If $f\in L^1(\mathbb{R})$ and $M>0$ is it true that $\mathcal{F}\left( \chi_{[-M,M]}\mathcal{F}(f) \right)\in L^1(\mathbb{R})$?

Question

If $f\in L^1(\mathbb{R})$ and $M>0$ is it true that $\mathcal{F}\left( \chi_{[-M,M]}\mathcal{F}(f) \right)\in L^1(\mathbb{R})$, where $\mathcal{F}$ is the Fourier transform? It is pretty clear that $f\in L^p(\mathbb{R})$ for every $p>1$ (e.g. via Young's convolution inequality) but I can't see if the same holds true for $p=1$.

Any suggestion?

Answer 1

Fix a function $g$ from Schwarz class such that $g = 1$ on $|x| \leq 1$, and define $f$ to be the inverse Fourier transform of $g$, namely $f = \mathcal{F}^{-1}(g)$. Then $f$ is again from Schwarz class, in particular from $L^1(\mathbb{R})$, and taking $M=1$ we get $$ \mathcal{F}(\chi_{[-1,1]}\mathcal{F}(f)) = \mathcal{F}(\chi_{[-1,1]}g) = \mathcal{F}(\chi_{[-1,1]}) $$ which is not $L^1$.