TITLE: Lipschitz-like function which is nowhere differentiable function
QUESTION [5 upvotes]: I've been trying this problem from Stein, but with no luck.
Consider the function
$$f_{1}(x)=\sum_{n=0}^{\infty}{2^{-n} e^{2\pi i 2^{n} x} }.$$
a) Prove that $f_{1}$ satisfies $|f_{1}(x)-f_{1}(y)| \leq A_{\alpha}|x-y|^{\alpha}$ for each $\alpha \in (0,1)$.
b) $f_{1}$ is nowhere differentiable hence not of bounded variation.
It sounds beautiful and I was wondering if there's any nice proof. A friend tells me there's a more general theory about some so-called Hilbert functions which justify this, but I'm interested in something easier!
Thanks!

REPLY [6 votes]: Triangle Inequality and Mean Value Theorem yield
$$
\begin{align}
|f_1(x)-f_1(y)|
&\le\sum_{n=0}^\infty2^{-n}\left|e^{2\pi i2^nx}-e^{2\pi i2^ny}\right|\\
&=\sum_{n=0}^\infty2^{-n}\left|e^{2\pi i2^nx}-e^{2\pi i2^ny}\right|^{1-\alpha}\left|e^{2\pi i2^nx}-e^{2\pi i2^ny}\right|^\alpha\\
&\le\sum_{n=0}^\infty2^{-n}2^{1-\alpha}(2\pi2^n)^\alpha|x-y|^\alpha\\
&=2\pi^\alpha\frac{1}{1-2^{\alpha-1}}|x-y|^\alpha\\
&=A_\alpha|x-y|^\alpha
\end{align}
$$
Note that as $\alpha\to1^-$, $A_\alpha\to\infty$.
Hardy proves in Theorem $1.31$ of Weierstrass's Non-Differentiable Function that $f_1$ is nowhere differentiable.