Let $\Omega\subset\mathbb{R}^d$ be a compact subset of $\mathbb{R}^d$. For a Schwartz function $f:\mathbb{R}^d\to\mathbb{R}$ the Fourier Transform can be written as $$ \hat{f}(\xi) = \int_{\mathbb{R}^d}f(x)e^{ix^\intercal\xi}d\xi $$

If we look at an Schwartz arbitrary function $g:\Omega\to\mathbb{R}$, then this does not work directly. This is because the Fourier Transform is not defined for a compact subset of $\mathbb{R}^d$. According to two other posts, (1) and (2), it is possible to use the Paley Wiener theorem or some other approach to extend $g$ to a function from $\mathbb{R}^d\to\mathbb{R}$ when $g\in L^1$.

I am interested in the value of $$ \gamma(f) = \int_{\mathbb{R}^d}||\xi||^{s}_2|\hat{f}(\xi)e^{ix^\intercal\xi}|d\xi=\int_{\mathbb{R}^d}||\xi||^{s}_2|\hat{f}(\xi)|d\xi $$ for a function $f:\Omega\to\mathbb{R}$, in particular when $\gamma(f)$ is finite and well defined. What assumptions on $f$ and $\Omega$, aside from $f$ being a Schwartz function, are needed to make $\gamma$ finite and well defined? Does the boundary of $\Omega$ for example need to be smooth?

Dirier mentioned in the comments that $C^{\infty}$ and compactly supported on $\Omega$ should work. What if $f$ is less smooth or not compactly supported?

(1) Analog of the Fourier transform on a bounded domain?

(2) Fourier transform of function defined on subset of $\mathbb{R}^n$

  • $\begingroup$ What is the meaning of "a Schwartz function" when it is only defined on a compact set $\Omega$? $\endgroup$ May 4, 2021 at 12:33
  • $\begingroup$ If by Schwartz function you mean $C^{\infty}$ and compactly supported then you can extend $g$ to the entire space by $0$ and keep the same regularity. If the function was let regular then extending it by 0 you would lose some regularity but it is hardly relevant here. $\endgroup$ May 4, 2021 at 12:53
  • $\begingroup$ Adding to the above, a $C^\infty$ function on a sufficiently nice set has a $C^\infty_c$ extension, similarly with $W^{k,p}$. Details on how nice is "sufficiently nice" I would expect to find in Adams and Fournier. The smoothness of the extension is strongly linked to the decay of the Fourier transform (and therefore strongly linked to the convergence of the integral in question) $\endgroup$ May 4, 2021 at 13:35


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