Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 67 / 5

This addresses the general interpolation bound in part (I); the subsequent (a)–(c) address the cubic case in part (II). Write the interpolant in the B-spline basis asThe stipulated uniquely defined spline interpolation operator presupposes invertibility of the B-spline collocation matrix. For the standard increasing distinct sites, the conditions $N_i(x_i)>0$ give it by the Schoenberg–Whitney theorem. Thus $c=A_{\mathbf{x}}^{-1}Rf$.

Use the standard partition-normalized B-splines of the question: they are nonnegative and satisfy the subpartition of unity for B-splines, ${\sum }_j{N}_j(t)\le 1$. This also holds outside the basic knot interval; one may extend the knots and regard the finite family as a subfamily of a full partition of unity. The recurrence proof of the full partition is given in Question 6(c). Repeated knots retain the same nonnegativity and subpartition properties, with zero-denominator recurrence terms interpreted as zero or by knot limits. ConsequentlySampling has norm at most one, $\Vert Rf{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$, and soHere the induced matrix norm is the maximum absolute row sum. The label $L^{\infty }$-normalized means the usual partition normalization, not rescaling each spline to have maximum value one. Such a rescaling would invalidate the specific coefficient bound and the cardinal values used below.

Let $C_k(u)$ be the order-$k$ Cardinal B-spline supported on $[0,k]$. The Cox-de Boor recurrence on unit-spaced knots readsStart with $C_1=1_{[0,1)}$. The order-two hat satisfies $C_2(1)=1$ and vanishes at its support endpoints. Substituting these values into the recurrence gives $C_3(1)=C_3(2)=1/2$ and thenThe continuous cubic spline is zero at $0,4$ and outside its support. Since $N_j(t)=C_4(t-j)$ and $x_i=i+2$, this yields the midpoint cubic spline collocation valuesOnly indices $1\le j\le n$ are retained. In particular no nonexistent spline is added to an endpoint row.

The B-spline collocation matrix in the cubic case is the finite tridiagonal matrixFor $n=1$, this is simply the one-entry matrix $(2/3)$.

To evaluate the inverse norm exactly, let $D=\mathrm{diag}((-1{)}^1,\dots ,(-1{)}^n)$. Then $B_n=D{A}_{n}D=(4I-C)/6$, where $C$ has ones immediately above and below the diagonal and zeros elsewhere. Since $\Vert C/4{\Vert }_{{\ell }^{\infty }}\le 1/2$, the convergent Neumann serieshas nonnegative entries and proves invertibility. Since $A_n^{-1}=D{B}_n^{-1}D$, its absolute entries equal those of $B_n^{-1}$. The absolute row sums are therefore the entries of $u=B_n^{-1}1$.

The equation $B_{n}u=1$ becomesIts constant particular solution is $3$; the two characteristic roots of the homogeneous recurrence are $\rho =2-\sqrt{3}$ and ${\rho }^{-1}=2+\sqrt{3}$. Imposing the two boundary values givesThis can also be checked by direct substitution, since $\rho +{\rho }^{-1}=4$. The numerator is smallest at the central index or central pair of indices: its successive difference is $(1-\rho )({\rho }^{n-i}-{\rho }^i)$. Thus, with $j=\lfloor (n+1)/2\rfloor$, the exact inverse norm of midpoint cubic spline collocation isEquivalently, if $n=2m-1$ the fraction is $2{\rho }^m/(1+{\rho }^{2m})$, and if $n=2m$ it is $({\rho }^m+{\rho }^{m+1})/(1+{\rho }^{2m+1})$. The finite norm is not exactly three: it equals $3/2$ for $n=1$, two for $n=2$, and tends to three as $n$ grows. Hence three is the sharp dimension-independent bound for these inverse matrix norms.

Apply the general B-spline interpolation operator norm estimate in part (I) to the inverse norm from (b):and hence in particular $\Vert P_{\mathbf{x}}{\Vert }_{L^{\infty }}\le 3$, independently of $n$. The strict finite-dimensional estimate uses the actual finite collocation matrix; it does not require periodic closure, natural boundary conditions or endpoint knot repetitions.

There is also a direct shorter verification of the uniform bound. For an index $i$ maximizing $|z_i|$, the triangle inequality giveswith an even larger margin when a neighbor is missing. This is the inverse infinity-norm bound from diagonal dominance, and again gives $\Vert A_n^{-1}{\Vert }_{{\ell }^{\infty }}\le 3$.