art. rev 2REZ. 135KB Jun 04 2011 12:08:21 AM

ACTA UNIVERSITATIS APULENSIS

UNIVALENCE FOR INTEGRAL OPERATORS
by
Daniel Breaz, Nicoleta Breaz
Abstract: We study in this paper two integral operators and we determine the conditions for
univalence for these integral operators.

Let A be the class of the functions f, which are analytic in the unit disc
U = {z ∈ C ; z < 1} and f (0) = f ' (0) − 1 = 0 and denote by S the class of univalent
functions.
Teorema 1 [4] If f ∈ S , α ∈ C and α ≤ 1 / 4 then the function:

Gα ( z ) = ∫ [ f ′(t )] dt
α

z

0

is univalent in U.

Teorema 2 [3] If the function g ∈ S , α ∈ C , α ≤ 1 / 4n , then the function defined by

[

]

Gα ,n ( z ) = ∫ f ′(t n ) dt
z

is univalent in U for n ∈ N .
Teorema 3 [2]
Let α , a ∈ C , Re α > 0 and f ∈ A .
If

0

α

*

1− z

2 Re α

zf " ( z )
≤ 1 (∀) z ∈ U
f ' ( z)
Re α
then (∀) β ∈ C , Re β ≥ Reα , the function
 z

Fβ ( z ) =  β ∫ t β −1 f ' (t )
 0


is univalent.

51

1/ β

ACTA UNIVERSITATIS APULENSIS
Teorema 4 [1] If the function g is holomorphic in U and g ( z ) < 1 in U then
(∀)ξ ∈ U and z ∈ U the following inequalities hold:

g (ξ ) − g ( z )

1 − g ( z )g (ξ )
g ' ( z) ≤
the equalities hold in case g ( z ) =

ξ −z
,
1 − zξ

≤

1 − g ( z)

ε (z + t)

1 + tz

1− z

, where ε = 1 and

g (ξ ) − g (0)

1 − g (0)g (ξ )
g (ξ ) =

2

2

Remark 5 [1]
For z = 0, the inequalities (1) are the form:

and

(1)

u < 1.

≤ξ

ξ + g ( 0)
.
1 + g ( 0) ⋅ ξ

Considering g(0) = a and ξ = z ⇒ g ( z ) ≤

z+a

(∀) z ∈ U .

1+ a ⋅ z

Lemma 6 (Schwartz)
If the function g is holomorphic in U, g(0) = 0 and g (z ) ≤ 1 (∀) z ∈ U , then:

g ( z) ≤ z
(∀)z ∈ U and g ' (0) ≤ 1
the equalities hold in case g(z) = z, where =1.

Theorem 7. Let α, , , , complex number with the properties Reδ = a >0, f,g ∈ A
f = z + a 2 z 2 + ... , g = z + b2 z 2 + ...,

f "( z) 1
≤ , (∀) z ∈ U ,
f ' ( z) n

(2)

g" ( z ) 1
≤ , (∀) z ∈ U ,
g ' ( z) n
n + 2a  n + 2a  2 a
⋅
αβ ≤


2
 n 
n

52

(3)

ACTA UNIVERSITATIS APULENSIS

and
+

α

1

Then (∀)γ ∈ C , Re γ ≥ a the function:

β

1

< 1.

[

(4)

] [

]


 z
α
β
Dα , β ,γ , n ( z ) = γ ∫ t γ −1 ⋅ f ' (t n ) ⋅ g ' (t n ) dt 


 0

1/ γ

is univalent (∀)n ∈ N * \ {1} .

Proof: We consider the function:

h( z ) = ∫0 [ f ′(t n )]α [ g ′(t n )] β dt
z

(5)

The function
p( z ) =

αβ
1

⋅

h" ( z )
h' ( z )

(6)

where αβ satisfies the conditions (3).
We obtain:

{

h′′( z )
1 [ f ′(t n )]α ⋅ [ g ′(t n )] β
⋅
=
⋅
p( z ) =
αβ h ′( z ) αβ [ f ′(t n )]α ⋅ [ g ′(t n )] β
1

=

=

αβ
1

⋅

(

α ⋅ n ⋅ z n −1 ⋅ f ′( z n )

)

α −1

}′ =

[ ] + [ f ′( z )]
[ f ′( z )] ⋅ [g ′( z )]

⋅ f ′′( z n ) ⋅ g ′( z n )
n

β

α

α α ⋅ n ⋅ z n −1 ⋅ f ′′( z n )
β n ⋅ z n −1 ⋅ g ′′( z n )
⋅
+
⋅
αβ
αβ
f ′( z n )
g ′( z n )

We respect the conditions (2), (4) and (7) we have:

53

n

n

α

β

(

⋅ β ⋅ n ⋅ z n −1 g ′( z n )

(7)

)

β −1

⋅ g ′′( z n )

=

ACTA UNIVERSITATIS APULENSIS

p( z ) =

≤
≤

α nz n−1 ⋅ f " ( z n ) β nz n−1 ⋅ g" ( z n )
⋅
+
⋅
≤
αβ
αβ
f '(z n )
g'(z n )

α nz n −1 ⋅ f " ( z n )
β n ⋅ z n −1 ⋅ g " ( z n )
⋅
+
⋅
≤
αβ
αβ
f '(z n )
g'(z n )

α
β
f "(z n )
g" ( z n )
n −1
n
z
⋅ n ⋅ z n −1 ⋅
+
⋅
⋅
⋅
≤
α β
f '(z n ) α β
g'(z n )
≤

⋅n⋅

β

1

1 1
1 1
1
+ ⋅n⋅ =
+
< 1.
n α
n α
β

p (0) = 0
By Schwartz lemma we have:

αβ
1

⋅

h" ( z )
h" ( z )
n −1
≤ z
≤ z ⇔
≤ αβ ⋅ z n −1 ⇔
h' ( z )
h' ( z )

 1 − z 2a
⇔
 a


 zh" ( z )
⋅
≤ αβ
 h' ( z )


 1 − z 2a
⋅
 a


 n
⋅ z



Let’s the function Q : [0,1] → R, Q( x) =

(8)

1 − x 2a n
⋅x ,x = z .
a

We have
2
 n  2a
Q( x) ≤
⋅
 , (∀) x ∈ [0,1] .
n + 2 a  n + 2a 
n

.
Respect the conditions (3) and (8) we obtain:

54

ACTA UNIVERSITATIS APULENSIS
 1 − z 2a

 a


 z ⋅ h" ( z )
⋅
≤ 1.
 h' ( z )


In this conditions applying the theorem 3 we obtain:

[

] [

]


 z
α
β
Dα , β ,γ , n ( z ) = γ ∫ t γ −1 ⋅ f ' (t n ) ⋅ g ' (t n ) dt 

 0

1/ γ

is univalent (∀)n ∈ N * \ {1} .
Corollary 8 [5]. Let α , γ ∈ C , Re α = a > 0, g ∈ A
If
g" ( z ) 1
≤
(∀) z ∈ U
g ' ( z) n
and
n + 2 a  n + 2a 


2  n 
then (∀) β ∈ C , Re β ≥ a, the function

γ ≤

[

n / 2a

]

 z
γ 
G β ,γ ,n ( z ) =  β ∫ t β −1 g ' (t n ) 
 0


1/ β

is univalent (∀)n ∈ N * \ {1} .

Theorem 9. Let α, , , complex number, Re δ = c > 0 , and the functions f , g ∈ A

f = z + a 2 z 2 + ... , g = z + b2 z 2 + ..., .
If f and g satisfy the conditions
f "( z)
g" ( z )