Find coefficients of $F(x) = \frac{1}{\pi} \int_{-\pi}^{\pi}f(t)\,f(x+t)\,dt$ with periodic $f$. 
Consider $f(x)$ is continuous function and it has a period $2\pi$ and has Fourier transform 
  $$f(x) = \frac{a_{0}}{2} + \sum_{n>0} a_{n}\cos(nx)+b_{n}\sin(nx)$$
  Now consider $\displaystyle F(x) = \frac{1}{\pi} \int_{-\pi}^{\pi}f(t)\,f(x+t)\,dt$. Show that it could be represented as a Fourier series. 

My attempt was considering some special cases (for example $F(0)$ and that gives us Parseval's equation). But I guess using this function I can prove the latter. 
Any ideas?
 A: Hint:
Do it with simplifying
$$f(t)f(x+t)= \left(\frac{a_{0}}{2} + \sum_{n>0} a_{n}\cos(nt)+b_{n}\sin(nt)\right) \left(\frac{a_{0}}{2} + \sum_{n>0} a_{n}\cos(nx+nt)+b_{n}\sin(nx+nt)\right)$$
and integration on terms of 
$$a_{0},~\cos(nt),~\sin(nt)$$
Update:
Exponential form is better for discussion, but with series we have
\begin{align}
f(t)f(x+t)
&= \left(\frac{a_{0}}{2} + \sum_{m>0} a_m\cos mt+b_m\sin mt\right) \left(\frac{a_{0}}{2} + \sum_{n>0} \{a_n\cos nx+b_n\sin nx\}\cos nt + \{-a_n\sin nx+b_n\cos nx\}\sin nt \right)\\
&= \left(\frac{a_{0}}{2}\right)^2\\
&+\frac{a_{0}}{2} \left(\sum_{n>0} \{a_n+a_n\cos nx+b_n\sin nx\}\cos nt + \{b_n-a_n\sin nx+b_n\cos nx\}\sin nt \right)\\
&+\sum_{m>0}\sum_{n>0}\left(a_m\{a_n\cos nx+b_n\sin nx\} \cos mt\cos nt\right)\\
&+\sum_{m>0}\sum_{n>0}\left(a_m\{-a_n\sin nx+b_n\cos nx\}\ \cos mt\sin nt\right)\\
&+\sum_{m>0}\sum_{n>0}\left(b_m\{a_n\cos nx+b_n\sin nx\} \sin mt\cos nt\right)\\
&+\sum_{m>0}\sum_{n>0}\left(b_m\{-a_n\sin nx+b_n\cos nx\}\ \sin mt\sin nt\right)
\end{align}
using these formulas
\begin{eqnarray*}
&& \int_{-L}^{L}\sin\frac{m\pi x}{L}\cdot\sin\frac{n\pi x}{L}\cdot dx=\left\lbrace\begin{array}{c l}L&m=n,\\0&m\neq n.\end{array}\right. \\
&& \int_{-L}^{L}\cos\frac{m\pi x}{L}\cdot\cos\frac{n\pi x}{L}\cdot dx=\left\lbrace\begin{array}{c l}L&m=n,\\0&m\neq n.\end{array}\right. \\
&& \int_{-L}^{L}\cos\frac{m\pi x}{L}\cdot\sin\frac{n\pi x}{L}\cdot dx=0    ~~~~,~~~~\forall m,n   \\
\end{eqnarray*}
we cancel many terms during integration on $[-\pi,\pi]$, remains
\begin{align}
\int_{-\pi}^{\pi}f(t)f(x+t)
&= \int_{-\pi}^{\pi}\left(\frac{a_{0}}{2}\right)^2\\
&+\int_{-\pi}^{\pi}\frac{a_{0}}{2} \left(\sum_{n>0} \{a_n+a_n\cos nx+b_n\sin nx\}\cos nt + \{b_n-a_n\sin nx+b_n\cos nx\}\sin nt \right)\\
&+2\pi\sum_{n>0}\left(a_n^2+b_n^2\right)\cos nt\\
&+\cdots\sin nt
\end{align}
and you find the series!
A: I shall work wit the complex Fourier coefficients. For $n\in {\mathbb Z}$ we have
$$C_n={1\over2\pi}\int_TF(x)e^{-inx}\>dx={1\over2\pi^2}\int_T f(t)\left(\int_T f(x+t)e^{-inx}\>dx\right)\>dt\ .$$
Here the inner integral is equal to
$$e^{int}\int_T f(x+t)e^{-in(x+t)}\>dx=e^{int}\cdot2\pi c_n\ ,$$
where $c_n$ denotes the $n$th complex Fourier coefficient of $f$. It follows that
$$C_n={2\pi c_n\over 2\pi^2}\int_T f(t)e^{int}\>dt={2\pi c_n\over 2\pi^2}\cdot2\pi c_{-n}=2c_nc_{-n}\ .$$
This can easily be converted to a formula in terms of $A_k$, $B_k$, $a_k$, $b_k$ using
$$a_k=c_k+c_{-k},\qquad b_k=i(c_k-c_{-k})\qquad(k\geq0)\ .$$
