---
tag: 02CYB184RC
kind: statement
role: theorem
name: Mean value inequality
msc: [26B05]
uses: [0135XZZQ8M, 0XPR1E85W0]
sources:
  - url: https://en.wikipedia.org/wiki/Mean_value_theorem
    locator: Mean value inequality
  - url: https://en.wikipedia.org/wiki/Mean_value_theorem
    locator: Mean value theorem for vector-valued functions
---

## Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^m$ be [[differentiable|0135XZZQ8M]] 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|03HGK7EW6R]]. 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|0TKG7T3K0E]].

## 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.