ICTAMI 2005. 110KB Jun 04 2011 12:08:47 AM

ACTA UNIVERSITATIS APULENSIS

No 10/2005

Proceedings of the International Conference on Theory and Application of
Mathematics and Informatics ICTAMI 2005 - Alba Iulia, Romania

AN INTEGRAL UNIVALENT OPERATOR II
Daniel Breaz and Nicoleta Breaz
Abstract.In this paper we present the univalence conditions for the operator


Gα,n (z) = (n (α − 1) + 1)

in the unit disc.

Zz
0



1
n(α−1)+1

g1α−1 (t) . . . gnα−1 (t) dt

2000 Mathematics Subject Classification: 30C45
Keywords and phrases:Operator, unit disc, univalent functions, univalent
operator, holomorphic function.
Introduction
We consider U be the unit disc and denote by H (U ) the class of holomorphic functions in U . Let the set of analytic functions
n

An = f ∈ H (U ) : f (z) = z + an+1 z n+1 + . . .

o

(1)

For n = 1 obtain A1 = {f ∈ H (U ) : f (z) = z + a2 z 2 + . . .}, and denote
A1 = A. Let S the class of regular and univalent functions f (z) = z +a2 z 2 +. . .

in U, which satisfy the condition f (0) = f ′ (0) − 1 = 0.
Ozaki and Nunokawa proved in [3] the following results:
Theorem 1.If we assume that g ∈ A satisfies the condition



 z 2 f ′ (z)

 2
− 1 < 1, z ∈ U
 f (z)


(2)

then f is univalent in U .
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
and equality holds only if g (z) = εz, where |ε| = 1.

117

(3)

D. Breaz and N. Breaz-An integral univalent operator II
Theorem 2.Let α be a complex number with Re α > 0, and let f = z +
a2 z 2 + . . . be a regular function on U .
If


1 − |z|2Re α  zf ′′ (z) 
 ′
 ≤ 1, ∀z ∈ U
(4)
 f (z) 
Re α

then for any complex number β with Re β ≥ Re α the function


Fβ (z) = β

is regular and univalent in U .

Zz
0

1

β

tβ−1 f ′ (t) dt = z + . . .

(5)

Theorem 3.Assume that g ∈ A satisfies condition (2), and let α be a
complex number with
Re α
|α − 1| ≤
.

(6)
3
If
|g (z)| ≤ 1, ∀z ∈ U
(7)
then the function


Gα (z) = α

is of class S.

Zz
0

1

α

g (α−1) (t) dt

(8)

Main results
Theorem 4.Let gi ∈ A, for all i = 1, n, n ∈ N ∗ , satisfy the properties

and α ∈ C, with




 z 2 g ′ (z)

i
− 1 < 1, ∀z ∈ U, ∀i = 1, n
 2

 gi (z)

(9)

Re α
.
3n
If |gi (z)| ≤ 1, ∀z ∈ U, ∀i = 1, n, then the function

(10)

|α − 1| ≤



Gα,n (z) = (n (α − 1) + 1)

Zz
0



g1α−1 (t) . . . gnα−1 (t) dt

118

1
n(α−1)+1

(11)

D. Breaz and N. Breaz-An integral univalent operator II
is univalent.
Proof. From (11) Gα,n can be written:


Gα,n (z) = (n (α − 1) + 1)

Zz

tn(α−1)

0

g1 (t)
t

!α−1

...

gn (t)
t

!α−1



dt

1
n(α−1)+1

(12)

We consider the function
f (z) =

Zz
0

g1 (t)
t

!α−1

gn (t)
...
t

!α−1

dt.

(13)

The function f is regular in U , and from (13) we obtain
′

f (z) =

g1 (z)
z

!α−1

gn (z)
...
z

!α−1

(14)

and
f ′′ (z) = (α − 1)

n
X

Ak Bk

(15)

zgk′ (z) − gk (z)
z2

(16)

!α−1

(17)

k=1

where Ak and Bk has the next form:
Ak =

gk (z)
z

!α−2

and
z
Bk = f (z) ·
gk (z)
′

Next we calculate the expresion

.

zf ′′
.
f′

n
X
zgk′ (z) − 1
zf ′′ (z)
=
(α
−
1)
·
f ′ (z)
gk (z)
k=1

The modulus



 zf ′′ 


 ′ 
 f 

119

(18)

(19)

D. Breaz and N. Breaz-An integral univalent operator II
can then be evaluated as






n 
n
 zf ′′ (z) 

X
zgk′ (z) − 1 
zgk′ (z) − 1  X




 ′
(α − 1) ·
 = (α − 1) ·
≤
.

 f (z) 
gk (z)  k=1 
gk (z) 
k=1

Multiplying the first and the last terms of (20) with
1 − |z|2Re α
Re α

1−|z|2Re α
Re α



n
 zf ′′ (z) 
X
1 − |z|2Re α


|α − 1|
 ′
≤
 f (z) 
Re α

> 0, we obtain



!
 zg ′ (z) 

 k
+1

 gk (z) 

k=1

(20)

(21)

Applying the Schwartz Lemma and using (21), we obtain
1 − |z|2Re α
Re α



n
 zf ′′ (z) 
1 − |z|2Re α X


 ′
 ≤ |α − 1|
 f (z) 
Re α

k=1



!

 z 2 g ′ (z)


k
− 1 + 2
 2
 g (z)

(22)

k

Since gi satisfies the condition (2) ∀i = 1, n, then from (22) we obtain:
1 − |z|2Re α
Re α
But |α − 1| ≤



 zf ′′ (z) 
3n |α − 1|
1 − |z|2Re α


≤
.
 ′
 ≤ 3n |α − 1|
 f (z) 
Re α
Re α

Re α
3n

(23)

so from (10) we obtain that
1 − |z|2Re α
Re α



 zf ′′ (z) 


 ′
 ≤ 1,
 f (z) 

(24)

for all z ∈ U and the Theorem 2 implies that the function Gα,n is in the class
S.
Corollary 5.Let g ∈ A satisfy (2) and α a complex number such that
|α − 1| ≤

Re α
, k ∈ N∗
3k

(25)

If |g (z)| ≤ 1, ∀z ∈ U, then the function


Gkα (z) = (k (α − 1) + 1)
is univalent.
Proof.
120

Zz
0



g k(α−1) (t) dt

1
k(α−1)+1

(26)

D. Breaz and N. Breaz-An integral univalent operator II
We consider the functions


Gkα (z) = (k (α − 1) + 1)
and
f (z) =

Zz

tk(α−1)

0

Zz
0

g (t)
t

g (t)
t

!k(α−1)

!k(α−1)



1
k(α−1)+1

dt

dt.

(27)

(28)

The function f is regular in U. From (28) we obtain
′

f (z) =

g (z)
z

and
g (z)
f (z) = k (α − 1)
z
′′

!k(α−1)

!k(α−1)−1

zg ′ (z) − g (z)
.
z2

Next we have
1 − |z|2Re α
Re α



 zf ′′ (z) 
1 − |z|2Re α


 ′
≤
k |α − 1|
 f (z) 
Re α



!
 zg ′ (z) 



 + 1 , ∀z ∈ U.
 g (z) 

(29)

Applying the Schwartz Lemma and using (29) we obtain
1 − |z|2Re α
Re α



 zf ′′ (z) 
1 − |z|2Re α


 ′
 ≤ |α − 1|
 f (z) 
Re α



!
 z 2 g ′ (z)



 2
− 1 + 2 .
 g (z)


(30)

Since g satisfies conditions (2) then from (30) and (25) we obtain
1 − |z|2Re α
Re α



 zf ′′ (z) 

3k |α − 1|
3k |α − 1| 


1 − |z|2Re α ≤
 ′
≤
≤ 1.
 f (z) 
Re α
Re α

(31)

Now Theorem 2 and (31) imply that Gkα ∈ S.
Corollary 6.Let f, g ∈ A, satisfy (1) and α the complex number with the
property
Re α
|α − 1| ≤
.
(32)
6

121

D. Breaz and N. Breaz-An integral univalent operator II
If |f (z)| ≤ 1, ∀z ∈ U and |g (z)| ≤ 1, ∀z ∈ U , then the function


Gα (z) = (2α − 1)

Zz
0



1
2α−1

f (α−1) (t) g (α−1) (t) dt

(33)

is univalent.
Proof. In Theorem 4 we set n = 2, g1 = f , g2 = g.
Remark. Theorem 4 is a generalization of Theorem 3.
Remark. From Corollary 5, for k = 1, we obtain Theorem 3.
References
[1]Nehari, Z., Conformal Mapping, Mc Graw-Hill Book Comp., New York,
1952 Dover. Publ. Inc. 1975.
[2]Nunokava, M., On the theory of multivalent functions, Tsukuba J. Math.
11 (1987), no. 2, 273-286.
[3]Ozaki, S. Nunokawa, M., The Schwartzian derivative and univalent functions, Proc. Amer. Math. Soc. 33(2), (1972), 392-394.
[4]Pascu, N.N., On univalent criterion II, Itinerant seminar on functional
equations approximation and convexity, Cluj-Napoca, Preprint nr. 6, (1985),
153-154.
[5]Pascu, N.N., An improvement of Beker’s univelence criterion, Proceedings of the Commemorative Session Cimion Stoilow, Brasov, (1987), 43-48.
[6]Pescar, V., New criteria for univalence of certain integral operators,
Demonstratio Mathematica, XXXIII, (2000), 51-54.
D. Breaz and N. Breaz
Department of Mathematics
“1 Decembrie 1918” University
Alba Iulia, Romania
E-mail: dbreaz@uab.ro
E-mail: nbreaz@uab.ro

122