在最近一些年, 各种各样的连分数展式被赋予了极大的研究兴趣. 例如正规连分数展式^[1]、向后连分数展式^[2]、偶数连分数展式^[3]、Engel连分数展式、$ \alpha- $连分数展式^[4]等. 关于正规连分数展式, 在文献[5], [6], [7], [8]中, 他们讨论其部分商的增长速度并刻画相关例外集的Hausdorff维数. L.L.Fang和M. Wu^[9], T. Zhong和L. Tang^[10]讨论了Engel连分数展式中部分商的增长速度以及相关例外集的Hausdorff维数. 更多连分数的相关信息, 可参见文献[11].

A.H.Fan, B.W.Wang和J.Wu于2007年在文献[12]中引入了一种新的连分数展式. 对于任意一个$ x\in(0, 1] $, 定义算法:

$ \begin{align} x_{1}=x;\ d_{j}=\left[\frac{1}{x_{j}}\right];\ x_{j+1}=\frac{1}{d_{j}\left(d_{j}-1\right)+1}\left(\frac{1}{x_{j}}-d_{j}\right). \end{align} $

(1.1)

这个算法可以诱导出$ x\in(0, 1] $的一个无穷级数展式:

$ \begin{align} x=\frac{1}{d_{1}\left(x\right)+\frac{d_{1}\left(x\right)\left(d_{1}\left(x\right)-1\right)+1}{d_{2}\left(x\right) +\cdots+\frac{1}{d_{n-1}\left(x\right)+\frac{d_{n-1}\left(x\right)\left(d_{n-1}\left(x\right)-1\right)+1}{d_{n}\left(x\right)+\ddots}}}}. \end{align} $

(1.2)

该展式称为$ x $的Sylvester连分数展式, 并将$ x $的Sylvester连分数展式简记为$ x=\left[d_{1}(x), d_{2}(x), \cdots, d_{n}(x)\right] $. 其中$ \{d_{j}(x)\}_{j\geq1} $称为其Sylvester连分数展式的部分商. 由上述算法可得,

$ \begin{align} d_{n}(x)\in \mathbb{N}, d_{j+1} \geq d_{j}\left(d_{j}-1\right)+1, j \geq 1. \end{align} $

(1.3)

在文献[12]中, A.H.Fan, B.W.Wang和J.Wu研究了Sylvester部分商序列$ \{d_{j}(x)\}_{j\geq1} $的算术性质和度量性质, 包括弱大数定律、强大数定律以及中心极限定理等. 他们证明了如下引理.

引理1.1^[12]  对Lebesgue测度意义下几乎所有的$ x\in(0, 1] $都有$ \lim\limits_{n\to\infty}\left(\frac{b_{n+1}(x)}{b_1(x)b_2(x)\cdots b_n(x)}\right)^{1/n}=e $. 在文献[13]中, Z.L.Zhang研究了该极限的例外集, 即

$ \begin{align*} E(\alpha)=\left\{x \in(0, 1): \lim _{n \rightarrow \infty}\left(\frac{d_{n+1}(x)}{d_{1}(x) d_{2}(x) \cdots d_{n}(x)}\right)^{\frac{1}{n}}=\alpha\right\}, \end{align*} $

并证明了该集合$ E(\alpha) $的Hausdorff维数为1.

对于Sylvester部分商的增长速度, 他们证明了以下引理.

引理1.2^[12]  对Lebesgue测度意义下几乎所有的$ x\in(0, 1] $都有

$ \begin{align*} \lim _{n \rightarrow \infty} \frac{1}{2^{n}} \log d_{n}(x)=\lambda_{0}(x). \end{align*} $

这里的$ \lambda_{0}(x)=\frac{1}{2}\left\{\log d_1(x)+\sum_{n=1}^{\infty}2^{-n}\log \frac{d_{n+1}(x)}{d_n(x)}\right\} $是一个与$ x $有关的常数.

在此基础之上, 考虑$ x $的Sylvester连分数展式部分商的增长速度的刻画. 令$ \psi(n) $是给定的$ \mathbb{N} $上的正函数, 并考虑集合:

$ \begin{align*} A(\psi)=\left\{x \in(0, 1]: \lim _{n \rightarrow \infty} \frac{\log d_{n}(x)}{\psi(n)}=1\right\}. \end{align*} $

两个定义在$ \mathbb{N} $上的正函数$ \psi(n) $与$ \bar{\psi}(n) $称为是等价的, 如果$ n \rightarrow \infty $时, 有$ \frac{\bar{\psi}(n)}{\psi(n)} \rightarrow 1 $. 为方便, 记Sylvester连分数展式部分商全是$ 1 $的点为$ e_{0}=[1, 1, 1, \cdots] $. 计算的主要结果如下

定理1.1  令$ \psi(n) $为定义在$ \mathbb{N} $上的正函数, 则$ A(\psi)\setminus \{e_0\} $是非空的当且仅当函数$ \psi $等价于一个函数$ \bar{\psi} $且函数$ \bar{\psi} $满足: 存在一个$ n_{0} \in \mathbb{N} $, 使得$ \left[e^{\overline\psi{\left(n_{0}\right)}}\right]>e $, 当$ n\geq n_{0} $时, 有$ \left[e^{\overline\psi{\left(n+1\right)}}\right]\geq\left[e^{\overline\psi{\left(n\right)}}\right] \left(\left[e^{\overline\psi{\left(n\right)}}\right]-1\right)+1 $, 这时有

$ \begin{align*} \dim_{H} A(\psi)=\liminf _{n \rightarrow \infty} \frac{\psi(1)+\psi(2)+\cdots+\psi(n)}{\psi(n+1)} . \end{align*} $

由定理1.1的结果可以得到两个推论.

推论1.2  当$ \liminf _{n \rightarrow \infty} \frac{\psi(n+1)}{\psi(n)}<2 $时, 集合$ A(\psi) \backslash\{e_{0}\}=\emptyset $. 当$ \liminf _{n \rightarrow \infty} \frac{\psi(n+1)}{\psi(n)}>2 $时, 集合$ A(\psi) \backslash\{e_{0}\}\neq \emptyset $.

推论1.3  令$ \psi $为定义在$ \mathbb{N} $上的正函数, 且满足$ \lim _{n \rightarrow \infty} \frac{\psi(n+1)}{\psi(n)}=b \in[2, +\infty] $和$ A(\psi) \backslash\{e_{0}\}\neq\emptyset $.则有$ \operatorname{dim}_{H} A(\psi)=\frac{1}{b-1}. $与此同时, 考虑Sylvester展式部分商的增长速度是给定函数高阶无穷大和高阶无穷小的特殊情况, 即集合:

$ \begin{align*} A^{*}(\psi)=\left\{x \in(0, 1]: \lim _{n \rightarrow \infty} \frac{\log d_{n}(x)}{\psi(n)}=\infty\right\} \end{align*} $

和

$ \begin{align*} A_{*}(\psi)=\left\{x \in(0, 1]: \lim _{n \rightarrow \infty} \frac{\log d_{n}(x)}{\psi(n)}=0\right\}. \end{align*} $

对此, 可以得到以下结果.

定理1.4  令$ \psi(n) $为定义在$ \mathbb{N} $上的正函数且满足$ \limsup\limits _{n \rightarrow \infty} \frac{\log \psi(n)}{n}=\log B \in[0, +\infty] $则$ \operatorname{dim}_{H} A^{*}(\psi)=\min \left\{1, \frac{1}{B-1}\right\} $.

定理1.5  令$ \psi(n) $为定义在$ \mathbb{N} $上的正函数且满足$ \liminf\limits _{n \rightarrow \infty} \frac{\psi(n)}{2^{n}}=C \in[0, +\infty] $则$ A_{*}(\psi) $的维数为:

(1) 若$ 0 \leq C<\infty $, 则$ \operatorname{dim}_{H} A_{*}(\psi)=0 $.

(2) 若$ C=\infty $, 则$ \operatorname{dim}_{H} A_{*}(\psi)=1 $.

在这一节中罗列了一些Sylvester连分数展式的基本性质和相关结论. 记$ x $的Sylvester连分数展式的$ n $阶收敛因子为:

$ \begin{equation*} \frac{p_{n}\left(x\right)}{q_{n}\left(x\right)}=\frac{1}{d_{1}\left(x\right)+ \frac{d_{1}\left(x\right)\left(d_{1}\left(x\right)-1\right)+1}{d_{2}\left(x\right) +\cdots+\frac{1}{d_{n-1}\left(x\right)+\frac{d_{n-1}\left(x\right)\left(d_{n-1}\left(x\right)-1\right)+1}{d_{n}\left(x\right)}}}} :=[d_{1}\left(x\right), d_{2}\left(x\right), \cdots, d_{n}\left(x\right)], \end{equation*} $

则有以下命题.

命题2.1^[12]  对任意的$ x\in (0, 1] $, 有$ \lim\limits_{n\to\infty}\frac{p_{n}\left(x\right)}{q_{n}\left(x\right)}=x. $

命题2.2^[12]  在$ p_{0}\left(x\right)=0, q_{0}\left(x\right)=1;p_{1}\left(x\right)=1;q_{1}\left(x\right)=d_{1}\left(x\right) $的基础之上, 有:

$ \begin{equation} p_{n}\left(x\right)=d_{n}\left(x\right)p_{n-1}\left(x\right)+\left(d_{n-1}\left(x\right)\left(d_{n-1}\left(x\right)-1\right)+1\right)p_{n-2}\left(x\right), \qquad n\geq 2; \end{equation} $

(2.1)

$ \begin{equation} q_{n}\left(x\right)=d_{n}\left(x\right)q_{n-1}\left(x\right)+\left(d_{n-1}\left(x\right)\left(d_{n-1}\left(x\right)-1\right)+1\right)q_{n-2}\left(x\right), \qquad n\geq 2. \end{equation} $

(2.2)

对此进行进一步的推导可得:

$ \begin{equation} p_{n}\left(x\right)q_{n-1}\left(x\right)-p_{n-1}\left(x\right)q_{n}\left(x\right)=\left(-1\right)^{n-1}\prod\limits_{j=1}^{n-1}\left(d_{j}\left(x\right)\left(d_{j}\left(x\right)-1\right)+1\right). \end{equation} $

(2.3)

$ \begin{equation} q_{n}^{2}<q_{n}\left(q_{n}+q_{n-1}\right)<2q_{n}^{2}\qquad \qquad d_{1}d_{2}\cdots d_{n}\leq q_{n}\leq 2^{n}d_{1}d_{2}\cdots d_{n}. \end{equation} $

(2.4)

定义2.3^[12]  一个正整数向量$ \left(d_{1}, d_{2}, \cdots, d_{n}\right) $称为Sylvester连分数展式可允许的, 如果存在一个$ x\in (0, 1] $, 使得对$ 1\leq j\leq n $都有$ d_j(x)=d_j $. 一个正整数序列$ \left(d_{1}, d_{2}, \cdots, d_{n}, \cdots\right) $称为Sylvester连分数展式可允许的, 如果存在一个$ x\in (0, 1] $, 使得对$ j\geq1 $都有$ d_j(x)=d_j $.

命题2.4^[12]  一个正整数序列$ \left(d_{1}, d_{2}, \cdots, d_{n}, \cdots\right) $是Sylvester连分数展式可允许, 当且仅当对所有的$ j\geq 1 $, 有$ d_{n}\in \mathbb{N}, \ d_{j+1}\geq d_{j}\left(d_{j}-1\right)+1. $

记所有长度为$ n $的Sylvester连分数展式可允许序列的集合为

$ \Sigma_{n}=\{\left(d_{1}, d_{2}, \cdots, d_{n}\right)\in\mathbb{N}^{n}:d_{n}\in \mathbb{N}, \ d_{j+1}\geq d_{j}\left(d_{j}-1\right)+1, 1\leq j\leq n-1\}. $

对$ \left(d_{1}, d_{2}, \cdots, d_{n}\right)\in \Sigma_{n} $, 记

$ I\left(d_{1}, d_{2}, \cdots, d_{n}\right)\in \mathbb{N}^{n}:=\{x\in (0, 1]:d_{1}\left(x\right)=d_{1}, d_{2}\left(x\right)=d_{2}, \cdots, d_{n}\left(x\right)=d_{n}\}, $

将其称为$ n $阶柱集, 关于该柱集的长度有如下结论.

命题2.5^[12]  对$ \left(d_{1}, d_{2}, \cdots, d_{n}\right)\in \Sigma_{n} $,

$ \left|I\left(d_{1}, d_{2}, \cdots, d_{n}\right)\right|=\frac{\prod\limits_{j=1}^{n-1}\left(d_{j}\left(x\right)\left(d_{j}\left(x\right)-1\right)+1\right)}{q_{n}\left(q_{n}+q_{n-1}\right)}. $

引理2.6^[14]  令$ \mu $为一个测度, $ E\subset(0, 1] $为一个Borel集且$ \mu\left(E\right)>0 $.若对任意的$ x\in E $, 有$ \liminf\limits_{r\to\ 0 }\frac{\log\mu\left(B\left(x, r\right)\right)}{\log r}\geq s, $其中$ B\left(x, r\right) $表示一个以$ x $为中心, $ r $为半径的开球, 则可得$ \dim_{H}E\geq s. $

引理2.7    对$ N \geq 0 $, 令

$ A(\psi)=\left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_n(x)}{\psi(n)}=1\right\} $

且

$ A_N(\psi)=\left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_n(x)}{\psi(n+N)}=1\right\}. $

若$ A(\psi) \backslash\{e_{o}\} \neq \emptyset $且$ A_N(\psi) \backslash\{e_{o}\} \neq \emptyset $, 则可得:

$ \operatorname{dim}_H A(\psi)=\operatorname{dim}_H A_N(\psi). $

证  记$ A(\psi)=\bigcup\limits_{\left(d_{1}, d_{2}, \ldots, d_{N}\right) \in \Sigma_{n}} A_{\left(d_{1}, d_{2}, \ldots, d_{N}\right)}(\psi) $其中$ A_{\left(d_{1}, d_{2}, \ldots, d_{N}\right)}(\psi)=I\left(d_{1}, d_{2}, \ldots, d_{N}\right) \cap A(\psi) $结合$ \left(d_{1}, d_{2}, \ldots, d_{N}\right) \in \Sigma_{n} $, 由算法(1.1)可知, 在$ A_{\left(d_{1}, d_{2}, \ldots, d_{N}\right)}(\psi) $和$ A_{N}(\psi) $之间存在一个双bi-Lipschitz对应. 因此

$ \dim_{H} A_{\left(d_{1}, d_{2}, \ldots, d_{N}\right)}(\psi)=\dim_{H} A_{N}(\psi). $

进而可得

$ \dim_{H} A(\psi)=\sup\limits_{\left(d_{1}, d_{2}, \ldots, d_{N}\right) \in \Sigma_{N}} \dim_{H} A_{\left(d_{1}, d_{2}, \ldots, d_{N}\right)}(\psi)=\operatorname{dim}_{H} A_{N}(\psi). $

在证明该定理之前, 需要先证明一些在后续证明过程中需要的引理.

引理3.1  $ A(\psi)\setminus \{e_0\} $是非空的当且仅当函数$ \psi $等价于一个函数$ \bar{\psi} $且函数$ \bar{\psi} $满足: 存在一个$ n_{0} \in \mathbb{N} $, 使得$ \left[e^{\overline\psi{\left(n_{0}\right)}}\right]>e $, 当$ n\geq n_{0} $时, 有$ \left[e^{\overline\psi{\left(n+1\right)}}\right]\geq\left[e^{\overline\psi{\left(n\right)}}\right] \left(\left[e^{\overline\psi{\left(n\right)}}\right]-1\right)+1 $.

证  必要性: 由$ A(\psi)\backslash\{e_{0}\}\not=\emptyset $可知, 存在$ x_0 \in A(\psi)\backslash\{e_{0}\}\subset(0, 1) $, 由$ e_0 $的定义, 存在一个$ n_0 \in \mathbb{N} $, 当$ n\geq n_0 $时有$ d_{n_0}\left(x_0\right)\geq 3 $. 令:

$ \begin{align*} \bar{\psi}(n) = \begin{cases} 1 & \text { 若 } n< n_0 , \\ \log d_n\left(x_0\right) & \text { 若 } n\geq n_0. \end{cases} \end{align*} $

由算式(1.3), 可以得到$ \left[ e^{\bar{\psi}\left(n_0\right)}\right]=d_{n_0}\left(x_0\right) $且

$ \begin{align*} \left[ e^{\bar{\psi}(n+1)}\right]=d_{n+1}\left(x_0\right) \geq d_n\left(x_0\right)\left(d_n\left(x_0\right)-1\right)+1 =\left[ e^{\bar{\psi}(n)}\right]\left(\left[ e^{\bar{\psi}(n)}\right]-1\right)+1. \end{align*} $

显然$ \psi $与$ \bar{\psi} $是等价的.

充分性: 若$ \bar{\psi} $等价于$ \psi $, 且存在某一$ n_{0}\in\mathbb{N} $, 使得$ \left[e^{\overline\psi{\left(n_{0}\right)}}\right]>e $以及当$ n> n_{0} $, 有$ \left[e^{\overline\psi{\left(n+1\right)}}\right]\geq \left[e^{\overline\psi{\left(n\right)}}\right]\left(\left[e^{\overline\psi{\left(n\right)}}\right]-1\right)+1 $. 令:

$ \begin{align*} d_n(x)= \begin{cases} 1 & \text { 若 } n< n_0, \\ \left[ e^{\bar{\psi}(n)}\right] & \text { 若 } n\geq n_0. \end{cases} \end{align*} $

则$ x \in A(\bar{\psi})\backslash\{e_{0}\}=A(\psi)\backslash\{e_{0}\} $.

注意, 若$ \psi $与$ \bar{\psi} $等价, 则$ A(\psi)\backslash\{e_{0}\} $意味着$ A(\bar{\psi})\backslash\{e_{0}\} $, 因此在证明定理1.1时不妨假设存在一个$ n_{0}\in\mathbb{N} $, 使得$ \left[e^{\psi{\left(n_{0}\right)}}\right]>e $且当$ n>n_{0} $, 有$ \left[e^{\psi{\left(n+1\right)}}\right]\geq \left[e^{\psi{\left(n\right)}}\right]\left(\left[e^{\psi{\left(n\right)}}\right]-1\right)+1 $.

引理3.2  令正实数列$ \{u_{n}\}_{n\geq1} $与$ \{v_{n}\}_{n\geq1} $满足: 对任意的$ n\geq1 $, 有$ u_{n}\geq2 $, $ u_{n}+2<v_{n}u_{n} $, $ u_{n+1}\geq v_{n}u_{n}\left(v_{n}u_{n}-1\right)+1 $, 且有

$ \begin{align} \lim\limits_{n\to\infty}\frac{\log \prod\limits_{j=1}^n\left(v_{j}-1\right)}{\log u_{n}}=0 ; \lim\limits_{n\to\infty}\frac{\log v_{n}}{u_{n}}=0. \end{align} $

(3.1)

令$ F=\{x\in(0, 1]:u_{n}\leq d_{n}\left(x\right)<v_{n}u_{n}, n\geq1\}, $则

$ \dim_{H}F=\liminf\limits_{n\to\infty}\frac{\log\left(u_{1}u_{2}\cdots u_{n}\right)}{\log u_{n+1}}. $

证  令$ s_{0}=\liminf\limits_{n\to\infty}\frac{\log\left(u_{1}u_{2}\cdots u_{n}\right)}{\log u_{n+1}} $.

首先证明$ \dim_{H}F\geq s_0 $.

由已知, 对任意的$ n\geq1 $, 有$ u_{n}\geq2 $, $ u_{n}+2<v_{n}u_{n} $, $ u_{n+1}\geq v_{n}u_{n}\left(v_{n}u_{n}-1\right)+1 $, 故可得$ \lim_{n\to\infty}\frac{\log 2^{3n}}{\log u_{n}}=0 $. 由定义可得:

$ F=\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{\lceil u_{k}\rceil\leq d_{k}\left(x\right)\leq\lceil v_{k}u_{k}\rceil-1, 1\leq k\leq n}I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right), $

这里的$ \lceil\cdot\rceil $表示不小于该数的最小整数, $ I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right)=Cl\left(I\left(d_{1}, d_{2}, \cdots, d_{n}\right)\right) $称为一个$ n $阶基本区间, 其中$ Cl $表示其闭包. 对任意的$ n\geq 1 $和$ \lceil u_{k}\rceil\leq d_{k}\left(x\right)\leq\lceil v_{k}u_{k}\rceil-1\left(1\leq k\leq n\right) $, 由算式(2.4)和命题2.5, 可得

$ \begin{align*} |I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right)| =\frac{\prod\limits_{j=1}^{n-1}\left(d_{j}\left(x\right)\left(d_{j}\left(x\right)-1\right)+1\right)} {q_{n}\left(q_{n}+q_{n-1}\right)} \geq \frac{\left(\frac{d_{1}}{2}\right)^{2}\cdots \left(\frac{d_{n-1}}{2}\right)^{2}}{2q_{n}^{2}} \geq \frac{d_{1}^{2}\cdots d_{n-1}^{2}}{2^{n}q_{n}^{2}}\geq \frac{1}{2^{3n}v_{n}^{2}u_{n}^{2}} \end{align*} $

和

$ \begin{align*} |I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right)| =\frac{\prod\limits_{j=1}^{n-1}\left(d_{j}\left(x\right)\left(d_{j}\left(x\right)-1\right)+1\right)} {q_{n}\left(q_{n}+q_{n-1}\right)} \leq \frac{d_{1}^{2}\cdots d_{n-1}^{2}}{q_{n}^{2}} \leq \frac{1}{d_{n}^{2}}\leq \frac{1}{u_{n}^{2}}, \end{align*} $

即

$ \begin{equation} \frac{1}{2^{3n}v_{n}^{2}u_{n}^{2}}\leq |I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right)|\leq \frac{1}{u_{n}^{2}}. \end{equation} $

(3.2)

取$ s<s_{0} $. 由算式(3.1), 存在一个$ n_{0} $使得对所有的$ n\geq n_{0} $有

$ \begin{equation} \frac{u_{1}u_{2}\cdots u_{n}\prod\limits_{j=1}^n\left(v_{j}-1\right)}{2^{3n}} \geq \left(\frac{v_{n+1}^{2}u_{n+1}}{v_{n+1}-1}\right)^{s}. \end{equation} $

(3.3)

定义一个支撑在$ F $上的质量分布$ \mu $使得对任意的$ n\geq 1 $, $ \lceil u_{j}\rceil\leq d_{j}\left(x\right) \leq\lceil v_{j}u_{j}\rceil-1\left(1\leq j\leq n\right) $, 有

$ \begin{equation*} \mu\left(I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right)\right) =\prod\limits_{j=1}^n\frac{1}{\lceil v_{j}u_{j}\rceil-\lceil u_{j}\rceil}. \end{equation*} $

显然,

$ \begin{equation} \frac{1}{2^{n}u_{1}u_{2}\cdots u_{n}\prod\limits_{j=1}^n\left(v_{j}-1\right)} \leq \mu\left(I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right)\right) \leq \frac{2^{n}}{u_{1}u_{2}\cdots u_{n}\prod\limits_{j=1}^n\left(v_{j}-1\right)}. \end{equation} $

(3.4)

对任意的$ x\in F $, 都存在一个数列$ \{d_{n}\}_{n\geq 1} $, 其满足当$ n\geq1 $时, 有$ \lceil u_{n}\rceil\leq d_{n}\left(x\right)\leq \lceil v_{n}u_{n}\rceil $, 使得对所有的$ n $, 有$ x\in I^{*}\left(d_{1}, d_{2}, \cdots, d_{n}\right) $.

令$ r_{0}=\frac{1}{2^{3n_{0}}v_{n_{0}}^{2}u_{n_{0}}^{2}} $, 对任意的$ r<r_{0} $, 存在一个$ n\geq n_{0} $使得

$ \begin{equation} \frac{1}{2^{3(n+1)}v_{n+1}^{2}u_{n+1}^{2}}\leq r<\frac{1}{2^{3n}v_{n}^{2}u_{n}^{2}}. \end{equation} $

(3.5)

由算式(3.2)–(3.5), 开球$ B\left(x, r\right) $可能与最多两个$ n $阶基本区间相交, 或者可能与最多$ 2^{3(n+1)+1}rv_{n+1}^{2}u_{n+1}^{2} $个$ n+1 $阶基本区间相交. 综上, 可以得到

$ \begin{align*} \mu \left(B\left(x, r\right)\right) &\leq \min\{2\prod\limits_{j=1}^n\frac{1}{\lceil v_{j}u_{j}\rceil-\lceil u_{j}\rceil}, 2r2^{3(n+1)}v_{n+1}^{2}u_{n+1}^{2}\prod\limits_{j=1}^{n+1}\frac{1}{\lceil v_{j}u_{j}\rceil-\lceil u_{j}\rceil}\}\\ &=\prod\limits_{j=1}^n\frac{1}{\lceil v_{j}u_{j}\rceil-\lceil u_{j}\rceil} \min\{2, 2r2^{3(n+1)}v_{n+1}^{2}u_{n+1}^{2}\frac{1}{\lceil v_{n+1}u_{n+1}\rceil-\lceil u_{n+1}\rceil}\}\\ &\leq \frac{2^{3n}}{u_{1}u_{2}\cdots u_{n}\prod\limits_{j=1}^n\left(v_{j}-1\right)}2^{1-s} \left(2r2^{3}v_{n+1}^{2}u_{n+1}^{2}\frac{2}{\left({v_{n+1}-1}\right)u_{n+1}}\right)^{s}\\ &\leq 64 r^{s}. \end{align*} $

由命题2.6, 可得$ \dim_{H}F\geq s $. 由$ s<s_0 $的任意性, 可知$ \dim_{H}F\geq s_{0} $.

下面证明$ \dim_{H}F\leq s_{0} $.

结合$ s>s_0 $.由算式(3.1)和$ s_0 $的定义可知, 存在一个无穷数列$ \left\{n_l: l \geq 1\right\} \subset \mathbb{N} $, 满足: 对任意的$ l \geq 1 $, 有

$ \begin{equation} u_1 u_2 \cdots u_{n_l} \leq\left(\frac{u_{n_l+1}}{2}\right)^{\frac{s+s_0}{2}}, 2^{n_l} \prod\limits_{k=1}^{n_l}\left(v_k-1\right) \leq\left(\frac{u_{n_l+1}}{2}\right)^{\frac{s-s_0}{2}} . \end{equation} $

(3.6)

记集合$ F $为

$ \begin{equation} F=\bigcap\limits_{n=1}^{\infty} \bigcup\limits_{\left\lceil u_k\right\rceil \leq d_k \leq\left\lceil v_k u_k\right\rceil-1, 1 \leq k \leq n} J\left(d_1, d_2, \ldots, d_n\right), \end{equation} $

(3.7)

其中

$ \begin{equation*} J\left(d_1, d_2, \ldots, d_n\right)=\bigcup\limits_{d_{n+1}=\left\lceil u_{n+1}\right\rceil}^{\left\lceil v_{n+1} u_{n+1}\right\rceil-1} I^*\left(d_1, d_2, \ldots, d_n, d_{n+1}\right). \end{equation*} $

由命题2.5, 可得

$ \begin{align} \begin{aligned} \left|J\left(d_1, d_2, \cdots, d_n\right)\right| & =\sum\limits_{d_{n+1} \geqslant u_{n+1}}\left|I_{n+1}\left(d_1, \cdots, d_n, d_{n+1}\right)\right| \\ & =\sum\limits_{d_{n+1} \geqslant u_{n+1}} \frac{\prod_{j=1}^n\left(d_j\left(d_j-1\right)+1\right)}{q_{n+1}\left(q_{n+1}+q_n\right)} \\ & \leq \sum\limits_{d_{n+1} \geqslant u_{n+1}} \frac{\prod_{j=1}^n d_j^2}{\left(d_1 d_2 \cdots d_{n+1}\right)^2} \leq \sum\limits_{d_{n+1} \geqslant u_{n+1}} \frac{1}{d_{n+1}^2} \leq \frac{2}{u_{n+1}}. \end{aligned} \end{align} $

(3.8)

结合算式(3.6), (3.7)和(3.8), 集合$ F $的$ s $维Hausdorff测度满足

$ \begin{align*} \begin{aligned} \mathcal{H}^s(F) & \leq \liminf _{l \rightarrow \infty} \sum\limits_{\left\lceil u_k\right\rceil \leq d_k \leq\left\lceil v_k u_k\right\rceil-1, 1 \leq k \leq n_l}\left|J\left(d_1, d_2, \ldots, d_{n_l}\right)\right|^s \\ & \leq \liminf _{l \rightarrow \infty} \sum\limits_{\left\lceil u_k\right\rceil \leq d_k \leq\left[v_k u_k\right\rceil-1, 1 \leq k \leq n_l} \frac{2^s}{\left(u_{n_l+1}\right)^s} \\ & =2^s \liminf _{l \rightarrow \infty} \frac{1}{\left(u_{n_l+1}\right)^s} \prod\limits_{k=1}^{n_l}\left(\left\lceil v_k u_k\right\rceil-\left\lceil u_k\right\rceil\right) \\ & \leq 2^s \liminf _{l \rightarrow \infty} \frac{u_1 u_2 \cdots u_{n_l} 2^{n_l} \prod_{k=1}^{n_l}\left(v_k-1\right)}{\left(u_{n_l+1}\right)^s}\\ & \leq 2^s\left(\frac{u_{n_l+1}}{2}\right)^{\frac{s+s_0}{2}}\left(\frac{u_{n_l+1}}{2}\right)^{\frac{s-s_0}{2}}(u_{n_l+1})^{-s}\\ & \leq 2^{s}\frac{1}{2^{s}}=1 . \end{aligned} \end{align*} $

即$ \operatorname{dim}_H F \leq s_0 $.

引理3.3  对于$ \beta \geq 1 $, $ c\geq 0 $以及$ b\geq 0 $, 令

$ \overline{C}(c, b, \beta)=\left\{x \in(0, 1]: d_{n+1}(x) \geq c^n \left(d_1(x) d_2(x) \cdots d_n(x)\right)^\beta+b, n\in \mathbb{N}\right\} $

和

$ C(c, b, \beta)=\left\{x \in(0, 1]: d_{n+1}(x) \geq c^n \left(d_1(x) d_2(x) \cdots d_n(x)\right)^\beta+b \text { 对无穷多个 } n \in \mathbb{N}\right\}. $

则有

$ \operatorname{dim}_H \bar{C}(c, b, \beta)=\operatorname{dim}_H C(c, b, \beta)=\frac{1}{\beta}. $

证  首先证明$ \operatorname{dim}_H C(c, b, \beta) \leq \frac{1}{\beta} $.

令$ m^*(x)=\min\{n\geq1\colon d_n(x)\neq1\} $, 且$ \Delta_N=\{x\in(0, 1)\colon m^*(x)=N\} $. 可知

$ C(c, b, \beta)=\bigcup\limits_{N=1}^{\infty}\left(C(c, b, \beta)\cap \Delta_N\right). $

由Hausdorff维数的可列稳定性, 只需证明对任意的$ N\geq1 $, $ \dim_H \left(C(c, b, \beta)\cap \Delta_N\right)\leq \frac{1}{\beta} $. 由定义, 可得

$ C(c, b, \beta)\cap \Delta_N \subset \bigcap\limits_{m=1}^{\infty} \bigcup\limits_{n=m}^{\infty} \bigcup\limits_{\left(d_1, d_2, \ldots, d_n\right) \in \Sigma_{n}} J\left(d_1, d_2, \ldots, d_n\right)\cap \Delta_N, $

其中

$ J\left(d_1, d_2, \ldots, d_n\right)=\bigcup\limits_{d_{n+1} \geq\left\lfloor c^n\left(d_1 d_2 \cdots d_n\right)^\beta+b\right\rfloor} I\left(d_1, d_2, \ldots, d_{n+1}\right). $

因此, 通过算式(2.4), 可推导出:

$ \begin{align*} \left|J\left(d_1, d_2, \ldots, d_n\right)\right| &=\sum\limits_{d_{n+1}\geq\left\lfloor c^n\left(d_1 d_2 \cdots d_n\right)^\beta+b\right\rfloor} \frac{\prod\limits_{j=1}^{n}\left(d_{j}\left(x\right)\left(d_{j}\left(x\right)-1\right)+1\right)}{q_{n+1}\left(q_{n+1}+q_{n}\right)} \\ &\leq \sum\limits_{d_{n+1} \geq\left\lfloor c^n\left(d_1 d_2 \cdots d_n\right)^\beta+b\right\rfloor} \frac{d_{1}^{2}\cdots d_{n}^{2}}{q_{n+1}^{2}}\\ &\leq \sum\limits_{d_{n+1} \geq\left\lfloor c^n\left(d_1 d_2 \cdots d_n\right)^\beta+b\right\rfloor} \frac{d_{1}^{2}\cdots d_{n}^{2}}{d_{1}^{2}\cdots d_{n+1}^{2}}\\ &\leq \sum\limits_{d_{n+1} \geq\left\lfloor c^n\left(d_1 d_2 \cdots d_n\right)^\beta+b\right\rfloor}\frac{1}{d_{n+1}^{2}}\\ &\leq \frac{2}{c^n\left(d_1 d_2 \cdots d_n\right)^\beta}. \end{align*} $

取$ s>\frac{1}{\beta} $, 当$ n $充分大时, 令$ \Sigma_{n, N}=\{\left(d_1, d_2, \ldots, d_n\right) \in \Sigma_{n}\colon 1=d_1=\cdots=d_{N-1}<d_N\} $, 则

$ \begin{aligned} &\mathcal{H}^s(C(c, b, \beta)\cap \Delta_N)\\ \leq& \liminf _{m \rightarrow \infty} \sum\limits_{n=m}^{\infty} \sum\limits_{\left(d_1, d_2, \ldots, d_n\right) \in \Sigma_{n, N}} \left|J\left(d_1, d_2, \ldots, d_n\right)\right|^s \\ \leq& 2^s \liminf _{m \rightarrow \infty} \sum\limits_{n=m}^{\infty} \sum\limits_{\left(d_1, d_2, \ldots, d_n\right) \in \Sigma_{n, N}} \frac{1}{c^{n s}\left(d_1 d_2 \cdots d_n\right)^{\beta s}} \\ \leq& 2^s \liminf _{m \rightarrow \infty} \sum\limits_{n=m}^{\infty} \sum\limits_{1 = d_1=\cdots=d_{N-1}<d_N<\cdots<d_n} \frac{1}{\left(d_1 d_2 \cdots d_n\right)^{\beta s}} \frac{1}{c^{n s}} \\ \leq& 2^s \liminf _{m \rightarrow \infty} \sum\limits_{n=m}^{\infty} \left(\frac{1}{(n-N)!(\beta s-1)^{n-N}}+\frac{1}{(n-N-1) !(\beta s-1)^{n-N-1}}\right) \frac{1}{c^{n s}}=0 . \end{aligned} $

因此,

$ \operatorname{dim}_H C(c, b, \beta) \leq \frac{1}{\beta} . $

其中, 在计算得到: 当$ \alpha>1 $时, 有$ \sum_{1 = d_1=\cdots=d_{N-1}<d_N<\cdots<d_n}\frac{1}{\left(d_1 d_2 \cdots d_n\right)^{\beta s}} =\frac{1}{(n-N)!(\beta s-1)^{n-N}}+\frac{1}{(n-N-1) !(\beta s-1)^{n-N-1}} $的过程中, 使用了一个已知的结论: $ \sum_{d+1}^{\infty} \frac{1}{t^\alpha} \leq \int_d^{\infty} \frac{1}{t^\alpha} d t=\frac{1}{\alpha-1} \frac{1}{d^{\alpha-1}} $.

由于$ \bar{C}(c, b, \beta)\subset C(c, b, \beta) $, 故下面只需证明$ \operatorname{dim}_H \bar{C}(c, b, \beta) \geq \frac{1}{\beta} $成立.

选择一个足够大的$ n_0 \in \mathbb{N} $使得对于每一个$ n \geq n_0 $, 都有:

$ \begin{gathered} \frac{e^{(\beta+1)^n}}{n+1} \geq 2, \frac{e^{(\beta+1)^{n+1}}}{n+2} \geq c\left(\frac{e^{(\beta+1)^n}}{n}\right)^{\beta+1}+b, \\ \frac{e^{(\beta+1)^{n+1}}}{n+2} \geq \frac{e^{(\beta+1)^n}}{n}\left(\frac{e^{(\beta+1)^n}}{n}-1\right)+1, \frac{e^{(\beta+1)^n}}{n+1}+2<\frac{e^{(\beta+1)^n}}{n}, \end{gathered} $

并定义

$ F=\left\{x \in(0, 1]: \frac{e^{(\beta+1)^{n+n_0}}}{n+n_0+1} \leq d_n(x)<\frac{n+n_0+1}{n+n_0} \frac{e^{(\beta+1)^{n+n_0}}}{n+n_0+1}, n \geq 1\right\}, $

则

$ F \subset\left\{x \in(0, 1]: d_{n+1}(x) \geq c d_n(x)^{\beta+1}+b, n \geq 1\right\} \subset \bar{C}(c, b, \beta) . $

通过引理3.2, 可得$ \operatorname{dim}_H \bar{C}(c, b, \beta) \geq \operatorname{dim}_H F=\frac{1}{\beta} $.

因为存在$ n_{0}\in\mathbf{N} $, 使得$ \left[e^{\overline\psi{\left(n_{0}\right)}}\right]>e $且当$ n>n_{0} $, 有$ \left[e^{\overline\psi{\left(n+1\right)}}\right] \geq\left[e^{\overline\psi{\left(n\right)}}\right]\left(\left[e^{\overline\psi{\left(n\right)}}\right]-1\right)+1 . $取$ N>n_0 $满足：$ n \geq N $时, 有

$ \frac{e^{\psi(n)}}{n+1} \geq 2, \frac{e^{\psi(n)}}{n+1}+2<\frac{e^{\psi(n)}}{n} ;\ \frac{e^{\psi(n+1)}}{n+2} \geq \frac{e^{\psi(n)}}{n}\left(\frac{e^{\psi(n)}}{n}-1\right)+1. $

对$ n \geq 1 $, 令

$ u_n=\frac{e^{\psi(n+N)}}{n+N+1} ;\ v_n=\frac{n+N+1}{n+N}, $

取

$ A'=\left\{x \in(0, 1]: u_n \leq d_n(x)<v_n u_n, n \geq 1\right\}. $

显然$ A' $符合引理3.2的条件, 且有

$ A'\subset A_N(\psi)=\left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_n(x)}{\psi(n+N)}=1\right\}. $

由引理3.1, 可以得到$ A(\psi) \backslash\{e_{0}\} \neq \emptyset $和$ A_N(\psi) \backslash\{e_{0}\} \neq \emptyset $. 结合引理2.7和引理3.2, 可得

$ \begin{align*} \operatorname{dim}_H A(\psi)=\operatorname{dim}_H A_N(\psi) &\geq \operatorname{dim}_H A'\\ &=\liminf _{n \rightarrow \infty} \frac{\log \left(u_1u_2 \cdots u_n\right)}{\log u_{n+1}}\\ &=\liminf _{n \rightarrow \infty} \frac{\psi(1)+\psi(2)+\cdots+\psi(n)}{\psi(n+1)}. \end{align*} $

令

$ \alpha=\limsup\limits_{n \rightarrow \infty} \frac{\psi(n+1)}{\psi(1)+\psi(2)+\cdots+\psi(n)}. $

因为$ A(\psi) \backslash\{e_{0}\} \neq \emptyset $, 由引理3.1可以知道$ \alpha \geq 1 $. 因此只需要证明$ \operatorname{dim}_H E(\psi) \leq \frac{1}{\alpha} $.

当$ \alpha=1 $, 结论显然成立.

当$ \alpha>1 $, 对任意的$ \epsilon>0 $, 有

$ A(\psi) \subset\left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{1}{\psi(1)+\psi(2)+\cdots+\psi(n)} \sum\limits_{j=1}^n \log d_j(x)=1\right\} \subset \bigcup\limits_{N=1}^{\infty} B_N(\psi), $

其中

$ B_N(\psi)=\left\{x \in(0, 1]: 1-\epsilon<\frac{1}{\psi(1)+\psi(2)+\cdots+\psi(n)} \sum\limits_{j=1}^n \log d_j(x)<1+\epsilon, n \geq N\right\}. $

当$ N \geq 1 $, 选择一个数列$ \left\{n_k: k \geq 1\right\} \subset \mathbb{N} $, 满足$ n_1 \geq N $时有

$ \begin{equation} \psi\left(n_k+1\right) \geq \alpha(1-\epsilon)\left(\psi(1)+\psi(2)+\cdots+\psi\left(n_k\right)\right), k \geq 1 . \end{equation} $

(3.9)

对$ x \in B_N(\psi) $, 可以得到:

$ \begin{aligned} \log d_{n_k+1}(x) & =\sum\limits_{j=1}^{n_k+1} \log d_j(x)-\sum\limits_{j=1}^{n_k} \log d_j(x)>(1-\epsilon) \left(\psi(1)+\cdots+\psi\left(n_k+1\right)\right)-\sum\limits_{j=1}^{n_k} \log d_j(x) \\ & \geq(1-\epsilon)(1+\alpha(1-\epsilon))\left(\psi(1)+\cdots+\psi\left(n_k\right)\right)-\sum\limits_{j=1}^{n_k} \log d_j(x) \quad(\text { by }(3.9)) \\ & \geq \frac{(1-\epsilon)(1+\alpha(1-\epsilon))}{1+\epsilon} \sum\limits_{j=1}^{n_k} \log d_j(x)-\sum\limits_{j=1}^{n_k} \log d_j(x) \\ & =\left(\frac{(1-\epsilon)(1+\alpha(1-\epsilon))}{1+\epsilon}-1\right) \sum\limits_{j=1}^{n_k} \log d_j(x)\\ &=\lambda \sum\limits_{j=1}^{n_k} \log d_j(x)\\ &=\log\left( d_1(x) d_2(x) \cdots d_{n_{k}}(x)\right)^\lambda, \end{aligned} $

这里的$ \lambda=\frac{(1-\epsilon)(1+\alpha(1-\epsilon))}{1+\epsilon}-1 $. 由此可知

$ B_{N}(\psi) \subset\left\{x \in(0, 1]: d_{n+1}(x) \geq \left(d_1(x) d_2(x) \cdots d_{n}(x)\right)^{\lambda} \text { 对无穷多个 } n \in \mathbb{N}\right\}=C(1, 0, \lambda), $

当$ \epsilon \rightarrow 0 $, 由引理3.3, 可以得到

$ \dim_{H} E(\psi) \leq \sup\limits_{N \geq 1} \dim_{H} A_{N}(\psi) \leq \dim_{H} C(1, 0, \lambda) =\frac{1}{\lambda}\rightarrow \frac{1}{\alpha}. $

在证明定理1.4前, 先给出两个引理.

引理4.1  对$ a>1, b \geq 2 $, 令$ A(a, b)=\left\{x \in(0, 1]: d_{n}(x) \geq a^{b^{n}} \text { 对无穷多个 } n \in \mathbb{N}\right\}. $则有

$\operatorname{dim}_{H} A(a, b)=\frac{1}{b-1}.$

证  首先证明$ \operatorname{dim}_{H} A(a, b)\geq \frac{1}{b-1} $. 选择$ N>1 $满足

$ \begin{equation} a^{b^{n}-1} \geq 2, a^{b^{n}-1}+2<\frac{n+1}{n} a^{b^{n}-1} \text { 和 }\left(\frac{n+1}{n}\right)^{2} \leq a, \quad(n \geq N) . \end{equation} $

(4.1)

由引理3.2, 令$ u_{n}=a^{b^{n+1}-1}\ ;\ v_{n}=\frac{n+1}{n}, $因此由算式(4.1), 可以得到

$a^{b^{n+1}-1} \geq\left(\frac{n+1}{n} a^{b^{n}-1}\right)^{2} \geq \frac{n+1}{n} a^{b^{n}-1}\left(\frac{n+1}{n} a^{b^{n}-1}-1\right)+1, \quad(n \geq N) .$

令$ \bar{A}(a, b)=\left\{x \in(0, 1]: a^{b^{n+N}-1} \leq d_{n}(x)<\frac{n+1}{n} a^{b^{n+N}-1}, n \geq 1\right\} . $显然$ \bar{A}(a, b) \subset A(a, b) $.由引理3.2, 可得

$\operatorname{dim}_{H} A(a, b) \geq \operatorname{dim}_{H} \bar{A}(a, b)=\frac{1}{b-1}.$

接下来证明$ \operatorname{dim}_{H} A(a, b)\leq \frac{1}{b-1}. $当$ b=2 $时, $ \operatorname{dim}_{H} A(a, b)=1 $. 当$ b>2 $时, 取$ \lambda $满足$ 2<\lambda<b $.可知:

$ A\left(a, b\right) \subset \{x\in(0, 1]:d_{1}d_{2} \cdots d_{n+1}(x) >\left(d_1(x) d_2(x) \cdots d_n(x)\right)^{\lambda} \text {对无穷多的} n \in \mathbb{N}\}. $

实际上, 对任意的$ x \in A(a, b) $, 如果存在$ N \in \mathbb{N} $满足: 当$ n \geq N $时, 有

$ d_{1}d_{2} \cdots d_{n+1}(x)\leq\left(d_1(x) d_2(x) \cdots d_n(x)\right)^{\lambda}, $

则

$ \begin{align*} a^{b^{n}} \left(d_1(x) d_2(x) \cdots d_{n-1}(x)\right)&\leq d_{n}(x) \left(d_1(x) d_2(x) \cdots d_{n-1}(x)\right)\\ &=\left(d_1(x) d_2(x) \cdots d_n(x)\right)\\ &\leq \left(d_1(x) d_2(x) \cdots d_N(x)\right)^{\lambda^{n-N}}. \end{align*} $

对于无穷多个$ n \geq N $成立. 但是因为$ b>\lambda $, 可得: 当$ n $足够大时, 有$ a^{b^{n}}>\left(d_1(x) d_2(x) \cdots d_N(x)\right)^{\lambda^{n}} $, 与上式矛盾. 因此可得:

$ \begin{align*} A(a, b) &\subset \{x\in(0, 1]: d_1(x) d_2(x) \cdots d_{n+1}(x) >\left(d_1(x) d_2(x) \cdots d_n(x)\right)^{\lambda} \text {对无穷多的} n \in \mathbb{N}\} \\ &\subset\{x\in(0, 1]: d_{n+1}(x)>\left(d_1(x) d_2(x) \cdots d_n(x)\right)^{\lambda-1} \text {对无穷多的} n \in \mathbb{N}\} \\ &=C\left(1, 0, \lambda-1\right). \end{align*} $

结合引理3.3可得

$ \operatorname{dim}_{H} A(a, b) \leq \lim\limits_{\lambda \rightarrow b} \frac{1}{\lambda-1}=\frac{1}{b-1}. $

引理4.2  对$ b \geq 2 $, 令$ \hat{A}^{*}(b)=\left\{x \in(0, 1]: \lim _{n \rightarrow \infty} \frac{\log d_{n}(x)}{b^{n}}=\infty\right\}, $则有

$ \operatorname{dim}_{H} \hat{A}^{*}(b)=\frac{1}{b-1}. $

证  对于$ a>b $, 有

$ \left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_{n}(x)}{a^{n}}=1\right\} \subset \hat{A}^{*}(b) \subset\left\{x \in(0, 1]: d_{n}(x) \geq e^{b^{n}} \text { 对无穷多个 } n \in \mathbb{N}\right\}. $

由定理1.1和引理4.1, 可以得出:

$ \frac{1}{b-1}=\lim\limits_{a \rightarrow b} \frac{1}{a-1} \leq \operatorname{dim}_{H} \hat{A}^{*}(b) \leq \frac{1}{b-1}. $

下面证明定理1.4.

证  首先回顾集合

$ A^{*}(\psi)=\left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_{n}(x)}{\psi(n)} =\infty\right\}, \quad \log B=\limsup\limits_{n \rightarrow \infty} \frac{\log \psi(n)}{n}. $

证明将分三种情况来进行.

(1) $ 1 \leq B \leq 2 $.

在这种情况下, 对于任意的$ \varepsilon>0 $, 存在$ N \in \mathbb{N} $满足: 当$ n \geq N $时, 有$ \psi(n) \leq(2+\varepsilon)^{n} $. 可以得到:

$ A^{*}(\psi) \supset\left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_{n}(x)}{(2+\varepsilon)^{n}}=\infty\right\}, $

又由引理4.2可知: $ \operatorname{dim}_{H} A^{*}(\psi) \geq \frac{1}{1+\varepsilon} $.

(2) $ 2<B<\infty $.

在这种情况下, 对于任意的$ \varepsilon>0 $，$ \psi(n) \geq(B-\varepsilon)^{n} $对任意多的$ n $都成立, 且存在$ N \in \mathbb{N} $满足当$ n \geq N $时, 有$ \psi(n) \leq(B+\varepsilon)^{n} $. 因此

$ \left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_{n}(x)}{(B+\varepsilon)^{n}} =\infty\right\} \subset A^{*}(\psi) \subset\left\{x \in(0, 1]: d_{n}(x) \geq e^{(B-\varepsilon)^{n}} \text {对无穷多个} n \in \mathbb{N}\right\}. $

由引理4.1和引理4.2, 可得

$ \frac{1}{B-1+\varepsilon} \leq \operatorname{dim}_{H} A^{*}(\psi) \leq \frac{1}{B-1-\varepsilon}. $

(3) $ B=\infty $.

在这种情况下, 对任意的$ M>2 $, $ A^{*}(\psi) \subset\left\{x \in(0, 1]: d_{n}(x) \geq e^{M^{n}} \text { 对无穷多个 } n \in \mathbb{N}\right\}. $由引理4.1, 可得

$\operatorname{dim}_{H} A^{*}(\psi) \leq \frac{1}{M-1}.$

证  对集合

$A_{*}(\psi)=\left\{x\in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_{n}(x)}{\psi(n)}=0\right\}, C=\liminf\limits_{n \rightarrow \infty} \frac{\psi(n)}{2^{n}}.$

考虑两种情况:

(1) $ 0\leq C<\infty $.

在这种情况下, 这里存在一个数列$ \left\{n_{k}\right\}_{k \geq 1} $满足$ \lim _{k \rightarrow \infty} \frac{\psi\left(n_{k}\right)}{2^{n_{k}}}=C $. 因此

$ A_{*}(\psi) \subset\left\{x \in(0, 1]: \lim\limits_{k \rightarrow \infty} \frac{\log d_{n_{k}}(x)}{2^{n_{k}}}=0\right\}. $

由命题2.4, 可以得出

$ \begin{equation} d_{n}\left(x\right)\geq 2^{2^{n-2}}, n\in \mathbb{N}. \end{equation} $

(5.1)

由算式(5.1), 对所有的$ n\geq 1 $, 有$ \frac{\log d_{n}(x)}{2^{n}} \geq \frac{\log 2}{4} $. 因此$ A_{*}(\psi)=\emptyset $.

(2) $ C=\infty $.

在这种情况下, 有$ \lim _{n\rightarrow \infty}\frac{\psi(n)}{2^{n}}=\infty $, 因此

$ \left\{x \in(0, 1]: \lim\limits_{n \rightarrow \infty} \frac{\log d_{n}(x)}{2^{n}}=1\right\} \subset A_{*}(\psi). $

因此, 由定理1.1可以得到$ \operatorname{dim}_{H} A_{*}(\psi) \geq 1 $.