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

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.

Use the Fourier transform convention $f(t)=\hat{\varphi }(t)=\int \varphi (x)e^{-ixt}dx$ and the Plancherel theorem normalization $\Vert \varphi {\Vert }_2^2=(2\pi {)}^{-1}\Vert f{\Vert }_2^2$. For a general $L_2$ scaling function, the transform means its $L_2$ extension; an ordinary absolutely convergent integral is not required.

A change of variables gives $\widehat{\varphi (2\cdot -n)}(t)=\frac{1}{2}{e}^{-int/2}f(t/2)$. Thus the scaling refinement equation transforms to $f(t)=m(t/2)f(t/2)$, or $f(2t)=m(t)f(t)$, with the $2\pi$-periodic MRA low-pass filter $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The periodicity is necessary; allowing an arbitrary quotient $f(2t)/f(t)$ would not express refinement by integer translates.

For orthonormality, let $P(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. This is a $2\pi$-periodic $L_1$ function, since $f\in L_2$. The Plancherel theorem givesThese inner products equal ${\delta }_{j\ell }$ precisely when the Fourier coefficients of $P$ agree with those of one. The uniqueness of Fourier coefficients in L1 proves the equivalenceFor completeness, the converse refinement implication is also valid with the natural $L_2$ series interpretation. Suppose this periodization identity and $f(2t)=m(t)f(t)$ hold for a measurable $2\pi$-periodic $m$. Splitting the periodization at $2t$ into even and odd translates givesHence $m$ is bounded and belongs to $L_2$ of a period. Write $m=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$ in $L_2$ and let $m_N$ denote the symmetric partial sums. Since $P=1$,After substituting $t/2$, this proves convergence of the transformed refinement sums to $f$ in $L_2$. Inverting the Fourier transform proves $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$ in $L_2$. Thus the two spatial conditions and the two Fourier conditions are equivalent, with the periodic mask and convergence conventions made explicit.

First use the usual real-valued, measurable interpretation of the transition data. The complementary squares then imply $|f|\le 1$, and the compact support gives $f\in L_2$. Consider the periodization on $[-\pi ,\pi ]$. For $|t|<2\pi /3$, only $f(t)=1$ contributes. For $2\pi /3\le t\le \pi$, only $f(t)$ and $f(t-2\pi )$ can contribute, and their squares sum to one. For $-\pi \le t\le -2\pi /3$, apply the complementary identity at $t+2\pi$ to obtain $f(t{)}^2+f(t+2\pi {)}^2=1$. Therefore the orthonormal translates and Fourier periodization identity holds almost everywhere.

Define a periodic MRA low-pass filter by prescribing it on $[-\pi ,\pi ]$:In the central band, $f(t)=1$, so $f(2t)=m(t)f(t)$. In the rest of this fundamental interval, both $m(t)$ and $f(2t)$ vanish. If $\pi <|t|<4\pi /3$, reducing $t$ modulo $2\pi$ puts it in an outer band where $m=0$, and again $f(2t)=0$. For $|t|\ge 4\pi /3$, both $f(t)$ and $f(2t)$ vanish. Endpoint choices affect only a null set. Thus the complementary transition bands for a scaling function give both Fourier conditions, with this explicitly constructed periodic mask.

A concrete real-valued choice, illustrating that the transition constraints are consistent, isOn the positive transition band put $u=3t/(2\pi )-1\in [0,1]$. The other translate has parameter $1-u$, so the sum is ${\cos }^2(\pi u/2)+{\sin }^2(\pi u/2)=1$.

There is a genuine qualification in the printed formulation: it uses ordinary squares without explicitly declaring $f$ real-valued. If arbitrary complex values are allowed, the claimed verification is false. Take $f=1$ for $|t|\le 2\pi /3$, $f=0$ for $|t|\ge 4\pi /3$, and setThe printed complementary-square rule holds, including its endpoints: in the interior $i^2+(\sqrt{2}{)}^2=1$, while the endpoints pair one with zero. Nevertheless, for $2\pi /3<t<\pi$,This compactly supported $L_2$ function is the Fourier transform of an $L_2$ function, so it is a counterexample within the natural complex Hilbert space. The verification is valid for real-valued $f$, or with absolute squares in the transition rule; it is not valid for unrestricted complex $f$ with ordinary squares.