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Gabriel’s problem for harmonic Hardy spaces

Suman Das Department of Mathematics with Computer Science, Guangdong Technion - Israel Institute of Technology, Shantou, Guangdong 515063, P. R. China. suman.das@gtiit.edu.cn
Abstract.

We obtain inequalities of the form

C|f(z)|p|dz|A(p)𝕋|f(z)|p|dz|,(p>1)\int_{C}|f(z)|^{p}\,|dz|\leq A(p)\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|,\quad(p>1)

where ff is harmonic in the unit disk 𝔻{\mathbb{D}}, 𝕋{\mathbb{T}} is the unit circle, and CC is any convex curve in 𝔻{\mathbb{D}}. Such inequalities were originally studied for analytic functions by R. M. Gabriel [Proc. London Math. Soc. 28(2), 1928]. We show that these results, unlike in the case of analytic functions, cannot be true in general for 0<p10<p\leq 1. Therefore, we produce an inequality of a slightly different type, which deals with the case 0<p<10<p<1. An example is given to show that this result is “best possible”, in the sense that an extension to p=1p=1 fails. Then we consider the special case when CC is a circle, and prove a refined result which surprisingly holds for p=1p=1 as well. We conclude with a maximal theorem which has potential applications.

Key words and phrases:
Hardy space, Harmonic functions, Subharmonic functions, Gabriel’s problem, Riesz-Fejér inequality
2020 Mathematics Subject Classification:
31A05, 30H10

1. Introduction and Main Result

Suppose 𝔻={z:|z|<1}{\mathbb{D}}=\{z\in{\mathbb{C}}:\,|z|<1\} is the open unit disk in the complex plane {\mathbb{C}}, and 𝕋={z:|z|=1}\mathbb{T}=\{z\in{\mathbb{C}}:|z|=1\} is the unit circle. For a function ff analytic in 𝔻{\mathbb{D}}, the integral mean Mp(r,f)M_{p}(r,f) is defined as

Mp(r,f)=(12π02π|f(reiθ)|p𝑑θ)1p,p>0.M_{p}(r,f)=\left(\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{i\theta})|^{p}\,d\theta\right)^{\frac{1}{p}},~{}~{}~{}p>0.

The function ff is said to be in the Hardy space HpH^{p} if Mp(r,f)M_{p}(r,f) remains bounded as r1r\to 1^{-}. It is well known that every function fHpf\in H^{p} has a radial limit almost everywhere. We denote by f(eiθ)f(e^{i\theta}) the radial limit of ff on 𝕋{\mathbb{T}}. For a detailed survey on Hardy spaces, we refer to the books of Duren [5] and Koosis [17].

In [11], R. M. Gabriel considered the integral of |f|p|f|^{p} along certain curves, and proved the following result.

Theorem A.

[11] Let fHpf\in H^{p}, p>0p>0. If CC is any convex curve in 𝔻{\mathbb{D}}, then

C|f(z)|p|dz|2𝕋|f(z)|p|dz|.\int_{C}|f(z)|^{p}\,|dz|\leq 2\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|.

The constant 22 on the right hand side is best possible.

Indeed, the theorem was proved for an arbitrary circle Γ\Gamma (instead of 𝕋{\mathbb{T}}) and analyticity was assumed on and inside Γ\Gamma. However, that hypothesis is readily replaced by the weaker condition fHpf\in H^{p}.

This problem originated through a famous inequality of Riesz and Fejér, where CC is the diameter 1x1-1\leq x\leq 1 and the constant on the right hand side is 1/21/2 (see Theorem 3.13 of [5]). The study of similar inequalities continued in a series of individual papers by Gabriel and Frazer, including but possibly not limited to, [10, 11, 12, 13, 6, 7, 8, 9]. Later, Granados termed these as “Gabriel’s problem” in her review [14], which contains the background and subsequent developments, as well as open problems.

In this paper, we study Gabriel’s problem for complex-valued harmonic functions. A function f=u+ivf=u+iv is harmonic in 𝔻{\mathbb{D}}, if uu and vv are real-valued harmonic functions in 𝔻{\mathbb{D}}. Every such function has a decomposition f=h+g¯f=h+\overline{g}, where hh and gg are analytic functions in 𝔻\mathbb{D}. This decomposition is unique up to an additive constant. We say that the function ff belongs to the harmonic Hardy space hph^{p} (p>0)(p>0) if

sup0<r<1Mp(r,f)<.\sup_{0<r<1}M_{p}(r,f)<\infty.

In recent times, this problem resurfaced when Kayumov, Ponnusamy and Sairam Kaliraj [16] obtained the harmonic analogue of the Riesz-Fejér inequality for the space hph^{p}, p(1,2]p\in(1,2].

Theorem B.

[16] If fhpf\in h^{p} for 1<p21<p\leq 2, then

11|f(x)|p𝑑x12secp(π2p)02π|f(eiθ)|p𝑑θ.\int_{-1}^{1}|f(x)|^{p}\,dx\leq\frac{1}{2}\sec^{p}\left(\frac{\pi}{2p}\right)\int_{0}^{2\pi}|f(e^{i\theta})|^{p}\,d\theta.

The inequality is sharp.

Later Melentijević and Božin [19] showed that this sharp result holds for p>2p>2 as well. In [2], a slight variant of this inequality was established by the present author and Sairam Kaliraj. Very recently, Chen and Hamada [1] have produced a Riesz-Fejér type inequality for pluriharmonic functions in 2n{\mathbb{R}}^{2n}, thereby extending the problem to higher dimensions.

Continuing in this direction, in this paper, we prove Gabriel’s result (Theorem A) for the harmonic Hardy space hph^{p}. The main theorem is as follows.

Theorem 1.

Let fhpf\in h^{p} for some p>1p>1, and let CC be any convex curve in 𝔻{\mathbb{D}}. Then

C|f(z)|p|dz|4𝕋|f(z)|p|dz|,\int_{C}|f(z)|^{p}\,|dz|\leq 4\int_{\mathbb{T}}|f(z)|^{p}\,|dz|,

if p2p\geq 2, and

C|f(z)|p|dz|2secp(π2p)𝕋|f(z)|p|dz|,\int_{C}|f(z)|^{p}\,|dz|\leq 2\sec^{p}\left(\frac{\pi}{2p}\right)\int_{\mathbb{T}}|f(z)|^{p}\,|dz|,

if 1<p<21<p<2.

We give the proof of this result, followed by its consequences, in Section 2. In Section 3, we consider the special case when CC is a circle, and produce a refinement of Theorem 1.

2. Proof and Further Implications

2.1. The case p2p\geq 2:

We start with the simple case p=2p=2. If f=h+g¯h2f=h+\bar{g}\in h^{2}, we observe that

𝕋|f(z)|2|dz|=𝕋|h(z)|2|dz|+𝕋|g(z)|2|dz|,\displaystyle\int_{{\mathbb{T}}}|f(z)|^{2}\,|dz|=\int_{{\mathbb{T}}}|h(z)|^{2}\,|dz|+\int_{{\mathbb{T}}}|g(z)|^{2}\,|dz|,

so that h,gH2h,g\in H^{2}. Therefore, Theorem A gives

C|f(z)|2|dz|\displaystyle\int_{C}|f(z)|^{2}\,|dz| C(|h(z)|+|g(z)|)2|dz|\displaystyle\leq\int_{C}(|h(z)|+|g(z)|)^{2}\,|dz|
2[C|h(z)|2|dz|+C|g(z)|2|dz|]\displaystyle\leq 2\left[\int_{C}|h(z)|^{2}\,|dz|+\int_{C}|g(z)|^{2}\,|dz|\right]
4[𝕋|h(z)|2|dz|+𝕋|g(z)|2|dz|]\displaystyle\leq 4\left[\int_{{\mathbb{T}}}|h(z)|^{2}\,|dz|+\int_{{\mathbb{T}}}|g(z)|^{2}\,|dz|\right]
=4𝕋|f(z)|2|dz|.\displaystyle=4\int_{{\mathbb{T}}}|f(z)|^{2}\,|dz|.

Now assume that fhpf\in h^{p}, p>2p>2. Let us recall that ff has the Poisson integral representation

f(reiθ)=12π02π1r212rcos(θt)+r2f(eit)𝑑t.f(r{e^{i\theta}})=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}f(e^{it})\,dt.

Also, it is known that the Poisson integral of a function φL1(𝕋)\varphi\in L^{1}({\mathbb{T}}) is a harmonic function in 𝔻{\mathbb{D}}. For z=reiθz=r{e^{i\theta}}, an appeal to Jensen’s inequality shows that

C|f(z)|p|dz|\displaystyle\int_{C}|f(z)|^{p}\,|dz| C(12π02π1r212rcos(θt)+r2|f(eit)|𝑑t)p2.2|dz|\displaystyle\leq\int_{C}\left(\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}|f(e^{it})|\,dt\right)^{\frac{p}{2}.2}\,|dz|
C(12π02π1r212rcos(θt)+r2|f(eit)|p/2𝑑t)2|dz|.\displaystyle\leq\int_{C}\left(\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}|f(e^{it})|^{p/2}\,dt\right)^{2}\,|dz|.

Let us write

U(z)=12π02π1r212rcos(θt)+r2|f(eit)|p/2𝑑t,z=reiθ.U(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}|f(e^{it})|^{p/2}\,dt,\quad z=r{e^{i\theta}}.

Then UU is non-negative and harmonic in 𝔻{\mathbb{D}} such that U(eiθ)=|f(eiθ)|p/2U({e^{i\theta}})=|f({e^{i\theta}})|^{p/2}. Therefore, from what we have already proved,

C|f(z)|p|dz|\displaystyle\int_{C}|f(z)|^{p}\,|dz| CU2(z)|dz|\displaystyle\leq\int_{C}U^{2}(z)\,|dz|
4𝕋U2(z)|dz|\displaystyle\leq 4\int_{{\mathbb{T}}}U^{2}(z)\,|dz|
=4𝕋|f(z)|p|dz|,\displaystyle=4\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|,

as desired.

2.2. The case 1<p<21<p<2:

This requires little bit more technicalities. A positive real-valued function uu is called log-subharmonic, if logu\log u is subharmonic. To prove this part of the theorem, we make use of the following classical result of Lozinski [18].

Theorem C.

[18] Suppose Φ\Phi is a log-subharmonic function from 𝔻{\mathbb{D}} to {\mathbb{R}}, such that

sup0<r<102πΦp(reiθ)𝑑θ<,p>0.\sup_{0<r<1}\int_{0}^{2\pi}\Phi^{p}(re^{i\theta})\,d\theta<\infty,\quad p>0.

Then there exists fHpf\in H^{p} such that Φ(z)|f(z)|\Phi(z)\leq|f(z)| for z𝔻z\in{\mathbb{D}}, and Φ(eiθ)=|f(eiθ)|\Phi({e^{i\theta}})=|f({e^{i\theta}})| for almost every θ\theta.

This allows us to prove the next lemma which is essential for our purpose.

Lemma 1.

Let f=h+g¯hpf=h+\bar{g}\in h^{p} for some p>1p>1. Then

C(|h(z)|+|g(z)|)p|dz|2𝕋(|h(z)|+|g(z)|)p|dz|.\int_{C}(|h(z)|+|g(z)|)^{p}\,|dz|\leq 2\int_{\mathbb{T}}(|h(z)|+|g(z)|)^{p}\,|dz|.

The coefficient 22 on the right-hand side is sharp.

Proof.

If fhpf\in h^{p} for p>1p>1, an easy argument shows that h,gHph,g\in H^{p} (see, for example, Lemma 1 of [3] for the more general result). This ensures that the integral on the right-hand side is bounded. Therefore, in order to apply Theorem C, it is enough to show that log(|h(z)|+|g(z)|)\log(|h(z)|+|g(z)|) is subharmonic.

Let z0𝔻z_{0}\in{\mathbb{D}} be such that h(z0)0h(z_{0})\neq 0. Let us write

log(|h(z)|+|g(z)|)=log|h(z)|+log(1+|g(z)h(z)|)\log(|h(z)|+|g(z)|)=\log|h(z)|+\log\left(1+\left|\frac{g(z)}{h(z)}\right|\right)

in a neighbourhood of z0z_{0}. The function log|h(z)|\log|h(z)| is subharmonic. The second function is of the form log(1+eG)\log(1+e^{G}), where G=log|g/h|G=\log|g/h| is subharmonic in a neighbourhood of z0z_{0}. The function φ(x)=log(1+ex)\varphi(x)=\log(1+e^{x}) is convex and increasing, this can be checked by taking the derivatives. It is well-known that the composition of a convex, increasing function with a subharmonic function is subharmonic. Therefore, so is φG=log(1+|g(z)/h(z)|)\varphi\circ G=\log(1+|g(z)/h(z)|). So, log(|h(z)|+|g(z)|)\log(|h(z)|+|g(z)|) is the sum of two subharmonic functions, hence subharmonic.

Now, let z1𝔻z_{1}\in{\mathbb{D}} be a zero of hh. If g(z1)0g(z_{1})\neq 0, the above argument applies with hh and gg interchanged. If g(z1)=0g(z_{1})=0, then log(|h(z1)|+|g(z1)|)=\log(|h(z_{1})|+|g(z_{1})|)=-\infty, so that the local sub-mean-value property readily holds at z1z_{1}. This shows that log(|h(z)|+|g(z)|)\log(|h(z)|+|g(z)|) is subharmonic in the whole 𝔻{\mathbb{D}}.

Therefore, Theorem C implies that there exists FHpF\in H^{p} such that

|h(z)|+|g(z)||F(z)| in 𝔻,|h(eiθ)|+|g(eiθ)|=|F(eiθ)| a.e.|h(z)|+|g(z)|\leq|F(z)|\,\text{ in }{\mathbb{D}},\quad|h({e^{i\theta}})|+|g({e^{i\theta}})|=|F({e^{i\theta}})|\,\text{ a.e.}

It follows that

C(|h(z)|+|g(z)|)p|dz|\displaystyle\int_{C}(|h(z)|+|g(z)|)^{p}\,|dz| C|F(z)|p|dz|\displaystyle\leq\int_{C}|F(z)|^{p}\,|dz|
2𝕋|F(z)|p|dz|(Theorem A)\displaystyle\leq 2\int_{{\mathbb{T}}}|F(z)|^{p}\,|dz|\quad(\text{Theorem \ref{theGab}})
=2𝕋(|h(z)|+|g(z)|)p|dz|\displaystyle=2\int_{\mathbb{T}}(|h(z)|+|g(z)|)^{p}\,|dz|

and the proof of the lemma is complete. The sharpness follows from Theorem A. ∎

We also require the following inequality of Kalaj, the proof of which employs a method of plurisubharmonic functions.

Theorem D.

[15] Let 1<p<1<p<\infty and let f=h+g¯hpf=h+\overline{g}\in h^{p} with Re(h(0)g(0))=0{\operatorname{Re}\,}(h(0)g(0))=0. Then

𝕋(|h(z)|2+|g(z)|2)p/2|dz|1(1|cosπp|)p/2𝕋|f(z)|p|dz|.\int_{\mathbb{T}}(|h(z)|^{2}+|g(z)|^{2})^{p/2}\,|dz|\leq\frac{1}{\left(1-|\cos\frac{\pi}{p}|\right)^{p/2}}\int_{\mathbb{T}}|f(z)|^{p}\,|dz|.

The inequality is sharp.

Now we are ready to resume the proof of the case 1<p<21<p<2. Suppose, without any loss of generality, g(0)=0g(0)=0. We see that

C|f(z)|p|dz|\displaystyle\int_{C}|f(z)|^{p}\,|dz| C(|h(z)|+|g(z)|)p|dz|\displaystyle\leq\int_{C}(|h(z)|+|g(z)|)^{p}\,|dz|
2𝕋(|h(z)|+|g(z)|)p|dz|(Lemma 1)\displaystyle\leq 2\int_{{\mathbb{T}}}(|h(z)|+|g(z)|)^{p}\,|dz|\quad(\text{Lemma \ref{Dlem}})
2p2+1𝕋(|h(z)|2+|g(z)|2)p/2|dz|\displaystyle\leq 2^{\frac{p}{2}+1}\int_{{\mathbb{T}}}(|h(z)|^{2}+|g(z)|^{2})^{p/2}\,|dz|
2p2+1(1|cosπp|)p/2𝕋|f(z)|p|dz|,\displaystyle\leq\frac{2^{\frac{p}{2}+1}}{\left(1-|\cos\frac{\pi}{p}|\right)^{p/2}}\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|,

where the last inequality follows by Theorem D. Finally, we see that, for 1<p<21<p<2,

2p2+1(1|cosπp|)p/2=2secp(π2p),\frac{2^{\frac{p}{2}+1}}{\left(1-|\cos\frac{\pi}{p}|\right)^{p/2}}=2\sec^{p}\left(\frac{\pi}{2p}\right),

which completes the proof of the theorem.

2.3. Sharpness and further ramifications

This constant is best possible, in the sense that one cannot get rid of the factor secp(π2p)\sec^{p}(\frac{\pi}{2p}). This can be easily seen from Theorem B, which is contained in Theorem 1 (i.e., CC to be taken as the diameter 1x1-1\leq x\leq 1). Let us mention that the function

f(z)=Re{(1z2)1p}(z𝔻)f(z)={\operatorname{Re}\,}\left\{(1-z^{2})^{-\frac{1}{p}}\right\}\quad(z\in{\mathbb{D}})

works as an extremal function in this case (see [19]).

Also, we observe that

secp(π2p)+ as p1.\sec^{p}\left(\frac{\pi}{2p}\right)\to+\infty\ \textit{ as }\ p\to 1.

This allows us to show that, unlike in the case of analytic functions, results of this type are not true in general for 0<p10<p\leq 1.

Theorem 2.

The inequality in Theorem 1 does not generally hold for the case 0<p10<p\leq 1.

Proof.

Suppose, on the contrary, that there is p1(0,1]p_{1}\in(0,1] such that the inequality

C|f(z)|p1|dz|K(p1)𝕋|f(z)|p1|dz|\int_{C}|f(z)|^{p_{1}}\,|dz|\leq K(p_{1})\int_{\mathbb{T}}|f(z)|^{p_{1}}\,|dz|

holds, where K(p1)K(p_{1}) is a constant depending only on p1p_{1}. For p>p1p>p_{1} and z=reiθz=r{e^{i\theta}}, we use Poisson integral and Jensen’s inequality to deduce that

C|f(z)|p|dz|\displaystyle\int_{C}|f(z)|^{p}\,|dz| C(12π02π1r212rcos(θt)+r2|f(eit)|𝑑t)pp1p1|dz|\displaystyle\leq\int_{C}\left(\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}|f(e^{it})|\,dt\right)^{\frac{p}{p_{1}}\cdot{p_{1}}}\,|dz|
C(12π02π1r212rcos(θt)+r2|f(eit)|p/p1𝑑t)p1|dz|\displaystyle\leq\int_{C}\left(\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}|f(e^{it})|^{p/p_{1}}\,dt\right)^{p_{1}}\,|dz|
K(p1)𝕋|f(z)|pp1p1|dz|\displaystyle\leq K(p_{1})\int_{\mathbb{T}}|f(z)|^{\frac{p}{p_{1}}\cdot{p_{1}}}\,|dz|
=K(p1)𝕋|f(z)|p|dz|.\displaystyle=K(p_{1})\int_{\mathbb{T}}|f(z)|^{p}\,|dz|.

We can now choose p=1+ϵp=1+\epsilon, where ϵ>0\epsilon>0 is sufficiently small, to arrive at a contradiction. ∎

However, we produce a workaround in the form of the next result, which deals with the case 0<p<10<p<1.

Theorem 3.

Let fh1f\in h^{1} and CC be any convex curve in 𝔻{\mathbb{D}}. Then, for 0<p<10<p<1,

C|f(z)|p|dz|A(p)(𝕋|f(z)||dz|)p,\int_{C}|f(z)|^{p}\,|dz|\leq A(p)\left(\int_{{\mathbb{T}}}|f(z)|\,|dz|\right)^{p},

where

A(p)=2(2π)1p[1+sec(πp2)].A(p)=2(2\pi)^{1-p}\left[1+\sec\left(\frac{\pi p}{2}\right)\right].
Proof.

Let U(z)U(z) be the Poisson integral of |f(eit)||f(e^{it})|. Then U(z)U(z) is a non-negative harmonic function in 𝔻{\mathbb{D}} with U(eit)=|f(eit)|U(e^{it})=|f(e^{it})|. Since |f(z)||f(z)| is subharmonic in 𝔻{\mathbb{D}}, it follows (see [5, p. 7]) that

|f(z)|U(z),z𝔻.|f(z)|\leq U(z),\quad z\in{\mathbb{D}}.

Suppose V(z)V(z) is the harmonic conjugate of UU such that V(0)=0V(0)=0. Let us write F(z)=U(z)+iV(z)F(z)=U(z)+iV(z). Then, we see that

C|f(z)|p|dz|CUp(z)|dz|\displaystyle\int_{C}|f(z)|^{p}\,|dz|\leq\int_{C}U^{p}(z)\,|dz| C|F(z)|p|dz|\displaystyle\leq\int_{C}|F(z)|^{p}\,|dz|
2𝕋|F(z)|p|dz|(Theorem A)\displaystyle\leq 2\int_{{\mathbb{T}}}|F(z)|^{p}\,|dz|\quad(\text{Theorem \ref{theGab}})
=2𝕋[U2(z)+V2(z)]p/2|dz|\displaystyle=2\int_{{\mathbb{T}}}\left[U^{2}(z)+V^{2}(z)\right]^{p/2}\,|dz|
2[𝕋Up(z)|dz|+𝕋|V(z)|p|dz|].\displaystyle\leq 2\left[\int_{{\mathbb{T}}}U^{p}(z)\,|dz|+\int_{{\mathbb{T}}}|V(z)|^{p}\,|dz|\right].

As 1/p>11/p>1, Jensen’s inequality implies

(12π𝕋Up(z)|dz|)1/p12π𝕋U(z)|dz|,\left(\frac{1}{2\pi}\int_{{\mathbb{T}}}U^{p}(z)\,|dz|\right)^{1/p}\leq\frac{1}{2\pi}\int_{{\mathbb{T}}}U(z)\,|dz|,

so that

𝕋Up(z)|dz|(2π)1p(𝕋U(z)|dz|)p.\int_{{\mathbb{T}}}U^{p}(z)\,|dz|\leq(2\pi)^{1-p}\left(\int_{{\mathbb{T}}}U(z)\,|dz|\right)^{p}.

Also, a well-known theorem of Kolmogorov (see [5, p. 57]) leads to the inequality

𝕋|V(z)|p|dz|(2π)1psec(πp2)(𝕋U(z)|dz|)p.\int_{{\mathbb{T}}}|V(z)|^{p}\,|dz|\leq(2\pi)^{1-p}\sec\left(\frac{\pi p}{2}\right)\left(\int_{{\mathbb{T}}}U(z)\,|dz|\right)^{p}.

Combining these, we find that

C|f(z)|p|dz|\displaystyle\int_{C}|f(z)|^{p}\,|dz| 2(2π)1p[1+sec(πp2)](𝕋U(z)|dz|)p\displaystyle\leq 2(2\pi)^{1-p}\left[1+\sec\left(\frac{\pi p}{2}\right)\right]\left(\int_{{\mathbb{T}}}U(z)\,|dz|\right)^{p}
=A(p)(𝕋|f(z)||dz|)p,\displaystyle=A(p)\left(\int_{{\mathbb{T}}}|f(z)|\,|dz|\right)^{p},

which completes the proof. ∎

We are unable to check the sharpness of this result. Nevertheless, we show that A(p)A(p) cannot be refined to remove the term sec(πp2)\sec\left(\frac{\pi p}{2}\right). Let

f(z)=Re[(1+z1z)p],f(z)={\operatorname{Re}\,}\left[\left(\frac{1+z}{1-z}\right)^{p}\right],

and suppose CC is the diameter 1x1-1\leq x\leq 1. It is obvious that fh1f\in h^{1}, since the function

L(z)=1+z1zHpL(z)=\frac{1+z}{1-z}\in H^{p}

for every p<1p<1. Now,

C|f(z)|p|dz|=11(1+x1x)p2𝑑x.\int_{C}|f(z)|^{p}\,|dz|=\int_{-1}^{1}\left(\frac{1+x}{1-x}\right)^{p^{2}}\,dx.

This integral converges by the Riesz-Fejér inequality. On the other hand,

𝕋|f(z)||dz|\displaystyle\int_{{\mathbb{T}}}|f(z)|\,|dz| =02π|1+eit1eit|p|cos(parg1+eit1eit)|𝑑t\displaystyle=\int_{0}^{2\pi}\left|\frac{1+e^{it}}{1-e^{it}}\right|^{p}\left|\cos\left(p\arg\frac{1+e^{it}}{1-e^{it}}\right)\right|\,dt
=cos(πp2)02π|1+eit1eit|p𝑑t.\displaystyle=\cos\left(\frac{\pi p}{2}\right)\int_{0}^{2\pi}\left|\frac{1+e^{it}}{1-e^{it}}\right|^{p}\,dt.

The last integral is finite since LHpL\in H^{p}. It follows that

C|f(z)|p|dz|𝕋|f(z)||dz|=Ksec(πp2),\frac{\displaystyle\int_{C}|f(z)|^{p}\,|dz|}{\displaystyle\int_{{\mathbb{T}}}|f(z)|\,|dz|}=K\sec\left(\frac{\pi p}{2}\right),

where KK is an absolute constant. Thus, A(p)A(p) always contains a factor of sec(πp2)\sec\left(\frac{\pi p}{2}\right).

Remark 1.

This example further confirms that no general result of this type is true for p=1p=1, i.e., inequalities of the form

C|f(z)||dz|B𝕋|f(z)||dz|(B>0 constant)\int_{C}|f(z)|\,|dz|\leq B\int_{{\mathbb{T}}}|f(z)|\,|dz|\quad(B>0\text{ constant})

are typically not possible.

3. A Special Case

In [9], Frazer produced the “nicest” result of this type: when CC is a circle. In this case, the sharp constant turns out to be 11.

Theorem E.

[9] If fHpf\in H^{p} for some p>0p>0 and CC is a circle in 𝔻{\mathbb{D}}, then

C|f(z)|p|dz|𝕋|f(z)|p|dz|.\int_{C}|f(z)|^{p}\,|dz|\leq\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|.

Clearly the inequality is sharp.

Here we obtain the harmonic analogue of this inequality, and show that the constant on the right hand side is always less than 2. Curiously, this result also holds for p=1p=1. We do not know if this remains true for 0<p<10<p<1, as the reasoning of Theorem 2 fails.

Theorem 4.

Let fhpf\in h^{p} for some p1p\geq 1, and let CC be any circle in 𝔻{\mathbb{D}}. Then

C|f(z)|p|dz|𝕋|f(z)|p|dz|,\int_{C}|f(z)|^{p}\,|dz|\leq\int_{\mathbb{T}}|f(z)|^{p}\,|dz|,

if p2p\geq 2, and

C|f(z)|p|dz|<(1+r)𝕋|f(z)|p|dz|,\int_{C}|f(z)|^{p}\,|dz|<(1+r)\int_{\mathbb{T}}|f(z)|^{p}\,|dz|,

if 1p<2,1\leq p<2, where rr is the distance of the centre of CC from the origin. These inequalities are best possible.

Proof.

The case p2p\geq 2 will be handled as before. If f=h+g¯h2f=h+\bar{g}\in h^{2}, we see that

C|f(z)|2|dz|=C|h(z)|2|dz|+C|g(z)|2|dz|+2ReCh(z)g(z)|dz|.\int_{C}|f(z)|^{2}\,|dz|=\int_{C}|h(z)|^{2}\,|dz|+\int_{C}|g(z)|^{2}\,|dz|+2{\operatorname{Re}\,}\int_{C}h(z)g(z)\,|dz|.

The last integral is 0 since hghg is analytic in 𝔻{\mathbb{D}}. Therefore,

C|f(z)|2|dz|\displaystyle\int_{C}|f(z)|^{2}\,|dz| =C|h(z)|2|dz|+C|g(z)|2|dz|\displaystyle=\int_{C}|h(z)|^{2}\,|dz|+\int_{C}|g(z)|^{2}\,|dz|
𝕋|h(z)|2|dz|+𝕋|g(z)|2|dz|(Theorem E)\displaystyle\leq\int_{{\mathbb{T}}}|h(z)|^{2}\,|dz|+\int_{{\mathbb{T}}}|g(z)|^{2}\,|dz|\quad(\text{Theorem \ref{Frazer}})
=𝕋|f(z)|2|dz|.\displaystyle=\int_{{\mathbb{T}}}|f(z)|^{2}\,|dz|.

The result for p>2p>2 can be easily obtained using Poisson integral and Jensen’s inequality. The details are omitted.

The case 1p<21\leq p<2 is more delicate. Suppose z0=reiθz_{0}=r{e^{i\theta}} is the centre of CC and the radius is ρ<1|z0|=1r\rho<1-|z_{0}|=1-r. It is known that for p1p\geq 1, the function |f(z)|p|f(z)|^{p} is subharmonic in 𝔻{\mathbb{D}}. Let U(z)U(z) be the Poisson integral of |f(eit)|p|f(e^{it})|^{p}. Then UU is non-negative and harmonic in 𝔻{\mathbb{D}}, with U(eit)=|f(eit)|pU(e^{it})=|f(e^{it})|^{p}. It follows, like in the proof of Theorem 3, that

|f(z)|pU(z),z𝔻.|f(z)|^{p}\leq U(z),\quad z\in{\mathbb{D}}.

The mean value property implies

CU(z)|dz|=2πρU(z0).\int_{C}U(z)\,|dz|=2\pi\rho U(z_{0}).

Therefore, using these and the Poisson integral representation of U, we get

C|f(z)|p|dz|\displaystyle\int_{C}|f(z)|^{p}\,|dz| CU(z)|dz|\displaystyle\leq\int_{C}U(z)\,|dz|
=2πρ[12π02π1r212rcos(θt)+r2U(eit)𝑑t]\displaystyle=2\pi\rho\left[\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}U(e^{it})\,dt\right]
=ρ02π1r212rcos(θt)+r2|f(eit)|p𝑑t\displaystyle=\rho\int_{0}^{2\pi}\frac{1-r^{2}}{1-2r\cos(\theta-t)+r^{2}}|f(e^{it})|^{p}\,dt
<(1r)02π1r2(1r)2|f(eit)|p𝑑t\displaystyle<(1-r)\int_{0}^{2\pi}\frac{1-r^{2}}{(1-r)^{2}}|f(e^{it})|^{p}\,dt
=(1+r)02π|f(eit)|p𝑑t\displaystyle=(1+r)\int_{0}^{2\pi}|f(e^{it})|^{p}\,dt
=(1+r)𝕋|f(z)|p|dz|.\displaystyle=(1+r)\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|.

The sharpness can be seen by letting r0r\to 0, as CC and 𝕋{\mathbb{T}} become concentric. This completes the proof. ∎

We conclude this paper with another inequality which is closely related to the special case of circle. For this purpose, we recall the following result from [2], which is the consequence of a Riesz-Fejér type inequality.

Theorem F.

[2] Let {an}\{a_{n}\}, {bn}\{b_{n}\} be sequences of real numbers and θ\theta be any acute angle. Then

k=0l=0(akal+bkbl)cos(kl)θ2+2akblcos(k+l)θ2k+l+12πsinθ2+cosθ2n=0(an2+bn2).\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{(a_{k}a_{l}+b_{k}b_{l})\cos(k-l)\frac{\theta}{2}+2a_{k}b_{l}\cos(k+l)\frac{\theta}{2}}{k+l+1}\leq\frac{2\pi}{\sin\frac{\theta}{2}+\cos\frac{\theta}{2}}\sum_{n=0}^{\infty}(a_{n}^{2}+b_{n}^{2}).

It is worthwhile to mention that for an=bna_{n}=b_{n} and θ=0\theta=0, this reduces to Hilbert’s inequality. As an application of this, we obtain the next result which is somewhat along the line of, but far less intricate than, the Hardy-Littlewood maximal theorems.

Theorem 5.

If fhpf\in h^{p} for some p2p\geq 2, then

01max|z|=r|f(z)|pdr𝕋|f(z)|p|dz|.\int_{0}^{1}\max_{|z|=r}|f(z)|^{p}\,dr\leq\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|.
Proof.

It is enough to prove the theorem for p=2p=2, as the the case p>2p>2 can be deduced using our earlier techniques. Let us write

f(z)=h(z)+g(z)¯=k=0akzk+k=0bkzk¯.f(z)=h(z)+\overline{g(z)}=\sum_{k=0}^{\infty}a_{k}z^{k}+\overline{\sum_{k=0}^{\infty}b_{k}z^{k}}.

It is easy to see that

|f(reiθ)|k=0(|ak|+|bk|)rk,|f(r{e^{i\theta}})|\leq\sum_{k=0}^{\infty}(|a_{k}|+|b_{k}|)r^{k},

and therefore

01max|z|=r|f(z)|2dr\displaystyle\int_{0}^{1}\max_{|z|=r}|f(z)|^{2}\,dr 01[k=0(|ak|+|bk|)rk]2𝑑r\displaystyle\leq\int_{0}^{1}\left[\sum_{k=0}^{\infty}(|a_{k}|+|b_{k}|)r^{k}\right]^{2}\,dr
=01[k=0l=0(|ak||al|+|bk||bl|+2|ak||bl|)rk+l]𝑑r\displaystyle=\int_{0}^{1}\left[\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}(|a_{k}||a_{l}|+|b_{k}||b_{l}|+2|a_{k}||b_{l}|)r^{k+l}\right]\,dr
=k=0l=0|ak||al|+|bk||bl|+2|ak||bl|k+l+1.\displaystyle=\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{|a_{k}||a_{l}|+|b_{k}||b_{l}|+2|a_{k}||b_{l}|}{k+l+1}.

It follows from Theorem F, with θ=0\theta=0, that

k=0l=0|ak||al|+|bk||bl|+2|ak||bl|k+l+12πn=0(|an|2+|bn|2)=𝕋|f(z)|2|dz|\displaystyle\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{|a_{k}||a_{l}|+|b_{k}||b_{l}|+2|a_{k}||b_{l}|}{k+l+1}\leq 2\pi\sum_{n=0}^{\infty}(|a_{n}|^{2}+|b_{n}|^{2})=\int_{{\mathbb{T}}}|f(z)|^{2}\,|dz|

and the proof is complete. ∎

Remark 2.

This result can be particularly useful in finding the lower bounds of the hph^{p}-norms of functions, defined as

fp=(12π𝕋|f(z)|p|dz|)1/p.\|f\|_{p}=\left(\frac{1}{2\pi}\int_{{\mathbb{T}}}|f(z)|^{p}\,|dz|\right)^{1/p}.

This is often more challenging than finding the upper bounds. For example, in the case of univalent harmonic functions, several upper bounds are given in [3] and [4]. But no such lower bounds are available in the literature, to the best of our knowledge.

Acknowledgement

The research was partially supported by Natural Science Foundation of Guangdong Province (Grant no. 2024A1515010467), and Li Ka Shing Foundation STU-GTIIT Joint-Research Grant (2024LKSFG06). The author is indebted to Antti Rasila and Anbareeswaran Sairam Kaliraj for useful suggestions that improved the quality of the paper.

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