Prove that the Fourier transform of a test function has not compact support

Question

Let $f \in C^\infty_{c}(\mathbb{R})$ such that $\mathrm{supp}(f) \subset B(0,R)$. Prove that

$$ \widehat{f}(\xi)=\int_{-R}^R e^{i x \xi} f(x) dx $$ with $\xi \in \mathbb{C}$ is a holomorphic function in $\mathbb{C}$. Moreover, $|\widehat{f}(\xi)| \leq C e^{R |\mathrm{Im} \xi}|$. In particular, $\widehat{f}(\xi)$, $\xi \in \mathbb{R}$, is real analytic and $\mathrm{supp}(\widehat{f})$ is not compact unless $\widehat{f}(\xi)=0$.

is a bit of time that I do not touch complex analysis, can you help me?

thank you

That answers the part about the compact support (when you know $\hat{f}$ is holomorphic). The growth restriction is an ordinary estimate (bound $\lvert e^{ix\xi}\rvert$). For the holomorphicity, use Morera or differentiation under the integral, whatever you prefer. — Daniel Fischer, Sep 29 '16 at 20:24

syzygy · Answer 1 · 2016-09-29T20:35:05.243

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Hint: Differentiation under the integral sign. Check that the integrand is analytic and that the assumptions of the theorem on differentiation under the integral sign are satisfied.. For your estimate just estimate the modulus of the integral by the integral of the modulus of the integrand.

edited Sep 29 '16 at 20:35

answered Sep 29 '16 at 20:17

syzygy

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@haqnatural better now? XD – syzygy Sep 29 '16 at 20:35
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This may be of use. @JohnMartin – Sergio Charles Sep 29 '16 at 21:24

Prove that the Fourier transform of a test function has not compact support

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