Fourier transform and differentiability Let $f $ be $L^1 (R) $, and $x\mathcal {F}(f)(x) $ be also $L^1 (R) $. Prove that there exists $g $ differentiable which holds $f=g $ almost everywhere.
I know that has some relation with the fact that the Fourier transform of $f'$ is $L^1$, but I dont see how to conclude.

HINT:
It is convenient to work with tempered distributions. The inverse Fourier transform of $x \mathcal{F}(f)$ is the distributional derivative of $f$. However, this is also the Fourier transform of a function from $L^1$, so a function $h$ continuous and vanishing at infinity. We conclude that $f'=h$ in distribution sense, hence also classically. 
