if convolution of $f$ with itself remains same, then $f=0$ a.e? I'm trying to answer the question above.. But I'm not certain in either way. I tried to prove it by giving counter examples.. But it always failed.. Then i also tried to draw contradictions  But that's not successful as well.
Please give me some suggestion or ideas!
p.s I forgot the condition that  $f$ is in $L^1(\Bbb R)$.

Using the property $\widehat{f\star g}=\widehat f\widehat g$ for $f$ and $g$ integrable, we get $(\widehat f)²=\widehat f$, hence for all $x$, $\widehat f\in\{0,1\}$. By the dominated convergence theorem, $\widehat f$ is continuous, so either $\widehat f=1$ or $\widehat f=0$. By Riemann-Lebesgue lemma, $\widehat f(x)\to 0$ as $x\to +\infty$, so $f=0$ almost everywhere.
