Cantor function $f$ is defined by
$f: \Delta \to [0,1]$ by $$f(\sum_{n=1}^{\infty} \frac{2b_n}{3^n}) = \sum_{n=1}^{\infty} \frac{b_n}{2^n}, b_n \in \{0,1\}$$, where $\Delta$ is a Cantor set and each $x\in\Delta$ can be written as $x=\sum_{n=1}^{\infty} \frac{2b_n}{3^n}$, where $b_n = 0$ or $1$.
I read "Cantor function is uniformly continuous" on wikipedia https://en.wikipedia.org/wiki/Cantor_function and wiki also claims that it is Holder continuous of exponent $\alpha = \frac {\log2}{\log3}$. So how to prove it is uniformly continuous and even how to get $\alpha = \frac {\log2}{\log3}$?
Best Answer
It is enough to show that it is Holder continuous. Suppose $0 < \alpha < \frac{\ln 2}{\ln 3} < 1$. Take any two points $x, y \in [0,1]$. There exists some $n$ such that $|x - y| > 3^{-n}$ (the Archimedian property).
Then $|f(x) - f(y)| \leq \left|f(\frac{k}{3^{n}}) - f(\frac{k +1}{3^{n}})\right|\leq \left|\frac{m}{2^{n}} - \frac{m +1}{2^{n}}\right| \leq \frac{1}{2^{n}}\leq \left(\frac{1}{3^{n}} \right)^{\alpha} < M|x-y|^{\alpha}$, for an appropriate choice of M.