changing order of integration and sum

Question

Suppose I have a function that is represented as follows $$ f(x) = \sum_{n=1}^{\infty} a_n(x) $$ for $- \pi \leq x \leq \pi$.
Then is it always the case that $$ \int_{-\pi}^{\pi}f^2(x) dx = \sum_{n=1}^{\infty} \sum_{m = 1}^{\infty} \int_{-\pi}^{\pi} a_n(x) a_m(x) dx? $$ I'm trying to understand the proof of Parseval's identity and it seems to do this.... but I'm confused because it involves infinite sums and I thought one can not always switch the orders around when the sum/integral has infinite range.. any clarification appreciated. thank you

Answer 1

When you say "... is represented as follows", you need to be more specific. I'm guessing what is meant is that $f=\sum_{n=1}^{\infty}a_n$, where the convergence is with respect to the $L^2([-\pi,\pi])$ norm, i.e as $N\to\infty$, \begin{align} \left\|f-\sum_{n=1}^Na_n\right\|_{L^2([-\pi,\pi])}\to 0. \end{align} Even more explicitly, \begin{align} \sqrt{\int_{-\pi}^{\pi}\left|f(x)-\sum_{n=1}^Na_n(x)\right|^2\,dx}\to 0, \end{align} as $N\to\infty$. The equality you wrote $f(x)=\sum_{n=1}^{\infty}a_n(x)$ in general does not mean a pointwise equality on $[-\pi,\pi]$. If you only have pointwise convergence, you can't swap series with integrals. Now, keeping in mind that convergence here means with respect to the $L^2$ norm, and the fact that the inner product $\langle\cdot,\cdot\rangle$ on $L^2$ is continuous, do you see why you can swap the series with the inner product?