I want to show that $\{x \mapsto e^{2\pi i k x} \mid k \in \mathbb{N}\}$ is orthonormal basis of $L^2((0,1); \mathbb{C}) =: X$. Of course the only problem is to show completeness.

In our lecture we have shown that $\{x \mapsto \sqrt{2} \sin(\pi k x) \mid k \in \mathbb{N}\}$ (which will be referred as “the other orthonormal system”) as is an orthonormal basis of $L^2((0,1); \mathbb{R})$ and I hope I can reduce the proof to that fact.

Observations I have made:

  • $e^{2\pi i k x} = \cos(2 \pi k x) + i \sin(2 \pi k x)$. (This one made me believe that it might be possible to reduce the proof to the completeness of the other orthonormal system.)
  • $\{e_k: x \mapsto \sqrt{2} \sin(\pi k x) + \sqrt{2} i\sin(\pi k x) \mid k \in \mathbb{N}\}$ is an orthonormal basis of $X$. (This can be easily seen: Take $f \in X$ with $\langle f, e_k \rangle = 0$ for all $k \in \mathbb{N}$, split $f$ in $\Re f + i \Im f$, use the completeness of the other orthonormal system, get $\Re f - \Im f = 0$ and $\Re f + \Im f = 0$ and therefore $f = 0$.)
  • The main problem seems to be the $2$ in the exponent.

Is this orthonormal system complete? If yes, can the completeness be proven by using the completeness of the other orthonormal system? How?


1 Answer 1


This, as you've stated, is not true: the function $x \mapsto e^{-2\pi i x}$ is of course orthogonal to all the elements on the aforementioned set, and is nonzero.

In order to obtain an orthonormal basis, you should've included $\{ e^{-2\pi i k x}; k \in \Bbb{N} \}$ in the definition of the set above.

To prove this set is really an orthonormal basis, you can maybe mimic the proof of the fact for the other orthonormal basis. I say this because I do not know what was your proof.

A possible approach is as follows: we'll prove that the closure of

$$ \text{span} \{ \sqrt{2} \sin (2\pi i k x), \sqrt{2} \cos (2 \pi i k x), k \in \Bbb{N} $$

Contains the other orthonormal system. In fact, this follows by a simple computation of Fourier Coefficients, but one might also use the Stone-Weierstrass Theorem. With this powerful tool in hands, the density of the first orthonormal system comes quickly.

If the OP can provide the proof he knows that the other orthonormal system is an orthonormal basis, we might adapt it in order to finish the problem properly.


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