Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / P (Numerical Analysis)

A best approximation to $x$ from $U_n$ is an element $u_{\ast }\in U_n$ satisfying $\boxed{\Vert x-u_{\ast }\Vert =\underset{u\in U_n}{\inf }\Vert x-u\Vert }$. This is a minimum over the specified approximation subspace, not over all of $H$.

Expand using the inner product. The two cross terms cancel:Adding proves $\boxed{\Vert x-y{\Vert }^2+\Vert x+y{\Vert }^2=2(\Vert x{\Vert }^2+\Vert y{\Vert }^2)}$, for real or complex inner products.

Suppose $u,v\in U_n$ both attain the minimum distance $d$. Apply (b) to $x-u$ and $x-v$:The midpoint belongs to $U_n$, so its squared distance is at least $d^2$. Thus $\Vert u-v{\Vert }^2\le 0$, forcing $\boxed{u=v}$. Existence will also follow constructively in (d)(i).

Set $u={\sum }_{k=1}^n⟨x,e_k⟩e_k$, with the convention that the inner product is linear in its first argument. Orthonormality gives $⟨x-u,e_j⟩=0$, hence $x-u\perp U_n$. For any $v\in U_n$ the Pythagorean identity yieldsThus this $u$ is the best approximation, and uniqueness gives $\boxed{P_{n}x=\sum _{k=1}^n⟨x,e_k⟩e_k}$. No completeness assumption on $H$ is necessary for this finite-dimensional construction.

Take the squared norm of the orthonormal expansion from (i). All mixed inner products vanish and every diagonal inner product is one, so

The decomposition $x=P_{n}x+(x-P_{n}x)$ is orthogonal by (i). Therefore $\Vert x{\Vert }^2=\Vert P_{n}x{\Vert }^2+\Vert x-P_{n}x{\Vert }^2$. Substitute (ii) and rearrange to obtainIn particular nonnegativity of the left side gives the finite Bessel inequality.

Assume the intended nondegenerate interval $a<b$ and real-valued continuous functions. The integral is bilinear and symmetric. Since $w>0$, it gives $⟨f,f⟩\ge 0$. If a continuous $f$ is nonzero at some point, continuity makes $f^{2}w$ strictly positive on a subinterval of positive length, so its integral is positive. Hence $⟨f,f⟩=0$ forces $f=0$, proving positive definiteness. Orthogonality means $\boxed{{\int }_a^{b}f(x)g(x)w(x)dx=0}$. If $a=b$, the displayed integral is identically zero and is not an inner product on the stated function space; that degenerate case must be excluded.

An arbitrary orthonormal basis does not have the printed property. For example on $[-1,1]$ with $w=1$, take the usual normalized constant and linear Legendre polynomials and swap their labels $p_0,p_1$. The resulting family is still an orthonormal basis but $p_0$ has one zero, rather than zero. The necessary intended hypothesis is that $p_k$ has degree $k$, or equivalently that the basis is obtained by orthogonalizing $1,x,x^2,\dots$ in that order.

Under this hypothesis, $p_k$ is orthogonal to every polynomial of degree less than $k$. Let $t_1,\dots ,t_s$ be its sign-changing zeros in $(a,b)$ and put $q(x)={\prod }_{j=1}^s(x-t_j)$. If $s<k$, then $p_{k}q$ has a constant nonzero sign except at finitely many points: each interior sign change has been canceled, and all remaining zeros have even multiplicity or occur at endpoints. Since $w>0$, its integral cannot vanish. But orthogonality to $q$, of degree $s<k$, says that integral is zero, a contradiction. Thus $s\ge k$. Since the polynomial has degree $k$, it follows thatThis proves the requested conclusion in its standard intended setting and explains why it does not follow from the literal basis statement alone.

Use the degree-ordered orthogonal polynomials identified in (b); otherwise the printed nodes need not exist. Let ${\ell }_j$ be the degree-$n$ Lagrange interpolation polynomial with ${\ell }_j(x_i)={\delta }_{ij}$ and set $w_j={\int }_a^b{\ell }_j(x)w(x)dx$.

For any $f\in {\Pi }_{2n+1}$, polynomial division gives $f=p_{n+1}s+r$ with $\mathrm{deg}s\le n$ and $\mathrm{deg}r\le n$. Orthogonality gives $\int p_{n+1}sw=0$, while this product also vanishes at every node. The interpolation formula for $r$ then provesApply this exactness to ${\ell }_j^2$, of degree at most $2n$. It gives $w_j={\int }_a^b{\ell }_j(x{)}^{2}w(x)dx>0$, because ${\ell }_j$ is not the zero polynomial. Thus all quadrature weights are actually strictly positive, stronger than the requested nonnegativity. Here exactness is for the weighted integral defined at the start of the question.