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 $\varphi$, and ${\{\varphi (x-k)\}}_{k\in \mathbb{\mathbb{Z}}}$ is an orthonormal basis for $V_0$

Properties

Orthonormal Bases for Subspaces

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

Orthogonality:

$\begin{array}{rl}\int 2^{\frac{j}{2}}\varphi (2^{j}x-k)2^{\frac{j}{2}}\varphi (2^{j}x-\ell )\mathrm{d}x& =2^j\int \varphi (2^{j}x-k)\varphi (2^{j}x-k)\mathrm{d}x\\ & =\int \varphi (u-k)\varphi (u-\ell )\mathrm{d}x\\ & ={\delta }_{\ell ,k}\end{array}$

$f(x)\in V_j$

$f(2^{-j}x)\in V_0$

$f(2^{-j}x)=\sum _{k\in \mathbb{\mathbb{Z}}}{c}_k\varphi (x-k)$

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

Scaling / Two-Scale Relation

$\varphi (x)\in V_0\subset V_j$. Therefore $\varphi (x)=\sum _{k\in \mathbb{\mathbb{Z}}}\frac{p_k}{\sqrt{2}}\sqrt{2}\varphi (2x-k)$, where $\frac{p_k}{\sqrt{2}}=⟨\varphi ,\sqrt{2}\varphi (2x-k)⟩$ by the definition of vector space projection.

$\begin{array}{r}⟨\varphi (x),\sqrt{2}\varphi (2x-k)⟩=\frac{p_k}{\sqrt{2}}\\ & =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (x)\sqrt{2}\varphi (2x-k)\mathrm{d}x\\ & =\sqrt{2}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (x)\varphi (2x-k)\mathrm{d}x\\ p_k=2\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\varphi (x)\varphi (2x-k)\mathrm{d}x\end{array}$

Therefore

$\varphi (x)=\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k\varphi (2x-k)$

Example: Haar MRA

$\varphi (x)=\varphi (2x)+\varphi (2x-1)=p_0\varphi (2x)+p_1\varphi (2x+1)$

Hence $p_k=\{\begin{array}{ll}1& k\in \{0,1\}\\ 0& \text{otherwise}\end{array}$

Example: The Shannon MRA

$V_j=\{f\in L^2\mid \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}(\hat{f})\subseteq [-2^j\pi ,2^j\pi ]\}$ and $\varphi (x)=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(x)$

If $f(x)$ has $\hat{f}(\omega )$ with $\hat{f}(\omega )=0$ for $\omega \in ̸[-\pi ,\pi ]$, we have

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

I'm confused now...

The sampling theorem states

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

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

Therefore

$\varphi (x)=\sum _{k\in \mathbb{\mathbb{Z}}}\varphi \left(\frac{k}{2}\right)\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(2x-k)$, where $p_k=\varphi \left(\frac{k}{2}\right)=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left(\frac{k}{2}\right)$, and $\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(2x-k)=\varphi (2x-k)$.

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

Proof:

$\begin{array}{rl}\varphi (x-\ell )& =\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k\varphi (2(x-\ell )-k)\\ & =\sum _{k\in \mathbb{\mathbb{Z}}}{p}_k\varphi (2x-(k+2\ell )\\ & =\sum _{k\in \mathbb{\mathbb{Z}}}{p}_{k-2\ell }\varphi (2x-k)\\ ⟨\varphi (x),\varphi (x-\ell )⟩& =\sum _{k\in \mathbb{\mathbb{Z}}}\frac{p_{k-2\ell }}{\sqrt{2}}\cdot \sqrt{2}\varphi (2x-k)\\ & =\sum _{k\in \mathbb{\mathbb{Z}}}\frac{p_{k-2\ell }}{\sqrt{2}}\cdot \frac{p_k}{\sqrt{2}}\\ & =2{\delta }_{\ell ,0}\end{array}$

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

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