# Fourier inversion formula for compactly supported distributions

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.

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