Orthogonality of Haar wavelet functions I'm reading about wavelets and I bumped into the follwing:
$\text{Haar wavelet is a step function}\; \psi(x), \text{which takes values 1 and -1, when}\; x \;\text{is in the ranges}\; [0, \frac{1}{2}) \;\text{and}\; [\frac{1}{2}, 1).$
$\text{Dilations and translations of the Haar wavelet are defined as:}$
$$\psi_{jk}(x) = \text{const} \cdot \psi(2^jx-k) $$
Now here is what I need help with :
$\text{It is apparent that:}$
$$\int\psi_{jk}(x)\cdot\psi_{j'k'}(x) = 0$$
$\text{whenever}\; j=j'\; \text{and}\; k=k'\; \text{is not satisfied simultaneously}$.
My question is: Why is the integral above true?
P.S.
here is my reference: http://gtwavelet.bme.gatech.edu/wp/kidsA.pdf (page 4)
 A: When $(j,k)\ne(j',k')$ then the two supports are either disjoint, or one of the two wavelets is constant on the support of the other.
A: To simplify, first look at the case where $j'=0$, $k'=0$.
Clearly, for all nonzero integers $k$, $\int_{-\infty}^\infty \psi(x) \psi(x-k) dx = 0$, since $\psi(x)$ is nonzero only on $[0,1]$ and $\psi(x-k)$ is nonzero only on $[k,k+1]$.
For all $n \in \mathbb{Z}$, $\int_n^{n+1} \psi(x) dx = 0$, so if $a(x)$ is a function which is constant on the intervals $\{[n,n+1]\}_{n \in \mathbb{Z}}$, we have:
$$
\int_{-\infty}^\infty \psi(x)a(x)dx = \sum_{n=-\infty}^\infty \int_{n}^{n+1} \psi(x)a(x)dx= \sum_{k=-\infty}^\infty a(n+1/2) \int_{n}^{n+1} \psi(x)dx=0
$$
Note that $\psi(2^j x - k)$ is constant on the intervals $\{[n2^{-1-j},(n+1)2^{-1-j}]\}_{n \in \mathbb{Z}}$, so if $j<0$, it is constant on the unit intervals $\{[n,n+1]\}_{n \in \mathbb{Z}}$ as well, and so by the above equation we have $\int_{-\infty}^\infty \psi(2^j x - k) \psi(x) dx = 0$.
For the general case, we can shift and rescale to reduce it to the base case. Clearly, if $j>j'$ we can reverse the roles of $(j,k)$ and $(j',k')$, so let's assume $j\le j'$. To transform the integral, we make the substitution $2^{j'} x - k' = y \Leftrightarrow x = 2^{-j'} y + 2^{-j'}k'$.
$$\int_{-\infty}^\infty \psi(2^j x - k) \psi(2^{j'} x - k') dx =\frac{1}{2^{j'}} \int_{-\infty}^\infty \psi(2^{j-j'} y - (k-2^{j-j'}k')) \psi(y) dy$$
So by the base case, the above integral equals $0$ if $j-j'=0$ and $k-k' \ne 0$ or if $j-j'<0$.
A: The wavelets are similar to clipped versions of the Rademacher functions, and 'orthogonality' follows in a similar manner.
Note that $\psi^{-1} \{0\}^c = [0,1)$, and hence $\psi_{jk}^{-1} \{0\}^c = S_{jk} = \frac{1}{2^j}[k,k+1)$.
The keys to the proof are :(i) Any two sets $S_{jk}, S_{j'k'}$ satisfy one of $S_{jk} \subset S_{j'k'}$, the other way around, or $S_{jk} \cap S_{j'k'} = \emptyset$. (ii) If $S_{jk}, S_{j'k'}$ overlap, then either $S_{jk} = S_{j'k'}$, or one is contained in 'half the length' of the other (to be made precise below).
We have $\int \psi = \int_{S_{00}} \psi = 0$, and hence $\int \psi_{jk} = \int_{S_{jk}} \psi_{jk} = 0$.
Now suppose $j \le j'$. Then either $S_{jk} \supset S_{j'k'}$ or $S_{jk} \cap S_{j'k'} = \emptyset$.
Suppose $j=j'$ and $k \neq k'$. Then we have $S_{jk} \cap S_{jk'} = \emptyset$, and it follows that $\int \psi_{jk} \psi_{jk'} = 0$.
Now suppose $j < j'$. If $S_{jk} \cap S_{j'k'} = \emptyset$, we have $\int \psi_{jk} \psi_{jk'} = 0$. If $S_{jk} \supset S_{j'k'}$, then we must have (since $\frac{1}{2^j} \ge \frac{1}{2} \frac{1}{2^{j'}}  $) either
$\frac{1}{2^{j'}}[k',k'+1) \subset \frac{1}{2^j}[k,k+\frac{1}{2})$ or $\frac{1}{2^{j'}}[k',k'+1) \subset \frac{1}{2^j}[k+\frac{1}{2},k+1)$, and hence $\psi_{jk}$ is constant on $ S_{j'k'}$. It follows that $\int \psi_{jk} \psi_{jk'} = 0$.
