Bound on variation $|F(x)-F(y)|$ where $F$ is Cantor's function

Let $F$ be Cantor's function. Prove that there exists some $C>0$ s.t for all $x,y\in[0,1]$ $$|F(x)-F(y)|\leq C|x-y|^\alpha$$ Where $\alpha=\log_3 2$.

My direction of thought was to use the definition of F as

$$F(x):=\lim_{n\to\infty}\int_{0}^{x}g_n(t)dt$$ $$g_{n}(x):=\begin{cases} \left(\frac{3}{2}\right)^{n} & x\in C_{n}\\ 0 & x\notin C_{n} \end{cases}$$ and try to find some bound $A$ on the amount of sub-segments of $C_{n}$ contained in $[x,y]$, as then we could say $$\left|F\left(x\right)-F\left(y\right)\right|=\lim_{n\to\infty}\int_{x}^{y}g_{n}(t)dt\leq A\cdot\left(\frac{3}{2}\right)^{n}\cdot3^{-n}$$ because the length of each segment in $C_n$ is $3^{-n}$

Solution 1:

We'll show this holds for $C=2$. Clearly we should try to exploit the self-similarity of the Cantor set/function. And this can be done by induction.

Assume $x\not=y$ and let $n$ be the minimal positive integer such that $|x-y|\ge\frac{1}{3^n}$. We show the inequality holds by induction on $n$.

If $n=1$, then $|x-y|\ge\frac{1}{3}$ and $|x-y|^{\alpha}\ge \frac{1}{3^\alpha}=\frac{1}{2}$, and $C|x-y|^{\alpha}\ge 1 \ge |F(x)-F(y)|$ because of $0\le F(z)\le 1$ for all $z\in [0, 1]$.

For the induction step, assume $|x-y|>\frac{1}{3}$. By symmetry and relaxation (that is if one of $x,y$ sits in the middle $(\frac{1}{3}, \frac{2}{3})$, we can push it to one end of the interval without chaing $F(x)$ or $F(y)$.), we may assume $0\le x, y \le \frac{1}{3}$ and apply the identity $F(z)=\frac{F(3z)}{2}$ for all $0\le z\le\frac{1}{3}$. Suppose $\frac{1}{3^n}\le |x-y|<\frac{1}{3^{n-1}}$, then by the identity and induction hypothesis, $$|F(x)-F(y)|=\frac{1}{2}|F(3x)-F(3y)|\le \frac{C}{2} |3x-3y|^{\alpha}=2 |x-y|^{\alpha}$$