MATEMATIQKI VESNIK
originalni nauqni rad
research paper
65, 3 (2013), 394–402
September 2013
UNIVALENCE CONDITIONS OF GENERAL INTEGRAL OPERATOR
B. A. Frasin and D. Breaz
Abstract. In this paper, we obtain new univalence conditions for the integral operator
· Z
α ,β
Iξ i i (f1 , . . . , fn )(z)
z
= ξ
t
ξ−1
¡
¢
α1
f10 (t)
³
0
f1 (t)
t
´β1
···
¡
¢
αn
fn0 (t)
³
fn (t)
t
¸ ξ1
´βn
dt
of analytic functions defined in the open unit disc.
1. Introduction
Let A denote the class of functions of the form
∞
P
f (z) = z +
ak z k
k=2
which are analytic in the open unit disc U = {z : |z| < 1}. Further, by S we shall
denote the class of all functions in A which are univalent in U.
Very recently, Frasin [13] introduced and studied the following general integral
operator
Definition 1.1. Let αi , βi ∈ C for all i = 1, . . . , n, n ∈ N. We let Iξαi ,βi :
A → A to be the integral operator defined by
n
Iξαi ,βi (f1 , . . . , fn )(z)
" Z
= ξ
0
µ
z
t
ξ−1
α
(f10 (t)) 1
f1 (t)
t
# ξ1
¶β1
¶β
µ
fn (t) n
αn
0
· · · (fn (t))
dt ,
t
(1.1)
where ξ ∈ C\{0} and fi ∈ A for all i = 1, . . . , n.
Here and throughout in the sequel every many-valued function is taken with
the principal branch.
2010 Mathematics Subject Classification: 30C45
Keywords and phrases: Analytic function; univalent function; integral operator.
394
Univalence conditions of general integral operator
395
Remark 1.2. Note that the integral operator Iξαi ,βi (f1 , . . . , fn )(z) generalizes
the following operators introduced and studied by several authors:
(1) For ξ = 1, we obtain the integral operator
µ
¶β
µ
¶β
Z z
f1 (t) 1
fn (t) n
α
α
I αi ,βi (f1 , . . . , fn )(z) =
(f10 (t)) 1
· · · (fn0 (t)) n
dt
t
t
0
introduced and studied by Frasin [14].
(2) For ξ = 1 and αi = 0 for all i = 1, . . . , n, we obtain the integral operator
¶β
µ
¶β
Z zµ
fn (t) n
f1 (t) 1
Fn (z) =
...
dt
t
t
0
introduced and studied by Breaz and Breaz [3].
(3) For ξ = 1 and βi = 0 for all i = 1, . . . , n, we obtain the integral operator
Z z
α
α
Fα1 ,... ,αn (z) =
(f10 (t)) 1 · · · (fn0 (t)) n dt
0
introduced and studied by Breaz et al. [6].
(4) For ξ = 1, n = 1, α1 = α, β1 = β and f1 = f , we obtain the integral
operator
¶β
µ
Z z
f (t)
α
0
Fα,β (z) =
dt
(α, β ∈ R)
(f (t))
t
0
studied in [9] (see also [10]).
(5) For ξ = 1, n = 1, α1 = 0, β1 = β and f1 = f , we obtain the integral
operator
¶β
Z zµ
f (t)
Fβ (z) =
dt
t
0
studied in [7]. In particular, for β = 1, we obtain Alexander integral operator
introduced in [1]
Z z
f (t)
I(z) =
dt
t
0
(6) For ξ = 1, n = 1, β1 = 0, α1 = α and f1 = f , we obtain the integral
operator
Z z
α
Gα (z) =
(f 0 (t)) dt
0
studied in [21] (see also [25]).
Many authors studied the problem of integral operators which preserve the
class S (see, for example, [2, 4, 5, 7, 8, 12, 19, 22, 23, 24, 27]).
In particular, Pfaltzgraff [25] and Kim and Merkes [16], have obtained
the
Rz 0
R z ³ f (t) ´α
α
following univalence conditions for the functions 0 (f (t)) dt and 0
dt,
t
respectively.
396
B.A. Frasin, D. Breaz
1.3. [25] If f ∈ A and α ∈ C with |α| ≤ 1/4, then the function
R z Theorem
α
0
dt
is
in the class S.
(f
(t))
0
Theorem 1.4. [16] If f ∈ A and α ∈ C with |α| ≤ 1/4, then the function
R z ³ f (t) ´α
dt is in the class S.
t
0
Iξαi ,
In the present paper, we obtain univalence conditions for the integral operator
(f1 , . . . , fn )(z) defined by (1.1).
In order to derive our main results, we have to recall here the following lemmas.
βi
Lemma 1.5. [18] If f ∈ A satisfies
¯
¯
¯ 0 ¯
00
¯
¯
¯
¯
¯1 + zf (z) ¯ < 5 ¯ zf (z) ¯
¯
f 0 (z) ¯ 4 ¯ f (z) ¯
(z ∈ U ),
(1.2)
(z ∈ U ),
(1.3)
then f is univalent and starlike in U.
Lemma 1.6. [15, 26] If f ∈ A satisfies
¯
¯
0
00
¯
¯
¯1 + zf (z) − zf (z) ¯ < 1
¯
0
f (z)
f (z) ¯ 2
¯ 0
¯
¯ (z)
¯
then ¯ zff (z)
− 1¯ < 1 (z ∈ U ).
Lemma 1.7. [11] If f ∈ S then
¯ 0 ¯
¯ zf (z) ¯ 1 + |z|
¯
¯
¯ f (z) ¯ < 1 − |z|
(z ∈ U).
(1.4)
Lemma 1.8. [20] Let δ ∈ C with Re(δ) > 0. If f ∈ A satisfies
¯
2 Re(δ) ¯¯
00
¯
1 − |z|
¯ zf (z) ¯ ≤ 1,
¯
0
Re(δ)
f (z) ¯
for all z ∈ U, then, for any complex number ξ, with Re(ξ) ≥ Re(δ), the integral
operator
½ Z z
¾ ξ1
ξ−1 0
Fξ (z) = ξ
t f (t) dt
0
is in the class S.
2. Main results
We begin by proving the following theorem.
Theorem 2.1. Let αi , βi ∈ C for all i = 1, . . . , n and each fi ∈ A satisfies
the condition (1.2). If
½
n
P
4 Re δ, if Re δ ∈ (0, 1)
(29 |αi | + 16 |βi |) ≤
(2.1)
4,
if Re δ ∈ [1, ∞),
i=1
then, for any complex number ξ, with Re(ξ) ≥ Re(δ) > 0, the integral operator
Iξαi , βi (f1 , . . . , fn )(z) defined by (1.1) is in the class S.
Univalence conditions of general integral operator
397
Proof. Define a regular function h(z) by
Z
h(z) =
µ
n
zY
α
(fi0 (t)) i
0 i=1
fi (t)
t
¶βi
dt.
Then it is easy to see that
h0 (z) =
n
Y
µ
(fi0 (z))
αi
i=1
fi (z)
z
¶βi
(2.2)
and h(0) = h0 (0) − 1 = 0. Differentiating both sides of (2.2) logarithmically, we
obtain
¶
µ 00 ¶
µ 0
n
n
P
P
zfi (z)
zfi (z)
zh00 (z)
=
αi
+
βi
−1
h0 (z)
fi0 (z)
fi (z)
i=1
i=1
and so
n
P
zh00 (z)
=
αi
0
h (z)
i=1
µ
¶
µ 0
¶
n
n
P
P
zfi00 (z)
zfi (z)
+
1
−
α
+
β
−
1
.
i
i
fi0 (z)
fi (z)
i=1
i=1
From Lemma 1.5, it follows that
¯ 00 ¯
¯ 0 ¯
¯
¯ 0
n
n
n
¯ zh (z) ¯ 5 P
¯ zfi (z) ¯ P
¯ P
¯ zfi (z)
¯
¯≤
¯
¯
¯+
¯
|α
|
−
1
+
|β
|
|αi |
i ¯
i ¯
¯ h0 (z) ¯ 4
¯
¯
fi (z)
fi (z)
i=1
i=1
i=1
¯
¯
¯ 0
·¯ 0
¸
n
n
n
¯ zf (z)
¯
¯ P
¯ zf (z)
P
5 P
|αi | ¯¯ i
− 1¯¯ + 1 +
− 1¯¯ +
≤
|βi | ¯¯ i
|αi |
4 i=1
fi (z)
fi (z)
i=1
i=1
¯¸
¶¯ 0
·µ
n
n
¯ zf (z)
¯
P
9 P
5
|αi | + |βi | ¯¯ i
− 1¯¯ +
|αi |
≤
4
fi (z)
4 i=1
i=1
¶ µ¯ 0 ¯
·µ
¶¸
n
n
¯ zfi (z) ¯
P
P
5
¯
¯+1 + 9
≤
|αi | + |βi |
|αi | .
(2.3)
¯
¯
4
fi (z)
4 i=1
i=1
Multiplying both sides of (2.3) by
1−|z|2Re δ
,
Re δ
from (1.4), we get
¯ 00 ¯
¯ zh (z) ¯
¯
¯
¯ h0 (z) ¯
¶µ
¶
2Re δ Pn
2Re δ n µ
P 5
9(1 − |z|
) i=1 |αi |
2
1 − |z|
|αi | + |βi |
+
.
≤
Re δ
1 − |z|
4Re δ
i=1 4
2Re δ
1 − |z|
Re δ
(2.4)
Suppose that Re δ ∈ (0, 1). Define a function Φ : (0, 1) → R by
Φ(x) = 1 − a2x
(0 < a < 1).
Then Φ is an increasing function and consequently for |z| = a; z ∈ U , we obtain
2Re δ
1 − |z|
for all z ∈ U.
2
< 1 − |z|
(2.5)
398
B.A. Frasin, D. Breaz
We thus find from (2.4) and (2.5) that
2Re δ
1 − |z|
Re δ
¯ 00 ¯
¯ zh (z) ¯
¯
¯
¯ h0 (z) ¯ ≤
n
P
i=1
Re δ
n
P
=
i=1
9
(5 |αi | + 4 |βi |)
+
n
P
i=1
|αi |
4Re δ
(29 |αi | + 16 |βi |)
4Re δ
for all z ∈ U.
Using the hypothesis (2.1) for Re δ ∈ (0, 1), we readily get
¯
2Re δ ¯¯
00
¯
1 − |z|
¯ zh (z) ¯ ≤ 1.
¯
0
Re δ
h (z) ¯
Now if Re δ ∈ [1, ∞), we define a function Ψ : [1, ∞) → R by
Ψ(x) =
1 − a2x
x
(0 < a < 1).
We observe that the function Ψ is decreasing and consequently for |z| = a; z ∈ U,
we have
2Re δ
1 − |z|
2
≤ 1 − |z|
(2.6)
Re δ
for all z ∈ U. It follows from (2.4) and (2.6) that
¯
2Re δ ¯¯
00
n 29
¯ P
1 − |z|
¯ zh (z) ¯ ≤
( |αi | + 4 |βi |).
¯ h0 (z) ¯
Re δ
i=1 4
Using once again the hypothesis (2.1) when Re δ ∈ [1, ∞), we easily get
¯
2Re δ ¯¯
00
¯
1 − |z|
¯ zh (z) ¯ ≤ 1.
¯
0
Re δ
h (z) ¯
Finally by applying Lemma 1.8, we conclude that the integral operator
Iξαi , βi (f1 , . . . , fn )(z) defined by (1.2) is in the class S.
Letting n = 1, α1 = α, β1 = β and f1 = f in Theorem 2.1, we have
Corollary 2.2. Let α, β ∈ C and f ∈ A satisfy the condition (1.2). If
½
4Re δ, if Re δ ∈ (0, 1)
29 |α| + 16 |β| ≤
4,
if Re δ ∈ [1, ∞),
then, for any complex number ξ, with Re(ξ) ≥ Re(δ) > 0, the integral operator
· Z z
¸ ξ1
α
β
Iξα,β (z) = ξ
tξ−β−1 (f 0 (t)) (f (t)) dt
0
is in the class S.
399
Univalence conditions of general integral operator
Letting β = 0 in Corollary 2.2, we have
Corollary 2.3. Let α ∈ C and f ∈ A satisfy the condition (1.2). If
½ 4Re δ
29 , if Re δ ∈ (0, 1)
|α| ≤
4
if Re δ ∈ [1, ∞),
29 ,
then, for any complex number ξ, with Re(ξ) ≥ Re(δ) > 0, the integral operator
· Z
Iξα (z) = ξ
z
¸ ξ1
α
tξ−1 (f 0 (t)) dt
0
is in the class S.
Letting α = 0 in Corollary 2.2, we have
Corollary 2.4. Let β ∈ C and f ∈ A satisfy the condition (1.2). If
½ Re δ
, if Re δ ∈ (0, 1)
|β| ≤ 1 4
if Re δ ∈ [1, ∞),
4,
then, for any complex number ξ, with Re(ξ) ≥ Re(δ) > 0, the integral operator
Iξβ (z)
" Z
= ξ
µ
z
t
ξ−1
0
f (t)
t
¶β
# ξ1
dt
is in the class S.
Letting ξ = δ = 1 in Corollary 2.3, we have
Corollary
2.5. If f ∈ A and α ∈ C with |α| ≤ 4/29 ≈ 0.137, then the
Rz
α
function 0 (f 0 (t)) dt is in the class S.
Iξαi ,
Remark 2.6. If we let ξ = δ = 1 in Corollary 2.4, then we have Theorem 1.4.
Next, we obtain the following univalence condition for the integral operator
βi
(f1 , . . . , fn )(z) defined by (1.1) when βi = 1 − αi for all i = 1, . . . , n.
Theorem 2.7. Let αi , βi ∈ C for all i = 1, . . . , n and each fi ∈ A satisfy the
condition (1.3). If Re δ ≥ n, n ∈ N, δ ∈ C with
n
P
i=1
|αi | ≤ 2Re δ − 2n
(2.7)
then, for any complex number ξ, with Re(ξ) ≥ Re(δ) > 0, the integral operator
Iξαi (f1 , . . . , fn )(z)
is in the class S.
" Z
= ξ
z
t
0
ξ−1
¶α µ
¶ # ξ1
n µ
Y
fi0 (t) i fi (t)
t
dt
fi (t)
t
i=1
(2.8)
400
B.A. Frasin, D. Breaz
Proof. Define a regular function G(z) by
¶1−αi
µ
Z zY
n
fi (t)
α
G(z) =
(fi0 (t)) i
dt.
t
0 i=1
. Then it follows from (2.9) that
µ
¶
µ 0
¶
n
n
P
P
zG00 (z)
zfi00 (z) zfi0 (z)
zfi (z)
=
α
1
+
−
+
−
1
.
i
G0 (z)
fi0 (z)
fi (z)
fi (z)
i=1
i=1
Using Lemma 1.6, from (2.10), we have
¯ 00 ¯
n
¯ zG (z) ¯ 1 P
¯
¯
|αi | + n
¯ G0 (z) ¯ ≤ 2
i=1
(2.9)
(2.10)
(2.11)
2Re δ
Multiply both sides of (2.11) by 1−|z|
, we obtain
Re δ
¯
¶
2Re δ ¯¯
2Re δ µ
00
n
¯
1 − |z|
1 P
¯ zG (z) ¯ ≤ 1 − |z|
|α
|
+
n
i
¯ G0 (z) ¯
Re δ
Re δ
2 i=1
µ n
¶
P
1
≤
|αi | + 2n ,
2Re δ i=1
which, in the light of the hypothesis (2.7) yields
¯
2Re δ ¯¯
00
¯
1 − |z|
¯ zG (z) ¯ ≤ 1.
¯
0
Re δ
G (z) ¯
Finally by applying Lemma 1.8, we conclude that the integral operator
Iξαi (f1 , . . . , fn )(z) defined by (2.8) is in the class S.
Letting n = 1, α1 = α and f1 = f in Theorem 2.7, we have
Corollary 2.8. Let f ∈ A satisfies the condition (1.3), α, δ ∈ C and Re δ ≥
1. If
|α| ≤ 2Re δ − 2
then, for any complex number ξ, with Re(ξ) ≥ Re(δ), the integral operator
· Z z
¸ ξ1
α
1−α
α
ξ+α−2
0
Iξ (z) = ξ
t
(f (t)) (f (t))
dt
0
is in the class S.
Letting α = 0 in Corollary 2.8, we have
Corollary 2.9. Let f ∈ A satisfies the condition (1.3). If δ ∈ C, Re δ ≥ 1
then, for any complex number ξ, with Re(ξ) ≥ Re(δ), the integral operator
· Z z
¸ ξ1
ξ−2
Iξ (z) = ξ
t f (t) dt
0
is in the class S.
Univalence conditions of general integral operator
401
Letting α = 1 in Corollary 2.8, we have
Corollary 2.10. Let f ∈ A satisfies the condition (1.3). If δ ∈ C, Re δ ≥ 3/2
then, for any complex number ξ, with Re(ξ) ≥ Re(δ), the integral operator
· Z
Fξ (z) = ξ
z
t
¸ ξ1
f (t) dt
ξ−1 0
0
is in the class S.
Acknowledgement. The authors would like to thank the referee for his/her
helpful comments and suggestions.
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(received 23.08.2011; in revised form 21.02.2012; available online 10.06.2012)
B.A. Frasin, Faculty of Science, Department of Mathematics, Al al-Bayt University, P.O. Box:
130095 Mafraq, Jordan
E-mail: bafrasin@yahoo.com
D. Breaz, Department of Mathematics “1 Decembrie 1918”. University of Alba Iulia 510009 Alba
Iulia, Romania
E-mail: dbreaz@uab.ro