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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!

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    $\begingroup$ Can you say what paper/book you are reading? $\endgroup$ May 27 at 13:02
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    $\begingroup$ Thanks. I don't have an access to this book, though. How is $\|f \|_{1,\infty}$ defined? $\endgroup$ May 27 at 13:15
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    $\begingroup$ $\vert\vert f\vert\vert_{1,\infty} := \vert\vert f\vert\vert_{\infty} + L$, where the first summand is the supremum norm and $L$ is the Lipschitz constant. $\endgroup$ May 27 at 13:17
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    $\begingroup$ I guess, that the argument for "$I+f$ invertible" is that $x+f(x)=y$ has a unique solution, since the solutions are fixed points of $x\mapsto y-f(x)$ which is a contraction and Banach's fixed point theorem applies. $\endgroup$
    – Dirk
    May 27 at 13:50
  • $\begingroup$ Thank you Dirk, that makes sense! $\endgroup$ May 27 at 13:55
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$\newcommand{\R}{\mathbb R}$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 $|(I-f)^{-1}(y)|=|x|\ge|y|-|f(x)|\to\infty$.

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  • $\begingroup$ Thank you for your answer. A couple of questions come to my mind regarding your solution. 1) Why is $F_{0}$ a Banach-Algebra? Usually the multiplication has to be bilinear, but the composition of maps is not linear in the second argument. 2) Why is $I$ an element of $F_0$? It is not bounded and thus not an element of $F$. $\endgroup$ May 27 at 15:30
  • $\begingroup$ @OliverWatt : Thank you for your comment. This is now fixed. $\endgroup$ May 27 at 15:50
  • $\begingroup$ Ok, this fixes the second problem. But what about the first? I mean, we have $f\circ (g+h) \neq f\circ g + f\circ h$, because our functions are not necessarily linear. $\endgroup$ May 27 at 15:56
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    $\begingroup$ @OliverWatt : We don't need such an additivity. We need the facts that $L$ is a norm on $F_0$ and $L$ is sub-multiplicative with respect to the composition of functions. I have added a detail on the sub-multiplicativity. $\endgroup$ May 27 at 16:03
  • $\begingroup$ Thank you very much. I think I do understand the situation now :) $\endgroup$ May 27 at 16:07

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