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Let $\varphi$ defined on the real line be continuous and with compact support. What can we say about the summability of $\hat{\varphi}$? I've gone through some theorems such as Parseval's without success, and perhaps we cannot assert summability. I know that for continuous and compactly supported functions, their Fourier transforms are infinitely differentiable, but that's not enough since we need fast decay rather than smoothness (e.g., the constant function equal to $1$ is smooth but not summable). The context of this question is that I would like to have the following equality: $$ \int_0^{2\pi} \sum_{k\in \Bbb Z} \hat{\varphi}(t+2k\pi)e^{-ikt}\, dt=\sum_{k\in \Bbb Z}\int_0^{2\pi}\hat{\varphi}(t+2k\pi)e^{-ikt}\,dt. $$ Thanks in advance.

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  • $\begingroup$ Do you mean the fourier transform of $L^1(\mathbb R/2\pi\mathbb Z) \to l^\infty(\mathbb Z)$ or from $L^1(\mathbb R)\to L^\infty(\mathbb R)$ ? Because in the first case the functions are not defined on the real line. Or if they are they have to be $2\pi$-periodic in which case compact support doesn't make sense. In the second case summability doesn't make sense. $\endgroup$ – Tim B. May 22 '15 at 10:54
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    $\begingroup$ Oh, I just found this on MO mathoverflow.net/questions/3764/… $\endgroup$ – Orest Bucicovschi May 22 '15 at 12:38
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No. The function $$ f(x)=\begin{cases}0 & |x|>1\\ (1-\log(1-x^2))^{-1} & |x|\le1\end{cases} $$ is an explicit example. For more comments and a proof (due to Terry Tao) see this post.

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No, there exist continuous functions on $S^1$ whose Fourier coefficients are not in $l^{\beta}$ for every $\beta <2$ ( while certainly being in $l^2$). The first example seems to be given by T. Carleman in 1916. This should be discusses in some books on classical Fourier series. I learned about it from this paper of Hausdorff.

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    $\begingroup$ I think the OP is interested in the Fourier transform $L^1(\mathbb{R}) \to C_0(\mathbb{R})$, not in Fourier series. $\endgroup$ – Daniel Fischer May 22 '15 at 11:52
  • $\begingroup$ @DanielFischer: I think you are right, it was a bit confusing with the sums. OK, I would keep it as an example for the other group ($S^1$), if not too much opposition to it. $\endgroup$ – Orest Bucicovschi May 22 '15 at 12:03

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