Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 2 / 18D (Numerical Analysis)

For node $x_i$, let ${\ell }_i$ be the Lagrange interpolation polynomial with ${\ell }_i(x_j)={\delta }_{ij}$. It has degree $n$, so ${\ell }_i^2$ has degree $2n$. The Gaussian quadrature with $n+1$ nodes is exact through degree $2n+1$, henceThe integrand is nonnegative. Since ${\ell }_i$ is a nonzero polynomial, it is nonzero on some open subinterval; the positive weight has positive integral there. Consequently $\boxed{a_i>0}$ for every node. This proves the positive weights of Gaussian quadrature property without assuming any coefficient sign beforehand.

Write $W={\int }_{-1}^{1}w(x)dx$, finite and positive. Exactness on the constant polynomial gives ${\sum }_i{a}_i^{(n)}=W$. Positivity therefore implies the uniform operator bound $|I_n(h)|\le W\Vert h{\Vert }_{\infty }$, and the integral has the same bound.

For a continuous $f$ and $ϵ>0$, the Weierstrass approximation theorem gives a fixed polynomial $Q$ of degree $m$ with $\Vert f-Q{\Vert }_{\infty }<ϵ$. For every $n\ge m$, polynomial exactness gives $I_n(Q)=I(Q)$. ThusSince $ϵ$ is arbitrary, $\boxed{I_n(f)\to I(f)}$ for every continuous $f$. This is convergence of positive quadrature on continuous functions. The argument requires nodes in $[-1,1]$, implicit in a rule defined on all continuous functions there; otherwise their function values would not even be specified by the stated domain. Exactness alone would not provide the norm bound if arbitrarily signed weights were allowed.