ABSTRACT

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

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

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

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

is defined as:

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

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

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

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

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

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

It is natural to consider the class

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

$0<b<0.935449$

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

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

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

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

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

Now,

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

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

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

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 $M\left(I\mathrm{,}b\right)$ ,

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

we have that

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

which gives that

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

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

hence

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

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

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

References