I know how to prove that if $f,g \in \mathcal{S}$ (Schwartz space - or even ${L}^2$), then $$f*g \in \mathcal{S}$$ and $$\|f\|_2^2=\|\widehat{f}\|_2^2.$$ Here, $$\|f\|_2=\sqrt{<f,f>}= \left[ \int_{-\infty}^{+\infty}|f(x)|^2 dx\right]^{\frac{1}{2}}$$ and the identity is called Parseval's identity.
And I know that if $$S[f]= \frac{a_0}{2}+ \sum_{n=1}^{\infty} (a_n f_n(x)+b_ng_n(x)),$$ is the Fourier serie of $f$, in which $f_n(x)=\cos\frac{n \pi x}{L}$ and $g_n(x)=\sin\frac{n \pi x}{L}$, we have $$\frac{1}{2}a_0^2 + \sum_{n=1}^\infty (a_n^2 + b_n^2)= \frac{1}{L} \int_{-L}^L |f(x)|^2 dx$$ that is called Parseval's identity too.
I would like to know what is the connection between these two Parseval identities.
It is obvious that the norm of the Fourier Transform ends up having a connection with the coefficients of the Fourier series of $f$. Would it be just that?
Thaks a lot!
Cleto