Parseval's Theorem states that: If $$f(x)=\sum^\infty_{-\infty}c_ne^{inx}, g(x)=\sum^\infty_{-\infty}\alpha_ne^{inx}.$$ Then, $$\lim_{N\to\infty}\frac{1}{2\pi}\int^\pi_{-\pi}|f(x)-s_N(f;x)|^2dx=0,\frac{1}{2\pi}\int^\pi_{-\pi}f(x)g(x)dx=\sum^\infty_{-\infty}c_n\alpha_n.$$
where $f$ and $g$ are Riemann-integrable functions with a period of $2\pi$.
I don't really understand when it would hold, can somone prove this theorem for me?
Thanks!
Hints:
In general, when we have an orthonormal basis $\,\{u_i\}\,$ in a real (since this is, apparently, what you have) inner product linear space $\,V\,$, then
$$\forall\,v,w\in W\;,\;\;v=\sum_{n=1}^\infty a_nu_n\;,\;\;w=\sum_{n=1}^\infty b_nu_n\implies$$
$$\langle\,v\,,\,w\,\rangle=\left\langle\;\sum_{n=1}^\infty a_nu_n\,,\,\sum_{m=1}^\infty b_mu_m\;\right\rangle=\sum_{n=1}^\infty\sum_{m=1}^\infty a_nb_m\,\langle u_n\,,\,u_m\,\rangle =\sum_{n=1}^\infty\sum_{m=1}^\infty a_nb_m\delta_{nm}$$
and from here you get what you want. In case the space is a complex one you need the conjugate with the $\,b_n$'s and etc.
