1) The document presents theorems for determining the univalence of integral operators of certain forms. Specifically, it obtains conditions for the univalence of operators of the form F(z) and H(z), which involve integrals involving functions g(s) that satisfy certain constraints. 2) It proves Theorem 1, which provides conditions on the parameter λ for which the operator F(z) is univalent, if g(s) satisfies the given constraints. 3) It proves Theorem 2, which provides conditions on the parameters Reλ and Imλ for which the operator H(z) is univalent, if g(s) satisfies the given constraints.

1. Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
The Univalence of Some Integral Operators
Deborah O. Makinde
Department of Mathematics, Obafemi Awolowo University, Ile-Ife, Osun State, Nigeria
E-mail : domakinde.comp@gmail.com; domakinde@oauife.edu.ng
Abstract
In this paper, we obtain the conditions for univalence of the integral operators of
the form:
And 1/ 
 g (s) 
z k
F ( z )    t   i  1/𝜆
 1
ds
𝐻(𝑧) = {𝜆 𝑡 (
 ) 𝜆−1 𝑑𝑠}
0 ∏ 𝑘 i 1 𝑔 𝑖 (𝑠) s
𝑧 𝜆−1
∫0 𝑖=1 𝑠
AMS Mathematics Subject Classification: 30C45.
Key Words: Integral operator, Univalent.
1. Introduction
Let 𝐴 be the class of the functions in the unit disk 𝑈 = {𝑍 ∈ 𝐶: |𝑧| < 1 such that 𝑓(0) = 𝑓 ′ (0) − 1 = 0.
Let 𝑆 be the class of functions 𝑓 ∈ 𝐴 which are univalent in 𝑈. Ozaki and Nunokawa [2] showed the
condition for the univalence of the function 𝑓 ∈ 𝐴 as given in the lemma below.
Lemma 1 [3]: Let 𝑓 ∈ 𝐴 satisfy the condition
𝑧 2 𝑓 ′ (𝑧)
| − 1| ≤ 1 (1)
𝑓 2(𝑧)
For all 𝑧 ∈ 𝑈, then 𝑓 is univalent in 𝑈
Lemma 2 [2]: Let 𝛼 be a complex number, 𝛼 > 0 , and 𝑓 ∈ 𝐴. If
1 − |𝑧|2𝑅𝑒𝛼 𝑧𝑓 ′′ (𝑧)
| ′ |≤1
𝑅𝑒𝛼 𝑓 (𝑧)
𝑧 1⁄
𝛼
For all 𝑧 ∈ 𝑈, then the function 𝐹 𝛼 (𝑧) = [𝛼 ∫ 𝑈 𝛼−1 𝑓 ′ (𝑢)𝑑𝑢]
0
Is in the class 𝑆.
The Schwartz lemma [1] : Let the analytic function 𝑓 be regular in the unit disk and let 𝑓(0) = 0. If
|𝑓(𝑧)| ≤ 1, then
|𝑓(𝑧)| ≤ |𝑧|
For all 𝑧 ∈ 𝑈, where equality can hold only if |𝑓(𝑧)| ≡ 𝐾𝑧 and 𝑘 = 1
Lemma 3 [4]: Let 𝑔 ∈ 𝐴 satisfies (1) and 𝑎 + 𝑏 𝑖 a complex number, 𝑎, 𝑏 satisfies the conditions
3 3 1
𝑎 ∈ [ , ] ; 𝑏 ∈ [0, ] (2)
4 2 2√2
8𝑎2 + 𝑎9 𝑏 2 − 18𝑎 + 9 ≤ 0 (3)
1
𝑧 𝑔(𝑢) 𝑎+𝑏 𝑖 −1 𝑎+𝑏 𝑖
If |𝑔(𝑧) ≤ 1| for all 𝑧 ∈ 𝑈 then the function 𝐺(𝑧) = {𝑎 + 𝑏 𝑖 ∫ (
0
) 𝑑𝑢}
𝑢
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ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.2, No.4, 2012
Is univalent in 𝑈
1. MAIN RESULTS
Theorem 1: Let 𝑔 ∈ 𝐴 satisfies (1) and 𝜆 is a complex number, 𝜆 satisfies the conditions
𝑅𝑒𝜆 ∈ (0, √𝑘 + 2], 𝑘 ≥ 1 (4)
(𝑅𝑒𝜆)4 (𝑅𝑒𝜆)2 (𝐼𝑚𝜆)2 − (𝑘 + 2) ≥ 0 2
(5)
If |𝑔(𝑧)| ≤ 1 for all 𝑧 ∈ 𝑈 then the function
1/ 
 g (s) 
z k
 is 
 1
F ( z)    t ds Is univalent in 𝑈
0 i 1  
Proof: Let
z k 1/ 
g ( s)
 ( is )
 1
F ( z)    t ds
0 i 1
𝑘 𝑧 𝑔 𝑖 (𝑠) 1/𝜆
And let 𝑓(𝑧) = ∫ ∏ 𝑖=1 (
0
) 𝑑𝑠 (6)
𝑠
𝑘 𝑔 𝑖 (𝑧) 1/𝜆
Then 𝑓 ′ (𝑧) = ∏ 𝑖=1 ( )
𝑧
𝑔1 (𝑠) 1/𝜆 𝑔2 (𝑠) 1/𝜆 𝑔 𝑘 (𝑠) 1/𝜆
=( ) ( ) …( )
𝑠 𝑠 𝑠
1 1 ′ 1 1
𝑔1 (𝑠) 1/𝜆 𝑔2 (𝑠) 𝜆 𝑔 𝑘 (𝑠) 𝜆 𝑔1 (𝑠) 1/𝜆 𝑔2 (𝑠) 𝜆 𝑔 (𝑠) 𝜆
𝑓 ′′ (𝑧) = ( ) [( ) …( ) ] + [( ) ] ′ [( ) …( 𝑘 ) ]
𝑠 𝑠 𝑠 𝑠 𝑠 𝑠
𝑧𝑓 ′′ (𝑧) 1 𝑧𝑔′ 1 (𝑧) 1 𝑧𝑔′ 2 (𝑧) 1 𝑧𝑔′ 𝑘 (𝑧)
= ( − 1) + ( − 1) … ( − 1)
𝑓 ′ (𝑧) 𝜆 𝑔1 (𝑧) 𝜆 𝑔1 (𝑧) 𝜆 𝑔1 (𝑧)
1
𝑘 𝑧𝑔′ 𝑖 (𝑧)
= ∑ 𝑖=1 ( − 1)
𝜆 𝑔1 (𝑧)
This implies that
1−|𝑧|2𝑅𝑒𝜆 𝑧𝑓 ′′ (𝑧) 1−|𝑧|2𝑅𝑒𝜆 1 𝑘 𝑧𝑔′ 𝑖 (𝑧)
| |= |∑ 𝑖=1 − 1|
𝑅𝑒𝜆 𝑓 ′ (𝑧) 𝑅𝑒𝜆 |𝜆| 𝑔1 (𝑧)
1−|𝑧|2𝑅𝑒𝜆 1 𝑘 𝑧𝑔′ 𝑖 (𝑧)
≤ (|∑ 𝑖=1 | + 1)
𝑅𝑒𝜆 |𝜆| 𝑔1 (𝑧)
1−|𝑧|2𝑅𝑒𝜆 1 𝑘 𝑧 2 𝑔′ 𝑖 (𝑧) 𝑔 (𝑧)
≤ (|∑ 𝑖=1 || 𝑖 | + 1) (7)
𝑅𝑒𝜆 |𝜆| 𝑔2 1 (𝑧) 𝑧
Using Schwartz-lemma in (7), we have
1−|𝑧|2𝑅𝑒𝜆 𝑧𝑓 ′′ (𝑧) 1−|𝑧|2𝑅𝑒𝜆 1 𝑘 𝑧 2 𝑔′ 𝑖 (𝑧)
| |≤ (|∑ 𝑖=1 − 1| + 2)
𝑅𝑒𝜆 𝑓 ′ (𝑧) 𝑅𝑒𝜆 |𝜆| 𝑔2 1 (𝑧)
But 𝑔 satisfies (1), thus
𝑘 𝑧 2 𝑔′ 𝑖 (𝑧)
|∑ 𝑖=1 − 1| ≤ 𝑘
𝑔2 1 (𝑧)
92

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Vol.2, No.4, 2012
This implies that
𝑧𝑓 ′′ (𝑧) 1−|𝑧|2𝑅𝑒𝜆 1 𝑘+2
| |≤ (𝑘 + 2) ≤ (8)
𝑓 ′ (𝑧) 𝑅𝑒𝜆 |𝜆| 𝑅𝑒𝜆|𝜆|
From (4) and (5), we have
𝑘+2
≤1 (9)
𝑅𝑒𝜆|𝜆|
1−|𝑧|2𝑅𝑒𝜆 𝑧𝑓 ′′ (𝑧)
Using (9) in (8), we obtain | |≤1
𝑅𝑒𝜆 𝑓 ′ (𝑧)
𝑘 𝑔 𝑘 (𝑠) 1/𝜆
And from (6) 𝑓 ′ (𝑧) = ∏ 𝑖=1 ( )
𝑠
And by lemma 2 𝐹(𝑧) is univalent ∎
Theorem 2 Let 𝑔 ∈ 𝐴 satisfies (1) and let 𝜆 be a complex number with 𝑅𝑒𝜆, 𝐼𝑚𝜆 satisfying (2)
and (3). If |𝑔(𝑧)| ≤ 1 for all 𝑧 ∈ 𝑈 then the function
𝑘 1/𝜆
𝑧
𝜆−1
𝑔 𝑖 (𝑠) 𝜆−1
𝐻(𝑧) = {𝜆 ∫ 𝑡 ∏( ) }
0 𝑠
𝑖=1
Is univalent in 𝑈.
Proof: Let
𝑧 𝑔 𝑖 (𝑠) 𝜆−1 1/𝜆
𝑘
𝐻(𝑧) = {𝜆 ∫ 𝑡 𝜆−1 ∏ 𝑖=1 (
0
) } (10)
𝑠
Following the procedure of the proof of theorem 1, we obtain
𝑘
1 − |𝑧|2𝑅𝑒𝜆 𝑧ℎ′′ (𝑧) 1 − |𝑧|2𝑅𝑒𝜆 𝑧 2 𝑔′ (𝑧)
| ′ |≤ |𝜆 − 1| (|∑ 2 𝑖 − 1| + 2)
𝑅𝑒𝜆 ℎ (𝑧) 𝑅𝑒𝜆 𝑔 1 (𝑧)
𝑖=1
But 𝑔 satisfies (1), thus
1 − |𝑧|2𝑅𝑒𝜆 𝑧ℎ′′ (𝑧) 1 − |𝑧|2𝑅𝑒𝜆 |𝜆 − 1|(𝑘 + 2)
| ′ |≤ |𝜆 − 1|(𝑘 + 2) ≤
𝑅𝑒𝜆 ℎ (𝑧) 𝑅𝑒𝜆 𝑅𝑒𝜆
From (2) and (3) we have
|𝜆−1|(𝑘+2)
≤1
𝑅𝑒𝜆
This implies that
1−|𝑧|2𝑅𝑒𝜆 𝑧ℎ′′ (𝑧)
| | ≤ 1 for all 𝑧 ∈ 𝑈
𝑅𝑒𝜆 ℎ′ (𝑧)
𝑔(𝑧) 𝜆−1
𝑘
From (10), ℎ′ (𝑧) = ∏ 𝑖=1 ( )
𝑧
And by lemma 2, 𝐻(𝑧) is univalent in 𝑈 ∎
Remark: Theorem 1and theorem 2 gives a generalization of theorem 1[4] and lemma 3 respectively,.
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4. Mathematical Theory and Modeling www.iiste.org
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References
Mayer O, (1981). The function theory of one complex variable Bucuresti
Oaaki, S., & Nunokawa, M. (1972). The Schwarzian derivative and univalent function.Proc. Amer. Math.
Soc. 33(2), 392-394
Pescar, V., (2006). Univalence of certain integral operators. . Acta Universitatis Apulensis, 12, 43-48
Pescar, V., & Breaz, D. (2008). Some integral and their univalence. Acta Universitatis Apulensis, 15,
147-152
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