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I just came across with the following question.. suppose we are given a periodic function of period $2\pi$. We define $a_n$ and $b_n$ to be the Fourier coefficients of $f$. To be precise, we have $$a_n:=\frac 1\pi\int_0^{2\pi}f(x)\cos nx\mathrm dx\quad\;b_n:=\frac 1\pi\int_0^{2\pi}f(x)\sin nx\mathrm dx,$$ so that $$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left[a_n\cos(nx)+b_n\sin(nx)\right]$$ Now it is clear that if $f\in C^\infty(\mathbb R)$ then $$\lim_{n\to\infty}n^{2k}(a_n^2+b_n^2)=0,\:\forall k>0.$$ Is the converse also true?

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Yes. High decay of Fourier coefficients means that both the Fourier series and the series obtained by differentiating term-by-term converge uniformly. Hence the Fourier series for $f$ can be differentiated term by term. This process can be repeated as many times as we wish and so $f$ is infinitely differentiable. (Please check this assertion carefully, I'm answering in a hurry and have a high probability of being wrong.) – Giuseppe Negro Sep 4 '12 at 19:39
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Yes, the converse is true. To see that more easily, I will write the series with the exponential form, namely, $\sum_{n\in\Bbb Z}c_ne^{inx}$. The condition gives that $|c_n|^2n^{2k}\to 0$ for each $k\geq 1$. In particular, the sequence $\{|c_n|\}$ is summable, hence we can write $f(x)=\sum_{n\in\Bbb Z}c_ne^{inx}$. Now we can show by induction that $$f^{(d)}(x)=\sum_{n\in\Bbb Z}(in)^dc_ne^{inx}.$$ It's true for $d=0$, and to jump from $d$ to $d+1$, we can use Talor's formula and the fact that the series $\sum_{n\in\Bbb Z}|c_n|n^{d+1}$ is convergent.

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