Consider a function $f: X \to \mathbb{R}$ where $X$ is an interval in $\mathbb{R}$. Consider the set $$ S = \{ (x,f(x))\vert x \in X \}$$ According to Caratheodory's Theorem, any point in the convex hull of $S$ can be represented as a convex combination of three points in $S$ because $S \in \mathbb{R}^2$. But I feel that two points are enough (or maybe I missed some specific shape of the curve). Am I right? Does it have anything to do with the "dimension" of $S$ itself? (I feel that $S$ is "one-dimension", but I don't know how to precisely describe this.)
Thank @David Mitra for the counterexample. I feel that the argument fails because some interior points need more elements to be involved in the convex combination. So how about only considering points on the boundary? Formally,
$T(x) = \sup \{y \vert (x,y)\in co\{S\}\}$, where $\sup$ means supremum and co means convex hull. And $T=\{(x, T(x)\vert x \in X)\}$. Intuitively, $T$ is the "ceiling" of the convex hull of $S$. Is it true that every point in $T$ can be represented as a convex combination of two points in $S$?