Lipschitz-like function which is nowhere differentiable function 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!
 A: 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.
A: Lets divide the sum into two:
$$
\sum_{n=0}^{N}{2^{-n} e^{2\pi i 2^{n} x} }
+\sum_{n=N+1}^{\infty}{2^{-n} e^{2\pi i 2^{n} x} }=S_1(x)+S_2(x).
$$
The difference of the first sum can be estimated by the mean value theorem:
$$
|\Delta S_1(x)|\le \sum_{n=0}^{N}{2^{-n} (2\pi 2^{n} |\Delta x|)} =2\pi (N+1)|\Delta x|,
$$
and the second are marjorized  by the sum of an infinite geometric progression:
$$
|\Delta S_2(x)|\le \sum_{n=N+1}^{\infty}{2^{-n}}=2^{-N}.
$$
Now for given $\Delta x$ one can choose $N$ s.t. both summands satisfy the required estimate.
