1
$\begingroup$

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

  • 0
    as I wrote down I'm searching for a solution which does not involve the switching the order of integration,Than$k$s :)2012-01-08

1 Answers 1

4

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). } $

  • 1
    @DavidMitra: Could you please comment on why switching the order of integration is allowed here?2013-06-12