# Prove Parseval Identity for $f \in C(\Bbb T) 2\pi$ periodic continuous functions

Question: Prove Parseval Identity for $f \in C(\Bbb T)$ $2\pi$ periodic continuous functions

$$\frac{1}{2 \pi} \int_{-\pi}^\pi |f(x)|^2 dx =\sum_{n=-\infty}^\infty |\hat f(n)|^2$$

Thoughts:

We are somehow supposed to use the fact that exists $g\in C^1(\Bbb T)$ s.t. $f$ and $g$ are very close. Tried defining a function $h=f-g$, and then the identity is true for the $g$ part but not for the $f$ part...

Here is how,

$$\int_{-\pi}^{\pi}\left|\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{inx}\right|^{2}dx = \sum_{n=-\infty}^{\infty}\hat{f}(n)\sum_{m=-\infty}^{\infty}\overline {\hat{f}(m)} \int_{-\pi}^{\pi} e^{i(n-m)x}dx .$$

Now, see here for details and how to finish the problem.

Note:

$$\left|\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{inx}\right|^{2}= \sum_{n=-\infty}^{\infty}\hat{f}(n)e^{ix}\overline {\sum_{m=-\infty}^{\infty} {\hat{f}(m)}e^{ix}}.$$

• Hi. In the first step, did you mean the sum from -infty to infty? and also, why are you allowed to interchange the sum and the integral? Dec 25, 2013 at 13:35
• @user: Yes, it is a typo. Dec 25, 2013 at 13:42
• and the second question? Dec 25, 2013 at 13:47
• and also - I don't really understand where you start from. Do you start from the right side of the original equation? And if so- why are you allowed to replace f(x) with the sum? Dec 25, 2013 at 13:52
• @user: because your function admits Fourier series expansion. Dec 25, 2013 at 14:19
1. Do the computation when $f$ has the form $$\sum_{k=-n}^na_ke^{ikx}, n\in\mathbb N, a_k\in\mathbb R.$$

2. Use density in $C(\mathbb T)$ of such functions.

• are you saying in 1 that every contiunuous periodic function can be described as what you wrote? Dec 25, 2013 at 10:37
• Not exactly: a continuous $2\pi$-periodic function can be approximated by such functions, but for example $\sum_n 2^{—n}\sin(2^nx)$ is not of this form. Dec 25, 2013 at 10:39
• are you saying that the fourier series of f is so close to f (in this case) that we can just interchange them? Dec 25, 2013 at 14:53