Set of points at which a function coincides with its convexification is compact?

Question

Let $f:[0,1]\rightarrow\ \mathbb{\bar{R}}$, and let $\tilde{f}$ be the convexification of $f.$ (i.e., $\tilde{f}$ is the pointwise supremum of all affine functions that lie everywhere below $f$.) Let $X\subseteq[0,1]$ be the set of points at which $f(x)=\tilde{f}(x).$ Is $X$ a compact set?

Answer 1

If $f$ is general, then $X$ may not be compact. Pick an arbitrary, strictly convex function $g$ with $g(0.5) = 0$, and define $f(x) = g(x)$ for $x \neq 0.5$ and $f(0.5) = 1$. You have $\tilde f = g$ and $X = [0,1]\setminus \{0.5\}$ which is not compact.

If $f$ is continuous, then the result is true, because if $\tilde f(x)<f(x)$ then this happens on a whole neighborhood of $x$ (by the continuity of $f,\tilde f$). This means that $[0,1]\setminus X$ is open, and thus, $X$ is compact.