New Result for Strongly Starlike Functions

R. O. Ayinla¹, T. O. Opoola²

¹Department of Statistics and Mathematical Sciences, Kwara State University, Malete, Nigeria.
²Department of Mathematics, University of Ilorin, Ilorin, Nigeria.

DOI: 10.4236/am.2017.83027                

Abstract

In this paper, using Salagean differential operator, we define and investigate a new subclass of univalent functions . We also establish a characterization property for functions belonging to the class .

Keywords

Strongly Starlike Functions, Strongly Convex Functions, Salagean Differential Operator

1. Introduction

Let $A$ be the class of functions of the form

$f\left(z\right)=z+{\displaystyle \underset{k=2}{\overset{\infty }{\sum }}{a}_k{z}^k}$ (1)

which are analytic in the unit disk $U=\left\{z\in C:\left|z\right|<1\right\}$ . A function $f\left(z\right)\in A$ is said to be starlike of order $\alpha$ if and only if

$\mathrm{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>\alpha ,0\le \alpha <1\left(z\in U\right)$ (2)

We denote by $S^{\ast }\left(\alpha \right)$ the subclass of $A$ consisting of functions which are starlike of order $\alpha$ in $U$ .

Also, a function $f\left(z\right)\in A$ is said to be convex of order $\alpha$ if and only if

$\mathrm{Re}\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}>\alpha ,0\le \alpha <1\left(z\in U\right)$ (3)

We denote by $C\left(\alpha \right)$ the subclass of $A$ consisting of functions which are convex of order $\alpha$ in $U$ .

If $f\left(z\right)\in A$ satisfies

$\left|\arg \left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\alpha \right)\right|<\frac{\text{π}\beta }{2},0\le \alpha <1,0<\beta \le 1,\left(z\in U\right)$ (4)

then $f\left(z\right)$ is said to be strongly starlike of order $\beta$ and type $\alpha$ in $U$ , denoted by [1] .

If $f\left(z\right)\in A$ satisfies

$\left|\arg \left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\alpha \right)\right|<\frac{\text{π}\beta }{2},0\le \alpha <1,0<\beta \le 1,\left(z\in U\right)$ (5)

then $f\left(z\right)$ is said to be strongly convex of order $\beta$ and type $\alpha$ in $U$ , denoted by $C_{\alpha }\left(\beta \right)$ [1] .

The following lemma is needed to derive our result for class $S_{\alpha }^n\left(\beta \right)$ .

Lemma (1) [2] [3] [4] [5] . Let a function $p\left(z\right)$ be analytic in $U,p\left(0\right)=1\text{and}p\left(z\right)\ne 0\left(z\in U\right)$ , if there exists a point $z_0\in U$ such that

$\left|\arg \left(p\left(z\right)\right)\right|<\frac{\text{π}\beta }{2}\left(\left|z\right|<\left|z_0\right|\right)$ and $\left|\arg \left(p\left(z_0\right)\right)\right|=\frac{\text{π}\beta }{2}$ with $0<\beta \le 1$ , then

$\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}=ik\beta$ (6)

where

$k\ge \frac{1}{2}\left(a+\frac{1}{a}\right)\left(\text{when}\arg \left(p\left(z_0\right)\right)\right)=\frac{\text{π}\beta }{2}$

$k\le -\frac{1}{2}\left(a+\frac{1}{a}\right)\left(\text{when}\arg \left(p\left(z_0\right)\right)\right)=-\frac{\text{π}\beta }{2}$

And $p{\left(z_0\right)}^{\frac{1}{\beta }}=\pm ia\left(a>0\right)$ .

Definition 1. A function $f\left(z\right)\in A$ is said to be in the class $S_{\alpha }^n\left(\beta \right)$ if

$\left|\arg \left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}-\alpha \right)\right|<\frac{\text{π}\beta }{2},\left(z\in U\right)$ (7)

For some $\alpha ,0\le \alpha <1,n\in N_0=N\cup \left\{0\right\}0<\beta \le 1$ .

Remark

When $n=0$ then $S_{\alpha }^n\left(\beta \right)$ is the class studied by [1] .

Definition 2. For functions $f\left(z\right)\in A$ the Salagean differential operator [6] is $D^n:A\to A$

$D^{0}f\left(z\right)=f\left(z\right),D^{1}f\left(z\right)=z{f}^{\prime }\left(z\right),\cdots D^{n}f\left(z\right)=D\left[D^{n-1}f\left(z\right)\right],n=0,1,2,3,\cdots$

The main focus of this work is to provide a characterization property for the class of functions belonging to the class $S_{\alpha }^n\left(\beta \right)$ .

2. Main Result

Theorem 1. If $f\left(z\right)\in A$ satisfies

$\begin{array}{l}\left(i\right)\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\ne \frac{1}{2}\\ \left(ii\right)\left|\frac{D^{n+2}f\left(z\right)/D^{n+1}f\left(z\right)}{D^{n+1}f\left(z\right)/D^{n}f\left(z\right)}-1\right|<\frac{\beta }{2},\left(z\in U\right)\end{array}$

for some $\beta ,0<\beta \le 1,n\in N_0=N\cup \left\{0\right\},$ then $f\left(z\right)\in S_{\frac{1}{2}}^n\left(\beta \right)$

Proof. Let

$p\left(z\right)=2\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}-1,n\in N_{0}n=0,1,2,\cdots$ (8)

Taking the logarithmic differentiation in both sides of Equation (8), we have

$\begin{array}{c}\frac{p^{\prime }\left(z\right)}{p\left(z\right)}=\left[\frac{D^{n}f\left(z\right)2{\left(D^{n+1}f\left(z\right)\right)}^{\prime }-2D^{n+1}f\left(z\right){\left[D^{n}f\left(z\right)\right]}^{\prime }}{{\left[D^{n}f\left(z\right)\right]}^2}\right]\left[\frac{D^{n}f\left(z\right)}{2D^{n+1}f\left(z\right)-D^{n}f\left(z\right)}\right]\\ =\left[\frac{D^{n}f\left(z\right)2{\left(D^{n+1}f\left(z\right)\right)}^{\prime }-2D^{n+1}f\left(z\right){\left[D^{n}f\left(z\right)\right]}^{\prime }}{D^{n}f\left(z\right)}\right]\left[\frac{1}{2D^{n+1}f\left(z\right)-D^{n}f\left(z\right)}\right]\\ =\frac{2{\left(D^{n+1}f\left(z\right)\right)}^{\prime }}{D^{n}f\left(z\right)p\left(z\right)}-\frac{2D^{n+1}f\left(z\right){\left[D^{n}f\left(z\right)\right]}^{\prime }}{{\left[D^{n}f\left(z\right)\right]}^{2}p\left(z\right)}\end{array}$ (9)

Multiply Equation (9) through by $p\left(z\right)$ , to get

$p^{\prime }\left(z\right)=\frac{2{\left(D^{n+1}f\left(z\right)\right)}^{\prime }}{D^{n}f\left(z\right)}-\frac{2D^{n+1}f\left(z\right){\left(D^{n}f\left(z\right)\right)}^{\prime }}{{\left(D^{n}f\left(z\right)\right)}^2}$ (10)

Multiply Equation (10) by $z$ to obtain

$\begin{array}{c}z{p}^{\prime }\left(z\right)=\frac{2z{\left(D^{n+1}f\left(z\right)\right)}^{\prime }}{D^{n}f\left(z\right)}-\frac{2D^{n+1}f\left(z\right)z{\left(D^{n}f\left(z\right)\right)}^{\prime }}{{\left(D^{n}f\left(z\right)\right)}^2}\\ =\frac{2\left(D^{n+2}f\left(z\right)\right)}{D^{n}f\left(z\right)}-\frac{{\left(1+p\left(z\right)\right)}^2}{2}\end{array}$ (11)

Multiply Equation (11) through by 2 and divide through by ${\left(1+p\left(z\right)\right)}^2$ to give

$\frac{2z{p}^{\prime }\left(z\right)}{{\left(1+p\left(z\right)\right)}^2}=\frac{4\left(D^{n+2}f\left(z\right)\right)}{D^{n}f\left(z\right){\left(1+p\left(z\right)\right)}^2}-1$ (12)

Multiplying Equation (12) by $\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}=\frac{1+p\left(z\right)}{2}$ , and further simplifica-

tion, we obtain

$\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\left(1+\frac{2z{p}^{\prime }\left(z\right)}{{\left(1+p\left(z\right)\right)}^2}\right)=\frac{D^{n+2}f\left(z\right)}{D^{n+1}f\left(z\right)},z\in U,n\in N_0$ (13)

therefore,

$\frac{D^{n+2}f\left(z\right)/D^{n+1}f\left(z\right)}{D^{n+1}f\left(z\right)/D^{n}f\left(z\right)}=1+\frac{2z{p}^{\prime }\left(z\right)}{{\left(1+p\left(z\right)\right)}^2}$ (14)

If $\exists$ a point $z_0\in U$ which satisfies $\left|\arg p\left(z\right)\right|<\frac{\text{π}\beta }{2}\left(\left|z\right|<\left|z_0\right|\right)$ and

$\left|\arg p\left(z_0\right)\right|=\frac{\text{π}\beta }{2}$

then by lemma [2]

$\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}=ik\beta$

$k\ge \frac{1}{2}\left(a+\frac{1}{a}\right)$ and $p\left(z_0\right)=a^{\beta }{e}^{\frac{i\text{π}\beta }{2}}\text{or}p\left(z_0\right)=a^{\beta }{e}^{\frac{-i\beta }{2}}\left(a>0\right)$

Now,

$\begin{array}{c}\left|\frac{D^{n+2}f\left(z_0\right)/D^{n+1}f\left(z_0\right)}{D^{n+1}f\left(z_0\right)D^{n}f\left(z_0\right)}-1\right|=2k\beta \left|\frac{p\left(z_0\right)}{{\left(1+p\left(z_0\right)\right)}^2}\right|\\ \ge \frac{2\beta \frac{1}{2}\left(a+\frac{1}{a}\right)\left|p\left(z_0\right)\right|}{\left|{\left(1+p\left(z_0\right)\right)}^2\right|}\end{array}$ (15)

Since,

$\frac{1}{\left|{\left(1+p\left(z_0\right)\right)}^2\right|}\ge \frac{1}{1+2\left|p\left(z_0\right)\right|+\left|p{\left(z_0\right)}^2\right|}$ (16)

$\left|\frac{D^{n+2}f\left(z_0\right)/D^{n+1}f\left(z_0\right)}{D^{n+1}f\left(z_0\right)/D^{n}f\left(z_0\right)}-1\right|\ge \frac{\beta \left(a+\frac{1}{a}\right)\left|p\left(z_0\right)\right|}{1+2\left|p\left(z_0\right)\right|+{\left|p\left(z_0\right)\right|}^2}$ (17)

But $\begin{array}{c}p\left(z_0\right)=a^{\beta }{e}^{\frac{i\text{π}\beta }{2}},a>0\Rightarrow \left|p\left(z_0\right)\right|=a^{\beta }\\ =\frac{\beta \left(a+\frac{1}{a}\right)a^{\beta }}{1+2a^{\beta }+a^{2\beta }}\\ =\frac{\left(a+\frac{1}{a}\right)\beta }{a^{-\beta }+2+a^{\beta }}\end{array}$

Let

$S\left(a\right)=\frac{a+\frac{1}{a}}{a^{-\beta }+2+a^{\beta }}$

then

$S^{\prime }\left(a\right)=\frac{2\left(a^2-1\right)+\left(1-\beta \right)a^{-\beta }\left(a^2{}^{\left(1+\beta \right)}-1\right)+\left(1+\beta \right)a^{\beta }\left(a^{2\left(1-\beta \right)}-1\right)}{a^2{\left(a^{\beta }+2+a^{-\beta }\right)}^2}$ (18)

Hence, $S^{\prime }\left(a\right)=0\Rightarrow a=1$ .

It implies that

$S^{\prime }\left(a\right)<0\text{when}0<a<1\text{and}{S}^{\prime }\left(a\right)>0\text{when}a>1,\text{hence},a=1$ is a minimum

point of $S\left(a\right)\cdot S\left(1\right)=\frac{1}{2}$ .

Therefore, we have that

$\left|\frac{D^{n+2}f\left(z\right)f\left(z_0\right)/D^{n+1}f\left(z_0\right)}{D^{n+1}f\left(z_0\right)/D^{n}f\left(z_0\right)}-1\right|\ge \frac{\beta }{2},n\in N_0,z\in U$ (19)

which contradicts the condition of the theorem.

Hence, it is concluded from lemma [2] that

$\left|\arg p\left(z\right)\right|=\left|\arg \left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}-\frac{1}{2}\right)\right|<\frac{\text{π}\beta }{2},z\in U,n\in N_0$ (20)

so that

$f\left(z\right)\in S_{\frac{1}{2}}^n\left(\beta \right).$

Acknowledgements

The authors wish to thank the referees for their useful suggestions that lead to improvement of the quality of the work in this paper.

Conflicts of Interest

The authors declare no conflicts of interest.

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