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?
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!
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 $$a_k=c_k+c_{-k},\qquad b_k=i(c_k-c_{-k})\qquad(k\geq0)\ .$$
