Negative part of convex function is globally Lipschitz continuous?

Suppose $f:R^n\to R$ is a convex function. Define the the negative part $f^- (x) = |\min\{0, f(x) \}|$. Is $f^-(x)$ globally Lipschitz continuous in $x$?

I think it is since if it not then the epigraph of $f$ is not a convex set and we have a contradiction.

Edit: Let me clarify my thinking. I think $f^-$ is globally Lipschitz continuous because if it is not, then the epigraph of $f$ is not a convex set. But I don't know how to show this.

• The fact that the epigraph is not convex is not a contradiction -- convexity of it is not a necessary condition for Lipschitz continuity. Actually I'm rather sure that $f^-$ is (locally) Lipschitz, since sublevels of convex functions are convex regions, so $f^-$ equals $-f$ on a convex set and $0$ everywhere else.Globally Lipshitz is not necessarily true since it is not true for globally convex functions. – Thomas Jan 11 '18 at 16:45
• Let $n=1$ and $f(x)=-x^2$. – Andrew Jan 11 '18 at 17:06
• @Andrew: $f(x)=-x^2$ is not convex in x. No? – master_goon Jan 12 '18 at 2:30
• @Thomas: If the epigraph of $f$ is not a convex set, then $f$ is not a convex function. So this yields a contradiction. – master_goon Jan 12 '18 at 4:33
• @Thomas: I think maybe I wasn't clear in my question. Suppose $f$ is convex. I want to show that the negative part $f^-:=|\min\{ 0, f\}|$ is Lipschitz. – master_goon Jan 12 '18 at 9:00

Nice question - here is a partial answer: It is true when $n=1$.
Sketch for one case: Suppose $a$ is the smallest zero of $f$, and $b$ the largest. Then (if needed one-sided versions of) $f'$ at $a$ and $b$ are finite (since $f$ is locally Lipschitz) and since $f'$ is increasing, the set $f'([a,b])$ is bounded. Then $\sup \big|f'([a,b])\big|$ will be your global Lipschitz constant for $f^-$.
Sketch for another case: Suppose $a$ is the only zero of $f$ and say $f'(a)\leq 0$. Then $f'(x)\leq 0$ for $x\geq a$ for otherwise there would be another zero. Since $f'$ is increasing, we note that $|f'(a)| = \max_{x\geq a}|f'(a)|$ and thus $f^-$ is Lipschitz continuous with constant $|f'(a)|$.
I don't know about $n\geq 2$.
• Thanks. However, $f(x) = -x$ is convex but has only one zero. Also, the derivative $f'$ may not exist e.g. convex piecewise linear functions. – master_goon Jan 12 '18 at 2:54
• I wrote for simplicity $f'$.You can work with either the left or right directional derivative instead if you like. – max_zorn Jan 12 '18 at 2:59