Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 75 / 6

Write $r=x-u^{\ast }$. If $u^{\ast }$ is a best approximation, then for every $v\in {\mathcal{U}}_n$ and real $t$, the competitor $u^{\ast }+tv$ satisfiesThe linear coefficient must be zero, since otherwise choosing sufficiently small $t$ of the appropriate sign makes this expression negative. Hence $\mathrm{Re}(r,v)=0$. In a real inner product space this is the desired orthogonality; in a complex inner product space, apply the same argument to $iv$ to show that the imaginary part also vanishes. Thus $(r,v)=0$ for all $v\in {\mathcal{U}}_n$.

Conversely, suppose this orthogonality holds. Every competitor $u\in {\mathcal{U}}_n$ has $u-u^{\ast }\in {\mathcal{U}}_n$. The Pythagorean identity then givesEquality forces $u=u^{\ast }$. We have proved the characterization, as well as uniqueness:This is the orthogonal projection onto a finite-dimensional subspace. Completeness of the ambient space is not needed. Existence follows, for example, by solving the invertible Gram matrix system in a basis of the finite-dimensional linear subspace.

The dual vectors are the Riesz dual basis in an inner product space, and lie in ${\mathcal{U}}_n$ itself. Take the convention that the inner product is conjugate-linear in its first argument; in a real space the conjugations can simply be omitted. Since the $u_j$ form a basis, their Gram matrix $G$ is positive definite and invertible. Let $B=G^{-1}$ and putThe proposed vectors satisfyThey therefore give the unique dual family within ${\mathcal{U}}_n$. Their own Gram matrix isThe inverse of a Hermitian positive-definite matrix is Hermitian, so $B^{\ast }=B$. Using $GB=I$ gives $B^{\ast }GB=B$. ConsequentlyFor a real space this is the familiar identity $B^{T}GB=B$. If the opposite complex inner-product convention is adopted, conjugate the coefficients in the dual-vector formula; the stated identity remains the same. Requiring the dual vectors to lie in ${\mathcal{U}}_n$ is essential: arbitrary additions perpendicular to ${\mathcal{U}}_n$ preserve the pairings with $u_i$ but alter the Gram matrix of the dual vectors.

Use the standard nonnegative partition-normalized B-splines $N_j$; their sum is at most one, and is exactly one for the complete clamped basis on $[0,1]$. The printed question introduces $M_i$ only in its last matrix formula without defining it. The intended normalization is the unit-integral normalization of a B-spline:For supports wholly inside the interval, $c_i=(t_{i+k}-t_i)/k$. The positivity and partition convention, sometimes called $L_{\infty }$ normalization, should not be confused with individually rescaling every spline to have maximum exactly one.

The characterization in part 1(a) gives the normal equations for the orthogonal projection:Dividing each equation by $c_i$ yieldsThis is the mixed-normalization spline Gram matrix. It need not be symmetric, but $G=D^{-1}H$ with $D=\mathrm{diag}(c_i)$ and $H_{ij}=(N_i,N_j{)}_{L_2}$. The ordinary Gram matrix $H$ is positive definite because the splines are linearly independent, so $G$ is invertible and $a=G^{-1}d$.

Since each $M_i$ is nonnegative and has integral one,The induced matrix operator norm is the maximum absolute row sum, soFinally, nonnegativity and the subpartition of unity for B-splines giveTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the maximum-norm bound for spline projection:The formulas also establish linearity and boundedness of the projection on continuous inputs. Its range is exactly the finite-dimensional spline space $\mathcal{S}$, since it fixes every spline in that space. Thus, when the splines are continuous, it is an operator $C[0,1]\to C[0,1]$ with range $\mathcal{S}$; it is not onto all of $C[0,1]$ as the printed wording says. For example, $e^t$ cannot belong to a fixed finite-degree spline space because it is not a polynomial on any nonempty knot interval. The missing $M_i$ normalization and this range correction are needed for a precise reading of the result.