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$\newcommand{\span}{\operatorname{span}}$ Define $e_{0,0}\equiv 1$, and for all $n\in \mathbb{N}$ $$e_{n,k}=\begin{cases} 2^{n/2} &\text{if } \frac{k-1}{2^n}\leq x\lt \frac{k-\frac{1}{2}}{2^n}\\ -2^{n/2}&\text{if } \frac{k-\frac{1}{2}}{2^n}\leq x\lt \frac{k}{2^n}\\ 0 &\text{otherwise} \end{cases}$$ for $k=1,\ldots,2^n$. Let $$H:=\{e_{n,k}:n,k\in \mathbb{N}\}.$$

I want to prove that $H$ is a Hilbert's base for $L^2[0,1]$ with the usal inner product. In order to prove this we must show that $H$ is orthonormal and that $\span(H)$ is dense in $L^2[0,1]$. Here is a good place to begin to see the orthonormality. For the second thing I have the following exercise:

Let $f\in H^{\bot}$, i.e. $f$ is such that for all $n\in \mathbb{N}$ $$\int_0^1 f(x)e_{n,k}(x)dx=0,$$ for $k=1,\ldots,2^n$. Show that for all $n\in \mathbb{N}$ $$\int_0^1f\cdot 1_{[0,k/2^{n})}=0,$$ $k=1,\ldots,2^n$. Conclude that $f\equiv 0$.

The exercise show that $(\overline{\span(H)})^{\bot}=\{0\}$ and then the density follows. And here is where I'm stuck. I wish it $f$ were continuous function, but $f$ is square integrable only. If the notation is not clear, just tell me and I'll fix it. Thanks for your help.

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leo: could you follow me or do you need more info? As I said somewhere else, feel free to ask. By the way, concerning your doubt in a comment to ncmathsadist's answer: of course you can't find a sequence inside $V$ in general. Just think of the dense set $D = \mathbb{R}^2 \smallsetminus \mathbb{R}$ of $\mathbb{R}^2$ and $V = \mathbb{R}$. – t.b. Jul 28 '11 at 2:42
@Theo: Thank you, with the counterexample is clear now. I'm going to read your answer. – leo Jul 28 '11 at 2:53
leo: what is going on? Do you have problems with my answer? Is it insufficient? I thought we removed the doubts in our discussion in the chat room. If not, please let me know what I should clarify. – t.b. Aug 4 '11 at 19:07
oh no, it's just that some days ago that I do not visit the site. Excuse me @Theo. your answers, as always, are very useful to me. – leo Aug 4 '11 at 19:37
Thank you leo, that's nice to hear! – t.b. Aug 4 '11 at 19:43
up vote 11 down vote accepted

ncmathsadist's idea works with a minor change:

First of all, observe that $L^2[0,1] \subset L^1[0,1]$.

If $f \in L^1[0,1]$ satisfies $\int_{0}^{1} f \cdot e_{k,n} = 0$ for all $k,n \in \mathbb{N}$ then $f = 0$ a.e.

Note that the integral makes sense, as each $e_{k,n}$ is bounded.

Consider the function $\displaystyle F(t) = \int_{0}^{t} f(x)\,dx$ and note that it is absolutely continuous since $f$ is integrable. It is straightforward to show that $0 = F(1) = F(1/2) = F(1/4) = F(3/4) = \cdots $, in words, $F(r) = 0$ for each dyadic rational $r$. But this means $F \equiv 0$ on $[0,1]$.

On the other hand, by the Lebesgue differentiation theorem, we have $F'(t) = f(t)$ almost everywhere on $[0,1]$, so $f = 0$ a.e., as we wanted.

This is essentially Haar's original argument in his Ph.D. thesis, see III §1, pp.363-365. The first part of the thesis appeared as A. Haar, Zur Theorie der orthogonalen Funktionensysteme, Mathematische Annalen 69 (3) (1910), 331–371.

Here's a more hands-on but somewhat more laborious approach — I hope I've got the indices right, but most likely I haven't...

For $m = 2^{k} + n$ put $h_m = e_{k,n}$.

  1. For $f \in L^2 [0,1]$ put $P^M f = \sum_{m=0}^{2^{M}} \langle f, h_m \rangle h_m$. Note that $P^M$ is the orthogonal projection onto the space of functions that are constant on certain intervals with dyadic rational endpoints.

  2. If $f \in C[0,1]$ then $P^M f \to f$ uniformly on $[0,1]$.

  3. If $f \in L^2$ is arbitrary then $P^M f \to f$ in $L^2[0,1]$.

This proves that the subspace spanned by the Haar functions is dense in $L^2$.

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I think I prefer the first argument. It is cute. – Jonas Teuwen Jul 26 '11 at 22:08
I like that one more, too. But Lebesgue's differentiation theorem is not exactly easy to prove. – t.b. Jul 26 '11 at 22:12
Thanks! Nice job here. – ncmathsadist Jul 27 '11 at 0:43
@Theo: in the second point of your second argument, for what do you need it? The third point follows of general result? Or am I wrong? – leo Jul 28 '11 at 4:00
@leo: The second point is intended to help proving the third point. I don't know what general result you have in mind. – t.b. Jul 28 '11 at 4:03

I would suggest this tack. First assume $f$ is continuous (so it's in $L^2$). You should be able to show that for any two dyadic rationals $r, s\in[0,1]$, $\int_r^s f(x)\, dx = 0$. Use this to show that if $f$ is continuous, you must have $f = 0$. The continuous functions are dense in $L^2$. Chase some $\epsilon$s and it should work. Let me know if this is useful.

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Yes, that's what I was thinking. – leo Jul 26 '11 at 2:33
But I have some doubts. – leo Jul 26 '11 at 2:33
for example take a sequence of continuous functions $f_n$ such that $||f-f_n||_2\to 0$. If $f\in H^\bot$, why follows that $f_n\in H^\bot$? Or, can I assume that the sequence is in $H^\bot$? – leo Jul 26 '11 at 2:53
If $f \in L^2$ and $p$ is a Lebesgue point of $f$ with $f(x) \ne 0$, then there will be $s$ and $r$ near $p$ with $\int_r^s f(x) \, dx \ne 0$. – Robert Israel Jul 26 '11 at 3:17
@ncmathsadist: more general doubt: if $V$ is close vector subspace of a Hilbert space $H$ and $D$ is dense in $H$, given $x\in V$, there exists a sequence $x_n$ in V such that $x_n\to x$ consisting of elements of D? – leo Jul 26 '11 at 6:21

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