Analytic Functions Involving Complex Order

and Applied Analysis 3 Definition 1.1 Hadamard product or convolution . For functions f and g in the class A, where f z of the form 1.1 and g z is given by g z z ∞ ∑ k 2 ckz , 1.8 the Hadamard product or convolution f ∗ g z is defined by ( f ∗ g z z ∞ ∑ k 2 akbkz k ( g ∗ f z , z ∈ U. 1.9 Definition 1.2 subordination principle . For analytic functions g and hwith g 0 h 0 , g is said to be subordinate to h, denoted by g ≺ h, if there exists an analytic function w such that w 0 0, |w z | < 1, g z h w z , 1.10 for all z ∈ U. Definition 1.3 see 12 , subordinating factor sequence . A sequence {bk}k 1 of complex numbers is said to be a subordinating sequence if, whenever f z is of the form 1.1 is analytic, univalent, and convex in U, one has the subordination given by ∞ ∑ k 1 bkakz k ≺ f z ∞ ∑ k 1 akz , z ∈ U, a1 1. 1.11 Lemma 1.4 see 12 . The sequence {bk}k 1 is a subordinating factor sequence for the class K of convex univalent functions if and only if Re { 1 2 ∞ ∑ k 1 bkz k } > 0, z ∈ U. 1.12 2. Main Results Theorem 2.1. Let the function f of the form 1.1 satisfy the following condition: ∞ ∑ n 2 1 |B| 1 λ k − 1 k|ak| ≤ A − B |b|. 2.1 Then f ∈ Gn λ, b,A, B . 4 Abstract and Applied Analysis Proof. Suppose the inequality 2.1 holds. Then we have for z ∈ U ∣∣∣∣ 1 − λ Df z z λ ( Df z )′ − 1 ∣∣∣∣ − ∣∣∣∣ A − B b − B ( 1 − λ D f z z λ ( Df z )′ − 1 )∣∣∣∣ ∣∣∣∣ ∞ ∑ k 2 1 λ k − 1 knakzk−1 ∣∣∣∣ − ∣∣∣∣ A − B b − B ∞ ∑ k 2 1 λ k − 1 knakzk−1 ∣∣∣∣ ≤ ∞ ∑ k 2 1 λ k − 1 kn|ak||z| − { A − B |b| − |B| ∞ ∑ k 2 1 λ k − 1 kn|ak||z| } ≤ ∞ ∑ k 2 1 |B| 1 λ k − 1 k|ak| − A − B |b| ≤ 0 2.2 which shows that f belongs to the class Gn λ, b,A, B . In view of Theorem 2.1, we now introduce the subclass Gn λ, b,A, B which consists of functions f ∈ A whose Taylor-Maclaurin coefficients satisfy the inequality 2.1 . We note that Gn λ, b,A, B ⊂ Gn λ, b,A, B . In this work, we prove several subordination relationships involving the function class Gn λ, b,A, B employing the technique used earlier by Attiya 13 and Srivastava and Attiya 14 . Theorem 2.2. Let f ∈ Gn λ, b,A, B , and let g z be any function in the usual class of convex functions K, then 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| ( f ∗ g z ≺ g z 2.3 for every function g ∈ K. Further, Re { f z } > − 1 λ 1 |B| 2 n A − B |b| 1 λ 1 |B| 2n , z ∈ U. 2.4 The constant factor 1 λ 1 |B| 2/2 1 λ 1 |B| 2 A − B |b| in 2.3 cannot be replaced by a larger number. Proof. Let f ∈ Gn λ, b,A, B , and suppose that g z z ∑∞ k 2 ckz k ∈ K. Then 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| ( f ∗ g z 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| ( z ∞ ∑ k 2 ckakz k ) . 2.5 Abstract and Applied Analysis 5 Thus, by Definition 1.3, the subordination result holds true if { 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| ak }∞ k 1 2.6and Applied Analysis 5 Thus, by Definition 1.3, the subordination result holds true if { 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| ak }∞ k 1 2.6 is a subordinating factor sequence, with a1 1. In view of Lemma 1.4, this is equivalent to the following inequality: Re { 1 2 ∞ ∑ k 1 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| akz k } > 0, z ∈ U. 2.7 Since Ψ k 1 |B| 1 λ k − 1 k is an increasing function of k k ≥ 2 , we have, for |z| r < 1, Re { 1 ∞ ∑ k 1 2 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| akz k } Re { 1 1 λ 1 |B| 2 1 λ 1 |B| 2n A − B |b| z ∞ ∑ k 2 1 λ 1 |B| 2 1 λ 1 |B| 2n A − B |b| akz k } ≥ 1 − 1 λ 1 |B| 2 n 1 λ 1 |B| 2n A − B |b| r − 1 1 λ 1 |B| 2n A − B |b| × ∞ ∑ k 2 1 |B| 1 λ k − 1 k|ak|r ≥ 1 − 1 λ 1 |B| 2 n 1 λ 1 |B| 2n A − B |b| r − |b| A − B 1 λ 1 |B| 2n A − B |b| r 1 − r > 0, |z| r < 1, 2.8 where we have also made use of the assertion 2.1 of Theorem 2.1. This evidently proves the inequality 2.3 and hence also the subordination result 2.3 asserted by Theorem 2.2. The inequality 2.4 follows from 2.3 by taking g z z 1 − z z ∞ ∑ k 2 z ∈ K. 2.9 To prove the sharpness of the constant 1 λ 1 |B| 2/ 1 λ 1 |B| 2 A − B |b| , we consider the function F ∈ Gn λ, b,A, B defined by F z z − A − B |b| 1 λ 1 |B| 2n . 2.10 6 Abstract and Applied Analysis Thus, from 2.3 , we have 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| F z ≺ z 1 − z . 2.11 It is easily verified that min { Re ( 1 λ 1 |B| 2 2 1 λ 1 |B| 2n A − B |b| F z )} − 2 , z ∈ U. 2.12 This shows that the constant 1 λ 1 |B| 2/2 1 λ 1 |B| 2 A − B |b| cannot be replaced by any larger one. For the choices of A − 1 0 and B 1 0, we get the following corollary. Corollary 2.3. Let f ∈ Gn λ, b let and g z be any function in the usual class of convex functions K, then 1 λ 2 1 λ 1 |B| 2n |b| ( f ∗ g z ≺ g z , 2.13 where b ∈ C \ {0} and 0 ≤ λ < 1, Re { f z } > − 1 λ 2 n |b| 1 λ 2n , z ∈ U. 2.14 The constant factor 1 λ 2/2 1 λ 2 |b| in 2.13 cannot be replaced by a larger number. For the choices of A − 1 B 1 0 and n 0, one gets the following. Corollary 2.4. Let f ∈ G∗ λ, b , and let g z be any function in the usual class of convex functions K, then 1 λ 2 1 λ |b| ( f ∗ g z ≺ g z , 2.15 where b ∈ C \ {0} and 0 ≤ λ < 1, Re { f z } > − 1 λ |b| 1 λ , z ∈ U. 2.16 The constant factor 1 λ /2 1 λ |b| in 2.15 cannot be replaced by a larger number. For the choices of A − 1 B 1 0, n 0, and λ 0, one gets the following. Abstract and Applied Analysis 7 Corollary 2.5. Let f ∈ G∗ b , and g z let be any function in the usual class of convex functionsK, then 1 2 1 |b| ( f ∗ g z ≺ g z , 2.17and Applied Analysis 7 Corollary 2.5. Let f ∈ G∗ b , and g z let be any function in the usual class of convex functionsK, then 1 2 1 |b| ( f ∗ g z ≺ g z , 2.17 where b ∈ C \ {0} and 0 ≤ λ < 1, Re { f z } > − 1 |b| , z ∈ U. 2.18 The constant factor 1/2 1 |b| in 2.17 cannot be replaced by a larger number. For the choices ofA− 1 B 1 0, n 0, λ 0, and b 1− α, one gets the following. Corollary 2.6. Let f ∈ G∗ α , and let g z be any function in the usual class of convex functionsK, then 1 2 2 − α ( f ∗ g z ≺ g z , 2.19 where 0 ≤ α < 1, Re ( f z ) > − 2 − α , z ∈ U. 2.20 The constant factor 1/2 2 − α in 2.19 cannot be replaced by a larger number. For the choices of A − 1 B 1 0, n 0, λ 1, and b 1 − α 0 ≤ α < 1 , one gets the following. Corollary 2.7. Let f ∈ R∗ α , and g z let be any function in the usual class of convex functionsK, then 2 2 3 − α ( f ∗ g z ≺ g z 2.21