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Definition of smoothness

Definition of bandlimited: for simplicity, let's consider a function in $L^2(\mathbb{R})$ is bandlimited if the support of its Fourier transform lies in $[\Omega, \Omega]$ for some $\Omega\in\mathbb{R}$

Specifically, how smooth is this function given that I know it's bandlimited? $$Sinc(4\pi t)$$

tlaminator
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    The sinc function is $C^{\infty}$. There's a general principle that smoothness on one side corresponds to decay on the other, and band-limited functions have pretty nice decay. –  Dec 09 '15 at 05:58

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Yes, it is true that the Fourier transform of a compactly supported function is smooth. There is a cute proof using the fact that the Fourier transform is actually analytic. (Refer to the "Holomorphic Fourier transform" section.) A more direct proof (perhaps simpler) can be constructed using the general ideal that the Fourier transform trades off smoothness and decay. See Remark 9.3(a) in Rudin's Real and Complex Analysis, for example, but the idea is probably worked out more explicitly elsewhere.

In fact, more is true. Supposing the compactly supported function is smooth, its Fourier transform decays very rapidly and lies in the Schwarz space. Also, this result extends to distributions.

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