Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2008, Article ID 127645, 7 pages
doi:10.1155/2008/127645
Research Article
Sufficient Conditions for Univalence of
an Integral Operator
Georgia Irina Oros,1 Gheorghe Oros,1 and Daniel Breaz2
1
2
Department of Mathematics, University of Oradea, str. Universitatii nr. 1, 410087 Oradea, Romania
Department of Mathematics and Computer Science, 1 Decembrie 1918 University of Alba Iulia,
str. N. Iorga, no. 11–13, 510009 Alba Iulia, Romania
Correspondence should be addressed to Daniel Breaz, dbreaz@uab.ro
Received 27 February 2008; Accepted 24 May 2008
Recommended by Ulrich Abel
In this paper we have introduced an integral general operator. For this general operator which
is a generalization of more known integral operators we have demonstrated some univalence
properties.
Copyright q 2008 Georgia Irina Oros et al. 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 be the unit disk of the complex plane:
U z ∈ C : |z| < 1 .
Let HU be the space of holomorphic functions in U,
An f ∈ HU, fz z an1 zn1 · · · , z ∈ U
with A1 A, and
S f ∈ A : f is univalent in U .
1.1
1.2
1.3
Lemma 1.1 see 1. If the function f is regular in the unit disc U,
fz z a2 z2 · · · ,
zf z ≤ 1 ∀z ∈ U,
1 − |z|2 f z then the function f is univalent in U.
1.4
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Journal of Inequalities and Applications
Definition 1.2 St. Ruscheweyh 2. For f ∈ A, n ∈ N ∪ {0}, let Rn be the operator defined by
Rn : A → A,
R0 fz fz,
R1 fz zf z
..
.
n1
n 1R fz z Rn fz nRn fz,
1.5
z ∈ U.
Remark 1.3. If f ∈ A
fz z ∞
1.6
aj zj ,
j2
then
Rn fz z ∞
j1
n
Cnj−1
aj zj ,
z ∈ U,
1.7
with
Rn f0 0,
Rn f0 1.
1.8
Lemma 1.4 3, Schwarz’s lemma, 4, Lemma 4.26, page 103. If the analytic function fz is
regular in U with f0 0 and |fz| < 1 for all z ∈ U, then
fz ≤ |z|,
∀z ∈ U,
1.9
and |f 0| ≤ 1.
The equality holds if and only if fz cz, z ∈ U, |c| 1.
2. Main results
By using the Ruscheweyh differential operator given by Definition 1.2, we introduce the
following integral operator.
Definition 2.1. Let n, m ∈ N ∪ {0}, i ∈ {1, 2, 3, . . . , m}, αi ∈ C. Define the integral operator
If1 , f2 , . . . , fm : Am → A,
I f1 , f2 , . . . , fm z z 0
Rn f1 t
t
α1
Rn fm t
···
t
αm
dt,
z ∈ U,
2.1
where fi z ∈ A and Rn is the Ruscheweyh differential operator.
Remark 2.2. i For n 0, m 1, α1 1, α2 α3 · · · αm 0,
R0 fz fz ∈ A,
2.2
Georgia Irina Oros et al.
3
we obtain Alexander integral operator introduced in 1915 in 5:
Iz z
ft
dt,
t
0
z ∈ U.
2.3
ii For n 0, m 1, α1 α ∈ 0, 1, α2 α3 · · · αm 0, R0 fz fz ∈ S, and we
obtain the integral operator
Iα z z 0
ft
t
α
2.4
dt
studied in 6.
iii For n 1, m 1, α1 γ ∈ C, |γ| ≤ 1/4, α2 · · · αm 0, R1 fz zf z ∈ S, we
obtain the integral operator
Fγ z z
γ
f t dt
2.5
0
studied in 7, 8.
iv For n 0, m ∈ N ∪ {0}, αi ∈ C, i ∈ {1, 2, . . . , m}, R0 fz fz ∈ S, and we obtain
the integral operator
Fz z 0
f1 t
t
α1
···
fm t
t
αm
2.6
dt
studied in 9.
v For n, m ∈ N ∪ {0}, i ∈ {1, 2, . . . , m}, αi > 0, we obtain the integral operator Fm :
Am → A,
Fm f1 , f2 , . . . , fm z z 0
Rn f1 t
t
α1
···
Rn fm t
t
αm
dt
2.7
studied in 10.
vi For n 0, m 1, α1 γ, α2 · · · αm 0, R0 fz fz, and we obtain the
integral operator
Fγ z z 0
ft
t
γ
2.8
dt
studied in 11, 12.
Theorem 2.3. Let n, m ∈ N ∪ {0}, i ∈ {1, 2, . . . , m}, αi ∈ C, fi ∈ A. If
n
z R fi z ≤ 1,
−
1
Rn f z
i
α1 α2 · · · αm ≤ 1,
then If1 , f2 , . . . , fm z given by 2.1 is univalent.
z ∈ U,
2.9
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Journal of Inequalities and Applications
Proof. Since fi ∈ A, i ∈ {1, 2, . . . , m}, from Remark 1.3 we have
∞ n
j
∞
Rn fi z z j2 Cnj−1 aj,i z
n
aj,i zj−1 ,
1 Cnj−1
z
z
j2
Rn fi z
0,
/
z
2.10
z ∈ U.
For z 0, we have
Rn f1 z
z
α1
Rn fm z
···
z
αm
1.
2.11
By differentiating 2.1, we obtain
n
n
R fm z αm
R f1 z α1
···
,
I f1 , f2 , . . . , fm z z
z
I f1 , f2 , . . . , fm 0 1.
z ∈ U,
2.12
Using 2.12, we obtain
log I f1 , f2 , . . . , fm z α1 log Rn f1 z − log z · · · αm log Rn fm z − log z ,
z ∈ U.
2.13
By differentiating 2.13, we have
n
n
R f1 z
R fm z
I f1 , f2 , . . . , fm z
1
1
−
·
·
·
α
−
,
α
1
m
Rn f1 z
z
Rn fm z
z
I f1 , f2 , . . . , fm z
z∈U
and after a short calculus we obtain
z Rn fm z zI f1 , f2 , . . . , fm z z Rn f1 z
− 1 · · · αm
−1 ,
α1
Rn f1 z
Rn fm z
I f1 , f2 , . . . , fm z
2.14
z ∈ U.
2.15
We multiply the modulus of 2.15 by 1 − |z|2 and we obtain
zI f1 , f2 , . . . , fm z 2 1 − |z| I f1 , f2 , . . . , fm z n
n
z R f1 z
z R fm z
1 − |z| α1
−
1
·
·
·
α
−
1
m
n
n
R f z
R f z
2
1
m
z Rn fm z
z Rn f1 z
2 α1 − 1 · · · αm − 1
≤ 1 − |z|
n
n
R f1 z
R fm z
≤ α1 · · · αm 1 − z2 ≤ α1 · · · αm ≤ 1.
From Lemma A, we have If1 , f2 , . . . , fm z ∈ S.
2.16
Georgia Irina Oros et al.
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Remark 2.4. i For n 0, Rn fi z fi z ∈ S, we obtain Theorem 2.3 from 9.
ii For αi ∈ R, αi > 0, Theorem 2.3 can be rewritten as follows.
Corollary 2.5. Let n, m ∈ N ∪ {0}, i ∈ {1, 2, . . . , m}, αi > 0 with α1 α2 · · · αm ≤ 1. If fi ∈ A
satisfy
n
z R fi z ≤ 1,
−
1
Rn f z
i
z ∈ U,
2.17
then the integral operator given by 2.1 is univalent.
Theorem 2.6. Let n, m ∈ N ∪ {0}, i ∈ {1, 2, . . . , m}, αi ∈ C. If fi ∈ A satisfy
i |α1 | · · · |αm | ≤ 1/3,
ii |Rn fi z| ≤ 1,
iii |z2 Rn fi z /Rni fi z2 − 1| < 1
for all z ∈ U, then the integral operator given by 2.1 is univalent.
Proof. Using 2.14, we obtain
z I f1 , . . . , fm z zRn f1 z
z Rn fm z − 1 . . . αm − 1 .
α1 n
n
R f1 z
R fm z
I f1 , . . . , fm z We multiply 2.18 by 1 − |z|2 , use Schwarz’s lemma, and obtain
zT z 1 − |z| T z 2
zRn f1 z
z Rn fm z
2 1 − |z| α1 − 1 · · · 1 − |z| αm − 1
Rn f1 z
Rn f1 z
2
z Rn f1 z z Rn fm z 2 2 1 − |z| α1 1 − |z| α1 · · · 1 − |z| αm Rn f z Rn f z 2
1
1
1 − |z|2 αm n
z R fm z z Rn f1 z 1 − |z|2 α1 · · · αm 1 − |z| α1 ··· n
n
R f z
R f z
2
1
1
2 z2 Rn f1 z Rn f1 z2 Rn fm z Rn fm 1 − z α1 2 |z| · · · αm n
2 |z|
Rn f1 z
R fm z
1 − |z|2 α1 · · · αm 2.18
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Journal of Inequalities and Applications
z2 Rn fm z z2 Rn f1 z 2 1 − |z|2 α1 · · · αm · · · αm α1 ≤ 1 − |z|
2
2
Rn f1 z
Rn fm z
z2 Rn f1 z − α1 α1 α1 1 − |z|
2 Rn f1 z
2
z2 Rn fm z 2 −
· · · 1 − |z|
α
αm 1 − |z|2 α1 · · · αm α
m
m
2
Rn fm z
z2 Rn fm z z2 Rn f1 z
·
·
·
−
1
α
−
1
α1 1 − |z|
m
2
2
Rn f1 z
Rn fm z
1 − |z|2 α1 · · · αm 1 − |z|2 α1 · · · αm 2
≤ 1 − |z|2 α1 α1 · · · αm 2 1 − |z|2 α1 · · · αm 3 1 − |z|2 α1 · · · αm ≤ 3 α1 · · · αm .
2.19
From 2.19 and condition i, we have
zF z ≤1
1 − |z|2 F z 2.20
for all z ∈ U.
By Lemma A, it follows that the integral operator If1 , f2 , . . . , fm z is univalent.
Remark 2.7. For n 0, m 1, α1 α ∈ C, |α| ≤ 1/3, α2 · · · αm 0, the result was obtained
in 11, Theorem 1.
For αi ∈ R, αi > 0, Theorem 2.6 can be rewritten as follows.
Corollary 2.8. Let n, m ∈ N ∪ {0}, i ∈ {1, 2, . . . , m}, αi > 0. If fi ∈ A satisfy
i α1 α2 · · · αn ≤ 1/3,
ii |Rn fi z| ≤ 1,
iii |z2 Rn fi z /Rn fi z2 − 1| < 1
for all z ∈ U, then the integral operator given by 2.1 is univalent.
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