Schwarz Type Lemmas for Generalized Harmonic Functions

Abstract

Let ${\alpha },{\beta }\in (-1,\infty )$ such that ${\alpha }+{\beta }>-1$. Given two continuous functions $g \in \mathcal {C}(\overline{{\mathbb D}})$ and $f\in \mathcal {C}({\mathbb T})$, we provide various Schwarz type lemmas for mappings u satisfying the inhomogeneous $({\alpha },{\beta })$-harmonic equation $L_{{\alpha },{\beta }}u=g$ in ${\mathbb D}$ and $u=f$ in ${\mathbb T}$, where ${\mathbb D}$ is the unit disc of the complex plane ${\mathbb C}$ and ${\mathbb T}=\partial {\mathbb D}$ is the unit circle. The obtained results provide a significant improvement over previous research on the subject.

1 Introduction and Main Results

Let ${\mathbb D}$ be the open unit disc in the complex plane ${\mathbb C}$, let ${\mathbb T}= \partial {\mathbb D}$ be the unit circle, and denote by

$$\begin{aligned} \partial _z=\frac{1}{2}(\partial _x-i \partial _y)\quad \quad \text {and}\quad \quad \partial _{\overline{z}}=\frac{1}{2}(\partial _x+i \partial _y). \end{aligned}$$

For a differentiable function u on ${\mathbb D}$, we denote

$$\begin{aligned} \Vert Du(z)\Vert =|\partial _z u(z)|+|\partial _{\overline{z}}u(z)|. \end{aligned}$$

1.1 $({\alpha },{\beta })$-Harmonic Functions

We consider $L_{{\alpha },{\beta }}$, the family of differential operators on ${\mathbb D}$, defined by

$$\begin{aligned} L_{{\alpha },{\beta }}:=(1-|z|^2)^{-({\alpha }+{\beta }+1)}\bigg ((1-|z|^2)\partial _z\partial _{\overline{z}}+{\alpha }z\partial _z+{\beta }\overline{z}\partial _{\overline{z}}-{\alpha }{\beta }\bigg ), \end{aligned}$$

where ${\alpha },{\beta }$ are real numbers. There operators are essentially introduced by Geller [9], see also [15]. In fact, Geller introduced the following family of operators on ${\mathbb D}$

$$\begin{aligned} \Delta _{{\alpha },{\beta }}:=(1-|z|^2)\bigg ((1-|z|^2)\partial _z\partial _{\overline{z}}+{\alpha }z\partial _z+{\beta }\overline{z}\partial _{\overline{z}}-{\alpha }{\beta }\bigg )=(1-|z|^2)^{{\alpha }+{\beta }+2}L_{{\alpha },{\beta }}. \end{aligned}$$

We found that it is more convenient to work with the operators $L_{{\alpha },{\beta }}$ than $\Delta _{{\alpha },{\beta }}$ as for ${\alpha }={\beta }=0$, the operator $L_{0,0}$ is the classical Laplacian operator. In addition, it can be shown that

$$\begin{aligned} L_{{\alpha },0}=\partial _{\overline{z}} \omega _{{\alpha }}(z)^{-1}\partial _z, \end{aligned}$$

the operator for weighted harmonic functions with the standard weight $\omega _{{\alpha }}(z)=(1-|z|^2)^{\alpha }$ studied in [4, 8, 16, 22, 24]. Recall that a function u is ${\alpha }$-harmonic if $L_{{\alpha },0} u=0$.

A function $u\in \mathcal {C}^2({\mathbb D})$ is called $({\alpha },{\beta })$-harmonic if $L_{{\alpha },{\beta }}(u)=0.$

Let

$$\begin{aligned} T_\alpha := -\frac{{\alpha }^2}{4} \omega _{{\alpha }+1}^{-1}+ \frac{1}{2}( L_{{\alpha },0}+ \overline{L_{{\alpha },0}}), \end{aligned}$$

that is,

$$\begin{aligned} T_{\alpha }= & {} -\frac{{\alpha }^2}{4} (1-|z|^2)^{-{\alpha }-1}+\frac{{\alpha }}{2}(1-|z|^2)^{-{\alpha }-1}\overline{z} \partial _{\overline{z}}+\frac{{\alpha }}{2}(1-|z|^2)^{-{\alpha }-1}z\partial _z\\ {}{} & {} \quad +(1-|z|^2)^{-{\alpha }} \partial _{z}\partial _{\overline{z}}. \end{aligned}$$

Thus

$$\begin{aligned} T_{\alpha }= L_{\frac{{\alpha }}{2},\frac{{\alpha }}{2}}. \end{aligned}$$

Recall that a function u is $T_{\alpha }$-harmonic if $T_{\alpha }u=0$. Remark that $T_{\alpha }$-harmonic functions are exactly $(\frac{{\alpha }}{2},\frac{{\alpha }}{2})$-harmonic functions, for more details, we refer the reader to [6, 13, 14, 23].

Let $g\in \mathcal {C}({\mathbb D})$ and $u\in \mathcal {C}^2({\mathbb D})$, we consider the associated Dirichlet boundary value problem

$$\begin{aligned} \left\{ \begin{array}{ll} L_{{\alpha },{\beta }} \, u=g &{}\quad \text {in}\quad {\mathbb D},\\ u=f &{}\quad \text {on}\quad {\mathbb T}. \end{array}\right. \end{aligned}$$

(1.1)

If ${\beta }=0$, Eq. (1.1) is the inhomogeneous ${\alpha }$-harmonic equation studied in [4, 16, 17].

Theorem A

[9] The Dirichlet problem

$$\begin{aligned} L_{{\alpha },{\beta }}\, u=0,\quad u=f \text{ on } {\mathbb T},\quad f\in \mathcal {C}({\mathbb T}), \end{aligned}$$

has a solution for all $f \in \mathcal {C}({\mathbb T})$ if and only if ${\alpha }+{\beta }>-1$ and ${\alpha }, {\beta }\in {\mathbb R}\setminus {\mathbb Z}^-.$ In this case the solution is unique and is given by

$$\begin{aligned} u(z) := \mathcal {P}_{{\alpha },{\beta }}[f](z)= \frac{1}{2\pi }\int _0^{2\pi } P_{{\alpha },{\beta }}(ze^{-i\gamma })f(e^{i\gamma }) \textrm{d}\gamma , \end{aligned}$$

(1.2)

with

$$\begin{aligned} P_{{\alpha },{\beta }}(z)=c_{{\alpha },{\beta }}\frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{(1-z)^{{\alpha }+1}(1-\overline{z})^{{\beta }+1}}, \quad c_{{\alpha },{\beta }}=\frac{\Gamma ({\alpha }+1)\Gamma ({\beta }+1)}{\Gamma ({\alpha }+{\beta }+1)}. \end{aligned}$$

(1.3)

Denote

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

(1.4)

with

$$\begin{aligned} P_{\alpha }(z)=c_{\alpha }\frac{(1-|z|^2)^{{\alpha }+1}}{|1-z|^{{\alpha }+2}}, \quad c_{\alpha }=\frac{\Gamma ^2(\frac{{\alpha }}{2}+1)}{\Gamma ({\alpha }+1)}. \end{aligned}$$

(1.5)

The Hardy theory of $({\alpha },{\beta })$-harmonic functions is studied in detail in [1]. The authors proved the following

Theorem B

[1] Let ${\alpha },{\beta }\in {\mathbb R}\setminus {\mathbb Z}^-$ such that ${\alpha }+{\beta }>-1$ and let u be an $({\alpha },{\beta })$-harmonic function. Then:

• (i) $u=\mathcal {P}_{{\alpha },{\beta }}[f]$ for some $f\in L^p({\mathbb T})$, $1<p< \infty$ if and only if

  $$\begin{aligned} \sup _{0<r<1} \int _0^{2\pi } |u(re^{i\theta })|^p \textrm{d}\theta <+\infty . \end{aligned}$$

• (ii) $u=\mathcal {P}_{{\alpha },{\beta }}[\mu ]$ for some measure $\mu$ if and only if

  $$\begin{aligned} \sup _{0<r<1} \int _0^{2\pi } |u(re^{i\theta })| \textrm{d}\theta <+\infty . \end{aligned}$$

1.2 Hypergeometric Functions and Some Inequalities

Let us recall the Gaussian hypergeometric function defined by

$$\begin{aligned} F(a,b;c;z)=\sum _{n\ge 0}\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!},\ \ \ \ z\in {\mathbb D}, \end{aligned}$$

for $a,b,c\in {\mathbb C}$ such that $c\ne 0,-1,-2,\ldots$ where

$$\begin{aligned} (a)_n=\left\{ \begin{array}{ll} a(a+1)\ldots (a+n-1) &{}\quad \text{ for } \ \ n=1,2,3,\ldots \\ 1 &{}\quad \text{ for } \ \ n=0, \end{array}\right. \end{aligned}$$

which are called Pochhammer symbols.

We list few properties, see for instance ( [3], Chapter 2)

$$\begin{aligned} \lim \limits _{x\rightarrow 1}F(a,b;c;x)= & {} \frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)},\ \ \ \ \ \text {if}\ \ \ c-a-b>0, \end{aligned}$$

(1.6)

$$\begin{aligned} F(a,b;c;x)= & {} (1-x)^{c-a-b}F(c-a,c-b;c;x), \end{aligned}$$

(1.7)

$$\begin{aligned} \frac{d}{d x}F(a,b;c;x)= & {} \frac{ab}{c}F(1+a,1+b;1+c;x). \end{aligned}$$

(1.8)

$$\begin{aligned} F(a,b;a+b;x)\sim & {} -\frac{\Gamma (a+b)}{\Gamma (a)\Gamma (b)}\log (1-x) \quad (x\rightarrow 1^-), \end{aligned}$$

(1.9)

The asymptotic relation (1.9) is due to Gauss, and its refined form is due to Ramanujan [5, p. 71].

$$\begin{aligned} F(a,b;a+b;x)=\frac{\Gamma (a+b)}{\Gamma (a)\Gamma (b)}[R-\log (1-x)+O((1-x) \log (1-x))] \text{ as } x\rightarrow 1^{-}, \end{aligned}$$

where $R=-\psi (a)-\psi (b)-2\gamma , \psi (a)=\frac{\Gamma '(a)}{\Gamma (a)}$, and $\gamma$ denotes the Euler-Mascheroni constant.

For $\lambda >0$, let

$$\begin{aligned} I_{\lambda }(z):= \frac{1}{2\pi }\int _{0}^{2\pi } \frac{\textrm{d}\gamma }{|1-ze^{-i\gamma }|^\lambda }, \quad z\in {\mathbb D}. \end{aligned}$$

Using hypergeometric functions, one can see that

$$\begin{aligned} I_{\lambda }(z)\approx \left\{ \begin{array}{ll} 1, \ \ &{}\quad \lambda <1;\\ \log \frac{1}{1-|z|^2},&{}\quad \text{ if } \ \ \lambda =1;\\ (1-|z|^2)^{\lambda -1},&{}\quad \text{ if } \ \ \lambda >1. \end{array}\right. \end{aligned}$$

The notation $a(z) \approx b(z)$ means that the ratio a(z)/b(z) is bounded from above and below by two positive constants as $|z| \rightarrow 1^{-}$.

Sharp estimates of $I_\lambda$ are given in the following lemma.

Lemma 1.1

[13] Let $\lambda >0$. Then for all $z\in {\mathbb D}$, we have

• (i) If $\lambda >1$, then

  $$\begin{aligned} I_{\lambda }(z) \le \frac{\Gamma (\lambda -1)}{\Gamma (\lambda /2)^2}(1-|z|^2)^{1-\lambda }. \end{aligned}$$

• (ii) If $0<\lambda <1$, then

  $$\begin{aligned} I_{\lambda }(z) \le \frac{\Gamma (1-\lambda )}{\Gamma (1-\lambda /2)^2}. \end{aligned}$$

• (iii) If $\lambda =1$, then

  $$\begin{aligned} I_{1}(z)\le 1+ \frac{1}{\pi } \log \left( \frac{1}{1-|z|^2}\right) . \end{aligned}$$

For $t>-1,\ c\in {\mathbb R}$, let

$$\begin{aligned} J_{c,t}(z):=\int _{{\mathbb D}}\frac{(1-|\omega |^2)^t}{|1-z\overline{\omega }|^{2+t+c}}\, dA(\omega ). \end{aligned}$$

where $\textrm{d}A(w)$ is the normalized measure of the unit disc, given by $\displaystyle \textrm{d}A(w)=\frac{\textrm{d}u\textrm{d}v}{\pi }$, $w=u+iv$.

Recall that the well-known Forelli–Rudin [26] estimates states that

$$\begin{aligned} J_{c,t}(t)\approx \left\{ \begin{array}{ll} 1, &{}\quad \text{ if } \ \ c<0;\\ \log \frac{1}{1-|z|^2},&{}\quad \text{ if } \ \ c=0;\\ (1-|z|^2)^{-c},&{}\quad \text{ if } \ \ c>0. \end{array}\right. \end{aligned}$$

In [18], the author provided an estimate of the previous integral over the unit ball $\mathbb {B}_n$ in ${\mathbb C}^n$ for $n\ge 1$. By taking $n=1$, we get the following sharp estimates

Lemma 1.2

[18]

• (i) If $c<0$, then for all $z\in {\mathbb D}$,

  $$\begin{aligned} \frac{\Gamma (1+t)}{\Gamma (2+t)}\le J_{c,t}(z)\le \frac{\Gamma (1+t)\Gamma (-c)}{\Gamma ^2(\frac{2+t-c}{2})}. \end{aligned}$$

  (1.10)

• (ii) If $c>0$, then for all $z\in {\mathbb D}$,

  $$\begin{aligned} \frac{\Gamma (1+t)}{\Gamma (2+t)}\le (1-|z|^2)^cJ_{c,t}(z)\le \frac{\Gamma (1+t)\Gamma (c)}{\Gamma ^2(\frac{2+t+c}{2})}. \end{aligned}$$

• (iii) If $c=0$, then for all $z\in {\mathbb D}$,

  $$\begin{aligned} \frac{\Gamma (1+t)}{\Gamma ^2(1+\frac{t}{2})}\le |z|^2\bigg (\log \frac{1}{1-|z|^2}\bigg )^{-1}J_{0,t}(z)\le \frac{1}{1+t}. \end{aligned}$$

  (1.11)

The monotonicity of the hypergeometric function $F(a,b;a+b;x), a,b>0$ is studied in [2, Theorem 1.3], extending the complete elliptic integrals of the first kind. The authors proved the following

Lemma 1.3

[2, Theorem 1.3] For $a,b\in (0,\infty )$, the function

$$\begin{aligned} x \mapsto \frac{1-F(a,b;a+b,x)}{\log (1-x)} \end{aligned}$$

is increasing from (0, 1) into $(ab/(a+b),1/\textrm{B}(a,b)),$ where $\textrm{B}(a,b)$ is the Euler beta function.

1.3 Green’s Formula for $L_{{\alpha },{\beta }}$

Let

$$\begin{aligned} g_{{\alpha },{\beta }}(x):= \textrm{B}({\alpha }+1,{\beta }+1)(1-x)^{1+{\alpha }+{\beta }}F(1+{\alpha },1+{\beta };2+{\alpha }+{\beta };1-x), \end{aligned}$$

(1.12)

where $\textrm{B}$ is Euler beta function. In [1], the authors showed that $g_{{\alpha },{\beta }}(|z|^2)$ is a radial $({\alpha },{\beta })$-harmonic away from zero and playing the role of the Green’s function in the classical potential theory, and the weighted Green function $G_{{\alpha },{\beta }}$ of the differential operator $L_{{\alpha },{\beta }}$ could be written as

$$\begin{aligned} G_{{\alpha },{\beta }}(z,w):=(1-\overline{z}w)^{{\alpha }}(1-z\overline{w})^{{\beta }} g_{{\alpha },{\beta }}(|\varphi _z(w)|^2), \end{aligned}$$

(1.13)

where $\varphi _z$ is the Möbius transformation of the unit disc given by

$$\begin{aligned} \varphi _z(w)=\frac{z-w}{1-\overline{z}w}. \end{aligned}$$

Remark that

$$\begin{aligned} \displaystyle 1-|\varphi _z(w)|^2=\frac{(1-|z|^2)(1-|w|^2)}{|1-z\overline{w}|^2}. \end{aligned}$$

The weighted potential of a function g can be represented by

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

(1.14)

Following Riesz-type decomposition formula [1], we see that all solutions $u\in \mathcal {C}^2({\mathbb D}) \cap \mathcal {C}(\overline{{\mathbb D}})$ of (1.1) such that

$$\begin{aligned} \int _{\mathbb D}|g(w)| (1-|w|^2)^{{\alpha }+{\beta }+1} \textrm{d}A(w) <+\infty , \end{aligned}$$

(1.15)

are given by

$$\begin{aligned} u(z)=\mathcal {P}_{{\alpha },{\beta }}[f](z)-\mathcal {G}_{{\alpha },{\beta }}[g](z), \end{aligned}$$

(1.16)

where $\mathcal {P}_{{\alpha },{\beta }}[f]$ and $\mathcal {G}[g]$ are given, respectively, by (1.2) and (1.14). Clearly the condition (1.15) is satisfied if $u \in \mathcal {C}^2(\overline{{\mathbb D}})$.

In the case of $(0,\alpha )$-harmonic functions, Behm [4] showed that the weighted Green function $G_{{\alpha }}$ of $\Delta _{0,{\alpha }}$ could be written as

$$\begin{aligned} G_{{\alpha }}(z,w):=(1-\overline{z}w)^{{\alpha }}h \left( 1-|\varphi _z(w)|^2 \right) , \end{aligned}$$

where

$$\begin{aligned} h(x)=\int _0^x\frac{t^{\alpha }}{1-t}\textrm{d}t. \quad 0\le x<1. \end{aligned}$$

Using the zero-balanced Gauss’s hypergeometric function. One can see that

$$\begin{aligned} G_{{\alpha }}(z,w)= & {} \frac{1}{{\alpha }+1}\frac{(1-|z|^2)^{{\alpha }+1}(1-|w|^2)^{{\alpha }+1}}{(1 -\overline{z}w)(1-z\overline{w})^{{\alpha }+1}} \\{} & {} \times F\left( 1,{\alpha }+1;{\alpha }+2;\frac{(1-|z| ^2)(1-|w|^2)}{|1-z\overline{w}|^2}\right) . \end{aligned}$$

Hence $G_{\alpha }=G_{0,{\alpha }}$.

1.4 Schwarz and Schwarz–Pick Lemma

The Schwarz lemma for analytic functions plays a vital role in complex analysis, and it has been generalized in various settings; see [10, 13, 14, 16, 19,20,21] and the references therein.

Heinz [10] generalized it to the class of complex-valued harmonic functions. That is, if u is a complex-valued harmonic function from ${\mathbb D}$ into itself with $u (0) = 0$, then for $z \in {\mathbb D}$,

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

Moreover, this inequality is sharp for each point $z\in {\mathbb D}$.

Hethcote [11] and Pavlović [25] improved the above result of Heinz by removing the assumption $u (0) = 0$ and showed that for harmonic function u from ${\mathbb D}$ to ${\mathbb D}$, then

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

(1.17)

holds for all $z\in {\mathbb D}$.

Recently, Chen and Kalaj [7] established a Heinz-Hethcote type theorem for the solutions of the Dirichlet boundary value problem of the laplacian operator. In [13], we established a Heinz-Hethcote theorem for $T_\alpha$-harmonic functions.

Let ${\alpha }>-1$, define

$$\begin{aligned} U_\alpha (z):=\mathcal {P}_{\frac{{\alpha }}{2},\frac{{\alpha }}{2}} [\chi _{{\mathbb T}^r}- \chi _{{\mathbb T}^l}](z), \end{aligned}$$

(1.18)

where

$$\begin{aligned} {\mathbb T}^r=\{ z \in {\mathbb T}: {{\text {Re}}\,}{z}>0 \}, \text{ and } {\mathbb T}^l=\{ z \in {\mathbb T}: {{\text {Re}}\,}{z}<0 \}. \end{aligned}$$

Notice that $U_\alpha$ is a $T_\alpha$-harmonic function on ${\mathbb D}$ with values in $(-1,1)$ such that $U_\alpha (0)=0$.

Theorem C

Let $\alpha >-1$ and $u:{\mathbb D}\longrightarrow {\mathbb D}$ be a $T_\alpha$-harmonic function, then

$$\begin{aligned} \bigg |u(z)-\frac{(1-|z|^2)^{\alpha +1}}{(1+|z|^2)^{\frac{\alpha }{2}+1}}u(0)\bigg |\le U_\alpha (|z|), \end{aligned}$$

for all $z\in {\mathbb D}$, where $U_\alpha$ is the function defined in (1.18).

This theorem extends the estimate (1.17), indeed, for ${\alpha }=0$, we have $U_0(|z|)=\frac{4}{\pi }\arctan |z|$. Recently, a Heinz-Hethcote type theorem is proved for ${\alpha }$-harmonic functions, see [12].

Theorem D

[12] Let $\alpha >-1$ and $u:{\mathbb D}\rightarrow {\mathbb D}$ be an $\alpha$-harmonic function. Then

• (1) If $\alpha \ge 0$, then

  $$\begin{aligned} \left| u(z)-\frac{(1-|z|^2)^{\alpha +1}}{1+|z|^2}u(0)\right| \le \frac{2^{\alpha +2}}{\pi } \arctan |z|+ 2^{\alpha +1} (1-|z|) \left( 1-(1-|z|)^\alpha \right) . \end{aligned}$$

• (2) If $\alpha < 0$, then

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

Other variants of Schwarz lemma for ${\alpha }$-harmonic functions are considered in Li and Chen [16] for mappings u in ${\mathbb D}$ satisfying the $\alpha$-harmonic equation $L_{\alpha ,0}\, u = g$, extending previous results of Li et al. [17].

Here, we should point out that the inequalities obtained in the case ${\alpha }<0$ are not convenient due to the factor $(1-|z|)^{\alpha }$ which goes to infinity as $|z|\rightarrow 1$.

In this paper, we extend and improve the above estimates and obtain a Schwarz type lemma for solutions to the $({\alpha },{\beta })$-harmonic equation (1.1). Our first main result is the following theorem.

Theorem 1.1

Suppose that $g\in \mathcal {C}(\overline{{\mathbb D}})$ and $f\in \mathcal {C}({\mathbb T})$. If $u\in \mathcal {C}^2({\mathbb D})$ satisfies the $({\alpha },{\beta })$-harmonic equation (1.1) for ${\alpha },{\beta }\in (-1,\infty )$ such that ${\alpha }+{\beta }>-1$, for $z \in {\mathbb D}$,

• (i) If ${\alpha }+{\beta }>0$, then

  $$\begin{aligned}{} & {} \left| u(z)- \frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{1+|z|^2}\mathcal {P}_{{\alpha },{\beta }}[f](0)\right| \\{} & {} \quad \le 2^{{\alpha }+{\beta }+1}|c_{{\alpha },{\beta }}|\bigg [\frac{2}{\pi }\arctan |z| +\frac{|{\alpha }|+|{\beta }|}{{\alpha }+{\beta }} (1-|z|)\left( 1-(1-|z|)^{{\alpha }+{\beta }}\right) \bigg ]\Vert f\Vert _\infty \\{} & {} \qquad +d_{{\alpha },{\beta }}(1-|z|^2)^{{\alpha }+{\beta }+1}\Vert g\Vert _\infty . \end{aligned}$$

• (ii) If ${\alpha }+{\beta }=0$, then

  $$\begin{aligned} \left| u(z)- \frac{1-|z|^2}{1+|z|^2}\mathcal {P}_{{\alpha },{\beta }}[f](0)\right|\le & {} |c_{{\alpha },-{\alpha }}|\left[ \frac{4}{\pi }\arctan |z|+\frac{|{\alpha }|\pi }{4}(1-|z|)\right] \Vert f\Vert _\infty \\{} & {} +\, d_{{\alpha },-{\alpha }}(1-|z|^2)\Vert g\Vert _\infty . \end{aligned}$$

• (iii) If ${\alpha }+{\beta }<0$, then

  $$\begin{aligned}{} & {} \left| u(z)- \frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{(1+|z|^2)^{\frac{{\alpha }+{\beta }}{2}+1}}\mathcal {P}_{{\alpha },{\beta }}[f](0)\right| \\{} & {} \quad \le |c_{{\alpha },{\beta }}|\left[ \frac{U_{{\alpha }+{\beta }}(|z|)}{c_{{\alpha }+{\beta }}} + \frac{|{\alpha }-{\beta }|\pi }{4} (1-|z|^2)^{{\alpha }+{\beta }+1}\right] \Vert f\Vert _\infty \\{} & {} \qquad +d_{{\alpha },{\beta }}(1-|z|^2)^{{\alpha }+{\beta }+1}\Vert g\Vert _\infty , \end{aligned}$$

where $\Vert f\Vert _{\infty }=\sup _{\xi \in {\mathbb T}}|f(\xi )|$, $\Vert g\Vert _{\infty }=\sup _{z\in {\mathbb D}}|g(z)|$, $d_{{\alpha },{\beta }}:=2^{|{\alpha }+{\beta }|-2}+\frac{\Gamma ({\alpha }+1) \Gamma ({\beta }+1)}{\Gamma ^2(\frac{{\alpha }+{\beta }+4}{2})}$, $U_{{\alpha }+{\beta }}$ is defined by (1.18) and $c_{{\alpha },{\beta }}$, $c_{{\alpha }+{\beta }}$ are defined in (1.3) and (1.5).

Next, we give the Schwarz–Pick inequality for the solutions of the $({\alpha },{\beta })$-harmonic equation extending [16, Theorem 1.2]

Theorem 1.2

Suppose that $g\in \mathcal {C}({\mathbb D})$ and $f\in \mathcal {C}({\mathbb T})$. If $u\in \mathcal {C}^2({\mathbb D})$ satisfies the $({\alpha },{\beta })$-harmonic equation (1.1) for ${\alpha },{\beta }\in (-1,\infty )$ such that ${\alpha }+{\beta }>-1$, then for $z \in {\mathbb D}$,

$$\begin{aligned} \Vert Du(z)\Vert\le & {} \tau _{{\alpha },{\beta }}\frac{\Vert f\Vert _\infty }{1-|z|^2}+\sigma _{{\alpha },{\beta }}(1-|z|^2)^{{\alpha }+{\beta }}\Vert g\Vert _\infty , \end{aligned}$$

where

$$\begin{aligned} \tau _{{\alpha },{\beta }}:=(|{\alpha }|+|{\alpha }+1|+|{\beta }|+|{\beta }+1|)\frac{|c_{{\alpha },{\beta }}|}{c_{{\alpha }+{\beta }}}, \end{aligned}$$

and

$$\begin{aligned} \sigma _{{\alpha },{\beta }}:= & {} 5 \left( |{\alpha }|+|{\beta }|\right) 2^{|{\alpha }+{\beta }|-1}+2^{{\alpha }+{\beta }+2}({\alpha }+{\beta }+1))\textrm{B}({\alpha }+1,{\beta }+1)\\ {}{} & {} \quad + 2^{{\alpha }+{\beta }+2}({\alpha }+{\beta }+2). \end{aligned}$$

2 Schwarz Lemma for $({\alpha },{\beta })$-Harmonic Functions

To prove Schwarz lemma, we will distinguish two cases:

2.1 Case ${\alpha }+{\beta }\ge 0$

In this case, we write the generalized Poisson kernel $P_{{\alpha },{\beta }}$ in the following form:

$$\begin{aligned} P_{{\alpha },{\beta }}(z)=h_{{\alpha },{\beta }}(z)P(z), \end{aligned}$$

where

$$\begin{aligned} h_{{\alpha },{\beta }}(z):=c_{{\alpha },{\beta }}\frac{(1-|z|^2)^{{\alpha }+{\beta }}}{(1-z)^{\alpha }(1-\overline{z})^{\beta }}, \end{aligned}$$

and P is the Poisson kernel. As ${\alpha }+{\beta }\ge 0$, we have

$$\begin{aligned} \Vert h_{{\alpha },{\beta }}(r)\Vert _{\infty }:=\sup _{0\le \theta \le 2\pi }|h_{{\alpha },{\beta }}(re^{i\theta })|\le |c_{{\alpha },{\beta }}|2^{{\alpha }+{\beta }}. \end{aligned}$$

(2.1)

Let u be an $({\alpha },{\beta })$-harmonic mapping from the unit disc to itself. Then, we can write

$$\begin{aligned} u(z)=\frac{1}{2\pi }\int _{0}^{2\pi } P_{{\alpha },{\beta }}(ze^{-i\theta })f(e^{i\theta })\textrm{d}\theta , \end{aligned}$$

where f is the boundary function of u.

Let

$$\begin{aligned} H_{{\alpha },{\beta }}(z)=h_{{\alpha },{\beta }}[f](z)=\frac{1}{2\pi }\int _{0}^{2\pi }h_{{\alpha },{\beta }}(ze^{-i\theta }) f(e^{i\theta })\textrm{d}\theta . \end{aligned}$$

(2.2)

As in [12], we prove the following lemma.

Lemma 2.1

Let ${\alpha },{\beta }\in {\mathbb R}\setminus {\mathbb Z}^-$ such that ${\alpha }+{\beta }\ge 0$ and u be an $({\alpha },{\beta })$-harmonic function from the unit disc to itself. Then

$$\begin{aligned} \bigg |u(z)-\frac{1-|z|^2}{1+|z|^2}H_{{\alpha },{\beta }}(z)\bigg |\le \frac{ |c_{{\alpha },{\beta }}|2^{{\alpha }+{\beta }+2}}{\pi }\arctan |z|. \end{aligned}$$

Proof

We have

$$\begin{aligned} \bigg |u(z)-\frac{1-|z|^2}{1+|z|^2}H_{{\alpha },{\beta }}(z)\bigg |\le & {} \frac{1}{2\pi } \int ^{2\pi }_0\bigg |P(ze^{-i\theta })-\frac{1-|z|^2}{1+|z|^2}\bigg ||h_{{\alpha },{\beta }} (ze^{-i\theta })|f(e^{i\theta })|\textrm{d}\theta .\\\le & {} \frac{4}{\pi }\Vert h_{{\alpha },{\beta }}(|z|)\Vert _{\infty }\arctan |z|, \end{aligned}$$

and the conclusion follows from (2.1) and (1.17). $\square$

Next, we prove

Lemma 2.2

Let ${\alpha },{\beta }\in {\mathbb R}$ and $r \in [0, 1)$. Then

• (1) If ${\alpha }+{\beta }\not =0$, then

  $$\begin{aligned} \frac{1}{2\pi }\int ^{2\pi }_0|(1-re^{it})^{-{\alpha }}(1-re^{-it})^{-{\beta }}-1|\, \textrm{d}t \le \frac{|{\alpha }|+|{\beta }|}{|{\alpha }+{\beta }|}\left| (1-r)^{-{\alpha }-{\beta }}-1\right| . \end{aligned}$$

• (2) If ${\alpha }+{\beta }=0$, then

  $$\begin{aligned} \frac{1}{2\pi }\int ^{2\pi }_0|(1-re^{it})^{-{\alpha }}(1-re^{-it})^{-{\beta }}-1|\, \textrm{d}t\le \frac{|{\alpha }|\pi }{4}. \end{aligned}$$

Proof

Let

$$\begin{aligned} g(r,t)=(1-re^{it})^{-{\alpha }}(1-re^{-it})^{-{\beta }}. \end{aligned}$$

Differentiating g with respect r, we get

$$\begin{aligned} \partial _rg(r,t)=(1-re^{it})^{-{\alpha }-1}(1-re^{-it})^{-{\beta }-1}({\alpha }e^{it}+{\beta }e^{-it}-r({\alpha }+{\beta })). \end{aligned}$$

(1) For ${\alpha }+{\beta }\ne 0$, we have

$$\begin{aligned} |\partial _rg(r,t)|\le & {} (|{\alpha }|+|{\beta }|)|1-re^{it}|^{-({\alpha }+{\beta })-1}\le (|{\alpha }|+|{\beta }|)(1-r)^{-({\alpha }+{\beta })-1}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} g(r,t)-g(0,t)=\int _0^{r}\partial _xg(x,t)\textrm{d}x. \end{aligned}$$

Then

$$\begin{aligned} |g(r,t)-g(0,t)|\le & {} \int _0^{r}|\partial _xg(x,t)|\textrm{d}x\\\le & {} \int _0^{r}(|{\alpha }|+|{\beta }|)(1-x)^{-({\alpha }+{\beta })-1}\textrm{d}x\\= & {} \frac{|{\alpha }|+|{\beta }|}{{\alpha }+{\beta }}((1-r)^{-({\alpha }+{\beta })}-1). \end{aligned}$$

Hence

$$\begin{aligned} \frac{1}{2\pi }\int ^{2\pi }_0|(1-re^{it})^{-{\alpha }}(1-re^{-it})^{-{\beta }}-1|\textrm{d}t\le \frac{|{\alpha }|+|{\beta }|}{|{\alpha }+{\beta }|} |(1-r)^{-{\alpha }-{\beta }}-1|. \end{aligned}$$

(2) For ${\alpha }+{\beta }=0$, we have

$$\begin{aligned} |\partial _rg(r,t)|\le \frac{2|{\alpha }||\sin t|}{|1-re^{it}|^2}= \frac{2|{\alpha }||\sin t|}{1-2r \cos t+r^2}. \end{aligned}$$

Thus

$$\begin{aligned} |g(r,t)-g(0,t)|\le & {} 2|{\alpha }||\sin t|\int _0^{1}\frac{\textrm{d}x}{1-2 x\cos t +x^2}. \end{aligned}$$

One can check the following two integrals

$$\begin{aligned} \int _{0}^{2\pi } \frac{|\sin t| }{1-2x \cos t+x^2} \textrm{d}t= & {} \frac{1}{x}\ln \left( \frac{1+x}{1-x}\right) ,\\ \int _{0}^{1} \frac{1}{x}\ln \left( \frac{1+x}{1-x}\right) \textrm{d}x= & {} \textrm{Li}_2(1)-\textrm{Li}_2(-1)=\frac{\pi ^2}{4}, \end{aligned}$$

where $\textrm{Li}_2$ is the polylogarithm function. By Fubini theorem, it yields

$$\begin{aligned} \frac{1}{2\pi }\int ^{2\pi }_0|(1-re^{it})^{-{\alpha }}(1-re^{-it})^{{\alpha }}-1|\textrm{d}t\le \frac{|{\alpha }|\pi }{4}. \end{aligned}$$

$\square$

We establish the following Schwarz lemma for $({\alpha },{\beta })$-harmonic functions in the case ${\alpha }~+~{\beta }~\ge ~0$.

Theorem 2.1

Let ${\alpha },{\beta }\in {\mathbb R}\setminus {\mathbb Z}^-$ such that ${\alpha }+{\beta }\ge 0$ and u be an $({\alpha },{\beta })$-harmonic function from the unit disc ${\mathbb D}$ into itself, then

• (1) If ${\alpha }+{\beta }>0$, then

  $$\begin{aligned}{} & {} \left| u(z)- \frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{1+|z|^2}u(0)\right| \nonumber \\{} & {} \quad \le \frac{|c_{{\alpha },{\beta }}|2^{{\alpha }+{\beta }+2}}{\pi }\arctan |z| +\, \frac{2^{{\alpha }+{\beta }+1}|c_{{\alpha },{\beta }}|(|{\alpha }|+|{\beta }|)}{{\alpha }+{\beta }} (1-|z|) \left( 1-(1-|z|)^{{\alpha }+{\beta }}\right) .\nonumber \\ \end{aligned}$$

  (2.3)

• (2) If ${\alpha }+{\beta }=0$, then

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

Proof

Let

$$\begin{aligned} \Psi _{{\alpha },{\beta }}(z):=\frac{H_{{\alpha },{\beta }}(z)}{(1-|z|^2)^{{\alpha }+{\beta }}} =\frac{c_{{\alpha },{\beta }}}{2\pi }\int _0^{2\pi }\frac{f(e^{i\theta })}{(1-ze^{-i\theta })^{\alpha }(1-\overline{z}e^{i\theta })^{\beta }}\textrm{d}\theta , \end{aligned}$$

where $H_{{\alpha },{\beta }}$ is defined by (2.2) and

$$\begin{aligned} \Phi _{{\alpha },{\beta }}(z):=\Psi _{{\alpha },{\beta }}(z)-u(0)=\frac{c_{{\alpha },{\beta }}}{2\pi } \int _0^{2\pi }\big ((1-ze^{-i\theta })^{-{\alpha }}(1-\overline{z}e^{i\theta })^{-{\beta }}-1 \big )f(e^{i\theta })\textrm{d}\theta . \end{aligned}$$

(2.4)

It yields that

$$\begin{aligned} u(z)- \frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{1+|z|^2}u(0)=\bigg (u(z)-\frac{1-|z|^2}{1+|z|^2}H_{{\alpha },{\beta }}(z) \bigg )+\frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{1+|z|^2}\Phi _{{\alpha },{\beta }}(z). \end{aligned}$$

Since ${\alpha }+{\beta }\ge 0$, by Lemma 2.1, we get

$$\begin{aligned} \bigg |u(z)- \frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{1+|z|^2}u(0)\bigg |\le \frac{|c_{{\alpha },{\beta }}|2^{{\alpha }+{\beta }+2}}{\pi }\arctan |z|+(1-|z|^2)^{{\alpha }+{\beta }+1} |\Phi _{{\alpha },{\beta }}(z)|, \end{aligned}$$

and the conclusion follows from Lemma 2.2 to estimate $\Phi _{{\alpha },{\beta }}$. $\square$

2.2 Case ${\alpha }+{\beta }\in (-1,0)$

In the case $-1<{\alpha }+{\beta }<0$, we write the kernel $P_{{\alpha },{\beta }}$ in the following form

$$\begin{aligned} P_{{\alpha },{\beta }}(z)=\frac{k_{{\alpha },{\beta }}(z)}{c_{{\alpha }+{\beta }}} P_{{\alpha }+{\beta }}(z), \end{aligned}$$

(2.5)

where

$$\begin{aligned} k_{{\alpha },{\beta }}(z):= c_{{\alpha },{\beta }}(1-z)^{\frac{{\beta }-{\alpha }}{2}}(1-\overline{z})^{\frac{{\alpha }-{\beta }}{2}} \end{aligned}$$

(2.6)

and $P_{{\alpha }+{\beta }}$ is the Poisson kernel for $T_{{\alpha }+{\beta }}$-harmonic functions defined by equations (1.4) and (1.5). Remark that $|k_{{\alpha },{\beta }}(z)|=|c_{{\alpha },{\beta }}|$.

Let

$$\begin{aligned} K_{{\alpha },{\beta }}(z):=k_{{\alpha },{\beta }}[f](z)=\frac{1}{2\pi }\int _{0}^{2\pi } k_{{\alpha },{\beta }}(ze^{-i\theta })f(e^{i\theta })\textrm{d}\theta . \end{aligned}$$

First, we prove the following lemma.

Lemma 2.3

Let ${\alpha },{\beta }\in {\mathbb R}\setminus {\mathbb Z}^-$ such that ${\alpha }+{\beta }\in (-1,0)$ and u be an $({\alpha },{\beta })$-harmonic function from the unit disc to itself. Then

$$\begin{aligned} \bigg |u(z)-\frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{(1+|z|^2)^{\frac{{\alpha }+{\beta }}{2}+1}}K_{{\alpha },{\beta }}(z)\bigg |\le \frac{|c_{{\alpha },{\beta }}|}{c_{{\alpha }+{\beta }}} U_{{\alpha }+{\beta }}(|z|). \end{aligned}$$

We omit the proof as it is similar to Lemma 2.1 and uses Theorem C, a Heinz-Hethcote theorem of $T_{\alpha }$-harmonic functions.

Next, we show

Theorem 2.2

Let ${\alpha },{\beta }\in {\mathbb R}\setminus {\mathbb Z}^-$ such that ${\alpha }+{\beta }\in (-1,0)$ and u be an $({\alpha },{\beta })$-harmonic function from the unit disc to itself. Then

$$\begin{aligned}{} & {} \left| u(z)-\frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{(1+|z|^2)^{\frac{{\alpha }+{\beta }}{2}+1}}u(0) \right| \\{} & {} \quad \le |c_{{\alpha },{\beta }}|\left[ \frac{1}{c_{{\alpha }+{\beta }}} U_{{\alpha }+{\beta }}(|z|)+ \frac{|{\alpha }-{\beta }|\pi }{4} (1-|z|^2)^{{\alpha }+{\beta }+1}\right] . \end{aligned}$$

Proof

Using the triangle inequality, we have

$$\begin{aligned} \left| u(z)-\frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{(1+|z|^2)^{\frac{{\alpha }+{\beta }}{2}+1}}u(0) \right|\le & {} \bigg |u(z)-\frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{(1+|z|^2)^{\frac{{\alpha }+{\beta }}{2}+1}}K_{{\alpha },{\beta }}(z)\bigg |\nonumber \\{} & {} +\, (1-|z|^2)^{{\alpha }+{\beta }+1}\left| K_{{\alpha },{\beta }}(z)-u(0)\right| . \end{aligned}$$

(2.7)

We observe that

$$\begin{aligned} |\Phi _{\frac{{\beta }-{\alpha }}{2},\frac{{\alpha }-{\beta }}{2}}(z)|=\left| K_{{\alpha },{\beta }}(z)-u(0)\right| , \end{aligned}$$

where $\Phi _{{\alpha },{\beta }}$ is given by (2.4). Thus we can use the second case in Lemma 2.2 to obtain

$$\begin{aligned} \left| K_{{\alpha },{\beta }}(z)-u(0)\right| \le |c_{{\alpha },{\beta }}|\frac{|{\alpha }-{\beta }|\pi }{4}. \end{aligned}$$

(2.8)

Then, with an immediate consequence from Lemma 2.3 and the inequality (2.8) we obtain the desired result. $\square$

2.3 Estimates of $\mathcal {G}_{{\alpha },{\beta }}[g]$ and Its Derivatives

Lemma 2.4

Let $\gamma \in {\mathbb R}$ and z, w in ${\mathbb D}$. Then

$$\begin{aligned} \frac{(1-|z|^2)(1-|w|^2)^{\gamma +1}}{|1-\overline{z}w|^{\gamma +2}} \le 2^{|\gamma |}. \end{aligned}$$

Proof

Let us denote by $\displaystyle F_{\gamma }(z,w):= \frac{(1-|z|^2)(1-|w|^2)^{\gamma +1}}{|1-\overline{z}w|^{\gamma +2}}.$

If $\gamma \ge 0$, then

$$\begin{aligned} F_{\gamma } (z,w) = (1-|\varphi _w(z)|^2) \frac{(1-|w|^2)^{\gamma }}{|1-\overline{z}w|^{\gamma }}\le (1+|w|)^{\gamma } \le 2^\gamma . \end{aligned}$$

If $\gamma <0$, then

$$\begin{aligned} F_{\gamma } (z,w) = (1-|\varphi _w(z)|^2)^{\gamma +1} \frac{|1-\overline{z}w|^{\gamma }}{(1-|z|^2)^{\gamma }}\le (1+|z|)^{-\gamma } \le 2^{-\gamma }. \end{aligned}$$

$\square$

First, we estimate the Green functions $g_{{\alpha },{\beta }}$ and $G_{{\alpha },{\beta }}$.

Lemma 2.5

Let ${\alpha },{\beta }\in (-1,\infty )$ such that ${\alpha }+{\beta }>-1$. Then, the functions $g_{{\alpha },{\beta }}$ and $G_{{\alpha },{\beta }}$ satisfy the estimates

$$\begin{aligned} 0\le g_{{\alpha },{\beta }}(x)\le (1-x)^{{\alpha }+{\beta }+1}\bigg (\textrm{B}({\alpha }+1,{\beta }+1)+\log \frac{1}{x}\bigg ), \quad x\in (0,1]. \end{aligned}$$

(2.9)

and

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

(2.10)

The estimates (2.9) and (2.10) extend [16, Lemma B] and [16, Lemma 4.2], respectively.

Proof

From Lemma 1.3, we observe that the function

$$\begin{aligned} x \mapsto \frac{F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+2,x)-1}{\log \frac{1}{1-x}} \end{aligned}$$

is increasing from (0, 1) into $(\frac{({\alpha }+1)({\beta }+1)}{{\alpha }+{\beta }+2},\frac{1}{\textrm{B}({\alpha }+1,{\beta }+1)})$. Then

$$\begin{aligned} F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+2,x)\le \frac{1}{\textrm{B}({\alpha }+1,{\beta }+1)}\log \frac{1}{1-x}+1. \end{aligned}$$

(2.11)

Hence

$$\begin{aligned} g_{{\alpha },{\beta }}(x)\le (1-x)^{{\alpha }+{\beta }+1}\bigg ( \log \frac{1}{x}+\textrm{B}({\alpha }+1,{\beta }+1)\bigg ). \end{aligned}$$

The estimate of $G_{\ {\alpha },{\beta }}$ follows immediately from (2.9)

Remark

The inequalities (2.9) and (2.10) hold for ${\alpha },{\beta }\in {\mathbb R}{\setminus } {\mathbb Z}^-$ such that ${\alpha }+{\beta }>-1$ where the constant $\frac{1}{\textrm{B}({\alpha }+1,{\beta }+1)}$ in (2.11) should be replaced by

$$\begin{aligned} M_{{\alpha },{\beta }}:=\sup _{x\in (0,1)}\frac{F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+2,x)-1}{\log \frac{1}{1-x}}. \end{aligned}$$

$M_{{\alpha },{\beta }}$ is finite as the function $\displaystyle \frac{F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+2,x)-1}{\log \frac{1}{1-x}}$ is continuous on (0, 1) having finite limits at 0 and 1, see (1.9). Lemma 1.3 says simply that if ${\alpha },{\beta }>-1$ and ${\alpha }+{\beta }>-1$, then $M_{{\alpha },{\beta }}= \frac{1}{\textrm{B}({\alpha }+1,{\beta }+1)}$.

Proposition 2.1

Let ${\alpha },{\beta }\in (-1,\infty )$ such that ${\alpha }+{\beta }>-1$ and $g\in \mathcal {C}(\overline{{\mathbb D}})$. Then

$$\begin{aligned} \big |\mathcal {G}_{{\alpha },{\beta }}[g](z)\big |\le d_{{\alpha },{\beta }}(1-|z|^2)^{{\alpha }+{\beta }+1} \Vert g\Vert _\infty , \end{aligned}$$

where

$$\begin{aligned} d_{{\alpha },{\beta }}:=2^{|{\alpha }+{\beta }|-2}+\frac{\Gamma ({\alpha }+1) \Gamma ({\beta }+1)}{\Gamma ^2(\frac{{\alpha }+{\beta }+4}{2})}. \end{aligned}$$

Proof

Combining (2.10) and Lemma 2.4, we get

$$\begin{aligned} |G_{{\alpha },{\beta }}(z,w)|\le & {} 2^{|{\alpha }+{\beta }|} (1-|z|^2)^{{\alpha }+{\beta }}\bigg ( \log \bigg |\frac{1-\overline{z}w}{z-w}\bigg |^2\bigg ) \\{} & {} \quad +\, \textrm{B}({\alpha }+1,{\beta }+1) \frac{(1-|z|^2)^{{\alpha }+{\beta }+1}(1-|w|^2)^{{\alpha }+{\beta }+1}}{|1-\overline{z}w|^{{\alpha }+{\beta }+2}}. \end{aligned}$$

Hence

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

Let denote by

$$\begin{aligned} \mathcal {I}:=\int _{{\mathbb D}}\log \bigg |\frac{1-z\overline{w}}{z-w}\bigg |^2\textrm{d}A(w) \end{aligned}$$

and

$$\begin{aligned} \mathcal {J}:=\int _{{\mathbb D}}\frac{(1-|w|^2)^{{\alpha }+{\beta }+1}}{|1-\overline{z}w|^{{\alpha }+{\beta }+2}}\textrm{d}A(w). \end{aligned}$$

As $\mathcal {I}$ is the Green function of the Laplacian operator, we deduce that

$$\begin{aligned} \mathcal {I}= \frac{(1-|z|^2)}{4}. \end{aligned}$$

(2.12)

Now we estimate $\mathcal {J}$. By using the estimate (1.10) in Theorem 1.2 for $t={\alpha }+{\beta }+1$ and $c=-1$, we have

$$\begin{aligned} \mathcal {J}=J_{-1,{\alpha }+{\beta }+1}\le \frac{\Gamma ({\alpha }+{\beta }+2)}{\Gamma ^2(\frac{{\alpha }+{\beta }+4}{2})}, \end{aligned}$$

and, we reach our conclusion. $\square$

Next, we estimate $|\partial _z G_{{\alpha },{\beta }}(z,w)|$.

Lemma 2.6

Let ${\alpha },{\beta }\in (-1,\infty )$ such that ${\alpha }+{\beta }>-1$. Then

$$\begin{aligned} |\partial _z G_{{\alpha },{\beta }}(z,w)|\le & {} |{\beta }|2^{|{\alpha }+{\beta }|+1}(1-|z|^2)^{{\alpha }+{\beta }-1} \log \bigg |\frac{1-\overline{z}w}{z-w}\bigg |^2\\{} & {} \quad +\gamma _{{\alpha },{\beta }}\frac{(1-|z|^2)^{{\alpha }+{\beta }}}{|z-w|}. \end{aligned}$$

where

$$\begin{aligned} \gamma _{{\alpha },{\beta }}:=(|{\beta }|2^{|{\alpha }+{\beta }|}+2^{{\alpha }+{\beta }+1}({\alpha }+{\beta }+1))\textrm{B}({\alpha }+1,{\beta }+1)+ 2^{{\alpha }+{\beta }+1}({\alpha }+{\beta }+2). \end{aligned}$$

(2.13)

Proof

Using the chain rule and (1.8), we get

$$\begin{aligned} \partial _z G_{{\alpha },{\beta }}(z,w)=-\beta \overline{w}\frac{G_{{\alpha },{\beta }}(z,w)}{1-z\overline{w}}+l(z,w)H\left( 1-|\varphi _w(z)|^2\right) , \end{aligned}$$

where

$$\begin{aligned} l(z,w):=\frac{(1-|z|^2)^{{\alpha }+{\beta }}(1-|w|^2)^{{\alpha }+{\beta }+1}(\overline{w}-\overline{z})}{ (1-\overline{z}w)^{{\beta }+1}(1-z\overline{w})^{{\alpha }+2}}, \end{aligned}$$

and

$$\begin{aligned} H(x)= & {} \textrm{B}({\alpha }+1,{\beta }+1)\Big (({\alpha }+{\beta }+1)F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+2;x)\Big )\\{} & {} \quad +\, \textrm{B}({\alpha }+1,{\beta }+1)\Big (\frac{x({\alpha }+1)({\beta }+1)}{{\alpha }+{\beta }+2}F({\alpha }+2,{\beta }+2;{\alpha }+{\beta }+3;x)\Big ). \end{aligned}$$

Claim:

$$\begin{aligned} H(x)\le ({\alpha }+{\beta }+2)\frac{x}{1-x}+({\alpha }+{\beta }+1)\textrm{B}({\alpha }+1,{\beta }+1). \end{aligned}$$

(2.14)

Indeed, using (1.7), we have

$$\begin{aligned} F({\alpha }+2,{\beta }+2;{\alpha }+{\beta }+3;x)=\frac{1}{1-x}F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+3;x). \end{aligned}$$

As the function $F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+3;.)$ is increasing on (0, 1), then by (1.6), we have

$$\begin{aligned}{} & {} \frac{x({\alpha }+1)({\beta }+1)}{{\alpha }+{\beta }+2}F({\alpha }+2,{\beta }+2;{\alpha }+{\beta }+3;x)\nonumber \\{} & {} \qquad \le \frac{({\alpha }+1)({\beta }+1)}{{\alpha }+{\beta }+2}\frac{\Gamma ({\alpha }+{\beta }+3)}{\Gamma ({\alpha }+2)\Gamma ({\beta }+2)}\frac{x}{1-x}\nonumber \\{} & {} \qquad =\frac{1}{\textrm{B}({\alpha }+1,{\beta }+1)}\frac{x}{1-x}. \end{aligned}$$

(2.15)

On the other hand, by Lemma 1.3, we have

$$\begin{aligned} F({\alpha }+1,{\beta }+1;{\alpha }+{\beta }+2;x)\le \bigg (\frac{1}{\textrm{B}({\alpha }+1,{\beta }+1)}\log \frac{1}{1-x}+1\bigg ). \end{aligned}$$

(2.16)

Hence, by combining (2.15) and (2.16), we obtain

$$\begin{aligned} H(x)\le & {} ({\alpha }+{\beta }+1)\Big (\log \frac{1}{1-x}+\textrm{B}({\alpha }+1,{\beta }+1)\Big )+\frac{x}{1-x}. \end{aligned}$$

Using $\log (t) \le t-1$ for all $t\ge 1$, one can see that $\displaystyle \log \frac{1}{1-x}\le \frac{x}{1-x}$ for all $x\in [0,1).$ Thus the proof of the claim is complete.

It follows from the inequality (2.10) and Lemma 2.4 that

$$\begin{aligned} \bigg |-{\beta }\overline{w}\frac{G_{{\alpha },{\beta }}(z,w)}{1-z\overline{w}}\bigg |\le & {} |{\beta }|\frac{(1-|z|^2)^{{\alpha }+{\beta }+1}(1-|w|^2)^{{\alpha }+{\beta }+1}}{|1-z\overline{w}|^{{\alpha }+{\beta }+3}}\nonumber \\{} & {} \bigg ( \log \bigg |\frac{1-\overline{z}w}{z-w}\bigg |^2+\, \textrm{B}({\alpha }+1,{\beta }+1)\bigg )\nonumber \\\le & {} |{\beta }|2^{|{\alpha }+{\beta }|+1}(1-|z|^2)^{{\alpha }+{\beta }-1} \log \bigg |\frac{1-\overline{z}w}{z-w}\bigg |^2\nonumber \\{} & {} +\, |{\beta }|2^{|{\alpha }+{\beta }|}\textrm{B}({\alpha }+1,{\beta }+1)\frac{(1-|z|^2)^{{\alpha }+{\beta }}}{|1-z\overline{w}|}. \end{aligned}$$

(2.17)

Also, we have

$$\begin{aligned} |l(z,w)|= & {} \frac{(1-|z|^2)^{{\alpha }+{\beta }}(1-|w|^2)^{{\alpha }+{\beta }+1}|z-w|}{|1-z\overline{w}|^{{\alpha }+{\beta }+3}}\nonumber \\\le & {} \frac{(1-|z|^2)^{{\alpha }+{\beta }}}{|1-z\overline{w}|} \left( \frac{1-|w|^2}{|1-z\overline{w}|}\right) ^{{\alpha }+{\beta }+1} \left| \frac{z-w}{1-z\overline{w}}\right| \nonumber \\ {}\le & {} 2^{{\alpha }+{\beta }+1} \frac{(1-|z|^2)^{{\alpha }+{\beta }}}{|1-z\overline{w}|}. \end{aligned}$$

(2.18)

By the claim (2.14), we have

$$\begin{aligned} H\left( 1-|\varphi (z,w)|^2\right) \le ({\alpha }+{\beta }+2)\frac{(1-|z|^2)(1-|w|^2)}{|z-w|^2}+({\alpha }+{\beta }+1)\textrm{B}({\alpha }+1,{\beta }+1). \end{aligned}$$

Therefore,

$$\begin{aligned} |l(z,w)|H(1-|\varphi (z,w)|^2)\le & {} 2^{{\alpha }+{\beta }+1}({\alpha }+{\beta }+2)\frac{(1-|z|^2)^{{\alpha }+{\beta }}}{|z-w|}\\{} & {} +\, 2^{{\alpha }+{\beta }+1}({\alpha }+{\beta }+1)\textrm{B}({\alpha }+1,{\beta }+1)\frac{(1-|z|^2)^{{\alpha }+{\beta }}}{|1-z\overline{w}|}. \end{aligned}$$

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

Theorem E

[27] Suppose that X is an open subset of ${\mathbb R}$, and $\Omega$ is 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 $\frac{\partial F(x,w)}{\partial x}$ exists almost everywhere.]

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

  $$\begin{aligned} \int _a^b\int _{\Omega }\bigg |\frac{\partial }{\partial x}F(x,w)\bigg |\textrm{d}w\textrm{d}x<\infty . \end{aligned}$$

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

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

Proposition 2.2

Let ${\alpha },{\beta }\in (-1,\infty )$ such that ${\alpha }+{\beta }>-1$ and $g\in \mathcal {C}(\overline{{\mathbb D}})$. Then

$$\begin{aligned} \big |D{\mathcal {G}[g]}(z)\big |\le \delta _{{\alpha },{\beta }}(1-|z|^2)^{{\alpha }+{\beta }}\Vert g\Vert _\infty . \end{aligned}$$

where $\delta _{{\alpha },{\beta }}=2^{|{\alpha }+{\beta }|-1}\left( |{\alpha }|+|{\beta }|\right) +2(\gamma _{{\alpha },{\beta }}+\gamma _{{\beta },{\alpha }})$ and $\gamma _{{\alpha },{\beta }}$ is defined in Eq. (2.13).

Proof

Using Lemma (2.6), we have

$$\begin{aligned} \int _{{\mathbb D}}|\partial _z G_{{\alpha },{\beta }}(z,w)|\textrm{d}A(w)\le & {} |{\beta }|2^{|{\alpha }+{\beta }|+1}(1-|z|^2)^{{\alpha }+{\beta }-1}\int _{{\mathbb D}}\log \bigg |\frac{1-\overline{z}w}{z-w}\bigg |^2\textrm{d}A(w)\\{} & {} +\, \gamma _{{\alpha },{\beta }}(1-|z|^2)^{{\alpha }+{\beta }}\int _{{\mathbb D}}\frac{1}{|z-w|}\textrm{d}A(w). \end{aligned}$$

Using [28, proof of theorem 1.1], we have

$$\begin{aligned} \int _{\mathbb D}\frac{1}{|z-w|}\textrm{d}A(w)\le 2, \end{aligned}$$

and by (2.12), it yields

$$\begin{aligned} \int _{{\mathbb D}}\log \bigg |\frac{1-z\overline{w}}{z-w}\bigg |^2\textrm{d}A(w)=\frac{1-|z|^2}{4}. \end{aligned}$$

Thus $\partial _z G_{{\alpha },{\beta }}(z,w)$ is integrable on ${\mathbb D}\times {\mathbb D}$ and by Theorem E, we have

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

We conclude that

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

Similarly we obtain

$$\begin{aligned} \big |\partial _{\overline{z}}\mathcal {G}[g](z)\big |\le \left( |{\alpha }|2^{|{\alpha }+{\beta }|-1}+2\gamma _{{\beta },{\alpha }}\right) (1-|z|^2)^{{\alpha }+{\beta }}\Vert g\Vert _\infty . \end{aligned}$$

Thus, the proof is complete. $\square$

3 Proofs of Main Results

Proof of Theorem 1.1

The proof of Theorem 1.1 follows immediately from Theorems 2.1, 2.2 and Proposition 2.1.

Proof of Theorem 1.2

Differentiating $P_{{\alpha },{\beta }}$ with respect to z and $\overline{z}$, we get

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

(3.1)

and

$$\begin{aligned} \partial _{\overline{z}}P_{{\alpha },{\beta }}(z)=\bigg (-({\alpha }+{\beta }+1)\frac{z}{1-|z|^2}+({\beta }+1) \frac{1}{1-\overline{z}}\bigg )P_{{\alpha },{\beta }}. \end{aligned}$$

(3.2)

Therefore

$$\begin{aligned} \partial _z\mathcal {P}_{{\alpha },{\beta }}[f](z)=\frac{1}{2\pi }\int _{0}^{2\pi } \partial _zP_{{\alpha },{\beta }}(ze^{-i\theta })e^{-i\theta }f(e^{i\theta })\,\textrm{d}\theta , \end{aligned}$$

and

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

Hence, by using (3.1) and (3.2), we obtain

$$\begin{aligned} |\partial _z {\mathcal {P}_{{\alpha },{\beta }}}[f](z)|\le & {} \frac{1}{2\pi }\int _{0}^{2\pi } \bigg | \partial _zP_{{\alpha },{\beta }}(ze^{-i\theta })e^{-i\theta }f(e^{i\theta })\bigg |\textrm{d}\theta \nonumber \\\le & {} \frac{\Vert f\Vert _\infty }{2\pi }\int _{0}^{2\pi }\bigg |\frac{({\alpha }+1)(1 -\overline{z}e^{i\theta })-{\beta }\overline{z}e^{i\theta }(1-ze^{-i\theta })}{(1-ze^{-i\theta })(1-|z|^2)} |P_{{\alpha },{\beta }}(ze^{-i\theta })|\textrm{d}\theta \nonumber \\\le & {} \frac{\Vert f\Vert _\infty }{2\pi }\frac{1}{1-|z|^2}\int _{0}^{2\pi }(|{\alpha }+1|+|{\beta }|)|P_{{\alpha },{\beta }}(ze^{-i\theta }) |\textrm{d}\theta \nonumber \\\le & {} \frac{\Vert f\Vert _\infty \big ((|{\alpha }+1|+|{\beta }|\big )}{1-|z|^2}\frac{|c_{{\alpha },{\beta }}|}{2\pi }\int _{0}^{2\pi } \frac{(1-|z|^2)^{{\alpha }+{\beta }+1}}{|1-ze^{-i\theta }|^{{\alpha }+{\beta }+2}}\textrm{d}\theta . \end{aligned}$$

(3.3)

By using the first inequality in Lemma 1.1, we obtain

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

(3.4)

Similarly,

$$\begin{aligned} \Vert \partial _{\overline{z}}{\mathcal {P}_{{\alpha },{\beta }}}[f](z)\Vert\le & {} (|{\beta }+1|+|{\alpha }|)\frac{|c_{{\alpha },{\beta }}|}{c_{{\alpha }+{\beta }}}\frac{\Vert f\Vert _\infty }{1-|z|^2}.\nonumber \\ \Vert D u(z)\Vert\le & {} \Vert D{\mathcal {P}_{{\alpha },{\beta }}[f]}(z)\Vert +\Vert D{\mathcal {G}[g]}(z)\Vert .\nonumber \\ \end{aligned}$$

(3.5)

Combining Proposition 2.2 and (3.4) and (3.5), we get our conclusion and the proof of Theorem 1.2 is complete.