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11institutetext: Andrew Bakan 22institutetext: Institute of Mathematics, National Academy of Sciences of Ukraine, 01601 Kyiv, Ukraine
22email: andrew@bakan.kiev.ua
33institutetext: Håkan Hedenmalm 44institutetext: KTH Royal Institute of Technology, SE–10044 Stockholm, Sweden
44email: haakanh@math.kth.se
55institutetext: Alfonso Montes-Rodríguez 66institutetext: University of Sevilla, 4180 Sevilla, Spain
66email: amontes@us.es
77institutetext: Danylo Radchenko 88institutetext: ETH Zürich, Mathematics Department, 8092 Zürich, Switzerland
88email: danradchenko@gmail.com
99institutetext: Maryna Viazovska 1010institutetext: École Polytechnique Fédérale de Lausanne, 1015 Lausanne, Switzerland
1010email: viazovska@gmail.com

Hyperbolic Fourier series

  
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   5cm]Andrew Bakan    Håkan Hedenmalm   
0
   1cm] Alfonso Montes-Rodríguez   
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   12cm] Danylo Radchenko    Maryna Viazovska
Abstract

. In this article we explain the essence of the interrelation described in [PNAS 118, 15 (2021)] on how to write explicit interpolation formula for solutions of the Klein-Gordon equation by using the recent Fourier pair interpolation formula of Viazovska and Radchenko from [Publ Math-Paris 129, 1 (2019)]. By the words of Fourier pair interpolation, we mean the interpolation of a pair of functions of one variable, related to each other by the Fourier transform.

Hedenmalm and Montes-Rodríguez established in 2011 the weak-star completeness in L(L^{\infty}\!(\Bb{R}) of the sequence 11, exp(iπnx)\exp(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx), exp(iπn/x)\exp(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x), n{}n\in\Bb{Z}\setminus\{0\}, which is referred to as the hyperbolic trigonometric system. We construct explicitly the sequence in L1(L^{1}(\Bb{R}) which is biorthogonal to this system and show that it is complete in L1(L^{1}(\Bb{R}). The construction involves integrals with the kernel studied by Chibrikova in 1956, similar to what was used by Viazovska and Radchenko for the Fourier pair interpolation on the real line. We associate with each fL1(f\in L^{1}(\Bb{R},(1+x^{2})^{-1}{\rm{d}}x) its hyperbolic Fourier series

h0(f)+n{}(hn(f)eiπnx+mn(f)eiπn/x)\eurm{h}_{0}(f)+\sum_{n\in\Bb{Z}\setminus\{0\}}\left(\eurm{h}_{n}(f)\,{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}+\eurm{m}_{n}(f)\,{\rm{e}}^{{{\mbox{\footnotesize{$-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x$}}}}}\right)

and prove that it converges to ff in the space of tempered distributions on the real line. The integral transform of φL1(\varphi\in L^{1}(\Bb{R}) given by

Uφ(x,y):=φ(t)exp(ixt+iy/t)dt,(x,y)\displaystyle U_{\varphi}(x,y):=\textstyle\int_{\Bb{R}}\,\varphi(t)\exp\left(\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxt+{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty}/{t}\right){\rm{d}}t\,,\quad(x,y)\in\Bb{R}^{2}\,,

is continuous and bounded on \Bb{R}^{2}, vanishes at infinity, u=Uφu\!=\!U_{\varphi} solves the Klein-Gordon equation uxy+u=0u_{xy}\!+\!u\!=\!0, and, by the theorem of Hedenmalm and Montes-Rodríguez, it is determined uniquely by its values at {(πn,0),(0,πn)}n\{(\pi n,0),(0,\pi n)\}_{n\in\Bb{Z}}\!\subset\!\Bb{R}^{2}. Applied to the above mentioned biorthogonal system, this transform supplies interpolating functions {𝓃}𝓃0\{\eusm{R}_{n}\}_{n\geqslant 0} for the Klein-Gordon equation on the characteristics with 𝓃(π𝓂,0)=δ𝓃,𝓂\eusm{R}_{n}(\pi m,0)\!=\!\delta_{n,m}, 𝓃(0,π𝓂)=0\eusm{R}_{n}(0,\pi m)\!=\!0, mm\!\in\!\Bb{Z}, where δn,m\delta_{n,m} is the Kronecker delta symbol. Under additional decay conditions on UφU_{\varphi}, these interpolating functions allow us to restore UφU_{\varphi} from its values at {(πn,0),(0,πn)}n\{(\pi n,0),(0,\pi n)\}_{n\in\Bb{Z}}. The restriction of any smooth solution of the Klein-Gordon equation to a bounded rectangle can be represented in the form of UφU_{\varphi}, but this matter will be pursued elsewhere.

Keywords:
Theta functions Elliptic functions Gauss hypergeometric function
MSC:
81Q05 42C30 33C05 33E05
journal: PREPRINT

1 . Introduction

An important unsolved issue in mathematics is the problem of interpolating oscillatory processes. Uniqueness holds a special place in interpolation. For example, when gas fluctuates in a gas storage, it is necessary to select the position of the pressure gauges in order to fully know about the state of the gas at any point in the storage. Unfortunately, nothing but a few numerical methods, based on considering solutions with extreme value of entropy, are known (see, e.g., lio , ger ). There appear to have been no other approaches to this topic until 2011, when the second and the third authors hed considered uniformly bounded solutions of the Klein-Gordon equation Uxy+U=0U_{xy}+U=0 in the plane of the form

Uφ(x,y)=eixt+iy/tφ(t)dt,φL1(\displaystyle U_{\varphi}(x,y)=\int\nolimits_{\Bb{R}}\ {\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt+{\rm{i}}y/t$}}}}}\varphi(t){\rm{d}}t\ ,\quad\varphi\in L^{1}(\Bb{R})\,, (1.1)

On a compact subset, such solutions UφU_{\varphi} can approximate any continuous solution of this equation. They pioneered the study of the discretized Goursat problem which instead of prescribing the values of a solution on the two intersecting characteristics x=0x=0 and y=0y=0 assumes that these values are known only along the discrete subset of the characteristics consisting of equidistant points {(πn,0)}n\{(\pi n,0)\}_{n\in\Bb{Z}}, {(0,πn)}n\{(0,\pi n)\}_{n\in\Bb{Z}}. It was established in hed that the values Uφ(πn,0)U_{\varphi}(\pi n,0), nn\in\Bb{Z}, and Uφ(0,πn)U_{\varphi}(0,\pi n), n{}n\in\Bb{Z}\setminus\{0\}, determine the function UφU_{\varphi} uniquely. They obtained the following result, which we develop further in Section 9.

Theorem A ( (hed, , p.​ 1517, Theorem 3.1) )

. Let φL1(\varphi\in L^{1}(\Bb{R}) have

eiπnxφ(x)dx=eiπn/xφ(x)dx=0,n\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\varphi(x){\rm{d}}x=\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${-\rm{i}}\pi n/x$}}}}}\varphi(x){\rm{d}}x=0\,,\quad n\in\Bb{Z}\,.

Then φ=0\varphi=0.

In the present paper we explicitly construct a system of functions H0\eurm{H}_{0}, Hn\eurm{H}_{n}, MnL1(\eurm{M}_{n}\subset L^{1}(\Bb{R})\cap C^{\infty}(\Bb{R}), n{}n\in\Bb{Z}\setminus\{0\}, which is biorthogonal to the hyperbolic trigonometric system 11, exp(iπnx)\exp(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx), exp(iπn/x)L(\exp(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x)\subset L^{\infty}(\Bb{R}), n{}n\in\Bb{Z}\setminus\{0\}, such that

eiπnxHp(x)dx=δn,p,eiπm/xHp(x)dx=0,|Hq(x)|π6|q|21+x2,eiπnxMq(x)dx=0,eiπm/xMq(x)dx=δm,q,|Mq(x)|π6|q|21+x2,\displaystyle\begin{array}[]{lll}\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}\eurm{H}_{p}(x){\rm{d}}x=\delta_{n,p}\,,&\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi m}/{x}$}}}}}\eurm{H}_{p}(x){\rm{d}}x=0\,,&\ \left|\eurm{H}_{q}(x)\right|\leqslant\dfrac{\pi^{6}|q\,|^{2}}{1+x^{2}}\ ,\\[14.22636pt] \displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}\eurm{M}_{q}(x){\rm{d}}x=0\,,&\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi m}/{x}$}}}}}\eurm{M}_{q}(x){\rm{d}}x=\delta_{m,q}\,,&\ \left|\eurm{M}_{q}(x)\right|\leqslant\dfrac{\pi^{6}|q\,|^{2}}{1+x^{2}}\ ,\end{array} (1.4)

|H0(x)|3/(1+x2)\left|\eurm{H}_{0}(x)\right|\!\leqslant\!3/(1+x^{2}) for all xx\!\in\!\Bb{R}, n,pn,p\!\in\!\Bb{Z} and m,q{}m,q\!\in\!\Bb{Z}\setminus\{0\}, where δn,m=1\delta_{n,m}=1, if n=mn=m, and δn,m=0\delta_{n,m}=0, if nmn\neq m, for arbitrary n,mn,m\in\Bb{Z}. We prove that for each fL1(f\in L^{1}(\Bb{R},(1+x^{2})^{-1}{\rm{d}}x) its hyperbolic Fourier series

h0(f)+n{}(hn(f)eiπnx+mn(f)eiπn/x),\displaystyle\eurm{h}_{0}(f)+\sum_{n\in\Bb{Z}\setminus\{0\}}\left(\eurm{h}_{n}(f)\,{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}+\eurm{m}_{n}(f)\,{\rm{e}}^{{{\mbox{\footnotesize{$-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x$}}}}}\right), (1.5)
hn(f):=f(t)Hn(t)dt,n𝕄{}\displaystyle\eurm{h}_{n}(f):=\int\limits_{\Bb{R}}f(t)\eurm{H}_{-n}(t){\rm{d}}t\ ,\ n\in\Bb{Z}\ ;\quad\eurm{m}_{n}(f):=\int\limits_{\Bb{R}}f(t)\eurm{M}_{-n}(t){\rm{d}}t\ ,\ n\in\Bb{Z}\setminus\{0\}\ ,

converges to ff in the space S(\eurm{S}^{\,\prime}(\Bb{R}) of tempered distributions on the real line. We also show in Remark 4.4 that in case f(x)f(x) or f(1/x)f(-1/x) is 22-periodic and integrable on the period, the series (1.5) coincides with its usual Fourier series of the argument xx or 1/x-1/x, respectively. Using (1.12) below (see also Theorem 8.7), we explicitly restore UφU_{\varphi} in (1.1) from its values at the points {(πn,0),(0,πn)}n\{(\pi n,0),(0,\pi n)\}_{n\in\Bb{Z}} under the condition (1.11). The proof of the assertion that any smooth solution of the Klein-Gordon equation on a bounded rectangle can be represented in the form (1.1) will be supplied elsewhere. Another aspect of the theory which will be developed elsewhere is the series expansion of UφU_{\varphi} which is obtained by insertion of the series (1.5) for the function φ\varphi in the integral (1.1).

1.1 . Key proofs.

​​ Given the properties (1.4), it is easy to deduce from Theorem A that for every fL1(f\!\in\!L^{1}(\Bb{R},(1\!+\!x^{2})^{-1}{\rm{d}}x) the series (1.5) converges in S(\eurm{S}^{\,\prime}(\Bb{R}) and that the biorthogonal system {H0}{Hn,Mn}n{}\{\eurm{H}_{0}\}\cup\{\eurm{H}_{n},\eurm{M}_{n}\}_{n\in\Bb{Z}\setminus\{0\}} is complete in L1(L^{1}(\Bb{R}) as well. We indicate briefly the necessary arguments. The tempered test functions form the so-called Schwartz class S(\eurm{S}(\Bb{R}) on the real axis (see (vlad, , p.​ 74)), so that for all positive integers nn, kk and arbitrary φS(\varphi\in\eurm{S}(\Bb{R}) integration by part gives

φ(x)eiπnxdx=iπnφ(x)deiπnx=1(iπn)kφ(k)(x)eiπnxdx,\displaystyle\int\nolimits_{\Bb{R}}\varphi(x){\rm{e}}^{{{\mbox{\footnotesize{${-\rm{i}}\pi nx$}}}}}{\rm{d}}x=\dfrac{{\rm{i}}}{\pi n}\int\nolimits_{\Bb{R}}\varphi(x){\rm{d}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nx$}}}}}=\dfrac{1}{(i\,\pi\,n)^{k}}\int\nolimits_{\Bb{R}}\varphi^{(k)}(x){\rm{e}}^{{{\mbox{\footnotesize{${-\rm{i}}\pi nx$}}}}}{\rm{d}}x\,,
φ(x)eiπnxdx=iπnx2φ(x)deiπnx=1(iπn)k((x2ddx+2x)kφ(x))eiπnxdx.\displaystyle\int\nolimits_{\Bb{R}}\varphi(x){\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{x}$}}}}}{\rm{d}}x\!=\!\dfrac{{\rm{i}}}{\pi n}\int\nolimits_{\Bb{R}}x^{2}\varphi(x){\rm{d}}{\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{x}$}}}}}\!=\!\dfrac{1}{(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi\,n)^{k}}\int\nolimits_{\Bb{R}}\left(\left(x^{2}\dfrac{{\rm{d}}}{{\rm{d}}x}\!+\!2x\right)^{k}\!\!\varphi(x)\!\right){\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{x}$}}}}}{\rm{d}}x\,.

As the integrands here are in L1(L^{1}(\Bb{R}), the coefficients

hn(φ):=φ(x)eiπnxdx,nφφiπnx{}\displaystyle\eurm{h}_{n}^{\star}(\varphi)\!:=\!\int\limits_{\Bb{R}}\!\varphi(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${-\rm{i}}\pi nx$}}}}}{\rm{d}}x\,,\ n\!\in\!\Bb{Z}\,;\ \ \eurm{m}_{n}^{\star}(\varphi)\!:=\!\int\limits_{\Bb{R}}\!\varphi(x)\,{}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{x}$}}}}}{\rm{d}}x\,,\ n\!\in\!\Bb{Z}\setminus\{0\}, (1.6)

satisfy, for any positive integer kk,

hn(φ)=O(nk),mn(φ)=O(nk),n.\displaystyle\eurm{h}_{n}^{\star}(\varphi)\!=\!{\rm{O}}\left(n^{-k}\right)\ ,\quad\eurm{m}_{n}^{\star}(\varphi)\!=\!{\rm{O}}\left(n^{-k}\right)\,,\quad n\to\infty\,. (1.7)

Hence, for all {a0}{an,bn}n{}\{a_{0}\}\cup\{a_{n},b_{n}\}_{n\in\Bb{Z}\setminus\{0\}}\!\subset\!\Bb{C} and any positive integer NN we have

an,bn=O(nN),na0+n{}(aneiπnx+bneiπn/x)S(\displaystyle a_{n},b_{n}\!=\!{\rm{O}}(n^{N}),\ n\!\to\!\infty\ \Rightarrow\ a_{0}+\sum\nolimits_{n\in\Bb{Z}\setminus\{0\}}\!\!\left(a_{n}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}\!+\!b_{n}{\rm{e}}^{{{\mbox{\footnotesize{$-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x$}}}}}\right)\!\in\!\eurm{S}^{\,\prime}(\Bb{R}), (1.8)

as the series converges in the space S(\eurm{S}^{\,\prime}(\Bb{R}) (see (vlad, , p.​ 77))1{}^{\ref*{case012}}. In view of (1.4), we see that for each φS(\varphi\in\eurm{S}(\Bb{R}) the function

ρφ(x):=φ(x)h0(φ)H0(x)n{}(hn(φ)Hn(x)+mn(φ)Mn(x)),x\displaystyle\rho_{\varphi}(x)\!:=\!\varphi(x)\!-\!\eurm{h}_{0}^{\star}(\varphi)\eurm{H}_{0}(x)\!-\!\sum\nolimits_{n\in\Bb{Z}\setminus\{0\}}\!\Big{(}\eurm{h}_{n}^{\star}(\varphi)\eurm{H}_{n}(x)\!+\!\eurm{m}_{n}^{\star}(\varphi)\eurm{M}_{n}(x)\Big{)}\,,\ \ x\!\in\!\Bb{R}\,,

is in L1(L^{1}(\Bb{R}) and satisfies

eiπnxρφ(x)dx=eiπn/xρφ(x)dx=0,n\displaystyle\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\rho_{\varphi}(x){\rm{d}}x=\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi n/x$}}}}}\rho_{\varphi}(x){\rm{d}}x=0\ ,\quad n\in\Bb{Z}\,.

From Theorem A, it follows that ρφ=0\rho_{\varphi}=0 and consequently, for any φS(\varphi\in\eurm{S}(\Bb{R}),

φ(x)=h0(φ)H0(x)+n{}(hn(φ)Hn(x)+mn(φ)Mn(x)),x\displaystyle\varphi(x)\!=\!\eurm{h}_{0}^{\star}(\varphi)\eurm{H}_{0}(x)\!+\!\sum\nolimits_{n\in\Bb{Z}\setminus\{0\}}\!\Big{(}\eurm{h}_{n}^{\star}(\varphi)\eurm{H}_{n}(x)\!+\!\eurm{m}_{n}^{\star}(\varphi)\eurm{M}_{n}(x)\Big{)}\,,\ \ x\!\in\!\Bb{R}\,, (1.9)

where the series converges in the weighted uniform norm C2(\|\cdot\|_{C_{2}(\Bb{R})}, in view of (1.4) and (1.7). Here, fC2(:=supx(1+x2)|f(x)|\|f\|_{C_{2}(\Bb{R})}:=\sup_{x\in\Bb{R}}(1\!+\!x^{2})|f(x)|, fC(f\in C(\Bb{R}). Since S(\eurm{S}(\Bb{R}) is dense in L1(L^{1}(\Bb{R}) we deduce the completeness of the system {H0}{Hn,Mn}n{}\{\eurm{H}_{0}\}\cup\{\eurm{H}_{n},\eurm{M}_{n}\}_{n\in\Bb{Z}\setminus\{0\}} in L1(L^{1}(\Bb{R}), as claimed. As for the convergence of the series (1.5), we take an arbitrary fL1(f\in L^{1}(\Bb{R},(1+x^{2})^{-1}{\rm{d}}x) and for N1N\geqslant 1 we write

N[f](x):=h0(f)+n{}𝕜𝕜(hn(f)eiπnx+mn(f)eiπn/x),x\displaystyle\mathcal{F}_{N}[f](x):=\eurm{h}_{0}(f)+\sum_{n\in\Bb{Z}\setminus\{0\},\,|n|\leqslant N}\left(\eurm{h}_{n}(f)\,{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}+\eurm{m}_{n}(f)\,{\rm{e}}^{{{\mbox{\footnotesize{$-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x$}}}}}\right)\,,\ \ x\!\in\!\Bb{R}\,.

It follows from (1.4) and (1.8) that, as N+N\to+\infty, N[f]\mathcal{F}_{N}[f] converges in the space S(\eurm{S}^{\,\prime}(\Bb{R}) to an element [f]S(\mathcal{F}[f]\in\eurm{S}^{\,\prime}(\Bb{R}). For any test function φS(\varphi\in\eurm{S}(\Bb{R}) we have

N[f](x)φ(x)dx=h0(f)h0(φ)+n{}𝕜𝕜(hn(f)hn(φ)+mn(f)mn(φ))\displaystyle\int\limits_{\Bb{R}}\mathcal{F}_{N}[f](x)\varphi(x){\rm{d}}x=\eurm{h}_{0}(f)\eurm{h}_{0}^{\star}(\varphi)+\!\!\!\sum_{n\in\Bb{Z}\setminus\{0\},\,|n|\leqslant N}\left(\eurm{h}_{n}(f)\,\eurm{h}_{n}^{\star}(\varphi)+\eurm{m}_{n}(f)\,\eurm{m}_{n}^{\star}(\varphi)\right)
=f(x)[h0(φ)H0(x)+n{}𝕜𝕜(hn(φ)Hn(x)+mn(φ)Mn(x))]dx.\displaystyle=\int\limits_{\Bb{R}}f(x)\bigg{[}\eurm{h}_{0}^{\star}(\varphi)\eurm{H}_{0}(x)+\sum_{n\in\Bb{Z}\setminus\{0\},\,|n|\leqslant N}\Big{(}\eurm{h}_{n}^{\star}(\varphi)\eurm{H}_{n}(x)+\eurm{m}_{n}^{\star}(\varphi)\eurm{M}_{n}(x)\Big{)}\bigg{]}{\rm{d}}x\,.

From (1.9) combined with (1.4), (1.7) and the Lebesgue dominated convergence theorem (nat, , p.​ 161), we obtain, by passing to the limit as N+N\to+\infty,

[f](φ)=f(x)φ(x)dx,φS(\mathcal{F}[f]\big{(}\varphi\big{)}=\int_{\Bb{R}}f(x)\varphi(x){\rm{d}}x\,,\quad\varphi\in\eurm{S}(\Bb{R}).

This tells us that [f]=f\mathcal{F}[f]=f holds in the space S(\eurm{S}^{\,\prime}(\Bb{R}). In conclusion, the hyperbolic Fourier series (1.5) converges to ff in the space of tempered distributions on the real line.

1.2 . Interpolation formula.

​​ We proceed with the interpolation formula for the Klein-Gordon equation. Let φL1(\varphi\in L^{1}(\Bb{R}) be arbitrary and consider the solution UφU_{\varphi} of the Klein-Gordon equation given by (1.1). Then the coefficients of φ\varphi in (1.6) are called the conjugate hyperbolic Fourier coefficients and they equal the values of UφU_{\varphi} at the points {(πn,0),(0,πn)}n\{(\pi n,0),(0,\pi n)\}_{n\in\Bb{Z}},

hn(φ)=Uφ(πn,0),nφ𝕌φπ{}\displaystyle\eurm{h}_{n}^{\star}(\varphi)=U_{\varphi}(-\pi n,0)\,,\ n\!\in\!\Bb{Z}\,;\ \ \eurm{m}_{n}^{\star}(\varphi)=U_{\varphi}(0,\pi n)\,,\ n\!\in\!\Bb{Z}\setminus\{0\}\,. (1.10)

By the estimates (1.4) and manipulations similar to those employed in the proof of (1.9) we get that the conjugate hyperbolic Fourier series of φ\varphi in the right-hand side of the equality (1.9) converges to φ\varphi in the weighted uniform norm C2(\|\cdot\|_{C_{2}(\Bb{R})} on \Bb{R} provided that

n{}n2(|Uφ(πn,0)|+|Uφ(0,πn)|)<.\displaystyle\sum\nolimits_{n\in\Bb{Z}\setminus\{0\}}n^{2}\left(\,\left|U_{\varphi}(\pi n,0)\right|+\left|U_{\varphi}(0,\pi n)\right|\,\right)<\infty. (1.11)

If we multiply both sides of the equality (1.9) by exp(ixt+iy/t)\exp(\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxt+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty/t) and integrate with respect to tt over the real line, we restore UφU_{\varphi} from its values at the points {(πn,0),(0,πn)}n\{(\pi n,0),(0,\pi n)\}_{n\in\Bb{Z}} (see Theorem 8.7), by using the identities (1.15) below,

Uφ(x,y)=Uφ(0,0)0(𝓍,𝓎)+𝓃1[𝒰φ(π𝓃,0)𝓃(𝓍,𝓎)+𝒰φ(0,π𝓃)𝓃(𝓎,𝓍)]++n1[Uφ(πn,0)𝓃(𝓍,𝓎)+𝒰φ(0,π𝓃)𝓃(𝓎,𝓍)],(𝓍,𝓎)U_{\varphi}(x,y)\!=\!U_{\varphi}(0,0)\,\eusm{R}_{\hskip 0.71114pt0}(x,y)\!+\!\sum\limits_{n\geqslant 1}\!\Big{[}U_{\varphi}(\pi n,0)\,\eusm{R}_{\hskip 0.71114ptn}(x,y)+U_{\varphi}(0,-\pi n)\,\eusm{R}_{\hskip 0.71114ptn}(-y,-x)\Big{]}+\\[5.69046pt] \!+\!\sum\limits_{n\geqslant 1}\!\Big{[}\,U_{\varphi}(-\pi n,0)\,\eusm{R}_{\hskip 0.71114ptn}(-x,-y)+U_{\varphi}(0,\pi n)\,\eusm{R}_{\hskip 0.71114ptn}(y,x)\Big{]}\,,\quad(x,y)\in\Bb{R}^{2}\,. (1.12)

Here, {𝓃}𝓃0\{\eusm{R}_{\hskip 0.71114ptn}\}_{n\geqslant 0} are the interpolating functions for the Klein-Gordon equation given by

𝓃(𝓍,𝓎):=eixt+iy/tHn(t)dt,n0,x,y\displaystyle\eusm{R}_{n}(x,y)\!:=\!\int\nolimits_{\Bb{R}}\ {\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxt+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty/t$}}}}}\eurm{H}_{-n}(t){\rm{d}}t\,,\quad n\geqslant 0\,,\quad x,y\in\Bb{R}\,.

The biorthogonal system has the symmetry properties

Hn(x)=Hn(x),n0,x𝕄\displaystyle\eurm{H}_{-n}(x)\!=\!\eurm{H}_{n}(-x),\ n\!\geqslant\!0\,,\ x\!\in\!\Bb{R}\,;\ \ \eurm{M}_{n}(x)\!=\!\eurm{H}_{n}(-1/x)/x^{2}\,,\ n\!\geqslant\!1\,,\ x\!\in\!\Bb{R}_{\neq 0}\,,

which for arbitrary n1n\geqslant 1 and x,yx,y\in\Bb{R} lead to the identities,

eixt+iy/tHn(t)dt=𝓃(𝓍,𝓎),eixt+iy/tMn(t)dt=𝓃(𝓎,𝓍),eixt+iy/tHn(t)dt=𝓃(𝓍,𝓎),eixt+iy/tMn(t)dt=𝓃(𝓎,𝓍).\displaystyle\begin{array}[]{ll}\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxt\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty/t$}}}}}\eurm{H}_{n}(t){\rm{d}}t\!=\!\eusm{R}_{n}(-x,-y),&\displaystyle\ \int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxt\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty/t$}}}}}\eurm{M}_{n}(t){\rm{d}}t\!=\!\eusm{R}_{n}(y,x),\\ \displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxt\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty/t$}}}}}\eurm{H}_{-n}(t){\rm{d}}t\!=\!\eusm{R}_{n}(x,y),&\displaystyle\ \int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxt\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty/t$}}}}}\eurm{M}_{-n}(t){\rm{d}}t\!=\!\eusm{R}_{n}(-y,-x).\end{array}\vspace{-0,2cm} (1.15)

1.3 . Basic result.

Let 𝔻{ϝ𝕜𝕜ϝ𝕜}\Bb{D}:=\{z\in\Bb{C}\ |\ |z|<1\} and

F(z):=F(1/2,1/2;1;z)=1+1πn=1Γ(n+1/2)2n!2zn,z𝔻\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\!:=\!F(1/2,1/2;1;z)\!=\!1\!+\!\dfrac{1}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n\!+\!1/2)^{2}}{n!^{2}}z^{n}\ ,\quad z\!\in\!\Bb{D}\,, (1.16)

be the Gauss hypergeometric function which can be extended to a nonvanishing holomorphic function on the set ^\Bb{C}\!\setminus\![1,+\infty) (see (bh2, , p.​ 597, (1.19)(b))). In view of the Barnes expansion (bar, , p.​ 173)(1908) (see also (whi, , p.​ 299), (2.8)),

πF(1z)F(z)=(log16z)2πn=1Γ(n+1/2)2(n!)2[k=1n1(2k1)k]zn1+1πn=1Γ(n+1/2)2(n!)2zn,z𝔻~\displaystyle\dfrac{\pi F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1\!-\!z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}=\left(\log\dfrac{16}{z}\right)\!-\!\dfrac{\dfrac{2}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}\left[\sum\limits_{k=1}^{n}\dfrac{1}{(2k-1)k}\right]z^{n}}{1+\dfrac{1}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}z^{n}}\ ,\ z\!\in\!\Bb{D}\!\setminus\!(-1,0]\,,

for each positive integer nn, the function exp(nπF(1z)/F(z))\exp(n\pi F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)/F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)) is meromorphic in 𝔻\Bb{D}. Moreover, there exists an algebraic polynomial SnS^{{{\mbox{\tiny{$\triangle$}}}}}_{n} of degree nn with real coefficients and leading coefficient equal to 16n16^{n} such that Sn(0)=0S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(0)=0 and

enπF(1z)F(z)\displaystyle{\rm{e}}^{{{\mbox{\footnotesize{$n\pi\ \dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}$}}}}} =16nznexp(nz12+2πn=1Γ(n+3/2)2(n+1)!2[k=1n+11(2k1)k]zn1+1πn=1Γ(n+1/2)2(n!)2zn)\displaystyle=\dfrac{16^{n}}{z^{n}}\ \exp\left(-nz\ \dfrac{\dfrac{1}{2}+\dfrac{2}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+3/2)^{2}}{(n+1)!^{2}}\left[\sum\limits_{k=1}^{n+1}\dfrac{1}{(2k-1)k}\right]z^{n}}{1+\dfrac{1}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}z^{n}}\right)
=Sn(1/z)+ΔnS(z),z𝔻{}Δ𝕊𝔻\displaystyle=S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)+\Delta_{n}^{S}(z)\ ,\ z\in\Bb{D}\setminus\{0\}\ ,\ \Delta_{n}^{S}\in{\rm{Hol}}(\Bb{D})\,, (1.17)

where Hol(D){\rm{Hol}}(D) denotes the space of all holomorphic functions in a domain DD\!\subset\!\Bb{C}. In addition, all the Taylor coefficients of ΔnS\Delta_{n}^{S} are real (see (rud, , p.​ 199), (con, , p.​ 72)). Our main result follows.

Theorem 1.1

. The system {Hn(x)}n{Mn(x)}n{}L1(\left\{\eurm{H}_{n}(x)\right\}_{n\in\Bb{Z}}\cup\left\{\eurm{M}_{n}(x)\right\}_{n\in\Bb{Z}\setminus\{0\}}\subset L^{1}(\Bb{R}) given by

H0(x):=12π2+|F(1/2+it)|2dt(t2+1/4)(F(1/2it)2+x2F(1/2+it)2),\displaystyle\eurm{H}_{0}(x):=\dfrac{1}{2\pi^{2}}\int\limits_{-\infty}^{+\infty}\frac{\left|F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt)\right|^{2}{\rm{d}}t}{\big{(}t^{2}+1/4\big{)}\Big{(}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt)^{2}+x^{2}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt)^{2}\Big{)}}\,,
Hn(x)=14π3n+Sn(11/2+it)dt(t2+1/4)(F(1/2it)ixF(1/2+it))2,\displaystyle\eurm{H}_{n}(x)=\dfrac{1}{4\pi^{3}n}\int\limits_{-\infty}^{+\infty}\dfrac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\dfrac{1}{1/2+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt}\right){\rm{d}}t}{\big{(}t^{2}+1/4\big{)}\Big{(}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt)-\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxF_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt)\Big{)}^{2}}\ ,
Hn(x):=Hn(x),n1,x\displaystyle\eurm{H}_{-n}(x):=\eurm{H}_{n}(-x)\ ,\quad n\geqslant 1\ ,\ x\!\in\!\Bb{R}\,; (1.18)
Mn(x):=Hn(1/x)/x2,n{}{}\displaystyle\eurm{M}_{n}(x):=\eurm{H}_{n}(-1/x)/x^{2}\ ,\ n\in\Bb{Z}\setminus\{0\}\ ,\ x\!\in\!\Bb{R}\setminus\{0\}\,, (1.19)

is biorthogonal to the sequence 11, exp(iπnx)\exp({\rm{i}}\pi nx), exp(iπn/x)\exp({\rm{i}}\pi n/x), n{}n\in\Bb{Z}\setminus\{0\}, and satisfies the equalities (1.4). Here, all the integrals are absolutely convergent for every xx\!\in\!\Bb{R} and {Sn}n1\{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\}_{n\geqslant 1} are the algebraic polynomials defined in (1.17). This system is complete in L1(L^{1}(\Bb{R}) and enjoys the estimates

|H0(x)|31+x2,|Hn(x)|+|Mn(x)|π6|n|21+x2,n{}\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\frac{3}{1+x^{2}}\ ,\ \left|\eurm{H}_{n}(x)\right|+\left|\eurm{M}_{n}(x)\right|\leqslant\frac{\pi^{6}|n|^{2}}{1+x^{2}}\ ,\quad n\in\Bb{Z}\setminus\{0\}\ ,\ x\!\in\!\Bb{R}\,. (1.20)
Corrollary 1.2

. Suppose fL1(f\in L^{1}(\Bb{R},(1+x^{2})^{-1}{\rm{d}}x). Then the series

neiπnxf(t)Hn(t)dt+n{}eiπnxf(t)Mn(t)dt\displaystyle\sum\limits_{n\in\Bb{Z}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\int\limits_{\Bb{R}}f(t)\eurm{H}_{-n}(t){\rm{d}}t+\sum\limits_{n\in\Bb{Z}\setminus\{0\}}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{x}$}}}}}\int\limits_{\Bb{R}}f(t)\eurm{M}_{-n}(t){\rm{d}}t

represents a tempered distribution on \Bb{R} associated with the regular function ff.

Remark 1.3

. To prove Theorem 1.1 we first write down in (2.50) and (2.51) the integrands of the integrals (1.18) in terms of the Schwarz triangle function. Then using the change of variables we express in (2.52) these integrals through the elliptic modular function which is the inverse to the Schwarz triangle function. This allows us to prove in Theorem 4.3 the fact of biorthogonality indicated in Theorem 1.1.

To establish the estimates (1.20), we first find in (4.6) and (4.7) the generating functions of {Hn(x)}n\left\{\eurm{H}_{n}(x)\right\}_{n\in\Bb{N}} and {Mn(x)}n\left\{\eurm{M}_{n}(x)\right\}_{n\in\Bb{N}}. And then, having found in Section 6 a constructive description of the analytic extension of these generating functions, we get in (7.25) and (7.26) the desired estimates (1.20).

The relationship2{}^{\ref*{case12}}

F(12+it)=F(12+t4t2+1)+iF(12t4t2+1)(1+i)4t2+14,t\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt\right)=\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)}{(1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)\,\sqrt[4]{4t^{2}+1}}\ ,\quad t\in\Bb{R}\,, (1.21)

allows us3{}^{\ref*{case14}} to write the formulas (1.18) by using only the values of FF_{\!{{\mbox{\tiny{$\triangle$}}}}} on the interval (0,1)(0,1),

H0(x)\displaystyle\eurm{H}_{0}(x) =i2π21iy(t)1+iy(t)x2(1iy(t)1+iy(t))2dtt2+1/4,y(t):=F(12t4t2+1)F(12+t4t2+1),\displaystyle=\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}{2\pi^{2}}\int\limits_{\Bb{R}}\dfrac{\dfrac{1\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{y}(t)}{1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{y}(t)}}{x^{2}-\left(\dfrac{1\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{y}(t)}{1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{y}(t)}\right)^{2}}\,\dfrac{{\rm{d}}t}{t^{2}+1/4}\ ,\quad\eurm{y}(t)\!:=\!\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}\ ,\
Hn(x)\displaystyle\eurm{H}_{n}(x) =iπ3nSn(11/2+it)dtt2+1/4((1+x)F(12t4t2+1)+i(1x)F(12+t4t2+1))2.\displaystyle=\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}{\pi^{3}n}\int\limits_{\Bb{R}}\dfrac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\dfrac{1}{1/2+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt}\right)\dfrac{{\rm{d}}t}{\sqrt{t^{2}+1/4}}}{\left((1\!+\!x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt(1\!-\!x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\right)^{\!\!2}}\ .

Since F(0)=1F_{\!{{\mbox{\tiny{$\triangle$}}}}}(0)=1, FF_{\!{{\mbox{\tiny{$\triangle$}}}}} increases on [0,1)[0,1), πF(1x)ln(1/x)(0,3π)\pi F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-x)-\ln(1/x)\in(0,3\pi), x(0,1/3)x\in(0,1/3) (see (bh2, , p.​ 602, (2.4)), (bh1, , p.​ 30, (A.6i))), Sn(0)=0S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(0)=0, we obviously get an absolute convergence of these integrals for every xx\!\in\!\Bb{R} and n1n\geqslant 1. Here y:\eurm{y}:\Bb{R}\mapsto(0,+\infty) decreases from ++\infty to 0.

1.4 . Connection with the Perron-Frobenius-Ruelle operator.

It can be easily seen that for arbitrary fL1([1,1],dx)f\in L_{1}\big{(}[-1,1],{\rm{d}}x\big{)} in the space L1([1,1],dx)L_{1}\big{(}[-1,1],{\rm{d}}x\big{)} there exists the limit

𝐓1[f](x):=limn+k=nk0nf(12kx)(2kx)2L1([1,1],dx),\displaystyle{\mathbf{T}}_{1}[f](x)\ :=\ \lim\limits_{n\to+\infty}\sum_{\begin{subarray}{c}{{\mbox{\footnotesize{$k=-n$}}}}\\[1.42271pt] {{\mbox{\footnotesize{$k\neq 0$}}}}\end{subarray}}^{{{\mbox{\footnotesize{$n$}}}}}\dfrac{f\left(\dfrac{1}{2k-x}\right)}{(2k-x)^{2}}\in L_{1}\big{(}[-1,1],{\rm{d}}x\big{)}\,, (1.22)

and hence the mapping f𝐓1[f]f\mapsto{\mathbf{T}}_{1}[f] defines the operator 𝐓1:L1([1,1],dx)L1([1,1],dx){\mathbf{T}}_{1}:L_{1}\big{(}[-1,1],{\rm{d}}x\big{)}\mapsto L_{1}\big{(}[-1,1],{\rm{d}}x\big{)}, where dx:=dm(x){{\mathrm{d}}}x:={\rm{d}}m(x) and mm denotes the Lebesgue measure on the real line. This operator is known as the Perron-Frobenius-Ruelle operator corresponding to the even Gauss map G2:(1,1](1,1]G_{2}:(-1,1]\to(-1,1] defined by the formula G2(x)={1/x}2G_{2}(x)=\{-1/x\}_{2} if x0x\neq 0 and G2(0)=0G_{2}(0)=0. Here, {x}2\{x\}_{2} assigns to each real xx the unique number in the interval (1,1](-1,1] such that x{x}2x-\{x\}_{2} is an even integer (see (kes, , p.​ 81), (ios, , p.​ 57)). The properties stated in Lemma 4.1(4) and Lemma 4.2(3,4) below tell us that

(I+𝐓1)[Hn+](x)=eiπnx,(I𝐓1)[Hn](x)=eiπnx,n{}(I+𝐓1)[2H0](x)=1,x[1,1],\displaystyle\begin{array}[]{rcll}\displaystyle\left(I+{\mathbf{T}}_{1}\right)\left[\eurm{H}_{n}^{+}\right](x)&=&\displaystyle{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\,,&\displaystyle\quad\left(I-{\mathbf{T}}_{1}\right)\left[\eurm{H}_{n}^{-}\right](x)={\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\ ,\ \ n\!\in\!\Bb{Z}\setminus\{0\}\,,\\[5.69046pt] \displaystyle\left(I+{\mathbf{T}}_{1}\right)\left[2\eurm{H}_{0}\right](x)&=&1\,,&\quad x\in[-1,1]\,,\end{array} (1.25)

where

{Hn±(1/x)=±x2Hn±(x),H0(1/x)=H0(x)x2,Hn±:=2Hn±2Mn,n{}{}\displaystyle\left\{\begin{array}[]{ll}\eurm{H}_{n}^{\pm}(-1/x)=\pm x^{2}\eurm{H}_{n}^{\pm}(x)\ ,&\quad\eurm{H}_{0}(-1/x)\!=\!\eurm{H}_{0}(x)x^{2}\,,\\[5.69046pt] \eurm{H}_{n}^{\pm}:=2\eurm{H}_{n}\pm 2\eurm{M}_{n}\ ,&\quad n\in\Bb{Z}\setminus\{0\}\,,\ x\in\Bb{R}\setminus\{0\}\,.\end{array}\right. (1.28)

Hence another way to approach the functions H0\eurm{H}_{0}, Hn\eurm{H}_{n}, Mn\eurm{M}_{n}, n{}n\in\Bb{Z}\setminus\{0\}, is to solve the operator equations (1.25) and calculate the functions H0\eurm{H}_{0}, Hn±\eurm{H}^{\pm}_{n}, n{}n\in\Bb{Z}\setminus\{0\}, on the interval (1,1)(-1,1). The symmetry in (1.28) gives their values on ^~\Bb{R}\setminus[-1,1] and hence extends them to the real line \Bb{R}.

1.5 . Outline of the paper.

The paper is organized as follows. In Section 2 we list a few facts about the elliptic modular function λ\lambda following an approach suggested by the first and the second authors in bh2 . The polynomials introduced and investigated by the fourth and fifth author in rad have given reasons for us in Section 3 to introduce and explore a sequence of simpler polynomials, from which the polynomials in rad can be obtained by symmetrization. We devote Section 4 to various properties of the biorthogonal system. Two partitions of the upper half-plane are introduced and developed in Section 5. The results of Section 5 are needed in Section 6 to make a constructive description of the analytic extension of the generating function for the biorthogonal system. This extension is similar to what was proposed and implemented by the fourth and the fifth authors in rad . The results of Sections 5 and 6 are presented in the language of the even-integer continued fractions as suggested by the first author. Some estimates of biorthogonal functions are derived in Section 7. Applications to the Klein-Gordon equation of the obtained results are made in Section 8. In Section 9, Theorem A is extended with the purpose to apply it for partial solution of the issues raised in bhm . Section 10 elaborates on the proofs of the main results. Further explanatory notes are supplied in Section A.

1.6 . Notation.

We begin this section by listing the frequently used notations.

Complex​ plane, real​ line, all​ integers, positive​ integers, respectively.𝔻𝔻¯𝔻{ϝ𝕜ϝ𝕜}𝔻¯{ϝ𝕜ϝ𝕜}and{ϝ𝕀ϝ}^~{}{𝕜}{𝕜}ARe>a,ARe>a:={zA|Rez>a},a𝔸ARea,ARea:={zA|Reza},a𝔸ARe<aARe<a:={zA|Rez<a},a𝔸AReaARea:={zA|Reza},a𝔸A|Re|<a,A|Re|<a:={zA||Rez|<a},a𝔸A|Re|a,A|Re|a:={zA||Rez|a},a𝔸𝔻𝕀𝔻𝕀𝔻𝔻¯Im>0𝔻¯Im>0:=𝔻¯sign(x)sign(x)is equal to 1 if x<00 if x=0 and 1 if x>0.χA(x)χA(x)is equal to 1 if xA and 0 if xA for A.gcd(a,b)gcd(a,b)is the greatest common divisor of a,b andgcd(a,b):=gcd(|a|,|b|)gcd(0,b):=|b| for arbitrary a,b.intAintA:={aA|ε>0:a+ε𝔻𝔸}A.\displaystyle\begin{array}[]{ll}\Bb{C},\,\Bb{R},\,\Bb{Z},\,\Bb{N}&\ \mbox{Complex\! plane, real\! line, all\! integers, positive\! integers, respectively.}\\ \Bb{D},\ \overline{\Bb{D}},\ \Bb{H}&\ \Bb{D}\!:=\!\{z\!\in\!\Bb{C}\mid|z|\!<\!1\},\ \overline{\Bb{D}}\!:=\!\{z\!\in\!\Bb{C}\mid|z|\!\leqslant\!1\}\,\mbox{and}\,\Bb{H}\!:=\!\{z\!\in\!\Bb{C}\mid{\rm{Im}}\,z\!>\!0\}.\\ \Bb{R}_{>a},\,\Bb{R}_{\geqslant a}\,,&\ \Bb{R}_{>a}:=(a,+\infty),\quad\ \,\Bb{R}_{\geqslant a}:=[a,+\infty),\,\quad a\in\Bb{R}\,.\\ \Bb{R}_{<a},\,\Bb{R}_{\leqslant a}&\ \Bb{R}_{<a}:=(-\infty,a),\quad\ \,\Bb{R}_{\leqslant a}:=(-\infty,a],\quad\,a\in\Bb{R}\,.\\ \Bb{Z}_{\neq 0},\,\Bb{Z}_{\geqslant q}&\ \Bb{Z}_{\neq 0}\ :=\Bb{Z}\setminus\{0\},\ \ \ \ \ \Bb{Z}_{\geqslant q}\ :=\{\,n\in\Bb{Z}\,|\,n\geqslant q\,\},\ q\in\Bb{Z}\,.\\ \Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}},\,\Bb{Z}_{\leqslant q}&\ \Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\sqcup_{k\geqslant 1}\!\left(\Bb{Z}_{\neq 0}\right)^{k}\!,\ \Bb{Z}_{\leqslant q}\ :=\{\,n\in\Bb{Z}\,|\,n\leqslant q\,\},\ q\in\Bb{Z}\,.\\ A_{{\rm{Re}}>a},&\ A_{{\rm{Re}}>a}\ :=\,\{\,z\in A\ |\ {\rm{Re}}\,z>a\ \ \}\,,\ \,a\in\Bb{R}\,,\hskip 15.07993ptA\subset\Bb{C}\,.\\ A_{{\rm{Re}}\geqslant a},&\ A_{{\rm{Re}}\geqslant a}\ :=\,\{\,z\in A\ |\ {\rm{Re}}\,z\geqslant a\ \ \}\,,\ \,a\in\Bb{R}\,,\hskip 15.07993ptA\subset\Bb{C}\,.\\ A_{{\rm{Re}}<a}&\ A_{{\rm{Re}}<a}\ :=\,\{\,z\in A\ |\ {\rm{Re}}\,z<a\ \ \}\,,\ \,a\in\Bb{R}\,,\hskip 15.07993ptA\subset\Bb{C}\,.\\ A_{{\rm{Re}}\leqslant a}&\ A_{{\rm{Re}}\leqslant a}\ :=\,\{\,z\in A\ |\ {\rm{Re}}\,z\leqslant a\ \ \}\,,\ \,a\in\Bb{R}\,,\hskip 15.07993ptA\subset\Bb{C}\,.\\ A_{|{\rm{Re}}|<a},&\ A_{|{\rm{Re}}|<a}:=\{\,z\in A\ |\ |{\rm{Re}}\,z|<a\,\}\,,\ \,a\in\Bb{R}_{>0}\,,\ \,A\subset\Bb{C}\,.\\ A_{|{\rm{Re}}|\leqslant a},&\ A_{|{\rm{Re}}|\leqslant a}:=\{\,z\in A\ |\ |{\rm{Re}}\,z|\leqslant a\,\}\,,\ \,a\in\Bb{R}_{>0}\,,\ \,A\subset\Bb{C}\,.\\ \Bb{D}_{{\rm{Im}}>0}&\ \Bb{D}_{{\rm{Im}}>0}:=\Bb{D}\cap\Bb{H}\,.\\ \overline{\Bb{D}}_{{\rm{Im}}>0}&\ \overline{\Bb{D}}_{{\rm{Im}}>0}:=\overline{\Bb{D}}\cap\Bb{H}\,.\\ {\rm{sign}}(x)&\ {\rm{sign}}(x)\ \mbox{is equal to $-1$ if $x<0$, $0$ if $x=0$ and $1$ if $x>0$.}\\ \chi_{A}(x)&\ \chi_{A}(x)\ \mbox{is equal to $1$ if $x\in A$ and $0$ if $x\not\in A$ for $A\subset\Bb{C}$.}\\ {\rm{gcd}}(a,b)&\ {\rm{gcd}}(a,b)\in\Bb{N}\ \mbox{is the greatest common divisor of $a,b\in\Bb{N}$ and}\\ &\ \mbox{${\rm{gcd}}(a,b)\!:=\!{\rm{gcd}}(|a|,|b|)$, ${\rm{gcd}}(0,b)\!:=\!|b|$ for arbitrary $a,b\in\Bb{Z}_{\neq 0}$.}\\ {\rm{int}}A&\ \mbox{${\rm{int}}A\!:=\!\{a\in A\,|\,\exists\,\varepsilon\!>\!0:a\!+\!\varepsilon\Bb{D}\subset A\}$, \ $A\!\subset\!\Bb{C}$.}\end{array}

For any z1,z2z_{1},z_{2}\in\Bb{C}, z1z2z_{1}\neq z_{2} the straight line segment from z1z_{1} to z2z_{2} is denoted by [z1,z2][z_{1},z_{2}]. We extend interval notation on the reals to line segment notation, so that, e.g., for any two points z1,z2z_{1},z_{2}\in\Bb{C}, we write (z1,z2]:=[z1,z2]{z1}(z_{1},z_{2}]:=[z_{1},z_{2}]\setminus\{z_{1}\} and [z1,z2):=[z1,z2]{z2}[z_{1},z_{2}):=[z_{1},z_{2}]\setminus\{z_{2}\}.

Following the definitions of (sar, , pp.​ 6, 40), we denote by ln:\ln:\Bb{R}_{>0}\to\Bb{R} the real-valued logarithm defined on \Bb{R}_{>0}, and let Log(z)=ln|z|+iArg(z){\rm{Log}}\,(z)\!=\!\ln|z|\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{\rm{Arg}}\,(z) be the principal branch of the logarithm defined for z~z\!\in\!\Bb{C}\setminus(-\infty,0] with Arg(z)(π,π){\rm{Arg}}\,(z)\!\in\!(-\pi,\pi). Furthermore, for a simply connected domain DD\!\subset\!\Bb{C}, a point aDa\!\in\!D, and a function fHol(D)f\!\in\!{\rm{Hol}}(D) which is zero-free in DD with f(a)>0f(a)\!>\!0, we write logf(z)\log f(z) for the holomorphic function in DD such that exp(logf(z))=f(z)\exp(\log f(z))\!=\!f(z), zDz\!\in\!D, and logf(a)=lnf(a)\log f(a)\!=\!\ln f(a) (see (con, , p.​ 94)). Then Relogf(z)=ln|f(z)|{\rm{Re}}\,\log f(z)\!=\!\ln|f(z)| and argf(z):=Imlogf(z)\arg f(z)\!:=\!{\rm{Im}}\,\log f(z) for each zDz\!\in\!D.

As for topology, we denote by clos(A){\rm{clos}}(A) (or A¯\overline{A}), int(A)\mathrm{int}(A), and A\partial A the closure, interior, and boundary of a subset AA\subset\Bb{C}, respectively.

Let C(C(\Bb{R}) denote the linear space of all continuous complex-valued functions on \Bb{R}. For a non-negative Borel measure μ\mu on \Bb{R} we denote by L1(μL^{1}(\Bb{R},\mu) the standard normed space of integrable with respect to μ\mu complex-valued Borel functions. For arbitrary non-negative Borel measures μ\mu, ν\nu on \Bb{R} and gL1(μg\in L^{1}(\Bb{R},\mu), we write dν(x)=g(x)dμ(x){\rm{d}}\nu(x)=g(x){\rm{d}}\mu(x) if ν(A)=Ag(x)dμ(x)\nu(A)=\int_{A}g(x){\rm{d}}\mu(x) for arbitrary Borel subset AA of \Bb{R}. If here μ\mu is the Lebesgue measure mm on the real line and g(x)=1/(1+x2)g(x)=1/(1+x^{2}), xx\in\Bb{R}, then instead of L1(μL^{1}(\Bb{R},\mu) and L1(νL^{1}(\Bb{R},\nu) we write L1(L^{1}(\Bb{R},{\rm{d}}x) (or L1(L^{1}(\Bb{R})) and L1(L^{1}(\Bb{R},(1+x^{2})^{-1}{\rm{d}}x), respectively.

For 1p<+1\leqslant p<+\infty, we denote by H+p({\rm{H}}^{p}_{+}(\Bb{R}) the class of all nontangential limits on \Bb{R} of functions from the Hardy space (see (ko, , p.​ 112), (gar, , p.​ 57))

Hp({𝕜sup𝕜𝕜}\displaystyle{H}^{p}(\Bb{H}):=\left\{f\in{\rm{Hol}}(\Bb{H})\ \Big{|}\ \|f\|_{H^{p}_{+}}:=\sup\limits_{y>0}\int_{\Bb{R}}|f(x+\mathrm{i}\hskip 0.42677pty)|^{p}\,\mathrm{d}x<+\infty\right\}\,,

and Hp({𝕜}{\rm{H}}^{p}_{-}(\Bb{R})\!:=\{f(-x)\ |\ f\!\in\!{\rm{H}}^{p}_{+}(\Bb{R})\}.

2 . The Schwarz triangle function and its inverse

We denote γ(a,):=a+i(0,+)\gamma(a,\infty):=a+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt(0,+\infty), aa\in\Bb{R}, and

γ(a,b):={z𝕜𝕜ϝ𝕜𝕜𝕜},a,b\displaystyle\gamma(a,b):=\left\{\,z\in\Bb{H}\ \ \big{|}\ \ \left|z-(a+b)/2\right|=|b-a|/2\,\right\}\,,\hskip 6.25963pt\quad a,b\in\Bb{R}\,,\ a\neq b\,. (2.1)

Everywhere below we also use the notation γ(a,b)\gamma(a,b) for the contour of integration which is extended from aa to bb along the open semicircle γ(a,b)\gamma(a,b).

2.1 . The Gauss hypergeometric function.

Euler’s integral representation

F(z)=1π01dtt(1t)(1tz),z\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)=\dfrac{1}{\pi}\int_{0}^{1}\dfrac{{\mathrm{d}}t}{\sqrt{t(1-t)(1-tz)}}\ ,\quad z\in\Bb{C}\setminus\Bb{R}_{\geqslant 1}\,,\

of the Gauss hypergeometric function (1.16) gives its analytic extension from 𝔻\Bb{D} to \Bb{C}\setminus\Bb{R}_{\geqslant 1} satisfying

F(x)>0,x\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)>0\ ,\quad x\in\Bb{R}_{<\,1}\ ,\ (2.2)

and together with the Pfaff formula (see (and, , p.​ 79))

F(z)=11zF(zz1),z\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)=\dfrac{1}{\sqrt{1-z}}\ F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{z}{z-1}\right)\ ,\quad z\in\Bb{C}\setminus\Bb{R}_{\geqslant 1}\ ,\

it allows us to find out the boundary values of FF_{\!{{\mbox{\tiny{$\triangle$}}}}} on the both sides of the cut along \Bb{R}_{\geqslant 1} (see (olv, , p.​ 491, 19.7.3; p.​ 490, 19.5.1)),

F(x±i0)=1xF(1x)±ixF(11x),x\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x\!\pm\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)\!=\!\dfrac{1}{\sqrt{x}}\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{x}\right)\!\pm\!\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}{\sqrt{x}}\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1\!-\!\dfrac{1}{x}\right),\quad x\in\Bb{R}_{>1}\,.

Moreover, it also allows us to estimate FF_{\!{{\mbox{\tiny{$\triangle$}}}}} for x+iyx\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty\!\in\!\Bb{C}\setminus\Bb{R}_{\geqslant 1} as follows

|F(x+iy)|6+lnx2x1xχ(x)+χ[0,1)(x)ln211x+3+ln(1+|x|)1+|x|χ(x)\displaystyle\left|F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty)\right|\!\leq\!\dfrac{6\!+\!\ln\dfrac{x^{2}}{x\!-\!1}}{\sqrt{x}}\chi_{{{\mbox{\footnotesize{$\Bb{R}_{>1}$}}}}}(x)\!+\chi_{[0,1)}(x)\!\ln\dfrac{21}{1\!-\!x}\!+\!\dfrac{3\!+\!\ln\left(1\!+\!|x|\right)}{\sqrt{1+|x|}}\chi_{{{\mbox{\footnotesize{$\Bb{R}_{<0}$}}}}}(x)

(see (bh1, , p.​ 35, (A.9e))), and conclude that FF_{\!{{\mbox{\tiny{$\triangle$}}}}} belongs to the Hardy space H+p(H^{p}_{+}(\Bb{H}) for arbitrary 2<p<2<p<\infty. Then the Schwarz integral formula (see (ko, , p.​ 128))

F(z)=1π1F(11t)dt(tz)t,z\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)=\dfrac{1}{\pi}\int\nolimits_{1}^{\infty}F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1-\dfrac{1}{t}\right)\dfrac{{\mathrm{d}}t}{(t-z)\sqrt{t}}\ \ ,\quad z\in\Bb{C}\setminus\Bb{R}_{\geqslant 1}\,, (2.3)

can be applied to restore FF_{\!{{\mbox{\tiny{$\triangle$}}}}} on \Bb{C}\setminus\Bb{R}_{\geqslant 1} from the values of ImF{\rm{Im}}\,F_{\!{{\mbox{\tiny{$\triangle$}}}}} on \Bb{R} (see (bh2, , p.​ 604, (2.12))). The similar quadratic formula

F(z)2=2π1F(11t)F(1t)dt(tz)t,z\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}=\dfrac{2}{\pi}\int\nolimits_{1}^{\infty}F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1-\dfrac{1}{t}\right)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{t}\right)\dfrac{{\mathrm{d}}t}{(t-z)t}\ \ ,\quad z\in\Bb{C}\setminus\Bb{R}_{\geqslant 1}\,, (2.4)

also applies, because F2F_{\!{{\mbox{\tiny{$\triangle$}}}}}^{2} is in the Hardy space H+p(H^{p}_{+}(\Bb{H}) for all 1<p<1<p<\infty. For arbitrary z=x+iyz=x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty\in\Bb{C}\setminus\Bb{R}_{\geqslant 1} we use the relation

Imz(1z)1tz=ytx22x+1(1tx)2+t2y2,t[0,1],\displaystyle{\rm{Im}}\dfrac{z(1-z)}{1-tz}=y\dfrac{tx^{2}-2x+1}{(1-tx)^{2}+t^{2}y^{2}}\ ,\quad t\in[0,1]\,,\

to derive from (2.4) that

Im(z(1z)F(z)2)=2yπ01F(1t)F(t)(11+1tx)(1tx+1t)dt(1tx)2+t2y2,\displaystyle{\rm{Im}}\left(z(1\!-\!z)F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}\right)\!=\!\dfrac{2y}{\pi}\int\limits_{0}^{1}\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1\!-\!t\right)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(t\right)\left(\dfrac{1}{1\!+\!\sqrt{1\!-\!t}}\!-\!x\right)\left(1\!-\!tx\!+\!\sqrt{1\!-\!t}\right){\mathrm{d}}t}{(1-tx)^{2}+t^{2}y^{2}}\ ,

from which we conclude that

signIm(z(1z)F(z)2)=signIm(z),z\displaystyle{\rm{sign}}\,{\rm{Im}}\Big{(}z(1\!-\!z)F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}\Big{)}={\rm{sign}}\,{\rm{Im}}\left(z\right)\,,\quad z\in\Bb{C}_{{\rm{Re}}\,\leqslant 1/2}\,. (2.5)

It follows from (2.3) that FF_{\!{{\mbox{\tiny{$\triangle$}}}}} has property

ArgF(z)(π2,π2),ReF(z)>0,z\displaystyle{\rm{Arg}}\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\in\!\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\,,\ \ {{\rm{Re}}}\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\!>\!0\,,\quad z\!\in\!\Bb{C}\setminus\Bb{R}_{\geqslant 1}\,,

which reinforces (2.2), and allows us to form the logarithm of it and to obtain the integral representation

LogF(z)=1π201Log(11tz)t(1t)(F(t)2+F(1t)2)dt,z\displaystyle{\rm{Log}}\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)=\frac{1}{\pi^{2}}\!\int\nolimits_{0}^{1}\ \frac{{\rm{Log}}\,\left(\dfrac{1}{1-tz}\right)}{t(1-t)\left(\vphantom{\dfrac{a}{a}}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(t)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-t)^{2}\right)}\,{\mathrm{d}}t\,,\quad z\!\in\!\Bb{C}\setminus\Bb{R}_{\geqslant 1}\ ,\ (2.6)

(see (bh2, , p.​ 595, (1.15))), where

1π201dtt(1t)(F(t)2+F(1t)2)=12,\displaystyle\dfrac{1}{\pi^{2}}\int\nolimits_{0}^{1}\ \dfrac{{\rm{d}}t}{t(1-t)\left(\vphantom{\dfrac{a}{a}}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(t)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-t)^{2}\right)}=\dfrac{1}{2}\,,

because

ddxF(1x)F(x)=1πx(1x)F(x)2<0,x(0,1),\displaystyle\dfrac{{{\mathrm{d}}}}{{{\mathrm{d}}}x}\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-x)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)}=-\dfrac{1}{\pi x(1-x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)^{2}}<0\ ,\quad x\in(0,1)\ ,\ (2.7)

(see (bh2, , p.​ 606, (3.9))) implies

ddx(π2arctanF(1x)F(x))=1πx(1x)1F(x)2+F(1x)2,x(0,1).\displaystyle\dfrac{d}{dx}\left(\dfrac{\pi}{2}-\arctan\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1\!-\!x\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(x\right)}\right)=\dfrac{1}{\pi x(1-x)}\dfrac{1}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(x\right)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1-x\right)^{2}}\ ,\quad x\in(0,1)\,.

Here, in view of the Barnes expansion (see (bar, , p.​ 173), (whi, , p.​ 299)),

F(1z)=F(z)πLog16z2π2n=1Γ(n+1/2)2(n!)2(k=1n1(2k1)k)zn,\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)=\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}{\pi}{\rm{Log}}\,\dfrac{16}{z}-\dfrac{2}{\pi^{2}}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}\left(\sum\limits_{k=1}^{n}\dfrac{1}{(2k-1)k}\right)z^{n}\,,\ (2.8)

which is valid for z𝔻~z\in\Bb{D}\setminus(-1,0], we have (see (bh2, , p.​ 602, (2.5)))

F(1x)=1πln16x+O(xln1x),F(x)=1+O(x),x0,x(0,1).\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-x)\!=\!\dfrac{1}{\pi}\,\ln\dfrac{16}{x}\ +\!{\rm{O}}\Big{(}x\ln\dfrac{1}{x}\Big{)}\ ,\ \ F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)\!=\!1\!+{\rm{O}}(x)\ ,\ x\to 0,\,x\in(0,1)\,. (2.9)

2.2 . The Schwarz triangle function.

In 1873, Schwarz sch established the following fact (see (erd1, , p.​ 97)).

Theorem B

. The Schwarz triangle function

λ(z):=iF(1z)F(z),z(0,1)(),\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z):=\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\frac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}\,,\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), (2.10)

maps the set (0,1)()(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right) one-to-one onto the ideal hyperbolic quadrilateral

:={z𝕜ϝ𝕜ϝ𝕜𝕜ϝ𝕜}\displaystyle\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}:=\big{\{}\,z\in\Bb{H}\ \big{|}\,\ -1<{\rm{Re}}\,z<1\,,\,\ |2z-1|>1,\ |2z+1|>1\,\big{\}}\,, (2.11)

which will be referred to as Schwarz quadrilateral (see Fig. 1).

1-101/2-1/21/21/211i/2i/2\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}
Figure 1: . The Schwarz quadrilateral \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}.

Theorem B, (1.16), (2.9), (2.7) and the partial case

λ(x)=iF(1x)F(x)i\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(x)\!=\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\frac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-x)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)}\in{\rm{i}}\,\Bb{R}_{>0}\ ,\quad x\in(0,1)\ ,

of (2.10), imply

λ((0,1))=iλ\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\big{(}(0,1)\big{)}=\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0}\ ,\quad\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\big{(}1/2\big{)}=\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\ , (2.12)

and that the function λ(t)/i\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(t)/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt on the interval (0,1)(0,1) decreases from ++\infty to 0,

0<λ(t2)/i<λ(t1)/i<+,0<t1<t2<1.\displaystyle 0<\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(t_{2})/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt<\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(t_{1})/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt<+\infty\ ,\quad 0<t_{1}<t_{2}<1\,. (2.13)

To make other known properties of λ\lambda_{{{\mbox{\tiny{$\triangle$}}}}} obvious, it is necessary to use the integral representation (see (bh2, , p.​ 608, (3.13)))

Logλ(z)i=LogF(1z)F(z)=1π201Log1tz1t+tzt(1t)(F(t)2+F(1t)2)dt,\displaystyle{\rm{Log}}\,\dfrac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}={\rm{Log}}\,\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1\!-\!z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}\!=\!\dfrac{1}{\pi^{2}}\!\int\limits_{0}^{1}\!\!\dfrac{{\rm{Log}}\,\dfrac{1-tz}{1-t+tz}}{t(1-t)\left(F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(t\right)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1-t\right)^{2}\right)}\,{\mathrm{d}}t\ ,

which is immediate from (2.6) for every z(0,1)()z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right). This representation has the following convenient properties (see Section 10.1.1).

Lemma 2.1

. Denote Λ:=(0,1)()\Lambda:=(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right). The integrands in

ln|λ(z)|=ln|F(1z)||F(z)|=1π201ln|1tz||1t+tz|t(1t)(F(t)2+F(1t)2)dt,\displaystyle\ln\left|\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right|=\ln\dfrac{\left|F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1\!-\!z)\right|}{\left|F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\right|}\!=\!\dfrac{1}{\pi^{2}}\!\int\limits_{0}^{1}\!\!\dfrac{\ln\dfrac{|1-tz|}{|1-t+tz|}}{t(1-t)\left(F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(t\right)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1-t\right)^{2}\right)}\,{\mathrm{d}}t\ , (2.14)

and in

Argλ(z)i=ArgF(1z)F(z)=1π201Arg1tz1t+tzt(1t)(F(t)2+F(1t)2)dt\displaystyle{\rm{Arg}}\,\dfrac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}={\rm{Arg}}\,\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1\!-\!z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}\!=\!\dfrac{1}{\pi^{2}}\!\int\limits_{0}^{1}\!\!\dfrac{{\rm{Arg}}\,\dfrac{1-tz}{1-t+tz}}{t(1-t)\left(F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(t\right)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(1-t\right)^{2}\right)}\,{\mathrm{d}}t (2.15)

are of constant sign for each zΛz\in\Lambda. More precisely, if zΛz\in\Lambda then

|1tz||1t+tz|{>1,ifRez<1/2,=1,ifRez=1/2,<1,ifRez>1/2,Arg1tz1t+tz{(π,0),ifImz>0,=0,ifz(0,1),(0,π),ifImz<0,\displaystyle\dfrac{|1-tz|}{|1-t+tz|}\left\{\begin{array}[]{ll}>1\ ,&\hbox{if}\quad{\rm{Re}}\,z<1/2\,,\\ =1,&\hbox{if}\quad{\rm{Re}}\,z=1/2\,,\\ <1,&\hbox{if}\quad{\rm{Re}}\,z>1/2\,,\end{array}\right.\quad{\rm{Arg}}\,\dfrac{1-tz}{1-t+tz}\left\{\begin{array}[]{ll}\in(-\pi,0),&\hbox{if}\quad{\rm{Im}}\,z>0\,,\\ =0,&\hbox{if}\quad z\in(0,1)\,,\\ \in(0,\pi),&\hbox{if}\quad{\rm{Im}}\,z<0\,,\end{array}\right.

for arbitrary t(0,1)t\in(0,1).

The properties established in Lemma 2.1 allow us to see from (2.15) that

sign(Reλ(z))=sign(Imz),z(0,1)(),\displaystyle{\rm{sign}}\left({\rm{Re}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)={\rm{sign}}\left({\rm{Im}}\,z\right)\ ,\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\,, (2.16)

while (2.14) yields that |λ(z)|1|\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)|\lesseqqgtr 1 holds if and only of Rez1/2{\rm{Re}}\,z\lesseqqgtr 1/2, respectively, and according to Theorem B we obtain (see (bh1, , p.​ 41, (A11.i)))

(a)λ()=𝔻(b)λ()=𝔻¯,(c)λ((1/2)+i𝔻\displaystyle\begin{array}[]{l}{\rm{(a)}}\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\Bb{C}_{{\rm{Re}}>1/2}\setminus\Bb{R}_{\geqslant 1}\right)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\cap\Bb{D}\,,\\[2.84544pt] {\rm{(b)}}\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\Bb{C}_{{\rm{Re}}<1/2}\setminus\Bb{R}_{\leqslant 0}\right)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\,,\end{array}\qquad{\rm{(c)}}\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}\big{(}(1/2)+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}\big{)}=\Bb{H}\cap\partial\Bb{D}\,. (2.19)

2.3 . The elliptic modular function λ\lambda.

The function λ:(0,1)()\lambda\!:\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\to\!(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right) which is the inverse to λ\lambda_{{{\mbox{\tiny{$\triangle$}}}}}, i.e.,

(a)λ(λ(y))=y,y,(b)λ(λ(z))=z,z(0,1)(),\displaystyle{\rm{(a)}}\ \ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(\lambda(y))=y\,,\ \ y\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\ \ {\rm{(b)}}\ \ \lambda\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}=z\ ,\ \ z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), (2.20)

is called the (elliptic) modular function λ\lambda (see (erd1, , p.​ 99) and (olv, , p.​ 579)).

The modular function λ\lambda extends to a periodic nonvanishing holomorphic function in \Bb{H} with period 22 and (see (bh2, , p.​ 598, (1.29)))

λ(z)=Θ2(z)4/Θ3(z)40,λ(1/z)=1λ(z)=Θ4(z)4/Θ3(z)40,z\displaystyle\lambda(z)\!=\!{\Theta_{2}(z)^{4}}\big{/}{\Theta_{3}(z)^{4}}\!\neq\!0\,,\ \lambda(-1/z)\!=\!1\!-\!\lambda(z)\!=\!{\Theta_{4}(z)^{4}}\big{/}{\Theta_{3}(z)^{4}}\!\neq\!0\,,\ z\!\in\!\Bb{H}\,, (2.21)

where for zz\in\Bb{H} and u𝔻u\in\Bb{D} (see (bh2, , pp.​ 612–614, (6.1),(6.7),(6.8))),

Θ3(z)\displaystyle\Theta_{3}(z) :=θ3(eiπz),\displaystyle\!:=\!\theta_{3}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\,, Θ2(z)\displaystyle\Theta_{2}(z) :=2eiπz/4θ2(eiπz),\displaystyle\!:=\!2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/4}\theta_{2}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\,, Θ4(z)\displaystyle\Theta_{4}(z) :=θ4(eiπz),\displaystyle\!:=\!\theta_{4}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\,,
θ3(u)\displaystyle\theta_{3}(u) :=1+2n1un2,\displaystyle\!:=\!1\!+\!2\sum\limits_{n\geqslant 1}u^{n^{2}}\,, θ2(u)\displaystyle\theta_{2}(u) :=1+n1un2+n,\displaystyle\!:=\!1\!+\!\sum\limits_{n\geqslant 1}u^{n^{2}+n}\,, θ4(u)\displaystyle\theta_{4}(u) :=1+2n1(1)nun2.\displaystyle\!:=\!1\!+\!2\sum\limits_{n\geqslant 1}(-1)^{n}u^{n^{2}}\!.

Regarding these nonvanishing holomorphic functions in \Bb{H} and in 𝔻\Bb{D}, correspondingly (see (bh2, , p.​ 598, (1.26))), the main relationships between them can be written for arbitrary zz\in\Bb{H} as follows, by using the principal branch of the square root,

(a)Θ2(1/z)=(z/i)1/2Θ4(z),(b)Θ3(1/z)=(z/i)1/2Θ3(z),(c)Θ4(1/z)=(z/i)1/2Θ2(z),(d)Θ2(z+1)=eiπ/4Θ2(z),(e)Θ3(z+1)=Θ4(z),(g)Θ4(z+1)=Θ3(z),\displaystyle\begin{array}[]{llll}{\rm{(a)}}\ \Theta_{2}(-1/z)&=(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}\Theta_{4}(z)\,,&\quad{\rm{(b)}}\ \Theta_{3}(-1/z)&=(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}\Theta_{3}(z)\,,\\[2.84544pt] {\rm{(c)}}\ \Theta_{4}(-1/z)&=(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}\Theta_{2}(z)\,,&\quad{\rm{(d)}}\ \Theta_{2}(z+1)&={\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\Theta_{2}(z)\,,\\[2.84544pt] {\rm{(e)}}\ \Theta_{3}(z+1)&=\Theta_{4}(z)\,,&\quad{\rm{(g)}}\ \Theta_{4}(z+1)&=\Theta_{3}(z)\,,\end{array} (2.25)

(see (bh2, , p.​ 614, (6.8)), Re(z/i)>0{\rm{Re}}\,(z/i)>0). In addition, they are called the theta functions and meet the Jacobi identity

Θ3(z)4=Θ2(z)4+Θ4(z)4,zθθθ𝔻\displaystyle\Theta_{3}(z)^{4}\!=\!\Theta_{2}(z)^{4}\!+\!\Theta_{4}(z)^{4},\ z\!\in\!\Bb{H}\,;\ \ \theta_{3}(u)^{4}\!=\!16u\,\theta_{2}(u)^{4}\!+\!\theta_{3}(-u)^{4},\ u\!\in\!\Bb{D}\,, (2.26)

(see (bh2, , p.​ 599,​ (1.29))), which gives for u𝔻u\!\in\!\Bb{D} that (see (cha, , p.​ 157,​ (2.1)))

(a) 1+n=1r4(n)un:=θ3(u)4;(b) 1+n=1(1)nr4(n)un=θ4(u)4=θ3(u)4;(c)n=0r4(2n+1)u2n=8θ2(u)4.\displaystyle\begin{array}[]{lrcl}{\rm{(a)}}&\ \ 1+\sum\nolimits_{n=1}^{\infty}\ r_{4}(n)\,u^{n}&:=&\theta_{3}(u)^{4};\\[5.69046pt] {\rm{(b)}}&\ \ 1+\sum\nolimits_{n=1}^{\infty}(-1)^{n}r_{4}(n)u^{n}&=&\theta_{4}(u)^{4}=\theta_{3}(-u)^{4};\\[5.69046pt] {\rm{(c)}}&\ \ \sum\nolimits_{n=0}^{\infty}r_{4}(2n+1)u^{2n}&=&8\,\theta_{2}(u)^{4}.\end{array} (2.30)

It can be seen from (2.13), (2.21) and (2.41) that (see Section 10.1.1)

(a)λ(iλπyλπy\displaystyle{\rm{(a)}}\ \ \lambda\big{(}\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0}\big{)}\!=\!(0,1),\ \ \lambda\big{(}\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\big{)}\!=\!1/2,\ \ {\rm{(b)}}\ \ {}^{{{\mbox{\footnotesize{$-\pi y$}}}}}\!<\!\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty)\!<\!16\,{}^{{{\mbox{\footnotesize{$-\pi y$}}}}},\ \ 0\!<\!y\!<\!\infty\,, (2.31)

and that the function λ(it)\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt) on the interval (0,+)(0,+\!\infty) strictly decreases from 11 to 0,

λ(iy1)+λ(i/y1)=1,0<λ(iy2)<λ(iy1)<1,0<y1<y2<+.\displaystyle\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty_{1})+\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/y_{1})=1\ ,\quad 0<\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty_{2})<\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty_{1})<1\ ,\quad 0<y_{1}<y_{2}<+\infty\,. (2.32)

In addition, (2.19) and (2.16) can be written as follows ( γ(1,1):=𝔻\gamma(-1,1):=\Bb{H}\cap\partial\Bb{D} )

(a)λ(𝔻)=(b)λ(γ(1,1))=(1/2)+i(c)λ(𝔻¯)=(d)sign(Imλ(z))=sign(Rez),z.\displaystyle\begin{array}[]{ll}{\rm{(a)}}\ \lambda\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\cap\Bb{D}\right)\!=\!\Bb{C}_{{\rm{Re}}>1/2}\!\setminus\!\Bb{R}_{\geqslant 1}\,;&\ {\rm{(b)}}\ \lambda\big{(}\gamma(-1,1)\big{)}\!=\!(1/2)\!+\!i\Bb{R}\,;\\[2.84544pt] {\rm{(c)}}\ \lambda\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\right)\!=\!\Bb{C}_{{\rm{Re}}<1/2}\!\setminus\!\Bb{R}_{\leqslant 0}\,;&\ {\rm{(d)}}\ {\rm{sign}}\left({\rm{Im}}\,\lambda(z)\right)\!=\!{\rm{sign}}\left({\rm{Re}}\,z\right)\,,\ z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,.\end{array} (2.35)

The known relationships (see (bh2, , p.​ 598, (1.25); p.​ 599, (1.32)))4{}^{\ref*{case13}}

(a)Θ2(λ(z))=z1/4F(z)1/2,(b)Θ3(λ(z))=F(z)1/2,(c)Θ4(λ(z))=(1z)1/4F(z)1/2,z(0,1)(),\displaystyle\begin{array}[]{ll}{\rm{(a)}}\ \Theta_{2}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\!=\!z^{1/4}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}\!,&\qquad{\rm{(b)}}\ \Theta_{3}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\!=\!F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}\,,\\[2.84544pt] {\rm{(c)}}\ \Theta_{4}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\!=\!(1\!-\!z)^{1/4}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}\!,&\phantom{{\rm{(c)}}a}\qquad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right),\end{array}\ \ (2.38)

for the principal branches of the quadratic and of the fourth roots together with (2.20) and (2.21), show that the Schwarz triangle function λ\lambda_{{{\mbox{\tiny{$\triangle$}}}}} is a key that links the theta functions {Θk}k=24\{\Theta_{k}\}_{k=2}^{4} with the Gauss hypergeometric function FF_{\!{{\mbox{\tiny{$\triangle$}}}}}. This allows each property of FF_{\!{{\mbox{\tiny{$\triangle$}}}}} to be formulated as a property of theta functions and vice versa.

For instance, by virtue of (2.4), we have signImF(z)2=signImz{\rm{sign}}\,{\rm{Im}}\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}={\rm{sign}}\,{\rm{Im}}\,z for all zz\!\in\!\Bb{C}\!\setminus\!\Bb{R}_{\geqslant 1}, and as a consequence, if we take into account (2.16) and (2.38)(b), we get the result that

sign(ImΘ3(z)4)=sign(Rez),z,\displaystyle{\rm{sign}}\Big{(}{\rm{Im}}\,\Theta_{3}\big{(}z\big{)}^{4}\Big{)}={\rm{sign}}\left({\rm{Re}}\,z\right)\,,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,, (2.39)

which was established by the first and the second authors in Corollary 1.2 of (bh2, , p.​ 599). Our second example is the property

sign(Reλ(z))=sign(Rez),zclos(𝔻),\displaystyle{\rm{sign}}\big{(}{\rm{Re}}\,\lambda^{\,\prime}(z)\big{)}=-{\rm{sign}}\left({\rm{Re}}\,z\right)\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus{\rm{clos}}\left(\Bb{D}\right)\,, (2.40)

which follows from (2.5) and the identity

λ(z)=iπλ(z)(1λ(z))Θ3(z)4=iπΘ2(z)4Θ4(z)4Θ3(z)4,z\displaystyle\lambda^{\,\prime}(z)=\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi\,\lambda(z)\,\left(1-\lambda(z)\right)\,\Theta_{3}\left(z\right)^{4}={\rm{i}}\,\pi\,\dfrac{\Theta_{2}(z)^{4}\Theta_{4}(z)^{4}}{\Theta_{3}(z)^{4}}\ ,\ \ z\in\Bb{H} (2.41)

(see (bh2, , p.​ 599; (1.30))), by application to them (2.38), (2.19)(b) and (2.16), as with the help of (2.38), the identity (2.41) can be written in the form

λ(λ(z))iπ=Θ2(λ(z))4Θ4(λ(z))4Θ3(λ(z))4=z(1z)F(z)2,z(0,1)(),\displaystyle\dfrac{\lambda^{\,\prime}\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)}{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi}=\dfrac{\Theta_{2}\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)^{4}\Theta_{4}\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)^{4}}{\Theta_{3}\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)^{4}}=z(1\!-\!z)F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2},\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right),

and then, in a second step, using (2.5), (2.16) and (2.19)(b), we obtain

signReλ(λ(z))=signImλ(λ(z))iπ=signIm(z(1z)F(z)2)=\displaystyle-{\rm{sign}}\,{\rm{Re}}\,\lambda^{\,\prime}\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)={\rm{sign}}\,{\rm{Im}}\dfrac{\lambda^{\,\prime}\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)}{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi}={\rm{sign}}\,{\rm{Im}}\big{(}z(1\!-\!z)F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}\big{)}=
=signIm(z)=sign(Reλ(z)),λ(z)λ()=𝔻¯.\displaystyle={\rm{sign}}\,{\rm{Im}}\left(z\right)={\rm{sign}}\left({\rm{Re}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right),\ \ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\in\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\Bb{C}_{{\rm{Re}}<1/2}\setminus\Bb{R}_{\leqslant 0}\right)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\,.

As for conclusions going in the opposite direction, here we can mention that the combination of (2.38)(b), (2.20), (2.21) and the Landen transformation equations

(a) 2Θ2(2z)2=Θ3(z)2Θ4(z)2,(b) 2Θ3(2z)2=Θ3(z)2+Θ4(z)2,(c)Θ4(2z)2=Θ3(z)Θ4(z),z\displaystyle\begin{array}[]{ll}{\rm{(a)}}\ 2\Theta_{2}(2z)^{2}\!=\!\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2},&\quad{\rm{(b)}}\ 2\Theta_{3}(2z)^{2}\!=\!\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2},\\[5.69046pt] {\rm{(c)}}\ \Theta_{4}(2z)^{2}\!=\!\Theta_{3}(z)\Theta_{4}(z),&\quad\phantom{{\rm{(c)}}a}z\in\Bb{H}\,,\end{array} (2.44)

(see (law, , p.​ 18)5{}^{\ref*{case38}}), gives6{}^{\ref*{case1}} the quadratic transformation relation (3.1.10) with a=b=1a\!=\!b\!=\!1 of (and, , p.​ 128) for the hypergeometric function FF_{\!{{\mbox{\tiny{$\triangle$}}}}} and equality (1.21) as well. At the same time, each of the four nontrivial functional relationships for the modular function λ\lambda in the table of (cha, , p.​ 111) (except for the first and third, considered as trivial) can be written as the corresponding Kummer transformation rule for FF_{\!{{\mbox{\tiny{$\triangle$}}}}} (see (erd1, , p.​ 106)). For instance, in (bh1, , p.​ 33) the Kummer identity (27) of (erd1, , p.​ 106)) was derived from (3.38). It also follows7{}^{\ref*{case2}} from (2.44) that

λ(1+z1z)=12+iλ1(z)4λ2(z),\displaystyle\lambda\left(\dfrac{1+z}{1-z}\right)=\dfrac{1}{2}+\dfrac{i\lambda_{1}(z)}{4\sqrt{\lambda_{2}(z)}}\,, λ(z1z+1)=12iλ1(z)4λ2(z),\displaystyle\lambda\left(\dfrac{z-1}{z+1}\right)=\dfrac{1}{2}-\dfrac{i\lambda_{1}(z)}{4\sqrt{\lambda_{2}(z)}}\,, z,\displaystyle z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,, (2.45)
λ2(z):=λ(z)(1λ(z)),\displaystyle\lambda_{2}(z):=\lambda(z)\big{(}1-\lambda(z)\big{)}\,, λ1(z):=12λ(z),\displaystyle\lambda_{1}(z):=1-2\lambda(z)\,, z\displaystyle z\in\Bb{H}\,, (2.46)
λ2(1/z)=λ2(z),\displaystyle\lambda_{2}(-1/z)\!=\!\lambda_{2}(z)\,, λ1(1/z)=λ1(z),\displaystyle\lambda_{1}(-1/z)\!=\!-\!\lambda_{1}(z)\,, z\displaystyle z\in\Bb{H}\,.

Observe that the principal branch of the square root can be used in (2.45) because λ()=(0,1)()\lambda(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})=(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right) and the two inclusions zz\!\in\!\Bb{C}, z(1z)z(1-z)\!\in\!\Bb{R}_{<0} hold if and only if zz\!\in\!\Bb{R}_{<0}\!\cup\Bb{R}_{>1}. A combination of (2.31)(b) and (2.45) gives that8{}^{\ref*{case4}}

64|λ(t+iti)|=16/λ(it)λ(i/t)expπ2(t+1t),t>0,\displaystyle 64\left|\lambda\left(\dfrac{t+i}{t-i}\right)\right|=16\,\big{/}\!\sqrt{\lambda(it)\lambda(i/t)}\geqslant\exp\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)\ ,\quad t>0\,, (2.47)

while the relations (2.44) and (2.25) imply9{}^{\ref*{case3}} that

|Θ3(t+iti)|45(t+1t)exp(π4(t+1t)),t>0.\displaystyle\left|\Theta_{3}\left(\dfrac{t+i}{t-i}\right)\right|^{4}\leqslant 5\left(t+\dfrac{1}{t}\right)\exp\left(-\dfrac{\pi}{4}\left(t+\dfrac{1}{t}\right)\right)\ ,\quad t>0\,. (2.48)

The relation

iπλ(z)F(z)2=1z(1z),z(0,1)(),\displaystyle\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(z)\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}=\frac{1}{z(1-z)}\ ,\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), (2.49)

(see (bh2, , p.​ 597, (1.20))) allow us in Section 10.1.2 to show that the integral formulas of Theorem 1.1 for every xx\!\in\!\Bb{R} can be written as follows,

H0(x)\displaystyle\eurm{H}_{0}(x) =12πi1/2i1/2+iλ(y)λ(y)F(y)2x2λ(y)2𝑑y,\displaystyle=\dfrac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\!\!\!\!\int\limits_{1/2-{\rm{i}}\infty}^{1/2+{\rm{i}}\infty}\frac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(y)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(y)\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(y)^{2}}{x^{2}-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(y)^{2}}dy\,, (2.50)
Hn(x)\displaystyle\eurm{H}_{n}(x) =(1)4π2n1/2i1/2+iSn(1/y)λ(y)dy(x+λ(y))2,n1.\displaystyle=\dfrac{(-1)}{4\pi^{2}n}\!\!\!\!\int\limits_{1/2-{\rm{i}}\infty}^{1/2+{\rm{i}}\infty}\ \frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left({1}\big{/}{y}\right)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(y)dy}{\big{(}x+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(y)\big{)}^{2}}\,,\quad n\geqslant 1\,. (2.51)

In view of (2.38)(b), we can use (2.35) and Theorem B to change the variable z=λ(y)z=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(y) in the integrals (2.50), (2.51) and for arbitrary xx\!\in\!\Bb{R} obtain (see (3.14))

H0(x)=12πiγ(1,1)zΘ3(z)4x2z2dz,Hn(x)=(1)4π2nγ(1,1)Sn(1λ(z))dz(x+z)2,n1.\displaystyle\eurm{H}_{0}(x)\!=\dfrac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\frac{z\,\Theta_{3}\left(z\right)^{4}}{x^{2}-z^{2}}dz\,,\ \ \eurm{H}_{n}(x)\!=\dfrac{(-1)}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(z)}\right)dz}{(x+z)^{2}}\,,\quad n\geqslant 1\,. (2.52)

Here, the integrals are absolutely convergent for all xx\in\Bb{R} as follows from (3.14), (2.48), (2.47) and the parametrization γ(1,1)z=(t+i)/(ti)\gamma(-1,1)\ni z=(t+i)/(t-i), tt\in\Bb{R}_{>0}.

2.4 . Monotonicity properties of the modular function λ\lambda.

The equality (2.41) together with (2.39), the identities

ddxlog(|λ(iy+x)||1λ(iy+x)|)=πIm(Θ34(iy+x)2Θ34(iy+x1)),ddxlog|λ(iy+x)|=πImΘ3(iy+x1)4,x+iy\displaystyle\begin{array}[]{l}\displaystyle\dfrac{d}{dx}\log\left(\left|\lambda(iy\!+\!x)\right|\left|1\!-\!\lambda(iy\!+\!x)\right|\right)\!=\!\pi\,{\rm{Im}}\,\left(\Theta_{3}^{4}(iy\!+\!x)\!-\!2\Theta_{3}^{4}(iy\!+\!x\!-\!1)\right)\,,\\[5.69046pt] \displaystyle\dfrac{d}{dx}\log\left|\lambda(iy\!+\!x)\right|=-\pi\,{\rm{Im}}\,\Theta_{3}(iy\!+\!x\!-\!1)^{4}\,,\quad x+{\rm{i}}y\in\Bb{H}\,,\end{array} (2.55)

and (2.40) allow us to obtain in Section 10.1.3 the following monotonicity properties of λ\lambda and the estimates of Corollary 2.3 below as well.

Theorem 2.2

. Let a1/2a\geqslant 1/2 and λ2(z):=λ(z)(1λ(z))\lambda_{2}(z):=\lambda(z)\big{(}1-\lambda(z)\big{)}. Then

xddx|λ(x+iy)|>0,x+iy(i),\displaystyle x\dfrac{d}{dx}\ \left|\lambda(x+{\rm{i}}y)\right|\ \,>0,\hskip 10.52737ptx\!+\!iy\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\setminus\!\left({\rm{i}}\Bb{R}_{>0}\right)\!, (2.56)
xddx|λ2(x+iy)|>0,x+iy(i),\displaystyle x\dfrac{d}{dx}\,\left|\lambda_{2}(x+{\rm{i}}y)\right|\,>0,\hskip 10.52737ptx\!+\!iy\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\setminus\!\left({\rm{i}}\Bb{R}_{>0}\right)\!, (2.57)
xddxReλ(x+iy)<0,x+iy(𝔻),\displaystyle x\dfrac{d}{dx}\ {\rm{Re}}\,\lambda(x\!+\!iy)\,<0,\hskip 11.38092ptx\!+\!iy\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\setminus\!\left(\Bb{D}\cup{\rm{i}}\Bb{R}_{>0}\right), (2.58)
maxImza|λ(z)|=|λ(1+ia)|=λ(ia)1λ(ia),\displaystyle\max_{{{\mbox{\footnotesize{${\rm{Im}}z\!\geqslant\!a$}}}}}\left|\lambda(z)\right|\ =\left|\lambda(1+{\rm{i}}a)\right|\ =\frac{\lambda(ia)}{1-\lambda(ia)}\ \,, (2.59)
maxImza|λ2(z)|=|λ2(1+ia)|=λ(ia)(1λ(ia))2.\displaystyle\max_{{{\mbox{\footnotesize{${\rm{Im}}z\!\geqslant\!a$}}}}}\left|\lambda_{2}(z)\right|=\left|\lambda_{2}(1+{\rm{i}}a)\right|=\frac{\lambda(ia)}{\left(1-\lambda(ia)\right)^{2}}\ \,. (2.60)

Since 0<λ(iy)<1/2=λ(i)0\!<\!\lambda(iy)\!<\!1/2\!=\!\lambda(i) holds for y>1y\!>\!1, the next properties can be derived from Theorem 2.2, (2.41) and λ(γ(1,1))=(1/2)+i\lambda\big{(}\gamma(-1,1)\big{)}\!=\!(1/2)\!+\!{\rm{i}}\,\Bb{R} (see (2.31)–(2.35)).

Corrollary 2.3

. Suppose t>0t>0, x[1,1]x\in[-1,1], y>1y>1 and put

0:=1/,:=m𝔻¯𝔽𝔻𝔼\displaystyle\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}:=-1/\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}},\quad\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}:=\Bb{H}\setminus\underset{{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}{\cup}\left(2m+\overline{\Bb{D}}\right),\quad\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\cap\Bb{D}\subset\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}.

Then λ(iy)(0,1/2)\lambda(iy)\in(0,1/2) and

(a)λ(x+iy)λλλ𝔻(b)|λ(x+iy)|9,(c)4|λ(z)λ(x+iy)||λ(z)||12λ(iy)|,z0,(d)2|λ(z)λ(1+it)||λ(z)|+|λ(1+it)|,z0.\displaystyle\begin{array}[]{ll}\displaystyle{\rm{(a)}}\ \lambda(x\!+\!iy)\in\Bb{C}_{{\rm{Re}}\leqslant\lambda(iy)}\bigcap\frac{\lambda(iy)}{1-\lambda(iy)}\,{\rm{clos}}\left(\Bb{D}\right),&\ {\rm{(b)}}\ \left|\lambda^{\,\prime}(x\!+\!iy)\right|\!\leqslant\!9\ ,\\[14.22636pt] {\rm{(c)}}\ \hskip 8.5359pt4\left|\lambda(z)-\lambda(x+iy)\right|\geqslant\left|\lambda(z)\right|\,\left|1-2\lambda(iy)\right|\,,&\ z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\\[7.11317pt] {\rm{(d)}}\ \sqrt{2}\,\left|\lambda(z)-\lambda(1+it)\right|\geqslant\left|\lambda(z)\right|+\left|\lambda(1+it)\right|\,,&\ z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\,.\end{array} (2.64)

2.5 . Imaginary part of the Schwarz triangle function.

We note that λ\lambda_{{{\mbox{\tiny{$\triangle$}}}}} enjoys the functional relation

λ(z)λ(1z)=1,z(0,1)(),\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1-z)=-1,\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\,, (2.65)

(see (bh2, , p.​ 597; (1.21))) and for every x{}x\!\in\!\Bb{R}\setminus\!\{0,1\} there exist λ(x±i0):=limzϝλ(x±z)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\!(x\pm\!\mathrm{i}\hskip 0.42677pt0):=\lim_{\,z\in\Bb{H},\,z\to 0}\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(x\pm z) (see (bh2, , p.​ 604)). Moreover, there are relationships between the values of λ\lambda_{{{\mbox{\tiny{$\triangle$}}}}} on the two sides of the cut along \Bb{R}_{<0}:

λ(x+i0)=2+λ(xi0),x>0,\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=2+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)\ ,\quad x>0\,, (2.66)

(see (bh2, , p.​ 597; (1.22))), and along the other cut \Bb{R}_{>1} as well,

1λ(1+x+i0)=2+1λ(1+xi0),x>0,\displaystyle\dfrac{1}{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1+x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)}=2+\dfrac{1}{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1+x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)}\ ,\quad x>0\,, (2.67)

(see (bh2, , p.​ 597; (1.23))). It was also explained in (bh2, , p.​ 601; (4.1)) that the Pfaff formula (see (and, , p.​ 79)) gives that

λ(x±i0)=±1+iΔ(x),x>0,\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x\pm\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=\pm 1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)\,,\ \ x>0\,, (2.68)

where for x>0x>0 we write (see (bh2, , p.​ 599))

Δ(x):=F(1/(1+x))F(x/(1+x)),{Δ(0)=+,Δ(+)=0,dΔ(x)dx<0,Δ(x)Δ(1/x)=1.\displaystyle{{\mbox{\footnotesize{$\Delta$}}}}(x)\!:=\!\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({1}\big{/}{(1\!+\!x)}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({x}\big{/}{(1\!+\!x)}\right)}\ ,\quad\left\{\begin{array}[]{l}{{\mbox{\footnotesize{$\Delta$}}}}(0)\!=\!+\!\infty\ ,\\[5.69046pt] {{\mbox{\footnotesize{$\Delta$}}}}(+\infty)\!=\!0\ ,\end{array}\right.\ \ \ \dfrac{{{\mathrm{d}}}{{\mbox{\footnotesize{$\Delta$}}}}(x)}{{{\mathrm{d}}}x}\!<\!0,\ \ {{\mbox{\footnotesize{$\Delta$}}}}(x){{\mbox{\footnotesize{$\Delta$}}}}(1/x)\!=\!1. (2.71)

Then, by (2.65),

λ(1+x±i0)=1λ(xi0)=11+iΔ(x)=±1+iΔ(x)1+Δ(x)2,\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1+x\pm\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=-\frac{1}{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x\mp\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)}=-\frac{1}{\mp 1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}=\dfrac{\pm 1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\ , (2.72)

i.e., we have

λ(1+x+i0)=1+iΔ(x)1+Δ(x)2,λ(1+xi0)=1+iΔ(x)1+Δ(x)2,x>0.\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1+x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=\frac{1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\,,\ \ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1+x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=\frac{-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\,,\ \ x>0\,. (2.73)

Together with (2.68) this shows that the values of Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}} on the two sides of the cuts at \Bb{R}_{<0} and at \Bb{R}_{>1} coincide. Hence, we extend the function Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}, initially defined on (0,1)((0,1)\cup(\Bb{C}\setminus\Bb{R}), to \Bb{R}_{<0}\cup\Bb{R}_{>1} declaring its values to be given by

Imλ(1+x):=Imλ(1+x+i0)=Imλ(1+xi0)=Δ(x)1+Δ(x)2,Imλ(x):=Imλ(x+i0)=Imλ(xi0)=Δ(x),x>0.\displaystyle\begin{array}[]{lccclcl}{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x)&\!:=\!&{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)&\!=\!&{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)&\!=\!&\dfrac{{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1\!+\!{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\ ,\\[8.5359pt] {\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x)&\!:=\!&{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)&\!=\!&{\rm{Im}}\,\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)&\!=\!&{{\mbox{\footnotesize{$\Delta$}}}}(x)\ ,\ \ x>0\,.\end{array} (2.76)

We find that the resulting function Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}} is continuous on {}\Bb{C}\!\setminus\!\{0,1\}. Taking the limit in the relation (2.20)(b) as z(0,1)(),zz\in(0,1)\!\cup\!\left(\Bb{C}\!\setminus\!\Bb{R}\right),z\!\to\!\Bb{R}_{<0}\!\cup\Bb{R}_{>1}, we obtain from (2.68) and (2.72) that

{λ(λ(x±i0))=λ(±1+iΔ(x))=x,λ(λ(1+x±i0))=λ( 1/(±1iΔ(x)))=1+x,x>0,\displaystyle\left\{\begin{array}[]{rcccl}\lambda\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x\pm\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)\big{)}&=&\lambda\left(\pm 1+i{{\mbox{\footnotesize{$\Delta$}}}}(x)\right)&=&-x\,,\\[2.84544pt] \lambda\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1+x\pm\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)\big{)}&=&\lambda\left(\,1\big{/}\big{(}\pm 1-i{{\mbox{\footnotesize{$\Delta$}}}}(x)\hskip 0.42677pt\big{)}\hskip 0.42677pt\right)&=&1+x\,,\end{array}\right.\qquad x>0\,, (2.79)

where Δ\Delta maps \Bb{R}_{>0} onto itself in a one-to-one fashion, by (2.71). Together with Theorem B this means the following (see (2.1)).

Lemma 2.4

. The modular function λ\lambda maps each of the sets γ(1,)=1+i\gamma(1,\infty)=1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0} and γ(1,)=1+i\gamma(-1,\infty)=-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0} one-to-one onto \Bb{R}_{<0} and each of the set
γ(0,1)=1/(1i=1/2+(1/2)(𝔻γ(1,0)=1/(1i=1/2+(1/2)(𝔻\begin{array}[]{lcccr}\hskip 54.06006pt\gamma(0,1)&=&1/(1-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0})&=&1/2+(1/2)(\Bb{H}\cap\partial\Bb{D}),\\ \hskip 54.06006pt\gamma(-1,0)&=&1/(-1-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0})&=&-1/2+(1/2)(\Bb{H}\cap\partial\Bb{D})\end{array}

one-to-one onto \Bb{R}_{>1}. As a consequence, λ\lambda maps each of the sets

γ(1,)γ(0,1),γ(1,)γ(0,1),γ(1,)γ(1,0),γ(1,)γ(1,0),\displaystyle\begin{array}[]{ll}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\gamma(1,\infty)\sqcup\gamma(0,1)\ ,&\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\gamma(-1,\infty)\sqcup\gamma(0,1)\ ,\\[2.84544pt] \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\gamma(1,\infty)\sqcup\gamma(-1,0)\ ,&\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\gamma(-1,\infty)\sqcup\gamma(-1,0)\ ,\end{array} (2.82)

one-to-one onto {}\Bb{C}\setminus\{0,1\}.

By using the property Δ(x)+{{\mbox{\footnotesize{$\Delta$}}}}(x)\!\to\!+\infty as x>0x\!>\!0, x0x\!\to\!0, which comes from (2.71), we obtain from (2.76) that Imλ(x)+{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x)\!\to\!+\infty and Imλ(1+x)0{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x)\!\to\!0 as x>0x\!>\!0, x0x\!\to\!0, which together with the properties

limzΛ,z1λ(z)=0,limzΛ,z0Imλ(z)=+,Λ:=(0,1)(),\displaystyle\lim\nolimits_{{{\mbox{\footnotesize{$z\!\in\!\Lambda,z\!\to\!1$}}}}}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)=0\,,\ \lim\nolimits_{{{\mbox{\footnotesize{$z\!\in\!\Lambda,z\!\to\!0$}}}}}{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)=+\infty\,,\ \Lambda\!:=\!(0,1)\!\cup\!\left(\Bb{C}\!\setminus\!\Bb{R}\right),

(see (bh2, , p.​ 609; Lemmas 4.1, 4.2)) means that the function Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}, which is continuous on {}\Bb{C}\setminus\{0,1\}, can be continuously extended to the point 11 as well, with value Imλ(1)=0{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1)=0, while limz 0Imλ(z)=+\lim_{z\to\,0}{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)=+\infty. As a result, we get the following statement.

Lemma 2.5

. Let Δ\Delta be defined as in (2.71). The harmonic function Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}} on (0,1)()(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right) can be continuously extended to {}\Bb{C}\setminus\{0\} with values on \Bb{R}_{<0}\!\cup\Bb{R}_{\geqslant 1} given by (2.76) and Imλ(1):=0{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1)\!:=\!0. The extended function Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}} is positive on {}\Bb{C}\setminus\{0,1\} and satisfies

(a)limz 0Imλ(z)=+,(b)Imλ(λ(y))=Imy,y𝔽\displaystyle{\rm{(a)}}\ \lim_{z\to\,0}{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)=+\infty\,,\ \ {\rm{(b)}}\ {\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\big{(}\lambda(y)\big{)}\!=\!{\rm{Im}}\,y\,,\ \ y\!\in\!\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right). (2.83)

It follows from Theorem B, (2.79) and (3.39) that for every aa\in\Bb{R}_{<0}\cup\Bb{R}_{>1} the equation λ(z)=a\lambda(z)=a with z𝔽z\in\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right) has exactly two solutions z=λ(a±i0)z=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\pm{\rm{i}}0) with equal imaginary parts, by virtue of (2.76), while for every a(0,1)()a\in(0,1)\!\cup\!\left(\Bb{C}\!\setminus\!\Bb{R}\right) it has a unique solution given by z=λ(a)z=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a). In fact, Lemma 5.7 below implies the following stronger property, which we obtain later on in Section 10.1.4.

Theorem 2.6

. For every a{}a\in\Bb{C}\setminus\{0,1\} there exists a finite maximum

max{Imzλ(z)=a,z𝕜𝕜},𝕜𝕜{ϝϝ^~}\displaystyle\max\left\{\,{\rm{Im}}\,z\,\mid\,\lambda(z)=a\ ,\ z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,\right\},\ \ \Bb{H}_{|{\rm{Re}}|\leqslant 1}:=\left\{\,z\in\Bb{H}\,\mid\,{\rm{Re}}\,z\in[-1,1]\,\vphantom{\Bb{H}_{|{\rm{Re}}|\leqslant 1}}\right\},

attained at one or two points, which belong to 𝔽\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right). There is one such point λ(a)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a) if a(0,1)()a\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right) and two points λ(a±i0)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\pm{\rm{i}}0) if a(,0)(1,+)a\in(-\infty,0)\cup(1,+\infty). In particular

z{}𝕜𝕜𝔽λϝ𝕀𝕀λϝ\displaystyle z\in\Bb{C}\setminus\{0,1\},\ y\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),\ \lambda(y)=z\ \Rightarrow\ {\rm{Im}}\,y<{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\,, (2.84)

where Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}} is as in Lemma 2.5.

Corrollary 2.7

. For all pairs (y,z)(y,z) with y𝔽y\in\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right), z𝕜𝕜z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}, and Imz>Imy{\rm{Im}}\,z>{\rm{Im}}\,y, we have λ(z)λ(y)0\lambda(z)-\lambda(y)\neq 0.

3 . Schwarz triangle polynomials

We begin this section by recalling the following property of periodic functions.10{}^{\ref*{case711}}

Lemma 3.1

. Let aa\in\Bb{R} and f0Hol(ia+f_{0}\!\in\!{\rm{Hol}}({\rm{i}}a+\Bb{H}) be periodic with period 22. Then

z1z1+2f0(ζ)𝑑ζ=z2z2+2f0(ζ)𝑑ζ,z1,z2ia+\displaystyle\int_{z_{1}}^{z_{1}+2}f_{0}(\zeta)d\zeta=\int_{z_{2}}^{z_{2}+2}f_{0}(\zeta)d\zeta\ ,\quad z_{1},z_{2}\in{\rm{i}}a+\Bb{H}\,. (3.1)

If φ:𝔻{}\varphi:\Bb{D}\setminus\{0\}\mapsto\Bb{C} is such that φ(exp(iπz))Hol(\varphi(\exp({\rm{i}}\pi z))\in{\rm{Hol}}(\Bb{H}) then φHol(𝔻{}\varphi\in{\rm{Hol}}(\Bb{D}\setminus\{0\}).

3.1 . Definition and connection with Faber polynomials.

The notation (2.10) permits us for arbitrary positive integer nn to write the decomposition (1.17) in the form

enπiλ(z)=Sn(1/z)+ΔnS(z),z𝔻{}𝕊Δ𝕊𝔻\displaystyle{\rm{e}}^{{{\mbox{\footnotesize{$-n\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)$}}}}}\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)\!+\!\Delta_{n}^{S}(z)\ ,\ z\!\in\!\Bb{D}\setminus\{0\}\ ,\ S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(0)\!=\!0\,,\ \Delta_{n}^{S}\!\in\!{\rm{Hol}}(\Bb{D})\,, (3.2)

and we will call SnS^{{{\mbox{\tiny{$\triangle$}}}}}_{n} the nn-th Schwarz triangle polynomial.

A special symmetrization of algebraic polynomials occurs when we consider the expression Sn(1/λ(z))±Sn(1/λ(1/z))S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left({1}/{\lambda(z)}\right)\pm S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left({1}/{\lambda(-1/z)}\right) to make the function Sn(1/λ(z))S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left({1}/{\lambda(z)}\right) in (2.52) invariant up to a multiplier ±1\pm 1 with respect to the argument reversal z1/zz\mapsto-1/z. In view of (2.21), we have λ(1/z)=1λ(z)\lambda(-1/z)\!=\!1\!-\!\lambda(z), which suggests the following algebraic operations (see Section 10.2.1).

Lemma 3.2

. Let nn be a positive integer and PnP_{n} be an algebraic polynomial of degree nn with real coefficients. Then there exist two algebraic polynomials ±[𝒫𝓃]\eusm{R}^{\pm}\left[P_{n}\right] of degree nn with real coefficients such that for every x{}x\in\Bb{C}\setminus\{0,1\}, we have

Pn(1x)+Pn(11x)=+[𝒫𝓃](1𝓍(1𝓍)),\displaystyle P_{n}\left(\frac{1}{x}\right)+P_{n}\left(\frac{1}{1-x}\right)=\eusm{R}^{+}\!\!\left[P_{n}\right]\left(\frac{1}{x(1-x)}\right), +[𝒫𝓃](0)=2𝒫𝓃(0),\displaystyle\eusm{R}^{+}\!\left[P_{n}\right](0)=2P_{n}(0)\,,\
Pn(1x)Pn(11x)=(12x)[𝒫𝓃](1𝓍(1𝓍)),\displaystyle P_{n}\left(\frac{1}{x}\right)-P_{n}\left(\frac{1}{1-x}\right)=(1-2x)\,\eusm{R}^{-}\!\left[P_{n}\right]\left(\frac{1}{x(1-x)}\right)\,, [𝒫𝓃](0)=0.\displaystyle\eusm{R}^{-}\!\left[P_{n}\right](0)=0.

For each n1n\!\geqslant\!1, the symmetrized functions

Rn±(z):=Sn(1λ(z))±Sn(11λ(z)),Rn±(1/z)=±Rn±(z),z\displaystyle\eurm{R}^{\pm}_{n}(z)\!:=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\frac{1}{\lambda(z)}\right)\!\pm\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\frac{1}{1\!-\!\lambda(z)}\right),\ \eurm{R}^{\pm}_{n}(-1/z)\!=\!\pm\eurm{R}^{\pm}_{n}(z),\ \ z\!\in\!\Bb{H}, (3.3)

are holomorphic in \Bb{H} and 22-periodic. They were considered for the first time by the fourth and fifth authors rad in the context of Fourier pair interpolation on the real line.

In accordance with (2.21), for zz\in\Bb{H} and u𝔻u\in\Bb{D} we introduce the notation

λ(z)=16eiπzθ2(eiπz)4θ3(eiπz)4=λ𝔻(eiπz),λ𝔻(u):=16u(1+n1un2+n)4(1+2n1un2)4.\displaystyle\lambda(z)\!=\!\dfrac{16\hskip 0.56917pt\mathrm{e}\hskip 0.42677pt^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\theta_{2}\left({\hskip 0.56917pt\mathrm{e}\hskip 0.42677pt}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)^{4}}{\theta_{3}\left({\hskip 0.56917pt\mathrm{e}\hskip 0.42677pt}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)^{4}}\!=\!\lambda_{\Bb{D}}\left({\hskip 0.56917pt\mathrm{e}\hskip 0.42677pt}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\,,\ \ \lambda_{\Bb{D}}(u)\!:=\!\dfrac{16u\left(1\!+\!\sum\nolimits_{n\geqslant 1}u^{n^{2}+n}\right)^{4}}{\left(1\!+\!2\sum\nolimits_{n\geqslant 1}u^{n^{2}}\right)^{4}}. (3.4)

Then clearly, we have λ𝔻Hol(𝔻\lambda_{\Bb{D}}\in{\rm{Hol}}(\Bb{D}), λ𝔻(0)=0\lambda_{\Bb{D}}(0)=0, λ𝔻(0)=16\lambda^{\,\prime}_{\Bb{D}}(0)=16. As we substitute λ(z)\lambda(z) for zz in (3.2) and apply the left-hand side identity in (2.20), we find

ΔnS(λ𝔻(eiπz))=einπzSn(1λ(z)),zλ(𝔻~𝔽\displaystyle\Delta_{n}^{S}\left(\lambda_{\Bb{D}}\left({\rm{e}}^{i\pi z}\right)\right)\!=\!{\rm{e}}^{{{\mbox{\footnotesize{$-in\pi z$}}}}}\!-\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\frac{1}{\lambda(z)}\!\right),\ \ z\!\in\!\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(\Bb{D}\setminus(-1,0])\subset\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\ n\!\geqslant\!1\,. (3.5)

In view of (2.21), λ\lambda, 1/λHol(1/\lambda\in{\rm{Hol}}(\Bb{H}) and hence the right-hand side function in (3.5) is holomorphic on \Bb{H}. In view of Lemma 3.1 applied to the left-hand side function in (3.5) we see that ΔnS(λ𝔻)Hol(𝔻{}\Delta_{n}^{S}(\lambda_{\Bb{D}})\in{\rm{Hol}}(\Bb{D}\setminus\{0\}). But since ΔnS,λ𝔻Hol(𝔻\Delta_{n}^{S},\lambda_{\Bb{D}}\!\in\!{\rm{Hol}}(\Bb{D}) with λ𝔻(0)=0\lambda_{\Bb{D}}(0)=0, the composition ΔnS(λ𝔻)\Delta_{n}^{S}(\lambda_{\Bb{D}}) is holomorphic in a neighborhood of the origin, so that in fact the function ΔnSHol(𝔻\Delta_{n}^{S}\!\in\!{\rm{Hol}}(\Bb{D}) of (3.2) meets

ΔnS(λ𝔻)Hol(𝔻\displaystyle\Delta_{n}^{S}(\lambda_{\Bb{D}})\in{\rm{Hol}}(\Bb{D})\,,\quad n\geqslant 1\,,

and, in view of the uniqueness theorem for analytic functions (see (con, , p.​ 78)), the relationship (3.5) can be written as

Sn(1/λ𝔻(u))=un+ΔnS(λ𝔻(u)),u𝔻{}Δλ𝔻𝔻\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left({1}/{\lambda_{\Bb{D}}(u)}\right)\!=\!u^{-n}\!+\!\Delta_{n}^{S}\left(\lambda_{\Bb{D}}\left(u\right)\right)\,,\ u\!\in\!\Bb{D}\setminus\{0\}\,,\ \Delta_{n}(\lambda_{\Bb{D}})\!\in\!{\rm{Hol}}(\Bb{D})\,,\ n\!\geqslant\!1\,,

where Sn(0)=0S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(0)=0. According to the definition of Issai Schur (hur, , p.​ 34), if FnF_{n} is the nn-th Faber polynomial of 16/λ𝔻(1/u)16/\lambda_{\Bb{D}}(1/u) then Sn(x)=Fn(16x)Fn(0)S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(x)\!=\!F_{n}(16x)\!-\!F_{n}(0), n1n\!\geqslant\!1.

3.2 . The generating function.

The generating function of the Schwarz triangle polynomials is calculated in the next lemma.

Lemma 3.3

. We have

iπn=1Sn(1/z)eiπny=λ(y)zλ(y),Imy>Imλ(z),z{}\displaystyle{\rm{i}}\pi\sum\limits_{n=1}^{\infty}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(1/z\right){\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}=\frac{\lambda^{\,\prime}\left(y\right)}{z-\lambda\left(y\right)}\ ,\ \ {\rm{Im}}\,y>{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\ ,\ \ z\in\Bb{C}\setminus\{0\}, (3.6)

where the series converges absolutely and the function Imλ{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}} is continuously extended to {}\Bb{C}\setminus\{0\} in accordance with Lemma 2.5.

Proof of Lemma 3.3. Let z{}z\in\Bb{C}\setminus\{0\}. Then there exists β(0,1)\beta\in(0,1) such that |z|>β|z|>\beta. We calculate

12πiβ𝔻ζnζp1𝑑ζ=δn,p,n,pϝζζϝζβ𝔻\displaystyle\frac{1}{2\pi{\rm{i}}}\int\nolimits_{\beta\partial\Bb{D}}\,\zeta^{n}\zeta^{-p-1}d\zeta=\delta_{n,p}\ ,\ \ \ n,p\in\Bb{Z};\quad\dfrac{1}{z-\zeta}=\sum\nolimits_{n\geqslant 0}\ \frac{\zeta^{n}}{z^{n+1}}\ ,\quad\zeta\in\beta\partial\Bb{D},

and in view of (3.2), we have

Sn(1/z)=12πiβ𝔻enπiλ(ζ)dζzζ=12πiππenπiλ(βeiφ)d(βeiφ)zβeiφ,\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)=\frac{1}{2\pi{\rm{i}}}\int_{\beta\partial\Bb{D}}\frac{{\rm{e}}^{{{\mbox{\footnotesize{$-n\pi{\rm{i}}\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(\zeta)$}}}}}d\zeta}{z-\zeta}=\frac{1}{2\pi{\rm{i}}}\int_{-\pi}^{\pi}\frac{{\rm{e}}^{{{\mbox{\footnotesize{$-n\pi{\rm{i}}\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\beta{\rm{e}}^{{\rm{i}}\varphi}\right)$}}}}}d\left(\beta{\rm{e}}^{{\rm{i}}\varphi}\right)}{z-\beta{\rm{e}}^{{\rm{i}}\varphi}}\,, (3.7)

where the curve λ(βeiφ)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\beta{\rm{e}}^{{\rm{i}}\varphi}\right), π<φ<π-\pi<\varphi<\pi, lies in \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and connects two points λ(βi0)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\!\!\left(-\beta-{\rm{i}}0\right) and λ(β+i0)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\!\!\left(-\beta+{\rm{i}}0\right) which according to (see (2.68) and (2.71))

λ(β+i0)=2+λ(βi0)=1+iΔ(β),Δ(β)>1,\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-\beta+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=2+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-\beta-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(\beta),\quad{{\mbox{\footnotesize{$\Delta$}}}}(\beta)>1,

belong to the boundary of \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}. Making the change of variables ζ=λ(βeiφ)\zeta=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\beta{\rm{e}}^{{\rm{i}}\varphi}\right) for π<φ<π-\pi<\varphi<\pi in (3.7) and using (2.20), we obtain βeiφ=λ(ζ)\beta{\rm{e}}^{{\rm{i}}\varphi}=\lambda\left(\zeta\right) and

Sn(1/z)=12πi1+iΔ(β)1+iΔ(β)λ(ζ)enπiζzλ(ζ)𝑑ζ,|z|>β,β(0,1),\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)=\frac{1}{2\pi{\rm{i}}}\int\nolimits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{{\mbox{\footnotesize{$\Delta$}}}}(\beta)$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{{\mbox{\footnotesize{$\Delta$}}}}(\beta)$}}}}}\ \frac{\lambda^{\,\prime}\left(\zeta\right){\rm{e}}^{{{\mbox{\footnotesize{$-n\pi{\rm{i}}\zeta$}}}}}}{z-\lambda\left(\zeta\right)}\,d\zeta\,,\qquad|z|>\beta,\ \beta\in(0,1)\,, (3.8)

where the contour of integration {λ(βiφ)}π<φ<π\{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\!\left(\beta^{{\rm{i}}\varphi}\right)\}_{-\pi<\varphi<\pi} can be replaced by the straight line segment [1,1]+iΔ(β)[-1,1]\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{{\mbox{\footnotesize{$\Delta$}}}}(\beta) connecting the two points ±1+iΔ(β)\pm 1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{{\mbox{\footnotesize{$\Delta$}}}}(\beta), as by (2.56),

λ({z𝕜𝕜ϝ𝕜𝕀ϝΔβ}β𝔻β\displaystyle\lambda\Big{(}\big{\{}\ z\in\Bb{H}\ \big{|}\ |{\rm{Re}}\,z|\leqslant 1,\ {\rm{Im}}\,z\geqslant{{\mbox{\footnotesize{$\Delta$}}}}(\beta)\ \big{\}}\Big{)}\subset\beta\Bb{D}\ ,\quad\beta\in\left(0,1\right).

In view of (2.83), Corollary 2.7 tells us that the integrand in (3.8) is a holomorphic function of the variable ζ\zeta when Imζ>Imλ(z){\rm{Im}}\,\zeta>{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z), and hence the periodicity of the integrand in (3.8) allows us to use Lemma 3.1 to shift the segment of integration [1,1]+iΔ(β)[-1,1]\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{{\mbox{\footnotesize{$\Delta$}}}}(\beta) to [1,1]+ia[-1,1]\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta for any a>Imλ(z)a>{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z). This gives

Sn(1/z)=12πi1+ia1+iaλ(ζ)enπiζzλ(ζ)𝑑ζ,a>Imλ(z),z{}\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)\!=\!\frac{1}{2\pi{\rm{i}}}\int\nolimits_{{{\mbox{\footnotesize{$-1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}^{{{\mbox{\footnotesize{$1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}\frac{\lambda^{\,\prime}\left(\zeta\right){\rm{e}}^{{{\mbox{\footnotesize{$-n\pi{\rm{i}}\zeta$}}}}}}{z-\lambda\left(\zeta\right)}\,d\zeta\,,\ \ a>{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z),\ z\in\Bb{C}\setminus\{0\}. (3.9)

We conclude that the series

n=1Sn(1/z)eiπny=12πi1+ia1+iaλ(ζ)dζ(eπi(ζy)1)(zλ(ζ)),\displaystyle\sum\limits_{n=1}^{\infty}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(1/z\right){\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}=\frac{1}{2\pi{\rm{i}}}\int\nolimits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}\ \frac{\lambda^{\,\prime}\left(\zeta\right)d\zeta}{\left({\rm{e}}^{{{\mbox{\footnotesize{$\pi{\rm{i}}(\zeta-y)$}}}}}-1\right)\big{(}z-\lambda\left(\zeta\right)\big{)}}\,, (3.10)

converges absolutely if Imy>a>Imλ(z){\rm{Im}}\,y\!>\!a\!>\!{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z). By the periodicity of the integrand in (3.10), we can apply Lemma 3.1 again to shift the interval of integration horizontally by Rey{\rm{Re}}\,y. We then obtain from the Cauchy formula applied to the rectangle with vertices ±1+ia+Rey\pm 1+{\rm{i}}a+{\rm{Re}}\,y, ±1+iA+Rey\pm 1+{\rm{i}}A+{\rm{Re}}\,y with A>ImyA>{\rm{Im}}\,y that this integral equals

1iπλ(y)zλ(y)+12πi1+iA+Rey1+iA+Reyλ(ζ)dζ(eπi(ζy)1)(zλ(ζ)),A>Imy.\displaystyle\frac{1}{i\pi}\frac{\lambda^{\,\prime}\left(y\right)}{z-\lambda\left(y\right)}+\frac{1}{2\pi{\rm{i}}}\int\limits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}\frac{\lambda^{\,\prime}\left(\zeta\right)d\zeta}{\left({\rm{e}}^{{{\mbox{\footnotesize{$\pi{\rm{i}}(\zeta-y)$}}}}}-1\right)\big{(}z-\lambda\left(\zeta\right)\big{)}}\ ,\quad A>{\rm{Im}}\,y\,.

The integral expression tends to zero as A+A\to+\infty, because it follows from the expressions (2.41) for λ(ζ)\lambda^{\,\prime}\left(\zeta\right) and (3.4) for λ(ζ)\lambda\left(\zeta\right) that λ(ζ)0\lambda^{\,\prime}\left(\zeta\right)\to 0 and λ(ζ)0\lambda\left(\zeta\right)\to 0, as Imζ+{\rm{Im}}\,\zeta\to+\infty. This proves that the identity (3.6) holds. \square

3.3 . Asymptotic behavior for the large index.

The condition in (3.6) for the convergence of the series for the generating function of the Schwarz triangle polynomials is sharp. Indeed, it follows from Lemma 5.7 that for every a{}a\in\Bb{C}\setminus\{0,1\} the set {y𝕜𝕜𝕜λ}\{y\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,|\,\lambda(y)=a\} is countable and has no limit points in \Bb{H}. Then (2.84) implies that for every z{}z\!\in\!\Bb{C}\!\setminus\!\{0,1\} there exists β(z)(0,1)\beta(z)\!\in\!(0,1) such that Imy<β(z)Imλ(z){\rm{Im}}\,y\!<\!\beta(z){\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z) for all y𝕜𝕜𝔽y\!\in\!\Bb{H}_{|{\rm{Re}}|\!\leqslant 1}\!\setminus\!{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right) satisfying λ(y)=z\lambda(y)=z. This allows us to apply the Cauchy formula in (3.9) for the rectangle with vertices ±1+ia\pm\!1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta, ±1+iβ(z)Imλ(z)\pm 1\!+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\beta(z){\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z) and obtain

Sn(1/z)={yclos()|λ(y)=z}enπiy+O(enπβ(z)Imλ(z)),n+.\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)\!=\!\!\!\!\!\!\sum\limits_{{{\mbox{\footnotesize{$\left\{y\!\in\!{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)\ |\ \lambda(y)\!=\!z\right\}$}}}}}\!\!{\rm{e}}^{{{\mbox{\footnotesize{$-n\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty$}}}}}+{\rm{O}}\left({\rm{e}}^{{{\mbox{\footnotesize{$n\pi\beta(z){\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)$}}}}}\right)\,,\ n\to+\infty. (3.11)

where12{}^{\ref*{case9}} by Theorem 2.6, (2.68) and (2.73),

{yclos()|λ(y)=z}enπiy={exp(nπiλ(z)),ifz(0,1)();2enπΔ(x)1+Δ(x)2cosnπ1+Δ(x)2,ifz=1+x,x>0;2(1)nexp(nπΔ(x)),ifz=x,x>0.\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\left\{y\!\in\!{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)\ \big{|}\ \lambda(y)\!=\!z\right\}$}}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\rm{e}}^{{{\mbox{\footnotesize{$-n\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty$}}}}}=\left\{\begin{array}[]{ll}\exp\left(-n\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)\,,&\ \ \hbox{if}\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\,;\\[7.11317pt] 2{\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{\,n\pi{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}\cos\dfrac{n\pi}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\ ,&\ \ \hbox{if}\quad z=1+x\ ,\quad x>0\,;\\[14.22636pt] 2(-1)^{n}\exp\left(n\pi{{\mbox{\footnotesize{$\Delta$}}}}(x)\right),&\ \ \hbox{if}\quad z=-x\ ,\qquad x>0\,.\end{array}\right.

3.4 . Symmetry property.

The condition for the convergence of the series in (3.6) is at its weakest when z=1z=1 as Imλ(1)=0{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1)=0. In the next lemma we explain this fact by showing in (3.13) that Sn(1)=O(n2)S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)={\rm{O}}(n^{2}), as nn\to\infty, and derive in (3.12)(a) an important symmetric property of the Schwarz triangle polynomials.

Lemma 3.4

. The following identities hold

(a)(1)nSn(z)=Sn(1z)Sn(1),(b)S2n(1)=0,n1,\displaystyle{\rm{(a)}}\ \,(-1)^{n}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(z)=S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1-z)-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)\ ,\quad{\rm{(b)}}\ \,S^{{{\mbox{\tiny{$\triangle$}}}}}_{2n}(1)=0\ ,\qquad n\!\geqslant\!1\,, (3.12)
n=1Sn(1)un1=16θ2(u)4,u𝔻𝕊\displaystyle\sum\limits_{n=1}^{\infty}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(1\right)u^{n-1}\!=\!16\,\theta_{2}(u)^{4},\ u\!\in\!\Bb{D}\,;\ \ S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)\!=\!\left(1\!-\!(-1)^{n}\right)r_{4}(n),\ n\!\geqslant\!1\,, (3.13)
Sn(z):=k=1nsn,kzk,n1,S1(z)=16z,S2(z)=162z(z1),z\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(z)\!:=\!\sum\limits_{k=1}^{n}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}z^{k},\ n\!\geqslant\!1\,,\ S^{{{\mbox{\tiny{$\triangle$}}}}}_{1}(z)\!=\!16z,\ S^{{{\mbox{\tiny{$\triangle$}}}}}_{2}(z)\!=\!16^{2}z(z-1),\ \ z\!\in\!\Bb{C}\,, (3.14)
sn,n=16n,n1,sn,n1=8n16n1,n2,\displaystyle s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,n}=16^{n}\ ,\quad n\geqslant 1\,,\quad s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,n-1}=-8n16^{n-1}\ ,\quad n\geqslant 2\,, (3.15)
n=1sn,1un1=16θ2(u)4θ3(u)4θ3(u)4,u𝔻\displaystyle\sum\limits_{n=1}^{\infty}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,1}u^{n-1}\!=\!16\,\theta_{2}(u)^{4}\,\theta_{3}(-u)^{4}\,\theta_{3}(u)^{-4}\ ,\quad u\in\Bb{D}\,. (3.16)

We supply the proof of Lemma 3.4 in Section 10.2.1. By the symmetric property (3.12) and the change of variables z=1/zz^{\,\prime}=-1/z on the right-hand side integral of (2.52), for arbitrary xx\in\Bb{R} and n1n\geqslant 1 we obtain from (1.19), (2.21) and (10.4) that 11{}^{\ref*{case7}}

Mn(x)=14π2nγ(1,1)Sn(11λ(z))dz(x+z)2=(1)n4π2nγ(1,1)Sn(λ(z+1))Sn(1)(x+z)2dz.\displaystyle\eurm{M}_{n}(x)\!=\!\dfrac{1}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{1\!-\!\lambda(z)}\right)dz}{(x+z)^{2}}\!=\!\dfrac{(-1)^{n}}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\big{(}\lambda(z\!+\!1)\big{)}-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)}{(x+z)^{2}}dz\,. (3.17)

Here, in view of (2.25)(d),(e) and (3.4), we have

λ(z+1)=Θ2(z)4Θ4(z)4=λ𝔻(eiπz),z\displaystyle\lambda(z\!+\!1)=-\dfrac{\Theta_{2}(z)^{4}}{\Theta_{4}(z)^{4}}=\lambda_{\Bb{D}}\left(-{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right),\quad z\in\Bb{H},

where λ𝔻Hol(𝔻\lambda_{\Bb{D}}\in{\rm{Hol}}(\Bb{D}) with λ𝔻(0)=0\lambda_{\Bb{D}}(0)=0. Hence for arbitrary n1n\geqslant 1 and zz\in\Bb{H}, we obtain

(1)n\displaystyle(-1)^{n} Sn(1λ(1/z))=(1)nSn(11λ(z))\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(-1/z)}\right)\!=\!(-1)^{n}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{1\!-\!\lambda(z)}\right)
=Sn(λ(z+1))Sn(1)=Sn(1)16sn,1eiπz+k2bn,keiπkz,\displaystyle\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\big{(}\lambda(z\!+\!1)\big{)}-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)=-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)\!-\!16s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,1}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\!+\!\sum\nolimits_{k\geqslant 2}{\eurm{b}}_{n,k}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi kz}, (3.18)

where we have used the following notation for the Taylor series (see (3.14))

k2bn,kuk:=16sn,1u+k=1nsn,kλ𝔻(u)kHol(𝔻𝔻\displaystyle\sum\limits_{k\geqslant 2}{\eurm{b}}_{n,k}u^{k}:=16s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,1}u+\sum\limits_{k=1}^{n}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}\lambda_{\Bb{D}}\left(-u\right)^{k}\in{\rm{Hol}}(\Bb{D})\ ,\quad u\in\Bb{D}\,,\quad n\geqslant 1\,. (3.19)

Besides, by using (2.25)(b) and the change of variables z=1/zz^{\,\prime}=-1/z in the left-hand side integral of (2.52), we find 13{}^{\ref*{case8}}

H0(x)=H0(x),x{}\displaystyle\eurm{H}_{0}(x)=\eurm{H}_{0}(-x)\,,\ \ x\in\Bb{R}\,,\qquad\eurm{H}_{0}(-1/x)\!=\eurm{H}_{0}(x)x^{2}\,,\ \ x\in\Bb{R}\setminus\{0\}\,. (3.20)

3.5 . Characteristic behavior at the vertices of the Schwarz quadrilateral.

We note that there are only four distinct Möbius transformations (see (con, , p.​ 47))

z,1/z,(z1)/(z+1),(1+z)/(1z),\displaystyle z\ ,\ \ -{1}\big{/}{z}\ ,\ \ {(z-1)}\big{/}{(z+1)}\ ,\ \ {(1+z)}\big{/}{(1-z)}\ ,

which14{}^{\ref*{case11}} map \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} onto itself and interchange the vertices according to

z,z01/z,1/z,z,z1(1+z)/(1z),(1+z)/(1z),z,z1(z1)/(z+1),(z1)/(z+1).\displaystyle\begin{array}[]{lrclr}z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},&\ \ z\to 0&\Leftrightarrow&\ \ -1/z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},&\quad-1/z\to\infty\ ,\\ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},&\ \ z\to 1&\Leftrightarrow&\ \ {(1+z)}\big{/}{(1-z)}\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},&\quad{(1+z)}\big{/}{(1-z)}\to\infty\ ,\\ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},&\ \ z\to-1&\Leftrightarrow&\ \ {(z-1)}\big{/}{(z+1)}\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},&\quad{(z-1)}\big{/}{(z+1)}\to\infty\,.\end{array}

The nn-th Schwarz triangle polynomial SnS^{{{\mbox{\tiny{$\triangle$}}}}}_{n} can be fully characterized by the behavior of the function Sn(1/λ)S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/\lambda) from (3.5) at the vertices of the Schwarz quadrilateral. Note that (3.5) for arbitrary n1n\!\geqslant\!1 and zz\!\in\!\Bb{H} can also be written as

Sn(1λ(z))=Sn(11λ(1/z))=einπzΔnS(0)k1dn,keiπkz,\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\frac{1}{\lambda(z)}\!\right)\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{1\!-\!\lambda(-1/z)}\right)\!=\!{\rm{e}}^{{{\mbox{\footnotesize{$-in\pi z$}}}}}\!-\!\Delta_{n}^{S}(0)-\sum\nolimits_{k\geqslant 1}{\eurm{d}}_{n,k}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi kz}, (3.21)

where we use the Taylor series notation

k1dn,kuk:=ΔnS(λ𝔻(u))ΔnS(0)Hol(𝔻𝔻\displaystyle\sum\nolimits_{k\geqslant 1}{\eurm{d}}_{n,k}u^{k}:=\Delta_{n}^{S}(\lambda_{\Bb{D}}(u))-\Delta_{n}^{S}(0)\in{\rm{Hol}}(\Bb{D})\ ,\quad u\in\Bb{D}\,,\quad n\geqslant 1\,.

By (2.30)(b) and the same line of argument15{}^{\ref*{case10}} as in Section 3.2, we get

1+n=1ΔnS(0)un=θ3(u)4,u𝔻Δ𝕊\displaystyle 1\!+\!\sum\nolimits_{n=1}^{\infty}\Delta_{n}^{S}(0)u^{n}\!=\!\theta_{3}(-u)^{4},\ u\!\in\!\Bb{D}\,;\quad\Delta_{n}^{S}(0)\!=\!(-1)^{n}r_{4}(n),\ n\!\geqslant\!1\,, (3.22)

and, consequently, ΔnS(0)=O(n2)\Delta_{n}^{S}(0)\!=\!{\rm{O}}(n^{2}), as nn\!\to\!\infty. It follows from (2.21), from the identities

1λ(11/z)=λ(z+1),11λ(11/z)=λ(z),z\displaystyle\frac{1}{\lambda\left(1\!-\!{1}/{z}\right)}=\lambda(z\!+\!1)\ ,\quad\frac{1}{1-\lambda\left(1\!-\!{1}/{z}\right)}=\lambda(z)\ ,\quad z\in\Bb{H}, (3.23)

(see (cha, , p.​ 111)) and from (3.4) that

Sn(1λ(11/z))=Sn(λ𝔻(eiπz)),Sn(11λ(11/z))=Sn(λ𝔻(eiπz)),\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\frac{1}{\lambda\left(1\!-\!{1}/{z}\right)}\!\right)\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\Big{(}\!\lambda_{\Bb{D}}\left(-{\rm{e}}^{{\rm{i}}\pi z}\right)\!\Big{)}\ ,\ S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\frac{1}{1-\lambda\left(1\!-\!{1}/{z}\right)}\!\right)\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\Big{(}\!\lambda_{\Bb{D}}\left({\rm{e}}^{{\rm{i}}\pi z}\right)\!\Big{)}\,,

and thus for arbitrary n1n\geqslant 1 and zz\in\Bb{H}, we obtain, by (3.19),

Sn(1λ(11/z))\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\frac{1}{\lambda\left(1\!-\!{1}/{z}\right)}\!\right) =16sn,1eiπz+k2bn,keiπkz,\displaystyle\!=\!-16s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,1}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}+\sum\nolimits_{k\geqslant 2}{\eurm{b}}_{n,k}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi kz}, (3.24)
Sn(11λ(11/z))\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\frac{1}{1-\lambda\left(1\!-\!{1}/{z}\right)}\!\right) = 16sn,1eiπz+k2(1)kbn,keiπkz.\displaystyle\!=\ 16s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,1}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}+\sum\nolimits_{k\geqslant 2}(-1)^{k}{\eurm{b}}_{n,k}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi kz}. (3.25)

When zz\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} tends to one of the vertices α{1,1,0,}\alpha\in\{1,-1,0,\infty\} of \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, we assume that Imz1/2{\rm{Im}}\,z\geqslant 1/2 if α=\alpha=\infty and Imz<1/2{\rm{Im}}\,z<1/2 if α{1,1,0}\alpha\in\{1,-1,0\}, which imply that 2Imz|Rez|2\,{\rm{Im}}\,z\geqslant|{\rm{Re}}\,z| if α=\alpha=\infty, while Imz|Rezα|{\rm{Im}}\,z\geqslant|{\rm{Re}}\,z-\alpha| if α{0,1,1}\alpha\in\{0,1,-1\}. Hence it follows from (3.18), (3.21), (3.24) and (3.25) that for every n1n\geqslant 1 there exists 𝒟n{\eusm{D}}_{n}\in\Bb{R}_{>0} such that

(a)|Rn(z)|𝒟nexp(π2Imz),|z±1|1/2,z;(b)|Rn(z)exp(inπz)|𝒟n,Imz1/2,z;(c)|Rn(z)|𝒟n,|z|1/2,z;\displaystyle\hskip 5.69046pt\begin{array}[]{llll}{\rm{(a)}}&\ \ \left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\right|\!\leqslant\!{\eusm{D}}_{n}\exp\left(-\dfrac{\pi}{2\,{\rm{Im}}\,z}\right),&\ \ \left|z\pm 1\right|\leqslant 1/\sqrt{2}\,,&\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,;\\[8.5359pt] {\rm{(b)}}&\ \ \left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!-\!\exp\left(-\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptn\pi z\right)\right|\!\leqslant\!{\eusm{D}}_{n}\ ,&\ \ {\rm{Im}}\,z\geqslant 1/2\,,&\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,;\\[7.11317pt] {\rm{(c)}}&\ \ \left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\right|\!\leqslant\!{\eusm{D}}_{n}\ ,&\ \ \left|z\right|\leqslant 1/\sqrt{2}\,,&\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,;\end{array} (3.29)

and

(a)|Rn(1/z)|𝒟nexp(π2Imz),|z±1|1/2,z;(b)|Rn(1/z)|𝒟n,Imz1/2,z;(c)|Rn(1/z)exp(inπ/z)|𝒟n,|z|1/2,z;\displaystyle\begin{array}[]{llll}{\rm{(a)}}&\ \ \left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z)\right|\!\leqslant\!{\eusm{D}}_{n}\exp\left(-\dfrac{\pi}{2\,{\rm{Im}}\,z}\right),&\ \ \left|z\pm 1\right|\leqslant 1/\sqrt{2}\,,&\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,;\\[8.5359pt] {\rm{(b)}}&\ \ \left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z)\right|\!\leqslant\!{\eusm{D}}_{n}\ ,&\ \ {\rm{Im}}\,z\geqslant 1/2\,,&\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,;\\[7.11317pt] {\rm{(c)}}&\ \ \left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z)\!-\!\exp\left(\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptn\pi/z\right)\right|\!\leqslant\!{\eusm{D}}_{n}\ ,&\ \ \left|z\right|\leqslant 1/\sqrt{2}\,,&\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,;\end{array} (3.33)

where for every n1n\geqslant 1 the functions

(a)Rn(z):=Sn(1λ(z))=k=1nsn,kλ(z)k;(b)Rn(1/z)=Sn(11λ(z))=k=1nsn,k(1λ(z))k;\displaystyle\begin{array}[]{lrcccl}{\rm{(a)}}&\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!&:=&\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\dfrac{1}{\lambda(z)}\!\right)\!&=&\!\displaystyle\sum\nolimits_{k=1}^{n}\ \dfrac{s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}}{\lambda(z)^{k}}\ ;\\[11.38092pt] {\rm{(b)}}&\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z)\!&=&\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\!\dfrac{1}{1\!-\!\lambda(z)}\!\right)\!&=&\!\displaystyle\sum\nolimits_{k=1}^{n}\ \dfrac{s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}}{\big{(}1\!-\!\lambda(z)\big{)}^{k}}\ ;\end{array} (3.36)

are holomorphic in \Bb{H} and periodic with period 22. We observe that in this notation the formulas (2.52) and (3.17) can be written in the form

Hn(x)=(1)4π2nγ(1,1)Rn(z)dz(x+z)2,Mn(x)=14π2nγ(1,1)Rn(1/z)dz(x+z)2,n1,\displaystyle\eurm{H}_{n}(x)\!=\dfrac{(-1)}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z){\rm{d}}z}{(x+z)^{2}}\,,\quad\eurm{M}_{n}(x)\!=\!\dfrac{1}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z){\rm{d}}z}{(x+z)^{2}}\ ,\ \ n\geqslant 1\,, (3.37)

where the behavior of the both integrands in \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} is completely described by (3.29) and (3.33). In particular, it follows from (3.29)(a) and (3.33)(a) that the both integrals converge absolutely for all xx\in\Bb{R}. Furthermore, by writing (3.23) in the form (see (cha, , p.​ 111))

1λ(z)+1λ(z+1)=1,z\displaystyle\dfrac{1}{\lambda(z)}+\dfrac{1}{\lambda(z+1)}=1\ ,\quad z\in\Bb{H}\,, (3.38)

we conclude from (3.12) and (3.36) that for any zz\in\Bb{H} and n1n\geqslant 1 we have

R2n(z+1)=R2n(z),R2n1(z+1)=R2n1(z)+2r4(2n1).\displaystyle\eurm{R}_{2n}^{{{\mbox{\tiny{$\triangle$}}}}}(z+1)=\eurm{R}_{2n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\,,\quad\eurm{R}_{2n-1}^{{{\mbox{\tiny{$\triangle$}}}}}(z+1)=-\eurm{R}_{2n-1}^{{{\mbox{\tiny{$\triangle$}}}}}(z)+2\,r_{4}(2n-1)\,.

The following Liouville-type property holds for the Schwarz quadrilateral \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, where in the notation (2.11), =1/\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}=-1/\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and

:=𝔽γγγγ\displaystyle\partial_{\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!:=\!\Bb{H}\cap\partial\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!=\!\gamma(1,\infty)\cup\gamma(-1,\infty)\cup\gamma(0,1)\cup\gamma(0,-1). (3.39)
Lemma A ((bh2, , p.​ 597))

. Suppose that fHol()C(𝔽f\!\in\!{\rm{Hol}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})\cap C(\Bb{H}\,\cap\,{\rm{clos}}\,\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}) satisfies  f(z)=f(z+2)f(z)=f(z+2)   and  f(1/z)=f(1/(z+2))f(-1/z)=f\big{(}-1/(z+2)\big{)}\vphantom{\dfrac{A}{A}}   for each z1+iz\in-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0}. Suppose also that there exist nonnegative integers nn_{\infty}, n0n_{0}, and n1n_{1} such that

|f(z)|=o(exp(π(n+1)|z|)),\displaystyle\left|f(z)\right|=\mathrm{o}\big{(}\exp\left(\pi\left(n_{\infty}+1\right)|z|\right)\big{)}\ , z,z,\displaystyle z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\,\ \ z\to\infty\,,
|f(z)|=o(exp(π(n0+1)|z|1)),\displaystyle\left|f(z)\right|=\mathrm{o}\big{(}\exp\left(\pi\left(n_{0}+1\right)|z|^{-1}\right)\big{)}\ , z,z0,\displaystyle z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\,\ \ z\to 0\,,
|f(z)|=o(exp(π(n1+1)|zσ|1)),\displaystyle\left|f(z)\right|=\mathrm{o}\big{(}\exp\left(\pi\left(n_{1}+1\right)|z-\sigma|^{-1}\right)\big{)}\ , z,zσ,σ{1,1}.\displaystyle z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\,\ \ z\to\sigma\,,\ \sigma\in\{1,-1\}\,.

Then there exists an algebraic polynomial PP of degree n+n0+n1\leq n_{\infty}+n_{0}+n_{1} such that

f(z)=P(λ(z))λ(z)n(1λ(z))n0,z.\displaystyle f\big{(}z\big{)}=\frac{\vphantom{\frac{a}{b}}P\big{(}\lambda(z)\big{)}}{\vphantom{\dfrac{a}{b}}\lambda(z)^{\,{{\mbox{\footnotesize{$n_{\infty}$}}}}}\big{(}1-\lambda(z)\big{)}^{\,{{\mbox{\footnotesize{$n_{0}$}}}}}}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,.

The next lemma shows that the function Rn=Sn(1/λ)\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}=S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/\lambda) is uniquely determined by much weaker asymptotic conditions at the vertices of the Schwarz quadrilateral than those of (3.29).

Lemma 3.5

. For a given positive integer nn there exists a unique function Rn\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}} in C(𝔽𝔽C(\Bb{H}\,\cap\,{\rm{clos}}\,\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})\cap{\rm{Hol}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}) such that for any σ{1,1}\sigma\in\{1,-1\} we have

(a)Rn(z)=Rn(z+2),Rn(1/z)=Rn(1/(z+2)),z1+i(b)Rn(z)=O(1),z,z0,(c)Rn(z)=o(1),z,zσ,(d)Rn(z)=eiπnz+O(1),z,z.\displaystyle\begin{array}[]{lll}{\rm{(a)}}&\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!=\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z\!+\!2),\ \ \eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z)\!=\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\big{(}-1/(z\!+\!2)\big{)},&\ \ z\!\in\!-1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0}\,,\\[5.69046pt] {\rm{(b)}}&\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!=\!{\rm{O}}(1)\,,&\ \ z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\ z\!\to\!0\,,\\[5.69046pt] {\rm{(c)}}&\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!=\!\mathrm{o}\big{(}1\big{)}\ ,&\ \ z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\ z\!\to\!\sigma\,,\\[5.69046pt] {\rm{(d)}}&\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!=\!e^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nz$}}}}}\!+\!{\rm{O}}(1)\,,&\ \ z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\,z\!\to\!\infty\,.\end{array} (3.44)

Lemma 3.5 follows easily from Lemma A. Indeed, the conditions (3.44) are satisfied if (3.36)(a) holds, as follows from (3.29). As for the uniqueness, suppose that the function FnF_{n} has all the properties of (3.44) (with Rn\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}} replaced by FnF_{n}). Then the difference FnSn(1/λ)F_{n}-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/\lambda) meets the conditions of Lemma A with n=n0=n1=0n_{\infty}=n_{0}=n_{1}=0. By Lemma A, this difference is a constant function, which must equal zero, by property (3.44)(c). This completes the proof of Lemma 3.5.

The symmetrized functions Rn±(z):=Rn(z)±Rn(1/z)\eurm{R}^{\pm}_{n}(z)\!:=\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\pm\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z) introduced in (3.3) are characterized in a similar manner.

Lemma 3.6

. For every positive integer nn, there exist unique functions Rn+\eurm{R}_{n}^{+} and Rn\eurm{R}_{n}^{-} in the set C(𝔽𝔽C(\Bb{H}\,\cap\,{\rm{clos}}\,\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})\cap{\rm{Hol}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}) such that

(a)Rn±(z)=Rn±(z+2),z1+i(b)Rn±(z)=±Rn±(1/z),z,(c)Rn±(z)=o(1),z,z1,(d)Rn±(z)=eiπnz+O(1),z,z.\displaystyle\begin{array}[]{lll}{\rm{(a)}}&\eurm{R}_{n}^{\pm}(z)\!=\!\eurm{R}_{n}^{\pm}(z\!+\!2),&\ \ z\!\in\!-1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0}\,,\\[4.26773pt] {\rm{(b)}}&\eurm{R}_{n}^{\pm}(z)=\pm\,\eurm{R}_{n}^{\pm}(-1/z)\,,&\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\\[4.26773pt] {\rm{(c)}}&\eurm{R}_{n}^{\pm}(z)\!=\!\mathrm{o}\big{(}1\big{)}\ ,&\ \ z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\ \,z\!\to\!1\,,\\[4.26773pt] {\rm{(d)}}&\eurm{R}_{n}^{\pm}(z)\!=\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nz$}}}}}\!+\!{\rm{O}}(1)\,,&\ \ z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}},\ z\!\to\!\infty\,.\end{array}

On a side note, we observe that the functional properties (3.23) imply that for arbitrary n1n\geqslant 1 and zz\in\Bb{H}, we have

Rn+(11z)=Sn(λ(z))+Sn(λ(z+1)),Rn(11z)=Sn(λ(z))Sn(λ(z+1)).\displaystyle\eurm{R}^{+}_{n}\left(\!1\!-\!\dfrac{1}{z}\!\right)\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\lambda(z)\right)\!+\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\lambda(z\!+\!1)\right)\,,\ \ \eurm{R}^{-}_{n}\left(\!1\!-\!\dfrac{1}{z}\!\right)\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\lambda(z)\right)\!-\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\lambda(z\!+\!1)\right)\,.

4 . The biorthogonal sequence

This section is devoted to the study of the functions H0\eurm{H}_{0}, Hn\eurm{H}_{n}, Mn\eurm{M}_{n}, for nn\in\Bb{Z}_{\neq 0}, given in Theorem 1.1.

4.1 . Explicit expressions for the biorthogonal sequence.

It can easily be derived from (3.37), (4.5) and (2.47) that Hn,MnHol(\eurm{H}_{n},\eurm{M}_{n}\!\in\!{\rm{Hol}}(\Bb{H})\!\cap\!C(\Bb{H}\!\cup\!\Bb{R}) with16{}^{\ref*{case35}}

|Hn(z)|+|Mn(z)|8e2πn|z1|2+|z+1|2,z\displaystyle\left|\eurm{H}_{n}(z)\right|+\left|\eurm{M}_{n}(z)\right|\leqslant\dfrac{8\,{\rm{e}}^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{|z-1|^{2}+|z+1|^{2}}\ ,\quad z\in\Bb{H}\cup\Bb{R}\,,\ n\geqslant 1\,,

from which, in view of (1.18), we obtain

Hn,MnH+1(𝕄\displaystyle\eurm{H}_{n}\,,\ \eurm{M}_{n}\in{\rm{H}}^{1}_{+}(\Bb{R})\ ,\quad\eurm{H}_{-n}\,,\ \eurm{M}_{-n}\in{\rm{H}}^{1}_{-}(\Bb{R})\ ,\quad n\geqslant 1\,. (4.1)

The main results of this section are the following assertions, the proofs of which are deferred to Sections 10.3.1.

Lemma 4.1

. For every nn\in\Bb{Z}_{\neq 0} the functions Hn\eurm{H}_{n} and Mn\eurm{M}_{n} have the following properties:

(1)Hn,MnC(if𝕄if\displaystyle{\rm{(1)}}\ \eurm{H}_{n},\eurm{M}_{n}\!\in\!C^{\infty}(\Bb{R})\cap{\rm{H}}^{1}_{+}(\Bb{R}),\ \mbox{if}\ n\!\geqslant\!1,\quad\eurm{H}_{n},\ \eurm{M}_{n}\!\in\!C^{\infty}(\Bb{R})\cap{\rm{H}}^{1}_{-}(\Bb{R}),\ \mbox{if}\ n\!\leqslant\!-1;
(2)m𝕄𝕆𝕜𝕜𝕜𝕜\displaystyle{\rm{(2)}}\ \forall m\in\Bb{Z}_{\geqslant 0}:\,\,\eurm{H}_{n}^{(m)}(x),\eurm{M}_{n}^{(m)}(x)={\rm{O}}\left(|x|^{-2-m}\right),\ \ \ |x|\to+\infty\,;\
(3)Mn(x)=Hn(1/x)/x2,x{}\displaystyle{\rm{(3)}}\ \eurm{M}_{n}(x)=\eurm{H}_{n}(-1/x)/x^{2},\ \ \ x\!\in\!\Bb{R}\!\setminus\!\{0\};
(4) 2kHn(x+2k)=eiπnx,kMn(x+2k)=0,x\displaystyle{\rm{(4)}}\ 2\!\!\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{n}(x+2k)={\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\ ,\qquad\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{M}_{n}(x+2k)=0\ ,\quad x\in\Bb{R}\,.
Lemma 4.2

. The function H0\eurm{H}_{0} has the following properties:

(1)H 0C(\displaystyle{\rm{(1)}}\ \eurm{H}_{\,0}\!\in\!C^{\infty}(\Bb{R}),\ \eurm{H}_{0}(x)\!=\!\eurm{H}_{0}(-x),\ x\!\in\!\Bb{R}\,, (2)H 0(x)=O(x2),|x|+,\displaystyle{\rm{(2)}}\ \eurm{H}_{\,0}(x)={\rm{O}}\left(x^{-2}\right),\ |x|\to+\infty\,,
(3) 2kH 0(x+2k)=1,x\displaystyle{\rm{(3)}}\ 2\!\!\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{\,0}(x+2k)=1\ ,\ \ x\in\Bb{R}\,, (4)H0(1/x)=H0(x)x2,x{}\displaystyle{\rm{(4)}}\ \eurm{H}_{0}(-1/x)\!=\!\eurm{H}_{0}(x)x^{2}\!\!,\ \ \ x\!\in\!\Bb{R}\!\setminus\!\{0\}.

It can easily be obtained from Lemmas 4.1 and 4.2 that for every mm\in\Bb{Z} we have

eiπmxHn(x)dx=δm,n,\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\eurm{H}_{n}(x){\rm{d}}x=\delta_{m,n}\,, eiπm/xHn(x)dx=0,\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${{\rm{i}}\pi m}/{x}$}}}}}\eurm{H}_{n}(x){\rm{d}}x=0\,, n\displaystyle n\in\Bb{Z}\,,\ (4.2)
eiπmxMn(x)dx=0,\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\eurm{M}_{n}(x){\rm{d}}x=0\,, eiπm/xMn(x)dx=δm,n,\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${{\rm{i}}\pi m}/{x}$}}}}}\eurm{M}_{n}(x){\rm{d}}x=\delta_{m,n}\,, n\displaystyle n\in\Bb{Z}_{\neq 0}\,. (4.3)

Indeed, Lemma 4.1(2,3) together with Lemma 4.2(2,3) give that

eiπmxHn(x)dx\displaystyle\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\eurm{H}_{n}(x){\rm{d}}x =11eiπmx(kHn(x+2k))dx\displaystyle=\int\nolimits_{-1}^{1}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\left(\sum\nolimits_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{n}(x+2k)\right){\rm{d}}x
=1211eiπmxeiπnxdx=δn,m,m\displaystyle=\dfrac{1}{2}\int\nolimits_{-1}^{1}{\rm{e}}^{{{\mbox{\footnotesize{$-i\pi mx$}}}}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}{\rm{d}}x=\delta_{n,m}\ ,\quad m\in\Bb{Z}\,,\ n\in\Bb{Z}_{\geqslant 0}\,,

while (1.18), (1.19), Lemma 4.1(2),(4) and Lemma 4.2(1),(4) for any n1n\geqslant 1 and mm\in\Bb{Z} entail

eiπmxHn(x)dx\displaystyle\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\eurm{H}_{-n}(x){\rm{d}}x =eiπmxHn(x)dx=eiπmxHn(x)dx=δn,m,\displaystyle=\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\eurm{H}_{n}(-x){\rm{d}}x=\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi mx$}}}}}\eurm{H}_{n}(x){\rm{d}}x=\delta_{-n,m}\,,
eiπm/xH0(x)dx\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${{\rm{i}}\pi m}/{x}$}}}}}\eurm{H}_{0}(x){\rm{d}}x =eiπmxH0(x)dx=δ0,m,\displaystyle=\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\eurm{H}_{0}(x){\rm{d}}x=\delta_{0,m}\,,\
eiπm/xHn(x)dx\displaystyle\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${{\rm{i}}\pi m}/{x}$}}}}}\eurm{H}_{n}(x){\rm{d}}x =eiπmxMn(x)dx\displaystyle=\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\eurm{M}_{n}(x){\rm{d}}x
=11eiπmx(kMn(x+2k))dx=0.\displaystyle=\int\nolimits_{-1}^{1}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi mx$}}}}}\left(\sum\nolimits_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{M}_{n}(x+2k)\right){\rm{d}}x=0\,.

This concludes the verification of (4.2). By applying the change of the variables x=1/xx=-1/x^{\,\prime} in (4.2), the relations (4.3) are immediate, since the functions Hn,Mn\eurm{H}_{n},\eurm{M}_{n} are connected via the relation (1.19). Expressed differently, the following theorem holds, which contains part of the assertions made in Theorem 1.1.

Theorem 4.3

. The system consisting of the functions H0\eurm{H}_{0}, Hn\eurm{H}_{n}, MnL1(\eurm{M}_{n}\subset L^{1}(\Bb{R})\cap C^{\infty}(\Bb{R}), nn\in\Bb{Z}_{\neq 0}, is biorthogonal to the hyperbolic trigonometric system 11, exp(iπnx)\exp({\rm{i}}\pi nx), exp(iπn/x)\exp({\rm{i}}\pi n/x) L(\subset L^{\infty}(\Bb{R}), nn\in\Bb{Z}_{\neq 0}.

Remark 4.4

. Let fL1([0,2],dx)f\in L^{1}([0,2],{\rm{d}}x) be 22-periodic on \Bb{R}. Then clearly fL1(f\in L^{1}(\Bb{R},(1+x^{2})^{-1}{\rm{d}}x) and, in view of Lemma 4.1(4) and Lemma 4.2(3),

hn(f)=f(t)Hn(t)dt=02f(t)kHn(t+2k)dt=(1/2)02f(t)eiπntdt,\displaystyle\eurm{h}_{n}(f)=\int\limits_{\Bb{R}}f(t)\eurm{H}_{-n}(t){\rm{d}}t=\int\limits_{0}^{2}f(t)\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{-n}(t+2k){\rm{d}}t=(1/2)\int\limits_{0}^{2}f(t)\,{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nt$}}}}}{\rm{d}}t,
mm(f)=f(t)Mm(t)dt=02f(t)kMm(t+2k)dt=0,n\displaystyle\eurm{m}_{m}(f)=\int\limits_{\Bb{R}}f(t)\eurm{M}_{-m}(t){\rm{d}}t=\int\limits_{0}^{2}f(t)\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{M}_{-m}(t+2k){\rm{d}}t=0,\ n\in\Bb{Z}\,,\ m\in\Bb{Z}_{\neq 0}\,.

It follows that the hyperbolic Fourier series (1.5) of the 22-periodic function ff is the usual Fourier series nhn(f)exp(iπnx)\sum_{n\in\Bb{Z}}\eurm{h}_{n}(f)\exp({\rm{i}}\pi nx) of ff on the interval [0,2][0,2]. Next, assume that g(1/x)=f(x)g(-1/x)=f(x), xx\in\Bb{R}_{\neq 0}. Then, by (1.19) and the above identities for ff,

mn(g)=g(t)Mn(t)dt=f(t)Hn(t)dt=(1/2)02g(1/t)eiπntdt,\displaystyle\eurm{m}_{n}(g)=\int\nolimits_{\Bb{R}}g(t)\eurm{M}_{-n}(t){\rm{d}}t=\int\nolimits_{\Bb{R}}f(t)\eurm{H}_{-n}(t){\rm{d}}t=(1/2)\int\nolimits_{0}^{2}g(-1/t)\,{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nt$}}}}}{\rm{d}}t,
hn(g)=g(t)Hn(t)dt=f(t)Mn(t)dt=0,n\displaystyle\eurm{h}_{n}(g)=\int\nolimits_{\Bb{R}}g(t)\eurm{H}_{-n}(t){\rm{d}}t=\int\nolimits_{\Bb{R}}f(t)\eurm{M}_{-n}(t){\rm{}}dt=0,\ \ \ n\in\Bb{Z}_{\neq 0}\,,

while, as follows from Lemma 4.2(4) and the above equalities,

h0(g)=g(t)H0(t)dt=f(t)H0(t)dt=(1/2)02g(1/t)dt.\displaystyle\eurm{h}_{0}(g)=\int\nolimits_{\Bb{R}}g(t)\eurm{H}_{0}(t){\rm{d}}t=\int\nolimits_{\Bb{R}}f(t)\eurm{H}_{0}(t){\rm{d}}t=(1/2)\int\nolimits_{0}^{2}g(-1/t){\rm{d}}t\,.

It follows that the hyperbolic Fourier series (1.5) of gg equals the usual Fourier series h0(g)+nmn(g)exp(iπn/x)\eurm{h}_{0}(g)+\sum_{n\in\Bb{Z}_{\neq 0}}\eurm{m}_{n}(g)\exp(-i\pi n/x) of g(1/x)g(-1/x) on the interval [0,2][0,2] expressed in the variable 1/x-1/x.

4.2 . The generating function for the biorthogonal sequence.

In the notation of (3.36), it follows from (3.6) and (2.83) that

iπn=1Rn(z)eiπny=λ(y)λ(z)λ(y),Imy>1Imz,zγ(1,1),\displaystyle i\pi\sum\limits_{n=1}^{\infty}\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\,{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}\!=\!\frac{\lambda^{\,\prime}\left(y\right)}{\lambda(z)\!-\!\lambda\left(y\right)}\ ,\quad{\rm{Im}}\,y\!>\!1\!\geqslant\!{\rm{Im}}\,z,\ z\!\in\!\gamma(-1,1), (4.4)

where for arbitrary a(1,Imy)a\in(1,{\rm{Im}}\,y), in accordance with (3.9), we have

Rn(z)=12πi1+ia1+iaλ(ζ)enπiζλ(z)λ(ζ)dζ,Imy>a>1Imz,zγ(1,1).\displaystyle\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!=\!\frac{1}{2\pi i}\int\nolimits_{{{\mbox{\footnotesize{$-1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}^{{{\mbox{\footnotesize{$1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}\frac{\lambda^{\,\prime}\left(\zeta\right)e^{{{\mbox{\footnotesize{$-n\pi i\zeta$}}}}}}{\lambda(z)-\lambda\left(\zeta\right)}\,{\rm{d}}\zeta\,,\quad{\rm{Im}}\,y\!>\!a\!>\!1\!\geqslant\!{\rm{Im}}\,z,\ z\!\in\!\gamma(-1,1). (4.5)

In view of Corollary 2.3, we have |λ(ζ)|9|\lambda^{\,\prime}(\zeta)|\leqslant 9 and 4|λ(z)λ(ζ)||λ(z)||12λ(ia)|4|\lambda(z)-\lambda(\zeta)|\geqslant|\lambda(z)|\,|1-2\lambda(ia)| for all ζ[1+ia,1+ia]\zeta\in[-1+ia,1+ia] and zγ(1,1)z\!\in\!\gamma(-1,1). Hence we obtain

|n=1Rn(z)eiπny|n=112eπn(aImy)|12λ(ia)||λ(z)|=12(eπ((Imy)a)1)1|12λ(ia)||λ(z)|,\displaystyle\left|\sum\nolimits_{n=1}^{\infty}\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\,{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}\right|\!\leqslant\!\sum\nolimits_{n=1}^{\infty}\dfrac{12\,{\rm{e}}^{\pi n(a-{\rm{Im}}\,y)}}{|1\!-\!2\lambda(ia)|\,|\lambda(z)|}=\dfrac{12\,\left({\rm{e}}^{\pi(({\rm{Im}}\,y)-a)}-1\right)^{-1}}{|1\!-\!2\lambda(ia)|\,|\lambda(z)|}\,,

for every zγ(1,1)z\!\in\!\gamma(-1,1) and Imy>a>1{\rm{Im}}\,y\!>\!a\!>\!1. Together with (2.47), this inequality allows us to apply the Lebesgue dominated convergence theorem (see (nat, , p.​ 161)) in order to derive from (4.4) and (3.37) that

n1iπnHn(x)eiπny=14π2γ(1,1)λ(y)λ(y)λ(z)dz(x+z)2,Imy>1,\displaystyle\sum\limits_{n\geqslant 1}{\rm{i}}\pi\,n\,\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi ny$}}}}}=\frac{1}{4\pi^{2}}\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{\lambda(y)-\lambda(z)}\frac{{\rm{d}}z}{(x+z)^{2}}\,,\quad\ \ \ {\rm{Im}}\,y\!>\!1\,, (4.6)
n1iπnMn(x)eiπny=14π2γ(1,1)λ(y)1λ(y)λ(z)dz(x+z)2,Imy>1,\displaystyle\sum\limits_{n\geqslant 1}{\rm{i}}\pi\,n\,\eurm{M}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi ny$}}}}}=\frac{1}{4\pi^{2}}\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{1\!-\!\lambda(y)\!-\!\lambda(z)}\frac{{\rm{d}}z}{(x+z)^{2}}\,,\ \ \ {\rm{Im}}\,y\!>\!1\,, (4.7)

where both series converge absolutely and uniformly over all xx\in\Bb{R}.

5 . Partitions of the upper half-plane

  

In the Poincaré half-plane model of hyperbolic space the (generalized) semicircles γ(a,b)\gamma(a,b), γ(a,)\gamma(a,\infty), a,ba,b\in\Bb{R}, aba\neq b, (see (2.1)) are the hyperbolic straight lines connecting ideal points aa and bb, or aa and \infty, respectively. Moreover, given an arbitrary collection of four (three) points <a<b<c<d<+\!-\!\infty\!<\!a\!<\!b\!<\!c\!<\!d\!<\!+\!\infty (<a<b<c<+-\!\infty\!<\!a\!<\!b\!<\!c\!<\!+\!\infty) a bounded set AA\subset\Bb{H} is called an ideal hyperbolic quadrilateral (triangle), with four (three) vertices a,b,c,da,b,c,d (a,b,ca,b,c) if 𝔸\Bb{H}\cap\partial A is equal to the union of the three (two) lower arches γ(a,b),γ(b,c),γ(c,d)\gamma(a,b),\gamma(b,c),\gamma(c,d) (γ(a,b),γ(b,c)\gamma(a,b),\gamma(b,c)) and the roof (upper arch) γ(a,d)\gamma(a,d)  (γ(a,c)\gamma(a,c)). We will omit the word ”ideal” in the sequel, and note that clearly, there exists a unique bounded closed set satisfying this definition for any given collection of vertices lying on \Bb{R}.

Let Γ\Gamma be the group of Möbius transformations (called modular)
           ϕM(z):=(az+b)(cz+d)1\phi_{M}(z):=(az+b)(cz+d)^{-1}, M=(abcd)M=(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}), a,b,c,d\ a,b,c,d\in\Bb{Z}, adbc=1ad-bc=1,
on \Bb{H} with the superposition as a group operation (see (cha, , p.​ 11)). The theta subgroup Γϑ\Gamma_{\vartheta} of Γ\Gamma is defined as a collection of all ϕMΓ\phi_{M}\!\in\!\Gamma with MM, satisfying either M(1001)(mod 2)M\equiv(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix})({\rm{mod}}\,2) or M(0110)(mod 2)M\equiv(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix})({\rm{mod}}\,2), while the subgroup Γ(2)\Gamma(2) of Γϑ\Gamma_{\vartheta} as all ϕMΓϑ\phi_{M}\!\in\!\Gamma_{\vartheta} with M(1001)(mod 2)M\equiv(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix})({\rm{mod}}\,2).

Definition 5.1

. For (a,b){(1,),(1,1)}(a,b)\in\{(1,\infty),(-1,1)\}, we say that the set {ϕ(γ(a,b))}ϕΓϑ\{\phi(\gamma(a,b))\}_{\phi\in\Gamma_{\vartheta}} (called orbit of γ(a,b)\gamma(a,b) with respect to Γϑ\Gamma_{\vartheta}) generates a partition k1Ak=\sqcup_{k\geqslant 1}A_{k}=\Bb{H} of \Bb{H} if k1Ak=ϕΓϑϕ(γ(a,b))\cup_{k\geqslant 1}\partial A_{k}=\cup_{\phi\in\Gamma_{\vartheta}}\phi(\gamma(a,b)) and intAk{\rm{int}}A_{k}\neq\emptyset for all kk\in\Bb{N}.

In this section, we consider two partitions of \Bb{H} generated by {ϕ(γ(1,))}ϕΓϑ\{\phi(\gamma(1,\infty))\}_{\phi\in\Gamma_{\vartheta}} and {ϕ(γ(1,1))}ϕΓϑ\{\phi(\gamma(-1,1))\}_{\phi\in\Gamma_{\vartheta}}. Both partitions arise from the need to analyze integral operators with kernel of the form
                                  Gλ(z,ζ):=λ(z)/(λ(z)λ(ζ))G_{\lambda}(z,\zeta):=\lambda^{\,\prime}(z)/(\lambda(z)-\lambda(\zeta))
and integration over ζγ(1,1)\zeta\in\gamma(-1,1). An operator of this type (but with integration over zz) was first studied by Chibrikova in 1956 and later reproduced in Gakhov’s monograph (gah, , p.​ 513, (52.3)) from 1966 when considering the Riemann boundary value problem in a fundamental domain with respect to a Fuchsian group of linear fractional transformations. The importance of studying such integral operators was highlighted in the recent work of the fourth and fifth authors rad , where the following crucial observation was made. Given, e.g., a bounded function fHol(f\in{\rm{Hol}}(\Bb{H}), if we integrate of f(ζ)Gλ(z,ζ)f(\zeta)G_{\lambda}(z,\zeta) with respect to ζ\zeta along the semicircle γ(1,1)\gamma(-1,1), we obtain an analytic function in the domain 𝔻¯\Bb{H}\setminus(2\Bb{Z}+\overline{\Bb{D}}) which can be analytically extended to all of \Bb{H}, while preserving relevant growth control. The main goal of introducing the above mentioned partitions of the upper half-plane is to obtain explicit formulas for such an analytic extension, see Section 6 below. The results of Sections 5 and 6 are expressed in the language of the even-integer continued fractions which is the suggestion of the first author. This elaborates on the approach of the second and the third authors hed based on the connection between even-integer continued fractions and the even Gauss map, see, e.g., pio and lop .

5.1 . Even Gauss map in the upper half-plane.

For arbitrary NN\in\Bb{N} and n1,n2,,nNn_{1},n_{2},...,n_{N}\in\Bb{Z}_{\neq 0}\vphantom{A^{A^{A}}} let ϕnN,,n1\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}} be an even-integer continued fraction of the form (see sho and (10.67))

ϕnN,,n1(z):=12nN12nN112n212n1z,z[1,1]().\displaystyle\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(z):={{\mbox{\footnotesize{$\dfrac{1}{2n_{N}\!-\!\dfrac{1}{\begin{subarray}{l}\displaystyle\vphantom{A^{A}}2n_{N-1}\!-\!\\[-11.38092pt] \hphantom{2n_{k-1}11-}\ddots\\[-11.38092pt] \hphantom{2n_{k-1}11-\ddots}\displaystyle-\!\dfrac{1}{2n_{2}\!-\!\dfrac{1}{\vphantom{A^{A}}2n_{1}-z}}\end{subarray}}}$}}}}\ ,\ z\in[-1,1]\cup\left(\Bb{C}\setminus\Bb{R}\right)\,. (5.1)

The theta subgroup Γϑ\Gamma_{\vartheta} is generated by the Möbius transformations zz+2z\mapsto z+2 and z1/zz\mapsto-1/z (see (cha, , p.​ 112)). If we write, for zz\in\Bb{H}, \Bb{Z}_{\neq 0}^{k}:=\left(\Bb{Z}_{\neq 0}\right)^{k}, kk\in\Bb{N},

𝔫:=(𝔫𝔑,,𝔫1)ϕϕnN,,n1ϕϝϝ\displaystyle\eufm{n}\!:=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\sqcup_{k\geqslant 1}\Bb{Z}_{\neq 0}^{k},\ \phi_{\eufm{n}}\!:=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\,,\ \phi_{0}(z):=z, (5.2)

we can represent the theta subgroup as19{}^{\ref*{case17}}

{ϕ(z)}ϕΓϑ={ϕ𝔫(z),1/ϕ𝔫(z),ϕ𝔫(1/z),1/ϕ𝔫(1/z)}𝔫{}.\displaystyle\left\{\,\phi(z)\,\right\}_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}$}}}}}\!=\!\left\{\,\phi_{\eufm{n}}(z),\ -1/\phi_{\eufm{n}}(z),\ \phi_{\eufm{n}}(-1/z),\ -1/\phi_{\eufm{n}}(-1/z)\,\right\}_{\,{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\!\{0\}$}}}}}\,. (5.3)

Here, we see that

{1/ϕn1(z)=2n1+z,ifN=1,1/ϕnN,,n1(z)=2nN+ϕnN1,,n1(z),ifN2,\displaystyle\left\{\begin{array}[]{rlll}-1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(z)&=&-2n_{1}+z\,,&\ \mbox{if}\ N=1\,,\\ -1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(z)&=&-2n_{N}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-1},...,n_{1}$}}}}}(z)\,,&\ \mbox{if}\ N\geqslant 2\,,\end{array}\right. (5.6)

for every z[1,1]()z\in[-1,1]\cup\left(\Bb{C}\setminus\Bb{R}\right), NN\in\Bb{N} and (nN,,n1)(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{N}. Hence, if AA\subset\Bb{H} is invariant under the inversion z1/zz\mapsto-1/z, i.e., if A=1/A:={1/z|zA}A=-1/A:=\{-1/z\,|\,z\in A\}, then its orbit with respect to Γϑ\Gamma_{\vartheta} can be written in the form

{ϕ(A)}ϕΓϑ={ 2n0+ϕ𝔫(A)}n0,𝔫{},A𝔸𝔸\displaystyle\left\{\,\phi(A)\,\right\}_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}$}}}}}\!=\!\left\{\,2n_{0}+\phi_{\eufm{n}}(A)\,\right\}_{\,{{\mbox{\footnotesize{$n_{0}$}}}}\,{{\mbox{\footnotesize{$\in$}}}}\,{{\mbox{\footnotesize{$\Bb{Z}$}}}}\,,\ {{\mbox{\footnotesize{$\eufm{n}$}}}}\,{{\mbox{\footnotesize{$\in$}}}}\,{{\mbox{\footnotesize{$\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\!\{0\}$}}}}}\ ,\quad A\subset\Bb{H}\,,\ A=-1/A\,. (5.7)

This suggests the introduction of the following subset of Γϑ\Gamma_{\vartheta},

Γϑ||:={ϕ𝔫}𝔫,(aϕz+bϕ)(cϕz+dϕ)1:=ϕ(z),ϕΓϑ||,\displaystyle\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}:=\left\{\,\phi_{\eufm{n}}\,\right\}_{{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}$}}}}}\,,\quad(a_{\phi}z+b_{\phi})(c_{\phi}z+d_{\phi})^{-1}:=\phi(z)\,,\quad\phi\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\,, (5.8)

aϕ,bϕ,cϕ,dϕa_{\phi},b_{\phi},c_{\phi},d_{\phi}\!\in\!\Bb{Z}, with the basic properties (see Sections 10.4.1 and 1.6)

|aϕ|<|bϕ|<|dϕ|,|aϕ|<|cϕ|<|dϕ|,ϕ(𝕜𝕜)𝕜𝕜𝔽ϕΓϑ||\displaystyle|a_{\phi}|\!<\!|b_{\phi}|\!<\!|d_{\phi}|,\ |a_{\phi}|\!<\!|c_{\phi}|\!<\!|d_{\phi}|,\ \ \phi\left(\Bb{H}_{|{\rm{Re}}|\leqslant 1}\right)\!\subset\!\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\ \ \phi\!\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\,. (5.9)

Let {x}\{x\} and x\lfloor{x}\rfloor denote the usual fractional and integer parts of xx\in\Bb{R}, respectively, and put (see (bh2, , pp.​ 599, 600))

||:=(1+i\displaystyle\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}:=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup(-1+\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0})\sqcup(1+\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0})\,, :=m(2m+||).\displaystyle\mathcal{F}^{\infty}_{{{\mbox{\tiny{$\square$}}}}}:=\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\left(2m+\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\right). (5.10)

For arbitrary xx\!\in\!\Bb{R} we define its even integer part x22\rceil{x}\lceil_{\!2}\,\in\!2\Bb{Z} as the average of the endpoints of the interval that the point xx belongs to in the following interval partition of the real axis,

m1~^\displaystyle\Bb{R}=(-1,1)\ \sqcup\ \bigsqcup\limits_{{{\mbox{\footnotesize{$\ m\geqslant 1$}}}}}\ \bigg{(}\Big{(}-2m-1,-2m+1\Big{]}\ \sqcup\ \Big{[}2m-1,2m+1\Big{)}\bigg{)}, (5.11)

and we define its even fractional part {x}2[1,1]\{x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\in[-1,1] to be {x}2:=xx2\{x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}:=x-\rceil{x}\lceil_{\!2}. As such, they have the properties {x}2=(1+2{(1+|x|)/2})sign(x)\{x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\!=\!(-1+2\,\{(1+|x|)/2\})\,{\rm{sign}}(x), {x}2={x}2\{-x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\!=\!-\{x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}, x2=2(1+|x|)/2sign(x)\rceil{x}\lceil_{\!2}=\!2\,\lfloor{(1\!+\!|x|)/2}\rfloor\,{\rm{sign}}(x), x2=x2\rceil{-x}\lceil_{\!2}=\!-\rceil{x}\lceil_{\!2} for each xx\!\in\!\Bb{R}, and

2n+sign(n)2=2n,{2n+sign(n)}2=sign(n),n\displaystyle\rceil{-2n+{\rm{sign}}(n)}\lceil_{\!2}\,=-2n\,,\quad\{\hskip 1.9919pt-2n+{\rm{sign}}(n)\hskip 1.9919pt\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\!=\!{\rm{sign}}(n),\quad n\in\Bb{Z}\,. (5.12)
Definition 5.2

. We define the complex analogue 𝔾𝕜𝕜𝕜𝕜\Bb{G}_{2}:\Bb{H}_{|{\rm{Re}}|\leqslant 1}\to\Bb{H}_{|{\rm{Re}}|\leqslant 1} of the even Gauss map G2:[1,1][1,1]G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}:[-1,1]\to[-1,1], G2(0):=0G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}(0):=0, G2(x):={1/x}2G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}(x):=\{-1/x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}, x[1,1]{0}x\in[-1,1]\setminus\{0\}, associated with the even fractional part {x}2\{x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}, as follows

𝔾ϝ{ϝ}𝕀ϝϝ𝕜𝕜\displaystyle\Bb{G}_{2}(z):=\ \left\{{\rm{Re}}\left(-\dfrac{1}{z}\right)\right\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}+{\rm{i}}\,{\rm{Im}}\left(-\dfrac{1}{z}\right)\,,\ \ \ \ z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,. (5.13)

For z𝕜𝕜z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}, we have that

𝔾ϝϝϝ𝕜𝕜ϝ\displaystyle\Bb{G}_{2}(z)=-\dfrac{1}{z}-\left\rceil{{\rm{Re}}\left(-\dfrac{1}{z}\right)}\right\lceil_{2}\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\ ,\qquad\left\rceil{{\rm{Re}}\left(-\dfrac{1}{z}\right)}\right\lceil_{2}\in 2\,\Bb{Z}_{\neq 0}\ . (5.14)

We note that

𝔾ϝϝϝ𝕜𝕜ϝ𝕜𝕜𝕜𝕜𝔽||\displaystyle\vspace{-0,3cm}\Bb{G}_{2}(z)=-1/z\ ,\ z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\ \ \Longleftrightarrow\ \ z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\cap\left(-1\big{/}\Bb{H}_{|{\rm{Re}}|<1}\right)=\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}},\vspace{-0,1cm} (5.15)

and observe that it follows from (5.14) and (5.11) that22{}^{\ref*{case19}}

(a)𝔾γσ=γ(σ,),σ{1,1},(b)𝔾γσ=γ(σ,0),σ{1,1},(c)𝔾ϕnγσ=γ(σn,),σn:=sign(n),n\displaystyle\begin{array}[]{lrcll}{\rm{(a)}}&\Bb{G}_{2}(\gamma(\sigma,0))&=&\gamma(\sigma,\infty)\,,&\quad\sigma\in\{1,-1\}\,,\\[5.69046pt] {\rm{(b)}}&\Bb{G}_{2}(\gamma(\sigma,\infty))&=&\gamma(-\sigma,0)\,,&\quad\sigma\in\{1,-1\}\,,\\[2.84544pt] {\rm{(c)}}&\Bb{G}_{2}\Big{(}\phi_{{{\mbox{\footnotesize{$n$}}}}}\big{(}\hskip 1.9919pt\gamma(\sigma_{n},\infty)\hskip 1.9919pt\big{)}\Big{)}&=&\gamma(\sigma_{n},\infty)\,,&\quad\sigma_{n}:={\rm{sign}}(n)\,,\ n\in\Bb{Z}_{\neq 0}\,.\end{array} (5.19)

Estimating from above the modulus of zz lying on the union of semicircles γ(1,0)γ(1,0)\gamma(1,0)\cup\gamma(-1,0) we obtain from the identity Im(1/z)=(Imz)/|z|2{\rm{Im}}(-1/z)=({\rm{Im}}\,z)/|z|^{2}, zz\in\Bb{H}, that

Im𝔾ϝ𝕀ϝ𝕀ϝϝ𝕜𝕜𝔽||𝕀\displaystyle{\rm{Im}}\,\Bb{G}_{2}\left(z\right)\geqslant\dfrac{{\rm{Im}}\,z}{(1/2)+\sqrt{(1/4)-({\rm{Im}}\,z)^{2}}}\ ,\quad z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\subset\Bb{H}_{{\rm{Im}}\leqslant 1/2}\,. (5.20)

For arbitrary NN\in\Bb{N} and n1,n2,,nNn_{1},n_{2},...,n_{N}\in\Bb{Z}_{\neq 0} it follows from (5.6) and (5.9) that

{Re(1/ϕn1(z))2n1+(1,1),ifN=1,z𝕜𝕜Re(1/ϕnN,,n1(z))2nN+(1,1),ifN2,z𝕜𝕜\displaystyle\left\{\begin{array}[]{rlll}{\rm{Re}}\left(-1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(z)\right)&\in&-2n_{1}+(-1,1)\,,&\ \mbox{if}\ N=1\,,\ z\!\in\!\Bb{H}_{|{\rm{Re}}|<1}\,,\\ {\rm{Re}}\left(-1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(z)\right)&\in&-2n_{N}+(-1,1)\,,&\ \mbox{if}\ N\geqslant 2\,,\ z\!\in\!\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,,\end{array}\right. (5.23)

and hence we see from (5.14) and (5.6) that23{}^{\ref*{case20}}

(a)𝔾ϕnN,,n1ϝ=z,ifN1,z𝕜𝕜(b)𝔾ϕnN,,n1ϝ=ϕnNk,,n1(z),if 1kN1,z𝕜𝕜\displaystyle\begin{array}[]{lrlllc}{\rm{(a)}}&\Bb{G}^{N}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\big{)}&\!=\!&z\,,&\ \mbox{if}\ N\!\geqslant\!1\,,&\ \ z\!\in\!\Bb{H}_{|{\rm{Re}}|<1}\,;\\[2.84544pt] {\rm{(b)}}&\Bb{G}^{k}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\big{)}&\!=\!&\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)\,,&\ \mbox{if}\ 1\!\leqslant\!k\!\leqslant\!N\!-\!1\,,&\ \ z\!\in\!\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,.\end{array} (5.26)

5.2 . The Schwarz partition of the upper half-plane.

We introduce the notation

ϕ:=𝔫:=nN,,n1:=ϕnN,,n1(),𝔫=(𝔫𝔑,,𝔫1)\displaystyle\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\phi}\!:=\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}\!:=\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\!:=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)},\ \ \eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N},\ \ N\!\in\!\Bb{N}\,,

where ϕ:=ϕ𝔫\phi:=\phi_{\eufm{n}} is as in (5.1). We associate with an arbitrary NN-tuple 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, NN\!\in\!\Bb{N}, and, correspondingly, with each ϕ=ϕ𝔫Γϑ||\phi\!=\!\phi_{\eufm{n}}\!\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}, its ”sign”

σϕ:=σ𝔫:=σnN,,n1:=σn1:=sign(n1){+1,1},ϕ=ϕ𝔫Γϑ||.\displaystyle\sigma_{\phi}:=\sigma_{\eufm{n}}:=\sigma_{{}_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}}:=\sigma_{n_{1}}:={\rm{sign}}(n_{1})\in\{+1,-1\}\ ,\quad\phi=\phi_{\eufm{n}}\in\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\,. (5.27)

Depending on the sign of the transformation ϕ=ϕ𝔫Γϑ||\phi=\phi_{\eufm{n}}\in\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}, 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}, which we apply to the Schwarz quadrilateral \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, we add to \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} one of the rays ±1+i\pm 1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R}_{>0},

|σϕ:=|σ𝔫:=|σn1:=(σn1+i),\displaystyle\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\phi}}:=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\eufm{n}}}:=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}:=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\left(\sigma_{n_{1}}\!+\!i\Bb{R}_{>0}\right)\,, (5.28)

and denote the resulting image as follows

^ϕ:=^𝔫:=^nN,,n1:=ϕnN,,n1(|σn1)𝕜𝕜𝔽\displaystyle\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\phi}:=\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}:=\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}:=\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\big{)}\subset\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,. (5.29)

For each 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} the open set 𝔫\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}} is a hyperbolic quadrilateral with four vertices {ϕ𝔫(x)|x{1,0,1,}}[1,1]\{\,\phi_{\eufm{n}}(x)\,|\,x\in\{-1,0,1,\infty\}\,\}\subset[-1,1] whose position is completely determined by the value of σ𝔫\sigma_{\eufm{n}}, as follows from the relationships (see Section 10.4.1)

(a)ϕ𝔫(1)<ϕ𝔫(0)<ϕ𝔫(1),(b)σ𝔫ϕ𝔫()<σ𝔫ϕ𝔫(σ𝔫),(c)sign(ϕ𝔫(1))=sign(ϕ𝔫(0))=sign(ϕ𝔫(1)),𝔫\displaystyle\begin{array}[]{l}{\rm{(a)}}\ \phi_{\eufm{n}}(-1)<\phi_{\eufm{n}}(0)<\phi_{\eufm{n}}(1)\ ,\quad{\rm{(b)}}\ \sigma_{\eufm{n}}\phi_{\eufm{n}}(\infty)<\sigma_{\eufm{n}}\phi_{\eufm{n}}(-\sigma_{\eufm{n}})\,,\\[5.69046pt] {\rm{(c)}}\ \ {\rm{sign}}\big{(}\phi_{\eufm{n}}(-1)\big{)}={\rm{sign}}\big{(}\phi_{\eufm{n}}(0)\big{)}={\rm{sign}}\big{(}\phi_{\eufm{n}}(1)\big{)}\,,\quad\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,.\end{array} (5.32)

Moreover, for arbitrary 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} the properties (5.32) and (γ0(a,b):=γ(a,b)\gamma_{0}(a,b)\!:=\!\gamma(a,b))

γ𝔫(a,b):=ϕ𝔫(γ(a,b))=γ(ϕ𝔫(a),ϕ𝔫(b)),a[1,1],b[1,1]{},\displaystyle\gamma_{\eufm{n}}(a,b)\!:=\!\phi_{\eufm{n}}\big{(}\gamma(a,b)\big{)}\!=\!\gamma\big{(}\phi_{\eufm{n}}(a),\phi_{\eufm{n}}(b)\big{)}\,,\ \ a\in[-1,1]\,,\ b\in[-1,1]\cup\{\infty\}\,, (5.33)

aba\neq b, imply that 𝔽\Bb{H}\cap\partial\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}} is a hyperbolic polygon which consists of the three lower arches γ𝔫(σ𝔫,)\gamma_{\eufm{n}}(-\sigma_{\eufm{n}},\infty), γ𝔫(1,0)\gamma_{\eufm{n}}(-1,0), γ𝔫(0,1)\gamma_{\eufm{n}}(0,1), which do not belong to ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}, and of the roof γ𝔫(σ𝔫,)\gamma_{\eufm{n}}(\sigma_{\eufm{n}},\infty), which is in ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}(see Figure 3). Obviously, 𝔫=int(^𝔫)\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}={\rm{int}}(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}})  and  𝔫=^𝔫\partial\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}=\partial\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}.

Lemma 5.3

. Let NN\!\in\!\Bb{N} and 𝔫:=(𝔫𝔑,,𝔫1)\eufm{n}:=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N} (see also (5.1), (5.29)).

(a) For arbitrary 𝔫,𝔪\eufm{n},\eufm{m}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\cup_{k\geqslant 1}\Bb{Z}_{\neq 0}^{k}, 𝔫𝔪\eufm{n}\neq\eufm{m}, we have ^𝔫^𝔪=\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}\cap\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{m}}=\emptyset.

(b) The lower arches

ϕ𝔫(γ(1,0)),ϕ𝔫(γ(0,1)),ϕ𝔫(γ(σn1,)),\displaystyle\phi_{\eufm{n}}\big{(}\gamma(-1,0)\big{)}\,,\quad\phi_{\eufm{n}}\big{(}\gamma(0,1)\big{)}\,,\quad\phi_{\eufm{n}}\big{(}\gamma(-\sigma_{n_{1}},\infty)\big{)}\,,

of nN,,n1\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}} are the roofs of the respective quadrilaterals

nN,,n1,1,nN,,n1, 1,nN,,n1+σn1.\displaystyle\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1},-1}\,,\quad\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1},\,1}\,,\quad\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}+\sigma_{n_{1}}}\,.

(c) The roofs of 1\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271pt1} and 1\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271pt-1} are γ(0,1)\gamma(0,1) and γ(1,0)\gamma(-1,0), respectively.

(d) The roof of nN,,n1\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}} is the lower arch of

nN,,n1σn1,ifn1σn1,N1,nN,,n2,ifn1=σn1,N2.\displaystyle\begin{array}[]{lr}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}-\sigma_{n_{1}}}&,\ \mbox{if}\ \ n_{1}\!\neq\!\sigma_{n_{1}}\ ,\quad N\geqslant 1\,,\\[2.84544pt] \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{2}}&,\ \mbox{if}\ \ n_{1}\!=\!\sigma_{n_{1}}\ ,\quad N\geqslant 2\,.\end{array}

(e)  2max{Imz|z^nN,,n1}1/(1+k=2N(2|nk|1))1/N2\max\left\{\ {\rm{Im}}z\,\left|\,z\in\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\,\right\}\right.\leqslant 1\Big{/}\left(1+\sum\limits_{k=2}^{N}(2|n_{k}|-1)\right)\leqslant 1/N holds,
(e) where it is assumed that k=21:=0\sum\nolimits_{k=2}^{1}:=0.

The proof of Lemma 5.3 is supplied in Section 10.4.2. In the sequel, we agree to use the notation

ϕ𝔫,𝔪:=ϕnN,,n1,mM,,m1,𝔪=(𝔪𝔐,,𝔪1),𝔫=(𝔫𝔑,,𝔫1)\displaystyle\phi_{\eufm{n},\eufm{m}}\!:=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1},m_{M},...,m_{1}$}}}}}\,,\ \ \eufm{m}\!=\!(m_{M},...,m_{1}),\ \eufm{n}\!=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,. (5.34)

As we combine the properties (5.26)(b) and (5.19)(c) together with the definitions (5.29), (5.28) and (5.27), we find for every NN\!\in\!\Bb{N} and (nN,,n1)(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N} that24{}^{\ref*{case21}}

(a)𝔾𝔽^=|σn1=(σn1+i),ifN1;(b)𝔾𝔽^=^nNk,,n1=ϕnNk,,n1(|σn1),if 1kN1.\displaystyle\begin{array}[]{lrcccll}{\rm{(a)}}&\Bb{G}^{N}_{2}\left(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\right)&\!=\!&\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}&\!=\!&\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\left(\sigma_{n_{1}}\!+\!i\Bb{R}_{>0}\right)\,,&\ \mbox{if}\ N\!\geqslant\!1\,;\\[7.11317pt] {\rm{(b)}}&\Bb{G}^{k}_{2}\left(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\!\right)&\!=\!&\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N-k},...,n_{1}}&\!=\!&\phi_{{{\mbox{\footnotesize{$n_{N-k},...,n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\!\big{)}\,,&\ \mbox{if}\ 1\!\leqslant\!k\!\leqslant\!N\!-\!1\,.\end{array} (5.37)

It follows from Lemma 5.3(a) and (5.29) that the sets ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}, 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}, are disjoint subsets of 𝕜𝕜𝔽\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, and their union equals all of 𝕜𝕜𝔽\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}.

Lemma 5.4

. Let \Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\cup_{k\geqslant 1}\Bb{Z}_{\neq 0}^{k}. We have the partition

𝕜𝕜𝔫{}𝔽^𝔽||ϕΓϑ||ϕ𝔽|σϕ\displaystyle\Bb{H}_{|{\rm{Re}}|\leqslant 1}=\bigsqcup\limits_{\,{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\,\eufm{n}}=\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\bigsqcup\limits_{\,{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}$}}}}}\phi\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\phi}}\right), (5.38)

where for 𝔫=0\eufm{n}=0 we use the notation ^0:=||\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.13791pt0}:=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$||$}}}}}.

The proof of Lemma 5.4 is supplied in Section 10.4.2. It is clear that (see (5.10))

=n1((|+12n)(|1+2n)).\displaystyle\mathcal{F}^{\infty}_{{{\mbox{\tiny{$\square$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\bigsqcup\nolimits_{\,n\geqslant 1}\ \Big{(}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|\!+\!1$}}}}}-2n\big{)}\sqcup\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|\!-\!1$}}}}}+2n\big{)}\Big{)}\,. (5.39)

By combining (5.38) and (5.39), we arrive at the partition

𝔽n0𝔽ðn1,,nNN(2n0+ϕnN,,n1((sign(n1)+i\Bb{H}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ \sqcup\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}_{\neq 0}$}}}}}\bigg{(}-2n_{0}+\Big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\big{(}{\rm{sign}}(n_{0})+i\Bb{R}_{>0}\big{)}\Big{)}\bigg{)}\\[-2.84544pt] \sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\bigg{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\,n_{N},...,n_{1}$}}}}}\Big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\big{(}{\rm{sign}}(n_{1})+i\Bb{R}_{>0}\big{)}\Big{)}\bigg{)}\,. (5.40)

which we refer to as the Schwarz partition of \Bb{H}. The following property is obtained in Section 10.4.3.

Theorem 5.5

.​ The set {ϕ(1+i}ϕΓϑ\{\phi(1\!+\!{\rm{i}}\Bb{R}_{>0})\}_{\phi\in\!\Gamma_{\vartheta}}​ generates the Schwarz partition of \Bb{H}.

Remark 5.6

.​​ It is known that the fundamental domain (see (cha, , p.​ 15)) FΓ(2)F_{\Gamma(2)} for the subgroup Γ(2)Γ\Gamma(2)\!\subset\!\Gamma, (elements written below are linear fractional mappings)

Γ(2)\displaystyle\Gamma(2) =n0{2n0+z}n1,,n2NN{2n0+ϕn2N,n2N1,,n1(z)}\displaystyle=\bigsqcup\limits_{n_{0}\in\Bb{Z}}\left\{2n_{0}+z\right\}\ \ \sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{2N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\left\{2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{2N},n_{2N-1},...,n_{1}$}}}}}(z)\right\}
n1,,n2N1N{2n0+ϕn2N1,n2N2,,n1(1/z)},\displaystyle\hskip 46.37813pt\sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{2N-1}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\left\{2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{2N-1},n_{2N-2},...,n_{1}$}}}}}(-1/z)\right\}, (5.41)

generated by the Möbius transformations zz+2z\mapsto z+2 and zz/(12z)z\mapsto z/(1-2z) (see (cha, , p.​ 111)), can be chosen by any of the four sets written in (2.82), for instance, FΓ(2)=γ(1,)γ(1,0)F_{\Gamma(2)}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\gamma(-1,\infty)\sqcup\gamma(1,0), where intFΓ(2)={\rm{int}}F_{\Gamma(2)}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} (cp. (cha, , p.​ 115)). Then

ϕΓ(2)ϕ𝔽ΓϕΓ(2)ϕ𝔽ϕγϕγ\displaystyle\Bb{H}=\bigsqcup\limits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma(2)$}}}}}\phi\big{(}F_{\Gamma(2)}\big{)}=\bigsqcup\limits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma(2)$}}}}}\Big{(}\phi\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)}\sqcup\phi\big{(}\gamma(-1,\infty)\big{)}\sqcup\phi\big{(}\gamma(1,0)\big{)}\Big{)}\,,\ (5.42)

forms a partition of the upper half-plane corresponding to the subgroup Γ(2)\Gamma(2). But it follows from (5.41), 1/=-1/\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and (5.42) that

ϕΓ(2)ϕ()=n0(2n0+)n1,,nNN(2n0+ϕnN,,n1()),\displaystyle\bigsqcup\limits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma(2)$}}}}}\!\!\!\phi\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)}=\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}$}}}}}\left(2n_{0}+\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)\ \sqcup\!\!\!\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\!\!\!\!\!\!\left(2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})\right)\,,\ (5.43)

and therefore the partition (5.42) coincides with the Schwarz partition (5.40) on the set {ϕ}ϕΓϑϕΓϕ𝔽\Bb{H}\setminus\{\phi(1\!+\!{\rm{i}}\Bb{R}_{>0})\}_{\phi\in\Gamma_{\vartheta}}=\sqcup_{\phi\in\Gamma(2)}\phi(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}), because

ϕΓ(2)ϕ(γ(1,)γ(1,0))\displaystyle\bigsqcup\nolimits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma(2)$}}}}}\ \phi\big{(}\gamma(-1,\infty)\sqcup\gamma(1,0)\big{)} =ϕΓ(2)ϕ()=ϕΓϑϕ()\displaystyle=\bigcup\nolimits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma(2)$}}}}}\ \phi\big{(}\partial_{\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)}=\bigcup\nolimits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}$}}}}}\ \phi(\partial_{\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})
=ϕΓϑϕ(1+i\displaystyle=\bigcup\nolimits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}$}}}}}\ \phi(1\!+\!{\rm{i}}\Bb{R}_{>0}), (5.44)

as follows from γ(1,)=2+γ(1,)\gamma(1,\infty)=2+\gamma(-1,\infty), γ(1,0)=ϕ1(1/γ(1,0))\gamma(-1,0)=\phi_{-1}(-1/\gamma(1,0)), (3.39), 1/=-1/\partial_{\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}=\partial_{\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, {ϕ(z)}ϕΓϑ={ϕ(z),ϕ(1/z)}ϕΓ(2)\left\{\,\phi(z)\,\right\}_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}$}}}}}\!=\!\left\{\,\phi(z)\,,\ \phi(-1/z)\,\right\}_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma(2)$}}}}} (see (cha, , p.​ 115)) and ϕ1(z)=z/(12z)\phi_{-1}(z)=z/(1-2z), zz\in\Bb{H}, ϕ1Γ(2)\phi_{-1}\in\Gamma(2). But on the set

ϕΓϑϕ(1+iϕΓ(2)ϕ𝔽Γ\displaystyle\bigcup\limits_{{{\mbox{\footnotesize{$\phi\in\Gamma_{\vartheta}$}}}}}\phi\big{(}1+{\rm{i}}\Bb{R}_{>0}\big{)}=\Bb{H}\setminus\bigcup\limits_{{{\mbox{\footnotesize{$\phi\in\Gamma(2)$}}}}}\phi\big{(}\,{\rm{int}}F_{\Gamma(2)}\big{)} (5.45)

these two partitions are different. This difference is essential, since the sets that make up the Schwarz partition are much simpler than the sets of the partition (5.42). Actually, if the partition (5.42) adds to each open hyperbolic quadrilateral ϕnN,,n1()\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}) its two sides γnN,,n1(0,1)\gamma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(0,1) and γnN,,n1(1,)\gamma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(-1,\infty), which can be two lower arches or a lower arch and a roof, depending on the sign of n1n_{1}, then the Schwarz partition adds to such a quadrilateral its own roof γnN,,n1(sign(n1),)\gamma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}({\rm{sign}}(n_{1}),\infty), where n1,,nNn_{1},...,n_{N}\in\Bb{Z}_{\neq 0} and NN\in\Bb{N} (see the note before Lemma 5.3). Hence, the Schwarz partition (5.40) is an easy-to-use modification of the known partition (5.42) associated with the subgroup Γ(2)\Gamma(2). Moreover, the choice of the most convenient partition of the set (5.45) is a separate issue25{}^{\ref*{case42}}, which is no longer related to the structure of FΓ(2)FΓ(2)F_{\Gamma(2)}\cap\partial F_{\Gamma(2)}, but depends on the mutual arrangement of the sets {ϕ(intFΓ(2))}ϕΓ(2)\{\phi(\,{\rm{int}}F_{\Gamma(2)})\}_{\phi\in\Gamma(2)}.

5.3 . Even rational partition of the upper half-plane.

By virtue of Lemma 2 of (cha, , p.​ 112)20{}^{\ref*{case43}} and (5.9), the relationships (5.3) and (5.8) can be expressed in the form (the elements of the sets below are linear fractional mappings)26{}^{\ref*{case44}}

Γϑ={1z+00z+1,0z11z+0}\displaystyle\Gamma_{\vartheta}\!=\!\left\{\,{{\mbox{\footnotesize{$\dfrac{1\cdot z+0}{0\cdot z+1}$}}}}\,,\ {{\mbox{\footnotesize{$\dfrac{0\cdot z-1}{1\cdot z+0}$}}}}\,\right\}
{az+bcz+d,(c)z+(d)az+b,bz+(a)dz+(c),(d)z+cbz+(a)|az+bcz+dΓϑ||},\displaystyle\hskip 2.27626pt\bigsqcup\bigg{\{}\,{{\mbox{\footnotesize{$\dfrac{az+b}{cz+d}$}}}}\,,\,{{\mbox{\footnotesize{$\dfrac{(-c)z+(-d)}{az+b}$}}}}\,,\,{{\mbox{\footnotesize{$\dfrac{bz+(-a)}{dz+(-c)}$}}}}\,,\,{{\mbox{\footnotesize{$\dfrac{(-d)z+c}{bz+(-a)}$}}}}\ \bigg{|}\ {{\mbox{\footnotesize{$\dfrac{az+b}{cz+d}$}}}}\in\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\,\bigg{\}}\,, (5.46)
Γϑ||={az+bcz+d||a|<|b|<d,|a|<|c|<dabcd0(mod 2)adbc=1,a,b,c,d}\displaystyle\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\!=\!\bigg{\{}{{\mbox{\footnotesize{$\,\dfrac{az+b}{cz+d}$}}}}\ \bigg{|}\ {{\mbox{\footnotesize{$\begin{array}[]{l}|a|\!<\!|b|\!<\!d,\ |a|\!<\!|c|\!<\!d\\ ab\equiv cd\equiv 0({\rm{mod}}\,2)\\ ad-bc\!=\!1\,,\ a,b,c,d\!\in\!\Bb{Z}\end{array}$}}}}\,\bigg{\}} (5.50)
={12nN12nN112n212n1z|n1,n2,,nN}\displaystyle\hskip 5.40608pt=\!\!\Bigg{\{}\,{{\mbox{\footnotesize{$\dfrac{1}{2n_{N}\!-\!\dfrac{1}{\begin{subarray}{l}\displaystyle\vphantom{A^{A}}2n_{N-1}\!-\!\\[-11.38092pt] \hphantom{2n_{k-1}11-}\ddots\\[-11.38092pt] \hphantom{2n_{k-1}11-\ddots}\displaystyle-\!\dfrac{1}{2n_{2}\!-\!\dfrac{1}{\vphantom{A^{A}}2n_{1}-z}}\end{subarray}}}$}}}}\ \Bigg{|}\ n_{1},n_{2},...,n_{N}\in\Bb{Z}_{\neq 0}\,,\ N\in\Bb{N}\ \Bigg{\}}\,, (5.51)

where no repetitions of linear fractional maps occur in the listing of the sets on the right-hand sides of (5.51) and (5.46). That this is so follows from Lemma 5.3(a) and the fact that neither of the mappings zzz\mapsto z nor z1/zz\mapsto-1/z can be an element of the rightmost set of (5.46), where the listed four linear fractional maps are all different as the unique entry in the associated 2×22\times 2 matrix with biggest absolute value (denoted by dd) occupies four different positions. Consequently, it makes sense to define the ”order” of each element ϕΓϑ\phi\!\in\!\Gamma_{\vartheta} by

d(ϕ)={0,ifϕ(z)=ϕ0(z)=z;1,ifϕ(z)=1/z;N,ifϕ(z)=ϕ𝔫(z);N1,ifϕ(z)=1/ϕ𝔫(z);N+1,ifϕ(z)=ϕ𝔫(1/z);N,ifϕ(z)=1/ϕ𝔫(1/z),\displaystyle d(\phi)\!=\!\left\{\begin{array}[]{llll}0,&\mbox{if}\ \phi(z)=\phi_{0}(z)=z\,;&\ 1,&\mbox{if}\ \phi(z)=-1/z\,;\\ N,&\mbox{if}\ \phi(z)=\phi_{\eufm{n}}(z)\,;&\ N-1,&\mbox{if}\ \phi(z)=-1/\phi_{\eufm{n}}(z)\,;\\ N+1,&\mbox{if}\ \phi(z)=\phi_{\eufm{n}}(-1/z)\,;&\ N,&\mbox{if}\ \phi(z)=-1/\phi_{\eufm{n}}(-1/z)\,,\\ \end{array}\right. (5.55)

where 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{N}, NN\in\Bb{N}. Then the functional relations λ(z+2)=λ(z)\lambda(z+2)=\lambda(z), λ(1/z)=1λ(z)\lambda(-1/z)=1-\lambda(z), zz\in\Bb{H}, give that (cf. (2.21), (cha, , p.​ 111))

λ(ϕ(z))={λ(z),ifd(ϕ){0}21λ(z),ifd(ϕ)2ϕΓϑ,z\displaystyle\lambda\big{(}\phi(z)\big{)}=\left\{\begin{array}[]{rl}\lambda(z),&\ \mbox{if}\ d(\phi)\in\{0\}\cup 2\Bb{N}\,,\\[2.84544pt] 1-\lambda(z),&\ \mbox{if}\ d(\phi)\in 2\Bb{N}\!-\!1\,,\end{array}\right.\ \quad\phi\!\in\!\Gamma_{\vartheta}\,,\ z\in\Bb{H}\,. (5.58)

Moreover, since the semicircle γ(1,1)\gamma(-1,1) is invariant under the inversion z1/zz\mapsto-1/z, we realize that

λ(ϕ(γ(1,1)))=λ(γ(1,1))=(1/2)+iϕΓϑ\displaystyle\lambda\Big{(}\phi\big{(}\gamma(-1,1)\big{)}\Big{)}=\lambda\big{(}\gamma(-1,1)\big{)}=(1/2)+i\Bb{R}_{>0}\,,\quad\phi\!\in\!\Gamma_{\vartheta}\,,\ (5.59)

and, in the notation Λ:=(0,1)(\Lambda:=(0,1)\cup(\Bb{C}\setminus\Bb{R}),

λ(ϕ(|σϕ))={λ(γ(σϕ,))=Λifd(ϕ){0}2λ(γ(σϕ,0))=Λifd(ϕ)2\displaystyle\lambda\left(\phi\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\phi}}\big{)}\right)=\left\{\begin{array}[]{rl}\lambda\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\cup\gamma(\sigma_{\phi},\infty)\right)=\Bb{R}_{<0}\cup\Lambda,&\ \mbox{if}\ d(\phi)\in\{0\}\cup 2\Bb{N}\,,\\[2.84544pt] \lambda\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\cup\gamma(-\sigma_{\phi},0)\right)=\Bb{R}_{>1}\cup\Lambda,&\ \mbox{if}\ d(\phi)\in 2\Bb{N}\!-\!1\,,\end{array}\right. (5.62)

holds for every ϕΓϑ{ϕ0}\phi\!\in\!\Gamma_{\vartheta}\cup\{\phi_{0}\} (where ϕ0(z):=z\phi_{0}(z):=z). By Lemma 2.4, (5.62) means that λ\lambda is one-to-one on each set of the partition (5.38), with the exception of ||\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}, where λ(γ(1,))=λ(γ(1,))=\lambda(\gamma(1,\infty))=\lambda(\gamma(-1,\infty))=\Bb{R}_{<0}. This has one useful consequence, the proof of which is supplied in Section 10.4.4.

Lemma 5.7

. Let us write \Bb{Z}_{\neq 0}^{2\Bb{N}_{\eurm{f}}}\!:=\!\cup_{k\geqslant 1}\Bb{Z}_{\neq 0}^{2k} and \Bb{Z}_{\neq 0}^{2\Bb{N}_{\eurm{f}}-1}\!:=\!\cup_{k\geqslant 1}\Bb{Z}_{\neq 0}^{2k-1}. Then for arbitrary y𝔽y\in\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right) and σ{1,1}\sigma\in\{1,-1\} we have

{z𝕜𝕜𝕜λϝλ}\displaystyle\left\{z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\ |\ \lambda(z)=\lambda(y)\right\} (5.63)
={{y}{ϕ𝔫(y)|𝔫}{ϕ𝔫(1/y)|𝔫},ify;{y}{y2σ}{ϕ𝔫(yσ+σ𝔫)|𝔫},ifyγ(σ,);{ϕ𝔫(σ𝔫+σ(1/y))|𝔫},ifyγ(σ,0),\displaystyle=\left\{\begin{array}[]{ll}\left\{y\right\}\sqcup\left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(y)\ \left|\ \eufm{n}\in\Bb{Z}_{\neq 0}^{2\Bb{N}_{\eurm{f}}}\right\}\right.\sqcup\left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(-1/y)\ \left|\ \eufm{n}\in\Bb{Z}_{\neq 0}^{2\Bb{N}_{\eurm{f}}-1}\right\}\right.,&\ \hbox{if}\ \ y\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}};\\[4.26773pt] \left\{y\right\}\sqcup\left\{y-2\sigma\right\}\sqcup\left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(y-\sigma+\sigma_{\eufm{n}}\right)\ \left|\ \eufm{n}\in\Bb{Z}_{\neq 0}^{2\Bb{N}_{\eurm{f}}}\right\}\right.,&\ \hbox{if}\ \ y\in\gamma(\sigma,\infty);\\[4.26773pt] \left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\sigma_{\eufm{n}}+\sigma-{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\ \left|\ \eufm{n}\in\Bb{Z}_{\neq 0}^{2\Bb{N}_{\eurm{f}}-1}\right\}\right.,&\ \hbox{if}\ \ y\in\gamma(\sigma,0)\,,\end{array}\right. (5.67)

where each set is countable and has no limit points in \Bb{H}.

Lemma 5.7 can be used to characterize the sets

S||:={z𝕜𝕜𝕜λϝ},S:={z𝕜λϝ}.\displaystyle S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}:=\!\left\{\,z\!\in\!\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,\left|\,\lambda(z)\!\in\!(1/2)\!+\!i\Bb{R}\,\right\}\right.\!,\ S^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\!:=\!\left\{\,z\!\in\!\Bb{H}\,\left|\,\lambda(z)\!\in\!(1/2)\!+\!i\Bb{R}\,\right\}\right.\!. (5.68)

Indeed, the case yy\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} in (5.63) applied to the semicircle γ(1,1)\gamma(-1,1)\!\subset\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, gives

(a)S||=ϕΓϑ||{ϕ0}ϕ(γ(1,1)),(b)S=n(2n+S||),\displaystyle{\rm{(a)}}\ S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}\!=\!\bigsqcup\nolimits_{{{\mbox{\footnotesize{$\phi\in\Gamma_{\vartheta}^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}\cup\{\phi_{0}\}$}}}}}\ \phi\big{(}\gamma(-1,1)\big{)}\ ,\ \ {\rm{(b)}}\ S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}=\bigsqcup\nolimits_{n\in\Bb{Z}}\ (2n+S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}})\,, (5.69)

and it follows from 1/γ(1,1)=γ(1,1)-1/\gamma(-1,1)=\gamma(-1,1) and (5.7) that

S=ϕΓϑϕ(γ(1,1)),ϕ(S)=S,ϕ(𝕊)=𝕊ϕΓϑ\displaystyle S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\!=\!\!\bigcup\limits_{{{\mbox{\footnotesize{$\phi\in\Gamma_{\vartheta}$}}}}}\!\!\!\phi\big{(}\gamma(-1,1)\big{)}\ ,\ \ \phi\left(S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\right)\!=\!S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty},\ \phi\left(\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\right)\!=\!\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty},\ \ \ \phi\!\in\!\Gamma_{\vartheta}\,. (5.70)

Regarding the two triangular parts into which the open semicircle γ(1,1)\gamma(-1,1) splits the Schwarz quadrilateral, we introduce the notation (see (5.28))

:=𝔻¯,||:=||𝔻¯,:=𝔻𝔽𝔽|σ𝔽|σ𝔻¯\displaystyle\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}:=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\,,\ \ \mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$||$}}}}}:=\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\,,\ \ \mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}:=\Bb{D}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\ \ \mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma}:=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma}\setminus\overline{\Bb{D}}\,, (5.71)

where σ{+1,1}\sigma\in\{+1,-1\}. Then

=γ(1,1),|σ=|σγ(1,1),σ{+1,1},\displaystyle\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\sqcup\gamma(-1,1)\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\,,\ \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma}\sqcup\gamma(-1,1)\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\,,\ \ \sigma\in\{+1,-1\}\,, (5.72)

and for each 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} we can apply ϕ𝔫\phi_{\eufm{n}} to the equalities (5.72) with σ=σ𝔫\sigma=\sigma_{\eufm{n}}, to obtain

^𝔫=^𝔫γ𝔫(1,1)𝔫,^𝔫:=ϕ𝔫(|σ𝔫),𝔫:=ϕ𝔫(),𝔫=𝔫γ𝔫(1,1)𝔫,𝔫:=ϕ𝔫(),𝔫\displaystyle\begin{array}[]{lll}\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}=\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}\sqcup\gamma_{\eufm{n}}(-1,1)\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}\,,&\ \widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}:=\phi_{\eufm{n}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\eufm{n}}}\big{)},&\ \mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}:=\phi_{\eufm{n}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\big{)},\\[5.69046pt] \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}\!=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}\sqcup\gamma_{\eufm{n}}(-1,1)\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}\,,&\ \mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}:=\phi_{\eufm{n}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\big{)},&\ \ \eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,.\end{array} (5.75)

Here 𝔫\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}} and ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}} are hyperbolic triangles, the set {ϕ𝔫(x)|x{1,0,1}}\{\,\phi_{\eufm{n}}(x)\,|\,x\!\in\!\{-1,0,1\}\,\} forms the vertices of 𝔫\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}, and 𝔫\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}} is open, while γ𝔫(1,1)\gamma_{\eufm{n}}(-1,1) is the roof of 𝔫\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}. At the same time, {ϕ𝔫(x)|x{1,1,}}\{\,\phi_{\eufm{n}}(x)\,|\,x\!\in\!\{-1,1,\infty\}\,\} are also the vertices of ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}, γ𝔫(1,1)\gamma_{\eufm{n}}(-1,1) and γ𝔫(σ𝔫,)\gamma_{\eufm{n}}(-\sigma_{\eufm{n}},\infty) are the lower arches of ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}} not contained in ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}, while the roof γ𝔫(σ𝔫,)\gamma_{\eufm{n}}(\sigma_{\eufm{n}},\infty) of ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}} is contained in ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}.

A combination of (5.38) and (5.69), (5.72), (5.75) gives

𝕜𝕜𝕊||𝔽||𝔽n1,,nNN𝔽^𝔽\displaystyle\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$||$}}}}}\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\sqcup\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\,,\ {{\mbox{\footnotesize{$N\in\Bb{N}$}}}}}\left(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\right).

By regrouping the subsets involved, we find that

𝕜𝕜𝕊||𝔽||𝔫{}𝔼\displaystyle\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$||$}}}}}\sqcup\bigsqcup\nolimits_{{{\mbox{\footnotesize{$\ \eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\ (5.76)

where we use the notation

(a)0:=n0^𝓃0;𝓃𝒩,,𝓃1:=𝓃𝒩,,𝓃1n0^𝓃𝒩,,𝓃1,𝓃0,(b)𝓃𝒩,,𝓃1=ϕnN,,n1(0),𝓃𝒩,,𝓃1\displaystyle\begin{array}[]{rl}{\rm{(a)}}&\displaystyle\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!:=\!\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\sqcup\!\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\!\!\!\!\!\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{0}};\qquad\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\!:=\!\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}}\sqcup\!\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\!\!\!\!\!\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1},\,n_{0}},\\[17.07182pt] {\rm{(b)}}&\displaystyle\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)},\ \quad n_{N},...,n_{1}\!\in\!\Bb{Z}_{\neq 0}\,,\ \ N\in\Bb{N}\,,\end{array} (5.79)

and the identities (5.79)(b) follow directly from (5.75) and (5.29). Moreover,

1/0=n0((|+12𝓃0)(|1+2𝓃0))=m𝔻¯\displaystyle-1\big{/}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!=\!\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\sqcup\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\!\in\!\Bb{N}$}}}}}\!\!\!\Big{(}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$|$}}}}+1}-2n_{0}\big{)}\sqcup\big{(}\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$|$}}}}-1}+2n_{0}\big{)}\Big{)}=\Bb{H}\setminus\underset{{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}{\cup}\left(2m+\overline{\Bb{D}}\right),

which leads to (ϕ0(z):=z\phi_{0}(z):=z)

𝔫=ϕ𝔫(1/),𝔫{0}𝔼m𝔻¯\displaystyle\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\phi_{\eufm{n}}\big{(}-1/\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\ ,\quad\eufm{n}\in\{0\}\cup\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,,\quad\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}:=\Bb{H}\setminus\underset{{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}{\cup}\left(2m+\overline{\Bb{D}}\right). (5.80)

In particular, the set 0\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} is open and its boundary consists of the roof γ(1,1)\gamma(-1,1) and the infinitely many lower arches γ(1/(2n+1),1/(2n1))\gamma(1/(2n+1),1/(2n-1)), nn\!\in\!\Bb{Z}_{\neq 0}, which accumulate at the origin, i.e.,

co(0)=𝔻𝕀𝔼𝔻𝕀\nϕ𝔻¯𝕀\displaystyle{\rm{co}}\left(\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\!=\!\Bb{D}_{{\rm{Im}}>0}\,,\qquad\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!=\!\Bb{D}_{{\rm{Im}}>0}\Big{\backslash}\bigsqcup\limits_{{{\mbox{\footnotesize{$n\in\Bb{Z}_{\neq 0}$}}}}}\phi_{n}\big{(}\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)}, (5.81)

where co(A){\rm{co}}(A) denotes the convex hull of a given subset AA\subset\Bb{R}^{2}. For every 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} the associated subset 𝔫\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} has the similar structure, as follows from the relationships (5.80). To be specific, 𝔫\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} is open and its boundary consists of the roof γ(ϕ𝔫(1),ϕ𝔫(1))\gamma(\phi_{\eufm{n}}(-1),\phi_{\eufm{n}}(1)) and the infinitely many lower arches
                        γ(ϕ𝔫(1/(2n+1)),ϕ𝔫(1/(2n1)))\gamma\big{(}\phi_{\eufm{n}}(1/{{\mbox{\small{$($}}}}2n+1{{\mbox{\small{$)$}}}}),\phi_{\eufm{n}}(1/{{\mbox{\small{$($}}}}2n-1{{\mbox{\small{$)$}}}})\big{)},  nn\!\in\!\Bb{Z}_{\neq 0},
which accumulate at the point ϕ𝔫(0)\phi_{\eufm{n}}(0). The analogue of (5.81) reads (see (5.34))

co(𝔫)=ϕ𝔫(𝔻𝕀𝔼ϕ𝔻𝕀\n0ϕ𝔫,𝔫0𝔻¯𝕀\displaystyle\begin{array}[]{l}\ \,{\rm{co}}\left(\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\!=\!\phi_{\eufm{n}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)},\ \eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\phi_{\eufm{n}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\big{\backslash}\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}_{\neq 0}$}}}}}\phi_{{{\mbox{\footnotesize{$\eufm{n},n_{0}$}}}}}\big{(}\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)}\,.\end{array} (5.83)

It follows that (see (5.32), (5.33) and (2.1))

ϕ𝔫(𝔻𝕀γϕϕ1ϕ𝔫(1)<ϕ𝔫(1)1,ϕ𝔻𝕀γ𝔫\displaystyle\begin{array}[]{ll}\phi_{\eufm{n}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}={\rm{co}}\big{(}\,\gamma\left(\phi_{\eufm{n}}(-1),\phi_{\eufm{n}}(1)\right)\,\big{)}\,,&\quad-1\!\leqslant\!\phi_{\eufm{n}}(-1)\!<\!\phi_{\eufm{n}}(1)\!\leqslant\!1\ ,\\[7.11317pt] \Bb{H}\!\cap\partial\phi_{\eufm{n}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\!=\!\gamma_{\eufm{n}}(-1,1)\,,&\quad\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,.\end{array}

In particular, 𝔫\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} is simply connected for each 𝔫{}\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\!\{0\} (see (con, , p.​ 93)).

The rational number p/qp/q, pp\in\Bb{Z}, qq\in\Bb{Z}_{\neq 0}, gcd(p,q)=1{\rm{gcd}}(p,q)=1, is called even if pq0(mod 2)p\,q\,\equiv 0({\rm{mod}}\,2). It is known (see (lop, , p.​ 303))27{}^{\ref*{case39}} that each nonzero even rational number on (1,1)(-1,1) can be uniquely represented in the form ϕ𝔫(0)\phi_{\eufm{n}}(0) with some 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}, and conversely. This suggests the introduction of the following notions.

Definition 5.8

. Given mm\in\Bb{Z} and 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\sqcup_{k\geqslant 1}\!\left(\Bb{Z}_{\neq 0}\right)^{k}, the open sets
                            2m+02m\!+\!\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}},   \eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}  and   2m+𝔫2m+\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}
are called the even rational neighborhoods of 2m2m, \infty and the even rational number 2m+ϕ𝔫(0)(2m1,2m+1){2m}2m+\phi_{\eufm{n}}(0)\in(2m-1,2m+1)\setminus\{2m\}, respectively.

3-32-21-113-\dfrac{1}{3}\cdot\!\!\cdot\!\!\cdot0\cdot\!\!\cdot\!\!\cdot13\dfrac{1}{3}11223315-\dfrac{1}{5}15\dfrac{1}{5}0:=1/\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!:=\!-1/\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}:=m𝔻¯\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\!:=\!\Bb{H}\setminus\underset{{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}{\cup}\left(2m+\overline{\Bb{D}}\,\right)i\!i
Figure 2: . Even rational neighborhoods \eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} and 0\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} of \infty and of 0, respectively.

For each given 𝔫{0}\eufm{n}\in\{0\}\cup\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}, we add to the subset 𝔫\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} its corresponding roof ϕ𝔫(1,1)\phi_{\eufm{n}}(-1,1), and obtain the roofed subsets (see (5.33)),

^𝔫:=γ𝔫(1,1)𝔫,^𝔫=ϕ𝔫(𝔻¯Im>0)n0ϕ𝔫,𝔫0(𝔻¯Im>0),\displaystyle\widehat{\eusm{E}}^{\hskip 0.71114pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\!:=\!\gamma_{\eufm{n}}(-1,1)\sqcup\eusm{E}^{\hskip 1.42271pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}},\quad\widehat{\eusm{E}}^{\hskip 0.71114pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\!=\!\phi_{\eufm{n}}\big{(}\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)}\setminus\bigsqcup\nolimits_{\,{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}_{\neq 0}$}}}}}\ \phi_{{{\mbox{\footnotesize{$\eufm{n},n_{0}$}}}}}\big{(}\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)}, (5.84)

which in view of (5.76) and (5.69) gives us the following partition

𝔼n0𝔼^𝔫𝔼^\displaystyle\Bb{H}=\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}$}}}}}\Big{(}2n_{0}+\widehat{\eusm{E}}^{\hskip 0.71114pt0}_{\!{{\mbox{\tiny{$\frown$}}}}}\Big{)}\ \ \sqcup\!\!\!\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,,\ n_{0}\in\Bb{Z}$}}}}}\Big{(}2n_{0}+\widehat{\eusm{E}}^{\hskip 0.71114pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\Big{)}, (5.85)

which we refer to as the even rational partition of \Bb{H}. Here, we write =m𝔻¯\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\Bb{H}\setminus\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\ (2m+\overline{\Bb{D}}) and in accordance with (5.69)(b) and (5.76), it follows from (5.85) that (for notation, cf. (5.68))

\𝕊𝔼n0𝔼𝔫𝔼\displaystyle\Bb{H}\big{\backslash}S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}=\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}$}}}}}\Big{(}2n_{0}+\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\Big{)}\ \ \sqcup\!\!\!\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,,\ n_{0}\in\Bb{Z}$}}}}}\Big{(}2n_{0}+\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\Big{)}. (5.86)

The following property is obtained in Section 10.4.3.

Theorem 5.9

. The set {ϕ(𝔻}ϕΓϑ\{\phi(\Bb{H}\cap\partial\Bb{D})\}_{\phi\in\Gamma_{\vartheta}} generates the even rational partition (5.85) of \Bb{H}.

By intersecting the both sides of (5.85) with the closed unit disk 𝔻¯\overline{\Bb{D}}, we get

𝔻¯Im>0=^0𝔫^𝔫,\displaystyle\overline{\Bb{D}}_{{\rm{Im}}>0}=\widehat{\eusm{E}}^{\hskip 0.71114pt0}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \ \sqcup\ \ \bigsqcup\nolimits_{{{\mbox{\footnotesize{$\ \eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}$}}}}}\ \ \widehat{\eusm{E}}^{\hskip 0.71114pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,, (5.87)

and for each 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} we can apply ϕ𝔫\phi_{\eufm{n}} to both sides of (5.87), to obtain, in view of (5.34) and ϕ𝔫(𝔻¯Im>0)=γ𝔫(1,1)ϕ𝔫(𝔻𝕀\phi_{{{\mbox{\footnotesize{$\,\eufm{n}$}}}}}(\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)}\!=\!\gamma_{\eufm{n}}(-1,1)\sqcup\phi_{{{\mbox{\footnotesize{$\,\eufm{n}$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}),

ϕ𝔫(𝔻¯Im>0)=^𝔫𝔪^𝔫,𝔪,𝔫:=(𝔫𝔑,,𝔫1)\displaystyle\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)}=\widehat{\eusm{E}}^{\hskip 0.71114pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\ \bigsqcup\nolimits_{\ {{\mbox{\footnotesize{$\eufm{m}\in\Bb{Z}_{\neq 0}$}}}}}\ \widehat{\eusm{E}}^{\eufm{n},\eufm{m}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ ,\ \eufm{n}:=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\in\Bb{N}. (5.88)

This for arbitrary NN\in\Bb{N} shows that (cf. (bon, , p.​ 44))

𝔫ϕ𝔫(𝔻¯Im>0)=𝔫(^𝔫𝔪^𝔫,𝔪)=𝔪𝕄,MN^𝔪,\displaystyle\bigsqcup\limits_{{{\mbox{\footnotesize{$\eufm{n}\in\Bb{Z}^{N}_{\neq 0}$}}}}}\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)}=\bigsqcup\limits_{\ {{\mbox{\footnotesize{$\eufm{n}\in\Bb{Z}^{N}_{\neq 0}$}}}}}\Big{(}\widehat{\eusm{E}}^{\hskip 0.71114pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \ \sqcup\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$\eufm{m}\in\Bb{Z}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}_{\neq 0}$}}}}}\widehat{\eusm{E}}^{\eufm{n},\eufm{m}}_{\!{{\mbox{\tiny{$\frown$}}}}}\Big{)}=\bigsqcup\limits_{{{\mbox{\footnotesize{$\eufm{m}\in\Bb{Z}^{M}_{\neq 0}$}}}}\,,\ {{\mbox{\footnotesize{$M\geqslant N$}}}}}\widehat{\eusm{E}}^{\hskip 0.71114pt\eufm{m}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ ,

so that in view of (5.55),

d(ϕ)N,ϕΓϑ||{ϕ0}ϕ(^0)=d(ϕ)=N,ϕΓϑ||{ϕ0}ϕ(𝔻¯Im>0),N\displaystyle\bigsqcup\limits_{{{\mbox{\footnotesize{$d(\phi)\geqslant N$}}}}\,,\ {{\mbox{\footnotesize{$\phi\in\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}$}}}}\cup\{\phi_{0}\}}\phi\left(\widehat{\eusm{E}}^{\hskip 0.71114pt0}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)=\bigsqcup\limits_{{{\mbox{\footnotesize{$d(\phi)=N$}}}}\,,\ {{\mbox{\footnotesize{$\phi\in\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}$}}}}\cup\{\phi_{0}\}}\phi\big{(}\overline{\Bb{D}}_{{\rm{Im}}>0}\big{)},\ \ N\in\Bb{Z}_{\geqslant 0}\,,

where for N=0N\!=\!0 this splitting coincides with (5.87). In accordance with (5.55), (5.80) and (5.85), we can associate with each point zz\!\in\!\Bb{H} the corresponding even rational height

h(z):={0,ifz=m𝔻¯1+d(ϕ),ifzn0(2n0+ϕ(^0)),ϕΓϑ||{ϕ0}.\displaystyle\eurm{h}_{\eusm{E}}(z):=\left\{\begin{array}[]{ll}0\,,&\ \mbox{if}\ z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\Bb{H}\setminus\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\ \left(2m+\overline{\Bb{D}}\right)\,,\\[4.26773pt] 1+d(\phi)\,,&\displaystyle\ \mbox{if}\ z\in\bigsqcup\nolimits_{{\ {\mbox{\footnotesize{$n_{0}\in\Bb{Z}$}}}}}\ \Big{(}2n_{0}+\phi\big{(}\widehat{\eusm{E}}^{\hskip 0.71114pt0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\Big{)}\,,\ \phi\in\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\cup\{\phi_{0}\}\,.\end{array}\right. (5.91)

The results established in (bon, , p.​ 44) imply the asymptotics

11h(x+iy)dx=2π2log2(1/y)+O(log(1/y)),0<y0.\displaystyle\int\nolimits_{-1}^{1}\eurm{h}_{\eusm{E}}(x+iy)dx=2\pi^{-2}\log^{2}(1/y)+{\rm{O}}\big{(}\log(1/y)\big{)}\,,\qquad 0<y\to 0\,.

Finally, we notice that the inclusion 0𝕜𝕜\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!\subset\!\Bb{H}_{|{\rm{Re}}|<1}, taken together with the properties (5.79)(b) and (5.26), shows that28{}^{\ref*{case25}}

(a)𝔾𝔼=||𝔻¯,𝔾γ=γ(1,1);(b)𝔾𝔼=0,𝔾γ=γ(1,1),ifN1;(c)𝔾𝔼=𝔫𝔨,𝔾γ=γ𝔫𝔨(1,1),if 1kN1,N2.\displaystyle\begin{array}[]{lrllrcll}{\rm{(a)}}&\Bb{G}_{2}\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}&\!=\!&\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\,,&\Bb{G}_{2}\big{(}\gamma(-1,1)\big{)}&\!=\!&\gamma(-1,1)\,;&\\[8.5359pt] {\rm{(b)}}&\Bb{G}_{2}^{N}\big{(}\eusm{E}^{\hskip 1.42271pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}&\!=\!&\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\ ,&\Bb{G}_{2}^{N}\big{(}\gamma_{\eufm{n}}(-1,1)\big{)}&\!=\!&\gamma(-1,1),&\ \mbox{if}\ N\!\geqslant\!1\,;\\[8.5359pt] {\rm{(c)}}&\Bb{G}_{2}^{k}\big{(}\eusm{E}^{\hskip 1.42271pt\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}&\!=\!&\eusm{E}^{\hskip 1.42271pt\eufm{n}_{k}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,&\Bb{G}_{2}^{k}\big{(}\gamma_{\eufm{n}}(-1,1)\big{)}&\!=\!&\gamma_{\eufm{n}_{k}}(-1,1),&\ \mbox{if}\ 1\!\leqslant\!k\!\leqslant\!N\!-\!1,\,N\!\geqslant\!2\,.\end{array} (5.95)

Here, 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N} for a given NN\!\in\!\Bb{N}, and we write 𝔫𝔨:=(𝔫𝔑𝔨,,𝔫1)\eufm{n}_{k}\!:=\!(n_{N-k},...,n_{1}) for kk with 1kN11\!\leqslant\!k\!\leqslant\!N\!-\!1, N2N\!\geqslant\!2. We observe that it follows from (2.35), Lemma 2.4 and (2.21) that

(a)λ(||𝔻¯)=λ()={}(b)λ(0)=λ(1/)={}\displaystyle\begin{array}[]{lrcccl}{\rm{(a)}}&\,\lambda\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$||$}}}}}\setminus\overline{\Bb{D}}\,\right)&=&\lambda\left(\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)&=&\Bb{C}_{\,{\rm{Re}}<1/2}\setminus\{0\}\,;\\[8.5359pt] {\rm{(b)}}&\,\lambda\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}&=&\lambda\left(-1/\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)&=&\Bb{C}_{\,{\rm{Re}}>1/2}\setminus\{1\}\,.\end{array} (5.98)

6 . Analytic continuation of the generating function

  

For arbitrary NN\!\in\!\Bb{N} and 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N} let d(𝔫):=𝔡(ϕ𝔫)=𝔑d(\eufm{n})\!:=\!d(\phi_{\eufm{n}})\!=\!N (see (5.55)),

ψ𝔫(z):=ψnN,,n1(z):=1/ϕn1,,nN(1/z),ψ0(z):=z,z\displaystyle\psi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(z)\!:=\!\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\!:=\!{1}\big{/}{\phi_{{{\mbox{\footnotesize{$n_{1},...,n_{N}$}}}}}(1/z)}\,,\ \ \ \psi_{0}(z):=z\,,\ \ z\!\in\!\Bb{H}\,, (6.1)

and d(0):=0d(0):=0. It can easily be shown from (5.1) that

1/ϕn1,,nN(1/z)=1/ϕn1,,nN(1/z),(n1,,nN)ψ𝔫(ϕ𝔫(z))=ϕ𝔫(ψ𝔫(z))=z,zψ𝔫Γϑ{}\displaystyle\begin{array}[]{l}{1}\big{/}{\phi_{{{\mbox{\footnotesize{$n_{1},...,n_{N}$}}}}}(1/z)}=-{1}\big{/}{\phi_{{{\mbox{\footnotesize{$-n_{1},...,-n_{N}$}}}}}(-1/z)}\,,\quad\ \,(n_{1},...,n_{N})\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,,\\[4.26773pt] \psi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(z)\big{)}=\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\psi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(z)\big{)}=z\,,\qquad z\in\Bb{H}\,,\quad\psi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\!\in\!\Gamma_{\vartheta}\,,\quad\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}\,,\end{array} (6.4)

and the identities (5.79)(b) can be written as

ψnN,,n1(𝓃𝒩,,𝓃1)=0=1/,𝓃𝒩,,𝓃1\displaystyle\psi{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\big{(}\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}=\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}=-1/\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\ ,\quad n_{N},...,n_{1}\in\Bb{Z}_{\neq 0}\,,\ \ N\in\Bb{N}\,. (6.5)

The formulas (5.26) for arbitrary n1,n2,,nNn_{1},n_{2},...,n_{N}\in\Bb{Z}_{\neq 0} and NN\!\in\!\Bb{N} take the form29{}^{\ref*{case22}}

𝔾ψnN,,n1𝔾ϝψnN,,nNkϝ\displaystyle\Bb{G}^{N}_{2}(y)\!=\!\psi{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(y),\ N\!\geqslant\!1,\ \Bb{G}^{k+1}_{2}(z)\!=\!\psi{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}(z),\ 0\!\leqslant\!k\!\leqslant\!N\!-\!2\,, (6.6)

provided that yϕnN,,n1(𝕜𝕜)y\!\in\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\Bb{H}_{|{\rm{Re}}|<1}\right) and zϕnN,,n1(𝕜𝕜)z\!\in\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\Bb{H}_{|{\rm{Re}}|\leqslant 1}\right). In addition, according to (5.29), (5.75), (5.71) and (6.4), we have

ψ𝔫(^𝔫)=|σ𝔫,ψ𝔫(^𝔫)=|σ𝔫,ψ𝔫(𝔫)=,𝔫\displaystyle\psi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}\big{)}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\eufm{n}}}\,,\quad\psi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}\big{)}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\eufm{n}}}\,,\quad\psi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}\big{)}=\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\ ,\quad\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,.

Moreover, we observe that (5.58) gives

λ(z)=(1)d(ϕ)λ(ϕ(z))ϕ(z),ϕΓϑ,z\displaystyle\lambda^{\,\prime}(z)=(-1)^{d(\phi)}\lambda^{\,\prime}(\phi(z))\phi^{\,\prime}(z)\,,\ \quad\phi\!\in\!\Gamma_{\vartheta}\,,\ z\in\Bb{H}\,. (6.7)

We let xx\!\in\!\Bb{R} be fixed, and consider the following two functions of zz,

Φδ(x;z):=12πiγ(1,1)λ(z)dζ(λ(z)λ(ζ))(xδζ(x)1δ)2,δ{0,1}.\displaystyle\Phi^{\,\delta}_{\infty}(x;z):=\frac{1}{2\pi i}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\lambda^{\,\prime}(z){\rm{d}}\zeta}{\big{(}\lambda(z)-\lambda(\zeta)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}\ \,,\quad\delta\in\{0,1\}\,. (6.8)

By (5.59), (5.68) and (2.47), the functions Φ0(x;z)\Phi^{\hskip 1.42271pt0}_{\infty}(x;z) and Φ1(x;z)\Phi^{1}_{\infty}(x;z) are holomorphic at each point of 𝕊\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}, and, in view of (4.6) and (4.7),30{}^{\ref*{case24}}

n1nHn(x)eiπnz=Φ0(x;z)2π2,n1nMn(x)eiπnz=Φ1(x;z)2π2,z𝕀\displaystyle\sum\limits_{n\geqslant 1}\!n\,\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\Phi^{\hskip 0.56917pt0}_{\infty}(x;z)}{2\pi^{2}}\,,\ \ \sum\limits_{n\geqslant 1}\!n\,\eurm{M}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\Phi^{\hskip 0.56917pt1}_{\infty}(x;z)}{2\pi^{2}}\,,\ \ z\in\Bb{H}_{\,{\rm{Im}}>1}\,. (6.9)

Furthermore, (5.70) and (6.7) imply that31{}^{\ref*{case23}}

(1)d(ϕ)Φδ(x,z)={ϕ(z)Φδ(x;ϕ(z)),ifd(ϕ){0}2ϕ(z)Φ1δ(x;ϕ(z)),ifd(ϕ)2\displaystyle(-1)^{d(\phi)}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x,z)=\left\{\begin{array}[]{ll}\phi^{\,\prime}(z)\,\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;\phi(z))\,,&\quad\mbox{if}\ d(\phi)\in\{0\}\cup 2\Bb{N}\,,\\[7.11317pt] \phi^{\,\prime}(z)\,\Phi^{\hskip 0.56917pt1-\delta}_{\infty}(x;\phi(z))\,,&\quad\mbox{if}\ d(\phi)\in 2\Bb{N}\!-\!1\,,\end{array}\right. (6.12)

holds for all z𝕊z\in\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}, ϕΓϑ\phi\in\Gamma_{\vartheta} and δ{0,1}\delta\!\in\!\{0,1\}. It follows from (5.69) that Φδ(x;z)\Phi^{\hskip 0.71114pt\delta}_{\infty}(x;z) is holomorphic 22-periodic function on the set =m𝔻¯\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\Bb{H}\setminus\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\ (2m+\overline{\Bb{D}}). The goal of this section is to prove that for every δ{0,1}\delta\!\in\!\{0,1\} the function Φδ(x;z)\Phi^{\hskip 0.71114pt\delta}_{\infty}(x;z) can be analytically extended from this set to \Bb{H}. To do this, it suffices to show that such an extension is possible from the set \eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} to 𝕜𝕜𝔼\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}, because then the desired extension Φδ(x;z)\Phi^{\hskip 0.71114pt\delta}_{\Bb{H}}(x;z) on the remaining set n(2n+𝔻¯Im>0)\cup_{n\in\Bb{Z}_{\neq 0}}(2n+\overline{\Bb{D}}_{{\rm{Im}}>0}) can be constructed from the resulting extension Φδ||(x;z)\Phi^{\hskip 0.71114pt\delta}_{\,{{\mbox{\tiny{$||$}}}}}(x;z) by the formula

Φδ(x;z+2n):=Φδ||(x;z),z𝔻¯Im>0,nδ{}\displaystyle\Phi^{\hskip 0.71114pt\delta}_{\Bb{H}}(x;z\!+\!2n)\!:=\!\Phi^{\hskip 0.71114pt\delta}_{\,{{\mbox{\tiny{$||$}}}}}(x;z)\ ,\quad z\!\in\!\overline{\Bb{D}}_{{\rm{Im}}>0}\,,\ \ \ n\!\in\!\Bb{Z}_{\neq 0}\,,\quad\delta\in\{0,1\}\,. (6.13)

6.1 . Auxiliary lemmas.

In view of Lemma 2.4, for every xx\!\in\!\Bb{R}, zz\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, Imz2{\rm{Im}}\,z\!\neq\!2, we introduce the function Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z) as the integral (6.8), where γ(1,1)\gamma(-1,1) is replaced by the oriented contour Π(1,1)\Pi(-1,1), which passes from 1-1 to 11 along the polygonal contour (1,1+2i][1+2i,1+2i][1+2i,1)(-1,-1+2{\rm{i}}]\cup[-1+2{\rm{i}},1+2{\rm{i}}]\cup[1+2{\rm{i}},1).

Lemma 6.1

. For each xx\!\in\!\Bb{R} and δ{0,1}\delta\!\in\!\{0,1\} the function Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z) of zz can be analytically extended from =m𝔻¯\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\Bb{H}\setminus\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\ (2m+\overline{\Bb{D}}) to γ(1,1)0\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\sqcup\gamma(-1,1)\sqcup\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} such that the resulting extension Φ0δ(x,z)\Phi_{0}^{\hskip 0.56917pt\delta}(x,z) satisfies

Φ0δ(x,z)={Φδ(x;z),ifz;Φδ(x;z)(xδz(x)1δ)2,ifz0.\displaystyle\Phi_{0}^{\hskip 0.56917pt\delta}(x,z)\!=\!\left\{\begin{array}[]{ll}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\,,&\quad\ \mbox{if}\ \ z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\,;\\[5.69046pt] \Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)-\big{(}x^{\delta}z\!-\!(-x)^{1-\delta}\big{)}^{-2}\,,&\quad\ \mbox{if}\ \ z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\,.\end{array}\right. (6.16)
Proof

. By transforming the contour γ(1,1)\gamma(-1,1) of integration in (6.8) to Π(1,1)\Pi(-1,1) and using Lemma 2.4, we obtain from the residue theorem (con, , p.​ 112) that

Φδ(x;z)=Φδ(x;z),z𝕀Φδ(x;z)=Φδ(x;z)(xδz(x)1δ)2,z𝕀\displaystyle\begin{array}[]{ll}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\!=\!\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)\,,&\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\sqcup\Bb{H}_{{\rm{Im}}>2},\\[5.69046pt] \Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\!=\!\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)-\big{(}x^{\delta}z\!-\!(-x)^{1-\delta}\big{)}^{-2}\,,&\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\cap\Bb{H}_{{\rm{Im}}<2}\,.\end{array} (6.19)

This means that the function Φδ(x;z)(xδz(x)1δ)2\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)-(x^{\delta}z-(-x)^{1-\delta})^{-2}, being holomorphic on 𝕀𝔽\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, coincides on the set 𝕀𝔽𝕀𝔽\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\subset\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} with the function Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z), which is holomorphic on 𝕀𝔽\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\supset\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}. By the uniqueness theorem for analytic functions (see (con, , p.​ 78)), we find that the function Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z) can be analytically extended from \mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}} to =(𝕀𝔽\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\cup(\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}). Moreover, for z𝕀𝔽z\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\subset\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, the resulting extension equals the expression H(z):=Φδ(x;z)(xδz(x)1δ)2H(z):=\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)-(x^{\delta}z-(-x)^{1-\delta})^{-2} since Φδ(x;z)=Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)=\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z) holds for all zz\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}, in view of (6.19). But by (5.79)(a) and (5.86), we see that 0\𝕊\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\!\subset\!\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!\subset\!\Bb{H}\big{\backslash}S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}, and since 0\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} is simply connected it follows that the latter function H(z)H(z) is actually holomorphic on 0\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}. This proves (6.16) and completes the proof of Lemma 6.1. \square

Lemma 6.2

. Let xx\!\in\!\Bb{R}, NN\!\in\!\Bb{N}, 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{N}, δ{0,1}\delta\!\in\!\{0,1\} and the sets 𝔫\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}, 𝔫\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}} be defined as in (5.75). The function Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z) of zz can be analytically extended from 𝔫\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}} to 𝔫=𝔫γ𝔫(1,1)𝔫\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}\sqcup\gamma_{\eufm{n}}(-1,1)\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}} such that the resulting extension Φ𝔫δ(x,z)\Phi_{\eufm{n}}^{\hskip 0.56917pt\delta}(x,z) satisfies

Φ𝔫δ(x,z)={Φδ(x;z),ifz𝔫;Φδ(x;z)Δδ𝔫(x;z),ifz𝔫.\displaystyle\Phi_{\eufm{n}}^{\hskip 0.56917pt\delta}(x,z)\!=\!\left\{\begin{array}[]{ll}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\,,&\quad\ \mbox{if}\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}\,;\\[5.69046pt] \Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)-\Delta^{\delta}_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(x;z)\,,&\quad\ \mbox{if}\ \ z\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}\,.\end{array}\right. (6.22)

where the holomorphic at each zz\!\in\!\Bb{H} function Δδ𝔫(x;z)\Delta^{\delta}_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(x;z) is defined as

Δδ𝔫(x;z):={ψ𝔫(z)(xδψ𝔫(z)(x)1δ)2,N2ψ𝔫(z)(x1δψ𝔫(z)(x)δ)2,N2\displaystyle\Delta^{\delta}_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(x;z)\!:=\!\left\{\begin{array}[]{ll}\hphantom{-}\psi^{\,\prime}_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(z)\big{(}x^{\delta}\psi_{\eufm{n}}(z)-(-x)^{1-\delta}\big{)}^{-2}\,,&\quad N\in 2\Bb{N}\,,\\[5.69046pt] -\psi^{\,\prime}_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(z)\big{(}x^{1-\delta}\psi_{\eufm{n}}(z)-(-x)^{\delta}\big{)}^{-2}\,,&\quad N\!\in\!2\Bb{N}\!-\!1\,.\end{array}\right. (6.25)
Proof

. We put δN:=δ\delta_{N}:=\delta if NN is even, while δN:=1δ\delta_{N}:=1-\delta if NN is odd. Let the function zΦ0δN(x,z)Hol(γ(1,1)0)z\mapsto\Phi_{0}^{\hskip 0.56917pt\delta_{N}}(x,z)\in{\rm{Hol}}(\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\sqcup\gamma(-1,1)\sqcup\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}) be given as in Lemma 6.1 with δ=δN\delta=\delta_{N}. By (5.72), (5.75) and (6.4), we have ψ𝔫(𝔫)=\psi_{\eufm{n}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}})=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, and γ(1,1)0\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\subset\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\sqcup\gamma(-1,1)\sqcup\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}, in view of (5.79)(a). It now follows that zΦ0δN(x,ψ𝔫(z))z\mapsto\Phi_{0}^{\hskip 0.56917pt\delta_{N}}(x,\psi_{\eufm{n}}(z)) is in Hol(𝔫){\rm{Hol}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}). By (6.12), we obtain from (6.16) that, for z𝔫z\in\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}},

Φ0δN(x,ψ𝔫(z))={Φδ(x;ψ𝔫(z))=ψ𝔫(z)1Φδ(x;z),N2Φ1δ(x;ψ𝔫(z))=ψ𝔫(z)1Φδ(x;z),N2\displaystyle\Phi_{0}^{\hskip 0.56917pt\delta_{N}}(x,\psi_{\eufm{n}}(z))\!=\!\left\{\begin{array}[]{ll}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;\psi_{\eufm{n}}(z))=\psi_{\eufm{n}}^{\,\prime}(z)^{-1}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\,,&\quad N\in 2\Bb{N}\,,\\[2.84544pt] \Phi^{\hskip 0.56917pt1-\delta}_{\infty}(x;\psi_{\eufm{n}}(z))=-\psi_{\eufm{n}}^{\,\prime}(z)^{-1}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\,,&\quad N\!\in\!2\Bb{N}\!-\!1\,.\end{array}\right.

On the other hand, if z𝔫z\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}} then

Φ0δN(x,ψ𝔫(z))={ψ𝔫(z)1Φδ(x;z)(xδψ𝔫(z)(x)1δ)2,N2ψ𝔫(z)1Φδ(x;z)(x1δψ𝔫(z)(x)δ)2,N2\displaystyle\Phi_{0}^{\hskip 0.56917pt\delta_{N}}(x,\psi_{\eufm{n}}(z))\!=\!\left\{\begin{array}[]{lll}\psi_{\eufm{n}}^{\,\prime}(z)^{-1}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\!-\!(x^{\delta}\psi_{\eufm{n}}(z)\!-\!(-x)^{1-\delta})^{-2}\!,&\,N\in 2\Bb{N}\,,\\[2.84544pt] \!-\psi_{\eufm{n}}^{\,\prime}(z)^{-1}\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\!-\!(x^{1-\delta}\psi_{\eufm{n}}(z)\!-\!(-x)^{\delta})^{-2}\!,&\,N\!\in\!2\Bb{N}\!-\!1\,.\end{array}\right.

Hence the function Φ𝔫δ(x,z):=(1)Nψ𝔫(z)Φ0δN(x,ψ𝔫(z))\Phi_{\eufm{n}}^{\hskip 0.56917pt\delta}(x,z):=(-1)^{N}\psi_{\eufm{n}}^{\,\prime}(z)\Phi_{0}^{\hskip 0.56917pt\delta_{N}}(x,\psi_{\eufm{n}}(z)) is in Hol(𝔫){\rm{Hol}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}), where 𝔫=𝔫γ𝔫(1,1)𝔫\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}\sqcup\gamma_{\eufm{n}}(-1,1)\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}} and Φ𝔫δ(x,z)=Φδ(x;z)\Phi_{\eufm{n}}^{\hskip 0.56917pt\delta}(x,z)=\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z) for all z𝔫z\in\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}, while

Φ𝔫δ(x,z)=Φδ(x;z)(1)N(xδNψ𝔫(z)(x)1δN)2,z𝔫.\displaystyle\Phi_{\eufm{n}}^{\hskip 0.56917pt\delta}(x,z)=\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)-(-1)^{N}(x^{\delta_{N}}\psi_{\eufm{n}}(z)\!-\!(-x)^{1-\delta_{N}})^{-2}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}\,.

This shows that as a function of zz, Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z) can be analytically extended from 𝔫\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}} to 𝔫\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}, and the formulas (6.22), (6.25) hold as well. Lemma 6.2 follows.\square

The desired analytical extension Φδ||(x;z)\Phi^{\,\delta}_{\,{{\mbox{\tiny{$||$}}}}}(x;z) of the function Φδ(x;z)\Phi^{\,\delta}_{\infty}(x;z) from the set \eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} to 𝕜𝕜𝔼\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} can be easily constructed by combining the results of Lemmas 6.1 and 6.2.

Lemma 6.3

Let xx\in\Bb{R}, δ{0,1}\delta\!\in\!\{0,1\} and Φδ(x;z)\Phi^{\,\delta}_{\infty}(x;z) be given by (6.8). The function zΦδ(x;z)z\mapsto\Phi^{\,\delta}_{\infty}(x;z) extends analytically from the set \eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} to 𝕜𝕜𝔼\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} such that the resulting extension Φδ||(x;z)\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z) on the set

(𝕜𝕜𝔼)\S=0𝔫𝔫\displaystyle\left(\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\big{\backslash}S^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \ \sqcup\ \ \bigsqcup\nolimits_{{{\mbox{\footnotesize{$\ \eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}$}}}}}\ \ \eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} (6.26)

satisfies

Φδ||(x;z)=Φδ(x;z),\displaystyle\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)=\Phi^{\,\delta}_{\infty}(x;z)\,,\ z,\displaystyle z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\ (6.27)
Φδ||(x;z)=Φδ(x;z)Δδ0(x;z),\displaystyle\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)=\Phi^{\,\delta}_{\infty}(x;z)-\Delta^{\delta}_{0}(x;z)\,,\ z0,\displaystyle z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,,\ (6.28)
Φδ||(x;z)=Φδ(x;z)Δδ0(x;z)k=0N1ΔδnN,nN1,,nNk(x;z),\displaystyle\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)=\Phi^{\,\delta}_{\infty}(x;z)-\Delta^{\delta}_{0}(x;z)-\sum\limits_{k=0}^{N-1}\Delta^{\delta}_{n_{N},n_{N-1},...,n_{N-k}}(x;z)\ , (6.29)

when z𝓃𝒩,,𝓃1z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}} for arbitrary NN\!\in\!\Bb{N} and 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}. Here, for every 𝔫\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{N}, Δδ𝔫(x;z)\Delta^{\delta}_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(x;z) is defined as in (6.25) and

Δδ0(x;z):=(xδz(x)1δ)2,xδ{}ϝ\displaystyle\Delta^{\delta}_{0}(x;z):=(x^{\delta}z-(-x)^{1-\delta})^{-2}\ ,\quad x\in\Bb{R}\,,\ \delta\!\in\!\{0,1\}\,,\ z\in\Bb{H}\,. (6.30)

Proof. As follows from =m𝔻¯\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\Bb{H}\setminus\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\ (2m+\overline{\Bb{D}}) and (5.69), we have 𝒮=\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\cap S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}=\emptyset, and hence (6.26) is immediate from (5.76). Since Φδ(x;z)\Phi^{\,\delta}_{\infty}(x;z) is holomorphic at each point of 𝕊\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty} we conclude from (6.26) that Φδ||(x;z)\Phi^{\hskip 0.56917pt\delta}_{\,{{\mbox{\tiny{$||$}}}}}(x;z) is holomorphic on 𝔫\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} for every 𝔫{}\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}.

What remains to prove is that Φδ||(x;z)\Phi^{\hskip 0.56917pt\delta}_{\,{{\mbox{\tiny{$||$}}}}}(x;z) extends analytically across γ𝔫(1,1)\gamma_{\eufm{n}}(-1,1) for each 𝔫{}\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}. In the case when 𝔫=0\eufm{n}=0, this fact follows from Lemma 6.1 because, in accordance with (6.16), Φδ||(x;z)=Φ0δ(x,z)\Phi^{\hskip 0.56917pt\delta}_{\,{{\mbox{\tiny{$||$}}}}}(x;z)=\Phi_{0}^{\hskip 0.56917pt\delta}(x,z) for all z0z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\sqcup\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}. If NN\in\Bb{N} and 𝔫:=(𝔫𝔑,,𝔫1)\eufm{n}:=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N} then γ𝔫(1,1)\gamma_{\eufm{n}}(-1,1) is the roof of 𝔫\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}} and is the lower arch of 𝔫\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}, by virtue of (5.75). According to (5.79)(a), 𝔫𝔫\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}\subset\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} and nN,,n1^nN,,n1𝓃𝒩,,𝓃2\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}}\subset\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}}\subset\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{2}}_{\!{{\mbox{\tiny{$\frown$}}}}}, if N2N\geqslant 2, while n1^n10\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{1}}\subset\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{1}}\subset\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}, if N=1N=1. Hence, in this case, the required property follows from Lemma 6.2, because for each z𝔫𝔫z\in\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436pt\eufm{n}}\sqcup\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}^{\hskip 0.28436pt\eufm{n}}, the difference Φδ||(x;z)Φ𝔫δ(x,z)\Phi^{\hskip 0.56917pt\delta}_{\,{{\mbox{\tiny{$||$}}}}}(x;z)-\Phi_{\eufm{n}}^{\hskip 0.56917pt\delta}(x,z) equals the function

Δδ0(x;z)k=0N2ΔδnN,nN1,,nNk(x;z),k=01:=0,\displaystyle-\Delta^{\delta}_{0}(x;z)-\sum\nolimits_{k=0}^{N-2}\Delta^{\delta}_{n_{N},n_{N-1},...,n_{N-k}}(x;z)\,,\quad\sum\nolimits_{k=0}^{-1}:=0\,,

which is holomorphic on \Bb{H}, in view of (6.25) and (6.30). Lemma 6.3 follows. \square

6.2 . Main result.

By (5.84), we obtain from (5.87) and (5.88) that

𝔻𝕀𝕊||𝔼𝔪𝔼\displaystyle\Bb{D}_{\,{\rm{Im}}>0}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \ \sqcup\ \ \bigsqcup\limits_{{{\mbox{\footnotesize{$\ \eufm{m}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}$}}}}}\ \ \eusm{E}^{\eufm{m}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,, (6.31)
ϕ𝔫(𝔻𝕀𝕊||𝔼mM,,m1𝕄𝔼𝕄\displaystyle\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\Bb{D}_{\,{\rm{Im}}>0}\big{)}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\!\!\!\!\!\!\bigsqcup\limits_{\ {{\mbox{\footnotesize{$m_{M},...,\,m_{1}\in\Bb{Z}_{\neq 0}\,,\ M\in\Bb{N}$}}}}}\!\!\!\!\!\!\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1},\,m_{M},...,\,m_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ , (6.32)

where NN\in\Bb{N} and 𝔫:=(𝔫𝔑,,𝔫1)\eufm{n}:=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}. Consequently, for each xx\in\Bb{R} and δ{0,1}\delta\!\in\!\{0,1\}, the formula for Φδ||(x;z)\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z) in Lemma 6.3 can be written in the form

Φδ||(x;z)=Φδ(x;z)𝔫{}Δδ𝔫(x;z)χϕ𝔫(𝔻𝕀)(z),z𝕜𝕜𝕊||\displaystyle\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)\!=\!\Phi^{\,\delta}_{\infty}(x;z)\!-\!\!\!\!\!\!\sum\limits_{{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\Delta^{\delta}_{{{\eufm{n}}}}(x;z)\,\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{\,{\rm{Im}}>0}\right)$}}}}}(z),\quad z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}},

where, for every z𝕜𝕜𝕊||z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}, the sum has only finitely many nonzero terms, the number of which equals (see (5.91), (6.31) and (6.32))32{}^{\ref*{case5}}

𝔫{}χϕ𝔫(𝔻𝕀)(z)=h(z),z𝕜𝕜𝕊||\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{\,{\rm{Im}}>0}\right)$}}}}}(z)=\eurm{h}_{\eusm{E}}(z)\ ,\quad z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,. (6.33)

Let NN\!\in\!\Bb{Z}_{\geqslant 2}, 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, δ{0,1}\delta\!\in\!\{0,1\}, and put δ2n:=δ\delta_{2n}:=\delta, δ2n+1:=1δ\delta_{2n+1}:=1-\delta, nn\in\Bb{Z}_{\geqslant 0}. The equalities of sets in (6.5) and in (5.79)(b) lead to

ψnN,,nNk(𝓃𝒩,,𝓃1)\displaystyle\psi{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}\big{(}\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)} =ψnN,,nNk(ϕnN,,nNk(ϕnNk1,,n1(0)))\displaystyle=\psi{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}\big{(}\phi{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}\big{(}\phi{{{\mbox{\footnotesize{$n_{N-k-1},...,n_{1}$}}}}}\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\big{)}\big{)}
=ϕnNk1,,n1(0)=𝓃𝒩𝓀1,,𝓃1,0𝓀𝒩2,\displaystyle=\phi{{{\mbox{\footnotesize{$n_{N-k-1},...,n_{1}$}}}}}\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}=\eusm{E}^{\hskip 1.42271ptn_{N-k-1},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ ,\quad 0\!\leqslant\!k\!\leqslant\!N\!-\!2\,,

and hence, in view of (6.6), we find that

ψnN,,nNk(z)=𝔾ϝ 0kN1,z𝓃𝒩,,𝓃1,𝔾𝔼𝔼𝔾𝔼𝔼0kN2.\displaystyle\begin{array}[]{lll}\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}(z)\!=\!\Bb{G}^{k+1}_{2}(z),&\ \ \ 0\!\leqslant\!k\!\leqslant\!N\!-\!1\,,&\ z\!\in\!\eusm{E}^{n_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\\[8.5359pt] \Bb{G}^{k+1}_{2}\left(\eusm{E}^{n_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\!=\!\eusm{E}^{\hskip 1.42271ptn_{N-k-1},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,&\ \ \ \Bb{G}^{N}_{2}\left(\eusm{E}^{n_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\!=\!\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,&\ 0\!\leqslant\!k\!\leqslant\!N\!-\!2\,.\end{array} (6.36)

By applying the identity

|ϕ(z)|=Imϕ(z)Imz,ϕΓϑ,z\displaystyle\left|\phi^{\,\prime}(z)\right|=\dfrac{{\rm{Im}}\,\phi(z)}{{\rm{Im}}\,z}\,,\qquad\phi\in\Gamma_{\vartheta}\ ,\quad z\in\Bb{H}\,,

we see from (6.36) and (6.4) that

|ψnN,,nNk(z)|=Im𝔾ϝImz,0kN1,z𝓃𝒩,,𝓃1.\displaystyle\left|\psi^{\,\prime}_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}(z)\right|=\dfrac{{\rm{Im}}\,\Bb{G}^{k+1}_{2}(z)}{{\rm{Im}}\,z}\ ,\quad 0\leqslant k\leqslant\!N-1\,,\quad z\in\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,. (6.37)

In view of (6.36), for every 0kN10\leqslant k\leqslant N-1, xx\!\in\!\Bb{R} and z𝓃𝒩,,𝓃1z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}} we can write the function ΔδnN,nN1,,nNk\Delta^{\delta}_{n_{N},n_{N-1},...,n_{N-k}} from (6.29) in the form (see (6.25))

ΔδnN,nN1,,nNk(x,z)=(1)k+1ψnN,,nNk(z)(x1δk𝔾ϝδ)2,\displaystyle\Delta^{\delta}_{n_{N},n_{N-1},...,n_{N-k}}(x,z)=\dfrac{(-1)^{k+1}\psi_{n_{N},...,n_{N-k}}^{\,\prime}(z)}{\left(x^{1-\delta_{k}}\Bb{G}^{k+1}_{2}(z)-(-x)^{\delta_{k}}\right)^{2}}\ ,\ (6.38)

where δ{0,1}\delta\!\in\!\{0,1\} and the properties (6.36), (6.37) hold. On the other hand, in the formula (6.29) the values of the function Φδ(x;z)\Phi^{\,\delta}_{\infty}(x;z) for z𝓃𝒩,,𝓃1z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}} can be expressed with the help of (6.12) through its values on 0\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}:

(1)NΦδ(x;z)/ψnN,,n1(z)=ΦδN(x;ψnN,,n1(z))=ΦδN(x;𝔾ϝ\displaystyle(-1)^{N}\Phi^{\,\delta}_{\infty}(x;z)\big{/}\psi^{\,\prime}_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)=\Phi^{\,\delta_{N}}_{\infty}(x;\psi_{n_{N},...,n_{1}}(z))=\Phi^{\,\delta_{N}}_{\infty}(x;\Bb{G}^{N}_{2}(z)), (6.39)

because ψnN,,n1Γϑ\psi_{n_{N},...,n_{1}}\in\Gamma_{\vartheta}, by virtue of (6.4), and ψnN,,n1(z)=𝔾ϝ𝔼\psi_{n_{N},...,n_{1}}(z)=\Bb{G}^{N}_{2}(z)\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} for all z𝓃𝒩,,𝓃1z\in\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}. Moreover, the first equality in (6.19) actually holds for all z0z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}. Indeed, by (5.98)(b) we have λ(0)={}\lambda\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\!=\!\Bb{C}_{\,{\rm{Re}}>1/2}\leavevmode\nobreak\ \setminus\leavevmode\nobreak\ \{1\}, while λ(𝔻)=\lambda\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\Bb{D}\right)\!=\!\Bb{C}_{{\rm{Re}}\leqslant 1/2}\!\setminus\!\Bb{R}_{\leqslant 0}, as follows from (2.35)(c). It follows that λ(z)λ(ζ)0\lambda(z)-\lambda(\zeta)\neq 0 holds for all ζ𝔻\zeta\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\Bb{D} and z0z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}. Since 𝔻\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\Bb{D} is simply connected and γ(1,1)𝔻\gamma(-1,1)\subset\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\Bb{D}, Π(1,1)𝔻\Pi(-1,1)\subset\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus\Bb{D}, we can transform the contour γ(1,1)\gamma(-1,1) of integration in (6.8) to Π(1,1)\Pi(-1,1) for each z0z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} to get Φδ(x;z)=Φδ(x;z)\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\!=\!\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z) and hence

Φδ(x;𝔾ϝΦδ𝔾ϝϝ𝔼δ{}\displaystyle\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;\Bb{G}^{N}_{2}(z))\!=\!\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;\Bb{G}^{N}_{2}(z)),\ z\in\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\ x\in\Bb{R},\ \delta\!\in\!\{0,1\}\,. (6.40)

A similar change of the contour γ(1,1)\gamma(-1,1) can be made in (6.28) and in (6.27) after application (6.12) to (6.27) with ϕ(z)=1/z\phi(z)=-1/z. Combining this with (6.40), (6.39), (6.38), (6.36), (6.37) and Lemma 6.3 gives the main result of this section.

Let

Φδ(x;z)\displaystyle\Phi^{\,\delta}_{\infty}(x;z) :=Iγ(1,1)δ(x;z),z𝕊Φδϝ𝕀Π(1,1)δϝϝ𝔼\displaystyle:=I_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}^{\delta}(x;z),\ \,z\!\in\!\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\,;\ \ \Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z):=I_{{{\mbox{\footnotesize{$\Pi(-1,1)$}}}}}^{\delta}(x;z),\ \,z\!\in\!\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,,
IδΓ(x;z)\displaystyle I^{\,\delta}_{\Gamma}(x;z) :=12πiΓλ(z)dζ(λ(z)λ(ζ))(xδζ(x)1δ)2,xδ{}\displaystyle:={{\dfrac{1}{2\pi i}}}\int\limits_{{{\mbox{\footnotesize{$\Gamma$}}}}}{{\dfrac{\lambda^{\,\prime}(z)\,{\rm{d}}\zeta}{\big{(}\lambda(z)-\lambda(\zeta)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{\!2}}}}\ ,\quad x\in\Bb{R}\ ,\quad\delta\in\{0,1\}\,.

Here the contour Π(1,1)=(1,1+2i][1+2i,1+2i][1+2i,1)\Pi(-1,1)=(-1,-1+\!2{\rm{i}}]\cup[-1\!+\!2{\rm{i}},1+\!2{\rm{i}}]\cup[1+\!2{\rm{i}},1) passes from 1-1 to 11, λ(0)\lambda\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\!\subset\!\Bb{C}_{\,{\rm{Re}}>1/2}, λ(Π(1,1))\lambda\left(\Pi(-1,1)\right)\!\subset\!\Bb{C}_{{\rm{Re}}<1/2} and the convergence of both integrals is absolute, in view of (2.47) and (7.2).

Theorem 6.4

For each xx\in\Bb{R} and δ{0,1}\delta\in\{0,1\} the function Φδ(x;z)\Phi^{\,\delta}_{\infty}(x;z) of the variable zz can be analytically extended from the set =m𝔻¯\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\Bb{H}\setminus\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\ (2m+\overline{\Bb{D}}) to 𝕜𝕜𝔼\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} such that the resulting extension Φδ||(x;z)\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z) on the set (see (5.86))

(𝕜𝕜𝔼)\S||=0n1,,nN,N𝓃𝒩,,𝓃1\displaystyle\left(\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\big{\backslash}S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}\!=\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\sqcup\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\!\!\!\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$\,n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\,,\,{{\mbox{\footnotesize{$N\in\Bb{N}$}}}}}\,\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}} (6.41)

for δ2n:=δ\delta_{2n}\!:=\!\delta, δ2n+1:=1δ\delta_{2n+1}\!:=\!1\!-\!\delta, nn\!\in\!\Bb{Z}_{\geqslant 0}, and arbitrary NN\!\in\!\Bb{N}, nN,,n1n_{N},...,n_{1}\!\in\!\Bb{Z}_{\neq 0} satisfies

Φδ||(x;z)={z2Φ1δ(x;1/z),ifz,1(xδz(x)1δ)2+Φδ(x;z),ifz0,1(xδz(x)1δ)2+k=0N1(1)kψnN,,nNk(z)(x1δk𝔾ϝδ)2+(1)NψnN,,n1(z)ΦδN(x;𝔾ϝ),ifz𝓃𝒩,,𝓃1,\displaystyle\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)=\left\{\begin{array}[]{ll}-z^{-2}\Phi^{\hskip 0.56917pt1-\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;-1/z),&\mbox{if}\ z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\\[5.69046pt] -{{\dfrac{1}{(x^{\delta}z-(-x)^{1-\delta})^{2}}}}+\Phi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)\ ,&\mbox{if}\ z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,,\\[11.38092pt] -{{\dfrac{1}{(x^{\delta}z-(-x)^{1-\delta})^{2}}}}+\sum\limits_{k=0}^{N-1}{{\dfrac{(-1)^{k}\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}^{\,\prime}(z)}{\left(x^{1-\delta_{k}}\Bb{G}^{k+1}_{2}(z)-(-x)^{\delta_{k}}\right)^{2}}}}\\[19.91684pt] +(-1)^{N}\psi^{\,\prime}_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\,\Phi^{\hskip 0.56917pt\delta_{N}}_{{{\mbox{\tiny{$\sqcap$}}}}}\left(x;\Bb{G}^{N}_{2}(z)\right),&\mbox{if}\ z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}},\end{array}\right. (6.46)

where the sets in (6.41) and ψnN,,n1Γϑ\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\!\in\!\Gamma_{\vartheta} in (6.46) are defined as in (5.70), (5.79), (5.83), and as in (6.1), correspondingly. Here, in (6.46) we have

ψnN,,nNk(z)=𝔾ϝϝ𝔼\displaystyle\vspace{-0,25cm}\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}(z)\!=\!\Bb{G}^{k+1}_{2}(z)\,,\quad z\!\in\!\eusm{E}^{n_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}},\ \ 0\!\leqslant\!k\!\leqslant\!N\!-\!1\,,\vspace{-0,1cm}

𝔾𝔼𝔼\Bb{G}^{N}_{2}\left(\eusm{E}^{n_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\!=\!\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} and 1/=0-1/\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0} (see also (6.36) and (6.37)).

We observe that by the definition (5.14) of 𝔾ϝ\Bb{G}_{2}(z) and the identity Im(1/z)=(Imz)/|z|2{\rm{Im}}(-1/z)=({\rm{Im}}\,z)/|z|^{2}, zz\!\in\!\Bb{H}, we have Im𝔾ϝ𝕀ϝ{\rm{Im}}\,\Bb{G}_{2}(z)\!=\!{\rm{Im}}(-1/z) for every z𝕜𝕜z\!\in\!\Bb{H}_{|{\rm{Re}}|\leqslant 1} and thus

Im𝔾ϝ𝕀ϝ𝕀ϝϝ𝔻𝕀𝕀ϝ𝕀ϝϝ\displaystyle{\rm{Im}}\,\Bb{G}_{2}(z)\!=\!{\rm{Im}}(-1/z)\!\geqslant\!{\rm{Im}}\,z\,,\ \ z\!\in\!\Bb{D}_{{\rm{Im}}>0}\,;\ \ {\rm{Im}}\,(-1/z)\!\leqslant\!\dfrac{1}{{\rm{Im}}\,z}\,,\ \ z\!\in\!\Bb{H}\,. (6.47)

Since for every z𝓃𝒩,,𝓃1z\!\in\!\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}} we have 𝔾ϝ𝔻𝕀\Bb{G}_{2}^{k}(z)\in\Bb{D}_{{\rm{Im}}>0}, 0kN0\leqslant k\leqslant N, then

ImzIm𝔾ϝ𝕀𝔾ϝ𝕀𝔾ϝIm(1/𝔾ϝ)1/Imz,z𝓃𝒩,,𝓃1,𝓃𝒩,,𝓃1\displaystyle\begin{array}[]{l}{\rm{Im}}\,z\!\leqslant\!{\rm{Im}}\,\Bb{G}_{2}^{k}(z)\!\leqslant\!{\rm{Im}}\,\Bb{G}_{2}^{k+1}(z)\!=\!{\rm{Im}}\,\left(-1/\Bb{G}_{2}^{k}(z)\right)\!\leqslant\!1\ ,\quad 0\!\leqslant\!k\leqslant\!N\!-\!1\,,\\[8.5359pt] {\rm{Im}}\left(-1/\Bb{G}_{2}^{N}(z)\right)\!\leqslant\!1/{\rm{Im}}\,z\,,\quad z\!\in\!\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\ n_{N},...,n_{1}\!\in\!\Bb{Z}_{\neq 0},\ N\!\in\!\Bb{N}.\end{array} (6.50)

7 . Evaluation of biorthogonal functions for large index

The relations33{}^{\ref*{case28}}

Θ2(z)4=(σz)2Θ4(1/(σz))4,zIm(σz)11/2,|z|>1,σRez[0,1],Imz1,σ{1,1},\displaystyle\begin{array}[]{l}\Theta_{2}(z)^{4}=(\sigma-z)^{-2}\Theta_{4}\left({1}/{(\sigma-z)}\right)^{4},\ \ z\in\Bb{H}\,;\\[4.26773pt] {\rm{Im}}(\sigma-z)^{-1}\geqslant 1/2\,,\ |z|>1\,,\ \sigma{\rm{Re}}\,z\in[0,1]\,,\ {\rm{Im}}\,z\leqslant 1\,,\ \sigma\in\left\{1,-1\right\},\end{array}

and the Landen equations (2.44) imply that6{}^{\ref*{case1}}

|Θ2(z)|4Im2zθ3(eπ/2)45,z;λ(2𝓏)=(11λ(𝓏)1+1λ(𝓏))2,𝓏,\displaystyle\left|\Theta_{2}(z)\right|^{4}{\rm{Im}}^{2}z\!\leqslant\!\theta_{3}({\rm{e}}^{-\pi/2})^{4}\!\leqslant\!5,\ z\!\in\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\,;\ \ \lambda(2z)\!=\!\left(\dfrac{1\!-\!\sqrt{1\!-\!\lambda(z)}}{1\!+\!\sqrt{1\!-\!\lambda(z)}}\right)^{2}\!\!\!\!,\,z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,, (7.1)

where the value of θ3(eπ/2)\theta_{3}({\rm{e}}^{-\pi/2}) is taken from (ber1, , p.​ 325) and the principal branch of the square root is used. From (3.38) and (2.31)(b) we thus obtain34{}^{\ref*{case26}}

112λ(2i)=11+8221,|λ(±1+it)|1712216eπ/t,t(0,2].\displaystyle\dfrac{1}{1-2\lambda(2{\rm{i}})}=\dfrac{11+8\sqrt{2}}{21}\,,\quad\left|\lambda(\pm 1\!+\!{\rm{i}}t)\right|\geqslant\dfrac{17-12\sqrt{2}}{16}{\rm{e}}^{\pi/t}\ ,\quad t\in(0,2]\,. (7.2)

7.1 . Evaluation of the generating function.

Using the identity λ(z)=πiΘ4(z)4λ(z)\lambda^{\,\prime}(z)=\pi i\Theta_{4}\left(z\right)^{4}\lambda(z), zz\in\Bb{H}, written in (2.41), and the fact that all three values of the variable zz used in the formula (6.46) for the function Φδ(x;z)\Phi^{\,\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z) belong to 0\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}, this function can be estimated as

|Φδ(x;z)||Θ4(z)|4Iδ(z),Iδ(z):=Π(1,1)(|λ(z)|/|λ(z)λ(ζ)|)|dζ|2|xδζ(x)1δ|2,z0.\displaystyle\left|\Phi^{\,\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)\right|\!\leqslant\!\left|\Theta_{4}\left(z\right)\right|^{4}I_{\delta}(z)\,,\ I_{\delta}(z):=\!\!\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\Pi(-1,1)$}}}}}\!\!\!\!\!\dfrac{\left(\left|\lambda\left(z\right)\right|\big{/}\big{|}\lambda\left(z\right)\!-\!\lambda(\zeta)\big{|}\right)|{\rm{d}}\zeta|}{2\big{|}x^{\delta}\zeta-(-x)^{1-\delta}\big{|}^{2}}\ ,\quad z\!\in\!\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,.

For arbitrary z𝔻𝕀z\in\Bb{D}_{{\rm{Im}}>0} and xx\in\Bb{R} we obviously have

|z+x|Imz,|xz1|=|z||x(1/z)||z|Im(1/z)=Imz/|z|Imz,\displaystyle\left|z+x\right|\geqslant{\rm{Im}}z\,,\ \left|xz-1\right|=\left|z\right|\left|x-(1/z)\right|\geqslant\left|z\right|{\rm{Im}}(-1/z)={\rm{Im}}z/\left|z\right|\geqslant{\rm{Im}}z\,,

and hence

|xδz(x)1δ|(min{1,|z|1})Imz,zδ{}\displaystyle\left|x^{\delta}z-(-x)^{1-\delta}\right|\geqslant\left(\min\left\{1,|z|^{-1}\right\}\right){\rm{Im}}z\ ,\quad z\in\Bb{H}\,,\ \delta\!\in\!\{0,1\}\,,\ x\!\in\!\Bb{R}\,. (7.3)

Since the sets 0\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0} and 𝓃𝒩,,𝓃1\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271ptn_{N},...,\,n_{1}} in (6.46) are the subsets of 𝔻\Bb{D} and for any zz from these sets we have 𝔾ϝ𝔻\Bb{G}^{k}_{2}(z)\in\Bb{D}, 0kN0\leqslant k\leqslant N (N=0N=0, if z0z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}), in view of (6.36), (5.87) and (5.84), we deduce from (7.3) that

|1(xδ𝔾ϝδ|1Im2𝔾ϝ, 0kd(𝔫),𝔷𝔫,𝔫{0}\displaystyle\left|\dfrac{1}{(x^{\delta}\Bb{G}^{k}_{2}(z)-(-x)^{1-\delta})^{2}}\right|\leqslant\dfrac{1}{{\rm{Im}}^{2}\Bb{G}^{k}_{2}(z)}\,,\ 0\leqslant k\leqslant d(\eufm{n})\ ,\quad z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\eufm{n}}\,,\ \eufm{n}\in\{0\}\cup\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,,

where 𝔾ϝϝ\Bb{G}^{0}_{2}(z):=z. Together with (6.37), this gives the following estimate for the sum in (6.46)

|1(xδz(x)1δ)2k=0N1(1)kψnN,,nNk(z)(x1δk𝔾ϝδ)2|k=0NIm𝔾ϝ(Imz)Im2𝔾ϝ.\displaystyle\left|{{\dfrac{1}{(x^{\delta}z-(-x)^{1-\delta})^{2}}}}-\sum\limits_{k=0}^{N-1}{{\dfrac{(-1)^{k}\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}^{\,\prime}(z)}{\left(x^{1-\delta_{k}}\Bb{G}^{k+1}_{2}(z)-(-x)^{\delta_{k}}\right)^{2}}}}\right|\leqslant\sum\limits_{k=0}^{N}\dfrac{{\rm{Im}}\,\Bb{G}^{k}_{2}(z)}{({\rm{Im}}\,z)\,{\rm{Im}}^{2}\Bb{G}^{k}_{2}(z)}\ .

As the roof γ𝔫(1,1)\gamma_{\eufm{n}}(-1,1) of 𝔫\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} is also the roof of nN,,n1\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}} then, in accordance with Lemma 5.3(e), the property z𝔫z\in\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} implies that Imz1/(2N){\rm{Im}}\,z\leqslant 1/(2N), i.e., 2N1/(Imz)2N\leqslant 1/({\rm{Im}}\,z) and, by (6.50), we obtain for the latter sum the estimate

k=0N1(Imz)Im𝔾ϝN+1(Imz)21(Imz)3,z𝔫,𝔫\displaystyle\sum\limits_{k=0}^{N}\dfrac{1}{({\rm{Im}}\,z)\,{\rm{Im}}\,\Bb{G}^{k}_{2}(z)}\leqslant\dfrac{N+1}{({\rm{Im}}\,z)^{2}}\leqslant\dfrac{1}{({\rm{Im}}\,z)^{3}}\,,\quad z\!\in\!\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\ \eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{N}\,,\ N\in\Bb{N}\,. (7.4)

As a result, we obtain the following estimates for the functions from (6.46):

|Φδ||(x;z)|{|Θ4(1/z)|4|z|2Iδ(1/z),ifz;|Θ4(z)|4Iδ(z)+1(Imz)2,ifz0;|Θ4(𝔾ϝ|4(Imz)/Im𝔾ϝIδ(𝔾ϝ𝕀𝔾ϝ𝕀ϝifz𝔫.\displaystyle\left|\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)\right|\!\leqslant\!\left\{\begin{array}[]{ll}\dfrac{\left|\Theta_{4}\left(-1/z\right)\right|^{4}}{|z|^{2}}\,I_{\delta}(-1/z),&\ \ \hbox{if}\ \,z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\,;\\ \left|\Theta_{4}\left(z\right)\right|^{4}\,I_{\delta}(z)+\dfrac{1}{({\rm{Im}}\,z)^{2}},&\ \ \hbox{if}\ \,z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,;\\ \dfrac{\left|\Theta_{4}\big{(}\Bb{G}^{N}_{2}(z)\big{)}\right|^{4}}{\left({\rm{Im}}\,z\right)\big{/}{\rm{Im}}\,\Bb{G}^{N}_{2}(z)}\,I_{\delta}\big{(}\Bb{G}^{N}_{2}(z)\big{)}+\dfrac{\sum\limits_{k=0}^{N}\dfrac{1}{{\rm{Im}}\,\Bb{G}^{k}_{2}(z)}}{{\rm{Im}}\,z},&\ \ \hbox{if}\ \,z\in\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,.\end{array}\right. (7.8)

Here, NN\!\in\!\Bb{N}, 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}=(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, xx\!\in\!\Bb{R}, δ{0,1}\delta\!\in\!\{0,1\} and the equality (6.37) was used.

Since for arbitrary ζΠ(1,1)\zeta\in\Pi(-1,1), we have |ζ|5|\zeta|\leqslant\sqrt{5}, (7.3) gives that

Iδ(z)(5/2)Π(1,1)|λ(z)||dζ||λ(z)λ(ζ)|(Imζ)2=11|λ(z)|dt4|λ(z)λ(2i+t)|+022|λ(z)|dtt2|λ(z)λ(1+it)|.\displaystyle\dfrac{I_{\delta}(z)}{(5/2)}\!\leqslant\!\!\!\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\Pi(-1,1)$}}}}}\!\!\!\!\!\dfrac{\left|\lambda\left(z\right)\right||{\rm{d}}\zeta|}{\big{|}\lambda\left(z\right)-\lambda(\zeta)\big{|}({\rm{Im}}\,\zeta)^{2}}\!=\!\int\limits_{-1}^{1}\!\dfrac{\left|\lambda\left(z\right)\right|{\rm{d}}t}{4\big{|}\lambda\left(z\right)-\lambda(2{\rm{i}}\!+\!t)\big{|}}\!+\!\int\limits_{0}^{2}\!\dfrac{2\left|\lambda\left(z\right)\right|{\rm{d}}t}{t^{2}\big{|}\lambda\left(z\right)-\lambda(1\!+\!{\rm{i}}t)\big{|}}\,.

If we denote by Iδ(z)I^{-}_{\delta}(z) and I|δ(z)I^{|}_{\delta}(z) the first and the second integrals in the right-hand side of the above equality, correspondingly, we can apply (2.64)(c) and (2.64)(d) to get (2/5)Iδ(z)Iδ(z)+I|δ(z)(2/5)I_{\delta}(z)\!\leqslant\!I^{-}_{\delta}(z)+I^{|}_{\delta}(z) and

Iδ(z)212λ(2i),I|δ(z)0222|λ(z)|dtt2(|λ(z)|+|λ(1+it)|),z0.\displaystyle I^{-}_{\delta}(z)\leqslant\dfrac{2}{1-2\lambda(2{\rm{i}})}\,,\ I^{|}_{\delta}(z)\leqslant\int\limits_{0}^{2}\!\dfrac{2\sqrt{2}\left|\lambda\left(z\right)\right|{\rm{d}}t}{t^{2}\left(\big{|}\lambda\left(z\right)\big{|}+\big{|}\lambda(1\!+\!{\rm{i}}\,t)\big{|}\right)}\ ,\quad z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,. (7.9)

Let a:=|λ(z)|a:=\big{|}\lambda\left(z\right)\big{|}. Then λ(0)\lambda\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\!\subset\!\Bb{C}_{\,{\rm{Re}}>1/2} implies a>1/2a>1/2 and using (7.2) we get35{}^{\ref*{case27}}

I|δ(z)2a202dtt2(a+1712216eπ/t)=1πalog(1+16eπ/2a17122),\displaystyle\dfrac{I^{|}_{\delta}(z)}{2a\sqrt{2}}\leqslant\int\limits_{0}^{2}\!\dfrac{{\rm{d}}t}{t^{2}\left(a+\dfrac{17-12\sqrt{2}}{16}\,{\rm{e}}^{\pi/t}\right)}=\dfrac{1}{\pi a}\log\left(1+\dfrac{16\,{\rm{e}}^{-\pi/2}a}{17-12\sqrt{2}}\right)\,,\

from which

I|δ(z)22πlog(1+16eπ/217122|λ(z)|),z0,\displaystyle I^{|}_{\delta}(z)\leqslant\dfrac{2\sqrt{2}}{\pi}\log\left(1+\dfrac{16\,{\rm{e}}^{-\pi/2}}{17-12\sqrt{2}}\big{|}\lambda\left(z\right)\big{|}\right)\ ,\quad z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,,

and hence, we deduce from (7.9) and (7.2) that

Iδ(z)55+40221+52πlog(1+16eπ/217122|λ(z)|),z0.\displaystyle I_{\delta}(z)\!\leqslant\!\dfrac{55+40\sqrt{2}}{21}+\dfrac{5\sqrt{2}}{\pi}\log\left(1+\dfrac{16\,{\rm{e}}^{-\pi/2}}{17-12\sqrt{2}}\big{|}\lambda\left(z\right)\big{|}\right)\ ,\quad z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,\,. (7.10)

To obtain the final estimates of Φδ||\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}, we need to deal with the variable zz lying in =1/0\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=-1/\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0} but not in 0\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}. Hence, we transform (7.10), taking into account that due to 22-periodicity of λ\lambda we have 1λ(z)=λ(1/z)=λ(𝔾ϝ1-\lambda(z)=\lambda(-1/z)=\lambda(\Bb{G}_{2}(z)), where 𝔾ϝ𝔽||𝔻¯\Bb{G}_{2}(z)\in\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}} for all z0z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}, in view of (5.95)(a). Namely,

Iδ(z)55+40221+52πlog(1+16eπ/217122)+52πlog(1+|λ(1/z)|),\displaystyle I_{\delta}(z)\!\leqslant\!\dfrac{55+40\sqrt{2}}{21}+\dfrac{5\sqrt{2}}{\pi}\log\left(1+\dfrac{16\,{\rm{e}}^{-\pi/2}}{17-12\sqrt{2}}\right)+\dfrac{5\sqrt{2}}{\pi}\log\left(1+\big{|}\lambda\left(-1/z\right)\big{|}\right),

which after numerical calculations can be written as36{}^{\ref*{case29}}

Iδ(z)16+(9/4)log(1+|λ(𝔾ϝ|),𝔾ϝ𝔽||𝔻¯𝔼ϝ𝔼\displaystyle I_{\delta}(z)\leqslant 16+(9/4)\log\Big{(}1\!+\!\left|\lambda\big{(}\Bb{G}_{2}(z)\big{)}\right|\Big{)},\ \ \Bb{G}_{2}(z)\!\in\!\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\!\subset\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}},\ \ z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,. (7.11)

But for arbitrary x+iy||𝔻¯x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty\in\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}} by (2.56), (3.38) and (2.31)(b) we have

|λ(x+iy)||λ(1+iy)|=λ(iy)/λ(i/y)16eπy+π/y,\displaystyle\left|\lambda(x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty)\right|\leqslant\left|\lambda(1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty)\right|=\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty)\big{/}\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/y)\leqslant 16\,{\rm{e}}^{{{\mbox{\footnotesize{$-\pi y+\pi/y$}}}}}\,,

from which

log(1+|λ(x+iy)|)log(1+16eπy+π/y)π(1+1/y),\displaystyle\log\left(1+\left|\lambda(x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty)\right|\right)\leqslant\log\left(1+16\,{\rm{e}}^{{{\mbox{\footnotesize{$-\pi y+\pi/y$}}}}}\right)\leqslant\pi\left(1+{1}/{y}\right),

and hence36{}^{\ref*{case29}}

Iδ(z)147π20(1+1Im𝔾ϝ),𝔾ϝ𝔽||𝔻¯𝔼ϝ𝔼\displaystyle I_{\delta}(z)\leqslant\dfrac{147\pi}{20}\left(1+\dfrac{1}{{\rm{Im}}\,\Bb{G}_{2}(z)}\right),\ \ \Bb{G}_{2}(z)\!\in\!\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\!\subset\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}},\ \ z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,. (7.12)

By applying for the three expressions in (7.8) the obvious consequence Imz=Im𝔾ϝ{\rm{Im}}\,z={\rm{Im}}\,\Bb{G}_{2}(-1/z), zz\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}, of (5.13), (147/20)θ3(eπ/2)4<30(147/20)\theta_{3}({\rm{e}}^{-\pi/2})^{4}<30, the left-hand side inequality of (7.1), (2.25)(a),(c) and (7.12), we obtain in Section 10.5.1 the following properties.

Lemma 7.1

. Let xx\in\Bb{R}, δ{0,1}\delta\in\{0,1\}, z𝕜𝕜𝔼z\in\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} and the function Φδ||(x;z)\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z) be defined as in Theorem 6.4. Then

|Φδ||(x;z)|{30π(1+1Imz)Im2z,ifz;1Im2z+30π(1+1Im𝔾ϝ)(Imz)Im𝔾ϝ,ifz0;k=0N1Im𝔾ϝImz+30π(1+1Im𝔾ϝ)(Imz)Im𝔾ϝ,ifz𝔫;\displaystyle\left|\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)\right|\leqslant\left\{\begin{array}[]{ll}\dfrac{30\pi\left(1+\dfrac{1}{{\rm{Im}}\,z}\right)}{{\rm{Im}}^{2}z},&\quad\hbox{if}\ \ z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\,;\\[4.26773pt] \dfrac{1}{{\rm{Im}}^{2}\,z}+\dfrac{30\pi\,\left(1+\dfrac{1}{{\rm{Im}}\,\Bb{G}_{2}(z)}\right)}{({\rm{Im}}\,z)\,{\rm{Im}}\,\Bb{G}_{2}(z)},&\quad\hbox{if}\ \ z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,;\\[7.11317pt] \dfrac{\sum\limits_{k=0}^{N}\dfrac{1}{{\rm{Im}}\,\Bb{G}^{k}_{2}(z)}}{{\rm{Im}}\,z}+\dfrac{30\pi\,\left(1+\dfrac{1}{{\rm{Im}}\,\Bb{G}^{N+1}_{2}(z)}\right)}{({\rm{Im}}\,z)\,{\rm{Im}}\,\Bb{G}_{2}^{N+1}(z)},&\quad\hbox{if}\ \ z\in\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,;\end{array}\right. (7.16)

𝔾𝔼𝔾𝔼𝔽||𝔻¯\Bb{G}_{2}\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\!=\!\Bb{G}_{2}^{N+1}\big{(}\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\!=\!\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}, for arbitrary NN\!\in\!\Bb{N} and 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}.

By (6.47), Im𝔾ϝ𝕀ϝ{\rm{Im}}\,\Bb{G}_{2}(z)\geqslant{\rm{Im}}\,z for every z0𝔻𝕀z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\subset\Bb{D}_{{\rm{Im}}>0}. At the same time, for any 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} and z𝔫z\in\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} we have 𝔾ϝ𝔼𝔻𝕀\Bb{G}_{2}^{N}(z)\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\subset\Bb{D}_{{\rm{Im}}>0} and Im𝔾ϝ𝕀ϝ{\rm{Im}}\,\Bb{G}_{2}^{N}(z)\geqslant{\rm{Im}}\,z, in view of (5.95)(b) and (6.50), correspondingly. Applying (6.47) once more, we get Im𝔾ϝ𝕀ϝ{\rm{Im}}\,\Bb{G}_{2}^{N+1}(z)\geqslant{\rm{Im}}\,z. Since Φδ||\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}} of the variable zz is continuous on 𝕜𝕜\Bb{H}_{|{\rm{Re}}|\leqslant 1} we derive from (7.16) and (7.4) the next assertion.38{}^{\ref*{case31}}

Corrollary 7.2

. Let xx\in\Bb{R}, δ{0,1}\delta\in\{0,1\}, z𝕜𝕜𝔼z\in\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} and the function Φδ||(x;z)\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z) be defined as in Theorem 6.4. Then

|Φδ||(x;z)|{20π2Im3z,ifImz1,20π2Im2z,ifImz1,z𝕜𝕜δ{}\displaystyle\left|\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z)\right|\leqslant\left\{\begin{array}[]{ll}\dfrac{20\pi^{2}}{{\rm{Im}}^{3}z}\ ,&\mbox{if}\ \,{\rm{Im}}\,z\!\leqslant\!1,\\[8.5359pt] \dfrac{20\pi^{2}}{{\rm{Im}}^{2}z}\ ,&\mbox{if}\ \,{\rm{Im}}\,z\!\geqslant\!1,\end{array}\right.\ \quad\ z\!\in\!\Bb{H}_{|{\rm{Re}}|\leqslant 1},\ x\!\in\!\Bb{R},\ \delta\!\in\!\{0,1\}\,. (7.19)

7.2 . Main inequalities.

It follows from (6.9) that for arbitrary n1n\geqslant 1 and xx\in\Bb{R} we have

4π2nHn(x)=1+2i1+2ieiπnzΦ0(x;z)dz, 4π2nMn(x)=1+2i1+2ieiπnzΦ1(x;z)dz.\displaystyle 4\pi^{2}n\eurm{H}_{n}(x)=\!\!\!\int\limits_{-1+2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}^{1+2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\!\!\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nz$}}}}}\Phi^{\hskip 0.56917pt0}_{\infty}(x;z){\rm{d}}z,\ \ \ 4\pi^{2}n\eurm{M}_{n}(x)=\!\!\!\int\limits_{-1+2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}^{1+2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\!\!\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nz$}}}}}\Phi^{\hskip 0.56917pt1}_{\infty}(x;z){\rm{d}}z\,.

By Theorem 6.4, for every δ{0,1}\delta\!\in\!\{0,1\} the function Φδ\Phi^{\hskip 0.56917pt\delta}_{\infty} can be analytically extended to 𝕜𝕜𝔼\Bb{H}_{|{\rm{Re}}|<1}\cup\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} and in a second step, by (6.13), to 22-periodic holomorphic function Φδ(x;z)\Phi^{\hskip 0.71114pt\delta}_{\Bb{H}}(x;z) on \Bb{H}, which equals Φδ||(x;z)\Phi^{\,\delta}_{{{\mbox{\tiny{$||$}}}}}(x;z) on 𝕜𝕜\Bb{H}_{|{\rm{Re}}|\leqslant 1}. As a consequence, by Lemma 3.1, we obtain from the above formulas that

4π2nHn(x)=1+i/n1+i/neiπnzΦ 0||(x;z)dz, 4π2nMn(x)=1+i/n1+i/neiπnzΦ 1||(x;z)dz,\displaystyle 4\pi^{2}n\eurm{H}_{n}(x)=\!\!\!\int\limits_{-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/n}^{1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/n}\!\!\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nz$}}}}}\Phi^{\,0}_{{{\mbox{\tiny{$||$}}}}}(x;z){\rm{d}}z,\ \ \ 4\pi^{2}n\eurm{M}_{n}(x)=\!\!\!\int\limits_{-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/n}^{1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/n}\!\!\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nz$}}}}}\Phi^{\,1}_{{{\mbox{\tiny{$||$}}}}}(x;z){\rm{d}}z\,,

from which it follows that

{|Hn(x)||Mn(x)|}eπ4π2n11|Φ{01}||(x;in+t)|dt,,\displaystyle\left\{\begin{array}[]{c}\left|\eurm{H}_{n}(x)\right|\\[2.84544pt] \left|\eurm{M}_{n}(x)\right|\end{array}\right\}\!\leqslant\!\dfrac{{\rm{e}}^{{{\mbox{\footnotesize{$\pi$}}}}}}{4\pi^{2}n}\int\limits_{-1}^{1}\left|\Phi^{\,{{\mbox{\tiny{$\left\{\begin{array}[]{c}0\\ 1\end{array}\right\}$}}}}}_{\,{{\mbox{\tiny{$||$}}}}}\left(x;\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}{n}\!+\!t\right)\right|{\rm{d}}t,, (7.24)

for n1n\geqslant 1 and xx\in\Bb{R}. Since Im(t+i/n)=1/n1{\rm{Im}}\,(t+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/n)=1/n\leqslant 1 we can apply (7.19) to estimate the integral in the right-hand side of (7.24) as follows

|Hn(x)|,|Mn(x)|20π2n3eπ2π2n=10eπn2<π6n24,x\displaystyle\left|\eurm{H}_{n}(x)\right|\ ,\ \left|\eurm{M}_{n}(x)\right|\leqslant 20\pi^{2}n^{3}\dfrac{{\rm{e}}^{{{\mbox{\footnotesize{$\pi$}}}}}}{2\pi^{2}n}=10\,{\rm{e}}^{{{\mbox{\footnotesize{$\pi$}}}}}n^{2}<\dfrac{\pi^{6}n^{2}}{4}\,,\quad x\in\Bb{R}\,.

Here, we used39{}^{\ref*{case32}} that 40exp(π)<π640\exp(\pi)<\pi^{6}. Using (1.18) and (1.19), we obtain40{}^{\ref*{case33}}

|Hn(x)|,|Mn(x)|min{π6n24,π6n22(1+x2)},x\displaystyle\left|\eurm{H}_{n}(x)\right|,\left|\eurm{M}_{n}(x)\right|\leqslant\min\left\{\,\dfrac{\pi^{6}n^{2}}{4}\,,\ \ \dfrac{\pi^{6}n^{2}}{2(1+x^{2})}\,\right\}\,,\quad x\in\Bb{R}\,,\ n\in\Bb{Z}_{\neq 0}\,. (7.25)

The corresponding estimates of H0\eurm{H}_{0} can be derived from the explicit integral formula written after (1.21):41{}^{\ref*{case34}}

|H0(x)|min{32,31+x2},x\displaystyle\left|\eurm{H}_{\hskip 0.56917pt0}(x)\right|\leqslant\min\left\{\dfrac{3}{2}\ ,\ \dfrac{3}{1+x^{2}}\right\}\ ,\quad x\in\Bb{R}\ . (7.26)

The first immediate consequence of the obtained estimates and (6.9) is the possibility of expressing the functions Φδ\Phi^{\hskip 0.56917pt\delta}_{\Bb{H}}, δ{0,1}\delta\!\in\!\{0,1\}, for any xx\in\Bb{R} in the form

n1nHn(x)eiπnz=Φ0(x;z)2π2,n1nMn(x)eiπnz=Φ1(x;z)2π2,z\displaystyle\sum\limits_{n\geqslant 1}\!n\,\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\Phi^{\hskip 0.56917pt0}_{\Bb{H}}(x;z)}{2\pi^{2}}\,,\ \ \sum\limits_{n\geqslant 1}\!n\,\eurm{M}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\Phi^{\hskip 0.56917pt1}_{\Bb{H}}(x;z)}{2\pi^{2}}\,,\ \ z\in\Bb{H}\,, (7.27)

because, by (7.25) and (7.26), both series on the left-hand sides of the equalities in (6.9) turn out to be holomorphic on \Bb{H}. It follows from (6.8), the identity λ(z)=πiΘ4(z)4λ(z)\lambda^{\,\prime}(z)=\pi{\rm{i}}\Theta_{4}\left(z\right)^{4}\lambda(z), zz\in\Bb{H} (see (2.41)), and the asymptotics λ(it)0\lambda(\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptt)\to 0 as t+t\to+\infty, that for all δ{0,1}\delta\in\{0,1\} and xx\in\Bb{R}, the function

Ψδ(x;z):=12πiγ(1,1)(xδζ(x)1δ)2log(1λ(z)λ(ζ))dζ,\displaystyle\Psi^{\,\delta}_{\infty}(x;z):=\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{-2}\log\left(1-\dfrac{\lambda(z)}{\lambda(\zeta)}\right){\rm{d}}\zeta\ \,, (7.28)

of the variable zz\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} is the unique primitive of Φδ\Phi^{\,\delta}_{\infty} which has limit 0 as ziϝz\in{\rm{i}}\Bb{R}_{>1},z\to\infty. Then we derive from (6.9) that

n1Hn(x)eiπnz=iΨ0(x;z)2π,n1Mn(x)eiπnz=iΨ1(x;z)2π,z𝕀\displaystyle\sum\limits_{n\geqslant 1}\!\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Psi^{\hskip 0.56917pt0}_{\infty}(x;z)}{2\pi}\,,\ \ \sum\limits_{n\geqslant 1}\!\eurm{M}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Psi^{\hskip 0.56917pt1}_{\infty}(x;z)}{2\pi}\,,\ \ z\in\Bb{H}_{\,{\rm{Im}}>1}\,,

and since both series here for each xx\in\Bb{R} are holomorphic on \Bb{H}, in view of (7.25) and (7.26), we conclude that for any δ{0,1}\delta\in\{0,1\} and xx\in\Bb{R} the function Ψδ\Psi^{\hskip 0.56917pt\delta}_{\infty} has analytic extension from 𝕀\Bb{H}_{\,{\rm{Im}}>1} to \Bb{H} and the resulting extension Ψδ\Psi^{\hskip 0.56917pt\delta}_{\Bb{H}} satisfies

n1Hn(x)eiπnz=iΨ0(x;z)2π,n1Mn(x)eiπnz=iΨ1(x;z)2π,z\displaystyle\sum\limits_{n\geqslant 1}\!\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Psi^{\hskip 0.56917pt0}_{\Bb{H}}(x;z)}{2\pi}\,,\ \ \sum\limits_{n\geqslant 1}\!\eurm{M}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\!=\!\frac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Psi^{\hskip 0.56917pt1}_{\Bb{H}}(x;z)}{2\pi}\,,\ z\in\Bb{H}\,. (7.29)

It follows from (7.27) and (7.29) that

(/z)Ψδ(x;z)=Φδ(x;z),xϝδ{}\displaystyle({\partial}/{\partial z})\,\Psi^{\,\delta}_{\Bb{H}}(x;z)=\Phi^{\,\delta}_{\Bb{H}}(x;z)\ ,\quad x\in\Bb{R}\,,\ z\in\Bb{H}\,,\ \delta\in\{0,1\}\,. (7.30)

When we integrate by parts, (7.28) becomes (see (2.47))

Ψδ(x;z)=12πiγ(1,1)log(1λ(z)λ(ζ))dζδxδζ(x)1δ\displaystyle\Psi^{\,\delta}_{\infty}(x;z)=-\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\log\left(1-\dfrac{\lambda(z)}{\lambda(\zeta)}\right)\,{\rm{d}}\dfrac{\zeta^{\delta}}{x^{\delta}\zeta-(-x)^{1-\delta}}
=12πiγ(1,1)λ(z)λ(ζ)λ(ζ)(λ(ζ)λ(z))ζδdζ(xδζ(x)1δ),δ{0,1},x{}\displaystyle=\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\lambda(z)\lambda^{\,\prime}(\zeta)}{\lambda(\zeta)\left(\lambda(\zeta)-\lambda(z)\right)}\dfrac{\zeta^{\delta}{\rm{d}}\zeta}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}\,,\ \ \delta\in\{0,1\}\,,\ x\in\Bb{R}\setminus\{0\}\,,

and hence, in view of λ(z)=iπλ(z)Θ4(z)4\lambda^{\,\prime}(z)\!=\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi\lambda(z)\Theta_{4}(z)^{4}, zz\in\Bb{H} (see (2.41), (2.21)), we find

Ψδ(x;z)=12γ(1,1)λ(z)Θ4(ζ)4ζδdζ(λ(ζ)λ(z))(xδζ(x)1δ),δ{0,1}.\displaystyle\Psi^{\,\delta}_{\infty}(x;z)=\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\dfrac{\lambda(z)\Theta_{4}(\zeta)^{4}\,\zeta^{\delta}{\rm{d}}\zeta}{\big{(}\lambda(\zeta)-\lambda(z)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}\,,\ \ \delta\!\in\!\{0,1\}\,. (7.31)

In comparison with (7.28), the formula (7.31) determines the values of Ψδ(x;z)\Psi^{\,\delta}_{\infty}(x;z) for arbitrary xx\!\in\!\Bb{R} and z𝕊z\in\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}, and, it can be easily calculated that42{}^{\ref*{case40}}

(/z)Ψδ(x;z)=Φδ(x;z),z𝕊{}δ{}\displaystyle({\partial}/{\partial z})\,\Psi^{\,\delta}_{\infty}(x;z)=\Phi^{\,\delta}_{\infty}(x;z)\ ,\quad z\in\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\,,\ x\in\Bb{R}\setminus\{0\}\,,\ \delta\in\{0,1\}\,.

This relationship enables us to apply the reasons as in the proof of Lemma 6.1 to obtain for arbitrary δ{0,1}\delta\in\{0,1\} and xx\in\Bb{R} that43{}^{\ref*{case41}}

Ψδ(x;z)=Ψδ(x;z)+zδxδz(x)1δ,z0𝕊\displaystyle\Psi^{\hskip 0.56917pt\delta}_{\Bb{H}}(x;z)=\Psi^{\,\delta}_{\infty}(x;z)+\dfrac{z^{\delta}}{x^{\delta}z-(-x)^{1-\delta}}\ ,\quad z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\subset\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\,. (7.32)

8 . Interpolation formula for the Klein-Gordon equation

8.1 . Hyperbolic Fourier series in \Bb{H}.

In view of (6.13), (6.28) and (6.16), for arbitrary fixed xx\!\in\!\Bb{R} and δ{0,1}\delta\!\in\!\{0,1\} we have Φδ(x;z)=Φδ0(x;z)\Phi^{\hskip 0.71114pt\delta}_{\Bb{H}}(x;z)=\Phi^{\hskip 0.71114pt\delta}_{0}(x;z) for any z0z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\sqcup\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}, where 1/=0-1/\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}. Then for each zz\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} it follows from (6.12), (6.16), (6.27) and (6.13) that

Φδ(x;1/z)=Φδ0(x;1/z)=Φδ(x;1/z)(xδ(1/z)(x)1δ)2\displaystyle\Phi^{\hskip 0.71114pt\delta}_{\Bb{H}}(x;-1/z)\!=\!\Phi^{\hskip 0.71114pt\delta}_{0}(x;-1/z)\!=\!\Phi^{\hskip 0.56917pt\delta}_{\infty}(x;-1/z)-\big{(}x^{\delta}(-1/z)\!-\!(-x)^{1-\delta}\big{)}^{-2}
=z2Φ1δ(x;z)z2(xδ+z(x)1δ)2=z2Φ1δ(x;z)z2(x1δz(x)δ)2,\displaystyle=\!-z^{2}\Phi^{1-\delta}_{\infty}(x;z)-z^{2}\big{(}x^{\delta}\!+\!z(-x)^{1-\delta}\big{)}^{-2}\!=\!-z^{2}\Phi^{1-\delta}_{\Bb{H}}(x;z)-z^{2}\big{(}x^{1-\delta}z\!-\!(-x)^{\delta}\big{)}^{-2}\!\!,

and since the functions appearing on the two sides of this identity are holomorphic in \Bb{H} we obtain, by the uniqueness theorem for analytic functions (see (con, , p.​ 78, Theorem 3.7(c))), that

Φ1δ(x;z)+z2Φδ(x;1/z)=(x1δz(x)δ)2,zδ{}\displaystyle\Phi^{1-\delta}_{\Bb{H}}(x;z)\!+\!z^{-2}\Phi^{\hskip 0.71114pt\delta}_{\Bb{H}}(x;-1/z)\!=\!-\big{(}x^{1-\delta}z\!-\!(-x)^{\delta}\big{)}^{-2}\!\!,\ z\!\in\!\Bb{H},\ x\!\in\!\Bb{R},\ \delta\!\in\!\{0,1\}. (8.1)

In the case of δ=1\delta=1, we integrate (8.1) with respect to the variable zz and obtain, by (7.30),

Ψ0(x;z)+Ψ1(x;1/z)=(z+x)1+η(x),\displaystyle\Psi^{0}_{\Bb{H}}(x;z)+\Psi^{\hskip 0.71114pt1}_{\Bb{H}}(x;-1/z)=\big{(}z+x\big{)}^{-1}+\eta(x)\,,

where, for x{}x\in\Bb{R}\setminus\{0\}, we use (7.31), (7.32) and limt+λ(i/t)=1\lim_{t\to+\infty}\lambda({\rm{i}}/t)=1 to see that

η(x)\displaystyle\eta(x) =limt+Ψ1(x;i/t)=12γ(1,1)Θ3(ζ)4ζdζxζ1=12γ(1,1)Θ3(1/ζ)4dζζ2(x1/ζ)\displaystyle=\lim_{t\to+\infty}\Psi^{\hskip 0.56917pt1}_{\Bb{H}}(x;\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt/t)=\!-\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\Theta_{3}(\zeta)^{4}\zeta{\rm{d}}\zeta}{x\zeta-1}=-\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{-\Theta_{3}(-1/\zeta)^{4}{\rm{d}}\zeta}{\zeta^{2}\big{(}x-1/\zeta\big{)}}
=12γ(1,1)Θ3(ζ)4dζζ+x,\displaystyle=-\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\Theta_{3}(\zeta)^{4}{\rm{d}}\zeta}{\zeta+x}\ ,

where we have applied (2.21) and (2.25)(b). Thus,

Ψ0(x;z)+Ψ1(x;1z)=1z+x12γ(1,1)Θ3(ζ)4dζζ+x,z\displaystyle\Psi^{0}_{\Bb{H}}(x;z)+\Psi^{\hskip 0.71114pt1}_{\Bb{H}}\left(x;-\dfrac{1}{z}\right)=\dfrac{1}{z+x}-\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\Theta_{3}(\zeta)^{4}{\rm{d}}\zeta}{\zeta+x}\ ,\quad z\!\in\!\Bb{H},\ x\!\in\!\Bb{R}\,. (8.2)

By substituting the identities (7.27) and (7.29) into (8.1)(with δ=1\delta=1) and in (8.2), respectively, we see that for arbitrary zz\in\Bb{H} and x,tx,t\in\Bb{R} the following identities hold:

12πi(xz)2=n1iπnHn(x)eiπnz+1z2n1iπnMn(x)eiπnz,\displaystyle\dfrac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt(x-z)^{2}}=\sum\limits_{n\geqslant 1}\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n\,\eurm{H}_{-n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}+\dfrac{1}{z^{2}}\sum\limits_{n\geqslant 1}\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n\,\eurm{M}_{-n}(x)\,{\rm{e}}^{{\mbox{\footnotesize{${-\dfrac{{\rm{i}}\pi n}{z}}$}}}}\,, (8.3)

and

12πi(1tz12γ(1,1)Θ3(ζ)4dζtζ)=n1Hn(t)eiπnz+n1Mn(t)eiπnz.\displaystyle\dfrac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\bigg{(}\dfrac{1}{t-z}-\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\Theta_{3}(\zeta)^{4}d\zeta}{t-\zeta}\bigg{)}=\sum\limits_{n\geqslant 1}\eurm{H}_{-n}(t)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}+\sum\limits_{n\geqslant 1}\eurm{M}_{-n}(t)\,{\rm{e}}^{{\mbox{\footnotesize{${-\dfrac{{\rm{i}}\pi n}{z}}$}}}}\,. (8.4)

By multiplying the latter equality by a function FH1+(F\in{H}^{1}_{+}(\Bb{R}) we integrate it with respect to tt\in\Bb{R} and apply well-known property of functions in H1({H}^{1}(\Bb{H}) (see (ko, , p.​ 116)), together with the estimate (2.48) and the identity (see (2.52))

2πiF(t)H0(t)dt\displaystyle 2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\int\limits_{\Bb{R}}F(t)\eurm{H}_{0}(t){\rm{d}}t =F(t)(12γ(1,1)Θ3(ζ)4dζtζ12γ(1,1)Θ3(ζ)4dζt+ζ)dx\displaystyle=\int\limits_{\Bb{R}}F(t)\bigg{(}\ \frac{1}{2}\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\Theta_{3}(\zeta)^{4}{\rm{d}}\zeta}{t-\zeta}-\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\Theta_{3}(\zeta)^{4}{\rm{d}}\zeta}{t+\zeta}\ \bigg{)}{\rm{d}}x
=F(t)(12γ(1,1)Θ3(ζ)4dζtζ)dx,\displaystyle=\int\limits_{\Bb{R}}F(t)\bigg{(}\ \frac{1}{2}\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\Theta_{3}(\zeta)^{4}{\rm{d}}\zeta}{t-\zeta}\ \bigg{)}{\rm{d}}x\,,

to obtain that FF enjoys the representation (see (1.5))

F(z)=h0(F)+n1hn(F)eiπnz+n1mn(F)exp(iπnz),z\displaystyle F(z)=\eurm{h}_{0}(F)+\sum\limits_{n\geqslant 1}\eurm{h}_{n}(F)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}+\sum\limits_{n\geqslant 1}\eurm{m}_{n}(F)\,\exp\left({{\mbox{\footnotesize{${-\dfrac{{\rm{i}}\pi n}{z}}$}}}}\right),\ \ z\in\Bb{H}\,. (8.5)

We now extend this representation to a larger class of functions. According to (gar, , p.​ 60, Corollary 3.4), Φ(i(1z)/(1+z))H1(𝔻\Phi(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt(1-z)/(1+z))\in{H}^{1}(\Bb{D}) if and only if Φ(z)/(z+i)2H1(\Phi(z)/(z+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{2}\in{H}^{1}(\Bb{H}). Since then Fε(z):=Φ(z)/(i+εz)2H1(F_{\varepsilon}(z):=\Phi(z)/(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt+\varepsilon z)^{2}\in{H}^{1}(\Bb{H}) for any ε>0\varepsilon>0 we can expand FεF_{\varepsilon} as in (8.5). We then apply the estimates (7.25), (7.26) together with the Lebesgue dominated convergence theorem (nat, , p.​ 161) as 0<ε00<\varepsilon\to 0 to get the following representation for functions in a weighted Hardy class H1({H}^{1}(\Bb{H}) (cf. Theorem 3.1 in (bon, , p.​ 14)).

Theorem 8.1

. Let F(z)/(z+i)2H1(F(z)/(z+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{2}\in{H}^{1}(\Bb{H}). Then

F(z)=h0(F)+n1hn(F)eiπnz+n1mn(F)eiπnz,z\displaystyle F(z)=\eurm{h}_{0}(F)+\sum\limits_{n\geqslant 1}\eurm{h}_{n}(F)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}+\sum\limits_{n\geqslant 1}\eurm{m}_{n}(F)\,{\rm{e}}^{{\mbox{\footnotesize{${-\dfrac{{\rm{i}}\pi n}{z}}$}}}}\ ,\quad z\in\Bb{H}\,,

where

hn(F):=F(t)Hn(t)dt,n𝔽𝔽𝕄\displaystyle\eurm{h}_{n}(F)\!:=\!\int\limits_{\Bb{R}}\!F(t)\eurm{H}_{-n}(t){\rm{d}}t\,,\ n\!\in\!\Bb{Z}_{\geqslant 0}\,,\ \ \eurm{m}_{n}(F)\!:=\!\int\limits_{\Bb{R}}\!F(t)\eurm{M}_{-n}(t){\rm{d}}t\ ,\ n\!\in\!\Bb{N}\,,

and |hn(F)|,|mn(F)|(π6/4)n2F𝔻H1+|\eurm{h}_{n}(F)|,|\eurm{m}_{n}(F)|\leqslant(\pi^{6}/4)\,n^{2}\left\|F_{\Bb{D}}\right\|_{H^{1}_{+}}, n1n\geqslant 1, F𝔻(z):=F(z)/(z+i)2F_{\Bb{D}}(z):=F(z)/(z+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{2}, zz\in\Bb{H}.

In view of Jordan’s lemma, for any x,y>0x,y>0 the function zexp(ixziy/z)z\mapsto\exp(\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxz-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty/z) is bounded and holomorphic in \Bb{H}, and we can calculate that

eixtiy/tdttz\displaystyle\int\limits_{\Bb{R}}\dfrac{{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}{\rm{d}}t}{t-z} :=limA+AAeixtiy/tdttz\displaystyle:=\lim\limits_{A\to+\infty}\int\limits_{-A}^{A}\dfrac{{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}{\rm{d}}t}{t-z}
=2πieixziy/zχ(z),z\displaystyle=2\pi{\rm{i}}\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xz-{\rm{i}}y/z$}}}}}\chi_{\Bb{H}}(z)\,,\ \ z\in\Bb{C}\setminus\Bb{R}\ ,\ x>0\ ,\ y\geqslant 0\,. (8.6)

By multiplying (8.4) by this function and after integrating over tt\in\Bb{R}, we obtain as above,

eixziy/z=0(𝓍,𝓎)+𝓃1𝓃(𝓍,𝓎)eiπnz+𝓃1𝓃(𝓎,𝓍)eiπnz,\displaystyle{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xz-{\rm{i}}y/z$}}}}}=\eusm{R}_{0}(x,-y)+\sum\limits_{n\geqslant 1}\eusm{R}_{n}(x,-y)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}+\sum\limits_{n\geqslant 1}\eusm{R}_{n}(y,-x)\,{\rm{e}}^{{\mbox{\footnotesize{${-\dfrac{{\rm{i}}\pi n}{z}}$}}}},

for all zz\in\Bb{H} and x,y0x,y\geqslant 0, where

𝓃(𝓍,𝓎):=ixt+iy/tHn(t)dt,n\displaystyle\eusm{R}_{n}(x,y)\!:=\!\int\nolimits_{\Bb{R}}\ e{}^{{{\mbox{\footnotesize{${\rm{i}}xt+{\rm{i}}y/t$}}}}}\eurm{H}_{-n}(t){\rm{d}}t\,,\quad n\!\in\!\Bb{Z}_{\geqslant 0}\,,\quad x,y\in\Bb{R}\,. (8.7)

Here, the change of variable t=1/tt^{\,\prime}=-1/t and the symmetry property (1.19) entail that, for arbitrary x,yx,y\in\Bb{R},

𝓃(𝓎,𝓍)=eiytix/tHn(t)dt=eixtiy/tMn(t)dt,n\displaystyle\eusm{R}_{n}(y,-x)\!=\!\int\nolimits_{\Bb{R}}\,{\rm{e}}^{{{\mbox{\footnotesize{$i{\rm{}}yt-{\rm{{\rm{i}}}}x/t$}}}}}\eurm{H}_{-n}(t){\rm{d}}t=\int\nolimits_{\Bb{R}}\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}\eurm{M}_{-n}(t){\rm{d}}t\ ,\quad n\in\Bb{N}\,, (8.8)

while, as a consequence of (3.20), we have

0(𝓍,𝓎)=0(𝓎,𝓍)=0(𝓍,𝓎),𝓍,𝓎\displaystyle\eusm{R}_{0}(x,-y)=\eusm{R}_{0}(y,-x)=\eusm{R}_{0}(-x,y)\ ,\quad x,y\in\Bb{R}\,. (8.9)

8.2 . Conjugate hyperbolic Fourier series.

For arbitrary φL1(\varphi\in L^{1}(\Bb{R}), the series in the right-hand side of the equality (1.9) is called conjugate hyperbolic Fourier series of φ\varphi, where the coefficients are defined as in (1.7). Taking account the estimates (7.25), the result (1.9) can be improved as follows.

Theorem 8.2

. Let φL1(\varphi\in L^{1}(\Bb{R}), the numbers {hn(φ)}n\{\eurm{h}_{n}^{\star}(\varphi)\}_{n\in\Bb{Z}} , {mn(φ)}n\{\eurm{m}_{n}^{\star}(\varphi)\}_{n\in\Bb{Z}_{\neq 0}} be defined as in (1.6), and

n{}n2(|hn(φ)|+|mn(φ)|)<.\displaystyle\sum\nolimits_{n\in\Bb{Z}\setminus\{0\}}n^{2}\left(\left|\eurm{h}_{n}^{\star}(\varphi)\right|+\left|\eurm{m}_{n}^{\star}(\varphi)\right|\right)<\infty. (8.10)

Then φ\varphi can be expanded into the conjugate hyperbolic Fourier series (1.9), which converges absolutely and uniformly over all xx\in\Bb{R}.

We now apply Theorem 8.2 to obtain the conjugate hyperbolic Fourier series expansion of the Poisson kernel44{}^{\ref*{case36}}

1πy(tx)2+y2=H0(t)+n1(eπn(y+ix)Hn(t)+eπny+ixMn(t))\displaystyle\dfrac{1}{\pi}\dfrac{y}{(t-x)^{2}+y^{2}}=\eurm{H}_{0}(t)\!+\!\sum\limits_{n\geqslant 1}\!\Big{(}{\rm{e}}^{{{\mbox{\footnotesize{$-\pi n\,(y+{\rm{i}}x)$}}}}}\eurm{H}_{n}(t)\!+\!{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y+{\rm{i}}x}$}}}}}\eurm{M}_{n}(t)\Big{)}
+n1(eπn(yix)Hn(t)+eπnyixMn(t)),t,x\displaystyle+\sum\limits_{n\geqslant 1}\!\Big{(}{\rm{e}}^{{{\mbox{\footnotesize{$-\pi n\,(y-{\rm{i}}x)$}}}}}\eurm{H}_{-n}(t)\!+\!{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y-{\rm{i}}x}$}}}}}\eurm{M}_{-n}(t)\Big{)}\ ,\quad\ t,x\in\Bb{R},\ y>0\,. (8.11)

This allows us to expand the harmonic extension to the upper half-plane of a given fL1(f\in L^{1}(\Bb{R},(1+x^{2})^{-1}dx) given by convolution with the Poisson kernel in the form

1πy(tx)2+y2f(t)dt=h0(f)+n1(eπn(yix)hn(f)+eπnyixmn(f))\displaystyle\dfrac{1}{\pi}\int\limits_{\Bb{R}}\dfrac{y}{(t-x)^{2}+y^{2}}f(t){\rm{d}}t=\eurm{h}_{0}(f)\!+\!\sum\limits_{n\geqslant 1}\!\left({\rm{e}}^{{{\mbox{\footnotesize{$-\pi n\,(y-{\rm{i}}x)$}}}}}\eurm{h}_{n}(f)\!+\!{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y-{\rm{i}}x}$}}}}}\eurm{m}_{n}(f)\right)
+n1(eπn(y+ix)hn(f)+eπny+ixmn(f)),x\displaystyle+\sum\limits_{n\geqslant 1}\!\left({\rm{e}}^{{{\mbox{\footnotesize{$-\pi n\,(y+{\rm{i}}x)$}}}}}\eurm{h}_{-n}(f)\!+\!{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y+{\rm{i}}x}$}}}}}\eurm{m}_{-n}(f)\right)\ ,\quad\ x\in\Bb{R},\ y>0\,.

This formula can be understood as the regularization of the hyperbolic Fourier series (1.5) of ff found by considering the harmonic extensions of the basis functions.

8.3 . Density in Hardy classes.

By manipulations similar to those employed in the proof of Corollary 3.3 in (gar, , p.​ 59) we obtain in Section 10.6.1 the following property of the Hardy class H1+({\rm{H}}^{1}_{+}(\Bb{R}).

Lemma 8.3

. The linear subspace H1+(𝕊{\rm{H}}^{1}_{+}(\Bb{R})\cap\eurm{S}(\Bb{R}) is dense in H1+({\rm{H}}^{1}_{+}(\Bb{R}).

In view of (1.9), every φH1+(𝕊\varphi\in{\rm{H}}^{1}_{+}(\Bb{R})\cap\eurm{S}(\Bb{R}), can be expanded in an absolutely convergent conjugate hyperbolic Fourier series, all whose coefficients with non-positive indexes are zero, as follows from (1.6) and the known properties of the functions from the class H1+({\rm{H}}^{1}_{+}(\Bb{R}) (see (gar, , p.​ 88, 2(iv))). It now follows from (4.1) and (1.18) that the following property holds.

Theorem 8.4

. The function system {Hn(x)}n1{Mn(x)}n1H1+(\left\{\eurm{H}_{n}(x)\right\}_{n\geqslant 1}\!\cup\!\left\{\eurm{M}_{n}(x)\right\}_{n\geqslant 1}\!\subset\!{\rm{H}}^{1}_{+}(\Bb{R}) is complete in H1+({\rm{H}}^{1}_{+}(\Bb{R}), while the functions {Hn(x)}n1{Mn(x)}n1H1(\left\{\eurm{H}_{-n}(x)\right\}_{n\geqslant 1}\!\cup\!\left\{\eurm{M}_{-n}(x)\right\}_{n\geqslant 1}\!\subset\!{\rm{H}}^{1}_{-}(\Bb{R}) form a complete system in H1({\rm{H}}^{1}_{-}(\Bb{R}).

8.4 . Interpolating functions and interpolation formula.

We apply Jordan’s lemma and the residue theorem in the same way as for the proof of (8.6), and obtain

eixtiy/tdt(tz)2=2π(x+yz2)eixziy/zχ(z),z\displaystyle\int\limits_{\Bb{R}}\dfrac{{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}{\rm{d}}t}{(t-z)^{2}}=-2\pi\left(x+\dfrac{y}{z^{2}}\right){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xz-{\rm{i}}y/z$}}}}}\chi_{\Bb{H}}(z)\ ,\ \ z\in\Bb{C}\setminus\Bb{R}, (8.12)

for all x>0x>0, y0y\geqslant 0. We proceed and apply this identity together with (8.6) to the formulas (2.52) written in the form

H0(t)=14πiγ(1,1)Θ3(z)4(1tz1t+z)dz,\displaystyle\eurm{H}_{0}(t)\!=\dfrac{1}{4\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\Theta_{3}\left(z\right)^{4}\left(\dfrac{1}{t-z}-\dfrac{1}{t+z}\right){\rm{d}}z\,, t\displaystyle t\in\Bb{R}\,,
Hn(t)=Hn(t)=(1)4π2nγ(1,1)Sn(1/λ(z))dz(tz)2,\displaystyle\eurm{H}_{-n}(t)=\eurm{H}_{n}(-t)=\dfrac{(-1)}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\dfrac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left({1}/{\lambda(z)}\right){\rm{d}}z}{(t-z)^{2}}\ , n1,t\displaystyle n\geqslant 1\,,\ t\in\Bb{R}\,, (8.13)

and obtain from (8.7), by using (2.48), (3.36)(a) and (3.29)(a), that the integral representations

𝓃(𝓍,𝓎)=eixtiy/tHn(t)dt=12πnγ(1,1)(x+yz2)eixziy/zSn(1λ(z))dz,\displaystyle\eusm{R}_{n}(x,-y)=\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}\eurm{H}_{-n}(t){\rm{d}}t=\dfrac{1}{2\pi n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\left(x+\dfrac{y}{z^{2}}\right){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xz-{\rm{i}}y/z$}}}}}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(z)}\right){\rm{d}}z\,,
0(𝓍,𝓎)=eixtiy/tH0(t)dt=12γ(1,1)Θ3(z)4eixziy/zdz,n1,\displaystyle\eusm{R}_{0}(x,-y)=\!\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}\eurm{H}_{0}(t){\rm{d}}t=\dfrac{1}{2}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\Theta_{3}\left(z\right)^{4}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xz-{\rm{i}}y/z$}}}}}{\rm{d}}z\,,\ \ n\geqslant 1\,, (8.14)

hold for all x,y0x,y\geqslant 0, because 𝓃𝒞(\eusm{R}_{n}\in C(\Bb{R}^{2}) for all n0n\geqslant 0, as we see from the estimates (7.25) and (7.26). In addition to the symmetry property (8.9), we observe that by substituting (8.13) into (8.7), while taking into account (8.12) (interchange in orders of integration is justified by (3.36)(a) and (3.29)(a)), we arrive at

𝓃(𝓍,𝓎)\displaystyle\eusm{R}_{n}(-x,y) =eixt+iy/tHn(t)dt=eixtiy/tHn(t)dt\displaystyle=\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}xt+{\rm{i}}y/t$}}}}}\eurm{H}_{-n}(t){\rm{d}}t=\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}\eurm{H}_{n}(t){\rm{d}}t (8.15)
=(1)4π2nγ(1,1)Sn(1λ(z))eixtiy/tdt(t+z)2dz=0,x,y>0,n1,\displaystyle=\dfrac{(-1)}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(z)}\right)\int\limits_{\Bb{R}}\!\dfrac{{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt-{\rm{i}}y/t$}}}}}{\rm{d}}t}{(t+z)^{2}}\ {\rm{d}}z=0\ ,\ \ \ x,y>0\,,\ n\geqslant 1\,,

and hence, since 𝓃\eusm{R}_{n} is continuous on \Bb{R}^{2} for each n1n\geqslant 1, we find that

𝓃(𝓍,𝓎)=0,𝓍,𝓎0,𝓃1.\displaystyle\eusm{R}_{n}(-x,y)=0\,,\quad x,y\geqslant 0\,,\ n\geqslant 1\,. (8.16)

This property also follows directly from HnH1+(\eurm{H}_{n}\in{\rm{H}}^{1}_{+}(\Bb{R}) (see (4.1)) and the equality (8.15). In Section 10.6.1 we prove the following assertion.

Theorem 8.5

. Let {𝓃}𝓃0\{\eusm{R}_{\hskip 0.71114ptn}\}_{n\geqslant 0} be given by (8.7). Then 𝓃𝒞(\eusm{R}_{\hskip 0.71114ptn}\in C(\Bb{R}^{2}), and, in addition, for each n0n\geqslant 0, the restriction of the function 𝓃\eusm{R}_{\hskip 0.71114ptn} to the quadrant ×\Bb{R}_{\geqslant 0}\times\Bb{R}_{\leqslant 0} extends to all of \Bb{C}^{2} as an entire function of two variables. At the same time, we have

(a)0(π𝓂,0)=δ0𝓂,0(0,π𝓂)=δ0𝓂,m0,(b)𝓃(π𝓂,0)=δ𝓃𝓂,𝓃(0,π𝓂)=0,m0,n1.\displaystyle\begin{array}[]{lllll}{\rm{{(a)}}}&\ \ \eusm{R}_{\hskip 0.71114pt0}(\pi m,0)=\delta_{0m}\ ,&\ \ \eusm{R}_{\hskip 0.71114pt0}(0,-\pi m)=\delta_{0m}\ ,&\ m\geqslant 0\ ,&\\[5.69046pt] {\rm{{(b)}}}&\ \ \eusm{R}_{\hskip 0.71114ptn}(\pi m,0)=\delta_{nm}\ ,&\ \ \eusm{R}_{\hskip 0.71114ptn}(0,-\pi m)=0\ ,&\ m\geqslant 0\ ,&\ n\geqslant 1\ .\end{array} (8.19)

The function 0\eusm{R}_{\hskip 0.71114pt0} satisfies

0(𝓍,𝓎)=0(𝓎,𝓍)=0(𝓍,𝓎),|0(𝓍,𝓎)|3π,\displaystyle\eusm{R}_{0}(x,-y)=\eusm{R}_{0}(y,-x)=\eusm{R}_{0}(-x,y),\ \ \ \left|\eusm{R}_{0}(x,y)\right|\leqslant 3\pi\,, x,y\displaystyle x,y\in\Bb{R}\,; (8.20)
|0(𝓍,𝓎)|5K0(2π(x+y+1)),\displaystyle\left|\eusm{R}_{0}(x,-y)\right|\leqslant 5K_{0}\left(\sqrt{2\pi(x+y+1)}\right), x,y0,\displaystyle x,y\geqslant 0\,, (8.21)

while for any n1n\geqslant 1 we have

𝓃(𝓍,𝓎)=0,𝓍,𝓎0;|𝓃(𝓍,𝓎)|π7𝓃2/2,\displaystyle\eusm{R}_{\hskip 0.71114ptn}(-x,y)=0\,,\ \ \ \ x,y\geqslant 0\,;\ \ \ \ \ \ \ \ \ \ \left|\eusm{R}_{n}(x,y)\right|\leqslant\pi^{7}n^{2}/2\,, x,y\displaystyle x,y\in\Bb{R}\,; (8.22)
|𝓃(𝓍,𝓎)|2π3e2πn(x+y)x+y+1K1(2π(x+y+1)),\displaystyle\left|\eusm{R}_{n}(x,-y)\right|\leqslant 2\pi^{3}e^{{{\mbox{\footnotesize{$2\pi n$}}}}}\dfrac{(x+y)}{\sqrt{x+y+1}}\,K_{1}\left(2\sqrt{\pi(x+y+1)}\right), x,y0,\displaystyle x,y\geqslant 0\,, (8.23)

where K0(x)=1(t21)1/2exp(xt)dtK_{0}(x)=\int_{1}^{\infty}(t^{2}-1)^{-1/2}\exp(-xt)\,{\rm{d}}t, K1(x)=0exp(xt2+1)dtK_{1}(x)=\int_{0}^{\infty}\exp(-x\sqrt{t^{2}+1})\,{\rm{d}}t, x> 0x\leavevmode\nobreak\ >\leavevmode\nobreak\ 0, are the modified Hankel function satisfying Kj(x)x1/2exp(x)π/2K_{j}(x)\,x^{1/2}\,\exp(x)\to\sqrt{\pi/2} as x+x\to+\infty for each j{0,1}j\in\{0,1\}.

For φL1(\varphi\in L^{1}(\Bb{R}), let UφU_{\varphi} be defined as in (1.1), i.e.,

Uφ(x,y)=eixt+iy/tφ(t)dt,x,y\displaystyle U_{\varphi}(x,y)=\int\limits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt+{\rm{i}}y/t$}}}}}\varphi(t){\rm{d}}t\ ,\quad x,y\in\Bb{R}\,. (8.24)

Then u=Uφu=U_{\varphi} solves the Klein-Gordon equation uxy+u=0u_{xy}+u=0 in the sense of distribution theory on \Bb{R}^{2}. We recall that uu is a solution of the Klein-Gordon equation on a given open subset GG\!\subset\!\Bb{R}^{2} in the sense of distribution theory if u𝒟(G)u\!\in\!\mathcal{D}^{\,\prime}(G) and the equality uxy+u=0u_{xy}+u=0 holds in the sense of an equality for the linear functionals uxy,u𝒟(G)u_{xy},u\!\in\!\mathcal{D}^{\,\prime}(G) on test functions C0(G)C_{0}^{\infty}(G), the compactly supported CC^{\infty}-smooth functions mapping GG to \Bb{C} (cf. (hor, , pp.​ 14, 34)). Given that our primary interest is in solutions of the form UφU_{\varphi}, which are continuous on \Bb{R}^{2}, it is more convenient to use alternative definition which is equivalent to the definition above for continuous solutions of the Klein-Gordon equation.

Definition 8.6

. Let GG be an open convex subset of \Bb{R}^{2}. We say that UU is a continuous solution of the Klein-Gordon equation on GG if UC(G)U\in C\left(G\right) and

U(b,d)U(b,c)U(a,d)+U(a,c)+cdabU(t,s)dtds=0,\displaystyle U(b,d)\!-\!U(b,c)\!-\!U(a,d)\!+\!U(a,c)\!+\!\!\int\nolimits_{c}^{\,d}\int\nolimits_{a}^{\,b}U(t,s)\,{\rm{d}}t{\rm{d}}s=0\ , (8.25)

for all [a,b]×[c,d]G[a,b]\times[c,d]\subset G, where <a<b<+-\infty<a<b<+\infty and <c<d<+-\infty<c<d<+\infty.

We readily verify that (8.25) holds for U=UφU=U_{\varphi} and G=G=\Bb{R}^{2}, and, consequently, UφU_{\varphi} is a continuous solution of the Klein-Gordon equation on \Bb{R}^{2}.

Next, let us assume that φ\varphi meets the condition (8.10) of Theorem 8.2. In view of the identities (1.10), this is the same as requiring that (1.11) holds. Then the conjugate hyperbolic Fourier series (1.9) of φ\varphi can be substituted for φ\varphi into the integral in (8.24) to get the equality (1.12) by taking into account the identities (1.15). As a consequence, we obtain the following interpolation formula for the solutions of the Klein-Gordon equation of the type (8.24).

Theorem 8.7

. Suppose φL1(\varphi\in L^{1}(\Bb{R}) and that the continuous solution UφU_{\varphi} of the Klein-Gordon equation on \Bb{R}^{2} is given by (8.24). If UφU_{\varphi} satisfies (1.11) then

Uφ(x,y)=Uφ(0,0)0(𝓍,𝓎)+𝓃1[𝒰φ(π𝓃,0)𝓃(𝓍,𝓎)+𝒰φ(0,π𝓃)𝓃(𝓎,𝓍)]+n1[Uφ(πn,0)𝓃(𝓍,𝓎)+𝒰φ(0,π𝓃)𝓃(𝓎,𝓍)],(𝓍,𝓎)U_{\varphi}(x,y)\!=\!U_{\varphi}(0,0)\,\eusm{R}_{\hskip 0.71114pt0}(x,y)+\!\sum\limits_{n\geqslant 1}\!\Big{[}U_{\varphi}(\pi n,0)\eusm{R}_{\hskip 0.71114ptn}(x,y)+U_{\varphi}(0,-\pi n)\eusm{R}_{\hskip 0.71114ptn}(-y,-x)\Big{]}\\[5.69046pt] \!+\!\sum\limits_{n\geqslant 1}\Big{[}\,U_{\varphi}(-\pi n,0)\,\eusm{R}_{\hskip 0.71114ptn}(-x,-y)+U_{\varphi}(0,\pi n)\,\eusm{R}_{\hskip 0.71114ptn}(y,x)\Big{]}\,,\quad(x,y)\in\Bb{R}^{2}\,,

where the sequence {𝓃}𝓃0\{\eusm{R}_{\hskip 0.71114ptn}\}_{n\geqslant 0} is given by (8.7), and the series converges absolutely and uniformly over all (x,y)(x,y)\in\Bb{R}^{2}.

9 . Extension of Theorem A

In this section, we derive the following extension of Theorem A. We make an effort to present a direct proof which does not rely on ergodic theory, and which also covers the instance of Theorem A (when N=M=0N=M=0 in (9.3) below). This makes the approach direct, but we should mention that there are shortcuts available if we were to abandon this aim.

Theorem 9.1

. Let φL1(\varphi\in L^{1}(\Bb{R}), N,MN,M\in\Bb{Z}_{\geqslant 0}, δ0:=1\delta_{0}:=1, and δk{0,1}\delta_{k}\in\left\{0,1\right\}, kk\in\Bb{N}. Assume that

(a)φ(t)eiπntk=0N11k!(iπnt)k(it)N1+δNdt=0,(b)φ(t)(eiπntk=0M11k!(iπnt)k)(ti)M1+δMdt=0,n\displaystyle\begin{array}[]{l}{\rm{(a)}}\ \ \displaystyle\int\limits_{\Bb{R}}\varphi(t)\dfrac{{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}-\displaystyle\sum\limits_{k=0}^{N-1}\dfrac{1}{k\,!}\left({\rm{i}}\pi nt\right)^{k}}{\vphantom{A^{A^{A}}}({\rm{i}}t)^{N-1+\delta_{N}}}{\rm{d}}t=0\,,\\[19.91684pt] {\rm{(b)}}\ \ \displaystyle\int\limits_{\Bb{R}}\varphi(t)\left({\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{t}$}}}}}-\sum\limits_{k=0}^{M-1}\dfrac{1}{k\,!}\left(\dfrac{{\rm{i}}\pi n}{t}\right)^{k}\right)\left(\dfrac{t}{{\rm{i}}}\right)^{M-1+\delta_{M}}{\rm{d}}t=0\,,\end{array}\quad n\in\Bb{Z}\,, (9.3)

where k=01:=0\sum_{k=0}^{-1}:=0. Then φ=0\varphi=0.

In hed2 , the second and the third authors sharpened Theorem A as follows:

φL1(iπnxφiπn/xφφ\displaystyle\varphi\!\in\!L^{1}(\Bb{R})\,,\ \int\limits_{\Bb{R}}\!\!{}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\varphi(x){\rm{d}}x\!=\!\int\limits_{\Bb{R}}\!\!{}^{{{\mbox{\footnotesize{${-\rm{i}}\pi n/x$}}}}}\varphi(x){\rm{d}}x\!=\!0\,,\ n\!\geqslant\!0\ \Rightarrow\ \varphi\!\in\!{\rm{H}}^{1}_{+}(\Bb{R})\,.

The analogue of this assertion for the biorthogonal system H0\eurm{H}_{0}, Hn\eurm{H}_{n}, Mn\eurm{M}_{n}, nn\in\Bb{Z}_{\neq 0}, can be formulated as the following theorem.

Theorem 9.2

. Let φL1(\varphi\in L^{1}(\Bb{R}). Suppose that

φ(t)Hn(t)dt=φ(t)Mn(t)dt=0,n\displaystyle\int\limits_{\Bb{R}}\varphi(t)\eurm{H}_{n}(t){\rm{d}}t=\int\limits_{\Bb{R}}\varphi(t)\eurm{M}_{n}(t){\rm{d}}t=0\ ,\quad n\in\Bb{N}\,.

Then φH1+(\varphi\in{\rm{H}}^{1}_{+}(\Bb{R}).

Proof of Theorem 9.2. We apply (8.3) written in the form

12πi(x+z)2=n1iπnHn(x)eiπnz+1z2n1iπnMn(x)eiπnz,xϝ\displaystyle\dfrac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt(x+z)^{2}}=\sum\limits_{n\geqslant 1}\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n\,\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}+\dfrac{1}{z^{2}}\sum\limits_{n\geqslant 1}\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n\,\eurm{M}_{n}(x)\,{\rm{e}}^{{\mbox{\footnotesize{${-\dfrac{{\rm{i}}\pi n}{z}}$}}}}\ ,\quad x\in\Bb{R}\,,\ z\in\Bb{H}\,,

multiply it by the function φ(x)\varphi(x) and integrate over xx\in\Bb{R}, to obtain from the given assumptions that

0=φ(x)dx(x+z)2=ddzF(z),F(z):=φ(x)dxx+z,z\displaystyle 0=\int\limits_{\Bb{R}}\dfrac{\varphi(x){\rm{d}}x}{(x+z)^{2}}=\dfrac{{\rm{d}}}{{\rm{d}}z}\,F(z)\,,\quad F(z):=\int\limits_{\Bb{R}}\dfrac{\varphi(x){\rm{d}}x}{x+z}\ ,\qquad z\in\Bb{H}\,.

It follows that the function FF, which is holomorphic in \Bb{H}, must be constant. Since φL1(\varphi\in L^{1}(\Bb{R}) automatically implies that F(it)0F(it)\to 0 as t+t\to+\infty, we find that F(z)=0F(z)=0 for all zz\in\Bb{H}. By the known characterizations of the space H1+({\rm{H}}^{1}_{+}(\Bb{R}) (see (gar, , p.​ 88, 2(ii))), we obtain that φH1+(\varphi\in{\rm{H}}^{1}_{+}(\Bb{R}). Theorem 9.2 follows. \square

9.1 . Proof of Theorem 9.1.

We observe that an application of the finite difference Δr[f](x)=k=0r(rk)(1)rkf(x+k)\Delta^{r}[f](x)=\sum_{k=0}^{r}\binom{r}{k}(-1)^{r-k}f(x+k) over nn of order r=Nr=N and r=Mr=M to (9.3)(a) and to (9.3)(b), respectively, gives (see (mil, , p.​ 29))

(a)φ(t)eiπnt(eiπt1)N(it)N1+δNdt=0,(b)φ(t)eiπnt(eiπt1)M(ti)M1+δMdt=0,n\displaystyle\begin{array}[]{l}{\rm{(a)}}\ \ \displaystyle\int\limits_{\Bb{R}}\varphi(t){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}\dfrac{\left({\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi t$}}}}}-1\right)^{N}}{\vphantom{A^{A^{A}}}({\rm{i}}t)^{N-1+\delta_{N}}}\,{\rm{d}}t=0\,,\\[19.91684pt] {\rm{(b)}}\ \ \displaystyle\int\limits_{\Bb{R}}\varphi(t){\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{t}$}}}}}\left({\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi}{t}$}}}}}-1\right)^{M}\left(\dfrac{t}{{\rm{i}}}\right)^{M-1+\delta_{M}}{\rm{d}}t=0\,,\end{array}\quad n\in\Bb{Z}\,,

or, in the notation φJ(t):=φ(1/t)/t2\varphi^{J}(t):=\varphi(-1/t)/t^{2}, tt\in\Bb{R}_{\neq 0}, where φJL1(\varphi^{J}\in L^{1}(\Bb{R}) by a change-of-variables, the conditions may be written as follows:

φ(t)eiπnt(eiπt1)NtN1+δNdt=φJ(t)eiπnt(1eiπt)MtM1+δMdt=0,n\displaystyle\int\limits_{\Bb{R}}\varphi(t){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}\dfrac{\left({\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi t$}}}}}-1\right)^{N}}{\vphantom{A^{A^{A}}}t^{N-1+\delta_{N}}}\,{\rm{d}}t=\int\limits_{\Bb{R}}\varphi^{J}(t){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}\dfrac{\left(1-{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi t$}}}}}\right)^{M}}{\vphantom{A^{A^{A}}}t^{M-1+\delta_{M}}}\,{\rm{d}}t=0\,,\ \ n\in\Bb{Z}\,.

From this we obtain that

(a)φ(t)tN1+δN+kφ(t+2k)(t+2k)N1+δN=0,(b)φJ(t)tM1+δM+kφJ(t+2k)(t+2k)M1+δM=0,\displaystyle\begin{array}[]{l}{\rm{(a)}}\ \ \dfrac{\varphi(t)}{t^{N-1+\delta_{N}}}+\displaystyle\sum\limits_{k\in\Bb{Z}_{\neq 0}}\dfrac{\varphi(t+2k)}{(t+2k)^{N-1+\delta_{N}}}=0\,,\\[19.91684pt] {\rm{(b)}}\ \ \dfrac{\varphi^{J}(t)}{t^{M-1+\delta_{M}}}+\displaystyle\sum\limits_{k\in\Bb{Z}_{\neq 0}}\dfrac{\varphi^{J}(t+2k)}{(t+2k)^{M-1+\delta_{M}}}=0\,,\end{array} (9.6)

for almost all t[1,1]t\in[-1,1] because the system {exp(iπnx)}n\{\exp(i\pi nx)\}_{n\in\Bb{Z}} is complete in L([1,1])L^{\infty}([-1,1]). Since the right-hand side series in (9.6) represent integrable functions on [1,1][-1,1] we conclude that the left-hand side functions in (9.6), being obviously integrable on ^~\Bb{R}\setminus[-1,1], are also integrable on [1,1][-1,1], and hence

tpφ(t),tqφJ(t)L1(δ𝕄δ𝕄\displaystyle t^{-p}\varphi(t)\,,\ t^{-q}\varphi^{J}(t)\in L^{1}(\Bb{R})\ ,\ \ \ p\!:=\!N\!-\!1\!+\!\delta_{N}\!\geqslant\!0\,,\ q\!:=\!M\!-\!1\!+\!\delta_{M}\!\geqslant\!0\,.

If for arbitrary σ{1,1}\sigma\in\{1,-1\} we introduce

Fσ(t):=tpφ(t)+σtqφJ(t),t𝔽σ𝕃\displaystyle F_{\sigma}(t):=t^{-p}\varphi(t)+\sigma t^{-q}\varphi^{J}(t)\ ,\quad t\in\Bb{R}_{\neq 0}\ ,\quad F_{\sigma}\in L^{1}(\Bb{R})\,,\ (9.7)

then

Fσ(1/t)=(1)ptp+2φJ(t)+(1)qtq+2σφ(t),\displaystyle F_{\sigma}(-1/t)=(-1)^{p}t^{p+2}\varphi^{J}(t)+(-1)^{q}t^{q+2}\sigma\varphi(t)\,,\

from which

Fσ(1/t)/tp+q+2=(1)ptqφJ(t)+(1)qtpσφ(t)=σ(1)qFσ(1)p+q(t),\displaystyle F_{\sigma}(-1/t)/t^{p+q+2}=(-1)^{p}t^{-q}\varphi^{J}(t)+(-1)^{q}t^{-p}\sigma\varphi(t)=\sigma(-1)^{q}F_{\sigma(-1)^{p+q}}(t)\,,

and thus

Fσ(t)=σ(1)pFσ(1)p+q(1/t)/tp+q+2,σ{1,1},t\displaystyle F_{\sigma}(t)=\sigma(-1)^{p}F_{\sigma(-1)^{p+q}}(-1/t)/t^{p+q+2}\,,\quad\sigma\in\{1,-1\}\,,\ t\in\Bb{R}_{\neq 0}\,.

Then it follows from (9.6) that

Fσ(t)=kFσ(2k+t)=kFσ(2k+t)=σ(1)qkFσ(1)p+q(12kt)(2kt)p+q+2,\displaystyle-\!F_{\sigma}(t)\!=\!\!\sum\limits_{k\in\Bb{Z}_{\neq 0}}\!\!F_{\sigma}(2k\!+\!t)\!=\!\!\sum\limits_{k\in\Bb{Z}_{\neq 0}}\!\!F_{\sigma}(-2k\!+\!t)\!=\!\sigma(-1)^{q}\!\sum\limits_{k\in\Bb{Z}_{\neq 0}}\!\!\dfrac{F_{\sigma(-1)^{p+q}}\left(\dfrac{1}{2k\!-\!t}\right)}{(2k\!-\!t)^{p+q+2}}\ ,

i.e.,

Fσ(t)=σ(1)q+1𝐓p+q+1[Fσ(1)p+q](t),t[1,1]{0},σ{1,1},\displaystyle F_{\sigma}(t)\!=\!\sigma(-1)^{q+1}{\mathbf{T}}_{p+q+1}\left[F_{\sigma(-1)^{p+q}}\right](t)\,,\ \ t\!\in\![-1,1]\setminus\{0\}\,,\ \sigma\!\in\!\{1,-1\}\,, (9.8)

where it can be easily seen that for arbitrary fL1([1,1])f\in L_{1}\big{(}[-1,1]\big{)} in the space L1([1,1])L_{1}\big{(}[-1,1]\big{)} there exists the limit

𝐓p+q+1[f](x):=limn+n=Nn0Nf(1/(2nx))(2nx)p+q+2L1([1,1]).\displaystyle{\mathbf{T}}_{p+q+1}[f](x)\ :=\ \lim\limits_{n\to+\infty}\sum_{\begin{subarray}{c}{{\mbox{\footnotesize{$n=-N$}}}}\\[1.42271pt] {{\mbox{\footnotesize{$n\neq 0$}}}}\end{subarray}}^{{{\mbox{\footnotesize{$N$}}}}}\dfrac{f\big{(}{1}/{(2n-x)}\big{)}}{(2n-x)^{p+q+2}}\in L_{1}\big{(}[-1,1]\big{)}\,. (9.9)

In view of |2kx|1|2k-x|\geqslant 1, kk\in\Bb{Z}_{\neq 0}, x[1,1]x\in[-1,1], the operator (9.9) for arbitrary Borel set A(1,1)A\subset(-1,1) possesses the following essential property

A|𝐓p+q+1[f](x)|dxnA|f(1/(2nx))||2nx|p+q+2dxnA|f(1/(2nx))|(2nx)2dx\displaystyle\int\limits_{A}\left|{\mathbf{T}}_{p+q+1}[f](x)\right|\,{\rm{d}}x\leqslant\!\!\!\sum\limits_{n\in\Bb{Z}_{\neq 0}}\int\limits_{A}\dfrac{\left|f\big{(}{1}/{(2n-x)}\big{)}\right|}{|2n-x|^{p+q+2}}\,{\rm{d}}x\leqslant\!\!\!\sum\limits_{n\in\Bb{Z}_{\neq 0}}\int\limits_{A}\dfrac{\left|f\big{(}{1}/{(2n-x)}\big{)}\right|}{(2n-x)^{2}}\,{\rm{d}}x
=n1/(2nA)|f(x)|dx=n 1/(2nA)|f(x)|dx,\displaystyle=\sum\limits_{n\in\Bb{Z}_{\neq 0}}\ \ \int\limits_{1/(2n-A)}\!\!\!\!\!\!|f(x)|\,{\rm{d}}x=\int\limits_{{{\mbox{\footnotesize{$\sqcup_{{{\mbox{\footnotesize{$\,n\in\Bb{Z}_{\neq 0}$}}}}}\ 1/(2n-A)$}}}}}\!\!\!\!\!\!|f(x)|\,{\rm{d}}x\,,\

or, in the notation (5.1),

A|𝐓p+q+1[f](x)|dxω1(A)|f(x)|dx,A(1,1),ω1(A):=nϕ𝔸\displaystyle\int\limits_{A}\left|{\mathbf{T}}_{p+q+1}[f](x)\right|\,{\rm{d}}x\leqslant\!\!\!\!\!\int\limits_{\omega_{1}(A)}\!\!\!\!\!|f(x)|\,{\rm{d}}x\,,\ A\subset(-1,1)\,,\ \omega_{1}(A):=\underset{n\in\Bb{Z}_{\neq 0}}{\sqcup}\phi_{n}(A), (9.10)

because the sets 1/(2n(1,1))1/(2n-(-1,1)), nn\in\Bb{Z}_{\neq 0}, are disjoint. Since ω1(A)(1,1)\omega_{1}(A)\subset(-1,1) we can successively apply (9.10) to get

A|𝐓Np+q+1[f](x)|dxωN(A)|f(x)|dx,A(1,1),ωN(A):=n1,,nNϕn1,,nN(A),N\displaystyle\begin{array}[]{l}\displaystyle\int\limits_{A}\left|{\mathbf{T}}^{N}_{p+q+1}[f](x)\right|\,{\rm{d}}x\leqslant\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\omega_{N}(A)$}}}}}\!\!\!\!\!|f(x)|\,{\rm{d}}x\,,\ A\subset(-1,1)\,,\\[19.91684pt] \omega_{N}(A):=\underset{{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}}{\sqcup}\phi_{n_{1},...,n_{N}}(A)\,,\ \ N\in\Bb{N}\,,\end{array} (9.13)

where for the Lebesgue measure mm on the real line and a Borel set A(1,1)A\subset(-1,1) after the similar manipulations we obtain

m(ωN(A))=A𝐓N1[1](x)dx,A(1,1),N\displaystyle m\big{(}\omega_{N}(A)\big{)}=\int\limits_{A}{\mathbf{T}}^{N}_{1}[1](x)\,{\rm{d}}x\,,\qquad A\subset(-1,1)\,,\ \ N\in\Bb{N}\,. (9.14)

Here, in the notations (10.53), (10.57) and (10.62), for any x[1,1]x\!\in\![-1,1] we have

𝐓N1[1](x)=n1,n2,nN1(pN𝔫x+qN𝔫)2,𝔫=(𝔫𝔑,,𝔫1)\displaystyle{\mathbf{T}}^{N}_{1}[1](x)=\sum_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\frac{1}{\left(p_{N}^{\eufm{n}}x+q_{N}^{\eufm{n}}\right)^{2}}\ ,\quad\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\,,\ N\geqslant 1\,, (9.15)

because, in view of ϕn1,n2,nN1(ϕnN(x))ϕnN(x)=ϕn1,n2,nN(x)\phi^{\,\prime}_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N-1}$}}}}}(\phi_{n_{N}}(x))\phi^{\,\prime}_{n_{N}}(x)=\phi^{\,\prime}_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N}$}}}}}(x), N2N\geqslant 2\,,

𝐓N1[1](x)=n1,n2,nNϕn1,n2,nN(x),ϕn1,n2,nN(x)=zp𝔫N1+q𝔫N1zp𝔫N+q𝔫N,N1.\displaystyle{\mathbf{T}}^{N}_{1}[1](x)=\sum_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\phi^{\,\prime}_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N}$}}}}}(x)\,,\quad\phi_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N}$}}}}}(x)=\dfrac{zp^{\eufm{n}}_{N-1}+q^{\eufm{n}}_{N-1}}{zp^{\eufm{n}}_{N}+q^{\eufm{n}}_{N}}\,,\ N\geqslant 1\,.

But according to (10.77), |p𝔫N|<|q𝔫N|\left|p^{\eufm{n}}_{N}\right|<\left|q^{\eufm{n}}_{N}\right|, and hence, we obtain from

|pN𝔫x+qN𝔫||qN𝔫||x||pN𝔫|(1|x|)|qN𝔫|\displaystyle\left|p_{N}^{\eufm{n}}x+q_{N}^{\eufm{n}}\right|\geqslant\left|q_{N}^{\eufm{n}}\right|-\left|x\right|\left|p_{N}^{\eufm{n}}\right|\geqslant(1-|x|)|q_{N}^{\eufm{n}}|

and (9.15) that

𝐓2N[1](x)𝐓2N[1](0)(1|x|)2,x[1,1],\displaystyle{\bf{T}}_{2}^{N}[1](x)\leqslant\dfrac{{\bf{T}}_{2}^{N}[1](0)}{(1-|x|)^{2}}\ ,\quad x\in[-1,1]\,,

and, consequently, (9.14) implies that

m(ωN([α,β]))2𝐓2N[1](0)min{(1+α)2,(1β)2},1<α<β<1,N\displaystyle m\big{(}\omega_{N}([\alpha,\beta])\big{)}\leqslant\dfrac{2{\bf{T}}_{2}^{N}[1](0)}{\min\left\{(1+\alpha)^{2},\left(1-\beta\right)^{2}\right\}}\,,\ \ -1\!<\!\alpha\!<\!\beta\!<\!1\,,\ \ N\!\in\!\Bb{N}\,. (9.16)

Iterating (9.8) gives

Fσ(t)=𝐓2Np+q+1[Fσ](t),t[1,1]{0},σ{1,1},N\displaystyle F_{\sigma}(t)\!=\!{\mathbf{T}}^{2N}_{p+q+1}\left[F_{\sigma}\right](t)\,,\ \ t\!\in\![-1,1]\setminus\{0\}\,,\ \sigma\!\in\!\{1,-1\}\,,N\in\Bb{N}\,,

which together with (9.13) lead us to the conclusion that

αβ|Fσ(x)|dxω2N([α,β])|Fσ(x)|dx,σ{1,1},1<α<β<1,N\displaystyle\int\limits_{\alpha}^{\beta}\left|F_{\sigma}(x)\right|\,{\rm{d}}x\leqslant\int\limits_{{{\mbox{\footnotesize{$\omega_{2N}([\alpha,\beta])$}}}}}|F_{\sigma}(x)|\,{\rm{d}}x\,,\ \sigma\!\in\!\{1,-1\}\,,\ \ -1\!<\!\alpha\!<\!\beta\!<\!1\,,\ \ N\!\in\!\Bb{N}\,. (9.17)

But in view of (9.18) below and (9.16) we have lim¯Nm(ω2N([α,β]))=0\operatorname*{\underline{\lim}}\nolimits_{N\to\infty}m\big{(}\omega_{2N}([\alpha,\beta])\big{)}=0, which by FσL1([1,1])F_{\sigma}\in L^{1}([-1,1]) and (9.17) yields that Fσ(x)=0F_{\sigma}(x)=0 for almost all x[1,1]x\in[-1,1] and for any σ{1,1}\sigma\!\in\!\{1,-1\}. Definition (9.7) of the functions FσF_{\sigma} gives that φ(x)=0\varphi(x)=0 and φ(1/x)=0\varphi(-1/x)=0 for almost all x[1,1]x\in[-1,1]. Thus, φ(x)=0\varphi(x)=0 for almost all xx\in\Bb{R} and Theorem 9.1 follows. \square

We supply the following auxiliary lemma which was referred to in the proof of Theorem 9.1.

Lemma 9.3

. Let 𝐓1{\bf{T}}_{1} be defined as in (1.22). Then

lim¯N𝐓12N[1](0)=0.\displaystyle\operatorname*{\underline{\lim}}\limits_{N\to\infty}{\bf{T}}_{1}^{2N}[1](0)=0\ . (9.18)

Proof of Lemma 9.3. We first obtain that

lim¯N𝐓1N[1](0)=0.\displaystyle\operatorname*{\underline{\lim}}\limits_{N\to\infty}{\bf{T}}_{1}^{N}[1](0)=0\ . (9.19)

Assume that there exists ε>0\varepsilon>0 such that

𝐓1N[1](0)ε>0for allN1.\displaystyle{\bf{T}}_{1}^{N}[1](0)\geqslant\varepsilon>0\ \quad\ \mbox{for all}\ N\geqslant 1\ .

Since 𝐓1N[1](x){\bf{T}}_{1}^{N}[1](x) is nondecreasing on [0,1][0,1] (see (hed1, , p.​ 1715, Proposition 3.7.2)) then

𝐓1N[1](x)ε>0for allN1,x[0,1].\displaystyle{\bf{T}}_{1}^{N}[1](x)\geqslant\varepsilon>0\ \quad\ \mbox{for all}\ N\geqslant 1\,,\ \ x\in[0,1]\ .

For arbitrary x[0,1]x\in[0,1] and N1N\geqslant 1 we obviously have

𝐓1N[1](x)\displaystyle{\bf{T}}_{1}^{N}[1](x) =n1[𝐓1N1[1](12nx)(2nx)2+𝐓1N1[1](12n+x)(2n+x)2]\displaystyle=\sum_{n\geqslant 1}\left[\frac{{\bf{T}}_{1}^{N-1}[1]\left(\dfrac{1}{2n-x}\right)}{(2n-x)^{2}}+\frac{{\bf{T}}_{1}^{N-1}[1]\left(\dfrac{1}{2n+x}\right)}{(2n+x)^{2}}\right]
>𝐓1N1[1](12x)(2x)2+𝐓1N1[1](12+x)(2+x)2.\displaystyle>\frac{{\bf{T}}_{1}^{N-1}[1]\left(\dfrac{1}{2-x}\right)}{(2-x)^{2}}+\frac{{\bf{T}}_{1}^{N-1}[1]\left(\dfrac{1}{2+x}\right)}{(2+x)^{2}}\ . (9.20)

and, in view of the inequality

12xx 12xx2(x1)20,x[0,1],\displaystyle\dfrac{1}{2-x}\geqslant x\ \Leftrightarrow\ 1\geqslant 2x-x^{2}\ \Leftrightarrow\ (x-1)^{2}\geqslant 0\,,\quad x\in[0,1]\,,

and of nondecreasing property of 𝐓1N1[1](x){\bf{T}}_{1}^{N-1}[1](x) on [0,1][0,1], we get

𝐓1N1[1](12x)𝐓1N1[1](x),x[0,1].\displaystyle{\bf{T}}_{1}^{N-1}[1]\left(\dfrac{1}{2-x}\right)\geqslant{\bf{T}}_{1}^{N-1}[1](x)\,,\quad x\in[0,1]\ .

As a consequence, we derive from (9.20) that

𝐓1N[1](x)𝐓1N1[1](x)(2x)2+𝐓1N1[1](12+x)(2+x)2𝐓1N1[1](x)(2x)2+ε9,N1.\displaystyle{\bf{T}}_{1}^{N}[1](x)\geqslant\dfrac{{\bf{T}}_{1}^{N-1}[1](x)}{(2-x)^{2}}+\frac{{\bf{T}}_{1}^{N-1}[1]\left(\dfrac{1}{2+x}\right)}{(2+x)^{2}}\geqslant\dfrac{{\bf{T}}_{1}^{N-1}[1](x)}{(2-x)^{2}}+\dfrac{\varepsilon}{9}\ ,\ \ \ N\geqslant 1\,.

Iterating this procedure for N4N\geqslant 4, we obtain

𝐓1N[1](x)ε9+1(2x)2𝐓1N1[1](x)ε9+1(2x)2[ε9+1(2x)2𝐓1N2[1](x)]\displaystyle{\bf{T}}_{1}^{N}[1](x)\geqslant\dfrac{\varepsilon}{9}+\dfrac{1}{(2-x)^{2}}{\bf{T}}_{1}^{N-1}[1](x)\geqslant\dfrac{\varepsilon}{9}+\dfrac{1}{(2-x)^{2}}\left[\dfrac{\varepsilon}{9}+\dfrac{1}{(2-x)^{2}}{\bf{T}}_{1}^{N-2}[1](x)\right]
=ε9(1+1(2x)2)+1(2x)4𝐓1N2[1](x)\displaystyle=\dfrac{\varepsilon}{9}\left(1+\dfrac{1}{(2-x)^{2}}\right)+\dfrac{1}{(2-x)^{4}}{\bf{T}}_{1}^{N-2}[1](x)
ε9(1+1(2x)2)+1(2x)4[ε9+1(2x)2𝐓1N3[1](x)]\displaystyle\geqslant\dfrac{\varepsilon}{9}\left(1+\dfrac{1}{(2-x)^{2}}\right)+\dfrac{1}{(2-x)^{4}}\left[\dfrac{\varepsilon}{9}+\dfrac{1}{(2-x)^{2}}{\bf{T}}_{1}^{N-3}[1](x)\right]
=ε9(1+1(2x)2+1(2x)4)+1(2x)6𝐓1N3[1](x)\displaystyle=\dfrac{\varepsilon}{9}\left(1+\dfrac{1}{(2-x)^{2}}+\dfrac{1}{(2-x)^{4}}\right)+\dfrac{1}{(2-x)^{6}}{\bf{T}}_{1}^{N-3}[1](x)\geqslant\ldots
ε9(1+1(2x)2+1(2x)4++1(2x)2(m1))+1(2x)2m𝐓1Nm[1](x)\displaystyle\geqslant\dfrac{\varepsilon}{9}\left(1+\dfrac{1}{(2-x)^{2}}+\dfrac{1}{(2-x)^{4}}+\ldots+\dfrac{1}{(2-x)^{2(m-1)}}\right)+\dfrac{1}{(2-x)^{2m}}{\bf{T}}_{1}^{N-m}[1](x)
ε9k=0N11(2x)2k+1(2x)2N,\displaystyle\geqslant\ldots\geqslant\dfrac{\varepsilon}{9}\sum_{k=0}^{N-1}\dfrac{1}{(2-x)^{2k}}+\dfrac{1}{(2-x)^{2N}}\ ,

i.e.,

𝐓1N[1](x)ε9k=0N11(2x)2k+1(2x)2N,x[0,1].\displaystyle{\bf{T}}_{1}^{N}[1](x)\geqslant\dfrac{\varepsilon}{9}\sum_{k=0}^{N-1}\dfrac{1}{(2-x)^{2k}}+\dfrac{1}{(2-x)^{2N}}\ ,\qquad x\in[0,1]\,.

Hence,

01𝐓1N[1](x)dx\displaystyle\int_{0}^{1}{\bf{T}}_{1}^{N}[1](x)\,{\rm{d}}x ε9k=0N101dx(2x)2k+01dx(2x)2N\displaystyle\geqslant\dfrac{\varepsilon}{9}\sum_{k=0}^{N-1}\int_{0}^{1}\dfrac{{\rm{d}}x}{(2-x)^{2k}}+\int_{0}^{1}\dfrac{{\rm{d}}x}{(2-x)^{2N}}
=ε9(1+k=1N11122k12k1)+1122N12N1.\displaystyle=\dfrac{\varepsilon}{9}\left(1+\sum_{k=1}^{N-1}\dfrac{1-\dfrac{1}{2^{2k-1}}}{2k-1}\right)+\dfrac{1-\dfrac{1}{2^{2N-1}}}{2N-1}\ . (9.21)

In view of (9.13), for any N1N\geqslant 1 the set ωN((1,1))\omega_{N}((-1,1)) contains at least all irrational point of (1,1)(-1,1). Therefore its Lebesgue measure is equal to 22. Then, by virtue of (9.14),

11𝐓1N[1](x)dx=m(ωN((1,1)))=2,N1,\displaystyle\int_{-1}^{1}{\bf{T}}_{1}^{N}[1](x)\,{\rm{d}}x=m\Big{(}\omega_{N}\big{(}(-1,1)\big{)}\Big{)}=2\,,\quad N\geqslant 1\ ,\

and since 𝐓1N[1](x){\bf{T}}_{1}^{N}[1](x) is even on [0,1][0,1] (see (hed1, , p.​ 1715, Proposition 3.7.2)) we derive from (9.21) that

1\displaystyle 1 =1211𝐓1N[1](x)dx=01𝐓1N[1](x)dx\displaystyle=\dfrac{1}{2}\int_{-1}^{1}{\bf{T}}_{1}^{N}[1](x)\,{\rm{d}}x=\int_{0}^{1}{\bf{T}}_{1}^{N}[1](x)\,{\rm{d}}x
ε9k=1N112k1ε9k=1N1122k112k1ε9+ε9k=1N112k1+,\displaystyle\geqslant\dfrac{\varepsilon}{9}\sum_{k=1}^{N-1}\dfrac{1}{2k-1}-\dfrac{\varepsilon}{9}\sum_{k=1}^{N-1}\dfrac{1}{2^{2k-1}}\cdot\dfrac{1}{2k-1}\geqslant-\dfrac{\varepsilon}{9}+\dfrac{\varepsilon}{9}\sum_{k=1}^{N-1}\dfrac{1}{2k-1}\to+\infty\ ,

as NN\to\infty. This contradiction proves (9.19).

Next, to establish (9.18), we appeal to the formula (9.15) with N2N\geqslant 2. In view of (10.66), for arbitrary x(1,1)x\in(-1,1) we have

{yk=2nkyk1yk2, 1kN;yk:=pk𝔫x+qk𝔫,1kN;a:=y1=p1𝔫x+q1𝔫=x(1,1),b:=y0=p0𝔫x+q0𝔫=1,\displaystyle\left\{\begin{array}[]{ll}y_{k}=2n_{k}y_{k-1}-y_{k-2}\,,\ \ 1\leqslant k\leqslant N\,;&y_{k}:=p_{k}^{\eufm{n}}x+q_{k}^{\eufm{n}}\ ,\ \ -1\leqslant k\leqslant N\,;\\ a:=y_{-1}=p_{-1}^{\eufm{n}}x+q_{-1}^{\eufm{n}}=x\in(-1,1)\,,&b:=y_{0}=p_{0}^{\eufm{n}}x+q_{0}^{\eufm{n}}=1\,,\end{array}\right. (9.24)

where 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}. These relationships mean that we can apply conclusion (10.71) to the finite collection of numbers {yk}k=1Q\{y_{k}\}_{k=-1}^{Q} with Q=NQ=N, according to which,

1=|p0𝔫x+q0𝔫|<|p1𝔫x+q1𝔫|<<|pN𝔫x+qN𝔫|,x(1,1).\displaystyle 1=\left|p_{0}^{\eufm{n}}x+q_{0}^{\eufm{n}}\right|<\left|p_{1}^{\eufm{n}}x+q_{1}^{\eufm{n}}\right|<\ldots<\left|p_{N}^{\eufm{n}}x+q_{N}^{\eufm{n}}\right|\ ,\quad x\in(-1,1)\,.

In particular,

pN2𝔫x+qN2𝔫pN1𝔫x+qN1𝔫(1,1),x(1,1),𝔫=(𝔫𝔑,,𝔫1)\displaystyle\dfrac{p_{N-2}^{\eufm{n}}x+q_{N-2}^{\eufm{n}}}{p_{N-1}^{\eufm{n}}x+q_{N-1}^{\eufm{n}}}\in(-1,1)\,,\ x\in(-1,1)\,,\ \eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\,,\ N\geqslant 2\,.

Then we derive from (9.15) and (9.24) that

𝐓N1[1](x)=n1,n2,nN(pN𝔫x+qN𝔫)2=n1,n2,nN1(pN1𝔫x+qN1𝔫)2×\displaystyle{\mathbf{T}}^{N}_{1}[1](x)=\sum_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\left(p_{N}^{\eufm{n}}x+q_{N}^{\eufm{n}}\right)^{-2}=\sum_{{{\mbox{\footnotesize{$n_{1},n_{2},...n_{N-1}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\left(p_{N-1}^{\eufm{n}}x+q_{N-1}^{\eufm{n}}\right)^{-2}\times
×nN(2nNpN2𝔫x+qN2𝔫pN1𝔫x+qN1𝔫)2𝐓N11[1](x)maxy[1,1]n1(2n+y)2\displaystyle\times\sum_{{{\mbox{\footnotesize{$n_{N}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\left(2n_{N}-\dfrac{p_{N-2}^{\eufm{n}}x+q_{N-2}^{\eufm{n}}}{p_{N-1}^{\eufm{n}}x+q_{N-1}^{\eufm{n}}}\right)^{-2}\leqslant{\mathbf{T}}^{N-1}_{1}[1](x)\max_{{{\mbox{\footnotesize{$y\in[-1,1]$}}}}}\sum_{{{\mbox{\footnotesize{$n\in\!\Bb{Z}_{\neq 0}$}}}}}\frac{1}{(2n+y)^{2}}
=𝐓N11[1](x)maxy[1,1](π24sin2(πy/2)1y2)=(π241)𝐓N11[1](x),\displaystyle={\mathbf{T}}^{N-1}_{1}[1](x)\max_{{{\mbox{\footnotesize{$y\in[-1,1]$}}}}}\left(\frac{\pi^{2}}{4\sin^{2}(\pi y/2)}-\frac{1}{y^{2}}\right)=\left(\frac{\pi^{2}}{4}-1\right){\mathbf{T}}^{N-1}_{1}[1](x)\,,

for all x(1,1)x\in(-1,1). Hence,

0<𝐓N1[1](0)(π241)𝐓N11[1](0),N2.\displaystyle 0<{\mathbf{T}}^{N}_{1}[1](0)\leqslant\left(\frac{\pi^{2}}{4}-1\right){\mathbf{T}}^{N-1}_{1}[1](0)\,,\ N\geqslant 2\,. (9.25)

In view of (9.19), there exists an increasing sequence of positive integers {Nk}k1\{N_{k}\}_{k\geqslant 1} such that 𝐓1Nk[1](0)0{\bf{T}}_{1}^{N_{k}}[1](0)\to 0 as k+k\to+\infty. By keeping each even NkN_{k} and replacing each odd NkN_{k} by Nk+1N_{k}+1 we obtain a new sequence of even integers {2nk}k1\{2n_{k}\}_{k\geqslant 1} with the property that 𝐓12nk[1](0)0{\bf{T}}_{1}^{2n_{k}}[1](0)\to 0 as k+k\to+\infty, as follows from (9.25). This obtains (9.18) and establishes the validity of Lemma 9.3. \square

10 . Proofs

10.1 . Proofs for Section 2

10.1.1 . Proofs of Lemma 2.1 and (2.31)(b).

The statement of Lemma 2.1 is immediate from the relationship

|Arg(1tz)Arg(1t+tz)|<π,\displaystyle\big{|}{\rm{Arg}}\,(1\!-\!tz)\!-\!{\rm{Arg}}\,(1\!-\!t\!+\!tz)\big{|}\!<\!\pi\,,

proved in (bh2, , p.​ 608, (3.15)) and from the identities

|1t+tz|2|1tz|2=t(2x1)(2t),Im1tz1t+tz=yt(2t)(1t+tx)2+t2y2,\displaystyle|1\!-\!t\!+\!tz|^{2}\!-\!|1\!-\!tz|^{2}=t(2x\!-\!1)(2\!-\!t)\,,\ \ {\rm{Im}}\,\dfrac{1\!-\!tz}{1\!-\!t\!+\!tz}\!=\!-\dfrac{yt(2\!-\!t)}{(1\!-t\!+\!tx)^{2}\!+\!t^{2}y^{2}}\ ,

where t(0,1)t\in(0,1) and z=x+iy(0,1)()z=x+{\rm{i}}y\!\in\!(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right).

To prove (2.31)(b) we use (2.41), (2.21) and (2.26), to obtain, for any t>0t>0,

ddteπtλ(it)=eπt(πλ(it)+iλ(it))=eπt(πλ(it)+iλ(it))\displaystyle\frac{d}{dt}{\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\lambda({\rm{i}}t)={\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\left(\pi\lambda({\rm{i}}t)+{\rm{i}}\lambda^{\,\prime}({\rm{i}}t)\right)={\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\left(\pi\lambda({\rm{i}}t)+{\rm{i}}\lambda^{\,\prime}({\rm{i}}t)\right)
=eπt(πλ(it)πλ(it)(1λ(it))Θ3(it)4)=πλ(it)eπt(1(1λ(it))Θ3(it)4)\displaystyle={\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\left(\pi\lambda({\rm{i}}t)\!-\!\pi\,\lambda({\rm{i}}t)\,\left(1\!-\!\lambda({\rm{i}}t)\right)\,\Theta_{3}\left({\rm{i}}t\right)^{4}\right)=\pi\lambda({\rm{i}}t)\,{\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\left(1\!-\!\left(1\!-\!\lambda({\rm{i}}t)\right)\,\Theta_{3}\left({\rm{i}}t\right)^{4}\right)
=πλ(it)eπt(1Θ4(it)4),\displaystyle=\pi\lambda({\rm{i}}t)\,{\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\left(1-\Theta_{4}\left({\rm{i}}t\right)^{4}\right)\,,\

from which

ddteπtλ(it)=ddt16θ2(eπt)4θ3(eπt)4=16πθ2(eπt)4(1θ4(eπt)4)θ3(eπt)4>0,t>0,\displaystyle\frac{d}{dt}{\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\lambda({\rm{i}}t)=\frac{d}{dt}\dfrac{16\,\theta_{2}\left({\rm{e}}^{-\pi t}\right)^{4}}{\theta_{3}\left({\rm{e}}^{-\pi t}\right)^{4}}=\dfrac{16\pi\,\theta_{2}\left({\rm{e}}^{-\pi t}\right)^{4}\left(1-\theta_{4}\left({\rm{e}}^{-\pi t}\right)^{4}\right)}{\theta_{3}\left({\rm{e}}^{-\pi t}\right)^{4}}>0\ ,\quad t>0,

because

θ4(eπt)=12k0(eπ(2k+1)2teπ(2k+2)2t)<1,t>0.\displaystyle\theta_{4}\left({\rm{e}}^{-\pi t}\right)\!=\!1\!-2\sum\limits_{k\geqslant 0}\left({\rm{e}}^{-\pi(2k+1)^{2}t}\!-\!{\rm{e}}^{-\pi(2k+2)^{2}t}\right)<1\ ,\quad t>0.

Inequalities (2.31)(b) is now immediate from lim0<t0eπtλ(it)=1\lim_{0<t\to 0}{\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\lambda({\rm{i}}t)=1 and from

limt+eπtλ(it)=limt+16θ2(eπt)4θ3(eπt)4=16.\displaystyle\lim_{t\to+\infty}{\rm{e}}^{{{\mbox{\footnotesize{$\pi t$}}}}}\lambda({\rm{i}}t)=\lim_{t\to+\infty}\dfrac{16\,\theta_{2}\left({\rm{e}}^{-\pi t}\right)^{4}}{\theta_{3}\left({\rm{e}}^{-\pi t}\right)^{4}}=16\,.

10.1.2 . Proofs of (2.50) and (2.51).

For the integrand in the formula for H0\eurm{H}_{0} we have

I0:=F(1/2it)(t2+1/4)F(1/2+it)(x2F(1/2+it)2+F(1/2it)2)\displaystyle{\rm{I}}_{0}:=\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)}{\big{(}t^{2}+1/4\big{)}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\Big{(}x^{2}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)^{2}\Big{)}}
=F(1/2it)F(1/2+it)(1/2+it)(1/2it)F(1/2+it)2(x2+F(1/2it)2F(1/2+it)2),\displaystyle=\dfrac{\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)}}{\big{(}1/2+{\rm{i}}t\big{)}\big{(}1/2-{\rm{i}}t\big{)}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}\left(x^{2}+\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)^{2}}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}}\right)}\,,

where by (2.10) and (2.49), written for z=1/2+it(0,1)()z=1/2+{\rm{i}}t\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), tt\in\Bb{R},

F(1/2it)F(1/2+it)=λ(1/2+it)i,\displaystyle\frac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)}=\dfrac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)}{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\,, (10.1)
F(1/2it)2F(1/2+it)2=λ(1/2+it)2,\displaystyle\frac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)^{2}}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}}=\,-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}\,, (10.2)
1(1/2+it)(1/2it)F(1/2+it)2=iπλ(1/2+it),\displaystyle\dfrac{1}{\big{(}1/2+{\rm{i}}t\big{)}\big{(}1/2-{\rm{i}}t\big{)}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}}=\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(1/2+{\rm{i}}t)\,,\ (10.3)

we obtain

I0=πλ(1/2+it)λ(1/2+it)F(1/2+it)x2λ(1/2+it)2.\displaystyle{\rm{I}}_{0}=\ \pi\,\frac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(1/2+{\rm{i}}t)\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)}{x^{2}-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}}\ .

Hence,

H0(x)\displaystyle\eurm{H}_{0}(x) =12π2+F(1/2it)dt(t2+1/4)F(1/2+it)(x2F(1/2+it)2+F(1/2it)2)\displaystyle=\dfrac{1}{2\pi^{2}}\int\limits_{-\infty}^{+\infty}\frac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t){\rm{d}}t}{\big{(}t^{2}+1/4\big{)}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\Big{(}x^{2}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}+F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)^{2}\Big{)}}
=12π+λ(1/2+it)λ(1/2+it)F(1/2+it)dtx2λ(1/2+it)2\displaystyle=\dfrac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\frac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(1/2+{\rm{i}}t)\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t){\rm{d}}t}{x^{2}-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}}
=12πi+λ(1/2+it)λ(1/2+it)F(1/2+it)d(1/2+it)x2λ(1/2+it)2\displaystyle=\dfrac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{-\infty}^{+\infty}\frac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(1/2+{\rm{i}}t)\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t){\rm{d}}(1/2+{\rm{i}}t)}{x^{2}-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}}
=12πi1/2i1/2+iλ(y)λ(y)F(y)x2λ(y)2dy,\displaystyle=\dfrac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{1/2-{\rm{i}}\infty}^{1/2+{\rm{i}}\infty}\frac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(y)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(y)\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(y)}{x^{2}-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(y)^{2}}{\rm{d}}y\,,\

which proves (2.50).

Similarly, by using (10.1) and (10.3), we get

1(t2+1/4)(xF(1/2+it)+iF(1/2it))2=\displaystyle\dfrac{1}{\big{(}t^{2}+1/4\big{)}\Big{(}xF_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)+{\rm{i}}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)\Big{)}^{2}}=
=1(1/2+it)(1/2it)F(1/2+it)2(x+iF(1/2it)F(1/2+it))2\displaystyle=\dfrac{1}{\big{(}1/2+{\rm{i}}t\big{)}\big{(}1/2-{\rm{i}}t\big{)}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)^{2}\left(x+{\rm{i}}\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)}\right)^{2}}
=iπλ(1/2+it)(x+λ(1/2+it))2,\displaystyle=\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(1/2+{\rm{i}}t)}{\left(x+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\right)^{2}}\,,\

and therefore

Hn(x)\displaystyle\eurm{H}_{n}(x) =(1)4π3n+Sn(11/2+it)dt(t2+1/4)(xF(1/2+it)+iF(1/2it))2\displaystyle=\dfrac{(-1)}{4\pi^{3}n}\int\limits_{-\infty}^{+\infty}\dfrac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\dfrac{1}{1/2+{\rm{i}}t}\right){\rm{d}}t}{\big{(}t^{2}+1/4\big{)}\Big{(}xF_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)+{\rm{i}}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-{\rm{i}}t)\Big{)}^{2}}
=(1)4π3n+iπλ(1/2+it)Sn(11/2+it)dt(x+λ(1/2+it))2\displaystyle=\dfrac{(-1)}{4\pi^{3}n}\int\limits_{-\infty}^{+\infty}\ \dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(1/2+{\rm{i}}t)S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\dfrac{1}{1/2+{\rm{i}}t}\right){\rm{d}}t}{\left(x+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\right)^{2}}
=(1)4π2n+λ(1/2+it)Sn(11/2+it)d(1/2+it)(x+λ(1/2+it))2\displaystyle=\dfrac{(-1)}{4\pi^{2}n}\int\limits_{-\infty}^{+\infty}\ \dfrac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(1/2+{\rm{i}}t)S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\dfrac{1}{1/2+{\rm{i}}t}\right){\rm{d}}(1/2+{\rm{i}}t)}{\left(x+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/2+{\rm{i}}t)\right)^{2}}
=(1)4π2n1/2i1/2+iSn(1/y)λ(y)dy(x+λ(y))2,\displaystyle=\dfrac{(-1)}{4\pi^{2}n}\int\limits_{1/2-{\rm{i}}\infty}^{1/2+{\rm{i}}\infty}\ \frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left({1}\big{/}{y}\right)\lambda_{{{\mbox{\tiny{$\triangle$}}}}}^{\,\prime}(y){\rm{d}}y}{(x+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(y))^{2}}\,,\

which completes the proof of (2.51).

10.1.3 . Proofs of Theorem 2.2 and Corollary 2.3.

  

We first prove (2.55). By using (2.41), for arbitrary x+iyx+{\rm{i}}y\in\Bb{H} we get

ddx|λ(iy+x)|2=ddxλ(iy+x)λ(iyx)=λ(iy+x)λ(iyx)\displaystyle\dfrac{{\rm{d}}}{{\rm{d}}x}\left|\lambda({\rm{i}}y+x)\right|^{2}=\dfrac{{\rm{d}}}{{\rm{d}}x}\lambda({\rm{i}}y+x)\lambda({\rm{i}}y-x)=\lambda^{\,\prime}({\rm{i}}y+x)\lambda({\rm{i}}y-x)
λ(iy+x)λ(iyx)=iπ|λ(iy+x)|2(1λ(iy+x))Θ3(iy+x)4\displaystyle-\lambda({\rm{i}}y+x)\lambda^{\,\prime}({\rm{i}}y-x)={\rm{i}}\pi\left|\lambda({\rm{i}}y+x)\right|^{2}\big{(}1-\lambda({\rm{i}}y+x)\big{)}\Theta_{3}\left({\rm{i}}y+x\right)^{4}
iπ|λ(iy+x)|2(1λ(iyx))Θ3(iyx)4,\displaystyle-{\rm{i}}\pi\left|\lambda({\rm{i}}y+x)\right|^{2}\big{(}1-\lambda({\rm{i}}y-x)\big{)}\Theta_{3}\left({\rm{i}}y-x\right)^{4}\,,\

where, in accordance with (2.21) and (2.25)(g),

(1λ(z))Θ3(z)4=Θ4(z)4=Θ3(z1)4,z\displaystyle\left(1\!-\!\lambda(z)\right)\Theta_{3}(z)^{4}=\Theta_{4}(z)^{4}=\Theta_{3}(z-1)^{4}\,,\ \ z\!\in\!\Bb{H}\,,

and therefore

2ddxlog|λ(iy+x)|=iπ(Θ4(iy+x)4Θ4(iyx)4)=2πImΘ3(iy+x1)4,\displaystyle 2\dfrac{{\rm{d}}}{{\rm{d}}x}\log\left|\lambda({\rm{i}}y\!+\!x)\right|\!=\!{\rm{i}}\pi\left(\Theta_{4}({\rm{i}}y+x)^{4}-\Theta_{4}({\rm{i}}y-x)^{4}\right)=-2\pi{\rm{Im}}\Theta_{3}({\rm{i}}y+x-1)^{4}\,,\

which proves the second equality in (2.55). Similarly,

ddx|λ(iy+x)|2|1λ(iy+x)|2=\displaystyle\dfrac{{\rm{d}}}{{\rm{d}}x}\left|\lambda({\rm{i}}y+x)\right|^{2}\left|1-\lambda({\rm{i}}y+x)\right|^{2}=
=ddxλ(iy+x)(1λ(iy+x))λ(iyx)(1λ(iyx))=\displaystyle=\dfrac{{\rm{d}}}{{\rm{d}}x}\lambda({\rm{i}}y+x)\left(1-\lambda({\rm{i}}y+x)\right)\lambda({\rm{i}}y-x)\left(1-\lambda({\rm{i}}y-x)\right)=
=iπ|λ2(iy+x)|2[(12λ(iy+x))Θ3(iy+x)4(12λ(iyx))Θ3(iyx)4],\displaystyle={\rm{i}}\pi\left|\lambda_{2}({\rm{i}}y+x)\right|^{2}\left[\left(1-2\lambda({\rm{i}}y+x)\right)\Theta_{3}({\rm{i}}y+x)^{4}-\left(1-2\lambda({\rm{i}}y-x)\right)\Theta_{3}({\rm{i}}y-x)^{4}\right],

where, in accordance with (2.21) and (2.41),

(12λ(z))Θ34(z)=Θ32(z)2Θ22(z)=Θ44(z)Θ24(z)=2Θ44(z)Θ34(z)\displaystyle(1-2\lambda(z))\Theta_{3}^{4}(z)=\Theta_{3}^{2}(z)-2\Theta_{2}^{2}(z)=\Theta_{4}^{4}(z)-\Theta_{2}^{4}(z)=2\Theta_{4}^{4}(z)-\Theta_{3}^{4}(z)

for arbitrary zz\!\in\!\Bb{H}, and hence, by (2.25)(g), we obtain

2ddxlog|λ2(iy+x)|\displaystyle 2\dfrac{{\rm{d}}}{{\rm{d}}x}\log\left|\lambda_{2}({\rm{i}}y\!+\!x)\right| =2πIm(2Θ44(iy+x)Θ34(iy+x))\displaystyle\!=\!-2\pi{\rm{Im}}\left(2\Theta_{4}^{4}({\rm{i}}y+x)-\Theta_{3}^{4}({\rm{i}}y+x)\right)
=2πIm(Θ34(iy+x)2Θ34(iy+x1)),\displaystyle\!=\!2\pi{\rm{Im}}\left(\Theta_{3}^{4}({\rm{i}}y+x)-2\Theta_{3}^{4}({\rm{i}}y+x-1)\right)\,,\

which completes the proof of the first equality in (2.55).

Prove now (2.56) and (2.57). For x(0,1)x\in(0,1) it follows from x+iyx\!+\!{\rm{i}}y\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and (2.39) that ImΘ34(iy+x)>0{\rm{Im}}\Theta_{3}^{4}({\rm{i}}y+x)>0 and ImΘ34(iy+x1)<0{\rm{Im}}\Theta_{3}^{4}({\rm{i}}y+x-1)<0 because x1(1,0)x-1\in(-1,0). These two inequalities together with (2.55) yield the validity of (2.56) and (2.57) for x>0x>0. If x(1,0)x\in(-1,0) in (2.56) and in (2.57), we similarly get their validity by using inequalities ImΘ34(iy+x)<0{\rm{Im}}\Theta_{3}^{4}({\rm{i}}y+x)<0 and ImΘ34(iy+x1)=ImΘ34(iy+x+1)>0{\rm{Im}}\Theta_{3}^{4}({\rm{i}}y+x-1)={\rm{Im}}\Theta_{3}^{4}({\rm{i}}y+x+1)>0 which follow from (2.39), in view of x+1(0,1)x+1\in(0,1). The proof of (2.56) and (2.57) is completed.

To prove (2.59) and (2.60), observe that by using the relationship

λ(z+1)=λ(z1)=λ(z)λ(z)1,z\displaystyle\lambda(z+1)=\lambda(z-1)=\frac{\lambda(z)}{\lambda(z)-1}\ ,\quad z\in\Bb{H}, (10.4)

(see (cha, , p.​ 111)), we obtain from (2.56), (2.57) and (2.32) that

maxx[0,1]|λ(iy+x)|=|λ(iy±1)|=λ(iy)1λ(iy),maxx[0,1]|λ2(iy+x)|=|λ2(iy±1)|=λ(iy)(1λ(iy))2,y1/2,\displaystyle\begin{array}[]{l}\displaystyle\max_{x\in[0,1]}\left|\lambda({\rm{i}}y+x)\right|\ =\,\left|\lambda({\rm{i}}y\pm 1)\right|\ =\frac{\lambda({\rm{i}}y)}{1-\lambda({\rm{i}}y)}\ ,\\[14.22636pt] \displaystyle\max_{x\in[0,1]}\left|\lambda_{2}({\rm{i}}y+x)\right|=\left|\lambda_{2}({\rm{i}}y\pm 1)\right|=\dfrac{\lambda({\rm{i}}y)}{\big{(}1-\lambda({\rm{i}}y)\big{)}^{2}}\ ,\quad y\geqslant 1/2\,,\end{array} (10.7)

where, by virtue of (2.32), the both functions of yy in the right-hand sides of (10.7) strictly decrease on the interval (0,+)(0,+\infty) from ++\infty to 0. This proves (2.59) and (2.60).

The property (2.58) is immediate from (2.40). Furthermore, (2.64)(a) is a simple consequence of (2.58) and (2.59).

Prove now (2.64)(c). For arbitrary z0z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} and x[1,1],x\in[-1,1], y>1y>1, it follows from (2.64)(a) and (5.98)(b) that λ(x+iy)λ\lambda(x\!+\!{\rm{i}}y)\in\Bb{C}_{{\rm{Re}}\leqslant\lambda({\rm{i}}y)}, λ(z)\lambda(z)\!\in\!\Bb{C}_{{\rm{Re}}\geqslant 1/2} and

|λ(z)λ(x+iy)||12λ(iy)||12λ(iy)||λ(z)|2,\displaystyle\left|\lambda(z)-\lambda(x+{\rm{i}}y)\right|\geqslant\left|\dfrac{1}{2}-\lambda({\rm{i}}y)\right|\geqslant\left|\dfrac{1}{2}-\lambda({\rm{i}}y)\right|\dfrac{\left|\lambda(z)\right|}{2}\,,\ (10.8)

provided that |λ(z)|2\left|\lambda(z)\right|\leqslant 2. But if |λ(z)|>2\left|\lambda(z)\right|>2 then by (2.59) and λ(iy)(0,1/2)\lambda({\rm{i}}y)\in(0,1/2) we get |λ(x+iy)|λ(iy)/(1λ(iy))<1|\lambda(x+{\rm{i}}y)|\leqslant\lambda({\rm{i}}y)/(1-\lambda({\rm{i}}y))<1 and therefore

|λ(z)λ(x+iy)||λ(z)||λ(x+iy)||λ(z)|2|λ(z)|2|12λ(iy)|,\displaystyle\left|\lambda(z)-\lambda(x+{\rm{i}}y)\right|\geqslant\left|\lambda(z)\right|-\left|\lambda(x+{\rm{i}}y)\right|\geqslant\dfrac{\left|\lambda(z)\right|}{2}\geqslant\dfrac{\left|\lambda(z)\right|}{2}\left|\dfrac{1}{2}-\lambda({\rm{i}}y)\right|\,,

which together with (10.8) completes the proof of (2.64)(c). Next, by (2.41), (2.60) and (ber1, , p.​ 325, (i)), for x[1,1]x\in[-1,1] and y>1y>1 we obtain

|λ(x+iy)|=π|λ2(x+iy)||θ3(eiπ(x+iy))|4πλ(iy)(1λ(iy))2|θ3(eπy)|4\displaystyle\left|\lambda^{\,\prime}\left(x+{\rm{i}}y\right)\right|=\pi\left|\lambda_{2}(x+{\rm{i}}y)\right|\left|\theta_{3}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(x+{\rm{i}}y)}\right)\right|^{4}\leqslant\pi\dfrac{\lambda({\rm{i}}y)}{\big{(}1-\lambda({\rm{i}}y)\big{)}^{2}}\left|\theta_{3}\left({\rm{e}}^{-\pi y}\right)\right|^{4}\leqslant
2π|θ3(eπ)|4=2ππΓ(3/4)4<8,7537585<9,\displaystyle\leqslant 2\pi\left|\theta_{3}\left({\rm{e}}^{-\pi}\right)\right|^{4}=2\pi\dfrac{\pi}{\Gamma(3/4)^{4}}<8,7537585<9\,,

which proves (2.64)(b). Finally, to prove (2.64)(d) observe that for 1/2+η:=λ(z)1/2+\eta:=\lambda(z) we have u+iv:=ηu+{\rm{i}}v:=\eta\in\Bb{C}_{{\rm{Re}}>0}, in view of (5.98)(b) , and for a:=λ(1+it)a:=-\lambda(1+{\rm{i}}t) we have a>0a>0, by virtue of Lemma 2.4. Then

2|λ(z)λ(1+it)|2\displaystyle 2\left|\lambda(z)-\lambda(1+{\rm{i}}t)\right|^{2} =2|1/2+u+a+iv|2=2(1/2+u+a)2+2v2,\displaystyle=2\left|1/2+u+a+{\rm{i}}v\right|^{2}=2\left(1/2+u+a\right)^{2}+2v^{2}\,,
(|λ(z)|+|λ(1+it)|)2\displaystyle\left(\left|\lambda(z)\right|+\left|\lambda(1+{\rm{i}}t)\right|\right)^{2} =(u+1/2)2+v2+a2+2a(u+1/2)2+v2\displaystyle=(u+1/2)^{2}+v^{2}+a^{2}+2a\sqrt{(u+1/2)^{2}+v^{2}}
(u+1/2)2+v2+a2+(u+1/2)2+v2+a2\displaystyle\leqslant(u+1/2)^{2}+v^{2}+a^{2}+(u+1/2)^{2}+v^{2}+a^{2}
=2(u+1/2)2+2v2+2a2=2(1/2+u+a)2+2v2\displaystyle=2(u+1/2)^{2}+2v^{2}+2a^{2}=2\left(1/2+u+a\right)^{2}+2v^{2}
4a(u+1/2)<2|λ(z)λ(1+it)|2,\displaystyle-4a(u+1/2)<2\left|\lambda(z)-\lambda(1+{\rm{i}}t)\right|^{2}\,,

which proves (2.64)(d) and completes the proof of Theorem 2.2 and Corollary 2.3.

10.1.4 . Proofs of Theorem 2.6 and Corollary 2.7.

  

For nn\in\Bb{Z}_{\neq 0} and z=a+ibz=a+{\rm{i}}b with a[1,1]a\in[-1,1], b>0b>0 we have

Im12nz=b(2na)2+b2b1+b2<b.\displaystyle{\rm{Im}}\dfrac{1}{2n-z}=\dfrac{b}{(2n-a)^{2}+b^{2}}\leqslant\dfrac{b}{1+b^{2}}<b\ .\

Besides that,

Im12n+1/z=b(2na+1)2+4n2b2=b1+4n2(a2+an+b2),\displaystyle{\rm{Im}}\dfrac{1}{2n+{1}/{z}}=\dfrac{b}{(2na+1)^{2}+4n^{2}b^{2}}=\frac{b}{1+4n^{2}\left(a^{2}+\dfrac{a}{n}+b^{2}\right)}\ ,

which is strictly less than bb if a=0a=0, but if a0a\!\neq\!0 then σa:=sign(a){1,1}\sigma_{a}\!:=\!{\rm{sign}}(a)\!\in\!\{1,-1\} and

a2+an+b2=|a|(1+σan)+|zσa2|2140,\displaystyle a^{2}+\dfrac{a}{n}+b^{2}=|a|\left(1+\dfrac{\sigma_{a}}{n}\right)+\left|z-\frac{\sigma_{a}}{2}\right|^{2}-\frac{1}{4}\geqslant 0\ ,\

provided that zclos()z\!\in\!{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right) and the equality here is strict if |n|2|n|\geqslant 2 or nRez>0n\cdot{\rm{Re}}\,z>0, or zz\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}. Thus, the following properties hold.

{z|Re|1n\displaystyle\left\{\begin{array}[]{l}z\in\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}}\\[8.5359pt] n\in\Bb{Z}_{\neq 0}\end{array}\right. \displaystyle\ \Rightarrow\ {Im12nz<Imz,12nz𝔻|Re|1\displaystyle\left\{\begin{array}[]{l}{\rm{Im}}{{\mbox{\footnotesize{$\dfrac{1}{2n-z}$}}}}<{\rm{Im}}\,z\ ,\\[8.5359pt] {{\mbox{\footnotesize{$\dfrac{1}{2n-z}$}}}}\in\Bb{D}\cap\Bb{H}\subset\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}}\ ,\end{array}\right. (10.13)
{z𝔽n\displaystyle\left\{\begin{array}[]{l}z\!\in\!\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)\\[8.5359pt] n\in\Bb{Z}_{\neq 0}\end{array}\right. \displaystyle\ \Rightarrow\ {Im12n+(1/z)Imz,12n+(1/z)𝔻|Re|1Im12n+(1/z)<Imz,if{|n|2,ornRez>0,orz.\displaystyle\left\{\begin{array}[]{l}{\rm{Im}}{{\mbox{\footnotesize{$\dfrac{1}{2n+(1/z)}$}}}}\leqslant{\rm{Im}}\,z\ ,\\[8.5359pt] {{\mbox{\footnotesize{$\dfrac{1}{2n+(1/z)}$}}}}\in\Bb{D}\cap\Bb{H}\subset\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}}\ ,\\[5.69046pt] {\rm{Im}}{{\mbox{\footnotesize{$\dfrac{1}{2n+(1/z)}$}}}}<{\rm{Im}}\,z\ ,\ \mbox{if}\ \left\{\begin{array}[]{l}|n|\!\geqslant\!2\,,\ \mbox{or}\\[2.84544pt] n\cdot{\rm{Re}}\,z\!>\!0\,,\\[2.84544pt] \ \mbox{or}\ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,.\end{array}\right.\end{array}\right. (10.22)

According to the definition (5.1), for N1N\geqslant 1 and 𝔫=(𝔫𝔑,𝔫𝔑1,,𝔫1)\eufm{n}=(n_{N},n_{N-1},\ldots,n_{1})\in\Bb{Z}_{\neq 0}^{N} we have

ϕ𝔫(y)=ϕnN,nN1,,n1(y)=ϕnN(ϕnN1((ϕn1(y)))),y\displaystyle\phi_{\eufm{n}}(y)=\phi_{n_{N},n_{N-1},\ldots,n_{1}}(y)=\phi_{n_{N}}\Big{(}\phi_{n_{N-1}}\big{(}\ldots\big{(}\phi_{n_{1}}(y)\big{)}\ldots\big{)}\Big{)},\quad y\in\Bb{H}, (10.23)

where

ϕn(z)=12nz,ϕn(1/z)=12n+1/z,nϝ\displaystyle\ \phi_{n}(z)=\frac{1}{2n-z}\ ,\quad\phi_{n}(-1/z)=\frac{1}{2n+1/z}\ ,\quad n\in\Bb{Z}_{\neq 0}\,,\ z\in\Bb{H}. (10.24)

Applying successively (10.13) and (10.22) to (10.23) we obtain

(a)Imϕ𝔫(y)<Imy,𝔫𝔑1y|Re|1(b)Imϕ𝔫(1/z)<Imz,𝔫𝔑2z𝔽(c)Imϕ𝔫(1/z)<Imz,𝔫{}z𝔽(d)Imϕ𝔫(1/z)<Imz,𝔫=sign(Re𝔷){1,1},z𝔽(e)Imϕ𝔫(1/ζ)<Imζ,𝔫𝔑1ζ,\displaystyle\vspace{-0,4cm}\begin{array}[]{lrll}{\rm{(a)}}&{\rm{Im}}\,\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(y)\!<\!{\rm{Im}}\,y,&\ \ \eufm{n}\in\cup_{N\geqslant 1}\Bb{Z}_{\neq 0}^{N},&\ \ y\in\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}}\ ,\\[2.84544pt] {\rm{(b)}}&{\rm{Im}}\,\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(-1/z)\!<\!{\rm{Im}}\,z,&\ \ \eufm{n}\in\cup_{N\geqslant 2}\Bb{Z}_{\neq 0}^{N},&\ \ z\in\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),\\[2.84544pt] {\rm{(c)}}&{\rm{Im}}\,\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(-1/z)\!<\!{\rm{Im}}\,z,&\ \ \eufm{n}\in\Bb{Z}\setminus\{-1,0,1\},&\ \ z\in\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),\\[2.84544pt] {\rm{(d)}}&{\rm{Im}}\,\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(-1/z)\!<\!{\rm{Im}}\,z,&\ \ \eufm{n}\!=\!{\rm{sign}}({\rm{Re}}\,z)\!\in\!\{-1,1\},&\ \ z\in\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),\\[2.84544pt] {\rm{(e)}}&{\rm{Im}}\,\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(-1/\zeta)\!<\!{\rm{Im}}\,\zeta,&\ \ \eufm{n}\in\cup_{N\geqslant 1}\Bb{Z}_{\neq 0}^{N},&\ \ \zeta\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\end{array}\vspace{-0,1cm} (10.30)

where (10.30)(b) with 𝔫=(𝔫𝔑,𝔫𝔑1,,𝔫1)\eufm{n}=(n_{N},n_{N-1},\ldots,n_{1}) follows from (10.30)(a) with 𝔫=(𝔫𝔑,𝔫𝔑1,,𝔫2)\eufm{n}=(n_{N},n_{N-1},\ldots,n_{2}), applied to y=ϕn1(1/z)y=\phi_{n_{1}}(-1/z) which by (10.24) and (10.22) satisfies y|Re|1y\in\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}} and ImyImz{\rm{Im}}\,y\leqslant{\rm{Im}}\,z. Introduce the notation

λ(1)(a):={z𝔽λ(z)=a},λ(1)(a):={z𝕜𝕜𝔽λ(z)=a},a{}\displaystyle\begin{array}[]{rcccl}\lambda_{{{\mbox{\tiny{$\square$}}}}}^{(-1)}(a)&:=&\{\,z\in\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)&\mid&\,\lambda(z)=a\},\\ \lambda_{\,\sqcup}^{(-1)}(a)&:=&\{\,z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\setminus{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)&\mid&\,\lambda(z)=a\},\end{array}\qquad a\in\Bb{C}\setminus\{0,1\}\,. (10.33)

In view of Lemma 2.4, Theorem B and (2.79),

λ(1)(a)={λ(a)},a(0,1)(),λ(1)(a)={λ(a+i0),λ(ai0)},a\displaystyle\begin{array}[]{ll}\lambda_{{{\mbox{\tiny{$\square$}}}}}^{(-1)}(a)=\left\{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a)\right\}\subset\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,&\quad a\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right),\\ \lambda_{{{\mbox{\tiny{$\square$}}}}}^{(-1)}(a)=\left\{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a+{\rm{i}}0),\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a-{\rm{i}}0)\right\}\subset\partial\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,&\quad a\in\Bb{R}_{<0}\cup\Bb{R}_{>1},\end{array} (10.36)

where by (2.68), (2.72) and (2.66), (2.67),

λ(a±i0)(±1)+iλ(ai0)=λ(a+i0)2,aλ(a±i0)γ(0,±1),λ(ai0)=12+1/λ(a+i0),a\displaystyle\begin{array}[]{lll}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!\pm\!{\rm{i}}0)\!\in\!(\pm 1)\!+\!{\rm{i}}\Bb{R}_{>0}\,,&\ \ \ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!-\!{\rm{i}}0)\!=\!\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!+\!{\rm{i}}0)\!-\!2\,,&\ \ \ a\!\in\!\Bb{R}_{<0}\,,\\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!\pm\!{\rm{i}}0)\!\in\!\gamma(0,\pm 1)\ ,&\ \ \ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!-\!{\rm{i}}0)=\dfrac{1}{-2\!+\!1/\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!+\!{\rm{i}}0)}\ ,&\ \ \ a\!\in\!\Bb{R}_{>1}\,.\end{array} (10.39)

Furthermore, by Lemma 2.5,

Imλ(a)=Imλ(a+i0)=Imλ(ai0),a\displaystyle{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a)={\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!+\!{\rm{i}}0)={\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a\!-\!{\rm{i}}0)\ ,\ a\!\in\!\Bb{R}_{<0}\cup\Bb{R}_{>1}\,. (10.40)

It has been established in Lemma 5.7 that for each a{}λ𝔽a\in\Bb{C}\setminus\{0,1\}=\lambda(\Bb{H}\cap{\rm{clos}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})) the set λ(1)(a)\lambda_{\,\sqcup}^{(-1)}(a) is countable and cannot have the limit points in \Bb{H}. That’s why to prove the statement of Theorem 2.6 it suffices to show that the imaginary part of each number from λ(1)(a)\lambda_{\,\sqcup}^{(-1)}(a) is strictly less than Imλ(a){\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a), i.e.,

Imz<Imλ(a),zλ(1)(a),a{}\displaystyle{\rm{Im}}\,z<{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a)\ ,\quad z\in\lambda_{\,\sqcup}^{(-1)}(a)\ ,\quad a\in\Bb{C}\setminus\{0,1\}\,, (10.41)

which obviously coincides with the property (2.84).

Assume that the number aa in Theorem 2.6 belongs to (0,1)()(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right). In the notation y:=λ(a)y:=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a)\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, it follows from (10.36) that λ(1)(a)={y}\lambda_{{{\mbox{\tiny{$\square$}}}}}^{(-1)}(a)=\left\{y\right\} and therefore (5.63) yields

λ(1)(a)={ϕ𝔫(y)|𝔫}{ϕ𝔫(1/y)|𝔫},\displaystyle\lambda_{\sqcup}^{(-1)}(a)=\left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(y)\ \left|\ \eufm{n}\in\Bb{Z}_{0}^{2\Bb{N}}\right\}\right.\cup\left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}(-1/y)\ \left|\ \eufm{n}\in\Bb{Z}_{0}^{2\Bb{N}-1}\right\}\right., (10.42)

which satisfies (10.41), in view of (10.30)(a) and (10.30)(e). This proves Theorem 2.6 for a(0,1)()a\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right).

Let aa in Theorem 2.6 belong to \Bb{R}_{<0} and y:=λ(a+i0)γ(1,)y:=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a+{\rm{i}}0)\in\gamma(1,\infty). By virtue of (10.36) and (10.39), λ(1)(a)={y,y2}\lambda_{{{\mbox{\tiny{$\square$}}}}}^{(-1)}(a)=\{y,y-2\} and we deduce from (5.63) that

λ(1)(a)={ϕ𝔫(y1+σ𝔫)|𝔫},\displaystyle\lambda_{\sqcup}^{(-1)}(a)=\left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(y-1+\sigma_{\eufm{n}}\right)\ \left|\ \eufm{n}\in\Bb{Z}_{0}^{2\Bb{N}}\right\}\right., (10.43)

which satisfies (10.41), in view of (10.30)(a), Im(y1+σ𝔫)=Imy{\rm{Im}}\,(y-1+\sigma_{\eufm{n}})={\rm{Im}}\,y and y1+σ𝔫σ𝔫+i𝔽|Re|1y-1+\sigma_{\eufm{n}}\in\sigma_{\eufm{n}}+{\rm{i}}\Bb{R}_{>0}\subset\Bb{H}\cap{\rm{clos}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})\subset\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}}. This proves Theorem 2.6 for aa\!\in\!\Bb{R}_{<0}.

Let finally aa in Theorem 2.6 belong to \Bb{R}_{>1} and y:=λ(a+i0)γ(1,0)y:=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(a+{\rm{i}}0)\in\gamma(1,0). Then 1/yγ(1,)-1/y\in\gamma(-1,\infty),

σ+1(1/y)σ+i𝔽σ{}\displaystyle\sigma\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\in\sigma+{\rm{i}}\Bb{R}_{>0}\subset\Bb{H}\cap{\rm{clos}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})\ ,\quad\sigma\in\{1,-1\}\,,\ (10.44)

and (10.36), (10.39) and (10.40) imply that λ(1)(a)={y,1/(2+1/y)}\lambda_{{{\mbox{\tiny{$\square$}}}}}^{(-1)}(a)=\{y,1/(-2+1/y)\} and

12+1/yγ(1,0),\displaystyle\dfrac{1}{-2+1/y}\in\gamma(-1,0)\ , Im12+1/y=Imy,\displaystyle{\rm{Im}}\dfrac{1}{-2+1/y}={\rm{Im}}\,y\ ,\ (10.45)
y=ϕn1(σn1+11/y)|n1=1,\displaystyle y=\phi_{n_{1}}(\sigma_{n_{1}}+1-1/y)\Big{|}_{n_{1}=1}\ , 12+1/y=ϕn1(σn1+11/y)|n1=1\displaystyle\dfrac{1}{-2+1/y}=\phi_{n_{1}}(\sigma_{n_{1}}+1-1/y)\Big{|}_{n_{1}=-1}

(see (5.27)). In accordance with (5.63), we get

λ(1)(a)\displaystyle\lambda_{\sqcup}^{(-1)}(a)\! ={ϕ𝔫(σ𝔫+1(1/y))|𝔫}\displaystyle=\!\left\{\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\!\left(\sigma_{\eufm{n}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\ \left|\ \eufm{n}\!\in\!\Bb{Z}_{0}^{2\Bb{N}+1}\right\}\right.\cup
{ϕn1(σn1+1(1/y))}n1{},\displaystyle\,\cup\big{\{}\phi_{n_{1}}\!\!\left(\sigma_{n_{1}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\big{\}}_{n_{1}\!\in\,\Bb{Z}\setminus\{-1,0,1\}}, (10.46)

because y,1/(2+1/y)𝔽y,1/(-2+1/y)\in\Bb{H}\cap{\rm{clos}}(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}) and, in view of (10.24), (10.45), (10.44) and (10.13),

{ϕn1(σn1+1(1/y))}n1{}\displaystyle\left\{\phi_{n_{1}}\!\!\left(\sigma_{n_{1}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\right\}_{n_{1}\in\Bb{Z}\setminus\{0\}}
={y}{12+1/y}{ϕn1(σn1+1(1/y))}n1{}|Re|1\displaystyle=\!\left\{y\right\}\!\cup\!\left\{\dfrac{1}{-2\!+\!1/y}\right\}\!\cup\!\left\{\phi_{n_{1}}\!\!\left(\sigma_{n_{1}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\right\}_{n_{1}\in\Bb{Z}\setminus\{-1,0,1\}}\subset\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}}\,. (10.47)

Here ϕn1(σn1+1(1/y))=ϕn~1(1/y)\phi_{n_{1}}(\sigma_{n_{1}}\!+\!1\!-\!(1/y))=\phi_{\widetilde{n}_{1}}(-\!1/y), n~1=n1(σn1+1)/2{}\widetilde{n}_{1}=n_{1}-(\sigma_{n_{1}}\!+\!1)/2\in\Bb{Z}\setminus\{0,-1\} for every n1{}n_{1}\in\Bb{Z}\setminus\{-1,0,1\}, where n~1=1\widetilde{n}_{1}=1 if and only if n1=1=sign(Rey)n_{1}=1={\rm{sign}}({\rm{Re}}y). Applying (10.30)(c) and (10.30)(d) to ϕn~1(1/y)\phi_{\widetilde{n}_{1}}(-\!1/y) with yγ(1,0)𝔽y\in\gamma(1,0)\subset\Bb{H}\cap{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right), we obtain

Imϕn1(σn1+1(1/y))<Imy,n1{}γ\displaystyle{\rm{Im}}\,\phi_{n_{1}}(\sigma_{n_{1}}\!+\!1\!-\!(1/y))<{\rm{Im}}\,y\ ,\quad n_{1}\in\Bb{Z}\setminus\{-1,0,1\},\quad y\in\gamma(1,0). (10.48)

At the same time, if N1N\geqslant 1 and 𝔫=(𝔫2𝔑+1,𝔫2𝔑,,𝔫2,𝔫1)()2𝔑+1\eufm{n}=(n_{2N+1},n_{2N},\ldots,n_{2},n_{1})\in\left(\Bb{Z}_{\neq 0}\right)^{2N+1} then (10.23) means that in (10.46) we have

ϕ𝔫(σ𝔫+1(1/y))\displaystyle\phi_{\eufm{n}}\!\left(\sigma_{\eufm{n}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right) =ϕn2N+1,n2N,,n2,n1(σn1+1(1/y))\displaystyle=\phi_{n_{2N+1},n_{2N},\ldots,n_{2},n_{1}}\!\left(\sigma_{n_{1}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)
=ϕn2N+1,n2N,,n2(ϕn1(σn1+1(1/y))),\displaystyle=\phi_{n_{2N+1},n_{2N},\ldots,n_{2}}\big{(}\phi_{n_{1}}\!\!\left(\sigma_{n_{1}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\big{)}\,,\

where, by (10.47), ϕn1(σn1+1(1/y))|Re|1\phi_{n_{1}}\!\!\left(\sigma_{n_{1}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\in\Bb{H}_{{{\mbox{\footnotesize{$\left|{\rm{Re}}\right|\leqslant 1$}}}}}. Hence, in accordance with (10.30)(a), (10.13) and (10.47), (10.45), (10.48),

Imϕ𝔫(σ𝔫+1(1/y))<Imϕn1(σn1+1(1/y))Imy.\displaystyle{\rm{Im}}\,\phi_{\eufm{n}}\!\left(\sigma_{\eufm{n}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)<{\rm{Im}}\,\phi_{n_{1}}\!\!\left(\sigma_{n_{1}}\!+\!1\!-\!{{\mbox{\footnotesize{$($}}}}1/y{{\mbox{\footnotesize{$)$}}}}\hskip 1.42271pt\right)\leqslant{\rm{Im}}\,y\,.

This inequality together with (10.48) and (10.46) proves (10.41) for each aa\in\Bb{R}_{>1}. Thus, Theorem 2.6 holds in this case and its proof is completed.

If we assume the contrary in Corollary 2.7 we obtain the contradiction with (2.84) and therefore Corollary 2.7 holds as well.

10.2 . Proofs for Section 3

10.2.1 . Proofs of Lemmas 3.2 and 3.4.

We first prove Lemma 3.2. Let x{}x\in\Bb{C}\setminus\{0,1\} and Pn(x)=k=0npkxkP_{n}(x)=\sum_{k=0}^{n}p_{k}x^{k} satisfy the conditions of Lemma 3.2. Then

Pn(1x)±Pn(11x)=k=0npk(1xk±1(1x)k)=xn(1x)nRn±(x),\displaystyle P_{n}\left(\frac{1}{x}\right)\pm P_{n}\left(\frac{1}{1-x}\right)=\sum_{k=0}^{n}p_{k}\left(\dfrac{1}{x^{k}}\pm\dfrac{1}{(1-x)^{k}}\right)=x^{-n}(1-x)^{-n}R_{n}^{\pm}(x)\,,
Rn±(x):=k=0npk((1x)nxnk±xn(1x)nk),\displaystyle R_{n}^{\pm}(x):=\sum\nolimits_{k=0}^{n}\ p_{k}\left((1-x)^{n}x^{n-k}\pm x^{n}(1-x)^{n-k}\right)\,,\

where Rn±(1x)=±Rn±(x)R_{n}^{\pm}(1-x)=\pm R_{n}^{\pm}(x), and therefore, there exist the real numbers qkq_{k}, 0kn0\leqslant k\leqslant n, such that

Rn+(1/2+x)=0knq2kx2k,Rn(1/2+x)=0kn1q2k+1x2k+1,\displaystyle R_{n}^{+}(1/2+x)=\sum_{0\leqslant k\leqslant n}\!\!\!q_{2k}\,x^{2k}\,,\ R_{n}^{-}(1/2+x)=\sum_{0\leqslant k\leqslant n-1}\!\!\!q_{2k+1}\,x^{2k+1}\,,\

where kk\in\Bb{Z}_{\geqslant 0}. Then there exist the real numbers rk+r_{k}^{+}, 0kn0\leqslant k\leqslant n, satisfying

Rn+(x)\displaystyle R_{n}^{+}(x) =0knq2k(x1/2)2k=0kn22kq2k(2x1)2k\displaystyle=\sum_{0\leqslant k\leqslant n}\!\!\!q_{2k}\,(x-1/2)^{2k}=\sum_{0\leqslant k\leqslant n}\!\!\!2^{-2k}q_{2k}\,(2x-1)^{2k}
=0kn22kq2k(4x24x+1)k=0kn22kq2k(14x(1x))k\displaystyle=\sum_{0\leqslant k\leqslant n}\!\!\!2^{-2k}q_{2k}\,(4x^{2}-4x+1)^{k}=\sum_{0\leqslant k\leqslant n}\!\!\!2^{-2k}q_{2k}\,\big{(}1-4x(1-x)\big{)}^{k}
=0knr+nk(x(1x))k,\displaystyle=\sum_{0\leqslant k\leqslant n}r^{+}_{n-k}\big{(}x(1-x)\big{)}^{k}\,,\

and rkr_{k}^{-}, 1kn1\leqslant k\leqslant n, such that

Rn(x)=(x1/2)0kn1q2k+1(x1/2)2k=21(2x1)0kn122kq2k+1(2x1)2k\displaystyle R_{n}^{-}(x)\!=\!(x\!-\!1/2)\!\!\!\sum_{0\leqslant k\leqslant n-1}\!\!\!q_{2k+1}\,(x\!-\!1/2)^{2k}\!=\!2^{-1}(2x\!-\!1)\!\!\!\sum_{0\leqslant k\leqslant n-1}\!\!\!2^{-2k}q_{2k+1}\,(2x\!-\!1)^{2k}
=21(2x1)0kn122kq2k+1(14x(1x))k=(12x)0kn1rnk(x(1x))k.\displaystyle=\!2^{-1}(2x\!-\!1)\!\!\!\sum_{0\leqslant k\leqslant n-1}\!\!\!2^{-2k}q_{2k+1}\,\big{(}1\!-\!4x(1\!-\!x)\big{)}^{k}\!=\!(1\!-\!2x)\sum_{0\leqslant k\leqslant n-1}r^{-}_{n-k}\big{(}x(1-x)\big{)}^{k}\,.

Hence,

Pn(1x)+Pn(11x)=+[𝒫𝓃](1𝓍(1𝓍)),\displaystyle P_{n}\left(\frac{1}{x}\right)\!+\!P_{n}\left(\frac{1}{1-x}\right)=\eusm{R}^{+}\!\!\left[P_{n}\right]\left(\frac{1}{x(1-x)}\right), +[𝒫𝓃](𝓍):=0𝓀𝓃𝓇+𝓀𝓍𝓀,\displaystyle\!\!\eusm{R}^{+}\!\!\left[P_{n}\right](x)\!:=\!\sum_{0\leqslant k\leqslant n}r^{+}_{k}x^{k},\
Pn(1x)Pn(11x)=(12x)[𝒫𝓃](1𝓍(1𝓍)),\displaystyle P_{n}\left(\frac{1}{x}\right)\!-\!P_{n}\left(\frac{1}{1-x}\right)=(1\!-\!2x)\eusm{R}^{-}\!\!\left[P_{n}\right]\left(\frac{1}{x(1-x)}\right), [𝒫𝓃](𝓍):=1𝓀𝓃𝓇𝓀𝓍𝓀,\displaystyle\!\!\eusm{R}^{-}\!\!\left[P_{n}\right](x)\!:=\!\sum_{1\leqslant k\leqslant n}r^{-}_{k}x^{k},

where obviously, [𝒫𝓃](0)=0\eusm{R}^{-}\!\left[P_{n}\right](0)\!=\!0 and +[𝒫𝓃](0)=𝓇+0=2𝓅0=2𝒫(0)\eusm{R}^{+}\!\!\left[P_{n}\right](0)\!=\!r^{+}_{0}\!=\!2p_{0}\!=\!2P(0). Lemma 3.2 is proved.

Next we prove Lemma 3.4. For arbitrary z𝔻{}z\!\in\!\Bb{D}\!\setminus\!\{0\} we deduce from (1.17) that

enπF(1z)F(z)=16nznexp(nz12+2πn=1Γ(n+3/2)2(n+1)!2[k=1n+11(2k1)k]zn1+1πn=1Γ(n+1/2)2(n!)2zn)\displaystyle e^{{{\mbox{\footnotesize{$n\pi\ \dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}$}}}}}=\dfrac{16^{n}}{z^{n}}\ \exp\left(-nz\dfrac{\dfrac{1}{2}+\dfrac{2}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+3/2)^{2}}{(n+1)!^{2}}\left[\sum\limits_{k=1}^{n+1}\dfrac{1}{(2k-1)k}\right]z^{n}}{1+\dfrac{1}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}z^{n}}\right)
=16nznk0(nz)kk!(12+2πn=1Γ(n+3/2)2(n+1)!2[k=1n+11(2k1)k]zn1+1πn=1Γ(n+1/2)2(n!)2zn)k=16nzn\displaystyle=\dfrac{16^{n}}{z^{n}}\sum\limits_{k\geqslant 0}\dfrac{(-nz)^{k}}{k!}\left(\dfrac{\dfrac{1}{2}+\dfrac{2}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+3/2)^{2}}{(n+1)!^{2}}\left[\sum\limits_{k=1}^{n+1}\dfrac{1}{(2k-1)k}\right]z^{n}}{1+\dfrac{1}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}z^{n}}\right)^{k}=\dfrac{16^{n}}{z^{n}}
+16nzn(nz)12+2πn=1Γ(n+3/2)2(n+1)!2[k=1n+11(2k1)k]zn1+1πn=1Γ(n+1/2)2(n!)2zn\displaystyle+\dfrac{16^{n}}{z^{n}}(-nz)\dfrac{\dfrac{1}{2}+\dfrac{2}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+3/2)^{2}}{(n+1)!^{2}}\left[\sum\limits_{k=1}^{n+1}\dfrac{1}{(2k-1)k}\right]z^{n}}{1+\dfrac{1}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}z^{n}}
+16nznk2(nz)kk!(12+2πn=1Γ(n+3/2)2(n+1)!2[k=1n+11(2k1)k]zn1+1πn=1Γ(n+1/2)2(n!)2zn)k\displaystyle+\dfrac{16^{n}}{z^{n}}\sum\limits_{k\geqslant 2}\dfrac{(-nz)^{k}}{k!}\left(\dfrac{\dfrac{1}{2}+\dfrac{2}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+3/2)^{2}}{(n+1)!^{2}}\left[\sum\limits_{k=1}^{n+1}\dfrac{1}{(2k-1)k}\right]z^{n}}{1+\dfrac{1}{\pi}\sum\limits_{n=1}^{\infty}\dfrac{\Gamma(n+1/2)^{2}}{(n!)^{2}}z^{n}}\right)^{k}
=16nzn(1nz2)+16nznk=2βn,kzk=16nzn8n16n1zn1+16nk=2n1βn,kznk\displaystyle=\dfrac{16^{n}}{z^{n}}\left(1-\dfrac{nz}{2}\right)+\dfrac{16^{n}}{z^{n}}\sum\limits_{k=2}^{\infty}\beta_{n,k}z^{k}=\dfrac{16^{n}}{z^{n}}-8n\dfrac{16^{n-1}}{z^{n-1}}+16^{n}\sum\limits_{k=2}^{n-1}\dfrac{\beta_{n,k}}{z^{n-k}}
+16nk=nβn,kzkn,\displaystyle+16^{n}\sum\limits_{k=n}^{\infty}\beta_{n,k}z^{k-n}\,,\

where βn,k\beta_{n,k}, n1n\geqslant 1, k2k\geqslant 2, are the real numbers such that k=2βn,kzkHol(𝔻\sum_{k=2}^{\infty}\beta_{n,k}z^{k}\in{\rm{Hol}}(\Bb{D}) for each n1n\geqslant 1. Hence, in the notation Sn(z)=k=1nsn,kzkS^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(z)=\sum_{k=1}^{n}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}z^{k}, n1n\geqslant 1, of (3.14), it follows that

S1(z)=16z,S2(z)=162z(z1),Sn(z)=16nzn8n16n1zn1\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{1}(z)=16z\ ,\quad S^{{{\mbox{\tiny{$\triangle$}}}}}_{2}(z)=16^{2}z(z-1)\,,\ \ S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(z)=16^{n}z^{n}-8n16^{n-1}z^{n-1}
+k=2n1sn,nkznk,sn,n=16n,sn,n1=8n16n1,n3,\displaystyle+\sum\limits_{k=2}^{n-1}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,n-k}z^{n-k}\ ,\quad s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,n}=16^{n}\ ,\quad s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,n-1}=-8n16^{n-1}\ ,\quad n\geqslant 3\,,

which proves (3.14) and (3.15).

According to Lemma 2.5, Imλ(1)=0{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1)\!=\!0 and therefore (3.6) for z=1z\!=\!1 gives

n=1Sn(1)eiπny=λ(y)iπ(1λ(y)),Imy>Imλ(1)=0,\displaystyle\sum\limits_{n=1}^{\infty}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(1\right){\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}=\frac{\lambda^{\,\prime}\left(y\right)}{i\pi\left(1-\lambda\left(y\right)\right)}\ ,\quad{\rm{Im}}\,y>{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1)=0\,,\

where, in view of (2.41), (2.21) and (2.26), we have

λ(y)iπ(1λ(y))=Θ2(y)4Θ4(y)4Θ3(y)4Θ4(y)4Θ3(y)4=Θ2(y)4=16eiπyθ2(eiπy)4.\displaystyle\frac{\lambda^{\,\prime}\left(y\right)}{i\pi\left(1-\lambda\left(y\right)\right)}=\dfrac{\Theta_{2}(y)^{4}\Theta_{4}(y)^{4}\Theta_{3}(y)^{-4}}{\Theta_{4}(y)^{4}\Theta_{3}(y)^{-4}}=\Theta_{2}(y)^{4}=16{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi y}\theta_{2}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi y}\right)^{4}.

Together with (2.30)(c) this proves (3.13) and (3.12)(b).

To prove (3.16), observe that Imλ(1/z)0{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/z)\to 0 as zϝz\in\Bb{C}\!\setminus\!\Bb{R},z\to 0 (see (bh2, , p.​ 609, (4.6))), which yields, in view of Lemma 2.5, that Imλ(1/z)0{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/z)\to 0 as zϝz\in\Bb{C},z\to 0. So that for every a>0a>0 one can find δa>0\delta_{a}>0 such that 0<Imλ(1/z)<a0<{\rm{Im}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1/z)<a for all 0<|z|<δa0<|z|<\delta_{a} and (3.9) can be written as follows

Sn(z)z=k=1nsn,kzk1=12πi1+ia1+iaλ(ζ)enπiζ1zλ(ζ)dζ,z(δa𝔻){0}.\displaystyle\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(z)}{z}\!=\!\sum\limits_{k=1}^{n}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}z^{k-1}\!=\!\frac{1}{2\pi i}\int\nolimits_{{{\mbox{\footnotesize{$-1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}^{{{\mbox{\footnotesize{$1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}\frac{\lambda^{\,\prime}\left(\zeta\right){\rm{e}}^{{{\mbox{\footnotesize{$-n\pi{\rm{i}}\zeta$}}}}}}{1-z\lambda\left(\zeta\right)}\,d\zeta\,,\ \ z\in\left(\delta_{a}\Bb{D}\right)\setminus\{0\}\,. (10.49)

By choosing a positive number ηa<δa\eta_{a}<\delta_{a} less than (1/2)minζ[1+ia,1+ia]1/|λ(ζ)|(1/2)\min_{\zeta\in[-1+{\rm{i}}a,1+{\rm{i}}a]}1/|\lambda(\zeta)| we deduce from (10.49) that

|Sn(z)|/|z|(2/π)(maxζ[1+ia,1+ia]|λ(ζ)|)exp(nπa),zηa𝔻\displaystyle|S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(z)|/|z|\leqslant(2/\pi)\big{(}\max\nolimits_{\zeta\in[-1+{\rm{i}}a,1+{\rm{i}}a]}|\lambda^{\,\prime}(\zeta)|\big{)}\exp(n\pi a)\ ,\ z\in\eta_{a}\Bb{D}\,.

Therefore for every Imy>a{\rm{Im}}\,y>a the series in (3.6), written in the form,

iπn=1(k=1nsn,kzk1)eiπny=λ(y)/(1zλ(y)),z(δa𝔻){0},\displaystyle{\rm{i}}\pi\sum\nolimits_{n=1}^{\infty}\left(\sum\nolimits_{k=1}^{n}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}z^{k-1}\right){\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}={\lambda^{\,\prime}\left(y\right)}\big{/}{\left(1-z\lambda\left(y\right)\right)}\,,\quad z\in\left(\delta_{a}\Bb{D}\right)\setminus\{0\}\,,

converges uniformly over all zηa𝔻z\in\eta_{a}\Bb{D} and we can take the limit as z0z\to 0 to get, in view of (2.41) and arbitrariness of a>0a>0, the following equivalent form of (3.16)

n=1sn,1eiπny=λ(y)/(iπ)=Θ2(y)4Θ4(y)4Θ3(y)4,Imy>0.\displaystyle\sum\nolimits_{n=1}^{\infty}s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,1}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}=\lambda^{\,\prime}\left(y\right)\big{/}({\rm{i}}\pi)=\Theta_{2}(y)^{4}\Theta_{4}(y)^{4}\Theta_{3}(y)^{-4}\,,\ {\rm{Im}}\,y>0\ . (10.50)

It follows from the functional equation

λ(z)sign(Imz)=λ(z/(z1)),z\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)-{\rm{sign}}({\rm{Im}}\,z)=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left({z}\big{/}{(z-1)}\right)\ ,\quad z\in\Bb{C}\setminus\Bb{R}\,,

(see (bh1, , p.​ 48, (A.14n))) that

exp(iπnλ(z/(z1)))=(1)nexp(iπnλ(z)),z\displaystyle\exp\left(-{\rm{i}}\pi n\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left({z}\big{/}{(z\!-\!1)}\right)\right)\!=\!(-1)^{n}\exp\left(-{\rm{i}}\pi n\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right),\ z\!\in\!\Bb{C}\!\setminus\!\Bb{R},\ n\!\geqslant\!1. (10.51)

By using (3.2),

exp(iπnλ(z))=Sn(1/z)+ΔnS(z),z𝔻{}exp(iπnλ(z/(z1)))=Sn(11/z)+ΔnS(z/(z1)),z\displaystyle\begin{array}[]{rclcll}\exp\left(-{\rm{i}}\pi n\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)&=&S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)&+&\Delta_{n}^{S}(z)\ ,&\quad z\in\Bb{D}\setminus\{0\},\\[5.69046pt] \exp\left(-{\rm{i}}\pi n\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left({z}\big{/}{(z\!-\!1)}\right)\right)&=&S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1-1/z)&+&\Delta_{n}^{S}\left(z\big{/}(z-1)\right)\ ,&\quad z\in\Bb{C}_{{\rm{Re}}\,<1/2}\,,\end{array}

we obtain

Sn(11/z)=(1)nSn(1/z)+(1)nΔnS(z)ΔnS(z/(z1)),\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1-1/z)=(-1)^{n}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)\!+\!(-1)^{n}\Delta_{n}^{S}(z)-\Delta_{n}^{S}\left(z\big{/}(z-1)\right), (10.52)

for every z(𝔻{})z\in\left(\Bb{D}\setminus\{0\}\right)\cap\Bb{C}_{{\rm{Re}}\,<1/2}. Since Sn(0)=0S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(0)=0 and z/(z1)Hol(𝔻z/(z-1)\!\in\!{\rm{Hol}}(\Bb{D}) we have ΔnS(z/(z1))Hol(𝔻\Delta_{n}^{S}(z/(z-1))\!\in\!{\rm{Hol}}(\Bb{D}/2) and therefore (10.52) , (3.22) and (3.13) yield

Sn(11/z)(1)nSn(1/z)=((1)n1)ΔnS(0)=(1(1)n)r4(n)=Sn(1)\displaystyle S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1-1/z)-(-1)^{n}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)=\left((-1)^{n}-1\right)\Delta_{n}^{S}(0)=\left(1-(-1)^{n}\right)r_{4}(n)=S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)\,

for every z(𝔻{})/2z\in\left(\Bb{D}\setminus\{0\}\right)\big{/}2, which proves (3.12)(a) and completes the proof of Lemma 3.4.

10.3 . Proofs for Section 4

  

10.3.1 . Proofs of Lemmas 4.1 and 4.2.

We first prove Lemma 4.1. The properties Lemma 4.1(1),(2),(3) are immediate from (3.37), (3.29), (3.33) (4.1) and (1.19). To prove Lemma 4.1(4) for positive integer nn we change the contour γ(1,1)\gamma(-1,1) of integration in (3.37) to the contour which passes from 1-1 to 1+i-1+{\rm{i}} along [1,1+i][-1,-1+{\rm{i}}], from 1+i-1+{\rm{i}} to 1+i1+{\rm{i}} along [1+i,1+i][-1+{\rm{i}},1+{\rm{i}}] and from 1+i1+{\rm{i}} to 11 along [1+i,1][1+{\rm{i}},1]. By using the periodicity of Rn(z)\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z) and Rn(1/z)\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z), we obtain

4π2nHn(x)=11Rn(i+t)dt(x+i+t)2+i01Rn(1+it)[1(x1+it)21(x+1+it)2]dt,\displaystyle-4\pi^{2}n\eurm{H}_{n}(x)\!=\!\int\nolimits_{-1}^{1}\!\frac{\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}({\rm{i}}+t){\rm{d}}t}{(x+{\rm{i}}+t)^{2}}\!+\!{\rm{i}}\!\int\nolimits_{0}^{1}\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1\!+\!{\rm{i}}t)\!\left[\dfrac{1}{(x\!-\!1\!+\!{\rm{i}}t)^{2}}\!-\!\dfrac{1}{(x\!+\!1\!+\!{\rm{i}}t)^{2}}\right]\!{\rm{d}}t,
4π2nMn(x)=11Rn(1i+t)dt(x+2i+t)2+i01Rn(11it)[1(x1+it)21(x+1+it)2]dt,\displaystyle 4\pi^{2}n\eurm{M}_{n}(x)\!=\!\int\nolimits_{-1}^{1}\!\frac{\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{-1}{{\rm{i}}\!+\!t}\right){\rm{d}}t}{(x+2{\rm{i}}+t)^{2}}\!+\!{\rm{i}}\!\int\nolimits_{0}^{1}\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{1\!-\!{\rm{i}}t}\right)\!\left[\dfrac{1}{(x\!-\!1\!+\!{\rm{i}}t)^{2}}\!-\!\dfrac{1}{(x\!+\!1\!+\!{\rm{i}}t)^{2}}\right]\!{\rm{d}}t,

and therefore, for any positive integer N2N\geqslant 2 it follows that

4π2nk=NNHn(x+2k)\displaystyle-4\pi^{2}n\!\!\!\sum_{k=-N}^{N}\eurm{H}_{n}(x+2k) =11Rn(i+t)(k=NN1(x+2k+i+t)2)dt\displaystyle\!=\!\int\nolimits_{-1}^{1}\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}({\rm{i}}+t)\left(\sum_{k=-N}^{N}\frac{1}{(x+2k+{\rm{i}}+t)^{2}}\right){\rm{d}}t
+i01Rn(1+it)[1(x2N1+it)21(x+2N+1+it)2]dt,\displaystyle\!+\!{\rm{i}}\!\!\int\nolimits_{0}^{1}\!\!\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1\!+\!{\rm{i}}t)\!\left[\dfrac{1}{(x\!-\!2N\!-\!1\!+\!{\rm{i}}t)^{2}}\!-\!\dfrac{1}{(x\!+\!2N\!+\!1\!+\!{\rm{i}}t)^{2}}\right]\!{\rm{d}}t,
4π2nk=NNMn(x+2k)\displaystyle 4\pi^{2}n\!\!\!\sum_{k=-N}^{N}\eurm{M}_{n}(x+2k) =11Rn(1i+t)(k=NN1(x+2k+i+t)2)dt\displaystyle\!=\!\int\nolimits_{-1}^{1}\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(-\dfrac{1}{{\rm{i}}\!+\!t}\right)\left(\sum_{k=-N}^{N}\frac{1}{(x+2k+{\rm{i}}+t)^{2}}\right){\rm{d}}t
+i01Rn(11it)[1(x2N1+it)21(x+2N+1+it)2]dt.\displaystyle\!+\!{\rm{i}}\!\!\int\nolimits_{0}^{1}\!\!\!\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{1\!-\!{\rm{i}}t}\right)\!\left[\dfrac{1}{(x\!-2N\!\!-\!1\!+\!{\rm{i}}t)^{2}}\!-\!\dfrac{1}{(x\!+\!2N\!+\!1\!+\!{\rm{i}}t)^{2}}\right]\!{\rm{d}}t.

Letting NN\to\infty and using the identity (see (gra, , p.​ 44, 1.422.4))17{}^{\ref*{case15}}

k1(2k+z)2=π24sin2πz2=π2m1meiπmz,z\displaystyle\sum\limits_{k\in\Bb{Z}}\frac{1}{(2k+z)^{2}}=\frac{\pi^{2}}{4\sin^{2}\dfrac{\pi z}{2}}=-\pi^{2}\sum_{m\geqslant 1}m{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi mz$}}}}}\ ,\qquad z\in\Bb{H}\,,\

we derive from the first of the latter two equalities and (3.21) that

4π2nkHn(x+2k)=π2411Rn(i+t)dtsin2π(x+i+t)2=π2×\displaystyle-4\pi^{2}n\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{n}(x+2k)=\frac{\pi^{2}}{4}\int\limits_{-1}^{1}\!\frac{\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}({\rm{i}}+t)\,{\rm{d}}t}{\sin^{2}\dfrac{\pi(x+{\rm{i}}+t)}{2}}=-\pi^{2}\times
×m1meπmeiπmx11eiπmt(enπeinπt+ΔnS(0)+k1dn,keπkeiπkt)dt\displaystyle\times\!\sum_{m\geqslant 1}m{\rm{e}}^{-\pi m}{\rm{e}}^{{\rm{i}}\pi mx}\int\limits_{-1}^{1}{\rm{e}}^{{\rm{i}}\pi mt}\bigg{(}\!{\rm{e}}^{n\pi}{\rm{e}}^{-{\rm{i}}n\pi t}\!+\!\Delta_{n}^{S}(0)+\sum\limits_{k\geqslant 1}{\eurm{d}}_{n,k}{\rm{e}}^{-\pi k}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi kt}\!\bigg{)}{\rm{d}}t
=2π2neiπnx,\displaystyle=-2\pi^{2}n{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}\,,\

and from the second one and (3.18) that

4π2n(1)nkMn(x+2k)=(1)nπ2411Rn(1i+t)dtsin2π(x+i+t)2=π2×\displaystyle 4\pi^{2}n(-1)^{n}\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{M}_{n}(x+2k)\!=\!\frac{(-1)^{n}\pi^{2}}{4}\int\limits_{-1}^{1}\!\dfrac{\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(-\dfrac{1}{{\rm{i}}+t}\right)\,{\rm{d}}t}{\sin^{2}\dfrac{\pi(x+{\rm{i}}+t)}{2}}=-\pi^{2}\times
×m1meπmeiπmx11eiπmt(Sn(1)16sn,1eπeiπt+k2bn,keπkeiπkt)dt\displaystyle\times\!\sum_{m\geqslant 1}m{\rm{e}}^{-\pi m}{\rm{e}}^{{\rm{i}}\pi mx}\int\limits_{-1}^{1}{\rm{e}}^{{\rm{i}}\pi mt}\bigg{(}\!-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)\!-\!16s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,1}{\rm{e}}^{-\pi}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi t}\!+\!\sum\limits_{k\geqslant 2}{\eurm{b}}_{n,k}{\rm{e}}^{-\pi k}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi kt}\!\bigg{)}{\rm{d}}t
=0.\displaystyle=0\,.

This completes the proof of Lemma 4.1(4) for positive integer nn. But according to (1.18), for any n1n\geqslant 1 we have

kHn(x+2k)=kHn(x2k)=kHn(x+2k)=eiπnx2,x\displaystyle\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{-n}(x+2k)=\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{n}(-x-2k)=\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{n}(-x+2k)=\frac{{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nx$}}}}}}{2}\ ,\ x\!\in\!\Bb{R}\,,

and, by (1.19),

Mn(x):=Hn(1/x)/x2=Hn(1/x)/x2=Mn(x),x{}\displaystyle\eurm{M}_{-n}(x):=\eurm{H}_{-n}(-1/x)/x^{2}=\eurm{H}_{n}(1/x)/x^{2}=\eurm{M}_{n}(-x)\ ,\ x\!\in\!\Bb{R}\!\setminus\!\{0\}\ ,\

from which Mn(x)=Mn(x)\eurm{M}_{-n}(x)=\eurm{M}_{n}(-x) for all xx\!\in\!\Bb{R}, in view of the continuity of Mn\eurm{M}_{n}. Then

kMn(x+2k)=kMn(x2k)=kMn(x+2k)=0,x\displaystyle\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{M}_{-n}(x+2k)=\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{M}_{n}(-x-2k)=\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{M}_{n}(-x+2k)=0\ ,\ x\!\in\!\Bb{R}\,.

This proves Lemma 4.1(4) and completes the proof of Lemma 4.1.

Next, we prove Lemma 4.2. The properties Lemma 4.2(1),(2),(4) are simple consequences of (3.20) and (2.52). To prove Lemma 4.2(3) we change the contour γ(1,1)\gamma(-1,1) of integration in (2.52) similar to that of in Section 10.3.1 and by the periodicity of Θ3\Theta_{3} we get

2πiH0(x)=11(i+t)Θ3(i+t)4x2(i+t)2dt\displaystyle 2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{H}_{0}(x)\!=\!\int\limits_{-1}^{1}\!\frac{({\rm{i}}+t)\,\Theta_{3}\left({\rm{i}}+t\right)^{4}}{x^{2}-({\rm{i}}+t)^{2}}{\rm{d}}t
+i201Θ3(1+it)4[1x(1+it)1x+(1+it)1x(1+it)+1x+(1+it)]dt,\displaystyle+\dfrac{{\rm{i}}}{2}\int\limits_{0}^{1}\!\Theta_{3}\left(-1+{\rm{i}}t\right)^{4}\left[\frac{1}{x\!-\!(-1\!+\!{\rm{i}}t)}\!-\!\frac{1}{x\!+\!(-1\!+\!{\rm{i}}t)}\!-\!\frac{1}{x\!-\!(1\!+\!{\rm{i}}t)}\!+\!\frac{1}{x\!+\!(1\!+\!{\rm{i}}t)}\right]\!{\rm{d}}t,

and therefore, for any positive integer N2N\geqslant 2 this yields

2πik=NNH0(x+2k)=11Θ3(i+t)4(k=NN(i+t)(x+2k)2(i+t)2)dt\displaystyle 2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\sum_{k=-N}^{N}\eurm{H}_{0}(x+2k)\!=\!\int\limits_{-1}^{1}\!\Theta_{3}\left({\rm{i}}+t\right)^{4}\left(\sum_{k=-N}^{N}\frac{({\rm{i}}+t)\,}{(x+2k)^{2}-({\rm{i}}+t)^{2}}\right){\rm{d}}t
+i201Θ3(1+it)4[1x+2N+1it1x2N1it]dt\displaystyle+\dfrac{{\rm{i}}}{2}\int\limits_{0}^{1}\!\Theta_{3}\left(-1+{\rm{i}}t\right)^{4}\left[\frac{1}{x\!+2N+\!1\!-\!{\rm{i}}t}\!-\!\frac{1}{x\!-\!2N\!-\!1\!-\!{\rm{i}}t}\right]\!{\rm{d}}t
+i201Θ3(1+it)4[1x+2N+1+it1x2N1+it]dt.\displaystyle+\dfrac{{\rm{i}}}{2}\int\limits_{0}^{1}\!\Theta_{3}\left(-1+{\rm{i}}t\right)^{4}\left[\frac{1}{x\!+\!2N\!+\!1\!+\!{\rm{i}}t}\!-\!\frac{1}{x\!-\!2N\!-1\!+\!{\rm{i}}t}\right]\!{\rm{d}}t\,.

Letting NN\to\infty and using the identity18{}^{\ref*{case16}}

kz(2k+x)2z2\displaystyle\sum\limits_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\dfrac{z}{(2k+x)^{2}-z^{2}} =π4(cotπ(zx)2+cotπ(z+x)2)\displaystyle=-\dfrac{\pi}{4}\left(\cot\dfrac{\pi(z-x)}{2}+\cot\dfrac{\pi(z+x)}{2}\right)
=πi2+iπm1eiπmzcosπmx,z\displaystyle=\dfrac{\pi{\rm{i}}}{2}+{\rm{i}}\pi\sum\limits_{{{\mbox{\footnotesize{$m\geqslant 1$}}}}}{\rm{e}}^{{\rm{i}}\pi mz}\cos\pi mx\ ,\qquad z\in\Bb{H}\,,\ x\in\Bb{R}\,,\

we derive from the latter equality that

2πikH0(x+2k)=π411Θ3(i+t)4(cotπ(i+tx)2+cotπ(i+t+x)2)dt\displaystyle 2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\sum_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\eurm{H}_{0}(x+2k)=-\dfrac{\pi}{4}\int\limits_{-1}^{1}\!\Theta_{3}\left({\rm{i}}+t\right)^{4}\left(\cot\dfrac{\pi({\rm{i}}+t-x)}{2}+\cot\dfrac{\pi({\rm{i}}+t+x)}{2}\right){\rm{d}}t
=11(1+2n1eπn2eiπn2t)4(πi2+iπm1eπmeiπmtcosπmx)dt=iπ.\displaystyle=\int\limits_{-1}^{1}\!\left(1\!+\!2\sum\limits_{n\geqslant 1}{\rm{e}}^{-\pi n^{2}}{\rm{e}}^{{\rm{i}}\pi n^{2}t}\right)^{\!4}\left(\dfrac{\pi{\rm{i}}}{2}+{\rm{i}}\pi\sum\limits_{{{\mbox{\footnotesize{$m\geqslant 1$}}}}}{\rm{e}}^{-\pi m}{\rm{e}}^{{\rm{i}}\pi mt}\cos\pi mx\right){\rm{d}}t={\rm{i}}\pi\,.

This proves Lemma 4.2(3) and completes the proof of Lemma 4.2.

10.4 . Proofs for Section 5

  

10.4.1 . Proofs of (5.9) and (5.32).

We first prove (5.9). In the notation

I:=(1001),T:=(1101),S:=(0110),ϕ(abcd)(z):=az+bcz+d,\displaystyle I:=\begin{pmatrix}1&0\\ 0&1\\ \end{pmatrix}\,,\ \ T:=\begin{pmatrix}1&1\\ 0&1\\ \end{pmatrix}\,,\ \ S:=\begin{pmatrix}0&1\\ -1&0\\ \end{pmatrix}\,,\ \ \phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}(z):=\dfrac{az+b}{cz+d}\,, (10.53)

where zz\in\Bb{H} and a,b,c,da,b,c,d\!\in\!\Bb{Z} such that adbc0ad-bc\!\neq\!0, we obviously have

S2=I,T2n=(02n01),ST2n=(0112n),ϕT2n(z)=z2nϕST2nϝϝ\displaystyle\begin{array}[]{l}\displaystyle S^{2}=-I\,,\quad T^{-2n}=\begin{pmatrix}0&-2n\\ 0&1\\ \end{pmatrix}\,,\quad ST^{-2n}=\begin{pmatrix}0&1\\ -1&2n\\ \end{pmatrix}\,,\\[11.38092pt] \displaystyle\phi_{{{\mbox{\footnotesize{$T^{{{\mbox{\footnotesize{$-2n$}}}}}$}}}}}(z)=z-2n\in\Bb{H}\,,\quad\phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2n$}}}}}$}}}}}(z)=\dfrac{1}{2n-z}\in\Bb{H}\,,\quad n\in\Bb{Z}\,.\end{array} (10.56)

Let ϕΓϑ||\phi\!\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}. Then, by the definitions (5.1) and (5.2), we have

ϕ(z)=ϕ𝔫(z)=ϕnN,,n1(z)=ϕST2nNST2nN1ST2n1(z),\displaystyle\phi(z)\!=\!\phi_{\eufm{n}}(z)\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)=\phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2n_{N}$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}$}}}}}(z)\ , (10.57)

for some 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\cup_{k\geqslant 1}\Bb{Z}_{\neq 0}^{k}, NN\in\Bb{N} and every zz\in\Bb{H}.

To prove the right-hand side inclusion of (5.9), which by (5.8) can be written in the form

ϕ𝔫(𝕜𝕜)𝕜𝕜𝔽\displaystyle\phi_{\eufm{n}}\left(\Bb{H}_{|{\rm{Re}}|\leqslant 1}\right)\!\subset\!\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\ \ \eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,,\ (10.58)

observe that 𝕜𝕜𝕜𝕜\Bb{H}_{|{\rm{Re}}|\leqslant 1}-2n\subset\Bb{H}_{|{\rm{Re}}|\geqslant 1} for any nn\in\Bb{Z}_{\neq 0}, and therefore it follows from 1/𝕜𝕜𝕜𝕜𝔽-1/\Bb{H}_{|{\rm{Re}}|\geqslant 1}=\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} that

ϕST2n(𝕜𝕜)𝕜𝕜𝔽𝕜𝕜{}\displaystyle\phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2n$}}}}}$}}}}}\left(\Bb{H}_{|{\rm{Re}}|\leqslant 1}\right)\subset\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\subset\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,,\ \ \eufm{n}\!\in\!\Bb{Z}\!\setminus\!\{0\}\,. (10.59)

Successive applications of (10.59) to (10.57) prove (10.58) and the right-hand side inclusion of (5.9) as well.

For 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N} and NN\!\in\!\Bb{N} from (10.57) introduce the matrices

{(p𝔫1q𝔫1p𝔫0q𝔫0):=(1001),(p𝔫k1q𝔫k1p𝔫kq𝔫k):=(0112nk)(0112nk1)(0112n1), 1kN,\displaystyle\left\{\begin{array}[]{ll}\displaystyle\begin{pmatrix}p^{\eufm{n}}_{-1}&q^{\eufm{n}}_{-1}\\ p^{\eufm{n}}_{0}&q^{\eufm{n}}_{0}\\ \end{pmatrix}:=\begin{pmatrix}1&0\\ 0&1\\ \end{pmatrix}\ ,\\[14.22636pt] \displaystyle\begin{pmatrix}p^{\eufm{n}}_{k-1}&q^{\eufm{n}}_{k-1}\\ p^{\eufm{n}}_{k}&q^{\eufm{n}}_{k}\\ \end{pmatrix}\!:=\!\begin{pmatrix}0&1\\ -1&2n_{k}\\ \end{pmatrix}\begin{pmatrix}0&1\\ -1&2n_{k-1}\\ \end{pmatrix}\ldots\begin{pmatrix}0&1\\ -1&2n_{1}\\ \end{pmatrix},\ 1\!\leqslant\!k\!\leqslant\!N\,,\end{array}\right. (10.62)

whose elements obviously satisfy

p𝔫k1q𝔫kp𝔫kq𝔫k1=1,0kN,\displaystyle p^{\eufm{n}}_{k-1}q^{\eufm{n}}_{k}-p^{\eufm{n}}_{k}q^{\eufm{n}}_{k-1}=1\ ,\ \quad\quad 0\leqslant k\leqslant N\ , (10.63)

and

{p𝔫k=2nkp𝔫k1p𝔫k2,p𝔫0=0,p𝔫1=1,q𝔫k=2nkq𝔫k1q𝔫k2,q𝔫0=1,q𝔫1=0,1kN.\displaystyle\left\{\begin{array}[]{lll}p^{\eufm{n}}_{k}=2n_{k}p^{\eufm{n}}_{k-1}-p^{\eufm{n}}_{k-2}\ ,&\ p^{\eufm{n}}_{0}=0\ ,&\ p^{\eufm{n}}_{-1}=1\ ,\\ q^{\eufm{n}}_{k}=2n_{k}q^{\eufm{n}}_{k-1}-\,q^{\eufm{n}}_{k-2}\ ,&\ q^{\eufm{n}}_{0}=1\ ,&\ q^{\eufm{n}}_{-1}=0\ ,\end{array}\right.\quad\quad 1\leqslant k\leqslant N\,. (10.66)

In the notation (5.8) and (10.57), we have (see (bh1, , p.​ 63))

ϕ𝔫(z)=zp𝔫N1+q𝔫N1zp𝔫N+q𝔫N,𝔫=(𝔫𝔑,,𝔫1)ϝ\displaystyle\phi_{\eufm{n}}(z)\!=\!\dfrac{zp^{\eufm{n}}_{N-1}+q^{\eufm{n}}_{N-1}}{zp^{\eufm{n}}_{N}+q^{\eufm{n}}_{N}}\ ,\ \eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\!\in\!\Bb{N}\,,\ z\in\Bb{H}\,, (10.67)

and (ϕ=ϕ𝔫\phi\!=\!\phi_{\eufm{n}})

|aϕ|=|p𝔫N1|,|bϕ|=|q𝔫N1|,|cϕ|=|p𝔫N|,|dϕ|=|q𝔫N|.\displaystyle\left|a_{\phi}\right|=\left|p^{\eufm{n}}_{N-1}\right|\,,\ \left|b_{\phi}\right|=\left|q^{\eufm{n}}_{N-1}\right|\,,\ \left|c_{\phi}\right|=\left|p^{\eufm{n}}_{N}\right|\,,\ \left|d_{\phi}\right|=\left|q^{\eufm{n}}_{N}\right|\,. (10.68)

We prove that

(a)|q𝔫k|>|p𝔫k|>|p𝔫k1|,(b)|q𝔫k|>|q𝔫k1|>|p𝔫k1|,1kN.\displaystyle{\rm{(a)}}\ \ \big{|}q^{\eufm{n}}_{k}\big{|}>\big{|}p^{\eufm{n}}_{k}\big{|}>\big{|}p^{\eufm{n}}_{k-1}\big{|}\ ,\quad{\rm{(b)}}\ \ \big{|}q^{\eufm{n}}_{k}\big{|}>\big{|}q^{\eufm{n}}_{k-1}\big{|}>\big{|}p^{\eufm{n}}_{k-1}\big{|}\ ,\quad 1\leqslant k\leqslant N\,. (10.69)

For arbitrary positive integer QQ, real numbers a,ba,b satisfying |a|<|b||a|<|b| and nonzero integers {mk}k=1Q\{m_{k}\}_{k=1}^{Q}\subset\Bb{Z}_{\neq 0}, we define the following collection {yk}k=1Q\{y_{k}\}_{k=-1}^{Q} of real numbers

yk=2mkyk1yk2,y0=b,y1=a,1kQ,|a|<|b|.\displaystyle y_{k}=2m_{k}y_{k-1}-\,y_{k-2}\ ,\ \ y_{0}=b\ ,\ y_{-1}=a\ ,\ \quad 1\leqslant k\leqslant Q\,,\ \ |a|<|b|\ . (10.70)

We state that

0<|b|=|y0|<|y1|<<|yQ|.\displaystyle 0<|b|=|y_{0}|<|y_{1}|<\ldots<|y_{Q}|\ . (10.71)

We prove |yk1|<|yk||y_{k-1}|<|y_{k}|, 1kQ1\leqslant k\leqslant Q, by induction on kk. For k=1k=1 such an inequality holds because |y1|=|2m1ba|2|m1||b||a|>2|m1||b||b|=(2|m1|1)|b||b||y_{1}|=|2m_{1}b-\,a|\geqslant 2\cdot|m_{1}|\cdot|b|-|a|>2\cdot|m_{1}|\cdot|b|-|b|=(2\cdot|m_{1}|-1)|b|\geqslant|b|, and hence, |b|=|y0|<|y1||b|=|y_{0}|<|y_{1}|. This also completely proves (10.71) for Q=1Q=1. If Q2Q\geqslant 2 we assume that |yk2|<|yk1||y_{k-2}|<|y_{k-1}| for all 2kP2\leqslant k\leqslant P for some 2PQ2\leqslant P\leqslant Q. Then |yP|=|2mPyP1yP2|2|mP||yP1||yP2|>2|mP||yP1||yP1|=(2|mP|1)|yP1||yP1||y_{P}|=|2m_{P}y_{P-1}-\,y_{P-2}|\geqslant 2\cdot|m_{P}|\cdot|y_{P-1}|-|y_{P-2}|>2\cdot|m_{P}|\cdot|y_{P-1}|-|y_{P-1}|=(2\cdot|m_{P}|-1)|y_{P-1}|\geqslant|y_{P-1}|, i.e. |yP1|<|yP||y_{P-1}|<|y_{P}|. By induction, we conclude that (10.71) is true.

For k=1k=1 (10.69) holds because p𝔫0=0p^{\eufm{n}}_{0}=0, q𝔫0=1q^{\eufm{n}}_{0}=1, p𝔫1=1p^{\eufm{n}}_{1}=-1 and q𝔫1=2n1q^{\eufm{n}}_{1}=2n_{1}, in view of (10.62).

It remains to examine in (10.69) the case 2kN2\leqslant k\leqslant N. By setting first Q:=N1Q:=N-1, a:=0a:=0, b:=1b:=-1, yk:=p𝔫k+1y_{k}:=p^{\eufm{n}}_{k+1}, 1kN1-1\leqslant k\leqslant N-1, mk:=nk+1m_{k}:=n_{k+1}, 1kN11\leq k\leqslant N-1, in (10.70) and then Q:=NQ:=N, a:=0a:=0, b:=1b:=1, yk:=q𝔫ky_{k}:=q^{\eufm{n}}_{k}, 1kN-1\leq k\leqslant N, mk:=nkm_{k}:=n_{k}, 1kN1\leqslant k\leqslant N, we deduce from (10.66) and (10.71) that

|q𝔫k|>|q𝔫k1|1, 1kN;|p𝔫k|>|p𝔫k1|1, 2kN,|p𝔫1|>|p𝔫0|,\displaystyle|q^{\eufm{n}}_{k}|\!>\!|q^{\eufm{n}}_{k-1}|\!\geqslant\!1\,,\ 1\!\leqslant\!k\!\leqslant\!N\,;\ \ |p^{\eufm{n}}_{k}|\!>\!|p^{\eufm{n}}_{k-1}|\!\geqslant\!1\ ,\ 2\!\leqslant\!k\!\leqslant\!N\ ,\ \ |p^{\eufm{n}}_{1}|\!>\!|p^{\eufm{n}}_{0}|\,, (10.72)

where the last inequality follows from p𝔫0=0p^{\eufm{n}}_{0}=0 and p𝔫1=1p^{\eufm{n}}_{1}=-1.

For positive integer 2rN2\leqslant r\leqslant N let 𝔫(𝔯):=(𝔫1,𝔫2,,𝔫𝔯)\eufm{n}(r\!):=(-n_{1},-n_{2},\ldots,-n_{r})\!\in\!\Bb{Z}_{\neq 0}^{r}. Introducing by the formulas (10.62) matrices corresponding to 𝔫(𝔯)\eufm{n}(r\!), we get

Ak𝔫(𝔯):=(pk1𝔫(𝔯)qk1𝔫(𝔯)pk𝔫(𝔯)qk𝔫(𝔯))=(0112nrk+1)(0112nrk+2)(0112nr)\displaystyle A_{k}^{\eufm{n}(r)}:=\begin{pmatrix}p_{k-1}^{\eufm{n}(r)}&q_{k-1}^{\eufm{n}(r)}\\ p_{k}^{\eufm{n}(r)}&q_{k}^{\eufm{n}(r)}\\ \end{pmatrix}=\begin{pmatrix}0&1\\ -1&-2n_{r-k+1}\\ \end{pmatrix}\begin{pmatrix}0&1\\ -1&-2n_{r-k+2}\\ \end{pmatrix}\ldots\begin{pmatrix}0&1\\ -1&-2n_{r}\\ \end{pmatrix}

for every 1kr1\leqslant k\leqslant r, where, in view of (10.66) and (10.72),

|pk𝔫(𝔯)|>|pk1𝔫(𝔯)|,|qk𝔫(𝔯)|>|qk1𝔫(𝔯)|,1kr.\displaystyle\big{|}p_{k}^{\eufm{n}(r)}\big{|}>\big{|}p_{k-1}^{\eufm{n}(r)}\big{|}\,,\qquad\big{|}q_{k}^{\eufm{n}(r)}\big{|}>\big{|}q_{k-1}^{\eufm{n}(r)}\big{|}\,,\qquad 1\leqslant k\leqslant r\ . (10.73)

Applying the transpose operation to Ar𝔫(𝔯)A_{r}^{\eufm{n}(r)}, we obtain

(1)r(pr1𝔫(𝔯)qr1𝔫(𝔯)pr𝔫(𝔯)qr𝔫(𝔯))=(0112n1)(0112n2)(0112nr)=(p𝔫r1q𝔫r1p𝔫rq𝔫r)T=(p𝔫r1p𝔫rq𝔫r1q𝔫r),\displaystyle(-1)^{r}\!\begin{pmatrix}p_{r-1}^{\eufm{n}(r)}&q_{r-1}^{\eufm{n}(r)}\\ p_{r}^{\eufm{n}(r)}&q_{r}^{\eufm{n}(r)}\\ \end{pmatrix}\!=\!\begin{pmatrix}0&-1\\ 1&2n_{1}\\ \end{pmatrix}\begin{pmatrix}0&-1\\ 1&2n_{2}\\ \end{pmatrix}\ldots\begin{pmatrix}0&-1\\ 1&2n_{r}\\ \end{pmatrix}\!=\!\begin{pmatrix}p^{\eufm{n}}_{r-1}&q^{\eufm{n}}_{r-1}\\ p^{\eufm{n}}_{r}&q^{\eufm{n}}_{r}\\ \end{pmatrix}^{\!\!{\rm{T}}}\!\!\!=\!\begin{pmatrix}p^{\eufm{n}}_{r-1}&p^{\eufm{n}}_{r}\\ q^{\eufm{n}}_{r-1}&q^{\eufm{n}}_{r}\\ \end{pmatrix},

from which,

(1)rpr1𝔫(𝔯)=p𝔫r1,(1)rpr𝔫(𝔯)=q𝔫r1,(1)rqr1𝔫(𝔯)=p𝔫r,(1)rqr𝔫(𝔯)=q𝔫r,\displaystyle\begin{array}[]{rclrcl}(-1)^{r}p_{r-1}^{\eufm{n}(r)}&=&p^{\eufm{n}}_{r-1}\,,&\quad(-1)^{r}p_{r}^{\eufm{n}(r)}&=&q^{\eufm{n}}_{r-1}\,,\\[8.5359pt] (-1)^{r}q_{r-1}^{\eufm{n}(r)}&=&p^{\eufm{n}}_{r}\,,&\quad(-1)^{r}q_{r}^{\eufm{n}(r)}&=&q^{\eufm{n}}_{r}\,,\end{array} (10.76)

and therefore, by arbitrariness of 2rN2\!\leqslant\!r\!\leqslant\!N, we deduce from (10.73) that

|p𝔫k|<|q𝔫k|,1kN.\displaystyle\left|p^{\eufm{n}}_{k}\right|<\left|q^{\eufm{n}}_{k}\right|\ ,\quad 1\leqslant k\leqslant N\,. (10.77)

Together with (10.72) and (10.68) this proves the required inequalities |aϕ|<|bϕ|<|dϕ||a_{\phi}|\!<\!|b_{\phi}|\!<\!|d_{\phi}|,|aϕ|<|cϕ|<|dϕ||a_{\phi}|\!<\!|c_{\phi}|\!<\!|d_{\phi}| in the left-hand side of (5.9) (cp. (boc, , p.​ 4, Lemma 2)).

Besides that, the relationships (10.66) written for 𝔫(r){\eufm{n}}(r) and k=rk=r give

q𝔫r1=(1)rp𝔫(𝔯)r=(1)r(2n1p𝔫(𝔯)r1p𝔫(𝔯)r2)=2n1p𝔫r1p𝔫(𝔯)r2,|p𝔫(𝔯)r2|<|p𝔫r1|,q𝔫r=(1)rq𝔫(𝔯)r=(1)r(2n1q𝔫(𝔯)r1q𝔫(𝔯)r2)=2n1p𝔫rq𝔫(𝔯)r2,|q𝔫(𝔯)r2|<|p𝔫r|,\displaystyle\begin{array}[]{rccccclr}q^{\eufm{n}}_{r-1}\!&\!=\!&(-1)^{r}p^{\eufm{n}(r)}_{r}&\!=\!&(-1)^{r}\left(\!-\!2n_{1}p^{\eufm{n}(r)}_{r-1}\!-\!p^{\eufm{n}(r)}_{r-2}\right)&\!=\!&-\!2n_{1}p^{\eufm{n}}_{r-1}\!-\!p^{\eufm{n}(r)}_{r-2}\,,&|p^{\eufm{n}(r)}_{r-2}|<|p^{\eufm{n}}_{r-1}|,\\[8.5359pt] q^{\eufm{n}}_{r}&\!=\!&(-1)^{r}q^{\eufm{n}(r)}_{r}&\!=\!&(-1)^{r}\left(\!-\!2n_{1}q^{\eufm{n}(r)}_{r-1}\!-\!q^{\eufm{n}(r)}_{r-2}\right)&\!=\!&-\!2n_{1}p^{\eufm{n}}_{r}\!-\!q^{\eufm{n}(r)}_{r-2}\,,&|q^{\eufm{n}(r)}_{r-2}|<|p^{\eufm{n}}_{r}|,\end{array}

where the inequalities (10.73) have been used. Since rr is arbitrary integer satisfying 2rN2\!\leqslant\!r\!\leqslant\!N, we get

|q𝔫k+2n1p𝔫k||p𝔫k|1,1kN.\displaystyle\left|q^{\eufm{n}}_{k}+2n_{1}p^{\eufm{n}}_{k}\right|\leqslant\left|p^{\eufm{n}}_{k}\right|-1\ ,\quad 1\leqslant k\leqslant N\,. (10.78)

Next, we prove (5.32). Let NN\!\in\!\Bb{N} and 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}. Then in the notation (10.62) we have

(a)sign(q𝔫k)sign(q𝔫k1)=σk,(b)sign(p𝔫k)sign(q𝔫k)=σ1,σk:=sign(nk),1kN.\displaystyle\begin{array}[]{ll}{\rm{(a)}}\ \ {\rm{sign}}\,(q^{\eufm{n}}_{k})\cdot{\rm{sign}}\,(q^{\eufm{n}}_{k-1})=\sigma_{k}\ ,&\ \ {\rm{(b)}}\ \ {\rm{sign}}\,(p^{\eufm{n}}_{k})\cdot{\rm{sign}}\,(q^{\eufm{n}}_{k})=-\sigma_{1}\ ,\\ \phantom{{\rm{(a)}}\ \ }\sigma_{k}:={\rm{sign}}\,(n_{k})\ ,&\ \ \phantom{{\rm{(b)}}\ \ }1\leqslant k\leqslant N\,.\end{array} (10.81)

Indeed, (10.72) and (10.66) yield that q𝔫k/q𝔫k1(2nk1,2nk+1)q^{\eufm{n}}_{k}/q^{\eufm{n}}_{k-1}\in(2n_{k}-1,2n_{k}+1) for each 1kN1\leqslant k\leqslant N and hence, (10.81)(a) holds. To prove (10.81)(b) assume to the contrary that there exists 1kN1\leqslant k\leqslant N such that sign(p𝔫k)sign(q𝔫k)=σ1{\rm{sign}}\,(p^{\eufm{n}}_{k})\cdot{\rm{sign}}\,(q^{\eufm{n}}_{k})=\sigma_{1}. Then |q𝔫k+2n1p𝔫k|=|q𝔫k|+2|n1||p𝔫k|2|p𝔫k|+1>|p𝔫k|1|q^{\eufm{n}}_{k}+2n_{1}p^{\eufm{n}}_{k}|=|q^{\eufm{n}}_{k}|+2|n_{1}||p^{\eufm{n}}_{k}|\geqslant 2|p^{\eufm{n}}_{k}|+1>|p^{\eufm{n}}_{k}|-1 which contradicts (10.78). This contradiction proves (10.81)(b). Observe, that (10.81) for N2N\geqslant 2 yields

p𝔫0=0,p𝔫1=1,sign(p𝔫k)sign(p𝔫k1)=σk,2kN.\displaystyle p^{\eufm{n}}_{0}=0\ ,\quad p^{\eufm{n}}_{1}=-1\ ,\quad{\rm{sign}}\,(p^{\eufm{n}}_{k})\cdot{\rm{sign}}\,(p^{\eufm{n}}_{k-1})=\sigma_{k}\ ,\quad 2\leqslant k\leqslant N\,. (10.82)

Then we deduce from (10.67) and (10.81) that (in accordance with (5.27), we have σ𝔫=σ1\sigma_{\eufm{n}}=\sigma_{1})

ϕ𝔫(σ𝔫)=q𝔫N1σ1p𝔫N1q𝔫Nσ1p𝔫N=σN|q𝔫N1|+|p𝔫N1||q𝔫N|+|p𝔫N|,ϕ𝔫()=p𝔫N1p𝔫N,ϕ𝔫(σ𝔫)=q𝔫N1+σ1p𝔫N1q𝔫N+σ1p𝔫N=σN|q𝔫N1||p𝔫N1||q𝔫N||p𝔫N|,ϕ𝔫( 0)=q𝔫N1q𝔫N.\displaystyle\begin{array}[]{lcccclcl}\phi_{\eufm{n}}(-\sigma_{\eufm{n}})&=&\dfrac{q^{\eufm{n}}_{N-1}-\sigma_{1}p^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}-\sigma_{1}p^{\eufm{n}}_{N}}&=&\sigma_{N}\,\dfrac{|q^{\eufm{n}}_{N-1}|+|p^{\eufm{n}}_{N-1}|}{|q^{\eufm{n}}_{N}|+|p^{\eufm{n}}_{N}|}\ ,&\phi_{\eufm{n}}(\infty)&=&\dfrac{p^{\eufm{n}}_{N-1}}{p^{\eufm{n}}_{N}}\ ,\\[14.22636pt] \phi_{\eufm{n}}(\sigma_{\eufm{n}})&=&\dfrac{q^{\eufm{n}}_{N-1}+\sigma_{1}p^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}+\sigma_{1}p^{\eufm{n}}_{N}}&=&\sigma_{N}\,\dfrac{|q^{\eufm{n}}_{N-1}|-|p^{\eufm{n}}_{N-1}|}{|q^{\eufm{n}}_{N}|-|p^{\eufm{n}}_{N}|}\ ,&\phi_{\eufm{n}}(\,0\,)&=&\dfrac{q^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}}\ .\end{array} (10.85)

It follows from (10.77) that q𝔫N(q𝔫N±p𝔫N)>0q^{\eufm{n}}_{N}(q^{\eufm{n}}_{N}\pm p^{\eufm{n}}_{N})>0 and therefore we obtain the inequalities ϕ𝔫(1)<ϕ𝔫(0)<ϕ𝔫(1)\phi_{\eufm{n}}(-1)\!<\!\phi_{\eufm{n}}(0)\!<\!\phi_{\eufm{n}}(1) of (5.32)(a) , because (10.63) gives

ϕ𝔫(0)ϕ𝔫(1)=1/(q𝔫N(q𝔫Np𝔫N)),ϕ𝔫(1)ϕ𝔫(0)=1/(q𝔫N(q𝔫N+p𝔫N)).\displaystyle\phi_{\eufm{n}}(0)\!-\!\phi_{\eufm{n}}(-1)\!=\!{1}\big{/}\left(q^{\eufm{n}}_{N}(q^{\eufm{n}}_{N}\!-\!p^{\eufm{n}}_{N})\right),\ \phi_{\eufm{n}}(1)\!-\!\phi_{\eufm{n}}(0)\!=\!{1}\big{/}\left(q^{\eufm{n}}_{N}(q^{\eufm{n}}_{N}\!+\!p^{\eufm{n}}_{N})\right). (10.86)

At the same time, for N=1N=1 we have σ𝔫ϕ𝔫(σ𝔫)=σ1ϕ𝔫(σ𝔫)>0=ϕ𝔫()\sigma_{\eufm{n}}\phi_{\eufm{n}}(-\sigma_{\eufm{n}})=\sigma_{1}\phi_{\eufm{n}}(-\sigma_{\eufm{n}})>0=\phi_{\eufm{n}}(\infty), as follows from (10.85) and (10.66). This proves (5.32)(b) for N=1N=1. But if N2N\geqslant 2 then we deduce from (10.63) that the sign of the number ϕ𝔫(σ𝔫)ϕ𝔫()=1/(p𝔫N(q𝔫Nσ1p𝔫N))\phi_{\eufm{n}}(-\sigma_{\eufm{n}})-\phi_{\eufm{n}}(\infty)=-1/(p^{\eufm{n}}_{N}(q^{\eufm{n}}_{N}-\sigma_{1}p^{\eufm{n}}_{N})) is equal to sign(p𝔫Nq𝔫N)=σ1{\rm{sign}}(p^{\eufm{n}}_{N}q^{\eufm{n}}_{N})=-\sigma_{1}, in view of (10.77) and (10.81)(b). This proves (5.32)(b) for N2N\geqslant 2. Since (5.32)(c) is immediate from (10.69)(a) and (10.85), the proof of (5.32) is completed. Hence, for NN\!\in\!\Bb{N} and 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, the hyperbolic quadrilateral 𝔫\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}} has the shape shown in Figure 3, provided that n11n_{1}\geqslant 1 (in view of (10.81)(b), n11n_{1}\geqslant 1 yields p𝔫Nq𝔫N1p^{\eufm{n}}_{N}\,q^{\eufm{n}}_{N}\leqslant-1).

1-111p𝔫N1p𝔫N\dfrac{\vphantom{A^{A}}p^{\eufm{n}}_{N-1}}{p^{\eufm{n}}_{N}}\|ϕnN,,n1()\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(\infty)q𝔫N1p𝔫N1q𝔫Np𝔫N\dfrac{\vphantom{A^{A}}q^{\eufm{n}}_{N-1}-p^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}-p^{\eufm{n}}_{N}}\|ϕnN,,n1(1)\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(-1)q𝔫N1q𝔫N\dfrac{\vphantom{A^{A}}q^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}}\|ϕnN,,n1(0)\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(0)q𝔫N1+p𝔫N1q𝔫N+p𝔫N\dfrac{\vphantom{A^{A}}q^{\eufm{n}}_{N-1}+p^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}+p^{\eufm{n}}_{N}}\|ϕnN,,n1(1)\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(1)nN,,n1:=ϕnN,,n1()\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\!:=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)}γnN,,n1(1,1){{\gamma_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(-1,1)}}γnN,,n1(1,){{\gamma_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(1,\infty)}}n11n_{1}\!\geqslant\!1
Figure 3: . Image nN,,n1\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}} of \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} by ϕ𝔫(z)=zp𝔫N1+q𝔫N1zp𝔫N+q𝔫N\phi_{\eufm{n}}(z)\!=\!\dfrac{zp^{\eufm{n}}_{N-1}+q^{\eufm{n}}_{N-1}}{zp^{\eufm{n}}_{N}+q^{\eufm{n}}_{N}}, 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, NN\!\in\!\Bb{N}.

10.4.2 . Proofs of Lemmas 5.3 and 5.4.

We first prove Lemma 5.3.

(a) Assume the contrary, i.e., there exist N,MN,M\!\in\!\Bb{N}, NMN\!\geqslant\!M, 𝔫:=(𝔫𝔑,,𝔫1)\eufm{n}\!:=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, 𝔪:=(𝔪𝔑,,𝔪1)()𝔐\eufm{m}\!:=\!(m_{N},...,m_{1})\!\in\!\left(\Bb{Z}_{\neq 0}\right)^{M}, 𝔫𝔪\eufm{n}\neq\eufm{m}, z(σn1+i)z\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\left(\sigma_{n_{1}}\!+\!{\rm{i}}\Bb{R}_{>0}\right), y(σm1+i)y\!\in\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\sqcup\left(\sigma_{m_{1}}\!+\!{\rm{i}}\Bb{R}_{>0}\right) such that ϕnN,,n1(z)=ϕmM,,m1(y)\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\big{(}z\big{)}=\phi_{{{\mbox{\footnotesize{$m_{M},...,m_{1}$}}}}}\big{(}y\big{)}, which by (10.57) can be written as follows (see (5.27))

ϕST2nNST2nN1ST2n1(z)=ϕST2mMST2mM1ST2m1(y).\displaystyle\phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2n_{N}$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}$}}}}}\!(z)\!=\!\phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2m_{M}$}}}}}ST^{{{\mbox{\footnotesize{$-2m_{M-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2m_{1}$}}}}}$}}}}}\!(y). (10.87)

Without loss of generality one can consider in (10.87) that nNmMn_{N}\neq m_{M}, because otherwise, we can apply ϕT2nNS\phi_{T^{{{2n_{\!N}}}}S} to the both parts of (10.87) to get the similar relationship for 𝔫\eufm{n} and 𝔪\eufm{m} of lower dimension and to get 𝔫=𝔪\eufm{n}=\eufm{m}, if M=NM=N, or

𝕜𝕜𝔽ϕST2nNMST2n1ϝ𝔽|σif𝕄\displaystyle\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni\phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2n_{N-M}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}$}}}}}(z)=y\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{m_{1}}}\,,\ \ \mbox{if}\ N-M\geqslant 1\,, (10.88)

where the left-hand side inclusion follows from (5.9). But (10.88) cannot hold since |σm1(𝕜𝕜𝔽\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{m_{1}}}\cap(\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})=\emptyset. So that we assume everywhere below that nNmMn_{N}\neq m_{M}.

Let N=M=1N=M=1 in (10.87). Then zy=2n12m1z-y=2n_{1}-2m_{1}, where n1m1n_{1}\neq m_{1}. Since Rez,Rey[1,1]{\rm{Re}}\,z,{\rm{Re}}\,y\in[-1,1] then n1m1n_{1}-m_{1} can take only two values 11 or 1-1. If n1m1=σ{1,1}n_{1}-m_{1}=\sigma\in\{1,-1\} then there exists T>0T>0 such that z=σ+iTz=\sigma+{\rm{i}}T and y=σ+iTy=-\sigma+{\rm{i}}T, which yields σn1=σ\sigma_{n_{1}}=\sigma and σm1=σ\sigma_{m_{1}}=-\sigma, by virtue of (5.27) and (5.28). But then |n1m1|2|n_{1}-m_{1}|\geqslant 2 which contradicts |n1m1|=1|n_{1}-m_{1}|=1. Therefore N2N\geqslant 2, M1M\geqslant 1 and it follows from (10.87) that

ϕST2nN1ST2n1(z)=ϕT2(m1nN)(y),ifM=1,ϕST2(m2nN)ST2nN1ST2n1(z)=ϕT2m1(y),ifM=2,ϕST2m2T2mM1ST2(mMnN)ST2nN1ST2n1(z)=ϕT2m1(y),\displaystyle\begin{array}[]{l}\phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}$}}}}}(z)=\phi_{{{\mbox{\footnotesize{$T^{{{\mbox{\footnotesize{$-2(m_{1}-n_{N})$}}}}}$}}}}}(y)\,,\ \ \mbox{if}\ M=1\,,\\ \phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$-2(m_{2}-n_{N})$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}$}}}}}(z)=\phi_{{{\mbox{\footnotesize{$T^{{{\mbox{\footnotesize{$-2m_{1}$}}}}}$}}}}}(y)\,,\ \ \mbox{if}\ M=2\,,\\ \phi_{{{\mbox{\footnotesize{$ST^{{{\mbox{\footnotesize{$2m_{2}$}}}}}\ldots T^{{{\mbox{\footnotesize{$2m_{M-1}$}}}}}ST^{{{\mbox{\footnotesize{$-2(m_{M}-n_{N})$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}$}}}}}(z)=\phi_{{{\mbox{\footnotesize{$T^{{{\mbox{\footnotesize{$-2m_{1}$}}}}}$}}}}}(y)\,,\end{array}

if M3M\geqslant 3. But none of these equations can hold because their left-hand sides belong to 𝕜𝕜𝔽\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, in view of (5.9), while their right-hand sides belong to 𝕜𝕜\Bb{H}_{|{\rm{Re}}|\geqslant 1}. Lemma 5.3(a) is proved.

(d) Let 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{N} be such that N2N\geqslant 2 if n1=σn1=σ𝔫n_{1}\!=\!\sigma_{n_{1}}=\sigma_{\eufm{n}} (see (5.27)). As noted before Lemma 5.3, ϕ𝔫(γ(σ𝔫,))^𝔫\phi_{\eufm{n}}\big{(}\gamma(\sigma_{\eufm{n}},\infty)\big{)}\subset\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}} is the roof of ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}. But ϕ𝔫(γ(σ𝔫,))\phi_{\eufm{n}}\big{(}\gamma(\sigma_{\eufm{n}},\infty)\big{)} is also a part of the boundary of

ϕnN,,n1(+2σ𝔫)\displaystyle\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!+\!2\sigma_{\eufm{n}}\right) =ϕnN,,n2(12n12σ𝔫)\displaystyle\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\left(\dfrac{1}{2n_{1}\!-\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!-\!2\sigma_{\eufm{n}}}\right)
|1/=12n2+1/=12n2|\displaystyle\left|-1/\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!=\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,\Rightarrow\,\dfrac{1}{2n_{2}\!+\!1/\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}}\!=\!\dfrac{1}{2n_{2}\!-\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}}\right| ={ϕnN,,n2,n1σ𝔫(),ifn1σ𝔫,ϕnN,,n2(),ifn1=σ𝔫,\displaystyle\!=\!\left\{\begin{array}[]{ll}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2},n_{1}-\sigma_{\eufm{n}}$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),&\hbox{if}\ \,n_{1}\!\neq\!\sigma_{\eufm{n}},\\ \phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),&\hbox{if}\ \,n_{1}\!=\!\sigma_{\eufm{n}},\end{array}\right.

and since

ϕnN,,n1(γ(σ𝔫,))\displaystyle\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\gamma(\sigma_{\eufm{n}},\infty)\right) =ϕnN,,n2(12n1γ(σ𝔫,)2σ𝔫)\displaystyle=\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\left(\dfrac{1}{2n_{1}\!-\!\gamma(-\sigma_{\eufm{n}},\infty)\!-\!2\sigma_{\eufm{n}}}\right)
|12n2+1/γ(σ𝔫,)=12n2γ(σ𝔫,0)|\displaystyle\left|\dfrac{1}{2n_{2}\!+\!1/\gamma(-\sigma_{\eufm{n}},{{\mbox{\footnotesize{$\infty$}}}})}\!=\!\dfrac{1}{2n_{2}\!-\!\gamma(\sigma_{\eufm{n}},0)}\right| ={ϕnN,,n2,n1σ𝔫(γ(σ𝔫,)),ifn1σ𝔫,ϕnN,,n2(γ(σ𝔫,0)),ifn1=σ𝔫,\displaystyle\!=\!\left\{\begin{array}[]{ll}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2},n_{1}\!-\!\sigma_{\eufm{n}}$}}}}}\!\left(\gamma(-\sigma_{\eufm{n}},{{\mbox{\footnotesize{$\infty$}}}})\right),&\hbox{if}\,\,n_{1}\!\neq\!\sigma_{\eufm{n}},\\ \phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\!\left(\gamma(\sigma_{\eufm{n}},0)\right),&\hbox{if}\,\,n_{1}\!=\!\sigma_{\eufm{n}},\end{array}\right.

σnN,,n2,n1σ𝔫=sign(n1σ𝔫)=sign(n1)=σ𝔫\sigma_{\hskip 0.1424ptn_{N},...,n_{2},n_{1}\!-\sigma_{\eufm{n}}}\!=\!{\rm{sign}}(n_{1}\!-\sigma_{\eufm{n}})\!=\!{\rm{sign}}(n_{1})\!=\!\sigma_{\eufm{n}} for n1σ𝔫n_{1}\!\neq\!\sigma_{\eufm{n}}, we get that

ϕ𝔫(γ(σ𝔫,))={γnN,,n2,n1σ𝔫(σnN,,n2,n1σ𝔫,),ifn1σ𝔫,γnN,,n2(σ𝔫,0),σ𝔫{1,1},ifn1=σ𝔫.\displaystyle\phi_{\eufm{n}}\big{(}\gamma(\sigma_{\eufm{n}},\infty)\big{)}=\left\{\begin{array}[]{ll}\gamma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{2},n_{1}\!-\sigma_{\eufm{n}}$}}}}}(-\sigma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{2},n_{1}\!-\sigma_{\eufm{n}}$}}}}},\infty),&\hbox{if}\,\,n_{1}\!\neq\!\sigma_{\eufm{n}}\,,\\[5.69046pt] \gamma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{2}$}}}}}(\sigma_{\eufm{n}},0),\ \ \sigma_{\eufm{n}}\in\{1,-1\}\,,&\hbox{if}\,\,n_{1}\!=\!\sigma_{\eufm{n}}\,.\end{array}\right.

According to what was stated before Lemma 5.3, this means that ϕ𝔫(γ(σ𝔫,))\phi_{\eufm{n}}\big{(}\gamma(\sigma_{\eufm{n}},\infty)\big{)} is the lower arc of ^nN,,n1σn1\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}\!-\sigma_{n_{1}}}\!\!, if n1σ𝔫n_{1}\!\neq\!\sigma_{\eufm{n}}. At the same time, ϕ𝔫(γ(σ𝔫,))\phi_{\eufm{n}}\big{(}\gamma(\sigma_{\eufm{n}},\infty)\big{)} for n1=σ𝔫n_{1}\!=\!\sigma_{\eufm{n}} is the lower arc of ^nN,,n2\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{2}}. Lemma 5.3(d) is proved. Lemma 5.3(b) and (c) follow by similar arguments.21{}^{\ref*{case18}}

(e) As noted before Lemma 5.3, ϕ𝔫(γ(σ𝔫,))\phi_{\eufm{n}}\big{(}\gamma(\sigma_{\eufm{n}},\infty)\big{)} is the roof of ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}} and since ϕ𝔫(γ(σ𝔫,))=γ(ϕ𝔫(σ𝔫),ϕ𝔫())\phi_{\eufm{n}}\big{(}\gamma(\sigma_{\eufm{n}},\infty)\big{)}=\gamma\big{(}\phi_{\eufm{n}}(\sigma_{\eufm{n}}),\phi_{\eufm{n}}(\infty)\big{)}, by virtue of (5.33), we get

2max{Imz|z^nN,,n1}=|ϕ𝔫()ϕ𝔫(σ𝔫)|,\displaystyle 2\max\left\{\ {\rm{Im}}z\,\left|\,z\in\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\,\right\}\right.=\left|\phi_{\eufm{n}}(\infty)-\phi_{\eufm{n}}(\sigma_{\eufm{n}})\right|\,, (10.89)

where in accordance with (10.85), (10.63), (10.81) and (10.69),

|ϕ𝔫()ϕ𝔫(σ𝔫)|=|p𝔫N1p𝔫Nq𝔫N1+σ1p𝔫N1q𝔫N+σ1p𝔫N|=1|p𝔫N|(|q𝔫N||p𝔫N|)1|p𝔫N|.\displaystyle|\phi_{\eufm{n}}(\infty)\!-\!\phi_{\eufm{n}}(\sigma_{\eufm{n}})|\!=\!\left|\dfrac{p^{\eufm{n}}_{N-1}}{p^{\eufm{n}}_{N}}\!-\!\dfrac{q^{\eufm{n}}_{N-1}\!+\!\sigma_{1}p^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}\!+\!\sigma_{1}p^{\eufm{n}}_{N}}\right|\!=\!\dfrac{1}{\left|p^{\eufm{n}}_{N}\right|\left(\left|q^{\eufm{n}}_{N}\right|\!-\!\left|p^{\eufm{n}}_{N}\right|\right)}\leqslant\dfrac{1}{\left|p^{\eufm{n}}_{N}\right|}\ . (10.90)

We prove now

|p𝔫k|1+m=2k(2|nm|1)k,0kN,\displaystyle\left|p^{\eufm{n}}_{k}\right|\geqslant 1+\sum\nolimits_{m=2}^{k}\,(2|n_{m}|-1)\geqslant k\ ,\quad 0\leqslant k\leqslant N\,, (10.91)

by induction on kk, where m=20:=m=21:=0\sum_{m=2}^{0}:=\sum_{m=2}^{1}:=0 and {p𝔫k}k=1N\{p^{\eufm{n}}_{k}\}_{k=-1}^{N} are defined as in (10.66). In view of (10.66), p𝔫1=1p^{\eufm{n}}_{1}=-1, and therefore (10.91) holds for k=1k=1. In addition, (10.91) is proved for N=1N=1. Assume that N2N\geqslant 2, 2rN2\leqslant r\leqslant N and (10.66) is true for k=r1k=r-1. In view of (10.72), we have

0=|p𝔫0|<|p𝔫1|<|p𝔫2|<<|p𝔫r|.\displaystyle 0=|p^{\eufm{n}}_{0}|<|p^{\eufm{n}}_{1}|<|p^{\eufm{n}}_{2}|<\ldots<|p^{\eufm{n}}_{r}|\ .

Then |p𝔫r2||p𝔫r1|1|p^{\eufm{n}}_{r-2}|\leq|p^{\eufm{n}}_{r-1}|-1 and by (10.66) we get

|p𝔫r|\displaystyle|p^{\eufm{n}}_{r}| =|2nrp𝔫r1p𝔫r2|2|nr||p𝔫r1||p𝔫r2|1+(2|nr|1)|p𝔫r1|\displaystyle=|2n_{r}p^{\eufm{n}}_{r-1}-p^{\eufm{n}}_{r-2}|\geqslant 2|n_{r}||p^{\eufm{n}}_{r-1}|-|p^{\eufm{n}}_{r-2}|\geqslant 1+(2|n_{r}|-1)|p^{\eufm{n}}_{r-1}|
1+(2|nr|1)(1+m=1r1(2|nm|1))1+m=1r(2|nm|1),\displaystyle\geqslant 1+(2|n_{r}|-1)\left(1+\sum\limits_{m=1}^{r-1}(2|n_{m}|-1)\right)\geqslant 1+\sum\limits_{m=1}^{r}(2|n_{m}|-1),

which proves the validity of (10.91) for k=rk=r. By induction, we conclude that (10.91) is true for all 0kN0\leqslant k\leqslant N. Combining (10.91) for k=Nk=N, (10.90) and (10.89) gives Lemma 5.3(e) and completes the proof of Lemma 5.3.

Next we prove Lemma 5.4. Since ^𝔫𝕜𝕜𝔽\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}}\subset\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} for every 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} it suffices to prove that for any point in 𝕜𝕜𝔽\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} there exists 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} such that ^𝔫\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 0.28436pt\eufm{n}} contains this point.

Let z𝕜𝕜𝔽z\in\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, NN\in\Bb{N} and assume that 𝔾ϝ𝕜𝕜𝔽\Bb{G}^{k}_{2}(z)\in\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} for each 1kN1\leqslant k\leqslant N. We denote by Q=Q(z)Q=Q(z) the number (1/2)+((1/4)(Imz)2)1/2>1(1/2)+((1/4)-({\rm{Im}}\,z)^{2})^{1/2}>1. Then by (5.20), QNImz1/2Q^{N}{\rm{Im}}\,z\leqslant 1/2, because 𝕜𝕜𝕀𝔽||\Bb{H}_{|{\rm{Re}}|\leqslant 1}\cap\Bb{H}_{{\rm{Im}}>1/2}\subset\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}. Therefore for arbitrary z𝕜𝕜𝔽z\in\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} there exists a minimal positive integer r(z)\eurm{r}(z) such that

r(z):=min{k𝕜𝔾ϝ𝔽||},z𝕜𝕜𝔽\displaystyle\eurm{r}(z):=\min\left\{\,k\in\Bb{N}\,|\,\Bb{G}^{k}_{2}(z)\in\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\,\right\}\ ,\quad z\in\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,, (10.92)

where, in accordance with (5.14),

𝔾ϝϝ𝔾ϝ{𝔾ϝ}𝕀𝔾ϝjk(z):=12Re(1𝔾ϝ)2ϝ\Bb{G}_{2}^{k}(z)\!=\!2j_{k}(z)\!-\!\dfrac{1}{\Bb{G}_{2}^{k-1}(z)}=\left\{{\rm{Re}}\left(-\dfrac{1}{\Bb{G}_{2}^{k-1}(z)}\right)\right\}^{\!{{\mbox{\tiny{$\rceil\lceil$}}}}}_{\!2}+{\rm{i}}\,{\rm{Im}}\left(-\dfrac{1}{\Bb{G}_{2}^{k-1}(z)}\right),\\ j_{k}(z)\!:=\!-\dfrac{1}{2}\left\rceil{{\rm{Re}}\left(-\dfrac{1}{\Bb{G}_{2}^{k-1}(z)}\right)}\right\lceil_{2}\!\!\in\!\Bb{Z}_{\neq 0}\,,\quad 1\!\leqslant\!k\!\leqslant\!{\eurm{r}}(z). (10.93)

Since ϕjk(z)(𝔾ϝ𝔾ϝ\phi_{j_{k}(z)}(\Bb{G}_{2}^{k}(z))=\Bb{G}_{2}^{k-1}(z) for each 1kr(z)1\!\leqslant\!k\!\leqslant\!{\eurm{r}}(z) we deduce that

ϕj1(z),,jr(z)(z)(𝔾ϝϝϝ𝔾ϝϝ𝔽||ϝ𝕜𝕜𝔽\displaystyle\phi_{j_{1}(z),...,j_{{\eurm{r}}(z)}(z)}(\Bb{G}_{2}^{{\eurm{r}}(z)}(z))=z\ ,\quad\Bb{G}_{2}^{{\eurm{r}}(z)}(z)\in\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\ ,\quad z\in\Bb{H}_{|{\rm{Re}}|<1}\setminus\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ , (10.94)

and if 𝔾ϝϝγσ\Bb{G}_{2}^{{\eurm{r}}(z)}(z)\in\gamma(\sigma,\infty) for some σ{1,1}\sigma\in\{1,-1\}, then (10.93) yields

{Re(1/𝔾ϝ)}2=σ,σ{1,1}.\displaystyle\left\{{\rm{Re}}\left(-{1}\big{/}{\Bb{G}_{2}^{k-1}(z)}\right)\right\}^{\!{{\mbox{\tiny{$\rceil\lceil$}}}}}_{\!2}=\sigma\ ,\quad\sigma\in\{1,-1\}\,.

In view of (5.12), this means that Re(1/𝔾ϝ)σ(2{\rm{Re}}(-{1}\big{/}{\Bb{G}_{2}^{k-1}(z)})\in-\sigma(2\Bb{N}-1) and therefore Re(1/𝔾ϝ)22σ\rceil{{\rm{Re}}(-{1}\big{/}{\Bb{G}_{2}^{k-1}(z)})}\lceil_{2}\,\in-2\sigma\Bb{N}. Then sign(jr(z)(z))=σ{\rm{sign}}(j_{{\eurm{r}}(z)}(z))=\sigma, by virtue of (10.93). Hence, in addition to (10.94), we have

𝔾ϝϝγσσ{}σϝðϝϝσ\displaystyle\Bb{G}_{2}^{{\eurm{r}}(z)}(z)\in\gamma(\sigma,\infty)\,,\ \sigma\in\{1,-1\}\ \Rightarrow\ \sigma_{{\eurm{r}}}(z):={\rm{sign}}(j_{{\eurm{r}}(z)}(z))=\sigma\,. (10.95)

Thus,

zϕj1(z),,jr(z)(z)(|σr(z))=^j1(z),,jr(z)(z),\displaystyle z\in\phi_{j_{1}(z),...,j_{{\eurm{r}}(z)}(z)}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\,\sigma_{{\eurm{r}}}(z)}\right)=\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptj_{1}(z),...,j_{{\eurm{r}}(z)}(z)}\,, (10.96)

which completes the proof of Lemma 5.4.

10.4.3 . Proofs of Theorems 5.5 and 5.9.

  

We first prove Theorem 5.5. Let L:={ϕ(γ(1,))}ϕΓϑL:=\{\phi(\gamma(1,\infty))\}_{\phi\in\Gamma_{\vartheta}}. In accordance with (3.39),

:=𝔽γγγγ\displaystyle\partial_{\,\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!:=\!\Bb{H}\cap\partial\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!=\!\gamma(1,\infty)\sqcup\gamma(-1,\infty)\sqcup\gamma(-1,0)\sqcup\gamma(0,1), (10.97)

and since 1/()=-1/\left(\partial_{\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)=\partial_{\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} we deduce from (5.7) that the union SS of all boundary points of the sets composing the Schwarz partition (5.40) of \Bb{H}, satisfies

S=n(2n+{ϕ𝔫()}𝔫{0})={ϕ()}ϕΓϑL.\displaystyle S\!=\!\bigcup\limits_{n\in\Bb{Z}}\ \left(2n+\big{\{}\phi_{\eufm{n}}\big{(}\partial_{\,\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)}\big{\}}_{{{\mbox{\footnotesize{$\eufm{n}\in\{0\}\cup\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}$}}}}}\right)\!=\!\left\{\,\phi\big{(}\partial_{\,\Bb{H}}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)}\,\right\}_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}$}}}}}\supset L\,.

Inverse inclusion SLS\subset L is immediate from the invariance of LL under any transform ϕΓϑ\phi\!\in\!\Gamma_{\vartheta} and obvious properties γ(1,)=γ(1,)2L\gamma(-1,\infty)=\gamma(1,\infty)-2\in L, γ(1,0)=1/γ(1,)L\gamma(-1,0)=-1/\gamma(1,\infty)\in L and γ(1,0)=1/γ(1,)L\gamma(1,0)=-1/\gamma(-1,\infty)\in L. This completes the proof of Theorem 5.5.

Next, we prove Theorem 5.9. It follows from (5.81) and (5.83) that the union of all boundary points of the sets composing the even rational partition (5.86) of \Bb{H} is equal to the union EE of the roofs γ𝔫(1,1)\gamma_{\eufm{n}}(-1,1) of 𝔫\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} over all 𝔫{}\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\} and their shifts on any even integer. But by 1/γ(1,1)=γ(1,1)-1/\gamma(-1,1)=\gamma(-1,1) and (5.7) this yields that EE coincides with the orbit {ϕ(γ(1,1))}ϕΓϑ\{\phi(\gamma(-1,1))\}_{\phi\in\Gamma_{\vartheta}} of γ(1,1)\gamma(-1,1) with respect to Γϑ\Gamma_{\vartheta}. Theorem 5.9 is proved.

10.4.4 . Proof of Lemma 5.7.

According to the partition (5.38), we have

{z𝕜𝕜𝕜λϝλ}\displaystyle\big{\{}z\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\ |\ \lambda(z)=\lambda(y)\big{\}} (10.98)
={z|||λ(z)=λ(y)}ϕΓϑ||{zϕ(|σϕ)|λ(z)=λ(y)}.\displaystyle\hskip 28.45274pt=\left.\left\{z\in\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\ \right|\ \lambda(z)=\lambda(y)\right\}\sqcup\bigsqcup\nolimits_{\,{{\mbox{\footnotesize{$\phi\!\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}$}}}}}\left.\left\{z\in\phi\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{\phi}}\right)\ \right|\ \lambda(z)=\lambda(y)\right\}.

Let σ{1,1}\sigma\in\{1,-1\} be arbitrarily prescribed and fixed number. It follows from Lemma 2.4 that

{z|||λ(z)=λ(y)}={y,ify,{y,y2σ},ifyγ(σ,),,ifyγ(σ,0).\displaystyle\left.\left\{z\in\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\ \right|\ \lambda(z)=\lambda(y)\right\}=\left\{\begin{array}[]{ll}y\,,&\ \ \hbox{if}\ \ y\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\\ \left\{y,y-2\sigma\right\}\,,&\ \ \hbox{if}\ \ y\in\gamma(\sigma,\infty)\,,\\ \emptyset,&\ \ \hbox{if}\ \ y\in\gamma(\sigma,0)\,.\end{array}\right. (10.102)

Fix now an arbitrary ϕΓϑ||\phi\!\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}, which, in the notation (5.27) and (5.29), is associated with NN\in\Bb{N} and (nN,,n1)(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{N} such that ϕ=ϕnN,,n1\phi=\phi_{n_{N},...,n_{1}} and σϕ=σn1=sign(n1){1,1}\sigma_{\phi}=\sigma_{n_{1}}={\rm{sign}}(n_{1})\in\{1,-1\}, where ϕnN,,n1\phi_{n_{N},...,n_{1}} is the Möbius transformation defined as in (5.1). Since ϕnN,,n1\phi_{n_{N},...,n_{1}} is injective on \Bb{H}, we get

{zϕnN,,n1(|σn1)\displaystyle\big{\{}z\in\phi_{n_{N},...,n_{1}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\right)\ |λ(z)=λ(y)}\displaystyle\big{|}\ \lambda(z)=\lambda(y)\big{\}} (10.103)
={ϕnN,,n1(η)|η|σn1,λ(ϕnN,,n1(η))=λ(y)},\displaystyle=\left.\left\{\phi_{n_{N},...,n_{1}}\left(\eta\right)\ \right|\ \eta\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\ ,\ \lambda\left(\phi_{n_{N},...,n_{1}}\left(\eta\right)\right)=\lambda(y)\right\},

where, in view of (5.58),

λ(ϕnN,,n1(η))={λ(η),ifN2λ(1/η),ifN2ηγ(σn1,).\displaystyle\lambda\left(\phi_{n_{N},...,n_{1}}(\eta)\right)=\left\{\begin{array}[]{rl}\lambda(\eta),&\ \mbox{if}\ N\in 2\Bb{N}\,,\\[2.84544pt] \lambda(-1/\eta),&\ \mbox{if}\ N\in 2\Bb{N}\!-\!1\,,\end{array}\right.\ \ \eta\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\cup\gamma\left(\sigma_{n_{1}},\infty\right)\,.

By (2.21), the equation λ(y)=λ(1/η)=1λ(η)\lambda(y)=\lambda(-1/\eta)=1-\lambda(\eta) is equivalent to λ(η)=λ(1/y)\lambda(\eta)=\lambda(-1/y), and therefore (10.103) yields

EnN,,n1(y):={zϕnN,,n1(|σn1)|λ(z)=λ(y)}\displaystyle E_{n_{N},...,n_{1}}(y):=\left.\left\{z\in\phi_{n_{N},...,n_{1}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\right)\ \right|\ \lambda(z)=\lambda(y)\right\} (10.104)
={ϕnN,,n1(η)|ηγ(σn1,),λ(η)=λ(y),ifN2λ(η)=λ(1/y),ifN2}.\displaystyle=\left\{\phi_{n_{N},...,n_{1}}\left(\eta\right)\ \left|\ \eta\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\cup\gamma\left(\sigma_{n_{1}},\infty\right),\ \begin{array}[]{ll}\lambda(\eta)=\lambda(y),&\ \mbox{if}\ N\in 2\Bb{N}\,,\\[2.84544pt] \lambda(\eta)=\lambda(-1/y),&\ \mbox{if}\ N\in 2\Bb{N}\!-\!1\,,\end{array}\right.\right\}\,. (10.107)

If yy\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, then 1/y-1/y\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and, in view of Lemma 2.4, the solutions of the both equations in the right-hand side of the latter equality are unique. Hence,

EnN,,n1(y)={ϕnN,,n1(y),ifN2ϕnN,,n1(1/y),ifN2y.\displaystyle E_{n_{N},...,n_{1}}(y)=\left\{\begin{array}[]{ll}\phi_{n_{N},...,n_{1}}(y),&\ \mbox{if}\ N\in 2\Bb{N}\,,\\[2.84544pt] \phi_{n_{N},...,n_{1}}(-1/y),&\ \mbox{if}\ N\in 2\Bb{N}\!-\!1\,,\end{array}\right.\ \quad y\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,. (10.110)

If yγ(σ,)y\in\gamma(\sigma,\infty) then λ(y)\lambda(y)\in\Bb{R}_{<0}, while λ(1/η)\lambda(-1/\eta)\in\Bb{R}_{>1}\cup(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), by Lemma 2.4 and Theorem B. So that there is no solutions of the equation in the right-hand side of (10.104) for odd NN. But for even NN such an equation has the unique solution η=y\eta=y, if σ=σn1\sigma=\sigma_{n_{1}}, and η=y2σ\eta=y-2\sigma, if σ=σn1\sigma=-\sigma_{n_{1}}. Or, what is the same η=yσ+σn1\eta=y-\sigma+\sigma_{n_{1}}. Thus,

EnN,,n1(y)={ϕnN,,n1(yσ+σn1),ifN2,ifN2yγ(σ,).\displaystyle E_{n_{N},...,n_{1}}(y)\!=\!\left\{\begin{array}[]{ll}\phi_{n_{N},...,n_{1}}(y\!-\!\sigma\!+\!\sigma_{n_{1}}),&\ \mbox{if}\ N\!\in\!2\Bb{N}\,,\\[2.84544pt] \emptyset,&\ \mbox{if}\ N\!\in\!2\Bb{N}\!-\!1\,,\end{array}\right.\ \ y\!\in\!\gamma(\sigma,\infty)\,. (10.113)

If yγ(σ,0)y\in\gamma(\sigma,0) then λ(y)\lambda(y)\in\Bb{R}_{>1}, while λ(η)\lambda(\eta)\in\Bb{R}_{<0}\cup(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), by Lemma 2.4 and Theorem B. Therefore there is no solutions of the equation in the right-hand side of (10.104) for even NN. But for odd NN such an equation has the unique solution η=1/y\eta=-1/y, if σ=σn1\sigma=-\sigma_{n_{1}}, and η=1/y+2σ\eta=-1/y+2\sigma, if σ=σn1\sigma=\sigma_{n_{1}}. I.e., η=1/y+σ+σn1\eta=-1/y+\sigma+\sigma_{n_{1}}, and

EnN,,n1(y)={,ifN2ϕnN,,n1(1/y+σ+σn1),ifN2yγ(σ,0).\displaystyle E_{n_{N},...,n_{1}}(y)\!=\!\left\{\begin{array}[]{ll}\emptyset,&\ \mbox{if}\ N\in 2\Bb{N}\,,\\[2.84544pt] \phi_{n_{N},...,n_{1}}(-1/y\!+\!\sigma\!+\!\sigma_{n_{1}}),&\ \mbox{if}\ N\in 2\Bb{N}\!-\!1\,,\end{array}\right.\ \ y\!\in\!\gamma(\sigma,0)\,. (10.116)

The relationships (10.98), (10.102) and (10.104)–(10.116) prove (5.63). It follows from Lemma 5.3(e) that each set in (5.63) is countable and has no limit points in \Bb{H}. This completes the proof of Lemma 5.7.

10.5 . Proofs for Section 7

  

10.5.1 . Proof of Lemma 7.1.

If zz\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} in (7.8), then (5.13) yields Imz=Im𝔾ϝ{\rm{Im}}\,z={\rm{Im}}\,\Bb{G}_{2}(-1/z), zz\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}, and we deduce from (2.25)(c), (7.1) and (7.12) that

|Θ4(1/z)|4|z|2Iδ(1/z)|Θ2(z)|4Im2zIm2z147π20(1+1Imz)30πIm2z(1+1Imz),\displaystyle\dfrac{\left|\Theta_{4}\left(-1/z\right)\right|^{4}}{|z|^{2}}\,I_{\delta}(-1/z)\!\leqslant\!\dfrac{\left|\Theta_{2}\left(z\right)\right|^{4}{\rm{Im}}^{2}z}{{\rm{Im}}^{2}z}\dfrac{147\pi}{20}\left(1\!+\!\dfrac{1}{{\rm{Im}}\,z}\right)\!\leqslant\ \dfrac{30\pi}{{\rm{Im}}^{2}z}\left(1\!+\!\dfrac{1}{{\rm{Im}}\,z}\right),

because by (ber1, , p.​ 325), (147/20)θ3(eπ/2)4=(147/20)(π/2)(3+22)Γ(3/4)4<30(147/20)\theta_{3}(e^{-\pi/2})^{4}=(147/20)(\pi/2)(3+2\sqrt{2})\Gamma(3/4)^{-4}<30.37{}^{\ref*{case30}} This proves the first inequality of (7.16). If in (7.8) we have z0z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0} then (2.25)(a) implies

|Θ4(z)|4=|Θ2(1/z)|4|z|2=|Θ2(1/z)|4Im(1/z)Imz=|Θ2(𝔾ϝ)|4Im𝔾ϝImz,\displaystyle\left|\Theta_{4}\left(z\right)\right|^{4}=\dfrac{\left|\Theta_{2}\left(-1/z\right)\right|^{4}}{|z|^{2}}=\dfrac{\left|\Theta_{2}\left(-1/z\right)\right|^{4}{\rm{Im}}\,(-1/z)}{{\rm{Im}}\,z}=\dfrac{\left|\Theta_{2}\left(\Bb{G}_{2}(z)\right)\right|^{4}{\rm{Im}}\,\Bb{G}_{2}(z)}{{\rm{Im}}\,z}\ ,

and we derive from (7.1) and (7.12) that

|Θ4(z)|4Iδ(z)θ3(eπ/2)4Iδ(z)(Imz)Im𝔾ϝ30π(1+1/Im𝔾ϝ(Imz)Im𝔾ϝ,\displaystyle\left|\Theta_{4}\left(z\right)\right|^{4}\,I_{\delta}(z)\leqslant\dfrac{\theta_{3}(e^{-\pi/2})^{4}I_{\delta}(z)}{\left({\rm{Im}}\,z\right){\rm{Im}}\,\Bb{G}_{2}(z)}\leqslant\dfrac{30\pi\big{(}1+1/{\rm{Im}}\,\Bb{G}_{2}(z)\big{)}}{\left({\rm{Im}}\,z\right){\rm{Im}}\,\Bb{G}_{2}(z)}\ ,

which proves the second inequality of (7.16). Finally, if z𝔫z\in\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}} in (7.8), then (2.25)(a) yields

Im𝔾ϝ𝕜Θ𝔾ϝ𝕜𝕀𝔾ϝ𝕜𝔾ϝ𝕜𝕜Θ𝔾ϝ𝕜\displaystyle{\rm{Im}}\,\Bb{G}^{N}_{2}(z)\left|\Theta_{4}\big{(}\Bb{G}^{N}_{2}(z)\big{)}\right|^{4}={\rm{Im}}\,\Bb{G}^{N}_{2}(z)\left|\Bb{G}^{N}_{2}(z)\right|^{-2}\left|\Theta_{2}\big{(}-1/\Bb{G}^{N}_{2}(z)\big{)}\right|^{4}
=Im(1/𝔾ϝ)|Θ2(1/𝔾ϝ)|4=Im𝔾ϝ𝕜Θ𝔾ϝ𝕜\displaystyle={\rm{Im}}\,\left(-1/\Bb{G}^{N}_{2}(z)\right)\left|\Theta_{2}\left(-1/\Bb{G}^{N}_{2}(z)\right)\right|^{4}={\rm{Im}}\,\Bb{G}^{N+1}_{2}(z)\left|\Theta_{2}\left(\Bb{G}^{N+1}_{2}(z)\right)\right|^{4}

and we conclude, by (7.1) and (7.12), that the third inequality of (7.16) holds because

|Θ4(𝔾ϝ|4Iδ(𝔾ϝ(Imz)/Im𝔾ϝθ3(eπ/2)4Iδ(𝔾ϝ(Imz)Im𝔾ϝ30π(1+1/Im𝔾ϝ(Imz)Im𝔾ϝ.\displaystyle\dfrac{\left|\Theta_{4}\big{(}\Bb{G}^{N}_{2}(z)\big{)}\right|^{4}\,I_{\delta}\big{(}\Bb{G}^{N}_{2}(z)\big{)}}{\left({\rm{Im}}\,z\right)\big{/}{\rm{Im}}\,\Bb{G}^{N}_{2}(z)}\leqslant\dfrac{\theta_{3}(e^{-\pi/2})^{4}I_{\delta}\big{(}\Bb{G}^{N}_{2}(z)\big{)}}{\left({\rm{Im}}\,z\right){\rm{Im}}\,\Bb{G}^{N+1}_{2}(z)}\leqslant\dfrac{30\pi\big{(}1+1/{\rm{Im}}\,\Bb{G}^{N+1}_{2}(z)\big{)}}{\left({\rm{Im}}\,z\right){\rm{Im}}\,\Bb{G}^{N+1}_{2}(z)}\ .

10.6 . Proofs for Section 8

  

10.6.1 . Proofs of Lemma 8.3 and Theorem 8.5.

To prove Lemma 8.3 for arbitrary ε>0\varepsilon>0 we introduce the function

ωε(x):=χ(0,ε)(x)κεeε2x(εx)C(εωε\displaystyle\omega_{\varepsilon}(x):=\dfrac{\chi_{(0,\varepsilon)}(x)}{\kappa\cdot\varepsilon}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\varepsilon^{2}}{x(\varepsilon-x)}$}}}}}\in C^{\infty}(\Bb{R}),\quad x\in\Bb{R}\,,\qquad\int\limits_{0}^{\varepsilon}\omega_{\varepsilon}(x){\rm{d}}x=1\,,

where (see (bak, , p.​ 658, (1.17), (1.18))) maxt[0,1]exp(t1(1t)1)=1/e4\max_{t\in[0,1]}\exp(-t^{-1}(1-t)^{-1})={1}/{{\rm{e}}^{4}},

κ:=01e1t(1t)dt=2[K1(2)K0(2)]/e2(0.3838/e4, 0.38382/e4).\displaystyle\kappa:=\int\nolimits_{0}^{1}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{1}{t(1-t)}$}}}}}dt=2\left[K_{1}(2)-K_{0}(2)\right]/{\rm{e}}^{2}\in\left({0.3838}/{{\rm{e}}^{4}}\ ,\ {0.38382}/{{\rm{e}}^{4}}\right).

For an arbitrary φH1+(\varphi\in{\rm{H}}^{1}_{+}(\Bb{R}) let

φε(x)=ωε(tx)φ(t)dt,xε\displaystyle\varphi_{\varepsilon}(x)=\int\nolimits_{\Bb{R}}\omega_{\varepsilon}(t-x)\varphi(t){\rm{d}}t\ ,\quad x\in\Bb{R}\,,\ \varepsilon>0\,. (10.117)

Then for any nn\in\Bb{Z}_{\geqslant 0} the relationships

φε(n)(x)=(1)nωε(n)(tx)φ(t)dt,|φε(n)(x)|maxx[0,ε]|ωε(n)(x)||φ(t)|dt,\displaystyle\varphi_{\varepsilon}^{(n)}(x)=(-1)^{n}\int\limits_{\Bb{R}}\omega_{\varepsilon}^{(n)}(t-x)\varphi(t){\rm{d}}t\,,\ \left|\varphi_{\varepsilon}^{(n)}(x)\right|\leqslant\max\limits_{x\in[0,\varepsilon]}\left|\omega_{\varepsilon}^{(n)}(x)\right|\int\limits_{\Bb{R}}\left|\varphi(t)\right|{\rm{d}}t\,,
|φε(n)(x)|dx|ωε(n)(tx)||φ(t)|dtdx=|ωε(n)(x)|dx|φ(t)|dt,\displaystyle\int\limits_{\Bb{R}}\left|\varphi_{\varepsilon}^{(n)}(x)\right|{\rm{d}}x\leqslant\int\limits_{\Bb{R}}\int\limits_{\Bb{R}}\left|\omega_{\varepsilon}^{(n)}(t-x)\right|\left|\varphi(t)\right|{\rm{d}}t{\rm{d}}x=\int\limits_{\Bb{R}}\left|\omega_{\varepsilon}^{(n)}(x)\right|{\rm{d}}x\int\limits_{\Bb{R}}\left|\varphi(t)\right|{\rm{d}}t\,,

yield that

φ(n)εC(𝕃𝕃\displaystyle\varphi^{(n)}_{\,\varepsilon}\in C^{\infty}(\Bb{R})\cap L_{1}(\Bb{R})\cap L_{\infty}(\Bb{R})\ ,\quad n\in\Bb{Z}_{\geqslant 0}\,. (10.118)

Since by (gar, , p.​ 88, 2(iv)) we have φ(t)exp(ixt)dt=0\int_{\Bb{R}}\varphi(t)\exp(ixt){\rm{d}}t=0 for every x0x\geqslant 0 then the identity

eiyxφε(x)dx=eiyxφ(x)dxeiytωε(t)dt=0,y>0,\displaystyle\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}yx$}}}}}\varphi_{\varepsilon}(x){\rm{d}}x=\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}yx$}}}}}\varphi(x){\rm{d}}x\int\nolimits_{\Bb{R}}{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}yt$}}}}}\omega_{\varepsilon}(t){\rm{d}}t=0\,,\quad y>0\,,

implies

φεH1+(\displaystyle\varphi_{\varepsilon}\in{\rm{H}}^{1}_{+}(\Bb{R})\,,\ (10.119)

by virtue of (10.118) and (gar, , p.​ 88, 2(iv)). For arbitrary δ>0\delta>0 we introduce the Fourier transform of ωδ\omega_{\delta} as follows

hδ(x):=0δeitxωδ(t)dtH+(𝕜δϝ𝕜ϝ\displaystyle h_{\delta}(x):=\int\nolimits_{{{\mbox{\footnotesize{$0$}}}}}^{{{\mbox{\footnotesize{$\delta$}}}}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}tx$}}}}}\omega_{\delta}(t){\rm{d}}t\in{\rm{H}}^{\infty}_{+}(\Bb{R})\ ,\quad\left|h_{\delta}(z)\right|\leqslant 1\,,\ z\in\Bb{H}\cup\Bb{R}\,, (10.120)

where H+({\rm{H}}^{\infty}_{+}(\Bb{R}) denotes the class of all nontangential limits on \Bb{R} of the uniformly bounded and holomorphic on \Bb{H} functions. Integration by parts gives

hδ(n)(x)=0δeitx(it)nωδ(t)dt=imxm0δeitx((it)nωδ(t))(m)dt,n,m\displaystyle h_{\delta}^{(n)}(x)=\int\nolimits_{{{\mbox{\footnotesize{$0$}}}}}^{{{\mbox{\footnotesize{$\delta$}}}}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}tx$}}}}}({\rm{i}}t)^{n}\omega_{\delta}(t){\rm{d}}t=\dfrac{i^{m}}{x^{m}}\int\nolimits_{{{\mbox{\footnotesize{$0$}}}}}^{{{\mbox{\footnotesize{$\delta$}}}}}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}tx$}}}}}\big{(}({\rm{i}}t)^{n}\omega_{\delta}(t)\big{)}^{(m)}{\rm{d}}t\,,\quad n,m\in\Bb{Z}_{\geqslant 0}\,,

which proves that hδS(h_{\delta}\in\eurm{S}(\Bb{R}) and we conclude, by taking into account (10.118) and (10.119), that

hδφεH1+(𝕊εδ\displaystyle h_{\delta}\cdot\varphi_{\varepsilon}\in{\rm{H}}^{1}_{+}(\Bb{R})\cap\eurm{S}(\Bb{R})\ ,\quad\varepsilon,\delta>0\,. (10.121)

It follows from the inequality |ez1|<7|z|/4|{\rm{e}}^{z}-1|<7|z|/4, |z|<1|z|<1 (see (abr, , p.​ 70, 4.2.38)) that for arbitrary ρ>0\rho>0 the following inequality holds

|hδ(x)1|0δ|eitx1|ωδ(t)dt7δ4ρ0δωδ(t)dt=7δ4ρ,δρ,|x|1ρ.\displaystyle\left|h_{\delta}(x)\!-\!1\right|\!\leqslant\!\int\limits_{{{\mbox{\footnotesize{$0$}}}}}^{{{\mbox{\footnotesize{$\delta$}}}}}\left|{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}tx$}}}}}\!-\!1\right|\omega_{\delta}(t){\rm{d}}t\!\leqslant\!\dfrac{7\delta}{4\rho}\int\limits_{{{\mbox{\footnotesize{$0$}}}}}^{{{\mbox{\footnotesize{$\delta$}}}}}\omega_{\delta}(t){\rm{d}}t=\dfrac{7\delta}{4\rho}\,,\ \ \delta\leqslant\rho\,,\ |x|\leqslant\dfrac{1}{\rho}\,. (10.122)

We introduce the notation

{L(a,b):={x𝕜𝕜𝕜}=[b,a][a,b],G(a,b):=𝕃{𝕜𝕜𝕜}0<a<b<+.\displaystyle\left\{\begin{array}[]{l}L(a,b):=\left\{\ x\in\Bb{R}\ |\ a\leqslant|x|\leqslant b\ \right\}=[-b,-a]\cup[a,b]\,,\\[7.11317pt] G(a,b):=\Bb{R}\setminus L(a,b)=(-a,a)\cup\left\{\ x\in\Bb{R}\ |\ |x|>b\ \right\}\,,\end{array}\right.\quad 0<a<b<+\infty\,.

Now, for arbitrary ε(0,1/2)\varepsilon\in(0,1/2) and δ=ε2<ρ=ε\delta=\varepsilon^{2}<\rho=\varepsilon it follows from (10.120) that

|φ(x)hδ(x)φε(x)|dxL(ρ,1/ρ)|φ(x)hδ(x)φε(x)|dx+G(ρ,1/ρ)|φ(x)|dx+G(ρ,1/ρ)|φε(x)|dx,\displaystyle\int\limits_{\Bb{R}}\left|\varphi(x)\!-\!h_{\delta}(x)\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x\!\leqslant\int\limits_{L(\rho,1/\rho)}\left|\varphi(x)\!-\!h_{\delta}(x)\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x\!+\int\limits_{G(\rho,1/\rho)}\left|\varphi(x)\right|{\rm{d}}x\!+\int\limits_{G(\rho,1/\rho)}\left|\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x\,,

where

G(ρ,1/ρ)|φε(x)|dx|ωε(t)||φ(t+x)|χG(ρ,1/ρ)(x)χ[0,ε](t)dtdx=\displaystyle\int\limits_{G(\rho,1/\rho)}\left|\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x\leqslant\int\limits_{\Bb{R}}\int\limits_{\Bb{R}}\left|\omega_{\varepsilon}(t)\right|\left|\varphi(t+x)\right|\chi_{G(\rho,1/\rho)}(x)\chi_{[0,\varepsilon]}(t){\rm{d}}t{\rm{d}}x=
=|ωε(t)||φ(x)|χG(ρ,1/ρ)(xt)χ[0,ε](t)dxdtG(ρ+ε,(1/ρ)ε)|φ(x)|dx,\displaystyle=\int\limits_{\Bb{R}}\int\limits_{\Bb{R}}\left|\omega_{\varepsilon}(t)\right|\left|\varphi(x)\right|\chi_{G(\rho,1/\rho)}(x-t)\chi_{[0,\varepsilon]}(t){\rm{d}}x{\rm{d}}t\leqslant\int\limits_{G(\rho+\varepsilon,(1/\rho)-\varepsilon)}\left|\varphi(x)\right|{\rm{d}}x\,,\

and, in view of (10.122),

L(ρ,1/ρ)|φ(x)hδ(x)φε(x)|dx=L(ρ,1/ρ)|φ(x)φε(x)+(1hδ(x))φε(x)|dx\displaystyle\int\limits_{L(\rho,1/\rho)}\left|\varphi(x)-h_{\delta}(x)\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x=\int\limits_{L(\rho,1/\rho)}\Big{|}\varphi(x)-\varphi_{\,\varepsilon}(x)+\left(1-h_{\delta}(x)\right)\varphi_{\,\varepsilon}(x)\Big{|}{\rm{d}}x
L(ρ,1/ρ)|φ(x)φε(x)|dx+7δ4ρ|φ(x)|dx,\displaystyle\leqslant\int\limits_{L(\rho,1/\rho)}\left|\varphi(x)-\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x+\dfrac{7\delta}{4\rho}\int\limits_{{{\mbox{\footnotesize{$\Bb{R}$}}}}}\left|\varphi(x)\right|{\rm{d}}x\ ,\

while

L(ρ,1/ρ)|φ(x)φε(x)|dx|φ(x)φε(x)|dx=|0εωε(t)(φ(x)φ(t+x))dt|dx\displaystyle\int\limits_{L(\rho,1/\rho)}\!\left|\varphi(x)\!-\!\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x\!\leqslant\!\!\int\limits_{{{\mbox{\footnotesize{$\Bb{R}$}}}}}\!\!\left|\varphi(x)\!-\!\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x\!=\!\!\int\limits_{{{\mbox{\footnotesize{$\Bb{R}$}}}}}\!\left|\int\limits_{0}^{\varepsilon}\!\omega_{\varepsilon}(t)\left(\varphi(x)\!-\!\varphi(t+x)\right){\rm{d}}t\right|{\rm{d}}x
0εωε(t)|φ(x)φ(t+x)|dxdtmaxt[0,ε]|φ(x)φ(t+x)|dx.\displaystyle\leqslant\int\limits_{0}^{\varepsilon}\omega_{\varepsilon}(t)\int\limits_{{{\mbox{\footnotesize{$\Bb{R}$}}}}}\left|\varphi(x)-\varphi(t+x)\right|{\rm{d}}x{\rm{d}}t\leqslant\max\limits_{{{\mbox{\footnotesize{$t\in[0,\varepsilon]$}}}}}\int\limits_{{{\mbox{\footnotesize{$\Bb{R}$}}}}}\left|\varphi(x)-\varphi(t+x)\right|{\rm{d}}x\ .

So that

|φ(x)hε2(x)φε(x)|dx\displaystyle\int\limits_{\Bb{R}}\left|\varphi(x)\!-\!h_{\varepsilon^{2}}(x)\varphi_{\,\varepsilon}(x)\right|{\rm{d}}x\!
2G(2ε,1/(2ε))|φ(x)|dx+7ε4|φ(x)|dx+maxt[0,ε]|φ(x)φ(t+x)|dx,\displaystyle\leqslant 2\int\limits_{G(2\varepsilon,1/(2\varepsilon))}\left|\varphi(x)\right|{\rm{d}}x+\dfrac{7\varepsilon}{4}\int\limits_{{{\mbox{\footnotesize{$\Bb{R}$}}}}}\left|\varphi(x)\right|{\rm{d}}x+\max\limits_{{{\mbox{\footnotesize{$t\in[0,\varepsilon]$}}}}}\int\limits_{{{\mbox{\footnotesize{$\Bb{R}$}}}}}\left|\varphi(x)-\varphi(t+x)\right|{\rm{d}}x\,,\

where the right-hand side tends to zero as ε0\varepsilon\to 0 because φL1(\varphi\in L^{1}(\Bb{R}) and the translation is continuous in L1(L^{1}(\Bb{R}) (see (gar, , p.​ 16)). This together with (10.121) completes the proof of Lemma 8.3.

To prove Theorem 8.5 we notice that in accordance with (8.14), for each n1n\geqslant 1 the functions 0\eusm{R}_{0} and 𝓃\eusm{R}_{n} on the quadrant ×\Bb{R}_{\geqslant 0}\times\Bb{R}_{\leqslant 0} coincide, correspondingly, with the functions

12γ(1,1)Θ3(z)4eixz+iy/zdz=12i0πΘ3(eiφ)4eixeiφ+iyeiφ+iφdφ,\displaystyle\dfrac{1}{2}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\Theta_{3}\left(z\right)^{4}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xz+{\rm{i}}y/z$}}}}}{\rm{d}}z=\dfrac{1}{2{\rm{i}}}\int\limits_{0}^{\pi}\Theta_{3}\left({\rm{e}}^{{\rm{i}}\varphi}\right)^{4}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}x{\rm{e}}^{{\rm{i}}\varphi}+{\rm{i}}y{\rm{e}}^{-{\rm{i}}\varphi}+{\rm{i}}\varphi$}}}}}{\rm{d}}\varphi\ ,\
12πnγ(1,1)(xyz2)eixz+iy/zSn(1λ(z))dz\displaystyle\dfrac{1}{2\pi n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\left(x-\dfrac{y}{z^{2}}\right){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xz+{\rm{i}}y/z$}}}}}S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(z)}\right){\rm{d}}z
=12πin0π(xye2iφ)eixeiφ+iyeiφ+iφ(k=1nsn,kλ(eiφ)k)dφ,\displaystyle=\dfrac{1}{2\pi{\rm{i}}n}\int\limits_{0}^{\pi}\left(x-y{\rm{e}}^{-2{\rm{i}}\varphi}\right){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}x{\rm{e}}^{{\rm{i}}\varphi}+{\rm{i}}y{\rm{e}}^{-{\rm{i}}\varphi}+{\rm{i}}\varphi$}}}}}\left(\sum\limits_{k=1}^{n}\dfrac{s^{{{\mbox{\tiny{$\triangle$}}}}}_{n,k}}{\lambda({\rm{e}}^{{\rm{i}}\varphi})^{k}}\right){\rm{d}}\varphi\ ,\

which both are entire functions of two variables x,yx,y\in\Bb{C} because (2.48) and (2.47) can be written as

1|λ(eiφ)|64exp(πsinφ),|Θ3(eiφ)|410sinφexp(π2sinφ),\displaystyle\dfrac{1}{\left|\lambda\left({\rm{e}}^{{\rm{i}}\varphi}\right)\right|}\leqslant 64\exp\left(-\dfrac{\pi}{\sin\varphi}\right),\ \ \ \left|\Theta_{3}\left({\rm{e}}^{{\rm{i}}\varphi}\right)\right|^{4}\leqslant\dfrac{10}{\sin\varphi}\exp\left(-\dfrac{\pi}{2\sin\varphi}\right), (10.123)

for all 0<φ<π0<\varphi<\pi. This proves the first assertion of Theorem 8.5.

By setting y=0y=0 in (8.7) and in (8.8), we obtain, in view of (1.18), (1.19) and (8.9),

𝓃(𝓍,0)=eixtHn(t)dt=eixtHn(t)dt,n0;0(0,𝓍)=0(𝓍,0);\displaystyle\eusm{R}_{n}(x,0)\!=\!\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt$}}}}}\eurm{H}_{-n}(t){\rm{d}}t\!=\!\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}xt$}}}}}\eurm{H}_{n}(t){\rm{d}}t\,,\ n\geqslant 0\,;\ \ \ \eusm{R}_{0}(0,-x)=\eusm{R}_{0}(x,0)\,;
𝓃(0,𝓍)=eixtMn(t)dt=eixtMn(t)dt,n1,x\displaystyle\eusm{R}_{n}(0,-x)\!=\!\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}xt$}}}}}\eurm{M}_{-n}(t){\rm{d}}t=\int\limits_{\Bb{R}}\!{\rm{e}}^{{{\mbox{\footnotesize{$-{\rm{i}}xt$}}}}}\eurm{M}_{n}(t){\rm{d}}t\ ,\ \ n\geqslant 1\,,\ \ x\in\Bb{R}\,,

from which and (1.4) we derive (8.19).

The properties (8.20) and (8.22) are immediate of (8.9), (7.26), (8.16) and (7.25), respectively.

To prove (8.21) and (8.23) observe that the following estimates have been derived after (4.5) for a=2a=2,

|Sn(1λ(z))|=|Rn(z)|9e2πnπ|λ(z)|11+822110e2πnπ|λ(z)|,zγ(1,1),\displaystyle\left|S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(z)}\right)\right|=\left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\right|\leqslant\dfrac{9\,{\rm{e}}^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{\pi|\lambda(z)|}\dfrac{11+8\sqrt{2}}{21}\leqslant\dfrac{10\,{\rm{e}}^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{\pi|\lambda(z)|}\ ,\quad z\in\gamma(-1,1)\,, (10.124)

where n1n\geqslant 1 and the value of 12λ(2i)1-2\lambda(2{\rm{i}}) has been used from (7.2). Then, by using the estimates (2.47), (2.48), for the parametrization γ(1,1)z=(t+i)/(ti)\gamma(-1,1)\ni z=(t+{\rm{i}})/(t-{\rm{i}}), t>0t>0, we derive from (8.14) and (10.124) that

|Rn(t+iti)|640e2πnπeπ2(t+1t),|Θ3(t+iti)|45(t+1t)eπ4(t+1t),\displaystyle\left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{t+{\rm{i}}}{t-{\rm{i}}}\right)\right|\!\leqslant\!\dfrac{640{\rm{e}}^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{\pi}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)$}}}}}\,,\ \left|\Theta_{3}\left(\dfrac{t+{\rm{i}}}{t-{\rm{i}}}\right)\right|^{4}\leqslant 5\left(t+\dfrac{1}{t}\right){\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi}{4}\left(t+\dfrac{1}{t}\right)$}}}}}\ ,
𝓃(𝓍,𝓎)=1iπ𝓃0(𝓍+𝓎(𝓉i)2(𝓉+i)2)Rn(t+iti)eixt+itiiytit+idt(ti)2,\displaystyle\eusm{R}_{n}(x,-y)=\dfrac{1}{{\rm{i}}\pi n}\int\limits_{0}^{\infty}\left(x+y\dfrac{(t-{\rm{i}})^{2}}{(t+{\rm{i}})^{2}}\right)\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{t+{\rm{i}}}{t-{\rm{i}}}\right){\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}x\dfrac{t+{\rm{i}}}{t-{\rm{i}}}-{\rm{i}}y\dfrac{t-{\rm{i}}}{t+{\rm{i}}}$}}}}}\dfrac{{\rm{d}}t}{(t-{\rm{i}})^{2}}\ ,
0(𝓍,𝓎)=1i0Θ3(𝓉+i𝓉i)4eixt+itiiytit+id𝓉(𝓉i)2,𝓍,𝓎0,𝓉>0,𝓃1.\displaystyle\eusm{R}_{0}(x,-y)=\dfrac{1}{{\rm{i}}}\int\limits_{0}^{\infty}\Theta_{3}\left(\dfrac{t+{\rm{i}}}{t-{\rm{i}}}\right)^{4}{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}x\dfrac{t+{\rm{i}}}{t-{\rm{i}}}-{\rm{i}}y\dfrac{t-{\rm{i}}}{t+{\rm{i}}}$}}}}}\dfrac{{\rm{d}}t}{(t-{\rm{i}})^{2}}\,,\quad x,y\geqslant 0\,,\ t>0\,,\ n\geqslant 1\,.

So that

|𝓃(𝓍,𝓎)|640e2πnπ2n(x+y)0eπ2(t+1t)2t(x+y)t2+1dtt2+1,|0(𝓍,𝓎)|50eπ4(t+1t)2t(x+y)t2+1dtt,x,y0,n1,\displaystyle\begin{array}[]{l}\displaystyle\left|\eusm{R}_{n}(x,-y)\right|\leqslant\dfrac{640{\rm{e}}^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{\pi^{2}n}(x+y)\int\limits_{0}^{\infty}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)-\dfrac{2t(x+y)}{t^{2}+1}$}}}}}\dfrac{{\rm{d}}t}{t^{2}+1}\ ,\\ \displaystyle\left|\eusm{R}_{0}(x,-y)\right|\leqslant 5\int\limits_{0}^{\infty}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{\pi}{4}\left(t+\dfrac{1}{t}\right)-\dfrac{2t(x+y)}{t^{2}+1}$}}}}}\dfrac{{\rm{d}}t}{t}\ ,\quad x,y\geqslant 0\,,\ n\geqslant 1\,,\end{array} (10.127)

where for arbitrary a,b>0a,b>0 we have45{}^{\ref*{case37}}

0eb2(t+1t)2att2+1dtt2+1=12ba+1K1(2b(a+1)),0eb2(t+1t)2att2+1dtt=K0(2b(a+1)).\displaystyle\begin{array}[]{l}\displaystyle\int\limits_{0}^{\infty}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{b}{2}\left(t+\dfrac{1}{t}\right)-\dfrac{2at}{t^{2}+1}$}}}}}\dfrac{{\rm{d}}t}{t^{2}+1}=\dfrac{1}{2}\sqrt{\dfrac{b}{a+1}}K_{1}\left(2\sqrt{b(a+1)}\right)\ ,\\ \displaystyle\int\limits_{0}^{\infty}{\rm{e}}^{{{\mbox{\footnotesize{$-\dfrac{b}{2}\left(t+\dfrac{1}{t}\right)-\dfrac{2at}{t^{2}+1}$}}}}}\dfrac{{\rm{d}}t}{t}=K_{0}\left(2\sqrt{b(a+1)}\right)\,.\end{array} (10.130)

Taking into account that 320π/(π2n)<2π3320\sqrt{\pi}/(\pi^{2}n)<2\pi^{3} for all n1n\geqslant 1, we derive from (10.127) and (10.130) that

|𝓃(𝓍,𝓎)|2π3e2πn(x+y)x+y+1K1(2π(x+y+1)),|0(𝓍,𝓎)|5K0(2π(x+y+1)),x,y0,n1,\displaystyle\begin{array}[]{l}\displaystyle\left|\eusm{R}_{n}(x,-y)\right|\leqslant 2\pi^{3}{\rm{e}}^{{{\mbox{\footnotesize{$2\pi n$}}}}}\dfrac{(x+y)}{\sqrt{x+y+1}}K_{1}\left(2\sqrt{\pi(x+y+1)}\right)\ ,\\[14.22636pt] \displaystyle\left|\eusm{R}_{0}(x,-y)\right|\leqslant 5K_{0}\left(\sqrt{2\pi(x+y+1)}\right)\ ,\quad x,y\geqslant 0\,,\ n\geqslant 1\,,\end{array} (10.133)

which proves (8.21), (8.23) and completes the proof of Theorem 8.5.

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Remark to the References. The results of this preprint are based on several assertions which may be found in abr , and , bh2 , bh1 , (bak, , p.​ 665), ber1 , con , erd1 , gar , gra , (hed1, , p.​ 1715), ko , lav , nat , olv , rud , vlad and whi . The other cited books and articles are listed for completeness.

For example, the monograph law is cited on the page 2.44 in connection with the Landen transformation equations (2.44). However, these equations have self-contained proofs on the pages 56. Similarly, in connection with the reference to (lop, , p.​ 303) on the page 5.8 we supply a self-contained proof of (lop, , p.​ 303, Proposition 2) on the pages 2728. Finally, the references to Lemma 2 of (cha, , p.​ 112) on the pages 5.3 and 5.3 are complemented by a self-contained proof of this lemma on the pages 1921.

Moreover, in (hed1, , p.​ 1715) we use only Proposition 3.7.2 which establishes the elementary property of the operator 𝐓1{\bf{T}}_{1} that it preserves the properties of functions on (1,1)(-1,1) to be even and convex. Also, we need (bak, , p.​ 665) to write explicitly the value of the integral of the simplest test function. At last, when using a property of the theta functions needed in the preprint, we refer to bh2 and bh1 because all the basic properties find simple proofs there with the help of elementary integrals, Liouville’s theorem, Morera’s theorem, Riemann’s theorem about removable singularities, and elementary properties of the Schwarz triangle function λ\lambda_{{{\mbox{\tiny{$\triangle$}}}}}. All the other facts used in the preprint without proofs are from the fundamentals of real and complex analysis found in, e.g., abr , and , ber1 , con , erd1 , gar , gra , ko , lav , nat , olv , rud , vlad and whi .

A . Supplementary notes

A.1 . Notes for Section 1

  

1

\uparrow Let MM\in\Bb{N} and

𝒮M(x):=a0+n{}𝕜𝕜𝕄(aneiπnx+bneiπn/x),x\displaystyle\mathcal{S}_{M}(x):=a_{0}+\sum_{n\in\Bb{Z}\setminus\{0\},\,|n|\leqslant M}\left(a_{n}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}\!+\!b_{n}{\rm{e}}^{{{\mbox{\footnotesize{$-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x$}}}}}\right)\,,\ \ x\!\in\!\Bb{R}\,.

It follows from the two equalities preceding (1.6) that for arbitrary φS(\varphi\in\eurm{S}(\Bb{R}) and n{}n\in\Bb{Z}\setminus\{0\} we have

|anφ(x)eiπnxdx||an|(π|n|)N+2|φ(N+2)(x)|dx,\displaystyle\left|a_{n}\int\limits_{\Bb{R}}\!\varphi(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi nx$}}}}}{\rm{d}}x\right|\leq\dfrac{|a_{n}|}{(\pi\,|n|)^{N+2}}\int\nolimits_{\Bb{R}}\left|\varphi^{(N+2)}(x)\right|{\rm{d}}x\,,
|bnφ(x)eiπnxdx||bn|(π|n|)N+2|(x2ddx+2x)N+2φ(x)|dx.\displaystyle\left|b_{n}\int\limits_{\Bb{R}}\!\varphi(x)\,{\rm{e}}^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{x}$}}}}}{\rm{d}}x\right|\leq\dfrac{|b_{n}|}{(\pi\,|n|)^{N+2}}\int\nolimits_{\Bb{R}}\left|\left(x^{2}\dfrac{{\rm{d}}}{{\rm{d}}x}\!+\!2x\right)^{\!N+2}\!\!\varphi(x)\right|{\rm{d}}x\,.

Thus, the conditions an,bn=O(nN),n,a_{n},b_{n}\!=\!{\rm{O}}(n^{N}),\ n\!\to\!\infty, in (1.8) imply that the sequence

φ(x)𝒮M(x)dx,M\displaystyle\int\limits_{\Bb{R}}\!\varphi(x)\mathcal{S}_{M}(x){\rm{d}}x\ ,\quad M\in\Bb{N}\,,

converges and

𝒮(φ):=limMφ(x)𝒮M(x)dx,φS(\displaystyle\mathcal{S}_{\infty}\big{(}\varphi\big{)}:=\lim_{M\to\,\infty}\int\limits_{\Bb{R}}\!\varphi(x)\mathcal{S}_{M}(x){\rm{d}}x\,,\quad\varphi\in\eurm{S}(\Bb{R})\,,

satisfies

|𝒮(φ)|\displaystyle\left|\mathcal{S}_{\infty}\big{(}\varphi\big{)}\right|\leq |a0|+(n{}|an|(π|n|)N+2)|φ(N+2)(x)|dx+\displaystyle|a_{0}|+\left(\sum_{n\in\Bb{Z}\setminus\{0\}}\dfrac{|a_{n}|}{(\pi\,|n|)^{N+2}}\right)\int\nolimits_{\Bb{R}}\left|\varphi^{(N+2)}(x)\right|{\rm{d}}x\,+
(n{}|bn|(π|n|)N+2)|(x2ddx+2x)N+2φ(x)|dx.\displaystyle\left(\sum_{n\in\Bb{Z}\setminus\{0\}}\dfrac{|b_{n}|}{(\pi\,|n|)^{N+2}}\right)\int\nolimits_{\Bb{R}}\left|\left(x^{2}\dfrac{{\rm{d}}}{{\rm{d}}x}\!+\!2x\right)^{\!N+2}\!\!\varphi(x)\right|{\rm{d}}x\,.

This means that 𝒮\mathcal{S}_{\infty} defines a continuous linear functional on S(\eurm{S}(\Bb{R}) and therefore 𝒮S(\mathcal{S}_{\infty}\in\eurm{S}^{\,\prime}(\Bb{R}) (see (vlad, , p.​ 77)). In conclusion, the series

a0+n{}(aneiπnx+bneiπn/x)\displaystyle a_{0}+\sum\nolimits_{n\in\Bb{Z}\setminus\{0\}}\!\!\left(a_{n}{\rm{e}}^{{{\mbox{\footnotesize{$\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi nx$}}}}}\!+\!b_{n}{\rm{e}}^{{{\mbox{\footnotesize{$-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n/x$}}}}}\right)

converges to 𝒮S(\mathcal{S}_{\infty}\in\eurm{S}^{\,\prime}(\Bb{R}) in the space S(\eurm{S}^{\,\prime}(\Bb{R}).

2

\uparrow  We prove (1.21). It follows from (2.45), written in the form

λ(1+z1z)=12+i(12λ(z))4λ(z)(1λ(z)),z,\displaystyle\lambda\left(\dfrac{1+z}{1-z}\right)=\dfrac{1}{2}+\dfrac{i\left(1-2\lambda(z)\right)}{4\sqrt{\lambda(z)\left(1-\lambda(z)\right)}}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,

and (2.20), that

λ(1+λ(z)1λ(z))=12+i(12λ(λ(z)))4λ(λ(z))(1λ(λ(z)))=12+i(12z)4z(1z),zΛ,\displaystyle\lambda\left(\dfrac{1+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}{1-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}\right)=\dfrac{1}{2}+\dfrac{i\left(1-2\lambda(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z))\right)}{4\sqrt{\lambda(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z))\left(1-\lambda(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z))\right)}}=\dfrac{1}{2}+\dfrac{i\left(1-2z\right)}{4\sqrt{z\left(1-z\right)}}\ ,\quad z\in\Lambda\,,

where Λ:=(0,1)()\Lambda:=(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), and then, by using

ϕ(1111)()=,ϕ(1111)(z):=1+z1z,\displaystyle\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}1&1\\ -1&1\end{smallmatrix})$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ ,\quad\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}1&1\\ -1&1\end{smallmatrix})$}}}}}\left(z\right):=\dfrac{1+z}{1-z}\,, (A.14a)

we can apply to the latter equality the function λ\lambda_{{{\mbox{\tiny{$\triangle$}}}}} to get

1+λ(z)1λ(z)=λ(12+i(12z)4z(1z)),zΛ.\displaystyle\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni\dfrac{1+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}{1-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{i\left(1-2z\right)}{4\sqrt{z\left(1-z\right)}}\right)\ ,\quad z\in\Lambda\,. (A.2a)

Applying to (A.2a) the function Θ32\Theta_{3}^{2} and using the squared equality (2.38)(b), for any zΛz\in\Lambda we obtain

F(12+i(12z)4z(1z))=Θ3(λ(12+i(12z)4z(1z)))2=Θ3(1+λ(z)1λ(z))2\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{i\left(1-2z\right)}{4\sqrt{z\left(1-z\right)}}\right)=\Theta_{3}\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{i\left(1-2z\right)}{4\sqrt{z\left(1-z\right)}}\right)\right)^{2}=\Theta_{3}\left(\dfrac{1+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}{1-\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)}\right)^{2}

where, by virtue of (2.25) and (2.44)(c),

Θ3(1+z1z)2=Θ3(21z1)2=Θ4(21z)2=Θ3(11z)Θ4(11z),\displaystyle\Theta_{3}\left(\dfrac{1+z}{1-z}\right)^{2}=\Theta_{3}\left(\dfrac{2}{1-z}-1\right)^{2}=\Theta_{4}\left(\dfrac{2}{1-z}\right)^{2}=\Theta_{3}\left(\dfrac{1}{1-z}\right)\Theta_{4}\left(\dfrac{1}{1-z}\right)\,,
Θ3(1z1)=z1iΘ3(z1)=z1iΘ4(z),\displaystyle\Theta_{3}\left(-\dfrac{1}{z-1}\right)=\sqrt{\dfrac{z-1}{i}}\Theta_{3}(z-1)=\sqrt{\dfrac{z-1}{i}}\Theta_{4}(z)\,,
Θ4(1z1)=z1iΘ2(z1)=eiπ/4z1iΘ2(z),\displaystyle\Theta_{4}\left(-\dfrac{1}{z-1}\right)=\sqrt{\dfrac{z-1}{i}}\Theta_{2}(z-1)={\rm{e}}^{-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\sqrt{\dfrac{z-1}{i}}\Theta_{2}(z)\,,
Θ3(1+z1z)2=e3iπ/4(z1)Θ4(z)Θ2(z),z\displaystyle\Theta_{3}\left(\dfrac{1+z}{1-z}\right)^{2}={\rm{e}}^{-3\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}(z-1)\Theta_{4}(z)\Theta_{2}(z)\ ,\quad z\in\Bb{H}\,,

and therefore

F(12+i(12z)4z(1z))=e3iπ/4(λ(z)1)Θ4(λ(z))Θ2(λ(z)),zΛ.F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{i\left(1-2z\right)}{4\sqrt{z\left(1-z\right)}}\right)\\ ={\rm{e}}^{{{\mbox{\footnotesize{$-3\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4$}}}}}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)-1\big{)}\Theta_{4}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\Theta_{2}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\ ,\quad z\in\Lambda\,. (A.2b)

But according to (2.10), (2.38)(c) and (2.38)(a),

λ(z)=iF(1z)F(z),Θ4(λ(z))=(1z)1/4F(z)1/2,Θ2(λ(z))=z1/4F(z)1/2,\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)=\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}\,,\quad\Theta_{4}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\!=\!(1\!-\!z)^{1/4}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}\,,\quad\Theta_{2}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\!=\!z^{1/4}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}\,,

and, substituting these expressions in (A.2b), we get

F(12+i(12z)4z(1z))=e3iπ/4(iF(1z)F(z)1)z(1z)4F(z)\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{i\left(1-2z\right)}{4\sqrt{z(1-z)}}\right)={\rm{e}}^{-3\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\left(\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)}-1\right)\sqrt[4]{z(1\!-\!z)}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)
=e3iπ/4(iF(1z)F(z))z(1z)4=eiπ/4(F(1z)+iF(z))z(1z)4,\displaystyle={\rm{e}}^{-3\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\left(iF_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)-F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\right)\sqrt[4]{z(1\!-\!z)}={\rm{e}}^{-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\left(F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\right)\sqrt[4]{z(1\!-\!z)}\,,

from which

F(12+i(12z)4z(1z))=eiπ/4(F(1z)+iF(z))z(1z)4,zΛ.\displaystyle\vspace{-0,3cm}F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{i\left(1-2z\right)}{4\sqrt{z(1-z)}}\right)\!=\!{\rm{e}}^{-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\big{(}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1-z)\!+\!iF_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\sqrt[4]{z(1\!-\!z)}\,,\quad z\in\Lambda\,. (A.2c)

Since (0,1)Λ(0,1)\subset\Lambda, we can set z=τ(0,1)z=\tau\in(0,1) in (A.2c) and consider the equation

t=(12τ)4τ(1τ)\displaystyle t=\dfrac{\left(1-2\tau\right)}{4\sqrt{\tau(1-\tau)}}\ \Rightarrow\
16t2=(12τ)2τ(1τ)=14τ+4τ2τ(1τ)=1τ(1τ)41τ(1τ)=4(4t2+1)\displaystyle 16t^{2}=\dfrac{\left(1-2\tau\right)^{2}}{\tau(1-\tau)}=\dfrac{1-4\tau+4\tau^{2}}{\tau(1-\tau)}=\dfrac{1}{\tau(1-\tau)}-4\ \Rightarrow\ \dfrac{1}{\tau(1-\tau)}=4(4t^{2}+1)
1τ(1τ)=24t2+1,\displaystyle\ \Rightarrow\ \dfrac{1}{\sqrt{\tau(1-\tau)}}=2\sqrt{4t^{2}+1}\,,\

and therefore

t=(12τ)4τ(1τ)t=(12τ)24t2+1τ=12t4t2+1.\displaystyle t=\dfrac{\left(1-2\tau\right)}{4\sqrt{\tau(1-\tau)}}\ \Rightarrow\ t=\dfrac{\left(1-2\tau\right)}{2}\sqrt{4t^{2}+1}\ \Rightarrow\ \tau=\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\ .

Since

ddτ(12τ)τ(1τ)\displaystyle\dfrac{d}{d\tau}\dfrac{\left(1-2\tau\right)}{\sqrt{\tau(1-\tau)}} =2τ(1τ)(12τ)12τ2τ(1τ)τ(1τ)=\displaystyle=\dfrac{-2\sqrt{\tau(1-\tau)}-\left(1-2\tau\right)\dfrac{1-2\tau}{2\sqrt{\tau(1-\tau)}}}{\tau(1-\tau)}=
=4τ(1τ)+(12τ)22τ(1τ)τ(1τ)=4τ4τ2+14τ+4τ22τ(1τ)τ(1τ)=\displaystyle=-\dfrac{4\tau(1-\tau)+\left(1-2\tau\right)^{2}}{2\tau(1-\tau)\sqrt{\tau(1-\tau)}}=-\dfrac{4\tau-4\tau^{2}+1-4\tau+4\tau^{2}}{2\tau(1-\tau)\sqrt{\tau(1-\tau)}}=
=12τ(1τ)τ(1τ)<0,τ(0,1),\displaystyle=-\dfrac{1}{2\tau(1-\tau)\sqrt{\tau(1-\tau)}}<0\ ,\quad\tau\in(0,1)\,,

and

lim0<τ0(12τ)τ(1τ)=+,lim1>τ1(12τ)τ(1τ)=,\displaystyle\lim\limits_{0<\tau\to 0}\dfrac{\left(1-2\tau\right)}{\sqrt{\tau(1-\tau)}}=+\infty\,,\quad\lim\limits_{1>\tau\to 1}\dfrac{\left(1-2\tau\right)}{\sqrt{\tau(1-\tau)}}=-\infty\,,

we can replace zz in (A.2c) by

z=12t4t2+1 1z=12+t4t2+1,\displaystyle z=\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\ \Rightarrow\ 1-z=\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\,,
z(1z)=14t24t2+1=14(4t2+1),z(1z)4=124t2+14\displaystyle z(1-z)=\dfrac{1}{4}-\dfrac{t^{2}}{4t^{2}+1}=\dfrac{1}{4\left(4t^{2}+1\right)}\ ,\quad\sqrt[4]{z(1-z)}=\dfrac{1}{\sqrt{2}\sqrt[4]{4t^{2}+1}}

and to get

F(12+it)=eiπ/4F(12+t4t2+1)+iF(12t4t2+1)24t2+14\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!it\right)\!=\!{\rm{e}}^{-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)\!+\!iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)}{\sqrt{2}\sqrt[4]{4t^{2}+1}}

where

eiπ/42=1+i22=1+i,\displaystyle{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\sqrt{2}=\dfrac{1+i}{\sqrt{2}}\sqrt{2}=1+i\,,

and hence,

F(12+it)=F(12+t4t2+1)+iF(12t4t2+1)(1+i)4t2+14,t\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+it\right)=\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)}{(1+i)\,\sqrt[4]{4t^{2}+1}}\ ,\quad t\in\Bb{R}\,, (A.2d)

which proves (1.21).

3

\uparrow  We prove the new expressions for the integrals of Hn(x)\eurm{H}_{n}(x), n0n\geqslant 0. Obviously

H0(x)=12π2+F(1/2+it)F(1/2it)dt(t2+1/4)(F(1/2it)2+x2F(1/2+it)2)\displaystyle\eurm{H}_{0}(x)=\dfrac{1}{2\pi^{2}}\int\limits_{-\infty}^{+\infty}\frac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+it)F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-it)dt}{\big{(}t^{2}+1/4\big{)}\Big{(}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-it)^{2}+x^{2}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+it)^{2}\Big{)}}
=12π2F(1/2it)F(1/2+it)dt(t2+1/4)(F(1/2it)2F(1/2+it)2+x2),\displaystyle=\dfrac{1}{2\pi^{2}}\int\limits_{\Bb{R}}\frac{\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-it)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+it)}dt}{\big{(}t^{2}+1/4\big{)}\left(\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-it)^{2}}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+it)^{2}}+x^{2}\right)}\ ,
Hn(x)=14π3n+Sn(11/2+it)dt(t2+1/4)(F(1/2it)ixF(1/2+it))2,n1.\displaystyle\eurm{H}_{n}(x)=\dfrac{1}{4\pi^{3}n}\int\limits_{-\infty}^{+\infty}\dfrac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\dfrac{1}{1/2+it}\right)dt}{\big{(}t^{2}+1/4\big{)}\Big{(}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-it)-ixF_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+it)\Big{)}^{2}}\,,\quad n\geqslant 1\,.

But (A.2d) written in the form

F(12+it)(1+i)4t2+14=F(12+t4t2+1)+iF(12t4t2+1),\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+it\right)(1+i)\,\sqrt[4]{4t^{2}+1}=F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)\ ,
F(12it)(1+i)4t2+14=F(12t4t2+1)+iF(12+t4t2+1),\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-it\right)(1+i)\,\sqrt[4]{4t^{2}+1}=F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)\ ,

yield

F(12it)F(12+it)=F(12t4t2+1)+iF(12+t4t2+1)F(12+t4t2+1)+iF(12t4t2+1)=\displaystyle\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-it\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+it\right)}=\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)}=
=iF(12+t4t2+1)iF(12t4t2+1)F(12+t4t2+1)+iF(12t4t2+1),\displaystyle=i\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)-iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)}\ ,\

and

[F(1/2it)ixF(1/2+it)](1+i)4t2+14=F(12t4t2+1)\displaystyle\left[F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-it)-ixF_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+it)\right](1+i)\sqrt[4]{4t^{2}+1}=F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)
+iF(12+t4t2+1)ix(F(12+t4t2+1)+iF(12t4t2+1))\displaystyle\!+\!iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\!-\!ix\left(F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\!+\!iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\right)
=(1+x)F(12t4t2+1)+i(1x)F(12+t4t2+1),\displaystyle=(1+x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}\right)+i(1-x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}\right)\,,

i.e., for any t,xt,x\in\Bb{R} we have

F(1/2it)\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2-it)- ixF(1/2+it)\displaystyle ixF_{\!{{\mbox{\tiny{$\triangle$}}}}}(1/2+it)\!
=(1+x)F(12t4t2+1)+i(1x)F(12+t4t2+1)(1+i)2t2+1/44.\displaystyle=\!\dfrac{(1\!+\!x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\!+\!i(1\!-\!x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}{(1+i)\sqrt{2}\sqrt[4]{t^{2}+1/4}}\,.

Substituting these expressions in the above integral for H0\eurm{H}_{0} we get

H0(x)\displaystyle\eurm{H}_{0}(x) =i2π2F(12+t4t2+1)iF(12t4t2+1)F(12+t4t2+1)+iF(12t4t2+1)dtt2+1/4x2(F(12+t4t2+1)iF(12t4t2+1)F(12+t4t2+1)+iF(12t4t2+1))2\displaystyle=\dfrac{i}{2\pi^{2}}\int\limits_{\Bb{R}}\dfrac{\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)-iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)}\dfrac{dt}{t^{2}+1/4}}{x^{2}-\left(\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)-iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}+\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)+iF_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({{\mbox{\footnotesize{$\dfrac{1}{2}-\dfrac{t}{\sqrt{4t^{2}+1}}$}}}}\right)}\right)^{\!\!\!\!2}}
=i2π21iy(t)1+iy(t)x2(1iy(t)1+iy(t))2dtt2+1/4,y(t):=F(12t4t2+1)F(12+t4t2+1),\displaystyle=\dfrac{i}{2\pi^{2}}\int\limits_{\Bb{R}}\dfrac{\dfrac{1\!-\!i\eurm{y}(t)}{1\!+\!i\eurm{y}(t)}}{x^{2}-\left(\dfrac{1\!-\!i\eurm{y}(t)}{1\!+\!i\eurm{y}(t)}\right)^{2}}\dfrac{dt}{t^{2}+1/4}\ ,\quad\eurm{y}(t)\!:=\!\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}\ ,

and doing the same for the above integral for Hn\eurm{H}_{n} we obtain, taking account of ((1+i)2)2=4i((1+i)\sqrt{2})^{2}=4i,

Hn(x)\displaystyle\eurm{H}_{n}(x) =iπ3nSn(11/2+it)dtt2+1/4((1+x)F(12t4t2+1)+i(1x)F(12+t4t2+1))2\displaystyle=\dfrac{i}{\pi^{3}n}\int\limits_{\Bb{R}}\dfrac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\left(\dfrac{1}{1/2+it}\right)\dfrac{dt}{\sqrt{t^{2}+1/4}}}{\left((1\!+\!x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\!+\!i(1\!-\!x)F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\right)^{2}}\,\ \

for all xx\in\Bb{R} and n1n\geqslant 1.

A.2 . Notes for Section 2

4

\uparrow   We prove (2.38). It follows actually from (bh2, , p.​ 598, (1.25); p.​ 599, (1.32)) that

(a)Θ2(λ(z))4=zF(z)2,(b)Θ3(λ(z))2=F(z),(c)Θ4(λ(z))4=(1z)F(z)2,z(0,1)().\displaystyle\begin{array}[]{ll}{\rm{(a)}}\ \Theta_{2}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{4}\!=\!zF_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}\!,&\qquad{\rm{(b)}}\ \Theta_{3}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}\!=\!F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)\,,\\[2.84544pt] {\rm{(c)}}\ \Theta_{4}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{4}\!=\!(1\!-\!z)F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{2}\!,&\phantom{{\rm{(c)}}a}\qquad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right).\end{array}\ \ (A.4c)

Since (see (bh2, , p.​ 598, (1.26)))

Θ3(z)Θ4(z)Θ2(z)0,z\displaystyle\Theta_{3}(z)\Theta_{4}(z)\Theta_{2}(z)\neq 0\ ,\quad z\in\Bb{H}\,, (A.4d)

for the principal branches of the quadratic and of the fourth roots it makes possible to consider three functions

ω2(z):=z1/4F(z)1/2Θ2(λ(z)),ω4(z):=(1z)1/4F(z)1/2Θ4(λ(z)),ω3(z):=F(z)1/2Θ3(λ(z)),\displaystyle\omega_{2}(z):=\dfrac{z^{1/4}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}}{\Theta_{2}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}}\,,\ \omega_{4}(z):=\dfrac{(1-z)^{1/4}F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}}{\Theta_{4}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}}\,,\ \omega_{3}(z):=\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}(z)^{1/2}}{\Theta_{3}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}}\,,

which are holomorphic in Λ:=(0,1)()\Lambda:=(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right) and in accordance with (A.4c),

ω2(z),ω4(z){exp(πik2)}k=03,ω3(z){1,1},zΛ.\displaystyle\omega_{2}(z),\omega_{4}(z)\in\left\{\exp\left(\dfrac{\pi ik}{2}\right)\right\}_{k=0}^{3}\ ,\quad\omega_{3}(z)\in\left\{-1,1\right\}\ ,\quad z\in\Lambda\,. (A.4e)

Therefore

Λ=0k3Λkm,m{2,4},Λ=0k1Λk3,\displaystyle\Lambda=\underset{0\leqslant k\leqslant 3}{\sqcup}\Lambda_{k}^{m}\,,\ m\in\{2,4\}\ ,\quad\Lambda=\underset{0\leqslant k\leqslant 1}{\sqcup}\Lambda_{k}^{3}\,,
Λkm:={zΛ|ωm(z)=exp(πik2)},0k3,m{2,4},\displaystyle\Lambda_{k}^{m}:=\left\{z\in\Lambda\,\left|\,\omega_{m}(z)=\exp\left(\dfrac{\pi ik}{2}\right)\right\},\quad 0\leqslant k\leqslant 3\,,\ m\in\{2,4\}\right.,
Λk3:={zΛ|ω3(z)=(1)k},k{0,1},\displaystyle\Lambda_{k}^{3}:=\left\{z\in\Lambda\,\left|\,\omega_{3}(z)=(-1)^{k}\right\}\right.,\ k\in\left\{0,1\right\},

Here each of the set Λkm\Lambda_{k}^{m} is relatively closed in Λ\Lambda, i.e., KΛkmK\cap\Lambda_{k}^{m} is closed for every compact subset KK of Λ\Lambda, because ω2\omega_{2}, ω3\omega_{3}, ω4Hol(Λ)\omega_{4}\in{\rm{Hol}}(\Lambda). In view of the Baire Category Theorem (roy, , p.​ 159), for each m{2,3,4}m\in\{2,3,4\} there exists a number qmq_{m} such that q3{0,1}q_{3}\in\{0,1\}, qm{0,1,2,3}q_{m}\in\{0,1,2,3\}, if m{2,4}m\in\{2,4\}, and the set Λqmm\Lambda_{q_{m}}^{m} contains some neighborhood of at least one interior point of Λ\Lambda. Applying the uniqueness theorem for analytic functions (see (con, , p.​ 78)) we obtain Λqmm=Λ\Lambda_{q_{m}}^{m}=\Lambda for all m{2,3,4}m\in\{2,3,4\} and therefore

ωm(z)=exp(πiqm2),m{2,4},ω3(z)=(1)q3,zΛ.\displaystyle\omega_{m}(z)=\exp\left(\dfrac{\pi iq_{m}}{2}\right)\,,\ m\in\{2,4\}\ ,\quad\omega_{3}(z)=(-1)^{q_{3}}\ ,\quad z\in\Lambda\,. (A.4f)

By setting here z(0,1)z\in(0,1) we deduce that qm=0q_{m}=0 for each m{2,3,4}m\in\{2,3,4\}, because, in view of (2.12), λ((0,1))=i\lambda_{{{\mbox{\tiny{$\triangle$}}}}}((0,1))=i\Bb{R}, and Θm(i\Theta_{m}(i\Bb{R})\subset\Bb{R}_{>0}, F((0,1))F_{\!{{\mbox{\tiny{$\triangle$}}}}}((0,1))\subset\Bb{R}_{>0}, z,1z(0,1)z,1-z\in(0,1) for arbitrary z(0,1)z\in(0,1). Therefore all equalities (2.38) hold and the proof of (2.38) is completed.

5

\uparrow  We prove (2.44). In view of the uniqueness theorem for analytic functions (see (con, , p.​ 78)), it is enough to prove that

Ψ1(z):=Θ3(z)2Θ4(z)22Θ2(2z)2=1,Ψ2(z):=Θ3(z)2+Θ4(z)22Θ3(2z)2=1,Ψ3(z):=Θ4(2z)2Θ3(z)Θ4(z)=1,z,\displaystyle\begin{array}[]{ll}\Psi_{1}(z)\!:=\!\dfrac{\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2}}{2\Theta_{2}(2z)^{2}}\!=\!1\,,&\ \Psi_{2}(z)\!:=\!\dfrac{\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2}}{2\Theta_{3}(2z)^{2}}\!=\!1\,,\\[14.22636pt] \Psi_{3}(z)\!:=\!\dfrac{\Theta_{4}(2z)^{2}}{\Theta_{3}(z)\Theta_{4}(z)}\!=\!1\,,&\ z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\end{array} (A.5c)

because ΨjHol(\Psi_{j}\in{\rm{Hol}}(\Bb{H}), 1j31\leqslant j\leqslant 3, as follows from (A.4d).

We apply the approach suggested in (bh1, , p.​ 24). According to this approach, it is necessary to introduce the following three functions

Φj(z):=Ψj(λ(z)), 1j3,zΛ:=(0,1)(),\displaystyle\Phi_{j}(z)\!:=\!\Psi_{j}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\ ,\ 1\leqslant j\leqslant 3\ ,\ \ z\in\Lambda:=(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right), (A.5d)

which are holomorphic in Λ\Lambda and λ(Λ)=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(\Lambda)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}. And then to study the values of {Φj}j=13\{\Phi_{j}\}_{j=1}^{3} on the two sides of the cuts along \Bb{R}_{<0} and along \Bb{R}_{>1}, and their behavior near the points {0,1,}\{0,1,\infty\}.

We first look on their values on the two sides of the cut along \Bb{R}_{<0}. Since by (bh2, , p.​ 19, (6.8)) we have

Θ4(z+1)=Θ3(z),Θ3(z+1)=Θ4(z),Θ2(z+1)=eiπ/4Θ2(z),z\displaystyle\Theta_{4}(z+1)=\Theta_{3}(z)\,,\ \Theta_{3}(z+1)=\Theta_{4}(z)\,,\ \Theta_{2}(z+1)={\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi/4}\Theta_{2}(z)\ ,\quad z\in\Bb{H}\,,\

which yield (see (bh2, , p.​ 56, (A.18a)))

Θ4(z+2)=Θ4(z),Θ3(z+2)=Θ3(z),Θ2(z+2)=iΘ2(z),Θ4(z+4)=Θ4(z),Θ3(z+4)=Θ3(z),Θ2(z+4)=Θ2(z),z\displaystyle\begin{array}[]{llll}\Theta_{4}(z+2)=\Theta_{4}(z)\,,&\ \Theta_{3}(z+2)=\Theta_{3}(z)\,,&\ \Theta_{2}(z+2)=i\Theta_{2}(z)\,,&\\ \Theta_{4}(z+4)=\Theta_{4}(z)\,,&\ \Theta_{3}(z+4)=\Theta_{3}(z)\,,&\ \Theta_{2}(z+4)=-\Theta_{2}(z)\,,&z\in\Bb{H}\,,\end{array}

then

Φ1(2+z)\displaystyle\Phi_{1}(2+z) =Θ3(2+z)2Θ4(2+z)22Θ2(4+2z)2=Θ3(z)2Θ4(z)22Θ2(2z)2=Φ1(z),\displaystyle=\dfrac{\Theta_{3}\big{(}2+z\big{)}^{2}\!-\!\Theta_{4}\big{(}2+z\big{)}^{2}}{2\Theta_{2}\big{(}4+2z\big{)}^{2}}=\dfrac{\Theta_{3}\big{(}z\big{)}^{2}\!-\!\Theta_{4}\big{(}z\big{)}^{2}}{2\Theta_{2}\big{(}2z\big{)}^{2}}=\Phi_{1}(z)\,,\
Φ2(2+z)\displaystyle\Phi_{2}(2+z) =Θ3(2+z)2+Θ4(2+z)22Θ3(4+2z)2=Θ3(z)2+Θ4(z)22Θ3(2z)2=Φ2(z),\displaystyle=\dfrac{\Theta_{3}\big{(}2+z\big{)}^{2}\!+\!\Theta_{4}\big{(}2+z\big{)}^{2}}{2\Theta_{3}\big{(}4+2z\big{)}^{2}}=\dfrac{\Theta_{3}\big{(}z\big{)}^{2}\!+\!\Theta_{4}\big{(}z\big{)}^{2}}{2\Theta_{3}\big{(}2z\big{)}^{2}}=\Phi_{2}(z)\,,\
Φ3(2+z)\displaystyle\Phi_{3}(2+z) =Θ4(4+2z)2Θ3(2+z)Θ4(2+z)=Θ4(2z)2Θ3(z)Θ4(z)=Φ3(z),z\displaystyle=\dfrac{\Theta_{4}\big{(}4+2z\big{)}^{2}}{\Theta_{3}\big{(}2+z\big{)}\Theta_{4}\big{(}2+z\big{)}}=\dfrac{\Theta_{4}\big{(}2z\big{)}^{2}}{\Theta_{3}\big{(}z\big{)}\Theta_{4}\big{(}z\big{)}}=\Phi_{3}(z)\,,\ \quad z\in\Bb{H}\,,

and we derive from (2.66),

λ(x+i0)=2+λ(xi0),x>0,\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)=2+\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(-x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)\ ,\quad x>0\,,

that for every x>0x>0 we have

Φ1(x+i0)=Φ1(xi0),Φ2(x+i0)=Φ2(xi0),\displaystyle\Phi_{1}\big{(}-x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{1}\big{(}-x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\ \ \Phi_{2}\big{(}-x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{2}\big{(}-x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,, (A.5e)
Φ3(x+i0)=Φ3(xi0).\displaystyle\Phi_{3}\big{(}-x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{3}\big{(}-x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,. (A.5f)

Applying to these relationships the Morera theorem (see (lav, , p.​ 96)) we obtain that

ΦjHol({}^\displaystyle\Phi_{j}\in{\rm{Hol}}\big{(}\Bb{C}\setminus\left(\{0\}\cup[1,+\infty\right)\big{)},\ \ 1\leqslant j\leqslant 3, (A.5g)

and therefore 0 could be a point of an isolated singularity for all functions {Φj}j=13\{\Phi_{j}\}_{j=1}^{3} (see (con, , p.​ 103)). We prove that actually 0 is a point of a removable singularity for these functions. To prove this, we apply the Riemann theorem about removable singularities (see (con, , p.​ 103)), according to which 0 is a point of a removable singularity of Φj\Phi_{j}, where 1j31\leqslant j\leqslant 3, if and only if limz0zΦj(z)=0\lim_{z\to 0}z\Phi_{j}(z)=0. For any 1j31\leqslant j\leqslant 3 the latter equality obviously follows from more strong property, sup2z𝔻{}|Φj(z)|=sup2z𝔻~|Φj(z)|<+\sup_{2z\in\Bb{D}\setminus\{0\}}|\Phi_{j}(z)|=\sup_{2z\in\Bb{D}\setminus(-1,0]}|\Phi_{j}(z)|<+\infty, and this in turn is equivalent to lim¯z𝔻~ϝ|Φj(z)|<+\operatorname*{\overline{\lim}}_{z\in\Bb{D}\setminus(-1,0]\,,\ z\to 0}|\Phi_{j}(z)|<+\infty, because ΦjHol(𝔻{}\Phi_{j}\in{\rm{Hol}}(\Bb{D}\setminus\{0\}) and therefore Φj\Phi_{j} is continuous on 𝔻{}\Bb{D}\setminus\{0\}. According to (bh2, , p.​ 609, (4.2), (4.5), (4.6)),

(0,1)()z0λ(z)Imλ(z)+,\displaystyle(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\ni z\to 0\ \Rightarrow\ \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\to\infty\ \Rightarrow\ {\rm{Im}}\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\to+\infty\,,

and since

Θ3(z)2Θ4(z)2=(1+2n1eiπn2z12n1(1)neiπn2z)×\displaystyle\Theta_{3}(z)^{2}-\Theta_{4}(z)^{2}=\left(1\!+\!2\sum\limits_{n\geqslant 1}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n^{2}z}\!-\!1\!-\!2\sum\limits_{n\geqslant 1}(-1)^{n}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n^{2}z}\right)\times
×(1+2n1eiπn2z+1+2n1(1)neiπn2z)\displaystyle\times\left(1\!+\!2\sum\limits_{n\geqslant 1}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n^{2}z}\!+\!1\!+\!2\sum\limits_{n\geqslant 1}(-1)^{n}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi n^{2}z}\right)
=(4eiπz+4n1eiπ(2n+1)2z)(2+4n1e4πin2z)=8eiπz+O(e4πiz),\displaystyle=\left(4{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}+4\sum\limits_{n\geqslant 1}{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(2n+1)^{2}z}\right)\left(2+4\sum\limits_{n\geqslant 1}{\rm{e}}^{4\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptn^{2}z}\right)=8{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}+{\rm{O}}\left({\rm{e}}^{4\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz}\right),
Θ3(z)=1+O(eiπz),Θ4(z)=1+O(eiπz),Θ2(z)=2eiπz/4+O(e5πiz/4),\displaystyle\Theta_{3}(z)=1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right),\ \Theta_{4}(z)=1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right),\ \Theta_{2}(z)=2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/4}+{\rm{O}}\left({\rm{e}}^{5\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz/4}\right),

as Imz+{\rm{Im}}\,z\to+\infty, we obtain the existence of the following limits

lim𝔻~ϝΦ1(z)=lim𝔻~ϝΘ3(2λ(z))2Θ4(2λ(z))22Θ2(2λ(z))2\displaystyle\lim_{{{\Bb{D}\setminus(-1,0]\ni z\to 0}}}\Phi_{1}(z)=\lim_{{{\Bb{D}\setminus(-1,0]\ni z\to 0}}}\dfrac{\Theta_{3}\big{(}2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}\!-\!\Theta_{4}\big{(}2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}}{2\Theta_{2}\big{(}2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}}
=limzΘ3(z)2Θ4(z)22Θ2(2z)2=limImz+8eiπz+O(e4πiz)2(2eiπz/2+O(e5πiz/2))2=1,\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z\to\infty}}}\dfrac{\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2}}{2\Theta_{2}(2z)^{2}}=\lim_{{{{\rm{Im}}\,z\to+\infty}}}\dfrac{8{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}+{\rm{O}}\left({\rm{e}}^{4\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz}\right)}{2\left(2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/2}+{\rm{O}}\left({\rm{e}}^{5\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz/2}\right)\right)^{2}}=1\,,
lim𝔻~ϝΦ2(z)=lim𝔻~ϝΘ3(2λ(z))2+Θ4(2λ(z))22Θ3(2λ(z))2\displaystyle\lim_{{{\Bb{D}\setminus(-1,0]\ni z\to 0}}}\Phi_{2}(z)=\lim_{{{\Bb{D}\setminus(-1,0]\ni z\to 0}}}\dfrac{\Theta_{3}\big{(}2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}\!+\!\Theta_{4}\big{(}2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}}{2\Theta_{3}\big{(}2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}}
=limzΘ3(z)2+Θ4(z)22Θ3(2z)2=limImz+(1+O(eiπz))2+(1+O(eiπz))22(1+O(eiπz))2=1,\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z\to\infty}}}\dfrac{\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2}}{2\Theta_{3}(2z)^{2}}=\lim_{{{{\rm{Im}}\,z\to+\infty}}}\dfrac{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right)^{2}+\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right)^{2}}{2\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right)^{2}}=1\,,
lim𝔻~ϝΦ3(z)=lim𝔻~ϝΘ4(2λ(z))2Θ3(λ(z))Θ4(λ(z))\displaystyle\lim_{{{\Bb{D}\setminus(-1,0]\ni z\to 0}}}\Phi_{3}(z)=\lim_{{{\Bb{D}\setminus(-1,0]\ni z\to 0}}}\dfrac{\Theta_{4}\big{(}2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}^{2}}{\Theta_{3}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\Theta_{4}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}}
=limzΘ4(2z)2Θ3(z)Θ4(z)=limImz+(1+O(e2πiz))2(1+O(eiπz))(1+O(eiπz))=1.\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z\to\infty}}}\dfrac{\Theta_{4}(2z)^{2}}{\Theta_{3}(z)\Theta_{4}(z)}=\lim_{{{{\rm{Im}}\,z\to+\infty}}}\dfrac{\left(1+{\rm{O}}\left({\rm{e}}^{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz}\right)\right)^{2}}{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right)\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right)}=1\,.

Thus,

ΦjHol(^Φ\displaystyle\Phi_{j}\in{\rm{Hol}}\big{(}\Bb{C}\setminus[1,+\infty)\big{)},\ \ \Phi_{j}(0)=1\,,\ \ ,\quad 1\leqslant j\leqslant 3, (A.5h)

We now look on the values of {Φj}j=13\{\Phi_{j}\}_{j=1}^{3} on the two sides of the cut along \Bb{R}_{>1}. For arbitrary zz\in\Bb{H} and σ{1,1}\sigma\in\{1,-1\}, by using the identities,

Θ3(z)=iΘ3(z12z)(2z1)1/2,Θ4(z)=Θ4(z12z)(2z1)1/2,Θ2(z)=iΘ2(z12z)(2z1)1/2,Θ3(z)2=Θ3(z12z)22z1,Θ4(z)2=Θ4(z12z)22z1,Θ2(z)2=Θ2(z12z)212z,Θ3(z)2=iΘ2(1σz)2zσ,Θ4(z)2=iΘ3(1σz)2zσ,Θ2(z)2=Θ4(1σz)2σz,\displaystyle\begin{array}[]{lll}\Theta_{3}(z)\!=\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)}{(2z\!-\!1)^{1/2}}\,\,,&\ \Theta_{4}(z)\!=\!\,\,\dfrac{\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)}{(2z\!-\!1)^{1/2}}\,\,,&\ \Theta_{2}(z)\!=\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(\dfrac{z}{1\!-\!2z}\right)}{(2z\!-\!1)^{1/2}}\,,\\ \Theta_{3}(z)^{2}\!=\!\,\!-\!\,\dfrac{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{2z\!-\!1}\,\,,&\ \Theta_{4}(z)^{2}\!=\!\,\,\dfrac{\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{2z\!-\!1}\,\,,&\ \Theta_{2}(z)^{2}\!=\!\dfrac{\Theta_{2}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{1\!-\!2z}\,,\\ \Theta_{3}(z)^{2}\!=\!\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(\dfrac{1}{\sigma\!-\!z}\right)^{2}}{z\!-\!\sigma}\,\,,&\ \Theta_{4}(z)^{2}\!=\!\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{3}\left(\dfrac{1}{\sigma\!-\!z}\right)^{2}}{z\!-\!\sigma}\,\,,&\ \Theta_{2}(z)^{2}\!=\!\dfrac{\Theta_{4}\left(\dfrac{1}{\sigma\!-\!z}\right)^{2}}{\sigma\!-\!z}\ ,\end{array}

(see (bh1, , p.​ 57, (A.18b)(b), (A.18b)(c); p.58, (A.18e)(c))) we obtain, for every zz\in\Bb{H},

Ψ1(z)\displaystyle\Psi_{1}(z) =Θ3(z)2Θ4(z)22Θ2(2z)2=Θ3(z12z)2(2z1)Θ4(z12z)2(2z1) 2Θ4(112z)22z1\displaystyle\!=\!\dfrac{\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2}}{2\Theta_{2}(2z)^{2}}\!=\!\dfrac{\,\!-\!\,\dfrac{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{(2z\!-\!1)}\!-\!\dfrac{\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{(2z\!-\!1)}}{\,\!-\!\,2\dfrac{\Theta_{4}\left(\dfrac{1}{1\!-\!2z}\right)^{2}}{2z\!-\!1}}
=Θ3(z12z)2Θ4(z12z)2 2Θ4(11+112z)2=Θ3(z12z)2Θ4(z12z)2 2Θ3(1+112z)2\displaystyle\!=\!\dfrac{\,\!-\!\,\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}\!-\!\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{\,\!-\!\,2\Theta_{4}\left(1\!-\!1+\dfrac{1}{1\!-\!2z}\right)^{2}}\!=\!\dfrac{\,\!-\!\,\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}\!-\!\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{\,\!-\!\,2\Theta_{3}\left(\!-\!1+\dfrac{1}{1\!-\!2z}\right)^{2}}
=Θ3(z12z)2+Θ4(z12z)22Θ3(2z12z)2=Ψ2(z12z);\displaystyle\!=\!\dfrac{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}+\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{2\Theta_{3}\left(\dfrac{2z}{1\!-\!2z}\right)^{2}}\!=\!\Psi_{2}\left(\dfrac{z}{1\!-\!2z}\right)\,;\
Ψ2(z)\displaystyle\Psi_{2}(z) =Θ3(z)2+Θ4(z)22Θ3(2z)2=Θ3(z12z)2(2z1)+Θ4(z12z)2(2z1) 2iΘ2(112z)22z1\displaystyle\!=\!\dfrac{\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2}}{2\Theta_{3}(2z)^{2}}\!=\!\dfrac{\,\!-\!\,\dfrac{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{(2z\!-\!1)}\!+\!\dfrac{\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{(2z\!-\!1)}}{\,2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(\dfrac{1}{1\!-\!2z}\right)^{2}}{2z\!-\!1}}
=Θ3(z12z)2+Θ4(z12z)2 2iΘ2(11+112z)2=Θ3(z12z)2+Θ4(z12z)22iiΘ2(1+112z)2\displaystyle\!=\!\dfrac{\,\!-\!\,\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}\!+\!\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{\,2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\Theta_{2}\left(1\!-\!1+\dfrac{1}{1\!-\!2z}\right)^{2}}\!=\!\dfrac{\,\!-\!\,\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}\!+\!\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\cdot\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\Theta_{2}\left(\!-\!1+\dfrac{1}{1\!-\!2z}\right)^{2}}
=Θ3(z12z)2Θ4(z12z)22Θ2(2z12z)2=Ψ1(z12z);\displaystyle\!=\!\dfrac{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)^{2}\!-\!\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)^{2}}{2\Theta_{2}\left(\dfrac{2z}{1\!-\!2z}\right)^{2}}\!=\!\Psi_{1}\left(\dfrac{z}{1\!-\!2z}\right)\,;\
Ψ3(z)\displaystyle\Psi_{3}(z) =Θ4(2z)2Θ3(z)Θ4(z)=iΘ3(112z)22z1iΘ3(z12z)(2z1)1/2Θ4(z12z)(2z1)1/2=Θ3(112z)2Θ3(z12z)Θ4(z12z)\displaystyle\!=\!\dfrac{\Theta_{4}(2z)^{2}}{\Theta_{3}(z)\Theta_{4}(z)}\!=\!\dfrac{i\,\dfrac{\Theta_{3}\left(\dfrac{1}{1\!-\!2z}\right)^{2}}{2z\!-\!1}}{i\,\dfrac{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)}{(2z\!-\!1)^{1/2}}\dfrac{\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)}{(2z\!-\!1)^{1/2}}}\!=\!\dfrac{\Theta_{3}\left(\dfrac{1}{1\!-\!2z}\right)^{2}}{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)}
=Θ3(11+112z)2Θ3(z12z)Θ4(z12z)=Θ4(1+112z)2Θ3(z12z)Θ4(z12z)\displaystyle\!=\!\dfrac{\Theta_{3}\left(1\!-\!1+\dfrac{1}{1\!-\!2z}\right)^{2}}{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)}\!=\!\dfrac{\Theta_{4}\left(\!-\!1+\dfrac{1}{1\!-\!2z}\right)^{2}}{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)}
=Θ4(2z12z)2Θ3(z12z)Θ4(z12z)=Ψ3(z12z),\displaystyle\!=\!\dfrac{\Theta_{4}\left(\dfrac{2z}{1\!-\!2z}\right)^{2}}{\Theta_{3}\left(\dfrac{z}{1\!-\!2z}\right)\Theta_{4}\left(\dfrac{z}{1\!-\!2z}\right)}\!=\!\Psi_{3}\left(\dfrac{z}{1\!-\!2z}\right)\ ,\

and derive from (2.67),

λ(1+xi0)=λ(1+x+i0)12λ(1+x+i0), 2λ(1+xi0)=2λ(1+x+i0)12λ(1+x+i0),x>0,\displaystyle\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)\!=\!\dfrac{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)}{1\!-\!2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)}\,,\ 2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)\!=\!\dfrac{2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)}{1\!-\!2\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0)}\,,\ x\!>\!0\,,

that

(1)Φ1(1+xi0)=Φ2(1+x+i0),Φ2(1+xi0)=Φ1(1+x+i0),x>0,\displaystyle{\rm{(1)}}\ \,\Phi_{1}\big{(}1\!+\!x\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\!=\!\Phi_{2}\big{(}1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\ \Phi_{2}\big{(}1\!+\!x\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\!=\!\Phi_{1}\big{(}1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\ x>0, (A.5i)
(2)Φ3(1+xi0)=Φ3(1+x+i0),x>0.\displaystyle{\rm{(2)}}\ \,\Phi_{3}\big{(}1\!+\!x\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\!=\!\Phi_{3}\big{(}1\!+\!x\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\ x>0\,. (A.5j)

Applying the Morera theorem (see (lav, , p.​ 96)) to these relationships we obtain

Φ3Hol({}Φ{}\displaystyle\Phi_{3}\in{\rm{Hol}}\big{(}\Bb{C}\setminus\{1\}\big{)},\ \ \Phi_{0}\in{\rm{Hol}}\big{(}\Bb{C}\setminus\{0\}\big{)}, (A.5k)

where the function

Φ0(z):={Φ1(1+z2),zΦ2(1+z2),z\displaystyle\Phi_{0}(z):=\left\{\begin{array}[]{ll}\Phi_{1}\big{(}1+z^{2}\big{)}\,,&z\in\Bb{H}\,,\\ \Phi_{2}\big{(}1+z^{2}\big{)}\,,&z\in-\Bb{H}\,,\end{array}\right. (A.5n)

is holomorphic in \Bb{C}\setminus\Bb{R} as follows from (A.5h), theorem about analyticity of the composition of two holomorphic functions (see (con, , p.​ 34)), and the fact that 1+z21+z^{2} maps conformally ±\pm\Bb{H} onto ^\Bb{C}\setminus[1,+\infty). The latter property is the consequence of the fact that the function ±i1ζ\pm i\sqrt{1-\zeta} maps conformally ^\Bb{C}\setminus[1,+\infty) onto ±\pm\Bb{H} (see (con, , p.​ 46)). Therefore the inverse mapping 1+z21+z^{2} maps ±\pm\Bb{H} one-to-one onto ^\Bb{C}\setminus[1,+\infty). Then 1+z2=1+(x+iy)2=1+x2y2+2ixy1+z^{2}=1+(x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty)^{2}=1+x^{2}-y^{2}+2\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptxy, z:=x+iyz:=x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pty, x,yx,y\in\Bb{R} and (A.5n), for arbitrary x>0x>0 imply

{Φ0(x+i0)=Φ1(1+x2+i0),Φ0(xi0)=Φ2(1+x2i0),{Φ0(x+i0)=Φ1(1+x2i0),Φ0(xi0)=Φ2(1+x2+i0),\displaystyle\left\{\begin{array}[]{l}\Phi_{0}\big{(}x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{1}\big{(}1+x^{2}+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\\ \Phi_{0}\big{(}x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{2}\big{(}1+x^{2}-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\end{array}\right.\ \ \left\{\begin{array}[]{l}\Phi_{0}\big{(}-x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{1}\big{(}1+x^{2}-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\\ \Phi_{0}\big{(}-x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{2}\big{(}1+x^{2}+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,,\end{array}\right.

which by (A.5i) means that

Φ0(x+i0)=Φ1(1+x2+i0)=Φ2(1+x2i0)=Φ0(xi0),\displaystyle\Phi_{0}\big{(}x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{1}\big{(}1+x^{2}+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{2}\big{(}1+x^{2}-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{0}\big{(}x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,, x>0,\displaystyle\ \ x>0,
Φ0(x+i0)=Φ1(1+x2i0)=Φ2(1+x2+i0)=Φ0(xi0),\displaystyle\Phi_{0}\big{(}-x+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{1}\big{(}1+x^{2}-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{2}\big{(}1+x^{2}+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}=\Phi_{0}\big{(}-x-\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt0\big{)}\,, x>0,\displaystyle\ \ x>0,

and hence the Morera theorem can be applied to Φ0Hol(\Phi_{0}\in{\rm{Hol}}(\Bb{C}\setminus\Bb{R}) to get Φ0Hol({}\Phi_{0}\in{\rm{Hol}}(\Bb{C}\setminus\{0\}). The right-hand side inclusion of (A.5k) is completely proved.

We prove now that 11 and 0 are the points of a removable singularity for Φ3\Phi_{3} and Φ0\Phi_{0}, respectively. According to (bh2, , p.​ 609, (4.2), (4.4)),

(0,1)()z1λ(z)0Im(1/λ(z))+,\displaystyle(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\!\ni\!z\!\to\!1\ \Rightarrow\ \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\!\to\!0\ \Rightarrow\ {\rm{Im}}\big{(}-1/\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}\!\to\!+\!\infty\,, (A.5o)

because in view of 1/=-1/\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\subset\Bb{H}_{{\rm{Re}}<1}, it follows from λ(z)0\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\to 0 that 1/λ(z)-1/\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and |1/λ(z)|+|1/\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)|\to+\infty, which means that Im(1/λ(z))|1/λ(z)|1+{\rm{Im}}(-1/\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z))\geqslant|1/\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)|-1\to+\infty. Then the asymptotic equalities (see (2.25)(a),(b),(c)),

Θ3(z)=Θ3(1/z)(z/i)1/2,Θ2(z)=Θ4(1/z)(z/i)1/2,Θ4(z)=Θ2(1/z)(z/i)1/2,\displaystyle\Theta_{3}(z)=\dfrac{\Theta_{3}(-1/z)}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}}\ ,\ \Theta_{2}(z)=\dfrac{\Theta_{4}(-1/z)}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}}\ ,\ \Theta_{4}(z)=\dfrac{\Theta_{2}(-1/z)}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}}\ ,\
Θ3(1/z)=1+O(eiπ(1/z)),Θ4(1/z)=1+O(eiπ(1/z)),\displaystyle\Theta_{3}(-1/z)=1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\,,\ \Theta_{4}(-1/z)=1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\,,
Θ2(1/z)=2eiπ(1/z)/4(1+O(eiπ(1/z))),\displaystyle\Theta_{2}(-1/z)=2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)/4}\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\right)\,,

as zz\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, z0z\to 0 (and hence, Im(1/z)+{\rm{Im}}(-1/z)\to+\infty) yield

limz0Ψ1(z)=limz0Θ3(z)2Θ4(z)22Θ2(2z)2=limz0Θ3(1/z)2(z/i)Θ2(1/z)2(z/i)2Θ4(1/(2z))2(2z/i)\displaystyle\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\Psi_{1}(z)\!=\!\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2}}{2\Theta_{2}(2z)^{2}}\!=\!\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\dfrac{\Theta_{3}(-1/z)^{2}}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)}\!-\!\dfrac{\Theta_{2}(-1/z)^{2}}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)}}{2\dfrac{\Theta_{4}(-1/(2z))^{2}}{(2z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)}}
=limz0Θ3(1/z)2Θ2(1/z)2Θ4(1/(2z))2\displaystyle=\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\Theta_{3}(-1/z)^{2}\!-\!\Theta_{2}(-1/z)^{2}}{\Theta_{4}(-1/(2z))^{2}}
=limIm(1/z)+(1+O(eiπ(1/z)))24eiπ(1/z)/2(1+O(eiπ(1/z)))2(1+O(eiπ(1/(2z))))2=1,\displaystyle=\lim_{{{{\rm{Im}}(-1/z)\to+\infty}}}\dfrac{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\right)^{2}-4{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)/2}\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\right)^{2}}{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/(2z))}\right)\right)^{2}}=1\,,
limz0Ψ2(z)=limz0Θ3(z)2+Θ4(z)22Θ3(2z)2=limz0Θ3(1/z)2(z/i)+Θ2(1/z)2(z/i)2Θ3(1/(2z))2(2z/i)\displaystyle\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\Psi_{2}(z)\!=\!\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2}}{2\Theta_{3}(2z)^{2}}\!=\!\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\dfrac{\Theta_{3}(-1/z)^{2}}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)}\!+\!\dfrac{\Theta_{2}(-1/z)^{2}}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)}}{2\dfrac{\Theta_{3}(-1/(2z))^{2}}{(2z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)}}
=limz0Θ3(1/z)2+Θ2(1/z)2Θ3(1/(2z)2\displaystyle=\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\Theta_{3}(-1/z)^{2}\!+\!\Theta_{2}(-1/z)^{2}}{\Theta_{3}(-1/(2z)^{2}}
=limIm(1/z)+(1+O(eiπ(1/z)))2+4eiπ(1/z)/2(1+O(eiπ(1/z)))2(1+O(eiπ(1/(2z))))2=1,\displaystyle=\lim_{{{{\rm{Im}}(-1/z)\to+\infty}}}\dfrac{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\right)^{2}+4{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)/2}\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\right)^{2}}{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/(2z))}\right)\right)^{2}}=1\,,\
limz0Ψ3(z)=limz0Θ4(2z)2Θ3(z)Θ4(z)=limz0Θ2(1/(2z))2(2z/i)Θ3(1/z)(z/i)1/2Θ2(1/z)(z/i)1/2\displaystyle\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\Psi_{3}(z)\!=\!\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\Theta_{4}(2z)^{2}}{\Theta_{3}(z)\Theta_{4}(z)}\!=\!\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\dfrac{\Theta_{2}(-1/(2z))^{2}}{(2z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)}}{\dfrac{\Theta_{3}(-1/z)}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}}\dfrac{\Theta_{2}(-1/z)}{(z/\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt)^{1/2}}}
=limz0Θ2(1/(2z))22Θ3(1/z)Θ2(1/z)\displaystyle=\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\dfrac{\Theta_{2}(-1/(2z))^{2}}{2\Theta_{3}(-1/z)\Theta_{2}(-1/z)}
=limIm(1/z)+4eiπ(1/2z)/2(1+O(eiπ(1/(2z))))22(1+O(eiπ(1/z)))2eiπ(1/z)/4(1+O(eiπ(1/z)))=1.\displaystyle=\lim_{{{{\rm{Im}}(-1/z)\to+\infty}}}\dfrac{4{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/2z)/2}\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/(2z))}\right)\right)^{2}}{2\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\right)2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)/4}\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi(-1/z)}\right)\right)}=1\,.

Hence, by (A.5o), we have, for Λ:=(0,1)()\Lambda:=(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right),

limΛz1Φj(z)=limΛz1Ψj(λ(z))=limz0Ψj(z)=1, 1j3.\displaystyle\lim_{{{\mbox{\footnotesize{$\Lambda\!\ni\!z\!\to\!1$}}}}}\Phi_{j}(z)=\lim_{{{\mbox{\footnotesize{$\Lambda\!\ni\!z\!\to\!1$}}}}}\Psi_{j}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}=\lim_{{{\mbox{\footnotesize{$\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!\ni\!z\!\to\!0$}}}}}\Psi_{j}\big{(}z\big{)}=1\,,\ 1\leqslant j\leqslant 3. (A.5p)

In view of the continuity of Φ0\Phi_{0} on {}\Bb{C}\setminus\{0\}, as follows from (A.5k), the limit of Φ0\Phi_{0} as {}ϝ\Bb{C}\setminus\{0\}\ni z\to 0 exists if and only if there exists the limit of Φ0\Phi_{0} as ϝ\Bb{C}\setminus\Bb{R}\ni z\to 0, and we derive from (A.5p), \Bb{C}\setminus\Bb{R}=(-\Bb{H})\sqcup\Bb{H} and (1+z2):±Λ(1+z^{2}):\pm\Bb{H}\to\Lambda, the existence of the following limit

lim{}ϝΦ0(z)=limϝΦ0(z)\displaystyle\lim_{{{\mbox{\footnotesize{$\Bb{C}\!\setminus\!\{0\}\!\ni\!z\!\to\!0$}}}}}\Phi_{0}(z)=\lim_{{{\mbox{\footnotesize{$\Bb{C}\!\setminus\!\Bb{R}\!\ni\!z\!\to\!0$}}}}}\Phi_{0}(z)
=limϝ[Φ1(1+z2)χ(z)+Φ2(1+z2)χ(z)]=limΛz1{Φ1(z)Φ2(z)}=1.\displaystyle=\lim_{{{\mbox{\footnotesize{$\Bb{C}\!\setminus\!\Bb{R}\!\ni\!z\!\to\!0$}}}}}\big{[}\Phi_{1}\big{(}1+z^{2}\big{)}\chi_{\Bb{H}}(z)+\Phi_{2}\big{(}1+z^{2}\big{)}\chi_{-\Bb{H}}(z)\big{]}=\lim\limits_{{{\mbox{\footnotesize{$\Lambda\!\ni\!z\!\to\!1$}}}}}\left\{\begin{array}[]{l}\Phi_{1}(z)\\ \Phi_{2}(z)\end{array}\right\}=1.

Together with (A.5p) for j=3j=3 and (A.5k) this relationship yields that 11 and 0 are the points of a removable singularity for Φ3\Phi_{3} and Φ0\Phi_{0}, respectively, and so Φ3\Phi_{3} and Φ0\Phi_{0} are entire functions, i.e.,

Φ3,Φ0Hol(ΦΦ\displaystyle\Phi_{3},\ \Phi_{0}\in{\rm{Hol}}\big{(}\Bb{C}\big{)}\,,\quad\Phi_{0}(0)=\Phi_{3}(1)=1. (A.5q)

We prove now that the modulus of Φ3\Phi_{3} and Φ0\Phi_{0} are uniformly bounded on \Bb{C} by establishing the existence of the finite limits of Φ3(z)\Phi_{3}(z) and Φ0(z)\Phi_{0}(z) as ϝ\Bb{C}\ni z\to\infty.

Let Λz\Lambda\ni z\!\to\!\infty approaching from one of the half-planes σ:=sign(Imz){1,1}\sigma\!:=\!{\rm{sign}}({\rm{Im}}z)\!\in\!\{1,-1\}. Then λ(z)σ\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\to\sigma (see (bh2, , p.​ 609, (4.2), (4.3))) and, in view of (2.16), sign(Reλ(z))=σ{\rm{sign}}({\rm{Re}}\,\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z))=\sigma. The latter equality yields λ(z)σ0\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)-\sigma\!\to\!0 (see (bh1, , p.​ 24, item 2)) and by manipulations similar to those employed in the proof of (A.5o), for arbitrary σ{1,1}\sigma\!\in\!\{1,-1\} we get

σϝ𝔽λϝσ𝕀λϝσ\displaystyle\sigma\,\Bb{H}\ni z\!\to\!\infty\ \Rightarrow\ \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)-\sigma\!\to\!0\ \Rightarrow\ {\rm{Im}}\big{(}-1/\left(\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)-\sigma\right)\big{)}\!\to\!+\!\infty\,. (A.5r)

By using the relationships (see (bh1, , p.​ 58, (A.18e), (A.18d)),

Θ3(z)2=iΘ2(1σz)2zσ,Θ4(z)2=iΘ3(1σz)2zσ,Θ2(z)2=Θ4(1σz)2σz,\displaystyle\Theta_{3}(z)^{2}\!=\!\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(\dfrac{1}{\sigma\!-\!z}\right)^{2}}{z\!-\!\sigma}\,\,,\ \Theta_{4}(z)^{2}\!=\!\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{3}\left(\dfrac{1}{\sigma\!-\!z}\right)^{2}}{z\!-\!\sigma}\,\,,\ \Theta_{2}(z)^{2}\!=\!\dfrac{\Theta_{4}\left(\dfrac{1}{\sigma\!-\!z}\right)^{2}}{\sigma\!-\!z}\ ,
Θ3(z)2=iΘ3(1z)2z,Θ4(z)2=iΘ2(1z)2z,Θ2(z)2=iΘ4(1z)2z,\displaystyle\Theta_{3}(z)^{2}\!=\!\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{3}\left(-\dfrac{1}{z}\right)^{2}}{z}\,\,,\hskip 11.38092pt\Theta_{4}(z)^{2}\!=\!\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(-\dfrac{1}{z}\right)^{2}}{z}\,\,,\hskip 11.38092pt\Theta_{2}(z)^{2}\!=\!\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{4}\left(-\dfrac{1}{z}\right)^{2}}{z}\,\,,
Θ3(z)Θ4(z)=iΘ2(1σz)Θ3(1σz)zσ,zσ{}\displaystyle\Theta_{3}(z)\Theta_{4}(z)=\,\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(\dfrac{1}{\sigma\!-\!z}\right)\Theta_{3}\left(\dfrac{1}{\sigma\!-\!z}\right)}{z\!-\!\sigma}\ ,\quad z\in\Bb{H}\,,\ \ \sigma\in\{1,-1\}\,,

we obtain, for arbitrary σ{1,1}\sigma\!\in\!\{1,-1\}, the existence of the following limits

limσϝΦ1(z)=limσϝΨ1(λ(z))=limzσ0Ψ1(z)=limzσ0Θ3(z)2Θ4(z)22Θ2(2z)2\displaystyle\lim_{{{\sigma\Bb{H}\ni z\to\infty}}}\Phi_{1}(z)=\!\!\!\!\!\lim_{{{\sigma\Bb{H}\ni z\to\infty}}}\Psi_{1}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}=\!\!\!\!\!\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\Psi_{1}\big{(}z\big{)}=\!\!\!\!\!\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2}}{2\Theta_{2}(2z)^{2}}
=limzσ0Θ3(z)2Θ4(z)22Θ2(2(zσ)+2σ)2=limzσ0Θ3(z)2Θ4(z)22Θ2(2(zσ))2\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2}}{2\Theta_{2}\big{(}2(z\!-\!\sigma)+2\sigma\big{)}^{2}}=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{3}(z)^{2}\!-\!\Theta_{4}(z)^{2}}{-2\Theta_{2}\big{(}2(z\!-\!\sigma)\big{)}^{2}}
=limzσ0iΘ2(1zσ)2zσiΘ3(1zσ)2zσ2iΘ4(12(zσ))22(zσ)=limzΘ3(z)2Θ2(z)2Θ4(z/2)2\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(-\dfrac{1}{z\!-\!\sigma}\right)^{2}}{z\!-\!\sigma}\!-\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{3}\left(-\dfrac{1}{z\!-\!\sigma}\right)^{2}}{z\!-\!\sigma}}{-2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{4}\left(-\dfrac{1}{2(z\!-\!\sigma)}\right)^{2}}{2(z\!-\!\sigma)}}=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z\to\infty}}}\dfrac{\Theta_{3}(z)^{2}-\Theta_{2}(z)^{2}}{\Theta_{4}(z/2)^{2}}
=limImz+(1+O(eiπz))2(2eiπz/4+O(e5πiz/4))2(1+O(eiπz/2))2=1;\displaystyle=\lim_{{{{\rm{Im}}\,z\to+\infty}}}\dfrac{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right)^{2}-\left(2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/4}+{\rm{O}}\left({\rm{e}}^{5\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz/4}\right)\right)^{2}}{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/2}\right)\right)^{2}}=1\,;
limσϝΦ2(z)=limσϝΨ2(λ(z))=limzσ0Ψ2(z)=limzσ0Θ3(z)2+Θ4(z)22Θ3(2z)2\displaystyle\lim_{{{\sigma\Bb{H}\ni z\to\infty}}}\Phi_{2}(z)=\!\!\!\!\!\lim_{{{\sigma\Bb{H}\ni z\to\infty}}}\Psi_{2}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}=\!\!\!\!\!\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\Psi_{2}\big{(}z\big{)}=\!\!\!\!\!\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2}}{2\Theta_{3}(2z)^{2}}
=limzσ0Θ3(z)2+Θ4(z)22Θ3(2(zσ)+2σ)2=limzσ0Θ3(z)2+Θ4(z)22Θ3(2(zσ))2\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2}}{2\Theta_{3}\big{(}2(z\!-\!\sigma)+2\sigma\big{)}^{2}}=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{3}(z)^{2}\!+\!\Theta_{4}(z)^{2}}{2\Theta_{3}\big{(}2(z\!-\!\sigma)\big{)}^{2}}
=limzσ0iΘ2(1zσ)2zσ+iΘ3(1zσ)2zσ2iΘ3(12(zσ))22(zσ)=limzΘ2(z)2+Θ3(z)2Θ3(z/2)2\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(-\dfrac{1}{z\!-\!\sigma}\right)^{2}}{z\!-\!\sigma}\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{3}\left(-\dfrac{1}{z\!-\!\sigma}\right)^{2}}{z\!-\!\sigma}}{2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{3}\left(-\dfrac{1}{2(z\!-\!\sigma)}\right)^{2}}{2(z\!-\!\sigma)}}=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z\to\infty}}}\dfrac{\Theta_{2}(z)^{2}+\Theta_{3}(z)^{2}}{\Theta_{3}(z/2)^{2}}
=limImz+(2eiπz/4+O(e5πiz/4))2+(1+O(eiπz))2(1+O(eiπz/2))2=1;\displaystyle=\lim_{{{{\rm{Im}}\,z\to+\infty}}}\dfrac{\left(2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/4}+{\rm{O}}\left({\rm{e}}^{5\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz/4}\right)\right)^{2}+\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right)^{2}}{\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/2}\right)\right)^{2}}=1\,;
limσϝΦ3(z)=limσϝΨ3(λ(z))=limzσ0Ψ3(z)=limzσ0Θ4(2z)2Θ3(z)Θ4(z)\displaystyle\lim_{{{\sigma\Bb{H}\ni z\to\infty}}}\Phi_{3}(z)=\!\!\!\!\!\lim_{{{\sigma\Bb{H}\ni z\to\infty}}}\Psi_{3}\big{(}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\big{)}=\!\!\!\!\!\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\Psi_{3}\big{(}z\big{)}=\!\!\!\!\!\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{4}(2z)^{2}}{\Theta_{3}(z)\Theta_{4}(z)}
=limzσ0Θ4(2(zσ)+2σ)2Θ3(z)Θ4(z)=limzσ0Θ4(2(zσ))2Θ3(z)Θ4(z)\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{4}\big{(}2(z\!-\!\sigma)+2\sigma\big{)}^{2}}{\Theta_{3}(z)\Theta_{4}(z)}=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\Theta_{4}\big{(}2(z\!-\!\sigma)\big{)}^{2}}{\Theta_{3}(z)\Theta_{4}(z)}
=limzσ0iΘ2(12(zσ))22(zσ)iΘ2(1σz)Θ3(1σz)zσ=limzΘ2(z/2)22Θ2(z)Θ3(z)\displaystyle=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z-\sigma\to 0}}}\dfrac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(-\dfrac{1}{2(z\!-\!\sigma)}\right)^{2}}{2(z\!-\!\sigma)}}{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\dfrac{\Theta_{2}\left(\dfrac{1}{\sigma\!-\!z}\right)\Theta_{3}\left(\dfrac{1}{\sigma\!-\!z}\right)}{z\!-\!\sigma}}=\lim_{{{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ni z\to\infty}}}\dfrac{\Theta_{2}(z/2)^{2}}{2\Theta_{2}(z)\Theta_{3}(z)}
=limImz+(2eiπz/8+O(e5πiz/8))22(2eiπz/4+O(e5πiz/4))(1+O(eiπz/2))=1.\displaystyle=\lim_{{{{\rm{Im}}\,z\to+\infty}}}\dfrac{\left(2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/8}+{\rm{O}}\left({\rm{e}}^{5\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz/8}\right)\right)^{2}}{2\left(2{\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/4}+{\rm{O}}\left({\rm{e}}^{5\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptz/4}\right)\right)\left(1+{\rm{O}}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z/2}\right)\right)}=1\,.

Hence, for arbitrary σ{1,1}\sigma\!\in\!\{1,-1\} there exist the limits

limσϝΦj(z)=1,1j3.\displaystyle\lim_{{{\sigma\Bb{H}\ni z\to\infty}}}\Phi_{j}(z)=1\ ,\quad 1\leqslant j\leqslant 3\,. (A.5s)

Since by (A.5q) we have Φ3Hol(\Phi_{3}\in{\rm{Hol}}(\Bb{C}) then it follows from (A.5s) with j=3j=3 that Φ3(z)1\Phi_{3}(z)\to 1 as ϝ\Bb{H}\ni z\to\infty and we obtain that the entire function Φ3\Phi_{3} is bounded on \Bb{C} while Φ3(1)=1\Phi_{3}(1)=1, also according to (A.5q). The Liouville theorem (con, , p.​ 77) yields Φ3(z)=1\Phi_{3}(z)=1 for every zz\in\Bb{C} and, in particular, Φ3(z)=1\Phi_{3}(z)=1 for every zΛz\in\Lambda. Then λ(Λ)=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(\Lambda)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and (A.5d) imply the validity of (A.5c) for Ψ3\Psi_{3} and completes the proof of (2.44)(c).

Since 1+z21+z^{2}\in\Bb{R} if and only if z(iz\in(\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R})\cup\Bb{R} and, in accordance with (A.5q), we have Φ0Hol(\Phi_{0}\in{\rm{Hol}}(\Bb{C}), then the limit of Φ0(z)\Phi_{0}(z) as ϝ\Bb{C}\ni z\to\infty exists if and only if there exists the limit of Φ0(z)\Phi_{0}(z) as ϝ\Bb{C}\setminus((\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\Bb{R})\cup\Bb{R})\ni z\to\infty. But in the latter case 1+z21+z^{2}\in\Bb{C}\setminus\Bb{R} and 1+z21+z^{2}\to\infty. In view of the definition (A.5n) of Φ0\Phi_{0} and the properties (A.5s), we conclude that Φ0(z)1\Phi_{0}(z)\to 1 as ϝ\Bb{C}\ni z\to\infty and therefore the entire function Φ0\Phi_{0} is bounded on \Bb{C} while Φ0(1)=1\Phi_{0}(1)=1, according to (A.5q). The Liouville theorem (con, , p.​ 77) gives Φ0(z)=1\Phi_{0}(z)=1 for every zz\in\Bb{C}, which by (A.5n) yield that Φ1(1+z2)=1\Phi_{1}\big{(}1+z^{2}\big{)}=1 for every zz\in\Bb{H} and Φ2(1+z2)=1\Phi_{2}\big{(}1+z^{2}\big{)}=1 for every zz\in-\Bb{H}. This implies that Φ1(z)=Φ2(z)=1\Phi_{1}(z)=\Phi_{2}(z)=1 for every zΛz\in\Lambda because 1+z21+z^{2} maps conformally ±\pm\Bb{H} onto Λ:=^\Lambda:=\Bb{C}\setminus[1,+\infty). Then λ(Λ)=\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(\Lambda)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and (A.5d) yield the validity of (A.5c) for Ψ1\Psi_{1} and Ψ2\Psi_{2} which completes the proof of (2.44)(a) and (2.44)(b). The Landen transformation equations (2.44) have been completely proved.

6

\uparrow  \uparrow  Let zz\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}. Since λ()=1λ()=(0,1)()\lambda(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})=1-\lambda(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}})=(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right) it is possible to use the principal branch of the square root for λ(z)\lambda(z) and 1λ(z)1-\lambda(z) as well. Then by (2.21) and (2.26) we get

λ(z)=Θ2(z)2Θ3(z)2,1λ(z)=Θ4(z)2Θ3(z)2,z.\displaystyle\sqrt{\lambda(z)}=\dfrac{\Theta_{2}(z)^{2}}{\Theta_{3}(z)^{2}}\,,\ \sqrt{1-\lambda(z)}=\dfrac{\Theta_{4}(z)^{2}}{\Theta_{3}(z)^{2}}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,. (A.6a)

Then, by virtue of (2.44)(b), we get

2Θ3(2z)2=Θ3(z)2+Θ4(z)2=Θ3(z)2(1+Θ4(z)2Θ3(z)2)=Θ3(z)2(1+1λ(z)),\displaystyle 2\Theta_{3}(2z)^{2}=\Theta_{3}(z)^{2}+\Theta_{4}(z)^{2}=\Theta_{3}(z)^{2}\left(1+\dfrac{\Theta_{4}(z)^{2}}{\Theta_{3}(z)^{2}}\right)=\Theta_{3}(z)^{2}\left(1+\sqrt{1-\lambda(z)}\right),

from which

Θ3(z)2=2Θ3(2z)21+1λ(z),z.\displaystyle\Theta_{3}(z)^{2}=\dfrac{2\Theta_{3}(2z)^{2}}{1+\sqrt{1-\lambda(z)}}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,. (A.6b)

Substituting here (2.38)(b) written in the form

Θ3(z)2=F(λ(z)),z,\displaystyle\Theta_{3}\big{(}z\big{)}^{2}\!=\!F_{\!{{\mbox{\tiny{$\triangle$}}}}}\big{(}\lambda(z)\big{)}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,, (A.6c)

we obtain

F(λ(z))=2F(λ(2z))1+1λ(z),z.\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\lambda(z)\right)=\dfrac{2F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\lambda(2z)\right)}{1+\sqrt{1-\lambda(z)}}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,. (A.6d)

But if 2z2z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} and zz\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} then it follows from (2.44)(a), (2.44)(b) and (A.6a) that

λ(2z)\displaystyle\sqrt{\lambda(2z)} =Θ2(2z)2Θ3(2z)2=Θ3(z)2Θ4(z)2Θ3(z)2+Θ4(z)2=Θ3(z)2Θ4(z)21Θ3(z)2Θ4(z)2+1=11λ(z)111λ(z)+1\displaystyle=\dfrac{\Theta_{2}(2z)^{2}}{\Theta_{3}(2z)^{2}}=\dfrac{\Theta_{3}(z)^{2}-\Theta_{4}(z)^{2}}{\Theta_{3}(z)^{2}+\Theta_{4}(z)^{2}}=\dfrac{\dfrac{\Theta_{3}(z)^{2}}{\Theta_{4}(z)^{2}}-1}{\dfrac{\Theta_{3}(z)^{2}}{\Theta_{4}(z)^{2}}+1}=\dfrac{\dfrac{1}{\sqrt{1-\lambda(z)}}-1}{\dfrac{1}{\sqrt{1-\lambda(z)}}+1}
=11λ(z)1+1λ(z),\displaystyle=\dfrac{1-\sqrt{1-\lambda(z)}}{1+\sqrt{1-\lambda(z)}}\,,\

and after squaring this identity the constraint 2z2z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} can be dropped,

λ(2z)=(11λ(z)1+1λ(z))2,z,\displaystyle\lambda(2z)=\left(\dfrac{1-\sqrt{1-\lambda(z)}}{1+\sqrt{1-\lambda(z)}}\right)^{2}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,, (A.6e)

and we deduce from (A.6d),

F(λ(z))=21+1λ(z)F((11λ(z)1+1λ(z))2),z.\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\lambda(z)\right)=\dfrac{2}{1+\sqrt{1-\lambda(z)}}\ F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\left(\dfrac{1-\sqrt{1-\lambda(z)}}{1+\sqrt{1-\lambda(z)}}\right)^{2}\right)\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,. (A.6f)

Or, what is the same,

F(z)=21+1zF((11z1+1z)2),z(0,1)(),\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(z\right)=\dfrac{2}{1+\sqrt{1-z}}\ F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\left(\dfrac{1-\sqrt{1-z}}{1+\sqrt{1-z}}\right)^{2}\right)\ ,\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\,, (A.6g)

which coincides with the quadratic transformation (3.1.10) with a=b=1a=b=1 of (and, , p.​ 128) for the hypergeometric function FF_{\!{{\mbox{\tiny{$\triangle$}}}}}.

7

\uparrow   The Landen relationships (2.44)(a)–(c),

2Θ2(2z)2=Θ3(z)2Θ4(z)2,2Θ3(2z)2=Θ3(z)2+Θ4(z)2,Θ4(2z)2=Θ3(z)Θ4(z),z\displaystyle\begin{array}[]{rcl}2\Theta_{2}(2z)^{2}&=&\Theta_{3}(z)^{2}-\Theta_{4}(z)^{2}\ ,\\ 2\Theta_{3}(2z)^{2}&=&\Theta_{3}(z)^{2}+\Theta_{4}(z)^{2}\ ,\\ \Theta_{4}(2z)^{2}&=&\Theta_{3}(z)\Theta_{4}(z)\,,\end{array}\ \quad z\in\Bb{H}\,, (A.7d)

can obviously written as follows

Θ3(z)2=Θ3(2z)2+Θ2(2z)2,Θ4(z)2=Θ3(2z)2Θ2(2z)2,Θ4(2z)2=Θ3(z)Θ4(z),z\displaystyle\begin{array}[]{rcl}\Theta_{3}(z)^{2}&=&\Theta_{3}(2z)^{2}+\Theta_{2}(2z)^{2}\ ,\\ \Theta_{4}(z)^{2}&=&\Theta_{3}(2z)^{2}-\Theta_{2}(2z)^{2}\ ,\\ \Theta_{4}(2z)^{2}&=&\Theta_{3}(z)\Theta_{4}(z)\,,\end{array}\ \quad z\in\Bb{H}\,, (A.7h)

from which, by (2.26), we get

Θ2(z)4=(Θ3(2z)2+Θ2(2z)2)2(Θ3(2z)2Θ2(2z)2)2=4Θ3(2z)2Θ2(2z)2,\displaystyle\Theta_{2}(z)^{4}=\left(\Theta_{3}(2z)^{2}+\Theta_{2}(2z)^{2}\right)^{2}-\left(\Theta_{3}(2z)^{2}-\Theta_{2}(2z)^{2}\right)^{2}=4\Theta_{3}(2z)^{2}\Theta_{2}(2z)^{2}\,,\

i.e.,

Θ2(z)4=4Θ3(2z)2Θ2(2z)2,z\displaystyle\Theta_{2}(z)^{4}=4\Theta_{3}(2z)^{2}\Theta_{2}(2z)^{2}\ ,\quad z\in\Bb{H}\,. (A.7i)

Therefore

λ(z)=Θ2(z)4Θ3(z)2=4Θ3(2z)2Θ2(2z)2Θ3(2z)4+Θ2(2z)4+2Θ3(2z)2Θ2(2z)2\displaystyle\lambda(z)=\dfrac{\Theta_{2}(z)^{4}}{\Theta_{3}(z)^{2}}=\dfrac{4\Theta_{3}(2z)^{2}\Theta_{2}(2z)^{2}}{\Theta_{3}(2z)^{4}+\Theta_{2}(2z)^{4}+2\Theta_{3}(2z)^{2}\Theta_{2}(2z)^{2}}

and then, in view of (2.25) and (2.26),

Θ2(z1)2=iΘ2(z)2,Θ3(z1)2=Θ4(z)2,λ(z)1λ(z)=Θ2(z)4Θ4(z)4,z\displaystyle\Theta_{2}(z-1)^{2}=i\Theta_{2}(z)^{2}\ ,\quad\Theta_{3}(z-1)^{2}=\Theta_{4}(z)^{2}\ ,\quad\dfrac{\lambda(z)}{1-\lambda(z)}=\dfrac{\Theta_{2}(z)^{4}}{\Theta_{4}(z)^{4}}\,,\quad z\in\Bb{H}\,,

we obtain

λ(z12)\displaystyle\lambda\left(\dfrac{z-1}{2}\right) =Θ2(z12)4Θ3(z12)4=4Θ3(z1)2Θ2(z1)2Θ3(z1)4+Θ2(z1)4+2Θ3(z1)2Θ2(z1)2=\displaystyle=\dfrac{\Theta_{2}\left(\dfrac{z-1}{2}\right)^{4}}{\Theta_{3}\left(\dfrac{z-1}{2}\right)^{4}}=\dfrac{4\Theta_{3}(z-1)^{2}\Theta_{2}(z-1)^{2}}{\Theta_{3}(z-1)^{4}+\Theta_{2}(z-1)^{4}+2\Theta_{3}(z-1)^{2}\Theta_{2}(z-1)^{2}}=
=4iΘ4(z)2Θ2(z)2Θ4(z)4Θ2(z)4+2iΘ4(z)2Θ2(z)2,\displaystyle=\dfrac{4i\Theta_{4}(z)^{2}\Theta_{2}(z)^{2}}{\Theta_{4}(z)^{4}-\Theta_{2}(z)^{4}+2i\Theta_{4}(z)^{2}\Theta_{2}(z)^{2}}\,,\

from which, taking account of (2.21), for arbitrary zz\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} we get

λ(1+z1z)=λ(21z1)=λ(21z)λ(21z)1=λ(z12)1λ(z12)=\displaystyle\lambda\left(\dfrac{1+z}{1-z}\right)=\lambda\left(\dfrac{2}{1-z}-1\right)=\dfrac{\lambda\left(\dfrac{2}{1-z}\right)}{\lambda\left(\dfrac{2}{1-z}\right)-1}=\dfrac{\lambda\left(\dfrac{z-1}{2}\right)-1}{\lambda\left(\dfrac{z-1}{2}\right)}=
=11λ(z12)=1Θ4(z)4Θ2(z)4+2iΘ4(z)2Θ2(z)24iΘ4(z)2Θ2(z)2=\displaystyle=1-\dfrac{1}{\lambda\left(\dfrac{z-1}{2}\right)}=1-\dfrac{\Theta_{4}(z)^{4}-\Theta_{2}(z)^{4}+2i\Theta_{4}(z)^{2}\Theta_{2}(z)^{2}}{4i\Theta_{4}(z)^{2}\Theta_{2}(z)^{2}}=
=1214i(Θ4(z)2Θ2(z)2Θ2(z)2Θ4(z)2)=12+14i(λ(z)1λ(z)1λ(z)λ(z))=\displaystyle=\dfrac{1}{2}-\dfrac{1}{4i}\left(\dfrac{\Theta_{4}(z)^{2}}{\Theta_{2}(z)^{2}}-\dfrac{\Theta_{2}(z)^{2}}{\Theta_{4}(z)^{2}}\right)=\dfrac{1}{2}+\dfrac{1}{4i}\left(\sqrt{\dfrac{\lambda(z)}{1-\lambda(z)}}-\sqrt{\dfrac{1-\lambda(z)}{\lambda(z)}}\right)=
=12+2λ(z)14i(1λ(z))λ(z)=12+i12λ(z)4(1λ(z))λ(z),\displaystyle=\dfrac{1}{2}+\dfrac{2\lambda(z)-1}{4i\sqrt{(1-\lambda(z))\lambda(z)}}=\dfrac{1}{2}+i\dfrac{1-2\lambda(z)}{4\sqrt{(1-\lambda(z))\lambda(z)}}\,,

i.e., in the notation (2.46),

λ(1+z1z)=12+iλ1(z)4λ2(z),z,\displaystyle\lambda\left(\dfrac{1+z}{1-z}\right)=\dfrac{1}{2}+\dfrac{i\lambda_{1}(z)}{4\sqrt{\lambda_{2}(z)}}\ ,\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,, (A.7j)

which proves the validity of the left-hand side equality in (2.45). The right-hand side equality in (2.45) is obtained from (A.7j) by replacing zz by 1/z-1/z and by using (2.21).

8

\uparrow  The right-hand side equality (2.45) written for z=itz=it

z1z+1=it1it+1=tit+i=t+iti,\displaystyle\dfrac{z-1}{z+1}=\dfrac{it-1}{it+1}=\dfrac{-t-i}{-t+i}=\dfrac{t+i}{t-i}\ ,\
λ(z1z+1)=12iλ1(z)4λ2(z)\displaystyle\lambda\left(\dfrac{z-1}{z+1}\right)=\dfrac{1}{2}-\dfrac{i\lambda_{1}(z)}{4\sqrt{\lambda_{2}(z)}}\ \Rightarrow\
λ(t+iti)=12iλ1(it)4λ2(it),t>0,\displaystyle\lambda\left(\dfrac{t+i}{t-i}\right)=\dfrac{1}{2}-\dfrac{i\lambda_{1}(it)}{4\sqrt{\lambda_{2}(it)}}\ ,\quad t>0,

where λ(it)(0,1)\lambda(it)\in(0,1) for all t>0t>0, makes it possible to write

|λ(t+iti)|2=14+(12λ(it))216λ(it)(1λ(it))\displaystyle\left|\lambda\left(\dfrac{t+i}{t-i}\right)\right|^{2}=\dfrac{1}{4}+\dfrac{\left(1-2\lambda(it)\right)^{2}}{16\lambda(it)\left(1-\lambda(it)\right)}
=4λ(it)(1λ(it))+(12λ(it))216λ(it)(1λ(it))\displaystyle=\dfrac{4\lambda(it)\left(1-\lambda(it)\right)+\left(1-2\lambda(it)\right)^{2}}{16\lambda(it)\left(1-\lambda(it)\right)}
=4λ(it)4λ(it)2+14λ(it)+4λ(it)216λ(it)(1λ(it))=116λ(it)(1λ(it)),\displaystyle=\dfrac{4\lambda(it)-4\lambda(it)^{2}+1-4\lambda(it)+4\lambda(it)^{2}}{16\lambda(it)\left(1-\lambda(it)\right)}=\dfrac{1}{16\lambda(it)\left(1-\lambda(it)\right)}\ ,\

i.e., by (2.21),

|λ(t+iti)|2=116λ(it)(1λ(it))=116λ(it)λ(i/t),t>0,\displaystyle\left|\lambda\left(\dfrac{t+i}{t-i}\right)\right|^{2}=\dfrac{1}{16\lambda(it)\left(1-\lambda(it)\right)}=\dfrac{1}{16\lambda(it)\lambda(i/t)}\ ,\quad t>0\ ,\ (A.8a)

that proves equality in (2.47).

By using (2.31)(b) written in the forms,

0<λ(it)<16eπt, 0<λ(i/t)<16eπ/t, 0<t<,\displaystyle 0\!<\!\lambda(it)\!<\!16{\rm{e}}^{{{\mbox{\footnotesize{$-\pi t$}}}}},\ \ 0\!<\!\lambda(i/t)\!<\!16{\rm{e}}^{{{\mbox{\footnotesize{$-\pi/t$}}}}},\ \ 0\!<\!t\!<\!\infty,

we derive from (A.8a) that

16|λ(t+iti)|2116eπt16eπ/t=1162expπ(t+1t),\displaystyle 16\left|\lambda\left(\dfrac{t+i}{t-i}\right)\right|^{2}\geqslant\dfrac{1}{16\cdot{\rm{e}}^{{{\mbox{\footnotesize{$-\pi t$}}}}}\cdot 16{\rm{e}}^{{{\mbox{\footnotesize{$-\pi/t$}}}}}}=\frac{1}{16^{2}}\exp\pi\left(t+\dfrac{1}{t}\right)\ ,\

from which we obtain the inequality in (2.47),

64|λ(t+iti)|expπ2(t+1t),t>0.\displaystyle 64\left|\lambda\left(\dfrac{t+i}{t-i}\right)\right|\geqslant\exp\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)\ ,\quad t>0\ . (A.8b)
9

\uparrow   The Landen relationships (2.44)(c) together with (2.25)(g) and (A.18e) of (bh1, , p.​ 58) for any zz\in\Bb{H} give

Θ3(11z)2\displaystyle\Theta_{3}\left(\dfrac{1}{1-z}\right)^{2} =i(1z)Θ4(z)2,Θ4(11z)2=(1z)Θ2(z)2,\displaystyle=i(1-z)\Theta_{4}(z)^{2}\ ,\ \Theta_{4}\left(\dfrac{1}{1-z}\right)^{2}=(1-z)\Theta_{2}(z)^{2}\ ,
Θ3(1+z1z)4\displaystyle\Theta_{3}\left(\dfrac{1+z}{1-z}\right)^{4} =Θ3(21z1)4=Θ4(21z)4=Θ3(11z)2Θ4(11z)2\displaystyle=\Theta_{3}\left(\dfrac{2}{1-z}-1\right)^{4}=\Theta_{4}\left(\dfrac{2}{1-z}\right)^{4}=\Theta_{3}\left(\dfrac{1}{1-z}\right)^{2}\Theta_{4}\left(\dfrac{1}{1-z}\right)^{2}
=i(1z)2Θ2(z)2Θ4(z)2,\displaystyle=i(1-z)^{2}\Theta_{2}(z)^{2}\Theta_{4}(z)^{2}\ ,\

which by (2.25)(a), (c),

Θ2(1/z)Θ4(1/z)=(z/i)Θ2(z)Θ4(z)\displaystyle\Theta_{2}(-1/z)\Theta_{4}(-1/z)=(z/i)\Theta_{2}(z)\Theta_{4}(z)\ \Rightarrow\
Θ2(1/z)2Θ4(1/z)2=z2Θ2(z)2Θ4(z)2,\displaystyle\Theta_{2}(-1/z)^{2}\Theta_{4}(-1/z)^{2}=-z^{2}\Theta_{2}(z)^{2}\Theta_{4}(z)^{2}\ ,

lead to

i(1z)2Θ2(z)2Θ4(z)2=i(11/z)2Θ2(1/z)2Θ4(1/z)2,\displaystyle i(1-z)^{2}\Theta_{2}(z)^{2}\Theta_{4}(z)^{2}=-i(1-1/z)^{2}\Theta_{2}(-1/z)^{2}\Theta_{4}(-1/z)^{2}\ ,\

and therefore

Θ3(1+z1z)4={i(1z)2Θ2(z)2Θ4(z)2,i(11/z)2Θ2(1/z)2Θ4(1/z)2,z\displaystyle\Theta_{3}\left(\dfrac{1+z}{1-z}\right)^{4}=\left\{\begin{array}[]{l}i(1-z)^{2}\Theta_{2}(z)^{2}\Theta_{4}(z)^{2}\ ,\\[14.22636pt] -i(1-1/z)^{2}\Theta_{2}(-1/z)^{2}\Theta_{4}(-1/z)^{2}\ ,\end{array}\right.\quad z\in\Bb{H}\,. (A.9c)

Making here the change of variables z=1/zz=-1/z^{\,\prime}, we obtain

Θ3(z1z+1)4={i(1+z)2Θ2(z)2Θ4(z)2,i(1+1/z)2Θ2(1/z)2Θ4(1/z)2,z\displaystyle\Theta_{3}\left(\dfrac{z-1}{z+1}\right)^{4}=\left\{\begin{array}[]{l}-i(1+z)^{2}\Theta_{2}(z)^{2}\Theta_{4}(z)^{2}\ ,\\[14.22636pt] i(1+1/z)^{2}\Theta_{2}(-1/z)^{2}\Theta_{4}(-1/z)^{2}\,,\end{array}\right.\quad z\in\Bb{H}\,. (A.9f)

When z=itz=it, t>0t>0,

z1z+1=it1it+1=t+iti,\displaystyle\dfrac{z-1}{z+1}=\dfrac{it-1}{it+1}=\dfrac{t+i}{t-i}\ ,\

and (A.9f) yields

Θ3(t+iti)4={4i(ti)2eπt/2θ2(eπt)2θ4(eπt)2,4i(ti)2t2eπ/(2t)θ2(eπ/t)2θ4(eπ/t)2,t>0.\displaystyle\Theta_{3}\left(\dfrac{t+i}{t-i}\right)^{4}=\left\{\begin{array}[]{l}4i(t-i)^{2}e^{-\pi t/2}\theta_{2}\left(e^{-\pi t}\right)^{2}\theta_{4}\left(e^{-\pi t}\right)^{2}\ ,\\[14.22636pt] 4i\dfrac{(t-i)^{2}}{t^{2}}e^{-\pi/(2t)}\theta_{2}\left(e^{-\pi/t}\right)^{2}\theta_{4}\left(e^{-\pi/t}\right)^{2}\ ,\end{array}\right.\quad t>0\,.

Hence,

Φ(t):=|Θ3(t+iti)|4={4(1+t2)eπt/2θ2(eπt)2θ4(eπt)2,ift1,41+t2t2eπ/(2t)θ2(eπ/t)2θ4(eπ/t)2,if 0<t1,\displaystyle\Phi(t):=\left|\Theta_{3}\left(\dfrac{t+i}{t-i}\right)\right|^{4}=\left\{\begin{array}[]{ll}4(1+t^{2})e^{-\pi t/2}\theta_{2}\left(e^{-\pi t}\right)^{2}\theta_{4}\left(e^{-\pi t}\right)^{2}\ ,&\mbox{if}\ \ t\geqslant 1\ ,\\[14.22636pt] 4\dfrac{1+t^{2}}{t^{2}}e^{-\pi/(2t)}\theta_{2}\left(e^{-\pi/t}\right)^{2}\theta_{4}\left(e^{-\pi/t}\right)^{2}\ ,&\mbox{if}\ \ 0<t\leqslant 1\ ,\end{array}\right.

Here

θ4(u):=1+2n1(1)nun2=12n1(un2u(n+1)2)<1,u(0,1),\displaystyle\theta_{4}(u)\!:=\!1\!+\!2\sum\limits_{n\geqslant 1}(-1)^{n}u^{n^{2}}=1\!-\!2\sum\limits_{n\geqslant 1}\left(u^{n^{2}}-u^{(n+1)^{2}}\right)<1\,,\quad u\in(0,1),

while, by (ber1, , p.​ 325, (xii)),

θ2(u)=1+n1un2+nθ2(eπ), 0<u<eπ,\displaystyle\theta_{2}\left(u\right)=\!1\!+\!\sum\limits_{n\geqslant 1}u^{n^{2}+n}\leqslant\theta_{2}\left(e^{-\pi}\right)\,,\ 0<u<e^{-\pi}\,,
ψ(q)=n0qn(n+1)/2=θ2(q),θ2(eπ)=ψ(e2π)=π1/4eπ/425/4Γ(3/4),\displaystyle\psi(q)=\sum\limits_{n\geqslant 0}q^{n(n+1)/2}=\theta_{2}\left(\sqrt{q}\right)\ ,\quad\theta_{2}\left(e^{-\pi}\right)=\psi\left(e^{-2\pi}\right)=\dfrac{\pi^{1/4}e^{\pi/4}}{2^{5/4}\Gamma(3/4)}\ ,\

and therefore

Φ(t)=|Θ3(t+iti)|4{4(1+t2)eπt/2π1/2eπ/225/2Γ(3/4)2,ift1,41+t2t2eπ/(2t)π1/2eπ/225/2Γ(3/4)2,if 0<t1.\displaystyle\Phi(t)=\left|\Theta_{3}\left(\dfrac{t+i}{t-i}\right)\right|^{4}\leqslant\left\{\begin{array}[]{ll}4(1+t^{2})e^{-\pi t/2}\dfrac{\pi^{1/2}e^{\pi/2}}{2^{5/2}\Gamma(3/4)^{2}}\ ,&\mbox{if}\ \ t\geqslant 1\ ,\\[14.22636pt] 4\dfrac{1+t^{2}}{t^{2}}e^{-\pi/(2t)}\dfrac{\pi^{1/2}e^{\pi/2}}{2^{5/2}\Gamma(3/4)^{2}}\ ,&\mbox{if}\ \ 0<t\leqslant 1\ .\end{array}\right.

But

Φ(1/t)=|Θ3(1/t+i1/ti)|4=|Θ3(1+it1it)|4=|Θ3(iti+t)|4\displaystyle\Phi(1/t)=\left|\Theta_{3}\left(\dfrac{1/t+i}{1/t-i}\right)\right|^{4}=\left|\Theta_{3}\left(\dfrac{1+it}{1-it}\right)\right|^{4}=\left|\Theta_{3}\left(\dfrac{i-t}{i+t}\right)\right|^{4}
=|θ3(eiπiti+t)|4=|θ3(eiπiti+t)|4=|θ3(eiπt+iti)|4\displaystyle=\left|\theta_{3}\left(e^{{{\mbox{\footnotesize{$i\pi\dfrac{i-t}{i+t}$}}}}}\right)\right|^{4}=\left|\theta_{3}\left(e^{{{\mbox{\footnotesize{$-i\pi\dfrac{-i-t}{-i+t}$}}}}}\right)\right|^{4}=\left|\theta_{3}\left(e^{{{\mbox{\footnotesize{$i\pi\dfrac{t+i}{t-i}$}}}}}\right)\right|^{4}
=|Θ3(t+iti)|4=Φ(t),\displaystyle=\left|\Theta_{3}\left(\dfrac{t+i}{t-i}\right)\right|^{4}=\Phi(t)\ ,\

and so

Φ(t)=Φ(t)Φ(1/t)π1/2eπ/221/2Γ(3/4)2(t+1t)eπ4(t+1t),t>0,\displaystyle\Phi(t)=\sqrt{\Phi(t)\Phi(1/t)}\leqslant\dfrac{\pi^{1/2}e^{\pi/2}}{2^{1/2}\Gamma(3/4)^{2}}\left(t+\dfrac{1}{t}\right)e^{{{\mbox{\footnotesize{$-\dfrac{\pi}{4}\left(t+\dfrac{1}{t}\right)$}}}}}\ ,\quad t>0,

where

Γ(34)=1,2254167024,Γ(34)2=1,5016460,\displaystyle\Gamma\left(\dfrac{3}{4}\right)=1,2254167024\ldots\ ,\quad\Gamma\left(\dfrac{3}{4}\right)^{2}=1,5016460\ldots\,,
2=1,41421356,π=1,772453850,eπ/2=4,810477380,\displaystyle\sqrt{2}=1,41421356\ldots\,,\ \sqrt{\pi}=1,772453850\ldots\,,\quad e^{{{\mbox{\footnotesize{$\pi/2$}}}}}=4,810477380\ldots\,,
πeπ/2=8,526349152518913,2Γ(34)2=2,1236481390831,\displaystyle\sqrt{\pi}e^{{{\mbox{\footnotesize{$\pi/2$}}}}}=8,526349152518913\ldots\,,\quad\sqrt{2}\Gamma\left(\dfrac{3}{4}\right)^{2}=2,1236481390831\ldots\,,
π1/2eπ/221/2Γ(3/4)2=4,0149537937108354094530647660142<5,\displaystyle\dfrac{\pi^{1/2}e^{\pi/2}}{2^{1/2}\Gamma(3/4)^{2}}=4,0149537937108354094530647660142\ldots<5\ ,\

which completes the proof of (2.48).

A.3 . Notes for Section 3

10

\uparrow   We prove the statements of Lemma 3.1.

Suppose for the moment the assertion of Lemma 3.1 holds with a=0a\!=\!0. Then we would apply that statement to the new function g(z):=f0(ia+z)g(z)\!:=\!f_{0}(ia\!+\!z) and derive that Lemma 3.1 holds in fact for any aa\!\in\!\Bb{R}. Hence it suffices to obtain (3.1) in the case a=0a=0 only. Then our assumptions are that the functions f0f_{0} and f1(z):=φ(exp(iπz))f_{1}(z)\!:=\!\varphi(\exp({\rm{i}}\pi z)) are both holomorphic in \Bb{H} and 22-periodic.

For z=βexp(ib)𝔻~z=\beta\exp(ib)\in\Bb{D}\setminus(-1,0], β(0,1)\beta\in(0,1), π<b<π-\pi<b<\pi, we have

Logziπ=Log(βeib)iπ=iπln1β+bπ(1,1)+i\displaystyle\dfrac{{\rm{Log}}\,z}{{\rm{i}}\pi}=\dfrac{{\rm{Log}}\,\left(\beta{\rm{e}}^{{\rm{i}}b}\right)}{{\rm{i}}\pi}=\frac{{\rm{i}}}{\pi}\ln\frac{1}{\beta}+\frac{b}{\pi}\in(-1,1)+{\rm{i}}\Bb{R}_{>0}\subset\Bb{H}\,, (A.10a)

and since LogHol(){\rm{Log}}\,\in{\rm{Hol}}\left(\Bb{C}\setminus\Bb{R}_{\leqslant 0}\right) we obtain that ΦδHol(𝔻~\Phi^{\delta}\in{\rm{Hol}}(\Bb{D}\setminus(-1,0]) for every δ{0,1}\delta\in\{0,1\}, where

Φδ(z):=fδ(Logziπ),Φ1(z)=φ(z),z𝔻~δ{}\displaystyle\Phi^{\delta}(z):=f_{\delta}\left(\dfrac{{\rm{Log}}\,z}{i\pi}\right)\,,\quad\Phi^{1}(z)=\varphi(z)\,,\ \quad z\in\Bb{D}\setminus(-1,0]\,,\ \delta\in\{0,1\}\,.

Here, if we write Φδ±(x):=limz±ϝΦδ(z)\Phi^{\delta}_{\pm}(-x):=\lim_{z\in\pm\Bb{H},\,z\to-x}\ \Phi^{\delta}(z) for each δ{0,1}\delta\in\{0,1\}, we find

Φδ+(x)=fδ(iπln1x+1)=fδ(iπln1x1)=Φδ(x),x(0,1),\displaystyle\Phi^{\delta}_{+}(-x)=f_{\delta}\left(\frac{i}{\pi}\ln\frac{1}{x}+1\right)=f_{\delta}\left(\frac{i}{\pi}\ln\frac{1}{x}-1\right)=\Phi^{\delta}_{-}(-x)\ ,\ x\in(0,1)\,,

and consequently, Φδ\Phi^{\delta} extends continuously across (1,0)(-1,0). From Morera’s theorem (see (lav, , p.​ 96)) we conclude that ΦδHol(𝔻{}\Phi^{\delta}\in{\rm{Hol}}(\Bb{D}\setminus\{0\}). For δ=1\delta=1 this completes the proof of the second assertion of Lemma 3.1. As for δ=0\delta=0, we obtain that Φ0\Phi^{0} admits the Laurent expansion (see (con, , p.​ 107)) Φ0(z)=nanzn\Phi^{0}(z)=\sum\nolimits_{n\in\Bb{Z}}\ a_{n}z^{n}, z𝔻{}z\in\Bb{D}\setminus\{0\}, which is absolutely convergent in 𝔻{}\Bb{D}\setminus\{0\}. The change of variables z=eiπζz={\rm{e}}^{{\rm{i}}\pi\zeta}, ζ=x+iy\zeta=x+{\rm{i}}y, x(1,1)x\in(-1,1), y>0y>0, gives, by using (A.10a), that Logexp(iπζ)=iπζ{\rm{Log}}\,\exp(i\pi\zeta)={\rm{i}}\pi\zeta, and, moreover, that

f0(ζ)=naneiπnζ,Imζ>0,Reζ(1,1).\displaystyle f_{0}(\zeta)=\sum\nolimits_{n\in\Bb{Z}}\ a_{n}{\rm{e}}^{{\rm{i}}\pi n\zeta}\ ,\ \ \ {\rm{Im}}\,\zeta>0\ ,\ {\rm{Re}}\,\zeta\in(-1,1)\,.

Both f0f_{0} and the Fourier series on the right-hand side are continuous on \Bb{H} and periodic with period 22. Hence they must be equal for all ζ\zeta\in\Bb{H} and, consequently,

zz+2f0(ζ)dζ=2a0,\displaystyle\int_{z}^{z+2}f_{0}(\zeta)d\zeta=2a_{0}\,,

for arbitrary zz\in\Bb{H}. This proves the first assertion of Lemma 3.1 and completes the proof of Lemma 3.1.

11

\uparrow  We have, in view of (2.52),

4π2nMn(x)=4π2nHn(1/x)/x2=γ(1,1)Sn(1λ(z))dz(zx1)2=γ(1,1)Sn(1λ(z))(x1/z)2dzz2\displaystyle-4\pi^{2}n\eurm{M}_{n}(x)=-4\pi^{2}n\eurm{H}_{n}(-1/x)/x^{2}\!=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(z)}\right)dz}{(zx-1)^{2}}=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(z)}\right)}{(x-1/z)^{2}}\,\dfrac{dz}{z^{2}}
|z=1z,dz=dz(z)2|=γ(1,1)Sn(1λ(1/z))dz(x+z)2=γ(1,1)Sn(11λ(z))dz(x+z)2,\displaystyle\left|z=-\dfrac{1}{z^{\,\prime}}\,,\ dz=\dfrac{dz^{\,\prime}}{\left(z^{\,\prime}\right)^{2}}\right|=-\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{\lambda(-1/z)}\right)dz}{(x+z)^{2}}=-\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{1-\lambda(z)}\right)dz}{(x+z)^{2}}\,,\

from which

Mn(x)=14π2nγ(1,1)Sn(11λ(z))dz(x+z)2,\displaystyle\eurm{M}_{n}(x)=\dfrac{1}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{1-\lambda(z)}\right)dz}{(x+z)^{2}}\,,

and therefore by (3.12) and (10.4),

(1)n4π2nMn(x)=γ(1,1)Sn(11λ(z))dz(x+z)2=γ(1,1)Sn(111λ(z))Sn(1)(x+z)2dz\displaystyle(-1)^{n}4\pi^{2}n\eurm{M}_{n}(x)=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{1}{1-\lambda(z)}\right)dz}{(x+z)^{2}}=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(1-\dfrac{1}{1-\lambda(z)}\right)-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)}{(x+z)^{2}}dz
=γ(1,1)Sn(λ(z)λ(z)1)Sn(1)(x+z)2dz=γ(1,1)Sn(λ(z+1))Sn(1)(x+z)2dz\displaystyle=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\dfrac{\lambda(z)}{\lambda(z)-1}\right)-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)}{(x+z)^{2}}dz=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\lambda(z+1)\right)-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)}{(x+z)^{2}}dz

from which

Mn(x)=(1)n4π2nγ(1,1)Sn(λ(z+1))Sn(1)(x+z)2dz.\displaystyle\eurm{M}_{n}(x)=\dfrac{(-1)^{n}}{4\pi^{2}n}\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\frac{S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}\!\left(\lambda(z+1)\right)-S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1)}{(x+z)^{2}}dz\ .
12

\uparrow  According with Theorem 2.6, (2.68) and (2.73),

{yclos()|λ(y)=z}\displaystyle\left\{y\!\in\!{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)\ \big{|}\ \lambda(y)\!=\!z\right\}
={λ(z),ifz(0,1)();{λ(z+i0),λ(zi0)},ifz=1+x,x>0,{λ(z+i0),λ(zi0)},ifz=x,x>0,\displaystyle=\left\{\begin{array}[]{ll}\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\,,&\ \ \hbox{if}\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\,;\\[7.11317pt] \left\{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z+i0),\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z-i0)\right\}\ ,&\ \ \hbox{if}\quad z=1+x\ ,\quad x>0\,,\\[7.11317pt] \left\{\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z+i0),\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z-i0)\right\},&\ \ \hbox{if}\quad z=-x\ ,\qquad x>0\,,\end{array}\right. (A.12d)
{=λ(z),ifz(0,1)();=(2.73){1+iΔ(x)1+Δ(x)2,1+iΔ(x)1+Δ(x)2},ifz=1+x,x>0,=(2.68){1+iΔ(x),1+iΔ(x)},ifz=x,x>0.\displaystyle\left\{\begin{array}[]{cll}=&\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\,,&\ \ \hbox{if}\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\,;\\[7.11317pt] \stackrel{{\scriptstyle{{\mbox{\footnotesize{$\eqref{f7intthA}$}}}}}}{{\vphantom{A}=}}&\left\{\dfrac{-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}},\dfrac{1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\right\}\ ,&\ \ \hbox{if}\quad z=1+x\ ,\quad x>0\,,\\[14.22636pt] \stackrel{{\scriptstyle{{\mbox{\footnotesize{$\eqref{f6intthA}$}}}}}}{{\vphantom{A}=}}&\left\{-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x),1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)\right\},&\ \ \hbox{if}\quad z=-x\ ,\qquad x>0\,.\end{array}\right. (A.12h)

Therefore

{yclos()|λ(y)=z}enπiy\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\left\{y\!\in\!{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)\ \big{|}\ \lambda(y)\!=\!z\right\}$}}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!e^{{{\mbox{\footnotesize{$-n\pi iy$}}}}}

is equal to

enπiλ(z),\displaystyle e^{{{\mbox{\footnotesize{$-n\pi i\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)$}}}}}\,,\

if z(0,1)()z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right),

enπi1+iΔ(x)1+Δ(x)2+enπi1+iΔ(x)1+Δ(x)2=enπΔ(x)1+Δ(x)2(enπi1+Δ(x)2+enπi1+Δ(x)2)\displaystyle e^{{{\mbox{\footnotesize{$-n\pi i\dfrac{-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}+e^{{{\mbox{\footnotesize{$-n\pi i\dfrac{1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}=e^{{{\mbox{\footnotesize{$\dfrac{n\pi\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}\left(e^{{{\mbox{\footnotesize{$\dfrac{n\pi i}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}+e^{{{\mbox{\footnotesize{$-\dfrac{n\pi i}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}\right)
=2enπΔ(x)1+Δ(x)2cosnπ1+Δ(x)2,\displaystyle=2\,e^{{{\mbox{\footnotesize{$\dfrac{n\pi\,{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}\cos\dfrac{n\pi}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\,,\

if z=1+xz=1+x, x>0x>0, and

enπi(1+iΔ(x))+enπi(1+iΔ(x))=2(1)nenπΔ(x),\displaystyle e^{{{\mbox{\footnotesize{$-n\pi i\left(-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)\right)$}}}}}+e^{{{\mbox{\footnotesize{$-n\pi i\left(1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,{{\mbox{\footnotesize{$\Delta$}}}}(x)\right)$}}}}}=2(-1)^{n}e^{{{\mbox{\footnotesize{$n\pi{{\mbox{\footnotesize{$\Delta$}}}}(x)$}}}}}\,,\

if z=xz=-x, x>0x>0. Thus,

{yclos()|λ(y)=z}enπiy={exp(nπiλ(z)),ifz(0,1)();2enπΔ(x)1+Δ(x)2cosnπ1+Δ(x)2,ifz=1+x,x>0,2(1)nexp(nπΔ(x)),ifz=x,x>0,\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\left\{y\!\in\!{\rm{clos}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)\ \big{|}\ \lambda(y)\!=\!z\right\}$}}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!e^{{{\mbox{\footnotesize{$-n\pi iy$}}}}}=\left\{\begin{array}[]{ll}\exp\left(-n\pi i\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)\right)\,,&\ \ \hbox{if}\quad z\in(0,1)\cup\left(\Bb{C}\setminus\Bb{R}\right)\,;\\[7.11317pt] 2e^{{{\mbox{\footnotesize{$\dfrac{\,n\pi{{\mbox{\footnotesize{$\Delta$}}}}(x)}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}$}}}}}\cos\dfrac{n\pi}{1+{{\mbox{\footnotesize{$\Delta$}}}}(x)^{2}}\ ,&\ \ \hbox{if}\quad z=1+x\ ,\quad x>0\,,\\[14.22636pt] 2(-1)^{n}\exp\left(n\pi{{\mbox{\footnotesize{$\Delta$}}}}(x)\right),&\ \ \hbox{if}\quad z=-x\ ,\qquad x>0\,,\end{array}\right.

where

Δ(x):=F(1/(1+x))F(x/(1+x)),{Δ(0)=+,Δ(+)=0,dΔ(x)dx<0,Δ(x)Δ(1/x)=1,\displaystyle{{\mbox{\footnotesize{$\Delta$}}}}(x)\!:=\!\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({1}\big{/}{(1\!+\!x)}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left({x}\big{/}{(1\!+\!x)}\right)}\ ,\quad\left\{\begin{array}[]{l}{{\mbox{\footnotesize{$\Delta$}}}}(0)\!=\!+\!\infty\ ,\\[5.69046pt] {{\mbox{\footnotesize{$\Delta$}}}}(+\infty)\!=\!0\ ,\end{array}\right.\ \ \ \dfrac{{{\mathrm{d}}}{{\mbox{\footnotesize{$\Delta$}}}}(x)}{{{\mathrm{d}}}x}\!<\!0,\ \ {{\mbox{\footnotesize{$\Delta$}}}}(x){{\mbox{\footnotesize{$\Delta$}}}}(1/x)\!=\!1,

what has been asserted after (3.11).

13

\uparrow  It follows from (2.52)

2πiH0(1/x)/x2=γ(1,1)zΘ3(z)41z2x2dz=γ(1,1)zΘ3(z)4z2x2dzz2=γ(1,1)(1/z)Θ3(1/z)4z2x2dz,\displaystyle 2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{H}_{0}(1/x)/x^{2}\!=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\!\frac{z\,\Theta_{3}\left(z\right)^{4}}{1-z^{2}x^{2}}dz=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\!\frac{z\,\Theta_{3}\left(z\right)^{4}}{z^{-2}-x^{2}}\dfrac{dz}{z^{2}}=-\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\!\frac{(-1/z)\Theta_{3}\left(-1/z\right)^{4}}{z^{2}-x^{2}}dz\,,\

where by (2.25)(b),

Θ3(1/z)4=z2Θ3(z)4,\displaystyle\Theta_{3}(-1/z)^{4}=-z^{2}\Theta_{3}(z)^{4}\,,\

and therefore

2πiH0(1/x)/x2=γ(1,1)(1/z)(z2Θ3(z)4)z2x2dz=γ(1,1)zΘ3(z)4x2z2dz=2πiH0(x).\displaystyle 2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{H}_{0}(-1/x)/x^{2}\!=-\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\!\frac{(-1/z)\left(-z^{2}\Theta_{3}(z)^{4}\right)}{z^{2}-x^{2}}dz=\!\!\!\!\!\!\int\limits_{\gamma(-1,1)}\!\!\!\!\!\!\frac{z\Theta_{3}(z)^{4}}{x^{2}-z^{2}}dz=2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\eurm{H}_{0}(x)\,.

Thus,

H0(x)=H0(x),x{}\displaystyle\eurm{H}_{0}(x)=\eurm{H}_{0}(-x)\,,\ x\in\Bb{R}\,,\quad\eurm{H}_{0}(-1/x)\!=\eurm{H}_{0}(x)x^{2}\,,\ x\in\Bb{R}\setminus\{0\}\,.
14

\uparrow  We prove the next assertion.

Lemma A.1

. Let a,b,c,da,b,c,d\in\Bb{C}, adbc0ad-bc\neq 0 and ϕ(abcd)\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}} be the Möbius transformation (con, , p.​ 47, Definition 3.5) such that

ϕ(abcd)()=,ϕ(abcd)(z):=az+bcz+d.\displaystyle\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\ ,\quad\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}\left(z\right):=\dfrac{az+b}{cz+d}\ . (A.14a)

Then

ϕ(abcd)(z){z,1z,z1z+1,z+1z1}.\displaystyle\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}(z)\in\left\{\ z\ ,\ \ -\dfrac{1}{z}\ ,\ \ \dfrac{z-1}{z+1}\ ,\ \ -\dfrac{z+1}{z-1}\ \right\}\,. (A.14b)

Proof. The boundary of the image of \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} under the Möbius transformation

ϕ(abcd)(z)=az+bcz+d,\displaystyle\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}\left(z\right)=\dfrac{az+b}{cz+d}\ ,

according to the known property of the Möbius transformations to map circles onto circles (con, , p.​ 49, Theorem 3.14) and to the known open mapping theorem (con, , p.​ 99, Theorem 7.5) of nonconstant analytic functions to map open set to open set, consists of four circles and therefore four vertices of \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} are transformed into vertices, i.e.,

ϕ(abcd)({,0,1,1})={,0,1,1}.\displaystyle\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}\left(\left\{\infty,0,1,-1\right\}\right)=\left\{\infty,0,1,-1\right\}\,.

Since 1/=-1/\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} then if ϕ(abcd)()=\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} then

1/ϕ(abcd)(z),1/ϕ(abcd)(1/z),1/ϕ(abcd)(1/z),\displaystyle-1/\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}(z)\ ,\quad-1/\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}(-1/z)\ ,\quad-1/\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}(-1/z)\ , (A.14c)

also map \mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} onto itself. Therefore it suffices to consider two cases

1)f()=,2)f()=1,f(z):=ϕ(abcd)(z)=az+bcz+d.\displaystyle 1)\ \ f(\infty)=\infty\ ,\quad 2)\ \ f(\infty)=1\ ,\quad f(z):=\phi_{{{\mbox{\normalsize{$(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})$}}}}}\left(z\right)=\dfrac{az+b}{cz+d}\,.

In the first case we get c=0c=0 and therefore one may assume that f(z)=az+bf(z)=az+b. Then f(0)=b{0,1,1}f(0)=b\in\{0,-1,1\}, i.e. it should be

{f(1),f(1)}={0,1,1}{b},f(z)=az+b,b{0,1,1}.\displaystyle\{f(1),f(-1)\}=\{0,-1,1\}\setminus\{b\}\ ,\quad f(z)=az+b\ ,\quad b\in\{0,-1,1\}\,.

we get three cases f(z)=azf(z)=az, f(z)=az1f(z)=az-1, f(z)=az+1f(z)=az+1, where correspondingly

b=0{f(1),f(1)}={a,a}={1,1}a{1,1},\displaystyle b=0\ \Rightarrow\ \{f(1),f(-1)\}=\left\{a,-a\right\}=\{1,-1\}\ \Rightarrow\ a\in\{1,-1\}\ ,
b=1{f(1),f(1)}={a1,a1}={0,1}{a,a}={1,2}\displaystyle b=-1\ \Rightarrow\ \{f(1),f(-1)\}=\left\{a-1,-a-1\right\}=\{0,1\}\ \Rightarrow\ \left\{a,-a\right\}=\{1,2\}
impossible ,\displaystyle\Rightarrow\ \mbox{impossible },
b=1{f(1),f(1)}={a+1,a+1}={0,1}{a,a}={1,2}\displaystyle b=1\ \Rightarrow\ \{f(1),f(-1)\}=\left\{a+1,-a+1\right\}=\{0,-1\}\ \Rightarrow\ \left\{a,-a\right\}=\{-1,-2\}
impossible ,\displaystyle\Rightarrow\ \mbox{impossible },

and since f(z)=zf(z)=-z is not acceptable because -\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\neq\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, we obtain f(z)=zf(z)=z.

For the second case we have a=ca=c, i.e.

f(z)=z+bz+d,\displaystyle f(z)=\dfrac{z+b}{z+d}\,,

and it follows from f({1,1,0})\infty\in f(\{1,-1,0\}) that

d{1,1,0}.\displaystyle d\in\{1,-1,0\}\,.

and therefore it should be

f({0,1,1}{d})={1,0},f(z)=z+bz+d,d{1,1,0}.\displaystyle f\left(\{0,-1,1\}\setminus\{-d\}\right)=\{-1,0\}\ ,\quad f(z)=\dfrac{z+b}{z+d}\ ,\quad d\in\{1,-1,0\}\,.

We get three cases

d=0f({1,1})={1+b,1b}={1,0}{b,b}={2,1}\displaystyle d=0\ \Rightarrow\ f\left(\{-1,1\}\right)=\left\{1+b,1-b\right\}=\{-1,0\}\ \Rightarrow\ \left\{b,-b\right\}=\{-2,-1\}
impossible ,\displaystyle\Rightarrow\ \mbox{impossible },
d=1f({1,0})={1+b1+1,b1}={1+b2,b}={1,0}\displaystyle d=1\ \Rightarrow\ f\left(\{1,0\}\right)=\left\{\dfrac{1+b}{1+1},\dfrac{b}{1}\right\}=\left\{\dfrac{1+b}{2},b\right\}=\{-1,0\}
{1+b,2b}={2,0}b=1,\displaystyle\Rightarrow\ \left\{1+b,2b\right\}=\{-2,0\}\ \Rightarrow\ b=-1\ ,
d=1f({1,0})={1+b11,b1}={1b2,b}={1,0}\displaystyle d=-1\ \Rightarrow\ f\left(\{-1,0\}\right)=\left\{\dfrac{-1+b}{-1-1},\dfrac{b}{-1}\right\}=\left\{\dfrac{1-b}{2},-b\right\}=\{-1,0\}
b=1f(z)=z+1z1=1+2z1f(impossible \displaystyle\ \Rightarrow\ b=1\ \Rightarrow\ f(z)=\dfrac{z+1}{z-1}=1+\dfrac{2}{z-1}\ \Rightarrow\ f(\Bb{H})\subset-\Bb{H}\Rightarrow\ \mbox{impossible }.

Thus, f(z)=(z1)/(z+1)f(z)=(z-1)/(z+1) and taking account of (A.14c) we complete the proof.

15

\uparrow  To calculate ΔnS(0)\Delta_{n}^{S}(0) in (3.2), written as

enπiλ(z)=Sn(1/z)+ΔnS(z),z𝔻{}\displaystyle e^{{{\mbox{\footnotesize{$-n\pi i\lambda_{{{\mbox{\tiny{$\triangle$}}}}}(z)$}}}}}\!=\!S^{{{\mbox{\tiny{$\triangle$}}}}}_{n}(1/z)\!+\!\Delta_{n}^{S}(z)\ ,\ z\!\in\!\Bb{D}\setminus\{0\}\ ,\

we use the formula

12πiβ𝔻ζ1ζpdζ=δ0,p,pβ\displaystyle\frac{1}{2\pi i}\int\nolimits_{\beta\partial\Bb{D}}\,\zeta^{-1}\zeta^{p}d\zeta=\delta_{0,p}\ ,\quad p\in\Bb{Z}\ ,\quad\beta\in(0,1)\,,\

and similarly to (3.8), by applying Lemma 3.1 to the periodic integrands, obtain

ΔnS(0)=12πiβ𝔻enπiλ(ζ)dζζ=12πiππenπiλ(βeiφ)d(βeiφ)βeiφ=\displaystyle\Delta_{n}^{S}(0)=\frac{1}{2\pi i}\int_{\beta\partial\Bb{D}}\frac{e^{{{\mbox{\footnotesize{$-n\pi i\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}(\zeta)$}}}}}d\zeta}{\zeta}=\frac{1}{2\pi i}\int_{-\pi}^{\pi}\frac{e^{{{\mbox{\footnotesize{$-n\pi i\ \lambda_{{{\mbox{\tiny{$\triangle$}}}}}\left(\beta e^{i\varphi}\right)$}}}}}d\left(\beta e^{i\varphi}\right)}{\beta e^{i\varphi}}=
=12πi1+iΔ(β)1+iΔ(β)λ(ζ)enπiζλ(ζ)dζ=12πi1+ia1+iaλ(ζ)enπiζλ(ζ)dζ,\displaystyle=\frac{1}{2\pi i}\int\nolimits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{{\mbox{\footnotesize{$\Delta$}}}}(\beta)$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt{{\mbox{\footnotesize{$\Delta$}}}}(\beta)$}}}}}\ \frac{\lambda^{\,\prime}\left(\zeta\right)e^{{{\mbox{\footnotesize{$-n\pi i\zeta$}}}}}}{\lambda\left(\zeta\right)}\,d\zeta=\frac{1}{2\pi i}\int\nolimits_{{{\mbox{\footnotesize{$-1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}^{{{\mbox{\footnotesize{$1\!+\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}\frac{\lambda^{\,\prime}\left(\zeta\right)e^{{{\mbox{\footnotesize{$-n\pi i\zeta$}}}}}}{\lambda\left(\zeta\right)}\,d\zeta\ ,

where Δ(β)>1{{\mbox{\footnotesize{$\Delta$}}}}(\beta)>1, while aa is arbitrary positive real number. Then for any Imy>a{\rm{Im}}\,y>a we can apply Lemma 3.1 once more for arbitrary A>ImyA>{\rm{Im}}\,y to get similarly to the transform of (3.10),

n=1ΔnS(0)eiπny=12πi1+ia1+iaλ(ζ)dζ(eπi(ζy)1)λ(ζ)=\displaystyle\sum\limits_{n=1}^{\infty}\Delta_{n}^{S}(0){\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}=\frac{1}{2\pi i}\int\nolimits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677pta$}}}}}\ \frac{\lambda^{\,\prime}\left(\zeta\right)d\zeta}{\left(e^{{{\mbox{\footnotesize{$\pi i(\zeta-y)$}}}}}-1\right)\lambda\left(\zeta\right)}=
=1iπλ(y)λ(y)+12πi1+iA+Rey1+iA+Reyλ(ζ)dζ(eπi(ζy)1)λ(ζ),\displaystyle=\frac{1}{i\pi}\frac{\lambda^{\,\prime}\left(y\right)}{\lambda\left(y\right)}+\frac{1}{2\pi i}\int\limits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}\frac{\lambda^{\,\prime}\left(\zeta\right)d\zeta}{\left(e^{{{\mbox{\footnotesize{$\pi i(\zeta-y)$}}}}}-1\right)\lambda\left(\zeta\right)}\,,\

from which by letting A+A\to+\infty it follows from

1iπλ(y)λ(y)\displaystyle\frac{1}{i\pi}\frac{\lambda^{\,\prime}\left(y\right)}{\lambda\left(y\right)} =1iπiπλ(y)(1λ(y))Θ3(y)4λ(y)=(1λ(y))Θ3(y)4=\displaystyle=\frac{1}{i\pi}\frac{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\,\pi\,\lambda(y)\,\left(1-\lambda(y)\right)\,\Theta_{3}\left(y\right)^{4}}{\lambda\left(y\right)}=\left(1-\lambda(y)\right)\Theta_{3}\left(y\right)^{4}=
=(1Θ2(y)4Θ3(y)4)Θ3(y)4=Θ4(y)4,y\displaystyle=\left(1-\dfrac{\Theta_{2}\left(y\right)^{4}}{\Theta_{3}\left(y\right)^{4}}\right)\Theta_{3}\left(y\right)^{4}=\Theta_{4}\left(y\right)^{4}\ ,\quad y\in\Bb{H},

and Θ4(y)1\Theta_{4}\left(y\right)\to 1, as 0<Imy+0<{\rm{Im}}\,y\to+\infty, that

limA+12πi1+iA+Rey1+iA+Reyλ(ζ)dζ(eπi(ζy)1)λ(ζ)=limA+121+iA+Rey1+iA+ReyΘ4(ζ)4dζ1eπi(ζy)=1,\lim\limits_{A\to+\infty}\frac{1}{2\pi i}\!\!\!\int\limits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}\!\!\!\frac{\lambda^{\,\prime}\left(\zeta\right)d\zeta}{\left(e^{{{\mbox{\footnotesize{$\pi i(\zeta-y)$}}}}}-1\right)\lambda\left(\zeta\right)}\\ =-\lim\limits_{A\to+\infty}\frac{1}{2}\!\!\!\int\limits_{{{\mbox{\footnotesize{$-1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}^{{{\mbox{\footnotesize{$1+\hskip 0.56917pt\mathrm{i}\hskip 0.42677ptA+{\rm{Re}}\,y$}}}}}\!\!\!\frac{\Theta_{4}\left(\zeta\right)^{4}d\zeta}{1-e^{{{\mbox{\footnotesize{$\pi i(\zeta-y)$}}}}}}=-1,

and therefore

n=1ΔnS(0)eiπny=1iπλ(y)λ(y)1=Θ4(y)41,\displaystyle\sum\limits_{n=1}^{\infty}\Delta_{n}^{S}(0){\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi ny}=\frac{1}{i\pi}\frac{\lambda^{\,\prime}\left(y\right)}{\lambda\left(y\right)}-1=\Theta_{4}\left(y\right)^{4}-1,

i.e.,

1+n=1ΔnS(0)un=(1+2n1(1)nun2)4=θ3(u)4,u𝔻\displaystyle 1+\sum\limits_{n=1}^{\infty}\Delta_{n}^{S}(0)u^{n}=\left(1\!+\!2\sum\nolimits_{n\geqslant 1}(-1)^{n}u^{n^{2}}\right)^{4}=\theta_{3}(-u)^{4},\ u\in\Bb{D}\,,

which together with (2.30)(b) proves (3.22).

A.4 . Notes for Section 4

  

16

\uparrow  In accordance with (4.5) for a=2a=2 we have

Rn(z)=12πi1+2i1+2iλ(ζ)enπiζλ(z)λ(ζ)dζ,zγ(1,1),\displaystyle\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\!=\!\frac{1}{2\pi i}\int\nolimits_{{{\mbox{\footnotesize{$-1\!+\!2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt$}}}}}^{{{\mbox{\footnotesize{$1\!+\!2\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt$}}}}}\frac{\lambda^{\,\prime}\left(\zeta\right)e^{{{\mbox{\footnotesize{$-n\pi i\zeta$}}}}}}{\lambda(z)-\lambda\left(\zeta\right)}\,d\zeta\,,\quad z\!\in\!\gamma(-1,1), (A.16a)

where |λ(ζ)|9|\lambda^{\,\prime}(\zeta)|\leqslant 9 and 4|λ(z)λ(ζ)||λ(z)||12λ(2i)|4|\lambda(z)-\lambda(\zeta)|\geqslant|\lambda(z)|\,|1-2\lambda(2i)| for all ζ[1+2i,1+2i]\zeta\in[-1+2i,1+2i] and zγ(1,1)z\!\in\!\gamma(-1,1), in view of Corollary 2.3. Applying (7.2), we get

|Rn(z)|9e2πnπ|λ(z)|11+822110e2πnπ|λ(z)|,zγ(1,1).\displaystyle\left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\right|\leqslant\dfrac{9e^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{\pi|\lambda(z)|}\dfrac{11+8\sqrt{2}}{21}\leqslant\dfrac{10e^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{\pi|\lambda(z)|}\ ,\quad z\in\gamma(-1,1)\,. (A.16b)

For the parametrization γ(1,1)z=(t+i)/(ti)\gamma(-1,1)\ni z=(t+i)/(t-i), tt\in\Bb{R}_{>0}, taking into account

1z=tit+i=iti+t=1+it1it=1/t+i1/ti,\displaystyle-\dfrac{1}{z}=-\dfrac{t-i}{t+i}=\dfrac{i-t}{i+t}=\dfrac{1+it}{1-it}=\dfrac{1/t+i}{1/t-i}\,,

and (2.47),

64|λ(t+iti)|expπ2(t+1t),t>0\displaystyle 64\left|\lambda\left(\dfrac{t+i}{t-i}\right)\right|\geqslant\exp\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)\ ,\quad t>0\ \Rightarrow\ (2.47)
64|λ(1/(t+iti))|=64|λ(1/t+i1/ti)|expπ2(t+1t),t>0,\displaystyle 64\left|\lambda\left(-1/\left(\dfrac{t+i}{t-i}\right)\right)\right|=64\left|\lambda\left(\dfrac{1/t+i}{1/t-i}\right)\right|\geqslant\exp\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)\ ,\quad t>0\,,\

we obtain

|Rn(z)|,|Rn(1/z)|640e2πnπeπ2(t+1t),z=t+iti,t>0.\displaystyle\left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(z)\right|,\ \left|\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}(-1/z)\right|\!\leqslant\!\dfrac{640e^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{\pi}e^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)$}}}}}\,,\ z\!=\!\dfrac{t\!+\!i}{t\!-\!i}\ ,\ t\!>\!0\,. (A.16c)

Then we deduce from (3.37) and

z=t+iti=1+2iti,dz=2idt(ti)2,\displaystyle z=\dfrac{t+i}{t-i}=1+\dfrac{2i}{t-i}\ ,\quad dz=\dfrac{2idt}{(t-i)^{2}}\,,\

that for z=x+iyz=x+iy\in\Bb{H}

4π2nHn(z)=02iRn(t+iti)dt(ti)2(z+t+iti)2=02iRn(t+iti)dt(z(ti)+t+i)2,\displaystyle-4\pi^{2}n\eurm{H}_{n}(z)=\int\limits_{0}^{\infty}\frac{2i\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{t+i}{t-i}\right)dt}{(t-i)^{2}\left(z+\dfrac{t+i}{t-i}\right)^{2}}=\int\limits_{0}^{\infty}\frac{2i\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{t+i}{t-i}\right)dt}{\left(z(t-i)+t+i\right)^{2}}\ , (A.16d)
4π2nMn(z)=02iRn(1/t+i1/ti)dt(ti)2(z+t+iti)2=02iRn(1/t+i1/ti)dt(z(ti)+t+i)2,\displaystyle 4\pi^{2}n\eurm{M}_{n}(z)=\int\limits_{0}^{\infty}\frac{2i\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1/t+i}{1/t-i}\right)dt}{(t-i)^{2}\left(z+\dfrac{t+i}{t-i}\right)^{2}}=\int\limits_{0}^{\infty}\frac{2i\eurm{R}_{n}^{{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1/t+i}{1/t-i}\right)dt}{\left(z(t-i)+t+i\right)^{2}}\,,\ (A.16e)

where

z(ti)+t+i=(x+iy)(ti)+t+i=xtix+iyt+y+t+i\displaystyle z(t-i)+t+i=(x+iy)(t-i)+t+i=xt-ix+iyt+y+t+i
=xt+y+t+i(ytx+1)=t(x+1)+y+i(ty(x1)),\displaystyle=xt+y+t+i(yt-x+1)=t(x+1)+y+i\big{(}ty-(x-1)\big{)}\,,
|z(ti)+t+i|2=(t(x+1)+y)2+(ty(x1))2\displaystyle\left|z(t-i)+t+i\right|^{2}=\left(t(x+1)+y\right)^{2}+\left(ty-(x-1)\right)^{2}
=2ty((x+1)(x1))+(1+t2)y2+t2(x+1)2+(x1)2\displaystyle=2ty\big{(}(x+1)-(x-1)\big{)}+(1+t^{2})y^{2}+t^{2}(x+1)^{2}+(x-1)^{2}
=(1+t2)y2+t2(x+1)2+(x1)2+4ty.\displaystyle=(1+t^{2})y^{2}+t^{2}(x+1)^{2}+(x-1)^{2}+4ty\,.

Thus,

|z(ti)+t+i|2=(1+t2)y2+t2(x+1)2+(x1)2+4ty\displaystyle\left|z(t-i)+t+i\right|^{2}=(1+t^{2})y^{2}+t^{2}(x+1)^{2}+(x-1)^{2}+4ty
t2(y2+(x+1)2)+y2+(x1)2=|z1|2+t2|z+1|2\displaystyle\geqslant t^{2}(y^{2}+(x+1)^{2})+y^{2}+(x-1)^{2}=|z-1|^{2}+t^{2}|z+1|^{2}
|z1|2+|z+1|21+1t2,\displaystyle\geqslant\dfrac{|z-1|^{2}+|z+1|^{2}}{1+\dfrac{1}{t^{2}}}\,,\

from which

n|Hn(z)|,n|Mn(z)|320π3e2πn|z1|2+|z+1|20(1+1t2)eπ2(t+1t)dt,\displaystyle n\left|\eurm{H}_{n}(z)\right|,n\left|\eurm{M}_{n}(z)\right|\leqslant\dfrac{320\pi^{-3}e^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{|z\!-\!1|^{2}\!+\!|z\!+\!1|^{2}}\int\limits_{0}^{\infty}\left(1\!+\!\dfrac{1}{t^{2}}\right)e^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)$}}}}}dt, (A.16f)

where

0(1+1t2)eπ2(t+1t)dt=21(1+1t2)eπ2(t+1t)dt=\displaystyle\int\limits_{0}^{\infty}\left(1\!+\!\dfrac{1}{t^{2}}\right)e^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)$}}}}}dt=2\int\limits_{1}^{\infty}\left(1\!+\!\dfrac{1}{t^{2}}\right)e^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)$}}}}}dt\boxed{=}
=|2t=t+1t, 2dt=(11t2)dt,1+1t211t2=t+1tt1t|\displaystyle=\left|2t^{\,\prime}=t+\dfrac{1}{t}\,,\ 2dt^{\,\prime}=\left(1-\dfrac{1}{t^{2}}\right)dt\,,\ \dfrac{1\!+\!\dfrac{1}{t^{2}}}{1\!-\!\dfrac{1}{t^{2}}}=\dfrac{t+\dfrac{1}{t}}{t-\dfrac{1}{t}}\right|
|t22tt+1=0,t=t+(t)21,1t=t(t)21|\displaystyle\left|t^{2}-2tt^{\,\prime}+1=0\,,\ t=t^{\,\prime}+\sqrt{\left(t^{\,\prime}\right)^{2}-1}\,,\ \dfrac{1}{t}=t^{\,\prime}-\sqrt{\left(t^{\,\prime}\right)^{2}-1}\right|
= 211+1t211t2eπ2(t+1t)(11t2)dt=21t+1tt1teπ2(t+1t)(11t2)dt\displaystyle\boxed{=}\ 2\int\limits_{1}^{\infty}\dfrac{1\!+\!\dfrac{1}{t^{2}}}{1\!-\!\dfrac{1}{t^{2}}}e^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)$}}}}}\left(1-\dfrac{1}{t^{2}}\right)dt=2\int\limits_{1}^{\infty}\dfrac{t+\dfrac{1}{t}}{t-\dfrac{1}{t}}e^{{{\mbox{\footnotesize{$-\dfrac{\pi}{2}\left(t+\dfrac{1}{t}\right)$}}}}}\left(1-\dfrac{1}{t^{2}}\right)dt
=41tt21eπtdt=211t1eπtdt=201teπt+1dt\displaystyle=4\int\limits_{1}^{\infty}\dfrac{t}{\sqrt{t^{2}-1}}e^{-\pi t}dt=2\int\limits_{1}^{\infty}\dfrac{1}{\sqrt{t-1}}e^{-\pi\sqrt{t}}dt=2\int\limits_{0}^{\infty}\dfrac{1}{\sqrt{t}}e^{-\pi\sqrt{t+1}}dt
=40eπt2+1dt40eπtdt=4π\displaystyle=4\int\limits_{0}^{\infty}e^{-\pi\sqrt{t^{2}+1}}dt\leqslant 4\int\limits_{0}^{\infty}e^{-\pi t}dt=\dfrac{4}{\pi}

where

0eπt2+1dt=n0nn+1eπt2+1dtn0eπn2+1=\displaystyle\int\limits_{0}^{\infty}e^{-\pi\sqrt{t^{2}+1}}dt=\sum\limits_{n\geqslant 0}\int\limits_{n}^{n+1}e^{-\pi\sqrt{t^{2}+1}}dt\leqslant\sum\limits_{n\geqslant 0}e^{-\pi\sqrt{n^{2}+1}}=
=eπ+n1eπn=1eπ+1eπ+1=\displaystyle=e^{-\pi}+\sum\limits_{n\geqslant 1}e^{-\pi n}=\dfrac{1}{e^{\pi}}+\dfrac{1}{e^{\pi}+1}=
=0,04321391+0,04142383216636282681=0,08463775043\displaystyle=0,04321391+0,04142383216636282681=0,08463775043

But actually (erd, , p.​ 82, (19)), (abr, , p.​ 376, 9.6.27)

K0(z)=11t21eztdt1tt21eztdt=K0(z)\displaystyle K_{0}(z)=\int\limits_{1}^{\infty}\dfrac{1}{\sqrt{t^{2}-1}}e^{-zt}dt\ \Rightarrow\ \int\limits_{1}^{\infty}\dfrac{t}{\sqrt{t^{2}-1}}e^{-zt}dt=-K_{0}^{\,\prime}(z)\ \Rightarrow\
0eπt2+1dt=1tt21eπtdt=K0(π)=K1(π).\displaystyle\int\limits_{0}^{\infty}e^{-\pi\sqrt{t^{2}+1}}dt=\int\limits_{1}^{\infty}\dfrac{t}{\sqrt{t^{2}-1}}e^{-\pi t}dt=-K_{0}^{\,\prime}(\pi)=K_{1}(\pi)\,.

So that

n|Hn(z)|,n|Mn(z)|320π3e2πn|z1|2+|z+1|24K1(π),\displaystyle n\left|\eurm{H}_{n}(z)\right|,n\left|\eurm{M}_{n}(z)\right|\leqslant\dfrac{320}{\pi^{3}}\dfrac{e^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{|z\!-\!1|^{2}\!+\!|z\!+\!1|^{2}}4K_{1}(\pi)\ ,\
1280K1(π)π3<1280π3(1eπ+1eπ+1)<4,\displaystyle\dfrac{1280K_{1}(\pi)}{\pi^{3}}<\dfrac{1280}{\pi^{3}}\left(\dfrac{1}{e^{\pi}}+\dfrac{1}{e^{\pi}+1}\right)<4\ ,\

and hence,

n|Hn(z)|,n|Mn(z)|4e2πn|z1|2+|z+1|2,z\displaystyle n\left|\eurm{H}_{n}(z)\right|,n\left|\eurm{M}_{n}(z)\right|\leqslant\dfrac{4e^{{{\mbox{\footnotesize{$2\pi n$}}}}}}{|z\!-\!1|^{2}\!+\!|z\!+\!1|^{2}}\ ,\ z\in\Bb{H}\cup\Bb{R}\,. (A.16g)
17

\uparrow  Using (gra, , p.​ 44, 1.422.4), we obtain the validity of the left-hand side equality in

k1(2k+z)2=π24sin2πz2=π2m1meiπmz,z\displaystyle\sum\limits_{k\in\Bb{Z}}\frac{1}{(2k+z)^{2}}=\frac{\pi^{2}}{4\sin^{2}\dfrac{\pi z}{2}}=-\pi^{2}\sum_{m\geqslant 1}me^{{{\mbox{\footnotesize{$i\pi mz$}}}}}\ ,\qquad z\in\Bb{H}\,,\

while the validity of the right-hand side equality here follows from the following identities

sin2πz2=(eiπz2eiπz22i)2=eiπz4(1eiπz)2,\displaystyle\sin^{2}\dfrac{\pi z}{2}=\left(\dfrac{e^{{{\mbox{\footnotesize{$\dfrac{i\pi z}{2}$}}}}}-e^{{{\mbox{\footnotesize{${{\mbox{\normalsize{$-$}}}}\dfrac{i\pi z}{2}$}}}}}}{2i}\right)^{2}=-\dfrac{e^{{{\mbox{\footnotesize{$i\pi z$}}}}}}{4}\left(1-e^{{{\mbox{\footnotesize{$-i\pi z$}}}}}\right)^{2}\ ,\
1sin2πz2=4eiπz(1eiπz)2,\displaystyle\dfrac{1}{\sin^{2}\dfrac{\pi z}{2}}=-\dfrac{4e^{{{\mbox{\footnotesize{$-i\pi z$}}}}}}{\left(1-e^{{{\mbox{\footnotesize{$-i\pi z$}}}}}\right)^{2}}\ ,
11u=m0um,1(1u)2=m1mum1,u𝔻\displaystyle\dfrac{1}{1-u}=\sum_{m\geqslant 0}u^{m}\ ,\ \dfrac{1}{(1-u)^{2}}=\sum_{m\geqslant 1}mu^{m-1}\ ,\ u\in\Bb{D},
1sin2πz2=4eiπz(1eiπz)2=4eiπzm1meiπ(m1)z=4m1meiπmz.\displaystyle\dfrac{1}{\sin^{2}\dfrac{\pi z}{2}}=-\dfrac{4e^{{{\mbox{\footnotesize{$i\pi z$}}}}}}{\left(1-e^{{{\mbox{\footnotesize{$i\pi z$}}}}}\right)^{2}}=-4e^{{{\mbox{\footnotesize{$i\pi z$}}}}}\sum_{m\geqslant 1}me^{{{\mbox{\footnotesize{$i\pi(m-1)z$}}}}}=-4\sum_{m\geqslant 1}me^{{{\mbox{\footnotesize{$i\pi mz$}}}}}\ .
18

\uparrow  For any zz\in\Bb{C}\setminus\Bb{R} it follows from (gra, , p.​ 44, 1.421.3) that

π2cotπz2=1x+k1(1z+2k+1z2k)=limN+NkN12k+z,\displaystyle\frac{\pi}{2}\cot\dfrac{\pi z}{2}=\frac{1}{x}+\sum\limits_{k\geqslant 1}\left(\dfrac{1}{z+2k}+\dfrac{1}{z-2k}\right)=\lim\limits_{N\to+\infty}\sum\limits_{{{\mbox{\footnotesize{$-N\leqslant k\leqslant N$}}}}}\frac{1}{2k+z}\,,

and if xx\in\Bb{R} then

k2z(2k+x)2z2=k(12k+xz12k+x+z)\displaystyle\sum\limits_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\dfrac{2z}{(2k+x)^{2}-z^{2}}=\sum\limits_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\left(\frac{1}{2k+x-z}-\frac{1}{2k+x+z}\right)
=limN+NkN(12k+xz12k+x+z)\displaystyle=\lim\limits_{N\to+\infty}\sum\limits_{{{\mbox{\footnotesize{$-N\leqslant k\leqslant N$}}}}}\left(\frac{1}{2k+x-z}-\frac{1}{2k+x+z}\right)
=π2cotπ(xz)2π2cotπ(x+z)2=π2cotπ(zx)2π2cotπ(x+z)2,\displaystyle=\frac{\pi}{2}\cot\dfrac{\pi(x-z)}{2}-\frac{\pi}{2}\cot\dfrac{\pi(x+z)}{2}=-\frac{\pi}{2}\cot\dfrac{\pi(z-x)}{2}-\frac{\pi}{2}\cot\dfrac{\pi(x+z)}{2}\ ,

where for arbitrary zz\in\Bb{H} we have

cotπz2=ieiπz/2+eiπz/2eiπz/2eiπz/2=ieiπz+1eiπz1=i1+eiπz1eiπz=i2im1eiπmz,\displaystyle\cot\dfrac{\pi z}{2}=i\dfrac{e^{i\pi z/2}+e^{-i\pi z/2}}{e^{i\pi z/2}-e^{-i\pi z/2}}=i\dfrac{e^{i\pi z}+1}{e^{i\pi z}-1}=-i\dfrac{1+e^{i\pi z}}{1-e^{i\pi z}}=-i-2i\sum\limits_{{{\mbox{\footnotesize{$m\geqslant 1$}}}}}e^{i\pi mz}\ ,
4πkz(2k+x)2z2=cotπ(zx)2cotπ(z+x)2\displaystyle\dfrac{4}{\pi}\sum\limits_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\dfrac{z}{(2k+x)^{2}-z^{2}}=-\cot\dfrac{\pi(z-x)}{2}-\cot\dfrac{\pi(z+x)}{2}
=2i+2im1(eiπmzeiπmx+eiπmzeiπmx)=2i+4im1eiπmzcosπmx,\displaystyle=2i+2i\sum\limits_{{{\mbox{\footnotesize{$m\geqslant 1$}}}}}\left(e^{i\pi mz}e^{-i\pi mx}+e^{i\pi mz}e^{i\pi mx}\right)=2i+4i\sum\limits_{{{\mbox{\footnotesize{$m\geqslant 1$}}}}}e^{i\pi mz}\cos\pi mx\ ,\

which proves the identity

kz(2k+x)2z2\displaystyle\sum\limits_{{{\mbox{\footnotesize{$k\in\Bb{Z}$}}}}}\dfrac{z}{(2k+x)^{2}-z^{2}} =π4(cotπ(zx)2+cotπ(z+x)2)\displaystyle=-\dfrac{\pi}{4}\left(\cot\dfrac{\pi(z-x)}{2}+\cot\dfrac{\pi(z+x)}{2}\right)
=πi2+iπm1eiπmzcosπmx,z\displaystyle=\dfrac{\pi i}{2}+i\pi\sum\limits_{{{\mbox{\footnotesize{$m\geqslant 1$}}}}}e^{i\pi mz}\cos\pi mx\ ,\qquad z\in\Bb{H}\,,\ x\in\Bb{R}.

A.5 . Notes for Section 5

19

\uparrow   Let ϕΓϑ\phi\!\in\!\Gamma_{\vartheta}. Then by Lemma 2 of (cha, , p.​ 112)20{}^{\ref*{case43}} the transform ϕ\phi can be represented as a superposition of a finite number of the degrees of the transformations z+2z+2 and 1/z-1/z (uniqueness of such representation has not been proved by Chandrasekharan in cha ). In the notation of (10.53) and in view of (10.56), this means that either ϕ(z){z,1/z}={ϕ0(z),ϕ0(1/z)}={ϕ0(z),1/ϕ0(z),ϕ0(1/z),1/ϕ0(1/z)}\phi(z)\in\{z,-1/z\}=\{\phi_{0}(z),\phi_{0}(-1/z)\}=\{\phi_{0}(z),\ -1/\phi_{0}(z),\ \phi_{0}(-1/z),\ -1/\phi_{0}(-1/z)\} or there exist α,β{0,1}\alpha,\beta\in\{0,1\}, NN\in\Bb{N} and n1,,nNn_{1},\ldots,n_{N}\in\Bb{Z}_{\neq 0} such that (see (bh1, , p.​ 63))

ϕ(z)=ϕSαST2n1ST2nN1ST2nNSβ(z)=ϕSα(ϕnN,,n1(ϕSβ(z))),\displaystyle\phi(z)\!=\!\phi_{{{\mbox{\footnotesize{$S^{{{\mbox{\footnotesize{$\,\alpha$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N}$}}}}}S^{{{\mbox{\footnotesize{$\,\beta$}}}}}$}}}}}(z)=\phi_{{{\mbox{\footnotesize{$S^{{{\mbox{\footnotesize{$\,\alpha$}}}}}$}}}}}\left(\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\phi_{{{\mbox{\footnotesize{$S^{{{\mbox{\footnotesize{$\,\beta$}}}}}$}}}}}(z)\right)\right)\!,

where (10.57) is used for 𝔫:=(𝔫𝔑,,𝔫1)\eufm{n}\!:=\!(n_{N},...,n_{1})\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\cup_{k\geqslant 1}\Bb{Z}_{\neq 0}^{k} and zz\in\Bb{H}. In view of (5.8) and (5.2),

ϕ𝔫:=ϕnN,,n1Γϑ||,\displaystyle\phi_{\eufm{n}}\!:=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\in\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\,,

and by setting in the above equalities to (α,β)(\alpha,\beta) each of four possible values (0,0),(1,0),(0,1),(1,1)(0,0),(1,0),(0,1),(1,1) we obtain that

ϕ(z){ϕ𝔫(z),1/ϕ𝔫(z),ϕ𝔫(1/z),1/ϕ𝔫(1/z)},\displaystyle\phi(z)\in\left\{\phi_{\eufm{n}}(z),\ -1/\phi_{\eufm{n}}(z),\ \phi_{\eufm{n}}(-1/z),\ -1/\phi_{\eufm{n}}(-1/z)\right\}\,,\

which completes the proof of (5.3).

20

\uparrow  \uparrow   Lemma 2 of (cha, , p.​ 112) can be sharpened as follows.

Theorem A.1

. For arbitrary ASL2(ϑ,{𝕀𝕀𝕊𝕊}A\in{\rm{SL}}_{2}(\vartheta,\Bb{Z})\setminus\{I,-I,S,-S\} there exists a unique collection of numbers σ{1,1}\sigma\in\{1,-1\}, α,β{0,1}\alpha,\beta\in\{0,1\}, NN\in\Bb{N} and n1,,nNn_{1},\ldots,n_{N}\in\Bb{Z}_{\neq 0} such that

A=σSα(ST2n1ST2nN1ST2nN)Sβ,\displaystyle A=\sigma\cdot S^{{{\mbox{\footnotesize{$\,\alpha$}}}}}\Big{(}ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N}$}}}}}\Big{)}S^{{{\mbox{\footnotesize{$\,\beta$}}}}}, (A.20a)

where

SL2(ϑ,{(abcd)𝕊𝕃𝕜(abcd)(1001)or(abcd)(0110)}\displaystyle{\rm{SL}}_{2}(\vartheta,\Bb{Z})\!=\!\left\{\,{{\mbox{\footnotesize{$\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}$}}}}\!\in\!{\rm{SL}}_{2}(\Bb{Z})\ \left|\ {{\mbox{\footnotesize{$\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}$}}}}\!\equiv\!{{\mbox{\footnotesize{$\begin{pmatrix}1&0\\ 0&1\\ \end{pmatrix}$}}}}({\rm{mod}}\,2)\ \ \mbox{or}\ {{\mbox{\footnotesize{$\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}$}}}}\!\equiv\!{{\mbox{\footnotesize{$\begin{pmatrix}0&1\\ 1&0\\ \end{pmatrix}$}}}}({\rm{mod}}\,2)\,\right\}\right.\!,
T=(1101)SL2(ϑ,𝕋±(1±201)𝕊𝕃ϑ𝕊(0110)𝕊𝕃ϑ\displaystyle T={{\mbox{\footnotesize{$\begin{pmatrix}1&1\\ 0&1\\ \end{pmatrix}$}}}}\not\in{\rm{SL}}_{2}(\vartheta,\Bb{Z}),\ T^{\pm 2}={{\mbox{\footnotesize{$\begin{pmatrix}1&\pm 2\\ 0&1\\ \end{pmatrix}$}}}}\in{\rm{SL}}_{2}(\vartheta,\Bb{Z}),\ S={{\mbox{\footnotesize{$\begin{pmatrix}0&1\\ -1&0\\ \end{pmatrix}$}}}}\in{\rm{SL}}_{2}(\vartheta,\Bb{Z}),

S2=IS^{2}=-I, I:=(1001)I:=(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix}), and SL2({\rm{SL}}_{2}(\Bb{Z}) denotes the set of all 2×22\times 2 matrices (abcd)(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}) with integer coefficients a,b,c,da,b,c,d\!\in\!\Bb{Z} satisfying adbc=1ad-bc\!=\!1.

The fact that SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) is invariant under multiplication of matrices follows from the possibility of a termwise multiplying the congruences (see (apo, , p.​ 107, Theorem 5.2(b))) and (0110)(0110)=(1001)(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix})(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix})=(\begin{smallmatrix}1&0\\ 0&1\end{smallmatrix}), while belonging to SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) of each its inverse matrix is a consequence of the general formula

M=(abcd)SL2(𝕄𝕊𝕃\displaystyle M=\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}\in\mathrm{SL}_{2}(\Bb{Z})\ \ \Rightarrow\ \ M^{-1}=\begin{pmatrix}d&-b\\ -c&a\\ \end{pmatrix}\in\mathrm{SL}_{2}(\Bb{Z})\ . (A.20b)

As well as SL2({\rm{SL}}_{2}(\Bb{Z}), the set SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) is also invariant with respect to the transpose operation

SL2(ϑ,𝕄𝕄𝕊𝕃ϑ\displaystyle{\rm{SL}}_{2}(\vartheta,\Bb{Z})\ni M=\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}\ \ \ \Rightarrow\ \ \ M^{\ast}=\begin{pmatrix}a&c\\ b&d\\ \end{pmatrix}\in{\rm{SL}}_{2}(\vartheta,\Bb{Z})\ . (A.20c)

Everywhere below we use the notation {±I,±S}:={I,I,S,S}\{\pm I,\pm S\}:=\{I,-I,S,-S\}.

Before proving Theorem A.1, we first establish several special properties of matrices from the set SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}). Each column and each row of any matrix from the set SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) contains two numbers of different parity and therefore they cannot be equal to each other. In fact, a much stronger property of these matrices takes place.

Lemma A.2

. Let A=(abcd)SL2(ϑ,{𝕀𝕀𝕊𝕊}A=(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})\in{\rm{SL}}_{2}(\vartheta,\Bb{Z})\setminus\{I,-I,S,-S\}. Then the value of min{|a|,|b|,|c|,|d|}\min\{|a|,|b|,|c|,|d|\} is attained on only one element.

Proof

. Suppose to the contrary that min{|a|,|b|,|c|,|d|}\min\{|a|,|b|,|c|,|d|\} is attained on more than one element. Since the pairs (a,b)(a,b), (a,c)(a,c), (c,d)(c,d) and (b,d)(b,d) consist of numbers with different parity then the only one of two following cases can take place

(1)|a|=|d|=min{|a|,|b|,|c|,|d|}<min{|b|,|c|},(2)|b|=|c|=min{|a|,|b|,|c|,|d|}<min{|a|,|d|}.\displaystyle\begin{array}[]{l}{\rm{(1)}}\quad|a|=|d|=\min\{|a|,|b|,|c|,|d|\}<\min\{|b|,|c|\}\ ,\\ {\rm{(2)}}\quad|b|=|c|=\min\{|a|,|b|,|c|,|d|\}<\min\{|a|,|d|\}\ .\end{array} (A.20f)

If (A.20f)(1) holds then

|a|=|d|=min{|a|,|b|,|c|,|d|}min{|b|,|c|}1,\displaystyle|a|=|d|=\min\{|a|,|b|,|c|,|d|\}\leqslant\min\{|b|,|c|\}-1\ ,\ (A.20g)

from which |ad|(min{|b|,|c|}1)2(min{|b|,|c|})22min{|b|,|c|}+1|bc|+12min{|b|,|c|}|ad|\leqslant\left(\min\{|b|,|c|\}-1\right)^{2}\leqslant\left(\min\{|b|,|c|\}\right)^{2}-2\min\{|b|,|c|\}+1\leq|bc|+1-2\min\{|b|,|c|\} and therefore 2min{|b|,|c|}1|bc||ad||bcad|=12\min\{|b|,|c|\}-1\leqslant|bc|-|ad|\leqslant|bc-ad|=1, i.e.,

min{|b|,|c|}1.\displaystyle\min\{|b|,|c|\}\leqslant 1\ . (A.20h)

Substituting this inequality in (A.20g) we get a=d=0a=d=0, while (A.20h) and adbc=1ad-bc=1 mean that bc=1bc=-1 and |b|=|c|=1|b|=|c|=1. These conditions are satisfied only when A=±SA=\pm S that contradicts A{±I,±S}A\not\in\{\pm I,\pm S\}. Thus, (A.20f) cannot be valid.

If (A.20f)(2) holds then the matrix AS=(badc)AS=(\begin{smallmatrix}-b&a\\ -d&c\\ \end{smallmatrix}) satisfies the condition (A.20f)(1) and using the above reasoning we conclude that either AS=SAS=S or AS=SAS=-S, i.e., either A=IA=I or A=IA=-I. This also contradicts A{±I,±S}A\not\in\{\pm I,\pm S\} and finishes the proof of Lemma A.2. \square

We say that A=(abcd)SL2(ϑ,A=(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})\in{\rm{SL}}_{2}(\vartheta,\Bb{Z}) is modulo isotonic if

|a|<|b|<|d| and|a|<|c|<|d|.\displaystyle|a|<|b|<|d|\quad\mbox{ and}\quad|a|<|c|<|d|\ . (A.20i)

Lemma A.2 can be essentially sharpened as follows.

Lemma A.3

. Let A=(abcd)SL2(ϑ,{±𝕀±𝕊}A=(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})\in{\rm{SL}}_{2}(\vartheta,\Bb{Z})\setminus\{\pm I,\pm S\}. Then among four matrices

A:=(abcd),AS=(badc),SA=(cdab),SAS=(dcba),\displaystyle A:={{\mbox{\footnotesize{$\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}$}}}}\ ,\ AS={{\mbox{\footnotesize{$\begin{pmatrix}-b&a\\ -d&c\\ \end{pmatrix}$}}}}\ ,\ SA={{\mbox{\footnotesize{$\begin{pmatrix}c&d\\ -a&-b\\ \end{pmatrix}$}}}}\ ,\ SAS={{\mbox{\footnotesize{$\begin{pmatrix}-d&c\\ b&-a\\ \end{pmatrix}$}}}}\ , (A.20j)

each of which belongs to SL2(ϑ,{±𝕀±𝕊}{\rm{SL}}_{2}(\vartheta,\Bb{Z})\setminus\{\pm I,\pm S\}, there exists exactly one modulo isotonic matrix.

Proof

. Four matrices {±I,±S}\{\pm I,\pm S\} form the group with the matrix multiplication as a group operation, and therefore one of the matrices from (A.20j) can belong to {±I,±S}\{\pm I,\pm S\} if and only if {±I,±S}\{\pm I,\pm S\} contains the matrix AA. It readily follows from Lemma A.2 and (A.20j) that there exists a unique ordered pair (α,β)(\alpha,\beta) of numbers α,β{0,1}\alpha,\beta\in\{0,1\} such that the matrix SαASβS^{\alpha}AS^{\beta} at the intersection of its first row and first column contains exactly that element of AA which has the smallest absolute value. By designating the elements of the transformed matrix SαASβS^{\alpha}AS^{\beta} by the same letters, we have

B:=SαASβ=(abcd),|a||b|1,|a||c|1,|a||d|1.\displaystyle B:=S^{\alpha}AS^{\beta}=\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}\ ,\ |a|\leqslant|b|-1\ ,\ |a|\leqslant|c|-1\ ,\ |a|\leqslant|d|-1\ . (A.20k)

To prove that BB is modulo isotonic, it is sufficient to show that |b|<|d||b|<|d| and |c|<|d||c|<|d|.

Assume first that

|d||b|.\displaystyle|d|\leqslant|b|\ . (A.20l)

By virtue of the different parity of |b||b| and |d||d|, we have |d||b|1|d|\leqslant|b|-1 and therefore |ad||a||b||a|(|c|1)|b||a|=|b||c||b||a||ad|\leqslant|a||b|-|a|\leqslant(|c|-1)|b|-|a|=|b||c|-|b|-|a|, i.e., |a|+|b||b||c||ad||a|+|b|\leqslant|b||c|-|ad| and 1=|bcad||b||c||a||d||a|+|b|2|a|+11=|bc-ad|\geqslant|b||c|-|a||d|\geqslant|a|+|b|\geqslant 2|a|+1, which yields a=0a=0. Then adbc=1ad-bc=1 implies bc=1bc=-1 and therefore |c|=|b|=1|c|=|b|=1. Our assumption |d||b|1=0|d|\leqslant|b|-1=0 gives d=0d=0. Finally, we obtain a=d=0a=d=0 and |c|=|b|=1|c|=|b|=1. This holds if and only if B=σSB=\sigma\cdot S for some σ{1,1}\sigma\in\{1,-1\}. By solving the equation SαASβ=σSS^{\alpha}AS^{\beta}=\sigma\cdot S we get A=(1)α+βσSα+β+1{I,I,S,S}A=(-1)^{\alpha+\beta}\sigma\cdot S^{\alpha+\beta+1}\in\{I,-I,S,-S\} which contradicts A{±I,±S}A\not\in\{\pm I,\pm S\}. This contradiction proves |b|<|d||b|<|d|.

If we assume that |d||c||d|\leqslant|c| then the conjugate matrix B=(acbd)B^{*}=(\begin{smallmatrix}a&c\\ b&d\end{smallmatrix}) also belongs to SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) and satisfies the conditions (A.20k) and (A.20l). By application of the above reasoning we conclude that B=σSB^{*}=\sigma\cdot S for some σ{1,1}\sigma\in\{1,-1\}. Then B=(σ)SB=(-\sigma)\cdot S and similarly to the above A=(1)α+β+1σSα+β+1{I,I,S,S}A=(-1)^{\alpha+\beta+1}\sigma\cdot S^{\alpha+\beta+1}\in\{I,-I,S,-S\}, which contradicts A{±I,±S}A\not\in\{\pm I,\pm S\}. This contradiction proves |c|<|d||c|<|d| and that BB is modulo isotonic. But if one of the matrices in (A.20j) is modulo isotonic then no other one in (A.20j) can have this property because all other matrices contain the unique element of AA with the smallest absolute value at the wrong place. Lemma A.3 is proved. \square

The next assertion is immediate from Lemma A.3.

Corollary 1

. Let A=(abcd)SL2(ϑ,A=(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})\in{\rm{SL}}_{2}(\vartheta,\Bb{Z}) satisfy |a||b||d||a|\leqslant|b|\leqslant|d| and |a||c||d||a|\leqslant|c|\leqslant|d|. Then |a|<|b|<|d||a|<|b|<|d| and |a|<|c|<|d||a|<|c|<|d|, i.e., AA is modulo isotonic.

The following result shows that any modulo isotonic matrix from the set SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) is uniquely determined by its second column.

Lemma A.4

. Let two matrices A1=(a1b1c1d1)A_{1}\!=\!(\begin{smallmatrix}a_{1}&b_{1}\\ c_{1}&d_{1}\end{smallmatrix}) and A2=(a2b2c2d2)A_{2}\!=\!(\begin{smallmatrix}a_{2}&b_{2}\\ c_{2}&d_{2}\end{smallmatrix}) belong to SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) and be modulo isotonic. Suppose that b1=b2b_{1}\!=\!b_{2} and d1=d2d_{1}\!=\!d_{2}. Then A1=A2A_{1}\!=\!A_{2}.

Proof

. Assume that two different modulo isotonic matrices in SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) satisfy

A1=(a1bc1d),A2=(a2bc2d);(1)(a1a2)2+(c1c2)2>0,\displaystyle A_{1}=\begin{pmatrix}a_{1}&b\\ c_{1}&d\\ \end{pmatrix}\ ,\ A_{2}=\begin{pmatrix}a_{2}&b\\ c_{2}&d\\ \end{pmatrix}\ ;\quad{\rm{(1)}}\ \ (a_{1}-a_{2})^{2}+(c_{1}-c_{2})^{2}>0\ , (A.20m)

In view of (A.20i), |d|>|b|>0|d|>|b|>0 and it follows from a1dbc1=1a_{1}d-bc_{1}=1, a2dbc2=1a_{2}d-bc_{2}=1 that

(1)(a1a2)d=(c1c2)b;(2)gcd(d,b)=1;(3)d,b{}\displaystyle{\rm{(1)}}\ \,(a_{1}-a_{2})d=(c_{1}-c_{2})b\ ;\quad{\rm{(2)}}\ \,{\rm{gcd}}(d,b)=1\ ;\quad{\rm{(3)}}\ \,d,b\in\Bb{Z}\setminus\{0\}\ . (A.20n)

By virtue of (A.20n)(1) and (A.20n)(3), the property (A.20m)(1) yields that

a1a20,c1c20.\displaystyle a_{1}-a_{2}\neq 0\ ,\ c_{1}-c_{2}\neq 0\ . (A.20o)

But (A.20n)(1) and (A.20n)(2) implies that dd divides (c1c2)b(c_{1}-c_{2})b and since gcd(d,b)=1{\rm{gcd}}(d,b)=1 we get that dd divides (c1c2)(c_{1}-c_{2}). Hence, taking account of (A.20o) and (A.20n)(3), there exists mm\in\Bb{Z}_{\neq 0} such that c1c2=mdc_{1}-c_{2}=md. Substituting this in (A.20n)(1) we obtain (a1a2)d=mdb(a_{1}-a_{2})d=mdb, where we can divide by nonzero number dd and derive from (A.20i) that

(1)a1=a2+mb,|a1|,|a2||b|1;(2)c1=c2+md,|c1|,|c2||d|1;\displaystyle{\rm{(1)}}\ \ a_{1}=a_{2}+mb\ ,\ |a_{1}|,|a_{2}|\leqslant|b|-1\ ;\quad{\rm{(2)}}\ \ c_{1}=c_{2}+md\ ,\ |c_{1}|,|c_{2}|\leqslant|d|-1\ ;
(3)m{}\displaystyle{\rm{(3)}}\ \ m\in\Bb{Z}\setminus\{0\}\ . (A.20p)

It follows from (A.20p)(1) that |mb|2|b|2|mb|\leq 2|b|-2 and since by (A.20n)(3) bb is nonzero we conclude that the condition (A.20p)(3) can be replaced by

m{1,1}.\displaystyle m\in\{1,-1\}\ . (A.20q)

According to the definition SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}), we have only two possibilities

(1)(a2bc2d)(a1bc1d)(1001)(mod 2)or(2)(a2bc2d)(a1bc1d)(0110)(mod 2).\displaystyle\begin{array}[]{l}{\rm{(1)}}\ \ \begin{pmatrix}a_{2}&b\\ c_{2}&d\\ \end{pmatrix}\!\equiv\!\begin{pmatrix}a_{1}&b\\ c_{1}&d\\ \end{pmatrix}\!\equiv\!\begin{pmatrix}1&0\\ 0&1\\ \end{pmatrix}({\rm{mod}}\,2)\quad\mbox{or}\\[14.22636pt] {\rm{(2)}}\ \ \begin{pmatrix}a_{2}&b\\ c_{2}&d\\ \end{pmatrix}\!\equiv\!\begin{pmatrix}a_{1}&b\\ c_{1}&d\\ \end{pmatrix}\!\equiv\!\begin{pmatrix}0&1\\ 1&0\\ \end{pmatrix}({\rm{mod}}\,2)\ .\end{array} (A.20t)

Assume that (A.20t)(1) holds. But (A.20p)(2) and (A.20q) yield c1c2+md1(mod 2)c_{1}\equiv c_{2}+md\equiv 1({\rm{mod}}\,2) which contradicts (A.20t)(1).

Suppose now that (A.20t)(2) holds. Then (A.20p)(1) and (A.20q) yield a1a2+mb1(mod 2)a_{1}\equiv a_{2}+mb\equiv 1({\rm{mod}}\,2) which contradicts (A.20t)(2).

These contradictions together with (A.20m) complete the proof of Lemma A.4. \square

Lemma A.5

. Let B=(abcd)SL2(ϑ,B=(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix})\in{\rm{SL}}_{2}(\vartheta,\Bb{Z}) be modulo isotonic (see (A.20i)). Then there exists a unique collection of numbers γ{0,1}\gamma\in\{0,1\}, NN\in\Bb{N} and n1,,nNn_{1},\ldots,n_{N}\in\Bb{Z}_{\neq 0} such that

(1)γB=ST2nNST2nN1ST2n1.\displaystyle(-1)^{\gamma}B\!=\!ST^{{{\mbox{\footnotesize{$-2n_{N}$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}. (A.20u)
Proof

. By (A.20i) and the definition of SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}), bb and dd are nonzero integers of different parity, b/d(1,1){0}b/d\in(-1,1)\setminus\{0\} and it follows from adbc=1ad-bc\!=\!1 that gcd(b,d)=1{\rm{gcd}}(b,d)=1. By applying Lemma A.6 to the rational number b/db/d we obtain the existence of the unique NN\!\in\!\Bb{N} and (nN,,n1)(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N} such that

bd=ϕnN,,n1(0):=12nN12nN112n212n1,(nN,,n1)\displaystyle\dfrac{b}{d}=\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(0):={{\mbox{\footnotesize{$\dfrac{1}{2n_{N}\!-\!\dfrac{1}{\begin{subarray}{l}\displaystyle\vphantom{A^{A}}2n_{N-1}\!-\!\\[-11.38092pt] \hphantom{2n_{k-1}11-}\ddots\\[-11.38092pt] \hphantom{2n_{k-1}11-\ddots}\displaystyle-\!\dfrac{1}{2n_{2}\!-\!\dfrac{1}{\vphantom{A^{A}}2n_{1}}}\end{subarray}}}$}}}}\ ,\quad(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\!\in\!\Bb{N}\,,

where

ϕ𝔫(z)=zp𝔫N1+q𝔫N1zp𝔫N+q𝔫N=12nN12nN112n212n1z,z[1,1](\displaystyle\phi_{\eufm{n}}(z)\!=\!\dfrac{zp^{\eufm{n}}_{N-1}+q^{\eufm{n}}_{N-1}}{zp^{\eufm{n}}_{N}+q^{\eufm{n}}_{N}}={{\mbox{\footnotesize{$\dfrac{1}{2n_{N}\!-\!\dfrac{1}{\begin{subarray}{l}\displaystyle\vphantom{A^{A}}2n_{N-1}\!-\!\\[-11.38092pt] \hphantom{2n_{k-1}11-}\ddots\\[-11.38092pt] \hphantom{2n_{k-1}11-\ddots}\displaystyle-\!\dfrac{1}{2n_{2}\!-\!\dfrac{1}{\vphantom{A^{A}}2n_{1}-z}}\end{subarray}}}$}}}}\,,\ \ z\in[-1,1]\cup(\Bb{C}\setminus\Bb{R})\,,

in view of (10.67) and (5.1). Therefore there exists a unique γ{0,1}\gamma\in\{0,1\} such that b=(1)γq𝔫N1b=(-1)^{\gamma}q^{\eufm{n}}_{N-1} and d=(1)γq𝔫Nd=(-1)^{\gamma}q^{\eufm{n}}_{N}. By virtue of (10.62), (10.69) and (10.77), two modulo isotonic matrices

(p𝔫N1q𝔫N1p𝔫Nq𝔫N)=(0112nN)(0112nN1)(0112n1)\displaystyle\begin{pmatrix}p^{\eufm{n}}_{N-1}&q^{\eufm{n}}_{N-1}\\ p^{\eufm{n}}_{N}&q^{\eufm{n}}_{N}\\ \end{pmatrix}\!=\!\begin{pmatrix}0&1\\ -1&2n_{N}\\ \end{pmatrix}\begin{pmatrix}0&1\\ -1&2n_{N-1}\\ \end{pmatrix}\ldots\begin{pmatrix}0&1\\ -1&2n_{1}\\ \end{pmatrix}

and (1)γB(-1)^{\gamma}B belong to SL2(ϑ,{\rm{SL}}_{2}(\vartheta,\Bb{Z}) and have the same second column. In accordance with Lemma A.4 and (10.56), this means that (A.20u) holds and that Lemma A.5 is proved. \square

Proof of Theorem A.1. Let ASL2(ϑ,{±𝕀±𝕊}A\in{\rm{SL}}_{2}(\vartheta,\Bb{Z})\setminus\{\pm I,\pm S\}. According to Lemma A.3, there exists a unique ordered pair (α,β)(\alpha,\beta) of numbers α,β{0,1}\alpha,\beta\in\{0,1\} such that the matrix B:=(abcd):=SαASβB:=(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}):=S^{\alpha}AS^{\beta} is modulo isotonic. Then, obviously,

A=(1)α+βSαBSβ,α,β{0,1}.\displaystyle A=(-1)^{\alpha+\beta}S^{\alpha}BS^{\beta}\ ,\quad\alpha,\beta\in\{0,1\}\,. (A.20v)

By using (A.20u) and (A.20v), we obtain

A=(1)α+β+γSα(ST2nNST2nN1ST2n1)Sβ.\displaystyle A=(-1)^{\alpha+\beta+\gamma}S^{\alpha}\big{(}ST^{{{\mbox{\footnotesize{$-2n_{N}$}}}}}ST^{{{\mbox{\footnotesize{$-2n_{N-1}$}}}}}\ldots ST^{{{\mbox{\footnotesize{$-2n_{1}$}}}}}\big{)}S^{\beta}\,.

This proves (A.20a) for {1,0}σα+β+γ(mod 2)\{1,0\}\ni\sigma\equiv\alpha+\beta+\gamma({\rm{mod}}\,2) and the uniqueness of such representation as well. Theorem A.1 is proved. \square

21

\uparrow  We prove Lemma 5.3(b). The lower arc ϕ𝔫(γ(σn1,))\phi_{\eufm{n}}\big{(}\gamma(-\sigma_{n_{1}},\infty)\big{)} of ^nN,,n1\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}} is a part of the boundary of

ϕnN,,n1(2σn1)\displaystyle\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!-\!2\sigma_{n_{1}}\right) =ϕnN,,n2(12n1+2σn1)=ϕnN,,n2,n1+σn1(),\displaystyle\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\left(\dfrac{1}{2n_{1}\!-\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!+\!2\sigma_{n_{1}}}\right)\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2},n_{1}+\sigma_{n_{1}}$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),

and also

ϕnN,,n1(γ(σn1,))=ϕnN,,n2(12n1γ(σn1,)+2σn1)\displaystyle\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\gamma(-\sigma_{n_{1}},\infty)\right)=\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\left(\dfrac{1}{2n_{1}-\gamma(\sigma_{n_{1}},\infty)+2\sigma_{n_{1}}}\right)
=ϕnN,,n2,n1+σn1(γ(σn1,))=γnN,,n2,n1+σn1(σnN,,n2,n1+σn1,).\displaystyle=\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2},n_{1}+\sigma_{n_{1}}$}}}}}\left(\gamma(\sigma_{n_{1}},\infty)\right)=\gamma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{2},n_{1}+\sigma_{n_{1}}$}}}}}(\sigma_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{2},n_{1}+\sigma_{n_{1}}$}}}}},\infty).

As stated before Lemma 5.3, this means that ϕ𝔫(σn1+i\phi_{\eufm{n}}\big{(}-\sigma_{n_{1}}+i\Bb{R}_{>0}\big{)} is the roof of ^nN,,n1+σn1\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}+\sigma_{n_{1}}}.

Let σ{1,1}\sigma\in\{1,-1\}. Then the lower arc ϕ𝔫(γ(σ,0))\phi_{\eufm{n}}\big{(}\gamma(\sigma,0)\big{)} of ^nN,,n1\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}} is a part of the boundary of

ϕnN,,n1(12σ)=ϕnN,,n2(12n112σ)=ϕnN,,n2,n1,σ(),\displaystyle\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(-\dfrac{1}{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\!-\!2\sigma}\right)\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\left(\dfrac{1}{2n_{1}\!-\!\dfrac{1}{2\sigma\!-\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}}}\right)\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2},n_{1},\sigma$}}}}}\left(\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\right),

and

ϕnN,,n1(γ(σ,0))=ϕnN,,n1(1γ(σ,))=ϕnN,,n1(1γ(σ,)2σ)\displaystyle\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\gamma(\sigma,0)\right)\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(-\dfrac{1}{\gamma(-\sigma,\infty)}\right)\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(-\dfrac{1}{\gamma(\sigma,\infty)-2\sigma}\right)
=ϕnN,,n2(12n112σγ(σ,))=ϕnN,,n2,n1,σ(γ(σ,)).\displaystyle\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2}$}}}}}\left(\dfrac{1}{2n_{1}\!-\!\dfrac{1}{2\sigma\!-\!\gamma(\sigma,\infty)}}\right)\!=\!\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{2},n_{1},\sigma$}}}}}\big{(}\gamma(\sigma,\infty)\big{)}\,.

According to what was stated before Lemma 5.3, this means that ϕnN,,n1(γ(σ,0))\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\gamma(\sigma,0)\right) is the roof of ^nN,,n1,σ\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1},\sigma} for every σ{1,1}\sigma\in\{1,-1\}. Lemma 5.3(b) is proved. Lemma 5.3(c) follows from the obvious fact that γ(0,1)\gamma(0,1) and γ(1,0)\gamma(-1,0), correspondingly, are the roofs of

^1=12|+1=1|+12and ^1=12|1=1|1+2.\displaystyle\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271pt1}=\dfrac{1}{2-\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|\!+\!1$}}}}}}=-\dfrac{1}{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|\!+\!1$}}}}}-2}\ \ \mbox{and }\ \ \widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271pt-1}=\dfrac{1}{-2-\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|\!-\!1$}}}}}}=-\dfrac{1}{\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|\!-\!1$}}}}}+2}\ .
22

\uparrow  We prove (5.19). (a) We have zγ(σ,0)=1/γ(σ,)z\in\gamma(\sigma,0)=-1/\gamma(-\sigma,\infty) if and only if there exists T>0T>0 such that

z=1σ+iT1z=σ+iT.\displaystyle z=-\dfrac{1}{-\sigma+iT}\ \Rightarrow\ -\dfrac{1}{z}=-\sigma+iT\,.

By using (5.12), according to which

Re(1z)2=σ2=2σ,\displaystyle\left\rceil{{\rm{Re}}\left(-\dfrac{1}{z}\right)}\right\lceil_{2}=\left\rceil{-\sigma}\right\lceil_{2}=-2\sigma\,,

we obtain by (5.14),

𝔾ϝϝϝσ𝕋σσ𝕋γσ\displaystyle\Bb{G}_{2}(z)=-\dfrac{1}{z}-\left\rceil{{\rm{Re}}\left(-\dfrac{1}{z}\right)}\right\lceil_{2}=-\sigma+iT+2\sigma=\sigma+iT\in\gamma(\sigma,\infty)\,.

Since T>0T>0 is arbitrary, (5.19)(a) is proved.

(b) Since γ(σ,)||\gamma(\sigma,\infty)\subset\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}, (5.19)(b) is immediate from (5.15) and evident relationship 1/γ(σ,)=γ(σ,0)-1/\gamma(\sigma,\infty)=\gamma(-\sigma,0).

(c) In view of (5.6),

1ϕn(γ(σn,))=2n+γ(σn,)=2n+σn+i\displaystyle-\dfrac{1}{\phi_{{{\mbox{\footnotesize{$n$}}}}}\big{(}\hskip 1.9919pt\gamma(\sigma_{n},\infty)\hskip 1.9919pt\big{)}}=-2n+\gamma(\sigma_{n},\infty)=-2n+\sigma_{n}+{\rm{i}}\,\Bb{R}_{>0}\,.

By (5.13) and (5.12),

𝔾ϕnγσ{ϕnγσ}𝕀ϕnγσ\displaystyle\Bb{G}_{2}\Big{(}\phi_{{{\mbox{\footnotesize{$n$}}}}}\big{(}\hskip 1.9919pt\gamma(\sigma_{n},\infty)\hskip 1.9919pt\big{)}\Big{)}=\left\{{\rm{Re}}\left(-\dfrac{1}{\phi_{{{\mbox{\footnotesize{$n$}}}}}\big{(}\hskip 1.9919pt\gamma(\sigma_{n},\infty)\hskip 1.9919pt\big{)}}\right)\right\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}+{\rm{i}}\,{\rm{Im}}\left(-\dfrac{1}{\phi_{{{\mbox{\footnotesize{$n$}}}}}\big{(}\hskip 1.9919pt\gamma(\sigma_{n},\infty)\hskip 1.9919pt\big{)}}\right)
={2n+σn}2+iσγσ\displaystyle=\left\{-2n+\sigma_{n}\right\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}+{\rm{i}}\,\Bb{R}_{>0}=\sigma_{n}+{\rm{i}}\,\Bb{R}_{>0}=\gamma(\sigma_{n},\infty)\,,\

which proves (5.19)(c) and completes the proof of (5.19).

23

\uparrow  We prove (5.26). It follows from

{1/ϕn1(z)=2n1+z,ifN=1,1/ϕnN,,n1(z)=2nN+ϕnN1,,n1(z),ifN2,\displaystyle\left\{\begin{array}[]{rlll}-1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(z)&=&-2n_{1}+z\,,&\ \mbox{if}\ N=1\,,\\ -1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(z)&=&-2n_{N}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-1},...,n_{1}$}}}}}(z)\,,&\ \mbox{if}\ N\geqslant 2\,,\end{array}\right.

and from

ϕ(𝕜𝕜)𝕜𝕜𝔽ϕΓϑ||\displaystyle\phi\left(\Bb{H}_{|{\rm{Re}}|\leqslant 1}\right)\!\subset\!\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\ \ \phi\!\in\!\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\,, (5.9)

that (5.23) holds because

Re(1/ϕn1(z))=Re(2n1+z)2n1+(1,1),ifN=1,z𝕜𝕜\displaystyle{\rm{Re}}\left(-1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(z)\right)={\rm{Re}}\left(-2n_{1}+z\right)\in-2n_{1}+(-1,1)\,,\ \mbox{if}\ N=1\,,\ z\!\in\!\Bb{H}_{|{\rm{Re}}|<1}\,,
Re(1/ϕnNk+1,,n1(z))=Re(2nNk+1+ϕnNk,,n1(z))\displaystyle{\rm{Re}}\left(-1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k+1},...,n_{1}$}}}}}(z)\right)={\rm{Re}}\left(-2n_{N-k+1}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)\right)
2nNk+1+(1,1),ifN2,z𝕜𝕜ϕnNk,,n1ϝ𝕜𝕜\displaystyle\in-2n_{N-k+1}+(-1,1)\,,\ \mbox{if}\ N\geqslant 2\,,\ z\!\in\!\Bb{H}_{|{\rm{Re}}|\leqslant 1}\ \Rightarrow\ \phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)\!\in\!\Bb{H}_{|{\rm{Re}}|<1}\,.

Therefore

𝔾ϕn1ϝ\displaystyle\Bb{G}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{1}$}}}}}(z)\big{)} =1ϕn1(z)Re(1ϕn1(z))2\displaystyle=-\dfrac{1}{\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(z)}-\left\rceil{{\rm{Re}}\left(-\dfrac{1}{\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(z)}\right)}\right\lceil_{2}
=2n1+z2n1+Rez2=z,z𝕜𝕜\displaystyle=-2n_{1}+z-\left\rceil{-2n_{1}+{\rm{Re}}\,z}\right\lceil_{2}=z\ ,\quad z\!\in\!\Bb{H}_{|{\rm{Re}}|<1}\,,

and

𝔾ϕnNk+1,,n1ϝϕnNk+1,,n1ϝϕnNk+1,,n1ϝ\displaystyle\Bb{G}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{N-k+1},...,n_{1}$}}}}}(z)\big{)}=-\dfrac{1}{\phi_{{{\mbox{\footnotesize{$n_{N-k+1},...,n_{1}$}}}}}(z)}-\left\rceil{{\rm{Re}}\left(-\dfrac{1}{\phi_{{{\mbox{\footnotesize{$n_{N-k+1},...,n_{1}$}}}}}(z)}\right)}\right\lceil_{2}
=2nNk+1+ϕnNk,,n1(z)2nNk+1+ReϕnNk,,n1(z)2\displaystyle=-2n_{N-k+1}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)-\left\rceil{-2n_{N-k+1}+{\rm{Re}}\,\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)}\right\lceil_{2}
=ϕnNk,,n1(z),ϕnNk,,n1(z)𝕜𝕜\displaystyle=\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)\ ,\quad\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)\in\Bb{H}_{|{\rm{Re}}|<1}\,,

what was to be proved in (5.26).

24

\uparrow  For 1kN11\!\leqslant\!k\!\leqslant\!N\!-\!1 in (5.37)(b) we have

𝔾𝔽^𝔾ϕnN,,n1𝔽|σ\displaystyle\Bb{G}^{k}_{2}\left(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N},...,n_{1}}\right)=\Bb{G}^{k}_{2}\left(\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\big{)}\right)
=ϕnNk,,n1(|σn1)=^nNk,,n1,\displaystyle=\phi_{{{\mbox{\footnotesize{$n_{N-k},...,n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\big{)}=\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{N-k},...,n_{1}}\,,\

by virtue of

𝔾ϕnN,,n1ϝϕnNk,,n1ϝif\displaystyle\Bb{G}^{k}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\big{)}\!=\!\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(z)\,,\ \mbox{if}\ 1\!\leqslant\!k\!\leqslant\!N\!-\!1\,, z𝕜𝕜\displaystyle\ \ z\!\in\!\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,. (5.26(b))

And

𝔾𝔽^𝔾ϕn1𝔽|σ𝔾ϕn1𝔽𝔾ϕn1γσ\displaystyle\Bb{G}_{2}\left(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\square$}}}}}^{\hskip 1.42271ptn_{1}}\right)=\Bb{G}_{2}\left(\phi_{{{\mbox{\footnotesize{$n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\big{)}\right)=\Bb{G}_{2}\left(\phi_{{{\mbox{\footnotesize{$n_{1}$}}}}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\big{)}\right)\bigcup\Bb{G}_{2}\Big{(}\phi_{{{\mbox{\footnotesize{$n_{1}$}}}}}\big{(}\hskip 1.9919pt\gamma(\sigma_{n_{1}},\infty)\hskip 1.9919pt\big{)}\Big{)}
=γ(σn1,)=|σn1,\displaystyle=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\bigcup\gamma(\sigma_{n_{1}},\infty)=\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}^{\,{{\mbox{\tiny{$|$}}}}\sigma_{n_{1}}}\,,\

by virtue (5.26)(a) since 𝕜𝕜\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\subset\Bb{H}_{|{\rm{Re}}|<1}, and by virtue of (5.19)(c).

25

\uparrow    It follows from (5.40) and Theorem 5.5 that

ϕΓϑϕ(1+in0ðn1,,nNN(2n0+ϕnN,,n1(sign(n1)+i\bigcup\limits_{\phi\in\Gamma_{\vartheta}}\phi(1+{\rm{i}}\Bb{R}_{>0})=\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}_{\neq 0}$}}}}}\Big{(}-2n_{0}+\big{(}{\rm{sign}}(n_{0})+i\Bb{R}_{>0}\big{)}\Big{)}\\ \sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\,n_{N},...,n_{1}$}}}}}\big{(}{\rm{sign}}(n_{1})+i\Bb{R}_{>0}\big{)}\Big{)}\,,\ (A.25a)

while (5.44) and (5.43) for A=γ(1,)γ(1,0)A=\gamma(-1,\infty)\sqcup\gamma(1,0) give

ϕΓϑϕ(1+in0γγn1,,nNN(2n0+ϕnN,,n1(γ(1,)γ(1,0))),\bigcup\limits_{\phi\in\Gamma_{\vartheta}}\phi(1+{\rm{i}}\Bb{R}_{>0})=\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\in\Bb{Z}_{\neq 0}$}}}}}\Big{(}-2n_{0}+\big{(}\gamma(-1,\infty)\sqcup\gamma(1,0)\big{)}\Big{)}\\ \sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\,n_{N},...,n_{1}$}}}}}\big{(}\gamma(-1,\infty)\sqcup\gamma(1,0)\big{)}\Big{)}\,,\ (A.25b)

We prove directly that the right-hand side set in (A.25a) equals to the right-hand side set in (A.25b). Everywhere below we use the notation

σn:=sign(n),n\displaystyle\sigma_{n}:={\rm{sign}}(n)\ ,\quad n\in\Bb{Z}_{\neq 0}\,,

and the following identities

(1)γ(0,1)=ϕ1(γ(1,)),(2)ϕn(γ(1,))={ϕn+1(γ(σn,)),ifnϕn(γ(σn,)),ifn(3) 2n+γ(1,)={2(n)+γ(σn,),ifn2(1n)+γ(σ1n,),ifn\displaystyle\begin{array}[]{l}{\rm{(1)}}\ \ \gamma(0,1)=\phi_{1}\big{(}\gamma(1,\infty)\big{)}\ ,\\[14.22636pt] {\rm{(2)}}\ \ \phi_{n}\big{(}\gamma(-1,\infty)\big{)}=\left\{\begin{array}[]{ll}\phi_{n+1}\big{(}\gamma(\sigma_{n},\infty)\big{)}\,,&\ \ \mbox{if}\ n\in\Bb{Z}_{\geqslant 1}\,;\\[8.5359pt] \phi_{n}\big{(}\gamma(\sigma_{n},\infty)\big{)},&\ \ \mbox{if}\ n\in\Bb{Z}_{\leqslant-1}\,,\end{array}\right.\\[21.33955pt] {\rm{(3)}}\ \ 2n+\gamma(-1,\infty)=\left\{\begin{array}[]{ll}-2(-n)+\gamma(\sigma_{-n},\infty)\,,&\ \ \mbox{if}\ n\in\Bb{Z}_{\geqslant 1}\,;\\[8.5359pt] -2(1-n)+\gamma(\sigma_{1-n},\infty)\,,&\ \ \mbox{if}\ n\in\Bb{Z}_{\leqslant 0}\,,\end{array}\right.\end{array} (A.25j)

which hold because

ϕ1(γ(1,))=12(1+i)=11i=γ(0,1),\displaystyle\phi_{1}\left(\gamma(1,\infty)\right)=\dfrac{1}{2-\left(1+i\Bb{R}\right)}=\dfrac{1}{1-i\Bb{R}}=\gamma(0,1)\,,\

and

ϕn(γ(1,))={12n(1+i)=12n+21i=12(n+1)(σn+i)=ϕn+1(γ(σn,)),ifn12n(1+i)=12n(σn+i)=ϕn(γ(σn,)),ifn\displaystyle\phi_{n}\left(\gamma(-1,\infty)\right)=\left\{\begin{array}[]{ll}\dfrac{1}{2n-\left(-1+i\Bb{R}\right)}=\dfrac{1}{2n+2-1-i\Bb{R}}&\\[8.5359pt] =\dfrac{1}{2(n+1)-\left(\sigma_{n}+i\Bb{R}\right)}=\phi_{n+1}\left(\gamma(\sigma_{n},\infty)\right),&\ \ \mbox{if}\ n\in\Bb{Z}_{\geqslant 1}\,;\\[8.5359pt] \dfrac{1}{2n-\left(-1+i\Bb{R}\right)}=\dfrac{1}{2n-\left(\sigma_{n}+i\Bb{R}\right)}&\\[11.38092pt] =\phi_{n}\left(\gamma(\sigma_{n},\infty)\right),&\ \ \mbox{if}\ n\in\Bb{Z}_{\leqslant-1}\,.\end{array}\right.

Applying first identity (A.25j)(1) and then identities (A.25j)(2),(3) to the set on the right-hand side of the equality (A.25b), we obtain that it equals to

n0(2n0+γ(1,))n1,,nNN(2n0+ϕnN,,n1(γ(1,)))\displaystyle\bigsqcup\limits_{n_{0}\in\Bb{Z}}\big{(}2n_{0}+\gamma(-1,\infty)\big{)}\sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}\big{(}\gamma(-1,\infty)\big{)}\Big{)}
n0(2n0+ϕ1(γ(1,)))n2,,nNN(2n0+ϕnN,,n2,1(γ(1,)))\displaystyle\sqcup\bigsqcup\limits_{n_{0}\in\Bb{Z}}\Big{(}2n_{0}+\phi_{1}\big{(}\gamma(1,\infty)\big{)}\Big{)}\sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{2},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{Z}_{\geqslant 2}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{2},1$}}}}}\big{(}\gamma(1,\infty)\big{)}\Big{)}
=n0(2(n0)+γ(σn0,))n0(2(1n0)+γ(σ1n0,))\displaystyle=\bigsqcup\limits_{n_{0}\in\Bb{Z}_{\geqslant 1}}\big{(}-2(-n_{0})+\gamma(\sigma_{-n_{0}},\infty)\big{)}\sqcup\bigsqcup\limits_{n_{0}\in\Bb{Z}_{\leqslant 0}}\big{(}-2(1-n_{0})+\gamma(\sigma_{1-n_{0}},\infty)\big{)}
n0[(2n0+ϕ1(γ(1,)))n1(2n0+ϕn1(γ(σn1,)))\displaystyle\sqcup\bigsqcup\limits_{n_{0}\in\Bb{Z}}\bigg{[}\Big{(}2n_{0}+\phi_{1}\big{(}\gamma(1,\infty)\big{)}\Big{)}\sqcup\bigsqcup\limits_{n_{1}\in\Bb{Z}_{\geqslant 2}}\Big{(}2n_{0}+\phi_{n_{1}}\big{(}\gamma(\sigma_{n_{1}},\infty)\big{)}\Big{)}
n1(2n0+ϕn1(γ(σn1,)))]\displaystyle\sqcup\bigsqcup\limits_{n_{1}\in\Bb{Z}_{\leqslant-1}}\Big{(}2n_{0}+\phi_{n_{1}}\big{(}\gamma(\sigma_{n_{1}},\infty)\big{)}\Big{)}\bigg{]}
n2,,nNN[(2n0+ϕnN,,n2,1(γ(1,)))\displaystyle\sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{2},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{Z}_{\geqslant 2}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\bigg{[}\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{2},1$}}}}}\big{(}\gamma(1,\infty)\big{)}\Big{)}
n1(2n0+ϕnN,,n1(γ(σn1,)))n1(2n0+ϕnN,,n1(γ(σn1,)))]\displaystyle\sqcup\bigsqcup\limits_{n_{1}\in\Bb{Z}_{\geqslant 2}}\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}\big{(}\gamma(\sigma_{n_{1}},\infty)\big{)}\Big{)}\sqcup\bigsqcup\limits_{n_{1}\in\Bb{Z}_{\leqslant-1}}\!\!\!\!\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}\big{(}\gamma(\sigma_{n_{1}},\infty)\big{)}\Big{)}\bigg{]}
=n(2n+γ(σn,))n1n0(2n0+ϕn1(γ(σn1,)))\displaystyle=\bigsqcup\limits_{n\in\Bb{Z}_{\neq 0}}\big{(}-2n+\gamma(\sigma_{n},\infty)\big{)}\sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$n_{0}\in\Bb{Z}$}}}}\end{subarray}}\Big{(}2n_{0}+\phi_{n_{1}}\big{(}\gamma(\sigma_{n_{1}},\infty)\big{)}\Big{)}
n1,,nNN(2n0+ϕnN,,n1(γ(σn1,)))\displaystyle\sqcup\bigsqcup\limits_{\begin{subarray}{c}{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{Z}_{\geqslant 2}\,,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}\big{(}\gamma(\sigma_{n_{1}},\infty)\big{)}\Big{)}
=n0(2n0+(σn0+in1,,nNNϕnN,,n1γσ\displaystyle=\bigsqcup\limits_{n_{0}\in\Bb{Z}_{\neq 0}}\big{(}-2n_{0}+(\sigma_{n_{0}}+i\Bb{R}_{>0})\big{)}\sqcup\bigsqcup\limits_{\begin{subarray}{{\mbox{\footnotesize{$n_{1},...,n_{N}\in\Bb{Z}_{\neq 0}$}}}}\\[2.13387pt] {{\mbox{\footnotesize{$N\in\Bb{N},,\ \ n_{0}\in\Bb{Z}$}}}}\end{subarray}}\!\!\!\!\Big{(}2n_{0}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}\big{(}\gamma(\sigma_{n_{1}},\infty)\big{)}\Big{)},

where the latter set coincides with the right-hand set in (A.25a). The proof of the equality between the right-hand sets in (A.25a) and in (A.25b) is finished. So that these two sets represent two different partitions of the set

ϕΓϑϕ(1+iϕΓ(2)ϕ𝔽Γ\displaystyle\bigcup\limits_{{{\mbox{\footnotesize{$\phi\in\Gamma_{\vartheta}$}}}}}\phi(1+{\rm{i}}\Bb{R}_{>0})=\bigcup\limits_{{{\mbox{\footnotesize{$\phi\!\in\!\Gamma(2)$}}}}}\phi\big{(}\Bb{H}\cap\partial F_{\Gamma(2)}\big{)}\,.
26

\uparrow    We prove (5.51). The set in the right-hand side of (5.51) equals to Γϑ||\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}, in view of the definition (5.8). The relationship (5.3) yields that Γϑ||Γϑ\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\subset\Gamma_{\vartheta} and since

Γϑ={az+bcz+d|abcd0(mod 2)adbc=1,a,b,c,d},\displaystyle\Gamma_{\vartheta}=\bigg{\{}{{\mbox{\footnotesize{$\,\dfrac{az+b}{cz+d}$}}}}\ \bigg{|}\ {{\mbox{\footnotesize{$\begin{array}[]{l}ab\equiv cd\equiv 0({\rm{mod}}\,2)\\[2.84544pt] ad-bc\!=\!1\,,\ a,b,c,d\!\in\!\Bb{Z}\end{array}$}}}}\,\bigg{\}}\,, (A.26c)

we deduce from (5.9) that

Γϑ||{az+bcz+d||a|<|b|<d,|a|<|c|<dabcd0(mod 2)adbc=1,a,b,c,d}.\displaystyle\Gamma_{\vartheta}^{\hskip 0.42677pt{{\mbox{\tiny{$||$}}}}}\subset\bigg{\{}{{\mbox{\footnotesize{$\,\dfrac{az+b}{cz+d}$}}}}\ \bigg{|}\ {{\mbox{\footnotesize{$\begin{array}[]{l}|a|\!<\!|b|\!<\!d,\ |a|\!<\!|c|\!<\!d\\ ab\equiv cd\equiv 0({\rm{mod}}\,2)\\ ad-bc\!=\!1\,,\ a,b,c,d\!\in\!\Bb{Z}\end{array}$}}}}\,\bigg{\}}\,. (A.26g)

The inverse inclusion in (A.26g) follows from (A.26c), Lemma A.5, (10.57) and (5.8). The equalities (5.51) are proved.

27

\uparrow    We show that this fact can be easily deduced from the properties of the even integer part x22\rceil{x}\lceil_{\!2}\,\in\!2\Bb{Z} and of the even fractional part {x}2[1,1]\{x\}^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\in[-1,1] of the real number xx\!\in\!\Bb{R}. Moreover, our further reasoning does not depend on how these two functions are defined on the odd integers.

We first prove that ϕ𝔫(0)\phi_{\eufm{n}}(0) for arbitrary 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} is an even rational number lying in (1,1){0}(-1,1)\setminus\{0\}. Observe that

ϕ𝔫(z)=zp𝔫N1+q𝔫N1zp𝔫N+q𝔫N,𝔫=(𝔫𝔑,,𝔫1)ϝ\displaystyle\phi_{\eufm{n}}(z)\!=\!\dfrac{zp^{\eufm{n}}_{N-1}+q^{\eufm{n}}_{N-1}}{zp^{\eufm{n}}_{N}+q^{\eufm{n}}_{N}}\ ,\ \eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\!\in\!\Bb{N}\,,\ z\in\Bb{H}\,, (10.67)

and

(a)|q𝔫k|>|p𝔫k|>|p𝔫k1|,(b)|q𝔫k|>|q𝔫k1|>|p𝔫k1|,1kN.\displaystyle{\rm{(a)}}\ \ \big{|}q^{\eufm{n}}_{k}\big{|}>\big{|}p^{\eufm{n}}_{k}\big{|}>\big{|}p^{\eufm{n}}_{k-1}\big{|}\ ,\quad{\rm{(b)}}\ \ \big{|}q^{\eufm{n}}_{k}\big{|}>\big{|}q^{\eufm{n}}_{k-1}\big{|}>\big{|}p^{\eufm{n}}_{k-1}\big{|}\ ,\quad 1\leqslant k\leqslant N\,. (10.69)

yield that

ϕ𝔫Hol([1,1]()),𝔫=(𝔫𝔑,,𝔫1)\displaystyle\phi_{\eufm{n}}\in{\rm{Hol}}\big{(}\,[-1,1]\cup\left(\Bb{C}\setminus\Bb{R}\right)\big{)}\,,\quad\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\!\in\!\Bb{N}\,. (A.27a)

So that, by setting z=0z=0 in (5.6) and (10.67), we obtain

{1/ϕn1(0)=2n1,ifN=1,1/ϕnN,,n1(0)=2nN+ϕnN1,,n1(0),ifN2,\displaystyle\left\{\begin{array}[]{rlll}-1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(0)&=&-2n_{1}\,,&\ \mbox{if}\ N=1\,,\\ -1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(0)&=&-2n_{N}+\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-1},...,n_{1}$}}}}}(0)\,,&\ \mbox{if}\ N\geqslant 2\,,\end{array}\right. (A.27d)

and, in view of (10.69),

ϕ𝔫(0)=q𝔫N1q𝔫N(1,1){0},𝔫=(𝔫𝔑,,𝔫1)\displaystyle\phi_{\eufm{n}}(0)\!=\!\dfrac{q^{\eufm{n}}_{N-1}}{q^{\eufm{n}}_{N}}\in\left(-1,1\right)\setminus\{0\}\,,\quad\eufm{n}\!=\!(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\!\in\!\Bb{N}\,. (A.27e)

It follows from (A.27e) that ϕ𝔫(0)\phi_{\eufm{n}}(0) is a rational number lying in (1,1){0}(-1,1)\setminus\{0\}. Furthermore,

p𝔫k1q𝔫kp𝔫kq𝔫k1=1,0kN,\displaystyle p^{\eufm{n}}_{k-1}q^{\eufm{n}}_{k}-p^{\eufm{n}}_{k}q^{\eufm{n}}_{k-1}=1\ ,\ \quad\quad 0\leqslant k\leqslant N\ , (10.63)

for k=Nk=N implies that

gcd(q𝔫N1,q𝔫N)=1,\displaystyle{\rm{gcd}}\left(q^{\eufm{n}}_{N-1},q^{\eufm{n}}_{N}\right)=1\,,\ (A.27f)

while (10.66) written in the form

q𝔫k=2nkq𝔫k1q𝔫k2,q𝔫0=1,q𝔫1=0,1kN,\displaystyle q^{\eufm{n}}_{k}=2n_{k}q^{\eufm{n}}_{k-1}-\,q^{\eufm{n}}_{k-2}\ ,\ \ q^{\eufm{n}}_{0}=1\ ,\ q^{\eufm{n}}_{-1}=0\ ,\quad 1\leqslant k\leqslant N\,, (10.66)

means that

q2n10(mod 2),q2n1(mod 2),0n<(N+1)/2,\displaystyle q_{2n-1}\equiv 0({\rm{mod}}\,2)\ ,\quad q_{2n}\equiv 1({\rm{mod}}\,2)\ ,\quad 0\leqslant n<(N+1)/2\ ,

and therefore q𝔫N1q𝔫N0(mod 2)q^{\eufm{n}}_{N-1}q^{\eufm{n}}_{N}\equiv 0({\rm{mod}}\,2). This property together with (A.27f) proves that ϕ𝔫(0)\phi_{\eufm{n}}(0) is an even rational number lying in (1,1){0}(-1,1)\setminus\{0\} for arbitrary 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}.

By applying the even integer and the even fractional parts to (A.27d), we obtain, taking into account of (A.27e),

{G2(ϕn1(0))=0,1/ϕn1(0)2=2n1,ifN=1,G2(ϕnN,,n1(0))=ϕnN1,,n1(0),1/ϕnN,,n1(0)2=2nN,ifN2.\displaystyle\left\{\begin{array}[]{rllrlll}G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\left(\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(0)\right)&=&0\,,&\ \left\rceil{1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(0)}\right\lceil_{2}&=&2n_{1}\,,&\ \mbox{if}\ N=1\,,\\[5.69046pt] G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\left(\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(0)\right)&=&\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-1},...,n_{1}$}}}}}(0)\,,&\ \left\rceil{1/\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(0)}\right\lceil_{2}&=&2n_{N}\,,&\ \mbox{if}\ N\geqslant 2\,.\end{array}\right.

Hence, we can easily calculate the coefficients of the continued fraction

ϕnN,,n1(0):=12nN12nN112n212n1,(nN,,n1)\displaystyle\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(0):={{\mbox{\footnotesize{$\dfrac{1}{2n_{N}\!-\!\dfrac{1}{\begin{subarray}{l}\displaystyle\vphantom{A^{A}}2n_{N-1}\!-\!\\[-11.38092pt] \hphantom{2n_{k-1}11-}\ddots\\[-11.38092pt] \hphantom{2n_{k-1}11-\ddots}\displaystyle-\!\dfrac{1}{2n_{2}\!-\!\dfrac{1}{\vphantom{A^{A}}2n_{1}}}\end{subarray}}}$}}}}\ ,\quad(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\!\in\!\Bb{N}\,,

by the formulas,

2nNk=1/(G2)k(ϕnN,,n1(0))2,0kN1,\displaystyle 2n_{N-k}=\big{\rceil}1\big{/}\big{(}G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\big{)}^{k}\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(0)\big{)}\big{\lceil}_{\!2}\ ,\quad 0\leqslant k\leqslant N-1\,, (A.27g)

and also,

{(G2)k(ϕnN,,n1(0))=ϕnNk,,n1(0),0kN1,(G2)N(ϕnN,,n1(0))=0,(nN,,n1)\displaystyle\left\{\begin{array}[]{ll}\big{(}G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\big{)}^{k}\!\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(0)\big{)}\!=\!\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k},...,n_{1}$}}}}}(0)\,,&\ 0\!\leqslant\!k\!\leqslant\!N\!-\!1\,,\\[2.84544pt] \big{(}G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\big{)}^{N}\!\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(0)\big{)}\!=\!0\,,&\ (n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}\ ,\ N\!\in\!\Bb{N}\,.\end{array}\right. (A.27j)

It follows from (A.27g) that

ϕ𝔫(0)=ϕ𝔪(0),𝔫,𝔪\displaystyle\phi_{\eufm{n}}(0)=\phi_{\eufm{m}}(0)\,,\ \eufm{n},\eufm{m}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\ \ \Longleftrightarrow\ \ \eufm{n}=\eufm{m}\,. (A.27k)

We prove now that the following analogue of (A.27j) for an arbitrary even rational number p/q(1,1)p/q\in(-1,1), p,qp,q\in\Bb{Z}_{\neq 0}, gcd(p,q)=1{\rm{gcd}}(p,q)=1, holds

pq(1,1){0}is even{G2(pq)=rp(1,1){0}is even,if|p|2;G2(pq)=0,if|p|=1;\displaystyle\dfrac{p}{q}\!\in\!(-1,1)\!\setminus\!\{0\}\ \mbox{is even}\Rightarrow\ \left\{\begin{array}[]{ll}G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\left(\dfrac{p}{q}\right)\!=\!\dfrac{r}{p}\!\in\!(-1,1)\!\setminus\!\{0\}\ \mbox{is even}\,,&\ \mbox{if}\ |p|\!\geqslant\!2;\\[8.5359pt] G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\left(\dfrac{p}{q}\right)=0\,,&\ \mbox{if}\ |p|=1\,;\end{array}\right. (A.27n)
(a)gcd(r,p)=gcd(p,q)=1;(b)rq(mod 2),(c) 1|r||p|1,\displaystyle{\rm{(a)}}\ {\rm{gcd}}(r,p)={\rm{gcd}}(p,q)=1;\ {\rm{(b)}}\ r\equiv q({\rm{mod}}\,2),\ {\rm{(c)}}\ 1\leqslant|r|\leqslant|p|-1, (A.27o)

where rr\in\Bb{Z}_{\neq 0}. Since for |p|=1|p|=1 the integer qq is even, then p/q=ϕn1(0)p/q=\phi_{n_{1}}(0) with n1:=(q/2)sign(p)n_{1}:=(q/2)\,{\rm{sign}}(p)\in\Bb{Z}_{\neq 0} and G2(ϕn1(0))=0G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\left(\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{1}$}}}}}(0)\right)=0, as follows from (A.27j) for N=1N=1. It remains to prove (A.27o) for |p|2|p|\!\geqslant\!2. Since qq and pp are coprime then qq belongs to the set

(|p|,|p|)m(2m|p||p|, 2m|p|+|p|)=m{}\displaystyle\big{(}-|p|,|p|\big{)}\ \sqcup\bigsqcup_{{{\mbox{\footnotesize{$m\in\Bb{Z}_{\neq 0}$}}}}}\Big{(}2m|p|-|p|,\ 2m|p|+|p|\Big{)}=\Bb{R}\setminus\bigsqcup_{{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\Big{\{}\big{(}2m+1\big{)}\cdot p\Big{\}}\ ,

and does not lie on the interval (|p|,|p|)(-|p|,|p|) because |q|>|p||q|>|p|. Thus, there exists a unique m1m_{1}\in\Bb{Z}_{\neq 0} such that

q(2m1|p||p|,2m1|p|+|p|){𝕜𝕜}k=|p|+1k0|p|1{𝕜𝕜}\displaystyle q\in\Big{(}2m_{1}|p\,|-|p\,|,2m_{1}|p\,|+|p\,|\Big{)}\cap\Bb{Z}=\big{\{}2m_{1}|p\,|\big{\}}\ \sqcup\bigsqcup_{\begin{subarray}{{\mbox{\footnotesize{$k=-|p\,|+1$}}}}\\[2.84544pt] {{\mbox{\footnotesize{$k\neq 0$}}}}\end{subarray}}^{{{\mbox{\footnotesize{$|p\,|-1$}}}}}\big{\{}2m_{1}|p\,|-k\big{\}}\ ,

and since q2m1|p|q\neq 2m_{1}|p\,|, there also exists a unique k1k_{1}\in\Bb{Z}_{\neq 0} satisfying

q=2m1|p|k1,|p|+1k1|p|1,k10.\displaystyle q=2m_{1}|p\,|-k_{1}\ ,\quad-|p\,|+1\leq k_{1}\leq|p\,|-1\ ,\ k_{1}\neq 0\ .

By setting n:=m1sign(p)n:=m_{1}\cdot{\rm{sign}}(p) and r:=k1r:=k_{1} we obtain the existence of the integers nn and rr satisfying

q=2npr,n,r𝕜𝕜𝕜𝕜\displaystyle q=2np-r\ ,\quad n,r\in\Bb{Z}_{\neq 0}\,,\ 1\leqslant|r|\leqslant|p|-1\,. (A.27p)

Here, the property (A.27o)(b) is immediate, while (A.27o)(a) follows from (jon, , p.​ 5, Lemma 1.5), which states that

a,cðð\displaystyle a,c\in\Bb{Z}\ ,\ \ b,d\in\Bb{Z}_{\neq 0}\ ,\quad a=db-c\ \Longrightarrow\ {\rm{gcd}}(a,b)={\rm{gcd}}(c,b). (A.27q)

Finally, q/p=2n+r/p-q/p=-2n+r/p and r/p(1,1){0}r/p\in(-1,1)\setminus\{0\} yield G2(p/q)=r/pG^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}({p}/{q})\!=\!{r}/{p} and completes the proof of (A.27o).

We prove that an arbitrary even rational number p/qp/q, p,qp,q\in\Bb{Z}_{\neq 0}, gcd(p,q)=1{\rm{gcd}}(p,q)=1, lying on the set (1,1)(-1,1), can be represented as ϕnN,,n1(0)\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}(0) with (nN,,n1)(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, NN\!\in\!\Bb{N}, by employing induction on |p||p|. It has already been written above that for |p|=1|p|=1 the integer qq is even, and therefore p/q=ϕn1(0)p/q=\phi_{n_{1}}(0) with n1:=(q/2)sign(p)n_{1}:=(q/2)\,{\rm{sign}}(p)\in\Bb{Z}_{\neq 0}. Assume that PP\in\Bb{Z}_{\geqslant 2} and this statement is proved for all nonzero even rational numbers from (1,1)(-1,1) the modulus of whose nominators is less than or equal P1P-1. Let p/q(1,1)p/q\in(-1,1), p,qp,q\in\Bb{Z}_{\neq 0}, gcd(p,q)=1{\rm{gcd}}(p,q)=1 and |p|=P|p|=P. Invoking the induction hypothesis on the even rational number r/pr/p satisfying (A.27p), we obtain the existence of NN\in\Bb{Z}_{\geqslant 2} and n1,,nN1n_{1},...,n_{N-1}\in\Bb{Z}_{\neq 0} such that r/p=ϕnN1,,n1(0)r/p=\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-1},...,n_{1}$}}}}}(0). But then (A.27o) and (A.27p) yield

pq=1qp=12nrp=12nϕnN1,,n1(0)=ϕnN,nN1,,n1(0),nN:=n\displaystyle\dfrac{p}{q}=\dfrac{1}{\dfrac{q}{p}}\!=\!\dfrac{1}{2n\!-\!\dfrac{r}{p}}\!=\!\dfrac{1}{2n\!-\!\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-1},...,n_{1}$}}}}}(0)}\!=\!\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},n_{N-1},...,n_{1}$}}}}}(0)\ ,\ n_{N}\!:=\!n\!\in\!\Bb{Z}_{\neq 0}\,.

The induction is complete. In view of (A.27g) and (A.27j), we have proved the following assertion.

Lemma A.6

. Each nonzero even rational number p/qp/q in (1,1)(-1,1) with nonzero coprime integers pp and qq can be uniquely represented in the form ϕnN,,n1(0)\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(0) with (nN,,n1)(n_{N},...,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}, NN\!\in\!\Bb{N}, satisfying

2nNk=1(G2)k(p/q)2, 0kN1,(G2)N(p/q)=0.\displaystyle 2n_{N-k}=\Bigg{\rceil}\dfrac{1}{\big{(}G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\big{)}^{k}\big{(}p/q\big{)}}\Bigg{\lceil}_{\!2}\ ,\ \ 0\leqslant k\leqslant N-1\,,\ \ \big{(}G^{{{\mbox{\tiny{$\rceil\lceil$}}}}}_{2}\big{)}^{N}\big{(}p/q\big{)}=0\,. (A.27r)

Conversely, ϕ𝔫(0)\phi_{\eufm{n}}(0) for arbitrary 𝔫\eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\!:=\!\sqcup_{k\geqslant 1}\!\left(\Bb{Z}_{\neq 0}\right)^{k} is an even rational number lying in (1,1){0}(-1,1)\setminus\{0\}. For different 𝔫\eufm{n} and 𝔪\eufm{m} from \Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}} the numbers ϕ𝔫(0)\phi_{\eufm{n}}(0) and ϕ𝔪(0)\phi_{\eufm{m}}(0) are different.

28

\uparrow    We prove (5.95)(a). Since γ(1,1)||\gamma(-1,1)\subset\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}} we can apply (5.15) to get 𝔾γγγ\Bb{G}_{2}\big{(}\gamma(-1,1)\big{)}=-1/\gamma(-1,1)=\gamma(-1,1). To prove the left-hand side equality

𝔾𝔼𝔽||γ\displaystyle\Bb{G}_{2}\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}\!=\!\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\gamma(-1,1)\,, (5.95(a))

of (5.95)(a), we use (5.79)(a),

0:=n0^𝓃0𝔾𝔼𝔾𝔽n0𝔾𝔽^\displaystyle\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!:=\!\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\sqcup\!\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\!\!\!\!\!\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{0}}\ \Rightarrow\ \Bb{G}_{2}\left(\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\right)\!:=\!\Bb{G}_{2}\left(\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\right)\sqcup\!\!\!\!\bigsqcup\limits_{{{\mbox{\footnotesize{$n_{0}\!\in\!\Bb{Z}_{\neq 0}$}}}}}\!\!\!\!\!\Bb{G}_{2}\left(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{0}}\right)\,,\

where by (5.71) =𝔻𝔽𝔽||\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}=\Bb{D}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\subset\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}} we apply (5.15) to get

𝔾𝔽𝔽\displaystyle\Bb{G}_{2}\left(\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\right)=-1/\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\,,\

while for every n0n_{0}\!\in\!\Bb{Z}_{\neq 0} we can apply to (see (5.75))

^n0=ϕn0(γ(σn0,))=ϕn0()ϕn0(γ(σn0,)),𝕜𝕜\displaystyle\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 0.28436ptn_{0}}=\phi_{n_{0}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\sqcup\gamma(\sigma_{n_{0}},\infty)\big{)}=\phi_{n_{0}}\big{(}\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\big{)}\sqcup\phi_{n_{0}}\big{(}\gamma(\sigma_{n_{0}},\infty)\big{)}\ ,\quad\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\subset\Bb{H}_{|{\rm{Re}}|<1}\,,\

the equality (5.26)(a)

𝔾ϕn0ϝϝϝ𝕜𝕜\displaystyle\Bb{G}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{0}$}}}}}(z)\big{)}=z\ ,\quad\ z\!\in\!\Bb{H}_{|{\rm{Re}}|<1}\,, (5.26(a))

and the equality (5.19)(c)

𝔾ϕnγσγσ\displaystyle\Bb{G}_{2}\Big{(}\phi_{{{\mbox{\footnotesize{$n$}}}}}\big{(}\hskip 1.9919pt\gamma(\sigma_{n},\infty)\hskip 1.9919pt\big{)}\Big{)}=\gamma(\sigma_{n},\infty)\ ,\quad n\!\in\!\Bb{Z}_{\neq 0}\,, (5.19(c))

to obtain

𝔾𝔽^𝔽γσ\displaystyle\Bb{G}_{2}\left(\widehat{\mathcal{F}}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}^{\hskip 1.42271ptn_{0}}\right)=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\sqcup\gamma(\sigma_{n_{0}},\infty)\ ,\quad n_{0}\!\in\!\Bb{Z}_{\neq 0}\,.

Finally,

𝔾𝔼\displaystyle\Bb{G}_{2}\left(\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\right) =(1/)γ(1,)γ(1,)\displaystyle=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\sqcup\left(-1/\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\right)\sqcup\gamma(1,\infty)\sqcup\gamma(-1,\infty)
=γ(1,)γ(1,)=||𝔻¯.\displaystyle=\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\sqcup\gamma(1,\infty)\sqcup\gamma(-1,\infty)=\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\,.

This finishes the proof of (5.95)(a).

29

\uparrow  We prove (6.6). Since zϕnN,,n1(𝕜𝕜)z\in\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}\left(\Bb{H}_{|{\rm{Re}}|\leqslant 1}\right) then there exists z0𝕜𝕜z_{0}\in\Bb{H}_{|{\rm{Re}}|\leqslant 1} such that

z=ϕnN,,n1(z0)(5.9)𝕜𝕜𝔽ϝ𝕜𝕜\displaystyle z=\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z_{0})\stackrel{{\scriptstyle{{\mbox{\footnotesize{$\eqref{f8contgen}$}}}}}}{{\vphantom{A}\in}}\Bb{H}_{|{\rm{Re}}|<1}\!\setminus\!\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\,,\quad z_{0}\in\Bb{H}_{|{\rm{Re}}|\leqslant 1}\,, (A.29a)

and by (6.4),

z0=ψnN,,n1(z)=1ϕn1,,nN(1/z)={2n1ϕn2,,nN(1/z),ifN2;2n11z,ifN=1.\displaystyle z_{0}\!=\!\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\!=\!\dfrac{1}{\phi_{{{\mbox{\footnotesize{$n_{1},...,n_{N}$}}}}}(1/z)}\!=\!\left\{\begin{array}[]{ll}2n_{1}\!-\!\phi_{{{\mbox{\footnotesize{$n_{2},...,n_{N}$}}}}}(1/z),&\hbox{if}\ N\!\geqslant\!2;\\ 2n_{1}\!-\!\dfrac{1}{z},&\hbox{if}\ N\!=\!1\,.\end{array}\right. (A.29d)

If N2N\!\geqslant\!2 the application of (5.26)(b) to (A.29a) gives

𝔾ϝ𝔾ϕnN,,n1ϝϕnNk1,,n1ϝif\displaystyle\Bb{G}^{k+1}_{2}(z)=\Bb{G}^{k+1}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z_{0})\big{)}\!=\!\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k-1},...,n_{1}$}}}}}(z_{0})\,,\quad\mbox{if}\ \ \ 0\!\leqslant\!k\!\leqslant\!N\!-\!2\,,

where, by virtue of (A.29d) and (6.4),

ϕnNk1,,n1(z0)\displaystyle\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k-1},...,n_{1}$}}}}}(z_{0}) =ϕnNk1,,n1(ψnN,,n1(z))\displaystyle=\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k-1},...,n_{1}$}}}}}\big{(}\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\big{)}
=ψnN,,nNk(ϕnN,,nNk(ϕnNk1,,n1(ψnN,,n1(z))))\displaystyle=\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}\left(\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}\Big{(}\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N-k-1},...,n_{1}$}}}}}\big{(}\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\big{)}\Big{)}\right)
=ψnN,,nNk(ϕnN,,n1(ψnN,,n1(z)))=ψnN,,nNk(z),\displaystyle=\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}\Big{(}\phi_{{{\mbox{\footnotesize{$\hskip 0.1424ptn_{N},...,n_{1}$}}}}}\big{(}\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(z)\big{)}\Big{)}=\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}(z)\,,\

and therefore

𝔾ϝψnN,,nNkϝif\displaystyle\Bb{G}^{k+1}_{2}(z)=\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{N-k}$}}}}}(z)\,,\quad\mbox{if}\ \ \ 0\!\leqslant\!k\!\leqslant\!N\!-\!2\,, (A.29e)

which proves the right-hand side equalities in (6.6).

Then for y=ϕnN,,n1(y0)y=\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(y_{0}), y0𝕜𝕜y_{0}\in\Bb{H}_{|{\rm{Re}}|<1} and N2N\geqslant 2 we have

𝔾𝔾𝔾𝔾𝔾ϕnN,,n1𝔾ϕn1\displaystyle\Bb{G}^{N}_{2}(y)=\Bb{G}_{2}\left(\Bb{G}^{N-1}_{2}(y)\right)=\Bb{G}_{2}\left(\Bb{G}^{N-1}_{2}\big{(}\phi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(y_{0})\big{)}\right)=\Bb{G}_{2}\left(\phi_{{{\mbox{\footnotesize{$n_{1}$}}}}}(y_{0})\right)
=𝔾ψnN,,n1\displaystyle=\Bb{G}_{2}\left(\dfrac{1}{2n_{1}-y_{0}}\right)=y_{0}-2n_{1}-\left\rceil{{\rm{Re}}\left(y_{0}-2n_{1}\right)}\right\lceil_{2}=y_{0}\!=\!\psi_{{{\mbox{\footnotesize{$n_{N},...,n_{1}$}}}}}(y)\,,\

while, if N=1N\!=\!1, then similarly 𝔾𝔾ϕn1ψn1\Bb{G}^{N}_{2}(y)=\Bb{G}_{2}\left(\phi_{{{\mbox{\footnotesize{$n_{1}$}}}}}(y_{0})\right)=y_{0}\!=\!\psi_{{{\mbox{\footnotesize{$n_{1}$}}}}}(y). This proves the left-hand side equalities in (6.6) and finishes the proof of (6.6).

A.6 . Notes for Section 6

  

30

\uparrow   We prove (6.9). In view of

n1nHn(x)eiπny=14π3iγ(1,1)λ(y)λ(y)λ(z)dz(x+z)2,Imy>1,\displaystyle\sum\limits_{n\geqslant 1}\,n\,\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi ny$}}}}}=\frac{1}{4\pi^{3}{\rm{i}}}\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{\lambda(y)-\lambda(z)}\frac{dz}{(x+z)^{2}}\,,\quad\ \ \ {\rm{Im}}\,y\!>\!1\,, (4.6)
n1nMn(x)eiπny=14π3iγ(1,1)λ(y)1λ(y)λ(z)dz(x+z)2,Imy>1,\displaystyle\sum\limits_{n\geqslant 1}\,n\,\eurm{M}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi ny$}}}}}=\frac{1}{4\pi^{3}{\rm{i}}}\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{1\!-\!\lambda(y)\!-\!\lambda(z)}\frac{dz}{(x+z)^{2}}\,,\ \ \ {\rm{Im}}\,y\!>\!1\,, (4.7)

and (6.8)

Φ0(x;y):=12πiγ(1,1)λ(y)dz(λ(y)λ(z))(z+x)2,\displaystyle\Phi^{0}_{\infty}(x;y):=\frac{1}{2\pi i}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\lambda^{\,\prime}(y)dz}{\big{(}\lambda(y)-\lambda(z)\big{)}\big{(}z+x\big{)}^{2}}\ \,,
Φ1(x;y):=12πiγ(1,1)λ(y)dz(λ(y)λ(z))(xz1)2,\displaystyle\Phi^{1}_{\infty}(x;y):=\frac{1}{2\pi i}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\lambda^{\,\prime}(y)dz}{\big{(}\lambda(y)-\lambda(z)\big{)}\big{(}xz-1\big{)}^{2}}\ \,,

we get

n1nHn(x)eiπny=Φ0(x;y)2π2,Imy>1,\displaystyle\sum\limits_{n\geqslant 1}\,n\,\eurm{H}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi ny$}}}}}=\frac{\Phi^{0}_{\infty}(x;y)}{2\pi^{2}}\,,\quad\ \ \ {\rm{Im}}\,y\!>\!1\,, (A.30a)

and since

γ(1,1)λ(y)1λ(y)λ(z)dz(x+z)2=γ(1,1)λ(y)λ(1/z)λ(y)dz(x+z)2\displaystyle\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{1\!-\!\lambda(y)\!-\!\lambda(z)}\frac{dz}{(x+z)^{2}}=\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{\lambda(-1/z)\!-\!\lambda(y)}\frac{dz}{(x+z)^{2}}
=|z=1z,dz=dz(z)2,1γ(1,1)=γ(1,1)|\displaystyle=\left|z=-\dfrac{1}{z^{\,\prime}}\,,\ dz=\dfrac{dz^{\,\prime}}{\left(z^{\,\prime}\right)^{2}}\,,\ -\dfrac{1}{\gamma(-1,1)}=\gamma(1,-1)\right|
=γ(1,1)λ(y)λ(z)λ(y)dzz2(x1/z)2=γ(1,1)λ(y)λ(y)λ(z)dz(xz1)2=2πiΦ1(x;y),\displaystyle=-\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{\lambda(z)\!-\!\lambda(y)}\frac{dz}{z^{2}(x-1/z)^{2}}=\int\limits_{\gamma(-1,1)}\dfrac{\lambda^{\,\prime}(y)}{\lambda(y)\!-\!\lambda(z)}\frac{dz}{(xz-1)^{2}}=2\pi i\Phi^{1}_{\infty}(x;y)\,,\

we obtain

n1nMn(x)eiπny=Φ1(x;y)2π2,Imy>1.\displaystyle\sum\limits_{n\geqslant 1}\,n\,\eurm{M}_{n}(x)\,{\rm{e}}^{{{\mbox{\footnotesize{${\rm{i}}\pi ny$}}}}}=\frac{\Phi^{1}_{\infty}(x;y)}{2\pi^{2}}\,,\quad\ \ \ {\rm{Im}}\,y\!>\!1\,. (A.30b)

The equalities (A.30a) and (A.30b) finishes the proof of (6.9).

31

\uparrow    We prove (6.12). We introduce the parity indicator p:{}\eurm{p}:\Bb{Z}\mapsto\{0,1\} such that pn=0\eurm{p}_{n}=0 if the integer nn is even and pn=1\eurm{p}_{n}=1 if nn is odd. Let ϕΓϑ\phi\in\Gamma_{\vartheta} and ψΓϑ\psi\in\Gamma_{\vartheta} be inverse to ϕ\phi, i.e., ϕ(ψ(z))=ψ(ϕ(z))=z\phi(\psi(z))=\psi(\phi(z))=z for all zz\in\Bb{H}. Denote α:=pd(ϕ){0,1}\alpha:=\eurm{p}_{d(\phi)}\in\{0,1\}. Then by (5.58) and (6.7),

λ(ϕ(z))=α+(1)αλ(z),λ(z)=(1)αλ(ϕ(z))ϕ(z),z\displaystyle\lambda\big{(}\phi(z)\big{)}=\alpha+(-1)^{\alpha}\lambda(z)\ ,\quad\lambda^{\,\prime}(z)=(-1)^{\alpha}\lambda^{\,\prime}(\phi(z))\phi^{\,\prime}(z)\,,\ \quad z\in\Bb{H}\,.

Then

Φδ(x;ϕ(z))\displaystyle\Phi_{\infty}^{\,\delta}(x;\phi(z)) :=12πiγ(1,1)λ(ϕ(z))dζ(λ(ϕ(z))λ(ζ))(xδζ(x)1δ)2\displaystyle\!:=\!\frac{1}{2\pi i}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}(\phi(z))d\zeta}{\big{(}\lambda(\phi(z))\!-\!\lambda(\zeta)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}
=(1)α2πiϕ(z)γ(1,1)λ(z)dζ(α+(1)αλ(z)λ(ζ))(xδζ(x)1δ)2,\displaystyle=\frac{(-1)^{\alpha}}{2\pi i\phi^{\,\prime}(z)}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}\big{(}z\big{)}d\zeta}{\big{(}\alpha+(-1)^{\alpha}\lambda\big{(}z\big{)}\!-\!\lambda(\zeta)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}\,,\

from which

(1)αΦδ(x;ϕ(z))ϕ(z)=12πiγ(1,1)λ(z)dζ(α+(1)αλ(z)λ(ζ))(xδζ(x)1δ)2\displaystyle(-1)^{\alpha}\Phi^{\,\delta}_{\infty}(x;\phi(z))\phi^{\,\prime}(z)=\frac{1}{2\pi i}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}\big{(}z\big{)}d\zeta}{\big{(}\alpha+(-1)^{\alpha}\lambda\big{(}z\big{)}\!-\!\lambda(\zeta)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}

where the right-hand side is Φδ(x;z)\Phi^{\,\delta}_{\infty}(x;z) if α=0\alpha=0 and (6.12) is proved for that case. But if α=1\alpha=1 then

Φδ(x;ϕ(z))ϕ(z)\displaystyle-\Phi^{\,\delta}_{\infty}(x;\phi(z))\phi^{\,\prime}(z) =12πiγ(1,1)λ(z)dζ(1λ(z)λ(ζ))(xδζ(x)1δ)2\displaystyle=\frac{1}{2\pi i}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}\big{(}z\big{)}d\zeta}{\big{(}1-\lambda\big{(}z\big{)}\!-\!\lambda(\zeta)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}
=12πiγ(1,1)λ(z)dζ(λ(1/ζ)λ(z))(xδζ(x)1δ)2\displaystyle=\frac{1}{2\pi i}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}\big{(}z\big{)}d\zeta}{\big{(}\lambda(-1/\zeta)-\lambda\big{(}z\big{)}\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}
=|1/ζ:γ(1,1)γ(1,1),d(1/ζ)=1/ζ2|\displaystyle=\left|-1/\zeta:\gamma(1,-1)\mapsto\gamma(-1,1),\ d(-1/\zeta)=1/\zeta^{2}\right|
=12πiγ(1,1)λ(z)dζ(λ(ζ)λ(z))ζ2(xδζ(x)1δ)2\displaystyle=-\frac{1}{2\pi i}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}\big{(}z\big{)}d\zeta}{\big{(}\lambda(\zeta)-\lambda(z)\big{)}\zeta^{2}\left(-\dfrac{x^{\delta}}{\zeta}-(-x)^{1-\delta}\right)^{2}}
=12πiγ(1,1)λ(z)dζ(λ(z)λ(ζ))(xδ+(x)1δζ)2\displaystyle=\frac{1}{2\pi i}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}\big{(}z\big{)}d\zeta}{\big{(}\lambda(z)-\lambda(\zeta)\big{)}\left(x^{\delta}+(-x)^{1-\delta}\zeta\right)^{2}}
=12πiγ(1,1)λ(z)dζ(λ(z)λ(ζ))(x1δζ(x)δ)2=Φ1δ(x;z),\displaystyle=\frac{1}{2\pi i}\int\limits_{\gamma(1,-1)}\dfrac{\lambda^{\,\prime}\big{(}z\big{)}d\zeta}{\big{(}\lambda(z)-\lambda(\zeta)\big{)}\big{(}x^{1-\delta}\zeta-(-x)^{\delta}\big{)}^{2}}=\Phi^{1-\delta}_{\infty}(x;z)\,,\

because

(xδ+(x)1δζ)2={(1xζ)2=(xζ1)2,ifδ=0,(x+ζ)2,ifδ=1,=(x1δζ(x)δ)2.\displaystyle\left(x^{\delta}+(-x)^{1-\delta}\zeta\right)^{2}=\left\{\begin{array}[]{ll}\left(1-x\zeta\right)^{2}=\left(x\zeta-1\right)^{2},&\hbox{if}\ \ \delta=0\,,\\ \left(x+\zeta\right)^{2},&\hbox{if}\ \ \delta=1\,,\end{array}\ \ \ =\big{(}x^{1-\delta}\zeta-(-x)^{\delta}\big{)}^{2}\,.\right.

This finished the proof of (6.12).

32

\uparrow  We prove (6.33). If z=m𝔻¯z\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}=\Bb{H}\setminus\cup_{\,{{\mbox{\footnotesize{$m\in\Bb{Z}$}}}}}\ (2m+\overline{\Bb{D}}) then zSz\not\in S^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} and

zϕ𝔫(𝔻𝕀),𝔫{}\displaystyle z\not\in\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)\ ,\quad\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}\,.

Therefore, in this case,

χϕ𝔫(𝔻𝕀)(z)=0,𝔫{}\displaystyle\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)=0\ ,\quad\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}\,,

while h(z)=0\eurm{h}_{\eusm{E}}(z)=0, as follows from (5.91). So that (6.33) is proved for zz\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}.

Let

z𝔻𝕀𝕊||𝔼𝔫𝔼\displaystyle z\in\Bb{D}_{{\rm{Im}}>0}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \ \sqcup\ \ \bigsqcup\nolimits_{{{\mbox{\footnotesize{$\ \eufm{n}\in\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}$}}}}}\ \ \eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,.

In view of (5.91), written for z𝔻𝕀𝕊||z\in\Bb{D}_{{\rm{Im}}>0}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}, zz\not\in\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}} in the form,

h(z)=1+d(ϕ𝔫),ifzϕ𝔫(0)=𝔫,𝔫{}\displaystyle\eurm{h}_{\eusm{E}}(z)=1+d\left(\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\right)\,,\ \ \mbox{if}\ z\in\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\big{)}=\eusm{E}^{\eufm{n}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\ \eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}\,, (5.91)

we have h(z)=1\eurm{h}_{\eusm{E}}(z)=1, if z0z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}. Together with (ϕ0(z)=z\phi_{0}(z)=z),

χϕ0(𝔻𝕀)(z)=1,χϕ𝔫(𝔻𝕀)(z)=0,𝔫\displaystyle\chi_{{{\mbox{\footnotesize{$\phi_{0}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)=1\ ,\quad\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)=0\ ,\quad\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\,,

this yields the validity of (6.33) for z0z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}},

𝔫{}χϕ𝔫(𝔻𝕀)(z)=h(z),z0.\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)=\eurm{h}_{\eusm{E}}(z)\ ,\quad z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\,.

At the same time, h(z)=1+N\eurm{h}_{\eusm{E}}(z)=1+N, if z𝓃𝒩,,𝓃1z\in\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}, NN\in\Bb{N}, 𝔫=(𝔫𝔑,,𝔫1)\eufm{n}=(n_{N},...,\,n_{1})\!\in\!\Bb{Z}_{\neq 0}^{N}. But for such zz, in view of (6.32),

ϕ𝔫(𝔻𝕀𝕊||𝔼kK,,k1𝕂𝔼𝕂\displaystyle\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\!\!\!\!\!\!\bigsqcup\limits_{\ {{\mbox{\footnotesize{$k_{K},...,\,k_{1}\in\Bb{Z}_{\neq 0}\,,\ K\in\Bb{N}$}}}}}\!\!\!\!\!\!\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1},k_{K},...,\,k_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ , (6.32)

we have z𝔻𝕀𝕊||ϕ 0𝔻𝕀𝕊||z\in\Bb{D}_{{\rm{Im}}>0}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}=\phi_{{{\mbox{\footnotesize{$\,0$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}, and it follows from

χϕ𝔪(𝔻𝕀)(z)0,M𝕄𝕄\displaystyle\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{m}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)\neq 0\ ,\quad M\in\Bb{N}\,,\ \ \eufm{m}=(m_{M},...,\,m_{1})\!\in\!\Bb{Z}_{\neq 0}^{M}\ \Leftrightarrow (A.32a)
𝓃𝒩,,𝓃1ϕ𝔪(𝔻𝕀𝕊||𝔼𝕄kK,,k1𝕂𝔼𝕄𝕂\displaystyle\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\!\subset\!\phi_{{{\mbox{\footnotesize{$\eufm{m}$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}\!=\!\eusm{E}^{\hskip 1.42271ptm_{M},...,\,m_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \sqcup\!\!\!\!\!\!\!\!\bigsqcup\limits_{\ {{\mbox{\footnotesize{$k_{K},...,\,k_{1}\in\Bb{Z}_{\neq 0}\,,\ K\in\Bb{N}$}}}}}\!\!\!\!\!\!\eusm{E}^{\hskip 1.42271ptm_{M},...,\,m_{1},\,k_{K},...,\,k_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}

that

zϕnN,,n1(𝔻𝕀ϝϕnN,,n2𝔻𝕀zϕnN(𝔻𝕀ϝϕ 0𝔻𝕀z\in\phi_{{{\mbox{\footnotesize{$\,n_{N},...,n_{1}$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\,,\ z\in\phi_{{{\mbox{\footnotesize{$\,n_{N},...,n_{2}$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\,,\ \ldots\,,\\[5.69046pt] z\in\phi_{{{\mbox{\footnotesize{$\,n_{N}$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\,,\ z\in\phi_{{{\mbox{\footnotesize{$\,0$}}}}}\big{(}\Bb{D}_{{\rm{Im}}>0}\big{)}\,,

while

χϕ𝔫(𝔻𝕀)(z)=0,𝔫{}\displaystyle\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)=0\ ,\quad\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\setminus\{(n_{N}),(n_{N},n_{N-1}),\ldots,(n_{N},...,n_{1})\}\,,

which yields the validity of (6.33) for z𝓃𝒩,,𝓃1z\in\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}},

𝔫{}χϕ𝔫(𝔻𝕀)(z)=h(z),z𝓃𝒩,,𝓃1.\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)=\eurm{h}_{\eusm{E}}(z)\ ,\quad z\in\eusm{E}^{\hskip 1.42271ptn_{N},...,\,n_{1}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,.

This completes the proof of (6.33). Besides, we have proved that

𝔫{}χϕ𝔫(𝔻𝕀)(z)=h(z),z𝕊||\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\Bb{D}_{{\rm{Im}}>0}\right)$}}}}}(z)=\eurm{h}_{\eusm{E}}(z)\ ,\quad z\in\Bb{H}_{{\rm{Re}}\leqslant 1}\setminus S^{\hskip 0.56917pt{{\mbox{\tiny{$||$}}}}}_{\!{{\mbox{\tiny{$\frown$}}}}}\,,\ (A.32b)

and

𝔫{}χϕ𝔫(𝔻¯Im>0)(z)=h(z),z\displaystyle\sum\limits_{{{\mbox{\footnotesize{$\eufm{n}\!\in\!\Bb{Z}_{\neq 0}^{\hskip 0.56917pt\Bb{N}_{\eurm{f}}}\cup\{0\}$}}}}}\chi_{{{\mbox{\footnotesize{$\phi_{{{\mbox{\footnotesize{$\eufm{n}$}}}}}\left(\overline{\Bb{D}}_{{\rm{Im}}>0}\right)$}}}}}(z)=\eurm{h}_{\eusm{E}}(z)\ ,\quad z\in\Bb{H}_{{\rm{Re}}\leqslant 1}\,. (A.32c)

A.7 . Notes for Section 7

  

33

\uparrow  We prove the left-hand side inequality of (7.1).

Since x+iy𝓍+𝒾𝓎x+iy\!\in\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\ \Rightarrow\ -x+iy\!\in\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}, and Θ2(x+iy)=Θ2(x+iy)¯\Theta_{2}(-x+iy)=\overline{\Theta_{2}(x+iy)}, xx\in\Bb{R}, y>0y>0, it is enough to consider the case Rez0{\rm{Re}}\,z\geqslant 0 for which it is stated that

|Θ2(z)|4Im2zθ3(eπ/2)45,z\displaystyle\left|\Theta_{2}(z)\right|^{4}{\rm{Im}}^{2}z\!\leqslant\!\theta_{3}(e^{-\pi/2})^{4}\!\leqslant\!5\ ,\quad z\!\in\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\cap\Bb{C}_{{\rm{Re}}\geqslant 0}\,. (A.33a)

We first assume that zz\!\in\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\cap\Bb{C}_{{\rm{Re}}\geqslant 0}, Imz1{\rm{Im}}\,z\leqslant 1, and prove that

Im(1z)11/2,|z|>1,Rez[0,1],Imz1,z\displaystyle{\rm{Im}}(1-z)^{-1}\geqslant 1/2\,,\quad|z|>1\,,\quad{\rm{Re}}\,z\in[0,1]\,,\quad{\rm{Im}}\,z\leqslant 1\ ,\quad z\in\Bb{H}\,. (A.33b)

For any such that there exists a(0,π/2)a\in(0,\pi/2) such that z=cosa+iyz=\cos a+iy, sinay1\sin a\leqslant y\leqslant 1. Then it is necessary to obtain that

12Im11z=Imz|1z|2=y(1cosa)2+y2,sinay1,\displaystyle\dfrac{1}{2}\leqslant{\rm{Im}}\dfrac{1}{1-z}=\dfrac{{\rm{Im}}\,z}{|1-z|^{2}}=\dfrac{y}{\left(1-\cos a\right)^{2}+y^{2}}\ ,\ \ \sin a\leqslant y\leqslant 1\,,\

i.e.,

(1cosa)2+y22y,sinay1(1cosa)2y(2y),sinay1.\displaystyle\left(1-\cos a\right)^{2}+y^{2}\leqslant 2y\,,\ \sin a\leqslant y\leqslant 1\ \Leftrightarrow\ \left(1-\cos a\right)^{2}\leqslant y(2-y)\,,\ \sin a\leqslant y\leqslant 1\,.

But on the interval [0,1][0,1] the function y(2y)y(2-y) increases from 0 to 11. Therefore the above inequality is equivalent to

(1cosa)2(2sina)sina 1+cos2a2cosa2sinasin2a\displaystyle\left(1-\cos a\right)^{2}\leqslant(2-\sin a)\sin a\ \Leftrightarrow\ 1+\cos^{2}a-2\cos a\leqslant 2\sin a-\sin^{2}a
1cosasinacosa+sina1sin(a+π4)sinπ4,\displaystyle 1-\cos a\leqslant\sin a\ \Leftrightarrow\ \cos a+\sin a\geqslant 1\ \Leftrightarrow\ \sin\left(a+\dfrac{\pi}{4}\right)\geqslant\sin\dfrac{\pi}{4}\,,

which is true because a(0,π/2)a\in(0,\pi/2). Thus, (A.33b) is proved.

Then for such zz we deduce from the following consequence of (2.25),

Θ2(z)4=(1z)2Θ4(1/(1z))4,z\displaystyle\Theta_{2}(z)^{4}=(1-z)^{-2}\Theta_{4}\left({1}/{(1-z)}\right)^{4},\ \ z\in\Bb{H}\,,

and (A.33b) that

|Θ2(z)|4Im2z\displaystyle\left|\Theta_{2}(z)\right|^{4}{\rm{Im}}^{2}z |1z|2|Θ2(z)|4=|Θ4(11z)|4\displaystyle\leqslant|1-z|^{2}\left|\Theta_{2}(z)\right|^{4}=\left|\Theta_{4}\left(\dfrac{1}{1-z}\right)\right|^{4}
=|1+2n1(1)nun2|4u=eiπ1zθ3(eπ/2)4,\displaystyle=\left|1\!+\!2\sum\limits_{n\geqslant 1}(-1)^{n}u^{n^{2}}\right|^{4}_{{{\mbox{\normalsize{$u=e^{{{\mbox{\footnotesize{$\dfrac{i\pi}{1-z}$}}}}}$}}}}}\leqslant\theta_{3}\left(e^{-\pi/2}\right)^{4}\ ,

where by (ber1, , p.​ 325),

θ3(eπ/2)4=πΓ(3/4)4(1+2)22=πΓ(3/4)43+222,\displaystyle\theta_{3}\left(e^{-\pi/2}\right)^{4}=\dfrac{\pi}{\Gamma(3/4)^{4}}\dfrac{(1+\sqrt{2})^{2}}{2}=\dfrac{\pi}{\Gamma(3/4)^{4}}\dfrac{3+2\sqrt{2}}{2}\,,
2,914213562373095<3+222<2,9142135623730951,\displaystyle 2,914213562373095<\dfrac{3+2\sqrt{2}}{2}<2,9142135623730951\,,
1,393203929652002<πΓ(3/4)4<1,393203929652003\displaystyle 1,393203929652002<\dfrac{\pi}{\Gamma(3/4)^{4}}<1,393203929652003\ \Rightarrow\
4<4,06009<θ3(eπ/2)4<4,060094<5.\displaystyle 4<4,06009<\theta_{3}\left(e^{{{\mbox{\footnotesize{$-\pi/2$}}}}}\right)^{4}<4,060094<5\,. (A.33c)

It remains to prove (A.33a) for zz\!\in\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}}\cap\Bb{C}_{{\rm{Re}}\geqslant 0}, Imz>1{\rm{Im}}\,z>1, where due to 22-periodicity of Θ2\Theta_{2} it is possible to consider that Rez[1,1]{\rm{Re}}\,z\in[-1,1]. Then

|Θ2(z)|416eπImz|θ2(eiπz)|416eπImzθ2(eπ)4,\displaystyle\left|\Theta_{2}(z)\right|^{4}\leqslant 16e^{-\pi{\rm{Im}}\,z}\left|\theta_{2}\left({\rm{e}}^{\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi z}\right)\right|^{4}\leqslant 16e^{-\pi{\rm{Im}}\,z}\theta_{2}\left(e^{-\pi}\right)^{4}\,,

and therefore

|Θ2(z)|4Im2z16(Im2z)eπImzθ2(eπ)4,\displaystyle\left|\Theta_{2}(z)\right|^{4}{\rm{Im}}^{2}z\leqslant 16\left({\rm{Im}}^{2}z\right)e^{-\pi{\rm{Im}}\,z}\theta_{2}\left(e^{-\pi}\right)^{4}\,,\

where by (ber1, , p.​ 325),

θ2(eπ)4=eπ32πΓ(3/4)4,\displaystyle\theta_{2}\left(e^{-\pi}\right)^{4}=\dfrac{e^{\pi}}{32}\dfrac{\pi}{\Gamma(3/4)^{4}}\,,\

and the function f(y)=y2eπyf(y)=y^{2}e^{-\pi y} has the derivative, satisfying

f(y)=y2eπy=2yeπyπy2eπy=yeπy(2πy)<0,y>2/π,\displaystyle f^{\,\prime}(y)=y^{2}e^{-\pi y}=2ye^{-\pi y}-\pi y^{2}e^{-\pi y}=ye^{-\pi y}\left(2-\pi y\right)<0\,,\ y>2/\pi\,,

and therefore f(y)f(1)=eπf(y)\leqslant f(1)=e^{-\pi}, y1y\geqslant 1. Thus,

|Θ2(z)|4Im2z16(Im2z)eπImzθ2(eπ)416eπeπ32πΓ(3/4)4=π2Γ(3/4)4,\displaystyle\left|\Theta_{2}(z)\right|^{4}{\rm{Im}}^{2}z\leqslant 16\left({\rm{Im}}^{2}z\right)e^{-\pi{\rm{Im}}\,z}\theta_{2}\left(e^{-\pi}\right)^{4}\leqslant 16e^{-\pi}\dfrac{e^{\pi}}{32}\dfrac{\pi}{\Gamma(3/4)^{4}}=\dfrac{\pi}{2\Gamma(3/4)^{4}}\,,

where

πΓ(3/4)4<1,3932039296520022532040318760206,\displaystyle\dfrac{\pi}{\Gamma(3/4)^{4}}<1,3932039296520022532040318760206\,,\

and hence,

|Θ2(z)|4Im2z<0,69660196482600112660201593801<θ3(eπ/2)4,\displaystyle\left|\Theta_{2}(z)\right|^{4}{\rm{Im}}^{2}z<0,69660196482600112660201593801<\theta_{3}\left(e^{{{\mbox{\footnotesize{$-\pi/2$}}}}}\right)^{4}\,,\

by (A.33c). This completes the proof of (A.33a) and of the left-hand side inequality of (7.1) as well.

34

\uparrow  In view of (2.31)(a) and (A.6e), λ(i)=1/2\lambda(i)=1/2,

λ(2i)=(1121+12)2=3223+22=17122=(0.029437251,0.029437252),\displaystyle\lambda(2i)=\left(\dfrac{1-\dfrac{1}{\sqrt{2}}}{1+\dfrac{1}{\sqrt{2}}}\right)^{2}=\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}=17-12\sqrt{2}=(0.029437251,0.029437252)\,,\
12λ(2i)=123223+22=3+226+423+22=6233+22=32213+22,\displaystyle 1-2\lambda(2i)=1-2\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}=\dfrac{3+2\sqrt{2}-6+4\sqrt{2}}{3+2\sqrt{2}}=\dfrac{6\sqrt{2}-3}{3+2\sqrt{2}}=3\dfrac{2\sqrt{2}-1}{3+2\sqrt{2}}\ ,
112λ(2i)=133+22221=(22+1)(3+22)21=62+8+3+2221\displaystyle\dfrac{1}{1-2\lambda(2i)}=\dfrac{1}{3}\cdot\dfrac{3+2\sqrt{2}}{2\sqrt{2}-1}=\dfrac{(2\sqrt{2}+1)(3+2\sqrt{2})}{21}=\dfrac{6\sqrt{2}+8+3+2\sqrt{2}}{21}
=11+8221(22.31370,22.31371)21(1,062557547570,1,062557547571),\displaystyle=\dfrac{11+8\sqrt{2}}{21}\in\dfrac{(22.31370,22.31371)}{21}\in(1,062557547570,1,062557547571)\,,
15(12λ(2i))=11+82105(0.21251150,0.21251151).\displaystyle\dfrac{1}{5(1-2\lambda(2i))}=\dfrac{11+8\sqrt{2}}{105}\in(0.21251150,0.21251151)\,.

At the same time, by virtue of (3.38) and (2.31)(b),

1λ(1+it)+1λ(it)=11λ(1+it)=λ(it)1λ(it)\displaystyle\dfrac{1}{\lambda(1\!+\!it)}+\dfrac{1}{\lambda(it)}=1\ \Rightarrow\ \dfrac{1}{\lambda(1\!+\!it)}=\dfrac{\lambda(it)-1}{\lambda(it)}\ \Rightarrow\
|λ(1+it)|=λ(it)1λ(it)=λ(it)λ(i(1/t))λ(2i)16eπ/t=1712216eπ/t,t(0,2].\displaystyle\left|\lambda(1\!+\!it)\right|=\dfrac{\lambda(it)}{1-\lambda(it)}=\dfrac{\lambda(it)}{\lambda(i(1/t))}\geqslant\dfrac{\lambda(2i)}{16}e^{\pi/t}=\dfrac{17-12\sqrt{2}}{16}e^{\pi/t}\ ,\quad t\in(0,2]\,.

Here

1617122=163+22322=16(3+22)2=16(17+122)\displaystyle\dfrac{16}{17-12\sqrt{2}}=16\cdot\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}=16\cdot(3+2\sqrt{2})^{2}=16(17+12\sqrt{2})
=272+1922(543.52900,543.52901)555.\displaystyle=272+192\sqrt{2}\in(543.52900,543.52901)\leqslant 555\,.

Since λ\lambda is 22-periodic, we obtain

|λ(±1+it)|1712216eπ/teπ/t555,t(0,2],\displaystyle\left|\lambda(\pm 1\!+\!it)\right|\geqslant\dfrac{17-12\sqrt{2}}{16}e^{\pi/t}\geqslant\dfrac{e^{\pi/t}}{555}\ ,\quad t\in(0,2]\,, (A.34a)
15(12λ(2i))=11+821050.21251151.\displaystyle\dfrac{1}{5(1-2\lambda(2i))}=\dfrac{11+8\sqrt{2}}{105}\leqslant 0.21251151\,. (A.34b)
35

\uparrow  The expression for the integral follows from the following transforms

02dtt2(a+1712216eπ/t)\displaystyle\int\limits_{0}^{2}\!\dfrac{dt}{t^{2}\left(a+\dfrac{17-12\sqrt{2}}{16}e^{\pi/t}\right)} =1/2dta+1712216eπt=|t=eπt,t=logtπ|\displaystyle=\int\limits_{1/2}^{\infty}\dfrac{dt}{a+\dfrac{17-12\sqrt{2}}{16}e^{\pi t}}=\left|t^{\,\prime}=e^{\pi t}\,,\ t=\dfrac{\log t^{\,\prime}}{\pi}\right|
=16π(17122)eπ/2dtt(t+b)=|b:=16a17122|\displaystyle=\dfrac{16}{\pi\left(17-12\sqrt{2}\right)}\int\limits_{e^{\pi/2}}^{\infty}\dfrac{dt}{t(t+b)}=\left|b:=\dfrac{16a}{17-12\sqrt{2}}\right|
=16πb(17122)eπ/2(1t1t+b)dt\displaystyle=\dfrac{16}{\pi b\left(17-12\sqrt{2}\right)}\int\limits_{e^{\pi/2}}^{\infty}\left(\dfrac{1}{t}-\dfrac{1}{t+b}\right)dt
=16πb(17122)logtt+b|t=eπ/2t=\displaystyle=\dfrac{16}{\pi b\left(17-12\sqrt{2}\right)}\left.\log\dfrac{t}{t+b}\right|_{t=e^{\pi/2}}^{t=\infty}
=16πb(17122)log(1+beπ/2)\displaystyle=\dfrac{16}{\pi b\left(17-12\sqrt{2}\right)}\log\left(1+be^{-\pi/2}\right)
=1πalog(1+16eπ/217122a).\displaystyle=\dfrac{1}{\pi a}\log\left(1+\dfrac{16e^{-\pi/2}}{17-12\sqrt{2}}a\right)\,.
36

\uparrow  We use the following results of the numerical calculations

55+40221<5,313,\displaystyle\dfrac{55+40\sqrt{2}}{21}<5,313\,,
1+16eπ/217122<113,9884,\displaystyle 1+\dfrac{16e^{-\pi/2}}{17-12\sqrt{2}}<113,9884\,,
52π<2,251<94,\displaystyle\dfrac{5\sqrt{2}}{\pi}<2,251<\dfrac{9}{4}\,,
log(1+16eπ/217122)<4,7361,\displaystyle\log\left(1+\dfrac{16e^{-\pi/2}}{17-12\sqrt{2}}\right)<4,7361\,,
52πlog(1+16eπ/217122)<10,6609536465591,\displaystyle\dfrac{5\sqrt{2}}{\pi}\log\left(1+\dfrac{16e^{-\pi/2}}{17-12\sqrt{2}}\right)<10,6609536465591\,,
55+40221+52πlog(1+16eπ/217122)<15,974\displaystyle\dfrac{55+40\sqrt{2}}{21}+\dfrac{5\sqrt{2}}{\pi}\log\left(1+\dfrac{16e^{-\pi/2}}{17-12\sqrt{2}}\right)<15,974

which yield that

Iδ(z)16+(9/4)log(1+|λ(1/z)|),z0.\displaystyle I_{\delta}(z)\!\leqslant\!16+(9/4)\log\left(1+\left|\lambda(-1/z)\right|\right)\ ,\quad z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,.

To prove (7.12), observe that

log(1+|λ(x+iy)|)\displaystyle\log\left(1+\left|\lambda(x+iy)\right|\right) log(1+16eπy+π/y)\displaystyle\leqslant\log\left(1+16\,{\rm{e}}^{{{\mbox{\footnotesize{$-\pi y+\pi/y$}}}}}\right)
πy+log(eπ/y+16eπy)πy+log17π(1+1/y),\displaystyle\dfrac{\pi}{y}+\log\left({\rm{e}}^{{{\mbox{\footnotesize{$-\pi/y$}}}}}+16{\rm{e}}^{{{\mbox{\footnotesize{$-\pi y$}}}}}\right)\leqslant\dfrac{\pi}{y}+\log 17\leqslant\pi\left(1+{1}/{y}\right)\,,\

and therefore it follows from (7.11)

Iδ(z)16+(9/4)log(1+|λ(𝔾ϝ|),𝔾ϝ𝔽||𝔻¯𝔼ϝ𝔼\displaystyle I_{\delta}(z)\!\leqslant\!16\!+\!(9/4)\log\Big{(}1\!+\!\left|\lambda\big{(}\Bb{G}_{2}(z)\big{)}\right|\Big{)},\ \Bb{G}_{2}(z)\!\in\!\mathcal{F}^{\,{{\mbox{\tiny{$||$}}}}}_{{{\mbox{\tiny{$\square$}}}}}\setminus\overline{\Bb{D}}\!\subset\!\eusm{E}^{\infty}_{\!{{\mbox{\tiny{$\frown$}}}}},\ z\in\eusm{E}_{\!{{\mbox{\tiny{$\frown$}}}}}^{\hskip 1.42271pt0}\,, (7.11)

that

Iδ(z)16+9π4+9π4Im𝔾ϝ147π20(1+1Im𝔾ϝ),\displaystyle I_{\delta}(z)\!\leqslant\!16\!+\!\dfrac{9\pi}{4}+\dfrac{9\pi}{4{\rm{Im}}\,\Bb{G}_{2}(z)}\leqslant\dfrac{147\pi}{20}\left(1+\dfrac{1}{{\rm{Im}}\,\Bb{G}_{2}(z)}\right)\,,

i.e., (7.12) holds, because

16+9π4<23,068583471<23,09<147π20.\displaystyle 16+\dfrac{9\pi}{4}<23,068583471<23,09<\dfrac{147\pi}{20}\ .
37

\uparrow   In view of (ber1, , p.​ 325), we have

9π4<7,068583471,\displaystyle\dfrac{9\pi}{4}<7,068583471\,,
16+9π4<23,0685834705771,\displaystyle 16+\dfrac{9\pi}{4}<23,0685834705771\,,
16π+94<7,342958178941<14720=7,35,\displaystyle\dfrac{16}{\pi}+\dfrac{9}{4}<7,342958178941<\dfrac{147}{20}=7,35\ ,
θ3(eπ/2)4=πΓ(34)4(1+2)22<4,0600937869433563,\displaystyle\theta_{3}(e^{-\pi/2})^{4}=\dfrac{\pi}{\Gamma\left(\dfrac{3}{4}\right)^{4}}\dfrac{\left(1+\sqrt{2}\right)^{2}}{2}<4,0600937869433563\,,
14720θ3(eπ/2)4<147204,0600937869433563<29,8416893341<30.\displaystyle\dfrac{147}{20}\cdot\theta_{3}(e^{-\pi/2})^{4}<\dfrac{147}{20}\cdot 4,0600937869433563<29,8416893341<30\,.
38

\uparrow  Here we use the inequality 1+60π<20π21+60\pi<20\pi^{2} which holds because

60π<188,49555921538759430775860299677,\displaystyle 60\pi<188,49555921538759430775860299677\,,
1+60π<189,49555921538759430775860299677,\displaystyle 1+60\pi<189,49555921538759430775860299677\,,
20π2=197,39208802178717237668981999752>190>1+60π.\displaystyle 20\pi^{2}=197,39208802178717237668981999752...>190>1+60\pi\,.
39

\uparrow   Actually, we have

eπ<23,140692632779269005729086367949,\displaystyle e^{\pi}<23,140692632779269005729086367949\,,
40eπ<925,6277053111707602291634547178,\displaystyle 40e^{\pi}<925,6277053111707602291634547178\,,
π6=961,38919357530443703021944365242>961>40eπ.\displaystyle\pi^{6}=961,38919357530443703021944365242...>961>40e^{\pi}\,.
40

\uparrow  We prove (7.25). In view of (1.18),

|Hn(x)|,|Mn(x)|π6n24,x\displaystyle\left|\eurm{H}_{n}(x)\right|\ ,\ \left|\eurm{M}_{n}(x)\right|\leqslant\dfrac{\pi^{6}n^{2}}{4}\,,\quad x\in\Bb{R}\ ,\ n\in\Bb{Z}_{\neq 0}\,.

By using (1.19), we get

|Hn(x)|,|Mn(x)|π6n24x2π6n22(1+x2),|x|1,\displaystyle\left|\eurm{H}_{n}(x)\right|\ ,\ \left|\eurm{M}_{n}(x)\right|\leqslant\dfrac{\pi^{6}n^{2}}{4x^{2}}\leqslant\dfrac{\pi^{6}n^{2}}{2(1+x^{2})}\,,\quad|x|\geqslant 1\,,

while

|Hn(x)|,|Mn(x)|π6n24π6n22(1+x2),x[1,1].\displaystyle\left|\eurm{H}_{n}(x)\right|\ ,\ \left|\eurm{M}_{n}(x)\right|\leqslant\dfrac{\pi^{6}n^{2}}{4}\leqslant\dfrac{\pi^{6}n^{2}}{2(1+x^{2})}\,,\quad x\in[-1,1]\,.

This proves (7.25).

41

\uparrow  We prove (7.26). The explicit integral formula for H0\eurm{H}_{0} written after (1.21),

H0(x)=i2π21iy(t)1+iy(t)x2(1iy(t)1+iy(t))2dtt2+1/4\displaystyle\eurm{H}_{0}(x)=\dfrac{i}{2\pi^{2}}\int\limits_{\Bb{R}}\dfrac{\dfrac{1\!-\!i\eurm{y}(t)}{1\!+\!i\eurm{y}(t)}}{x^{2}-\left(\dfrac{1\!-\!i\eurm{y}(t)}{1\!+\!i\eurm{y}(t)}\right)^{2}}\,\dfrac{dt}{t^{2}+1/4}
=i4π21x1iy(t)1+iy(t)dtt2+1/4i4π21x+1iy(t)1+iy(t)dtt2+1/4\displaystyle=\dfrac{i}{4\pi^{2}}\int\limits_{\Bb{R}}\dfrac{1}{x-\dfrac{1\!-\!i\eurm{y}(t)}{1\!+\!i\eurm{y}(t)}}\,\dfrac{dt}{t^{2}+1/4}-\dfrac{i}{4\pi^{2}}\int\limits_{\Bb{R}}\dfrac{1}{x+\dfrac{1\!-\!i\eurm{y}(t)}{1\!+\!i\eurm{y}(t)}}\,\dfrac{dt}{t^{2}+1/4}

yields

|H0(x)|12π21Imiy(t)1iy(t)+1dtt2+1/4=14π21+y(t)2y(t)dtt2+1/4,\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\dfrac{1}{2\pi^{2}}\int\limits_{\Bb{R}}\dfrac{1}{{\rm{Im}}\dfrac{i\eurm{y}(t)\!-\!1}{i\eurm{y}(t)\!+\!1}}\,\dfrac{dt}{t^{2}+1/4}=\dfrac{1}{4\pi^{2}}\int\limits_{\Bb{R}}\dfrac{1\!+\!\eurm{y}(t)^{2}}{\eurm{y}(t)}\dfrac{dt}{t^{2}+1/4}\ ,\

where y(t)=1/y(t)\eurm{y}(-t)=1/\eurm{y}(t) and therefore

1+y(t)2y(t)dtt2+1/4=(y(t)+y(t))dtt2+1/4=20+(y(t)+y(t))dtt2+1/4\displaystyle\int\limits_{\Bb{R}}\dfrac{1\!+\!\eurm{y}(t)^{2}}{\eurm{y}(t)}\dfrac{dt}{t^{2}+1/4}=\int\limits_{\Bb{R}}\left(\eurm{y}(t)+\eurm{y}(-t)\right)\dfrac{dt}{t^{2}+1/4}=2\int\limits_{0}^{+\infty}\left(\eurm{y}(t)+\eurm{y}(-t)\right)\dfrac{dt}{t^{2}+1/4}
=20+(y(t)+1y(t))dtt2+1/4.\displaystyle=2\int\limits_{0}^{+\infty}\left(\eurm{y}(t)+\dfrac{1}{\eurm{y}(t)}\right)\dfrac{dt}{t^{2}+1/4}\ .

Thus, by using 1F(x)21\leqslant F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)\leqslant 2, 0x1/20\leqslant x\leqslant 1/2 (see (bh1, , p.​ 35, (A.9c))), which actually gives F(x)2F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)\geqslant 2, 1/2x<11/2\leqslant x<1, we get

|H0(x)|12π20+(y(t)+1y(t))dtt2+1/4\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\dfrac{1}{2\pi^{2}}\int\limits_{0}^{+\infty}\left(\eurm{y}(t)+\dfrac{1}{\eurm{y}(t)}\right)\dfrac{dt}{t^{2}+1/4}
=12π20+(F(12t4t2+1)F(12+t4t2+1)+F(12+t4t2+1)F(12t4t2+1))dtt2+1/4\displaystyle=\dfrac{1}{2\pi^{2}}\int\limits_{0}^{+\infty}\left(\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}+\dfrac{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}{F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)}\right)\dfrac{dt}{t^{2}+1/4}
12π20+(1+F(12+t4t2+1))dtt2+1/4\displaystyle\leqslant\dfrac{1}{2\pi^{2}}\int\limits_{0}^{+\infty}\left(1+F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\right)\dfrac{dt}{t^{2}+1/4}
12π20+dtt2+1/4+12π20+F(12+t4t2+1)dtt2+1/4.\displaystyle\leqslant\dfrac{1}{2\pi^{2}}\int\limits_{0}^{+\infty}\dfrac{dt}{t^{2}+1/4}+\dfrac{1}{2\pi^{2}}\int\limits_{0}^{+\infty}F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)\dfrac{dt}{t^{2}+1/4}\ .

Applying F(x)3+log(1x)1F_{\!{{\mbox{\tiny{$\triangle$}}}}}(x)\leqslant 3+\log(1-x)^{-1}, 0x<10\leqslant x<1 (see (bh1, , p.​ 35, (A.9d))), we deduce that

F(12+t4t2+1)3log112t4t2+1=log24t2+14t2+12t\displaystyle F_{\!{{\mbox{\tiny{$\triangle$}}}}}\left(\dfrac{1}{2}\!+\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}\right)-3\leqslant\log\dfrac{1}{\dfrac{1}{2}\!-\!\dfrac{t}{\sqrt{4t^{2}\!+\!1}}}=\log\dfrac{2\sqrt{4t^{2}\!+\!1}}{\sqrt{4t^{2}\!+\!1}-2t}
=log2+log4t2+1+log(4t2+1+2t)log2+log4t2+1\displaystyle=\log 2+\log\sqrt{4t^{2}\!+\!1}+\log\left(\sqrt{4t^{2}\!+\!1}+2t\right)\leqslant\log 2+\log\sqrt{4t^{2}\!+\!1}
+log(4t2+1+4t2+1)=2log2+2log4t2+1,\displaystyle+\log\left(\sqrt{4t^{2}\!+\!1}+\sqrt{4t^{2}\!+\!1}\right)=2\log 2+2\log\sqrt{4t^{2}\!+\!1}\,,\

and therefore from the above inequality we obtain

|H0(x)|4+2log22π20+dtt2+1/4+1π20+log4t2+1t2+1/4dt\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\dfrac{4+2\log 2}{2\pi^{2}}\int\limits_{0}^{+\infty}\dfrac{dt}{t^{2}+1/4}+\dfrac{1}{\pi^{2}}\int\limits_{0}^{+\infty}\dfrac{\log\sqrt{4t^{2}\!+\!1}}{t^{2}+1/4}dt
=4+2log2π20+dtt2+1+2π20+logt2+1t2+1dt=2+log2π\displaystyle=\dfrac{4+2\log 2}{\pi^{2}}\int\limits_{0}^{+\infty}\dfrac{dt}{t^{2}+1}+\dfrac{2}{\pi^{2}}\int\limits_{0}^{+\infty}\dfrac{\log\sqrt{t^{2}\!+\!1}}{t^{2}+1}dt=\dfrac{2+\log 2}{\pi}
+1π20+log(t2+1)t2+1dt=2+log2π+12π20+log(1+t)(t+1)tdt=\displaystyle+\dfrac{1}{\pi^{2}}\int\limits_{0}^{+\infty}\dfrac{\log(t^{2}\!+\!1)}{t^{2}+1}dt=\dfrac{2+\log 2}{\pi}+\dfrac{1}{2\pi^{2}}\int\limits_{0}^{+\infty}\dfrac{\log(1\!+\!t)}{(t+1)\sqrt{t}}dt\ \boxed{=}
0+log(1+t)(t+1)tdt=1+logtt(t1)dt=01log(1/t)(1t)tdt\displaystyle\int\limits_{0}^{+\infty}\dfrac{\log(1\!+\!t)}{(t+1)\sqrt{t}}dt=\int\limits_{1}^{+\infty}\dfrac{\log t}{t\sqrt{(t-1)}}dt=\int\limits_{0}^{1}\dfrac{\log(1/t)}{\sqrt{\left(1-t\right)t}}dt
=(pru, , p.​ 490, 2.6.5.4)=B(1/2,1/2)[ψ(1/2)ψ(1)]\displaystyle=\ \mbox{\cite[cite]{(\@@bibref{AuthorsPhrase1Year}{pru}{\@@citephrase{, }}{}, p.\! 490, 2.6.5.4)}}\ =-B(1/2,1/2)\left[\psi(1/2)-\psi(1)\right]
=B(1/2,1/2)(ψ(1)ψ(1/2))=π((γ)(γlog4))=πlog4,\displaystyle=B(1/2,1/2)\left(\psi(1)-\psi(1/2)\right)=\pi\left((-\gamma)-(-\gamma-\log 4)\right)=\pi\log 4\,,\
=2+log2π+log42π=2+log2π+log2π=21+log2π<1,077891.\displaystyle\boxed{=}\ \dfrac{2+\log 2}{\pi}+\dfrac{\log 4}{2\pi}=\dfrac{2+\log 2}{\pi}+\dfrac{\log 2}{\pi}=2\,\dfrac{1+\log 2}{\pi}<1,077891\,.

So that

|H0(x)|32,x\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\dfrac{3}{2}\ ,\quad x\in\Bb{R}\ .

By using (3.20), we get

|H0(x)|32x231+x2,|x|1,\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\dfrac{3}{2x^{2}}\leqslant\dfrac{3}{1+x^{2}}\,,\quad|x|\geqslant 1\,,

while

|H0(x)|3231+x2,x[1,1].\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\dfrac{3}{2}\leqslant\dfrac{3}{1+x^{2}}\,,\quad x\in[-1,1]\,.

This proves

|H0(x)|31+x2,x\displaystyle\left|\eurm{H}_{0}(x)\right|\leqslant\dfrac{3}{1+x^{2}}\ ,\quad x\in\Bb{R}\ ,

and completes the proof of (7.26).

42

\uparrow  Actually, let xx\!\in\!\Bb{R}, δ{0,1}\delta\in\{0,1\}, ζγ(1,1)\zeta\in\gamma(-1,1) and z𝕊z\in\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}. Since

ddζζδxδζ(x)1δ=1(xδζ(x)1δ)2,\displaystyle\dfrac{d}{d\zeta}\ \ \dfrac{\zeta^{\delta}}{x^{\delta}\zeta-(-x)^{1-\delta}}=-\dfrac{1}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}\,, (A.42a)

and

ddzλ(z)λ(ζ)λ(ζ)(λ(ζ)λ(z))=λ(ζ)λ(ζ)ddz(1+λ(ζ)λ(ζ)λ(z))\displaystyle\dfrac{d}{dz}\ \dfrac{\lambda(z)\lambda^{\,\prime}(\zeta)}{\lambda(\zeta)\left(\lambda(\zeta)-\lambda(z)\right)}=\dfrac{\lambda^{\,\prime}(\zeta)}{\lambda(\zeta)}\ \dfrac{d}{dz}\ \left(-1+\dfrac{\lambda(\zeta)}{\lambda(\zeta)-\lambda(z)}\right)
=λ(ζ)λ(ζ)λ(ζ)λ(z)(λ(ζ)λ(z))2=λ(z)λ(ζ)(λ(ζ)λ(z))2,\displaystyle=\dfrac{\lambda^{\,\prime}(\zeta)}{\lambda(\zeta)}\ \dfrac{\lambda(\zeta)\lambda^{\,\prime}(z)}{\big{(}\lambda(\zeta)-\lambda(z)\big{)}^{2}}=\lambda^{\,\prime}(z)\dfrac{\lambda^{\,\prime}(\zeta)}{\big{(}\lambda(\zeta)-\lambda(z)\big{)}^{2}}\,,\

we deduce from (7.31) that

ddzΨδ(x;z)\displaystyle\dfrac{d}{dz}\ \ \Psi^{\,\delta}_{\infty}(x;z) =ddz12πiγ(1,1)λ(z)λ(ζ)λ(ζ)(λ(ζ)λ(z))ζδdζ(xδζ(x)1δ)\displaystyle=\dfrac{d}{dz}\ \ \frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\lambda(z)\lambda^{\,\prime}(\zeta)}{\lambda(\zeta)\left(\lambda(\zeta)-\lambda(z)\right)}\dfrac{\zeta^{\delta}{\rm{d}}\zeta}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}
=12πiγ(1,1)λ(z)λ(ζ)(λ(ζ)λ(z))2ζδdζ(xδζ(x)1δ)\displaystyle=\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\lambda^{\,\prime}(z)\lambda^{\,\prime}(\zeta)}{\big{(}\lambda(\zeta)-\lambda(z)\big{)}^{2}}\dfrac{\zeta^{\delta}{\rm{d}}\zeta}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}
=12πiγ(1,1)ζδ(xδζ(x)1δ)dλ(z)λ(z)λ(ζ)=|λ(±1)=|\displaystyle=\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\zeta^{\delta}}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}d\dfrac{\lambda^{\,\prime}(z)}{\lambda(z)-\lambda(\zeta)}=\left|\lambda(\pm 1)=\infty\right|
=12πiγ(1,1)λ(z)λ(z)λ(ζ)dζδ(xδζ(x)1δ)\displaystyle=-\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\lambda^{\,\prime}(z)}{\lambda(z)-\lambda(\zeta)}d\dfrac{\zeta^{\delta}}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}
=(A.42a)12πiγ(1,1)λ(z)λ(z)λ(ζ)dζ(xδζ(x)1δ)2=(6.8)Φδ(x;z),\displaystyle\stackrel{{\scriptstyle\eqref{f1case40}}}{{=}}\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\dfrac{\lambda^{\,\prime}(z)}{\lambda(z)-\lambda(\zeta)}\dfrac{{\rm{d}}\zeta}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}^{2}}\stackrel{{\scriptstyle\eqref{f1bsdenac}}}{{=}}\Phi^{\,\delta}_{\infty}(x;z)\,,

i.e.,

ddzΨδ(x;z)=Φδ(x;z),z𝕊{}δ{}\displaystyle\dfrac{d}{dz}\ \Psi^{\,\delta}_{\infty}(x;z)=\Phi^{\,\delta}_{\infty}(x;z)\ ,\quad z\in\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\,,\ x\in\Bb{R}\setminus\{0\}\,,\ \delta\in\{0,1\}\,. (A.42b)
43

\uparrow  We prove (7.32). Let xx\!\in\!\Bb{R}, z𝕊z\in\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty} and

Ψδ(x;z)=12γ(1,1)λ(z)Θ4(ζ)4ζδdζ(λ(ζ)λ(z))(xδζ(x)1δ),δ{0,1},\displaystyle\Psi^{\,\delta}_{\infty}(x;z)=\frac{1}{2}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\dfrac{\lambda(z)\Theta_{4}(\zeta)^{4}\,\zeta^{\delta}{\rm{d}}\zeta}{\big{(}\lambda(\zeta)-\lambda(z)\big{)}\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}\,,\ \ \delta\!\in\!\{0,1\}\,, (7.31)

or, by the identity λ(z)=iπλ(z)Θ4(z)4\lambda^{\,\prime}(z)\!=\!\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt\pi\lambda(z)\Theta_{4}(z)^{4}, zz\in\Bb{H} (see (2.41), (2.21)), it can be written as follows

Ψδ(x;z)=12πiγ(1,1)λ(z)λ(ζ)λ(ζ)(λ(ζ)λ(z))ζδdζ(xδζ(x)1δ),δ{0,1}.\displaystyle\Psi^{\,\delta}_{\infty}(x;z)\!=\!\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\gamma(-1,1)$}}}}}\!\!\!\!\!\dfrac{\lambda(z)\lambda^{\,\prime}(\zeta)}{\lambda(\zeta)\left(\lambda(\zeta)-\lambda(z)\right)}\dfrac{\zeta^{\delta}{\rm{d}}\zeta}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}\,,\ \delta\!\in\!\{0,1\}\,. (A.43a)

For z[1+2i,1+2i]z\in\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}\setminus[-1\!+\!2{\rm{i}},1+\!2{\rm{i}}] introduce

Ψδ(x;z)=12πiΠ(1,1)λ(z)λ(ζ)λ(ζ)(λ(ζ)λ(z))ζδdζ(xδζ(x)1δ),δ{0,1},\displaystyle\Psi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)\!=\!\frac{1}{2\pi\hskip 0.56917pt\mathrm{i}\hskip 0.42677pt}\!\!\!\!\!\int\limits_{{{\mbox{\footnotesize{$\Pi(-1,1)$}}}}}\!\!\!\!\!\dfrac{\lambda(z)\lambda^{\,\prime}(\zeta)}{\lambda(\zeta)\left(\lambda(\zeta)-\lambda(z)\right)}\dfrac{\zeta^{\delta}{\rm{d}}\zeta}{\big{(}x^{\delta}\zeta-(-x)^{1-\delta}\big{)}}\,,\ \delta\!\in\!\{0,1\}\,, (A.43b)

where the contour Π(1,1)=(1,1+2i][1+2i,1+2i][1+2i,1)\Pi(-1,1)=(-1,-1+\!2{\rm{i}}]\cup[-1\!+\!2{\rm{i}},1+\!2{\rm{i}}]\cup[1+\!2{\rm{i}},1) passes from 1-1 to 11.

By transforming the contour γ(1,1)\gamma(-1,1) of integration in (A.43a) to Π(1,1)\Pi(-1,1) and using Lemma 2.4, we obtain from the residue theorem (con, , p.​ 112) that

Ψδ(x;z)=Ψδ(x;z),z𝕀Ψδ(x;z)=Ψδ(x;z)+zδ(xδz(x)1δ),z𝕀\displaystyle\begin{array}[]{ll}\Psi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\!=\!\Psi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)\,,&\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\sqcup\Bb{H}_{{\rm{Im}}>2},\\[5.69046pt] \Psi^{\hskip 0.56917pt\delta}_{\infty}(x;z)\!=\!\Psi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)+\dfrac{z^{\delta}}{\big{(}x^{\delta}z-(-x)^{1-\delta}\big{)}}\,,&\quad z\in\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\cap\Bb{H}_{{\rm{Im}}<2}\,.\end{array} (A.43e)

This means that the function Ψδ(x;z)+zδ(xδz(x)1δ)1\Psi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)+z^{\delta}(x^{\delta}z-(-x)^{1-\delta})^{-1}, being holomorphic on 𝕀𝔽\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, coincides on the set 𝕀𝔽𝕀𝔽\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\subset\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}} with the function Ψδ(x;z)\Psi^{\hskip 0.56917pt\delta}_{\infty}(x;z), which is holomorphic on 𝕀𝔽\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}\supset\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}}.

By the uniqueness theorem for analytic functions (see (con, , p.​ 78)), we find that for z𝕀𝔽z\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\subset\Bb{H}_{{\rm{Im}}<2}\cap\mathcal{F}_{{{\mbox{\tiny{$\square$}}}}}, the analytic extension Ψδ(x;z)\Psi^{\,\delta}_{\Bb{H}}(x;z) of the function Ψδ(x;z)\Psi^{\hskip 0.56917pt\delta}_{\infty}(x;z) from \mathcal{F}_{{{\mbox{\tiny{$\bigtriangledown$}}}}} to \Bb{H} (see (7.29)) equals the expression H(z):=Ψδ(x;z)+zδ(xδz(x)1δ)1H(z):=\Psi^{\hskip 0.56917pt\delta}_{\infty}(x;z)+z^{\delta}(x^{\delta}z-(-x)^{1-\delta})^{-1} since Ψδ(x;z)=Ψδ(x;z)\Psi^{\hskip 0.56917pt\delta}_{{{\mbox{\tiny{$\sqcap$}}}}}(x;z)=\Psi^{\hskip 0.56917pt\delta}_{\infty}(x;z) holds for all zz\in\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}, in view of (A.43e).

But by (5.79)(a) and (5.86), we see that 0\𝕊\mathcal{F}_{{{\mbox{\tiny{$\triangle$}}}}}\!\subset\!\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\!\subset\!\Bb{H}\big{\backslash}S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}, and since 0\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}} is simply connected it follows that the latter function H(z)H(z) is actually holomorphic on 0\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}.

So that the equality

Ψδ(x;z)=Ψδ(x;z)+zδxδz(x)1δ,z0𝕊\displaystyle\Psi^{\hskip 0.56917pt\delta}_{\Bb{H}}(x;z)=\Psi^{\,\delta}_{\infty}(x;z)+\dfrac{z^{\delta}}{x^{\delta}z-(-x)^{1-\delta}}\ ,\quad z\in\eusm{E}^{0}_{\!{{\mbox{\tiny{$\frown$}}}}}\subset\Bb{H}\setminus S_{\!{{\mbox{\tiny{$\frown$}}}}}^{\infty}\,, (7.32)

is proved for arbitrary δ{0,1}\delta\in\{0,1\} and xx\in\Bb{R}.

44

\uparrow  We prove (8.11). For z=x+iyz=x+iy\in\Bb{H}, z¯:=x+iy\overline{z}:=x+iy and t{}t\in\Bb{R}\setminus\{0\} the Poisson kernel

2πiPz(t):=1tz1tz¯=1txiy1tx+iy=2iy(tx)2+y2,\displaystyle 2\pi iP_{z}(t):=\dfrac{1}{t-z}-\dfrac{1}{t-\overline{z}}=\dfrac{1}{t-x-iy}-\dfrac{1}{t-x+iy}=\dfrac{2iy}{(t-x)^{2}+y^{2}}\,,

possesses the following property

2πiPz(1/t)t2\displaystyle\dfrac{2\pi iP_{z}(-1/t)}{t^{2}} =1t2(1(1/t)z1(1/t)z¯)=1t(1zt)1t(1tz¯)\displaystyle=\dfrac{1}{t^{2}}\left(\dfrac{1}{(-1/t)-z}-\dfrac{1}{(-1/t)-\overline{z}}\right)=\dfrac{1}{t\left(-1-zt\right)}-\dfrac{1}{t\left(-1-t\overline{z}\right)}
=t(1tz¯)t(1zt)t2(1zt)(1tz¯)=tt2z¯+t+t2zt2(1zt)(1tz¯)\displaystyle=\dfrac{t\left(-1-t\overline{z}\right)-t\left(-1-zt\right)}{t^{2}\left(-1-zt\right)\left(-1-t\overline{z}\right)}=\dfrac{-t-t^{2}\overline{z}+t+t^{2}z}{t^{2}\left(-1-zt\right)\left(-1-t\overline{z}\right)}
=zz¯|z|2(1zt)(1z¯t)=zz¯zz¯(t(1z))(t(1z¯))\displaystyle=\dfrac{z-\overline{z}}{|z|^{2}\left(\dfrac{-1}{z}-t\right)\left(\dfrac{-1}{\overline{z}}-t\right)}=\dfrac{\dfrac{z-\overline{z}}{z\overline{z}}}{\left(t-\left(\dfrac{-1}{z}\right)\right)\left(t-\left(\dfrac{-1}{\overline{z}}\right)\right)}
=(1z)(1z¯)(t(1z))(t(1z¯))=1t(1z)1t(1z¯)\displaystyle=\dfrac{\left(\dfrac{-1}{z}\right)-\left(\dfrac{-1}{\overline{z}}\right)}{\left(t-\left(\dfrac{-1}{z}\right)\right)\left(t-\left(\dfrac{-1}{\overline{z}}\right)\right)}=\dfrac{1}{t-\left(\dfrac{-1}{z}\right)}-\dfrac{1}{t-\left(\dfrac{-1}{\overline{z}}\right)}
=2πiP1/z(t),\displaystyle=2\pi iP_{-1/z}(t)\,,\

i.e.,

Pz(1/t)t2=P1/z(t),z{}\displaystyle\dfrac{P_{z}(-1/t)}{t^{2}}=P_{-1/z}(t)\ ,\quad z\in\Bb{H}\,,\ \ t\in\Bb{R}\setminus\{0\}\,. (A.44a)

Since for every zz\in\Bb{H} the Poisson kernel Pz(t)P_{z}(t) of the variable tt belongs to L1(L^{1}(\Bb{R}), by (1.6) and Jordan’s lemma (8.6), for every n1n\geqslant 1 we have

2πihn(Pz):=2πiPz(t)eiπntdt=eiπnt(1tz1tz¯)dt\displaystyle 2\pi i\eurm{h}_{-n}^{\star}(P_{z})\!:=\!2\pi i\int\limits_{\Bb{R}}\!P_{z}(t)\,e^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}dt=\int\limits_{\Bb{R}}e^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}\left(\dfrac{1}{t-z}-\dfrac{1}{t-\overline{z}}\right)dt
=limA+AAeiπntdttzlimA+AAeiπntdttz¯=2πieiπnz,\displaystyle=\lim\limits_{A\to+\infty}\int\limits_{-A}^{A}\dfrac{e^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}dt}{t-z}-\lim\limits_{A\to+\infty}\int\limits_{-A}^{A}\dfrac{e^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}dt}{t-\overline{z}}=2\pi ie^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}\,,

from which

hn(Pz)=Pz(t)eiπntdt=eiπnz=eiπn(x+iy),n1,z\displaystyle\eurm{h}_{-n}^{\star}(P_{z})=\int\limits_{\Bb{R}}\!P_{z}(t)\,e^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}dt=e^{{{\mbox{\footnotesize{${\rm{i}}\pi nz$}}}}}=e^{{{\mbox{\footnotesize{${\rm{i}}\pi n(x+iy)$}}}}}\ ,\quad n\geqslant 1\ ,\quad z\in\Bb{H}\,. (A.44b)

Applying complex conjugation, taking into account that Pz(t)P_{z}(t)\in\Bb{R} for arbitrary zz\in\Bb{H} and tt\in\Bb{R}, we get

hn(Pz)=Pz(t)eiπntdt=eiπnz¯=eiπn(xiy),n1,z\displaystyle\eurm{h}_{n}^{\star}(P_{z})=\int\limits_{\Bb{R}}\!P_{z}(t)\,e^{{{\mbox{\footnotesize{$-{\rm{i}}\pi nt$}}}}}dt=e^{{{\mbox{\footnotesize{$-{\rm{i}}\pi n\overline{z}$}}}}}=e^{{{\mbox{\footnotesize{$-{\rm{i}}\pi n(x-iy)$}}}}}\ ,\quad n\geqslant 1\ ,\ \ z\in\Bb{H}\,, (A.44c)

while

h0(Pz)\displaystyle\eurm{h}_{0}^{\star}(P_{z}) =Pz(t)dt=1πydt(tx)2+y2=1πarctantxy|t=t=+\displaystyle=\int\limits_{\Bb{R}}\!P_{z}(t)\,dt=\dfrac{1}{\pi}\int\limits_{\Bb{R}}\dfrac{ydt}{(t-x)^{2}+y^{2}}=\dfrac{1}{\pi}\left.\arctan\dfrac{t-x}{y}\right|_{t=-\infty}^{t=+\infty}
=1,z\displaystyle=1\ ,\quad z\in\Bb{H}\,. (A.44d)

Next,

mn(Pz)\displaystyle\eurm{m}_{-n}^{\star}(P_{z}) :=Pz(t)eiπntdt=|t=1/t|=Pz(1/t)t2eiπntdt\displaystyle:=\int\limits_{\Bb{R}}\!P_{z}(t)\,e^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{t}$}}}}}dt=\left|t^{\,\prime}=-1/t\right|=\int\limits_{\Bb{R}}\!\dfrac{P_{z}(-1/t)}{t^{2}}\,e^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}dt
=P1/z(t)eiπntdt=hn(P1/z)=eiπnz=eiπnx+iy,\displaystyle=\int\limits_{\Bb{R}}\!P_{-1/z}(t)\,e^{{{\mbox{\footnotesize{${\rm{i}}\pi nt$}}}}}dt=\eurm{h}_{-n}^{\star}(P_{-1/z})=e^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{z}$}}}}}=e^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{x+iy}$}}}}}\ ,

i.e.,

mn(Pz)=Pz(t)eiπntdt=eiπnz=eiπnx+iy,n1,z\displaystyle\eurm{m}_{-n}^{\star}(P_{z})=\int\limits_{\Bb{R}}\!P_{z}(t)\,e^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{t}$}}}}}dt=e^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{z}$}}}}}=e^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{x+iy}$}}}}}\ ,\quad n\geqslant 1\ ,\quad z\in\Bb{H}\,. (A.44e)

Applying complex conjugation, as above ,we get

mn(Pz)=Pz(t)eiπntdt=eiπnz¯=eiπnxiy,n1,z=x+iy\displaystyle\eurm{m}_{n}^{\star}(P_{z})=\int\limits_{\Bb{R}}\!P_{z}(t)\,e^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{t}$}}}}}dt=e^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{\overline{z}}$}}}}}=e^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{x-iy}$}}}}}\ ,\ \ n\geqslant 1\ ,\ \ z\!=x\!+\!iy\in\Bb{H}\,. (A.44f)

Finally,

hn(Pz)=eiπn(xiy)=eπn(y+ix),mn(Pz)=eiπnxiy=eπny+ix,hn(Pz)=eiπn(x+iy)=eπn(yix),mn(Pz)=eiπnx+iy=eπnyix,\displaystyle\begin{array}[]{ll}\eurm{h}_{n}^{\star}(P_{z})\!=\!e^{{{\mbox{\footnotesize{$-{\rm{i}}\pi n(x-iy)$}}}}}\!=\!e^{{{\mbox{\footnotesize{$-\pi n(y+ix)$}}}}}\,,&\ \ \eurm{m}_{n}^{\star}(P_{z})\!=\!e^{{{\mbox{\footnotesize{$\dfrac{{\rm{i}}\pi n}{x-iy}$}}}}}\!=\!e^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y+ix}$}}}}}\,,\\ \eurm{h}_{-n}^{\star}(P_{z})\!=\!e^{{{\mbox{\footnotesize{${\rm{i}}\pi n(x+iy)$}}}}}\!=\!e^{{{\mbox{\footnotesize{$-\pi n(y-ix)$}}}}}\,,&\ \ \eurm{m}_{-n}^{\star}(P_{z})\!=\!e^{{{\mbox{\footnotesize{$-\dfrac{{\rm{i}}\pi n}{x+iy}$}}}}}\!=\!e^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y-ix}$}}}}}\,,\end{array} (A.44i)

for all n1n\geqslant 1, which together with (A.44c) and Theorem 8.2 give

1πy(tx)2+y2\displaystyle\dfrac{1}{\pi}\dfrac{y}{(t-x)^{2}+y^{2}} =Pz(t)=h0(Pz)H0(t)+n(hn(Pz)Hn(t)+mn(Pz)Mn(t))\displaystyle=P_{z}(t)\!=\!\eurm{h}_{0}^{\star}(P_{z})\eurm{H}_{0}(t)\!+\!\sum\nolimits_{n\in\Bb{Z}_{\neq 0}}\!\Big{(}\eurm{h}_{n}^{\star}(P_{z})\eurm{H}_{n}(t)\!+\!\eurm{m}_{n}^{\star}(P_{z})\eurm{M}_{n}(t)\Big{)}
=H0(t)+n1(eπn(y+ix)Hn(t)+eπny+ixMn(t))\displaystyle=\eurm{H}_{0}(t)\!+\!\sum\limits_{n\geqslant 1}\!\Big{(}e^{{{\mbox{\footnotesize{$-\pi n(y+ix)$}}}}}\eurm{H}_{n}(t)\!+\!e^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y+ix}$}}}}}\eurm{M}_{n}(t)\Big{)}
+n1(eπn(yix)Hn(t)+eπnyixMn(t)),t,x\displaystyle+\sum\limits_{n\geqslant 1}\!\Big{(}e^{{{\mbox{\footnotesize{$-\pi n(y-ix)$}}}}}\eurm{H}_{-n}(t)\!+\!e^{{{\mbox{\footnotesize{$-\dfrac{\pi n}{y-ix}$}}}}}\eurm{M}_{-n}(t)\Big{)}\ ,\quad\ t,x\in\Bb{R},\ y>0\,,

which completes the proof of (8.11).

A.8 . Notes for Section 8

  

45

\uparrow  For arbitrary a,b>0a,b>0 we calculate the integrals (10.130). Obviously,

0eb2(t+1t)2att2+1dtt2+1=0ebt+1t2at+1t2dtt2+1=\displaystyle\int\limits_{0}^{\infty}e^{{{\mbox{\footnotesize{$-\dfrac{b}{2}\left(t+\dfrac{1}{t}\right)-\dfrac{2at}{t^{2}+1}$}}}}}\dfrac{dt}{t^{2}+1}=\int\limits_{0}^{\infty}e^{{{\mbox{\footnotesize{$-b\dfrac{t+\dfrac{1}{t}}{2}-\dfrac{a}{\dfrac{t+\dfrac{1}{t}}{2}}$}}}}}\dfrac{dt}{t^{2}+1}\boxed{=}
2t=t+1t, 2dt=(11t2)dt=(t1t)dtt,dtt=2dtt1t,\displaystyle 2t^{\,\prime}=t+\dfrac{1}{t}\,,\ 2dt^{\,\prime}=\left(1-\dfrac{1}{t^{2}}\right)dt=\left(t-\dfrac{1}{t}\right)\dfrac{dt}{t}\,,\ \dfrac{dt}{t}=\dfrac{2dt^{\,\prime}}{t-\dfrac{1}{t}}\,,
t22tt+1=0,t=t+(t)21,1t=t(t)21\displaystyle t^{2}-2tt^{\,\prime}+1=0\,,\ t=t^{\,\prime}+\sqrt{\left(t^{\,\prime}\right)^{2}-1}\,,\ \dfrac{1}{t}=t^{\,\prime}-\sqrt{\left(t^{\,\prime}\right)^{2}-1}
t1t=2(t)21,dtt2+1=1t+1tdtt=12t2dt2(t)21\displaystyle t-\dfrac{1}{t}=2\sqrt{\left(t^{\,\prime}\right)^{2}-1}\,,\ \dfrac{dt}{t^{2}+1}=\dfrac{1}{t+\dfrac{1}{t}}\dfrac{dt}{t}=\dfrac{1}{2t^{\,\prime}}\dfrac{2dt^{\,\prime}}{2\sqrt{\left(t^{\,\prime}\right)^{2}-1}}
=121ebtatdttt21,\displaystyle\boxed{=}\ \dfrac{1}{2}\int\limits_{1}^{\infty}e^{{{\mbox{\footnotesize{$-bt-\dfrac{a}{t}$}}}}}\dfrac{dt}{t\sqrt{t^{2}-1}}\,,\

and

0eb2(t+1t)2att2+1dtt=0ebt+1t2at+1t2dtt=\displaystyle\int\limits_{0}^{\infty}e^{{{\mbox{\footnotesize{$-\dfrac{b}{2}\left(t+\dfrac{1}{t}\right)-\dfrac{2at}{t^{2}+1}$}}}}}\dfrac{dt}{t}=\int\limits_{0}^{\infty}e^{{{\mbox{\footnotesize{$-b\dfrac{t+\dfrac{1}{t}}{2}-\dfrac{a}{\dfrac{t+\dfrac{1}{t}}{2}}$}}}}}\dfrac{dt}{t}\boxed{=}
2t=t+1t, 2dt=(11t2)dt=(t1t)dtt,dtt=2dtt1t,\displaystyle 2t^{\,\prime}=t+\dfrac{1}{t}\,,\ 2dt^{\,\prime}=\left(1-\dfrac{1}{t^{2}}\right)dt=\left(t-\dfrac{1}{t}\right)\dfrac{dt}{t}\,,\ \dfrac{dt}{t}=\dfrac{2dt^{\,\prime}}{t-\dfrac{1}{t}}\,,
t22tt+1=0,t=t+(t)21,1t=t(t)21\displaystyle t^{2}-2tt^{\,\prime}+1=0\,,\ t=t^{\,\prime}+\sqrt{\left(t^{\,\prime}\right)^{2}-1}\,,\ \dfrac{1}{t}=t^{\,\prime}-\sqrt{\left(t^{\,\prime}\right)^{2}-1}
t1t=2(t)21,dtt=2dt2(t)21\displaystyle t-\dfrac{1}{t}=2\sqrt{\left(t^{\,\prime}\right)^{2}-1}\,,\ \dfrac{dt}{t}=\dfrac{2dt^{\,\prime}}{2\sqrt{\left(t^{\,\prime}\right)^{2}-1}}
=1ebtatdtt21,\displaystyle\boxed{=}\ \int\limits_{1}^{\infty}e^{{{\mbox{\footnotesize{$-bt-\dfrac{a}{t}$}}}}}\dfrac{dt}{\sqrt{t^{2}-1}}\,,\

where (see (erd, , p.​ 82, (19)), (abr, , p.​ 376, 9.6.27)) for x,y>0x,y>0 we have

K0(x)=1extdtt21K0(2x(y+1))=1extytdtt21\displaystyle K_{0}(x)=\int\limits_{1}^{\infty}\dfrac{e^{-xt}dt}{\sqrt{t^{2}-1}}\ \Rightarrow\ K_{0}\left(2\sqrt{x(y+1)}\right)=\int\limits_{1}^{\infty}\dfrac{e^{{{\mbox{\footnotesize{$-xt-\dfrac{y}{t}$}}}}}dt}{\sqrt{t^{2}-1}}\ \Rightarrow\
1extytdttt21=yK0(2x(y+1))=xy+1K1(2x(y+1)),\displaystyle-\int\limits_{1}^{\infty}\dfrac{e^{{{\mbox{\footnotesize{$-xt-\dfrac{y}{t}$}}}}}dt}{t\sqrt{t^{2}-1}}=\dfrac{\partial}{\partial y}K_{0}\left(2\sqrt{x(y+1)}\right)=-\sqrt{\dfrac{x}{y+1}}K_{1}\left(2\sqrt{x(y+1)}\right)\,,\

i.e.,

0eb2(t+1t)2att2+1dtt2+1=12ba+1K1(2b(a+1)),\displaystyle\int\limits_{0}^{\infty}e^{{{\mbox{\footnotesize{$-\dfrac{b}{2}\left(t+\dfrac{1}{t}\right)-\dfrac{2at}{t^{2}+1}$}}}}}\dfrac{dt}{t^{2}+1}=\dfrac{1}{2}\sqrt{\dfrac{b}{a+1}}K_{1}\left(2\sqrt{b(a+1)}\right)\,,
0eb2(t+1t)2att2+1dtt=K0(2b(a+1)),a,b>0,\displaystyle\int\limits_{0}^{\infty}e^{{{\mbox{\footnotesize{$-\dfrac{b}{2}\left(t+\dfrac{1}{t}\right)-\dfrac{2at}{t^{2}+1}$}}}}}\dfrac{dt}{t}=K_{0}\left(2\sqrt{b(a+1)}\right)\,,\ a,b>0\,,

which proves (10.130).

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