Bounded derivative on a convex set implies Lipschitz

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^m$ be differentiable on $U$, and let $C \subseteq U$ be a convex set. Suppose that $M \ge 0$ and $\|Df(z)\| \le M$ for each $z \in C$, with the operator norm. Then $f$ is Lipschitz on $C$ with constant $M$: $\|f(y) - f(x)\| \le M \, \|y - x\|$ for all $x, y \in C$.

Remarks

Because $C$ is convex, the segment between two points of $C$ lies in $C$. So this follows from the mean value inequality. Convexity matters. Take the annulus $1 < \|x\| < 2$ 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 $2\pi$. A function of class $C^1$ on $U$ is Lipschitz on each closed ball in $U$, because $\|Df\|$ is bounded on the compact ball.

Explanations

L1 human

A Lipschitz function cannot change faster than a fixed rate $M$ per unit of distance. A 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.