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Under the isomorphism of Hilbert spaces $L^2(S^1)\to\ell^2(\mathbb Z),\quad e^{2\pi i n t}\mapsto e_n$, smooth functions on the circle are mapped to rapidly decaying sequences (see wikipedia).

Is the converse also true? That is, does every rapidly decaying sequence in $\ell^2(\mathbb Z)$ represent a smooth function?

Motivation: For certain arguments, it would be really nice to read off smoothness from the Fourier coefficients.

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When you say a sequence $c_n$ is rapidly decaying, do you mean it is $O(n^{-k})$ for all $k$ ? – TCL Nov 18 '10 at 16:42
@TCL: Yes, I do. – Rasmus Nov 18 '10 at 21:01
up vote 8 down vote accepted

Yes. Smoothness is equivalent to the Fourier coefficients forming a sequence that decays rapidly (faster than any polynomial). To see the direction you asked about, note that if $\{c_n\}$ is a rapidly decaying sequence, then the sum $\sum c_n e^{in x}$ will converge uniformly as will every derivative. So the sum will represent a smooth function. (The continuous analog of this is that the Fourier transform induces an isomorphism of the Schwartz space onto itself.)

In general, one can use the Fourier coefficients to give an $L^2$ definition of differentiability. Morally, a function ought to be $C^k$ if its Fourier coefficients decay like $|n|^{-k}$. This is not quite accurate, but one can use this to define the Sobolev spaces on the circle (more generally on the torus, or using the Fourier transform on euclidean space, and by an extension process on compact manifolds) that describe how many "weak" derivatives a function has. This sort of argument shows that $L^2$ differentiability is roughly comparable with normal differentiability, and using such Sobolev spaces turns out to be integral when one tries to prove explicit bounds about things such as elliptic regularity. There, working with Fourier coefficients (or transforms) and a multiplication operator is significantly easier than working in the original function space with the correspondingly more complicated differential operator. (I am using the fact that constant-coefficient differential operators correspond under the Fourier transform to multiplication by a polynomial.)

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Sorry for bringing up this old topic again, but I just came across this question and answer. If you have the time, could you elaborate a little on how the relationship between degree of differentiability and decay of Fourier coefficients is "not quite accurate"? – Rahul Mar 22 '11 at 5:13

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