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Abstract and Applied Analysis
Volume 2012, Article ID 140924, 10 pages
doi:10.1155/2012/140924
Research Article
Univalence Conditions Related to a General
Integral Operator
Nicoleta Breaz1 and Virgil Pescar2
1
Department of Mathematics, Faculty of Science, “1 Decembrie 1918” University of Alba Iulia,
Nicolae Iorga Street No. 11-13, 510009 Alba Iulia, Romania
2
Department of Mathematics, Faculty of Mathematics and Computer Science,
Transilvania University of Braşov, 500091 Braşov, Romania
Correspondence should be addressed to Nicoleta Breaz, nicoletabreaz@yahoo.com
Received 28 June 2012; Revised 29 October 2012; Accepted 31 October 2012
Academic Editor: Naseer Shahzad
Copyright q 2012 N. Breaz and V. Pescar. 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.
We consider a general integral operator based on two types of analytic functions, namely, regular
functions and, respectively, functions having a positive real part. Some univalence conditions for
this integral operator are obtained.
1. Introduction
Let A be the class of functions of the form
fz z ∞
ai zi ,
1.1
i2
which are analytic in the open unit disk, U {z ∈ C : |z| < 1} and normalized by f0 f 0 − 1 0. Also, let S denote the subclass of A consisting of univalent functions, in the
open unit disk. We denote by P, the class of functions p which are analytic in U, Re pz > 0,
for all z ∈ U and p0 1, the so-called Caratheodory functions.
In 1, Pescar introduced and studied the following integral operator:
αj
z n
fj u
β
gj u j du,
En z 1.2
u
0 j1
where αj , βj are complex numbers, fj ∈ A, gj ∈ P, j 1, n.
2
Abstract and Applied Analysis
In this paper, we generalize this integral operator, by considering the general integral
operator defined as follows:
⎛
Kn z ⎝δ
z
0
uδ−1
n
j1
fj u
u
αj
gj u
β j
⎞1/δ
du⎠
,
1.3
0, fj ∈ A, gj ∈ P, j 1, n.
where δ, αj , βj are complex numbers, δ /
Remarks. This integral operator extends many other integral operators from related works
on this field as those given by Srivastava, Mocanu, Owa, Pescar, Orhan, Breaz, and others.
Using this new operator and the related univalence conditions that are going to be proved
here, one can study some other already known operators in a unified perspective. Thus, for
different particular cases of the parameters δ, αj , βj , our integral operator Kn is shown to be
an extension of the following integral operators.
i For δ 1, the integral operator Kn is the operator En from 1.2, introduced by
Pescar in 1.
ii For n 1, αj δ, βj 0, the integral operator Kn was studied in 2, by Miller and
Mocanu.
iii For βj 0, j 1, n, the integral operator Kn was studied in 3, by Pescar and Breaz.
iv For βj 0, j 1, n, αj α − 1, and δ nα − 1 1, the integral operator Kn was
studied in 4, by Breaz et al. and also in 5, by Srivastava et al.
v For δ 1, βj 0, j 1, n, we get the integral operator defined by D. Breaz and N.
Breaz, in the paper 6.
vi For δ 1, αj 0, βj > 0, and gj fj , the operator was studied by Breaz et al., in the
paper 7.
More precisely, if we are interested to study various properties of two or more different
operators reminded above and other similar operators which are not mentioned here, we
can do this in an integrated manner by simply allowing the parameters involved in the
definition, to be more general and consequently by studying only one operator, having the
form 1.3. In this paper, we will study some univalence criteria for this new operator.
The following known results will be used in order to prove our results.
Lemma 1.1 see 8. Let γ be a complex number, Re γ > 0 and f ∈ A. If
1 − |z|2 Re γ zf z ≤ 1,
Re γ f z 1.4
for all z ∈ U, then for any complex number δ, Re δ ≥ Re γ,
z
1/δ
δ−1 Fδ z δ u f udu
∈ S.
0
1.5
Abstract and Applied Analysis
3
Lemma 1.2 see 9. Let f be a regular function in the open disk UR {z ∈ C : |z| < R} with
|fz| < M, M fixed. If f has in z 0 one zero with multiplicity ≥ m, then
fz ≤ M |z|m ,
Rm
z ∈ UR ,
1.6
the equality for (z /
0) can hold only if
fz eiθ
M m
z ,
Rm
1.7
where θ is constant.
Lemma 1.3 see 10. For each f ∈ S,
1 − r zf z 1 r
≤
≤
,
1 r fz 1 − r
Lemma 1.4 see 11. If gz 1 ∞
n1
|z| r < 1.
1.8
bn zn is regular in |z| < 1 and Regz > 0, then
zg z gz ≤
2|z|
1 − |z|2
.
1.9
Also, for the statement of our main results, we need to define the following classes:
AM zf z
− 1 ≤ M, M ≥ 1 ,
f ∈ A : fz
PL zf z ≤ L, L > 0 .
f ∈ P : fz 1.10
2. Main Results
Theorem 2.1. Let δ, αj , βj be complex numbers, Re δ ≥ 1, Mj , Lj positive real numbers, Mj ≥ 1,
j 1, n, and the functions fj ∈ AMj , gj ∈ PLj , j 1, n.
If
√
n αj Mj βj Lj ≤ 3 3 ,
2
j1
then the general integral operator Kn , defined by 1.3, is in the class S.
2.1
4
Abstract and Applied Analysis
Proof. We consider the regular function
hn z z 0
f1 u
u
α1
···
fn u
u
αn
β
β
· g1 u 1 · · · gn u n du.
2.2
After some calculus, we have
n zf z
zgj z zh z j
n
2
≤ 1 − |z|
αj 1 − |z| − 1 βj ,
f
g
z
z
hn z j
j
j1
2
2.3
for all z ∈ U.
Since fj ∈ AMj , by applying Lemma 1.2 for m 1, we get
zfj z
− 1 ≤ Mj |z|,
fj z
j 1, n; z ∈ U .
2.4
Also since gj ∈ PLj , from Lemma 1.2, we have
zgj z ≤ Lj |z|,
gj z j 1, n; z ∈ U .
2.5
If we put these last two inequalities in 2.3, together with the inequality from the
hypothesis, we get
√
zh z 3
3
n
2
≤ 1 − |z| |z| ·
.
1 − |z| 2
hn z
2
2.6
Now we take into account the fact that
max
|z|<1
2
1 − |z|2 |z| √ ,
3 3
2.7
thus obtaining
zh z ≤ 1,
1 − |z|2 n
h z z ∈ U.
2.8
n
Further from Lemma 1.1, for Re δ ≥ Re γ 1, it is obvious that Kn ∈ S.
Corollary 2.2. Let δ, αj be complex numbers, Re δ ≥ 1, Mj positive real numbers, Mj ≥ 1, fj ∈
AMj , j 1, n. If
√
n αj Mj ≤ 3 3
2
j1
2.9
Abstract and Applied Analysis
5
then the function
⎡
z
Kn z ⎣δ
uδ−1
0
n
j1
fj u
u
αj
⎤1/δ
du⎦
,
2.10
is in the class S.
Proof. In Theorem 2.1, we take βj 0, j 1, n.
Corollary 2.3. Let δ, βj be complex numbers, Re δ ≥ 1, Lj positive real numbers and gj ∈ PLj ,
j 1, n. If
√
n βj Lj ≤ 3 3 ,
2
j1
2.11
then the function
⎡
Kn z ⎣δ
z
u
δ−1
0
n
gj u
β j
⎤1/δ
du⎦
2.12
j1
belongs to the class S.
Proof. In Theorem 2.1, we take αj 0, j 1, n.
Theorem 2.4. Let γ, δ, αj , βj be complex numbers, j 1, n, Re δ ≥ Re γ > 0, and the functions
fj ∈ Aμ , gj ∈ Pμ , μ ≥ 1, with
11/2 Re γ
2 Re γ 1
.
μ
2
2.13
If
n αj βj ≤ 1,
2.14
j1
then the general integral operator Kn belongs to the class S.
Proof. Let consider again the regular function:
hn z z 0
f1 u
u
α1
···
fn u
u
αn
β
β
· g1 u 1 · · · gn u n du.
2.15
6
Abstract and Applied Analysis
After some calculus, we have
n zf z
zgj z 1 − |z|2 Re γ zhn z 1 − |z|2 Re γ j
αj ≤
− 1 βj ,
Re γ hn z Re γ
f
g
z
z
j
j
j1
z ∈ U.
2.16
Since fj ∈ Aμ and gj ∈ Pμ , for all j 1, n, from Lemma 1.2, we obtain
zfj z
j 1, n; z ∈ U ,
− 1 ≤ μ|z|,
fj z
zgj z j 1, n; z ∈ U .
≤ μ|z|,
gj z 2.17
Using these last two inequalities, the inequality from the hypothesis, and the fact that
⎤
1 − |z|2 Re γ |z|
⎥ 1
⎢
max⎣
⎦ ,
Re γ
μ
|z|<1
⎡
2.18
from 2.16 we get
1 − |z|2 Re γ zhn z ≤ 1,
Re γ hn z 2.19
for all z ∈ U.
Hence, by Lemma 1.1, we have that Kn ∈ S.
Corollary 2.5. Let γ, δ, αj be complex numbers, Re δ ≥ Re γ > 0, and fj ∈ Aμ , j 1, n, μ ≥ 1, with
11/2 Re γ
2 Re γ 1
.
μ
2
2.20
If
n αj ≤ 1,
2.21
j1
then the integral operator defined by
⎡
Kn z ⎣δ
z
0
is in the class S.
uδ−1
n
j1
fj u
u
αj
⎤1/δ
du⎦
,
2.22
Abstract and Applied Analysis
7
Proof. In Theorem 2.4, we take β1 · · · βn 0.
Corollary 2.6. Let γ, δ, βj be complex numbers, j 1, n, Re δ ≥ Re γ > 0, and gj ∈ Pμ , with
11/2 Re γ
2 Re γ 1
μ
.
2
2.23
If
n βj ≤ 1,
2.24
j1
then the integral operator defined by
⎡
Kn z ⎣δ
z
uδ−1
0
n
gj u
βj
⎤1/δ
du⎦
,
2.25
j1
is in the class S.
Proof. We put in Theorem 2.4, α1 · · · αn 0.
Theorem 2.7. Let γ, δ, αj , βj be complex numbers, j 1, n, Re δ ≥ Re γ > 0 and fj ∈ S, gj ∈ P. If
n n Re γ 1
,
,
2 αj βj ≤ min
2 2
j1
j1
2.26
then the general integral operator Kn ∈ S.
Proof. We consider the same regular function hn as in the proof of the previous theorems, and
after some calculus we get
n zf z zgj z 1 − |z|2 Re γ zhn z 1 − |z|2 Re γ j
αj ≤
1 βj ,
fj z gj z Re γ hn z Re γ
j1
2.27
for all z ∈ U.
Since fj ∈ S, from Lemma 1.3, we get
zfj z 1 |z|
,
≤
fj z 1 − |z|
2.28
8
Abstract and Applied Analysis
j 1, n, z ∈ U, and since gj ∈ P , from Lemma 1.4, we get
zgj z 2|z|
≤
gj z 1 − |z|2
2.29
j 1, n, z ∈ U.
Now if we use 2.28 and 2.29 in 2.27, we obtain
⎤
⎡
n n 2|z|
1 − |z|2 Re γ zhn z 1 − |z|2 Re γ ⎣ 2 αj βj ⎦,
≤
Re γ hn z Re γ
1 − |z| j1
1 − |z|2 j1
2.30
for all z ∈ U.
We consider two cases.
1 If min{Re γ/2, 1/2} Re γ/2, we have
1 − |z|2 Re γ ≤ 1 − |z|2
2.31
and further if we use this inequality together with the inequality from the hypothesis, from
2.30, we get
1 − |z|2 Re γ zhn z ≤ 1,
Re γ hn z 2.32
for all z ∈ U, and with Lemma 1.1 the proof is complete.
2 If min{Re γ/2, 1/2} 1/2, we obtain
1 − |z|2 Re γ
≤ 1 − |z|2 ,
Re γ
2.33
z ∈ U and further if we put this last inequality in 2.30 we get
n n 1 − |z|2 Re γ zhn z αj 2 βj ,
≤
4
Re γ hn z j1
j1
2.34
for all z ∈ U.
Now by applying first the inequality condition from the hypothesis and then
Lemma 1.1 for hn , the proof is complete.
Corollary 2.8. Let γ, δ, αj be complex numbers, j 1, n, Re δ ≥ Re γ > 0 and fj ∈ S. If
n αj ≤ min Re γ , 1 ,
4 4
j1
2.35
Abstract and Applied Analysis
9
then the integral operator
⎡
Kn z ⎣δ
z
u
δ−1
0
n
j1
fj u
u
αj
⎤1/δ
du⎦
2.36
is in the class S.
Proof. In Theorem 2.7, we take β1 · · · βn 0.
Corollary 2.9. Let γ, δ, βj be complex numbers, j 1, n, Re δ ≥ Re γ > 0 and gj ∈ P. If
n βj ≤ min Re γ , 1
2 2
j1
2.37
then the integral operator
⎡
Kn z ⎣δ
z
u
0
δ−1
n
gj u
β j
⎤1/δ
du⎦
2.38
j1
is in the class S.
Proof. In Theorem 2.7 we take α1 · · · αn 0.
Acknowledgment
The authors wish to thank the reviewers for their constructive comments which helped them
to improve the quality of the paper.
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Abstract and Applied Analysis
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