# A Hausdorff dimension analysis of sets with the product of consecutive vs
single partial quotients in continued fractions

Mumtaz Hussain  Mumtaz Hussain, Department of Mathematical and Physical
Sciences, La Trobe University, Bendigo 3552, Australia.
<EMAIL_ADDRESS>,  Bixuan Li  Bixuan Li, School of Mathematics and
Statistics, Huazhong University of Science and Technology, Wuhan 430074, P. R.
China<EMAIL_ADDRESS>and  Nikita Shulga  Nikita Shulga, Department of
Mathematical and Physical Sciences, La Trobe University, Bendigo 3552,
Australia<EMAIL_ADDRESS>
###### Abstract.

We present a detailed Hausdorff dimension analysis of the set of real numbers
where the product of consecutive partial quotients in their continued fraction
expansion grow at a certain rate but the growth of the single partial quotient
is at a different rate. We consider the set

$\mathcal{F}(\Phi_{1},\Phi_{2})\stackrel{{\scriptstyle\text{def}}}{{=}}\mathcal{E}(\Phi_{1})\backslash\mathcal{E}(\Phi_{2})=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq\Phi_{1}(n)\text{\,\,
for infinitely many }n\in\mathbb{N}\\\ a_{n+1}(x)&<\Phi_{2}(n)\text{\,\, for
all sufficiently large }n\in\mathbb{N}\end{split}\right\\},$

where $\Phi_{i}:\mathbb{N}\to(0,\infty)$ are any functions such that
$\lim\limits_{n\to\infty}\Phi_{i}(n)=\infty$. We obtain some surprising
results including the situations when $\mathcal{F}(\Phi_{1},\Phi_{2})$ is
empty for various non-trivial choices of $\Phi_{i}$’s. Our results contribute
to the metrical theory of continued fractions by generalising several known
results including the main result of [Nonlinearity, 33(6):2615–2639, 2020]. To
obtain some of the results, we consider an alternate generalised set, which
may be of independent interest, and calculate its Hausdorff dimension. One of
the main ingredients is in the usage of the classical mass distribution
principle; specifically a careful distribution of the mass on the Cantor
subset by introducing a new idea of two different types of probability
measures.

##  1\.  Introduction

It is well-known that every irrational number $x\in(0,1)$ has a unique
infinite continued fraction expansion. This expansion can be induced by the
Gauss map $T:[0,1)\to[0,1)$ defined by

$T(0)=0,~{}T(x)=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor\textmd{ for
}x\in(0,1),$

where $\lfloor x\rfloor$ denotes the integer part of $x$. We write
$x:=[a_{1}(x),a_{2}(x),a_{3}(x),\ldots]$ for the continued fraction of $x$
where $a_{1}(x)=\lfloor 1/x\rfloor$, $a_{n}(x)=a_{1}(T^{n-1}(x))$ for $n\geq
2$ are called the partial quotients of $x$.

The metric theory of continued fractions concerns the quantitative study of
properties of partial quotients for almost all $x\in(0,1)$. This area of
research is closely connected with metric Diophantine approximation, for
example, fundamental theorems in this field by Khintchine (1924) and Jarník
(1931) can be represented in terms of the growth of partial quotients. The
classical Borel-Bernstein theorem (1912) states that the Lebesgue measure of
the set

$\mathcal{E}_{1}(\Phi):=\left\\{x\in[0,1):a_{n}(x)\geq\Phi(n)\ {\rm for\
infinitely\ many}\ n\in\mathbb{N}\right\\}$

is either zero or one depending upon the convergence or divergence of the
series $\sum_{n=1}^{\infty}\Phi(n)^{-1}$ respectively. Here and throughout
$\Phi:\mathbb{N}\to[1,\infty)$ will be an arbitrary function such that
$\Phi(n)\to\infty$ as $n\to\infty$. For rapidly increasing functions $\Phi$,
the Borel-Bernstein (1911, 1912) theorem gives no further information than the
zero Lebesgue measure of $\mathcal{E}_{1}(\Phi)$. To distinguish between the
sets of zero Lebesgue measure, the Hausdorff dimension is an appropriate tool.
When the function $\Phi(n)$ tends to infinity at a polynomial rate $n^{a}$,
the Hausdorff dimension of $\mathcal{E}_{1}(\Phi)$ was calculated by Good [7],
and when $\Phi(n)$ tends to infinity at a super-exponential rate $a^{b^{n}},$
the Hausdorff dimension was established by Łuczak [16]. For an arbitrary
function $\Phi$ the dimension of $\mathcal{E}_{1}(\Phi)$ was computed by Wang-
Wu [23].

######  Theorem 1.1 ([23, Wang-Wu]).

Let $\Phi:\mathbb{N}\rightarrow\mathbb{[}1,\infty)$. Suppose

$\log B=\liminf\limits_{n\rightarrow\infty}\frac{\log\Phi(n)}{n}\ {and}\ \log
b=\liminf\limits_{n\rightarrow\infty}\frac{\log\log\Phi(n)}{n}.$

  * (i)

When $B=1$, $\dim_{H}\mathcal{E}_{1}(\Phi)=1.$

  * (ii)

When $B=\infty$, $\dim_{H}\mathcal{E}_{1}(\Phi)=1/(1+b).$

  * (iii)

When $1<B<\infty$, $\dim_{H}\mathcal{E}_{1}(\Phi)=s_{B}:=\inf\\{s\geq
0:\mathsf{P}(T,-s(\log B+\log|T^{\prime}|))\leq 0\\},$

where $T^{\prime}$ denotes the derivative of the Gauss map $T$, and
$\mathsf{P}$ represents the pressure function defined in the Subection 2.3.

Recently, Kleinbock-Wadleigh [13] observed that the growth of the product of
consecutive partial quotients gives a characterisation for a real number to be
$\Phi$-Dirichlet improvable or not. In fact, in [13] the Lebesgue measure of
the set of $\Phi$-Dirichlet non-improvable numbers

$\mathcal{E}_{2}(\Phi):=\left\\{x\in[0,1):a_{n}(x)a_{n+1}(x)\geq\Phi(n)\ {\rm
for\ infinitely\ many}\ n\in\mathbb{N}\right\\}$

was established. Soon after that the Hausdorff dimension of this set (in a
generalised form) was established in [9, 10].

######  Theorem 1.2 ([9, Huang-Wu-Xu]).

Let $\Phi:\mathbb{N}\rightarrow\mathbb{[}1,\infty)$. Suppose

$\log B=\liminf\limits_{n\rightarrow\infty}\frac{\log\Phi(n)}{n}\ {and}\ \log
b=\liminf\limits_{n\rightarrow\infty}\frac{\log\log\Phi(n)}{n}.$

  * (i)

When $B=1$, $\dim_{H}\mathcal{E}_{2}(\Phi)=1.$

  * (ii)

When $B=\infty$, $\dim_{H}\mathcal{E}_{2}(\Phi)=1/(1+b).$

  * (iii)

When $1<B<\infty$, $\dim_{H}\mathcal{E}_{2}(\Phi)=s_{0}:=\inf\\{s\geq
0:\mathsf{P}(T,-s^{2}\log B-s\log|T^{\prime}|))\leq 0\\}.$

Note that the set $\mathcal{E}_{1}(\Phi)$ is properly contained in
$\mathcal{E}_{2}(\Phi)$ raising a natural question of the size of the set

$\mathcal{F}(\Phi):=~{}\mathcal{E}_{2}(\Phi)\setminus\mathcal{E}_{1}(\Phi)=\left\\{x\in[0,1):\begin{array}[]{r}a_{n+1}(x)a_{n}(x)\geq\Phi(n)\text{
for infinitely many }n\in\mathbb{N}\text{ and}\\\ a_{n+1}(x)<\Phi(n)\text{ for
all sufficiently large }n\in\mathbb{N}\end{array}\right\\}.$

The Hausdorff dimension of $\mathcal{F}(\Phi)$ was established in [1].

######  Theorem 1.3 ([1, Bakhtawar-Bos-Hussain]).

Let $\Phi:\mathbb{N}\rightarrow\mathbb{[}1,\infty)$. Suppose

$\log B=\liminf\limits_{n\rightarrow\infty}\frac{\log\Phi(n)}{n}\ {\rm and}\
\log b=\liminf\limits_{n\rightarrow\infty}\frac{\log\log\Phi(n)}{n}.$

Then

$\dim_{H}\mathcal{F}(\Phi)=\left\\{\begin{array}[]{ll}s_{0}&\mathrm{if}\ \
1<B<\infty;\\\\[12.91663pt] \frac{1}{1+b}&\mathrm{if}\ \
B=\infty,\end{array}\right.$

where $s_{0}$ is from Theorem 1.2.

Note that the Hausdorff dimension proved in Theorem 1.3 is the same as in
Theorem 1.2, that is,
$\dim_{H}\mathcal{F}(\Phi)=\dim_{H}\mathcal{E}_{2}(\Phi)$. The main reason
that the subset $\mathcal{F}(\Phi)$ has the same dimension as the superset
$\mathcal{E}_{2}(\Phi)$, is that the set $\mathcal{E}_{2}(\Phi)$ is much
larger than $\mathcal{E}_{1}(\Phi)$. However, if we consider different
approximation functions then the problem becomes more intricate and
challenging. The complications arises in constructing a suitable Cantor type
subset supporting a probability measure. To overcome this difficulty we
introduce two types of probability measures – an idea that we believe can lend
itself in resolving problems of similar nature. Indeed our results below
highlight situations when $\mathcal{F}(\Phi)$ and $\mathcal{E}_{2}(\Phi)$ have
different dimensions for different $\Phi$. To be precise, define the set

$\mathcal{F}(\Phi_{1},\Phi_{2})\stackrel{{\scriptstyle\text{def}}}{{=}}\mathcal{E}_{2}(\Phi_{1})\backslash\mathcal{E}_{1}(\Phi_{2})=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq\Phi_{1}(n)\text{\,\,
for infinitely many }n\in\mathbb{N}\\\ a_{n+1}(x)&<\Phi_{2}(n)\text{\,\, for
all sufficiently large }n\in\mathbb{N}\end{split}\right\\}.$

We fully characterise the Hausdorff dimension of
$\mathcal{F}(\Phi_{1},\Phi_{2})$ as well as prove some more results regarding
general sets, in some way connected to $\mathcal{F}(\Phi_{1},\Phi_{2})$, which
are also of their own independent interest.

It is worth noting that when $\Phi$ is a function of the denominator of the
$n^{\rm th}$ convergent $q_{n}$, the Hausdorff dimension (and measure) study
is carried out in a sequence of papers [2, 4, 14]. In particular, these papers
give insight into comparing the uniform Diophantine approximation (Dirichlet
non-improvable numbers) to the asymptotic Diophantine approximation (Jarník
set).

###  1.1. Main results

The first result of this paper gives a full description of the Hausdorff
dimension of $\mathcal{F}(\Phi_{1},\Phi_{2})$.

######  Theorem 1.4.

Let $\Phi_{1}$ and $\Phi_{2}$ be positive functions such that

$\displaystyle\log B_{1}=\liminf\limits_{n\to\infty}\frac{\log\Phi_{1}(n)}{n}$
$\displaystyle\text{ \,\,\,\,\,\, and \,\,\,\,\,}\log
b_{1}=\liminf\limits_{n\to\infty}\frac{\log\log\Phi_{1}(n)}{n}.$
$\displaystyle\log B_{2}=\lim\limits_{n\to\infty}\frac{\log\Phi_{2}(n)}{n}$
$\displaystyle\text{ \,\,\,\,\,\, and \,\,\,\,\,}\log
b_{2}=\lim\limits_{n\to\infty}\frac{\log\log\Phi_{2}(n)}{n}.$

Then

$\dim_{H}\mathcal{F}(\Phi_{1},\Phi_{2})=\left\\{\begin{array}[]{ll}s_{0}&{\rm
if}\ \ B_{1}^{s_{0}}\leq B_{2}\ {\rm and}\ B_{1}\ {\rm is\
finite};\\\\[12.91663pt] g_{B_{1},B_{2}}&{\rm if}\ \ B_{1}^{s_{0}}\geq
B_{2}>B_{1}^{1/2};\\\\[12.91663pt] {\diameter}&{\rm if}\ \ B_{1}^{1/2}>B_{2}\
{\rm and}\ B_{2}\ {\rm is\ finite};\\\\[12.91663pt] \frac{1}{1+b_{1}}&{\rm
if}\ \ B_{1}=B_{2}=\infty,\ 1\leq b_{1}<b_{2},\ {\rm and}\ b_{1}\ {\rm is\
finite};\\\\[12.91663pt] {\diameter}&{\rm if}\ \ B_{1}=B_{2}=\infty,\
b_{1}>b_{2}\geq 1,\ {\rm and}\ b_{2}\ {\rm is\ finite};\\\\[12.91663pt]
{\diameter}\text{\,\, or\,\, }0&{\rm if}\ \ B_{1}=B_{2}=\infty,\
b_{1}=b_{2}=\infty,\par\end{array}\right.$

where $s_{0}$ is from Theorem 1.1 and

$g_{B_{1},B_{2}}\stackrel{{\scriptstyle\text{def}}}{{=}}\inf\\{s\geq
0:P(T,-s\log|T^{\prime}|-s\log B_{1}+(1-s)\log B_{2})\leq 0\\}.$

The reason behind the slight variation in the definitions of $B_{2}$ and
$b_{2}$ compared to $B_{1}$ and $b_{1}$ respectively, is in the nature of the
conditions of our set. To get the proper classification, we need to bound
$\Phi_{2}(n)$ from both sides for all $n$, while bounding $\Phi_{1}(n)$ from
above for infinitely many $n$ is sufficient. Reader might also notice that
this theorem does not cover two cases, namely $B_{1}=B_{2}^{2}<\infty$ and
$b_{1}=b_{2}<\infty$. Their omission qualifies them to be highlighted as
standalone propositions.

######  Proposition 1.5.

Denote

$E_{1}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
B_{2}^{2n}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<B_{2}^{n}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\},$

and

$E_{2}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
B_{2}^{2n-1}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<3B_{2}^{n+1}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

Then $B_{1}=B_{2}^{2}<\infty$ for both of these sets, but

$E_{1}={\diameter},\,\,\,\dim_{\mathrm{H}}E_{2}=g_{B_{1},B_{2}}.$

######  Proposition 1.6.

Denote

$P_{1}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
e^{b_{1}^{n}}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<e^{b_{1}^{n}}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\},$

and

$P_{2}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
e^{5b_{1}^{n}}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<e^{b_{1}^{n-1}}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

Then $B_{1}=B_{2}=\infty$, $b_{1}=b_{2}<\infty$ for both of these sets, but

$P_{2}={\diameter},\,\,\,\dim_{\mathrm{H}}P_{1}=\frac{1}{1+b_{1}}.$

The key ingredient in proving Theorem 1.4 is to consider a supporting set.
Define

$E(A_{1},A_{2})\stackrel{{\scriptstyle\text{def}}}{{=}}\left\\{x\in[0,1):\,c_{1}A_{1}^{n}\leq
a_{n}(x)<2c_{1}A_{1}^{n},\,\,c_{2}A_{2}^{n}\leq
a_{n+1}(x)<2c_{2}A_{2}^{n},\,\,\text{for i.m. }n\in\mathbb{N}\right\\}.$

Here and throughout, ‘i.m.’ stands for ‘infinitely many’. Then we have

######  Theorem 1.7.

For any $A_{1}>1$,

$\dim_{H}E(A_{1},A_{2})=\min\left\\{s_{A_{1}},g_{(A_{1}A_{2}),A_{1}}\right\\},$

where $s_{A_{1}}$ is from Theorem 1.1 and $g_{(A_{1}A_{2}),A_{1}}$ was defined
in Theorem 1.4.

There is one more supporting set we need to consider to calculate the
Hausdorff dimension of $\mathcal{F}(\Phi_{1},\Phi_{2})$. Define

$\mathcal{F}_{B_{1},B_{2}}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
B_{1}^{n}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<B_{2}^{n}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

This set is a special case of $\mathcal{F}(\Phi_{1},\Phi_{2})$ with
$\Phi_{i}(n)=B_{i}^{n},i=1,2$. However, dealing with this set is crucial for
the general case. We prove the following theorem explicitly.

######  Theorem 1.8.

For any $B_{1},B_{2}>1$,

  * •

when $B_{1}^{s_{0}}\leq B_{2}$,

$\dim_{H}\mathcal{F}_{B_{1},B_{2}}=s_{0};$

  * •

when $B_{1}^{s_{0}}\geq B_{2}>B_{1}^{1/2}$,

$\dim_{H}\mathcal{F}_{B_{1},B_{2}}=g_{B_{1},B_{2}};$

  * •

when $B_{1}^{1/2}\geq B_{2}$,

$\mathcal{F}_{B_{1},B_{2}}={\diameter}.$

The paper is structured as follows. In Section 2 we provide some auxiliary
results and group together various definitions that we shall appeal to in the
course of proving our theorems in the subsequent sections. In Section 3, which
serves as the main subsection of this paper, we prove Theorem 1.7, that is, we
find the Hausdorff dimension for the set $E(A_{1},A_{2})$. In Section 4 we use
this set to prove Theorem 1.8. In Section 5, we prove Theorem 1.4 using
Theorem 1.8 as well as some auxiliary results. In Section 6, we prove both
Propositions 1.5 and 1.6.

Acknowledgements. The research of Mumtaz Hussain and Nikita Shulga is
supported by the Australian Research Council Discovery Project (200100994).
The authors thank Professor Baowei Wang for useful discussions.

##  2\. Preliminaries and auxiliary results

For completeness we give a brief introduction to Hausdorff measures and
dimension. For further details we refer to the beautiful texts [3, 5].

###  2.1. Hausdorff measure and dimension

Let $0<s\in\mathbb{R}^{n}$ let $E\subset\mathbb{R}^{n}$. Then, for any
$\rho>0$ a countable collection $\\{B_{i}\\}$ of balls in $\mathbb{R}^{n}$
with diameters $\mathrm{diam}(B_{i})\leq\rho$ such that
$E\subset\bigcup_{i}B_{i}$ is called a $\rho$-cover of $E$. Let

$\mathcal{H}_{\rho}^{s}(E)=\inf\sum_{i}\mathrm{diam}(B_{i})^{s},$

where the infimum is taken over all possible $\rho$-covers $\\{B_{i}\\}$ of
$E$. It is easy to see that $\mathcal{H}_{\rho}^{s}(E)$ increases as $\rho$
decreases and so approaches a limit as $\rho\rightarrow 0$. This limit could
be zero or infinity, or take a finite positive value. Accordingly, the
$s$-Hausdorff measure $\mathcal{H}^{s}$ of $E$ is defined to be

$\mathcal{H}^{s}(E)=\lim_{\rho\to 0}\mathcal{H}_{\rho}^{s}(E).$

It is easy to verify that Hausdorff measure is monotonic and countably sub-
additive, and that $\mathcal{H}^{s}(\varnothing)=0$. Thus it is an outer
measure on $\mathbb{R}^{n}$.

For any subset $E$ one can verify that there exists a unique critical value of
$s$ at which $\mathcal{H}^{s}(E)$ ‘jumps’ from infinity to zero. The value
taken by $s$ at this discontinuity is referred to as the Hausdorff dimension
of $E$ and is denoted by $\dim_{\mathrm{H}}E$; i.e.,

$\dim_{\mathrm{H}}E:=\inf\\{s\in\mathbb{R}_{+}\;:\;\mathcal{H}^{s}(E)=0\\}.$

When $s=n$, $\mathcal{H}^{n}$ coincides with standard Lebesgue measure on
$\mathbb{R}^{n}$.

Computing Hausdorff dimension of a set is typically accomplished in two steps:
obtaining the upper and lower bounds separately. Upper bounds often can be
handled by finding appropriate coverings. When dealing with a limsup set, one
usually applies the Hausdorff measure version of the famous Borel–Cantelli
lemma (see Lemma 3.10 of [3]):

######  Proposition 2.1.

Let $\\{B_{i}\\}_{i\geq 1}$ be a sequence of measurable sets in $\mathbb{R}$
and suppose that,

$\sum_{i}\mathrm{diam}(B_{i})^{s}\,<\,\infty.$

Then

$\mathcal{H}^{s}({\limsup_{i\to\infty}B_{i}})=0.$

The main tool in establishing the lower bound for the dimension of
$E(A_{1},A_{2})$ will be the following well-known mass distribution principle
[5].

######  Proposition 2.2 (Mass Distribution Principle [5]).

Let $\mu$ be a probability measure supported on a measurable set $F$. Suppose
there are positive constants $c$ and $r_{0}$ such that

$\mu(B(x,r))\leq cr^{s}$

for any ball $B(x,r)$ with radius $r\leq r_{0}$ and center $x\in F$. Then
$\dim_{\mathrm{H}}F\geq s$.

###  2.2. Continued fractions and Diophantine approximation

Recall that the Gauss map $T:[0,1)\to[0,1)$ is defined by

$T(0)=0,~{}T(x)=\frac{1}{x}-\left\lfloor\frac{1}{x}\right\rfloor\textmd{ for
}x\in(0,1),$

where $\lfloor x\rfloor$ denotes the integer part of $x$. We write
$x:=[a_{1}(x),a_{2}(x),a_{3}(x),\ldots]$ for the continued fraction of $x$
where $a_{1}(x)=\lfloor 1/x\rfloor$, $a_{n}(x)=a_{1}(T^{n-1}(x))$ for $n\geq
2$ are called the partial quotients of $x$. The sequences $p_{n}=p_{n}(x)$,
$q_{n}=q_{n}(x)$, referred to as $n^{\rm th}$ convergents, has the recursive
relation

(2.1) $p_{n+1}=a_{n+1}(x)p_{n}+p_{n-1},\ \ q_{n+1}=a_{n+1}(x)q_{n}+q_{n-1},\ \
n\geq 0.$

Thus $p_{n}=p_{n}(x),q_{n}=q_{n}(x)$ are determined by the partial quotients
$a_{1},\dots,a_{n}$, so we may write
$p_{n}=p_{n}(a_{1},\dots,a_{n}),q_{n}=q_{n}(a_{1},\dots,a_{n})$. When it is
clear which partial quotients are involved, we denote them by $p_{n},q_{n}$
for simplicity.

For any integer vector $(a_{1},\dots,a_{n})\in\mathbb{N}^{n}$ with $n\geq 1$,
write

$I_{n}(a_{1},\dots,a_{n}):=\left\\{x\in[0,1):a_{1}(x)=a_{1},\dots,a_{n}(x)=a_{n}\right\\}$

for the corresponding ‘cylinder of order $n$’, i.e. the set of all real
numbers in $[0,1)$ whose continued fraction expansions begin with
$(a_{1},\dots,a_{n}).$

We will frequently use the following well known properties of continued
fraction expansions. They are explained in the standard texts [11, 12].

######  Proposition 2.3.

For any positive integers $a_{1},\dots,a_{n}$, let
$p_{n}=p_{n}(a_{1},\dots,a_{n})$ and $q_{n}=q_{n}(a_{1},\dots,a_{n})$ be
defined recursively by (2.1). Then:

  1. ($\rm P_{1}$)

$\displaystyle
I_{n}(a_{1},a_{2},\dots,a_{n})=\left\\{\begin{array}[]{ll}\left[\frac{p_{n}}{q_{n}},\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}\right)&{\rm
if}\ \ n\ {\rm{is\ even}};\\\
\left(\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}},\frac{p_{n}}{q_{n}}\right]&{\rm if}\
\ n\ {\rm{is\ odd}}.\end{array}\right.$

Thus, its length is given by

$\frac{1}{2q_{n}^{2}}\leq|I_{n}(a_{1},\ldots,a_{n})|=\frac{1}{q_{n}(q_{n}+q_{n-1})}\leq\frac{1}{q_{n}^{2}},$

since

$p_{n-1}q_{n}-p_{n}q_{n-1}=(-1)^{n},\ {\rm for\ all}\ n\geq 1.$

  2. ($\rm P_{2}$)

For any $n\geq 1$, $q_{n}\geq 2^{(n-1)/2}$ and

$1\leq\frac{q_{n+m}(a_{1},\cdots,a_{n},b_{1},\cdots,b_{m})}{q_{n}(a_{1},\cdots,a_{n})\cdot
q_{m}(b_{1},\cdots,b_{m})}\leq 2.$

  3. ($\rm P_{3}$)

$\prod_{i=1}^{n}a_{i}\leq q_{n}\leq\prod_{i=1}^{n}(a_{i}+1)\leq
2^{n}\prod_{i=1}^{n}a_{i}.$

  4. ($\rm P_{4}$)

$\frac{1}{3a_{n+1}(x)q^{2}_{n}(x)}\,<\,\Big{|}x-\frac{p_{n}(x)}{q_{n}(x)}\Big{|}=\frac{1}{q_{n}(x)(q_{n+1}(x)+T^{n+1}(x)q_{n}(x))}\,<\,\frac{1}{a_{n+1}q^{2}_{n}(x)}.$

  5. ($\rm P_{5}$)

there exists a constant $K>1$ such that for almost all $x\in[0,1)$,

$q_{n}(x)\leq K^{n},\ {\text{for all $n$ sufficiently large}}.$

Let $\mu$ be the Gauss measure given by

$d\mu=\frac{1}{(1+x)\log 2}dx.$

It is clear that $\mu$ is $T$-invariant and equivalent to Lebesgue measure
$\mathcal{L}$.

The next proposition concerns the position of a cylinder in $[0,1)$.

######  Proposition 2.4 ([12], Khintchine).

Let $I_{n}=I_{n}(a_{1},\dots,a_{n})$ be a cylinder of order $n$, which is
partitioned into sub-cylinders
$\\{I_{n+1}(a_{1},\dots,a_{n},a_{n+1}):a_{n+1}\in\mathbb{N}\\}$. When $n$ is
odd, these sub-cylinders are positioned from left to right, as $a_{n+1}$
increases from 1 to $\infty$; when $n$ is even, they are positioned from right
to left.

The following result is due to Łuczak [16].

######  Lemma 2.5 ([16], Łuczak).

For any $b,c>1$, the sets

$\displaystyle\left\\{x\in[0,1):a_{n}(x)\geq c^{b^{n}}\ {\text{for infinitely
many}}\ n\in\mathbb{N}\right\\},$ $\displaystyle\left\\{x\in[0,1):a_{n}(x)\geq
c^{b^{n}}\ {\text{for all }}\ n\geq 1\right\\},$

have the same Hausdorff dimension $\frac{1}{b+1}$.

We will also need the following lemma from [6, Lemma 3.2].

######  Lemma 2.6 ([6], Fan-Liao-Wang-Wu).

Let $\\{s_{n}\\}_{n\geq 1}$ be a sequence of positive integers tending to
infinity with $s_{n}\geq 3$ for all $n\geq 1$. Then for any positive number
$N\geq 2$, we have

$\dim_{\mathrm{H}}\\{x\in[0,1):s_{n}\leq a_{n}(x)<Ns_{n}\,\,\forall n\geq
1\\}=\liminf_{n\to\infty}\frac{\log(s_{1}s_{2}\cdots
s_{n})}{2\log(s_{1}s_{2}\cdots s_{n})+\log s_{n+1}}.$

###  2.3. Pressure function

When dealing with dimension problems in non-linear dynamical system, pressure
function and other concepts from thermodynamics are good tools. The concept of
a general pressure function was introduced by Walters [22] as a generalization
of entropy. It describes exponential growth rate of ergodic sum which
determine the whole system in some sense. We are interested in a way of
calculating Hausdorff dimensions using pressure functions.

A method in [21, Theorem 2.2.1] can be used to calculate the Hausdorff
dimension of self-similar sets for linear system. As for the non-linear
setting, the relation between Hausdorff dimension and pressure functions is
given in [21] as the corresponding generalization of Moran [20]. The main
required ingredient is the following

######  Definition 2.7.

Given any continuous function $f:X\rightarrow\mathbb{R}$ we define its
_pressure_ $\mathsf{P}\left(f\right)$, with respect to $T$, as

$\mathsf{P}\left(f\right):=\underset{n\rightarrow\infty}{\lim\sup}\frac{1}{n}\log{{\left(\sum\limits_{\begin{subarray}{c}T^{n}x=x\\\
x\in
X\end{subarray}}e^{f\left(x\right)+f\left(Tx\right)+\cdots+f\left(T^{n-1}x\right)}\right)}}$

where the summation is over all periodic points.

We note that we can write $``\lim"$ instead of $``\limsup"$ since the above
limit actually exists. Combined this with dimension theory, we are concerned
with a family of functions
$f_{t}\left(x\right)=-t\log\left|T^{\prime}\left(x\right)\right|$, $x\in X$
and $0\leq t\leq d$. Substituting these functions into the above definition
leads to a real-valued map

$\begin{array}[]{l}\left[0,d\right]\rightarrow\mathbb{R},\qquad
t\mapsto\mathsf{P}\left(f_{t}\right)=\underset{n\rightarrow\infty}{\lim\sup}\frac{1}{n}\log\left(\sum\limits_{\begin{subarray}{c}T^{n}x=x\\\
x\in
X\end{subarray}}\frac{1}{\left|\left(T^{n}\right)^{\prime}\left(x\right)\right|^{t}}\right).\end{array}$

We observe that a function $t\mapsto\mathsf{P}\left(f_{t}\right)$ satisfies
interesting properties:

  1. (i)

$\mathsf{P}\left(0\right)=\log k$;

  2. (ii)

$t\mapsto\mathsf{P}\left(f_{t}\right)$ is strictly decreasing;

  3. (iii)

$t\mapsto\mathsf{P}\left(f_{t}\right)$ is analytic on $\left[0,d\right]$.

Proof can be found on page 32 of [21]. Moreover, property (i) is easily
obtained by the definition.

Bowen and Ruelle had given key results for calculating the Hausdorff dimension
from the pressure function. Bowen showed the result in the context of quasi-
circles, and Bowen-Ruelle developed the method for the case of hyperbolic
Julia sets. More precisely,

######  Theorem 2.8 (Bowen-Ruelle).

Let $T:X\rightarrow X$ be a $C^{1+\alpha}$ conformal expanding map, for some
$\alpha>0$. There is a unique solution $0\leq s\leq d$ to

$\mathsf{P}\left(-s\log\left|T^{\prime}\right|\right)=0,$

which occurs precisely at $s=\dim_{H}X$.

More details and context on pressure functions can be found in [17, 18, 19].
We use the fact that the pressure function with a continuous potential can be
approximated by the pressure function restricted to the sub-systems in
continued fractions.

Let $\mathcal{A}\subset\mathbb{N}$ be a finite or infinite set and define

$X_{\mathcal{A}}=\\{x\in[0,1):{\text{for all}}\ n\geq
1,a_{n}(x)\in\mathcal{A}\\}.$

Then $(X_{\mathcal{A}},T)$ is a subsystem of $([0,1),T)$ where $T$ is a Gauss
map. Given any real function $\psi:[0,1)\rightarrow\mathbb{R},$ the pressure
function restricted to the system $(X_{\mathcal{A}},T)$ is defined by

(2.3)
$\mathsf{P}_{\mathcal{A}}(T,\psi):=\lim_{n\rightarrow\infty}\frac{1}{n}\log\sum_{a_{1},\cdots,a_{n}\in\mathcal{A}}\sup_{x\in
X_{\mathcal{A}}}e^{S_{n}\psi([a_{1},\cdots,a_{n}+x])},$

where $S_{n}\psi(x)$ denotes the ergodic sum $\psi(x)+\cdots+\psi(T^{n-1}x)$.
For simplicity, we denote $\mathsf{P}_{\mathbb{N}}(T,\psi)$ by
$\mathsf{P}(T,\psi)$ when $\mathcal{A}=\mathbb{N}$. We note that the supremum
in equation (2.3) can be removed if $\psi$ satisfy the continuity property.

For each $n\geq 1$, the $n^{\rm th}$ variation of $\psi$ is denoted by

${\mathop{\rm{Var}}}_{n}(\psi):=\sup\Big{\\{}|\psi(x)-\psi(y)|:I_{n}(x)=I_{n}(y)\Big{\\}}.$

The following results [15, Proposition 2.4] ensure the existence of the limit
in the equation (2.3).

######  Proposition 2.9 ([15], Li-Wang-Wu-Xu).

Let $\psi:[0,1)\to\mathbb{R}$ be a real function with
$\mathrm{Var}_{1}(\psi)<\infty$ and $\mathrm{Var}_{n}(\psi)\to 0$ as
$n\to\infty$. Then the limit defining $\mathsf{P}_{\mathcal{A}}(T,\psi)$
exists and the value of $\mathsf{P}_{\mathcal{A}}(T,\psi)$ remains the same
even without taking supremum over $x\in X_{\mathcal{A}}$ in (2.3).

The system $([0,1),T)$ is approximated by its subsystems $(X_{\mathcal{A}},T)$
then the pressure function has a continuity property in the system of
continued fractions. A detailed proof can be seen in [8, Proposition 2] or
[15].

######  Proposition 2.10 ([8], Hanus-Mauldin-Urbański).

Let $\psi:[0,1)\to\mathbb{R}$ be a real function with
$\mathrm{Var}_{1}(\psi)<\infty$ and $\mathrm{Var}_{n}(\psi)\to 0$ as
$n\to\infty$. We have

$\mathsf{P}_{\mathbb{N}}(T,\psi)=\sup\\{\mathsf{P}_{\mathcal{A}}(T,\psi):\mathcal{A}\
\mathrm{is\ a\ finite\ subset\ of}\ \mathbb{N}\\}.$

Now we consider the specific potentials,

$\displaystyle\psi_{1}(x)$ $\displaystyle=-s(\log B+\log|T^{\prime}(x)|)$
$\displaystyle\psi_{2}(x)$ $\displaystyle=-s\log|T^{\prime}(x)|-s\log
B_{1}+(1-s)\log B_{2}$

where $1<B,B_{1},B_{2}<\infty,$ and $s\geq 0$.It is clear that $\psi_{1}$ and
$\psi_{2}$ satisfy the variation condition and then Proposition 2.10 holds.

Thus, the pressure function (2.3) with potential $\psi_{1}$ is represented by

$\displaystyle\mathsf{P}_{\mathcal{A}}(T,-s(\log B+\log|T^{\prime}(x)|))$
$\displaystyle=\lim_{n\to\infty}\frac{1}{n}\log\sum_{a_{1},\ldots,a_{n}\in\mathcal{A}}e^{S_{n}(-s(\log
B+\log|T^{\prime}(x)|))}$
$\displaystyle=\lim_{n\to\infty}\frac{1}{n}\log\sum_{a_{1},\ldots,a_{n}\in\mathcal{A}}\left(\frac{1}{B^{n}q_{n}^{2}}\right)^{s}.$

The last equality holds by

${S_{n}(-s(\log B+\log|T^{\prime}(x)|))}={-ns\log B-s\log q_{n}^{2}}.$

which is easy to check by Proposition 2.3.

As before, we obtain the pressure function with potential $\psi_{2}$ by

$\displaystyle\mathsf{P}_{\mathcal{A}}(T,-s\log|T^{\prime}(x)|-s\log
B_{1}+(1-s)\log B_{2})$
$\displaystyle=\lim_{n\to\infty}\frac{1}{n}\log\sum_{a_{1},\ldots,a_{n}\in\mathcal{A}}e^{S_{n}(-s\log|T^{\prime}(x)|-s\log
B_{1}+(1-s)\log B_{2})}$
$\displaystyle=\lim_{n\to\infty}\frac{1}{n}\log\sum_{a_{1},\ldots,a_{n}\in\mathcal{A}}\left(\frac{1}{B_{1}^{n}q_{n}^{2}}\right)^{s}B_{2}^{(1-s)n}.$

For any $n\geq 1$ and $s\geq 0,$ we write

$\displaystyle f_{n}^{(1)}\left(s\right)$
$\displaystyle:=\sum_{a_{1},\ldots,a_{n}\in\mathcal{A}}\frac{1}{\left(B^{n}q_{n}^{2}\right)^{s}}$
$\displaystyle f_{n}^{(2)}\left(s\right)$
$\displaystyle:=\sum_{a_{1},\ldots,a_{n}\in\mathcal{A}}\left(\frac{1}{B_{1}^{n}q_{n}^{2}}\right)^{s}B_{2}^{(1-s)n}.$

and denote

$s_{n,B}\left(\mathcal{A}\right)=\inf\left\\{s\geq
0:f_{n}^{(1)}\left(s\right)\leq
1\right\\},g_{n,B_{1},B_{2}}\left(\mathcal{A}\right)=\inf\left\\{s\geq
0:f_{n}^{(2)}\left(s\right)\leq 1\right\\},$ $\displaystyle
s_{B}(\mathcal{A})=\inf\\{s\geq 0:\mathsf{P}_{\mathcal{A}}(T,\psi_{1})\leq
0\\},$ $\displaystyle g_{B_{1},B_{2}}(\mathcal{A})=\inf\\{s\geq
0:\mathsf{P}_{\mathcal{A}}(T,\psi_{2})\leq 0\\},$ $\displaystyle
s_{B}(\mathbb{N})=\inf\\{s\geq 0:\mathsf{P}(T,\psi_{1})\leq 0\\},$
$\displaystyle g_{B_{1},B_{2}}(\mathbb{N})=\inf\\{s\geq
0:\mathsf{P}(T,\psi_{2})\leq 0\\}.$

If $\mathcal{A}\in\mathbb{N}$ is a finite set, then by [23] it is
straightforward to check that both $f_{n}^{(i)}\left(s\right)$ and
$\mathsf{P}_{\mathcal{A}}(T,\psi_{i})$ for $i=1,2$ are monotonically
decreasing and continuous with respect to $s$. Thus,
$s_{n,B}\left(\mathcal{A}\right)$, $s_{B}(\mathcal{A})$,
$g_{n,B_{1},B_{2}}\left(\mathcal{A}\right)$ and $g_{B_{1},B_{2}}(\mathcal{A})$
are, respectively, the unique solutions of $f_{n}^{(1)}\left(s\right)=1$,
$\mathsf{P}_{\mathcal{A}}(T,\psi_{1})=0$, $f_{n}^{(2)}\left(s\right)=1$ and
$\mathsf{P}_{\mathcal{A}}(T,\psi_{2})=0.$

For simplicity, when $\mathcal{A}=\left\\{1,2,\ldots,M\right\\}$ for some
$M>0$, we write $s_{n,B}\left({M}\right)$ for
$s_{n,B}\left(\mathcal{A}\right)$, $s_{B}\left({M}\right)$ for
$s_{B}\left(\mathcal{A}\right)$, $g_{n,B_{1},B_{2}}\left({M}\right)$ for
$g_{n,B_{1},B_{2}}\left(\mathcal{A}\right)$ and
$g_{B_{1},B_{2}}\left({M}\right)$ for
$g_{B_{1},B_{2}}\left(\mathcal{A}\right)$. When $\mathcal{A}=\mathbb{N}$, we
write $s_{n,B}$ for $s_{n,B}(\mathbb{N})$, $s_{B}$ for $s_{B}(\mathbb{N})$,
$g_{n,B_{1},B_{2}}$ for $g_{n,B_{1},B_{2}}(\mathbb{N})$ and $g_{B_{1},B_{2}}$
for $g_{B_{1},B_{2}}(\mathbb{N})$.

As a consequence, we have

######  Corollary 2.11.

For any integer $M\in\mathbb{N},$

$\lim_{n\to\infty}s_{n,B}(M)=s_{B}(M),\ \ \lim_{M\to\infty}s_{B}(M)=s_{B},\ \
\lim_{n\to\infty}g_{n,B_{1},B_{2}}(M)=g_{B_{1},B_{2}}(M),\ \
\lim_{M\to\infty}g_{B_{1},B_{2}}(M)=g_{B_{1},B_{2}}.$

Note that, $s_{B}$ and $g_{B_{1},B_{2}}$ are continuous respectively as a
function of $B$ and $B_{1},B_{2}$. Moreover,

$\lim_{B\rightarrow 1}s_{B}=1,\ \ \lim_{B\rightarrow\infty}s_{B}=1/2.$

###### Proof.

The last two equations can be proved by following similar steps as for $s_{B}$
in [23] and others are consequences of Proposition 2.10. ∎

##  3\. Hausdorff dimension of $E(A_{1},A_{2})$.

The proof of Theorem 1.7 consists of two parts, the upper bound and the lower
bound. For notational simplicity, we take $c_{1}=c_{2}=1$ and the other case
can be done with obvious modifications. That is we will deal with the set

$E(A_{1},A_{2})\stackrel{{\scriptstyle\text{def}}}{{=}}\left\\{x\in[0,1):\,A_{1}^{n}\leq
a_{n}(x)<2A_{1}^{n},\,\,A_{2}^{n}\leq a_{n+1}(x)<2A_{2}^{n},\,\,\text{for i.m.
}n\in\mathbb{N}\right\\}.$

###  3.1. Upper bound.

It is known that the covering of an upper limit set is natural, so the natural
covering of $E(A_{1},A_{2})$ is found routinely as follows.

$\displaystyle E(A_{1})$
$\displaystyle=\bigcap\limits_{N=1}^{\infty}\bigcup\limits_{n=N}^{\infty}\left\\{x\in[0,1):\,A_{1}^{n}\leq
a_{n}(x)<2A_{1}^{n},\,\,A_{2}^{n}\leq a_{n+1}(x)<2A_{2}^{n}\right\\}$
$\displaystyle\subset\bigcup\limits_{n=N}^{\infty}\left\\{x\in[0,1):\,A_{1}^{n}\leq
a_{n}(x)<2A_{1}^{n},\,\,A_{2}^{n}\leq
a_{n+1}(x)<2A_{2}^{n}\right\\}:=\bigcup\limits_{n=N}^{\infty}E_{n}$
$\displaystyle=\bigcup\limits_{n=N}^{\infty}\bigcup\limits_{a_{1},\ldots,a_{n-1}\in\mathbb{N}}\left\\{x\in[0,1)\,:\,a_{i}(x)=a_{i},1\leq
i<n,\ A_{1}^{n}\leq a_{n}(x)<2A_{1}^{n},\ A_{2}^{n}\leq
a_{n+1}(x)<2A_{2}^{n}\right\\}.$

There are two potential optimal covers for $E_{n}$ for each $n\geq N$. For any
integers $a_{1},\ldots,a_{n-1}\in\mathbb{N}$, define

$J_{n-1}(a_{1},\ldots,a_{n-1})=\bigcup\limits_{A_{1}^{n}\leq
a_{n}<2A_{1}^{n}}I_{n}(a_{1},\ldots,a_{n}).$

That is one type of suitable coverings of $E(A_{1})$, which is specifically
represented as

$E(A_{1})\subset\bigcup\limits_{n=N}^{\infty}\bigcup\limits_{a_{1},\ldots,a_{n-1}\in\mathbb{N}}J_{n-1}(a_{1},\ldots,a_{n-1}).$

Then, by using Proposition 2.3 and Proposition 2.4 recursively, we obtain

$|J_{n-1}(a_{1},\ldots,a_{n-1})|=\sum\limits_{A_{1}^{n}\leq
a_{n}<2A_{1}^{n}}\left|\frac{p_{n}}{q_{n}}-\frac{p_{n}+p_{n-1}}{q_{n}+q_{n-1}}\right|\asymp\frac{1}{A_{1}^{n}q_{n-1}^{2}}.$

Therefore, for any $\varepsilon>0$, an $(s_{A_{1}}+2\varepsilon)$-dimensional
Hausdorff dimension of $E(A_{1},A_{2})$ can be estimated as

$\begin{split}\mathcal{H}^{s_{A_{1}}+2\varepsilon}(E(A_{1},A_{2}))&\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\sum\limits_{a_{1},\ldots,a_{n-1}}|J_{n-1}(a_{1},\ldots,a_{n-1})|^{s_{A_{1}}+2\varepsilon}\\\
&\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\frac{1}{2^{(n-1)\varepsilon}}\sum\limits_{a_{1},\ldots,a_{n-1}}\left(\frac{1}{A_{1}^{n}q_{n-1}^{2}}\right)^{s_{A_{1}}}\\\
&\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\frac{1}{2^{(n-1)\varepsilon}}<\infty.\end{split}$

Hence, from the definition of Hausdorff dimension, it follows that

$\dim_{H}E(A_{1},A_{2})\leq s_{A_{1}}.$

As for another covering type of $E(A_{1})$, similar with the former, for any
integers $a_{1},\ldots,a_{n-1}\in\mathbb{N}$ and $A_{1}^{n}\leq
a_{n}(x)<2A_{1}^{n}$, define

$J_{n}(a_{1},\ldots,a_{n})=\bigcup\limits_{A_{2}^{n}\leq
a_{n+1}<2A_{2}^{n}}I_{n+1}(a_{1},\ldots,a_{n+1})$

Then

$|J_{n}(a_{1},\ldots,a_{n})|\asymp\frac{1}{A_{2}^{n}q_{n}^{2}},$

and

$E(A_{1})\subset\bigcup_{n=N}^{\infty}\bigcup\limits_{a_{1},\ldots,a_{n-1}}\bigcup\limits_{A_{1}^{n}\leq
a_{n}<2A_{1}^{n}}J_{n}(a_{1},\ldots,a_{n}).$

Therefore, for any $\varepsilon>0$, a
$(g_{(A_{1}A_{2}),A_{1}}+2\varepsilon)$-dimensional Hausdorff dimension of
$E(A_{1},A_{2})$ can be estimated as

$\begin{split}\mathcal{H}^{g_{(A_{1}A_{2}),A_{1}}+2\varepsilon}(E(A_{1},A_{2}))&\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\sum\limits_{a_{1},\ldots,a_{n-1}}\sum\limits_{A_{1}^{n}\leq
a_{n}<2A_{1}^{n}}|J_{n}(a_{1},\ldots,a_{n})|^{g_{(A_{1}A_{2}),A_{1}}+2\varepsilon}\\\
&\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\sum\limits_{a_{1},\ldots,a_{n-1}}\sum\limits_{A_{1}^{n}\leq
a_{n}<2A_{1}^{n}}\left(\frac{1}{A_{2}^{n}q_{n}^{2}}\right)^{g_{(A_{1}A_{2}),A_{1}}+2\varepsilon}\\\
&\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\sum\limits_{a_{1},\ldots,a_{n-1}}A_{1}^{n}\left(\frac{1}{(A_{2}A_{1}^{2})^{n}q_{n-1}^{2}}\right)^{g_{(A_{1}A_{2}),A_{1}}+2\varepsilon}\\\
&\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\frac{1}{2^{(n-1)\varepsilon}}<\infty.\end{split}$

The third inequality holds by recursive relation (2.1). Thus, the upper bound
is obtained immediately by

$\dim_{H}E(A_{1},A_{2})\leq g_{(A_{1}A_{2}),A_{1}}.$

Hence,

$\dim_{H}E(A_{1},A_{2})\leq\min\left\\{s_{A_{1}},g_{(A_{1}A_{2}),A_{1}}\right\\}.$

###  3.2. Lower bound.

In this subsection we will determine the lower bound for the dimension of
$E(A_{1},A_{2})$ by using the mass distribution principle (Proposition 2.2).

For convenience, let us define some dimensional numbers in first. For any
integers $N,M$, define the dimensional number $s=s_{N}(M)$ and $g=g_{N}(M)$
repectively as the solution to

$\sum\limits_{1\leq a_{1},\ldots,a_{N}\leq
M}\left(\frac{1}{A_{1}^{N}q_{N}^{2}}\right)^{s}=1$

and

$\sum\limits_{1\leq a_{1},\ldots,a_{N}\leq
M}\frac{A_{1}^{N}}{((A_{1}^{2}A_{2})^{N}q_{N}^{2})^{g}}=1.$

More specifically, each equation has a unique solution and, by Corollary 2.11,

$\lim\limits_{M\to\infty}\lim\limits_{N\to\infty}s_{N}(M)=s_{A_{1}},$

and

$\lim\limits_{M\to\infty}\lim\limits_{N\to\infty}g_{N}(M)=g_{(A_{1}A_{2}),A_{1}}.$

Take a sequence of large sparse integers $\\{\ell_{k}\\}_{k\geq 1}$, say,
$\ell_{k}\gg e^{\ell_{1}+\dots+\ell_{k-1}}.$ For any $\varepsilon>0$, choose
integers $N,M$ sufficiently large such that

$s>s_{A_{1}}-\varepsilon,\quad
g>g_{(A_{1}A_{2}),A_{1}}-\varepsilon,\qquad\left(2^{(N-1)/2})\right)^{\varepsilon/2}\geq
2^{100}.$

Without loss of generality, we assume that $s<g$. Let

$n_{k}-n_{k-1}=\ell_{k}N+1,\,\forall k\geq 1,$

such that

$\left(2^{\ell_{k}(N-1)/2}\right)^{\frac{\varepsilon}{2}}\geq\prod\limits_{t=1}^{k-1}(M+1)^{\ell_{t}N}(A_{1}A_{2})^{\sum_{i=1}^{t}\ell_{i}N+t}.$

At this point, define a subset of $E(A_{1},A_{2})$ as

(3.1) $\displaystyle E=\\{x\in[0,1):\,A_{1}^{n_{k}}\leq
a_{n_{k}}(x)<2A_{1}^{n_{k}},\ $ $\displaystyle A_{2}^{n_{k}}\leq
a_{n_{k}+1}<2A_{2}^{n_{k}}\text{ for all }k\geq 1$ $\displaystyle\text{ and
}a_{n}(x)\in\\{1,\ldots,M\\}\text{ for other }n\in\mathbb{N}\\}.$

Next we proceed to make use of a symbolic space. Define
$D_{0}=\\{{\diameter}\\}$, and for any $n\geq 1$, define

$\displaystyle
D_{n}=\Bigg{\\{}(a_{1},\cdots,a_{n})\in\mathbb{N}^{n}:A_{i}^{n_{k}}\leq
a_{n_{k}+i}$ $\displaystyle<2A_{i}^{n_{k}},\ {\text{for all}}\ 0\leq i\leq
1,k\geq 1\ {\text{with}}\ {n_{k}+i}\leq n;$ $\displaystyle{\text{and}}\
a_{j}\in\\{1,\cdots,M\\},\ {\text{for other $j\leq n$}}\Bigg{\\}}.$

This set is just the collection of the prefixes of the points in $E$.
Moreover, the collection of finite words of length $N$ is denoted by

$\mathcal{U}=\\{w=(\sigma_{1},\dots,\sigma_{N}):1\leq\sigma_{i}\leq M,1\leq
i\leq N\\}$

and, in the rest of the paper we always use $w$ to denote a generic word in
$\mathcal{U}$.

####  3.2.1. Cantor structure of $E$.

In this subsection, we depict the structure of $E$ with the help of symbolic
space as mentioned above. For any $(a_{1},\cdots,a_{n})\in D_{n}$, define

$J_{n}(a_{1},\cdots,a_{n})=\bigcup_{a_{n+1}:(a_{1},\cdots,a_{n},a_{n+1})\in
D_{n+1}}I_{n+1}(a_{1},\cdots,a_{n},a_{n+1})$

and call it a basic cylinder of order $n$. More precisely, for any $k\geq 0$

  * •

when $n_{k}+1\leq n<n_{k+1}-1$ (by viewing $n_{0}=0$),

$J_{n}(a_{1},\cdots,a_{n})=\bigcup_{1\leq a_{n+1}\leq
M}I_{n+1}(a_{1},\cdots,a_{n},a_{n+1}).$

  * •

when $n=n_{k+1}-1$ and $n=n_{k+1}$, for $i=1,2$,

$J_{n}(a_{1},\cdots,a_{n})=\bigcup_{A_{i}^{n_{k+1}}\leq
a_{n+1}<2A_{i}^{n_{k+1}}}I_{n+1}(a_{1},\cdots,a_{n},a_{n+1}).$

Then we define the level $n$ of the Cantor set $E$ as

$\mathcal{F}_{n}=\bigcup_{(a_{1},\cdots,a_{n})\in
D_{n}}J_{n}(a_{1},\cdots,a_{n}).$

Consequently, the Cantor structure of $E$ is described as follows

$E=\bigcap_{n=1}^{\infty}\mathcal{F}_{n}=\bigcap_{n=1}^{\infty}\bigcup_{(a_{1},\cdots,a_{n})\in
D_{n}}J_{n}(a_{1},\cdots,a_{n}).$

We observe that every element $x\in E$ can be written as

$\displaystyle
x=[w_{1}^{(1)},\cdots,w_{\ell_{1}}^{(1)},a_{n_{1}},a_{n_{1}+1},$
$\displaystyle
w_{1}^{(2)},\cdots,w_{\ell_{2}}^{(2)},a_{n_{2}},a_{n_{2}+1},\cdots,w_{1}^{(k)},\cdots,w_{\ell_{k}}^{(k)},a_{n_{k}},a_{n_{k}+1},\cdots],$

where

$w^{(p)}_{k}\in\mathcal{U},\ {\text{and}}\ \ A_{i+1}^{n_{k}}\leq
a_{n_{k}+i}\leq 2A_{i+1}^{n_{k}},\ \text{ for all }k,p\in\mathbb{N}\ \ 0\leq
i\leq 1.$

Then the length of cylinder set can be estimated as follows.

######  Lemma 3.1 (Length estimation).

Let $x\in E$ and $n_{k}+1\leq n<n_{k+1}-1$.

  * •

for $n=n_{k}-1$,

(3.2) $\displaystyle|J_{n_{k}-1}(x)|$
$\displaystyle\geq\frac{1}{2^{3}A_{1}^{n_{k}}}\cdot\left(\frac{1}{2^{\ell_{k}}}\cdot\prod_{i=1}^{\ell_{k}}\frac{1}{q_{N}(w_{i}^{(k)})}\cdot\frac{1}{q_{n_{k-1}+1}}\right)^{2}\geq\frac{1}{A_{1}^{n_{k}}}\left(\prod_{i=1}^{\ell_{k}}\frac{1}{q_{N}(w_{i}^{(k)})}\right)^{2(1+\varepsilon)}.$

  * •

for $n=n_{k}$,

(3.3)
$\displaystyle|J_{n_{k}}(x)|\geq\frac{1}{2^{11}}\cdot\frac{1}{A_{1}^{n_{k}}A_{2}^{n_{k}}}\cdot|J_{n_{k}-1}(x)|.$

  * •

for $n=n_{k}+1$,

(3.4)
$\displaystyle|J_{n_{k}+1}(x)|\geq\frac{1}{2^{11}}\cdot\frac{1}{A_{1}^{n_{k}}A_{2}^{2n_{k}}}\cdot|J_{n_{k}-1}(x)|.$

  * •

For each $1\leq\ell<\ell_{k+1}$,

(3.5) $\displaystyle|J_{n_{k}+1+\ell N}(x)|$
$\displaystyle\geq\frac{1}{2^{3}}\cdot\left(\frac{1}{2^{2\ell}}\cdot\prod_{i=1}^{\ell}\frac{1}{q_{N}^{2}(w_{i}^{(k+1)})}\right)\cdot\frac{1}{q^{2}_{{n_{k}}+1}}\geq\left(\prod_{i=1}^{\ell}\frac{1}{q_{N}^{2}(w_{i}^{(k+1)})}\right)^{1+\varepsilon}\cdot\frac{1}{q^{2}_{{n_{k}}+1}}.$

  * •

for $n_{k}+1+(\ell-1)N<n<n_{k}+1+\ell N$ with $1\leq\ell\leq\ell_{k+1}$,

(3.6) $\displaystyle|J_{n}(x)|\geq c\cdot|J_{n_{k}+1+(\ell-1)N}(x)|,$

where $c=c(M,N)$ is an absolute constant.

###### Proof.

The proof of this lemma is based on the following inequality. Let

$V_{a}:=\max{a_{n+1}},\qquad v_{a}:=\min{a_{n+1}}.$

Then

$\displaystyle|J_{n}(x)|$
$\displaystyle=\left|\frac{v_{a}p_{n}+p_{n-1}}{v_{a}q_{n}+q_{n-1}}-\frac{(V_{a}+1)p_{n}+p_{n-1}}{(V_{a}+1)q_{n}+q_{n-1}}\right|=\frac{V_{a}+1-v_{a}}{(v_{a}q_{n}+q_{n-1})((V_{a}+1)q_{n}+q_{n-1})},$
$\Longrightarrow\frac{1}{2^{3}}\cdot\frac{V_{a}-v_{a}}{V_{a}v_{a}q_{n}^{2}}\leq|J_{n}(x)|\leq\frac{V_{a}+1-v_{a}}{V_{a}v_{a}q_{n}^{2}}.$

The cases of $n=n_{k}-1$, $n=n_{k}$ and any others follows by replacing
$V_{a}$ with $2A_{1}^{n_{k}}$, $2A_{2}^{n_{k}}$, $M$ and $v_{a}$ with
$A_{1}^{n_{k}}$, $A_{2}^{n_{k}}$, $1$ respectively. Then using subtle scale by
the choice of $\ell_{k}$ is enough. More precisely, we give the main terms
here and omit straightforward substitutions.

$\displaystyle\frac{1}{A_{1}^{n_{k}}q_{n_{k}-1}^{2}}>$
$\displaystyle|J_{n_{k}-1}(x)|\geq\frac{1}{8}\frac{1}{A_{1}^{n_{k}}q_{n_{k}-1}^{2}},$
$\displaystyle\frac{1}{(A_{2}A^{2}_{1})^{n_{k}}q_{n_{k}-1}^{2}}\geq\frac{1}{A_{2}^{n_{k}}q_{n_{k}}^{2}}>$
$\displaystyle|J_{n_{k}}(x)|\geq\frac{1}{8}\frac{1}{A_{2}^{n_{k}}q_{n_{k}}^{2}}\geq\frac{1}{2^{7}}\frac{1}{(A_{2}A^{2}_{1})^{n_{k}}q_{n_{k}-1}^{2}}.$

For $n\neq n_{k}-1,n_{k}$,

$\frac{1}{q_{n}^{2}}>|J_{n}(x)|\geq\frac{1}{8q_{n}^{2}}.$

∎

###  3.3. Mass distribution

In this subsection, we define two mass distributions along the basic intervals
$J_{n}(x)$ containing $x$. These mass distributions then can be extended
respectively into probability measure supported on $E$ by the Carathéodory
extension theorem. Let us begin this idea by induction. For $n\leq n_{1}+1$,

  1. (1)

when $n=\ell N$ for each $1\leq\ell\leq\ell_{1}$, define

$\mu_{1}(J_{\ell
N}(x))=\prod_{i=1}^{\ell}\frac{1}{q_{N}(w_{i}^{(1)})^{2s}\cdot A_{1}^{sN}}$

and

$\mu_{2}(J_{\ell
N}(x))=\prod_{i=1}^{\ell}\frac{A_{1}^{N}}{q_{N}(w_{i}^{(1)})^{2g}\cdot(A_{2}A_{1}^{2})^{gN}}.$

We note that measures $\mu_{1}$ and $\mu_{2}$ can be defined on all basic
cylinders of order $\ell N$ since $x$ is arbitrary.

  2. (2)

when $(\ell-1)N<n<\ell N$ for some $1\leq\ell\leq\ell_{1}$ and for all $1\leq
j\leq 2$, define

$\mu_{j}(J_{n}(x))=\sum_{J_{\ell N}\subset J_{n}(x)}\mu_{j}(J_{\ell N}(x))$

The consistency property as mentioned above fulfills the measure of other
basic intervals of order less than $n_{1}-1$.

  3. (3)

when $n=n_{1}+i$ for each $0\leq i\leq 1$ and $1\leq j\leq 2$, define

$\mu_{j}(J_{n_{1}+i}(x))=\prod_{k=0}^{i}\frac{1}{A_{k}^{n}}\cdot\mu_{j}(J_{n_{1}-1}(x)).$

Assume the measure of all basic intervals of order $n_{k}+1$ has been defined
when $n_{k}+1<n\leq n_{k+1}+1$.

  1. (1)

When $n=n_{k}+1+\ell N$ for each $1\leq\ell\leq\ell_{k+1}$, define

(3.7) $\displaystyle\mu_{1}(J_{n_{k}+1+\ell N}(x))$
$\displaystyle=\prod_{i=1}^{\ell}\frac{1}{q_{N}(w_{i}^{(k+1)})^{2s}\cdot
A_{1}^{sN}}\cdot\mu_{1}(J_{n_{k}+1}(x))$

and

(3.8) $\displaystyle\mu_{2}(J_{n_{k}+1+\ell
N}(x))=\prod_{i=1}^{\ell}\frac{A_{1}^{N}}{q_{N}(w_{i}^{(k+1)})^{2g}\cdot(A_{2}A_{1}^{2})^{gN}}\mu_{2}(J_{n_{k}+1}(x)).$

  2. (2)

When $n_{k}+1+(\ell-1)N<n<n_{k}+1+\ell N$ for some $1\leq\ell\leq\ell_{1}$ and
for $1\leq j\leq 2$, define

$\mu_{j}(J_{n}(x))=\sum_{J_{n_{k}+1+\ell N}\subset
J_{n}(x)}\mu_{j}(J_{n_{k}+1+\ell N}).$

Furthermore, for each measure, compared with the measure of a basic cylinder
of order $n_{k}+1+(\ell-1)N$ and its offsprings of order $n_{k}+1+\ell N$,
there is only a multiplier between them. More precisely, that is the term

$\frac{1}{q_{N}^{2s}(w_{\ell}^{(k+1)})A_{1}^{sN}}$

in the case of $\mu_{1}$ and

$\frac{A_{1}^{N}}{q_{N}^{2g}(w_{\ell}^{(k+1)})(A_{2}A_{1}^{2})^{gN}}$

in the case of $\mu_{2}$. Thus for each $1\leq j\leq 2$, there is an absolute
constant $c>0$ such that

$\mu_{j}(J_{n}(x))\geq c\cdot\mu_{j}\Big{(}J_{n_{k}+1+(\ell-1)N}(x)\Big{)}$

since the above two terms are uniformly bounded.

  3. (3)

when $n=n_{k}+1+i$ for each $0\leq i\leq 1$ and $1\leq j\leq 2$, define

(3.9) $\displaystyle\mu_{j}(J_{n_{k}+1+i}(x))$
$\displaystyle=\prod_{j=0}^{i}\frac{1}{A_{j}^{n_{k}}}\cdot\mu_{j}(J_{n_{k}-1}(x)).$

  4. (4)

As for other orders of measure, to ensure the consistency property, let their
measure equal to the summation of the measure of their offsprings. Moreover,
for each integer $n$, the relation between measures of a basic cylinder and
its predecessor acts like the case $n_{k}+1+(\ell-1)N<n<n_{k}+1+\ell N$, for
each $1\leq j\leq 2$, there is a constant $c>0$ such that

(3.10) $\displaystyle\mu_{j}(J_{n+1}(x))\geq c\cdot\mu_{j}(J_{n}(x)).$

###  3.4. Hölder exponent of $\mu$ for basic cylinders

We compare the measure and length of $J_{n}(x)$. However, different measures
for different values of parameters are considered as the following three
cases:

  1. i)

$A_{1}\geq(A_{1}A_{2})^{s}$ .

  2. ii)

$(A_{1}A_{2})^{g}<A_{1}<(A_{1}A_{2})^{s}$.

  3. iii)

$A_{1}\leq(A_{1}A_{2})^{g}$.

We will use measure $\mu_{1}$ for the first and the second cases, while we use
$\mu_{2}$ for the third case.

  1. (1)

When $n={n_{k}-1}$. Recall (3.7) and (3.2), it follows that for $\mu_{1}$,

$\displaystyle\mu_{1}(J_{n_{k}-1}(x))$
$\displaystyle\leq\frac{1}{A_{1}^{sn_{k}}}\prod_{i=1}^{\ell_{k}}\frac{1}{q_{N}(w_{i}^{(k)})^{2s}}\leq|J_{n_{k}-1}(x)|^{\frac{s}{1+\varepsilon}}\leq\left(\frac{1}{A_{1}^{n_{k}}q_{n_{k}-1}^{2}}\right)^{\frac{s}{1+\varepsilon}}\leq\left(\frac{1}{A_{1}^{n_{k}}q_{n_{k}-1}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

For the measure $\mu_{2}$ recall (3.8) and (3.2). It follows that

$\displaystyle\mu_{2}(J_{n_{k}-1}(x))\leq\frac{A_{1}^{n_{k}}}{(A_{1}^{2}A_{2})^{gn_{k}}}\prod_{i=1}^{\ell_{k}}\frac{1}{q_{N}(w_{i}^{(k)})^{2g}}$
$\displaystyle\leq\frac{1}{A_{1}^{gn_{k}}}\prod_{i=1}^{\ell_{k}}\frac{1}{q_{N}(w_{i}^{(k)})^{2g}}$
$\displaystyle\leq|J_{n_{k}-1}(x)|^{\frac{g}{1+\varepsilon}}\leq|J_{n_{k}-1}(x)|^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

  2. (2)

When $n={n_{k}}$. Recall (3.9) and (3.3), for $\mu_{1}$ we get

$\displaystyle\mu_{1}(J_{n_{k}}(x))=\frac{1}{A_{1}^{n_{k}}}\cdot\mu_{1}(J_{n_{k}-1}(x))$
$\displaystyle\leq\frac{1}{A_{1}^{n_{k}}}\cdot\left(\frac{1}{A_{1}^{n_{k}}q_{n_{k}-1}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}\leq\left(\frac{1}{(A_{2}A^{2}_{1})^{n_{k}}q_{n_{k}-1}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}$
$\displaystyle\leq c|J_{n_{k}}(x)|^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}\leq
c\cdot\left(\frac{1}{A_{2}^{n_{k}}q_{n_{k}}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

For the measure $\mu_{2}$ we have

$\displaystyle\mu_{2}(J_{n_{k}}(x))=\frac{1}{A_{1}^{n_{k}}}\cdot\mu_{2}(J_{n_{k}-1}(x))$
$\displaystyle\leq\frac{1}{A_{1}^{n_{k}}}\cdot\frac{A_{1}^{n_{k}}}{(A_{1}^{2}A_{2})^{gn_{k}}}\prod_{i=1}^{\ell_{k}}\frac{1}{q_{N}(w_{i}^{(k)})^{2g}}\leq\left(\frac{1}{(A_{2}A^{2}_{1})^{n_{k}}q_{n_{k}-1}^{2}}\right)^{\frac{g}{1+\varepsilon}}$
$\displaystyle\leq c|J_{n_{k}}(x)|^{\frac{g}{1+\varepsilon}}\leq
c\cdot\left(\frac{1}{A_{2}^{n_{k}}q_{n_{k}}^{2}}\right)^{\frac{g}{1+\varepsilon}}\leq
c\cdot\left(\frac{1}{A_{2}^{n_{k}}q_{n_{k}}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

  3. (3)

When $n=n_{k}+1$. Recall (3.9) and (3.4). Note that $0\leq\min\\{s,g\\}\leq
1$, for each $1\leq j\leq 2$,

$\displaystyle\mu_{j}(J_{n_{k}+1}(x))=\frac{1}{A_{2}^{n_{k}}}\cdot\mu_{j}(J_{n_{k}}(x))$
$\displaystyle\leq\frac{1}{A_{2}^{n_{k}}}\cdot
c\cdot\left(\frac{1}{A_{2}^{n_{k}}q_{n_{k}}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}\leq
c\left(\frac{1}{A_{2}^{2n_{k}}q_{n_{k}}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}$
$\displaystyle\leq
c_{2}|J_{n_{k}+1}(x)|^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}\leq
c_{2}\left(\frac{1}{q_{n_{k}+1}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

  4. (4)

When $n=n_{k}+1+\ell N$ for some $1\leq\ell<\ell_{k+1}$. Recall (3.7), (3.8)
and (3.5), for each $1\leq j\leq 2$,

$\displaystyle\mu_{j}(J_{n_{k}+1+\ell N}(x))\leq
c_{2}\prod_{i=1}^{\ell}\frac{1}{q_{N}(w_{i}^{(k+1)})^{2s}}\left(\frac{1}{q_{n_{k}+1}^{2}}\right)^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}\leq
c_{2}|J_{n_{k}+1+\ell N}(x)|^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

  5. (5)

For other $n$, let $1\leq\ell\leq\ell_{k}$ be the integer such that

$n_{k}+1+(\ell-1)N\leq n<n_{k}+1+\ell N.$

Recall (3.6). Then for each $1\leq j\leq 2$,

$\displaystyle\mu_{j}(J_{n}(x))$
$\displaystyle\leq\mu_{j}(J_{n_{k}+1+(\ell-1)N}(x))\leq
c_{2}\cdot\big{|}J_{n_{k}+2+(\ell-1)N}(x)\big{|}^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}\leq
c_{2}\cdot
c\cdot\big{|}J_{n}(x)\big{|}^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

In a summary, we have shown that for some absolute constant $c_{3}$, for any
$n\geq 1$ and $x\in E$,

(3.11) $\displaystyle\mu_{j}(J_{n}(x))\leq
c_{3}\cdot|J_{n}(x)|^{\frac{{\min\\{s,g\\}}}{1+\varepsilon}}.$

###  3.5. Hölder exponent for a general ball

For simplicity, write

$\tau=\frac{{\min\\{s,g\\}}}{1+\varepsilon}.$

The next lemma gives a minimum gap between two adjacent fundamental cylinders.

######  Lemma 3.2 (Gap estimation).

Denote by $G_{n}(a_{1},\cdots,a_{n})$ the gap between
$J_{n}(a_{1},\cdots,a_{n})$ and other basic cylinders of order $n$. Then

$G_{n}(a_{1},\cdots,a_{n})\geq\frac{1}{M}\cdot|J_{n}(a_{1},\cdots,a_{n})|.$

###### Proof.

The proof of this lemma is derived from the positions of the cylinders in
Proposition 2.4. We omit the details and refer the reader to its analogous
proof in [9, Lemma 5.3]. ∎

Then for any $x\in E$ and $r$ small enough, there is a unique integer $n$ such
that

$G_{n+1}(x)\leq r<G_{n}(x).$

This implies that the ball $B(x,r)$ can only intersect one basic cylinder
$J_{n}(x)$, and so all the basic cylinders of order $n+1$ for which $B(x,r)$
can intersect are all contained in $J_{n}(x)$. We discuss the measure of the
ball with several cases. Note that $n_{k-1}+1\leq n<n_{k}+1$.

  1. (1)

For $n_{k-1}+1\leq n<n_{k}-1$, by (3.10) and (3.11), it follows that for each
$1\leq j\leq 2$,

$\displaystyle\mu_{j}(B(x,r))$ $\displaystyle\leq\mu_{j}(J_{n}(x))\leq
c\cdot\mu_{j}(J_{n+1}(x))\leq c\cdot
c_{3}\cdot\big{|}J_{n+1}(x)\big{|}^{\tau}$ $\displaystyle\leq c\cdot
c_{3}\cdot M\cdot(G_{n+1}(x))^{\tau}\leq c\cdot c_{3}\cdot M\cdot r^{\tau}.$

  2. (2)

For $n=n_{k}-1$, the ball $B(x,r)$ can only intersect one basic cylinder
$J_{n_{k}-1}(x)$ of order $n_{k}-1$. Next, the number of basic cylinders of
order $n_{k}$ which are contained in $J_{n_{k}-1}(x)$ and intersect the ball
can be calculated as follows.

We write $x=[a_{1}(x),a_{2}(x),\ldots]$ and observe that any basic cylinder
$J_{n_{k}}(a_{1},\ldots,a_{n_{k}})$ is contained in the cylinder
$I_{n_{k}}(a_{1},\ldots,a_{n_{k}})$. Note that $A_{1}^{n_{k}}\leq
a_{n_{k}}\leq 2A_{1}^{n_{k}}$, the length of cylinder $I_{n_{k}}$ is

$\frac{1}{q_{n_{k}}(q_{n_{k}}+q_{n_{k}-1})}\geq\frac{1}{2^{5}}\cdot\frac{1}{q_{n_{k}-1}^{2}A_{1}^{2n_{k}}}.$

We also note that radius $r$ is sometimes too small to cover a whole cylinder
of order $n_{k}$. The exposition needs to split into two parts. When

$r<\frac{1}{2^{5}}\frac{1}{q_{n_{k}-1}^{2}(u)A_{1}^{2n_{k}}},$

then the ball $B(x,r)$ can intersect at most three cylinders
$I_{n_{k}}(a_{1},\ldots,a_{n_{k}})$ and so three basic cylinders
$J_{n_{k}}(a_{1},\ldots,a_{n_{k}})$. Since each measure has the same
distribution on these intervals, for $1\leq j\leq 2$,

$\displaystyle\mu_{j}(B(x,r))$ $\displaystyle\leq 3\mu_{j}(J_{n_{k}}(x))\leq
3\cdot c_{3}\cdot|J_{n_{k}}(x)|^{\tau}$ $\displaystyle\leq 3\cdot c_{3}\cdot
M\cdot G_{n+1}(x)^{\tau}\leq 3\cdot c_{1}\cdot M\cdot r^{\tau}.$

When

$r\geq\frac{1}{2^{5}}\frac{1}{q_{n_{k}-1}^{2}A_{1}^{2n_{k}}},$

then the number of basic cylinders of order $n_{k}$ for which the ball
$B(x,r)$ can intersect is at most

${2^{6}r}\cdot q_{n_{k}-1}^{2}A_{1}^{2n_{k}}+2\leq 2^{7}\cdot{r}\cdot
q_{n_{k}-1}^{2}A_{1}^{2n_{k}}.$

Thus, for $1\leq j\leq 2$,

$\displaystyle\mu_{j}(B(x,r))$
$\displaystyle\leq\min\Big{\\{}\mu_{j}(J_{n_{k}-1}(x)),\ \ 2^{7}\cdot{r}\cdot
q_{n_{k}-1}^{2}(u)A_{1}^{2n_{k}}\cdot\frac{1}{A_{1}^{n_{k}}}\cdot\mu_{j}(J_{n_{k}-1}(x))\Big{\\}}$
$\displaystyle\leq
c_{3}\cdot|J_{n_{k}-1}|^{\tau}\cdot\min\Big{\\{}1,2^{7}\cdot{r}\cdot
q_{n_{k}-1}^{2}(u)A_{1}^{n_{k}}\Big{\\}}$ $\displaystyle\leq
c_{3}\cdot\left(\frac{1}{q_{n_{k}-1}(u)^{2}A_{1}^{n_{k}}}\right)^{\tau}\cdot
1^{1-\tau}\cdot\Big{(}2^{7}\cdot{r}\cdot
q_{n_{k}-1}^{2}(u)A_{1}^{n_{k}}\Big{)}^{\tau(1-\epsilon)}$
$\displaystyle=c_{4}\cdot r^{\tau}.$

  3. (3)

When $n=n_{k}$. Replaced $n_{k}-1$ and $A_{1}$ in case (2) by $n_{k}$ and
$A_{2}$ respectively can arrive the same conclusion with the same argument as
in case (2).

###  3.6. Conclusion

Thus by applying the mass distribution principle (Proposition 2.2), it yields
that

$\dim_{H}E(A_{1},A_{2})\geq\frac{{\min\\{s,g\\}}}{1+\varepsilon}.$

Since $\varepsilon>0$ is arbitrary, by letting $N\to\infty$ as then
$M\to\infty$, we arrive at

$\dim_{H}E(A_{1},A_{2})\geq\min\\{s_{A_{1}},\ g_{(A_{1}A_{2}),A_{1}}\\}.$

This finishes the proof.

##  4\.  Hausdorff dimension of $\mathcal{F}_{B_{1},B_{2}}$

In this section we prove Theorem 1.8. There are different possible cases for
$B_{1}$ and $B_{2}$.

###  4.1.  Case $B_{1}^{s_{0}}\leq B_{2}$

To get the upper bound for dimension of $\mathcal{F}_{B_{1},B_{2}}$ recall
that for every $B_{2}$

$\mathcal{F}_{B_{1},B_{2}}\subset\mathcal{E}_{2}=\\{x:a_{n}(x)a_{n+1}(x)\geq
B_{1}^{n}\text{\,\, for i.m. \,\,}n\in\mathbb{N}\\}.$

Hence, by the Theorem 1.2, we have that

$\dim_{H}\mathcal{F}_{B_{1},B_{2}}\leq s_{0}.$

To get a lower bound, recall a set (3.1) from Theorem 1.7 and let
$A_{1}=B_{1}^{s_{0}},A_{1}A_{2}=B_{1}$, so that we have a set

$E=\left\\{x\in[0,1):\begin{split}&\begin{split}&c_{1}B_{1}^{ns_{0}}\leq
a_{n}(x)<2c_{1}B_{1}^{ns_{0}},\\\ &c_{2}B_{1}^{n(1-s_{0})}\leq
a_{n+1}(x)<2c_{2}B_{1}^{n(1-s_{0})},\end{split}&\text{ for i.m.
}n\in\mathbb{N}\\\ &1\leq a_{n}(x)\leq M&\text{for all
other\,\,}n\in\mathbb{N}\end{split}\right\\}.$

Since we have $B_{1}^{s_{0}}\leq B_{2}$ and $s_{0}\geq 1/2$ by Propositon 2.11
in [9], one can easily see that $E$ is indeed a subset of
$\mathcal{F}_{B_{1},B_{2}}$ and by the proof of Theorem 1.7 we know that

$\dim_{H}E\geq\min\\{s_{B_{1}^{s_{0}}},\
g_{B_{1},B_{1}^{s_{0}}}\\}=s_{B_{1}^{s_{0}}}=s_{0},$

so $\dim_{H}\mathcal{F}_{B_{1},B_{2}}\geq s_{0}.$ Hence, we get that when
$B_{1}^{s_{0}}\leq B_{2}$, we have $\dim_{H}\mathcal{F}_{B_{1},B_{2}}=s_{0}.$

###  4.2.  Case $B_{1}^{s_{0}}\geq B_{2}>B_{1}^{1/2}$

For the upper bound a subset

$U_{1}=\left\\{x:1\leq a_{n}(x)\leq B_{2}^{n},\
a_{n+1}(x)\geq\frac{B_{1}^{n}}{a_{n}(x)}\text{\,\, for i.m.
\,}n\in\mathbb{N}\right\\}$

of $\mathcal{F}_{B_{1},B_{2}}.$ This set can be covered by collections of
basic cylinders $J_{n}$ of order $n$:

$\displaystyle U_{1}$
$\displaystyle=\bigcup_{a_{1},\ldots,a_{n-1}\in\mathbb{N}}\left\\{x\in[0,1):a_{k}(x)=a_{k},1\leq
k\leq n-1,1\leq a_{n}(x)\leq
B_{2}^{n},a_{n+1}(x)\geq\frac{B_{1}^{n}}{a_{n}(x)}\right\\}$
$\displaystyle=\bigcup_{a_{1},\ldots,a_{n-1}\in\mathbb{N}}\bigcup_{1\leq
a_{n}\leq
B_{2}^{n}}\bigcup_{a_{n+1}\geq\frac{B_{1}^{n}}{a_{n}}}I_{n+1}(a_{1},\ldots,a_{n+1})$
$\displaystyle=\bigcup_{\begin{subarray}{c}a_{1},\ldots,a_{n-1}\in\mathbb{N}\\\
1\leq a_{n}\leq B_{2}^{n}\end{subarray}}J_{n}(a_{1},\ldots,a_{n}).$

Note that

$J_{n}(a_{1},\ldots,a_{n})=\bigcup_{a_{n+1}\geq\frac{B_{1}^{n}}{a_{n}}}I_{n+1}(a_{1},\ldots,a_{n+1}),$

and

$|J_{n}(a_{1},\ldots,a_{n})|\asymp\frac{1}{B_{1}^{n}a_{n}q_{n-1}^{2}}.$

Now consider the $s$-volume of the cover of $U_{1}$:

$\sum\limits_{a_{1},\ldots,a_{n-1}\in\mathbb{N}}\sum\limits_{1\leq a_{n}\leq
B_{2}^{n}}\left(\frac{1}{B_{1}^{n}a_{n}q_{n-1}^{2}}\right)^{s}\asymp\sum\limits_{a_{1},\ldots,a_{n-1}\in\mathbb{N}}B^{n(1-s)}_{2}\left(\frac{1}{B_{1}^{n}q_{n-1}^{2}}\right)^{s}.$

Therefore, an $s$-dimensional Hausdorff measure of $U_{1}$ can be estimated as

$\mathcal{H}^{s}(E(A_{1},A_{2}))\leq\liminf_{N\to\infty}\sum\limits_{n=N}^{\infty}\sum\limits_{a_{1},\ldots,a_{n-1}\in\mathbb{N}}B^{n(1-s)}_{2}\left(\frac{1}{B_{1}^{n}q_{n-1}^{2}}\right)^{s}.$

Hence, we have that

$\dim_{H}\mathcal{F}_{B_{1},B_{2}}\leq\dim_{H}U_{1}\leq g_{B_{1},B_{2}}.$

To get a lower bound, recall a set (3.1) from Theorem 1.7 and let
$A_{1}=B_{2},\ A_{1}A_{2}=B_{1}$, so that we have

$E=\left\\{x\in[0,1):\begin{split}&\begin{split}&c_{1}B_{2}^{n}\leq
a_{n}(x)<2c_{1}B_{2}^{n},\\\ &c_{2}\frac{B_{1}^{n}}{B_{2}^{n}}\leq
a_{n+1}(x)<2c_{2}\frac{B_{1}^{n}}{B_{2}^{n}},\end{split}&\text{ for i.m.
}n\in\mathbb{N}\\\ &1\leq a_{n}(x)\leq M&\text{for all
other\,\,}n\in\mathbb{N}\end{split}\right\\}.$

Notice that if $B_{2}<B_{1}^{s_{0}}$, we have

$\min\\{s_{B_{2}},g_{B_{1},B_{2}}\\}=g_{B_{1},B_{2}}.$

So in this case we have

$\dim_{H}E\geq g_{B_{1},B_{2}}.$

Since $B_{2}>B_{1}^{1/2}$, or $B_{2}^{2}>B_{1}$, one can easily see that $E$
is indeed a subset of $\dim_{H}\mathcal{F}_{B_{1},B_{2}}$. Hence

$\dim_{H}\mathcal{F}_{B_{1},B_{2}}\geq g_{B_{1},B_{2}}.$

Hence, together with the upper bound, when $B_{1}^{s_{0}}\geq
B_{2}>B_{1}^{1/2}$, we have that
$\dim_{H}\mathcal{F}_{B_{1},B_{2}}=g_{B_{1},B_{2}}$.

###  4.3.  Case $B_{1}^{1/2}\geq B_{2}$

In this case let us investigate the conditions on partial quotients in the
definition of $\mathcal{F}_{B_{1},B_{2}}$. We know that $a_{n+1}<B_{2}^{n}$
and $a_{n}<B_{2}^{n-1}$.
Hence we get

$a_{n}a_{n+1}<B_{2}^{2n-1}.$

But $B_{1}^{1/2}\geq B_{2}$, we come to a contradiction with $a_{n}a_{n+1}\geq
B_{1}^{n}$. Hence in this case the set $\mathcal{F}_{B_{1},B_{2}}$ is empty.

##  5\. Hausdorff dimension of $\mathcal{F}(\Phi_{1},\Phi_{2})$

To make it easier to follow, let us recall the statement of the result. Let us
consider arbitrary positive functions $\Phi_{1}$ and $\Phi_{2}$. Suppose

$\displaystyle\log B_{1}$
$\displaystyle=\liminf\limits_{n\to\infty}\frac{\log\Phi_{1}(n)}{n}\quad\text{and}\quad\log
b_{1}=\liminf\limits_{n\to\infty}\frac{\log\log\Phi_{1}(n)}{n}.$
$\displaystyle\log B_{2}$
$\displaystyle=\lim\limits_{n\to\infty}\frac{\log\Phi_{2}(n)}{n}\quad\ \
\text{and}\quad\log
b_{2}=\lim\limits_{n\to\infty}\frac{\log\log\Phi_{2}(n)}{n}.$

Then

  1. (1)

if $B_{1}^{s_{0}}\leq B_{2}$ and $B_{1}$ is finite, then

$\dim_{H}\mathcal{F}(\Phi_{1},\Phi_{2})=s_{0}.$

  2. (2)

if $B_{1}^{s_{0}}\geq B_{2}>B_{1}^{1/2}$, then

$\dim_{H}\mathcal{F}(\Phi_{1},\Phi_{2})=g_{B_{1},B_{2}}.$

  3. (3)

if $B_{1}^{1/2}>B_{2}$ and $B_{2}$ is finite, then

$\mathcal{F}(\Phi_{1},\Phi_{2})={\diameter}.$

  4. (4)

if $B_{1}=B_{2}=\infty$, $1\leq b_{1}<b_{2}$ and $b_{1}$ is finite, then

$\dim_{H}\mathcal{F}(\Phi_{1},\Phi_{2})=\frac{1}{1+b_{1}}.$

  5. (5)

if $B_{1}=B_{2}=\infty$, $b_{1}>b_{2}\geq 1$ and $b_{2}$ is finite, then

$\mathcal{F}(\Phi_{1},\Phi_{2})={\diameter}.$

  6. (6)

if $B_{1}=B_{2}=\infty$, $b_{1}=b_{2}=\infty$, then

$\text{either\,\, }\mathcal{F}(\Phi_{1},\Phi_{2})={\diameter}\text{\,\, or\,\,
}\dim_{H}\mathcal{F}(\Phi_{1},\Phi_{2})=0.$

###### Proof.

  1. (1)

$B_{1}^{s_{0}}\leq B_{2}$ and $B_{1}$ is finite.
For any $\epsilon>0$, one has

$\Phi_{1}(n)\geq(B_{1}-\epsilon)^{n},\text{\,\, for all \,}n\gg 1.$

Thus

$\mathcal{F}(\Phi_{1},\Phi_{2})\subset\left\\{x\in[0,1):a_{n}(x)a_{n+1}(x)\geq(B_{1}-\epsilon)^{n},\,\,\text{
i.m. n }\in\mathbb{N}\right\\}.$

By Theorem 1.2 and continuity of dimensional number established in [9], we
have

$\dim_{H}\mathcal{F}(\Phi_{1},\Phi_{2})\leq s_{0}.$

As for the lower bound, fix $\widetilde{B_{1}}>B_{1}$ and
$\widetilde{B_{2}}<B_{2}$, where we take finite $\widetilde{B_{2}}$
independently of whether $B_{2}$ is finite or not. Then by the definition of
$B_{1}$, we can choose a largely sparse integer sequence $\\{n_{j}\\}_{j\geq
1}$ such that for all $j\geq 1$,

$\Phi_{1}(n_{j})\leq\widetilde{B_{1}}^{n_{j}}.$

By the definition of $B_{2}$ one has

$\Phi_{2}(n)\geq\widetilde{B_{2}}^{n},\text{\,\, for all \,}n\gg 1.$

So we can see that

$\mathcal{F}(\Phi_{1},\Phi_{2})\supset\left\\{x\in[0,1):\begin{split}a_{n_{j}}(x)a_{n_{j}+1}(x)&\geq\widetilde{B_{1}}^{n_{j}}&\text{for
all }j\in\mathbb{N}\\\ a_{n+1}(x)&<\widetilde{B_{2}}^{n}&\text{for all
sufficiently large }n\in\mathbb{N}\end{split}\right\\}.$

In the definition of (3.1) we required that $n_{k+1}-n_{k}-2$ is a multiple of
some large integer $N$. Here we cannot guarantee this. However, we can define
a slightly different subset as $\widetilde{E}$.
Still fix a large integer $N$ and an integer $M$. Let

$n_{k+1}-n_{k}-2=\ell_{k+1}N+i_{k+1},\text{\,\, for some }0\leq i_{k+1}<N.$

Denote

$y_{k}=n_{k-1}+1+\ell_{k}N.$

Now we consider the following set, which is the subset of our set under
consideration:

$\displaystyle\widetilde{E}=\Bigg{\\{}x\in[0,1):\,$ $\displaystyle
c_{1}\widetilde{B_{1}}^{s_{0}n_{j}}\leq
a_{n_{j}}(x)<2c_{1}\widetilde{B_{1}}^{s_{0}n_{j}},c_{2}\left(\frac{\widetilde{B_{1}}}{\widetilde{B_{1}}^{s_{0}}}\right)^{n_{j}}\leq
a_{n_{j}+1}(x)<2c_{2}\left(\frac{\widetilde{B_{1}}}{\widetilde{B_{1}}^{s_{0}}}\right)^{n_{j}},$
$\displaystyle\alpha_{n}(x)=2\text{\,\, for all }y_{j}<n<n_{j},j\geq
1,\text{\,\, and }1\leq a_{n}(x)\leq M\text{\,\, for other
}n\in\mathbb{N}\Bigg{\\}}.$

The rest of the argument can be carried out with no more changes to conclude
that

$\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})\geq
s_{\widetilde{B_{1}}^{s_{0}}}.$

By the continuity of $s_{B}$ and using the upper bound, we have
$\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})=s_{0}$.

  2. (2)

$B_{1}^{s_{0}}\geq B_{2}>B_{1}^{1/2}$.
To get the upper bound for dimension of $\mathcal{F}(\Phi_{1},\Phi_{2})$ in
this case, we can take the following set $U_{2}$, which is a superset of
$\mathcal{F}(\Phi_{1},\Phi_{2})$:

$U_{2}=\left\\{x:1\leq a_{n}(x)\leq(B_{2}+\epsilon_{2})^{n},\
a_{n+1}(x)\geq\frac{(B_{1}-\epsilon_{1})^{n}}{a_{n}(x)}\text{\,\, for i.m.
\,\,}n\in\mathbb{N}\right\\}.$

Now one can repeat the argument from the subsection (4.2). By the continuity
of $g_{B_{1},B_{2}}$ with respect to $B_{1}$ and $B_{2}$ we get that

$\dim_{H}\mathcal{F}(\Phi_{1},\Phi_{2})\leq g_{B_{1},B_{2}}.$

For the lower bound, we can completely repeat the argument from the lower
bound in last case, but for a slightly different set, namely

$\displaystyle\widetilde{E}=\Bigg{\\{}x\in[0,1):$
$\displaystyle\widetilde{B_{2}}^{n_{j}}c_{1}\leq
a_{n_{j}}(x)<2c_{1}\widetilde{B_{2}}^{n_{j}},c_{2}\left(\frac{\widetilde{B_{1}}}{\widetilde{B_{2}}}\right)^{n_{j}}\leq
a_{n_{j}+1}(x)<2c_{2}\left(\frac{\widetilde{B_{1}}}{\widetilde{B_{2}}}\right)^{n_{j}},$
$\displaystyle a_{n}(x)=2\text{\,\, for all }y_{j}<n<n_{j},j\geq 1,\text{\,\,
and }1\leq a_{n}(x)\leq M\text{\,\, for other }n\in\mathbb{N}\Bigg{\\}}.$

By the continuity of $g_{B_{1},B_{2}}$ in corollary 2.11 and using the upper
bound, we have
$\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})=g_{B_{1},B_{2}}$.

  3. (3)

$B_{1}^{1/2}>B_{2}$ and $B_{2}$ is finite.
If $B_{1}=+\infty$, then obviously our set is empty. Hence we can assume that
$B_{1}<\infty$. By definition of $B_{1}$ and $B_{2}$ we know that
$\Phi_{1}(n)\geq(B_{1}-\varepsilon_{1})^{n}$ for $n$ large enough with any
$\varepsilon_{1}>0$ and $\Phi_{2}(n)\leq(B_{2}+\varepsilon_{2})^{n}$ for $n$
large enough with any $\varepsilon_{2}>0$.
By conditions of our set we know that

$a_{n+1}<\Phi_{2}(n)\leq(B_{2}+\varepsilon_{2})^{n}\text{\,\,\, and
\,\,\,}a_{n}<\Phi_{2}(n)\leq(B_{2}+\varepsilon_{2})^{n-1}.$

Combining these together, we have that
$a_{n}a_{n+1}<(B_{2}+\varepsilon_{2})^{2n-1}$. On the other hand, the second
condition of our set is

$a_{n+1}a_{n}\geq\Phi_{1}(n)\geq(B_{1}-\varepsilon_{1})^{n}.$

In our case we know that $B_{1}>B_{2}^{2}$, or we can rewrite it as
$B_{1}=B^{2}_{2}+\delta$ for some fixed $\delta>0$. Let
$\delta^{\prime}=\sqrt{B_{2}^{2}+\delta/2}-B_{2}$ and take
$\varepsilon_{1}<\delta/2$ and $\varepsilon_{2}<\delta^{\prime}$. Now we
easily come to a contradiction between conditions, since

$(B_{1}-\varepsilon_{1})^{n}=(B_{2}^{2}+\delta-\varepsilon_{1})^{n}>(B_{2}^{2}+\delta/2)^{n}=(B_{2}+\delta^{\prime})^{2n}>(B_{2}+\varepsilon_{2})^{2n-1},$

which contradicts our conditions. Hence our set is empty.

  4. (4)

$B_{1}=B_{2}=\infty$, $1\leq b_{1}<b_{2}$ and $b_{1}$ is finite.
First, let us assume that $b_{1}=1$. Note that $B=\infty$, therefore for any
$C>1$, $\Phi_{1}(n)\geq C^{n}$ for all sufficiently large $n\in\mathbb{N}$.
Thus one can see that

$\mathcal{F}(\Phi_{1},\Phi_{2})\subset\\{x\in[0,1):a_{n}(x)\geq C^{n}\text{
\,\, i.m. }n\in\mathbb{N}\\}.$

Then by Theorem 1.1 we have

$\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})\leq\lim_{C\to\infty}s_{C}=\frac{1}{2}.$

Now for the lower bound notice that $b_{2}=1+\delta$ for some fixed $\delta>0$
and that for every $0<\epsilon<\frac{\delta}{2}$ one has

$\Phi_{1}(n)\leq e^{(1+\epsilon)^{n}}\text{\,\, and \,\, }\Phi_{2}(n)\geq
e^{(1+\delta-\epsilon)^{n}}.$

Now it is obvious that

$\mathcal{F}(\Phi_{1},\Phi_{2})\supset\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
e^{(1+\epsilon)^{n}}&\text{for i.m. }n\in\mathbb{N}\\\
a_{n+1}(x)&<e^{(1+\delta-\epsilon)^{n}}&\text{for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

At this point we will make use of Lemma 2.6. Set
$s_{n}=\frac{1}{2}e^{(1+\epsilon)^{n}}$. One can easily check that

$U_{3}=\\{x\in[0,1):s_{n}\leq a_{n+1}<2s_{n}\,\,\,\forall n\in\mathbb{N}\\}$

is a subset of our original set.
Applying Lemma 2.6 to this choice of the sequence $s_{n}$, we get

$\displaystyle\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})$
$\displaystyle\geq\lim_{\epsilon\to
0}\left(\liminf_{n\to\infty}\frac{\log(s_{1}s_{2}\cdots
s_{n})}{2\log(s_{1}s_{2}\cdots s_{n})+\log s_{n+1}}\right)$
$\displaystyle=\lim_{\epsilon\to
0}\left(\liminf_{n\to\infty}\frac{1-\frac{1}{(1+\epsilon)^{n}}}{2+\epsilon-\frac{2}{(1+\epsilon)^{n}}}\right)$
$\displaystyle=\lim_{\epsilon\to 0}\frac{1}{2+\epsilon}=\frac{1}{2}.$

Now let $b_{1}>1$. By the definition of $b_{1}$, for any
$\epsilon>0,\frac{\log\log\Phi_{1}(n)}{n}\leq\log(b_{1}+\epsilon)$ that is
$\Phi_{1}(n)\leq e^{(b_{1}+\epsilon)^{n}}$ for infinitely many
$n\in\mathbb{N}$, whereas $\Phi_{1}(n)\geq e^{(b_{1}-\epsilon)^{n}}$ for all
sufficiently large $n\in\mathbb{N}$. Note that

$\mathcal{F}(\Phi_{1},\Phi_{2})\subset\left\\{x\in[0,1):a_{n}(x)\geq
e^{\frac{1}{2(b_{1}-\epsilon)}(b_{1}-\epsilon)^{n}}\text{ \,\, for i.m.
}n\in\mathbb{N}\right\\}.$

Therefore, by using Lemma 2.5 we get an upper bound

$\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})\leq\lim_{\epsilon\to
0}\frac{1}{1+b_{1}-\epsilon}=\frac{1}{1+b_{1}}.$

Now for the lower bound we use Lemma 2.6 once again. Set
$s_{n}=\frac{1}{2}e^{(b_{1}+\epsilon)^{n}}$. One can easily check that

$U_{4}=\\{x\in[0,1):s_{n}\leq a_{n+1}(x)<2s_{n}\,\,\,\forall
n\in\mathbb{N}\\}$

is a subset of our original set. Applying Lemma 2.6 to this choice of the
sequence $s_{n}$, we get

$\displaystyle\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})$
$\displaystyle\geq\lim_{\epsilon\to
0}\left(\liminf_{n\to\infty}\frac{\log(s_{1}s_{2}\cdots
s_{n})}{2\log(s_{1}s_{2}\cdots s_{n})+\log s_{n+1}}\right)$
$\displaystyle=\lim_{\epsilon\to
0}\left(\liminf_{n\to\infty}\frac{1-\frac{1}{(b_{1}+\epsilon)^{n}}}{b_{1}+\epsilon+1-\frac{2}{(b_{1}+\epsilon)^{n}}}\right)$
$\displaystyle=\lim_{\epsilon\to 0}\frac{1}{1+b_{1}+\epsilon}$
$\displaystyle=\frac{1}{1+b_{1}}.$

  5. (5)

if $B_{1}=B_{2}=\infty$, $b_{1}>b_{2}\geq 1$ and $b_{2}$ is finite.
We want to show that in this case our set is empty. Let us first assume that
$b_{1}$ is finite. Then $b_{1}=b_{2}+\delta$ for some fixed $\delta>0$. Next,
by the conditions of our set, for every $\epsilon_{2}>0$ one has

$a_{n}<\Phi_{2}(n)\leq e^{(b_{2}+\epsilon_{2})^{n-1}},\
a_{n+1}<\Phi_{2}(n)\leq e^{(b_{2}+\epsilon_{2})^{n}}.$

Combining these together, we have that

$a_{n}a_{n+1}<e^{(b_{2}+\epsilon_{2})^{n-1}+(b_{2}+\epsilon_{2})^{n}}.$

On the other hand, the second condition of our set is that for every
$\epsilon_{1}>0$ one has

$a_{n}a_{n+1}\geq\Phi_{1}(n)\geq
e^{(b_{1}-\epsilon_{1})^{n}}=e^{(b_{2}+\delta-\epsilon_{1})^{n}}.$

We can take $\epsilon_{1}$ and $\epsilon_{2}$ small enough, so that

$\displaystyle\log a_{n}a_{n+1}$
$\displaystyle\geq(b_{2}+\delta-\epsilon_{1})^{n}>(b_{2}+\delta/2)^{n}$
$\displaystyle>2(b_{2}+\epsilon_{2})^{n}>(b_{2}+\epsilon_{2})^{n-1}+(b_{2}+\epsilon_{2})^{n}>\log
a_{n}a_{n+1}.$

This clearly gives us a contradiction for $n$ large enough, because $\delta$
is a fixed positive number. Hence our set is empty. Now if $b_{1}=+\infty$,
then simply for some finite $b_{1}^{\prime}>b_{2}^{2}$ one has

$a_{n}a_{n+1}\geq\Phi_{1}(n)\geq e^{(b_{1}^{\prime}-\epsilon_{1})^{n}},$

and then proceed like for finite $b_{1}$ earlier.

  6. (6)

$B_{1}=B_{2}=\infty$, $b_{1}=b_{2}=\infty$.

From the definition of $b_{1}$, for every $C>1$ we have $\Phi_{1}(n)\geq
e^{C^{n}}$. So we can consider the following superset of our set
$\mathcal{F}(\Phi_{1},\Phi_{2})$:

$\mathcal{F}(\Phi_{1},\Phi_{2})\subset\left\\{x\in[0,1):a_{n}(x)\geq
e^{C^{n}}\text{ \,\, i.m. }n\in\mathbb{N}\right\\}.$

Then by Lemma 2.5

$\dim_{\mathrm{H}}\mathcal{F}(\Phi_{1},\Phi_{2})\leq\lim_{C\to\infty}\frac{1}{C+1}=0.$

So if the set is non-empty, then its dimension is 0. However, it can happen
that under conditions of this case the set $\mathcal{F}(\Phi_{1},\Phi_{2})$ is
empty. Recall the conditions of our set,

$a_{n+1}<\Phi_{2}(n),\,a_{n}<\Phi_{2}(n-1)\text{\,\, and \,\,
}a_{n}a_{n+1}\geq\Phi_{1}(n).$

So if for all $n$ large enough

$\Phi_{2}(n)\Phi_{2}(n-1)\leq\Phi_{1}(n),$

then the set $\mathcal{F}(\Phi_{1},\Phi_{2})$ is empty because conditions are
incompatible.

∎

##  6\. Proofs of the propositions

In this section we prove Propositions 1.5 and 1.6.

###  6.1. Proof of Proposition 1.5

Note that this proposition concerns the case $B_{1}=B_{2}^{2}$. Recall that

$E_{1}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
B_{2}^{2n}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<B_{2}^{n}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

Then simply by Theorem 1.8, this set is empty.
Our second set is

$E_{2}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
B_{2}^{2n-1}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<3B_{2}^{n+1}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

To get the lower bound on the Hausdorff dimension, we take the subset

$\displaystyle F=\\{x\in[0,1)\,:\,B_{2}^{n}\leq a_{n}(x)<2B_{2}^{n},$
$\displaystyle\ B_{2}^{n}\leq a_{n+1}(x)<2B_{2}^{n},\,\,\text{i.m.
}n\in\mathbb{N};$ $\displaystyle\text{ and }1\leq a_{n}(x)\leq M\text{\,\, for
all other\,\,}n\in\mathbb{N}\\}.$

It is indeed a subset of $E_{2}$, because for the first condition of $E_{2}$
we have

$a_{n}(x)a_{n+1}(x)\geq B_{2}^{n}B_{2}^{n}=B_{2}^{2n}>B_{2}^{2n-1},$

and for the second one we have

$a_{n+1}(x)<2B_{2}^{n}<3B_{2}^{n+1},$ $a_{n}(x)<2B_{2}^{n}<3B_{2}^{n}.$

We dealt with set $F$ in Theorem 1.7, which tells us that its dimension is
$g_{B_{1},B_{2}}$. So, we have that

$\dim_{\mathrm{H}}E_{2}\geq g_{B_{1},B_{2}}.$

The upper bound of dimension for the set $E_{2}$ is clear by the result for
the case $B_{1}^{1/2}<B_{2}$ in Theorem 1.8, since $E_{2}$ is becoming larger
when we decrease the value of $B_{1}$. And so
$\dim_{\mathrm{H}}E_{2}=g_{B_{1},B_{2}}.$

###  6.2. Proof of Proposition 1.6

Note that this proposition concerns the case $B_{1}=B_{2}=\infty$,
$b_{1}=b_{2}<\infty$. The first set in consideration is

$P_{1}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
e^{b_{1}^{n}}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<e^{b_{1}^{n}}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

This set is a particular case of the set $\mathcal{F}(\Phi)$ from Theorem 1.3
and its Hausdorff dimension by this theorem is equal to $\frac{1}{b_{1}+1}$.
The second set is

$P_{2}=\left\\{x\in[0,1):\begin{split}a_{n}(x)a_{n+1}(x)&\geq
e^{5b_{1}^{n}}\text{\,\, for infinitely many }n\in\mathbb{N}\\\
a_{n+1}(x)&<e^{b_{1}^{n-1}}\text{\,\, for all sufficiently large
}n\in\mathbb{N}\end{split}\right\\}.$

This set is empty, because the conditions are incompatible. Indeed, on the one
hand we have

$a_{n}(x)a_{n+1}(x)<e^{b_{1}^{n-1}+b_{1}^{n-2}}.$

On the other hand,

$a_{n}(x)a_{n+1}(x)\geq e^{5b_{1}^{n}}.$

Combining this together, we get $5b_{1}^{2}<b_{1}+1$, which can happen only
when $b_{1}$ is less than 1. But we know that $b_{1}\geq 1$, hence the set
$P_{2}$ is empty.

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