Simplify $\langle f, S_{n}(f) \rangle$

Question

I am trying to simplify $$\langle f, S_{n}(f)\rangle$$ where $S_{n}(f)$ is the Fourier partial sum of $f$, and $$\langle f,g\rangle=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)g(x)\,\mathrm dx$$ I believe $\langle f, S_{n}(f)\rangle$ $=$ $||S_{n}(f)||_{2}^{2},$ where $||f||_{2}=\Big(\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)^{2}\mathrm dx\Big)^{\frac{1}{2}}$however, have not been able to show this. Any help is appreciated.

Answer 1

Let $f = S_n(f) + T_n(f)$, where we use $T$ to be the Fourier tail after the cutoff at $n$ terms. You should be able to show $\langle S_n(f), T_n(f) \rangle = 0$ (since the basis elements for the $T_n$ part are not in the span of the first $n$ basis elements).

\begin{align*} \langle f, S_n(f) \rangle &= \langle S_n(f) + T_n(f), S_n(f) \rangle \\ &= \langle S_n(f), S_n(f) \rangle + \langle T_n(f), S_n(f) \rangle \\ &= \dots \end{align*}

Answer 2

$\langle f, S_n(f) \rangle = \sum_n \overline{f_n}\langle f, e_n \rangle = \sum_n |f_n|^2$.