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

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.