# Univalence Conditions Related to a General Integral Operator

Authors: Nicoleta Breaz; Virgil Pescar
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We consider a general integral operator based on two types of analytic functions, namely, regular functions and, respectively, functions having a positive real part. Some univalence conditions for this integral operator are obtained.


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## Body

## 1. Introduction

Let𝒜 be the class of functions of the form
(1.1)f(z)=z+∑i=2∞‍aizi,
which are analytic in the open unit disk, 𝒰={z∈ℂ:|z|<1} and normalized by f(0)=f′(0)-1=0. Also, let 𝒮 denote the subclass of 𝒜 consisting of univalent functions, in the open unit disk. We denote by 𝒫, the class of functions p which are analytic in 𝒰, Rep(z)>0, for all z∈𝒰 and p(0)=1, the so-called Caratheodory functions.In [1], Pescar introduced and studied the following integral operator:
(1.2)En(z)=∫0z‍∏j=1n‍(fj(u)u)αj(gj(u))βjdu,
where αj,βj are complex numbers, fj∈𝒜, gj∈𝒫, j=1,n¯.In this paper, we generalize this integral operator, by considering the general integral operator defined as follows:(1.3)Kn(z)=(δ∫0z‍uδ-1∏j=1n‍(fj(u)u)αj(gj(u))βjdu)1/δ,
where δ, αj,βj are complex numbers, δ≠0, fj∈𝒜, gj∈𝒫, j=1,n¯.Remarks.
This integral operator extends many other integral operators from related works on this field as those given by Srivastava, Mocanu, Owa, Pescar, Orhan, Breaz, and others. Using this new operator and the related univalence conditions that are going to be proved here, one can study some other already known operators in a unified perspective. Thus, for different particular cases of the parametersδ, αj, βj, our integral operator Kn is shown to be an extension of the following integral operators.(i)
Forδ=1, the integral operator Kn is the operator En from (1.2), introduced by Pescar in [1].(ii)
Forn=1, αj=δ, βj=0, the integral operator Kn was studied in [2], by Miller and Mocanu.(iii)
Forβj=0, j=1,n¯, the integral operator Kn was studied in [3], by Pescar and Breaz.(iv)
Forβj=0, j=1,n¯, αj=α-1, and δ=n(α-1)+1, the integral operator Kn was studied in [4], by Breaz et al. and also in [5], by Srivastava et al.(v)
Forδ=1, βj=0, j=1,n¯, we get the integral operator defined by D. Breaz and N. Breaz, in the paper [6].(vi)
Forδ=1, αj=0, βj>0, and gj=fj′, the operator was studied by Breaz et al., in the paper [7].More precisely, if we are interested to study various properties of two or more different operators reminded above (and other similar operators which are not mentioned here), we can do this in an integrated manner by simply allowing the parameters involved in the definition, to be more general and consequently by studying only one operator, having the form (1.3). In this paper, we will study some univalence criteria for this new operator.The following known results will be used in order to prove our results.Lemma 1.1 (see [8]).
Letγ be a complex number, Reγ>0 and f∈𝒜. If
(1.4)1-|z|2ReγReγ|zf′′(z)f′(z)|≤1,
for all z∈𝒰, then for any complex number δ, Reδ≥Reγ,
(1.5)Fδ(z)=[δ∫0z‍uδ-1f′(u)du]1/δ∈𝒮.Lemma 1.2 (see [9]).
Letf be a regular function in the open disk 𝒰R={z∈ℂ:|z|<R} with |f(z)|<M, M fixed. If f has in z=0 one zero with multiplicity ≥m, then
(1.6)|f(z)|≤MRm|z|m,(z∈𝒰R),
the equality for (z≠0) can hold only if
(1.7)f(z)=eiθMRmzm,
where θ is constant.Lemma 1.3 (see [10]).
For eachf∈𝒮,
(1.8)1-r1+r≤|zf′(z)f(z)|≤1+r1-r,|z|=r<1.Lemma 1.4 (see [11]).
Ifg(z)=1+∑n=1∞‍bnzn is regular in |z|<1 and Re(g(z))>0, then
(1.9)|zg′(z)g(z)|≤2|z|1-|z|2.Also, for the statement of our main results, we need to define the following classes:(1.10)𝒜M={f∈𝒜:|zf′(z)f(z)-1|≤M,M≥1},𝒫L={f∈𝒫:|zf′(z)f(z)|≤L,L>0}.


## 2. Main Results

Theorem 2.1.
Letδ,αj,βj be complex numbers, Reδ≥1, Mj,Lj positive real numbers, Mj≥1, j=1,n¯, and the functions fj∈𝒜Mj, gj∈𝒫Lj, j=1,n¯.
If(2.1)∑j=1n‍[|αj|Mj+|βj|Lj]≤332,
then the general integral operator Kn, defined by (1.3), is in the class 𝒮.Proof.
We consider the regular function(2.2)hn(z)=∫0z‍(f1(u)u)α1⋯(fn(u)u)αn·(g1(u))β1⋯(gn(u))βndu.
After some calculus, we have
(2.3)(1-|z|2)|zhn′′(z)hn′(z)|≤(1-|z|2)∑j=1n‍[|αj||zfj′(z)fj(z)-1|+|βj||zgj′(z)gj(z)|],
for all z∈𝒰.
Sincefj∈AMj, by applying Lemma 1.2 for m=1, we get
(2.4)|zfj′(z)fj(z)-1|≤Mj|z|,(j=1,n¯;z∈𝒰).
Also sincegj∈PLj, from Lemma 1.2, we have
(2.5)|zgj′(z)gj(z)|≤Lj|z|,(j=1,n¯;z∈𝒰).
If we put these last two inequalities in (2.3), together with the inequality from the hypothesis, we get
(2.6)(1-|z|2)|zhn′′(z)hn′(z)|≤(1-|z|2)|z|·332.
Now we take into account the fact that(2.7)max|z|<1[(1-|z|2)|z|]=233,
thus obtaining
(2.8)(1-|z|2)|zhn′′(z)hn′(z)|≤1,(z∈𝒰).
Further from Lemma1.1, for Reδ≥Reγ=1, it is obvious that Kn∈𝒮.Corollary 2.2.
Letδ, αj be complex numbers, Reδ≥1, Mj positive real numbers, Mj≥1, fj∈𝒜Mj, j=1,n¯. If
(2.9)∑j=1n‍[|αj|Mj]≤332
then the function(2.10)Kn(z)=[δ∫0z‍uδ-1∏j=1n‍(fj(u)u)αjdu]1/δ,
is in the class 𝒮.Proof.
In Theorem2.1, we take βj=0,j=1,n¯.Corollary 2.3.
Letδ, βj be complex numbers, Reδ≥1, Lj positive real numbers and gj∈𝒫Lj, j=1,n¯. If
(2.11)∑j=1n‍[|βj|Lj]≤332,
then the function
(2.12)Kn(z)=[δ∫0z‍uδ-1∏j=1n‍(gj(u))βjdu]1/δ
belongs to the class 𝒮.Proof.
In Theorem2.1, we take αj=0,j=1,n¯.Theorem 2.4.
Letγ,δ,αj,βj be complex numbers, j=1,n¯, Reδ≥Reγ>0, and the functions fj∈𝒜μ, gj∈𝒫μ, μ≥1, with
(2.13)μ=(2Reγ+1)1+(1/2Reγ)2.
If(2.14)∑j=1n‍[|αj|+|βj|]≤1,
then the general integral operator Kn belongs to the class 𝒮.Proof.
Let consider again the regular function:(2.15)hn(z)=∫0z‍(f1(u)u)α1⋯(fn(u)u)αn·(g1(u))β1⋯(gn(u))βndu.
After some calculus, we have(2.16)1-|z|2ReγReγ|zhn′′(z)hn′(z)|≤1-|z|2ReγReγ∑j=1n‍[|αj||zfj′(z)fj(z)-1|+|βj||zgj′(z)gj(z)|],(z∈𝒰).
Sincefj∈Aμ and gj∈Pμ, for all j=1,n¯, from Lemma 1.2, we obtain
(2.17)|zfj′(z)fj(z)-1|≤μ|z|,(j=1,n¯;z∈𝒰),|zgj′(z)gj(z)|≤μ|z|,(j=1,n¯;z∈𝒰).
Using these last two inequalities, the inequality from the hypothesis, and the fact that(2.18)max|z|<1[(1-|z|2Reγ)|z|Reγ]=1μ,
from (2.16) we get
(2.19)1-|z|2ReγReγ|zhn′′(z)hn′(z)|≤1,
for all z∈𝒰.
Hence, by Lemma1.1, we have that Kn∈𝒮.Corollary 2.5.
Letγ,δ,αj be complex numbers, Reδ≥Reγ>0, and fj∈𝒜μ, j=1,n¯, μ≥1, with
(2.20)μ=(2Reγ+1)1+(1/2Reγ)2.
If
(2.21)∑j=1n‍|αj|≤1,
then the integral operator defined by
(2.22)Kn(z)=[δ∫0z‍uδ-1∏j=1n‍(fj(u)u)αjdu]1/δ,
is in the class 𝒮.Proof.
In Theorem2.4, we take β1=⋯=βn=0.Corollary 2.6.
Letγ,δ,βj be complex numbers, j=1,n¯, Reδ≥Reγ>0, and gj∈𝒫μ, with
(2.23)μ=(2Reγ+1)1+(1/2Reγ))2.
If
(2.24)∑j=1n‍|βj|≤1,
then the integral operator defined by
(2.25)Kn(z)=[δ∫0z‍uδ-1∏j=1n‍(gj(u))βjdu]1/δ,
is in the class 𝒮.Proof.
We put in Theorem2.4, α1=⋯=αn=0.Theorem 2.7.
Letγ,δ,αj,βj be complex numbers, j=1,n¯, Reδ≥Reγ>0 and fj∈𝒮, gj∈𝒫. If
(2.26)2∑j=1n‍|αj|+∑j=1n‍|βj|≤min{Reγ2,12},
then the general integral operator Kn∈𝒮.Proof.
We consider the same regular functionhn as in the proof of the previous theorems, and after some calculus we get
(2.27)1-|z|2ReγReγ|zhn′′(z)hn′(z)|≤1-|z|2ReγReγ∑j=1n‍[|αj|(|zfj′(z)fj(z)|+1)+|βj||zgj′(z)gj(z)|],
for all z∈𝒰.
Sincefj∈𝒮, from Lemma 1.3, we get
(2.28)|zfj′(z)fj(z)|≤1+|z|1-|z|,j=1,n¯, z∈𝒰, and since gj∈P, from Lemma 1.4, we get
(2.29)|zgj′(z)gj(z)|≤2|z|1-|z|2j=1,n¯, z∈𝒰.
Now if we use (2.28) and (2.29) in (2.27), we obtain
(2.30)1-|z|2ReγReγ|zhn′′(z)hn′(z)|≤1-|z|2ReγReγ[21-|z|∑j=1n‍|αj|+2|z|1-|z|2∑j=1n‍|βj|],
for all z∈𝒰.
We consider two cases.
(1) Ifmin{Reγ/2,1/2}=Reγ/2, we have
(2.31)1-|z|2Reγ≤1-|z|2
and further if we use this inequality together with the inequality from the hypothesis, from (2.30), we get
(2.32)1-|z|2ReγReγ|zhn′′(z)hn′(z)|≤1,
for all z∈𝒰, and with Lemma 1.1 the proof is complete.
(2) Ifmin{Reγ/2,1/2}=1/2, we obtain
(2.33)1-|z|2ReγReγ≤1-|z|2,z∈𝒰 and further if we put this last inequality in (2.30) we get
(2.34)1-|z|2ReγReγ|zhn′′(z)hn′(z)|≤4∑j=1n‍|αj|+2∑j=1n‍|βj|,
for all z∈𝒰.
Now by applying first the inequality condition from the hypothesis and then Lemma1.1 for hn, the proof is complete.Corollary 2.8.
Letγ,δ,αj be complex numbers, j=1,n¯, Reδ≥Reγ>0 and fj∈𝒮. If
(2.35)∑j=1n‍|αj|≤min{Reγ4,14},
then the integral operator
(2.36)Kn(z)=[δ∫0z‍uδ-1∏j=1n‍(fj(u)u)αjdu]1/δ
is in the class 𝒮.Proof.
In Theorem2.7, we take β1=⋯=βn=0.Corollary 2.9.
Letγ,δ,βj be complex numbers, j=1,n¯, Reδ≥Reγ>0 and gj∈𝒫. If
(2.37)∑j=1n‍|βj|≤min{Reγ2,12}
then the integral operator
(2.38)Kn(z)=[δ∫0z‍uδ-1∏j=1n‍(gj(u))βjdu]1/δ
is in the class 𝒮.Proof.
In Theorem2.7 we take α1=⋯=αn=0.



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*Source: 140924-2012-12-24.xml*