# The Univalence Conditions of Some Integral Operators

Authors: Laura Stanciu
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

We introduce new integral operators of analytic functionsf and g defined in the open unit disk 𝕌. For these operators, we discuss some univalence conditions.


---

## Body

## 1. Introduction and Preliminaries

Let𝒜 denote  the class of all functions of the form(1.1)f(z)=z+∑n=2∞anzn,
which are analytic in the open unit disk(1.2)U={z∈C:|z|<1}
and satisfy the following usual normalization condition:(1.3)f(0)=f′(0)-1=0.
Also, let 𝒮 denote the subclass of 𝒜 consisting of functions f, which are univalent in 𝕌 (see, for details [1]; see also [2, 3]).In [4, 5], Pescar gave the following univalence conditions for the functions f∈𝒜.Theorem 1.1 (see [4]).
Letα be a complex number,  Reα>0, and  c a complex number,  |c|≤1,c≠-1, and  f(z)=z+⋯ a regular function in  𝕌. If
(1.4)|c|z|2α+(1-|z|2α)zf′′(z)αf′(z)|≤1,
for all z∈𝕌, then the function
(1.5)Fα(z)=(α∫0ztα-1f′(t)dt)1/α=z+⋯
is regular and univalent in 𝕌.Theorem 1.2 (see [5]).
Letα be a complex number,  Reα>0, and  c a complex number,  |c|≤1,c≠-1, and  f∈𝒜. If
(1.6)1-|z|2ReαReα|zf′′(z)f′(z)|≤1-|c|
for all z∈𝕌, then for any complex number β,Reβ≥Reα, the function
(1.7)Fβ(z)=(β∫0ztβ-1f′(t)dt)1/β
is in the class 𝒮.On the other hand, for the functionsf∈𝒜, Ozaki and Nunokawa [6] proved another univalence condition asserted by Theorem 1.3.Theorem 1.3 (see [6]).
Letf∈𝒜 satisfy the condition
(1.8)|z2f′(z)[f(z)]2-1|<1(z∈U).
Then f is univalent in 𝕌.In the paper [7], Pescar determined some univalence conditions for the following integral operators.Theorem 1.4 (see [7]).
Let the functiong satisfy (1.8),  M a positive real number fixed, and  c a complex number. If  α∈[(2M+1)/(2M+2),(2M+1)/2M],
(1.9)|c|≤1-|α-1α|(2M+1),c≠-1,|g(z)|≤M
for all z∈𝕌, then the function
(1.10)Gα(z)=(α∫0z[g(t)]α-1dt)1/α
is in the class 𝒮.Theorem 1.5 (see [7]).
Letg∈𝒜,α a real number,  α≥1, and  c a complex number,  |c|≤1/α,c≠-1. If
(1.11)|g′′(z)g′(z)|≤1(z∈U),
then the function
(1.12)Hα(z)=(α∫0z[tg′(t)]α-1dt)1/α
is in the class 𝒮.Theorem 1.6 (see [7]).
Letg∈𝒜 satisfies (1.8),  α a complex number,  M>1 fixed,  Reα>0, and  c a complex number,  |c|<1. If  |g(z)|≤M for all  z∈𝕌, then for any complex number  β(1.13)Reβ≥Reα≥2M+1|α|(1-|c|)
the function
(1.14)Hβ(z)=(β∫0ztβ-1(g(t)t)1/αdt)1/β
is in the class 𝒮.In this paper, we introduce the following integral operators as follows:(1.15)F1(f,g)(z)=(α∫0z(f(t)eg(t))α-1dt)1/α(f,g∈A;α∈C),(1.16)G1(f,g)(z)=(α∫0z(tf′(t)eg(t))α-1dt)1/α(f,g∈A;α∈C),(1.17)H1(f,g)(z)=(β∫0ztβ-1(f(t)teg(t))1/αdt)1/β(f,g∈A;α,β∈C-{0}).Remark 1.7.
Foreg(z)=1 and f(z)=g(z), the integral operators (1.15), (1.16), and (1.17) would reduce to the integral operators (1.10), (1.12), and (1.14).In this paper, we generalize the integral operators given by Pescar [7], and we study the univalence conditions for the integral operators defined by (1.15), (1.16), and (1.17).For this purpose, we need the following result.Lemma 1.8 (General Schwarz Lemma [8]).
Let the functionf be regular in the disk  𝕌R={z∈ℂ:|z|<R}, with  |f(z)|<M for fixed  M. If  f has one zero with multiplicity order bigger than  m for  z=0, then
(1.18)|f(z)|≤MRm|z|m(z∈UR).
The equality can hold only if
(1.19)f(z)=eiθMRmzm,
where θ is constant.


## 2. Main Results

Theorem 2.1.
Letf,g∈𝒜, where g satisfies the condition (1.8), M1 and M2 are real positive numbers, and α a complex number, Reα>0. If
(2.1)|f′(z)f(z)|≤M1(z∈U),|g(z)|≤M2(z∈U),(2.2)|c|≤1-|α-1α|(M1+2M22+1),c∈C,c≠-1,
then the integral operator F1(f,g)(z) defined by (1.15) is in the class 𝒮.Proof.
From (1.15), we have
(2.3)F1(f,g)(z)=(α∫0ztα-1(f(t)teg(t))α-1dt)1/α.
Let us consider the function
(2.4)h(z)=∫0z(f(t)teg(t))α-1dt.
The function h is regular in 𝕌. From (2.4), we get
(2.5)h′(z)=(f(z)zeg(z))α-1,h′′(z)=(α-1)(f(z)zeg(z))α-2(zf′(z)-f(z)z2eg(z)+f(z)zg′(z)eg(z)).
Also, a simple computation yields
(2.6)zh′′(z)h′(z)=(α-1)[(zf′(z)f(z)-1)+zg′(z)],
which readily shows that
(2.7)|c|z|2α+(1-|z|2α)zh′′(z)αh′(z)|=|c|z|2α+(1-|z|2α)α-1α((zf′(z)f(z)-1)+zg′(z))|≤|c|+|α-1α|((|zf′(z)f(z)|+1)+|z2g′(z)[g(z)]2||[g(z)]2z|)(z∈U).
From the hypothesis of Theorem 2.1, we have
(2.8)|f′(z)f(z)|≤M1(z∈U),|g(z)|≤M2(z∈U),
then by General Schwarz Lemma for the function g, we obtain
(2.9)|g(z)|≤M2|z|(z∈U).
Using the inequality (2.7), we have
(2.10)|c|z|2α+(1-|z|2α)zh′′(z)αh′(z)|≤|c|+|α-1α|(M1+1+(|z2g′(z)[g(z)]2-1|+1)M22).
From (2.10) and since g satisfies the condition (1.8), we have
(2.11)|c|z|2α+(1-|z|2α)zh′′(z)αh′(z)|≤|c|+|α-1α|(M1+2M22+1),
from which, by (2.2), we get
(2.12)|c|z|2α+(1-|z|2α)zh′′(z)αh′(z)|≤1(z∈U).
Applying Theorem 1.1, we conclude that the integral operator F1(f,g)(z) defined by (1.15) is in the class 𝒮.SettingM1=1 and M2=1 in Theorem 2.1, we immediately arrive at the following application of Theorem 2.1.Corollary 2.2.
Letf,g∈𝒜, where  g satisfies the condition (1.8) and  α a complex number,  Reα>0. If
(2.13)|f′(z)f(z)|≤1(z∈U),|g(z)|≤1(z∈U),|c|≤1-|4(α-1)α|,c∈C,c≠-1,
then the integral operator F1(f,g)(z) defined by (1.15) is in the class 𝒮.Theorem 2.3.
Letf,g∈𝒜, where  g satisfies the inequality  |g(z)|≤M,M≥1. Also, let  α be a real number,  α≥1, and  c a complex number with
(2.14)|c|≤1-|α-1α|(M+1),c≠-1.
If
(2.15)|f′′(z)f′(z)|≤1(z∈U),|g′(z)g(z)|≤1(z∈U),
then the integral operator G1(f,g)(z) defined by (1.16) is in the class 𝒮.Proof.
We observe that(2.16)G1(f,g)(z)=(α∫0ztα-1(f′(t)eg(t))α-1dt)1/α.
Let us consider the function
(2.17)h(z)=∫0z(f′(t)eg(t))α-1dt.
The function h is regular in 𝕌. From (2.17), we have
(2.18)h′(z)=(f′(z)eg(z))α-1,zh′′(z)h′(z)=(α-1)(zf′′(z)f′(z)+zg′(z)),
which readily shows that
(2.19)|c|z|2α+(1-|z|2α)zh′′(z)αh′(z)|≤|c|+|α-1α|(|zf′′(z)f′(z)|+|zg′(z)g(z)||g(z)|)(z∈U).
From (2.19) and the conditions of Theorem 2.3, we get
(2.20)|c|z|2α+(1-|z|2α)zh′′(z)αh′(z)|≤|c|+|α-1α|(1+M)≤1(z∈U).
Applying Theorem 1.1, we conclude that the integral operator G1(f,g)(z) defined by (1.16) is in the class 𝒮.SettingM=1 in Theorem 2.3, we obtain the following consequence of Theorem 2.3.Corollary 2.4.
Letf,g∈𝒜, where  g satisfies the condition  |g′(z)/g(z)|≤1,α a real number,  α≥1, and  c a complex number with  |c|≤1-|2(α-1)/α|,c≠-1. If
(2.21)|f′′(z)f′(z)|≤1(z∈U),|g(z)|≤1(z∈U),
then the integral operator G1(f,g)(z) defined by (1.16) is in the class 𝒮.Theorem 2.5.
Letf,g∈𝒜, where  g satisfies the condition (1.8),  α a complex number,  Reα>0,M1 and  M2 are real positive numbers, and  c a complex number,  |c|<1. If
(2.22)|f′(z)f(z)|≤M1(z∈U),|g(z)|≤M2(z∈U),
then for any complex number β,
(2.23)Reβ≥Reα≥M1+2M22+1|α|(1-|c|),
the integral operator H1(f,g)(z) defined by (1.17) is in the class 𝒮.Proof.
Let us consider the function(2.24)h(z)=∫0z(f(t)teg(t))1/αdt.
The function h is regular in 𝕌. From (2.24), we have
(2.25)h′(z)=(f(z)zeg(z))1/α,zh′′(z)h′(z)=1α((zf′(z)f(z)-1)+zg′(z)),
which readily shows that
(2.26)1-|z|2ReαReα|zh′′(z)h′(z)|≤1-|z|2Reα|α|Reα(|zf′(z)f(z)|+1+|z2g′(z)[g(z)]2||[g(z)]2z|)≤1-|z|2Reα|α|Reα(|zf′(z)f(z)|+1+(|z2g′(z)[g(z)]2-1|+1)|[g(z)]2z|).
By the General Schwarz Lemma for the function g, we obtain
(2.27)|g(z)|≤M2|z|(z∈U),
and using the inequality (2.26), we have
(2.28)1-|z|2ReαReα|zh′′(z)h′(z)|≤1-|z|2Reα|α|Reα(M1+1+(|z2g′(z)[g(z)]2-1|+1)M22).
From (2.28) and since g satisfies the condition (1.8), we get
(2.29)1-|z|2ReαReα|zh′′(z)h′(z)|≤1-|z|2ReαReα(M1+2M22+1)|α|≤M1+2M22+1|α|Reα.
From (2.23), we have
(2.30)M1+2M22+1|α|Reα≤1-|c|,
and using (2.29), we obtain
(2.31)1-|z|2ReαReα|zh′′(z)h′(z)|≤1-|c|(z∈U).
Applying Theorem 1.2, we conclude that the integral operator H1(f,g)(z) defined by (1.17) is in the class 𝒮.SettingM1=1 and M2=1 in Theorem 2.5, we obtain the following corollary.Corollary 2.6.
Letf,g∈𝒜, where  g satisfies the condition (1.8),  α a complex number,  Reα>0, and  c a complex number,  |c|<1. If
(2.32)|f′(z)f(z)|≤1(z∈U),|g(z)|≤1(z∈U),
then for any complex number β,
(2.33)Reβ≥Reα≥4|α|(1-|c|),
the integral operator H1(f,g)(z) defined by (1.17) is in the class 𝒮.



---

*Source: 924645-2012-05-10.xml*