Here is the problem:
Use the complex Fourier Series on $[-L,L] $ with complex coefficients to find a representation of
$\frac{1}{2L} \int_{-L}^{L} |f(x)|^{2} dx$
Here is my attempt:
The complex Fourier Series is represented as
$f(x) = \sum_{n=0}^{\infty} C_n e^{\frac{n \pi x}{L}}$
Initializing the absolute value function would imply
$|f(x)|^{2} = |\sum_{n=0}^{\infty} C_n e^{\frac{n \pi x}{L}}|^{2}$ $ \leq |\sum_{n=0}^{\infty} C_n|^{2}*|e^{\frac{n \pi x}{L}}|^{2}$ = $|\sum_{n=0}^{\infty} C_n|^{2}$
This is by Cauchy-Schwarz. Am I on the right track about this?
I appreciate all of your help.
Note that \begin{align*} \frac{1}{2L}\int_{-L}^L \mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}} \mathrm{e}^{\mathrm{i} \frac{-m \pi x}{L}} \mathrm{d} x = \delta_{n,m} \end{align*}
I.e. the basis functions $\mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}}$ are orthogonal.
So note that for $N$ finite, we have
\begin{align*} & \left \| \sum_{n = 0}^N C_n \mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}} \right \| \\ & = \sum_{n = 0}^N \vert C_n \vert^2 \end{align*} With the inner product defined above, we have
\begin{align*} & \frac{1}{2L} \int_{-L}^L \left \vert \sum_{n = 0} ^\infty C_n \mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}} \right \vert^2 \mathrm{d} x \\ & = \frac{1}{2L} \int_{-L}^L \left \vert \lim_{N \rightarrow \infty} \sum_{n = 0} ^N C_n \mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}} \right \vert^2 \mathrm{d} x \\ & =\left \| \lim_{N \rightarrow \infty} \sum_{n = 0} ^N C_n \mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}} \right \| \\ & = \lim_{N \rightarrow \infty} \left \| \sum_{n = 0} ^N C_n \mathrm{e}^{\mathrm{i} \frac{n \pi x}{L}} \right \| \\ & = \sum_{n = 0}^\infty \vert C_n \vert^2 \end{align*}
You may move the limit in front of the norm because norms are continuous.