Schwarz Lemma for Solutions of the $\alpha$-harmonic Equation

Abstract

In this paper, we study Schwarz lemma for solutions of the $\alpha$-harmonic equation. Applying the method given by Burgeth, we build a new form of Schwarz lemma for solutions of the $\alpha$-harmonic equation, which is suitable for $\alpha >-1$. Moreover, we generalize the Schwarz–Pick inequality for solutions of the $\alpha$-harmonic equation from the case $\alpha \ge 0$ to $\alpha >-1$.

1 Introduction

Let $\mathbb {D}$ denote the unit disc of the complex plane $\mathbb {C}$, and $\mathbb {T}=\partial {\mathbb {D}}$, the boundary of $\mathbb {D}$. Let $\mathbb {R}^n$ be the standard Euclidean space, with the norm $|x|=\sqrt{\sum _{i=1}^n{x_i}^2}$, where $x=(x_1,\dots x_n)$. We write $\mathbb {B}^n=\{x\in \mathbb {R}^n: |x|< 1\}$, and $\mathbb {B}^2=\mathbb {D}$. Let $S^n$ denote the boundary of $\mathbb {B}^n$, and $S^2=\mathbb {T}$. Denote by $\mathcal {X}_A$ the characteristic function of a set A of $S^{n}$, and $\sigma$ the usual surface measure on $S^{n}.$ Write

$$\begin{aligned} e_x=\left\{ \begin{array}{rl} x/|x|,&{} x\ne 0, \\ 1,&{} x=0. \end{array} \right. \end{aligned}$$

1.1 $\alpha$-harmonic Function and Related Definitions

Let $\Delta _\alpha =\partial z(1-|z|^2)^{-\alpha }\bar{\partial }z$ denote the weighted Laplacian operator in $\mathbb {D}$, and the standard weight by $\omega _\alpha =(1-|z|^2)^{-\alpha }$, where $\alpha >-1$. If $\alpha =0$, then $\Delta _\alpha$ reduces to the classical Laplacian operator. One can refer to [1] for the original study of weighted Laplacian operators. Let $g \in \mathcal {C}(\mathbb {D})$ and $f \in \mathcal {C}^2(\mathbb {D})$, we write an inhomogeneous $\alpha$-harmonic equation by

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

(1)

and the associated Dirichlet boundary value problem by

$$\begin{aligned} \left\{ \begin{array}{rl} \Delta _\alpha (f)&{} =g \quad \ in\ \mathbb {D},\\ f&{}=f^*\quad on\ \mathbb {T} , \end{array} \right. \end{aligned}$$

(2)

where the boundary data $f^* \in L^1({\mathbb {T}})$ , and the boundary condition is understood as $f_r \rightarrow f^*$ for $r \rightarrow 1^-$, where $f_r=f(re^{i\theta }), 0<r<1$. If $g=0$, a Dirichlet solution of the Eq. (2) is called an $\alpha$-harmonic mapping.

Olofsson and Wittsten [2] proved that if the boundary data $f^* \in L^1({\mathbb {T}})$, then an $\alpha$-harmonic mapping f had the form of a Poisson type integral

$$\begin{aligned} f(z)=\mathcal {P}_\alpha [f^*](z)=\frac{1}{2\pi }\int ^{2\pi }_0 P_\alpha (ze^{-i\theta })f^*(e^{i\theta })\mathrm {d}\theta \quad (\alpha >-1), \end{aligned}$$

where

$$\begin{aligned} P_\alpha =\frac{(1-|z|^2)^{\alpha +1}}{(1-z)(1-\overline{z})^{\alpha +1}} \end{aligned}$$

was called the $\alpha$-harmonic Poisson kernel. We note that $P_{\alpha }$ is real if and only if $\alpha =0$, for this special case, $P_\alpha$ reduces to the classical Poisson kernel

$$\begin{aligned} P(z)=\frac{1-|z|^2}{|1-z|^2} \end{aligned}$$

(3)

and the convolution type integral

$$\begin{aligned} \mathcal {P}[f^*](z)=\frac{1}{2\pi }\int ^{2\pi }_0 P(ze^{-i\theta })f^*(e^{i\theta })\mathrm {d}\theta \end{aligned}$$

is the classical harmonic mapping in ${\mathbb {D}}$ with boundary data $f^*\in L^1({\mathbb {T}})$. One can refer to [3] and the references therein for further study of harmonic mappings.

Behm [4] found that the weighted Green function $G_\alpha$ of the weighted Laplacian operator $\Delta _\alpha$ could be written by

$$\begin{aligned} G_\alpha (z,w)=-(1- \bar{w}z)^\alpha h \circ \omega (z,w), \ \ \ \ \ \ z\ne w \end{aligned}$$

(4)

where

$$\begin{aligned} h(s)=\int ^s_0 \frac{t^\alpha }{1-t}\mathrm {d}t=\sum ^{\infty }_{n=0} \frac{s^{\alpha +1+n}}{\alpha +1+n},\quad 0\le s<1, \end{aligned}$$

(5)

and

$$\begin{aligned} \omega (z,w)=1-\bigg |\frac{z-w}{1-\bar{z}w}\bigg |^2=\frac{(1-|z|^2)(1-|w|^2)}{|1-z\bar{w}|^2}. \end{aligned}$$

(6)

The weighted potential of a function $g\in \mathcal {C}(\mathbb {D})$ can be represented by

$$\begin{aligned} \mathcal {G}[g](z)=\int _{\mathbb {D}}G_\alpha (z,w)g(w)\mathrm {d}A(w). \end{aligned}$$

Where $\mathrm {d}A(w)=\frac{1}{\pi }\mathrm {d}u\mathrm {d}v, w=u+iv$.

The combination of the results of Olofsson–Wittsten [2] and Behm [4] says that a solution of the $\alpha$-harmonic equation (2) has the form

$$\begin{aligned} f(z)=\mathcal {P}_\alpha [f^*](z)+\mathcal {G}[g](z). \end{aligned}$$

(7)

For the latest research on $\alpha$-harmonic mappings one can see the papers [5,6,7,8,9,10,11,12,13,14,15].

1.2 Schwarz lemma and Schwarz–Pick lemma

The Schwarz lemma for analytic functions plays a vital role in complex analysis and has been generalized to various spaces of functions.

Heinz [16] generalized it to the class of complex-valued harmonic functions. That is, if f is a complex-valued harmonic function from $\mathbb {D}$ into itself with $f(0)=0$, then for $z \in \mathbb {D}$,

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

Hethcote [17] and Pavlović [18] improved Heinz’s result, by removing the assumption $f(0)=0$, i.e., let f be a harmonic function from ${\mathbb {D}}$ to ${\mathbb {D}}$, then

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

(8)

Colonna [19] obtained a Schwarz–Pick lemma for complex-valued harmonic functions which was stated that if f was a complex-valued harmonic function from $\mathbb {D}$ into itself, then for $z\in \mathbb {D}$,

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

where $\Vert D_f(z)\Vert =\sup \{|D_f(z)\zeta |: |\zeta |=1\}=|f_z(z)|+|f_{\bar{z}}(z)|.$

For $\alpha$-harmonic function, Li et al. [9] stated a Schwarz type lemma for solutions of the $\alpha$-harmonic equation with the condition $\alpha \ge 0$.

Theorem A

Suppose that $g\in \mathcal {C}(\overline{\mathbb {D}})$, and $f^*\in \mathcal {C}^1(\mathbb {T})$. If $f\in \mathcal {C}^2(\mathbb {D})$ satisfies (2) with $\alpha \ge 0$ and $\mathcal {P}_\alpha [f^*](0)=0$, then for $z\in \mathbb {D},$

$$\begin{aligned} |f(z)|\le 2^\alpha \bigg [\frac{4}{\pi }\Vert f^*\Vert _\infty \arctan |z|+\Vert g\Vert _\infty (1-|z|^2)^{\alpha +1}\bigg ], \end{aligned}$$

where $\Vert f^*\Vert _\infty =\sup _{z\in \mathbb {T}}\{|f^*(z)|\},$ and $\Vert g\Vert _\infty =\sup _{z\in \mathbb {D}}\{|g(z)|\}.$

Li et al. [9] also established a Schwarz–Pick type inequality for the solutions of the Eq. (2) for the case $\alpha \ge 0$.

Theorem B

Suppose that $g\in \mathcal {C}(\bar{\mathbb {D}})$, and $f\in \mathcal {C}^2(\mathbb {D})$ satisfies (2) with $f^*\in \mathcal {C}(\mathbb {T})$ and $\alpha \ge 0$. Then for any $z \in \mathbb {D},$

$$\begin{aligned} \Vert D_f(z)\Vert \le (\alpha +1)2^{\alpha +1}\Vert f^*\Vert _\infty \frac{1}{1-|z|^2}+(\alpha +\frac{4}{3})2^{\alpha +1}\Vert g\Vert _\infty , \end{aligned}$$

where $\Vert f^*\Vert _\infty =\sup _{z\in \mathbb {T}}\{|f^*(z)|\},$ and $\Vert g\Vert _\infty =\sup _{z\in \mathbb {D}}\{|g(z)|\}.$

One can refer to the papers [19,20,21,22,23,24,25,26,27,28] for recent progress on the Schwarz lemma and the Schwarz–Pick lemma.

1.3 Statement of main results

In this paper, we continue to study the Schwarz lemma and the Schwarz–Pick lemma for solutions of the $\alpha$-harmonic equation (2). The method to estimate an $\alpha$-Poisson kernel by the classical Poisson kernel is key to complete the proof of Theorem A. To reach their estimate, one needs to make sure that the condition $\mathcal {P}_\alpha [f^*](0)=0$ imply that $\mathcal {P}_\alpha [|f^*|](0)=0$. However, Example 2.1 shows that this is not necessarily true. Hethcote’s inequality may overcome this difficulty to reach an estimate for an $\alpha$-harmonic mapping when $\alpha \ge 0$ as follows

$$\begin{aligned} |f(z)|\le 2^\alpha \bigg [\frac{4}{\pi }\arctan |z|+\frac{1-|z|^2}{1+|z|^2}\mathcal {P}_\alpha [|f|](0)\bigg ]\Vert f^*\Vert _\infty . \end{aligned}$$

(9)

We aim to give another form of the above estimate and obtain a Schwarz type lemma for solutions to the $\alpha$-harmonic equation (2) under a more general condition, which includes the case that $-1<\alpha <0$.

Theorem 1.1

Suppose that $g \in \mathcal {C}(\overline{{\mathbb {D}}})$. If non-constant $f\in \mathcal {C}^2(\mathbb {D})$ with $f^*\in \mathcal {C}^1(\mathbb {T})$ satisfies the Eq. (2), then for $\alpha >-1$ and $z \in \mathbb {D}$,

$$\begin{aligned} |f(z)|\le \left\{ \begin{array}{rl} 2^{1+\alpha }\frac{\Vert f^*\Vert _\infty }{\pi }&{}\arctan (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2})+2^{\alpha }(1-|z|^2)^\alpha \big (\frac{2+\alpha }{1+\alpha }-|z|^2\big )\Vert g\Vert _\infty ,\ \\ &{}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \alpha \ge 0,\\ &{}\quad \\ 2^{1+|\alpha |}\frac{\Vert f^*\Vert _\infty }{\pi }&{}(1-|z|^2)^\alpha \arctan (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2})+2^{|\alpha |}(1-|z|^2)^\alpha \big (\frac{2+\alpha }{1+\alpha }- \\ &{}\qquad \qquad \qquad \qquad \qquad \qquad \qquad |z|^2\big )\Vert g\Vert _\infty ,\ -1<\alpha <0 . \end{array} \right. \end{aligned}$$

where $\Vert f^*\Vert _\infty =\sup _{z\in \mathbb {T}}\{|f^*(z)|\}$, $\Vert g\Vert _\infty =\sup _{z\in \mathbb {D}}\{|g(z)|\}$, $c=\frac{\mathcal {P}_{\alpha }[|f^*|](0)}{\Vert f^*\Vert _\infty }.$

Note that the above theorem is an application of the growth estimate for positive harmonic functions, which is given by Burgeth at [29]. We also note that for a harmonic mapping, that is, $\alpha =0$ and $g=0$, the estimate at Theorem 1.1 attains equality if $c=0$ or $c=1$.

We next give the Schwarz–Pick inequality for the solution of the $\alpha$-harmonic equation when $\alpha >-1$, which is a corresponding result given by Theorem B.

Theorem 1.2

Suppose that $g \in \mathcal {C}(\overline{{\mathbb {D}}})$. If non-constant $f\in \mathcal {C}^2(\mathbb {D})$ with $f^*\in \mathcal {C}^1(\mathbb {T})$ satisfies the Eq. (2) with $\alpha >-1$, then for $z\in \mathbb {D},$

$$\begin{aligned} \Vert D_f(z) \Vert \le \left\{ \begin{array}{rl} (\alpha +1)2^{\alpha +1}&{}\Vert f^*\Vert _\infty \frac{1}{1-|z|^2}+2^{\alpha +3}(1-|z|^2)^{\alpha -1}\big (\frac{\alpha }{1+\alpha }+(\alpha +1)(1- \\ &{}\qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad |z|^2)\big )\Vert g\Vert _\infty ,\ \alpha \ge 0,\\ &{}\quad \\ 2^{1-\alpha }\Vert f^*\Vert _\infty &{}\frac{1}{(1-|z|^2)^{1+|\alpha |}}+2^{\alpha +3}(1-|z|^2)^{\alpha -1}\big (\frac{|\alpha |}{1+\alpha }+(1+|\alpha |)(1-\\ &{}\qquad \qquad \qquad \qquad \qquad \qquad \quad |z|^2)\big )\Vert g\Vert _\infty , -1<\alpha <0.\\ \end{array} \right. \end{aligned}$$

where $\Vert f^*\Vert _\infty =\sup _{z\in \mathbb {T}}\{|f^*(z)|\}$, $\Vert g\Vert _\infty =\sup _{z\in \mathbb {D}}\{|g(z)|\}.$

We note that both two estimates in Theorem 1.1 and Theorem 1.2 are continuous in $\alpha$ for all $\alpha >-1$.

2 Auxiliary Examples

In [9], Li et al. used the Heinz inequality of $\mathcal {P}[|f^*|](z)$ to obtain an upper estimate of $\mathcal {P}_\alpha [f^*](z)$. In fact, to reach their result, they need to require that the normalization that $\mathcal {P}_{\alpha }[f^*](0)=0$ should imply that $\mathcal {P}_{\alpha }[|f^*|](0)=0$. However, the following example shows that it will not be true in general.

Example 2.1

Let

$$\begin{aligned} f(z)=z+\frac{1}{2}\overline{z}^2. \end{aligned}$$

Then

$$\begin{aligned} f^*=e^{i\theta }+\frac{1}{2}e^{-2i\theta }. \end{aligned}$$

Thus the Poisson integral

$$\begin{aligned} \mathcal {P}_{\alpha }[f^*](z)=\frac{1}{2\pi }\int ^{2\pi }_{0}P_\alpha (ze^{-i\theta })f^*(e^{i\theta })\mathrm {d}\theta \end{aligned}$$

implies that

$$\begin{aligned} \mathcal {P}_{\alpha }[f^*](0)=\frac{1}{2\pi }\int ^{2\pi }_{0}(e^{i\theta }+\frac{1}{2}e^{-2i\theta } )\mathrm {d}\theta =0 , \end{aligned}$$

but

$$\begin{aligned} \mathcal {P}_{\alpha }[|f^*|](0)=\frac{1}{2\pi }\int ^{2\pi }_{0}\frac{|1+2e^{3i\theta }|}{2}\mathrm {d}\theta \ne 0. \end{aligned}$$

In fact, the inequality

$$\begin{aligned} |1+2e^{3i\theta }|=|1+2\cos 3\theta +2i\sin 3\theta |=\sqrt{5+4\cos 3\theta }\ge 1 \end{aligned}$$

implies that

$$\begin{aligned} \mathcal {P}_{\alpha }[|f^*|](0)=\frac{1}{2\pi }\int ^{2\pi }_{0}\frac{|1+2e^{3i\theta }|}{2}\mathrm {d}\theta \ge \frac{1}{2}>0. \end{aligned}$$

3 Schwarz Lemma for Real Harmonic Functions

Write

$$\begin{aligned} \mathcal {H}^d : =\{f\ is\ a\ harmonic\ function\ on\ \mathbb {B}^n:\ f(0)=d,\ a \le f \le b \} , \end{aligned}$$

and

$$\begin{aligned} \mathcal {K}^d :=\{ f \in L^\infty (S^{n}) :\int _{S^{n}}f\mathrm {d}\sigma =\sigma (S^{n})\cdot c, c=\frac{d-a}{b-a},\ a \le f \le b \} , \end{aligned}$$

where $L^\infty (S^{n})$ denotes the Lebesgue space of essentially bounded functions on $S^{n}$, $\sigma$ denotes the sphere measure. Let $P_x$ denote the Poisson kernel of the harmonic function on ${\mathbb {B}}^n$ as follows

$$\begin{aligned} P_x(x,y)=\frac{1}{\sigma (S^n)}\frac{1-|x|^2}{|x-y|^n},\ x\in {\mathbb {B}}^n,\ y\in S^{n}. \end{aligned}$$

The Poisson kernel $P_x$ satisfies a normalization that $\int _{S^n}P_x(x,y)\mathrm {d}\sigma =1$.

For convenience, we set

$$\begin{aligned} M^n_c(|x|) = (b-a)\int _{S^{n}}\mathcal {X}_{S_{(c,e_x)}}\cdot P_x \mathrm {d}\sigma +a \end{aligned}$$

(10)

and

$$\begin{aligned} m^n_c(|x|) = (b-a)\int _{S^{n}}\mathcal {X}_{S_{(c,-e_x)}}\cdot P_x \mathrm {d}\sigma +a, \end{aligned}$$

(11)

where $S_{(c,e_x)}$ denote the polar cap centered at $e_x$ with $\sigma$-measure c.

After introducing spherical coordinates, we can write $\sigma (S^n)$ as

$$\begin{aligned} \sigma (S^{n})&=\int _0^\pi \dots \int _0^\pi \int _0^{2\pi } \sin ^{n-2}\varphi _1\dots \sin \varphi _{n-2}\mathrm {d}{\varphi _{n-1}}\mathrm {d}{\varphi _{n-2}}\dots \mathrm {d}{\varphi _1}. \end{aligned}$$

So,

$$\begin{aligned}&\ \ \ \ \int _{S^n}\mathcal {X}_{S_{(c,e_x)}}\cdot P_x \mathrm {d}\sigma ,\\&=\int _0^{\alpha (c)}\frac{1-|x|^2}{\sigma (S^{n})(1-2|x|\cos \varphi +|x|^2)^{n/2}}\sin ^{n-2}\varphi _1\mathrm {d}\varphi _1\int _0^\pi \dots \int _0^{2\pi } \\&\qquad \qquad \qquad \qquad \quad \qquad \ \sin ^{n-3}\varphi _2\dots \sin \varphi _{n-2}\mathrm {d}\varphi _{n-1}\mathrm {d}{\varphi _{n-2}}\dots \mathrm {d}{\varphi _2},\\&=\frac{\sigma (S^{n-1})}{\sigma (S^{n})}\int _0^{\alpha (c)}\frac{(1-|x|^2)\sin ^{n-2}\varphi }{(1-2|x|\cos \varphi +|x|^2)^{n/2}} \mathrm {d}\varphi . \end{aligned}$$

By the fact that

$$\begin{aligned} \sigma (S^{n})=\frac{2\pi ^{\frac{n}{2}}}{\Gamma (\frac{n}{2})} \end{aligned}$$

and

$$\begin{aligned} \sigma (S^{n-1})=\frac{2\pi ^{\frac{n-1}{2}}}{\Gamma (\frac{n-1}{2})} \end{aligned}$$

we have

$$\begin{aligned} \frac{\sigma (S^{n-1})}{\sigma (S^{n})}=\frac{1}{\sqrt{\pi }}\cdot \frac{\Gamma (\frac{n}{2})}{\Gamma (\frac{n-1}{2})}. \end{aligned}$$

So, we can rewrite (10) and (11) as

$$\begin{aligned} M^n_c(|x|) =\frac{b-a}{\sqrt{\pi }}\cdot \frac{\Gamma (\frac{n}{2})}{\Gamma (\frac{n-1}{2})}(1-|x|^2)\int _0^{\alpha (c)}\frac{\sin ^{n-2}\varphi }{(1-2|x|\cos \varphi +|x|^2)^{n/2}} \mathrm {d}\varphi +a \end{aligned}$$

(12)

and

$$\begin{aligned} m^n_c(|x|) = \frac{b-a}{\sqrt{\pi }}\cdot \frac{\Gamma (\frac{n}{2})}{\Gamma (\frac{n-1}{2})}(1-|x|^2)\int _{\pi -\alpha (c)}^\pi \frac{ \sin ^{n-2}\varphi }{(1-2|x|\cos \varphi +|x|^2)^{n/2}} \mathrm {d}\varphi +a , \end{aligned}$$

(13)

where $\alpha (c)$ is the spherical angle of $S_{(c,e_x)}$. Li and Chen [30] used the method of Theorem 1 at [4] to obtain the following theorem.

Theorem C

Let h be a harmonic function $\mathbb {B}^n$, a, b are two real numbers with $a<b$. If $h(0)=d$ and $a<h(x)<b$, then for $\displaystyle c=\frac{d-a}{b-a}$ and any $z\in \mathbb {B}^n$,

$$\begin{aligned} m_c^n(|x|)\le h(x)\le M_c^n(|x|). \end{aligned}$$

Since $\mathcal {P}[|f^*|](z)$ is a real harmonic function on the unit disk ${\mathbb {D}}$, the above theorem will imply the following corollary.

Corollary 1

Let $\mathcal {P}[|f^*|](z)$ be a positive harmonic function with $\mathcal {P}[|f^*|](0)=\mathcal {P}_{\alpha }[|f^*|](0)=d$. If $\mathcal {P}[|f^*|](z) \le \Vert f^*\Vert _{\infty }<\infty$ and $\displaystyle c=\frac{d}{\Vert f^*\Vert _{\infty }}$, then for all $z\in \mathbb {D}$

$$\begin{aligned} \mathcal {P}[|f^*|](z)\le \frac{2\Vert f^{*}\Vert _{\infty }}{\pi }\arctan \bigg (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2}\bigg ). \end{aligned}$$

(14)

Proof

By Theorem C, we obtain that

$$\begin{aligned} {\mathcal {P}}[|f^*|](x)\le M_c^2(|x|). \end{aligned}$$

By the formula (12), we have

$$\begin{aligned} M_c^2(|x|)&=\frac{\Vert f^*\Vert _\infty }{\sqrt{\pi }}\frac{\Gamma (1)}{\Gamma (\frac{1}{2})}(1-|x|^2)\int _0^{\alpha (c)}\frac{1}{1-2|x|\cos \varphi +|x|^2}\mathrm {d}\varphi \\&=\frac{\Vert f^*\Vert _\infty }{\pi }\int _0^{\alpha (c)}\frac{1-|x|^2}{1-2|x|\cos \varphi +|x|^2}\mathrm {d}\varphi \\&=\frac{\Vert f^*\Vert _\infty }{\pi }\int _0^{\alpha (c)}\frac{1-|x|^2}{(1-|x|)^2 \cos ^2\frac{\varphi }{2}+(1+|x|)^2 \sin ^2\frac{\varphi }{2}}\mathrm {d}\varphi \\&=\frac{2\Vert f^*\Vert _\infty }{\pi }\int _0^{\alpha (c)}\frac{\frac{1+|x|}{1-|x|}\sec ^2\frac{\varphi }{2}}{1+(\frac{1+|x|}{1-|x|}\tan \frac{\varphi }{2})^2}\mathrm {d}\frac{\varphi }{2} \\&=\frac{2\Vert f^*\Vert _\infty }{\pi }\arctan \left( \frac{1+|x|}{1-|x|}\tan \frac{\varphi }{2}\right) \bigg \vert _0^{\alpha (c)}. \end{aligned}$$

The fact that $\displaystyle \alpha (c)=\pi \cdot c$, we have

$$\begin{aligned} M_c^2(|x|)=\frac{2\Vert f^{*}\Vert _{\infty }}{\pi }\arctan \bigg (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2}\bigg ). \end{aligned}$$

So,

$$\begin{aligned} \mathcal {P}[|f^*|](z)\le \frac{2\Vert f^{*}\Vert _{\infty }}{\pi }\arctan \bigg (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2}\bigg ). \end{aligned}$$

Thus, the proof of Corollary 1 is complete. $\square$

One can refer to [31, 32] for more knowledge of the growth estimate for real harmonic functions.

4 Some Basic Lemmas

In [9], Li et al. obtained an estimate for the function h(s) as follows.

Lemma A

For $\alpha \ge 0$, the function h(s) satisfies the estimate

$$\begin{aligned} h(s)\le s^\alpha \log {\frac{1}{1-s}}. \end{aligned}$$

We note the above estimate of h(s) is not suitable for the case $-1<\alpha <0$. In fact, in [4], Behm obtained an estimate of h(s) in the case that $\alpha >-1$.

Lemma B

For $\alpha >-1$, the function h(s) satisfies the estimate

$$\begin{aligned} h(s)< s^{1+\alpha }\bigg (\frac{1}{1+\alpha }+\log \frac{1}{1-s}\bigg ). \end{aligned}$$

Especially, for $-1<\alpha <0$, it holds

$$\begin{aligned} s^\alpha \log {\frac{1}{1-s}}<h(s)< s^{1+\alpha }\bigg (\frac{1}{1+\alpha }+\log \frac{1}{1-s}\bigg ). \end{aligned}$$

(15)

For completeness, we give the proof of Lemma B.

Proof

By the definition of h(s) given by (5) and the fact that $1+n+\alpha >n$ for all $n\ge 1$ and $\alpha >-1$, we obtain that

$$\begin{aligned} h(s)=s^{\alpha +1}\bigg [\frac{1}{1+\alpha }+\sum _{n=1}^\infty \frac{s^n}{1+n+\alpha }\bigg ]<s^{\alpha +1}\bigg [\frac{1}{1+\alpha }+\log \bigg (\frac{1}{1-s}\bigg )\bigg ] \end{aligned}$$

When $-1<\alpha <0$, it follows that $0<n+\alpha <n$ for all $n\ge 1$ and hence, we have

$$\begin{aligned} h(s)=s^{\alpha }\sum _{n=1}^\infty \frac{s^n}{n+\alpha }> s^\alpha \log {\frac{1}{1-s}}. \end{aligned}$$

Thus, the proof of Lemma B is complete. $\square$

Remark

• (1) Two estimates of the upper bound of h are two infinitesimals of the same order when s tends to 0; Furthermore,

• (2) For $\alpha >0$, we have

  $$\begin{aligned} \left\{ \begin{array}{rl} \displaystyle s^\alpha \log \frac{1}{1-s}\ge s^{1+\alpha }(\frac{1}{1+\alpha }+\log \frac{1}{1-s}),&{}0<s\le 1-e^{-\frac{\alpha }{\alpha +1}},\\ &{}\qquad \\ \displaystyle s^\alpha \log \frac{1}{1-s}<s^{1+\alpha }(\frac{1}{1+\alpha }+\log \frac{1}{1-s}),&{}1-e^{-\frac{\alpha }{\alpha +1}}<s<1. \end{array}\right. \end{aligned}$$

Proof

• (1) Since

  $$\begin{aligned} \lim _{s\rightarrow 0}\frac{\log \frac{1}{1-s}}{s(\frac{1}{1+\alpha }+\log \frac{1}{1-s})}=1+\alpha , \end{aligned}$$

  the proof of Remarks (1) is complete.

• (2) By a direct calculation, we obtain that

  $$\begin{aligned} s^\alpha \log \frac{1}{1-s}-s^{1+\alpha }(\frac{1}{1+\alpha }+\log \frac{1}{1-s})=s^\alpha \big ((1-s)\log \frac{1}{1-s}-\frac{s}{1+\alpha }\big ). \end{aligned}$$

  Let $\displaystyle A_\alpha (s)=(1-s)\log \frac{1}{1-s}-\frac{s}{1+\alpha }$.

Since

$$\begin{aligned} A'_\alpha (s)=\frac{\alpha }{1+\alpha }-\log \frac{1}{1-s}, \end{aligned}$$

we have the unique zero of the $A'_\alpha (s)$, which is denoted by $B(\alpha )=1-e^{-\frac{\alpha }{1+\alpha }}$.

Moreover,

$$\begin{aligned} A_\alpha (B(\alpha ))=e^{-\frac{\alpha }{\alpha +1}}\log e^{\frac{\alpha }{\alpha +1}}-\frac{1-e^{-\frac{-\alpha }{\alpha +1}}}{1+\alpha }=e^{-1}e^{\frac{1}{\alpha +1}}-\frac{1}{1+\alpha }. \end{aligned}$$

By a change of variable $\displaystyle x=\frac{1}{1+\alpha }$, $A_\alpha (B(\alpha ))$ can be represented by $e^{-1}e^x-x$.

Let $H(x)=e^{-1}e^x-x$. The fact that

$$\begin{aligned} H(0)>0, H(1)=0 \end{aligned}$$

and

$$\begin{aligned} H'(x)=\frac{e^x}{e}-1<\frac{e}{e}-1=0 \end{aligned}$$

imply that

$$\begin{aligned} A_\alpha (B(\alpha ))=e^{-1}e^{\frac{1}{\alpha +1}}-\frac{1}{1+\alpha }>0. \end{aligned}$$

Thus, Remarks (2) is proved. $\square$

Li et al. used Lemma A to obtain an inequality of $\Vert D_{{\mathcal {P}}_\alpha [f^*]}(z)\Vert$ for $\alpha \ge 0$ in [9].

Lemma C

Assume that $f^*\in \mathcal {C}(\mathbb {T})$. Then for $\alpha \ge 0$

$$\begin{aligned} \Vert D _{{\mathcal {P}}_\alpha [f^*]}(z) \Vert \le (\alpha +1)2^{\alpha +1}\Vert f^*\Vert _\infty \frac{1}{1-|z|^2}. \end{aligned}$$

Zhu and Kalaj used the hypergeometric function to obtain the maximum of the $K_p(|z|)$ in [33] for $z\in {\mathbb {D}}$, here $\displaystyle K_p(|z|)=\int _{{\mathbb {D}}}\frac{1}{|w-z|^{q}}\mathrm {d}A(w).$

Lemma D

For $p>2$ and $q\in {\mathbb {R}}$ such that $\frac{1}{p}+\frac{1}{q}=1$, $K_p(|z|)$ has its maximum

$$\begin{aligned} K_p(0)=\frac{2}{2-q}. \end{aligned}$$

Next, we give Green’s formula of complex form.

Lemma 4.1

The complex form of Green’s formula can be expressed as

$$\begin{aligned} \iint _{\mathbb {D}} f_z\mathrm {d}x\mathrm {d}y=\frac{1}{2}\oint _{\partial {\mathbb {D}}}{\mathscr {R}}(f\bar{v})\mathrm {d}s+\frac{i}{2}\oint _{\partial {\mathbb {D}}}{\mathscr {R}}(f\bar{u})\mathrm {d}s, \end{aligned}$$

(16)

where $z=x+iy$, $u=\cos \alpha +i\cos \beta$, $v=\cos \beta -i\cos \alpha$, and $(\cos \alpha , \cos \beta )$ is the unit tangent vector of the curve $\partial {\mathbb {D}}$.

Proof

Let $(\cos \alpha , \cos \beta )$ be the unit tangent vector of the curve $\partial {\mathbb {D}}$, then the unit exterior normal vector of $\partial {\mathbb {D}}$ is $(\cos \beta , -\cos \alpha )$.

By the classic Green’s formula, it follows,

$$\begin{aligned} \frac{1}{2} \iint _{{\mathbb {D}}}(P_x+Q_y)\mathrm {d}x\mathrm {d}y=\frac{1}{2}\oint _{\partial {\mathbb {D}}}P\mathrm {d}y-Q\mathrm {d}x=\frac{1}{2} \oint _{\partial {\mathbb {D}}}(P\cos \beta -Q\cos \alpha )\mathrm {d}s, \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2} \iint _{{\mathbb {D}}}(Q_x-P_y)\mathrm {d}x\mathrm {d}y=\frac{1}{2}\oint _{\partial {\mathbb {D}}} P\mathrm {d}x+Q\mathrm {d}y=\frac{1}{2} \oint _{\partial {\mathbb {D}}}(P\cos \alpha +Q\cos \beta )\mathrm {d}s. \end{aligned}$$

Write

$$\begin{aligned} v=\cos \beta -i\cos \alpha ,\ \ \ u=\cos \alpha +i\cos \beta . \end{aligned}$$

Let $f=P+iQ$, then we have

$$\begin{aligned} P\cos \beta -Q\cos \alpha =\mathscr {R}(f\bar{v}), \end{aligned}$$

and

$$\begin{aligned} P\cos \alpha +Q\cos \beta =\mathscr {R}(f\bar{u}). \end{aligned}$$

So

$$\begin{aligned} \iint _{\mathbb {D}} f_z\mathrm {d}x\mathrm {d}y&=\frac{1}{2}\iint _{\mathbb {D}} (P_x+Q_y)\mathrm {d}x\mathrm {d}y+\frac{i}{2}\iint _{\mathbb {D}} (Q_x-P_y)\mathrm {d}x\mathrm {d}y\\&=\frac{1}{2}\oint _{\partial {\mathbb {D}}}-Q\mathrm {d}x+P\mathrm {d}y+iP\mathrm {d}x+iQ\mathrm {d}y\\&=\frac{1}{2}\oint _{\partial {\mathbb {D}}}if\mathrm {d}x+f\mathrm {d}y=\frac{i}{2}\oint _{\partial {\mathbb {D}}}f\mathrm {d}\bar{z}\\&=\frac{1}{2}\oint _{\partial {\mathbb {D}}}\mathscr {R}(f\bar{v})\mathrm {d}s+\frac{i}{2}\oint _{\partial {\mathbb {D}}}\mathscr {R}(f\bar{u})\mathrm {d}s. \end{aligned}$$

Therefore, the proof of Lemma 4.1 is complete. $\square$

Next, we will give an estimate of the weighted Green function $G_\alpha (z,w)$.

Lemma 4.2

Assume that $\alpha >-1$. Then the weighted Green function $G_\alpha (z,w)$ satisfies the estimate

$$\begin{aligned} |G_\alpha (z,w)|\le 2^{|\alpha |}(1-|z|^2)^\alpha \bigg (\frac{1}{1+\alpha }+\log |\frac{ 1-\bar{z}w}{z-w}|^2\bigg ). \end{aligned}$$

Proof

By the definition of $G_\alpha (z,w)$ given by (4), we obtain that

$$\begin{aligned} |G_\alpha (z,w)|\le |(1-\bar{w}z)^\alpha ||h\circ \omega (z,w)|, \end{aligned}$$

where h(s)and $\omega (z,w)$ are given by (5) and (6).

Moreover, we have from Lemma B that

$$\begin{aligned} |G_\alpha (z,w)|&\le |(1-z\bar{w})^\alpha |\frac{(1-|z|^2)^{1+\alpha }(1-|w|^2)^{1+\alpha }}{|1-z\bar{w}|^{2+2\alpha }}\bigg (\frac{1}{1+\alpha }+\log |\frac{1-\bar{z}w}{z-w}|^2\bigg ) \\&=\frac{(1-|z|^2)^{1+\alpha }(1-|w|^2)^{1+\alpha }}{|1-z\bar{w}|^{2+\alpha }}\bigg (\frac{1}{1+\alpha }+\log |\frac{1-\bar{z}w}{z-w}|^2\bigg ) \\&\le 2^{|\alpha |}(1-|z|^2)^\alpha \bigg (\frac{1}{1+\alpha }+\log |\frac{1-\bar{z}w}{z-w}|^2\bigg ). \end{aligned}$$

Thus, the proof of Lemma 4.2 is complete. $\square$

The following Lemma will be used to give the form of the Schwarz–Pick inequality for ${\mathcal {P}}_\alpha [f^*]$ with $\alpha >-1$.

Lemma 4.3

Assume that $f^*\in {\mathcal {C}}(\mathbb {T})$, then

$$\begin{aligned} \Vert D _{{\mathcal {P}}_\alpha [f^*]}(z) \Vert \le \left\{ \begin{array}{rl} &{}2^{1-\alpha }\Vert f^*\Vert _\infty (1-|z|^2)^{\alpha -1},\qquad -1<\alpha <0,\\ &{}\ \\ &{}\displaystyle (1+\alpha )2^{1+\alpha }\Vert f^*\Vert _\infty \frac{1}{1-|z|^2},\ \ \ \ \ \alpha \ge 0. \end{array}\right. \end{aligned}$$

Two estimates coincides with each other when $\alpha =0$.

Proof

For the case $\alpha \ge 0$, one can see Lemma 3.3 of [9] for details. We next prove the case $-1<\alpha <0$. By an elementary calculation of P(z) given by (3), we obtain that

$$\begin{aligned} \frac{\partial }{\partial z} P (ze^{-i\theta })=\frac{e^{-i\theta }}{(1-ze^{-i\theta })^2}, \end{aligned}$$

and

$$\begin{aligned} \frac{\partial }{\partial \bar{z}} P (ze^{-i\theta })=\frac{e^{i\theta }}{(1-\bar{z}e^{i\theta })^2} . \end{aligned}$$

Then

$$\begin{aligned} \frac{\partial }{\partial z} P _\alpha (ze^{-i\theta })&=\frac{(1-|z|^2)^\alpha \big [e^{-i\theta }(1-|z|^2)-(1+\alpha )\bar{z}(1-ze^{-i\theta })\big ]}{(1-ze^{-i\theta })^2(1-\bar{z}e^{i\theta })^{1+\alpha }}\nonumber \\&=\frac{(1-|z|^2)^\alpha }{(1-\bar{z}e^{i\theta })^{1+\alpha }}\bigg [1-|z|^2-(1+\alpha )\bar{z}e^{i\theta }(1-ze^{-i\theta })\bigg ]\frac{\partial }{\partial z} P (ze^{-i\theta }) , \end{aligned}$$

(17)

and

$$\begin{aligned} \frac{\partial }{\partial \bar{z}} P _\alpha (ze^{-i\theta }) =\frac{(\alpha +1)(1-|z|^2)^\alpha e^{i\theta }}{(1-\bar{z}e^{i\theta })^{\alpha +2}} =\frac{(\alpha +1)(1-|z|^2)^\alpha }{(1-\bar{z}e^{i\theta })^\alpha }\frac{\partial }{\partial \bar{z}} P (ze^{-i\theta }) . \end{aligned}$$

(18)

By the equality

$$\begin{aligned} |1-|z|^2-(1+\alpha )\bar{z}e^{i\theta }(1-ze^{-i\theta })|=|(1-ze^{-i\theta })(-\alpha \bar{z}e^{i\theta })+(1-\bar{z}e^{i\theta })|\end{aligned}$$

and the fact that $-1<\alpha <0$, we have

$$\begin{aligned} |1-|z|^2-(1+\alpha )\bar{z}e^{i\theta }(1-ze^{-i\theta })|&\le |1-ze^{-i\theta }||-\alpha \bar{z}e^{i\theta }|+|1-\bar{z}e^{i\theta }|\\&\le (1-\alpha )|1-\bar{z}e^{i\theta }|, \end{aligned}$$

and

$$\begin{aligned} \bigg |\frac{(1-|z|^2)^\alpha }{(1-\bar{z}e^{i\theta })^{1+\alpha }}\bigg |=\frac{1}{|1-\bar{z}e^{i\theta }|}\bigg |\frac{1-|z|^2}{1- \bar{z}e^{i\theta }}\bigg |^\alpha \le \frac{1}{|1-\bar{z}e^{i\theta }|}\bigg (\frac{1+|z|}{1-|z|^2}\bigg )^{-\alpha }=\frac{(1-|z|)^\alpha }{|1-\bar{z}e^{i\theta }|}. \end{aligned}$$

We obtain from (17) that

$$\begin{aligned} \bigg |\frac{\partial }{\partial z} P _\alpha (ze^{-i\theta })\bigg |\le (1-\alpha )(1-|z|)^\alpha \bigg |\frac{\partial }{\partial z} P (ze^{-i\theta })\bigg |, \end{aligned}$$

and from (18) that

$$\begin{aligned} \bigg |\frac{\partial }{\partial \bar{z}} P _\alpha (ze^{-i\theta })\bigg |\le (\alpha +1)(1-|z|)^\alpha \bigg |\frac{\partial }{\partial \bar{z}} P (ze^{-i\theta })\bigg |. \end{aligned}$$

Thus, the identities

$$\begin{aligned} \frac{1}{2\pi }\int _0^{2\pi }\bigg |\frac{\partial }{\partial z} P (ze^{-i\theta })\bigg |\mathrm {d}\theta =\frac{1}{2\pi }\int _0^{2\pi }\bigg |\frac{\partial }{\partial \bar{z}} P (ze^{-i\theta })\bigg |\mathrm {d}\theta =\frac{1}{1-|z|^2}, \end{aligned}$$

imply

$$\begin{aligned} \Vert D _{{\mathcal {P}}_\alpha [f^*]}(z) \Vert&=\bigg |\frac{\partial }{\partial z}{\mathcal {P}}_{\alpha [f^*]}(ze^{-i\theta })\bigg |+\bigg |\frac{\partial }{\partial \bar{z}}{\mathcal {P}}_{\alpha [f^*]}(ze^{-i\theta })\bigg |\nonumber \\&=\bigg |\frac{\partial }{\partial z}\frac{1}{2\pi }\int _0^{2\pi }P_\alpha (ze^{-i\theta })f^*(e^{i\theta })\mathrm {d}\theta \bigg |\nonumber \\&\quad +\bigg |\frac{\partial }{\partial \bar{z}}\frac{1}{2\pi }\int _0^{2\pi }P_\alpha (ze^{-i\theta })f^*(e^{i\theta })\mathrm {d}\theta \bigg |\nonumber \\&=\bigg |\frac{1}{2\pi }\int _0^{2\pi }\frac{\partial }{\partial z}P_\alpha (ze^{-i\theta })f^*(e^{i\theta })\mathrm {d}\theta \bigg |\nonumber \\&\quad +\bigg |\frac{1}{2\pi }\int _0^{2\pi }\frac{\partial }{\partial \bar{z}}P_\alpha (ze^{-i\theta })f^*(e^{i\theta })\mathrm {d}\theta \bigg |\nonumber \\&\le \Vert f^*\Vert _\infty \bigg [\frac{1}{2\pi }\int _0^{2\pi }\bigg |\frac{\partial }{\partial z} P_\alpha (ze^{-i\theta })\bigg |\mathrm {d}\theta +\frac{1}{2\pi }\int _0^{2\pi }\bigg |\frac{\partial }{\partial \bar{z}} P_\alpha (ze^{-i\theta })\bigg |\mathrm {d}\theta \bigg ]\nonumber \\&\le \Vert f^*\Vert _\infty \bigg [(1-\alpha )+(1+\alpha )\bigg ](1-|z|)^\alpha \frac{1}{1-|z|^2}\nonumber \\&\le 2^{1-\alpha }\Vert f^*\Vert _\infty (1-|z|^2)^{\alpha -1} . \end{aligned}$$

(19)

Therefore, the proof of Lemma 4.3 is complete. $\square$

We also need some inequalities of the weighted Green potential ${\mathcal {G}}[g](z)$ as the following lemma.

Lemma 4.4

For $\alpha >-1$ and $g\in {\mathcal {C}} ({\overline{\mathbb {D}} })$, the potential ${\mathcal {G}}[g](z)$ satisfies the following inequalities:

• (a) $\displaystyle \big |{\mathcal {G}}[g](z)\big |\le 2^{|\alpha |}(1-|z|^2)^\alpha \bigg [\frac{2+\alpha }{1+\alpha }-|z|^2\bigg ]\Vert g\Vert _\infty ;$

• (b) For fixed $w\in \mathbb {D}$,

  $$\begin{aligned} |\frac{\partial {\mathcal {G}}[g](z)}{\partial z}|\le 2^{\alpha +2} \bigg [\bigg |\frac{2\alpha }{1+\alpha }\bigg |+(2|\alpha |+1)(1-|z|^2)\bigg ](1-|z|^2)^{\alpha -1}\Vert g\Vert _\infty . \end{aligned}$$

  (20)

• (c) For fixed $w\in \mathbb {D}$,

  $$\begin{aligned} |\frac{\partial {\mathcal {G}}[g](z)}{\partial \bar{z}}|\le 2^{\alpha +2}(1-|z|^2)^{\alpha }\Vert g\Vert _\infty . \end{aligned}$$

  (21)

Proof

We first give the proof of (a). By the Lemma 4.2, we have

$$\begin{aligned} |\mathcal {G}[g](z)|&\le \int _{\mathbb {D}} |G_\alpha (z,w)g(w)|\mathrm {d}A(w)\le \Vert g\Vert _\infty \int _{\mathbb {D}} |G_\alpha (z,w)|\mathrm {d}A(w)\\&\le 2^{|\alpha |}\Vert g\Vert _\infty (1-|z|^2)^\alpha \int _{\mathbb {D}}\bigg (\frac{1}{1+\alpha }+\log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\bigg )\mathrm {d}A(w)\\&\le 2^{|\alpha |}(1-|z|^2)^\alpha \bigg [\frac{1}{1+\alpha }+\int _{\mathbb {D}} \log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\mathrm {d}A(w)\bigg ]\Vert g\Vert _\infty . \end{aligned}$$

Let $\displaystyle I=\int _{\mathbb {D}} \log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\mathrm {d}A(w)$ and $\displaystyle \eta =\varphi (w)=\frac{z-w}{1-w\bar{z}}=re^{iv}$, so,

$$\begin{aligned} \mathrm {d}A(w)=|(\varphi ^{-1})'(\eta )|^2\mathrm {d}A(\eta )=\frac{(1-|z|^2)^2}{|1-\eta \bar{z}|^4}\mathrm {d}A(\eta ). \end{aligned}$$

Consequently,

$$\begin{aligned} I&=\int _{\mathbb {D}}\frac{(1-|z|^2)^2}{|1-\eta \bar{z}|^4}\log \frac{1}{|\eta |^2}\mathrm {d}A(\eta )\\&=\frac{(1-|z|^2)^2}{\pi }\int _0^1\int _0^{2\pi }\frac{r}{|1-\bar{z}re^{iv}|^4}\log \frac{1}{r^2}\mathrm {d}v\mathrm {d}r. \end{aligned}$$

Using the technique of power series expansion, we obtain that

$$\begin{aligned} I=2(1-|z|^2)^2\sum _{n=0}^\infty (n+1)^2|z|^{2n}\int _0^1r^{2n+1}\log \frac{1}{r^2}\mathrm {d}r=1-|z|^2, \end{aligned}$$

(22)

which implies that

$$\begin{aligned} |\mathcal {G}[g](z)|\le 2^{|\alpha |}(1-|z|^2)^\alpha \bigg [\frac{2+\alpha }{1+\alpha }-|z|^2\bigg ]\Vert g\Vert _\infty . \end{aligned}$$

(23)

Next, we will give the proof of (b).

For $\varepsilon >0$, let $D_\varepsilon =\mathbb {D}(z,\varepsilon )$ and $\varphi \in {\mathcal {C}}_0^\infty ({\mathbb {D}})$ be a test function. By Lebesgue’s dominated convergence theorem we get

$$\begin{aligned} \int _{\mathbb {D}} G_\alpha (z,w) \varphi _w (w) \mathrm {d}A(w)=\lim _{\varepsilon \rightarrow 0} \int _{{\mathbb {D}} \setminus D_\varepsilon } G_\alpha (z,w)\varphi _w (w) \mathrm {d}A(w), \end{aligned}$$

where $G_\alpha (z,w) \varphi _w (w)$ is the dominant function.

By Lemma 4.1, we have

$$\begin{aligned}&\bigg |\int _{{\mathbb {D}} \backslash D_\varepsilon } G_\alpha (z,w)\varphi _w (w) \mathrm {d}A(w)+\int _{{\mathbb {D}} \backslash D_\varepsilon } \frac{\partial G_\alpha (z,w)}{\partial w}\varphi (w) \mathrm {d}A(w)\bigg |\\&\ \ \ \ \ =\bigg |\frac{1}{2}\oint _{ \partial D_\varepsilon }\mathscr {R}[G_\alpha (z,w)\varphi (w)\bar{v}]\mathrm {d}s+\frac{i}{2}\oint _{ \partial D_\varepsilon }{\mathscr {R}}[G_\alpha (z,w)\varphi (w)\bar{u}]\mathrm {d}s\bigg |\\&\ \ \ \ \ \le \frac{1}{2}\oint _{ \partial D_\varepsilon } |G_\alpha (z,w)\varphi (w)|\mathrm {d}s+\frac{1}{2}\oint _{ \partial D_\varepsilon } |G_\alpha (z,w)\varphi (w)|\mathrm {d}s\\&\ \ \ \ \ =\oint _{ \partial D_\varepsilon } |G_\alpha (z,w)\varphi (w)|\mathrm {d}s \\&\ \ \ \ \ \le C_\varphi \oint _{ \partial D_\varepsilon } |G_\alpha (z,w)|\mathrm {d}s \\&\ \ \ \ \ \le C_\varphi 2\pi \varepsilon 2^{|\alpha |}(1-|z|^2)^\alpha (\frac{1}{1+\alpha }+\log \frac{4}{\varepsilon ^2}), \end{aligned}$$

where $C_\varphi$ is a constant depending only on $\Vert \varphi \Vert _\infty$. Therefore, let $\varepsilon \rightarrow 0$, it follows

$$\begin{aligned} \int _{{\mathbb {D}} } G_\alpha (z,w)\varphi _w (w) \mathrm {d}A(w)=-\int _{{\mathbb {D}}} \frac{\partial G_\alpha (z,w)}{\partial w}\varphi (w) \mathrm {d}A(w). \end{aligned}$$

From the proof of (a), we obtain that

$$\begin{aligned} \int _{\mathbb {D}}\int _{\mathbb {D}}|G_{\alpha }(z,w)|\mathrm {d}A(w)\mathrm {d}A(z)&\le 2^{|\alpha |}\frac{2+\alpha }{1+\alpha } \int _{\mathbb {D}}(1-|z|^2)^\alpha \frac{r}{\pi }\mathrm {d}r\mathrm {d}v\\&=2^{|\alpha |}\frac{2+\alpha }{1+\alpha }\int _0^{2\pi }\mathrm {d}\theta \int _0^1(1-r^2)^\alpha \frac{\mathrm {d}(r^2)}{2\pi }\\&=2^{|\alpha |}\frac{2+\alpha }{(1+\alpha )^2}. \end{aligned}$$

Hence, $G_{\alpha }(z,w)\in L^1({\mathbb {D}}\times {\mathbb {D}}).$ Therefore, the assumption that $\varphi (z)\in {\mathcal {C}}_0^\infty ({\mathbb {D}})$ and $g(w)\in {\mathcal {C}}(\overline{{\mathbb {D}}})$ shows that the integral $\int _{\mathbb {D}}\int _{\mathbb {D}}G_\alpha (z,w) g(w)\varphi _z(z)\mathrm {d}A(w)\mathrm {d}A(z)$ is absolutely convergent. Moreover, we also have that ${\mathcal {G}}[g](z)\in L^1(\mathbb {D})$ for $g(w)\in {\mathcal {C}}(\overline{{\mathbb {D}}})$.

The definition of distributional partial derivatives says that for $f\in L_{loc}^1(\mathbb {D})$ and $g\in \mathcal {C}_0^\infty (\mathbb {D})$, it follows

$$\begin{aligned} \langle f_z,g\rangle =-\int _{\mathbb {D}}fg_z\mathrm {d}A. \end{aligned}$$

Hence, the fact that ${\mathcal {G}}[g](z)\in L^1(\mathbb {D})$ and the absolute convergence of the integral $\int _{\mathbb {D}}\int _{\mathbb {D}}G_\alpha (z,w) g(w)\varphi _z(z)\mathrm {d}A(w)\mathrm {d}A(z)$ shows that

$$\begin{aligned} \int _{\mathbb {D}}\frac{\partial {\mathcal {G}}[g(w)](z)}{\partial z} \varphi (z) \mathrm {d}A(z) =-\int _{\mathbb {D}}\int _{\mathbb {D}}G_\alpha (z,w) \varphi _z(z)g(w)\mathrm {d}A(z)\mathrm {d}A(w). \end{aligned}$$

From (a) and the proof of Proposition 4 of [9], for any fixed $w\in \mathbb {D}$ we know that $G_\alpha (z,w)\in L^1({\mathbb {D}})$. Thus, for a test function $\varphi \in {\mathcal {C}}_0^\infty ({\mathbb {D}})$ it follows

$$\begin{aligned} \bigg \langle \frac{\partial G_\alpha (z,w)}{\partial z},\varphi (z) \bigg \rangle =-\int _{\mathbb {D}} G_\alpha (z,w)\varphi _z(z)\mathrm {d}A(z). \end{aligned}$$

To complete the proof of (b), we also need the absolute convergence of the integral $\int _{{\mathbb {D}}}\int _{{\mathbb {D}}} \frac{\partial G_\alpha (z,w)}{\partial z}\mathrm {d}A(w)\mathrm {d}A(z)$. By some direct calculations, we obtain that for $z\ne w$

$$\begin{aligned} \frac{\partial G_\alpha (z,w)}{\partial z} = \alpha \bar{w}(1-z\bar{w})^{\alpha -1}h\circ \omega (z,w)+ \frac{(1-|z|^2)^\alpha (1-|w|^2)^{\alpha +1}}{(1-\bar{z}w)^\alpha (1-z\bar{w})}\frac{1}{z-w}. \end{aligned}$$

Hence,

$$\begin{aligned} \bigg |\frac{\partial G_\alpha (z,w)}{\partial z}\bigg |&=\bigg |\alpha \bar{w}(1-z\bar{w})^{\alpha -1}h\circ \omega (z,w)+ \frac{(1-|z|^2)^\alpha (1-|w|^2)^{\alpha +1}}{(1-\bar{z}w)^\alpha (1-z\bar{w})}\frac{1}{z-w}\bigg |\\&\le |\alpha \bar{w}(1-z\bar{w})^{\alpha -1}h\circ \omega (z,w)|+\bigg |\frac{(1-|z|^2)^\alpha (1-|w|^2)^{\alpha +1}}{(1-\bar{z}w)^\alpha (1-z\bar{w})}\frac{1}{z-w}\bigg |. \end{aligned}$$

Moreover, we have

$$\begin{aligned} |\alpha \bar{w}(1-z\bar{w})^{\alpha -1}h\circ \omega (z,w)|&\le \bigg |\alpha \bar{w}(1-z\bar{w})^{\alpha -1}\bigg (\frac{\omega ^{1+\alpha }}{1+\alpha }+\omega ^{\alpha +1} \log \frac{1}{1-\omega }\bigg )\bigg |\\&=\bigg |\alpha \bar{w}(1-z\bar{w})^{\alpha -1}\bigg (\frac{(1-|z|^2)^{\alpha +1}(1-|w|^2)^{\alpha +1}}{|1-z\bar{w}|^{2\alpha +2}(\alpha +1)}\\&+\frac{(1-|z|^2)^{\alpha +1}(1-|w|^2)^{\alpha +1}}{|1-z\bar{w}|^{2\alpha +2}}\log |\frac{1-\bar{z}w}{z-w}|^2\bigg )\bigg |\\&=\bigg |\frac{\alpha }{\alpha +1} \bar{w}\frac{(1-|z|^2)^{\alpha +1}(1-|w|^2)^{\alpha +1}}{(1-z\bar{w})^2(1- \bar{z}w)^{\alpha +1}} \\&+\alpha \bar{w}\frac{( 1-|z|^2)^{\alpha +1}(1-|w|^2)^{\alpha +1}}{(1-z\bar{w})^2(1-\bar{z}w)^{\alpha +1}}\log |\frac{1-\bar{z}w}{z-w}|^2\bigg |. \end{aligned}$$

So,

$$\begin{aligned} |\alpha \bar{w}(1-z\bar{w})^{\alpha -1}h\circ \omega (z,w)|\le 2^{\alpha +1}\frac{(1-|z|^2)^{\alpha +1}}{|1-z\bar{w}|^2}\bigg [\bigg |\frac{\alpha }{\alpha +1}\bigg |+|\alpha |\log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\bigg ] \end{aligned}$$

and

$$\begin{aligned} \bigg |\frac{(1-|z|^2)^\alpha (1-|w|^2)^{\alpha +1}}{(1-\bar{z}w)^\alpha (1-z\bar{w})}\frac{1}{z-w}\bigg |&=\frac{(1-|z|^2)^\alpha (1-|w|^2)^{\alpha +1}}{|1-\bar{z}w|^{\alpha +1}}\frac{1}{|z-w|}\\&\le 2^{\alpha +1}(1-|z|^2)^\alpha \frac{1}{|z-w|}. \end{aligned}$$

Therefore, we obtain that

$$\begin{aligned} \int _{\mathbb {D}}&\bigg |\frac{\partial G_\alpha (z,w)}{\partial z}\bigg |\mathrm {d}A(w)\le 2^{\alpha +1}\bigg |\frac{\alpha }{\alpha +1}\bigg |(1-|z|^2)^{\alpha +1}\int _{\mathbb {D}} \frac{\mathrm {d}A(w)}{|1-z\bar{w}|^2}\\&+|\alpha |2^{3+\alpha }(1-|z|^2)^{\alpha -1}\int _{\mathbb {D}}\log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\mathrm {d}A(w) \\&\quad +2^{\alpha +1}(1-|z|^2)^\alpha \int _{{\mathbb {D}} } \frac{1}{|z-w|}\mathrm {d}A(w)\\&=I_1(z)+I_2(z)+I_3(z). \end{aligned}$$

By the Lemma D we can obtain

$$\begin{aligned} \int _{{\mathbb {D}}}I_3(z)\mathrm {d}A(z)\le 2^{\alpha +2}\int _{{\mathbb {D}}}(1-|z|^2)^\alpha \mathrm {d}A(z)=\frac{2^{\alpha +2}}{1+\alpha }<\infty \end{aligned}$$

(24)

By the equality (22) we have

$$\begin{aligned} \int _{{\mathbb {D}}}I_2(z)\mathrm {d}A(z)=\frac{|\alpha |2^{3+\alpha }}{1+\alpha }<\infty . \end{aligned}$$

Using Parseval’s theorem, we can obtain that

$$\begin{aligned} \int _{\mathbb {D}} \frac{1}{|1-z\bar{w}|^2}\mathrm {d}A(w)=-\frac{\ln (1-|z|^2)}{|z|^2}. \end{aligned}$$

Since $\displaystyle \lim _{x\rightarrow 0}\frac{-\ln (1-x)}{x}=1,$ there exists a constant $M_1$ such that

$$\begin{aligned} \int _{{\mathbb {D}}_{1/2}}\int _{{\mathbb {D}}} \frac{(1-|z|)^{1+\alpha }}{|1-z\bar{w}|^2}\mathrm {d}A(w)\mathrm {d}A(z)\le M_1. \end{aligned}$$

Thus, from the fact

$$\begin{aligned} \int _{\frac{1}{2}}^{1}(1-r^2)^{1+\alpha }\ln (1-r^2)r\mathrm {d}r<\infty \end{aligned}$$

we have that $\int _{{\mathbb {D}}} I_1(z)\mathrm {d}A(z)$ is also finite. So we have completed the absolute convergence of the integral $\int _{{\mathbb {D}}}\int _{{\mathbb {D}}} \frac{\partial G_\alpha (z,w)}{\partial z}\mathrm {d}A(w)\mathrm {d}A(z)$.

Therefore, we complete the proof of each steps of the following relation

$$\begin{aligned} \bigg \langle \frac{\partial }{\partial z}{\mathcal {G}}[g(w)](z),\varphi (z) \bigg \rangle&=\int _{\mathbb {D}}\frac{\partial {\mathcal {G}}[g(w)](z)}{\partial z} \varphi (z) \mathrm {d}A(z)\\&=-\int _{\mathbb {D}}\bigg (\int _{\mathbb {D}}G_\alpha (z,w) g(w) \mathrm {d}A(w)\bigg )\varphi _z(z)\mathrm {d}A(z)\\&=-\int _{\mathbb {D}}\bigg (\int _{\mathbb {D}}G_\alpha (z,w) \varphi _z(z) \mathrm {d}A(z)\bigg )g(w)\mathrm {d}A(w)\\&=\int _{\mathbb {D}}\bigg \langle \frac{\partial G_\alpha (z,w)}{\partial z},\varphi (z) \bigg \rangle g(w)\mathrm {d}A(w)\\&=\int _{\mathbb {D}}\bigg (\int _{\mathbb {D}} \frac{\partial G_\alpha (z,w)}{\partial z} \varphi (z) \mathrm {d}A(z)\bigg )g(w)\mathrm {d}A(w)\\&=\int _{\mathbb {D}}\bigg (\int _{\mathbb {D}} \frac{\partial G_\alpha (z,w)}{\partial z}g(w) \mathrm {d}A(w)\bigg )\varphi (z)\mathrm {d}A(z)\\&=\bigg \langle \int _{\mathbb {D}} \frac{\partial }{\partial z}G_\alpha (z,w)g(w)\mathrm {d}A(w),\varphi (z) \bigg \rangle . \end{aligned}$$

Thus, we can conclude that

$$\begin{aligned} \frac{\partial {\mathcal {G}}[g](z)}{\partial z}=\int _{\mathbb {D}} \frac{\partial G_\alpha (z,w)}{\partial z}g(w)\mathrm {d}A(w). \end{aligned}$$

The following estimate

$$\begin{aligned} |\alpha \bar{w}(1-z\bar{w})^{\alpha -1}h\circ \omega (z,w)|&\le 2^{\alpha +1}\frac{(1-|z|^2)^{\alpha +1}}{|1-z\bar{w}|^2}\bigg [\bigg |\frac{\alpha }{\alpha +1}\bigg |+|\alpha |\log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\bigg ]\\&\le 2^{\alpha +3}(1-|z|^2)^{\alpha -1}\bigg [\bigg |\frac{\alpha }{\alpha +1}\bigg |+|\alpha |\log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\bigg ] \end{aligned}$$

implies that

$$\begin{aligned} \bigg |\frac{\partial G_\alpha (z,w)}{\partial z}\bigg |&\le 2^{\alpha +3}(1-|z|^2)^{\alpha -1}\bigg [\bigg |\frac{\alpha }{\alpha +1}\bigg |+|\alpha |\log \bigg |\frac{1-\bar{z}w}{z-w}\bigg |^2\bigg ]\\&\quad +2^{\alpha +1}(1-|z|^2)^\alpha \frac{1}{|z-w|}. \end{aligned}$$

Hence, we conclude that

$$\begin{aligned} \bigg |\frac{\partial {\mathcal {G}}[g](z)}{\partial z}\bigg |&\le \Vert g \Vert _\infty \int _{\mathbb {D}} \bigg |\frac{\partial G_\alpha (z,w)}{\partial z}\bigg |\mathrm {d}A(w) \\&\le 2^{\alpha +2} \bigg [\bigg |\frac{2\alpha }{1+\alpha }\bigg |+(2|\alpha |+1)(1-|z|^2)\bigg ](1-|z|^2)^{\alpha -1}\Vert g\Vert _\infty .\end{aligned}$$

Finally, we will give the proof of (c). In fact, by (4), we obtain that for $z\ne w$

$$\begin{aligned} \frac{\partial G_\alpha (z,w)}{\partial \bar{z}} =\frac{(1-|z|^2)^\alpha (1-|w|^2)^{\alpha +1}}{(1-\bar{z}w)^{\alpha +1}(\bar{z}-\bar{w})} . \end{aligned}$$

(25)

By a similar method as above, Lemma D implies that

$$\begin{aligned} \int _{\mathbb {D}} \bigg |\frac{\partial G_\alpha (z,w)}{\partial \bar{z}}\bigg |\mathrm {d}A(w)\le 2^{\alpha +2}(1-|z|^2)^{\alpha }. \end{aligned}$$

Hence, we conclude that

$$\begin{aligned} \bigg |\frac{\partial {\mathcal {G}}[g](z)}{\partial \bar{z}}\bigg |\le \int _{\mathbb {D}}\bigg |\frac{\partial G_\alpha (z,w)}{\partial \bar{z}}g(w)\bigg |\mathrm {d}A(w)\le 2^{\alpha +2}(1-|z|^2)^{\alpha }\Vert g\Vert _\infty . \end{aligned}$$

Thus, the proof of the Lemma 4.4 is complete. $\square$

5 Proof of Theorem 1.1

Proof

By the formula (7) we have

$$\begin{aligned} |f(z)|\le |\mathcal {P}_\alpha [f^*](z)|+|{\mathcal {G}}[g](z)|. \end{aligned}$$

For $\alpha \ge 0$, we have

$$\begin{aligned} |\mathcal {P}_\alpha [f^*](z)|&=\bigg |\frac{1}{2\pi }\int ^{2\pi }_{0}\frac{1-|z|^2}{|1-ze^{-i\theta }|^2}\cdot \frac{(1-|z|^2)^\alpha }{(1-\overline{z}e^{i\theta })^\alpha }f^*(e^{i\theta })\mathrm {d}\theta \bigg |\\&\le (1+|z|)^\alpha \frac{1}{2\pi }\int ^{2\pi }_{0}\frac{1-|z|^2}{|1-ze^{-i\theta }|^2}\big |f^*(e^{i\theta })\big |\mathrm {d}\theta \\&\le 2^\alpha \mathcal {P}[|f^*|](z) \end{aligned}$$

and $\mathcal {P}[|f^*|](z)$ is a positive harmonic function in $\mathbb {D}.$

By the inequality (14), we know that, for $z\in \mathbb {D}$

$$\begin{aligned} |\mathcal {P}_\alpha [f^*](z)|\le 2^\alpha \mathcal {P}[|f^*|](z) \le 2^{\alpha +1} \frac{\Vert f^*\Vert _\infty }{\pi }\arctan \bigg (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2}\bigg ) . \end{aligned}$$

On the other hand, for $\alpha \ge 0$, by the inequality (23), we get

$$\begin{aligned} |\mathcal {G}[g](z)|\le 2^{\alpha }(1-|z|^2)^\alpha \bigg (\frac{2+\alpha }{1+\alpha }-|z|^2\bigg )\Vert g\Vert _\infty . \end{aligned}$$

Hence, for $\alpha \ge 0$,

$$\begin{aligned} |f(z)|\le 2^{\alpha +1} \frac{\Vert f^*\Vert _\infty }{\pi }\arctan \bigg (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2}\bigg )+2^{\alpha }(1-|z|^2)^\alpha \bigg (\frac{2+\alpha }{1+\alpha }-|z|^2\bigg )\Vert g\Vert _\infty . \end{aligned}$$

For $-1<\alpha <0$, we have

$$\begin{aligned} |\mathcal {P}_\alpha [f^*](z)|&=\bigg |\frac{1}{2\pi }\int ^{2\pi }_{0}\frac{1-|z|^2}{|1-ze^{-i\theta }|^2}\cdot \frac{(1-|z|^2)^\alpha }{(1-\overline{z}e^{i\theta })^\alpha }f^*(e^{i\theta })\mathrm {d}\theta \bigg |\\&\le (1-|z|)^\alpha {\mathcal {P}}[|f^*|](z). \end{aligned}$$

So,

$$\begin{aligned} |\mathcal {P}_\alpha [f^*](z)|\le 2^{1-\alpha }(1-|z|^2)^\alpha \frac{\Vert f^*\Vert _\infty }{\pi }\arctan \bigg (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2}\bigg ). \end{aligned}$$

By the formula (23), we obtain that

$$\begin{aligned} |\mathcal {G}[g](z)|\le 2^{|\alpha |}(1-|z|^2)^\alpha \bigg (\frac{2+\alpha }{1+\alpha }-|z|^2\bigg )\Vert g\Vert _\infty . \end{aligned}$$

Hence, for $-1<\alpha <0$,

$$\begin{aligned} |f(z)|&\le 2^{1-\alpha }(1-|z|^2)^\alpha \frac{\Vert f^*\Vert _\infty }{\pi }\arctan \bigg (\frac{1+|z|}{1-|z|}\tan \frac{c\pi }{2}\bigg )\\&\quad +2^{|\alpha |}(1-|z|^2)^\alpha \bigg (\frac{2+\alpha }{1+\alpha }-|z|^2\bigg )\Vert g\Vert _\infty . \end{aligned}$$

Thus, the proof of Theorem 1.1 is complete. $\square$

6 Proof of Theorem 1.2

Proof

For $-1<\alpha <0$, by the inequalities (19), (20) and (21) imply that

$$\begin{aligned} \Vert D _f(z) \Vert&\le \bigg |\frac{\partial D _{{\mathcal {P}}_\alpha [f^*]}(z)}{\partial z}\bigg |+\bigg |\frac{\partial D _{{\mathcal {P}}_\alpha [f^*]}(z)}{\partial \bar{z}} \bigg |+\bigg |\frac{\partial {\mathcal {G}}[g](z)}{\partial z} \bigg |+\bigg |\frac{\partial {\mathcal {G}}[g](z)}{\partial \bar{z}}\bigg |\\&=\Vert D _{{\mathcal {P}}_\alpha [f^*]}(z)\Vert +\Vert D _{{\mathcal {G}}[g]}(z) \Vert \\&\le 2^{\alpha +2}(1-|z|^2)^{\alpha -1}\bigg (\bigg |\frac{2\alpha }{\alpha +1}\bigg |+(2|\alpha |+2)(1-|z|^2)\bigg )\Vert g\Vert _\infty +2^{1-\alpha }\\&\qquad \Vert f^*\Vert _\infty (1-|z|^2)^{\alpha -1}\\&=2^{\alpha +3}(1-|z|^2)^{\alpha -1}\bigg (\frac{-\alpha }{\alpha +1}+(1-\alpha )(1-|z|^2)\bigg )\Vert g\Vert _\infty +2^{1-\alpha }\Vert f^*\Vert _\infty \\&\qquad (1-|z|^2)^{\alpha -1}. \end{aligned}$$

Hence, for $\alpha \ge 0$, by the inequalities (20), (21) and Lemma C, we obtain that

$$\begin{aligned} \Vert D _f(z) \Vert&\le \bigg |\frac{\partial D _{{\mathcal {P}}_\alpha [f^*]}(z)}{\partial z}\bigg |+\bigg |\frac{\partial D _{{\mathcal {P}}_\alpha [f^*]}(z)}{\partial \bar{z}} \bigg |+\bigg |\frac{\partial {\mathcal {G}}[g](z)}{\partial z} \bigg |+\bigg |\frac{\partial {\mathcal {G}}[g](z)}{\partial \bar{z}}\bigg |\\&\le (\alpha +1)2^{\alpha +1}\Vert f^*\Vert _\infty \frac{1}{1-|z|^2}+ 2^{\alpha +2}(1-|z|^2)^{\alpha -1}\\&\qquad \bigg (\bigg |\frac{2\alpha }{\alpha +1}\bigg |+(2|\alpha |+2) (1-|z|^2)\bigg )\Vert g\Vert _\infty \\&=(\alpha +1)2^{\alpha +1}\Vert f^*\Vert _\infty \frac{1}{1-|z|^2}+2^{\alpha +3}(1-|z|^2)^{\alpha -1}\\&\qquad \bigg (\frac{\alpha }{1+\alpha }+(\alpha +1)(1-|z|^2)\bigg )\Vert g\Vert _\infty . \end{aligned}$$

Thus, the proof of the Theorem 1.2 is complete. $\square$