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I'd like to prove that Fourier coefficient of two continuous and periodic functions, with a period of $1$, the following equality holds:

$$\widehat {f*g}(n)=\hat f(n)\cdot \hat g(n)$$

( $\hat x(n)$- Fourier coefficient of $x$, and $y*z$- Convolution of $y$ and $z$)

If I knew that $f, g$ are continually differentiable I could have used the fact that Fourier series $f,g$ converge to the actual functions, and I could use it to prove the claim. How can I prove it with these conditions and without using the order of the integrals, cause I am able to prove with that, but I am not allowed to.

Thanks a lot

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Write down the definition for the $n$th Fourier coefficient for $f*g$; then switch the order of integration. The inner integral can be seen to be the Fourier coefficient of $g$, then the outer integral gives the Fourier coefficient of $f$. –  David Mitra Jan 8 '12 at 10:16
    
Possible duplicate of math.stackexchange.com/q/96913/15941 ? –  Dilip Sarwate Jan 8 '12 at 15:33
    
as I wrote down I'm searching for a solution which does not involve the switching the order of integration,Thanks :) –  Jozef Jan 8 '12 at 16:54

1 Answer 1

up vote 4 down vote accepted

We have $$\eqalign{ (\widehat{f\star g })(n) &={1\over T}\int_{-T/2}^{T/2} (f\star g)(t) e^{-in\omega t}\,dt\cr &={1\over T}\int_{-T/2}^{T/2} \int_{-T/2}^{T/2} f(u)g(t-u)\,du \ e^{-in\omega t}\,dt \cr &={1\over T}\int_{-T/2}^{T/2} \int_{-T/2}^{T/2} f(u)g(t-u) \ e^{-in\omega t} \,dt\,\,du \cr &={1\over T}\int_{-T/2}^{T/2}\Bigl[ \int_{-T/2}^{T/2} g(t-u) e^{-in\omega (t-u)}\,dt \Bigr] e^{-in\omega u}f(u) \,du \cr &=T\cdot {1\over T} \Bigl[ \int_{-T/2}^{T/2} g(t-u) e^{-in\omega (t-u)}\,dt \Bigr] {1\over T} \Bigl[\int_{-T/2}^{T/2}e^{-in\omega u}f(u) \,du \Bigr]\cr &=T\cdot\hat f(n)\hat g(n). } $$

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What does $T$ stands for? I mean why did you chose to integrate between $ -T / 2 $ to $ T / 2 $? –  Jozef Jan 8 '12 at 17:04
    
@Josef Sorry, the period of the function. –  David Mitra Jan 8 '12 at 17:04
    
So basically you showed that it works for every $T$? –  Jozef Jan 8 '12 at 17:06
    
@Josef Yes but notice the factor of $T$ at the end... –  David Mitra Jan 8 '12 at 17:10
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@DavidMitra: Could you please comment on why switching the order of integration is allowed here? –  Leo Jun 12 '13 at 1:03

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