Journal of Inequalities in Pure and
Applied Mathematics
UNIVALENCE CONDITIONS FOR CERTAIN INTEGRAL
OPERATORS
volume 7, issue 4, article 147,
2006.
VIRGIL PESCAR
"Transilvania" University of Braşov
Faculty of Mathematics and Computer Science
Department of Mathematics
2200 Braşov
ROMANIA
EMail: virgilpescar@unitbv.ro
Received 19 June, 2006;
accepted 04 September, 2006.
Communicated by: H. Silverman
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
In this paper we consider some integral operators and we determine conditions
for the univalence of these integral operators.
2000 Mathematics Subject Classification: Primary 30C45.
Key words: Univalent function, Integral operator.
Univalence Conditions for
Certain Integral Operators
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Preliminary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
5
6
Virgil Pescar
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1.
Introduction
Let U = {z ∈ C : |z| < 1} be the unit disc in the complex plane. The class
A and the class S are defined in [2]: let A be the class of functions f (z) =
z + a2 z 2 + · · · , which are analytic in the unit disk normalized with f (0) =
f 0 (0) − 1 = 0; let S the class of the functions f ∈ A which are univalent in U.
In [7] is defined the class S(α). For 0 < α ≤ 2, let S(α) denote the class of
functions f ∈ A which satisfy the conditions:
(1.1)
f (z) 6= 0
Univalence Conditions for
Certain Integral Operators
for 0 < |z| < 1
Virgil Pescar
and
(1.2)
00 z
f (z) ≤ α
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for all z ∈ U.
In [7] is proved the next result. For 0 < α ≤ 2, the functions f ∈ S(α) are
univalent.
In this work, we consider the integral operators
(1.3)
Z
Gα (z) = α
z
g α−1 (u) du
α1
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0
and
(1.4)
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Z
Hα, γ (z) = α
0
z
u
α−1
h(u)
u
α1
γ
du
J. Ineq. Pure and Appl. Math. 7(4) Art. 147, 2006
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for g(z) ∈ S, h(z) ∈ S and for some α, γ ∈ C.
Kim - Merkes [1] studied the integral operator
γ
Z z
h(u)
(1.5)
Fγ (z) =
du
u
0
and obtained the following result
Theorem 1.1. If the function h(z) belongs to the class S, then for any complex
number γ, |γ| ≤ 41 , the function Fγ (z) defined by (1.5) is in the class S.
Univalence Conditions for
Certain Integral Operators
Virgil Pescar
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2.
Preliminary Results
In order to prove our main results we will use the lemma due to N.N. Pascu [4]
presented in this section.
Lemma 2.1. Let the function f ∈ A and α a complex number, Re α > 0. If
1 − |z|2 Re α zf 00 (z) (2.1)
f 0 (z) ≤ 1,
Re α
for all z ∈ U , then for all complex numbers β, Re β ≥ Re α the function
(2.2)
Z
Fβ (z) = β
0
is regular and univalent in U .
z
uβ−1 f 0 (u)du
β1
Univalence Conditions for
Certain Integral Operators
Virgil Pescar
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3.
Main Results
Theorem 3.1. Let α be a complex number, Re α ≥ 0 and the function g ∈
S, g(z) = z + a2 z 2 + · · · . If
(j1 )
|α − 1| ≤
Re α
4
for Re α ∈ (0, 1)
or
(j2 )
|α − 1| ≤
1
4
Univalence Conditions for
Certain Integral Operators
for Re α ∈ [1, ∞),
Virgil Pescar
then the function
(3.1)
Z
Gα (z) = α
α1
z
g
α−1
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(u) du
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is in the class S.
Proof. From (3.1) we have
(3.2)
" Z
Gα (z) = α
0
z
uα−1
g(u)
u
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# α1
α−1
du
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.
The function g(z) is regular and univalent, hence g(z)
6= 0 for all z ∈ U. We
h izα−1
can choose the regular branch of the function g(z)
to be equal to 1 at the
z
origin.
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Let us consider the regular function in U , given by
α−1
Z z
g(u)
du.
(3.3)
p(z) =
u
0
Because g ∈ S, we obtain
0 z g (z) 1 + |z|
g(z) ≤ 1 − |z|
(3.4)
Univalence Conditions for
Certain Integral Operators
for all z ∈ U.
We have
(3.5)
Virgil Pescar
1 − |z|2 Re α
Re α
00 z p (z) 1 − |z|2 Re α z g 0 (z)
=
−
1
p0 (z) g(z)
Re α
zg 0 (z) 1 − |z|2 Re α
|α − 1| +1 .
≤
Re α
g(z) From (3.5) and (3.4) we obtain
1 − |z|2 Re α z p00 (z) 1 − |z|2 Re α
2
(3.6)
≤
|α − 1|
.
0
Re α
p (z)
Re α
1 − |z|
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Now, we consider the cases
i1 ) 0 < Re α < 1.
The function
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s : (0, 1) → <,
s(x) = 1 − a2x
(0 < a < 1)
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is a increasing function and for a = |z|, z ∈ U, we obtain
(3.7)
1 − |z|2 Re α ≤ 1 − |z|2
for all z ∈ U.
From (3.6) and (3.7), we have
(3.8)
1 − |z|2 Re α
Re α
00 zp (z) 4 |α − 1|
p0 (z) ≤ Re α
for all z ∈ U.
Using the condition (j1 ) and (3.8) we get
1 − |z|2 Re α zp00 (z) (3.9)
p0 (z) ≤ 1
Re α
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q(x) =
1 − a2x
(0 < a < 1)
x
is a decreasing function, and that, if we take a = |z|, z ∈ U, then
(3.10)
Virgil Pescar
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for all z ∈ U.
i2 ) Re α ≥ 1.
We observe that the function
q : [1, ∞) → <,
Univalence Conditions for
Certain Integral Operators
1 − |z|2 Re α
≤ 1 − |z|2
Re α
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J. Ineq. Pure and Appl. Math. 7(4) Art. 147, 2006
for all z ∈ U.
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From (3.6) and (3.10) we obtain
1 − |z|2 Re α zp00 (z) (3.11)
p0 (z) ≤ 4 |α − 1|.
Re α
From (3.11) and (j2 ), we have
(3.12)
1 − |z|2 Re α
Re α
00 zp (z) p0 (z) ≤ 1
for all z ∈ U.
α−1
Using (3.9), (3.12) and because p0 (z) = g(z)
, from Lemma 2.1 for
z
α = β it results that the function Gα (z) is in the class S.
Theorem 3.2. If α is a real number, α ∈ 45 , 54 and the function g ∈ S(α),
then the function
Z z
α1
α−1
(3.13)
Gα (z) = α
g (u) du
0
is in the class S.
Proof. If g ∈ S(α), then g ∈ S and by Theorem 3.1 for α ∈
the function Gα (z) in the class S.
4
5
,
5
4
, we obtain
Theorem 3.3. Let α, γ be a complex numbers and the function h ∈ S, h(z) =
z + a2 z 2 + · · · .
If
(p1 )
|γ| ≤
Re α
4
for
Re α ∈ (0, 1)
Univalence Conditions for
Certain Integral Operators
Virgil Pescar
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or
|γ| ≤
(p2 )
1
4
Re α ∈ [1, ∞)
for
then the function
Z
Hα, γ (z) = α
(3.14)
0
z
u
α−1
h(u)
u
α1
γ
du
is regular and univalent in U.
Proof. Let us consider the regular function in U, defined by
γ
Z z
h(u)
(3.15)
f (z) =
du.
u
0
For the function h ∈ S, we obtain
0 z h (z) 1 + |z|
(3.16)
h(z) ≤ 1 − |z|
for all z ∈ U.
We obtain
(3.17)
1 − |z|2 Re α
Re α
Univalence Conditions for
Certain Integral Operators
Virgil Pescar
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00 0 z f (z) 1 − |z|2 Re α
zh (z) ≤
|γ| +1 .
f 0 (z) Re α
h(z) From (3.17) and (3.16), we have
1 − |z|2 Re α z f 00 (z) 1 − |z|2 Re α
2
(3.18)
≤
|γ|
0
Re α
f (z)
Re α
1 − |z|
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We consider the cases
j1 ) 0 < Re α < 1.
In this case we obtain
1 − |z|2 Re α ≤ 1 − |z|2
(3.19)
for all z ∈ U.
From (3.18) and (3.19), we get
(3.20)
1 − |z|2 Re α
Re α
00 zf (z) 4 |γ|
f 0 (z) ≤ Re α
Virgil Pescar
for all z ∈ U.
By (3.20) and (p1 ) we have
1 − |z|2 Re α
Re α
(3.21)
Univalence Conditions for
Certain Integral Operators
00 zf (z) f 0 (z) ≤ 1
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for all z ∈ U.
j2 ) Re α ≥ 1.
For this case we obtain
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2 Re α
(3.22)
1 − |z|
Re α
≤ 1 − |z|2
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for all z ∈ U.
From (3.18) and (3.22) we have
(3.23)
1 − |z|2 Re α
Re α
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00 zf (z) f 0 (z) ≤ 4 |γ|.
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From (3.23) and (p2 ), we get
(3.24)
1 − |z|2 Re α zf 00 (z) f 0 (z) ≤ 1
Re α
for all z ∈ U.
γ
h(z)
0
From (3.21), (3.24) and because f (z) = z
, from Lemma 2.1 for α =
β it results that the function Hα,γ (z) is in the class S.
Remark 1. For α = 1, from Theorem 3.3 we obtain Theorem 1.1, the result due
to Kim-Merkes.
Univalence Conditions for
Certain Integral Operators
Virgil Pescar
Theorem 3.4. Let γ be a complex number and the function h ∈ S(a).
If
(3.25)
|γ| ≤
α
4
for
α ∈ (0, 1)
|γ| ≤
1
4
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or
(3.26)
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for
α ∈ [1, 2]
then the function Hα, γ (z) defined by (3.14) is in the class S.
Proof. Because h(z) ∈ S(α), 0 < α ≤ 2, then h(z) ∈ S and by Theorem 3.3
the function Hα, γ (z) belongs to the class S.
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References
[1] Y.J. KIM AND E.P. MERKES, On an integral of powers of a spirallike function, Kyungpook Math. J., 12(2) (1972), 249–253.
[2] P.T. MOCANU, T. BULBOACĂ AND G.St. SĂLĂGEAN, The Geometric
Theory of Univalent Functions, Cluj, 1999.
[3] Z. NEHARI, Conformal Mapping, Mc Graw-Hill Book Comp., New York,
1952 (Dover. Publ. Inc., 1975).
[4] N.N. PASCU, An improvement of Becker’s univalence criterion, Proceedings of the Commemorative Session Simion Stoilov, Braşov, (1987), 43–48.
[5] V. PESCAR, New univalence criteria, "Transilvania" University of Braşov,
Braşov, 2002.
[6] C. POMMERENKE, Univalent Functions, Vanderhoeck Ruprecht, Gőttingen, 1975.
[7] D. YANG AND J. LIU, On a class of univalent functions, IJMMS, 22(3)
(1999), 605–610.
Univalence Conditions for
Certain Integral Operators
Virgil Pescar
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