1
$\begingroup$

I know that the Fourier transform of a compactly support distribution $u\in \mathscr{E}'(\mathbb{R}^{n})$ is smooth and also satisfies $$ |\hat{u}(\xi)|\leqslant C_{N}(1+|\xi|)^N,\label{1}\tag{1} $$ where the constants $C_N$ are positives, because $\hat{u}$ maps $\mathscr{S}(\mathbb{R}^{n})$ into itself.

But I can't understand why this converse is true:

If $u\in \mathscr{E}'(\mathbb{R}^{n})$ is a compactly support distribution satisfying \eqref{1}, then by Fourier inversion formula it implies $\hat{u}$ is smooth. That's my first question, I don't know how to use the inversion formula in order to do that.

Moreover since $u\in \mathscr{E}'(\mathbb{R}^{n})$ we can conclude $u\in \mathcal{C}_{0}^{\infty}(\mathbb{R}^{n})$. Why can we concude this last assertion?

Many thanks in advance.

$\endgroup$
  • $\begingroup$ What do textbooks say about it? $\endgroup$ – user64494 Jan 30 at 18:39

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.