Lüroth展式可由如下变换$ T:\left[ {0, 1} \right) \to \left[ {0, 1} \right) $导出,

$ T(x) = \left\{ {\begin{array}{*{20}{c}} {n\left( {n + 1} \right)x - n, }&{x \in \left[ \frac{1}{n + 1}, \frac{1}{n} \right)}, \\ {0, }&{x = 0.} \end{array}} \right. $

对任意的$ x \in (0, 1) $, 定义$ d_1(x)\geq 2 $为使得下式成立的唯一的正整数

$ \frac{1}{d_1(x)} \le x < \frac{1}{d_1(x) - 1}. $

对$ n\ge 2 $, 如果$ T^{n-1}(x)\neq 0 $, 则定义$ d_n(x) = d_1(T^{n-1}(x)) $. 从而对任意的无理数$ x\in [0, 1) $, 可以得到由$ T $所诱导的如下无穷级数

$ x = \frac{1}{d_1(x)} + \frac{1}{d_1(x)(d_1(x) - 1)d_2(x)} + \cdots + \frac{1}{{{d_1}\left( x \right) \cdots {d_{n - 1}}\left( x \right)\left( {{d_{n - 1}}\left( x \right) - 1} \right){d_n}\left( x \right)}} + \cdots . $

我们称这一级数为$ x $的Lüroth展式, 它由Lüroth^[1]1883年首次引入. 其中$ d_n(x) $称为$ x $的Lüroth展式的第$ n $个字符. 对于有理数$ x\in [0, 1) $, 其Lüroth展式只含有限项. Lüroth展式与度量数论、动力系统和分形几何的研究密切相关, 已有很多的研究成果. 例如变换$ T $相对勒贝格测度$ \lambda $是不变的和遍历的, 字符列$ \{d_n(x): n\geq 1\} $相对于$ \lambda $是独立同分布的^[2]. 关于单个Lüroth展式字符的增长速度及相关维数问题的研究, 可参见文献[3]. 在文献[4]和[5]中, 研究了Lüroth展式两个相邻字符乘积的增长速度, 集合$ \{x\in(0, 1): d_n(x)d_{n+1}(x) \geq \varphi(n){无穷多次成立}\} $的勒贝格测度与Hausdorff维数得到了解决.

在本文中, 我们将研究Lüroth展式相邻字符乘积的部分和的分布. 令

$ {S_n}\left( x \right) = \sum\limits_{i = 1}^n {{d_i}(x){d_{i + 1}}(x)} . $

首先考虑对适当的增长速度$ \varphi(n) $, $ S_n(x)/\varphi(n) $的收敛问题. 具体地, 我们有下面的结果.

定理1.1  比值$ \frac{S_n(x)}{n\log^2n} $依勒贝格测度$ \lambda $收敛到$ \frac{1}{2} $, 即对任意的$ \varepsilon >0 $,

$ \lim\limits_{n \to \infty } \lambda \left\{ {x \in \left( {0, 1} \right):\left| \frac{S_n(x)}{n\log^2 n} - \frac{1}{2} \right| > \varepsilon } \right\} = 0. $

需要指出的是, 由文献[5]知, 对勒贝格几乎处处的$ x\in(0, 1) $, $ d_n(x)d_{n+1}(x)\ge n\log^2 n $对于无穷多的$ n $成立. 从而$ S_n(x)/n\log^2n $并不是几乎处处收敛到$ 1/2 $. 另一个很自然的问题就是考虑$ S_n(x) $的相关例外集

$ G\left( \psi \right): = \left\{ {x \in \left( {0, 1} \right):\mathop {\lim }\limits_{n \to \infty } \frac{{{S_n}\left( x \right)}}{{\psi \left( n \right)}} = 1} \right\} $

的Hausdorff维数, 其中函数$ \psi: \mathbb{N} \to (1, \infty) $单调递增且$ \lim\limits_{n\to \infty}\psi(n) = \infty $.

定理1.2  若$ \psi :\mathbb{N} \to (1, \infty) $单调递增且满足

$ \mathop {\lim }\limits_{n \to \infty } \frac{{\psi (n)}}{n} = \infty , \mathop {\lim }\limits_{n \to \infty } \frac{{\psi (n + 1)}}{{\psi (n)}} = 1, \mathop {\lim \sup }\limits_{n \to \infty } \frac{{\log \log \psi (n)}}{{\log n}} < \frac{1}{2}, $

则有$ {\dim _H}G(\psi ) = 1. $特别地, 对任意的$ \alpha > 0 $, $ {\dim _H}\left\{ {x \in \left( {0, 1} \right):\mathop {\lim }\limits_{n \to \infty } \frac{{{S_n}\left( x \right)}}{{n{{\log }^2}n}} = \alpha } \right\} = 1. $

对任意的$ n\geq 1 $和正整数$ d_1, d_2, \ldots, d_n $(其中$ d_i \geq 2, 1\leq i\leq n $), 集合

$ I(d_1, d_2, \ldots, d_n) = \{x\in [0, 1): d_i(x) = d_i, 1\leq i\leq n\} $

称为一个$ n $阶柱集. $ I(d_1, d_2, \ldots, d_n) $是一个区间, 长度为$ |I(d_1, d_2, \ldots, d_n)| = \prod \limits_{i = 1}^n\frac{1}{d_i(d_i-1)} $.

下文将用到Borel-Cantelli引理的收敛部分. 设$ \left\{ {{A_n}, n \ge 1} \right\} $是一列事件, $ P $是一个概率测度. 若$ \sum \limits_{n \geq 1} P\left( {{A_n}} \right) < \infty , $则$ P\{A_n \text{无穷多次发生}\} = 0 $, 其中

$ \{A_n \text{无穷多次发生}\} = \bigcap \limits_{k = 1}^\infty \bigcup \limits_{n = k}^\infty {A_n}. $

命题2.1^[6]  设$ E \subset \mathbb{R}^n $, 若映射$ f:E \to \mathbb{R}^m $满足$ \alpha $阶-Hölder条件, 即

$ \left| {f\left( x \right) - f\left( y \right)} \right| \le c|x - y{|^\alpha } \ \left( {x, y \in E} \right), $

则有$ \dim_H f(E) \le \frac{1}{\alpha }{\rm{di}}{{\rm{m}}_H}E. $

这一节我们先介绍三个引理, 再给出定理1.1的证明.

引理3.1  设$ \phi: \mathbb{N} \to (1, \infty ) $是一个正值函数且$ \lim\limits_{n \to \infty } \phi (n) = \infty $. 对任意的$ n \ge 1 $, 令

$ A_n = \{ x \in (0, 1):{d_1}(x){d_2}(x) \ge \phi (n)\} , $

其中$ {d_1}(x), {d_2}(x) $分别是$ x $的Lüroth展式的第一和第二个字符. 则我们有

$ \lambda ({A_n}) = \frac{{{\rm{log}}\phi \left( n \right)}+O(1)}{{\phi \left( n \right)}}. $

证  因为对任意的$ x\in (0, 1) $和$ i\geq 1 $, 都有$ {d_i}(x) \ge 2 $, 所以不妨设$ \phi \left( n \right)\geq 4 $. 因为

$ \begin{aligned} {A_n} & = \left\{ {x \in (0, 1):{d_1}(x){d_2}\left( x \right) \ge \phi \left( n \right)} \right\} = B_1\cup B_2, \end{aligned} $

其中

$ B_1 = \{ {x \in (0, 1):{d_1}\left( x \right) \ge \frac{\phi(n)}{2}} \}, B_2 = \{ x \in (0, 1):{d_1}\left( x \right) < \frac{\phi(n)}{2}, {d_1}(x){d_2}\left( x \right) \ge \phi \left( n \right)\}, $

所以$ \lambda \left( {{A_n}} \right) = \lambda \left( {{B_1}} \right) + \lambda \left( {{B_2}} \right) $. 此外,

$ \begin{equation} \lambda \left( {{B_1}} \right) = \mathop \sum \limits_{k = \left\lceil {\phi(n)/2} \right\rceil }^{ + \infty } \lambda \left\{ {x \in (0, 1):{d_1}\left( x \right) = k} \right\} = \mathop \sum \limits_{k = \left\lceil {\phi(n)/2} \right\rceil }^{ + \infty } \frac{1}{{k\left( {k - 1} \right)}} = \frac{{O(1)}}{{\phi (n)}}. \end{equation} $

(3.1)

$ \begin{aligned} \lambda \left( {{B_2}} \right) & = \mathop \sum \limits_{k = 2}^{\left\lceil {\phi(n)/2} \right\rceil - 1} \lambda \left\{ {x \in (0, 1):{d_1}\left( x \right) = k} \right\}\lambda \left\{ {x \in (0, 1):{d_2}\left( x \right) \ge \frac{{\phi \left( n \right)}}{k}} \right\}\\ & = \mathop \sum \limits_{k = 2}^{\left\lceil {\phi(n)/2} \right\rceil - 1} \frac{1}{{k\left( {k - 1} \right)}}\frac{1}{{\left\lceil {\frac{{\phi \left( n \right)}}{k}} \right\rceil - 1}}, \end{aligned} $

这里记号$ \left\lceil \xi \right\rceil $表示不小于$ \xi $的最小整数. 接下来继续估计$ \lambda \left( {{B_2}} \right) $. 一方面,

$ \lambda \left( {{B_2}} \right) \ge \mathop \sum \limits_{k = 2}^{\left\lceil {\phi \left( n \right)}/2 \right\rceil - 1} \frac{1}{{k\left( {k - 1} \right)}}\frac{1}{{\frac{{\phi \left( n \right)}}{k}}} \ge \frac{{\log \phi \left( n \right) + O(1)}}{{\phi \left( n \right)}}. $

另一方面,

$ \lambda \left( {{B_2}} \right) \le \frac{1}{{\phi \left( n \right) - 1}}\mathop \sum \limits_{k = 2}^{\lceil \phi(n)/2\rceil -1} \left( {\frac{1}{k-1} + \frac{1}{{\phi \left( n \right) - k}}} \right) \le \frac{{\log \phi \left( n \right) + O(1)}}{{\phi \left( n \right)}}. $

所以

$ \begin{equation} \lambda \left( {{B_2}} \right) = \frac{{\log \phi \left( n \right) + O(1)}}{{\phi \left( n \right)}}. \end{equation} $

(3.2)

由关系式(3.1), (3.2)知该引理成立.

对任意无理数$ x\in (0, 1) $及$ i\geq 1 $, 令$ {b_i}\left( x \right) = {d_i}\left( x \right){d_{i + 1}}\left( x \right) $. 对任给$ \varepsilon > 0 $和正整数$ N \ge 1 $, 令$ \varphi \left( N \right) = \left\lfloor {{\rm{ \mathsf{ ε} }}N{\rm{lo}}{{\rm{g}}^2}N} \right\rfloor+ 1 $. 对任意$ 1 \le i \le N $, 令

$ b_i^*\left( x \right) = \left\{ \begin{aligned} &{b_i}\left( x \right), \;{b_i}\left( x \right) \le \varphi \left( N \right), \\ &0, \quad \quad \texttt{其它}, \end{aligned} \right. $

其中记号$ \left\lfloor \eta \right\rfloor $表示不大于$ \eta $的最大整数. 令$ S_N^*\left( x \right) = \mathop \sum \limits_{i = 1}^N b_i^*\left( x \right) $. 下面分别给出对$ S_N^*\left( x \right) $的期望和方差的估计.

引理3.2  $ \lim \limits_{N \to \infty } \frac{\mathbb{E}\left( {S_N^*\left( x \right)} \right)}{{N{\rm{lo}}{{\rm{g}}^2}N}} = \frac{1}{2}, $其中$ \mathbb{E} $表示数学期望.

证  先计算$ S_N^*\left( x \right) $的数学期望. 因为$ \mathbb{E}\left( {S_N^*\left( x \right)} \right) = N\mathbb{E}\left( {b_1^*\left( x \right)} \right) $, 而

$ \begin{aligned} & \mathbb{E}\left( {b_1^*\left( x \right)} \right) = \mathop \sum \limits_{4\leq k\leq \varphi \left( N \right)} k \cdot \lambda \left\{ {x:{b_1}\left( x \right) = k} \right\}\\ = & \sum\limits_{4\leq k\leq \varphi \left( N \right)} {k \cdot \left( {\lambda \left\{ {x:{b_1}\left( x \right) \ge k} \right\} - \lambda \left\{ {x:{b_1}\left( x \right) \ge k + 1} \right\}} \right)} \\ = & 3\lambda \left\{ {x:{b_1}\left( x \right) \ge 4} \right\} + \mathop \sum \limits_{4 \le k \le \varphi \left( N \right)} \lambda \left\{ {x:{b_1}\left( x \right) \ge k} \right\} - \varphi \left( N \right) \cdot \lambda \left\{ {x:{b_1}\left( x \right) \ge \varphi \left( N \right) + 1} \right\}. \end{aligned} $

由引理3.1  有$ \mathbb{E}\left( {{b_1}^*\left( x \right)} \right) = \frac{1}{2}{\rm{lo}}{{\rm{g}}^2}\varphi \left( N \right)\left( {1 + o\left( 1 \right)} \right) $. 又因为$ \mathbb{E}\left( {S_N^*\left( x \right)} \right) = N\mathbb{E}\left( {b_1^*\left( x \right)} \right) $和$ \varphi \left( N \right) = \left\lfloor {\varepsilon N{{\log }^2 N}} \right\rfloor $, 所以$ \mathop {\lim }\limits_{N \to \infty } \frac{{\mathbb{E}({S_N}^*\left( x \right))}}{N{\log }^2 N} = \frac{1}{2} $. 该引理得证.

引理3.3  $ {\rm{Var}}\left( {S_N^*\left( x \right)} \right) \ll N\varphi \left( N \right){\rm{log}}\varphi \left( N \right) $, 其中$ \rm{Var}( \cdot ) $表示方差, 记号$ a \ll b $表示存在常数$ c $使得$ a < cb $.

证  我们先估计$ {S_N^*}^2\left( x \right) $的期望.

$ \begin{equation*} \begin{aligned} &\mathbb{E}\left( {S_N^{{*^2}}\left( x \right)} \right) = \mathop \sum \limits_{1 \le i, j \le N} \mathbb{E}\left( {b_i^*\left( x \right)b_j^*\left( x \right)} \right)\\ = & \mathop \sum \limits_{1 \le i = j \le N} \mathbb{E}\left( {b_i^*\left( x \right)b_j^*\left( x \right)} \right) + \mathop \sum \limits_{{1 \le i, j \le N} \atop {\left| {i - j} \right| = 1}} \mathbb{E}\left( {b_i^*\left( x \right)b_j^*\left( x \right)} \right) + \mathop \sum \limits_{{1 \le i, j \le N} \atop {\left| {i - j} \right| \ge 2}} \mathbb{E}\left( {b_i^*\left( x \right)b_j^*\left( x \right)} \right). \end{aligned} \end{equation*} $

当$ i = j $时, 有

$ \begin{equation*} \begin{aligned} & \mathbb{E}\left( {b_i^*\left( x \right)b_j^*\left( x \right)} \right) = \mathbb{E}\left( {{b{{_i^*}}^2}\left( x \right)} \right) = \mathbb{E}\left( {b{{_1^*}^2}\left( x \right)} \right)\\ = & \mathop \sum \limits_{4 \le k \le \varphi \left( N \right)} {k^2} \cdot \lambda \left\{ {x:{b_i}\left( x \right) = k} \right\} \ll \varphi \left( N \right)\log \varphi \left( N \right). \end{aligned} \end{equation*} $

当$ \left| {i - j} \right| = 1 $时, 不妨设$ j = i + 1 $, 则有

$ \begin{equation*} \begin{aligned} & \mathbb{E}\left( {b_i^*\left( x \right)b_j^*\left( x \right)} \right) = \mathbb{E}\left( {b_i^*\left( x \right)b_{i + 1}^*\left( x \right)} \right) = \mathop \sum \limits_{4 \le s, t \le \varphi \left( N \right)} st\lambda \left\{ {x:{b_i}\left( x \right) = s, \;{b_{i + 1}}\left( x \right) = t} \right\}\\ = & \mathop \sum \limits_{4 \le s, t \le \varphi \left( N \right)} st\lambda \left\{ {x:{d_1}\left( x \right){d_2}\left( x \right) = s, \;{d_2}\left( x \right){d_3}\left( x \right) = t} \right\}\\ = & \mathop \sum \limits_{2 \le k \le \frac{\varphi (N)}{2}} \mathop \sum \limits_{2 \le m \le \frac{{\varphi \left( N \right)}}{k}} \mathop \sum \limits_{2 \le n \le \frac{{\varphi \left( N \right)}}{k}} {k^2}mn\lambda \left\{ {x:{d_2}\left( x \right) = k, {d_1}\left( x \right) = m, {d_3}\left( x \right) = n} \right\}\\ \ll & \mathop \sum \limits_{2 \le k \le \frac{\varphi (N)}{2}} {\left( {1 + \log \varphi \left( N \right) - \log k} \right)^2} \ll \varphi \left( N \right). \end{aligned} \end{equation*} $

当$ \left| {i - j} \right| \ge 2 $时, 有

$ \begin{equation*} \begin{aligned} & \mathbb{E}\left( {b_i^*\left( x \right)b_j^*\left( x \right)} \right) = \mathop \sum \limits_{4 \le m, n \le \varphi \left( N \right)} mn\lambda \left\{ {x:{b_i}\left( x \right) = m, {b_j}\left( x \right) = n} \right\}\\ = & \mathop \sum \limits_{4 \le m, n \le \varphi \left( N \right)} mn\lambda \left\{ {x:{b_1}\left( x \right) = m} \right\} \cdot \lambda \left\{ {x:{b_1}\left( x \right) = n} \right\} = { \mathbb{E}^2}\left( {b_1^*\left( x \right)} \right). \end{aligned} \end{equation*} $

由上面三种情况的估计, $ \mathbb{E}\left( {S_N^{{*^2}}\left( x \right)} \right) \ll N\varphi \left( N \right)\log \varphi \left( N \right) + N\varphi \left( N \right) + { \mathbb{E}^2}\left( {S_N^*\left( x \right)} \right). $所以$ \rm{Var}\left( {S_N^*\left( x \right)} \right) \ll N\varphi \left( N \right){\rm{log}}\varphi \left( N \right) $. 该引理得证.

定理1.1的证明  由引理3.2 和引理3.3 可知, 当$ N $充分大时, 有

$ \begin{equation*} \begin{aligned} & \lambda \left\{ {x \in \left( {0, 1} \right):\left| {S_N^*\left( x \right) - \frac{1}{2}N\log ^2N}\right| > \varepsilon N\log ^2N} \right\}\\ \le & \lambda \left\{ {x \in \left( {0, 1} \right):\left| {S_N^*\left( x \right) - \mathbb{E}\left( {S_N^*\left( x \right)} \right)} \right| > \frac{\varepsilon }{2}N\log ^2N} \right\}\\ \le & \frac{{{\rm{Var}}\left( {S_N^*\left( x \right)} \right)}}{{{{\left( {\frac{\varepsilon }{2}N\log ^2N} \right)}^2}}} \ll \frac{N\varphi \left( N \right)\log\varphi \left( N \right)}{\left( \frac{\varepsilon }{2} \right)^2N^2\log ^4N} \ll \frac{1}{\varepsilon \log N}. \end{aligned} \end{equation*} $

由$ S_N^*(x) $定义知, 对任意的$ x \in \left( {0, 1} \right) $, 如果对所有的$ 1 \le i \le N $都有$ b_i(x) \le \varphi(N) $, 则$ {S_N}\left( x \right) = S_N^*\left( x \right) $. 因此

$ \begin{aligned} &\left\{ {x \in \left( {0, 1} \right):\left| {\frac{{{S_N}\left( x \right)}}{{N{\rm{lo}}{{\rm{g}}^{\rm{2}}}N}} - \frac{1}{{\rm{2}}}} \right| > \varepsilon } \right\}\\ \subset & \left\{ {x \in \left( {0, 1} \right):\left| {S_N^*\left( x \right) - \frac{1}{{\rm{2}}}N\log ^2N} \right| > \frac{\varepsilon }{2}N\log ^2N} \right\}\cup\bigcup\limits_{i = 1}^{N}\left\{ {x \in \left( {0, 1} \right):{b_i}\left( x \right) \ge \varphi(N)+1 }\right\}. \end{aligned} $

从而

$ \begin{equation*} \begin{aligned} & \lambda \left\{ {x \in \left( {0, 1} \right):\left| {\frac{{{S_N}\left( x \right)}}{{N\log ^2N}} - \frac{1}{{\rm{2}}}} \right| > \varepsilon } \right\}\\ \ll & \frac{1}{\varepsilon \log N} + \sum \limits_{1 \le i \le N} \lambda \left\{ x \in \left( {0, 1} \right):b_i(x) \ge \varphi(N)+1 \right\}. \end{aligned} \end{equation*} $

由引理3.1 可知,

$ \sum \limits_{1 \le i \le N} \lambda\left\{ x \in (0, 1):b_i(x) \ge \varphi(N)+1 \right\} \ll \frac{1}{\varepsilon \log N}. $

由此$ \lambda \{ x \in(0, 1):\left| \frac{S_N(x)}{N\log^2N} - \frac{1}{2} \right| > \varepsilon\} \ll \frac{1}{\varepsilon\log N}\to 0\;\left( {N \to \infty } \right). $定理得证.

本节将先介绍四个引理, 之后再给出定理1.2的证明, 其中集合

$ G\left( \psi \right) = \left\{ {x \in \left( {0, 1} \right):\mathop {\lim }\limits_{n \to \infty } \frac{{{S_n}\left( x \right)}}{{\psi \left( n \right)}} = 1} \right\}. $

引理4.1  对任意正整数$ M \ge 3 $, 令

$ F\left( M \right): = \left\{ {x \in \left( {0, 1} \right):{d_i}\left( x \right) \le M, \;i \ge 1} \right\}. $

则

(1) $ {\dim _H}F\left( M \right) = {s_M} $, 其中$ s_M $是方程$ \mathop \sum \limits_{j = 2}^M \frac{1}{{{j^s}{{\left( {j - 1} \right)}^s}}} = 1 $的唯一解.

(2) $ \mathop {\lim }\limits_{M \to \infty } {s_M} = 1. $

证  (1) 对任意$ 2 \le j \le M $, 令

$ {f_{j - 1}}\left( x \right) = \frac{1}{j} + \frac{1}{{j\left( {j - 1} \right)}}x, \ x \in \left( {0, 1} \right). $

则$ F\left( M \right) $是迭代函数系$ {\left\{ {{f_{j - 1}}} \right\}_{2 \le j \le M}} $的吸引子. 从而$ F\left( M \right) $是压缩比为$ r_j = \frac{1}{{j\left( {j - 1} \right)}} $ $ \left( {2 \le j \le M} \right) $的自相似集, 所以$ {\dim _H}F\left( M \right) $是方程$ \mathop \sum \limits_{j = 2}^M \frac{1}{{{r_j}^s}} = 1 $的唯一解.

(2) 由$ {s_M} $的定义可知, $ {s_M} $关于$ M $是单调递增的. 一方面, 对任意的$ M \ge 3 $, 有$ {s_M} < 1 $. 另一方面, 对任意的$ s < 1 $, 有$ \mathop \sum \limits_{j = 2}^{ + \infty } \frac{1}{{{j^s}{{\left( {j - 1} \right)}^s}}} > \mathop \sum \limits_{j = 2}^{ + \infty } \frac{1}{{j\left( {j - 1} \right)}} = 1, $从而存在整数$ {M_0} = {M_0}\left( s \right) $, 使得$ \mathop \sum \limits_{j = 2}^{{M_0}} \frac{1}{{{j^s}{{\left( {j - 1} \right)}^s}}} \ge 1 $, 因此$ {s_{{M_0}}} \ge s $. 从而对任意$ M \ge {M_0} $, 有$ {s_M} \ge {s_{{M_0}}} \ge s $. 所以$ \mathop {\lim }\limits_{M \to \infty } {s_M} = 1 $, 即此引理得证.

取定常数$ 0 < \tau < \frac{1}{2} $, 使条件$ \mathop {\lim \sup }\limits_{n \to \infty } \frac{{\log \log \psi \left( n \right)}}{{\log n}} < \frac{1}{2} - \tau $成立. 另取$ 0 < \delta < 1 $使得不等式$ \left( {1 + \frac{1}{{1 - \delta }}} \right)\left( {\frac{1}{2} - \tau } \right) < 1 $成立. 对任意$ k \ge 1 $, 令$ {\varepsilon _k} = {k^{ - \delta }} $. 下面归纳定义自然数的子序列$ \{n_k\} $. 设$ {n_1} \ge 3 $是使得$ \log \psi \left( n \right) < {n^{\frac{1}{2} - \tau }} $对所有$ n \ge {n_1} $都成立的最小正整数. 对任意$ k \ge 2 $, 令$ {n_k} $是满足条件$ {n_k} \ge {n_{k - 1}} + 4 $和$ \psi \left( {{n_k}} \right) \ge \left( {1 + {\varepsilon _{k - 1}}} \right)\psi \left( {{n_{k - 1}}} \right) $的最小正整数.

定义

$ G\left( {\psi , M} \right): = \left\{ \begin{array}{l} x \in \left( {0, 1} \right):{d_{{n_1}}}\left( x \right) = \left\lfloor {\frac{1}{4}\left( {1 + {\varepsilon _1}} \right)\psi \left( {{n_1}} \right)} \right\rfloor + 1, \;\\ \quad \;\;\, {d_{{n_{k + 1}}}}\left( x \right) = \left\lfloor {\frac{1}{4}\left(( {1 + {\varepsilon _{k + 1}}} )\psi \left( {{n_{k + 1}}} \right) - \left( {1 + {\varepsilon _k}} \right)\psi \left( {{n_k}} \right)\right)} \right\rfloor + 1\left( {k \ge 1} \right), \\ \quad \;\, \;{d_{{n_k} - 1}}\left( x \right) = {d_{{n_k} + 1}}\left( x \right) = 2\left( {k \ge 1} \right), \;\\ \quad \, \, \;2 \le {d_i}\left( x \right) \le M,i\not \in { \cup _{k \ge 1}}\{ {n_k},{n_k} - 1,{n_k} + 1\}. \end{array} \right\} $

下证集合$ G\left( {\psi , M} \right) $是集合$ G\left( \psi \right) $的子集.

引理4.2  $ G\left( {\psi , M} \right) \subset G\left( \psi \right). $

证  任取$ x \in G\left( {\psi , M} \right) $. 对充分大的$ n $, 存在$ k\geq 1 $使得$ {n_k} \le n < {n_{k + 1}} $. 一方面,

$ {S_n}\left( x \right) \ge {S_{{n_{_k}}}}\left( x \right) \ge \left( {1 + {\varepsilon _k}} \right)\psi \left( {{n_k}} \right). $

另一方面,

$ \begin{aligned} {S_n}\left( x \right) \le & {S_{{n_{_{k + 1}}}}}\left( x \right) \le \left( {{n_{k + 1}} - 2k} \right){M^2} + \sum\limits_{i = 1}^{k + 1} {\left( {{d_{{n_i} - 1}}(x){d_{{n_i}}}(x) + {d_{{n_i}}}(x){d_{{n_i} + 1}}(x)} \right)} \\ \le &\left( {{n_{k + 1}} - 2k} \right){M^2} + \left( {1 + {\varepsilon _{k + 1}}} \right)\psi \left( {{n_{k + 1}}} \right) + 4\left( {k + 1} \right). \end{aligned} $

所以

$ \begin{equation} \left( {1 + {\varepsilon _k}} \right)\psi \left( {{n_k}} \right) \le {S_n}\left( x \right) \le \left( {{n_{k + 1}} - 2k} \right){M^2} + \left( {1 + {\varepsilon _{k + 1}}} \right)\psi \left( {{n_{k + 1}}} \right) + 4\left( {k + 1} \right). \end{equation} $

(4.1)

由$ n_k(k\ge2) $定义, 不等式$ \psi \left( {{n_k} - 1} \right) < \left( {1 + {\varepsilon _{k - 1}}} \right)\psi \left( {{n_{k - 1}}} \right) $, $ \psi \left( {{n_k} - 4} \right) < \left( {1 + {\varepsilon _{k - 1}}} \right)\psi \left( {{n_{k - 1}}} \right) $中总有一个是成立的. 又因为$ \mathop {\lim }\limits_{n \to \infty } \frac{{\psi \left( {n + 1} \right)}}{{\psi \left( n \right)}} = 1 $, 所以

$ \begin{equation} \mathop {\lim }\limits_{n \to \infty } \frac{{\psi \left( {{n_k}} \right)}}{{\psi \left( {{n_{k - 1}}} \right)}} = 1. \end{equation} $

(4.2)

再结合(4.1), (4.2), $ \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\psi \left( n \right)}} = 0 $以及

$ \frac{{{S_n}\left( x \right)}}{{\psi \left( {{n_{k + 1}}} \right)}} \le \frac{{{S_n}\left( x \right)}}{{\psi \left( n \right)}} \le \frac{{{S_n}\left( x \right)}}{{\psi \left( {{n_k}} \right)}}, $

可知$ \mathop {\lim }\limits_{n \to \infty } \frac{{{S_n}\left( x \right)}}{{\psi \left( n \right)}} = 1. $即$ x \in G\left( \psi \right) $. 因此$ G\left( {\psi , M} \right) \subset G\left( \psi \right). $该引理得证.

对$ n \ge 1 $, 令$ r\left( n \right) = \sharp \left\{ {k:{n_k}\leq n} \right\}. $由集合$ G\left( {\psi , M} \right) $的定义可知, 给定$ k \ge 1 $, 对任意的$ x \in G\left( {\psi , M} \right) $, $ {d_{{n_k}}}\left( x \right) $的值都是相等的. 为了简化记号, 记$ {d_{{n_k}}}\left( x \right) $为$ {d_{{n_k}}} $.

引理4.3  数列$ \left\{ {{n_k}} \right\} $与$ \left\{ {{d_k}} \right\} $的定义如前所述. 则

$ \mathop {\lim }\limits_{n \to \infty } \frac{{r\left( n \right)}}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{\log \left( {{d_{{n_1}}}{d_{{n_{_2}}}} \cdots {d_{{n_{r\left( n \right)}}}}} \right)}}{n} = 0. $

证  由[7]中关系式（10）（12）可知, 如果$ \psi \left( {{n_k}} \right) \ge \left( {1 + {\varepsilon _{k - 1}}} \right)\psi \left( {{n_{k - 1}}} \right) $, 那么$ r(n) \ll {n^\frac{{\frac{1}{2} - \tau }}{{1 - \delta }}}. $从而由$ \frac{{\frac{1}{2} - \tau }}{{1 - \delta }} < 1 $可知$ \mathop {\lim }\limits_{n \to \infty } \frac{{r\left( n \right)}}{n} = 0. $此外,

$ \begin{aligned} \log \left( {{d_{{n_1}}}{d_{{n_2}}} \cdots {d_{{n_{r\left( n \right)}}}}} \right) & \ll \sum \limits_{i = 1}^{r(n)}\log\left((1 + \varepsilon _i)\psi (n_i)\right)\\ & \ll r\left( n \right)\log\psi \left( n \right) + \left( {{\varepsilon _1} + \cdots + {\varepsilon _{r\left( n \right)}}} \right)\\ & \ll r\left( n \right) \cdot {n^{\frac{1}{2} - \tau }} + r{\left( n \right)^{1 - \delta }}. \end{aligned} $

所以$ \mathop {\lim }\limits_{n \to \infty } \frac{{\log \left( {{d_{{n_1}}}{d_{{n_2}}} \cdots {d_{{n_{r\left( n \right)}}}}} \right)}}{n} = 0 $. 引理得证.

令$ \Lambda = \bigcup\limits_{k \ge 1} {\left\{ {{n_k} - 1, {n_k}, {n_k} + 1} \right\}}. $定义映射$ f:G\left( {\psi , M} \right) \to F\left( M \right) $如下,

$ f(x) = \sum \limits_{{i \ge 1} \atop {i \notin \Lambda }} \frac{1}{d_i(x)}\prod\limits_{{1\le k<i} \atop {k \notin \Lambda }} \frac{1}{d_k(x)(d_{k}(x)-1)}, $

即$ f(x) $的Lüroth展式的字符依次为$ d_1(x), \cdots, $$ d_{n_1-2}(x), d_{n_1+2}(x), \cdots, $$ d_{n_2-2}(x), d_{n_2+2}(x), $ $ \cdots. $因此$ f(G(\psi , M)) = F(M) $.

引理4.4  对任意$ \varepsilon > 0 $, 映射$ f $满足$ \frac{1}{{1 + \varepsilon }} $阶-Hölder条件.

证  任取$ x, y \in G\left( {\psi , M} \right) $且距离充分小, 则存在$ n \ge 0 $, 使得当$ 1 \le i \le n $时, $ {d_i}\left( x \right) = {d_i}\left( y \right) $; 但$ {d_{n + 1}}\left( x \right) \ne {d_{n + 1}}\left( y \right) $. 从而$ n + 1 \notin \Lambda $并且$ 2 \le {d_{n + 1}}\left( x \right), \;{d_{n + 1}}\left( y \right) \le M $. 一方面, $ I\left( {{d_1}\left( x \right), \cdots , {d_n}\left( x \right), {d_{n + 1}}\left( x \right), M + 1} \right) $或$ I\left( {{d_1}\left( y \right), \cdots , {d_n}\left( y \right), {d_{n + 1}}\left( y \right), M + 1} \right) $一定落在$ x $与$ y $之间的空隙里. 从而

$ \left| {x - y} \right| \ge \min \left\{ {\left| {I\left( {{d_1}\left( x \right), \cdots , {d_{n + 1}}\left( x \right), M + 1} \right)} \right|, \left| {I\left( {{d_1}\left( y \right), \cdots , {d_{n + 1}}\left( y \right), M + 1} \right)} \right|\;} \right\}. $

由[8]中引理4.1知, 存在某个常数$ c_M $使得

$ \left| {I\left( {{d_1}\left( x \right), \cdots , {d_{n + 1}}\left( x \right), M + 1} \right)} \right| \ge {c_M}\left| {I\left( {{d_1}\left( x \right), \cdots , {d_n}\left( x \right)} \right)} \right| $

和

$ \begin{aligned} \left| {I\left( {{d_1}\left( y \right), \cdots , {d_{n + 1}}\left( y \right), M + 1} \right)} \right| & \ge {c_M}\left| {I\left( {{d_1}\left( y \right), \cdots , {d_n}\left( y \right)} \right)} \right|\\ & = {c_M}\left| {I\left( {{d_1}\left( x \right), \cdots , {d_n}\left( x \right)} \right)} \right|. \end{aligned} $

所以$ \left| {x - y} \right| \ge {c_M}\left| {I\left( {{d_1}\left( x \right), \cdots , {d_n}\left( x \right)} \right)} \right|. $由$ f $的定义知, $ f(I\left( {{d_1}\left( x \right), \ldots , {d_n}\left( x \right)} \right) $为$ n-3r(n) $阶柱集, 记作$ I\left( {{b_1}, \cdots , {b_{n - 3r\left( n \right)}}} \right) $. 从而

$ \begin{aligned} &\left| {x - y} \right| \ge {c_M}\left| {I\left( {{d_1}\left( x \right), \cdots , {d_n}\left( x \right)} \right)} \right|\\ = & {c_M}|I({b_1}, \cdots , {b_{n - 3r(n)}})|\mathop \prod \limits_{i = 1}^{r(n)} \prod\limits_{k = 0}^2\left(d_{n_i-1+k}(x)(d_{n_i-1+k}(x)-1)\right)^{-1}\\ = & {c_M}{2^{ - 2r(n)}}|I({b_1}, \cdots , {b_{n - 3r\left( n \right)}})| \prod\limits_{i = 1}^{r(n)}\left(d_{n_i}(x)(d_{n_i}(x)-1)\right)^{-1}. \end{aligned} $

由引理4.3可知, 如果$ n \ge {n_0} $, 则

$ \left| {x - y} \right| \ge {c_M}{\left| {I({b_1}, \cdots , {b_{n - 3r\left( n \right)}})} \right|^{1 + \varepsilon }} \ge {c_M}{\left| {f(x) - f(y)} \right|^{1 + \varepsilon }}, $

其中$ {n_0} $是跟$ M $和$ \varepsilon $有关的整数. 从而引理得证.

定理1.2的证明  由引理4.1知$ {\dim _H}F(M) = {s_M} $, 且$ \lim \limits_{M\to \infty} s_M = 1 $. 由引理4.2和引理4.3知,

$ {\dim _H}G(\psi ) \ge {\dim _H}G(\psi , M) \ge \frac{1}{{1 + \varepsilon }}{\dim _H}F(M). $

所以$ {\dim _H}G(\psi ) \ge \frac{1}{{1 + \varepsilon }}{s_M} $. 令$ \varepsilon \to 0, M \to \infty $, 则定理结论得证.