0
$\begingroup$

If $f\in L^{1}(\mathbb{R}^n)$, and $f$ has compact support, then how can I show that $\hat{f}$ cannot satisfy $\hat{f}(x)=O(e^{-\epsilon|x|})$ for any $\epsilon>0$?

This is similar in the spirit of the Paley-Wiener-Schwartz theorem, which gives a characterization of a smooth function with compact support of a distribution with compact support in terms of the decay property of their Fourier transform. So similarly, I'm also interested to know if there is a characterization for $L^p$ functions with compact support.

Improve this question
$\endgroup$
1
  • 9
    $\begingroup$ $\widehat{f}\lesssim e^{-\epsilon|x|}$ makes $f$ holomorphic on a strip including $\mathbb R$ (just allow complex values $z$ in $f(z)=\int\widehat{f}(x)e^{ixz}\, dx$), so $f\equiv 0$ if it has compact support. $\endgroup$
    – Christian Remling
    Commented Jun 27, 2014 at 18:51

1 Answer 1

Reset to default
1
$\begingroup$

Unni addresses $L^p$ Paley Wiener type theorems for Hankel transforms in (MR0174941). In the introduction, he also lists references for $L^p$ Paley Wiener theorems for Fourier transforms.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .