Let $f\in C[a,b]$ and $q\in Q_n$ which minimizes $||f-p||^2=\int_a^b(f(x)-p(x))^2\ dx \forall p\in P_n$ . Show that there exists $a<x_0<x_1<\cdots<x_n<b$ such that $f(x_i) = q(x_i), i = 0,\cdots,n$
I'm studying the least squares method and I came upon this question. I think this can be solved in multiple ways but one that involves orthogonal families of polynomials is the way my teacher wants.
I have no idea on how to solve it.
Representing $p(x) = c_0 + c_1x + \cdots + c_nx^n$ and $\textbf{c} = (c_0, \cdots, c_n)$, we can express \begin{align*} \frac{d}{d\textbf{c}} \|f - p\|^2 &= -2 \int_a^b(f(x) - p(x)) \begin{pmatrix} 1 \\ x \\ \vdots \\ x^n \end{pmatrix} dx \overset{\text{set}}{=} 0 \\ \frac{d}{d\textbf{c}d\textbf{c}^\intercal} \|f - p\|^2 &= 2 \int_a^b\begin{pmatrix} 1 \\ x \\ \vdots \\ x^n \end{pmatrix}\begin{pmatrix} 1 \\ x \\ \vdots \\ x^n \end{pmatrix}^\intercal dx \succ 0 \end{align*} Since the Hessian is positive-definite everywhere, the solution, $q$, to the vector of first-derivatives is a unique global minimum. This implies that \begin{align*} \int_{a}^{b}(f(x) - q(x)) x^k = 0 \end{align*} for all $k = 0, \cdots, n$. By the Intermediate Value theorem, there exists a $x_k \in (a, b)$ such that \begin{align*} (f(x_k) - q(x_k))x_k^k(b-a) = \int_{a}^{b}(f(x) - q(x)) x^k = 0 \implies f(x_k) = q(x_k) \end{align*} And since $\{1, \cdots, x^n\}$ are linearly independent, these $\{x_k\}$ will all be distinct as well.