Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2008, Article ID 246909, 12 pages
doi:10.1155/2008/246909
Research Article
On Certain Subclasses of Meromorphic
Close-to-Convex Functions
Georgia Irina Oros, Adriana Cătaş, and Gheorghe Oros
Department of Mathematics, University of Oradea, 1, Universităţii street, 410087 Oradea, Romania
Correspondence should be addressed to Georgia Irina Oros, georgia oros ro@yahoo.co.uk
Received 20 February 2008; Accepted 31 March 2008
Recommended by Narendra Kumar Govil
By using the operator Dλn fz, z ∈ U, Definition 2.1, we introduce a class of meromorphic functions
denoted by Σα, λ, n and we obtain certain differential subordinations.
Copyright q 2008 Georgia Irina Oros et al. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction and preliminaries
Denote by U the unit disc of the complex plane:
U z ∈ C : |z| < 1 ,
U̇ U − {0}.
1.1
Let HU be the space of holomorphic function in U.
Let
An f ∈ HU, fz z an1 zn1 · · · , z ∈ U
1.2
with A1 A.
For a ∈ C and n ∈ N, we let
Ha, n f ∈ HU, fz a an zn an1 zn1 · · · , z ∈ U .
1.3
Let
K
f ∈ A, Re
zf z
1 > 0, z ∈ U
f z
denote the class of normalized convex functions in U.
1.4
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Journal of Inequalities and Applications
If f and g are analytic functions in U, then we say that f is subordinate to g, written
f ≺ g, if there is a function w analytic in U, with w0 0, |wz| < 1, for all z ∈ U such
that fz gwz for z ∈ U. If g is univalent, then f ≺ g if and only if f0 g0 and
fU ⊆ gU.
A function f, analytic in U, is said to be convex if it is univalent and fU is convex.
Let ψ : C3 × U → C and let h be univalent in U. If p is analytic in U and satisfies the
second-order differential subordination,
i ψpz, zp z, z2 p z; z ≺ hz, z ∈ U,
then p is called a solution of the differential subordination.
The univalent function q is called a dominant of the solutions of the differential
subordination, or more simply a dominant, if p ≺ q for all p satisfying i.
A dominant q that satisfies q ≺ q for all dominants q of i is said to be the best dominant
of i. Note that the best dominant is unique up to a rotation of U. In order to prove the
original results, we use the following lemmas.
Lemma 1.1 see 1, Theorem 3.1.6, page 71, and the references therein. Let h be a convex
function with h0 a, and let γ ∈ C∗ be a complex number with Re γ ≥ 0. If p ∈ Ha, n and
1
pz zp z ≺ hz,
γ
z ∈ U,
1.5
then
pz ≺ qz ≺ hz,
where
qz γ
nzγ/n
z
z ∈ U,
httγ/n−1 dt,
1.6
z ∈ U.
1.7
0
The function q is convex and the best dominant.
Lemma 1.2 see 2, Lemma 13.5.1, page 375, and the references therein. Let g be a convex
function in U, and let
hz gz nαzg z,
z ∈ U,
1.8
where α > 0, and n is a positive integer.
If
pz g0 pn zn pn1 zn1 · · · ,
z∈U
1.9
is holomorphic in U, and
pz αzp z ≺ hz,
z ∈ U,
1.10
then
pz ≺ gz,
and this result is sharp.
1.11
Georgia Irina Oros et al.
3
Lemma 1.3 see 1, Corollary 2.6.g.2, page 66. Let f ∈ A and
2 z
Fz ftdt, z ∈ U.
z 0
1.12
If
Re
zf z
1
f z
1
>− ,
2
1.13
then
F ∈ K.
1.14
Lemma 1.4 see 3, Lemma 1.5. Let Re c > 0, and let
k 2 |c|2 − k 2 − c2 .
w
4kRe c
Let h be an analytic function in U with h0 1, and suppose that
zh z
Re
1
> −w, z ∈ U.
h z
1.15
1.16
If pz 1 pk zk · · · (k ≥ 1 integer) is analytic in U and
1
pz zp z ≺ hz,
c
z ∈ U,
1.17
then
pz ≺ qz,
z ∈ U,
1.18
where q is the solution of the differential equation:
qz k zq z hz,
c
qz 1,
1.19
given by
c
qz kzc/k
z
tc/k−1 htdt.
1.20
0
Moreover, q is the best dominant.
Definition 1.5 see 4. For f ∈ A, n ∈ N∗ ∪ {0}, the operator Sn f is defined by Sn : A → A
S0 fz fz,
S1 fz zf z,
1.21
...
Sn1 fz z Sn fz ,
z ∈ U.
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Journal of Inequalities and Applications
Remark 1.6. If f ∈ A,
fz z ∞
1.22
aj zj ,
j2
then
Sn fz z ∞
j n aj zj ,
z ∈ U.
1.23
j2
Definition 1.7 see 5. For f ∈ A, n ∈ N∗ ∪ {0}, the operator Rn f is defined by Rn : A → A
R0 fz fz,
R1 fz zf z,
1.24
...
n 1Rn1 fz z Rn fz nRn fz,
z ∈ U.
Remark 1.8. If f ∈ A,
fz z ∞
1.25
aj zj ,
j2
then
Rn fz z ∞
n
Cnj−1
aj zj ,
z ∈ U.
1.26
j2
2. Main results
Definition 2.1. Let n ∈ N∗ ∪ {0} and λ ≥ 0. Let Dλn f denote the operator defined by Dλn : A → A
Dλn fz 1 − λSn fz λRn fz,
z ∈ U,
2.1
where the operators Sn f and Rn f are given by Definitions 1.5 and 1.7, respectively.
Remark 2.2. We observe that Dλn is a linear operator and for
fz z ∞
aj zj ,
2.2
∞
j
n
1 − λj n λCnj−1
aj z .
2.3
j2
we have
Dλn fz z j2
Also, it is easy to observe that if we consider λ 1 in Definition 2.1, we obtain the
Ruscheweyh differential operator, and if we consider λ 0 in Definition 2.1, we obtain the
Sălăgean differential operator.
Georgia Irina Oros et al.
5
Remark 2.3. For n 0,
Dλ0 fz 1 − λS0 fz λR0 fz fz S0 fz R0 fz,
2.4
Dλ1 fz 1 − λS1 fz λR1 fz zf z S1 fz R1 fz.
2.5
and for n 1,
Remark 2.4. If f ∈ Σ,
fz 1
a0 a1 z a2 z2 · · · ,
z
2.6
and we let
gz z2 fz z a0 z2 a1 z3 · · · ,
z ∈ U.
2.7
Definition 2.5. If 0 ≤ α < 1, λ ≥ 0, and n ∈ N, let Σα, λ, n 1 denote the class of functions
f ∈ Σ which satisfy the inequality,
n1
λzn Rn gz
Re Dλ gz > α,
n1
2.8
where Dλn1 g is given by Definition 2.1, g is given by 2.7, and Rn g is given by Definition 1.7.
Theorem 2.6. If 0 ≤ α < 1, λ ≥ 0, and n ∈ N, then
Σα, λ, n 1 ⊂ Σδ, λ, n 1,
2.9
δ δα 2α − 1 21 − α ln 2.
2.10
where
Proof. Let f ∈ Σα, λ, n 1,
gz z2 fz z a0 z2 a1 z3 · · · ,
g ∈ A.
2.11
Since f ∈ Σα, λ, n 1 by using Definition 2.5, we deduce
λnz Rn gz
n1
Re Dλ gz > α,
n1
z ∈ U,
2.12
which is equivalent to
λnz Rn gz
n1
1 2α − 1z
≺
hz,
Dλ gz n1
1z
z ∈ U.
2.13
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Journal of Inequalities and Applications
By using the properties of the operators Dλn g, Sn g, and Rn g, we have
λnz Rn gz
1 − λSn1 gz λRn1 gz n1
n
z R gz nRn gz
λnz Rn gz
n
1 − λ z S gz
λ
n1
n1
n
n
R gz z R gz n Rn gz
λnz Rn gz
n
n
1 − λ S gz z S gz
λ
n1
n1
n
n
n
n
1 − λ S gz λ R gz z 1 − λ S gz λ R gz , z ∈ U.
2.14
Using 2.14 in 2.13, we obtain
1 2α − 1z
,
1 − λ Sn gz λ Rn gz z 1 − λ Sn gz λ Rn gz
≺
1z
z ∈ U.
2.15
Let
pz Dλn gz
1 − λ Sn gz λ Rn gz
∞
∞
n
1 − λ z j n aj zj λ z Cnj−1
aj zj
j2
j2
∞
∞
n
1 − λ 1 j n1 aj zj−1 λ 1 jCnj−1
aj zj−1
j2
1
∞
2.16
j2
j−1
n
1 − λj n1 λjCnj−1
aj z
j2
1 b1 z b 2 z 2 · · · ,
z ∈ U.
We have that p ∈ H1, 1. From 2.16, we have
p z 1 − λ Sn gz λ Rn gz .
2.17
Using 2.16 and 2.17 in 2.15, we obtain
pz zp z ≺
1 2α − 1z
hz,
1z
z ∈ U.
2.18
By using Lemma 1.1, we have
pz ≺ qz ≺ hz,
z ∈ U,
2.19
Georgia Irina Oros et al.
7
where
qz 1
z
z
htdt 0
1
z
z
0
ln1 z
1 2α − 1t
dt 2α − 1 21 − α
,
1t
z
z ∈ U.
2.20
The function q is convex and best dominant.
Since q is convex and qU is symmetric with respect to the real axis, we deduce
Re pz > Re q1 δ δα 2α − 1 21 − α ln 2,
2.21
from which we deduce that Σα, λ, n 1 ⊂ Σδ, λ, n 1.
Example 2.7. If n 0, α 1/2, λ ≥ 0, then δ1/2 ln 2, and we deduce for f ∈ Σ that
1
Re 4zfz 5z2 f z z3 f z > ,
2
z∈U
2.22
implies
Re 2zfz z2 f z > ln 2,
z ∈ U.
2.23
Theorem 2.8. Let r be a convex function, r0 1, and let h be a function such that
hz rz zr z,
z ∈ U.
2.24
If f ∈ Σ, g is given by 2.7, and the following differential subordination holds
Dλn1 gz
λnz Rn gz
≺ hz rz zr z,
n1
z ∈ U,
2.25
then
n
Dλ gz ≺ rz,
z ∈ U,
2.26
and this result is sharp.
Proof. By using the properties of the operator Dλn g, we have
Dλn1 gz 1 − λSn1 gz λRn1 gz.
2.27
By using the properties of operators Sn gz, Rn gz, and by differentiating 2.27, we
obtain
n1
Dλ gz 1 − λSn1 gz λRn1 gz
n 1 Rn gz z Rn gz
n
n
1 − λ S gz z S gz
λ
.
n1
2.28
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Journal of Inequalities and Applications
Using 2.28 in 2.25 and relations 2.16 and 2.17, after a simple calculation, Subordination 2.25 becomes
pz zp z ≺ rz zr z,
z ∈ U.
2.29
By using Lemma 1.2, we have
pz ≺ rz,
that is,
z ∈ U,
n
Dλ gz ≺ rz,
2.30
z ∈ U.
2.31
Example 2.9. If n 0, λ ≥ 0, rz 1 z/1 − z, from Theorem 2.8, we deduce that if f ∈ Σ
and
1 2z − z2
4zfz 5z2 f z z3 f z ≺
, z ∈ U,
2.32
1 − z2
then
1z
,
1−z
Theorem 2.10. Let r be a convex function, r0 1, and
2zfz z2 f z ≺
hz rz zr z,
z ∈ U.
z ∈ U.
2.33
2.34
If f ∈ Σ, g is given by 2.7, and the following differential subordination holds
n
Dλ gz ≺ hz rz zr z, z ∈ U,
2.35
then
Dλn gz
≺ rz,
z
z ∈ U,
2.36
Dλn gz
,
z
z ∈ U.
2.37
and this result is sharp.
Proof. We let
pz By differentiating 2.37, we obtain
n
Dλ gz pz zp z,
z ∈ U.
2.38
Using 2.38, Subordination 2.35 becomes
pz zp z ≺ rz zr z hz,
z ∈ U.
2.39
By using Lemma 1.2, we have
pz ≺ rz,
2.40
that is,
Dλn gz
≺ rz,
z
and this result is sharp.
z ∈ U,
2.41
Georgia Irina Oros et al.
9
Example 2.11. If we let rz 1/1 − z, n 1, λ ≥ 0, then
hz 1
1 − z2
2.42
,
and from Theorem 2.10, we deduce that if f ∈ Σ, and
4zfz 5z2 f z z3 f z ≺
1
1 − z2
,
z ∈ U,
2.43
then
2fz zf z ≺
1
,
1−z
z ∈ U,
2.44
and this result is sharp.
0 which verifies the inequality:
Theorem 2.12. Let h ∈ HU, with h0 1, h 0 /
zh z
1
Re 1 > − , z ∈ U.
h z
2
2.45
If f ∈ Σ, g is given by 2.7 and the following differential subordination holds
n
Dλ gz ≺ hz,
z ∈ U,
2.46
then
Dλn gz
≺ qz,
z
where
qz 1
z
z
htdt,
z ∈ U,
2.47
z ∈ U.
2.48
0
Function q is convex and the best dominant.
Proof. In order to prove Theorem 2.12, we will use Lemmas 1.3 and 1.4. We deduce the value
of w from Lemma 1.4 by using the conditions of Theorem 2.12. From 2.37, Definition 2.1 and
Remark 2.2, we have
pz Dλn gz
z
j
n
n
z ∞
j2 1 − λj λCnj−1 aj z
1
z
∞
n
1 − λj n λCnj−1
aj zj−1
j2
1 b1 z b 2 z · · · ,
z ∈ U.
2.49
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Journal of Inequalities and Applications
Using Lemma 1.4, we deduce from 2.49 that k 1. Using 2.38 in Subordination 2.46, we
have
pz zp z ≺ hz,
z ∈ U.
2.50
From Subordination 2.50, by using Lemma 1.4, we deduce that c 1. Then,
w
k 2 c2 − |k 2 − c2 | 1 1 − |1 − 1| 1
.
4kRe c
4
2
Applying Lemma 1.4, from Subordination 2.50, we obtain
1 z
htdt, z ∈ U,
pz ≺ qz z 0
2.51
2.52
that is,
Dλn gz
1
≺ qz z
z
z
htdt,
z ∈ U,
2.53
0
where q is the best dominant.
Since the function h verifies the relation 2.45, from Lemma 1.3, we deduce that q is a
convex function.
Example 2.13. If n 0, λ ≥ 0, hz e3/2z − 1, from Theorem 2.12, we deduce for f ∈ Σ that if
4zfz 5z2 f z z3 f z ≺ e3/2z − 1,
z ∈ U,
2.54
then
2fz zf z ≺
2 3/2z 2
−
e
− 1,
3z
3z
z ∈ U.
0 which verifies the inequality:
Theorem 2.14. Let h ∈ HU, with h0 1, h 0 /
1
zh z
> − , z ∈ U.
Re 1 h z
2
2.55
2.56
If f ∈ Σ, g is given by 2.7, and the following differential subordination holds
Dλn1 gz
λnz Rn gz
≺ hz,
n1
z ∈ U,
2.57
then
Dλn gz ≺ qz,
where
qz is convex and the best dominant.
1
z
z ∈ U,
z
htdt
0
2.58
2.59
Georgia Irina Oros et al.
11
Proof. In order to prove Theorem 2.14, we will use Lemmas 1.3 and 1.4. The value of w is
obtained using the conditions of Theorem 2.14.
Using 2.16 and 2.17, Subordination 2.49 becomes
pz zp z ≺ hz,
z ∈ U.
2.60
From Subordination 2.60, by using Lemma 1.4, we deduce that c 1; and from the
relation 2.16, Definition 2.1, and Remark 2.2, we obtain
pz Dλn gz
∞
j
n
n
z
1 − λj λCnj−1 aj z
j2
∞
n
1
1 − λj n λCnj−1
·j·aj ·zj−1
2.61
j2
1 c1 z c2 z 2 · · · ,
z ∈ U.
From 2.61, by using Lemma 1.4, we deduce that k 1, then
w
k 2 |c|2 − |k − c|2 1 1 − |1 − 1|2 1
.
4kRe c
4
2
2.62
Applying Lemma 1.4, from Subordination 2.60, we obtain
pz ≺ qz 1
z
z
htdt,
z ∈ U,
2.63
0
that is,
Dλn gz
1
≺ qz z
z
htdt,
z ∈ U,
2.64
0
where q is the best dominant.
Since the function h verifies the inequality 2.45, from Lemma 1.3, we deduce that q is a
convex function.
Example 2.15. If n 0, λ ≥ 0, f ∈ Σ, hz 2z z2 /1 z2 , from Theorem 2.14, we deduce
that if
4zfz 5z2 f z z3 f z ≺
2z z2
21 z2
,
z ∈ U,
2.65
then
1
1 1
2fz zf z ≺ z 1,
2
21z
z ∈ U.
2.66
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Journal of Inequalities and Applications
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