volume 7, issue 4, article 134, 2006.
Received 21 June, 2005;
accepted 01 June, 2006.
Communicated by:A. Sofo
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Journal of Inequalities in Pure and Applied Mathematics
SUBORDINATION RESULTS FOR A CLASS OF ANALYTIC FUNCTIONS DEFINED BY A LINEAR OPERATOR
B.A. FRASIN
Department of Mathematics Al al-Bayt University P.O. Box: 130095 Mafraq, Jordan
EMail:bafrasin@yahoo.com
2000c Victoria University ISSN (electronic): 1443-5756 189-05
Subordination Results for a Class of Analytic Functions Defined by a Linear Operator
B.A. Frasin
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Abstract
In this paper, we derive several interesting subordination results for certain class of analytic functions defined by the linear operatorL(a, c)f(z)which in- troduced and studied by Carlson and Shaffer [2].
2000 Mathematics Subject Classification: Primary 30C45; Secondary 30A10, 30C80.
Key words: Analytic functions, Hadamard product, Subordinating factor sequence.
Contents
1 Introduction and Definitions . . . 3 2 Main Theorem. . . 8
References
Subordination Results for a Class of Analytic Functions Defined by a Linear Operator
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1. Introduction and Definitions
LetAdenote the class of functions of the form:
(1.1) f(z) =z+
∞
X
n=2
anzn
which are analytic in the open unit disc ∆ = {z :|z|<1}.For two functions f(z)andg(z)given by
(1.2) f(z) =z+
∞
X
n=2
anzn and g(z) = z+
∞
X
n=2
cnzn
their Hadamard product (or convolution) is defined by
(1.3) (f ∗g)(z) :=z+
∞
X
n=2
ancnzn.
Define the functionφ(a, c;z)by (1.4) φ(a, c;z) :=
∞
X
n=0
(a)n
(c)nzn+1 (c /∈Z−0 :={0,−1,−2, . . .}, z∈∆), where(λ)nis the Pochhammer symbol given, in terms of Gamma functions,
(λ)n:= Γ(λ+n) (1.5) Γ(λ)
=
( 1, n= 0,
λ(λ+ 1)(λ+ 2). . .(λ+n−1), n∈N:{1,2, . . .}.
Subordination Results for a Class of Analytic Functions Defined by a Linear Operator
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Corresponding to the functionφ(a, c;z),Carlson and Shaffer [2] introduced a linear operatorL(a, c) :A → Aby
(1.6) L(a, c)f(z) := φ(a, c;z)∗f(z), or, equivalently, by
L(a, c)f(z) :=z+
∞
X
n=1
(a)n
(c)nan+1zn+1 (z ∈∆).
Note thatL(1,1)f(z) =f(z),L(2,1)f(z) =zf0(z)andL(3,1)f(z) =zf0(z)+
1
2z2f00(z).
For−1 ≤ α < 1, β ≥ 0, we let L(a, c;α, β) consist of functions f in A satisfying the condition
(1.7) Re
aL(a+ 1, c)f(z)
L(a, c)f(z) −(a−1)
> β
aL(a+ 1, c)f(z) L(a, c)f(z) −a
+α, (z ∈∆) The family L(a, c;α, β) is of special interest for it contains many well- known as well as many new classes of analytic univalent functions. ForL(1,1;α,0), we obtain the family of starlike functions of orderα(0≤α <1)andL(2,1;α,0) is the family of convex functions of orderα(0≤ α <1).ForL(1,1; 0, β)and L(2,1; 0, β), we obtain the class of uniformly β- starlike functions and uni- formlyβ- convex functions, respectively, introduced by Kanas and Winsiowska
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([3],[4]) (see also the work of Kanas and Srivastava [5], Goodman ([7],[8]), Rønning ([10],[11]), Ma and Minda [9] and Gangadharan et al. [6]).
Before we state and prove our main result we need the following definitions and lemmas.
Definition 1.1 (Subordination Principle). Let g(z)be analytic and univalent in∆.Iff(z)is analytic in∆, f(0) =g(0),andf(∆) ⊂g(∆),then we see that the functionf(z)is subordinate tog(z)in∆,and we writef(z)≺g(z).
Definition 1.2 (Subordinating Factor Sequence). A sequence{bn}∞n=1of com- plex numbers is called a subordinating factor sequence if, whenever f(z) is analytic , univalent and convex in∆, we have the subordination given by (1.8)
∞
X
n=2
bnanzn ≺f(z) (z ∈∆, a1 = 1).
Lemma 1.1 ([14]). The sequence{bn}∞n=1is a subordinating factor sequence if and only if
(1.9) Re
( 1 + 2
∞
X
n=1
bnzn )
>0 (z ∈∆).
Lemma 1.2. If (1.10)
∞
X
n=2
σn(a, c;α, β)|an| ≤1−α
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where, for convenience,
σn(a, c;α, β) := (1 +β)(a)n+ [1−α−a(1 +β)](a)n−1
(c)n−1
(1.11)
(−1≤α <1; β ≥0, n≥2), thenf(z)∈ L(a, c;α, β).
Proof. It suffices to show that β
aL(a+ 1, c)f(z) L(a, c)f(z) −a
−Re
aL(a+ 1, c)f(z) L(a, c)f(z) −a
≤1−α.
We have β
aL(a+ 1, c)f(z) L(a, c)f(z) −a
−Re
aL(a+ 1, c)f(z) L(a, c)f(z) −a
≤(1 +β)
aL(a+ 1, c)f(z) L(a, c)f(z) −a
≤
(1 +β)P∞ n=2
a(a+1)n−1−a(a)n−1
(c)n−1
|an| |z|n−1
1−P∞ n=2
(a)n−1
(c)n−1 |an| |z|n−1
≤
(1 +β)P∞ n=2
(a)n−a(a)n−1
(c)n−1
|an|
1−P∞ n=2
(a)n−1
(c)n−1 |an| . The last expression is bounded above by1−αif
∞
X
n=2
(1 +β)(a)n+ [1−α−a(1 +β)](a)n−1 (c)n−1
|an| ≤1−α
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and the proof is complete.
LetL?(a, c;α, β)denote the class of functionsf(z)∈ Awhose coefficients satisfy the condition (1.10). We note thatL?(a, c;α, β)⊆ L(a, c;α, β).
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2. Main Theorem
Employing the techniques used earlier by Srivastava and Attiya [13], Attiya [1]
and Singh [12], we state and prove the following theorem.
Theorem 2.1. Let the functionf(z)defined by (1.1) be in the classL?(a, c;α, β) where−1≤α < 1 ; β ≥0;a >0;c >0.Also let Kdenote the familiar class of functionsf(z)∈ Awhich are also univalent and convex in∆. Then
(2.1) σ2(a, c;α, β)
2[1−α+σ2(a, c;α, β)](f ∗g)(z)≺g(z) (z ∈∆; g ∈ K), and
(2.2) Re(f(z))>−1−α+σ2(a, c;α, β)
σ2(a, c;α, β) , (z ∈∆).
The constant 2[1−α+σσ2(a,c;α,β)
2(a,c;α,β)] is the best estimate.
Proof. Letf(z)∈ L?(a, c;α, β)and letg(z) = z+P∞
n=2cnzn∈ K. Then (2.3) σ2(a, c;α, β)
2[1−α+σ2(a, c;α, β)](f∗g)(z)
= σ2(a, c;α, β)
2[1−α+σ2(a, c;α, β)] z+
∞
X
n=2
ancnzn
! .
Thus, by Definition1.2, the assertion of our theorem will hold if the sequence (2.4)
σ2(a, c;α, β)
2[1−α+σ2(a, c;α, β)]an ∞
n=1
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is a subordinating factor sequence, witha1 = 1.In view of Lemma1.1, this will be the case if and only if
(2.5) Re (
1 + 2
∞
X
n=1
σ2(a, c;α, β)
2[1−α+σ2(a, c;α, β)]anzn )
>0 (z ∈∆).
Now Re
(
1 + σ2(a, c;α, β) 1−α+σ2(a, c;α, β)
∞
X
n=1
anzn )
= Re
1 + σ2(a, c;α, β) 1−α+σ2(a, c;α, β)z
+ 1
1−α+σ2(a, c;α, β)
∞
X
n=1
σ2(a, c;α, β)anzn )
≥1−
σ2(a, c;α, β) 1−α+σ2(a, c;α, β)r
− 1
1−α+σ2(a, c;α, β)
∞
X
n=1
σn(a, c;α, β)anrn )
.
Sinceσn(a, c;α, β)is an increasing function ofn(n ≥2) 1−
σ2(a, c;α, β) 1−α+σ2(a, c;α, β)r
− 1
1−α+σ2(a, c;α, β)
∞
X
n=1
σn(a, c;α, β)anrn )
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>1− σ2(a, c;α, β)
1−α+σ2(a, c;α, β)r− 1−α
1−α+σ2(a, c;α, β)r (|z|=r)
>0.
Thus (2.5) holds true in∆.This proves the inequality (2.1). The inequality (2.2) follows by taking the convex functiong(z) = 1−zz =z+P∞
n=2zn in (2.1). To prove the sharpness of the constant 2[1−α+σσ2(a,c;α,β)
2(a,c;α,β)], we consider the function f0(z)∈ L?(a, c;α, β)given by
(2.6) f0(z) = z− 1−α
σ2(a, c;α, β)z2 (−1≤α <1; β ≥0).
Thus from (2.1), we have
(2.7) σ2(a, c;α, β)
2[1−α+σ2(a, c;α, β)]f0(z)≺ z 1−z. It can easily verified that
(2.8) min
Re
σ2(a, c;α, β)
2[1−α+σ2(a, c;α, β)]f0(z)
=−1
2 (z ∈∆), This shows that the constant 2[1−α+σσ2(a,c;α,β)
2(a,c;α,β)] is best possible.
Corollary 2.2. Let the functionf(z)defined by (1.1) be in the classL?(1,1;α, β) and satisfy the condition
(2.9)
∞
X
n=2
[n(1 +β)−(α+β)]|an| ≤1−α
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then
β+ 2−α
2(β+ 3−2α)(f ∗g)(z)≺g(z) (2.10)
(−1≤α <1; β ≥0; z ∈∆; g ∈ K) and
(2.11) Re(f(z))>−β+ 3−2α
β+ 2−α , (z ∈∆).
The constant 2(β+3−2α)β+2−α is the best estimate.
Corollary 2.3. Let the functionf(z)defined by (1.1) be in the classL?(1,1;α,0) and satisfy the condition
(2.12)
∞
X
n=2
(n−α)|an| ≤1−α, then
(2.13) 2−α
6−4α(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and
(2.14) Re(f(z))>−3−2α
2−α , (z ∈∆).
The constant 6−4α2−α is the best estimate.
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Puttingα= 0in Corollary2.3, we obtain
Corollary 2.4 ([12]). Let the function f(z) defined by (1.1) be in the class L?(1,1; 0,0)and satisfy the condition
(2.15)
∞
X
n=2
n|an| ≤1 then
(2.16) 1
3(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and
(2.17) Re(f(z))>−3
2, (z ∈∆).
The constant1/3 is the best estimate.
Corollary 2.5. Let the functionf(z)defined by (1.1) be in the classL?(2,1;α, β) and satisfy the condition
(2.18)
∞
X
n=2
n[n(1 +β)−(α+β)]|an| ≤1−α, then
β+ 2−α
2β+ 5−3α(f ∗g)(z)≺g(z) (2.19)
(−1≤α <1; β ≥0; z ∈∆; g ∈ K)
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and
(2.20) Re(f(z))>−2β+ 5−3α
2(β+ 2−α), (z ∈∆).
The constant 2β+5−3αβ+2−α is the best estimate.
Corollary 2.6. Let the functionf(z)defined by (1.1) be in the classL?(2,1;α,0) and satisfy the condition
(2.21)
∞
X
n=2
n(n−α)|an| ≤1−α, then
(2.22) 2−α
5−3α(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and
(2.23) Re(f(z))>− 5−3α
2(2−α), (z ∈∆).
The constant 5−3α2−α is the best estimate.
Puttingα= 0in Corollary2.6, we obtain
Corollary 2.7. Let the functionf(z)defined by (1.1) be in the classL?(2,1; 0,0) and satisfy the condition
(2.24)
∞
X
n=2
n2|an| ≤1
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then
(2.25) 2
5(f∗g)(z)≺g(z) (z ∈∆; g ∈ K) and
(2.26) Re(f(z))> −5
4 , (z ∈∆).
The constant2/5is the best estimate.
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