Bounded derivative on a convex set implies Lipschitz

Builds on Mean value inequality Differentiable function of several variables

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.

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Euclidean space object
Function concept
Function of several variables concept
Limit of a function of several variables concept
Differentiable function of several variables definition
Mean value inequality statement
Bounded derivative on a convex set implies Lipschitz statement

Tag	`0TKG7T3K0E`
URI	`m2://0TKG7T3K0E`
Kind	statement, corollary
Writing	W0 stub
MSC	26B35
Status	This copy of the site has no build data. Run `m2 status 0TKG7T3K0E`.
Uses	Mean value inequality, Differentiable function of several variables

Bounded derivative on a convex set implies Lipschitz

Statement

Remarks

Reading path

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