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I want to show that the Haar functions in $L^2([0,1])$ forms an orthonormal basis:

Let $$f = 1_{[0, 1/2)} - 1_{[1/2,0)} \ \ \mbox{,} \ \ f_{j,k}(t) = 2^{j/2}f(2^jt - k).$$ Let $\mathscr{A} = \{(j.k) : j \geq 0, k = 0, 1, 2, ..., 2^j -1\}.$ I can prove that $\ A := \{1_{[0,1]}\} \cup \{f_{j,k}: (j,k) \in \mathscr{A}\}$ is an orthonormal system in $L^2([0,1])$.

(using the fact that each of them is supported on $[2^{-j}k, 2^{-j}(k+1))$, and each different pairs $i, j$ either has disjoint support or contained in each other support)

I want to show that $A$ is complete.

Let $g \in L^2([0,1])$ with $\left<g,f_{i,j}\right> = 0$ and $\left<g, 1_{[0,1]}\right> = 0$ for all $(i, j) \in A.$ I will show that $g = 0 $ a.e. Let $$I^l_{j,k} = [2^{-j},2^{-j}k + 2^{-j-1}), I^r_{j,k} = [2^{-j}k + 2^{-j-1}, 2^{-j}(k+1)).$$ Then $$f_{i,j} = 2^{-j}(1_{I^l_{i,j}} - 1_{I^r_{i,j}}).$$ So I see that $$\int_{I^l_{i,j}} f = \int _{I^r_{i,j}} f$$ for all $(i,j) \in A $ and $$\int_{[0.1]} f = 0.$$

It just "seems" that $f$ should be $0$ a.e., but I cannot think of rigorous reasons for this to happen (how to clearly show that it is true).

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  • $\begingroup$ you forgot to periodize your mother wavelet. proving the orthogonaly is straightforward. I think the simplest way for proving it is a basis of $L^2([0;1])$ is by showing that the discrete Haar wavelet transform is invertible in $\mathbb{R}^{N}$ with $N = 2^n$, thus the Haar wavelet transform is invertible for functions which are constant on $ [k2^{-n}, (k+1)2^{-n}[$ for every $k$. lettting $n \to \infty$ you get that it is invertible for any piecewise continuous function, which are dense in $L^2([0;1])$ $\endgroup$ – reuns Feb 12 '16 at 8:02
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One approach is to show that every continuous function on $[0,1]$ is a uniform limit of linear combination of Haar functions. Suppose $\phi$ is continuous, $n$ is a positive integer, and let $$\psi = \sum_{j\le n, k}\langle \phi, f_{j,k}\rangle f_{j,k}$$ By construction, $\psi$ is a piecewise constant function; it is constant on subintervals of the form $[2^{-n-1}k, 2^{-n-1}(k+1))$, $k=0,\dots, 2^{n+1}$. I claim that on each such subinterval, the value of $\psi$ is equal to the average of $\phi$ on that subinterval: this implies uniform convergence, thanks to the uniform continuity of $\phi$.

The proof is by induction. The base case is $$\int_0^1 \psi = \int_0^1 \langle \phi, 1_{[0,1]}\rangle 1_{[0,1]} = \int_0^1 \phi$$ which holds because all functions except $1_{[0,1]}$ have zero mean on $[0,1]$. Once it has been established that on some dyadic interval $I_{j,k}$ the averages of $\phi$ and $\psi$ are equal, consider its halves $I_-$ and $I_+$: then $$ \int_{I_+}\psi -\int_{I_-}\psi = 2^{-j/2}\int_0^1 f_{j,k}\psi = 2^{-j/2} \int_0^1 f_{j,k}\phi = \int_{I_+}\phi -\int_{I_-}\phi $$ which together with $$ \int_{I_+}\psi + \int_{I_-}\psi = \int_I \psi = \int_I \phi = \int_{I_+}\phi + \int_{I_-}\phi $$ imply that $\int_{I_+}\psi = \int_{I_+}\phi$ and $\int_{I_-}\psi = \int_{I_-}\phi$.


This proof is just a rephrasing of the construction of Haar martingale, which is a dyadic martingale converging to $f$.

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