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 [1] 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 $|h^{\prime }(z)|>|g^{\prime }(z)|$ (see also [2–4] and [5]).

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

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

(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 $S_{\overline{H}}$ denote the subclass of $S_H$ consisting of functions $f=h+\overline{g}$ such that the functions h and g are of the form

$$h(z)=z-\sum _{n=2}^{\mathrm{\infty }}|a_n|z^n,g(z)=\sum _{n=1}^{\mathrm{\infty }}|b_n|z^n,|b_1|<1.$$

(1.2)

Recently Kanas and Wisniowska [6] (see also Kanas and Srivastava [7]) studied the class of k-uniformly convex analytic functions, denoted by $k-UCV$, $k\ge 0$, so that $\varphi \in k-UCV$ if and only if

$$Re\{1+\frac{(z-\zeta ){\varphi }^{″}(z)}{{\varphi }^{\prime }(z)}\}\ge 0(|\zeta |\le k;z\in U).$$

(1.3)

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

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

(1.4)

Kim et al. [8] introduced and studied the class $HCV(k,\alpha )$ consisting of functions $f=h+\overline{g}$, such that h and g are given by (1.1), and satisfying the condition

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

(1.5)

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

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

(1.6)

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

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

(1.7)

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

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

(1.8)

The convolution of two functions of the form

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{z}^n\text{and}F(z)=z+\sum _{n=2}^{\mathrm{\infty }}{A}_n{z}^n$$

is defined as

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

(1.9)

while the integral convolution is defined by

$$(f♢F)(z)=z+\sum _{n=2}^{\mathrm{\infty }}\frac{a_n{A}_n}{n}{z}^n.$$

(1.10)

From (1.9) and (1.10), we have

$$(f♢F)(z)={\int }_0^z\frac{(f\ast F)(t)}{t}dt.$$

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

$$Re\{(1+k{e}^{i\theta })\frac{h(z)\ast \phi (z)-\overline{g(z)\ast \chi (z)}}{h(z)♢\phi (z)+\overline{g(z)♢\chi (z)}}-k{e}^{i\theta }\}\ge \alpha (0\le \alpha <1;k\ge 0;\theta \text{ real}),$$

(1.11)

where

$$\phi (z)=z+\sum _{n=2}^{\mathrm{\infty }}{\lambda }_n{z}^n({\lambda }_n\ge 0)\text{and}\chi (z)=z+\sum _{n=2}^{\mathrm{\infty }}{\mu }_n{z}^n({\mu }_n\ge 0).$$

(1.12)

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

We note that

• (i) $\overline{\mathit{HST}}(\varphi ,\chi ,0,\alpha )=\overline{HS}(\varphi ,\chi ,\alpha )$ (see Dixit et al. [9]);

• (ii) $\overline{\mathit{HST}}(\frac{z}{{(1-z)}^2},\frac{z}{{(1-z)}^2},1,\alpha )=G_{\overline{H}}(\alpha )$ (see Rosy et al. [10]);

• (iii) $\overline{\mathit{HST}}(\frac{z+z^2}{{(1-z)}^3},\frac{z+z^2}{{(1-z)}^3},k,\alpha )=\overline{H}CV(k,\alpha )$ (see Kim et al. [8]);

• (iv) $\overline{\mathit{HST}}(\frac{z}{{(1-z)}^2},\frac{z}{{(1-z)}^2},0,\alpha )=T_H^{\ast }(\alpha )$ (see Jahangiri [3], see also Joshi and Darus [11]);

• (v) $\overline{\mathit{HST}}(\frac{z+z^2}{{(1-z)}^3},\frac{z+z^2}{{(1-z)}^3},0,\alpha )=C_H(\alpha )$ (see Jahangiri [3], see also Joshi and Darus [11]).

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

2 Coefficient characterization and distortion theorem

Unless otherwise mentioned, we assume throughout this paper that $\phi (z)$ and $\chi (z)$ 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 $\mathit{HST}(\varphi ,\chi ,k,\alpha )$.

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

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

(2.1)

where

$$\begin{array}{c}{n}^2(1-\alpha )\le {\lambda }_n[(1+k)n-(k+\alpha )]\mathit{\text{and}}{n}^2(1-\alpha )\le {\mu }_n[(1+k)n+(k+\alpha )]\hfill \\ \mathit{\text{for}}n\ge 2.\hfill \end{array}$$

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

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

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

To show that f is univalent in U, suppose $z_1,z_2\in U$ so that $z_1\ne z_2$, then

$$\begin{array}{rcl}\left|\frac{f(z_1)-f(z_2)}{h(z_1)-h(z_2)}\right|& \ge & 1-\left|\frac{g(z_1)-g(z_2)}{h(z_1)-h(z_2)}\right|=1-\left|\frac{{\sum }_{n=1}^{\mathrm{\infty }}{b}_n(z_1^n-z_2^n)}{(z_1-z_2)+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n(z_1^n-z_2^n)}\right|\\ \ge & 1-\frac{{\sum }_{n=1}^{\mathrm{\infty }}n|b_n|}{1-{\sum }_{n=2}^{\mathrm{\infty }}n|a_n|}>1-\frac{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}(\frac{(1+k)n+(k+\alpha )}{(1-\alpha )})|b_n|}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}(\frac{(1+k)n-(k+\alpha )}{(1-\alpha )})|a_n|}\ge 0.\end{array}$$

Now, we prove that $f\in \mathit{HST}(\varphi ,\psi ,k,\alpha )$, 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

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

(2.2)

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

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

and

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

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

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

The harmonic functions

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

(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 $\mathit{HST}(\varphi ,\chi ,k,\alpha )$ because

$$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}[\frac{{\lambda }_n}{n}\frac{(1+k)n-(k+\alpha )}{(1-\alpha )}|a_n|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}\frac{(1+k)n+(k+\alpha )}{(1-\alpha )}|b_n|]\hfill \\ =\sum _{n=2}^{\mathrm{\infty }}|x_n|+\sum _{n=1}^{\mathrm{\infty }}|y_n|=1.\hfill \end{array}$$

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 $f\in \overline{\mathit{HST}}(\varphi ,\chi ,k,\alpha )$ if and only if

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

(2.4)

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

$$Re\{(1+k{e}^{i\theta })\frac{h(z)\ast \phi (z)-\overline{g(z)\ast \chi (z)}}{h(z)♢\phi (z)+\overline{g(z)♢\chi (z)}}-k{e}^{i\theta }\}\ge \alpha .$$

This is equivalent to

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

which implies that

$$\begin{array}{r}Re\{\frac{(1-\alpha )z-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}[(1+k{e}^{i\theta })n-(k{e}^{i\theta }+\alpha )]|a_n|z^n}{z-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}|a_n|z^n+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}|b_n|{\overline{z}}^n}\\ -\frac{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}[(1+k{e}^{i\theta })n+(k{e}^{i\theta }+\alpha )]|b_n|{\overline{z}}^n}{z-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}|a_n|z^n+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}|b_n|{\overline{z}}^n}\}\\ =Re\{\frac{(1-\alpha )-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}[(1+k{e}^{i\theta })n-(k{e}^{i\theta }+\alpha )]|a_n|z^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}|a_n|z^{n-1}+(\frac{\overline{z}}{z}){\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}|b_n|{\overline{z}}^{n-1}}\\ -\frac{(\frac{\overline{z}}{z}){\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}[(1+k{e}^{i\theta })n+(k{e}^{i\theta }+\alpha )]|b_n|{\overline{z}}^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}|a_n|z^n+(\frac{\overline{z}}{z}){\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}|b_n|{\overline{z}}^{n-1}}\}>0,\end{array}$$

(2.5)

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

$$\begin{array}{r}\{\frac{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}\frac{(1+k)n-(k+\alpha )}{(1-\alpha )}|a_n|r^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}|a_n|r^{n-1}+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}|b_n|r^{n-1}}-\frac{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}\frac{(1+k)n+(k+\alpha )}{(1-\alpha )}|b_n|r^{n-1}}{1-{\sum }_{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}|a_n|r^{n-1}+{\sum }_{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}|b_n|r^{n-1}}\}\\ \ge 0.\end{array}$$

(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 $(0,1)$ for which the quotient in (2.6) is negative. This contradicts the required condition for $f\in \overline{\mathit{HST}}(\varphi ,\chi ,k,\alpha )$, and so the proof of Theorem 2 is completed. □

Theorem 3 Let $f\in \overline{\mathit{HST}}(\varphi ,\chi ,k,\alpha )$. Then, for $|z|=r<1$, $|b_1|<\frac{1-\alpha }{2k+\alpha +1}$ and

$$D_n\le \frac{{\lambda }_n}{n},E_n\le \frac{{\mu }_n}{n}\mathit{\text{for}}n\ge 2\mathit{\text{and}}C=min\{D_2,E_2\},$$

(2.7)

we have

$$|f(z)|\le (1+|b_1|)r+\{\frac{(1-\alpha )}{C(2+k-\alpha )}-\frac{2k+1+\alpha }{C(2+k-\alpha )}|b_1|\}{r}^2$$

and

$$|f(z)|\ge (1-|b_1|)r-\{\frac{(1-\alpha )}{C(2+k-\alpha )}-\frac{2k+1+\alpha }{C(2+k-\alpha )}|b_1|\}{r}^2.$$

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

$$\begin{array}{rl}|f(z)|=& |z-\sum _{n=2}^{\mathrm{\infty }}|a_n|z^n+\sum _{n=1}^{\mathrm{\infty }}|b_n|{\overline{z}}^n|\\ \ge & r-|b_1|r-\sum _{n=2}^{\mathrm{\infty }}(|a_n|+|b_n|)r^2\\ \ge & r-|b_1|r\\ -\frac{(1-\alpha )}{C(2+k-\alpha )}\sum _{n=2}^{\mathrm{\infty }}\frac{C((1+k)n-(k+\alpha ))}{(1-\alpha )}(|a_n|+|b_n|)r^2\\ \ge & r-|b_1|r\\ -\frac{(1-\alpha )}{C(2+k-\alpha )}\sum _{n=2}^{\mathrm{\infty }}\{\frac{C((1+k)n-(k+\alpha ))}{(1-\alpha )}|a_n|\\ +\frac{C((1+k)n+(k+\alpha ))}{(1-\alpha )}|b_n|\}{r}^2\\ \ge & (1-|b_1|)r-\frac{(1-\alpha )}{C(2+k-\alpha )}\{1-\frac{2k+1+\alpha }{(1-\alpha )}|b_1|\}{r}^2\\ \ge & (1-|b_1|)r-\{\frac{(1-\alpha )}{C(2+k-\alpha )}-\frac{2k+1+\alpha }{C(2+k-\alpha )}|b_1|\}{r}^2.\end{array}$$

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

$$f(z)=z+|b_1|\overline{z}+(\frac{(1-\alpha )}{C(2+k-\alpha )}-\frac{2k+1+\alpha }{C(2+k-\alpha )}|b_1|){\overline{z}}^2$$

and

$$f(z)=(1-|b_1|)z-(\frac{(1-\alpha )}{C(2+k-\alpha )}-\frac{2k+1+\alpha }{C(2+k-\alpha )}|b_1|)z^2.$$

 □

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

Corollary 1 Let $f\in \overline{\mathit{HST}}(\varphi ,\chi ,k,\alpha )$, then for $|b_1|<\frac{1-\alpha }{2k+\alpha +1}$ the set

$$\{w:|w|<1-\frac{(1-\alpha )}{C(2+k-\alpha )}-(1-\frac{2k+1+\alpha }{C(2+k-\alpha )})|b_1|\}$$

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

3 Extreme points

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

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

$$f(z)=\sum _{n=1}^{\mathrm{\infty }}[X_n{h}_n(z)+Y_n{g}_n(z)],$$

(3.1)

where

$$\begin{array}{c}{h}_1(z)=z,\hfill \\ h_n(z)=z-\frac{n(1-\alpha )}{{\lambda }_n((1+k)n-(k+\alpha ))}{z}^n(n\ge 2),\hfill \\ g_n(z)=z+\frac{n(1-\alpha )}{{\mu }_n((1+k)n+(k+\alpha ))}{\overline{z}}^n(n\ge 1),\hfill \\ X_n\ge 0,Y_n\ge 0,\sum _{n=1}^{\mathrm{\infty }}[X_n+Y_n]=1.\hfill \end{array}$$

In particular, the extreme points of the class $\overline{\mathit{HST}}(\varphi ,\chi ,k,\alpha )$ are $\{h_n\}$ and $\{g_n\}$, respectively.

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

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

Then

$$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n((1+k)n-(k+\alpha ))}{n(1-\alpha )}\left(\frac{n(1-\alpha )}{{\lambda }_n((1+k)n-(k+\alpha ))}\right)X_n\hfill \\ +\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_n((1+k)n+(k+\alpha ))}{n(1-\alpha )}\left(\frac{n(1-\alpha )}{{\mu }_n((1+k)n+(k+\alpha ))}\right)Y_n\hfill \\ =\sum _{n=2}^{\mathrm{\infty }}{X}_n+\sum _{n=1}^{\mathrm{\infty }}{Y}_n=1-X_1\le 1,\hfill \end{array}$$

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

$$X_n=\frac{{\lambda }_n((1+k)n-(k+\alpha ))}{n(1-\alpha )}|a_n|(n\ge 2)$$

and

$$Y_n=\frac{{\mu }_n((1+k)n+(k+\alpha ))}{n(1-\alpha )}|b_n|(n\ge 1),$$

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

$$f(z)=\sum _{n=1}^{\mathrm{\infty }}[X_n{h}_n(z)+Y_n{g}_n(z)].$$

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

4 Convolution result

For harmonic functions of the form

$$f(z)=z-\sum _{n=2}^{\mathrm{\infty }}|a_n|z^n+\sum _{n=1}^{\mathrm{\infty }}|b_n|{\overline{z}}^n$$

(4.1)

and

$$G(z)=z-\sum _{n=2}^{\mathrm{\infty }}{A}_n{z}^n+\sum _{n=1}^{\mathrm{\infty }}{B}_n{\overline{z}}^n(A_n,B_n\ge 0),$$

(4.2)

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

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

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

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

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

$$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}\frac{(1+k)n-(k+\alpha )}{(1-\alpha )}|a_n|A_n+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}\frac{(1+k)n+(k+\alpha )}{(1-\alpha )}|b_n|B_n\hfill \\ \le \sum _{n=2}^{\mathrm{\infty }}[\frac{{\lambda }_n}{n}\frac{(1+k)n-(k+\alpha )}{(1-\alpha )}|a_n|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}\frac{(1+k)n+(k+\alpha )}{(1-\alpha )}|b_n|],\hfill \end{array}$$

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

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

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

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

$$f_i(z)=z-\sum _{n=2}^{\mathrm{\infty }}|a_{n,i}|z^n+\sum _{n=1}^{\mathrm{\infty }}|b_{n,i}|{\overline{z}}^n.$$

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

$$\sum _{i=1}^{\mathrm{\infty }}{t}_i{f}_i(z)=z-\sum _{n=2}^{\mathrm{\infty }}(\sum _{i=1}^{\mathrm{\infty }}{t}_i|a_{n,i}|)z^n+\sum _{n=1}^{\mathrm{\infty }}(\sum _{i=1}^{\mathrm{\infty }}{t}_i|b_{n,i}|){\overline{z}}^n,$$

then by (2.4) we have

$$\begin{array}{c}\sum _{n=2}^{\mathrm{\infty }}\frac{{\lambda }_n}{n}\frac{(1+k)n-(k+\alpha )}{(1-\alpha )}\sum _{i=1}^{\mathrm{\infty }}{t}_i|a_{n,i}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}\frac{(1+k)n+(k+\alpha )}{(1-\alpha )}\sum _{i=1}^{\mathrm{\infty }}{t}_i|b_{n,i}|\hfill \\ =\sum _{i=1}^{\mathrm{\infty }}{t}_i\left\{\sum _{n=2}^{\mathrm{\infty }}[\frac{{\lambda }_n}{n}\frac{(1+k)n-(k+\alpha )}{(1-\alpha )}|a_{n,i}|+\sum _{n=1}^{\mathrm{\infty }}\frac{{\mu }_n}{n}\frac{(1+k)n+(k+\alpha )}{(1-\alpha )}|b_{n,i}|]\right\}\hfill \\ \le \sum _{i=1}^{\mathrm{\infty }}{t}_i=1.\hfill \end{array}$$

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

Remarks

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

• (ii) Putting $\phi (z)=\chi (z)=\frac{z}{{(1-z)}^2}$ and $k=1$ in our results, we obtain the results obtained by Rosy et al. [10];

• (iii) Putting $\phi (z)=\chi (z)=\frac{z+z^2}{{(1-z)}^3}$ in our results, we obtain the results obtained by Kim et al. [8];

• (iv) Putting $\phi (z)=\chi (z)=\frac{z}{{(1-z)}^2}$ and $k=0$ in our results, we obtain the results obtained by Jahangiri [3];

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