Fourier series and convolution Let $f$ and $g$ be $2\pi$-periodic, piece-wise smooth functions having Fourier series $f(x)=\sum_n\alpha_ne^{inx}$ and $g(x)=\sum_n\beta_ne^{inx}$, and define the convolution of $f$ and $g$ to be $f\star g(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)g(x-t)dt$. Show that the complex form of the Fourier series for $f\star g$ is $$f\star g(x)=\sum_{n=-\infty}^{\infty}\alpha_n\beta_ne^{inx}.$$
I have been trying different approaches to this for a while but I haven't been able to figure it out. I think that I am supposed to use the fact that
$$\frac{1}{2\pi}\langle f,g\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt=\sum_{n=-\infty}^{\infty}\alpha_n\overline{\beta_n}$$
but I can't figure out how to manipulate the algebra to show the correct result. I may be forgetting some key property of the inner product or something.
 A: Let the Fourier Series representation for $f$ and $g$ be written respectively by
$$f(x)=\sum_{n=-\infty}^\infty \alpha_ne^{inx}$$
$$g(x)=\sum_{n=-\infty}^\infty \beta_ne^{inx}$$
with coefficients $\alpha_n$ and $\beta_n$ given by 
$$\alpha_n=\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{-int}\,dt$$
$$\beta_n=\frac1{2\pi}\int_{-\pi}^\pi g(t)e^{-int}\,dt$$
The convolution of $f$ and $g$ is defined as
$$\begin{align}
(f*g)(x)&\equiv \frac1{2\pi}\int_{-\pi}^\pi f(t)g(x-t)\,dt\\\\
&=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{n=-\infty}^\infty \beta_n e^{in(x-t)}\right)\,dt\\\\
&=\sum_{n=-\infty}^\infty \beta_n e^{inx}\left(\frac1{2\pi}\,\int_{-\pi}^\pi f(t)e^{-int}\,dt\right)\\\\
&=\sum_{n=-\infty}^\infty \alpha_n\,\beta_n e^{inx}
\end{align}$$
and we are done!
A: Since Dr. MV already posted an answer, I'll post the one I alluded to in my comment:
Convolving $f$ with $g$ gives
\begin{align}
(f*g)(x) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(\sum_m \alpha_m e^{imt}\right) \left(\sum_n \beta_n e^{in(x-t)}\right)\,dt \\
&= \frac{1}{2\pi} \sum_m\sum_n \alpha_m\beta_n e^{inx} \int_{-\pi}^{\pi} e^{i(m-n)t}\,dt.
\end{align}
This integral evaluates to $2\pi$ if $m=n$ and $0$ otherwise. (Check this yourself by breaking it into these two cases.) Meaning that the double sum is zero if $m\neq n$ so really you can drop down to one sum to get
$$(f*g)(x) = \sum_n \alpha_n \beta_n e^{inx}.$$
