doi: 10.2478/v10062-010-0012-z
A N N A L E S
U N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K A
L U B L I N – P O L O N I A
VOL. LXIV, NO. 2, 2010 SECTIO A 29–35
ERHAN DEN
˙
IZ and HALIT ORHAN
An extension of the univalence criterion
for a family of integral operators
Abstract. The main object of the present paper is to extend the univalence
condition for a family of integral operators. Relevant connections of some of
the results obtained in this paper with those in earlier works are also provided.
1. Introduction and preliminaries. Let A denote the class of functions
f normalized by
(1.1) f(z) = z +
∞
X
n=2
a
n
z
n
,
which are analytic in the open unit disk U = {z : z ∈ C and |z| < 1} and
satisfy the conditions f(0) = f
0
(0) − 1 = 0.
Consider S = {f ∈ A : f is a univalent function in U}.
A function f (z) ∈ A is said to be a member of the class B(γ) if and only
if
(1.2)
z
2
f
0
(z)
f
2
(z)
− 1
< 1 − γ, 0 ≤ γ < 1.
Recently, Frasin and Darus (see [6]) defined and studied the class B(γ).
In his paper Frasin (see [4]) obtained some results for functions belonging
2000 Mathematics Subject Classification. 30C45.
Key words and phrases. Integral operator, analytic functions, univalent functions, open
unit disk, univalence criterion.
30 E. Deniz and H. Orhan
to this class and also he showed that if f (z) ∈ B(γ) then f (z) satisfies the
following inequality
(1.3)
zf
00
(z)
f
0
(z)
≤
(1 − γ)(2 + |z|)
1 − |z|
(z ∈ U).
For γ = 0 the class B(0) = T was studied by Ozaki and Nunokawa
(see [8]).
We denote by W the class of functions w which are analytic in U satisfying
the conditions |w(z)| < 1 and w(0) = w
0
(0) = 0 for all z ∈ U.
Now, by Schwarz’s lemma, it follows that
(1.4) |w(z)| < |z| .
In [7], we see that if w(z) ∈ W, then w(z) satisfies
(1.5)
w
0
(z)
≤
1 − |w(z)|
2
1 − |z|
2
(z ∈ U).
In [11], N. Seenivasagan and D. Breaz considered the following family of
integral operators F
α
1
,α
2
,...,α
n
,β
(z) defined as follows
(1.6) F
α
1
,α
2
,...,α
n
,β
(z)
:
=
β
z
Z
0
t
β−1
n
Y
i=1
f
i
(t)
t
1
α
i
dt
1
β
,
where f
i
∈ A, f
00
i
(0) = 0 and α
1
, α
2
, . . . , α
n
, β ∈ C for all i ∈ {1, 2, . . . , n}.
When α
i
= α for all i ∈ {1, 2, . . . , n}, F
α
1
,α
2
,...,α
n
,β
(z) becomes the inte-
gral operator F
α,β
(z) considered in (see [1]).
We begin by recalling each of the following theorems dealing with univa-
lence criterion, which will be required in our present paper.
In [10], Pascu proved the following theorem.
Theorem 1 (Pascu [10]). Let β be a complex number with Re(β) > 0 and
f ∈ A. If
1 − |z|
2 Re(β)
Re(β)
zf
00
(z)
f
0
(z)
≤ 1,
for z ∈ U, then the function
F
β
(z)
:
=
β
z
Z
0
t
β−1
f
0
(t)dt
1
β
belongs to S.
In [9], Pascu and Pescar obtained the next result.
An extension of the univalence criterion for a family... 31
Theorem 2 (Pascu and Pescar [9]). Let β and µ be complex numbers, and
g ∈ S. If Re(β) > 0 and |µ| ≤ min
1
2
Re(β);
1
4
, then the function
G
β,µ
(z)
:
=
β
z
Z
0
t
β−1
g(t)
t
µ
dt
1
β
belongs to S.
Note that Theorem 2 includes the special case of Pascu and Pescar’s
theorem (see [9]) when Re(α) = Re(β).
In the present paper, we propose to investigate further univalence condi-
tion involving the general a family of integral operators defined by (1.6).
2. Main results. In this section, we first state an inclusion for f (z) ∈
B(γ), then we give the main univalence condition involving the general in-
tegral operator given by (1.6).
Theorem 3. If f (z) ∈ B(γ), then the inequality is satisfied
(2.1)
zf
0
(z)
f(z)
− 1
≤
(1 − γ)(1 + |z|)
1 − |z|
for all z ∈ U.
Proof. Let f(z) ∈ B(γ). Then we have
(2.2)
z
2
f
0
(z)
f
2
(z)
= 1 + (1 − γ)w(z),
where w(z) ∈ W. By applying the logarithmic differentiation, we obtain
from (2.2) that
zf
00
(z)
f
0
(z)
=
(1 − γ)zw
0
(z)
1 + (1 − γ)w(z)
+
2zf
0
(z)
f(z)
− 2
and
zf
0
(z)
f(z)
− 1 =
1
2
zf
00
(z)
f
0
(z)
−
(1 − γ)zw
0
(z)
1 + (1 − γ)w(z)
,
thereby, it follows that
zf
0
(z)
f(z)
− 1
=
1
2
zf
00
(z)
f
0
(z)
−
(1 − γ)zw
0
(z)
1 + (1 − γ)w(z)
≤
1
2
zf
00
(z)
f
0
(z)
+
(1 − γ)zw
0
(z)
1 + (1 − γ)w(z)
≤
1
2
zf
00
(z)
f
0
(z)
+
(1 − γ) |z| |w
0
(z)|
1 − (1 − γ) |w(z)|
.
32 E. Deniz and H. Orhan
From (1.3) and (1.5), we have
(2.3)
zf
0
(z)
f(z)
− 1
≤
1
2
(1 − γ)(2 + |z|)
1 − |z|
+
(1 − γ) |z|
1 − (1 − γ) |w(z)|
1 − |w(z)|
2
1 − |z|
2
!
and for 0 ≤ γ < 1, it is easy to show that
(2.4)
1 − |w(z)|
1 − (1 − γ) |w(z)|
≤ 1 (z ∈ U).
From (1.4), (2.3) and (2.4), we obtain that
(2.5)
zf
0
(z)
f(z)
− 1
≤
(1 − γ)(1 + |z|)
1 − |z|
.
This evidently completes the proof of Theorem 3.
Next we prove the following main theorem.
Theorem 4. Let f
i
(z) ∈ B(γ) for i ∈ {1, 2, . . . , n}. Let β be a complex
number with Re(β) > 0. If
(2.6)
n
X
i=1
1
|α
i
|
≤ min
1
2(1 − γ)
Re β;
1
4(1 − γ)
for all z ∈ U, then the function
F
α
1
,α
2
,...,α
n
,β
(z)
:
=
β
z
Z
0
t
β−1
n
Y
i=1
f
i
(t)
t
1
α
i
dt
1
β
belongs to S.
Proof. Define function
h(z) =
z
Z
0
n
Y
i=1
f
i
(t)
t
1
α
i
dt.
We have h(0) = h
0
(0) − 1 = 0. Also, a simple computation yields
(2.7) h
0
(z) =
n
Y
i=1
f
i
(z)
z
1
α
i
.
Making use of logarithmic differentiation in (2.7), we obtain
(2.8)
zh
00
(z)
h
0
(z)
=
n
X
i=1
1
α
i
zf
0
i
(z)
f
i
(z)
− 1
.
We thus have from (2.8) that
zh
00
(z)
h
0
(z)
≤
n
X
i=1
1
|α
i
|
zf
0
i
(z)
f
i
(z)
− 1
.
An extension of the univalence criterion for a family... 33
By using the Theorem 3, we get the inequality
(2.9)
zh
00
(z)
h
0
(z)
≤
n
X
i=1
1
|α
i
|
(1 − γ)(1 + |z|)
1 − |z|
.
From (2.9), we obtain
(2.10)
1 − |z|
2 Re(β)
Re(β)
zh
00
(z)
h
0
(z)
≤
1 − |z|
2 Re(β)
Re(β)
(1 − γ)(1 + |z|)
1 − |z|
n
X
i=1
1
|α
i
|
≤
1 − |z|
2 Re(β)
1 − |z|
2(1 − γ)
Re(β)
n
X
i=1
1
|α
i
|
for all z ∈ U.
Let us denote |z| = x, x ∈ [0, 1), Re(β) = a > 0 and ψ(x) =
1−x
2a
1−x
. It is
easy to prove that
(2.11) ψ(x) ≤
(
1, if 0 < a <
1
2
2a, if
1
2
< a < ∞.
From (2.6), (2.10) and (2.11), we have
1 − |z|
2 Re(β)
Re(β)
zh
00
(z)
h
0
(z)
≤
2(1−γ)
Re(β)
n
P
i=1
1
|α
i
|
, if 0 < Re(β) <
1
2
4(1 − γ)
n
P
i=1
1
|α
i
|
, if
1
2
< Re(β) < ∞
≤ 1
for all z ∈ U.
Finally, by applying Theorem 1, we conclude that the function
F
α
1
,α
2
,...,α
n
,β
(z) defined by (1.6) is in the function class S. This evidently
completes the proof of Theorem 4.
3. Some applications of Theorem 4. In this section, we give some re-
sults of Theorem 4.
First of all, upon setting α
i
= α, for all i ∈ {1, 2, . . . , n} in Theorem 4,
we immediately arrive at the following application of Theorem 4.
Corollary 1. Let f
i
(z) ∈ B(γ) for i ∈ {1, 2, . . . , n}. Let β be a complex
number with Re(β) > 0. If
(3.1)
1
|α|
≤ min
1
2n(1 − γ)
Re β;
1
4n(1 − γ)
holds for all z ∈ U, then the function
F
α,β
(z)
:
=
β
z
Z
0
t
β−1
n
Y
i=1
f
i
(t)
t
1
α
dt
1
β