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From Paley winner theorem, if we have compactly supported distribution (not even function) still its Fourier transform is smooth. Is same thing valid for inverse Fourier transform. i.e. given compactly supported function then its inverse Fourier transform is smooth or not?

I thought this because of following theorem

If $m \in L_{\text {comp }}^{1}\left(\mathbb{R}^{n}\right)$, then the operator $$ S: f \mapsto \mathscr{F}^{-1}\{m(\xi) \hat{f}\} $$ is smoothing in the sense that it maps $\mathscr{E}^{\prime}\left(\mathbb{R}^{n}\right)$ to $C^{\infty}\left(\mathbb{R}^{n}\right)$.

Proof: If $f \in \mathscr{E}^{\prime}\left(\mathbb{R}^{n}\right)$ then $\hat{f} \in C^{\infty}\left(\mathbb{R}^{n}\right) $(Schwartz's Paley winner theorem). Consequently $F(\xi):=m(\xi) \hat{f}(\xi)$ is in $L^{1}\left(\mathbb{R}^{n}\right)$ by the assumption on $m$. Moreover, $F$ is compactly supported, which implies that $S f=\mathscr{F}^{-1} F$ is $C^{\infty}$.

I do not know how last step follows.

Any help or hint will be greatly appreciated.

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Since the inverse Fourier transform operator differs from the forward one merely by a complex conjugation in the integrand, ...

  • If $f$ has compact support, then $\mathscr Ff$ is smooth

implies

  • If $g$ has compact support, then $\mathscr F^{-1}g=\overline {\mathscr F\overline g}$ is smooth
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