I am studying the Christoffel-Darboux kernel polynomials (as a follow up of orthogonal polynomials), defined as $K_n(y,x)=\sum_{k=0}^n p_k(y)p_k(x)$ and i am having some doubts on how to prove the following theorem my teacher gave me:
Theorem Let $x_0 \in \mathbb{R}$. We can say that $\frac{1}{K_n(x_0,x_0)}= \underset{q_n}{min} \int_a^b (q_n(x))^2 w(x) dx$, where $q_n$ is a polynomial of degree at most $n$, normalized by the condition $q_n(x_0)=1$.
My attempt so far: I have written $q_n(x)$ in the orthonormal basis $\{p_n\}$ (for some context this is such that $\int_a^b p_n(x)p_m(x)w(x)dx $ is 0 when $m \neq n$ and 1 when $n=m$), this is, $q_n=\sum_{k=0}^n a_k p_k(x)$ and replaced it in the integral and obtained that $\int_a^b (q_n(x))^2 w(x)dx = \sum_{k=0}^n (a_k)^2$. I don't really know where to use the normalized condition my teacher gave, do you have any tips? Thanks for the help!
Edit (My proposed solution): I tried the following, can anyone tell me if it is correct: I started by analyzing $(q_n(x))^2=(\sum_{k=0}^n a_k p_k(x))^2$. Applying the Cauchy-Schwarz inequality we can see that: $(q_n(x))^2=(\sum_{k=0}^n a_k p_k(x))^2 \leq (\sum_{k=0}^n a_k^2)(\sum_{k=0}^np_k^2(x))$. Then I applied the normalized condition by replacing $x \rightarrow x_0$ in the previous inequality and we have that, $1=q_n(x_0))^2\leq (\sum_{k=0}^n a_k^2)(\sum_{k=0}^np_k^2(x_0))$. As i observed previously we have that $\int_a^b (q_n(x))^2 w(x)dx = \sum_{k=0}^n (a_k)^2$. Substituting it we finally have that:
$\frac{1}{(\sum_{k=0}^np_k^2(x_0))} \leq \sum_{k=0}^n (a_k)^2 \Leftrightarrow \frac{1}{K_n(x_0,x_0)}\leq \int_a^b (q_n(x))^2 w(x)dx$.
Therefore,we have that the minimum value of that integral it hit at $\frac{1}{K_n(x_0,x_0)}$.