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!



Classical fixed-point argument and invertible function
$\newcommand{\R}{\mathbb R}$Let $F:=W^{1,\infty}(\R^n)$, with $\|f\|\_{1,\infty}:=\|f\|\_\infty+L(f)$ for $f\in F$, where $L(f)$ is the Lipschitz constant of $f$.


Take any $f\in F$ with $q:=\|f\|\_{1,\infty}<1$.
Let us show that then $I-f$ is invertible. Consider the Banach space
\begin{equation}
 F\_0:=\{h\in\R^\R\colon h(0)=0,L(h)<\infty\}
\end{equation}
with the norm $L$. Note that $F$ is closed with respect to the composition of functions, since $L(u\circ v)\le L(u)L(v)$ for $u$ and $v$ in $F\_0$. The functions $I$ and $g:=f-f(0)$ are in $F\_0$, and also $L(g)=L(f)\le q<1$, so that $L(g^{\circ k})\le L(g)^k\le q^k$ for natural $k$, where $g^{\circ k}$ denotes the $k$-fold composition of $g$ with itself.


So, we have the inverse $(I-g)^{-1}=\sum\_{k=0}^\infty g^{\circ k}\in F\_0$, so that $L((I-g)^{-1})<\infty$.


Note that $I-f=I-g-f(0)=s\_{f(0)}\circ(I-g)$, where $s\_a(x):=x-a$ for $x\in\R^n$. The shifts $s\_a$ are obviously invertible. So, we have the inverse $(I-f)^{-1}=(I-g)^{-1}\circ s\_{f(0)}^{-1}$, and $L((I-f)^{-1})=L((I-g)^{-1}\circ s\_{f(0)}^{-1})=L((I-g)^{-1})<\infty$.


However, $(I-f)^{-1}\notin F$, since the conditions $(I-f)(x)=y$ and $|y|\to\infty$ for $f\in F$ and $x,y$ in $\R^n$ imply $|(I-f)^{-1}(y)|=|x|\ge|y|-|f(x)|\to\infty$.


