# Classical fixed-point argument and invertible function

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!

• Can you say what paper/book you are reading? May 27 at 13:02
• Thanks. I don't have an access to this book, though. How is $\|f \|_{1,\infty}$ defined? May 27 at 13:15
• $\vert\vert f\vert\vert_{1,\infty} := \vert\vert f\vert\vert_{\infty} + L$, where the first summand is the supremum norm and $L$ is the Lipschitz constant. May 27 at 13:17
• 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.
– Dirk
May 27 at 13:50
• Thank you Dirk, that makes sense! May 27 at 13:55

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

• 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 $F$. May 27 at 15:30
• @OliverWatt : Thank you for your comment. This is now fixed. May 27 at 15:50
• 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. May 27 at 15:56
• @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. May 27 at 16:03
• Thank you very much. I think I do understand the situation now :) May 27 at 16:07