Hindawi Publishing Corporation Journal of Function Spaces Volume 2016, Article ID 7123907, 6 pages http://dx.doi.org/10.1155/2016/7123907
Research Article A Subclass of Harmonic Functions Related to a Convolution Operator Saqib Hussain,1 Akhter Rasheed,1 and Maslina Darus2 1
Department of Mathematics, COMSATS Institute of Information Technology, Abbottabad Campus, Abbottabad, Pakistan School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600 Bangi, Selangor, Malaysia
2
Correspondence should be addressed to Saqib Hussain; saqib
[email protected] Received 23 May 2016; Accepted 21 July 2016 Academic Editor: Wilfredo Urbina Copyright © 2016 Saqib Hussain et al. 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. We introduce a new subclass of harmonic functions by using a certain linear operator. For this class we derive coefficient bounds, extreme points, and inclusion results and also show that this class is closed under an integral operator.
1. Introduction
Hence
Harmonic functions have long been used in the representation of minimal surface; for example, Heinz [1] in 1952 used such mappings in the study of the Gaussian curvature of nonparametric minimal surface over the unit disc. Such mappings have vast application in the field of engineering, physics, electronics, medicine, operations research, aerodynamics, and other branches of applied mathematics (see [2]). 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖V(𝑥, 𝑦) is said to be complex valued harmonic function if both 𝑢 and V are continuous and real harmonic; that is, 𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0 and V𝑥𝑥 + V𝑦𝑦 = 0. In simply connected domain [3], we can write 𝑓(𝑧) = ℎ(𝑧) + 𝑔(𝑧), where ℎ and 𝑔 are analytic in 𝐸 = {𝑧: |𝑧| < 1}. We call ℎ the analytic part and 𝑔 the coanalytic part of 𝑓. The necessary and sufficient condition for a function 𝑓(𝑧) = ℎ(𝑧) + 𝑔(𝑧) to be univalent and sense preserving in 𝐸 is |ℎ (𝑧)| > |𝑔 (𝑧)| (see [3]). A function 𝑓(𝑧) = ℎ(𝑧) + 𝑔(𝑧) is in class SH if it is harmonic, univalent, and sense preserving in 𝐸, where ℎ and 𝑔 have the following series: ∞
ℎ (𝑧) = 𝑧 + ∑ 𝑎𝑘 𝑧 , 𝑘=2
∞
(1)
𝑔 (𝑧) = ∑ 𝑏𝑘 𝑧𝑘 , 𝑘=1
𝑏1 < 1.
∞
𝑘=2
𝑘=1
𝑏1 < 1.
(2)
Note that if the coanalytic part 𝑔 is identically zero, then SH reduces to well-known class of normalized univalent analytic functions S. For this class, the function 𝑓(𝑧) may be expressed as ∞
𝑓 (𝑧) = 𝑧 + ∑ 𝑎𝑘 𝑧𝑘 .
(3)
𝑘=2
A function 𝑓(𝑧) given by (2) is said to be harmonic starlike of order 𝛽 (0 ≤ 𝛽 < 1) for |𝑧| = 𝑟 < 1, if 𝜕 (arg (𝑓 (𝑟𝑒𝑖𝜃 ))) > 𝛽, 𝜕𝜃
(4)
or equivalently Re (
𝑘
∞
𝑓 (𝑧) = 𝑧 + ∑ 𝑎𝑘 𝑧𝑘 + ∑ 𝑏𝑘 𝑧𝑘
𝑧ℎ (𝑧) − 𝑧𝑔 (𝑧) ) > 𝛽, ℎ (𝑧) + 𝑔 (𝑧)
𝑧 = 𝑟𝑒𝑖𝜃 .
(5)
The class of all harmonic starlike functions of order 𝛽 is denoted by S∗H (𝛽). This class was studied by Jahangiri [4]. The cases 𝛽 = 0 and 𝑏1 = 0 were studied by Silverman and Silvia [5] and Silverman [6].
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Journal of Function Spaces
∞ 𝑘 𝑘 For 𝐹(𝑧) = 𝑧 + ∑∞ 𝑘=2 𝐴 𝑘 𝑧 + ∑𝑘=1 𝐵𝑘 𝑧 and 𝑓(𝑧) given by (2), the convolution is denoted by 𝑓 ∗ 𝐹 and defined as ∞
∞
𝑘=2
𝑘=1
𝑓 ∗ 𝐹 = 𝑧 + ∑ 𝑎𝑘 𝐴 𝑘 𝑧𝑘 + ∑ 𝑏𝑘 𝐵𝑘 𝑧𝑘 .
(6)
Let 𝛼1 , 𝐴 1 , . . . , 𝛼𝑝 , 𝐴 𝑝 and 𝛽1 , 𝐵1 , . . . , 𝛽𝑞 , 𝐵𝑞 (𝑝, 𝑞 ∈ 𝑁 = {1, 2, 3, . . .}) be positive real parameter such that ∞
∞
𝑘=1
𝑘=1
1 + ∑ 𝐵𝑘 − ∑ 𝐴 𝑘 ≥ 0.
𝑊 [(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ] 𝑓 (𝑧)
where 𝑝 Φ𝑞 [(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ; 𝑧] is given by 𝑝 Φ𝑞
[(𝛼1 , 𝐴 1 ) , . . . , (𝛼𝑝 , 𝐴 𝑝 ) ; (𝛽1 , 𝐵1 ) , . . . , (𝛽𝑞 , 𝐵𝑞 ) ; 𝑧]
[(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ; 𝑧]
(13)
= Ω𝑧𝑝 Ψ𝑞 [(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ; 𝑧] .
(8)
= 𝑝 Ψ𝑞 [(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ; 𝑧]
(12)
=𝑝 Φ𝑞 [(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ; 𝑧] ∗ 𝑓 (𝑧) ,
(7)
The Wright generalized hypergeometric function [7] is defined as 𝑝 Ψ𝑞
Dziok and Srivastava [8] introduced a linear operator which is generalization of Dziok-Srivastava operator [8–10], Carlson-Shaffer operator [11], and the generalized Bernardi’s integral operator [12]. Dziok and Raina [13] considered the linear operator 𝑊[(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ] : S → S defined by
For 𝑓(𝑧) ∈ S, we have
which is defined by 𝑝 Ψ𝑞
𝑊 [(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ] 𝑓 (𝑧)
[(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ; 𝑧]
=
𝑝 Π𝑖=1 Γ (𝛼𝑖 + 𝑘𝐴 𝑖 ) ∑ 𝑞 𝑘=0 Π𝑖=1 Γ (𝛽𝑖 + 𝑘𝐵𝑖 ) ∞
𝑘
𝑧 𝑘!
∞
𝑧 ∈ 𝐸.
𝑘=2
If 𝐴 𝑖 = 1 (𝑖 = 1, 2, . . . , 𝑝) and 𝐵𝑖 = 1 (𝑖 = 1, 2, . . . , 𝑞), we have the relationship Ω𝑝 Ψ𝑞 [(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ; 𝑧] = 𝑝 𝐹 (𝛼1 , . . . , 𝛼𝑝 ; 𝛽1 , . . . , 𝛽𝑞 ; 𝑧) 𝑞 ∞
=∑
(𝛼1 )𝑘 ⋅ ⋅ ⋅ (𝛼𝑝 )𝑘 𝑧𝑘
𝑘=0 (𝛽1 )𝑘
⋅ ⋅ ⋅ (𝛽𝑞 )𝑘 𝑘!
(14)
= 𝑧 + ∑ Ω𝜎𝑘 (𝛼1 ) 𝑎𝑘 𝑧𝑘 ,
(9)
(10)
.
(𝑝 ≤ 𝑞 + 1; 𝑝, 𝑞 ∈ 𝑁0 = 𝑁 ∪ {0}; 𝑧 ∈ 𝐸) is the generalized hypergeometric function [8], where 𝑁 denotes the set of all positive integers and (𝛼)𝑘 is the Pochhammer symbol and
where 𝜎𝑘 (𝛼1 ) is given by 𝜎𝑘 (𝛼1 ) =
Γ (𝛼1 + 𝐴 1 (𝑘 − 1)) ⋅ ⋅ ⋅ Γ (𝛼𝑝 + 𝐴 𝑝 (𝑘 − 1)) Γ (𝛽1 + 𝛽1 (𝑘 − 1)) ⋅ ⋅ ⋅ Γ (𝛽𝑞 + 𝛽𝑞 (𝑘 − 1)) (𝑘 − 1)!
.
(15)
For our convenience we write 𝑊[(𝛼𝑖 , 𝐴 𝑖 )1,𝑝 ; (𝛽𝑖 , 𝐵𝑖 )1,𝑞 ] = 𝑊[𝛼1 ]. Motivated by the work of [4, 14–21], we extend the work of Chandrashekar et al. [22] by introducing some new subclasses of SH using the generalized hypergeometric function.
𝑝
Ω=
Π𝑖=1 Γ (𝛼𝑖 ) 𝑞
Π𝑖=1 Γ (𝛽𝑖 )
.
(11)
Definition 1. A function 𝑓 ∈ SH is in class GH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) if
[𝑧2 (𝑊 [𝛼1 ] ℎ (𝑧)) + 2 (𝑊 [𝛼1 ] 𝑔 (𝑧)) + 𝑧2 (𝑊 [𝛼1 ] 𝑔 (𝑧)) ] } { } { 𝑖𝜙 Re {1 + (1 + 𝑒 ) ≥ 𝛾. } } { 𝑧 (𝑊 [𝛼1 ] ℎ (𝑧)) − 𝑧 (𝑊𝑔 (𝑧)) } {
Definition 2. Let TH denote the class of functions of the form ∞
∞
𝑘=2
𝑘=1
𝑓 (𝑧) = 𝑧 − ∑ 𝑎𝑘 𝑧𝑘 − ∑ 𝑏𝑘 𝑧𝑘 ,
𝑏1 < 1,
(17)
and TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) = GH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) ∩ TH .
(16)
Throughout this paper, we shall assume 0 ≤ 𝑟 < 1, 𝑓 = ℎ + 𝑔, Ω, and 𝜎𝑘 (𝛼1 ) as given in (11) and (15), respectively, unless otherwise mentioned.
2. Main Results In Theorem 3, we shall present a sufficient condition for 𝑓 ∈ 𝑆𝐻 to be in class GH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾).
Journal of Function Spaces
3
Theorem 3. Let 𝑓 = ℎ + 𝑔 be given by (2). If ∞
∑𝑘 ( 𝑘=1
(2𝑘 − 1 − 𝛾) (2𝑘 + 1 + 𝛾) 𝑎𝑘 + 𝑏𝑘 ) 𝜎𝑘 (𝛼1 ) Ω 1−𝛾 1−𝛾 (18)
Proof. When inequality (18) holds for the coefficients of 𝑓 = ℎ + 𝑔 given in (2), we have to show that inequality (16) is satisfied. Arranging the left side inequality (16), we have
≤ 2,
then 𝑓 ∈ GH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾).
{ 𝑧 (𝑊 [𝛼1 ] ℎ (𝑧)) + (1 + 𝑒𝑖𝜙 ) 𝑧2 (𝑊 [𝛼1 ] ℎ (𝑧)) + (1 + 2𝑒𝑖𝜙 ) 𝑧 (𝑊 [𝛼1 ] 𝑔 (𝑧)) + (1 + 𝑒𝑖𝜙 ) 𝑧2 (𝑊 [𝛼1 ] 𝑔 (𝑧)) } Re { }≥𝛾 𝑧 (𝑊 [𝛼1 ] ℎ (𝑧)) − 𝑧 (𝑊 [𝛼1 ] 𝑔 (𝑧)) (19) } { 𝐴 (𝑧) . = Re 𝐵 (𝑧) ∞
As we know Re(𝑤) ≥ 𝛾 if and only if |1−𝛾+𝑤| ≥ |1+𝛾−𝑤|, it is sufficient to show that 𝐴 (𝑧) + (1 − 𝛾) 𝐵 (𝑧) − 𝐴 (𝑧) − (1 + 𝛾) 𝐵 (𝑧) ≥ 0.
− (∑𝑘 𝑘=1
(20)
⋅ 𝜎𝑘 (𝛼1 ) Ω} ≥ 0
Substituting the values 𝐴(𝑧) and 𝐵(𝑧) in left side of (20), we obtain 𝐴 (𝑧) + (1 − 𝛾) 𝐵 (𝑧) − 𝐴 (𝑧) − (1 + 𝛾) 𝐵 (𝑧) ≥ (2 ∞
𝑘=2
∞
(1)
𝑘=1 ∞ 𝑘=2 ∞
∞
∑𝑘 {
− ∑ 𝑘 (2𝑘 + 2 + 𝛾) 𝑎𝑘 𝜎𝑘 (𝛼1 ) Ω |𝑧|𝑘 ≥ 2 (1 − 𝛾)
𝑘=1
𝑘=1
∞
⋅ {1 − ∑ 𝑘 𝑘=2 ∞
𝑘=1
Now we obtain the necessary and sufficient condition for the function 𝑓 = ℎ + 𝑔 given by (16) to be in TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾). Theorem 4. Let 𝑓 = ℎ + 𝑔 be given by (16). Then 𝑓 ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) if and only if
− ∑ 𝑘 (2𝑘 − 2 − 𝛾) 𝑎𝑘 𝜎𝑘 (𝛼1 ) Ω |𝑧|𝑘
+ ∑𝑘
(21) by hypothesis in (18), which implies that 𝑓 ∈ GH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾).
− 𝛾) |𝑧| − ∑ 𝑘 (2𝑘 − 𝛾) 𝑎𝑘 𝜎𝑘 (𝛼1 ) Ω |𝑧|𝑘 − ∑ 𝑘 (2𝑘 + 𝛾) 𝑏𝑘 𝜎𝑘 (𝛼1 ) Ω |𝑧|𝑘 − 𝛾 |𝑧|
(2𝑘 − 1 − 𝛾) ∞ (2𝑘 + 1 + 𝛾) 𝑎𝑘 + ∑ 𝑘 𝑏𝑘 ) 1−𝛾 1−𝛾 𝑘=1
(2𝑘 − 1 − 𝛾) (2𝑘 + 1 + 𝛾) 𝑎𝑘 + 𝑏𝑘 } 𝜎𝑘 (𝛼1 ) Ω 1−𝛾 1−𝛾 (22)
≤ 2.
(2𝑘 − 1 − 𝛾) 𝑎𝑘 𝜎𝑘 (𝛼1 ) |Ω| 1−𝛾
Proof. Since TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) ⊂ GH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾), we only have to prove the necessary part of theorem. Assume that 𝑓 ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾), and then by virtue of (16), we obtain
(2𝑘 + 1 + 𝛾) 𝑏𝑘 𝜎𝑘 (𝛼1 ) Ω} = 2 (1 − 𝛾) {2 1−𝛾
[𝑧2 (𝑊 [𝛼1 ] ℎ (𝑧)) + 2 (𝑊 [𝛼1 ] 𝑔 (𝑧)) + 𝑧2 (𝑊 [𝛼1 ] 𝑔 (𝑧)) ] } { } { 𝑖𝜙 Re {(1 − 𝛾) + (1 + 𝑒 ) ≥ 0. } } { 𝑧 (𝑊 [𝛼1 ] ℎ (𝑧)) − 𝑧 (𝑊 [𝛼1 ] 𝑔 (𝑧)) } {
(23)
This is equivalent to 𝑘 ∞ ∞ { (1 − 𝛾) 𝑧 − (∑𝑘=2 𝑘 [𝑘 (1 + 𝑒𝑖𝜙 ) − 𝛾 − 𝑒𝑖𝜙 ] 𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 𝑧𝑘 + ∑𝑘=1 𝑘 [𝑘 (1 + 𝑒𝑖𝜙 ) + 𝛾 + 𝑒𝑖𝜙 ] 𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 𝑧 ) } Re { } 𝑘 𝑘 ∞ 𝑧 − ∑∞ 𝑘=2 𝑘𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 𝑧 − ∑𝑘=1 𝑘𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 𝑧 } { 𝑘−1 𝑘−1 ∞ ∞ 𝑖𝜙 𝑖𝜙 𝑖𝜙 𝑖𝜙 } (24) { (1 − 𝛾) − ∑𝑘=2 𝑘 [𝑘 (1 + 𝑒 ) − 𝛾 − 𝑒 ] 𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 𝑧 − (𝑧/𝑧) ∑𝑘=1 𝑘 [𝑘 (1 + 𝑒 ) + 𝛾 + 𝑒 ] 𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 𝑧 = Re { } 𝑘−1 𝑘−1 ∞ ∞ 1 − ∑𝑘=2 𝑘𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 𝑧 + (𝑧/𝑧) ∑𝑘=1 𝑘𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 𝑧 } { ≥ 0.
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Journal of Function Spaces
This condition must hold for all values of 𝑧 ∈ 𝐸 and for real 𝜙, so that, by taking 𝑧 = 𝑟 < 1 and 𝜙 = 0, the above inequality reduces to 𝑘−1 𝑘−1 + (𝑧/𝑧) ∑∞ ] (1 − 𝛾) − [∑∞ 𝑘=2 𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 𝑟 𝑘=1 𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 𝑟 ≥ 0. 𝑘−1 𝑘−1 ∞ ∞ 1 − ∑𝑘=2 𝑘𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 𝑟 + (𝑧/𝑧) ∑𝑘=1 𝑘𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 𝑟 Letting 𝑟 → 1− through real values, we obtain condition (22). This completes the proof.
(25)
Using (22) for the coefficients in (29), we have ∞
𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝐴𝑘 1−𝛾 𝑘=2 ∑
We determine the extreme points of closed convex hulls of TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) denoted by clcoTH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾). Theorem 5. A function 𝑓(𝑧) ∈ 𝑐𝑙𝑐𝑜TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) if and only if ∞
𝑓 (𝑧) = ∑ (𝑋𝑘 ℎ𝑘 (𝑧) + 𝑌𝑘 𝑔𝑘 (𝑧)) ,
(26)
𝑘=1
ℎ1 (𝑧) = 𝑧,
𝑔𝑘 (𝑧) = 𝑧 −
∞ ∞ 𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝐵𝑘 = ∑ 𝑋𝑘 + ∑ 𝑌𝑘 1−𝛾 𝑘=1 𝑘=2 𝑘=1
1−𝛾 𝑧𝑘 ; 𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 (𝛼1 ) Ω 1−𝛾 𝑧𝑘 ; 𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 Ω
(𝑘 ≥ 2) ,
(30)
= 1 − 𝑥1 ≤ 1, and hence 𝑓(𝑧) ∈ clcoTH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾). Conversely, suppose that 𝑓(𝑧) ∈ clcoTH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾), and set 𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝐴 𝑘, 1−𝛾
(𝑘 ≥ 2) ,
𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝑌𝑘 = 𝐵𝑘 , 1−𝛾
(𝑘 ≥ 1) ,
𝑋𝑘 =
where
ℎ𝑘 (𝑧) = 𝑧 −
∞
+∑
(31)
where ∑∞ 𝑘=1 (𝑋𝑘 + 𝑌𝑘 ) = 1. Then (27)
(𝑘 ≥ 2) ,
∞
∞
𝑘=2
𝑘=1
𝑓 (𝑧) = 𝑧 − ∑ 𝐴 𝑘 𝑧𝑘 − ∑ 𝐵𝑘 𝑧𝑘 , 𝐴 𝑘 ≥ 0, 𝐵𝑘 ≥ 0 ∞
∑∞ 𝑘=1 (𝑋𝑘
+ 𝑌𝑘 ) = 1, 𝑋𝑘 ≥ 0, and 𝑌𝑘 ≥ 0. In particular, the extreme points of TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) are {ℎ𝑘 } and {𝑔𝑘 }.
1−𝛾 𝑋𝑘 𝑧𝑘 𝑘 (2𝑘 − 1 − 𝛾) 𝜎 (𝛼 ) Ω 𝑘 1 𝑘=2
=𝑧−∑ ∞
1−𝛾 𝑌𝑘 𝑧𝑘 𝑘 (2𝑘 + 1 + 𝛾) 𝜎 (𝛼 ) Ω 𝑘 1 𝑘=1
−∑
Proof. First, we consider ∞
∞
∞
𝑘=2
𝑘=1
(32)
= 𝑧 + ∑ (ℎ𝑘 (𝑧) − 𝑧) 𝑋𝑘 + ∑ (𝑔𝑘 (𝑧) − 𝑧)𝑘 𝑌𝑘
𝑓 (𝑧) = ∑ (𝑋𝑘 ℎ𝑘 (𝑧) + 𝑌𝑘 𝑔𝑘 (𝑧)) 𝑘=1
∞
∞
= ∑ (𝑋𝑘 ℎ𝑘 (𝑧) + 𝑌𝑘 𝑔𝑘 (𝑧)) ,
= ∑ (𝑋𝑘 + 𝑌𝑘 ) 𝑧
𝑘=1
𝑘=1
∞
1−𝛾 𝑋𝑘 𝑧𝑘 𝑘 (2𝑘 − 1 − 𝛾) 𝜎 (𝛼 ) Ω 𝑘 1 𝑘=2
(28)
−∑
Next we show that TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) is close under convex combinations of its members.
∞
1−𝛾 𝑌𝑘 𝑧𝑘 , 𝑘 (2𝑘 + 1 + 𝛾) 𝜎 (𝛼 ) Ω 𝑘 1 𝑘=1
−∑
∞
𝑘=1
Theorem 6. The family TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) is closed under convex combination.
∞
𝑓 (𝑧) = 𝑧 − ∑ 𝐴 𝑘 𝑧𝑘 − ∑ 𝐵𝑘 𝑧𝑘 ,
which is the required result.
(29)
𝑘=1
where 𝐴 𝑘 = ((1 − 𝛾)/𝑘(2𝑘 − 1 − 𝛾)𝜎𝑘 (𝛼1 )Ω)𝑋𝑘 , and 𝐵𝑘 = ((1 − 𝛾)/𝑘(2𝑘 + 1 + 𝛾)𝜎𝑘 (𝛼1 )Ω)𝑌𝑘 .
Proof. For 𝑖 = 1, 2, . . ., suppose that 𝑓𝑖 ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾), where ∞
∞
𝑘=2
𝑘=2
𝑓𝑖 (𝑧) = 𝑧 − ∑ 𝑎𝑖,𝑘 𝑧𝑘 − ∑ 𝑏𝑖,𝑘 𝑧𝑘 .
(33)
Journal of Function Spaces
5
For coefficients in (22) the relation given in (33) takes the form
We note that |𝐴 𝑘 | ≤ 1 and |𝐵𝑘 | ≤ 1. Therefore ∞
𝑘 (2𝑘 − 1 − 𝛿) 𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 𝐴 𝑘 1−𝛿 𝑘=2 ∑
∞
∞ 𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 Ω 𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 Ω 𝑎𝑖,𝑘 + ∑ 𝑏𝑖,𝑘 ∑ 1−𝛾 1−𝛾 𝑘=2 𝑘=1 (34)
∞
𝑘 (2𝑘 + 1 + 𝛿) 𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 𝐵𝑘 1−𝛿 𝑘=1
+∑
≤ 1. For ∑∞ 𝑖=1 𝑡𝑖 = 1, 0 ≤ 𝑡𝑖 ≤ 1, the convex combination of 𝑓𝑖 may be written as ∞
∞
∞
∞
∞
𝑖=1
𝑘=2
𝑖=1
𝑘=2
𝑖=1
∞
𝑘 (2𝑘 − 1 − 𝛿) 𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 1−𝛿 𝑘=2
≤∑ ∞
𝑘 (2𝑘 + 1 + 𝛿) 𝜎𝑘 (𝛼1 ) Ω +∑ 𝑏𝑘 1−𝛿 𝑘=1
∑𝑡𝑖 𝑓𝑖 (𝑧) = 𝑧 − ∑ (∑𝑡𝑖 𝑎𝑖,𝑘 ) 𝑧𝑘 − ∑ (∑𝑡𝑖 𝑏𝑖,𝑘 )𝑧𝑘 . (35)
(40)
∞
𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝑎𝑘 1−𝛾 𝑘=2
≤∑
From inequality (22) for (35), ∞
𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 (𝛼1 ) Ω ∞ (∑𝑡𝑖 𝑎𝑖,𝑘 ) 1−𝛾 𝑖=1 𝑘=2
∞
𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝑏𝑘 ≤ 1, 1−𝛾 𝑘=1
∑
𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 (𝛼1 ) Ω ∞ (∑𝑡𝑖 𝑎𝑖,𝑘 ) 1−𝛾 𝑖=1 𝑘=1
+∑
∞
+∑ ∞
∞
𝑘 (2𝑘 − 1 − 𝛾) 𝜎𝑘 (𝛼1 ) Ω ⋅ ∑𝑡𝑖 ( ∑ 𝑎𝑖,𝑘 1−𝛾 𝑖=1 𝑘=2
(36)
by using (22), since 𝑓(𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) and 0 ≤ 𝛿 ≤ 𝛾 ≤ 1. This proves that 𝑓(𝑧) ∗ 𝐹(𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛿). Now for the class TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) the closure property under the generalized Bernardi-Livingston integral operator 𝐿 𝑐 (𝑓) is examined which is defined by
∞
∞ 𝑘 (2𝑘 + 1 + 𝛾) 𝜎𝑘 (𝛼1 ) Ω 𝑏𝑖,𝑘 ) ≤ ∑𝑡𝑖 = 1, 1−𝛾 𝑡=1 𝑘=1
+∑
𝐿 𝑐 (𝑓) =
and therefore ∑∞ 𝑡=1 𝑡𝑖 𝑓𝑖 (𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾). Theorem 7. For 0 ≤ 𝛿 ≤ 𝛾 ≤ 1, let 𝑓(𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) and 𝐹(𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛿). Then 𝑓 (𝑧) ∗ 𝐹 (𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ] , 𝛾) ⊂ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ] , 𝛿) .
∞
𝑘=2
𝑘=1
(37) 𝐿 𝑐 (𝑓) =
𝑓 (𝑧) = 𝑧 − ∑ 𝑎𝑘 𝑧𝑘 − ∑ 𝑏𝑘 𝑧𝑘
∞
−∫
0
(38)
𝐹 (𝑧) = 𝑧 − ∑ 𝐴 𝑘 𝑧𝑘 − ∑ 𝐵𝑘 𝑧𝑘 𝑘=2
𝑐 + 1 𝑧 𝑐−1 ∫ 𝑡 [ℎ (𝑡) + 𝑔 (𝑡)] 𝑑𝑡 𝑧𝑐 0
∞ 𝑧 𝑐+1 𝑐−1 (∫ 𝑡 (𝑡 − 𝑎𝑘 𝑡𝑘 ) 𝑑𝑡 ∑ 𝑧𝑐 0 𝑘=2 𝑧
∞
𝑡𝑐−1
∞
∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ] , 𝛿) ,
( ∑ 𝑏𝑘 𝑘=1
∞
∞
𝑘=2
𝑘=1
where
and then
𝐴𝑘 = ∞
𝑘=2
∞
𝑘=1
(42) 𝑡𝑘 )) ,
= 𝑧 − ∑ 𝐴 𝑘 𝑧𝑘 − ∑ 𝐵𝑘 𝑧𝑘 ,
𝑘=1
𝑓 (𝑧) ∗ 𝐹 (𝑧) = 𝑧 + ∑ 𝑎𝑘 𝐴 𝑘 𝑧𝑘 + ∑ 𝑏𝑘 𝐵𝑘 𝑧𝑘 .
(41)
Proof. Consider the generalized Bernardi-Livingston integral operator 𝐿 𝑐 (𝑓) given in (41):
=
∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ] , 𝛾) ,
𝑐 > −1.
Theorem 8. Let 𝑓(𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾) and then 𝐿 𝑐 (𝑓) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾).
Proof. Let ∞
𝑐 + 1 𝑧 𝑐−1 ∫ 𝑡 𝑓 (𝑡) 𝑑𝑡, 𝑧𝑐 0
(39)
𝑐+1 𝑎; 𝑐+𝑛 𝑘
𝑐+1 𝐵𝑘 = 𝑏. 𝑐+𝑛 𝑘
(43)
6
Journal of Function Spaces Therefore ∞
∑𝑘 ( 𝑘=1
+
(2𝑘 − 1 − 𝛾) 𝑐 + 1 ( ) 𝑎 1−𝛾 𝑐 + 𝑛 𝑘
(2𝑘 + 1 + 𝛾) 𝑐 + 1 ( ) 𝑏 ) 𝜎 (𝛼 ) Ω 1−𝛾 𝑐 + 𝑛 𝑘 𝑘 1 ∞
≤ ∑𝑘 ( 𝑘=1
(44)
(2𝑘 − 1 − 𝛾) (2𝑘 + 1 + 𝛾) 𝑎𝑘 + 𝑏𝑘 ) 1−𝛾 1−𝛾
⋅ 𝜎𝑘 (𝛼1 ) Ω ≤ 2 (1 − 𝛾) . Since 𝑓(𝑧) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾), therefore, by Theorem 4, 𝐿 𝑐 (𝑓) ∈ TH ([𝛼1 , 𝐴 1 , 𝛽1 , 𝐵1 ], 𝛾).
Competing Interests The authors declare that they have no competing interests.
Acknowledgments This work is supported by AP-2013-009 and GP-K006392, Universiti Kebangsaan Malaysia.
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