MATH 414 Lecture 27

« previous | Wednesday, March 26, 2014 | next »

Wavelets

Haar scaling function: $\varphi (x)=\{\begin{array}{ll}1& x\in [0,1)\\ 0& \text{otherwise}\end{array}$

Shifts: $\varphi (x-k)$

"Sampling" spaces

• Integers: $V_0=\{f\in L^2\mid f\text{is constant on}k\le x<k+1,k\in \mathbb{\mathbb{Z}}\}$ (sampling interval has unit length 1)
• Half-Integers: $V_1=\{f\in L^2\mid f\text{is constant on}\frac{k}{2}\le x<\frac{k+1}{2}\}$ (sampling interval has length $\frac{1}{2}$)
• $V_j=\{f\in L^2\mid f\text{is constant on}\frac{k}{2^j}\le x<\frac{k+1}{2^j}\}$ (sampling interval has length $2^{-j}$)

Note: As $j$ increases, the "sampling" occurs at finer scales.

Properties

The following 5 properties institute a multi-resolution analysis.

Definition: The $V_j$'s are called approximation spaces (or scaling spaces)

Nesting

Theorem. [Nesting.] The spaces $V_j$ are nested:

$\dots \subset V_{j-1}\subset V_j\subset V_{j+1}\subset \dots$

Proof by induction. $V_0\subset V_1$.

If $f\in V$, then $f$ is constant on $2^{-j}k\le x<2^{-j}(k+1)$. This implies that $f$ is constant on $2^{-(j+1)}k\le x<2^{-(j+1)}(k+1)$ and on $2^{-(j+1)}(k+1)\le x<2^{-(j+1)}(k+2)$. Hence $V_j\subset V_{j+1}$

The hypothesis follows by induction on $j$.

quod erat demonstrandum

Density

One can approximate any $f$ in $L^2$ arbitrarily well by functions in $V_j$ if $j$ is large enough.

$\bigcup _{j\in \mathbb{\mathbb{Z}}}{V}_j=L^2$

Separation

$\bigcap _{j\in \mathbb{\mathbb{Z}}}{V}_j=\left\{0\right\}$

Orthonormal Basis

Theorem. The set ${\left\{2^{\frac{j}{2}}\varphi (2^{j}x-k)\right\}}_{k=-\mathrm{\infty }}^{\mathrm{\infty }}$ is an orthonormal basis for $V_j$

Proof. We know from last time that $⟨\varphi (x-k),\varphi (x-\ell )⟩={\delta }_{k,\ell }$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{2}^{\frac{j}{2}}\varphi (2^{j}x-k)\varphi (2^j-\ell )\cdot 2^{\frac{j}{2}}\mathrm{d}x=2^j\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2^{j}x-k)\varphi (2^{j}x-\ell )\mathrm{d}x$

If we let $u=2^{j}x$, then

$2^j\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2^{j}x-k)\varphi (2^{j}x-\ell )\mathrm{d}x=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (u-k)\varphi (u-\ell )\mathrm{d}u={\delta }_{k,\ell }$

quod erat demonstrandum

Scaling Property

$f(x)\in V_j$ if and only if $f(2^{-j}x)\in V_0$

We know if $f\in V_j$, then $f\in V_{j+1}$ (nesting).

$f=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{⟨2^{\frac{j}{2}}\varphi (2^j(\cdot )-k),f⟩}_{L^2}{2}^{\frac{j}{2}}\varphi (2^{j}x-k)$

Let $a_k^j=2^{\frac{j}{2}}{⟨\varphi (2^j(\cdot )-k),f⟩}_{L^2}$

$\begin{array}{rl}{a}_k^j& =2^j\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (2^{j}x-k)f(x)\mathrm{d}x\\ & =2^j\underset{2^{-j}k}{\overset{2^{-j}(k+1)}{\int }}f(x)\mathrm{d}x\\ & =\frac{1}{2^{-j}}\underset{2^{-j}k}{\overset{2^{-j}(k+1)}{\int }}f(x)\mathrm{d}x\end{array}$

This integral represents the average value over the interval $[2^{-j}k,2^{-j}(k+1))$