p(x) is a n-degree polynomial and it is orthogonal to all polynomial that has degree m (m<n) or less with respect to the inner product $<f,g>=\int_a^bf(x)g(x)w(x)dx$, where w(x)>0, continuous on [a,b]. Take ${x_1,\cdots, x_m}$ be the roots of p(x)=0. Prove that for the quadrature rule $$\int_a^bf(x)w(x)dx=\sum_{i=1}^mA_if(x_i)$$ is exact for polynomial of degree m+n or less. Here $A_i$'s are chosen to make this quadrature rule exact for polynomials of degree m-1 or less.
I have proved that p(x) does have at least m roots by the following way:
Assume that p(x) only has m-1 roots, then we can take a polynomial in the form of $g(x)=(x-x_1)(x-x_2)\cdots(x-x_{m-1})$, which has degree less than m, but $g(x)p(x)w(x)$ will always keep nonnegative or nonpositive on [a,b]. So it cannot be orthogonal.
Then I tried to take any f(x) with degree k, $m-1<k\leq m+n$, and prove this rule is exact for f(x). There exist q(x) and r(x) such that f(x)=p(x)q(x)+r(x). Here deg $q(x)\leq m$, deg r(x)<n. It's easy to notice that the quadrature rule is exact for p(x)q(x) part, but I'm stuck with how to prove the rule is exact for r(x) as the degree of r(x) may be higher than m-1.
Any help will be appreciated!