TITLE: convolution of a function with itself equals itself
QUESTION [9 upvotes]: In a homework question,  I was asked to show: (1) in $L^1(R)$, if $f*f = f$, then $f$ must be a zero function. (2) In $L^2(R)$, find a function $f*f=f$. I don't know how to proceed. 
for (1), $f*f=f$ gives $\widehat{f*f}=\hat{f}$, which is equal to $\hat{f}\cdot\hat{f}=\hat{f}$, but this does not guarantee the result. I tried to prove by contradiction, no success. 
for (2), I don't know where to proceed. Is there any help that I could get? Thanks.

REPLY [3 votes]: For $(i)$, $\hat{f}\cdot\hat{f}=\hat{f}$ implies that $\hat{f}(\xi) \in \{ 0,1 \}$ for all $\xi$.
Now use the fact that $\hat{f}$ is continuous. 
For $2$ try to solve the problem backwards. Try to find some $g \in L^2(\mathbb{R})$ so that
$g(x) \in \{ 0,1 \}$ and whose FT is real valued.