Let $a_n=\mathcal{O}(\lvert n\rvert^{-\alpha})$ where $\alpha>\frac{3}{2}$ then \begin{equation} f(x):=\sum_{n\in\mathbb{Z}}a_n e^{inx}, \end{equation} is differentiable and its derivative is in $L^2([0,2\pi])$.
I do not understand how to prove that the limit: \begin{equation} \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} \end{equation} Exists finite for any $x\in [0,2\pi]$. The fact that that limit exists in $L^2([0,2\pi])$ is rather obvious (prove that it is a Cauchy sequence), however I cannot figure out how to proceed to pointwise existence. Any help is welcome. This is an exercise from a Fourier Analysis course, therefore I would really appreciate if you can clear my mind.
this seems possible with more restrictions:
If ${a_{n}+\overline{a_{n}}}=\large{\frac{2}{n^{l/m}}}$, where $l,m$ are integers and $\large{\frac{l}{m}\ > \frac{3}{2}}$ and the sum is taken over $A=\{k^m| k\in\mathbb{Z}\}$ then one gets a sum of the form $\sum\limits_{n\in A} \frac{1}{n^{l/k}} \cos{nt}=\sum\limits_{k=0}^{\infty} \frac{1}{k^{l}} \cos{k^m t}$ which when differentiated term-wise gives an absolutely convergent series converging pointwise everywhere to $f'$.
No counterexample comes to mind in the general case.