An. Şt. Univ. Ovidius Constanţa
Vol. 17(1), 2009, 37–50
Univalence preserving integral operators
defined by generalized Al-Oboudi differential
operators
Serap BULUT
Abstract
In this paper, we investigate sufficient conditions for the univalence
of an integral operator defined by generalized Al-Oboudi differential
operator.
1
Introduction
Let A denote the class of all functions of the form
f (z) = z +
∞
X
ak z k
(1.1)
k=2
which are analytic in the open unit disk U = {z ∈ C : |z| < 1}, and S =
{f ∈ A : f is univalent in U}.
The following definition of fractional derivative by Owa [8] (also by Srivastava and Owa [14]) will be required in our investigation.
Key Words: Analytic function; Univalent function; Differential operator; Integral operator.
Mathematics Subject Classification: 30C45.
Received: January, 2009
Accepted: April, 2009
37
38
Serap BULUT
The fractional derivative of order γ is defined, for a function f , by
Z z
d
f (ξ)
1
dξ (0 ≤ γ < 1),
Dzγ f (z) =
Γ(1 − γ) dz 0 (z − ξ)γ
(1.2)
where the function f is analytic in a simply connected region of the complex
z-plane containing the origin, and the multiplicity of (z − ξ)−γ is removed by
requiring log(z − ξ) to be real when z − ξ > 0.
It readily follows from (1.2) that
Dzγ z k =
Γ(k + 1)
z k−γ
Γ(k + 1 − γ)
(0 ≤ γ < 1, k ∈ N = {1, 2, . . .}).
Using Dzγ f , Owa and Srivastava [9] introduced the operator Ωγ : A → A,
which is known as an extension of fractional derivative and fractional integral,
as follows:
Ωγ f (z)
= Γ (2 − γ) z γ Dzγ f (z), γ 6= 2, 3, 4, . . .
∞
X
Γ(k + 1)Γ (2 − γ)
= z+
ak z k .
Γ(k + 1 − γ)
(1.3)
k=2
Note that
Ω0 f (z) = f (z).
In [3], Al-Oboudi and Al-Amoudi defined the linear multiplier fractional
differential operator Dλn,γ as follows:
D0 f (z) = f (z),
Dλ1,γ f (z) =
(1 − λ) Ωγ f (z) + λz (Ωγ f (z))
′
= Dλγ (f (z)) , λ ≥ 0, 0 ≤ γ < 1,
Dλ2,γ f (z) = Dλγ Dλ1,γ f (z) ,
(1.4)
Dλn,γ f (z) = Dλγ Dλn−1,γ f (z) ,
(1.5)
..
.
n ∈ N.
If f is given by (1.1), then by (1.3), (1.4) and (1.5), we see that
Dλn,γ f (z) = z +
∞
X
k=2
Ψk,n (γ, λ) ak z k ,
n ∈ N0 = N∪ {0} ,
(1.6)
39
Univalence preserving integral operators
where
Ψk,n (γ, λ) =
n
Γ(k + 1)Γ (2 − γ)
(1 + (k − 1) λ) .
Γ(k + 1 − γ)
(1.7)
Remark 1.1. (i) When γ = 0, we get Al-Oboudi differential operator [2].
(ii) When γ = 0 and λ = 1, we get Sălăgean differential operator [13].
(iii) When n = 1 and λ = 0, we get Owa-Srivastava fractional differential
operator [9].
By using the generalized Al-Oboudi differential operator Dλn,γ , we introduce the following integral operator:
Definition 1.1 Let n ∈ N0 , m ∈ N, β ∈ C with ℜ(β) > 0 and αi ∈ C
(i ∈ {1, . . . , m}). We define the integral operator
Iβn,γ (f1 , . . . , fm ) : Am → A,
Iβn,γ (f1 , . . . , fm )(z)
( Z
= β
0
z
t
β−1
α
m n,γ
Y
D fi (t) i
λ
t
i=1
dt
) β1
(z ∈ U), (1.8)
where Dλn,γ is the generalized Al-Oboudi differential operator.
Remark 1.2. (i) For m ∈ N, β ∈ C, ℜ(β) > 0, αi ∈ C and Dλ0,γ fi (z) =
D01,0 fi (z) = fi (z) ∈ S (i ∈ {1, . . . , m}), we have the integral operator
( Z
Iβ (f1 , . . . , fm )(z) = β
0
z
t
β−1
α
m Y
fi (t) i
i=1
t
dt
) β1
which was introduced in [4].
(ii) For m ∈ N, β = 1, αi ∈ C and Dλ0,γ fi (z) = D01,0 fi (z) = fi (z) ∈ S
(i ∈ {1, . . . , m}), we have the integral operator
α
α
Z z
f1 (t) 1
fm (t) m
···
dt
I(f1 , . . . , fm )(z) =
t
t
0
which was studied in [4].
(iii) For n ∈ N0 , m ∈ N, β = 1, αi ∈ C and Dλn,0 fi (z) = Dn fi (z) (i ∈ {1, . . . , m}),
we have the integral operator
α
α
n
Z z n
D f1 (t) 1
D fm (t) m
n
I (f1 , . . . , fm )(z) =
···
dt
t
t
0
40
Serap BULUT
which was studied in [5].
(iv) For n = 0, m = 1, β = 1, α1 = 1, α2 = α3 = · · · = αm = 0 and
Dλ0,γ f1 (z) = D01,0 f1 (z) = f (z) ∈ A, we have Alexander integral operator
Z z
f (t)
I(f )(z) =
dt
t
0
which was introduced in [1].
(v) For n = 0, m = 1, β = 1, α1 = α ∈ [0, 1], α2 = α3 = · · · = αm = 0 and
Dλ0,γ f1 (z) = D01,0 f1 (z) = f (z) ∈ S, we have the integral operator
α
Z z
f (t)
I(f )(z) =
dt
t
0
which was studied in [6].
To discuss our problems, we have to recall here the following results.
General Schwarz Lemma [7]. Let the function f be regular in the disk
UR = {z ∈ C : |z| < R}, with |f (z)| < M for fixed M . If f (z) has one zero
with multiplicity order bigger than m for z = 0, then
|f (z)| ≤
M
m
|z|
Rm
(z ∈ UR ).
The equality can hold only if
f (z) = eiθ (M/Rm ) z m ,
where θ is constant.
Theorem A [10]. Let α be a complex number with ℜ(α) > 0 and f ∈ A.
If f (z) satisfies
2ℜ(α) ′′
1 − |z|
zf (z) ≤ 1,
′
ℜ(α)
f (z) for all z ∈ U, then the integral operator
Z
Fα (z) = α
0
is in the class S.
z
t
α−1 ′
f (t)dt
α1
41
Univalence preserving integral operators
Theorem B [11]. Let α be a complex number with ℜ(α) > 0 and f ∈ A.
If f (z) satisfies
2ℜ(α)
1 − |z|
ℜ(α)
′′ zf (z) f ′ (z) ≤ 1
(z ∈ U),
then, for any complex number β with ℜ(β) ≥ ℜ(α), the integral operator
Z
Fβ (z) = β
z
t
β−1 ′
f (t)dt
0
β1
is in the class S.
Theorem C [12]. Let β be a complex number with ℜ(β) > 0, c a complex
number with |c| ≤ 1, c 6= −1, and f (z) given by (1.1) an analytic function in
U. If
′′
c |z|2β + (1 − |z|2β ) zf (z) ≤ 1
′
βf (z) for all z ∈ U, then the function
Z
Fβ (z) = β
z
t
β−1 ′
f (t)dt
0
β1
= z + ···
is analytic and univalent in U.
2
Main Results
Theorem 2.1 Let α1 , . . . , αm , β ∈ C and each of the functions fi ∈ A (i ∈ {1, . . . , m}).
If
z (Dn,γ f (z))′
i
λ
− 1 ≤ 1 (z ∈ U, n ∈ N0 )
n,γ
Dλ fi (z)
and
ℜ(β) ≥
m
X
|αi | > 0,
i=1
then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
42
Serap BULUT
Proof. Since fi ∈ A (i ∈ {1, . . . , m}), by (1.6), we have
∞
X
Dλn,γ fi (z)
Ψk,n (γ, λ) ak,i z k−1
=1+
z
(n ∈ N0 )
k=2
and
Dλn,γ fi (z)
6= 0
z
for all z ∈ U.
Let us define
h(z) =
Z
m zY
0 i=1
so that, obviously,
′
h (z) =
Dλn,γ f1 (z)
z
Dλn,γ fi (t)
t
α1
···
αi
dt,
Dλn,γ fm (z)
z
αm
for all z ∈ U. This equality implies that
ln h′ (z) = α1 ln
Dλn,γ f1 (z)
Dn,γ fm (z)
+ · · · + αm ln λ
z
z
or equivalently
ln h′ (z) = α1 [ln Dλn,γ f1 (z) − ln z] + · · · + αm [ln Dλn,γ fm (z) − ln z] .
By differentiating above equality, we get
"
#
m
′
(Dλn,γ fi (z))
1
h′′ (z) X
.
αi
=
−
h′ (z)
Dλn,γ fi (z)
z
i=1
Hence, we obtain from this equality that
′′ m
z (Dn,γ f (z))′
zh (z) X
i
λ
≤
−
1
|α
|
.
i
n,γ
h′ (z) Dλ fi (z)
i=1
So by the conditions of the Theorem 2.1, we find
m
2ℜ(β) X
2ℜ(β) z (Dn,γ f (z))′
′′
1
−
|z|
zh
(z)
1 − |z|
i
λ
≤
− 1
|αi | n,γ
′
Dλ fi (z)
ℜ(β)
h (z)
ℜ(β)
i=1
m
≤
1 X
|αi | ≤ 1.
ℜ(β) i=1
43
Univalence preserving integral operators
Finally, applying Theorem A for the function h(z), we prove that Iβn,γ (f1 , . . . , fm ) ∈
S.
Remark 2.1. If we set β = 1 and γ = 0 in Theorem 2.1, then we have
Theorem 2.3 in [5].
Corollary 2.2 Let αi > 0, β ∈ C and each of the functions fi ∈ A (i ∈ {1, . . . , m}).
If
z (Dn,γ f (z))′
i
λ
− 1 ≤ 1 (z ∈ U, n ∈ N0 )
n,γ
Dλ fi (z)
and
ℜ(β) ≥
m
X
αi ,
i=1
then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
Remark 2.2. If we set β = 1 and γ = 0 in Corollary 2.2, then we have
Corollary 2.5 in [5].
Theorem 2.3 Let Mi ≥ 1 and suppose that each of the functions fi ∈ A
(i ∈ {1, . . . , m} , m ∈ N) satisfies the inequality
z 2 (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
(Dn,γ fi (z))2
λ
Also let α1 , . . . , αm , α ∈ C with
ℜ(α) ≥
m
X
|αi | (2Mi + 1) > 0.
i=1
If
|Dλn,γ fi (z)| ≤ Mi (z ∈ U; i ∈ {1, . . . , m}),
then, for any complex number β with ℜ(β) ≥ ℜ(α), the integral operator
Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent function class S.
Proof. We know from the proof of Theorem 2.1 that
′′ m
z (Dn,γ f (z))′
zh (z) X
i
λ
≤
|α
|
−
1
.
i
n,γ
h′ (z) D fi (z)
i=1
λ
44
Serap BULUT
So, by the imposed conditions, we find
m
2ℜ(α) 2ℜ(α) X
′′
1 − |z|
zh (z) ≤ 1 − |z|
|αi |
h′ (z) ℜ(α)
ℜ(α)
i=1
2ℜ(α)
≤
1 − |z|
ℜ(α)
≤
1 − |z|
ℜ(α)
2ℜ(α)
m
≤
!
z (Dn,γ f (z))′ i
λ
+1
Dλn,γ fi (z) !
m
z 2 (Dn,γ f (z))′ Dn,γ f (z) X
λ i i
λ
|αi | +1
(Dn,γ fi (z))2 z
λ
i=1
!
m
z 2 (Dn,γ f (z))′
X
i
λ
− 1 Mi + Mi + 1
|αi | (Dn,γ fi (z))2
λ
i=1
1 X
|αi | (2Mi + 1) ≤ 1
ℜ(α) i=1
By applying Theorem B for the function h(z), we prove that Iβn,γ (f1 , . . . , fm ) ∈
S.
Corollary 2.4 Let Mi ≥ 1, αi > 0 and suppose that each of the functions
fi ∈ A (i ∈ {1, . . . , m}) satisfies the inequality
z 2 (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
(Dn,γ fi (z))2
λ
Also let α ∈ C with
ℜ(α) ≥
m
X
αi (2Mi + 1) .
i=1
If
|Dλn,γ fi (z)| ≤ Mi (z ∈ U; i ∈ {1, . . . , m}),
then, for any complex number β with ℜ(β) ≥ ℜ(α), the integral operator
Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent function class S.
Corollary 2.5 Let M ≥ 1 and suppose that each of the functions fi ∈ A
(i ∈ {1, . . . , m} , m ∈ N) satisfies the inequality
z 2 (Dn f (z))′
i
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
(Dn fi (z))2
Also let α1 , . . . , αm , α ∈ C with
ℜ(α) ≥ (2M + 1)
m
X
i=1
|αi | > 0.
45
Univalence preserving integral operators
If
|Dn fi (z)| ≤ M (z ∈ U; i ∈ {1, . . . , m}),
then, for any complex number β with ℜ(β) ≥ ℜ(α), the integral operator
Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent function class S.
Proof. In Theorem 2.3, we consider M1 = · · · = Mm = M .
Corollary 2.6 Suppose that each of the functions fi ∈ A (i ∈ {1, . . . , m})
satisfies the inequality
z 2 (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
(Dn,γ fi (z))2
λ
Also let α1 , . . . , αm , α ∈ C with
ℜ(α) ≥ 3
m
X
|αi | > 0.
i=1
If
|Dλn,γ fi (z)| ≤ 1 (z ∈ U; i ∈ {1, . . . , m}),
then, for any complex number β with ℜ(β) ≥ ℜ(α), the integral operator
Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent function class S.
Proof. In Corollary 2.5, we consider M = 1.
Remark 2.3. In Corollary 2.6, if we set
(i) β = 1 and γ = 0, then we have Theorem 2.6,
(ii) β = 1, γ = 0 and αi > 0 (i ∈ {1, . . . , m}), then we have Corollary 2.8
in [5].
Theorem 2.7 Suppose that each of the functions fi ∈ A (i ∈ {1, . . . , m})
satisfies the inequality
z (Dn,γ f (z))′
i
λ
− 1 ≤ 1 (z ∈ U, n ∈ N0 ).
n,γ
Dλ fi (z)
Also let α1 , . . . , αm , β ∈ C with
ℜ(β) ≥
m
X
i=1
|αi | > 0
46
Serap BULUT
and let c ∈ C be such that
m
1 X
|αi | .
|c| ≤ 1 −
ℜ(β) i=1
Then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
Proof. We know from the proof of Theorem 2.1 that
"
#
m
′
z (Dλn,γ fi (z))
zh′′ (z) X
αi
=
−1 .
h′ (z)
Dλn,γ fi (z)
i=1
So we obtain
′′
c |z|2β + (1 − |z|2β ) zh (z) ′
βh (z) "
#
m
′
X
z (Dλn,γ fi (z))
2β
2β 1
= c |z| + (1 − |z| )
−
1
αi
β i=1
Dλn,γ fi (z)
m
1 − |z|2β X
z (Dn,γ f (z))′
i
λ
− 1
|αi | ≤ |c| + n,γ
β
Dλ fi (z)
i=1
m
≤ |c| +
1 X
|αi |
|β| i=1
m
≤ |c| +
1 X
|αi | ≤ 1.
ℜ(β) i=1
Finally, applying Theorem C for the function h(z), we prove that Iβn,γ (f1 , . . . , fm ) ∈
S.
Corollary 2.8 Suppose that each of the functions fi ∈ A (i ∈ {1, . . . , m})
satisfies the inequality
z (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
Dλn,γ fi (z)
Also let αi > 0, β ∈ C with
ℜ(β) ≥
m
X
αi
i=1
and let c ∈ C be such that
m
|c| ≤ 1 −
1 X
αi .
ℜ(β) i=1
Univalence preserving integral operators
47
Then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
Theorem 2.9 Let Mi ≥ 1 and suppose that each of the functions fi ∈ A
(i ∈ {1, . . . , m}) satisfies the inequality
z 2 (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
(Dn,γ fi (z))2
λ
Also let α1 , . . . , αm , β ∈ C with
ℜ(β) ≥
m
X
|αi | (2Mi + 1) > 0,
i=1
c ∈ C be such that
m
|c| ≤ 1 −
1 X
|αi | (2Mi + 1)
ℜ(β) i=1
and
|Dλn,γ fi (z)| ≤ Mi (z ∈ U; i ∈ {1, . . . , m}).
Then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
Proof. We know from the proof of Theorem 2.1 that
"
#
m
′
z (Dλn,γ fi (z))
zh′′ (z) X
αi
=
−1 .
h′ (z)
Dλn,γ fi (z)
i=1
So we obtain
′′
c |z|2β + (1 − |z|2β ) zh (z) ′
βh (z) =
≤
≤
≤
"
#
m
′
X
z (Dλn,γ fi (z))
2β
2β 1
−
1
αi
c |z| + (1 − |z| )
β i=1
Dλn,γ fi (z)
m
!
1 − |z|2β X
z (Dn,γ f (z))′ i
λ
|c| + |αi | +1
Dλn,γ fi (z) β
i=1
!
m
z 2 (Dn,γ f (z))′ Dn,γ f (z) 1 X
λ i i
λ
|αi | |c| +
+1
(Dn,γ fi (z))2 |β| i=1
z
λ
!
m
z 2 (Dn,γ f (z))′
1 X
i
λ
|αi | − 1 Mi + Mi + 1
|c| +
(Dn,γ fi (z))2
|β| i=1
λ
48
Serap BULUT
m
≤ |c| +
1 X
|αi | (2Mi + 1) ≤ 1.
ℜ(β) i=1
Finally, applying Theorem C for the function h(z), we prove that Iβn,γ (f1 , . . . , fm ) ∈
S.
Corollary 2.10 Let Mi ≥ 1, αi > 0 and suppose that each of the functions
fi ∈ A (i ∈ {1, . . . , m}) satisfies the inequality
z 2 (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
(Dn,γ fi (z))2
λ
Also let β ∈ C with
ℜ(β) ≥
m
X
αi (2Mi + 1) ,
i=1
c ∈ C be such that
m
|c| ≤ 1 −
1 X
αi (2Mi + 1)
ℜ(β) i=1
and
|Dλn,γ fi (z)| ≤ Mi (z ∈ U; i ∈ {1, . . . , m}).
Then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
Corollary 2.11 Let M ≥ 1 and suppose that each of the functions fi ∈ A
(i ∈ {1, . . . , m}) satisfies the inequality
z 2 (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
(Dn,γ fi (z))2
λ
Also let α1 , . . . , αm , β ∈ C with
ℜ(β) ≥ (2M + 1)
m
X
|αi | > 0,
i=1
c ∈ C be such that
m
|c| ≤ 1 −
2M + 1 X
|αi |
ℜ(β) i=1
49
Univalence preserving integral operators
and
|Dλn,γ fi (z)| ≤ M (z ∈ U; i ∈ {1, . . . , m}).
Then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
Proof. In Theorem 2.9, we consider M1 = · · · = Mm = M .
Corollary 2.12 Suppose that each of the functions fi ∈ A (i ∈ {1, . . . , m})
satisfies the inequality
z 2 (Dn,γ f (z))′
i
λ
−
1
≤ 1 (z ∈ U, n ∈ N0 ).
2
(Dn,γ fi (z))
λ
Also let α1 , . . . , αm , β ∈ C with
ℜ(β) ≥ 3
m
X
|αi | > 0,
i=1
c ∈ C be such that
m
|c| ≤ 1 −
3 X
|αi |
ℜ(β) i=1
and
|Dλn,γ fi (z)| ≤ 1 (z ∈ U; i ∈ {1, . . . , m}).
Then the integral operator Iβn,γ (f1 , . . . , fm ) defined by (1.8) is in the univalent
function class S.
Proof. In Corollary 2.11, we consider M = 1.
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50
Serap BULUT
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