Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 69 / 6

Put $e=x-u^{\ast }$. If $u^{\ast }$ minimizes distance, then for every $v\in {\mathcal{U}}_n$ and every real $t$,Both signs of arbitrarily small $t$ force $\mathrm{Re}(e,v)=0$. In a complex inner product space, apply the same argument to $iv$ to obtain the imaginary part too. Conversely, if $e$ is orthogonal to the subspace, the Pythagorean theorem for inner product spaces givesThus orthogonality characterizes the unique best approximation:Finite dimension also guarantees existence by solving the invertible Gram matrix normal equations; completeness of the ambient inner product space is unnecessary.

Use the convention that the inner product is conjugate-linear in its first argument; the real case is the same without conjugates. Linear independence makes the Gram matrix $G$ positive definite, so $B=G^{-1}$ exists and is Hermitian. Define the Riesz dual basis in an inner product space within ${\mathcal{U}}_n$ byThen $(u_i,{\hat{u}}_k)={\sum }_l{G}_{il}{b}_{lk}={\delta }_{ik}$. Its own Gram entries areHence the inverse Gram matrix is the Gram matrix of the dual vectors:The dual vectors are required to lie in ${\mathcal{U}}_n$; adding arbitrary vectors orthogonal to that subspace would preserve the displayed biorthogonality but would destroy this conclusion.

Use the standard partition normalization of B-splines, not rescaling each spline to have supremum exactly one. Let $h_i=t_{i+k}-t_i>0$. Then $M_i=k{N}_i/h_i$. In the ordinary real inner product, the mixed-normalization spline Gram matrix is $G=D^{-1}H$, where $D=\mathrm{diag}(h_i/k)$ and $H_{ij}=(N_i,N_j)$. Since the $N_i$ form a basis, $H$ is positive definite; $D$ is invertible, so $G$ is invertible. The undeclared upper index $h$ in the printed matrix should be $n$.

We need the standard positivity and normalization properties with their justification. The Cox-de Boor recurrence iswith zero-width terms interpreted as zero. It starts from nonnegative interval indicators. On each lower-order spline's support, its recurrence coefficients are nonnegative, so induction gives $N_i\ge 0$. Extend the finite knots outside their endpoints to a full knot sequence. In the full sum, coefficients of each lower-order spline add to one, so the recurrence preserves partition of unity. Our finite collection is a subset of nonnegative B-splines; therefore ${\sum }_{i=1}^n{N}_i(t)\le 1$ everywhere. The Marsden identity gives equality on the basic knot interval.

The divided-difference definition also gives support in $[t_i,t_{i+k}]$: below the left endpoint the knot-data function is a polynomial of degree $k-1$, whose order-$k$ divided difference vanishes; above the right endpoint the knot data vanish. Choose $a<t_i$ and $b>t_{i+k}$. Integrating in $t$ and then taking the divided difference in its knot variable givesbecause an order-$k$ divided difference of a monic degree-$k$ polynomial is one. The support lies inside $[0,1]$, so ${\int }_0^1{M}_i=1$.

Orthogonality of $f-s^{\ast }$ to the spline space gives $Ga=r$, with $r_i=(M_i,f)$. Since $M_i\ge 0$ and has unit integral, $|r_i|\le \Vert f{\Vert }_{\infty }$. The induced matrix maximum norm is the absolute row-sum norm, soTaking the supremum over nonzero $f$ proves the maximum-norm bound for spline projection:The range is $\mathcal{S}$, so the printed phrase “onto $C[0,1]$” must mean into $C[0,1]$, and onto $\mathcal{S}$. Continuity of the range requires continuous B-splines, for example $k\ge 2$ and internal multiplicities at most $k-1$. For order-one or discontinuous B-splines, the same bound is valid with codomain $L^{\infty }[0,1]$ rather than $C[0,1]$.

Fix $i$. If $t<t_i$, all knot values of the function $x\mapsto (x-t{)}_{+}^{k-1}$ agree with a polynomial of degree $k-1$. Its order-$k$ divided difference vanishes. If $t>t_{i+k}$, all the knot values are zero, so the same divided difference again vanishes. Thus the support is finite:The support and Gram bandwidth of B-splines also holds at repeated knots: the same conclusion follows with the standard limiting convention. If $j\ge i+k$, then $t_j\ge t_{i+k}$, so the two supports have no positive-length overlap. An endpoint intersection contributes zero to the integral defining $G_{ij}$. The case $i\ge j+k$ is identical. HenceFor strictly increasing knots, B-splines whose indices differ by $k-1$ overlap on a positive-length interval and their product is positive there, so this universal bound is sharp. In standard matrix terminology the half-bandwidth is $k-1$ and the full bandwidth is $2k-1$; the paper's integer $d$ is $k$.