# On Univalence Criteria for a General Integral Operator

Authors: Vasile Marius Macarie; Daniel Breaz
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We consider a new general integral operator, and we give sufficient conditions for the univalence of this integral operator in the open unit disk of the complex plane. Several consequences of the main results are also shown.


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## Body

## 1.  Introduction

LetU={z∈ℂ:|z|<1} be the unit disk of the complex plane and let 𝒜 be the class of functions f of the form
(1.1)f(z)=z+∑m=2∞‍amzm,
that are analytic in U and satisfy the usual normalization conditions f(0)=f′(0)-1=0. We denote by S the subclass of 𝒜 consisting of all univalent functions f in U and consider the class P of functions h that are analytic in U and that satisfy h(0)=1 and Reh(z)>0 for all z∈U.In the present paper we obtain sufficient conditions for the following general integral operator to be in the classS (The univalent functions are of importance in geometric functions theory and may have some applications in fluid mechanics and physics):
(1.2)Hα1,…,αn,β,g1,…,gn,h1,…,hp(z)={β∫0z‍uβ-1∏i=1n‍gi′(u)αi∏j=1p‍hj(u)du}1/β,
where αi,β are complex numbers, β≠0, the functions gi(u)∈𝒜 for all i=1,2,…,n and hj(u)∈P for all j=1,2,…,p, where n,p are positive integers.For proving our main results we need the following theorems.Theorem 1.1 (see [1]).
Letα be a complex number, let Reα>0; and let f∈𝒜. If
(1.3)1-|z|2ReαReα|zf′′(z)f′(z)|≤1,
for all z∈U, then for any complex number β with Reβ≥Reα, the function
(1.4)Fβ(z)={β∫0z‍uβ-1f′(u)du}1/β
is in the class S.Theorem 1.2 (see [2]).
If the functiong(z) is regular in U and |g(z)|<1 in U, then for all ξ∈U and z∈U, the following inequalities hold:
(1.5)|g(ξ)-g(z)1-g(z)¯⋅g(ξ)|≤|ξ-z1-z¯⋅ξ|,(1.6)|g′(z)|≤1-|g(z)|21-|z|2.
These are equalities if and only if g(z)=ɛ(z+u)/(1+¯uz), where |ɛ|=1 and |u|<1.Remark 1.3 (see [2]).
Forz=0, from inequality (1.5) we have
(1.7)|g(ξ)-g(0)1-g(0)¯⋅g(ξ)|≤|ξ|,
and hence,
(1.8)|g(ξ)|≤|ξ|+|g(0)|1+|g(0)|⋅|ξ|.
Writing g(0)=a and letting ξ=z, we get
(1.9)|g(z)|≤|z|+|a|1+|a|⋅|z|,
for all z∈U.


## 2.  Main Results

Theorem 2.1.
Letδ>0. For i=1,2,…,n, let αi be a complex number, let Mi>0, and let gi∈𝒜. For j=1,2,…,p, let Nj>0 and hj∈P. If
(2.1)|zgi′′(z)gi′(z)|≤Mi,∀i=1,2,…,n(z∈U),(2.2)|zhj′(z)hj(z)|≤Nj,∀j=1,2,…,p(z∈U),(2.3)∑i=1n‍|αi|⋅Mi+∑j=1p‍Nj≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.Proof.
Let us define the function:(2.4)f(z)=∫0z‍∏i=1n‍gi′(u)αi∏j=1p‍hj(u)du,
with gi∈𝒜, for all i=1,2,…,n and hj∈P, for all j=1,2,…,p, and, thus, we obtain
(2.5)f′(z)=∏i=1n‍gi′(z)αi∏j=1p‍hj(z).
The function f is regular in U and f(0)=f′(0)-1=0. We have
(2.6)zf′′(z)f′(z)=∑i=1n‍αizgi′′(z)gi′(z)+∑j=1p‍zhj′(z)hj(z)(z∈U).
From (2.1), (2.2), and (2.6) we obtain
(2.7)1-|z|2δδ⋅|zf′′(z)f′(z)|≤1-|z|2δδ(∑i=1n‍|αi|⋅Mi+∑j=1p‍Nj)(z∈U),
and by (2.3), we have
(2.8)1-|z|2δδ⋅|zf′′(z)f′(z)|≤1,∀z∈U.
Using (2.8), by Theorem 1.1, it results that the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.Lettingp=1 in Theorem 2.1, we have the following.Corollary 2.2.
Letδ>0,  let N>0, and let h∈P. For i=1,2,…,n, let αi be a complex number, let Mi>0, and let gi∈𝒜. If
(2.9)|zgi′′(z)gi′(z)|≤Mi,∀i=1,2,…,n(z∈U),|zh′(z)h(z)|≤N(z∈U)∑i=1n‍|αi|⋅Mi+N≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator
(2.10)Fα1,…,αn,β,g1,…,gn,h(z)={β∫0z‍uβ-1∏i=1n‍gi′(u)αih(u)du}1/β
is in the class S.Lettingn=1 in Theorem 2.1, we have the following.Corollary 2.3.
Letδ>0, let M>0, let g∈𝒜, and let α be a complex number. For j=1,2,…,p, let Nj>0 and let hj∈P. If
(2.11)|zg′′(z)g′(z)|≤M(z∈U),|zhj′(z)hj(z)|≤Nj,∀j=1,2,…,p(z∈U),|α|⋅M+∑j=1p‍Nj≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator
(2.12)Gα,β,g,h1,…,hp(z)={β∫0z‍uβ-1g′(u)α∏j=1p‍hj(u)du}1/β
is in the class S.ForM1=M2=⋯=Mn=M and N1=N2=⋯=Np=N in Theorem 2.1, we have the following.Corollary 2.4.
Letδ>0, let M>0, and let N>0. For i=1,2,…,n, let αi be a complex number and let gi∈𝒜. For j=1,2,…,p, let hj∈P. If
(2.13)|zgi′′(z)gi′(z)|≤M,∀i=1,2,…,n(z∈U),|zhj′(z)hj(z)|≤N,∀j=1,2,…,p(z∈U),M∑i=1n‍|αi|+pN≤δ,
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.Theorem 2.5.
Letδ>0. For i=1,2,…,n, let αi be a complex number, let Mi>0, and let gi∈𝒜,gi(z)=z+a2iz2+a3iz3+⋯. For j=1,2,…,p, let Nj>0 and hj∈P with hj′(0)=0, and c=∑i=1n‍αia2i/|∏i=1n‍αi|. If
(2.14)|gi′′(z)gi′(z)|<Mi,∀i=1,2,…,n(z∈U),(2.15)|hj′(z)hj(z)|<Nj,∀j=1,2,…,p(z∈U),(2.16)∑i=1n‍|αi|Mi+∑j=1p‍Nj|∏i=1n‍αi|<1,(2.17)|∏i=1n‍αi|≤1max⁡|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|c|)/(1+2|c|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.Proof.
We define the function:(2.18)f(z)=∫0z‍∏i=1n‍gi′(u)αi∏j=1p‍hj(u)du(z∈U),
with gi∈𝒜, for all i=1,2,…,n and hj∈P, for all j=1,2,…,p.
We consider the function(2.19)K(z)=1|∏i=1n‍αi|⋅f′′(z)f′(z),∀z∈U.
We have
(2.20)f′′(z)f′(z)=∑i=1n‍αigi′′(z)gi′(z)+∑j=1p‍hj′(z)hj(z)(z∈U).
From (2.14), (2.15), and (2.20) we obtain
(2.21)1|∏i=1n‍αi|⋅|f′′(z)f′(z)|≤∑i=1n‍|αi|Mi+∑j=1p‍Nj|∏i=1n‍αi|(z∈U).
From (2.16), (2.19), and (2.21) we obtain |K(z)|<1 for all z∈U.
We haveK(0)=2∑i=1n‍αia2i/|∏i=1n‍αi|=2c, and, using Remark 1.3 we get
(2.22)|K(z)|≤|z|+2|c|1+2|c|⋅|z|(z∈U).
From (2.19) and (2.22), we obtain
(2.23)1-|z|2δδ⋅|zf′′(z)f′(z)|≤|∏i=1n‍αi|⋅max⁡|z|<1[1-|z|2δδ⋅|z|⋅|z|+2|c|1+2|c|⋅|z|],
for all z∈U.
From (2.17) and (2.23), we have
(2.24)1-|z|2δδ⋅|zf′′(z)f′(z)|≤1(z∈U).
So, applying Theorem 1.1, we obtain that the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.Lettingp=1 in Theorem 2.5, we have the following.Corollary 2.6.
Letδ>0, let N>0, and let h∈P with h′(0)=0. For i=1,2,…,n, let αi be a complex number, Mi>0 and gi∈𝒜,gi(z)=z+a2iz2+a3iz3+⋯, and c=∑i=1n‍αia2i/|∏i=1n‍αi|. If
(2.25)|gi′′(z)gi′(z)|<Mi,∀i=1,2,…,n(z∈U),|h′(z)h(z)|<N(z∈U),∑i=1n‍|αi|Mi+N|∏i=1n‍αi|<1,|∏i=1n‍αi|≤1max⁡|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|c|)/(1+2|c|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator
(2.26)Fα1,…,αn,β,g1,…,gn,h(z)={β∫0z‍uβ-1∏i=1n‍gi′(u)αih(u)du}1/β
is in the class S.Lettingn=1 in Theorem 2.5, we have the following.Corollary 2.7.
Letδ>0, M>0,  g∈𝒜,g(z)=z+a2z2+a3z3+⋯ and let α be a complex number. For j=1,2,…,p, let Nj>0 and hj∈P with hj′(0)=0. If
(2.27)|g′′(z)g′(z)|<M(z∈U),|h′j(z)hj(z)|<Nj,∀j=1,2,…,p(z∈U),|α|⋅M+∑j=1pNj<|α|≤1max⁡|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|a2|)/(1+2|a2|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator
(2.28)Gα,β,g,h1,…,hp(z)={β∫0z‍uβ-1g′(u)α∏j=1p‍hj(u)du}1/β
is in the class S.ForM1=M2=⋯=Mn=M and N1=N2=⋯=Np=N in Theorem 2.5, we have the following.Corollary 2.8.
Letδ>0, M>0, N>0. For i=1,2,…,n, let αi be a complex number and gi∈𝒜,gi(z)=z+a2iz2+a3iz3+⋯. For j=1,2,…,p, let hj∈P with hj′(0)=0, and c=∑i=1n‍αia2i/|∏i=1n‍αi|.
If(2.29)|gi′′(z)gi′(z)|<M∀i=1,2,…,n(z∈U),|hj′(z)hj(z)|<N∀j=1,2,…,p(z∈U),M∑i=1n‍|αi|+pN|∏i=1n‍αi|<1,|∏i=1n‍αi|≤1max⁡|z|<1[((1-|z|2δ)/δ)⋅|z|⋅((|z|+2|c|)/(1+2|c|⋅|z|))],
then for every complex number β with Reβ≥δ>0, the integral operator Hα1,…,αn,β,g1,…,gn,h1,…,hp(z) given by (1.2) is in the class S.



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*Source: 207410-2012-07-03.xml*