I am trying to prove the following theorem:
Theorem Let $x_0 \in \mathbb{R}$ and $q_n(x)$ a polynomial of degree at most n, normalized by the following condition:
$\int_a^b (q_n(x))^2 w (x)dx$=1.
The maximum of $q_n(x_0)$ is hit at the polynomial $q(x)=K_n(x_0,x_0)^{- \frac{1}{2}} K_n(x,x_0)$.
For some more context, $K_n(y,x)$ is the kernel polynomial given by $K_n(y,x)=\sum_{k=0}^n p_k(y)p_k(x)$, where $p_k(x)$ is a polynomial of degree at most $k$ associated to a weight function $w(x)$, such that $\int_a^b p_k(x)p_m(x)w(x) dx$ is equal to 0 when $m<k$ or $m>k$; or is equal to 1 when $m=k$.
My approach so far: I have expanded $q_n(x)$ is the orthonormal basis $p_k(x)$, such that $q_n(x)=\sum_{k=0}^n a_kp_k(x)$ and replacing this in the normality condition i obtained that $\sum_k a_k^2 =1$.
Next, I applied the Cauchy-Schwarz inequality to obtain the following:
$(q_n(x_0))^2 = (\sum_{k=0}^n a_k p_k(x_0))^2 \leq \sum_k a_k^2 \sum_k (p_k(x_0))^2 =\sum_k (p_k(x_0))^2 = K_n(x_0,x_0)$.
Now, however, I don't know what to do next, i feel like i am going in the right direction but i don't know to obtain the rest of the expression. Any tips?
Thank you for all the help in advance!