IJMMS 2004:47, 2489–2494
PII. S0161171204401392
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
SOME CRITERIA FOR UNIVALENCE OF CERTAIN
INTEGRAL OPERATORS
VIRGIL PESCAR and SHIGEYOSHI OWA
Received 21 January 2004 and in revised form 25 March 2004
We derive some criteria for univalence of certain integral operators for analytic functions
in the open unit disk.
2000 Mathematics Subject Classification: 30C45.
1. Introduction. Let Ꮽ be the class of the functions f (z) which are analytic in the
open unit disk U = {z ∈ C : |z| < 1} and f (0) = f (0) − 1 = 0.
We denote by ᏿ the subclass of Ꮽ consisting of functions f (z) ∈ Ꮽ which are univalent in U. Miller and Mocanu [1] have considered many integral operators for functions
f (z) belonging to the class Ꮽ. In this paper, we consider the integral operators
Fα (z) =
1
α
z
1/α −1
f (u)
u du
α
0
(z ∈ U)
(1.1)
for f (z) ∈ Ꮽ and for some α ∈ C. It is well known that Fα (z) ∈ ᏿ for f (z) ∈ ᏿∗ and
α > 0, where ᏿∗ denotes the subclass of ᏿ consisting of all starlike functions f (z) in U.
2. Preliminary results. To discuss our integral operators, we need the following
theorems.
Theorem 2.1 [3]. Let α be a complex number with Re(α) > 0 and f (z) ∈ Ꮽ. If f (z)
satisfies
1 − |z|2 Re(α) zf (z) 1,
Re(α)
f (z) (2.1)
for all z ∈ U, then the integral operator
z
1/α
Gα (z) = α uα−1 f (u)du
0
is in the class ᏿.
(2.2)
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V. PESCAR AND S. OWA
Theorem 2.2 [4]. Let α be a complex number with Re(α) > 0 and f (z) ∈ Ꮽ. If f (z)
satisfies (2.1) for all z ∈ U, then, for any complex number β with Re(β) Re(α), the
integral operator
z
1/β
Gβ (z) = β uβ−1 f (u)du
(2.3)
0
is in the class ᏿.
Example 2.3. Defining the function f (z) by
z
f (z) =
0
1 + uRe(α)
1 − uRe(α)
1/2
du
(2.4)
with Re(α) > 0, we have that
1 − z2 Re(α) zf (z)
= zRe(α) .
Re(α)
f (z)
(2.5)
Thus the function f (z) satisfies the condition of Theorem 2.2. Therefore, for Re(β) Re(α),
z
1/β
1 + uRe(α) 1/2
du
Gβ (z) = β uβ−1
1 − uRe(α)
0
(2.6)
is in the class ᏿.
Theorem 2.4 [2]. If the function g(z) is regular in U, then, for all ξ ∈ U and z ∈ U,
g(z) satisfies
g(ξ) − g(z) ξ − z 1 − zξ ,
1 − g(z)g(ξ)
2
g (z) 1 − g(z) .
2
1 − |z|
(2.7)
(2.8)
The equalities hold only in the case g(z) = ε((z+u)/(1+uz)), where |ε| = 1 and |u| < 1.
Remark 2.5 [2]. For z = 0, from inequality (2.7),
g(ξ) − g(0) |ξ|
1 − g(0)g(ξ)
(2.9)
and, hence
g(ξ) |ξ|+ g(0)
.
1 + g(0)|ξ|
(2.10)
Considering g(0) = a and ξ = z, we see that
g(z) |z| + |a|
1 + |a||z|
for all z ∈ U.
(2.11)
SOME CRITERIA FOR UNIVALENCE OF CERTAIN INTEGRAL OPERATORS
2491
Schwarz lemma [2]. If the function g(z) is regular in U, g(0) = 0, and |g(z)| 1,
for all z ∈ U, then
g(z) |z|
(2.12)
for all z ∈ U, and |g (0)| 1. The equality in (2.12) for z ≠ 0 holds only in the case
g(z) = z, where || = 1.
3. Main results
Theorem 3.1. Let α be a complex number with Re(1/α) = a > 0 and the function
g(z) ∈ Ꮽ satisfying
zg (z)
g(z) − 1 1
(z ∈ U).
(3.1)
Then, for
2
,
(2a + 1)(2a+1)/2a
|α| (3.2)
the integral operator
Fα (z) =
z
1
α
1/α −1
u du
g(u)
α
(3.3)
0
is in the class ᏿.
Proof. Let 1/α = β. Then we have
z
1/β
g(u) β
du
.
F1/β (z) = β uβ−1
u
0
(3.4)
We consider the function
z
f (z) =
0
g(u)
u
β
du.
(3.5)
Then the function
h(z) =
1 zf (z)
|β| f (z)
(3.6)
is regular in U and the constant |β| satisfies the inequality
|β| (2a + 1)
2
(2a+1)/2a
.
(3.7)
From (3.5) and (3.6), we have that
h(z) =
β zg (z)
−1 .
|β| g(z)
(3.8)
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V. PESCAR AND S. OWA
Using (3.8) and (3.1), we obtain
h(z) 1
(z ∈ U).
(3.9)
Noting that h(0) = 0 and applying the Schwarz lemma for h(z), we get
1 zf (z) |z| (z ∈ U)
|β| f (z) (3.10)
2a 1 − |z|2a zf (z) |β| 1 − |z|
|z| (z ∈ U).
f (z) a
a
(3.11)
and hence we obtain
Because
max
|z|≤1
2
1 − |z|2a
|z| =
,
a
(2a + 1)(2a+1)/2a
(3.12)
from (3.11) and (3.7), we have
1 − |z|2a zf (z) 1
a
f (z) (3.13)
for z ∈ U. From (3.13) and Theorem 2.1, it follows that (2.4) belongs to the class ᏿.
By means of (2.4) and (3.5), we have that the integral operator F1/β (z) is in the class
᏿, and hence we conclude that the integral operator Fα (z) is in the class ᏿.
Example 3.2. If we take the function g(z) = zez and α = 1/a > 0, then
g(z) = z + a2 z2 + a3 z3 + · · ·
(3.14)
is analytic in U and
zg (z)
= |z| < 1
−
1
g(z)
(z ∈ U).
(3.15)
Since the function g(z) satisfies the condition of Theorem 3.1, we have
Tα (z) =
1
α
z
α
eu/α u1/α−1 du
0
∈ ᏿.
(3.16)
Theorem 3.3. Let α, β be complex numbers with Re(β) Re(α) > 0 and the function
g(z) ∈ Ꮽ satisfying
zg (z) − g(z) 1
zg(z)
(z ∈ U).
(3.17)
Then, for
|α| max
|z|1
|z| + a2 1 − |z|2 Re(α)
|z|
Re(α)
1 + a2 |z|
,
(3.18)
SOME CRITERIA FOR UNIVALENCE OF CERTAIN INTEGRAL OPERATORS
2493
the integral operator
z
1/β
1/α β−1/α−1
Fα,β (z) = β
u
du
g(u)
(3.19)
0
is in the class ᏿.
Proof. We have
z
1/β
g(u) 1/α
du
.
Fα,β (z) = β uβ−1
u
0
(3.20)
We consider the function
z
f (z) =
0
g(u)
u
1/α
du,
(3.21)
which is regular in U. The function
p(z) = |α|
f (z)
,
f (z)
(3.22)
where the constant |α| satisfies inequality (3.18), is regular in U. From (3.22) and (3.21),
we obtain
p(z) =
|α| zg (z) − g(z)
α
zg(z)
(3.23)
and using (3.17), we have
p(z) < 1
(z ∈ U)
(3.24)
and |p(0)| = |a2 |. Applying Remark 2.5, we obtain
2
f (z) ≤ |z|+ a
α
f (z) 1 + a 2 |z|
(z ∈ U).
(3.25)
It follows that
|z| + a2 1 − |z|2 Re(α)
1 − |z|2 Re(α) zf (z) 1
|z|
f (z) Re(α)
|α|
Re(α)
1 + a2 |z|
(3.26)
for all z ∈ U. We consider the function
Q(x) =
x + a2 1 − x 2 Re(α)
x
Re(α)
1 + a2 x
x = |z|; x ∈ [0, 1] .
(3.27)
Because Q(1/2) > 0, Q(x) satisfies
max Q(x) > 0.
x∈[0,1]
(3.28)
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V. PESCAR AND S. OWA
Using this fact, (3.26) gives us that
1 − |z|2 Re(α)
|z| + a2 1 − |z|2 Re(α) z f (z) 1 max
|z|
f (z) |α| |z|1
Re(α)
Re(α)
1 + a2 |z|
.
(3.29)
From (3.29) and (3.18), we obtain (2.1). Using (2.1) and Theorem 2.2, we obtain that
the integral operator (2.4) belongs to the class ᏿. Therefore, it follows from (2.4) and
(3.21) that Fα,β (z) is in the class ᏿.
Corollary 3.4. Let α be a complex number with Re(α) > 0 and the function g(z) ∈
Ꮽ satisfying (3.18). Then, for
max
|z|1
|z| + a2 1 − |z|2 Re(α)
|z|
Re(α)
1 + a2 |z|
|α| 1,
(3.30)
the integral operator (3.3) is in the class ᏿.
Proof. From Theorem 3.3 for β = 1/α, the condition Re(β) Re(α) > 0 is identical
with |α| < 1 and we have Fα,β (z) = Fα (z).
Acknowledgment. The authors thankfully acknowledge the kind and helpful advice of Professor H. M. Srivastava on the paper.
References
[1]
[2]
[3]
[4]
S. S. Miller and P. T. Mocanu, Differential Subordinations. Theory and applications, Monographs and Textbooks in Pure and Applied Mathematics, vol. 225, Marcel Dekker,
New York, 2000.
Z. Nehari, Conformal Mapping, McGraw-Hill, New York, 1952.
N. N. Pascu, On a univalence criterion. II, Itinerant Seminar on Functional Equations, Approximation and Convexity (Cluj-Napoca, 1985), Preprint, vol. 85, Universitatea “BabeşBolyai”, Cluj-Napoca, 1985, pp. 153–154.
, An improvement of Becker’s univalence criterion, Proceedings of the Commemorative Session: Simion Stoïlow (Braşov, 1987), University of Braşov, Braşov, 1987, pp. 43–
48.
Virgil Pescar: Department of Mathematics, Faculty of Mathematics and Computer Science,
Transilvania University of Braşov, 2200 Braşov, Romania
E-mail address: virgilpescar@hotmail.com
Shigeyoshi Owa: Department of Mathematics, Kinki University, Higashi-Osaka, Osaka 5778502, Japan
E-mail address: owa@math.kindai.ac.jp