A convex function on an open interval is continuous

Statement

Let $I$ be an open interval and let $f \colon I \to \mathbb{R}$ be convex. Then $f$ is continuous on $I$.

Remarks

Fix $y \in I$ and points $x < y < z$ in $I$. By the slope inequalities, the slopes of chords from $y$ are bounded for points near $y$. So $|f(w) - f(y)| \le C |w - y|$ near $y$. On a closed interval this fails. The function on $[0, 1]$ with $f(0) = 1$ and $f(x) = 0$ for $x > 0$ is convex but not continuous at $0$. A convex function on an open interval has one-sided derivatives at every point. Mathlib states continuity as ConvexOn.continuousOn, for open sets.

Explanations

L1 human

A convex function cannot jump inside an open interval. A jump would break the condition that the chords lie above the graph.