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.

  • $\begingroup$ what did you try so far? $\endgroup$ – mm-aops Jun 22 '14 at 14:34
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    $\begingroup$ @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
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    $\begingroup$ you might want to take a look here: math.stackexchange.com/questions/785439/… $\endgroup$ – mm-aops Jun 22 '14 at 15:34
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    $\begingroup$ 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

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 $l \in \mathbb{Z}$.


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