Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 33 / 3

Use half-open intervals and setThe Haar scaling functions and Haar wavelets areFor fixed $j$, each scaling function has squared integral $2^j{2}^{-j}=1$, and different $k$ have disjoint supports, proving that this family is orthonormal. Each wavelet also has norm one and integral zero. At a fixed level, distinct wavelets have disjoint supports. At different levels, their dyadic supports are either disjoint or the finer support lies in a half on which the coarser wavelet is constant. The finer wavelet has integral zero, so their inner product vanishes. Finally, a wavelet with $l\ge 0$ is supported inside a unit interval; it is orthogonal to its containing ${\phi }_{0,k}$ because its integral is zero, and to all other unit scaling functions because their supports are disjoint. Thus both the scaling family at each fixed level and the complete stated Haar family are orthonormal sets.

The Haar refinement identity isIt shows that the closed scaling spaces $V_j$ satisfy $V_{j+1}=V_j\oplus W_j$, where $W_j$ is the closed span of the level-$j$ wavelets. The union of the $V_j$ is dense in $L^2(\mathbb{R})$: continuous compactly supported functions can be approximated in $L^2$ by their dyadic cell averages, and such continuous functions are dense in $L^2$. Hence the orthonormal Haar family is also a complete orthonormal basis.

For a locally integrable function, define the compact-support coefficient integralsThese exist without requiring $f\in L^2$. The two formulas for the Haar approximation areFor $j=0$ the wavelet sum is empty. For every fixed $j$, these sums are locally finite, so they make sense pointwise and locally in $L^1$. The refinement identities and their orthogonal two-by-two coefficient transformation establish equality by induction, even for merely locally integrable $f$. If $f\in L^2$, this is also the orthogonal projection onto $V_j$.

In particular, for $\delta =2^{-j}$ and $x\in I_k=[k\delta ,(k+1)\delta )$,Now impose the symmetry and monotonicity conditions. The limit zero and decrease on $[0,\infty )$ imply $0\le f(x)\le f(0)$ for $x\ge 0$. For $k\ge 0$, the cell average and every value inside $I_k$ lie between $f((k+1)\delta )$ and $f(k\delta )$. ThereforeSumming over the positive half-line telescopes and gives a bound $\delta f(0)$. Reflection maps each dyadic cell to another dyadic cell up to endpoints. Since $f$ is even, its Haar approximation is even almost everywhere as well, so the negative half-line has the same error. ConsequentlyThe argument proves integrability of the difference, even if $f$ itself is not integrable on the whole line. It is the symmetric monotone case of the Haar approximation error for a function of bounded variation.