convolution of a function with itself equals itself

Question

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.

Answer 1

Hints: For (1), you're on the right track.. also use that $\hat{f}$ is continuous. For (2), try to define $\hat{f}$ instead of $f$. So you need a nonzero $L^2$ function equal to its square....

Answer 2

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.