advertisement

General Mathematics Vol. 17, No. 4 (2009), 83–95 On a Certain Differential Sandwich Theorem Associated with a New Generalized Derivative Operator Adriana Cătaş Abstract The purpose of this paper is to derive certain subordinations and superordinations results involving a new differential operator. By means of the new introduced operator, I m (λ, β, l)f (z), for certain normalized analytic functions in the open unit disc, we establish differential sandwich-type theorems. These results extend corresponding previously known results. 2000 Mathematics Subject Classification: Primary 30C80, 30C45. Key words and phrases: differential subordination, differential superordination, dominant, differential operator, subordinant. 83 84 1 Adriana Cătaş Introduction and definitions Let H(U ) be the class of analytic functions in the open unit disc U = {z ∈ C : |z| < 1}. For a ∈ C and n ∈ N let H[a, n] be the subclass of H(U ) consisting of functions of the form f (z) = a + an z n + an+1 z n+1 + . . . . Let An = {f ∈ H(U ), f (z) = z + an+1 z n+1 + . . . } with A1 := A. 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, written symbolically as f ≺g or f (z) ≺ g(z), z∈U if there exists a Schwarz function w analytic in U such that f (z) = g(w(z)), z ∈ U . In particular, if the function g is univalent in U , the above subordination is equivalent to f (0) = g(0) and f (U ) ⊂ g(U ). Let p, h ∈ H(U ) and let ψ(r, s, t; z) : C3 × U → C. If p and ψ(p(z), zp0 (z), z 2 p00 (z); z) are univalent and if p satisfies the second order differential superordination (1) h(z) ≺ ψ(p(z), zp0 (z), z 2 p00 (z); z), z∈U On a Certain Differential Sandwich Theorem ... 85 then p is a solution of the differential superordination (1). If f is subordinate to g, then g is superordinate to f . An analytic function q is called a subordinant of the differential superordination, or more simply a subordinant if q ≺ p for all p satisfying (1). A univalent subordinant qe that satisfies q ≺ qe for all subordinants q of (1) is said to be the best subordinant. The best subordinant is unique up to a rotation of U . Recently Miller and Mocanu [7] obtained conditions on h, q and ψ for which the following implication holds: h(z) ≺ ψ(p(z), zp0 (z), z 2 p00 (z); z) =⇒ q(z) ≺ p(z), z ∈ U. In order to prove our subordination and superordination results, we make use of the following definition and lemmas. Definition 1 [7] Denote by Q, the set of all functions f that are analytic and injective on U − E(f ), where E(f ) = {ζ ∈ ∂U : lim f (z) = ∞} z→ζ and are such that f 0 (ζ) 6= 0 for ζ ∈ ∂U − E(f ). Lemma 1 [8] Let the function q be univalent in the unit disc U and θ and φ be analytic in a domain D containing q(U ) with φ(w) 6= 0 when w ∈ q(U ). Set Q(z) = zq 0 (z)φ(q(z)) and h(z) = θ(q(z)) + Q(z). Suppose that (1) Q(z)½is starlike ¾ univalent in U and zh0 (z) (2) Re > 0 for z ∈ U . Q(z) 86 Adriana Cătaş If p is analytic with p(0) = q(0), p(U ) ⊆ D and θ(p(z)) + zp0 (z)φ(p(z)) ≺ θ(q(z)) + zq 0 (z)φ(q(z)) then p(z) ≺ q(z) and q is the best dominant. Lemma 2 [4] Let q be convex univalent in the unit disc U and ν and ϕ be analytic in ½ a domain ¾ D containing q(U ). Suppose that ν 0 (q(z)) (1) Re > 0 for z ∈ U and ϕ(q(z)) (2) ψ(z) = zq 0 (z)ϕ(q(z)) is starlike univalent in U . If p(z) ∈ H[q(0), 1] ∩ Q with p(U ) ⊆ D and ν(p(z)) + zp0 (z)ϕ(p(z)) is univalent in U and ν(q(z)) + zq 0 (z)ϕ(q(z)) ≺ ν(p(z)) + zp0 (z)ϕ(p(z)) then q(z) ≺ p(z) and q is the best subordinant. 2 Main results Definition 2 Let the function f be in the class An . For m, β ∈ N0 = {0, 1, 2, . . . }, λ ≥ 0, l ≥ 0, we define the following differential operator ¸m ∞ · X 1 + λ(k − 1) + l m (2) I (λ, β, l)f (z) := z + C(β, k)ak z k 1 + l k=n+1 On a Certain Differential Sandwich Theorem ... where µ C(β, k) := and 87 ¶ k+β−1 (β + 1)k−1 = β (k − 1)! 1, n=0 (a)n := a(a + 1) . . . (a + n − 1), n ∈ N = N − {0} 0 is Pochhamer symbol. Using simple computation one obtains the next result. Proposition 1 For m, β ∈ N0 , λ ≥ 0, l ≥ 0 (3) (l+1)I m+1 (λ, β, l)f (z) = (1−λ+l)I m (λ, β, l)f (z)+λz(I m (λ, β, l)f (z))0 and (4) z(I m (λ, β, l)f (z))0 = (1 + β)I m (λ, β + 1, l)f (z) − βI m (λ, β, l)f (z). Remark 1 Special cases of this operator includes the Ruscheweyh derivative operator I 0 (1, β, 0)f (z) ≡ Dβ defined in [9], the Sălăgean derivative operator I m (1, 0, 0)f (z) ≡ Dm , studied in [10], the generalized Sălăgean operator I m (λ, 0, 0) ≡ Dλm introduced by Al-Oboudi in [1], the generalized Ruscheweyh derivative operator I 1 (λ, β, 0)f (z) ≡ Dλ,β introduced in [6], the m operator I m (λ, β, 0) ≡ Dλ,β introduced by K. Al-Shaqsi and M. Darus in [3] and finally the operator I m (λ, 0, l) ≡ I1 (m, λ, l) introduced in [5]. The main object of the present paper is to find sufficient conditions for certain normalized analytic functions f to satisfy q1 (z) ≺ I m+1 (λ, β, l)f (z) ≺ q2 (z), I m (λ, β, l)f (z) 88 Adriana Cătaş where m, β ∈ N0 , λ ≥ 0 and q1 , q2 are given univalent functions in U . Also, we obtain the number of known results as their special cases. Theorem 1 Let m, β ∈ N0 , λ > 0 and q be convex univalent in U with q(0) = 1. Further, assume that ½ ¾ 2(δ + α)q(z) zq 00 (z) (5) Re +1+ 0 > 0. δ q (z) Let (6) ψ(m, λ, β, δ, α; z) = + δ[1 − λ(1 + β) + l] I m+1 (λ, β, l)f (z) · m + λ I (λ, β, l)f (z) δλ(β + 1)(β + 2) I m (λ, β + 2, l)f (z) · + l+1 I m (λ, β, l)f (z) δ(1 + β)[1 − λ(β + 2) + l] I m (λ, β + 1, l) · + l+1 I m (λ, β, l) · µ ¶¸ µ m+1 ¶2 I (λ, β, l)f (z) l+1 + α+δ 1− . λ I m (λ, β, l)f (z) + If f ∈ An satisfies ψ(m, λ, β, δ, α; z) ≺ δzq 0 (z) + (δ + α)(q(z))2 (7) then I m+1 (λ, β, l)f (z) ≺ q(z) I m (λ, β, l)f (z) (8) and q is the best dominant. Proof. Define the function p(z) by (9) p(z) = I m+1 (λ, β, l)f (z) , I m (λ, β, l)f (z) z ∈ U. On a Certain Differential Sandwich Theorem ... 89 Then the function p(z) is analytic in U and p(0) = 1. Therefore, by making use of (3) and (4) we have δ[1 − λ(1 + β) + l] I m+1 (λ, β, l)f (z) · m + λ I (λ, β, l)f (z) (10) δλ(β + 1)(β + 2) I m (λ, β + 2, l)f (z) + · + l+1 I m (λ, β, l)f (z) δ(1 + β)[1 − λ(β + 2) + l] I m (λ, β + 1, l) · + l+1 I m (λ, β, l) · µ ¶¸ µ m+1 ¶2 l+1 I (λ, β, l)f (z) + α+δ 1− = λ I m (λ, β, l)f (z) + = δzp0 (z) + (δ + α)(p(z))2 . By using (10) in (7) we get δzp0 (z) + (δ + α)(p(z))2 ≺ δzq 0 (z) + (δ + α)(q(z))2 . By setting θ(w) = (δ + α)w2 and φ(w) = δ are analytic in C \ {0} and that φ(w) 6= 0. Hence the result follows by an application of Lemma 1. Remark 2 Similar results were obtained earlier in [6] for the operator defined in [2]. Let q(z) = 1 + Az , 1 + Bz −1 ≤ B < A ≤ 1 in Theorem 1. One obtains the following result. Corollary 1 Let m, β ∈ N0 , λ > 0. Assume that (5) holds. If f ∈ An , then, differential subordination (11) δ(A − B)z ψ(m, λ, β, δ, α; z) ≺ + (δ + α) (1 + Bz)2 µ 1 + Az 1 + Bz ¶2 90 Adriana Cătaş implies 1 + Az I m+1 (λ, β, l)f (z) ≺ m I (λ, β, l)f (z) 1 + Bz and 1 + Az is the best dominant. 1 + Bz Corollary 2 Let m, β ∈ N0 , λ > 0. Assume that (5) holds. If f ∈ An , then differential subordination 2δz ψ(m, λ, β, δ, α; z) ≺ + (δ + α) (1 − z)2 (12) µ 1+z 1−z ¶2 implies 1+z I m+1 (λ, β, l)f (z) ≺ m I (λ, β, l)f (z) 1−z and 1+z is the best dominant. 1−z Corollary 3 Let m, β ∈ N0 , λ > 0, 0 < µ ≤ 1. Assume that (5) holds. If f ∈ An , then differential subordination (13) 2δµz ψ(m, λ, β, δ, α; z) ≺ (1 − z)2 µ 1+z 1−z ¶µ−1 µ + (α + δ) implies I m+1 (λ, β, l)f (z) ≺ I m (λ, β, l)f (z) µ and 1+z 1−z ¶µ is the best dominant. µ 1+z 1−z ¶µ 1+z 1−z ¶2µ On a Certain Differential Sandwich Theorem ... 91 Theorem 2 Let q be convex univalent in U with q(0) = 1. Assume that ½ ¾ 2(δ + α)q(z)q 0 (z) (14) Re > 0. δ I m+1 (λ, β, l)f (z) ∈ H[q(0), 1] ∩ Q. I m (λ, β, l)f (z) If function ψ(m, λ, β, δ, α; z), given by (6), is univalent in U and Let f ∈ A, (15) (δ + α)(q(z))2 + δzq 0 (z) ≺ ψ(m, λ, β, δ, α; z) then q(z) ≺ I m+1 (λ, β, l)f (z) I m (λ, β, l)f (z) and q is the best subordinant. Proof. Theorem 2 follows by using the same technique to prove Theorem 1 and by an application of Lemma 2. By using Theorem 2 we obtain the following corollaries. Corollary 4 Let q(z) = 1 + Az , −1 ≤ B < A ≤ 1, f ∈ A and 1 + Bz I m+1 (λ, β, l)f (z) ∈ H[q(0), 1] ∩ Q. I m (λ, β, l)f (z) Assume that (14) holds. If µ (16) then (δ + α) 1 + Az 1 + Bz ¶2 + δ(A − B)z ≺ ψ(m, λ, β, δ, α; z) (1 + Bz)2 92 Adriana Cătaş 1 + Az I m+1 (λ, β, l)f (z) ≺ m 1 + Bz I (λ, β, l)f (z) and 1 + Az is the best subordinant. 1 + Bz Corollary 5 Let q(z) = 1+z , f ∈ A and 1−z I m+1 (λ, β, l)f (z) ∈ H[q(0), 1] ∩ Q. I m (λ, β, l)f (z) Assume that (14) holds. If 2δz + (δ + α) (1 − z)2 (17) µ 1+z 1−z ¶2 ≺ ψ(m, λ, β, δ, α; z) then 1+z I m+1 (λ, β, l)f (z) ≺ m 1−z I (λ, β, l)f (z) 1+z is the best subordinant. 1−z µ ¶µ 1+z Corollary 6 Let q(z) = , 0 < µ ≤ 1, f ∈ A and 1−z and I m+1 (λ, β, l)f (z) ∈ H[q(0), 1] ∩ Q. I m (λ, β, l)f (z) Assume that (14) holds. If µ ¶µ−1 µ ¶2µ 2δµz 1+z 1+z (18) + (α + δ) ≺ ψ(m, λ, β, δ, α; z) (1 − z)2 1 − z 1−z then µ µ and 1+z 1−z ¶µ 1+z 1−z ¶µ ≺ I m+1 (λ, β, l)f (z) I m (λ, β, l)f (z) is the best subordinant. On a Certain Differential Sandwich Theorem ... 93 Combining the results of differential subordination and superordination we state the following Sandwich Theorems. Theorem 3 Let q1 and q2 be convex univalent in U and satisfy (14) and (5) respectively. I m+1 (λ, β, l)f (z) If f ∈ A, m ∈ H[q(0), 1] ∩ Q and ψ(m, λ, β, δ, α; z) given I (λ, β, l)f (z) in (6) is univalent in U and (19) δzq10 (z) + (δ + α)(q1 (z))2 ≺ ψ(m, λ, β, δ, α; z) ≺ ≺ δzq20 (z) + (δ + α)(q2 (z))2 , then q1 (z) ≺ I m+1 (λ, β, l)f (z) ≺ q2 (z) I m (λ, β, l)f (z) and q1 and q2 are the best subordinant and best dominant respectively. 1 + A1 z 1 + A2 z , q2 (z) = , where −1 ≤ B2 < B1 < A1 ≤ 1 + B1 z 1 + B2 z A2 ≤ 1 we have the following corollary. For q1 (z) = Corollary 7 If f ∈ A, I m+1 (λ, β, l)f (z) ∈ H[q(0), 1] ∩ Q and I m (λ, β, l)f (z) δ(A1 − B1 )z + (δ + α) (1 + B1 z)2 µ 1 + A1 z 1 + B1 z ¶2 δ(A2 − B2 )z ≺ + (δ + α) (1 + B2 z)2 ≺ ψ(m, λ, β, δ, α; z) ≺ µ 1 + A2 z 1 + B2 z ¶2 , then I m+1 (λ, β, l)f (z) 1 + A2 z 1 + A1 z ≺ m ≺ 1 + B1 z I (λ, β, l)f (z) 1 + B2 z 1 + A2 z 1 + A1 z and are the best subordinant and the best dominant Hence 1 + B1 z 1 + B2 z respectively. 94 Adriana Cătaş References [1] Al-Oboudi, F.M.: On univalent functions defined by a generalized Sălăgean operator. Int. J. Math. Math. Sci. 27, 1429-1436, (2004). [2] Al-Shaqsi, K., Darus, M.: Differential subordination with generalized derivative operator. (submitted). [3] Al-Shaqsi, K., Darus, M.: On univalent functions with respect to ksymmetric points defined by a generalized Ruscheweyh derivatives operator. (submitted). [4] Bulboacă, T.: Classes of first order differential superordinations. Demonstratio Math. 35(2), 287-292, (2002). [5] Cătaş, A.: On certain class of p-valent functions defined by a new multiplier transformations. Proceedings Book of the International Symposium G.F.T.A., Istanbul Kultur University, Turkey, 241-250, (2007). [6] Darus, M., Al-Shaqsi, K.: Differential sandwich theorems with generalized derivative operator. Int. J. of Computational and Mathematical Sciences. Vol 2, No. 2 Spring, 75-78, (2008). [7] Miller, S.S., Mocanu, P.T.: Subordinants of differential superordinations. Complex Variables, Theory and Applications. 48(10), 815-826, (2003). [8] Miller, S.S., Mocanu, P.T.: Differential Subordinations: Theory and Applications. Pure and Applied Mathematics. No. 225, Marcel Dekker, New York, (2000). On a Certain Differential Sandwich Theorem ... 95 [9] Ruscheweyh, S.: New criteria for univalent functions. Proc. Amer. Math. Soc. 49, 109-115, (1975). [10] Sălăgean, G. Şt.: Subclasses of univalent functions. Lecture Note in Math. (Springer-Verlag), 1013, 362-372, (1983). Adriana Cătaş Faculty of Sciences, University of Oradea Department of Mathematics and Computer Science 1 University Street, 410087 Oradea, Romania E-mail: [email protected]