Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 69 / 7 / a / Solution

The orthogonal projection condition is $(f-s^{\ast },N_i)=0$ for all $i$. Since $M_i=k{N}_i/(t_{i+k}-t_i)$, it is equivalentlyThe mixed-normalization spline Gram matrix is invertible: $G=D^{-1}H$, where $D_{ii}=(t_{i+k}-t_i)/k>0$ and $H_{ij}=(N_i,N_j)$ is the positive-definite ordinary Gram matrix of the basis. Nonnegativity and the unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }{\int }_0^1{M}_i=\Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines gives, on all of $[0,1]$,ConsequentlysoThis is the maximum-norm bound for spline projection. Using a subpartition matters when the basic knot interval is a proper subinterval of $[0,1]$. As in the paper's assertion that the image lies in $C[0,1]$, the spline space here must consist of continuous functions; order-one step splines or full-multiplicity interior knots do not have that property.