Zeros of Gram-Schmidt derived polynomials in weighted integral space

Question

This is a problem out of Chapter 3 of Luenberger's Optimization by Vector Space Methods that I have been having trouble with. Any guidance would be appreciated.

Let $w(t)$ be a positive (weight) function defined on the interval $[a,b]$ of the real line. Assume the integrals $$\int_a^bt^nw(t)\text{ d}t$$ exist for $n=1,2,\ldots$. Define the inner product of two polynomials $p$ and $q$ as $$\langle p,q\rangle=\int_a^bp(t)q(t)w(t)\text{ d}t.$$ Beginning with the sequence $\{1,t,t^2,\ldots\}$, we can employ the Gram-Schmidt procedure to produce a sequence of orthonormal polynomials with respect to this weight function. Show that the zeros of the real orthonormal polynomials are real, simple, and located on the interior of $[a,b]$.

Answer 1

If we call $\{P_0,P_1,P_2,\ldots\}$ the polynomials given by the Gram-Schmidt process, you can easily show that $P_n$ is of degree $n$ and orthogonal to $\mathbb R_{n-1}[X]$ (the space of polynomials of degree $<n$).

Then there is a trick. If you call $\omega_1,\ldots,\omega_p$ those among the roots of $P_n$ which are in $[a,b]$ and have odd multiplicity, and write $P_n = (X-\omega_1)\cdots(X-\omega_p)Q$, then $Q(t)$ is of constant sign for $t\in [a,b]$, and hence : $$ \langle (X-\omega_1)\cdots(X-\omega_p), P_n\rangle = \int_a^b (t-\omega_1)^2\cdots(t-\omega_p)^2Q(t)w(t)dt \neq 0 $$ as the integral of a constant sign non zero function. But since $P_n \perp \mathbb R_{n-1}[X]$, that must mean that $(X-\omega_1)\cdots(X-\omega_p)$ is at least of degree $n$, ie that $p=n$.

Hence $P_n$ has $n$ roots of odd multiplicity within $[a,b]$, and furthermore they are all simple because $\deg P_n = n$.