MATH 414 Lecture 35

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Wavelet Review

$\left\{2^{\frac {j}{2}}\,\phi (2^{j}\,x-k)\right\}_{k\in \mathbb {Z} }$ is an orthonormal basis for $V_{j}$ $\left\{2^{\frac {j}{2}}\,\psi (2^{j}\,x-k)\right\}_{k\in \mathbb {Z} }$ is an orthonormal basis for $W_{j}$

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

Decomposition and Reconstruction

For $f_{j}\in V_{j}$, $f_{j}={\begin{cases}\sum _{k=-\infty }^{\infty }a_{k}^{j}\,\phi (2^{j}\,x-k)\\\underbrace {\sum _{k=-\infty }^{\infty }a_{k}^{j-1}\,\phi (2^{j-1}\,x-k)} _{f_{j-1}}+\underbrace {\sum _{k=-\infty }^{\infty }b_{k}^{j-1}\psi (2^{j-1}\,x-k)} _{w_{j-1}}\end{cases}}$

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

${\begin{aligned}p_{k}&=2\,\int _{-\infty }^{\infty }\phi (2x-k)\,\phi (x)\,\mathrm {d} x\\q_{k}&=2\,\int _{-\infty }^{\infty }\phi (2x-k)\,\psi (x)\,\mathrm {d} x\end{aligned}}$

${\begin{aligned}w_{j-1}=\mathrm {proj} _{W_{j-1}}(f_{j})&=\sum _{k\in \mathbb {Z} }b_{k}^{j-1}\,\psi (2^{j-1}\,x-k)\\&=\sum _{k\in \mathbb {Z} }{\frac {b_{k}^{j-1}}{2^{\frac {j-1}{2}}}}\,\underbrace {2^{\frac {j-1}{2}}\,\psi (2^{j-1}\,x-k)} _{\mbox{orthonormal}}\\{\frac {b_{k}^{j-1}}{2^{\frac {j-1}{2}}}}&=\left(\int _{-\infty }^{\infty }f_{j}(x)\,\psi (2^{j-1}\,x-k)\,\mathrm {d} x\right)\,2^{\frac {j-1}{2}}\\&=\left\langle f_{j},\psi _{j-1,k}\right\rangle \\b_{k}^{j-1}=2^{j-1}\,\int _{-\infty }^{\infty }f_{j}(x)\,\psi (2^{j-1}\,x-k)\,\mathrm {d} x\end{aligned}}$

We already have $f_{j}=\sum _{\ell \in \mathbb {Z} }a_{\ell }^{j}\,\phi (2^{j}\,x-\ell )$, so

${\begin{aligned}b_{k}^{j-1}&=2^{j-1}\,\int _{-\infty }^{\infty }\psi (2^{j-1}\,x-k)\,f_{j}(x)\,\mathrm {d} x\\&=2^{j-1}\,\int _{-\infty }^{\infty }\psi (2^{j-1}\,x-k)\,\left(\sum _{\ell \in \mathbb {Z} }a_{\ell }^{j}\,\phi (2^{j}\,x-\ell )\right)\\&=2^{j-1}\,\sum _{\ell \in \mathbb {Z} }a_{\ell }^{j}\,\int _{-\infty }^{\infty }\psi (2^{j-1}\,x-k)\,\phi (2^{j}\,x-\ell )\,\mathrm {d} x\end{aligned}}$

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 $q_{m}=2\int _{-\infty }^{\infty }\psi (x)\,\phi (2x-m)\,\mathrm {d} x$.

${\begin{aligned}b_{k}^{j-1}&=2^{j-1}\,\sum _{\ell \in \mathbb {Z} }a_{\ell }^{j}\,\underbrace {\int _{-\infty }^{\infty }\psi (u)\,\phi (2u+2k-\ell )\,\mathrm {d} x} _{2}\\&={\frac {1}{2}}\,\sum _{\ell \in \mathbb {Z} }a_{\ell }^{j}\,q_{\ell -2k}\\\end{aligned}}$

Therefore, we are left with

${\begin{aligned}a_{k}^{j-1}&={\frac {1}{2}}\,\sum _{\ell \in \mathbb {Z} }a_{\ell }^{j}\,p_{2k-\ell }\\b_{k}^{j-1}&={\frac {1}{2}}\,\sum _{\ell \in \mathbb {Z} }a_{\ell }^{j}\,q_{2k-\ell }\end{aligned}}$

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

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

Therefore, we can downsample:

$a_{k}^{j-1}=D\,L(a^{j})$

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

Likewise, $b_{k}^{j-1}=D\,H(a^{j})$

Projection

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

${\begin{aligned}a_{k}^{j}&=2^{j}\,\int _{-\infty }^{\infty }f(x)\,\phi (2^{j}\,x-k)\\&=\int _{-\infty }^{\infty }f(2^{-j}\,u+2^{-j}\,k)\,\phi (u)\,\mathrm {d} u\end{aligned}}$

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

$a_{k}^{j}=\int _{a}^{b}f(2^{-j}\,u+2^{-j}\,u)\,\phi (u)\,\mathrm {d} u$

If $j$ is large, then $2^{-j}\,a\leq 2^{-j}\,u\leq 2^{-j}\,b$, 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:

$a_{k}^{j}\approx f(2^{-j}\,k)\,\underbrace {\int _{a}^{b}\phi (u)\,\mathrm {d} u} _{1}$

(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