Let $f \in L^1(\mathbb{R})$ and suppose that there is $g\in L^1(\mathbb{R})$ such that $f*g=f$. Can we conclude that the Fourier transform of $f$ has compact support?
I believe yes, in fact, the converse of this is true. A simple/elegant proof can be constructed by considering the approximate identity given by the de la Vallee-Poussin kernel. How can one proceed in the other direction? Any hints/help is appreciated!
Thanks :)!
True: $\hat f (t)=\hat {f*g}=\hat f \hat g$. By Riemann Lebesgue Lemma $|\hat g (t) | <1$ for $|t|$ sufficiently large. Hence, we get $\hat f (t)=0$ for $|t|$ sufficiently large.