Daniel acta 9. 99KB Jun 04 2011 12:08:35 AM

ACTA UNIVERSITATIS APULENSIS

No 9/2005

THE UNIVALENT CONDITION FOR AN INTEGRAL
OPERATOR
ON THE CLASSES S (α) AND T2

Daniel Breaz and Nicoleta Breaz
Abstract.In this paper we present a few conditions of univalency for the
operator Fα,β on the classes of the univalent functions S (p) and T2 . These are
actually generalizations(extensions) of certain results published in the paper
[1].
AMS 2000 Subject Classification: 30C45.
Keywords and phrases: Unit disk, analytical function, univalent function,integral operator.
1.Introduction
Let be the class of analytical functions, A = {f : f = z + a2 z 2 + ...} , z ∈ U ,
where U is the unit disk , U = {z : |z| < 1} and denote S the class of univalent
functions in the unit disk.
Let p the real number with the property 0 < p ≤ 2. We define the class
S (p) as the class

 of the functions f ∈ A, that satisfy the conditions f (z) 6= 0

′′ 
and (z/f (z))  ≤ p, z ∈ U and if f ∈ S (p) then the following property is true

 2 ′

 z f (z)

2
 f 2 (z) − 1 ≤ p |z| , z ∈ U, relation proved in [3].

We denote
 2 ′ with T2 the class of the univalent functions that satisfy the


(z)
condition  zf 2f(z)
− 1 < 1, z ∈ U , and also have the property f ′′ (0) = 0.
These functions have the form f = z + a3 z 3 + a4 z 4 + .... For 0 < µ < 1 we have

a subclass of functions
 by Tµ,2 , containing the functions f ∈ T2 that
 2 denoted
 z f ′ (z)
satisfy the property  f 2 (z) − 1 < µ < 1, z ∈ U .
Next we present a few well known results related to these classes, results
on which shall rely in this paper.
63

Daniel Breaz and Nicoleta Breaz - The univalent condition for an integral ...
The Schwartz Lemma. Let the analytic function g be regular in the unit
disc U and g (0) = 0. If |g (z)| ≤ 1, ∀z ∈ U, then
|g (z)| ≤ |z| , ∀z ∈ U

(1)

and equality holds only if g (z) = εz, where |ε| = 1.
Theorem 1.[2]. Let α ∈ C, Reα > 0 and f ∈ A. If
1 − |z|2Reα
Reα



 zf ′′ (z) 


 ′
 ≤ 1, ∀z ∈ U
 f (z) 

(2)

then ∀β ∈, Reβ ≥ Reα, the function


Fβ (z) = β

Zz
0

1/β

tβ−1 f ′ (t) dt

(3)

is univalent.
Theorem 2.[1]. Let fi ∈T2 , fi (z) = z + ai3 z 3 + ai4 z 4 + ..., ∀i = 1, 2, so that
α, β ∈ C, Reα ≥

6
, Reβ ≥ Reα.
|α|

(4)

If |fi (z)| ≤ 1, ∀z ∈ U, i = 1, 2 then ∀β ∈ C the function
 z
 Z

Fα,β (z) = β

tβ−1

0

f1 (t)
t

!1

α

·

f2 (t)
t

!1

1
β

α

dt ∈ S.

(5)

Theorem 3.[1]. Let fi ∈T2,µ , fi (z) = z + ai3 z 3 + ai4 z 4 + ..., ∀i = 1, 2, so that
α, β ∈ C, Reα ≥

2 (µ + 2)
, Reβ ≥ Reα.
|α|

(6)

If |fi (z)| ≤ 1, ∀z ∈ U, i = 1, 2 then ∀β ∈ C the function
 z

 Z

Fα,β (z) = β

0

tβ−1

f1 (t)
t

!1

α

·

f2 (t)
t

!1

α

1
β

dt ∈ S.

(7)

Theorem 4.[1]. Let fi ∈S (p) , 0 < p < 2, fi (z) = z + ai3 z 3 + ai4 z 4 + ..., ∀i =
1, 2, so that
64

Daniel Breaz and Nicoleta Breaz - The univalent condition for an integral ...

2 (p + 2)
, Reβ ≥ Reα.
|α|

α, β ∈ C, Reα ≥

(8)

If |fi (z)| ≤ 1, ∀z ∈ U, i = 1, 2 then ∀β ∈ C the function
 z
 Z

Fα,β (z) = β

tβ−1

0

f1 (t)
t

!1

α

·

f2 (t)
t

!1

1
β

α

2.Main results

dt ∈ S.

(9)

Theorem 5. Let M ≥ 1,fi ∈T2 , fi (z) = z + ai3 z 3 + ai4 z 4 + ..., ∀i = 1, 2, so
that
α, β ∈ C, Reα ≥

(4M + 2)
, Reβ ≥ Reα.
|α|

(10)

If |fi (z)| ≤ M, ∀z ∈ U, i = 1, 2 then ∀β ∈ C the function
Fα,β (z) =
Proof.

 z
 Z

β



tβ−1

0

f1 (t)
t

We consider the function h (z) =

!1

α

f2 (t)
·
t

Rz  f1 (t)  α1
t

0

·



!1

1
β

α

f2 (t)
t

dt

1

α

∈ S.

(11)



dt.

We observe that h (0) = h′ (0) − 1 = 0.
By calculating the derivatives of the order I and II for this function we
obtain:
′

h (z) =

f1 (t)
t

!1

α

f2 (t)
·
t

!1

α

(12)

respectively
h′′ (z) =

1
1 ′
h (z) · B1 + h′ (z) · B2
α
α

where

65

(13)

Daniel Breaz and Nicoleta Breaz - The univalent condition for an integral ...

Bk =

z
zfk′ (z) − fk (z)
·
, k = 1, 2.
fk (z)
z2
!

We calculate the fraction
′′

zh (z)
=
h′ (z)

zh′′ (z)
h′ (z)

z · α1 h′ (z) ·

(14)

and we obtain:
2
P

k=1

h′ (z)

Bk

2
1 X
Bk , ∀z ∈ U.
=z· ·
α k=1

(15)

Replacing Bk , k = 1, 2, in formula (15) we obtain:
1
zh′′ (z)
=
′
h (z)
α

zf1′ (z)
1
−1 +
f1 (z)
α
!

zf2′ (z)
−1 .
f2 (z)
!

(16)

We evaluate the modulus and we multiply in both terms of the relation
2Reα
(16) with 1−|z|
, obtain:
Reα
1 − |z|2Reα
Reα





!
2
 zh′′ (z) 
 z 2 f ′ (z)  |f (z)|
1 − |z|2Reα X
i




i
 ′
≤

+1 .

 h (z) 
|α| Reα i=1  fi2 (z)  |z|

(17)

Because |fi (z)| ≤ M, ∀z ∈ U , ∀i = 1, 2 and applying Schwarz Lemma we
obtain that
|fi (z)|
≤ M, ∀z ∈ U, ∀i = 1, 2.
|z|

(18)

We apply this relation in the above inequality and we obtain:
1 − |z|2Reα
Reα
But





!
2
 z 2 f ′ (z) 
 zh′′ (z) 
1 − |z|2Reα X




i

 ′
≤
M + 1 .
 h (z) 
|α| Reα i=1  fi2 (z) 







 z 2 f ′ (z)
 z 2 f ′ (z)
 z 2 f ′ (z) 







i
i
i
= 2
 2
− 1 + 1 ≤  2
− 1 + 1, ∀z ∈ U, ∀i = 1, 2.
 fi (z)


 fi (z)
 fi (z) 


2 ′



(z)
Because f ∈ T2 , so  zf 2f(z)
− 1 < 1.
Applying this property and (20) in (19), obtain that:

66

(19)

(20)

Daniel Breaz and Nicoleta Breaz - The univalent condition for an integral ...

2Reα

1 − |z|
Reα





2Reα
 zh′′ (z) 
(4M + 2)
1
−
|z|
(4M + 2)


 ′
≤
≤
, ∀z ∈ U.
 h (z) 
|α| Reα
|α| Reα

Because Reα >

(4M +2)
,
|α|

(21)

we obtain that

1 − |z|2Reα
Reα



 zh′′ (z) 


 ′
 ≤ 1, ∀z ∈ U.
 h (z) 

(22)

Applying Theorem 1 we obtain that Fα,β ∈ S.

Remark. Theorem 5 is a generalization of the Theorem 2.
Theorem 6. Let M ≥ 1, fi ∈T2,µ , fi (z) = z + ai3 z 3 + ai4 z 4 + ..., ∀i = 1, 2,
so that
α, β ∈ C, Reα ≥

2 (µM + M + 1)
, Reβ ≥ Reα.
|α|

(23)

If |fi (z)| ≤ M, ∀z ∈ U, i = 1, 2 then ∀β ∈ C the function
 z
 Z

Fα,β (z) = β

0

tβ−1

f1 (t)
t

!1

α

·

f2 (t)
t

!1

α

1
β

dt ∈ S.

(24)

Proof. Considering the same steps as in the above proof we obtain:
1 − |z|2Reα
Reα





!
2
 zh′′ (z) 

 z 2 f ′ (z)
1 − |z|2Reα X




i
− 1 M + M + 1 .
 ′
≤

 h (z) 
|α| Reα i=1  fi2 (z)

(25)

2 ′ (z)
− 1 < µ, ∀z ∈ U.
But f ∈ T2,µ , which implies that  zf 2f(z)
In these conditions we obtain:



1 − |z|2Reα
Reα





 zh′′ (z) 
2 (µM + M + 1)


, ∀z ∈ U.
 ′
≤
 h (z) 
|α| Reα

By applying the relation (23) we obtain that
67

1−|z|2Reγ
Reγ

(26)

 ′′ 
 zh (z) 
 h′ (z)  ≤ 1, ∀z ∈ U.

Daniel Breaz and Nicoleta Breaz - The univalent condition for an integral ...
So according to the Theorem 1 the function Fα,β is univalent.
Remark. Theorem 6 is a generalization of the Theorem 3.
Theorem 7. Let M ≥ 1, fi ∈S (p) , 0 < p < 2, fi (z) = z + ai3 z 3 + ai4 z 4 +
..., ∀i = 1, 2, so that
α, β ∈ C, Reα ≥

2 (pM + M + 1)
, Reβ ≥ Reα.
|α|

(27)

If |fi (z)| ≤ M, ∀z ∈ U, i = 1, 2 then ∀β ∈ C the function
Fα,β (z) =

 z
 Z

β



tβ−1

0

f1 (t)
t

!1

α

f2 (t)
·
t

!1

α

1
β

dt

∈ S.

(28)



Proof. Considering the same steps as in the above proof we obtain:
1 − |z|2Reα
Reα





!
2
 z 2 f ′ (z)
 zh′′ (z) 

1 − |z|2Reα X




i
− 1 M + M + 1 .

 ′
≤
 h (z) 
|α| Reα i=1  fi2 (z)

(29)

But f ∈ S (p) , so




 z 2 f ′ (z)

− 1 ≤ p |z|2 , ∀z ∈ U.
 2

 f (z)

(30)

In these conditions we obtain:
1 − |z|2Reα
Reα
so,



2 
 zh′′ (z) 

1 − |z|2Reα X


p |z|2 M + M + 1 , ∀z ∈ U,
 ′
≤
 h (z) 
|α| Reα i=1

1 − |z|2Reα
Reα



 zh′′ (z) 
2 (pM + M + 1)


 ′
≤
.
 h (z) 
|α| Reα

By applying in (32) the relation (27) we obtain that
1, ∀z ∈ U.
68

(31)

(32)
1−|z|2Reγ
Reγ

 ′′ 
 zh (z) 
 h′ (z)  ≤

Daniel Breaz and Nicoleta Breaz - The univalent condition for an integral ...
So according to the Theorem 1 the function Fα,β is univalent.
Remark. Theorem 7 is a generalization of the Theorem 4.
References
[1] Breaz, D.,Breaz, N., The univalent conditions for an integral operator
on the classes S (p) and T2 , Preprint,2004.
[2] Pascu, N.N., An improvement of Becker’s univalence criterion, Proceedings of the Commemorative Session Simion Stoilow, Brasov, (1987), 43-48.
[3] Singh, V., On class of univalent functions, Internat. J. Math &Math.
Sci. 23(2000), 12, 855-857. [4] Yang, D., Liu, J., On a class of univalent
functions, Internat. J. Math &Math. Sci. 22(1999), 3, 605-610.
D. Breaz:
Department of Mathematics
”‘1 Decembrie 1918”’ University of Alba Iulia
Alba Iulia, str. N. Iorga, No. 11-13, 510009, Romania
E-mail address: dbreaz@uab.ro
N. Breaz:
Department of Mathematics
”‘1 Decembrie 1918”’ University of Alba Iulia
Alba Iulia, str. N. Iorga, No. 11-13, 510009, Romania
E-mail address: nbreaz@uab.ro

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