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.

DOI: 10.4236/apm.2017.711038                

Abstract

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

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 $U=\left\{z\in ℂ:\left|z\right|<1\right\}$ and let S be the subclass of A consisting of functions univalent in U and have the form

$f\left(z\right)=z+{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}{a}_k{z}^k,$ (1.1)

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

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

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

$g\left(z\right)=z+{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}{b}_k{z}^k,$

is defined as:

$\left(f\ast g\right)\left(z\right)=z+{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}{a}_k{b}_k{z}^k=\left(g\ast f\right)\left(z\right),$ (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:

$f\left(z\right)\prec g\left(z\right)\mathrm{,\; <mi>\; </mi>\; <mi>\; </mi>}z\in U$

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

$f\left(z\right)=g\left(\omega \left(z\right)\right),\mathrm{<mi>\; </mi>\; <mi>\; </mi>}z\in U$ (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

$f\left(0\right)=g\left(0\right),\mathrm{<mi>\; </mi>\; <mi>\; </mi>\; <mi>\; </mi>}f\left(u\right)\subset g\left(u\right)$ (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

${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{a}_k{c}_k{z}^k\prec f\left(z\right),\mathrm{<mi>\; </mi>\; <mi>\; </mi>}z\in U,a_1=1$

We have the following theorem:

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

$Re\left\{1+2{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{c}_k{z}^k\right\}>0,\mathrm{<mi>\; </mi>\; <mi>\; </mi>}z\in U$ (1.5)

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

$\left|P\left(z\right)-1\right|<b,\mathrm{<mi>\; </mi>\; <mi>\; </mi>}b>0,\mathrm{<mi>\; </mi>}z\in U.$

The class $P\left(\mathrm{1,}b\right)$ was studied by Janwoski [2] . The family $P\left(\mathrm{1,}b\right)$ contains many interesting classes of functions. For example, for $f\left(z\right)\in A$ , if

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

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

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

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

Let $F\left(\mathrm{1,}b\right)$ be the subclass of $P\left(\mathrm{1,1}-\alpha \right)$ 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. [3] Let $P\left(f\right)$ be given by Equation (1.6) with $f\left(z\right)=z+{\displaystyle \sum }{a}_k{z}^k$ . If

${\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}\left(k^2+b-1\right)\left|a_k\right|<b,\mathrm{<mi>\; </mi>}b>0$

then $P\left(f\right)\in F\left(\mathrm{1,}b\right)$ , $0<b<0.935449$ .

It is natural to consider the class

$M\left(1,b\right)=\left\{f\in A:{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}\left(k^2+b-1\right)\left|a_k\right|<b,\mathrm{<mi>\; </mi>}b>0\right\}$

$0<b<0.935449$

Remark 1.1. [4] If $b=1-\alpha$ , then $M\left(\mathrm{1,1}-\alpha \right)$ consists of starlike functions of order $\alpha$ , $0\le \alpha <1$ since

${\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}\left(k-\alpha \right)\left|a_k\right|<{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}\left(k^2-\alpha \right)\left|a_k\right|$

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

2. Main Results

2.1. Theorem

Let $f\left(z\right)\in M\left(\mathrm{1,}b\right)$ , 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 $0<b<0.935449$ and $g\left(z\right)$ is convex function.

Proof:

Let

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

and suppose that

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

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

By definition (1.1) we have

$\begin{array}{l}\frac{3+b}{2\left(3+2b\right)}\left(f\ast g\right)\left(z\right)\\ =\frac{3+b}{2\left(3+2b\right)}\left(z+{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}{a}_k{b}_k{z}^k\right)\\ ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}\frac{3+b}{2\left(3+2b\right)}{a}_k{b}_k{z}^k,\mathrm{<mi>\; </mi>}{a}_1=1,\mathrm{<mi>\; </mi>}{b}_1=1\end{array}$ (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

$Re\left\{1+2{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}\frac{3+b}{2\left(3+2b\right)}{a}_k{z}^k\right\}>\mathrm{0,\; <mi>\; </mi>\; <mi>\; </mi>}z\in U$ (2.4)

Now,

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

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

$Re\left\{1+2{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}\frac{3+b}{2\left(3+2b\right)}{a}_k{z}^k\right\}>\mathrm{0,\; <mi>\; </mi>\; <mi>\; </mi>}z\in U$

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

$Re\left(\frac{z}{1-z}\right)>-\frac{1}{2}\mathrm{,\; <mi>\; </mi>\; <mi>\; </mi>}\left|z\right|=r$ (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

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

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

This implies that

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

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

$f\left(z\right)=z+{\displaystyle {\sum }_{k=2}^{\infty }}{a}_k{z}^k\in M\left(1,b\right)$ , $0<b<0.935449$

then $\left|a_k\right|\le \frac{1}{2}$ .

Proof:

Let

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

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

${\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}\left(k^2+b-1\right)\left|a_k\right|\le b,0<b<0.935449$

we have that

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

which gives that

${\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}k\left|a_k\right|\le \frac{k^2+b-1}{b}\left|a_k\right|\le 1$

$i.e{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}}k\left|a_k\right|\le 1$

hence

$2{\displaystyle \sum }\left|a_k\right|\le 1$

$\left|a_k\right|\le \frac{1}{2}$

Conflicts of Interest

The authors declare no conflicts of interest.

References

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