SOME SUBORDINATION CRITERIA CONCERNING THE SǍLǍGEAN OPERATOR Subordination Criteria KAZUO KUROKI AND SHIGEYOSHI OWA Department of Mathematics Kinki University Higashi-Osaka, Osaka 577-8502, Japan EMail: freedom@sakai.zaq.ne.jp owa@math.kindai.ac.jp Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page Contents Received: 14 September, 2008 Accepted: 11 January, 2009 Communicated by: H.M. Srivastava 2000 AMS Sub. Class.: 26D15. Key words: Janowski function, Starlike, Convex, Univalent, Subordination. Abstract: Applying Sǎlǎgean operator, for the class A of analytic functions f (z) in the open unit disk U which are normalized by f (0) = f 0 (0) − 1 = 0, the generalization of an analytic function to discuss the starlikeness is considered. Furthermore, from the subordination criteria for Janowski functions generalized by some complex parameters, some interesting subordination criteria for f (z) ∈ A are given. JJ II J I Page 1 of 23 Go Back Full Screen Close Contents 1 Introduction, Definition and Preliminaries 3 2 Subordinations for the Class Defined by the Sǎlǎgean Operator 11 3 Subordination Criteria for Other Analytic Functions 17 Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page Contents JJ II J I Page 2 of 23 Go Back Full Screen Close 1. Introduction, Definition and Preliminaries Let A denote the class of functions f (z) of the form: (1.1) f (z) = z + ∞ X an z n n=2 Subordination Criteria which are analytic in the open unit disk Kazuo Kuroki and Shigeyoshi Owa U = {z : z ∈ C |z| < 1}. and vol. 10, iss. 2, art. 36, 2009 Also, let P denote the class of functions p(z) of the form: (1.2) p(z) = 1 + ∞ X Title Page pn z n Contents n=1 which are analytic in U. If p(z) ∈ P satisfies Re p(z) > 0 (z ∈ U), then we say that p(z) is the Carathéodory function (cf. [1]). By the familiar principle of differential subordination between analytic functions f (z) and g(z) in U, we say that f (z) is subordinate to g(z) in U if there exists an analytic function w(z) satisfying the following conditions: w(0) = 0 and |w(z)| < 1 (z ∈ U), JJ II J I Page 3 of 23 Go Back Full Screen Close such that f (z) = g w(z) (z ∈ U). We denote this subordination by f (z) ≺ g(z) (z ∈ U). In particular, if g(z) is univalent in U, then it is known that f (z) ≺ g(z) (z ∈ U) ⇐⇒ f (0) = g(0) and f (U) ⊂ g(U). For the function p(z) ∈ P, we introduce the following function p(z) = (1.3) 1 + Az 1 + Bz (−1 5 B < A 5 1) Subordination Criteria which has been investigated by Janowski [3]. Thus, the function p(z) given by (1.3) is said to be the Janowski function. And, as a generalization of the Janowski function, Kuroki, Owa and Srivastava [2] have discussed the function p(z) = 1 + Az 1 + Bz (ii) |A| 5 1, |B| = 1, A 6= B, and 1 − AB > 0. Here, for some complex numbers A and B which satisfy condition (i), the function p(z) is analytic and univalent in U and p(z) maps the open unit disk U onto the open disk given by 1 − AB |A − B| (1.4) p(z) − 1 − |B|2 < 1 − |B|2 . Thus, it is clear that (1.5) Re(1 − AB) − |A − B| Re p(z) > =0 1 − |B|2 vol. 10, iss. 2, art. 36, 2009 Title Page for some complex parameters A and B which satisfy one of following conditions (i) |A| 5 1, |B| < 1, A 6= B, and Re 1 − AB = |A − B| Kazuo Kuroki and Shigeyoshi Owa (z ∈ U). Contents JJ II J I Page 4 of 23 Go Back Full Screen Close Also, for some complex numbers A and B which satisfy condition (ii), the function p(z) is analytic and univalent in U and the domain p(U) is the right half-plane satisfying (1.6) 1 − |A|2 Re p(z) > = 0. 2(1 − AB) Hence, we see that the generalized Janowski function maps the open unit disk U onto some domain which is on the right half-plane. Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Remark 1. For the function 1 + Az 1 + Bz defined with the condition (i), the inequalities (1.4) and (1.5) give us that p(z) = Title Page Contents p(z) 6= 0 namely, 1 + Az 6= 0 (z ∈ U). Since, after a simple calculation, we see the condition |A| 5 1, we can omit the condition |A| 5 1 in (i). Hence, the condition (i) is newly defined by the following conditions (1.7) |B| < 1, A 6= B, and Re 1 − AB = |A − B|. A function f (z) ∈ A is said to be starlike of order α in U if it satisfies 0 zf (z) (1.8) Re >α (z ∈ U) f (z) for some α (0 5 α < 1). We denote by S ∗ (α) the subclass of A consisting of all functions f (z) which are starlike of order α in U. JJ II J I Page 5 of 23 Go Back Full Screen Close Similarly, if f (z) ∈ A satisfies the following inequality zf 00 (z) >α (z ∈ U) (1.9) Re 1 + 0 f (z) for some α (0 5 α < 1), then f (z) is said to be convex of order α in U. We denote by K(α) the subclass of A consisting of all functions f (z) which are convex of order α in U. As usual, in the present investigation, we write ∗ S (0) ≡ S ∗ and K(0) ≡ K. D0 f (z) = f (z) = z + (1.10) an z n , n=2 (1.11) 1 0 D f (z) = Df (z) = zf (z) = z + ∞ X Title Page Contents JJ II J I Page 6 of 23 nan z n , n=2 (1.12) Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 The classes S ∗ (α) and K(α) were introduced by Robertson [7]. We define the following differential operator due to Sǎlǎgean [8]. For a function f (z) and j = 1, 2, 3, . . . , ∞ X Subordination Criteria ∞ X j j−1 D f (z) = D D f (z) = z + nj an z n . n=2 Also, we consider the following differential operator Z z ∞ X f (ζ) −1 dζ = z + n−1 an z n , (1.13) D f (z) = ζ 0 n=2 Go Back Full Screen Close (1.14) ∞ X D−j f (z) = D−1 D−(j−1) f (z) = z + n−j an z n n=2 for any negative integers. Then, for f (z) ∈ A given by (1.1), we know that (1.15) Dj f (z) = z + ∞ X Subordination Criteria n j an z n (j = 0, ±1, ±2, . . . ). Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 n=2 We consider the subclass Sjk (α) as follows: k D f (z) k >α Sj (α) = f (z) ∈ A : Re Dj f (z) Title Page (z ∈ U ; 0 5 α < 1) . Sjj+1 (α) In particular, putting k = j + 1, we also define j+1 D f (z) j+1 Sj (α) = f (z) ∈ A : Re >α Dj f (z) by (z ∈ U ; 0 5 α < 1) . II J I Page 7 of 23 Full Screen 0 z zf 0 (z) D2 f (z) zf 00 (z) = = 1 + , D1 f (z) zf 0 (z) f 0 (z) we see that S01 (α) ≡ S ∗ (α), JJ Go Back Remark 2. Noting D1 f (z) zf 0 (z) = , D0 f (z) f (z) Contents S12 (α) ≡ K(α) (0 5 α < 1). Close Furthermore, by applying subordination, we consider the following subclass Dk f (z) 1 + Az k Pj (A, B) = f (z) ∈ A : j ≺ (z ∈ U ; A 6= B, |B| 5 1) . D f (z) 1 + Bz In particular, putting k = j + 1, we also define Dj+1 f (z) 1 + Az j+1 Pj (A, B) = f (z) ∈ A : ≺ Dj f (z) 1 + Bz (z ∈ U; A 6= B, |B| 5 1) . Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa Remark 3. Noting vol. 10, iss. 2, art. 36, 2009 Dk f (z) 1 + (1 − 2α)z ≺ Dj f (z) 1−z ⇐⇒ Re Dk f (z) Dj f (z) >α (z ∈ U; 0 5 α < 1), Title Page Contents we see that P01 (1 − 2α, −1) ≡ S ∗ (α), P12 (1 − 2α, −1) ≡ K(α) (0 5 α < 1). In our investigation here, we need the following lemma concerning the differential subordination given by Miller and Mocanu [5] (see also [6, p. 132]). Lemma 1.1. Let the function q(z) be analytic and univalent in U. Also let φ(ω) and ψ(ω) be analytic in a domain C containing q(U), with ψ(ω) 6= 0 ω ∈ q(U) ⊂ C . Set Q(z) = zq 0 (z)ψ q(z) and h(z) = φ q(z) + Q(z), and suppose that (i) Q(z) is starlike and univalent in U; JJ II J I Page 8 of 23 Go Back Full Screen Close and (ii) Re zh0 (z) Q(z) = Re ! φ0 q(z) zQ0 (z) + >0 Q(z) ψ q(z) (z ∈ U). If p(z) is analytic in U, with p(0) = q(0) and p(U) ⊂ C, Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa and vol. 10, iss. 2, art. 36, 2009 φ p(z) + zp0 (z)ψ p(z) ≺ φ q(z) + zq 0 (z)ψ q(z) =: h(z) (z ∈ U), Title Page then p(z) ≺ q(z) (z ∈ U) Contents and q(z) is the best dominant of this subordination. JJ II By making use of Lemma 1.1, Kuroki, Owa and Srivastava [2] have investigated some subordination criteria for the generalized Janowski functions and deduced the following lemma. J I Lemma 1.2. Let the function f (z) ∈ A be chosen so that f (z) 6= 0 (z ∈ U). z Also, let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of following conditions: (i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be such that β(1 − α) (1 + β) Re 1 − AB − |A − B| 1−β 1+β + + + −1 = 0, 2 α 1 − |B| 1 + |A| 1 + |B| Page 9 of 23 Go Back Full Screen Close (ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be such that β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|) + + = 0. α 2(1 + |A|) 2(1 − AB) If (1.16) zf 0 (z) f (z) β zf 00 (z) 1+α 0 f (z) ≺ h(z) (z ∈ U), Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 where h(z) = Subordination Criteria 1 + Az 1 + Bz β−1 1 + Az α(1 + Az)2 + α(A − B)z (1 − α) + , 1 + Bz (1 + Bz)2 Title Page Contents then zf 0 (z) 1 + Az ≺ f (z) 1 + Bz (z ∈ U). JJ II J I Page 10 of 23 Go Back Full Screen Close 2. Subordinations for the Class Defined by the Sǎlǎgean Operator First of all, by applying the Sǎlǎgean operator for f (z) ∈ A, we consider the following subordination criterion in the class Pjk (A, B) for some complex parameters A and B. j Theorem 2.1. Let the function f (z) ∈ A be chosen so that D fz (z) 6= 0 (z ∈ U). Also, let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of following conditions: (i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that 1−β 1+β β(1 − α) (1 + β) Re 1 − AB − |A − B| + + −1 = 0, + 1 − |B|2 1 + |A| 1 + |B| α (ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|) + + = 0. α 2(1 + |A|) 2(1 − AB) If Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page Contents JJ II J I Page 11 of 23 Go Back Full Screen k β k D f (z) D f (z) Dk+1 f (z) Dj+1 f (z) (2.1) (1−α)+α + k − j ≺ h(z), Dj f (z) Dj f (z) D f (z) D f (z) where h(z) = 1 + Az 1 + Bz β−1 1 + Az α(1 + Az)2 + α(A − B)z (1 − α) + , 1 + Bz (1 + Bz)2 Close then Dk f (z) 1 + Az ≺ Dj f (z) 1 + Bz (z ∈ U). Proof. If we define the function p(z) by Dk f (z) p(z) = j D f (z) (z ∈ U), Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa then p(z) is analytic in U with p(0) = 1. Further, since k k+1 D f (z) Dj+1 f (z) D f (z) 0 zp (z) = − , Dj f (z) Dk f (z) Dj f (z) the condition (2.1) can be written as follows: β β−1 p(z) (1 − α) + αp(z) + αzp0 (z) p(z) ≺ h(z) vol. 10, iss. 2, art. 36, 2009 Title Page Contents (z ∈ U). We also set q(z) = 1 + Az , 1 + Bz φ(z) = z β (1 − α + αz), and ψ(z) = αz β−1 for z ∈ U. Then, it is clear that the function q(z) is analytic and univalent in U and has a positive real part in U for the conditions (i) and (ii). Therefore, φ and ψ are analytic in a domain C containing q(U), with ψ(ω) 6= 0 ω ∈ q(U) ⊂ C . Also, for the function Q(z) given by α(A − B)z(1 + Az)β−1 , Q(z) = zq 0 (z)ψ q(z) = (1 + Bz)β+1 JJ II J I Page 12 of 23 Go Back Full Screen Close we obtain 1−β 1+β zQ0 (z) = + − 1. Q(z) 1 + Az 1 + Bz (2.2) Furthermore, we have h(z) = φ q(z) + Q(z) β 1 + Az 1 + Az α(A − B)z(1 + Az)β−1 = 1−α+α + 1 + Bz 1 + Bz (1 + Bz)β+1 Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 and zh0 (z) β(1 − α) zQ0 (z) = + (1 + β)q(z) + . Q(z) α Q(z) (2.3) Title Page Contents Hence, (i) For the complex numbers A and B such that JJ II |B| < 1, A 6= B, J I and Re(1 − AB) = |A − B|, Page 13 of 23 it follows from (2.2) and (2.3) that 0 zQ (z) 1−β 1+β Re > + − 1 = 0, Q(z) 1 + |A| 1 + |B| Go Back Full Screen and Close Re zh0 (z) Q(z) β(1 − α) (1 + β) Re 1 − AB − |A − B| > + α 1 − |B|2 1+β 1−β + + −1=0 1 + |A| 1 + |B| (z ∈ U). (ii) For the complex numbers A and B such that |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0, from (2.2) and (2.3), we get 0 zQ (z) 1−β 1 (1 − β)(1 − |A|) Re > + (1 + β) − 1 = = 0, Q(z) 1 + |A| 2 2(1 + |A|) Subordination Criteria and 0 zh (z) β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|) Re > + + =0 Q(z) α 2(1 + |A|) 2(1 − AB) Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 (z ∈ U). Since all the conditions of Lemma 1.1 are satisfied, we conclude that Dk f (z) 1 + Az ≺ j D f (z) 1 + Bz Title Page Contents (z ∈ U), which completes the proof of Theorem 2.1. Remark 4. We know that a function f (z) satisfying the conditions in Theorem 2.1 belongs to the class Pjk (A, B). Letting k = j + 1 in Theorem 2.1, we obtain the following theorem. j Theorem 2.2. Let the function f (z) ∈ A be chosen so that D fz (z) 6= 0 (z ∈ U). Also, let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of following conditions (i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that 1−β 1+β β(1 − α) (1 + β) Re 1 − AB − |A − B| + + − 1 = 0, + 2 1 − |B| 1 + |A| 1 + |B| α JJ II J I Page 14 of 23 Go Back Full Screen Close (ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|) + = 0. + α 2(1 + |A|) 2(1 − AB) If (2.4) Dj+1 f (z) Dj f (z) β Dj+2 f (z) 1 − α + α j+1 D f (z) ≺ h(z), where Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 h(z) = then Subordination Criteria β−1 1 + Az α(1 + Az)2 + α(A − B)z 1 + Az (1 − α) + , 1 + Bz 1 + Bz (1 + Bz)2 Dj+1 f (z) 1 + Az ≺ j D f (z) 1 + Bz Contents (z ∈ U). Remark 5. A function f (z) satisfying the conditions in Theorem 2.2 belongs to the class Pjj+1 (A, B). Setting j = 0 in Theorem 2.2, we obtain Lemma 1.2 proven by Kuroki, Owa and Srivastava [2]. Also, if we assume that 1 − µ iθ α = 1, β = A = 0, and B = e (0 5 µ < 1, 0 5 θ < 2π), 1+µ Theorem 2.2 becomes the following corollary. j Corollary 2.3. If f (z) ∈ A D fz (z) 6= 0 in U satisfies Dj+2 f (z) 1 + µ − (1 − µ)eiθ z ≺ Dj+1 f (z) 1 + µ + (1 − µ)eiθ z Title Page (z ∈ U; 0 5 θ < 2π) JJ II J I Page 15 of 23 Go Back Full Screen Close for some µ (0 5 µ < 1), then Dj+1 f (z) 1+µ ≺ j D f (z) 1 + µ + (1 − µ)eiθ z (z ∈ U). From the above corollary, we have j+2 j+1 D f (z) D f (z) 1+µ > µ =⇒ Re > Re j+1 j D f (z) D f (z) 2 In particular, making j = 0, we get 0 zf 00 (z) zf (z) 1+µ > µ =⇒ Re > Re 1 + 0 f (z) f (z) 2 (z ∈ U; 0 5 µ < 1). Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 (z ∈ U; 0 5 µ < 1), Title Page Contents namely f (z) ∈ K(µ) =⇒ f (z) ∈ S ∗ 1+µ 2 (z ∈ U; 0 5 µ < 1). And, taking µ = 0, we find that every convex function is starlike of order 21 . This fact is well-known as the Marx-Strohhäcker theorem in Univalent Function Theory (cf. [4, 9]). JJ II J I Page 16 of 23 Go Back Full Screen Close 3. Subordination Criteria for Other Analytic Functions In this section, by making use of Lemma 1.1, we consider some subordination critej ria concerning the analytic function D fz (z) for f (z) ∈ A. Theorem 3.1. Let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B which satisfy one of following conditions Subordination Criteria (i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that β 1−β 1+β + + − 1 = 0, α 1 + |A| 1 + |B| Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page (ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that β (1 − β)(1 − |A|) + = 0. α 2(1 + |A|) If f (z) ∈ A satisfies β j Dj+1 f (z) D f (z) 1−α+α j ≺ h(z), (3.1) z D f (z) Contents JJ II J I Page 17 of 23 Go Back Full Screen where h(z) = then 1 + Az 1 + Bz β + α(A − B)z(1 + Az)β−1 , (1 + Bz)β+1 Dj f (z) 1 + Az ≺ z 1 + Bz (z ∈ U). Close Proof. If we define the function p(z) by p(z) = Dj f (z) z (z ∈ U), then p(z) is analytic in U with p(0) = 1 and the condition (3.1) can be written as follows: β β−1 p(z) + αzp0 (z) p(z) ≺ h(z) (z ∈ U). We also set Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa 1 + Az q(z) = , 1 + Bz vol. 10, iss. 2, art. 36, 2009 β φ(z) = z , and ψ(z) = αz β−1 for z ∈ U. Then, the function q(z) is analytic and univalent in U and satisfies Re q(z) > 0 (z ∈ U) for the condition (i) and (ii). Thus, the functions φ and ψ satisfy the conditions required by Lemma 1.1. Further, for the functions Q(z) and h(z) given by Q(z) = zq 0 (z)ψ q(z) and h(z) = φ q(z) + Q(z), we have Title Page Contents JJ II J I Page 18 of 23 Go Back Full Screen 1−β 1+β zQ0 (z) = + − 1 and Q(z) 1 + Az 1 + Bz zh0 (z) β zQ0 (z) = + . Q(z) α Q(z) Then, similarly to the proof of Theorem 2.1, we see that 0 0 zQ (z) zh (z) Re > 0 and Re >0 Q(z) Q(z) (z ∈ U) Close for the conditions (i) and (ii). Thus, by applying Lemma 1.1, we conclude that p(z) ≺ q(z) The proof of the theorem is completed. (z ∈ U). Letting j = 0 in Theorem 3.1, we obtain the following theorem. Theorem 3.2. Let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of following conditions: (i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that β 1−β 1+β + + − 1 = 0, α 1 + |A| 1 + |B| Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page (ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that β (1 − β)(1 − |A|) + = 0. α 2(1 + |A|) If f (z) ∈ A satisfies (3.2) then f (z) z β−1 Contents JJ II J I Page 19 of 23 f (z) 0 (1 − α) + αf (z) z β 1 + Az α(A − B)z(1 + Az)β−1 ≺ + , 1 + Bz (1 + Bz)β+1 f (z) 1 + Az ≺ z 1 + Bz (z ∈ U). Go Back Full Screen Close Also, taking α = 1, β = A = 0, 1 − ν iθ e and B = ν 1 5 ν < 1, 0 5 θ < 2π 2 in Theorem 3.2, we have Corollary 3.3. If f (z) ∈ A satisfies for some ν 1 2 zf 0 (z) ν ≺ f (z) ν + (1 − ν)eiθ z 5 ν < 1 , then Subordination Criteria (z ∈ U; 0 5 θ < 2π) f (z) ν ≺ z ν + (1 − ν)eiθ z Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page (z ∈ U). Contents Further, making α = β = 1, A = 0, and B = 1 − ν iθ e ν 1 5 ν < 1, 0 5 θ < 2π 2 in Theorem 3.2, we get Corollary 3.4. If f (z) ∈ A satisfies 2 ν 0 f (z) ≺ ν + (1 − ν)eiθ z for some ν 21 5 ν < 1 , then ν f (z) ≺ z ν + (1 − ν)eiθ z JJ II J I Page 20 of 23 Go Back Full Screen (z ∈ U ; 0 5 θ < 2π) (z ∈ U). Close The above corollaries give: 0 zf (z) (3.3) Re > ν =⇒ f (z) Re f (z) z >ν 1 z ∈ U; 5 ν < 1 , 2 and (3.4) Re p f 0 (z) >ν =⇒ Re f (z) z >ν 1 z ∈ U; 5 ν < 1 . 2 1 , 2 Here, taking ν = we find some results that are known as the Marx-Strohhäcker theorem in Univalent Function Theory (cf. [4], [9]). Setting j = 1 in Theorem 3.1, we obtain the following theorem. Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page Theorem 3.5. Let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of following conditions (i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that 1−β 1+β β + + − 1 = 0, α 1 + |A| 1 + |B| Contents JJ II J I Page 21 of 23 Go Back (ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that β (1 − β)(1 − |A|) + = 0. α 2(1 + |A|) If f (z) ∈ A satisfies β β 1 + Az α(A − B)z(1 + Az)β−1 zf 00 (z) 0 (3.5) f (z) 1+α 0 ≺ + , f (z) 1 + Bz (1 + Bz)β+1 Full Screen Close then f 0 (z) ≺ 1 + Az 1 + Bz (z ∈ U). Here, making α = 1, β = A = 0, 1 − ν iθ and B = e ν 1 5 ν < 1, 0 5 θ < 2π 2 in Theorem 3.5, we have: Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Corollary 3.6. If f (z) ∈ A satisfies zf 00 (z) ν 1+ 0 ≺ f (z) ν + (1 − ν)eiθ z for some ν 12 5 ν < 1 , then f 0 (z) ≺ ν ν + (1 − ν)eiθ z (z ∈ U ; 0 5 θ < 2π) Title Page Contents (z ∈ U). JJ II J I Page 22 of 23 Go Back Also, from Corollary 3.6 we have: zf 00 (z) (3.6) Re 1 + 0 > ν =⇒ Re (f 0 (z)) > ν f (z) Full Screen 1 z ∈ U; 5 ν < 1 . 2 Close References [1] P.L. DUREM, Univalent Functions, Springer-Verlag, New York, Berlin, Heidelberg, Tokyo, 1983. [2] K. KUROKI, S. OWA AND H.M. SRIVASTAVA, Some subordination criteria for analytic functions, Bull. Soc. Sci. Lett. Lodz, Vol., 52 (2007), 27–36. [3] W. JANOWSKI, Extremal problem for a family of functions with positive real part and for some related families, Ann. Polon. Math., 23 (1970), 159–177. [4] A. MARX, Untersuchungen uber schlichte Abbildungen, Math. Ann., 107 (1932/33), 40–67. Subordination Criteria Kazuo Kuroki and Shigeyoshi Owa vol. 10, iss. 2, art. 36, 2009 Title Page [5] S.S. MILLER AND P.T. MOCANU, On some classes of first-order differential subordinations, Michigan Math. J., 32 (1985), 185–195. [6] S.S. MILLER AND P.T. MOCANU, Differential Subordinations, Pure and Applied Mathematics 225, Marcel Dekker, 2000. [7] M.S. ROBERTSON, On the theory of univalent functions, Ann. Math., 37 (1936), 374–408. [8] G.S. SǍLǍGEAN, Subclass of univalent functions, Complex Analysis-Fifth Romanian-Finnish Seminar, Part 1(Bucharest, 1981), Lecture Notes in Math., vol. 1013, Springer, Berlin, 1983, pp. 362–372. [9] E. STROHHÄCKER, Beitrage zur Theorie der schlichten Funktionen, Math. Z., 37 (1933), 356–380. Contents JJ II J I Page 23 of 23 Go Back Full Screen Close