On extreme points and product properties of a new subclass of p-harmonic functions

Abstract

In this paper, we introduce a new subclass of p-harmonic functions and investigate the univalence and sense-preserving, extreme points, distortion bounds, convex combination, neighborhoods of mappings belonging to the subclass. Relevant connections of the results presented here with the results of previous research are briefly indicated. Finally, we also prove new properties of the Hadamard product of these classes.

1 Introduction

Let $\mathcal{H}$ denote the class of all complex-valued harmonic functions $f = h+\bar{g}$ in $\mathbb{U}=\{z:|z|<1\}$, where h and g are analytic in $\mathbb{U}$ and normalized such that

$$ f(z)=h(z)+\overline{g(z)}=z+\sum_{j=2}^{\infty }a_{j}z^{j}+ \overline{ \sum_{j=1}^{\infty }b_{j}z^{j}}. $$

(1.1)

A necessary and sufficient condition for f to be locally univalent and sense-preserving in $\mathbb{U}$ is that $J_{f}=|f_{z}|^{2}-|f _{\bar{z}}|^{2}>0$ in $\mathbb{U}$ (see [1, 2]). Let $S_{H}$ denote the subclass of $\mathcal{H}$ consisting of sense-preserving univalent functions in $\mathbb{U}$. Then the function $f \in S_{H}$ of the form (1.1) satisfies the condition $|b_{1}|<1$.

A 2p-times continuously differentiable complex-valued function $F =u+iv$ in a domain $\mathbb{U}$ is p-harmonic if F satisfies the p-harmonic equation $\bigtriangleup ^{p}F=\bigtriangleup ( \bigtriangleup ^{p-1}F)=0\ (p=1,2,\ldots )$, where Δ represents the complex Laplacian operator:

$$ \Delta =4\frac{\partial ^{2}}{\partial z \,\partial \bar{z}}:=\frac{ \partial ^{2}}{\partial {x^{2}}}+\frac{\partial ^{2}}{\partial {y^{2}}}. $$

Obviously, if we take $p=1$ and $p=2$, then F is harmonic and biharmonic, respectively.

A function F is p-harmonic in a simply connected domain $\mathbb{U}$ if and only if F has the following representation:

$$ F(z)=\sum_{k=1}^{p} \vert z \vert ^{2(k-1)}f_{p-k+1}(z)\quad \bigl(k\in \{1,2,\ldots,p\}\bigr), $$

(1.2)

where each $f_{p-k+1}(z)$ is harmonic (or $\triangle f_{p-k+1}=0$) (see [3]) and $f_{p-k+1}(z)$ has the form

$$ f_{p-k+1}=h_{p-k+1}+\bar{g}_{p-k+1}, $$

(1.3)

where

$$\begin{aligned} &h_{p-k+1}(z)=\sum_{j=1}^{\infty }a_{j,p-k+1}z^{j}\quad (a_{1,p}=1, k\geq 1), \end{aligned}$$

(1.4)

$$\begin{aligned} &g_{p-k+1}(z)=\sum_{j=1}^{\infty }b_{j,p-k+1}z^{j}\quad (k\geq 1). \end{aligned}$$

(1.5)

Denote by $SH_{p}$ the class of functions F of the form (1.2) that are p-harmonic, univalent, and sense-preserving in the unit disk. Recently, there has been significant interest in results about the class $SH_{p}$ (see, for details, [4,5,6,7,8,9]).

Denote by $\mathit{HL}_{p}(\alpha,\lambda )\ (0\leq \alpha <1,\lambda \geq 0)$ the class of all mappings of the form (1.2) which satisfy the condition

$$\begin{aligned} &\sum_{k=1}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda }(j- \alpha )}{1-\alpha } \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \\ &\quad \leq 1- \vert b_{1,p} \vert -\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \end{aligned}$$

(1.6)

with

$$ 0\leq \vert b_{1,p} \vert +\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr)< 1. $$

(1.7)

Clearly, inequality (1.6) implies that

$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \leq 1, $$

(1.8)

where $a_{1,p}=1,k\in \{1,\ldots,p\}$.

It is easy to see that various subclasses of $SH_{p}$ consisting of mappings $F(z)$ of the form (1.2) and (1.3) can be represented as $\mathit{HL}_{p}(\alpha,\lambda )(b_{1,p}=a_{1,p-k+1}=b_{1,p-k+1}=0,k=2, \ldots,p)$ for suitable choices of $p,\alpha$, and λ in the earlier studies by various authors.

• (i) $\mathit{HL}_{p}(0,0)=\mathit{HS}_{p}$ and $\mathit{HL}_{p}(0,1)=\mathit{HC}_{p}$ (see Qiao and Wang [4]);

• (ii) $\mathit{HL}_{p}(\alpha,0)=\mathit{HS}_{p}(\alpha )$ and $\mathit{HL}_{p}(\alpha,1)=\mathit{HC} _{p}(\alpha )$ (see Saurabh Porwal and Dixit [5]);

• (iii) $\mathit{HL}_{1}(\alpha,0)=\mathit{HS}(\alpha )$ and $\mathit{HL}_{1}(\alpha,1)=\mathit{HC}( \alpha )$ (see Öztürk and Yalcin [10]);

• (vi) $\mathit{HL}_{1}(0,0)=\mathit{HS}$ and $\mathit{HL}_{1}(0,1)=\mathit{HC}$ (see Avci and Zlotkiewicz [11]).

For $\lambda \in \mathbb{N}=\{1,2,\ldots \}\cup \{0\}$, we have the following inclusion relation:

$$ \mathit{HL}_{p}(\alpha,\lambda )\subset \mathit{HL}_{p}(\alpha,\lambda -1) \subset \cdots \subset \mathit{HL}_{p}(\alpha,2)\subset \mathit{HC}_{p}(\alpha ) \subset \mathit{HS}_{p}( \alpha ). $$

Suppose that F is a p-harmonic mapping with expression (1.2). Following Ruscheweyh [12], we use $N_{\lambda,\alpha }^{\delta }(F)$ to denote the δ-neighborhood of F in p-harmonic mappings:

$$\begin{aligned} N_{\lambda,\alpha }^{\delta }(F)={}& \Biggl\{ \tilde{F}: \vert b_{1,p}-B_{1,p} \vert + \sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1}-A_{1,p-k+1} \vert + \vert b_{1,p-k+1}-B _{1,p-k+1} \vert \bigr) \\ &{}+\sum_{k=1}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda }(j- \alpha )}{1-\alpha } \biggr) \bigl( \vert a_{j,p-k+1}-A_{j,p-k+1} \vert \\ &{}+ \vert b_{j,p-k+1}-B _{j,p-k+1} \vert \bigr)\leq \delta \Biggr\} , \end{aligned}$$

where

$$ \tilde{F}=z+\sum_{j=2}^{\infty }A_{j,p}z^{j}+ \sum_{j=1}^{\infty } \bar{B}_{j,p} \bar{z}^{j} +\sum_{k=2}^{p} \vert z \vert ^{2(k-1)} \Biggl(\sum_{j=1} ^{\infty }A_{j,p-k+1}z^{j}+\sum_{j=1}^{\infty } \bar{B}_{j,p-k+1} \bar{z}^{j} \Biggr). $$

If $F, G\in SH_{p}$ satisfy

$$ F=\sum_{k=1}^{p} \vert z \vert ^{2(k-1)} \Biggl(\sum_{j=1}^{\infty }a_{j,p-k+1}z ^{j}+\sum_{j=1}^{\infty } \bar{b}_{j,p-k+1}\bar{z}^{j} \Biggr) $$

and

$$ G=\sum_{k=1}^{p} \vert z \vert ^{2(k-1)} \Biggl(\sum_{j=1}^{\infty }A_{j,p-k+1}z ^{j}+\sum_{j=1}^{\infty } \bar{B}_{j,p-k+1}\bar{z}^{j} \Biggr), $$

then the convolution $F\ast G$ of F and G is defined to be the mapping

$$ F\ast G =\sum_{k=1}^{p} \vert z \vert ^{2(k-1)} \Biggl(\sum_{j=1}^{\infty }a_{j,p-k+1}A _{j,p-k+1}z^{j}+\sum_{j=1} ^{\infty }\bar{b}_{j,p-k+1}\bar{B}_{j,p-k+1} \bar{z}^{j} \Biggr). $$

Let

$$ \mathit{TH}_{p}= \bigl\{ F(z):F\in SH_{p} \text{ with } a_{1,p}=1,a_{j,p-k+1} \geq 0,b _{j,p-k+1}\geq 0\text{ for }j\geq 1,k=1,\ldots,p \bigr\} $$

and denote $\overline{\mathit{HL}}_{p}(\alpha,\lambda )=\mathit{HL}_{p}(\alpha,\lambda )\cap \mathit{TH}_{p}$.

The main objective of the paper is to introduce a new subclass of p-harmonic mappings and investigate the univalence and sense-preserving, extreme points, neighborhoods and Hadamard product of mappings for the above subclass. Relevant connections of the results presented here with the results of Qiao et al. [4] and Porwal et al. [5] are briefly indicated. Finally, we also prove new properties of the Hadamard product of these classes.

2 Main results

Firstly, we discuss the inclusion relation of $\mathit{HL}_{p}(\alpha,\lambda )$.

Theorem 2.1

Let $\lambda _{2}\geq \lambda _{1}\geq 0, 1> \alpha _{2}\geq \alpha _{1}\geq 0$, then $\mathit{HL}_{p}(\alpha _{2},\lambda _{2}) \subseteq \mathit{HL}_{p}(\alpha _{1},\lambda _{1})$.

Proof

Let $F\in \mathit{HL}_{p}(\alpha _{2},\lambda _{2})$, then using (1.6), we have

$$\begin{aligned} &\sum_{k=1}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda _{1}}(j- \alpha _{1})}{1-\alpha _{1}} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \\ &\quad\leq \sum_{k=1}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda _{2}}(j-\alpha _{2})}{1-\alpha _{2}} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \\ &\quad \leq 1- \vert b_{1,p} \vert -\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr), \end{aligned}$$

therefore $F\in \mathit{HL}_{p}(\alpha _{1},\lambda _{1})$, and so $\mathit{HL}_{p}(\alpha _{2},\lambda _{2})\subseteq \mathit{HL}_{p}(\alpha _{1},\lambda _{1})$. □

Next, we prove that the mapping in $\mathit{HL}_{p}(\alpha,\lambda )$ is univalent and sense-preserving.

Theorem 2.2

Each mapping in $\mathit{HL}_{p}(\alpha,\lambda )$ is univalent and sense-preserving.

Proof

Let $F\in \mathit{HL}_{p}(\alpha,\lambda )$ and $z_{1},z _{2}\in \mathbb{U}$ with $z_{1}\neq z_{2}$, so that $|z_{1}|\leq |z _{2}|$:

$$\begin{aligned} &\bigl\vert F(z_{1})-F(z_{2}) \bigr\vert \\ &\quad = \Biggl\vert \sum_{k=1}^{p} \bigl( \vert z_{1} \vert ^{2(k-1)}f_{p-k+1}(z _{1})- \vert z_{2} \vert ^{2(k-1)}f_{p-k+1}(z_{2}) \bigr) \Biggr\vert \\ &\quad\geq \vert z_{1}-z_{2} \vert \Biggl\{ 1- \Biggl\vert \sum_{j=2}^{\infty }a_{j,p} \frac{z_{1} ^{j}-z_{2}^{j}}{z_{1}-z_{2}}+\sum_{j=1}^{\infty } \bar{b}_{j,p}\frac{ \bar{z}_{1}^{j}-\bar{z}_{2}^{j}}{z_{1}-z_{2}} \Biggr\vert \Biggr\} \\ &\qquad{}- \Biggl\vert \sum_{k=2}^{p} \Biggl(\sum _{j=1}^{\infty }a_{j,p-k+1}\frac{ \vert z _{1} \vert ^{2(k-1)}z_{1}^{j} - \vert z_{2} \vert ^{2(k-1)}z_{2}^{j}}{z_{1}-z_{2}} \\ &\qquad{}+\sum_{j=1}^{\infty }\bar{b}_{j,p-k+1} \frac{ \vert z_{1} \vert ^{2(k-1)}\bar{z} _{1}^{j}- \vert z_{2} \vert ^{2(k-1)}\bar{z}_{2}^{j}}{z_{1}-z_{2}} \Biggr) \Biggr\vert \\ &\quad\geq \vert z_{1}-z_{2} \vert \Biggl(1- \vert b_{1,p} \vert - \vert z_{2} \vert \sum _{j=2}^{\infty }j \bigl( \vert a_{j,p} \vert + \vert b_{j,p} \vert \bigr) \Biggr) \\ &\qquad{}- \vert z_{2} \vert \sum_{k=2}^{p} \sum_{j=1}^{\infty }\bigl(2(k-1)+j\bigr) \bigl( \vert a_{j,p-k+1} \vert + \vert b _{j,p-k+1} \vert \bigr) \\ &\quad\geq \vert z_{1}-z_{2} \vert \Biggl(1- \vert b_{1,p} \vert - \vert z_{2} \vert \sum _{j=2}^{\infty }\frac{j ^{\lambda }(j-\alpha )}{1-\alpha } \bigl( \vert a_{j,p} \vert + \vert b_{j,p} \vert \bigr) \Biggr) \\ &\qquad{}- \vert z_{2} \vert \sum_{k=2}^{p} \sum_{j=1}^{\infty } \biggl(2(k-1)+ \frac{j^{ \lambda }(j-\alpha )}{1-\alpha } \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \\ &\quad \geq \vert z_{1}-z_{2} \vert \bigl(1- \vert b_{1,p} \vert \bigr) \bigl(1- \vert z_{2} \vert \bigr) \\ &\quad >0, \end{aligned}$$

which proves univalence.

In order to prove that F is sense-preserving, we need to show that $J_{F}=|F_{z}|^{2}-|F_{\bar{z}}|^{2}>0$:

$$\begin{aligned} J_{F}(z)={}& \vert F_{z} \vert ^{2}- \vert F_{\bar{z}} \vert ^{2} =\bigl( \vert F_{z} \vert + \vert F_{\bar{z}} \vert \bigr) \bigl( \vert F_{z} \vert - \vert F_{\bar{z}} \vert \bigr) \\ ={}&\bigl( \vert F_{z} \vert + \vert F_{\bar{z}} \vert \bigr) \Biggl\{ \Biggl\vert 1+\sum_{j=2}^{\infty }ja_{j,p}z ^{j-1}+\sum_{k=2}^{p}\sum _{j=2}^{\infty }|z | ^{2(k-1)}ja_{j,p-k+1}z ^{j-1} \\ &{}+\sum_{k=2}^{p} \vert z \vert ^{2(k-1)}ja_{1,p-k+1}+\sum_{k=2}^{p}(k-1) \vert z \vert ^{2(k-1)} \\ &{}\times\Biggl(\sum_{j=1}^{\infty }a_{j,p-k+1}z^{j-1}+ \frac{\bar{z}}{z}\sum_{j=1}^{\infty } \bar{b}_{j,p-k+1}\bar{z}^{j-1} \Biggr)\Biggr\vert \\ &{}- \Biggl\vert \sum_{j=1}^{\infty }j \bar{b}_{j,p}\bar{z}^{j-1}+\sum_{k=2}^{p} \sum_{j=2}^{\infty } \vert z \vert ^{2(k-1)}j\bar{b}_{j,p-k+1}\bar{z}^{j-1} \\ &{}+\sum_{k=2}^{p} \vert z \vert ^{2(k-1)}\bar{b}_{1,p-k+1}+\sum_{k=2}^{p}(k-1) \vert z \vert ^{2(k-1)}\\ &{}\times \Biggl(\frac{z}{\bar{z}}\sum _{j=1}^{\infty }a_{j,p-k+1}z^{j-1}+ \frac{ \bar{z}}{z}\sum_{j=1}^{\infty } \bar{b}_{j,p-k+1}\bar{z}^{j-1} \Biggr) \Biggr\vert \Biggr\} \\ \geq{}& \bigl( \vert F_{z} \vert + \vert F_{\bar{z}} \vert \bigr) \Biggl[1- \vert b_{1,p} \vert -\sum _{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \\ &{}- \vert z \vert \sum_{k=1}^{p}\sum _{j=2}^{\infty }\bigl(2(k-1)+j\bigr) \bigl( \vert a_{j,p-k+1} \vert + \vert b _{j,p-k+1} \vert \bigr) \Biggr] \\ \geq{}& \bigl( \vert F_{z} \vert + \vert F_{\bar{z}} \vert \bigr) \Biggl[1- \vert b_{1,p} \vert -\sum _{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \\ &{}- \vert z \vert \sum_{k=1}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda }(j- \alpha )}{1-\alpha } \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \Biggr] \\ \geq{}& \bigl( \vert F_{z} \vert + \vert F_{\bar{z}} \vert \bigr) \bigl(1- \vert b_{1,p} \vert \bigr) \bigl(1- \vert z \vert \bigr) \\ >{}&0. \end{aligned}$$

From $z\neq 0$ and the obvious fact $J_{F}(0)>0$, we thus complete the proof. □

Example 2.1

Let $F(z)=z+\frac{1}{(2p-1)}|z|^{2(p-1)} \bar{z}$. Then $F(z)$ is a p-harmonic function and

$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda +1}}{2} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr)< 1, $$

using (1.8), we get $F\in \mathit{HL}_{p}(0,\lambda )$.

Also, we determine the extreme points of $\overline{\mathit{HL}}_{p}(\alpha, \lambda )$.

Theorem 2.3

Let F be given by (1.2). Then $F\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$ if and only if

$$ F(z)=\sum_{k=1}^{p}\sum _{j=1}^{\infty } \bigl(X_{j,p-k+1}h_{j,p-k+1}(z)+Y _{j,p-k+1}g_{j,p-k+1}(z) \bigr), $$

(2.1)

where

$$\begin{aligned} &h_{j,p-k+1}(z)=z+ \vert z \vert ^{2(k-1)}\frac{z^{j}}{(k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )}}\quad (2\leq k\leq p;j\geq 1), \\ &g_{j,p-k+1}(z)=z+ \vert z \vert ^{2(k-1)}\frac{\bar{z}^{j}}{(k-1)+\frac{j^{\lambda }(j-\alpha )}{2(1-\alpha )}}\quad (2\leq k\leq p;j\geq 1), \\ &h_{1,p}(z)=z,\qquad h_{1,j}(z)=z+\frac{z^{j}}{\frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )}}\quad(j\geq 2), \\ &g_{1,j}(z)=z+\frac{\bar{z}^{j}}{\frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )}}\quad (j\geq 1) \end{aligned}$$

and

$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } (X_{j,p-k+1}+Y_{j,p-k+1} )=1\quad (X _{j,p-k+1}\geq 0,Y_{j,p-k+1}\geq 0). $$

In particular, the extreme points of $\overline{\mathit{HL}}_{p}(\alpha, \lambda )$ are $\{h_{j,p-k+1}(z)\}$ and $\{g_{j,p-k+1}(z)\}$, where $j\geq 1$ and $1\leq k \leq p$.

Proof

Since

$$\begin{aligned} F(z)={}&\sum_{k=1}^{p}\sum _{j=1}^{\infty } \bigl(X_{j,p-k+1}h_{j,p-k+1}(z)+Y _{j,p-k+1}g_{j,p-k+1}(z) \bigr) \\ ={}&z+\sum_{k=2}^{p} \vert z \vert ^{2(k-1)}\sum_{j=1}^{\infty } \biggl( \frac{X_{j,p-k+1}}{(k-1)+\frac{j ^{\lambda }(j-\alpha )}{2(1-\alpha )}}z^{j}+\frac{Y_{j,p-k+1}}{(k-1)+\frac{j ^{\lambda }(j-\alpha )}{2(1-\alpha )}}\bar{z}^{j} \biggr) \\ &{}+\sum_{j=1}^{\infty }\frac{X_{j,p}}{\frac{j^{\lambda (j-\alpha )}}{1- \alpha }}z^{j} +\sum_{j=1}^{\infty }\frac{Y_{j,p}}{\frac{j^{\lambda (j- \alpha )}}{2(1-\alpha )}} \bar{z}^{j} \end{aligned}$$

and

$$\begin{aligned} &\sum_{k=1}^{p}\sum _{j=2}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \biggl( \biggl\vert \frac{X_{j,p-k+1}}{(k-1)+\frac{j ^{\lambda }(j-\alpha )}{2(1-\alpha )}} \biggr\vert + \biggl\vert \frac{Y_{j,p-k+1}}{(k-1)+\frac{j ^{\lambda }(j-\alpha )}{2(1-\alpha )}} \biggr\vert \biggr) \\ &\qquad{}+ \vert Y_{1,p} \vert +\sum_{k=2}^{p} \frac{(2k-1)}{2} \biggl( \biggl\vert \frac{X_{1,p-k+1}}{ \frac{2k-1}{2}} \biggr\vert + \biggl\vert \frac{Y_{1,p-k+1}}{\frac{2k-1}{2}} \biggr\vert \biggr) \\ &\quad\leq \sum_{k=1}^{p}\sum _{j=2}^{\infty } (X_{j,p-k+1}+Y_{j,p-k+1} )+\sum _{k=2}^{p} (X_{1,p-k+1}+Y_{1,p-k+1} )+Y_{1,p} \\ &\quad \leq 1-X_{1,p} \\ &\quad \leq 1, \end{aligned}$$

we see that $F\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$.

Conversely, assuming that $F\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$ and setting

$$\begin{aligned} &X_{j,p-k+1}= \biggl((k-1)+\frac{j^{\lambda }(j-\alpha )}{2(1-\alpha )} \biggr) a_{j,p-k+1}\quad (2\leq k \leq p, j\geq 1), \\ &X_{j,p}=\frac{j^{\lambda }(j-\alpha )}{2(1-\alpha )}a_{j,p}\quad (j\geq 2), \\ &Y_{j,p-k+1}= \biggl((k-1)+\frac{j^{\lambda }(j-\alpha )}{2(1-\alpha )} \biggr) b_{j,p-k+1}\quad (1\leq k \leq p, j\geq 1) \end{aligned}$$

and

$$ X_{1,p}=1-\sum_{k=1}^{p}\sum _{j=2}^{\infty } (X_{j,p-k+1}+Y_{j,p-k+1} )-\sum_{k=2}^{p} (X_{1,p-k+1}+Y_{1,p-k+1} )-Y_{1,p}, $$

where $X_{1,p}\geq 0$. Then, as required, we obtain

$$ F(z)=\sum_{k=1}^{p}\sum _{j=1}^{\infty } \bigl(X_{j,p-k+1}h_{j,p-k+1}(z)+Y _{j,p-k+1}g_{j,p-k+1}(z) \bigr). $$

 □

Example 2.2

Let $F(z)=z+\frac{1}{(2p-1)}|z|^{2(p-1)}z+ \frac{1}{(2p-1)}|z|^{2(p-1)}\bar{z}$. Then $F(z)$ is a p-harmonic function, and using Theorem 2.3, we have $F\in \overline{\mathit{HL}}_{p}(0, \lambda )$. Here, we give the figures for $p=4$ and $p=10$, respectively (see Fig. 1 and Fig. 2).

Figure 1

When $F(z)$ is a 4-harmonic function

Figure 2

When $F(z)$ is a 10-harmonic function

Theorem 2.4

Let F be given by (1.2) and $F\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$. Then, for $|z|=r<1$, we have

$$\begin{aligned} \bigl\vert F(z) \bigr\vert \leq{}& \Biggl(\sum_{k=1}^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r \\ &{}+\frac{1}{\psi _{2,1 }(\lambda,\alpha )} \Biggl(1- \vert b_{1,p} \vert -\sum _{k=2} ^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2} \end{aligned}$$

(2.2)

and

$$\begin{aligned} \bigl\vert F(z) \bigr\vert \geq{}& \Biggl(1- \vert b_{1,p} \vert - \sum_{k=2}^{p}\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r \\ &{}-\frac{1}{\psi _{2,1 }(\lambda,\alpha )} \Biggl(1- \vert b_{1,p} \vert -\sum _{k=2} ^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2}, \end{aligned}$$

(2.3)

where

$$ \psi _{j,k }(\lambda,\alpha )=(k-1)+\frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )}. $$

(2.4)

Proof

Let $F\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$. Taking the absolute value of $F(z)$, we have

$$\begin{aligned} \bigl\vert F(z) \bigr\vert \leq{}& \Biggl(\sum_{k=1}^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r+ \Biggl(\sum_{k=1}^{p}\sum _{j=2}^{\infty }\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2} \\ \leq{}& \Biggl(\sum_{k=1}^{p}\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r \\ &{}+ \Biggl(\frac{1}{\psi _{2,1 }(\lambda,\alpha )}\sum_{k=1}^{p} \sum_{j=2} ^{\infty }\psi _{j,k }(\lambda, \alpha ) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2} \\ \leq{}& \Biggl(\sum_{k=1}^{p}\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r \\ &{}+\frac{1}{\psi _{2,1 }(\lambda,\alpha )} \Biggl(1- \vert b_{1,p} \vert -\sum _{k=2} ^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2} \end{aligned}$$

and

$$\begin{aligned} \bigl\vert F(z) \bigr\vert \geq{}& \Biggl(\sum_{k=1}^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r- \Biggl(\sum_{k=1}^{p}\sum _{j=2}^{\infty }\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2} \\ \geq{}& \Biggl(\sum_{k=1}^{p}\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r \\ &{}- \Biggl(\frac{1}{\psi _{2,1 }(\lambda,\alpha )}\sum_{k=1}^{p} \sum_{j=2} ^{\infty }\psi _{j,k }(\lambda, \alpha ) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2} \\ \geq{}& \Biggl(1- \vert b_{1,p} \vert -\sum _{k=2}^{p}\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r \\ &{}-\frac{1}{\psi _{2,1 }(\lambda,\alpha )} \Biggl(1- \vert b_{1,p} \vert -\sum _{k=2} ^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr)r^{2}. \end{aligned}$$

 □

Corollary 2.5

Let F be given by (1.2) and $F\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$. Then

$$ \bigl\{ \omega: \vert \omega \vert < \rho \bigr\} \subset F(\mathbb{U}), $$

where

$$ \rho =\frac{1+\psi _{2,1 }(\lambda,\alpha )}{\psi _{2,1 }(\lambda, \alpha )} -\frac{1-\psi _{2,1 }(\lambda,\alpha )}{\psi _{2,1 }(\lambda ,\alpha )} \Biggl( \vert b_{1,p} \vert +\sum_{k=2}^{p}\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr) $$

and $\psi _{j,k }(\lambda,\alpha )$ is given by (2.4).

Theorem 2.6

The class $F\in \overline{\mathit{HL}}_{p}( \alpha,\lambda )$ is closed under combination.

Proof

For $i=1,2,\ldots$ , let $F_{i}\in \overline{\mathit{HL}} _{p}(\alpha,\lambda )$, where

$$ F_{i}(z)=z+\sum_{j=2}^{\infty }a_{ij,p}z^{j}+ \sum_{j=1}^{\infty }b _{ij,p} \bar{z}^{j} +\sum_{k=2}^{p} \vert z \vert ^{2(k-1)}\sum_{j=1}^{\infty } \bigl( \vert a_{j,p-k+1} \vert z^{j}+ \vert \bar{b}_{j,p-k+1} \vert \bar{z}^{j} \bigr). $$

Then, by (1.6) and (2.4), we get

$$ \sum_{k=1}^{p}\sum _{j=2}^{\infty }\psi _{j,k }(\lambda,\alpha ) \bigl( \vert a _{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr)\leq 1- \vert b_{1,p} \vert -\sum_{k=2}^{p} \bigl( \vert a _{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr). $$

(2.5)

For $\sum_{i=1}^{\infty }t_{i}=1,0\leq t_{i}\leq 1$, the convex combination of $F_{i}$ may be written as

$$\begin{aligned} \sum_{i=1}^{\infty }t_{i}F_{i}={}&z- \sum_{j=2}^{\infty } \Biggl(\sum _{i=1} ^{\infty }t_{i}\bigl[ \vert a_{ij,p} \vert z^{j} + \vert b_{ij,p} \vert \bar{z}^{j}\bigr] \Biggr)\\ &{}-\sum_{k=2}^{p} \vert z \vert ^{2(k-1)} \sum_{j=1}^{\infty } \Biggl(\sum_{i=1}^{\infty }t_{i}\bigl[ \vert a_{ij,p-k+1} \vert z^{j}+ \vert b_{ij,p-k+1} \vert \bar{z}^{j}\bigr] \Biggr). \end{aligned}$$

Then, by (2.5), we obtain

$$\begin{aligned} &\sum_{k=1}^{p}\sum _{j=2}^{\infty }\psi _{j,k }(\lambda,\alpha ) \Biggl(\sum_{i=1}^{\infty }t_{i}\bigl[ \vert a_{ij,p-k+1}+|b_{ij,p-k+1} \vert \bigr] \Biggr) \\ &\quad=\sum_{i=1}^{\infty }t_{i} \Biggl[\sum _{k=1}^{p}\sum_{j=2}^{\infty } \psi _{j,k }(\lambda,\alpha )\cdot \bigl( \vert a_{ij,p-k+1} \vert + \vert b_{ij,p-k+1} \vert \bigr) \Biggr] \\ &\quad \leq \Biggl[(1- \vert b_{1,p} \vert -\sum _{k=2}^{p}\bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr) \Biggr]\sum_{i=1}^{\infty }t_{i} \\ &\quad =1- \vert b_{1,p} \vert -\sum_{k=2}^{p} \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr). \end{aligned}$$

Therefore, using (1.6), we obtain $\sum_{i=1}^{\infty }t_{i}F_{i} \in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$. □

Theorem 2.7

Let

$$ F_{1}(z)=z+\sum_{j=2}^{\infty }a_{j,p}z^{j}+ \sum_{j=2}^{p} \vert z \vert ^{2(k-1)} \Biggl(\sum_{j=1}^{\infty }a_{j,p-k+1}z^{j}+ \sum_{j=1}^{\infty } \bar{b}_{j,p-k+1} \bar{z}^{j} \Biggr) $$

belong to $\overline{\mathit{HL}}_{p}(\alpha,\lambda _{2})$. If $\lambda _{2}> \lambda _{1}\geq 0$ and

$$ \delta \leq (1-c_{0}) \bigl(1- \vert b_{1,p} \vert \bigr)-\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b _{1,p-k+1} \vert \bigr), $$

(2.6)

then $N_{\lambda _{1},\alpha }^{\delta }(F_{1})\subset \mathit{HL}_{p}(\alpha, \lambda _{1})$, where

$$ c_{0}=\frac{2(p-1)(1-\alpha )+2^{\lambda _{1}}(2-\alpha )}{2(p-1)(1- \alpha )+2^{\lambda _{2}}(2-\alpha )}. $$

(2.7)

Proof

The δ-neighborhood of $F_{1}$ is the set

$$\begin{aligned} N_{\lambda _{1},\alpha }^{\delta }(F_{1})= {}&\Biggl\{ F_{2}:\sum _{k=1}^{p} \sum_{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda _{1}}(j-\alpha )}{1- \alpha } \biggr) \bigl( \vert a_{j,p-k+1}-A_{j,p-k+1} \vert \\ &{}+ \vert b_{j,p-k+1}-B_{j,p-k+1} \vert \bigr)+ \vert b_{1,p}+B_{1,p} \vert +\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1}-A_{1,p-k+1} \vert \\ &{}+ \vert b_{1,p-k+1}-B_{1,p-k+1} \vert \bigr)\leq \delta \Biggr\} , \end{aligned}$$

where

$$ F_{2}(z)=z+\sum_{j=2}^{\infty }A_{j,p}z^{j}+ \sum_{j=1}^{\infty } \bar{B}_{j,p} \bar{z}^{j} +\sum_{k=2}^{\infty } \vert z \vert ^{2(k-1)} \Biggl(\sum_{j=1}^{\infty }A_{j,p-k+1}z^{j} +\sum_{j=1}^{\infty }\bar{B}_{j,p-k+1} \bar{z}^{j} \Biggr). $$

If

$$ \delta \leq (1-c_{0}) \bigl(1- \vert b_{1,p} \vert \bigr)-\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b _{1,p-k+1} \vert \bigr), $$

then we have

$$\begin{aligned} &\sum_{j=2}^{\infty }\frac{j^{\lambda _{1}}(j-\alpha )}{1-\alpha } \vert A _{j,p} \vert +\sum_{j=1}^{\infty } \frac{j^{\lambda _{1}}(j-\alpha )}{1- \alpha } \vert B_{j,p} \vert \\ &\qquad{} +\sum _{k=2}^{p}\sum_{j=2}^{\infty } \biggl(2(k-1)+\frac{j ^{\lambda _{1}} (j-\alpha )}{1-\alpha } \biggr) \bigl( \vert A_{j,p-k+1} \vert + \vert B _{j,p-k+1} \vert \bigr) \\ &\quad\leq \sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1}-A_{1,p-k+1} \vert + \vert b_{1,p-k+1}-B _{1,p-k+1} \vert + \vert b_{1,p}-B_{1,p} \vert \bigr) \\ &\qquad{}+\sum_{k=2}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda _{1}}(j- \alpha )}{1-\alpha } \biggr) ( \bigl( \vert a_{j,p-k+1}-A_{1,p-k+1} \vert + \vert b_{j,p-k+1}-B _{j,p-k+1} \vert \bigr) \\ &\qquad{}+\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr)+ \vert b_{1,p} \vert \\ &\qquad{}+\sum_{k=2}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda _{1}}(j- \alpha )}{1-\alpha } \biggr) ( \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \\ &\quad\leq \delta +\sum_{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b_{1,p-k+1} \vert \bigr)+ \vert b_{1,p} \vert \\ &\qquad{}+c_{0}\sum_{k=2}^{p}\sum _{j=2}^{\infty } \biggl(2(k-1)+\frac{j^{\lambda _{2}}(j-\alpha )}{1-\alpha } \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \\ &\quad\leq \delta +c_{0}+(1-c_{0}) \Biggl(\sum _{k=2}^{p}(2k-1) \bigl( \vert a_{1,p-k+1} \vert + \vert b _{1,p-k+1} \vert \bigr)+ \vert b_{1,p} \vert \Biggr) \\ &\quad \leq 1. \end{aligned}$$

Hence $F_{2}\in \overline{\mathit{HL}}_{p}(\alpha,\lambda _{1})$. □

Remark 2.8

• 1. If $\alpha =0,\lambda =0$ and $\alpha =0,\lambda =1$, then Theorem 2.2, Theorem 2.4, and Theorem 2.7, respectively, coincide with Theorem 3.1, Theorem 4.3, Theorem 4.4, Lemma 4.1, and Theorem 5.1 in [4].

• 2. If $\lambda =0$ and $\lambda =1$, then Theorem 2.2, Theorem 2.3, and Theorem 2.7, respectively, coincide with Theorem 3.1, Theorem 3.6, Theorem 3.7, and Theorem 4.1 in [5].

At last, we discuss the Hadamard product of $\overline{\mathit{HL}}_{p}(\alpha ,\lambda )$.

Theorem 2.9

Let $\lambda \geq 0,~0\leq \alpha <1,~p \in \{1,2,\ldots \}$. If $F,~G\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$, then $F\ast G \in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$, where

$$ 2^{\lambda -1}(2-\alpha )\geq (1-\alpha )p^{2}~. $$

(2.8)

Proof

Let $F,G\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$, then, from (1.8), we know that, in order to prove $F\ast G \in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$, we need to show that

$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b _{j,p-k+1} \vert \bigr)\leq 1. $$

(2.9)

Since $F,G\in \overline{\mathit{HL}}_{p}(\alpha,\lambda )$, using (1.8), we have

$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr) \leq 1 $$

(2.10)

and

$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr) \leq 1. $$

(2.11)

From (2.10) and (2.11), we obtain

$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr)\leq 1 $$

(2.12)

and

$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr) \bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr)\leq 1. $$

(2.13)

Using the Cauchy–Schwarz inequations, from (2.12) and (2.13), we get

$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr) \sqrt{\bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr) \bigl( \vert a_{j,p-k+1} \vert + \vert b _{j,p-k+1} \vert \bigr)}\leq 1, $$

(2.14)

because

$$\begin{aligned} &\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr) \\ &\quad\leq \bigl( \vert A_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \bigr) \bigl( \vert a_{j,p-k+1} \vert + \vert b_{j,p-k+1} \vert \bigr)\quad (1 \leq k\leq 1,j\in \mathbb{N}). \end{aligned}$$

(2.15)

So from (2.14) and (2.15), we have

$$ \sum_{j=1}^{\infty } \biggl((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1- \alpha )} \biggr)\sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b _{j,p-k+1} \vert \bigr)}\leq 1, $$

and hence

$$ \sum_{k=1}^{p}\sum _{j=1}^{\infty } \biggl((k-1)+\frac{j^{\lambda }(j- \alpha )}{2(1-\alpha )} \biggr) \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B _{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)}\leq p, $$

(2.16)

which implies that

$$ \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)} \leq \frac{p}{ ((k-1)+ \frac{j^{\lambda }(j-\alpha )}{2(1-\alpha )} )}. $$

(2.17)

In addition, if

$$ \bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)\leq \frac{1}{p} \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)}, $$

that is,

$$ \sqrt{\bigl( \vert A_{j,p-k+1} \vert \vert a_{j,p-k+1} \vert + \vert B_{j,p-k+1} \vert \vert b_{j,p-k+1} \vert \bigr)} \leq \frac{1}{p}, $$

(2.18)

then we obtain the conditions of satisfaction (2.9). Again, combining (2.17) and (2.18) with $k=1$ and $j=2$, we can get

$$ \frac{p}{ (\frac{2^{\lambda }(2-\alpha )}{2(1-\alpha )} )} \leq \frac{1}{p}, $$

which deduces condition (2.8). The proof is completed. □

Taking $\lambda =0$ and $\lambda =1$ in Theorem 2.9, respectively, we obtain the following corollaries.

Corollary 2.10

Let $0\leq \alpha <1,2-\alpha \geq 2(1-\alpha )p^{2}(p\geq 1)$. If $F,G\in \mathit{HS}_{p}(\alpha )$, then $F\ast G\in \mathit{HS}_{p}(\alpha )$.

Corollary 2.11

Let $0\leq \alpha <1,2-\alpha \geq (1-\alpha )p^{2}(p\geq 1)$. If $F,G\in \mathit{HC}_{p}(\alpha )$, then $F\ast G\in \mathit{HC}_{p}(\alpha )$.

3 Conclusions

In this paper, we mainly introduce a new subclass of p-harmonic mappings and investigate the univalence and sense-preserving, extreme points, distortion bounds, convex combination, neighborhoods of mappings belonging to the subclass. Relevant connections of the results presented here with the results of Qiao et al. [4] and Porwal et al. [5] are briefly indicated. Finally, we also prove new properties of the Hadamard product of these classes.