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(1)

ON AN INTEGRAL OPERATOR WHICH PRESERVE THE

UNIVALENCE

by

Maria E. Gageonea and Silvia Moldoveanu

1. Introduction and preliminaries

We denote by Ur the disc {z∈C : z <r}, 0 <r≤ 1, U1 = U.

Let A be the class of functions f which are analytic in U and f (0) = f ’(0) – 1 = 0.

Let S be the class of the functions fA which are univalent in U.

Definition 1. Let f and g be two analytic functions in U. We say that f is subordinate to g, fpg, if there exists a function ϕ analytic in U, which satisfies ϕ(0) = 0, |ϕ(z)| < 1 and f(z)= g(ϕ(z)) in U.

Definition 2. A function L : U x I → C, I = [o, ∞), L(z, t) is a Loewner chain, or a subordination chain if L is analytic and univalent in U for all zU and for all t1, t2∈

I, 0 ≤ t1<t2, L(z, t1) pL(z, t2).

Lemma 1 [4]. Let r0 ∈ (0, 1] and let L(z, t) = a1(t) z + …, a1(t) ≠ 0 be analytic in 0

r

U

for all tI and locally absolutely continuous in I, locally uniform with respect to

0 r

U

.

For almost all tI suppose:

t

t

z,

L

t

z,

p

z

t

z,

L

z

=

(

)

)

(

)

(

, z ∈ 0 r

U

where p is analytic in U and satisfies Re p(z, t) > 0, zU, t∈ I. If |a1(t)| → ∞ for t → ∞ and

)

(

)

(

t

a

t

z,

L

1

forms a normal family in

0 r

U then for each tI, L has an analytic and univalent extension to the whole disc U.

In the theory of univalent functions an interesting problem is to find those integral operators, which preserve the univalence, respectively certain classes of

(2)

univalent functions. The integral operators which transform the class S into S are presented in the theorems A, B and C which follow.

The integral operators studied by Kim and Merkes is that from the theorem: Theorem A [1]. If fS, then for α∈C,

4

1

α

the function Fα defined by:

(1)

( )

u

u

u

f

z

F

z

(

)

d

0

α

α

=

belongs to the class S.

A similar result, for other integral operator has been obtained by Pfaltzgraff in: Theorem B [3]. If fS, then for α∈C,

4

1

α

, the function Gα defined by:

(2)

G

( )

z

z

[

f

'

u

]

u

0

α α

=

(

)

d

belongs to the class S.

An integral operator different of (1) and (2) is obtained by Silvia Moldoveanu and N.N. Pascu in the next theorem:

Theorem C [2]. If fS, then for α∈C,

4

1

1

α

, the function Iα defined by:

(3) I

( )

z z f u u α

  

 α

=

α−

α

1 d ) (

0 1

belongs to the class S.

In this note, using the subordination chains method, we obtain sufficient conditions for the regularity and univalence of the integral operator:

(4)

(

)

α

  

 

⋅ α

= λ λ λ α−

α

1 2

1( ) ( ) ( ) d

)

( 1

0 f1 u f2 u ... f u u

z

H z n n

where fkA,

k

=

1

,

n

, 0 ≤λk≤ 1,

1

1

=

λ

=

n

k

(3)

2. Main results

Theorem 1. Let f1, f2, …, fn A, α∈C, |α – 1| < 1 and λ1, λ2, …, λn ∈R, 0 ≤λk ≤ 1,

n

k

=

1

,

,

1

1

=

λ

=

n

k

k . If:

(5)

(

)

1

)

(

)

(

)

1

(

1

2

α

'

z

f

z

f

z

z

k

k , () zU,

k

=

1

,

n

,

then the function Hα defined by (4) is analytic and univalent in U.

Proof. Because f1λ1(z)⋅ f2λ2(z) ... fnλn(z) = z + a2z2 + … is analytic in U, there

exists a number r1 ∈ (0, 1] such that

(

)

(

)

(

)

0

2

1

2

λ λ

z

z

f

...

z

f

z

f

n

n for any z

1 r

U

. Then for the function

1 n

2

1

(

)

1

(

)

2

(

)

n

− α λ λ

λ

z

z

f

....

z

z

f

z

z

f

we can choose the uniform branch equal to 1 at the origin, analytic in

1 r

U

: (6)

n

z

z

f

....

z

z

f

z

z

f

z

g

n

λ λ

λ

=

(

)

(

)

(

)

)

(

1 1 2 2

1 = 1 + b1z +

... + bnzn + ..., and we have:

(7) 0 u 1g1(u)du z g2(z,t ) z

e t α α

=

− ,

where:

(8) 2

(

)

=

α

1

tα

+

...

+

α

+

n

e

−(α+n)t

z

n

+

...

n

b

e

t

z,

g

Let we consider the function:

(9)

g

3

(

z,

t

)

=

α

f

2

(

z,

t

)

+

α

(

e

t

e

t

)

e

t(α−1)

g

1

(

e

t

z

)

Since |α− 1| < 1 we have g3(0,t )=e−αt (1−α+αe2t)≠0 for any tI and it

(4)

function [g3(z, t)]1/α we can choose an uniform branch, analytic in

U

r0 for any tI. It

results that the function:

(10) L(z, t) = z [g3(z, t)]1/α = [g4(z, t)]1/α,

where:

(11)

g

(

z,

t

)

e tz

(

f

(

u

)

f

(

u

)

...

f

nn

(

u

)

)

d

u

1

0 1 2

4 1 2

− α λ λ

λ

α

=

+

z

e

e

t t

)

(

α

+

(

)

1

1λ1(ez) ... fλ (ez) α−

f t nn t

is analytic in

0 r

U .

Using Lemma 1 we will prove that L is a subordination chain. We observe that L(z, t) = a1(t) z + …, where:

(12) a1(t) = et (1 −α + αe2t)1/α.

Because |α− 1| < 1 we have a1(t) ≠ 0 for all tI and

(13) = α

[

α+α

]

α

α

1

2 2

1( ) (1 ) t

t

e e

t a and

(14) = α α α =∞

α − ∞ → ∞

1

2 Re

1( ) lim

lim a t e t

t

t if

α

α

2

Re

> 1 or |α− 1| < 1.

It follows that

)

(

)

(

1

t

a

t

z,

L

forms a normal family of analytic functions in

2 r

U

,

2

0 2

r

r

=

.

Let p :

0 r

U

×I,

t

t

z,

L

z

t

z,

L

z

t

z,

p

=

)

(

)

(

)

(

.

In order to prove that p has an analytic extension with positive real part in U, for all tI it is sufficient to prove that the function:

(5)

(15)

1

)

(

1

)

(

)

(

+

=

t

z,

p

t

z,

p

t

z,

w

is analytic in U for tI and (16) |w (z, t)| < 1, (∀) zU, tI.

But

(

)

max

(

)

(

)

1

w

z,

t

w

e

t,

t

z,

w

i

z

θ

=

=

<

, θ∈R and it is sufficient that (17)

w

(

e

iθ

t,

)

1

, () t> 0, or

(18)

(

)

f

f

u

u

u

f

u

f

u

f

u

f

u

u

n n

n

+

+

+

α

)

(

)

(

'

λ

...

)

(

)

(

'

λ

)

(

)

(

'

λ

)

1

(

1

1

2 2 2 1

1 1 2

(

)

1

) (

) ( ' ) 1 ( 1

1 2

1

= λ ≤ ⋅

− α −

λ

n k

k k n

k

u f

u f u

u ,

from (5) where u = eteiθ, uU, |u| = et. It results (16).

Hence the function L is a subordination chain and L(z, t) = Hα(z) from (4) is

analytic and univalent in U.

Theorem 2. If f1, f2, …, fnS, 0 ≤λk≤ 1,

1

1

=

λ

=

n

k

k , then for α∈C, |α – 1|

4

1

, the function Hα defined by (4) belongs to the class S.

Proof. Because fk S,

k

=

1

,

n

we have:

z z z f

z f z

k k

− + ≤ ⋅

1 1 )

( ) ( '

, (∀) zU, then

(

)

(

1

)

4

) (

) ( '

1− 2 ⋅ ≤ + z 2<

z f

z f z z

k k

and

(

)

1

(

1

)

4 1 1

) (

) ( ' ) 1 (

1 2 z 2

z f

z f z

z

k

k α + < α

⋅ − α −

because |α – 1|

4

1

(6)

Example. Let 1 2

)

1

(

)

(

z

z

z

f

=

,

z

z

z

f

=

1

)

(

2 and f3(z) = z.

Then, for

3

1

1

=

λ

,

4

1

2

=

λ

,

12

5

3

=

λ

we have:

) ( ) ( )

( 2 3

1

3 2

1 z f z f z

fλ ⋅ λ ⋅ λ

=

=

λ λ 3

2 2 1

1

λ λ

2 λ

)

1

(

)

1

(

z

z

z

z

z

) λ ( 1 2

λ

2 1 2

1 2 1

)

1

(

λ + − λ + λ λ +

=

z

z

z

2 1

2

)

1

(

λ +λ

=

z

z

1211

)

1

(

z

z

=

α

    

    

    

  

− α

=

− α α

1

1211

0

1

d )

1 ( )

( z u

z z z

H .

For

α

=

2

3

3 2 2 1

12 11 2

3

d

)

1

(

2

3

)

(

0

=

u

z

z

z

H

z .

References

[1]

Y.J., Kim, E.P., Merkes:

On an integral of powers of a spirallike function. Kyungpook Math. J., vol. 12, nr. 2, (1972), p. 191-210.

[2]

Silvia Moldoveanu, N.N., Pascu: Integral operators which preserve the univalence. Mathematica (Cluj), 32 (55), nr. 2, (1990), p. 159-166.

[3]

J. Pfaltzgraff: Univalence of the integral

(

f

'

(

z

)

)

c. Bull. London Math. Soc. 7, (1975), nr. 3, p. 254-256.

[4]

Ch., Pommerenke: Uber die subordinaction analytischer Functionen. J. Reine Angew. Math 218 (1965), p. 159-173.

Authors:

Maria E. Gageonea and Silvia Moldoveanu - Transilvania University of Braşov Departament of Mathematics Braşov

References

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