0
$\begingroup$

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.

| cite | improve this question | |
$\endgroup$
  • 3
    $\begingroup$ Take for simplicity $C\subset \mathbb R^1$, where $C=[-1,1]$ and $f$ to be the lower semicircle of the unit sphere in $\mathbb R^2$, i.e $f=-\sqrt{1-x^2}$. This function is convex and continuous, but not Lipschitz. The problem arises on the boundary of the set, where the derivatives can be unbounded. $\endgroup$ – Svetoslav Mar 3 '16 at 18:18
  • $\begingroup$ This is correct, thank you. $\endgroup$ – user284439 Mar 3 '16 at 19:48
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.