Mean value inequality (multivariable)

Mean value inequality (multivariable): Let $U\subseteq\mathbb{R}^n$ be open and let $f:U\to\mathbb{R}^m$ be differentiable . Suppose $x,y\in U$ and the line segment $[x,y]=\{x+t(y-x):0\le t\le 1\}$ is contained in $U$. If there is a constant $M$ such that $\|Df(z)\| \le M \quad \text{for all } z\in[x,y]$ (where $\|Df(z)\|$ is the operator norm ), then $\|f(y)-f(x)\|\le M\|y-x\|.$

This inequality is the multivariable analogue of the one-dimensional mean value estimate and is used to prove Lipschitz bounds , uniqueness results, and the inverse /implicit function theorems .

Proof sketch: Define $\gamma(t)=x+t(y-x)$ and consider $F(t)=f(\gamma(t))$. Then $F'(t)=Df(\gamma(t))(y-x)$ by the chain rule , and $\|F'(t)\|\le \|Df(\gamma(t))\|\,\|y-x\|\le M\|y-x\|.$ Integrate from $0$ to $1$: $f(y)-f(x)=F(1)-F(0)=\int_0^1 F'(t)\,dt,$ and bound the integral norm by $M\|y-x\|$.