Find function $f(x)$ so $x\hat{f}(x)\in L^1(\mathbb{R})$

Question

Suppose we have $f\in L^1(\mathbb{R})$ so $\xi\mapsto \xi\hat{f}(\xi)$ is in $L^1(\mathbb{R})$, where $\hat{f}$ is the Fourier Transform for the function $f$. I'm trying to show that there exists some $g\in C^1(\mathbb{R})$ so $g=f$ almost everywhere. I've already shown that the condition implies $\hat{f}$ is in $ L^1(\mathbb{R})$. I've tried using this to perform the inverse Fourier transform on $\text{id}\cdot f$, but so far haven't achieved anything.

Answer 1

Hi (too long for a comment). Since $\hat f\in L^1$, then $g:={(\hat f)}\check\space$ is continuous and goes to zero at infinity (you probably know this from your book). By the Fourier inversion formula (you can apply it since $f$ and $\hat f$ are in $L^1$), $g=f$ a.e. So this is the reason you need to be careful about a.e. equality.

So, $g$ is crearly the right candidate. Now to conclude, you need to use one of the properties of the Fourier transform by which $\widehat{(\partial_x f)}=i\xi \hat f$. If you are not allowed to do this (I suspect this is part of the proof of that property), then you need to prove that the function is $C^1$ by hand.

In principle, note that $h:={(i\xi \hat f)}\check\space$ is well-defined and continuous since $\xi\hat f\in L^1$, and this should be the ansatz for the derivative of $g$.

So, fix a point $x_0\in\mathbb R$; you need to formally differentiate the function $g$ (which remember is defined using the Fourier inversion formula) at the point $x_0$, so you need to prove that the derivative commutes with an integral, i.e., $$ \frac{d}{dx}g(x_0)=\frac{d}{dx}\left.\int e^{ix\xi}\hat f(\xi)d\xi\right|_{x=x_0}\overset{?}{=} \left.\int\left[ \frac{d}{dx}e^{ix\xi}\hat f(\xi)\right]\right|_{x=x_0}d\xi=\left.\int\left[ i\xi e^{ix\xi}\hat f(\xi)\right]\right|_{x=x_0}d\xi=h(x_0). $$ To prove the "?" equality, you need to use one of the theorems from the real analysis course that let you commute a derivative with an integral. Can you continue the proof?

I am afraid I am missing a lot of $2\pi$ factors in the above formulas, but I hope it is still clear