Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 11H (Topics in Analysis)

Bilinearity and symmetry are immediate. Since $r\ge 0$, $⟨u,u⟩\ge 0$. If this is zero, continuity of $u^{2}r$ makes it identically zero. The points where $r>0$ are dense because only finitely many are excluded, so $u$ vanishes on a dense set and hence everywhere. Thus the form is an inner product.

Apply the Gram-Schmidt process to $1,x,\dots ,x^n$. The last resulting vector $P_n$ is nonzero, has degree exactly $n$, and is orthogonal to every polynomial in $\mathrm{span}(1,x,\dots ,x^{n-1})$.

Let $x_1,\dots ,x_k$ be the distinct zeros in $(-1,1)$ at which $P_n$ changes sign, and put $Q(x)={\prod }_{j=1}^k(x-x_j)$. Then $P_{n}Q$ has constant sign and is nonzero except at finitely many points, soIf $k<n$, this contradicts the orthogonality of $P_n$ to the lower-degree polynomial $Q$. Hence $k\ge n$; degree $n$ forces exactly $n$ distinct, simple roots, all in $(-1,1)$.

Let ${\ell }_j$ be the Lagrange interpolation polynomial for the roots ${\omega }_j$, and defineFor $\mathrm{deg}Q\le 2n-1$, polynomial division gives $Q=S{P}_n+R$ with $\mathrm{deg}S,\mathrm{deg}R<n$. Orthogonality kills the first term, while interpolation gives $R={\sum }_{j}R({\omega }_j){\ell }_j$. ThereforeTaking $Q={\ell }_j$ proves uniqueness. Taking $Q={\ell }_j^2$ also showsThis is Gaussian quadrature.

Given $ϵ>0$, the Weierstrass approximation theorem supplies a polynomial $Q$ with $\Vert f-Q{\Vert }_{\infty }<ϵ$. For all sufficiently large $n$, the quadrature is exact on $Q$. Positivity of the weights and ${\sum }_j{A}_j=\int r$ then giveLetting $ϵ\downarrow 0$ proves convergence.

Write $f=Q_{2n-1}+E_n$, where $Q_{2n-1}$ contains the terms through degree $2n-1$. On $[-1,1]$,The quadrature error of $Q_{2n-1}$ is zero. Bounding the integral and quadrature errors of $E_n$ separately, using positive weights with total $\int r$, gives exactly

With $x=\cos \theta$, the weight becomes ${\sin }^2\theta d\theta$. Sincethe sine orthogonality relations show that $P_n$ is a scalar multiple of the Chebyshev polynomial of the second kind $U_n$. Its roots areThe printed normalization $P_n(0)=1$ is possible only for even $n$; then $U_n(0)=(-1{)}^{n/2}$ andFor odd $n$, every such orthogonal polynomial is odd and vanishes at zero, so that normalization is inconsistent.