3
$\begingroup$

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...

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

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:

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

$\endgroup$
9
  • $\begingroup$ 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? $\endgroup$
    – jreing
    Dec 25, 2013 at 13:35
  • $\begingroup$ @user: Yes, it is a typo. $\endgroup$
    – Mhenni Benghorbal
    Dec 25, 2013 at 13:42
  • $\begingroup$ and the second question? $\endgroup$
    – jreing
    Dec 25, 2013 at 13:47
  • $\begingroup$ 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? $\endgroup$
    – jreing
    Dec 25, 2013 at 13:52
  • $\begingroup$ @user: because your function admits Fourier series expansion. $\endgroup$
    – Mhenni Benghorbal
    Dec 25, 2013 at 14:19
1
$\begingroup$

  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.

$\endgroup$
3
  • $\begingroup$ are you saying in 1 that every contiunuous periodic function can be described as what you wrote? $\endgroup$
    – jreing
    Dec 25, 2013 at 10:37
  • $\begingroup$ 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. $\endgroup$
    – Davide Giraudo
    Dec 25, 2013 at 10:39
  • $\begingroup$ are you saying that the fourier series of f is so close to f (in this case) that we can just interchange them? $\endgroup$
    – jreing
    Dec 25, 2013 at 14:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .