Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page
Contents
JJ II
J I
Page1of 21 Go Back Full Screen
Close
SOME PROPERTIES 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
Received: 29 August, 2006
Accepted: 03 April, 2007
Communicated by: A. Sofo 2000 AMS Sub. Class.: 30C45.
Key words: Analytic functions, Hadamard product, Linear operator.
Abstract: The object of the present paper is to derive some properties of analytic functions defined by the Carlson - Shaffer linear operatorL(a, c)f(z).
Acknowledgements: The author would like to thank the referee for his valuable suggestions.
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page2of 21 Go Back Full Screen
Close
Contents
1 Introduction and Definitions 3
2 Main Results 6
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page3of 21 Go Back Full Screen
Close
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 U ={z :|z|<1}.For two functionsf(z) andg(z)given by
(1.2) f(z) = z+
∞
X
n=2
anzn and g(z) =z+
∞
X
n=2
bnzn,
their Hadamard product (or convolution) is defined by
(1.3) (f∗g)(z) :=z+
∞
X
n=2
anbnzn.
Define the functionφ(a, c;z)by
φ(a, c;z) :=
∞
X
n=0
(a)n (c)n
zn+1, (1.4)
(a∈R; c∈R\Z−0; Z−0 :={0,−1,−2, . . .}, z∈ U), where(λ)nis the Pochhammer symbol given in terms of Gamma functions,
(λ)n:= Γ(λ+n) Γ(λ)
=
1, n= 0,
λ(λ+ 1)(λ+ 2). . .(λ+n−1), n∈N:{1,2, . . .}.
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page4of 21 Go Back Full Screen
Close
Corresponding to the function φ(a, c;z), Carlson and Shaffer [1] introduced a linear operatorL(a, c) :A → Aby
(1.5) L(a, c)f(z) :=φ(a, c;z)∗f(z), or, equivalently, by
(1.6) L(a, c)f(z) :=z+
∞
X
n=1
(a)n
(c)nan+1zn+1 (z ∈ U).
It follows from (1.6) that
(1.7) z(L(a, c)f(z))0 =aL(a+ 1, c)f(z)−(a−1)L(a, c)f(z), and
L(1,1)f(z) =f(z), L(2,1)f(z) =zf0(z), L(3,1)f(z) =zf0(z) + 1
2z2f00(z).
Many properties of analytic functions defined by the Carlson-Shaffer linear op- erator were studied by (among others) Owa and Srivastava [10], Ding [5], Kim and Lee [6], Ravichandran et al. ([8,9]), Shanmugam et al. [7] and Frasin ([2,3]).
In this paper we shall derive some properties of analytic functions defined by the linear operatorL(a, c)f(z).
In order to prove our main results, we recall the following lemma:
Lemma 1.1 ([4]). LetΦ(u, v)be a complex valued function,
Φ :D→C, (D⊂C2;Cis the complex plane),
and letu=u1+iu2andv =v1+iv2.Suppose that the functionΦ(u, v)satisfies:
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page5of 21 Go Back Full Screen
Close
(i) Φ(u, v)is continuous inD;
(ii) (1,0)∈DandRe(Φ(1,0))>0;
(iii) Re(Φ(iu2, v1))≤0for all(iu2, v1)∈Dand such thatv1 ≤ −(1 +u22)/2.
Letp(z) = 1 +p1z+p2z2+· · · be regular inU such that(p(z), zp0(z))∈D for allz ∈ U.IfRe(Φ(p(z), zp0(z)))>0 (z ∈ U), thenRe(p(z))>0 (z ∈ U).
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page6of 21 Go Back Full Screen
Close
2. Main Results
Theorem 2.1. Letα ∈C, β ∈C,(Re(β)≥0),(α−β)∈R, and suppose that
(2.1) Re
L(a, c)f(z) L(a+ 1, c)f(z)
α−β(a+ 1)L(a+ 2, c)f(z) L(a+ 1, c)f(z)
+aα
> γ
for someγ(γ <(α−β)(a+ 1))and2(α−β) + Re(β)6= 0,then
(2.2) Re
L(a, c)f(z) L(a+ 1, c)f(z)
> 2γ−2a(α−β) + Re(β)
2(α−β) + Re(β) (z∈ U).
Proof. Define the functionp(z)by
(2.3) L(a, c)f(z)
L(a+ 1, c)f(z) :=δ+ (1−δ)p(z), where
(2.4) δ= 2γ−2a(α−β) + Re(β)
2(α−β) + Re(β) <1.
Thenp(z) = 1 +b1z+b2z+· · · is regular inU. It follows from (2.3) that (2.5) z(L(a, c)f(z))0
L(a, c)f(z) − z(L(a+ 1, c)f(z))0
L(a+ 1, c)f(z) = (1−δ)zp0(z) δ+ (1−δ)p(z) by making use of the familiar identity (1.7) in (2.5), we get
(2.6) (a+ 1)L(a+ 2, c)f(z)
L(a+ 1, c)f(z) = 1 + a
δ+ (1−δ)p(z) − (1−δ)zp0(z) δ+ (1−δ)p(z) or, equivalently,
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page7of 21 Go Back Full Screen
Close
(2.7) L(a, c)f(z) L(a+ 1, c)f(z)
α−β(a+ 1)L(a+ 2, c)f(z) L(a+ 1, c)f(z)
+aα
= (δ+a)(α−β) + (1−δ)(α−β)p(z) +β(1−δ)zp0(z).
Therefore, we have (2.8) Re
L(a, c)f(z) L(a+ 1, c)f(z)
α−β(a+ 1)L(a+ 2, c)f(z) L(a+ 1, c)f(z)
+aα−γ
= Re{(δ+a)(α−β)−γ+ (1−δ)(α−β)p(z) +β(1−δ)zp0(z)}>0.
If we define the functionΦ(u, v)by
(2.9) Φ(u, v) = (δ+a)(α−β)−γ+ (1−δ)(α−β)u+β(1−δ)v withu=u1+iu2 andv =v1+iv2,then
(i)Φ(u, v)is continuous inD=C2;
(ii)(1,0)∈DandRe(Φ(1,0)) = (α−β)(a+ 1)−γ >0;
(iii) For all(iu2, v1)∈Dand such thatv1 ≤ −(1 +u22)/2, Re(Φ(iu2, v1)) = Re{(δ+a)(α−β)−γ+β(1−δ)v1}
≤(δ+a)(α−β)−γ− (1−δ)(1 +u22) Re(β) 2
=−(1−δ)u22Re(β)
2 ≤0.
Therefore, the functionΦ(u, v)satisfies the conditions in Lemma1.1. Thus we have Re(p(z))>0 (z ∈ U),that is
Re
L(a, c)f(z) L(a+ 1, c)f(z)
> 2γ−2a(α−β) + Re(β) 2(α−β) + Re(β) .
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page8of 21 Go Back Full Screen
Close
Lettingβ=−α−in Theorem2.1, we have
Corollary 2.2. Letα ∈C,(Re(α)<0),and suppose that (2.10) Re
L(a, c)f(z) L(a+ 1, c)f(z)
α+−α(a+ 1)L(a+ 2, c)f(z) L(a+ 1, c)f(z)
+aα
> γ
for someγ(γ <2(a+ 1) Re(α)),then
(2.11) Re
L(a, c)f(z) L(a+ 1, c)f(z)
> 2γ−(4a+ 1) Re(α)
3 Re(α) (z ∈ U).
Lettinga=c= 1in Theorem2.1, we have
Corollary 2.3. Letα, β ∈C,(Re(β)≥0),(α−β)∈R, and suppose that
(2.12) Re
f(z) zf0(z)
α−2β
1 + zf00(z) 2f0(z)
+α
> γ
for someγ(γ <2(α−β))and2(α−β) + Re(β)6= 0,then
(2.13) Re
f(z) zf0(z)
> 2γ−2(α−β) + Re(β)
2(α−β) + Re(β) (z ∈ U).
Theorem 2.4. Letα ∈C, β ∈C,(Re(β)≥0),(α−β)∈R, and suppose that (2.14) Re
L(a, c)f(z) L(a+ 1, c)f(z)
α−β(a+ 1)L(a+ 2, c)f(z) L(a+ 1, c)f(z)
+aα
< γ
for someγ(γ >(α−β)(a+ 1))and2(α−β) + Re(β)6= 0,then
(2.15) Re
L(a, c)f(z) L(a+ 1, c)f(z)
< 2γ−2a(α−β) + Re(β)
2(α−β) + Re(β) (z ∈ U).
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page9of 21 Go Back Full Screen
Close
Proof. Define the functionp(z)by
(2.16) L(a, c)f(z)
L(a+ 1, c)f(z) :=δ+ (1−δ)p(z), where
(2.17) δ= 2γ−2a(α−β) + Re(β)
2(α−β) + Re(β) >1.
Thenp(z) = 1 +b1z+b2z+· · · is regular inU. It follows from (2.16) that (2.18) Re
γ− L(a, c)f(z) L(a+ 1, c)f(z)
α−β(a+ 1)L(a+ 2, c)f(z) L(a+ 1, c)f(z)
−aα
= Re{γ−(δ+a)(α−β)−(1−δ)(α−β)p(z)−β(1−δ)zp0(z)}>0.
If we define the functionΦ(u, v)by
(2.19) Φ(u, v) =γ−(δ+a)(α−β)−(1−δ)(α−β)u−β(1−δ)v withu=u1+iu2 andv =v1+iv2,then
(i)Φ(u, v)is continuous inD=C2;
(ii)(1,0)∈DandRe(Φ(1,0)) =γ−(α−β)(a+ 1)>0;
(iii) For all(iu2, v1)∈Dand such thatv1 ≤ −(1 +u22)/2, Re(Φ(iu2, v1)) = Re{γ−(δ+a)(α−β)−β(1−δ)v1}
≤γ−(δ+a)(α−β) + (1−δ)(1 +u22) Re(β) 2
= (1−δ)u22Re(β)
2 ≤0,
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page10of 21 Go Back Full Screen
Close
applying Lemma1.1, we haveRe(p(z))>0 (z ∈ U),that is Re
L(a, c)f(z) L(a+ 1, c)f(z)
< 2γ−2a(α−β) + Re(β) 2(α−β) + Re(β) .
Theorem 2.5. Leta > −1,µ >0, α∈ C, β ∈C,(Re(β)≥0),(α+β)∈ R, and suppose that
(2.20) Re
(
L(a+ 1, c)f(z) z
µ
α+βL(a+ 2, c)f(z) L(a+ 1, c)f(z)
)
> γ
for someγ(γ < α+β)and2µ(α+β)(a+ 1) + Re(β)6= 0,then (2.21) Re
(
L(a+ 1, c)f(z) z
µ)
> 2µγ(a+ 1) + Re(β)
2µ(α+β)(a+ 1) + Re(β) (z ∈ U).
Proof. Define the functionp(z)by (2.22)
L(a+ 1, c)f(z) z
µ
:=δ+ (1−δ)p(z), where
δ = 2µγ(a+ 1) + Re(β)
2µ(α+β)(a+ 1) + Re(β) >1.
Thenp(z) = 1 +b1z+b2z+· · · is regular inU. Also, by a simple computation and by making use of the familiar identity (1.7) we find from (2.22) that
(2.23) βL(a+ 2, c)f(z)
L(a+ 1, c)f(z) = β(1−δ)zp0(z)
µ(a+ 1)(δ+ (1−δ)p(z))+β,
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page11of 21 Go Back Full Screen
Close
and by using (2.22) and (2.23), we get (2.24)
L(a+ 1, c)f(z) z
µ
βL(a+ 2, c)f(z) L(a+ 1, c)f(z) +α
= β(1−δ)zp0(z)
µ(a+ 1) + (α+β)(δ+ (1−δ)p(z)).
Therefore, we have (2.25) Re
(
L(a+ 1, c)f(z) z
µ
βL(a+ 2, c)f(z) L(a+ 1, c)f(z) +α
−γ )
= Re
(α+β)δ−γ+ (1−δ)(α+β)p(z) + β(1−δ) µ(a+ 1)zp0(z)
>0.
If we define the functionΦ(u, v)by
(2.26) Φ(u, v) = (α+β)δ−γ+ (1−δ)(α+β)u+β(1−δ) µ(a+ 1)v withu=u1+iu2 andv =v1+iv2,then
(i)Φ(u, v)is continuous inD=C2;
(ii)(1,0)∈DandRe(Φ(1,0)) = (α+β)−γ >0;
(iii) For all(iu2, v1)∈Dand such thatv1 ≤ −(1 +u22)/2, Re(Φ(iu2, v1)) = Re{(α+β)δ−γ+ β(1−δ)
µ(a+ 1)v1}
≤(α+β)δ−γ− (1−δ)(1 +u22) Re(β) 2µ(a+ 1)
=−(1−δ)u22Re(β) 2µ(a+ 1) ≤0.
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page12of 21 Go Back Full Screen
Close
Therefore, the functionΦ(u, v)satisfies the conditions in Lemma1.1. Thus we haveRe(p(z))>0 (z∈ U),that is,
Re (
L(a+ 1, c)f(z) z
µ)
> 2µγ(a+ 1) + Re(β)
2µ(α+β)(a+ 1) + Re(β) (z ∈ U).
Lettingα=
−
β in Theorem2.5, we have
Corollary 2.6. Leta >−1,µ >0, β ∈C,(Re(β)>0), and suppose that
(2.27) Re
L(a+ 1, c)f(z) z
µ
α+βL(a+ 2, c)f(z) L(a+ 1, c)f(z)
> γ
for someγ(γ <2 Re(β)),then (2.28) Re
(
L(a+ 1, c)f(z) z
µ)
> 2µγ(a+ 1) + Re(β)
(4µ(a+ 1) + 1) Re(β) (z ∈ U).
Lettinga=c= 1in Theorem2.5, we have,
Corollary 2.7. Letµ >0, α∈C, β ∈C, (Re(β)≥0),(α+β)∈R, and suppose that
(2.29) Re
(f0(z))µ
α+β
1 + zf00(z) 2f0(z)
> γ
for someγ(γ < α+β)and4µ(α+β) + Re(β)6= 0,then
(2.30) Ren
(f0(z))µo
> 4µγ+ Re(β)
4µ(α+β) + Re(β) (z ∈ U).
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page13of 21 Go Back Full Screen
Close
Employing the same manner as in the proofs of Theorems2.4and2.5, we have:
Theorem 2.8. Leta > −1,µ >0, α∈ C, β ∈C,(Re(β)≥0),(α+β)∈ R, and suppose that
(2.31) Re
(
L(a+ 1, c)f(z) z
µ
α+βL(a+ 2, c)f(z) L(a+ 1, c)f(z)
)
< γ
for someγ(γ > α+β)and2µ(α+β)(a+ 1) + Re(β)6= 0,then (2.32) Re
(
L(a+ 1, c)f(z) z
µ)
< 2µγ(a+ 1) + Re(β)
2µ(α+β)(a+ 1) + Re(β) (z ∈ U).
Theorem 2.9. Letλ > 0, α∈ C, β ∈ C,(Re(β)≥ 0),(α+β) ∈ R, and suppose that
(2.33) Re (
L(a+ 1, c)f(z) L(a, c)f(z)
λ
×
λβ(a+ 1)L(a+ 2, c)f(z)
L(a+ 1, c)f(z)−aλβL(a+ 1, c)f(z) L(a, c)f(z) +λα
> γ
for someγ(γ < λ(α+β))and2λ(α+β) + Re(β)6= 0, then
(2.34) Re
L(a+ 1, c)f(z) L(a, c)f(z)
λ
> 2γ+ Re(β)
2λ(α+β) + Re(β) (z ∈ U).
Proof. Define the functionp(z)by (2.35)
L(a+ 1, c)f(z) L(a, c)f(z)
λ
:=δ+ (1−δ)p(z),
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page14of 21 Go Back Full Screen
Close
where
(2.36) δ= 2γ+ Re(β)
2λ(α+β) + Re(β) <1.
Thenp(z) = 1 +b1z +b2z+· · · is regular inU . Also, by a simple computation and by making use of the familiar identity (1.7), we find from (2.35) that
(2.37)
L(a+ 1, c)f(z) L(a, c)f(z)
λ
×
λβ(a+ 1)L(a+ 2, c)f(z)
L(a+ 1, c)f(z) −aλβL(a+ 1, c)f(z) L(a, c)f(z) +λα
=λ(α+β)(δ+ (1−δ)p(z)) +β(1−δ)zp0(z) Therefore, we have
(2.38) Re (
L(a+ 1, c)f(z) L(a, c)f(z)
λ
×
λβ(a+ 1)L(a+ 2, c)f(z)
L(a+ 1, c)f(z) −aλβL(a+ 1, c)f(z) L(a, c)f(z) +λα
−γ
= Re{λ(α+β)(δ+ (1−δ)p(z)) +β(1−δ)zp0(z)−γ}>0.
If we define the functionΦ(u, v)by
(2.39) Φ(u, v) =λδ(α+β)−γ+λ(1−δ)(α+β)u+β(1−δ)v withu=u1+iu2 andv =v1+iv2,then
(i)Φ(u, v)is continuous inD=C2;
(ii)(1,0)∈DandRe(Φ(1,0)) =λ(α+β)−γ >0;
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page15of 21 Go Back Full Screen
Close
(iii) For all(iu2, v1)∈Dand such thatv1 ≤ −(1 +u22)/2, Re(Φ(iu2, v1)) = Re{λδ(α+β)−γ+β(1−δ)v1}
≤λδ(α+β)−γ− (1−δ)(1 +u22)
2 Re(β)
=−(1−δ)u22Re(β)
2 ≤0.
Therefore, the functionΦ(u, v)satisfies the conditions in Lemma1.1. Thus we have Re
L(a+ 1, c)f(z) L(a, c)f(z)
λ
> 2γ+ Re(β)
2λ(α+β) + Re(β) (z ∈ U).
Lettingα=
−
β in Theorem2.9, we have:
Corollary 2.10. Letλ >0, α∈C, β ∈C,(Re(β)>0),(α+β)∈R, and suppose that
(2.40) Re (
L(a+ 1, c)f(z) L(a, c)f(z)
λ
×
λβ(a+ 1)L(a+ 2, c)f(z)
L(a+ 1, c)f(z) −aλβL(a+ 1, c)f(z) L(a, c)f(z) +λ
−
β
> γ
for someγ(γ < λ(α+β)),then
(2.41) Re
L(a+ 1, c)f(z) L(a, c)f(z)
λ
> 2γ+ Re(β)
(4λ+ 1) Re(β) (z ∈ U).
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page16of 21 Go Back Full Screen
Close
Lettinga=c=λ= 1in Theorem2.9, we have:
Corollary 2.11. Letα ∈C, β ∈C,(Re(β)≥0),(α+β)∈R, and suppose that:
(2.42) Re
zf0(z)
f(z) (2β+α) +βz
f00(z)
f0(z) − f0(z) f(z)
> γ
for someγ(γ < α+β)and2(α+β) + Re(β)6= 0, thenf(z)is starlike of orderδ, whereδ = 2(α+β)+Re(β)2γ+Re(β) .
Employing the same manner as in the proofs of Theorems2.4and2.9, we have:
Theorem 2.12. Letλ >0, α∈ C, β ∈C,(Re(β)≥ 0),(α+β)∈R, and suppose that:
(2.43) Re (
L(a+ 1, c)f(z) L(a, c)f(z)
λ
×
λβ(a+ 1)L(a+ 2, c)f(z)
L(a+ 1, c)f(z)−aλβL(a+ 1, c)f(z) L(a, c)f(z) +λα
< γ
for someγ(γ > λ(α+β))and2λ(α+β) + Re(β)6= 0, then
(2.44) Re
L(a+ 1, c)f(z) L(a, c)f(z)
λ
< 2γ+ Re(β)
2λ(α+β) + Re(β) (z ∈ U).
Theorem 2.13. Leta >−1, α∈C, β∈C,(Re(β)≥0),(α−β)∈R, and suppose that
(2.45) Re
z
L(a+ 1, c)f(z)
(a+ 1)α−βL(a+ 2, c)f(z) L(a+ 1, c)f(z)
> γ
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page17of 21 Go Back Full Screen
Close
for someγ(γ <(α−β)(a+ 1)),then (2.46) Re
z
L(a+ 1, c)f(z)
> 2γ+ Re(β)
2(a+ 1)(α−β) + Re(β) (z ∈ U).
Proof. Define the functionp(z)by
(2.47) z
L(a+ 1, c)f(z) :=δ+ (1−δ)p(z), where
(2.48) δ = 2γ+ Re(β)
2(a+ 1)(α−β) + Re(β) >1.
Thenp(z) = 1 +b1z+b2z+· · · is regular inU. Also, by a simple computation and by making use of the familiar identity (1.7), we find from (2.47) that
(2.49) z
L(a+ 1, c)f(z)
(a+ 1)α−βL(a+ 2, c)f(z) L(a+ 1, c)f(z)
= (a+ 1)(α−β)(δ+ (1−δ)p(z)) +β(1−δ)zp0(z).
Therefore, we have (2.50) Re
z
L(a+ 1, c)f(z)
(a+ 1)α−βL(a+ 2, c)f(z) L(a+ 1, c)f(z)
−γ
= Re{(a+ 1)(α−β)(δ+ (1−δ)p(z)) +β(1−δ)zp0(z)−γ}>0.
If we define the functionΦ(u, v)by
(2.51) Φ(u, v) = δ(a+ 1)(α−β)−γ+ (a+ 1)(α−β)(1−δ)u+β(1−δ)v
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page18of 21 Go Back Full Screen
Close
withu=u1+iu2 andv =v1+iv2,then (i)Φ(u, v)is continuous inD=C2;
(ii)(1,0)∈DandRe(Φ(1,0)) = (α−β)(a+ 1)−γ >0;
(iii) For all(iu2, v1)∈Dand such thatv1 ≤ −(1 +u22)/2, Re(Φ(iu2, v1)) = Re{δ(a+ 1)(α−β)−γ+β(1−δ)v1}
≤δ(a+ 1)(α−β)−γ−(1−δ)(1 +u22)
2 Re(β)
=−(1−δ)u22Re(β)
2 ≤0.
Therefore, the functionΦ(u, v)satisfies the conditions in Lemma1.1. Thus we have Re
z
L(a+ 1, c)f(z)
> 2γ+ Re(β)
2(a+ 1)(α−β) + Re(β) (z ∈ U).
Lettingβ=−α−in Theorem2.13, we have:
Corollary 2.14. Leta >−1, a6= 0, α∈C,(Re(α)<0),and suppose that
(2.52) Re
z
L(a+ 1, c)f(z)
(a+ 1)α+α−L(a+ 2, c)f(z) L(a+ 1, c)f(z)
> γ
for someγ(γ <2(a+ 1) Re(α)),then
(2.53) Re
z
L(a+ 1, c)f(z)
> 2γ−Re(α)
(4a+ 3) Re(α) (z ∈ U).
Lettinga=c= 1in Theorem2.13, we have:
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page19of 21 Go Back Full Screen
Close
Corollary 2.15. Let a > −1, α ∈ C, β ∈ C, (Re(β) ≥ 0), (α −β) ∈ R, and suppose that
(2.54) Re
1 f0(z)
2α−β−zf00(z) 2f0(z)β
> γ
for someγ(γ <2(α−β))and4(α−β) + Re(β)6= 0, then
(2.55) Re
1 f0(z)
> 2γ+ Re(β)
4(α−β) + Re(β) (z ∈ U).
Employing the same manner as in the proofs of Theorem2.4and2.13, we have:
Theorem 2.16. Leta >−1, α∈C, β∈C,(Re(β)≥0),(α−β)∈R, and suppose that
(2.56) Re
z
L(a+ 1, c)f(z)
(a+ 1)α−βL(a+ 2, c)f(z) L(a+ 1, c)f(z)
< γ
for someγ(γ >(α−β)(a+ 1))and2(a+ 1)(α−β) + Re(β)6= 0, then (2.57) Re
z
L(a+ 1, c)f(z)
< 2γ+ Re(β)
2(a+ 1)(α−β) + Re(β) (z ∈ U).
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page20of 21 Go Back Full Screen
Close
References
[1] B.C. CARLSONANDD.B. SHAFFER, Starlike and prestarlike hypergeomet- ric functions, SIAM J. Math. Anal., 15(4) (1984), 737–745.
[2] B.A. FRASIN, Subordinations results for a class of analytic functions defined by linear operator, J. Inequal. Pure Appl. Math., 7(4) (2006), Art. 134 [ON- LINE:http://jipam.vu.edu.au/article.php?sid=754].
[3] B.A. FRASIN, Subclasses of analytic functions defined by Carlson-Shaffer lin- ear operator, to appear in Tamsui Oxford J. Math.
[4] S. S. MILLER AND P.T. MOCANU, Second order differential inequalities in the complex plane, J. Math. Anal. Appl., 65 (1978), 289–305.
[5] S. DING, Some properties of a class of analytic functions, J. Math. Anal. Appl., 195 (1995), 71–81.
[6] Y.C. KIM AND K.S. LEE, Some applications of fractional integral operators and Ruscheweyh derivatives, J. Math. Anal. Appl., 197 (1996), 505–517.
[7] T.N. SHANMUGAM, V. RAVICHANDRANANDS. SIVASUBRAMANIAN, Differential sandwich theorems for some subclasses of analytic functions, Aust.
J. Math. Anal. Appl., 3(1) (2006), 1–11.
[8] V. RAVICHANDRAN, H. SELVERMAN, S.S KUMAR AND K.G. SUBRA- MANIAN, On differential subordinations for a class of analytic functions de- fined by a linear operator, International Journal of Mathematics and Mathe- matical Sciences, 42 (2004), 2219–2230.
[9] V. RAVICHANDRAN, N. SEENIVASAGANANDH.M. SRIVASTAVA, Some inequalities associated with a linear operator defined for a class of multivalent
Some Properties of Analytic Functions
B.A. Frasin vol. 8, iss. 2, art. 53, 2007
Title Page Contents
JJ II
J I
Page21of 21 Go Back Full Screen
Close
functions, J. Inequal. Pure Appl. Math., 4(4) (2003), Art.70 [ONLINE:http:
//jipam.vu.edu.au/article.php?sid=311].
[10] S. OWAAND H.M. SRIVASTAVA, Univalent and starlike generalized hyper- geometric functions, Canad. J. Math., 39 (1987), 1057–1077.