Consider the function defined by the Fourier series
$$ f(x;\alpha) = \sum_{n=1}^\infty \frac{1}{n^\alpha} \exp(i n^2 x ) , $$
where $\alpha >1 $.
For what values of $\alpha $ is $f$ differentiable? Based on numerics, it is conjectured that $\alpha = 2$ is a critical value. For $\alpha <2 $, the function is nowhere differentiable; while for $\alpha >2 $, the function is differentiable almost everywhere.
Best Answer
Edit: fedja has pointed out the previous result (first theorem) stated was weaker than the present purposes require. This has now been addressed with the addition of references and the result of Luther below.
The fact that $f(x)$ is everywhere non-differentiable for $f$ with lacunary Fourier series follows by Theorem 2.1 here with earlier versions known to Freud and Hardy
Theorem [Hardy]: An integrable periodic function $f$ with Fourier series $\sum a_k \sin(n_k x)$, satisfying $\frac{n_k+1}{n_k} >\lambda > 1$ is differentiable at a point only if $\lim\limits_{k\rightarrow\infty} a_{k}n_k = 0$.
The particular function that is provided (known as Weirstrass' function) is not lacunary, but many results are known. For $\alpha=2$ one has a mix of non-differentiability and differentiability results known, one interesting result being the following of Gerver
Theorem [Gerver, 1970]: $f(x)=\sum\limits_{n=1}^{\infty} \frac{\sin{n^2 x}}{n^2}$ is differentiable at points $k\pi$ if $k=\frac{2p+1}{2q+1}$, $p,q\in\mathbb{Z}$.
A literature search shows that the question for varying $\alpha$ was considered even earlier by Hardy here (see second paragraph on the fifth page) who showed that this function has no derivative for $x$ irrational and $\alpha<\frac{5}{2}$. A more recent result of Luther (Theorem 2, here) shows more, namely that the above series is nowhere differentiable for $1\leq \alpha\leq \frac{3}{2}$ and that for $\frac{3}{2}<\alpha<3$ the function is differentiable at $x=\frac{p}{q}$, $(p,q)=1$, only for $pq \equiv\ 1\ (\text{mod }2)$.
This last result shows that the mix between differentiability and non-differentibility happens for $\alpha$ smaller than two ($\alpha=\frac{3}{2}$) while for $f(x;\alpha)$ to be differentiable it is necessary that $\alpha \geq 3$.