MATH 414 Lecture 33

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Multiresolution Analysis (MRA)

1. $\dots \subset V_{j-1}\subset V_{j}\subset V_{j+1}\subset \dots$
2. Density, approximating everything in $L^{2}$
3. Separation: $\bigcap V_{j}=\left\{0\right\}$
4. Scaling: $f(x)\in V_{j}$ if and only if $f\left(2^{-j}\,x\right)\in V_{0}$
5. Scaling unctions $\phi$, and $\left\{\phi (x-k)\right\}_{k\in \mathbb {Z} }$ is an orthonormal basis for $V_{0}$

Properties

Orthonormal Bases for Subspaces

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

Orthogonality:

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

$f(x)\in V_{j}$

$f(2^{-j}\,x)\in V_{0}$

$f(2^{-j}\,x)=\sum _{k\in \mathbb {Z} }c_{k}\,\phi (x-k)$

Let $u=2^{-j}\,x$, then $f(u)=\sum _{k\in \mathbb {Z} }c_{k}\,\phi (2^{j}\,x-k)$

Scaling / Two-Scale Relation

$\phi (x)\in V_{0}\subset V_{j}$. Therefore $\phi (x)=\sum _{k\in \mathbb {Z} }{\frac {p_{k}}{\sqrt {2}}}\,{\sqrt {2}}\,\phi (2x-k)$, where ${\frac {p_{k}}{\sqrt {2}}}=\left\langle \phi ,{\sqrt {2}}\,\phi (2x-k)\right\rangle$ by the definition of vector space projection.

${\begin{aligned}\left\langle \phi (x),{\sqrt {2}}\,\phi (2x-k)\right\rangle ={\frac {p_{k}}{\sqrt {2}}}\\&=\int _{-\infty }^{\infty }\phi (x)\,{\sqrt {2}}\,\phi (2x-k)\,\mathrm {d} x\\&={\sqrt {2}}\,\int _{-\infty }^{\infty }\phi (x)\,\phi (2x-k)\,\mathrm {d} x\\p_{k}=2\,\int _{-\infty }^{\infty }\phi (x)\,\phi (2x-k)\,\mathrm {d} x\end{aligned}}$

Therefore

$\phi (x)=\sum _{k\in \mathbb {Z} }p_{k}\,\phi (2x-k)$

Example: Haar MRA

$\phi (x)=\phi (2x)+\phi (2x-1)=p_{0}\,\phi (2x)+p_{1}\,\phi (2x+1)$

Hence $p_{k}={\begin{cases}1&k\in \left\{0,1\right\}\\0&{\mbox{otherwise}}\end{cases}}$

Example: The Shannon MRA

$V_{j}=\left\{f\in L^{2}~\mid ~\mathrm {supp} ({\hat {f}})\subseteq \left[-2^{j}\,\pi ,2^{j}\,\pi \right]\right\}$ and $\phi (x)=\mathrm {sinc} (x)$

If $f(x)$ has ${\hat {f}}(\omega )$ with ${\hat {f}}(\omega )=0$ for $\omega \not \in \left[-\pi ,\pi \right]$, we have

• $\Omega$, the Nyquist frequency
• $2\Omega$, the Nyquist rate
• $\nu _{nat}=\nu _{freq}={\frac {\Omega }{2\pi }}$, (the natural frequency (also called Nyquist frequency)
• $\nu _{rate}={\frac {\Omega }{\pi }}$, (also called Nyquist rate)

I'm confused now...

The sampling theorem states

$f(x)=\sum _{k\in \mathbb {Z} }f\left({\frac {\pi \,k}{\Omega }}\right)\,\mathrm {sinc} \left(\nu _{freq}\,x-k\right)$

Suppose $\Omega =2\pi$. then $\nu _{nat}={\frac {\Omega }{2\pi }}=1$, so $\nu _{rate}=2$

Therefore

$\phi (x)=\sum _{k\in \mathbb {Z} }\phi \left({\frac {k}{2}}\right)\,\mathrm {sinc} \left(2x-k\right)$, where $p_{k}=\phi \left({\frac {k}{2}}\right)=\mathrm {sinc} \left({\frac {k}{2}}\right)$, and $\mathrm {sinc} \left(2x-k\right)=\phi (2x-k)$.

What are the $p_{k}$'s?

In the even case, $p_{2k}=0$, and in the odd case, $p_{2k+1}={\frac {2(-1)^{k}}{\pi (2\ell +1)}}$

Properties

1. $\sum _{k\in \mathbb {Z} }p_{k-2\ell }\,p_{k}=2\delta _{\ell ,0}$
2. In the above case if $\ell =0$, we have $\sum _{k\in \mathbb {Z} }p_{k}^{2}=2$

Proof:

${\begin{aligned}\phi (x-\ell )&=\sum _{k\in \mathbb {Z} }p_{k}\,\phi (2(x-\ell )-k)\\&=\sum _{k\in \mathbb {Z} }p_{k}\,\phi (2x-(k+2\ell )\\&=\sum _{k\in \mathbb {Z} }p_{k-2\ell }\phi (2x-k)\\\left\langle \phi (x),\phi (x-\ell )\right\rangle &=\sum _{k\in \mathbb {Z} }{\frac {p_{k-2\ell }}{\sqrt {2}}}\cdot {\sqrt {2}}\,\phi (2x-k)\\&=\sum _{k\in \mathbb {Z} }{\frac {p_{k-2\ell }}{\sqrt {2}}}\cdot {\frac {p_{k}}{\sqrt {2}}}\\&=2\delta _{\ell ,0}\end{aligned}}$

The proof of the latter property comes from taking the Fourier series expansion of the sawtooth wave function:

$F(0)={\frac {\pi }{4}}-\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}$