---
abstract: 'We extend monotone quasiconformal mappings from dimension $n$ to while preserving both monotonicity and quasiconformality. The extension is given explicitly by an integral operator. In the case $n=1$ it yields a refinement of the Beurling-Ahlfors extension.'
address:
- |
    Department of Mathematics\
    Syracuse University\
    Syracuse, NY 13244, USA
- |
    Department of Mathematics\
    Syracuse University\
    Syracuse, NY 13244, USA
author:
- 'Leonid V. Kovalev'
- Jani Onninen
date: 'April 17, 2008'
title: |
    An $N$-dimensional version of\
    the Beurling-Ahlfors extension
---

[^1]

[^2]

Introduction
============

Given a mapping $f \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ of class $\mathscr A$, find $F \colon {\mathbb{R}}^{n+1} \to {\mathbb{R}}^{n+1}$ of class $\mathscr A$ such that the restriction of $F$ to ${\mathbb{R}}^n$ agrees with $f$.

Let us introduce coordinate notation $x=(x^1, \dots ,x^n)$ and $f=(f^1, \dots , f^n)$. By setting $F^i=f^i$ for $i=1, \dots , n$ and $F^{n+1}=x^{n+1}$ one immediately obtains a solution to the extension problem for many classes $\mathscr A$ such as continuous ($\mathscr A=C^0$), smooth ($\mathscr A=C^{k}$), homeomorphic, diffeomorphic, and (bi-)Lipschitz mappings.

When $\mathscr A={\mathcal{QC}}$, the class of quasiconformal mappings, the extension problem is much more difficult. It was solved

-   for $n=1$ by Beurling and Ahlfors [@BA] in 1956,

-   for $n=2$ by Ahlfors [@Ah] in 1964,

-   for $n\le 3$ by Carleson [@Ca] in 1974, and

-   for all $n \ge 1$ by Tukia and Väisälä [@TV] in 1982.

The Tukia-Väisälä extension uses, among other things, Sullivan’s theory [@Su] of deformations of Lipschitz embeddings. Our goal is to give an explicit extension for a subclass of ${\mathcal{QC}}$. Quasiconformal mappings can be defined as orientation-preserving quasisymmetric mappings [@Heb; @Vab].

\[qs\] A homeomorphism $f \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ is quasisymmetric if there is a homeomorphism $\eta \colon [0,\infty ) \to [0, \infty )$ such that $$\label{Kav24}
 \frac{|f(x)-f(z)|}{|f(y)-f(z)|} \le \eta \left(\frac{|x-z|}{|y-z|}\right) .$$ for $x,y, z \in {\mathbb{R}}^n$, $z \ne y$.

One can say that quasisymmetry is a three-point condition. But there are two subclasses of ${\mathcal{QC}}$ that are defined by *two-point* conditions, namely bi-Lipschitz class ${\mathcal{BL}}$ and the class of nonconstant delta-monotone mappings [@AIMb Chapter 3]. Recall that a mapping $f\colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ is [*monotone*]{} if $$\label{mon}
{\langlef(x)-f(y), x-y \rangle} \ge  0 \qquad \textnormal{ for all } x, y\in {\mathbb{R}}^n.$$ We called $f$ [*delta-monotone*]{} if there exists $\delta >0$ such that $$\label{demon}
{\langlef(x)-f(y), x-y \rangle} \ge   \delta{\lvertf(x)-f(y)\rvert} {\lvertx-y\rvert}  \qquad \textnormal{ for all } x, y\in {\mathbb{R}}^n.$$ The class of nonconstant delta-monotone mappings is denoted by ${ \mathcal{DM}}$. When we want to specify the value of $\delta$ we write that $f$ is $\delta$-monotone.

In contrast to the bi-Lipschitz case, the extension problem for the class ${ \mathcal{DM}}$ cannot be solved by means of the trivial extension. For example, the mapping $f(x)={\lvertx\rvert}^p x$, $p>-1$, belongs to ${ \mathcal{DM}}$ but its trivial extension does not (unless $p=0$).

Let $n \ge 2$. For any mapping $f \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ of class ${ \mathcal{DM}}$ there exists $F \colon {\mathbb{R}}^{n+1} \to {\mathbb{R}}^{n+1}$ of class ${ \mathcal{DM}}$ such that the restriction of $F$ to ${\mathbb{R}}^n$ agrees with $f$.

Our proof is by an explicit construction that can be viewed as an $n$-dimensional version of the Beurling-Ahlfors extension. Suppose $f\in { \mathcal{DM}}$. Let ${\mathbb{R}}^{n+1}_+={\mathbb{R}}^n\times[0,\infty)$ and $$\label{gauss}
 \phi(x) =(2 \pi)^{-\frac{n}{2}} e^{-{\lvertx\rvert}^2/2}, \qquad x\in {\mathbb{R}}^n.$$ We define $F \colon {\mathbb{R}}^{n+1}_+ \to  {\mathbb{R}}^{n+1}_+ $ by $$\begin{aligned}
F^i(x,t)&= &\int_{{\mathbb{R}}^n} f^i(x+ty)\,\phi (y)\, d y  \qquad i=1, \dots , n \label{Fi}\\
F^{n+1}(x,t)&= &\int_{{\mathbb{R}}^n}  {\langlef(x+ty),y\rangle}\, \phi(y)\, dy \label{Fn1} 
 \end{aligned}$$ where $x\in{\mathbb{R}}^n$, $t\ge 0$ (see §\[pfth\] for the convergence of these integrals). Observe that $F(x,0)=(f(x),0)$. Furthermore, $F^{n+1}(x,t) \ge 0$ because $$\int_{{\mathbb{R}}^n}  {\langlef(x+ty),y\rangle}\, \phi(y)\, dy =  \int_{{\mathbb{R}}^n}  {\langlef(x+ty) -f(x),y\rangle}  \, \phi(y)\, dy \ge 0$$ due to the monotonicity of $f$. Finally, we extend $F$ to ${\mathbb{R}}^{n+1}$ by reflection $$F^i (x,t) = F^i(x,-t) \quad i=1,\dots , n \quad \mbox{ and } \quad  F^{n+1}(x,t)= - F^{n+1}(x,-t).$$

\[Thm\] Let $n \ge 2$. If $f \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ is $\delta$-monotone, then $F \colon {\mathbb{R}}^{n+1} \to  {\mathbb{R}}^{n+1} $ is $\delta_1$-monotone where $\delta_1$ depends only on $\delta$ and $n$. In addition, $F \colon \mathbb H^{n+1} \to \mathbb H^{n+1}$ is bi-Lipschitz in the hyperbolic metric.

Here $\mathbb H^{n+1} ={\mathbb{R}}^n\times(0,\infty) $ and the hyperbolic metric on $\mathbb H^{n+1}$ is ${\lvertd x\rvert}/x^{n+1}$. Theorem \[Thm\] can be also formulated for $n=1$, in which case it becomes a refinement of the Beurling-Ahlfors extension theorem.

\[n1\] If $f \colon {\mathbb{R}}\to {\mathbb{R}}$ is increasing and quasisymmetric, then $F \colon {\mathbb{R}}^{2} \to  {\mathbb{R}}^{2} $ is $\delta_1$-monotone where $\delta_1$ depends only on $\eta$ in Definition \[qs\]. Furthermore, $F \colon \mathbb H^{2} \to \mathbb H^{2}$ is bi-Lipschitz in the hyperbolic metric.

Fefferman, Kenig and Pipher [@FKP Lemma 4.4] proved that $F$ in Proposition \[n1\] is quasiconformal. Proposition \[n1\] was originally proved in [@Ko] using their result. In this paper we give a direct proof.

Theorem \[Thm\] has an application to mappings with a convex potential [@Ca92], i.e., those of the form $f=\nabla u$ with $u$ convex. The basic properties and examples of quasiconformal mappings with a convex potential are given in [@KM].

\[convex\] Suppose that $f\colon {\mathbb{R}}^n\to{\mathbb{R}}^n$, $n\ge 2$, is a $K$-quasiconformal mapping with a convex potential. Then $f$ can be extended to a $K_1$-quasiconformal mapping $F\colon {\mathbb{R}}^{n+1}\to{\mathbb{R}}^{n+1}$ with a convex potential, where $K_1$ depends only on $K$ and $n$.

Preliminaries {#secdef}
=============

Let $e_1,\dots,e_{n+1}$ be the standard basis of ${\mathbb{R}}^{n+1}$. All vectors are treated as column vectors. The transpose of a vector $v$ is denoted by $v^T$. We use the operator norm ${\lVert\cdot\rVert}$ for matrices. A Borel measure $\mu$ on ${\mathbb{R}}^n$ is *doubling* if there exists $\mathscr D_\mu$, called the doubling constant of $\mu$, such that $$\mu (2B) \le \mathscr D_\mu \, \mu (B)$$ for all balls $B=B(x,r)$. Here $2B=B(x,2r)$.

The geometric definition of class ${\mathcal{QC}}$ given in the introduction is equivalent to the following analytic definition [@Heb; @Vab].

\[defKqc\] A homeomorphism $f \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ ($n \ge 2$) is quasiconformal if $f \in W^{1,n}_{\operatorname{loc}} ({\mathbb{R}}^n , {\mathbb{R}}^n)$ and there exists a constant $K$ such that the differential matrix $Df(x)$ satisfies the distortion inequality $${\lVertDf(x)\rVert}^n \le K \det Df(x) \quad \mbox{ a.e. in } {\mathbb{R}}^n.$$

Delta-monotone mappings also have an analytic definition.

\[deltas\] Let $\Omega$ be a convex domain in ${\mathbb{R}}^n$, $n \ge 2$. Suppose $f\in W^{1,1}_{\operatorname{loc}} (\Omega , {\mathbb{R}}^n)$ is continuous. The following are equivalent:

(i) \[delta1\] $f$ is $\delta$-monotone in $\Omega$ for some $\delta >0$; that is,  holds for all $x,y\in\Omega$;

(ii) \[delta2\] there exists $\delta>0$ such that for a.e. $x\in \Omega$ the matrix $Df(x)$ satisfies $$v^TDf(x)v \ge \delta {\lvertDf(x)v\rvert} {\lvertv\rvert}\qquad \text{for every vector $v\in {\mathbb{R}}^n$;}$$

(iii) \[delta3\] there exists $\gamma>0$ such that for a.e. $x\in\Omega$ the matrix $Df(x)$ satisfies $$v^TDf(x)v \ge \gamma {\lVertDf(x)\rVert} {\lvertv\rvert}^2\qquad \text{for every vector $v\in {\mathbb{R}}^n$.}$$

The constants $\delta$ and $\gamma$ depend only on each other.

The equivalence of  and , with the same constant $\delta$, was proved in [@Ko p. 397]. It is obvious that  implies  with $\delta=\gamma$. It remains to establish the converse implication  $\implies$ . To this end we need the following\
[**Claim:**]{} if a real square matrix $A$ satisfies $$v^TAv \ge \delta {\lvertAv\rvert} {\lvertv\rvert}\qquad \text{for every $v\in {\mathbb{R}}^n$}$$ then $$\label{something}
{\lvertAv\rvert} \ge c{\lVertA\rVert}{\lvertv\rvert}\qquad c=c(\delta)>0.$$ Although this claim is known, even with a sharp constant [@AIM], we give a proof for the sake of completeness. It suffices to estimate ${\lvertAv\rvert}$ from below under the assumptions that $Av \ne 0$ and ${\lVertA\rVert}=1={\lvertv\rvert}$. Let $u$ be a unit vector in ${\mathbb{R}}^n$ such that ${\lvertAu\rvert}=1$. Replacing $u$ by $-u$ if necessary we may assume that $u^TAv+v^TAu\le0$. Let $\lambda = \sqrt{{\lvertAv\rvert}}$. On one hand we have $$\label{no1}
(\lambda u +v)^T A (\lambda u +v) \le \lambda^2 u^T Au + v^T Av \le \lambda^2 + \lambda^2=2\lambda^2.$$ On the other hand $$\label{no2}
(\lambda u +v)^T A (\lambda u +v) \ge  \delta {\lvert\lambda Au + Av\rvert} {\lvert\lambda u +v\rvert} \ge \delta (\lambda - \lambda^2)(1-\lambda).$$ Combining  and  we obtain $2 \lambda \ge    \delta (1-\lambda)^2$, hence $$\lambda \ge \delta^{-1}+1- \sqrt{(\delta^{-1}+1)^2-1}>0.$$ This proves the claim.

Delta-monotone mappings and doubling measures
=============================================

The following result shows that ${ \mathcal{DM}}\subset {\mathcal{QC}}$. In particular, $f\in { \mathcal{DM}}$ implies that $f$ is a continuous Sobolev mapping, and therefore – of Lemma \[deltas\] hold.

\[dmqs\] [@Ko Theorem 6] Every nonconstant $\delta$-monotone mapping is $\eta$-quasisymmetric where $\eta$ depends only on $\delta$.

It is well-known that quasisymmetric mappings are closely related to doubling measures [@Heb]. The following lemma is another instance of this relation.

\[Dfdoub\] For any nonconstant $\delta$-monotone mapping $f \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ ($n\ge 2$) the measure $\mu= {\lVertDf(x)\rVert} \, dx$ is doubling. The doubling constant $\mathscr D_\mu$ depends only on $\delta$ and $n$.

Recall that $f$ is quasisymmetric. Lemma 3.2 in [@KMW] implies the existence of a constant $C=C(\delta, n)$ such that $$\label{box}
 C^{-1} \frac{\operatorname{diam}f(B)}{\operatorname{diam}B} \le \frac{1}{|B|} \int_B {\lVertDf\rVert} \, d x \le C \frac{\operatorname{diam}f(B)}{\operatorname{diam}B}$$ for all balls $B \subset {\mathbb{R}}^n$. Since $\operatorname{diam}f(2B) \le C \operatorname{diam}f(B)$ with $C=C(\eta)$, the lemma follows.

Recall that $\phi\colon{\mathbb{R}}^n\to (0,\infty)$ is the Gaussian kernel . Let ${\mathbb{B}}=B(0,1)$ be the open unit ball in ${\mathbb{R}}^n$.

\[doublem\] Let $\mu$ be a doubling measure in ${\mathbb{R}}^n$ and $p \ge 0$. Let $\Omega$ be either ${\mathbb{R}}^n$ or the half space $\{y \colon {\langley, \xi\rangle} \ge 0\}$ for some $\xi \in {\mathbb{R}}^n$. Then $$\label{doubineq}
C^{-1} \mu \big( {\mathbb{B}}\big) \le \int_{\Omega} {\lverty\rvert}^p \phi(y)\, d \mu (y) \le C   \mu \big( {\mathbb{B}}\big)$$ where the constant $C$ depends only on $\mathscr D_\mu$, $p$ and $n$.

We begin by estimating the integral in  from above as follows $$\int_{{\mathbb{R}}^n} {\lverty\rvert}^p \phi(y)\, d \mu (y)  = \int_{{\mathbb{B}}}  {\lverty\rvert}^p  \phi(y)\, d \mu (y)  + \sum_{k=0}^\infty    \int_{2^{k} < |y| \le 2^{k+1}} {\lverty\rvert}^p  \phi(y)\, d \mu (y),$$ where $$\int_{{\mathbb{B}}}  {\lverty\rvert}^p \phi(y)\, d \mu (y) \le \phi(0) \mu({\mathbb{B}}) = (2 \pi)^{-\frac{n}{2}} \mu({\mathbb{B}})$$ and $$\begin{split}
 \int_{2^{k} < |y| \le 2^{k+1}}  {\lverty\rvert}^p \phi(y)\, d \mu (y) & \le  2^{p(k+1)}(2 \pi)^{-\frac{n}{2}} e^{-2^{2k-1}}  \mu(B(0,2^{k+1})) \\
 & \le 2^{p(k+1)} (2 \pi)^{-\frac{n}{2}} e^{-2^{2k-1}} \mathscr D_\mu^{k+1} \mu({\mathbb{B}}). 
  \end{split}$$ Summing over $k=0,1, 2 \dots$ we obtain $$\int_{{\mathbb{R}}^n}  \phi(y)\, d \mu (y) \le    {C} \,  \mu({\mathbb{B}})$$ where $C=C(\mathscr D_\mu, p, n)>0$.

We turn to the left side of . The inequality $${\lverty\rvert}^p \phi(y) \ge  \frac{e^{-1/2}}{2^p (2 \pi)^{n/2}} \qquad \mbox{ for } \quad \frac{1}{2}\le {\lverty\rvert} \le 1$$ implies $$\int_{\Omega} {\lverty\rvert}^p \phi(y)\, d \mu (y) \ge \frac{e^{-1/2}}{2^p (2 \pi)^{n/2}} \mu (\Omega \cap \{  1/2\le {\lverty\rvert} \le 1 \}).$$ Since $ \mu (\Omega \cap \{  1/2\le {\lverty\rvert} \le 1 \})  \ge \mathscr D_\mu^{-3} \mu({\mathbb{B}})$, the left side of  follows.

Proof of main results {#pfth}
=====================

Since $f$ is quasisymmetric by Proposition \[dmqs\], it satisfies the growth condition ${\lvertf(x)\rvert} \le \alpha {\lvertx\rvert}^p + \beta$ for some constants $\alpha, \beta,  p $, see [@Heb Theorem 11.3]. Therefore, the integrals  and  converge and $F$ is $C^\infty$-smooth in $\mathbb H^{n+1}$. Let $\gamma=\gamma(\delta)>0$ be as in part  of Lemma \[deltas\].

Our first step is to prove that for $(x,t)\in\mathbb H^{n+1}$ the matrix $\mathscr B:=DF(x,t)$ satisfies the condition $$\label{final}
w^T\mathscr B w \ge \gamma_1 {\lVert\mathscr B \rVert} {\lvertw\rvert}^2\qquad \text{for every vector $w\in {\mathbb{R}}^{n+1}$}$$ where $\gamma_1=\gamma_1(\delta,n)>0$. Fix $x\in {\mathbb{R}}^n$ and $t>0$. We compute the partial derivatives of $F$ at $(x,t)\in \mathbb H^{n+1}$ as follows. $$\begin{split}
\frac{\partial F^i}{\partial x_j}  &= \int_{{\mathbb{R}}^n} f_j^i (x+ty) \phi(y) d y ,\quad 1 \le i,j \le n; \\
\frac{\partial F^i}{\partial t} &= \int_{{\mathbb{R}}^n}  \sum_{j=1}^n  f_j^i (x+ty) y^i \phi(y) d y ,\quad 1 \le i \le n; \\
\frac{\partial F^{n+1}}{\partial x_j}   &=  \int_{{\mathbb{R}}^n} \sum_{i=1}^n   f_j^i (x+ty) y^j \phi(y) d y ,\quad 1 \le j \le n; \\
\frac{\partial F^{n+1}}{\partial t}  &= \int_{{\mathbb{R}}^n}  \sum_{i=1}^n \sum_{j=1}^n  f_j^i (x+ty) y^iy^j \phi(y) d y .
 \end{split}$$ To simplify formulas we write $A(y)=Df(x+ty)$ and let $B(y)$ be the $(n+1)\times (n+1)$ matrix written in block form below. $$\label{Bmatrix} B(y)=\left(
   \begin{BMAT}(@,50pt,20pt){c.c}{c.c}
      A(y) & A(y)y \\ y^T A(y)& y^T A(y) y
   \end{BMAT}
   \right) .$$ With this notation we have $$\label{afterBmatrix}
DF(x,t) = \int_{{\mathbb{R}}^n} B(y) \phi(y) \, dy .$$ First we show that the norm of $\mathscr B$ is dominated by the quantity $$\alpha := \int_{B(0,1)} {\lVertA(y)\rVert}   \, dy .$$ Indeed, $${\lVert\mathscr B\rVert} \le  \int_{{\mathbb{R}}^n} {\lVertB(y)\rVert} \phi(y) \, dy \le    \int_{{\mathbb{R}}^n} {\lVertA(y)\rVert} (1+{\lverty\rvert})^2\phi(y) \, dy.$$ By Lemma \[Dfdoub\] the measure $\mu = {\lVertA(y)\rVert}\,  dy$ is doubling. Applying Lemma \[doublem\] we obtain $$\label{DF1}
{\lVert\mathscr B\rVert} \le  C \alpha, \qquad C=C(\delta, n) .$$ Next we estimate the quadratic form $w\mapsto w^T\mathscr B w$ generated by $\mathscr B$ from below. For this we fix a vector $w \in {\mathbb{R}}^{n+1}$, written as $w= v+se_{n+1}$ with $v \in {\mathbb{R}}^n$ and $s\in {\mathbb{R}}$. It is easy to see that $$w^T B(y) w  = (v+sy)^T A(y) (v+sy).$$ Let $\Omega = \{y \in {\mathbb{R}}^n \colon {\langlev,sy\rangle} \ge 0\}$. Then $$\begin{split}
w^T\mathscr B w& = \int_{{\mathbb{R}}^n} \left\{  (v+sy)^T A(y) (v+sy)  \right\} \phi(y) \, dy \\
& \ge \gamma \int_{{\mathbb{R}}^n} {\lVertA(y)\rVert} {\lvertv+sy\rvert}^2 \phi(y) \, dy \\
& \ge \gamma \int_{\Omega} {\lVertA(y)\rVert} {\lvertv+sy\rvert}^2 \phi(y) \, dy \\
& \ge \gamma {\lvertv\rvert}^2 \int_{\Omega} {\lVertA(y)\rVert}   \phi(y) \, dy +   \gamma s^2 \int_{\Omega} {\lVertA(y)\rVert} {\lverty\rvert}^2  \phi(y) \, dy. 
\end{split}$$ Applying Lemma \[doublem\] with $\mu = {\lVertA(y)\rVert}\,  dy$ we obtain $$\label{DF2}
w^T \mathscr B w \ge c\, \alpha  \gamma  ({\lvertv\rvert}^2 + s^2)  = c \,  \alpha \gamma {\lvertw\rvert}^2,  \qquad c=c(\delta, n) .$$ Combining  and  we obtain  with $\gamma_1=(c/C) \gamma $. By virtue of Lemma \[deltas\] $F$ is $\delta_1$-monotone in the upper half-space $\mathbb H^{n+1}$ where $\delta_1=\delta_1(\delta,n)$. By symmetry, $F$ is also $\delta_1$-monotone in the lower half-space.

To prove that $F$ is $\delta_1$-monotone in the entire space ${\mathbb{R}}^{n+1}$, we consider two points $a,b\in{\mathbb{R}}^{n+1}$ such that the line segment $[a,b]$ crosses the hyperplane ${\mathbb{R}}^n$ at some point $c$. We have $$\begin{split}
{\langleF(a)-F(b), a-b\rangle}& = {\langlef(a)-f(c), a-b\rangle} + {\langleF(c)-F(b), a-b\rangle}\\
& \ge \delta_1 {\lvertF(a)-F(c)\rvert} {\lverta-b\rvert} + \delta_1 {\lvertF(c)-F(b)\rvert} {\lverta-b\rvert}\\
& \ge \delta_1 {\lvertF(a)-F(b)\rvert} {\lverta-b\rvert}
\end{split}$$ Therefore, $F\in { \mathcal{DM}}$.

It remains to show that $F \colon \mathbb H^{n+1} \to \mathbb H^{n+1}$ is bi-Lipschitz in the hyperbolic metric. Since $F\in {\mathcal{QC}}$ and $\mathbb H^{n+1}$ is a geodesic space, it suffices to prove that $$\label{name}
 {\lVertDF(x,t)\rVert} \approx \frac{F^{n+1} (x,t)}{t}.$$ Here $X \approx Y$ means that $X$ and $Y$ are comparable, i.e., $C^{-1} Y \le X \le CY$ where $C=C(\delta, n)$. It follows from  and  that ${\lVertDF(x,t)\rVert}$ is comparable to the integral average of ${\lVertDf\rVert}$ over the ball $B(x,t)$. By  this average is comparable to $t^{-1}\operatorname{diam}f(B(x,t))$. The quasisymmetry of $F$ implies (cf. [@Heb 11.18]) $$\operatorname{diam}f(B(x,t)) \approx {\lvertF(x,t)- F(x, t/2)\rvert} \approx F^{n+1}(x,t) .$$ This proves .

The proof of Theorem \[Thm\] also works in the case $n=1$ with the following interpretation. Since quasisymmetric mappings on the line need not be absolutely continuous [@BA], the derivative $f'$ must be understood in the sense of distributions. In fact, $\mu:=f'$ is a positive doubling measure with $\mathscr D_{\mu}=\mathscr D_{\mu}(\eta)$ [@Heb 13.20]. Lemma \[Dfdoub\] is not needed in this case. The rest of the proof carries over with $\gamma=1$ and $\gamma_1=\gamma_1(\mathscr D_{\mu})$.

According to [@Ko Lemma 18], a $K$-quasiconformal mapping with a convex potential is also $\delta$-monotone with $\delta=\delta(K,n)$. Let $F$ be the $\delta_1$-monotone extension of $f$ provided by Theorem \[Thm\]. Since the differential matrix $Df$ is symmetric, the formulas  and  show that $DF$ is symmetric as well. In addition, $DF$ is positive semidefinite by Lemma \[deltas\]. Thus, $F=\nabla U$ for some convex function $U\colon {\mathbb{R}}^{n+1}\to {\mathbb{R}}$.

Concluding remarks
==================

Both classes ${\mathcal{QC}}$ (quasiconformal) and ${\mathcal{BL}}$ (bi-Lipschitz) are groups under composition. However, the class of delta-monotone mappings ${ \mathcal{DM}}$ is not closed under composition (consider the rotation of the complex plane given by $z \mapsto e^{i \theta} z$ where ${\lvert\theta\rvert} < \pi /2$). Let ${\mathcal{QC}}_d \subset {\mathcal{QC}}$ be the group generated by ${\mathcal{BL}}$ and ${ \mathcal{DM}}$. In other words, $f$ belongs to ${\mathcal{QC}}_d$ if it can be decomposed into bi-Lipschitz and delta-monotone mappings. This should be compared with the notion of polar factorization of mappings introduced by Brenier [@Br].

Theorem \[Thm\] together with the trivial extension of bi-Lipschitz mappings yield a solution to the extension problem for ${\mathcal{QC}}_d$.

\[cor\] Let $n \ge 2$. For any mapping $f \colon {\mathbb{R}}^n \to {\mathbb{R}}^n$ of class ${\mathcal{QC}}_d $ there exists $F \colon {\mathbb{R}}^{n+1} \to {\mathbb{R}}^{n+1}$ of class ${\mathcal{QC}}_d $ such that the restriction of $F$ to ${\mathbb{R}}^n$ agrees with $f$.

It seems likely that ${\mathcal{QC}}_d $ is a proper subset of ${\mathcal{QC}}$. This motivates the following question:

\[which\] Which quasiconformal mappings are decomposable?

Both bi-Lipschitz and delta-monotone mappings take smooth curves into rectifiable curves [@AIMb Theorem 3.11.7]. This is no longer true for their composition. More precisely, for any $1<\alpha<2$ one can construct a mapping $f\colon {\mathbb{R}}^2\to{\mathbb{R}}^2$ such that $f \in {\mathcal{QC}}_d$ and $f({\mathbb{R}})$ has Hausdorff dimension at least $\alpha$. To this end, one first finds a bi-Lipschitz mapping $g\colon {\mathbb{R}}^2\to{\mathbb{R}}^2$ such that $g({\mathbb{R}})$ contains a planar Cantor set $E$ of dimension $0<\beta<1$ (see Lemma 3.1[@Bi] and the comment after its proof). Second, there is a delta-monotone mapping $h\colon {\mathbb{R}}^2\to{\mathbb{R}}^2$ such that the Hausdorff dimension of $h(E)$ is equal to $\alpha$ (see the construction in [@GV Theorem 5]). Finally, let $f=h\circ g$.

Acknowledgments {#acknowledgments .unnumbered}
===============

We thank Mario Bonk and Jang-Mei Wu for conversations related to the subject of this paper.

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[^1]: Kovalev was supported by the NSF grant DMS-0913474.

[^2]: Onninen was supported by the NSF grant DMS-0701059.
