# Orthonormal basis for $\mathcal{L}^2([0,1])$

$$\textbf{Theorem:}$$ The orthonormal family $$\{e_n(x):\, n\in\mathbb{N}\}$$, where $$e_n(x)=e^{2\pi inx}$$, is a basis for $$\mathcal{L}^2([0,1])$$.

In this case, $$\{e_n(x):\, n\in\mathbb{N}\}$$ being a basis would mean that any $$f\in\mathcal{L}^2([0,1])$$ can be written in the form

$$f=\sum^\infty_{k=0} \hat{f}(k)e_k(x)$$

where $$\hat{f}(k)=\langle f,e_k\rangle =\int_{[0,1]} f(x)\overline{e_k(x)} \ \text{d}x$$

I am attempting to get a solution in which we can say

$$\left\vert\left\vert f-\sum^k_{k=0}\hat{f}(k)e_k(x)\right\vert\right\vert\rightarrow 0 \ \ \text{as} \ \ n\rightarrow \infty$$

via Parsevals and Plancheral Identities, but I have been unable to do so.

• The lower limit in the sum in the first displayed equation should be $0$, not $-\infty$? Commented Jul 1, 2015 at 19:53
• @joriki yes you're correct - my apologies, I shall edit it now
– Naji
Commented Jul 1, 2015 at 19:56
• @Naji : Change the first sum to $\sum_{k=-\infty}^{\infty}$ and the finite sum to $\sum_{k=-n}^{n}$ or something like that. Commented Jul 3, 2015 at 11:04
• The meaning of $f = \sum_{k=-\infty}^{\infty}\hat{f}(k)e_{k}(x)$ in this context would normally be taken to mean $L^{2}$ convergence, which would mean $\|f - \sum_{k=-N}^{N}\hat{f}(k)e_{k}\|\rightarrow 0$ as $N\rightarrow\infty$. In fact, it would mean unordered convergence (i.e., independent of order). Commented Jul 7, 2015 at 20:54

This is an old question, but it's not an uncommon exercise, so I think that it's worth having an approach sketched.

The orthonormal basis for $$L^2([0,1])$$ is given by elements of the form $$e_n=e^{2\pi i nx},$$ with $$n\in\mathbb{Z}$$ (not in $$\mathbb{N}$$). Clearly, this family is an orthonormal system with respect to $$L^2$$, so let's focus on the basis part. One of the easiest ways to do this is to appeal to the Stone-Weierstrass theorem. Here are the general steps:

Consider the set $$\mathcal{A}$$ consisting of all finite linear combinations of elements $$e_n$$, where $$n$$ ranges over $$\mathbb{Z}$$ (i.e. the linear span). This is clearly a subspace of the continuous functions on $$[0,1]$$. Show that

1. $$\mathcal{A}$$ forms a unital sub-algebra over $$\mathbb{C}$$. That is, show that if $$f,g\in\mathcal{A}$$ and $$c\in\mathbb{C},$$ then $$cf,\ f+g,\ fg, 1\in \mathcal{A}$$.

2. $$\mathcal{A}$$ separates points. That is, show that if $$x,y\in [0,1]$$ with $$x\neq y,$$ then there exists $$f_{xy}\in \mathcal{A}$$ so that $$f_{xy}(x)\neq f_{xy}(y).$$

3. $$\mathcal{A}$$ is self-adjoint. That is, show that if $$f\in\mathcal{A}$$, then $$\bar{f}\in \mathcal{A}.$$

None of these steps are particularly hard, using the properties $$e^{2\pi i nx}e^{2\pi i mx}=e^{2\pi i (n+m)x}$$ and $$\overline{e^{2\pi i nx}}=e^{-2\pi i n x}.$$ Stone-Weierstrass will tell you that the linear span is dense in continuous functions, which you can extend to density in $$L^2$$ (using that continuous functions are dense in $$L^2$$). This tells you that this orthonormal system is, indeed, an orthonormal basis for $$L^2.$$