Is the Fourier transform of a continuous and compactly supported function summable?

Question

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.

Answer 1

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.

Answer 2

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.