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If I have a function $f:\Omega \to \mathbb{R}$ in $H^k(\Omega)$ where $\Omega \subset \mathbb{R}^n$ is compact, then what is known about the Fourier transform $\hat{f}$? What space does it lie in? I want to say something like it's in $L^2(K)$ or $H^k(K)$ where $K$ is some other compact subset of $\mathbb{R}^n$.

I don't know if this makes sense... but essentially I want to bound $|(1+|\xi|^2)^k|$ where $\xi$ is the variable in $\hat{f}$, so I need to know how big $\xi$ can get..

Does this hold? Thanks for any help.

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How are you extending $f$ to $\mathbb{R}^n$? – Jose27 Dec 10 '12 at 0:55
@Jose27 I don't know, all I know if $f$ has domain $\Omega$. I guess extension by 0? – maximumtag Dec 10 '12 at 9:11
If you extend as $0$, and $\Omega$ is bounded, then $\hat{f}$ has an extension to $\mathbb{C}^n$ as an entire function. – Jose27 Dec 11 '12 at 6:13
up vote 2 down vote accepted

Extending a function in $H^k$ by $0$, you lose the Sobolev regularity: the result is just an $L^2$ function of compact support ($L^p$ with $p>2$ if we use Sobolev embedding, but this does not help). In particular, if the extended $f$ is discontinuous, then $\hat f$ has a substantial tail: it's in $L^2(\mathbb R^n)\setminus L^1(\mathbb R^n)$. This is rather far from being compactly supported.

In order to have $\hat f$ compactly supported, a necessary condition for $f$ is to be real-analytic on $\mathbb R^n$. If $f$ is not real analytic on $\Omega$, then no matter how you extend it, the Fourier transform will not be compactly supported.

The fundamental result in this direction is the Paley-Wiener theorem.

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