# On quasiconformal extensions of harmonic mappings associated with pre-
Schwarzian derivative

Xiao-Yuan Wang, Jin-Hua Fan, Zhen-Yong Hu and Zhi-Gang Wang∗ Xiao-Yuan Wang
School of Science, Nanjing University of Science and Technology, Nanjing
210094, Jiangsu, P. R. China<EMAIL_ADDRESS>Jin-Hua FanSchool of
Science, Nanjing University of Science and Technology, Nanjing 210094,
Jiangsu, P. R. China<EMAIL_ADDRESS>Zhen-Yong HuSchool of Science,
Nanjing University of Science and Technology, Nanjing 210094, Jiangsu, P. R.
China<EMAIL_ADDRESS>Zhi-Gang WangSchool of Mathematics and
Statistics, Hunan First Normal University, Changsha 410205, Hunan, P. R.
China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we extend Ahlfors’s univalent criteria and Ahlfors’s
quasiconformal extension for analytic functions to harmonic mappings defined
in the unit disk. Moreover, we give a general quasiconformal extension of
harmonic Teichmüller mappings, whose maximal dilatation estimate is
asymptotically sharp.

###### Key words and phrases:

Quasiconformal mapping, harmonic mapping, Teichmüller mapping, pre-Schwarzian
derivative.

###### 2010 Mathematics Subject Classification:

Primary 30C62, 30C55; Secondary 31A05.

∗Corresponding author.

##  1\. Introduction and statements of the main results

Let $\triangle:=\\{z:|z|<1\\}$ be the unit disk in the complex plane
$\mathbb{C}$. For a locally univalent analytic function $\phi$ in $\triangle$,
we denote by $P_{\phi}$ the pre-Schwarzian derivative and $S_{\phi}$ the
Schwarzian derivative of $\phi$ as follows:

$\displaystyle P_{\phi}=\frac{\phi^{\prime\prime}}{\phi^{\prime}}\ \ {\rm
and}\ \
S_{\phi}=\left(P_{\phi}\right)^{\prime}-\frac{1}{2}\left(P_{\phi}\right)^{2}.$

For $P_{\phi}$ and $S_{\phi}$, we have the following two norms:

$\|P_{\phi}\|=\sup_{z\in\triangle}|P_{\phi}|(1-|z|^{2}),$

and

$\left\|S_{\phi}\right\|=\sup_{z\in\triangle}\left|S_{\phi}\right|(1-|z|^{2})^{2}.$

In 1972, Becker [3] stated that if

$\|P_{\phi}\|\leq 1,$ (1.1)

then $\phi$ is univalent in $\triangle$. Moreover, the constant 1 is sharp
(see [4]). Becker [3] also proved that if

$\|P_{\phi}\|\leq k<1,$ (1.2)

then not only $\phi$ is univalent, but also it exists a continuous extension
$\widetilde{\phi}$ to $\overline{\triangle}$ and
$\widetilde{\phi}(\partial\triangle)$ is a quasicircle.

Indeed, Becker [3] proved that $\phi$ has a $K$-quasiconformal extension to
$\mathbb{C}$ if (1.2) holds. We call (1.1) and (1.2) are Becker’s univalence
criterion and Becker’s extended univalence criterion, respectively. In 1974,
Ahlfors gave a generalized result expressed as follows.

######  Theorem A.

Let $\phi$ be a locally univalent analytic function in $\triangle$. $\sigma$
is a continuous function in $\triangle$, which satisfies the following
conditions: (i) $\sigma_{z}$ and $\sigma_{\overline{z}}$ exist in $\triangle$
a.e.; (ii) $\frac{1}{\sigma}=0$ on $\partial\triangle$; (iii)
$\frac{\sigma_{\overline{z}}}{\sigma^{2}}\neq 0$ in $\overline{\triangle}$.
Then the inequality

$\left|\sigma
P_{\phi}+\sigma^{2}-\sigma_{z}\right|\leq\left|\sigma_{\overline{z}}\right|\ \
(z\in\triangle)$ (1.3)

is sufficient to imply that $\phi$ is univalent in $\triangle$.

Moreover, Ahlfors [1] gave a generalized quasiconformal extension formulated
as below.

######  Theorem B.

Let $\phi$ be a locally univalent analytic function in $\triangle$. $\sigma$
is a continuous function in $\triangle$, which satisfies conditions given by
Theorem A. Then the inequality

$\left|\sigma P_{\phi}+\sigma^{2}-\sigma_{z}\right|\leq
k\left|\sigma_{\overline{z}}\right|\ \ (0\leq k<1)$ (1.4)

is sufficient to imply that $\phi$ has an explicit homeomorphic extension

$\Phi(z)=\begin{cases}\widetilde{\phi}(z)&(|z|\leq 1),\\\
\phi\left(\frac{1}{\overline{z}}\right)+u\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$

where $u(z)=\phi^{\prime}(z)/{\sigma(z)}$ for
$z\in\triangle\backslash\\{0\\}$. Also, the mapping $\Phi$ is
$K$-quasiconformal in $\mathbb{C}$, where $K=\frac{1+k}{1-k}$.

A special case of interest, suggested by Ahlfors [1], arises from taking

$\sigma(z)=\frac{(c+1)\overline{z}}{1-|z|^{2}}$ (1.5)

in (1.4). Then the principal condition for quasiconformal extension displayed
as follows:

######  Theorem C.

If

$\left|c|z|^{2}+\left(1-|z|^{2}\right)zP_{\phi}\right|\leq k<1\quad(|c|\leq
k;\,z\in\triangle),$ (1.6)

then $\phi$ has an explicit homeomorphic extension

$\Phi(z)=\begin{cases}\widetilde{\phi}(z)&(|z|\leq 1),\\\
\phi\left(\frac{1}{\overline{z}}\right)+u\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$
(1.7)

where

$u(z)=\frac{\phi^{\prime}(z)\left(1-|z|^{2}\right)}{\overline{z}}\quad(z\in\triangle\backslash\\{0\\}).$

Also, the mapping $\Phi$ is a $K$-quasiconformal extension in $\mathbb{C}$,
where $K=\frac{1+k}{1-k}$.

We remark that even though (1.4) appears to depend on $\phi$ only through
$P_{\phi}$, one may, by choosing $\sigma$ to depend on $\phi$, obtain
conditions such as the classical Becker criterion which involve $\phi$ through
quantities other than $P_{\phi}$. Instead of $P_{\phi}$ in Theorem B, Nehari
[18] used non-oscillating solutions of differential equations, established the
following criterion with Schwarzian derivative.

######  Theorem D.

For a locally univalent analytic function $\phi$ in $\triangle$, the condition

$\left\|S_{\phi}(z)\right\|\leq 2$ (1.8)

implies that $\phi$ is univalent.

We call Theorem D is the classical Nehari’s univalence criterion. At the same
time, as a supplement note, Hille showed that the bound 2 is the best
possible. Later, a remarkable feature of the explicit extension was provided
by Ahlfors and Weill [2].

######  Theorem E.

Let $\phi$ be a locally univalent analytic function in $\triangle$. For a
constant $t\;(0\leq t<1)$, if $\|S_{\phi}\|\leq 2t$, then the mapping

$F(z)\,=\,\left\\{\begin{array}[]{rl}\phi(z)&\quad(|z|\leq 1),\\\
E_{\phi}(1/\overline{z})&\quad(|z|>1),\end{array}\right.$ (1.9)

is a quasiconformal extension of $\phi$ to $\mathbb{C}$, where

$E_{\phi}(\xi)=\phi(\xi)+\frac{(1-|\xi|^{2})\phi^{\prime}(\xi)}{\overline{\xi}-\tfrac{1}{2}(1-|\xi|^{2})P_{\phi}(\xi)}\quad(\xi\in\triangle).$

######  Remark 1.1.

One can see that Theorem E can be obtained by taking

$\sigma(z)=\frac{\overline{z}}{1-|z|^{2}}-\frac{1}{2}P_{\phi}$

in Theorem B. Indeed, Ahlfors and Weill [2] applied a quasiconformal extension
to obtain the Nehari condition (1.8) for the half plane.

Ahlfors [1] also showed general formulas on the problem of univalence and
quasiconformal extensibility through the following two results.

######  Theorem F.

Let $\phi$ be a locally univalent analytic function in $\triangle$. Then the
inequality

$\left|\frac{1}{2}S_{\phi}+v^{2}-v_{z}\right|\leq
k\left|v_{\overline{z}}\right|\ \ (0<k<1)$ (1.10)

is sufficient to imply that there exists a $\frac{1+k}{1-k}$-quasiconformal
extension of $\phi$ to $\mathbb{C}$, where the continue function $v$ satisfies
$\frac{v_{\overline{z}}}{v^{2}}\neq 0$ and $v\rightarrow\infty$ for
$|z|\rightarrow 1$.

######  Theorem G.

Let $\phi$ be a locally univalent analytic function in $\triangle$. Then the
inequality

$\displaystyle\left|\frac{1}{2}S_{\phi}(1-|z|^{2})^{2}-c(1-c)\overline{z}\right|\leq
k|c|\quad(0<k<1;\,|c-1|\leq k)$ (1.11)

is sufficient to imply that there exists a $\frac{1+k}{1-k}$-quasiconformal
extension of $\phi$ to $\mathbb{C}$.

We call (1.3) is the Ahlfors’s univalent criterion and (1.4), (1.6), (1.10),
(1.11) are the Ahlfors’s quasiconformal extension criteria. Note that the
Becker’s extension criterion arises from taking

$\sigma(z)=\frac{\overline{z}}{1-|z|^{2}}$ (1.12)

in (1.4) or $c=0$ in (1.6).

######  Remark 1.2.

By setting $\sigma(z)=v-1/2P_{\phi}$ and
$\sigma(z)=c\overline{z}/(1-|z|^{2})-1/2P_{\phi}$ in (1.4), one can get
Theorem F and Theorem G, respectively.

######  Remark 1.3.

By putting $v(z)=c\overline{z}/(1-|z|^{2})$ in (1.10), one can get Theorem G.

######  Remark 1.4.

By taking $v(z)=\overline{z}/(1-|z|^{2})$ in (1.10) or $c=1$ in (1.11), one
can obtain Theorem E.

A complex-valued function $f$ in $\triangle$ is harmonic if $\Delta
f=4f_{z\overline{z}}=0$. It is well-known that $f$ has a canonical
representation $f=h+\overline{g}$, where $h$ and $g$ are analytic in
$\triangle$. Lewy [17] proved that a harmonic mapping $f$ is locally univalent
if and only if its Jacobian
$J_{f}=\left|h^{\prime}\right|^{2}-\left|g^{\prime}\right|^{2}\neq 0$. If
$J_{f}>0\,\left(J_{f}<0\right)$, then $f$ is sense-preserving (sense-
reserving). Denote by $\omega={g^{\prime}}/{h^{\prime}}$ the second complex
dilatation of $f=h+\overline{g}$. If a harmonic mapping has the representation
$f=h+\alpha\overline{h}$, where $h$ is a conformal mapping and $\alpha$ is a
constant such that $0<|\alpha|<1$, then it is called a harmonic Teichmüller
mapping (see [6]).

As natural generalizations of conformal mappings, quasiconformal extensions
for harmonic mappings have been investigated by many researchers (see [9, 10]
and the references therein).

In 2003, Chuaqui $et\ al.$ [8] generalized the idea of Schwarzian derivatives
of $f=h+\overline{g}$ into the case of locally univalent harmonic mappings
with $\omega=q^{2}$, where $q$ is an analytic function. The pre-Schwarzian
derivative $P_{f}$ and Schwarzian derivative $S_{f}$ for sense-preserving
harmonic mappings are defined as follows:

$\displaystyle P_{f}=\frac{\partial}{\partial
z}\log\left|J_{f}\right|=P_{h}-\frac{\overline{\omega}\omega^{\prime}}{1-|\omega|^{2}},$
(1.13)

and

$\displaystyle
S_{f}=S_{h}+\frac{\overline{\omega}}{1-|\omega|^{2}}\left(\frac{h^{\prime\prime}}{h^{\prime}}\omega^{\prime}-\omega^{\prime\prime}\right)-\frac{3}{2}\left(\frac{\overline{\omega}\omega^{\prime}}{1-|\omega|^{2}}\right)^{2},$
(1.14)

which were introduced and investigated by Hernández and Martín [13]. In the
same paper, Hernández and Martín proved following result.

######  Theorem H.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
with the second dilatation $\omega$. If $f$ satisfies

$\left(1-|z|^{2}\right)|P_{f}|+\frac{\left|\omega^{\prime}(z)\left(1-|z|^{2}\right)\right|}{1-|\omega(z)|^{2}}\leq
1\ \ (z\in\triangle),$ (1.15)

then $f$ is univalent in $\triangle$. The constant 1 is sharp.

Moreover, Hernández and Martín [11] obtained a quasiconformal extension as
follows.

######  Theorem I.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
with $\|\omega\|_{\infty}<1$. If $f$ satisfies the condition

$\left(1-|z|^{2}\right)|P_{f}|+\frac{\left|\omega^{\prime}(z)\left(1-|z|^{2}\right)\right|}{1-|\omega(z)|^{2}}\leq
k<1\quad(z\in\triangle),$ (1.16)

then $f$ has a continuous and injective extension $\widetilde{f}$ to
$\overline{\triangle}$, and the mapping

$F(z)=\begin{cases}\widetilde{f}(z)&(|z|\leq 1),\\\
f\left(\frac{1}{\overline{z}}\right)+U\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$
(1.17)

is a homeomorphism of $\mathbb{C}$ onto itself, where

$U(z)=\frac{h^{\prime}(z)}{\overline{z}}\left(1-|z|^{2}\right)+\frac{\overline{g^{\prime}(z)}}{z}\left(1-|z|^{2}\right)\quad(z\in\triangle\backslash\\{0\\}).$

Moreover, if $f$ satisfies the condition (1.16), then
$\widetilde{f}(\partial\triangle)$ is a quasicircle and $f$ can be extended to
a quasiconformal mapping in $\mathbb{C}$. Under the additional condition

$k<\frac{1-\|\omega\|_{\infty}}{1+\|\omega\|_{\infty}},$ (1.18)

the mapping $F$ defined by (1.17) is an explicit $K$-quasiconformal mapping of
$\mathbb{C}$ onto itself, where

$K=\frac{1+k+\|\omega\|_{\infty}(1-k)}{1-k-\|\omega\|_{\infty}(1+k)}.$

The following result was derived by Bravo $et\ al.$ [5], which generalized
Theorem H. It also can be regarded as a generalization of harmonic analogue of
Theorem A.

######  Theorem J.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
with the second dilatation $\omega$. If $f$ satisfies

$\left|c|z|^{2}+(1-|z|^{2})z{P_{f}}\right|+\frac{|z\omega^{\prime}(z)(1-|z|^{2})|}{1-|\omega(z)|^{2}}\leq
1\ \ (z\in\triangle),$ (1.19)

then $f$ is univalent in $\triangle$.

Recently, Chen and Que [7] generalized Theorem I as follows.

######  Theorem K.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
with $\|\omega\|_{\infty}<1$. If $f$ satisfies (1.16), then for all
$|\lambda|\leq 1,f_{\lambda}=h+\lambda\overline{g}$ has a continuous and
injective extension $\widetilde{f_{\lambda}}$ to $\overline{\triangle}$ and
the mapping

$F_{\lambda}(z)=\begin{cases}\widetilde{f}_{\lambda}(z)&(|z|\leq 1),\\\
f_{\lambda}\left(\frac{1}{\bar{z}}\right)+U_{\lambda}\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$
(1.20)

is a homeomorphism of $\mathbb{C}$ onto itself, where

$U_{\lambda}(z)=\frac{h^{\prime}(z)}{\overline{z}}\left(1-|z|^{2}\right)+\lambda\frac{\overline{g^{\prime}(z)}}{z}\left(1-|z|^{2}\right)\quad(z\in\triangle\backslash\\{0\\}).$

Moreover, if $k$ satisfies (1.18), then the family of mappings
$F_{\lambda}(z)$ are $K$-quasiconformal mappings in $\mathbb{C}$, where

$K=\frac{1+k+|\lambda|\|\omega\|_{\infty}(1-k)}{1-k-|\lambda|\|\omega\|_{\infty}(1+k)}.$
(1.21)

We observe that the following result obtained by Chen and Que [7] is a
quasiconformal extension of harmonic Teichmüller mappings, whose maximal
dilatation estimate is asymptotically sharp.

######  Theorem L.

Let $f$ be a sense-preserving harmonic mapping in $\triangle$ with a
representation $f=h+\alpha\overline{h}$, where $h$ is a locally univalent
analytic function in $\triangle$ and $\alpha$ is a constant with $|\alpha|<1$.
Assume that

$\left|P_{h}(z)\left(1-|z|^{2}\right)\right|\leq k<1\quad(z\in\triangle).$
(1.22)

Then $f$ is a harmonic Teichmüller mappings of $\triangle$ and has a
continuous and homeomorphic extension $\widetilde{f}$ to
$\overline{\triangle}$. Moreover, the mapping

$F(z)=\begin{cases}\widetilde{f}(z)&\quad(|z|\leq 1),\\\
f\left(\frac{1}{\overline{z}}\right)+U_{\alpha}\left(\frac{1}{\overline{z}}\right)&\quad(|z|>1),\end{cases}$
(1.23)

is a $K$-quasiconformal mapping of $\mathbb{C}$ with

$K=\frac{(1+k)(1+|\alpha|)}{(1-k)(1-|\alpha|)},$ (1.24)

where

$U_{\alpha}(z)=\frac{h^{\prime}(z)\left(1-|z|^{2}\right)}{\bar{z}}+\frac{\alpha\overline{h^{\prime}(z)}\left(1-|z|^{2}\right)}{z}\quad(z\in\triangle\backslash\\{0\\}).$

The maximal dilatation estimate of the quasiconformal extension $F$ is
asymptotically sharp in $k$, and extremal mappings are of the form
$f(z)=az+b\overline{z}$, where $a$ and $b$ are two nonvanishing constants.

In a recent paper, Hu and Fan [14] generalized Theorem K by posing one of
harmonic version of the classical Becker’s criteria. We present it here in a
slightly modified form.

######  Theorem M.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$.
Assume that

$\left|c|z|^{2}+(1-|z|^{2})z{P_{f}}\right|+\frac{|z\omega^{\prime}(z)(1-|z|^{2})|}{1-|\omega(z)|^{2}}\leq
k<1\quad(|c|\leq k;\,z\in\triangle).$ (1.25)

Then $f_{\lambda}=h+\lambda\overline{g}\;(|\lambda|\leq 1)$ has a continuous
and injective extension $\widetilde{f_{\lambda}}$ to $\overline{\triangle}$
and the family of mappings $F_{\lambda}(z)$ defined by (1.20) are
$K$-quasiconformal mappings of $\mathbb{C}$ onto themselves. Moreover, if $k$
satisfies the conditions (1.18), then the family of mappings $F_{\lambda}(z)$
are $K$-quasiconformal mappings of $\mathbb{C}$, where $K$ is given by (1.21).

######  Remark 1.5.

In Theorem K, the condition (1.16) for $P_{f}$ and (1.18) for $k$, imply that
$\widetilde{f}_{\lambda}(\partial\triangle)$ is a quasicircle. In Theorem M,
if inequality (1.25) holds for $P_{f}$, it shows that
$\widetilde{f}_{\lambda}(\partial\triangle)$ is a quasicircle. We would like
to draw reader’s attention to Theorem 1.3, it is shown that
$\widetilde{f}_{\lambda}(\partial\triangle)$ is a quasicircle, if inequality
(1.27) holds for $P_{f}$.

Motivated essentially by the above discussions, we aim at deriving general
harmonic analogues of Theorems A and B by using the pre-Schwarzian derivative
of harmonic mappings.

######  Theorem 1.1.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
and let $\omega$ be its second complex dilatation. Let $\sigma$ be a
continuous function in $\triangle$, which satisfies the following conditions:
(i) $\sigma_{z}$ and $\sigma_{\overline{z}}$ exist in $\triangle$ a.e.; (ii)
$\frac{1}{\sigma}=0$ on $\partial\triangle$; (iii)
$\frac{\sigma_{\overline{z}}}{\sigma^{2}}\neq 0$ in $\overline{\triangle}$. If
$f$ satisfies the condition

$\left|\sigma
P_{f}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma\omega^{\prime}|}{1-|\omega|^{2}}\leq\left|\sigma_{\overline{z}}\right|\
\ (z\in\triangle),$ (1.26)

then $f$ is univalent in $\triangle$.

######  Remark 1.6.

By setting $\sigma(z)=\frac{(c+1)\overline{z}}{1-|z|^{2}}$ in (1.26) of
Theorem 1.1, we get Theorem J due to Bravo $et\ al.$ [5]; By putting
$\sigma(z)=\frac{\overline{z}}{1-|z|^{2}}$ in (1.26) of Theorem 1.1, we obtain
Theorem H due to Hernández and Martín [13].

######  Theorem 1.2.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
and let $\omega$ be its second complex dilatation with
$\|\omega\|_{\infty}<1$. Let $\sigma$ be a continuous function in $\triangle$,
which satisfies conditions given in Theorem 1.1. Assume that

$\left|\sigma
P_{f}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma\omega^{\prime}|}{1-|\omega|^{2}}\leq
k\left|\sigma_{\overline{z}}\right|\ \ (0\leq k<1;\,z\in\triangle).$ (1.27)

Then the harmonic mapping $f_{\lambda}=h+\lambda\overline{g}\;(|\lambda|\leq
1)$ has a continuous and injective extension $\widetilde{f}$ to
$\overline{\triangle}$. Moreover, the function

$F_{\lambda}(z)=\begin{cases}\widetilde{f}_{\lambda}(z)&(|z|\leq 1),\\\
f_{\lambda}\left(\frac{1}{\overline{z}}\right)+U_{\lambda}\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$
(1.28)

is a homeomorphic extension of $f$ to $\mathbb{C}$ onto itself. The function
$U_{\lambda}$ that appears in (1.28) is defined by

$U_{\lambda}(z)=\frac{h^{\prime}(z)}{\sigma(z)}+\lambda\frac{\overline{g^{\prime}(z)}}{\overline{\sigma(z)}}\quad(z\in\triangle\backslash\\{0\\}).$

Furthermore, if $k$ satisfies the condition (1.18), then the family of
mappings $F_{\lambda}(z)$ are $K$-quasiconformal mappings of $\mathbb{C}$,
where $K$ is given by (1.21).

######  Remark 1.7.

By setting $\lambda=0$ in Theorem 1.2, we get Theorem B exactly given by
Ahlfors [1]; By putting $\sigma(z)=\frac{\overline{z}}{1-|z|^{2}}$ in Theorem
1.2, we obtain the result in Theorem K due to Chen and Que [7] and Theorem M
due to Hu and Fan[14]. By setting $\sigma(z)=\frac{\overline{z}}{1-|z|^{2}}$
and $\lambda=1$ in Theorem 1.2, we get the result just as Theorem I due to
Hernández and Martín [11].

######  Theorem 1.3.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
with $\|\omega\|_{\infty}<1$. If $f$ satisfies the condition (1.27), then
$\widetilde{f}(\partial{\triangle})$ is a quasicircle and $f$ can be extended
to a quasiconformal mapping in $\mathbb{C}$.

By means of the pre-Schwazian derivative and Schwazian derivative of harmonic
mappings, by setting

$\sigma=v-\frac{1}{2}P_{f},$

and

$\sigma=\frac{c\overline{z}}{1-|z|^{2}}-\frac{1}{2}P_{f}$

in Theorem 1.2, we get following two corollaries, respectively.

######  Corollary 1.1.

For a sense-preserving harmonic mapping $f$ in $\triangle$, the inequality

$\left|\frac{1}{2}S_{f}+v^{2}-v_{z}\right|+\left|(v-P_{f})\frac{\omega^{\prime}}{1-|\omega|^{2}}\right|\leq
k\left|v_{\overline{z}}-\overline{P_{f}}\right|\ \ (0\leq k<1)$

together with $v\rightarrow\infty$ for $|z|\rightarrow 1$ and
$\frac{v_{\overline{z}}}{v^{2}}\neq 0$ is sufficient to imply the existence of
a $\frac{1+k}{1-k}$-quasiconformal extension of $f$.

######  Corollary 1.2.

For a sense-preserving harmonic mapping $f$ in $\triangle$, the inequality

$\displaystyle\left|\frac{1}{2}S_{f}(1-|z|^{2})^{2}-c(1-c)\overline{z}\right|$
$\displaystyle+\left|\left(c\overline{z}(1-|z|^{2})-\frac{1}{2}P_{f}(1-|z|^{2})^{2}\right)\frac{\omega^{\prime}}{1-|\omega|^{2}}\right|$
(1.29) $\displaystyle\qquad\leq
k\left|c-\overline{P_{f}}(1-|z|^{2})^{2}\right|\ \ (0\leq k<1;\,|c-1|\leq k)$

is sufficient to imply the existence of a $\frac{1+k}{1-k}$-quasiconformal
extension of $f$.

######  Remark 1.8.

Corollary 1.1 and Corollary 1.2 extended Theorem F and Theorem G to harmonic
cases, respectively.

######  Theorem 1.4.

Let $f$ be a sense-preserving harmonic mapping in $\triangle$ with a
representation $f=h+\alpha\overline{h}$, where $h$ is a locally univalent
analytic function in $\triangle$. Let $\sigma$ satisfy the conditions in
Theorem 1.1 and $\alpha$ be a constant with $|\alpha|<1$. Assume that $h$
satisfies the condition

$\left|\sigma P_{h}+\sigma^{2}-\sigma_{z}\right|\leq
k\left|\sigma_{\overline{z}}\right|\ \ (0\leq k<1).$ (1.30)

Then $f$ is a harmonic Teichmüller mappings of $\triangle$ and has a
continuous and homeomorphic extension $\tilde{f}$ to $\overline{\triangle}$.
Moreover, the mapping

$F_{\alpha}(z)=\begin{cases}\widetilde{f}(z)&(|z|\leq 1),\\\
f\left(\frac{1}{\overline{z}}\right)+U_{\alpha}\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$
(1.31)

is a $K$-quasiconformal mapping of the complex plane $\mathbb{C}$ with $K$
given by (1.24), where

$U_{\alpha}(z)=\frac{h^{\prime}(z)}{\sigma(z)}+\alpha\frac{\overline{h^{\prime}(z)}}{\overline{\sigma(z)}}\quad(z\in\triangle\backslash\\{0\\}).$

The maximal dilatation estimate of the quasiconformal extension $F_{\alpha}$
is asymptotically sharp in $k$, and extremal mappings are of the form
$f(z)=mz+n\overline{z}$, where $m$ and $n$ are two nonvanishing constants.

######  Remark 1.9.

By setting $\sigma(z)=\frac{\overline{z}}{1-|z|^{2}}$ in Theorem 1.4, we get
the corresponding result obtained by Chen and Que [7] (see Theorem L).

##  2\. Proof of Theorem 1.1

###### Proof of Theorem 1.1.

If $f=h+\overline{g}$ satisfied the condition (1.26), we know that

$\displaystyle\left|{\sigma}\frac{h^{\prime\prime}}{h^{\prime}}+{\sigma^{2}}-{\sigma_{z}}\right|$
$\displaystyle=\left|{\sigma}\left(\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\overline{\omega}\omega^{\prime}}{1-|\omega|^{2}}\right)+{\sigma^{2}}-{\sigma_{z}}+\frac{{\sigma}\overline{\omega}{\omega}^{\prime}}{1-|\omega|^{2}}\right|$
$\displaystyle\leq\left|{\sigma}P_{f}+{\sigma^{2}}-{\sigma_{z}}\right|+\left|\frac{{\sigma}\overline{\omega}{\omega}^{\prime}}{1-|\omega|^{2}}\right|$
(2.1)
$\displaystyle\leq\left|{\sigma}P_{f}+{\sigma^{2}}-{\sigma_{z}}\right|+\left|\frac{{\sigma}{\omega}^{\prime}}{1-|\omega|^{2}}\right|$
$\displaystyle\leq|\sigma_{\overline{z}}|.$

Hence, from Theorem A, we see that $h$ is univalent.

For $a\in\triangle$, we let the function $f_{a}$ has the canonical
decomposition

$\displaystyle
f_{a}=f+a\overline{f}=h+{a}g+\overline{g}+a\overline{h}=h_{a}+\overline{g_{a}}\quad(|a|<1).$
(2.2)

By noting that $f_{a}$ is also a sense-preserving harmonic mapping and its the
second complex dilatation is $\omega_{a}$ with

$\omega_{a}=\varphi_{a}\circ\omega,$

where $\varphi_{a}$ is the automorphism of $\triangle$ defined by

$\varphi_{a}(z)=\frac{\overline{a}+z}{1+az}\quad(z\in\triangle).$

It follows that

$\frac{\left|z\omega_{a}^{\prime}\right|}{1-\left|\omega_{a}\right|^{2}}=\frac{\left|z\varphi_{a}^{\prime}(\omega)\omega^{\prime}\right|}{1-\left|\varphi_{a}(\omega)\right|^{2}}=\frac{\left|z\omega^{\prime}\right|}{1-|\omega|^{2}}\cdot\frac{\left|\varphi_{a}^{\prime}(\omega)\right|\left(1-|\omega|^{2}\right)}{1-\left|\varphi_{a}(\omega)\right|^{2}}=\frac{\left|z\omega^{\prime}\right|}{1-|\omega|^{2}}.$

Since the Jacobian $J_{f_{a}}$ of $f_{a}$ is $(1-|a|^{2})J_{f}$, we see that
$P_{f_{a}}=P_{f}$. Therefore, $f_{a}$ satisfies the hypothesis of Theorem A.
Then for any $a\in\triangle$, the function $h_{a}=h+ag$ is univalent. By
Hurwitz’s theorem, the functions $h+\varepsilon g$ are univalent for all
$|\varepsilon|=1$, which implies that $h+\varepsilon\overline{g}$ is univalent
(see [12]). We thus conclude that $f$ is univalent in $\triangle$. ∎

##  3\. Proof of Theorem 1.2

###### Proof of Theorem 1.2.

For every $\lambda\in\overline{\triangle}$, let
$f_{\lambda}=h+\lambda\overline{g}$. Then the second complex dilatation of
$f_{\lambda}$ is $\omega_{\lambda}=\overline{\lambda}\omega$. Since
$(1-|\lambda|)(1-|\omega|)\geq 0$, we obtain

$\displaystyle\left|\sigma
P_{f_{\lambda}}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma\omega_{\lambda}^{\prime}|}{1-|\omega_{\lambda}|^{2}}$
$\displaystyle=\left|\sigma\left(\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\overline{\omega_{\lambda}}\omega_{\lambda}^{\prime}}{1-\left|\omega_{\lambda}\right|^{2}}\right)+\sigma^{2}-\sigma_{z}\right|+\frac{\left|\sigma\omega^{\prime}_{\lambda}\right|}{1-|\omega_{\lambda}|^{2}}$
$\displaystyle\leq\left|\sigma\left(\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\overline{\omega}\omega^{\prime}}{1-\left|\omega\right|^{2}}\right)+\sigma^{2}-\sigma_{z}\right|+\left|\sigma\left(\frac{\overline{\omega}\omega^{\prime}}{1-\left|\omega\right|^{2}}-\frac{\overline{\omega_{\lambda}}\omega_{\lambda}^{\prime}}{1-\left|\omega_{\lambda}\right|^{2}}\right)\right|+\frac{\left|\sigma\omega^{\prime}_{\lambda}\right|}{1-|\omega_{\lambda}|^{2}}$
(3.1) $\displaystyle=\left|\sigma
P_{f}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma||\omega||\omega^{\prime}|\left(1-|\lambda|^{2}\right)}{\left(1-|\omega|^{2}\right)\left(1-|\lambda|^{2}|\omega|^{2}\right)}+\frac{|\sigma||\lambda|\left|\omega^{\prime}\right|}{1-|\lambda|^{2}|\omega|^{2}}$
$\displaystyle=\left|\sigma
P_{f}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma||\omega^{\prime}|}{1-|\omega|^{2}}\frac{|\lambda|+|\omega|}{1+|\lambda||\omega|}$
$\displaystyle\leq\left|\sigma
P_{f}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma\omega^{\prime}|}{1-|\omega|^{2}}.$

Therefore, for all $\lambda\in\overline{\triangle}$, if $f$ satisfies the
condition (1.4), we have

$\left|\sigma
P_{f_{\lambda}}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma\omega_{\lambda}^{\prime}|}{1-|\omega_{\lambda}|^{2}}\leq
k|\sigma_{\overline{z}}|\quad(0\leq k<1).$

If the second complex dilatation $\omega$ of $f$ satisfies
$\|\omega\|_{\infty}<1$, then for each $\lambda\in\overline{\triangle}$, by
[11, Theorem 1], the mapping $\widetilde{f}_{\lambda}$ can be continuously
extended to a homeomorphism of $\overline{\triangle}$. Moreover,
$F_{\lambda}(z)$ given by (1.28) is a homeomorphism extension of
${f}_{\lambda}$.

Next, we shall estimate the maximal complex dilatation of the mapping
$F_{\lambda}$. We divide it into the following two cases.

${\bf C{ase}\ 1.}$ If $|z|<1$, then
$|\mu_{{F}_{\lambda}}|=|\lambda||\omega(z)|\leq|\lambda|\|\omega\|_{\infty}<1$
for $z\in\triangle$.

${\bf C{ase}\ 2.}$ If $|z|>1$, we make a reciprocal transformation
$w=1/\overline{z}$ for some $\left|z\right|<1$, then

$\displaystyle\left|\mu_{F_{\lambda}}(w)\right|$
$\displaystyle=\left|\frac{h^{\prime}(z)+(U_{\lambda})_{z}(z)}{\lambda\overline{g^{\prime}(z)}+(U_{\lambda})_{\overline{z}}(z)}\right|$
$\displaystyle=\left|\frac{h^{\prime}+\frac{h^{\prime\prime}\sigma-h^{\prime}\sigma_{z}}{\sigma^{2}}-\lambda\frac{\overline{g^{\prime}}\overline{\sigma_{\overline{z}}}}{\overline{\sigma}^{2}}}{\lambda\overline{g^{\prime}}+\lambda\frac{\overline{g^{\prime\prime}}\overline{\sigma}-\overline{g^{\prime}}\overline{\sigma_{z}}}{\overline{\sigma}^{2}}-\frac{h^{\prime}\sigma_{\overline{z}}}{\sigma^{2}}}\right|$
$\displaystyle=\left|\frac{\sigma^{2}+\sigma\frac{h^{\prime\prime}}{h^{\prime}}-\sigma_{z}-\lambda{\overline{\omega}}\overline{\sigma_{\overline{z}}}\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{h^{\prime}}}{h^{\prime}}}{\overline{\lambda}\sigma^{2}\omega+\overline{\lambda}\sigma\frac{g^{\prime\prime}}{h^{\prime}}-\overline{\lambda}\sigma_{z}\omega-\overline{{\sigma_{\overline{z}}}}\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{h^{\prime}}}{h^{\prime}}}\right|$
(3.2)
$\displaystyle=\left|\frac{\frac{\sigma^{2}}{\sigma_{\overline{z}}}+\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}-\lambda\overline{\omega}\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{\sigma_{\overline{z}}}}{\sigma_{\overline{z}}}\frac{\overline{h^{\prime}}}{h^{\prime}}}{\overline{\lambda}\frac{\sigma^{2}}{\sigma_{\overline{z}}}\omega+\overline{\lambda}\frac{\sigma}{\sigma_{\overline{z}}}\frac{g^{\prime\prime}}{h^{\prime}}-\overline{\lambda}\frac{\sigma_{z}}{\sigma_{\overline{z}}}\omega-\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{\sigma_{\overline{z}}}}{\sigma_{\overline{z}}}\frac{\overline{h^{\prime}}}{h^{\prime}}}\right|$
$\displaystyle\leq\frac{|\frac{\sigma^{2}}{\sigma_{\overline{z}}}+\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}|+|\lambda|\|\omega\|_{\infty}}{1-|\lambda|\left|\frac{\sigma^{2}}{\sigma_{\overline{z}}}\omega-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\omega+\frac{\sigma}{\sigma_{\overline{z}}}\frac{g^{\prime\prime}}{h^{\prime}}\right|}.$

On the one hand, since $f$ satisfies the condition (1.27), by the triangle
inequality, we get

$\displaystyle\left|\frac{\sigma^{2}}{\sigma_{\overline{z}}}+\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\sigma_{z}}{\sigma_{{\overline{z}}}}\right|$
$\displaystyle=\left|\frac{\sigma}{\sigma_{\overline{z}}}\left(\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\overline{\omega}\omega^{\prime}}{1-|\omega|^{2}}\right)+\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}+\frac{\sigma}{\sigma_{\overline{z}}}\frac{\overline{\omega}{\omega}^{\prime}}{(1-|\omega|^{2})}\right|$
$\displaystyle\leq\left|\frac{\sigma}{\sigma_{\overline{z}}}\left(\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\overline{\omega}\omega^{\prime}}{1-|\omega|^{2}}\right)+\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\right|+\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\overline{\omega}{\omega}^{\prime}}{(1-|\omega|^{2})}\right|$
$\displaystyle\leq\left|\frac{\sigma}{\sigma_{\overline{z}}}P_{f}+\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\right|+\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\overline{\omega}{\omega}^{\prime}}{(1-|\omega|^{2})}\right|$
(3.3) $\displaystyle\leq
k-\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{{\omega}^{\prime}}{1-|\omega|^{2}}\right|+\|\omega\|_{\infty}\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{{\omega}^{\prime}}{(1-|\omega|^{2})}\right|$
$\displaystyle=k-\left(1-\|\omega\|_{\infty}\right)\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{{\omega}^{\prime}}{(1-|\omega|^{2})}\right|.$

On the other hand, we have

$g^{\prime\prime}=\left(\omega
h^{\prime}\right)^{\prime}=\omega^{\prime}h^{\prime}+\omega h^{\prime\prime}.$

Therefore, for all $z\in\triangle$, we obtain

$\displaystyle\left|\frac{\sigma^{2}}{\sigma_{\overline{z}}}\omega-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\omega+\frac{\sigma}{\sigma_{\overline{z}}}\frac{g^{\prime\prime}}{h^{\prime}}\right|$
$\displaystyle=\left|\frac{\sigma^{2}}{\sigma_{\overline{z}}}\omega-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\omega+\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}+\frac{\sigma}{\sigma_{\overline{z}}}\omega^{\prime}\right|$
$\displaystyle=\left|\frac{\sigma}{\sigma_{\overline{z}}}\left(\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\overline{\omega}\omega^{\prime}}{1-|\omega|^{2}}\right)+\frac{\sigma^{2}}{\sigma_{\overline{z}}}\omega-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\omega+\frac{\sigma}{\sigma_{\overline{z}}}\frac{{\omega}^{\prime}}{(1-|\omega|^{2})}\right|$
$\displaystyle\leq\|\omega\|_{\infty}\left|\frac{\sigma}{\sigma_{\overline{z}}}P_{f}+\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\right|+\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{(1-|\omega|^{2})}\right|$
(3.4) $\displaystyle\leq
k\|\omega\|_{\infty}-\|\omega\|_{\infty}\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{(1-|\omega|^{2})}\right|+\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{(1-|\omega|^{2})}\right|$
$\displaystyle=k\|\omega\|_{\infty}+\left(1-\|\omega\|_{\infty}\right)\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{(1-|\omega|^{2})}\right|.$

By substituting (3) and (3) into (3), we know that

$\displaystyle\left|\mu_{F_{\lambda}}(w)\right|$
$\displaystyle\leq\frac{k-\left(1-\|\omega\|_{\infty}\right)\left|\omega^{*}(z)\right|+|\lambda|\|\omega\|_{\infty}}{1-\left(k|\lambda|\|\omega\|_{\infty}+\left(1-\|\omega\|_{\infty}\right)\left|\omega^{*}(z)\right|\right)}$
(3.5)
$\displaystyle=\frac{k+|\lambda|\|\omega\|_{\infty}-\left(1-|\lambda|\|\omega\|_{\infty}\right)\left|\omega^{*}(z)\right|}{1-k|\lambda|\|\omega\|_{\infty}-\left(1-\|\omega\|_{\infty}\right)\left|\omega^{*}(z)\right|},$

where

$\omega^{*}(z)=\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{(1-|\omega|^{2})}.$

Since $f$ satisfies the condition (1.27), we deduce that $|\omega^{*}(z)|\leq
k$.

Define the function $\rho(x):[0,k]\rightarrow\mathbb{R}$ by

$\rho(x)=\frac{k+|\lambda|\|\omega\|_{\infty}-\left(1-\|\omega\|_{\infty}\right)x}{1-k|\lambda|\|\omega\|_{\infty}-\left(1-\|\omega\|_{\infty}\right)x}.$

Bearing in mind that (1.18) holds, we see that $\rho^{\prime}(x)<0$ for all
$x\in[0,k].$ Thus,

$\rho(x)\leq\rho(0)=\frac{k+|\lambda|\|\omega\|_{\infty}}{1-k|\lambda|\|\omega\|_{\infty}}=\frac{K-1}{K+1}.$

We then find from (3.5) that the inequality

$\left|\mu_{F_{\lambda}}(w)\right|\leq\frac{K-1}{K+1}:=k_{1}$

holds for all $|w|>1$. Since the assumption (1.18) shows that

$\left\|\omega_{\lambda}\right\|_{\infty}=|\lambda|\|\omega\|_{\infty}\leq\|\omega\|_{\infty}<\frac{1-k}{1+k}$

holds for all $\lambda\in\overline{\triangle}$, which implies that $k_{1}<1$.
Therefore, $F_{\lambda}(w)$ is a quasiconformal mapping in
$\mathbb{C}\backslash\overline{\triangle}$.

By [15, Lemma 6.1], we see that

$\left|\mu_{F_{\lambda}}(z)\right|\leq\frac{K-1}{K+1}$

for all $|z|\neq 1$ in $\mathbb{C}$. Thus, the mapping $F_{\lambda}$ defined
by (1.28) is $K$-quasiconformal whenever (3.5) holds. Moreover, by noting that
$|\lambda|\|\omega\|_{\infty}\leq k_{1}<1$, we deduce that $F_{\lambda}$ is a
$K$-quasiconformal mapping of $f_{\lambda}$ in $\mathbb{C}$, where $K$ is
given by (1.21). ∎

######  Remark 3.1.

The method to prove $F_{\lambda}(z)$ given by (1.28) is a homeomorphism
extension of $\mathbb{C}$ due to [11]. Here, we give the main key points.

Firstly, we shall prove that for all $a\in\overline{\triangle}$, the functions
$h_{a}=h+ag$ have a continuous and injective extension to $\mathbb{C}$. It
follows from (1.27) that for all $z\in\triangle$,

$\displaystyle k|\sigma_{\overline{z}}|$ $\displaystyle\geq\left|\sigma
P_{f_{a}}+\sigma^{2}-\sigma_{z}\right|+\frac{|\sigma\omega_{a}^{\prime}(z)|}{1-|\omega_{a}(z)|^{2}}$
(3.6)
$\displaystyle=\left|\sigma\left(\frac{h^{\prime\prime}(z)+ag^{\prime\prime}(z)}{h^{\prime}(z)+ag^{\prime}(z)}-\frac{\overline{\omega_{a}(z)}\omega_{a}^{\prime}(z)}{1-\left|\omega_{a}(z)\right|^{2}}\right)+\sigma^{2}-\sigma_{z}\right|+\frac{\left|\sigma\omega^{\prime}_{a}(z)\right|}{1-|\omega_{a}(z)|^{2}}.$

In view of (3.6) and the triangle inequality, we see that

$\displaystyle\left|\sigma\frac{h_{a}^{\prime\prime}(z)}{h_{a}^{\prime}(z)}+\sigma^{2}-\sigma_{z}\right|\leq$
$\displaystyle\left|\sigma\left(\frac{h^{\prime\prime}(z)+ag^{\prime\prime}(z)}{h^{\prime}(z)+ag^{\prime}(z)}-\frac{\overline{\omega_{a}(z)}\omega_{a}^{\prime}(z)}{1-\left|\omega_{a}(z)\right|^{2}}\right)+\sigma^{2}-\sigma_{z}\right|+\left|\frac{\sigma\overline{\omega_{a}(z)}\omega^{\prime}_{a}(z)}{1-|\omega_{a}(z)|^{2}}\right|$
(3.7) $\displaystyle\leq$ $\displaystyle\left|\sigma
P_{f_{a}}(z)+\sigma^{2}-\sigma_{z}\right|+\left|\frac{\sigma\omega^{\prime}_{a}(z)}{1-|\omega_{a}(z)|^{2}}\right|$
$\displaystyle\leq$ $\displaystyle k|\sigma_{\overline{z}}|.$

By virtue of (1.4), it implies that for each $a\in\triangle$, the function
$h_{a}$ is univalent and can be extended to a continuous and injective mapping
$\widetilde{h}_{a}$ in $\overline{\triangle}$. Moreover, in [1], it was shown
that the function

$H_{a}(z)=\begin{cases}\widetilde{h}_{a}(z)&(|z|\leq 1),\\\
h_{a}\left(\frac{1}{\overline{z}}\right)+u_{a}\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$
(3.8)

is a $K$-quasiconformal extension of $\mathbb{C}$ onto itself with
$K=(1+k)/(1-k)$, where

$u_{a}(z)=\frac{h_{a}^{\prime}(z)}{{\sigma(z)}}\quad(z\in\triangle\backslash\\{0\\}).$

Hence, $H_{a}$ is continuous and univalent in $\mathbb{C}$.

Define

$H_{\varepsilon}(z)=\lim_{r\rightarrow
1^{-}}H_{re^{i\theta}}(z)\quad(z\in\mathbb{C};\,\varepsilon=e^{i\theta}\in\partial\triangle).$

According to [11, Theorem 1], we show that $H_{\varepsilon}$ is continuous and
one-to-one in $\mathbb{C}$ for all $|\varepsilon|\leq 1$.

Next, we assume that $H=H_{0}$ and $H_{1}$ are the corresponding extension of
$h_{0}=h$ and $h_{1}=h+g$, respectively. By virtue of (3.8), we define the
function

$G(z)=H_{1}(z)-H(z)$

to obtain a continuous extension $G$ of $g$ to $\mathbb{C}$, which was given
by

$G(z)=\begin{cases}g(z)&(|z|<1),\\\
\widetilde{g}(z)=\widetilde{h}_{1}(z)-\widetilde{h}(z)&(|z|=1),\\\
g\left(\frac{1}{\overline{z}}\right)+V\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$

where

$V(z)=\frac{g^{\prime}(z)}{\sigma(z)}\quad(z\in\triangle\backslash\\{0\\}).$

Since $H_{a}$ are univalent in $\mathbb{C}$ for all
$a\in\overline{\triangle}$, for each $z\in\mathbb{C}$, we see that
$H(z)+aG(z)$ is univalent in $\mathbb{C}$. Now, we construct an explicit
candidate for a continuous and injective extension of $f$ to $\mathbb{C}$
defined by

$F_{\lambda}(z)=H(z)+\lambda\overline{G(z)}=\begin{cases}f_{\lambda}(z)&(|z|<1),\\\
\widetilde{h}(z)+\lambda\overline{\widetilde{g}}(z)&(|z|=1),\\\
f_{\lambda}\left(\frac{1}{\overline{z}}\right)+U\left(\frac{1}{\overline{z}}\right)&(|z|>1),\end{cases}$

where $U$ is given by

$U(z)=\frac{h^{\prime}(z)}{\sigma(z)}+\lambda\frac{\overline{g^{\prime}(z)}}{\overline{\sigma(z)}}\quad(z\in\triangle\backslash\\{0\\}).$

Finally, by using the similar method as in [11], we can prove that
$F_{\lambda}$ is univalent in $\mathbb{C}$, and $F_{\lambda}$ is a
homeomorphism of $\mathbb{C}$ onto itself.

##  4\. Proof of Theorem 1.3

The following three lemmas are crucial to prove Theorem 1.3.

######  Lemma 4.1.

([7]) Let $\varepsilon\in\triangle$ and
$T(z)=(z+|\varepsilon|)/(1+|\varepsilon|z)$. Then $T(z)$ is a Möbius
transformation of the unit disk $\triangle$ onto itself and

$\frac{\|\varepsilon|-|z||}{1-|\varepsilon||z|}\leq|T(z)|\leq\frac{|\varepsilon|+|z|}{1+|\varepsilon||z|}.$

######  Lemma 4.2.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in $\triangle$
with complex dilatation $\omega\not\equiv 0.$ Assume that
$\|\omega\|_{\infty}<1$ and that $f$ satisfies (1.27). Then, the analytic
functions $h_{a}=h+ag$ are univalent in $\triangle$ for all $0\leq|a|<\delta$,
where

$1<\delta=\frac{1+k\|\omega\|_{\infty}}{k+\|\omega\|_{\infty}}\leq\frac{1}{\|\omega\|_{\infty}}.$

Moreover, $h_{a}$ has a continuous and injective extension $\widetilde{h}_{a}$
to $\overline{\triangle}$.

###### Proof.

In Remark 3.1, we have found that $h_{a}$ has a continuous and injective
extension $\widetilde{h}_{a}$ to $\overline{\triangle}$ for all $|a|\leq 1$.
Next, we prove that $h_{a}$ has a continuous and injective extension
$\widetilde{f_{a}}$ to $\overline{\triangle}$ for $1<|a|<\delta$. By noting
that

$\frac{h_{a}^{\prime\prime}}{h_{a}^{\prime}}=\frac{h^{\prime\prime}}{h^{\prime}}+\frac{a\omega^{\prime}}{1+a\omega},$
(4.1)

in view of (4.1) and the formula (1.13) for pre-Schwarzian derivative of $f$,
we get

$\displaystyle\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{h_{a}^{\prime\prime}}{h_{a}^{\prime}}+\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\right|$
$\displaystyle\leq\left|\frac{\sigma}{\sigma_{\overline{z}}}P_{f}+\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}\right|+\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{1-|\omega|^{2}}\frac{\overline{\omega}+a}{1+a\omega}\right|$
(4.2) $\displaystyle\leq
k-\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{1-|\omega|^{2}}\right|+\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{1-|\omega|^{2}}\right|\left|\frac{\overline{\omega}+a}{1+a\omega}\right|$
$\displaystyle=k+\left|\frac{\sigma}{\sigma_{\overline{z}}}\frac{\omega^{\prime}}{1-|\omega|^{2}}\right|\left(\left|\frac{\overline{\omega}+a}{1+a\omega}\right|-1\right)$
$\displaystyle\leq k\left|\frac{\overline{\omega}+a}{1+a\omega}\right|.$

From Lemma 4.1, we have

$\sup_{z\in\mathbf{D}}\left|\frac{\omega+\bar{a}}{1+a\omega}\right|\leq\frac{|a|-\|\omega\|_{\infty}}{1-\|\omega\|_{\infty}|a|}<\frac{1}{k}$
(4.3)

for

$1<|a|<\delta=\frac{1+k\|\omega\|_{\infty}}{k+\|\omega\|_{\infty}}\leq\frac{1}{\|\omega\|_{\infty}}.$

Combining (4.2), (4.3) with Theorem B, we conclude that $h_{a}$ is univalent
in $\triangle$ and $h_{a}$ has a continuous and injective extension
$\widetilde{f}_{a}$ to $\overline{\triangle}$ for all $0\leq|a|<\delta$, where

$1<\delta=\frac{1+k\|\omega\|_{\infty}}{k+\|\omega\|_{\infty}}\leq\frac{1}{\|\omega\|_{\infty}}.$

We thus complete the proof of Lemma 4.2. ∎

######  Lemma 4.3.

Let $f=h+\overline{g}$ satisfy the hypothesis of Lemma 4.2. Assume in addition
that both $h$ and $g$ are analytic functions in $\overline{\triangle}$. Then,

$\sup_{\alpha,\beta\in\overline{\triangle},\alpha\neq\beta}\left|\frac{g(\alpha)-g(\beta)}{h(\alpha)-h(\beta)}\right|=\sup_{\alpha,\beta\in\overline{\triangle},\alpha\neq\beta}\left|\frac{\lambda\overline{(g(\alpha)-g(\beta))}}{h(\alpha)-h(\beta)}\right|\leq\frac{1}{\delta}=\frac{k+\|\omega\|_{\infty}}{1+k\|\omega\|_{\infty}}<1.$
(4.4)

###### Proof.

By using the similar method as in [14, Lemma 2], we can prove Lemma 4.3. The
details are omitted here. ∎

###### Proof of Theorem 1.3.

Define

$(f_{\lambda})_{r}(z)=f_{\lambda}(rz)=h(rz)+\lambda\overline{g(rz)}\quad(0<r<1),$

where both $h$ and $g$ are analytic in $\overline{\triangle}$. Each of these
functions $f_{\lambda}$ with $|\lambda||\omega_{r}|=|\lambda||\omega(rz)|$, so
that

$|\lambda|\|\omega_{r}\|_{\infty}=\|\omega_{r}\|_{\infty}\leq\|\omega\|_{\infty}<1.$

If we prove that the mapping $(f_{\lambda})_{r}$ can be extended to a
$K$-quasiconformal mapping in $\mathbb{C}$, where $K$ does not depend on $r$.
Then from [16, Theorem 5.3], we conclude that $f_{\lambda}$ can be extended to
a $K$-quasiconformal mappings in $\mathbb{C}$.

We note that if $f$ satisfies (1.27), then

$\displaystyle\sup_{z\in\triangle}\left\\{|\sigma
P_{(f_{\lambda})_{r}}+\sigma^{2}-\sigma_{z}|+\frac{|\sigma\omega^{\prime}|}{1-|\omega|^{2}}\right\\}$
$\displaystyle\leq r\sup_{z\in\triangle}|\sigma
P_{f_{\lambda}}+\sigma^{2}-\sigma_{z}|+r\sup_{z\in\triangle}\frac{|\sigma\omega^{\prime}|}{1-|\omega|^{2}}$
$\displaystyle\leq r\sup_{z\in\triangle}|\sigma
P_{f}+\sigma^{2}-\sigma_{z}|+r\sup_{z\in\triangle}\frac{|\sigma\omega^{\prime}|}{1-|\omega|^{2}}$
$\displaystyle\leq rk|\sigma_{\overline{z}}|\leq k|\sigma_{\overline{z}}|.$

Thus, $(f_{\lambda})_{r}$ satisfies the condition (1.27).

By Lemmas 4.2 and 4.3, and use the similar method as in [14, Theorem 3], we
get the assertion $(f_{\lambda})_{r}$ admits a $K$-quasiconformal reflection.
By allowing $r\rightarrow 1$, we deduce that $f_{\lambda}$ has a
quasiconformal extension to $\mathbb{C}$. This completes the proof of Theorem
1.3. ∎

##  5\. Proof of Theorem 1.4

###### Proof of Theorem 1.4.

Step 1. We shall prove that for any $\alpha$ with
$|\alpha|<1,f=h+\alpha\overline{h}$ is a harmonic Teichmüller mapping in
$\triangle$.

Assume that $f$ is a harmonic mapping in $\triangle$ with a representation
$f(z)=h(z)+\alpha\overline{h(z)}$, where $h$ is a locally univalent analytic
function and $|\alpha|<1$. It follows that

$\frac{|\sigma\omega^{\prime}|}{1-|\omega|^{2}}=0.$

The Jacobian $J_{f}$ of $f$ is equal to $\left(1-|\alpha|^{2}\right)J_{f}$.
Hence,

$P_{f}=\frac{\partial}{\partial
z}\log\left(\left(1-|a|^{2}\right)\left|h^{\prime}(z)\right|^{2}\right)=\frac{h^{\prime\prime}}{h^{\prime}}=P_{h}.$
(5.1)

By (1.3) and (1.4), the Ahlfors’s univalent and extension criterion for a
locally univalent analytic function shows that $h$ is univalent, and it can be
extended to a continuous and injective mapping $\widetilde{h}(z)$ in
$\overline{\triangle}$. Thus, for any $\alpha$ with
$|\alpha|<1,f=h+\alpha\overline{h}$ is a harmonic Teichmüller mapping in
$\triangle$.

Step 2. We will show that ${F}(z)$ is a homeomorphic extension of $f$ in
$\mathbb{C}$ onto itself. By virtue of [11, Theorem 1] and (5.1), we have

$\left|\sigma P_{f}+\sigma^{2}-\sigma_{z}\right|\leq
k\left|\sigma_{\overline{z}}\right|.$ (5.2)

Hence, $f=h+\alpha\overline{h}$ can be extended to a homeomorphism of
$\mathbb{C}$ onto itself. Moreover, the homeomorphism can be constructed as
$F_{\alpha}(z)$ given by (1.31).

Step 3. We shall estimate the maximal dilatation of $F_{\alpha}$, and then
show that $F_{\alpha}$ is a quasiconformal mapping in $\mathbb{C}$. By the
assumption $|\alpha|<1$, we now separate the argument into two cases:

Case 1. If $|z|<1$, then $\left|\mu_{F_{\alpha}}(z)\right|=|\alpha|<1.$

Case 2. If $|z|>1$, we make a reciprocal transformation $\zeta=1/\overline{z}$
for $|z|<1$, then

$\displaystyle\left|\mu_{F_{\alpha}}(\zeta)\right|$
$\displaystyle=\left|\frac{(F_{\alpha})_{\overline{\zeta}}}{(F_{\alpha})_{\zeta}}\right|=\left|\frac{(F_{\alpha})_{z}z_{\overline{\zeta}}+(F_{\alpha})_{\overline{z}}\overline{z_{\zeta}}}{(F_{\alpha})_{z}z_{\zeta}+(F_{\alpha})_{\overline{z}}\overline{z_{\overline{\zeta}}}}\right|=\left|\frac{(F_{\alpha})_{z}}{(F_{\alpha})_{\overline{z}}}\right|$
$\displaystyle=\left|\frac{h^{\prime}+\frac{h^{\prime\prime}\sigma-h^{\prime}\sigma_{z}}{\sigma^{2}}-\alpha\frac{\overline{h^{\prime}}\overline{\sigma_{\overline{z}}}}{\overline{\sigma}^{2}}}{\alpha\overline{h^{\prime}}+\alpha\frac{\overline{h^{\prime\prime}}\overline{\sigma}-\overline{h^{\prime}}\overline{\sigma_{z}}}{\overline{\sigma}^{2}}-\frac{h^{\prime}\sigma_{\overline{z}}}{\sigma^{2}}}\right|$
$\displaystyle=\left|\frac{\sigma^{2}+\sigma\frac{h^{\prime\prime}}{h^{\prime}}-\sigma_{z}-\alpha\overline{\sigma_{\overline{z}}}\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{h^{\prime}}}{h^{\prime}}}{\overline{\alpha}\sigma^{2}+\overline{\alpha}\sigma\frac{h^{\prime\prime}}{h^{\prime}}-\overline{\alpha}\sigma_{z}-\overline{{\sigma_{\overline{z}}}}\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{h^{\prime}}}{h^{\prime}}}\right|$
(5.3)
$\displaystyle=\left|\frac{\frac{\sigma^{2}}{\sigma_{\overline{z}}}+\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}-\frac{\sigma_{z}}{\sigma_{\overline{z}}}-\alpha\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{\sigma_{\overline{z}}}}{\sigma_{\overline{z}}}\frac{\overline{h^{\prime}}}{h^{\prime}}}{\overline{\alpha}\frac{\sigma^{2}}{\sigma_{\overline{z}}}+\overline{\alpha}\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}-\overline{\alpha}\frac{\sigma_{z}}{\sigma_{\overline{z}}}-\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{\sigma_{\overline{z}}}}{\sigma_{\overline{z}}}\frac{\overline{h^{\prime}}}{h^{\prime}}}\right|$
$\displaystyle=\left|\frac{\alpha+\frac{-\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}+\frac{\sigma_{z}}{\sigma_{\overline{z}}}}{\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{\sigma_{\overline{z}}}}{\sigma_{\overline{z}}}\frac{\overline{h^{\prime}}}{h^{\prime}}}}{1+\overline{\alpha}\frac{-\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}+\frac{\sigma_{z}}{\sigma_{\overline{z}}}}{\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{\sigma_{\overline{z}}}}{\sigma_{\overline{z}}}\frac{\overline{h^{\prime}}}{h^{\prime}}}}\right|.$

We now suppose that

$\eta(z)=\frac{-\frac{\sigma^{2}}{\sigma_{\overline{z}}}-\frac{\sigma}{\sigma_{\overline{z}}}\frac{h^{\prime\prime}}{h^{\prime}}+\frac{\sigma_{z}}{\sigma_{\overline{z}}}}{\frac{\sigma^{2}}{\overline{\sigma}^{2}}\frac{\overline{\sigma_{\overline{z}}}}{\sigma_{\overline{z}}}\frac{\overline{h^{\prime}}}{h^{\prime}}}.$

From (1.27), we see that

$|\eta(z)|\leq k\quad(z\in\triangle).$ (5.4)

By setting $\theta=\overline{\alpha}/|\alpha|$, it follows from (5) that

$\left|\mu_{F_{\alpha}}(\zeta)\right|=\left|\frac{\alpha+\eta(z)}{1+\overline{\alpha}\eta(z)}\right|=\left|\frac{\theta\alpha+\theta\eta(z)}{1+\overline{\theta}\overline{\alpha}\theta\eta(z)}\right|=\left|\frac{|\alpha|+\theta\eta(z)}{1+|\alpha|\theta\eta(z)}\right|\quad(\zeta\in\mathbb{C}\backslash\overline{\triangle}).$

By Lemma 4.1 and (5.4), we know that

$\left|\mu_{F_{\alpha}}(\zeta)\right|\leq\frac{|\alpha|+|\eta(z)|}{1+|\alpha\eta(z)|}\leq\frac{|\alpha|+k}{1+|\alpha|k}=k_{2}\quad(\zeta\in\mathbb{C}\backslash\overline{\triangle}).$
(5.5)

Therefore, $F_{\alpha}(\zeta)$ is a quasiconformal mapping in
$\mathbb{C}\backslash\overline{\triangle}$.

By taking

$f=mz+n\overline{z}\quad(|n|<|m|),$ (5.6)

then $f$ satisfies the condition (1.30) with $k=0$. Its quasiconformal
extension given by (1.31) is just of the form (5.6), and its maximal
dilatation is

$k_{2}=\left|\frac{\overline{n}}{m}\right|=|\alpha|=\frac{K-1}{K+1},$

which shows that the estimate (5.5) is asymptotically sharp for $k$. The proof
of Theorem 1.4 is thus completed. ∎

Acknowledgments

The present investigation was supported by the Key Project of Education
Department of Hunan Province under Grant no. 19A097, and the Natural Science
Foundation of Hunan Province under Grant no. 2018JJ2074 of the P. R. China.

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