= Solution
An orthonormal <multiresolution analysis> consists of closed subspaces $(V_j)_{j\in\mathbb Z}$ of the <L2 space> $L^2(\mathbb R)$ with $V_j\subset V_{j+1}$, dense union, intersection $\{0\}$, and the dilation rule $f\in V_j\iff f(2\cdot)\in V_{j+1}$. A <scaling function> $\phi$ has integer translates forming an <orthonormal basis> of $V_0$. Equivalently,
$$
\phi_{j,n}(x)=2^{j/2}\phi(2^jx-n),\qquad n\in\mathbb Z,
$$
form an <orthonormal basis> of $V_j$. Let $W_j=V_{j+1}\ominus V_j$. Density and the trivial intersection imply $L^2(\mathbb R)=\bigoplus_{j\in\mathbb Z}W_j$.
Nesting gives the <scaling refinement equation> $\phi(x)=\sum_na_n\phi(2x-n)$, with $h_n=a_n/\sqrt2$ the coefficients in the normalized finer-scale basis. <Orthogonality> of the translates gives $\sum_na_n\overline{a_{n-2r}}=2\delta_{r0}$. Its Fourier form is the <quadrature mirror filter> identity $|m(t)|^2+|m(t+\pi)|^2=1$, where $m(t)=\tfrac12\sum_na_ne^{-int}$.
Set $b_n=(-1)^n\overline{a_{1-n}}$ and define $\psi(x)=\sum_nb_n\phi(2x-n)$ in $L^2$. Its high-pass symbol is $q(t)=-e^{-it}\overline{m(t+\pi)}$. The matrix
$$
\begin{pmatrix}m(t)&m(t+\pi)\\q(t)&q(t+\pi)\end{pmatrix}
$$
is unitary almost everywhere. Under the identification of $V_1$ with its finer-scale coefficient space, this two-channel unitary transform splits that space into the translates of $\phi$ and the translates of $\psi$. The latter are therefore an <orthonormal basis> of $W_0$. Dilating gives \b[an <orthonormal wavelet> basis]
$$
\boxed{\{2^{j/2}\psi(2^jx-n):j,n\in\mathbb Z\}\text{ of }L^2(\mathbb R).}
$$
This <wavelet completion of a multiresolution filter> explains how the <multiresolution analysis> supplies a single <orthonormal wavelet>. The argument uses coefficient-space completeness, not just pairwise <orthogonality>.
Back to article page
