So I am stuck in the middle of this proof:
Let $f,g \in E[-\pi,\pi]$ be periodic functions with periods of $2\pi$, where
$f(x)$ ~ $\sum_{n=-\inf}^{\inf}a_ne^{inx}$
$g(x)$ ~ $\sum_{n=-\inf}^{\inf}b_ne^{inx}$
$\forall x\in R: h(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-t)g(t)dt$ | $h(x)$ ~ $\sum_{n=-\inf}^{\inf}c_ne^{inx}$
Prove that $c_n=a_nb_n$
($\forall n \in Z$)
Here is what I've got so far:
By definition, $c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}h(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-t)g(t)dte^{-inx}dx$
$c_n=\frac{1}{(2\pi)^2}\int_{-\pi}^{\pi}(\int_{-\pi}^{\pi}f(x-t)e^{-inx}dx)g(t)dt$
However, in order to finish the proof, I need to, presumably using periodicity, say that
$c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx \cdot \frac{1}{2\pi}\int_{-\pi}^{\pi}g(t)dt = a_nb_n$
Any help would be appreciated!
Almost there
\begin{eqnarray} c_n&=&\frac{1}{2\pi}\int_{-\pi}^{\pi}h(x)e^{-inx}dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left\{\frac{1}{2\pi}\int_{-\pi}^{\pi}\color{blue}{f(x-t)}\color{red}{g(t)}dt\right\}~e^{-inx}~dx \\ &=& \frac{1}{(2\pi)^2} \int_{-\pi}^{\pi}\int_{-\pi}^{\pi} \color{blue}{\sum_{k} a_ke^{ik (x-t)}}\color{red}{\sum_l b_l e^{il t}}e^{-inx} ~dx dt \\ &=& \frac{1}{(2\pi)^2} \int_{-\pi}^{\pi}\int_{-\pi}^{\pi} \sum_{k,l} a_k b_l e^{ikx}e^{-i(k-l)t}dt dx \\ &=& \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{k,l} a_k b_l e^{-i(n - k)x} \color{orange}{\left\{\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-i(k-l)t}dt\right\}} dx \\ &=& \frac{1}{2\pi} \int_{-\pi}^{\pi} \sum_{k,l} a_k b_l e^{-i(n - k)x} \color{orange}{\delta_{kl}} dx \\ &=& \frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_k a_kb_k e^{-i(n - k)x}dx \\ &=&\sum_k a_k b_k \color{magenta}{\left\{\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-i(n-k)t}dx\right\}} \\ &=& \sum_k a_k b_k \color{magenta}{\delta_{nk}} = a_n b_n \end{eqnarray}