 Solution

Use the degree-ordered orthogonal polynomials identified in (b); otherwise the printed nodes need not exist. Let ${\ell }_j$ be the degree-$n$ Lagrange interpolation polynomial with ${\ell }_j(x_i)={\delta }_{ij}$ and set $w_j={\int }_a^b{\ell }_j(x)w(x)dx$.

For any $f\in {\Pi }_{2n+1}$, polynomial division gives $f=p_{n+1}s+r$ with $\mathrm{deg}s\le n$ and $\mathrm{deg}r\le n$. Orthogonality gives $\int p_{n+1}sw=0$, while this product also vanishes at every node. The interpolation formula for $r$ then provesApply this exactness to ${\ell }_j^2$, of degree at most $2n$. It gives $w_j={\int }_a^b{\ell }_j(x{)}^{2}w(x)dx>0$, because ${\ell }_j$ is not the zero polynomial. Thus all quadrature weights are actually strictly positive, stronger than the requested nonnegativity. Here exactness is for the weighted integral defined at the start of the question.