Subordination Properties of Multivalent Functions Defined by

Hindawi Publishing Corporation Journal of Complex Analysis Volume 2014, Article ID 187482, 12 pages http://dx.doi.org/10.1155/2014/187482

Research Article Subordination Properties of Multivalent Functions Defined by Generalized Multiplier Transformation M. P. Jeyaraman1 and T. K. Suresh2 1 2

Department of Mathematics, L. N. Government College, Ponneri, Chennai, Tamil Nadu 601 204, India Department of Mathematics, Easwari Engineering College, Chennai, Tamil Nadu 600 089, India

Correspondence should be addressed to M. P. Jeyaraman; jeyaraman [email protected] Received 31 August 2013; Accepted 11 November 2013; Published 25 February 2014 Academic Editor: Janne Heittokangas Copyright © 2014 M. P. Jeyaraman and T. K. Suresh. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. The main object of the present paper is to investigate several interesting subordination properties and a sharp inclusion relationship for certain subclass of multivalent analytic functions, which are defined here by the generalized multiplier transformation. Relevant connections of the results which are presented in this paper with various known results are also considered.

1. Introduction and Definitions Let A denote the class of analytic functions in the open unit disk U := {𝑧 : 𝑧 ∈ C , |𝑧| < 1} .

(1)

If 𝑓 and 𝑔 are in A, we say that the function 𝑓 is said to be subordinate to 𝑔 or (equivalently) 𝑔 is said to be superordinate to 𝑓, written symbolically as 𝑓≺𝑔

in U or 𝑓 (𝑧) ≺ 𝑔 (𝑧)

(𝑧 ∈ U) ,

(2)

if there exists a Schwarz function 𝑤 analytic in U, with 𝑤(0) = 0 and |𝑤(𝑧)| < 1, for all 𝑧 ∈ U, such that 𝑓 (𝑧) = 𝑔 (𝑤 (𝑧))

(𝑧 ∈ U) .

(3)

In particular, if the function 𝑔 is univalent in U, then we have the following equivalence (cf. [1, 2]): 𝑓 ≺ 𝑔 ⇐⇒ 𝑓 (0) = 𝑔 (0) ,

𝑓 (U) ⊂ 𝑔 (U) .

𝑓 (𝑧) = 𝑧𝑝 + ∑ 𝑎𝑘 𝑧𝑘 𝑘=𝑝+1

(𝑧 ∈ U) ,

∞

𝑔 (𝑧) = 𝑧𝑝 + ∑ 𝑏𝑘 𝑧𝑘 ,

(5)

(6)

𝑘=𝑝+1

the Hadamard product (or convolution) of 𝑓 and 𝑔 is given by ∞

(𝑓 ∗ 𝑔) (𝑧) := 𝑧𝑝 + ∑ 𝑎𝑘 𝑏𝑘 𝑧𝑘 =: (𝑔 ∗ 𝑓) (𝑧) . 𝑘=𝑝+1

(7)

Recently, C˘atas¸ [3] defined the generalized multiplier transformation 𝐼𝑝𝑚 (𝜆, 𝑙) on A(𝑝) by the following infinite series: ∞

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) = 𝑧𝑝 + ∑ ( 𝑘=𝑝+1

(4)

Let A(𝑝) (𝑝 ∈ N = {1, 2, . . .}) be the subclass of A consisting of functions 𝑓 defined by ∞

which are analytic and 𝑝-valent in the open unit disk U. We note that A(1) ≡ A. For a function 𝑓 ∈ A(𝑝) given by (5) and 𝑔 ∈ A(𝑝) defined by

𝑚

𝑝 + 𝜆 (𝑘 − 𝑝) + 𝑙 ) 𝑎𝑘 𝑧𝑘 𝑝+𝑙

(8)

(𝜆 > 0, 𝑙 ≥ 0, 𝑝 ∈ N, 𝑚 ∈ N0 = N ∪ {0} , 𝑧 ∈ U) . We note that 𝐼𝑝0 (1, 0) 𝑓 (𝑧) = 𝑓 (𝑧) ,

𝐼𝑝1 (1, 0) 𝑓 (𝑧) =

𝑧𝑓󸀠 (𝑧) . 𝑝

(9)

2

Journal of Complex Analysis

It is easily verified from (8) that 󸀠

𝜆𝑧(𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓) (𝑧) = (𝑝 + 𝑙) 𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓 (𝑧) − [𝑝 (1 − 𝜆) + 𝑙] 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) . (10) The generalized multiplier transformation 𝐼𝑝𝑚 (𝜆, 𝑙) reduces several familiar operators by specializing the parameters 𝑚, 𝜆, 𝑙, and 𝑝. (1) For the choice of 𝜆 = 1, the operator defined by (8) reduces the operator 𝐼𝑝 (𝑚, 𝑙), studied by Srivastava et al. [4] and Sivaprasad Kumar et al. [5]. (2) By taking 𝜆 = 𝑝 = 1, the generalized multiplier transformation 𝐼𝑝𝑚 (𝜆, 𝑙) yields the operator 𝐼𝑙𝑚 , which was investigated by Cho et al. [6, 7]. (3) For 𝜆 = 1 and 𝑙 = 0, the operator 𝐼𝑝𝑚 (𝜆, 𝑙) reduces the differential operator 𝐷𝑝𝑚 studied by Kamali and Orhan [8] and Orhan and Kiziltunc¸ [9] and also, for 𝜆 = 𝑝 = 1 and 𝑙 = 0, it yields the differential operator 𝐷𝑚 introduced by S˘al˘agean [10]. (4) As a special case of this operator 𝐼𝑝𝑚 (𝜆, 𝑙) for 𝑙 = 0 and 𝑝 = 1, it reduces the generalized S˘al˘agean operator 𝐷𝜆𝑚 studied by Al-Oboudi [11] and also earlier, for 𝜆 = 𝑙 = 𝑝 = 1, it gives the operator 𝐼𝑚 investigated by Uralegaddi and Somanatha [12]. Now, we introduce a new subclass of functions in A(𝑝), by making use of the generalized multiplier transformation 𝐼𝑝𝑚 (𝜆, 𝑙) as follows. Definition 1. Let 𝑚 ∈ N0 , 𝐴, 𝐵, 𝛼, 𝜆, 𝑙 be arbitrary fixed real numbers such that −1 ≤ 𝐵 < 𝐴 ≤ 1, 0 ≤ 𝛼 < 𝑝, 𝜆 > 0, 𝑝 ∈ N, and 𝑙 ≥ 0. A function 𝑓 ∈ A(𝑝) is said to be in the class R𝑚 𝑝 (𝜆, 𝑙, 𝛼; 𝐴, 𝐵), if it satisfies the following subordination condition:

the multivalent analytic function class R𝑚 𝑝 (𝜆, 𝑙, 𝛼; 𝐴, 𝐵). We also derive a number of sufficient conditions for functions belonging to the subclass R𝑚 𝑝 (𝜆, 𝑙, 𝛼) which satisfy certain subordination properties. Relevant connections of the results presented in this paper with earlier sequels are also pointed out.

2. Preliminaries To prove our results, we will need the following lemmas. Lemma 2 (see [1, 2]). Let a function ℎ be analytic and convex (univalent) in U, with ℎ(0) = 1. Suppose also that the function 𝜑 given by 𝜑 (𝑧) = 1 + 𝑏1 𝑧 + 𝑏2 𝑧2 + ⋅ ⋅ ⋅

(13)

is analytic in U. If 𝜑 (𝑧) +

𝑧𝜑󸀠 (𝑧) ≺ ℎ (𝑧) 𝑐

(Re 𝑐 ≥ 0, 𝑐 ≠ 0) ,

(14)

𝑐 𝑧 𝑐−1 ∫ 𝑡 ℎ (𝑡) 𝑑𝑡 ≺ ℎ (𝑧) , 𝑧𝑐 0

(15)

then 𝜑 (𝑧) ≺ 𝜓 (𝑧) =

where 𝜓 is the best dominant of (14). We denote by 𝑃(𝛾) the class of functions 𝜑 given by (13) which are analytic in U and satisfy the following inequality: Re 𝜑 (𝑧) > 𝛾,

(0 ≤ 𝛾 < 1, 𝑧 ∈ U) .

(16)

Lemma 3 (see [15]). Let the function 𝜑 given by (13) be in the class 𝑃(𝛾). Then

󸀠

𝑧(𝐼𝑝𝑚 (𝜆, 𝑙)𝑓) (𝑧) 1 1 + 𝐴𝑧 ( 𝑚 − 𝛼) ≺ , 𝑝−𝛼 𝐼𝑝 (𝜆, 𝑙) 𝑓 (𝑧) 1 + 𝐵𝑧

(𝑧 ∈ U) .

Re 𝜑 (𝑧) ≥ 2𝛾 − 1 +

2 (1 − 𝛾) 1 + |𝑧|

(0 ≤ 𝛾 < 1, 𝑧 ∈ U) .

(17)

(11) In particular, for 𝐴 = 1 and 𝐵 = −1, we write R𝑚 𝑝 (𝜆, 𝑙, 𝛼; 1, −1) = R𝑚 (𝜆, 𝑙, 𝛼), where 𝑝

Lemma 4 (see [16]). For 0 ≤ 𝛾1 < 𝛾2 < 1, 𝑃 (𝛾1 ) ∗ 𝑃 (𝛾2 ) ⊂ 𝑃 (𝛾3 ) ,

R𝑚 𝑝 (𝜆, 𝑙, 𝛼)

𝑤ℎ𝑒𝑟𝑒 𝛾3 = 1 − 2 (1 − 𝛾1 ) (1 − 𝛾2 ) .

(18)

󸀠

𝑧(𝐼𝑝𝑚 (𝜆, 𝑙)𝑓) (𝑧) { } = {𝑓 ∈ A (𝑝) : Re ( 𝑚 ) > 𝛼, 𝑧 ∈ U} . 𝐼𝑝 (𝜆, 𝑙) 𝑓 (𝑧) { } (12) Motivated by the recent work of Bulboac˘a et al. [13] and Patel and Mishra [14], we investigate the subordination properties of the generalized multiplier transformation 𝐼𝑝𝑚 (𝜆, 𝑙) defined by (8) and obtain a sharp inclusion relationship for

The result is the best possible. For any complex numbers 𝑎, 𝑏, and 𝑐 (𝑐 ∉ Z−0 := {0, −1, −2, . . .}), the Gaussian hypergeometric function is defined by 2 𝐹1

(𝑎, 𝑏; 𝑐; 𝑧) = 1 +

𝑎𝑏 𝑧 𝑎 (𝑎 + 1) 𝑏 (𝑏 + 1) 𝑧2 + + ⋅⋅⋅ . 𝑐 1! 𝑐 (𝑐 + 1) 2! (19)

Journal of Complex Analysis

3

Lemma 5 (see [17]). For any complex numbers 𝑎, 𝑏, c (𝑐 ∉ Z−0 ), one has

then Re (

1

∫ 𝑡𝑏−1 (1 − 𝑡)𝑐−𝑏−1 (1 − 𝑧𝑡)−𝑎 𝑑𝑡 0

Re 𝑐 > Re 𝑏 > 0, (𝑎, 𝑏; 𝑐; 𝑧) = (1 − 𝑧)−𝑎 2 𝐹1 (𝑎, 𝑐 − 𝑏; 𝑐;

𝑔 (𝑧) ≺ 1 +

𝑧 ), 𝑧−1

(𝑏 + 1) 2 𝐹1 (1, 𝑏; 𝑏+1; 𝑧) = (𝑏 + 1) + 𝑏𝑧 2 𝐹1 (1, 𝑏 + 1; 𝑏 + 2; 𝑧) . (20) Lemma 6 (see [18]). If −1 ≤ 𝐵 < 𝐴 ≤ 1, 𝛽 > 0, and the complex number 𝛾 is constrained by Re(𝛾) ≥ −𝛽(1−𝐴)/(1−𝐵), then the following differential equation: (21)

If the function 𝜑(𝑧) given by (13) is analytic in U and satisfies the following subordination: (𝑧 ∈ U) ,

(23)

(𝑧 ∈ U) ,

Lemma 9 (see [2]). Suppose that the function Ψ : C2 × U → C satisfies the following condition: (31)

for all 𝑥 ∈ R and 𝑦 ≤ −(1/2)(1 + 𝑥2 ) and for all 𝑧 ∈ U. If the function 𝜑 of the form (13) is analytic in U and Re Ψ (𝜑 (𝑧) , 𝑧𝜑󸀠 (𝑧) ; 𝑧) > 𝜀,

Re (𝜑 (𝑧)) > 0

(𝑧 ∈ U) .

Unless otherwise mentioned, we assume throughout this paper that 𝑙 ≥ 0,

𝑚 ∈ N0 ,

0 ≤ 𝛼 < 𝑝,

𝑝 ∈ N,

−1 ≤ 𝐵 < 𝐴 ≤ 1.

(34)

Theorem 10. Let 𝛽 > 0 and −1 ≤ 𝐵𝑗 < 𝐴 𝑗 ≤ 1, 𝑗 = 1, 2. If the functions 𝑓𝑗 ∈ A(𝑝) satisfy the following subordination condition: (1 − 𝛽)

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓𝑗 (𝑧) 𝑧𝑝

+𝛽

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓𝑗 (𝑧) 𝑧𝑝

≺

1 + 𝐴 𝑗𝑧 1 + 𝐵𝑗 𝑧

,

𝑗 = 1, 2, (35) then

If the function 𝐻(𝑧) is defined by

0

(33)

3. Subordination Properties of 𝐼𝑝𝑚 (𝜆,𝑙)

(|𝑧| ≤ 𝑟 < 1; 𝑡 ∈ [0, 1]) . (25)

𝐻 (𝑧) = ∫ ℎ (𝑧, 𝑡) 𝑑V (𝑡) ,

(32)

then

(24)

Lemma 7 (see [19]). Let V be a positive measure on the interval [0, 1]. Let ℎ(𝑧, 𝑡) be a complex-valued function defined on U × [0, 1] such that ℎ(⋅, 𝑡) is analytic in U for each 𝑡 ∈ [0, 1] and ℎ(𝑧, ⋅) is V-integrable on [0, 1] for each 𝑧 ∈ U. In addition, suppose that Re(ℎ(𝑧, 𝑡)) > 0, ℎ(−𝑟, 𝑡) is real, and

1

(30)

then Re 𝑃(𝑧) > 0 for 𝑧 ∈ U.

𝜆 > 0,

and 𝑞(𝑧) is the best dominant of (23).

1 1 )≥ Re ( ℎ (𝑧, 𝑡) ℎ (−𝑟, 𝑡)

(29)

𝑔 (𝑧) {1 − 𝜆 + 𝜆 [(1 − 𝛽) 𝑃 (𝑧) + 𝛽]} ≺ 1 + 𝑀𝑧,

then 1 + 𝐴𝑧 𝜑 (𝑧) ≺ 𝑞 (𝑧) ≺ 1 + 𝐵𝑧

(28)

Re Ψ (𝑖𝑥, 𝑦; 𝑧) ≤ 𝜀,

𝛾 𝑧𝛽+𝛾 (1 + 𝐵𝑧)𝛽(𝐴−𝐵)/𝐵 { { − , 𝑖𝑓 𝐵 ≠ 0, { 𝑧 𝛽+𝛾−1 𝛽(𝐴−𝐵)/𝐵 { {𝛽 ∫ 𝑡 𝑑𝑡 𝛽 (1 + 𝐵𝑡) 𝑞 (𝑧) = { 0 𝛽+𝛾 { 𝑧 exp (𝛽𝐴𝑧) 𝛾 { { − , 𝑖𝑓 𝐵 = 0. { 𝑧 𝛽+𝛾−1 𝛽 exp (𝛽𝐴𝑡) 𝑑𝑡 { 𝛽 ∫0 𝑡 (22)

𝑧𝜑󸀠 (𝑧) 1 + 𝐴𝑧 ≺ 𝛽𝜑 (𝑧) + 𝛾 1 + 𝐵𝑧

(𝑛 ∈ N) ,

If 𝑃(𝑧) = 1 + 𝑑𝑛 𝑧𝑛 + 𝑑𝑛+1 𝑧𝑛+1 + ⋅ ⋅ ⋅ is analytic in U and satisfies the subordination relation

has a univalent solution in U given by

𝜑 (𝑧) +

𝑎𝑀𝑧 𝑛𝜆 + 𝑎

(1 − 𝛽) |𝜆| (1 + 𝑛𝜆/𝑎) . 󵄨 󵄨󵄨 2 󵄨󵄨1 − 𝜆 + 𝜆𝛽󵄨󵄨󵄨 + √1 + (1 + 𝑛𝜆/𝑎)

𝑀=

2 𝐹1 (𝑎, 𝑏; 𝑐; 𝑧) = 2 𝐹1 (𝑏, 𝑎; 𝑐; 𝑧) ,

(𝑧 ∈ U)

(27)

where

𝑧 ∉ (1, ∞) ,

𝑧𝑞󸀠 (𝑧) 1 + 𝐴𝑧 𝑞 (𝑧) + = 𝛽𝑞 (𝑧) + 𝛾 1 + 𝐵𝑧

(|𝑧| ≤ 𝑟 < 1) .

Lemma 8 (see [20]). Let 𝜆 ≠ 0 be a real number, 𝑎/𝜆 > 0, and 0 ≤ 𝛽 < 1. Let 𝑔(𝑧) = 1 + 𝑐𝑛 𝑧𝑛 + 𝑐𝑛+1 𝑧𝑛+1 + ⋅ ⋅ ⋅ be analytic in U and

Γ (𝑏) Γ (𝑐 − 𝑏) = 2 𝐹1 (𝑎, 𝑏; 𝑐; 𝑧) , Γ (𝑐)

2 𝐹1

1 1 )≥ 𝐻 (𝑧) 𝐻 (−𝑟)

(1 − 𝛽) (26)

𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹 (𝑧) 𝑧𝑝

+𝛽

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝐹 (𝑧) 𝑧𝑝

≺

1 + (1 − 2𝛿) 𝑧 , 1−𝑧 (36)

4

Journal of Complex Analysis

where 𝐹 = 𝐼𝑝𝑚 (𝜆, 𝑙)(𝑓1 ∗ 𝑓2 ) and

Since 𝜑𝑗 ∈ 𝑃(𝛾𝑗 ), 𝑗 = 1, 2, it follows from Lemma 4 that 𝜑1 ∗ 𝜑2 ∈ 𝑃 (𝛾3 ) ,

4 (𝐴 1 − 𝐵1 ) (𝐴 2 − 𝐵2 ) (1 − 𝐵1 ) (1 − 𝐵2 )

𝛿=1−

𝑝+𝑙 1 1 × (1 − 2 𝐹1 (1, 1; + 1; )) . 2 𝜆𝛽 2

(37)

and the bound 𝛾3 is the best possible. Hence, by using Lemma 3 in (43), we deduce that Re 𝜑0 (𝑧)

The result is the best possible when 𝐵1 = 𝐵2 = −1. Proof. Let the functions 𝑓𝑗 ∈ A(𝑝), 𝑗 = 1, 2, satisfy the subordination condition (35). Then, by setting

𝜑𝑗 (𝑧) = (1 − 𝛽) ≺

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓𝑗 (𝑧) 𝑧𝑝

1 + 𝐴 𝑗𝑧 1 + 𝐵𝑗 𝑧

,

+𝛽

=

𝑝 + 𝑙 1 (𝑝+𝑙)/𝜆𝛽−1 Re (𝜑1 ∗ 𝜑2 ) (𝑢𝑧) 𝑑𝑢 ∫ 𝑢 𝜆𝛽 0

≥

2 (1 − 𝛾3 ) 𝑝 + 𝑙 1 (𝑝+𝑙)/𝜆𝛽−1 (2𝛾3 − 1 + ) 𝑑𝑢 ∫ 𝑢 𝜆𝛽 0 1 + 𝑢 |𝑧|

>

2 (1 − 𝛾3 ) 𝑝 + 𝑙 1 (𝑝+𝑙)/𝜆𝛽−1 (2𝛾3 − 1 + ) 𝑑𝑢 ∫ 𝑢 𝜆𝛽 0 1+𝑢

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓𝑗 (𝑧) 𝑧𝑝

(38)

𝑗 = 1, 2,

=1−

we have 𝛾𝑗 =

1 − 𝐴𝑗 1 − 𝐵𝑗

,

𝑗 = 1, 2.

(39)

By making use of (10) and (38), we obtain 𝐼𝑝𝑚

𝑝 + 𝑙 𝑝−(𝑝+𝑙)/𝜆𝛽 𝑧 (𝑝+𝑙)/𝜆𝛽−1 𝜑𝑗 (𝑡) 𝑑𝑡, 𝑧 ∫ 𝑡 (𝜆, 𝑙) 𝑓𝑗 (𝑧) = 𝜆𝛽 0

(45)

4 (𝐴 1 − 𝐵1 ) (𝐴 2 − 𝐵2 ) (1 − 𝐵1 ) (1 − 𝐵2 )

× (1 − 𝜑𝑗 ∈ 𝑃 (𝛾𝑗 ) ,

where 𝛾3 = 1 − 2 (1 − 𝛾1 ) (1 − 𝛾2 ) , (44)

𝑝 + 𝑙 1 𝑢(𝑝+𝑙)/𝜆𝛽−1 𝑑𝑢) = 𝛿 ∫ 𝜆𝛽 0 1+𝑢

(𝑧 ∈ U) ,

where 𝛿 is given by (37). When 𝐵1 = 𝐵2 = −1, we consider the functions 𝑓𝑗 ∈ A(𝑝) (𝑗 = 1, 2) which satisfy the hypothesis (35) and are given by 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓𝑗 (𝑧) =

𝑝 + 𝑙 𝑝−(𝑝+𝑙)/𝜆𝛽 𝑧 (𝑝+𝑙)/𝜆𝛽−1 1 + 𝐴 𝑗 𝑡 ( 𝑧 ) 𝑑𝑡, ∫ 𝑡 𝜆𝛽 1−𝑡 0 𝑗 = 1, 2. (46)

𝑗 = 1, 2. (40) Since Now, if we let 𝐹 = 𝐼𝑝𝑚 (𝜆, 𝑙)(𝑓1 ∗ 𝑓2 ), then by using (40) and the fact that 𝐼𝑝𝑚

(𝜆, 𝑙) 𝐹 (𝑧) =

𝐼𝑝𝑚

(𝜆, 𝑙) (𝐼𝑝𝑚

(𝜆, 𝑙) (𝑓1 ∗ 𝑓2 ) (𝑧))

= 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓1 (𝑧) ∗ 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓2 (𝑧) ,

(41)

1 + 𝐴 2𝑧 1 + 𝐴 1𝑧 )∗( ) 1−𝑧 1−𝑧 (1 + 𝐴 1 ) (1 + 𝐴 2 ) = 1 − (1 + 𝐴 1 ) (1 + 𝐴 2 ) + , 1−𝑧

𝜑0 (𝑧) =

𝑝 + 𝑙 𝑝−(𝑝+𝑙)/𝜆𝛽 𝑧 (𝑝+𝑙)/𝜆𝛽−1 𝜑0 (𝑡) 𝑑𝑡, (42) 𝑧 ∫ 𝑡 𝜆𝛽 0

𝑝+𝑙 𝜆𝛽 1

× ∫ 𝑢(𝑝+𝑙)/𝜆𝛽−1 (1 − (1 + 𝐴 1 ) (1 + 𝐴 2 ) 0

+

where 𝜑0 (𝑧) = (1 − 𝛽) =

𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹 (𝑧) 𝑧𝑝 𝑧

+𝛽

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝐹 (𝑧) 𝑧𝑝

𝑝 + 𝑙 −(𝑝+𝑙)/𝜆𝛽 𝑧 ∫ 𝑡(𝑝+𝑙)/𝜆𝛽−1 (𝜑1 ∗ 𝜑2 ) (𝑡) 𝑑𝑡. 𝜆𝛽 0

(47)

it follows from (43) that

a simple computation shows that 𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹 (𝑧) =

(

(1 + 𝐴 1 ) (1 + 𝐴 2 ) ) 𝑑𝑢 1 − 𝑢𝑧

= 1 − (1 + 𝐴 1 ) (1 + 𝐴 2 ) (43)

+

(1 + 𝐴 1 ) (1 + 𝐴 2 ) 𝑝+𝑙 𝑧 + 1; ). 2 𝐹1 (1, 1; 𝜆𝛽 𝑧−1 (1 − 𝑧) (48)

Journal of Complex Analysis

5 Then, the function 𝑔 is of the form (13). Differentiating (55) with respect to 𝑧 and using the identity (10), we obtain

Therefore 𝜑0 (𝑧) 󳨀→ 1 − (1 + 𝐴 1 ) (1 + 𝐴 2 ) +

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓 (𝑧)

1 (1 + 𝐴 1 ) (1 + 𝐴 2 ) 2

× 2 𝐹1 (1, 1;

𝑧𝑝

(49)

𝑝+𝑙 1 + 1; ) 𝜆𝛽 2

Corollary 11. If the functions 𝑓𝑗 ∈ A(𝑝) satisfy the following subordination condition: 1 + (1 − 2𝛿𝑗 ) 𝑧

≺

1−𝑧

,

𝑗 = 1, 2,

(50)

then Re (

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝐹 (𝑧) 𝑧𝑝

) (51)

> 1 − 2 (1 − 𝛿1 ) (1 − 𝛿2 ) × [2 − 2 𝐹1 (1, 1; where 𝐹 =

𝐼𝑝𝑚 (𝜆, 𝑙)(𝑓1

𝑝+𝑙 1 + 1; )] 𝜆 2

(𝑧 ∈ U) ,

𝑔 (𝑧) ≺ 𝑄 (𝑧) =

By using Lemma 5, we get 𝐴 𝐴 + (1 − ) (1 + 𝐵𝑧)−1 { { { 𝐵 𝐵 { { { 𝑝+𝑙 𝐵𝑧 { + 1; ) , if 𝐵 ≠ 0, 𝑄 (𝑧) = { × 2 𝐹1 (1, 1; 𝜆𝛽 𝐵𝑧 +1 { { { { { {1 + 𝑝 + 𝑙 𝐴𝑧, if 𝐵 = 0. 𝑝 + 𝑙 + 𝜆𝛽 { (59) Now, we will show that

In Theorem 12, we have determined the sufficient condition for the functions 𝐼𝑝𝑚 (𝜆, 𝑙)𝑓(𝑧)/𝑧𝑝 to be a member of the class 𝑃(𝜂).

We have

Theorem 12. If 𝑓 ∈ A(𝑝) satisfy the following subordination condition:

and setting

(1 − 𝛽)

𝑧𝑝

+𝛽

𝑧𝑝

Re

Re (

𝑧𝑝

)>𝜂

(𝑧 ∈ U) ,

(53)

1 + 𝐴𝑧 1 − 𝐴𝑟 ≥ 1 + 𝐵𝑧 1 − 𝐵𝑟

ℎ (𝑠, 𝑧) =

1 + 𝐴𝑧 (52) ≺ , 1 + 𝐵𝑧

1 + 𝐴𝑧𝑠 1 + 𝐵𝑧𝑠

|𝑧| = 𝑟 < 1,

1

𝑄 (𝑧) = ∫ ℎ (𝑠, 𝑧) 𝑑𝜇 (𝑠) , 0

(54)

1

Re 𝑄 (𝑧) ≥ ∫

0

𝑖𝑓 𝐵 = 0.

1 − 𝐴𝑠𝑟 𝑑𝜇 (𝑠) = 𝑄 (−𝑟) , 1 − 𝐵𝑠𝑟

Re ( (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

(63)

for 𝑓 ∈ A (𝑝) .

|𝑧| = 𝑟 < 1. (64)

As 𝑟 → 1− in (64), we obtain the assertion (60). Now, by using (59) and (60), we get

Proof. Let 𝑔 (𝑧) =

(62)

so that 𝑖𝑓 𝐵 ≠ 0,

The result is the best possible.

𝐼𝑝𝑚

(61)

which is a positive measure on the closed interval [0, 1], we get

where 𝐴 𝐴 + (1 − ) (1 − 𝐵)−1 { { { 𝐵 𝐵 { { { 𝑝+𝑙 𝐵 { + 1; ), 𝜂 = { × 2 𝐹1 (1, 1; 𝜆𝛽 𝐵−1 { { { { { {1 − 𝑝 + 𝑙 𝐴, 𝑝 + 𝑙 + 𝜆𝛽 {

(60)

(0 ≤ 𝑠 ≤ 1) ,

𝑝 + 𝑙 (𝑝+𝑙)/𝜆𝛽 𝑑𝑠, 𝑑𝜇 (𝑠) = 𝑠 𝜆𝛽

then 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

(57)

𝑝 + 𝑙 −(𝑝+𝑙)/𝜆𝛽 𝑧 (𝑝+𝑙)/𝜆𝛽−1 1 + 𝐴𝑡 𝑧 ( ) 𝑑𝑡. ∫ 𝑡 𝜆𝛽 1 + 𝐵𝑡 0 (58)

inf {Re 𝑄 (𝑧) : |𝑧| < 1} = 𝑄 (−1) .

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓 (𝑧)

(56)

Now, by applying Lemma 2, we have

∗ 𝑓2 ).

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

𝜆𝛽 1 + 𝐴𝑧 𝑧𝑔󸀠 (𝑧) ≺ . 𝑝+𝑙 1 + 𝐵𝑧

𝑔 (𝑧) +

By setting 𝛽 = 1, 𝐵𝑗 = −1, and 𝐴 𝑗 = 1 − 2𝛿𝑗 , 𝑗 = 1, 2, in Theorem 10, we have the following corollary.

𝑧𝑝

𝜆 𝑧𝑔󸀠 (𝑧) . 𝑝+𝑙

By using (52), (55), and (56), we get

as 𝑧 󳨀→ −1,

which evidently completes our proof of Theorem 10.

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓𝑗 (𝑧)

= 𝑔 (𝑧) +

(55)

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

where 𝜂 is given by (54).

𝑧𝑝

) > 𝜂,

(65)

6

Journal of Complex Analysis

To show that the estimate (54) is the best possible, we consider the function 𝑓 ∈ A(𝑝) defined by 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

=

𝑝 + 𝑙 1 (𝑝+𝑙)/𝜆𝛽−1 1 + 𝐴𝑢𝑧 ( ) 𝑑𝑢. (66) ∫ 𝑢 𝜆𝛽 0 1 + 𝐵𝑢𝑧

For the above function, we find that 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

(1 − 𝛽)

𝑧𝑝

+𝛽

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

Theorem 14. If 𝑓 ∈ A(𝑝) and 𝐹𝜇,𝑝 𝑓 is given by (70), satifies the subordination condition: (1 − 𝛽)

1 + 𝐴𝑧 = , 1 + 𝐵𝑧

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹𝜇,𝑝 𝑓 (𝑧)

1

𝑝+𝑙 1 − 𝐴𝑢 ) 𝑑𝑢 ∫ 𝑢(𝑝+𝑙)/𝜆𝛽−1 ( 𝜆𝛽 0 1 − 𝐵𝑢

Re (

𝐴 𝐴 + (1 − ) (1 − 𝐵)−1 { { { 𝐵 𝐵 { { { 𝑝+𝑙 𝐵 { + 1; ) , if 𝐵 ≠ 0, = { × 2 𝐹1 (1, 1; 𝜆𝛽 𝐵 −1 { { { { { {1 − 𝑝 + 𝑙 𝐴, if 𝐵 = 0 𝑝 + 𝑙 + 𝜆𝛽 { (67) as 𝑧 → −1, and the proof of the Theorem 12 is completed. In its special case when 𝐴 = 1 − 2𝛾, 𝐵 = −1 and 𝛽 = 1, Theorem 12 yields the following corollary.

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

≺

1 + 𝐴𝑧 , 1 + 𝐵𝑧 (72)

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

≺

1 + (1 − 2𝛾) 𝑧 1−𝑧

(𝑧 ∈ U) ,

𝑧𝑝

(68)

)

× [ 2 𝐹1 (1, 1;

(𝑧 ∈ U) ,

(73)

The result is the best possible. Proof. Let 𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹𝜇,𝑝 𝑓 (𝑧) 𝑧𝑝

.

(75)

Then by using the hypothesis (72) together with (71) and (75), we obtain 𝛽 𝑧ℎ󸀠 (𝑧) 𝜇+𝑝

= (1 − 𝛽) (69) 𝑝+𝑙 1 + 1; ) − 1] 𝜆 2

) > 𝜌0

𝐴 𝐴 { + (1 − ) (1 − 𝐵)−1 { { 𝐵 𝐵 { { { { 𝜇+𝑝 𝐵 + 1; ) , 𝑖𝑓 𝐵 ≠ 0, (74) 𝜌0 = { × 2 𝐹1 (1, 1; 𝛽 𝐵 −1 { { { { { {1 − 𝜇 + 𝑝 𝐴, 𝑖𝑓 𝐵 = 0. 𝑝+𝑙+𝛽 {

ℎ (𝑧) +

> 𝛾 + (1 − 𝛾)

𝑧𝑝

ℎ (𝑧) =

then 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹𝜇,𝑝 𝑓 (𝑧)

where

Corollary 13. If 𝑓 ∈ A(𝑝) satisfy the following condition:

Re (

𝑧𝑝

+𝛽

then

𝑧𝑝 󳨀→

In the next Theorem 14, by using the integral operator defined in (70), we established the sufficient condition for the functions 𝐼𝑝𝑚 (𝜆, 𝑙)𝐹𝜇,𝑝 𝑓(𝑧)/𝑧𝑝 belongs to 𝑃(𝜌0 ).

𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹𝜇,𝑝 𝑓 (𝑧) 𝑧𝑝

+𝛽

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

≺

1 + 𝐴𝑧 . 1 + 𝐵𝑧 (76)

The remaining part of the proof of Theorem 14 is similar to that of Theorem 12 and hence we omit the details.

(𝑧 ∈ U) .

The result is the best possible.

4. Inclusion Relationship for the Class R𝑚 𝑝 (𝜆,𝑙,𝛼;𝐴,𝐵)

For a function 𝑓 ∈ A(𝑝), the integral operator 𝐹𝜇,𝑝 𝑓 (𝑧) =

𝜇 + 𝑝 𝑧 𝜇−1 ∫ 𝑡 𝑓 (𝑡) 𝑑𝑡 𝑧𝜇 0

= 𝑧𝑝 2 𝐹1 (1, 𝜇 + 𝑝; 𝜇 + 𝑝 + 1; 𝑧) ∗ 𝑓 (𝑧)

Theorem 15. If 𝑓(𝑧) ∈ R𝑚+1 𝑝 (𝜆, 𝑙, 𝛼; 𝐴, 𝐵) and (70)

(𝜇 > −𝑝, 𝑧 ∈ U) .

then 󸀠

Also, it is easily verified from (70) that 󸀠

𝑧(𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹𝜇,𝑝 𝑓) (𝑧) = (𝜇 + 𝑝) 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) − 𝜇𝐼𝑝𝑚 (𝜆, 𝑙) 𝐹𝜇,𝑝 𝑓 (𝑧) .

𝜆 (𝑝 − 𝛼) (1 − 𝐴) + [𝛼𝜆 + 𝑝 (1 − 𝜆) + 𝑙] (1 − 𝐵) ≥ 0, (77)

(71)

𝑧(𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓) (𝑧) 1 1 + 𝐴𝑧 − 𝛼) ≺ 𝑞 (𝑧) ≺ ( 𝑚 , 𝑝−𝛼 𝐼𝑝 (𝜆, 𝑙) 𝑓 (𝑧) 1 + 𝐵𝑧 (𝑧 ∈ U) ,

(78)

Journal of Complex Analysis

7 is analytic in |𝑧| < 𝑟1 and 𝜙(0) = 1. Using the identity (10) in (84) and logarithmic differentiation of the resulting equation yields the following:

where 𝑞 (𝑧) =

1 1 − [𝛼𝜆 + 𝑝 (1 − 𝜆) + 𝑙]) , ( 𝜆 (𝑝 − 𝛼) 𝑄 (𝑧)

𝜆(𝑝−𝛼)(𝐴−𝐵)/𝐵 { {∫1 𝑡𝑝+𝑙−1 ( 1 + 𝐵𝑡𝑧 ) 𝑑𝑡, 𝑄 (𝑧) = { 0 1 + 𝐵𝑧 { 1 𝑝+𝑙−1 ∫ 𝑡 exp (𝜆 (𝑝 − 𝛼) 𝐴𝑧 (𝑡 − 1)) 𝑑𝑡, { 0

󸀠

𝑖𝑓 𝐵 ≠ 0,

𝑧(𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓) (𝑧) 1 − 𝛼) ( 𝑚+1 𝑝−𝛼 𝐼𝑝 (𝜆, 𝑙) 𝑓 (𝑧)

𝑖𝑓 𝐵 = 0 (79)

= 𝜙 (𝑧) +

and 𝑞(𝑧) is the best dominant of (78). If, in addition to (77), 𝐴>

−𝐵 [𝛼𝜆 + 𝑝 (1 − 𝜆) + 𝑙 + 1] 𝜆 (𝑝 − 𝛼)

(80)

𝜙 (𝑧) ≺

then

Re (

× ((𝑝 + 𝑙) 𝜆 (𝑝 − 𝛼) (𝐵 − 𝐴) 𝐵 ; 𝑝 + 𝑙 + 1; )] 𝐵 𝐵−1

−1

− [𝛼𝜆 + 𝑝 (1 − 𝜆) + 𝑙] ) . (82)

𝑧𝑝

𝜙 (𝑧) =

(83)

≺

1 + 𝐴𝑧 1 + 𝐵𝑧

(87)

(𝑧 ∈ U) .

(88)

This proves the assertion (78) of Theorem 15. In order to establish (81), we have to find the greatest lower bound of 𝜌 (0 < 𝜌 < 1) such that 1 + (1 − 2𝜌) 𝑧 1−𝑧

(𝑧 ∈ U) .

(89)

By (86), we have to show that 𝜌 = inf Re (𝑞 (𝑧)) = 𝑞 (−1) .

(90)

To prove (90), we need to show that inf Re (

𝑧∈U

1 1 )= . 𝑄 (𝑧) 𝑄 (−1)

(91)

From (79), we see that, for 𝐵 ≠ 0, 1

𝑄 (𝑧) = (1 + 𝐵𝑧)𝑎 ∫ 𝑡𝑏−1 (1 − 𝑡)𝑐−𝑏−1 (1 + 𝐵𝑧𝑡)−𝑎 𝑑𝑡 0

󸀠

𝑧𝑔 (𝑧) 1 = ( 𝑚 𝑔 (𝑧) 𝑝−𝛼 𝐼𝑝 (𝜆, 𝑙) 𝑓 (𝑧)

1 + 𝐴𝑧 1 + 𝐵𝑧

𝑧∈U

)

𝑧(𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓) (𝑧)

(|𝑧| < 𝑟1 ) ,

(−1 ≤ 𝐵 < 𝐴 ≤ 1; 𝑧 ∈ U) ,

𝜙 (𝑧) ≺ 𝑞 (𝑧) ≺

1/(𝑝−𝛼)

and 𝑟1 = sup{𝑟 : 𝑔(𝑧) ≠ 0, 0 < |𝑧| < 𝑟 < 1}. Then 𝑔(𝑧) is single-valued and analytic function in |𝑧| < 𝑟1 . By logarithmic differentiation in (83), it follows that the function 𝜙(𝑧) given by 󸀠

1 + 𝐴𝑧 )>0 1 + 𝐵𝑧

𝜙 (𝑧) ≺

Proof. Let 𝑓(𝑧) ∈ R𝑚+1 𝑝 (𝜆, 𝑙, 𝛼; 𝐴, 𝐵). Define the function 𝑔 by (𝜆, 𝑙) 𝑓 (𝑧)

(86)

by (86), we have Re(𝜙(𝑧)) > 0 (|𝑧| < 𝑟1 ). Now, (84) shows that 𝑔(𝑧) is starlike (univalent) in |𝑧| < 𝑟1 . Thus, it is not possible that 𝑔(𝑧) vanishes on |𝑧| = 𝑟1 if 𝑟1 < 1. So, we conclude that 𝑟1 = 1, and, therefore, 𝜙(𝑧) is analytic in U. Hence, (86) implies that

The bound on 𝜌 is the best possible.

𝑔 (𝑧) = 𝑧(

1 1 − [𝛼𝜆 + 𝑝 (1 − 𝜆) + 𝑙]) ( 𝜆 (𝑝 − 𝛼) 𝑄 (𝑧)

where 𝑞(𝑧) is the best dominant of (86) and 𝑄(𝑧) is given by (79). Since

1 𝜌= 𝜆 (𝑝 − 𝛼)

𝐼𝑝𝑚

(|𝑧| < 𝑟1 ) .

1 + 𝐴𝑧 = 𝑞 (𝑧) ≺ 1 + 𝐵𝑧

(81)

where

× [ 2 𝐹1 (1,

1 + 𝐴𝑧 1 + 𝐵𝑧

≺

(85)

Hence, by using Lemma 6 with 𝛽 = 𝜆(𝑝 − 𝛼) and 𝛾 = 𝛼𝜆 + 𝑝(1 − 𝜆) + 𝑙, we find that

𝑤𝑖𝑡ℎ − 1 ≤ 𝐵 < 0,

R𝑚+1 (𝜆, 𝑙, 𝛼; 𝐴, 𝐵) ⊂ R𝑚 𝑝 𝑝 (𝜆, 𝑙, 𝛼; 1 − 2𝜌, −1) ,

𝜆𝑧𝜙󸀠 (𝑧) 𝜆 (𝑝 − 𝛼) 𝜙 (𝑧) + 𝛼𝜆 + 𝑝 (1 − 𝜆) + 𝑙

(92)

(𝑧 ∈ U) ,

− 𝛼) (84)

where 𝑎=

𝜆 (𝑝 − 𝛼) (𝐵 − 𝐴) , 𝐵

𝑏 = 𝑝 + 𝑙,

𝑐 = 𝑝 + 𝑙 + 1.

(93)

8

Journal of Complex Analysis

Since 𝑐 > 𝑏 > 0, by using Lemma 5, we get following: 𝑄 (𝑧) = (1 + 𝐵𝑧)𝑎

𝐴 = −𝐵[𝛼𝜆 + 𝑝(1 − 𝜆) + 𝑙 + 1]/𝜆(𝑝 − 𝛼) and using (78), we get (81). The result is the best possible as the function 𝑞(𝑧) is the best dominant of (78). This completes the proof of Theorem 15.

Γ (𝑏) Γ (𝑐 − 𝑏) 2 𝐹1 (𝑎, 𝑏; 𝑐; −𝐵𝑧) Γ (𝑐)

=

Γ (𝑏) 𝐵𝑧 ) 2 𝐹1 (𝑎, 𝑐 − 𝑏; 𝑐; Γ (𝑐) 𝐵𝑧 + 1

=

𝐵𝑧 Γ (𝑏) ). 2 𝐹1 (1, 𝑎; 𝑐; Γ (𝑐) 𝐵𝑧 + 1

(94)

5. Sufficient Conditions for the Class R𝑚 𝑝 (𝜆, 𝑙, 𝛼)

Since 𝐴>

−𝐵 [𝛼𝜆 + 𝑝 (1 − 𝜆) + 𝑙 + 1] 𝜆 (𝑝 − 𝛼)

In the following section, we obtain the sufficient condition for the function 𝑓 to be a member of the class R𝑚 𝑝 (𝜆, 𝑙, 𝛼).

Theorem 16. If 𝑓 ∈ A(𝑝) satisfy the following subordination condition:

with − 1 ≤ 𝐵 < 0 (95)

(1 − 𝛽)

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

implies that 𝑐 > 𝑎 > 0, by using Lemma 5, we find from (94) that 𝑄 (𝑧) = ∫ ℎ (𝑧, 𝑡) 𝑑V (𝑡) , 0

(96)

𝑀1 =

where 1 + 𝐵𝑧 1 + (1 − 𝑡) 𝐵𝑧

(0 ≤ 𝑡 ≤ 1) ,

Γ (𝑏) 𝑡𝑎−1 (1 − 𝑡)𝑐−𝑎−1 𝑑𝑡, 𝑑V (𝑡) = Γ (𝑎) Γ (𝑐 − 𝑎)

(97)

inf Re (

𝑧∈U

1 1 )≥ 𝑄 (𝑧) 𝑄 (−𝑟)

≺ 1 + 𝑀1 𝑧, (100)

𝜆𝛽 (𝑝 − 𝛼) (1 + 𝜆𝛽/ (𝑝 + 𝑙)) , 2 2 󵄨󵄨 √ 󵄨󵄨 + 𝑝 + 𝑙 − 𝜆𝛽 (𝑝 − 𝛼) (𝑝 + 𝑙) + (𝑝 + 𝑙 + 𝜆𝛽) 󵄨󵄨 󵄨󵄨 (101)

then 𝑓 ∈ R𝑚 𝑝 (𝜆, 𝑙, 𝛼).

𝑔 (𝑧) =

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

.

(102)

Then, the function 𝑔 is of the form (13) and is analytic in U. From Theorem 12 with 𝐴 = 𝑀1 and 𝐵 = 0, we have

(|𝑧| ≤ 𝑟 < 1) ,

𝑔 (𝑧) ≺ 1 +

1 1 1 ) = inf = 1 −1 0

(𝑧 ∈ U) .

(108)

then (109) holds true if Ψ(𝑥) ≥ 0, for any 𝑥 ∈ U. Since (𝑢2 + V2 )(𝜆𝛽(𝑝 − 𝛼)/(𝑝 + 𝑙))2 > 0, the inequality Ψ(𝑥) ≥ 0 holds true if the discriminant Δ ≤ 0; that is, 2

Suppose that this is false. Since 𝑃(0) = 1, there exists a point 𝑧0 ∈ U such that 𝑃(𝑧0 ) = 𝑖𝑥 for some 𝑥 ∈ R. Therefore, in order to show that (108), it is sufficient to obtain the contradiction from the inequality

Δ = 4(

𝜆𝛽 (𝑝 − 𝛼) ) 𝑝+𝑙

× {V2 − (𝑢2 + V2 ) 2 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 󵄨 󵄨 × [( − 󵄨󵄨1 − 󵄨 𝑁) 𝑝+𝑙 𝑝 + 𝑙 󵄨󵄨󵄨 󵄨󵄨

󵄨󵄨 󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 𝜆𝛽 (𝑝 − 𝛼) 󵄨 󵄨 ) 𝑔 (𝑧0 )+ 𝑃 (𝑧0 ) 𝑔 (𝑧0 ) − 1󵄨󵄨󵄨󵄨 𝐸 = 󵄨󵄨󵄨󵄨(1− 𝑝+𝑙 𝑝+𝑙 󵄨󵄨 󵄨󵄨 ≥ 𝑀1 .

− 𝑁2 (

(109) If we let 𝑔(𝑧0 ) = 𝑢 + 𝑖V, then, by using (104) and the triangle inequality, we obtain that 󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 󵄨 ) 𝑔 (𝑧0 ) 𝐸 = 󵄨󵄨󵄨󵄨(1 − 𝑝+𝑙 󵄨󵄨

= (𝑢2 + V2 ) (

2

−𝑁2 (

(110)

≤(

If we let

2 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 󵄨󵄨 𝑁) − 󵄨󵄨󵄨1 − 𝑝+𝑙 𝑝 + 𝑙 󵄨󵄨󵄨 󵄨󵄨

− 𝑁2 (

Ψ (𝑥) = 𝐸2 − 𝑀12 2

𝜆𝛽𝑥 (𝑝 − 𝛼) 2V𝜆𝛽𝑥 (𝑝 − 𝛼) ) + 𝑝+𝑙 𝑝+𝑙

𝑝 + 𝑙 + 𝜆𝛽 2 ) 𝑝+𝑙

× (1 − [(

2 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 󵄨󵄨 𝑁) − 󵄨󵄨󵄨1 − 𝑝+𝑙 𝑝 + 𝑙 󵄨󵄨󵄨 󵄨󵄨

− 𝑁2 (

𝑝 + 𝑙 + 𝜆𝛽 2 ) ]. 𝑝+𝑙

𝜌2 V2 𝑁2 ≤ ≤ 2 2 𝑢 1−𝜌 1 − 𝑁2

2 𝜆𝛽(𝑝 − 𝛼) 𝜆𝛽(𝑝 − 𝛼) +( − (1 − ) 𝑁) . 𝑝+𝑙 𝑝+𝑙

+(

(113)

After a simple computation, by using (104), we obtain the inequality

2

𝜆𝛽𝑥 (𝑝 − 𝛼) 2V𝜆𝛽𝑥 (𝑝 − 𝛼) ≥ (𝑢2 + V2 ) ( ) + 𝑝+𝑙 𝑝+𝑙

≥ (𝑢2 + V2 ) (

𝑝 + 𝑙 + 𝜆𝛽 2 ) ]} 𝑝+𝑙

2 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 󵄨󵄨 𝑁) ≤ 𝑢2 [( − 󵄨󵄨󵄨1 − 𝑝+𝑙 𝑝 + 𝑙 󵄨󵄨󵄨 󵄨󵄨

𝜆𝛽𝑥 (𝑝 − 𝛼) 2V𝜆𝛽𝑥 (𝑝 − 𝛼) ) + 𝑝+𝑙 𝑝+𝑙

󵄨󵄨 󵄨󵄨2 𝜆𝛽 (𝑝 − 𝛼) 󵄨 󵄨 + 󵄨󵄨󵄨󵄨(1 − ) 𝑔 (𝑧0 ) − 1󵄨󵄨󵄨󵄨 𝑝+𝑙 󵄨󵄨 󵄨󵄨

2 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 󵄨󵄨 𝑁) − 󵄨󵄨󵄨1 − 𝑝+𝑙 𝑝 + 𝑙 󵄨󵄨󵄨 󵄨󵄨

+𝑁2 (

󵄨󵄨2 𝜆𝛽 (𝑝 − 𝛼) 󵄨 + 𝑃 (𝑧0 ) 𝑔 (𝑧0 ) − 1󵄨󵄨󵄨󵄨 𝑝+𝑙 󵄨󵄨

𝑝 + 𝑙 + 𝜆𝛽 2 ) ]} ≤ 0, 𝑝+𝑙

which is equivalent to

V2 {1 − [(

2

(112)

(114)

2 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 𝜆𝛽 (𝑝 − 𝛼) 󵄨󵄨󵄨󵄨 󵄨󵄨 𝑁) − 󵄨󵄨󵄨1 − 𝑝+𝑙 𝑝 + 𝑙 󵄨󵄨󵄨 󵄨󵄨 −1

𝑝 + 𝑙 + 𝜆𝛽 2 +𝑁 ( ) ]) , 𝑝+𝑙 2

𝑝 + 𝑙 + 𝜆𝛽 2 ), 𝑝+𝑙 (111)

which yields Δ ≤ 0. Therefore 𝐸 ≥ 𝑀1 , which contradicts (107). It follows that Re 𝑃(𝑧) > 0, and 𝑓 ∈ R𝑚 𝑝 (𝜆, 𝑙, 𝛼).

10

Journal of Complex Analysis

Theorem 17. Let 𝛽 > 0 and if 𝑓 ∈ A(𝑝) such that 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓(𝑧)/𝑧𝑝 ≠ 0, 𝑧 ∈ U, satisfies the following differential subordination: (1 − 𝛽) (

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

𝜎

where 𝑃 is given by (105). Therefore, by Lemma 8, we find that Re 𝑃 (𝑧) > 0

(122)

that is

)

󸀠

󸀠

(𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓) (𝑧) 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) +𝛽 ( ) 𝑝𝑧𝑝−1 𝑧𝑝

𝜎−1

(115)

Re (

𝑧(𝐼𝑝𝑚 (𝜆, 𝑙)𝑓) (𝑧) 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧)

)>𝛼

(0 ≤ 𝛼 < 𝑝, 𝑧 ∈ U) , (123)

which completes the proof of Theorem 17

≺ 1 + 𝑀2 𝑧, where the powers are understood as the principle value, and (𝑝 − 𝛼) 𝛽 (1 + 𝛽/𝜎𝑝) { { , 𝑖𝑓 𝜎 ≠ 0, { { { { 󵄨󵄨󵄨󵄨𝑝 − (𝑝 − 𝛼) 𝛽󵄨󵄨󵄨󵄨 + √𝑝2 + (𝑝 + 𝛽/𝜎)2 𝑀2 = { { { (𝑝 − 𝛼) 𝛽 { { { , 𝑖𝑓 𝜎 = 0, 𝑝 { (116)

By taking 𝑚 = 0 in Theorem 17, we get the following corollary due to Patel et al. [21]. Corollary 18. Let 𝛽 > 0, 0 ≤ 𝛼 < 𝑝, 𝑝 ∈ N, and 𝜎 ≥ 0. If 𝑓 ∈ A(𝑝) such that 𝑓(𝑧)/𝑧𝑝 ≠ 0 (𝑧 ∈ U), and satisfies the following subordination condition: (1 − 𝛽) (

where the powers are understood as the principal value and 𝑀2 is given by (116), then 𝑓 ∈ 𝑆𝑝∗ (𝛼).

Proof. If 𝜎 = 0, then the condition (115) is equivalent to 󵄨󵄨 𝑚 󵄨󵄨 󸀠 󵄨󵄨 𝑧(𝐼𝑝 (𝜆, 𝑙)𝑓) (𝑧) 󵄨󵄨 󵄨󵄨 󵄨 − 𝑝󵄨󵄨󵄨 < 𝑝 − 𝛼 󵄨󵄨 𝑚 󵄨󵄨 𝐼𝑝 (𝜆, 𝑙) 𝑓 (𝑧) 󵄨󵄨 󵄨󵄨 󵄨󵄨

(𝑧 ∈ U) .

(117)

𝐼𝑝𝑚 (𝜆, 𝑙)𝑓(𝑧) 𝑧𝑝

𝜎

)

(𝑧 ∈ U) .

(118)

Choosing the principal value in (118), we note that 𝑔 is of the form (13) and is analytic in U. Differentiating (118) with respect to 𝑧, we obtain 𝛽 󸀠 𝑧𝑔 (𝑧) 𝑔 (𝑧) + 𝜎𝑝 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) 𝑧𝑝

)

󸀠

(119) 𝜎−1

,

𝜎𝑝 𝑀 𝑧. 𝜎𝑝 + 𝛽 2

(120)

Also, with the aid of (118), (115) can be written as follows: 𝑔 (𝑧) {1 − 𝛽 + 𝛽 [(1 −

Remark 22. Taking 𝑚 = 𝛼 = 0, and 𝑝 = 𝛽 = 𝜎 = 1 in Theorem 17, we obtain the result of Mocanu [23, with 𝑛 = 1]. Taking 𝑚 = 0 and 𝛽 = 1/(𝑝−𝛼), 0 ≤ 𝛼 < 𝑝 in Theorem 17, we obtain the following corollary.

󵄨󵄨 󵄨󵄨 𝑓 (𝑧) 𝜎 𝑓󸀠 (𝑧) 𝑓 (𝑧) 𝜎−1 󵄨󵄨 󵄨 󵄨󵄨(𝑝 − 𝛼 − 1) ( 𝑝 ) + 𝑝−1 ( 𝑝 ) + 𝛼 − 𝑝󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 𝑧 𝑝𝑧 𝑧 󵄨 󵄨
0, 0 ≤ 𝛼 < 𝑝, and 𝑝 ∈ N. If 𝑓 ∈ A(𝑝) such that 𝑓(𝑧)/𝑧𝑝 ≠ 0 (𝑧 ∈ U) satisfies the inequality:

𝜎

(𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓) (𝑧) 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑓 (𝑧) +𝛽 ( ) 𝑝𝑧𝑝−1 𝑧𝑝

Remark 19. Taking 𝑚 = 0 and 𝑝 = 1 in Theorem 17, we get the result of Liu [20, Theorem 2.2, with 𝑛 = 1]. Remark 20. Putting 𝑚 = 0 and 𝛽 = 𝑝 = 1 in Theorem 17, we obtain the result of Liu [20, Corollary 2.1, with 𝑛 = 1].

The above equation (117) implies that 𝑓 ∈ R𝑚 𝑝 (𝜆, 𝑙, 𝛼). If we consider 𝜎 > 0 and suppose that 𝑔 (𝑧) = (

𝑓 (𝑧) 𝜎 𝑓󸀠 (𝑧) 𝑓 (𝑧) 𝜎−1 ) + 𝛽 ( ) ≺ 1 + 𝑀2 𝑧 𝑧𝑝 𝑝𝑧𝑝−1 𝑧𝑝 (124) (𝑧 ∈ U) ,

then 𝑓 ∈ R𝑚 𝑝 (𝜆, 𝑙, 𝛼).

= (1 − 𝛽) (

(𝑧 ∈ U) ;

𝛼 𝛼 ) 𝑃 (𝑧) + ]} ≺ 1 + 𝑀2 𝑧, (121) 𝑝 𝑝

((𝑝 − 𝛼) /𝑝) [𝑝𝜎 (𝑝 − 𝛼) + 1] 2 2 󵄨 󵄨 𝜎 (𝑝 − 𝛼) 󵄨󵄨󵄨𝑝 − 1󵄨󵄨󵄨 + √[𝑝𝜎 (𝑝 − 𝛼)] + [𝑝𝜎 (𝑝 − 𝛼) + 1] (𝑧 ∈ U) , (125)

where the powers are the principal value ones, then 𝑓 ∈ 𝑆𝑝∗ (𝛼). Remark 24. Taking 𝑝 = 1 in Corollary 23, we obtain the result of Liu [20, Corollary 2.2, with 𝑛 = 1].

Journal of Complex Analysis

11

Remark 25. Putting 𝑝 = 𝜎 = 1 in Corollary 23, we get the result of Mocanu and Oros [22, Corollary 2.4, with 𝑛 = 1]. In Theorem 26, we obtain the necessary condition for functions belonging to class 𝑃(𝛾𝑗 ). Theorem 26. Let 𝑓𝑗 ∈ A(𝑝) (𝑗 = 1, 2). If the functions 𝐼𝑝𝑚+1 (𝜆, 𝑙)𝑓𝑗 (𝑧)/𝑧𝑝 ∈ 𝑃(𝛾𝑗 ) (0 ≤ 𝛾𝑗 < 1), then the function 𝑔 = 𝐼𝑝𝑚 (𝜆, 𝑙), (𝑓1 ∗ 𝑓2 ) satisfies the following inequality:

Re (

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑔 (𝑧) 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑔 (𝑧)

)>0

(𝑧 ∈ U) ,

(126)

𝑧𝑝 =

𝜆 + 2 (𝑝 + 𝑙)

. 2 𝜆[ 2 𝐹1 (1, 1; (𝑝 + 𝑙) /𝜆 + 1; 1/2) − 1] + 2 (𝑝 + 𝑙) (127)

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓1 (𝑧) 𝑧𝑝 = Re (

∗

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑓2 (𝑧)

𝑧𝑝

𝑧𝑝

[𝜑2 (𝑧) +

󸀠

)

𝑧𝜆𝜑󸀠 (𝑧) ] 𝑝+𝑙

(131)

where Ψ(𝑢, V; 𝑧) = (𝐼𝑝𝑚 (𝜆, 𝑙)𝑔(𝑧)/𝑧𝑝 )(𝑢2 + 𝜆V/(𝑝 + 𝑙)). Thus, by using (128), (131) can be written as follows:

𝑧𝑝

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑔 (𝑧) 𝑧𝑝

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑔 (𝑧) 𝜆𝑦 2 =( ) − 𝑥 ) Re ( 𝑝+𝑙 𝑧𝑝 ≤−

)

𝑚+1

𝜆𝑧 𝐼𝑝 + ( 𝑝+𝑙

Re (Ψ (𝑖𝑥, 𝑦; 𝑧))

(𝜆, 𝑙) 𝑔 (𝑧) 𝑧𝑝

󸀠

))

≤−

) > 1 + 2 (1 − 𝛾1 ) (1 − 𝛾2 ) × [ 2 𝐹1 (1, 1;

𝑝+𝑙 1 + 1; ) − 1] . 𝜆 2 (129)

Again, from (129) and Corollary 13, we obtain 𝐼𝑝𝑚 (𝜆, 𝑙) 𝑔 (𝑧) 𝑧𝑝

) > 1 − 2 (1 − 𝛾1 ) (1 − 𝛾2 ) × [ 2 𝐹1 (1, 1;

𝑧𝑝

(133)

)

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑔 (𝑧) 𝜆 ) Re ( 𝑧𝑝 2 (𝑝 + 𝑙) (𝑧 ∈ U) ,

where we used (127) and (130). Thus, by Lemma 9, we get

Then, by using Lemma 2 with 𝑐 = (𝑝 + 𝑙)/𝜆, 𝐴 = −1 + 4(1 − 𝛾1 )(1 − 𝛾2 ), and 𝐵 = −1, we have

𝑧𝑝

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑔 (𝑧)

< 1 − 2 (1 − 𝛾1 ) (1 − 𝛾2 )

(128)

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑔 (𝑧)

2 (𝑝 + 𝑙) 𝜆 [1 + (1 + ) 𝑥2 ] 𝜆 2 (𝑝 + 𝑙)

× Re (

> 1 − 2 (1 − 𝛾1 ) (1 − 𝛾2 ) .

Re (

𝐼𝑝𝑚 (𝜆, 𝑙) 𝑔 (𝑧)

(𝜆, 𝑙) 𝑔 (𝑧)

For all real 𝑥 and 𝑦 ≤ −(1/2)(1 + 𝑥2 ), we have

Proof. Let 𝑔 = 𝐼𝑝𝑚 (𝜆, 𝑙)(𝑓1 ∗ 𝑓2 ). By using Lemma 4 and (10), we obtain

Re (

𝑚+1

𝜆𝑧 𝐼𝑝 + ( 𝑝+𝑙

Re (Ψ (𝜑 (𝑧) , 𝑧𝜑󸀠 (𝑧) ; 𝑧)) > 1 − 2 (1 − 𝛾1 ) (1 − 𝛾2 ) . (132)

2 (1 − 𝛾1 ) (1 − 𝛾2 )

Re (

𝐼𝑝𝑚+1 (𝜆, 𝑙) 𝑔 (𝑧)

= Ψ (𝜑 (𝑧) , 𝑧𝜑󸀠 (𝑧) ; 𝑧) ,

provided that


0

(𝑧 ∈ U) ,

(134)

which completes the proof of Theorem 26.

Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment The authors thank the referees for their insightful suggestions.

References [1] S. S. Miller and P. T. Mocanu, “Differential subordinations and univalent functions,” The Michigan Mathematical Journal, vol. 28, no. 2, pp. 157–172, 1981. [2] S. S. Miller and P. T. Mocanu, Differential Subordinations: Theory and Applications, vol. 225 of Monographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 2000.

12 [3] A. C˘atas¸, “On certain classes of p-valent functions defined by multiplier transformations,” in Proceedings of the International Symposium on Geometric Function Theory and Applications (GFTA ’07), S. Owa and Y. Polatoglu, Eds., vol. 91, pp. 241–250, TC ˙Istanbul K¨ult¨ur University Publications, Istanbul, Turkey, August 2007. [4] H. M. Srivastava, K. Suchithra, B. A. Stephen, and S. Sivasubramanian, “Inclusion and neighborhood properties of certain subclasses of analytic and multivalent functions of complex order,” Journal of Inequalities in Pure and Applied Mathematics, vol. 7, no. 5, pp. 1–8, 2006. [5] S. Sivaprasad Kumar, H. C. Taneja, and V. Ravichandran, “Classes of multivalent functions defined by Dziok-Srivastava linear operator and multiplier transformation,” Kyungpook Mathematical Journal, vol. 46, no. 1, pp. 97–109, 2006. [6] N. E. Cho and T. H. Kim, “Multiplier transformations and strongly close-to-convex functions,” Bulletin of the Korean Mathematical Society, vol. 40, no. 3, pp. 399–410, 2003. [7] N. E. Cho and H. M. Srivastava, “Argument estimates of certain analytic functions defined by a class of multiplier transformations,” Mathematical and Computer Modelling, vol. 37, no. 1-2, pp. 39–49, 2003. [8] M. Kamali and H. Orhan, “On a subclass of certain starlike functions with negative coefficients,” Bulletin of the Korean Mathematical Society, vol. 41, no. 1, pp. 53–71, 2004. [9] H. Orhan and H. Kiziltunc¸, “A generalization on subfamily of p-valent functions with negative coefficients,” Applied Mathematics and Computation, vol. 155, no. 2, pp. 521–530, 2004. [10] G. S. S˘al˘agean, “Subclasses of univalent functions,” in Complex Analysis—Fifth Romanian-Finnish Seminar, vol. 1013 of Lecture Notes in Mathematics, pp. 362–372, Springer, Berlin, Germany, 1983. [11] F. M. Al-Oboudi, “On univalent function defined by a generalized S˘al˘agean operator,” International Journal of Mathematics and Mathematical Sciences, vol. 27, pp. 1429–1436, 2004. [12] B. A. Uralegaddi and C. Somanatha, “Certain classes of univalent functions,” in Current Topics in Analytic Function Theory, pp. 371–374, World Scientific, Singapore, 1992. [13] T. Bulboac˘a, M. K. Aouf, and R. M. El-Ashwah, “Subordination properties of multivalent functions defined by certain integral operetor,” Banach Journal of Mathematical Analysis, vol. 6, no. 2, pp. 69–85, 2012. [14] J. Patel and A. K. Mishra, “On certain subclasses of multivalent functions associated with an extended fractional differintegral operator,” Journal of Mathematical Analysis and Applications, vol. 332, no. 1, pp. 109–122, 2007. [15] D. Z. Pashkouleva, “The starlikeness and spiral-convexity of certain subclasses of analytic functions,” in Current Topics in Analytic Function Theory, pp. 266–273, World Scientific, Singapore, 1992. [16] J. Stankiewicz and Z. Stankiewicz, “Some applications of the Hadamard convolution in the theory of functions,” Annales Universitatis Mariae Curie-Skłodowska A, vol. 40, pp. 251–265, 1986. [17] E. T. Whittaker and G. N. Watson, A Course of Modern Analysis: An Introduction to the General Theory of Infinite Processes and of Analytic Functions, Cambridge Mathematical Library, Cambridge University Press, Cambridge, UK, 1996. [18] S. S. Miller and P. T. Mocanu, “Univalent solutions of BriotBouquet differential equations,” Journal of Differential Equations, vol. 56, no. 3, pp. 297–309, 1985.

Journal of Complex Analysis [19] D. R. Wilken and J. Feng, “A remark on convex and starlike functions,” The Journal of the London Mathematical Society, vol. 21, no. 2, pp. 287–290, 1980. [20] M. Liu, “On certain sufficient condition for starlike functions,” Soochow Journal of Mathematics, vol. 29, no. 4, pp. 407–412, 2003. [21] J. Patel, A. K. Mishra, and H. M. Srivastava, “Classes of multivalent analytic functions involving the Dziok-Srivastava operator,” Computers & Mathematics with Applications, vol. 54, no. 5, pp. 599–616, 2007. [22] P. T. Mocanu and Gh. Oros, “A sufficient condition for starlikeness of order 𝛼,” International Journal of Mathematics and Mathematical Sciences, vol. 28, no. 9, pp. 557–560, 2001. [23] P. T. Mocanu, “Some simple criteria for starlikeness and convexity,” Libertas Mathematica, vol. 13, pp. 27–40, 1993.

Advances in

Operations Research Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Advances in

Decision Sciences Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Journal of

Applied Mathematics

Algebra

Hindawi Publishing Corporation http://www.hindawi.com

Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Journal of

Probability and Statistics Volume 2014

The Scientific World Journal Hindawi Publishing Corporation http://www.hindawi.com

Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

International Journal of

Differential Equations Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Volume 2014

Submit your manuscripts at http://www.hindawi.com International Journal of

Advances in

Combinatorics Hindawi Publishing Corporation http://www.hindawi.com

Mathematical Physics Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Journal of

Complex Analysis Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

International Journal of Mathematics and Mathematical Sciences

Mathematical Problems in Engineering

Journal of

Mathematics Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Volume 2014

Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Discrete Mathematics

Journal of

Volume 2014

Hindawi Publishing Corporation http://www.hindawi.com

Discrete Dynamics in Nature and Society

Journal of

Function Spaces Hindawi Publishing Corporation http://www.hindawi.com

Abstract and Applied Analysis

Volume 2014

Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

International Journal of

Journal of

Stochastic Analysis

Optimization

Hindawi Publishing Corporation http://www.hindawi.com

Hindawi Publishing Corporation http://www.hindawi.com

Volume 2014

Volume 2014