Inverse Fourier transform of a function generated from a compactly supported function

Question

Suppose $u$ belongs to a Sobolev space $H^s$ ($s\ge 0$). Assume $u$ has compact support. Does $f$ which is defined by $f=\mathcal{F}^{-1}(\hat{u}(\xi^2))$ have also compact support? Or more generally, $f=\mathcal{F}^{-1}(p(\xi)\hat{u}(\xi^2))$ where $p(\xi)$ is polynomial or rational.

Answer 1

For $u\ne 0\in L^2$ supported on $[-r,r]$ then its Fourier transform $$\hat{u}(z)=\int_\Bbb{R} u(t)e^{-i z x}dx$$ is entire and $u(z)=O(e^{r|\Im(z)|})$ as $|z|\to \infty$.

$\hat{u}(z^2)$ is $L^2(\Bbb{R})$ and entire but it isn't $O(e^{R|\Im(z)|})$ for any $R$, whence $\mathcal{F}^{-1}(\hat{u}(z^2))$ is $L^2$ but not compactly supported.