Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 70 / 6

An orthonormal multiresolution analysis consists of closed subspaces of a Hilbert space $(V_j{)}_{j\in \mathbb{Z}}$ of the Hilbert space $L^2(\mathbb{R})$ such thatThere is a scaling function $\varphi$ whose integer function translations form an orthonormal basis of $V_0$. Consequently ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ form an orthonormal basis of $V_j$; in particular $V_0$ is invariant under integer function translations.

Let $W_j=V_{j+1}\ominus V_j$ be the orthogonal complement of $V_j$ in $V_{j+1}$. Nesting, density and the trivial intersection imply the orthogonal direct sumFor example, telescoping $V_{J+1}=V_L\oplus {⨁}_{j=L}^J{W}_j$ and letting $L\to -\infty$, $J\to \infty$ gives this decomposition; the corresponding orthogonal projections converge strongly to zero and the identity by the intersection and density properties.

To construct the orthonormal wavelet, use $\varphi \in V_1$ to write the scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, with $a\in {\ell }^2$ and ${\sum }_n|a_n{|}^2=2$. Put $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The orthonormal translates and Fourier periodization criterion proved in part (b), applied to the scaling refinement equation, yieldsThis is the quadrature mirror filter identity. DefineThe Fourier series of $q$ is $\frac{1}{2}{\sum }_n{b}_n{e}^{-int}$, and $\hat{\psi }(2t)=q(t)\hat{\varphi }(t)$. The two rows $(m(t),m(t+\pi ))$ and $(q(t),q(t+\pi ))$ form a unitary $2\times 2$ matrix almost everywhere: both have squared norm one and their inner product is zero. This wavelet completion of a multiresolution filter proves not just orthogonality but completeness. Splitting a finer-scale coefficient Fourier series into its values at $t$ and $t+\pi$, this invertible two-channel matrix resolves it into a coarse-scale channel and its complementary channel. Thus the integer function translations of $\psi$ form an orthonormal basis of $W_0$.

The dilated functions ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ therefore form an orthonormal basis of each $W_j$. The orthogonal direct sum proves the conclusion: every orthonormal multiresolution analysis produces an orthonormal wavelet basis $\{{\psi }_{j,k}:j,k\in \mathbb{Z}\}$ of $L^2(\mathbb{R})$.

Use the printed Fourier transform convention $\hat{\varphi }(t)=\int \varphi (x)e^{-ixt}dx$, interpreted by the Plancherel theorem when $\varphi$ is only square integrable. By the scaling and function translation rules,Taking the Fourier transform of the scaling refinement equation gives $f(t)=m(t/2)f(t/2)$, or equivalentlyConversely this identity, the same scaling rule and injectivity of the Fourier transform recover the scaling refinement equation.

For infinite filters, these statements are equalities in $L^2$, with Fourier series understood in $L^2[0,2\pi ]$. Under the accompanying orthonormality condition, $\{\sqrt{2}\varphi (2\cdot -n)\}$ is an orthonormal sequence, so the refinement series converges in $L^2$ precisely for $a\in {\ell }^2$. Moreover the condition in part (b)(2) makes multiplication by $f$ an isometry from periodic $L^2$ symbols into $L^2(\mathbb{R})$:Applying this to finite filter truncations justifies passage to the limit in both directions. Thus the two paired assertions have a rigorous equivalent interpretation without requiring absolute integrability of $\varphi$.

Let $F(t)={\sum }_{k\in \mathbb{Z}}|f(t+2\pi k){|}^2$. By the Tonelli theorem, $F$ is integrable on $[0,2\pi ]$, since its integral is ${\int }_{\mathbb{R}}|f{|}^2<\infty$. The Plancherel theorem and periodization giveConsequently, if $F=1$ almost everywhere these inner products are ${\delta }_{mn}$, proving orthonormality. Conversely orthonormality says that every Fourier coefficient of $F-1$ is zero. The uniqueness of Fourier coefficients in L1 implies $F-1=0$ almost everywhere. HenceThis proves orthonormal translates and Fourier periodization, including both directions and the normalization constant associated with the printed Fourier transform.

The half-open translates $[-\pi ,\pi )+2\pi k$ partition the real line. Exactly one summand is therefore nonzero in the periodization of $|f{|}^2$, giving ${\sum }_k|f(t+2\pi k){|}^2=1$.

Choose the Shannon scaling mask to be the $2\pi$-periodic extension of $1_{[-\pi /2,\pi /2)}$. On the support of $f$, its product with $f$ is $1_{[-\pi /2,\pi /2)}(t)=f(2t)$; outside that support both sides vanish. Thus the scaling refinement equation holds in frequency. The Fourier coefficients give its precise filter normalization:These coefficients are square summable, so the interpretation in part (b) applies.

The Fourier inversion theorem gives the Shannon scaling functionThis is the normalized sinc function. Its Fourier transform is taken in the Plancherel theorem sense: the Shannon scaling function belongs to $L^2$ but not $L^1$, so its Fourier integral is not absolutely convergent. The inverse integral above is an ordinary integral because the rectangular transform belongs to $L^1$. The resulting refinement spaces are exactly the $L^2$ functions with Fourier transform supported in $[-2^j\pi ,2^j\pi ]$; their union is dense and their intersection is zero. Thus this example also gives a full multiresolution analysis.