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.
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 $n\geq 2$.