$\newcommand{\span}{\operatorname{span}}$ 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.