Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 70 / 5

The orthogonal projection has range $\mathcal{S}$, not all of $C[0,1]$: the latter is an infinite-dimensional vector space whereas $\mathcal{S}$ is a finite-dimensional vector space. For a concrete counterexample, $e^x$ cannot belong to any finite-degree piecewise polynomial space, since all its derivatives are nonzero on every open interval. For $k\ge 2$, the B-splines are continuous, so the orthogonal projection is a map $C[0,1]\to \mathcal{S}\subset C[0,1]$. For order one it instead has a generally discontinuous piecewise constant range. For example, a single interval indicator on $[1/4,3/4)$ is the projection of the constant function one onto its span, and is not continuous.

Write $s^{\ast }={\sum }_j{a}_j{N}_j$. Since $M_i$ is a positive scalar multiple of $N_i$, the operator normal equations $⟨f-s^{\ast },N_i⟩=0$ are equivalent toThe mixed-normalization spline Gram matrix is invertible: $G=DH$ where $D_{ii}=k/(t_{i+k}-t_i)>0$ and $H_{ij}=⟨N_i,N_j⟩$ is the ordinary positive-definite matrix of inner products of a linearly independent basis. The positivity and unit-integral normalization of a B-spline give $|b_i|\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines givesHere the operator norm of a matrix on ${\ell }^{\infty }$ is its maximum absolute row sum. Consequently,This proves the maximum-norm bound for spline projection.

For $k=2$ on an equidistant spline knot sequence, $N_i$ is the triangular hat with value one at $t_{i+1}$ and support $[t_i,t_{i+2}]$. Its slopes on the two pieces are $1/h$ and $-1/h$, and $M_i=N_i/h$. Direct integration givesNonadjacent hats have disjoint interiors of their supports, so their inner product vanishes. Dividing by $h$ therefore givesThis linear-spline mixed Gram matrix is tridiagonal. Every diagonal entry is $2/3$, including the first and last: the distinct-knot hats have their full supports inside $[0,1]$. End rows simply lack one off-diagonal neighbor; there are no repeated endpoint spline knots here.

A strict diagonal dominance argument proves the estimate even on a nonuniform strictly increasing spline knot sequence. Put $h_i=t_{i+1}-t_i>0$. Then $N_i$ has a rising piece of length $h_i$ and a falling piece of length $h_{i+1}$, while $M_i=2N_i/(h_i+h_{i+1})$. The same integrations as in part (b) give the linear-spline mixed Gram matrixOnly entries with indices in $1,\dots ,n$ are present. Thus ${\sum }_{j\ne i}|g_{ij}|\le 1/3$ in every row.

For $Ga=b$, choose an index $i$ with $|a_i|=\Vert a{\Vert }_{{\ell }^{\infty }}$. The reverse triangle inequality then yieldsHence $\Vert a{\Vert }_{{\ell }^{\infty }}\le 3\Vert b{\Vert }_{{\ell }^{\infty }}$. Applying this to every $b$ provesThis is the inverse infinity-norm bound from diagonal dominance; no total positivity theorem is needed. The proof also gives invertibility directly by taking $b=0$.