Subordination and Superordination Results for Analytic Functions Associated With Convolution Structure

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1 Int. J. Open Problems Complex Analysis, Vol. 2, No. 2, July 2010 ISSN ; Copyright c ICSRS Publication, Subordination and Superordination Results for Analytic Functions Associated With Convolution Structure N. Magesh, G. Murugusundaramoorthy, T. Rosy and K. Muthunagai Department of Mathematics, Government Arts College (Men) Krishnagiri , Tamilnadu, India. nmagi 2000@yahoo.co.in School of Advanced Sciences, VIT University Vellore , Tamilnadu, India. gmsmoorthy@yahoo.com Department of Mathematics, Madras Christian College Chennai , Tamilnadu, India. thomas.rosy@gmail.com Abstract In the present investigation, we obtain some subordination and superordination results involving Hadamard product for certain normalied analytic functions in the open unit disk. Relevant connections of the results, which are presented in this paper, with various other known results are also pointed out. Keywords: Univalent functions, starlike functions, convex functions, differential subordination, differential superordination, Hadamard product (convolution) Mathematics Subject Classification: Primary 30C45; Secondary 30C80. 1 Introduction Let H be the class of analytic functions in U := { : < 1} and H[a, n] be the subclass of H consisting of functions of the form f() = a + a n n + a n+1 n

2 68 N. Magesh, et al. Let A be the subclass of H consisting of functions of the form f() = + a (1) Let p, h H and let φ(r, s, t; ) : C 3 U C. If p and φ(p(), p (), 2 p (); ) are univalent and if p satisfies the second order superordination h() φ(p(), p (), 2 p (); ), (2) p is a solution of the differential superordination (2). (If f is subordinate to F, F is superordinate to f.) An analytic function q is called a subordinant if q p for all p satisfying (2). A univalent subordinant q that satisfies q q for all subordinants q of (2) is said to be the best subordinant. Recently Miller and Mocanu[10] obtained conditions on h, q and φ for which the following implication holds: h() φ(p(), p (), 2 p (); ) q() p(). For two functions f() = + n=2 a n n and g() = + n=2 b n n, the Hadamard product (or convolution) of f and g is defined by (f g)() := + a n b n n =: (g f)(). (3) n=2 Recently Bulboacă [2] (see also [1]) considered certain classes of first order differential superordinations as well as superordination-preserving integral operators by using the results of Miller and Mocanu[10]. Further, using the results in [1] and [10], Magesh and Murugusundaramoorthy [7], Magesh et al., [8], Murugusundaramoorthy and Magesh [11, 12, 13] and Shanmugam et al., [18] have obtained sandwich results for certain classes of analytic functions. The main object of the present paper is to find sufficient condition for certain normalied analytic functions f in U to satisfy ( α(f Φ)() + β(f Ψ)() q 1 () q 2 (), (4) where q 1, q 2 are given univalent functions in U with q 1 (0) = 1, q 2 (0) = 1 and Φ() = + λ n n, Ψ() = + µ n n are analytic functions in U with n=2 n=2 λ n 0, µ n 0 and λ n µ n. Also, we obtain the number of known results as their special cases. 2 Preliminary Results For our present investigation, we shall need the following:

3 Subordination and Superordination Results 69 Lemma 2.1 [15, p.159, Theorem 6.2] The function L(, t) = a 1 (t) + a 2 (t) with a 1 (t) 0 for t 0 and lim a 1 (t) = +, is a subordination chain t if Re L(,t) L(,t) t > 0, U, t 0. Definition 2.2 [10, p.817, Definition 2] Denote by Q, the set of all functions f that are analytic and injective on U E(f), where E(f) = {ζ U : lim ζ f() = } and are such that f (ζ) 0 for ζ U E(f). Lemma 2.3 [9, p.132, Theorem 3.4h] Let q be univalent in the unit disk U and θ and φ be analytic in a domain D containing q(u) with φ(w) 0 when w q(u). Set Suppose that Q() := q ()φ(q()) and h() := θ(q()) + Q(). 1. Q() is starlike univalent in U and 2. Re { } h () Q() > 0 for U. If p is analytic with p(0) = q(0), p(u) D and θ(p()) + p ()φ(p()) θ(q()) + q ()φ(q()), (5) and q is the best dominant. p() q() Lemma 2.4 [2, p.289, Corollary 3.2] Let q be convex univalent in the unit disk U and ϑ and ϕ be analytic in a domain D containing q(u). Suppose that 1. Re {ϑ (q())/ϕ(q())} > 0for U and 2. ψ() = q ()ϕ(q()) is starlike univalent in U. If p() H[q(0), 1] Q, with p(u) D, and ϑ(p())+p ()ϕ(p()) is univalent in U and ϑ(q()) + q ()ϕ(q()) ϑ(p()) + p ()ϕ(p()), (6) q() p() and q is the best subordinant.

4 70 N. Magesh, et al. 3 Subordination results Using Lemma 2.3, we first prove the following theorem. Theorem 3.1 Let Φ, Ψ A, γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, and assume that { } γ2 Re q() q () γ 3 q () q () > 0 ( U). (7) q() If f A satisfies (γ i) 3 1 (f; Φ, Ψ, α, β) = (f, Φ, Ψ, γ1, γ 2, γ 3, α, β) γ 1 + γ 2 q() + γ 3 q () q(), (8) where (γ i) 3 1 (f; Φ, Ψ, α, β) := ( ) γ 1 + γ α(f Φ)()+β(f Ψ)( 2 +γ 3 µ ( α(f Φ) ()+β(f ψ) () 1 ), α(f Φ)()+β(f ψ)() ( α(f Φ)() + β(f Ψ)() q() (9) and q is the best dominant. Proof: Define the function p by p() := ( α(f Φ)() + β(f Ψ)() ( U). (10) Then the function p is analytic in U and p(0) = 1. Therefore, by making use of (10), we obtain By using (11) in (8), we have ( α(f Φ)() + β(f Ψ)() γ 1 + γ 2 ( α(f Φ) () + β(f ψ) ) () +γ 3 µ 1 α(f Φ)() + β(f ψ)() = γ 1 + γ 2 p() + γ 3 p () p(). (11) γ 1 + γ 2 p() + γ 3 p () p() γ 1 + γ 2 q() + γ 3 q () q(). (12)

5 Subordination and Superordination Results 71 By setting θ(w) := γ 1 + γ 2 ω and φ(ω) := γ 3 w, it can be easily observed that θ(w), φ(w) are analytic in C {0} and φ(w) 0. Also we see that Q() := q q () ()φ(q()) = γ 3 q() and h() := θ(q()) + Q() = γ 1 + γ 2 q() + γ 3 q () q(). It is clear that Q() is starlike univalent in U and { h } { } () γ2 Re = Re q() q () Q() γ 3 q () q () q() > 0. By the hypothesis of Theorem 3.1, the result now follows by an application of Lemma 2.3. By fixing Φ() = following corollary. 1 and Ψ() = (1 ) 2 in Theorem 3.1, we obtain the Corollary 3.2 Let γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, and (7) holds true. If f A satisfies γ 1 + γ 2 ( αf() + βf () + γ 3 µ γ 1 + γ 2 q() + γ 3 q () q(), and q is the best dominant. ( αf() + βf () q() ( α(f () f()) + β 2 f ) () αf() + βf () Putting α = 1 and β = 1 in Corollary 3.2, we obtain the following corollary. Corollary 3.3 Let γ i C (i = 1, 2, 3)(γ 3 0), 0 µ C, q be convex univalent with q(0) = 1, and (7) holds true. If f A satisfies ( f() + f ( () 2 f () + f ) () f() γ 1 + γ 2 + γ 3 µ f() + f () q () γ 1 + γ 2 q() + γ 3 q(),

6 72 N. Magesh, et al. ( f() + f () q() and q is the best dominant. Putting α = 1 and β = 0 in Corollary 3.2, we obtain the following corollary. Corollary 3.4 Let γ i C (i = 1, 2, 3) (γ 3 0), 0 µ C, q be convex univalent with q(0) = 1, and (7) holds true. If f A satisfies ( ( f() f ) () γ 1 + γ 2 + γ 3 µ f() 1 q () γ 1 + γ 2 q() + γ 3 q(), and q is the best dominant. ( f() q() Putting α = 0 and β = 1 in Corollary 3.2, we obtain the following corollary. Corollary 3.5 Let γ i C (i = 1, 2, 3)(γ 3 0), 0 µ C, q be convex univalent with q(0) = 1, and (7) holds true. If f A satisfies γ 1 + γ 2 (f () + γ 3 µ f () f () γ 1 + γ 2 q() + γ 3 q () q(), and q is the best dominant. (f () q() By taking q() = 1+A 1+B following corollary. ( 1 B < A 1) in Theorem 3.1, we have the Corollary 3.6 Let Φ, Ψ A, γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, assume that (7) holds. If f A and and 1+A 1+B (γ 1 + A i) 3 1 (f; Φ, Ψ, α, β) γ1 + γ B + γ (A B) 3 (1 + A)(1 + B), ( α(f Φ)() + β(f Ψ)() is the best dominant. 1 + A 1 + B

7 Subordination and Superordination Results 73 Putting α = 1 and β = 0 in Corollary 3.6, we obtain the following corollary. Corollary 3.7 Let Φ A, γ i C (i = 1, 2, 3)(γ 3 0), 0 µ C and q be convex univalent with q(0) = 1, assume that (7) holds. If f A and and 1+A 1+B ( ( (f Φ)() (f Φ) ) () γ 1 + γ 2 + γ 3 µ (f Φ)() A γ 1 + γ B + γ (A B) 3 (1 + A)(1 + B), is the best dominant. ( (f Φ)() 1 + A 1 + B Putting γ 1 = 1, γ 2 = 0 and γ 3 = 1 in Corollary 3.7, we obtain the following corollary. Corollary 3.8 Let Φ, A, 0 µ C and q be convex univalent with q(0) = 1, assume that (7) holds. If f A and and 1+A 1+B 1 + µ ( (f Φ) ) () (f Φ)() 1 (A B) 1 + (1 + A)(1 + B), is the best dominant. ( (f Φ)() 1 + A 1 + B By setting γ 1 = 1, γ 2 = 0, γ 3 = 1, α = 1, β = 0, Φ() = (1 + B (A B) B 1 in Theorem 3.1, we obtain the following corollary. and q() = Corollary 3.9 Let 0 µ C, 1 B < A 1 and q be convex univalent with q(0) = 1, assume that (7) holds. If f A and 1 + µ ( f ) () f() A 1 + B, ( f() (1 + B (A B) B and (1 + B (A B) B is the best dominant.

8 74 N. Magesh, et al. We note that q() = (1+B (A B) B 1 or µ(a B) B By setting γ 1 = 1, γ 2 = 0, α = 1, β = 0, Φ() = C {0}), µ = 1 and γ 3 = 1 b as stated in [19]. is univalent if and only if µ(a B) 1 B, q() = 1 (b 1 (1 ) 2b in Theorem 3.1, we obtain the following corollary Corollary 3.10 Let 0 b C and q be convex univalent with q(0) = 1, assume that (7) holds. If f A and ( f ) () b f() , and 1 (1 ) 2b is the best dominant. f() 1 (1 ) 2b By setting γ 1 = 1, γ 2 = 0, α = 1, β = 0, Φ() =, q() = 1 (b (1 ) 2 (1 ) 2b C {0}), µ = 1 and γ 3 = 1 in Theorem 3.1, we obtain the following corollary b as stated in [19]. Corollary 3.11 Let 0 b C and q be convex univalent with q(0) = 1, assume that (7) holds. If f A and and 1 (1 ) 2b b is the best dominant. f () f () 1 + 1, f () 1 (1 ) 2b By setting γ 1 = 1, γ 2 = 0, α = 1, β = 0, Φ() =, q() = 1 (b 1 (1 ) 2ab C {0}), µ = a and γ 3 = 1 in Theorem 3.1, we obtain the following corollary b as stated in [14]. Corollary 3.12 Let 0 b C and q be convex univalent with q(0) = 1, assume that (7) holds. If f A and ( f ) () b f() , and 1 (1 ) 2ab ( ) a f() is the best dominant. 1 (1 ) 2ab

9 Subordination and Superordination Results 75 By taking q() = ( ) δ 1+ 1 (0 < δ 1), in Theorem 3.1, we have the following corollary. Corollary 3.13 Let Φ, Ψ A, γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, assume that (7) holds. If f A and ( ) (γ i) δ 2δ (f; Φ, Ψ, α, β) γ1 + γ 2 + γ 3 1 (1 2 ), ( α(f Φ)() + β(f Ψ)() ( ) 1 + δ 1 and ( ) δ 1+ 1 is the best dominant. Putting q() = e µa ( µa π), in Theorem 3.1, we have the following corollary. Corollary 3.14 Let Φ, Ψ A, γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, assume that (7) holds. If f A and (γ i) 3 1 (f; Φ, Ψ, α, β) γ1 + γ 2 e µa + γ 3 µa, ( α(f Φ)() + β(f Ψ)() e µa and e µa is the best dominant. 3.1 Superordination results Now, by applying Lemma 2.4, we prove the following theorem. Theorem 3.15 Let Φ, Ψ A, γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, and assume that } { γ2 Re q() 0. (13) γ 3 If f A, ( ) α(f Φ)()+β(f Ψ)( H[q(0), 1] Q. Let (γ i ) 3 1 (f; Φ, Ψ, α, β) be univalent in U and γ 1 + γ 2 q() + γ 3 q () q() (γ i)3 1(f; Φ, Ψ, α, β), (14)

10 76 N. Magesh, et al. where (γ i) 3 1 (f; Φ, Ψ, α, β) is given by (9), ( α(f Φ)() + β(f Ψ)() q() and q is the best subordinant. Proof Define the function p by p() := Simple computation from (15), we get, ( α(f Φ)() + β(f Ψ)(). (15) (γ i) 3 1 (f; Φ, Ψ, α, β) = γ1 + γ 2 p() + γ 3 p () p(), γ 1 + γ 2 q() + γ 3 q () q() γ 1 + γ 2 p() + γ 3 p () p(). By setting ϑ(w) = γ 1 + γ 2 w and φ(w) = γ 3, it is easily observed that ϑ(w) w is analytic in C. Also, φ(w) is analytic in C {0} and φ(w) 0. If we let L(, t) = ϑ(q()) + φ(q())tq () = γ 1 + γ 2 q() + γ 3 t q () q() = a 1 (t) +... (16) Differentiating (16) with respect to and t, we have and Also, L(, t) [ q = γ 2 q () () + tγ 3 q() = a 1 (t) +... L(0, t) L(, t) t = γ 3 q (0) = γ 3 q () q(). [ γ2 + q () q() (q ()) 2 ] q 2 () ] 1 + t γ 3 q(0) From the univalence of q we have q (0) 0 and q(0) = 1, it follows that a 1 (t) 0 for t 0 and lim t a 1 (t) = +.

11 Subordination and Superordination Results 77 A simple computation yields, L(,t) { ( )} Re L(,t) = Re γ2 q() + t 1 + q () γ 3 q () q (). q() t Using the fact that q is convex univalent function in U and γ 4 0, we have, L(,t) { } Re L(,t) > 0 if R γ2 q() > 0, U, t 0. γ 3 t Now Theorem 3.15 follows by applying Lemma 2.4. By fixing Φ() = following corollary. 1 and Ψ() = (1 ) 2 in Theorem 3.15, we obtain the Corollary 3.16 Let γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, and (13) hold true. If f A, ( ) αf()+βf ( ( ) H[q(0), 1] Q. Let γ1 + γ αf()+βf ( 2 + γ 3 µ ( ) α(f () f())+β 2 f () αf()+βf (), be univalent in U and γ 1 + γ 2 q() + γ 3 q () q() γ 1 + γ 2 ( αf() + βf () q() and q is the best subordinant. ( α(f () f()) + β 2 f ) () + γ 3 µ, αf() + βf () ( αf() + βf () Putting α = 1 and β = 1 in Corollary 3.16, we obtain the following corollary. Corollary 3.17 Let γ i C (i = 1, 2, 3)(γ 3 0), 0 µ C, q be convex univalent with q(0) = 1, and (13) hold true. If f A, ( ) f()+f ( ( ) H[q(0), 1] Q. Let γ 1 + γ f()+f ( 2 + γ3 µ ( ) 2 f ()+f () f() f()+f (), be univalent in U and q () γ 1 + γ 2 q() + γ 3 q() ( f() + f ( () 2 f () + f ) () f() γ 1 + γ 2 + γ 3 µ, f() + f () q() and q is the best subordinant. ( f() + f ()

12 78 N. Magesh, et al. By taking q() = (1 + A)/(1 + B) ( 1 B < A 1) in Theorem 3.15, we obtain the following corollary. Corollary 3.18 Let γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, q be convex univalent with q(0) = 1, and (13) hold true. If f A, ( ) α(f Φ)()+β(f Ψ)( H[q(0), 1] Q. Let (γ i ) 3 1 (f; Φ, Ψ, α, β) be univalent in U and 1 + A γ 1 + γ B + γ (A B) 3 (1 + A)(1 + B) (γ i)3 1(f; Φ, Ψ, α, β), and 1+A 1+B 1 + A 1 + B is the best subordinant. 4 Sandwich results ( α(f Φ)() + β(f Ψ)() There is a complete analog of Theorem 3.1 for differential subordination and Theorem 3.15 for differential superordination. We can combine the results of Theorem 3.1 with Theorem 3.15 and obtain the following sandwich theorem. Theorem 4.1 Let q 1 and q 2 be convex univalent in U, γ i C (i = 1, 2, 3)(γ 3 0), µ, α, β C such that µ 0 and α + β 0, and let q 2 satisfy (7) and q 1 satisfy (13). For f, Φ, Ψ A, let ( ) α(f Φ)()+β(f Ψ)( H[1, 1] Q and (γ i) 3 1 (f; Φ, Ψ, α, β) defined by (9) be univalent in U satisfying γ 1 + γ 2 q 1 () + γ 3 q 1() q 1 () (γ i)3 1(f; Φ, Ψ, α, β) γ1 + γ 2 q 2 () + γ 3 q 2() q 2 (), q 1 () ( α(f Φ)() + β(f Ψ)() q 2 () and q 1, q 2 are respectively the best subordinant and best dominant. By taking q 1 () = 1+A 1 ( 1 B 1+B 1 1 < A 1 1) and q 2 () = 1+A 2 ( 1 1+B 2 B 2 < A 2 1) in Theorem 4.1 we obtain the following result. Corollary 4.2 For f, Φ, Ψ A, let ( ) α(f Φ)()+β(f Ψ)( H[1, 1] Q and (γ i) 3 1 (f; Φ, Ψ, α, β) defined by (9) be univalent in U satisfying 1 + A 1 γ 1 + γ B 1 + γ (A 1 B 1 ) 3 (1 + A 1 )(1 + B 1 ) (γ i) 3 1 (f; Φ, Ψ, α, β) 1 + A 2 γ 1 + γ B 2 + γ (A 2 B 2 ) 3 (1 + A 2 )(1 + B 2 )

13 Subordination and Superordination Results A B 1 ( α(f Φ)() + β(f Ψ)() 1 + A B 2 and 1+A 1 1+B 1, 1+A 2 1+B 2 are respectively the best subordinant and best dominant. 5 Conclusion and Open Problem We conclude this paper by remarking that in view of the function class defined by the subordination relation (4) and expressed in terms of the convolution (3) involving arbitrary coefficients, the main results would lead to additional new results. In fact, by appropriately selecting the arbitrary sequences (Φ() and Ψ()), the results presented in this paper would find further applications for the classes which incorparate generalied forms of linear operators illustrated below 1. and Φ() = + Ψ() = + n=2 n=2 (α 1 ) n 1... (α l ) n 1 n (β 1 ) n 1... (β m ) n 1 (n 1)! n (α 1) n 1... (α l ) n 1 n (β 1 ) n 1... (β m ) n 1 (n 1)! where α 1,..., α l and β 1,..., β m, (l, m N = 1, 2, 3,...) are complex parameters β j / {0, 1, 2,...} for j = 1, 2,... m, l m + 1 (results for Diok-Srivastava operator [6]) 2. Φ() = + (a) n 1 n n=2 (c) n 1 and Ψ() = + (n 1)! n=2 n (a) n 1 n, where (c) n 1 (n 1)! c 0, 1, 2,..., ( results for Carlson and Shaffer operator [3]) 3. Φ() =, λ 1, and Ψ() = (1 ) λ+1 Ruscheweyh derivative operator [16]), λ 1, (results for (1 ) λ+2 4. Φ() = + n=2 n k n, and Ψ() = + n=2 n k+1 n, k 0, ( results for Salagean operator [17]) 5. Φ() = + ( ) k n+λ n=2 1+λ n, and Ψ() = + n=2 n ( ) k n+λ 1+λ n, where λ 0 k Z,( results for Multiplier transformation [4, 5])

14 80 N. Magesh, et al. and specialiing the parameters α, β, µ, γ 1, γ 2, γ 3 and γ 4 and the function q() Theorem 3.1, Theorem 3.15 and Theorem 4.1 would eventually lead further new results for the classes of functions defined analogously by associating in the process of Φ() and Ψ(). These considerations can fruitfully be worked out and we skip the details in this regard. ACKNOWLEDGEMENTS. The authors would like to thank the referee for his useful comments. References [1] T. Bulboacă, A class of superordination-preserving integral operators, Indag. Math., New Ser., 13(3) (2002), pp [2] T. Bulboacă, Classes of first order differential superordinations, Demonstratio Math., 35(2) (2002), pp [3] B. C. Carlson and S. B. Shaffer, Starlike and prestarlike hypergeometric functions, SIAM J. Math. Anal., 15 (1984), pp [4] N. E. Cho, and T. H. Kim, Multiplier transformations and strongly closeto-convex functions, Bull. Korean Math. Soc., 40(3) (2003), pp [5] N. E. Cho, and H. M. Srivastava, Argument estimates of certain analytic functions defined by a class of multiplier transformations, Math. Comput. Modelling, 37 (2003) no. 12, pp [6] J. Diok and H. M. Srivastava, Certain subclasses of analytic functions associated with the generalied hypergeometric function, Intergral Transform Spec. Funct., 14 (2003), pp [7] N. Magesh and G. Murugusundaramoorthy, Differential subordinations and superordinations for comprehensive class of analytic functions, SUT J. Math., 44(2) (2008), pp [8] N. Magesh, T. Rosy and K. Muthunagi, Subordinations and superordinations results for certain class of analytic functions, Inter. J. Math. Sci. Engg. Appls., 3(1) (2009), pp [9] S. S. Miller and P. T. Mocanu, Differential Subordinations: Theory and Applications, Marcel Dekker Inc., New York, [10] S. S. Miller and P. T. Mocanu, Subordinants of differential superordinations, Complex Variables, 48(10) (2003), pp

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