The supremum norm is given by \(||f-p||_{\infty}=\max_{a\le x\le b}|f(x)-p(x)|\). It is denoted by \(L^{\infty}\).
Given a function \(f\), we are looking for a polynomial of up to a certain degree that minimizes this norm. This polynomial exists and is unique.
Weierstrass’s Theorem: Let \(f\) be a continuous function in \([a,b]\). For any \(\epsilon>0\), there exists a polynomial such that \(||p-f||_{\infty}\le\epsilon\).
(I did not prove this theorem).
Chebyshev also showed the following. \(p\) is the degree-n minimax approximation of \(f\) if and only if there are at least \(n+2\) values \(a\le x_{0}\le\cdots\le x_{n+1}\le b\) that satisfy:
In words, \(x_{0}\) is such that the difference of the two equals the norm in magnitude, and all the other points also equal the norm in magnitude, but alternate signs.
Keep in mind that the theorem only requires at least \(n+2\) values - there could be more!
(No proof of this theorem).