Past exam of the options II course of the University of Oxford / 2010 / P / P4

Assume the usual nondegenerate interval $a<b$. The integral is bilinear and symmetric because the functions and weight are real. It is nonnegative on $(f,f)$. If $f$ is nonzero at a point, continuity makes $|f|$ bounded below by a positive number on a nontrivial relative interval nearby, and the positive continuous weight is also bounded below there. Hence ${\int }_a^b{f}^{2}w>0$. Thus the formula is positive definite and defines an inner product.

The functions are orthogonal precisely whenA singleton interval would make this integral zero for every function, so the condition $a<b$ is necessary.

The intended orthogonal-polynomial convention is $\mathrm{deg}{p}_k=k$. An arbitrary orthonormal basis without this degree ordering does not satisfy the claim: permuting a standard basis can place the nonzero constant polynomial last, where it has no roots rather than $n+1$. The following proof uses the required degree ordering.

It makes $p_k$ orthogonal to every polynomial of degree less than $k$, since $p_0,\dots ,p_{k-1}$ form a basis of that lower-degree space. Let $c_1,\dots ,c_r$ be the distinct roots in $(a,b)$ at which $p_k$ changes sign, and put $q(x)={\prod }_{j=1}^r(x-c_j)$, with $q=1$ if $r=0$. The product $p_{k}q$ has constant sign on $(a,b)$ except at its finitely many zeros: each sign change of $p_k$ is canceled by that of $q$.

If $r<k$, then $q$ has degree below $k$, so orthogonality gives ${\int }_a^b{p}_{k}qw=0$. But a nonzero constant-sign continuous product against the positive weight has a nonzero integral, a contradiction. Thus $r\ge k$. Since $\mathrm{deg}{p}_k=k$, there are exactly $k$ such roots, all simple, with no additional endpoint roots. In particularThis also includes $k=0$, where the polynomial is nonzero constant and has no roots.

Use the intended degree ordering from part (b). Let ${\ell }_j$ be the degree-$n$ Lagrange cardinal polynomial at the distinct roots $x_0,\dots ,x_n$ of $p_{n+1}$:For any $f\in {\Pi }_{2n+1}$, polynomial division gives $f=s{p}_{n+1}+r$ with $\mathrm{deg}s\le n$ and $\mathrm{deg}r\le n$. Orthogonality of $p_{n+1}$ to ${\Pi }_n$ makes the integral of $s{p}_{n+1}$ zero, and that product also vanishes at every quadrature node. Meanwhile $r={\sum }_{j}r(x_j){\ell }_j$ by interpolation. ConsequentlyThus the rule is exact up to degree $2n+1$. To determine the sign of a weight, apply exactness to ${\ell }_j^2$, of degree $2n$:The inequality is strict because ${\ell }_j$ is a nonzero polynomial and the weight is positive. This is the Gaussian quadrature construction; the source asks only for nonnegative weights, and the proof gives positivity.