Undergrad Fourier coefficients of convolution

[psie]

TL;DR

I'm trying to verify to myself that the Fourier coefficients of a convolution are the products of the coefficients of the convoluted functions, but I get stuck.

Let $h(x)=(f*g)(x)=\frac1{2\pi}\int_{-\pi}^\pi f(x-y)g(y)dy$ be the convolution. Then its Fourier coefficients are given by $$ {1\over2\pi}\int_{-\pi}^\pi (f*g)(x)e^{-inx}dx={1\over4\pi^2}\int_{-\pi}^\pi\left(\int_{-\pi}^\pi f(x-y)g(y)dy\right)e^{-inx}\ dx\ . $$
Changing the order of integration, we get $${1\over4\pi^2}\int_{-\pi}^\pi g(y) \left(\int_{-\pi}^\pi f(x-y)e^{-inx} dx\right)\,dy\ .$$ Now here I'd like to do the substitution $t=x-y$ in the inner integral, but this makes the limits of integration depend on $y$, which I do not want. How can I go about this issue?

EDIT: I know that $h(t)$ is periodic with period $2\pi$. I don't know if this can be helpful in any way.

 

[psie]

I guess, after the substitution $t=x-y$, we get $${1\over4\pi^2}\int_{-\pi}^\pi g(y)e^{-iny} \left(\int_{-\pi-y}^{\pi-y} f(t)e^{-int} dt\right)dy .$$ So here the interval $[-\pi-y,\pi-y]$ is still an interval over a whole period, so we can safely replace it $[-\pi,\pi]$ since $y$ is kept constant in the inner integral anyway. Therefor we get $${1\over4\pi^2}\int_{-\pi}^\pi g(y) e^{-iny}\left(\int_{-\pi}^\pi f(t) e^{-int} dt\right)\,dy\ ,$$ which proves the result.