Let $C_k$ be the set obtained in the $k-$th stage of building the Cantor set, where
$$C_1=[0,\frac{1}{3}]\cup[\frac{1}{3},\frac{2}{3}]$$
$$C_2=[0,\frac{1}{9}]\cup[\frac{2}{9},\frac{1}{3}]\cup[\frac{2}{3},\frac{7}{9}]\cup[\frac{8}{9},1]$$
$$\dots$$
So we have
$$C_k=\bigcup_{j=1}^{2^k}I_{k,j}$$
where $I_{k,j}$ are pairwise disjoint closed intervals of length $3^{-k}$ in $[0,1]$, let
$$[0,1]\setminus C_k=\bigcup_{j=1}^{2^k-1}J_{k,j}$$
where $J_{k,j}$ are pairwise disjoint open intervals and each $J_{k,j}$ are located between intervals $I_{k,j},I_{k,j+1}$
define
$$F_k:[0,1]\to[0,1]$$
where
$$F_k(0)=0,F_k(1)=1,F_k(x)=\frac{j}{2^k}\quad x\in J_{k,j}$$
and $F_k$ is linear on each subinterval of $C_k$. We can show that $F_k$ converges to the Cantor function $F:[0,1]\to[0,1]$. Moreover, it can be shown that
$$|F_k(x)-F_k(y)|\leq\frac{3^k}{2^k}|x-y|$$
How can we show that $F$ is Hölder continuous of order $\log_32$ ?
Best Answer
Using $\|F-F_k\|_\infty \le 2^{-k+1} \|F_0-F_1\|_\infty = 2^{-k}/3$ you get $$ |F(y)-F(x)| \le 2/3\cdot 2^{-k} + |F_k(y)-F_k(x)| \le (2/3 + 3^k |x-y|) 2^{-k}. $$ Now use $3^{-k-1} \le |x-y| < 3^{-k}$ for some $k$ to obtain $$ |F(y)-F(x)| < \frac{5}{3} 2^{-k} \le \frac{10}{3}|x-y|^{\log_3(2)} $$
Problem 4.12 in my lecture notes contains a hint for an alternate proof:
http://www.mat.univie.ac.at/~gerald/ftp/book-ra/ra.pdf