Mappings of Finite Inner Distortion: Global Homeomorphism Theorem

Abstract

We show that if $f\in W_{loc}^{1,n-1}(\mathbb {R}^n,\mathbb {R}^n)$, $n\ge 3$, is a local homeomorphism of $K$-bounded 1-mean inner distortion, then $f$ is a homeomorphism onto $\mathbb {R}^n$. We also establish the $K_I$-weighted Väisälä’s inequality for mappings of finite inner distortion under minimal assumptions.

Appendix: Proof of the Regularity of Generalized Inverse

Let $f\in W_{loc}^{1,n-1}(\Omega ,\mathbb {R}^n)$ be a continuous, discrete and open mapping of finite inner distortion with $K_I(\cdot ,f)\in L_{loc}^1(\Omega )$. Moreover, assume that $|f(B_f)|=0$. Let $U\subset \subset \Omega$ be a normal domain such that $f(U)=V$. We set $m=\mu (f,U)$ and define the generalized inverse mapping $g_U:V\rightarrow \mathbb {R}^n$ of $f$ by

$$\begin{aligned} g_U(y)=\frac{1}{m}\sum _{x\in f^{-1}(y)\cap U}i(x,f)x. \end{aligned}$$

(5.1)

We prove the following regularity result for the generalized inverse in this Appendix.

Proposition 5.1

Let $f$ and $U$ be as above. Then $g_U\in W^{1,n}(V,\mathbb {R}^n)$.

Proposition 5.1 can be viewed as a generalization of the main results of [19]. To be more precise, consider the case where $f$ is a homeomorphism. Then $g_U=f^{-1}|_V$ and hence $f^{-1}\in W^{1,n}(V,\mathbb {R}^n)$. On the other hand, Proposition 5.1 also generalizes [14, Theorem 2.1]. We will mainly follow the proof used in the homeomorphic case [1].

We will use the notation $\pi _r(x)=|x|$ for the radial projection and $\pi _S(x)=x/|x|$ for the projection to the unit sphere.

Lemma 5.2

Let $B(x,r_0)\subset U$. Then for almost every $r\in (0,r_0)$ the mapping $f:S(x,r)\rightarrow \mathbb {R}^n$ satisfies condition $N$, i.e.,

$$\begin{aligned} \fancyscript{H}^{n-1}(f(A))=0 \quad \text {for every}\ A\subset S(x,r)\ \text {such that}\ \fancyscript{H}^{n-1}(A)=0. \end{aligned}$$

Proof

This follows immediately from [26, Proposition 3.3].

The next lemma is quite similar to [1, Lemma 4.2].

Lemma 5.3

Let $h=\pi _S\circ f$ and let $E\subset \Omega$ be a measurable set. Then

$$\begin{aligned} \int \limits _{\partial B(0,1)}\fancyscript{H}^1(\pi _r(\{x\in E:h(x)=z\}))d\fancyscript{H}^{n-1}(z)\le \int \limits _E |D^{\#}h(x)|dx. \end{aligned}$$

Proof

If $f$ is Lipschitz, we can use the coarea formula from [16, Theorem 1.1] to obtain

$$\begin{aligned}&\int \limits _{\partial B(0,1)}\fancyscript{H}^1(\pi _r(\{x\in E:h(x)=z\}))d\fancyscript{H}^{n-1}(z)\\&\quad \le \int \limits _{\partial B(0,1)}\fancyscript{H}^1(\{x\in E:h(x)=z\})d\fancyscript{H}^{n-1}(z)=\int \limits _E |D^{\#}h(x)|dx. \end{aligned}$$

In the general case, we cover the domain of $f$ up to a set of measure zero by countably many sets of the type $\{f=f_j\}$ with $f_j$ Lipschitz.

It remains to consider the case where $E=N$ with $|N|=0$. For $z\in S(0,1)$ we denote $S_z=\pi _S^{-1}(z)$. To obtain a contradiction suppose that there is a set $P\subset S(0,1)$ such that $\fancyscript{H}^{n-1}(P)>0$ and for every $z\in P$ we have $\fancyscript{H}^1(\pi _r(f^{-1}(S_z)\cap E))>0$. Consider the set $A\subset (0,\infty )\times S(0,1)$:

$$\begin{aligned} (r,z)\in A \quad \Leftrightarrow \quad z\in P\ \text {and}\ r\in \pi _r(f^{-1}(S_z)\cap E). \end{aligned}$$

By Fubini’s theorem we have

$$\begin{aligned} |A|=\int \limits _P \fancyscript{H}^1(\pi _r(f^{-1}(S_z)\cap E))d\fancyscript{H}^{n-1}(z)>0. \end{aligned}$$

Set $E_r=E\cap S(x,r)$. For almost every $r$ we have $\fancyscript{H}^{n-1}(E_r)=0$ and therefore we obtain $\fancyscript{H}^{n-1}(\pi _S\circ f(E_r))=0$ for almost every $r$ by Lemma 5.2. Now the Fubini theorem implies that

$$\begin{aligned} |A|=\int \limits _0^\infty \fancyscript{H}^{n-1}(\pi _S\circ f(E_r))dr=0, \end{aligned}$$

which gives us a contradiction. $\square$

The next lemma will give us the regularity of $g_U$. The proof is similar to [1, Proof of Lemma 4.3]; see also [7, Proof of Lemma 3.2].

Lemma 5.4

The following Poincaré type inequality holds:

$$\begin{aligned} \int \limits _B |g_U(y)-c|dy\le Cr_0\int \limits _{f^{-1}(B)}|D^{\#}f(x)|dx, \end{aligned}$$

(5.2)

for each ball $B=B(y_0,r_0)\subset V$, where and $C=C(m,n)$.

Proof

We fix a point $y'\in B$ and set

$$\begin{aligned} h(x)=\frac{f(x)-y'}{|f(x)-y'|}. \end{aligned}$$

Let $y''\in B$ and $\beta$ be the line segment joining $y'$ and $y''$. Let $\{x_i'\}=f^{-1}(y')$ and $\{x_i''\}=f^{-1}(y'')$, $i=1,\ldots ,m$.  [25, Corollary II 3.4] implies that there exist paths $\alpha _i\subset U\cap f^{-1}(B)$ connecting $x_i'$ and $x_i''$ such that $f\circ \alpha _i=\beta$, $\text {Card}\{i:\alpha _i(t)=x\}=i(x,f)$ for all $x\in U\cap f^{-1}(\beta (t))$, and $|\alpha _1|\cup \ldots \cup |\alpha _m|=U\cap f^{-1}(|\beta |)$. Choose $x_{i_0}'$ and $x_{i_0}''$ such that

$$\begin{aligned} |x_{i_0}'-x_{i_0}''|=\max \{|x_i'-x_i''|:i=1,\ldots ,m \}. \end{aligned}$$

Without loss of generality, we may assume that $x_{i_0}'=0$. It is clear that

$$\begin{aligned} |x_i'-x_i''|\le \fancyscript{H}^1(\pi _r\circ \alpha _{i_0}), \quad \text {for all}\quad i=1,\ldots ,m. \end{aligned}$$

It follows that

$$\begin{aligned} |g_U(y')-g_U(y'')|\le C(m)\fancyscript{H}^1(\pi _r\circ f^{-1}(\beta )). \end{aligned}$$

On the other hand,

$$\begin{aligned} y\in \beta \quad \Rightarrow \quad \frac{y-y'}{|y-y'|}=\frac{y''-y'}{|y''-y'|}. \end{aligned}$$

Hence, if $r=|y''-y'|$, then

$$\begin{aligned} |g_U(y')-g_U(y'')|&\le C(m)\fancyscript{H}^1(\pi _r\circ f^{-1}(\beta ))\\&\le C(m)\fancyscript{H}^1(\pi _r(\{x\in f^{-1}(B):h(x)=\frac{y''-y'}{r}\})). \end{aligned}$$

Given $r>0$, using Lemma 5.3 for the mapping $f(x)-y'$ we estimate

$$\begin{aligned}&\int \limits _{B\cap \partial B(y',r)}|g_U(y'')-g_U(y')|d\fancyscript{H}^{n-1}(y'')\\&\quad \le C(m)\int \limits _{B\cap \partial B(y',r)}\fancyscript{H}^1(\pi _r(\{x\in f^{-1}(B):h(x)=\frac{y''-y'}{r}\}))d\fancyscript{H}^{n-1}(y'')\\&\quad \le C(m)r^{n-1}\int \limits _{B(0,1)}\fancyscript{H}^1(\pi _r(\{x\in f^{-1}(B):h(x)=z\}))d\fancyscript{H}^{n-1}(z)\\&\quad \le C(m)r^{n-1}\int \limits _{f^{-1}(B)}|D^{\#}h(x)|dx\\&\quad \le C(n,m)r^{n-1}\int \limits _{f^{-1}(B)}\frac{|D^{\#}f(x)|}{|f(x)-y'|^{n-1}}dx, \end{aligned}$$

where the last inequality follows from the chain rule, the formula

$$\begin{aligned} |D^{\#}(AB)|\le C|D^{\#}A||D^{\#}B|, \end{aligned}$$

and the estimate

$$\begin{aligned} \left| D^{\#}\left( \frac{z-y'}{z-y'}\right) \right| \le \frac{C}{|z-y'|^{n-1}}. \end{aligned}$$

Hence

$$\begin{aligned} |B||g_U(y')-c|&\le \int \limits _B |g_U(y'')-g_U(y')|dy''\\&=\int \limits _0^{2r_0}\Big (\int \limits _{B\cap \partial B(y',r)}|g_U(y'')-g_U(y')|d\fancyscript{H}^{n-1}(y'') \Big )dr\\&\le C\int \limits _0^{2r_0}r^{n-1}\left( \int \limits _{f^{-1}(B)}\frac{|D^{\#}f(x)|}{|f(x)-y'|^{n-1}}dx \right) dr\\&\le Cr_0^n \int \limits _{f^{-1}(B)}\frac{|D^{\#}f(x)|}{|f(x)-y'|^{n-1}}dx. \end{aligned}$$

Integrating with respect to $y'$ and then using Fubini’s theorem we obtain

$$\begin{aligned} \int \limits _B |g_U(y')-c|dy'&\le C\int \limits _{f^{-1}(B)}|D^{\#}f(x)|\left( \int \limits _B\frac{dy'}{|f(x)-y'|^{n-1}} \right) dx\\&\le Cr_0\int \limits _{f^{-1}(B)}|D^{\#}f(x)|dx. \end{aligned}$$

$\square$

Proof of Proposition 5.1

We claim that there is a function $g\in L^n(V)$ such that

$$\begin{aligned} \int \limits _{f^{-1}(B)}|D^{\#}f|\le m\int \limits _B g. \end{aligned}$$

(5.3)

This and Lemma 5.4 imply that the pair $g_U$, $g$ satisfies a $1$-Poincaré inequality in $V$. From [3, Theorem 9] or [5, Theorem 9.4.2], we then deduce that $g_U\in W^{1,n}(V,\mathbb {R}^n)$.

By Theorem 3.3, $f$ is a mapping of finite distortion. Since $K_I(\cdot ,f)\in L^1(U)$ and by the last inequality in [26, p. 18]

$$\begin{aligned} |Df(x)|^n\le C(n)J_f(x)K_I(x,f)^{n-1} \end{aligned}$$

for a.e. $x\in U$, we know that $K(\cdot ,f)\in L^1(U)$. It follows from [13, Theorem 1.2] that $f$ satisfies condition $N^{-1}$ and hence $|f^{-1}(f(B_f))|=0$. Since $U$ is compact, we may decompose $U\backslash f^{-1}(f(B_f))=\cup _{i=1}^k U_i$ such that $f|_{U_i}$ is a homeomorphism and $U_i\cap U_j=\emptyset$ if $i\ne j$ (here we have used the fact that $f$ is a covering mapping on $U\backslash f^{-1}(f(B_f))$, which is true since $U$ is a normal domain; see, e.g., [27]). Let $V_i=f(U_i)$. Then $V=\cup _{i=1}^k V_i\cup f(B_f)$. On the other hand, since the area formula holds on a full measure set $U_i'\subset U_i$ by Lemma 2.2, we define a function $g_i:V_i\rightarrow \mathbb {R}^n$ by setting

$$\begin{aligned} g_i(f(x))={\left\{ \begin{array}{ll} \frac{|D^{\#}f(x)|}{J_f(x)} &{} \quad \text {if } \quad x\in U_i',\ J_f(x)>0 \\ 0 &{} \quad \text {otherwise}. \end{array}\right. } \end{aligned}$$

Note that $J(\cdot ,f)>0$ a.e. again by [13, Theorem 1.2], we have

$$\begin{aligned} |D^{\#}f(x)|=g_i(f(x))J_f(x)\qquad \text {a.e.}\ \text {in}\ U_i. \end{aligned}$$

The desired mapping $g$ is then defined to be $g_i$ on each $V_i$ and $0$ on $f(B_f)$. Now

$$\begin{aligned} \int \limits _V g(y)^ndy&=\sum _{i=1}^k \int \limits _{V_i}g_i(y)^ndy\le \sum _{i=1}^k\int \limits _{V_i} g_i(y)^nN(y,f,U_i')dy\\&=\sum _{i=1}^k\int \limits _{f^{-1}(V_i)\cap U_i'}g_i(f(x))^nJ_f(x)dx=\int \limits _{U}K_I(x)dx \end{aligned}$$

and

$$\begin{aligned} \int \limits _{f^{-1}(B)}|D^{\#}f(x)|dx&\le \int \limits _{f^{-1}(B)\cap U'}g(f(x))J_f(x)dx\\&\le m\int \limits _B g(y)dy. \end{aligned}$$

The proof is then complete.