TITLE: Show that $g(x)=x + f(x)$ is surjective
QUESTION [5 upvotes]: Let $f: {\mathbb{R}}^n \rightarrow {\mathbb{R}}^n$ be continuously differentiable and $C \in (0,1)$ a constant, so that ${||Df(x)||}_{op} \leq C$ $\forall x \in {\mathbb{R}}^n$ with $op$ being a operator norm.
Show that $g: {\mathbb{R}}^n \rightarrow {\mathbb{R}}^n$, $g(x)=x+f(x)$ is surjective.
I tried following:
$g(x)=x+f(x)$
$\Leftrightarrow Dg(x)=Dx+Df(x)$
$\Leftrightarrow {||Df(x)||}_{op}={||Dg(x)-Dx||}_{op} \leq C$
I'm not realy sure how to get on from here. Is it possible to use the sub-additive of matrix norms even if there is a minus in the equation?
I'm thankfull for every hint.

REPLY [3 votes]: For any fixed $Y \in \mathbb{R}^n$, consider the function $x \mapsto Y - f(x)$.  By the given hypothesis, you get that this function satisfies the conditions of the Banach fixed-point theorem.  Therefore, this map has a unique fixed point, which is equivalent to a solution of $x + f(x) = Y$.