General Mathematics ,Vol. 16, Nr. 1 (2008), 11–17
On some integral classes of integral
operators1
Virgil Pescar
Abstract
Let A be the class of the functions f which are analytic in the unit
disk U = {z ∈ C; |z| < 1} and f (0) = f ′ (0) − 1 = 0. The object of
the present paper is to derive univalence conditions of certain integral
∞
X
ak z k .
operators for f (z) ∈ A and f (z) has the form: f (z) = z +
k=3
2000 Mathematics Subject Classification: 30C45.
Key words: Univalent, integral operator.
1
Introduction
Let A be the class of the functions f (z) which are analytic in the unit disk
U = {z ∈ C : |z| < 1} and f (0) = f ′ (0) − 1 = 0.
We denote by S the class of the functions f (z) ∈ A which are univalent
in U.
In this paper we consider the integral operators
(1.1)
Fα (z) =
Z
z
α
[f ′ (u)] du
0
1
Received 29 August 2007
Accepted for publication (in revised form) 4 January 2008
11
12
(1.2)
Virgil Pescar
½ Z
Hβ,γ (z) = β
z
· Z
Lβ (z) = β
z
β−1
¾ β1
[f (u)] du
β−1
¸ β1
[f (u)] du
u
0
(1.3)
u
0
2
′
γ
′
Preliminary Results
We need the following theorems.
Lemma 2.1. [1]. If f (z) ∈ A satisfies
¯
¯
¡
¢ ¯ ′′ ¯
2 ¯ z f (z) ¯
(2.1)
1 − |z| ¯ ′
≤ 1, z ∈ U
f (z) ¯
then f (z) ∈ S.
Theorem 2.2. [3]. Let α be a complex number, Re α > 0 and f (z) ∈ A. If
(2.2)
1 − |z|2Re α
Re α
¯ ′′ ¯
¯ zf (z) ¯
¯
¯
¯ f ′ (z) ¯ ≤ 1,
for all z ∈ U , then for any complex number β, Re β ≥ Re α the function
(2.3)
· Z
Fβ (z) = β
0
z
u
β−1
¸ β1
f (u)du
′
is in the class S.
Theorem 2.3. [2 ]. If the function g(z) is regular in U and |g(z)| < 1 in
U, then for all ξ ∈ U and z ∈ U the following inequalities hold:
¯
¯
¯ g(ξ) − g(z) ¯ ¯¯ ξ − z ¯¯
¯ ¯
¯
¯,
(2.4)
¯≤
¯
¯ 1 − g(z)g(ξ) ¯ ¯ 1 − zξ ¯
(2.5)
1 − |g(z)|2
,
|g (z)| ≤
1 − |z|2
′
On some integral classes of integral operators
13
, where |ǫ| = 1 and |u| < 1.
the equalities hold only in the case g(z) = ǫ(z+u)
1+uz
Remark 2.4. [2] For z = 0, from inequality (2.4). We have
¯
¯
¯ g(ξ) − g(0) ¯
¯
¯
(2.6)
¯ ≤ |ξ|
¯
¯ 1 − g(0)g(ξ) ¯
and, hence
(2.7)
|g(ξ)| ≤
|ξ| + |g(0)|
.
1 + |g(0)| |ξ|
Considering g(0) = a and ξ = z,
(2.8)
|g(z)| ≤
|z| + |a|
1 + |a||z|
for all z ∈ U.
3
Main Results
Theorem 3.1. Let α be a complex number and f (z) ∈ A,
∞
X
ak z k . If
f (z) = z +
k=3
and
¯ ′′ ¯
¯ f (z) ¯
¯
¯
¯ f ′ (z) ¯ < 1, z ∈ U
(3.2)
|α| ≤ 4
(3.1)
then the function
(3.3)
Fα (z) =
Z
z
α
[f ′ (u)] du
0
is in the class S.
Proof. The function Fα (z) is regular in U. Let us consider the function
(3.4)
1 Fα′′ (z)
p(z) =
|α| Fα′ (z)
14
Virgil Pescar
where the constant |α| satisfies the inequality (3.2).
The function p(z) is regular in U. From (3.4) and (3.3) we obtain
(3.5)
p(z) =
α f ′′ (z)
.
|α| f ′ (z)
Using (3.1) and (3.5) we get
(3.6)
|p(z)| ≤ 1, z ∈ U
and we have p(0) = 0.
By Remark 2.4 we have
(3.7)
|p(z)| ≤ |z|, z ∈ U.
From (3.4) and (3.7) we obtain
¯
¯
1 ¯¯ Fα′′ (z) ¯¯
≤ |z|, z ∈ U
|α| ¯ Fα′ (z) ¯
(3.8)
and
¡
(3.9)
¯
¯
¢ ¯ zFα′′ (z) ¯
¡
¢
¯
¯ ≤ |α| max 1 − |z|2 |z|2 .
1 − |z| ¯ ′
¯
|z|<1
Fα (z)
2
¡
¢
1
Because max 1 − |z|2 |z|2 = , from (3.9) and (3.2) we get
|z|<1
4
¯ ′′ ¯
¡
¢ ¯ zF (z) ¯
(3.10)
1 − |z|2 ¯¯ ′α ¯¯ ≤ 1, z ∈ U.
Fα (z)
By Lemma 2.1 it results that the function Fα (z) ∈ S.
Theorem 3.2. Let γ be a complex number and the function f (z) ∈ A,
∞
X
f (z) = z +
ak z k . If
k=3
(3.11)
¯ ′′ ¯
¯ f (z) ¯
¯
¯
¯ f ′ (z) ¯ < 1, z ∈ U
15
On some integral classes of integral operators
and
(3.12)
|γ| ≤ 4
then for any complex number β, Re β ≥ 1 the function
½ Z
Hβ,γ (z) = β
(3.13)
z
u
0
β−1
¾ β1
[f (u)] du
′
γ
is in the class S.
Proof. Let us consider the function
Z z
γ
(3.14)
g(z) =
[f ′ (u)] du.
0
The function
(3.15)
p(z) =
1 g ′′ (z)
,
|γ| g ′ (z)
where the constant |γ| satisfies the inequality (3.12), is regular in U .
From (3.15) and (3.14) we obtain
(3.16)
p(z) =
γ f ′′ (z)
.
|γ| f ′ (z)
and using (3.11) we have
|p(z)| ≤ 1, z ∈ U
Remark 2.4 applied to the function p(z) give
¯
¯
1 ¯¯ g ′′ (z) ¯¯
≤ |z|, z ∈ U
(3.17)
|γ| ¯ g ′ (z) ¯
and, hence
(3.18)
¡
¯ ′′ ¯
¯ zg (z) ¯
¢
¡
¢
¯ ≤ |γ| max 1 − |z|2 |z|2 .
1 − |z|2 ¯¯ ′
|z|<1
g (z) ¯
16
Virgil Pescar
From (3.18) and (3.12) we obtain
¯
¯
¡
¢ ¯ ′′ ¯
2 ¯ zg (z) ¯
(3.19)
1 − |z| ¯ ′
≤ 1, z ∈ U.
g (z) ¯
By Theorem 2.2 for Re α = 1, it results that Hβ,γ (z) ∈ S.
Theorem 3.3. Let β a complex number, Re β ≥ 1 and f (z) ∈ A,
6= 0, z ∈ U. If
f (z) = z + a3 z 3 + . . . , f (z)
z
¯ ′′ ¯
¯ f (z) ¯
¯
¯
(3.20)
¯ f ′ (z) ¯ ≤ 4, z ∈ U
then the function
· Z
Lβ (z) = β
(3.21)
z
u
0
β−1
¸ β1
[f (u)] du
′
is in the class S.
Proof. Let us consider the function
g(z) =
1 f ′′ (z)
4 f ′ (z)
which is regular in U. Remark 2.4 applied to the function g(z) give
¯
¯
1 ¯¯ f ′′ (z) ¯¯
≤ |z|, z ∈ U
(3.22)
4 ¯ f ′ (z) ¯
and, hence, we obtain
(3.23)
¡
¯
¯
¢ ¯ zf ′′ (z) ¯
¡
¢
¯
¯ ≤ 4 max 1 − |z|2 |z|2 , z ∈ U
1 − |z| ¯ ′
¯
|z|<1
f (z)
2
¡
¢
1
Since max 1 − |z|2 |z|2 = , from (3.23) we have
|z|<1
4
¯
¯
¡
¢ ¯ ′′ ¯
2 ¯ zf (z) ¯
(3.24)
1 − |z| ¯ ′
≤ 1, z ∈ U.
f (z) ¯
From (3.24) and Theorem 2.2 for Re α = 1, we obtain Fβ (z) ∈ S.
On some integral classes of integral operators
17
References
[1] J. Becker, Lőwnersche Differentialgleichung und quasikonform fortsetzbare schlichte Functionen, J. Reine Angew. Math., 255 (1972), 23-43.
[2] Z. Nehari, Conformal mapping, Mc Graw-Hill Book Co., Inc., New York,
Toronto, London, 1952 .
[3] N. N. Pascu, An improvement of Beckerş univalence criterion, Proceedings of the Commemorative Session Simion Stoilow (Braşov), 1987, University of Braşov, pp. 43-48.
[4] V. Pescar, New univalence criteria, Transilvania University of Braşov,
2002.
[5] C. Pommerenke, Univalent functions, Vanderhoeck Ruprecht in
Gőttingen, 1975.
”Transilvania” University of Braşov
Faculty of Mathematics and Computer Science
Department of Mathematics
2200 Braşov, Romania
E-mail: virgilpescar@unitbv.ro