2.1 Continued fraction

It is well known that if $x \in (0,1)$ is a rational number, the expansion of x is finite; if $x \in (0,1)$ is an irrational number, the expansion of x is infinite. A finite truncation on the expansion of x gives rational fraction $p_n(x)/q_n(x): =[a_1(x), a_2(x), \ldots , a_n(x)]$ , which is called the nth convergents of x. With the conventions

$$ \begin{align*} p_{-1}=1,\quad q_{-1}=0,\quad p_0=0\quad\text{and}\quad q_0=1, \end{align*} $$

the sequence $p_n=p_n(x)$ and $q_n =q_n(x)$ can be given by the following recursive relations:

(2.1) $$ \begin{align} p_{n+1}=a_{n+1}(x)p_n+p_{n-1},\quad q_{n+1}=a_{n+1}(x)q_n+q_{n-1}. \end{align} $$

Clearly, $q_n(x)$ is determined by $a_1(x), \ldots , a_n(x)$ . So we may write $q_n(a_1(x),\ldots , a_n(x))$ . When no confusion is likely to arise, we write $a_n$ and $ q_n$ , respectively, in place of $a_n(x)$ and $q_n(x)$ for simplicity.

For an integer vector $(a_1,a_2,\ldots ,a_n) \in \mathbb {N}^n$ with $ n\geq 1$ , denote

(2.2) $$ \begin{align} I_n (a_1, a_2, \ldots,a_n):=\{x \in [0,1): a_1(x)=a_1, a_2(x)=a_2, \ldots, a_n(x)=a_n\} \end{align} $$

for the corresponding nth level cylinder, that is, the set of all real numbers in $[0,1)$ whose continued fraction expansions begin with $(a_1, \ldots , a_n)$ .

We will frequently use the following well-known properties of continued fraction expansion. They are explained in the standard texts [Reference Iosifescu and Kraaikamp16, Reference Khintchine17].

The next proposition describes the positions of cylinders $I_{n+1}$ of level $n+1$ inside the nth level cylinder $I_n$ .

2.2 Pressure function

Pressure function is an appropriate tool in dealing with dimension problems in infinite conformal iterated function systems. We recall that the pressure function with a continuous potential can be approximated by the pressure functions restricted to the sub-systems in continued fractions. For more information on pressure functions, we refer the readers to [Reference Hanus, Mauldin and Urbański9, Reference Mauldin and Urbański24, Reference Mauldin and Urbański25].

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

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

Then, with the Gauss map T restricted to it, $X_{\mathbb A}$ forms a dynamical system. The pressure function restricted to this sub-system $(X_{\mathbb A}, T)$ with potential $\phi : [0,1) \to \mathbb {R}$ is defined as

(2.5) $$ \begin{align} P_{\mathbb A} (T, \phi) = \lim_{n \to \infty} \frac{1}{n} \log \sum_{(a_1, \ldots, a_n ) \in {\mathbb A}^n } \sup_{x \in X_{\mathbb A}} e^{S_n \phi([a_1, \ldots, a_n +x])}, \end{align} $$

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

The nth variation $\mathrm {Var}_n(\phi )$ of $\phi $ is defined as

$$ \begin{align*} \mathrm{Var}_n (\phi) := \sup \{ |\phi(x)-\phi(y)|: I_n (x) =I_n (y)\}. \end{align*} $$

The existence of the limit in equation (2.5) is due to the following result.

The following proposition states a continuity of the pressure function when the continued fraction system $( [0,1), T) $ is approximated by its sub-systems $(X_{\mathbb A}, T)$ .

Proposition 2.4. [Reference Hanus, Mauldin and Urbański9, Proposition 2]

Let $\phi : [0,1) \to \mathbb {R}$ be a real function with $\mathrm {Var}_1(\phi ) < \infty $ and $\mathrm {Var}_n(\phi ) \to 0$ as $n \to \infty $ . We have

(2.6) $$ \begin{align} P(T, \phi) = P_{\mathbb{N}} (T, \phi)=\sup \{ P_{\mathbb A}(T, \phi) : {\mathbb A} \text{ is a finite subset of } \mathbb{N}\}. \end{align} $$

The potential functions related to the dimension of $E_2(\{z_n\}_{n\ge 1}, B)$ will be taken as the following forms:

1. (1) $\phi _1(x)=-s\log |T^{\prime }(x)|-s^2\log B$ corresponding to the pre-dimensional number $s_{n,1}$ (see equation (4.5));

2. (2) $\phi _2(x)=-s\log |T^{\prime }(x)|-s\log B+(1-s)\alpha $ for some $\alpha>0$ corresponding to the pre-dimensional number $s_{n,2}$ (see equation (4.22));

3. (3) $\phi _3 (x)=-s\log |T^{\prime }(x)|-(s/2)\log B-s\beta $ for some $\beta \ge 0$ corresponding to the pre-dimensional number $s_{n,3}$ (see equation (4.31)).

2.3 Basic facts

Some lemmas will be established in this subsection for future use. The first one involves a summation that naturally appearing in the proof of the upper bound of $E_2(\{z_n\}_{n\ge 1},B)$ .

Lemma 2.5. For any $a\in {\mathbb {N}}^{+}$ and $t>1/2$ , the following holds:

(2.7) $$ \begin{align} \sum_{b \in {\mathbb{N}}^{+}\setminus\{a\}}\frac{a^t }{b^t|a-b|^t} \asymp a^{1-t}. \end{align} $$

Proof. We are going to decompose $\{b\in {\mathbb {N}}: b\ne a\}$ appearing in the left summation into four blocks, and in each block using a particular estimate. Write

$$ \begin{align*}\{b\in{\mathbb{N}}:b\ne a\}=A+B+C+D,\end{align*} $$

where

$$ \begin{align*} A&=\{b\in{\mathbb{N}}:1\le b\le a/2\},\quad B=\{b\in{\mathbb{N}}:a/2\le b<a\},\\ C&=\{b\in{\mathbb{N}}:a< b\le 2a\},\quad D=\{b\in{\mathbb{N}}:b>2a\}. \end{align*} $$

For the block A, since $1 \leq b \leq a/2$ for any $b\in A$ , one has $a/2 \le |a-b| < a$ . Thus,

$$ \begin{align*}\sum_{b\in A}\frac{a^t }{b^t|a-b|^t} \asymp \sum_{b\in A}\frac{1 }{b^t}\asymp \int_1^{a/2} \frac{1}{x^t }\, dx\asymp a^{1-t}.\end{align*} $$

For the block B, using $ a/2\leq b< a$ and both a and b are integers, we have

$$ \begin{align*} \sum_{b\in B}\frac{a^t }{b^t|a-b|^t} \asymp \sum_{b\in B}\frac{1 }{|a-b|^t}=\sum_{1\le c\le a/2}\frac{1}{c^t}\asymp a^{1-t}.\end{align*} $$

The estimation for the block C is similar to B, we omit the details.

For the block D, since $b>2a$ for any $b\in D$ , it follows that $ b/2 \leq |a-b|< b$ . So, by the condition $t>1/2$ ,

$$ \begin{align*}\sum_{b\in D}\frac{a^t }{b^t|a-b|^t} \asymp a^t\cdot\sum_{b\in D}\frac{1 }{b^{2t}}\asymp a^t\cdot\int_{2a}^{\infty} \frac{1}{x^{2t}}\,dx\asymp a^{1-t}.\end{align*} $$

Combine the above four estimations and the proof is completed.

Next, we explore some properties of the pre-dimension numbers $s_{n,i}$ and their relationship to $a_1(z_n)$ . Recall the definitions of $s_{n,i}$ in equation (1.3).

Lemma 2.6. Let $n\in {\mathbb {N}}$ and $f(s)=\sum _{a_1,\ldots ,a_n\in {\mathbb {N}}}q_n(a_1,\ldots ,a_n)^{-2s}$ . We have

$$ \begin{align*}f(s)<\infty \;\Longleftrightarrow\; s>1/2.\end{align*} $$

Moreover, $f(s)$ is continuous on $s>1/2$ and goes to infinity as $s\downarrow 1/2$ . Consequently:

1. (1) $s_{n,i}>1/2 \text { for } i=1,2,3;$

2. (2) $s_{n,1}$ satisfies

   $$ \begin{align*} \sum_{a_1,\ldots,a_n\in{\mathbb{N}}} \frac{1}{q_n(a_1,\ldots,a_n)^{2s_{n,1}}B^{ns_{n,1}^2}}=1.\end{align*} $$
   Similar arguments apply to $s_{n,2}$ and $s_{n,3}$ , with the summations being replaced in the obvious way according to their definitions.

Proof. The proofs follow the ideas in [Reference Wang and Wu26, Lemma 2.6] and [Reference Hanus, Mauldin and Urbański9, Lemma 3.2].

By Proposition 2.1(1), for any $s>0$ ,

(2.8) $$ \begin{align} f(s)&=\sum_{a_1,\ldots,a_n\in{\mathbb{N}}}q_n(a_1,\ldots,a_n)^{-2s}\asymp \sum_{a_1,\ldots,a_n\in{\mathbb{N}}}\prod_{i=1}^{n}a_i^{-2s}=\bigg(\sum_{a\in{\mathbb{N}}} a^{-2s}\bigg)^n\notag\\ &\asymp\bigg(\int_1^\infty x^{-2s}\,dx\bigg)^n=(1-2s)^{-n}, \end{align} $$

which clearly implies the first point of the lemma.

By Hölder inequality, $\log f(s)$ is convex on $s>1/2$ . Hence, $\log f(s)$ is continuous on $s>1/2$ , so is $f(s)$ . By equation (2.8), it is easily seen that $f(s)$ goes to infinity as $s\downarrow 1/2$ .

Items (1) and (2) follow from the continuity of $f(s)$ and the definitions of $s_{n,i}$ ( $i=1,2,3$ ) directly.

Proof. (1) Since $s_{n,1}\le s_{n,2}$ , by Lemma 2.6(2) and the definitions of $s_{n,1}$ and $s_{n,2}$ ,

$$ \begin{align*}\begin{aligned} &\sum_{a_1,\ldots,a_n\in{\mathbb{N}}}\frac{1}{q_n(a_1,\ldots,a_n)^{2s_{n,1}}B^{ns_{n,1}^2}} = 1\le \sum_{a_1,\ldots,a_n\in{\mathbb{N}}}\frac{a_1(z_n)^{1-s_{n,1}}}{q_n(a_1,\ldots,a_n)^{2s_{n,1}}B^{ns_{n,1}}}, \end{aligned}\end{align*} $$

which is equivalent to

$$ \begin{align*}a_1(z_n)\ge B^{ns_{n,1}}.\end{align*} $$

The proofs of the items (2)–(4) follow the same lines as item (1).

3.1 Optimal cover of $F_n$ when n is large enough and $s_{n,1}\le s_{n,2}$

Suppose that n is large enough so that equation (3.2) is satisfied. Since $s_{n,1}\le s_{n,2}$ , we have

(3.3) $$ \begin{align} s_n=s_{n,1}<s-\epsilon \end{align} $$

with s and $\epsilon $ given in equation (3.2), and by Lemma 2.7

$$ \begin{align*}a_1(z_n)\ge B^{ns_{n,1}}.\end{align*} $$

Now, consider the intersection of $F_n$ with $(n+1)$ th level cylinders. For any $ (\boldsymbol {a},a_{n+1})\in {\mathbb {N}}^{n+1} $ with $\boldsymbol {a}\in {\mathbb {N}}^n$ and $ a_{n+1}\le B^{ns_{n,1}}/2$ , let

$$ \begin{align*}J_{n+1}(\boldsymbol{a},a_{n+1}):=F_n\cap I_{n+1}(\boldsymbol{a},a_{n+1}).\end{align*} $$

Let $x\in I_{n+1}(\boldsymbol {a},a_{n+1})$ . Then, we have $a_{n+1}(x)=a_{n+1}$ , and so $|a_{n+1}(x)-a_1(z_n)|\ge B^{ns_{n,1}}/2$ , which is much larger than $|Tz_n-T^{n+1}x|$ . Applying the identity (3.1) and using $0\le T^{n+1}x, Tz_n\le 1$ , we get

(3.4) $$ \begin{align} \frac{|a_1(z_n)-a_{n+1}|}{8a_1(z_n)a_{n+1}}\le |T^nx-z_n|\le \frac{2|a_1(z_n)-a_{n+1}|}{a_1(z_n)a_{n+1}}. \end{align} $$

If x also belongs to $J_{n+1}(\boldsymbol {a},a_{n+1})$ , then it satisfies the above inequality and $ |T^nx-z_n||T^{n+1}x-Tz_n|\le B^{-n} $ . Thus,

$$ \begin{align*} |T^{n+1}x-Tz_n|\le \frac{8a_1(z_n)a_{n+1}}{|a_1(z_n)-a_{n+1}|B^n},\end{align*} $$

which implies that

$$ \begin{align*}\begin{aligned} J_{n+1}(\boldsymbol{a},a_{n+1}) &\subset\bigg\{x\in I_{n+1}(\boldsymbol{a},a_{n+1}):|T^{n+1}x-Tz_n|\le \frac{8a_1(z_n)a_{n+1}}{|a_1(z_n)-a_{n+1}|B^n}\bigg\}\\ &=\bigg\{x\in I_{n+1}(\boldsymbol{a},a_{n+1}):T^{n+1}x\in B\bigg(Tz_n,\frac{8a_1(z_n)a_{n+1}}{|a_1(z_n)-a_{n+1}|B^n}\bigg)\bigg\}. \end{aligned}\end{align*} $$

Since for any $x\in I_{n+1}(\boldsymbol {a},a_{n+1})$ ,

(3.5) $$ \begin{align} q_{n+1}(\boldsymbol{a},a_{n+1})^2\le (T^{n+1})^{\prime}(x)\le2q_{n+1}(\boldsymbol{a},a_{n+1})^2, \end{align} $$

it follows that

(3.6) $$ \begin{align} |J_{n+1}(\boldsymbol{a},a_{n+1})|&\le \frac{16 a_1(z_n)a_{n+1}}{q_{n+1}(\boldsymbol{a},a_{n+1})^2|a_1(z_n)-a_{n+1}|B^n}\notag\\ &\le\frac{16 a_1(z_n)}{q_{n}(\boldsymbol{a})^2a_{n+1}|a_1(z_n)-a_{n+1}|B^n}. \end{align} $$

We take this opportunity to infer that $J_{n+1}(\boldsymbol {a},a_{n+1})$ contains an interval of length comparable to equation (3.6), which will be needed in the subsequent proof of the lower bound of $\mathrm {dim}_{\mathrm {H}} E_2(\{z_n\}_{n\ge 1}, B)$ . Note that equation (3.4) holds for any $x \in I_{n+1}(\boldsymbol {a},a_{n+1})$ . If x further satisfies

$$ \begin{align*} |T^{n+1}x-Tz_n| \le \frac{a_1(z_n)a_{n+1}}{2|a_1(z_n)-a_{n+1}|B^n},\end{align*} $$

then by the second inequality in equation (3.4), one has $ |T^nx-z_n||T^{n+1}x-Tz_n|\le B^{-n} $ , which implies

$$ \begin{align*}J_{n+1}(\boldsymbol{a},a_{n+1})\supset \bigg\{x\in I_{n+1}(\boldsymbol{a},a_{n+1}):T^{n+1}x\in B\bigg(Tz_n,\frac{a_1(z_n)a_{n+1}}{2|a_1(z_n)-a_{n+1}|B^n}\bigg)\bigg\}.\end{align*} $$

By equation (3.5), $J_{n+1}(\boldsymbol {a},a_{n+1})$ contains an interval with length greater than

(3.7) $$ \begin{align} \frac{a_1(z_n)a_{n+1}}{4q_{n+1}(\boldsymbol{a},a_{n+1})^2|a_1(z_n)-a_{n+1}|B^n}\ge \frac{ a_1(z_n)}{16q_{n}(\boldsymbol{a})^2 a_{n+1}|a_1(z_n)-a_{n+1}|B^n}. \end{align} $$

Note that $F_n$ can be covered by the union of the interval $ \bigcup _{a_{n+1}>B^{ns_{n,1}}/2}I_{n+1}(\boldsymbol {a},a_{n+1}) $ and the sets $ J_{n+1}(\boldsymbol {a},a_{n+1})$ with $a_{n+1}\le B^{ns_{n,1}}/2$ . Therefore, by the previous discussion, the s-volume of optimal cover of $F_n$ can be estimated as follows:

$$ \begin{align*}\begin{aligned} &\sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\bigg(\frac{1}{q_n(\boldsymbol{a})^{2s}(B^{ns_{n,1}}/2)^s}+\sum_{a_{n+1}\le B^{ns_{n,1}}/2 }\frac{16^{s}a_1(z_n)^{s}}{q_{n}(\boldsymbol{a})^{2s}a_{n+1}^{s}|a_1(z_n)-a_{n+1}|^{s}B^{ns}}\bigg)\\ &\quad =\sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\bigg(\frac{1}{q_n(\boldsymbol{a})^{2s}(B^{ns_{n,1}}/2)^s}+\frac{16^s}{q_n(\boldsymbol{a})^{2s}B^{ns}}\cdot\sum_{a_{n+1}\le B^{ns_{n,1}}/2 }\frac{a_1(z_n)^{s}}{a_{n+1}^{s}|a_1(z_n)-a_{n+1}|^{s}}\bigg)\\ &\quad \asymp \sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\bigg(\frac{B^{-n s s_{n,1}}}{q_n(\boldsymbol{a})^{2s}}+\frac{B^{ns_{n,1}(1-s)-ns}}{q_{n}(\boldsymbol{a})^{2s}}\bigg) \asymp\sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\frac{B^{-n s s_{n,1}}}{q_n(\boldsymbol{a})^{2s}}, \end{aligned}\end{align*} $$

where we use the main ideas of the proof of Lemma 2.5 with $t=s>s_{n,1}+\epsilon >1/2$ (by Lemma 2.6(1)), $a=a_1(z_n)$ and $b=a_{n+1}$ for the inner-most summation in the second formula. By the definition of $s_{n,1}$ and the fact that $s>s_{n,1}+\epsilon $ , we see that

(3.8) $$ \begin{align} \sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\frac{B^{-n s s_{n,1}}}{q_n(\boldsymbol{a})^{2s}}\le 2^{-n\delta_1} \end{align} $$

for some $\delta _1>0$ depending on s only.

3.2 Optimal cover of $F_n$ when n is large enough and $s_{n,1}> s_{n,2}$

Suppose that n is large enough so that equation (3.2) is satisfied. Since $s_{n,1}> s_{n,2}$ , we have

(3.9) $$ \begin{align} s_n=\max \{ s_{n,2},s_{n,3} \}<s-\epsilon \end{align} $$

with s and $\epsilon $ given in equation (3.2), and by Lemma 2.7,

$$ \begin{align*}a_1(z_n)< B^{ns_{n,2}}.\end{align*} $$

Following the same notation in the last section, still let $J_{n+1}(\boldsymbol {a},a_{n+1}):=F_n\cap I_{n+1}(\boldsymbol {a},a_{n+1})$ with $ (\boldsymbol {a},a_{n+1})\in {\mathbb {N}}^{n+1} $ . Let $x\in I_{n+1}(\boldsymbol {a},a_{n+1})$ , and so $a_{n+1}(x)=a_{n+1}$ . The discussion is split into three subcases.

Subcase (1A): $ |a_1(z_n)-a_{n+1}|>1 $ . Since $|Tz_n-T^{n+1}x|<1$ , applying the identity (3.1), we can get

$$ \begin{align*}\frac{|a_1(z_n)-a_{n+1}|}{8a_1(z_n)a_{n+1}}\le |T^nx-z_n|\le \frac{2|a_1(z_n)-a_{n+1}|}{a_1(z_n)a_{n+1}}.\end{align*} $$

Since $x\in I_{n+1}(\boldsymbol {a},a_{n+1})$ is arbitrary, by the same reason as equations (3.6) and (3.7), we get that $J_{n+1}(\boldsymbol {a},a_{n+1})$ is contained in an interval with length at most

(3.10) $$ \begin{align} \frac{16 a_1(z_n)}{q_{n}(\boldsymbol{a})^2a_{n+1}|a_1(z_n)-a_{n+1}|B^n} \end{align} $$

and contains an interval with length at least

(3.11) $$ \begin{align} \frac{a_1(z_n)}{16q_{n}(\boldsymbol{a})^2a_{n+1}|a_1(z_n)-a_{n+1}|B^n}. \end{align} $$

Subcase (1B): $ |a_1(z_n)-a_{n+1}|=1 $ . Applying the identity (3.1) again, we obtain

$$ \begin{align*} |T^nx-z_n|=\bigg|\frac{1-(Tz_n-T^{n+1}x)}{(a_{n+1}+T^{n+1}x)(a_1(z_n)+Tz_n)}\bigg|\ge \frac{|1-(Tz_n-T^{n+1}x)|}{4a_1(z_n)a_{n+1}}.\end{align*} $$

Denote

$$ \begin{align*}J_{n+1}^{(1)}(\boldsymbol{a},a_{n+1}):=\{x\in I_{n+1}(\boldsymbol{a},a_{n+1}):|T^{n+1}x-Tz_n|\le 8a_1(z_n)a_{n+1}B^{-n}\}\end{align*} $$

and

$$ \begin{align*}J_{n+1}^{(2)}(\boldsymbol{a},a_{n+1}):=\{x\in I_{n+1}(\boldsymbol{a},a_{n+1}):|T^{n+1}x-Tz_n|\ge 1-8a_1(z_n)a_{n+1}B^{-n}\}.\end{align*} $$

If $ 8a_1(z_n)a_{n+1} B^{-n} \ge 1/2 $ , then the union of the above two sets is $ I_{n+1}(\boldsymbol {a},a_{n+1}) $ , which can obviously cover $J_{n+1}(\boldsymbol {a},a_{n+1})$ .

Now suppose that $ 8a_1(z_n)a_{n+1} B^{-n}< 1/2$ . Let $y \notin J_{n+1}^{(1)}(\boldsymbol {a},a_{n+1})\cup J_{n+1}^{(2)}(\boldsymbol {a},a_{n+1})$ . There are two cases, one is $ 1/2\le |T^{n+1}y-Tz_n|<1-8a_1(z_n)a_{n+1} B^{-n} $ , and the other is $ 8a_1(z_n)a_{n+1} B^{-n}<|T^{n+1}y-Tz_n|<1/2 $ . For the first case,

$$ \begin{align*}\begin{aligned} |T^ny-z_n||T^{n+1}y-Tz_n|&\geq \frac{|1-(Tz_n-T^{n+1}y)|}{4a_1(z_n)a_{n+1}}\cdot|T^{n+1}y-Tz_n|\\ &> \frac{8a_1(z_n)a_{n+1} B^{-n}}{4a_1(z_n)a_{n+1}}\cdot\frac{1}{2}=B^{-n}. \end{aligned}\end{align*} $$

By employing the same strategy on the other case, one has for $y\notin J_{n+1}^{(1)}(\boldsymbol {a},a_{n+1})\cup J_{n+1}^{(2)}(\boldsymbol {a},a_{n+1})$ ,

$$ \begin{align*}|T^ny-z_n||T^{n+1}y-Tz_n|>B^{-n},\end{align*} $$

which implies that $y\notin J_{n+1}(\boldsymbol {a},a_{n+1})$ . Summarizing,

$$ \begin{align*}J_{n+1}(\boldsymbol{a},a_{n+1})\subset J_{n+1}^{(1)}(\boldsymbol{a},a_{n+1})\cup J_{n+1}^{(2)}(\boldsymbol{a},a_{n+1}),\end{align*} $$

and therefore by the same reason as equation (3.6), $J_{n+1}(\boldsymbol {a},a_{n+1})$ can be covered by two intervals whose length is at most

(3.12) $$ \begin{align} \frac{16a_1(z_n)B^{-n}}{q_{n}(\boldsymbol{a})^2a_{n+1}|a_1(z_n)-a_{n+1}|}. \end{align} $$

Subcase (1C): $ a_1(z_n)=a_{n+1} $ . Using the identity (3.1), one has

$$ \begin{align*}\frac{|T^{n+1}x-Tz_n|^2}{4a_1(z_n)^2}\le |T^nx-z_n||T^{n+1}x-Tz_n|\le \frac{|T^{n+1}x-Tz_n|^2}{a_1(z_n)^2}.\end{align*} $$

Clearly, the following holds:

$$ \begin{align*}\begin{aligned} &\{x\in I_{n+1}(\boldsymbol{a},a_{n+1}):|T^{n+1}x-Tz_n| \le a_1(z_n)B^{-n/2}\}\\ &\quad \subset J_{n+1}(\boldsymbol{a},a_{n+1}) \\ &\quad \subset \{x\in I_{n+1}(\boldsymbol{a},a_{n+1}):|T^{n+1}x-Tz_n|\le 2 a_1(z_n)B^{-n/2}\}. \end{aligned} \end{align*} $$

This together with equation (3.5) gives

(3.13) $$ \begin{align} |J_{n+1}(\boldsymbol{a},a_{n+1})|\le \frac{2a_1(z_n) B^{-n/2}}{q_{n+1}(\boldsymbol{a},a_1(z_n))^2}\le\frac{2 B^{-n/2}}{q_n(\boldsymbol{a})^2a_1(z_n)} \end{align} $$

and $J_{n+1}(\boldsymbol {a},a_{n+1})$ contains an interval with length greater than

(3.14) $$ \begin{align} \frac{a_1(z_n) B^{-n/2}}{q_{n+1}(\boldsymbol{a},a_1(z_n))^2}\ge\frac{B^{-n/2}}{4q_n(\boldsymbol{a})^2a_1(z_n)}. \end{align} $$

Combining the estimations (3.10), (3.12) and (3.13), the s-volume of optimal cover of $F_n$ can be estimated as follows:

$$ \begin{align*}\begin{aligned} &\ll\sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\bigg(\sum_{a_{n+1}\ne a_1(z_n)}\frac{a_1(z_n)^s B^{-ns}}{q_{n}(\boldsymbol{a})^{2s}a_{n+1}^s|a_1(z_n)-a_{n+1}|^s}+\frac{B^{-ns/2}}{q_n(\boldsymbol{a})^{2s}a_1(z_n)^{s}}\bigg)\\ &\quad \asymp \sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\bigg(\frac{a_1(z_n)^{1-s} B^{-ns}}{q_{n}(\boldsymbol{a})^{2s}}+\frac{B^{-ns/2}}{q_n(\boldsymbol{a})^{2s}a_1(z_n)^{s}}\bigg), \end{aligned}\end{align*} $$

where we have used Lemma 2.5 with $t=s>\max \{ s_{n,2},s_{n,3} \}+\epsilon >1/2$ (by Lemma 2.6(1)), $a=a_1(z_n)$ and $b=a_{n+1}$ for the inner-most summation. By the definitions of $s_{n,2}$ and $s_{n,3}$ and $s>\max \{ s_{n,2},s_{n,3} \}+\epsilon $ (see equation (3.9)), we have

(3.15) $$ \begin{align} \sum_{\boldsymbol{a}\in{\mathbb{N}}^n}\bigg(\frac{a_1(z_n)^{1-s} B^{-ns}}{q_{n}(\boldsymbol{a})^{2s}}+\frac{B^{-ns/2}}{q_n(\boldsymbol{a})^{2s}a_1(z_n)^{s}}\bigg)\le 2^{-n\delta_2} \end{align} $$

for some $\delta _2>0$ depending on s only.

3.3 Completing the proof of the upper bound of $\mathrm {dim}_{\mathrm {H}} E_2(\{z_n\}_{n\ge 1},B)$

By the previous two subsections, we see that for given $s>\limsup _{n \to \infty } s_n$ , by equations (3.8) and (3.15), there exists a cover of $F_n$ for which the corresponding s-volume

$$ \begin{align*}\ll 2^{-n\min \{ \delta_1,\delta_2 \}},\end{align*} $$

where $\delta _1>0$ and $\delta _2>0$ are respectively given in (3.8) and (3.15) and depend on s only. Note that for any $N\in {\mathbb {N}}$ ,

$$ \begin{align*}E_2(\{z_n\}_{n\ge 1}, B)\subset \bigcup_{n=N}F_n.\end{align*} $$

By the definition of Hausdorff measure, one has

$$ \begin{align*}\mathcal H^s(E_2(\{z_n\}_{n\ge 1}, B))\ll\lim_{N\to\infty}\sum_{n=N}^{\infty}2^{-n\min \{ \delta_1,\delta_2 \}}=0,\end{align*} $$

which implies that $ \mathrm {dim}_{\mathrm {H}} E_2(\{z_n\}_{n\ge 1}, B)\le s $ . By the arbitrariness of $ s $ ( $>\limsup _{n \to \infty } s_n$ ), we have

$$ \begin{align*}\mathrm{dim}_{\mathrm{H}} E_2(\{z_n\}_{n\ge 1}, B)\le \limsup_{n\to\infty} s_n,\end{align*} $$

which is what we want.

4.1 Case I: $ s^*=\limsup _{n \to \infty } s_n$ is obtained along a subsequence of $\{s_{n,1}\}_{n\ge 1}$

Note that $t<s^*$ . There exist infinitely many n such that

(4.3) $$ \begin{align} s_n=s_{n,1}>t. \end{align} $$

For such n, by the definition of $s_n$ ,

$$ \begin{align*}s_{n,1}\le s_{n,2},\end{align*} $$

which by Lemma 2.7(1) gives

(4.4) $$ \begin{align} a_1(z_n)\ge B^{ns_{n,1}}. \end{align} $$

Following the same argument as [Reference Wang and Wu26, Lemma 2.4] with some obvious modification, we have

(4.5) $$ \begin{align} s^*=\lim_{n\text{ satisfies equation~}(4.3)\atop n\to\infty}s_{n,1}=\inf\{s>0:P(T,-s\log|T^{\prime}|-s^2\log B)\le 0\}. \end{align} $$

Fix an $\epsilon <s^*-t$ . By Proposition 2.4 and Lemma 2.6(2), there exist integers $ \ell>\max \{\log _B4/\epsilon , 2t/\epsilon +1 \} $ and $ M>0 $ such that the unique positive number $s=s(\ell ,M)$ satisfying the equation

(4.6) $$ \begin{align} \sum_{\boldsymbol{a}\in\{1,\ldots,M \}^\ell}\frac{1}{q_{\ell}(\boldsymbol{a})^{2s}B^{\ell s^2}}=1 \end{align} $$

is greater than $t+\epsilon $ . It should be emphasized that $ \ell>\max \{ \log _B4/\epsilon , 2t/\epsilon +1 \} $ implies that

(4.7) $$ \begin{align} B^{\ell\epsilon}\ge 4 \end{align} $$

and that for any $\boldsymbol {a}\in {\mathbb {N}}^\ell $ ,

(4.8) $$ \begin{align} q_\ell(\boldsymbol{a})^{-2s}\le q_\ell(\boldsymbol{a})^{-2(t+\epsilon)}\le q_\ell(\boldsymbol{a})^{-2t}2^{-(\ell-1)\epsilon}\le q_\ell(\boldsymbol{a})^{-2t}2^{-2t}. \end{align} $$

Choose $n\in {\mathbb {N}}$ so that $n-k\ge \ell kt/\epsilon $ and equation (4.3) is satisfied, and write $n-k=m\ell +\ell _0$ , where $0\le \ell _0<\ell $ . By the choice of n, one has $m\ge kt/\epsilon $ .

From now on, let $\ell , M $ and n be fixed. Let $1^{\ell _0}$ be the word consisting only of $1$ and of length $\ell _0$ . Let $\tilde {\boldsymbol {u}}=(\boldsymbol {u},1^{\ell _0})\in {\mathbb {N}}^{k+\ell _0}$ and consider the following auxiliary set defined by $(n+1)$ th level cylinders:

(4.9) $$ \begin{align} \begin{aligned} \{I_{n+1}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b): \ & \boldsymbol{a}_i\in\{1,\ldots,M\}^\ell, 1\le i\le m,\\ & B^{nt}\le b\le 2B^{nt} \text{ and } b \text{ is even}\}. \end{aligned} \end{align} $$

The additional requirement here, that b be even, is that any two cylinders in the above set are well separated (see Lemma 4.4 below). By equation (3.7), there is an interval, denoted by ${\mathcal I}_1(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ (given that there are three cases (§4.1–§4.3) to consider, each requiring the construction of subsets and measures, we will use subscripts 1, 2, 3 or superscripts (1), (2), (3) to distinguish and avoid burdening of notation, where there is no risk of ambiguity) such that

(4.10) $$ \begin{align} {\mathcal I}_1(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)\subset F_n\cap I_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b). \end{align} $$

It is easy to see that ${\mathcal I}_1(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)\subset F_n\cap I_{k+\ell _0}(\tilde { \boldsymbol {u}})\subset F_n\cap I_k(\boldsymbol {u})$ . Moreover, the length of this interval can be estimated as

(4.11) $$ \begin{align} |{\mathcal I}_1(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b) | & \geq \frac{a_1(z_n)}{16 q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2b|a_1(z_n)-b|B^n} \notag\\ & \geq \frac{1}{16 q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2 b B^{n}}\notag\\ & \geq \frac{1}{32 q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2 B^{n(1+t)}}, \end{align} $$

where we use $a_1(z_n)\ge B^{ns_{n,1}}>B^{nt}B^{n\epsilon }\ge 4B^{nt}$ (see equations (4.4) and (4.7)) and $b \leq 2B^{nt}$ in the second and third inequalities. In what follows, we call ${\mathcal I}_1(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ in $F_n\cap I_{k+\ell _0}(\tilde {\boldsymbol {u}})$ as fundamental interval, and the cylinders $I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ and $I_{k+\ell _0+p\ell }(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_p)$ with $0\le p\le m$ as basic cylinders.

Define a probability measure $\mu _1$ supported on $F_n\cap I_{k+\ell _0}(\tilde {\boldsymbol {u}})\subset F_n\cap I_k(\boldsymbol {u})$ as follows:

(4.12) $$ \begin{align} \mu_1=\sum_{\boldsymbol{a}_1\in\{1,\ldots,M \}^\ell}\cdots\sum_{\boldsymbol{a}_m\in\{1,\ldots,M \}^\ell}\sum_{B^{nt}\le b\le 2 B^{nt}\atop b\text{ is even}}\bigg(\prod_{i=1}^m\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s^2}}\bigg)\cdot \frac{2}{B^{nt}}\cdot{\mathcal L}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b}^{(1)}, \end{align} $$

where

$$ \begin{align*}{\mathcal L}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b}^{(1)}:=\frac{{\mathcal L}|_{{\mathcal I}_1(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)}}{{\mathcal L}({\mathcal I}_1(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b))} \end{align*} $$

denotes the normalized Lebesgue measure on the fundamental interval ${\mathcal I}_1(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots , \boldsymbol {a}_m,b)$ . Roughly speaking, we assign each word $(\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ a different weight: the weights corresponding to $\boldsymbol {a}_i$ and b are respectively

$$ \begin{align*}\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s^2}}\quad\text{and}\quad \frac{2}{B^{nt}}.\end{align*} $$

The definition of s (see equation (4.6)) ensures that $\mu _1$ is a probability measure.

The next two lemmas describe the gap between fundamental intervals defined in equation (4.10) and their $\mu _1$ -measures, respectively.

Proof. (1) Bear in mind that ${\mathcal I}_1$ and ${\mathcal I}_1^{\prime }$ are two fundamental intervals contained in the $(n+1)$ th level cylinders $I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ and $I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1^{\prime },\ldots ,\boldsymbol {a}_m^{\prime },b^{\prime })$ , respectively. For further discussion, we write $ (\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)=(c_1,c_2,\ldots , c_{m\ell },c_{m\ell +1})$ and $ (\boldsymbol {a}_1^{\prime },\ldots ,\boldsymbol {a}_m^{\prime },b^{\prime }) = (c_1^{\prime },c_2^{\prime },\ldots , c_{m\ell }^{\prime },c_{m\ell +1}^{\prime }) $ for the moment. Assume that $1\le i\le m\ell +1$ is the smallest integer such that $c_i \ne c_i^{\prime }$ . By the assumption in item (1), we have $(p-1) \ell < i \le p \ell \le m\ell $ . Therefore, the distance between ${\mathcal I}_1$ and ${\mathcal I}_1^{\prime }$ is majorized by

(4.13) $$ \begin{align} \mathrm{dist} (I_{k+\ell_0+i+1}(\tilde{\boldsymbol{u}},c_1,\ldots,c_{i-1},c_i,c_{i+1}),I_{k+\ell_0+i+1}(\tilde{\boldsymbol{u}},c_1,\ldots,c_{i-1},c_i^{\prime},c_{i+1}^{\prime})). \end{align} $$

Now, we consider two cases.

Case 1: $1\le i<m\ell $ . Since $i<m\ell $ , we have $c_{i+1},c_{i+1}^{\prime } \leq M$ . By the distribution of cylinders (see Proposition 2.2), either $I_{k+\ell _0+i+1}(\tilde {\boldsymbol {u}}, c_1, \ldots , c_{i-1},c_i, M+1)$ or $I_{k+\ell _0+i+1}(\tilde {\boldsymbol {u}}, c_1, \ldots , c_{i-1},c_i^{\prime }, M+1)$ lies between two cylinders listed in equation (4.13). Without loss of generality, assume that this is satisfied by the former one. Then, by equation (4.13) and using Proposition 2.1(2) repeatedly, we have

(4.14) $$ \begin{align} \mathrm{dist}({\mathcal I}_1,{\mathcal I}_2)&\ge| I_{k+\ell_0+i+1}(\tilde{\boldsymbol{u}}, c_1, \ldots, c_{i-1},c_i, M+1) | \notag\\ &\geq \frac{1}{2 q_{k+\ell_0+i+1} (\tilde{\boldsymbol{u}}, c_1, \ldots, c_{i-1},c_i, M+1)^2 } \notag\\ &\geq \frac{1}{2 (M+2)^2 (c_i+1)^2 q_{k+\ell_0+i-1} (\tilde{\boldsymbol{u}}, c_1, \ldots, c_{i-1})^2 } \notag\\ &\geq \frac{|I_{k+\ell_0+i-1} (\tilde{\boldsymbol{u}}, c_1, \ldots, c_{i-1})|}{2 (M+2)^4} \notag \\ &\geq \frac{|I_{k+\ell_0+p\ell} (\tilde{\boldsymbol{u}}, \boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)|}{2 (M+2)^4} \end{align} $$

as desired.

Case 2: $i=m\ell $ . In this case, $c_{i+1}=b$ and $c_{i+1}^{\prime }=b^{\prime }$ . By the distribution of cylinders (see Proposition 2.2), we see that either $I_{n+1}(\tilde {\boldsymbol {u}},c_1,\ldots ,c_{m\ell -1},c_{m\ell },1)$ or $I_{n+1}(\tilde {\boldsymbol {u}},c_1,\ldots ,c_{m\ell -1},c_{m\ell }^{\prime },1)$ lies between two cylinders listed in equation (4.13). By the same reason as equation (4.14), we can obtain the same conclusion.

(2) Without loss of generality, assume that $b < b^{\prime } $ . Note that by equation (4.9), both b and $b^{\prime }$ are even. Then, for any $b^{\prime \prime }$ with $b<b^{\prime \prime }<b^{\prime }$ , the $(n+1)$ th level cylinder $I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b^{\prime \prime })$ lies between ${\mathcal I}_1$ and ${\mathcal I}_1^{\prime }$ . This means the distance between ${\mathcal I}_1$ and ${\mathcal I}_1^{\prime }$ is at least

$$ \begin{align*}\begin{aligned} |I_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b^{\prime\prime})|&\ge \frac{1}{2q_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b^{\prime\prime})^2}\\ &\geq \frac{1}{2 \cdot 4^2 q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2B^{2nt}}\\ & \geq \frac{1}{2 \cdot 4^2 q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2 b^{2}}\\ &\ge \frac{1}{32}|I_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)|, \end{aligned}\end{align*} $$

where we use $B^{nt}\le b^{\prime \prime },b\le 2B^{nt}$ in the second and third inequalities.

Proof. (1) The conclusion is clear if $p=0$ . Now suppose that $p\ne 0$ . By the definition of $ \mu _1 $ , one has

(4.15) $$ \begin{align} &\mu_1(I_{k+\ell_0+p\ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p))\notag\\ &\quad =\prod_{i=1}^p\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s^2}}\le \prod_{i=1}^p\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}}\overset{\text{equation}~(4.8)}{\le} \prod_{i=1}^p\frac{1}{(2q_\ell(\boldsymbol{a}_i))^{2t}}\notag\\ &\quad \le \frac{1}{2^{2t} q_{p\ell}(\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)^{2t}}= \frac{q_{k+\ell_0}(\tilde{ \boldsymbol{u}})^{2t}}{q_{k+\ell_0}(\tilde {\boldsymbol{u}})^{2t}}\cdot\frac{1}{2^{2t} q_{p\ell}(\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)^{2t}}\notag\\ &\quad \le \frac{q_{k+\ell_0}(\tilde {\boldsymbol{u}})^{2t}}{q_{k+\ell_0+p \ell}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)^{2t}}\notag \\ &\quad \le\frac{q_{k+\ell_0+p \ell}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)^{-2t}}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}. \end{align} $$

(2) Recall that $n=k+\ell _0+m\ell $ . It should be noticed that equation (4.15) is actually an upper bound of $\prod _{i=1}^pq_\ell (\boldsymbol {a}_i)^{-2s}$ . In the spirit of equation (4.15), by the definition of $ \mu _1$ ,

$$ \begin{align*} \mu_1({{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)) &=\bigg(\prod_{i=1}^m\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s^2}}\bigg)\cdot \frac{2}{B^{nt}}\\ &\le \frac{q_{k+\ell_0+m \ell}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{-2t}}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot \frac{1}{B^{m\ell s^2} }\cdot \frac{2}{B^{nt}}. \end{align*} $$

Since $m\ge kt/\epsilon $ (which follows from the choice of n), one has

(4.16) $$ \begin{align} m\ell s^2&\ge m\ell(t+\epsilon)^2 \ge m\ell t^2+2m\ell t\epsilon\ge m\ell t^2+2 \ell kt^2 \nonumber\\&\ge m\ell t^2+(\ell+k)t^2\ge nt^2.\end{align} $$

Therefore,

(4.17) $$ \begin{align} \mu_1({{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)) & \le \frac{q_{k+\ell_0+m \ell}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{-2t}}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot \frac{1}{B^{n t^2} }\cdot \frac{2}{B^{nt}}\notag\\& =\frac{1}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot\frac{2}{q_{n}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\dots,\boldsymbol{a}_m)^{2t}B^{nt(1+t)}} \\ & \overset{\text{equation}~(4.11)}{\le} \frac{64 |{\mathcal I}_1(\tilde{\mathbf{u}},\mathbf{a}_1,\ldots,\mathbf{a}_m,b) |^t}{|I_{k+\ell_0}(\tilde {\mathbf{u}})|^t}. \notag\end{align} $$

Lemma 4.6. Let $ \mu _1$ be as in equation (4.12). For any $ r>0 $ and $ x\in [0,1] $ , we have

$$ \begin{align*}\mu_1(B(x,r))\le \frac{16(M+2)^4(M+1)^{2\ell}r^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}.\end{align*} $$

Proof. Without loss of generality, assume that $ x\in {{\mathcal I}_1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b) $ . Obviously, if $ r\ge |I_{k+\ell _0}(\tilde {\boldsymbol {u}})| $ , then

$$ \begin{align*}\mu_1 (B(x,r))=1\le \frac{r^{t}}{|I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^{t}}.\end{align*} $$

Hence, it is sufficient to focus on the case $ r<|I_{k+\ell _0}(\tilde {\boldsymbol {u}})| $ . Lemma 4.4 suggests that we need to consider three cases.

Case 1: $r\le |I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)|/32$ . By Lemma 4.4, we see that the distance between ${{\mathcal I}_1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ and other fundamental intervals contained in $F_n\cap I_{k+\ell _0}(\tilde {\boldsymbol {u}})$ is at least r. So $B(x,r)$ only intersects the fundamental interval ${{\mathcal I}_1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ to which x belongs. It follows that

$$ \begin{align*}\begin{aligned} \mu_1(B(x,r))&=\bigg(\prod_{i=1}^m\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s^2}}\bigg)\cdot \frac{2}{B^{nt}}\cdot {\mathcal L}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b}^{(1)}(B(x,r))\\ &=\mu_1({{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b))\cdot {\mathcal L}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b}^{(1)}(B(x,r)). \end{aligned}\end{align*} $$

Since

$$ \begin{align*}\begin{aligned} {\mathcal L}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b}^{(1)}(B(x,r))&\le\min\biggl\{1,\frac{2r}{|{{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)|}\biggr\}\le \frac{2r^t}{|{{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)|^t}, \end{aligned}\end{align*} $$

where the last inequality follows from $\min \{a, c\} \leq a^t c ^{1-t}$ for any $t \in [0,1]$ . By Lemma 4.5(2), we have

$$ \begin{align*}\begin{aligned} \mu_1(B(x,r)) &\leq \frac{64|{{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)|^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot \frac{2r^t}{|{{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)|^t}\\ &= \frac{128r^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}. \end{aligned} \end{align*} $$

Case 2: $|I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)|/32< r\le |I_{n}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m)|/(2(M+2)^4)$ . In this case, the ball $B(x,r)$ intersects exactly the nth level basic cylinder $I_{n}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m)$ , but may intersect multiple $(n+1)$ th level basic cylinders inside it. Therefore, by the definition of $\mu _1$ and equation (4.17),

(4.18) $$ \begin{align} \mu_1(B(x,r)) &\leq \#\Delta_1(x)\cdot\max_{B^{nt}\le b\le 2B^{nt}\atop b\text{ is even}} \mu_1({{\mathcal I}_1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b))\notag\\ &\le \#\Delta_1(x)\cdot \frac{1}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot\frac{2}{q_n(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2t}B^{nt(1+t)}}, \end{align} $$

where

$$ \begin{align*}\Delta_1(x)=\{B^{nt}\le b\le 2B^{nt}:~ b \text{ is even and } I_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)\cap B(x,r)\ne\emptyset\}.\end{align*} $$

To get the best upper bound for $\mu_1(B(x, r))$ , we need to use two methods to bound $ \#\Delta _1(x) $ from above. First, there are at most $B^{nt}/2$ choices for b and so

(4.19) $$ \begin{align} \#\Delta_1(x)\le B^{nt}/2. \end{align} $$

On the other hand, each $(n+1)$ th level basic cylinder $I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,b)$ is of length at least

$$ \begin{align*}2^{-1}q_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,b)^{-2}\ge 32^{-1}q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{-2}B^{-2nt},\end{align*} $$

which means that

$$ \begin{align*}\#\Delta_1(x)\le 2r\cdot 32 q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2}B^{2nt}.\end{align*} $$

This together with equation (4.19) gives

$$ \begin{align*}\begin{aligned} \# \Delta_1(x)&\le\min\{B^{nt}/2, 2r\cdot 32 q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2B^{2nt}\}\\ &\leq 64 B^{nt(1-t)}\cdot( r\cdot q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2}B^{2nt})^t\\ &=64 B^{nt(1+t)}\cdot r^t\cdot q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2t}. \end{aligned}\end{align*} $$

Substituting this upper bound for $\#\Delta _1(x)$ into equation (4.18), we get

$$ \begin{align*}\begin{aligned} \mu_1(B(x,r)) &\le 64 B^{nt(1+t)}\cdot r^t\cdot q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2t}\cdot \frac{1}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\\ &\quad\cdot \frac{2}{q_n(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2t}B^{nt(1+t)}}\\ &= 128\frac{r^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}. \end{aligned}\end{align*} $$

Case 3: $|I_{k+\ell _0+(p+1) \ell }(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_{p+1})|/(2(M+2)^4) \le r< | I_{k+\ell _0+p\ell }(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots, \boldsymbol {a}_p)|/ (2(M+2)^4)$ for some $1\le p\le m-1$ . In this case, the ball $B(x,r)$ only intersects one $(k+\ell _0+p \ell )$ th level basic cylinder, that is, $I_{ k+\ell _0+p \ell }(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_p)$ . Hence, by Lemma 4.5(1), we get

$$ \begin{align*}\begin{aligned} \mu_1(B(x,r))&\le\mu_1(I_{k+\ell_0+p\ell}(\tilde{ \boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)) \\ & \le \frac{q_{k+\ell_0+p \ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_{p})^{-2t}}{| I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^t} \\ & = \frac{q_{k+\ell_0+p \ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_{p})^{-2t} }{| I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^t}\cdot \frac{q_{\ell} (\boldsymbol{a}_{p+1})^{-2t}}{q_{\ell} (\boldsymbol{a}_{p+1})^{-2t}}\\ & \le \frac{4q_{k+\ell_0+(p+1) \ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_{p+1})^{-2t}}{| I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^t }\cdot (M+1)^{2\ell}\\ & \le \frac{ 16(M+2)^4(M+1)^{2\ell} r^t}{| I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^t }. \end{aligned}\end{align*} $$

Combining the estimation given in Cases 1–3, we arrive at the conclusion.

Completing the proof of Theorem 1.3 .

Recall that $\tilde { \boldsymbol {u}}=(\boldsymbol {u}, 1^{\ell _0}) \in {\mathbb {N}}^{k+\ell _0}$ . By Proposition 2.1(1), a simple calculation shows that

$$ \begin{align*}\frac{|I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^t}{|I_k(\boldsymbol{u})|^t}\ge\frac{ q_{k+\ell_0}(\tilde{ \boldsymbol{u}})^{-2t}}{2^{2t}\cdot q_{k}({ \boldsymbol{u}})^{-2t}}\ge \frac{1}{2^{4t}\cdot q_{\ell_0}(1^{\ell_0})^{2t}}\ge \frac{1}{2^{4t}\cdot 2^{t\ell_0}}\ge \frac{1}{2^{\ell+4}}.\end{align*} $$

Therefore, for any $\boldsymbol {u}\in {\mathbb {N}}^k$ , by Lemma 4.6 and mass distribution principle, we have

$$ \begin{align*}\begin{aligned} \mathcal H_\infty^t(F_n\cap I_{k}(\boldsymbol{u}))&\ge \mathcal H_\infty^t(F_n\cap I_{k+\ell_0}(\tilde{\boldsymbol{u}}))\\ &\ge \frac{1}{16(M+2)^4(M+1)^{2\ell}} |I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^t \mu_1 (F_n\cap I_{k+\ell_0}(\tilde{\boldsymbol{u}}))\\ &\ge \frac{1}{2^{\ell+8}(M+2)^4(M+1)^{2\ell}} |I_{k}(\boldsymbol{u})|^t\\ &\ge \frac{1}{2^{\ell+8}(M+2)^4(M+1)^{2\ell}} |I_{k}(\boldsymbol{u})|, \end{aligned}\end{align*} $$

where the last inequality follows from $t<s^*\le 1$ . Since $\ell $ and M depend on t only, and since the above Hausdorff content bound holds for infinitely many $n\in {\mathbb {N}}$ , by Lemma 4.1, we have

$$ \begin{align*} \mathrm{dim}_{\mathrm{H}} E_2(\{z_n\}_{n\ge 1},B)\ge s^*.\\[-34pt] \end{align*} $$

4.2 Case II: $s^*=\limsup _{n \to \infty } s_n$ is obtained along a subsequence of $\{s_{n,2}\}_{n\ge 1}$

In this case, there exist infinitely many n such that

(4.20) $$ \begin{align} s_n=s_{n,2}>t. \end{align} $$

For such n, by the definition of $s_n$ ,

$$ \begin{align*}s_{n,1}> s_{n,2}\ge s_{n,3},\end{align*} $$

which by Lemma 2.7(2) and (4) gives

(4.21) $$ \begin{align} B^{ns_{n,2}/2}\le a_1(z_n) < B^{ns_{n,2}}. \end{align} $$

By taking a subsequence, assume that integers satisfying equation (4.20) will ensure the existence of the following limit:

$$ \begin{align*}(s^* \log B)/2\le\lim_{n\text{ satisfies equation~}(4.20)\atop n\to\infty}\frac{\log a_1(z_n)}{n}=: \alpha\le s^* \log B.\end{align*} $$

Recall the definition of $s_{n,2}$ and using the continuity of pressure functions, the above discussion gives

(4.22) $$ \begin{align} s^*=\inf\{s\in [0,1]:P(T, -s\log|T^{\prime}|-s\log B+(1-s)\alpha)\le 0\}. \end{align} $$

Since most of the arguments in this subsection are quite identical to the last subsection, we will follow the same notation when there is no risk of ambiguity. In addition, to keep the paper of a manageable length, some proofs will not be detailed here if they are similar to those in §4.1. Instead, we will present only the main ideas.

Fix an $\epsilon <s^*-t$ . By Proposition 2.4, there exist integers $ \ell \ge 2t/\epsilon +1$ and $ M $ such that the unique positive number $s=s(\ell ,M)$ satisfying the equation

(4.23) $$ \begin{align} \sum_{\boldsymbol{a}\in\{1,\ldots,M \}^\ell}\frac{e^{\alpha\ell (1-s)}}{q_\ell(\boldsymbol{a})^{2s}B^{\ell s}}=1 \end{align} $$

is greater than $t+\epsilon $ .

Choose a large integer $n\in {\mathbb {N}}$ so that equation (4.20) is satisfied and

(4.24) $$ \begin{align} n-k\ge \ell,\quad e^{n(\alpha-\epsilon)}\le a_1(z_n)\le e^{n(\alpha+\epsilon)}. \end{align} $$

Write $n-k=m\ell +\ell _0$ , where $0\le \ell _0<\ell $ .

Analogously, let $\tilde {\boldsymbol {u}}=(\boldsymbol {u},1^{\ell _0})\in {\mathbb {N}}^{k+\ell _0}$ and consider the following set defined by $(n+1)$ th level cylinders:

$$ \begin{align*}\begin{aligned} \{I_{n+1}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c): ~&\boldsymbol{a}_i\in\{1,\ldots,M\}^\ell, 1\le i\le m,\\ & 2e^{(\alpha+\epsilon)n}\le c \le 3e^{(\alpha+\epsilon)n} \text{ and } c \text{ is even}\}. \end{aligned}\end{align*} $$

Similar to §4.1, we restrict c to be even for the only reason that any two cylinders in the above set are well separated (see Lemma 4.7 below). By equation (3.11), there is an interval, denoted by ${\mathcal I}_2(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,c)$ such that

(4.25) $$ \begin{align} {\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c)\subset F_n\cap I_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c). \end{align} $$

Moreover, the length of this interval can be estimated as

(4.26) $$ \begin{align} |{\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c)| &\ge \frac{a_1(z_n)}{16q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2 c |a_1(z_n)-c|B^n} \notag\\ & \ge \frac{1}{16\cdot 3 e^{2n\epsilon} q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2cB^{n}} , \end{align} $$

where the last inequality follows from $c\le 3e^{(\alpha +\epsilon )n}$ and $ e^{n(\alpha -\epsilon )}\le a_1(z_n)\le e^{n(\alpha +\epsilon )} $ (see equation (4.24)).

Define a probability measure $\mu _2$ supported on $F_n\cap I_{k+\ell _0}(\tilde {\boldsymbol {u}})\subset F_n\cap I_k(\boldsymbol {u})$ as follows:

$$ \begin{align*} \mu_2=\!\!\!\!\sum_{\boldsymbol{a}_1\in\{1,\ldots,M \}^\ell}\!\cdots\!\!\sum_{\boldsymbol{a}_m\in\{1,\ldots,M \}^\ell}\sum_{2e^{n(\alpha+\epsilon)}\le c\le 3e^{n(\alpha+\epsilon)}\atop c\text{ is even}} \!\!\bigg(\!\prod_{i=1}^m\frac{e^{\ell \alpha(1-s)}}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s}}\bigg)\cdot \frac{2}{e^{n(\alpha+\epsilon)}}\cdot\!{\mathcal L}^{(2)}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c}, \end{align*} $$

where

$$ \begin{align*}{\mathcal L}^{(2)}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c}:=\frac{{\mathcal L}|_{{\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c)}}{{\mathcal L}({\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c))}.\end{align*} $$

The next two lemmas describe the gap between fundamental intervals defined in equation (4.25) and their $\mu _2$ -measures, respectively. The first one follows the same lines as the proof of Lemma 4.4.

Instead of giving a complete proof of the following lemma, we merely point out which changes have to be made.

Sketch proof.

(1) Note that

$$ \begin{align*}\mu_2(I_{k+\ell_0+p\ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p))=\prod_{i=1}^p\frac{e^{\ell \alpha (1-s)} }{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s}}.\end{align*} $$

By equations (4.21) and (4.24),

$$ \begin{align*}e^{n\alpha}\le e^{n\epsilon}a_1(z_n)\le e^{n\epsilon}B^{ns_{n,2}}\le e^{n\epsilon}B^{n(s+O(\epsilon))},\end{align*} $$

which is equivalent to

(4.27) $$ \begin{align} e^{\alpha}\le e^\epsilon B^{s+O(\epsilon)}. \end{align} $$

By decreasing $\epsilon $ if necessary, we have

$$ \begin{align*}e^{\alpha(1-s)}\le B^{s}.\end{align*} $$

Therefore,

(4.28) $$ \begin{align} \mu_2(I_{k+\ell_0+p\ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p))\le \prod_{i=1}^p\frac{1 }{q_\ell(\boldsymbol{a}_i)^{2s}} \le \frac{q_{k+\ell_0+p \ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)^{-2t}}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^{t}}, \end{align} $$

where the last inequality follows the same argument identical to equation (4.15).

(2) By the definition of $\mu _2$ ,

$$ \begin{align*} \mu_2({\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c)) &=\bigg(\prod_{i=1}^m\frac{e^{ \ell \alpha (1-s)}}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s}}\bigg)\cdot \frac{2}{e^{n(\alpha+\epsilon)}}\\ &= \bigg(\prod_{i=1}^m\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s} }\bigg) \cdot \frac{e^{m \ell \alpha (1-s)}} {B^{m\ell s}}\cdot \frac{2}{e^{n(\alpha+\epsilon)}}. \end{align*} $$

Although the setting is slightly different, one can follow the proof of equation (4.16) and show that whenever n is large enough,

(4.29) $$ \begin{align} n(1-\epsilon) \le m\ell \le n \end{align} $$

since $n=k+m\ell +\ell _0$ and $\ell $ is fixed. Hence, using $2e^{n(\alpha +\epsilon )}\le c \le 3e^{n(\alpha +\epsilon )}$ , we get

$$ \begin{align*}\begin{aligned} \frac{e^{m \ell \alpha (1-s)}} {B^{m\ell s}}\cdot \frac{2}{e^{n(\alpha+\epsilon)}}&= \frac{e^{n\alpha(1-s)}} {B^{ns}}\cdot \frac{1}{e^{n\alpha}} \cdot e^{O(n\epsilon)}=\frac{1} {e^{n\alpha s}B^{ns}}\cdot e^{O(n\epsilon)}\\ &=\frac{1} {c^sB^{ns}}\cdot e^{O(n\epsilon)}. \end{aligned}\end{align*} $$

Again, by decreasing $\epsilon $ and increasing s if necessary, we have

$$ \begin{align*}\frac{1} {c^sB^{ns}}\cdot e^{O(n\epsilon)}\le \frac{1} {e^{2nt\epsilon}c^tB^{nt}}.\end{align*} $$

By equations (4.28) and (4.26), it follows that

$$ \begin{align*}\begin{aligned} \mu_2({\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c))&\le\frac{q_{k+\ell_0+m \ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{-2t}}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^{t}}\cdot \frac{1} {e^{2nt\epsilon}c^tB^{nt}}\\ &\ll \frac{|{\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c)|^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t }. \end{aligned}\end{align*} $$

Remark 4.9. We present an inequality that will be also used in a later discussion:

$$ \begin{align*} &e^{n(\alpha+\epsilon)(1+t)}\cdot q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2t}\cdot \bigg(\prod_{i=1}^m\frac{e^{ \ell \alpha (1-s)}}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s}}\bigg)\cdot \frac{2}{e^{n(\alpha+\epsilon)}}\\&\quad \ll e^{n(\alpha+\epsilon)(1+t)} \cdot\frac{1}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot \frac{e^{m\ell\alpha(1-s)}}{B^{m\ell s}}\cdot \frac{2}{e^{n(\alpha+\epsilon)}}\hspace{1.5em}\text{by equation~}(4.28)\\&\quad =\frac{1}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot\frac{e^{n\alpha(1+t-s)}} {B^{ns}}\cdot e^{O(n\epsilon)}\hspace{7.55em}\text{by equation~}(4.29)\\&\quad \le \frac{1}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}\cdot\frac{B^{ns(1+t-s)}} {B^{ns}}\cdot e^{O(n\epsilon)}\hspace{7.45em}\text{by equation~}(4.27)\\&\quad \le \frac{1}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}, \end{align*} $$

where the last inequality follows from the fact that the error term $e^{O(n\epsilon )}$ can be made smaller than $B^{ns(s-t)}$ .

Next, the following lemma gives the estimation of the $\mu _2$ -measure of any ball $B(x, r)$ with $x\in [0,1] $ and $r>0$ .

Lemma 4.10. For any $ r>0 $ and $ x\in [0,1] $ , we have

$$ \begin{align*}\mu_2(B(x,r))\ll (M+2)^4(M+1)^{2\ell}\cdot\frac{r^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t},\end{align*} $$

where the unspecified constant is absolute.

Sketch proof.

It suffices to focus on the case $ r<|I_{k+\ell _0}(\tilde {\boldsymbol {u}})| $ . We need to consider three cases according to Lemma 4.7.

Case 1: $r\le |I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,c)|/18$ . In view of Lemma 4.7(2), $B(x,r)$ only intersects the fundamental interval ${\mathcal I}_2(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,c)$ to which x belongs. By the same reason as Case 1 in Lemma 4.6, and using Lemma 4.8(2) instead of Lemma 4.5(2), we deduce that

$$ \begin{align*}\begin{aligned} \mu_2(B(x,r))\ll\frac{r^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}. \end{aligned}\end{align*} $$

Case 2: $|I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,c)|/18< r\le |I_{n}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m)|/(2(M+2)^4)$ . In this case, the ball $B(x,r)$ intersects exactly the nth level basic cylinder $I_{n}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m)$ , but may intersect multiple $(n+1)$ th level basic cylinders inside this cylinder. Let

$$ \begin{align*}\Delta_2(x)=\{2e^{n(\alpha+\epsilon)}\le c\le 3e^{n(\alpha+\epsilon)}:~c \text{ is even and } I_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c)\cap B(x,r)\ne\emptyset\}.\end{align*} $$

In comparison to Case 2 in Lemma 4.6, here the choices for c are at most $e^{n(\alpha +\epsilon )}/2$ and each $(n+1)$ th level basic cylinder $I_{n+1}(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,c)$ is of length at least

$$ \begin{align*}\ge 2^{-1}q_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c)^{-2}\ge 2^ {-1} 2^{-2} 3^{-2}q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{-2}e^{-2n(\alpha+\epsilon)}.\end{align*} $$

This gives

$$ \begin{align*}\begin{aligned} \# \Delta_2(x)&\ll\min\{e^{n(\alpha+\epsilon)}/2, r\cdot q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2}e^{2n(\alpha+\epsilon)}\}\\ &\le e^{n(\alpha+\epsilon)(1-t)}\cdot(r\cdot q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2}e^{2n(\alpha+\epsilon)})^t\\ &=e^{n(\alpha+\epsilon)(1+t)}\cdot r^t\cdot q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2t}. \end{aligned}\end{align*} $$

By the definition of $\mu _2$ and the inequality presented in Remark 4.9,

$$ \begin{align*}\begin{aligned} \mu_2(B(x,r))\le& \#\Delta_2(x)\cdot\max_{2e^{n(\alpha+\epsilon)}\le c\le 3e^{n(\alpha+\epsilon)}\atop c\text{ is even}} \mu_2({\mathcal I}_2(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,c))\\ \ll &e^{n(\alpha+\epsilon)(1+t)}\cdot r^t\cdot q_{n}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^{2t}\cdot \bigg(\prod_{i=1}^m\frac{e^{ \ell \alpha (1-s)}}{q_\ell(\boldsymbol{a}_i)^{2s}B^{\ell s}}\bigg)\cdot \frac{2}{e^{n(\alpha+\epsilon)}}\\ \le & \frac{r^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}. \end{aligned}\end{align*} $$

Case 3: $|I_{k+\ell _0+(p+1) \ell }(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_{p+1})|/(2(M+2)^4) \le r< | I_{k+\ell _0+p\ell }(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots , \boldsymbol {a}_{p})|/ (2(M+2)^4) $ for some $1\le p\le m-1$ . In this case, the ball $B(x,r)$ only intersects one $(k+\ell _0+p \ell )$ th level basic cylinder, that is, $I_{ k+\ell _0+p \ell }(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_p)$ . Hence, following the same line as Case 3 in Lemma 4.6, we have

$$ \begin{align*}\begin{aligned} \mu_2(B(x,r))&\ll(M+2)^4(M+1)^{2\ell}\cdot\frac{r^t}{| I_{k+\ell_0}(\tilde{\boldsymbol{u}})|^t}. \end{aligned}\end{align*} $$

Combining the estimation given in Cases 1–3, we arrive at the conclusion.

4.3 Case III: $s^*=\limsup _{n \to \infty } s_n$ is obtained along a subsequence of $\{s_{n,3}\}_{n\ge 1}$

There exist infinitely many n such that

(4.30) $$ \begin{align} s_n=s_{n,3}>t. \end{align} $$

Fix an $\epsilon <s^*-t$ . By the definition of $s_n$ ,

$$ \begin{align*}s_{n,3}> s_{n,2}.\end{align*} $$

With Lemma 2.7(3), it follows that

$$ \begin{align*}a_1(z_n)\le B^{ns_{n,3}/2}<B^{n/2}.\end{align*} $$

By taking a subsequence, assume that integers satisfying equation (4.30) will ensure the existence of the following limit:

$$ \begin{align*}\lim_{n\text{ satisfies equation~}(4.20)\atop n\to\infty}\frac{\log a_1(z_n)}{n}=: \beta \le \frac{\log B}{2}.\end{align*} $$

Recall the definition of $s_{n,3}$ . By the continuity of the pressure functions, we can infer from the above discussion that

(4.31) $$ \begin{align} s^*=\inf\{s\in [0,1]:P(T, -s\log|T^{\prime}|-(s/2)\log B-s\beta)\le 0\}. \end{align} $$

By Proposition 2.4, there exist integers $ \ell \ge 2t/\epsilon +1$ and $ M $ such that the unique positive number $s=s(\ell ,M)$ satisfying the equation

(4.32) $$ \begin{align} \sum_{\boldsymbol{a}\in\{1,\ldots,M \}^\ell}\frac{B^{-\ell s/2}}{q_\ell(\boldsymbol{a})^{2s}e^{\beta\ell s}}=1 \end{align} $$

is greater than $t+\epsilon $ .

Choose a large integer $n\in {\mathbb {N}}$ so that equation (4.30) is satisfied and

(4.33) $$ \begin{align} n-k\ge \ell,\quad e^{n(\beta-\epsilon)}\le a_1(z_n)\le e^{n(\beta+\epsilon)}. \end{align} $$

Write $n-k=m\ell +\ell _0$ , where $0\le \ell _0<\ell $ .

Let $\tilde {\boldsymbol {u}}=(\boldsymbol {u},1^{\ell _0})\in {\mathbb {N}}^{k+\ell _0}$ and consider the following set defined by $(n+1)$ th level cylinders:

(4.34) $$ \begin{align} \{I_{n+1}(\tilde {\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n)): ~\boldsymbol{a}_i\in\{1,\ldots,M\}^\ell, 1\le i\le m\}. \end{align} $$

We stress that for the $(n+1)$ th position, there is only one choice, that is, $a_1(z_n)$ . This makes the discussion much easier than the previous two. By equation (3.14), there is an interval, denoted by ${\mathcal I}_3(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,a_1(z_n))$ such that

(4.35) $$ \begin{align} {\mathcal I}_3(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n))\subset F_n\cap I_{n+1}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n)). \end{align} $$

Additionally, since $a_1(z_n)\le B^{n/2}$ , by equation (3.14),

$$ \begin{align*}|{\mathcal I}_3(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n))|\ge\frac{B^{-n/2}}{4q_n(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m)^2a_1(z_n)}.\end{align*} $$

Define a probability measure $\mu _3$ supported on $F_n\cap I_{k+\ell _0}(\tilde {\boldsymbol {u}})$ as follows:

$$ \begin{align*} \mu_3=\sum_{\boldsymbol{a}_1\in\{1,\ldots,M \}^\ell}\cdots\sum_{\boldsymbol{a}_m\in\{1,\ldots,M \}^\ell}\bigg(\prod_{i=1}^m\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}e^{\beta\ell s} B^{\ell s/2}}\bigg)\cdot {\mathcal L}^{(3)}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n)}, \end{align*} $$

where

$$ \begin{align*}{\mathcal L}^{(3)}_{\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n)}:=\frac{{\mathcal L}|_{{\mathcal I}_3(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n))}}{{\mathcal L}({\mathcal I}_3(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n)))}.\end{align*} $$

The next two lemmas describe the gap between fundamental intervals defined in equation (4.25) and the $\mu _3$ -measures of these fundamental intervals, respectively. The first one follows the same lines as the proof of Lemma 4.4.

Lemma 4.11. Let $ {\mathcal I}_3={\mathcal I}_3(\tilde {\boldsymbol {u}},\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_m,a_1(z_n)) $ and $ {\mathcal I}_3^{\prime }={\mathcal I}_3(\tilde {\boldsymbol {u}},\boldsymbol {a}_1^{\prime },\ldots ,\boldsymbol {a}_m^{\prime },a_1(z_n))$ be two intervals defined in equation (4.25). If $ (\boldsymbol {a}_1,\ldots ,\boldsymbol {a}_{p-1})= (\boldsymbol {a}_1^{\prime },\ldots ,\boldsymbol {a}_{p-1}^{\prime }) $ but $ \boldsymbol {a}_p \ne \boldsymbol {a}_p^{\prime } $ for some $ 1\le p\le m $ , then

$$ \begin{align*} \mathrm{dist}({\mathcal I}_3,{\mathcal I}_3^{\prime})\ge\frac{|I_{k+\ell_0+p \ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)|}{2(M+2)^4}. \end{align*} $$

Instead of giving the complete proof of the following lemma, we merely point out which changes have to be made.

Sketch proof.

(1) Following the same lines as in Lemma 4.5(1), one has

$$ \begin{align*}\begin{aligned} \mu_3(I_{k+\ell_0+p\ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p))&=\prod_{i=1}^p\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}e^{\beta\ell s} B^{\ell s/2}}\le \prod_{i=1}^p\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}}\\ &\ll \frac{q_{k+\ell_0+p \ell}(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_p)^{-2t}}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^{t}}. \end{aligned}\end{align*} $$

(2) By the definition of $\mu _3$ ,

$$ \begin{align*} \mu_3({\mathcal I}_3(\tilde{\boldsymbol{u}},\boldsymbol{a}_1,\ldots,\boldsymbol{a}_m,a_1(z_n))) &=\prod_{i=1}^m\frac{1}{q_\ell(\boldsymbol{a}_i)^{2s}e^{\beta\ell s} B^{\ell s/2}}. \end{align*} $$

In view of item (2), we only need to estimate $\prod _{i=1}^m(e^{\beta \ell s} B^{\ell s/2})^{-1}=e^{-m\ell \beta s} B^{-m\ell s/2}$ . Although the setting is slightly different, one can follow the proof of equation (4.16) and show that whenever n is large enough,

$$ \begin{align*} n(1-\epsilon) \le m\ell\le n,\end{align*} $$

since $n=k+m\ell +\ell _0$ and $\ell $ is fixed. Hence, by equation (4.33),

$$ \begin{align*}\begin{aligned} e^{-m\ell\beta s}B^{-m\ell s/2}=e^{-n\beta s}B^{-ns/2}\cdot e^{O(n\epsilon)}=a_1(z_n)^{-s}B^{-ns/2}\cdot e^{O(\epsilon)}. \end{aligned}\end{align*} $$

Since $a_1(z_n)=e^{n(\beta +O(\epsilon ))}$ , by decreasing $\epsilon $ if necessary, we have

$$ \begin{align*}a_1(z_n)^{-s}B^{-ns/2}\cdot e^{O(\epsilon)}\le a_1(z_n)^{-t}B^{-nt/2}.\end{align*} $$

This together with item (1) yields the conclusion.

Next, the following lemma gives the estimation of the $\mu _3$ -measure of arbitrary ball $B(x, r)$ with $x \in [0,1]$ and $r>0$ .

Lemma 4.13. For any $ r>0 $ and $ x\in [0,1] $ , we have

$$ \begin{align*} \mu_3(B(x, r))\ll (M+2)^4(M+1)^{2\ell}\cdot \frac{r^t}{|I_{k+\ell_0}(\tilde {\boldsymbol{u}})|^t}.\end{align*} $$