Mean value estimate lemma (differentiable maps)

Let $U\subseteq\mathbb{R}^n$ be open and let $f:U\to\mathbb{R}^m$ be of class $C^1$ . Fix $a\in U$.

Mean value estimate lemma: For every $\varepsilon>0$ there exists $\delta>0$ such that if $x,y\in U$ satisfy $\|x-a\|<\delta$, $\|y-a\|<\delta$, and the line segment $[x,y]\subseteq U$, then $\|f(x)-f(y)-Df(a)(x-y)\|\le \varepsilon\,\|x-y\|.$ In particular, $\|f(x)-f(y)\|\le \bigl(\|Df(a)\|+\varepsilon\bigr)\,\|x-y\|.$

This estimate is a standard quantitative form of differentiability used in proofs of the inverse and implicit function theorems : it says that on sufficiently small scales, $f$ behaves like the linear map $Df(a)$ with a uniformly small relative error.

Proof sketch: Using the fundamental theorem of calculus along the segment $\gamma(t)=x+t(y-x)$, $f(y)-f(x)=\int_0^1 Df(\gamma(t))(y-x)\,dt.$ Subtract $Df(a)(y-x)$ and take norms: $\|f(y)-f(x)-Df(a)(y-x)\| \le \int_0^1 \|Df(\gamma(t))-Df(a)\|\,dt\;\|y-x\|.$ Continuity of $Df$ at $a$ gives $\delta$ such that $\|Df(z)-Df(a)\|<\varepsilon$ whenever $\|z-a\|<\delta$, and $\gamma(t)$ stays within that ball when $x,y$ do.