Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 62 / 4 / b

Use $s_j=t_{i+j}$ again and take $k\ge 2$. Set $g(s)=(s-t{)}_{+}^{k-2}$, so $(s-t)g(s)=(s-t{)}_{+}^{k-1}$. LetThe Leibniz rule for divided differences, applied to the linear factor $s-t$, readsMultiply by $s_k-s_0$ and substitute $G$ to get $(t-s_0)H+(s_k-t)K$. Converting $H$ and $K$ to the lower-order normalized B-splines proves the Cox-de Boor recurrence:It starts with $N_{i,1}=1_{[t_i,t_{i+1})}$. The coefficients are nonnegative wherever the corresponding lower-order B-spline is nonzero, so induction proves nonnegativity.

For later use, extend the finite knot sequence to a strictly increasing sequence in both directions. The full order-one sum is one. When the Cox-de Boor recurrence is summed, the two coefficients of each fixed lower-order B-spline add to one. Induction therefore gives a full partition of unity. Removing terms proves the subpartition of unity for B-splines, ${\sum }_{i=1}^n{N}_{i,k}(t)\le 1$. For repeated knots, the same statement follows by a limit of strictly increasing knot sequences, with the usual convention that a recurrence term with zero denominator is zero; the bound holds on each open knot interval and for the standard one-sided endpoint values. The terminology “$L_{\infty }$-normalized” refers to this standard basis normalization; it does not claim that each individual B-spline has supremum norm exactly one.