If $\int_0^1\lvert f\rvert^2=\sum\limits_{n\in\mathbf Z}\lvert\hat f_n\rvert^2$ then how can I derive $\int_0^1f\bar g =\sum\limits_{n\in\mathbf Z}\hat f_n\overline{\hat g_n}$
$\hat f_n$ is the fourier transform of $f$ and I try to show the parseval identity, the problem is, one does not know if the function admits a fourier series, so I have to derive it only from the knowledge that LHS holds
I know normally the converse is true (Identitiy on the right implies the one on the left) but is there a trick to show the other direction If I take for example $f+g$ then
$\int_0^1\lvert f+g\rvert^2=\int_0^1\lvert f\rvert^2+\int_0^1\lvert g\rvert^2+\int_0^1f\bar g+\int_0^1g\bar f$
$\sum\limits_{n\in\mathbf Z}\lvert\hat f_n+\hat g_n\rvert^2=\sum\limits_{n\in\mathbf Z}\lvert\hat f_n\rvert^2+\sum\limits_{n\in\mathbf Z}\lvert\hat g_n\rvert^2+\sum\limits_{n\in\mathbf Z}\hat f_n\overline{\hat g_n}+\sum\limits_{n\in\mathbf Z}\overline{\hat f_n}\hat g_n$
At least I have
$\int_0^1f\bar g+\int_0^1g\bar f=\sum\limits_{n\in\mathbf Z}\hat f_n\overline{\hat g_n}+\sum\limits_{n\in\mathbf Z}\overline{\hat f_n}\hat g_n$