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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? The coefficients tend to have order 0? Ie the series is finite? Am I right? Thanks in advance!

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up vote 4 down vote accepted

The result

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

becomes for infinitely differentiable function

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

And the converse is true: if $\displaystyle f(x)=\sum_{n=-\infty}^{+\infty}c_ne^{inx}$ with $c_n=O(n^{d+1})$ 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 $|a|\leq 1$ and $a\neq 0$.

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