Classical fixed-point argument and invertible function

Question

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!

Answer 1

$\newcommand{\R}{\mathbb R}$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 $|(I-f)^{-1}(y)|=|x|\ge|y|-|f(x)|\to\infty$.