Several properties of $\alpha$-harmonic functions in the unit disk

Abstract

The aim of this paper is to obtain the Schwarz–Pick type inequality for $\alpha$-harmonic functions f in the unit disk and get estimates on the coefficients of f. As an application, a Landau type theorem of $\alpha$-harmonic functions is established.

1 Introduction and main results

Let $\mathbb {C}$ be the complex plane. For $a\in \mathbb {C}$, let $\mathbb {D}(a,r)=\{z:|z-a|<r\}$ $(r>0)$ and $\mathbb {D}(0,r)=\mathbb {D}_{r}$. Also, we use the notations $\mathbb {D}=\mathbb {D}_1$ and $\mathbb {T}=\partial \mathbb {D}$, the boundary of $\mathbb {D}$.

Let

$$\begin{aligned} A=\left( \begin{array}{cc} a &{} b \\ c &{} d \\ \end{array} \right) \in \mathbb {R}^{2\times 2}. \end{aligned}$$

We will consider the matrix norm

$$\begin{aligned} |A|=\sup \{|Az|:\; z\in \mathbb {C},\;|z|=1\} \end{aligned}$$

and the matrix function

$$\begin{aligned} l(A)=\inf \{|Az|:\;z\in \mathbb {C},\;|z|=1\}. \end{aligned}$$

Let D and $\Omega$ be domains in $\mathbb {C}$, and let $f=u+iv$: $D\rightarrow \Omega$ be a function that has both partial derivatives at $z=x+iy$ in D, where u and v are real functions. The Jacobian matrix of f at z is denoted by

$$\begin{aligned} D f(z) = \left( \begin{array}{cc} u_x &{} u_y \\ v_x &{} v_y \\ \end{array} \right) . \end{aligned}$$

Set

$$\begin{aligned} \frac{\partial }{\partial {z}}=\frac{1}{2}\Big (\frac{\partial }{\partial {x}}-i\frac{\partial }{\partial {y}}\Big )\quad \hbox {and}\quad \frac{\partial }{\partial {\overline{z}}}=\frac{1}{2}\Big (\frac{\partial }{\partial {x}}+i\frac{\partial }{\partial {y}}\Big ). \end{aligned}$$

Then

$$\begin{aligned} |D f(z)|= & {} \sup \{|D f(z)\varsigma |:\; |\varsigma |=1\}=|f_z(z)|+|f_{\overline{z}}(z)|, \end{aligned}$$

(1.1)

$$\begin{aligned} l(D f(z))= & {} \inf \{|D f(z)\varsigma |:\; |\varsigma |=1\}=\big ||f_z(z)|-|f_{\overline{z}}(z)|\big | \end{aligned}$$

(1.2)

and

$$\begin{aligned} |J_{f}(z)|=|D f(z)|\cdot l(D f(z)), \end{aligned}$$

where $J_{f}(z)$ stands for the Jacobian of f at z.

We denote by $\Delta _{\alpha }$ the weighted Laplace operator corresponding to the so-called standard weight $w_{\alpha }=(1-|z|^{2})^{\alpha }$, that is,

$$\begin{aligned} \Delta _{\alpha ,z}=\frac{\partial }{\partial z}(w_{\alpha })^{-1}\frac{\partial }{\partial \overline{z}}=\frac{\partial }{\partial z}(1-{|{z}|}^{2})^{-\alpha }\frac{\partial }{\partial \overline{z}} \end{aligned}$$

in $\mathbb {D}$, where $\alpha > -1$ (see [20, Proposition 1.5] for the reason for this constraint).

Olofsson and Wittsten [20] introduced this operator $\Delta _{\alpha }$ and a counterpart of the classical Poisson integral formula was given.

We remark that in the study of Bergman spaces of $\mathbb {D}$, one often considers the weights $w_{\alpha }$ in $\mathbb {D}$ $(\alpha >-1)$. For an account of recent developments in Bergman space theory, we mention the monograph by Hedenmalm et al. [13]. The case $\alpha = 0$ is commonly referred to as the unweighted case, whereas the case $\alpha = 1$ has attracted special attention recently with contributions by Hedenmalm, Shimorin and others (see for instance [14,15,16, 22] etc).

Of particular interest to us is the following $\alpha$-harmonic equation in $\mathbb {D}$:

$$\begin{aligned} \Delta _{\alpha }(f)=0. \end{aligned}$$

(1.3)

Denote the associated Dirichlet boundary value problem of functions f satisfying the Eq. (1.3) by

$$\begin{aligned} \left\{ \begin{array}{ll} \Delta _{\alpha }(f)=0&{}\quad \text {in}\;\mathbb {D},\\ f=f^{*}&{}\quad \text {on}\;\mathbb {T}. \end{array} \right. \end{aligned}$$

(1.4)

Here the boundary data $f^{*}$ is a distribution on $\mathbb {T}$, i.e. $f^{*}\in \mathcal {D}^{\prime }(\mathbb {T})$, and the boundary condition in the Eq. (1.4) is to be understood as $f_{r}\rightarrow f^{*}\in \mathcal {D}^{\prime }(\mathbb {T})$ as $r \rightarrow 1^{-}$, where

$$\begin{aligned} f_{r}(e^{i\theta })=f(re^{i\theta }) \end{aligned}$$

(1.5)

for $\theta \in [0, 2\pi ]$ and $r\in [0,1)$.

For simplicity, we introduce the following definition.

Definition 1.1

For $\alpha >-1$, a complex-valued function f is said to be $\alpha$-harmonic if f is twice continuously differentiable in $\mathbb {D}$ and satisfies the condition (1.3).

Olofsson and Wittsten [20] showed that if an $\alpha$-harmonic function f satisfies

$$\begin{aligned} \lim _{r\rightarrow 1^{-}}f_{r}=f^{*} \in \mathcal {D}^{\prime }(\mathbb {T})\;\; (\alpha > -1), \end{aligned}$$

then it has the form of a Poisson type integral

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

(1.6)

in $\mathbb {D}$, where

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

In the following, we always assume that any $\alpha$-harmonic function has such a representation which plays a key role in the discussions of this paper.

Obviously, $\alpha$-harmonicity coincides with harmonicity when $\alpha = 0$. See [12] and the references therein for the properties of harmonic mappings. Particularly, Colonna proved the following Schwarz–Pick type inequality.

Theorem A

[11, Theorems 3 and 4] Let f be a harmonic function of $\mathbb {D}$ into $\mathbb {D}$. Then for $z\in \mathbb {D}$,

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

This estimate is sharp and all the extremal functions are

$$\begin{aligned} f(z)=\frac{2\delta }{\pi }\arg \Big (\frac{1+\psi (z)}{1-\psi (z)}\Big ), \end{aligned}$$

where $\delta \in \mathbb {C}$, $|\delta |=1$ and $\psi$ is a conformal automorphism of $\mathbb {D}$.

For the related discussions on this topic, see [2, 4, 7, 10, 17, 21] etc.

As the first aim of this paper, we shall generalize Theorem A to the case of $\alpha$-harmonic functions. Our first result is as follows.

Theorem 1.1

Suppose that f is an $\alpha$-harmonic function in $\mathbb {D}$ with $\alpha >-1$, that $f^{*}\in C(\mathbb {T})$ and that $\sup _{z\in \overline{\mathbb {D}}}|f(z)|\le M$, where M is a constant. Then for $z\in \mathbb {D}$,

$$\begin{aligned} |D{f}(z)|\le \frac{M(\alpha +2)}{\mathrm {c}_{\alpha }}\cdot \frac{1}{1-|z|}\le \frac{2M(\alpha +2)}{\mathrm {c}_{\alpha }}\cdot \frac{1}{1-|z|^2}, \end{aligned}$$

where $\mathrm {c}_{\alpha }=\frac{(\Gamma (\frac{\alpha }{2}+1))^{2}}{\Gamma (\alpha +1)}$ and $\Gamma (s)=\int ^{\infty }_{0}t^{s-1}e^{-t}dt$ $(s>0)$ is the Gamma function.

In particular, if f maps $\mathbb {D}$ into $\mathbb {D}$, then

$$\begin{aligned} |D{f}(z)|\le \frac{2(\alpha +2)}{\mathrm {c}_{\alpha }}\cdot \frac{1}{1-|z|^2}. \end{aligned}$$

Let $\lambda _{D}(z)|dz|$ be the hyperbolic metric of the domain D having constant Gaussian curvature $-1$. The hyperbolic distance $d_{h_D}(z_1, z_2)$ between two points $z_1$ and $z_2$ in D is defined by

$$\begin{aligned} \inf _{\gamma }\Big \{\int _{\gamma }\lambda _{D}(z)|dz|\Big \}, \end{aligned}$$

where the infimum is taken over all rectifiable curves $\gamma$ in D connecting $z_1$ and $z_2$.

We have known that if $D=\mathbb {D}$, then (cf. [1])

$$\begin{aligned} \lambda _{\mathbb {D}}(z)=\frac{2}{1-|z|^2}\;\;\hbox {and}\;\;d_{h_\mathbb {D}}(z_1, z_2)=\log \frac{|1-z_1\overline{z}_2|+|z_1-z_2|}{|1-z_1\overline{z}_2|-|z_1-z_2|}. \end{aligned}$$

As a consequence of Theorem 1.1, we have

Corollary 1.1

Under the assumptions of Theorem 1.1, if f maps $\mathbb {D}$ into $\mathbb {D}$, then for $z_1$ and $z_2\in \mathbb {D}$,

$$\begin{aligned} |f(z_1)-f(z_2)|\le \frac{\alpha +2}{\mathrm {c}_{\alpha }}d_{h_\mathbb {D}}(z_1, z_2). \end{aligned}$$

In [20], the authors got the following homogeneous expansion of $\alpha$-harmonic functions (see [20, Theorem 1.2]):

A function f in $\mathbb {D}$ is $\alpha$-harmonic if and only if it has the following convergent power series expansion:

$$\begin{aligned} f(z)=\sum ^{\infty }_{k=0}c_{k}z^{k}+\sum ^{\infty }_{k=1}c_{-k}P_{\alpha ,k}(|z|^2)\overline{z}^{k}, \end{aligned}$$

(1.7)

where $P_{\alpha ,k}(x)=\int ^{1}_{0}t^{k-1}(1-tx)^{\alpha }dt$ $(-1<x<1)$ and $\{c_{k}\}^{\infty }_{k=-\infty }$ denotes a sequence of complex numbers with $\lim _{|k|\rightarrow \infty }\sup |c_{k}|^{\frac{1}{|k|}}\le 1.$

The second aim of this paper is to prove the following estimates on coefficients $c_{k}$ and $c_{-k}$.

Theorem 1.2

Suppose that f is an $\alpha$-harmonic function in $\mathbb {D}$ with $\alpha >-1$ and that $\sup _{z\in \mathbb {D}}|f(z)|\le M$, where M is a constant. If f has the series expansion (1.7), then for $k\in \{0,1,2,...\},$

$$\begin{aligned} |c_{k}|\le M, \end{aligned}$$

(1.8)

and for $k\in \{1,2,...\},$

$$\begin{aligned} |c_{k}|+|c_{-k}|B(k, \alpha +1)\le \frac{4M}{\pi }, \end{aligned}$$

(1.9)

where B(p, q) denotes the Beta function.

By [20, Definition 2.1], we find that

$$\begin{aligned} \mathcal {P}_{\alpha }(ze^{-i\theta })=\sum ^{\infty }_{k=0}e^{-ik\theta }z^{k} +\sum ^{\infty }_{k=1}\frac{\Gamma (k+\alpha +1)}{\Gamma (k)\Gamma (\alpha +1)}P_{\alpha ,k}(|z|^2)e^{ik\theta }\overline{z}^{k}. \end{aligned}$$

If $|f^{*}(z)|\le M$, then by (1.6), we get

$$\begin{aligned} |c_{-k}|=\left| \frac{\Gamma (k+\alpha +1)}{\Gamma (k)\Gamma (\alpha +1)}\frac{1}{2\pi }\int ^{2\pi }_{0}e^{ik\theta }f^{*}(e^{i\theta })d\theta \right| \le M\frac{\Gamma (k+\alpha +1)}{\Gamma (k)\Gamma (\alpha +1)}\rightarrow \infty \end{aligned}$$

as $k\rightarrow \infty$.

Moreover, from the proof of [20, Theorem 1.2], we see that

$$\begin{aligned} (1-{|{z}|}^{2})^{-\alpha }\frac{\partial }{\partial \overline{z}}f(z)=\overline{h(z)}, \end{aligned}$$

where $h(z)=\sum ^{\infty }_{k=0}a_{k}z^k$, $z\in \mathbb {D}$ and $c_{-k}=\overline{a_{k-1}}$ for $k\ge 1$. Note that if h(z) is a normalized (in the sense that $h(0)=h'(0)-1=0$) univalent analytic function in $\mathbb {D}$, then by Louis de Branges’s theorem it is well-known that $|a_{k}|\le k$ for all $k\ge 2$ so that

$$\begin{aligned} c_{-1}=0,\quad c_{-2}=1\quad \hbox {and}\quad |c_{-k}|=|a_{k-1}|\le k-1\quad \hbox {for all}\;\;k\ge 3. \end{aligned}$$

(1.10)

The classical Landau theorem says that there is a $\rho =\frac{1}{M+\sqrt{M^{2}-1}}$ such that every function f, analytic in $\mathbb {D}$ with $f(0)=f^{'}(0)-1=0$ and $|f(z)|< M$, is univalent in the disk $\mathbb {D}_{\rho }$. Moreover, the range $f(\mathbb {D}_{\rho })$ contains a disk of radius $M\rho ^{2}$, where $M\ge 1$ is a constant (see [18]). Recently, many authors considered Landau type theorem for $\alpha$-harmonic functions f when $\alpha =0$ (see [3, 5,6,7,8,9] etc).

As an application of Theorems 1.1 and 1.2, we get the following Landau type theorem for $\alpha$-harmonic functions.

Theorem 1.3

Suppose that f is an $\alpha$-harmonic function in $\mathbb {D}$ with $\alpha \ge 0$, that $f^{*}\in C(\mathbb {T})$, that $\sup _{z\in \overline{\mathbb {D}}}|f(z)|\le M$, where M is a constant, and that $f(0)=|J_{f}(0)|-\beta =0$. If f satisfies (1.10), then we have the following:

• (1) f is univalent in $\mathbb {D}_{\rho _{0}}$, where $\rho _{0}$ satisfies the following equation

  $$\begin{aligned} \frac{\beta \mathrm {c}_{\alpha }}{M(\alpha +2)}-(M+5)\,\frac{\rho _{0}(2-\rho _{0})}{(1-\rho _{0})^{2}}=0; \end{aligned}$$

  (1.11)

• (2) $f(\mathbb {D}_{\rho _{0}})$ contains a univalent disk $\mathbb {D}_{R_{0}}$ with

  $$\begin{aligned} R_{0}\ge (M+5)\,\Big (\frac{\rho _{0}}{1-\rho _{0}}\Big )^2. \end{aligned}$$

The arrangement of the rest of this paper is as follows. In Sect. 2, we shall prove Theorem 1.1 and Corollary 1.1. Section 3 will be devoted to the proof of Theorem 1.2. In Sect. 4, Theorem 1.3 will be demonstrated.

2 Schwarz–Pick type inequality

The aim of this section is to prove Theorem 1.1 and Corollary 1.1. The proofs need a result from [19]. Before the statement of this result, we do some preparation.

In [19], the author considered the following integral means:

$$\begin{aligned} \mathcal {M}_{\alpha }(r)= \frac{1}{2\pi } \int _{0}^{2\pi }K_{\alpha }(re^{i\theta })d\theta , \end{aligned}$$

(2.1)

where $r\in [0, 1)$ and

$$\begin{aligned} K_{\alpha }(z)=\mathrm {c}_{\alpha }|\mathcal {P}_{\alpha }(z)|=\mathrm {c}_{\alpha }\frac{(1-|z|^{2})^{\alpha +1}}{|1-z|^{\alpha +2}} \end{aligned}$$

in $\mathbb {D}$.

Let us recall the following result from [19].

Theorem B

[19, Theorem 3.1] Let $\alpha >-1$. The integral means function $\mathcal {M}_{\alpha }(r)$ given by (2.1) satisfies the following assertions.

• (1) $\lim _{r\rightarrow 1^{-}}\mathcal {M}_{\alpha }(r)=1;$

• (2) $\mathcal {M}_{\alpha }^{(n)}(r)\ge 0$ for $r\in [0,1)$ and $n\ge 0$.

The following result also plays a key role in the proof of Theorem 1.1.

Lemma 2.1

If $\alpha >-1$ and $f^{*}\in C(\mathbb {T})$, then

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

and

$$\begin{aligned} \frac{\partial }{\partial \overline{z}}\int ^{2\pi }_{0}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta =\int ^{2\pi }_{0}\frac{\partial }{\partial \overline{z}}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta . \end{aligned}$$

Proof

By elementary calculations we see that the following equalities hold:

$$\begin{aligned} \frac{\partial }{\partial {z}}\mathcal {P}_{\alpha }(ze^{-i\theta }) =\frac{(1-|z|^{2})^{\alpha }\left[ e^{-i\theta }(1-|z|^{2})-(\alpha +1)\overline{z}(1-ze^{-i\theta })\right] }{(1-ze^{-i\theta })^{2}(1-\overline{z}e^{i\theta })^{\alpha +1}} \end{aligned}$$

(2.2)

and

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

(2.3)

Then we know that functions

$$\begin{aligned} \frac{\partial }{\partial z}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })\quad \hbox {and}\quad \frac{\partial }{\partial \overline{z}}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta }) \end{aligned}$$

are continuous on $\overline{\mathbb {D}}_r\times [0, 2\pi ]$, where $r\in [0, 1)$.

Let $z=\rho e^{i\varphi }\in \overline{\mathbb {D}}_r$. It follows from

$$\begin{aligned} \frac{\partial }{\partial \rho } \mathcal {P}_{\alpha }(ze^{-i\theta })=\frac{\partial }{\partial z} \mathcal {P}_{\alpha }(ze^{-i\theta })e^{i\varphi } +\frac{\partial }{\partial \overline{z}}\mathcal {P}_{\alpha }(ze^{-i\theta })e^{-i\varphi } \end{aligned}$$

and

$$\begin{aligned} \frac{\partial }{\partial \varphi }\mathcal {P}_{\alpha }(ze^{-i\theta })=\frac{\partial }{\partial z} \mathcal {P}_{\alpha }(ze^{-i\theta })iz -\frac{\partial }{\partial \overline{z}}\mathcal {P}_{\alpha }(ze^{-i\theta })i\overline{z} \end{aligned}$$

that both

$$\begin{aligned} \frac{\partial }{\partial \rho } \mathcal {P}_{\alpha }(ze^{-i\theta })f(e^{i\theta })\;\;\hbox { and} \;\;\frac{\partial }{\partial \varphi }\mathcal {P}_{\alpha }(ze^{-i\theta })f(e^{i\theta }) \end{aligned}$$

are continuous in $\overline{\mathbb {D}}_r\times [0, 2\pi ]$. Hence

$$\begin{aligned} \int ^{\rho }_{0}\int ^{2\pi }_{0}\frac{\partial }{\partial \rho } \mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta d\rho= & {} \int ^{2\pi }_{0}\int ^{\rho }_{0}\frac{\partial }{\partial \rho } \mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\rho d\theta \\= & {} \int ^{2\pi }_{0}\big (\mathcal {P}_{\alpha }(ze^{-i\theta })-\mathcal {P}_{\alpha }(0)\big )f^{*}(e^{i\theta })d\theta \end{aligned}$$

and

$$\begin{aligned} \int ^{\varphi }_{0}\int ^{2\pi }_{0}\frac{\partial }{\partial \varphi }\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta d\varphi= & {} \int ^{2\pi }_{0}\int ^{\varphi }_{0}\frac{\partial }{\partial \varphi }\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\varphi d\theta \\= & {} \int ^{2\pi }_{0}\big (\mathcal {P}_{\alpha }(ze^{-i\theta })-\mathcal {P}_{\alpha }(\rho e^{-i\theta })\big )f^{*}(e^{i\theta })d\theta . \end{aligned}$$

By differentiating with respect to $\rho$ and $\varphi$, respectively, we get

$$\begin{aligned} \int ^{2\pi }_{0}\frac{\partial }{\partial \rho } \mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta =\frac{\partial }{\partial \rho }\int ^{2\pi }_{0}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta \end{aligned}$$

(2.4)

and

$$\begin{aligned} \int ^{2\pi }_{0}\frac{\partial }{\partial \varphi }\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta =\frac{\partial }{\partial \varphi }\int ^{2\pi }_{0}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta . \end{aligned}$$

(2.5)

Since

$$\begin{aligned} \frac{\partial }{\partial z} \mathcal {P}_{\alpha }(ze^{-i\theta })=\frac{e^{-i\varphi }}{2}\left( \frac{\partial }{\partial \rho } \mathcal {P}_{\alpha }(ze^{-i\theta })-\frac{i}{\rho }\frac{\partial }{\partial \varphi }\mathcal {P}_{\alpha }(ze^{-i\theta })\right) \end{aligned}$$

and

$$\begin{aligned} \frac{\partial }{\partial \overline{z}}\mathcal {P}_{\alpha }(ze^{-i\theta })=\frac{e^{i\varphi }}{2}\left( \frac{\partial }{\partial \rho } \mathcal {P}_{\alpha }(ze^{-i\theta })+\frac{i}{\rho }\frac{\partial }{\partial \varphi }\mathcal {P}_{\alpha }(ze^{-i\theta })\right) , \end{aligned}$$

it follows from (2.4) and (2.5) that the proof of the lemma is complete. $\square$

Now, we are ready to present the proofs of Theorem 1.1 and Corollary 1.1.

Proof of Theorem 1.1

From (2.2) and (2.3), we can easily get

$$\begin{aligned} \left| \frac{\partial }{\partial {z}}\mathcal {P}_{\alpha }(ze^{-i\theta })\right| \le \frac{1}{\mathrm {c}_{\alpha }}\cdot \frac{(\alpha +2)|z|+1}{1-|z|^{2}}K_{\alpha }(ze^{-i\theta }) \end{aligned}$$

and

$$\begin{aligned} \left| \frac{\partial }{\partial {\overline{z}}}\mathcal {P}_{\alpha }(ze^{-i\theta })\right| = \frac{\alpha +1}{\mathrm {c}_{\alpha }}\cdot \frac{1}{1-|z|^{2}}K_{\alpha }(ze^{-i\theta }). \end{aligned}$$

In the first inequality above, the fact “$1-|z|\le |1-ze^{-i\theta }|$” is applied. By (1.1), (1.6) and Lemma 2.1 yield

$$\begin{aligned} |D{f}(z)|=\left| \frac{1}{2\pi }\int _{0}^{2\pi }\frac{\partial }{\partial {z}}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta \right| +\left| \frac{1}{2\pi }\int _{0}^{2\pi }\frac{\partial }{\partial {\overline{z}}}\mathcal {P}_{\alpha }(ze^{-i\theta })f^{*}(e^{i\theta })d\theta \right| , \end{aligned}$$

we see from (2.1) and Theorem B that

$$\begin{aligned} |D{f}(z)| \le \frac{M(\alpha +2)}{\mathrm {c}_{\alpha }}\cdot \frac{1}{1-|z|}\mathcal {M}_{\alpha }(|z|) \le \frac{M(\alpha +2)}{\mathrm {c}_{\alpha }}\cdot \frac{1}{1-|z|}, \end{aligned}$$

and so the proof of Theorem 1.1 is complete. $\square$

Proof of Corollary 1.1

For any $z_1$ and $z_2\in \mathbb {D}$, let $\gamma$ be the hyperbolic geodesic connecting $z_1$ and $z_2$. It follows from Theorem 1.1 that

$$\begin{aligned} |f(z_1)-f(z_2)|\le \int _{\gamma }|D{f}(z)|\cdot |dz| \le \frac{\alpha +2}{\mathrm {c}_{\alpha }}\int _{\gamma }\frac{2}{1-|z|^2}|dz| = \frac{\alpha +2}{\mathrm {c}_{\alpha }}d_{h_\mathbb {D}}(z_1, z_2), \end{aligned}$$

as required. $\square$

3 Estimates on coefficients

The aim of this paper is to prove Theorem 1.2. We start with a lemma.

Lemma 3.1

Under the assumptions of Theorem 1.2, if f has the series expansion (1.7), then

• (1) $|c_k|\le M$ for $k\ge 0$;

• (2) $(|c_{k}|+|c_{-k}|P_{\alpha ,k}(r^{2}))r^{k}\le \frac{4}{\pi }M$ for $k>0$ and $r\in (0, 1)$.

Proof

If $k\not =0$, let $z=re^{i\theta }\in {\mathbb {D}}$. Then by (1.7), we have

$$\begin{aligned} c_{k}r^{k}=\frac{1}{2\pi }\int _{0}^{2\pi }f(re^{i\theta })e^{-ik\theta }d{\theta }\quad \hbox {and}\quad c_{-k}P_{\alpha ,k}(r^{2})r^{k}=\frac{1}{2\pi }\int _{0}^{2\pi }f(re^{i\theta })e^{ik\theta }d{\theta }. \end{aligned}$$

Letting $c_k=|c_k|e^{i\mu _k}$ and $c_{-k}=|c_{-k}|e^{i\nu _k}$ leads to

$$\begin{aligned} (|c_{k}|+|c_{-k}|P_{\alpha ,k}(r^{2}))r^{k}= & {} \frac{1}{2\pi }\int ^{2\pi }_{0}f(re^{i\theta })\left( e^{-i(k\theta +\mu _k)}+e^{i(k\theta -\nu _k)}\right) d{\theta } \nonumber \\\le & {} \frac{1}{2\pi }\int ^{2\pi }_{0}|f(re^{i\theta })|\cdot \left| e^{-i(k\theta +\mu _k)}+e^{i(k\theta -\nu _k)}\right| d{\theta } \nonumber \\\le & {} \frac{M}{\pi }\int ^{2\pi }_{0}\left| \cos \left( k\theta +\frac{\mu _k-\nu _k}{2}\right) \right| d{\theta }, \end{aligned}$$

and so [4, Lemma 1] gives

$$\begin{aligned} (|c_{k}|+|c_{-k}|P_{\alpha ,k}(r^{2}))r^{k} \le \frac{4M}{\pi }. \end{aligned}$$

Thus the assertion (2) in the lemma is true.

To prove the assertions (1), we first recall from [20, Definition 2.1] that

$$\begin{aligned} \mathcal {P}_{\alpha }(ze^{-i\theta })=\sum ^{\infty }_{k=0}e^{-ik\theta }z^{k} +\sum ^{\infty }_{k=1}\frac{\Gamma (k+\alpha +1)}{\Gamma (k)\Gamma (\alpha +1)}P_{\alpha ,k}(|z|^2)e^{ik\theta }\overline{z}^{k}. \end{aligned}$$

Then by (1.6), we get

$$\begin{aligned} f(z)= & {} \sum ^{\infty }_{k=0}z^k\frac{1}{2\pi }\int ^{2\pi }_{0}e^{-ik\theta }f^{*}(e^{i\theta })d{\theta }\\&+\sum ^{\infty }_{k=1}\frac{\Gamma (k+\alpha +1)}{\Gamma (k)\Gamma (\alpha +1)}P_{\alpha ,k}(|z|^2)\overline{z}^{k} \frac{1}{2\pi }\int ^{2\pi }_{0}e^{ik\theta }f^{*}(e^{i\theta })d{\theta }, \end{aligned}$$

which implies

$$\begin{aligned} |c_k|=\left| \frac{1}{2\pi }\int ^{2\pi }_{0}e^{-ik\theta }f^{*}(e^{i\theta })d{\theta }\right| \le M \end{aligned}$$

as required. $\square$

Proof of Theorem 1.2

To prove this theorem, by Lemma 3.1, we only need to check (1.9) in the theorem. By letting $r\rightarrow 1^{-}$ in Lemma 3.1(2), we see that the inequalities (1.9) easily follows. $\square$

4 Landau type theorem

This section consists of two subsections. In the first subsection, we shall prove an auxiliary result. In the second subsection, Theorem 1.3 will be checked.

4.1 A lemma

Lemma 4.1

For constants $\alpha >-2$, $\beta >0$ and $M>0$, let

$$\begin{aligned} \varphi (x)=\frac{\beta \mathrm {c}_{\alpha }}{M(\alpha +2)}+(M+5)\,\frac{x(x-2)}{(1-x)^{2}} \end{aligned}$$

in [0, 1). Then

• (1) $\varphi$ is continuous in [0, 1) and strictly decreasing in (0, 1);

• (2) there is a unique $x_{0}\in (0,1)$ such that $\varphi (x_{0})=0$.

Proof

For $x\in [0,1)$, obviously,

$$\begin{aligned} \varphi '(x)=-\frac{2(M+5)}{(1-x)^{3}}<0. \end{aligned}$$

Hence $\varphi (x)$ is continuous and strictly decreasing in [0, 1). It follows from

$$\begin{aligned} \varphi (0)=\frac{\beta \mathrm {c}_{\alpha }}{M(\alpha +2)}>0 \quad \text {and}\quad \lim _{x\rightarrow 1^{-}}\varphi (x)=-\infty <0 \end{aligned}$$

that there is a unique $x_{0}\in (0,1)$ such that $\varphi (x_{0})=0$. The proof of this lemma is complete. $\square$

4.2 Proof of Theorem 1.3

To prove this theorem, we need estimates on two quantities $|f_{z}(z)-f_{z}(0)|+|f_{\overline{z}}(z)-f_{\overline{z}}(0)|$ and l(D f(0)). First, we estimate $|f_{z}(z)-f_{z}(0)|+|f_{\overline{z}}(z)-f_{\overline{z}}(0)|$. Obviously, by (1.7), we see that

$$\begin{aligned} f_{z}(z)-f_{z}(0)=\sum _{k=2}^{\infty }kc_{k}z^{k-1}+\sum _{k=2}^{\infty }c_{-k}\frac{d}{dw}P_{\alpha , k}(w)\overline{z}^{k+1} \end{aligned}$$

and

$$\begin{aligned} f_{\overline{z}}(z)-f_{\overline{z}}(0)=\sum _{k=2}^{\infty }kc_{-k}P_{\alpha , k}(w)\overline{z}^{k-1}+\sum _{k=2}^{\infty }c_{-k}\frac{d}{dw}P_{\alpha , k}(w)z\overline{z}^{k}, \end{aligned}$$

where $w=|z|^{2}$.

Since

$$\begin{aligned} \frac{d}{dw}P_{\alpha ,k}(w)=-\int ^{1}_{0}t^{k}\alpha (1-tw)^{\alpha -1}dt\le 0, \end{aligned}$$

we get that

$$\begin{aligned} P_{\alpha ,k}(w)\le P_{\alpha ,k}(0)=\frac{1}{k}. \end{aligned}$$

(4.1)

Moreover, since

$$\begin{aligned} P_{\alpha , k}(w)=\frac{1}{w^{k}}\int ^{w}_{0}x^{k-1}(1-x)^{\alpha }dx, \end{aligned}$$

we easily get

$$\begin{aligned} \frac{d}{dw}P_{\alpha ,k}(w)=-\frac{k}{w}P_{\alpha ,k}(w)+\frac{(1-w)^\alpha }{w}. \end{aligned}$$

(4.2)

Then (1.9), (1.10), (4.1) and (4.2) guarantee that

$$\begin{aligned} |f_{z}(z)-f_{z}(0)|+|f_{\overline{z}}(z)-f_{\overline{z}}(0)|\le & {} \sum _{k=2}^{\infty }k\big (|c_{k}|+|c_{-k}|P_{\alpha ,k}(w)\big )|z|^{k-1}\nonumber \\&+2\sum _{k=2}^{\infty }|c_{-k}|\big (kP_{\alpha ,k}(w)+1\big )|z|^{k-1} \nonumber \\\le & {} (M+5)\sum _{k=2}^{\infty }k|z|^{k-1} \nonumber \\= & {} (M+5)\,\frac{|z|(2-|z|)}{(1-|z|)^{2}}, \end{aligned}$$

(4.3)

which is what we want.

Next, we estimate l(D f(0)). Applying Theorem 1.1 leads to

$$\begin{aligned} \beta =|J_{f}(0)|=|D{f}(0)|\;l(D{f}(0))\le \frac{M(\alpha +2)}{\mathrm {c}_{\alpha }}\;l(D{f}(0)), \end{aligned}$$

which gives

$$\begin{aligned} l(D{f}(0))\ge \frac{\beta \mathrm {c}_{\alpha }}{M(\alpha +2)}. \end{aligned}$$

(4.4)

Now, we are ready to finish the proof of the theorem. First, we demonstrate the univalence of f in $\mathbb {D}_{\rho _{0}}$, where $\rho _{0}$ is determined by Eq. (1.11). For this, let $z_{1}$, $z_{2}$ be two points in $\mathbb {D}_{\rho _{0}}$ with $z_{1}\ne z_{2}$, and denote the segment from $z_{1}$ to $z_{2}$ with the endpoints $z_{1}$ and $z_{2}$ by $[z_{1},z_{2}]$. Since

$$\begin{aligned} |f(z_{2})-f(z_{1})|= & {} \left| \int _{[z_{1},z_{2}]}f_{z}(z)dz+f_{\overline{z}}(z)d\overline{z}\right| \nonumber \\\ge & {} \left| \int _{[z_{1},z_{2}]}f_{z}(0)dz+f_{\overline{z}}(0)d\overline{z}\right| \nonumber \\&-\left| \int _{[z_{1},z_{2}]}[f_{z}(z)-f_{z}(0)]dz+[f_{\overline{z}}(z)-f_{\overline{z}}(0)]d\overline{z}\right| , \end{aligned}$$

we see from (4.3), (4.4) and Lemma 4.1 that

$$\begin{aligned} |f(z_{2})-f(z_{1})|\ge & {} l(D{f}(0))\cdot |z_{2}-z_{1}|-(M+5)\int _{0}^{|z_{2}-z_{1}|}\frac{|z|(2-|z|)}{(1-|z|)^{2}}|dz| \nonumber \\ {}> & {} \left[ \frac{\beta \mathrm {c}_{\alpha }}{M(\alpha +2)}-(M+5)\,\frac{\rho _{0}(2-\rho _{0})}{(1-\rho _{0})^{2}}\right] |z_{2}-z_{1}| \nonumber \\= & {} 0. \end{aligned}$$

Thus, for arbitrary $z_{1}$ and $z_{2}\in \mathbb {D}_{\rho _{0}}$ with $z_{1}\ne z_{2}$, we have

$$\begin{aligned} f(z_{1})\ne f(z_{2}), \end{aligned}$$

which implies the univalence of f in $\mathbb {D}_{\rho _{0}}$.

Next, we prove Theorem 1.3(2). For any $\zeta =\rho _{0}e^{i\theta }\in \partial \mathbb {D}_{\rho _{0}}$, we obtain that

$$\begin{aligned} |f(\zeta )-f(0)|= & {} \left| \int _{[0,\zeta ]}f_{z}(z)dz+f_{\overline{z}}(z)d\overline{z}\right| \nonumber \\\ge & {} \left| \int _{[0,\zeta ]}f_{z}(0)dz+f_{\overline{z}}(0)d\overline{z}\right| \nonumber \\&-\left| \int _{[0,\zeta ]}[f_{z}(z)-f_{z}(0)]dz+[f_{\overline{z}}(z)-f_{\overline{z}}(0)]d\overline{z}\right| \nonumber \\\ge & {} l(D{f}(0))\rho _{0}-(M+5)\int _{0}^{\rho _0}\frac{|z|(2-|z|)}{(1-|z|)^{2}}|dz| \quad (\text {by }4.3) \nonumber \\= & {} \frac{\beta \mathrm {c}_{\alpha }\rho _{0}}{M(\alpha +2)}-(M+5)\,\frac{\rho _{0}^2}{1-\rho _{0}}. \nonumber \\= & {} (M+5)\,\Big (\frac{\rho _{0}}{1-\rho _{0}}\Big )^2. \quad (\text {by }1.11) \end{aligned}$$

Hence $f(\mathbb {D}_{\rho _{0}})$ contains a univalent disk $\mathbb {D}_{R_{0}}$, where

$$\begin{aligned} R_{0}\ge (M+5)\,\Big (\frac{\rho _{0}}{1-\rho _{0}}\Big )^2. \end{aligned}$$

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