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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 $L^1(\Bbb R)$.

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up vote 5 down vote accepted

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.

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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.

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I have merged your "blaber" account into this one. Try registering your account to prevent it "splitting" like the present case. – Willie Wong Oct 17 '12 at 9:07

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

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We use the L^2 condition to ensure that the Fourier transform of g exists, and the L^1 condition to ensure that the Fourier transform of f exists. If either of these is violated probably you can construct counter-examples... – ajja Oct 17 '12 at 8:13
I don't understand this answer. You claim that $f = 0$ in a proof that concludes with $f = 0$. – Christopher A. Wong Oct 17 '12 at 8:21
Sorry it was a typo... It should have been "since g = 0" – ajja Oct 17 '12 at 8:25
Why is $g = 0$? Isn't the problem $f \star f = f$? – Christopher A. Wong Oct 17 '12 at 8:25
This is still not right, a Gaussian convolved with itself gives a different Gaussian. – Christopher A. Wong Oct 17 '12 at 8:34

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