Finite sum $\sum_{r,k} p_kP_r(x_k)f(x_k)P_r(x_m)=f(x_m)$

Question

Let $x_0,\ldots,x_n\in\mathbb{R}$ be $n+1$ arbitrary real points and $p_0,...,p_n>0$ be positive real numbers. Let $P_0,P_1,\ldots,P_n$ be polinomials such that $$\sum_{k=0}^n p_kP_r(x_k)P_s(x_k)=\delta_{rs}$$where $\delta_{rs}$ is the Kronecker delta (this means that they are orthogonal in $L^2(\mathbb{R},\mu)$ where $\mu$ is defined by $\mu(E)=\sum_{x_k\in E}p_k$). Given function $f:\mathbb{R}\to\mathbb{C}$ such that $\forall x\notin\{x_0,\ldots,x_n\}\quad f(x)=0$ let us define $$c_r:=\sum_{k=0}^n p_kP_r(x_k)f(x_k)$$ (i.e. $c_r:=\langle f,P_r\rangle$). I read (p. 404 here) that, for any $m=0,\ldots,n$,$$\sum_{r=0}^n c_r P_r(x_m)=f(x_m).$$I am convinced that it is an easy thing that can be proved with some manipulation of the sum, but I land nowhere. Could anybody explain how to prove this equality?

Answer 1

This works because the polynomials form an orthonormal basis when restricted to those points. $f(x_m)$ is of the form $$f(x_m)=c_1P_1(x_m)+c_2P_2(x_m)+\cdots+c_nP_n(x_m)$$ for all $m$. where we don't a priori know what the coefficients $c_i$ are. Let $$g(x)=c_1P_1(x)+c_2P_2(x)+\cdots+c_nP_n(x)$$ Then $g(x_m)=f(x_m)$ for all $m$. Take $$\langle f(x),P_i\rangle=\langle g(x),P_i\rangle=c_1\langle P_1,P_i\rangle +c_2\langle P_2,P_i\rangle+\cdots+c_n\langle P_n,P_i\rangle=c_i\langle P_i,P_i\rangle=c_i$$ to get what you want.