Mean value inequality

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^m$ be differentiable on $U$, and let $x, y \in U$ be points such that the segment $[x, y] = \{x + t (y - x) : 0 \le t \le 1\}$ lies in $U$. Suppose that $M \ge 0$ and $\|Df(z)\| \le M$ for each $z \in [x, y]$, where $\|Df(z)\| = \sup_{\|h\| \le 1} \|Df(z) h\|$ is the operator norm. Then $$\|f(y) - f(x)\| \le M \, \|y - x\|.$$

Remarks

For $m = 1$, the inequality follows from the mean value theorem in several variables. For $m \ge 2$ there is in general no equality form, but the inequality still holds. A standard proof applies the one-variable mean value theorem to $\varphi(t) = (f(y) - f(x)) \cdot f(x + t(y - x))$. On a convex set $C \subseteq U$ with $\|Df\| \le M$ on $C$, the inequality holds for all $x, y \in C$, so $f$ is Lipschitz on C.

Explanations

L1 human

Suppose that the speed of change of $f$ is at most $M$ everywhere on a straight path. Then $f$ changes by at most $M$ times the length of the path. This is like saying that a car that never goes faster than $100$ km/h cannot cover more than $200$ km in two hours.