18_864_frasin.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 3, No. 6, 2010, 1141-1149 ISSN 1307-5543 – www.ejpam.com SPECIAL ISSUE ON COMPLEX ANALYSIS: THEORY AND APPLICATIONS DEDICATED TO PROFESSOR HARI M. SRIVASTAVA, ON THE OCCASION OF HIS 70TH BIRTHDAY Certain Sufficient Conditions for Univalence of Two Integral Operators B.A. Frasin Faculty of Science, Department of Mathematics, Al al-Bayt University, P.O. Box: 130095 Mafraq, Jordan Abstract. In this paper, we obtain new sufficient conditions for two general integral operators to be univalent in the open unit disc. A number of new univalent conditions would follow upon specializing the parameters involved in our main results. 2000 Mathematics Subject Classifications: 30C45 Key Words and Phrases: Analytic and univalent functions, Integral operators 1. Introduction and Definitions LetA denote the class of functions of the form : f (z) = z + ∞ ∑ n=2 anzn which are analytic in the open unit disc U = {z : |z| < 1} . Further, by S we shall denote the class of all functions inA which are univalent in U . In [10] Ozaki and Nunokawa showed that if f ∈A and � � � � � z2 f ′(z) f 2(z) − 1 � � � � � ≤ |z|2 , for all z ∈ U , (1) then the function f is univalent in U . Email address: bafrasin�yahoo. om http://www.ejpam.com 1141 c© 2010 EJPAM All rights reserved. B. Frasin / Eur. J. Pure Appl. Math, 3 (2010), 1141-1149 1142 Making use of the univalence criteria (1), several authors (e.g., see [1, 3, 4, 6, 8, 14, 16, 17]), obtained many sufficient conditions for the univalency of the integral operators Fα1,α2,...,αn,β(z) =    z ∫ 0 β tβ−1 � f1(t) t � 1 α1 . . . � fn(t) t � 1 αn d t    1 β (2) and Gn,β (z) =    [n(β − 1) + 1] z ∫ 0 � f1(t) �β−1 . . . � fn(t) �β−1 d t    1 n(β−1)+1 (3) where the functions f1, f2, . . . , fn belong to the classA and the parameters α1,α2, . . . ,αn and β are complex numbers such that the integrals in (2) and (3) exist. Here and throughout in the sequel every many-valued function is taken with the principal branch. The integral operator in (2) was introduced and studied by Seenivasagan and Breaz [15], and the integral operator in (3) was introduced and studied by Breaz and Breaz [2]. In this paper we are mainly interested on some integral operators of the type (2) and (3). More precisely, we obtain new sufficient conditions for this operators to be univalent in the open unit disc U . In the proofs of our main results we need the following univalence criteria. The first result, i.e. Lemma 1 is a generalization of Ozaki - Nunokawa’s criterion (1) obtained by Raducanu et al. [13], while the second, i.e. Lemma 2 is a generalization of Ahlfors’ and Becker’s univalence criterion [11]. Finally, we need the well-known general Schwarz Lemma. Lemma 1 ([13]). Let f ∈ A and m > 0 such that � � � � � � z2 f ′(z) f 2(z) − 1 � − m− 1 2 |z|m+1 � � � � � ≤ m+ 1 2 |z|m+1 , (4) for all z ∈ U . Then the function f is analytic and univalent in U . Lemma 2 ([11]). Let β ∈ C with Re(β)> 0, c ∈ C with |c| ≤ 1, c 6= −1. If h ∈A satisfies � � � � c |z|2β + (1− |z|2β ) zh′′(z) βh′(z) � � � � ≤ 1, for all z ∈ U , then the integral operator Fβ (z) =    β z ∫ 0 tβ−1h′(t)d t    1 β is analytic and univalent in U . B. Frasin / Eur. J. Pure Appl. Math, 3 (2010), 1141-1149 1143 Lemma 3 ([9]). Let the function f be regular in the disk UR = {z : |z| < R}, with � � f (z) � � < M for fixed M . If f (z) has one zero with multiplicity order bigger than m for z = 0, then � � f (z) � �≤ M Rm |z|m (z ∈ UR). The equality can hold only if f (z) = eiθ (M/Rm) zm where θ is constant. 2. Univalence Conditions for Fα1,α2,...,αn,β(z) We first prove Theorem 1. Let αi ∈ C, Mi ≥ 1, mi > 0 for all i = 1, . . . , n and β ∈ C with Re(β)≥ n ∑ i=1 (mi + 1)Mi + 1 � �αi � � . (5) and let c ∈ C be such that |c| ≤ 1− 1 Re(β) n ∑ i=1 (mi + 1)Mi + 1 � �αi � � . (6) If fi ∈A (i=1, . . . , n) satisfies the inequality (4) and � � fi(z) � �≤ Mi (z ∈ U , i = 1, . . . , n), then the integral operator Fα1 ,α2,...,αn,β(z) defined by (2) is analytic and univalent in U . Proof. Define h(z) = z ∫ 0 n ∏ i=1 � fi(t) t � 1 αi d t we observe that h(0) = h′(0)− 1= 0. On the other hand, it is easy to see that h′(z) = n ∏ i=1 � fi(z) z � 1 αi . (7) Now we differentiate (7) logarithmically and multiply by z on both sides, we obtain zh′′(z) h′(z) = n ∑ i=1 1 αi � z fi ′(z) fi(z) − 1 � . (8) B. Frasin / Eur. J. Pure Appl. Math, 3 (2010), 1141-1149 1144 Since � � fi(z) � � ≤ Mi (z ∈ U , i = 1, . . . , n), then by the general Schwarz Lemma, we obtain � � fi(z) � � ≤ Mi |z| for all z ∈ U and i = 1, . . . , n, we thus from (4) and (8) find that � � � � zh′′(z) h′(z) � � � � ≤ n ∑ i=1 1 � �αi � � �� � � � z fi ′(z) fi(z) � � � � + 1 � = n ∑ i=1 1 � �αi � � � � � � � z2 fi ′(z) [ fi(z)] 2 � � � � � � � � � fi(z) z � � � � + 1 ! ≤ n ∑ i=1 1 � �αi � � � � � � � � z2 fi ′(z) [ fi(z)] 2 − 1 � − mi − 1 2 |z|mi+1 � � � � � Mi + � 1+ mi − 1 2 |z|mi+1 � Mi + 1 ! ≤ n ∑ i=1 1 � �αi � � � mi + 1 2 |z|mi+1 Mi + � 1+ mi − 1 2 |z|mi+1 � Mi + 1 � ≤ n ∑ i=1 (mi + 1)Mi + 1 � �αi � � Therefore, we have � � � � c |z|2β + (1− |z|2β) zh′′(z) βh′(z) � � � � ≤ |c|+ 1 � �β � � � � � � zh′′(z) h′(z) � � � � ≤ |c|+ 1 � �β � � n ∑ i=1 (mi + 1)Mi + 1 � �αi � � ≤ |c|+ 1 Re(β) n ∑ i=1 (mi + 1)Mi + 1 � �αi � � , which, in the light of the hypothesis (6), yields � � � � c |z|2β + (1− |z|2β ) zh′′(z) βh′(z) � � � � ≤ 1. Finally, by applying Lemma 2, we conclude that Fα1 ,α2,...,αn,β(z) ∈ S . Letting m1 = m2 = · · ·= mn = m in Theorem 1, we have Corollary 1. Let αi ∈ C Mi ≥ 1 for all i = 1, . . . , n, m > 0 and β ∈ Cwith Re(β)≥ n ∑ i=1 (m+ 1)Mi + 1 � �αi � � . (9) and let c ∈ C be such that |c| ≤ 1− n ∑ i=1 (m+ 1)Mi + 1 � �αi � � . (10) B. Frasin / Eur. J. Pure Appl. Math, 3 (2010), 1141-1149 1145 If fi ∈A (i = 1, . . . , n) satisfies the inequality (4) and � � fi(z) � �≤ Mi (z ∈ U , i = 1, . . . , n), then the integral operator Fα1 ,α2,...,αn,β(z) defined by (2) is analytic and univalent in U . Remark 1. If we put m= 1 in Corollary 1, we obtain Theorem 2.1 in [5]. Letting α1 = α2 = · · ·= αn = α and M1 = M2 = · · · = Mn = M in Corollary 1, we have Corollary 2. Let α ∈ C, M ≥ 1 , m > 0 and β ∈ C with Re(β)≥ n(m+ 1)M + n |α| . (11) and let c ∈ C be such that |c| ≤ 1− n(m+ 1)M + n |α|Re(β) . (12) If fi ∈A (i = 1, . . . , n) satisfies the inequality (4) and � � fi(z) � � ≤ M (z ∈ U , i = 1, . . . , n), then the integral operator Fα,β (z) =    z ∫ 0 β tβ−1 n ∏ i=1 � fi(t) t � 1 α d t    1 β is analytic and univalent in U . Letting n= 1, α1 = α, M1 = M , m1 = m and f1 = f in Theorem 1, we have Corollary 3. Let α ∈ C, M ≥ 1, m > 0 and β ∈ C with Re(β)≥ (m+ 1)M + 1 |α| . (13) and let c ∈ C be such that |c| ≤ 1− (m+ 1)M + 1 |α|Re(β) . (14) If f ∈ A satisfies the inequality (4) and � � f (z) � � ≤ M (z ∈ U ), then the integral operator Fα,β(z) =    z ∫ 0 β tβ−1 � f (t) t � 1 α d t    1 β defined by (2) is analytic and univalent in U . B. Frasin / Eur. J. Pure Appl. Math, 3 (2010), 1141-1149 1146 3. Univalence Conditions for Gn,β(z) Next, we prove Theorem 2. Let Mi ≥ 1, mi > 0 for all i = 1, . . . , n and β ≥ 1 with � β − 1 β � n ∑ i=1 [(mi + 1)Mi + 1]≤ 1. (15) and let c ∈ C be such that |c| ≤ 1+ � 1− β β � n ∑ i=1 [(mi + 1)Mi + 1]. (16) If fi ∈A (i=1, . . . , n) satisfies the inequality (4) and � � fi(z) � �≤ Mi (z ∈ U , i = 1, . . . , n), then the integral operator Gn,β (z) defined by (2) is analytic and univalent in U . Proof. Setting h(z) = z ∫ 0 n ∏ i=1 � fi(t) t �β−1 d t so that, h′(z) = n ∏ i=1 � fi(z) z �β−1 . (17) and h′′(z) = (β − 1). n ∑ i=1 � fi(z) z �β−2�z fi − fi z2 � . n ∏ k=1(k 6=i) � fk(z) z �β−1 . (18) It follows from (17) and (18) that zh′′(z) h′(z) = n ∑ i=1 (β − 1) � z fi ′(z) fi(z) − 1 � . (19) Since � � fi(z) � � ≤ Mi (z ∈ U , i = 1, . . . , n), then by the general Schwarz Lemma, we obtain � � fi(z) � � ≤ Mi |z| for all z ∈ U and i = 1, . . . , n, we thus from (4) and (19) find that � � � � zh′′(z) h′(z) � � � � ≤ n ∑ i=1 (β − 1) �� � � � z fi ′(z) fi(z) � � � � + 1 � ≤ n ∑ i=1 (β − 1) � � � � � z2 fi ′(z) [ fi(z)] 2 � � � � � � � � � fi(z) z � � � � + 1 ! B. Frasin / Eur. J. Pure Appl. Math, 3 (2010), 1141-1149 1147 ≤ n ∑ i=1 (β − 1) � � � � � � z2 fi ′(z) [ fi(z)] 2 − 1 � − mi − 1 2 |z|mi+1 � � � � � Mi + � 1+ mi − 1 2 |z|mi+1 � Mi + 1 ! ≤ n ∑ i=1 (β − 1) � mi + 1 2 |z|mi+1 Mi + � 1+ mi − 1 2 |z|mi+1 � Mi + 1 � ≤ n ∑ i=1 (β − 1)[(mi + 1)Mi + 1]. Therefore, we have � � � � c |z|2β + (1− |z|2β ) zh′′(z) βh′(z) � � � � ≤ |c|+ 1 β � � � � zh′′(z) h′(z) � � � � ≤ |c|+ � β − 1 β � n ∑ i=1 [(mi + 1)Mi + 1], which, in the light of the hypothesis (16), yields � � � � c |z|2β + (1− |z|2β ) zh′′(z) βh′(z) � � � � ≤ 1. Finally, by applying Lemma 2, we conclude that Gn,β(z) ∈ S . Letting m1 = m2 = . . . = mn = m and M1 = M2 = . . . = Mn = M in Theorem 2, we have Corollary 4. Let M ≥ 1, m > 0 and β ∈ R with β ∈ � 1, ((m+ 1)M + 1)n ((m+ 1)M + 1)n− 1 � . (20) and let c ∈ C be such that |c| ≤ 1+ � 1− β β � ((m+ 1)M + 1)n. 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