11.the univalence of some integral operators

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11.the univalence of some integral operators

  1. 1. Mathematical Theory and Modeling www.iiste.orgISSN 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.ngAbstractIn this paper, we obtain the conditions for univalence of the integral operators ofthe form:And 1/   g (s)  z k F ( z )    t   i  1/𝜆  1 ds𝐻(𝑧) = {𝜆 𝑡 (  ) 𝜆−1 𝑑𝑠} 0 ∏ 𝑘 i 1 𝑔 𝑖 (𝑠) s 𝑧 𝜆−1 ∫0 𝑖=1 𝑠AMS Mathematics Subject Classification: 30C45.Key Words: Integral operator, Univalent. 1. IntroductionLet 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 thecondition 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 𝑈Lemma 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⁄ 𝛼For all 𝑧 ∈ 𝑈, then the function 𝐹 𝛼 (𝑧) = [𝛼 ∫ 𝑈 𝛼−1 𝑓 ′ (𝑢)𝑑𝑢] 0Is 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 = 1Lemma 3 [4]: Let 𝑔 ∈ 𝐴 satisfies (1) and 𝑎 + 𝑏 𝑖 a complex number, 𝑎, 𝑏 satisfies the conditions 3 3 1𝑎 ∈ [ , ] ; 𝑏 ∈ [0, ] (2) 4 2 2√28𝑎2 + 𝑎9 𝑏 2 − 18𝑎 + 9 ≤ 0 (3) 1 𝑧 𝑔(𝑢) 𝑎+𝑏 𝑖 −1 𝑎+𝑏 𝑖If |𝑔(𝑧) ≤ 1| for all 𝑧 ∈ 𝑈 then the function 𝐺(𝑧) = {𝑎 + 𝑏 𝑖 ∫ ( 0 ) 𝑑𝑢} 𝑢 91
  2. 2. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012Is univalent in 𝑈 1. MAIN RESULTSTheorem 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/𝜆And let 𝑓(𝑧) = ∫ ∏ 𝑖=1 ( 0 ) 𝑑𝑠 (6) 𝑠 𝑘 𝑔 𝑖 (𝑧) 1/𝜆Then 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 that1−|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) 𝑧Using 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
  3. 3. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012This 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/𝜆And from (6) 𝑓 ′ (𝑧) = ∏ 𝑖=1 ( ) 𝑠And 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=1Is univalent in 𝑈.Proof: Let 𝑧 𝑔 𝑖 (𝑠) 𝜆−1 1/𝜆 𝑘 𝐻(𝑧) = {𝜆 ∫ 𝑡 𝜆−1 ∏ 𝑖=1 ( 0 ) } (10) 𝑠Following 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=1But 𝑔 satisfies (1), thus 1 − |𝑧|2𝑅𝑒𝜆 𝑧ℎ′′ (𝑧) 1 − |𝑧|2𝑅𝑒𝜆 |𝜆 − 1|(𝑘 + 2) | ′ |≤ |𝜆 − 1|(𝑘 + 2) ≤ 𝑅𝑒𝜆 ℎ (𝑧) 𝑅𝑒𝜆 𝑅𝑒𝜆From (2) and (3) we have|𝜆−1|(𝑘+2) ≤1 𝑅𝑒𝜆This implies that1−|𝑧|2𝑅𝑒𝜆 𝑧ℎ′′ (𝑧) | | ≤ 1 for all 𝑧 ∈ 𝑈 𝑅𝑒𝜆 ℎ′ (𝑧) 𝑔(𝑧) 𝜆−1 𝑘From (10), ℎ′ (𝑧) = ∏ 𝑖=1 ( ) 𝑧And 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,. 93
  4. 4. Mathematical Theory and Modeling www.iiste.orgISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)Vol.2, No.4, 2012ReferencesMayer O, (1981). The function theory of one complex variable BucurestiOaaki, S., & Nunokawa, M. (1972). The Schwarzian derivative and univalent function.Proc. Amer. Math. Soc. 33(2), 392-394Pescar, V., (2006). Univalence of certain integral operators. . Acta Universitatis Apulensis, 12, 43-48Pescar, V., & Breaz, D. (2008). Some integral and their univalence. Acta Universitatis Apulensis, 15, 147-152 94
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