Convex and continuous function on compact set implies Lipschitz

Question

Let the function $f: C \rightarrow \mathbb{R}$ be convex and continuous, where $C \subset \mathbb{R}^n$ is a compact set.

Prove or disprove that $f$ is Lipschitz continuous on $C$.

Comments: If $f$ were defined on an open set $O$, then I can show that it is Lipschitz continuous on every compact subset of $O$, see Proving that a convex function is Lipschitz for the scalar case. In this case, however, I need to exploit the assumption that $f$ is continuous also on the boundary of $C$ - continuity of $f$ in the interior of $C$ already follows from the convexity assumption.

Answer 1

As noted by Svetoslav in a comment, the statement is false: a counterexample is given by $f(x) = -\sqrt{1-x^2}$ on $[-1,1]$.

A slightly simpler counterexample is $f(x) = -\sqrt{x}$ on $[0,1]$.

In both cases, the issue is that $f'$ is not bounded. A function with unbounded derivative cannot be Lipschitz.

Answer 2

This statement is true if instead of dealing with arbitrary compact subsets, you deal with compact subsets of the interior of the domain. This is Theorem 3.3.1 in these notes.