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  "body": "\n## Proof\n\nWe apply the mean value inequality to each pair of points of $C$.\n\n1. Let $U$, $f$, $C$ and $M$ be as in the statement, and let $x, y \\in C$.\n2. For $0 \\le t \\le 1$, the point $x + t (y - x)$ equals $(1 - t) x + t y$.\n3. Since $C$ is [[convex|0K06X9QXDX]], step 2 shows that the segment $[x, y] = \\{x + t (y - x) : 0 \\le t \\le 1\\}$ lies in $C$. As $C \\subseteq U$, it lies in $U$.\n4. By hypothesis, $\\|Df(z)\\| \\le M$ for each $z \\in C$, so in particular for each $z \\in [x, y]$.\n5. The map $f$ is [[differentiable|0135XZZQ8M]] on the open set $U$. By steps 3 and 4 and the [[mean value inequality|02CYB184RC]], $\\|f(y) - f(x)\\| \\le M \\, \\|y - x\\|$.\n6. Since $x, y \\in C$ were arbitrary, $f$ is [[Lipschitz on C|055VVSN00V]] with constant $M$.\n\n## Idea\n\nConvexity puts the whole segment between two points of $C$ inside $C$. On that segment the derivative is bounded by $M$, so the mean value inequality bounds the change of $f$ by $M$ times the distance.\n",
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