Fourier coefficients for an infinitely differentiable function

Question

I hope I have understood this coreectly: A Fourier series has coefficients of order $O(n^{d+1})$ for a $d$ times differentiable function. But what if the function is infinitely differentiable? Do the coefficients tend to have order 0? That is, is the series finite? Am I right? Thanks in advance!

Answer 1

The result

A periodic function which has Fourier coefficients of order $O\left(1/n^{d+1}\right)$ is $d$ times differentiable.

becomes for infinitely differentiable function

A periodic function which has Fourier coefficients of order $O\left(1/n^{d+1}\right)$ for all integer $d$ is infinitely differentiable.

And the converse is true: if $\displaystyle f(x)=\sum_{n=-\infty}^{+\infty}c_ne^{inx}$ with $c_n=O\left(1/n^{d+1}\right)$ then for all $d$ the series $\displaystyle\sum_{n=-\infty}^{+\infty}c_n(in)^de^{inx}$ is normally convergent, and therefore we can take the derivative under the sum. Note that we may have that the coefficients $c_n$ are all different from $0$, for example taking $c_n=e^{-n^2}$, or $a^{-n^2}$ where $|a|\gt 1$.