Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $L^1(\Bbb R)$.

share|improve this question

3 Answers 3

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.

share|improve this answer

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.

share|improve this answer
    
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.

share|improve this answer
    
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
4  
This is still not right, a Gaussian convolved with itself gives a different Gaussian. –  Christopher A. Wong Oct 17 '12 at 8:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.