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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function, with $|f(x)| \le e^{-|x|}$ a.e.

Then how can we prove that its Fourier transform, $\hat{f}$, cannot have compact support (unless $f = 0$ a.e.).

I have a hint which says to show that $\hat{f} \in C^{\infty}$; I can do this using the differentiation rules for Fourier transforms, but am unsure of how to proceed from here. Can we use this to show that $\hat{f}$ is analytic in some neighbourhood of $\mathbb{R}$ (in which case the result follows easily)? I have another hint which says to then consider a suitable Taylor expansion. I am not too sure what to make of the second hint, but answers that involve the hints would be preferred.

John Don
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  • How does the result follow easily assuming $f$ is analytic in some nbhd? – mathworker21 Dec 29 '18 at 13:01
  • @mathworker21 Essentially by the identity theorem. Alternatively, see this question (if $f$ is analytic in a neighbourhood of $\mathbb{R}$, then in particular, it is $\mathbb{R}$-analytic). – John Don Dec 29 '18 at 13:11
  • Going back to the definition of the Fourier transform, you can show directly that $\hat{f}$ extends to an analytic function in ${\Im(z) >-1}$ (or ${\Im(z) < 1}$ depending on which convention you use). @JohnDon – Michh Dec 29 '18 at 13:12
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    This is what you are looking for with the roles of $f$ and $\hat{f}$ interchanged. – Michh Dec 29 '18 at 13:31
  • @JohnDon do you mean $\hat{f}$ is analytic in some nbhd? – mathworker21 Dec 29 '18 at 13:34
  • @Michh Thanks - this looks good! Just one point of confusion - the answer in the link shows that the Taylor series converges in a nbhd of each $x \in \mathbb{R}$, but why does this mean that it is $\mathbb{R}$-analytic? – John Don Dec 29 '18 at 13:35
  • @mathworker21 Sorry, I do mean $\hat{f}$ is analytic! – John Don Dec 29 '18 at 13:35

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You can show that $\hat{f}(s)$ is continuous in the strip $|\Im s| < 1$. Dominated convergence will do the job. Then you can use Morera's theorem to show that $\hat{f}$ is holomorphic in this strip by showing that $\int_{\Delta}\hat{f}(s)ds=0$ for every triangle in the same strip; all this requires is interchanging orders of integration.

Disintegrating By Parts
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