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Research ArticleConvolution Properties for Some Subclasses of
MeromorphicFunctions of Complex Order
M. K. Aouf,1 A. O. Mostafa,1 and H. M. Zayed2
1Department of Mathematics, Faculty of Science, Mansoura
University, Mansoura 35516, Egypt2Department of Mathematics,
Faculty of Science, Menofia University, Shebin El Kom 32511,
Egypt
Correspondence should be addressed to H. M. Zayed; hanaa
[email protected]
Received 3 June 2015; Accepted 5 July 2015
Academic Editor: Allan Peterson
Copyright © 2015 M. K. Aouf 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.
Making use of the operator L𝜐for functions of the form 𝑓(𝑧) =
1/𝑧 + ∑∞
𝑘=1𝑎𝑘𝑧𝑘−1, which are analytic in the punctured unit disc
U∗ = {𝑧 : 𝑧 ∈ C and 0 < |𝑧| < 1} = U \ {0}, we introduce
two subclasses of meromorphic functions and investigate
convolutionproperties, coefficient estimates, and containment
properties for these subclasses.
1. Introduction
Let Σ denote the class of meromorphic functions of the form
𝑓 (𝑧) =1𝑧+∞
∑𝑘=1𝑎𝑘𝑧𝑘−1, (1)
which are analytic in the punctured unit discU∗ = {𝑧 : 𝑧 ∈ Cand
0 < |𝑧| < 1} = U \ {0}. Let 𝑔(𝑧) ∈ Σ be given by
𝑔 (𝑧) =1𝑧+∞
∑𝑘=1𝑏𝑘𝑧𝑘−1; (2)
then, the Hadamard product (or convolution) of 𝑓(𝑧) and𝑔(𝑧) is
given by
(𝑓 ∗ 𝑔) (𝑧) =1𝑧+∞
∑𝑘=1𝑎𝑘𝑏𝑘𝑧𝑘 = (𝑔 ∗ 𝑓) (𝑧) . (3)
We recall some definitions which will be used in our paper.
Definition 1. For two functions 𝑓(𝑧) and 𝑔(𝑧), analytic in U,we
say that the function 𝑓(𝑧) is subordinate to 𝑔(𝑧) in Uand written
𝑓(𝑧) ≺ 𝑔(𝑧), if there exists a Schwarz function𝑤(𝑧), analytic in U
with 𝑤(0) = 0 and |𝑤(𝑧)| < 1 such that𝑓(𝑧) = 𝑔(𝑤(𝑧)) (𝑧 ∈ U).
Furthermore, if the function 𝑔(𝑧) is
univalent in U, then we have the following equivalence
(see[1]):
𝑓 (𝑧) ≺ 𝑔 (𝑧)
⇐⇒ 𝑓 (0) = 𝑔 (0) ,
𝑓 (U) ⊂ 𝑔 (U) .
(4)
Now, consider Bessel’s function of the first kind of order
𝜐where 𝜐 is an unrestricted (real or complex) number, definedby
(see Watson [2, page 40]) (see also Baricz [3, page 7])
𝐽𝜐(𝑧) =
∞
∑𝑘=0
(−1)𝑘
Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)(𝑧
2)2𝑘+]
, (5)
which is a particular solution of the second order
linearhomogenous Bessel differential equation (see, e.g.,Watson
[2,page 38]) (see also Baricz [3, page 7])
𝑧2𝑤 (𝑧) + 𝑧𝑤
(𝑧) + (𝑧2 − 𝜐2)𝑤 (𝑧) = 0. (6)
Also, let us define
L𝜐(𝑧) =
2𝜐Γ (𝜐 + 1)𝑧𝜐/2+1
𝐽𝜐(𝑧1/2)
=1𝑧+∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑧𝑘−1
(𝑧 ∈ U∗) .
(7)
Hindawi Publishing CorporationAbstract and Applied
AnalysisVolume 2015, Article ID 973613, 6
pageshttp://dx.doi.org/10.1155/2015/973613
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2 Abstract and Applied Analysis
The operatorL𝜐is a modification of the operator introduced
by Szász and Kupán [4] for analytic functions.By using the
Hadamard product (or convolution), we
define the operatorL𝜐as follows:
(L𝜐𝑓) (𝑧) = L
𝜐(𝑧) ∗ 𝑓 (𝑧)
=1𝑧+∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘−1.
(8)
It is easy to verify from (8) that
𝑧 ((L𝜐+1𝑓) (𝑧))
= (𝜐 + 1) (L𝜐𝑓) (𝑧)
− (𝜐 + 2) (L𝜐+1𝑓) (𝑧) .
(9)
Definition 2. For 0 ≤ 𝜆 < 1, −1 ≤ 𝐵 < 𝐴 ≤ 1, and 𝑏 ∈C∗ = C
\ {0}, let ΣS∗
𝜆[𝑏; 𝐴, 𝐵] be the subclass of Σ consisting
of function 𝑓(𝑧) of the form (1) and satisfying the
analyticcriterion
1+ 1𝑏[
−𝑧𝑓 (𝑧)
(1 − 𝜆) 𝑓 (𝑧) − 𝜆𝑧𝑓 (𝑧)− 1] ≺ 1 + 𝐴𝑧
1 + 𝐵𝑧. (10)
Also, let ΣK𝜆[𝑏; 𝐴, 𝐵] be the subclass of Σ consisting of
function 𝑓(𝑧) of the form (1) and satisfying the
analyticcriterion
1+ 1𝑏[
[
−𝑧 (𝑧𝑓 (𝑧))
(1 − 𝜆) 𝑧𝑓 (𝑧) − 𝜆𝑧 (𝑧𝑓 (𝑧))− 1]
]
≺1 + 𝐴𝑧1 + 𝐵𝑧
.
(11)
It is easy to verify from (10) and (11) that
𝑓 (𝑧) ∈ ΣK𝜆[𝑏; 𝐴, 𝐵]
⇐⇒ −𝑧𝑓 (𝑧) ∈ ΣS∗
𝜆[𝑏; 𝐴, 𝐵] .
(12)
We note that
(i) ΣS∗0 [𝑏; 𝐴, 𝐵] = ΣS∗[𝑏; 𝐴, 𝐵] and ΣK0[𝑏; 𝐴, 𝐵] =
ΣK[𝑏; 𝐴, 𝐵] (see Bulboacă et al. [5]);(ii) ΣS∗0 [𝑏; 1, −1] =
ΣS(𝑏) and ΣK0[𝑏; 1, −1] = ΣK(𝑏)
(see Aouf [6]);(iii) ΣS∗0 [(1 − 𝛼)𝑒
−𝑖𝜇cos𝜇; 1, −1] = ΣS𝜇(𝛼) and ΣK0[(1 −𝛼)𝑒−𝑖𝜇cos𝜇; 1, −1] = ΣK𝜇(𝛼)
(𝜇 ∈ R, |𝜇| ≤ 𝜋/2, 0 ≤𝛼 < 1) (see Ravichandran et al. [7, with 𝑝
= 1]).
Definition 3. For 0 ≤ 𝜆 < 1, −1 ≤ 𝐵 < 𝐴 ≤ 1, 𝑏 ∈ C∗ and
𝜐is an unrestricted (real or complex) number, let
ΣS∗𝜆,𝜐[𝑏; 𝐴, 𝐵]
= {𝑓 (𝑧) ∈ Σ : (L𝜐𝑓) (𝑧) ∈ ΣS
∗
𝜆[𝑏; 𝐴, 𝐵]} ,
ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵]
= {𝑓 (𝑧) ∈ Σ : (L𝜐𝑓) (𝑧) ∈ ΣK
𝜆[𝑏; 𝐴, 𝐵]} .
(13)
It is easy to show that
𝑓 (𝑧) ∈ ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵]
⇐⇒ −𝑧𝑓 (𝑧) ∈ ΣS∗
𝜆,𝜐[𝑏; 𝐴, 𝐵] .
(14)
The object of the present paper is to investigatesome
convolution properties, coefficient estimates, and con-tainment
properties for the subclasses ΣS∗
𝜆,𝜐[𝑏; 𝐴, 𝐵] and
ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵].
2. Main Results
Unless otherwise mentioned, we assume throughout thispaper that
0 ≤ 𝜆 < 1, −1 ≤ 𝐵 < 𝐴 ≤ 1, 𝑏 ∈ C∗ and 𝜐 isan unrestricted
(real or complex) number.
Theorem 4. If 𝑓(𝑧) ∈ Σ, then 𝑓(𝑧) ∈ ΣS∗𝜆[𝑏; 𝐴, 𝐵] if and
only
if
𝑧 [𝑓 (𝑧) ∗1 − [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧
𝑧 (1 − 𝑧)2] ̸= 0
𝑓𝑜𝑟 𝑧 ∈ U,
(15)
where𝑀 = 𝑀𝜃= (𝑒−𝑖𝜃 + 𝐵)/(𝐴 − 𝐵)𝑏, 𝜃 ∈ [0, 2𝜋), and also
𝑀 = 0.
Proof. It is easy to verify that
𝑓 (𝑧) ∗1
𝑧 (1 − 𝑧)= 𝑓 (𝑧) ,
𝑓 (𝑧) ∗ [1
𝑧 (1 − 𝑧)2−
2(1 − 𝑧)2
] = − 𝑧𝑓 (𝑧)
∀𝑧 ∈ U∗; 𝑓 ∈ Σ.
(16)
(i) In view of (10), 𝑓(𝑧) ∈ ΣS∗𝜆[𝑏; 𝐴, 𝐵] if and only if
(10)
holds. Since the function (1 + [𝐵 + (𝐴 − 𝐵)𝑏]𝑧)/(1 + 𝐵𝑧)
isanalytic in U, it follows that (1 − 𝜆)𝑓(𝑧) − 𝜆𝑧𝑓(𝑧) ̸= 0 for𝑧 ∈
U∗ or 𝑧[(1 − 𝜆)𝑓(𝑧) − 𝜆𝑧𝑓(𝑧)] ̸= 0 for 𝑧 ∈ U; this isequivalent to
(15) holdding for 𝑀 = 0. To prove (15) for all𝑀 ̸= 0, we write (10)
by using the principle of subordinationas
−𝑧𝑓 (𝑧)
(1 − 𝜆) 𝑓 (𝑧) − 𝜆𝑧𝑓 (𝑧)=1 + [𝐵 + (𝐴 − 𝐵) 𝑏]𝑤 (𝑧)
1 + 𝐵𝑤 (𝑧), (17)
where 𝑤(𝑧) is Schwarz function, analytic in U with 𝑤(0) = 0and
|𝑤(𝑧)| < 1; hence,
𝑧 [−𝑧𝑓 (𝑧) (1+𝐵𝑒𝑖𝜃) − [(1−𝜆) 𝑓 (𝑧) − 𝜆𝑧𝑓 (𝑧)]
⋅ [1+ [𝐵+ (𝐴−𝐵) 𝑏] 𝑒𝑖𝜃]] ̸= 0
for 𝑧 ∈ U, 𝜃 ∈ [0, 2𝜋) .
(18)
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Abstract and Applied Analysis 3
Using (16), (18) may be written as
𝑧 [𝑓 (𝑧)
∗1 − [(𝜆 − 1) ((𝑒−𝑖𝜃 + 𝐵) / (𝐴 − 𝐵) 𝑏) + (𝜆 + 1)] 𝑧
𝑧 (1 − 𝑧)2]
̸= 0 for 𝑧 ∈ U.
(19)
Thus, the first part of Theorem 4 was proved.(ii) Reversely,
because assumption (15) holds for𝑀 = 0,
it follows that 𝑧[(1 − 𝜆)𝑓(𝑧) − 𝜆𝑧𝑓(𝑧)] ̸= 0 for 𝑧 ∈ U.
Thisimplies that𝜑(𝑧) = −𝑧𝑓(𝑧)/((1−𝜆)𝑓(𝑧)−𝜆𝑧𝑓(𝑧)) is analyticin U
(i.e., it is regular in 𝑧 = 0, with 𝜑(0) = 1).
Since it was shown in the first part of the proof thatassumption
(18) is equivalent to (15), we obtain that
−𝑧𝑓 (𝑧)
(1 − 𝜆) 𝑓 (𝑧) − 𝜆𝑧𝑓 (𝑧)̸=1 + [𝐵 + (𝐴 − 𝐵) 𝑏] 𝑒𝑖𝜃
1 + 𝐵𝑒𝑖𝜃
for 𝑧 ∈ U, 𝜃 ∈ [0, 2𝜋) .
(20)
Assume that
𝜓 (𝑧) =1 + [𝐵 + (𝐴 − 𝐵) 𝑏] 𝑒𝑖𝜃
1 + 𝐵𝑒𝑖𝜃. (21)
Relation (20) means that 𝜑(U) ∩ 𝜓(𝜕U) = 0. Thus, the
simplyconnected domain is included in a connected component ofC \
𝜓(𝜕U). From this, using the fact that 𝜑(0) = 𝜓(0) andthe univalence
of the function 𝜓, it follows that 𝜑(𝑧) ≺ 𝜓(𝑧);this implies that
𝑓(𝑧) ∈ ΣS∗
𝜆,𝜐[𝑏; 𝐴, 𝐵]. Thus, the proof of
Theorem 4 is completed.
Remark 5. (i) Putting 𝜆 = 0 in Theorem 4, we obtain theresult
obtained by Bulboacă et al. [5, Theorem 1].
(ii) Putting 𝜆 = 0, 𝑏 = 1, and 𝑒𝑖𝜃 = 𝑥 in Theorem 4, weobtain
the result obtained by Ponnusamy [8, Theorem 2.1].
(iii) Putting 𝜆 = 0, 𝑏 = (1 − 𝛼)𝑒−𝑖𝜇cos𝜇 (𝜇 ∈ R, |𝜇| ≤𝜋/2, 0 ≤ 𝛼
< 1),𝐴 = 1, 𝐵 = −1, and 𝑒𝑖𝜃 = 𝑥 inTheorem 4, weobtain the result
obtained by Ravichandran et al. [7,Theorem1.2 with 𝑝 = 1].
Theorem 6. If 𝑓(𝑧) ∈ Σ, then 𝑓(𝑧) ∈ ΣK𝜆[𝑏; 𝐴, 𝐵] if and
only if
𝑧 [𝑓 (𝑧) ∗1 − 3𝑧 + 2 [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧2
𝑧 (1 − 𝑧)3] ̸= 0
𝑓𝑜𝑟 𝑧 ∈ U,
(22)
where𝑀 = 𝑀𝜃= (𝑒−𝑖𝜃 + 𝐵)/(𝐴 − 𝐵)𝑏, 𝜃 ∈ [0, 2𝜋), and also
𝑀 = 0.
Proof. Putting
𝑔 (𝑧) =1 − [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧
𝑧 (1 − 𝑧)2, (23)
then
− 𝑧𝑔 (𝑧) =1 − 3𝑧 + 2 [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧2
𝑧 (1 − 𝑧)3. (24)
From (12) and using the identity
[−𝑧𝑓 (𝑧)] ∗ 𝑔 (𝑧) = 𝑓 (𝑧) ∗ [−𝑧𝑔
(𝑧)] , (25)
we obtain the required result fromTheorem 4.
Remark 7. (i) Putting𝜆 = 0 inTheorem6,we obtain the
resultobtained by Bulboacă et al. [5, Theorem 2].
(ii) Putting 𝜆 = 0, 𝑏 = 1, and 𝑒𝑖𝜃 = 𝑥 in Theorem 4, weobtain
the result obtained by Ponnusamy [8, Theorem 2.2].
Theorem 8. If 𝑓(𝑧) ∈ Σ, then 𝑓(𝑧) ∈ ΣS∗𝜆,𝜐[𝑏; 𝐴, 𝐵] if and
only if
1+∞
∑𝑘=1
(−1)𝑘 (1 − 𝑘𝜆) Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘 ̸= 0, (26)
1+∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 1)
⋅ [(1 − 𝑘𝜆) (𝐴 − 𝐵) 𝑏
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
−−𝑘 (𝜆 − 1) (𝑒−𝑖𝜃 + 𝐵)
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)] 𝑎𝑘𝑧𝑘 ̸= 0,
(27)
for all 𝜃 ∈ [0, 2𝜋).
Proof. If 𝑓(𝑧) ∈ Σ, from Theorem 4, we have 𝑓(𝑧) ∈ΣS∗𝜆,𝜐[𝑏; 𝐴,
𝐵] if and only if
𝑧 [(L𝜐𝑓) (𝑧) ∗
1 − [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧𝑧 (1 − 𝑧)2
] ̸= 0
for 𝑧 ∈ U,(28)
where𝑀 = 𝑀𝜃= (𝑒−𝑖𝜃 + 𝐵)/(𝐴 − 𝐵)𝑏, 𝜃 ∈ [0, 2𝜋), and also
𝑀 = 0. Since
1 − (𝜆 + 1) 𝑧𝑧 (1 − 𝑧)2
=1𝑧+∞
∑𝑘=1
(1− 𝑘𝜆) 𝑧𝑘−1, 𝑧 ∈ U∗, (29)
it is easy to show that (28) holds for𝑀 = 0 if and only if
(26)holds. Also,
1 − [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧𝑧 (1 − 𝑧)2
=1𝑧+∞
∑𝑘=1
[(1− 𝑘𝜆) − 𝑘 (𝜆 − 1)𝑀] 𝑧𝑘−1, 𝑧 ∈ U∗;(30)
we may easily check that (28) is equivalent to (27).
Thiscompletes the proof of Theorem 8.
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4 Abstract and Applied Analysis
Theorem 9. If 𝑓(𝑧) ∈ Σ, then 𝑓(𝑧) ∈ ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵] if and
only if
1+∞
∑𝑘=1
(−1)𝑘 (𝑘𝜆 − 1) (𝑘 − 1) Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘 ̸= 0, (31)
1+∞
∑𝑘=1
(−1)𝑘 (𝑘 − 1) Γ (𝜐 + 1)
⋅ [(𝑘𝜆 − 1) (𝐴 − 𝐵) 𝑏
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
+𝑘 (𝜆 − 1) (𝑒−𝑖𝜃 + 𝐵)
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)] 𝑎𝑘𝑧𝑘 ̸= 0,
(32)
for all 𝜃 ∈ [0, 2𝜋).
Proof. If 𝑓(𝑧) ∈ Σ, from Theorem 6, we have 𝑓(𝑧) ∈ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵]
if and only if
𝑧 [(L𝜐𝑓) (𝑧) ∗
1 − 3𝑧 + 2 [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧2
𝑧 (1 − 𝑧)3]
̸= 0 for 𝑧 ∈ U,
(33)
where𝑀 = 𝑀𝜃= (𝑒−𝑖𝜃 + 𝐵)/(𝐴 − 𝐵)𝑏, 𝜃 ∈ [0, 2𝜋), and also
𝑀 = 0. Since
1 − 3𝑧 + 2 (𝜆 + 1) 𝑧2
𝑧 (1 − 𝑧)3=1𝑧+∞
∑𝑘=1
(𝑘𝜆 − 1) (𝑘 − 1) 𝑧𝑘−1,
𝑧 ∈ U∗,
(34)
it is easy to show that (33) holds for𝑀 = 0 if and only if
(31)holds. Also,
1 − 3𝑧 + 2 [(𝜆 − 1)𝑀 + (𝜆 + 1)] 𝑧2
𝑧 (1 − 𝑧)3
=1𝑧+∞
∑𝑘=1
(𝑘 − 1) [𝑘 (𝜆 − 1)𝑀+ (𝑘𝜆 − 1)] 𝑧𝑘−1;(35)
for 𝑧 ∈ U∗, we may easily check that (33) is equivalent to
(32).This completes the proof of Theorem 9.
Unless otherwise mentioned, we assume throughout theremainder of
this section that 𝜐 is a real number (𝜐 > −1).
Theorem 10. If 𝑓(𝑧) ∈ Σ satisfies inequalities∞
∑𝑘=1
|𝑘𝜆 − 1| Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘 < 1, (36)
∞
∑𝑘=1
[(|1 − 𝑘𝜆|) (𝐴 − 𝐵) |𝑏| + 𝑘 (1 − 𝜆) (1 + |𝐵|)] Γ (𝜐 + 1)4𝑘Γ (𝑘 +
1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘
< (𝐴−𝐵) |𝑏| ,
(37)
then 𝑓(𝑧) ∈ ΣS∗𝜆,𝜐[𝑏; 𝐴, 𝐵].
Proof. We have
1−∞
∑𝑘=1
(−1)𝑘 (𝑘𝜆 − 1) Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘
≥ 1−
∞
∑𝑘=1
(−1)𝑘 (𝑘𝜆 − 1) Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘
≥ 1−∞
∑𝑘=1
|𝑘𝜆 − 1| Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘
≥ 1−∞
∑𝑘=1
|𝑘𝜆 − 1| Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘
> 0, for 𝑧 ∈ U,
(38)
which implies inequality (36). Also,
1+∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 1)
⋅ [(1 − 𝑘𝜆) (𝐴 − 𝐵) 𝑏
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
−𝑘 (𝜆 − 1) (𝑒−𝑖𝜃 + 𝐵)
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)] 𝑎𝑘𝑧𝑘
≥ 1
−
∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 1)
⋅ [(1 − 𝑘𝜆) (𝐴 − 𝐵) 𝑏
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
−𝑘 (𝜆 − 1) (𝑒−𝑖𝜃 + 𝐵)
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)] 𝑎𝑘𝑧𝑘
≥ 1
−∞
∑𝑘=1Γ (𝜐 + 1)
⋅ [|1 − 𝑘𝜆| (𝐴 − 𝐵) |𝑏|
4𝑘 (𝐴 − 𝐵) |𝑏| Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
+𝑘 (1 − 𝜆) (1 + |𝐵|)
4𝑘 (𝐴 − 𝐵) |𝑏| Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)]𝑎𝑘𝑧𝑘
≥ 1−∞
∑𝑘=1Γ (𝜐 + 1)
⋅ [|1 − 𝑘𝜆| (𝐴 − 𝐵) |𝑏|
4𝑘 (𝐴 − 𝐵) |𝑏| Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
+𝑘 (1 − 𝜆) (1 + |𝐵|)
4𝑘 (𝐴 − 𝐵) |𝑏| Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)]𝑎𝑘 > 0,
for 𝑧 ∈ U,
(39)
-
Abstract and Applied Analysis 5
which implies inequality (37). Thus, the proof of Theorem 10is
completed.
Using similar arguments to those in the proof of Theo-rem 10, we
obtain the following theorem.
Theorem 11. If 𝑓(𝑧) ∈ Σ satisfies inequalities∞
∑𝑘=1
|𝑘𝜆 − 1| (𝑘 − 1) Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘 < 1,
∞
∑𝑘=1
(𝑘 − 1) Γ (𝜐 + 1) [ (|𝑘𝜆 − 1|) (𝐴 − 𝐵) |𝑏|4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 +
1)
+𝑘 (1 − 𝜆) (1 + |𝐵|)
4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)]𝑎𝑘 < (𝐴−𝐵) |𝑏| ,
(40)
then 𝑓(𝑧) ∈ ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵].
Now, using themethod due toAhuja [9], wewill prove
thecontainment relations for the subclasses ΣS∗
𝜆,𝜐[𝑏; 𝐴, 𝐵] and
ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵].
Theorem 12. For 𝜐 > −1, we have ΣS∗𝜆,𝜐+1[𝑏; 𝐴, 𝐵] ⊂
ΣS∗𝜆,𝜐[𝑏; 𝐴, 𝐵].
Proof. Since 𝑓(𝑧) ∈ ΣS∗𝜆,𝜐+1[𝑏; 𝐴, 𝐵], we see fromTheorem 8
that
1+∞
∑𝑘=1
(−1)𝑘 (1 − 𝑘𝜆) Γ (𝜐 + 2)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 2)
𝑎𝑘𝑧𝑘 ̸= 0,
1+∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 2)
⋅ [(1 − 𝑘𝜆) (𝐴 − 𝐵) 𝑏
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 2)
−𝑘 (𝜆 − 1) (𝑒−𝑖𝜃 + 𝐵)
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 2)] 𝑎𝑘𝑧𝑘 ̸= 0.
(41)
We can write (41) as
[1+∞
∑𝑘=1
𝜐 + 1𝑘 + 𝜐 + 1
𝑧𝑘] ∗ [1
+∞
∑𝑘=1
(−1)𝑘 (1 − 𝑘𝜆) Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘] ̸= 0,
[1+∞
∑𝑘=1
𝜐 + 1𝑘 + 𝜐 + 1
𝑧𝑘] ∗ [1+∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 1)
⋅ [(1 − 𝑘𝜆) (𝐴 − 𝐵) 𝑏
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
−𝑘 (𝜆 − 1) (𝑒−𝑖𝜃 + 𝐵)
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)] 𝑎𝑘𝑧𝑘] ̸= 0,
(42)
since
[1+∞
∑𝑘=1
𝜐 + 1𝑘 + 𝜐 + 1
𝑧𝑘] ∗ [1+∞
∑𝑘=1
𝑘 + 𝜐 + 1𝜐 + 1
𝑧𝑘]
= 1+∞
∑𝑘=1𝑧𝑘.
(43)
By using the property, if𝑓 ̸= 0 and𝑔∗ℎ ̸= 0, then𝑓∗(𝑔∗ℎ) ̸=0;
(42) can be written as
1+∞
∑𝑘=1
(−1)𝑘 (1 − 𝑘𝜆) Γ (𝜐 + 1)4𝑘Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
𝑎𝑘𝑧𝑘 ̸= 0,
1+∞
∑𝑘=1
(−1)𝑘 Γ (𝜐 + 1)
⋅ [(1 − 𝑘𝜆) (𝐴 − 𝐵) 𝑏
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)
−𝑘 (𝜆 − 1) (𝑒−𝑖𝜃 + 𝐵)
4𝑘 (𝐴 − 𝐵) 𝑏Γ (𝑘 + 1) Γ (𝑘 + 𝜐 + 1)] 𝑎𝑘𝑧𝑘 ̸= 0,
(44)
which means that 𝑓(𝑧) ∈ ΣS∗𝜆,𝜐[𝑏; 𝐴, 𝐵]. This completes the
proof of Theorem 12.
Using the same arguments as in the proof of Theorem 12,we obtain
the following theorem.
Theorem 13. For 𝜐 > −1, we have ΣK𝜆,𝜐+1[𝑏; 𝐴, 𝐵] ⊂
ΣK𝜆,𝜐[𝑏; 𝐴, 𝐵].
Conflict of Interests
The authors declare that there is no conflict of
interestsregarding the publication of this paper.
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