2.1 Properties of continued fractions

This section is devoted to recalling some elementary properties in continued fractions. For more information on the continued fraction expansion, the readers are referred to [Reference Hardy and Wright12, Reference Khintchine14, Reference Schmidt22]. We also introduce some basic techniques for estimating the Hausdorff dimension of a fractal set (see [Reference Falconer8, Reference Schweiger23]).

For any irrational number $x\in [0,1)$ with continued fraction expansion in equation (1.3), we write ${p_{n}(x)}/{q_{n}(x)} = [a_{1}(x),\ldots ,a_{n}(x)]$ and call it the nth convergent of $x.$ With the conventions $p_{-1}(x)=1, q_{-1}(x)=0, p_{0}(x)=0$ , and $q_{0}(x)=1,$ we know that $p_{n}(x)$ and $q_{n}(x)$ satisfy the recursive relations [Reference Khintchine14]:

(2.1) $$ \begin{align}\begin{aligned} p_{n+1}(x)&= a_{n+1}(x)p_{n}(x)+p_{n-1}(x),\\q_{n+1}(x) &= a_{n+1}(x)q_{n}(x)+q_{n-1}(x),\quad n\geq 0. \end{aligned}\end{align} $$

Clearly, $q_{n}(x)$ is determined by $a_{1}(x),\ldots ,a_{n}(x),$ so we also write $q_{n}(a_{1}(x),\ldots ,a_{n}(x))$ instead of $q_n(x)$ . We write $a_{n}$ and $q_n$ in place of $a_{n}(x)$ and $q_n(x)$ for simplicity when no confusion can arise.

Proof. For the convenience of readers, we give the proof.

(1) By the recursive relations in equation (2.1), we readily check that

$$ \begin{align*} \prod\limits_{k=1}^{n}a_{k}\leq q_{n}\leq \prod\limits_{k=1}^{n}(a_{k}+1). \end{align*} $$

Since $a_n\ge 1$ for $n\ge 1,$ we have

$$ \begin{align*} 1=q_0\le q_1<q_2<\cdots q_{n-1}<q_n. \end{align*} $$

By induction, $q_n\ge 2^{({n-1})/{2}}$ for all $n\ge 1$ ; similarly $p_n\ge 2^{({n-1})/{2}}$ .

(2) Induction on k: assuming that

$$ \begin{align*} 1\leq \frac{q_{n+k}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{n+k})}{q_{n}(a_{1},\ldots,a_{n})q_{k}(a_{n+1},\ldots,a_{n+k})} \leq 2 \end{align*} $$

holds for all $k\in \{1,\ldots ,m\}$ , we prove that the above inequality holds for $k=m+1.$ Indeed, this is the case because

$$ \begin{align*} &q_{n+m+1}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{n+m+1})\\ &\quad =a_{n + m+1}q_{n+m}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{n+m})\\ & \qquad+ q_{n + m-1}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{n+m-1})\\ &\quad \ge a_{n+m+1}q_{n}(a_{1},\ldots,a_{n})q_{m}(a_{n+1},\ldots,a_{n+m})\\ &\qquad+q_{n}(a_{1},\ldots,a_{n})q_{m-1}(a_{n+1},\ldots,a_{n+m-1})\\ & \quad =q_{n}(a_{1},\ldots,a_{n})q_{m+1}(a_{n+1},\ldots,a_{n+m+1}), \end{align*} $$

and

$$ \begin{align*} &q_{n+m+1}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{n+m+1})\\&\quad = a_{n+m+1}q_{n+m}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{n+m})\\&\qquad +q_{n+m-1}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{n+m-1})\\&\quad \le 2a_{n+m+1}q_{n}(a_{1},\ldots,a_{n})q_{m}(a_{n+1},\ldots,a_{n+m})\\&\qquad +2q_{n}(a_{1},\ldots,a_{n})q_{m-1}(a_{n+1},\ldots,a_{n+m-1})\\&\quad =2q_{n}(a_{1},\ldots,a_{n})q_{m+1}(a_{n+1},\ldots,a_{n+m+1}). \end{align*} $$

(3) By the recursive relations in equation (2.1), we deduce that

$$ \begin{align*} \left(\!\! \begin{array}{cc} p_{n+1} & p_{n} \\ q_{n+1} & q_{n} \\ \end{array}\!\!\right)&=\left(\!\! \begin{array}{cc} p_{n} & p_{n-1} \\ q_{n} & q_{n-1} \\ \end{array}\!\!\right)\left(\!\!\begin{array}{cc} a_{n+1} & 1 \\ 1 & 0 \\ \end{array}\!\!\right)\\ & =\left(\!\!\begin{array}{cc} p_0 & p_{-1} \\ q_0 & q_{-1} \\ \end{array}\!\!\right)\left(\!\!\begin{array}{cc} a_{1} & 1 \\ 1 & 0 \\ \end{array}\!\!\right)\cdots\left(\!\!\begin{array}{cc} a_{n+1} & 1 \\ 1 & 0 \\ \end{array}\!\!\right)\\ &= \left(\!\!\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\!\!\right)\left(\!\! \begin{array}{cc} a_{1} & 1 \\ 1 & 0 \\ \end{array}\!\!\right)\cdots\left(\!\!\begin{array}{cc} a_{n+1} & 1 \\ 1 & 0 \\ \end{array}\!\!\right). \end{align*} $$

Taking $a_1=\cdots =a_n=a_{n+1}=i$ yields that

$$ \begin{align*}\left(\!\! \begin{array}{cc} p_{n+1} & p_{n} \\ q_{n+1} & q_{n} \\ \end{array}\!\!\right)=\left(\!\!\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\!\!\right)\left(\!\! \begin{array}{cc} i & 1 \\ 1 & 0 \\ \end{array}\!\!\right)\cdots\left(\!\!\begin{array}{cc} i & 1 \\ 1 & 0 \\ \end{array}\!\!\right). \end{align*} $$

The symmetric matrix $A=(\begin {smallmatrix} i & 1 \\ 1 & 0 \\ \end {smallmatrix})$ is diagonalizable:

$$ \begin{align*}P^{-1}AP=\left(\!\! \begin{array}{cc} \tau(i) & 0 \\ 0 & \zeta(i) \\ \end{array}\!\!\right) \end{align*} $$

with $P=\big (\begin {smallmatrix} \tau (i) & \zeta (i) \\ 1 & 1 \\ \end {smallmatrix}\big )$ .

A direct calculation yields that

$$ \begin{align*} q_n(i,\ldots,i) =\frac{(\tau(i))^{n+1}- (\zeta(i))^{n+1}}{\tau(i)-\zeta(i)}. \end{align*} $$

Also,

$$ \begin{align*} \frac{(\tau(i))^{n+1}- (\zeta(i))^{n+1}}{\tau(i)-\zeta(i)}\le\frac{2(\tau(i))^{n+1}}{\tau(i)}=2(\tau(i))^{n}, \end{align*} $$

and, if n is even,

$$ \begin{align*} \frac{(\tau(i))^{n+1} - (\zeta(i))^{n+1}}{\tau(i)-\zeta(i)}\ge\frac{(\tau(i))^{n+1}}{2\tau(i)}=\frac{(\tau(i))^{n}}{2}; \end{align*} $$

if n is odd (since $\zeta (i)\cdot \tau (i)=-1$ ),

$$ \begin{align*} \frac{(\tau(i))^{n+1} - (\zeta(i))^{n+1}}{\tau(i)-\zeta(i)} =\frac{(\tau(i))^{2(n+1)}-1}{(\tau(i))^{n+2}+(\tau(i))^{n}} \ge\frac{(\tau(i))^{n}}{2}. \end{align*} $$

This completes the proof.

For $n\geq 1$ and $(a_{1},\ldots ,a_{n})\in \mathbb {N}^{n}$ , we write

$$ \begin{align*} I_{n}(a_{1},\ldots,a_{n})=\{x\in[0,1)\colon a_{k}(x)=a_{k}, 1\leq k\leq n\}, \end{align*} $$

and call it a basic interval of order n. The basic interval of order n which contains x will be denoted by $I_{n}(x)$ , that is, $I_{n}(x)=I_{n}(a_{1}(x),\ldots ,a_{n}(x))$ .

Lemma 2.2. [Reference Khintchine14]

For $n\geq 1$ and $(a_{1},\ldots ,a_{n})\in \mathbb {N}^{n},$ we have

(2.2) $$ \begin{align} \frac{1}{2q_{n}^{2}}\leq|I_{n}(a_{1},\ldots,a_{n})|=\frac{1}{q_{n}(q_{n}+q_{n+1})}\leq\frac{1}{q_{n}^{2}}. \end{align} $$

Here and hereafter, $|\cdot |$ denotes the length of an interval.

The next lemma describes the distribution of basic intervals $I_{n+1}$ of order $n+1$ inside an nth basic interval $I_{n}.$

The following lemma displays the relationship between the ball $B(x,|I_{n}(x)|)$ and the basic interval $I_{n}(x)$ .

2.2 Hausdorff dimension

The following two properties, namely, Hölder property and the mass distribution principle, are often used to estimate the Hausdorff dimension of a fractal set.

Lemma 2.5. [Reference Falconer8]

If $f\colon X \to Y$ is an $\alpha $ -Hölder mapping between metric spaces, that is, there exists $c>0$ such that for all $x_{1},x_{2}\in X$ ,

$$ \begin{align*}d(f(x_{1}),f(x_{2}))\leq cd(x_{1},x_{2})^{\alpha},\end{align*} $$

then $\dim _{H}f(X)\leq ({1}/{\alpha }) \dim _{H}X.$

Lemma 2.6. [Reference Falconer8]

Let $E\subseteq [0,1]$ be a Borel set and $\mu $ be a measure with $\mu (E)> 0.$ If for every $x \in E$ ,

$$ \begin{align*}\liminf_{r \to 0}\frac{\log \mu(B(x,r))}{\log r} \geq s,\end{align*} $$

then $ \dim _{H}E\geq s.$

We conclude this subsection by quoting a dimensional result related to continued fractions, which will be used in the proof of Theorem 1.3.

Let $\mathbf {K}=\{k_{n}\}_{n= 1}^{\infty }$ be a subsequence of $\mathbb {N}$ which is not cofinite. Let $x=[a_{1},a_{2},\ldots ]$ be an irrational number in $[0,1)$ . Eliminating all the terms $a_{k_n}$ from the sequence $a_1,a_2,\ldots $ , we obtain an infinite subsequence $c_1,c_2,\ldots $ , and put $\phi _{\mathbf {K}}(x)=y$ with $y=[c_1,c_2,\ldots ]$ . In this way, we define a mapping $\phi _{\mathbf {K}}\colon [0,1)\cap \mathbb {Q}^{c}\to [0,1)\cap \mathbb {Q}^{c}$ .

Let $\{M_{n}\}_{n\geq 1}$ be a sequence with $M_{n}\in \mathbb {N}$ , $n\geq 1$ . Set

$$ \begin{align*} S(\{M_{n}\})= \{x\in [0,1)\cap \mathbb{Q}^{c}\colon 1\leq a_{n}(x)\leq M_{n} \text{ for all } n\geq 1\}. \end{align*} $$

Lemma 2.7. [Reference Chang and Chen6]

Suppose that $\{M_{n}\}_{n=1}^{\infty }$ is a bounded sequence. If the sequence $\mathbf {K}=\{k_{n}\}_{n=1}^{\infty }$ is of density zero in $\mathbb {N}$ , then

$$ \begin{align*}\dim_{H}S(\{M_{n}\})=\dim_{H}\phi_{\mathbf{K}}S(\{M_{n}\}).\end{align*} $$

2.3 Pressure function and pre-dimensional number

We now introduce the notions of the pressure function and pre-dimensional number in the continued fraction dynamical system. For more details, we refer the reader to [Reference Hanus, Mauldin and Urbański11].

For $\mathcal {A}$ a finite or infinite subset of $\mathbb {N},$ we set

$$ \begin{align*} X_{\mathcal{A}}= \{x\in[0,1)\colon a_{n}(x)\in \mathcal{A} \text{ for all } n\geq 1\}. \end{align*} $$

The pressure function restricted to the subsystem $(X_{\mathcal {A}},T)$ with potential $\phi \colon [0,1)\to \mathbb {R} $ is defined as

(2.3) $$ \begin{align} P_{\mathcal{A}}(T,\phi)=\lim_{n\to\infty} \frac{\log\sum\limits_{(a_{1},\ldots,a_{n})\in \mathcal{A}^{n}}\sup\limits_{x\in X_{\mathcal{A}}}\exp{S_n \phi([a_{1},\ldots,a_{n}+x])}}{n}, \end{align} $$

where $S_{n}\phi (x)=\phi (x)+\cdots +\phi (T^{n-1}(x))$ denotes the ergodic sum of $\phi $ . When $\mathcal {A}=\mathbb {N}$ , we write $P(T,\phi )$ for $P_{\mathbb {N}}(T,\phi )$ .

The nth variation $\textrm {Var}_{n}(\phi )$ of $\phi $ is defined as

$$ \begin{align*} \textrm{Var}_{n}(\phi)=\sup \{|\phi(x)-\phi(y)|\colon I_{n}(x)=I_{n}(y)\}. \end{align*} $$

The following lemma shows the existence of the limit in equation (2.3).

For $0<\alpha <1$ and $i\in \mathbb {N}$ , we define

$$ \begin{align*} \widehat{s}_{n}(\mathcal{A},\alpha,{\tau(i)}) = \inf \bigg\{\rho \geq 0\colon \sum\limits_{a_{1},\ldots,a_{n}\in \mathcal{A}} \bigg(\frac{1}{(\tau(i))^{{n\alpha}/({1-\alpha})}q_{n}(a_{1},\ldots,a_{n})}\bigg)^{2\rho}\leq 1\bigg\}. \end{align*} $$

Following [Reference Wang and Wu31], we call $\widehat {s}_{n}(\mathcal {A},\alpha ,{\tau (i)})$ the nth pre-dimensional number with respect to $\mathcal {A}$ and $\alpha $ . The properties of pre-dimensional numbers are presented in the following lemmas; the original ideas for the proofs date back to Good [Reference Good10] (see also [Reference Mauldin and Urbański20]).

By equation (2.2) and the definition of $\widehat {s}_{n}(\mathcal {A},\alpha ,{\tau (i)}),$ we know $0\le \widehat {s}_{n}(\mathcal {A},\alpha ,{\tau (i)})\le 1.$ Furthermore, Lemma 2.9 implies that $0\le s(\mathcal {A},\alpha ,{\tau (i)})\le 1.$

Similarly to pre-dimensional numbers $\{\widehat {s}_{n}(\mathcal {A},\alpha ,{\tau (i)}{)}\},$ we define

$$ \begin{align*} s_{n}(\mathcal{A},\alpha,{\tau(i)})= \inf \bigg\{\rho \geq 0\!\colon \sum\limits_{a_{1},\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\! \bigg(\frac{1}{q_{n}(a_{1},\ldots,a_{n-\lfloor na\rfloor},i,\ldots,i)}\bigg)^{2\rho}\leq 1\bigg\}. \end{align*} $$

Remark 2.11. We remark that

$$ \begin{align*}\sum\limits_{a_{1},\ldots,a_{n}\in \mathcal{A}} \bigg(\frac{1}{(\tau(i))^{{n\alpha}/({1-\alpha}}) q_{n}(a_{1},\ldots,a_{n})}\bigg)^{2\widehat{s}_{n}(\mathcal{A},\alpha,{\tau(i)})}\leq 1\end{align*} $$

and

$$ \begin{align*}\sum_{a_{1},\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}} \bigg(\frac{1}{q_{n}(a_{1},\ldots,a_{n-\lfloor n\alpha\rfloor},i,\ldots,i)}\bigg)^{2s_{n}(\mathcal{A},\alpha,{\tau(i)})}\leq1,\end{align*} $$

with equalities holding when $\mathcal {A}$ is finite.

By Lemmas 2.9 and 2.10, we have the following result.

Lemma 2.12. Let $\mathcal {A}$ be a finite or infinite subset of $\mathbb {N}.$ For $0<\alpha <1$ and $i\in \mathbb {N},$ we have

$$ \begin{align*}\lim_{n\to\infty}s_{n}(\mathcal{A},\alpha,{\tau(i)})=s(\mathcal{A},\alpha,{\tau(i)}).\end{align*} $$

In particular, if $\mathcal {A}=\mathbb {N},$ then

$$ \begin{align*}\lim_{n\to\infty}s_{n}(\mathbb{N},\alpha,{\tau(i)})=s(\mathbb{N},\alpha,{\tau(i)}).\end{align*} $$

Proof. For $\varepsilon>0$ and n large enough, we have

(2.4) $$ \begin{align} 2^{(({n-\lfloor n\alpha\rfloor})/{2}) \varepsilon}>64, \end{align} $$

(2.5) $$ \begin{align} \frac{3}{(1-\alpha)(n\alpha-1)}+\frac{\log 4}{n\alpha-1}<\varepsilon, \end{align} $$

(2.6) $$ \begin{align} |\widehat{s}_{n}(\mathcal{A},\alpha,{\tau(i)})-s(\mathcal{A},\alpha,{\tau(i)})|<\frac{\varepsilon}{2}. \end{align} $$

On the one hand, by Remark 2.11, we deduce that

$$ \begin{align*} 1 \ge&\! \sum_{a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\!\!\bigg(\frac{1}{q_{n}(a_1,\ldots,a_{n-\lfloor n\alpha\rfloor},i,\ldots,i)}\bigg)^{2s_{n}(\mathcal{A},\alpha,{\tau(i)})} \\ \ge &\! \sum_{a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\!\! \bigg(\frac{1}{2q_{n-\lfloor n\alpha\rfloor}(a_1,\ldots,a_{n-\lfloor n\alpha\rfloor})q_{\lfloor n\alpha\rfloor}(i,\ldots,i)}\bigg)^{2s_{n}(\mathcal{A},\alpha,{\tau(i)})}\\ \ge &\! \sum_{a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\!\!\bigg(\frac{1}{4q_{n-\lfloor n\alpha\rfloor}(a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}) (\tau(i))^{({\alpha}/({1-\alpha})) (n-\lfloor n\alpha\rfloor)}}\bigg)^{\!2s_{n}(\mathcal{A},\alpha,{\tau(i)})}\\ \ge &\! \sum_{a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\!\!\bigg(\frac{1}{q_{n-\lfloor n\alpha\rfloor}(a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}) (\tau(i))^{({\alpha}/({1-\alpha}))(n-\lfloor n\alpha\rfloor)}}\bigg)^{\!2s_{n}(\mathcal{A},\alpha,{\tau(i)})+\varepsilon}, \end{align*} $$

where the second inequality holds by Lemma 2.1(2); the third inequality is right by Lemma 2.1(3) and the fact that $({\alpha }/({1-\alpha }))(n-\lfloor n\alpha \rfloor )\ge \lfloor n\alpha \rfloor $ for $n\in \mathbb {N};$ the last inequality is true by Lemma 2.1(1) and equation (2.4). This means that

$$ \begin{align*}s_{n}(\mathcal{A},\alpha,{\tau(i)})+\frac{\varepsilon}{2}\ge \widehat{s}_{n-\lfloor n\alpha\rfloor}(\mathcal{A},\alpha,{\tau(i)}).\end{align*} $$

On the other hand, we have

$$ \begin{align*} 1 \ge &\! \sum_{a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\!\!\bigg(\frac{1}{q_{n-\lfloor n\alpha\rfloor}(a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}) (\tau(i))^{({\alpha}/({1-\alpha})) (n-\lfloor n\alpha\rfloor)}}\bigg)^{\!2\widehat{s}_{n-\lfloor n\alpha\rfloor}(\mathcal{A},\alpha,{\tau(i)})}\\ \ge &\! \sum_{a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\!\!\bigg(\frac{1}{q_{n-\lfloor n\alpha\rfloor}(a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}) (\tau(i))^{\lfloor n\alpha\rfloor+{1}/{(1-\alpha)}}}\bigg)^{\!2\widehat{s}_{n-\lfloor n\alpha\rfloor}(\mathcal{A},\alpha,{\tau(i)})}\\ \ge &\! \sum_{a_1,\ldots,a_{n-\lfloor n\alpha\rfloor}\in \mathcal{A}}\!\!\bigg(\frac{1}{q_{n}(a_1,\ldots,a_{n-\lfloor n\alpha\rfloor},i,\ldots,i)}\bigg)^{\!2\widehat{s}_{n-\lfloor n\alpha\rfloor}(\mathcal{A},\alpha,{\tau(i)})+\varepsilon}, \end{align*} $$

where the second inequality is obtained by $({\alpha }/({1-\alpha }))(n-\lfloor n\alpha \rfloor )\le \lfloor n\alpha \rfloor + {1}/({1-\alpha })$ for $n\in \mathbb {N}$ ; the last inequality holds by Lemma 2.1(3) and equation (2.5). This implies that

$$ \begin{align*}s_{n}(\mathcal{A},\alpha,{\tau(i)})\le \widehat{s}_{n-\lfloor n\alpha\rfloor}(\mathcal{A},\alpha,{\tau(i)})+\frac{\varepsilon}{2}.\end{align*} $$

Thus, by equation (2.6), we obtain that

$$ \begin{align*}|s_{n}(\mathcal{A},\alpha,{\tau(i)})-s(\mathcal{A},\alpha,{\tau(i)})|<\varepsilon\end{align*} $$

for n large enough. This completes the proof.

For simplicity, write $s_{n}(\alpha ,{\tau (i)})$ for $s_{n}(\mathbb {N},\alpha ,{\tau (i)}), s(\alpha ,{\tau (i)})$ for $s(\mathbb {N},\alpha ,{\tau (i)}).$

From a point of view of a dynamical system, $s(\alpha ,{\tau (i)})$ can be regarded as the solution to the pressure function [Reference Wang, Wu and Xu32]

$$ \begin{align*}P\bigg(T,-s\bigg(\log |T'|+\frac{\alpha}{1-\alpha}\log{\tau(i)}\bigg)\bigg)=0.\end{align*} $$

Furthermore, by Lemma 2.13, we may extend $s(\alpha ,{\tau (i)})$ to $[0,1]$ as follows:

(2.7) $$ \begin{align} s(\alpha,{\tau(i)})=\left\{\! \begin{array}{ll} 1, &\quad \alpha=0, \\ s(\alpha,{\tau(i)}), &\quad 0<\alpha <1, \\ \frac{1}{2}, &\quad \alpha=1. \end{array} \right. \end{align} $$

4.1 Lower bound of $\dim _{H}E(B)$

Before proceeding, we cite an analogous definition of the pre-dimensional numbers. Let $l_{k}=m_{k}-m_{k-1}$ for $k\geq 1$ ( $m_{0}=0$ by convention). Let

$$ \begin{align*}\widetilde{f}_{k}(s,{\tau(i)}) =\sum\limits_{1\leq a_{m_{k-1}+1},\ldots,a_{n_{k}}\leq B} \bigg(\frac{1}{q_{l_{k}}(a_{m_{k-1}+1},\ldots,a_{n_{k}}, i,\ldots,i)}\bigg)^{2s}.\end{align*} $$

Recall $\xi = {\nu ^{2}}/({(1+\nu )(\nu -\hat {\nu })})$ . We define $\widetilde {s}_{l_{k}}(\mathcal {A}_{B},\xi ,\tau (i))$ to be the solution of the equation $\widetilde {f}_{k}(s,\tau (i))=1$ .

Proof. By Lemma 2.12 and the fact $l_{k}\to \infty $ as $k\rightarrow \infty $ (cf. condition (1)), we deduce that, for any $\varepsilon>0,$ when $k\gg 1,$

(4.1) $$ \begin{align} |s_{l_{k}}(\mathcal{A}_{B},\xi+\varepsilon,\tau(i))-s(\mathcal{A}_{B},\xi+\varepsilon,\tau(i))|<\frac{\varepsilon}{2}. \end{align} $$

Further, from conditions (2) and (3), we have that

$$ \begin{align*} \lim_{k\to\infty}\frac{m_{k}-n_{k}}{l_{k}} &=\lim_{k\to\infty}\frac{ (({m_k-n_k})/{m_k})\cdot ({m_k}/{n_k}) }{({m_k}/{n_k}) - (({m_{k-1}-n_{k-1}})/{n_k})\cdot ({m_{k-1}})/({m_{k-1} -n_{k-1}})} =\xi, \end{align*} $$

and thus for $k\gg 1$ ,

(4.2) $$ \begin{align} \lfloor l_{k}(\xi-\varepsilon)\rfloor \leq m_{k}-n_{k}\leq \lfloor l_{k}(\xi+\varepsilon)\rfloor. \end{align} $$

Hence, by equations (4.1) and (4.2), we obtain that

$$ \begin{align*} &\sum\limits_{1\leq a_{1},\ldots,a_{n_{k}-m_{k-1}}\leq B} \bigg(\frac{1}{q_{l_{k}}(a_{1},\ldots,a_{n_{k}-m_{k-1}}, i,\ldots,i)}\bigg)^{2 (s(\mathcal{A}_{B}, \xi+\varepsilon,\tau(i))-\varepsilon)} \\ &\quad\geq\sum\limits_{1\leq a_{1},\ldots,a_{n_{k}-m_{k-1}}\leq B} \bigg(\frac{1}{q_{l_{k}}(a_{1},\ldots,a_{n_{k}-m_{k-1}}, i,\ldots,i)}\bigg)^{2s_{l_{k}}(\mathcal{A}_{B}, \xi+\varepsilon,\tau(i))-\varepsilon}\\ &\quad\geq\sum\limits_{1\leq a_{1},\ldots,a_{l_{k}-\lfloor l_{k}(\xi+\epsilon)\rfloor}\leq B} \bigg(\frac{1}{q_{l_{k}}(a_{1},\ldots,a_{l_{k}-\lfloor l_{k}(\xi+\varepsilon)\rfloor}, i,\ldots,i)}\bigg)^{2s_{l_{k}}(\mathcal{A}_{B}, \xi+\varepsilon,{\tau(i)})-\varepsilon}\\ &\quad\geq\bigg(\frac{1}{q_{l_{k}}(B,\ldots,B)}\bigg)^{-\varepsilon}\geq\tau(B)^{l_{k}\varepsilon}\geq 1, \end{align*} $$

where $\tau (B)= ({B+\sqrt {B^{2}+4}})/{2}.$ Moreover, we get that

$$ \begin{align*} &\sum\limits_{1\leq a_{1},\ldots,a_{n_{k}-m_{k-1}}\leq B} \bigg(\frac{1}{q_{l_{k}}(a_{1},\ldots,a_{n_{k}-m_{k-1}}, i,\ldots,i)}\bigg)^{2 (s(\mathcal{A}_{B}, \xi-\varepsilon,\tau(i))+\varepsilon)} \\ &\quad\leq\sum\limits_{1\leq a_{1},\ldots,a_{l_{k}-\lfloor l_{k}(\xi-\varepsilon)\rfloor}\leq B} \bigg(\frac{1}{q_{l_{k}}(a_{1},\ldots,a_{l_{k}-\lfloor l_{k}(\xi-\varepsilon)\rfloor}, i,\ldots,i)}\bigg)^{2s_{l_{k}}(\mathcal{A}_{B}, \xi-\varepsilon,\tau(i))+\varepsilon}\\ &\quad\leq\bigg(\frac{1}{q_{l_{k}}(1,\ldots,1)}\bigg)^{\varepsilon}<1. \end{align*} $$

By the monotonicity of $\widetilde {f}_{k}(s,{\tau (i)})$ with respect to s, we have

$$ \begin{align*}s(\mathcal{A}_{B},\xi+\varepsilon,\tau(i))-\varepsilon \le\widetilde{s}_{l_{k}}(\mathcal{A}_{B},\xi,{\tau(i)}) \le s(\mathcal{A}_{B},\xi-\varepsilon,\tau(i))+\varepsilon,\end{align*} $$

which completes the proof.

4.1.1 Supporting measure

We define a probability measure $\mu $ on $E(B)$ by distributing mass among the basic intervals. We introduce the symbolic space to code these basic intervals: write $\mathcal {A}_{B}=\{1,\ldots ,B\}$ ; for $n\geq 1,$ set

$$ \begin{align*} \mathcal{B}_{n} = \{(\sigma_{1},\ldots,\sigma_{n})\in \mathcal{A}_{B}^{n}\colon\sigma_{j}=i \text{ for } n_{k}<j\leq m_{k} \text{ with some } k\geq 1\}. \end{align*} $$

Step I: For $(a_{1},\ldots ,a_{m_{1}})\in \mathcal {B}_{m_{1}}$ , we define

$$ \begin{align*} \mu (I_{m_{1}}(a_{1},\ldots,a_{m_{1}})) = \bigg(\frac{1}{q_{l_{1}}(a_{1},\ldots,a_{m_{1}})}\bigg)^{2\widetilde{s}_{l_{1}}(\mathcal{A}_{B},\xi,{\tau(i)})}\end{align*} $$

and for $1\leq n<m_{1}$ , set

$$ \begin{align*} \mu (I_{n}(a_{1},\ldots,a_{n})) = \sum\limits_{a_{n+1},\ldots,a_{m_{1}}}\mu (I_{m_{1}}(a_{1},\ldots,a_{n},a_{n+1},\ldots,a_{m_{1}})), \end{align*} $$

where the summation is taken over all $(a_{n+1},\ldots ,a_{m_{1}})$ with $(a_{1},\ldots ,a_{m_{1}})\in \mathcal {B}_{m_{1}}$ .

Step II: Assuming that $\mu (I_{m_{k}}(a_{1},\ldots ,a_{m_{k}}))$ is defined for some $k\geq 1$ , we define

$$ \begin{align*} &\mu(I_{m_{k+1}}(a_{1},\ldots,a_{m_{k+1}}))\\ &\quad = \mu(I_{m_{k}}(a_{1},\ldots,a_{m_{k}}))\cdot \bigg(\frac{1}{q_{l_{k+1}}(a_{m_{k}+1},\ldots,a_{m_{k+1}})}\bigg)^{2\widetilde{s}_{l_{k+1}}(\mathcal{A}_{B},\xi,{\tau(i)})} \end{align*} $$

and for $m_{k}<n<m_{k+1}$ , set

$$ \begin{align*} \mu (I_{n}(a_{1},\ldots,a_{n})) = \sum\limits_{a_{n+1},\ldots,a_{m_{k+1}}} \mu (I_{m_{k+1}}(a_{1},\ldots,a_{n},a_{n+1},\ldots, a_{m_{k+1}})). \end{align*} $$

Likewise, the last summation is taken under the restriction that $(a_{1},\ldots ,a_{m_{k+1}})\in \mathcal {B}_{m_{k+1}}.$

Step III: We have distributed the measure among basic intervals. By the definition of $\widetilde {s}_{l_{k}}(\mathcal {A}_{B},\xi ,{\tau (i)})$ , we readily check the consistency: for $n\geq 1$ and $(a_{1},\ldots ,a_{n})\in \mathcal {B}_{n}$ ,

$$ \begin{align*} \mu (I_{n}(a_{1},\ldots,a_{n})) = \sum\limits_{a_{n+1}}\mu (I_{n+1}(a_{1},\ldots,a_{n},a_{n+1})). \end{align*} $$

We then extend the measure to all Borel sets by Kolmogorov extension theorem. The extension measure is also denoted by $\mu $ .

From the construction, we know that $\mu $ is supported on $E(B)$ and

$$ \begin{align*} \mu (I_{m_{k}}(a_{1},\ldots,a_{m_{k}}))=&\prod\limits_{j=1}^{k} \bigg(\frac{1}{q_{l_{j}}(a_{m_{j-1}+1},\ldots,a_{m_{j}})}\bigg)^{2\widetilde{s}_{l_{j}}(\mathcal{A}_{B},\xi,{\tau(i)})},\\ & \sum\limits_{a_{1}\in \mathcal{B}_{1}}\mu (I_{1}(a_{1})) = 1. \end{align*} $$

4.1.2 Hölder exponent of $\boldsymbol \mu $

We shall start with the study of a basic interval.

For $0<\varepsilon <{s(\mathcal {A}_{B},\xi ,{\tau (i)})}/{4}$ , by Lemmas 2.12, 4.1, and the fact that $m_k$ grows exponentially, we can find $K\in \mathbb {N}$ such that for any $k, j\geq K,$

(4.3) $$ \begin{align} |\widetilde{s}_{l_{k}}(\mathcal{A}_{B},\xi,{\tau(i)})-s(\mathcal{A}_{B},\xi,{\tau(i)})|<\varepsilon, \end{align} $$

(4.4) $$ \begin{align} |\widetilde{s}_{l_{k}}(\mathcal{A}_{B},\xi,{\tau(i)})-s_j(\mathcal{A}_{B},\xi,{\tau(i)})|<\frac{\varepsilon\log 2}{2\log (B+1)}:=\varepsilon', \end{align} $$

and

(4.5) $$ \begin{align} \max \{(B+1)^K,2^k\}\le \frac14 (q_{m_k}(a_1,\ldots,a_{m_k}))^\varepsilon. \end{align} $$

Lemma 4.2. Let $n\ge m_K$ . For $(a_{1}, \ldots , a_{n})\in \mathcal {B}_{n}$ , we have

$$ \begin{align*} \mu (I_{n}(a_{1},\ldots,a_{n}))\leq C_0\cdot |I_{n}(a_1,\ldots,a_n)|^{s(\mathcal{A}_{B},\xi,{\tau(i)})-3\varepsilon}, \end{align*} $$

where $C_0=(B+1)^{2(l_1+\cdots +l_{K-1})}.$

Proof. To shorten notation, we will write $\widetilde {s}_{l_{k}}, s_k$ and s instead of $\widetilde {s}_{l_{k}}(\mathcal {A}_{B},\xi ,{\tau (i)}), s_k (\mathcal {A}_{B},\xi ,{\tau (i)})$ and $s(\mathcal {A}_{B},\xi ,{\tau (i)})$ , respectively. Fixing $(a_{1}, \ldots , a_{n})\in \mathcal {B}_{n}$ , we also write $I_n$ for $I_n(a_1,\ldots ,a_n)$ , $q_n$ for $q_n(a_1,\ldots ,a_n)$ , and $q_{l_j}$ for ${q_{l_{j}}(a_{m_{j-1}+1},\ldots ,a_{m_{j}})}$ when no confusion can arise. The proof falls naturally into three parts according to the range of n.

Case 1: $n=m_{k}$ for $k\ge K$ . By Lemmas 2.2 and 2.3, equations (4.3), (4.5), and the fact $q_{l_j}\le (B+1)^{l_j}$ , we obtain

$$ \begin{align*} \mu(I_{m_{k}})&=\prod\limits_{j=1}^{k} q_{l_j}^{-2\widetilde{s}_{l_j}}=\prod\limits_{j=1}^{K}q_{l_j}^{-2\widetilde{s}_{l_j}}\cdot\prod\limits_{j=K+1}^{k}q_{l_j}^{-2\widetilde{s}_{l_j}}\le C_0\prod\limits_{j=1}^{K}q_{l_j}^{-2(s-\varepsilon)}\cdot\prod\limits_{j=K+1}^{k}q_{l_j}^{-2(s-\varepsilon)}\\ &\leq C_{0}2^{2(k-1)}(q_{m_k})^{-2(s-\varepsilon)}\le \frac{C_{0}}4(q_{m_k})^{-2(s-2\varepsilon)}\leq C_{0}|I_{m_{k}}|^{s-2\varepsilon}. \end{align*} $$

Case 2: $m_{k}<n< n_{k+1}$ for $k\ge K$ . In this case, we have

$$ \begin{align*} \mu(I_n)&=\sum_{a_{n+1},\ldots,a_{m_{k+1}}}\mu(I_{m_{k+1}})=\sum_{{a_{n+1},\ldots,a_{m_{k+1}}}}\prod_{j=1}^{k+1}(q_{l_j})^{-2\widetilde{s}_{l_{j}}} \\ &= \prod_{j=1}^{k}(q_{l_j})^{-2\widetilde{s}_{l_{j}}}\cdot\sum_{{a_{n+1},\ldots,a_{m_{k+1}}}}(q_{l_{k+1}})^{-2\widetilde{s}_{l_{k+1}}}. \end{align*} $$

We have already seen in Case 1 that $\prod _{j=1}^{k}(q_{l_j})^{-2\widetilde {s}_{l_{j}}}\le C_{0}2^{2(k-1)}(q_{m_k})^{-2(s-\varepsilon )}$ . Additionally,

$$ \begin{align*} \sum_{{a_{n+1},\ldots,a_{m_{k+1}}}} (q_{l_{k+1}})^{-2\widetilde{s}_{l_{k+1}}}&\le ({q_{n-m_{k}}(a_{m_{k}+1},\ldots,a_{n})})^{-2(s-\varepsilon)}\\ &\quad\cdot\sum_{{a_{n+1},\ldots,a_{n_{k+1}}}} ({q_{m_{k+1}-n}(a_{n+1},\ldots,a_{n_{k+1}},i,\ldots,i)})^{-2\widetilde{s}_{l_{k+1}}}. \end{align*} $$

We then obtain that

$$ \begin{align*} \mu(I_n)\le C_02^{2k}(q_n)^{-2(s-\varepsilon)}\cdot \sum_{a_{n+1},\ldots,a_{m_{k+1}}} ({q_{m_{k+1}-n}(a_{n+1},\ldots,a_{n_{k+1}},i,\ldots,i)})^{-2\widetilde{s}_{l_{k+1}}}. \end{align*} $$

Now we need an upper estimate of the last sum. By the definition of $\widetilde {s}_{l_{k+1}}$ , we have that

$$ \begin{align*} &\sum\limits_{1\leq a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_n,a_{n+1},\ldots,a_{n_{k+1}}\leq B} (q_{l_{k+1}}( a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_n,a_{n+1},\ldots,a_{n_{k+1}},i,\ldots,i))^{-2\widetilde{s}_{l_{k+1}}}=1. \end{align*} $$

This yields that

$$ \begin{align*} &\sum_{{a^{\prime}_{m_k+1},\ldots,a^{\prime}_{n}}} ( q_{n-m_{k}}(a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n}))^{-2\widetilde{s}_{l_{k+1}}}\\ &\qquad\times \sum_{{a_{n+1},\ldots,a_{n_{k+1}}}} (q_{m_{k+1}-n}(a_{n+1},\ldots,a_{n_{k+1}},i,\ldots,i))^{-2\widetilde{s}_{l_{k+1}}}\le 4. \end{align*} $$

We will bound the first sum from below to reach the desired upper estimate of the second sum. We consider two cases.

(1) If $n-m_{k}< K$ ,

$$ \begin{align*} \sum_{a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n}} (q_{n-m_k}(a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n}))^{-2\widetilde{s}_{l_{k+1}}} \geq (q_{n-m_k}(B,\ldots,B))^{-2}\geq( {B+1})^{-2K}. \end{align*} $$

And thus, by equation (4.5), we reach that

$$ \begin{align*} \mu(I_{n}) \leq C_{0}2^{2k+2}(B+1)^{2K} ({q_{n}} )^{-2(s-\varepsilon)}\le\frac{C_{0}}4(q_{m_k})^{-2(s-3\varepsilon)} \leq C_{0} |I_{n}|^{s-3\varepsilon}. \end{align*} $$

(2) If $n-m_{k}\geq K$ , then, by equations (4.4), (4.5), and Remark 2.11, we have

$$ \begin{align*} &\sum_{a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n}} (q_{n-m_k}(a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n}))^{-2\widetilde{s}_{l_{k+1}}} \ge \sum_{a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n}} (q_{n-m_k}(a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n}))^{-2{s}_{n-m_k}-\varepsilon'}\\&\quad\ge\sum_{a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n-\lfloor(n-m_k)\xi\rfloor}}({q_{n-m_{k}}(a^{\prime}_{m_{k}+1},\ldots,a^{\prime}_{n-\lfloor(n-m_k)\xi\rfloor}, i,\ldots,i)})^{-2s_{n-m_{k}}-\varepsilon'}\\&\quad\ge ({q_{n-m_{k}}(B,\ldots,B)})^{-\varepsilon'} \ge({B+1})^{-(n-m_{k})\varepsilon'} \ge({B+1})^{-n\varepsilon'} \ge2^{{-n\varepsilon}/{2}}. \end{align*} $$

Therefore,

$$ \begin{align*}\mu(I_{n}) \leq C_{0}2^{2k+2}2^{{n\varepsilon}/{2}} ({q_{n}} )^{-2(s-2\varepsilon)} \leq \frac{C_{0}}2({q_{n}})^{-2(s-3\varepsilon)} \leq C_{0}|I_{n}|^{s-3\varepsilon}. \end{align*} $$

Case 3: $n_{k+1}\leq n<m_{k+1}$ for $k\ge K$ . In this case, since $(a_{n+1},\ldots ,a_{m_{k+1}})=(i,\ldots ,i),$ we have

$$ \begin{align*}\mu(I_{n}(a_{1},\ldots,a_{n}))=\mu(I_{m_{k+1}}(a_{1},\ldots,a_{m_{k+1}})),\end{align*} $$

then

$$ \begin{align*} \mu(I_{n}) \leq C_{0}|I_{m_{k+1}}|^{s-2\varepsilon} \leq C_{0}|I_{n}|^{s-2\varepsilon}. \end{align*} $$

These conclude the verification of the lemma.

Now we study the Hölder exponent for the measure of a general ball $B(x,r)$ .

Lemma 4.3. For $x\in E(B)$ and $r>0$ small enough, we have

$$ \begin{align*}\mu(B(x,r))\leq C_{0}\cdot r^{s(\mathcal{A}_{B},\xi,{\tau(i)})-4\varepsilon}.\end{align*} $$

Proof. Let $x=[a_{1},a_{2},\ldots ]$ be its continued fraction expansion. Let $n\ge K+2$ be the integer such that

$$ \begin{align*}|I_{n+1}(a_{1},\ldots,a_{n+1})|\leq r<|I_{n}(a_{1},\ldots,a_{n})|.\end{align*} $$

Therefore, it follows from Lemmas 2.4 and 4.2 that

$$ \begin{align*} \mu(B(x,r))&\leq \mu(I_{n-2}(a_{1},\ldots,a_{n-2})) \leq C_{0}\cdot|I_{n-2}(a_{1},\ldots,a_{n-2})|^{s(\mathcal{A}_{B},\xi,{\tau(i)})-3\varepsilon} \\& \le C_{0}(B+1)^{6}\cdot |I_{n+1}(a_{1},\ldots,a_{n+1})|^{s(\mathcal{A}_{B},\xi,{\tau(i)})-3\varepsilon} \\& \le C_{0} \cdot |I_{n+1}(a_{1},\ldots,a_{n+1})|^{s(\mathcal{A}_{B},\xi,{\tau(i)})-4\varepsilon} \le C_{0}\cdot r^{s(\mathcal{A}_{B},\xi,{\tau(i)})-4\varepsilon}.\\[-3pc] \end{align*} $$

Applying mass distribution principle (see Lemma 2.6), letting $\varepsilon \to 0$ , we conclude that

$$ \begin{align*}\dim_{H}E(B)\geq s(\mathcal{A}_{B},\xi,{\tau(i)}).\end{align*} $$

4.2 Lower bound of $\dim _{H}E(\hat{\nu}, \nu)$

We build a mapping f from $E(B)$ to $ E(\hat{\nu} ,\nu )$ and prove that f is dimension-preserving.

Fix an integer $d>B$ . For $x=[a_{1},a_{2},\ldots ]$ in $E(B)$ , we remark that the continued fraction of x is the concatenation of $\mathbb B_0=[a_1,\ldots ,a_{n_1}]$ and the blocks

$$ \begin{align*}\mathbb B_k=[\,\underbrace{ i,\ldots, i}_{m_k-n_k}, a_{m_{k}+1}, \ldots, a_{n_{k+1}}] \quad (k=1,2,\ldots).\end{align*} $$

In the block $\mathbb B_k$ , from the beginning, we insert a digit d after each $m_k-n_k$ digits to obtain a new block $\mathbb B_k'$ , that is,

$$ \begin{align*}\mathbb B^{\prime}_k=[ d, i,\ldots, i, d, a_{m_{k}+1}, \ldots, a_{m_{k}+(m_k-n_k)}, d, \ldots, a_{n_{k+1}}].\end{align*} $$

Concatenating the blocks $\mathbb B_0, \mathbb B^{\prime }_1, \mathbb B^{\prime }_2,\ldots $ , we get $[\mathbb B_0, \mathbb B^{\prime }_1, \mathbb B^{\prime }_2,\ldots ]$ , which is a continued fraction expansion of some $\bar {x}$ . We then define $f(x)=\bar {x}.$ Let ${\mathbf {K}}=\{k_n\}\subset \mathbb N$ be the collection of the occurrences of the digit d in the continued expansion of $\bar x$ . It is trivially seen that $\mathbf K$ is independent of the choice of $x\in E(B)$ , and, in the notation of Lemma 2.7, $\phi _{\mathbf {K}}(\bar x)=x$ for $x\in E(B)$ .

Let $h_k$ be the length of the block $\mathbb B^{\prime }_k.$ Noting that the number of the inserted digit d is at most $({n_{k+1}-m_{k}})/({m_{k}-n_{k}}) + 1=o(h_k)$ in the block $\mathbb B^{\prime }_k$ , we readily check that $\mathbf K$ is a subset of $\mathbb N$ of density zero. Hence, by Lemma 2.7, we have

$$ \begin{align*} \dim_{H}f (E(B))=\dim_{H}E(B). \end{align*} $$

It remains to prove that $f (E(B))$ is a Cantor subset of $E(\hat {\nu },\nu ).$

Proof. Fix $\bar {x}\in f (E(B))$ .

For $\varepsilon>0$ and n large enough, there exists some k such that $(\sum _{j=0}^{k-1}h_j)\leq n<(\sum _{j=0}^{k}h_j)$ . From the construction, we deduce that if $n=(\sum _{j=0}^{k-1}h_j)+1$ , then

$$ \begin{align*}|T^{n}(\bar{x})-y|<|I_{m_{k}-n_{k}}(y)|\le|I_{n}(y)|^{\nu-\varepsilon},\end{align*} $$

where the last inequality holds by the fact $\lim _k ({\sum _{j=0}^{k-1}h_j}/{n_k})=1;$ if $(\sum _{j=0}^{k-1}h_j)< n<(\sum _{j=0}^{k}h_j),$ then

$$ \begin{align*}|T^{n}(\bar{x})-y|\ge \frac{1}{2(d+2)^{2}}|I_{m_{k}-n_{k}}(y)|\ge|I_n(y)|^{\nu+\varepsilon}.\end{align*} $$

We then prove that $\nu (\bar {x})=\nu $ by the arbitrariness of $\varepsilon .$

However, for $(\sum _{j=0}^{k-1}h_j)\leq N<(\sum _{j=0}^{k}h_j)$ with k large enough, we pick $n=(\sum _{j=0}^{k-1}h_j)+1$ to obtain that

$$ \begin{align*}|T^{n}(\bar{x})-y|<|I_{m_{k}-n_{k}}(y)|\le|I_{\sum_{j=0}^{k}h_j}(y)|^{\hat{\nu}-\varepsilon}<|I_{N}(y)|^{\hat{\nu}-\varepsilon}.\end{align*} $$

When $N=(\sum _{j=0}^{k}h_j)$ , for all $n\in [1,N],$ we have that

$$ \begin{align*}|T^{n}(\bar{x})-y|\ge\frac{1}{2(d+2)^{2}}|I_{m_{k}-n_{k}}(y)|\ge|I_{N}(y)|^{\hat{\nu}+\varepsilon}.\end{align*} $$

We prove that $\hat {\nu }(\bar {x})=\hat {\nu }.$

Hence, $\bar {x}\in E(\hat {\nu },\nu ),$ as desired.

Consequently, for $0\le \hat {\nu }\le {\nu }/({1+\nu }) <\nu <\infty ,$ we have $\dim _{H}E(\hat {\nu },\nu )\geq s(\mathcal {A}_{B}, \xi ,{\tau (i)})$ .

Letting $B\rightarrow \infty $ yields $\dim _{H}E(\hat {\nu },\nu )\geq s(\xi ,{\tau (i)})$ , where $\xi = {\nu ^{2}}/({(1+\nu )(\nu -\hat {\nu })})$ .

We conclude this section by determining the lower bound of $\dim _{H}E(\hat {\nu }, +\infty ).$ We first study the case $0<\hat {\nu }<1$ . Let

$$ \begin{align*} n_{1}=2,\quad n_{k+1}=n_{k}^{k}+2n_{k},\quad m_0=0,\quad m_{k}= \lfloor\hat{\nu}n_{k}^{k}\rfloor + n_{k},\quad B_{k}= \lfloor m_{k}\log m_{k}\rfloor. \end{align*} $$

Thus,

$$ \begin{align*} \lim_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}}=\hat{\nu},\quad \lim_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k}}=\infty,\quad \lim_{k\to\infty}\frac{m_{k}-n_{k}}{m_{k}}=1. \end{align*} $$

Define

$$ \begin{align*} E=\{x\in[0,1)\colon a_{n}(x)\le B_k \text{ if } m_{k}<n\le n_{k+1} \text{ for some } k; \ a_n(x)=i \text{ otherwise}\}. \end{align*} $$

As before, for any $x=[a_1,a_2,\ldots ]$ in $E,$ we construct an element $\bar {x}:=f(x)\colon $ insert a digit $B_k+1$ after positions $n_{k}$ and $m_{k}+i(m_{k}-n_{k}), 0\leq i\leq t_{k}$ in the continued fraction expansion of x, where $t_{k}=\max \{t\in \mathbb {N}\colon m_{k}+t(m_{k}-n_{k})< n_{k+1}\}$ ; the resulting sequence is the continued fraction of $\bar x$ .

The method establishing the lower bound of $\dim _{H}E(B)$ applies to show that ${\dim _{H}E\geq 1/2 .}$ Moreover, $f(E)$ is a subset of $E(\hat {\nu }, +\infty ).$ It remains to prove that the Hausdorff dimension of $f(E)$ coincides with the one of E. To this end, we shall show that $f^{-1}$ is a $(1-\varepsilon )$ -Hölder mapping for any $\varepsilon>0$ . We remark that Lemma 2.7 may not apply directly here since $\{B_{k}\}$ is an unbounded sequence.

Proof. We write

$$ \begin{align*} m_{k}'=m_{k}+\sum_{l=1}^{k-1}(t_{l}+2),\quad n_{k}'=n_{k}+\sum_{l=1}^{k-1}(t_{l}+2), \end{align*} $$

and define the marked set

$$ \begin{align*} \mathbf{K}= \{m_{k}'+i(m_{k}-n_{k})+1\colon 0\leq i \leq t_{k}, k\geq 1\}\cup \{ n_{k}'\colon k\geq 1\}. \end{align*} $$

Let $\Delta _{n}=\sharp \{i\leq n\colon i\in \mathbf {K}\}$ , where $\sharp $ denotes the cardinality of a finite set. Let $k\in \mathbb {N}$ such that $m_{k}'\leq n<m_{k+1}'$ . We have

$$ \begin{align*} \frac{\Delta_{n}\log B_{k}}{n} &\leq\frac{(\sum_{l=1}^{k-1}(t_{l}+2)+ ({n-m_{k}'})/({m_{k}-n_{k}}) +1)\log B_{k}}{n}\\ &\leq \frac{(\sum_{l=1}^{k-1}(t_{l}+2)+1)\log B_{k}}{m_{k}'}+\frac{\log B_{k}}{m_{k}-n_{k}}+\frac{\log B_{k}}{m_{k}}\to 0. \end{align*} $$

So there exists $K\in \mathbb {N}$ , such that for $k\geq K$ and $n\geq m_{K}',$

(4.6) $$ \begin{align} (B_{k}+2)^{2\Delta_{n}+4}<2^{(n-1)\varepsilon}. \end{align} $$

For $\overline {x_{1}}=f(x_{1})$ and $\overline {x_{2}}=f(x_{2})$ in $f(E),$ we assume without loss of generality that

$$ \begin{align*} |\overline{x_{1}}-\overline{x_{2}}|<\frac{1}{2(B_{K}+2)^{2}q^{2}_{m_{K}'}(\overline{x_{1}})}; \end{align*} $$

otherwise, $|f^{-1}(\overline {x_{1}})-f^{-1}(\overline {x_{2}})|<C|\overline {x_{1}}-\overline {x_{2}}|^{1-\varepsilon }$ for some $C,$ as desired. Let

$$ \begin{align*}n=\min\{j\geq 1\colon a_{j+1}(\overline{x_{1}})\neq a_{j+1}(\overline{x_{2}})\}.\end{align*} $$

By Lemma 2.2, we have $m_{k}'\leq n<m_{k+1}'$ for some $k\geq K$ and $n+1<n_{k+1}'.$ Assume that $\overline {x_{1}}>\overline {x_{2}}$ and n is even (the same conclusion can be drawn for the remaining cases). There exist $1\leq \tau _{n+1}(\overline {x_{1}})<\sigma _{n+1}(\overline {x_{2}}) \leq B_{k}+1$ such that $\overline {x_{1}}\in I_{n+1}(a_{1},\ldots ,a_{n},\tau _{n+1}(\overline {x_{1}}))$ , $\overline {x_{2}}\in I_{n+1}(a_{1},\ldots ,a_{n},\sigma _{n+1}(\overline {x_{2}})).$ Combining Lemma 2.3 and the construction yields that $\overline {x_{1}}-\overline {x_{2}}$ is greater than the length of basic interval $I_{n+2}(a_{1},\ldots ,a_{n},\sigma _{n+1}(\overline {x_{2}}),B_{k}+1).$ This implies that

$$ \begin{align*} \overline{x_{1}}-\overline{x_{2}}\geq |I_{n+2}(a_{1},\ldots,a_{n},\sigma_{n+1}(\overline{x_{2}}),B_{k}+1)| \geq \frac{1}{2(B_{k}+2)^{4}q^{2}_{n}}. \end{align*} $$

Furthermore, noting that $f^{-1}(\overline {x_{1}}), f^{-1}(\overline {x_{2}})\in I_{n-\Delta _{n}}(c_{1},\ldots ,c_{n-\Delta _{n}}),$ where $(c_{1},\ldots , c_{n-\Delta _{n}})$ is obtained by eliminating all the terms $a_i$ with $i\in \mathbf {K}$ from $(a_1,\ldots ,a_n)$ , we conclude that

$$ \begin{align*} |f^{-1}(\overline{x_{1}})-f^{-1}(\overline{x_{2}})|&\leq |I_{n-\Delta_{n}}(a_{1},\ldots,a_{n-\Delta_{n}})| \\&\leq \frac{1}{q^{2}_{n-\Delta_{n}}} \leq(B_{k}+2)^{2\Delta_{n}}\frac{1}{q^{2}_{n}} \leq2|\overline{x_{1}}-\overline{x_{2}}|^{1-\varepsilon}, \end{align*} $$

where the penultimate inequality follows by equation (4.6). This completes the proof.

Now we deduce that $\dim _{H}E(\hat {\nu },\infty )\geq ({1-\varepsilon })/{2}$ by Lemma 2.5. Letting $\varepsilon \to 0$ , we establish that $\dim _{H}E(\hat {\nu },\infty )\geq \tfrac 12$ when $0<\hat \nu <1$ . A slight change in the proof actually shows that the estimate $\dim _{H}E(\hat {\nu },\infty )\geq \tfrac 12$ also works for $\hat \nu =0$ or $1$ . Indeed, when $\hat {\nu }=0,$ we may take

$$ \begin{align*} n_{k}=2^{2^{2k}},\quad m_{k}=n_{k}^{2},\quad B_{k}=2^{n_{k}}; \end{align*} $$

when $\hat {\nu }=1,$ we may take

$$ \begin{align*} m_{k}=(k+1)!,\quad n_{1}=1,\quad n_{k+1}=m_{k}+\frac{m_{k}}{\log m_{k}},\quad B_{k}=\lfloor2^{\sqrt{m_{k}}}\rfloor. \end{align*} $$

Remark 4.6. Combining Remark 3.5 and the fact

$$ \begin{align*} E(\hat{\nu},\nu)\subseteq E_{\ast}(\hat{\nu},\nu)=\{x\in[0,1]\colon \hat{\nu}(x)\ge\hat{\nu},~\nu(x)=\nu\}, \end{align*} $$

we have $\dim _{H}E_{\ast }(\hat {\nu },\nu )=s(\xi ,{\tau (i)}).$

Remark 4.7. By Lagrange’s theorem, any quadratic irrational number $y_1$ is represented by a periodic continued fraction expansion, that is,

$$ \begin{align*}{y_1=[a_1(y_1),a_2(y_1),\ldots,a_{k_0}(y_1),\overline{a_{k_{0}+1}(y_1),\ldots,a_{k_{0}+h}(y_1)}]}\end{align*} $$

for some positive integers $k_{0}$ and $h.$ By Lemma 2.1(2), we readily check that the limit $\lim _n (({\log q_{n}(y_1)})/{n})$ exists, denoted by $\log g(y_1).$ However, unlike in the special $y=[i,i,\ldots ]$ , we cannot obtain a closed-form expression for $\log g(y_1)$ .

In Theorem 1.3, we replace $y=[i,i,\ldots ]$ by a general quadratic irrational number $y_1$ and consider the Hausdorff dimension of the corresponding set $E(\hat {\nu },\nu ).$ We obtain that

$$ \begin{align*} \dim_{H}E(\hat{\nu},\nu) =\left\{\!\begin{array}{ll} 1 & \ \ \ \text{if }~ \nu=0,\\ s\bigg(\dfrac{\nu^{2}}{(1+\nu)(\nu-\hat{\nu})},g(y_1)\bigg) & \ \ \ \text{if }~ 0\le\hat{\nu}\le\dfrac{\nu}{1+\nu}<\nu\leq\infty, \\ 0 & \ \ \ \text{otherwise,} \end{array} \right. \end{align*} $$

where $s ({\nu ^{2}}/({(1+\nu )(\nu -\hat {\nu })}),g(y_1))$ is the solution to

$$ \begin{align*}{P\bigg(T,-s\bigg(\log |T'|+\frac{\alpha}{1-\alpha}\log{g(y_1)}\bigg)\bigg)=0}\end{align*} $$

with $\alpha = {\nu ^{2}}/({(1+\nu )(\nu -\hat {\nu })}).$ The proof can be established following the same line as for the original theorem, with some crucial modifications as follows.

1. (1) In the proof of Lemma 3.2: two sequences $\{n_{k}'\}_{k\geq 1}$ and $\{m_{k}'\}_{k\geq 1}$ are modified as

   $$ \begin{align*} m_{0}'=0,~ n_k'&=\min\{n\ge m_{k-1}'\colon a_{n+1}(x)=a_1(y_1)\},\\ m_k'&=\max\{n\ge n_k'\colon (a_{n_k'+1}(x),\ldots,a_{n}(x))=(a_1(y_1),\ldots,a_{n-n_k'}(y_1))\}. \end{align*} $$

   The partial quotient $a_{n}(y_1)$ of $y_1$ is bounded uniformly in $n,$ which guarantees that equations (3.1) and (3.2) hold.

   The limit

   $$ \begin{align*}{\lim_{n\to\infty}\frac{-\log|I_n(y_1)|}{2n}=\lim_{n\to\infty}\frac{\log q_n(y_1)}{n}=\log g(y_1).}\end{align*} $$

2. (2) For $B> \max \{a_1(y_1),\ldots ,a_{k_0+h}(y_1)\},$ the cantor-like subset $E(B)$ is modified as

   $$ \begin{align*} \begin{aligned} &\{x\in[0,1)\colon 1\leq a_{n}(x)\leq B, (a_{n_{k}+1}(x),\ldots,a_{m_{k}}(x))\\ &\quad =(a_1(y_1),\ldots,a_{m_{k}-n_{k}}(y_1)), n\geq 1, k\geq 1\}. \end{aligned} \end{align*} $$

6.1 Upper bound of $\dim _{H} (F(\alpha )\cap G(\beta ))$

For $x=[a_{1}(x),a_{2}(x),\ldots ]\in F(\alpha )\cap G(\beta )$ with non-periodic continued fraction expansion, we associate x with two sequences $\{n_k\}$ and $\{m_k\}$ that satisfy the following properties:

1. (1) $n_k<m_k<n_{k+1}<m_{k+1}$ for $k\ge 1$ ;

2. (2) $a_{n_k}(x)=\cdots =a_{m_k}(x)$ for $k\ge 1;$

3. (3) $\liminf _{k\to \infty } (({m_k-n_k})/{n_{k+1}}) = {\alpha }/({1-\alpha }), \limsup _{k\to \infty }(({m_k-n_k})/{m_k})=\beta ;$

4. (4) the sequence $\{m_k\}$ grows exponentially;

5. (5) write $\xi = ({\beta ^{2}(1-\alpha )})/({\beta -\alpha })$ . For any $\varepsilon>0$ , there exist infinitely many k such that

   $$ \begin{align*} \sum_{i=1}^{k}(m_{i}-n_{i}+1)\ge m_{k}(\xi-\varepsilon). \end{align*} $$

   To this end, we define two ascending sequences $\{n_{k}'\}$ and $\{m_{k}'\}$ as follows:

   $$ \begin{align*} n_{1}'=1,\quad &a_{n_{k}'}(x)= a_{n_{k}'+1}(x)= \cdots =a_{m_{k}'}(x)\neq a_{m_{k}'+1}(x), n_{k+1}'=m_{k}'+1. \end{align*} $$

   Since $\beta =\limsup R_n(x)/n>0$ , we have that $\limsup _{k\to \infty }(m_{k}'-n_{k}')= +\infty $ , which enables us to pick a non-decreasing subsequence of $\{(n_k', m_k')\}_{k\geq 1}\colon $ , put $(n_{1},m_{1})=(n^{\prime }_{1},m^{\prime }_{1});$ having chosen $(n_{k}, m_k)=(n_{j_{k}}', m_{j_{k}}')$ for $k\ge 1$ , we set $j_{k+1}=\min \{j>j_k\colon m_{j}'-n_{j}'> m_{k}-n_{k}\},$ and put $(n_{k+1},m_{k+1})=(n_{j_{k+1}}', m_{j_{k+1}}')$ .

We readily check the following properties.

1. (a) The sequence $\{m_{k}-n_{k}\}_{k\geq 1}$ is non-decreasing and $\lim _{k\to \infty } (m_{k}-n_{k})= +\infty .$

2. (b) If $m_{k}\leq n\leq n_{k+1}+(m_{k}-n_{k})$ for $k\geq 1$ , then $R_{n}(x)=m_{k}-n_{k}+1$ and

   $$ \begin{align*} \frac{m_{k}-n_{k}+1}{n_{k+1}+(m_{k}-n_{k})}\leq \frac{R_{n}(x)}{n}\leq \frac{m_{k}-n_{k}+1}{m_{k}}. \end{align*} $$

3. (c) If $n_{k+1}+(m_{k}-n_{k})<n< m_{k+1}$ for $k\geq 1$ , then $R_{n}(x)=n-n_{k+1}+1$ , and

   $$ \begin{align*} \frac{m_{k}-n_{k}+1}{n_{k+1}+(m_{k}-n_{k})}\leq \frac{R_{n}(x)}{n}\leq \frac{m_{k+1}-n_{k+1}+1}{m_{k+1}}. \end{align*} $$

Properties (a) and (b) imply

(6.1) $$ \begin{align} \alpha=\liminf_{n\to\infty}\frac{R_{n}(x)}{n} =\liminf_{k\rightarrow\infty}\frac{m_{k}-n_{k}+1}{n_{k+1}+(m_{k}-n_{k})} \end{align} $$

and

(6.2) $$ \begin{align} \beta=\limsup_{n\to\infty}\frac{R_{n}(x)}{n} =\limsup_{k\rightarrow\infty}\frac{m_{k+1}-n_{k+1}+1}{m_{k+1}}. \end{align} $$

From equation (6.1), we obtain that

(6.3) $$ \begin{align} \liminf_{k\to\infty}\frac{R_{n_{k+1}}(x)}{n_{k+1}}=\liminf_{k\to\infty}\frac{m_{k}-n_{k}+1}{n_{k+1}} =\frac{\alpha}{1-\alpha}, \end{align} $$

which combined with equation (6.2) yields $\alpha /({1-\alpha })\le \beta $ , or equivalently $\alpha \leq {\beta }/({1+\beta })$ . Thus, the set $F(\alpha )\cap G(\beta )$ is at most countable when $\alpha> {\beta }/({1+\beta })$ . Moreover, equation (6.3) implies that $\{m_{k}\}_{k\geq 1}$ grows at least exponentially, namely, there exists $C>0$ , independent of $x,$ such that $k \leq C\log {m_{k}}$ for k large enough. Further, by equations (6.2) and (6.3), we also have

$$ \begin{align*} \liminf_{k\to\infty}\frac{m_{k}}{n_{k+1}} \ge \liminf_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}} \cdot\liminf_{k\to\infty}\frac{m_{k}}{m_{k}-n_{k}} =\frac{\alpha}{\beta(1-\alpha)}, \end{align*} $$

which combined with the Stolz–Cesàro theorem implies that

$$ \begin{align*} \begin{split} \liminf_k\frac{\sum_{i=1}^{k-1}(m_i-n_i+1)}{m_{k}} \ge \frac{\alpha\beta(1-\beta)}{\beta-\alpha}, \end{split} \end{align*} $$

and thus, for $\varepsilon>0$ and k large enough,

$$ \begin{align*} \sum_{i=1}^{k}(m_{i}-n_{i}+1)\geq \bigg(\frac{\alpha\beta(1-\beta)}{\beta-\alpha}-\frac{\varepsilon}{2}\bigg)m_{k}+(m_{k}-n_{k}+1). \end{align*} $$

Since there exist infinitely many k such that

(6.4) $$ \begin{align} m_{k}-n_{k}+1\ge m_{k}\bigg(\beta-\frac{\varepsilon}{2}\bigg), \end{align} $$

property (5) holds for such k.

6.2 Covering of $F(\alpha )\cap G(\beta )$

We collect all sequences $(\{n_{k}\}, \{m_{k}\})$ associated with $x\in F(\alpha )\cap G(\beta )$ as above to form a set

$$ \begin{align*} \Omega= \{(\{n_{k}\},\{m_{k}\})\colon \text{Properties } (1) ~\text{and} ~(3) \text{ are fulfilled}\}. \end{align*} $$

For $(\{n_{k}\},\{m_{k}\})\in \Omega $ and $\{b_{k}\}\subset \mathbb N$ , write

$$ \begin{align*} H(\{n_{k}\},\{m_{k}\})&= \{x\in[0,1)\colon \text{Property } (2) \text{ is fulfilled}\},\\ \Lambda_{k, m_k} &= \{(n_1,m_1; \ldots; n_{k-1},m_{k-1};n_{k})\colon n_{1}<m_{1}<\cdots<m_{k-1}<n_{k},\\ &\qquad \text{equation~(6.4)}\text{ holds}\},\\ \mathcal{D}_{n_1,m_1;\ldots;n_{k}, m_{k}}(\{b_{k}\})&= \{(\sigma_{1},\ldots,\sigma_{m_{k}})\in \mathbb{N}^{m_{k}}\colon \sigma_{n_{j}}=\cdots=\sigma_{m_{j}}=b_{j}\\ &\qquad\text{for all }1\leq j\leq k\}. \end{align*} $$

We obtain a covering of $ F(\alpha )\cap G(\beta )$ :

$$ \begin{align*} \begin{aligned} F(\alpha)\cap G(\beta) &\subseteq \bigcup_{(\{n_{k}\},\{m_{k}\})\in \Omega}H(\{n_{k}\},\{m_{k}\}) \\ &\subseteq\bigcap_{K=1}^{\infty} \bigcup_{k=K}^{\infty}\bigcup_{m_{k}= e^{{k}/{C}}}^{\infty} \bigcup_{(n_{1},m_{1},\ldots,m_{k-1},n_{k})\in\Lambda_{k, m_k}} \bigcup_{(b_{1},\ldots,b_{k})\in \mathbb{N}^{k}} \\ &\quad\times\bigcup_{(a_{1},\ldots,a_{m_{k}})\in \mathcal{D}_{n_1,m_1;\ldots;n_{k}, m_{k}}(\{b_{k}\})}I_{m_{k}}(a_{1},\ldots,a_{m_{k}}). \end{aligned} \end{align*} $$

Writing $t=s(\xi -\varepsilon ,{\tau (1)})+ ({\varepsilon }/{2})$ , we estimate the $(t+ ({\varepsilon }/{2}))$ -dimensional Hausdorff measure of $F(\alpha )\cap G(\beta )$ . Setting $\psi (m_{k})=m_{k}-\sum _{i=1}^{k}(m_{i}-n_{i}+1),$ we have $M=512\sum _{i=1}^{\infty } (\tau (i))^{-2t}.$ For sufficiently large k, we first have the following estimate:

(6.5) $$ \begin{align} & \sum\limits_{b_{k}=1}^{\infty}\cdots\sum\limits_{b_{1}=1}^{\infty}\sum\limits_{(a_{1},\ldots,a_{m_{k}})\in \mathcal{D}_{n_{k}, m_{k}}(\{b_{k}\})}|I_{m_{k}}(a_{1},\ldots,a_{m_{k}})|^{t+ ({\varepsilon}/{2})}\nonumber\\[-1.5pt]&\quad\le\sum\limits_{b_{k}=1}^{\infty}\cdots\sum\limits_{b_{1}=1}^{\infty}\sum\limits_{a_{1},\ldots,a_{\psi(m_{k})}\in\mathbb{N}} 4^{k(t+ ({\varepsilon}/{2}))} \bigg(\frac{1}{q_{\psi(m_{k})}(a_{1},\ldots, a_{\psi(m_{k})})}\bigg)^{2t+\varepsilon}\nonumber\\[-1.5pt]&\qquad\times\prod_{j=1}^{k}\bigg(\frac{1}{q_{m_j-n_j+1}(b_j,\ldots,b_j)}\bigg)^{2t+\varepsilon}\nonumber\\[-1.5pt]&\quad\le4^{k(t+ ({\varepsilon}/{2}))} \sum\limits_{a_{1},\ldots,a_{\psi(m_{k})}\in\mathbb{N}}\bigg(\frac{1}{q_{\psi(m_{k})}(a_{1},\ldots,a_{\psi(m_{k})})}\bigg)^{2t+\varepsilon}\nonumber\\[-1.5pt]&\qquad\times\prod_{j=1}^{k} \bigg(\sum_{i=1}^{\infty}\bigg({1}/({q_{m_j-n_j+1}) (i,\ldots,i)}\bigg)^{2t+\varepsilon}\bigg)\nonumber\\[-1.5pt]&\quad\le(4^{t+({\varepsilon}/{2})}M)^{k} \sum\limits_{a_{1},\ldots,a_{\psi(m_{k})}\in\mathbb{N}}\bigg(\frac{1}{q_{\psi(m_{k})}(a_{1},\ldots,a_{\psi(m_{k})})}\bigg)^{2t+\varepsilon}\nonumber\\[-1.5pt]& \qquad \times \prod_{j=1}^{k} \bigg(\frac{1}{q_{m_j-n_j+1}(1,\ldots,1)}\bigg)^{2t+\varepsilon}\nonumber\\&\quad\le(16^{t+ ({\varepsilon}/{2})}M)^{k}\sum\limits_{a_{1},\ldots,a_{\psi(m_{k})}\in \mathbb{N}}\bigg(\frac{1}{q_{m_{k}}(a_{1},\ldots,a_{\psi(m_{k})},1,\ldots,1)}\bigg)^{2t+\varepsilon}\nonumber\\ &\quad\le(16^{t+ ({\varepsilon}/{2})}M)^{k}\nonumber\\ & \qquad \times \sum\limits_{a_{1},\ldots,a_{m_{k}-\lfloor m_{k}(\xi-\varepsilon)\rfloor}\in \mathbb{N}}\bigg(\frac{1}{q_{m_{k}}(a_{1},\ldots,a_{m_{k}-\lfloor m_{k}(\xi-\delta)\rfloor},1,\ldots,1)}\bigg)^{2s_{m_{k}}(\xi-\varepsilon,{\tau(1)})+\varepsilon}\nonumber\\&\quad\le(16^{t+ ({\varepsilon}/{2})}M)^{k}\bigg(\frac{1}{2}\bigg)^{(({m_{k}-1})/{2})\varepsilon}, \end{align} $$

where the third inequality holds since

$$ \begin{align*} \begin{split} \sum\limits_{i=1}^{\infty} \bigg(\frac{1}{q_{n}(i,\ldots,i)}\bigg)^{2t+\varepsilon} &=\bigg(\frac{1}{q_{n}(1,\ldots,1)}\bigg)^{2t+\varepsilon} \sum\limits_{i=1}^{\infty}\bigg(\frac{q_{n}(1,\ldots,1)}{q_{n}(i,\ldots,i)}\bigg)^{2t+\varepsilon}\\ &\le\bigg(\frac{1}{q_{n}(1,\ldots,1)}\bigg)^{2t+\varepsilon} \sum\limits_{i=1}^{\infty}\bigg(\frac{4\tau(1)}{\tau(i)}\bigg)^{2t+\varepsilon}; \end{split} \end{align*} $$

the penultimate one follows from equations (6.4) and (3.7), and the last one is by Remark 2.11 and Lemma 2.1(1).

Hence,

$$ \begin{align*} \begin{split} &\mathcal{H}^{t+ ({\varepsilon}/{2})} (F(\alpha)\cap G(\beta)) \leq \liminf_{K\to\infty}\sum\limits_{k=K}^{\infty}\sum_{m_{k}= e^{{k}/{C}}}^{\infty} \sum_{(n_{1},m_{1},\ldots,m_{k-1},n_{k})\in\Lambda_{k, m_k}}\sum\limits_{b_{k}=1}^{\infty}\cdots \sum\limits_{b_{1}=1}^{\infty}\\ &\qquad\times\sum\limits _{(a_{1},\ldots,a_{m_{k}})\in\mathcal{D}_{n_{k}, m_{k}}(\{b_{k}\})}|I_{m_{k}}(a_{1},\ldots,a_{m_{k}})|^{t+ ({\varepsilon}/{2})}\\ &\quad\overset{({6.5})}{\leq}\liminf_{K\to\infty}\sum\limits_{k=K}^{\infty}\sum\limits_{m_{k}=e^{{k}/{C}}}^{\infty} \sum_{n_{k}=1}^{m_{k}}\sum_{m_{k-1}=1}^{n_{k}}\cdots\sum\limits_{m_{1}=1}^{n_{2}} \sum\limits_{n_{1}=1}^{m_{1}}(16^{t+ ({\varepsilon}/{2})}M)^{k}\bigg(\frac{1}{2}\bigg)^{(({m_{k}-1})/{2})\varepsilon}\\ &\quad\leq\liminf_{K\to\infty}\sum\limits_{k=K}^{\infty} \sum\limits_{m_{k}=e^{{k}/{C}}}^{\infty}(16^{t + ({\varepsilon}/{2})}Mm_{k})^{2C\log m_{k}}\bigg(\frac{1}{2}\bigg)^{(({m_{k}-1})/{2})\varepsilon}\\ &\quad\leq\liminf_{K\to\infty}\sum\limits_{k=K}^{\infty} \sum\limits_{m_{k}=e^{{k}/{C}}}^{\infty}\bigg(\frac{1}{2}\bigg)^{(({m_{k}-1})/{4})\varepsilon} \leq \frac{1}{1-(\frac{1}{2})^{{\varepsilon}/{4}}}\sum\limits_{k=1}^{\infty} \bigg(\frac{1}{2^{\varepsilon}}\bigg)^{({e^{{k}/{C}}-1})/{4}} < +\infty, \end{split} \end{align*} $$

where the penultimate one holds since $(16^{t+ ({\varepsilon }/{2})}Mk)^{2C\log k}<2^{(({k-1})/{4})\varepsilon }$ for k large enough.

6.3 Lower bound of $\dim _{H} (F(\alpha )\cap G(\beta ))$

Note that $F(\alpha )\cap F(\beta )$ is at most countable for $\alpha> {\beta }/({1+\beta })$ ; we assume that $\alpha \le {\beta }/({1+\beta }).$ Let $\{n_{k}\}$ and $\{m_{k}\}$ be two strictly increasing sequences satisfying the following conditions:

1. (1) $(m_{k}-n_{k})\leq (m_{k+1}-n_{k+1})$ and $n_{k}< m_{k}< n_{k+1}$ for $k\geq 1$ ;

2. (2) $\lim _{k\to \infty }(({m_{k}-n_{k}})/{n_{k+1}}) = {\alpha }/({1-\alpha })$ ;

3. (3) $\lim _{k\to \infty } (({m_{k}-n_{k}})/{m_{k}}) = \beta $ .

With the help of these sequences, we construct a Cantor subset of $F(\alpha )\cap G(\beta )$ to provide a lower bound estimate of its dimension. The set $E(B)$ is defined in much the same way as in §4.2, the only difference being that the digit i is replaced by the digit $1$ ; the mapping f is defined in exactly the same way. It remains to verify that the set $f (E(B))$ is a subset of $F(\alpha )\cap G(\beta ).$

Proof. Recall the definitions of

$$ \begin{align*}t_{k}=\max \{t\in \mathbb{N}\colon m_{k}+t(m_{k}-n_{k})< n_{k+1}\},\end{align*} $$

$$ \begin{align*}m_{k}'=m_{k}+\sum_{l=1}^{k-1}(t_{l}+2),\quad n_{k}'=n_{k}+\sum_{l=1}^{k-1}(t_{l}+2).\end{align*} $$

We know that

$$ \begin{align*} \lim_{k\to\infty}\frac{\sum_{l=1}^{k}(t_{l}+2)}{n_{k+1}}=0,\quad \lim_{k\to\infty}\frac{n_k}{n_{k}'}=\lim_{k\to\infty}\frac{m_k}{m_k'}=1. \end{align*} $$

For $x\in f(E(B))$ , and $m_{k}'\leq n<m_{k+1}'$ with $k\in \mathbb {N},$ we have that

$$ \begin{align*} R_{n}(x)=\left\{\!\begin{array}{ll} m_{k}-n_{k}& \ \text{if } m_{k}'\leq n\leq n_{k+1}'+m_{k}-n_{k}, \\ n-n_{k+1}' & \ \text{if } n_{k+1}'+m_{k}-n_{k}< n< m_{k+1}' .\end{array} \right. \end{align*} $$

Observing that for $n_{k+1}'+m_{k}-n_{k}< n< m_{k+1}',$

$$ \begin{align*} \frac{m_{k}-n_{k}}{n_{k+1}'+m_{k}-n_{k}}\leq\frac{R_{n}(x)}{n}=\frac{n-n_{k+1}'}{n}\leq\frac{m_{k+1}-n_{k+1}}{m_{k+1}'}, \end{align*} $$

we deduce that

$$ \begin{align*} \liminf_{n\to\infty}\frac{R_{n}(x)}{n} =\lim_{k\to\infty}\frac{R_{n_{k+1}'+m_{k}-n_{k}}(x)}{n_{k+1}'+m_{k}-n_{k}} =\lim_{k\to\infty}\frac{m_{k}-n_{k}}{n_{k+1}+m_{k}-n_{k}} =\alpha \end{align*} $$

and

$$ \begin{align*} \limsup_{n\to\infty}\frac{R_{n}(x)}{n} = \lim_{k\to\infty}\frac{m_{k+1}-n_{k+1}}{m_{k+1}'} = \lim_{k\to\infty}\frac{m_{k+1}-n_{k+1}}{m_{k+1}} =\beta. \end{align*} $$

Hence, $x\in F(\alpha )\cap G(\beta )$ .