6
$\begingroup$

Are the functions $e_n := e^{i\cdot(2n+1)\cdot\pi\cdot x}$, $n \in \mathbb{Z}$ an orthonormal basis of $L^{2}(0,1)$? I suppose it is true, but I haven't been able to prove it myself yet.

$\endgroup$
  • $\begingroup$ what did you try so far? $\endgroup$ – mm-aops Jun 22 '14 at 14:34
  • 1
    $\begingroup$ @mm-aops: well, my first thought was to try to apply the Stone-Weierstraß theorem, which doesn't work out. Then I tried to deduce it from the fact that $e^{2n\cdot i \cdot \pi x}$ is an orthonormal basis (after norming), but that did not seem to work out either. $\endgroup$ – user159517 Jun 22 '14 at 14:42
  • 1
    $\begingroup$ you might want to take a look here: math.stackexchange.com/questions/785439/… $\endgroup$ – mm-aops Jun 22 '14 at 15:34
  • 1
    $\begingroup$ Thank you, that seems helpful. Sadly, I don't understand the argument why the function F is supposed to extend to an entire Function. $\endgroup$ – user159517 Jun 22 '14 at 19:15
  • $\begingroup$ the singularities are removable because $n$ is a simple zero of the fraction in front of the integral and the integral term is $0$ at $n$ because of the assumption $\endgroup$ – mm-aops Jun 22 '14 at 19:21
6
$\begingroup$

I have found a proof. It makes use of the following facts:

  • $\{e^{i\cdot 2\pi nx}: n \in \mathbb{Z}\}$ is an orthonormal basis of $L^{2}(0,1)$.
  • Let $\{e_k : k \in I \}$ be an orthonormal set in a Hilbert Space H and let M denote the closure of its span. Then, for $ x \in H$, the following two statements are equivalent:
    1. $ x \in M$
    2. $\sum_{k\in I}|(x,e_k )|^2 = \|x\|^2$
  • $\sum_{n\in \mathbb{N}}\frac{1}{n^2} = \frac{\pi^{2}}{6}$

Let M denote the closure of the span of the set $\{e^{i\cdot (2n+1)\pi x}: n \in \mathbb{Z}\}$. Let $\tilde e_{l} := e^{i\cdot 2\pi lx}$ for $l \in \mathbb{Z}$. We show that M is dense in H using the statements above: It suffices to check $\sum_{k \in \mathbb{Z}} |(\tilde e_l,e_k)|^2 = 1 $ for all $l \in \mathbb{Z}$. For all $l \in \mathbb{Z}$, a straightforward calculation yields

$$\sum_{k \in \mathbb{Z}} |(\tilde e_l,e_k)|^2 = \frac{4}{\pi^2} \sum_{k \in \mathbb{Z}} \frac{1}{(2k+1)^2} $$

We can now find the value of this series by observing

$$\sum_{n\in \mathbb{N}} \frac{1}{n^2} = \sum_{n\in \mathbb{N}}^{} \frac{1}{(2n)^2} + \sum_{n\in \mathbb{N}} \frac{1}{(2n - 1)^2}$$

Therefore, $$ \sum_{n\in \mathbb{N}} \frac{1}{(2n - 1)^2} = \frac{3}{4}\cdot \sum_{n\in \mathbb{N}} \frac{1}{n^2} = \frac{\pi^2}{8}$$

Using the identity $\sum_{k \in \mathbb{Z}} \frac{1}{(2k+1)^2} = 2 \cdot \sum_{n\in \mathbb{N}} \frac{1}{(2n - 1)^2}$ we obtain $\sum_{k \in \mathbb{Z}} |(\tilde e_l,e_k)|^2 = 1 $ for all $l \in \mathbb{Z}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.