1_540_aouf.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 4, No. 1, 2011, 1-13 ISSN 1307-5543 – www.ejpam.com Differential Sandwich Theorems of Analytic Functions Defined by Linear Operators M. K. Aouf 1,∗, Tamer M. Seoudy 2 1 Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt 2 Department of Mathematics, Faculty of Science, Fayoum University, Fayoum 63514, Egypt Abstract. In this paper, we obtain some applications of first order differential subordination and super- ordination results involving a linear operator and other linear operators for certain normalized analytic functions. Some of our results generalize previously known results. 2000 Mathematics Subject Classifications: 30C45 Key Words and Phrases: Analytic function, Hadamard product, differential subordination, superordi- nation, linear operator 1. Introduction Let H (U) be the class of analytic functions in the open unit disk U = {z ∈ C : |z| < 1} and let H[a, k] be the subclass of H (U) consisting of functions of the form: f (z) = a+ akzk + ak+1zk+1 . . . (a ∈ C). (1) For simplicity H[a] = H[a, 1]. Also, letA be the subclass of H (U) consisting of functions of the form: f (z) = z + ∞ ∑ k=2 akzk. (2) If f , g ∈ H (U), we say that f is subordinate to g or f is superordinate to g, written f (z) ≺ g(z) if there exists a Schwarz function ω, which (by definition) is analytic in U with ω(0) = 0 and |ω(z)| < 1 for all z ∈ U , such that f (z) = g(ω(z)), z ∈ U . Furthermore, if the function g is univalent in U , then we have the following equivalence, [cf., e.g., 6, 16, 17]: f (z)≺ g(z)⇔ f (0) = g(0) and f (U)⊂ g(U). ∗Corresponding author. Email addresses: mkaouf127�yahoo. om (M. Aouf), tms00�fayoum.edu.eg (T. Seoudy) http://www.ejpam.com 1 c© 2010 EJPAM All rights reserved. M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 2 Let φ : C2 × U → C and h(z) be univalent in U . If p (z) is analytic in U and satisfies the first order differential subordination: φ � p (z) , zp ′ (z) ; z � ≺ h(z) , (3) then p (z) is a solution of the differential subordination (3). The univalent function q (z) is called a dominant of the solutions of the differential subordination (3) if p (z) ≺ q (z) for all p (z) satisfying (3). A univalent dominant q̃ that satisfies q̃ ≺ q for all dominants of (3) is called the best dominant. If p (z) and φ � p (z) , zp ′ (z) ; z � are univalent in U and if p (z) satisfies first order differential superordination: h(z) ≺ φ � p (z) , zp ′ (z) ; z � , (4) then p (z) is a solution of the differential superordination (4). An analytic function q (z) is called a subordinant of the solutions of the differential superordination (4) if q (z) ≺ p (z) for all p (z) satisfying (4). A univalent subordinant q̃ that satisfies q ≺ q̃ for all subordinants of (4) is called the best subordinant.Using the results of Miller and Mocanu [17], Bulboaca [5] con- sidered certain classes of first order differential superordinations as well as superordination- preserving integral operators [6]. Ali et al. [1], have used the results of Bulboaca [5] to obtain sufficient conditions for normalized analytic functions to satisfy: q1(z) ≺ z f ′(z) f (z) ≺ q2(z), where q1 and q2 are given univalent functions in U with q1(0) = q2(0) = 1. Also, Tuneski [25] obtained a sufficient condition for starlikeness of f in terms of the quantity f ′′(z) f (z) ( f ′(z))2 . Recently, Shanmugam et al. [24] obtained sufficient conditions for the normalized analytic function f to satisfy q1(z)≺ f (z) z f ′(z) ≺ q2(z) and q1(z) ≺ z2 f ′(z) { f (z)}2 ≺ q2(z). They [24] also obtained results for functions defined by using Carlson-Shaffer operator [7], Ruscheweyh derivative [20] and Sălăgean operator [22]. For functions f given by (1) and g ∈ A given by g(z) = z + ∞ ∑ k=2 bkzk, the Hadamard product (or convolution) of f and g is defined by ( f ∗ g)(z) = z + ∞ ∑ k=2 ak bkzk = (g ∗ f )(z). (5) For functions f , g ∈A , we define the linear operator Dn λ :A →A (λ≥ 0, n ∈ N0 = N∪ {0},N = {1,2, . . .}) by: D0 λ( f ∗ g)(z) = ( f ∗ g)(z) , M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 3 D1 λ( f ∗ g)(z) = Dλ( f ∗ g)(z) = (1−λ )( f ∗ g)(z) +λz ( ( f ∗ g)(z))′ , (6) and ( in general ) Dn λ( f ∗ g)(z) = Dλ(D n−1 λ ( f ∗ g)(z)) = z + ∞ ∑ k=2 [1+λ(k− 1)]nak bkzk � λ≥ 0; n ∈ N0 � . (7) From (7), we can easily deduce that λz � Dn λ( f ∗ g)(z) �′ = Dn+1 λ ( f ∗ g)(z)− (1−λ)Dn λ( f ∗ g)(z) (λ > 0). (8) The linear operator Dn λ ( f ∗ g)(z) was introduced by Aouf and Seoudy [3] and we observe that Dn λ ( f ∗ g)(z) reduces to several interesting many other linear operators considered earlier for different choices of n, λ and the function g (z) : (i) For bk = 1 (or g(z) = z 1− z ), we have Dn λ ( f ∗ g)(z) = Dn λ f (z), where Dn λ is the gener- alized Sălăgean operator ( or Al-Oboudi operator [2] which yield Sălăgean operator Dn for λ= 1 introduced and studied by Sălăgean [21]; (ii) For n= 0 and g(z) = z + ∞ ∑ k=2 (a1)k−1 . . . (al)k−1 (b1)k−1 . . . (bm)k−1(1)k−1 zk (9) � ai ∈ C; i = 1, . . . , l; b j ∈ C\Z − 0 = {0,−1,−2, . . .} ; j = 1, . . . , m; l ≤ m+ 1; l, m ∈ N0; z ∈ U � , where (x)k = ¨ 1 (k = 0; x ∈ C∗ = C\{0}) x(x + 1) . . . (x + k− 1) (k ∈ N ; x ∈ C), we have D0 λ ( f ∗ g)(z) = ( f ∗ g)(z) = Hl ,m � a1; b1 � f (z), where the operator Hl ,m � a1; b1 � is the Dziok-Srivastava operator introduced and studied by Dziok and Srivastava [10] ([see also 11, 12]). The operator Hl ,m � a1; b1 � , contains in turn many interesting opera- tors such as, Hohlov linear operator (see [13]), the Carlson-Shaffer linear operator (see [7, 21]), the Ruscheweyh derivative operator (see [20]), the Bernardi-Libera-Livingston operator ( see [4, 14, 15]) and Owa-Srivastava fractional derivative operator (see [19]); (iii) For n= 0 and g(z) = z + ∞ ∑ k=2 � 1+ l +λ(k− 1) 1+ l �s zk (λ≥ 0; l, s ∈ N0), (10) we see that D0 λ ( f ∗ g)(z) = ( f ∗ g)(z) = I(s,λ, l) f (z), where I(s,λ, l) is the generalized multiplier transformations which was introduced and studied by Cătaş et al. [8]. The operator I(s,λ, l), contains as special cases, the multiplier transformation I(s, l) (see [9]) for λ = 1, the generalized Sălăgean operator Dn λ introduced and studied by Al- Oboudi [2] which in turn contains as special case the Sălăgean operator Dn (see [21]); M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 4 (iv) For g(z) of the form (9), the operator Dn λ ( f ∗ g)(z) = Dn λ (a1, b1) f (z), introduced and studied by Selvaraj and Karthikeyan [23]. In this paper, we will derive several subordination results, superordination results and sandwich results involving the operator Dn λ ( f ∗ g)(z) and some of its special operators by some choices of n,λ and the function g(z). 2. Preliminaries In order to prove our subordinations and superordinations, we need the following defini- tion and lemmas. Definition 1. [17] Denote by Q, the set of all functions f that are analytic and injective on U\E( f ), where E( f ) = � ζ ∈ ∂ U : lim z→ζ f (z) =∞ � , and are such that f ′ (ζ) 6= 0 for ζ ∈ ∂ U\E � f � . Lemma 1. [17] Let q (z) be univalent in the unit disk U and θ and ϕ be analytic in a domain D containing q(U) with ϕ (w) 6= 0 when w ∈ q(U). Set ψ (z) = zq ′ (z)ϕ � q (z) � and h(z) = θ � q (z) � +ψ (z) . (11) Suppose that (i) ψ (z) is starlike univalent in U, (ii) ℜ ( zh ′ (z) ψ (z) ) > 0 for z ∈ U. If p (z) is analytic with p(0) = q(0), p(U)⊂ D and θ � p (z) � + zp ′ (z)ϕ � p (z) � ≺ θ � q (z) � + zq ′ (z)ϕ � q (z) � , (12) then p(z) ≺ q(z) and q (z) is the best dominant. Taking θ (w) = αw and ϕ (w) = γ in Lemma 1, Shanmugam et al. [24] obtained the following lemma. Lemma 2. [24] Let q (z) be univalent in U with q(0) = 1. Let α ∈ C; γ ∈ C∗, further assume that ℜ ( 1+ zq ′′ (z) q ′ (z) ) >max � 0,−ℜ � α γ �� . (13) If p (z) is analytic in U, and αp (z) + γzp ′ (z) ≺ αq (z) + γzq ′ (z) , then p (z)≺ q (z) and q (z) is the best dominant. M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 5 Lemma 3. [5] Let q (z) be convex univalent in U and ϑ and φ be analytic in a domain D containing q(U). Suppose that (i) ℜ � ϑ ′ (q(z)) φ(q(z)) � > 0 for z ∈ U, (ii) Ψ(z) = zq ′ (z)φ � q (z) � is starlike univalent in U. If p(z) ∈ H[q(0), 1]∩Q, with p(U) ⊆ D, and ϑ � p (z) � + zp ′ (z)φ � p (z) � is univalent in U and ϑ � q (z) � + zq ′ (z)φ � q (z) � ≺ ϑ � p (z) � + zp ′ (z)φ � p (z) � , (14) then q(z)≺ p(z) and q (z) is the best subordinant. Taking ϑ (w) = αw and φ (w) = γ in Lemma 3, Shanmugam et al. [24] obtained the following lemma. Lemma 4. [24] Let q (z) be convex univalent in U , q(0) = 1. Let α ∈ C; γ ∈ C∗ and ℜ � α γ � > 0. If p(z) ∈ H[q(0), 1]∩Q, αp (z) + γzp ′ (z) is univalent in U and αq (z) + γzq ′ (z)≺ αp (z) + γzp ′ (z) , then q (z) ≺ p (z) and q (z) is the best subordinant. 3. Sandwich Results Unless otherwise mentioned, we assume throughout this paper that λ > 0 and n ∈ N0. Theorem 1. Let q (z) be univalent in U with q(0) = 1, and γ ∈ C∗. Further, assume that ℜ ( 1+ zq ′′ (z) q ′ (z) ) >max � 0,−ℜ � 1 γ �� . (15) If f , g ∈A satisfy the following subordination condition: � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    ≺ q (z) + γzq ′ (z) , (16) then zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ≺ q (z) and q (z) is the best dominant. M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 6 Proof. Define a function p (z) by p (z) = zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 (z ∈ U) . (17) Then the function p (z) is analytic in U and p(0) = 1. Therefore, differentiating (17) logarith- mically with respect to z and using the identity (8) in the resulting equation, we have � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    = p (z)+γzp ′ (z) , that is, p (z) + γzp ′ (z) ≺ q (z) + γzq ′ (z) . Therefore, Theorem 1 now follows by applying Lemma 2. Putting q(z) = 1+ Az 1+ Bz (−1≤ B < A≤ 1) in Theorem 1, we obtain the following corollary. Corollary 1. Let γ ∈ C∗ and ℜ � 1− Bz 1+ Bz � >max � 0,−ℜ � 1 γ �� . If f , g ∈A satisfy the following subordination condition: � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    ≺ 1+A z 1+ Bz + γ (A− B) z (1+ Bz)2 , then zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ≺ 1+ Az 1+ Bz and the function 1+ Az 1+ Bz is the best dominant. Remark 1. Taking g(z) = z 1− z in Theorem 1, we obtain the subordination result of Nechita [18, Theorem 14]. Remark 2. Taking λ = 1 and g(z) = z 1− z in Theorem 1, we obtain the subordination result for Sălăgean operator which was obtained by Shanmugam et al. [24, Theorem 5.4] and also obtained by Nechita [18, Corollary 16]. M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 7 Taking n = 0,λ = 1 and g (z) of the form (9) in Theorem 1, we obtain the following subordination result for Dziok-Srivastava operator. Corollary 2. Let q (z) be univalent in U with q(0) = 1, and γ ∈ C∗. Further assume that (15) holds. If f ∈A satisfies the following subordination condition: z2 � Hl ,m � a1; b1 � f (z) �′ � Hl ,m � a1; b1 � f (z) �2 − γz2 z � Hl ,m � a1; b1 � f (z) � !′′ ≺ q (z) + γzq ′ (z) , then z2 � Hl ,m � a1; b1 � f (z) �′ � Hl ,m � a1; b1 � f (z) �2 ≺ q (z) and q (z) is the best dominant. Taking g (z) of the form (9) in Theorem 1, we obtain the following subordination result for the operator Dn λ (a1; b1). Corollary 3. Let q (z) be univalent in U with q(0) = 1, and γ ∈ C∗. Further assume that [15] holds. If f ∈A satisfies the following subordination condition: � 1+ γ λ � zDn+1 λ (a1; b1) f (z)) � Dn λ (a1; b1) f (z) �2 + γ λ    zDn+2 λ (a1; b1) f (z) � Dn λ (a1; b1) f (z) �2 − 2 z � Dn+1 λ (a1; b1) f (z) �2 � Dn λ (a1; b1) f (z) �3    ≺ q (z) + γzq ′ (z) , then zDn+1 λ (a1; b1) f (z)) � Dn λ (a1; b1) f (z) �2 ≺ q (z) and q (z) is the best dominant. Taking n= 0,λ= 1 and g(z) = z + ∞ ∑ k=2 � l + k 1+ l �s zk (l, s ∈ N0), (18) in Theorem 1, we obtain the following subordination result for the multiplier transformations I(s, l). Corollary 4. Let q (z) be univalent in U with q(0) = 1, and γ ∈ C∗. Further assume that (15) holds. If f ∈A satisfies the following subordination condition: z2 � I(s, l) f (z) �′ � I(s, l) f (z) �2 − γz2 � z I(s, l) f (z) �′′ ≺ q (z) + γzq ′ (z) , M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 8 then z2 � I(s, l) f (z) �′ � I(s, l) f (z) �2 ≺ q (z) and q (z) is the best dominant. Remark 3. Taking n = 0,λ = 1 and g(z) = z 1− z in Theorem 1, we obtain the subordination result of Shanmugam et al. [24, Theorem 3.4] and also obtained by Nechita [18, Corollary 17]. Now, by appealing to Lemma 4 it can be easily prove the following theorem. Theorem 2. Let q (z) be convex univalent in U with q (0) = 1. Let γ ∈ C with ℜ � γ � > 0. If f , g ∈A , zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ∈ H [1,1]∩Q, � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    is univalent in U, and the following superordination condition q (z)+γzq ′ (z) ≺ � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    holds, then q (z) ≺ zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 and q (z) is the best subordinant. Taking q(z) = 1+ Az 1+ Bz (−1≤ B < A≤ 1) in Theorem 2, we have the following corollary. Corollary 5. Let γ ∈ C with ℜ � γ � > 0. If f , g ∈A , zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ∈ H [1,1]∩Q, � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    is univalent in U, and the following superordination condition 1+ Az 1+ Bz +γ (A− B) z (1+ Bz)2 ≺ � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 9 holds, then 1+ Az 1+ Bz ≺ zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 and q (z) is the best subordinant. Remark 4. Taking g(z) = z 1− z in Theorem 2, we obtain the superordination result of Nechita [18, Theorem 19]. Remark 5. Taking λ = 1 and g(z) = z 1− z in Theorem 2, we obtain the following superordina- tion result for Sălăgean operator which is obtained Shanmugam et al. [24, Theorem 5.5]. Taking n = 0,λ = 1 and g (z) of the form (9) in Theorem 2, we obtain the following superordination result for Dziok-Srivastava operator. Corollary 6. Let q (z) be convex univalent in U with q (0) = 1. Let γ ∈ C with ℜ � γ � > 0. If f ∈A , z2 � Hl ,m � a1; b1 � f (z) �′ � Hl ,m � a1; b1 � f (z) �2 ∈ H [1,1]∩Q, z2 � Hl ,m � a1; b1 � f (z) �′ � Hl ,m � a1; b1 � f (z) �2 − γz2 z � Hl ,m � a1; b1 � f (z) � !′′ is univalent in U, and the following superordination condition q (z) + γzq ′ (z)≺ z2 � Hl ,m � a1; b1 � f (z) �′ � Hl ,m � a1; b1 � f (z) �2 − γz2 z � Hl ,m � a1; b1 � f (z) � !′′ holds, then q (z) ≺ z2 � Hl ,m � a1; b1 � f (z) �′ � Hl ,m � a1; b1 � f (z) �2 and q (z) is the best subordinant. Taking g (z) of the form (9) in Theorem 2, we obtain the following superordination result for the operator Dn λ (a1; b1). Corollary 7. Let q (z) be convex univalent in U with q (0) = 1. Let γ ∈ C with ℜ � γ � > 0. If f , g ∈A , zDn+1 λ (a1; b1) f (z) � Dn λ (a1; b1) f (z) �2 ∈ H [1,1]∩Q, � 1+ γ λ � zDn+1 λ (a1; b1) f (z) � Dn λ (a1; b1) f (z) �2 + γ λ    zDn+2 λ (a1; b1) f (z) � Dn λ (a1; b1) f (z) �2 − 2 z � Dn+1 λ (a1; b1) f (z) �2 � Dn λ (a1; b1) f (z) �3    M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 10 is univalent in U, and the following superordination condition q (z)+γzq ′ (z) ≺ � 1+ γ λ � zDn+1 λ (a1; b1) f (z) � Dn λ (a1; b1) f (z) �2 + γ λ    zDn+2 λ (a1; b1) f (z) � Dn λ (a1; b1) f (z) �2 − 2 z � Dn+1 λ (a1; b1) f (z) �2 � Dn λ (a1; b1) f (z) �3    holds, then q (z) ≺ zDn+1 λ (a1; b1) f (z) � Dn λ (a1; b1) f (z) �2 and q (z) is the best subordinant. Taking n = 0,λ = 1 and g(z) of the form (18) in Theorem 2, we obtain the following supordination result for the multiplier transformations I(s, l). Corollary 8. Let q (z) be convex univalent in U with q (0) = 1. Let γ ∈ C with ℜ � γ � > 0. If f ∈A , z2 � I(s, l) f (z) �′ � I(s, l) f (z) �2 ∈ H [1,1]∩Q, z2 � I(s, l) f (z) �′ � I(s, l) f (z) �2 − γz2 � z I(s, l) f (z) �′′ is univalent in U, and the following superordination condition q (z) + γzq ′ (z)≺ z2 � I(s, l) f (z) �′ � I(s, l) f (z) �2 − γz2 � z I(s, l) f (z) �′′ holds, then q (z) ≺ z2 � I(s, l) f (z) �′ � I(s, l) f (z) �2 and q (z) is the best subordinant. Remark 6. Taking n= 0,λ= 1 and g(z) = z 1− z in Theorem 2, we obtain the superordination result of Shanmugam et al. [24, Theorem 3.5]. Combining Theorem 1 and Theorem 2, we get the following sandwich theorem for the linear operator Dn λ ( f ∗ g). Theorem 3. Let q1 (z) be convex univalent in U with q1 (0) = 1, γ ∈ C with ℜ � γ � > 0,q2 (z) be univalent in U with q2 (0) = 1, and satisfies (15). If f , g ∈A , zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ∈ H [1,1]∩Q, � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    M. Aouf, T. Seoudy / Eur. J. Pure Appl. Math, 4 (2011), 1-13 11 is univalent in U, and q1 (z) + γzq ′ 1 (z) ≺ � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    ≺ q2 (z) + γzq ′ 2 (z) holds, then q1 (z) ≺ zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ≺ q2 (z) and q1 (z) and q2 (z) are, respectively, the best subordinant and the best dominant. Taking qi(z) = 1+ Aiz 1+ Biz � i = 1,2;−1≤ B2 ≤ B1 < A1 ≤ A2 ≤ 1 � in Theorem 3, we have the following corollary. Corollary 9. Let γ ∈ C with ℜ � γ � > 0. If f , g ∈A , zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ∈ H [1,1]∩Q, � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    is univalent in U, and 1+ A1z 1+ B1z + γ � A1 − B1 � z � 1+ B1z �2 ≺ � 1+ γ λ � zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 + γ λ    zDn+2 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 − 2 z � Dn+1 λ ( f ∗ g)(z) �2 � Dn λ ( f ∗ g)(z) �3    ≺ 1+ A2z 1+ B2z + γ � A2 − B2 � z � 1+ B2z �2 holds, then 1+ A1z 1+ B1z ≺ zDn+1 λ ( f ∗ g)(z) � Dn λ ( f ∗ g)(z) �2 ≺ 1+ A2z 1+ B2z and 1+ A1z 1+ B1z and 1+A2z 1+ B2z are, respectively, the best subordinant and the best dominant. REFERENCES 12 Remark 7. Taking g(z) = z 1− z in Theorem 3, we obtain sandwich result of Nechita [18, Theorem 19]. Remark 8. Taking λ = 1 and g(z) = z 1− z in Theorem 3, we obtain sandwich result of Shan- mugam et al. [24, Theorem 5.6]. Remark 9. 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