 Negative part of a finite convex function is Lipschitz

For a finite convex function $f:\mathbb{R}\to \mathbb{R}$, its negative part $f^{-}=\max (-f,0)$ is globally Lipschitz continuous. The negative sublevel set is an interval. If bounded, local Lipschitz continuity of finite convex functions bounds the slopes on its closure. If it is a left ray, any negative slope there would force $f$ positive far to the left by the supporting-line inequality; all slopes on that ray are therefore nonnegative and bounded above near its finite endpoint. For a right ray the analogous slopes are nonpositive and bounded below. If $f$ is negative everywhere, convexity and boundedness above force it to be constant. These cases bound the slopes of $f^{-}$ globally, including its zero extension outside the interval. In particular, integrability of $f(Z)$ implies integrability of the negative part of every fixed translate $f(Z+x)$.