# Starlikeness Properties of a New Integral Operator for Meromorphic Functions

Authors: Aabed Mohammed; Maslina Darus
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

We define here an integral operatorℋγ1,…,γn for meromorphic functions
in the punctured open unit disk. Several starlikeness conditions for the integral operator ℋγ1,…,γn are derived.


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## Body

## 1. Introduction

LetΣ denotes the class of functions of the form(1.1)f(z)=1z+∑n=0∞anzn,
which are analytic in the punctured open unit disk(1.2)U*={z∈C:0<|z|<1}=U∖{0},
where 𝕌 is the open unit disk 𝕌={z∈ℂ:|z|<1}.We say that a functionf∈Σ is meromorphic starlike of order α(0≤α<1), and belongs to the class Σ*(α), if it satisfies the inequality(1.3)-R(zf′(z)f(z))>α.A functionf∈Σ is a meromorphic convex function of order α(0≤α<1), if f satisfies the following inequality(1.4)-R(1+zf′′(z)f′(z))>α,
and we denote this class by Σk(α).Analogous to the integral operator defined by Breaz et al. [1] on the normalized analytic functions, we now define the following integral operator on the space meromorphic functions in the class ∑.Definition 1.1.
Letn∈ℕ,γi>0,i∈{1,2,3,…,n}. We define the integral operator ℋγ1,…,γn(f1,f2,…,fn):Σn→Σ by
(1.5)Hγ1,…,γn(f1,…,fn)(z)=1z2∫0z(-u2f1′(u))γ1⋯(-u2fn′(u))γndu.
For the sake of simplicity, from now on we will write ℋγ1,…,γn(z) instead of ℋγ1,…,γn(f1,…,fn)(z).
ByΣkp(β)(-1≤β<1), we denote the class of functions f∈Σ such that
(1.6)|zf′′(z)f′(z)+2|<-R(zf′′(z)f′(z)+β)-1.
In order to derive our main results, we have to recall here the following preliminary results.Lemma 1.2 (see [2]).
Suppose that the functionΨ:ℂ2→ℂ satisfies the following condition:
(1.7)R{Ψ(is,t)}≤0,(s,t∈R;t≤-(1+s2)2).
If the function p(z)=1+p1z+⋯ is analytic in 𝕌 and
(1.8)R{Ψ(p(z),zp′(z))}>0,(z∈U),
then,
(1.9)R{p(z)}>0(z∈U).Proposition 1.3 (see [3]).
Iff∈Σ satisfying
(1.10)-R{z(zf′′(z)+3f′(z))zf′(z)+2f(z)}>α,0≤α<1,|zf′(z)f(z)+1|<1,
then,
(1.11)-R{zf′(z)f(z)}>α.


## 2. Starlikeness of the Operatorℋγ1,…,γn(z)

In this section, we investigate sufficient conditions for the integral operatorℋγ1,…,γn(z) which is defined in Definition 1.1, to be in the class Σ*(α),0≤α<1.Theorem 2.1.
Letfi∈∑,γi>0 for alli∈{1,…,n}. If
(2.1)-R(zfi′′(z)fi′(z))>-1nγi+2,
then ℋγ1,…,γn(z) belongs to Σ*(0).Proof.
On successive differentiation ofℋγ1,…,γn(z), which is defined in (1.5), we get
(2.2)2zHγ1,…,γn(z)+z2Hγ1,…,γn′(z)=(-z2f1′(z))γ1⋯(-z2fn′(z))γn,z2Hγ1,…,γn′′(z)+4zHγ1,…,γn′(z)+2Hγ1,…,γn(z)=∑i=1nγi(-z2fi′(z))γi-1(-z2fi′′(z)-2zfi′(z))∏j=1,j≠in(-z2fj′(z))γj.
Then from (2.2), we obtain
(2.3)z2Hγ1,…,γn′′(z)+4zHγ1,…,γn′(z)+2Hγ1,…,γn(z)z2Hγ1,…,γn′(z)+2zHγ1,…,γn(z)=∑i=1nγi(fi′′(z)fi′(z)+2z).
By multiplying (2.3) with z yield,
(2.4)z2Hγ1,…,γn′′(z)+4zHγ1,…,γn′(z)+2Hγ1,…,γn(z)zHγ1,…,γn′(z)+2Hγ1,…,γn(z)=∑i=1nγi(zfi′′(z)fi′(z)+2).
That is equivalent to
(2.5){z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}+1=∑i=1nγi(zfi′′(z)fi′(z)+2).
Or
(2.6)-{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}=∑i=1nγi(-zfi′′(z)fi′(z))-2∑i=1nγi+1.
We can write the left-hand side of (2.6), as the following:
(2.7)-(zHγ1,…,γn′(z)/Hγ1,…,γn(z))((zHγ1,…,γn′′(z)/Hγ1,…,γn′(z))+3)(zHγ1,…,γn′(z)/Hγ1,…,γn(z))+2=∑i=1nγi(-zfi′′(z)fi′(z))-2∑i=1nγi+1.
We define the regular function p in 𝕌 by
(2.8)p(z)=-zHγ1,…,γn′(z)Hγ1,…,γn(z),
and p(0)=1. Differentiating p(z) logarithmically, we obtain
(2.9)-p(z)+zp′(z)p(z)=1+zHγ1,…,γn′′(z)Hγ1,…,γn′(z).
From (2.7), (2.8), and (2.9), we obtain
(2.10)p(z)+zp′(z)-p(z)+2=∑i=1nγi(-zfi′′(z)fi′(z))-2∑i=1nγi+1.
Let us put
(2.11)Ψ(u,v)=u+v-u+2.
From (2.1), (2.10), and (2.11), we obtain
(2.12)R{Ψ(p(z),zp′(z))}=γ1(-Rzf1′′(z)f1′(z))+⋯+(-Rzfn′′(z)fn′(z))-2(γ1+⋯+γn)+1>γ1(-1nγ1+2)+⋯+γn(-1nγn+2)-2(γ1+⋯+γn)+1=0.
Now, we proceed to show that
(2.13)R{Ψ(is,t)}≤0,(s,t∈R;t≤-(1+s2)2).
Indeed, from (2.11), we have
(2.14)R{Ψ(is,t)}=R{is+t-is+2}=2t4+s2≤-1+s24+s2<0.
Thus, from (2.12), (2.14), and by using Lemma 1.2, we conclude that ℜ{p(z)}>0, and so
(2.15)-R{zHγ1,…,γn′(z)Hγ1,…,γn(z)}>0
that is, ℋγ1,…,γn(z) is starlike of order 0.Theorem 2.2.
Fori∈{1,…,n}, let γi>0 and fi∈Σk(αi)(0≤αi<1). If 0<∑i=1nγi(1-αi)≤1,ℋγ1,…,γn(z) be the integral operator given by (1.5) and
(2.16)|zHγ1,…,γn′(z)Hγ1,…,γn(z)+1|<1.
Then ℋγ1,…,γn(z) belong to Σ*(μ), where μ=1-∑i=1nγi(1-αi).Proof.
Following the same steps as in Theorem2.1, we obtain
(2.17)-{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}=∑i=1nγi{-(zfi′′(z)fi′(z)+1)}+1-∑i=1nγi.
Taking the real part of both terms of the last expression, we have
(2.18)-R{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}=∑i=1nγi{-R(zfi′′(z)fi′(z)+1)}+1-∑i=1nγi.
Since fi∈Σk(αi), for i∈{1,…,n}, we receive
(2.19)-R{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}>∑i=1nγiαi+1-∑i=1nγi.
Therefore,
(2.20)-R{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}>1-∑i=1nγi(1-αi).
Using (2.16), (2.20), and applying Proposition 1.3, we get ℋγ1,…,γn(z)∈Σ*(μ), where μ=1-∑i=1nγi(1-αi).Lettingαi=α,i∈{1,…,n} in Theorem 2.2, we get the following.Corollary 2.3.
Fori∈{1,…,n}, let γi>0 and fi∈Σk(α)(0≤α<1). If
(2.21)0<∑i=1nγi≤11-α,ℋγ1,…,γn be the integral operator given by (1.5) and
(2.22)|zHγ1,…,γn′(z)Hγ1,…,γn(z)+1|<1.
Then ℋγ1,…,γn(z) is starlike of order 1-(1-α)∑i=1nγi.Theorem 2.4.
Fori∈{1,…,n}, let γi>0 and fi∈Σkp(βi)(-1≤βi<1). If
(2.23)0<∑i=1nγi(1-βi)≤1,ℋγ1,…,γn(z) be the integral operator given by (1.5) and
(2.24)|zHγ1,…,γn′(z)Hγ1,…,γn(z)+1|<1.
Then ℋγ1,…,γn(z) is starlike of order 1-∑i=1nγi(1-βi).Proof.
Following the same steps as in Theorem2.1, we obtain
(2.25)-{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}=-∑i=1nγi(zfi′′(z)fi′(z)+2)+1=∑i=1nγi{-(zfi′′(z)fi′(z)+βi)-1}+1-∑i=1nγi+∑i=1nγiβi=∑i=1nγi{-(zfi′′(z)fi′(z)+βi)-1}+1-∑i=1nγi(1-βi).
We calculate the real part from both terms of the above equality and obtain
(2.26)-R{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}=∑i=1nγi{-R(zfi′′(z)fi′(z)+βi)-1}+1-∑i=1nγi(1-βi).
Since fi∈Σkp(βi) for all i∈{1,…,n}, the above relation then yields
(2.27)-R{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}>∑i=1nγi|zfi′′(z)fi′(z)+2|+1-∑i=1nγi(1-βi).
Because ∑i=1nγi|zfi′′(z)/fi′(z)+2|≥0, we obtain that
(2.28)-R{z(zHγ1,…,γn′′(z)+3Hγ1,…,γn′(z))zHγ1,…,γn′(z)+2Hγ1,…,γn(z)}>1-∑i=1nγi(1-βi).
Using (2.24), (2.28) and applying Proposition 1.3, we get ℋγ1,…,γn(z) is a starlike function of order 1-∑i=1nγi(1-βi).Lettingβi=β,i∈{1,…,n} in Theorem 2.4, we get the following.Corollary 2.5.
Fori∈{1,…,n}, let γi>0 and fi∈Σkp(β)(-1≤β<1). If
(2.29)0<∑i=1nγi≤11-β,
ℋγ1,…,γn be the integral operator given by (1.5) and
(2.30)|zHγ1,…,γn′(z)Hγ1,…,γn(z)+1|<1.
Then ℋγ1,…,γn(z) is starlike of order 1-(1-β)∑i=1nγi.Lettingn=1,γ1=γ and f1=f in Corollary 2.5, we get the following.Corollary 2.6.
Letγ>0, and f∈Σkp(β)(-1≤β<1). If
(2.31)0<γ≤11-β,ℋγ(z) be the integral operator,
(2.32)Hγ(z)=1z2∫0z(-u2f′(u))γdu,|zHγ′(z)Hγ(z)+1|<1.
Then ℋγ(z) is starlike of order 1-(1-β)γ.Other work related to integral operator for different studies can also be found in [4–6].



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