# Metric results for the eventually always hitting points and level sets in
subshift with specification

Bo Wang and Bing Li
Department of Mathematics, South China University of Technology,
Guangzhou 510640, P.R. China
E-mails<EMAIL_ADDRESS>
<EMAIL_ADDRESS>Corresponding author

Abstract. We study the set of eventually always hitting points for symbolic
dynamics with specification. The measure and Hausdorff dimension of such sets
are obtained. Moreover, we establish the stronger metric results by
introducing a new quantity $L_{N}(\omega)$ which describes the maximal length
of string of zeros of the prefix among the first $N$ iterations of $\omega$ in
symbolic space. The Hausdorff dimensions of the level sets for this quantity
are also completely determined.

Key words and phrases. Eventually always hitting points, Subshift,
Specification, Hausdorff dimension, Ruelle operator ††footnotetext: MSC
(2010): 11J83, 37A50, 37B10 (primary), 28A80, 28D05, 37C30 (secondary).

##  1 Introduction

###  1.1 The set of eventually always hitting points

Let $(X,d,\mathcal{B},\mu,T)$ be a metric measure preserving dynamical system.
Namely $\mu$ is a probability measure on $(X,\mathcal{B})$ and $T$-invariant,
$(X,d)$ is a metric space. Here the measure and metric structure are
compatible meaning that all Borel sets are $\mu$-measurable. Let
$\\{E_{n}\\}_{n=1}^{\infty}$ be a sequence of subsets in $\mathcal{B}$ and let

$H_{io}(T,\\{E_{n}\\}):=\\{x\in X:T^{n}x\in E_{n}\ {\rm for\ i.o.}\\}.$

Here and throughout, “i.o.” stands for “infinitely often”. The sets $E_{n}$
can be thought of as targets that the orbit under $T$ of points in $X$ have to
hit. The interesting situation is usually, when the diameters of $E_{n}$ tend
to zero as $n$ increases. It is thus natural to refer to
$H_{io}(T,\\{E_{n}\\})$ as the corresponding shrinking target set associated
with the given dynamical system and target sets. The classical “shrinking
target problem” is to consider the $\mu$-measure and the Hausdorff dimension
of $H_{io}(T,\\{E_{n}\\})$, which was formulated in [15]. There are many
results about shrinking target problem, we refer the reader to [7, 11, 18, 26]
for the $\mu$-measure of $H_{io}(T,\\{E_{n}\\})$ and [15, 16, 17, 25, 26, 27,
32, 33] for the Hausdorff dimension of $H_{io}(T,\\{E_{n}\\})$. In this paper
we will investigate a set which is closely related to $H_{io}(T,\\{E_{n}\\})$,
namely the set of eventually always hitting points. This is the set of points
$x$ satisfies that for all large enough $N$, there exists a
$n\in\\{1,2,\cdots,N\\}$ such that $T^{n}x\in E_{N}$. That is,

$\displaystyle H_{ea}(T,\\{E_{n}\\}):$ $\displaystyle=\\{x\in X:\forall\ N\gg
1,1\leq\exists n\leq N,\ {\rm such\ that}\ T^{n}x\in E_{N}\\}$
$\displaystyle=\bigcup_{m=1}^{\infty}\bigcap_{N=m}^{\infty}\bigcup_{n=1}^{N}T^{-n}E_{N}.$

Here and throughout, “$N\gg 1$” stands for “$N$ large enough”. The name
eventually always hitting points was introduced by Kelmer [19]. We remark that
Kelmer [19] studied a slightly different definition of eventually always
hitting points, and required that for all $N\gg 1$, we have $T^{n}x\in E_{N}$
for some $0\leq n\leq N-1$, whereas we require $1\leq n\leq N$. For the result
discussed in this paper, it is insignificant which definition we use.

Note that $H_{io}(T,\\{E_{n}\\})$ is a limsup set and $H_{ea}(T,\\{E_{n}\\})$
is a liminf set. If $\mu(E_{n})\to 0$ and $\\{E_{n}\\}_{n=1}^{\infty}$ is a
nested sequence, that is, $E_{n+1}\subset E_{n}$ holds for all $n$, then
$H_{ea}(T,\\{E_{n}\\})\subset H_{io}(T,\\{E_{n}\\})$ except for a null measure
set [22]. In this sense, $H_{ea}(T,\\{E_{n}\\})$ is smaller than
$H_{io}(T,\\{E_{n}\\})$. Similar with the shrinking target problem, we have
the following two problems.
(P1) What is the $\mu$-measure of $H_{ea}(T,\\{E_{n}\\})$?
(P2) What is the Hausdorff dimension of $H_{ea}(T,\\{E_{n}\\})$ if
$\mu(H_{ea}(T,\\{E_{n}\\}))=0$?

There have recently been several results about the $\mu$-measure of the set
$H_{ea}(T,\\{E_{n}\\})$. If $\mu$ is ergodic and $\\{E_{n}\\}$ is a nested
sequence, the $\mu$-measure of $H_{ea}(T,\\{E_{n}\\})$ is either zero or one
(see [22]).

  1. •

Kelmer gave necessary and sufficient conditions for
$\mu(H_{ea}(T,\\{E_{n}\\}))=1$ in the setting of flows on hyperbolic manifolds
[19], and for flows on homogeneous spaces together with Yu [21].

  2. •

In the general setting, Kelmer showed [19] that if there exists $c<1$ such
that $\mu(E_{n})\leq\frac{c}{n}$ for infinitely many $n$, then
$\mu(H_{ea}(T,\\{E_{n}\\}))=0$.

  3. •

Kleinbock, Konstantoulas and Richter gave sufficient conditions for
$\mu(H_{ea}(T,\\{E_{n}\\}))$ $=0$ ($\mu(H_{ea}(T,\\{E_{n}\\}))=1$)
respecitively in some mixing systems [23], such as Gauss map, Product systems
and Bernoulli schemes. For the Gauss map and balls $E_{n}$ with centers in 0,
they proved that if there exists $c>\frac{1}{\log 2}$, such that
$\mu(E_{n})\geq\frac{c\log\log n}{n}$ for all but finitely many $n$, then
$\mu(H_{ea}(T,\\{E_{n}\\}))=1$, and if there exists $c<\frac{1}{\log 2}$, such
that $\mu(E_{n})\leq\frac{c\log\log n}{n}$ for all but finitely many $n$, then
$\mu(H_{ea}(T,\\{E_{n}\\}))=0$.

  4. •

Kirseborn, Kunde and Persson [22] obtained results similar to those of
Kleinbock, Konstantoulas and Richter. Their results hold for a wider class of
dynamical systems, but they need to add stronger conditions on $\mu(E_{n})$ in
order to be able to conclude that the measure of $H_{ea}(T,\\{E_{n}\\})$ is
one or zero. For piecewise expanding interval map including the doubling map,
Gauss map, and for some quadratic maps, they proved that
$\mu(H_{ea}(T,\\{E_{n}\\}))=1$ when $\mu(E_{n})\geq\frac{c(\log n)^{2}}{n}$
for some $c>0$ and all sufficiently large $n$. Furthermore, they gave a
sufficient condition for $\mu(H_{ea}(T,\\{E_{n}\\}))=0$ under additional
assumptions, they prove that if there exists $c>0$, such that
$\mu(E_{n})\leq\frac{c}{n}$ for infinitely many $n$, then
$\mu(H_{ea}(T,\\{E_{n}\\}))=0$.

  5. •

Ganotaki and Persson [13] gave sufficient conditions for
$\mu(H_{ea}(T,\\{E_{n}\\}))=1$ in the systems which have exponential decay of
correlations for either Hölder continuous functions or functions of bounded
variation. They also gave an asymptotic estimate as $N\to\infty$ on the number
of $n\leq N$ with $T^{n}x\in E_{N}$.

  6. •

Other recent results can be found in [20, 24].

Unfortunately, the sufficient condition and necessary condition above does not
coincide.

There are also some results about the Hausdorff dimension of the set
$H_{ea}(T,\\{E_{n}\\})$ for $\beta$-transformation $T_{\beta}(x)=\beta x({\rm
mod}\ 1)$ on $X=[0,1]$.

  1. •

Bugeaud and Liao [6] proved that $\dim_{\rm
H}H_{ea}(T,\\{E_{n}\\})=\left(\frac{1-v}{1+v}\right)^{2}$ when
$E_{n}=[0,\beta^{-vn})$, where $v\in[0,1]$. Here and below, $\dim_{\rm H}$
stands for Hausdorff dimension.

  2. •

Wu [35] extended the result of Bugeaud and Liao to every points $x_{0}$ in the
unit interval.

  3. •

Wu and Zheng [36] gave the upper bound and lower bound of $\dim_{\rm
H}H_{ea}(T,\\{E_{n}\\})$ when $E_{n}=[0,\psi(n))$, where $\psi$ is a general
positive function.

###  1.2 Main results

In this paper, we will consider the set of eventually always hitting points
for symbolic dynamics. Given a positive integer $m\geq 2$, let
$\mathcal{A}:=\\{0,1,\cdots,m-1\\}$. We denote

$\mathcal{A}^{\mathbb{N}}=\\{\omega_{1}\cdots\omega_{n}\cdots:\omega_{i}\in\mathcal{A},\forall\
i\geq 1\\}.$

The set $\mathcal{A}^{\mathbb{N}}$ is compact in the metric
$d(u,\omega)=m^{-\min\\{i:u_{i}\neq\omega_{i}\\}}$ if $u\neq\omega$ and
$d(u,\omega)=0$ if $u=\omega$. Furthermore, the shift map
$\sigma:\mathcal{A}^{\mathbb{N}}\to\mathcal{A}^{\mathbb{N}}$ defined by
$(\sigma u)_{i}=u_{i+1}$ is continuous. We call
$(\mathcal{A}^{\mathbb{N}},d,\sigma)$ the symbolic space. A subshift on
$\mathcal{A}$ is a closed set $\Sigma\subset\mathcal{A}^{\mathbb{N}}$ with
$\sigma(\Sigma)\subset\Sigma$. If $\Sigma$ has specification, by the theory of
Ruelle operator, there exists a unique $\sigma$-invariant Borel probability
measure $\mu$ satisfies the Gibbs property. We refer the reader to Section 3
for more details.

In this paper, we are dedicated to answer (P1) and (P2) when the targets are
balls with center $0^{\infty}$ in a subshift with specification. Assume that
$0^{\infty}\in\Sigma$, for function $\psi:\mathbb{R}^{+}\to\mathbb{R}^{+}$
with $\lim\limits_{x\to\infty}\psi(x)=0$, let $E_{n}=B(0^{\infty},\psi(n))$,
where
$B(0^{\infty},\psi(n)):=\\{\omega\in\Sigma:d(\omega,0^{\infty})<\psi(n)\\}$.
For simplicity, we denote $H_{ea}(\sigma,\\{E_{n}\\})$ by
$H_{ea}(\sigma,\psi)$. Denote by $h(\Sigma)$ the topological entropy of
$\Sigma$, the definition of topological entropy can be found in Section 3.
Throughout this paper, let $A=e^{h(\Sigma)}$.

We first answer the problems (P1) and (P2) by the two following theorems.
Theorem 1.1 describes the $\mu$-measure of $H_{ea}(\sigma,\psi)$. Theorem 1.3
gives the Hausdorff dimension of $H_{ea}(\sigma,\psi)$ when
$\lim\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{N}$ exists.

######  Theorem 1.1.

Suppose that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$. We have

$\mu(H_{ea}(\sigma,\psi))=\begin{cases}0,&{\rm if}\
\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}>1,\\\ 1,&{\rm
if}\
\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}<1.\end{cases}$

######  Remark 1.2.

By Lemma 4.2, we know that the system $(\Sigma,\sigma,\mu)$ is exponentially
mixing, which implies $\sigma$ is ergodic. Therefore, the first part of
Theorem 1.1 can be deduced from the result of Kirseborn, Kunde and Persson
[22, Corollary 1]. However, their sufficient condition on $\mu(H_{ea})=1$
requires that $X\subset\mathbb{R}$.

In view of Theorem 1.1, (P2) can be regarded as what is the Hausdorff
dimension of $H_{ea}(\sigma,\psi)$ if
$\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}>1$? From
Corollary 1.5 below, we show that

$\dim_{\rm H}H_{ea}(\sigma,\psi)=\dim_{\rm H}\Sigma$

if $\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}<+\infty$.
The remaining case is
$\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}=+\infty$, which
means that the growth rate of $-\log_{m}\psi(N)$ is larger than $\log N$. Then
the natural candidate for comparing rate is $N$ instead of $\log N$. In the
following we assume that $\lim\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{N}$
exists.

######  Theorem 1.3.

Assume that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$. Suppose
$\lim\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{N}$ exists, denoted by $\tau$
($\tau$ can be $+\infty$).
(i) If $\tau>1$, then

$H_{ea}(\sigma,\psi)=\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\}).$

Here and below, $\sigma^{-1}(\\{\omega\\}):=\\{u\in\Sigma:\sigma(u)=\omega\\}$
for all $\omega\in\Sigma$.
(ii) If $0\leq\tau\leq 1$, then

$\dim_{\rm
H}H_{ea}(\sigma,\psi)=\left(\frac{1-\tau}{1+\tau}\right)^{2}\dim_{\rm
H}\Sigma.$

######  Remark 1.4.

It is known that

$\dim_{\rm H}\Sigma=\frac{h(\Sigma)}{\log m}$

for any subshift $\Sigma$ (for example, see [12, Proposition iii@.1]). Thus
the dimension formula for Theorem 1.3 can be reformulated as

$\dim_{\rm
H}H_{ea}(\sigma,\psi)=\left(\frac{1-\tau}{1+\tau}\right)^{2}\frac{h(\Sigma)}{\log
m}.$

By Theorem 1.3, we immediately obtain the following corollary.

######  Corollary 1.5.

Assume that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$. For any $\psi$ with
$\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}<+\infty$, we
have $\dim_{\rm H}H_{ea}(\sigma,\psi)=\dim_{\rm H}\Sigma$.

Moreover, we can establish stronger results, which deduce Theorem 1.1 and 1.3
directly due to the quantity $L_{N}(\omega)$ describing the longest string of
zeros. Given $\omega\in\Sigma$, $n\geq 1$ and $N\geq 1$, define

$l_{n}(\omega)=\sup\\{k\geq 0:\omega_{n+i}=0,\forall\ 1\leq i\leq k\\}$

and

$L_{N}(\omega)=\max\limits_{1\leq n\leq N}l_{n}(\omega).$

The quantity $l_{n}(\omega)$ is the maximal length of string of zeros of the
prefix of $\sigma^{n}\omega$, which was introduced by Li and Wu [29]. Li,
Persson, Wang and Wu [28] gave a kind of classification of $\beta>1$ according
to $\\{l_{n}\\}_{n\geq 1}$ and calculated the Lebesgue measure and Hausdorff
dimension of each class. Fang, Wu and Li [10] studied the approximation orders
of real numbers by $\beta$-expansion in terms of investigating the behaviour
of $l_{n}(\omega)$. We introduce a new quantity $L_{N}(\omega)$ describing the
maximal length of string of zeros of the prefix among
$\\{\sigma\omega,\sigma^{2}\omega,\cdots,\sigma^{N}\omega\\}$. Recall that

$\displaystyle H_{ea}(\sigma,\psi)$ $\displaystyle=\\{\omega\in\Sigma:\forall
N\gg 1,1\leq\exists n\leq N,\ {\rm such\ that}\ \sigma^{n}\omega\in E_{N}\\}$
$\displaystyle=\left\\{\omega\in\Sigma:\forall N\gg 1,\min_{1\leq n\leq
N}d(\sigma^{n}\omega,0^{\infty})<\psi(N)\right\\}$
$\displaystyle=\left\\{\omega\in\Sigma:\forall N\gg
1,L_{N}(\omega)>-\log_{m}\psi(N)-1\right\\}.$

We have

$\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}>1\right\\}\subset
H_{ea}(\sigma,\psi)\subset\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}\geq
1\right\\}.$ (1.1)

So it suffices to consider
$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}$ for the
study of the set $H_{ea}(\sigma,\psi)$. The typical result for the limit of
$L_{N}(\omega)$ is the following.

######  Theorem 1.6.

Suppose that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$. For $\mu$-almost all $\omega\in\Sigma$, we have

$\lim_{N\to\infty}\frac{L_{N}(\omega)}{\log_{A}N}=1.$

Theorem 1.6 tells us that for $\mu$-almost all $\omega\in\Sigma$,
$L_{N}(\omega)$ increases to infinity with the speed $\log_{A}N$ as $N$ goes
to infinity. Generally, for any $a\geq 0$, let

$\mathcal{U}_{\psi}(a):=\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}\geq
a\right\\}.$

By $\eqref{18}$, the study of the Hausdorff dimension of $H_{ea}(\sigma,\psi)$
can be transformed into the the study of the Hausdorff dimension of
$\mathcal{U}_{\psi}(a)$. We choose $\psi_{0}$ with $\psi_{0}(N)=m^{-N}$, that
is, $-\log_{m}\psi_{0}(N)=N$. So

$\mathcal{U}_{\psi_{0}}(a)=\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{N}\geq
a\right\\}.$

The Hausdorff dimension of $\mathcal{U}_{\psi_{0}}(a)$ is given in Corollary
1.9 below. Indeed, we establish the stronger results. That is, let $0\leq
a\leq b$, we can further study the level sets

$\mathcal{U}_{\psi}(a,b):=\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}\geq
a,\limsup_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=b\right\\}$

and

$\mathcal{U}^{\ast}_{\psi}(a,b):=\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=a,\limsup_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=b\right\\}.$

Obviously, $\mathcal{U}^{\ast}_{\psi}(a,b)\subset\mathcal{U}_{\psi}(a,b)$.
Theorem 1.7 describes the Hausdorff dimensions of $\mathcal{U}_{\psi}(a,b)$
and $\mathcal{U}^{\ast}_{\psi}(a,b)$ completely when
$\lim\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{N}$ exists.

######  Theorem 1.7.

Assume that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$. Suppose
$\lim\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{N}$ exists, denoted by $\tau$
($\tau$ can be $+\infty$).
Case i $0<\tau<+\infty$
(1) $a\geq\tau^{-1}$

  1. •

If $b<+\infty$, then

$\mathcal{U}^{\ast}_{\psi}(a,b)=\mathcal{U}_{\psi}(a,b)=\emptyset.$

  2. •

If $b=+\infty$, then

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi}(a,b)=0.$

(2) $0\leq a<\tau^{-1}$

  1. •

If $b<\frac{a}{1-\tau a}$, then

$\mathcal{U}^{\ast}_{\psi}(a,b)=\mathcal{U}_{\psi}(a,b)=\emptyset.$

  2. •

If $b\geq\frac{a}{1-\tau a}$, then

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi}(a,b)=\frac{b(1-\tau a)-a}{(1+\tau b)(b-a)}\dim_{\rm
H}\Sigma.$

Case ii $\tau=+\infty$

  1. •

If $b>0$, then

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi}(a,b)=0.$

  2. •

If $b=0$, then

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi}(a,b)=\dim_{\rm H}\Sigma.$

Case iii $\tau=0$
Additionally, suppose $\psi(x)$ is strictly decreasing and
$\lim\limits_{N\to\infty}\frac{\log_{m}\psi(N-\log_{m}\psi(N))}{\log_{m}\psi(N)}=1$,
then

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi}(a,b)=\dim_{\rm H}\Sigma.$

######  Remark 1.8.

(i) In (2) of Theorem 1.7 Case i, we have

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi}(a,b)=\frac{b(1-\tau a)-a}{(1+\tau b)(b-a)}\dim_{\rm
H}\Sigma$

for all $b\geq\frac{a}{1-\tau a}$. Note that $b(1-\tau a)-a=(1+\tau b)(b-a)=0$
when $a=b=0$. However, we know that the denominator can not be zero. In fact,
in the proof of Theorem 1.7, we show that

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi}(0,0)=\dim_{\rm
H}\mathcal{U}_{\psi}(0,0)=\dim_{\rm H}\Sigma.$

Thus, we stipulate that $\frac{b(1-\tau a)-a}{(1+\tau b)(b-a)}=1$ when
$a=b=0$.
(ii) More precisely, in (1) of Theorem 1.7 Case i, we have

  1. •

$a>\tau^{-1}$ and $b=+\infty$ $\Rightarrow$
$\mathcal{U}_{\psi}(a,b)=\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\})$.

  2. •

$\tau^{-1}<a<+\infty$ $\Rightarrow$
$\mathcal{U}^{\ast}_{\psi}(a,b)=\emptyset$.

In Theorem 1.7 Case ii, we have

  1. •

$a>0$ $\Rightarrow$
$\mathcal{U}_{\psi}(a,b)\subset\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\})$.

######  Corollary 1.9.

Suppose that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$. We have
(i) If $a>1$, then

$\mathcal{U}_{\psi_{0}}(a)=\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\}).$

(ii) If $0\leq a\leq 1$, then

$\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a)=\left(\frac{1-a}{1+a}\right)^{2}\dim_{\rm
H}\Sigma.$ (1.2)

The remainder of the paper is organized as follows. In Section 2, we give the
proof of Theorem 1.1 (Theorem 1.3) modulo Theorem 1.6 (Corollary 1.9)
respectively. In Section 3, we introduce some preliminary concepts, including
symbolic dynamics and Ruelle operator. The proof of Theorem 1.6 is given in
Section 4. In Section 5, we calculate the Hausdorff dimensions of
$\mathcal{U}_{\psi_{0}}(a,b)$ and $\mathcal{U}_{\psi_{0}}^{\ast}(a,b)$.
Section 6 is dedicated to the proof of Theorem 1.7 and Corollary 1.9.

##  2 Establishing Theorems 1.1 and 1.3

######  Proof of Theorem 1.1 modulo Theorem 1.6.

By $\eqref{18}$, it suffices to show that

$\mu\left(\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}\geq
1\right\\}\right)=0\ {\rm if}\
\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}>1$ (2.1)

and

$\mu\left(\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}>1\right\\}\right)=1\
{\rm if}\ \limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}<1.$
(2.2)

Denote
$\Gamma=\left\\{\omega\in\Sigma:\lim\limits_{N\to\infty}\frac{L_{N}(\omega)}{\log_{A}N}=1\right\\}$.
Note that for any $\omega\in\Gamma$, we have

$\displaystyle\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}$
$\displaystyle=\liminf_{N\to\infty}\frac{L_{N}(\omega)}{\log_{A}N}\cdot\frac{\log_{A}N}{-\log_{m}\psi(N)}$
(2.3) $\displaystyle=\liminf_{N\to\infty}\frac{\log_{A}N}{-\log_{m}\psi(N)}$
$\displaystyle=\left(\limsup\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{\log_{A}N}\right)^{-1}.$

The combination of Theorem 1.6 and $\eqref{116}$ gives $\eqref{114}$ and
$\eqref{115}$. ∎

######  Proof of Theorem 1.3 modulo Corollary 1.9.

We use two important facts.
Fact 1: When $\tau>0$, for any $\omega\in\mathcal{A}^{\mathbb{N}}$ with
$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}>0$, we have

$\liminf_{N\to\infty}\frac{L_{N}(\omega)}{N}=\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}\cdot\frac{-\log_{m}\psi(N)}{N}=\tau\liminf_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}$
(2.4)

Furthermore, $\eqref{117}$ holds for $0<\tau<+\infty$ and any
$\omega\in\mathcal{A}^{\mathbb{N}}$.
Fact 2: When $\tau=0$, for any $\omega\in\mathcal{A}^{\mathbb{N}}$ with
$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{N}>0$, we have

$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=\liminf_{N\to\infty}\frac{L_{N}(\omega)}{N}\cdot\frac{N}{-\log_{m}\psi(N)}=+\infty.$
(2.5)

(i) If $\tau>1$, by Corollary 1.9 (i), we have

$\mathcal{U}_{\psi_{0}}(\tau)=\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\}).$
(2.6)

The upshot of $\eqref{18}$, $\eqref{117}$ and $\eqref{118}$ is that
$H_{ea}(\sigma,\psi)\subset\mathcal{U}_{\psi_{0}}(\tau)=\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\})$.
On the other hand, for any
$\omega\in\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\})$, there exists
$n_{0}\geq 0$, such that $\sigma^{n_{0}}\omega=0^{\infty}$. Thus,
$L_{N}(\omega)=+\infty$ for every $N\geq n_{0}$, which implies that $\omega\in
H_{ea}(\sigma,\psi)$. Thus,
$\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\})\subset
H_{ea}(\sigma,\psi)$. Therefore,

$H_{ea}(\sigma,\psi)=\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\}).$

(ii) If $0<\tau\leq 1$, the upshot of $\eqref{18}$ and $\eqref{117}$ is that

$\mathcal{U}_{\psi_{0}}(\tau(1+\epsilon))\subset
H_{ea}(\sigma,\psi)\subset\mathcal{U}_{\psi_{0}}(\tau)$ (2.7)

for any $\epsilon>0$. When $\tau=1$, in view of $\eqref{120}$ and Corollary
1.9 (ii), we obtain that $\dim_{\rm H}H_{ea}(\sigma,\psi)=0$. When $0<\tau<1$,
for any $\epsilon\in(0,\tau^{-1}-1)$, the combination of $\eqref{120}$ and
Corollary 1.9 (ii) gives

$\left(\frac{1-\tau(1+\epsilon)}{1+\tau(1+\epsilon)}\right)^{2}\dim_{\rm
H}\Sigma\leq\dim_{\rm
H}H_{ea}(\sigma,\psi)\leq\left(\frac{1-\tau}{1+\tau}\right)^{2}\dim_{\rm
H}\Sigma.$

Letting $\epsilon\to 0$, we have

$\dim_{\rm
H}H_{ea}(\sigma,\psi)=\left(\frac{1-\tau}{1+\tau}\right)^{2}\dim_{\rm
H}\Sigma.$

The remaining case is $\tau=0$, by $\eqref{18}$ and $\eqref{118}$, we have
$\mathcal{U}_{\psi_{0}}(a)\subset H_{ea}(\sigma,\psi)$ for all $0<a<1$. It
follows from Corollary 1.9 (ii) that $\dim_{\rm
H}H_{ea}(\sigma,\psi)\geq\left(\frac{1-a}{1+a}\right)^{2}\dim_{\rm H}\Sigma$
for all $0<a<1$, which implies that $\dim_{\rm
H}H_{ea}(\sigma,\psi)\geq\dim_{\rm H}\Sigma$. Thus

$\dim_{\rm H}H_{ea}(\sigma,\psi)=\dim_{\rm H}\Sigma.$

∎

##  3 Preliminary

###  3.1 Symbolic dynamics

Let $(\mathcal{A}^{\mathbb{N}},d,\sigma)$ be the symbolic space. For any
$n\geq 1$, define

$\mathcal{A}^{n}=\\{\omega_{1}\cdots\omega_{n}:\omega_{i}\in\mathcal{A},\forall\
1\leq i\leq n\\}.$

The element in $\mathcal{A}^{\mathbb{N}}$ and $\mathcal{A}^{n}$ is called the
infinite word and the word of length $n$, respectively. We also denote
$\mathcal{A}^{\ast}$ by the collection of all finite words, i.e.
$\mathcal{A}^{\ast}=\cup_{n\geq 1}\mathcal{A}^{n}$. For any
$\omega,u\in\mathcal{A}^{\ast}$, we write $\omega u$ as the concatenation of
words $\omega$ and $u$. In particular, $\omega^{n}$ denotes the $n$ times
self-concatenation of $\omega$ and $\omega^{\infty}$ stands for the infinite
times self-concatenation of $\omega$. The lexicographical order $\prec$ on
$\mathcal{A}^{\mathbb{N}}$ is defined as follows:

$\varepsilon_{1}\varepsilon_{2}\cdots\varepsilon_{n}\cdots\prec\omega_{1}\omega_{2}\cdots\omega_{n}\cdots$

if there exists an integer $n\geq 1$ such that $\varepsilon_{k}=\omega_{k}$
for all $1\leq k<n$ but $\varepsilon_{n}<\omega_{n}$. Similarly, for each
$n\in\mathbb{N}$, the lexicographical order $\prec$ on $\mathcal{A}^{n}$ is
defined as follows:

$\varepsilon_{1}\varepsilon_{2}\cdots\varepsilon_{n}\prec\omega_{1}\omega_{2}\cdots\omega_{n}$

if there exists an integer $i\leq n$ such that $\varepsilon_{k}=\omega_{k}$
for all $1\leq k<i$ but $\varepsilon_{i}<\omega_{i}$. Given
$u\in\mathcal{A}^{\mathbb{N}}$ and positive integers $i\leq j$, let
$u_{[i,j]}=u_{i}u_{i+1}\cdots u_{j}$. Let
$\Sigma\subset\mathcal{A}^{\mathbb{N}}$ be a subshift. For each $n\geq 1$,
define

$\Sigma^{n}=\\{\omega_{1}\cdots\omega_{n}\in\mathcal{A}^{n}:\exists\
u\in\Sigma,{\rm such\ that}\ u_{[1,n]}=\omega_{1}\cdots\omega_{n}\\}$

which is the set of all finite word of length $n$ appears in some elements of
$\Sigma$. Similar with $\mathcal{A}^{\ast}$, we denote $\Sigma^{\ast}$ the
collection of all finite words in $\Sigma$, that is,
$\Sigma^{\ast}=\cup_{n\geq 1}\Sigma^{n}$. For $u\in\Sigma\cup\Sigma^{\ast}$
and $n\in\mathbb{N}$, denote $I_{n}(u)$ the set of all $\omega\in\Sigma$ such
that $\omega_{[1,n]}=u_{1}\cdots u_{n}$. We call $I_{n}(u)$ the $n$-th
cylinder containing $u$. Furthermore, for any $\omega\in\Sigma$ and $r>0$, we
denote $B(\omega,r):=\\{u\in\Sigma:d(u,\omega)<r\\}$, which is the ball with
center $\omega$ and radius $r$. Let $\sharp$ denote the cardinality of a
finite set. Note that

$\sharp\Sigma^{i+j}\leq\sharp\Sigma^{i}\cdot\sharp\Sigma^{j}$ (3.1)

for all $i,j\in\mathbb{N}$, which is a simple but important property.

We introduce two important definitions, which appears many times in our paper.
The first definition is the topological entropy, which can be found in [34,
Chapter 7].

######  Definition 3.1.

The topological entropy of a subshift $\Sigma$ is

$h(\Sigma):=\lim_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}=\inf_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n},$
(3.2)

where existence of the limit and the second equality follow from $\eqref{11}$
and a standard lemma regarding subadditive sequences.

In fact, the above definition is not the original one [34, Definition 7.6],
but in view of [34, Theorem 7.13], for subshift, the above definition is
equivalent to the original one. The below definition of specification is
specialised for the symbolic setting and is slightly weaker than Bowen’s
original one [4].

######  Definition 3.2.

A subshift $\Sigma$ has the specification property or simply has specification
if there exists $M\in\mathbb{N}$ such that for any $u,v\in\Sigma^{\ast}$,
there exists $\omega\in\Sigma^{M}$ such that $u\omega v\in\Sigma^{\ast}$.

There are many examples of symbolic dynamics with specification, we refer the
reader to [2, 3].

###  3.2 Ruelle operator

An important tool to find $\sigma$-invariant measure on $\Sigma$ is Ruelle
operator. Denote by $\mathcal{C}=\mathcal{C}(\Sigma,\mathbb{R})$ the space of
all continuous functions $\phi:\Sigma\to\mathbb{R}$ with the supremum norm
$\|\phi\|:=\max\limits_{\omega\in\Sigma}|\phi(\omega)|$. We define a linear
operator $\mathcal{L}$ as

$\mathcal{L}\phi(\omega):=\sum_{u:\sigma
u=\omega}\phi(u),\quad\phi\in\mathcal{C}.$ (3.3)

The above operator is called Ruelle operator. Here we choose a special
potential, that is, $\varphi(\omega)=1$, $\forall\ \omega\in\Sigma$. Note that
$d(\sigma u,\sigma v)=m\cdot d(u,v)$ for any $u,v\in\Sigma$ with $d(u,v)\leq
m^{-2}$. In view of [9, Proposition 1], there exists $0<\kappa\leq m^{-2}$,
such that $\sharp\sigma^{-1}(u)=\sharp\sigma^{-1}(v)$ for all $u,v\in\Sigma$
with $d(u,v)\leq\kappa$. Furthermore, we can arrange
$\sigma^{-1}(u)=\\{x_{1},\cdots,x_{n}\\}$ and
$\sigma^{-1}(v)=\\{y_{1},\cdots,y_{n}\\}$ such that $d(x_{i},y_{i})\leq
m^{-1}\cdot d(u,v)$ for all $1\leq i\leq n$. Hence, we have
$\mathcal{L}\phi\in\mathcal{C}$ for every $\phi\in\mathcal{C}$. Let
$\mathcal{M}$ be the dual space of $\mathcal{C}$, by the Riesz representation
theorem, $\mathcal{M}$ is actually the space of all Borel measures on
$\Sigma$. Let $\mathcal{L}^{\ast}:\mathcal{M}\to\mathcal{M}$ be the dual
operator of $\mathcal{L}:\mathcal{C}\to\mathcal{C}$. That is, for any
$\phi\in\mathcal{C}$ and any $\nu\in\mathcal{M}$,

$\int\phi\mathrm{d}\mathcal{L}^{\ast}\nu=\int\mathcal{L}\phi\mathrm{d}\nu.$

For simplicity, we denote $\int\phi\mathrm{d}\nu$ by $\langle\nu,\phi\rangle$
for each $\phi\in\mathcal{C}$ and $\nu\in\mathcal{M}$. Due to the result of
Fan and Jiang [9, Theorem 1], there are a strictly positive number $\rho$ and
a strictly positive function $h$ such that $\mathcal{L}h=\rho h$. Furthermore,
there exists a probability measure $\nu$ satsifying
$\mathcal{L}^{\ast}\nu=\rho\nu$. Choose $h$ such that $\langle\nu,h\rangle=1$.
Let $\mu:=h\nu$, i.e. $\frac{\mathrm{d}\mu}{\mathrm{d}\nu}=h$. The measure
$\mu$ has the Gibbs property: there exists a constant $\gamma>1$ such that

$\gamma^{-1}\rho^{-n}\leq\mu(I_{n}(\omega))\leq\gamma\rho^{-n}$ (3.4)

holds for all $\omega\in\Sigma$ and $n\geq 1$. What is more, by the result of
Fan and Jiang [9, Theorem 1], $\mu$ is the unique $\sigma$-invariant Borel
probability measure which satisfies the Gibbs property. Take
$\widetilde{\varphi}$ as

$\widetilde{\varphi}(\omega)=\frac{h(\omega)}{\rho\cdot h(\sigma\omega)}>0,\
\forall\ \omega\in\Sigma.$

Consider the Ruelle operator associated to
$\widetilde{\mathcal{L}}=\mathcal{L}_{\widetilde{\varphi}}$. The important
feature of $\widetilde{\mathcal{L}}$ is that

$\widetilde{\mathcal{L}}1=1.$ (3.5)

We call it the normalized Ruelle operator. Denote
$\widetilde{\mathcal{L}}^{\ast}=\mathcal{L}_{\widetilde{\varphi}}^{\ast}$,
which is the dual operator of $\widetilde{\mathcal{L}}$. We have the following
relation between $\mathcal{L}^{\ast}$ and $\widetilde{\mathcal{L}}^{\ast}$:

$\mathcal{L}^{\ast n}\nu=\rho^{n}h^{-1}\widetilde{\mathcal{L}}^{\ast
n}(h\nu),\ \forall n\geq 1.$

Thus, we have that $\widetilde{\mathcal{L}}^{\ast}\mu=\mu$.

##  4 Proof of Theorem 1.6

The following lemma implies that the eigenvalue $\rho$ is equal to
$e^{h(\Sigma)}$ under the potential $\varphi$ with $\varphi(\omega)=1$,
$\forall\ \omega\in\Sigma$.

######  Lemma 4.1.

Assume that $\Sigma$ has specification and $h(\Sigma)>0$. Then for any
$\omega\in\Sigma$ and all $n\in\mathbb{N}$, we have

$\gamma^{-1}e^{-n\cdot h(\Sigma)}\leq\mu(I_{n}(\omega))\leq\gamma e^{-n\cdot
h(\Sigma)},$ (4.1)

where $\gamma>1$ is the constant that appears in $\eqref{15}$.

######  Proof.

The proof is standard, we give the proof here for completeness. By
$\eqref{15}$, we only need to show that $\rho=e^{h(\Sigma)}$. In view of [9,
Theorem 1], we know that

$\lim_{n\to\infty}\|\mathcal{L}^{n}1\|^{\frac{1}{n}}=\rho.$ (4.2)

Then we show that
$\lim\limits_{n\to\infty}\|\mathcal{L}^{n}1\|^{\frac{1}{n}}=e^{h(\Sigma)}$. On
the one hand, since

$\mathcal{L}^{n}1(\omega)=\sum_{u:\sigma^{n}u=\omega}1\leq\sharp\Sigma^{n}$

for every $\omega\in\Sigma$, we have
$\|\mathcal{L}^{n}1\|\leq\sharp\Sigma^{n}$. On the other hand, in view of
$\Sigma$ has specification, for any $n>M$ and any $v\in\Sigma^{n-M}$, there
exists $u\in\Sigma^{M}$, such that $vu\omega\in\Sigma$, which implies that
$vu\omega\in\sigma^{-n}\omega$. Therefore,

$\sum_{u:\sigma^{n}u=\omega}1\geq\sharp\Sigma^{n-M}.$

Hence, $\|\mathcal{L}^{n}1\|\geq\sharp\Sigma^{n-M}$. Therefore

$\lim_{n\to\infty}\|\mathcal{L}^{n}1\|^{\frac{1}{n}}=\lim_{n\to\infty}(\sharp\Sigma^{n})^{\frac{1}{n}}=e^{h(\Sigma)}.$
(4.3)

The upshot of $\eqref{22}$ and $\eqref{23}$ is that $\rho=e^{h(\Sigma)}$. ∎

The following statement shows that the system $(\Sigma,\sigma,\mu)$ is
exponentially mixing when $\Sigma$ has specification.

######  Lemma 4.2.

Suppose that $\Sigma$ has specification. The system $(\Sigma,\sigma,\mu)$ is
exponentially mixing, i.e. there exists $0<\theta<1$, such that for any
cylinder $E$ and any measurable set $F$, we have

$\mu(E\cap\sigma^{-n}F)=\mu(E)\mu(F)+\mu(F)\cdot O(\theta^{n}),$

for all $n\geq 1$, where the constant involved by the $O$ is absolute.

######  Proof.

Given $\alpha>0$, denote by $\mathcal{C}^{\alpha}$ the set of all
$\alpha$-Hölder continuous function. We prove that there exists $c>0$ and
$\theta\in(0,1)$, such that for any $f\in L^{1}(\Sigma,\mu)$ and
$g\in\mathcal{C}^{\alpha}$, we have

$\left|\int f\circ\sigma^{n}\cdot g\mathrm{d}\mu-\int f\mathrm{d}\mu\cdot\int
g\mathrm{d}\mu\right|\leq c\cdot\int|f|\mathrm{d}\mu\cdot\theta^{n}.$

In view of $\widetilde{\mathcal{L}}^{\ast}\mu=\mu$ and
$\langle\widetilde{\mathcal{L}}^{\ast}\mu,f\rangle=\langle\mu,\widetilde{\mathcal{L}}f\rangle$,
we have that

$\displaystyle\int f\circ\sigma^{n}\cdot g\mathrm{d}\mu-\int
f\mathrm{d}\mu\cdot\int g\mathrm{d}\mu$ $\displaystyle=\int
f\circ\sigma^{n}\cdot
g\mathrm{d}((\widetilde{\mathcal{L}}^{\ast})^{n}\mu)-\int
f\mathrm{d}\mu\cdot\int g\mathrm{d}\mu$
$\displaystyle=\int\widetilde{\mathcal{L}}^{n}(f\circ\sigma^{n}\cdot
g)\mathrm{d}\mu-\int f\mathrm{d}\mu\cdot\int g\mathrm{d}\mu.$

Note that $\widetilde{\mathcal{L}}^{n}(f\circ\sigma^{n}\cdot
g)=f\cdot\widetilde{\mathcal{L}}^{n}g$. Therefore,

$\displaystyle\int f\circ\sigma^{n}\cdot g\mathrm{d}\mu-\int
f\mathrm{d}\mu\cdot\int g\mathrm{d}\mu$ $\displaystyle=\int
f\cdot\widetilde{\mathcal{L}}^{n}g\mathrm{d}\mu-\int f\mathrm{d}\mu\cdot\int
g\mathrm{d}\mu$ $\displaystyle=\int
f\cdot(\widetilde{\mathcal{L}}^{n}g-\langle\mu,g\rangle)\mathrm{d}\mu.$

Thus,

$\displaystyle\left|\int f\circ\sigma^{n}\cdot g\mathrm{d}\mu-\int
f\mathrm{d}\mu\cdot\int g\mathrm{d}\mu\right|$ $\displaystyle=\left|\int
f\cdot(\widetilde{\mathcal{L}}^{n}g-\langle\mu,g\rangle)\mathrm{d}\mu\right|$
$\displaystyle\leq\int|f|\mathrm{d}\mu\cdot\|\widetilde{\mathcal{L}}^{n}g-\langle\mu,g\rangle\|$
$\displaystyle=\int|f|\mathrm{d}\mu\cdot\|\widetilde{\mathcal{L}}^{n}g-\langle\mu,g\rangle\cdot
1\|.$

Due to $\eqref{13}$, 1 is the strictly positive eigenfunction of
$\widetilde{\mathcal{L}}$. By [30, Theorem 2.2], there exists $c>0$ and
$\theta\in(0,1)$ such that for any $g\in\mathcal{C}^{\alpha}$,

$\|\widetilde{\mathcal{L}}^{n}g-\langle\mu,g\rangle\cdot 1\|\leq
c\cdot\theta^{n}$

for all $n\geq 1$. Hence,

$\left|\int f\circ\sigma^{n}\cdot g\mathrm{d}\mu-\int f\mathrm{d}\mu\cdot\int
g\mathrm{d}\mu\right|\leq c\cdot\int|f|\mathrm{d}\mu\cdot\theta^{n}$ (4.4)

for any $n\geq 1$. For any cylinder $E$ and measurable set $F$, let
$f=\chi_{F}$ and $g=\chi_{E}$. Here and below, $\chi$ is the characteristic
function. Then $f\in L^{1}(\Sigma,\mu)$ and $g\in\mathcal{C}^{\alpha}$, it
follows from $\eqref{25}$ that for each $n\geq 1$,

$\left|\int\chi_{F}\circ\sigma^{n}\cdot\chi_{E}\mathrm{d}\mu-\int\chi_{F}\mathrm{d}\mu\cdot\int\chi_{E}\mathrm{d}\mu\right|\leq
c\cdot\int|\chi_{F}|\mathrm{d}\mu\cdot\theta^{n}.$ (4.5)

That is, for all $n\geq 1$,

$|\mu(E\cap\sigma^{-n}F)-\mu(E)\mu(F)|\leq c\cdot\mu(F)\theta^{n}$

∎

Let $\mathcal{D}$ be the set of all cylinders. When $\Sigma$ has
specification, it follows from Lemma 4.2 that $(\Sigma,\sigma,\mu)$ is
exponentially mixing. Thus, $\mu$ is summable-mixing with respect to
$(\sigma,\mathcal{D})$ (see [26, Definition 1]). By [26, Theorem 1], we
immediately obtain the below lemma.

######  Lemma 4.3.

Suppose that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$. Let $I_{n}=I_{\left\lfloor\frac{\log
n}{h(\Sigma)}\right\rfloor+1}(0^{\infty})$, where $\lfloor x\rfloor$ stands
for the integer part of $x$. For $\omega\in\Sigma$, denote by $R(N,\omega)$
the number of positive integer $n\leq N$ such that $\sigma^{n}\omega\in
I_{n}$. Then for any given $\epsilon>0$, we have

$R(N,\omega)=F(N)+O\left(F^{\frac{1}{2}}(N)(\log
F(N))^{\frac{3}{2}+\epsilon}\right)$

for $\mu$-almost all $\omega\in\Sigma$, where

$F(N)=\sum_{n\leq N}\mu(I_{n}).$

######  Proof of Theorem 1.6.

For simplicity, we denote “almost all” by “a.a.”. Firstly, we prove that

$\limsup_{N\to\infty}\frac{L_{N}(\omega)}{\log N}\leq\frac{1}{h(\Sigma)}\ {\rm
for}\ \mu-{\rm a.a.}\ \omega\in\Sigma.$ (4.6)

It suffices to show that

$\mu\left(\left\\{\omega\in\Sigma:\limsup_{N\to\infty}\frac{L_{N}(\omega)}{\log
N}>\frac{1+\epsilon}{h(\Sigma)}\right\\}\right)=0$

for each $\epsilon>0$. Note that

$\displaystyle\left\\{\omega\in\Sigma:\limsup_{N\to\infty}\frac{L_{N}(\omega)}{\log
N}>\frac{1+\epsilon}{h(\Sigma)}\right\\}$
$\displaystyle\subset\left\\{\omega\in\Sigma:\frac{L_{N}(\omega)}{\log
N}>\frac{1+\epsilon}{h(\Sigma)}\quad{\rm i.o.}\right\\}$ (4.7)
$\displaystyle\subset\left\\{\omega\in\Sigma:l_{n}(\omega)>(1+\epsilon)\frac{\log
n}{h(\Sigma)}\quad{\rm i.o.}\right\\}$
$\displaystyle=\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}\left\\{\omega\in\Sigma:l_{n}(\omega)>(1+\epsilon)\frac{\log
n}{h(\Sigma)}\right\\}$
$\displaystyle=\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}\sigma^{-n}I^{\prime}_{n},$

where $I^{\prime}_{n}=I_{\left\lfloor(1+\epsilon)\frac{\log
n}{h(\Sigma)}\right\rfloor+1}(0^{\infty})$. In view of the $\sigma$-invariance
of $\mu$, by $\eqref{31}$, we have that

$\mu(\sigma^{-n}I^{\prime}_{n})=\mu(I^{\prime}_{n})\leq\gamma\cdot
e^{-\left(\left\lfloor(1+\epsilon)\frac{\log
n}{h(\Sigma)}\right\rfloor+1\right)h(\Sigma)}<\gamma\cdot n^{-(1+\epsilon)}.$

So $\sum_{n=1}^{\infty}\mu(\sigma^{-n}I^{\prime}_{n})<+\infty$, applying the
first Borel-Cantelli lemma [14, Lemma 1.2], we have
$\mu\left(\bigcap\limits_{N=1}^{\infty}\bigcup\limits_{n=N}^{\infty}\sigma^{-n}I^{\prime}_{n}\right)=0$.
By $\eqref{32}$,

$\mu\left(\left\\{\omega\in\Sigma:\limsup_{N\to\infty}\frac{L_{N}(\omega)}{\log
N}>\frac{1+\epsilon}{h(\Sigma)}\right\\}\right)=0.$

Secondly, we show that

$\liminf_{N\to\infty}\frac{L_{N}(\omega)}{\log N}\geq\frac{1}{h(\Sigma)}\ {\rm
for}\ \mu-{\rm a.a.}\ \omega\in\Sigma.$ (4.8)

It suffices to prove that

$\mu\left(\left\\{\omega\in\Sigma:\liminf_{N\to\infty}\frac{L_{N}(\omega)}{\log
N}\geq\frac{1-\epsilon}{h(\Sigma)}\right\\}\right)=1$

for any $\epsilon>0$. So we need to verify that for $\mu$-a.a.
$\omega\in\Sigma$, when $N$ is large enough, there exists a $n$ such that
$1\leq n\leq N$ and

$l_{n}(\omega)\geq(1-\epsilon)\frac{\log N}{h(\Sigma)}.$

Our idea is to prove that for $\mu$-a.a. $\omega\in\Sigma$, when $N$ is large
enough, there exists a $n$ such that $N^{1-\epsilon}\leq n\leq N$ and
$\sigma^{n}\omega\in I_{n}$, where $I_{n}=I_{\left\lfloor\frac{\log
n}{h(\Sigma)}\right\rfloor+1}(0^{\infty})$. That is, we need to show that for
$\mu$-a.a. $\omega\in\Sigma$, when $N$ is large enough,

$\sum_{n=\lfloor
N^{1-\epsilon}\rfloor+1}^{N}\chi_{\sigma^{-n}I_{n}}(\omega)\geq 1.$

Note that

$\sum_{n=\lfloor
N^{1-\epsilon}\rfloor+1}^{N}\chi_{\sigma^{-n}I^{\prime}_{n}}(\omega)=R(N,\omega)-R(\lfloor
N^{1-\epsilon}\rfloor,\omega).$

It follows from Lemma 4.3 that

$R(N,\omega)=F(N)+O\left(F^{\frac{1}{2}}(N)(\log F(N))^{2}\right)$ (4.9)

for $\mu$-a.a. $\omega\in\Sigma$, where

$F(N)=\sum_{n\leq N}\mu(I_{n}).$

By Lemma 4.1 and the $\sigma$-invariance of $\mu$, we have that
$\gamma_{1}^{-1}n^{-1}\leq\mu(I_{n})\leq\gamma_{1}n^{-1}$, where
$\gamma_{1}=\gamma\cdot e^{h(\Sigma)}$. Thus, $\gamma_{1}^{-1}\log N\leq
F(N)\leq\gamma_{1}(1+\log N)$. Therefore,

$R(N,\omega)=F(N)+O\left(\log^{\frac{2}{3}}N\right)\ {\rm for}\ \mu-{\rm
a.a.}\ \omega\in\Sigma.$ (4.10)

Combining $\eqref{33}$ and $\eqref{34}$, for $\mu$-a.a. $\omega\in\Sigma$, we
have

$\displaystyle\sum_{n=\lfloor
N^{1-\epsilon}\rfloor+1}^{N}\chi_{\sigma^{-n}I_{n}}(\omega)$
$\displaystyle=F(N)-F(\lfloor
N^{1-\epsilon}\rfloor)+O\left(\log^{\frac{2}{3}}N\right)$
$\displaystyle=\sum_{n=\lfloor
N^{1-\epsilon}\rfloor+1}^{N}\mu(I_{n})+O\left(\log^{\frac{2}{3}}N\right)$
$\displaystyle\geq\gamma_{1}^{-1}\frac{\epsilon}{2}\log
N+O\left(\log^{\frac{2}{3}}N\right)\geq 1$

for all $N$ large enough, which implies $\eqref{28}$. By $\eqref{27}$ and
$\eqref{28}$, the conclusion follows. ∎

##  5 The Hausdorff dimensions of $\mathcal{U}_{\psi_{0}}(a,b)$ and
$\mathcal{U}_{\psi_{0}}^{\ast}(a,b)$

Recall that $\psi_{0}(N)=m^{-N}$, that is, $-\log_{m}\psi_{0}(N)=N$. In this
section, we calculate the Hausdorff dimensions of
$\mathcal{U}_{\psi_{0}}(a,b)$ and $\mathcal{U}_{\psi_{0}}^{\ast}(a,b)$, which
plays an important role in the proof of Theorem 1.7. For simplicity, denote
$\underline{L}(\omega)=\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{N}$ and
$\overline{L}(\omega)=\limsup\limits_{N\to\infty}\frac{L_{N}(\omega)}{N}$.

######  Lemma 5.1.

Assume that $\Sigma$ has specification, $h(\Sigma)>0$ and
$0^{\infty}\in\Sigma$.
(i) If $a<1$ and $b<\frac{a}{1-a}$, then

$\mathcal{U}_{\psi_{0}}(a,b)=\emptyset.$

(ii) If $a<1$ and $b\geq\frac{a}{1-a}$, then

$\dim_{\rm H}\mathcal{U}^{\ast}_{\psi_{0}}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a,b)=\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm H}\Sigma.$

Furthermore, for any positive number $\varrho$, we have that

$\dim_{\rm H}(\\{\omega\in\Sigma:\underline{L}(\omega)\geq
a,b\leq\overline{L}(\omega)\leq
b+\varrho\\})\leq\left(\frac{b(1-a)-a}{(1+b)(b-a)}+\frac{\varrho(1-a)^{2}}{a^{3}}\right)\dim_{\rm
H}\Sigma.$

###  5.1 Proof of Lemma 5.1 (i)

Lemma 5.1 (i) follows from Propositions 5.2 and 5.3.

######  Proposition 5.2.

For $\omega=\omega_{1}\cdots\omega_{n}\cdots\in\mathcal{A}^{\mathbb{N}}$, we
have $\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{N}>1$ if and only if
$\omega_{n}=0$ for all $n$ large enough.

######  Proof.

The “if” part is obviously, we only need to show the “only if” part. Since
$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{N}>1$, there exists
$\epsilon>0$ such that
$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{N}>1+\epsilon$. So there
exists $N_{0}\in\mathbb{N}$, such that $\frac{L_{N}(\omega)}{N}>1+\epsilon$
for all $N\geq N_{0}$. It means that for every $N\geq N_{0}$, there exists
$1\leq n\leq N$, satisfies $l_{n}(\omega)\geq\lfloor(1+\epsilon)N\rfloor+1$.
Because of $n+1\leq N+1$ and $n+l_{n}(\omega)\geq
n+\lfloor(1+\epsilon)N\rfloor+1\geq N+2$, we obtain that $\omega_{N+1}=0$.
Thus, $\omega_{n}=0$ for every $n\geq N_{0}+1$. ∎

######  Proposition 5.3.

For any $\omega\in\mathcal{A}^{\mathbb{N}}$, we have

$\overline{L}(\omega)=+\infty\ {\rm when}\ \underline{L}(\omega)=1$ (5.1)

and

$\overline{L}(\omega)\geq\frac{\underline{L}(\omega)}{1-\underline{L}(\omega)}\
{\rm when}\ \underline{L}(\omega)<1.$ (5.2)

######  Proof.

By Proposition 5.2, we know that $\underline{L}(\omega)\leq 1$ is equivalent
to $\omega_{n}\neq 0$ for infinitely many $n\in\mathbb{N}$. Without loss of
generality, we assume that $\underline{L}(\omega)>0$. Let
$\mathcal{F}=\\{n:\omega_{n+1}=0\\}$. Since
$\overline{L}(\omega)\geq\underline{L}(\omega)>0$, $\mathcal{F}$ is a infinite
set. So we can write $\mathcal{F}=\\{n^{\prime}_{k}\\}_{k=1}^{\infty}$, where
$n^{\prime}_{1}<n^{\prime}_{2}<\cdots$. Since $\omega_{n}\neq 0$ for
infinitely many $n$, for every $k\geq 1$, there exists $m^{\prime}_{k}\geq
n^{\prime}_{k}+1$, satisfies

$\omega_{n^{\prime}_{k}+1}=\cdots=\omega_{m^{\prime}_{k}}=0,\omega_{m^{\prime}_{k}+1}\neq
0.$

That is, $l_{n^{\prime}_{k}}(\omega)=m^{\prime}_{k}-n^{\prime}_{k}$. In view
of $\overline{L}(\omega)>0$, we have

$\limsup_{k\to\infty}l_{n^{\prime}_{k}}(\omega)=+\infty.$ (5.3)

Now, we take the maximal subsequences $\\{n_{k}\\}_{k=1}^{\infty}$ and
$\\{m_{k}\\}_{k=1}^{\infty}$ of $\\{n^{\prime}_{k}\\}_{k=1}^{\infty}$ and
$\\{m^{\prime}_{k}\\}_{k=1}^{\infty}$, respectively, in such a way that the
sequence $\\{m_{k}-n_{k}\\}$ is strictly increasing. More precisely, let
$n_{1}=n^{\prime}_{1}$ and $m_{1}=m^{\prime}_{1}$. Let $k\geq 1$ be such that
$n_{k}=n^{\prime}_{i_{k}}$ and $m_{k}=m^{\prime}_{i_{k}}$ have been defined.
Denote

$i_{k+1}=\min\\{i>i_{k}:m^{\prime}_{i}-n^{\prime}_{i}>m_{k}-n_{k}\\}.$

Then, define

$n_{k+1}=n^{\prime}_{i_{k+1}}\ {\rm and}\ m_{k+1}=m^{\prime}_{i_{k+1}}.$

By $\eqref{43}$, the sequence $\\{i_{k}\\}_{k=1}^{\infty}$ is well defined.
Furthermore, by construction, we have

$\overline{L}(\omega)=\limsup_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k}}=\limsup_{k\to\infty}\frac{m_{k}}{n_{k}}-1$
(5.4)

and

$\underline{L}(\omega)=\liminf_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}}\leq\liminf_{k\to\infty}\frac{m_{k}-n_{k}}{m_{k}}=1-\limsup_{k\to\infty}\frac{n_{k}}{m_{k}}.$
(5.5)

If $\underline{L}(\omega)=1$, it follows from $\eqref{45}$ that
$\limsup\limits_{k\to\infty}\frac{n_{k}}{m_{k}}=0$. Thus,
$\limsup\limits_{k\to\infty}\frac{m_{k}}{n_{k}}=+\infty$. It follows from
$\eqref{44}$ that $\overline{L}(\omega)=+\infty$. If
$\underline{L}(\omega)<1$, we assume that $\overline{L}(\omega)<+\infty$,
otherwise, $\eqref{42}$ is trivial. Then by $\eqref{44}$, we have
$\limsup\limits_{k\to\infty}\frac{m_{k}}{n_{k}}<+\infty$. Hence,

$\limsup_{k\to\infty}\frac{m_{k}}{n_{k}}\cdot\limsup_{k\to\infty}\frac{n_{k}}{m_{k}}\geq
1.$ (5.6)

The upshot of $\eqref{44}$, $\eqref{45}$ and $\eqref{46}$ is that
$(\overline{L}(\omega)+1)\cdot(1-\underline{L}(\omega))\geq 1$. It follows
that
$\overline{L}(\omega)\geq\frac{\underline{L}(\omega)}{1-\underline{L}(\omega)}$.
∎

######  Proof of Lemma 5.1 (i).

For any $\omega\in\mathcal{A}^{\mathbb{N}}$ with $\underline{L}(\omega)\geq
a$, it follows from Propositions 5.2 and 5.3 that
$\overline{L}(\omega)\geq\frac{a}{1-a}$. Thus,
$\mathcal{U}_{\psi_{0}}(a,b)=\emptyset$ when $a<1$ and $b<\frac{a}{1-a}$. ∎

###  5.2 Proof of Lemma 5.1 (ii)

####  5.2.1 Upper bound

In this section, we prove that

$\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a,b)\leq\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm
H}\Sigma,\ \forall\ a<1\ {\rm and}\ b\geq\frac{a}{1-a}.$

######  Proof.

An easy cover argument shows that $\dim_{\rm
H}(\\{\omega\in\Sigma:\overline{L}(\omega)=+\infty\\})=0$. It follows from
Proposition 5.3 and Proposition 5.2 that
$\mathcal{U}_{\psi_{0}}(1)\subset\\{\omega\in\Sigma:\overline{L}(\omega)=+\infty\\}$.
Thus, we can assume that $0<b<+\infty$ and we only need to consider the set
$\mathcal{U}_{\psi_{0}}(a,b)\setminus\mathcal{U}_{\psi_{0}}(1)$. For any
$\omega\in\mathcal{U}_{\psi_{0}}(a,b)\setminus\mathcal{U}_{\psi_{0}}(1)$, that
is, $\omega\in\Sigma$ with $a\leq\underline{L}(\omega)<1$ and
$\overline{L}(\omega)=b>0$. By the proof of Proposition 5.3, we know that
there exists two strictly increasing sequences $\\{n_{k}\\}_{k=1}^{\infty}$,
$\\{m_{k}\\}_{k=1}^{\infty}$ with $\\{m_{k}-n_{k}\\}_{k=1}^{\infty}$ is
strictly increasing and

$\omega_{n_{k}+1}=\cdots=\omega_{m_{k}}=0,\omega_{m_{k}+1}\neq 0$

for each $k\in\mathbb{N}$. Furthermore,

$\limsup_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k}}=\overline{L}(\omega)=b$ (5.7)

and

$\liminf_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}}=\underline{L}(\omega)\geq a.$
(5.8)

Take a subsequence $\\{k_{j}\\}_{j=1}^{\infty}$ along which the supremum of
$\eqref{48}$ is obtained. For simplicity, we still denote
$\\{n_{k_{j}}\\}_{j=1}^{\infty}$, $\\{m_{k_{j}}\\}_{j=1}^{\infty}$ by
$\\{n_{k}\\}_{k=1}^{\infty}$, $\\{m_{k}\\}_{k=1}^{\infty}$. We remark that
when passing to the subsequence, the first equality in $\eqref{49}$ becomes an
inequality. Let $\epsilon\in(0,\frac{a}{2})$, there exists
$K_{1}\in\mathbb{N}$, such that for all $k\geq K_{1}$, we have

$(b-\epsilon)n_{k}<m_{k}-n_{k}<(b+\epsilon)n_{k}$ (5.9)

and

$m_{k}-n_{k}>(a-\epsilon)n_{k+1}.$ (5.10)

Combining the second inequality of $\eqref{410}$ and $\eqref{411}$, we obtain
that

$m_{k-1}-n_{k-1}>\frac{a-\epsilon}{b+\epsilon}(m_{k}-n_{k}),\ \forall\ k\geq
K_{1}+1.$ (5.11)

Since $0<\frac{a-\epsilon}{b+\epsilon}<1$, for all $k$ large enough, we have
$(\frac{a-\epsilon}{b+\epsilon})^{k-K_{1}}<\epsilon$. So when $k$ is large
enough, the sum of all the lengths of the blocks of 0 in the prefix of length
$n_{k}$ in the infinite sequence $\omega_{1}\omega_{2}\cdots$ is at least
equal to

$\displaystyle\sum_{j=K_{1}}^{k-1}(m_{j}-n_{j})$
$\displaystyle\geq\sum_{j=K_{1}}^{k-1}\left(\frac{a-\epsilon}{b+\epsilon}\right)^{k-j}(m_{k}-n_{k})>(b-\epsilon)n_{k}\sum_{j=K_{1}}^{k-1}\left(\frac{a-\epsilon}{b+\epsilon}\right)^{k-j}$
(5.12)
$\displaystyle>(b-\epsilon)n_{k}\frac{a-\epsilon}{b-a+2\epsilon}(1-\epsilon)=\left(\frac{ba}{b-a}-\epsilon^{\prime}\right)n_{k},$

where
$\epsilon^{\prime}=\frac{ba}{b-a}-\frac{(b-\epsilon)(a-\epsilon)}{b-a+2\epsilon}+\epsilon\frac{(b-\epsilon)(a-\epsilon)}{b-a+2\epsilon}$.
The first inequality is due to $\eqref{412}$, the second inequality is due to
the first inequality of $\eqref{410}$, the third inequality is due to the
arithmetic series summation formula and
$(\frac{a-\epsilon}{b+\epsilon})^{k-K_{1}}<\epsilon$. It follows from the
first inequality of $\eqref{410}$ that
$m_{k}>(b-\epsilon+1)n_{k}\geq(b-\epsilon+1)m_{k-1}$. Thus,
$\\{m_{k}\\}_{k=1}^{\infty}$ increases at least exponentially. Since
$n_{k}\geq m_{k-1}$, the sequence $\\{n_{k}\\}_{k=1}^{\infty}$ also increases
at least exponentially. It follows that there exists $c_{1}>0$, such that

$k\leq c_{1}\log n_{k},\quad\forall\ k\gg 1.$ (5.13)

Now, for any $\delta>0$, let us construct a $\delta$-cover. Let
$\Gamma_{n,\epsilon}$ be the set of all
$\omega_{1}\cdots\omega_{\lfloor(b-\epsilon+1)n\rfloor}\in\Sigma^{\lfloor(b-\epsilon+1)n\rfloor}$
satisfies $\omega_{n+1}=\cdots=\omega_{\lfloor(b-\epsilon+1)n\rfloor}=0$ and
there exists $i$ blocks of 0 of $\omega_{1}\cdots\omega_{n}$, with $i\leq
c_{1}\log n$ and the total length of these $i$ blocks is at least equal to
$\left(\frac{ba}{b-a}-\epsilon^{\prime}\right)n$. In view of $\eqref{413}$,
$\eqref{414}$ and the first inequality of $\eqref{410}$, for all $N$ large
enough,

$\bigcup_{n=N}^{\infty}\bigcup_{\omega\in\Gamma_{n,\epsilon}}I_{\lfloor(b-\epsilon+1)n\rfloor}(\omega)$

is a $\delta$-cover of the set
$\mathcal{U}_{\psi_{0}}(a,b)\setminus\mathcal{U}_{\psi_{0}}(1)$. Thus, for any
$s\geq 0$, we have

$\mathcal{H}_{\delta}^{s}(\mathcal{U}_{\psi_{0}}(a,b)\setminus\mathcal{U}_{\psi_{0}}(1))\leq\sum_{n=N}^{\infty}\sum_{\omega\in\Gamma_{n,\epsilon}}|I_{\lfloor(b-\epsilon+1)n\rfloor}(\omega)|^{s}\leq\sum_{n=N}^{\infty}m^{-(b-\epsilon+1)ns}\cdot\sharp\Gamma_{n,\epsilon}.$
(5.14)

So, we need to estimate $\sharp\Gamma_{n,\epsilon}$. Note that for any $i$
with $1\leq i\leq c_{1}\log n$ and every $l$ with
$\left(\frac{ba}{b-a}-\epsilon^{\prime}\right)n\leq l\leq n$, there are at
most

$n(n-1)\cdots(n-i+1)\binom{l-1}{i-1}$

choices of $i$ blocks of 0 of $\omega_{1}\cdots\omega_{n}$, with the total
length of these $i$ blocks is equal to $l$. Denote by
$B=h(\Sigma):=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$. We
claim that the remaining $n-l$ digits has at most

$e^{\left(n-\left(\frac{ba}{b-a}-\epsilon^{\prime}\right)n+iM\right)(B+\epsilon)}$

choices. In fact, if $n-l<\log n$, then the choices of the remaining $n-l$
digits is at most equal to

$m^{n-l}<m^{\log
n}<e^{\left(n-\left(\frac{ba}{b-a}-\epsilon^{\prime}\right)n+iM\right)(B+\epsilon)}.$

If $n-l\geq\log n$, since the remaining $n-l$ digits could be $i-1$ or $i$ or
$i+1$ blocks of $\omega_{1}\cdots\omega_{n}$, we consider three cases.
Case (i@). The remaining $n-l$ digits are $i-1$ blocks of
$\omega_{1}\cdots\omega_{n}$, we assume that these $i-1$ blocks is
$u_{1},\cdots,u_{i-1}$. Since $\Sigma$ has specification, there exists
$v_{1},\cdots,v_{i-2}\in\Sigma^{M}$, such that $u_{1}v_{1}\cdots
u_{i-2}v_{i-2}u_{i-1}\in\Sigma^{n-l+(i-2)M}$. Furthermore, $v_{j}$ ($1\leq
j\leq i-2$) is the minimal word (in the sense of lexicographical order) such
that $u_{1}v_{1}\cdots u_{j}v_{j}u_{j+1}\in\Sigma^{\ast}$.
Case (ii@). The remaining $n-l$ digits are $i$ blocks of
$\omega_{1}\cdots\omega_{n}$, we assume that these $i$ blocks is
$u_{1},\cdots,u_{i}$. Since $\Sigma$ has specification, there exists
$v_{1},\cdots,v_{i-1}\in\Sigma^{M}$, such that $u_{1}v_{1}\cdots
u_{i-1}v_{i-1}u_{i}\in\Sigma^{n-l+(i-1)M}$. Furthermore, $v_{j}$ ($1\leq j\leq
i-1$) is the minimal word (in the sense of lexicographical order) such that
$u_{1}v_{1}\cdots u_{j}v_{j}u_{j+1}\in\Sigma^{\ast}$.
Case (iii@). The remaining $n-l$ digits are $i+1$ blocks of
$\omega_{1}\cdots\omega_{n}$, we assume that these $i+1$ blocks is
$u_{1},\cdots,u_{i+1}$. Since $\Sigma$ has specification, there exists
$v_{1},\cdots,v_{i}\in\Sigma^{M}$, such that $u_{1}v_{1}\cdots
u_{i}v_{i}u_{i+1}\in\Sigma^{n-l+iM}$. Furthermore, $v_{j}$ ($1\leq j\leq i$)
is the minimal word (in the sense of lexicographical order) such that
$u_{1}v_{1}\cdots u_{j}v_{j}u_{j+1}\in\Sigma^{\ast}$.
Thus, for all $n$ large enough, the choices of the remaining $n-l$ digits is
at most equal to

$\max\left\\{\sharp\Sigma^{n-l+(i-2)M},\sharp\Sigma^{n-l+(i-1)M},\sharp\Sigma^{n-l+iM}\right\\}<e^{(n-l+iM)(B+\epsilon)}\leq
e^{\left(n-\left(\frac{ba}{b-a}-\epsilon^{\prime}\right)n+iM\right)(B+\epsilon)}.$
(5.15)

The first inequality of $\eqref{416}$ is due to
$B=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$ and $n-l\geq\log
n$. The second inequality of $\eqref{416}$ is due to
$l\geq\left(\frac{ba}{b-a}-\epsilon^{\prime}\right)n$. Therefore,
$\sharp\Gamma_{n,\epsilon}$ is not larger than

$n^{2c_{1}\log n}c_{1}\log n\cdot e^{\left(n\left(\frac{b-a-
ba}{b-a}+\epsilon^{\prime}\right)+Mc_{1}\log n\right)(B+\epsilon)}.$

Combining this and $\eqref{415}$, we have to consider the series

$\sum_{n=1}^{\infty}n^{2c_{1}\log n}c_{1}\log n\cdot e^{n(-(b-\epsilon+1)s\log
m+\left(\frac{b-a-ba}{b-a}+\epsilon^{\prime}\right)(B+\epsilon))+Mc_{1}\log
n(B+\epsilon)}.$ (5.16)

The critical exponent $s_{0}$ such that $\eqref{417}$ converges if $s>s_{0}$
and diverges if $s<s_{0}$ is given by

$s_{0}=(b-\epsilon+1)^{-1}\left(\frac{b(1-a)-a}{b-a}+\epsilon^{\prime}\right)\frac{B+\epsilon}{\log
m}$

It then follows from $\dim_{\rm H}\Sigma=\frac{h(\Sigma)}{\log m}$ ([12,
Proposition<EMAIL_ADDRESS>that

$\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a,b)\leq\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm
H}\Sigma.$

Actually, we have proved that, for every $b\geq\frac{a}{1-a}$ and for every
$\varrho>0$, we have

$\dim_{\rm H}(\\{\omega\in\Sigma:\underline{L}(\omega)\geq
a,b\leq\overline{L}(\omega)\leq
b+\varrho\\})\leq\left(\frac{b(1-a)-a}{(1+b)(b-a)}+\frac{\varrho(1-a)^{2}}{a^{3}}\right)\dim_{\rm
H}\Sigma.$ (5.17)

∎

####  5.2.2 Lower bound

In this section, we show that

$\dim_{\rm
H}\mathcal{U}_{\psi_{0}}^{\ast}(a,b)\geq\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm
H}\Sigma,\ \forall\ a<1\ {\rm and}\ b\geq\frac{a}{1-a}.$ (5.18)

The following proposition enables us to consider only the measure of
cylinders, which is helpful in the estimate of the lower bound of Hausdorff
dimension. For any $U\subset\mathcal{A}^{\mathbb{N}}$, let
$|U|=\sup\\{d(u,v):u,v\in U\\}$, which is the diameter of $U$.

######  Proposition 5.4.

(MDP) Let $E\subset\Sigma$ be a Borel measurable set and $\lambda$ be a Borel
measure with $\lambda(E)>0$. If there exists some $s\in[0,\log m]$ such that
for any $\omega\in E$,

$\liminf_{n\to\infty}\frac{\log\lambda(I_{n}(\omega))}{\log|I_{n}(\omega)|}\geq
s,$

then $\dim_{\rm H}E\geq s$.

######  Proof.

The proof is standard, we give the proof here for completeness. For any
$\omega\in E$ and $r\in(0,m^{-1})$, there exists $n\geq 1$ such that
$m^{-n-1}\leq r<m^{-n}$, hence

$I_{n}(\omega)=B(\omega,r).$

Therefore,

$\frac{\log\lambda(B(\omega,r))}{\log
2r}=\frac{-\log\lambda(B(\omega,r))}{-\log
2r}=\frac{-\log\lambda(I_{n}(\omega))}{-\log
2r}\geq\frac{-\log\lambda(I_{n}(\omega))}{-\log|I_{n}(\omega)|}=\frac{\log\lambda(I_{n}(\omega))}{\log|I_{n}(\omega)|}.$

Thus,

$\liminf_{r\to 0}\frac{\log\lambda(B(\omega,r))}{\log
r}\geq\liminf_{n\to\infty}\frac{\log\lambda(I_{n}(\omega))}{\log|I_{n}(\omega)|}\geq
s.$

By [8, Proposition 2.3], we have that $\dim_{\rm H}E\geq s$. ∎

The following statement shows that if $\Sigma$ has specification and
$h(\Sigma)>0$, then the diameter of of $n$th cylinder is comparable to
$m^{-n}$.

######  Proposition 5.5.

Suppose that $\Sigma$ has specification and $h(\Sigma)>0$. There exists
$c^{\ast}>0$, which only depend on $M$, such that for any $\omega\in\Sigma$
and any $n\in\mathbb{N}$,

$c^{\ast}m^{-n}\leq|I_{n}(\omega)|\leq m^{-n-1}.$

######  Proof.

By the definition of $I_{n}(\omega)$, we immediately obtain that
$|I_{n}(\omega)|\leq m^{-n-1}$. Since $\Sigma$ has specification, for any
$\omega_{1},\omega_{2}\in\Sigma^{\ast}$, there exists
$\varepsilon\in\Sigma^{M}$, such that
$\omega_{1}\varepsilon\omega_{2}\in\Sigma^{\ast}$. Because of

$h(\Sigma):=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}>0,$

there exists $K\in\mathbb{N}$, such that
$\sharp\Sigma^{K}>m^{M}\geq\sharp\Sigma^{M}$. By pigeon-hole principle, there
exists $v_{1}\neq v_{2}\in\Sigma^{K}$ and $\varepsilon\in\Sigma^{M}$,
satisfies $\omega_{[1,n]}\varepsilon v_{1},\omega_{[1,n]}\varepsilon
v_{2}\in\Sigma^{\ast}$. Note that for any $x\in
I_{n+M+K}(\omega_{[1,n]}\varepsilon v_{1})$ and any $y\in
I_{n+M+K}(\omega_{[1,n]}\varepsilon v_{2})$, we have $d(x,y)\geq m^{-n-M-K}$.
Thus, $|I_{n}(\omega)|\geq m^{-n-M-K}$. ∎

Now, we start proving $\eqref{426}$.

######  Proof.

If $0<a<1$ and $b=\frac{a}{1-a}$, the inequality is trivial. If $0<a<1$ and
$b>\frac{a}{1-a}$, we choose an integer $k_{0}$ with
$(a^{-1}b)^{k_{0}}>\max\left\\{\frac{a(2+b)}{b(1-a)-a},\frac{a(1+b)}{b(b-a)}\right\\}$
and let $n_{k}=\left\lfloor(a^{-1}b)^{k+k_{0}}\right\rfloor+1$,
$m_{k}=\lfloor(1+b)n_{k}\rfloor+1$. If $a=0$, we choose integers $k_{0}\geq
b+1$ and $n_{1}\geq 4(\sqrt{2}+1)^{2}$. Let $n_{k+1}=(k+k_{0}+1)n_{k}$ and
$m_{k}=\lfloor(1+b)n_{k}\rfloor+\lfloor\sqrt{n_{k}}\rfloor+1$. It is easy to
check that $m_{k}<n_{k+1}$ and $\\{m_{k}-n_{k}\\}_{k=1}^{\infty}$ is strictly
increasing. Furthermore,

$\lim_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}}=a$

and

$\lim_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k}}=b.$

Let $t_{k}$ be the largest integer such that
$m_{k}+t_{k}(m_{k}-n_{k})<n_{k+1}$. Since $B>0$, we have
$\Sigma^{1}\setminus\\{0\\}\neq\emptyset$. Fix
$\omega\in\Sigma^{1}\setminus\\{0\\}$, for any
$u_{k}^{(1)},\cdots,u_{k}^{(t_{k})}\in\Sigma^{m_{k}-n_{k}-2M-1}$ and any
$v_{k}\in\Sigma^{n_{k+1}-m_{k}-t_{k}(m_{k}-n_{k})}$, since $\Sigma$ has
specification, there exists
$\omega_{k}^{(1)},\cdots,\omega_{k}^{(2t_{k}+3)}\in\Sigma^{M}$, such that

$\omega\omega_{k}^{(1)}0^{m_{k}-n_{k}-2M-1}\omega_{k}^{(2)}\omega\omega_{k}^{(3)}u_{k}^{(1)}\omega_{k}^{(4)}\cdots\omega\omega_{k}^{(2t_{k}+1)}u_{k}^{(t_{k})}\omega_{k}^{(2t_{k}+2)}\omega\omega_{k}^{(2t_{k}+3)}v_{k}\in\Sigma^{\ast}.$

Furthermore, we choose
$\omega_{k}^{(1)},\cdots,\omega_{k}^{(2t_{k}+3)}\in\Sigma^{M}$ that satisfies

  1. (1)

$\omega_{k}^{(1)}$ is the minimal word (in the sense of lexicographical order)
such that
$\omega\omega_{k}^{(1)}0^{m_{k}-n_{k}-2M-1}\in\Sigma^{\ast}$;

  2. (2)

For any $1\leq i\leq t_{k}+1$, $\omega_{k}^{(2i)}$ is the minimal word (in the
sense of lexicographical order) such that
$\omega\omega_{k}^{(1)}0^{m_{k}-n_{k}-2M-1}\cdots\omega_{k}^{(2i)}\omega\in\Sigma^{\ast}$;

  3. (3)

For any $1\leq i\leq t_{k}$, $\omega_{k}^{(2i+1)}$ is the minimal word (in the
sense of lexicographical order) such that
$\omega\omega_{k}^{(1)}0^{m_{k}-n_{k}-2M-1}\cdots\omega\omega_{k}^{(2i+1)}u_{k}^{(i)}\in\Sigma^{\ast}$;

  4. (4)

$\omega_{k}^{(2t_{k}+3)}$ is the minimal word (in the sense of lexicographical
order) such that
$\omega\omega_{k}^{(1)}0^{m_{k}-n_{k}-2M-1}\omega_{k}^{(2)}\cdots\omega\omega_{k}^{(2t_{k}+1)}u_{k}^{(t_{k})}\omega_{k}^{(2t_{k}+2)}\omega\omega_{k}^{(2t_{k}+3)}v_{k}\in\Sigma^{\ast}$.

According to the above rules, for every
$u_{k}^{(1)},\cdots,u_{k}^{(t_{k})}\in\Sigma^{m_{k}-n_{k}-2M-1}$ and every
$v_{k}\in\Sigma^{n_{k+1}-m_{k}-t_{k}(m_{k}-n_{k})}$, the choice of
$\omega_{k}^{(1)},\cdots,\omega_{k}^{(2t_{k}+3)}\in\Sigma^{M}$ is determined,
we put

$\displaystyle\mathcal{W}_{k}=\left\\{\omega\omega_{k}^{(1)}0^{m_{k}-n_{k}-2M-1}\omega_{k}^{(2)}\cdots\omega\omega_{k}^{(2t_{k}+1)}u_{k}^{(t_{k})}\omega_{k}^{(2t_{k}+2)}\omega\omega_{k}^{(2t_{k}+3)}v_{k}\in\Sigma^{\ast}:\right.$
$\displaystyle\left.v_{k}\in\Sigma^{n_{k+1}-m_{k}-t_{k}(m_{k}-n_{k})},u_{k}^{(i)}\in\Sigma^{m_{k}-n_{k}-2M-1},\
\forall 1\leq i\leq t_{k}\right\\}.$

Note that every element of $\mathcal{W}_{k}$ is of length $n_{k+1}-n_{k}+M+1$.
Again, since $\Sigma$ has specification, for any $k\geq 1$ and any
$u_{i}\in\mathcal{W}_{i}$ $(1\leq i\leq k)$, there exists
$\varepsilon_{1},\cdots,\varepsilon_{k}\in\Sigma^{M}$, such that
$0^{n_{1}}\varepsilon_{1}u_{1}\varepsilon_{2}u_{2}\cdots\varepsilon_{k}u_{k}\in\Sigma^{\ast}$.
Here we choose $\varepsilon_{1},\cdots,\varepsilon_{k}\in\Sigma^{M}$, which
satisfies for any $1\leq i\leq k$, $\varepsilon_{i}$ is the minimal word (in
the sense of lexicographic order) such that
$0^{n_{1}}\varepsilon_{1}u_{1}\cdots\varepsilon_{i}u_{i}\in\Sigma^{\ast}$. Let

$G_{k}=\\{0^{n_{1}}\varepsilon_{1}u_{1}\cdots\varepsilon_{k}u_{k}\in\Sigma^{\ast}:u_{i}\in\mathcal{W}_{i},\forall\
1\leq i\leq k\\}.$

Note that every element of $G_{k}$ is of length $n_{k+1}+(2M+1)k$. Define

$E_{0}=I_{n_{1}}(0^{n_{1}})$

and

$E_{k}=\bigcup_{w\in G_{k}}I_{n_{k+1}+(2M+1)k}(w)$

for each $k\geq 1$. Put

$E=\bigcap_{k=0}^{\infty}E_{k},$

we show that $E\subset\mathcal{U}_{\psi_{0}}^{\ast}(a,b)$ and $\dim_{\rm
H}E\geq\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm H}\Sigma$. Denote
$N_{k}:=n_{k+1}+(2M+1)k$. For any $\omega\in E$, we prove that
$\underline{L}(\omega)=a$ and $\overline{L}(\omega)=b$. Since the element of
$E$ are of the form
$0^{n_{1}}\varepsilon_{1}u_{1}\cdots\varepsilon_{k}u_{k}\cdots$ with
$u_{k}\in\mathcal{W}_{k}$ $(k\geq 1)$ and $\varepsilon_{k}\in\Sigma^{M}$ is
the minimal word such that
$0^{n_{1}}\varepsilon_{1}u_{1}\cdots\varepsilon_{k}u_{k}\in\Sigma^{\ast}$, by
the definition of $\mathcal{W}_{k}$, we have the following two facts:
(i) If $N_{k}\leq N\leq N_{k}+2M$, then

$m_{k}-n_{k}-2M-1\leq L_{N}(\omega)\leq m_{k}-n_{k}+2M.$

(ii) If $N_{k}+2M<N<N_{k+1}$, then

$m_{k+1}-n_{k+1}-2M-1\leq L_{N}(\omega)\leq m_{k+1}-n_{k+1}+2M.$

Thus, we have

$\underline{L}(\omega):=\liminf_{N\to\infty}\frac{L_{N}(\omega)}{N}\geq\liminf_{k\to\infty}\frac{m_{k}-n_{k}}{N_{k}}=\lim_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}}=a$

and

$\overline{L}(\omega):=\limsup_{N\to\infty}\frac{L_{N}(\omega)}{N}\leq\limsup_{k\to\infty}\frac{m_{k+1}-n_{k+1}}{N_{k}}=\lim_{k\to\infty}\frac{m_{k+1}-n_{k+1}}{n_{k+1}}=b.$

Note that

$\lim_{k\to\infty}\frac{L_{N_{k}}(\omega)}{N_{k}}=\lim_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}}=a$

and

$\lim_{k\to\infty}\frac{L_{N_{k}+2M+1}(\omega)}{N_{k}+2M+1}=\lim_{k\to\infty}\frac{m_{k+1}-n_{k+1}}{n_{k+1}}=b.$

Therefore, $\underline{L}(\omega)=a$ and $\overline{L}(\omega)=b$. It remain
to prove that $\dim_{\rm H}E\geq\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm
H}\Sigma$. Firstly, we assign a mass distribution on $E$. We first assign a
mass distribution on all cylinders. Let $\lambda(I_{N}(0^{N}))=1$ for any
$1\leq N\leq n_{1}$. For any cylinder $I_{N_{1}}\in E_{1}$, we define

$\lambda(I_{N_{1}})=\frac{1}{\sharp\mathcal{W}_{1}}\lambda(I_{n_{1}}(0^{n_{1}}))=\frac{1}{\sharp\mathcal{W}_{1}}.$

Suppose that the measure on $I_{N_{k}}\in E_{k}$ $(k\geq 1)$ is well defined,
then the measure on $I_{N_{k+1}}\in E_{k+1}$ is defined as

$\lambda(I_{N_{k+1}})=\frac{1}{\sharp\mathcal{W}_{k+1}}\lambda(I_{N_{k}})=\prod_{i=1}^{k+1}\frac{1}{\sharp\mathcal{W}_{i}}.$

For any $n_{1}<N<N_{1}$, define

$\lambda(I_{N})=\sum_{I_{N_{1}}:I_{N_{1}}\cap
I_{N}\neq\emptyset}\lambda(I_{N_{1}}).$

For any $N_{k}<N<N_{k+1}$, define

$\lambda(I_{N})=\sum_{I_{N_{k+1}}:I_{N_{k+1}}\cap
I_{N}\neq\emptyset}\lambda(I_{N_{k+1}}).$

Note that

$\lambda(I_{N})=\sum_{I_{N+1}:I_{N+1}\cap
I_{N}\neq\emptyset}\lambda(I_{N+1}),$

thus, $\lambda$ can be extended to a measure on $E$. Secondly, in view of
Proposition 5.4, we only need to show that for any $\omega\in E$,

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm
H}\Sigma,$

where $I_{N}(\omega)$ is the $N$-th cylinder contain $\omega$. It follows from
Proposition 5.5 that

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}=\liminf_{N\to\infty}\frac{-\log\lambda(I_{N}(\omega))}{N\log
m}.$ (5.19)

If $N_{k}\leq N\leq M_{k}:=m_{k+1}+(2M+1)(k+1)$, then
$\lambda(I_{N}(\omega))=\lambda(I_{N_{k}}(\omega))=\prod_{i=1}^{k}\frac{1}{\sharp\mathcal{W}_{i}}$.
Thus,

$\displaystyle\frac{-\log\lambda(I_{N}(\omega))}{N}\geq\frac{\sum_{i=1}^{k}\log\sharp\mathcal{W}_{i}}{M_{k}}.$
(5.20)

Note that

$\liminf_{k\to\infty}\frac{\sum_{i=1}^{k}\log\sharp\mathcal{W}_{i}}{M_{k}}=\liminf_{k\to\infty}\frac{\sum_{i=1}^{k}\log\sharp\mathcal{W}_{i}}{m_{k+1}}\geq\liminf_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k}}{m_{k+1}-m_{k}}.$
(5.21)

The equality follows from $\lim\limits_{k\to\infty}\frac{k}{m_{k}}=0$. The
inequality follows from $\lim\limits_{k\to\infty}m_{k}=+\infty$ and the
definition of limit inferior. By the definition of $\mathcal{W}_{k}$, we have
that

$\sharp\mathcal{W}_{k}=\left(\sharp\Sigma^{m_{k}-n_{k}-2M-1}\right)^{t_{k}}\cdot\sharp\Sigma^{n_{k+1}-m_{k}-t_{k}(m_{k}-n_{k})}.$

Since $B=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$, we obtain
that
$\lim\limits_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k}}{n_{k+1}-m_{k}}=B$.
Hence,

$\lim_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k}}{m_{k+1}-m_{k}}=B\cdot\lim_{k\to\infty}\frac{n_{k+1}-m_{k}}{m_{k+1}-m_{k}}=\frac{b(1-a)-a}{(1+b)(b-a)}h(\Sigma).$
(5.22)

The last equality follows from our choice of sequences $\\{n_{k}\\}$,
$\\{m_{k}\\}$. For any $\delta>0$, the combination of $\eqref{421}$,
$\eqref{422}$ and $\eqref{423}$ gives

$\frac{-\log\lambda(I_{N}(\omega))}{N\log m}>(\log
m)^{-1}\left(\frac{b(1-a)-a}{(1+b)(b-a)}h(\Sigma)-\delta\right)$ (5.23)

for all $N$ large enough with $N_{k}\leq N\leq M_{k}$.
Recall that $N_{k+1}=n_{k+2}+(2M+1)(k+1)$ and $M_{k}=m_{k+1}+(2M+1)(k+1)$.
Thus,

$N_{k+1}=M_{k}+n_{k+2}-m_{k+1}\leq M_{k}+(t_{k+1}+1)(m_{k+1}-n_{k+1}).$

If $M_{k}<N<N_{k+1}$, then $M_{k}<N<M_{k}+(t_{k+1}+1)(m_{k+1}-n_{k+1})$. So
there exists $0\leq q\leq t_{k+1}$ and $0\leq r<m_{k+1}-n_{k+1}$, such that
$N=M_{k}+q(m_{k+1}-n_{k+1})+r$.
Case 1. If $r\leq M$, then $N\leq M_{k}+q(m_{k+1}-n_{k+1})+M$. So

$\displaystyle\lambda(I_{N}(\omega))$
$\displaystyle\leq\lambda(I_{M_{k}+q(m_{k+1}-n_{k+1})}(\omega))$
$\displaystyle=\left(\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1}\right)^{-q}\lambda(I_{M_{k}}(\omega)).$

Thus,

$\frac{-\log\lambda(I_{N}(\omega))}{N}\geq\frac{-\log\lambda(I_{M_{k}}(\omega))+q\log\sharp\Sigma^{m_{k}-n_{k}-2M-1}}{M_{k}+q(m_{k+1}-n_{k+1})+M}.$

In view of $\eqref{423}$, $\lim\limits_{k\to\infty}m_{k}=+\infty$ and the
definition of limit superior, we have

$\displaystyle\limsup_{k\to\infty}\frac{-\log\lambda(I_{M_{k}}(\omega))}{M_{k}}$
$\displaystyle=\limsup_{k\to\infty}\frac{\sum_{i=1}^{k}\log\sharp\mathcal{W}_{i}}{m_{k+1}}\leq\limsup_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k}}{m_{k+1}-m_{k}}$
$\displaystyle=\lim_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k}}{m_{k+1}-m_{k}}=\frac{b(1-a)-a}{(1+b)(b-a)}B\leq
B.$

Thus, for every sufficiently small $\delta>0$, for all $k$ large enough, we
have

$(B+\delta)^{-1}(-\log\lambda(I_{M_{k}}(\omega)))<M_{k}$

This together with $B=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$
gives

$\displaystyle\frac{-\log\lambda(I_{N}(\omega))}{N}$
$\displaystyle\geq\frac{-\log\lambda(I_{M_{k}}(\omega))+(B-\delta)q(m_{k+1}-n_{k+1})}{M_{k}+q(m_{k+1}-n_{k+1})+M}$
$\displaystyle\geq\frac{B-\delta}{1+\delta}\cdot\frac{(B+\delta)^{-1}(-\log\lambda(I_{M_{k}}(\omega)))+q(m_{k+1}-n_{k+1})}{M_{k}+q(m_{k+1}-n_{k+1})}$
$\displaystyle\geq\frac{B-\delta}{(B+\delta)(1+\delta)}\cdot\frac{-\log\lambda(I_{M_{k}}(\omega))}{M_{k}}$
$\displaystyle>\frac{B-\delta}{(B+\delta)(1+\delta)}\left(\frac{b(1-a)-a}{(1+b)(b-a)}h(\Sigma)-\delta\right)$

for all $N$ large enough. The third inequality follows from the inequality

$\frac{y+s}{z+s}\geq\frac{y}{z},0\leq y<z,s\geq 0.$

The last inequality follows from $\eqref{424}$.
Case 2. If $0\leq q\leq t_{k+1}-1$ and $r\geq m_{k+1}-n_{k+1}-2M-1$, we have

$\lambda(I_{N}(\omega))=\left(\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1}\right)^{-(q+1)}\lambda(I_{M_{k}}(\omega)).$

Thus,

$\frac{-\log\lambda(I_{N}(\omega))}{N}>\frac{-\log\lambda(I_{M_{k}}(\omega))+(q+1)\log\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1}}{M_{k}+(q+1)(m_{k+1}-n_{k+1})}.$

Similar with Case 1, for every sufficiently small positive number $\delta$,
for $N$ large enough, we have

$\frac{-\log\lambda(I_{N}(\omega))}{N}>\frac{B-\delta}{B+\delta}\cdot\left(\frac{b(1-a)-a}{(1+b)(b-a)}h(\Sigma)-\delta\right).$

Case 3. If $0\leq q\leq t_{k+1}-1$ and $M<r<m_{k+1}-n_{k+1}-2M-1$, then

$\displaystyle\lambda(I_{N}(\omega))$
$\displaystyle\leq\lambda(I_{M_{k}}(\omega))\cdot(\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1})^{-q-1}\cdot\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1-r},$
$\displaystyle=\lambda(I_{M_{k}}(\omega))\cdot(\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1})^{-q}\cdot\frac{\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1-r}}{\Sigma^{m_{k+1}-n_{k+1}-2M-1}}.$

Since $\Sigma$ has specification, we have

$\sharp\Sigma^{i}\geq\sharp\Sigma^{i-j}\cdot\sharp\Sigma^{j-M}$ (5.24)

for all $i>j>M$. Hence,

$\lambda(I_{N}(\omega))\leq\lambda(I_{M_{k}}(\omega))\cdot(\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1})^{-q}\cdot(\sharp\Sigma^{r-M})^{-1}.$

Thus, for every sufficiently small $\delta>0$, for all $N$ large enough, we
have

$\displaystyle\frac{\lambda(I_{N}(\omega))}{N}$
$\displaystyle\geq\frac{-\log\lambda(I_{M_{k}}(\omega))+q\cdot\log\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1}+\log\sharp\Sigma^{r-M}}{M_{k}+q(m_{k+1}-n_{k+1})+r}$
$\displaystyle\geq\frac{-\log\lambda(I_{M_{k}}(\omega))+q\cdot(1-\delta)\log\sharp\Sigma^{m_{k+1}-n_{k+1}}+\log\sharp\Sigma^{r-M}}{M_{k}+q(m_{k+1}-n_{k+1})+r}$
$\displaystyle\geq\frac{-\log\lambda(I_{M_{k}}(\omega))+(1-\delta)\log\sharp\Sigma^{q(m_{k+1}-n_{k+1})+r-M}}{M_{k}+q(m_{k+1}-n_{k+1})+r}$
$\displaystyle\geq\frac{1}{1+\delta}\cdot\frac{-\log\lambda(I_{M_{k}}(\omega))+(1-\delta)\log\sharp\Sigma^{q(m_{k+1}-n_{k+1})+r-M}}{M_{k}+q(m_{k+1}-n_{k+1})+r-M}.$

The second inequality follows from
$\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$ exists and
$\lim\limits_{k\to\infty}(m_{k+1}-n_{k+1})=+\infty$. The third inequality
follows from $\eqref{11}$ and $\sharp\Sigma^{r-M}\geq 1$. Similar with Case 1,
for all $N$ large enough, we have

$\frac{\lambda(I_{N}(\omega))}{N}>\frac{(1-\delta)(B-\delta)}{(1+\delta)(B+\delta)}\left(\frac{b(1-a)-a}{(1+b)(b-a)}h(\Sigma)-\delta\right).$

Case 4. If $q=t_{k+1}$ and $M<r<n_{k+2}-m_{k+1}-t_{k+1}(m_{k+1}-n_{k+1})$,
then

$\displaystyle\lambda(I_{N}(\omega))$
$\displaystyle\leq\sharp\Sigma^{n_{k+2}-m_{k+1}-t_{k+1}(m_{k+1}-n_{k+1})-r}\cdot\lambda(I_{N_{k+1}}(\omega))$
$\displaystyle=\lambda(I_{M_{k}}(\omega))\cdot(\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1})^{-t_{k+1}}\cdot\frac{\sharp\Sigma^{n_{k+2}-m_{k+1}-t_{k+1}(m_{k+1}-n_{k+1})-r}}{\sharp\Sigma^{n_{k+2}-m_{k+1}-t_{k+1}(m_{k+1}-n_{k+1})}}$
$\displaystyle\leq\lambda(I_{M_{k}}(\omega))\cdot(\sharp\Sigma^{m_{k+1}-n_{k+1}-2M-1})^{-t_{k+1}}\cdot(\sharp\Sigma^{r-M})^{-1}.$

The last inequality follows from $\eqref{425}$. Similar with Case 3, when $N$
is large enough, we have

$\frac{\lambda(I_{N}(\omega))}{N}>\frac{(1-\delta)(B-\delta)}{(1+\delta)(B+\delta)}\left(\frac{b(1-a)-a}{(1+b)(b-a)}h(\Sigma)-\delta\right).$

For every sufficient small $\delta$, the combination of $\eqref{424}$ and
Cases 1-4 gives

$\frac{\lambda(I_{N}(\omega))}{N\log m}>(\log
m)^{-1}\frac{(1-\delta)(B-\delta)}{(1+\delta)(B+\delta)}\left(\frac{b(1-a)-a}{(1+b)(b-a)}h(\Sigma)-\delta\right)$

for all $N$ large enough. Thus, we have

$\liminf_{N\to\infty}\frac{\lambda(I_{N}(\omega))}{N\log
m}\geq\frac{b(1-a)-a}{(1+b)(b-a)}\cdot\frac{h(\Sigma)}{\log
m}=\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm H}\Sigma.$ (5.25)

The equality follows from $\dim_{\rm H}\Sigma=\frac{h(\Sigma)}{\log m}$ (see
[12, Proposition iii@.1]). The combination of $\eqref{420}$, $\eqref{427}$ and
Proposition 5.4 gives $\eqref{426}$. ∎

##  6 Proof of Theorem 1.7 and Corollary 1.9

In this section, we give the proof of Theorem 1.7 and Corollary 1.9.

###  6.1 Proof of Theorem 1.7 Case iii

######  Proof.

We consider six cases.
Case 1: $a=b=\infty$. Since
$\lim\limits_{N\to\infty}\frac{-\log_{m}\psi(N)}{N}=0$, we have
$\overline{L}(\omega)>0$ ($\underline{L}(\omega)>0$) implies
$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=+\infty$
($\limsup\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=+\infty$)
respectively. Thus, for any $0<a^{\prime}<1$ and
$b^{\prime}\geq\frac{a^{\prime}}{1-a^{\prime}}$, we have
$\mathcal{U}_{\psi_{0}}^{\ast}(a^{\prime},b^{\prime})\subset\mathcal{U}_{\psi}^{\ast}(\infty,\infty)$.
It follows from Lemma 5.1 that

$\dim_{\rm
H}(\mathcal{U}_{\psi}^{\ast}(\infty,\infty))\geq\frac{b^{\prime}-b^{\prime}a^{\prime}-a^{\prime}}{(1+b^{\prime})(b^{\prime}-a^{\prime})}\dim_{\rm
H}\Sigma,\ \forall\ 0<a^{\prime}<1\ {\rm and}\
b^{\prime}\geq\frac{a^{\prime}}{1-a^{\prime}}.$

Therefore, $\dim_{\rm
H}(\mathcal{U}_{\psi}^{\ast}(\infty,\infty))\geq\dim_{\rm H}\Sigma$. Thus,
$\dim_{\rm H}(\mathcal{U}_{\psi}^{\ast}(\infty,\infty))=\dim_{\rm H}\Sigma$.
Case 2: $0<a\leq b<\infty$. Firstly, we construct a Cantor-type subset of
$\mathcal{U}_{\psi}^{\ast}(a,b)$. Denote by $\Phi(x)=-\log_{m}\psi(x)$. Fix a
integer $P\geq 3$. We define the following two sequences $\\{n_{k}\\}_{k\geq
0}$ and $\\{d_{k}\\}_{k\geq 0}$ of integers as follows: choose $n_{0}$ such
that

$\lfloor b\Phi(n_{0})\rfloor\geq P+2+3M$

and then take $d_{0}=\lfloor b\Phi(n_{0})\rfloor$, for any $k\geq 1$,
$\\{n_{k}\\}$ and $\\{d_{k}\\}$ are recursively defined as

$n_{k}:=\left\lfloor\Phi^{-1}\left(\frac{b}{a}\Phi(n_{k-1})\right)\right\rfloor+Pd_{k-1}$

and

$d_{k}:=\lfloor b\Phi(n_{k})\rfloor.$

Here, $\\{n_{k}\\}$ is well defined because of the strictly increasing
property of the function $\Phi$. Note that

$\lim_{k\to\infty}\frac{\Phi(n_{k})}{\Phi(n_{k-1})}=\frac{b}{a}.$ (6.1)

In fact, since $a\leq b$ and $\Phi$ is strictly increasing, we have
$\Phi^{-1}\left(\frac{b}{a}\Phi(n_{k-1})\right)\geq n_{k-1}$. Thus,
$\left\lfloor\Phi^{-1}\left(\frac{b}{a}\Phi(n_{k-1})\right)\right\rfloor\geq
n_{k-1}$. Therefore,

$n_{k}\leq\left\lfloor\Phi^{-1}\left(\frac{b}{a}\Phi(n_{k-1})\right)\right\rfloor+Pb\cdot\Phi\left(\left\lfloor\Phi^{-1}\left(\frac{b}{a}\Phi(n_{k-1})\right)\right\rfloor\right).$
(6.2)

Combining $\eqref{62}$ and
$\lim\limits_{N\to\infty}\frac{\Phi(N+\Phi(N))}{\Phi(N)}=1$, we obtain that

$\lim_{k\to\infty}\frac{\Phi(n_{k})}{\Phi\left(\left\lfloor\Phi^{-1}\left(\frac{b}{a}\Phi(n_{k-1})\right)\right\rfloor\right)}=1.$

Thus,

$\lim_{k\to\infty}\frac{\Phi(n_{k})}{\Phi(n_{k-1})}=\frac{b}{a}.$

In the following, we construct a Cantor-type subset of
$\mathcal{U}_{\psi}^{\ast}(a,b)$. Denote by
$B=h(\Sigma):=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$. Since
$B>0$, we have $\Sigma^{1}\setminus\\{0\\}\neq\emptyset$. Fix
$\omega\in\Sigma^{1}\setminus\\{0\\}$. For each $k\in\mathbb{N}$, we write

$n_{k}-n_{k-1}=d_{k-1}(l_{k}+1)+r_{k},$

where

$l_{k}=\left\lfloor\frac{n_{k}-n_{k-1}}{d_{k-1}}\right\rfloor-1,\quad 0\leq
r_{k}<d_{k-1}.$

Let

$v_{k}=\begin{cases}0^{r_{k}},&{\rm if}\ 0\leq r_{k}\leq P,\\\ \omega
0^{r_{k}-1},&{\rm if}\ P<r_{k}<d_{k-1}.\end{cases}$

Since $\Sigma$ has specification, for any
$u_{k}^{(1)},u_{k}^{(2)},\cdots,u_{k}^{(l_{k})}\in\Sigma^{d_{k-1}-1-2M}$,
there exists $\varepsilon_{k}^{(1)},\varepsilon_{k}^{(2)},$
$\cdots,\varepsilon_{k}^{(2l_{k}+2)}\in\Sigma^{M}$, such that

$\omega\varepsilon_{k}^{(1)}0^{d_{k-1}-1-3M}\varepsilon_{k}^{(2)}\omega\varepsilon_{k}^{(3)}u_{k}^{(1)}\varepsilon_{k}^{(4)}\omega\varepsilon_{k}^{(5)}u_{k}^{(2)}\varepsilon_{k}^{(6)}\cdots\omega\varepsilon_{k}^{(2l_{k}+1)}u_{k}^{(l_{k})}\varepsilon_{k}^{(2l_{k}+2)}v_{k}\in\Sigma^{\ast}.$

Similar with the proof of Lemma 5.1, we choose
$\varepsilon_{k}^{(1)},\cdots,\varepsilon_{k}^{(2l_{k}+2)}\in\Sigma^{M}$ that
satisfies

  1. (i)

For any $0\leq i\leq l_{k}$, $\varepsilon_{k}^{(2i+1)}$ is the minimal word
(in the sense of lexicographical order) such that
$\omega\varepsilon_{k}^{(1)}0^{d_{k-1}-1-3M}\varepsilon_{k}^{(2)}\omega\varepsilon_{k}^{(3)}\cdots\varepsilon_{k}^{(2i)}\omega\varepsilon_{k}^{(2i+1)}u_{k}^{(i)}\in\Sigma^{\ast}$;

  2. (ii)

For any $1\leq i\leq l_{k}$, $\varepsilon_{k}^{(2i)}$ is the minimal word (in
the sense of lexicographical order) such that
$\omega\varepsilon_{k}^{(1)}0^{d_{k-1}-1-3M}\varepsilon_{k}^{(2)}\omega\varepsilon_{k}^{(3)}u_{k}^{(1)}\varepsilon_{k}^{(4)}\cdots\varepsilon_{k}^{(2i-1)}u_{k}^{(i-1)}\varepsilon_{k}^{(2i)}\omega\in\Sigma^{\ast}$;

  3. (iii)

$\varepsilon_{k}^{(2l_{k}+2)}$ is the minimal word (in the sense of
lexicographical order) such that
$\omega\varepsilon_{k}^{(1)}0^{d_{k-1}-1-3M}\varepsilon_{k}^{(2)}\omega\varepsilon_{k}^{(3)}u_{k}^{(1)}\varepsilon_{k}^{(4)}\omega\varepsilon_{k}^{(5)}u_{k}^{(2)}\varepsilon_{k}^{(6)}\cdots\omega\varepsilon_{k}^{(2l_{k}+1)}u_{k}^{(l_{k})}\varepsilon_{k}^{(2l_{k}+2)}v_{k}\in\Sigma^{\ast}$.

According to the above rules, for every
$u_{k}^{(1)},\cdots,u_{k}^{(l_{k})}\in\Sigma^{d_{k-1}-1-2M}$, the choice of
$\varepsilon_{k}^{(1)},\cdots,\varepsilon_{k}^{(2l_{k}+2)}\in\Sigma^{M}$ is
determined. Put

$\displaystyle\mathcal{W}_{k}=\left\\{\omega\varepsilon_{k}^{(1)}0^{d_{k-1}-1-3M}\varepsilon_{k}^{(2)}\omega\varepsilon_{k}^{(3)}u_{k}^{(1)}\varepsilon_{k}^{(4)}\omega\varepsilon_{k}^{(5)}u_{k}^{(2)}\varepsilon_{k}^{(6)}\cdots\omega\varepsilon_{k}^{(2l_{k}+1)}u_{k}^{(l_{k})}\varepsilon_{k}^{(2l_{k}+2)}v_{k}\in\Sigma^{\ast}:\right.$
$\displaystyle\left.u_{k}^{(i)}\in\Sigma^{d_{k-1}-1-2M},\forall\ 1\leq i\leq
l_{k}\right\\}.$

Note that every element of $\mathcal{W}_{k}$ is of length $n_{k}-n_{k-1}-M$.
Furthermore, since $\Sigma$ has specification, for any $k\geq 1$ and any
$u_{i}\in\mathcal{W}_{i}(1\leq i\leq k)$, there exists
$\varepsilon_{1},\cdots,\varepsilon_{k}\in\Sigma^{M}$, such that
$0^{n_{0}}\varepsilon_{1}u_{1}\varepsilon_{2}u_{2}\cdots\varepsilon_{k}u_{k}\in\Sigma^{\ast}$.
Here we choose $\varepsilon_{1},\cdots,\varepsilon_{k}\in\Sigma^{M}$, which
satisfies for any $1\leq i\leq k$, $\varepsilon_{i}$ is the minimal word (in
the sense of lexicographical order) such that
$0^{n_{0}}\varepsilon_{1}u_{1}\cdots\varepsilon_{i}u_{i}\in\Sigma^{\ast}$. Let

$G_{k}=\\{0^{n_{0}}\varepsilon_{1}u_{1}\varepsilon_{2}u_{2}\cdots\varepsilon_{k}u_{k}:u_{i}\in\mathcal{W}_{i},\forall\
1\leq i\leq k\\}.$

Note that every element of $G_{k}$ is of length $n_{k}$. Define

$E_{0}=I_{n_{0}}(0^{n_{0}})$

and

$E_{k}=\bigcup_{w\in G_{k}}I_{n_{k}}(w),\quad\forall\ k\geq 1.$

The desired Cantor-type subset of $\mathcal{U}_{\psi}^{\ast}(a,b)$ is defined
as

$\mathcal{U}_{P}^{\ast}(a,b):=\bigcap_{k=0}^{\infty}E_{k}.$

Now, we show that $\mathcal{U}_{P}^{\ast}(a,b)$ is the subset of
$\mathcal{U}_{\psi}^{\ast}(a,b)$. For any
$\omega\in\mathcal{U}_{P}^{\ast}(a,b)$, since the elements of
$\mathcal{U}_{P}^{\ast}(a,b)$ are of the form
$0^{n_{0}}\varepsilon_{1}u_{1}\cdots\varepsilon_{k}u_{k}\cdots$ with
$u_{k}\in\mathcal{W}_{k}$ $(k\geq 1)$ and $\varepsilon_{k}\in\Sigma^{M}$ is
the minimal word (in the sense of lexicographical order) such that
$0^{n_{0}}\varepsilon_{1}u_{1}\cdots\varepsilon_{k}u_{k}\in\Sigma^{\ast}$, by
the definition of $\mathcal{W}_{k}$, we have the following two facts:

  1. (i)

If $n_{k}<N\leq n_{k}+2M$, then

$d_{k-1}-1-3M\leq L_{N}(\omega)\leq\max\\{d_{k-1}-1+F+M,d_{k}-1-M\\}.$

  2. (ii)

If $n_{k}+2M<N\leq n_{k+1}$, then

$d_{k}-1-3M\leq L_{N}(\omega)\leq d_{k}-1+F+M.$

Combining these and $\eqref{61}$, we obtain that

$\liminf_{N\to\infty}\frac{L_{N}(\omega)}{\Phi(N)}\geq\lim_{k\to\infty}\frac{d_{k-1}}{\Phi(n_{k})}=a$

and

$\limsup_{N\to\infty}\frac{L_{N}(\omega)}{\Phi(N)}\leq\lim_{k\to\infty}\frac{d_{k}}{\Phi(n_{k})}=b.$

Note that

$d_{k-1}-1-3M\leq L_{n_{k}+1}(\omega)\leq d_{k-1}-1+F+M,$

thus

$\lim_{k\to\infty}\frac{L_{n_{k}+1}(\omega)}{\Phi(n_{k}+1)}=\lim_{k\to\infty}\frac{d_{k-1}}{\Phi(n_{k})}=a.$

Furthermore,

$\lim_{k\to\infty}\frac{L_{n_{k}+2M+1}(\omega)}{\Phi(n_{k}+2M+1)}=\lim_{k\to\infty}\frac{d_{k}}{\Phi(n_{k})}=b.$

Therefore, we have
$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{\Phi(N)}=a$ and
$\limsup\limits_{N\to\infty}\frac{L_{N}(\omega)}{\Phi(N)}=b$. Thus,
$\mathcal{U}_{P}^{\ast}(a,b)\subset\mathcal{U}_{\psi}^{\ast}(a,b)$. Similar
with the proof of Lemma 5.1, we assign a mass distribution on
$\mathcal{U}_{P}^{\ast}(a,b)$. We first assign a mass distribution on all
cylinders. Let $\lambda(I_{k}(0^{k}))=1$ for any $1\leq k\leq n_{0}$. For any
$I_{n_{1}}\in E_{1}$, let

$\lambda(I_{n_{1}})=\frac{1}{\sharp\mathcal{W}_{1}}\lambda(I_{n_{0}}(0^{n_{0}}))=\frac{1}{\sharp\mathcal{W}_{1}}.$

Assume that the measure on $I_{n_{k}}\in E_{k}$ $(k\geq 1)$ is well defined,
then the measure on $I_{n_{k+1}}\in E_{k+1}$ is defined as

$\lambda(I_{n_{k+1}})=\frac{1}{\sharp\mathcal{W}_{k+1}}\lambda(I_{n_{k}})=\prod_{j=1}^{k+1}\frac{1}{\sharp\mathcal{W}_{j}}.$

For any $n_{k}<N<n_{k+1}$, let

$\lambda(I_{N})=\sum_{I_{n_{k+1}}\in E_{k+1}:I_{N}\cap
I_{n_{k+1}}\neq\emptyset}\lambda(I_{n_{k+1}}).$

Note that

$\lambda(I_{N})=\sum_{I_{N+1}:I_{N+1}\cap I_{N}\neq\emptyset}\lambda(I_{N+1})$

for all $N\geq 1$, thus, $\lambda$ can be extended to a Borel probability
measure on $\mathcal{U}_{P}^{\ast}(a,b)$. Lastly we estimate the Hausdorff
dimension of $\mathcal{U}_{P}^{\ast}(a,b)$. We show that

$\dim_{\rm
H}\mathcal{U}_{P}^{\ast}(a,b)\geq\left(1-\frac{2}{P}\right)\dim_{\rm
H}\Sigma.$ (6.3)

Then we have $\dim_{\rm
H}\mathcal{U}_{\psi}^{\ast}(a,b)\geq(1-\frac{2}{P})\dim_{\rm H}\Sigma$ for
every $P\geq 3$. Letting $P\to\infty$, we obtain that $\dim_{\rm
H}\mathcal{U}_{\psi}^{\ast}(a,b)=\dim_{\rm H}\Sigma$. To prove $\eqref{63}$,
in view of Proposition 5.4, we only need to prove that for any
$\omega\in\mathcal{U}_{P}^{\ast}(a,b)$,

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\left(1-\frac{2}{P}\right)\dim_{\rm
H}\Sigma,$

where $I_{N}(\omega)$ is the $N$-th cylinder containing $\omega$. By the
definition of $\lambda$, we consider two cases.
Case i@: $n_{k}\leq N\leq n_{k}+d_{k}$. Then

$\lambda(I_{N}(\omega))=\lambda(I_{n_{k}}(\omega))=\prod_{j=1}^{k}\frac{1}{\sharp\mathcal{W}_{j}}.$

By Proposition 5.5, we have

$\displaystyle\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}$
$\displaystyle=\liminf_{N\to\infty}\frac{-\log\lambda(I_{N}(\omega))}{N\log
m}\geq\frac{1}{\log
m}\liminf_{k\to\infty}\frac{\sum_{j=1}^{k}\log\sharp\mathcal{W}_{j}}{n_{k}+d_{k}}$
$\displaystyle=\frac{1}{\log
m}\liminf_{k\to\infty}\frac{\sum_{j=1}^{k}\log\sharp\mathcal{W}_{j}}{n_{k}}\geq\frac{1}{\log
m}\liminf_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k+1}}{n_{k+1}-n_{k}}.$

Since
$\sharp\mathcal{W}_{k+1}=\left(\sharp\Sigma^{d_{k}-2M-1}\right)^{l_{k+1}}$ and
$B=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$, we obtain that

$\lim_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k+1}}{l_{k+1}d_{k}}=B.$ (6.4)

Thus,

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\frac{h(\Sigma)}{\log
m}\liminf_{k\to\infty}\frac{l_{k+1}d_{k}}{n_{k+1}-n_{k}}.$

In view of

$n_{k+1}-n_{k}=d_{k}(l_{k+1}+1)+r_{k+1}<d_{k}(l_{k+1}+2)$

and

$l_{k+1}:=\left\lfloor\frac{n_{k+1}-n_{k}}{d_{k}}\right\rfloor-1\geq P-2,$

we obtain that

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\frac{h(\Sigma)}{\log
m}\cdot\liminf_{k\to\infty}\frac{l_{k+1}}{l_{k+1}+2}\geq\left(1-\frac{2}{P}\right)\frac{h(\Sigma)}{\log
m}.$

Case ii@: $n_{k}+d_{k}<N<n_{k+1}$. Since
$n_{k+1}=n_{k}+(l_{k+1}+1)d_{k}+r_{k+1}<n_{k}+(l_{k+1}+2)d_{k}$, there exists
$1\leq l\leq l_{k+1}+1$, such that $n_{k}+ld_{k}\leq N<n_{k}+(l+1)d_{k}$.
Therefore,

$\lambda(I_{N}(\omega))\leq\lambda(I_{n_{k}+ld_{k}}(\omega))=\left(\sharp\Sigma^{d_{k}-1-2M}\right)^{-(l-1)}\lambda(I_{n_{k}}(\omega)).$

By Proposition 5.5, we have

$\displaystyle\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}$
$\displaystyle\geq\liminf_{k\to\infty}\frac{-\log\lambda(I_{n_{k}}(\omega))+(l-1)\log\sharp\Sigma^{d_{k}-1-2M}}{N\log
m}$ $\displaystyle\geq\frac{1}{\log
m}\liminf_{k\to\infty}\frac{-\log\lambda(I_{n_{k}}(\omega))+(l-1)\log\sharp\Sigma^{d_{k}-1-2M}}{n_{k}+(l+1)d_{k}}$
$\displaystyle=\frac{1}{\log
m}\liminf_{k\to\infty}\frac{-\log\lambda(I_{n_{k}}(\omega))+(l+1)\log\sharp\Sigma^{d_{k}-1-2M}}{n_{k}+(l+1)d_{k}}.$

The above equality follows from
$\lim\limits_{k\to\infty}\frac{\log\sharp\Sigma^{d_{k}-1-2M}}{d_{k}}\leq\log
m$ and $\lim\limits_{k\to\infty}\frac{d_{k}}{n_{k}}=0$. Since
$B=\lim\limits_{n\to\infty}\frac{\log\sharp\Sigma^{n}}{n}$ and
$\lim\limits_{k\to\infty}d_{k}=+\infty$, for any $\delta\in(0,B)$, we have
$\log\sharp\Sigma^{d_{k}-1-2M}>(B-\delta)d_{k}$ for all $k$ large enough.
Thus,

$\displaystyle\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}$
$\displaystyle\geq\frac{1}{\log
m}\liminf_{k\to\infty}\frac{-\log\lambda(I_{n_{k}}(\omega))+(B-\delta)(l+1)d_{k}}{n_{k}+(l+1)d_{k}}$
(6.5) $\displaystyle=\frac{B-\delta}{\log
m}\liminf_{k\to\infty}\frac{(B-\delta)^{-1}(-\log\lambda(I_{n_{k}}(\omega)))+(l+1)d_{k}}{n_{k}+(l+1)d_{k}}$
$\displaystyle\geq\frac{B-\delta}{\log
m}\liminf_{k\to\infty}\frac{(B+\delta)^{-1}(-\log\lambda(I_{n_{k}}(\omega)))+(l+1)d_{k}}{n_{k}+(l+1)d_{k}}.$

Note that

$\displaystyle\limsup_{k\to\infty}\frac{-\log\lambda(I_{n_{k}}(\omega))}{n_{k}}=\limsup_{k\to\infty}\frac{\sum_{j=1}^{k}\log\sharp\mathcal{W}_{j}}{n_{k}}$
$\displaystyle\leq\limsup_{k\to\infty}\frac{\log\sharp\mathcal{W}_{k+1}}{n_{k+1}-n_{k}}$
$\displaystyle=\limsup_{k\to\infty}\frac{Bl_{k+1}d_{k}}{n_{k+1}-n_{k}}\leq B,$

the second equality is due to $\eqref{64}$ and the last inequality is due to
$n_{k+1}-n_{k}\geq(l_{k+1}+1)d_{k}$. Therefore, for all $k$ large enough, we
have

$(B+\delta)^{-1}(-\log\lambda(I_{n_{k}}(\omega)))<n_{k}.$ (6.6)

Combining $\eqref{65}$, $\eqref{66}$ and the inequality
$\frac{y+s}{z+s}\geq\frac{y}{z},0\leq y<z,s\geq 0$, we obtain that

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\frac{B-\delta}{\log
m}\liminf_{k\to\infty}\frac{(B+\delta)^{-1}(-\log\lambda(I_{n_{k}}(\omega)))}{n_{k}}.$

Furthermore, by the proof of Case i@, we know that

$\liminf_{k\to\infty}\frac{-\log\lambda(I_{n_{k}}(\omega))}{n_{k}}\geq\left(1-\frac{2}{P}\right)h(\Sigma).$

Therefore,

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\frac{B-\delta}{B+\delta}\left(1-\frac{2}{P}\right)\frac{h(\Sigma)}{\log
m}.$

Letting $\delta\to 0$, we obtain that

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\left(1-\frac{2}{P}\right)\frac{h(\Sigma)}{\log
m}.$

The combination of Case i@ and Case ii@ gives

$\liminf_{N\to\infty}\frac{\log\lambda(I_{N}(\omega))}{\log|I_{N}(\omega)|}\geq\left(1-\frac{2}{P}\right)\frac{h(\Sigma)}{\log
m}=\left(1-\frac{2}{P}\right)\dim_{\rm H}\Sigma.$

The equality follows from $\dim_{\rm H}\Sigma=\frac{h(\Sigma)}{\log m}$ (see
[12, Proposition iii@.1]). Similar arguments apply to the remaining cases. We
only give the constructions for the sequences $\\{n_{k}\\}_{k\geq 0}$ and
$\\{d_{k}\\}_{k\geq 0}$.
Case 3: $0<a<b=\infty$. Fix a integer $P\geq 3$. We choose $n_{0}$ such that

$\left\lfloor a\Phi(n_{0})\log\frac{n_{0}}{\Phi(n_{0})}\right\rfloor\geq
P+2+2M$

and then take $d_{0}=\lfloor
a\Phi(n_{0})\log\frac{n_{0}}{\Phi(n_{0})}\rfloor$, for any $k\geq 1$,
$\\{n_{k}\\}$ and $\\{d_{k}\\}$ are recursively defined as

$n_{k}:=\left\lfloor\Phi^{-1}\left(\Phi(n_{k-1})\log\frac{n_{k-1}}{\Phi(n_{k-1})}\right)\right\rfloor+Pd_{k-1}\
{\rm and}\ d_{k}:=\left\lfloor
a\Phi(n_{k})\log\frac{n_{k}}{\Phi(n_{k})}\right\rfloor.$

Case 4: $0=a<b<\infty$. Fix a integer $P\geq 3$. We choose $n_{0}$ such that

$\lfloor b\Phi(n_{0})\rfloor\geq P+2+2M$

and then take $d_{0}=\lfloor b\Phi(n_{0})\rfloor$, for any $k\geq 1$,
$\\{n_{k}\\}$ and $\\{d_{k}\\}$ are recursively defined as

$n_{k}:=\lfloor\Phi^{-1}((k-1)\Phi(n_{k-1}))\rfloor+Pd_{k-1}\ {\rm and}\
d_{k}:=\lfloor b\Phi(n_{k})\rfloor.$

Case 5: $0=a<b=\infty$. Fix a integer $P\geq 3$. We choose $n_{0}$ such that

$\lfloor\sqrt{n_{0}\Phi(n_{0})}\rfloor\geq P+2+2M$

and then take $d_{0}=\lfloor\sqrt{n_{0}\Phi(n_{0})}\rfloor$, for any $k\geq
1$, $\\{n_{k}\\}$ and $\\{d_{k}\\}$ are recursively defined as

$n_{k}:=\lfloor\Phi^{-1}(n_{k-1}\Phi(n_{k-1}))\rfloor+Pd_{k-1}\ {\rm and}\
d_{k}:=\lfloor\sqrt{n_{k}\Phi(n_{k})}\rfloor.$

Case 6: $a=b=0$. Fix a integer $P\geq 3$. We choose $n_{0}$ such that

$\lfloor\sqrt{\Phi(n_{0})}\rfloor\geq P+2+2M$

and then take $d_{0}=\lfloor\sqrt{\Phi(n_{0})}\rfloor$, for any $k\geq 1$,
$\\{n_{k}\\}$ and $\\{d_{k}\\}$ are recursively defined as

$n_{k}:=n_{k-1}+Pd_{k-1}+1\ {\rm and}\
d_{k}:=\lfloor\sqrt{\Phi(n_{k})}\rfloor.$

∎

###  6.2 Proof of Theorem 1.7 Cases i and ii

######  Proof.

Case i Since $0<\tau<+\infty$, we have

$\limsup\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=\tau^{-1}\overline{L}(\omega)\
{\rm and}\
\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}=\tau^{-1}\underline{L}(\omega).$
(6.7)

Thus,

$\mathcal{U}_{\psi}(a,b)=\mathcal{U}_{\psi_{0}}(\tau a,\tau b)$ (6.8)

and

$\mathcal{U}_{\psi}^{\ast}(a,b)=\mathcal{U}_{\psi_{0}}^{\ast}(\tau a,\tau b).$
(6.9)

Recall that $\psi_{0}(N)=m^{-N}$. In view of $\eqref{612}$, $\eqref{68}$,
$\eqref{69}$ and Lemma 5.1, we immediately obtain (2) of Theorem 1.7 Case i.
The remaining case is $a\geq\tau^{-1}$. If $b<+\infty$, the upshot of
$\eqref{41}$, $\eqref{68}$ and Proposition 5.2 is that
$\mathcal{U}_{\psi}(a,b)=\emptyset$. Therefore,
$\mathcal{U}_{\psi}^{\ast}(a,b)=\mathcal{U}_{\psi}(a,b)=\emptyset$.
If $b=+\infty$, in view of $\eqref{68}$, we have
$\mathcal{U}_{\psi}(a,b)\subset\\{\omega\in\Sigma:\overline{L}(\omega)=+\infty\\}$.
Since $\dim_{\rm H}(\\{\omega\in\Sigma:\overline{L}(\omega)=+\infty\\})=0$, we
obtain that $\dim_{\rm H}\mathcal{U}_{\psi}(a,b)=0$. Therefore, $\dim_{\rm
H}\mathcal{U}_{\psi}^{\ast}(a,b)=\dim_{\rm H}\mathcal{U}_{\psi}(a,b)=0$.
What is more, by Proposition 5.2, we know that $\underline{L}(\omega)>1$ is
equivalent to $\underline{L}(\omega)=+\infty$. Combining this and
$\eqref{69}$, we have $\mathcal{U}_{\psi}^{\ast}(a,b)=\emptyset$ when
$\tau^{-1}<a<+\infty$. Lastly, if $a>\tau^{-1}$ and $b=+\infty$, in view of
$\eqref{68}$ and Proposition 5.2, we obtain that
$\mathcal{U}_{\psi}(a,b)=\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\})$.
Case ii Since $\tau=+\infty$, we have

$\limsup\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}>0\ {\rm
implies}\ \overline{L}(\omega)=+\infty.$ (6.10)

and

$\liminf\limits_{N\to\infty}\frac{L_{N}(\omega)}{-\log_{m}\psi(N)}>0\ {\rm
implies}\ \underline{L}(\omega)=+\infty.$ (6.11)

If $b>0$, it follows from $\eqref{610}$ that
$\mathcal{U}_{\psi}(a,b)\subset\\{\omega\in\Sigma:\overline{L}(\omega)=+\infty\\}$.
Since $\dim_{\rm H}(\\{\omega\in\Sigma:\overline{L}(\omega)=+\infty\\})=0$, we
obtain that $\dim_{\rm H}\mathcal{U}_{\psi}(a,b)=0$. Hence,

$\dim_{\rm H}\mathcal{U}_{\psi}^{\ast}(a,b)=\dim_{\rm
H}\mathcal{U}_{\psi}(a,b)=0.$

The remaining case is $b=0$, more precisely $a=b=0$ (we assume that $0\leq
a\leq b$). Note that
$\mathcal{U}_{\psi_{0}}^{\ast}(a^{\prime},b^{\prime})\subset\mathcal{U}^{\ast}_{\psi}(0,0)$
for all $0\leq a^{\prime}<1$ and $\frac{a^{\prime}}{1-a^{\prime}}\leq
b^{\prime}<+\infty$, it follows from Proposition 5.1 that $\dim_{\rm
H}\mathcal{U}^{\ast}_{\psi}(0,0)\geq\frac{b^{\prime}-b^{\prime}a^{\prime}-a^{\prime}}{(1+b^{\prime})(b^{\prime}-a^{\prime})}\dim_{\rm
H}\Sigma$ for all $0\leq a^{\prime}<1$ and
$\frac{a^{\prime}}{1-a^{\prime}}\leq b^{\prime}<+\infty$. Thus, $\dim_{\rm
H}\mathcal{U}^{\ast}_{\psi}(0,0)=\dim_{\rm H}\Sigma.$ Furthermore, when $a>0$,
it follows from $\eqref{611}$ and Proposition 5.2 that

$\mathcal{U}_{\psi}(a,b)\subset\bigcup_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\}).$

∎

###  6.3 Proof of Corollary 1.9

######  Proof.

Recall that $\psi_{0}(N)=m^{-N}$, thus
$\tau=\lim\limits_{N\to\infty}\frac{-\log_{m}\psi_{0}(N)}{N}=1$.
(i) If $a>1$, by Proposition 5.2, we have
$\mathcal{U}_{\psi_{0}}(a)=\bigcup\limits_{n=0}^{\infty}\sigma^{-n}(\\{0^{\infty}\\})$.
(ii) Since $\mathcal{U}_{\psi_{0}}(0)=\Sigma$, we only need to consider
$0<a\leq 1$. If $0<a<1$, it follows from Lemma 5.1 (ii) that

$\dim_{\rm H}\mathcal{U}_{\psi_{0}}(a)\geq\frac{b(1-a)-a}{(1+b)(b-a)}\dim_{\rm
H}\Sigma,\ \forall\ b\geq\frac{a}{1-a}.$

An easy calculation shows that the maximum of the right hand side of above
inequality is attained for $b=\frac{2a}{1-a}$. Thus, $\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a)\geq\left(\frac{1-a}{1+a}\right)^{2}\dim_{\rm
H}\Sigma$. On the other hand, for any $\varrho>0$, in view of Lemma 5.1 (i)
and the denseness of the set of rational numbers, we have that

$\displaystyle\quad\mathcal{U}_{\psi_{0}}(a)\cap\\{\omega\in\Sigma:\overline{L}(\omega)<+\infty\\}$
$\displaystyle\subset\mathcal{U}_{\psi_{0}}(a,\frac{a}{1-a})\cup\bigcup_{b\in[\frac{a}{1-a},+\infty)\cap\mathbb{Q}}\\{\omega\in\Sigma:\underline{L}(\omega)\geq
a,b\leq\overline{L}(\omega)\leq b+\varrho\\}.$

By Lemma 5.1 (ii) and the countable stability of Hausdorff dimension, we
obtain that

$\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a)\leq\sup_{b\in[\frac{a}{1-a},+\infty)}\left(\frac{b(1-a)-a}{(1+b)(b-a)}+\frac{\varrho(1-a)^{2}}{a^{3}}\right)\dim_{\rm
H}\Sigma.$

Similar to the above proof, the maximum of the right hand side of above
inequality is attained for $b=\frac{2a}{1-a}$. Therefore, $\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a)\leq\left(\frac{1-a}{1+a}\right)^{2}\dim_{\rm
H}\Sigma$. Therefore,

$\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(a)=\left(\frac{1-a}{1+a}\right)^{2}\dim_{\rm
H}\Sigma.$

The remaining case is $a=1$, it follows from Propositions 5.2 and 5.3 that
$\mathcal{U}_{\psi_{0}}(1)=\mathcal{U}_{\psi_{0}}(1,+\infty)$. An easy cover
argument shows that $\dim_{\rm
H}(\\{\omega\in\Sigma:\overline{L}(\omega)=+\infty\\})=0$, hence $\dim_{\rm
H}\mathcal{U}_{\psi_{0}}(1)=0$. ∎

### Acknowledgements

It was supported by NSFC12271176 and Guangdong Natural Science Foundation
2023A1515010691.

## References

  * [1] B. Bárány and M. Rams, _Shrinking targets on Bedford-McMullen carpets_ , Proc. Lond. Math. Soc. (3), 117 (2018), no.5, 951-995.
  * [2] R. A. Barrera, _Topological and symbolic dynamics of the doubling map with a hole_ , PhD thesis University of Manchester, 2014.
  * [3] A. Bertrand-Mathis, _Développement en base $\theta$; répartition modulo un de la suite $(x\theta^{n})_{n\geq 0}$; langages codés et $\theta$-shift_, Bull. Soc. Math. France, 114 (1986), no.3, 271-323.
  * [4] R. Bowen, _Some systems with unique equilibrium states_ , Math. Syst. Theory, 8 (1974), no.3, 193-202.
  * [5] R. Bowen, _Equilibrium states and the ergodic theory of Anosov diffeomorphisms_ , LNM 470, Berlin: Springer, 1975.
  * [6] Y. Bugeaud and L. Liao, _Uniform Diophantine approximation related to $b$-ary and $\beta$-expansions_, Ergodic Theory Dynam. Systems, 36 (2016), no.1, 1-22.
  * [7] N. Chernov and D. Kleinbock, _Dynamical Borel-Cantelli lemmas for Gibbs measures_ , Israel J. Math., 122 (2001), 1-27.
  * [8] K. Falconer, _Techniques in fractal geometry_ , John Wiley & Sons, Ltd., Chichester, 1997.
  * [9] A. Fan and Y. Jiang, _On Ruelle-Perron-Frobenius operators<EMAIL_ADDRESS>Ruelle Theorem_ , Comm. Math. Phys., 223 (2001), no.1, 125-141.
  * [10] L. Fang, M. Wu and B. Li, _Approximation orders of real numbers by $\beta$-expansions_, Math. Z., 296 (2020), no.1-2, 13-40.
  * [11] J. L. Fernández, M. V. Melián and D. Pestana, _Quantitative mixing results and inner functions_ , Math. Ann., 337 (2007), no.1, 233-251.
  * [12] H. Furstenberg, _Disjointness in ergodic theory, minimal sets, and a problem in Diophantine approximation_ , Math. Systems Theory, 1 (1967), 1-49.
  * [13] C. Ganotaki and T. Persson, _On eventually always hitting points_ , Monatsh. Math., 196 (2021), no.4, 763-784.
  * [14] G. Harman, _Metric number theory_ , London Mathematical Society Monographs. New Series, 18. The Clarendon Press, Oxford University Press, New York, 1998.
  * [15] R. Hill and S. L. Velani, _The ergodic theory of shrinking targets_ , Invent. Math., 119 (1995), 175-198.
  * [16] R. Hill and S. L. Velani, _Metric Diophantine approximation in Julia sets of expanding rational maps_ , Inst. Hautes Études Sci. Publ. Math., (1997), no.85, 193-216.
  * [17] R. Hill and S. L. Velani, _The shrinking target problem for matrix transformations of tori_ , Proc. Lond. Math. Soc., 60 (1999), 381-398.
  * [18] R. Hill and S. L. Velani, _A zreo-infinity law for well-approximable points in Julia sets_ , Ergodic Theory Dynam. Systems, 22 (2002), no.6, 1773-1782.
  * [19] D. Kelmer, _Shrinking targets for discrete time flows on hyperbolic manifolds_ , Geom. Funct. Anal., 27 (2017), no.5, 1257-1287.
  * [20] D. Kelmer and H. Oh, _Shrinking targets for the geodesic flow on geometrically finite hyperbolic manifolds_ , J. Mod. Dyn., 17 (2021), 401-434.
  * [21] D. Kelmer and S. Yu, _Shrinking targets problems for flows on homogeneous spaces_ , Trans. Am. Math. Soc., 372 (2019), no.9, 6283-6314.
  * [22] M. Kirsebom, P. Kunde and T. Persson, _Shrinking targets and eventually always hitting points for interval maps_ , Nonlinearity, 33 (2020), no.2, 892-914.
  * [23] D. Kleinbock, I. Konstantoulas and F. K. Richter, _Zero-one laws for eventually always hitting points in mixing systems_ , (2019), arXiv:1904.08584.
  * [24] D. Kleinbock and N. Wadleigh, _An inhomogeneous Dirichlet Theorem via shrinking targets_ , Compos. Math., 155 (2019), no.7, 1402-1423.
  * [25] H. Koivusalo and F. A. Ramírez, _Recurrence to shrinking targets on typical self-affine fractals_ , Proc. Edinb. Math. Soc. (2), 61 (2018), no.2, 387-400.
  * [26] B. Li, L. Liao, S. L. Velani and E. Zorin, _The shrinking target peoblem for matrix transformations of tori: revisiting the standard problem_ , Adv. Math., 421 (2023).
  * [27] B. Li, B.-W. Wang, J. Wu and J. Xu, _The shrinking target peoblem in the dynamical system of continued fractions_ , Proc. Lond. Math. Soc., 108 (2014), no.1, 159-186.
  * [28] B. Li, T. Persson, B.-W. Wang and J. Wu, _Diophantine approximation of the orbit of 1 in the dynamical system of beta expansions_ , Math. Z., 276 (2014), no.3-4, 799-827.
  * [29] B. Li and J. Wu, _Beta-expansion and continued fraction expansion_ , J. Math. Anal. Appl., 339 (2008), no.2, 1322-1331.
  * [30] W. Parry and M. Pollicott, _Zeta functions and the periodic orbits structure of hyperbolic dynamics_ , Astérisque (1990).
  * [31] W. Philipp, _Some metrical theorems in number theory_ , Pacific J. Math., 20 (1967), 109-127.
  * [32] L.-M. Shen and B.-W. Wang, _Shrinking target problems for beta-dynamical system_ , Sci. China Math., 56 (2013), no.1, 91-104.
  * [33] M. Urbański, _Diophantine analysis of conformal iterated function systems_ , Monatsh. Math., 137 (2002), no.4, 325-340.
  * [34] P. Walters, _An Introduction to Ergodic Theory_ , Graduate Texts in Mathematics, vol. 79, Springer-Verlag, New York-Berlin, 1982.
  * [35] W. Wu, _Uniform Diophantine approximation approximation related to beta-transformation_ , (2020), arXiv:2005.14538.
  * [36] W. Wu and L. Zheng, _Dimension theory of uniform Diophantine approximation related to $\beta$-transformations_, Acta Math. Sci. Ser. A (Chinese Ed.), 42 (2022), no.4, 978-1002.