MATH 414 Lecture 27

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Wavelets

Haar scaling function: $\phi (x)={\begin{cases}1&x\in [0,1)\\0&{\mbox{otherwise}}\end{cases}}$

Shifts: $\phi (x-k)$

"Sampling" spaces

• Integers: $V_{0}=\left\{f\in L^{2}~\mid ~f\ {\mbox{is constant on}}\ k\leq x<k+1,k\in \mathbb {Z} \right\}$ (sampling interval has unit length 1)
• Half-Integers: $V_{1}=\left\{f\in L^{2}~\mid ~f\ {\mbox{is constant on}}\ {\frac {k}{2}}\leq x<{\frac {k+1}{2}}\right\}$ (sampling interval has length ${\frac {1}{2}}$)
• $V_{j}=\left\{f\in L^{2}~\mid ~f\ {\mbox{is constant on}}\ {\frac {k}{2^{j}}}\leq x<{\frac {k+1}{2^{j}}}\right\}$ (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 $2^{-j}\,k\leq x<2^{-j}\,\left(k+1\right)$. This implies that $f$ is constant on $2^{-(j+1)}\,k\leq x<2^{-(j+1)}\,\left(k+1\right)$ and on $2^{-(j+1)}\,\left(k+1\right)\leq x<2^{-(j+1)}\,\left(k+2\right)$. 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.

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

Separation

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

Orthonormal Basis

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

Proof. We know from last time that $\left\langle \phi (x-k),\phi (x-\ell )\right\rangle =\delta _{k,\ell }$

$\int _{-\infty }^{\infty }2^{\frac {j}{2}}\,\phi \left(2^{j}\,x-k\right)\,\phi \left(2^{j}-\ell \right)\cdot 2^{\frac {j}{2}}\,\mathrm {d} x=2^{j}\,\int _{-\infty }^{\infty }\phi \left(2^{j}\,x-k\right)\,\phi \left(2^{j}\,x-\ell \right)\,\mathrm {d} x$

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

$2^{j}\,\int _{-\infty }^{\infty }\phi \left(2^{j}\,x-k\right)\,\phi \left(2^{j}\,x-\ell \right)\,\mathrm {d} x=\int _{-\infty }^{\infty }\phi \left(u-k\right)\,\phi \left(u-\ell \right)\,\mathrm {d} u=\delta _{k,\ell }$

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).

$f=\sum _{k=-\infty }^{\infty }\left\langle 2^{\frac {j}{2}}\,\phi \left(2^{j}\,\left(\cdot \right)-k\right),f\right\rangle _{L^{2}}\,2^{\frac {j}{2}}\,\phi \left(2^{j}\,x-k\right)$

Let $a_{k}^{j}=2^{\frac {j}{2}}\,\left\langle \phi \left(2^{j}\,\left(\cdot \right)-k\right),f\right\rangle _{L^{2}}$

${\begin{aligned}a_{k}^{j}&=2^{j}\int _{-\infty }^{\infty }\phi \left(2^{j}\,x-k\right)\,f(x)\,\mathrm {d} x\\&=2^{j}\,\int _{2^{-j}\,k}^{2^{-j}\,(k+1)}\,f(x)\,\,\mathrm {d} x\\&={\frac {1}{2^{-j}}}\,\int _{2^{-j}\,k}^{2^{-j}\,(k+1)}\,f(x)\,\,\mathrm {d} x\end{aligned}}$

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