Show that $g(x)=x + f(x)$ is surjective

Question

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.

Answer 1

Hint: By the reverse triangle inequality, we have $$ \|Dg(x)\|_{op} = \|Dx + Df(x)\|_{op} = \|\operatorname{id}_{\Bbb R^n} + Df(\cdot)\|_{op} \\ \geq \|\operatorname{id}_{\Bbb R^n}\|_{op} - \|Df(\cdot)\|_{op} = 1-C > 0 $$

Answer 2

Hint: Show that the image of $g$ is open and closed.

For openness, use the inverse function theorem locally.

For closedness, bound $\|f(x)\|$ and use Bolzano Weierstrass.

Answer 3

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 $x + f(x) = Y$.