Proving parseval identity for trigonometric polynomials

Question

Show that $$||P||_2^2=\sum_{k=-N}^N \langle P,e_k\rangle^2,$$ where $e_k$ are the Fourier basic functions, and $P$ is a trigonometric polynomial of degree $N.$

I am not sure how to link trigonometric polynomial with Fourier basic function. I know that fourier basic function has a similar form to trigonometric polynomial of degree N. However, fourier series is a summation from 1 to $\infty$, while trigonometric polynomial is a summation of 1 to N. How do I proceed from there?

Answer 1

$ \begin{align*} ||P||_2^2&= ||\sum^{k=N}_{k=-N}\langle P, e_k \rangle e_k||^2\\ &=\langle \sum^{k=N}_{k=-N}\langle P, e_k \rangle e_k, \sum^{k=N}_{k=-N}\langle P, e_k \rangle e_k\rangle\\ &=\sum^{k=N}_{k=-N}\langle P, e_k \rangle^2 \cdot 1 \hspace{2mm}\text{ Since }\langle e_k, e_j\rangle = \begin{cases}1 \text{ when } k=j \\ 0 \text{ otherwise}\end{cases}\\ &=\sum^{k=N}_{k=-N}\langle P, e_k \rangle^2 \end{align*}$

Answer 2

For any square integrable periodic function $f$ we have $\|f\|_2^{2}=\sum\limits_{k=-\infty}^{\infty} |\langle f, e_k \rangle|^{2}$. For a trigonometric polynomials $f$ of dergree $N$ the coefficients $\langle f, e_k \rangle$ are $0$ for $|k| >N$