Let $f\in L^1(\mathbb{R})$ and continuous on $\mathbb{R}$ such that its Fourier transform $\hat f$ equals zero in a neighborhood of zero.


Let $F$ be function such that $\hat F$ exists and


$$\hat f(x) =x\hat F(x),\quad \forall x\in \mathbb{R}$$


Prove that $F\in L^1(\mathbb{R})$.


Any hints on how to prove that?


I already asked [this question on MSE](https://math.stackexchange.com/questions/3646361/prove-that-f-in-l1-mathbbr?noredirect=1#comment7493238_3646361), I hope it will have a chance here



Integrability of a function under a condition on its Fourier transform
We have $\widehat{f}=0$ near zero, so we can write $\widehat{F}=\widehat{g}\widehat{f}$ with a function $\widehat{g}\in C^{\infty}$, $\widehat{g}(x)=1/x$ for $|x|\ge a>0$.


Since $\widehat{g}'' = -(t^2 g)\,\widehat{}\in L^1$, we have $|g(t)|\lesssim 1/t^2$. Moreover, $g\in L^2\subseteq L^1\_{\textrm{loc}}$. So $g\in L^1$, and thus also $F=g\*f\in L^1$, since $f\in L^1$ by assumption.


