if convolution of $f$ with itself remains same, then $f=0$ a.e?

Question

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 $L^1(\Bbb R)$.

Answer 1

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.

Answer 2

A Gaussian has the property that it is equal to a convolution with itself.

Answer 3

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.