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Increasing rate of weighted product of partial quotients in continued fractions

Ayreena Bakhtawar a.bakhtawar@unsw.edu.au Jing Feng jingfeng0425@gmail.com School of Mathematics and Statistics, University of New South Wales, Sydney NSW 2052, Australia. School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan, 430074 PR China and LAMA UMR 8050, CNRS, Université Paris-Est Créteil, 61 Avenue du Général de Gaulle, 94010 Créteil Cedex, France
Abstract

Let [a1(x),a2(x),,an(x),][a_{1}(x),a_{2}(x),\cdots,a_{n}(x),\cdots] be the continued fraction expansion of x[0,1)x\in[0,1). In this paper, we study the increasing rate of the weighted product ant0(x)an+1t1(x)an+mtm(x)a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x) ,where ti+(0im)t_{i}\in\mathbb{R}_{+}\ (0\leq i\leq m) are weights. More precisely, let φ:+\varphi:\mathbb{N}\to\mathbb{R}_{+} be a function with φ(n)/n\varphi(n)/n\to\infty as nn\to\infty. For any (t0,,tm)+m+1(t_{0},\cdots,t_{m})\in\mathbb{R}^{m+1}_{+} with ti0t_{i}\geq 0 and at least one ti0(0im)t_{i}\neq 0\ (0\leq i\leq m), the Hausdorff dimension of the set

E¯({ti}i=0m,φ)={x[0,1):lim infnlog(ant0(x)an+1t1(x)an+mtm(x))φ(n)=1}\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi)=\left\{x\in[0,1):\liminf\limits_{n\to\infty}\dfrac{\log\left(a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\right)}{\varphi(n)}=1\right\}

is obtained. Under the condition that (t0,,tm)+m+1(t_{0},\cdots,t_{m})\in\mathbb{R}^{m+1}_{+} with 0<t0t1tm0<t_{0}\leq t_{1}\leq\cdots\leq t_{m}, we also obtain the Hausdorff dimension of the set

E¯({ti}i=0m,φ)={x[0,1):lim supnlog(ant0(x)an+1t1(x)an+mtm(x))φ(n)=1}.\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi)=\left\{x\in[0,1):\limsup\limits_{n\to\infty}\dfrac{\log\left(a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\right)}{\varphi(n)}=1\right\}.
keywords:
Continued fractions, Hausdorff dimension, Product of partial quotients
MSC:
[2010] Primary 11K50, Secondary 37E05, 28A80

1 Introduction

Each irrational number x[0,1)x\in[0,1) admits a unique continued fraction expansion of the from

x=1a1(x)+1a2(x)+1a3(x)+,x=\frac{1}{a_{1}(x)+\displaystyle{\frac{1}{a_{2}(x)+\displaystyle{\frac{1}{a_{3}(x)+{\ddots}}}}}}, (1.1)

where for each n1n\geq 1, the positive integers an(x)a_{n}(x) are known as the partial quotients of x.x. The partial quotients can be generated by using the Gauss transformation T:[0,1)[0,1)T:[0,1)\rightarrow[0,1) defined as

T(0):=0andT(x)=1x(mod 1), for x(0,1).T(0):=0\ {\rm}\ {\rm and}\ T(x)=\frac{1}{x}\ ({\rm mod}\ 1),\text{ for }x\in(0,1). (1.2)

In fact, let a1(x)=1xa_{1}(x)=\big{\lfloor}\frac{1}{x}\big{\rfloor} (where \lfloor\cdot\rfloor denotes the integral part of real number). Then an(x)=1Tn1(x)a_{n}(x)=\big{\lfloor}\frac{1}{T^{n-1}(x)}\big{\rfloor} for n2n\geq 2. Sometimes (1.1) is written as x=[a1(x),a2(x),a3(x),].x=[a_{1}(x),a_{2}(x),a_{3}(x),\ldots]. Further, the nn-th convergent pn(x)/qn(x)p_{n}(x)/q_{n}(x) of xx is defined by pn(x)/qn(x)=[a1(x),a2(x),,an(x)].p_{n}(x)/q_{n}(x)=[a_{1}(x),a_{2}(x),\ldots,a_{n}(x)].

The metrical aspect of the theory of continued fractions has been very well studied due to its close connections with Diophantine approximation. For example, for any τ>0\tau>0 the famous Jarník-Besicovitch set

Jτ:={x[0,1):|xpq|<1qτ+2forinfinitelymany(p,q)×},{J}_{\tau}:=\left\{x\in[0,1):\left|x-\frac{p}{q}\right|<\frac{1}{q^{\tau+2}}\ \ \mathrm{for\ infinitely\ many\ }(p,q)\in\mathbb{Z}\times\mathbb{N}\right\},

can be described by using continued fractions. In fact,

Jτ={x[0,1):an+1(x)qnτ(x)forinfinitelymanyn}.{J}_{\tau}=\left\{x\in[0,1):a_{n+1}(x)\geq q^{\tau}_{n}(x)\ \ \mathrm{for\ infinitely\ many\ }n\in\mathbb{N}\right\}. (1.3)

For further details about this connection we refer to [11]. Thus the growth rate of the partial quotients reveals how well a real number can be approximated by rationals.

The well-known Borel-Bernstein Theorem [3, 4] states that for Lebesgue almost all x[0,1),x\in[0,1), an(x)φ(n)a_{n}(x)\geq\varphi(n) holds for finitely many nsn^{\prime}s or infinitely many nsn^{\prime}s according to the convergence or divergence of the series n=11/φ(n)\sum_{n=1}^{\infty}{1}/{\varphi(n)} respectively. However, for rapidly growing function φ,{\varphi}, the Borel-Bernstein Theorem does not give any conclusive information other than Lebesgue measure zero. To distinguish the sizes of zero Lebesgue measure sets, Hausdorff dimension is considered as an appropriate conception and has gained much importance in the metrical theory of continued fractions. Jarník [14] proved that the set of real numbers with bounded partials quotients has full Hausdorff dimension. Later on, Good [11] showed that the Hausdorff dimension of the set of numbers whose partial quotients tend to infinity is one half.

After that, a lot of work has been done in the direction of improving Borel-Bernstein Theorem, for example, the Hausdorff dimension of sets when partial quotients an(x)a_{n}(x) obeys different conditions has been obtained in [6, 7, 8, 17, 18, 19, 20].

Motivation for studying the growth rate of the products of consecutive partial quotients aroses from the works of Davenport-Schmidt [5] and Kleinbock-Wadleigh [16] where they considered improvements to Dirichlet’s theorem. Let ψ:[t0,)+\psi:[t_{0},\infty)\rightarrow\mathbb{R}_{+} be a monotonically decreasing function, where t01t_{0}\geq 1 is fixed. Denote by D(ψ)D(\psi) the set of all real numbers xx for which the system

|qxp|ψ(t)and|q|<t|qx-p|\leq\psi(t)\ \rm{and}\ |q|<t

has a nontrivial integer solution for all large enough tt. A real number xD(ψ)x\in D(\psi) (resp. xD(ψ)cx\in D(\psi)^{c}) will be referred to as ψ\psi-Dirichlet improvable (resp. ψ\psi-Dirichlet non-improvable) number.

The starting point for the work of Davenport-Schmidt [5] and Kleinbock-Wadleigh [16, Lemma 2.2] is an observation that Dirichlet improvability is equivalent to a condition on the growth rate of the products of two consecutive partial quotients. Precisely, they observed that

xD(ψ)\displaystyle x\in D(\psi) |qn1xpn1|ψ(qn), for all n large\displaystyle\Longleftrightarrow|q_{n-1}x-p_{n-1}|\leq\psi(q_{n}),\text{ for all }n\text{ large }
[an+1,an+2,][an,an1,,a1]((qnψ(qn))11), for all n large.\displaystyle\Longleftrightarrow[a_{n+1},a_{n+2},\cdots]\cdot[a_{n},a_{n-1},\cdots,a_{1}]\geq((q_{n}\psi(q_{n}))^{-1}-1),\text{ for all }n\text{ large. }

Then

{x[0,1):an(x)an+1(x)((qn(x)ψ(qn(x)))11)1for i.m.n}D𝖼(ψ){x[0,1):an(x)an+1(x)41((qn(x)ψ(qn(x)))11)1for i.m.n},\Big{\{}x\in[0,1)\colon a_{n}(x)a_{n+1}(x)\geq((q_{n}(x)\psi(q_{n}(x)))^{-1}-1)^{-1}\ {\text{for i.m.}}\ n\in\mathbb{N}\Big{\}}\subset D^{\mathsf{c}}(\psi)\\ \subset\Big{\{}x\in[0,1)\colon a_{n}(x)a_{n+1}(x)\geq 4^{-1}((q_{n}(x)\psi(q_{n}(x)))^{-1}-1)^{-1}\ {\text{for i.m.}}\ n\in\mathbb{N}\Big{\}},

where i.m. stands for infinitely many.

In other words, a real number x[0,1)x\in[0,1)\setminus\mathbb{Q} is ψ\psi-Dirichlet improvable if and only if the products of consecutive partial quotients of xx do not grow quickly. We refer the reader to [1, 2, 10, 12, 13] for more metrical results related with the set of Dirichlet non-improvable numbers.

As a consequence of Borel-Bernstein Theorem, for almost all x[0,1)x\in[0,1) there exists a subsequence of partial quotients tending to infinity with a linear speed. In other words, for Lebesgue almost every x[0,1)x\in[0,1)

lim supnlogan(x)logn=1.\limsup_{n\to\infty}\frac{\log a_{n}(x)}{\log n}=1.

Taking inspirations from the study of the growth rate of the products of consecutive partial quotients for the real numbers, in this paper we consider the growth rate of the products of the consecutive weighted partial quotients. More precisely, by [2, Theorem 1.4], we have for Lebesgue almost all x[0,1)x\in[0,1)

lim supnlogant0(x)an+1t1(x)an+mtm(x)logntmax=1,\limsup_{n\to\infty}\frac{\log a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)}{\log n^{t_{\max}}}=1, (1.4)

where tmax=max{ti:0im}t_{\max}=\max\{t_{i}:0\leq i\leq m\}. This paper is concerned with Hausdorff dimension of some exceptional sets of (1.4). Let φ:+\varphi:\mathbb{N}\to\mathbb{R}_{+} be a function satisfying φ(n)/n\varphi(n)/n\to\infty as nn\to\infty and let ti+(0im)t_{i}\in\mathbb{R}_{+}\ (0\leq i\leq m). Define the sets

E¯({ti}i=0m,φ)={x[0,1):lim supnlog(ant0(x)an+1t1(x)an+mtm(x))φ(n)=1},\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi)=\left\{x\in[0,1):\limsup\limits_{n\to\infty}\dfrac{\log\left(a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\right)}{\varphi(n)}=1\right\},

and

E¯({ti}i=0m,φ)={x[0,1):lim infnlog(ant0(x)an+1t1(x)an+mtm(x))φ(n)=1}.\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi)=\left\{x\in[0,1):\liminf\limits_{n\to\infty}\dfrac{\log\left(a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\right)}{\varphi(n)}=1\right\}.

The study of the level sets about the growth rate of {an(x)an+1(x):n1}\{a_{n}(x)a_{n+1}(x):n\geq 1\} relative to that of {qn(x):n1}\{q_{n}(x):n\geq 1\} was discussed in [12]. Let m2m\geq 2 be an integer and Ψ:+\Psi:\mathbb{N}\to\mathbb{R}_{+} be a positive function. The Lebesgue measure and the Hausdorff dimension of the set

{x[0,1):an(x)an+m(x)Ψ(n)for infinitely manyn},\left\{x\in[0,1):{a_{n}(x)\cdots a_{n+m}(x)}\geq\Psi(n)\ \text{for infinitely many}\ n\in\mathbb{N}\right\}, (1.5)

have been comprehensively determined by Huang-Wu-Xu [13]. Very recently the results of [13] were generalized by Bakhtawar-Hussain-Kleinbock-Wang [2] to a weighted generalization of the set (1.5). For more details we refer the reader to [2, 13].

Our main results are as follows.

Theorem 1.1.

Let φ:+\varphi:\mathbb{N}\to\mathbb{R}_{+} be a function satisfying φ(n)/n\varphi(n)/n\to\infty as nn\to\infty. Write

logB=lim supnlogφ(n)n.\log B=\limsup\limits_{n\rightarrow\infty}\frac{\log\varphi(n)}{n}.

Assume B[1,]B\in[1,\infty]. Then for any (t0,,tm)+m+1(t_{0},\cdots,t_{m})\in\mathbb{R}^{m+1}_{+} with ti0t_{i}\geq 0 and at least one ti0(0im)t_{i}\neq 0\ (0\leq i\leq m),

dimHE¯({ti}i=0m,φ)=11+B.\dim_{\mathrm{H}}\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi)=\frac{1}{1+B}.
Theorem 1.2.

Let φ:+\varphi:\mathbb{N}\to\mathbb{R}_{+} be a function satisfying φ(n)/n\varphi(n)/n\to\infty as nn\to\infty. Write

logb=lim infnlogφ(n)n.\log b=\liminf\limits_{n\rightarrow\infty}\frac{\log\varphi(n)}{n}.

Assume b[1,]b\in[1,\infty]. Then for any (t0,t1,,tm)+m+1(t_{0},t_{1},\cdots,t_{m})\in\mathbb{R}_{+}^{m+1} with 0<t0t1tm0<t_{0}\leq t_{1}\leq\cdots\leq t_{m}, we have

dimHE¯({ti}i=0m,φ)=11+b.\dim_{\mathrm{H}}\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi)=\frac{1}{1+b}.
Remark 1.1.

Note that in Theorem 1.2 we are only able to treat the case when the sequence {ti}i=0m\{t_{i}\}_{i=0}^{m} is nondecreasing. We would like to drop this monotonic condition. Indeed, our method for the upper bound is true for all sequences {ti}i=0m\{t_{i}\}_{i=0}^{m}. However, when dealing with the lower bound, the sequence {cn}n1\{c_{n}\}_{n\geq 1} we construct (see the proof for details) might not be bounded away from 0 once we drop the monotonic condition, which is important in constructing a suitable subset of E¯({ti}i=0m,φ)\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi).

2 Preliminaries

In this section, we fix some notations and recall some known results in theory of continued fraction expansions.

For an irrational number x[0,1)x\in[0,1), recall an(x)a_{n}(x) is the nn-th partial quotient of xx in its continued fraction expansion. The sequences {pn(x)}n1,\{p_{n}(x)\}_{n\geq 1}, {qn(x)}n1\{q_{n}(x)\}_{n\geq 1} are the numerator and denominator of the nn-th convergent of xx. It is well-known that {pn(x)}n1\{p_{n}(x)\}_{n\geq 1} and {qn(x)}n1\{q_{n}(x)\}_{n\geq 1} can be obtain by the following recursive relations (see [15]):

{pn(x)=an(x)pn1(x)+pn2(x),for any n 1 ;qn(x)=an(x)qn1(x)+qn2(x),for any n 1 ,\begin{cases}&p_{n}(x)=a_{n}(x)p_{n-1}(x)+p_{n-2}(x),\ \text{for any $n\geq$ 1 };\\ &q_{n}(x)=a_{n}(x)q_{n-1}(x)+q_{n-2}(x),\ \text{for any $n\geq$ 1 },\end{cases} (2.1)

with the conventions p1=1,q1=0,p0=0p_{-1}=1,\ q_{-1}=0,\ p_{0}=0 and q0=1q_{0}=1.

For any nn-tuple (a1,,an)n(a_{1},\cdots,a_{n})\in\mathbb{N}^{n} with n1n\geq 1, we call

In(a1,,an)={x[0,1):a1(x)=a1,a2(x)=a2,,an(x)=an},I_{n}(a_{1},\cdots,a_{n})=\big{\{}x\in[0,1)\colon a_{1}(x)=a_{1},\ a_{2}(x)=a_{2},\cdots,a_{n}(x)=a_{n}\big{\}},

a cylinder of order nn.

Note that pn(x)p_{n}(x) and qn(x)q_{n}(x) are determined by the first nn partial quotients of xx. So all points in In(a1,,an)I_{n}(a_{1},\cdots,a_{n}) determine the same pn(x)p_{n}(x) and qn(x)q_{n}(x). Hence for simplicity, if there is no confusion, we write ana_{n}, pnp_{n} and qnq_{n} to denote an(x)a_{n}(x), pn(x)p_{n}(x) and qn(x)q_{n}(x) for xIn(a1,,an)x\in I_{n}(a_{1},\cdots,a_{n}) respectively.

The following lemma is a collection of basic facts on continued fractions which can be found in the book of Khintchine [15].

Lemma 2.1.

For any (a1,,an)n(a_{1},\cdots,a_{n})\in\mathbb{N}^{n}, let qnq_{n} and pnp_{n} be given recursively by (2.1). Then

(1)

In(a1,,an)={[pnqn,pn+pn1qn+qn1), if n is even ,[pn+pn1qn+qn1,pnqn), if n is odd ;I_{n}(a_{1},\cdots,a_{n})=\begin{cases}&\Big{[}\dfrac{p_{n}}{q_{n}},\dfrac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}\Big{)},\text{ if }n\text{ is even },\\ &\Big{[}\dfrac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}},\dfrac{p_{n}}{q_{n}}\Big{)},\text{ if }n\text{ is odd };\end{cases}

(2) qn2n12,k=1nakqn2nk=1nak;q_{n}\geq 2^{\frac{n-1}{2}},\ \prod_{k=1}^{n}a_{k}\leq q_{n}\leq 2^{n}\prod_{k=1}^{n}a_{k};

(3)

13an+1qn2<|xpnqn|=1qn(qn+1+Tn+1(x)qn)<1an+1qn2,\frac{1}{3a_{n+1}q_{n}^{2}}\,<\,\Big{|}x-\frac{p_{n}}{q_{n}}\Big{|}=\frac{1}{q_{n}(q_{n+1}+T^{n+1}(x)q_{n})}<\,\frac{1}{a_{n+1}q_{n}^{2}},

and for any n1n\geq 1 the derivative of TnT^{n} is given by

(Tn)(x)=(1)n(xqn1pn1)2.(T^{n})^{\prime}(x)=\frac{(-1)^{n}}{(xq_{n-1}-p_{n-1})^{2}}.

The next theorem, known as Legendre’s Theorem, connects 1-dimensional Diophantine approximation with continued fractions.

Theorem 2.1 (Legendre).

Let pq\frac{p}{q} be an irreducible rational number. Then

|xpq|<12q2pq=pn(x)qn(x),forsomen1.\Big{|}x-\frac{p}{q}\Big{|}<\frac{1}{2q^{2}}\Longrightarrow\frac{p}{q}=\frac{p_{n}(x)}{q_{n}(x)},\quad\mathrm{for\ some\ }n\geq 1.

According to Legendre’s theorem if an irrational xx is “well” approximated by a rational p/qp/q, then this rational must be a convergent of xx. So, the continued fraction expansions is a quick and efficient tool for finding good rational approximations to real numbers. For more basic properties of continued fraction expansions, one can refer to [15]. We also give some auxiliary results on the Hausdorff dimension theory of continued fractions that will be used later.

Lemma 2.2 ([6]).

Let {sn}n1\{s_{n}\}_{n\geq 1} be a sequence of positive integers tending to infinity, then for any positive integer number N2N\geq 2,

dimH{x[0,1):snan(x)<Nsn,for alln1}\displaystyle\dim_{\mathrm{H}}\big{\{}x\in[0,1):s_{n}\leq a_{n}(x)<Ns_{n},\ \text{for all}\ n\geq 1\big{\}}
=\displaystyle= lim infnlog(s1s2sn)2log(s1s2sn)+logsn+1\displaystyle\liminf_{n\to\infty}\frac{\log(s_{1}s_{2}\cdots s_{n})}{2\log(s_{1}s_{2}\cdots s_{n})+\log s_{n+1}}
=\displaystyle= 12+lim supnlogsn+1log(s1s2sn).\displaystyle\frac{1}{2+\limsup\limits_{n\to\infty}\frac{\log s_{n+1}}{\log(s_{1}s_{2}\cdots s_{n})}}.
Lemma 2.3 ([9, 17]).

For any a,c>1a,c>1,

dimH{x[0,1):an(x)can,for alln1}\displaystyle\dim_{\mathrm{H}}\left\{x\in[0,1):a_{n}(x)\geq c^{a^{n}},\ \text{for all}\ n\geq 1\right\}
=\displaystyle= dimH{x[0,1):an(x)can,for infinitely manyn}\displaystyle\dim_{\mathrm{H}}\left\{x\in[0,1):a_{n}(x)\geq c^{a^{n}},\ \text{for infinitely many}\ n\in\mathbb{N}\right\}
=\displaystyle= 11+a.\displaystyle\frac{1}{1+a}.

Applying Lemma 2.3, we obtain the follwing corollary which will be useful for the upper bound estimation on dimHE¯({ti}i=0m,φ).\dim_{\mathrm{H}}\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi).

Corollary 2.1.

For any a,c>1a,c>1 and (t0,,tm)+m+1(t_{0},\cdots,t_{m})\in\mathbb{R}^{m+1}_{+} with ti0t_{i}\geq 0 and at least one ti0(0im)t_{i}\neq 0\ (0\leq i\leq m),

dimH{x[0,1):ant0(x)an+1t1(x)an+mtm(x)can,for alln1}\displaystyle\dim_{\mathrm{H}}\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq c^{a^{n}},\ \text{for all}\ n\geq 1\right\}
=\displaystyle= dimH{x[0,1):ant0(x)an+1t1(x)an+mtm(x)can,for infinitely manyn1}\displaystyle\dim_{\mathrm{H}}\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq c^{a^{n}},\ \text{for infinitely many}\ n\geq 1\right\}
=\displaystyle= 11+a.\displaystyle\frac{1}{1+a}.
Proof.

Denote k=min{0im:ti0}k=\min\{0\leq i\leq m:\ t_{i}\neq 0\}. It is clear that for some tj0(0jm),t_{j}\neq 0\ (0\leq j\leq m),

{x[0,1):an+ktk(x)can,for alln1}\displaystyle\left\{x\in[0,1):a^{t_{k}}_{n+k}(x)\geq c^{a^{n}},\ \text{for all}\ n\geq 1\right\}
\displaystyle\subset {x[0,1):ant0(x)an+1t1(x)an+mtm(x)can, for alln1}\displaystyle\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq c^{a^{n}},\ \text{ for all}\ n\geq 1\right\}
\displaystyle\subset {x[0,1):ant0(x)an+1t1(x)an+mtm(x)can, for infinitely manyn1}\displaystyle\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq c^{a^{n}},\ \text{ for infinitely many}\ n\geq 1\right\}
\displaystyle\subset {x[0,1):an+jtj(x)canm+1, for infinitely manyn1}.\displaystyle\left\{x\in[0,1):a^{t_{j}}_{n+j}(x)\geq c^{\frac{a^{n}}{m+1}},\ \text{ for infinitely many}\ n\geq 1\right\}.

From Lemma 2.3, we deduce that for some tj0(0jm),t_{j}\neq 0\ (0\leq j\leq m),

dimH{x[0,1):an+ktk(x)can, for alln1}\displaystyle\dim_{\mathrm{H}}\left\{x\in[0,1):a^{t_{k}}_{n+k}(x)\geq c^{a^{n}},\ \text{ for all}\ n\geq 1\right\}
=\displaystyle= dimH{x[0,1):an+jtj(x)can,for infinitely manyn}\displaystyle\dim_{\mathrm{H}}\left\{x\in[0,1):a^{t_{j}}_{n+j}(x)\geq c^{a^{n}},\ \text{for infinitely many}\ n\in\mathbb{N}\right\}
=\displaystyle= 11+a.\displaystyle\frac{1}{1+a}.

Then the desired results are obtained. ∎

3 Proof of theorem 1.1

Let φ:+\varphi:\mathbb{N}\to\mathbb{R}_{+} be a positive function with φ(n)\varphi(n)\rightarrow\infty as nn\rightarrow\infty. For any (t0,,tm)+m+1(t_{0},\cdots,t_{m})\in\mathbb{R}^{m+1}_{+} with ti0t_{i}\geq 0 and at least one ti0(0im)t_{i}\neq 0\ (0\leq i\leq m), we introduce the sets

ND(φ)={x[0,1):ant0(x)an+1t1(x)an+mtm(x)φ(n), for all n1},\displaystyle ND(\varphi)=\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq\varphi(n),\text{ for all }\ n\geq 1\right\},

and

ND(φ)={x[0,1):ant0(x)an+1t1(x)an+mtm(x)φ(n), for n large enough }.\displaystyle{ND}^{\prime}(\varphi)=\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq\varphi(n),\text{ for $n$ large enough }\right\}.

In order to prove Theorem 1.1, we first give a complete characterization on the size of the sets ND(φ)ND(\varphi) and ND(φ){ND}^{\prime}(\varphi) in terms of Hausdorff dimension.

Proposition 3.1.

For any (t0,,tm)+m+1(t_{0},\cdots,t_{m})\in\mathbb{R}^{m+1}_{+} with ti0t_{i}\geq 0 and at least one ti0(0im)t_{i}\neq 0~{}(0\leq i\leq m),

dimHND(φ)=dimHND(φ)=11+A,wherelogA=lim supnloglogφ(n)n.\dim_{\mathrm{H}}ND(\varphi)=\dim_{\mathrm{H}}{ND}^{\prime}(\varphi)=\frac{1}{1+A},\ \text{where}\ \log A=\limsup\limits_{n\rightarrow\infty}\frac{\log\log\varphi(n)}{n}.

We remark that recently Zhang ([21]) obtained the Hausdorff dimension results of ND(φ)ND(\varphi) and ND(φ){ND}^{\prime}(\varphi) for the special case t0=t1=1t_{0}=t_{1}=1, ti=0(i2).t_{i}=0\ (i\geq 2).

3.1 Proof of Proposition 3.1

To prove Proposition 3.1, we start with the following lemma.

Lemma 3.1.

For any (t0,,tm)+m+1(t_{0},\cdots,t_{m})\in\mathbb{R}^{m+1}_{+} with ti0t_{i}\geq 0 and at least one ti0(0im)t_{i}\neq 0~{}(0\leq i\leq m),

dimH{x[0,1):ant0(x)an+1t1(x)an+mtm(x),asn}=12.\dim_{\mathrm{H}}\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\rightarrow\infty,\ \text{as}\ n\rightarrow\infty\right\}=\frac{1}{2}.
Proof.

Denote by C()C(\infty) the set above and k=min{0im:ti0}k=\min\{0\leq i\leq m:\ t_{i}\neq 0\}. It is evident that

{x[0,1):an+ktk(x),asn}\displaystyle\left\{x\in[0,1):a^{t_{k}}_{n+k}(x)\rightarrow\infty,\ \text{as}\ n\rightarrow\infty\right\}
\displaystyle\subset {x[0,1):ant0(x)an+1t1(x)an+mtm(x),asn}.\displaystyle\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\rightarrow\infty,\ \text{as}\ n\rightarrow\infty\right\}.

So, dimHC()12.\dim_{\mathrm{H}}C(\infty)\geq\frac{1}{2}.

In the following, we give the upper bound for dimHC()\dim_{\mathrm{H}}C(\infty).

Step I. We find a cover for C()C(\infty). For any M>0M>0,

C()\displaystyle C(\infty) {x[0,1):ant0(x)an+1t1(x)an+mtm(x)M,for n large enough}\displaystyle\subset\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq M,\ \text{for $n$ large enough}\right\}
=N=1{x[0,1):ant0(x)an+1t1(x)an+mtm(x)M,fornN}\displaystyle=\bigcup_{N=1}^{\infty}\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq M,\ \text{for}\ n\geq N\right\}
:=N=1EM(N).\displaystyle:=\bigcup_{N=1}^{\infty}E_{M}(N).

It is clear that dimHC()dimHEM(1)\dim_{\mathrm{H}}C(\infty)\leq\dim_{\mathrm{H}}E_{M}(1), since dimHC()dimHsupN1EM(N)\dim_{\mathrm{H}}C(\infty)\leq\dim_{\mathrm{H}}\sup\limits_{N\geq 1}E_{M}(N), and by [11, Lemma 1], we have dimHEM(N)=dimHEM(1)\dim_{\mathrm{H}}E_{M}(N)=\dim_{\mathrm{H}}E_{M}(1) for any N1N\geq 1. So it is sufficient to estimate the upper bound for EM(1)E_{M}(1). For any n1n\geq 1, set

Dn(M)={(a1,,an)n:akt0ak+1t1ak+mtmM,for 1knm}.D_{n}(M)=\left\{(a_{1},\cdots,a_{n})\in\mathbb{N}^{n}:\ a^{t_{0}}_{k}a^{t_{1}}_{k+1}\cdots a^{t_{m}}_{k+m}\geq M,\ \text{for}\ 1\leq k\leq n-m\right\}.

Hence,

EM(1)\displaystyle E_{M}(1)\subset (a1,,an)Dn(M)In(a1,,an).\displaystyle\bigcup_{(a_{1},\cdots,a_{n})\in D_{n}(M)}I_{n}(a_{1},\cdots,a_{n}). (3.1)

Step II. We construct a family of Bernoulli measures {μt}t>1\{\mu_{t}\}_{t>1} on [0,1)[0,1). For each t>1t>1 and any (a1,,an)n,(a_{1},\cdots,a_{n})\in\mathbb{N}^{n}, put

μt(In(a1,,an))=enP(t)tΣj=1nlogaj,\mu_{t}(I_{n}(a_{1},\cdots,a_{n}))=e^{-nP(t)-t\Sigma_{j=1}^{n}\log a_{j}},

where eP(t)=k=1kt.e^{P(t)}=\sum_{k=1}^{\infty}k^{-t}. It is easy to see that

an+1μt(In(a1,,an,an+1))=μt(In(a1,,an))\sum_{a_{n+1}}\mu_{t}(I_{n}(a_{1},\cdots,a_{n},a_{n+1}))=\mu_{t}(I_{n}(a_{1},\cdots,a_{n}))

and

(a1,,an)nμt(In(a1,,an))=1.\sum_{(a_{1},\cdots,a_{n})\in\mathbb{N}^{n}}\mu_{t}(I_{n}(a_{1},\cdots,a_{n}))=1.

So the measures {μt}t>1\{\mu_{t}\}_{t>1} are well defined by Kolmogorov’s consistency theorem.

Fix s>12s>\frac{1}{2} and set t=s+12>1t=s+\frac{1}{2}>1. Choose MM sufficiently large such that

(s12)logMtmax12mP(s+12),\left(s-\frac{1}{2}\right)\cdot\frac{\log M^{t^{-1}_{\max}}}{2m}\geq P(s+\frac{1}{2}), (3.2)

where tmax=max{ti:0im}t_{\max}=\max\{t_{i}:0\leq i\leq m\}.

We claim that for any (a1,,an)Dn(M)(a_{1},\cdots,a_{n})\in D_{n}(M),

qn2sμs+12(In(a1,,an)).q_{n}^{-2s}\leq\mu_{s+\frac{1}{2}}(I_{n}(a_{1},\cdots,a_{n})). (3.3)

More precisely, for any (a1,,an)Dn(M)(a_{1},\cdots,a_{n})\in D_{n}(M), by the fact that for 1lnm1\leq l\leq n-m,

alt0al+1t1al+mtmM,a^{t_{0}}_{l}a^{t_{1}}_{l+1}\cdots a^{t_{m}}_{l+m}\geq M,

then for 1lnm1\leq l\leq n-m,

alal+1al+mMtmax1,a_{l}a_{l+1}\cdots a_{l+m}\geq M^{t^{-1}_{\max}},

where tmax=max{ti:0im}t_{\max}=\max\{t_{i}:0\leq i\leq m\}. Then we have

e2sj=1nlogaje(s+12)j=1nlogaj(s12)nmlogMtmax1.e^{-2s\sum_{j=1}^{n}\log a_{j}}\leq e^{-(s+\frac{1}{2})\sum_{j=1}^{n}\log a_{j}-(s-\frac{1}{2})\lfloor\frac{n}{m}\rfloor\log M^{t^{-1}_{\max}}}.

Thus, by qni=1naiq_{n}\geq\prod\limits_{i=1}^{n}a_{i} and then (3.2), we get

qn2se2sj=1nlogaje(s+12)j=1nlogajnP(s+12)q_{n}^{-2s}\leq e^{-2s\sum_{j=1}^{n}\log a_{j}}\leq e^{-(s+\frac{1}{2})\sum_{j=1}^{n}\log a_{j}-nP(s+\frac{1}{2})}

Therefore, by (3.1) and (3.3),

s(EM(1))\displaystyle\mathcal{H}^{s}(E_{M}(1)) lim infn(a1,,an)Dn(M)|In(a1,,an)|s\displaystyle\leq\liminf_{n\to\infty}\sum_{(a_{1},\cdots,a_{n})\in D_{n}(M)}{\big{|}I_{n}(a_{1},\cdots,a_{n})\big{|}}^{s}
lim infn(a1,,an)Dn(M)1qn2s\displaystyle\leq\liminf_{n\to\infty}\sum_{(a_{1},\cdots,a_{n})\in D_{n}(M)}\frac{1}{q_{n}^{2s}}
lim infn(a1,,an)Dn(M)μs+12(In(a1,,an))=1.\displaystyle\leq\liminf_{n\to\infty}\sum_{(a_{1},\cdots,a_{n})\in D_{n}(M)}\mu_{s+\frac{1}{2}}(I_{n}(a_{1},\cdots,a_{n}))=1.

Hence dimHEM(1)s\dim_{\mathrm{H}}E_{M}(1)\leq s, and then dimHC()s\dim_{\mathrm{H}}C(\infty)\leq s. Consequently, dimHC()12\dim_{\mathrm{H}}C(\infty)\leq\frac{1}{2} by the arbitrariness of s>12s>\frac{1}{2}. This completes the proof of Lemma 3.1. ∎

Now we are ready to prove Proposition 3.1.

Proof of Proposition 3.1: We see that dimHND(φ)=dimHND(φ).\dim_{\mathrm{H}}{ND}^{\prime}(\varphi)=\dim_{\mathrm{H}}ND(\varphi). The proof is divided into two cases according to A=1A=1 or A>1A>1.

(1) If A=1,A=1, then for any ϵ>0\epsilon>0, φ(n)e(1+ϵ)nholds for n large enough,\varphi(n)\leq e^{{(1+\epsilon)}^{n}}\ \text{holds for $n$ large enough}, and we have

{x[0,1):an(x)e(1+ϵ)nfor n large enough}ND(φ).\big{\{}x\in[0,1):\ a_{n}(x)\geq e^{{(1+\epsilon)}^{n}}\ \text{for $n$ large enough}\big{\}}\subset{ND}^{\prime}(\varphi).

By Lemma 2.3, we obtain

dimHND(φ)12.\dim_{\mathrm{H}}{ND}^{\prime}(\varphi)\geq\frac{1}{2}.

On the other hand,

ND(φ){x[0,1):ant0(x)an+1t1(x)an+mtm(x),asn}.ND(\varphi)\subset\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\rightarrow\infty,\ \text{as}\ n\rightarrow\infty\right\}.

Thus,

dimHND(φ)12.\dim_{\mathrm{H}}ND(\varphi)\leq\frac{1}{2}.

(2) If A>1A>1, by the definition of limsup, for any ϵ>0\epsilon>0,

{φ(n)e(Aϵ)n,for infinitely manyn,φ(n)e(A+ϵ)n,for all sufficiently largen.\begin{cases}&\varphi(n)\geq e^{{(A-\epsilon)}^{n}},\;\text{for infinitely many}\ n\in\mathbb{N},\\ &\varphi(n)\leq e^{{(A+\epsilon)}^{n}},\;\text{for all sufficiently large}\;n.\end{cases}

Therefore

{x[0,1):ant0(x)an+1t1(x)an+mtm(x)e(A+ϵ)n,for any n1}ND(φ).\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq e^{{(A+\epsilon)}^{n}},\ \text{for any $n\geq 1$}\right\}\subset{ND}^{\prime}(\varphi).

Applying Corollary 2.1, we obtain

dimHND(φ)11+A+ϵ.\dim_{\mathrm{H}}{ND}^{\prime}(\varphi)\geq\frac{1}{1+A+\epsilon}.

By the arbritrary of ϵ\epsilon, we have

dimHND(φ)11+A.\dim_{\mathrm{H}}{ND}^{\prime}(\varphi)\geq\frac{1}{1+A}.

On the other hand,

ND(φ){x[0,1):ant0(x)an+1t1(x)an+mtm(x)e(Aϵ)n,for infinitely manyn}.ND(\varphi)\subset\left\{x\in[0,1):a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}(x)\geq e^{{(A-\epsilon)}^{n}},\ \text{for infinitely many}\ n\in\mathbb{N}\right\}.

From Corollary 2.1, we obtain

dimHND(φ)11+Aϵ.\dim_{\mathrm{H}}ND(\varphi)\leq\frac{1}{1+A-\epsilon}.

Taking ϵ0\epsilon\to 0, we conclude

dimHND(φ)11+A.\dim_{\mathrm{H}}ND(\varphi)\leq\frac{1}{1+A}.

∎ 

Let us give a proof of Theorem 1.1.

3.2 Proof of Theorem 1.1

Upper bound: For xE¯({ti}i=0m,φ)x\in\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi), for any ϵ>0\epsilon>0, we have

ant0(x)an+1t1(x)an+mtme(1ϵ)φ(n),for n large enough.a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}\geq e^{(1-\epsilon)\varphi(n)},\ \text{for $n$ large enough}.

Then it follows from Proposition 3.1 that

dimHE¯({ti}i=0m,φ)11+B.\dim_{\mathrm{H}}\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi)\leq\frac{1}{1+B}.

Lower bound: It is trivial for B=B=\infty, so we only need to consider the case 1B<1\leq B<\infty.

We construct a suitable Cantor subset of E¯({ti}i=0m,φ)\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi) in two steps.

Step I. Since logB=lim supnlogφ(n)n,\log B=\limsup\limits_{n\rightarrow\infty}\frac{\log\varphi(n)}{n}, for any ϵ>0\epsilon>0, we have φ(n)(B+ϵ2)n\varphi(n)\leq{(B+\frac{\epsilon}{2})}^{n} for nn large enough. Hence,

φ(n)(B+ϵ)jn(B+ϵ2)n(B+ϵ)jn0,asn.\varphi(n){(B+\epsilon)}^{j-n}\leq{(B+\frac{\epsilon}{2})}^{n}{(B+\epsilon)}^{j-n}\to 0,\ \text{as}\ n\to\infty.

We define a sequence {Lj}j1\{L_{j}\}_{j\geq 1}: For j,k1j,k\geq 1, let

cj,k={exp(φ(k)), 1kj;exp(φ(k)(B+ϵ)jk),kj+1.c_{j,k}=\begin{cases}&\exp({\varphi(k)}),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1\leq k\leq j;\\ &\exp(\varphi(k){(B+\epsilon)}^{j-k}),\ k\geq j+1.\end{cases}

Define Lj=supk1{cj,k}L_{j}=\sup\limits_{k\geq 1}\{c_{j,k}\}. Clearly,

LjLj+1LjB+ϵandLjeφ(j)for anyj1.L_{j}\leq L_{j+1}\leq L_{j}^{B+\epsilon}\ \text{and}\ L_{j}\geq e^{\varphi(j)}\ \text{for any}\ j\geq 1. (3.4)

By the first part of (3.4),

logLj+1logLj(B+ϵ1)logLj.\log L_{j+1}-\log L_{j}\leq(B+\epsilon-1)\log L_{j}.

Hence

logLn+1logL1(B+ϵ1)j=1nlogLj.\log L_{n+1}-\log L_{1}\leq(B+\epsilon-1)\sum\limits_{j=1}^{n}\log L_{j}. (3.5)

We claim that

lim infnlogLnφ(n)=1.\liminf_{n\to\infty}\dfrac{\log L_{n}}{\varphi(n)}=1. (3.6)

In fact, on the one hand, in view of the second part of (3.4), we see at once that

lim infnlogLnφ(n)1.\liminf_{n\to\infty}\dfrac{\log L_{n}}{\varphi(n)}\geq 1.

For the opposite inequality, let tj:=min{k1:cj,k=Lj}t_{j}:=\min\{k\geq 1:c_{j,k}=L_{j}\}. Notice that for many consecutive jj^{\prime}s, the number tjt_{j} will be the same. More precisely, if tj<jt_{j}<j, ttj=ttj+1==tj;t_{t_{j}}=t_{t_{j}+1}=\cdots=t_{j}; if tjjt_{j}\geq j, tj=tj+1==ttj.t_{j}=t_{j+1}=\cdots=t_{t_{j}}. Let {li}\{l_{i}\} be the sequence of all ttj{t_{t_{j}}}^{\prime}s in the strictly increasing order. Then we obtain Lli=expφ(li)L_{l_{i}}=\exp{\varphi(l_{i})} and thus

lim infnlogLnφ(n)lim infilogLliφ(li)=1.\liminf_{n\to\infty}\dfrac{\log L_{n}}{\varphi(n)}\leq\liminf_{i\to\infty}\dfrac{\log L_{l_{i}}}{\varphi(l_{i})}=1.

Let

Z:=lim infnlog(Lnt0Ln+1t1Ln+mtm+1)φ(n).Z:=\liminf\limits_{n\to\infty}\dfrac{\log\left(L_{n}^{t_{0}}L_{n+1}^{t_{1}}\cdots L_{n+m}^{t_{m+1}}\right)}{\varphi(n)}.

We claim that

tkZ<,t_{k}\leq Z<\infty,

where k=min{0im:ti0}k=\min\{0\leq i\leq m:\ t_{i}\neq 0\}. In fact, by the first part of (3.4) and (3.6), we can check that

Zlim infnlogLn+ktkφ(n)lim infnlogLntkφ(n)=tk.Z\geq\liminf\limits_{n\to\infty}\dfrac{\log L_{n+k}^{t_{k}}}{\varphi(n)}\geq\liminf\limits_{n\to\infty}\dfrac{\log L_{n}^{t_{k}}}{\varphi(n)}=t_{k}.

On the other hand,

lim infnlog(Lnt0Ln+1t1Ln+mtm+1)φ(n)\displaystyle\liminf_{n\to\infty}\dfrac{\log\left(L_{n}^{t_{0}}L_{n+1}^{t_{1}}\cdots L_{n+m}^{t_{m+1}}\right)}{\varphi(n)} (3.4)lim infn(t0+t1(B+ϵ)++tm+1(B+ϵ)m)logLnφ(n)\displaystyle\overset{\eqref{L j+1-L_j}}{\leq}\liminf\limits_{n\to\infty}\dfrac{\left(t_{0}+t_{1}(B+\epsilon)+\cdots+t_{m+1}{(B+\epsilon)}^{m}\right)\log L_{n}}{\varphi(n)}
=(3.6)t0+t1(B+ϵ)++tm+1(B+ϵ)m<.\displaystyle\overset{\eqref{L j+1-L_j-2}}{=}t_{0}+t_{1}(B+\epsilon)+\cdots+t_{m+1}{(B+\epsilon)}^{m}<\infty.

Step II. We use the sequence {Lj}j1\{L_{j}\}_{j\geq 1} and ZZ to construct a subset of E¯({ti}i=0m,φ).\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi). Define

E({Ln}n1)={x[0,1):Ln1Zan(x)<2Ln1Z,for anyn1}.E(\{L_{n}\}_{n\geq 1})=\left\{x\in[0,1):\lfloor L^{\frac{1}{Z}}_{n}\rfloor\leq a_{n}(x)<2\lfloor L^{\frac{1}{Z}}_{n}\rfloor,\ \text{for any}\ n\geq 1\right\}.

Then

lim infnlog(ant0an+1t1an+mtm+1)φ(n)\displaystyle\liminf\limits_{n\to\infty}\dfrac{\log\left(a^{t_{0}}_{n}a^{t_{1}}_{n+1}\cdots a^{t_{m+1}}_{n+m}\right)}{\varphi(n)} =lim infnlog(Lnt0ZLn+1t1ZLn+mtm+1Z)φ(n)\displaystyle=\liminf\limits_{n\to\infty}\dfrac{\log\left(L^{\frac{t_{0}}{Z}}_{n}L^{\frac{t_{1}}{Z}}_{n+1}\cdots L^{\frac{t_{m+1}}{Z}}_{n+m}\right)}{\varphi(n)}
=1Zlim infnlog(Lnt0Ln+1t1Ln+mtm+1)φ(n)=1.\displaystyle=\frac{1}{Z}\liminf\limits_{n\to\infty}\dfrac{\log\left(L^{t_{0}}_{n}L^{t_{1}}_{n+1}\cdots L^{t_{m+1}}_{n+m}\right)}{\varphi(n)}=1.

Hence E({Ln}n1)E¯({ti}i=0m,φ).E(\{L_{n}\}_{n\geq 1})\subset\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi). Since φ(n)/n\varphi(n)/n\to\infty as nn\to\infty, by the second part of (3.4), we see that

limnlog(L1L2Ln)n=.\lim\limits_{n\to\infty}\dfrac{\log(L_{1}L_{2}\cdots L_{n})}{n}=\infty.

By Lemma 2.2, we obtain

dimHE¯({ti}i=0m,φ)dimHE({Ln}n1)=12+lim supnlogLn+1log(L1L2Ln)(3.5)1B+1+ϵ.\dim_{\mathrm{H}}\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi)\geq\dim_{\mathrm{H}}E(\{L_{n}\}_{n\geq 1})=\dfrac{1}{2+\limsup\limits_{n\to\infty}\frac{\log L_{n+1}}{\log(L_{1}L_{2}\cdots L_{n})}}\overset{\eqref{for limsup set}}{\geq}\frac{1}{B+1+\epsilon}.

Taking ϵ0\epsilon\to 0, we conclude

dimHE¯({ti}i=0m,φ)1B+1.\dim_{\mathrm{H}}\underline{E}(\{t_{i}\}_{i=0}^{m},\varphi)\geq\frac{1}{B+1}.

4 Proof of Theorem 1.2

In this section, we give a proof of Theorem 1.2. We adopt the strategies in [19]. The proof of the theorem splits into two parts: finding the upper bound and the lower bound separately.

Upper bound: For xE¯({ti}i=0m,φ)x\in\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi), for any ϵ>0\epsilon>0, there exist infinitely many nn such that

ant0(x)an+1t1(x)an+mtme(1ϵ)φ(n).a^{t_{0}}_{n}(x)a^{t_{1}}_{n+1}(x)\cdots a^{t_{m}}_{n+m}\geq e^{(1-\epsilon)\varphi(n)}.

Then by [2, Theorem 1.5],

dimHE¯m({ti}i=0m,φ)11+b.\dim_{\mathrm{H}}\overline{E}_{m}(\{t_{i}\}_{i=0}^{m},\varphi)\leq\frac{1}{1+b}.

Lower bound: We construct a suitable Cantor subset of E¯({ti}i=0m,φ)\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi) in two steps.

Step I. We will construct a sequence {cn}n1\{c_{n}\}_{n\geq 1} of positive real numbers such that

lim supnlog(cnt0cn+1t1cn+mtm)φ(n)=1,\limsup\limits_{n\to\infty}\dfrac{\log\left(c^{t_{0}}_{n}c^{t_{1}}_{n+1}\cdots c^{t_{m}}_{n+m}\right)}{\varphi(n)}=1,

and

lim supnlogcn+1log(c1c2cn)b+ϵ1.\limsup\limits_{n\to\infty}\dfrac{\log c_{n+1}}{\log\left(c_{1}c_{2}\cdots c_{n}\right)}\leq b+\epsilon-1.

For all nn\in\mathbb{N}, let Φ(n)=minknφ(k).\Phi(n)=\min\limits_{k\geq n}\varphi(k). Since φ(n)\varphi(n)\to\infty, as nn\to\infty, Φ(n)\Phi(n) is well defined. Thus, Φ(n)φ(n),\Phi(n)\leq\varphi(n), Φ(n)Φ(n+1)\Phi(n)\leq\Phi(n+1) for all nn\in\mathbb{N}. We claim that

Φ(n)=φ(n),infinitely manyn.\Phi(n)=\varphi(n),\ \text{infinitely many}\ n\in\mathbb{N}.

If not, there exists NN\in\mathbb{N} such that for any nNn\geq N, Φ(n)<φ(n).\Phi(n)<\varphi(n). Then for nNn\geq N, Φ(n)<minknφ(k),\Phi(n)<\min\limits_{k\geq n}\varphi(k), which contradicts to the definition of Φ(n)\Phi(n).

We define a sequence {cn}n1\{c_{n}\}_{n\geq 1} as follows:

c1=c2==cm=1,cm+1tm=eΦ(1),\displaystyle c_{1}=c_{2}=\cdots=c_{m}=1,\ c^{t_{m}}_{m+1}=e^{\Phi(1)}, (4.1)
cn+mtm=min{eΦ(n)cnt0cn+1t1cn+m1tm1,(c1c2cn+m1)tm(b+ϵ1)},forn2.\displaystyle c^{t_{m}}_{n+m}=\min\left\{\dfrac{e^{\Phi(n)}}{c^{t_{0}}_{n}c^{t_{1}}_{n+1}\cdots c^{t_{m-1}}_{n+m-1}},{(c_{1}c_{2}\cdots c_{n+m-1})}^{t_{m}(b+\epsilon-1)}\right\},\ \text{for}\ n\geq 2.

Since (t0,t1,,tm)+m+1(t_{0},t_{1},\cdots,t_{m})\in\mathbb{R}_{+}^{m+1} with 0<t0t1tm0<t_{0}\leq t_{1}\leq\cdots\leq t_{m} and Φ\Phi is nondecreasing, we have cn1c_{n}\geq 1 for all n1n\geq 1. Thus

lim supnlogcn+1log(c1c2cn)lim supnlog(c1c2cn)b+ϵ1log(c1c2cn)=b+ϵ1.\limsup\limits_{n\to\infty}\frac{\log c_{n+1}}{\log(c_{1}c_{2}\cdots c_{n})}\leq\limsup\limits_{n\to\infty}\frac{\log{(c_{1}c_{2}\cdots c_{n})}^{b+\epsilon-1}}{\log(c_{1}c_{2}\cdots c_{n})}=b+\epsilon-1. (4.2)

We also claim that

cn+mtm=eφ(n)cnt0cn+1t1cn+m1tm1for infinitely manyn.c^{t_{m}}_{n+m}=\frac{e^{\varphi(n)}}{c^{t_{0}}_{n}c^{t_{1}}_{n+1}\cdots c^{t_{m-1}}_{n+m-1}}\ \text{for infinitely many}\ n. (4.3)

In order to prove (4.3), we first show that

cn+mtm=eΦ(n)cnt0cn+1t1cn+m1tm1for infinitely manyn.c^{t_{m}}_{n+m}=\frac{e^{\Phi(n)}}{c^{t_{0}}_{n}c^{t_{1}}_{n+1}\cdots c^{t_{m-1}}_{n+m-1}}\ \text{for infinitely many}\ n. (4.4)

If not, there exists NN\in\mathbb{N} such that for any nNn\geq N,

{cn+m=(c1c2cn+m1)b+ϵ1eΦ(n)cnt0cn+1t1cn+m1tm1>(c1c2cn+m1)tm(b+ϵ1).\begin{cases}&c_{n+m}={\left(c_{1}c_{2}\cdots c_{n+m-1}\right)}^{b+\epsilon-1}\\ &\frac{e^{\Phi(n)}}{c^{t_{0}}_{n}c^{t_{1}}_{n+1}\cdots c^{t_{m-1}}_{n+m-1}}>{\left(c_{1}c_{2}\cdots c_{n+m-1}\right)}^{t_{m}(b+\epsilon-1)}.\end{cases} (4.5)

Then

{cn+m=cn+m1b+ϵeΦ(n)>(c1c2cn+m1)b+ϵ1c˙nt0cn+1t1cn+m1tm1>(c1c2cn+m1)tm(b+ϵ1).\begin{cases}&c_{n+m}=c^{b+\epsilon}_{n+m-1}\\ &e^{\Phi(n)}>{\left(c_{1}c_{2}\cdots c_{n+m-1}\right)}^{b+\epsilon-1}\dot{c}^{t_{0}}_{n}c^{t_{1}}_{n+1}\cdots c^{t_{m-1}}_{n+m-1}>{\left(c_{1}c_{2}\cdots c_{n+m-1}\right)}^{t_{m}(b+\epsilon-1)}.\end{cases} (4.6)

Therefore, by (4.5) and (4.6)

k=1nck=\displaystyle\prod_{k=1}^{n}c_{k}= (k=1N+m1ck)cN+mcN+m+1cn\displaystyle\left(\prod_{k=1}^{N+m-1}c_{k}\right)\cdot c_{N+m}\cdot c_{N+m+1}\cdots c_{n} (4.7)
=\displaystyle= (k=1N+m1ck)(k=1N+m1ck)b+ϵ1cN+m+1cn\displaystyle\left(\prod_{k=1}^{N+m-1}c_{k}\right)\cdot{\left(\prod_{k=1}^{N+m-1}c_{k}\right)}^{b+\epsilon-1}\cdot c_{N+m+1}\cdots c_{n}
=\displaystyle= (k=1N+m1ck)(k=1N+m1ck)b+ϵ1(k=1N+m1ck)(b+ϵ1)(b+ϵ)\displaystyle\left(\prod_{k=1}^{N+m-1}c_{k}\right)\cdot{\left(\prod_{k=1}^{N+m-1}c_{k}\right)}^{b+\epsilon-1}\cdot{\left(\prod_{k=1}^{N+m-1}c_{k}\right)}^{(b+\epsilon-1)(b+\epsilon)}
\displaystyle\cdots (k=1N+m1ck)(b+ϵ1)(b+ϵ)nNm\displaystyle{\left(\prod_{k=1}^{N+m-1}c_{k}\right)}^{(b+\epsilon-1){(b+\epsilon)}^{n-N-m}}
=\displaystyle= (k=1N+m1ck)(b+ϵ)nNm+1.\displaystyle{\left(\prod_{k=1}^{N+m-1}c_{k}\right)}^{{(b+\epsilon)}^{n-N-m+1}}.

Combining (4.6) with (4.7), we obtain

lim infnΦ(n+1)n+1\displaystyle\liminf\limits_{n\to\infty}\frac{\Phi(n+1)}{n+1} >lim infnloglog(c1c2cn+m)tm(b+ϵ1)n+1\displaystyle>\liminf\limits_{n\to\infty}\dfrac{{\log\log\left(c_{1}c_{2}\cdots c_{n+m}\right)}^{t_{m}(b+\epsilon-1)}}{n+1}
=lim infnloglog(k=1N+m1ck)tm(b+ϵ1)(b+ϵ)nN+1n+1=log(b+ϵ).\displaystyle=\liminf\limits_{n\to\infty}\dfrac{\log\log{\left(\prod\limits_{k=1}^{N+m-1}c_{k}\right)}^{t_{m}(b+\epsilon-1){(b+\epsilon)}^{n-N+1}}}{n+1}=\log(b+\epsilon).

Then

lim infnφ(n+1)n+1lim infnΦ(n+1)n+1>log(b+ϵ)>logb,\liminf\limits_{n\to\infty}\frac{\varphi(n+1)}{n+1}\geq\liminf\limits_{n\to\infty}\frac{\Phi(n+1)}{n+1}>\log(b+\epsilon)>\log b,

which contradicts to logb=lim infnφ(n)n.\log b=\liminf\limits_{n\to\infty}\frac{\varphi(n)}{n}.

Now we begin to prove (4.3). If the equality (4.4) holds for some nn such that Φ(n)φ(n)\Phi(n)\neq\varphi(n), then Φ(n)=Φ(n+1)\Phi(n)=\Phi(n+1), and the equality (4.4) holds for n+1n+1, since

eΦ(n+1)cn+1t0cn+2t1cn+mtm1=eΦ(n)cn+1t0cn+2t1cn+mtm1=cnt0cn+1t1t0cn+mtmtm1\displaystyle\frac{e^{\Phi(n+1)}}{c^{t_{0}}_{n+1}c^{t_{1}}_{n+2}\cdots c^{t_{m-1}}_{n+m}}=\frac{e^{\Phi(n)}}{c^{t_{0}}_{n+1}c^{t_{1}}_{n+2}\cdots c^{t_{m-1}}_{n+m}}=c^{t_{0}}_{n}c^{t_{1}-t_{0}}_{n+1}\cdots c^{t_{m}-t_{m-1}}_{n+m}
\displaystyle\leq (c1c2cn1)t0(b+ϵ1)(c1c2cn)(t1t0)(b+ϵ1)\displaystyle{\left(c_{1}c_{2}\cdots c_{n-1}\right)}^{t_{0}(b+\epsilon-1)}{\left(c_{1}c_{2}\cdots c_{n}\right)}^{(t_{1}-t_{0})(b+\epsilon-1)}
(c1c2cn+m1)(tmtm1)(b+ϵ1)\displaystyle\cdots{\left(c_{1}c_{2}\cdots c_{n+m-1}\right)}^{(t_{m}-t_{m-1})(b+\epsilon-1)}
=\displaystyle= (c1c2cn1)tm(b+ϵ1)cn(tmt0)(b+ϵ1)cn+m1(tmtm1)(b+ϵ1)\displaystyle{(c_{1}c_{2}\cdots c_{n-1})}^{t_{m}(b+\epsilon-1)}c_{n}^{(t_{m}-t_{0})(b+\epsilon-1)}\cdots c_{n+m-1}^{(t_{m}-t_{m-1})(b+\epsilon-1)}
<\displaystyle< (c1c2cn+m1)tm(b+ϵ1).\displaystyle{(c_{1}c_{2}\cdots c_{n+m-1})}^{t_{m}(b+\epsilon-1)}.

By the fact that Φ(n)=φ(n)\Phi(n)=\varphi(n) for infinitely many nn\in\mathbb{N}, we can repeat this argument until we get to some n+kn+k such that Φ(n+k)=φ(n+k)\Phi(n+k)=\varphi(n+k). Then the desired result is obtained.

Combining (4.1) with (4.3), we have

lim supnlog(cnt0cn+1t1cn+mtm)φ(n)=1.\limsup\limits_{n\to\infty}\dfrac{\log\left(c^{t_{0}}_{n}c^{t_{1}}_{n+1}\cdots c^{t_{m}}_{n+m}\right)}{\varphi(n)}=1. (4.8)

Step II. We use the sequence {cn}n1\{c_{n}\}_{n\geq 1} to construct a subset of E¯({ti}i=0m,φ).\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi).

By φ(n)/n\varphi(n)/n\to\infty as nn\to\infty, we choose an increasing sequence {nk}k=1\{n_{k}\}_{k=1}^{\infty} such that for each k1k\geq 1

φ(n)nk2,whennnk.\frac{\varphi(n)}{n}\geq k^{2},\ \text{when}\ n\geq n_{k}.

Let αn=2\alpha_{n}=2 if 1n<n11\leq n<n_{1} and

αn=k+1,whennkn<nk+1.\alpha_{n}=k+1,\ \text{when}\ n_{k}\leq n<n_{k+1}.

For any n1n\geq 1, there exists k(n)k(n) such that nk(n)n+m<nk(n)+1n_{k(n)}\leq n+m<n_{k(n)+1}. Then

limnlog(αnt0αn+1t1αn+mtm)φ(n)limn(t0++tm)log(k(n)+1)nk(n)2=0,\lim\limits_{n\to\infty}\dfrac{\log\left(\alpha^{t_{0}}_{n}\alpha^{t_{1}}_{n+1}\cdots\alpha^{t_{m}}_{n+m}\right)}{\varphi(n)}\leq\lim\limits_{n\to\infty}\dfrac{(t_{0}+\cdots+t_{m})\log(k(n)+1)}{n{k(n)}^{2}}=0, (4.9)

and

limnlogαn+1log(α1α2αn)limnlog(n+1)nlog2=0.\lim\limits_{n\to\infty}\dfrac{\log\alpha_{n+1}}{\log(\alpha_{1}\alpha_{2}\cdots\alpha_{n})}\leq\lim\limits_{n\to\infty}\dfrac{\log(n+1)}{n\log 2}=0. (4.10)

For any n1,n\geq 1, take sn=cn+αns_{n}=c_{n}+\alpha_{n}. Then we have sns_{n}\to\infty as nn\to\infty.

Define

E({sn}n1)={x[0,1):snan(x)<2sn,for anyn1}.E(\{s_{n}\}_{n\geq 1})=\left\{x\in[0,1):\lfloor s_{n}\rfloor\leq a_{n}(x)<2\lfloor s_{n}\rfloor,\ \text{for any}\ n\geq 1\right\}.

Since cn1c_{n}\geq 1 and αn2\alpha_{n}\geq 2 for all n1n\geq 1, we can check that for any n1n\geq 1

logcnlogsnlogcn+2logαn.\log c_{n}\leq\log s_{n}\leq\log c_{n}+2\log\alpha_{n}.

Combining (4.8), (4.9), (4.10), we get

lim supnlog(snt0sn+1t1sn+mtm)φ(n)=1.\limsup\limits_{n\to\infty}\dfrac{\log\left(s^{t_{0}}_{n}s^{t_{1}}_{n+1}\cdots s^{t_{m}}_{n+m}\right)}{\varphi(n)}=1.

So E({sn}n1)E¯({ti}i=0m,φ).E(\{s_{n}\}_{n\geq 1})\subset\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi). Applying Lemma 2.2, we obtain

dimHE¯({ti}i=0m,φ)dimHE({sn}n1)=12+lim supnlogsn+1log(s1s2sn)(4.2)1b+1+ϵ.\dim_{\mathrm{H}}\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi)\geq\dim_{\mathrm{H}}E(\{s_{n}\}_{n\geq 1})=\dfrac{1}{2+\limsup\limits_{n\to\infty}\frac{\log s_{n+1}}{\log(s_{1}s_{2}\cdots s_{n})}}\overset{\eqref{limsupb-1}}{\geq}\frac{1}{b+1+\epsilon}.

Therefore,

dimHE¯({ti}i=0m,φ)1b+1.\dim_{\mathrm{H}}\overline{E}(\{t_{i}\}_{i=0}^{m},\varphi)\geq\frac{1}{b+1}.

Acknowledgements

A. Bakhtawar is supported by the Australian Research Council Discovery Project (ARC Grant DP180100201) and J. Feng is supported by the National Natural Science Foundation of China (NSFC Grant No. 11901204). J. Feng would like to thank China Scholarship Council financial support (No. 202106160053). The authors are grateful to Professor Lingmin Liao for helpful discussions.

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