Let $\mathcal H$ be a (separable) Hilbert space on $\mathbb C$.
Suppose $f\in L^2([-\pi,\pi), \mathcal H)$ and $c_n(f)$ is n-th Fourier coefficient of $f$, i.e., $c_n(f)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{-in x}\ f(x)dx.$
Then, I want to see why $$\sum_{n\in\mathbb Z}\|c_n(f)\|_{\mathcal H}^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}\|f(x)\|_{\mathcal H}^2dx.$$
(This is parallel with the Parseval's identity for complex-valued functions.)
My one idea is using Parseval for complex valued fct.
Let $\tilde f(x):=\|f(x)\|.$ Use Parseval to $\tilde f$, I get $$\sum_{n}|c_n(\tilde f)|^2=\frac{1}{2\pi}\int_{-\pi}^\pi |\tilde f(x)|^2dx.$$ RHS=$\int_{-\pi}^\pi\|f(x)\|^2dx$ but I cannot see how I should handle LHS. If $|c_n(\tilde f)|=\|c_n(f)\|$ holds, I will come to conclusion but I cannot see $|c_n(\tilde f)|=\|c_n(f)\|$.
So parhaps this method doesn't work.
I would like you tell me how I can get the Parseval identity : $\sum_{n\in\mathbb Z}\|c_n(f)\|_{\mathcal H}^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}\|f(x)\|_{\mathcal H}^2dx$, or tell me any effective reference.
Let $\{v_k\}_{k=1}^\infty$ be an orthonormal basis in $\mathcal{H}.$ For $f\in L^2([-\pi,\pi),\mathcal{H})$ consider the functionz $f_k$ defined by $$f_k(x)=\langle f(x),v_k\rangle,\ -\pi\le x<\pi$$ Then $f_k\in L^2([-\pi,\pi)).$ Moreover $c_n(f_k)=\langle c_n(f),v_k\rangle.$ By the Parseval identity we have $$\sum_{n=-\infty}^\infty|\langle c_n(f),v_k\rangle|^2=\sum_{n=-\infty}^\infty |c_n(f_k)|^2\\ ={1\over 2\pi}\int\limits_{-\pi}^\pi |f_k(x)|^2\,dx={1\over 2\pi}\int\limits_{-\pi}^\pi |\langle f(x),v_k\rangle |^2\,dx$$ Thus, by the Parseval identity in $\mathcal{H}$ we get $$\sum_{n=-\infty}^\infty\|c_n(f)\|^2=\sum_{n=-\infty}^\infty\sum_{k=1}^\infty |\langle c_n(f),v_k\rangle |^2\\ =\sum_{k=1}^\infty {1\over 2\pi}\int\limits_{-\pi}^\pi |\langle f(x),v_k\rangle |^2\,dx = {1\over 2\pi}\int\limits_{-\pi}^\pi \|f(x)\|^2\,dx $$