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# 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 $DâŠ‚\mathbb{C}$ if both u and v are real harmonic in D. It was shown by Clunie and Sheil-Small [1] that such a harmonic function can be represented by $f=h+\stackrel{Â¯}{g}$, 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 $|{h}^{â€²}\left(z\right)|>|{g}^{â€²}\left(z\right)|$ (see also [2â€“4] and [5]).

Denote by ${S}_{H}$ the class of functions f that are harmonic univalent and sense-preserving in the open unit disc $U=\left\{zâˆˆ\mathbb{C}:|z|<1\right\}$, for which $f\left(0\right)=h\left(0\right)={f}_{z}^{â€²}\left(0\right)âˆ’1=0$. Then for $f=h+\stackrel{Â¯}{g}âˆˆ{S}_{H}$ we may express the analytic functions h and g as

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

Clunie and Sheil-Small [1] investigated the class ${S}_{H}$ as well as its geometric subclasses and obtained some coefficient bounds.

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

$h\left(z\right)=zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}|{a}_{n}|{z}^{n},\phantom{\rule{2em}{0ex}}g\left(z\right)=\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}|{b}_{n}|{z}^{n},\phantom{\rule{1em}{0ex}}|{b}_{1}|<1.$
(1.2)

Recently Kanas and Wisniowska [6] (see also Kanas and Srivastava [7]) studied the class of k-uniformly convex analytic functions, denoted by $kâˆ’UCV$, $kâ‰¥0$, so that $\mathrm{Ï•}âˆˆkâˆ’UCV$ if and only if

$Re\left\{1+\frac{\left(zâˆ’\mathrm{Î¶}\right){\mathrm{Ï•}}^{â€³}\left(z\right)}{{\mathrm{Ï•}}^{â€²}\left(z\right)}\right\}â‰¥0\phantom{\rule{1em}{0ex}}\left(|\mathrm{Î¶}|â‰¤k;zâˆˆU\right).$
(1.3)

For $\mathrm{Î¸}âˆˆ\mathbb{R}$, if we let $\mathrm{Î¶}=âˆ’kz{e}^{i\mathrm{Î¸}}$, then condition (1.3) can be written as

$Re\left\{1+\left(1+k{e}^{i\mathrm{Î¸}}\right)\frac{z{\mathrm{Ï•}}^{â€³}\left(z\right)}{{\mathrm{Ï•}}^{â€²}\left(z\right)}\right\}â‰¥0.$
(1.4)

Kim et al. [8] introduced and studied the class $HCV\left(k,\mathrm{Î±}\right)$ consisting of functions $f=h+\stackrel{Â¯}{g}$, such that h and g are given by (1.1), and satisfying the condition

$Re\left\{1+\left(1+k{e}^{i\mathrm{Î¸}}\right)\frac{{z}^{2}{h}^{â€³}\left(z\right)+\stackrel{Â¯}{2z{g}^{â€²}\left(z\right)+{z}^{2}{g}^{â€³}\left(z\right)}}{z{h}^{â€²}\left(z\right)âˆ’\stackrel{Â¯}{z{g}^{â€²}\left(z\right)}}\right\}â‰¥\mathrm{Î±}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î±}<1;\mathrm{Î¸}âˆˆ\mathbb{R};kâ‰¥0\right).$
(1.5)

Also, the class of $kâˆ’UST$ uniformly starlike functions is defined by using (1.4) as the class of all functions $\mathrm{Ïˆ}\left(z\right)=z{\mathrm{Ï•}}^{â€²}\left(z\right)$ such that $\mathrm{Ï•}âˆˆkâˆ’UCV$, then $\mathrm{Ïˆ}\left(z\right)âˆˆkâˆ’UST$ if and only if

$Re\left\{\left(1+k{e}^{i\mathrm{Î¸}}\right)\frac{z{\mathrm{Ïˆ}}^{â€²}\left(z\right)}{\mathrm{Ïˆ}\left(z\right)}âˆ’k{e}^{i\mathrm{Î¸}}\right\}â‰¥0.$
(1.6)

Generalizing the class $kâˆ’UST$ to include harmonic functions, we let $\mathit{HST}\left(k,\mathrm{Î±}\right)$ denote the class of functions $f=h+\stackrel{Â¯}{g}$, such that h and g are given by (1.1), which satisfies the condition

$Re\left\{\left(1+k{e}^{i\mathrm{Î¸}}\right)\frac{z{f}^{â€²}\left(z\right)}{{z}^{â€²}f\left(z\right)}âˆ’k{e}^{i\mathrm{Î¸}}\right\}â‰¥\mathrm{Î±}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î±}<1;\mathrm{Î¸}âˆˆ\mathbb{R};kâ‰¥0\right).$
(1.7)

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

$Re\left\{\left(1+k{e}^{i\mathrm{Î¸}}\right)\frac{z{h}^{â€²}\left(z\right)âˆ’\stackrel{Â¯}{z{g}^{â€²}\left(z\right)}}{h\left(z\right)+\stackrel{Â¯}{g\left(z\right)}}âˆ’k{e}^{i\mathrm{Î¸}}\right\}â‰¥\mathrm{Î±}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î±}<1;\mathrm{Î¸}âˆˆ\mathbb{R};kâ‰¥0\right).$
(1.8)

The convolution of two functions of the form

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

is defined as

$\left(fâˆ—F\right)\left(z\right)=z+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{n}{A}_{n}{z}^{n},$
(1.9)

while the integral convolution is defined by

$\left(f\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}F\right)\left(z\right)=z+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{a}_{n}{A}_{n}}{n}{z}^{n}.$
(1.10)

From (1.9) and (1.10), we have

$\left(f\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}F\right)\left(z\right)={âˆ«}_{0}^{z}\frac{\left(fâˆ—F\right)\left(t\right)}{t}\phantom{\rule{0.2em}{0ex}}dt.$

Now we consider the subclass $\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ïˆ},k,\mathrm{Î±}\right)$ consisting of functions $f=h+\stackrel{Â¯}{g}$, such that h and g are given by (1.1), and satisfying the condition

(1.11)

where

$\mathrm{Ï†}\left(z\right)=z+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathrm{Î»}}_{n}{z}^{n}\phantom{\rule{1em}{0ex}}\left({\mathrm{Î»}}_{n}â‰¥0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{Ï‡}\left(z\right)=z+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{\mathrm{Î¼}}_{n}{z}^{n}\phantom{\rule{1em}{0ex}}\left({\mathrm{Î¼}}_{n}â‰¥0\right).$
(1.12)

We further consider the subclass $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ of $\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ for h and g given by (1.2).

We note that

1. (i)

$\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},0,\mathrm{Î±}\right)=\stackrel{Â¯}{HS}\left(\mathrm{Ï•},\mathrm{Ï‡},\mathrm{Î±}\right)$ (see Dixit et al. [9]);

2. (ii)

$\stackrel{Â¯}{\mathit{HST}}\left(\frac{z}{{\left(1âˆ’z\right)}^{2}},\frac{z}{{\left(1âˆ’z\right)}^{2}},1,\mathrm{Î±}\right)={G}_{\stackrel{Â¯}{H}}\left(\mathrm{Î±}\right)$ (see Rosy et al. [10]);

3. (iii)

$\stackrel{Â¯}{\mathit{HST}}\left(\frac{z+{z}^{2}}{{\left(1âˆ’z\right)}^{3}},\frac{z+{z}^{2}}{{\left(1âˆ’z\right)}^{3}},k,\mathrm{Î±}\right)=\stackrel{Â¯}{H}CV\left(k,\mathrm{Î±}\right)$ (see Kim et al. [8]);

4. (iv)

$\stackrel{Â¯}{\mathit{HST}}\left(\frac{z}{{\left(1âˆ’z\right)}^{2}},\frac{z}{{\left(1âˆ’z\right)}^{2}},0,\mathrm{Î±}\right)={T}_{H}^{âˆ—}\left(\mathrm{Î±}\right)$ (see Jahangiri [3], see also Joshi and Darus [11]);

5. (v)

$\stackrel{Â¯}{\mathit{HST}}\left(\frac{z+{z}^{2}}{{\left(1âˆ’z\right)}^{3}},\frac{z+{z}^{2}}{{\left(1âˆ’z\right)}^{3}},0,\mathrm{Î±}\right)={C}_{H}\left(\mathrm{Î±}\right)$ (see Jahangiri [3], see also Joshi and Darus [11]).

In this paper, we extend the results of the above classes to the classes $\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ and $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, we also obtain some basic properties for the class $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$.

## 2 Coefficient characterization and distortion theorem

Unless otherwise mentioned, we assume throughout this paper that $\mathrm{Ï†}\left(z\right)$ and $\mathrm{Ï‡}\left(z\right)$ are given by (1.12), $0â‰¤\mathrm{Î±}<1$, $kâ‰¥0$ and Î¸ is real. We begin with a sufficient condition for functions in the class $\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$.

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

$\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\left(\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{a}_{n}|+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{b}_{n}|â‰¤1,$
(2.1)

where

$\begin{array}{c}{n}^{2}\left(1âˆ’\mathrm{Î±}\right)â‰¤{\mathrm{Î»}}_{n}\left[\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right]\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{n}^{2}\left(1âˆ’\mathrm{Î±}\right)â‰¤{\mathrm{Î¼}}_{n}\left[\left(1+k\right)n+\left(k+\mathrm{Î±}\right)\right]\hfill \\ \phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.25em}{0ex}}nâ‰¥2.\hfill \end{array}$

Then f is sense-preserving, harmonic univalent in U and $fâˆˆ\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$.

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

$\begin{array}{rcl}|{h}^{â€²}\left(z\right)|& â‰¥& 1âˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}n|{a}_{n}|{r}^{nâˆ’1}>1âˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}n|{a}_{n}|â‰¥1âˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\left(\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{a}_{n}|\\ â‰¥& \underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{b}_{n}|â‰¥\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}n|{b}_{n}|â‰¥\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}n|{b}_{n}|{r}^{kâˆ’1}>|{g}^{â€²}\left(z\right)|.\end{array}$

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

$\begin{array}{rcl}|\frac{f\left({z}_{1}\right)âˆ’f\left({z}_{2}\right)}{h\left({z}_{1}\right)âˆ’h\left({z}_{2}\right)}|& â‰¥& 1âˆ’|\frac{g\left({z}_{1}\right)âˆ’g\left({z}_{2}\right)}{h\left({z}_{1}\right)âˆ’h\left({z}_{2}\right)}|=1âˆ’|\frac{{âˆ‘}_{n=1}^{\mathrm{âˆž}}{b}_{n}\left({z}_{1}^{n}âˆ’{z}_{2}^{n}\right)}{\left({z}_{1}âˆ’{z}_{2}\right)+{âˆ‘}_{n=2}^{\mathrm{âˆž}}{a}_{n}\left({z}_{1}^{n}âˆ’{z}_{2}^{n}\right)}|\\ â‰¥& 1âˆ’\frac{{âˆ‘}_{n=1}^{\mathrm{âˆž}}n|{b}_{n}|}{1âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}n|{a}_{n}|}>1âˆ’\frac{{âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{b}_{n}|}{1âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}\left(\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{a}_{n}|}â‰¥0.\end{array}$

Now, we prove that $fâˆˆ\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ïˆ},k,\mathrm{Î±}\right)$, 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

$\begin{array}{r}Re\left\{\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)\left(h\left(z\right)âˆ—\mathrm{Ï†}\left(z\right)âˆ’\stackrel{Â¯}{g\left(z\right)âˆ—\mathrm{Ï‡}\left(z\right)}\right)âˆ’\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)\left(h\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï†}\left(z\right)+\stackrel{Â¯}{g\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï‡}\left(z\right)}\right)}{h\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï†}\left(z\right)+\stackrel{Â¯}{g\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï‡}\left(z\right)}}\right\}\\ \phantom{\rule{1em}{0ex}}â‰¥0.\end{array}$
(2.2)

Substituting for h, g, Ï† and Ï‡ in (2.2) and dividing by $\left(1âˆ’\mathrm{Î±}\right)z$, we obtain $Re\frac{A\left(z\right)}{B\left(z\right)}â‰¥0$, where

$\begin{array}{rcl}A\left(z\right)& =& 1+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)nâˆ’\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}{a}_{n}{z}^{nâˆ’1}\\ âˆ’\left(\frac{\stackrel{Â¯}{z}}{z}\right)\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)n+\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}{b}_{n}{\stackrel{Â¯}{z}}^{nâˆ’1}\end{array}$

and

$B\left(z\right)=1+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}{a}_{n}{z}^{nâˆ’1}+\left(\frac{\stackrel{Â¯}{z}}{z}\right)\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}{b}_{n}{\stackrel{Â¯}{z}}^{nâˆ’1}.$

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

$\begin{array}{c}|A\left(z\right)+B\left(z\right)|âˆ’|A\left(z\right)âˆ’B\left(z\right)|\hfill \\ \phantom{\rule{1em}{0ex}}=|2+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)nâˆ’\left(k{e}^{i\mathrm{Î¸}}+2\mathrm{Î±}âˆ’1\right)}{\left(1âˆ’\mathrm{Î±}\right)}{a}_{n}{z}^{nâˆ’1}\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\left(\frac{\stackrel{Â¯}{z}}{z}\right)\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)n+\left(k{e}^{i\mathrm{Î¸}}+2\mathrm{Î±}âˆ’1\right)}{\left(1âˆ’\mathrm{Î±}\right)}{b}_{n}{\stackrel{Â¯}{z}}^{nâˆ’1}|\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’|\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)nâˆ’\left(1+k{e}^{i\mathrm{Î¸}}\right)}{\left(1âˆ’\mathrm{Î±}\right)}{a}_{n}{z}^{nâˆ’1}\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\left(\frac{\stackrel{Â¯}{z}}{z}\right)\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)n+\left(1+k{e}^{i\mathrm{Î¸}}\right)}{\left(1âˆ’\mathrm{Î±}\right)}{b}_{n}{\stackrel{Â¯}{z}}^{nâˆ’1}|\hfill \\ \phantom{\rule{1em}{0ex}}â‰¥2âˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+2\mathrm{Î±}âˆ’1\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}||z{|}^{nâˆ’1}\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+2\mathrm{Î±}âˆ’1\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}||z{|}^{nâˆ’1}\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(1+k\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n+1}||z{|}^{nâˆ’1}\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(1+k\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}||z{|}^{nâˆ’1}\hfill \\ \phantom{\rule{1em}{0ex}}â‰¥2\left\{1âˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}|âˆ’\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}|\right\}\hfill \\ \phantom{\rule{1em}{0ex}}â‰¥0\phantom{\rule{1em}{0ex}}\text{by (2.1)}.\hfill \end{array}$

The harmonic functions

$\begin{array}{rl}f\left(z\right)=& z+\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{n}{{\mathrm{Î»}}_{n}}\frac{\left(1âˆ’\mathrm{Î±}\right)}{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{x}_{n}{z}^{n}\\ +\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{n}{{\mathrm{Î¼}}_{n}}\frac{\left(1âˆ’\mathrm{Î±}\right)}{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\stackrel{Â¯}{y}}_{n}{\stackrel{Â¯}{z}}^{n},\end{array}$
(2.3)

where ${âˆ‘}_{n=2}^{\mathrm{âˆž}}|{x}_{n}|+{âˆ‘}_{n=1}^{\mathrm{âˆž}}|{y}_{n}|=1$, show that the coefficient bound given by (2.1) is sharp. The functions of the form (2.3) are in the class $\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ because

$\begin{array}{c}\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\left[\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}|+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}|\right]\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}|{x}_{n}|+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}|{y}_{n}|=1.\hfill \end{array}$

This completes the proof of Theorem 1.â€ƒâ–¡

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

Theorem 2 Let $f=h+\stackrel{Â¯}{g}$ be such that h and g are given by (1.2). Then $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ if and only if

$\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\left(\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{a}_{n}|+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\left(\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\right)|{b}_{n}|â‰¤1.$
(2.4)

Proof Since $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)âŠ‚\mathit{HST}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, we only need to prove the â€˜only ifâ€™ part of the theorem. To this end, we notice that the necessary and sufficient condition for $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ is that

$Re\left\{\left(1+k{e}^{i\mathrm{Î¸}}\right)\frac{h\left(z\right)âˆ—\mathrm{Ï†}\left(z\right)âˆ’\stackrel{Â¯}{g\left(z\right)âˆ—\mathrm{Ï‡}\left(z\right)}}{h\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï†}\left(z\right)+\stackrel{Â¯}{g\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï‡}\left(z\right)}}âˆ’k{e}^{i\mathrm{Î¸}}\right\}â‰¥\mathrm{Î±}.$

This is equivalent to

$Re\left\{\frac{\left(1+k{e}^{i\mathrm{Î¸}}\right)\left(h\left(z\right)âˆ—\mathrm{Ï†}\left(z\right)âˆ’\stackrel{Â¯}{g\left(z\right)âˆ—\mathrm{Ï‡}\left(z\right)}\right)âˆ’\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)\left(h\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï†}\left(z\right)+\stackrel{Â¯}{g\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï‡}\left(z\right)}\right)}{h\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï†}\left(z\right)+\stackrel{Â¯}{g\left(z\right)\phantom{\rule{0.2em}{0ex}}â™¢\phantom{\rule{0.2em}{0ex}}\mathrm{Ï‡}\left(z\right)}}\right\}>0,$

which implies that

$\begin{array}{r}Re\left\{\frac{\left(1âˆ’\mathrm{Î±}\right)zâˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}\left[\left(1+k{e}^{i\mathrm{Î¸}}\right)nâˆ’\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)\right]|{a}_{n}|{z}^{n}}{zâˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}|{a}_{n}|{z}^{n}+{âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}|{b}_{n}|{\stackrel{Â¯}{z}}^{n}}\\ \phantom{\rule{2em}{0ex}}âˆ’\frac{{âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}\left[\left(1+k{e}^{i\mathrm{Î¸}}\right)n+\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)\right]|{b}_{n}|{\stackrel{Â¯}{z}}^{n}}{zâˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}|{a}_{n}|{z}^{n}+{âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}|{b}_{n}|{\stackrel{Â¯}{z}}^{n}}\right\}\\ \phantom{\rule{1em}{0ex}}=Re\left\{\frac{\left(1âˆ’\mathrm{Î±}\right)âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}\left[\left(1+k{e}^{i\mathrm{Î¸}}\right)nâˆ’\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)\right]|{a}_{n}|{z}^{nâˆ’1}}{1âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}|{a}_{n}|{z}^{nâˆ’1}+\left(\frac{\stackrel{Â¯}{z}}{z}\right){âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}|{b}_{n}|{\stackrel{Â¯}{z}}^{nâˆ’1}}\\ \phantom{\rule{2em}{0ex}}âˆ’\frac{\left(\frac{\stackrel{Â¯}{z}}{z}\right){âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}\left[\left(1+k{e}^{i\mathrm{Î¸}}\right)n+\left(k{e}^{i\mathrm{Î¸}}+\mathrm{Î±}\right)\right]|{b}_{n}|{\stackrel{Â¯}{z}}^{nâˆ’1}}{1âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}|{a}_{n}|{z}^{n}+\left(\frac{\stackrel{Â¯}{z}}{z}\right){âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}|{b}_{n}|{\stackrel{Â¯}{z}}^{nâˆ’1}}\right\}>0,\end{array}$
(2.5)

since $Re\left({e}^{i\mathrm{Î¸}}\right)â‰¤|{e}^{i\mathrm{Î¸}}|=1$, the required condition (2.5) is equivalent to

$\begin{array}{r}\left\{\frac{1âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}|{r}^{nâˆ’1}}{1âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}|{a}_{n}|{r}^{nâˆ’1}+{âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}|{b}_{n}|{r}^{nâˆ’1}}âˆ’\frac{{âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}|{r}^{nâˆ’1}}{1âˆ’{âˆ‘}_{n=2}^{\mathrm{âˆž}}\frac{{\mathrm{Î»}}_{n}}{n}|{a}_{n}|{r}^{nâˆ’1}+{âˆ‘}_{n=1}^{\mathrm{âˆž}}\frac{{\mathrm{Î¼}}_{n}}{n}|{b}_{n}|{r}^{nâˆ’1}}\right\}\\ \phantom{\rule{1em}{0ex}}â‰¥0.\end{array}$
(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 $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, and so the proof of Theorem 2 is completed.â€ƒâ–¡

Theorem 3 Let $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$. Then, for $|z|=r<1$, $|{b}_{1}|<\frac{1âˆ’\mathrm{Î±}}{2k+\mathrm{Î±}+1}$ and

${D}_{n}â‰¤\frac{{\mathrm{Î»}}_{n}}{n},\phantom{\rule{2em}{0ex}}{E}_{n}â‰¤\frac{{\mathrm{Î¼}}_{n}}{n}\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.25em}{0ex}}nâ‰¥2\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}C=min\left\{{D}_{2},{E}_{2}\right\},$
(2.7)

we have

$|f\left(z\right)|â‰¤\left(1+|{b}_{1}|\right)r+\left\{\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}âˆ’\frac{2k+1+\mathrm{Î±}}{C\left(2+kâˆ’\mathrm{Î±}\right)}|{b}_{1}|\right\}{r}^{2}$

and

$|f\left(z\right)|â‰¥\left(1âˆ’|{b}_{1}|\right)râˆ’\left\{\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}âˆ’\frac{2k+1+\mathrm{Î±}}{C\left(2+kâˆ’\mathrm{Î±}\right)}|{b}_{1}|\right\}{r}^{2}.$

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 $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, then we have

$\begin{array}{rl}|f\left(z\right)|=& |zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}|{a}_{n}|{z}^{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}|{b}_{n}|{\stackrel{Â¯}{z}}^{n}|\\ â‰¥& râˆ’|{b}_{1}|râˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(|{a}_{n}|+|{b}_{n}|\right){r}^{2}\\ â‰¥& râˆ’|{b}_{1}|r\\ âˆ’\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{C\left(\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right)}{\left(1âˆ’\mathrm{Î±}\right)}\left(|{a}_{n}|+|{b}_{n}|\right){r}^{2}\\ â‰¥& râˆ’|{b}_{1}|r\\ âˆ’\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\left\{\frac{C\left(\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}|\\ +\frac{C\left(\left(1+k\right)n+\left(k+\mathrm{Î±}\right)\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}|\right\}{r}^{2}\\ â‰¥& \left(1âˆ’|{b}_{1}|\right)râˆ’\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}\left\{1âˆ’\frac{2k+1+\mathrm{Î±}}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{1}|\right\}{r}^{2}\\ â‰¥& \left(1âˆ’|{b}_{1}|\right)râˆ’\left\{\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}âˆ’\frac{2k+1+\mathrm{Î±}}{C\left(2+kâˆ’\mathrm{Î±}\right)}|{b}_{1}|\right\}{r}^{2}.\end{array}$

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

$f\left(z\right)=z+|{b}_{1}|\stackrel{Â¯}{z}+\left(\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}âˆ’\frac{2k+1+\mathrm{Î±}}{C\left(2+kâˆ’\mathrm{Î±}\right)}|{b}_{1}|\right){\stackrel{Â¯}{z}}^{2}$

and

$f\left(z\right)=\left(1âˆ’|{b}_{1}|\right)zâˆ’\left(\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}âˆ’\frac{2k+1+\mathrm{Î±}}{C\left(2+kâˆ’\mathrm{Î±}\right)}|{b}_{1}|\right){z}^{2}.$

â€ƒâ–¡

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

Corollary 1 Let $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, then for $|{b}_{1}|<\frac{1âˆ’\mathrm{Î±}}{2k+\mathrm{Î±}+1}$ the set

$\left\{w:|w|<1âˆ’\frac{\left(1âˆ’\mathrm{Î±}\right)}{C\left(2+kâˆ’\mathrm{Î±}\right)}âˆ’\left(1âˆ’\frac{2k+1+\mathrm{Î±}}{C\left(2+kâˆ’\mathrm{Î±}\right)}\right)|{b}_{1}|\right\}$

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 $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, denoted by $clco\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$.

Theorem 4 Let $f=h+\stackrel{Â¯}{g}$ be such that h and g are given by (1.2). Then $fâˆˆclco\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ if and only if f can be expressed as

$f\left(z\right)=\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left[{X}_{n}{h}_{n}\left(z\right)+{Y}_{n}{g}_{n}\left(z\right)\right],$
(3.1)

where

$\begin{array}{c}{h}_{1}\left(z\right)=z,\hfill \\ {h}_{n}\left(z\right)=zâˆ’\frac{n\left(1âˆ’\mathrm{Î±}\right)}{{\mathrm{Î»}}_{n}\left(\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right)}{z}^{n}\phantom{\rule{1em}{0ex}}\left(nâ‰¥2\right),\hfill \\ {g}_{n}\left(z\right)=z+\frac{n\left(1âˆ’\mathrm{Î±}\right)}{{\mathrm{Î¼}}_{n}\left(\left(1+k\right)n+\left(k+\mathrm{Î±}\right)\right)}{\stackrel{Â¯}{z}}^{n}\phantom{\rule{1em}{0ex}}\left(nâ‰¥1\right),\hfill \\ {X}_{n}â‰¥0,\phantom{\rule{2em}{0ex}}{Y}_{n}â‰¥0,\phantom{\rule{1em}{0ex}}\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left[{X}_{n}+{Y}_{n}\right]=1.\hfill \end{array}$

In particular, the extreme points of the class $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ 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

$f\left(z\right)=\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left[{X}_{n}+{Y}_{n}\right]zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{n\left(1âˆ’\mathrm{Î±}\right)}{{\mathrm{Î»}}_{n}\left(\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right)}{X}_{n}{z}^{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{n\left(1âˆ’\mathrm{Î±}\right)}{{\mathrm{Î¼}}_{n}\left(\left(1+k\right)n+\left(k+\mathrm{Î±}\right)\right)}{Y}_{n}{\stackrel{Â¯}{z}}^{n}.$

Then

$\begin{array}{c}\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}\left(\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right)}{n\left(1âˆ’\mathrm{Î±}\right)}\left(\frac{n\left(1âˆ’\mathrm{Î±}\right)}{{\mathrm{Î»}}_{n}\left(\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right)}\right){X}_{n}\hfill \\ \phantom{\rule{2em}{0ex}}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}\left(\left(1+k\right)n+\left(k+\mathrm{Î±}\right)\right)}{n\left(1âˆ’\mathrm{Î±}\right)}\left(\frac{n\left(1âˆ’\mathrm{Î±}\right)}{{\mathrm{Î¼}}_{n}\left(\left(1+k\right)n+\left(k+\mathrm{Î±}\right)\right)}\right){Y}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{X}_{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{Y}_{n}=1âˆ’{X}_{1}â‰¤1,\hfill \end{array}$

and so $f\left(z\right)âˆˆclco\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$. Conversely, suppose that $f\left(z\right)âˆˆclco\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$. Set

${X}_{n}=\frac{{\mathrm{Î»}}_{n}\left(\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)\right)}{n\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}|\phantom{\rule{1em}{0ex}}\left(nâ‰¥2\right)$

and

${Y}_{n}=\frac{{\mathrm{Î¼}}_{n}\left(\left(1+k\right)n+\left(k+\mathrm{Î±}\right)\right)}{n\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}|\phantom{\rule{1em}{0ex}}\left(nâ‰¥1\right),$

then note that by Theorem 2, $0â‰¤{X}_{n}â‰¤1$ ($nâ‰¥2$) and $0â‰¤{Y}_{n}â‰¤1$ ($nâ‰¥1$).

Consequently, we obtain

$f\left(z\right)=\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left[{X}_{n}{h}_{n}\left(z\right)+{Y}_{n}{g}_{n}\left(z\right)\right].$

Using Theorem 2, it is easily seen that the class $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ is convex and closed and so $clco\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)=\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$.â€ƒâ–¡

## 4 Convolution result

For harmonic functions of the form

$f\left(z\right)=zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}|{a}_{n}|{z}^{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}|{b}_{n}|{\stackrel{Â¯}{z}}^{n}$
(4.1)

and

$G\left(z\right)=zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{A}_{n}{z}^{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{B}_{n}{\stackrel{Â¯}{z}}^{n}\phantom{\rule{1em}{0ex}}\left({A}_{n},{B}_{n}â‰¥0\right),$
(4.2)

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

$\left(fâˆ—G\right)\left(z\right)=f\left(z\right)âˆ—G\left(z\right)=zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{n}{A}_{n}{z}^{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{b}_{n}{B}_{n}{\stackrel{Â¯}{z}}^{n}.$

Using this definition, we show that the class $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ is closed under convolution.

Theorem 5 For $0â‰¤\mathrm{Î±}<1$, let $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ and $Gâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$. Then $f\left(z\right)âˆ—G\left(z\right)âˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$.

Proof Let the functions $f\left(z\right)$ defined by (4.1) be in the class $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, and let the functions $G\left(z\right)$ defined by (4.2) be in the class $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$. Obviously, the coefficients of f and G must satisfy a condition similar to inequality (2.4). So, for the coefficients of $f\left(z\right)âˆ—G\left(z\right)$, we can write

$\begin{array}{c}\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}|{A}_{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}|{B}_{n}\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\left[\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n}|+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n}|\right],\hfill \end{array}$

the right-hand side of this inequality is bounded by 1 because $fâˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$. Then $f\left(z\right)âˆ—G\left(z\right)âˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$.â€ƒâ–¡

Finally, we show that $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ is closed under convex combinations of its members.

Theorem 6 The class $\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$ is closed under convex linear combination.

Proof For $i=1,2,3,â€¦$â€‰, let ${f}_{i}âˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$, where the functions ${f}_{i}$ are given by

${f}_{i}\left(z\right)=zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}|{a}_{n,i}|{z}^{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}|{b}_{n,i}|{\stackrel{Â¯}{z}}^{n}.$

For ${âˆ‘}_{i=1}^{\mathrm{âˆž}}{t}_{i}=1$; $0â‰¤{t}_{i}â‰¤1$, the convex linear combination of ${f}_{i}$ may be written as

$\underset{i=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{t}_{i}{f}_{i}\left(z\right)=zâˆ’\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{i=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{t}_{i}|{a}_{n,i}|\right){z}^{n}+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\left(\underset{i=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{t}_{i}|{b}_{n,i}|\right){\stackrel{Â¯}{z}}^{n},$

then by (2.4) we have

$\begin{array}{c}\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\underset{i=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{t}_{i}|{a}_{n,i}|+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}\underset{i=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{t}_{i}|{b}_{n,i}|\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{i=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{t}_{i}\left\{\underset{n=2}{\overset{\mathrm{âˆž}}{âˆ‘}}\left[\frac{{\mathrm{Î»}}_{n}}{n}\frac{\left(1+k\right)nâˆ’\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{a}_{n,i}|+\underset{n=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\mathrm{Î¼}}_{n}}{n}\frac{\left(1+k\right)n+\left(k+\mathrm{Î±}\right)}{\left(1âˆ’\mathrm{Î±}\right)}|{b}_{n,i}|\right]\right\}\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{i=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{t}_{i}=1.\hfill \end{array}$

This condition is required by (2.4) and so ${âˆ‘}_{i=1}^{\mathrm{âˆž}}{t}_{i}{f}_{i}\left(z\right)âˆˆ\stackrel{Â¯}{\mathit{HST}}\left(\mathrm{Ï•},\mathrm{Ï‡},k,\mathrm{Î±}\right)$. This completes the proof of Theorem 6.â€ƒâ–¡

Remarks

1. (i)

Putting $k=0$ in our results, we obtain the results obtained by Dixit et al. [9];

2. (ii)

Putting $\mathrm{Ï†}\left(z\right)=\mathrm{Ï‡}\left(z\right)=\frac{z}{{\left(1âˆ’z\right)}^{2}}$ and $k=1$ in our results, we obtain the results obtained by Rosy et al. [10];

3. (iii)

Putting $\mathrm{Ï†}\left(z\right)=\mathrm{Ï‡}\left(z\right)=\frac{z+{z}^{2}}{{\left(1âˆ’z\right)}^{3}}$ in our results, we obtain the results obtained by Kim et al. [8];

4. (iv)

Putting $\mathrm{Ï†}\left(z\right)=\mathrm{Ï‡}\left(z\right)=\frac{z}{{\left(1âˆ’z\right)}^{2}}$ and $k=0$ in our results, we obtain the results obtained by Jahangiri [3];

5. (v)

Putting $\mathrm{Ï†}\left(z\right)=\mathrm{Ï‡}\left(z\right)=\frac{z+{z}^{2}}{{\left(1âˆ’z\right)}^{3}}$ and $k=0$ in our results, we obtain the results obtained by Jahangiri [2].

## References

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

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

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

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

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

6. Kanas S, Wisniowska A: Conic regions and k -uniform convexity. J. Comput. Appl. Math. 1999, 105: 327â€“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):121â€“132. 10.1080/10652460008819249

8. Kim YC, Jahangiri JM, Choi JH: Certain convex harmonic functions. Int. J. Math. Math. Sci. 2002, 29(8):459â€“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):255â€“264.

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

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

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

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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

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