Is it true that the Fourier coefficient of convolution is the product of the coefficients?

Question

what I mean by the title is the following: if we define the convolution between two $2\pi$-periodic, $C^1$ functions as $f*g(x) = (2\pi)^{-1}\int_{-\pi}^\pi f(x-y)g(y)dy$, is it true that the Fourier coefficients of $f*g(x)$ equal $\hat{f}(n)\hat{g}(n)$ where $\hat{f}(n),\hat{g}(n)$ are the Fourier coefficients of $f,g$ respectively?

I think I have a correct proof of this, but I am just unsure, but the result seems so beautiful, so I wanted to hear your expert opinions. Please do not post the proof of this...

Thanks

Answer 1

It is true that Fourier coefficient of convolution of two functions is the product of Fourier coefficients of the individual functions. This is called convolution theorem in Fourier theory . You may be interested to look at http://en.wikipedia.org/wiki/Convolution_theorem.

Answer 2

Given the assumed smoothness of $f$ and $g$ this is a simple exercise about double integrals. I write $\int_T$ for the integral over a full period. Then $$\widehat{f*g}(n)={1\over2\pi}\int_T f*g(x)\>e^{-inx}\>dx={1\over4\pi^2}\int_T\left(\int_Tf(x-y)g(y)\>dy\right)e^{-inx}\ dx\ .$$ Interchanging the order of integration gives $$\widehat{f*g}(n)={1\over4\pi^2}\int_T g(y) \left(\int_Tf(x-y)\>e^{-inx}\> dx\right)\,dy\ .$$ In the inner integral we substitute $x:=y+t$ $(t\in T)$ and obtain $$\widehat{f*g}(n)={1\over4\pi^2}\int_T g(y) \left(\int_Tf(t)\>e^{-in(y+t)}\> dt\right)\,dy={1\over4\pi^2}\int_T g(y) e^{-iny}\left(\int_Tf(t)\>e^{-int}\> dt\right)\,dy\ .$$ On the right hand side the inner integral is independent of $y$ and essentially $=\hat f(n)$. We therefore can write $$\widehat{f*g}(n)=\hat f(n)\>{1\over2\pi}\int_T g(y) e^{-iny}\>dy=\hat f(n)\cdot\hat g(n)\ .$$