I tried many ways to solve this problem but none seems to be conclusive :( any help will be great.
Suppose $f\in L^1(\mathbb{T})$ with Fourier coefficients $\hat{f}(n)=O(|n|^{-k})$ where $k$ isn't necessarily an integer. Then for any integer $m<k-1/2$ , $f$ is $m$ times differentiable with $f^{(m)}\in L^2(\mathbb{T})$.
I know that as long as $m<k-1$, then $\sum \hat{f}(n)n^m$ converges absolutely, so $f$ is $m$ times continuously differentiable. With this, if $m$ is the greatest integer such that $m<k-1/2$, then $f$ has continuous $m-1$ derivative, which is given by the absolutely and uniformly converging series $\sum \hat{f}(n)(in)^{m-1}e^{inx}$. I suppose now we need to calculate the derivative directly so we write the quotients
$$\frac{f(x+h)-f(h)}{h}=\sum\hat{f}(n)(im)^{m-1}e^{inx}\frac{e^{inh}-1}{h}$$
I thought about using dominating convergence in $\ell^1$, but that would be equivalent to asking $\sum |\hat{f}(n)||n|^m<\infty$ implying $f\in C^m$ which is too much; dominating convergence in $\ell^2$ will give $L^2$ convergence for the LHS by the Parserval isometry.
Thanks in advance.
Update: I noticed that $f(x)=x^2$ has Fourier expansion $$f(x)=\frac{\pi^2}{3}+\sum_{n\neq0}\frac{2(-1)^n}{n^2}e^{inx}$$ so $\hat{f}(n)=O(|n|^{-2})$, yet $f$ isn't differentiable at $x=$, so we have a contradiction, can anyone second this?