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.

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

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