1 Introduction
Let
$T:[0,1)\to [0,1)$
be the Gauss map defined by
$$ \begin{align*} T(0)= 0, \quad T(x)= \frac{1}{x} \pmod1 \quad\text{for } x \in (0,1).\end{align*} $$
Every irrational number
$x\in [0,1)$
can be uniquely expanded into an infinite form
$$ \begin{align} x:=\frac{1}{a_{1}(x)+\dfrac{1}{a_{2}(x)+\dfrac{1}{\ddots+\dfrac{1}{a_{n}+T^{n}(x)}}}} =\frac{1}{a_{1}(x)+\dfrac{1}{a_{2}(x)+\dfrac{1}{a_{3}(x)+\dfrac{1}{\ddots}}}}, \end{align} $$
where
$a_{n}(x)=\lfloor 1/T^{n-1}(x)\rfloor $
are called the partial quotients of
$x.$
(Here
$\lfloor \cdot \rfloor $
denotes the greatest integer less than or equal to a real number and
$T^0$
denotes the identity map.) For simplicity of notation, we write (1.1) as
$$ \begin{align*} x=[a_{1}(x),a_{2}(x),\ldots,a_{n}(x)+T^{n}(x)]=[a_{1}(x),a_{2}(x),a_{3}(x),\ldots]. \end{align*} $$
It is clear that the Gauss transformation T acts as the shift map on the continued fraction system. That is, for each
$x=[a_1(x),a_2(x),a_3(x),\ldots ]\in [0,1)\cap \mathbb {Q}^{c},$
$$ \begin{align*}T(x)=T([a_{1}(x),a_{2}(x),a_3(x),\ldots])=[a_{2}(x),a_{3}(x),\ldots].\end{align*} $$
Gauss observed that T is measure-preserving and ergodic with respect to the Gauss measure
$\mu $
defined by
$$ \begin{align*}d\mu=\frac{1}{\log 2}\frac{1}{x+1}dx.\end{align*} $$
For more information on the continued fraction expansion, see [Reference Khintchine3].
The metrical theory of continued fractions, which concerns the properties of the partial quotients for almost all
$x\in [0,1),$
is one of the major themes in the study of continued fractions. Wang and Wu [Reference Wang and Wu7] considered the metrical properties of the maximal run-length function
$$ \begin{align*}R_{n}(x)=\max\{l\in \mathbb{N}: a_{i+1}(x)=\cdots= a_{i+l}(x) \ \text{for some}\ i \mbox{ with } 0\leq i \leq n-l\},\end{align*} $$
which counts the longest run of the same symbol among the first n partial quotients of x. They proved that, for
$\mu $
almost all
$x\in [0,1),$
$$ \begin{align*}\lim_{n\rightarrow\infty}\frac{R_{n}(x)}{\log n}=\frac{1}{2\log({(\sqrt{5}+1)}/{2})}.\end{align*} $$
Song and Zhou [Reference Song and Zhou6] gave a more subtle characterisation of the function
$R_{n}(x).$
In this paper, we continue the study by considering the shortest distance function
$$ \begin{align*} M_{n,m}(x_{1},\ldots,x_{m})=\max\{k\in \mathbb{N}: a_{i+j}(x_{1}) = & \cdots= a_{i+j}(x_{m}) \ \text{for}~ j=1,\ldots,k, \\&\ \text{and some}\ i \mbox{ with } 0\leq i \leq n-k\}. \end{align*} $$
This is motivated by the behaviour of the shortest distance between two orbits,
$$ \begin{align*}S_{n,2}(x,y)=\min_{i=0,\ldots,n-1}(d(T^{i}(x),T^{i}(y)))\end{align*} $$
in the continued fraction system. Shi et al. [Reference Shi, Tan and Zhou5] proved that, for
$\mu ^{2}$
almost all
$(x,y)\in [0,1)\times [0,1),$
$$ \begin{align*}{H_{2}}\cdot\lim_{n\rightarrow\infty}\frac{M_{n,2}(x,y)}{\log n}=\lim_{n\rightarrow\infty}\frac{-\log S_{n,2}(x,y)}{\log n},\end{align*} $$
where
$H_{2}$
is the Rényi entropy defined by (1.2). Investigating the shortest distance between two orbits amounts to estimating the longest common substrings between two sequences of partial quotients. In fact, [Reference Shi, Tan and Zhou5] focused on the asymptotics of the length of the longest common substrings in two sequences of partial quotients.
For
$n\geq 1$
and
$(a_{1},\ldots ,a_{n})\in \mathbb {N}^{n}$
, we call
$$ \begin{align*}I_{n}(a_{1},\ldots,a_{n})=\{x\in[0,1): a_{1}(x)=a_{1},\ldots, a_{n}(x)=a_{n}\}\end{align*} $$
an nth cylinder. For
$m\geq 2$
, we define the generalised Rényi entropy with respect to the Gauss measure
$\mu $
by
$$ \begin{align} {H}_{m}=\lim_{n\rightarrow\infty}\frac{-\log \sum_{(a_1,\ldots,a_n)\in \mathbb{N}^{n}} \mu(I_{n}(a_1,\ldots,a_n))^{m}}{(m-1)n}. \end{align} $$
The existence of the limit (1.2) for the Gauss measure
$\mu $
was established in [Reference Haydn and Vaienti2].
Theorem 1.1. For
$\mu ^{m}$
-almost all
$(x_{1},\ldots ,x_{m})\in [0,1)^{m},$
$$ \begin{align*}\lim_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\frac{1}{(m-1)H_{m}}.\end{align*} $$
Here we use the convention that
${1}/{0}=\infty $
and
${1}/{\infty }=0.$
It is natural to study the exceptional set in this limit theorem. We define the exceptional set as
$$ \begin{align*}\widetilde{E}=\bigg\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: \liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n} <\limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\bigg\}\end{align*} $$
and the level set as
$$ \begin{align*}E(\alpha)=\bigg\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: \lim_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\alpha\bigg\}.\end{align*} $$
Throughout the paper,
$\dim _{H}A$
denotes the Hausdorff dimension of the set
$A.$
Theorem 1.2. For any
$\alpha $
with
$0\leq \alpha \leq \infty ,$
$\dim _{H}\widetilde {E}=\dim _{H}E(\alpha )=m.$
In fact, Theorem 1.2 follows immediately from the following more general result. For any
$0\leq \alpha \leq \beta \leq \infty ,$
set
$$ \begin{align*} E(\alpha,\beta)=\bigg\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: & \liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\alpha, \\ & \limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}=\beta\bigg\}. \end{align*} $$
Theorem 1.3. For any
$\alpha ,\beta $
with
$0\leq \alpha \leq \beta \leq \infty ,$
$\dim _{H}E(\alpha ,\beta )=m.$
2 Preliminaries
In this section, we fix some notation and recall some basic properties of continued fraction expansions. A detailed account of continued fractions can be found in Khintchine’s book [Reference Khintchine3].
For any irrational number
$x\in [0,1)$
with continued fraction expansion (1.1), we denote by
$$ \begin{align*}\frac{p_{n}(x)}{ q_{n}(x)} :=[a_{1}(x),\ldots,a_{n}(x)]\end{align*} $$
the nth convergent of
$x.$
With the conventions
$$ \begin{align*}p_{-1}(x)=1, \quad q_{-1}(x)=0, \quad p_{0}(x)=0, \quad q_{0}(x)=1,\end{align*} $$
we have, for any
$n\ge 0,$
$$ \begin{align*}p_{n+1}(x)=a_{n+1}(x) p_{n}(x)+p_{n-1}(x), \quad q_{n+1}(x)=a_{n+1}(x) q_{n}(x)+q_{n-1}(x).\end{align*} $$
Obviously,
$q_{n}(x)$
is determined by the first n partial quotients
$a_{1}(x),\ldots ,a_{n}(x).$
So we also write
$q_{n}(a_{1}(x),\ldots ,a_{n}(x))$
in place of
$q_{n}(x).$
If no confusion is likely to arise, we write
$a_{n}$
and
$q_{n}$
in place of
$a_{n}(x)$
and
$q_{n}(x),$
respectively.
Proposition 2.1 [Reference Khintchine3].
For
$n\geq 1$
and
$(a_{1},\ldots ,a_{n})\in \mathbb {N}^{n}$
:
-
(1)
$ q_n\ge 2^{(n-1)/2}$
and
$$ \begin{align*} \prod^{n}_{k=1}a_{k}\leq q_n \leq \prod^{n}_{k=1}(a_{k}+1)\leq 2^{n}\prod^{n}_{k=1}a_{k};\end{align*} $$
-
(2) the length of
$I_{n}(a_{1},\ldots ,a_{n})$
satisfies
$$ \begin{align*} \frac{1}{2q_n^2} \leq |I_{n}(a_{1},\ldots,a_{n})|=\frac{1}{(q_n+q_{n-1})q_n} \leq \frac{1}{q_n^2}. \end{align*} $$
The following
$\psi $
-mixing property is essential in proving Theorem 1.1.
Lemma 2.2 [Reference Philipp4].
For any
$k\geq 1,$
let
$\mathbb {B}^{k}_{1}=\sigma (a_{1},\ldots ,a_{k})$
and let
$\mathbb {B}^{\infty }_{k}=\sigma (a_{k},a_{k+1},\ldots )$
denote the
$\sigma $
-algebras generated by the random variables
$(a_{1},\ldots ,a_{k})$
and
$(a_{k},a_{k+1},\ldots )$
respectively. Then, for any
$E\in \mathbb {B}^{k}_{1}$
and
$F\in \mathbb {B}^{\infty }_{k+n},$
$$ \begin{align*}\mu(E\cap F)=\mu(E)\cdot \mu(F)(1+\theta \rho^n),\end{align*} $$
where
$|\theta |\leq K, \rho <1$
and
$K, \rho $
are positive constants independent of
$E, F, n$
and
$k.$
To estimate the measure of a limsup set in a probability space, the following lemma is widely used.
Lemma 2.3 (Borel–Cantelli lemma).
Let
$(\Omega , \mathcal {B}, \nu )$
be a finite measure space and let
$\{A_n\}_{n\ge 1}$
be a sequence of measurable sets. Define
$A=\bigcap _{N=1}^{\infty }\bigcup _{n=N}^{\infty }A_n.$
Then
$$ \begin{align*} \nu(A)=\begin{cases} 0 & \text{if } \displaystyle\sum_{n= 1}^{\infty}\nu(A_n)<\infty, \\[3pt] \nu(\Omega) & \displaystyle \text{if }\sum_{n= 1}^{\infty}\nu(A_n)=\infty \text{ and } \{A_n\}_{n\ge 1}\text{ are pairwise independent}. \end{cases} \end{align*} $$
Let
$\mathbb {K}=\{k_{n}\}_{n\geq 1}$
be a subsequence of
$\mathbb {N}$
that is not cofinite. Define a mapping
$\phi _{\mathbb {K}}:[0,1)\cap \mathbb {Q}^{c}\rightarrow [0,1)\cap \mathbb {Q}^{c}$
as follows. For each
$x=[a_1,a_2,\ldots ]\in [0,1)\cap \mathbb {Q}^{c},$
put
$\phi _{\mathbb {K}}(x)=\overline {x}=[c_1,c_2,\ldots ],$
where
$[c_1,c_2,\ldots ]$
is obtained by eliminating all the terms
$a_{k_{n}}$
from the sequence
$a_1, a_2,\ldots .$
Let
$\{b_{n}\}_{n\geq 1}$
be a sequence with
$b_{n}\in \mathbb {N},n\geq 1.$
Write
$$ \begin{align*}E(\mathbb{K},\{b_{n}\})=\{x\in[0,1)\cap \mathbb{Q}^{c}: a_{k_{n}}(x)= b_{n} \ \text{for all}~ n\geq1\}.\end{align*} $$
Lemma 2.4 [Reference Song and Zhou6].
Assume that
$\{b_{n}\}_{n\geq 1}$
is bounded. If the sequence
$\mathbb {K}$
is of density zero in
$\mathbb {N},$
that is,
$$ \begin{align*}\lim_{n\rightarrow\infty}\frac{\sharp\{i\leq n: i\in\mathbb{K}\}}{n}=0,\end{align*} $$
where
$\sharp $
denotes the number of elements in a set, then
$$ \begin{align*}\dim_{H}E(\mathbb{K},\{b_{n}\})=\dim_{H}\phi_{\mathbb{K}}(E(\mathbb{K},\{b_{n}\}))=1.\end{align*} $$
We close this section by citing Marstrand’s product theorem.
Lemma 2.5 [Reference Falconer1].
If
$E,F\subset \mathbb {R}^{d}$
for some
$d,$
then
$\dim _{H}(E\times F)\geq \dim _{H}E+\dim _{H}F.$
3 Proof of Theorem 1.1
Theorem 1.1 can be proved from the following two propositions.
Proposition 3.1. For
$\mu ^{m}$
-almost all
$(x_{1},\ldots ,x_{m})\in [0,1)^{m},$
$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\leq\frac{1}{(m-1)H_{m}}.\end{align*} $$
Proof. We can assume that
$H_{m}>0$
(the case
$H_m=0$
is obvious). Fix
$s_1< s_2< (m-1)H_{m}.$
By the definition of the
$H_{m},$
$$ \begin{align} \sum_{(a_{1},\ldots,a_{n})\in\mathbb{N}^{n}}\mu(I_{n}(a_{1},\ldots,a_{n}))^{m}<\exp\bigg\{-\frac{s_1+s_2}{2}n\bigg\} \end{align} $$
for sufficiently large n. Set
$u_{n}=\lfloor {\log n}/{s_1}\rfloor .$
Note that, for any
$(x_{1},\ldots ,x_{m})\in [0,1)^{m}$
with
$M_{n,m}(x_{1},\ldots ,x_{m})=k$
, there exists i with
$0\leq i\leq n-k$
such that
$$ \begin{align*} a_{i+j}(x_{1})=\cdots=a_{i+j}(x_{m})\end{align*} $$
for
$j=1,\ldots ,k.$
We deduce
$$ \begin{align*} &\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots,x_{m})>u_{n}\})\\ &\quad =\sum_{k=u_{n}+1}^{\infty}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots,x_{m})=k\})\\ &\quad \leq\sum_{k=u_{n}+1}^{\infty}\sum_{i=0}^{n-k}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: a_{i+j}(x_{1})=\cdots=a_{i+j}(x_{m}),\ j=1,\ldots,k\}). \end{align*} $$
By the invariance of
$\mu $
under
$T,$
it follows that
$$ \begin{align*} &\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots,x_{m})>u_{n}\})\\&\quad\leq n \sum_{k=u_{n}+1}^{\infty}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: a_{j}(x_{1})=\cdots=a_{j}(x_{m}),\ j=1,\ldots,k\})\\&\quad= n\sum_{k=u_{n}+1}^{\infty}\sum_{(a_{1},\ldots,a_{k})\in \mathbb{N}^{k}}\mu(I_{k}(a_{1},\ldots,a_{k}))^{m}\\&\quad\leq Cn\cdot\exp\bigg\{-\frac{s_1+s_2}{2}(u_{n}+1)\bigg\} \qquad (\text{by } (3.1))\\&\quad\leq Cn^{-{(s_2-s_1)}/{2s_1}}, \end{align*} $$
where
$C=\sum _{k=1}^{\infty }\exp \{-{(s_1+s_2)k}/{2}\}.$
Choose an infinite subsequence of integers
$\{n_{k}\}_{k\geq 1},$
where
$n_{k}=k^{L}$
and
$L\cdot {(s_2-s_1)}/{2s_1}>1.$
Then
$$ \begin{align*} \sum_{k=1}^{\infty}\mu^{m}(\{(x_{1},\ldots,x_{m})\in [0,1)^{m}: M_{n_{k},m}(x_{1},\ldots,x_{m})>u_{n_{k}}\})<\infty. \end{align*} $$
From the Borel–Cantelli Lemma 2.3, for almost all
$(x_{1},\ldots ,x_{m})\in [0,1)^{m},$
$$ \begin{align*}M_{n_{k},m}(x_{1},\ldots,x_{m})\leq u_{n_{k}} \end{align*} $$
for sufficiently large k. Thus,
$$ \begin{align*} \limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n} \leq&\limsup_{k\rightarrow\infty}\frac{M_{n_{k+1},m}(x_{1},\ldots,x_{m})}{\log n_{k}}\\ \leq&\limsup_{k\rightarrow\infty}\frac{M_{n_{k+1},m}(x_{1},\ldots,x_{m})}{\log n_{k+1}}\cdot\limsup_{k\rightarrow\infty}\frac{n_{k+1}}{n_{k}} \leq\frac{1}{s_{1}}. \end{align*} $$
Therefore, by the arbitrariness of
$s_{1},$
$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\leq\frac{1}{(m-1)H_{m}}.\end{align*} $$
This completes the proof.
Proposition 3.2. For
$\mu ^{m}$
-almost all
$(x_{1},\ldots ,x_{m})\in [0,1)^{m},$
$$ \begin{align*}\liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\geq\frac{1}{(m-1)H_{m}}.\end{align*} $$
Proof. We can assume that
$H_{m}<\infty $
(the case
$H_m=\infty $
is obvious). For
$1\le d<n$
, set
$$ \begin{align*} M_{[d,n]}(x_{1},\ldots,x_{m})=\max\{k\in \mathbb{N}: a_{i+j}(x_{1}) & =\cdots= a_{i+j}(x_{m}) \ \text{for}~ j=1,\ldots,k, \\&\ \text{and for some } i \mbox{ with } d-1\leq i \leq n-k\}. \end{align*} $$
We denote
$\{(x_{1},\ldots ,x_{m})\in [0,1)^{m}: M_{n,m}(x_{1},\ldots ,x_{m})<k\}$
by
$\{M_{n,m}<k\}$
for brevity.
For any
$s>(m-1)H_{m},$
by the definition of the
$H_{m},$
$$ \begin{align} \sum_{(a_{1},\ldots,a_{n}) \in\mathbb{N}^{n}}\mu(I_{n}(a_{1},\ldots,a_{n}))^{m}>\exp\bigg\{-\frac{s+(m-1)H_{m}}{2}n\bigg\} \end{align} $$
for sufficiently large n. Let
$u_{n}=\lfloor {\log n}/{s}\rfloor $
and
$l_{n}=\lfloor {n}/{u_{n}^{2}}\rfloor .$
Then
$$ \begin{align*} \begin{aligned} \{M_{n,m}<u_{n}\}&\subset \{M_{[iu_{n}^{2}+1,iu_{n}^{2}+u_{n}]}<u_{n} : 0\leq i <l_{n}\}\\& \subset\{M_{u_{n},m}<u_{n}\}\cap(\underbrace{T\times\cdots \times T}_{m \text{ times }})^{-u_{n}^{2}}\{M_{[iu_{n}^{2}+1,i\mu_{n}^{2}+u_{n}]}<u_{n} : 0\leq i <l_{n}-1\}. \end{aligned} \end{align*} $$
By Lemma 2.2, it follows that
$$ \begin{align*} & \mu^{m}(\{M_{n,m}<u_{n}\}) \\& \quad \leq \mu^{m}(\{M_{u_{n},m}<u_{n}\}\cap(\underbrace{T\times\cdots \times T}_{m \text{ times }})^{-u_{n}^{2}}\{M_{[iu_{n}^{2}+1,i\mu_{n}^{2}+u_{n}]}<u_{n}, 0\leq i <l_{n}-1\})\\& \quad \leq\mu^{m}(\{M_{u_{n},m}<u_{n}\})^{l_{n}}(1+\theta \rho^{u_{n}^{2}-u_{n}})^{m\cdot l_{n}}\\& \quad \leq\bigg(1-\sum_{(a_{1},\ldots,a_{u_{n}})\in \mathbb{N}^{u_{n}}} \mu(I_{u_{n}}(a_{1},\ldots,a_{u_{n}}))^{m}\bigg)^{l_{n}}(1+\theta \rho^{u_{n}^{2}-u_{n}})^{m\cdot l_{n}}\\& \quad \leq\exp\bigg\{-\frac{n}{u_{n}^{2}}\cdot n^{-{(s+(m-1)H_{m})}/{2s}}\bigg\}\cdot\exp\bigg\{\theta \rho^{u_{n}^{2}-u_{n}}\frac{m\cdot n}{u_{n}^{2}}\bigg\}\\& \quad \leq M\exp\bigg\{-\frac{n^{{(s-(m-1)H_{m})}/{2s}}}{u_{n}^{2}}\bigg\}, \end{align*} $$
where the penultimate inequality follows from (3.2) and the two facts
$ (1-x)<\exp (-x)$
for
$0<x<1$
and
$\lim _{n\to \infty }(1+1/n)^{n}=e$
, and the last inequality follows because
$\theta \rho ^{u_{n}^{2}-u_{n}}{m\cdot n}/{u_{n}^{2}}\rightarrow 0$
as
$n\rightarrow \infty .$
Thus,
$$ \begin{align*} \sum_{n=1}^{\infty}\mu^{m}(\{M_{n,m}<u_{n}\})<\infty. \end{align*} $$
From the Borel–Cantelli Lemma, for
$\mu ^{m}$
-almost all
$(x_{1},\ldots ,x_{m})\in [0,1)^{m},$
$$ \begin{align*}\liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}\geq\frac{1}{(m-1)H_{m}}.\end{align*} $$
This completes the proof of Proposition 3.2 and of Theorem 1.1.
4 Proof of Theorem 1.3
This section is devoted to the proof of Theorem 1.3. Our strategy is to construct Cantor-like subsets with full Hausdorff dimension. The proof is divided into several cases according to the values of
$\alpha $
and
$\beta .$
We give a detailed proof for the case
$0<\alpha <\beta <\infty $
and a sketch of the proof for the remaining cases.
Case 1:
$0<\alpha <\beta <\infty .$
Choose two positive integer sequences
$\{n_{k}\}_{k\geq 1}$
and
$\{s_{k}\}_{k\geq 1}$
such that, for each
$k\geq 1,$
$$ \begin{align} n_{1}=2,\quad n_{k+1}=\lfloor n_{k}^{{\beta}/{\alpha}}\rfloor, \quad s_{k}=\lfloor \,\beta \log n_{k}\rfloor. \end{align} $$
We readily check that
$$ \begin{align} \lim_{k\rightarrow\infty}\frac{s_{k}}{n_{k+1}-n_{k}}=0. \end{align} $$
Without loss of generality, we assume that
$n_{k+1}-n_{k}>s_{k}$
for all
$k\geq 1.$
Otherwise, we consider only sufficiently large k. Put
$$ \begin{align*}n_{k+1}-n_{k}=s_{k}\cdot\iota_{k}+\theta_{k},\end{align*} $$
where
$$ \begin{align*}\iota_{k}=\bigg\lfloor\frac{n_{k+1}-n_{k}}{s_{k}}\bigg\rfloor \quad \text{for } 0\leq\theta_{k}<s_{k}.\end{align*} $$
Define a marked set
$\mathbb {K}$
of positive integers by
$$ \begin{align*}\mathbb{K}:=\mathbb{K}(\{n_{k}\}, \{s_{k}\})=\bigcup_{k\ge1}\{n_{k},n_{k}+1,n_{k}+2,\ldots,n_{k}+s_{k}, n_{k}+2s_{k},n_{k}+3s_{k},\ldots, n_{k}+\iota_ks_{k}\}.\end{align*} $$
Now we define m sequences as follows.
-
• For
$i=1,$
$$ \begin{align*}a_{n_{k}}^{(1)}=1,\ a_{n_{k}+1}^{(1)}=\cdots=a_{n_{k}+s_{k}-1}^{(1)}=1,\ a_{n_{k}+s_{k}}^{(1)}=a_{n_{k}+2s_{k}}^{(1)}= \cdots=a_{n_{k}+\iota_ks_{k}}^{(1)}=1.\end{align*} $$
-
• For
$2\leq i\leq m,$
$$ \begin{align*}a_{n_{k}}^{(i)}=i,\ a_{n_{k}+1}^{(i)}=\cdots=a_{n_{k}+s_{k}-1}^{(i)}=1,\ a_{n_{k}+s_{k}}^{(i)}=a_{n_{k}+2s_{k}}^{(i)}= \cdots=a_{n_{k}+\iota_ks_{k}}^{(i)}=i.\end{align*} $$
Then, for
$i=1,2,\ldots ,m,$
write
$$ \begin{align*}E(\mathbb{K},\{a_{n}^{(i)}\}_{n\geq1})=\{x\in[0,1)\cap \mathbb{Q}^{c}: a_{n}(x)= a_{n}^{(i)} \ \text{for all}~ n\in\mathbb{K}\}.\end{align*} $$
Now we prove
$\prod _{i=1}^{m}E(\mathbb {K},\{a_{n}^{(i)}\}_{n\geq 1})\subset E(\alpha ,\beta ).$
Fix
$(x_{1},\ldots ,x_{m})\in \prod _{i=1}^{m}E(\mathbb {K},\{a_{n}^{(i)}\}_{n\geq 1})$
for any
$n\geq n_{1}$
and let k be the integer such that
$n_{k}\leq n< n_{k+1}.$
From the construction of the set
$\prod _{i=1}^{m}E(\mathbb {K},\{a_{n}^{(i)}\}_{n\geq 1})$
, we see that
$$ \begin{align*} M_{n,m}(x_{1},\ldots,x_{m})=\begin{cases} s_{k-1}-1 & \text{if}\ n_{k}\leq n<n_{k}+s_{k-1},\\ n-n_{k} & \text{if}\ n_{k}+s_{k-1}\leq n<n_{k}+s_{k},\\ s_{k}-1 & \text{if}\ n_{k}+s_{k}\leq n<n_{k+1}. \end{cases} \end{align*} $$
Further, by (4.1), we deduce that
$$ \begin{align*} \begin{aligned} \liminf_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}&= \liminf_{k\rightarrow\infty} \min\bigg\{\frac{M_{n_{k}+s_{k-1}-1,m}(x_{1},\ldots,x_{m})}{\log(n_{k}+s_{k-1}-1)},\frac{M_{n_{k+1}-1,m}(x_{1},\ldots,x_{m})}{\log(n_{k+1}-1)}\bigg\}\\& =\liminf_{k\rightarrow\infty} \min\bigg\{\frac{s_{k-1}-1}{\log(n_{k}+s_{k-1}-1)},\frac{s_{k}-1}{\log(n_{k+1}-1)}\bigg\}\\& =\alpha \end{aligned} \end{align*} $$
and
$$ \begin{align*} \begin{aligned} \limsup_{n\rightarrow\infty}\frac{M_{n,m}(x_{1},\ldots,x_{m})}{\log n}&= \limsup_{k\rightarrow\infty} \max\bigg\{\frac{M_{n_{k},m}(x_{1},\ldots,x_{m})}{\log(n_{k})},\frac{M_{n_{k}+s_{k}-1,m}(x_{1},\ldots,x_{m})}{\log(n_{k}+s_{k}-1)}\bigg\}\\& =\limsup_{k\rightarrow\infty} \max\bigg\{\frac{s_{k-1}-1}{\log(n_{k})},\frac{s_{k}-1}{\log(n_{k}+s_{k}-1)}\bigg\}\\& =\beta. \end{aligned} \end{align*} $$
Hence,
$(x_{1},\ldots ,x_{m})\in E(\alpha ,\beta ).$
It remains to prove that the density of
$\mathbb {K}\subset \mathbb {N}$
is zero. For
$n_{k}\leq n<n_{k+1}$
with some
$k\geq 1{:}$
-
• if
$n_{k}\leq n< n_{k}+s_{k},$
then
$\sharp \{i\leq n : i\in \mathbb {K}\}=\sum _{j=1}^{k-1}(m_{j}+\iota _{j})+n-n_{k}+1;$
-
• if
$n_{k}+ls_{k}\leq n<n_{k}+(l+1)s_{k}$
for some
$l$
with
$0< l<\iota _{k},$
then we see that
${\sharp \{i\leq n : i\in \mathbb {K}\}=\sum _{j=1}^{k-1} (s_{j}+\iota _{j})+s_{k}+l};$
-
• if
$n_{k}+\iota _{k}s_{k}\leq n<n_{k+1},$
then
$\sharp \{i\leq n : i\in \mathbb {K}\}=\sum _{j=1}^{k}(s_{j}+\iota _{j}).$
Consequently,
$$ \begin{align*} \begin{aligned} \limsup_{n\rightarrow\infty}\frac{\sharp\{i\leq n : i\in\mathbb{K}\}}{n}&\leq \limsup_{k\rightarrow\infty} \max_{0\leq l<\iota_{k}}\bigg\{\frac{\sum_{j=1}^{k-1}(s_{j}+\iota_{j})+s_{k}+l}{n_{k}+ls_{k}}\bigg\}\\& \leq\limsup_{k\rightarrow\infty}\bigg\{\frac{\sum_{j=1}^{k-1}(s_{j}+\iota_{j})+s_{k}+\iota_{k})}{n_{k}}\bigg\}\\& =0, \end{aligned} \end{align*} $$
where the last equality follows by the Stolz–Cesàro theorem and (4.2). By Lemmas 2.4 and 2.5,
$$ \begin{align*}\dim_{H}E_{\alpha,\beta}\geq\dim_{H}\bigg(\prod\limits_{i=1}^{m}E(\mathbb{K},\{a_{n}^{(i)}\}_{n\geq1})\bigg)\geq\sum_{i=1}^{m}\dim_{H}E(\mathbb{K},\{a_{n}^{(i)}\}_{n\geq1})=m.\end{align*} $$
Similar arguments apply to the remaining cases. We only give the constructions for the proper sequences
$\{n_k\}_{k\geq 1}$
and
$\{s_k\}_{k\geq 1}.$
Case 2:
$0<\alpha =\beta <\infty .$
Take
$n_{k}=2^{k}$
and
$s_{k}=\lfloor \alpha \log n_{k}\rfloor \text { for } k\geq 1.$
Case 3:
$\alpha =0<\beta <\infty .$
Take
$n_{k}=2^{2^{2^{k}}}$
and
$s_{k}=\lfloor \,\beta \log n_{k}\rfloor \text { for } k\geq 1.$
Case 4:
$\alpha =0, \beta =\infty .$
Take
$n_{k}=2^{2^{2^{k}}}$
and
$s_{k}=\lfloor k\log n_{k}\rfloor ~\text { for } k\geq 1.$
Case 5:
$0<\alpha <\beta =\infty .$
Take
$n_{k}=2^{k!}$
and
$s_{k}=\lfloor \alpha k\log n_{k}\rfloor \text { for } k\geq 1.$
Case 6:
$\alpha =\beta =0.$
Take
$n_{k}=2^{k}$
and
$s_{k}=\lfloor \log \log n_{k}\rfloor \text { for } k\geq 1.$
Case 7:
$\alpha =\beta =\infty .$
Take
$n_{k}=2^{k}$
and
$s_{k}=\lfloor k\log n_{k}\rfloor \text { for } k\geq 1.$
