The Schwarz Type Lemmas and the Landau Type Theorem of Mappings Satisfying Poisson’s Equations

Abstract

For a given continuous function $g:~\overline{\mathbb {D}}\rightarrow {\mathbb {C}}$ and a given continuous function $\psi :~{\mathbb {T}}\rightarrow {\mathbb {C}}$, we establish some Schwarz type Lemmas for mappings f satisfying the PDE: $\Delta f=g$ in ${\mathbb {D}}$, and $f=\psi$ in ${\mathbb {T}}$, where ${\mathbb {D}}$ is the unit disk of the complex plane ${\mathbb {C}}$ and ${\mathbb {T}}=\partial {\mathbb {D}}$ is the unit circle. Then we apply these results to obtain a Landau type theorem, which is a partial answer to the open problem in Chen and Ponnusamy (Bull Aust Math Soc 97: 80–87, 2018).

1 Preliminaries and Main Results

Let ${\mathbb {C}} \cong {\mathbb {R}}^{2}$ be the complex plane. For $a\in {\mathbb {C}}$ and $r>0$, we let ${\mathbb D}(a,r)=\{z:\, |z-a|<r\}$ so that ${\mathbb {D}}_r:={\mathbb {D}}(0,r)$ and thus, ${\mathbb {D}}:={\mathbb D}_1$ denotes the open unit disk in the complex plane ${\mathbb {C}}$. Let ${\mathbb {T}}=\partial {\mathbb {D}}$ be the boundary of ${\mathbb {D}}$. We denote by ${\mathcal {C}}^{m}(\Omega )$ the set of all complex-valued m-times continuously differentiable functions from $\Omega$ into ${\mathbb {C}}$, where $\Omega$ is a subset of ${\mathbb {C}}$ and $m\in {\mathbb {N}}_0:={\mathbb {N}}\cup \{0\}$. In particular, let ${\mathcal {C}}(\Omega ):={\mathcal {C}}^{0}(\Omega )$, the set of all continuous functions defined in $\Omega$.

For a real $2\times 2$ matrix A, we use the matrix norm $\Vert A\Vert =\sup \{|Az|:\,|z|=1\}$ and the matrix function $\lambda (A)=\inf \{|Az|:\,|z|=1\}$. For $z=x+iy\in {\mathbb {C}}$, the formal derivative of the complex-valued functions $f=u+iv$ is given by

$$\begin{aligned} D_{f}=\left( \begin{array}{cccc} \displaystyle u_{x}\;~~ u_{y}\\ \displaystyle v_{x}\;~~ v_{y} \end{array}\right) , \end{aligned}$$

so that

$$\begin{aligned} \Vert D_{f}\Vert =|f_{z}|+|f_{\overline{z}}| \quad \text{ and }\quad \lambda (D_{f})=\big | |f_{z}|-|f_{\overline{z}}|\big |, \end{aligned}$$

where

$$\begin{aligned} f_{z}=\frac{\partial f}{\partial z}=\frac{1}{2}\big ( f_x-if_y\big )\quad \text{ and }\quad f_{\overline{z}}=\frac{\partial f}{\partial \overline{z}}=\frac{1}{2}\big (f_x+if_y\big ). \end{aligned}$$

We use

$$\begin{aligned} J_{f}:=\det D_{f} =|f_{z}|^{2}-|f_{\overline{z}}|^{2} \end{aligned}$$

to denote the Jacobian of f and

$$\begin{aligned} \Delta f:=\frac{\partial ^{2}f}{\partial x^{2}}+\frac{\partial ^{2}f}{\partial y^{2}}=4f_{z \overline{z}} \end{aligned}$$

is the Laplacian of f.

For $t\in {\mathbb {R}}$ and $z, w\in {\mathbb {D}}$ with $z\ne w$ and $|z|+|w|\ne 0$, let

$$\begin{aligned} G(z,w)=\log \left| \frac{1-z\overline{w}}{z-w}\right| \quad \text{ and } \quad P(z,e^{it})=\frac{1-|z|^{2}}{|1-ze^{-it}|^{2}} \end{aligned}$$

be the Green function and Poisson kernel, respectively.

Let $\psi :~{\mathbb {T}}\rightarrow {\mathbb {C}}$ be a bounded integrable function and let $g\in {\mathcal {C}}({\mathbb {D}})$. For $z\in {\mathbb {D}}$, the solution to the Poisson’s equation

$$\begin{aligned} \Delta f(z)=g(z) \end{aligned}$$

satisfying the boundary condition $f|_{{\mathbb {T}}}=\psi \in L^{1}({\mathbb {T}})$ is given by

$$\begin{aligned} f(z)={\mathcal {P}}_{\psi }(z)-{\mathcal {G}}_{g}(z), \end{aligned}$$

(1.1)

where

$$\begin{aligned} {\mathcal {G}}_{g}(z)=\frac{1}{2\pi }\int _{{\mathbb {D}}}G(z,w)g(w)dA(w),~ ~{\mathcal {P}}_{\psi }(z)=\frac{1}{2\pi }\int _{0}^{2\pi }P(z,e^{it})\psi (e^{it})dt,\nonumber \\ \end{aligned}$$

(1.2)

and dA(w) denotes the Lebesgue measure in ${\mathbb {D}}$. It is well known that if $\psi$ and g are continuous in ${\mathbb {T}}$ and in $\overline{{\mathbb {D}}}$, respectively, then $f={\mathcal {P}}_{\psi }-{\mathcal {G}}_{g}$ has a continuous extension $\tilde{f}$ to the boundary, and $\tilde{f}=\psi$ in ${\mathbb {T}}$ (see [18, pp. 118–120] and [2, 19, 20, 22]).

Heinz in his classical paper [17] proved the following result, which is called the Schwarz Lemma of complex-valued harmonic functions: If f is a complex-valued harmonic function from ${\mathbb {D}}$ into itself satisfying the condition $f(0)=0$, then, for $z\in {\mathbb {D}}$,

$$\begin{aligned} |f(z)|\le \frac{4}{\pi }\arctan |z|. \end{aligned}$$

(1.3)

Later, Pavlović [30, Theorem 3.6.1] removed the assumption $f(0)=0$ and improved (1.3) into the following sharp form

$$\begin{aligned} \left| f(z)-\frac{1-|z|^{2}}{1+|z|^{2}}f(0)\right| \le \frac{4}{\pi }\arctan |z|, \end{aligned}$$

(1.4)

where f is a complex-valued harmonic function from ${\mathbb {D}}$ to itself.

The first aim of this paper is to extend (1.4) into mappings satisfying the Poisson’s equation as follows.

Theorem 1

For a given $g\in {\mathcal {C}}(\overline{{\mathbb {D}}})$ and a given $\psi \in {\mathcal {C}}({\mathbb {T}})$, if a complex-valued function f satisfies $\Delta f=g$ in ${\mathbb {D}}$ and $f=\psi$ in ${\mathbb {T}}$, then, for $z\in \overline{{\mathbb {D}}}$,

$$\begin{aligned} \left| f(z)-\frac{1-|z|^{2}}{1+|z|^{2}}{\mathcal {P}}_{\psi }(0)\right| \le \frac{4\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }}{\pi }\arctan |z|+\frac{\Vert g\Vert _{\infty }}{4}(1-|z|^{2}), \end{aligned}$$

(1.5)

where

$$\begin{aligned} {\mathcal {P}}_{\psi }(z)=\frac{1}{2\pi }\int _{0}^{2\pi }\!\! P(z,e^{it})\psi (e^{it})dt,~\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }= \sup _{z\in {\mathbb {D}}}|{\mathcal {P}}_{\psi }(z)|~\text{ and }~\Vert g\Vert _{\infty }=\sup _{z\in {\mathbb {D}}}|g(z)|. \end{aligned}$$

If we take $g(z)=-4M$ and $f(z)=M(1-|z|^{2})$ for $z\in \overline{{\mathbb {D}}}$, where M is a positive constant, then the inequality (1.5) is sharp in $\overline{{\mathbb {D}}}$.

The following result is a classical Schwarz Lemma at the boundary.

Theorem A

(see [15]) Let f be a holomorphic function from ${\mathbb {D}}$ into itself. If f is holomorphic at $z=1$ with $f(0)=0$ and $f(1)=1$, then $f'(1)\ge 1$. Moreover, the inequality is sharp.

Theorem A has attracted much attention and has been generalized in various forms (See [6, 23, 26, 27] for holomorphic functions, and see [21] for harmonic functions). In the following result, applying Theorem 1, we establish a Schwarz Lemma at the boundary for mappings satisfying the Poisson’s equation, which is a generalization of Theorem A.

Theorem 2

For a given $g\in {\mathcal {C}}(\overline{{\mathbb {D}}})$, let $f\in {\mathcal {C}}^{2}({\mathbb {D}})\cap {\mathcal {C}}({\mathbb {T}})$ be a function of ${\mathbb {D}}$ into itself satisfying $\Delta f=g,$ where $\Vert g\Vert _{\infty }<\frac{8}{3\pi }.$ If $f(0)=0$ and, for some $\zeta \in {\mathbb {T}}$, $\lim _{r\rightarrow 1^{-}}|f(r\zeta )|=1$, then

$$\begin{aligned} \liminf _{r\rightarrow 1^{-}}\frac{|f(\zeta )-f(r\zeta )|}{1-r}\ge \frac{2}{\pi }-\frac{3\Vert g\Vert _{\infty }}{4}. \end{aligned}$$

(1.6)

In particular, if $\Vert g\Vert _{\infty }=0$, then the estimate of (1.6) is sharp.

In [14], Colonna proved a sharp Schwarz-Pick type Lemma of complex-valued harmonic functions, which is as follows: If f is a complex-valued harmonic function from ${\mathbb {D}}$ into itself, then, for $z\in {\mathbb {D}}$,

$$\begin{aligned} \Vert D_{f}(z)\Vert \le \frac{4}{\pi }\frac{1}{1-|z|^{2}}. \end{aligned}$$

(1.7)

We extend (1.7) into the following form.

Theorem 3

For a given $g\in {\mathcal {C}}(\overline{{\mathbb {D}}})$ and a given $\psi \in {\mathcal {C}}({\mathbb {T}})$, if a complex-valued function f satisfies $\Delta f=g$ in ${\mathbb {D}}$ and $f=\psi$ in ${\mathbb {T}}$, then, for $z\in {\mathbb {D}}\backslash \{0\},$

$$\begin{aligned} \Vert D_{f}(z)\Vert \le \frac{4\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }}{\pi }\frac{1}{1-|z|^{2}}+2\mu (|z|), \end{aligned}$$

(1.8)

where

$$\begin{aligned} \frac{\Vert g\Vert _{\infty }}{4}\le \mu (|z|)=\frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{8|z|^{2}} \bigg [\frac{1+|z|^{2}}{1-|z|^{2}}-\frac{(1-|z|^{2})}{2|z|}\log \frac{1+|z|}{1-|z|}\bigg ]\le \frac{\Vert g\Vert _{\infty }}{3} \end{aligned}$$

and $\mu (|z|)$ is decreasing on $|z|\in (0,1)$. In particular, if $z=0$, then

$$\begin{aligned} \Vert D_{f}(0)\Vert \le \lim _{|z|\rightarrow 0^{+}}\bigg (\frac{4\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }}{\pi }\frac{1}{1-|z|^{2}}+2\mu (|z|)\bigg )= \frac{4}{\pi }\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }+\frac{2}{3}\Vert g\Vert _{\infty }. \end{aligned}$$

(1.9)

Moreover, if $\Vert g\Vert _{\infty }=0$, then the extremal functions

$$\begin{aligned} f(z)=\frac{2M\alpha }{\pi }\arg \left( \frac{1+\phi (z)}{1-\phi (z)}\right) \end{aligned}$$

show that the estimate of (1.8) and (1.9) are sharp, where $|\alpha |=1$ and $M>0$ are constants, and $\phi$ is a conformal automorphism of ${\mathbb {D}}$.

We remark that if $\Vert g\Vert _{\infty }=0$ and $\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }=1$ in Theorem 3, then (1.8) and (1.9) coincide with (1.7).

Let ${\mathcal {A}}$ denote the set of all analytic functions f defined in ${\mathbb {D}}$ satisfying the standard normalization: $f(0)=f'(0)-1=0$. In the early 20th century, Landau [24] showed that there is a constant $r>0$, independent of $f\in {\mathcal {A}}$, such that $f({\mathbb {D}})$ contains a disk of radius r. Let $L_{f}$ be the supremum of the set of positive numbers r such that $f({\mathbb {D}})$ contains a disk of radius r, where $f\in {\mathcal {A}}$. Then we call $\inf _{f\in {\mathcal {A}}}L_{f}$ the Landau–Bloch constant. One of the long standing open problems in geometric function theory is to determine the precise value of the Landau–Bloch constant. It has attracted much attention, see [4, 25, 28, 29, 32] and references therein. The Landau theorem is an important tool in geometric function theory of one complex variable (cf. [5, 33]). Unfortunately, for general class of functions, there is no Landau type theorem (see [7, 32]). In order to obtain some analogs of the Landau type theorem for more general classes of functions, it is necessary to restrict the class of functions considered (cf. [1, 3, 7,8,9,10,11, 13, 16, 32]). Let’s recall some known results as follows.

Theorem B

([7, Theorem 2]) Let f be a harmonic mapping in ${\mathbb {D}}$ such that $f(0)=J_{f}(0)-1=0$ and $|f(z)|<M$ for $z\in {\mathbb {D}}$, where M is a positive constant. Then f is univalent in ${\mathbb {D}}_{\rho _{0}}$ with $\rho _{0}=\pi ^{3}/(64mM^{2})$, and $f({\mathbb {D}}_{\rho _{0}})$ contains a univalent disk ${\mathbb {D}}_{R_{0}}$ with

$$\begin{aligned} R_{0}=\frac{\pi }{8M}\rho _{0}=\frac{\pi ^{4}}{512mM^{3}}, \end{aligned}$$

where $m\approx 6.85$ is the minimum of the function $(3-r^{2})/[r(1-r^{2})]$ for $r\in (0,1)$.

Theorem C

([1, Theorem 1]) Let $f(z)=|z|^{2}G(z)+K(z)$ be a biharmonic mapping, that is $\Delta (\Delta f)=0$, in ${\mathbb {D}}$ such that $f(0)= K(0)=J_{f}(0)-1=0$, where G and K are harmonic satisfying $|G(z)|, ~|K(z)|<M$ for $z\in {\mathbb {D}}$, where M is a positive constant. Then there is a constant $\rho _{2}\in (0,1)$ such that f is univalent in ${\mathbb {D}}_{\rho _{2}}$. Specifically, $\rho _{2}$ satisfies

$$\begin{aligned} \frac{\pi }{4M}-2\rho _{2}M-2M\left[ \frac{\rho _{2}^{2}}{(1-\rho _{2})^{2}}+\frac{1}{(1-\rho _{2})^{2}}-1\right] =0 \end{aligned}$$

and $f({\mathbb {D}}_{\rho _{2}})$ contains a disk ${\mathbb {D}}_{R_{2}}$, where

$$\begin{aligned} R_{2}=\frac{\pi }{4M}\rho _{2}-2M\frac{\rho _{2}^{3}+\rho _{2}^{2}}{1-\rho _{2}}. \end{aligned}$$

For some $g\in {\mathcal {C}}(\overline{{\mathbb {D}}})$, let ${\mathcal {F}}_{g}(\overline{{\mathbb {D}}})$ denote the class of all complex-valued functions $f\in {\mathcal {C}}^{2}({\mathbb {D}})\cap {\mathcal {C}}({\mathbb {T}})$ satisfying $\Delta f=g$ and $f(0)=J_{f}(0)-1=0$. We extend Theorems B and C into the following from.

Theorem 4

For a given $g\in {\mathcal {C}}(\overline{{\mathbb {D}}})$, let $f\in {\mathcal {F}}_{g}(\overline{{\mathbb {D}}})$ satisfying $\Vert g\Vert _{\infty }\le M_{1}$ and $\Vert f\Vert _{\infty }\le M_{2}$, where $M_{1}\ge 0$ and $M_{2}>0$ are constants. Then f is univalent in ${\mathbb {D}}_{r_{0}}$, where $r_{0}$ satisfies the following equation

$$\begin{aligned} \frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}}{\pi }\frac{r_{0}(2-r_{0})}{(1-r_{0})^{2}} -2M_{1}\big [\log 4(1+r_{0})- \log r_{0}\big ](2+r_{0})r_{0}=0. \end{aligned}$$

Moreover, $f({\mathbb {D}}_{r_{0}})$ contains an univalent disk ${\mathbb {D}}_{R_{0}}$ with

$$\begin{aligned} R_{0}\ge \frac{2M_{2}}{\pi }\frac{r_{0}^{2}(2-r_{0})}{(1-r_{0})^{2}}. \end{aligned}$$

Remark 1.1

Theorem 4 gives an affirmative answer to the open problem of [13] for the u-gradient mapping$f\in {\mathcal {C}}^{2}({\mathbb {D}})$. If g is harmonic, then all $f\in {\mathcal {F}}_{g}(\overline{{\mathbb {D}}})$ are biharmonic. Furthermore, if $\Vert g\Vert _{\infty }=0$, then all $f\in {\mathcal {F}}_{g}(\overline{{\mathbb {D}}})$ are harmonic. Hence, Theorem 4 is also a generalization of a series of known results, such as [1, Theorem 2], [7, Theorems, 3, 4, 5 and 6], [8, Theorems 2 and 3].

In the following two Examples, we will show that there is no Landau type Theorem for $f\in {\mathcal {F}}_{g}(\overline{{\mathbb {D}}})$ without the boundedness hypothesis of $\Vert f\Vert _{\infty }$.

Example 1.10

For $g\equiv 1$ and $z=x+iy\in {\mathbb {D}},$ let $f_{k}(z)=kx+|z|^{2}/4+i\frac{y}{k},$ where $k\in \{1,2,\ldots \}$. Then, for all $k\in \{1,2,\ldots \}$, $f_{k}$ is univalent. For all $k\in \{1,2,\ldots \}$, by simple calculations, we see that $J_{f_{k}} (0)-1=f_{k}(0)=0$, and there is no an absolute constant $\rho _{0}>0$ such that ${\mathbb {D}}_{\rho _{0}}$ is contained in $f_{k}({\mathbb {D}})$.

Example 1.11

For $\Vert g\Vert _{\infty }=0$ and $z=x+iy\in {\mathbb {D}},$ let $f_{k}(z)=kx+i\frac{y}{k},$ where $k\in \{1,2,\ldots \}$. For all $k\in \{1,2,\ldots \}$, it is not difficult to see that $f_{k}$ is univalent and $J_{f_{k}} (0)-1=f_{k}(0)=0$. Moreover, for all $k\in \{1,2,\ldots \}$, $f_{k}({\mathbb {D}})$ contains no disk with radius bigger than 1 / k. Hence, for all $k\in \{1,2,\ldots \}$, there is no an absolute constant $r_{0}>0$ such that ${\mathbb {D}}_{r_{0}}$ is contained in $f_{k}({\mathbb {D}})$.

Corollary 1

Under the same hypothesis of Theorem 4, there is a $r_{0}\in (0,1)$ such that f is bi-Lipschitz in ${\mathbb {D}}_{r_{0}}$.

The proofs of Theorems 1, 2, 3, 4 and Corollary 1 will be presented in Sect. 2.

2 Proofs of the Main Results

Proof of Theorem 1

For a given $g\in {\mathcal {C}}({\mathbb {D}})$, by (1.1), we have

$$\begin{aligned} f(z)={\mathcal {P}}_{\psi }(z)-{\mathcal {G}}_{g}(z), \quad z\in {\mathbb {D}}, \end{aligned}$$

(2.1)

where ${\mathcal {P}}_{\psi }$ and ${\mathcal {G}}_{g}$ are defined in (1.2). Since ${\mathcal {P}}_{\psi }$ is harmonic in ${\mathbb {D}}$, by (1.4), we see that, for $z\in {\mathbb {D}}$,

$$\begin{aligned} \left| {\mathcal {P}}_{\psi }(z)-\frac{1-|z|^{2}}{1+|z|^{2}}{\mathcal {P}}_{\psi }(0)\right| \le \frac{4\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }}{\pi }\arctan |z|. \end{aligned}$$

(2.2)

On the other hand, for a fixed $z\in {\mathbb {D}}$, let

$$\begin{aligned} \zeta =\frac{z-w}{1-\overline{z}w}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} w=\frac{z-\zeta }{1-\overline{z}\zeta }. \end{aligned}$$

Then

$$\begin{aligned} \big |{\mathcal {G}}_{g}(z)\big |= & {} \bigg |\frac{1}{2\pi }\int _{{\mathbb {D}}}\left( \log \frac{1}{|\zeta |}\right) g\left( \frac{z-\zeta }{1-\overline{z}\zeta }\right) \frac{(1-|z|^{2})^{2}}{|1-\overline{z}\zeta |^{4}}dA(\zeta )\bigg |\nonumber \\\le & {} \frac{\Vert g\Vert _{\infty }}{2\pi }\bigg |\int _{{\mathbb {D}}}\left( \log \frac{1}{|\zeta |}\right) \frac{(1-|z|^{2})^{2}}{|1-\overline{z}\zeta |^{4}}dA(\zeta )\bigg |\nonumber \\= & {} (1-|z|^{2})^{2}\Vert g\Vert _{\infty }\int _{0}^{1}\left[ \bigg (\frac{1}{2\pi }\int _{0}^{2\pi }\frac{dt}{|1-\overline{z}re^{it}|^{4}}\bigg )r\log \frac{1}{r}\right] dr\nonumber \\= & {} (1-|z|^{2})^{2}\Vert g\Vert _{\infty }\int _{0}^{1}\left[ \bigg (\frac{1}{2\pi }\int _{0}^{2\pi }\frac{dt}{|(1-\overline{z}re^{it})^{2}|^{2}}\bigg )r\log \frac{1}{r}\right] dr\nonumber \\= & {} (1-|z|^{2})^{2}\Vert g\Vert _{\infty }\int _{0}^{1}\left[ \left( \frac{1}{2\pi }\int _{0}^{2\pi }\left| \sum _{n=0}^{\infty }(n+1)(r\overline{z})^{n}e^{int}\right| ^{2}dt\right) r\log \frac{1}{r}\right] dr \nonumber \\= & {} (1-|z|^{2})^{2}\Vert g\Vert _{\infty }\int _{0}^{1}\bigg (r\log \frac{1}{r}\bigg )\sum _{n=0}^{\infty }(n+1)^{2}|z|^{2n}r^{2n}dr\nonumber \\= & {} (1-|z|^{2})^{2}\Vert g\Vert _{\infty }\sum _{n=0}^{\infty }(n+1)^{2}|z|^{2n}\int _{0}^{1}r^{2n+1}\bigg (\log \frac{1}{r}\bigg )dr \nonumber \\= & {} \frac{(1-|z|^{2})^{2}\Vert g\Vert _{\infty }}{4}\sum _{n=0}^{\infty }|z|^{2n}\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }}{4}(1-|z|^{2}). \end{aligned}$$

(2.3)

Hence, by (2.2) and (2.3), we conclude that

$$\begin{aligned} \left| f(z)-\frac{1-|z|^{2}}{1+|z|^{2}}{\mathcal {P}}_{\psi }(0)\right|\le & {} \left| {\mathcal {P}}_{\psi }(z)-\frac{1-|z|^{2}}{1+|z|^{2}}{\mathcal {P}}_{\psi }(0)\right| +\big |{\mathcal {G}}_{g}(z)\big |\\\le & {} \frac{4\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }}{\pi }\arctan |z|+\frac{\Vert g\Vert _{\infty }}{4}(1-|z|^{2}). \end{aligned}$$

Now we prove the sharpness part. For $z\in \overline{{\mathbb {D}}}$, let

$$\begin{aligned} g(z)=-4M\quad \text{ and }\quad f(z)=M(1-|z|^{2}), \end{aligned}$$

where M is a positive constant. Then $\psi \equiv 0$ in ${\mathbb {T}}$ and

$$\begin{aligned} \left| f(z)-\frac{1-|z|^{2}}{1+|z|^{2}}{\mathcal {P}}_{\psi }(0)\right|= & {} |f(z)|=\left| \frac{1}{2\pi }\int _{{\mathbb {D}}}G(z,w)g(w)dA(w)\right| \\= & {} \frac{\Vert g\Vert _{\infty }}{4}(1-|z|^{2}), \end{aligned}$$

which shows (1.5) is sharp in $\overline{{\mathbb {D}}}$. The proof of this theorem is complete. $\square$

Proof of Theorem 2

For a given $g\in {\mathcal {C}}(\overline{{\mathbb {D}}})$, by (1.1) with f in place of $\psi$, we have

$$\begin{aligned} f(z)={\mathcal {P}}_{f}(z)-{\mathcal {G}}_{g}(z), \quad z\in {\mathbb {D}}, \end{aligned}$$

where ${\mathcal {P}}_{f}$ and ${\mathcal {G}}_{g}$ are defined in (1.2). Since $f(0)=0$, we see that

$$\begin{aligned} |{\mathcal {P}}_{f}(0)|= & {} |{\mathcal {G}}_{g}(0)|=\left| \frac{1}{2\pi }\int _{{\mathbb {D}}}\log \frac{1}{|w|}g(w)dA(w)\right| \nonumber \\\le & {} \frac{\Vert g\Vert _{\infty }}{2\pi }\int _{0}^{2\pi }dt\int _{0}^{1}r\log \frac{1}{r}dr\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }}{4}. \end{aligned}$$

(2.4)

Let $z=r\zeta \in {\mathbb {D}}$, where $\zeta \in {\mathbb {T}}$ is as in the statement of the theorem. Then, by (2.4) and Theorem 1, we have

$$\begin{aligned} |f(\zeta )-f(r\zeta )|= & {} \left| f(\zeta )+{\mathcal {P}}_{f}(0)\frac{1-|z|^{2}}{1+|z|^{2}}-{\mathcal {G}}_{g}(0)\frac{1-|z|^{2}}{1+|z|^{2}}-f(r\zeta )\right| \\\ge & {} 1-\left| f(r\zeta )-{\mathcal {P}}_{f}(0)\frac{1-|z|^{2}}{1+|z|^{2}}\right| -|{\mathcal {G}}_{g}(0)|\frac{1-|z|^{2}}{1+|z|^{2}}\\\ge & {} 1-\frac{4}{\pi }\arctan |z|-\frac{\Vert g\Vert _{\infty }}{4}(1-|z|^{2})-|{\mathcal {G}}_{g}(0)|\frac{1-|z|^{2}}{1+|z|^{2}}\\\ge & {} 1-\frac{4}{\pi }\arctan |z|-\frac{\Vert g\Vert _{\infty }}{4}(1-|z|^{2})-\frac{\Vert g\Vert _{\infty }}{4}\frac{(1-|z|^{2})}{1+|z|^{2}}, \end{aligned}$$

which, together with L’Hospital’s rule, gives that

$$\begin{aligned} \liminf _{r\rightarrow 1^{-}}\frac{|f(e^{i\theta })-f(re^{i\theta })|}{1-r}\ge & {} \lim _{r\rightarrow 1^{-}}\frac{1-\frac{4}{\pi }\arctan r-\frac{\Vert g\Vert _{\infty }}{4}(1-r^{2})-\frac{\Vert g\Vert _{\infty }}{4}\frac{(1-r^{2})}{1+r^{2}}}{1-r}\\= & {} \lim _{r\rightarrow 1^{-}}\left[ \frac{4}{\pi }\frac{1}{1+r^{2}}-\Vert g\Vert _{\infty }\frac{r}{2}-\Vert g\Vert _{\infty }\frac{r}{(1+r^{2})^{2}}\right] \\= & {} \frac{2}{\pi }-\frac{3\Vert g\Vert _{\infty }}{4}. \end{aligned}$$

Now we prove the sharpness part. For $z\in {\mathbb {D}}$, let

$$\begin{aligned} f(z)=\frac{2}{\pi }\arctan \frac{2\text{ Re }(z)}{1-|z|^{2}}. \end{aligned}$$

Then f is harmonic in ${\mathbb {D}}$ with $f(0)=f(1)-1=0$, and

$$\begin{aligned} f(\rho )=\frac{4}{\pi }\arctan \rho , \end{aligned}$$

where $\rho \in (-1,1)$. Elementary calculations show that

$$\begin{aligned} \liminf _{\rho \rightarrow 1^{-}}\frac{|f(1)-f(\rho )|}{1-\rho }=\frac{2}{\pi }, \end{aligned}$$

which implies that (1.6) is sharp for $\Vert g\Vert _{\infty }=0$. The proof of this theorem is complete. $\square$

Theorem D

([31] $\text{ or }$ [19, Proposition 2.4]) Suppose that X is an open subset of ${\mathbb {R}}$, and $\Omega$ a measure space. Suppose, further, that a function F :  $X\times \Omega \rightarrow {\mathbb {R}}$ satisfies the following conditions:

• (1) F(x, w) is a measurable function of x and w jointly, and is integrable with respect to w for almost every $x\in X.$

• (2) For almost every $w\in \Omega$, F(x, w) is an absolutely continuous function with respect to x. [This guarantees that $\partial F(x,w)/\partial x$ exists almost everywhere.]

• (3) $\partial F/\partial x$ is locally integrable, that is, for all compact intervals [a, b] contained in X:

  $$\begin{aligned} \int _{a}^{b}\int _{\Omega }\left| \frac{\partial }{\partial x}F(x,w)\right| dwdx<\infty . \end{aligned}$$

Then, $\int _{\Omega }F(x,w)dw$ is an absolutely continuous function with respect to x, and for almost every $x\in X$, its derivative exists, which is given by

$$\begin{aligned} \frac{d}{dx}\int _{\Omega }F(x,w)dw=\int _{\Omega }\frac{\partial }{\partial x}F(x,w)dw. \end{aligned}$$

Proof of Theorem 3

For a given $g\in {\mathcal {C}}(\overline{{\mathbb {D}}})$, by (2.1), we have

$$\begin{aligned} f(z)={\mathcal {P}}_{\psi }(z)-{\mathcal {G}}_{g}(z), \quad z\in {\mathbb {D}}, \end{aligned}$$

where ${\mathcal {P}}_{\psi }$ and ${\mathcal {G}}_{g}$ are the same as in (2.1). Applying [19, Lemma 2.3] and Theorem D, we have

$$\begin{aligned} \frac{\partial }{\partial z}{\mathcal {G}}_{g}(z)= & {} \frac{1}{2\pi }\int _{{\mathbb {D}}}\frac{\partial }{\partial z}G(z,w)g(w)dA(w)\nonumber \\= & {} \frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{(z-w)(z\overline{w}-1)}g(w)dA(w)\in {\mathcal {C}}({\mathbb {D}}) \end{aligned}$$

(2.5)

and

$$\begin{aligned} \frac{\partial }{\partial \overline{z}}{\mathcal {G}}_{g}(z)= & {} \frac{1}{2\pi }\int _{{\mathbb {D}}}\frac{\partial }{\partial \overline{z}}G(z,w)g(w)dA(w)\\= & {} \frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{(\overline{z}-\overline{w})(w\overline{z}-1)}g(w)dA(w)\in {\mathcal {C}}({\mathbb {D}}). \end{aligned}$$

For a fixed $z\in {\mathbb {D}}\backslash \{0\}$, let

$$\begin{aligned} \zeta =\frac{z-w}{1-\overline{z}w} \end{aligned}$$

(2.6)

which implies that

$$\begin{aligned} w=\frac{z-\zeta }{1-\overline{z}\zeta },\quad 1-\overline{z}w=\frac{1-|z|^{2}}{1-\overline{z}\zeta }\quad \text{ and }\quad 1-|w|^{2}=\frac{(1-|\zeta |^{2})(1-|z|^{2})}{|1-\overline{z}\zeta |^{2}}. \nonumber \\ \end{aligned}$$

(2.7)

Then, by (2.5), (2.7) and the change of variables (2.6), we have

$$\begin{aligned} \left| \frac{\partial }{\partial z}{\mathcal {G}}_{g}(z)\right|\le & {} \frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{|z-w||z\overline{w}-1|}|g(w)|dA(w)\nonumber \\\le & {} \frac{\Vert g\Vert _{\infty }}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{|z-w||z\overline{w}-1|}dA(w)\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{|\zeta ||1-\overline{z}w|^{2}}\frac{(1-|z|^{2})^{2}}{|1-\overline{z}\zeta |^{4}}dA(\zeta ) \nonumber \\= & {} \frac{\Vert g\Vert _{\infty }}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|z|^{2})(1-|\zeta |^{2})}{|\zeta ||1-\overline{z}\zeta |^{4}}dA(\zeta )\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{2}\int _{0}^{1}\left[ (1-r^{2})\bigg (\frac{1}{2\pi }\int _{0}^{2\pi }\frac{dt}{|1-\overline{z}re^{it}|^{4}}\bigg )\right] dr\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{2}\int _{0}^{1} \left[ (1-r^{2})\left( \frac{1}{2\pi }\int _{0}^{2\pi }\left| \sum _{n=0}^{\infty }(n+1)(r\overline{z})^{n}e^{int}\right| ^{2}dt\right) \right] dr\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{2}\int _{0}^{1} (1-r^{2})\left[ \sum _{n=0}^{\infty }(n+1)^{2}|z|^{2n}r^{2n}\right] dr\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{2}\int _{0}^{1} \frac{(1-r^{2})(1+|z|^{2}r^{2})}{(1-|z|^{2}r^{2})^{3}}dr\nonumber \\= & {} \frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{2}\bigg [-\frac{1}{|z|^{2}}I_{1}+ \left( \frac{3}{|z|^{2}}-1\right) I_{2} +2\left( 1-\frac{1}{|z|^{2}}\right) I_{3}\bigg ], \end{aligned}$$

(2.8)

where

$$\begin{aligned} I_{1}= & {} \int _{0}^{1}\frac{dr}{1-r^{2}|z|^{2}}=\frac{1}{|z|}\log \frac{1+|z|r}{\sqrt{1-|z|^{2}r^{2}}}\bigg |_{0}^{1}=\frac{1}{|z|}\log \frac{1+|z|}{\sqrt{1-|z|^{2}}}, \end{aligned}$$

(2.9)

$$\begin{aligned} I_{2}= & {} \int _{0}^{1}\frac{dr}{(1-r^{2}|z|^{2})^{2}}=\frac{1}{2|z|}\left( \log \frac{1+|z|r}{\sqrt{1-|z|^{2}r^{2}}}+ \frac{|z|r}{1-|z|^{2}r^{2}}\right) \bigg |_{0}^{1}\nonumber \\= & {} \frac{1}{2|z|}\log \frac{1+|z|}{\sqrt{1-|z|^{2}}}+\frac{1}{2(1-|z|^{2})} \end{aligned}$$

(2.10)

and

$$\begin{aligned} I_{3}= & {} \int _{0}^{1}\frac{dr}{(1-r^{2}|z|^{2})^{3}}=\frac{1}{4|z|}\left( \frac{|z|r}{(1-r^{2}|z|^{2})^{2}} +\frac{3}{2}\frac{|z|r}{1-r^{2}|z|^{2}}+\frac{3}{2}\log \frac{1+|z|r}{\sqrt{1-|z|^{2}r^{2}}}\right) \bigg |_{0}^{1}\nonumber \\= & {} \frac{1}{4(1-|z|^{2})^{2}}+\frac{3}{8(1-|z|^{2})}+\frac{3}{8|z|}\log \frac{1+|z|}{\sqrt{1-|z|^{2}}}. \end{aligned}$$

(2.11)

By (2.9), (2.10) and (2.11), we get

$$\begin{aligned}&-\frac{1}{|z|^{2}}I_{1}+ \left( \frac{3}{|z|^{2}}-1\right) I_{2} +2\left( 1-\frac{1}{|z|^{2}}\right) I_{3}\\&\quad =\frac{1}{4|z|^{2}}\bigg [\frac{1+|z|^{2}}{1-|z|^{2}}-\frac{(1-|z|^{2})}{2|z|}\log \frac{1+|z|}{1-|z|}\bigg ], \end{aligned}$$

which, together with (2.8), yields that

$$\begin{aligned} \left| \frac{\partial }{\partial z}{\mathcal {G}}_{g}(z)\right| \le \mu (|z|), \end{aligned}$$

(2.12)

where

$$\begin{aligned} \mu (|z|)=\frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{8|z|^{2}} \bigg [\frac{1+|z|^{2}}{1-|z|^{2}}-\frac{(1-|z|^{2})}{2|z|}\log \frac{1+|z|}{1-|z|}\bigg ]. \end{aligned}$$

By a similar proof process of (2.12), we have

$$\begin{aligned} \left| \frac{\partial }{\partial \overline{z}}{\mathcal {G}}_{g}(z)\right| \le \mu (|z|). \end{aligned}$$

(2.13)

By direct calculation (or by [19, Lemma 2.3]), we obtain

$$\begin{aligned}&\lim _{|z|\rightarrow 0^{+}}\frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{8|z|^{2}} \bigg [\frac{1+|z|^{2}}{1-|z|^{2}}-\frac{(1-|z|^{2})}{2|z|}\log \frac{1+|z|}{1-|z|}\bigg ]=\frac{\Vert g\Vert _{\infty }}{3},\\&\lim _{|z|\rightarrow 1^{-}}\frac{\Vert g\Vert _{\infty }(1-|z|^{2})}{8|z|^{2}} \bigg [\frac{1+|z|^{2}}{1-|z|^{2}}-\frac{(1-|z|^{2})}{2|z|}\log \frac{1+|z|}{1-|z|}\bigg ]=\frac{\Vert g\Vert _{\infty }}{4}\nonumber \end{aligned}$$

(2.14)

and $\mu (|z|)$ is decreasing on $|z|\in (0,1)$.

On the other hand, since ${\mathcal {P}}_{\psi }$ is harmonic in ${\mathbb {D}}$, by [14, Theorem 3] (see also [11, 12]), we see that, for $z\in {\mathbb {D}}$,

$$\begin{aligned} \Vert D_{{\mathcal {P}}_{\psi }}(z)\Vert \le \frac{4\Vert {\mathcal {P}}_{\psi }\Vert _{\infty }}{\pi }\frac{1}{1-|z|^{2}}. \end{aligned}$$

(2.15)

Hence (1.8) follows from (2.12), (2.13) and (2.15). Furthermore, applying (1.8) and (2.14), we get (1.9). The proof of this theorem is complete. $\square$

Now we formulate the following well-known result.

Lemma 1

The improper integral

$$\begin{aligned} \int _{0}^{\frac{\pi }{2}}\log \sin x dx=\int _{0}^{\frac{\pi }{2}}\log \cos x dx=-\frac{\pi }{2}\log 2. \end{aligned}$$

Lemma 2

For $z\in {\mathbb {D}}\backslash \{0\}$, the improper integral

$$\begin{aligned} \int _{{\mathbb {D}}}\frac{dA(w)}{|w||z-w|}= & {} \int _{0}^{2\pi }\log \big (1-r\cos t+\sqrt{1+r^{2}-2r\cos t}\big )dt\\&-\,2\pi \log r+2\pi \log 2\\\le & {} 2\pi \log 4(1+r)-2\pi \log r, \end{aligned}$$

where $r=|z|$.

Proof

Let $z=re^{i\alpha }$ and $w=\rho e^{i\theta }$. Then

$$\begin{aligned} \int _{{\mathbb {D}}}\frac{dA(w)}{|w||z-w|}= & {} \int _{0}^{1}d\rho \int _{0}^{2\pi }\frac{d\theta }{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos (\theta -\alpha )}}\nonumber \\= & {} \int _{0}^{1}d\rho \int _{0}^{2\pi }\frac{dt}{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}}\nonumber \\= & {} \int _{0}^{2\pi }dt\int _{0}^{1}\frac{d\rho }{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}}\nonumber \\= & {} \int _{0}^{2\pi }\bigg \{\frac{1}{2r \cos t}\bigg [\int _{0}^{1}\frac{2\rho d\rho }{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}}\nonumber \\&-\int _{0}^{1}\frac{d(r^{2}+\rho ^{2}-2\rho r\cos t)}{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}}\bigg ]\bigg \}dt\nonumber \\= & {} \int _{0}^{2\pi }\bigg [\frac{1}{r\cos t}\int _{0}^{1}\frac{\rho d\rho }{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}} \nonumber \\&-\,\frac{1}{r\cos t}\big (\sqrt{1+r^{2}-2r\cos t}-r\big )\bigg ]dt\nonumber \\= & {} \int _{0}^{2\pi }\bigg [\frac{1}{r\cos t}\int _{0}^{1}\frac{\rho d\rho }{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}}\nonumber \\&-\,\frac{\sqrt{1+r^{2}-2r\cos t}}{r\cos t}+\frac{1}{\cos t}\bigg ]dt. \end{aligned}$$

(2.16)

By calculations, we get

$$\begin{aligned} \int _{0}^{1}\frac{\rho d\rho }{\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}}= & {} H(\rho )|_{0}^{1}\nonumber \\= & {} \sqrt{1+r^{2}-2r\cos t}\nonumber \\&+\,r\cos t \log \left( 1-r\cos t+\sqrt{1+r^{2}-2r\cos t}\right) \nonumber \\&-\, r-r\cos t\log r(1-\cos t), \end{aligned}$$

(2.17)

where

$$\begin{aligned} H(\rho )=\sqrt{\rho ^{2}+r^{2}-2r\rho \cos t}+r \cos t\log \left( \rho -r\cos t+\sqrt{r^{2}+\rho ^{2}-2\rho r\cos t}\right) . \end{aligned}$$

By (2.16), (2.17) and Lemma 1, we see that

$$\begin{aligned} \int _{{\mathbb {D}}}\frac{dA(w)}{|w||z-w|}= & {} \int _{0}^{2\pi }\log \big (1-r\cos t+\sqrt{1+r^{2}-2r\cos t}\big )dt\nonumber \\&-\int _{0}^{2\pi }\log r(1-\cos t)dt\nonumber \\= & {} \int _{0}^{2\pi }\log \big (1-r\cos t+\sqrt{1+r^{2}-2r\cos t}\big )dt \nonumber \\&-\,2\pi \log r-\int _{0}^{2\pi }\log \left( 2\sin ^{2}\frac{t}{2}\right) dt \nonumber \\= & {} \int _{0}^{2\pi }\log \big (1-r\cos t+\sqrt{1+r^{2}-2r\cos t}\big )dt\nonumber \\&-\,2\pi \log 2r-8\int _{0}^{\frac{\pi }{2}}\log (\sin t) dt \nonumber \\= & {} \int _{0}^{2\pi }\log \big (1-r\cos t+\sqrt{1+r^{2}-2r\cos t}\big )dt \nonumber \\&-\,2\pi \log r+2\pi \log 2\nonumber \\\le & {} 2\pi \log 4(1+r)-2\pi \log r. \end{aligned}$$

(2.18)

The proof of this lemma is complete. $\square$

Lemma E

([10, Lemma 1]) Let f be a harmonic mapping of ${\mathbb {D}}$ into ${\mathbb {C}}$ such that $|f(z)|\le M$ and $f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}+\sum _{n=1}^{\infty }\overline{b}_{n}\overline{z}^{n}$. Then $|a_{0}|\le M$ and for all $n\ge 1,$

$$\begin{aligned} |a_{n}|+|b_{n}|\le \frac{4M}{\pi }. \end{aligned}$$

Lemma 3

For $x\in (0,1)$, let

$$\begin{aligned} \phi (x)=\frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}}{\pi }\frac{x(2-x)}{(1-x)^{2}} -2M_{1}\big [\log 4(1+x)- \log x\big ](2+x)x, \end{aligned}$$

where $M_{2}>0$ and $M_{1}\ge 0$ are constant. Then $\phi$ is strictly decreasing and there is an unique $x_{0}\in (0,1)$ such that $\phi (x_{0})=0.$

Proof

For $x\in (0,1)$, let

$$\begin{aligned} f_{1}(x)=\frac{4M_{2}}{\pi }\frac{x(2-x)}{(1-x)^{2}} \end{aligned}$$

and

$$\begin{aligned} f_{2}(x)=2M_{1}\big [\log 2(1+x)- \log x+\log 2\big ](2+x)x. \end{aligned}$$

Since, for $x\in (0,1)$,

$$\begin{aligned} f_{1}'(x)=\frac{8M_{2}}{\pi }\frac{1}{(1-x)^{3}}>0 \end{aligned}$$

and

$$\begin{aligned} f_{2}'(x)= & {} 2M_{1}\left[ 2(x+1)\log \frac{4(1+x)}{x}-\frac{2+x}{1+x}\right] \\= & {} 2M_{1}\left\{ 2(x+1)\bigg [\log 4+\log \Big (1+\frac{1}{x}\Big )\bigg ]-\frac{2+x}{1+x}\right\} \\\ge & {} 2M_{1}\left\{ 2(x+1)\bigg [1+\frac{1}{1+x}\bigg ]-\frac{2+x}{1+x}\right\} \\= & {} 2M_{1}\frac{(2+x)(2x+1)}{1+x}\ge 0, \end{aligned}$$

we see that $f_{1}+f_{2}$ is continuous and strictly increasing in (0, 1). Then $\phi$ is continuous and strictly decreasing in (0, 1), which, together with

$$\begin{aligned} \lim _{x\rightarrow 0^{+}}\phi (x)=\frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}\quad \text{ and }\quad \lim _{x\rightarrow 1^{-}}\phi (x)=-\infty , \end{aligned}$$

implies that there is an unique $x_{0}\in (0,1)$ such that $\phi (x_{0})=0.$$\square$

Lemma 4

For $x\in (0,1]$, let

$$\begin{aligned} \tau _{1}(x)=\frac{2-r_{0}x}{(1-r_{0}x)^{2}}\quad \text{ and }\quad \tau _{2}(x)=x\big [\log 4(1+r_{0}x)-\log (r_{0}x)\big ], \end{aligned}$$

where $r_{0}\in (0,1)$ is a constant. Then $\tau _{1}$ and $\tau _{2}$ are increasing functions in (0, 1].

Proof of Theorem 4

As before, by (2.1) with f in place of $\psi$, we have

$$\begin{aligned} f(z)={\mathcal {P}}_{f}(z)-{\mathcal {G}}_{g}(z), \quad z\in {\mathbb {D}}, \end{aligned}$$

where ${\mathcal {P}}_{f}$ and ${\mathcal {G}}_{g}$ are defined in (2.1). By [19, Lemma 2.3], Theorem D and Lemma 2, we have

$$\begin{aligned} \left| \frac{\partial {\mathcal {G}}_{g}(z)}{\partial z}-\frac{\partial {\mathcal {G}}_{g}(0)}{\partial z}\right|= & {} \bigg |\frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{(z-w)(z\overline{w}-1)}g(w)dA(w)\nonumber \\&-\,\frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{w}g(w)dA(w)\bigg |\nonumber \\= & {} \bigg |\frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{z(1-|w|^{2})(1+|w|^{2}-z\overline{w})}{w(z-w)(z\overline{w}-1)}g(w)dA(w)\bigg |\nonumber \\\le & {} \frac{M_{1}|z|}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})\big |1+|w|^{2}-z\overline{w}\big |}{|w||z-w||1-z\overline{w}|}dA(w)\nonumber \\\le & {} \frac{|z|(2+|z|)M_{1}}{4\pi }\int _{{\mathbb {D}}}\frac{(1+|w|)}{|w||z-w|}dA(w)\nonumber \\\le & {} \frac{|z|(2+|z|)M_{1}}{2\pi }\int _{{\mathbb {D}}}\frac{1}{|w||z-w|}dA(w)\nonumber \\\le & {} M_{1}\big [\log 4(1+|z|)- \log |z|\big ]|z|(2+|z|). \end{aligned}$$

(2.19)

By a similar argument, we get

$$\begin{aligned} \left| \frac{\partial {\mathcal {G}}_{g}(z)}{\partial \overline{z}}-\frac{\partial {\mathcal {G}}_{g}(0)}{\partial \overline{z}}\right|= & {} \bigg |\frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{(\overline{z}-\overline{w})(w\overline{z}-1)}g(w)dA(w)\nonumber \\&-\,\frac{1}{4\pi }\int _{{\mathbb {D}}}\frac{(1-|w|^{2})}{\overline{w}}g(w)dA(w)\bigg |\nonumber \\\le & {} M_{1}\big [\log 4(1+|z|)- \log |z|\big ]|z|(2+|z|). \end{aligned}$$

(2.20)

On the other hand, ${\mathcal {P}}_{f}$ can be written by

$$\begin{aligned} {\mathcal {P}}_{f}(z)=\sum _{n=0}^{\infty }a_{n}z^{n}+\sum _{n=1}^{\infty }\overline{b}_{n}\overline{z}^{n} \end{aligned}$$

because ${\mathcal {P}}_{f}$ is harmonic in ${\mathbb {D}}$.

Since $|{\mathcal {P}}_{f}(z)|\le M_{2}$ for $z\in {\mathbb {D}}$, by Lemma E, we have

$$\begin{aligned} |a_{n}|+|b_{n}|\le \frac{4M_{2}}{\pi } \end{aligned}$$

(2.21)

for $n\ge 1$.

By (2.21), we see that

$$\begin{aligned} \left| \frac{\partial {\mathcal {P}}_{f}(z)}{\partial z}- \frac{\partial {\mathcal {P}}_{f}(0)}{\partial z}\right| +\left| \frac{\partial {\mathcal {P}}_{f}(z)}{\partial \overline{z}}- \frac{\partial {\mathcal {P}}_{f}(0)}{\partial \overline{z}}\right|= & {} \left| \sum _{n=2}^{\infty }na_{n}z^{n-1}\right| +\left| \sum _{n=2}^{\infty }nb_{n}\overline{z}^{n-1}\right| \nonumber \\\le & {} \sum _{n=2}^{\infty }n\big (|a_{n}|+|b_{n}|\big )|z|^{n-1} \nonumber \\\le & {} \frac{4M_{2}}{\pi }\sum _{n=2}^{\infty }n|z|^{n-1}\nonumber \\= & {} \frac{4M_{2}}{\pi }\frac{|z|(2-|z|)}{(1-|z|)^{2}}. \end{aligned}$$

(2.22)

Applying Theorem 3, we obtain

$$\begin{aligned} 1=J_{f}(0)=\Vert D_{f}(0)\Vert \lambda (D_{f}(0))\le \lambda (D_{f}(0))\left( \frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}\right) , \end{aligned}$$

which gives that

$$\begin{aligned} \lambda (D_{f}(0))\ge \frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}. \end{aligned}$$

(2.23)

In order to prove the univalence of f in ${\mathbb {D}}_{r_{0}}$, we choose two distinct points $z_{1}, z_{2}\in {\mathbb {D}}_{r_{0}}$ and let $[z_{1},z_{2}]$ denote the segment from $z_{1}$ to $z_{2}$ with the endpoints $z_{1}$ and $z_{2}$, where $r_{0}$ satisfies the following equation

$$\begin{aligned} \frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}}{\pi }\frac{r_{0}(2-r_{0})}{(1-r_{0})^{2}}-2M_{1}\big [\log 4(1+r_{0})- \log r_{0}\big ](2+r_{0})r_{0}=0. \end{aligned}$$

By (2.19), (2.20), (2.22), (2.23), Lemmas 3 and 4, we have

$$\begin{aligned} f(z_{2})-f(z_{1})|= & {} \left| \int _{[z_{1},z_{2}]}f_{z}(z)dz+f_{\overline{z}}(z)d\overline{z}\right| \nonumber \\\ge & {} \left| \int _{[z_{1},z_{2}]}f_{z}(0)dz+f_{\overline{z}}(0)d\overline{z}\right| \nonumber \\&-\left| \int _{[z_{1},z_{2}]}\big (f_{z}(z)-f_{z}(0)\big )dz+\big (f_{\overline{z}}(z)-f_{\overline{z}}(0)\big )d\overline{z}\right| \nonumber \\\ge & {} \lambda (D_{f}(0))|z_{2}-z_{1}|\nonumber \\&-\int _{[z_{1},z_{2}]}\big (|f_{z}(z)-f_{z}(0)|+|f_{\overline{z}}(z)-f_{\overline{z}}(0)|\big )|dz|\nonumber \\\ge & {} \lambda (D_{f}(0))|z_{2}-z_{1}|\nonumber \\&-\int _{[z_{1},z_{2}]}\left( \bigg |\frac{\partial {\mathcal {G}}_{g}(z)}{\partial z}-\frac{\partial {\mathcal {G}}_{g}(0)}{\partial z}\bigg |+\bigg |\frac{\partial {\mathcal {G}}_{g}(z)}{\partial \overline{z}}-\frac{\partial {\mathcal {G}}_{g}(0)}{\partial \overline{z}}\bigg |\right) |dz|\nonumber \\&-\int _{[z_{1},z_{2}]}\left( \bigg |\frac{\partial {\mathcal {P}}_{f}(z)}{\partial z}- \frac{\partial {\mathcal {P}}_{f}(0)}{\partial z}\bigg | +\bigg |\frac{\partial {\mathcal {P}}_{f}(z)}{\partial \overline{z}}- \frac{\partial {\mathcal {P}}_{f}(0)}{\partial \overline{z}}\bigg |\right) |dz|\nonumber \\> & {} |z_{2}-z_{1}|\big \{\lambda (D_{f}(0))-\frac{4M_{2}}{\pi }\frac{r_{0}(2-r_{0})}{(1-r_{0})^{2}}\nonumber \\&-\,2M_{1}\big [\log 4(1+r_{0})- \log r_{0}\big ](2+r_{0})r_{0}\big \}\nonumber \\\ge & {} |z_{2}-z_{1}|\bigg \{\frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}}{\pi }\frac{r_{0}(2-r_{0})}{(1-r_{0})^{2}}\nonumber \\&-\,2M_{1}\big [\log 4(1+r_{0})- \log r_{0}\big ](2+r_{0})r_{0}\bigg \}\nonumber \\= & {} 0, \end{aligned}$$

(2.24)

which yields that $f(z_{2})\ne f(z_{1})$. The univalence of f follows from the arbitrariness of $z_{1}$ and $z_{2}$.

Now, for all $\zeta =r_{0}e^{i\theta }\in \partial {\mathbb {D}}_{r_{0}}$, by (2.19), (2.20), (2.22), (2.23), Lemmas 3 and 4, we obtain

$$\begin{aligned} |f(\zeta )-f(0)|= & {} \left| \int _{[0,\zeta ]}f_{z}(z)dz+f_{\overline{z}}(z)d\overline{z}\right| \\= & {} \left| \int _{[0,\zeta ]}f_{z}(0)dz+f_{\overline{z}}(0)d\overline{z}\right| \\&-\left| \int _{[0,\zeta ]}\big (f_{z}(z)-f_{z}(0)\big )dz+\big (f_{\overline{z}}(z)-f_{\overline{z}}(0)\big )d\overline{z}\right| \\\ge & {} \lambda (D_{f}(0))r_{0}\\&-\int _{[0,\zeta ]}\big (|f_{z}(z)-f_{z}(0)|+|f_{\overline{z}}(z)-f_{\overline{z}}(0)|\big )|dz|\\\ge & {} \lambda (D_{f}(0))r_{0}\\&-\int _{[0,\zeta ]}\left( \bigg |\frac{\partial {\mathcal {G}}_{g}(z)}{\partial z}-\frac{\partial {\mathcal {G}}_{g}(0)}{\partial z}\bigg |+\bigg |\frac{\partial {\mathcal {G}}_{g}(z)}{\partial \overline{z}}-\frac{\partial {\mathcal {G}}_{g}(0)}{\partial \overline{z}}\bigg |\right) |dz|\\&-\int _{[0,\zeta ]}\left( \bigg |\frac{\partial {\mathcal {P}}_{f}(z)}{\partial z}- \frac{\partial {\mathcal {P}}_{f}(0)}{\partial z}\bigg | +\bigg |\frac{\partial {\mathcal {P}}_{f}(z)}{\partial \overline{z}}- \frac{\partial {\mathcal {P}}_{f}(0)}{\partial \overline{z}}\bigg |\right) |dz| \\\ge & {} \frac{r_{0}}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}}{\pi }\int _{[0,\zeta ]}\frac{|z|(2-|z|)}{(1-|z|)^{2}}|dz|\\&-\,2M_{1}\int _{[0,\zeta ]}\big [\log 4(1+|z|)- \log |z|\big ]|z|(2+|z|)|dz|\nonumber \\= & {} \frac{r_{0}}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}r_{0}^{2}}{\pi }\int _{0}^{1}\frac{t(2-r_{0}t)}{(1-r_{0}t)^{2}}dt\\&-\,2M_{1}r_{0}^{2}\int _{0}^{1}\big [\log 4(1+r_{0}t)- \log (r_{0}t)\big ]t(2+r_{0}t)dt\\\ge & {} \frac{r_{0}}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}r_{0}^{2}}{\pi }\frac{(2-r_{0})}{(1-r_{0})^{2}}\int _{0}^{1}t dt\\&-\,2M_{1}r_{0}^{2}(2+r_{0})\int _{0}^{1}\big [\log 4(1+r_{0}t)- \log (r_{0}t)\big ]t dt\\\ge & {} r_{0}\bigg \{\frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{2M_{2}}{\pi }\frac{r_{0}(2-r_{0})}{(1-r_{0})^{2}}\\&-\,2M_{1}r_{0}(2+r_{0})\big [\log 4(1+r_{0})- \log r_{0}\big ]\bigg \}\\= & {} \frac{2M_{2}}{\pi }\frac{r_{0}^{2}(2-r_{0})}{(1-r_{0})^{2}}. \end{aligned}$$

Hence $f({\mathbb {D}}_{r_{0}})$ contains an univalent disk ${\mathbb {D}}_{R_{0}}$ with

$$\begin{aligned} R_{0}\ge \frac{2M_{2}}{\pi }\frac{r_{0}^{2}(2-r_{0})}{(1-r_{0})^{2}}. \end{aligned}$$

The proof of this theorem is complete. $\square$

Proof of Corollary 1

For $z_{1}, z_{2}\in {\mathbb {D}}_{r_{0}}$, by (2.24), we see that there is a positive constant $L_{1}$ such that

$$\begin{aligned} L_{1}|z_{1}-z_{2}|\le |f(z_{1})-f(z_{2})|, \end{aligned}$$

where $r_{0}$ satisfies the following equation

$$\begin{aligned} \frac{1}{\frac{4}{\pi }M_{2}+\frac{2}{3}M_{1}}-\frac{4M_{2}}{\pi }\frac{r_{0}(2-r_{0})}{(1-r_{0})^{2}}-2M_{1}\big [\log 4(1+r_{0})- \log r_{0}\big ](2+r_{0})r_{0}=0. \end{aligned}$$

On the other hand, for $z_{1}, z_{2}\in {\mathbb {D}}_{r_{0}}$, we use Theorem 3 to get

$$\begin{aligned} |f(z_{2})-f(z_{1})|= & {} \left| \int _{[z_{1},z_{2}]}df(z)\right| \\\le & {} \int _{[z_{1},z_{2}]}\Vert D_{f}(z)\Vert |dz|\\\le & {} \int _{[z_{1},z_{2}]}\left( \frac{4M_{2}}{\pi }\frac{1}{1-r_{0}^{2}}+\frac{2}{3}M_{1}\right) |dz|\\= & {} \left( \frac{4M_{2}}{\pi }\frac{1}{1-r_{0}^{2}}+\frac{2}{3}M_{1}\right) |z_{1}-z_{2}|, \end{aligned}$$

where $[z_{1},z_{2}]$ is the segment from $z_{1}$ to $z_{2}$ with the endpoints $z_{1}$ and $z_{2}$. Therefore, f is bi-Lipschitz in ${\mathbb {D}}_{r_{0}}$. $\square$