Dimension theory of Diophantine approximation related to $\beta$-transformations

TL;DR

This paper investigates the Hausdorff dimension of sets related to Diophantine approximation under $eta$-transformations, providing explicit dimension calculations for sets defined by approximation conditions.

Contribution

It introduces a novel analysis of Diophantine approximation sets for $eta$-transformations and computes their Hausdorff dimensions, extending understanding of orbit approximation properties.

Findings

Calculated Hausdorff dimension of $igcap ext{sets}$

Established dimension estimates for approximation sets

Extended results to general $eta$-transformations

Abstract

Let $T_\beta$ be the $\beta$-transformation on $[0,1)$ defined by $$T_\beta(x)=\beta x\text{ mod }1.$$ We study the Diophantine approximation of the orbit of a point $x$ under $T_\beta$. Precisely, for given two positive functions $\psi_1,~\psi_2: \mathbb{N} \rightarrow \mathbb{R}^+$, define $$\mathcal{L}(\psi_1):=\left\{x\in[0,1]:T_\beta^n x<\psi_1(n),\text{ for infinitely many $n\in\mathbb{N}$}\right\},$$ $$\mathcal{U}(\psi_2):=\left\{x\in [0,1]:\forall~N\gg1,~\exists~n\in[0,N],\ s.t.\ T^n_\beta x<\psi_2(N)\right\},$$ where $\gg$ means large enough. We compute the Hausdorff dimension of the set $\mathcal{L}(\psi_1)\cap\mathcal{U}(\psi_2)$. As a corollary, we estimate the Hausdorff dimension of the set $\mathcal{U}(\psi_2)$.

Equations

$$T_{\beta }(x)=\beta x\text{ mod }1.$$

$$\mathcal{L}({\psi }_1):=\left\{x\in [0,1]:T_{\beta }^{n}x<{\psi }_1(n),\text{ for infinitely many n∈N}\right\},$$

$$\mathcal{U}({\psi }_2):=\left\{x\in [0,1]:\forall N\gg 1,\exists n\in [0,N],s.t.T_{\beta }^{n}x<{\psi }_2(N)\right\},$$

$$\Vert n\theta \Vert <Q^{-1},$$

$$\Vert n\theta \Vert <n^{-1}.$$

$${\mathcal{L}}_{\psi }:=\left\{\theta \in \mathbb{R}:\Vert n\theta \Vert <\psi (n),\text{ for infinitely many }n\in \mathbb{N}\right\}$$

$$\left\{\theta \in \mathbb{R}:\Vert n\theta \Vert <\frac{1}{n^{\tau }},\text{ for infinitely many }n\in \mathbb{N}\right\}$$

$$\left\{x\in M:T^{n}x\in B(n),\text{ for infinitely many n∈N}\right\},$$

$$T_{\beta }(x):=\beta x-\lfloor \beta x\rfloor ,$$

$$\mathcal{L}(\psi ,x_0):=\{x\in [0,1]:|T_{\beta }^{n}x-x_0|<\psi (n),\text{ for infinitely many n∈N}\}.$$

$${\dim }_H\left(\mathcal{L}(\psi ,x_0)\right)=\frac{1}{1+v},\text{where v:=n→∞liminfn−logβψ(n)}.$$

$${\nu }_{\beta }(x):=\sup \left\{v\ge 0:T_{\beta }^{n}x<({\beta }^n{)}^{-v},\text{ for infinitely many n∈N}\right\},$$

$${\hat{\nu }}_{\beta }(x):=\sup \left\{v\ge 0:\forall N\gg 1,T_{\beta }^{n}x<({\beta }^N{)}^{-v}\text{ has a solution n∈[0,N]}\right\}.$$

$$\left\{x\in [0,1]:{\nu }_{\beta }(x)=v\right\}\cap \left\{x\in [0,1]:{\hat{\nu }}_{\beta }(x)\ge \hat{v}\right\}$$

$${\dim }_H\left(\{x\in [0,1]:{\nu }_{\beta }(x)=v\}\cap \{x\in [0,1]:{\hat{\nu }}_{\beta }(x)=\hat{v}\}\right)=\frac{v-\hat{v}-v\hat{v}}{(1+v)(v-\hat{v})}.$$

$$\mathcal{L}({\psi }_1):=\left\{x\in [0,1]:T_{\beta }^{n}x<{\psi }_1(n),\text{ for infinitely many n∈N}\right\},$$

$$\mathcal{U}({\psi }_2):=\left\{x\in [0,1]:\forall N\gg 1,T_{\beta }^{n}x<{\psi }_2(N)\text{ has a solution n∈[0,N]}\right\}.$$

$${\underline{v}}_1:=\underset{n\to \infty }{\mathrm{liminf}}\frac{-{\log }_{\beta }{\psi }_1(n)}{n},{\overline{v}}_1:=\underset{n\to \infty }{\mathrm{limsup}}\frac{-{\log }_{\beta }{\psi }_1(n)}{n};$$

$${\underline{v}}_2:=\underset{n\to \infty }{\mathrm{liminf}}\frac{-{\log }_{\beta }{\psi }_2(n)}{n},{\overline{v}}_2:=\underset{n\to \infty }{\mathrm{limsup}}\frac{-{\log }_{\beta }{\psi }_2(n)}{n}.$$

$$\underset{j\to \infty }{\lim }\frac{-{\log }_{\beta }{\psi }_1(n_j)}{n_j}={\underline{v}}_1.$$

$$1<{\beta }^{-n_j({\underline{v}}_1+\epsilon )}\le {\psi }_1(n_j),\text{for any j≥j0}.$$

$$[0,1)\subseteq \mathcal{L}({\psi }_1).$$

$${\psi }_2(n_i)>1,\text{for i=1,2,3⋯}$$

$${\tilde{\psi }}_2(n)=\{\begin{array}{ll}{\psi }_2(n),& \text{ if n=ni}\\ 1,& \text{ if n=ni}\end{array},i=1,2,\cdots$$

$${\dim }_H\left(\mathcal{L}({\psi }_1)\cap \mathcal{U}({\psi }_2)\right)=1;$$

$$m\left(\mathcal{L}({\psi }_1)\cap \mathcal{U}({\psi }_2)\right)=0,$$

$$m\left(\mathcal{L}({\psi }_1)\cap \mathcal{U}({\psi }_2)\right)=1.$$

$$0\le {\dim }_H\left(\mathcal{L}({\psi }_1)\cap \mathcal{U}({\psi }_2)\right)\le \min \left\{\frac{1}{1+{\overline{v}}_2},{\left(\frac{1-{\underline{v}}_2}{1+{\underline{v}}_2}\right)}^2\right\}.$$

$${\left(\frac{1-{\overline{v}}_2}{1+{\overline{v}}_2}\right)}^2\le {\dim }_H\left(\mathcal{L}({\psi }_1)\cap \mathcal{U}({\psi }_2)\right)\le \min \left\{\frac{1}{1+{\overline{v}}_2},{\left(\frac{1-{\underline{v}}_2}{1+{\underline{v}}_2}\right)}^2\right\}.$$

$${\left(\frac{1-{\overline{v}}_2}{1+{\overline{v}}_2}\right)}^2\le {\dim }_H\left(\mathcal{L}({\psi }_1)\cap \mathcal{U}({\psi }_2)\right)\le \frac{{\underline{v}}_1-{\underline{v}}_2-{\underline{v}}_1\cdot {\underline{v}}_2}{(1+{\underline{v}}_1)({\underline{v}}_1-{\underline{v}}_2)}.$$

Full text

[TABLE]

We study the Diophantine approximation of the orbit of a point $x$ under $T_{\beta}$. Precisely, for given two positive functions $\psi_{1},~{}\psi_{2}:\mathbb{N}\rightarrow\mathbb{R}^{+}$, define

[TABLE]

[TABLE]

where $\gg$ means large enough. We compute the Hausdorff dimension of the set $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$. As a corollary, we estimate the Hausdorff dimension of the set $\mathcal{U}(\psi_{2})$.

1. Introduction

Diophantine approximation, which originaly asks how closely can a given irrational number be approximated by a rational number $p/q$ with denominator $q$ no larger than a fixed positive integer $q_{0}$, has been widely studied by mathematicians. In 1842, Dirichlet [5] proved the following theorem.

Dirichlet Theorem Given two real numbers $\theta,~{}Q$ with $Q\geq 1$, there is an integer $n$ with $1\leq n\leq Q$ such that

[TABLE]

where $\lVert\xi\rVert$ denotes the distance from $\xi$ to the nearest integer.

Dirichlet Theorem is called a uniform approximation theorem in [18, pp.2]. A weak form of Dirichlet Theorem, called an asymptotic approximation theorem in [18, pp.2], which was often refered to as a corollary of Dirichlet Theorem in the litterature has already existed in the book of Legendre [12, 1808, pp.18-19] (using a continued fraction fact): for any real number $\theta$, there are infinitely many $n\in\mathbb{N}$ such that

[TABLE]

For the general case, Khintchine in 1924 [11] showed that for a positive function $\psi:\mathbb{N}\rightarrow\mathbb{R}^{+}$, if $x\mapsto x\psi(x)$ is non-increasing, then

[TABLE]

has Lebesgue measure zero if the series $\sum\psi(n)$ converges and has full Lebesgue measure otherwise. In the case where the set has Lebesgue measure zero, it is natural to calculate the Hausdorff dimension of $\mathcal{L}_{\psi}$. The first result on the Hausdorff dimension of $\mathcal{L}_{\psi}$ dates back to Jarník-Bosicovitch Theorem [2, 10]. It was shown that the set

[TABLE]

has Hausdorff deminsion $\dfrac{2}{1+\tau}$, for any $\tau>1$.

In analogy with the classical Diophantine approximation, Hill and Velani [9] studied the approximation properties of the orbits of a dynamical system and introduced the so called shrinking target problems: for a measure preserving transformation $T:M\rightarrow M$ on a manifold $M$, what is the size (Lebesgue measure, Hausdorff dimension) of the set

[TABLE]

where $B(n)=B(x_{0},r(n))$ is a ball centred at $x_{0}$ with radius $r(n)$($r(n)\rightarrow 0$)? They answered the case where $T$ is an expanding rational map of the Riemann sphere $\overline{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$.

In this papper, we are interested in the approximation properties of the orbits of $\beta$-transformations. The $\beta$-transformation $T_{\beta}~{}(\beta>1)$ on $[0,1)$ is defined by

[TABLE]

where $\lfloor\cdot\rfloor$ is the integer part function. For any positive function $\psi:\mathbb{N}\rightarrow\mathbb{R}^{+}$, define the set of $\psi$-well asymptotically approximable points by $x_{0}$ as

[TABLE]

By [15, Theorem 2A, B, C], the set $\mathcal{L}(\psi,x_{0})$ has Lebesgue measure zero if and only if the series $\sum\psi(n)$ converges. Shen and Wang [17, Theorem 1.1] established the following result on the Hausdorff dimension of $\mathcal{L}(\psi,x_{0})$.

Theorem SW([17, Theorem 1.1]) For any real number $\beta>1$ and any point $x_{0}\in[0,1]$, one has

[TABLE]

Parallel to the asymptotic approximation theorem, it is also worth of studying the uniform approximation properties as in Dirichlet Theorem. The uniform Diophantine approximation related to $\beta$-transformations was studied by Bugeaud and Liao [4]. For $x\in[0,1)$, let

[TABLE]

[TABLE]

Bugeaud and Liao [4] proved the following theorem.

Theorem BL ([4, Theorem 1.4]) For any $v\in(0,+\infty)$ and any $\hat{v}\in(0,1)$, if $v<\hat{v}/(1-\hat{v})$, then the set

[TABLE]

is empty. Otherwise,

[TABLE]

The exponents $\nu_{\beta}$ and $\hat{\nu}_{\beta}$ were introduced in [1](see also [3, Ch.7]). They are strongly related to the run-length function of $\beta$-expansions (see [19]). The aim of this paper is to study the Diophantine approximation sets in [4] when the approximation speed function $n\mapsto\beta^{-nv}$ is replaced by a general positive function. More precisely, fix two positive functions $\psi_{1},~{}\psi_{2}:\mathbb{N}\rightarrow\mathbb{R}^{+}$, and define

[TABLE]

[TABLE]

We will estimate the Hausdorff dimension of the sets $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ and $\mathcal{U}(\psi_{2})$. Let

[TABLE]

[TABLE]

If $\underline{v}_{1}<0$, by the definition of $\underline{v}_{1}$, there is a sequence $\{n_{j}\}$ such that

[TABLE]

Then, for $\varepsilon>0$ small enough, there exists an integer $j_{0}$ such that

[TABLE]

By the fact $T^{n}_{\beta}x<1$, for any $x\in[0,1)$ and any $n\in\mathbb{N}$, for any $x\in[0,1)$, we have $T^{n_{j}}_{\beta}x<1<\psi_{1}(n_{j})$. This implies

[TABLE]

On the other hand, if we take all the integers $n_{i}$ with the following property

[TABLE]

then for any $x\in[0,1)$ and any integer $n\in[1,n_{i}]$, we have $T^{n}_{\beta}x<1<\psi_{2}(n_{i})$. Thus, we can replace $\psi_{2}(n)$ by the function

[TABLE]

The size (Lebesgue measure, Hausdorff dimension) of the sets $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ and $\mathcal{U}(\psi_{2})$ are the same as that of the sets $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\widetilde{\psi}_{2})$ and $\mathcal{U}(\widetilde{\psi}_{2})$. Therefore, in this paper, we always assume $\underline{v}_{1}\geq 0$ and $\underline{v}_{2}\geq 0$. We establish the following theorems on the Hausdorff dimension of the set $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$.

Theorem A**.**

1. $(1)$

If $\underline{v}_{1}=\overline{v}_{1}=\underline{v}_{2}=\overline{v}_{2}=0$, then

[TABLE] 2. $(2)$

If $\underline{v}_{2}=\infty$ and $0\leq\underline{v}_{1}\leq\overline{v}_{1}\leq\infty$, then $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is countable. 3. $(3)$

If $\underline{v}_{1}=\infty$ and $0\leq\underline{v}_{2}\leq\overline{v}_{2}\leq\infty$, then ${\rm dim}_{H}\left(\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})\right)=0$.

Remark**.**

For Item (1), the set $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is not necessary of full Lebesgue measure. In fact, if the series $\sum\psi_{1}(n)$ converges, by [15, Theorem 2A, B, C],

[TABLE]

where $m(A)$ denotes the Lebesgue measure of $A$. However, the set $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ can also be of full Lebesgue measure. For example, if $\psi_{1}(n)=\psi_{2}(n)=1/n$, according to Dmitry, Konstantoulas, and Florian [6, Theorem 1.1],

[TABLE]

For Item $(3)$, if $\underline{v}_{2}=\infty$, then $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is countable. If $1<\underline{v}_{2}<\infty$, then $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is empty (see Lemma 3.5). If $0<\underline{v}_{2}\leq 1$, then $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is uncountable (see Proposition 3.6).

Theorem B**.**

If $\underline{v}_{2}>1$, then $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is countable. If $\overline{v}_{1}/(2+\overline{v}_{1})\leq\underline{v}_{2}\leq 1<\overline{v}_{2}$, then

[TABLE]

If $\underline{v}_{1}/(2+\underline{v}_{1})<\underline{v}_{2}\leq\overline{v}_{2}\leq 1$ and $\overline{v}_{1}/(2+\overline{v}_{1})<\overline{v}_{2}$, then

[TABLE]

If $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$ and $\overline{v}_{1}/(2+\overline{v}_{1})<\overline{v}_{2}\leq 1$, then

[TABLE]

If $\underline{v}_{1}/(2+\underline{v}_{1})<\underline{v}_{2}\leq\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, then

[TABLE]

If $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$ and $\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, then

[TABLE]

We remark that Theorems A and B give all the cases. We also estimate the Hausdorff dimension of $\mathcal{U}(\psi_{2})$.

Theorem C**.**

If $\underline{v}_{2}>1$, then $\mathcal{U}(\psi_{2})$ is countable. If $\underline{v}_{2}\leq 1<\overline{v}_{2}$, then

[TABLE]

If $\overline{v}_{2}\leq 1$, then

[TABLE]

We will show in Examples 5.1, 5.2, 5.3, 5.4, 5.5, 5.6 and 5.7, that the upper and lower bound of the Hausdorff dimension in Theorems B and C can be all reached. When $\underline{v}_{1}=\overline{v}_{1}=0$, we have the result as Theorem D.

Theorem D**.**

Assume $\underline{v}_{1}=\overline{v}_{1}=0$. If $\underline{v}_{2}>0$, then $\mathcal{U}(\psi_{2})\subseteq\mathcal{L}(\psi_{1})$.

Our paper is organized as follows. We recall some classical results of the theory of $\beta$-expansion in Section $2$. Theorems A and B are proved in Section $3$. Section $4$ establishes Theorems C and D. In Section $5$, we give examples to show that the estimations in Theorems B and C are sharp.

2. $\beta$-expansions

The notion of $\beta$-expansion was introduced by Rényi [16] in 1957. For any $\beta>1$, the $\beta$-transformation $T_{\beta}$ on $[0,1)$ is defined by

[TABLE]

where $\lfloor\xi\rfloor$ denotes the largest integer less than or equal to $\xi$. Let

[TABLE]

Definition 2.1**.**

The $\beta$-expansion of a number $x\in[0,1)$ is the sequence $\{\varepsilon_{n}\}_{n\geq 1}:=\{\varepsilon_{n}(x,\beta)\}_{n\geq 1}$ of integers from $\{0,1,\cdots,\lceil\beta\rfloor\}$ such that

[TABLE]

where

[TABLE]

We also write $d_{\beta}(x)=\left(\varepsilon_{1},\cdots,\varepsilon_{n},\cdots\right)$.

We can extend the definition of the $\beta$-transformation to the point $1$ as:

[TABLE]

One can obtain

[TABLE]

where \varepsilon_{1}(1,\beta)=\lfloor\beta\rfloor,~{}\varepsilon_{n}=\lfloor\beta T^{n-1}_{\beta}1\rfloor,~{}\text{for all n\geq 2}. We also write

[TABLE]

If $d_{\beta}(1)$ is finite, i.e., there is an integer $m>0$ such that $\varepsilon_{m}(1,\beta)\neq 0$ and $\varepsilon_{i}(1,\beta)=0$ for all $i>m$, then $\beta$ is called a simple Parry number. In this case, the infinite $\beta$-expansion of $1$ is defined as:

[TABLE]

where $(\omega)^{\infty}$ denotes the periodic sequence $(\omega,\omega,\cdots)$. If $d_{\beta}(1)$ is infinite, then we define

[TABLE]

Endow the set $\{0,1,\cdots,\lceil\beta\rfloor\}^{\mathbb{N}}$ with the product topology and define the one-sided shift operator $\sigma$ as:

[TABLE]

for any infinite sequence $(\omega_{n})_{n\geq 1}$ in $\{0,1,\cdots,\lceil\beta\rfloor\}^{\mathbb{N}}$. The lexicographical order $<_{lex}$ on $\{0,1,\cdots,\lceil\beta\rfloor\}^{\mathbb{N}}$ is defined as:

[TABLE]

if $\omega_{1}<\omega^{\prime}_{1}$ or if there is an integer $k\geq 2$ such that for all $1\leq i<k$, $\omega_{i}=\omega^{\prime}_{i}$ but $\omega_{k}<\omega^{\prime}_{k}$. Denote by $\omega\leq_{lex}\omega^{\prime}$ if $\omega<_{lex}\omega^{\prime}$ or $\omega=\omega^{\prime}$.

Definition 2.2**.**

A finite word $(\omega_{1},\omega_{2},\cdots,\omega_{n})$ is called $\beta$-admissible, if there is $x\in[0,1]$ such that the $\beta$-expansion of $x$ begins with $(\omega_{1},\omega_{2},\cdots,\omega_{n}).$ An infinite sequence $(\omega_{1},\omega_{2},\cdots,\omega_{n},\cdots)$ is called $\beta$-admissible, if there is $x\in[0,1]$ such that the $\beta$-expansion of $x$ is $(\omega_{1},\omega_{2},\cdots,\omega_{n},\cdots).$

Denote by $\Sigma_{\beta}$ the set of all infinite $\beta$-admissible sequences and $\Sigma^{n}_{\beta}$ the set of all $\beta$-admissible sequences with length $n$. The $\beta$-admissible sequences are characterized by Parry [14] and Rényi [16].

Theorem 2.3**.**

Let $\beta>1$,

1. (1)

(**[14, Lemma 1]**) A word $\omega=(\omega_{n})_{n\geq 1}\in\Sigma_{\beta}$ if and only if

[TABLE] 2. (2)

(**[14, Lemma 3]**) For any $x_{1},~{}x_{2}\in[0,1]$, $x_{1}<x_{2}$ if and only if

[TABLE] 3. (3)

(**[14, Lemma 4]**) For any $\beta_{2}>\beta_{1}>1$, one has

[TABLE]

Theorem 2.4**.**

([16, Theorem 2]) For any $\beta>1$, one has

[TABLE]

where $\sharp$ denotes the cardinality of a finite set.

For every $(\omega_{1},\cdots,\omega_{n})\in\Sigma^{n}_{\beta}$, we call

[TABLE]

an $n$-th order basic interval with respect to $\beta$. Denote by $I_{n}(x)$ the $n$-th order basic interval containing $x$. The basic intervals are also called cylinders by some authors. It is crucial to estimate the lengths of the basic intervals. We will use the key notion of “full basic interval”as follows (see [13, 8]).

Definition 2.5**.**

For any $(\omega_{1},\cdots,\omega_{n})\in\Sigma^{n}_{\beta}$, a basic interval $I_{n}(\omega_{1},\cdots,\omega_{n})$ is said to be full if its length is $\beta^{-n}$.

Proposition 2.6**.**

([8, Lemma 3.1] and [17, Lemma 2.5])

For any $(\omega_{1},\cdots,\omega_{n})\in\Sigma^{n}_{\beta}$, the following statements are equivalent:

1. (1)

$I_{n}(\omega_{1},\cdots,\omega_{n})$* is a full basic interval;* 2. (2)

$T^{n}_{\beta}I_{n}(\omega_{1},\cdots,\omega_{n})=[0,1)$; 3. (3)

For any $\omega^{\prime}=(\omega^{\prime}_{1},\cdots,\omega^{\prime}_{m})\in\Sigma^{m}_{\beta}$, the concatenation

[TABLE]

Proposition 2.7**.**

([17, Corollary 2.6])

1. (1)

If $(\omega_{1},\cdots,\omega_{n+1})$ is a $\beta$-admissible sequence with $\omega_{n+1}\neq 0$, then

[TABLE]

is full for any $0\leq\omega^{\prime}_{n+1}<\omega_{n+1}$. 2. (2)

For every $\omega\in\Sigma^{n}_{\beta}$, if $I_{n}(\omega)$ is full, then for any $\omega^{\prime}\in\Sigma^{m}_{\beta}$, one has

[TABLE] 3. (3)

For any $\omega\in\Sigma^{n}_{\beta}$, if $I_{n+m}(\omega,\omega^{\prime})$ is a full basic interval contained in $I_{n}(\omega)$ with the smallest order, then

[TABLE]

Next, we define a sequence of numbers $\beta_{N}$ approaching to $\beta$ as follows. Let $\{\varepsilon^{\ast}_{k}(\beta):k\geq 1\}$ be the infinite $\beta$-expansion of $1$. Let $\beta_{N}$ be the unique real solution of the equation

[TABLE]

Therefore, $\beta_{N}<\beta$ and the sequence $\{\beta_{N}:N\geq 1\}$ increases and converges to $\beta$ when $N$ tends to infinity.

Lemma 2.8**.**

([17, Lemma 2.7]) For every $\omega\in\Sigma^{n}_{\beta_{N}}$ viewed as an element of $\Sigma^{n}_{\beta}$, one has

[TABLE]

3. Proofs of Theorems A and B

First, we give an easy fact which is useful for the proofs of Theorems A and B.

Fact 3.1**.**

For any $x\in[0,1)$, if there is an integer $n_{0}$ such that $T^{n_{0}}_{\beta}x=0$, then

[TABLE]

Now, we prove that if $\overline{v}_{2}=\infty$, then the Hausdorff dimensions of the sets $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ and $\mathcal{U}(\psi_{2})$ are zero.

Proposition 3.2**.**

If $\overline{v}_{2}=\infty$, then $\mathcal{U}(\psi_{2})\subseteq\{x\in[0,1]:\nu_{\beta}(x)=\infty\}$. Thus,

[TABLE]

Proof.

For every $x\in\mathcal{U}(\psi_{2})$, we distinguish two cases:

Case 1: There is an integer $n_{0}$ such that $T^{n_{0}}_{\beta}x=0$. By Fact 3.1, we have

[TABLE]

Case 2: For any $n\in\mathbb{N}$, we always have $T^{n}_{\beta}x>0$. Since $x\in\mathcal{U}(\psi_{2})$, there is $N_{0}\geq 1$ such that for any $N\geq N_{0}$, there is an integer $n\in[0,N]$ such that

[TABLE]

Since $\limsup\limits_{n\rightarrow\infty}\dfrac{-\log_{\beta}\psi_{2}(n)}{n}=\overline{v}_{2}=\infty$, for any $L>0$ large enough, there is a sequence $\{n_{i}\}$ such that $\psi_{2}(n_{i})\leq\beta^{-n_{i}L}$. Let $m_{1}:=\min\{n_{i}:~{}n_{i}\geq N_{0}\}$, there is an integer $j_{1}\in[0,m_{1}]$ such that

[TABLE]

Take $m_{2}:=\min\left\{n_{i}>m_{1}:\beta^{-n_{i}L}<T_{\beta}^{j_{1}}x\right\}$. There is $j_{2}\in[0,m_{2}]$ such that

[TABLE]

Since $T_{\beta}^{j_{2}}x<\psi_{2}(m_{2})\leq\beta^{-m_{2}L}<T_{\beta}^{j_{1}}x$, $j_{2}\neq j_{1}$. Repeat this process, one can get a sequence of pairwise disjoint integers $\{j_{i}:i\geq 1\}$ such that

[TABLE]

Therefore, $\nu_{\beta}(x)\geq L$. By the arbitrariness of $L$, we have $\nu_{\beta}(x)=\infty$.

Hence, in all cases, we have

[TABLE]

By Theorem SW,

[TABLE]

∎

We discuss the relation between $\nu_{\beta}(x)$, $\hat{\nu}_{\beta}(x)$ and $\underline{v}_{1}$, $\overline{v}_{1}$, $\underline{v}_{2}$, $\overline{v}_{2}$, which are important to the proof of Theorem B.

Lemma 3.3**.**

For $0\leq\underline{v}_{1}\leq\overline{v}_{1}<\infty$ and $0\leq\underline{v}_{2}\leq\overline{v}_{2}<\infty$, one has

1. $(1)$

$\{x\in[0,1]:\nu_{\beta}(x)>\overline{v}_{1}\}\cap\{x\in[0,1]:\hat{\nu}_{\beta}(x)>\overline{v}_{2}\}\subseteq\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2});$** 2. $(2)$

$\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})\subseteq\{x\in[0,1]:\nu_{\beta}(x)\geq\underline{v}_{1}\}\cap\{x\in[0,1]:\hat{\nu}_{\beta}(x)\geq\underline{v}_{2}\}.$**

Proof.

$(1)$ For any $x$ with $\nu_{\beta}(x)>\overline{v}_{1}$ and any $\varepsilon>0$ small enough, there is a sequence $\{n_{i}\}$ such that $T_{\beta}^{n_{i}}x<\beta^{-n_{i}(\overline{v}_{1}+\varepsilon)}$. By the definition of $\overline{v}_{1}$, for the above $\varepsilon$, there is an integer $i_{0}$ such that

[TABLE]

Then,

[TABLE]

Thus, $x\in\mathcal{L}(\psi_{1})$. Therefore,

[TABLE]

By similar discussion, $\{x\in[0,1]:\hat{\nu}_{\beta}(x)>\overline{v}_{2}\}\subseteq\mathcal{U}(\psi_{2}).$ Thus,

[TABLE]

$(2)$ For any $x\in\mathcal{L}(\psi_{1})$, there is a sequence $\{n_{i}\}$ such that

[TABLE]

By the definition of $\underline{v}_{1}$, for any $\varepsilon>0$, there is an integer $i_{0}$ such that

[TABLE]

Thus, $\nu_{\beta}(x)\geq\underline{v}_{1}-\varepsilon$. Therefore,

[TABLE]

By the arbitrariness of $\varepsilon$, one can obtain

[TABLE]

By similar discussion, $\mathcal{U}(\psi_{2})\subseteq\{x\in[0,1]:\hat{\nu}_{\beta}(x)\geq\underline{v}_{2}\}.$ Thus,

[TABLE]

∎

To prove Theorem A, we characterize the set of all points $x$ with $\nu_{\beta}(x)=\infty$ and $\hat{\nu}_{\beta}(x)=\infty$.

Lemma 3.4**.**

If $\nu_{\beta}(x)=\infty$ and $\hat{\nu}_{\beta}(x)=\infty$, then one has

[TABLE]

[TABLE]

Proof.

By Fact 3.1, the statement (1) and the inclusion

[TABLE]

are obvious. What is left is to show

[TABLE]

By contrary, for any $x$ with $\hat{\nu}_{\beta}(x)=\infty$, we suppose

[TABLE]

Then $T^{n}_{\beta}x>0$ for every $n\in\mathbb{N}$. Denote the $\beta$-expansion of $x$ by

[TABLE]

where $a_{i}\in\{0,\cdots,\lceil\beta\rfloor\}$, for all $i\geq 1$. We can take two increasing sequences $\left\{n^{\prime}_{i}:i\geq 1\right\}$ and $\left\{m^{\prime}_{i}:i\geq 1\right\}$ with the following properties:

1. (1)

For every $i\geq 1$, one has

[TABLE] 2. (2)

For every $a_{n}=0$, there is an integer $i$ such that $n^{\prime}_{i}<n<m^{\prime}_{i}$.

By the choices of $\left\{n^{\prime}_{i}:i\geq 1\right\}$ and $\left\{m^{\prime}_{i}:i\geq 1\right\}$, for every $i\geq 1$, one has $n^{\prime}_{i}<m^{\prime}_{i}<n^{\prime}_{i+1}$. Since $\hat{\nu}_{\beta}(x)>0$, one has

[TABLE]

Take $n_{1}=n^{\prime}_{1}$ and $m_{1}=m^{\prime}_{1}$. Suppose that $m_{k},~{}n_{k}$ have been defined. Let $i_{1}=1$ and $i_{k+1}:=\min\{i>i_{k}:m^{\prime}_{i}-n^{\prime}_{i}>m_{k}-n_{k}\},~{}{\rm for}~{}k\geq 1.$ Then, define

[TABLE]

Note that $\limsup_{i\rightarrow\infty}(m^{\prime}_{i}-n^{\prime}_{i})=\infty$, the sequence $\{i_{k}:k\geq 1\}$ is well defined. By this way, we obtain the subsequences $\{n_{k}:k\geq 1\}$ and $\{m_{k}:k\geq 1\}$ of $\left\{n^{\prime}_{i}:i\geq 1\right\}$ and $\left\{m^{\prime}_{i}:i\geq 1\right\}$, respectively, such that the sequence $\{m_{k}-n_{k}:k\geq 1\}$ is non-decreasing. As the similar discussion in [4], one has

[TABLE]

This contradicts our assumption $\hat{\nu}_{\beta}(x)=\infty$. Thus, we have proved

[TABLE]

Therefore,

[TABLE]

which implies that the set $\{x\in[0,1]:\hat{\nu}_{\beta}(x)=\infty\}$ is countable. ∎

Lemma 3.5**.**

The set $\{x\in[0,1]:1<\hat{\nu}_{\beta}(x)<\infty\}$ is empty.

Proof.

This follows from the proof of Item $(2)$ of Lemma 3.4. ∎

Proposition 3.6**.**

The sets

[TABLE]

are uncountable.

Proof.

For any real number $a>1$, we give a correspondence:

[TABLE]

The infinite string is a $\beta$-expansion of some $x\in[0,1)$. Denote this $x$ by $x_{a}$. Then, we can obtain a correspondence:

[TABLE]

from $\{a:a>1\}$ to $\{x_{a}\}_{a>1}$. One can check $\nu_{\beta}(x_{a})=\infty$ and $\hat{\nu}_{\beta}(x_{a})=1$. Therefore,

[TABLE]

For different $a_{1}>1$ and $a_{2}>1$, there is a $k_{0}\in\mathbb{N}^{+}$ such that

[TABLE]

Thus, $\Psi(a_{1})\neq\Psi(a_{2})$. Then, $\Phi(a_{1})\neq\Phi(a_{2})$. Hence, the cardinality of $\{a:a>1\}$ is less than or equal to that of $\{x_{a}\}_{a>1}$. Similarly, the cardinality of $\{x_{a}:a>1\}$ is less than or equal to that of $\{x\in[0,1]:\nu_{\beta}(x)=\infty\}$ ($\{x\in[0,1]:\hat{\nu}_{\beta}(x)=1\}$). Since $\{a:a>1\}$ is uncountable, $\{x\in[0,1]:\nu_{\beta}(x)=\infty\}$ and $\{x\in[0,1]:\hat{\nu}_{\beta}(x)=1\}$ are uncountable. ∎

Proof of Theorem A.

$(1)$ If $\underline{v}_{1}=\overline{v}_{1}=\underline{v}_{2}=\overline{v}_{2}=0$, then

[TABLE]

Thus, for any positive integer $m$ large enough, there is an integer $n_{0}>0$ such that

[TABLE]

Therefore,

[TABLE]

By the fact $\{x\in[0,1]:\hat{\nu}_{\beta}(x)=1/m\}\subseteq\{x\in[0,1]:\nu_{\beta}(x)\geq 1/m\}$, one has

[TABLE]

According to [4, Theorem 1.5], we have

[TABLE]

Thus, ${\rm dim}_{H}\left(\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})\right)=1.$

$(2)$ If $\underline{v}_{2}=\infty$, then

[TABLE]

For any $L>0$ large enough, there is an integer $n_{0}$ such that

[TABLE]

Therefore, for any $x\in\mathcal{U}(\psi_{2})$, one has $\hat{\nu}_{\beta}(x)\geq L$. By the arbitrariness of $L$, $\hat{\nu}_{\beta}(x)=\infty$. Thus, $\mathcal{U}(\psi_{2})\subseteq\{x\in[0,1]:\hat{\nu}_{\beta}(x)=\infty\}$. By Lemma 3.4 (2),

[TABLE]

According to Lemma 3.4 (1),

[TABLE]

Since $\{x\in[0,1]:\hat{\nu}_{\beta}(x)=\infty\}$ is countable, $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is countable.

$(3)$ If $\underline{v}_{1}=\infty$, by Theorem SW,

[TABLE]

∎

Now, we prove Theorem B. We divide the proof into three propositions as Propositions 3.7, 3.8, 3.9.

Proposition 3.7**.**

For any $0\leq\underline{v}_{2}\leq\overline{v}_{2}<\infty$, one has

[TABLE]

Proof.

By the definition of $\overline{v}_{2}$, take a subsequence $\{n_{k}:k\geq 1\}$ such that

[TABLE]

Then, for any $\varepsilon>0$, there is an integer $k_{0}$ such that

[TABLE]

For every $x\in\mathcal{U}(\psi_{2})$, by the same argument as Proposition 3.2, we have

[TABLE]

By [17, Theorem 1.1], ${\rm dim}_{H}\left(\mathcal{U}(\psi_{2})\right)\leq\dfrac{1}{1+\overline{v}_{2}}$. Then,

[TABLE]

∎

Proposition 3.8**.**

If $\underline{v}_{2}>1$, then the set $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is countable. If $\underline{v}_{1}/(2+\underline{v}_{1})<\underline{v}_{2}\leq 1$, then

[TABLE]

If $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$, then

[TABLE]

Proof.

By Lemma 3.3 $(2)$,

[TABLE]

The argument on the upper bound of the Hausdorff dimension of the set $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ can be obtained by a natural covering of the set

[TABLE]

According to Theorem A $(2)$ and $(3)$, we only need to consider the case $\nu_{\beta}(x)\in[\underline{v}_{1},\infty)$ and $\hat{\nu}_{\beta}(x)\in[\underline{v}_{2},\infty)$. For any $x\in\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$, there is a number $v_{\beta}\in[\underline{v}_{1},\infty)$ such that

[TABLE]

Denote its $\beta$-expansion by

[TABLE]

where $a_{i}\in\{0,\cdots,\lceil\beta\rfloor\}$, for all $i\geq 1$. Since $\nu_{\beta}(x)<\infty$, $T_{\beta}^{n}x>0$ for all $n\geq 0$. By the same way as Lemma 3.4, we take the maximal subsequences $\{n_{k}:k\geq 1\}$ and $\{m_{k}:k\geq 1\}$ of $\left\{n^{\prime}_{i}:i\geq 1\right\}$ and $\left\{m^{\prime}_{i}:i\geq 1\right\}$, respectively. Notice $\beta^{n_{k}-m_{k}}<T^{n_{k}}_{\beta}x<\beta^{n_{k}-m_{k}+1}$. We have the following claim.

Claim**.**

$v_{\beta}=\limsup\limits_{k\rightarrow\infty}\dfrac{m_{k}-n_{k}}{n_{k}},\quad\underline{v}_{2}\leq\liminf\limits_{k\rightarrow\infty}\dfrac{m_{k}-n_{k}}{n_{k+1}}.$**

Proof of Claim.

Without loss of generality, we assume

[TABLE]

First, we show $v_{\beta}=c_{1}$. For any $\varepsilon>0$, there is an integer $k_{0}>0$ such that

[TABLE]

Since $T^{n_{k}}_{\beta}x>\beta^{n_{k}-m_{k}}$, we have $T^{n_{k}}_{\beta}x>\beta^{n_{k}-m_{k}}\geq\beta^{-n_{k}(c_{1}+\varepsilon)}$. In general, for any $n\geq n_{k_{0}}$, there is an integer $k\geq k_{0}$ such that $n_{k}\leq n<n_{k+1}$. By the choice of $\{n_{k}\}$, we have

[TABLE]

It means $v_{\beta}=v_{\beta}(x)<c_{1}+\varepsilon$. On the other hand, by the definition of $c_{1}$, taking subsequence $\{n_{k_{i}}\}$ and $\{m_{k_{i}}\}$ such that

[TABLE]

one has $T^{n_{k_{i}}}_{\beta}x<\beta^{n_{k_{i}}-m_{k_{i}}+1}\leq\beta^{-n_{k_{i}}(c_{1}-\varepsilon)+1}$. Thus, $v_{\beta}=\nu_{\beta}(x)\geq c_{1}-\varepsilon$. By the arbitrariness of $\varepsilon$, $v_{\beta}=c_{1}$.

Next, we will prove $\underline{v}_{2}\leq c_{2}$. By the definition of $\underline{v}_{2}$, for any $\varepsilon>0$, there is an integer $n_{0}=n_{0}(\varepsilon)>0$ such that

[TABLE]

By the definition of $c_{2}$, one can take a subsequence $\{k_{i}:i\geq 1\}$ such that

[TABLE]

For the above $\varepsilon>0$, there is an integer $i_{0}=i_{0}(\varepsilon)>0$ such that

[TABLE]

By contrary, suppose $c_{2}<\underline{v}_{2}$. Then, for any $\varepsilon\in(0,(\underline{v}_{2}-c_{2})/4)$ and any integer $J\geq K:=\max\{n_{0}(\varepsilon),~{}n_{i_{0}(\varepsilon)}\}$, there is an integer $n_{k_{i+1}}>J$ such that for any integer $n\in[1,n_{k_{i+1}}]$, one has

[TABLE]

This contracts the fact $x\in\mathcal{U}(\psi_{2})$. Therefore,

[TABLE]

∎

Now, we consider

[TABLE]

[TABLE]

Since $\left(\limsup\dfrac{n_{k}}{m_{k}}\right)\cdot\left(\limsup\dfrac{m_{k}}{n_{k}}\right)\geq 1$, one has

[TABLE]

If $\underline{v}_{2}>1$, then $\underline{v}_{2}>v_{\beta}/(1+v_{\beta})$, for any $v_{\beta}\geq v_{1}$. This contradicts (3.3). Thus, $\mathbb{B}$ is empty. By Lemma 3.4,

[TABLE]

Hence, $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is countable.

If $\underline{v}_{2}\leq 1$, by the inequality (3.3), for any $v_{\beta}<\underline{v}_{2}/(1-\underline{v}_{2})$, the set $\mathbb{B}$ is empty. Therefore, we consider the case $v_{\beta}\geq\underline{v}_{2}/(1-\underline{v}_{2})$. Take a subsequence $\{k_{i}:i\geq 1\}$ such that the supremum of (3.1) is obtained. For abbreviation, we continue to write $\{n_{k}:k\geq 1\}$ and $\{m_{k}:k\geq 1\}$ for the subsequence $\{n_{k_{i}}:i\geq 1\}$ and $\{m_{k_{i}}:i\geq 1\}$, respectively. Given $0<\varepsilon<\underline{v}_{2}/2$, for $k$ large enough, one has

[TABLE]

[TABLE]

By inequality (3.4), one has

[TABLE]

Therefore, the sequence $\{m_{k}:k\geq 1\}$ increases at least exponentially. Since $n_{k}\geq m_{k-1}$ for every $k\geq 2$, the sequence $\{n_{k}:k\geq 1\}$ also increases at least exponentially. Thus, there is a positive constant $C$ such that $k\leq C\log_{\beta}n_{k}$. Combining (3.4) and (3.5), one obtains

[TABLE]

Thus, for $k$ large enough, there is an integer $n_{0}$ and a postive real number $\varepsilon_{1}$ small enough such that the sum of all lengths of the blocks of [math] in the prefix of length $n_{k}$ of the infinite sequence $a_{1}a_{2}\cdots$ is at least equal to

[TABLE]

Among the digits $a_{1}\cdots a_{m_{k}}$, there are $k$ blocks of digits which are ‘free’. Denote their lengths by $l_{1},\cdots,l_{k}$. Let $\varepsilon_{2}=\dfrac{(v_{\beta}-\underline{v}_{2}-v_{\beta}\cdot\underline{v}_{2})\varepsilon_{1}}{v_{\beta}-\underline{v}_{2}}$, one has

[TABLE]

By Theorem 2.4, there are at most $\beta\cdot\beta^{l_{i}}/(\beta-1)$ ways to choose the block with length $l_{i}$. Thus, one has in total at most

[TABLE]

possible choices of the digits $a_{1}\cdots a_{m_{k}}$. On the other hand, there are at most $k(k\leq C\log_{\beta}n_{k})$ blocks of [math] in the prefix of length $n_{k}$ of the infinite sequence $a_{1}a_{2}\cdots$. Since there are at most $n_{k}$ possible choices for their first index, one has in total at most $(n_{k})^{C\log_{\beta}n_{k}}$ possible choices. Consequently, the set of those $x\in\mathbb{B}$ is covered by

[TABLE]

basic intervals of length at most $\beta^{-m_{k}}$. Moreover, by (3.4) and by letting $\varepsilon_{3}=\varepsilon/(1+v_{\beta})$, we have $\beta^{-m_{k}}\leq\beta^{-(1+v_{\beta})(1-\varepsilon_{3})n_{k}}$. Set $\varepsilon^{\prime}=\max\{\varepsilon_{2},~{}\varepsilon_{3}\}$. The set of those $x$ is covered by

[TABLE]

basic intervals of length at most $\beta^{-(1+v_{\beta})(1-\varepsilon^{\prime})n_{k}}$. We consider the series

[TABLE]

The critical exponent $s_{0}$ such that the series converges if $s>s_{0}$ and diverges if $s<s_{0}$ is given by

[TABLE]

By a standard covering argument and the arbitrariness of $\varepsilon^{\prime}$, the Hausdorff dimension of $\mathbb{B}^{\prime}:=\{x\in[0,1]:\nu_{\beta}(x)=\nu_{\beta}\}\cap\mathcal{U}(\psi_{2})$ is at most equal to

[TABLE]

For $v_{\beta}\geq\underline{v}_{2}/(1-\underline{v}_{2})$, fix $L$ large enough. We consider the set

[TABLE]

Repeat the above discussion, if $\underline{v}_{2}<1$, then

[TABLE]

If $\underline{v}_{2}=1$, then $v_{\beta}=+\infty$. By Theorem SW, ${\rm dim}_{H}\left(\mathbb{D}\right)=0$. If $\underline{v}_{2}<1$, since $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is a subset of

[TABLE]

let $L$ tend to $+\infty$, one has

[TABLE]

Regard the right side as a function of $v_{\beta}$, if $\underline{v}_{1}/(2+\underline{v}_{1})<\underline{v}_{2}$, then the maximum is attained for $v_{\beta}=2\underline{v}_{2}/(1-\underline{v}_{2})$. Therefore,

[TABLE]

If $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$, then the maximum is attained for $v_{\beta}=\underline{v}_{1}$. Thus,

[TABLE]

∎

Proposition 3.9**.**

If $\overline{v}_{1}/(2+\overline{v}_{1})<\overline{v}_{2}\leq 1$, then

[TABLE]

If $\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, then

[TABLE]

Proof.

By Lemma 3.3 $(1)$,

[TABLE]

If $\overline{v}_{2}=1$, then ${\rm dim}_{H}\left(\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})\right)\geq 0$ always holds. If $\overline{v}_{2}<1$, then we fix $\delta>0$ with $\overline{v}_{2}+\delta<1$, we consider the lower bound of the Hausdorff dimension of the set

[TABLE]

where $v_{\beta}\geq\overline{v}_{1}$ is a real number. By Theorem BL, if $\dfrac{\overline{v}_{1}+\delta}{\overline{v}_{2}+\delta}<\dfrac{1}{1-(\overline{v}_{2}+\delta)}$, then $\mathbb{F}$ is empty. Therefore, we consider the case $\dfrac{\overline{v}_{1}+\delta}{\overline{v}_{2}+\delta}\geq\dfrac{1}{1-(\overline{v}_{2}+\delta)}$. If $\overline{v}_{2}>0$, then there is $\delta_{0}>0$ such that for any $\delta\in(0,\delta_{0}]$, one has

[TABLE]

For any $\delta\in(0,\delta_{0}]$, we will construct a Cantor subset $E_{\delta}$ of $\mathbb{F}$. Let

[TABLE]

If $\overline{v}_{2}=0$, let

[TABLE]

Making an adjustment, we can choose two subsequences $\{n_{k}\}$ and $\{m_{k}\}$ with $n_{k}<m_{k}<n_{k+1}$ for every $k\geq 1$ such that $\{m_{k}-n_{k}\}$ is a non-decreasing sequence and

[TABLE]

Consider the set of real numbers $x\in[0,1)$ whose $\beta$-expansion

[TABLE]

satisfies that for all $k\geq 1$,

[TABLE]

[TABLE]

where $t_{k}$ is the largest integer such that $m_{k}+t_{k}(m_{k}-n_{k})<n_{k+1}$. Then,

[TABLE]

for $k$ large enough. Therefore, the sequence $\{t_{k}:k\geq 1\}$ is bounded. Fix $N$, let $\beta_{N}$ be the real number defined by the infinite $\beta$-expansion of $1$ as equality (2.2). We replace the digit $1$ for $a_{n_{k}},~{}a_{m_{k}}$ and $a_{m_{k}+i(m_{k}-n_{k})}$ for any $1\leq i\leq t_{k}$ by the block $0^{N}10^{N}$. Fill other places by blocks belonging to $\Sigma_{\beta_{N}}$. Thus, we have constructed the Cantor type subset $E_{\delta}$. Since $\{t_{k}\}$ is bound, one has

[TABLE]

[TABLE]

According to the construction, the sequence $d_{\beta}(x)$ is in $\Sigma_{\beta_{N}}.$

Claim**.**

$E_{\delta}\subseteq\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$.

Proof of Claim.

Given $\varepsilon>0$, by (3.7), there exists an integer $k_{0}$ such that

[TABLE]

By the definitions of $\overline{v}_{1}$ and $\overline{v}_{2}$, there is an integer $n_{0}$ such that

[TABLE]

Let $N_{0}=\max\{n_{k_{0}},~{}n_{0}\}$, for any $x\in E_{\delta}$ and any $n_{k}\geq N_{0}$, one has

[TABLE]

It means $x\in\mathcal{L}(\psi_{1})$. On the other hand, for $N\geq N_{0}$, there is an integer $i$ such that $n_{k+i}\leq N<n_{k+i+1}$. Therefore,

[TABLE]

It means $x\in\mathcal{U}(\psi_{2})$. Then, $x\in\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$. Therefore,

[TABLE]

∎

We distribute the mass uniformly when meet a block in $\Sigma_{\beta_{N}}$ and keep the mass when go through the positions where the digits are determined by construction of $E_{\delta}$. The Bernoulli measure $\mu$ on $E_{\delta}$ is defined as follows.

If $n<n_{1}$, define $\mu(I_{n})=1/\sharp\Sigma^{n}_{\beta_{N}}$. If $n_{1}\leq n\leq m_{1}+4N$, define $\mu(I_{n})=1/\sharp\Sigma^{n_{1}-1}_{\beta_{N}}$. If there is an integer $t$ with $0\leq t\leq t_{1}-1$ such that

[TABLE]

define

[TABLE]

If there is an integer $t$ with $0\leq t\leq t_{1}$ such that

[TABLE]

where $c:=\min\{n_{2}+4N+2Nt_{1},m_{1}+4N+(t+1)(m_{1}-n_{1})+2Nt\}$, define

[TABLE]

For $k\geq 2$, let

[TABLE]

[TABLE]

If $l_{k}\leq n\leq h_{k}$, define

[TABLE]

If there is an integer $t$ with $0\leq t\leq t_{k}-1$ such that

[TABLE]

define

[TABLE]

If there is an integer $t$ with $0\leq t\leq t_{k}$ such that

[TABLE]

define

[TABLE]

By the construction and Proposition 2.6, $I_{h_{k}}$ is full. For calculating the local dimension of $\mu$, we discuss different cases as follows.

Case $A$: If $n=h_{k}$, then

[TABLE]

Recall that $\{t_{k}:k\geq 1\}$ is bounded and $\{m_{k}:k\geq 1\}$ grows exponentially fast in terms of $k$, therefore,

[TABLE]

By equalities (3.7), one has

[TABLE]

According to Stolz-Cesàro Theorem,

[TABLE]

Thus,

[TABLE]

Case $B$: For an integer $n$ large enough, if there is $k\geq 2$ such that $l_{k}\leq n\leq h_{k}$, then

[TABLE]

Case $C$: For $n$, if there is an integer $t$ with $0\leq t\leq t_{k}-1$ such that

[TABLE]

then one has

[TABLE]

Since $I_{h_{k}}$ is full, by Proposition 2.7, $\lvert I_{n}\rvert=\lvert I_{h_{k}}\rvert\cdot\lvert I_{n-h_{k}}(\omega^{\prime})\rvert$, where $\omega^{\prime}$ is an admissible block in $\Sigma^{n-h_{k}}_{\beta_{N}}$. By Lemma 2.8,

[TABLE]

Hence,

[TABLE]

where $\varphi(N)<1$ and $\varphi(N)$ tends to $1$ as $N$ tends to infinity. If there is an integer $t$ with $0\leq t\leq t_{k}$ such that

[TABLE]

then letting $l:=n-(h_{k}+t(m_{k}-n_{k})+2Nt)$, one has

[TABLE]

Since $I_{h_{k}}$ is full, by Proposition 2.7, $\lvert I_{n}\rvert=\lvert I_{h_{k}}\rvert\cdot\lvert I_{n-h_{k}}(\omega^{\prime})\rvert$, where $\omega^{\prime}$ is an admissible block in $\Sigma^{n-h_{k}}_{\beta_{N}}$. By Lemma 2.8, $\lvert I_{n-h_{k}}(\omega^{\prime})\rvert\geq\beta^{-(n-h_{k}+N)}.$ Therefore,

[TABLE]

Hence,

[TABLE]

Therefore, in all cases,

[TABLE]

Given a point $x\in E_{\delta}$, let $r$ be a number with $\lvert I_{n+1}(x)\rvert\leq r<\lvert I_{n}(x)\rvert$. We consider the ball $B(x,r)$. By Lemma 2.8, every $n$-th order basic interval $I_{n}$ satisfies $\lvert I_{n}\rvert\geq\beta^{-(n+N)}$. Hence, the ball $B(x,r)$ interests at most $\lfloor 2\beta^{N}\rfloor+2$ basic intervals of order $n$. On the other hand,

[TABLE]

Therefore,

[TABLE]

By the arbitrariness of $\delta\in(0,\delta_{0}]$, one has

[TABLE]

Let $N$ tend to infinity, by Mass Distribution Principle [7, pp.60], one has

[TABLE]

Regarding the right side as a function of $v_{\beta}$ with $v_{\beta}\geq\underline{v}_{2}/(1-\underline{v}_{2})$, if $\overline{v}_{1}/(2+\overline{v}_{1})<\overline{v}_{2}$, then the maximum is attained for $v_{\beta}=2\overline{v}_{2}/(1-\overline{v}_{2})$. Therefore,

[TABLE]

If $\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, then the maximum is attained for $v_{\beta}=\overline{v}_{1}$. Thus,

[TABLE]

∎

Proof of Theorem B.

If $\underline{v}_{2}>1$, by Proposition 3.8, $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is countable. If $\underline{v}_{1}/(2+\underline{v}_{1})\leq\underline{v}_{2}\leq 1<\overline{v}_{2}$, by Proposition 3.7,

[TABLE]

By Proposition 3.8 and the definition of Hausdorff dimension, we have

[TABLE]

If $\overline{v}_{1}/(2+\overline{v}_{1})<\underline{v}_{2}\leq\overline{v}_{2}\leq 1$ and $\overline{v}_{1}/(2+\overline{v}_{1})<\overline{v}_{2}$, by Propositions 3.7 and 3.8, we also have

[TABLE]

Since $\overline{v}_{1}/(2+\overline{v}_{1})<\overline{v}_{2}$, according to Propositions 3.9,

[TABLE]

Then,

[TABLE]

If $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$ and $\overline{v}_{1}/(2+\overline{v}_{1})<\overline{v}_{2}\leq 1$, by Propositions 3.7, 3.8 and 3.9,

[TABLE]

If $\underline{v}_{1}/(2+\underline{v}_{1})<\underline{v}_{2}\leq\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, combining Proposition 3.7 with Proposition 3.8, one has

[TABLE]

According to Proposition 3.9, we have

[TABLE]

Therefore,

[TABLE]

If $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$ and $\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, by Propositions 3.7, 3.8 and 3.9,

[TABLE]

∎

4. Proofs of Theorems C and D

In this section, we will give the proofs of Theorems C and D.

Proof of Theorem C.

By Lemma 3.3, one has

[TABLE]

Replace the role of $\underline{v}_{1}$ by $v_{\beta}$, for $L$ large enough, we consider the set

[TABLE]

By the similar discussions in Proposition 3.8, if $\underline{v}_{2}>1$, then $\underline{v}_{2}\geq(v_{\beta}+1/L)/(1+v_{\beta}+1/L)$ for any $v_{\beta}$. Therefore, $\mathbb{D}$ is empty. Thus,

[TABLE]

By Proposition 3.8 and Lemma 3.4, $\mathcal{U}(\psi_{2})$ is countable.

If $\underline{v}_{2}=1$, then $v_{\beta}=+\infty$. By Theorem SW, ${\rm dim}_{H}\left(\mathbb{D}\right)=0$. If $\underline{v}_{2}<1$, then for $v_{\beta}\geq\underline{v}_{2}/(1-\underline{v}_{2})$, one has

[TABLE]

Since $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$ is a subset of

[TABLE]

let $L$ tend to $+\infty$, one has

[TABLE]

Regarding the right side as a function of $v_{\beta}$ with $v_{\beta}\geq\underline{v}_{2}/(1-\underline{v}_{2})$, we obtain the maximum at $v_{\beta}=2\underline{v}_{2}/(1-\underline{v}_{2})$. Therefore,

[TABLE]

Combining with Proposition 3.7, one has

[TABLE]

If $\overline{v}_{2}\leq 1$, then $\underline{v}_{2}\leq\overline{v}_{2}\leq 1$, one also has

[TABLE]

To obtain the lower bound of the Hausdor dimension of $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$, we will construct a Cantor type subset $E$ of $\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$. By Lemma 3.3,

[TABLE]

We replace the role of $\overline{v}_{1}$ by $v_{\beta}$ with $v_{\beta}\geq\overline{v}_{2}/(1-\overline{v}_{2})$. Fix $\delta>0$, consider

[TABLE]

If $\overline{v}_{2}=0$, then let

[TABLE]

If $\overline{v}_{2}>0$, then, let

[TABLE]

Arguing as in the proof of Proposition 3.9, one has

[TABLE]

Regarding the right side as a function of $v_{\beta}$ and taking $v_{\beta}\geq\overline{v}_{2}/(1-\overline{v}_{2})$ into account, we obtain the maximum at $v_{\beta}=2\overline{v}_{2}/(1-\overline{v}_{2})$. Then,

[TABLE]

∎

Proof of Theorem D.

If $\underline{v}_{1}=\overline{v}_{1}=0$, then

[TABLE]

If $\underline{v}_{2}>0$, by the definitions of $\underline{v}_{2}$ and $\underline{v}_{1}=\overline{v}_{1}=0$, for any $\varepsilon\in(0,\underline{v}_{2}/2)$, there is an integer $n_{0}$ such that for any $n\geq n_{0}$, one has

[TABLE]

For any $x\in\mathcal{U}(\psi_{2})$, by the same argument as Proposition 3.2, we have

[TABLE]

∎

5. Examples

In this section, we will show that the upper and lower bounds of Theorems B and C can be all reached. Examples 5.1, 5.2 and 5.3 explain that the upper bound estimation $\dfrac{1}{1+\overline{v}_{2}}$, $\left(\dfrac{1-\underline{v}_{2}}{1+\underline{v}_{2}}\right)^{2}$ and $\dfrac{\underline{v}_{1}-\underline{v}_{2}-\underline{v}_{1}\underline{v}_{2}}{(1+\underline{v}_{1})(\underline{v}_{1}-\underline{v}_{2})}$ are reachable, respectively.

Example 5.1**.**

Let $\psi_{1}(n)=1$, for $n=1,2,\cdots$ and

[TABLE]

Then, $\underline{v}_{1}=\overline{v}_{1}=0$, $\underline{v}_{2}=0$, $\overline{v}_{2}=3$ and

[TABLE]

Proof.

By Proposition 3.7, we only need to show that

[TABLE]

Now, we construct a Cantor subset $E$ of $\mathcal{L}(\psi_{1})\bigcap\mathcal{U}(\psi_{2})$ such that

[TABLE]

Let $n_{k}=k^{k}$ and $m_{k}=4k^{k}$, for $k=1,2,\cdots$, we construct a Cantor subset $E$ by the same way as Proposition 3.9, then

[TABLE]

∎

Example 5.2**.**

Let $\psi_{1}(n)=1$, for $n=1,2,\cdots$ and

[TABLE]

Then, $\underline{v}_{1}=\overline{v}=0$, $\underline{v}_{2}=1/2$, $\overline{v}_{2}=2$ and

[TABLE]

Proof.

Since $\underline{v}_{1}/(2+\underline{v}_{1})<\underline{v}_{2}$, by Proposition 3.8, the proof is completed by showing

[TABLE]

Let $n_{k}=4^{k}$ and $m_{k}=3\cdot 4^{k}$, for $k=1,2,\cdots$, we construct a Cantor subset $E$ by the same way as Proposition 3.9, then

[TABLE]

∎

Example 5.3**.**

For $k=1,2,\cdots$, let

[TABLE]

Then, $\underline{v}_{1}=1/2$, $\overline{v}_{1}=1$, $\underline{v}_{2}=1/6$, $\overline{v}_{2}=1/2$ and

[TABLE]

Proof.

Since $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$, by Proposition 3.8, one has

[TABLE]

It suffices to show ${\rm dim}_{H}\left(\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})\right)\geq 1/2$. Let $n_{k}=4^{k}$ and $m_{k}=3\cdot 4^{k}$, for $k=1,2,\cdots$, we construct a Cantor subset $E$ by the same way as Proposition 3.9, then

[TABLE]

∎

Examples 5.4 and 5.5 explain that the lower bound estimation [math] and $\left(\dfrac{1-\overline{v}_{2}}{1+\overline{v}_{2}}\right)^{2}$ are reachable, respectively.

Example 5.4**.**

Let $\psi_{1}(n)=1$ for $n=1,2,\cdots$ and

[TABLE]

Then, $\underline{v}_{1}=\overline{v}_{1}=0$, $\underline{v}_{2}=1$, $\overline{v}_{2}=3$ and

[TABLE]

Proof.

It remains to prove ${\rm dim}_{H}\left(\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})\right)\leq 0$. In fact, by Proposition 3.8, one has

[TABLE]

∎

Example 5.5**.**

For $k=1,2,\cdots$, let

[TABLE]

Then, $\underline{v}_{1}=3$, $\overline{v}_{1}=10/3$, $\underline{v}_{2}=21/32$ and $\overline{v}_{2}=2/3$. One has

[TABLE]

Proof.

Since $\underline{v}_{1}/(2+\underline{v}_{1})<\overline{v}_{2}$, by Proposition 3.9, the proof is completed by showing

[TABLE]

In fact, since $\underline{v}_{2}\leq\underline{v}_{1}/(2+\underline{v}_{1})$, by Proposition 3.8, we have

[TABLE]

∎

The following two examples explain that the lower bound estimation $\dfrac{\overline{v}_{1}-\overline{v}_{2}-\overline{v}_{1}\overline{v}_{2}}{(1+\overline{v}_{1})(\overline{v}_{1}-\overline{v}_{2})}$ is reachable. We consider the cases of $\underline{v}_{1}=\overline{v}_{1}$ and $\underline{v}_{2}=\overline{v}_{2}$, respectively.

Example 5.6**.**

Let $\psi_{1}(n)=\beta^{-n}$ for $n=1,2,\cdots$ and

[TABLE]

Then, $\underline{v}_{1}=\overline{v}_{1}=1$, $\underline{v}_{2}=0$, $\overline{v}_{2}=1/4$ and

[TABLE]

Proof.

Since $\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, by Proposition 3.9, what is left is to show

[TABLE]

For any $x\in[0,1]$, denote its $\beta$-expansion by

[TABLE]

where $a_{i}\in\{0,\cdots,\lceil\beta\rfloor\}$, for all $i\geq 1$. Let

[TABLE]

If $x\in\mathcal{L}(\psi_{1})\cap\mathcal{U}(\psi_{2})$, since $v_{1}>0$, then one has

[TABLE]

Arguing as in the proof of Proposotion 3.8, we take the maximal subsequences $\{n_{k}:k\geq 1\}$ and $\{m_{k}:k\geq 1\}$ of $\left\{n^{\prime}_{i}:i\geq 1\right\}$ and $\left\{m^{\prime}_{i}:i\geq 1\right\}$, respectively, such that the sequence $\{m_{k}-n_{k}:k\geq 1\}$ is non-decreasing. Notice $\beta^{n_{k}-m_{k}}<T^{n_{k}}_{\beta}x<\beta^{n_{k}-m_{k}+1}$, one has

[TABLE]

Since $x\in\mathcal{U}(\psi_{2})$, there is an integer $k_{0}$ such that for any $k\geq k_{0}$, one has $m_{k}-n_{k}\geq\lfloor n_{k+1}/4\rfloor$. If not, for any $j\geq 1$, there is an integer $k_{j}$ such that $m_{k_{j}}-n_{k_{j}}<n_{k_{j+1}}/4$. Since one of $n_{k_{j+1}}$ and $n_{k_{j+1}}+1$ is even, denote it by $l_{k_{j+1}}$, for any integer $n\in[1,l_{k_{j+1}}]$, one has

[TABLE]

It contradicts $x\in\mathcal{U}(\psi_{2})$.

Choose the subsequence $\{n_{k_{i}}:i\geq 1\}$ and $\{m_{k_{i}}:i\geq 1\}$ of $\{n_{k}:k\geq 1\}$ and $\{m_{k}:k\geq 1\}$, respectively, such that

[TABLE]

For simplicity, let $\{n_{k}:k\geq 1\}$ and $\{m_{k}:k\geq 1\}$ stand for $\{n_{k_{i}}:i\geq 1\}$ and $\{m_{k_{i}}:i\geq 1\}$, respectively. For any $\varepsilon>0$, there is an integer $k^{\prime}$ such that for any $k\geq k^{\prime}$, one has

[TABLE]

Arguing as in the proof of Proposition 3.8, one has

[TABLE]

∎

Example 5.7**.**

Let

[TABLE]

and $\psi_{2}(n)=\beta^{-2n/11}$ for $n=1,2,\cdots$. Then, $\underline{v}_{1}=1/3$, $\overline{v}_{1}=2/3$, $\underline{v}_{2}=\overline{v}_{2}=2/11$ and

[TABLE]

Proof.

Since $\overline{v}_{2}\leq\overline{v}_{1}/(2+\overline{v}_{1})$, by Proposition 3.9, what is left is to show

[TABLE]

In fact, since $\underline{v}_{2}=\dfrac{2}{11}\leq\dfrac{1}{7}=\underline{v}_{1}/(2+\underline{v}_{1})$, by Proposition 3.8,

[TABLE]

∎