DIFFERENTIAL SUBORDINATION RESULTS FOR
NEW CLASSES OF THE FAMILY E(Φ, Ψ)
RABHA W. IBRAHIM AND MASLINA DARUS
School of Mathematical Sciences
Faculty of Science and Technology
Universiti Kebangsaan Malaysia
Bangi 43600, Selangor Darul Ehsan
Malaysia
EMail: rabhaibrahim@yahoo.com
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
maslina@ukm.my
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Contents
Received:
05 July, 2008
Accepted:
02 December, 2008
Communicated by:
JJ
II
S.S. Dragomir
J
I
2000 AMS Sub. Class.:
34G10, 26A33, 30C45.
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Key words:
Fractional calculus; Subordination; Hadamard product.
Abstract:
We define new classes of the family E(Φ, Ψ), in a unit disk U := {z ∈
C, |z| < 1}, as follows: for analytic functions F (z), Φ(z) and Ψ(z) so that
(z)∗Φ(z)
<{ FF (z)∗Ψ(z)
} > 0, z ∈ U, F (z) ∗ Ψ(z) 6= 0 where the operator ∗ denotes the
convolution or Hadamard product. Moreover, we establish some subordination
results for these new classes.
Acknowledgement:
The work presented here was supported by SAGA: STGL-012-2006, Academy
of Sciences, Malaysia.
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Contents
1
Introduction and preliminaries.
3
2
Main Results
7
3
Applications
14
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
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1.
Introduction and preliminaries.
Let Bα+ be the class of all analytic functions F (z) in the open disk U := {z ∈
C, |z| < 1}, of the form
F (z) = 1 +
∞
X
an z n+α−1 ,
0 < α ≤ 1,
n=1
satisfying F (0) = 1. And let
open disk U of the form
F (z) = 1 −
∞
X
Bα−
be the class of all analytic functions F (z) in the
an z n+α−1 ,
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
0 < α ≤ 1,
an ≥ 0;
n = 1, 2, 3, . . . ,
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n=1
satisfying F (0) = 1. With a view to recalling the principle of subordination between
analytic functions, let the functions f and g be analytic in U. Then we say that the
function f is subordinate to g if there exists a Schwarz function w(z), analytic in U
such that
f (z) = g(w(z)), z ∈ U.
We denote this subordination by f ≺ g or f (z) ≺ g(z), z ∈ U. If the function g is
univalent in U the above subordination is equivalent to
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f (0) = g(0)
and f (U ) ⊂ g(U ).
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3
Let φ : C × U → C and let h be univalent in U. Assume that p, φ are analytic
and univalent in U and p satisfies the differential superordination
(1.1)
h(z) ≺ φ(p(z)), zp0 (z), z 2 p00 (z); z),
then p is called a solution of the differential superordination.
An analytic function q is called a subordinant if q ≺ p for all p satisfying (1.1).
A univalent function q such that p ≺ q for all subordinants p of (1.1) is said to be the
best subordinant.
Let B + be the class of analytic functions of the form
f (z) = 1 +
∞
X
an z n ,
an ≥ 0.
n=1
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
Given two functions f, g ∈ B + ,
f (z) = 1 +
∞
X
an z n
and g(z) = 1 +
n=1
∞
X
vol. 10, iss. 1, art. 8, 2009
bn z n
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n=1
their convolution or Hadamard product f (z) ∗ g(z) is defined by
f (z) ∗ g(z) = 1 +
∞
X
an b n z n ,
an ≥ 0, bn ≥ 0,
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z ∈ U.
n=1
Φ(z) = z +
∞
X
n=2
ϕn z n
and
Ψ(z) = z +
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Juneja et al. [1] define the family E(Φ, Ψ), so that
f (z) ∗ Φ(z)
<
> 0, z ∈ U
f (z) ∗ Ψ(z)
where
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∞
X
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ψn z n
n=2
are analytic in U with the conditions ϕn ≥ 0, ψn ≥ 0, ϕn ≥ ψn for n ≥ 2 and
f (z) ∗ Ψ(z) 6= 0.
Definition 1.1. Let F (z) ∈ Bα+ , we define the family Eα+ (Φ, Ψ) so that
F (z) ∗ Φ(z)
(1.2)
<
> 0, z ∈ U,
F (z) ∗ Ψ(z)
where
Φ(z) = 1 +
∞
X
ϕn z n+α−1
and
Ψ(z) = 1 +
n=1
∞
X
ψn z n+α−1
n=1
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
are analytic in U under the conditions ϕn ≥ 0, ψn ≥ 0, ϕn ≥ ψn for n ≥ 1 and
F (z) ∗ Ψ(z) 6= 0.
vol. 10, iss. 1, art. 8, 2009
Definition 1.2. Letting F (z) ∈ Bα− , we define the family Eα− (Φ, Ψ) which satisfies
(1.2) where
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Φ(z) = 1 −
∞
X
ϕn z n+α−1
and
Ψ(z) = 1 −
n=1
∞
X
ψn z n+α−1
n=1
are analytic in U under the conditions ϕn ≥ 0, ψn ≥ 0, ϕn ≥ ψn for n ≥ 1 and
F (z) ∗ Ψ(z) 6= 0.
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In the present paper, we establish some sufficient conditions for functions F ∈
Bα+ and F ∈ Bα− to satisfy
(1.3)
F (z) ∗ Φ(z)
≺ q(z),
F (z) ∗ Ψ(z)
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z ∈ U,
where q(z) is a given univalent function in U such that q(0) = 1. Moreover, we
give applications for these results in fractional calculus. We shall need the following
known results.
Lemma 1.3 ([2]). Let q(z) be convex in the unit disk U with q(0) = 1 and <{q} >
1
, z ∈ U. If 0 ≤ µ < 1, p is an analytic function in U with p(0) = 1 and if
2
(1 − µ)p2 (z) + (2µ − 1)p(z) − µ + (1 − µ)zp0 (z)
≺ (1 − µ)q 2 (z) + (2µ − 1)q(z) − µ + (1 − µ)zq 0 (z),
then p(z) ≺ q(z) and q(z) is the best dominant.
Lemma 1.4 ([3]). Let q(z) be univalent in the unit disk U and let θ(z) be analytic in
a domain D containing q(U ). If zq 0 (z)θ(q) is starlike in U, and
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
zp0 (z)θ(p(z)) ≺ zq 0 (z)θ(q(z))
then p(z) ≺ q(z) and q(z) is the best dominant.
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2.
Main Results
In this section, we verify some sufficient conditions of subordination for analytic
functions in the classes Bα+ and Bα− .
Theorem 2.1. Let the function q(z) be convex in the unit disk U such that q(0) = 1
(z)∗Φ(z)
and <{q} > 12 . If F ∈ Bα+ and FF (z)∗Ψ(z)
an analytic function in U satisfies the
subordination
2
F (z) ∗ Φ(z)
F (z) ∗ Φ(z)
+ (2µ − 1)
(1 − µ)
−µ
F (z) ∗ Ψ(z)
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z) z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
+ (1 − µ)
−
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
≺ (1 − µ)q 2 (z) + (2µ − 1)q(z) − µ + (1 − µ)zq 0 (z),
then
F (z) ∗ Φ(z)
≺ q(z)
F (z) ∗ Ψ(z)
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
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and q(z) is the best dominant.
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Proof. Let the function p(z) be defined by
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F (z) ∗ Φ(z)
p(z) :=
,
F (z) ∗ Ψ(z)
z ∈ U.
It is clear that p(0) = 1. Then straightforward computation gives us
(1 − µ)p2 (z) + (2µ − 1)p(z) − µ + (1 − µ)zp0 (z)
Close
2
F (z) ∗ Φ(z)
F (z) ∗ Φ(z)
= (1 − µ)
+ (2µ − 1)
−µ
F (z) ∗ Ψ(z)
F (z) ∗ Ψ(z)
0
F (z) ∗ Φ(z)
+ (1 − µ)z
F (z) ∗ Ψ(z)
2
F (z) ∗ Φ(z)
F (z) ∗ Φ(z)
= (1 − µ)
+ (2µ − 1)
−µ
F (z) ∗ Ψ(z)
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z) z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
+ (1 − µ)
−
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
2
0
≺ (1 − µ)q (z) + (2µ − 1)q(z) − µ + (1 − µ)zq (z).
By the assumption of the theorem we have that the assertion of the theorem follows
by an application of Lemma 1.3.
(z)∗Φ(z)
Corollary 2.2. If F ∈ Bα+ and FF (z)∗Ψ(z)
is an analytic function in U satisfying the
subordination
2
F (z) ∗ Φ(z)
F (z) ∗ Φ(z)
(1 − µ)
+ (2µ − 1)
−µ
F (z) ∗ Ψ(z)
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z) z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
−
+ (1 − µ)
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
2
1 + Az
1 + Az
≺ (1 − µ)
+ (2µ − 1)
1 + Bz
1 + Bz
(A − B)z
1 + Az
− µ + (1 − µ)
,
1 + Bz (1 + Az)(1 + Bz)
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
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then
F (z) ∗ Φ(z)
≺
F (z) ∗ Ψ(z)
1 + Az
1 + Bz
−1 ≤ B < A ≤ 1
,
1+Az
) is the best dominant.
and ( 1+Bz
Proof. Let the function q(z) be defined by
1 + Az
q(z) :=
,
1 + Bz
It is clear that q(0) = 1 and <{q} >
Theorem 2.1 we obtain the result.
1
2
z ∈ U.
for arbitrary A, B, z ∈ U, then in view of
(z)∗Φ(z)
Corollary 2.3. If F ∈ Bα+ and FF (z)∗Ψ(z)
is an analytic function in U satisfying the
subordination
2
F (z) ∗ Φ(z)
F (z) ∗ Φ(z)
(1 − µ)
+ (2µ − 1)
−µ
F (z) ∗ Ψ(z)
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z) z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
−
+ (1 − µ)
F (z) ∗ Ψ(z)
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
2
1+z
1+z
≺ (1 − µ)
+ (2µ − 1)
−µ
1−z
1−z
2z
1+z
,
+ (1 − µ)
1−z
1 − z2
then
F (z) ∗ Φ(z)
≺
F (z) ∗ Ψ(z)
1+z
and ( 1−z
) is the best dominant.
1+z
1−z
,
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
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Define the function ϕα (a, c; z) by
ϕα (a, c; z) := 1 +
∞
X
(a)n
n=1
(c)n
z n+α−1 ,
(z ∈ U ; a ∈ R, c ∈ R\{0, −1, −2, . . . }),
where (a)n is the Pochhammer symbol defined by

1,
(n = 0);
Γ(a + n) 
(a)n :=
=
 a(a + 1)(a + 2) · · · (a + n − 1), (n ∈ N).
Γ(a)
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
Corresponding to the function ϕα (a, c; z), define a linear operator Lα (a, c) by
Lα (a, c)F (z) := ϕα (a, c; z) ∗ F (z),
F (z) ∈ Bα+
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or equivalently by
Lα (a, c)F (z) := 1 +
∞
X
(a)n
n=1
(c)n
an z n+α−1 .
For details see [4]. Hence we have the following result:
Corollary 2.4. Let the function q(z) be convex in the unit disk U such that q(0) = 1
(a,c)Φ(z)
and <{q} > 21 . If LLαα(a,c)Ψ(z)
is an analytic function in U satisfying the subordination
2
Lα (a, c)Φ(z)
Lα (a, c)Φ(z)
(1 − µ)
+ (2µ − 1)
−µ
Lα (a, c)Ψ(z)
Lα (a, c)Ψ(z)
Lα (a, c)Φ(z) z(Lα (a, c)Φ(z))0 z(Lα (a, c)Ψ(z))0
−
+ (1 − µ)
Lα (a, c)Ψ(z)
Lα (a, c)Φ(z)
Lα (a, c)Ψ(z)
2
≺ (1 − µ)q (z) + (2µ − 1)q(z) − µ + (1 − µ)zq 0 (z),
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then
Lα (a, c)Φ(z)
≺ q(z)
Lα (a, c)Ψ(z)
and q(z) is the best dominant.
Theorem 2.5. Let the function q(z) be univalent in the unit disk U such that q 0 (z) 6=
0 (z)
0 and zqq(z)
is starlike in U. If F ∈ Bα− satisfies the subordination
z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
zq 0 (z)
a
−
≺a
,
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
q(z)
then
F (z) ∗ Φ(z)
≺ q(z),
F (z) ∗ Ψ(z)
z ∈ U,
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
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and q(z) is the best dominant.
Proof. Let the function p(z) be defined by
F (z) ∗ Φ(z)
p(z) :=
,
F (z) ∗ Ψ(z)
By setting
z ∈ U.
a
θ(ω) := , a 6= 0,
ω
it can easily observed that θ(ω) is analytic in C − {0}. Then we obtain
zp0 (z)
z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
a
=a
−
p(z)
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
0
zq (z)
.
≺a
q(z)
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By the assumption of the theorem we have that the assertion of the theorem follows
by an application of Lemma 1.4.
Corollary 2.6. If F ∈ Bα− satisfies the subordination
z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
(A − B)z
a
−
≺a
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
(1 + Az)(1 + Bz)
then
and
F (z) ∗ Φ(z)
≺
F (z) ∗ Ψ(z)
1+Az
( 1+Bz
)
1 + Az
1 + Bz
,
−1 ≤ B < A ≤ 1
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
is the best dominant.
Corollary 2.7. If F ∈ Bα− satisfies the subordination
z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
2z
a
−
≺a
,
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
1 − z2
then
F (z) ∗ Φ(z)
≺
F (z) ∗ Ψ(z)
1+z
1−z
,
n=1
(a)n n+α−1
z
,
(c)n
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Define the function φα (a, c; z) by
φα (a, c; z) := 1 −
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1+z
and ( 1−z
) is the best dominant.
∞
X
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(z ∈ U ; a ∈ R, c ∈ R\{0, −1, −2, . . . }),
where (a)n is the Pochhammer symbol. Corresponding to the function φα (a, c; z),
define a linear operator Lα (a, c) by
Lα (a, c)F (z) := φα (a, c; z) ∗ F (z),
F (z) ∈ Bα−
Close
or equivalently by
Lα (a, c)F (z) := 1 −
∞
X
(a)n
n=1
(c)n
an z n+α−1 .
Hence we obtain the next result.
Corollary 2.8. Let the function q(z) be univalent in the unit disk U such that q 0 (z) 6=
0 (z)
0 and zqq(z)
is starlike in U. If F ∈ Bα− satisfies the subordination
z(Lα (a, c)Φ(z))0 z(Lα (a, c)Ψ(z))0
zq 0 (z)
a
−
≺a
,
Lα (a, c)Φ(z)
Lα (a, c)Ψ(z)
q(z)
then
Lα (a, c)Φ(z)
≺ q(z),
Lα (a, c)Ψ(z)
and q(z) is the best dominant.
Differential Subordination Results
Rabha W. Ibrahim and
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vol. 10, iss. 1, art. 8, 2009
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z ∈ U,
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3.
Applications
In this section, we introduce some applications
of Section 2 containing fractional inP∞
tegral operators. Assume that f (z) = n=1 ϕn z n and let us begin with the following
definitions
Definition 3.1 ([5]). The fractional integral of order α for the function f (z) is defined by
Z z
1
α
f (ζ)(z − ζ)α−1 dζ; 0 ≤ α < 1,
Iz f (z) :=
Γ(α) 0
where the function f (z) is analytic in a simply-connected region of the complex
z−plane (C) containing the origin. The multiplicity of (z − ζ)α−1 ishremoved
i by
α−1
requiring log(z − ζ) to be real when(z − ζ) > 0. Note that, Izα f (z) = zΓ(α) f (z),
for z > 0 and 0 for z ≤ 0 (see [6]).
From Definition 3.1, we have
α−1 ∞
∞
X
z α−1 X
z
n
α
ϕn z =
an z n+α−1
Iz f (z) =
f (z) =
Γ(α)
Γ(α) n=1
n=1
ϕn
where an := Γ(α)
for all n = 1, 2, 3, . . . , thus 1 + Izα f (z) ∈ Bα+ and 1 − Izα f (z) ∈
Bα− (ϕn ≥ 0). Then we have the following results.
Theorem 3.2. Let the assumptions of Theorem 2.1 hold. Then
(1 + Izα f (z)) ∗ Φ(z)
≺ q(z), z 6= 0, z ∈ U
(1 + Izα f (z)) ∗ Ψ(z)
and q(z) is the best dominant.
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
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Proof. Let the function F (z) be defined by
F (z) := 1 + Izα f (z),
z ∈ U.
Theorem 3.3. Let the assumptions of Theorem 2.5 hold. Then
(1 − Izα f (z)) ∗ Φ(z)
≺ q(z), z ∈ U
(1 − Izα f (z)) ∗ Ψ(z)
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
and q(z) is the best dominant.
Proof. Let the function F (z) be defined by
F (z) := 1 − Izα f (z),
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z ∈ U.
Let F (a, b; c; z) be the Gauss hypergeometric function (see [7]) defined, for z ∈
U, by
∞
X
(a)n (b)n n
F (a, b; c; z) =
z .
(c)n (1)n
n=0
We need the following definitions of fractional operators in Saigo type fractional
calculus (see [8, 9]).
α,β,η
Definition 3.4. For α > 0 and β, η ∈ R, the fractional integral operator I0,z
is
defined by
Z
z −α−β z
ζ
α,β,η
α−1
I0,z f (z) =
(z − ζ) F α + β, −η; α; 1 −
f (ζ)dζ,
Γ(α) 0
z
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where the function f (z) is analytic in a simply-connected region of the z−plane
containing the origin, with order
f (z) = O(|z| )(z → 0),
> max{0, β − η} − 1
and the multiplicity of (z − ζ)α−1 is removed by requiring log(z − ζ) to be real when
z − ζ > 0.
From Definition 3.4, with β < 0, we have
Z
ζ
z −α−β z
α,β,η
I0,z f (z) =
(z − ζ)α−1 F (α + β, −η; α; 1 − )f (ζ)dζ
Γ(α) 0
z
n
Z
∞
X (α + β)n (−η)n z −α−β z
ζ
α−1
(z − ζ)
1−
f (ζ)dζ
=
(α)n (1)n
Γ(α) 0
z
n=0
Z
∞
X
z −α−β−n z
=
Bn
(z − ζ)n+α−1 f (ζ)dζ
Γ(α)
0
n=0
=
∞
X
Bn
n=0
=
B
Γ(α)
z −β−1
f (ζ)
Γ(α)
∞
X
n=1
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and B :=
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X
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Bn .
n=0
for all n = 1, 2, 3, . . . , and let α = −β. Thus,
α,β,η
1 + I0,z
f (z) ∈ Bα+
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ϕn z n−β−1 ,
(α + β)n (−η)n
Bn :=
(α)n (1)n
Bϕn
Γ(α)
vol. 10, iss. 1, art. 8, 2009
Page 16 of 18
where
Denote an :=
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
and
α,β,η
1 − I0,z
f (z) ∈ Bα− (ϕn ≥ 0),
andn we have the following results
Theorem 3.5. Let the assumptions of Theorem 2.1 hold. Then
"
#
α,β,η
(1 + I0,z
f (z)) ∗ Φ(z)
≺ q(z), z ∈ U
α,β,η
(1 + I0,z
f (z)) ∗ Ψ(z)
and q(z) is the best dominant.
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
Proof. Let the function F (z) be defined by
vol. 10, iss. 1, art. 8, 2009
F (z) := 1 +
α,β,η
I0,z
f (z),
z ∈ U.
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Theorem 3.6. Let the assumptions of Theorem 2.5 hold. Then
#
"
α,β,η
(1 − I0,z
f (z)) ∗ Φ(z)
≺ q(z), z ∈ U
α,β,η
(1 − I0,z
f (z)) ∗ Ψ(z)
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JJ
II
J
I
Page 17 of 18
and q(z) is the best dominant.
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Proof. Let the function F (z) be defined by
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F (z) := 1 −
α,β,η
I0,z
f (z),
z ∈ U.
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References
[1] O. JUNEJA, T. REDDY AND M. MOGRA, A convolution approach for analytic
functions with negative coefficients, Soochow J. Math., 11 (1985), 69–81.
[2] M .OBRADOVIC, T. YAGUCHI AND H. SAITOH, On some conditions for
univalence and starlikeness in the unit disc, Rendiconti di Math. Series VII, 12
(1992), 869–877.
[3] S.S. MILLER AND P.T. MOCANU, Differential Subordinations: Theory and
Applications, Pure and Applied Mathematics, No.225, Dekker, N.Y., (2000).
[4] J. LIU AND H.M. SRIVASTAVA, A linear operator and associated families of
meromorphically multivalent functions, J. Math. Anal. Appl., 259 (2001), 566–
581.
[5] H.M. SRIVASTAVA AND S. OWA, Univalent Functions, Fractional Calculus,
and Their Applications, Halsted Press, John Wiley and Sons, New York, Chichester, Brisbane, and Toronto, 1989.
[6] K.S. MILLER AND B. ROSS, An Introduction to the Fractional Calculus and
Fractional Differential Equations, John-Wiley and Sons, Inc., 1993.
[7] H.M. SRIVASTAVA AND S. OWA (Eds.), Current Topics in Analytic Function
Theory, World Scientific Publishing Company, Singapore, New Jersey, London
and Hong Kong, 1992.
[8] R.K. RAINA AND H.M. SRIVASTAVA, A certain subclass of analytic functions associated with operators of fractional calculus, Comput. Math. Appl.,
32(7) (1996), 13–19.
[9] R.K. RAINA, On certain class of analytic functions and applications to fractional calculus operator, Integral Transf. and Special Func., 5 (1997), 247–260.
Differential Subordination Results
Rabha W. Ibrahim and
Maslina Darus
vol. 10, iss. 1, art. 8, 2009
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Contents
JJ
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Page 18 of 18
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