General Mathematics Vol. 18, No. 3 (2010), 41–50
Differential subordination for classes of normalized
analytic functions 1
Rabha W. Ibrahim, Maslina Darus
Abstract
We determine the sufficient conditions for subordination for new
classes of normalized analytic functions with applications in fractional
calculus in complex domain.
2010 Mathematics Subject Classification: 34G10, 26A33, 30C45.
Key words and phrases: Fractional calculus, Subordination.
1
Introduction and preliminaries.
Let A+
α be the class of all normalized analytic functions F (z) in the open disk
U := {z ∈ C, |z| < 1}, take the form
F (z) = z +
∞
X
an,α z n+α−1 , 0 < α ≤ 1,
n=2
where a0,1 = 0, a1,1 = 1 satisfying F (0) = 0 and F 0 (0) = 1. And let A−
α be
the class of all normalized analytic functions F (z) in the open disk U take the
form
∞
X
F (z) = z −
an,α z n+α−1 , an,α ≥ 0; n = 2, 3, ...,
n=2
1
Received 17 February, 2009
Accepted for publication (in revised form) 25 March, 2009
41
42
R.W. Ibrahim, M. Darus
satisfying F (0) = 0 and F 0 (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
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 if p satisfies the differential superordination
(1)
h(z) ≺ φ(p(z)), zp0 (z), z 2 p00 (z); z),
then p is called a solution of the differential superordination.(If f is subordinate
to g, then g is called to be superordinate to f. )An analytic function q is called
a subordinant if q ≺ p for all p satisfying (1). An univalent function q such
that p ≺ q for all subordinants p of (1) is said to be the best subordinant.
Let A be the class of analytic functions of the form f (z) = z + a2 z 2 + ... .
Obradović and Owa [1] obtained sufficient conditions for certain normalized
analytic functions f (z) ∈ A to satisfy
q1 (z) ≺ [
f (z) µ
] ≺ q2 (z)
z
where q1 and q2 are given univalent functions in U. The main object of the
present work is to apply a method based on the differential subordination in
−
order to derive sufficient conditions for functions F ∈ A+
α and F ∈ Aα to
satisfy
(2)
[
F (z) µ
] ≺ q(z)
z
where q(z) is a given univalent function in U such that q(z) 6= 0. Moreover,
we give applications for these results in fractional calculus. We shall need the
following known results.
Differential subordination...
43
Lemma 1 [2] Let q(z) be univalent in the unit disk U and θ and φ be analytic
in a domain D containing q(U ) with φ(w) 6= 0 when w ∈ q(U ). Set Q(z) :=
zq 0 (z)φ(q(z)), h(z) := θ(q(z)) + Q(z). Suppose that
1. Q(z) is starlike univalent in U , and
0 (z)
2. < zh
Q(z) > 0 for z ∈ U.
If θ(p(z)) + zp0 (z)φ(p(z)) ≺ θ(q(z)) + zq 0 (z)φ(q(z)) then p(z) ≺ q(z) and q(z)
is the best dominant.
Lemma 2 [3] Let q(z) be convex univalent in the unit disk U and ψ and γ ∈ C
00 (z)
+ ψγ } > 0. If p(z) is analytic in U and ψp(z) + γzp0 (z) ≺
with <{1 + zqq0 (z)
ψq(z) + γzq 0 (z), then p(z) ≺ q(z) and q is the best dominant.
2
Main results.
In this section, we study sufficient subordination normalized analytic functions
−
in the classes A+
α and Aα .
Theorem 1 Let the function q(z) be univalent in the unit disk U such that
0 (z)
is starlike univalent in U and
q(z) 6= 0, zqq(z)
(3) <{1 + (
a
zq 00 (z) zq 0 (z)
+ 1)( 0
−
)} > 0, b 6= 0, z 6= 0, q 0 (z) 6= 0, z ∈ U.
bz
q (z)
q(z)
If F ∈ A+
α satisfies the subordination
µ zF 0 (z)
q 0 (z)
(a + bz) (
− 1) ≺ (a + bz)
, F (z) 6= 0, z ∈ U.
z F (z)
q(z)
Then
F (z) µ
) ≺ q(z), z 6= 0, z ∈ U,
z
and q(z) is the best dominant.
(
Proof. Let the function p(z) be defined by
p(z) := (
F (z) µ
) , z 6= 0, z ∈ U.
z
By setting
θ(ω) :=
aω 0
b
and φ(ω) := , b 6= 0,
ω
ω
44
R.W. Ibrahim, M. Darus
it can easily be observed that θ(ω) is analytic in C − {0}, φ(ω) is analytic in
C − {0} and that φ(ω) 6= 0, ω ∈ C − {0}. Also we obtain
Q(z) = zq 0 (z)φ(q(z)) =
bzq 0 (z)
q 0 (z)
and h(z) = θ(q(z)) + Q(z) = (a + bz)
.
q(z)
q(z)
It is clear that Q(z) is starlike univalent in U,
<{
a
zq 00 (z) zq 0 (z)
zh0 (z)
= <{1 + ( + 1)( 0
−
)} > 0.
Q(z)
bz
q (z)
q(z)
Straightforward computation, we have
(a + bz)
p0 (z)
µ zF 0 (z)
= (a + bz) (
− 1)
p(z)
z F (z)
q 0 (z)
≺ (a + bz)
q(z)
Then by the assumption of the theorem we have that the assertion of the
theorem follows by an application of Lemma 1.
Corollary 1 Assume that (3) holds and q is convex univalent in U. If F ∈ A+
α
and
A−B
µ zF 0 (z)
− 1) ≺ µ(a + bz)
,
(a + bz) (
z F (z)
(1 + Az)(1 + Bz)
then
(
F (z) µ
1 + Az µ
) ≺(
) , −1 ≤ B < A ≤ 1
z
1 + Bz
1+Az µ
) is the best dominant.
and q(z) = ( 1+Bz
Corollary 2 Assume that (3) holds and q is convex univalent in U. If F ∈ A+
α
and
2µ
µ zF 0 (z)
− 1) ≺ (a + bz)
,
(a + bz) (
z F (z)
(1 + z)(1 − z)
for z ∈ U, µ 6= 0, then
(
F (z) µ
1+z µ
) ≺(
)
z
1−z
µ
and q(z) = ( 1+z
1−z ) is the best dominant.
Differential subordination...
45
Corollary 3 Assume that (3) holds and q is convex univalent in U. If F ∈ A+
α
and
µ zF 0 (z)
(a + bz) (
− 1) ≺ µA(a + bz)
z F (z)
for z ∈ U, µ 6= 0, then
(
F (z) µ
) ≺ eµAz
z
and q(z) = eµAz is the best dominant.
Theorem 2 Let the function q(z) be convex univalent in the unit disk U such
that q 0 (z) 6= 0 and
(4)
<{1 +
zq 00 (z) 1
+ } > 0, γ 6= 0.
q 0 (z)
γ
µ
−
Suppose that ( F (z)
z ) is analytic in U. If F ∈ Aα satisfies the subordination
(
zF 0 (z)
F (z) µ
) [1 + γµ(
− 1)] ≺ q(z) + γzq 0 (z), F (z) 6= 0.
z
F (z)
Then
(
F (z) µ
) ≺ q(z), z ∈ U, z 6= 0
z
and q(z) is the best dominant.
Proof. Let the function p(z) be defined by
p(z) := (
F (z) µ
) , z 6= 0, , z ∈ U.
z
By setting ψ = 1, it can easily be observed that
p(z) + γzp0 (z) = (
F (z) µ
zF 0 (z)
) [1 + γµ(
− 1)]
z
F (z)
≺ q(z) + γzq 0 (z).
Then by the assumption of the theorem we have that the assertion of the
theorem follows by an application of Lemma 2.
46
R.W. Ibrahim, M. Darus
Corollary 4 Assume that (4) holds and q is convex univalent in U. If F ∈ A−
α
and
(
F (z) µ
zF 0 (z)
1 + Az µ
(1 + Az)µ−1
) [1 + γµ(
− 1)] ≺ (
) + µγz(A − B)
z
F (z)
1 + Bz
(1 + Bz)µ+1
then
(
F (z) µ
1 + Az µ
) ≺(
) , −1 ≤ B < A ≤ 1
z
1 + Bz
1+Az µ
and q(z) = ( 1+Bz
) is the best dominant.
Corollary 5 Assume that (4) holds and q is convex univalent in U. If F ∈ A−
α
and
zF 0 (z)
1+z µ
2γµz
F (z) µ
) [1 + γµ(
− 1)] ≺ [
] {1 +
}
(
z
F (z)
1−z
1 − z2
for z ∈ U, µ 6= 0, then
(
F (z) µ
1+z µ
) ≺(
)
z
1−z
µ
and q(z) = ( 1+z
1−z ) is the best dominant.
Corollary 6 Assume that (4) holds and q is convex univalent in U. If F ∈ A−
α
and
F (z) µ
zF 0 (z)
(
) [1 + γµ(
− 1)] ≺ eµAz (1 + µγAz)
z
F (z)
for z ∈ U, µ 6= 0, then
(
F (z) µ
) ≺ eµAz
z
and q(z) = eµAz is the best dominant.
3
Applications.
In this section, we introduce some applications of section (2) containing fracP
n
tional integral operators. Assume that f (z) = ∞
n=2 ϕn z and let us begin
with the following definitions
Definition 1 [4] The fractional integral of order α is defined, for a function
f, by
Z z
1
Izα f (z) :=
f (ζ)(z − ζ)α−1 dζ; α > 0,
Γ(α) 0
Differential subordination...
47
where the function f (z) is analytic in simply-connected region of the complex
z-plane (C) containing the origin and the multiplicity of (z − ζ) α−1 is removed
by requiring log(z − ζ) to be real when(z − ζ) > 0.
α
From Definition 1 and see ([5]), thus z + Izα f (z) ∈ A+
α and z − Iz f (z) ∈
−
Aα (ϕn ≥ 0), then we have the following results
Theorem 3 Let the assumptions of Theorem 1 hold, then
(
z + Izα f (z) µ
) ≺ q(z),
z
and q(z) is the best dominant.
Proof. Let the function F (z) be defined by
F (z) := z + Izα f (z),
z ∈ U, z 6= 0.
Theorem 4 Let the assumptions of Theorem 2 hold, then
(
z − Izα f (z) µ
) ≺ q(z),
z
and q(z) is the best dominant.
Proof. Let the function F (z) be defined by
F (z) := z − Izα f (z),
z ∈ U, z 6= 0.
Let F (a, b; c; z) be the Gauss hypergeometric function (see [6]) defined, for
z ∈ U, by
∞
X
(a)n (b)n n
z ,
F (a, b; c; z) =
(c)n (1)n
n=0
where is the Pochhammer symbol defined by
(
Γ(a + n)
1,
(n = 0);
(a)n :=
=
Γ(a)
a(a + 1)(a + 2)...(a + n − 1), (n ∈ N).
We need the following definitions of fractional operators in the Saigo type
fractional calculus (see [7],[8]).
48
R.W. Ibrahim, M. Darus
α,β,η
Definition 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 the 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 2, with β < 0, we have
Z
ζ
z −α−β z
α,β,η
(z − ζ)α−1 F (α + β, −η; α; 1 − )f (ζ)dζ
I0,z
f (z) =
Γ(α) 0
z
Z
∞
X (α + β)n (−η)n z −α−β z
ζ
=
(z − ζ)α−1 (1 − )n 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
n=0
Bn
z −β−1
f (ζ)
Γ(α)
∞
B X
:=
ϕn z n−β−1
Γ(α)
n=2
where B :=
Bϕn
n=0 Bn . Denote an := Γ(α) ,
α,β,η
∈ A+
α and z − I0,z f (z)
P∞
α,β,η
I0,z
f (z)
z+
following results
∀ n = 2, 3, ..., and let α = −β thus
∈ A−
α (ϕn ≥ 0), then we have the
Theorem 5 Let the assumptions of Theorem 1 hold, then
(
α,β,η
z + I0,z
f (z)
and q(z) is the best dominant.
z
)µ ≺ q(z),
Differential subordination...
49
Proof. Let the function F (z) be defined by
α,β,η
F (z) := z + I0,z
f (z),
z ∈ U, z 6= 0.
Theorem 6 Let the assumptions of Theorem 2 hold, then
(
α,β,η
z − I0,z
f (z)
z
)µ ≺ q(z),
and q(z) is the best dominant.
Proof. Let the function F (z) be defined by
α,β,η
f (z),
F (z) := z − I0,z
z ∈ U, z 6= 0.
Acknowledgement The work here is supported by UKM-ST-06-FRGS01072009, MOHE, MALAYSIA
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50
R.W. Ibrahim, M. Darus
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1
Rabha W. Ibrahim, 2 Maslina Darus
School of Mathematical Sciences
Faculty of Science and Technology
Universiti Kebangsaan Malaysia
UKM Bangi 43600, Selangor D. Ehsan,Malaysia
e-mail: 1 rabhaibrahim@yahoo.com , 2 maslina@ukm.my