# convolution of a function with itself equals itself

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. –  Ahriman Aug 26 '12 at 15:49
For 2), take a convenable function $g$ in $L^2$ which satisfies your identity $g = g.g$ and consider its inverse by Fourier transform. –  Ahriman Aug 26 '12 at 15:51
@Ahriman, we cannot take the value $-1$. –  sos440 Aug 26 '12 at 15:52
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 $$f(x) = \frac{\sin \pi x}{\pi x}.$$ –  sos440 Aug 26 '12 at 15:52
You know, this is a homework problem, so maybe you should leave a little for him to do himself. –  Zarrax Aug 26 '12 at 16:02
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Hints: For (1), you're on the right track.. also use that $\hat{f}$ is continuous. For (2), try to define $\hat{f}$ instead of $f$. So you need a nonzero $L^2$ function equal to its square....

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Hi , I've added some working above , still need more help though.. –  studenthp Aug 26 '12 at 18:11

For $(i)$, $\hat{f}\cdot\hat{f}=\hat{f}$ implies that $\hat{f}(\xi) \in \{ 0,1 \}$ for all $\xi$.

Now use the fact that $\hat{f}$ is continuous.

For $2$ try to solve the problem backwards. Try to find some $g \in L^2(\mathbb{R})$ so that $g(x) \in \{ 0,1 \}$ and whose FT is real valued.

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For part (2) , now my thinking is : $f(x) =\int\hat{f(w)}e^{2\pi iwx} = lim_{B\to \infty}\int_{-B}^B\hat{f(w)}e^{2\pi iwx}$ . After a few lines of change order of integral , this is equal to $lim_{B\to \infty}\int_Rf(t)\frac{sin(B2\pi(x-t))}{\pi(x-t)}dt$ . –  studenthp Aug 26 '12 at 18:03
Now If I can put the limit under the integral and show $lim_{B\to \infty}\frac{sin(B2\pi(x-t))}{\pi(x-t)}$ is in both L1(R) and L2(R), then by replacing $f(x)$ by $lim_{B\to \infty}\frac{sin(B2\pi(x-t))}{\pi(x-t)}$, can I get the conclusion ? The reason that I need to show the function is in L1(R) is because to get the first equality , I assumed $f(x)$ is absolutely integrable. –  studenthp Aug 26 '12 at 18:04
I cant see the relation between $\hat{f}$ is continuous and f is zero function .. –  studenthp Aug 26 '12 at 18:07
Can a continuous function on $\mathbb{R}$ take only 2 values? –  N. S. Aug 26 '12 at 18:21
sorry , what I want to ask is : how do you get $\hat{f}$ goes to zero at infinity ? and why fourier transform maps a L1 funciton to a contiuous one ? is that some propositon ? Is it possible for you to show the solution ? –  studenthp Aug 26 '12 at 18:31
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