Mean value theorem in several variables

Statement

Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}$ be differentiable on $U$, and let $x, y \in U$ be points such that the segment $\{x + t (y - x) : 0 \le t \le 1\}$ lies in $U$. Then there is a number $c \in (0, 1)$ such that $$f(y) - f(x) = \nabla f\big(x + c (y - x)\big) \cdot (y - x).$$

Remarks

The proof applies the one-variable mean value theorem to $g(t) = f(x + t(y - x))$ on $[0, 1]$, with $g'(t) = \nabla f(x + t(y - x)) \cdot (y - x)$ by the chain rule for a curve. The theorem is false for maps into $\mathbb{R}^m$ with $m \ge 2$. For $f(t) = (\cos t, \sin t)$ on $[0, 2\pi]$, $f(2\pi) - f(0) = 0$, but $f'(t) \ne 0$ for each $t$. The mean value inequality replaces it. By the Cauchy–Schwarz inequality, it gives $|f(y) - f(x)| \le \sup_{z} \|\nabla f(z)\| \, \|y - x\|$, with $z$ on the segment.

Explanations

L1 human

On a straight path from $x$ to $y$, the average slope of $f$ equals the actual slope at some point between them. This reduces the problem to one variable, by looking at $f$ only along the segment.