General Mathematics Vol. 18, No. 2 (2010), 85–93
A new univalent integral operator defined by
Al-Oboudi differential operator 1
Serap Bulut
Abstract
In [3], Breaz and Breaz gave an univalence condition of the integral
operator Gn,α introduced in [2]. The purpose of this paper is to give
univalence condition of the generalized integral operator Gn,m,α defined
in [4]. Our results generalize the results of [3].
2000 Mathematics Subject Classification: 30C45.
Key words and phrases:Analytic functions, Univalent functions, Integral
operator, Differential operator, Schwarz lemma.
1
Introduction
Let A denote the class of all functions of the form
(1)
f (z) = z +
∞
X
ak z k
k=2
1
Received 20 November, 2008
Accepted for publication (in revised form) 3 December, 2008
85
86
S. Bulut
which are analytic in the open unit disk U = {z ∈ C : |z| < 1}, and
S = {f ∈ A : f is univalent in U} .
For f ∈ A, Al-Oboudi [1] introduced the following operator:
D0 f (z) = f (z),
(2)
(3)
D1 f (z) = (1 − δ)f (z) + δzf 0 (z) = Dδ f (z),
(4)
Dn f (z) = Dδ (Dn−1 f (z)),
δ≥0
(n ∈ N := {1, 2, 3, . . .}).
If f is given by (1), then from (3) and (4) we see that
(5)
Dn f (z) = z +
∞
X
[1 + (k − 1)δ]n ak z k ,
(n ∈ N0 := N ∪ {0}),
k=2
with Dn f (0) = 0.
Remark 1 When δ = 1, we get Sălăgean’s differential operator [9].
The following results will be required in our investigation.
General Schwarz Lemma. [5] Let the function f be regular in the disk
UR = {z ∈ C : |z| < R}, with |f (z)| < M for fixed M . If f has one zero with
multiplicity order bigger than m for z = 0, then
|f (z)| ≤
M
|z|m
Rm
(z ∈ UR ).
The equality can hold only if
f (z) = eiθ
where θ is constant.
M m
z ,
Rm
A new univalent integral operator ...
87
Theorem A. [7] Let α be a complex number with Reα > 0 and f ∈ A. If f
satisfies
1 − |z|2Reα
Reα
¯ 00 ¯
¯ zf (z) ¯
¯
¯
¯ f 0 (z) ¯ ≤ 1
(z ∈ U) ,
then, for any complex number β with Reβ ≥ Reα, the integral operator
½ Z z
¾1
β
β−1 0
Fβ (z) = β
t
f (t)dt
0
is in the class S.
Theorem B. [6] Let f ∈ A satisfy the following inequality:
¯
¯ 2 0
¯ z f (z)
¯
¯
¯
(6)
¯ (f (z))2 − 1¯ ≤ 1 (z ∈ U).
Then f is univalent in U.
Theorem C. [8] Assume that g ∈ A satisfies condition (6), and let α be a
complex number with
|α − 1| ≤
Reα
.
3
If
|g(z)| ≤ 1,
∀z ∈ U
then the function
(7)
Gα (z) =
½ Z
α
z
¾1
(g(t))α−1 dt
α
0
is of class S.
In [2], Breaz and Breaz considered the integral operator
½
Z
(8) Gn,α (z) := [n (α − 1) + 1]
0
z
α−1
(g1 (t))
α−1
· · · (gn (t))
¾ 1
n(α−1)+1
,
dt
(g1 , . . . , gn ∈ A), and proved that the function Gn,α is univalent in U.
Remark 2 Note that for n = 1, we obtain the integral operator Gα defined
by (7).
88
S. Bulut
Theorem D. [3] Let gi ∈ A,
∀i = 1, . . . , n, n ∈ N, satisfy the properties
¯ 2 0
¯
¯
¯ z gi (z)
¯
¯ < 1,
−
1
¯ (g (z))2
¯
i
∀z ∈ U,
∀i = 1, . . . , n
and α ∈ C with
|α − 1| ≤
Reα
.
3n
If
|gi (z)| ≤ 1, ∀z ∈ U, ∀i = 1, . . . , n,
then the function Gn,α defined by (8) is univalent.
In [4], the author introduced a new general integral operator by means of
the Al-Oboudi differential operator as follows.
Definition 1 [4] Let n ∈ N, m ∈ N0 and α ∈ C. We define the integral
operator Gn,m,α by
(9)

 1
Z zY
n

 n(α−1)+1
α−1
m
Gn,m,α (z) := [n (α − 1) + 1]
(D gj (t))
dt


0
(z ∈ U),
j=1
where g1 , . . . , gn ∈ A and Dm is the Al-Oboudi differential operator.
Remark 3 In the special case n = 1, we obtain the integral operator
(10)
½ Z
Gm,α (z) := α
z
m
α−1
(D g(t))
¾1
α
dt
(z ∈ U).
0
Remark 4 If we set m = 0 in (9) and (10), then we obtain the integral
operators defined in (8) and (7), respectively.
A new univalent integral operator ...
2
89
Main Results
Theorem 1 Let Mj ≥ 1, each of the functions gj ∈ A (j ∈ {1, . . . , n})
satisfies the inequality
¯
¯
¯ z 2 (Dm g (z))0
¯
¯
¯
j
(11)
−
1
¯≤1
¯
2
m
¯ (D gj (z))
¯
(z ∈ U; m ∈ N0 ).
and α ∈ C with
Reα
, Re (n (α − 1) + 1) ≥ Reα > 0.
j=1 (2Mj + 1)
|α − 1| ≤ Pn
If
|Dm gj (z)| ≤ Mj
(z ∈ U; j ∈ {1, . . . , n}),
then the integral operator Gn,m,α defined by (9) is in the univalent function
class S.
Proof. Since gj ∈ A (j ∈ {1, . . . , n}), by (5), we have
∞
X
Dm gj (z)
[1 + (k − 1)δ]m ak,j z k−1
=1+
z
(m ∈ N0 )
k=2
and
Dm gj (z)
6= 0
z
for all z ∈ U.
Also we note that

 1
¶α−1  n(α−1)+1
Z z
n µ m

Y
D gj (t)
Gn,m,α (z) = [n(α − 1) + 1]
tn(α−1)
dt
.


t
0
j=1
Define a function
Z
f (z) =
n µ
zY
0 j=1
Dm gj (t)
t
¶α−1
dt.
90
S. Bulut
Then we obtain
0
(12)
f (z) =
¶
n µ m
Y
D gj (z) α−1
z
j=1
.
It is clear that f (0) = f 0 (0) − 1 = 0.
The equality (12) implies that
0
ln f (z) = (α − 1)
n
X
ln
j=1
Dm gj (z)
z
or equivalently
0
ln f (z) = (α − 1)
n
X
(ln Dm gj (z) − ln z) .
j=1
By differentiating above equality, we get
n
X
f 00 (z)
=
(α
−
1)
f 0 (z)
µ
j=1
(Dm gj (z))0 1
−
Dm gj (z)
z
¶
.
Hence we obtain
n
X
zf 00 (z)
= (α − 1)
0
f (z)
µ
j=1
¶
z (Dm gj (z))0
−1 ,
Dm gj (z)
which readily shows that
¯
¯
1 − |z|2Reα ¯¯ zf 00 (z) ¯¯
¯ f 0 (z) ¯ ≤
Reα
≤
¯
¶
n µ¯
X
¯ z (Dm gj (z))0 ¯
1 − |z|2Reα
¯
¯
|α − 1|
¯ Dm gj (z) ¯ + 1
Reα
j=1
¯¯
ï
!
¯
n
¯
2
m g (z))0 ¯ ¯ D m g (z) ¯
X
z
(D
|α − 1|
¯
¯¯
j
j
¯+1 .
¯
¯
¯
¯ (Dm gj (z))2 ¯ ¯
Reα
z
j=1
From the hypothesis, we have |gj (z)| ≤ Mj (j ∈ {1, . . . , n} ; z ∈ U), then by
the General Schwarz Lemma, we obtain that
|gj (z)| ≤ Mj |z|
(j ∈ {1, . . . , n} ; z ∈ U).
A new univalent integral operator ...
Then we find
¯
¯
1 − |z|2Reα ¯¯ zf 00 (z) ¯¯
¯ f 0 (z) ¯ ≤
Reα
≤
since |α − 1| ≤
n
|α − 1| X
Reα
j=1
n
|α − 1| X
Pn Reα
.
j=1 (2Mj +1)
Reα
91
¯
ï
!
¯ z 2 (Dm g (z))0
¯
¯
¯
j
− 1¯ Mj + Mj + 1
¯
¯ (Dm gj (z))2
¯
(2Mj + 1) ≤ 1
j=1
Applying Theorem A, we obtain that Gn,m,α is
in the univalent function class S.
Corollary 1 Let M ≥ 1, each of the functions gj ∈ A (j ∈ {1, . . . , n})
satisfies the inequality (11) and α ∈ C with
|α − 1| ≤
Reα
, Re (n (α − 1) + 1) ≥ Reα > 0.
(2M + 1) n
If
|Dm gj (z)| ≤ M
(z ∈ U; j ∈ {1, . . . , n}),
then the integral operator Gn,m,α defined by (9) is in the univalent function
class S.
Proof. In Theorem 1, we consider M1 = · · · = Mn = M .
Corollary 2 Let each of the functions gj ∈ A (j ∈ {1, . . . , n}) satisfies the
inequality (11) and α ∈ C with
|α − 1| ≤
Reα
, Re (n (α − 1) + 1) ≥ Reα > 0.
3n
If
|Dm gj (z)| ≤ 1
(z ∈ U; j ∈ {1, . . . , n}),
then the integral operator Gn,m,α defined by (9) is in the univalent function
class S.
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S. Bulut
Proof. In Corollary 1, we consider M = 1.
Remark 5 If we set m = 0 in Corollary 2, then we have Theorem D.
Corollary 3 Let the function g ∈ A satisfies the inequality (11) and α ∈ C
with
|α − 1| ≤
Reα
, Reα > 0.
3
If
|Dm g(z)| ≤ 1
(z ∈ U),
then the integral operator Gm,α defined by (10) is in the univalent function
class S.
Proof. In Corollary 2, we consider n = 1.
Remark 6 If we set m = 0 in Corollary 3, then we have Theorem C.
References
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operator, Int. J. Math. Math. Sci. 2004, no. 25-28, 1429-1436.
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(Cluj) 47 (2005), no. 70, 35-38.
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Comenian. (N.S.) 76 (2007), no. 2, 137-142.
[4] S. Bulut, Univalence condition for a new generalization of the family of
integral operators, Acta Univ. Apulensis Math. Inform. No. 18 (2009), 7178.
A new univalent integral operator ...
93
[5] Z. Nehari, Conformal Mapping, Dover, New York, NY, USA, 1975.
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Serap Bulut
Kocaeli University
Civi Aviation College
Arslanbey Campus
41285 İzmit-Kocaeli, Turkey
E-mail: serap.bulut@kocaeli.edu.tr