Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 340 / 1

An orthonormal wavelet is a function whose integer translates and dyadic dilatesform an orthonormal basis of the Hilbert space $L^2(\mathbb{R})$. Thus their pairwise inner products are ${\delta }_{j,j^{\prime }}{\delta }_{k,k^{\prime }}$, and their closed linear span is the whole space. Equivalently, every $f\in L^2(\mathbb{R})$ has the norm-convergent expansionThe factor $2^{j/2}$ preserves the $L^2$ norm under dilation.

Let $P_j$ be the orthogonal projection onto $V_j$. The dilation property and the orthonormal basis property give the orthonormal basis ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$ of $V_j$. The periodized nonnegative functionhas period one and ${\int }_0^{1}F=\Vert \phi {\Vert }_2^2=1$, by Tonelli theorem. For a function $g\in L^2(\mathbb{R})$ with compact support contained in a bounded interval $I$, Cauchy-Schwarz inequality and the Parseval identity giveas $j\to -\infty$. The last step uses absolute continuity of the integral of a locally $L^1$ function, since the interval's length tends to zero. If $f\in {\bigcap }_j{V}_j$, thenFunctions with compact support are dense in $L^2$, so $f=0$. ConsequentlyThis proof needs neither the density axiom nor any compact support assumption on the scaling function.

Put $W_j=V_{j+1}\cap V_j^{\perp }$, the detail space of the multiresolution analysis. The Meyer-Mallat theorem gives a function $\psi \in W_0$ whose integer translates form an orthonormal basis of $W_0$. Dilations then give an orthonormal basis of each $W_j$, and the density and trivial-intersection axioms implyFor an explicit filter construction, write the scaling refinement equation $\phi (x)=\sqrt{2}{\sum }_k{h}_k\phi (2x-k)$. With a quadrature mirror filter one may take $g_k=(-1{)}^k\overline{h_{1-k}}$ and $\psi (x)=\sqrt{2}{\sum }_k{g}_k\phi (2x-k)$. A harmless constant sign or integer translation gives an equivalent orthonormal wavelet.

The answer is no with the integer-translation and dilation conventions of this multiresolution analysis. The orthonormal basis axiom forces $V_0$ to consist of $L^2$ functions constant on the cells $[k-1/2,k+1/2)$. Dilation forces $V_1$ to consist of functions constant on $[k/2-1/4,k/2+1/4)$. However, the proposed scaling function changes value at $1/2$, in the interior of the $V_1$ cell $[1/4,3/4)$. It is therefore not in $V_1$, contradicting nesting:Changing values at endpoints makes no difference in $L^2$. The usual Haar wavelet instead uses the scaling function ${\chi }_{[0,1)}$; an arbitrary half-unit translation does not preserve the required refinement grid.

With $N$ vanishing moments, a wavelet annihilates every polynomial of degree below $N$. A Taylor polynomial then explains small wavelet coefficients on smooth parts of a signal, and efficient best N-term approximation. Compact support localizes coefficients near a feature, limits the number of boundary interactions, and permits a finite filter implementation. Increasing the number of vanishing moments while keeping an orthonormal basis generally requires a larger support of a function: a finite orthonormal filter with $N$ vanishing moments needs at least $2N$ taps, and the minimal-support Daubechies wavelet has support length $2N-1$ in the standard normalization.

The Haar wavelet, $\psi ={\chi }_{[0,1/2)}-{\chi }_{[1/2,1)}$, has one vanishing moment, unit support length and discontinuities. It is inexpensive and particularly suitable for piecewise constant data with sharp jumps. A Daubechies wavelet of a larger order has more vanishing moments and a longer finite filter; sufficiently large orders also provide greater regularity. It is useful when smooth trends should yield small coefficients, although the wider support of a function can spread a jump across more coefficients. Choose Haar for compact jump localization; choose a higher-order Daubechies wavelet for smooth polynomial structure. More vanishing moments alone does not make every low-order wavelet highly differentiable.