Let $n\in\mathbb{N}$ and $W^{1,\infty}(\mathbb{R}^n)=\lbrace f:\mathbb{R}^n\rightarrow \mathbb{R}^n : \text{ f is bounded and Lipschitz continuous } \rbrace$. Suppose $f\in W^{1,\infty}(\mathbb{R}^n)$ with $\vert\vert f \vert\vert_{1,\infty}<1$ is given and $I:\mathbb{R}^n\rightarrow\mathbb{R}^n$ denotes the identity map.

By a **classical fixed-point argument** $I+f$ is invertible, $(I+f)^{-1} -I \in W^{1,\infty}(\mathbb{R}^n)$ and the following inequalities hold:

$$\vert\vert (I+f)^{-1} -I \vert\vert_{1,\infty} \leq \vert\vert f \vert\vert_{1,\infty} \cdot (1- \vert\vert f \vert\vert_{1,\infty})^{-1}, \\ \vert\vert (I+f)^{-1} -I + f \vert\vert_{\infty} \leq \vert\vert f \vert\vert_{1,\infty} \cdot \vert\vert I - (I+f)^{-1} \vert\vert_{\infty}.$$

I have problems understanding the reasoning because many details are left out. Does anyone understand how to obtain those results?

At least formally I have obtained both inequalities by using a Neumann series and setting $(I+f)^{-1} := \sum_{k=0}^{\infty} (-f)^k $. But I do not know exactly how to interpret the multiplication in the expression $f^k$, because problems appeared with any choice I could think of. For example, if we choose the composition of maps as multiplication, then the set $W^{1,\infty}$ does not become a Banach Algebra and the series is (probably?) only a left inverse. Maybe, if this idea could be made somehow precise, then I would appreciate if you can comment how to do it.

Best wishes and thank you for your help!