# Fourier Transform of Exponentially Decaying Function Cannot Have Compact Support

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.

• 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
• 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

## 1 Answer

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.