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

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

{

∈ ; <1

}

= z C z

U 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:

[ ]

∫

′ =z f t dt z

G

0

) ( )

( α

α

is univalent in U.

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

[ ]

∫

′ =z n

n z f t dt

G

0

, ( ) ( )

α α

is univalent in U for

n

∈

N

*. Teorema 3 [2]

Let α,a∈C ,Reα>0 and f ∈A. If

U z z

f z zf z

∈ ∀ ≤ −

) ( 1 ) ( '

) ( " Re

1 2Re α

α

then (∀)β∈C, Reβ ≥Reα , the function

( )

β β

β β

/ 1

0 1

) ( ' _

  

 

=

∫

z_t − _{f t}

z F

Teorema 4 [1] If the function g is holomorphic in U and g(z)<1 in U then

U

∈ ∀)ξ

( and z∈U the following inequalities hold:

2 2

1 ) ( 1 ) ( '

) 1 ( , 1 ) ( ) ( 1

) ( ) (

z z g z

g

z z g

z g

z g g

− − ≤

− − ≤ −

−

ξ ξ ξ ξ

the equalities hold in case , where 1 1

) ( )

( =

+ +

=ε ε

z t

t z z

g and u <1.

Remark 5 [1]

For z = 0, the inequalities (1) are the form: ξ ξ

ξ _≤

− (−0) ( ) 1

) 0 ( ) (

g g

g g

and

ξ ξ

ξ

⋅ +

+ =

) 0 ( 1

) 0 ( )

(

g g

g .

Considering g(0) = a and z U

z a

a z z g

z ∀ ∈

⋅ +

+ ≤ ⇒

= ( )

1 ) (

ξ .

Lemma 6 (Schwartz)

If the function g is holomorphic in U, g(0) = 0 and g

( )

z ≤1 (∀)z∈U, then: U

)z ( )

(z ≤ z ∀ ∈

g 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 ...,

z b z g z

a z

f = + ₂ 2 +... , = + ₂ 2 +

, ) ( , 1 ) ( '

) ( "

, ) ( , 1 ) ( '

) ( "

U z n z g

z g

U z n z f

z f

∈ ∀ ≤

∈ ∀ ≤

(2)

2 2

2 2a

n

n a n a n

   

  + ⋅ + ≤

and

1 1 1

< + _β

α . (4)

Then (∀)γ ∈C,Reγ ≥athe function:

[ ] [ ]

α β γ

γ γ

β

α γ

/ 1

0 1 ,

,

, ( ) '( ) '( )

   



 _{⋅ ⋅}

=

∫

z − n n

n z t f t g t dt

D

is univalent (∀)n∈N* \

{}

1 .

Proof: We consider the function:

∫

′ ′

= z _{n n}

dt t g t f z h

0[ ( )] [ ( )]

)

( α β (5)

The function

) ( '

) ( " 1 ) (

z h

z h z

p = ⋅

αβ (6)

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

{

}

₌

′ ⋅ ′

′ ′ ⋅ ′ ⋅ = ′′′ ⋅

= _{αβ αβ} α_α β_β

)] ( [ )] ( [

)] ( [ )] ( [ 1 ) (

) ( 1 ) (

n n

n n

t g t

f

t g t

f z

h z h z

p

(

)

[ ] [

]

(

)

[

′

] [ ]

⋅ ′ =

′′ ⋅ ′

⋅ ⋅ ⋅ ′ + ′

⋅ ′′ ⋅ ′

⋅ ⋅ ⋅ ⋅

= α − α− _αβ _βα β − β−

αβ ( ) ( )

) ( )

( )

( )

( ) ( )

(

1 1 1 1 1

n n

n n

n n

n n

n n

z g z f

z g z

g z n z

f z

g z f z

f z n

)

(

) ( )

(

)

( 1

1

n n n

n n n

z g

z g z n z

f

z f z n

′⋅ ′′ ⋅

⋅ + ′ ⋅ ′′ ⋅

⋅ ⋅

≤ ⋅

⋅ + ⋅

⋅

= − −

) ( '

) ( " )

( '

) ( " )

(

1 1

n n n

n n n

z g

z g nz z

f z f nz z

p

αββ αβα

≤ ⋅

⋅ ⋅ + ⋅

⋅

≤ − −

) ( '

) ( " )

( '

) (

" 1

1

n n n

n n n

z g

z g z n z

f z f nz

αββ αβα

≤ ⋅

⋅ ⋅ + ⋅

⋅ ⋅

≤ − −

) ( '

) ( " )

( '

) (

" 1

1

n n n

n n n

z g

z g z n z

f z f z n

β α

β β

α α

1 1 1 1 1 1 1

< + = ⋅ ⋅ + ⋅ ⋅

≤ _{β α α β}

n n n

n .

p(0) = 0

By Schwartz lemma we have:

⇔ ⋅

≤ ⇔

≤ ≤

⋅ −1 −1

) ( '

) ( " )

( '

) ( "

1 n n

z z

h z h z z z h

z h

αβ αβ

1

) ( '

) ( "

1 2 2 n

a a

z a

z

z h

z zh a

z

⋅     

   − ⋅ ≤ ⋅

    

   −

⇔ αβ (8)

Let’s the function x x z

a x x

Q R

Q n

a

= ⋅ − =

→ , ( ) 1 , ]

1 , 0 [ :

2

.

We have

[ ]

0,1 )

( , 2 2

2 )

( 2 ∀ ∈

  

 

+ ⋅ +

≤ x

a n

n a n x

Q a

n

.

1 ) ( '

) ( "

1 2

≤ ⋅

⋅     

   −

z h

z h z a

z a

.

In this conditions applying the theorem 3 we obtain:

[ ] [ ]

α β γ

γ γ

β

α γ

/ 1

0 1 ,

,

, ( ) '( ) '( )

   



 _{⋅ ⋅}

=

∫

z − n n

n z t f t g t dt

D

is univalent (∀)n∈N* \{1}.

Corollary 8 [5]. Let α,γ∈C ,Reα =a>0, g∈A

If

)

( 1 ) ( '

) ( "

U z n

z g

z g

∈ ∀ ≤

and

a n

n a n a

n 2 /2

2 2

   

  + +

≤

γ

then (∀)β∈C, Reβ ≥a, the function

[ ]

γ β

β γ

β β

/ 1

0 1 ,

, ( ) '( )

   

 

=

∫

z − n

n z t g t

G

is univalent (∀)n∈N* \

{}

1 .

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

z b z g z

a z

f = + ₂ 2 +... , = + ₂ 2 + .

If f and g satisfy the conditions 1 ) ( '

) ( "

<

z f

z f

and 1 ( ) , )

( '

) ( "

U z z

g z g

∈ ∀

< (9)

and

α

with

β

the conditions: 1 1 1

< + _β

2 1

2 1

max

1

2 2

2 2 2

1

    

 

    

 

+ ⋅ +

+ ⋅ + ⋅ ⋅ − ≤

≤

z b a

b a z

z c

z c

z

αβ β α

αββ α

αβ (11),

then (∀)γ∈C,Reγ ≥c the function

[ ] [ ]

α β γ γ

γ β

α γ

/ 1

0 1 ,

, ( ) '( ) '( )

   



 _{⋅ ⋅}

=

∫

z_t − _{f t g t}

z

F

is univalent. Proof: Let

[ ] [ ]

f t g t dt z

h

z

∫

⋅

=

0

) ( ' ) ( ' )

( α β .

Let

) ( '

) ( " 1 ) (

z h

z h z

p = ⋅

αβ .

Then:

( )

(

(

)

) (

(

)

)

(

(

)

) (

(

)

)

= ⋅

⋅ ⋅

⋅ ⋅

+ ⋅

⋅ ⋅ ⋅

= _αβ α α− _{α β} β _αβ α β _α β−_β

) ( ' ) ( '

) ( " )

( ' )

( ' 1 )

( ' ) ( '

) ( ' ) ( " )

( '

1 1 1

z g z f

z g z g z

f

z g z f

z g z f z f z

p

) ( '

) ( " )

( '

) ( "

z g

z g z

f z f

⋅ + ⋅

= _αβα _αββ .

The function p is olomorf , (∀)z∈U .

We respect the conditions (9) and (10) obtain:

1 1 1 ) ( '

) ( " )

( '

) ( " )

( '

) ( " )

( '

) ( " )

( = ⋅ + ⋅ ≤ ⋅ + ⋅ = + <

β α αββ

αβα αββ

αβα g z

z g z

f z f z

g z g z

f z f z

p

U z∈

∀)

( .

αβ β α αββ

αβα αββ

αβα 2 2 2 2 2 2 2

) 0 ( '

) 0 ( " )

0 ( '

) 0 ( " )

0

( a b a b

g g f

f

p = ⋅ + ⋅ = ⋅ + ⋅ = ⋅ +

U z

z b a

b a z

z

p ∀ ∈

⋅ + ⋅ +

+ ⋅ +

≤ , ( )

2 1

2 )

(

2 2

2 2

αβ β

α αβ

β α

⇔ ∈ ∀ +

+

+ ⋅ + ≤ ⋅

⇔ z U

z b a

b a z

z h

z h

) ( , 2

1 2 )

( '

) ( " 1

2 2

2 2

αββ

α αβ

β α

αβ

U z

z b a

b a z

z c

z

z h

z zh c

z c c

∈ ∀ ⋅

+ ⋅ +

+ ⋅ + ⋅ ⋅ − ⋅ ≤ ⋅

−

⇔ , ( )

2 1

2 1

) ( '

) ( " 1

2 2

2 2 2

2

αβ β

α αβ

β α

αβ .

The last inequality implies:

    

 

    

 

⋅ + ⋅ +

+ ⋅ + ⋅ ⋅ − ⋅

≤ ⋅

−

≤

z b a

b a z

z c

z

z h

z zh c

z c

z c

αβ β

α αβ

β α αβ

2 2

2 2 2

1 2

2 1

2 1

max )

( '

) ( " 1

The relation (11) and the last inequality implies:

U z z

h z zh c

z c

∈ ∀ ≤ ⋅

−

) ( , 1 ) ( '

) ( "

1 2

and by the theorem 3 we obtain

[ ] [ ]

α β γ γ

γ β

α γ

/ 1

0 1 ,

,

, ( ) '( ) '( )

   



 _{⋅ ⋅}

=

∫

z_t − _{f t g t}

z F

Corollary 10 [5] Let

α

,

γ

∈

C

,

Re

α

=

b

>

0

and function ...

) (

, = + ₂ 2 +

∈A g z z a z g

Dacă

U z z

g z g

∈ ∀ <1, ( ) )

( '

) ( "

and satisfy the condition:

    

  

  

 +

+ ⋅     

  

⋅ − ≤

≤ a z

a z z b

z b

z

2 2 2

1 _{1 2}

2 1

max

1 γ

then for allβ∈C, Reβ ≥b, the function

[ ]

_γ β β

γ

β β

1

0 1

, ( ) '( )

   

 

=

∫

z_t − _{g t}

z G

is univalent.

REFERENCES

[1] Z. Nehari- Conformal Mapping, McGraw-Hill Book Co., Inc., New York, Toronto, London, 1952.

[2] N.N. Pascu- An improvement of Becker’s univalence criterion, Proceedings of the Commemorative Session Simion Stoilow, Braşov, (1987), 43-48.

[3] N.N. Pascu, V. Pescar, On the integral operators of Kim-Merkens and Pfaltzgraff, Studia(Mathematica), Univ. Babeş-Bolyai, Cluj-Napoca, 32, 2(1990), 185-192.

[4] J. A. Pfaltzgraff, Univalence of the integral of f′

( )

z λ, Bull. London Math. Soc. 7(1975), no 3.254-256.

[5] V. Pescar and Shigeyoshi Owa- Univalence of certain integral operators, International J. Math. & Math. Sci. Vol. 23.No.10(2000), 697-701.

Authors:

Daniel Breaz - „ 1 Decembrie 1918” University of Alba Iulia, str. N.Iorga, No.13, Departament of Mathematics and Computer Science, dbreaz@lmm.uab.ro