paper 10-17-2009. 111KB Jun 04 2011 12:03:06 AM

ACTA UNIVERSITATIS APULENSIS

No 17/2009

SUFFICIENT CONDITIONS FOR UNIVALENCE OF A
GENERAL INTEGRAL OPERATOR

C.Selvaraj and K.R.Karthikeyan
Abstract. Using Dziok-Srivastava operator, we define a new family of
integral operators. For this general integral operator we study some interesting
univalence properties, to which a number of univalent conditions would follow
upon specializing the parameters involved.
2000 Mathematics Subject Classification: 30C45.
Keywords and Phrases: Analytic functions, univalent functions, Hadamard
product (or convolution), integral operators.
1. Introduction, Definitions And Preliminaries
Let A1 = A denote the class of functions of the form
f (z) = z +

∞
X

an z n

an ≥ 0,

(1)

n=2

which are analytic in the open disc U = {z : z ∈ C | z |< 1} and S be the
class of function f (z) ∈ A which are univalent in U.
For αj ∈ C (j = 1, 2, . . . , q) and βj ∈ C \ {0, −1, −2, . . .} (j = 1, 2, . . . , s),
the generalized hypergeometric function q Fs (α1 , . . . , αq ; β1 , . . . , βs ; z) is defined by the infinite series
q Fs (α1 , α2 , . . . , αq ; β1 , β2 , . . . , βs ; z) =

∞
X
(α1 )n . . . (αq )n z n
(β1 )n . . . (βs )n n!
n=0

(q ≤ s + 1; q, s ∈ N0 = {0, 1, 2, . . .}; z ∈ U),
87

C.Selvaraj, K.R.Karthikeyan - Sufficient Conditions For Univalence
where (x)n is the Pochhammer symbol defined by
(
1
if n = 0
(x)n =
x(x + 1)(x + 2) . . . (x + n − 1)
if n ∈ N = {1, 2, , . . .}.
Corresponding to a function h(α1 , . . . , αq ; β1 , . . . , βs ; z) defined by
h(α1 , . . . , αq ; β1 , . . . , βs ; z) := z q Fs (α1 , α2 , . . . , αq ; β1 , β2 , . . . , βs ; z),
the Dziok and Srivastava operator H(α1 , . . . , αq ; β1 , . . . , βs ; z)f (z) is defined
by the Hadamard product
H(α1 , . . . , αq ; β1 , . . . , βs ; z)f (z) := h(α1 , . . . , αq ; β1 , . . . , βs ; z) ∗ f (z)

For convenience, we write

∞
X
(α1 )n−1 . . . (αq )n−1 an z n
=z+
(β1 )n−1 . . . (βs )n−1 (n − 1)!
n=2

(2)

Hsq (α1 , β1 )f := H(α1 , . . . , αq ; β1 , . . . , βs ; z)f (z).
The linear operator Hsq (α1 , β1 )f includes (as its special cases) various other
linear operators which were introduced and studied by Hohlov, Carlson and
Shaffer, Ruscheweyh. For details see [1].
Using Dziok-Srivastava operator, we now introduce the following:
Definition 1. For γ ∈ C. We now define the integral operator Fγ (α1 , β1 ; z) :
An −→ A
Z

γ−1

z
Fγ (α1 , β1 ; z) =

Hsq (α1 , β1 )f1 (t)
...
(3)
1 + n(γ − 1)
0

1
! 1+n(γ−1)


γ−1
,
Hsq (α1 , β1 )fn (t)
dt

where fi ∈ A and Hsq (α1 , β1 )f (z) is the Dziok-Srivastava operator.

Remark 1. It is interesting to note that the integral operator Fγ (α1 , β1 ; z)
generalizes many operators which were introduced and studied recently. Here
we list a few of them
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C.Selvaraj, K.R.Karthikeyan - Sufficient Conditions For Univalence
1. Let q = 2, s = 1, α1 = β1 and α2 = 1, then the operator Fγ (α1 , α1 ; z)
reduces to an integral operator

Z
Fγ (z) = (n(γ − 1) + 1)

0

z

1
 (n(γ−1)+1)

γ−1

γ−1
f1 (t)
. . . fn (t)
dt.
, (4)

studied by D.Breaz and N.Breaz [2].

2. When q = 2, s = 1, α1 = 2, β1 = 1 and α2 = 1, the integral operator
Fγ (2, 1; z) reduces to an operator of the form

Z
Fγ (z) = (n(γ−1)+1)

z
n(γ−1)

t

0

1
 (n(γ−1)+1)

γ−1

γ−1

′
dt.
. . . fn (t)
f1 (t)
′

,
(5)

where γ ≥ 0.
We now state the following lemma which we need to establish our results
in the sequel.
Lemma 1.[3] If f ∈ A satisfies the condition
 2 ′

 z f (z)



 f 2 (z) − 1 ≤ 1, for all z ∈ U,

(6)

then the function f (z) is univalent in U.

Lemma 2.(Schwartz Lemma) Let f ∈ A satisfy the condition | f (z) |≤ 1,
for all z ∈ U, then
| f (z) |≤| z |, for all z ∈ U,
and equality holds only if f (z) = ǫz, where | ǫ |= 1.
Pascu [4, 5] has proved the following result
Lemma 3. Let β ∈ C, Reβ ≥ α > 0. If f ∈ A satisfies
 ′′ 
1− | z |2Re α  zf (z) 
 f ′ (z)  ≤ 1 (z ∈ U),
Re α

then the integral operator

 Z
Fβ (z) = β

z

0

89

β−1

t

 β1
f (t) dt
′

C.Selvaraj, K.R.Karthikeyan - Sufficient Conditions For Univalence
is univalent.
Lemma 4. Let g ∈ A satisfies the condition (6) and let α be a complex
number with | α − 1 |≤ Re3 α . If | g(z) |≤ 1, ∀ z ∈ U then the function
z

 Z
Gα (z) = α

g

(α−1)

0

 α1
(t)dt

belongs to S.
2.Main Results
Theorem 1. Let fi ∈ A, γ ∈ C. If

 2 q

 z (Hs (α1 , β1 )fi (z))′

 ≤ 1, | γ − 1 |≤ Re γ ,

−
1
q

 (Hs (α1 , β1 )fi (z))2
3n

and | Hsq (α1 , β1 )fi (z) |≤ 1

for all z ∈ U then Fγ (α1 , β1 ; z) given by (3) is univalent.
Proof. From the definition of the operator Hsq (α1 , β1 )f (z), it can be easily
seen that
Hsq (α1 , β1 )f (z)
6= 0 (z ∈ U)
z
and moreover for z = 0, we have
γ−1
γ−1
 q
 q
Hs (α1 , β1 )fn (z)
Hs (α1 , β1 )f1 (z)
...
=1
z
z
From (3), we have
Fγ (α1 , β1 ; z) =



1 + n(γ − 1)

Z

z
n(γ−1)

t

0





Hsq (α1 , β1 )f1 (t)
t

Hsq (α1 , β1 )fn (t)
t

γ−1

γ−1

...
1
! 1+n(γ−1)

dt

We consider the function
γ−1
γ−1
 q
Z z q
Hs (α1 , β1 )f1 (t)
Hs (α1 , β1 )fn (t)
h(z) =
...
dt
t
t
0
90

.

(7)

C.Selvaraj, K.R.Karthikeyan - Sufficient Conditions For Univalence
The function h is regular in U and from (7) we obtain
′

h (z) =



Hsq (α1 , β1 )f1 (z)
z

γ−1

...



Hsq (α1 , β1 )fn (z)
z

γ−1

and
 q

′
(Hs (α1 , β1 )f1 (z))
1 ′
−
h (z) + . . .
h (z) = (γ − 1)
Hsq (α1 , β1 )f1 (z)
z

 q
′
1 ′
(Hs (α1 , β1 )fn (z))
−
h (z)
+(γ − 1)
Hsq (α1 , β1 )fn (z)
z
′′

From the above inequalities we have



′
′′
z(Hsq (α1 , β1 )f1 (z))
zh (z)
= (γ − 1)
− 1 + ...
h′ (z)
Hsq (α1 , β1 )f1 (z)


′
z(Hsq (α1 , β1 )fn (z))
−1 .
+(γ − 1)
Hsq (α1 , β1 )fn (z)
On multiplying the modulus of equation (8) by

1−|z|2Re γ
Re γ

(8)

> 0, we obtain

"
 ′′ 



 z(Hsq (α1 , β1 )f1 (z))′ 
1− | z |2Re γ  zh (z)  1− | z |2Re γ


| γ −1 | 
+1 +
 h′ (z)  ≤
Re γ
Re γ
Hsq (α1 , β1 )f1 (z) 


#
 z(Hsq (α1 , β1 )fn (z))′ 
+1
. . . + | γ − 1 | 
Hsq (α1 , β1 )fn (z) 


′
1− | z |2Re γ  z 2 (Hsq (α1 , β1 )f1 (z))  | Hsq (α1 , β1 )f1 (z) |
+ 1 + ...
≤| γ − 1 |
 (Hsq (α1 , β1 )f1 (z))2 
Re γ
|z|


′
1− | z |2Re γ  z 2 (Hsq (α1 , β1 )fn (z))  | Hsq (α1 , β1 )fn (z) |
+|γ−1|
+1 .
 (Hsq (α1 , β1 )fn (z))2 
Re γ
|z|

Since Hsq (α1 , β1 )f (z) satisfies the conditions of the Schwartz Lemma, on ap-

91

C.Selvaraj, K.R.Karthikeyan - Sufficient Conditions For Univalence
plying the same on the above inequality, we obtain
 
 ′′ 

′

1− | z |2Re γ  zh (z) 
1− | z |2Re γ  z 2 (Hsq (α1 , β1 )f1 (z))
+2 +
−1
≤|
γ−1
|
 (Hsq (α1 , β1 )f1 (z))2
 h′ (z) 

Re γ
Re γ



′

1− | z |2Re γ  z 2 (Hsq (α1 , β1 )fn (z))
+2
−
1
... + | γ − 1 |
q

 (Hs (α1 , β1 )fn (z))2
Re γ
3|γ−1|
3n | γ − 1 |
3|γ−1|
+ ... +
=
.
≤
Re γ
Re γ
Re γ
But | γ − 1 |≤

Re γ
,
3n

so we obtain for all z ∈ U
 ′′ 
1− | z |2Re γ  zh (z) 
 h′ (z)  ≤ 1.
Re γ

(9)

It follows from Lemma 3 that



1 + n(γ − 1)

Z

1
! 1+n(γ−1)

z
′

tn(γ−1) h (t) dt

0

∈S

Since
Z



1 + n(γ − 1)


1 + n(γ − 1)

= Fγ (α1 , β1 ; z),

z
′

tn(γ−1) h (t) dt

0

Z

0

z

1
! 1+n(γ−1)

=

1
! 1+n(γ−1)




γ−1
γ−1
Hsq (α1 , β1 )f1 (t)
. . . Hsq (α1 , β1 )fn (t)
dt

hence Fγ (α1 , β1 ; z) ∈ S.
Putting q = 2, s = 1, α1 = β1 and α2 = 1, in Theorem 1, we have
Corollary 1. [2] Let fi ∈ A, γ ∈ C. If

 2 ′
 z fi (z)


 ≤ 1, | γ − 1 |≤ Re γ ,
−
1
 (fi (z))2

3n
92

and | fi (z) |≤ 1

C.Selvaraj, K.R.Karthikeyan - Sufficient Conditions For Univalence
for all z ∈ U then the operator
1
 (n(γ−1)+1)

γ−1

γ−1
f1 (t)
. . . fn (t)
dt

z


Z
Fγ (z) = (n(γ − 1) + 1)

0

is univalent.

Theorem 2. Let f ∈ A satisfy

 2 q

 z (Hs (α1 , β1 )f (z))′
 < 1,

−
1
q

 (Hs (α1 , β1 )f (z))2

and γ ∈ C with | γ − 1 |≤
function
Fγk (α1 , β1 ; z)

=



Re γ
.
3k

∀z ∈ U

If | Hsq (α1 , β1 )f (z) |≤ 1, ∀ z ∈ U, then the



k(γ − 1) + 1

Z

z

(Hsq (α1 , β1 )f (t))k(γ−1)

0

1
 k(γ−1)+1
dt

(10)

is univalent.
Proof. From (10) we have
Fγk (α1 , β1 ; z) =



k(γ − 1) + 1

and let
h(z) =

Z

0

z

Z



z

tk(γ−1)

0



Hsq (α1 , β1 )f (t)
t

Hsq (α1 , β1 )f (t)
t

k(γ−1)

!k(γ−1)

1
! k(γ−1)+1

dt

dt.

From here using arguments similar to those detailed in Theorem 1, we can
prove the assertion of the Theorem 2.
We note that on restating the Theorem 2 for the choice of q = 2, s =
1, α1 = β1 and α2 = 1, we have the result proved by D.Breaz and N.Breaz [2].
References
[1] R. Aghalary et al., Subordinations for analytic functions defined by the
Dziok-Srivastava linear operator, Appl. Math. Comput., 187, (2007), no. 1,
13–19.
93

C.Selvaraj, K.R.Karthikeyan - Sufficient Conditions For Univalence
[2] D. Breaz and N. Breaz, An integral univalent operator, Acta Math.
Univ. Comenian. (N.S.), 7, (2007), no. 2, 137–142.
[3] S. Ozaki and M. Nunokawa, The Schwarzian derivative and univalent
functions, Proc. Amer. Math. Soc., 33, (1972), 392–394.
[4] N. N. Pascu, On a univalence criterion. II, in Itinerant seminar on
functional equations, approximation and convexity (Cluj-Napoca, 1985), 153–
154, Univ. “Babe¸s-Bolyai”, Cluj.
[5] N. N. Pascu, An improvement of Becker’s univalence criterion, in Proceedings of the Commemorative Session: Simion Sto¨ılow (Bra¸sov, 1987), 43–
48, Univ. Bra¸sov, Bra¸sov.
Authors:
C.Selvaraj
Department of Mathematics,
Presidency College,
Chennai-600 005,
Tamilnadu, India
email: pamc9439@yahoo.co.in
K.R.Karthikeyan
R.M.K.Engineering College
R.S.M.Nagar,
Kavaraipettai-601206,
Thiruvallur District, Gummidipoondi Taluk,
Tamilnadu,India
email: kr karthikeyan1979@yahoo.com.

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