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

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.

For distinct ordered knots, put $h_i=t_{i+1}-t_i>0$. The linear B-spline $N_i$ is the triangular hat supported on $[t_i,t_{i+2}]$, equal to $(t-t_i)/h_i$ on its rising segment and $(t_{i+2}-t)/h_{i+1}$ on its falling segment. Its integral-normalized partner is $M_i=2N_i/(h_i+h_{i+1})$. ThusFor example, the left overlap is ${\int }_0^{h_i}(u/h_i)(1-u/h_i)du=h_i/6$. Nonadjacent hats have disjoint interiors of their supports, so their inner product is zero. The linear-spline mixed Gram matrix therefore has rowRetain only column indices in $1,\dots ,n$: the first or last row has no neighbor outside this range. Admissible repeated endpoint knots are obtained by a limit with zero spacings, provided $h_i+h_{i+1}>0$; a degenerate hat with zero support width is not a basis element.

The entries in (b) satisfyThus $G$ has strict diagonal dominance with margin at least $1/3$. To prove the inverse infinity-norm bound from diagonal dominance directly, choose $i$ with $|a_i|=\Vert a{\Vert }_{{\ell }^{\infty }}$. The reverse triangle inequality givesThis also proves that the kernel is zero and hence that $G$ is invertible. Substituting $a=G^{-1}b$ and taking the operator norm givesCombined with (a), it yields $\Vert P_S{\Vert }_{\infty }\le 3$ for linear splines, independently of the knot spacings. The diagonal-dominance argument already proves the required estimate.