Hoping you can help me out!
Theorem. Let $u \in Lip^1(\mathbb{R}^n, \mathbb{R}^n)$, such that $||u||\leq \frac{1}{2}$ where $||u||= \text{ess} \sup |Du(x)|$, then $I_d+ u$ is invertible. Also, we have $$ (I_d+u)^{-1}=I_d+v \quad \text{ for some } v \in Lip^1(\mathbb{R}^n, \mathbb{R}^n) $$ and $||v||\leq C ||u||$ where $C$ does not depend on u.
In the book is says that it follows immediately from the fact that $u$ is a contraction (this is clear, since $u$ is Lipschitz). The question is then, how can it be proved that way?
What I did was to prove it using Banach Algebras, since the ball $\{ x+I: ||x||< 1\}$ is a subset of $A^{\times}$, where $A^{\times}$ is the set of all invertible elements of A, in this case $A=Lip^1(\mathbb{R}^n, \mathbb{R}^n)$, but that only proves the invertible part, it isn't clear (at least to me) how the remaining follows.
Thanks!
$\|(I+u)-I\|=\|u\|<1\Rightarrow (I+u)^{-1}=\sum^{\infty}_{n=0}(-1)^nu^n=I+\sum^{\infty}_{n=1}(-1)^nu^n$
Set $v=\sum^{\infty}_{n=1}(-1)^nu^n$ and note that
$\sum^{\infty}_{n=1}(-1)^nu^n=u\sum^{\infty}_{n=1}(-1)^nu^{n-1}$ and
$\|\sum^{\infty}_{n=1}(-1)^nu^{n-1}\|\le 2$