Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 71 / 4 / 1

Let $y=(f(x_1),\dots ,f(x_n){)}^T$ and write the interpolating spline as $P_{x}f={\sum }_{j=1}^n{c}_j{N}_j$. The interpolation equations are $A_{x}c=y$, so $c=A_x^{-1}y$. Since $P_x$ is defined as a unique interpolant for arbitrary data, its B-spline collocation matrix is invertible. Ordered sites satisfying diagonal positivity ensure this by question 7. Positivity alone without admissible ordering does not: two equal sites in overlapping basis supports could give two identical rows. The norm argument below uses the existence of the operator stipulated here.

Standard partition-normalized B-splines satisfy $N_j\ge 0$ and ${\sum }_{j=1}^n{N}_j\le 1$. The latter is a subpartition of unity for B-splines: extending the knot sequence gives a full partition of unity, and discarding basis functions cannot increase its sum. HenceTaking the supremum over $t$ and then over unit-norm $f$ gives the B-spline interpolation operator norm estimateHere the matrix norm is the induced maximum-row-sum norm, not the maximum of its individual entries. The proof only needs nonnegative basis functions, their sum at most one, and invertible interpolation.