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 $L^1(\Bbb R)$.

Blaber here. Since $f \star f = f$ both $f$ and $f \star f$ are in $L^1$, therefore we can take their Fourier transform getting
$$
\hat{f}^2 = \hat{f}
$$
The function $\hat{f}(x)$ is continuous, and by the above relation $\hat{f}(x)$
can be only either $1$ or $0$ for all $x$. By the Riemann-Lebesgue lemma we know that $\hat{f}(x) \rightarrow 0$ as $x \rightarrow 0$. Therefore $\hat{f}(x) = 0$
for all $x$. By Fourier inversion it follows that $f = 0$ a.e.
