I have seen a proof of the claim that the Fourier coefficients of $f\ast g$ equals $\displaystyle \hat{f(n)}\cdot{\hat{g(n)}}$.
I can't understand a step of it.
The proof goes like this:
Let $f,g$ be $2\pi$ periodic functions that are continuous almost everywhere.
Let the Fourier series of $f,g$ be $\displaystyle \sum_{n=-\infty}^{\infty}c_{1,n}e^{inx} \ , \ \sum_{n=-\infty}^{\infty}c_{2,n}e^{inx}$, respectively.
Now, if $c_n$ are Fourier coefficients of $f\ast g$, then \begin{align*}c_n&=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}(f\ast g) \cdot e^{inx}\text{d}x=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(t)g(x-t)\text{d}t\cdot e^{inx}\text{d}x \\ &=\frac{1}{2\pi} \int\limits_{-\pi}^{\pi}f(t)\underbrace{\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}g(x-t)e^{in(t-x)}\text{d}x}_{c_{2,n}}\cdot{e^{int}}\text{d}t \\ &=c_{2,n}\underbrace{\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(t)e^{-int}}_{c_{1,n}}=c_{1,n}c_{2,n}\end{align*}I can't understand why we can change order of integration as done.
Please help me, thanks.