3.1 Proof of Theorem A assuming the Key Lemma

Proof. Fix $\unicode{x3bb} \in [\alpha _{\inf }, \alpha _{\sup }]$ . For simplicity, we denote by $\tau ,\tau _{\inf }$ and $\tau _{\sup }$ the Hausdorff dimension of the sets $E(\unicode{x3bb} ), E_{\inf }(\unicode{x3bb} )$ , and $E_{\sup }(\unicode{x3bb} )$ , respectively. Then,

$$ \begin{align*} \tau \leq \tau_{\inf}\quad \text{and} \quad \tau \leq \tau_{\sup}. \end{align*} $$

Hence, it suffices to prove that $\tau _{\inf } \leq \tau $ and $\tau _{\sup } \leq \tau $ . We prove only the first inequality, as the second follows analogously.

We first claim that for any $\varepsilon>0$ , there exist $\widetilde {a},\widetilde {b} \in \mathbb Q$ with $\widetilde {a}<\unicode{x3bb} <\widetilde {b}$ and $N \in \mathbb N$ such that for all $n \geq N$ and all $r\ge n$ ,

(13) $$ \begin{align} \sum_{\alpha \in Z_{n,r}(\widetilde{a},\widetilde{b})} |I_n(\alpha)|^{\tau+\varepsilon} < 1. \end{align} $$

In fact, if the claim is false, then there exists $\varepsilon _{0}>0$ such that for any $a,b\in \mathbb {Q}$ with ${a<\unicode{x3bb} <b}$ and any integer $n_{0}\geq 1$ , there exist integers $n\geq n_{0}$ and $r\ge n$ satisfying

$$ \begin{align*} \sum_{\alpha \in Z_{n,r}(a,b)} |I_n(\alpha)|^{\tau+\varepsilon_{0}}\geq 1. \end{align*} $$

Applying the Key Lemma with $\unicode{x3bb} = \xi $ , we obtain $\dim _{\mathrm {H}} E(\unicode{x3bb} ) \ge \tau + \varepsilon _0$ , which contradicts the definition of $\tau $ .

We now show that

(14) $$ \begin{align} E_{\inf}(\unicode{x3bb}) \subset \bigcap_{K=1}^\infty \bigcup^\infty_{n=K}\bigcup_{\alpha \in Z_n(\widetilde{a},\widetilde{b})} I_n(\alpha)=\bigcap_{K=1}^\infty \bigcup^\infty_{n=K}\bigcup^\infty_{r=n}\bigcup_{\alpha \in Z_{n,r}(\widetilde{a},\widetilde{b})} I_n(\alpha). \end{align} $$

The equality in (14) follows directly from the definitions of $Z_n(\widetilde {a},\widetilde {b})$ and $Z_{n,r}(\widetilde {a},\widetilde {b})$ . To prove the inclusion, take $a^\prime , b^\prime \in \mathbb {R}$ such that $\widetilde {a}<a^\prime <\unicode{x3bb} <b^\prime <\widetilde {b}$ , and define

$$ \begin{align*} \widetilde{\varepsilon}:= \min\{a^\prime -\widetilde{a},\; \widetilde{b}-b^\prime\}. \end{align*} $$

Since the potential $\psi $ satisfies the weakly tempered distortion property given in (6), there exists $N_0\in \mathbb {N}$ with $N_0 \geq N$ such that for all $n \geq N_0$ ,

(15) $$ \begin{align} \frac{1}{n}\mathrm{Var}_n(S_n\psi) <\widetilde{\varepsilon}. \end{align} $$

For any $x\in E_{\inf }(\unicode{x3bb} )$ , we have

$$ \begin{align*} \liminf_{n \to \infty} \frac{1}{n}S_n\psi(x) =\unicode{x3bb}. \end{align*} $$

Hence, for any $j \in \mathbb {N}$ , there exists $N_j\in \mathbb {N}$ with $N_j \geq N_0$ such that

(16) $$ \begin{align} a^\prime< \frac{1}{N_j}S_{N_j}\psi(x) <b^\prime. \end{align} $$

For any $j \in \mathbb {N}$ , let $\alpha _j := a_1(x)a_2(x)\cdots a_{N_j}(x)$ . Then, $x\in I_{N_j}(\alpha _j)$ . We next prove that $\alpha _j \in Z_{N_j}(\widetilde {a},\widetilde {b})$ for all $j \in \mathbb {N}$ . Fix $j \in \mathbb {N}$ . For any $y\in I_{N_j}(\alpha _j)$ , it follows from (15) that

$$ \begin{align*} \bigg|\frac{1}{N_j}S_{N_j}\psi(x) - \frac{1}{N_j}S_{N_j}\psi(y) \bigg| \leq \frac{1}{N_j}\mathrm{Var}_{N_j}(S_{N_j}\psi) <\widetilde{\varepsilon}. \end{align*} $$

Combining this with (16), we derive that

$$ \begin{align*} \frac{1}{N_j}S_{N_j}\psi(y)> \frac{1}{N_j}S_{N_j}\psi(x) -\widetilde{\varepsilon} >a^\prime- \widetilde{\varepsilon} \geq \widetilde{a} \end{align*} $$

and

$$ \begin{align*} \frac{1}{N_j}S_{N_j}\psi(y) < \frac{1}{N_j}S_{N_j}\psi(x) +\widetilde{\varepsilon} <b^\prime+ \widetilde{\varepsilon} \leq \widetilde{b}. \end{align*} $$

Thus,

$$ \begin{align*} \widetilde{a}< \frac{1}{N_j}S_{N_j}\psi(y) <\widetilde{b}, \end{align*} $$

which implies that $\alpha _j \in Z_{N_j}(\widetilde {a},\widetilde {b})$ . Therefore, the inclusion in (14) holds.

Since $|I_n(\alpha )| \leq 2^{-(r-1)}$ for all $\alpha \in Z_{n,r}(\widetilde {a},\widetilde {b})$ , for any $K \geq N$ , we deduce from (13) that

$$ \begin{align*} \sum^\infty_{n=K}\sum^\infty_{r=n} \sum_{\alpha \in Z_{n,r}(\widetilde{a},\widetilde{b})} |I_n(\alpha)|^{\tau+2\varepsilon} &\leq \sum^\infty_{n=K}\sum^\infty_{r=n} \bigg(\sum_{\alpha \in Z_{n,r}(\widetilde{a},\widetilde{b})} |I_n(\alpha)|^{\tau+\varepsilon}\bigg) 2^{-(r-1)\varepsilon} \\ &\leq \sum^\infty_{n=K}\sum^\infty_{r=n} 2^{-(r-1)\varepsilon}. \end{align*} $$

By (14), we conclude that

$$ \begin{align*} \mathcal{H}^{\tau+2\varepsilon}(E_{\inf}(\unicode{x3bb})) &\leq \liminf_{K\to \infty}\sum^\infty_{n=K}\sum^\infty_{r=n} 2^{-(r-1)\varepsilon}\\ &=(1-2^{-\varepsilon})^{-1}\liminf_{K\to \infty}\sum^\infty_{n=K} 2^{-(n-1)\varepsilon}=0. \end{align*} $$

Hence, $\dim _{\mathrm {H}} E_{\inf }(\unicode{x3bb} ) \le \tau + 2\varepsilon $ ; that is, $\tau _{\inf } \le \tau + 2\varepsilon $ . Letting $\varepsilon \to 0$ , we obtain $\tau _{\inf } \leq \tau $ . This completes the proof of Theorem A.

3.2 Proof of Theorem B assuming the Key Lemma

The proof of Theorem B proceeds similarly to that of Theorem A.

Proof. Let $\unicode{x3bb} ,\xi \in [\alpha _{\inf }, \alpha _{\sup }]$ with $\unicode{x3bb} \leq \xi $ be fixed. Define

$$ \begin{align*} \rho:=\dim_{\mathrm{H}} E(\unicode{x3bb},\xi). \end{align*} $$

Since $E(\unicode{x3bb} ,\xi ) \subset E_{\inf }(\unicode{x3bb} )$ and $E(\unicode{x3bb} ,\xi ) \subset E_{\sup }(\xi )$ , it follows from Theorem A that

$$ \begin{align*} \dim_{\mathrm{H}} E(\unicode{x3bb}) = \dim_{\mathrm{H}}E_{\inf}(\unicode{x3bb}) \geq \rho \quad \text{and}\quad \dim_{\mathrm{H}} E(\xi)= \dim_{\mathrm{H}}E_{\sup}(\xi) \geq \rho, \end{align*} $$

which implies that

$$ \begin{align*} \min\{\dim_{\mathrm{H}} E(\unicode{x3bb}),\ \dim_{\mathrm{H}} E(\xi)\} \geq \rho. \end{align*} $$

Hence, it remains to prove that

(17) $$ \begin{align} \min\{\dim_{\mathrm{H}} E(\unicode{x3bb}),\ \dim_{\mathrm{H}} E(\xi)\} \leq \rho. \end{align} $$

By the Key Lemma, we derive that for any $\varepsilon>0$ , there exist $a_0,b_0 \in \mathbb Q$ with ${a_0<\unicode{x3bb} <b_0}$ , $c_0,d_0 \in \mathbb {Q}$ with $c_0<\xi <d_0$ , and $\widetilde {N}\in \mathbb N$ such that, for all $n \geq \widetilde {N}$ and all $r\ge n$ ,

(18) $$ \begin{align} \sum_{\alpha \in Z_{n,r}(a_0,b_0)} |I_n(\alpha)|^{\rho+\varepsilon} < 1, \end{align} $$

or for all $\tilde n \geq \widetilde {N}$ and all $\tilde r\ge \tilde n$ ,

(19) $$ \begin{align} \sum_{\alpha \in Z_{\tilde n,\tilde r}(c_0,d_0)} |I_{\tilde n}(\alpha)|^{\rho+\varepsilon} < 1. \end{align} $$

Moreover, we claim that

(20) $$ \begin{align} E_{\inf}(\unicode{x3bb}) \subset \bigcap_{K=1}^\infty\bigcup^\infty_{n=K}\bigcup^\infty_{r=n}\bigcup_{\alpha \in Z_{n,r}(a_0,b_0)} I_n(\alpha) \end{align} $$

and

(21) $$ \begin{align} E_{\inf}(\xi) \subset \bigcap_{K=1}^\infty\bigcup^\infty_{\tilde n=K}\bigcup^\infty_{\tilde r=\tilde n}\bigcup_{\alpha \in Z_{\tilde n,\tilde r}(c_0,d_0)} I_{\tilde n}(\alpha). \end{align} $$

The justification of the inclusions in (20) and (21) is analogous to that of (14), and is therefore omitted.

We are now in a position to prove inequality (17). If (18) holds, then, by the inclusion in (20), we obtain

$$ \begin{align*} \mathcal{H}^{\rho+2\varepsilon}(E_{\inf}(\unicode{x3bb})) &\leq \liminf_{K\to \infty}\sum^\infty_{n=K}\sum^\infty_{r=n} \sum_{\alpha \in Z_{n,r}(a_0,b_0)} |I_n(\alpha)|^{\rho+2\varepsilon}\\& \leq \liminf_{K\to \infty}\sum^\infty_{n=K} \sum^\infty_{r=n}\bigg(\sum_{\alpha \in Z_{n,r}(a_0,b_0)} |I_n(\alpha)|^{\rho+\varepsilon}\bigg) 2^{-(r-1)\varepsilon} \\ &\leq \liminf_{K\to \infty}\sum^\infty_{n=K}\sum^\infty_{r=n} 2^{-(r-1)\varepsilon} = 0. \end{align*} $$

This implies that $\dim _{\mathrm {H}} E_{\inf }(\unicode{x3bb} ) \leq \rho +2\varepsilon $ . By Theorem A, it follows that

$$ \begin{align*} \dim_{\mathrm{H}} E(\unicode{x3bb}) \leq \rho +2\varepsilon. \end{align*} $$

Letting $\varepsilon \to 0$ , we have $\dim _{\mathrm {H}} E(\unicode{x3bb} ) \leq \rho $ . Similarly, if (19) holds, then, by the inclusion in (21), we conclude that $\dim _{\mathrm {H}} E(\xi ) \leq \rho $ . Therefore, inequality (17) follows. We complete the proof.

4.1 Step I: Construction of the sequence $\{\mathrm {E}_n\}_{n \geq 1}$

In the rest of this paper, when the order of a cylinder set is clear from the context or does not need to be emphasized, we omit the corresponding subscript. This convention is adopted mainly for typographical clarity, as the expressions appearing in these subscripts can be rather long.

Define $a_1=\lceil \unicode{x3bb} \rceil -1, b_1=\lfloor \unicode{x3bb} \rfloor +1$ , $c_2 = (\lceil 2\xi \rceil -1)/2$ , $d_2= (\lfloor 2\xi \rfloor +1)/2$ and, for any $k>1$ ,

$$ \begin{align*} a_{2k-1}:= \max\bigg\{a_{2k-3},\frac{\lceil(2k-1)\unicode{x3bb}\rceil-1}{2k-1}\bigg\}, \quad b_{2k-1}:= \min\bigg\{b_{2k-3},\frac{\lfloor(2k-1)\unicode{x3bb} \rfloor+1}{2k-1}\bigg\} \end{align*} $$

and

$$ \begin{align*} c_{2k}:= \max\bigg\{c_{2k-2},\frac{\lceil 2k\xi\rceil-1}{2k}\bigg\}, \quad d_{2k}:= \min\bigg\{d_{2k-2},\frac{\lfloor2k\xi\rfloor+1}{2k}\bigg\}. \end{align*} $$

Then, $a_{2k-1}, b_{2k-1},c_{2k},d_{2k} \in \mathbb Q$ and

$$ \begin{align*} \unicode{x3bb} -\frac{1}{2k-1} \leq a_{2k-1}\leq a_{2k+1} <\unicode{x3bb} <b_{2k+1}\leq b_{2k-1} \leq \unicode{x3bb} +\frac{1}{2k-1} \end{align*} $$

and

$$ \begin{align*} \xi -\frac{1}{2k} \leq c_{2k}\leq c_{2k+2} <\xi <d_{2k+2}\leq d_{2k} \leq \xi +\frac{1}{2k}. \end{align*} $$

By the condition of the Key Lemma, there exist strictly increasing sequences $\{n_k\}_{k\geq 1}$ and $\{r_k\}_{k\geq 1}$ of positive integers with $r_k \geq n_k$ such that, for all $k \in \mathbb {N}$ ,

(24) $$ \begin{align} \sum_{\alpha \in Z_{n_{2k-1},r_{2k-1}}(a_{2k-1},b_{2k-1})} |I_{n_{2k-1}}(\alpha)|^s \geq 1 \quad \text{and}\quad \sum_{\alpha \in Z_{n_{2k},r_{2k}}(c_{2k},d_{2k})} |I_{n_{2k}}(\alpha)|^s \geq 1. \end{align} $$

For notational convenience, let

$$ \begin{align*} \widetilde{a}_k:= \begin{cases} a_{2m-1} & \text{if } k=2m-1, \\ c_{2m} & \text{if } k=2m, \end{cases} \quad\text{and}\quad \widetilde{b}_k := \begin{cases} b_{2m-1} & \text{if } k=2m-1, \\ d_{2m} & \text{if } k=2m. \end{cases} \end{align*} $$

Under this notation, the inequalities in (24) can be rewritten as

(25) $$ \begin{align} \sum_{\alpha \in Z_{n_{k},r_{k}}(\widetilde{a}_{k},\widetilde{b}_{k})} |I_{n_{k}}(\alpha)|^s \geq 1\quad \text{for all } k \geq 1. \end{align} $$

The cardinality of $Z_{n_k, r_k}(\widetilde {a}_k,\widetilde {b}_k)$ is estimated as follows: since $|I_{n_k}(\alpha )|<(1/{2^{r_k-1}})$ for all $\alpha \in Z_{n_k, r_k}(\widetilde {a}_k,\widetilde {b}_k)$ , by (25), we have

$$ \begin{align*}\sharp Z_{n_k, r_k}(\widetilde{a}_k,\widetilde{b}_k)\cdot\frac1{2^{s(r_k-1)}}>1,\end{align*} $$

and therefore,

$$ \begin{align*}\sharp Z_{n_k, r_k}(\widetilde{a}_k,\widetilde{b}_k)>2^{s(r_k-1)}.\end{align*} $$

Let $0<\varepsilon <s$ be fixed. It follows that, for any sufficiently large k,

$$ \begin{align*} \sharp Z_{n_k, r_k}(\widetilde{a}_k,\widetilde{b}_k)> 2^{(s-\varepsilon)(r_k+1)}. \end{align*} $$

Hence, there exists $0<C \leq 1$ such that

(26) $$ \begin{align} \sharp Z_{n_k, r_k}(\widetilde{a}_k,\widetilde{b}_k)> C\cdot2^{(s-\varepsilon)(r_k+1)}\quad \text{for all } k \geq 1. \end{align} $$

For simplicity, let $X_k:= Z_{n_k,r_k}(\widetilde {a}_k,\widetilde {b}_k)$ .

Let

(27) $$ \begin{align} M_k:= \max\bigg\{\bigg|\sum_{k=0}^{j-1}\psi(G^k(x))\bigg|\colon x\in I_{n_k}(\alpha), \alpha\in X_k, 1\le j\le n_k\bigg\}. \end{align} $$

Notice that, since $\mathrm {Var}_1(\psi )<\infty $ , we derive that $\psi $ is bounded on the interval $I(w)$ for any finite sequence w of positive integers.

Choose a sequence $\{u_k\}_{k\geq 0}$ of integers such that $u_0=0$ and

(28) $$ \begin{align} u_k>(k+1)\bigg(r_{k+1}+M_{k+1}+\sum_{j=0}^{k-1}u_jn_j\bigg)\quad \text{for all } k\geq 1. \end{align} $$

We proceed to construct the sequence $\{\mathrm {E}_n\}_{n\geq 1}$ of subsets of $[0,1]$ . For this purpose, we first introduce some notation.

For any $k\in \mathbb N$ and $n \in \mathbb {N}$ with $\sum ^{k-1}_{i=0}u_i <n \leq \sum ^{k}_{i=0}u_i$ , define

$$ \begin{align*} \mathcal{A}^{*(n)}_k:= \bigg\{w^{(1)}_kw^{(2)}_k\cdots w^{(n-\sum^{k-1}_{i=0}u_i)}_k\colon w^{(j)}_k \in X_k\ \text{for all}\ 1\leq j \leq n-\sum^{k-1}_{i=0}u_i \bigg\}. \end{align*} $$

The length of a word in $\mathcal {A}^{*(n)}_k$ is $(n-\sum ^{k-1}_{i=0}u_i)n_k$ . In particular, when $n = \sum ^{k}_{i=0}u_i$ , we write $\mathcal {A}^{*(n)}_k$ as $\mathcal {A}_k$ :

$$ \begin{align*} \mathcal{A}_k:=\{w^{(1)}_kw^{(2)}_k\cdots w^{(u_k)}_k\colon w^{(j)}_k \in X_k\ \text{for all}\ 1\leq j \leq u_k \}. \end{align*} $$

Define

$$ \begin{align*} \mathcal{W}^{*(n)}_k:=\{w_1\cdots w_{k-1}w^{*(n)}_k\colon w_1 \in \mathcal{A}_1, \ldots, w_{k-1} \in \mathcal{A}_{k-1}, w^{*(n)}_k \in \mathcal{A}^{*(n)}_k \}. \end{align*} $$

The length of a word in $\mathcal {W}^{*(n)}_k$ is $\sum ^{k-1}_{j=1}u_jn_j + (n-\sum ^{k-1}_{i=0}u_i)n_k$ . When $n = \sum ^{k}_{i=0}u_i$ , we write $\mathcal {W}^{*(n)}_k$ as $\mathcal {W}_k$ :

$$ \begin{align*} \mathcal{W}_k:=\{w_1\cdots w_{k-1}w_k\colon w_1 \in \mathcal{A}_1, \ldots, w_{k-1} \in \mathcal{A}_{k-1}, w_k \in \mathcal{A}_k \}. \end{align*} $$

Let $v_k$ denote the maximum component among all vectors in $X_k$ for each $k \in \mathbb {N}$ . The sequence $\{\mathrm {E}_n\}_{n\geq 1}$ of sets is defined as follows.

1. (I-i) When $\sum _{i=0}^{k-1}u_i < n < \sum _{i=0}^{k}u_i$ for some $k\in \mathbb {N}$ , for any $w_1\cdots w_{k-1}w^{*(n)}_k \in \mathcal {W}^{*(n)}_k$ , we define the nth level basic interval as

   $$ \begin{align*} J(w_1\cdots w_{k-1}w^{*(n)}_k):=\bigcup^{v_k}_{j=1} \mathrm{cl}(I(w_1\cdots w_{k-1}w^{*(n)}_kj)), \end{align*} $$
   where $\mathrm {cl}(A)$ denotes the closure of the set A. We then set
   $$ \begin{align*} \mathrm{E}_n:= \bigcup_{w_1\cdots w_{k-1}w^{*(n)}_k \in \mathcal{W}^{*(n)}_k} J(w_1\cdots w_{k-1}w^{*(n)}_k). \end{align*} $$

2. (I-ii) When $n= \sum ^k_{i=0}u_i$ for some $k\in \mathbb {N}$ , for any $w_1\cdots w_{k-1}w_k \in \mathcal {W}_k$ , we define the nth level basic interval as

   $$ \begin{align*} J(w_1\cdots w_{k-1}w_k):=\bigcup^{v_{k+1}}_{j=1} \mathrm{cl}(I(w_1\cdots w_{k-1}w_kj)). \end{align*} $$
   We then set
   $$ \begin{align*} \mathrm{E}_n:= \bigcup_{w_1\cdots w_{k-1}w_k \in \mathcal{W}_k} J(w_1\cdots w_{k-1}w_k). \end{align*} $$

With this construction, the sequence $\{\mathrm {E}_n\}_{n\geq 1}$ is now fully defined.

4.2 Step II: Inclusion of the limit set of $\{\mathrm {E}_n\}_{n\geq 1}$ in $E(\unicode{x3bb} ,\xi )$

Since $\mathrm {E}_{n+1} \subseteq \mathrm {E}_n$ for all $n \in \mathbb {N}$ , we define $\mathrm {E}_0 := [0,1]$ and the limit set

$$ \begin{align*} \mathrm{E} := \bigcap_{n=0}^{\infty} \mathrm{E}_n. \end{align*} $$

We will prove that

(29) $$ \begin{align} \mathrm{E}\subset E(\unicode{x3bb},\xi). \end{align} $$

For any $x\in \mathrm {E}$ , the word consisting of the partial quotients of the continued fraction expansion of x is

$$ \begin{align*} \underbrace{w^{(1)}_1 \cdots w^{(u_1)}_1}_{=: w_1 \in \mathcal{W}_1} \underbrace{w^{(1)}_2 \cdots w^{(u_2)}_2}_{=: w_2 \in \mathcal{W}_2} \cdots \underbrace{w^{(1)}_{k-1} \cdots w^{(u_{k-1})}_{k-1}}_{=: w_{k-1} \in \mathcal{W}_{k-1}} \cdots. \end{align*} $$

For any $k \in \mathbb {N}$ and $0 \leq \ell \leq u_k-1$ , we have

$$ \begin{align*} G^{\sum_{i=1}^{k-1}u_in_i + \ell n_k}(x) \in I_{n_k}(w_k^{(\ell +1)}), \end{align*} $$

which, together with (12), implies that

(30) $$ \begin{align} n_k \widetilde{a}_k <\sum_{j=0}^{n_k-1}\psi(G^{\sum_{i=1}^{k-1}u_in_i + \ell n_k +j}(x)) < n_k \widetilde{b}_k. \end{align} $$

Thus,

(31) $$ \begin{align} u_kn_k \widetilde{a}_k < \sum_{j=0}^{u_kn_k-1}\psi(G^{\sum_{i=1}^{k-1}u_in_i + j}(x)) < u_kn_k \widetilde{b}_k. \end{align} $$

For any $N\in \mathbb {N}$ , there exist $k \in \mathbb N$ , $0\leq \ell _k <u_k$ , and $0 \leq t_k <n_k$ such that

$$ \begin{align*} N= u_1n_1 +\cdots +u_{k-1}n_{k-1} +\ell_k n_k +t_k. \end{align*} $$

Let us write

$$ \begin{align*} D(N):= \sum^{k-1}_{j=1}u_jn_j +\ell_k n_k. \end{align*} $$

By (30), (31), and the definition of $M_k$ in (27), we derive that

$$ \begin{align*} S_N\psi(x)> \sum^{k-1}_{j=1}u_jn_j \widetilde{a}_j + \ell_k n_k\widetilde{a}_k -M_k \end{align*} $$

and

$$ \begin{align*} S_N\psi(x) < \sum^{k-1}_{j=1}u_jn_j \widetilde{b}_j + \ell_k n_k\widetilde{b}_k + M_k. \end{align*} $$

By (28), we have

$$ \begin{align*} \sum^{k-1}_{j=1}u_jn_j +\ell_k n_k> u_{k-1} >kr_k \geq kn_k, \end{align*} $$

which implies that

$$ \begin{align*} D(N) {\kern-1pt}\leq{\kern-1pt} N {\kern-1pt}<{\kern-1pt} \sum^{k-1}_{j=1}u_jn_j +\ell_k n_k +n_k {\kern-1pt}<{\kern-1pt} \bigg(1+\frac{1}{k}\bigg)\bigg(\sum^{k-1}_{j=1}u_jn_j+\ell_k n_k\bigg) {\kern-1pt}={\kern-1pt} \bigg(\frac{k+1}{k}\bigg)D(N). \end{align*} $$

Hence,

$$ \begin{align*} &\frac{1}{N}S_N\psi(x) < \frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{b}_j + \ell_k n_k\widetilde{b}_k + M_k}{N}\\& \quad \leq \max\bigg\{\frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{b}_j + \ell_k n_k\widetilde{b}_k + M_k}{D(N)}, \frac{k}{k+1}\cdot \frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{b}_j + \ell_k n_k\widetilde{b}_k + M_k}{D(N)} \bigg\}\\& \quad < \max\bigg\{\frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{b}_j + \ell_k n_k\widetilde{b}_k }{D(N)}, \frac{k}{k+1}\cdot \frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{b}_j + \ell_k n_k\widetilde{b}_k}{D(N)} \bigg\} + \frac{M_k}{u_{k-1}} \end{align*} $$

and

$$ \begin{align*} \frac{1}{N}S_N\psi(x) &> \frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{a}_j + \ell_k n_k\widetilde{a}_k -M_k}{N}\\&>\min{\kern-1pt}\bigg\{\frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{a}_j + \ell_k n_k\widetilde{a}_k }{D(N)}, \frac{k}{k+1}\cdot \frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{a}_j + \ell_k n_k\widetilde{a}_k}{D(N)} \bigg\} {\kern-1pt}-{\kern-1pt} \frac{M_k}{u_{k-1}}. \end{align*} $$

Since $\liminf _{k\to \infty }\widetilde {a}_k = \unicode{x3bb} $ and $\limsup _{k\to \infty }\widetilde {b}_k = \xi $ , it follows that

$$ \begin{align*} \liminf_{k\to \infty} \min_{0\leq \ell_k <u_k}\bigg\{\frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{a}_j + \ell_k n_k\widetilde{a}_k}{D(N)}\bigg\} \geq \unicode{x3bb} \end{align*} $$

and

$$ \begin{align*} \limsup_{k\to \infty} \max_{0\leq \ell_k <u_k}\bigg\{\frac{\sum^{k-1}_{j=1}u_jn_j \widetilde{b}_j + \ell_k n_k\widetilde{b}_k}{D(N)}\bigg\} \leq \xi. \end{align*} $$

By the definition of $u_{k-1}$ defined in (28), we have ${M_k}/{u_{k-1}}<(1/k)$ for all $k\in \mathbb {N}$ . Therefore, we conclude that

$$ \begin{align*} \unicode{x3bb} \leq \liminf_{N\to \infty} \frac{1}{N}S_N\psi(x)\leq \limsup_{N\to \infty}\frac{1}{N}S_N\psi(x) \leq \xi. \end{align*} $$

Next, we show that

(32) $$ \begin{align} \liminf_{N\to \infty} \frac{1}{N}S_N\psi(x) =\unicode{x3bb} \quad \text{and} \quad \limsup_{N\to \infty} \frac{1}{N}S_N\psi(x)= \xi. \end{align} $$

For the lower limit, let $N_k:= \sum ^{2k-1}_{j=1} u_jn_j$ for all $k \in \mathbb {N}$ . Since $u_{2k-1}>\sum ^{2k-2}_{j=1} u_jn_j$ , we have

$$ \begin{align*}1\le\frac{N_k}{u_{2k-1}n_{2k-1}}<1+\frac1{n_{2k-1}},\end{align*} $$

which implies that $\lim _{k\to \infty }({N_k}/{u_{2k-1}n_{2k-1}})=1$ . Hence,

$$ \begin{align*}\liminf_{k\to \infty} \frac{1}{N_k} S_{N_k}\psi(x)=\liminf_{k\to \infty} \frac{1}{u_{2k-1}n_{2k-1}} S_{N_k}\psi(x).\end{align*} $$

By (31), we obtain

$$ \begin{align*} S_{N_k}\psi(x) < \sum^{2k-1}_{j=1} u_jn_j\widetilde{b}_j, \end{align*} $$

which implies that

$$ \begin{align*} \frac{1}{u_{2k-1}n_{2k-1}} S_{N_k}\psi(x)< \widetilde{b}_{2k-1} + \frac{\sum^{2k-2}_{j=1} u_jn_j\widetilde{b}_j}{u_{2k-1}n_{2k-1}}. \end{align*} $$

Note that $u_{2k-1}n_{2k-1}>n_{2k-1}\sum ^{2k-2}_{j=1} u_jn_j$ for all $k\in \mathbb {N}$ , while $\widetilde {b}_{2k-1} \to \unicode{x3bb} $ and $\widetilde {b}_{2k} \to \xi $ as $k\to \infty $ . It follows that $\{\widetilde {b}_{2k-1}\}_{k\geq 1}$ and $\{\widetilde {b}_{2k}\}_{k\geq 1}$ are bounded. Therefore,

$$ \begin{align*} \lim_{k\to \infty} \frac{\sum^{2k-2}_{j=1} u_jn_j\widetilde{b}_j}{u_{2k-1}n_{2k-1}}= 0, \end{align*} $$

and consequently,

$$ \begin{align*} \liminf_{k\to \infty} \frac{1}{N_k} S_{N_k}\psi(x) \leq \unicode{x3bb}. \end{align*} $$

Hence,

$$ \begin{align*} \liminf_{N\to \infty} \frac{1}{N} S_{N}\psi(x) =\unicode{x3bb}. \end{align*} $$

For the upper limit, let $\widetilde {N}_k:= \sum ^{2k}_{j=1} u_jn_j$ for all $k \in \mathbb {N}$ . Since $\sum ^{2k-1}_{j=1} u_jn_j<u_{2k}$ , it follows that $\lim _{k\to \infty }({\widetilde {N}_k}/{u_{2k}n_{2k}})=1$ . Therefore,

$$ \begin{align*}\limsup_{k\to \infty} \frac{1}{\widetilde{N}_k}S_{\widetilde{N}_k}\psi(x)=\limsup_{k\to \infty} \frac{1}{u_{2k}n_{2k}}S_{\widetilde{N}_k}\psi(x).\end{align*} $$

From (31), we deduce that

$$ \begin{align*} S_{\widetilde{N}_k}\psi(x)> \sum^{2k}_{j=1} u_jn_j\widetilde{a}_j, \end{align*} $$

and thus,

$$ \begin{align*} \frac{1}{u_{2k}n_{2k}}S_{\widetilde{N}_k}\psi(x)> \widetilde{a}_{2k}+\frac{\sum^{2k-1}_{j=1} u_jn_j\widetilde{a}_j}{u_{2k}n_{2k}}. \end{align*} $$

Since, as before, $\lim _{k\to \infty }({\sum ^{2k-1}_{j=1} u_jn_j\widetilde {a}_j}/{u_{2k}n_{2k}})=0$ , we deduce that

$$ \begin{align*} \limsup_{N\to \infty} \frac{1}{N}S_{N}\psi(x) \geq \limsup_{k\to \infty} \frac{1}{\widetilde{N}_k}S_{\widetilde{N}_k}\psi(x) \geq \xi, \end{align*} $$

where the last inequality follows from $\widetilde {a}_{2k} \to \xi $ as $k\to \infty $ . Hence,

$$ \begin{align*} \limsup_{N\to \infty} \frac{1}{N} S_{N}\psi(x)= \xi. \end{align*} $$

Therefore, (32) holds. That is, $\mathrm {E}$ is a subset of $E(\unicode{x3bb} ,\xi )$ .

By (29), we obtain

$$ \begin{align*} \dim_{\mathrm{H}} E(\unicode{x3bb},\xi) \geq \dim_{\mathrm{H}} \mathrm{E}. \end{align*} $$

To calculate the lower bound of $\dim _{\mathrm {H}}\mathrm {E}$ , by Lemma 4, it remains to estimate the values of $m_n$ and $\varepsilon _n$ .

4.3 Step III: Estimation of $m_n$ and $\varepsilon _n$ in Lemma 4

We first recall the meanings of $m_n$ and $\varepsilon _n$ in Lemma 4: $m_n$ are chosen such that every $(n-1)$ th level basic interval contains at least $m_n\, n$ th level basic intervals, and $\varepsilon _n$ is chosen such that the gaps between these nth level basic intervals have lengths at least $\varepsilon _n$ .

The estimates of the values of $m_n$ and $\varepsilon _n$ are as follows.

1. (II-i) When $\sum ^{k-1}_{i=0}u_i < n< \sum ^k_{i=0}u_i$ for some $k\in \mathbb {N}$ , equivalently,

   $$ \begin{align*} \sum^{k-1}_{i=0}u_i \leq n-1 < \sum^k_{i=0}u_i -1, \end{align*} $$
   the $(n-1)$ th level basic interval is of the form $J(w)$ for some $w \in \mathcal {W}^{*(n-1)}_k$ , and it contains the nth level basic intervals:
   $$ \begin{align*} J(ww^\prime):=\bigcup^{v_k}_{j=1} \mathrm{cl}(I(ww^\prime j)) \end{align*} $$
   for all $w^\prime \in X_k$ . Hence, we may take $m_n= \sharp X_k$ . In this case, the minimum length of the gaps between these nth level basic intervals is at least
   (33) $$ \begin{align} \widetilde{\varepsilon}_n :=\min\{|I(w(v_k+1))|\colon w \in \mathcal{W}^{*(n)}_k\}. \end{align} $$

2. (II-ii) When $n=\sum ^k_{i=0}u_i$ for some $k \in \mathbb {N}$ , equivalently,

   $$ \begin{align*} n-1= \sum^k_{i=0}u_i -1, \end{align*} $$
   the $(n-1)$ th level basic interval is of the form $J(w)$ for some $w \in \mathcal {W}^{*(n-1)}_k$ , and it contains the nth level basic intervals:
   $$ \begin{align*} J(ww^\prime):=\bigcup^{v_{k+1}}_{j=1} \mathrm{cl}(I(ww^\prime j)) \end{align*} $$
   for all $w^\prime \in X_k$ . Hence, we may take $m_n= \sharp X_k$ . In this case, the minimum length of the gaps between these nth level basic intervals is at least
   (34) $$ \begin{align} \widetilde{\varepsilon}_n:= \min\{|I(w(v_{k+1}+1))|\colon w \in \mathcal{W}_k\}. \end{align} $$

Therefore, for any $k\in \mathbb {N}$ and $n \in \mathbb {N}$ with $\sum ^{k-1}_{i=0}u_i < n \leq \sum ^k_{i=0}u_i$ , by (26), we have

(35) $$ \begin{align} m_n = \sharp X_k> C\cdot 2^{(s-\varepsilon)(r_k+1)}. \end{align} $$

For $\varepsilon _n$ , we may take

(36) $$ \begin{align} \varepsilon_n := \frac{1}{8\cdot 2^{r_{k}}}\cdot \min\{|I(w)|\colon w \in \mathcal{W}^{*(n)}_k\} \end{align} $$

for all $n \in \mathbb {N}$ with $\sum ^{k-1}_{i=0}u_i < n <\sum ^k_{i=0}u_i$ , and take

(37) $$ \begin{align} \varepsilon_n := \frac{1}{8\cdot 2^{r_{k+1}}}\cdot \min\{|I(w)|\colon w \in \mathcal{W}_k\} \end{align} $$

for all $n=\sum ^k_{i=0}u_i$ . Since $r_{k+1}> r_k$ , it follows that the sequence $\{\varepsilon _n\}_{n\geq 1}$ is strictly decreasing. In fact, when $\sum ^{k-1}_{i=0}u_i <n < \sum ^{k}_{i=0}u_i -1$ , by (36), we obtain

$$ \begin{align*} \varepsilon_n = \frac{1}{8\cdot 2^{r_{k}}}\cdot \min\{|I(u)|\colon u \in \mathcal{W}^{*(n)}_k\}, \quad \varepsilon_{n+1} = \frac{1}{8\cdot 2^{r_{k}}}\cdot \min\{|I(w)|\colon w \in \mathcal{W}^{*(n+1)}_k\}. \end{align*} $$

Since $|I(u)| \geq 2|I(w)|$ for any $u \in \mathcal {W}^{*(n)}_k$ and $w \in \mathcal {W}^{*(n+1)}_k$ , it follows that $\varepsilon _n> \varepsilon _{n+1}$ . When $n= \sum ^{k}_{i=0}u_i -1$ , we deduce from (36) and (37) that

$$ \begin{align*} \varepsilon_n = \frac{1}{8\cdot 2^{r_{k}}}\cdot \min\{|I(u)|\colon u \in \mathcal{W}^{*(n)}_k\}, \quad \varepsilon_{n+1} = \frac{1}{8\cdot 2^{r_{k+1}}}\cdot \min\{|I(w)|\colon w \in \mathcal{W}_k\}. \end{align*} $$

Since $r_{k+1}>r_k$ , we have $\varepsilon _n> \varepsilon _{n+1}$ . When $n= \sum ^{k}_{i=0}u_i$ , from (36) and (37), we derive that

$$ \begin{align*} \varepsilon_{n} = \frac{1}{8\cdot 2^{r_{k+1}}}\cdot \min\{|I(u)|\colon u \in \mathcal{W}_k\},\quad \varepsilon_{n+1} = \frac{1}{8\cdot 2^{r_{k+1}}}\cdot \min\{|I(w)|\colon w \in \mathcal{W}^{*(n+1)}_{k+1}\}, \end{align*} $$

and hence, $\varepsilon _{n}>\varepsilon _{n+1}$ .

We next show that

$$ \begin{align*} \widetilde{\varepsilon}_n>\varepsilon_n \quad \text{ for all } n \geq 1. \end{align*} $$

To this end, we will make use of the following fact: for any two finite words u and v,

(38) $$ \begin{align} \tfrac{1}{2}|I(u)||I(v)|<|I(uv)| \leq 2 |I(u)||I(v)|, \end{align} $$

see [Reference Moreira30, Lemma A.2] for details. We also note that $v^2_k < 2^{r_k}$ for all $k\geq 1$ . In fact, since $v_k$ is the maximum component among all vectors in $X_k$ , there exists $\omega ^\prime \in X_k$ having $v_k$ as one of its components. In this case, by the definition of $X_k$ , we have

$$ \begin{align*} \frac{1}{2^{r_k}} \leq |I(\omega^\prime)|< \frac{1}{v^2_k}, \end{align*} $$

which implies that $v^2_k < 2^{r_k}$ . When $\sum ^{k-1}_{i=0}u_i < n < \sum ^k_{i=0}u_i$ , for any $ w \in \mathcal {W}^{*(n)}_k$ , by (38), we derive that

$$ \begin{align*} |I(w(v_k+1))| \geq \frac{1}{2(v_k+1)(v_k+2)} |I(w)|> \frac{1}{8v^2_k} |I(w)| \geq \frac{1}{8\cdot 2^{r_k}} |I(w)|, \end{align*} $$

which, together with (33), implies that

$$ \begin{align*} \widetilde{\varepsilon}_n> \frac{1}{8\cdot 2^{r_{k}}}\cdot \min\{|I(w)|\colon w \in \mathcal{W}^{*(n)}_k\} = \varepsilon_n. \end{align*} $$

When $n=\sum ^k_{i=0}u_i$ , for any $w \in \mathcal {W}_k$ , it follows from (38) that

$$ \begin{align*} |I(w(v_{k+1}+1))|>\frac{1}{8\cdot 2^{r_{k+1}}} |I(w)|, \end{align*} $$

which, together with (34), implies that

$$ \begin{align*} \widetilde{\varepsilon}_n> \frac{1}{8\cdot 2^{r_{k+1}}}\cdot \min\{|I(w)|\colon w \in \mathcal{W}_k\} = \varepsilon_n. \end{align*} $$

4.4 Step IV: Computation of the lower limit in Lemma 4

We are ready to calculate the lower limit

(39) $$ \begin{align} \liminf_{n \to \infty} \frac{\log(m_1m_2 \cdots m_{n-1})}{-\log (m_{n}\varepsilon_n)}. \end{align} $$

Concerning the numerator in (39), for any $\sum ^{k-1}_{i=0}u_i < n \leq \sum ^k_{i=0}u_i$ , it follows from (35) that

$$ \begin{align*} m_1m_2\cdots m_{n-1} &\geq (\sharp X_1)^{u_1}\cdots(\sharp X_{k-1})^{u_{k-1}} (\sharp X_k)^{n-1-\sum^{k-1}_{i=0}u_i} \\ &>C^{n-1} \cdot \bigg(\prod^{k-1}_{j=1} 2^{(s-\varepsilon)(r_j+1)u_j}\bigg) 2^{(s-\varepsilon)(r_k+1)(n-1-\sum^{k-1}_{i=0}u_i)}. \end{align*} $$

Define

(40) $$ \begin{align} U_n:= \sum^{k-1}_{j=1} (r_j+1)u_j + (r_k+1)\bigg(n-1-\sum^{k-1}_{i=0}u_i\bigg). \end{align} $$

Then,

(41) $$ \begin{align} \log(m_1m_2 \cdots m_{n-1})> (s-\varepsilon)U_n\log 2 + (n-1)\log C. \end{align} $$

Turning to the denominator in (39), for any $w_j= w^{(1)}_j w^{(2)}_j\cdots w^{(u_j)}_j \in \mathcal {A}_j$ , by (38), we have

(42) $$ \begin{align} |I(w_j)| &\geq \frac{1}{2}|I(w^{(1)}_j)| \cdot \frac{1}{2}|I(w^{(2)}_j)| \cdots \frac{1}{2}|I(w^{(u_j-1)}_j)| \cdot |I(w^{(u_j)}_j)| \nonumber\\ &> \frac{1}{2^{u_j-1}} \bigg(\frac{1}{2^{r_j}}\bigg)^{u_j} \nonumber\\ & > \frac{1}{2^{(r_j+1)u_j}}. \end{align} $$

When $\sum ^{k-1}_{i=0}u_i < n < \sum ^k_{i=0}u_i$ , for any $w=w_1w_2\cdots w_{k-1}w^{*(n)}_k \in \mathcal {W}^{*(n)}_k$ , from (38), we derive that

(43) $$ \begin{align} |I(w)| &\geq \frac{1}{2}|I(w_1)| \cdot \frac{1}{2}|I(w_2)| \cdots \frac{1}{2}|I(w_{k-1})| \cdot |I(w^{*(n)}_k)| \nonumber\\ &= \frac{1}{2^{k-1}} |I(w_1)| \cdot |I(w_2)| \cdots |I(w_{k-1})| \cdot |I(w^{*(n)}_k)|. \end{align} $$

Note that

$$ \begin{align*} |I(w^{*(n)}_k)|> \frac{1}{2^{(r_k+1)(n-\sum^{k-1}_{i=0}u_i)}}. \end{align*} $$

Combining this with (42) and (43), we conclude that

$$ \begin{align*} |I(w)|> \frac{1}{2^{U_n+r_k+k}}, \end{align*} $$

which, together with (36), implies that

(44) $$ \begin{align} m_n\varepsilon_n> \varepsilon_n> \frac{1}{2^{U_n+ 2r_k +k + 3}}. \end{align} $$

When $n=\sum ^k_{i=0}u_i$ , for any $w=w_1w_2\cdots w_{k-1}w_k \in \mathcal {W}_k$ , we deduce from (38) and (42) that

$$ \begin{align*} |I(w)| &\geq \frac{1}{2}|I(w_1)| \cdot \frac{1}{2}|I(w_2)| \cdots \frac{1}{2}|I(w_{k-1})| \cdot |I(w_k)| \\ &> \frac{1}{2^{k-1}}\cdot \prod^k_{j=1} \frac{1}{2^{(r_j+1)u_j}} \\ &= \frac{1}{2^{U_n +r_k+k}}, \end{align*} $$

which, together with (37), implies that

(45) $$ \begin{align} m_n\varepsilon_n> \varepsilon_n> \frac{1}{2^{U_n+ r_{k+1}+r_k +k + 3}}. \end{align} $$

We are now in a position to estimate the lower limit in (39). When $\sum ^{k-1}_{i=0}u_i < n < \sum ^k_{i=0}u_i$ , by (28) and the definition of $U_n$ in (40), we see that

$$ \begin{align*} U_n> u_{k-1} >kr_k. \end{align*} $$

Combining this with (41) and (44), we derive that

(46) $$ \begin{align} \frac{\log(m_1m_2 \cdots m_{n-1})}{-\log (m_{n}\varepsilon_n)} &> \frac{(s-\varepsilon)U_n +(n-1)\log_2 C}{U_n+ 2r_k +k + 3} \nonumber \\ &> \frac{s-\varepsilon}{1+ ({2r_k +k + 3})/{kr_k}} + \frac{n}{U_n} \log_2 C, \end{align} $$

where the last inequality comes from $0<C\leq 1$ . When $n=\sum ^k_{i=0}u_i$ , it follows from (28) and (40) that

$$ \begin{align*} U_n>u_k -1 > (k+1)r_{k+1} - 1 > kr_{k+1}. \end{align*} $$

Combining this with (41) and (45), we conclude that

(47) $$ \begin{align} \frac{\log(m_1m_2 \cdots m_{n-1})}{-\log (m_{n}\varepsilon_n)} &> \frac{(s-\varepsilon)U_n+(n-1)\log_2 C}{U_n+ r_{k+1}+r_k +k + 3}\nonumber \\ &> \frac{s-\varepsilon}{1+ ({r_{k+1}+r_k +k + 3})/{kr_{k+1}}} + \frac{n}{U_n} \log_2 C. \end{align} $$

Since $\{r_k\}_{k\geq 1}$ is strictly increasing and $r_k \to \infty $ as $k\to \infty $ , we obtain

$$ \begin{align*} \lim_{k\to \infty}\frac{2r_k +k + 3}{kr_k}= \lim_{k\to \infty} \frac{r_{k+1}+r_k +k + 3}{kr_{k+1}} =0. \end{align*} $$

It follows from the definition of $U_n$ in (40) that

$$ \begin{align*} \lim_{n\to \infty} \frac{U_n}{n} = \infty. \end{align*} $$

By (46) and (47), we derive that

$$ \begin{align*} \liminf_{n\to \infty} \frac{\log(m_1m_2 \cdots m_{n-1})}{-\log (m_{n}\varepsilon_n)} \geq s-\varepsilon, \end{align*} $$

which, together with (29) and Lemma 4, implies that

$$ \begin{align*} \dim_{\mathrm{H}} E(\unicode{x3bb},\xi)\geq \dim_{\mathrm{H}}\mathrm{E} \geq s-\varepsilon. \end{align*} $$

Letting $\varepsilon \to 0$ , we have

$$ \begin{align*} \dim_{\mathrm{H}} E(\unicode{x3bb},\xi) \geq s. \end{align*} $$

Combining the above four steps, we complete the proof of the Key Lemma.