Orthogonal base-$b$ multiresolution and variance decomposition

On one parent interval, the unnormalized child coefficient vector for contrast $r$ is $(1,\dots ,1,-r,0,\dots ,0)$, with $r$ initial ones. Its entries sum to zero and its squared Euclidean norm is $r(r+1)$. The normalization in [F1] therefore makes each $h_{k,t,r}$ have integral zero and squared $L^2$ norm one. If $r<s$ are contrasts on the same parent, the $s$-contrast is constant on the first $r+1$ children, where the $r$-contrast coefficients sum to zero, so their inner product is zero. Distinct parents have disjoint interiors. At different levels, a finer parent is either disjoint from a coarser parent or lies in one of its children, where the coarser contrast is constant; the finer contrast has integral zero. Hence all one-dimensional contrasts are orthonormal and orthogonal to $1$.

On the partition into $b^J$ level-$J$ intervals, the constant and contrasts with $0\le k<J$ are orthonormal by step 1.1. Their number is $1+(b-1){\sum }_{k=0}^{J-1}{b}^k=b^J$, the dimension of the space of functions constant on those intervals. They therefore form an orthonormal basis of that finite-dimensional space.

Products of the one-dimensional bases from step 2.1 are orthonormal on rectangular grids by [F5]: each factor is bounded and supported on a finite union of intervals, hence integrable, and the cube is a finite-measure product space. Their inner products therefore factor coordinatewise; a coordinate with distinct basis functions gives zero. Their finite tensor products span each rectangular-grid step function because each one-dimensional basis in step 2.1 spans its level grid. Thus the family indexed by $(u,\kappa ,\tau ,\rho )$ in [F1], together with $1$, is orthonormal and its span contains every base-$b$ box step function.

By [F2] the closed span in step 3.1 is all $L^2([0,1{)}^d)$. Apply [F3] and [F4], using [A1], to obtain the orthogonal expansion of $f$. Grouping the finitely many translation and contrast terms for each fixed $(u,\kappa )$ gives exactly ${\nu }_{u,\kappa }(f)$ from [F1]; the empty-coordinate term is the constant $I(f)$. Since $U$ has uniform law on the unit cube, [F7] identifies $\mathbb{E}[f(U)]=I(f)$ and its centered second moment with $\mathrm{\parallel }f-I(f){\mathrm{\parallel }}_2^2$; [F6] identifies this with $\mathrm{Var}(f(U))$. Parseval then gives the displayed sum of squared component norms.