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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.

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    If $f$ belongs to $L^1$, then $\hat{f}$ is a continuous function, and by your identity, it can only takes three values : $0,-1, 1$. Conclusion ? Then use the fact that $\hat{f}$ goes to $0$ at infinity.2012-08-26
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    For 2), take a convenable function $g$ in $L^2$ which satisfies your identity $g = g.g$ and consider its inverse by Fourier transform.2012-08-26
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    @Ahriman, we cannot take the value $-1$.2012-08-26
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    Yes, I dunno why I say $-1$ ... Anyway, it doesn't change anything.2012-08-26
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    For (2), now the continuity restriction on $\hat{f}$ disappears and that we can choose the value of $\hat{f}$ piecewise. For example, if we let $\hat{f} = \chi_{[-1/2,1/2]}$, then $$f(x) = \frac{\sin \pi x}{\pi x}. $$2012-08-26
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    You know, this is a homework problem, so maybe you should leave a little for him to do himself.2012-08-26

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