Let $$f(e^{i\theta})\sim\sum_kc_ke^{ik\theta}$$, where $c_k=1/(|k|^3+1)$. Can we say anything about the derivatives $f',f'',f'''$ (like if they exist, continuous, and differentiable)?
From a theorem, the derivative of $f(e^{i\theta})$ is obtained by differentiating term by term: $$\frac{d}{d\theta}f(e^{i\theta})=\sum ikc_ke^{ik\theta} $$ However, I don't think directly applying this theorem alone could explain the existence, continuity, and differentiability of those derivatives. What else should I consider?
Thanks！
In this case one can show that $f'$ exists and the Fourier series is obtained by differentiation term by term, $f''$ exists except at $2k\pi$ where it is infinite, its Fourier series obtained by differentiation is convergent as usual (uniformly on compact sets avoiding $2k\pi$, but diverging to infinity at $2k\pi$), $f'''$ exists same as $f''$ but the trigonometric series obtained by differentiation is now only summable but not convergent since $f'''$ is not integrable any more near $2k\pi$ (so it is a generalized Fourier series or a Fourier-Stieltjes series rather than a "regular" Fourier series)
First we note that $f(e^{i\theta})=1+2\sum_{k\ge 1}\frac{\cos k\theta}{k^3+1}=2g(e^{i\theta})+h(e^{i\theta})$, where:
$g(e^{i\theta})=\sum_{k\ge 1}\frac{\cos k\theta}{k^3}$ and $h(e^{i\theta})=1-2\sum_{k\ge 1}\frac{\cos k\theta}{k^3(k^3+1)}$
By general Fourier series theory, we notice that $q(e^{i\theta})=\sum_{k\ge 1}\frac{k\cos k\theta}{k^3+1}$ is continuous (series absolutely convergent) and $h^{(4)}=-2q$ (we can integrate $q$ term by term bunch of times since the result is still periodic as there is no free term in the Fourier series etc), so we need just to analyse $g$ for our purposes of studying the derivatives of $f$ up to order $3$
But now it is well known that $\sum_{k\ge 1}\frac{\cos k\theta}{k}$ is the Fourier series of the integrable function $p(e^{i\theta})=-\log (2\sin \theta/2), 0<\theta<2\pi$ (extended by periodicity to the line if one so desires), so by the same results as before $g''(e^{i\theta})=-p(e^{i\theta}), \theta \ne 2k\pi$ and by writing explicitly $g'$ it is not hard to show that the result holds at the ends too ($g''(e^{i\theta}) \to \infty, \theta \to 2k\pi$).
In particular $g''$ is differentiable at $\theta \ne 2k\pi$ and $g'''(e^{i\theta})=\frac{1}{2}\cot \frac{\theta}{2}, 0<\theta<2\pi$ and the cotangent is clearly not integrable near $0,2\pi$ so $f'''$ is not either, hence it has no Fourier series in the classical sense which also is obvious from the fact that the formal derivative term by term of $\sum_{k\ge 1}\frac{\cos k\theta}{k}$ has coefficients which do not satisfy Riemann_Lebesgue for example; however it is not difficult to show summability of $\sum_{k \ge 1}\sin k\theta$ to $-\frac{1}{2}\cot \frac{\theta}{2}$ for $0<\theta<2\pi$ by considering the imaginary part of $\frac{1}{1-z}=\sum z^n, |z|<1$ and letting $|z| \to 1$ for example