# Kudriasov Type Univalence Criteria for Some Integral Operators

Authors: Virgil Pescar; Nicoleta Breaz
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider some integral operators defined by analytic functions in the open unit disk and derive new univalence criteria for these operators, using Kudriasov condition for a function to be univalent.


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## Body

## 1. Introduction

Let𝒜 be the class of functions f which are analytic in the open unit disk U={z∈ℂ:|z|<1} of the form
(1)f(z)=z+∑n=2∞‍anzn,
normalized by f(0)=f′(0)-1=0.We denote by𝒮 the subclass of 𝒜 consisting of functions f∈𝒜, which are univalent in U.We consider the integral operators(2)Hn(z)={β∫0zuβ-1(f1(u)u)γ1⋯(fn(u)u)γndu}1/β,Tn(z)={β∫0zuβ-1(f1′(u))γ1⋯(fn′(u))γndu}1/β,
for β, γj being complex numbers, β≠0, j=1,n¯, and the functions fj∈𝒜, j=1,n¯.Some univalence criteria for these integral operators were studied in [1]. Applying univalence conditions given by Kudriasov [2] and Pascu [3], we obtain new Kudriasov type univalence criteria for these two integral operators.


## 2. Preliminaries

Various generalizations of Becker’s univalence criteria for analytic functions given in [4] were obtained by many authors. For example, the result obtained by Pascu in [3] is also known as an improvement of Becker’s univalence criteria. This result or other similar generalizations of Becker’s univalence criteria have been used further to derive new univalence criteria for integral operators (see, e.g., some relatively recent works as [1, 5, 6]). In this paper we use Pascu improvement of Becker’s univalence criteria and also another univalence condition for a function to be univalent, given by Kudriasov in [2]. There are also some papers devoted to univalence criteria that use some Kudriasov type conditions (see, e.g., the work [1] containing a chapter dedicated to Kudriasov type univalence conditions and other papers as, e.g., [7–10]).The following univalence criteria are given by Kudriasov for a regular function.Lemma 1 (see [2]).
Letf be a regular function in U, f(z)=z+a2z2+⋯. If
(3)|f′′(z)f′(z)|≤K,z∈U,
for all z∈U, where K≅3.05, the function f is univalent in U.Remark 2.
The constantK is a solution of the equation 8[x(x-2)3]1/2-3(4-x)2=12. An approximation of this solution in MATLAB environment is 3.03902118847875. However, we call this constant in our further results like Kudriasov gave it, approximately equal to 3.05.The improvement of Becker’s univalence condition is given by Pascu for integral operators as follows.Lemma 3 (see [3]).
Letα be a complex number, Reα>0, and the function f∈𝒜. If
(4)1-|z|2ReαReα|zf′′(z)f′(z)|≤1,
for all z∈U, then for every complex number β, Reβ≥Reα, the function
(5)Fβ(z)=[β∫0zuβ-1f′(u)du]1/β
is regular and univalent in U.


## 3. Main Results

Theorem 4.
Letα, γj be complex numbers, Reα>0, j=1,n¯, the functions fj∈𝒜, fj=z+a2jz2+⋯, j=1,n¯, n∈ℕ-{0}, and K the positive real number K≅3.05.
If(6)|fj′′(z)fj′(z)|≤K,z∈U,j=1,n¯,(7)|γ1|+|γ2|+⋯+|γn|≤min{Reα4,14},
then fj∈𝒮, j=1,n¯, and for every complex number β, Reβ≥Reα, the integral operator Hn is in the class 𝒮.Proof.
Let us consider the function(8)hn(z)=∫0z(f1(u)u)γ1⋯(fn(u)u)γndu.
The functionhn is regular in U and hn(0)=hn′(0)-1=0. We have
(9)zhn′′(z)hn′(z)=γ1(zf1′(z)f1(z)-1)+⋯+γn(zfn′(z)fn(z)-1),
for all z∈U.
From (9), we obtain further
(10)|zhn′′(z)hn′(z)|≤|γ1|(|zf1′(z)f1(z)|+1)+⋯+|γn|(|zfn′(z)fn(z)|+1).
By (6) and Lemma 1, we have fj∈𝒮, j=1,n¯, and hence we obtain
(11)|zfj′(z)fj(z)|≤1+|z|1-|z|,z∈U,j=1,n¯.
From (10) and (11), we get
(12)1-|z|2ReαReα|zhn′′(z)hn′(z)|≤1-|z|2ReαReα21-|z|(|γ1|+⋯+|γn|)
for all z∈U.
Now we consider the following cases.
(1)0<Reα<1. The function s:(0,1)→ℝ, s(x)=1-a2x, x=Reα, a=|z|, (0≤a<1) is increasing and we obtain
(13)1-|z|2Reα≤1-|z|2,z∈U.
From (12) and (13), we obtain
(14)1-|z|2ReαReα|zhn′′(z)hn′(z)|≤4Reα(|γ1|+⋯+|γn|)
for all z∈U.
Using the hypothesis condition (7), from (14), we have
(15)1-|z|2ReαReα|zhn′′(z)hn′(z)|≤1,
for all z∈U.
(2)Reα≥1. We notice that the function
(16)q:[1,∞)⟶ℝ,q(x)=1-a2xx,x=Reα,a=|z|,(0≤a<1)
is decreasing function, and we obtain
(17)1-|z|2ReαReα≤1-|z|2,z∈U.
Using the last inequality in (12), we have
(18)1-|z|2ReαReα|zhn′′(z)hn′(z)|≤4(|γ1|+⋯+|γn|)
for all z∈U.
Now using the hypothesis condition (7), from (18), we get
(19)1-|z|2ReαReα|zhn′′(z)hn′(z)|≤1
for all z∈U.
Hence, based on the conditions obtained in (15) and in (19), applying Lemma 3, we have that Hn∈𝒮.Theorem 5.
Letα, γj be complex numbers, j=1,n¯, Reα>0, the functions fj∈𝒜, fj(z)=z+a2jz2+⋯, j=1,n¯, n∈ℕ-{0}, and K the positive real number, K≅3.05.
If(20)|fj′′(z)fj′(z)|≤K,z∈U,j=1,n¯,(21)|γ1|+|γ2|+⋯+|γn|≤(2Reα+1)(2Reα+1)/(2Reα)2K,
then fj∈𝒮, j=1,n¯, and for every complex number β, Reβ≥Reα, the integral operator Tn is in the class 𝒮.Proof.
By (20) and Lemma 1, we obtain that fj∈𝒮, j=1,n¯.
We consider the function(22)pn(z)=∫0z(f1′(u))γ1⋯(fn′(u))γndu.
The function pn is regular in U and pn(0)=pn′(0)-1=0.
We have(23)zpn′′(z)pn′(z)=γ1zf1′′(z)f1′(z)+⋯+γnzfn′′(z)fn′(z),
for all z∈U.
Further we obtain(24)1-|z|2ReαReα|zpn′′(z)pn′(z)|≤1-|z|2ReαReα|z|[|γ1||f1′′(z)f1′(z)|+⋯+|γn||fn′′(z)fn′(z)|].
From (24) and from the Kudriasov condition within the hypothesis, (20), we have
(25)1-|z|2ReαReα|zpn′′(z)pn′(z)|≤[1-|z|2ReαReα|z|]×(K|γ1|+⋯+K|γn|),
for all z∈U.
Let us consider the functionG:[0,1]→ℝ, G(x)=((1-x2a)/a)x, x=|z|, a=Reα. We have
(26)maxx∈[0,1]G(x)=2(2a+1)(2a+1)/(2a).
By (25), (26), and (21) we obtain
(27)1-|z|2ReαReα|zpn′′(z)pn′(z)|<1,
for all z∈U.
Now from (27) and Lemma 3, it results that Tn∈𝒮.


## 4. Corollaries

Corollary 1.
Letγ be complex number, Re[n(γ-1)+1]>0, the functions fj∈𝒜, j=1,n¯, n∈ℕ-{0}, and K the positive real number, K≅3.05.
If(28)|fj′′(z)fj′(z)|≤K,z∈U,j=1,n¯,n|γ-1|≤min{Re[n(γ-1)+1]4,14},
then fj∈𝒮, j=1,n¯, and the integral operator Iα,n defined by
(29)Iα,n(z)={[n(γ-1)+1]∫0zf1γ-1(u)⋯fnγ-1(u)du}1/(n(γ-1)+1)
is in the class 𝒮.Proof.
From (29), we have
(30)Iα,n(z)={[n(γ-1)+1]∫0zun(γ-1)(f1(u)u)γ-1iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii⋯(fn(u)u)γ-1du}1/(n(γ-1)+1),
and for α=β=n(γ-1)+1, γ1=γ2=⋯=γn=γ-1, from Theorem 4, we obtain Corollary 1.Corollary 2.
Letα, γj be complex numbers, j=1,n¯, 0<Reα≤1, the functions fj∈𝒜, fj(z)=z+a2jz2+⋯, j=1,n¯, n∈ℕ-{0}, and K the positive real number K≅3.05.
If(31)|fj′′(z)fj′(z)|≤K,z∈U,j=1,n¯,|γ1|+|γ2|+⋯+|γn|≤Reα4,
then fj∈𝒮, j=1,n¯ and the integral operator Ln defined by
(32)Ln(z)=∫0z(f1(u)u)γ1⋯(fn(u)u)γndu
belongs to the class 𝒮.Corollary 3.
Letα, γj be complex numbers, j=1,n¯, 0<Reα≤1, the functions fj∈𝒜, fj(z)=z+a2jz2+⋯, j=1,n¯, n∈ℕ-{0},  and K the positive real number, K≅3.05.
If(33)|fj′′(z)fj′(z)|≤K,z∈U,j=1,n¯,|γ1|+|γ2|+⋯+|γn|≤(2Reα+1)(2Reα+1)/(2Reα)2K,
then fj∈𝒮, j=1,n¯, and the integral operator Gn defined by
(34)Gn(z)=∫0z(f1′(u))γ1‍⋯(fn′(u))γndu
is in the class 𝒮.



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