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  "body": "\n## Statement\n\nLet $U \\subseteq \\mathbb{R}^n$ be open, let $f \\colon U \\to \\mathbb{R}^m$ be [[differentiable|0135XZZQ8M]] on $U$, and let $C \\subseteq U$ be a convex set.\nSuppose that $M \\ge 0$ and $\\|Df(z)\\| \\le M$ for each $z \\in C$, with the operator norm.\nThen $f$ is Lipschitz on $C$ with constant $M$: $\\|f(y) - f(x)\\| \\le M \\, \\|y - x\\|$ for all $x, y \\in C$.\n\n## Remarks\n\nBecause $C$ is convex, the segment between two points of $C$ lies in $C$. So this follows from the [[mean value inequality|02CYB184RC]].\nConvexity matters. Take the annulus $1 \u003c \\|x\\| \u003c 2$ in $\\mathbb{R}^2$ without the points on the negative $x$-axis, and the polar angle $\\theta \\in (-\\pi, \\pi)$ on it. Then $\\|\\nabla \\theta\\| = 1/\\|x\\| \\le 1$, but two points just above and just below the slit are close, and their angles differ by almost $2\\pi$.\nA function of class $C^1$ on $U$ is Lipschitz on each closed ball in $U$, because $\\|Df\\|$ is bounded on the compact ball.\n\n## Explanations\n\n### L1 human\n\nA Lipschitz function cannot change faster than a fixed rate $M$ per unit of distance.\nA bound on the derivative gives such a rate. For this, you must be able to walk between any two points along a straight line inside the domain.\n",
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