Haar Wavelets

**mathematicalbagpiper** · March 1st 2011, 11:27 AM

Show that the rescaled Haar wavelets $\psi_{jk}(x)=2^{j/2}\psi(2^jx-k)$ form an orthonormal basis basis for $L^2(\mathbb{R})$ .

So I know that what I have to show is that:

$\int_{\mathbb{R}}\psi_{jk}(x)\psi_{lm}(x)dx=\delta _{jl}\delta_{km}$

I'm just stuck as to how to simplify that integral.

**Ackbeet** · March 1st 2011, 07:07 PM

Hmm. Maybe this might help:

$\psi_{jk}(x)=2^{j/2}\begin{cases} 1,&\quad 0\le 2^{j}x-k<1/2\\ -1,&\quad 1/2\le 2^{j}x-k<1\\ 0,&\quad\text{otherwise} \end{cases}=2^{j/2}\begin{cases} 1,&\quad k\le 2^{j}x<1/2+k\\ -1,&\quad 1/2+k\le 2^{j}x<1+k\\ 0,&\quad\text{otherwise} \end{cases}$

$=2^{j/2}\begin{cases} 1,&\quad 2^{-j}k\le x<2^{-j}(1/2+k)\\ -1,&\quad 2^{-j}(1/2+k)\le x<2^{-j}(1+k)\\ 0,&\quad\text{otherwise} \end{cases}.$

Here I'm using the mother wavelet function $\psi(x)$ as defined in the wiki. So the only points at which this function is nonzero are in the half-open interval $[2^{-j}k,2^{-j}(k+1)).$

Now, what if you could show that $[2^{-j}k,2^{-j}(k+1))\cap[2^{-l}m,2^{-l}(m+1))=\varnothing$ if either $j\not= l$ or $k\not=m?$ That would certainly be sufficient to show the zero part of the delta functions, wouldn't it? Because then, under the integral sign, each function would drag the other one down to zero.

What happens when both $j=l$ and $k=m?$ What does the integrand do?

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