Weierstrass nowhere differentiable function is Lip$_\alpha$ for $\alpha < 1$

Question

Let $f(x) = \sum_{n=1}^{\infty}{(\frac{3}{4})^{n} \phi(4^nx) }$, where $\phi(x)$ is defined as follows.

1. $x-n$ if $n \leq x <n+1$ and $n$ is even.
2. $n + 1 - x$ if $ n \leq x < n+1 $ and n is odd.

Then it can be proved that $f$ is continuous(actually, uniformly continuous) but nowhere differentiable. How can I prove that $f$ is in Lip$_\alpha$ for any $\alpha < 1$? (It means there is a constant $C > 0$ such that for any $s, t$, it holds that $|f(s) - f(t)| \leq C|s - t|^\alpha$. I was able to prove that $f$ is not Lip$_1$.)

Answer 1

Actually it's not true that $f\in Lip_\alpha$ for every $\alpha<1$. It's true for some $\alpha$:

$\exists\alpha f\in Lip_\alpha$:

First note two things about $\phi$, namely that

(i) $|\phi(x)|\le1$

and

(ii) $|\phi(x)-\phi(y)|\le|x-y|$.

Write $$f(x)-f(y)=\sum\left(\frac34\right)^n(\phi(4^n x)-\phi(4^ny)).$$ Choose $N$ so $$\frac14\le4^N|x-y|\le1,$$and write $\sum=\sum_1^N+\sum_{N+1}^\infty=I+II$ above. Choose $\alpha$ so $4^{(-\alpha+1)}=3$, and use (ii) to estimate I: $$|I|\le\sum_1^N\left(\frac34\right)^n4^n|x-y|\sim 3^N|x-y|\sim4^{(-\alpha+1)N}4^{-N}=4^{-N\alpha}\sim|x-y|^\alpha.$$Similarly (i) shows that $|II|\le c|x-y|^\alpha$.

$f\notin\bigcap_{\beta<1}Lip_\beta$:

Since $\phi\ge0$ and $\phi(x)=x$ for $0<x<1$ we have $$f(4^{-N})-f(0) =f(4^{-N})\ge\sum_{n=1}^N\left(\frac34\right)^n4^{n-N}=c\left(\frac34\right)^N=c(4^{-N})^\alpha,$$where $\alpha$ is as above. Hence $f\notin Lip_\beta$ for $\beta>\alpha$.

Exercise. Let $g(x)=\sum4^{-n}\phi(4^nx)$. Then $g\in Lip_\alpha$ for every $\alpha\in(0,1)$.