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  "body": "\n## Statement\n\nLet $I$ be an open interval and let $f \\colon I \\to \\mathbb{R}$ be convex.\nThen $f$ is [[continuous|0P27W9RW0G]] on $I$.\n\n## Remarks\n\nFix $y \\in I$ and points $x \u003c y \u003c z$ in $I$. By the [[slope inequalities|0BY5FTHEEE]], the slopes of chords from $y$ are bounded for points near $y$. So $|f(w) - f(y)| \\le C |w - y|$ near $y$.\nOn a closed interval this fails. The function on $[0, 1]$ with $f(0) = 1$ and $f(x) = 0$ for $x \u003e 0$ is convex but not continuous at $0$.\nA convex function on an open interval has one-sided derivatives at every point. Mathlib states continuity as `ConvexOn.continuousOn`, for open sets.\n\n## Explanations\n\n### L1 human\n\nA convex function cannot jump inside an open interval.\nA jump would break the condition that the chords lie above the graph.\n",
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