Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 62 / 6 / a / Solution

An orthonormal multiresolution analysis consists of closed subspaces $V_j\subset L_2(\mathbb{R})$, $j\in \mathbb{Z}$, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L_2(\mathbb{R})$, ${\bigcap }_j{V}_j=\{0\}$, and the dilation equivalence $h\in V_j⟺h(2\cdot )\in V_{j+1}$. A scaling function $\varphi$ has integer translates forming an orthonormal basis of $V_0$. Henceis an orthonormal basis of $V_j$. Set $W_j=V_{j+1}\ominus V_j$, the orthogonal complement of $V_j$ inside $V_{j+1}$. A function $\psi$ whose integer translates form an orthonormal basis of $W_0$ produces a complete orthonormal wavelet system $\{2^{j/2}\psi (2^j\cdot -k):j,k\in \mathbb{Z}\}$: different $W_j$ are orthogonal, their sum contains every difference $V_b\ominus V_a$, and density together with the zero intersection exhausts $L_2(\mathbb{R})$.

Here is the filter construction that guarantees such a wavelet. Nesting puts $\varphi$ in $V_1$, so expansion in its orthonormal basis gives the scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, with $a\in {\ell }_2$. Define the MRA low-pass filter and its complementary quadrature mirror filter bySplitting the orthonormal translates and Fourier periodization identity at $2t$, as in part (b), gives $|m(t){|}^2+|m(t+\pi ){|}^2=1$ almost everywhere. Consequently the rows ofhave length one and are orthogonal; the matrix is unitary. Put $b_n=(-1{)}^n\overline{a_{1-n}}$ and $\psi ={\sum }_n{b}_n\varphi (2\cdot -n)$, so $q(t)=\frac{1}{2}{\sum }_n{b}_n{e}^{-int}$.

To see completeness as well as orthogonality, identify $V_1$ with its fine-scale coefficient sequences and take their Fourier series. Group the frequencies $t$ and $t+\pi$ over a half-period. The two rows of the displayed matrix are the normalized coarse and complementary channels; their unitarity is exactly an onto isometry from these two channels to the fine-scale coefficient space. Translates of $\varphi$ and $\psi$ correspond to the Fourier basis in the respective channels. They therefore jointly form an orthonormal basis of $V_1$. The orthogonal complement channel gives the wavelet completion of a multiresolution filter, namely the integer translates of $\psi$ span $W_0$. This proves the asserted relation with an orthonormal wavelet.