I was studying Fourier transform in one of my classes and I came across the following question
Let $f \in L^1$ such that $\hat{f}$ has compact support. Is it true that $f \in C^{\infty}$?
By inversion formula $f=(\hat{f}\check{)}$ almost everywhere. On the other hand, since the inverse Fourier transform maps functions with compact support to smooth functions, $(\hat{f}\check{)}\in C^{\infty}$. Hence $f$ is smooth almost everywhere. I dont know how to proceed from here nor if the statement is in fact true. I appreciate any help.
Best Answer
You have almost finished the argument. We cannot say that $f$ itself is smooth because the hypothesis holds if you replace it by any function equal to $f$ almost everywhere. The inversion formula tells you that there exists a smooth function (namely $\hat{f}\check{}$) which is equal to $f$ almost everywhere.