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I have the following problem. We defined $\mathbb{H}=\{f_0,\quad f_{j,n} \quad j=1,...,2^{n-1} \quad n=1,2,...\}$ where for all $t\in[0,1]$ we put $f_0(t)=1$ and setting $K=2j-1$, $$f_{j,n}(t)=\left\{ \begin{array}{ll} 2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K-1}{2^n},\frac{K}{2^n}\big)$};\\ -2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K}{2^n},\frac{K+1}{2^n}\big)$};\\ 0 &\mbox{otherwise} \end{array} \right. $$

Then we are asked to show that $\mathbb{H}$ is a complete orthonormal system in $L^2[0,1]$ (Lebesgue measure). In particular $\mathbb{H}^{\perp}_{L^2[0,1]}=\{0\}$.

I have never done this before, I know what to do for orthonormal, but I can not deal with the 2 varying indices. For the completeness I have no clue what to do.

I hope anybody can help me proving this, thanks in advance!

EDIT: for orthogonal I found this: Let $\{ \psi_{j,k}(t)\}$ haar system. How to prove that it is orthogonal?

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Mathoman
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There is actually a very nice way to prove completeness, which I learned in my stochastic calculus course. Just to simplify notation, let $J_{j,n}:=((j-1)2^{-n},j2^{-n}]$ for $j=1,\dots,2^n$. Then you can write $f_{n,j}(t):=2^{\frac{n}{2}}(\mathbf1_{J_{2j-1,n+1}}(t)-\mathbf1_{J_{2j,n+1}}(t))$, where $\mathbf1_{J_{2j,n+1}}(t)$ stands for the characteristic function of the set $J_{2j,n+1}$. Moreover, let

$$\mathcal{B}_n:=\sigma(J_{j,n+1};j=1\dots,2^{n+1})$$

then $\mathcal{B}_n$ increase to $\mathcal{B}[0,1]$. Let $f\in L^2[0,1]$, we have by the standard example of conditional expectation:

$$E[f|\mathcal{B}_n]=\sum_{j=1}^{2^{n+1}}2^{n+1}(\int f\mathbf1_{J_{j,n+1}}d\lambda)\mathbf1_{J_{j,n+1}}\tag{1}$$

where $\lambda$ denotes the Lebesgue measure. Now suppose $(f,f_{n,j})=0$ for all $n,j$ then one obtains by induction that $\int f\mathbf1_{J_{j,n+1}}d\lambda=0$ for all $n,j$. Hence the RHS in $(1)$ vanishes. But using the martingale convergence theorem, the LHS of $(1)$ converges to $f$ $\lambda$-a.s. on $[0,1]$. So we conclude that if $f$ is orthogonal in $L^2[0,1]$ to all $f_{n,j}$ we get $f\equiv 0$.

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