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Journal of Function Spaces and Applications
Volume 2012 (2012), Article ID 207410, 8 pages
doi:10.1155/2012/207410
Research Article

On Univalence Criteria for a General Integral Operator

Vasile Marius Macarie1 and Daniel Breaz2

1Department of Mathematics, University of Piteşti, Târgu din Vale Street, No. 1, 110040 Piteşti, Romania
2Department of Mathematics, “1 Decembrie 1918” University of Alba Iulia, Nicolae Iorga Street, No. 11–13, 510009 Alba Iulia, Romania

Received 5 February 2012; Revised 19 May 2012; Accepted 19 May 2012

Academic Editor: Gestur Ólafsson

Copyright © 2012 Vasile Marius Macarie and Daniel Breaz. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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.

1. Introduction

Let 𝑈 = { 𝑧 ∈ ℂ ∶ | 𝑧 | < 1 } be the unit disk of the complex plane and let 𝒜 be the class of functions 𝑓 of the form 𝑓 ( 𝑧 ) = 𝑧 + ∞ 𝑚 = 2 𝑎 𝑚 𝑧 𝑚 , ( 1 . 1 ) 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 R e ℎ ( 𝑧 ) > 0 for all 𝑧 ∈ 𝑈 .

In the present paper we obtain sufficient conditions for the following general integral operator to be in the class 𝑆 (The univalent functions are of importance in geometric functions theory and may have some applications in fluid mechanics and physics): 𝐻 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 1 , … , ℎ 𝑝 𝛽 ( 𝑧 ) = 𝑧 0 𝑢 𝑛 𝛽 − 1 𝑖 = 1 𝑔 𝑖 ( 𝑢 ) 𝛼 𝑖 𝑝 𝑗 = 1 ℎ 𝑗 ( 𝑢 ) 𝑑 𝑢 1 / 𝛽 , ( 1 . 2 ) where 𝛼 𝑖 , 𝛽 are complex numbers, 𝛽 ≠ 0 , the functions 𝑔 𝑖 ( 𝑢 ) ∈ 𝒜 for all 𝑖 = 1 , 2 , … , 𝑛 and ℎ 𝑗 ( 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]). Let 𝛼 be a complex number, let R e 𝛼 > 0 ; and let 𝑓 ∈ 𝒜 . If 1 − | 𝑧 | 2 R e 𝛼 | | | | R e 𝛼 𝑧 𝑓 ( 𝑧 ) 𝑓 | | | | ( 𝑧 ) ≤ 1 , ( 1 . 3 ) for all 𝑧 ∈ 𝑈 , then for any complex number 𝛽 with R e 𝛽 ≥ R e 𝛼 , the function 𝐹 𝛽 𝛽 ( 𝑧 ) = 𝑧 0 𝑢 𝛽 − 1 𝑓 ( 𝑢 ) 𝑑 𝑢 1 / 𝛽 ( 1 . 4 ) is in the class S.

Theorem 1.2 (see [2]). If the function 𝑔 ( 𝑧 ) is regular in 𝑈 and | 𝑔 ( 𝑧 ) | < 1 in 𝑈 , then for all 𝜉 ∈ 𝑈 and 𝑧 ∈ 𝑈 , the following inequalities hold: | | | | 𝑔 ( 𝜉 ) − 𝑔 ( 𝑧 ) 1 − | | | | ≤ | | | | 𝑔 ( 𝑧 ) ⋅ 𝑔 ( 𝜉 ) 𝜉 − 𝑧 1 − | | | | , | | 𝑔 𝑧 ⋅ 𝜉 ( 1 . 5 ) | | ≤ | | | | ( 𝑧 ) 1 − 𝑔 ( 𝑧 ) 2 1 − | 𝑧 | 2 . ( 1 . 6 ) These are equalities if and only if 𝑔 ( 𝑧 ) = 𝜀 ( 𝑧 + 𝑢 ) / ( 1 + ̄ 𝑢 𝑧 ) , where | 𝜀 | = 1 and | 𝑢 | < 1 .

Remark 1.3 (see [2]). For 𝑧 = 0 , from inequality (1.5) we have | | | | 𝑔 ( 𝜉 ) − 𝑔 ( 0 ) 1 − | | | | ≤ | | 𝜉 | | , 𝑔 ( 0 ) ⋅ 𝑔 ( 𝜉 ) ( 1 . 7 ) and hence, | | | | ≤ | | 𝜉 | | + | | | | 𝑔 ( 𝜉 ) 𝑔 ( 0 ) | | | | ⋅ | | 𝜉 | | 1 + 𝑔 ( 0 ) . ( 1 . 8 ) Writing 𝑔 ( 0 ) = 𝑎 and letting 𝜉 = 𝑧 , we get | | | | ≤ 𝑔 ( 𝑧 ) | 𝑧 | + | 𝑎 | , 1 + | 𝑎 | ⋅ | 𝑧 | ( 1 . 9 ) for all 𝑧 ∈ 𝑈 .

2. Main Results

Theorem 2.1. Let 𝛿 > 0 . For 𝑖 = 1 , 2 , … , 𝑛 , let 𝛼 𝑖 be a complex number, let 𝑀 𝑖 > 0 , and let 𝑔 𝑖 ∈ 𝒜 . For 𝑗 = 1 , 2 , … , 𝑝 , let 𝑁 𝑗 > 0 and ℎ 𝑗 ∈ 𝑃 . If | | | | 𝑧 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 | | | | ( 𝑧 ) ≤ 𝑀 𝑖 | | | | , ∀ 𝑖 = 1 , 2 , … , 𝑛 ( 𝑧 ∈ 𝑈 ) , ( 2 . 1 ) 𝑧 ℎ 𝑗 ( 𝑧 ) ℎ 𝑗 | | | | ( 𝑧 ) ≤ 𝑁 𝑗 , ∀ 𝑗 = 1 , 2 , … , 𝑝 ( 𝑧 ∈ 𝑈 ) , ( 2 . 2 ) 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | ⋅ 𝑀 𝑖 + 𝑝 𝑗 = 1 𝑁 𝑗 ≤ 𝛿 , ( 2 . 3 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐻 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 1 , … , ℎ 𝑝 ( 𝑧 ) given by (1.2) is in the class 𝑆 .

Proof. Let us define the function: 𝑓 ( 𝑧 ) = 𝑧 0 𝑛 𝑖 = 1 𝑔 𝑖 ( 𝑢 ) 𝛼 𝑖 𝑝 𝑗 = 1 ℎ 𝑗 ( 𝑢 ) 𝑑 𝑢 , ( 2 . 4 ) with 𝑔 𝑖 ∈ 𝒜 , for all 𝑖 = 1 , 2 , … , 𝑛 and ℎ 𝑗 ∈ 𝑃 , for all 𝑗 = 1 , 2 , … , 𝑝 , and, thus, we obtain 𝑓 ( 𝑧 ) = 𝑛 𝑖 = 1 𝑔 𝑖 ( 𝑧 ) 𝛼 𝑖 𝑝 𝑗 = 1 ℎ 𝑗 ( 𝑧 ) . ( 2 . 5 ) The function 𝑓 is regular in 𝑈 and 𝑓 ( 0 ) = 𝑓 ′ ( 0 ) − 1 = 0 . We have 𝑧 𝑓 ( 𝑧 ) 𝑓 = ( 𝑧 ) 𝑛 𝑖 = 1 𝛼 𝑖 𝑧 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 + ( 𝑧 ) 𝑝 𝑗 = 1 𝑧 ℎ 𝑗 ( 𝑧 ) ℎ 𝑗 ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) . ( 2 . 6 ) From (2.1), (2.2), and (2.6) we obtain 1 − | 𝑧 | 2 𝛿 𝛿 ⋅ | | | | 𝑧 𝑓 ( 𝑧 ) 𝑓 | | | | ≤ ( 𝑧 ) 1 − | 𝑧 | 2 𝛿 𝛿 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | ⋅ 𝑀 𝑖 + 𝑝 𝑗 = 1 𝑁 𝑗 ( 𝑧 ∈ 𝑈 ) , ( 2 . 7 ) and by (2.3), we have 1 − | 𝑧 | 2 𝛿 𝛿 ⋅ | | | | 𝑧 𝑓 ( 𝑧 ) 𝑓 | | | | ( 𝑧 ) ≤ 1 , ∀ 𝑧 ∈ 𝑈 . ( 2 . 8 ) Using (2.8), by Theorem 1.1, it results that the integral operator 𝐻 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 1 , … , ℎ 𝑝 ( 𝑧 ) given by (1.2) is in the class 𝑆 .

Letting 𝑝 = 1 in Theorem 2.1, we have the following.

Corollary 2.2. Let 𝛿 > 0 ,  let 𝑁 > 0 , and let ℎ ∈ 𝑃 . For 𝑖 = 1 , 2 , … , 𝑛 , let 𝛼 𝑖 be a complex number, let 𝑀 𝑖 > 0 , and let 𝑔 𝑖 ∈ 𝒜 . If | | | | 𝑧 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 | | | | ( 𝑧 ) ≤ 𝑀 𝑖 | | | | , ∀ 𝑖 = 1 , 2 , … , 𝑛 ( 𝑧 ∈ 𝑈 ) , 𝑧 ℎ ( 𝑧 ) | | | | ℎ ( 𝑧 ) ≤ 𝑁 ( 𝑧 ∈ 𝑈 ) 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | ⋅ 𝑀 𝑖 + 𝑁 ≤ 𝛿 , ( 2 . 9 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐹 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 𝛽 ( 𝑧 ) = 𝑧 0 𝑢 𝑛 𝛽 − 1 𝑖 = 1 𝑔 𝑖 ( 𝑢 ) 𝛼 𝑖 ℎ ( 𝑢 ) 𝑑 𝑢 1 / 𝛽 ( 2 . 1 0 ) is in the class 𝑆 .

Letting 𝑛 = 1 in Theorem 2.1, we have the following.

Corollary 2.3. Let 𝛿 > 0 , let 𝑀 > 0 , let 𝑔 ∈ 𝒜 , and let 𝛼 be a complex number. For 𝑗 = 1 , 2 , … , 𝑝 , let 𝑁 𝑗 > 0 and let ℎ 𝑗 ∈ 𝑃 . If | | | | 𝑧 𝑔 ( 𝑧 ) 𝑔 | | | | | | | | ( 𝑧 ) ≤ 𝑀 ( 𝑧 ∈ 𝑈 ) , 𝑧 ℎ 𝑗 ( 𝑧 ) ℎ 𝑗 | | | | ( 𝑧 ) ≤ 𝑁 𝑗 , ∀ 𝑗 = 1 , 2 , … , 𝑝 ( 𝑧 ∈ 𝑈 ) , | 𝛼 | ⋅ 𝑀 + 𝑝 𝑗 = 1 𝑁 𝑗 ≤ 𝛿 , ( 2 . 1 1 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐺 𝛼 , 𝛽 , 𝑔 , ℎ 1 , … , ℎ 𝑝 𝛽 ( 𝑧 ) = 𝑧 0 𝑢 𝛽 − 1 𝑔 ′ ( 𝑢 ) 𝛼 𝑝 𝑗 = 1 ℎ 𝑗 ( 𝑢 ) 𝑑 𝑢 1 / 𝛽 ( 2 . 1 2 ) is in the class 𝑆 .

For 𝑀 1 = 𝑀 2 = ⋯ = 𝑀 𝑛 = 𝑀 and 𝑁 1 = 𝑁 2 = ⋯ = 𝑁 𝑝 = 𝑁 in Theorem 2.1, we have the following.

Corollary 2.4. Let 𝛿 > 0 , let 𝑀 > 0 , and let 𝑁 > 0 . For 𝑖 = 1 , 2 , … , 𝑛 , let 𝛼 𝑖 be a complex number and let 𝑔 𝑖 ∈ 𝒜 . For 𝑗 = 1 , 2 , … , 𝑝 , let ℎ 𝑗 ∈ 𝑃 . If | | | | 𝑧 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 | | | | | | | | ( 𝑧 ) ≤ 𝑀 , ∀ 𝑖 = 1 , 2 , … , 𝑛 ( 𝑧 ∈ 𝑈 ) , 𝑧 ℎ 𝑗 ( 𝑧 ) ℎ 𝑗 | | | | 𝑀 ( 𝑧 ) ≤ 𝑁 , ∀ 𝑗 = 1 , 2 , … , 𝑝 ( 𝑧 ∈ 𝑈 ) , 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | + 𝑝 𝑁 ≤ 𝛿 , ( 2 . 1 3 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐻 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 1 , … , ℎ 𝑝 ( 𝑧 ) given by (1.2) is in the class 𝑆 .

Theorem 2.5. Let 𝛿 > 0 . For 𝑖 = 1 , 2 , … , 𝑛 , let 𝛼 𝑖 be a complex number, let 𝑀 𝑖 > 0 , and let 𝑔 𝑖 ∈ 𝒜 , 𝑔 𝑖 ( 𝑧 ) = 𝑧 + 𝑎 𝑖 2 𝑧 2 + 𝑎 𝑖 3 𝑧 3 + ⋯ . For 𝑗 = 1 , 2 , … , 𝑝 , let 𝑁 𝑗 > 0 and ℎ 𝑗 ∈ 𝑃 with ℎ 𝑗 ( 0 ) = 0 , and ∑ 𝑐 = 𝑛 𝑖 = 1 𝛼 𝑖 𝑎 𝑖 2 ∏ / | 𝑛 𝑖 = 1 𝛼 𝑖 | . If | | | | 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 | | | | ( 𝑧 ) < 𝑀 𝑖 | | | | ℎ , ∀ 𝑖 = 1 , 2 , … , 𝑛 ( 𝑧 ∈ 𝑈 ) , ( 2 . 1 4 ) 𝑗 ( 𝑧 ) ℎ 𝑗 | | | | ( 𝑧 ) < 𝑁 𝑗 ∑ , ∀ 𝑗 = 1 , 2 , … , 𝑝 ( 𝑧 ∈ 𝑈 ) , ( 2 . 1 5 ) 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | 𝑀 𝑖 + ∑ 𝑝 𝑗 = 1 𝑁 𝑗 | | ∏ 𝑛 𝑖 = 1 𝛼 𝑖 | | | | | | | < 1 , ( 2 . 1 6 ) 𝑛 𝑖 = 1 𝛼 𝑖 | | | | | ≤ 1 m a x | 𝑧 | < 1 1 − | 𝑧 | 2 𝛿 / 𝛿 ⋅ | 𝑧 | ⋅ ( ( | 𝑧 | + 2 | 𝑐 | ) / ( 1 + 2 | 𝑐 | ⋅ | 𝑧 | ) ) , ( 2 . 1 7 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐻 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 1 , … , ℎ 𝑝 ( 𝑧 ) given by (1.2) is in the class 𝑆 .

Proof. We define the function: 𝑓 ( 𝑧 ) = 𝑧 0 𝑛 𝑖 = 1 𝑔 𝑖 ( 𝑢 ) 𝛼 𝑖 𝑝 𝑗 = 1 ℎ 𝑗 ( 𝑢 ) 𝑑 𝑢 ( 𝑧 ∈ 𝑈 ) , ( 2 . 1 8 ) with 𝑔 𝑖 ∈ 𝒜 , for all 𝑖 = 1 , 2 , … , 𝑛 and ℎ 𝑗 ∈ 𝑃 , for all 𝑗 = 1 , 2 , … , 𝑝 .
We consider the function 1 𝐾 ( 𝑧 ) = | | ∏ 𝑛 𝑖 = 1 𝛼 𝑖 | | ⋅ 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) , ∀ 𝑧 ∈ 𝑈 . ( 2 . 1 9 ) We have 𝑓 ( 𝑧 ) 𝑓 = ( 𝑧 ) 𝑛 𝑖 = 1 𝛼 𝑖 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 + ( 𝑧 ) 𝑝 𝑗 = 1 ℎ 𝑗 ( 𝑧 ) ℎ 𝑗 ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) . ( 2 . 2 0 ) From (2.14), (2.15), and (2.20) we obtain 1 | | ∏ 𝑛 𝑖 = 1 𝛼 𝑖 | | ⋅ | | | | 𝑓 ( 𝑧 ) 𝑓 | | | | ≤ ∑ ( 𝑧 ) 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | 𝑀 𝑖 + ∑ 𝑝 𝑗 = 1 𝑁 𝑗 | | ∏ 𝑛 𝑖 = 1 𝛼 𝑖 | | ( 𝑧 ∈ 𝑈 ) . ( 2 . 2 1 ) From (2.16), (2.19), and (2.21) we obtain | 𝐾 ( 𝑧 ) | < 1 for all 𝑧 ∈ 𝑈 .
We have ∑ 𝐾 ( 0 ) = 2 𝑛 𝑖 = 1 𝛼 𝑖 𝑎 𝑖 2 ∏ / | 𝑛 𝑖 = 1 𝛼 𝑖 | = 2 𝑐 , and, using Remark 1.3 we get | | | | ≤ 𝐾 ( 𝑧 ) | 𝑧 | + 2 | 𝑐 | 1 + 2 | 𝑐 | ⋅ | 𝑧 | ( 𝑧 ∈ 𝑈 ) . ( 2 . 2 2 ) From (2.19) and (2.22), we obtain 1 − | 𝑧 | 2 𝛿 𝛿 ⋅ | | | | 𝑧 𝑓 ( 𝑧 ) 𝑓 | | | | ≤ | | | | | ( 𝑧 ) 𝑛 𝑖 = 1 𝛼 𝑖 | | | | | ⋅ m a x | 𝑧 | < 1 1 − | 𝑧 | 2 𝛿 𝛿 ⋅ | 𝑧 | ⋅ | 𝑧 | + 2 | 𝑐 | 1 + 2 | 𝑐 | ⋅ | 𝑧 | , ( 2 . 2 3 ) for all 𝑧 ∈ 𝑈 .
From (2.17) and (2.23), we have 1 − | 𝑧 | 2 𝛿 𝛿 ⋅ | | | | 𝑧 𝑓 ( 𝑧 ) 𝑓 | | | | ( 𝑧 ) ≤ 1 ( 𝑧 ∈ 𝑈 ) . ( 2 . 2 4 ) So, applying Theorem 1.1, we obtain that the integral operator 𝐻 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 1 , … , ℎ 𝑝 ( 𝑧 ) given by (1.2) is in the class 𝑆 .

Letting 𝑝 = 1 in Theorem 2.5, we have the following.

Corollary 2.6. Let 𝛿 > 0 , let 𝑁 > 0 , and let ℎ ∈ 𝑃 with ℎ ′ ( 0 ) = 0 . For 𝑖 = 1 , 2 , … , 𝑛 , let 𝛼 𝑖 be a complex number, 𝑀 𝑖 > 0 and 𝑔 𝑖 ∈ 𝒜 , 𝑔 𝑖 ( 𝑧 ) = 𝑧 + 𝑎 𝑖 2 𝑧 2 + 𝑎 𝑖 3 𝑧 3 + ⋯ , and ∑ 𝑐 = 𝑛 𝑖 = 1 𝛼 𝑖 𝑎 𝑖 2 ∏ / | 𝑛 𝑖 = 1 𝛼 𝑖 | . If | | | | 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 | | | | ( 𝑧 ) < 𝑀 𝑖 | | | | ℎ , ∀ 𝑖 = 1 , 2 , … , 𝑛 ( 𝑧 ∈ 𝑈 ) , ( 𝑧 ) | | | | ∑ ℎ ( 𝑧 ) < 𝑁 ( 𝑧 ∈ 𝑈 ) , 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | 𝑀 𝑖 + 𝑁 | | ∏ 𝑛 𝑖 = 1 𝛼 𝑖 | | | | | | | < 1 , 𝑛 𝑖 = 1 𝛼 𝑖 | | | | | ≤ 1 m a x | 𝑧 | < 1 1 − | 𝑧 | 2 𝛿 , / 𝛿 ⋅ | 𝑧 | ⋅ ( ( | 𝑧 | + 2 | 𝑐 | ) / ( 1 + 2 | 𝑐 | ⋅ | 𝑧 | ) ) ( 2 . 2 5 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐹 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 𝛽 ( 𝑧 ) = 𝑧 0 𝑢 𝑛 𝛽 − 1 𝑖 = 1 𝑔 𝑖 ( 𝑢 ) 𝛼 𝑖 ℎ ( 𝑢 ) 𝑑 𝑢 1 / 𝛽 ( 2 . 2 6 ) is in the class 𝑆 .

Letting 𝑛 = 1 in Theorem 2.5, we have the following.

Corollary 2.7. Let 𝛿 > 0 , 𝑀 > 0 ,   𝑔 ∈ 𝒜 , 𝑔 ( 𝑧 ) = 𝑧 + 𝑎 2 𝑧 2 + 𝑎 3 𝑧 3 + ⋯ and let 𝛼 be a complex number. For 𝑗 = 1 , 2 , … , 𝑝 , let 𝑁 𝑗 > 0 and ℎ 𝑗 ∈ 𝑃 with ℎ 𝑗 ( 0 ) = 0 . If | | | | 𝑔 ( 𝑧 ) 𝑔 | | | | | | | | ℎ ( 𝑧 ) < 𝑀 ( 𝑧 ∈ 𝑈 ) , 𝑗 ( 𝑧 ) ℎ 𝑗 | | | | ( 𝑧 ) < 𝑁 𝑗 , ∀ 𝑗 = 1 , 2 , … , 𝑝 ( 𝑧 ∈ 𝑈 ) , | 𝛼 | ⋅ M + 𝑝 ∑ 𝑗 = 1 𝑁 𝑗 1 < | 𝛼 | ≤ m a x | 𝑧 | < 1 1 − | 𝑧 | 2 𝛿 | | 𝑎 / 𝛿 ⋅ | 𝑧 | ⋅ | 𝑧 | + 2 2 | | / | | 𝑎 1 + 2 2 | | , ⋅ | 𝑧 | ( 2 . 2 7 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐺 𝛼 , 𝛽 , 𝑔 , ℎ 1 , … , ℎ 𝑝 𝛽 ( 𝑧 ) = 𝑧 0 𝑢 𝛽 − 1 𝑔 ′ ( 𝑢 ) 𝛼 𝑝 𝑗 = 1 ℎ 𝑗 ( 𝑢 ) 𝑑 𝑢 1 / 𝛽 ( 2 . 2 8 ) is in the class 𝑆 .

For 𝑀 1 = 𝑀 2 = ⋯ = 𝑀 𝑛 = 𝑀 and 𝑁 1 = 𝑁 2 = ⋯ = 𝑁 𝑝 = 𝑁 in Theorem 2.5, we have the following.

Corollary 2.8. Let 𝛿 > 0 , 𝑀 > 0 , 𝑁 > 0 . For 𝑖 = 1 , 2 , … , 𝑛 , let 𝛼 𝑖 be a complex number and 𝑔 𝑖 ∈ 𝒜 , 𝑔 𝑖 ( 𝑧 ) = 𝑧 + 𝑎 𝑖 2 𝑧 2 + 𝑎 𝑖 3 𝑧 3 + ⋯ . For 𝑗 = 1 , 2 , … , 𝑝 , let ℎ 𝑗 ∈ 𝑃 with ℎ 𝑗 ( 0 ) = 0 , and ∑ 𝑐 = 𝑛 𝑖 = 1 𝛼 𝑖 𝑎 𝑖 2 ∏ / | 𝑛 𝑖 = 1 𝛼 𝑖 | .
If | | | | 𝑔 𝑖 ( 𝑧 ) 𝑔 𝑖 | | | | | | | | ℎ ( 𝑧 ) < 𝑀 ∀ 𝑖 = 1 , 2 , … , 𝑛 ( 𝑧 ∈ 𝑈 ) , 𝑗 ( 𝑧 ) ℎ 𝑗 | | | | 𝑀 ∑ ( 𝑧 ) < 𝑁 ∀ 𝑗 = 1 , 2 , … , 𝑝 ( 𝑧 ∈ 𝑈 ) , 𝑛 𝑖 = 1 | | 𝛼 𝑖 | | + 𝑝 𝑁 | | ∏ 𝑛 𝑖 = 1 𝛼 𝑖 | | | | | | | < 1 , 𝑛 𝑖 = 1 𝛼 𝑖 | | | | | ≤ 1 m a x | 𝑧 | < 1 1 − | 𝑧 | 2 𝛿 , / 𝛿 ⋅ | 𝑧 | ⋅ ( ( | 𝑧 | + 2 | 𝑐 | ) / ( 1 + 2 | 𝑐 | ⋅ | 𝑧 | ) ) ( 2 . 2 9 ) then for every complex number 𝛽 with R e 𝛽 ≥ 𝛿 > 0 , the integral operator 𝐻 𝛼 1 , … , 𝛼 𝑛 , 𝛽 , 𝑔 1 , … , 𝑔 𝑛 , ℎ 1 , … , ℎ 𝑝 ( 𝑧 ) given by (1.2) is in the class 𝑆 .

References

  1. N. N. Pascu, “An improvement of Becker's univalence criterion,” in Proceedings of the Commemorative Session: Simion Stoilow, pp. 43–48, University of Brasov, 1987.
  2. Z. Nehari, Conformal Mapping, McGraw-Hill Book, New York, NY, USA, 1952, Dover, 1975. View at Zentralblatt MATH