paper 15-17-2009. 112KB Jun 04 2011 12:03:07 AM

ACTA UNIVERSITATIS APULENSIS

No 17/2009

CRITERIA FOR UNIVALENCE OF CERTAIN
INTEGRAL OPERATORS

V. Ravichandran
Abstract. In this paper, we determine conditions on β, αi and fi (z) so
n R
o1/β
Q
z
that the integral operator β 0 ξ β−1 ni=1 (fi (ξ)/ξ)1/αi dξ
is univalent in
the open unit disk. We also obtain similar results for the integral operator
 R z β−1
1
P
β 0 ξ
exp ( ni=1 αi fi (ξ)) dξ β .

2000 Mathematics Subject Classification: 30C45.
1. Introduction
An analytic function that maps different points in the open unit disk U :=
{z ∈ C : |z| < 1} to different points in C is a univalent function. Such functions
have been studied for long time. Let A be the class of all analytic functions
f (z) defined in U and normalized by the conditions f (0) = 0 = f ′ (0) − 1. Let
S be the subclass of A consisting of univalent functions. Pascu [2] has proved
the following theorem:
Theorem 1 ([1],[2]) Let β ∈ C, γ ∈ R and ℜβ ≥ γ > 0. If f ∈ A satisfies


1 − |z|2γ  zf ′′ (z) 
 f ′ (z)  ≤ 1 (z ∈ U ),
γ

then the integral operator

 Z
Fβ (z) = β

z

ξ

0

is in S.
141

β−1 ′

f (ξ)dξ

 β1

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
We denote by Sβ the class of functions f ∈ A satisfying
 2 ′

 z f (z)




 f 2 (z) − 1 < β (0 < β ≤ 1; z ∈ U )

while Sβ∗ denote the class of functions f ∈ A satisfying

 ′

 zf (z)
 < β (0 < β ≤ 1; z ∈ U ).

−
1

 f (z)

Using Theorem 1, Pescar [4] proved the following theorem.

Theorem 2 ([4]) Let α, β ∈ C and ℜβ ≥ ℜα ≥ 3/|α|. If f ∈ S1 satisfies the

condition
|f (z)| ≤ 1 (z ∈ U ),
then the integral operator
( Z
Hα,β (z) = β

z

ξ β−1

0



f (ξ)
ξ

 α1

dξ

) β1

(1)

is in S.
Using Theorem 1, Breaz and Breaz [1] extended Theorem 2 and obtained
the following theorem.
Theorem 3 (Theorem 1, p.260 [1]) Let α, β ∈ C and ℜβ ≥ ℜα > 3n/|α|. If
fi ∈ S1 (i = 1, 2, · · · , n) satisfies the conditions
|fi (z)| ≤ 1

(z ∈ U,

i = 1, 2, · · · , n),

then the integral operator
( Z
Fα,β (z) = β

1

z

ξ β−1

0


n 
Y
fi (ξ) α
i=1

ξ

dξ

) β1

is in S.
Using Theorem 1, Pescar [5] obtained the following theorem.
142

(2)

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
Theorem 4 (Theorem 1, p.452 [5]) Let f ∈ A and β ∈ C satisfies
√
3 3
1 ≤ ℜβ ≤ |β| ≤
.
2
If
then the function

|zf ′ (z)| ≤ 1
 Z
Tβ (z) = β

(z ∈ U ),

z

ξ β−1

0

is in S.


β
ef (ξ) dξ

 β1

By Schwarz’s Lemma (see below), the only function f ∈ A with |f (z)| ≤ 1
is f (z) = z and hence the hypothesis of Theorem 3–4 are satisfied only by a
single function f (z) = z.
In this paper, we extend Theorem 2 and Theorem 4 to obtain a sufficient
conditions for univalence of a more general integral operator. We also prove

some related results. The class of functions to which our theorems are applicable is non-trivial (when Mi 6= 1 for some i).
To prove our main results, we need the following lemma:
Lemma 1 (Schwarz’s Lemma) If the function w(z) is analytic in the unit
disk U , w(0) = 0 and |w(z)| ≤ 1 for all z ∈ U , then
|w(z)| ≤ |z| (z ∈ U ),

|w′ (0)| ≤ 1

and the either of these equalities holds if and only if w(z) = ǫz, where |ǫ| = 1.
2. Univalence Criteria
For fi ∈ A (i = 1, 2, · · · , n) and α1 , α2 , . . . , αn , β ∈ C, we define an integral
operator by
( Z
Fα1 ,α2 ,··· ,αn ;β (z) := β

1

z

ξ β−1

0


n 
Y
fi (ξ) αi
i=1

ξ

dξ

) β1

.

(3)

When α1 = α2 = · · · = αn = α, Fα1 ,α2 ,··· ,αn ;β (z) becomes the integral operator

Fα,β (z) considered in Theorem 3.
143

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
Theorem 5 Let 0 < βi ≤ 1. Let fi ∈ Sβi (i = 1, 2, · · · , n) satisfy the conditions
|fi (z)| ≤ Mi (Mi ≥ 1; z ∈ U, i = 1, 2, · · · , n).
If α1 , α2 , . . . , αn , β ∈ C, ℜβ ≥ γ and
γ :=

n
X
1 + Mi (1 + βi )

|αi |

i=1

,

(4)

then the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is in S.
Proof.

Define the function h(z) by
h(z) :=

Z

0

n 
zY
i=1

fi (ξ)
ξ

 α1

i

dξ.

Then we have h(0) = h′ (0) − 1 = 0. Also a simple computation yields
1

′

h (z) =


n 
Y
fi (z) αi
z

i=1

and

n

zh′′ (z) X 1
=
h′ (z)
αi
i=1




zfi′ (z)
−1 .
fi (z)

(5)

From equation (5), we have


 ′′ 


n
X
 zh (z) 
1  zfi′ (z) 

 ≤
+1
 h′ (z) 
|αi |  fi (z) 
i=1




n
X
1  z 2 fi′ (z)   fi (z) 
=
+1 .
|αi |  fi2 (z)   z 
i=1

From the hypothesis, we have |fi (z)| ≤ Mi
by Schwartz Lemma, yields
|fi (z)| ≤ Mi |z| (z ∈ U,

144

(z ∈ U,

(6)

i = 1, 2, · · · , n) which,

i = 1, 2, · · · , n).

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
Using this inequality in the inequality (6), we obtain
 ′′ 

  2 ′ 
n
X
 zh (z) 
 z fi (z) 
1

 ≤


+1
Mi  2
 h′ (z) 
|αi |
fi (z) 
i=1


  2 ′
n
X

 z fi (z)
1
Mi  2
− 1 + 1 + Mi .
≤
|α
|
f
(z)
i
i
i=1
Since fi ∈ Sβi , in view of (4), the equation (7) yields
 ′′ 
n
X
 zh (z) 
1 + Mi (1 + βi )
 <

= γ.
 h′ (z) 
|αi |
i=1

(7)

(8)

Multiply (8) by

1 − |z|2γ
,
γ

we obtain
1 − |z|2γ
γ

 ′′ 
 zh (z) 
2γ


 h′ (z)  ≤ 1 − |z| < 1 (z ∈ U ).

Since ℜβ ≥ γ > 0, it follows from Theorem 1 that
 β1
 Z z
β−1 ′
ξ h (ξ)dξ ∈ S.
β
0

Since
 Z
β

0

z

ξ β−1 h′ (ξ)dξ

 β1

( Z
= β

1

z

ξ β−1

0


n 
Y
fi (ξ) αi
i=1

ξ

dξ

) β1

= Fα1 ,α2 ,··· ,αn ;β (z),

the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is in S.
Theorem 6 Let α1 , α2 , . . . , αn , β ∈ C, 0 < βi ≤ 1 (i = 1, 2, . . . , n) and
ℜβ ≥ γ :=

n
X
βi
.
|α
|
i
i=1

(9)

If fi ∈ Sβ∗i (i = 1, 2, · · · , n), then the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is
in S.
145

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
Proof.

From (5), we get
 ′′ 
 ′

n
 zh (z)  X
 zfi (z)

1

≤


−
1
 h′ (z) 
 fi (z)

|α
|
i
i=1

and by using (9) and fi ∈ Sβ∗i , we obtain

 ′′ 
n
 zh (z)  X
βi
≤

= γ.
 h′ (z) 
|α
|
i
i=1

This, as in the proof of Theorem 5, shows that


1 − |z|2γ  zh′′ (z) 
 h′ (z)  < 1 (z ∈ U )
γ

and therefore, by Theorem 1,

 Z
β

z

ξ β−1 h′ (ξ)dξ

0

 β1

∈ S.

Therefore the function Fα1 ,α2 ,··· ,αn ;β (z) defined by (3) is in S.
Example 1 Let α, β ∈ C, 0 < βi ≤ 1 (i = 1, 2, . . . , n) satisfy
n

1 X
ℜβ ≥
βi .
|α| i=1
If fi ∈ Sβ∗i (i = 1, 2, · · · , n), then the function Fα,β (z) defined by (2) is in S.
β1
and f ∈ Sβ∗1 , then the
In particular, if α, β ∈ C, 0 < β1 ≤ 1 satisfy ℜβ ≥ |α|
function Hα,β (z) defined by (1) is in S.
For i = 1, 2, . . . , n, let αi , β ∈ C and fi (z) ∈ A. Define the integral operator
Tα1 ,...,αn ;β (z) by
( Z
Tα1 ,...,αn ;β (z) = β

z

ξ β−1 exp

0

n
X
i=1

146

!

αi fi (ξ) dξ

) β1

.

(10)

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
Theorem 7 Let fi ∈ A (i = 1, 2, · · · , n) satisfy
|zfi′ (z)| ≤ Mi

(Mi ≥ 1;

z ∈ U ).

Let αi , β ∈ C (i = 1, 2, . . . , n) and γ > 0 be the smallest number such that
2

n
X
i=1

Mi |αi | ≤ (2γ + 1)

2γ+1
2γ

.

(11)

Then, for β ∈ C, ℜβ ≥ γ, the function Tα1 ,...,αn ;β (z) defined (10) is in S.
Proof.

Define the function h(z) by
h(z) :=

Z

z

exp

0

n
X

!

αi fi (ξ) dξ.

i=1

Then we have h(0) = h′ (0) − 1 = 0. Also a simple computation yields
!
n
X
′
αi fi (z)
h (z) = exp
i=1

and

n

zh′′ (z) X
=
αi zfi′ (z).
′
h (z)
i=1

(12)

By Schwartz’s Lemma, we have |zfi′ (z)| ≤ |z|, (z ∈ U, i = 1, 2, · · · , n), and
therefore we obtain, from (12),
 ′′ 
n
n
X
 zh (z)  X
′

≤
|α
|
|zf
(z)|
≤
(
Mi |αi |)|z|.
i
i
 h′ (z) 
i=1
i=1

Thus



n
1 − |z|2γ  zh′′ (z)  |z|(1 − |z|2γ ) X
Mi |αi | (z ∈ U ).
 h′ (z)  ≤
γ
γ
i=1

For the function Q : [0, 1] → R defined by Q(t) = t(1 − t2γ ), γ > 0, the
maximum is attained at the point t = 1/(2γ + 1)1/2γ and thus we have
Q(t) ≤

2γ
(2γ + 1)
147

2γ+1
2γ

.

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
In view of this inequality and our assumption (11), we obtain


1 − |z|2γ  zh′′ (z) 
 h′ (z)  ≤ 1 (z ∈ U )
γ

and, by Theorem 1, Tα1 ,...,αn ;β (z) is univalent in U for β ∈ C with ℜβ ≥ γ.
Also we have the following result.

Theorem 8 Let αi , β ∈ C, 0 ≤ βi ≤ 1 (i = 1, 2, . . . , n). Let fi ∈ Sβ∗i (i =
1, 2, · · · , n) satisfy
|fi (z)| ≤ Mi

(Mi ≥ 1;

z ∈ U;

i = 1, 2, . . . , n).

and γ > 0 be the smallest number such that
2

n
X
i=1

Mi |αi |(1 + βi ) ≤ (2γ + 1)

2γ+1
2γ

.

(13)

Then, for β ∈ C, ℜβ ≥ γ, the function Tα1 ,...,αn ;β (z) defined (10) is in S.
Proof.

From (12), we have
 ′′  X
n
 zh (z) 

≤
|αi |
 h′ (z) 
i=1

 ′ 
 zfi (z) 


 fi (z)  |fi (z)|

 ′

n
X

 zfi (z)
− 1 |fi (z)|
≤
|αi | 1 + 
fi (z)
i=1
≤

n
X
i=1

Mi |αi |(1 + βi )|z|.

The remaining part of the proof is similar to the proof of Theorem 7.
Similarly we have the following result.
Theorem 9 Let fi ∈ Sβi (i = 1, 2, · · · , n) satisfy
|fi (z)| ≤ Mi

(Mi ≥ 1;

z ∈ U;

i = 1, 2, . . . , n).

Let αi , β ∈ C, 0 ≤ βi ≤ 1 (i = 1, 2, . . . , n), and γ > 0 be the smallest number
such that
n
X
2γ+1
2
Mi2 |αi |(1 + βi ) ≤ (2γ + 1) 2γ .
(14)
i=1

Then, for β ∈ C, ℜβ ≥ γ, the function Tα1 ,...,αn ;β (z) defined (10) is in S.
148

V. Ravichandran - Criteria for Univalence of Certain Integral Operators
References
[1] D. Breaz and N. Breaz, An univalent condition for an integral operator,
Nonlinear Funct. Anal. Appl. 11 (2006), no. 2, 259–263.
[2] N. N. Pascu, On a univalence criterion II, in Itinerant seminar on functional equations, approximation and convexity (Cluj-Napoca, 1985), 153–154,
Univ. “Babeş-Bolyai”, Cluj.
[3] N. N. Pascu, An improvement of Becker’s univalence criterion, in Proceedings of the Commemorative Session: Simion Stoı̈low (Braşov, 1987), 43–48,
Univ. Braşov, Braşov.
[4] V. Pescar, New criteria for univalence of certain integral operators,
Demonstratio Math. 33 (2000), no. 1, 51–54.
[5] V. Pescar, Certain sufficient conditions for univalence, Hokkaido Math.
J. 32(2) (2003), 451–455.
[6] V. Pescar and S. Owa, Some criteria for univalence of certain integral
operators, Int. J. Math. Math. Sci. 2004, no. 45-48, 2489–2494.
Author:
V. Ravichandran
Department of Mathematics
University of Delhi
Delhi 110 007 India
email: vravi@maths.du.ac.in

149