On the Generalization of a Class of Harmonic Univalent Functions Defined by Differential Operator

Abstract

In this article, a new class of harmonic univalent functions, defined by the differential operator, is introduced. Some geometric properties, like, coefficient estimates, extreme points, convex combination and convolution (Hadamard product) are obtained.

1. Introduction

A continuous function $f=u+iv$ is a complex-valued harmonic function in a complex domain $ℂ$ if both $u$ and $v$ are real harmonic. In any simply connected domain $\mathcal{B}\subset ℂ$, we can write $f=h+\overline{g}$, where $h$ and $g$ are analytic in $\mathcal{B}$. We call $h$ and $g$ are analytic part and co-analytic part of $f$ respectively. Clunie and Sheil-Small [1] observed that a necessary and sufficient condition for the harmonic functions $f=h+\overline{g}$ to be locally univalent and sense-preserving in $\mathcal{B}$ is that $\left|h^{\prime }\left(z\right)\right|>\left|g^{\prime }\left(z\right)\right|,\left(z\in \mathcal{B}\right).$

Denote by $S_H$ the family of harmonic functions $f=h+\overline{g}$, which are univalent and sense-preserving in the open unit disc $U=\left\{z\in ℂ:\left|z\right|<1\right\}$ where $h$ and $g$ are analytic in $\mathcal{B}$ and $f$ is normalized by $f\left(0\right)=h\left(0\right)=f_z\left(0\right)-1=0.$ Then for $f=h+\overline{g}\in S_H$, we may express the analytic functions $h$ and $g$ as

$$h\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{a}_n{z}^n,g\left(z\right)={\displaystyle \sum }_{n=1}^{\infty }{b}_n{z}^n,\left|b_1\right|<1.$$

(1)

Note that $S_H$ reduces to the class of normalized analytic univalent functions if the co-analytic part of its members equals to zero.

Also, denote by $S_{\overline{H}}$ the subclass of $S_H$ consisting of all functions $f_k\left(z\right)=h\left(z\right)+\overline{g_k\left(z\right)}$, where $h$ and $g$ are given by

$$h\left(z\right)=z-{\displaystyle \sum }_{n=2}^{\infty }\left|a_n\right|z^{n}and{g}_k\left(z\right)={\left(-1\right)}^k{\displaystyle \sum }_{n=1}^{\infty }\left|b_n\right|z^n,\left|b_1\right|<1.$$

(2)

In 1984 Clunie and Sheil-Small [1] investigated the class $S_H$, as well as its geometric subclass and obtained some coefficient bounds. Many authors have studied the family of harmonic univalent function (see References [2,3,4,5,6,7]).

In 2016 Makinde [8] introduced the differential operator $F^k$ such that

$$F^{k}f\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{C}_{nk}{z}^n,$$

(3)

where

$$C_{nk}=\frac{n!}{\left|n-k\right|!},F^{k}f\left(z\right)=z^k\left[z^{-\left(k-1\right)}+{\displaystyle \sum }_{n=2}^{\infty }{C}_{nk}{z}^n\right],k\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\},$$

and

$$F^{0}f\left(z\right)=f\left(z\right),F^{1}f\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{C}_{n1}{z}^n.$$

Thus, it implies that $F^{k}f\left(z\right)$ is identically the same as $f\left(z\right)$ when $k=0$. Also, it reduced the first differential coefficient of the Salagean differential operator when $k=1.$

For $f=h+\overline{g}$ given by Equation (1), Sharma and Ravindar [9] considered the differential operator which defined by Equation (3) of $f$ as

$$F^{k}f\left(z\right)=F^{k}h\left(z\right)+{\left(-1\right)}^k\overline{F^{k}g\left(z\right)},k\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\},z\in ℂ,$$

(4)

where

$$F^{k}h\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }{C}_{nk}{a}_n{z}^n,F^{k}g\left(z\right)={\displaystyle \sum }_{n=1}^{\infty }{C}_{nk}{b}_n{z}^n\mathrm{and}{C}_{nk}=\frac{n!}{\left|n-k\right|!}.$$

In this paper, motivated by study in [9], a new class $A_H\left(k,\alpha ,\gamma \right)\left(k\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\},0\le \gamma \le 1,0\le \alpha <1,\right)$ of harmonic univalent functions in $U=\left\{z\in ℂ:\left|z\right|<1\right\}$ is introduced and studied. Furthermore, coefficient conditions, distortion bounds, extreme points, convex combination and radii of convexity for this class are obtained.

2. Main Results

2.1. The Class $A_H\left(k,\alpha ,\gamma \right)$

Definition 1.

Let$f\left(z\right)=h\left(z\right)+\overline{g\left(z\right)}beaharmonicfunction,$where$h\left(z\right)$and$g\left(z\right)$are given by Equation (1). Then$f\left(z\right)\in A_H\left(k,\alpha ,\gamma \right)$it satisfies

$$Re\left\{\frac{F^{k+1}f\left(z\right)}{\left(1-\gamma \right)z+\gamma F^{k}f\left(z\right)}\right\}>\alpha ,$$

(5)

for$k\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\},0\le \gamma \le 1,0\le \alpha <1,z\in U,$and$F^{k}f\left(z\right)$defined by Equation (4)

Let $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$ be the subclass of $A_H\left(k,\alpha ,\gamma \right)$, where $A_{\overline{H}}\left(k,\alpha ,\gamma \right)=S_{\overline{H}}\cap A_H\left(k,\alpha ,\gamma \right).$

Remark 1.

The class$A_{\overline{H}}\left(k,\alpha ,\gamma \right)$reduces to the class$B_{\overline{H}}\left(k,\alpha \right)$[9], when$\gamma =1.$

Here, we give a sufficient condition for a function $f$ to be in the class $A_H\left(k,\alpha ,\gamma \right)$.

Theorem 1.

$Letf\left(z\right)=h\left(z\right)+\overline{g\left(z\right)}whereh\left(z\right)andg\left(z\right)weregivenby$(1). If

$${\displaystyle \sum }_{n=2}^{\infty }\varnothing \left(n,k,\alpha ,\gamma \right)\left|a_n\right|+{\displaystyle \sum }_{n=1}^{\infty }\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left|b_n\right|\le 1,$$

(6)

where

$$\varnothing \left(n,k,\alpha ,\gamma \right)=\frac{\left(\left|n-k\right|-\alpha \gamma \right)C_{nk}}{\left(1-\alpha \right)}$$

$$\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)=\frac{\left(\left|n-k\right|+\alpha \gamma \right)C_{nk}}{\left(1-\alpha \right)}$$

$$\left(k\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\},0\le \gamma \le 1,0\le \alpha <1,n\in \mathbb{N}\right),$$

then $f\left(z\right)$ is harmonic univalent and sense-preserving in $U$ and $f\left(z\right)\in A_H\left(k,\alpha ,\gamma \right).$

Proof. 

Firstly, to show that $f\left(z\right)$ is harmonic univalent in $U$, suppose that $z_1,z_2\in U\mathrm{for}\lfloor z_1\rfloor \le \lfloor z_2\rfloor <1$, we have by inequality so that $z_1\ne z_2$, then

$$\begin{array}{cc}\hfill \left|\frac{f\left(z_1\right)-f\left(z_2\right)}{h\left(z_1\right)-h\left(z_2\right)}\right|& \\ & \ge 1-\left|\frac{g\left(z_1\right)-g\left(z_2\right)}{h\left(z_1\right)-h\left(z_2\right)}\right|=1-\left|\frac{{\sum }_{n=1}^{\infty }{b}_n\left(z_1^n-z_2^n\right)}{\left(z_1-z_2\right)-{\sum }_{n=2}^{\infty }{a}_n\left(z_1^n-z_2^n\right)}\right|\hfill \\ & \ge 1-\frac{{\sum }_{n=1}^{\infty }n\left|b_n\right|}{1-{\sum }_{n=2}^{\infty }n\left|a_n\right|}\ge 1-\frac{{\sum }_{n=1}^{\infty }\frac{\left(\left|n-k\right|+\alpha \gamma \right)C_{nk}}{\left(1-\alpha \right)}\left|b_n\right|}{1-{\sum }_{n=2}^{\infty }\frac{\left(\left|n-k\right|-\alpha \gamma \right)C_{nk}}{\left(1-\alpha \right)}\left|a_n\right|}\ge 0.\hfill \end{array}$$

Thus $f$ is a univalent function in $U$.

Note that $f$ is sense-preserving in $U$. This is because

$$\begin{array}{c}\hfill \left|h^{\prime }\left(z\right)\right|\ge 1-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}n\left|a_n\right|{\left|z\right|}^{n-1}>1-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}n\left|a_n\right|\ge 1-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\frac{\left(\left|n-k\right|-\alpha \gamma \right)C_{nk}}{\left(1-\alpha \right)}\left|a_n\right|\\ \hfill \ge {\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\frac{\left(\left|n-k\right|+\alpha \gamma \right)C_{nk}}{\left(1-\alpha \right)}\left|b_n\right|\ge {\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}n\left|b_n\right|\ge {\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}n\left|b_n\right|{\left|z\right|}^{n-1}\ge \left|g^{\prime }\left(z\right)\right|.\end{array}$$

According to the condition of Equation (5), we only need to show that if Equation (6) holds, then

$$Re\left\{\frac{F^{k+1}f\left(z\right)}{\left(1-\gamma \right)z+\gamma F^{k}f\left(z\right)}\right\}=Re\left(w=\frac{A\left(z\right)}{B\left(z\right)}\right)>\alpha$$

where $z=r{e}^{i\theta },0\le \theta \le 2\pi ,0\le r<1\mathrm{and}0\le \alpha <1.$

Note that $A\left(z\right)=F^{k+1}f\left(z\right)\mathrm{and}B\left(z\right)=\left(1-\gamma \right)z+\gamma F^{k}f\left(z\right).$

Using the fact that $Re\left(w\right)>\alpha$ if and only if $\left|w-\left(1+\alpha \right)\right|\le \left|w+\left(1-\alpha \right)\right|,$ it suffices to show that

$$\left|A\left(z\right)-\left(1+\alpha \right)B\left(z\right)\right|-\left|A\left(z\right)+\left(1-\alpha \right)B\left(z\right)\right|\le 0$$

(7)

Substituting for $A\left(z\right)\mathrm{and}B\left(z\right)$ in $\left|A\left(z\right)-\left(1+\alpha \right)B\left(z\right)\right|$, we obtain

$$\begin{array}{c}\left|A\left(z\right)-\left(1+\alpha \right)B\left(z\right)\right|=\left|F^{k+1}f\left(z\right)-\left(1+\alpha \right)\left[\left(1-\gamma \right)z+\gamma F^{k}f\left(z\right)\right]\right|\hfill \\ =|\left[z+{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{C}_{n\left(k+1\right)}{a}_n{z}^n+{\left(-1\right)}^{\left(k+1\right)}{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{C}_{n\left(k+1\right)}\overline{b_n{z}^n}\right]\\ \hfill -\left(1+\alpha \right)\left[\left(1-\gamma \right)z+\gamma z+\gamma {\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{C}_{nk}{a}_n{z}^n+\gamma {\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{C}_{nk}\overline{b_n{z}^n}\right]|\\ \le \alpha \left|z\right|+{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|\left(\gamma \left(1+\alpha \right)\right)-\left|n-k\right|\right|C_{nk}\left|a_n\right|{\left|z\right|}^n\hfill \\ +{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|\left(\gamma \left(1+\alpha \right)\right)+\left|n-k\right|\right|C_{nk}\left|a_n\right|{\left|\overline{z}\right|}^n.\end{array}$$

(8)

Now, substituting for $A\left(z\right)\mathrm{and}B\left(z\right)$ in $\left|A\left(z\right)+\left(1-\alpha \right)B\left(z\right)\right|$, we obtain

$$\begin{array}{c}\left|A\left(z\right)+\left(1-\alpha \right)B\left(z\right)\right|=\left|F^{k+1}f\left(z\right)+\left(1-\alpha \right)\left[\left(1-\gamma \right)z+\gamma F^{k}f\left(z\right)\right]\right|\\ =|\left[z+{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{C}_{n\left(k+1\right)}{a}_n{z}^n+{\left(-1\right)}^{\left(k+1\right)}{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{C}_{n\left(k+1\right)}\overline{b_n{z}^n}\right]\hfill \\ \hfill +\left(1-\alpha \right)\left[\left(1-\gamma \right)z+\gamma z+\gamma {\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{C}_{nk}{a}_n{z}^n+\gamma {\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{C}_{nk}\overline{b_n{z}^n}\right]|\\ \ge \left(2-\alpha \right)\left|z\right|-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|\left(\gamma \left(\alpha -1\right)\right)-\left|n-k\right|\right|C_{nk}\left|a_n\right|{\left|z\right|}^n\hfill \\ -{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|\left|n-k\right|-\left(\gamma \left(1-\alpha \right)\right)\right|C_{nk}\left|a_n\right|{\left|\overline{z}\right|}^n.\end{array}$$

(9)

Substituting for Equations (8) and (9) in the inequality we obtain

$$\begin{array}{l}\left|A\left(z\right)-\left(1+\alpha \right)B\left(z\right)\right|-\left|A\left(z\right)+\left(1-\alpha \right)B\left(z\right)\right|\\ \le \alpha \left|z\right|+{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|\left(\gamma \left(1+\alpha \right)\right)-\left|n-k\right|\right|C_{nk}\left|a_n\right|{\left|z\right|}^n\\ +{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|\left(\gamma \left(1+\alpha \right)\right)+\left|n-k\right|\right|C_{nk}\left|b_n\right|{\left|\overline{z}\right|}^n\\ +\left(\alpha -2\right)\left|Z\right|+{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|\left(\gamma \left(\alpha -1\right)\right)-\left|n-k\right|\right|C_{nk}\left|a_n\right|{\left|z\right|}^n\\ +{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|\left|n-k\right|-\left(\gamma \left(1-\alpha \right)\right)\right|C_{nk}\left|b_n\right|{\left|\overline{z}\right|}^n.\\ =2{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left(\left|n-k\right|-\alpha \gamma \right)C_{nk}\left|a_n\right|+2{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left(\left|n-k\right|+\alpha \gamma \right)C_{nk}\left|b_n\right|-2\left(1-\alpha \right)\\ \le 0.\left(\mathrm{by}\mathrm{hypothesis}\right).\end{array}$$

Therefore, we have

$${\displaystyle \sum }_{n=2}^{\infty }\left(\left|n-k\right|-\alpha \gamma \right)C_{nk}\left|a_n\right|+{\displaystyle \sum }_{n=1}^{\infty }\left(\left|n-k\right|+\alpha \gamma \right)C_{nk}\left|b_n\right|\le \left(1-\alpha \right).$$

 □

The harmonic univalent function

$$f\left(z\right)=z+{\displaystyle \sum }_{n=2}^{\infty }\frac{1}{\varnothing \left(n,k,\alpha ,\gamma \right)}{\mathcal{X}}_n{z}^n+{\displaystyle \sum }_{n=1}^{\infty }\frac{1}{\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)}\overline{{\mathcal{Y}}_n{z}^n},$$

(10)

where $k\in {\mathbb{N}}_0$ and ${\sum }_{k=2}^{\infty }\left|{\mathcal{X}}_n\right|+{\sum }_{k=1}^{\infty }\left|{\mathcal{Y}}_n\right|=1$, shows that the coefficient bound given by Equation (6) is sharp. Since

$$\begin{array}{c}{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\left|a_n\right|+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left|b_n\right|\hfill \\ ={\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\frac{1}{\varnothing \left(n,k,\alpha ,\gamma \right)}\left|{\mathcal{X}}_n\right|+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\frac{1}{\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)}\left|{\mathcal{Y}}_n\right|\\ ={\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|{\mathcal{X}}_n\right|+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|{\mathcal{Y}}_n\right|=1.\end{array}$$

Now, we show that the condition of Equation (6) is also necessary for functions $f_k=h+\overline{g_k},$where $h$ and $g_n$ are given by Equation (6).

Theorem 2.

Let$f_k=h+\overline{g_k}be$given by Equation (6). Then$f_k\left(z\right)\in A_{\overline{H}}\left(k,\alpha ,\gamma \right)$if and only if the coefficient in condition of Equation (6) holds.

Proof. 

We only need to prove the “only if” part of the theorem because of $A_{\overline{H}}\left(k,\alpha ,\gamma \right)\subset A_H\left(k,\alpha ,\gamma \right).$ Then by Equation (5), we have

$$Re\left\{\frac{F^{k+1}f\left(z\right)}{\left(1-\gamma \right)z+\gamma F^{k}f\left(z\right)}\right\}>\alpha$$

or, equivalently

$$Re\left[\frac{\begin{array}{c}z-{\sum }_{n=2}^{\infty }{C}_{n\left(k+1\right)}\left|a_n\right|z^n+{\left(-1\right)}^{2k+1}{\sum }_{n=1}^{\infty }{C}_{n\left(k+1\right)}\left|b_n\right|{\overline{z}}^n\\ -\alpha \left\{\left(1-\gamma \right)z+\gamma z+\gamma {\sum }_{n=2}^{\infty }{C}_{nk}\left|a_n\right|z^n+\gamma {\left(-1\right)}^{2k}{\sum }_{n=1}^{\infty }{C}_{nk}\left|b_n\right|{\overline{z}}^n\right\}\end{array}}{\left(1-\gamma \right)z+\gamma z-\gamma {\sum }_{n=2}^{\infty }{C}_{nk}\left|a_n\right|z^n+\gamma {\left(-1\right)}^{2k}{\sum }_{n=1}^{\infty }{C}_{nk}\left|b_n\right|{\overline{z}}^n}\right]\ge 0$$

(11)

We observe that the above-required condition of Equation (11) must behold for all values of $zinU.$ If we choose $z$ to be real and $z\to 1^{-}$, we get

$$\frac{\begin{array}{c}\left(1-\alpha \right)-{\sum }_{n=2}^{\infty }\left(\left|n-k\right|-\alpha \gamma \right)C_{nk}\left|a_n\right|\\ +{\sum }_{n=1}^{\infty }\left(\left|n-k\right|+\alpha \gamma \right)C_{nk}\left|b_n\right|\end{array}}{1-\gamma {\sum }_{n=2}^{\infty }{C}_{nk}\left|a_n\right|z^{n-1}+\gamma {\sum }_{n=1}^{\infty }{C}_{nk}\left|b_n\right|{\overline{z}}^{n-1}}\ge 0$$

(12)

If the condition (6) does not hold, then the numerator in Equation (12) is negative for $r$ sufficiently closed to 1. Hence there exist $z_0=r_0$ in $\left(0,1\right)$ for which the quotient in Equation (12) is negative, therefore there is a contradicts the required condition for $f_k\in A_{\overline{H}}\left(k,\alpha ,\gamma \right).$ □

2.2. Extreme Points

Here, we determine the extreme points of the closed convex hull of $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$, denoted by $clco{A}_{\overline{H}}\left(k,\alpha ,\gamma \right).$

Theorem 3.

$Let{f}_{k}givenby\left(1.2\right).Then{f}_k\in A_{\overline{H}}\left(k,\alpha ,\gamma \right)ifandonlyif$

$$f_k\left(z\right)={\displaystyle \sum }_{n=1}^{\infty }\left({\mathcal{X}}_n{h}_n+{\mathcal{Y}}_n{g}_{kn}\right)$$

where

$$h_1\left(z\right)=z,h_n\left(z\right)=z-\frac{1}{\varnothing \left(n,k,\alpha ,\gamma \right)}{z}^n,n=2,3,\dots ,$$

$$g_{kn}\left(z\right)=z+{\left(-1\right)}^k\frac{1}{\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)}{\overline{z}}^n,n=1,2,\dots ,$$

and

$${\mathcal{X}}_n\ge 0,{\mathcal{Y}}_n\ge 0,{\mathcal{X}}_1=1-{\displaystyle \sum }_{n=2}^{\infty }\left({\mathcal{X}}_n+{\mathcal{Y}}_n\right)\ge 0$$

In particular the extreme points of $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$ are $\left\{h_n\right\}and\left\{g_{kn}\right\}.$

Proof. 

Suppose

$$\begin{array}{c}{f}_k\left(z\right)={\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left({\mathcal{X}}_n{h}_n+{\mathcal{Y}}_n{g}_{kn}\right)\\ ={\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left({\mathcal{X}}_n{h}_n+{\mathcal{Y}}_n{g}_{kn}\right)z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\frac{1}{\varnothing \left(n,k,\alpha ,\gamma \right)}{\mathcal{X}}_n{z}^n+{\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\frac{1}{\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)}{\mathcal{Y}}_n{\overline{z}}^n\\ =z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\frac{1}{\varnothing \left(n,k,\alpha ,\gamma \right)}{\mathcal{X}}_n{z}^n+{\left(-1\right)}^{k-1}{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\frac{1}{\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)}{\mathcal{Y}}_n{\overline{z}}^n\end{array}$$

Then

$$\begin{array}{c}{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\left|a_n\right|+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left|b_n\right|\\ ={\displaystyle {\displaystyle \sum }_{k=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\left(\frac{1}{\varnothing \left(n,k,\alpha ,\gamma \right)}{\mathcal{X}}_n\right)+{\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left(\frac{1}{\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)}{\mathcal{Y}}_n\right)\\ ={\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{\mathcal{X}}_n+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{\mathcal{Y}}_n=1-{\mathcal{X}}_1\le 1.\end{array}$$

Therefore $f_k\left(z\right)\in clco{A}_{\overline{H}}\left(k,\alpha ,\gamma \right).$

Conversely, if $f_k\left(z\right)\in clco{A}_{\overline{H}}\left(k,\alpha ,\gamma \right).$ Then

$$\begin{array}{c}\mathrm{Set}{\mathcal{X}}_n=\varnothing \left(n,k,\alpha ,\gamma \right)\left|a_n\right|,\left(n=2,3,\dots \right)\mathrm{and}{\mathcal{Y}}_n=\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left|b_n\right|,\\ \left(n=1,2,\dots \right)\mathrm{and}{\mathcal{X}}_1=1-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{\mathcal{X}}_n+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{\mathcal{Y}}_n\end{array}$$

The required representation is obtained as

$$\begin{array}{l}{f}_k\left(z\right)=z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|a_n\right|z^n+{\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|b_n\right|{\overline{z}}^n\\ =z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\frac{1}{\varnothing \left(n,k,\alpha ,\gamma \right)}{\mathcal{X}}_n{z}^n+{\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\frac{1}{\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)}{\mathcal{Y}}_n{\overline{z}}^n\\ =z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left[z-h_n\left(z\right)\right]{\mathcal{X}}_n+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left[z-g_{kn}\left(z\right)\right]{\mathcal{Y}}_n\\ =\left[1-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{\mathcal{X}}_n-{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{\mathcal{Y}}_n\right]z+{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{h}_n\left(z\right){\mathcal{X}}_n+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{g}_{kn}\left(z\right){\mathcal{Y}}_n={\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left({\mathcal{X}}_n{h}_n+{\mathcal{Y}}_n{g}_{kn}\right)\end{array}$$

□

2.3. Convex Combination

Here, we show that the class $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$ is closed under convex combination of its members.

Let the function $f_{k,i}\left(z\right)$ be defined, for $i=1,2,\dots ,m$ by

$$f_{k,i}\left(z\right)=z-{\displaystyle \sum }_{n=2}^{\infty }\left|a_{n,i}\right|z^n+{\left(-1\right)}^k{\displaystyle \sum }_{n=1}^{\infty }\left|b_{n,i}\right|{\overline{z}}^n$$

(13)

Theorem 4.

Let the functions$f_{k,i}\left(z\right),$defined by Equation (13) be in the class $A_{\overline{H}}\left(k,\alpha ,\gamma \right),$for every $i=1,2,\dots ,m.$ Then the functions $c_i\left(z\right)$ defined by

$$c_i\left(z\right)={\displaystyle \sum }_{i=1}^{\infty }{t}_i{f}_{k,i}\left(z\right),0\le t_i\le 1$$

are also in the class $A_{\overline{H}}\left(k,\alpha ,\gamma \right),where{\sum }_{i=1}^{\infty }{t}_i=1.$

Proof. 

According to the definition of $c_i\left(z\right)$, we can write

$$c_i\left(z\right)=z-{\displaystyle \sum }_{n=2}^{\infty }\left({\displaystyle \sum }_{i=1}^{\infty }{t}_i\left|a_{n,i}\right|\right)z^n+{\left(-1\right)}^k{\displaystyle \sum }_{n=1}^{\infty }\left({\displaystyle \sum }_{i=1}^{\infty }{t}_i\left|b_{n,i}\right|\right){\overline{z}}^n$$

Further, since $f_{k,i}\left(z\right)$ are in $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$ for every$i=1,2,\dots ,m$, then by Theorem 2, we obtain

$$\begin{array}{c}{\displaystyle {\displaystyle \sum }_{k=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\left({\displaystyle {\displaystyle \sum }_{i=1}^{\infty }}{t}_i\left|a_{n,i}\right|\right)+{\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left({\displaystyle {\displaystyle \sum }_{i=1}^{\infty }}{t}_i\left|b_{n,i}\right|\right)\hfill \\ \hfill ={\displaystyle {\displaystyle \sum }_{i=1}^{\infty }}{t}_i\left({\displaystyle {\displaystyle \sum }_{k=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\left|a_{n,i}\right|+{\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left|b_{n,i}\right|\right)\le {\displaystyle {\displaystyle \sum }_{i=1}^{\infty }}{t}_i=1,\end{array}$$

which is required coefficient condition. □

2.4. Convolution (Hadamard Product) Property

Here, we show that the class $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$ is closed under convolution.

The convolution of two harmonic functions

$$f_k\left(z\right)=z-{\displaystyle \sum }_{n=2}^{\infty }\left|a_n\right|z^n+{\left(-1\right)}^k{\displaystyle \sum }_{n=1}^{\infty }\left|b_n\right|{\overline{z}}^n,$$

(14)

and

$$Q_n\left(z\right)=z-{\displaystyle \sum }_{n=2}^{\infty }\left|L_n\right|z^n+{\left(-1\right)}^k{\displaystyle \sum }_{n=1}^{\infty }\left|M_n\right|{\overline{z}}^n$$

(15)

is defined as

$$\begin{array}{c}\left(f_n\ast Q_n\right)\left(z\right)=f_n\left(z\right)\ast Q_n\left(z\right)\\ =z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|a_n{L}_n\right|z^n+{\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|b_n{M}_n\right|{\overline{z}}^n\end{array}$$

(16)

Using Equations (12)–(14), we prove the following theorem.

Theorem 5.

For$0\le \mu \le \alpha <1,k\in {\mathbb{N}}_0$, let$f_n\in$$A_{\overline{H}}\left(k,\alpha ,\gamma \right)$and$Q_n\in$$A_{\overline{H}}\left(k,\mu ,\gamma \right)$. Then

$$f_n\ast Q_n\in A_{\overline{H}}\left(k,\alpha ,\gamma \right)\subset A_{\overline{H}}\left(k,\mu ,\gamma \right).$$

Proof. 

Let

$$f_k\left(z\right)=z-{\displaystyle \sum }_{n=2}^{\infty }\left|a_n\right|z^n+{\left(-1\right)}^k{\displaystyle \sum }_{n=1}^{\infty }\left|b_n\right|{\overline{z}}^n$$

be in the class $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$ and

$$Q_n\left(z\right)=z-{\displaystyle \sum }_{n=2}^{\infty }\left|L_n\right|z^n+{\left(-1\right)}^k{\displaystyle \sum }_{n=1}^{\infty }\left|M_n\right|{\overline{z}}^n,$$

be in $A_{\overline{H}}\left(k,\mu ,\gamma \right)$.

Then the convolution $f_n\ast Q_n$ is given by Equation (16), we want to show that the coefficients of $f_n\ast Q_n$ satisfy the required condition given in Theorem 1.

For $Q_n\in$ $A_{\overline{H}}\left(k,\mu ,\gamma \right)$, we note that $\left|L_n\right|<1$ and $\left|M_n\right|<1.$ Now consider convolution functions $f_n\ast Q_n$ as follows:

$$\begin{array}{c}{\displaystyle {\displaystyle \sum }_{k=2}^{\infty }}\varnothing \left(n,k,\mu ,\gamma \right)\left|a_n\right|\left|L_n\right|+{\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}\mathsf{\psi }\left(n,k,\mu ,\gamma \right)\left|b_n\right|\left|M_n\right|\\ \le {\displaystyle {\displaystyle \sum }_{k=2}^{\infty }}\varnothing \left(n,k,\mu ,\gamma \right)\left|a_n\right|+{\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}\mathsf{\psi }\left(n,k,\mu ,\gamma \right)\left|b_n\right|\le 1.\end{array}$$

$\mathrm{Since}0\le \mu \le \alpha <1\mathrm{and}{f}_n\in A_{\overline{H}}\left(k,\alpha ,\gamma \right).$ Therefore $f_n\ast Q_n\in$ $A_{\overline{H}}\left(k,\alpha ,\gamma \right)\subset A_{\overline{H}}\left(k,\mu ,\gamma \right)$. □

2.5. Integral Operator

Here, we examine the closure property of the class $A_{\overline{H}}\left(k,\alpha ,\gamma \right)$ under the generalized Bernardi-Libera-Livingston integral operator (see References [10,11]) ${\mathcal{L}}_u\left(f\right)$ which is defined by,

$${\mathcal{L}}_u\left(f\right)=\frac{u+1}{z^u}{{\displaystyle \int }}_0^z{t}^{u-1}f\left(t\right)dt,u>-1.$$

(17)

Theorem 6.

Let$f_k\left(z\right)\in$$A_{\overline{H}}\left(k,\alpha ,\gamma \right)$. Then

$${\mathcal{L}}_u\left(f_k\left(z\right)\right)\in A_{\overline{H}}\left(k,\alpha ,\gamma \right)$$

Proof. 

From definition of ${\mathcal{L}}_u\left(f_k\left(z\right)\right)$ given by Equation (17), it follows that

$$\begin{array}{c}{\mathcal{L}}_u\left(f_k\left(z\right)\right)=\frac{u+1}{z^u}{\displaystyle {{\displaystyle \int }}_0^z}{t}^{u-1}\left(t-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\left|a_n\right|t^n+{\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left|b_n\right|{\overline{t}}^n\right)dt\hfill \\ =z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\frac{u+1}{u+n}\left|a_n\right|z^n+{\left(-1\right)}^k{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\frac{u+1}{u+n}\left|b_n\right|z^n\\ =z-{\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}{G}_n{z}^n+{\left(-1\right)}^{n-1}{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{L}_n{z}^n\end{array}$$

where

$$\begin{array}{c}{G}_n=\frac{u+1}{u+n}\left|a_n\right|,\mathrm{and}\\ L_n=\frac{u+1}{u+n}\left|b_n\right|\end{array}$$

Hence

$$\begin{array}{c}{\displaystyle {\displaystyle \sum }_{k=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\frac{u+1}{u+n}\left|a_n\right|+{\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\frac{u+1}{u+n}\left|b_n\right|\hfill \\ \hfill \le {\displaystyle {\displaystyle \sum }_{n=2}^{\infty }}\varnothing \left(n,k,\alpha ,\gamma \right)\left|a_n\right|+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\mathsf{\psi }\left(n,k,\alpha ,\gamma \right)\left|b_n\right|\le 1.\end{array}$$

$\mathrm{by}\mathrm{Theorem}2.$

Therefore, we have ${\mathcal{L}}_u\left(f_k\left(z\right)\right)\in$ $A_{\overline{H}}\left(k,\alpha ,\gamma \right).$ □