DIFFERENTIAL SUBORDINATION RESULTS FOR NEW CLASSES OF THE FAMILY E(Φ, Ψ)

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1 DIFFERENTIAL SUBORDINATION RESULTS FOR NEW CLASSES OF THE FAMILY E(Φ, Ψ) Received: 05 July, 2008 RABHA W. IBRAHIM AND MASLINA DARUS School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 43600, Selangor Darul Ehsan Malaysia Accepted: 02 December, 2008 Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 34G10, 26A33, 30C45. Page 1 of 18 Key words: Fractional calculus; Subordination; Hadamard product. Abstract: We define new classes of the family E(Φ, Ψ), in a unit disk U := {z C, z < 1}, as follows: for analytic functions F (z), Φ(z) and Ψ(z) so that R{ F (z) Φ(z) } > 0, z U, F (z) Ψ(z) 0 where the operator denotes the F (z) Ψ(z) convolution or Hadamard product. Moreover, we establish some subordination results for these new classes. Acknowledgement: The work presented here was supported by SAGA: STGL , Academy of Sciences, Malaysia.

2 1 Introduction and preliminaries. 3 2 Main Results 7 3 Applications 14 Page 2 of 18

3 1. Introduction and preliminaries. Let B α + be the class of all analytic functions F (z) in the open disk U := {z C, z < 1}, of the form F (z) = 1 + a n z n+α 1, 0 < α 1, satisfying F (0) = 1. And let Bα be the class of all analytic functions F (z) in the open disk U of the form F (z) = 1 a n z n+α 1, 0 < α 1, a n 0; n = 1, 2, 3,..., satisfying F (0) = 1. With a view to recalling the principle of subordination between analytic functions, let the functions f and g be analytic in U. Then we say that the function f is subordinate to g if there exists a Schwarz function w(z), analytic in U such that f(z) = g(w(z)), z U. We denote this subordination by f g or f(z) g(z), z U. If the function g is univalent in U the above subordination is equivalent to f(0) = g(0) and f(u) g(u). Let φ : C 3 U C and let h be univalent in U. Assume that p, φ are analytic and univalent in U and p satisfies the differential superordination (1.1) h(z) φ(p(z)), zp (z), z 2 p (z); z), then p is called a solution of the differential superordination. Page 3 of 18

4 An analytic function q is called a subordinant if q p for all p satisfying (1.1). A univalent function q such that p q for all subordinants p of (1.1) is said to be the best subordinant. Let B + be the class of analytic functions of the form f(z) = 1 + Given two functions f, g B +, f(z) = 1 + a n z n, a n 0. a n z n and g(z) = 1 + b n z n their convolution or Hadamard product f(z) g(z) is defined by f(z) g(z) = 1 + a n b n z n, a n 0, b n 0, z U. Juneja et al. [1] define the family E(Φ, Ψ), so that { } f(z) Φ(z) R > 0, z U f(z) Ψ(z) where Φ(z) = z + ϕ n z n and Ψ(z) = z + n=2 ψ n z n are analytic in U with the conditions ϕ n 0, ψ n 0, ϕ n ψ n for n 2 and f(z) Ψ(z) 0. n=2 Page 4 of 18

5 Definition 1.1. Let F (z) B α +, we define the family E α + (Φ, Ψ) so that { } (1.2) R > 0, z U, where Φ(z) = 1 + ϕ n z n+α 1 and Ψ(z) = 1 + ψ n z n+α 1 are analytic in U under the conditions ϕ n 0, ψ n 0, ϕ n ψ n for n 1 and 0. Definition 1.2. Letting F (z) B α, we define the family E α (Φ, Ψ) which satisfies (1.2) where Φ(z) = 1 ϕ n z n+α 1 and Ψ(z) = 1 ψ n z n+α 1 are analytic in U under the conditions ϕ n 0, ψ n 0, ϕ n ψ n for n 1 and 0. In the present paper, we establish some sufficient conditions for functions F B + α and F B α to satisfy (1.3) q(z), z U, where q(z) is a given univalent function in U such that q(0) = 1. Moreover, we give applications for these results in fractional calculus. We shall need the following known results. Page 5 of 18

6 Lemma 1.3 ([2]). Let q(z) be convex in the unit disk U with q(0) = 1 and R{q} >, z U. If 0 µ < 1, p is an analytic function in U with p(0) = 1 and if 1 2 (1 µ)p 2 (z) + (2µ 1)p(z) µ + (1 µ)zp (z) (1 µ)q 2 (z) + (2µ 1)q(z) µ + (1 µ)zq (z), then p(z) q(z) and q(z) is the best dominant. Lemma 1.4 ([3]). Let q(z) be univalent in the unit disk U and let θ(z) be analytic in a domain D containing q(u). If zq (z)θ(q) is starlike in U, and zp (z)θ(p(z)) zq (z)θ(q(z)) then p(z) q(z) and q(z) is the best dominant. Page 6 of 18

7 2. Main Results In this section, we verify some sufficient conditions of subordination for analytic functions in the classes B + α and B α. Theorem 2.1. Let the function q(z) be convex in the unit disk U such that q(0) = 1 and R{q} > 1. If F 2 B+ α and F (z) Φ(z) an analytic function in U satisfies the F (z) Ψ(z) subordination [ ] 2 [ ] (1 µ) + (2µ 1) µ [ ] [ ] z() z() + (1 µ) (1 µ)q 2 (z) + (2µ 1)q(z) µ + (1 µ)zq (z), then and q(z) is the best dominant. q(z) Proof. Let the function p(z) be defined by p(z) :=, z U. Page 7 of 18 It is clear that p(0) = 1. Then straightforward computation gives us (1 µ)p 2 (z) + (2µ 1)p(z) µ + (1 µ)zp (z)

8 [ ] 2 [ ] = (1 µ) + (2µ 1) µ [ ] + (1 µ)z [ ] 2 [ ] = (1 µ) + (2µ 1) µ [ ] [ ] z() z() + (1 µ) (1 µ)q 2 (z) + (2µ 1)q(z) µ + (1 µ)zq (z). By the assumption of the theorem we have that the assertion of the theorem follows by an application of Lemma 1.3. Corollary 2.2. If F B α + and F (z) Φ(z) is an analytic function in U satisfying the F (z) Ψ(z) subordination [ ] 2 [ ] (1 µ) + (2µ 1) µ [ ] [ ] z() z() + (1 µ) ( ) 2 ( ) 1 + Az 1 + Az (1 µ) + (2µ 1) 1 + Bz 1 + Bz ( ) 1 + Az µ + (1 µ) 1 + Bz (A B)z (1 + Az)(1 + Bz), Page 8 of 18

9 then and ( 1+Az ) is the best dominant. 1+Bz Proof. Let the function q(z) be defined by q(z) := ( ) 1 + Az, 1 B < A Bz ( 1 + Az 1 + Bz ), z U. It is clear that q(0) = 1 and R{q} > 1 for arbitrary A, B, z U, then in view of 2 Theorem 2.1 we obtain the result. Corollary 2.3. If F B α + and F (z) Φ(z) is an analytic function in U satisfying the F (z) Ψ(z) subordination [ ] 2 [ ] (1 µ) + (2µ 1) µ [ ] [ ] z() z() + (1 µ) then (1 µ) and ( 1+z ) is the best dominant. 1 z ( 1 + z 1 z ) 2 ( ) 1 + z + (2µ 1) µ 1 z ( 1 + z + (1 µ) 1 z ( ) 1 + z, 1 z ) ( ) 2z, 1 z 2 Page 9 of 18

10 Define the function ϕ α (a, c; z) by (a) n ϕ α (a, c; z) := 1 + z n+α 1, (z U; a R, c R\{0, 1, 2,... }), (c) n where (a) n is the Pochhammer symbol defined by Γ(a + n) 1, (n = 0); (a) n := = Γ(a) a(a + 1)(a + 2) (a + n 1), (n N). Corresponding to the function ϕ α (a, c; z), define a linear operator L α (a, c) by or equivalently by L α (a, c)f (z) := ϕ α (a, c; z) F (z), L α (a, c)f (z) := 1 + F (z) B + α (a) n (c) n a n z n+α 1. For details see [4]. Hence we have the following result: Corollary 2.4. Let the function q(z) be convex in the unit disk U such that q(0) = 1 and R{q} > 1 Lα(a,c)Φ(z). If 2 L α(a,c)ψ(z) is an analytic function in U satisfying the subordination [ ] 2 [ ] Lα (a, c)φ(z) Lα (a, c)φ(z) (1 µ) + (2µ 1) µ L α (a, c)ψ(z) L α (a, c)ψ(z) [ ] [ Lα (a, c)φ(z) z(lα (a, c)φ(z)) + (1 µ) z(l ] α(a, c)ψ(z)) L α (a, c)ψ(z) L α (a, c)φ(z) L α (a, c)ψ(z) (1 µ)q 2 (z) + (2µ 1)q(z) µ + (1 µ)zq (z), Page 10 of 18

11 then and q(z) is the best dominant. L α (a, c)φ(z) L α (a, c)ψ(z) q(z) Theorem 2.5. Let the function q(z) be univalent in the unit disk U such that q (z) 0 and zq (z) is starlike in U. If F B q(z) α satisfies the subordination [ ] z() z() a a zq (z) q(z), then and q(z) is the best dominant. q(z), z U, Proof. Let the function p(z) be defined by [ ] p(z) :=, z U. By setting θ(ω) := a ω, a 0, it can easily observed that θ(ω) is analytic in C {0}. Then we obtain [ ] a zp (z) z() p(z) = a z() a zq (z) q(z). Page 11 of 18

12 By the assumption of the theorem we have that the assertion of the theorem follows by an application of Lemma 1.4. Corollary 2.6. If F Bα satisfies the subordination [ ] z() z() (A B)z a a (1 + Az)(1 + Bz) then and ( 1+Az ) is the best dominant. 1+Bz ( ) 1 + Az, 1 B < A Bz Corollary 2.7. If F Bα satisfies the subordination [ ] z() z() a then and ( 1+z ) is the best dominant. 1 z Define the function φ α (a, c; z) by φ α (a, c; z) := 1 ( ) 1 + z, 1 z ( ) 2z a, 1 z 2 (a) n (c) n z n+α 1, (z U; a R, c R\{0, 1, 2,... }), where (a) n is the Pochhammer symbol. Corresponding to the function φ α (a, c; z), define a linear operator L α (a, c) by L α (a, c)f (z) := φ α (a, c; z) F (z), F (z) B α Page 12 of 18

13 or equivalently by L α (a, c)f (z) := 1 Hence we obtain the next result. (a) n (c) n a n z n+α 1. Corollary 2.8. Let the function q(z) be univalent in the unit disk U such that q (z) 0 and zq (z) is starlike in U. If F B q(z) α satisfies the subordination [ z(lα (a, c)φ(z)) a z(l ] α(a, c)ψ(z)) a zq (z) L α (a, c)φ(z) L α (a, c)ψ(z) q(z), then and q(z) is the best dominant. L α (a, c)φ(z) L α (a, c)ψ(z) q(z), z U, Page 13 of 18

14 3. Applications In this section, we introduce some applications of Section 2 containing fractional integral operators. Assume that f(z) = ϕ nz n and let us begin with the following definitions Definition 3.1 ([5]). The fractional integral of order α for the function f(z) is defined by Iz α f(z) := 1 z f(ζ)(z ζ) α 1 dζ; 0 α < 1, Γ(α) 0 where the function f(z) is analytic in a simply-connected region of the complex z plane (C) containing the origin. The multiplicity of (z ζ) α 1 is[ removed ] by requiring log(z ζ) to be real when(z ζ) > 0. Note that, Iz α z f(z) = α 1 f(z), Γ(α) for z > 0 and 0 for z 0 (see [6]). From Definition 3.1, we have [ ] z Iz α α 1 f(z) = f(z) = zα 1 ϕ n z n = Γ(α) Γ(α) a n z n+α 1 where a n := ϕn for all n = 1, 2, 3,..., thus 1 + Γ(α) Iα z f(z) B α + and 1 Iz α f(z) (ϕ n 0). Then we have the following results. B α Theorem 3.2. Let the assumptions of Theorem 2.1 hold. Then [ ] (1 + I α z f(z)) Φ(z) q(z), z 0, z U (1 + Iz α f(z)) Ψ(z) Page 14 of 18 and q(z) is the best dominant.

15 Proof. Let the function F (z) be defined by F (z) := 1 + I α z f(z), z U. Theorem 3.3. Let the assumptions of Theorem 2.5 hold. Then [ ] (1 I α z f(z)) Φ(z) q(z), z U (1 Iz α f(z)) Ψ(z) and q(z) is the best dominant. Proof. Let the function F (z) be defined by F (z) := 1 I α z f(z), z U. Let F (a, b; c; z) be the Gauss hypergeometric function (see [7]) defined, for z U, by (a) n (b) n F (a, b; c; z) = z n. (c) n (1) n n=0 We need the following definitions of fractional operators in Saigo type fractional calculus (see [8, 9]). Definition 3.4. For α > 0 and β, η R, the fractional integral operator I α,β,η 0,z is defined by I α,β,η 0,z f(z) = z α β Γ(α) z 0 (z ζ) α 1 F ( α + β, η; α; 1 ζ ) f(ζ)dζ, z Page 15 of 18

16 where the function f(z) is analytic in a simply-connected region of the z plane containing the origin, with order f(z) = O( z ɛ )(z 0), ɛ > max{0, β η} 1 and the multiplicity of (z ζ) α 1 is removed by requiring log(z ζ) to be real when z ζ > 0. where From Definition 3.4, with β < 0, we have z I α,β,η 0,z f(z) = z α β Γ(α) 0 (α + β) n ( η) n = (α) n (1) n = = Denote a n := Bϕn Γ(α) n=0 n=0 (z ζ) α 1 F (α + β, η; α; 1 ζ z )f(ζ)dζ z z α β n B n Γ(α) z β 1 B n Γ(α) f(ζ) n=0 = B Γ(α) 0 ϕ n z n β 1, z α β Γ(α) z 0 (z ζ) n+α 1 f(ζ)dζ B n := (α + β) n( η) n (α) n (1) n and B := (z ζ) α 1 ( 1 ζ z ) n f(ζ)dζ B n. n=0 for all n = 1, 2, 3,..., and let α = β. Thus, 1 + I α,β,η 0,z f(z) B + α and 1 I α,β,η 0,z f(z) B α (ϕ n 0), Page 16 of 18

17 andn we have the following results Theorem 3.5. Let the assumptions of Theorem 2.1 hold. Then [ ] (1 + I α,β,η 0,z f(z)) Φ(z) q(z), z U (1 + I α,β,η 0,z f(z)) Ψ(z) and q(z) is the best dominant. Proof. Let the function F (z) be defined by F (z) := 1 + I α,β,η 0,z f(z), z U. Theorem 3.6. Let the assumptions of Theorem 2.5 hold. Then [ ] (1 I α,β,η 0,z f(z)) Φ(z) q(z), z U (1 I α,β,η 0,z f(z)) Ψ(z) and q(z) is the best dominant. Proof. Let the function F (z) be defined by F (z) := 1 I α,β,η 0,z f(z), z U. Page 17 of 18

18 References [1] O. JUNEJA, T. REDDY AND M. MOGRA, A convolution approach for analytic functions with negative coefficients, Soochow J. Math., 11 (1985), [2] M.OBRADOVIC, T. YAGUCHI AND H. SAITOH, On some conditions for univalence and starlikeness in the unit disc, Rendiconti di Math. Series VII, 12 (1992), [3] S.S. MILLER AND P.T. MOCANU, Differential Subordinations: Theory and Applications, Pure and Applied Mathematics, No.225, Dekker, N.Y., (2000). [4] J. LIU AND H.M. SRIVASTAVA, A linear operator and associated families of meromorphically multivalent functions, J. Math. Anal. Appl., 259 (2001), [5] H.M. SRIVASTAVA AND S. OWA, Univalent Functions, Fractional Calculus, and Their Applications, Halsted Press, John Wiley and Sons, New York, Chichester, Brisbane, and Toronto, [6] K.S. MILLER AND B. ROSS, An Introduction to the Fractional Calculus and Fractional Differential Equations, John-Wiley and Sons, Inc., [7] H.M. SRIVASTAVA AND S. OWA (Eds.), Current Topics in Analytic Function Theory, World Scientific Publishing Company, Singapore, New Jersey, London and Hong Kong, [8] R.K. RAINA AND H.M. SRIVASTAVA, A certain subclass of analytic functions associated with operators of fractional calculus, Comput. Math. Appl., 32(7) (1996), [9] R.K. RAINA, On certain class of analytic functions and applications to fractional calculus operator, Integral Transf. and Special Func., 5 (1997), Page 18 of 18