Bulletin of Mathematical Analysis and Applications ISSN: 1821-1291, URL:

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Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 2, Issue 3(2010), Pages 15-26.
SUBORDINATION AND SUPERORDINATION FOR FUNCTIONS
BASED ON DZIOK-SRIVASTAVA LINEAR OPERATOR
RABHA W. IBRAHIM AND MASLINA DARUS
Abstract. In this article, we obtain some subordination and superordination results involving Dziok-Srivastava linear operator and fractional integral
operator for certain normalized analytic functions in the open unit disk.
1. Introduction and Preliminaries.
Let β„‹(U) denote the class of analytic functions in the unit disk
π‘ˆ := {𝑧 ∈ β„‚, βˆ£π‘§βˆ£ < 1}.
For 𝑛 positive integer and π‘Ž ∈ β„‚, let
β„‹[π‘Ž, 𝑛] := {𝑓 ∈ 𝐻(π‘ˆ ) : 𝑓 (𝑧) = π‘Ž + π‘Žπ‘› 𝑧 𝑛 + π‘Žπ‘›+1 𝑧 𝑛+1 + ..., 𝑧 ∈ π‘ˆ },
and π’œπ‘› = {𝑓 ∈ 𝐻(π‘ˆ ) : 𝑓 (𝑧) = 𝑧 + π‘Žπ‘› 𝑧 𝑛 + π‘Žπ‘›+1 𝑧 𝑛+1 + ..., 𝑧 ∈ π‘ˆ } with π’œ1 = π’œ.
A function 𝑓 ∈ β„‹[π‘Ž, 𝑛] is convex in π‘ˆ if it is univalent and 𝑓 (π‘ˆ ) is convex. It is
well-known that 𝑓 is convex if and only if 𝑓 (0) βˆ•= 0 and
ℜ{1 +
𝑧𝑓 ′′ (𝑧)
} > 0, 𝑧 ∈ π‘ˆ.
𝑓 ′ (𝑧)
Definition 1.1. [1] Denote by Q the set of all functions 𝑓 (𝑧) that are analytic and
injective on π‘ˆ − 𝐸(𝑓 ) where
𝐸(𝑓 ) := {𝜁 ∈ ∂π‘ˆ : π‘™π‘–π‘šπ‘§→𝜁 𝑓 (𝑧) = ∞}
′
and are such that 𝑓 (𝜁) βˆ•= 0 for 𝜁 ∈ ∂π‘ˆ − 𝐸(𝑓 ).
Given two functions 𝐹 and 𝐺 in the unit disk π‘ˆ, the function 𝐹 is π‘ π‘’π‘π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’ to
𝐺, written 𝐹 β‰Ί 𝐺, if 𝐺 is univalent, 𝐹 (0) = 𝐺(0) and 𝐹 (π‘ˆ ) ⊂ 𝐺(π‘ˆ ). Alternatively,
given two functions 𝐹 and 𝐺, which are analytic in π‘ˆ, the function 𝐹 is said to be
subordinate to 𝐺 in π‘ˆ if there exists a function β„Ž, analytic in π‘ˆ with
β„Ž(0) = 0 π‘Žπ‘›π‘‘ βˆ£β„Ž(𝑧)∣ < 1 𝑓 π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑧 ∈ π‘ˆ
such that
𝐹 (𝑧) = 𝐺(β„Ž(𝑧)) 𝑓 π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑧 ∈ π‘ˆ.
2000 Mathematics Subject Classification. 34G10, 26A33,30C45.
Key words and phrases. Fractional calculus;
Univalent solution;
Subordination;
Superordination.
c
⃝2010
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted May 6, 2010. Published June 9, 2010.
The authors were partially supported by UKM-ST-06-FRGS0107-2009, MOHE Malaysia.
15
16
R. W. IBRAHIM, M. DARUS
Let πœ™ : β„‚2 → β„‚ and let β„Ž be univalent in π‘ˆ. If 𝑝 is analytic in π‘ˆ and satisfies
the differential subordination πœ™(𝑝(𝑧)), 𝑧𝑝′ (𝑧)) β‰Ί β„Ž(𝑧), then 𝑝 is called a solution of
the differential subordination. The univalent function π‘ž is called a dominant of the
solutions of the differential subordination, if 𝑝 β‰Ί π‘ž. If 𝑝 and πœ™(𝑝(𝑧)), 𝑧𝑝′ (𝑧)) are
univalent in π‘ˆ and satisfy the differential superordination β„Ž(𝑧) β‰Ί πœ™(𝑝(𝑧)), 𝑧𝑝′ (𝑧)),
then 𝑝 is called a solution of the differential superordination. An analytic function
π‘ž is called subordinant of the solution of the differential superordination if π‘ž β‰Ί 𝑝.
We shall need the following results:
Lemma 1.1. [2] Let π‘ž be univalent in the unit disk π‘ˆ , and let πœƒ and πœ™ be analytic
in a domain 𝐷 containing π‘ž(π‘ˆ ) with πœ™(𝑀) βˆ•= 0 when 𝑀 ∈ π‘ž(π‘ˆ ). Set 𝑄(𝑧) :=
π‘§π‘ž ′ (𝑧)πœ™(π‘ž(𝑧)), β„Ž(𝑧) := πœƒ(π‘ž(𝑧)) + 𝑄(𝑧). Suppose that
1. 𝑄(𝑧) is starlike univalent in π‘ˆ , and
′
(𝑧)
2. ℜ π‘§β„Ž
𝑄(𝑧) > 0 for 𝑧 ∈ π‘ˆ.
If πœƒ(𝑝(𝑧)) + 𝑧𝑝′ (𝑧)πœ™(𝑝(𝑧)) β‰Ί πœƒ(π‘ž(𝑧)) + π‘§π‘ž ′ (𝑧)πœ™(π‘ž(𝑧)), then 𝑝(𝑧) β‰Ί π‘ž(𝑧) and π‘ž is the
best dominant.
Lemma 1.2. [3] Let π‘ž be convex univalent in the unit disk π‘ˆ and πœ“ and 𝛾 ∈ β„‚ with
′′
(𝑧)
+ πœ“π›Ύ } > 0. If 𝑝(𝑧) is analytic in π‘ˆ and πœ“π‘(𝑧)+𝛾𝑧𝑝′ (𝑧) β‰Ί πœ“π‘ž(𝑧)+π›Ύπ‘§π‘ž ′ (𝑧),
ℜ{1+ π‘§π‘žπ‘ž′ (𝑧)
then 𝑝(𝑧) β‰Ί π‘ž(𝑧) and π‘ž is the best dominant.
Lemma 1.3. [4] Let π‘ž be convex univalent in the unit disk π‘ˆ and πœ— and πœ‘ be
analytic in a domain 𝐷 containing π‘ž(π‘ˆ ). Suppose that
1. π‘§π‘ž ′ (𝑧)πœ‘(π‘ž(𝑧)) is starlike univalent in π‘ˆ, and
′
(π‘ž(𝑧))
} > 0 for 𝑧 ∈ π‘ˆ.
2. ℜ{ πœ—πœ‘(π‘ž(𝑧))
If 𝑝(𝑧) ∈ β„‹[π‘ž(0), 1] ∩ Q, with 𝑝(π‘ˆ ) ⊆ 𝐷 and πœ—(𝑝(𝑧)) + 𝑧𝑝′ (𝑧)πœ‘(𝑧) is univalent in
π‘ˆ and πœ—(π‘ž(𝑧)) + π‘§π‘ž ′ (𝑧)πœ‘(π‘ž(𝑧)) β‰Ί πœ—(𝑝(𝑧)) + 𝑧𝑝′ (𝑧)πœ‘(𝑝(𝑧)) then π‘ž(𝑧) β‰Ί 𝑝(𝑧) and π‘ž is
the best subordinant.
Lemma 1.4. [1] Let π‘ž be convex univalent in the unit disk π‘ˆ and 𝛾 ∈ β„‚. Further,
assume that ℜ{𝛾} > 0. If 𝑝(𝑧) ∈ β„‹[π‘ž(0), 1] ∩ Q, with 𝑝(𝑧) + 𝛾𝑧𝑝′ (𝑧) is univalent
in π‘ˆ then π‘ž(𝑧) + π›Ύπ‘§π‘ž ′ (𝑧) β‰Ί 𝑝(𝑧) + 𝛾𝑧𝑝′ (𝑧) implies π‘ž(𝑧) β‰Ί 𝑝(𝑧) and π‘ž is the best
subordinant.
∑∞
∑∞
For two functions 𝑓 (𝑧) = 𝑧 + 𝑛=2 π‘Žπ‘› 𝑧 𝑛 and 𝑔(𝑧) = 𝑧 + 𝑛=2 𝑏𝑛 𝑧 𝑛 , the Hadamard
product (or convolution) of 𝑓 and 𝑔 defined by
∞
∑
(𝑓 ∗ 𝑔)(𝑧) := 𝑧 +
π‘Žπ‘› 𝑏𝑛 𝑧 𝑛 =: (𝑔 ∗ 𝑓 )(𝑧).
𝑛=2
For 𝛼𝑗 ∈ β„‚ (𝑗 = 1, 2, ..., 𝑙) and 𝛽𝑗 ∈ β„‚βˆ–{0, −1, −2, ...} (𝑗 = 1, 2, ..., π‘š), the generalized hypergeometric function 𝑙 πΉπ‘š (𝛼1 , ..., 𝛼𝑙 ; 𝛽1 , ...π›½π‘š ; 𝑧) is defined by the infinite
series
∞
∑
(𝛼1 )𝑛 ...(𝛼𝑙 )𝑛 𝑧 𝑛
𝑙 πΉπ‘š (𝛼1 , ..., 𝛼𝑙 ; 𝛽1 , ..., π›½π‘š ; 𝑧) :=
(𝛽1 )𝑛 ...(π›½π‘š )𝑛 𝑛!
𝑛=0
(𝑙 ≤ π‘š + 1 : 𝑙, π‘š ∈ β„•0 := {0, 1, 2, ...})
where (π‘Ž)𝑛 is the Pochhammer symbol defined by
{
Γ(π‘Ž + 𝑛)
1,
(𝑛 = 0);
(π‘Ž)𝑛 :=
=
π‘Ž(π‘Ž + 1)(π‘Ž + 2)...(π‘Ž + 𝑛 − 1), (𝑛 ∈ β„•).
Γ(π‘Ž)
SUBORDINATION AND SUPERORDINATION
17
Corresponding to the function
β„Ž(𝛼1 , ..., 𝛼𝑙 ; 𝛽1 , ...π›½π‘š ; 𝑧) := 𝑧𝑙 πΉπ‘š (𝛼1 , ..., 𝛼𝑙 ; 𝛽1 , ...π›½π‘š ; 𝑧),
𝑙
the Dziok-Srivastava operator (see [5-7]) π»π‘š (𝛼1 , ..., 𝛼𝑙 ; 𝛽1 , ...π›½π‘š ) is defined by the
Hadamard product
𝑙
π»π‘š
(𝛼1 , ..., 𝛼𝑙 ; 𝛽1 , ...π›½π‘š )𝑓 (𝑧) := β„Ž(𝛼1 , ..., 𝛼𝑙 ; 𝛽1 , ...π›½π‘š ; 𝑧) ∗ 𝑓 (𝑧)
=𝑧+
∞
∑
(𝛼1 )𝑛−1 ...(𝛼𝑙 )𝑛−1 π‘Žπ‘› 𝑧 𝑛
(𝛽1 )𝑛−1 ...(π›½π‘š )𝑛−1 (𝑛 − 1)!
𝑛=2
𝑙
:= π»π‘š
[𝛼1 ]𝑓 (𝑧).
We can verify that
𝑙
𝑙
𝑙
𝑧(π»π‘š
[𝛼1 ]𝑓 (𝑧))′ = 𝛼1 π»π‘š
[𝛼1 + 1]𝑓 (𝑧) − (𝛼1 − 1)π»π‘š
[𝛼1 ]𝑓 (𝑧).
Special cases of the Dziok-Srivastava linear operator include the Hohlov linear operator [8], the Carlson-Shaffer linear operator 𝐿(π‘Ž, 𝑐) [9], the Ruscheweyh derivative
operator π’Ÿπ‘› [10], the generalized Bernardi-Libera-Livingston linear integral operator [11] and the Srivastava-Owa fractional derivative operator [12]:
Definition 1.2. The fractional derivative of order 𝛼 is defined, for a function 𝑓 by
∫ 𝑧
𝑑
𝑓 (𝜁)
1
𝐷𝑧𝛼 𝑓 (𝑧) :=
π‘‘πœ; 0 ≤ 𝛼 < 1,
Γ(1 − 𝛼) 𝑑𝑧 0 (𝑧 − 𝜁)𝛼
where the function 𝑓 is analytic in simply-connected region of the complex z-plane
β„‚ containing the origin and the multiplicity of (𝑧 − 𝜁)−𝛼 is removed by requiring
π‘™π‘œπ‘”(𝑧 − 𝜁) to be real when(𝑧 − 𝜁) > 0.
Definition 1.3. The fractional integral of order 𝛼 is defined, for a function 𝑓, by
∫ 𝑧
1
𝐼𝑧𝛼 𝑓 (𝑧) :=
𝑓 (𝜁)(𝑧 − 𝜁)𝛼−1 π‘‘πœ; 𝛼 > 0,
Γ(𝛼) 0
where the function 𝑓 is analytic in simply-connected region of the complex 𝑧-plane
(β„‚) containing the origin and the multiplicity of (𝑧 − 𝜁)𝛼−1 is removed by requiring
π‘™π‘œπ‘”(𝑧 − 𝜁) to be real when(𝑧 − 𝜁) > 0.
Remark 1.1. [12]
𝐷𝑧𝛼 {𝑧 πœ‡ } =
Γ(πœ‡ + 1)
{𝑧 πœ‡−𝛼 }, πœ‡ > −1; 0 ≤ 𝛼 < 1
Γ(πœ‡ − 𝛼 + 1)
and
Γ(πœ‡ + 1)
{𝑧 πœ‡+𝛼 }, πœ‡ > −1; 𝛼 > 0.
Γ(πœ‡ + 𝛼 + 1)
The main object of the present paper is to find the sufficient conditions for certain
normalized analytic functions 𝑓, 𝑔 to satisfy
𝐼𝑧𝛼 {𝑧 πœ‡ } =
[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔1 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑔2 (𝑧) πœ‡
] β‰Ί[ 𝑧 π‘š
] β‰Ί[ 𝑧 π‘š
]
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
and
π‘ž1 (𝑧) β‰Ί [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
] β‰Ί π‘ž2 (𝑧), πœŒπ›Ό (𝑧) βˆ•= 0, 𝑧 ∈ π‘ˆ
πœŒπ›Ό (𝑧)
18
R. W. IBRAHIM, M. DARUS
where πœ‡ ≥ 1, π‘ž1 and π‘ž2 are given univalent functions in π‘ˆ. Also, we obtain the
results as special cases. Further, in this paper, we study the existence of univalent
solution for the fractional differential equation
𝑙
𝐷𝑧𝛼 πœŒπ›Ό (𝑧)𝑒(𝑧) = π»π‘š
[𝛼1 ]𝑓 (𝑧),
(1.1)
subject to the initial condition 𝑒(0) = 0, where 𝑒 : π‘ˆ → β„‚ is an analytic function
for all 𝑧 ∈ π‘ˆ, 𝜌 : π‘ˆ → β„‚βˆ–{0} is an analytic functions in 𝑧 ∈ π‘ˆ and 𝑓 : π‘ˆ → β„‚ is a
univalent function in π‘ˆ. The existence is obtained by applying Schauder fixed point
theorem. Moreover, we discuss some properties of this solution involving fractional
differential subordination. The following results are used in the sequel.
Theorem 1.1. (Arzela-Ascoli) (see [13]) Let 𝐸 be a compact metric space and
π’ž(𝐸) be the Banach space of real or complex valued continuous functions normed
by
βˆ₯𝑓 βˆ₯ := 𝑠𝑒𝑝𝑑∈𝐸 βˆ£π‘“ (𝑑)∣.
If 𝐴 = {𝑓𝑛 } is a sequence in π’ž(𝐸) such that 𝑓𝑛 is uniformly bounded and equicontinuous, then 𝐴 is compact.
Let 𝑀 be a subset of Banach space 𝑋 and 𝐴 : 𝑀 → 𝑀 an operator. The operator
𝐴 is called π‘π‘œπ‘šπ‘π‘Žπ‘π‘‘ on the set 𝑀 if it carries every bounded subset of 𝑀 into a
compact set. If 𝐴 is continuous on 𝑀 (that is, it maps bounded sets into bounded
sets ) then it is said to be π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘™π‘¦ π‘π‘œπ‘›π‘‘π‘–π‘›π‘’π‘œπ‘’π‘  on 𝑀.
Theorem 1.2. (Schauder) (see [14]) Let 𝑋 be a Banach space, 𝑀 ⊂ 𝑋 a nonempty
closed bounded convex subset and 𝑃 : 𝑀 → 𝑀 is compact. Then 𝑃 has a fixed
point.
Recently, the subordination and superordination containing the Dziok-Srivastava
linear operator are studied by many authors [15].
2. Subordination and superordination.
In this section, we study some important properties of the fractional differential
and integral operators 𝐷𝑧𝛼 , 𝐼𝑧𝛼 , given by the authors [16] which are useful in the
next results of the subordination and superordination.
Theorem 2.1[16] For 𝛼, ∈ (0, 1] and 𝑓 is a continuous function, then
(𝑧)𝛼−1
𝑑
𝑓 (0) + 𝐼𝑧𝛼 𝐷𝑓 (𝑧); 𝐷 =
.
Γ(𝛼)
𝑑𝑧
2 − 𝐼𝑧𝛼 𝐷𝑧𝛼 𝑓 (𝑧) = 𝐷𝑧𝛼 𝐼𝑧𝛼 𝑓 (𝑧) = 𝑓 (𝑧).
1 − 𝐷𝐼𝑧𝛼 𝑓 (𝑧) =
But, first we consider the subordination results involving Dziok-Srivastava linear
operator and fractional integral operator as the following:
Theorem 2.2. Let 𝑓, 𝑔 be analytic in π‘ˆ. [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧)
πœŒπ›Ό (𝑧)
that
subordination
[
βˆ•= 0 and
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧) πœ‡
]
πœŒπ›Ό (𝑧)
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑔(𝑧) πœ‡ ′
𝑧([ 𝑧 πœŒπ‘šπ›Ό (𝑧)
] )
be univalent in π‘ˆ such
be starlike univalent in π‘ˆ. If the
𝑙
𝐼𝑧𝛼 π»π‘š
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
[𝛼1 ]𝑓 (𝑧) πœ‡
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
SUBORDINATION AND SUPERORDINATION
β‰Ί[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑔(𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
𝜌(𝑧)
holds and
ℜ{
𝑧𝐺′ (𝑧)
𝑧𝐺(𝑧)πœŒπ›Ό (𝑧)
+ (πœ‡ − 1) 𝛼 𝑙
} > 0, 𝑧 ∈ π‘ˆ,
𝐺(𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
where
𝐺(𝑧) := [
Then
[
and [
19
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔 πœ‡
]
πœŒπ›Ό
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧) ′
].
πœŒπ›Ό (𝑧)
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑔(𝑧) πœ‡
] β‰Ί[ 𝑧 π‘š
]
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
is the best dominant.
Proof. Setting
𝑝(𝑧) := [
𝑙
[𝛼1 ]𝑓 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑔(𝑧) πœ‡
𝐼𝑧𝛼 π»π‘š
] , π‘ž(𝑧) := [ 𝑧 π‘š
] .
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
Our aim is to apply Lemma 1.1. First we show that ℜ{1 +
ℜ{1 +
π‘§π‘ž ′′ (𝑧)
π‘ž ′ (𝑧) }
> 0.
𝑧𝐺′ (𝑧)
𝐺(𝑧)πœŒπ›Ό (𝑧)
π‘§π‘ž ′′ (𝑧)
} = ℜ{1 +
+ (πœ‡ − 1) 𝛼 𝑙
} > 0.
′
π‘ž (𝑧)
𝐺(𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
Assume that
πœƒ(πœ”) := πœ” π‘Žπ‘›π‘‘ πœ™(πœ”) := 1,
it can easily be observed that πœƒ, πœ™ are analytic in β„‚. Also, we let
𝑄(𝑧) := π‘§π‘ž ′ (𝑧)πœ™(𝑧) = π‘§π‘ž ′ (𝑧),
β„Ž(𝑧) := πœƒ(π‘ž(𝑧)) + 𝑄(𝑧) = π‘ž(𝑧) + π‘§π‘ž ′ (𝑧).
By the assumptions of the theorem we find that 𝑄 is starlike univalent in π‘ˆ and
that
π‘§β„Ž′ (𝑧)
π‘§π‘ž ′′ (𝑧)
ℜ{
} = ℜ{2 + ′
} > 0.
𝑄(𝑧)
π‘ž (𝑧)
By using Theorem 2.1, a computation shows
𝑙
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑔(𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑔(𝑧)]′
π‘§πœŒ′ (𝑧)
β‰Ί[ 𝑧 π‘š
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
)}
)−
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
𝜌(𝑧)
= π‘ž(𝑧) + π‘§π‘ž ′ (𝑧).
𝑝(𝑧) + 𝑧𝑝′ (𝑧) = [
Thus in view of Lemma 1.1, 𝑝(𝑧) β‰Ί π‘ž(𝑧) and π‘ž is the best dominant.
Corollary 2.1. Let 𝑓, 𝑔 be analytic in π‘ˆ. [
𝐼 𝛼 𝐿(π‘Ž,𝑐)𝑔
𝑧([ 𝑧 πœŒπ›Ό ]πœ‡ )′
𝐼𝑧𝛼 𝐿(π‘Ž,𝑐)𝑔 πœ‡
]
πœŒπ›Ό
be univalent in π‘ˆ and
be starlike univalent in π‘ˆ. If the subordination
[
𝑧𝐼 𝛼 [𝐿(π‘Ž, 𝑐)𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
𝐼𝑧𝛼 𝐿(π‘Ž, 𝑐)𝑓 (𝑧) πœ‡
] {1 + πœ‡( 𝑧𝛼
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 𝐿(π‘Ž, 𝑐)𝑓 (𝑧)
𝜌(𝑧)
β‰Ί[
𝐼𝑧𝛼 𝐿(π‘Ž, 𝑐)𝑔(𝑧) πœ‡
𝑧𝐼 𝛼 [𝐿(π‘Ž, 𝑐)𝑔(𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡( 𝑧𝛼
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 𝐿(π‘Ž, 𝑐)𝑔(𝑧)
𝜌(𝑧)
20
R. W. IBRAHIM, M. DARUS
holds and
ℜ{
𝑧𝐺′ (𝑧)
𝑧𝐺(𝑧)πœŒπ›Ό (𝑧)
+ (πœ‡ − 1) 𝛼
} > 0, 𝑧 ∈ π‘ˆ,
𝐺(𝑧)
𝐼𝑧 𝐿(π‘Ž, 𝑐)𝑔(𝑧)
where
𝐺(𝑧) := [
Then
[
𝐼𝑧𝛼 𝐿(π‘Ž,𝑐)𝑔 πœ‡
]
πœŒπ›Ό
and [
𝐼𝑧𝛼 𝐿(π‘Ž, 𝑐)𝑔(𝑧) ′
].
πœŒπ›Ό (𝑧)
𝐼𝑧𝛼 𝐿(π‘Ž, 𝑐)𝑓 (𝑧) πœ‡
𝐼 𝛼 𝐿(π‘Ž, 𝑐)𝑔(𝑧) πœ‡
] β‰Ί[ 𝑧
]
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
is the best dominant.
Proof. By putting 𝑙 = 2, π‘š = 1, 𝛼1 = π‘Ž, 𝛼2 = 1 and 𝛽1 = 𝑐 in Theorem 2.2.
Corollary 2.2. Let 𝑓, 𝑔 be analytic in π‘ˆ, [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧)
πœŒπ›Ό (𝑧)
[
βˆ•= 0 and
𝐼𝑧𝛼 𝑓 (𝑧) πœ‡
πœŒπ›Ό (𝑧)
] {1 +
𝐼𝛼𝑔
𝑧([ πœŒπ‘§π›Ό ]πœ‡ )′
𝑧𝐼 𝛼 [𝑓 (𝑧)]′
πœ‡( 𝑧𝛼
𝐼𝑧 𝑓 (𝑧)
−
𝐼𝑧𝛼 𝑔 πœ‡
πœŒπ›Ό ]
be univalent in π‘ˆ such that
be starlike univalent in π‘ˆ. If the subordination
π‘§πœŒ′ (𝑧)
π‘§πœŒ′ (𝑧)
𝐼 𝛼 𝑔(𝑧) πœ‡
𝑧𝐼 𝛼 [𝑔(𝑧)]′
)} β‰Ί [ 𝑧
] {1 + πœ‡( 𝑧𝛼
−
)}
𝜌(𝑧)
πœŒπ›Ό (𝑧)
𝐼𝑧 𝑔(𝑧)
𝜌(𝑧)
holds and
𝑧𝐺′ (𝑧)
𝑧𝐺(𝑧)πœŒπ›Ό (𝑧)
𝐼 𝛼 𝑔(𝑧) ′
ℜ{
+ (πœ‡ − 1)
} > 0, 𝑧 ∈ π‘ˆ, π‘€β„Žπ‘’π‘Ÿπ‘’ 𝐺(𝑧) := [ 𝑧
].
𝛼
𝐺(𝑧)
𝐼𝑧 𝑔(𝑧)
πœŒπ›Ό (𝑧)
Then
[
and [
𝐼𝑧𝛼 𝑔 πœ‡
πœŒπ›Ό ]
𝐼 𝛼 𝑔(𝑧) πœ‡
𝐼𝑧𝛼 𝑓 (𝑧) πœ‡
] β‰Ί[ 𝑧
]
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
is the best dominant.
Proof. By putting 𝑙 = 1, π‘š = 0, 𝛼1 = 1, in Theorem 2.2.
Theorem 2.3. Let 𝑓, 𝑔 be analytic in π‘ˆ, π‘ž be convex univalent in π‘ˆ with ℜ{1 +
π‘§π‘ž ′′ (𝑧)
π‘ž ′ (𝑧)
[
+ 𝛾1 }, 𝛾 ∈ β„‚ and [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 πœ‡
]
πœŒπ›Ό
be analytic in π‘ˆ. If the subordination
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡π›Ύ( 𝑧𝛼 π‘™π‘š
−
)} β‰Ί π‘ž(𝑧) + π›Ύπ‘§π‘ž ′ (𝑧)
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
holds. Then
[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
] β‰Ί π‘ž(𝑧)
πœŒπ›Ό (𝑧)
and π‘ž is the best dominant.
Proof. Setting
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
] .
πœŒπ›Ό (𝑧)
Our aim is to applied Lemma 1.2. Let πœ“ := 1, since
𝑝(𝑧) := [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡ ′
] + 𝛾𝑧([ 𝑧 π‘š
] )
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
=[ 𝑧 π‘š
] {1 + πœ‡π›Ύ( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
β‰Ί π‘ž(𝑧) + π›Ύπ‘§π‘ž ′ (𝑧)
𝑝(𝑧) + 𝛾𝑧𝑝′ (𝑧) = [
SUBORDINATION AND SUPERORDINATION
21
then, in view of Lemma 1.2, 𝑝(𝑧) β‰Ί π‘ž(𝑧) and π‘ž is the best dominant.
1+𝐴𝑧 πœ‡
Corollary 2.3. Let 𝑓, 𝑔 be analytic in π‘ˆ, −1 ≤ 𝐡 ≤ 𝐴 ≤ 1, π‘ž(𝑧) := [ 1+𝐡𝑧
] with
ℜ{1 +
[
π‘§π‘ž ′′ (𝑧)
π‘ž ′ (𝑧)
+ 𝛾1 }, 𝛾 ∈ β„‚ and [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 πœ‡
]
πœŒπ›Ό
be analytic in π‘ˆ. If the subordination
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′ π‘§πœŒ′ (𝑧)
1 + 𝐴𝑧 πœ‡
πœ‡π›Ύπ‘§(𝐴 − 𝐡)
] {1+πœ‡( 𝑧𝛼 π‘™π‘š
−
)} β‰Ί [
] {1+
}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
1 + 𝐡𝑧
(1 + 𝐴𝑧)(1 + 𝐡𝑧)
holds. Then
[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
1 + 𝐴𝑧 πœ‡
] β‰Ί[
] , −1 ≤ 𝐡 < 𝐴 ≤ 1
πœŒπ›Ό (𝑧)
1 + 𝐡𝑧
1+𝐴𝑧 πœ‡
] is the best dominant.
and [ 1+𝐡𝑧
Next, applying Lemma 1.3 and Lemma 1.4 respectively, to obtain the following
theorems.
Theorem 2.4.
π‘ˆ such that
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧)
πœŒπ›Ό (𝑧)
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 πœ‡ ′
(𝑧[ 𝑧 π‘š
] )
πœŒπ›Ό
[
βˆ•= 0,
convex univalent in
univalent in π‘ˆ and
be univalent in π‘ˆ. If the subordination
𝑙
[𝛼1 ]𝑔(𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑔(𝑧)]′
π‘§πœŒ′ (𝑧)
𝐼𝑧𝛼 π»π‘š
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
𝜌(𝑧)
β‰Ί[
𝑙
[𝛼1 ]𝑓 (𝑧) πœ‡
𝐼𝑧𝛼 π»π‘š
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
𝛼−1
holds and [ 𝑧Γ(𝛼) ]πœ‡ [
𝑙
π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
]
πœŒπ›Ό (𝑧)
[
and [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔 πœ‡
] be
πœŒπ›Ό
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔 πœ‡ ′
] ) be starlike
𝑧([
πœŒπ›Ό
Let 𝑓, 𝑔 be analytic in π‘ˆ,[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔 πœ‡
]
πœŒπ›Ό
∈ β„‹[0, 1] ∩ Q . Then
𝑙
[𝛼1 ]𝑔(𝑧) πœ‡
𝐼𝑧𝛼 π»π‘š
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
] β‰Ί[ 𝑧 π‘š
]
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
is the best subordinant.
Proof. Setting
𝑝(𝑧) := [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑔(𝑧) πœ‡
] , π‘ž(𝑧) := [ 𝑧 π‘š
] .
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
Our aim is to apply Lemma 1.3. By taking
πœ—(πœ”) := πœ” π‘Žπ‘›π‘‘ πœ‘(πœ”) := 1,
it can easily observed that πœ—, πœ‘ are analytic in β„‚. Thus
ℜ{
πœ—′ (π‘ž(𝑧))
} = 1 > 0.
πœ‘(π‘ž(𝑧))
Now we must show that
π‘ž(𝑧) + π‘§π‘ž ′ (𝑧) β‰Ί 𝑝(𝑧) + 𝑧𝑝′ (𝑧).
22
R. W. IBRAHIM, M. DARUS
a computation shows that
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑔(𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
𝜌(𝑧)
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
−
β‰Ί[ 𝑧 π‘š
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
= 𝑝(𝑧) + 𝑧𝑝′ (𝑧).
π‘ž(𝑧) + π‘§π‘ž ′ (𝑧) = [
Thus in view of Lemma 1.3, π‘ž(𝑧) β‰Ί 𝑝(𝑧) and 𝑝 is the best subordinant.
Theorem 2.5. Let 𝑓, 𝑔 be analytic in π‘ˆ, π‘ž be convex univalent in π‘ˆ, [
β„‹[0, 1] ∩ Q and
[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
]
πœŒπ›Ό (𝑧)
∈
𝑙
π‘§πœŒ′ (𝑧)
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
] {1 + πœ‡π›Ύ( 𝑧𝛼 π‘™π‘š
−
)}, ℜ{𝛾} > 0,
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
be univalent in π‘ˆ. If the subordination
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
π‘ž(𝑧) + π›Ύπ‘§π‘ž ′ (𝑧) β‰Ί [ 𝑧 π‘š
] {1 + πœ‡π›Ύ( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
holds. Then
π‘ž(𝑧) β‰Ί [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
]
πœŒπ›Ό (𝑧)
and π‘ž is the best subordinant.
Proof. Setting
𝑙
[𝛼1 ]𝑓 (𝑧) πœ‡
𝐼𝑧𝛼 π»π‘š
] .
πœŒπ›Ό (𝑧)
Our aim is to apply Lemma 1.4. Since
𝑝(𝑧) := [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔(𝑧) πœ‡
π‘§πœŒ′ (𝑧)
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑔(𝑧)]′
−
] {1 + πœ‡π›Ύ( 𝑧𝛼 π‘™π‘š
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
𝜌(𝑧)
π‘§πœŒ′ (𝑧)
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
β‰Ί[ 𝑧 π‘š
] {1 + πœ‡π›Ύ( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
= 𝑝(𝑧) + 𝛾𝑧𝑝′ (𝑧)
π‘ž(𝑧) + π›Ύπ‘§π‘ž ′ (𝑧) = [
then, in view of Lemma 1.4, π‘ž(𝑧) β‰Ί 𝑝(𝑧) and π‘ž is the best subordinant.
Combining the results of differential subordination and superordination, we state
the following sandwich theorems.
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔1 πœ‡
] be convex univalent in π‘ˆ
πœŒπ›Ό
𝛼
𝑙
𝛼
𝑙
𝛼
𝑙
𝐼𝑧 π»π‘š [𝛼1 ]𝑔(𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔1 πœ‡ ′
𝐼𝑧 𝐻 π‘š
[𝛼1 ]𝑔2 πœ‡ ′
that
βˆ•= 0, 𝑧([
] ) , 𝑧([
] ) be starlike univalent
πœŒπ›Ό (𝑧)
πœŒπ›Ό
πœŒπ›Ό
𝛼
𝑙
𝛼
𝑙
𝐼 𝐻 [𝛼1 ]𝑓 πœ‡ ′ 𝐼𝑧 π»π‘š [𝛼1 ]𝑔2 πœ‡
and let (𝑧[ 𝑧 π‘š
] ),[
] be univalent in π‘ˆ. If the subordination
πœŒπ›Ό
πœŒπ›Ό
Theorem 2.6. Let 𝑓, 𝑔1 , 𝑔2 be analytic in π‘ˆ, [
such
in π‘ˆ
[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔1 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑔1 (𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔1 (𝑧)
𝜌(𝑧)
β‰Ί[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
SUBORDINATION AND SUPERORDINATION
β‰Ί[
𝛼−1
holds, [ 𝑧Γ(𝛼) ]πœ‡ [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔2 (𝑧) πœ‡
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑔2 (𝑧)]′
π‘§πœŒ′ (𝑧)
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
−
)}
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔2 (𝑧)
𝜌(𝑧)
𝑙
π»π‘š
[𝛼1 ]𝑓 (𝑧) πœ‡
]
πœŒπ›Ό (𝑧)
ℜ{
∈ β„‹[0, 1] ∩ Q and
𝑧𝐺′2 (𝑧)
𝑧𝐺2 (𝑧)πœŒπ›Ό (𝑧)
+ (πœ‡ − 1) 𝛼 𝑙
} > 0, 𝑧 ∈ π‘ˆ,
𝐺2 (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑔2 (𝑧)
where
𝐺2 (𝑧) := [
Then
[
23
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔2 (𝑧) ′
].
πœŒπ›Ό (𝑧)
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔1 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑔2 (𝑧) πœ‡
] β‰Ί[ 𝑧 π‘š
] β‰Ί[ 𝑧 π‘š
]
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
πœŒπ›Ό (𝑧)
𝑙
𝐼 𝛼 π»π‘š
[𝛼1 ]𝑔1 πœ‡
]
πœŒπ›Ό
and [ 𝑧
inant.
and [
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑔2 πœ‡
]
πœŒπ›Ό
are respectively the best subordinant and dom-
Theorem 2.7. Let 𝑓, 𝑔1 , 𝑔2 ∈ π’œ, π‘ž1 , π‘ž2 be convex univalent in π‘ˆ, with ℜ{1 +
π‘§π‘ž2′′ (𝑧)
π‘ž2′ (𝑧)
+ 𝛾1 }, 𝛾 ∈ β„‚, [
[
𝑙
𝐼𝑧𝛼 π»π‘š
[𝛼1 ]𝑓 πœ‡
]
πœŒπ›Ό
∈ β„‹[0, 1] ∩ Q, and analytic in π‘ˆ and
𝑙
[𝛼1 ]𝑓 πœ‡
𝐼𝑧𝛼 π»π‘š
π‘§πœŒ′
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 ]′
−
)}, ℜ{𝛾} > 0,
] {1 + πœ‡( 𝑧𝛼 π‘™π‘š
πœŒπ›Ό
𝐼𝑧 π»π‘š [𝛼1 ]𝑓
𝜌
be univalent in π‘ˆ. If the subordination
𝑧𝐼 𝛼 [𝐻 𝑙 [𝛼1 ]𝑓 (𝑧)]′ π‘§πœŒ′ (𝑧)
𝐼 𝛼 𝐻 𝑙 [𝛼1 ]𝑓 (𝑧) πœ‡
] {1+πœ‡π›Ύ( 𝑧𝛼 π‘™π‘š
−
)} β‰Ί π‘ž2 (𝑧)+π›Ύπ‘§π‘ž2′ (𝑧)
π‘ž1 (𝑧)+π›Ύπ‘§π‘ž1′ (𝑧) β‰Ί [ 𝑧 π‘š
πœŒπ›Ό (𝑧)
𝐼𝑧 π»π‘š [𝛼1 ]𝑓 (𝑧)
𝜌(𝑧)
holds. Then
𝑙
[𝛼1 ]𝑓 (𝑧) πœ‡
𝐼𝑧𝛼 π»π‘š
] β‰Ί π‘ž2 (𝑧)
πœŒπ›Ό (𝑧)
and π‘ž1 , π‘ž2 are respectively the best subordinant and the best dominant.
π‘ž1 (𝑧) β‰Ί [
3. Existence of univalent solution.
Let ℬ := π’ž[π‘ˆ, β„‚] be a Banach space of all continuous functions on π‘ˆ endowed
with the sup. norm
βˆ₯𝑒βˆ₯ := 𝑠𝑒𝑝𝑧∈π‘ˆ βˆ£π‘’(𝑧)∣.
By using the properties in Theorem 2.1, we can easily obtain the following result:
Lemma 3.1. If the function 𝑓 ∈ π’œ, then the initial value problem (1.1) is equivalent
to the nonlinear integral equation
∫ 𝑧
1
(𝑧 − 𝜁)𝛼−1 𝑙
𝑒(𝑧) =
π»π‘š [𝛼1 ]𝑓 (𝜁)π‘‘πœ.
(3.1)
πœŒπ›Ό (𝑧) 0
Γ(𝛼)
In other words, every solution of the equation (3.1) is also a solution of the initial
value problem (1.1) and vice versa.
Theorem 3.1.(Existence) Assume that βˆ£πœŒπ›Ό1(𝑧)∣ ≤ 𝑀 ; 𝑀 > 0. Then there exists a
univalent function 𝑒 : π‘ˆ → β„‚ solving the problem (1.1).
Proof. Define an operator 𝑃 : β„‚ → β„‚
∫ 𝑧
1
(𝑧 − 𝜁)𝛼−1 𝑙
π»π‘š [𝛼1 ]𝑓 (𝜁)π‘‘πœ.
(3.2)
(𝑃 𝑒)(𝑧) :=
πœŒπ›Ό (𝑧) 0
Γ(𝛼)
24
R. W. IBRAHIM, M. DARUS
(𝛼1 )𝑛−1 ...(𝛼𝑙 )𝑛−1
1
Denotes 𝐡𝑛 := (𝛽
. Our aim is to apply Theorem 2.1. First we
1 )𝑛−1 ...(π›½π‘š )𝑛−1 (𝑛−1)!
show that 𝑃 is bounded operator:
∫ 𝑧
(𝑧 − 𝜁)𝛼−1 𝑙
1
∣(𝑃 𝑒)(𝑧)∣ = ∣
π»π‘š [𝛼1 ]𝑓 (𝜁)π‘‘πœβˆ£
πœŒπ›Ό (𝑧) 0
Γ(𝛼)
∫ 𝑧
(𝑧 − 𝜁)𝛼−1 𝑙
1
∣∣
π»π‘š [𝛼1 ]𝑓 (𝜁)π‘‘πœβˆ£
≤∣
πœŒπ›Ό (𝑧) 0
Γ(𝛼)
∫ 𝑧
∞
∑
(𝑧 − 𝜁)𝛼−1
< 𝑀 (1 +
𝐡𝑛 βˆ£π‘Žπ‘› ∣)∣
π‘‘πœβˆ£
Γ(𝛼)
0
𝑛=2
= 𝑀 (1 +
∞
∑
𝐡𝑛 βˆ£π‘Žπ‘› ∣)
𝑛=2
βˆ£π‘§ 𝛼 ∣
Γ(𝛼 + 1)
∑∞
𝑀 (1 + 𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣)
<
Γ(𝛼 + 1)
Thus we obtain that
βˆ₯𝑃 βˆ₯ <
∑∞
𝑀 (1 + 𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣)
:= π‘Ÿ
Γ(𝛼 + 1)
that is 𝑃 : π΅π‘Ÿ → π΅π‘Ÿ . Then 𝑃 maps π΅π‘Ÿ into itself. Now we proceed to prove that 𝑃
is equicontinuous. For 𝑧1 , 𝑧2 ∈ π‘ˆ such that 𝑧1 βˆ•= 𝑧2 , βˆ£π‘§2 − 𝑧1 ∣ < 𝛿, 𝛿 > 0 Then for
all 𝑒 ∈ 𝑆, where
∑∞
𝑀 (1 + 𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣)
:= π‘Ÿ, π‘Ÿ > 0},
𝑆 := {𝑒 ∈ β„‚, : βˆ£π‘’βˆ£ ≤
Γ(𝛼 + 1)
we obtain
∣(𝑃 𝑒)(𝑧1 ) − (𝑃 𝑒)(𝑧2 )∣
∫𝑧
∑∞
≤ 𝑀 (1 + 𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣)∣ 0 1
∫𝑧
∑∞
≤ 𝑀 (1 + 𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣)∣ 0 1
=
<
<
∑∞
(𝑧1 −𝜁)𝛼−1
π‘‘πœ
Γ(𝛼)
∫ 𝑧2
0
(𝑧2 −𝜁)𝛼−1
π‘‘πœβˆ£
Γ(𝛼)
[(𝑧1 −𝜁)𝛼−1 −(𝑧2 −𝜁)𝛼−1 ]
π‘‘πœ
Γ(𝛼)
𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣)
∣[2(𝑧2 − 𝑧1 )𝛼
Γ(𝛼+1)
∑
2𝑀 (1+ ∞
𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣)
βˆ£π‘§2 − 𝑧1 βˆ£π›Ό
Γ(𝛼+1)
∑∞
2𝑀 (1+ 𝑛=2 𝐡𝑛 βˆ£π‘Žπ‘› ∣) 𝛼
𝛿 ,
Γ(𝛼+1)
𝑀 (1+
−
+
∫ 𝑧2
𝑧1
(𝑧2 −𝜁)𝛼−1
π‘‘πœβˆ£
Γ(𝛼)
+ 𝑧2𝛼 − 𝑧1𝛼 ]∣
which is independent on 𝑒. Hence 𝑃 is an equicontinuous mapping on 𝑆. By the
assumption of the theorem we can show that 𝑃 is a univalent function (see [17]).
The Arzela-Ascoli theorem yields that every sequence of functions from 𝑃 (𝑆) has
got a uniformly convergent subsequence, and therefore 𝑃 (𝑆) is relatively compact.
Schauder, s fixed point theorem asserts that 𝑃 has a fixed point. By construction,
a fixed point of 𝑃 is a univalent solution of the initial value problem (1.1).
The next theorems show the relation between univalent solutions and the subordination for a class of fractional differential problem.
Theorem 3.2. Let the assumptions of Theorem 2.6 be satisfied. Then univalent
solutions 𝑒1 , 𝑒, 𝑒2 , of the problem
𝐷𝑧𝛼 𝑒(𝑧) = 𝐹 (𝑧, 𝑒(𝑧)),
(3.3)
SUBORDINATION AND SUPERORDINATION
25
subject to the initial condition 𝑒(0) = 0, where 𝑒 : π‘ˆ → β„‚ is an analytic function
for all 𝑧 ∈ π‘ˆ and 𝐹 : π‘ˆ × β„‚ → β„‚, is an analytic functions in 𝑧 ∈ π‘ˆ, are satisfying
the subordination 𝑒1 β‰Ί 𝑒 β‰Ί 𝑒2 .
Proof. Setting πœ‡ = 1 and let 𝐹 (𝑧, 𝑒1 (𝑧)) :=
and 𝐹 (𝑧, 𝑒2 (𝑧)) :=
𝑙
π»π‘š
[𝛼1 ]𝑔2 (𝑧)
πœŒπ›Ό (𝑧)
𝑙
π»π‘š
[𝛼1 ]𝑔1 (𝑧)
,
πœŒπ›Ό (𝑧)
𝐹 (𝑧, 𝑒(𝑧)) :=
𝑙
π»π‘š
[𝛼1 ]𝑓 (𝑧)
,
πœŒπ›Ό (𝑧)
where πœŒπ›Ό (𝑧) βˆ•= 0, ∀ 𝑧 ∈ π‘ˆ.
Theorem 3.3. Let the assumptions of Theorem 2.7 be satisfied. Then every
univalent solution 𝑒(𝑧) of the problem (3.3) satisfies the subordination π‘ž1 (𝑧) β‰Ί
𝑒(𝑧) β‰Ί π‘ž2 (𝑧), where π‘ž1 (𝑧) and π‘ž2 (𝑧) are univalent function in π‘ˆ.
Proof. Setting πœ‡ = 1, 𝐹 (𝑧, 𝑒(𝑧)) :=
𝑙
π»π‘š
[𝛼1 ]𝑓 (𝑧)
.
πœŒπ›Ό (𝑧)
Acknowledgement: The authors would like to thank the anonymous referee for the
informative and creative comments given to the article.
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26
R. W. IBRAHIM, M. DARUS
Rabha W. Ibrahim and Maslina Darus
School of Mathematical Sciences
Faculty of Science and Technology
Universiti Kebangsaan Malaysia
Bangi 43600 Selangor D.E., Malaysia
E-mail address: saibahmath@yahoo.com
E-mail address: maslina@ukm.my (corresponding author)
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