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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.

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    $\begingroup$ 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, 2016 at 18:18
  • $\begingroup$ This is correct, thank you. $\endgroup$
    – user284439
    Mar 3, 2016 at 19:48

2 Answers 2

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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.

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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.

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  • $\begingroup$ Just the result I'm looking for¡ But the link is broken :( $\endgroup$
    – Psaro
    Nov 17, 2023 at 20:28

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