Volume 10 (2009), Issue 1, Article 8, 9 pp.
DIFFERENTIAL SUBORDINATION RESULTS FOR NEW CLASSES OF THE
FAMILY E(Φ, Ψ)
RABHA W. IBRAHIM AND MASLINA DARUS
S CHOOL OF M ATHEMATICAL S CIENCES
FACULTY OF S CIENCE AND T ECHNOLOGY
U NIVERSITI K EBANGSAAN M ALAYSIA
BANGI 43600, S ELANGOR DARUL E HSAN
M ALAYSIA
rabhaibrahim@yahoo.com
maslina@ukm.my
Received 05 July, 2008; accepted 02 December, 2008
Communicated by S.S. Dragomir
A BSTRACT. We define new classes of the family E(Φ, Ψ), in a unit disk U := {z ∈ C, |z| <
(z)∗Φ(z)
1}, as follows: for analytic functions F (z), Φ(z) and Ψ(z) so that <{ 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.
Key words and phrases: Fractional calculus; Subordination; Hadamard product.
2000 Mathematics Subject Classification. 34G10, 26A33, 30C45.
1. I NTRODUCTION 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
∞
X
F (z) = 1 +
an z n+α−1 , 0 < α ≤ 1,
n=1
Bα−
satisfying F (0) = 1. And let
be the class of all analytic functions F (z) in the open disk U
of the form
∞
X
F (z) = 1 −
an z n+α−1 , 0 < α ≤ 1, an ≥ 0; n = 1, 2, 3, . . . ,
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.
The work presented here was supported by SAGA: STGL-012-2006, Academy of Sciences, Malaysia.
188-08
2
R ABHA W. I BRAHIM AND M ASLINA DARUS
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
f (0) = g(0)
and
f (U ) ⊂ g(U ).
Let φ : C3 × 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
h(z) ≺ φ(p(z)), zp0 (z), z 2 p00 (z); z),
(1.1)
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
Given two functions f, g ∈ B + ,
f (z) = 1 +
∞
X
an z
n
and
g(z) = 1 +
n=1
∞
X
bn z n
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,
z ∈ U.
n=1
Juneja et al. [1] define the family E(Φ, Ψ), so that
f (z) ∗ Φ(z)
> 0,
<
f (z) ∗ Ψ(z)
where
Φ(z) = z +
∞
X
ϕn z
n
z ∈U
and Ψ(z) = z +
n=2
∞
X
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
are analytic in U under the conditions ϕn ≥ 0, ψn ≥ 0, ϕn ≥ ψn for n ≥ 1 and F (z)∗Ψ(z) 6= 0.
Definition 1.2. Letting F (z) ∈ Bα− , we define the family Eα− (Φ, Ψ) which satisfies (1.2) where
Φ(z) = 1 −
∞
X
ϕn z
n+α−1
n=1
and Ψ(z) = 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.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 8, 9 pp.
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D IFFERENTIAL S UBORDINATION R ESULTS
3
In the present paper, we establish some sufficient conditions for functions F ∈ Bα+ and
F ∈ Bα− to satisfy
F (z) ∗ Φ(z)
≺ q(z),
F (z) ∗ Ψ(z)
(1.3)
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 1.1 ([2]). Let q(z) be convex in the unit disk U with q(0) = 1 and <{q} > 12 , z ∈ U.
If 0 ≤ µ < 1, p is an analytic function in U with p(0) = 1 and if
(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 1.2 ([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
zp0 (z)θ(p(z)) ≺ zq 0 (z)θ(q(z))
then p(z) ≺ q(z) and q(z) is the best dominant.
2. M AIN R ESULTS
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 and
(z)∗Φ(z)
<{q} > 12 . If F ∈ Bα+ and FF (z)∗Ψ(z)
an analytic function in U satisfies the subordination
2
F (z) ∗ Φ(z)
+ (2µ − 1)
−µ
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 − µ)q (z) + (2µ − 1)q(z) − µ + (1 − µ)zq 0 (z),
F (z) ∗ Φ(z)
(1 − µ)
F (z) ∗ Ψ(z)
then
F (z) ∗ Φ(z)
≺ q(z)
F (z) ∗ Ψ(z)
and q(z) is the best dominant.
Proof. Let the function p(z) be defined by
p(z) :=
F (z) ∗ Φ(z)
,
F (z) ∗ Ψ(z)
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 8, 9 pp.
z ∈ U.
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4
R ABHA W. I BRAHIM AND M ASLINA DARUS
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)
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)
+ (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)
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 1.1.
(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)
+ (2µ − 1)
−µ
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
+ (2µ − 1)
≺ (1 − µ)
1 + Bz
1 + Bz
1 + Az
(A − B)z
− µ + (1 − µ)
,
1 + Bz (1 + Az)(1 + Bz)
F (z) ∗ Φ(z)
(1 − µ)
F (z) ∗ Ψ(z)
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
q(z) :=
It is clear that q(0) = 1 and <{q} >
we obtain the result.
1
2
1 + Az
1 + Bz
,
z ∈ U.
for arbitrary A, B, z ∈ U, then in view of Theorem 2.1
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 8, 9 pp.
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D IFFERENTIAL S UBORDINATION R ESULTS
5
(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)
+ (2µ − 1)
−µ
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
1+z
2z
+ (1 − µ)
,
1−z
1 − z2
F (z) ∗ Φ(z)
(1 − µ)
F (z) ∗ Ψ(z)
then
F (z) ∗ Φ(z)
≺
F (z) ∗ Ψ(z)
1+z
1−z
,
1+z
and ( 1−z
) is the best dominant.
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)
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+
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 and
(a,c)Φ(z)
<{q} > 12 . 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),
then
Lα (a, c)Φ(z)
≺ q(z)
Lα (a, c)Ψ(z)
and q(z) is the best dominant.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 8, 9 pp.
http://jipam.vu.edu.au/
6
R ABHA W. I BRAHIM AND M ASLINA DARUS
Theorem 2.5. Let the function q(z) be univalent in the unit disk U such that q 0 (z) 6= 0 and
zq 0 (z)
is starlike in U. If F ∈ Bα− satisfies the subordination
q(z)
zq 0 (z)
z(F (z) ∗ Φ(z))0 z(F (z) ∗ Ψ(z))0
a
−
≺a
,
F (z) ∗ Φ(z)
F (z) ∗ Ψ(z)
q(z)
then
F (z) ∗ Φ(z)
≺ q(z), z ∈ U,
F (z) ∗ Ψ(z)
and q(z) is the best dominant.
Proof. Let the function p(z) be defined by
F (z) ∗ Φ(z)
p(z) :=
,
z ∈ U.
F (z) ∗ Ψ(z)
By setting
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)
By the assumption of the theorem we have that the assertion of the theorem follows by an
application of Lemma 1.2.
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
F (z) ∗ Φ(z)
1 + Az
≺
, −1 ≤ B < A ≤ 1
F (z) ∗ Ψ(z)
1 + Bz
1+Az
and ( 1+Bz
) 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)
1+z
≺
,
F (z) ∗ Ψ(z)
1−z
1+z
) is the best dominant.
and ( 1−z
Define the function φα (a, c; z) by
∞
X
(a)n n+α−1
φα (a, c; z) := 1 −
z
,
(c)n
n=1
(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),
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 8, 9 pp.
F (z) ∈ Bα−
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D IFFERENTIAL S UBORDINATION R ESULTS
7
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 and
zq 0 (z)
is starlike in U. If F ∈ Bα− satisfies the subordination
q(z)
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)
z ∈ U,
and q(z) is the best dominant.
3. A PPLICATIONS
In this section, we introduceP
some applications of Section 2 containing fractional integral
n
operators. Assume that f (z) = ∞
n=1 ϕn z 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
Iz f (z) :=
f (ζ)(z − ζ)α−1 dζ; 0 ≤ α < 1,
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 (zh− ζ)α−1
i is removed by requiring log(z − ζ) to be
z α−1
α
real when(z − ζ) > 0. Note that, Iz f (z) = Γ(α) f (z), for z > 0 and 0 for z ≤ 0 (see [6]).
From Definition 3.1, we have
α−1 ∞
∞
X
z
z α−1 X
α
n
Iz f (z) =
f (z) =
ϕn z =
an z n+α−1
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 3.1. 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.
Proof. Let the function F (z) be defined by
F (z) := 1 + Izα f (z),
z ∈ U.
Theorem 3.2. 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)
and q(z) is the best dominant.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 8, 9 pp.
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8
R ABHA W. I BRAHIM AND M ASLINA DARUS
Proof. Let the function F (z) be defined by
F (z) := 1 − Izα f (z),
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)
(1)
n
n
n=0
We need the following definitions of fractional operators in Saigo type fractional calculus
(see [8, 9]).
α,β,η
Definition 3.2. 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
I0,z f (z) =
(z − ζ)α−1 F α + β, −η; α; 1 −
f (ζ)dζ,
Γ(α) 0
z
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),
α−1
and the multiplicity of (z − ζ)
> max{0, β − η} − 1
is removed by requiring log(z − ζ) to be real when z − ζ > 0.
From Definition 3.2, 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)
(1)
Γ(α)
z
n
n
0
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
z −β−1
f (ζ)
Γ(α)
∞
=
B X
ϕn z n−β−1 ,
Γ(α) n=1
where
(α + β)n (−η)n
Bn :=
(α)n (1)n
Denote an :=
Bϕn
Γ(α)
and B :=
∞
X
Bn .
n=0
for all n = 1, 2, 3, . . . , and let α = −β. Thus,
α,β,η
1 + I0,z
f (z) ∈ Bα+
α,β,η
and 1 − I0,z
f (z) ∈ Bα− (ϕn ≥ 0),
andn we have the following results
Theorem 3.3. 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.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 8, 9 pp.
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D IFFERENTIAL S UBORDINATION R ESULTS
9
Proof. Let the function F (z) be defined by
α,β,η
F (z) := 1 + I0,z
f (z),
z ∈ U.
Theorem 3.4. 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)
and q(z) is the best dominant.
Proof. Let the function F (z) be defined by
α,β,η
F (z) := 1 − I0,z
f (z),
z ∈ U.
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