Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2008, Article ID 263408, 10 pages
doi:10.1155/2008/263408
Research Article
A Univalence Preserving Integral Operator
Georgia Irina Oros
Department of Mathematics, University of Oradea, Universitatii street, No. 1, 410087 Oradea, Romania
Correspondence should be addressed to Georgia Irina Oros, georgia oros ro@yahoo.co.uk
Received 23 July 2008; Accepted 8 September 2008
Recommended by Vijay Gupta
We define an integral operator denoted by I, and we give sufficient conditions such that F If1 , f2 , . . . , fm is univalent.
Copyright q 2008 Georgia Irina Oros. 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
Let U denote the unit disc of the complex plane:
U {z ∈ C : |z| < 1}.
1.1
Let HU denote the space of holomorphic functions in U and let
An f ∈ HU, fz z an1 zn1 · · · , z ∈ U
1.2
S {f ∈ A : f is univalent in U}.
1.3
with A1 A.
Let
Definition 1.1 Ruscheweyh 1. For f ∈ A, n ∈ N ∪ {0}, let Rn be the operator defined by
Rn : A → A,
R0 fz fz,
n 1Rn1 fz z Rn fz nRn fz,
z ∈ U.
1.4
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Journal of Inequalities and Applications
Remark 1.2. If f ∈ A,
fz z ∞
aj zj ,
z ∈ U,
1.5
j2
then
Rn fz z ∞
z ∈ U.
n
Cnj−1
aj zj ,
1.6
j2
In order to prove our main results, we shall use the following lemmas.
Lemma A see 2. Let α be a complex number, Re α > 0, and f ∈ A. If
1 − |z|2Re α zf z ≤ 1,
Re α f z ∀z ∈ U,
1.7
then the function
z
1/α
Fα z α tα−1 f tdt
1.8
0
is in the class S.
Lemma B see 3. Let α be a complex number, Re α > 0, and let fz z a2 z2 · · · be a regular
function in U. If
1 − |z|2Re α zf z ≤ 1,
Re α f z ∀z ∈ U,
1.9
then, for any complex number β with Re β ≥ Re α, the function
β
1/β
Fβ z β tβ−1 f tdt
1.10
0
is in the class S.
2. Main results
By using the Ruscheweyh differential operator given by Definition 1.1, we introduce the
following integral operator.
Georgia Irina Oros
3
Definition 2.1. Let n, m ∈ N ∪ {0}, i ∈ {1, 2, 3, . . . , m}, αi ∈ C, α ∈ C, with Re α > 0, Am m
A × A × ··· × A
. We let I : A → A be the integral operator given by
m times
I f1 , f2 , . . . , fm
z
1/α
α1
n
n
R fm t αm
α−1 R f1 t
z Fz α t
···
dt
,
t
t
0
2.1
where fi ∈ A, i ∈ {1, 2, 3, . . . , n} and Rn is the Ruscheweyh differential operator.
Remark 2.2. i For n 0, m 1, α 1, α1 1, α2 α3 · · · αm 0, and fz ∈ A, we obtain
Alexander integral operator introduced in 1915 in 4:
Iz z
0
ft
dt,
t
z ∈ U.
2.2
ii For n 0, m 1, α 1, α1 β ∈ 0, 1, α2 α3 · · · αm 0, and fz ∈ S, we
obtain the integral operator
Iz z 0
ft
t
β
dt,
z ∈ U,
2.3
studied in 5.
For β ∈ C with |β| ≤ 1/4, this integral operator was studied in 6, 7 and for |β| ≤ 1/3,
in 8.
iii For n 1, m 1, α 1, α1 β ∈ C, |β| ≤ 1/4, α2 · · · αm 0, R1 fz zf z,
z ∈ U, f ∈ S, we obtain the integral operator
Iz z
β
f t dt,
z∈U
2.4
0
studied in 9.
iv For n 0, m ∈ N ∪ {0}, α 1, αi > 0, i ∈ {1, 2, . . . , m}, we obtain the integral
operator
Fz z 0
f1 t
t
α1
···
fm t
t
αm
dt,
2.5
studied in 10.
v For n 0, m ∈ N ∪ {0}, α ∈ R, Re α > 0, αi ∈ C, fi ∈ S, i ∈ {1, 2, . . . , n}, we obtain
the integral operator introduced in 10 by D. Breaz and N. Breaz:
z
1/α
α1
fn t αm
α−1 f1 t
Gz α t
···
dt
.
t
t
0
2.6
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Journal of Inequalities and Applications
vi For n, m ∈ N ∪ {0}, α 1, αi ∈ R, i ∈ {1, 2, . . . , m} with αi ≥ 0, we obtain the integral
operator studied in 11, 12
I f1 , f2 , . . . , fm z z 0
Rn f1 t
t
α1
Rn fm t
···
t
αm
dt.
2.7
vii For n 0, m 1, α1 1, α2 · · · αm 0, α ∈ C with Re α ≥ 3, we obtain the
integral operator studied in 7, 13
z
1/α
gu
.
Gα z α uα−1
du
u
0
2.8
viii For n 1, m 1, α ∈ C, Re α > 0, α1 1, α2 · · · αm 0, we obtain the integral
operator
z
1/α
Fα z α uα−1 f udu
2.9
0
studied in 14, 15.
We study the conditions for the integral operator introduced in Definition 2.1 to be
univalent.
Theorem 2.3. Let n, m ∈ N ∪ {0}, α ∈ C with Re α > 0, fi ∈ A, αi ∈ C, i ∈ {1, 2, . . . , m} with
|α1 | |α2 | · · · |αm | ≤ 1.
If
z Rn fi z
−
1
≤ 1,
Rn fi z
z ∈ U, i ∈ {1, 2, . . . , m},
2.10
then Fz given by 2.1 belongs to class S.
Proof. Let
fz z 0
Rn f1 t
t
α1
···
Rn fm t
t
αm
dt,
z ∈ U.
2.11
By differentiating 2.11, we obtain
f z z ∈ U.
Rn f1 z
z
α1
···
Rn fm z
z
αm
1 A2 z A3 z2 · · · ,
2.12
0, z ∈ U.
From 2.11, 2.12, and the condition in the theorem, we have that f z /
Georgia Irina Oros
5
Then using 2.12, we obtain
log f z α1 log Rn f1 z − log z · · · αm log Rn fm z − log z ,
z ∈ U.
2.13
z ∈ U.
2.14
By differentiating 2.13, after a short calculation, we have
n
n
z R f1 z
z R fm z
zf z
α1
− 1 · · · αm
−1 ,
f z
Rn f1 z
Rn fm z
Using the conditions given by the hypothesis of Theorem 2.3, we obtain
z Rn fm z 1 − |z|2Re α zf z 1 − |z|2Re α z Rn f1 z
α
·
·
·
−
1
−
1
α
1
m
Rn f1 z
Rn fm z
Re α f z Re α
≤
1 − |z|2Re α α1 α2 · · · αm Re α
≤
1
1 − |z|2Re α
≤
≤ 1.
Re α
Re α
2.15
Using 2.12, the conditions in Lemma A are satisfied, hence Fz belongs to the
class S.
Remark 2.4. For αi ∈ R, i ∈ {1, 2, . . . , m}, then Theorem 2.3 can be rewritten as following.
Corollary 2.5. Let n, m ∈ N∪{0}, α ∈ C, with Re α > 0, let αi ∈ R, αi ≥ 0 with α1 α2 · · ·αm ≤ 1
and let fi ∈ A, i ∈ {1, 2, . . . , m}.
If
z Rn fi z
−
1
≤ 1,
Rn fi z
z ∈ U, i ∈ {1, 2, . . . , m},
2.16
then Fz given by 2.1 belongs to the class S, where Rn is the Ruscheweyh differential operator.
√
Example 2.6. Let n√∈ N ∪ {0}, m 2, α 2 3i, α1 1/4 i 3/4, α2 1/5 − i2/5,
|α1 | |α2 | 5 2 5/10 < 1, f1 z z az2 , Rn f1 z z n 1az2 , f2 z z bz2 ,
Rn f2 z z n 1bz2 , a, b ∈ C, with |a| ≤ 1/2n 1, |b| ≤ 1/2n 1, z ∈ U. Then
z z n 1bz2 1 2n 1bz
n 1bz ,
− 1 − 1 z 1 n 1bz
1 n 1bz
1 n 1bz z ∈ U.
2.17
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Journal of Inequalities and Applications
But
n 1bz2
n 1bz 2
1 n 1bz 1 n 1bz2
n 1bzn 1bz
1 n 1bz 1 n 1bz
≤
n 12 |b|2 |z|2
1 n 1bz n 1bz n 12 |b|2 |z|2
n 12 |b|2 |z|2
2.18
1 2n 1Re bz n 12 |b|2 |z|2
n 12 |b|2 |z|2
1 − 2n 1|b||z| n 12 |b|2 |z|2
n 12 |b|2 |z|2
1 − n 1|b||z|
2
n 1|b|2
n 1 · 1/4n 12
≤
2 ≤ 2
1 − n 1|b|
1 − n 1 · 1/2n 1
1
1/4n 12
≤ 1.
1/4
n 12
It implies that
n 1bz 1 n 1bz ≤ 1,
z ∈ U.
2.19
Similarly, we obtain
z z n 1az2 n 1az ≤ 1,
− 1 z 1 n 1az
1 n 1az z ∈ U.
2.20
Using Theorem 2.3, we have
z
1/23i
√
I f1 , f2 Fz 2 3i t13i 1 at1/4i 3/4 1 bt1/5−i2/5 dt
∈ S,
2.21
0
for all z ∈ U.
Theorem 2.7. Let n, m ∈ N ∪ {0}, α, β ∈ C, with Re β ≥ Re α > 0, let fi ∈ A, and let αi ∈ C,
i ∈ {1, 2, . . . , m}, with |α1 | |α2 | · · · |αm | ≤ 1.
If
z Rn fi z
−
1
≤ 1,
Rn fi z
z ∈ U, i ∈ {1, 2, . . . , m},
2.22
Georgia Irina Oros
7
where Rn is the Ruscheweyh differential operator, then the function given by
z
1/β
n
n
α1
R fm t αm
β−1 R f1 t
Fβ z β t
···
dt
t
t
0
2.23
belongs to the class S.
Proof. Using 2.14 and 2.22, from the proof of Theorem 2.3, we obtain
1 − |z|2Re α zf z ≤ 1,
Re α f z z ∈ U.
2.24
Using 2.12, the conditions from Lemma B are satisfied and by applying it we have
that the function Fβ z given by 2.23 belongs to the class S.
√
Example 2.8. Let n ∈ N ∪ {0}, m 2, β√ 4 − i, α 2 3i, Re β > Re α > 0, α1 1/4 i 3/4,
α2 1/5 − i2/5, |α1 | |α2 | 5 2 5/10 < 1, f1 z z az2 , f2 z z bz2 , Rn f1 z z n 1az2 , Rn f2 z z n 1bz2 , a, b ∈ C, |a| ≤ 1/2n 1, |b| ≤ 1/2n 1, z ∈ U. Then
n 1az z z n 1az2 ≤ 1,
− 1 z n 1az2
1 n 1az z z n 1bz2 n 1bz ≤ 1,
− 1 z1 n 1bz
1 n 1bz z ∈ U.
2.25
Using Theorem 2.7, we have
z
1/4−i
√
Fβ z 4 − i t3−i 1 at1/4i 3/4 1 bt1/5−i2/5 dt
∈ S,
2.26
0
for all z ∈ U.
Theorem 2.9. Let n, m ∈ N ∪ {0}, μ > 0, α ∈ C with Re α > 0, let fi ∈ A and let αi ∈ C,
i ∈ {1, 2, . . . , m} with |α1 | |α2 | · · · |αm | ≤ 1/2μ 1.
If
i Rn fi z ≤ μ,
z2 Rn fi z
ii −
1
≤ 1,
Rn f z 2
z ∈ U, i ∈ {1, 2, . . . , m},
2.27
i
where Rn is the Ruscheweyh differential operator, then the function Fz given by 2.1 belongs to the
class S.
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Journal of Inequalities and Applications
Proof. Using 2.14, we have
z Rn fm z zf z z Rn f1 z α1 α
·
·
·
−
1
−
1
,
m
f z Rn f1 z
Rn fm z
z ∈ U.
2.28
Using the conditions from Theorem 2.9 and 2.28 we calculate
1 − |z|2Re α zf z Re α f z 1 − |z|2Re α z Rn fm z
1 − |z|2Re α z Rn f1 z
·
·
·
α1 −
1
α
−
1
m
Rn f1 z
Rn fm z
Re α
Re α
n
z R f1 z 1 − |z|2Re α z Rn fm z αm 1 ··· 1
Rn f1 z Rn fm z Re α
1 − |z|2Re α ≤
α1
Re α
1 − |z|2Re α z Rn fm z 1 − |z|2Re α z Rn f1 z α1 αm ··· Rn f1 z Rn fm z Re α
Re α
1 − |z|2Re α α1 · · · αm Re α
1 − |z|2Re α z2 Rn fm z Rn fm z
1 − |z|2Re α z2 Rn f1 z Rn f1 z
α1 ··· αm Rn f z2 Rn f z2 Re α
|z|
Re α
|z|
1
m
1 − |z|2Re α α1 · · · αm Re α
1 − |z|2Re α z2 Rn fm z 1 − |z|2Re α z2 Rn f1 z ≤μ
··· Re α Rn f1 z2 Re α Rn fm z2 1 − |z|2Re α α1 · · · αm Re α
1 − |z|2Re α z2 Rn fm z 1 − |z|2Re α z2 Rn f1 z − 1 1 ·μ ··· − 1 1 ·μ
Rn f z2 Rn f z2 Re α
Re α
1
m
1 − |z|2Re α α1 · · · αm Re α
1 − |z|2Re α z2 Rn fm z
1 − |z|2Re α z2 Rn f1 z
·
μ
·
·
·
α1 α
≤
−
1
−
1
·μ
m
Rn f z2
Rn f z2
Re α
Re
α
1
m
1 − |z|2Re α 1 − |z|2Re α · α1 · · · αm · μ α1 · · · αm Re α
Re α
Georgia Irina Oros
9
≤
1 − |z|2Re α 1 − |z|2Re α α1 · · · αm 2μ · α1 · · · αm Re α
Re α
≤
1 − |z|2Re α α1 · · · αm 2μ 1
Re α
≤
1
1 − |z|2Re α
≤
≤ 1,
Re α
Re α
z ∈ U.
2.29
By using 2.12 and 2.29 and by applying Lemma A, we obtain that the function Fz
given by 2.1 belongs to the class S.
Theorem 2.10. Let n, m ∈ N ∪ {0}, μ > 0, α, β ∈ C with Re β ≥ Re α > 0, and let αi ∈ C,
i ∈ {1, 2, 3, . . . , m}, with |α1 | |α2 | · · · |αm | ≤ 1/2μ 1.
If
i Rn fi z ≤ μ,
z2 Rn f z
i
ii −
1
≤ 1,
Rn f z 2
z ∈ U, i ∈ {1, 2, . . . , m},
2.30
i
where Rn is the Ruscheweyh differential operator, then function Fβ z given by 2.23 belongs to the
class S.
Proof. By using 2.30 and 2.12 and from the proof of Theorem 2.9, we have
1 − |z|2Re α zf z < 1,
Re α f z z ∈ U,
2.31
and applying Lemma B we obtain that the Fβ z given by 2.23 belongs to the class S.
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