# Subclass of univalent harmonic functions defined by dual convolution

## Abstract

In the present paper, we study a subclass of univalent harmonic functions defined by convolution and integral convolution. We obtain the basic properties such as coefficient characterization and distortion theorem, extreme points and convolution condition.

MSC:30C45, 30C50.

## 1 Introduction

A continuous function $f=u+iv$ is a complex-valued harmonic function in a simply connected complex domain $D\subset \mathbb{C}$ if both u and v are real harmonic in D. It was shown by Clunie and Sheil-Small  that such a harmonic function can be represented by $f=h+\overline{g}$, where h and g are analytic in D. Also, a necessary and sufficient condition for f to be locally univalent and sense-preserving in D is that $|{h}^{\prime }\left(z\right)|>|{g}^{\prime }\left(z\right)|$ (see also  and ).

Denote by ${S}_{H}$ the class of functions f that are harmonic univalent and sense-preserving in the open unit disc $U=\left\{z\in \mathbb{C}:|z|<1\right\}$, for which $f\left(0\right)=h\left(0\right)={f}_{z}^{\prime }\left(0\right)-1=0$. Then for $f=h+\overline{g}\in {S}_{H}$ we may express the analytic functions h and g as

$h\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n},\phantom{\rule{2em}{0ex}}g\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}{b}_{n}{z}^{n},\phantom{\rule{1em}{0ex}}|{b}_{1}|<1.$
(1.1)

Clunie and Sheil-Small  investigated the class ${S}_{H}$ as well as its geometric subclasses and obtained some coefficient bounds.

Also, let ${S}_{\overline{H}}$ denote the subclass of ${S}_{H}$ consisting of functions $f=h+\overline{g}$ such that the functions h and g are of the form

$h\left(z\right)=z-\sum _{n=2}^{\mathrm{\infty }}|{a}_{n}|{z}^{n},\phantom{\rule{2em}{0ex}}g\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}|{b}_{n}|{z}^{n},\phantom{\rule{1em}{0ex}}|{b}_{1}|<1.$
(1.2)

Recently Kanas and Wisniowska  (see also Kanas and Srivastava ) studied the class of k-uniformly convex analytic functions, denoted by $k-UCV$, $k\ge 0$, so that $\varphi \in k-UCV$ if and only if

$Re\left\{1+\frac{\left(z-\zeta \right){\varphi }^{″}\left(z\right)}{{\varphi }^{\prime }\left(z\right)}\right\}\ge 0\phantom{\rule{1em}{0ex}}\left(|\zeta |\le k;z\in U\right).$
(1.3)

For $\theta \in \mathbb{R}$, if we let $\zeta =-kz{e}^{i\theta }$, then condition (1.3) can be written as

$Re\left\{1+\left(1+k{e}^{i\theta }\right)\frac{z{\varphi }^{″}\left(z\right)}{{\varphi }^{\prime }\left(z\right)}\right\}\ge 0.$
(1.4)

Kim et al.  introduced and studied the class $HCV\left(k,\alpha \right)$ consisting of functions $f=h+\overline{g}$, such that h and g are given by (1.1), and satisfying the condition

$Re\left\{1+\left(1+k{e}^{i\theta }\right)\frac{{z}^{2}{h}^{″}\left(z\right)+\overline{2z{g}^{\prime }\left(z\right)+{z}^{2}{g}^{″}\left(z\right)}}{z{h}^{\prime }\left(z\right)-\overline{z{g}^{\prime }\left(z\right)}}\right\}\ge \alpha \phantom{\rule{1em}{0ex}}\left(0\le \alpha <1;\theta \in \mathbb{R};k\ge 0\right).$
(1.5)

Also, the class of $k-UST$ uniformly starlike functions is defined by using (1.4) as the class of all functions $\psi \left(z\right)=z{\varphi }^{\prime }\left(z\right)$ such that $\varphi \in k-UCV$, then $\psi \left(z\right)\in k-UST$ if and only if

$Re\left\{\left(1+k{e}^{i\theta }\right)\frac{z{\psi }^{\prime }\left(z\right)}{\psi \left(z\right)}-k{e}^{i\theta }\right\}\ge 0.$
(1.6)

Generalizing the class $k-UST$ to include harmonic functions, we let $\mathit{HST}\left(k,\alpha \right)$ denote the class of functions $f=h+\overline{g}$, such that h and g are given by (1.1), which satisfies the condition

$Re\left\{\left(1+k{e}^{i\theta }\right)\frac{z{f}^{\prime }\left(z\right)}{{z}^{\prime }f\left(z\right)}-k{e}^{i\theta }\right\}\ge \alpha \phantom{\rule{1em}{0ex}}\left(0\le \alpha <1;\theta \in \mathbb{R};k\ge 0\right).$
(1.7)

Replacing $h+\overline{g}$ for f in (1.7), we have

$Re\left\{\left(1+k{e}^{i\theta }\right)\frac{z{h}^{\prime }\left(z\right)-\overline{z{g}^{\prime }\left(z\right)}}{h\left(z\right)+\overline{g\left(z\right)}}-k{e}^{i\theta }\right\}\ge \alpha \phantom{\rule{1em}{0ex}}\left(0\le \alpha <1;\theta \in \mathbb{R};k\ge 0\right).$
(1.8)

The convolution of two functions of the form

$f\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}F\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{A}_{n}{z}^{n}$

is defined as

$\left(f\ast F\right)\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{A}_{n}{z}^{n},$
(1.9)

while the integral convolution is defined by

$\left(f\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}F\right)\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}\frac{{a}_{n}{A}_{n}}{n}{z}^{n}.$
(1.10)

From (1.9) and (1.10), we have

$\left(f\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}F\right)\left(z\right)={\int }_{0}^{z}\frac{\left(f\ast F\right)\left(t\right)}{t}\phantom{\rule{0.2em}{0ex}}dt.$

Now we consider the subclass $\mathit{HST}\left(\varphi ,\psi ,k,\alpha \right)$ consisting of functions $f=h+\overline{g}$, such that h and g are given by (1.1), and satisfying the condition

(1.11)

where

$\phi \left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{\lambda }_{n}{z}^{n}\phantom{\rule{1em}{0ex}}\left({\lambda }_{n}\ge 0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\chi \left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{\mu }_{n}{z}^{n}\phantom{\rule{1em}{0ex}}\left({\mu }_{n}\ge 0\right).$
(1.12)

We further consider the subclass $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ of $\mathit{HST}\left(\varphi ,\chi ,k,\alpha \right)$ for h and g given by (1.2).

We note that

1. (i)

$\overline{\mathit{HST}}\left(\varphi ,\chi ,0,\alpha \right)=\overline{HS}\left(\varphi ,\chi ,\alpha \right)$ (see Dixit et al. );

2. (ii)

$\overline{\mathit{HST}}\left(\frac{z}{{\left(1-z\right)}^{2}},\frac{z}{{\left(1-z\right)}^{2}},1,\alpha \right)={G}_{\overline{H}}\left(\alpha \right)$ (see Rosy et al. );

3. (iii)

$\overline{\mathit{HST}}\left(\frac{z+{z}^{2}}{{\left(1-z\right)}^{3}},\frac{z+{z}^{2}}{{\left(1-z\right)}^{3}},k,\alpha \right)=\overline{H}CV\left(k,\alpha \right)$ (see Kim et al. );

4. (iv)

$\overline{\mathit{HST}}\left(\frac{z}{{\left(1-z\right)}^{2}},\frac{z}{{\left(1-z\right)}^{2}},0,\alpha \right)={T}_{H}^{\ast }\left(\alpha \right)$ (see Jahangiri , see also Joshi and Darus );

5. (v)

$\overline{\mathit{HST}}\left(\frac{z+{z}^{2}}{{\left(1-z\right)}^{3}},\frac{z+{z}^{2}}{{\left(1-z\right)}^{3}},0,\alpha \right)={C}_{H}\left(\alpha \right)$ (see Jahangiri , see also Joshi and Darus ).

In this paper, we extend the results of the above classes to the classes $\mathit{HST}\left(\varphi ,\chi ,k,\alpha \right)$ and $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$, we also obtain some basic properties for the class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$.

## 2 Coefficient characterization and distortion theorem

Unless otherwise mentioned, we assume throughout this paper that $\phi \left(z\right)$ and $\chi \left(z\right)$ are given by (1.12), $0\le \alpha <1$, $k\ge 0$ and θ is real. We begin with a sufficient condition for functions in the class $\mathit{HST}\left(\varphi ,\chi ,k,\alpha \right)$.

Theorem 1 Let $f=h+\overline{g}$ be such that h and g are given by (1.1). Furthermore, let

$\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\left(\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{a}_{n}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{b}_{n}|\le 1,$
(2.1)

where

$\begin{array}{c}{n}^{2}\left(1-\alpha \right)\le {\lambda }_{n}\left[\left(1+k\right)n-\left(k+\alpha \right)\right]\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{n}^{2}\left(1-\alpha \right)\le {\mu }_{n}\left[\left(1+k\right)n+\left(k+\alpha \right)\right]\hfill \\ \phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.25em}{0ex}}n\ge 2.\hfill \end{array}$

Then f is sense-preserving, harmonic univalent in U and $f\in \mathit{HST}\left(\varphi ,\chi ,k,\alpha \right)$.

Proof First we note that f is locally univalent and sense-preserving in U. This is because

$\begin{array}{rcl}|{h}^{\prime }\left(z\right)|& \ge & 1-\sum _{n=2}^{\mathrm{\infty }}n|{a}_{n}|{r}^{n-1}>1-\sum _{n=2}^{\mathrm{\infty }}n|{a}_{n}|\ge 1-\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\left(\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{a}_{n}|\\ \ge & \sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{b}_{n}|\ge \sum _{n=1}^{\mathrm{\infty }}n|{b}_{n}|\ge \sum _{n=1}^{\mathrm{\infty }}n|{b}_{n}|{r}^{k-1}>|{g}^{\prime }\left(z\right)|.\end{array}$

To show that f is univalent in U, suppose ${z}_{1},{z}_{2}\in U$ so that ${z}_{1}\ne {z}_{2}$, then

$\begin{array}{rcl}|\frac{f\left({z}_{1}\right)-f\left({z}_{2}\right)}{h\left({z}_{1}\right)-h\left({z}_{2}\right)}|& \ge & 1-|\frac{g\left({z}_{1}\right)-g\left({z}_{2}\right)}{h\left({z}_{1}\right)-h\left({z}_{2}\right)}|=1-|\frac{{\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}\left({z}_{1}^{n}-{z}_{2}^{n}\right)}{\left({z}_{1}-{z}_{2}\right)+{\sum }_{n=2}^{\mathrm{\infty }}{a}_{n}\left({z}_{1}^{n}-{z}_{2}^{n}\right)}|\\ \ge & 1-\frac{{\sum }_{n=1}^{\mathrm{\infty }}n|{b}_{n}|}{1-{\sum }_{n=2}^{\mathrm{\infty }}n|{a}_{n}|}>1-\frac{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{b}_{n}|}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\left(\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{a}_{n}|}\ge 0.\end{array}$

Now, we prove that $f\in \mathit{HST}\left(\varphi ,\psi ,k,\alpha \right)$, by definition, we only need to show that if (2.1) holds, then condition (1.11) is satisfied. From (1.11), it suffices to show that

$\begin{array}{r}Re\left\{\frac{\left(1+k{e}^{i\theta }\right)\left(h\left(z\right)\ast \phi \left(z\right)-\overline{g\left(z\right)\ast \chi \left(z\right)}\right)-\left(k{e}^{i\theta }+\alpha \right)\left(h\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\phi \left(z\right)+\overline{g\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\chi \left(z\right)}\right)}{h\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\phi \left(z\right)+\overline{g\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\chi \left(z\right)}}\right\}\\ \phantom{\rule{1em}{0ex}}\ge 0.\end{array}$
(2.2)

Substituting for h, g, φ and χ in (2.2) and dividing by $\left(1-\alpha \right)z$, we obtain $Re\frac{A\left(z\right)}{B\left(z\right)}\ge 0$, where

$\begin{array}{rcl}A\left(z\right)& =& 1+\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k{e}^{i\theta }\right)n-\left(k{e}^{i\theta }+\alpha \right)}{\left(1-\alpha \right)}{a}_{n}{z}^{n-1}\\ -\left(\frac{\overline{z}}{z}\right)\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k{e}^{i\theta }\right)n+\left(k{e}^{i\theta }+\alpha \right)}{\left(1-\alpha \right)}{b}_{n}{\overline{z}}^{n-1}\end{array}$

and

$B\left(z\right)=1+\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}{a}_{n}{z}^{n-1}+\left(\frac{\overline{z}}{z}\right)\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}{b}_{n}{\overline{z}}^{n-1}.$

Using the fact that $Re\left(w\right)\ge 0$ if and only if $|1+w|\ge |1-w|$ in U, it suffices to show that $|A\left(z\right)+B\left(z\right)|-|A\left(z\right)-B\left(z\right)|\ge 0$. Substituting for $A\left(z\right)$ and $B\left(z\right)$ gives

$\begin{array}{c}|A\left(z\right)+B\left(z\right)|-|A\left(z\right)-B\left(z\right)|\hfill \\ \phantom{\rule{1em}{0ex}}=|2+\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k{e}^{i\theta }\right)n-\left(k{e}^{i\theta }+2\alpha -1\right)}{\left(1-\alpha \right)}{a}_{n}{z}^{n-1}\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\frac{\overline{z}}{z}\right)\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k{e}^{i\theta }\right)n+\left(k{e}^{i\theta }+2\alpha -1\right)}{\left(1-\alpha \right)}{b}_{n}{\overline{z}}^{n-1}|\hfill \\ \phantom{\rule{2em}{0ex}}-|\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k{e}^{i\theta }\right)n-\left(1+k{e}^{i\theta }\right)}{\left(1-\alpha \right)}{a}_{n}{z}^{n-1}\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\frac{\overline{z}}{z}\right)\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k{e}^{i\theta }\right)n+\left(1+k{e}^{i\theta }\right)}{\left(1-\alpha \right)}{b}_{n}{\overline{z}}^{n-1}|\hfill \\ \phantom{\rule{1em}{0ex}}\ge 2-\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+2\alpha -1\right)}{\left(1-\alpha \right)}|{a}_{n}||z{|}^{n-1}\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+2\alpha -1\right)}{\left(1-\alpha \right)}|{b}_{n}||z{|}^{n-1}\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(1+k\right)}{\left(1-\alpha \right)}|{a}_{n+1}||z{|}^{n-1}\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(1+k\right)}{\left(1-\alpha \right)}|{b}_{n}||z{|}^{n-1}\hfill \\ \phantom{\rule{1em}{0ex}}\ge 2\left\{1-\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}|{a}_{n}|-\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}|{b}_{n}|\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge 0\phantom{\rule{1em}{0ex}}\text{by (2.1)}.\hfill \end{array}$

The harmonic functions

$\begin{array}{rl}f\left(z\right)=& z+\sum _{n=2}^{\mathrm{\infty }}\frac{n}{{\lambda }_{n}}\frac{\left(1-\alpha \right)}{\left(1+k\right)n-\left(k+\alpha \right)}{x}_{n}{z}^{n}\\ +\sum _{n=1}^{\mathrm{\infty }}\frac{n}{{\mu }_{n}}\frac{\left(1-\alpha \right)}{\left(1+k\right)n+\left(k+\alpha \right)}{\overline{y}}_{n}{\overline{z}}^{n},\end{array}$
(2.3)

where ${\sum }_{n=2}^{\mathrm{\infty }}|{x}_{n}|+{\sum }_{n=1}^{\mathrm{\infty }}|{y}_{n}|=1$, show that the coefficient bound given by (2.1) is sharp. The functions of the form (2.3) are in the class $\mathit{HST}\left(\varphi ,\chi ,k,\alpha \right)$ because

$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}\left[\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}|{a}_{n}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}|{b}_{n}|\right]\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=2}^{\mathrm{\infty }}|{x}_{n}|+\sum _{n=1}^{\mathrm{\infty }}|{y}_{n}|=1.\hfill \end{array}$

This completes the proof of Theorem 1. □

In the following theorem, it is shown that condition (2.1) is also necessary for functions $f=h+\overline{g}$, where h and g are given by (1.2).

Theorem 2 Let $f=h+\overline{g}$ be such that h and g are given by (1.2). Then $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ if and only if

$\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\left(\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{a}_{n}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}\right)|{b}_{n}|\le 1.$
(2.4)

Proof Since $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)\subset \mathit{HST}\left(\varphi ,\chi ,k,\alpha \right)$, we only need to prove the ‘only if’ part of the theorem. To this end, we notice that the necessary and sufficient condition for $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ is that

$Re\left\{\left(1+k{e}^{i\theta }\right)\frac{h\left(z\right)\ast \phi \left(z\right)-\overline{g\left(z\right)\ast \chi \left(z\right)}}{h\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\phi \left(z\right)+\overline{g\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\chi \left(z\right)}}-k{e}^{i\theta }\right\}\ge \alpha .$

This is equivalent to

$Re\left\{\frac{\left(1+k{e}^{i\theta }\right)\left(h\left(z\right)\ast \phi \left(z\right)-\overline{g\left(z\right)\ast \chi \left(z\right)}\right)-\left(k{e}^{i\theta }+\alpha \right)\left(h\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\phi \left(z\right)+\overline{g\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\chi \left(z\right)}\right)}{h\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\phi \left(z\right)+\overline{g\left(z\right)\phantom{\rule{0.2em}{0ex}}♢\phantom{\rule{0.2em}{0ex}}\chi \left(z\right)}}\right\}>0,$

which implies that

$\begin{array}{r}Re\left\{\frac{\left(1-\alpha \right)z-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\left[\left(1+k{e}^{i\theta }\right)n-\left(k{e}^{i\theta }+\alpha \right)\right]|{a}_{n}|{z}^{n}}{z-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}|{a}_{n}|{z}^{n}+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}|{b}_{n}|{\overline{z}}^{n}}\\ \phantom{\rule{2em}{0ex}}-\frac{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\left[\left(1+k{e}^{i\theta }\right)n+\left(k{e}^{i\theta }+\alpha \right)\right]|{b}_{n}|{\overline{z}}^{n}}{z-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}|{a}_{n}|{z}^{n}+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}|{b}_{n}|{\overline{z}}^{n}}\right\}\\ \phantom{\rule{1em}{0ex}}=Re\left\{\frac{\left(1-\alpha \right)-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\left[\left(1+k{e}^{i\theta }\right)n-\left(k{e}^{i\theta }+\alpha \right)\right]|{a}_{n}|{z}^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}|{a}_{n}|{z}^{n-1}+\left(\frac{\overline{z}}{z}\right){\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}|{b}_{n}|{\overline{z}}^{n-1}}\\ \phantom{\rule{2em}{0ex}}-\frac{\left(\frac{\overline{z}}{z}\right){\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\left[\left(1+k{e}^{i\theta }\right)n+\left(k{e}^{i\theta }+\alpha \right)\right]|{b}_{n}|{\overline{z}}^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}|{a}_{n}|{z}^{n}+\left(\frac{\overline{z}}{z}\right){\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}|{b}_{n}|{\overline{z}}^{n-1}}\right\}>0,\end{array}$
(2.5)

since $Re\left({e}^{i\theta }\right)\le |{e}^{i\theta }|=1$, the required condition (2.5) is equivalent to

$\begin{array}{r}\left\{\frac{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}|{a}_{n}|{r}^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}|{a}_{n}|{r}^{n-1}+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}|{b}_{n}|{r}^{n-1}}-\frac{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}|{b}_{n}|{r}^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}|{a}_{n}|{r}^{n-1}+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}|{b}_{n}|{r}^{n-1}}\right\}\\ \phantom{\rule{1em}{0ex}}\ge 0.\end{array}$
(2.6)

If condition (2.4) does not hold, then the numerator in (2.6) is negative for $z=r$ sufficiently close to 1. Hence there exists ${z}_{0}={r}_{0}$ in $\left(0,1\right)$ for which the quotient in (2.6) is negative. This contradicts the required condition for $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$, and so the proof of Theorem 2 is completed. □

Theorem 3 Let $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. Then, for $|z|=r<1$, $|{b}_{1}|<\frac{1-\alpha }{2k+\alpha +1}$ and

${D}_{n}\le \frac{{\lambda }_{n}}{n},\phantom{\rule{2em}{0ex}}{E}_{n}\le \frac{{\mu }_{n}}{n}\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.25em}{0ex}}n\ge 2\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}C=min\left\{{D}_{2},{E}_{2}\right\},$
(2.7)

we have

$|f\left(z\right)|\le \left(1+|{b}_{1}|\right)r+\left\{\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}-\frac{2k+1+\alpha }{C\left(2+k-\alpha \right)}|{b}_{1}|\right\}{r}^{2}$

and

$|f\left(z\right)|\ge \left(1-|{b}_{1}|\right)r-\left\{\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}-\frac{2k+1+\alpha }{C\left(2+k-\alpha \right)}|{b}_{1}|\right\}{r}^{2}.$

The results are sharp.

Proof We prove the left-hand side inequality for $|f|$. The proof for the right-hand side inequality can be done by using similar arguments.

Let $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$, then we have

$\begin{array}{rl}|f\left(z\right)|=& |z-\sum _{n=2}^{\mathrm{\infty }}|{a}_{n}|{z}^{n}+\sum _{n=1}^{\mathrm{\infty }}|{b}_{n}|{\overline{z}}^{n}|\\ \ge & r-|{b}_{1}|r-\sum _{n=2}^{\mathrm{\infty }}\left(|{a}_{n}|+|{b}_{n}|\right){r}^{2}\\ \ge & r-|{b}_{1}|r\\ -\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}\sum _{n=2}^{\mathrm{\infty }}\frac{C\left(\left(1+k\right)n-\left(k+\alpha \right)\right)}{\left(1-\alpha \right)}\left(|{a}_{n}|+|{b}_{n}|\right){r}^{2}\\ \ge & r-|{b}_{1}|r\\ -\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}\sum _{n=2}^{\mathrm{\infty }}\left\{\frac{C\left(\left(1+k\right)n-\left(k+\alpha \right)\right)}{\left(1-\alpha \right)}|{a}_{n}|\\ +\frac{C\left(\left(1+k\right)n+\left(k+\alpha \right)\right)}{\left(1-\alpha \right)}|{b}_{n}|\right\}{r}^{2}\\ \ge & \left(1-|{b}_{1}|\right)r-\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}\left\{1-\frac{2k+1+\alpha }{\left(1-\alpha \right)}|{b}_{1}|\right\}{r}^{2}\\ \ge & \left(1-|{b}_{1}|\right)r-\left\{\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}-\frac{2k+1+\alpha }{C\left(2+k-\alpha \right)}|{b}_{1}|\right\}{r}^{2}.\end{array}$

The bounds given in Theorem 3 are respectively attained for the following functions:

$f\left(z\right)=z+|{b}_{1}|\overline{z}+\left(\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}-\frac{2k+1+\alpha }{C\left(2+k-\alpha \right)}|{b}_{1}|\right){\overline{z}}^{2}$

and

$f\left(z\right)=\left(1-|{b}_{1}|\right)z-\left(\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}-\frac{2k+1+\alpha }{C\left(2+k-\alpha \right)}|{b}_{1}|\right){z}^{2}.$

□

The following covering result follows from the left side inequality in Theorem 3.

Corollary 1 Let $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$, then for $|{b}_{1}|<\frac{1-\alpha }{2k+\alpha +1}$ the set

$\left\{w:|w|<1-\frac{\left(1-\alpha \right)}{C\left(2+k-\alpha \right)}-\left(1-\frac{2k+1+\alpha }{C\left(2+k-\alpha \right)}\right)|{b}_{1}|\right\}$

is included in $f\left(U\right)$, where C is given by (2.7).

## 3 Extreme points

Our next theorem is on the extreme points of convex hulls of the class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$, denoted by $clco\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$.

Theorem 4 Let $f=h+\overline{g}$ be such that h and g are given by (1.2). Then $f\in clco\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ if and only if f can be expressed as

$f\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\left[{X}_{n}{h}_{n}\left(z\right)+{Y}_{n}{g}_{n}\left(z\right)\right],$
(3.1)

where

$\begin{array}{c}{h}_{1}\left(z\right)=z,\hfill \\ {h}_{n}\left(z\right)=z-\frac{n\left(1-\alpha \right)}{{\lambda }_{n}\left(\left(1+k\right)n-\left(k+\alpha \right)\right)}{z}^{n}\phantom{\rule{1em}{0ex}}\left(n\ge 2\right),\hfill \\ {g}_{n}\left(z\right)=z+\frac{n\left(1-\alpha \right)}{{\mu }_{n}\left(\left(1+k\right)n+\left(k+\alpha \right)\right)}{\overline{z}}^{n}\phantom{\rule{1em}{0ex}}\left(n\ge 1\right),\hfill \\ {X}_{n}\ge 0,\phantom{\rule{2em}{0ex}}{Y}_{n}\ge 0,\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}\left[{X}_{n}+{Y}_{n}\right]=1.\hfill \end{array}$

In particular, the extreme points of the class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ are $\left\{{h}_{n}\right\}$ and $\left\{{g}_{n}\right\}$, respectively.

Proof For functions $f\left(z\right)$ of the form (3.1), we have

$f\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\left[{X}_{n}+{Y}_{n}\right]z-\sum _{n=2}^{\mathrm{\infty }}\frac{n\left(1-\alpha \right)}{{\lambda }_{n}\left(\left(1+k\right)n-\left(k+\alpha \right)\right)}{X}_{n}{z}^{n}+\sum _{n=1}^{\mathrm{\infty }}\frac{n\left(1-\alpha \right)}{{\mu }_{n}\left(\left(1+k\right)n+\left(k+\alpha \right)\right)}{Y}_{n}{\overline{z}}^{n}.$

Then

$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}\left(\left(1+k\right)n-\left(k+\alpha \right)\right)}{n\left(1-\alpha \right)}\left(\frac{n\left(1-\alpha \right)}{{\lambda }_{n}\left(\left(1+k\right)n-\left(k+\alpha \right)\right)}\right){X}_{n}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}\left(\left(1+k\right)n+\left(k+\alpha \right)\right)}{n\left(1-\alpha \right)}\left(\frac{n\left(1-\alpha \right)}{{\mu }_{n}\left(\left(1+k\right)n+\left(k+\alpha \right)\right)}\right){Y}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=2}^{\mathrm{\infty }}{X}_{n}+\sum _{n=1}^{\mathrm{\infty }}{Y}_{n}=1-{X}_{1}\le 1,\hfill \end{array}$

and so $f\left(z\right)\in clco\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. Conversely, suppose that $f\left(z\right)\in clco\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. Set

${X}_{n}=\frac{{\lambda }_{n}\left(\left(1+k\right)n-\left(k+\alpha \right)\right)}{n\left(1-\alpha \right)}|{a}_{n}|\phantom{\rule{1em}{0ex}}\left(n\ge 2\right)$

and

${Y}_{n}=\frac{{\mu }_{n}\left(\left(1+k\right)n+\left(k+\alpha \right)\right)}{n\left(1-\alpha \right)}|{b}_{n}|\phantom{\rule{1em}{0ex}}\left(n\ge 1\right),$

then note that by Theorem 2, $0\le {X}_{n}\le 1$ ($n\ge 2$) and $0\le {Y}_{n}\le 1$ ($n\ge 1$).

Consequently, we obtain

$f\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\left[{X}_{n}{h}_{n}\left(z\right)+{Y}_{n}{g}_{n}\left(z\right)\right].$

Using Theorem 2, it is easily seen that the class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ is convex and closed and so $clco\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)=\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. □

## 4 Convolution result

For harmonic functions of the form

$f\left(z\right)=z-\sum _{n=2}^{\mathrm{\infty }}|{a}_{n}|{z}^{n}+\sum _{n=1}^{\mathrm{\infty }}|{b}_{n}|{\overline{z}}^{n}$
(4.1)

and

$G\left(z\right)=z-\sum _{n=2}^{\mathrm{\infty }}{A}_{n}{z}^{n}+\sum _{n=1}^{\mathrm{\infty }}{B}_{n}{\overline{z}}^{n}\phantom{\rule{1em}{0ex}}\left({A}_{n},{B}_{n}\ge 0\right),$
(4.2)

we define the convolution of two harmonic functions f and G as

$\left(f\ast G\right)\left(z\right)=f\left(z\right)\ast G\left(z\right)=z-\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{A}_{n}{z}^{n}+\sum _{n=1}^{\mathrm{\infty }}{b}_{n}{B}_{n}{\overline{z}}^{n}.$

Using this definition, we show that the class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ is closed under convolution.

Theorem 5 For $0\le \alpha <1$, let $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ and $G\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. Then $f\left(z\right)\ast G\left(z\right)\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$.

Proof Let the functions $f\left(z\right)$ defined by (4.1) be in the class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$, and let the functions $G\left(z\right)$ defined by (4.2) be in the class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. Obviously, the coefficients of f and G must satisfy a condition similar to inequality (2.4). So, for the coefficients of $f\left(z\right)\ast G\left(z\right)$, we can write

$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}|{a}_{n}|{A}_{n}+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}|{b}_{n}|{B}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}\le \sum _{n=2}^{\mathrm{\infty }}\left[\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}|{a}_{n}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}|{b}_{n}|\right],\hfill \end{array}$

the right-hand side of this inequality is bounded by 1 because $f\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. Then $f\left(z\right)\ast G\left(z\right)\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. □

Finally, we show that $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ is closed under convex combinations of its members.

Theorem 6 The class $\overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$ is closed under convex linear combination.

Proof For $i=1,2,3,\dots$ , let ${f}_{i}\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$, where the functions ${f}_{i}$ are given by

${f}_{i}\left(z\right)=z-\sum _{n=2}^{\mathrm{\infty }}|{a}_{n,i}|{z}^{n}+\sum _{n=1}^{\mathrm{\infty }}|{b}_{n,i}|{\overline{z}}^{n}.$

For ${\sum }_{i=1}^{\mathrm{\infty }}{t}_{i}=1$; $0\le {t}_{i}\le 1$, the convex linear combination of ${f}_{i}$ may be written as

$\sum _{i=1}^{\mathrm{\infty }}{t}_{i}{f}_{i}\left(z\right)=z-\sum _{n=2}^{\mathrm{\infty }}\left(\sum _{i=1}^{\mathrm{\infty }}{t}_{i}|{a}_{n,i}|\right){z}^{n}+\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{i=1}^{\mathrm{\infty }}{t}_{i}|{b}_{n,i}|\right){\overline{z}}^{n},$

then by (2.4) we have

$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}\sum _{i=1}^{\mathrm{\infty }}{t}_{i}|{a}_{n,i}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}\sum _{i=1}^{\mathrm{\infty }}{t}_{i}|{b}_{n,i}|\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{i=1}^{\mathrm{\infty }}{t}_{i}\left\{\sum _{n=2}^{\mathrm{\infty }}\left[\frac{{\lambda }_{n}}{n}\frac{\left(1+k\right)n-\left(k+\alpha \right)}{\left(1-\alpha \right)}|{a}_{n,i}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_{n}}{n}\frac{\left(1+k\right)n+\left(k+\alpha \right)}{\left(1-\alpha \right)}|{b}_{n,i}|\right]\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\le \sum _{i=1}^{\mathrm{\infty }}{t}_{i}=1.\hfill \end{array}$

This condition is required by (2.4) and so ${\sum }_{i=1}^{\mathrm{\infty }}{t}_{i}{f}_{i}\left(z\right)\in \overline{\mathit{HST}}\left(\varphi ,\chi ,k,\alpha \right)$. This completes the proof of Theorem 6. □

Remarks

1. (i)

Putting $k=0$ in our results, we obtain the results obtained by Dixit et al. ;

2. (ii)

Putting $\phi \left(z\right)=\chi \left(z\right)=\frac{z}{{\left(1-z\right)}^{2}}$ and $k=1$ in our results, we obtain the results obtained by Rosy et al. ;

3. (iii)

Putting $\phi \left(z\right)=\chi \left(z\right)=\frac{z+{z}^{2}}{{\left(1-z\right)}^{3}}$ in our results, we obtain the results obtained by Kim et al. ;

4. (iv)

Putting $\phi \left(z\right)=\chi \left(z\right)=\frac{z}{{\left(1-z\right)}^{2}}$ and $k=0$ in our results, we obtain the results obtained by Jahangiri ;

5. (v)

Putting $\phi \left(z\right)=\chi \left(z\right)=\frac{z+{z}^{2}}{{\left(1-z\right)}^{3}}$ and $k=0$ in our results, we obtain the results obtained by Jahangiri .

## References

1. Clunie J, Sheil-Small T: Harmonic univalent functions. Ann. Acad. Sci. Fenn., Ser. A 1 Math. 1984, 9: 3–25.

2. Jahangiri JM: Coefficient bounds and univalent criteria for harmonic functions with negative coefficients. Ann. Univ. Marie-Curie Sklodowska Sect. A 1998, 52: 57–66.

3. Jahangiri JM: Harmonic functions starlike in the unit disc. J. Math. Anal. Appl. 1999, 235: 470–477. 10.1006/jmaa.1999.6377

4. Silverman H: Harmonic univalent function with negative coefficients. J. Math. Anal. Appl. 1998, 220: 283–289. 10.1006/jmaa.1997.5882

5. Silverman H, Silvia EM: Subclasses of harmonic univalent functions. N.Z. J. Math. 1999, 28: 275–284.

6. Kanas S, Wisniowska A: Conic regions and k -uniform convexity. J. Comput. Appl. Math. 1999, 105: 327–336. 10.1016/S0377-0427(99)00018-7

7. Kanas S, Srivastava HM: Linear operators associated with k -uniformly convex functions. Integral Transforms Spec. Funct. 2000, 9(2):121–132. 10.1080/10652460008819249

8. Kim YC, Jahangiri JM, Choi JH: Certain convex harmonic functions. Int. J. Math. Math. Sci. 2002, 29(8):459–465. 10.1155/S0161171202007585

9. Dixit KK, Pathak AL, Porwal S, Agarwal R: On a subclass of harmonic univalent functions defied by convolution and integral convolution. Int. J. Pure Appl. Math. 2011, 69(3):255–264.

10. Rosy T, Stephen BA, Subramanian KG, Jahangiri JM: Goodman-Ronning-type harmonic univalent functions. Kyungpook Math. J. 2001, 41: 45–54.

11. Joshi SB, Darus M: Unified treatment for harmonic univalent functions. Tamsui Oxford Univ. J. Math. Sci. 2008, 24(3):225–232.

## Acknowledgements

The author would like to express her sincere gratitude to Springer Open Accounts Team for their kind help.

## Author information

Authors

### Corresponding author

Correspondence to Rabha M El-Ashwah.

### Competing interests

The author declares that they have no competing interests.

## Rights and permissions

Reprints and Permissions

El-Ashwah, R.M. Subclass of univalent harmonic functions defined by dual convolution. J Inequal Appl 2013, 537 (2013). https://doi.org/10.1186/1029-242X-2013-537

• Accepted:

• Published:

• DOI: https://doi.org/10.1186/1029-242X-2013-537

### Keywords

• harmonic function
• univalent
• sense-preserving
• integral convolution 