﻿ On a Subordination Result of a Subclass of Analytic Functions

Vol.07 No.11(2017), Article ID:80546,6 pages
10.4236/apm.2017.711038

On a Subordination Result of a Subclass of Analytic Functions

Risikat Ayodeji Bello

Department of Mathematics and Statistics, College of Pure and Applied Science, Kwara State University, Malete, Nigeria    Received: March 13, 2017; Accepted: November 21, 2017; Published: November 24, 2017

ABSTRACT

In this paper, we investigate a subordination property and the coefficient inequality for the class $M\left(1,b\right)$ , The lower bound is also provided for the real part of functions belonging to the class $M\left(1,b\right)$ .

Keywords:

Analytic Function, Univalent Function, Hadamard Product, Subordination 1. Introduction

Let A denote the class of function $f\left(z\right)$ analytic in the open unit disk $U=\left\{z\in ℂ:|z|<1\right\}$ and let S be the subclass of A consisting of functions univalent in U and have the form

$f\left(z\right)=z+\sum _{k=2}^{\infty }\text{ }\text{ }{a}_{k}{z}^{k},$ (1.1)

The class of convex functions of order $\alpha$ in U, denoted as $K\left(\alpha \right)$ is given by

$K\left(\alpha \right)=\left\{f\in S:Re\left(1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)>\alpha ,0\le \alpha <1,z\in U\right\}$

Definition 1.1. The Hadamard product or convolution $f\ast g$ of the func- tion $f\left(z\right)$ and $g\left(z\right)$ , where $f\left(z\right)$ is as defined in (1.1) and the function $g\left(z\right)$ is given by

$g\left(z\right)=z+\sum _{k=2}^{\infty }\text{ }\text{ }{b}_{k}{z}^{k},$

is defined as:

$\left(f\ast g\right)\left(z\right)=z+\sum _{k=2}^{\infty }\text{ }\text{ }{a}_{k}{b}_{k}{z}^{k}=\left(g\ast f\right)\left(z\right),$ (1.2)

Definition 1.2. Let $f\left(z\right)$ and $g\left(z\right)$ be analytic in the unit disk $U$ . Then $f\left(z\right)$ is said to be subordination to $g\left(z\right)$ in $U$ and written as:

$f\left(z\right)\prec g\left(z\right),z\in U$

if there exist a Schwarz function $\omega \left(z\right)$ , analytic in U with $\omega \left(0\right)=0$ , $|\omega \left(z\right)|<1$ such that

$f\left(z\right)=g\left(\omega \left(z\right)\right),z\in U$ (1.3)

In particular, if the function $g\left(z\right)$ is univalent in U, then $f\left(z\right)$ is said to be subordinate to $g\left(z\right)$ if

$f\left(0\right)=g\left(0\right),f\left(u\right)\subset g\left(u\right)$ (1.4)

Definition 1.3. The sequence ${\left\{{c}_{k}\right\}}_{k=1}^{\infty }$ of complex numbers is said to be a subordinating factor sequence of the function $f\left(z\right)$ if whenever $f\left(z\right)$ in the form (1.1) is analytic, univalent and convex in the unit disk $U$ , the subordination is given by

$\sum _{k=1}^{\infty }\text{ }\text{ }{a}_{k}{c}_{k}{z}^{k}\prec f\left(z\right),z\in U,{a}_{1}=1$

We have the following theorem:

Theorem 1.1. (Wilf  ) The sequence ${\left\{{c}_{k}\right\}}_{k=1}^{\infty }$ is a subordinating factor sequence if and only if

$Re\left\{1+2\sum _{k=1}^{\infty }\text{ }\text{ }{c}_{k}{z}^{k}\right\}>0,z\in U$ (1.5)

Definition 1.4. A function $P\in A$ which is normalized by $P\left(0\right)=1$ is said to be in $P\left(1,b\right)$ if

$|P\left(z\right)-1|0,z\in U.$

The class $P\left(1,b\right)$ was studied by Janwoski  . The family $P\left(1,b\right)$ contains many interesting classes of functions. For example, for $f\left(z\right)\in A$ , if

$\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)\in P\left(1,1-\alpha \right),0\le \alpha <1$

Then $f\left(z\right)$ is starlike of order $\alpha$ in U and if

$\left(1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)\in P\left(1,1-\alpha \right),0\le \alpha <1$

Then $f\left(z\right)$ is convex of order $\alpha$ in U.

Let $F\left(1,b\right)$ be the subclass of $P\left(1,1-\alpha \right)$ consisting of functions $P\left(f\right)$ such that

$P\left(f\right)=\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\left(1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)$ (1.6)

we have the following theorem

Theorem 1.2.  Let $P\left(f\right)$ be given by Equation (1.6) with $f\left(z\right)=z+\sum \text{ }\text{ }{a}_{k}{z}^{k}$ . If

$\sum _{k=2}^{\infty }\left({k}^{2}+b-1\right)|{a}_{k}|0$

then $P\left(f\right)\in F\left(1,b\right)$ , $0 .

It is natural to consider the class

$M\left(1,b\right)=\left\{f\in A:\sum _{k=2}^{\infty }\left({k}^{2}+b-1\right)|{a}_{k}|0\right\}$

$0

Remark 1.1.  If $b=1-\alpha$ , then $M\left(1,1-\alpha \right)$ consists of starlike functions of order $\alpha$ , $0\le \alpha <1$ since

$\sum _{k=2}^{\infty }\left(k-\alpha \right)|{a}_{k}|<\sum _{k=2}^{\infty }\left({k}^{2}-\alpha \right)|{a}_{k}|$

Our main focus in this work is to provide a subordination results for functions belonging to the class $M\left(1,b\right)$

2. Main Results

2.1. Theorem

Let $f\left(z\right)\in M\left(1,b\right)$ , then

$\frac{3+b}{2\left(3+2b\right)}\left(f\ast g\right)\left(z\right)\prec g\left(z\right)$ (2.1)

where $0 and $g\left(z\right)$ is convex function.

Proof:

Let

$f\left(z\right)\in M\left(1,b\right)$

and suppose that

$g\left(z\right)=z+\sum \text{ }\text{ }{b}_{k}{z}^{k}\in C\left(\alpha \right)$

that is $g\left(z\right)$ is a convex function of order $\alpha$ .

By definition (1.1) we have

$\begin{array}{l}\frac{3+b}{2\left(3+2b\right)}\left(f\ast g\right)\left(z\right)\\ =\frac{3+b}{2\left(3+2b\right)}\left(z+\sum _{k=2}^{\infty }\text{ }\text{ }{a}_{k}{b}_{k}{z}^{k}\right)\\ =\sum _{k=1}^{\infty }\frac{3+b}{2\left(3+2b\right)}{a}_{k}{b}_{k}{z}^{k},{a}_{1}=1,{b}_{1}=1\end{array}$ (2.2)

Hence, by Definition 1.3…to show subordination (2.1) is by establishing that

${\left\{\frac{3+b}{2\left(3+2b\right)}{a}_{k}\right\}}_{k=1}^{\infty }$ (2.3)

is a subordinating factor sequence with ${a}_{1}=1$ . By Theorem 1.1, it is sufficient to show that

$Re\left\{1+2\sum _{k=1}^{\infty }\frac{3+b}{2\left(3+2b\right)}{a}_{k}{z}^{k}\right\}>0,z\in U$ (2.4)

Now,

$\begin{array}{l}Re\left\{1+2\sum _{k=1}^{\infty }\frac{3+b}{2\left(3+2b\right)}{a}_{k}{z}^{k}\right\}\\ =Re\left\{1+\frac{3+b}{3+2b}z+\sum _{k=2}^{\infty }\frac{3+b}{3+2b}{a}_{k}{z}^{k}\right\}\\ >Re\left\{1-\frac{3+b}{3+2b}r-\frac{3+b}{3+2b}\sum _{k=2}^{\infty }|{a}_{k}|{r}^{k}\right\}\\ >Re\left\{1-\frac{3+b}{3+2b}r-\frac{1}{3+2b}\sum _{k=2}^{\infty }\left({k}^{2}-b+1\right)|{a}_{k}|{r}^{k}\right\}\\ >Re\left\{1-\frac{3+b}{3+2b}r-\frac{br}{3+2b}\right\}=1-r>0\end{array}$

Since ( $|z|=r<1$ ), therefore we obtain

$Re\left\{1+2\sum _{k=1}^{\infty }\frac{3+b}{2\left(3+2b\right)}{a}_{k}{z}^{k}\right\}>0,z\in U$

which by Theorem 1.1 shows that $\frac{3+b}{2\left(3+2b\right)}{a}_{k}$ is a subordinating factor, hence, we have established Equation (2.5).

2.2. Theorem

Given $f\left(z\right)\in M\left(1,b\right)$ , then

$Ref\left(z\right)>-\frac{3+2b}{3+b}$ (2.6)

The constant factor $\frac{3+2b}{3+b}$ cannot be replaced by a larger one.

Proof:

Let

$g\left(z\right)=\frac{z}{1-z}$

which is a convex function, Equation (2.1) becomes

$\frac{3+b}{2\left(3+2b\right)}f\left(z\right)\ast \frac{z}{1-z}\prec \frac{z}{1-z}$

Since

$Re\left(\frac{z}{1-z}\right)>-\frac{1}{2},|z|=r$ (2.7)

This implies

$Re\left\{\frac{3+b}{2\left(3+2b\right)}f\left(z\right)\ast \frac{z}{1-z}\right\}>-\frac{1}{2}$ (2.8)

Therefore, we have

$Re\left(f\left(z\right)\right)>-\frac{3+2b}{3+b}$

which is Equation (2.6).

Now to show that sharpness of the constant factor

$\frac{3+b}{3+2b}$

We consider the function

${f}_{1}\left(z\right)=\frac{z\left(3+b\right)+b{z}^{2}}{3+b}$ (2.9)

Applying Equation (2.1) with $g\left(z\right)=\frac{z}{1-z}$ and $f\left(z\right)={f}_{1}\left(z\right)$ , we have

$\frac{z\left(3+b\right)+b{z}^{2}}{2\left(3+b\right)}\prec \frac{z}{1-z}$ (2.10)

Using the fact that

$|Re\left(z\right)|\le |z|$ (2.11)

We now show that the

$\underset{z\in U}{min}\left\{Re\left(\frac{z\left(3+b\right)+b{z}^{2}}{2\left(3+b\right)}\right)\right\}=-\frac{1}{2}$ (2.12)

we have

$\begin{array}{c}|Re\left(\frac{z\left(3+b\right)+b{z}^{2}}{2\left(3+b\right)}\right)|\le |\frac{z\left(3+b\right)+b{z}^{2}}{2\left(3+b\right)}|\le |z|\frac{|\left(3+b\right)+bz|}{|2\left(3+b\right)|}\\ \le \frac{|\left(3+b\right)+bz|}{2\left(3+b\right)}\le \frac{\left(3+b\right)+b}{2\left(3+2b\right)}\le \frac{3+2b}{2\left(3+2b\right)}=\frac{1}{2},\left(|z|=1\right)\end{array}$

This implies that

$|Re\left(\frac{z\left(3+b\right)+b{z}^{2}}{2\left(3+b\right)}\right)|\le \frac{1}{2}$

and therefore

$-\frac{1}{2}\le Re\left(\frac{z\left(3+b\right)+b{z}^{2}}{2\left(3+b\right)}\right)\le \frac{1}{2}$

Hence, we have that

$\underset{z\in U}{min}\left\{Re\left(\frac{z\left(3+b\right)+b{z}^{2}}{2\left(3+b\right)}\right)\right\}=-\frac{1}{2}$

That is

$\underset{z\in U}{min}\left\{Re\frac{3+b}{2\left(3+2b\right)}\left({f}_{1}\ast g\left(z\right)\right)\right\}=-\frac{1}{2}$

which shows the Equation (2.12).

2.3. Theorem

Let

$f\left(z\right)=z+{\sum }_{k=2}^{\infty }\text{ }\text{ }{a}_{k}{z}^{k}\in M\left(1,b\right)$ , $0

then $|{a}_{k}|\le \frac{1}{2}$ .

Proof:

Let

$f\left(z\right)=z+{\sum }_{k=2}^{\infty }\text{ }\text{ }{a}_{k}{z}^{k}\in M\left(1,b\right)$

then by definition of the class $M\left(I,b\right)$ ,

$\sum _{k=2}^{\infty }\left({k}^{2}+b-1\right)|{a}_{k}|\le b,\text{ }0

we have that

$\frac{{k}^{2}+b-1}{b}-k>0$

which gives that

$\sum _{k=2}^{\infty }\text{ }\text{ }k|{a}_{k}|\le \frac{{k}^{2}+b-1}{b}|{a}_{k}|\le 1$

$i.e\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sum _{k=2}^{\infty }\text{ }\text{ }k|{a}_{k}|\le 1$

hence

$2\sum |{a}_{k}|\le 1$

$|{a}_{k}|\le \frac{1}{2}$

Cite this paper

Bello, R.A. (2017) On a Subordination Result of a Subclass of Analytic Functions. Advances in Pure Mathematics, 7, 641-646. https://doi.org/10.4236/apm.2017.711038

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