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本帖最后由 hbghlyj 于 2022-10-14 18:38 编辑 找到了:cantor.pdf
Proposition 10.1. The function $G$ satisfies a Hölder condition of order $s_c = \lg 2/ \lg 3$, the Hölder coefficient being not greater than 1. In other words$$|G(x)-G(y)| \leqslant|x-y|^{s_{c}}\tag{10.2}$$The constants $s_c$ and 1 are the best possible in the sense that the inequality$$|G(x)-G(y)| \leqslant a|x-y|^{\beta}$$does not hold for all $x, y$ in $[0, 1]$ if either $β > s_c$ or $β = s_c$ and $a < 1$.
Proof. Using lemma from [32] we obtain the inequality
$$
\left(\frac{1}{2} t\right)^{s_{c}} \leqslant G(t) \leqslant t^{s_{c}}\tag{10.3}
$$
for all $t \in[0,1]$. Since $G$ is its first modulus of continuity (see Proposition 3.2), the last inequality implies (10.2). Moreover, it follows from (10.3) that $s_{c}$ is a sharp Hölder exponent in (10.2). By a simple calculation we have from formula (1.2) that
$$
G\left(3^{-n}\right)=3^{-n s_{c}}
$$
for all positive integers $n$. Thus, the Hölder coefficient 1 is the best possible in (10.2).□
Proposition 10.5. Let $x$ be an arbitrary point of $C$. Then we have$$\liminf _{\substack{y \rightarrow x \\ y \in C}} \frac{\lg |G(x)-G(y)|}{\lg |x-y|}=s_{c}$$Proof. It suffices to show that
$$
\lim _{\substack{y \rightarrow x \\ y \in C}} \inf \frac{\lg |G(x)-G(y)|}{\lg |x-y|} \leqslant s_{c}
$$
Choosing $|x-y|=\frac{2}{3^{n}}$ for $n \rightarrow \infty$ we obtain the result.□
Let $x$ be a point of the Cantor ternary set $C$. Then $x$ has a triadic representation
$$
x=\sum_{m=1}^{\infty} \frac{\alpha_{m}}{3^{m}},\tag{10.6}
$$
where $\alpha_{m} \in\{0,2\}$. Define a sequence $\left\{\mathscr{R}_{x}(n)\right\}_{n=1}^{\infty}$ by the rule
$$
\mathscr{R}_{x}(n):= \begin{cases}\inf \left\{m-n: \alpha_{m} \neq \alpha_{n}, m>n\right\} & \text { if } \exists m>n: \alpha_{m} \neq \alpha_{n}, \\ 0 & \text { if } \forall m>n: \alpha_{m}=\alpha_{n},\end{cases}
$$
i.e., $\quad \mathscr{R}_{x}(n)=1 \Longleftrightarrow\left(\alpha_{n} \neq \alpha_{n+1}\right)$;
$$
\mathscr{R}_{x}(n)=2 \Longleftrightarrow\left(\alpha_{n}=\alpha_{n+1}\right) \&\left(\alpha_{n+1} \neq \alpha_{n+2}\right)
$$
and so on. |
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