Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 2I (Topics in Analysis)

A degree-$n$ Legendre polynomial $P_n$ is orthogonal, for integration on $[-1,1]$, to every polynomial of degree at most $n-1$, and has $n$ distinct roots in $(-1,1)$. Divide $p$ by $P_n$:The Orthogonality of Legendre polynomials gives $\int p=\int r$. At every quadrature node $x_j$, $p(x_j)=r(x_j)$. The assumed exactness for $r$ therefore proves exactness up to degree $2n-1$, the characteristic property of Gaussian quadrature.

Use the Lagrange interpolation polynomialIts square has degree $2n-2$, so the exactness already proved givesThe positive weights of Gaussian quadrature follow because ${\ell }_j$ is a nonzero continuous polynomial. Thus these Gaussian quadrature weights are in fact positive, rather than merely nonnegative.

Apply the Gaussian quadrature rule to the constant polynomial $1$. Exactness gives ${\sum }_{j=1}^n{A}_j=2$.

To prove convergence of positive quadrature on continuous functions, for a continuous $f$, the Weierstrass approximation theorem supplies a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. For every sufficiently large $n$, the Gaussian quadrature rule is exact for this fixed polynomial. Positivity of its weights and their sum $2$ then giveSince $\epsilon$ is arbitrary, $Q_n(f)\to {\int }_{-1}^{1}f$. Notice that no differentiability of $f$ is needed; uniform polynomial approximation and positivity provide the whole argument.