 Solution

Let $x_1,\dots ,x_k$ be the distinct interior roots of $p_n$ where its sign changes, and put $q(x)={\prod }_{j=1}^k(x-x_j)$. If $k<n$, orthogonality gives ${\int }_a^{b}w{p}_{n}q=0$. But $p_{n}q$ has constant sign on $(a,b)$, apart from its isolated zeros, because every sign change has been canceled. It is not identically zero, and the positive weight makes its integral nonzero, a contradiction. Thus $k\ge n$. Degree $n$ forces exactly $n$ distinct simple roots, all inside $(a,b)$.

For distinct nodes, let ${\ell }_i(x)={\prod }_{j\ne i}(x-c_j)/(c_i-c_j)$. Lagrange interpolation polynomial gives $f={\sum }_{i}f(c_i){\ell }_i$ for every polynomial of degree at most $n-1$. Integrating this identity proves exactness with $b_i={\int }_a^{b}w{\ell }_i$.

For the Gaussian quadrature nodes, divide any $f\in {\mathbb{P}}_{2n-1}$ as $f=q{p}_n+r$, with $q,r\in {\mathbb{P}}_{n-1}$. The integral of $q{p}_n$ vanishes by orthogonality, and its nodal values vanish because $p_n(c_i)=0$. The remaining polynomial $r$ is integrated exactly by the preceding interpolation result. HenceIts weights are positive: exactness applied to ${\ell }_i^2$ gives $b_i={\int }_a^{b}w{\ell }_i^2>0$.

Define the error functional $L(f)={\int }_a^{b}w(x)f(x)dx-{\sum }_i{b}_{i}f(c_i)$. It annihilates polynomials of degree at most $2n-1$. For $f\in C^{2n}[a,b]$, the Peano kernel theorem giveswhere a formal Peano kernel with the normalization made explicit isNo evaluation of $K$ is needed. The supplied inner product presupposes an integrable positive weight on the finite interval, so the integral and point-evaluation error functional is bounded on continuous functions.