I need to prove the following property related to kernel polynomials:
$\int_a^b K_n(t,x)q_n(x)w(x)dx=q_n(t)$,
where $q_n(x)$ is a polynomial of degree less or equal to $n$, $w(x)$ is a weight function defined over the interval $[a,b]$ and the kernel polynomials are given by,
$K_n(t,x)=\sum_{k=0}^n p_k(t)p_k(x)$. I have seen on class that $\{p_k\}$ is an orthonormal sequence associated to the weight function $w(x)$.
I have no idea how to start, does anyone have a hint on how to begin?
All the help is appreciated, thank you in advance!
Your problem lacks information. I believe $w>0$ is continuous and that $(p_k)_{k\in\mathbb{N}}$ forms an orthonormal graduated (i.e. $\deg p_k = k$) basis of $\mathbb{R}[X]$ (which means the orthonormal polynomials for the weight $w$) for the inner product: $$\langle q,q' \rangle_w = \int_a^b wqq'$$
If so you have the parseval equality (not sure about the name): $$q(t) = \sum_{k=0}^\infty p_k(t) \langle q,p_k\rangle_w$$ don't worry this is in fact a finite sum.
If $\deg q \leqslant n$, you get $$q(t) = \sum_{k=0}^n p_k(t) \langle q,p_k\rangle_w = \sum_{k=0}^n p_k(t) \int_a^b p_k(x) q(x) w(x) dx = \int_a^b \left(\sum_{k=0}^n p_k(t)p_k(x)\right) q(x) w(x) dx$$ So $$q(t) = \int_a^b K_n(t,x) q(x) w(x) dx$$
My advise for next post: please add all the definitions so we don't have to guess and some personal tries.