Prove that $x+g$ is homeomorphism

Question

Problem: Assume we have $g:\mathbb{R}^n\longrightarrow \mathbb{R}^n$ of $C^1$ class with derivative bounded uniformly by some constant $M<1$. Consider $h=\mathcal{id}_{\mathbb{R}^n}+g:\mathbb{R}^n\longrightarrow \mathbb{R}^n$. Prove:

1. $h$ is homeomorphism

2. $h^{-1}$ is Lipschitz function

My solution: Clearly $h$ is continous, if $h(x)=h(y)$ than $|x-y|=|g(x)-g(y)|$ what contradicts increment theorem if $x\not = y$ so we have that $h$ is injective. From known fact from topology injective continous function between $\mathbb{R}^k$ is homeomorphism on image.

My question: I think that assumption of being of $C^1$ class can be ommited (if differentiability is kept) because all arguments work than, am I right ? Any ideas how to prove $2$?

Answer 1

Here is how I would prove questions 1. and 2.

1. $h$ is an homeomorphism

By the mean value theorem you have for $x,y \in \mathbb R^n$ $$\Vert g(x) - g(y) \Vert = \Vert (h(x) - h(y)) - (x-y)\Vert \le M \Vert x-y \Vert$$ hence by the reverse triangle inequality $$(1-M)\Vert x- y \Vert \le \Vert h(x)-h(y)\Vert$$ which implies that $h$ is injective. According to following post (which rely on Banach-Fixed point theorem) $h$ is also surjective. $h$ image is $\mathbb R^n$.

$h$ is differentiable as the sum of two differentiable maps. Also for all $x \in \mathbb R^n$, $h^\prime(x)$ is invertible. If not, there exists $k \in \mathbb R^n \setminus \{0\}$ with $0=h^\prime(x).k = k + g^\prime(x).k$ which leads to $\Vert k \Vert = \Vert g^\prime(x).k \Vert$ in contradiction with $\Vert g^\prime(x) \Vert \le M < 1$.

According to the inverse function theorem $h$ is invertible, the inverse being continuously differentiable. In particular $h$ is an homeomorphism from $\mathbb R^n$ onto $\mathbb R^n$.

2. $h^{-1}$ is Lipschitz function

For $x \in \mathbb R^n$ you have $$\Vert h^\prime(x) - \mathcal{id}_{\mathbb{R}^n} \Vert = \Vert g^\prime(x) \Vert \le M < 1$$ hence $$(h^\prime(x))^{-1} = \sum_{k=0}^\infty (\mathcal{id}_{\mathbb{R}^n}-h^\prime(x))^k$$ which implies $$\Vert (h^\prime(x))^{-1} \Vert \le \frac{1}{1-\Vert \mathcal{id}_{\mathbb{R}^n} - h^\prime(x) \Vert}= \frac{1}{1-\Vert g^\prime(x) \Vert} \le \frac{1}{1-M}$$

Using again mean value theorem, and the equality $$(h^{-1})^\prime(h(x))=(h^\prime(x))^{-1}$$ valid for all $x \in \mathbb R^n$ we get that $h$ is a Lipschitz function with coefficient $\frac{1}{1-M}$.