MATH 414 Lecture 35

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Wavelet Review

${\{2^{\frac{j}{2}}\varphi (2^{j}x-k)\}}_{k\in \mathbb{\mathbb{Z}}}$ is an orthonormal basis for $V_j$ ${\{2^{\frac{j}{2}}\psi (2^{j}x-k)\}}_{k\in \mathbb{\mathbb{Z}}}$ is an orthonormal basis for $W_j$

$V_j=V_{j-1}\oplus W_{j-1}$

Decomposition and Reconstruction

For $f_j\in V_j$, $f_j=\{\begin{array}{l}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k^j\varphi (2^{j}x-k)\\ \underset{f_{j-1}}{\underbrace{\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k^{j-1}\varphi (2^{j-1}x-k)}}+\underset{w_{j-1}}{\underbrace{\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_k^{j-1}\psi (2^{j-1}x-k)}}\end{array}$

Projecting an arbitrary function $f$ into function $f_j\in V_j$ involves sampling.

$f_j$ can then be decomposed into $f_{j-1}$ and $w_{j-1}$, and so on down to $f_0$, $w_0$, $w_1$, …

Inversely, $f_0$ and $w_0$ can be used to reconstruct $f_1$, $f_1$ and $w_1$ reconstruct $f_2$, and so on up to $f_j$

Questions:

1. How do we get from $a_k^j$'s to $a_k^{j-1}$'s and $b_k^{j-1}$'s?
2. How do we get from $a_k^{j-1}$'s and $b_k^{j-1}$'s back to $a_k^j$'s

Decomposition

$\begin{array}{rl}{p}_k& =2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2x-k)\varphi (x)\mathrm{d}x\\ q_k& =2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2x-k)\psi (x)\mathrm{d}x\end{array}$

$\begin{array}{rl}{w}_{j-1}={\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{j}}_{W_{j-1}}(f_j)& =\sum _{k\in \mathbb{\mathbb{Z}}}{b}_k^{j-1}\psi (2^{j-1}x-k)\\ & =\sum _{k\in \mathbb{\mathbb{Z}}}\frac{b_k^{j-1}}{2^{\frac{j-1}{2}}}\underset{\text{orthonormal}}{\underbrace{2^{\frac{j-1}{2}}\psi (2^{j-1}x-k)}}\\ \frac{b_k^{j-1}}{2^{\frac{j-1}{2}}}& =(\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_j(x)\psi (2^{j-1}x-k)\mathrm{d}x)2^{\frac{j-1}{2}}\\ & =⟨f_j,{\psi }_{j-1,k}⟩\\ b_k^{j-1}=2^{j-1}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_j(x)\psi (2^{j-1}x-k)\mathrm{d}x\end{array}$

We already have $f_j=\sum _{\ell \in \mathbb{\mathbb{Z}}}{a}_{\ell }^j\varphi (2^{j}x-\ell )$, so

$\begin{array}{rl}{b}_k^{j-1}& =2^{j-1}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (2^{j-1}x-k)f_j(x)\mathrm{d}x\\ & =2^{j-1}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (2^{j-1}x-k)(\sum _{\ell \in \mathbb{\mathbb{Z}}}{a}_{\ell }^j\varphi (2^{j}x-\ell ))\\ & =2^{j-1}\sum _{\ell \in \mathbb{\mathbb{Z}}}{a}_{\ell }^j\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (2^{j-1}x-k)\varphi (2^{j}x-\ell )\mathrm{d}x\end{array}$

Let $u=2^{j-1}x-k$, then $2^{j}x=2u+2k$, and $\mathrm{d}u=2^{j-1}\mathrm{d}x$

Also recall $q_m=2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (x)\varphi (2x-m)\mathrm{d}x$.

$\begin{array}{rl}{b}_k^{j-1}& =2^{j-1}\sum _{\ell \in \mathbb{\mathbb{Z}}}{a}_{\ell }^j\underset{2}{\underbrace{\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\psi (u)\varphi (2u+2k-\ell )\mathrm{d}x}}\\ & =\frac{1}{2}\sum _{\ell \in \mathbb{\mathbb{Z}}}{a}_{\ell }^j{q}_{\ell -2k}\\ \end{array}$

Therefore, we are left with

$\begin{array}{rl}{a}_k^{j-1}& =\frac{1}{2}\sum _{\ell \in \mathbb{\mathbb{Z}}}{a}_{\ell }^j{p}_{2k-\ell }\\ b_k^{j-1}& =\frac{1}{2}\sum _{\ell \in \mathbb{\mathbb{Z}}}{a}_{\ell }^j{q}_{2k-\ell }\end{array}$

If we let $\lambda$ be a filter sequence, we have

$a_k^{j-1}={(\lambda *a^j)}_{2k}$

Therefore, we can downsample:

$a_k^{j-1}=DL(a^j)$

Note that this does not depend on $j$!!

Likewise, $b_k^{j-1}=DH(a^j)$

Projection

We are interested in coefficients $a_k^j$ that are projections of $f(x)$ onto $V_j$. Therefore (let $u=2^{j}x-k$)

$\begin{array}{rl}{a}_k^j& =2^j\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)\varphi (2^{j}x-k)\\ & =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(2^{-j}u+2^{-j}k)\varphi (u)\mathrm{d}u\end{array}$

"All" interesting wavelets have support contained in some finite interval $[a,b]$. Hence

$a_k^j=\underset{a}{\overset{b}{\int }}f(2^{-j}u+2^{-j}u)\varphi (u)\mathrm{d}u$

If $j$ is large, then $2^{-j}a\le 2^{-j}u\le 2^{-j}b$, where $2^{-j}a$ and $2^{-j}b$ are small.

Since $f\in L^2$ is continuous, $f(2^{-j}u+2^{-j}k)\approx f(2^{-j}k)$ for small $2^{-j}u$. Now what happens next is crucial:

$a_k^j\approx f(2^{-j}k)\underset{1}{\underbrace{\underset{a}{\overset{b}{\int }}\varphi (u)\mathrm{d}u}}$

(the integral equality to 1 is the case in most MRAs)

This means that $a_k^j\approx f(2^{-j}k)$ is just a sampling of $f$ at $2^{-j}k$.

The Wavelet Crime

Given samples at any level $j$, we can decompose $a_j$ into component parts $a^{j-1}$, $b^{j-1}$ by using the same downsampling and convolution filters