TITLE: Differentiability of Fourier series
QUESTION [11 upvotes]: 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.

REPLY [5 votes]: The critical value is $a= 5/2$.
Chamizo and Cordoba showed that the fractal dimension of $f(x,a)$ is $2 + \frac{1}{2}(1/2 - a), \ \ \ 1 \leq a \leq \frac52$.Thus $f(x,a)$ is not differentiable for $a < 5/2$ (differentiable functions have dimension 1). On the other hand, for $\alpha > 5/2$, $f(x,a)$ is differentiable almost everywhere. They also include pictures for the $a = 3/2 case.
The point is that $\sum_{N \leq n \leq 2N} e^{itn^2}$ typically has size around $\sqrt{N}$ (for instance Weyl-differencing) and so at scale $N$ on the Fourier side, we see a decay of about $N^{1/2 - a}$. For differentiability, we want decay of $N^{-2}$ and so the critical value is $5/2$. 
See Jaffard's "The spectrum of singularities of Riemann's function" for precise results on the case $\alpha = 2$. He shows the set of points for which $f(x,2)$ is differentiable has Hausdorff dimension zero and gives a very precise formulation of the regularity at each point in terms of the continued fraction expansion of $x$.