Given a polynomial, can we construct an orthogonal sequence?

Question

Suppose we are given a polynomial $p(x)$ of degree $k$ which has all of its (distinct) roots in $[-1, 1]$. Is there is any way, generally, to choose a non-negative weight function $w(x)$ such that $\int_{-1}^{1} p(x) q(x) w(x) dx = 0$ for all $q$ with $\deg(q) < k$?

Answer 1

Yes. There is a theorem stating that in $\Bbb{R}^N$ with the standard inner product, if $V$ is a subspace such that $V \cap [0,\infty)^n = \{0\}$ then its orthogonal complement is such that $V^\perp \cap (0,\infty)^n \ne \emptyset$.

Given $p(x)$ your real polynomial of degree $k$ with $k$ simple zeros on $(-1,1)$.

Take $1/N$ smaller than the distance between the pairs of zeros and the distance between the zeros and $\pm 1$.

For each $a\in (0,1/N)$ let $V_a$ be the subspace of $\Bbb{R}^{2N}$ generated by the $k$ vectors $(v_n)_i=(\frac{i}N+a)^n p(\frac{i}N+a)$ with $i\in -N\ldots N-1,n\in 0\ldots k-1$.

Due to the number of sign changes of a polynomial of degree $<k$, $V_a$ doesn't contain any non-zero non-negative vector.

So by the theorem $V_a^\perp$ will contain a vector $u_a$ with strictly positive entries.

$V_a^\perp$ varies continuously with $a$. So we can choose the vectors $u_a$ to be continuous in $a$.

That's it, let $w(\frac{i}N + a) = (u_a)_i$.

By construction $w$ is non-negative and it is such that $$\forall n<k, \int_{-1}^1 w(x)x^np(x)dx=\int_0^{1/N}\sum_{i=-N}^{N-1}w(\frac{i}N+a)(\frac{i}N+a)^n p(\frac{i}N+a)da=0$$