Proof For any $\gamma \in (0, \infty )$ , let $\theta _{n}:=\left \lfloor \gamma \log n\right \rfloor +1$ for all $n \in \mathbb N$ . Then, $\left \{\theta _{n}\right \}_{n\geq 1}$ corresponds to a unique irrational number $x_{0} \in [0,1)$ such that $a_{n}\left (x_{0}\right )=\theta _{n}$ for all $n \in \mathbb N$ . Hence,

$$ \begin{align*}\lim _{n \rightarrow \infty} \frac{S_{n}(x_0)}{n \log n} = \gamma, \end{align*} $$

that is, $x_{0} \in S(\gamma )$ .

Set

$$ \begin{align*}D\left(x_{0}\right):=\bigcup_{N=1}^{\infty}\left\{x \in \mathbb{I}: a_{n}(x)=a_{n}\left(x_{0}\right), \forall n \geq N\right\}. \end{align*} $$

Then, $D\left (x_{0}\right ) \subseteq S(\gamma )$ is dense in $\mathbb {I}$ . To see this, for any $x \in \mathbb {I}$ and $n \in \mathbb N$ , define $x_{n}$ in terms of the continued fraction expansion as

$$ \begin{align*}x_{n}:=\left[a_{1}(x), \ldots, a_{n}(x), a_{n+1}\left(x_{0}\right), a_{n+2}\left(x_{0}\right), \ldots\right]. \end{align*} $$

Then, $x_{n} \in D\left (x_{0}\right )$ and $\left |x_{n}-x\right | \to 0$ as $n \to \infty $ , which implies that $D\left (x_{0}\right )$ is dense in $\mathbb {I}$ . Hence, $S(\gamma )$ is also dense in $\mathbb I$ .

Proof Let $\gamma \in (0,\infty )$ be fixed. It follows from Lemma 2.1 that $S_*(\gamma )$ and $S^*(\gamma )$ are dense. To prove that $S^*(\gamma )$ and $S_*(\gamma )$ are residual, it suffices to show that they are all $G_\delta $ sets.

For $S_*(\gamma )$ , we deduce that

$$\begin{align*}S_*(\gamma) =\bigcap_{k=1}^\infty\bigcap_{N=1}^\infty\bigcup_{n=N}^\infty E_n(\gamma, k), \end{align*}$$

where $E_n(\gamma , k)$ is given by

$$\begin{align*}E_n(\gamma, k):= \left\{x\in \mathbb{I}: S_n(x)<\left(\gamma +\dfrac{1}{k}\right)n\log n \right\}. \end{align*}$$

We observe that for sufficiently large n, $E_n(\gamma , k)$ is nonempty and can be written as a countable union of open sets in $\mathbb {I}$ . More precisely,

$$\begin{align*}E_n(\gamma, k) = \bigcup_{\sum_{i=1}^{n}k_i <(\gamma +1/k)n\log n }I_{n}(k_1,k_2,\dots,k_{n}) \cap\mathbb{I}, \end{align*}$$

where $(k_1,k_2,\dots ,k_{n})\in \mathbb {N}^{n}$ . Since $I_{n}\left (k_1,k_2,\dots ,k_{n}\right )$ is an open set in $\mathbb {I}$ , we have $E_{n}(\gamma , k)$ is open in $\mathbb {I}$ , and so $S_{*}(\gamma )$ is a $G_{\delta }$ set.

For $S^*(\gamma )$ , we conclude that

$$\begin{align*}S^*(\gamma) =\bigcap_{k=\lceil 1/\gamma \rceil}^\infty\bigcap_{N=1}^\infty\bigcup_{n=N}^\infty \left\{x\in \mathbb{I}: S_n(x)>\left(\gamma -\dfrac{1}{k}\right)n\log n \right\}, \end{align*}$$

and $\{x\in \mathbb {I}: S_n(x)>(\gamma -1/k)n\log n\} $ is a countable union of open sets in $\mathbb {I}$ . Hence, $S^{*}(\gamma )$ is a $G_{\delta }$ set.

Proof of Theorem 1.2

For any $K \in \mathbb {N}$ , it follows form Lemma 2.2 that $S_{*}(1 / K)$ and $S^{*}(K)$ are residual. By the definition of residual set, we see that the countable intersection of residual sets is also residual. Since

(2.1) $$ \begin{align} \left\{x \in \mathbb{I}: \liminf _{n \rightarrow \infty} \frac{S_{n}(x)}{n \log n}=0\right\}=\bigcap_{K=1}^{\infty} S_{*}(1 / K) \end{align} $$

and

(2.2) $$ \begin{align} \left\{x \in \mathbb{I}: \limsup _{n \rightarrow \infty} \frac{S_{n}(x)}{n \log n}=\infty\right\}=\bigcap_{K=1}^{\infty} S^{*}(K), \end{align} $$

we derive that the sets on the left-hand side of (2.1) and (2.2) are residual, and so their intersection $S(0, \infty )$ is residual.

Note that every subset of a set of first category is also of first category. For any $\alpha>0$ or $\beta <\infty $ , each $S(\alpha , \beta )$ is a subset of the complement of $S(0, \infty )$ , and so $S(\alpha , \beta )$ is of first category.

Proof of Theorem 1.3

The proof is divided into six cases. For convenience, we first introduce some notation. For any $n \in \mathbb N$ , let $t(n):= \max \{k \in \mathbb {N}:k^2 \leq n\}$ and ${\phi (n):= 2^{2^{n}}}$ . Then, $\sqrt {n}-1\leq t(n) \leq \sqrt {n} $ and $(\phi (n))^2 = \phi (n+1)$ .

In the following, we always take $n_k:=k^2$ for all $k \in \mathbb N$ . The definition of $\{u_k\}_{k\geq 1}$ depends on the values of $\alpha $ and $\beta $ .

Case 1: $ 0 < \alpha \leq \beta < \infty $ . Let $u_1 := 1$ . For all $ k \geq 2 $ , let

(3.3) $$ \begin{align} u_{k}:= \begin{cases} \lceil 4 \alpha k \log k \rceil, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is even}; \\ \lceil 4 \beta k \log k \rceil, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is odd}. \end{cases} \end{align} $$

Then, $\left \{n_{k}\right \}$ and $\left \{u_{k}\right \}$ satisfy condition (3.2). Moreover, we claim that the set $ S(\alpha , \beta )$ defined in (3.1) is a subset of $E_{M}(\{n_{k}\},\{u_{k}\})$ . In fact, for any $x\in E_{M}(\{n_{k}\},\{u_{k}\})$ and for all $k \in \mathbb N$ , we have $ 4 \alpha k \log k \leq a_{k^2}(x) < 4 \beta k \log k + 1 $ and $ 1 \leq a_{j}(x) \leq M $ for all $ j \neq k^2 $ . For any $n \in \mathbb N$ , we see that

$$ \begin{align*} S_n(x) \geq \sum_{k=2}^{t(n)} 4 \alpha k \log k \geq 4 \alpha \int_1^{t(n)} x \log x d x =\alpha (t(n))^2\big(2\log t(n)-1\big) \end{align*} $$

and

$$ \begin{align*} \begin{aligned} S_n(x) &< \sum_{k=2}^{t(n)}\left(4 \beta k \log k+1\right)+n M \\ &<4 \beta \int_1^{t(n)+1} x \log x d x +t(n)+nM \\ &= \beta(t(n)+1)^2\big(2\log \left(t(n)+1\right)-1\big)+ t(n)+nM. \end{aligned} \end{align*} $$

Combining this with the fact that $\sqrt {n}-1\leq t(n) \leq \sqrt {n}$ , we derive that

(3.4) $$ \begin{align} \alpha \leq \liminf_{n \to \infty} \frac{S_{n}(x)}{n \log n} \leq \limsup_{n \to \infty} \frac{S_{n}(x)}{n \log n} \leq \beta. \end{align} $$

Next, we show that the limit superior and the limit inferior in (3.4) are $\beta $ and $\alpha $ , respectively. To this end, consider

$$\begin{align*}\limsup_{m \to \infty} \frac{S_{\phi(2 m)}(x)}{\phi(2 m) \log \phi(2 m)} \end{align*}$$

and

$$\begin{align*}\liminf_{m \to \infty} \frac{S_{\phi(2 m+1)}(x)}{\phi(2 m+1) \log \phi(2 m+1)}. \end{align*}$$

Note that for all k,

1. (1) if $\phi (2m-2) < k \leq \phi (2m-1)$ , then $\lceil \log _2 \log _2 k \rceil $ is odd, and so $a_{k^2}(x) = \lceil 4 \beta k \log k \rceil $ ;

2. (2) if $\phi (2m-1) < k \leq \phi (2m)$ , then $\lceil \log _2 \log _2 k \rceil $ is even, and so $a_{k^2}(x) = \lceil 4 \alpha k \log k \rceil $ ,

and $ 1 \leq a_{j}(x) \leq M $ for all $ j \neq k^2$ . On the one hand, it follows from (1) that

$$ \begin{align*} \begin{aligned} S_{\phi(2 m)}(x)& \geq \sum_{k=\phi(2m-2)+1}^{\phi(2m-1)} 4\beta k\log k \\ & \geq 4 \beta \int_{\phi(2 m-2)}^{\phi(2 m-1)} x \log x\,\text{d} x\\ &= \beta (\phi(2 m-1))^2\left(2\log \phi(2 m-1) -1\right) \\ &\quad -\beta (\phi(2 m-2))^2\left(2\log \phi(2 m-2) -1\right). \end{aligned} \end{align*} $$

Note that $(\phi (n))^2 = \phi (n+1)$ , we have

$$\begin{align*}S_{\phi(2 m)}(x) \geq \beta \phi(2 m)(\log \phi(2 m)-1) - \beta \phi(2 m-1)(\log \phi(2 m-1)-1). \end{align*}$$

Since

$$\begin{align*}\lim_{m\to \infty} \frac{\phi(2 m-1)(\log \phi(2 m-1)-1)}{\phi(2 m)(\log \phi(2 m)-1)} =0, \end{align*}$$

we derive that

(3.5) $$ \begin{align} \limsup_{m \to \infty} \frac{S_{\phi(2 m)}(x)}{\phi(2 m) \log \phi(2 m)} \geq \beta. \end{align} $$

On the other hand, we deduce from (2) that

$$ \begin{align*} \begin{aligned} S_{\phi(2 m+1)}(x) & <\sum_{k=2}^{\phi(2 m)} (4 \alpha k \log k+1)+\sum_{k=2}^{\phi(2 m-1)} (4 \beta k \log k+1)\\ &\quad +\phi(2 m+1) M\\ & < 4 \alpha \int_{k=1}^{\phi(2m)+1} x \log x d x+4\beta\int_{k=1}^{\phi(2m-1)+1} x \log x d x\\ &\quad+\phi(2 m+1)(M+2)\\ & <2\alpha (\phi(2 m)+1)^2\log \left(\phi(2 m)+1\right)\\ &\quad +2\beta (\phi(2 m-1)+1)^2 \log \left(\phi(2 m-1)+1)\right)\\ &\quad +\phi(2 m+1)(M+2). \end{aligned} \end{align*} $$

By the definition of $\phi (n)$ , we see that

$$\begin{align*}\lim_{m\to \infty} \frac{2(\phi(2 m)+1)^2\log \left(\phi(2 m)+1\right)}{\phi(2 m+1) \log \phi(2 m+1)} =1 \end{align*}$$

and

$$\begin{align*}\lim_{m\to \infty} \frac{(\phi(2 m-1)+1)^2 \log \left(\phi(2 m-1)+1)\right)}{(\phi(2 m)+1)^2\log \left(\phi(2 m)+1\right)} =0, \end{align*}$$

and so

(3.6) $$ \begin{align} \liminf_{m \to \infty} \frac{S_{\phi(2 m+1)}(x)}{\phi(2 m+1) \log \phi(2 m+1)} \leq \alpha. \end{align} $$

Combining (3.4)–(3.6), we obtain

$$\begin{align*}\liminf_{n \rightarrow \infty} \frac{S_{n}(x)}{n \log n} = \alpha \quad \text{and} \quad \limsup_{n \rightarrow \infty} \frac{S_{n}(x)}{n \log n} = \beta, \end{align*}$$

i.e., $ x \in S(\alpha , \beta ) $ . Hence, $ \operatorname {dim}_{\mathrm {H}} S(\alpha , \beta ) \geq \operatorname {dim}_{\mathrm {H}} E_{M}(\{n_{k}\}, \{u_{k}\}) $ . Letting $ M \rightarrow \infty $ , we conclude from Lemma 3.1 that $ \operatorname {dim}_{\mathrm {H}} S(\alpha , \beta ) = 1 $ .

For other cases, when $ \alpha = 0 $ , replace $\lceil 4 \alpha k \log k \rceil $ in (3.3) with k; when $\beta =\infty $ , replace $\lceil 4 \beta k \log k \rceil $ in (3.3) with $k^2$ . More precisely,

Case 2: $\alpha = \beta =0$ . For all $ k \geq 1 $ , let $u_k := k$ .

Case 3: $\alpha =\beta = \infty $ . For all $ k \geq 1 $ , let $u_k := k^2 $ .

Case 4: $ \alpha =0$ and $\beta = \infty $ . Let $u_1 := 1$ . For all $ k \geq 2 $ , let

$$\begin{align*}u_{k} := \begin{cases} k, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is even}; \\ k^2, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is odd}. \end{cases} \end{align*}$$

Case 5: $\alpha = 0$ and $0< \beta < \infty $ . Let $u_1 := 1$ . For all $ k \geq 2 $ , let

$$\begin{align*}u_{k} := \begin{cases} k, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is even}; \\ \lceil 4 \beta k \log k \rceil, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is odd}. \end{cases} \end{align*}$$

Case 6: $ 0 <\alpha <\infty $ and $\beta = \infty $ . Let $u_1 := 1$ . For all $ k \geq 2 $ , let

$$\begin{align*}u_{k} := \begin{cases} \lceil 4 \alpha k \log k \rceil, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is even}; \\ k^2, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is odd}. \end{cases} \end{align*}$$

Following the proof in Case 1 step by step, we can similarly derive the same result. The details are omitted here for brevity.

Proof This result can be regarded as a consequence of Lemma 3.1. To see this, define $\{n_k\}_{k\geq 1}$ and $\{u_k\}_{k\geq 1}$ as follows: for any $k \in \mathbb N$ ,

$$\begin{align*}\begin{aligned} &\left\{ \begin{array}{@{}l} n_{2k-1}:= \omega_k, \\ n_{2k}:= \omega_k + 1, \end{array} \right. \quad \text{and} \quad \left\{ \begin{array}{@{}l} u_{2k-1}:= b_k, \\ u_{2k}:= c_k. \end{array} \right. \end{aligned} \end{align*}$$

Then, $E_{M}(\{\omega _{k}\},\{b_{k}\},\{c_{k}\}) = E_{M}(\{n_{k}\},\{u_{k}\})$ . Since $ \omega _k / k \to \infty $ as $k \rightarrow \infty $ , it follows that $ n_k / k \to \infty $ as $ k \to \infty $ . Moreover,

$$\begin{align*}\begin{aligned} \lim_{k \to \infty} \frac{1}{n_{2k}} \sum_{j=1}^{2k} \log u_j &= \lim_{k \to \infty} \frac{1}{\omega_k + 1} \sum_{j=1}^k \left(\log b_j + \log c_j\right) \\ &= \lim_{k \to \infty} \frac{\omega_k}{\omega_k + 1} \cdot \frac{1}{\omega_k} \sum_{j=1}^k \log \left(b_j c_j\right) \\ &= 0, \end{aligned} \end{align*}$$

and

$$\begin{align*}\begin{aligned} \lim_{k \to \infty} \frac{1}{n_{2k-1}} \sum_{j=1}^{2k-1} \log u_j &= \lim_{k \to \infty} \frac{1}{\omega_k} \left( \sum_{j=1}^{k-1} \left(\log b_j + \log c_j\right) + \log b_j \right) \\ &\leq \lim_{k \to \infty} \frac{1}{\omega_k} \sum_{j=1}^k \left(\log b_j + \log c_j\right). \\ &= 0. \end{aligned} \end{align*}$$

Thus, $\{n_k\}_{k\geq 1}$ and $\{u_k\}_{k\geq 1}$ satisfy condition (3.2). By Lemma 3.1, we derive that

$$\begin{align*}\lim _{M \rightarrow \infty} \operatorname{dim}_{\mathrm{H}}E_{M}(\{\omega_{k}\},\{b_{k}\},\{c_{k}\})= \lim _{M \rightarrow \infty}E_{M}(\{n_{k}\},\{u_{k}\}) =1. \end{align*}$$

The proof is completed.

Proof of Theorem 4.3

We only give the proof for the case where $0 < \alpha \leq \beta < \infty $ . Let $\omega _1 = b_1 = c_1 :=1$ . For all $ k \geq 2 $ , let $\omega _k := k^2$ ,

(4.4) $$ \begin{align} b_{k} := \begin{cases} \lceil 8 \alpha k \log k \rceil, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is even}; \\ \lceil 8 \beta k \log k \rceil, & \text{ if } \lceil \log_2 \log_2 k \rceil \text{ is odd}, \end{cases} \end{align} $$

and $c_k := \lfloor \log k \rfloor $ . Then, $\{\omega _{k}\}_{k\geq 1}, \{b_{k}\}_{k\geq 1}$ , and $\{c_{k}\}_{k\geq 1}$ satisfy condition (4.3). Moreover, we claim that $E_{M}(\{\omega _{k}\},\{b_{k}\},\{c_{k}\})$ is a subset of $ \widetilde {S}(\alpha , \beta )$ . In fact, for any $x\in E_{M}(\{\omega _{k}\},\{b_{k}\},\{c_{k}\})$ , we see that $ 8 \alpha k \log k \leq a_{k^2}(x) \leq 8 \beta k \log k + 1 $ , $ \log k -1 \leq a_{k^2+1}(x) \leq \log k $ , and $ 1 \leq a_{j}(x) \leq M $ for all $ j \notin \{k^2, k^2+1\}$ . Hence,

$$ \begin{align*} \widetilde{S}_n(x) \geq \sum_{k=2}^{t(n)}8 \alpha k \log k(\log k-1) \geq 8 \alpha\int_1^{t(n)}\left( x \log ^2 x d x - x\log x \right)d x, \end{align*} $$

and

$$ \begin{align*} \begin{aligned} \widetilde{S}_n(x) &\leq \sum_{k=2}^{t(n)} (8 \beta k \log k+1) \log k+M\sum_{k=2}^{t(n)}(8 \beta k \log k+\log k +1) +n M^2 \\ & \leq \int_1^{t(n)+1}\left(8 \beta x \log^2 x+8 M \beta x \log x +(M+1)\log x \right)d x + nM(M+1). \end{aligned} \end{align*} $$

Our analysis focuses exclusively on the main term of the integral, since the contributions from the integrals of the remaining terms are negligible, being infinitesimal of the order $ n \log ^2 n $ . Since $\sqrt {n}-1\leq t(n) \leq \sqrt {n}$ and

(4.5) $$ \begin{align} \int x \log ^2 x d x=\frac{1}{8} x^2\left( \log ^2 x^2-2 \log x^2+2\right)+C, \end{align} $$

we deduce that

(4.6) $$ \begin{align} \alpha \leq \liminf_{n \to \infty} \frac{\widetilde{S}_{n}(x)}{n \log^2 n} \leq \limsup_{n \to \infty} \frac{\widetilde{S}_{n}(x)}{n \log^2 n} \leq \beta. \end{align} $$

Note that for all $k \in \mathbb {N}$ ,

1. (1) if $\phi (2m-1) < k \leq \phi (2m)$ , then $\lceil \log _2 \log _2 k \rceil $ is even, and so $a_{k^2}(x) = \lceil 8 \alpha k \log k \rceil $ and $a_{k^2+1}(x) = \lfloor \log k \rfloor $ ;

2. (2) if $\phi (2m) < k \leq \phi (2m+1)$ , then $\lceil \log _2 \log _2 k \rceil $ is odd, and so $a_{k^2}(x) = \lceil 8 \beta k \log k \rceil $ and $a_{k^2+1}(x) = \lfloor \log k \rfloor $ ,

and $ 1 \leq a_{j}(x) \leq M $ for all $ j \notin \{k^2, k^2+1\}$ . Thus,

$$ \begin{align*} \begin{aligned} \widetilde{S}_{\phi(2 m)}(x)& \geq\sum_{k=\phi(2 m-2)+1}^{\phi(2 m-1)} 8 \beta k \log k (\log k-1)\\ & = 8\beta \int_{\phi(2 m-2)}^{\phi(2 m-1)} \left( x \log ^2 x d x- \log x \right) d x. \end{aligned} \end{align*} $$

By (4.5) and the definition of $\phi (n)$ , we obtain

(4.7) $$ \begin{align} \limsup_{m \to \infty} \frac{\widetilde{S}_{\phi(2 m)}(x)}{\phi(2 m) \log^2 \phi(2 m)} \geq \beta. \end{align} $$

Similarly, we can derive that

(4.8) $$ \begin{align} \liminf_{m \to \infty} \frac{\widetilde{S}_{\phi(2 m+1)}(x)}{\phi(2 m+1) \log^2 \phi(2 m+1)} \leq \alpha. \end{align} $$

Combining (4.6)–(4.8), we conclude that

$$\begin{align*}\liminf _{n \rightarrow \infty} \frac{\widetilde{S}_{n}(x)}{n \log^2 n}=\alpha\quad \text{and}\quad \limsup _{n \rightarrow \infty} \frac{\widetilde{S}_{n}(x)}{n \log^2 n}=\beta, \end{align*}$$

i.e., $x\in \widetilde {S}(\alpha ,\beta )$ . So $\operatorname {dim}_{\mathrm {H}} \widetilde {S}(\alpha ,\beta ) \geq \operatorname {dim}_{\mathrm {H}} E_{M}(\{\omega _{k}\},\{b_{k}\},\{c_{k}\})$ . Letting $ M \rightarrow \infty $ , it follows from Lemma 4.4 that $ \operatorname {dim}_{\mathrm {H}} \widetilde {S}(\alpha ,\beta ) = 1 $ .