On Certain Subclasses of Multivalent Functions Associated
with a Family of Linear Operators
Jae Ho Choi
Department of Mathematics Education, Daegu National University of Education, Daegu, South Korea E-mail: choijh@dnue.ac.kr
Received March 16,2011; revised April 7, 2011; accepted April 20, 2011
Abstract
Making use of a linear operator , which is defined here by means of the Hadamard product (or convolution), we introduce some new subclasses of multivalent functions and investigate various inclusion properties of these subclasses. Some radius problems are also discussed.
,p a c
Keywords: Multivalent Functions, Hadamard Product (or Convolution), Linear Operators, Radius Problem
1. Introduction and Definitions
Let
p denote the class of functions f z
of the form
=1
= p p k ( : {1, 2,3, })
p k k
f z z a z p
, (1)which are analytic in the open unit disk
= z z: and z < 1
.
We define the Hadamard product (or convolution) of two analytic functions
=0 =0
= k and = k
k k
k k
f z
a z g z
b z
,
as
=0
: k
k k k
f g z
a b z z .For a, c 0 ( ) H. Saitoh [13] introduced a linear operator 0
: , 2, 1,
0
, :
p a c p p
defined by
, :
, ;
( ;
p a c f z p a c z f z z f p
)
(2) where
=0
, ; : k k p ( )
p
k k
a
a c z z z
c
, (3)and
k is the Pochhammer symbol defined, in terms of the Gamma function, by
=
= 1
(1 1 (
= 0) )
k
k k
k k
.
The operator p
a c, is an extension of the Carlson- Shaffer operator (see [2]). In [3], Cho et al. introduced the following family of linear operators
, :
p a c p p
analogous to p
a c, (see also [14]):
†
0
, : , ;
, ; > ; ;
p a c f z p a c z f z
a c p z f p
. (4) where †
, ;
p a c z
is the function defined in terms of the Hadamard product (or convolution) by the following condition
†
, ; , ; =
1
p
p p p
z a c z a c z
z
, (5)
where p is given by (3). If f z
is given by (1),then from (3), (4) and (5), we deduce that
=1 ,
= .
!
p
p k k k
k p
k k
a c f z p c
z a z
a k
p z (6)
It is easily seen from (6) that
1 1,1 =
p p f z f
z and 1
,1 =
p
zf z p f z
p
229
.
1,
= , 1,
p
p p
z a c f z
a a c f z a p a c f z
(7)
and
1
,
= , ,
p
p p
z a c f z
p a c f z a c f z
(8)
Clearly, from (7) and (8), we have
1,
Re > 0
1,
,
Re > ( )
1,
p p
p p
z a c f z a c f z
a c f z a p
a p a
a c f z
(9)
and
1
1,
Re > 0
,
,
Re > 0 .
,
p p
p p
z a c f z a c f z
a c f z
p a c f z
(10)
When a n p n=
0:
0
and c== 1, the linear operator 1
1 ,1 =
p n p n p
, was introduced and studied by Liu and Noor [5] (see also [9] and [10]). Moreover, when , was first introduced and studied by Noor [8] which is known as Noor Integral operator.
= 1
p 1
1 n1,1 = n
Let k
be the class of functions analyticin the unit disk satisfying the properties
h z
h
0 = 1and
2
0 Re
d 1
h z
k
, (11)where z=rei, k2 and 0 < 1. For = 0 , the
class k
0 =k was introduced in [11]. For = 0, , we have the well known class of functions with and the class gives us the class= 2
k
= 2
Re h z
> 0 k of functions with Re h
z >. Also we can write, for h z
k
as
2
0
1 1 2 1
= d
2 1
it it
ze
h z t
ze
, (12)where is a function with bounded variation on such that
t
[0, 2 ]
2 2
0 d t = 2 and 0 d t
From (11) and (12) it can be seen that hk( ) if and only if there exist h h1, 2() such that
1
2
1 1
=
4 2 4 2
k k
h z h z h z
. (14)
It is known [7] that the class k
is a convex set. We also note that h z
k
if and only if there exists qk such that
= 1
h z q z
. (15) By using the linear operator , we now define some subclasses of
,
p a c
p
ap as follows:
Definition 1: Let , c 0
, >p, > 0, 0
, k2 and 0 < 1. A function f z
p
, , ,
is said to be in the class p k a c, , , if and only if it satisfies
1
1, (1 )
1,
, 1,
,
, 1,
p p
p p
k
p p
a c f z a c g z
a c f z a c f z a c g z a c g z
(16) where g z
p satisfies the condition
,
(0 < 1; ) 1,
p p
a c g z
z a c g z
. (17)
We note that g is starlike univalent in when = = = = 1
a c p in (16).
Definition 2: Let a c, 0 , 0, > 0, 0, and
2
k 0 < 1. A function f z
p is said to be in the class p
k a c, , , , , ,
if and only if it satisfies
1 1
1
, (1 )
,
, ,
,
, ,
p p
p p
k
p p
a c f z a c g z
a c f z a c f z a c g z a c g z
(18) where g z
p satisfies the condition
1 ,
0 < 1; ,
p p
a c f z
z
a c g z .
(19)
k
. (13)
In this manuscript, we investigate several inclusion and other properties of functions in the classes
, , , , , ,
p k a c
2. Main Results
In order to establish our results, we require the following lemmas.
Lemma 1: [6] Let u u= 1iu2 and v v= 1iv2 and let be a complex-valued function satisfying the conditions:
( , )u v
1) ( , )u v is continuous in a domain
1,0 2,
2) and
1,0 > 0. 0
3) Re ( iu2,v1) whenever
iu v2, 1
and
2
1 1 2 2
v u .
If p z
is analytic in , with , such that
,zp p
0 = 1
p z z
and Re
p z z
, p z
> 0 for z, then Rep z
> 0.Lemma 2: [12] If is analytic in with
, and if
h z
0 = 1p is a complex number satisfying Re0 (0), then
Re h z zh z >(0< 1) implies
1
Reh z > 1 2 1 , (20) where 1 is given by
1 1 0 R
d =
1 e
t t
which is an increasing function of Re and 1
1 2 < 1. The estimate (20) cannot be improved in general.
Lemma 3: [4] Let be analytic in with
and
. Then, for q zR > 0
0 = 1q e q z
(z) = <z r 1,
1 1
Re
1 1
r r
q z q z r
r
and
2 Re
1 z
q z q z
r
.
We begin by proving the following. Theorem 1:Let 0
, ,
. If , , , ,
p
f k a c , then
1, 1,
p
k p
a c f z a c g z
,
where
2 =
2 a
a
, (21)
and g
p satisfies the condition (16) and
0
0 2 0
, Re
= , =
1,
p p
a c g z h z
h z
a c g z h z
. (22)
Proof. Let f p
k a c, , , , , ,
and set
1,
= 1 1,
p p
a c f z
h z a c g z
, (23)
where h z
is analytic in with h
0 = 1 and we write
1
1 1
=
4 2 4 2
k k
h z h z h z
2 . (24)
A simple computation using (23) and (24) gives
1
1 1
0
2 2
0 1,
(1 )
1,
, 1,
, 1,
1 1
= 1
4 2
1
1 1 .
4 2
p p
p p
p p
a c f z a c g z
a c f z a c f z a c g z a c g z
zh z k
h z
a h z zh z
k h z
a h z
Now we form the functional
u v,
by choosing
1= i =
u h z u iu2 and v=zhi
z =v1iv2. Thus
0
1
, = 1 v
u v u
a h z
. (25)
The conditions 1) and 2) of Lemma 1 are clearly satisfied. Therefore, we show that the condition 3) of Lemma 1 is satisfied.
By virtue of (25), we have
0
2 1 2 1
0
1
1 Re
Re , =
1
= ,
h z
iu v v
a h z v a
where is given by (22). Thus, for
2
1 1 2 2v u ,
we obtain
22
222 1
1 1
Re , = ,
2 2
u A Bu iu v
a C
231
= 2 1
A a ,
= 1 and =
B C a
.
Since , and by (21), we get . Hence, by applying Lemma 1, it follows that which implies that
. The proof of Theorem 1 is thus completed. 0
B
2,v1
i
> 0 C
= 1 h i
0 A
z
Re iu 0
h
, 2;k
Remark: If we put a n p= and c== 1 in Theorem 1, we have the result due to Noor and Arif [9, Theorem 3.1].
Theorem 2: Let 0. If fp
k a c, , , , , ,
, then
, ,
p
k p
a c f a c g
z z
, where
2 =
2
p p
,
and g
p satisfies the condition (18) and
0 2 0Re =
| |
h z h z
,
1
0
, =
,
p p
a c f z h z
a c g z
.
Proof. Let fp
k a c, , , , , ,
and set
,
= 1 ,
p p
a c f z
h z a c g z
,
where is analytic in with . Then, by using same techniques as in the proof of Theorem 1, we obtain the desired result.
h z h
0 = 1We note that = when = 0 in Theorem 1. Corollary 1: Let 1. If f p
k a c, , , ,0,1,
, then
,
( )
,
p
k p
a c f z
z a c g z
.
Proof. It is clear that, for 1,
, ,
1, ,
= (1 )
1, ,
1,
( 1) .
1,
p p
p p
p p
p p
a c f z a c g z
a c f z a c f z a c g z a c g z
a c f z a c g z
This implies that
1 2, ,
1, ,
1
= 1
1, ,
1,
1 1
1 =
1,
p p
p p
p p
p p
a c f z a c g z
a c f z a c f z a c g z a c g z a c f z
P P
a c g z
1
1 .
Since k
is a convex set (see [7]), by usingTheorem 1 and Definition 1, we observe that
k
1, 2
P P and
, ,
p
k p
a c f z a c g z
,
which completes the proof of Corollary 1.
Making use of Theorem 2 and Definition 2, we can prove the following result.
Corollary 2: Let 1. If fp
k a c, , , ,0,1,
, then
1
1 ,
( )
,
p
k p
a c f z
z a c g z
.
Next, by using Lemma 2, we prove the following. Theorem 3: Let be a complex number satisfying Re> 0 and let a> 0, c 0
, 0 and > 0. If f
p satisfies the condition
1
1, 1
, 1,
,
p p
p p
k
p p
a c f z z
a c f z a c f z
z z
then
1,
( ) ( )
p
k p
a c f z
z z
,
where
1 Re 1
1 1 0
= 1 2 1 with = 1 t a dt
.(26) The value of is best possible and cannot be improved.
1
2
1,
1 1
= =
4 2 4 2
p p
a c f z z
k k
h z h z h z
,
then and is analytic in . By applying (7), we have
0 = 1
h h
1
1, 1
, 1,
= .
p p
p p
p p
k
a c f z z
a c f z a c f z
z z
h z zh z a
Therefore, by virtue of Lemma 2, we see that , where
( = 1, 2)i
h i is given by (26). Hence we conclude that hk
, which evidently provesTheorem 3.
By using (8) instead of (7) in Theorem 3, we have the following.
Theorem 4: Let be a complex number satisfying Re > 0 and let a c, 0 , 0 and > 0. If
f p satisfies the condition
1
1
1, 1
, ,
,
p p
p p
k
p p
a c f z z
a c f z a c f z
z z
then
,
( )
p
k p
a c f z
z z
,
where is given by (26) with 1
R 1 ( ) 1= 0 1 d
e p
t
t The value of is best possi- ble and cannot be improved.Theorem 5.Let 02<1
. Then
, , , ,0, , 1
, , , ,0, , p k a c p k a c 2
.
Proof. If 2= 0, then the proof is immediate from Theorem 1. Let 2> 0 and f p
k a c, , , ,0, , 1
1
. Then there exist two functions H , H2k
suchthat
1
1
1 1
1, 1
1,
, 1,
=
, 1,
p p
p p
p p
a c f z a c g z
a c f z a c f z
H z a c g z a c g z
and
2
1,
= 1,
p p
a c f z
H z a c g z
.
Then
2
1
2
2 2
1 2
1 1
1, 1
1,
, 1,
, 1,
= 1 .
p p
p p
p p
a c f z a c g z
a c f z a c f z a c g z a c g z
H z H z
(27)
Since k
is convex set (see [7]), it follows thatthe right hand side of (2.8) belongs to k
, whichproves Theorem 5.
Next, we consider the generalized Bernardi-Libera- Livingston integral operator defined by (cf. [1,8], and [15])
> p
1
0
= p z d ( ; > )
f z t f t t f p
z
p .
(28) Theorem 6: Let be a complex number satisfying Re> 0 and let f z
p and
f be given by (2.9). If
1
p , p p
,p k
a c f z a c f z
z z
then
,
p
k p
a c f z z
,
where is given by (26) with 1 Re
1= 0 1 d
z p
t t
.Proof. From (28), we obtain
,
= , ,
p
p p
z a c f z
p a c f z a c f
z .
233
Let p
a c, p f z =h z
z
. Then, by virtue of (29), we have
, ,
(1 )
= .
p p
p p
k
a c f z a c f z
z z
zh z h z
p
Hence, by using Lemma 2, we obtain the desired result.
Finally, we consider the converse case of Theorem 1 as follows.
Theorem 7: Let fp
k a c, , , , , ,0
, ,
. Then
p
, , , ,f k a c for z <R , where is given by
R
2 2
=
1 2
a R
a a a
1
. (30)
Proof. Let
1, =
1,
p p
a c f z H z
a c g z
and
0
, =
1,
p p
a c g z H z
a c g z
and since f p
k a c, , , ,0, ,0 , it follows that
k
H and H0k
h z. Proceeding as in Theorem 1, for
= 1H z and H z0
= 1
h z with hk, h0, we obtain
1
2
0
1 1
0
2 2
0 1, 1
1
1 1,
, 1,
, 1,
=
1
1 =
4 2 1
1
4 2 1
p p
p p
d
p p
a c f z a c g z a c f z a c f z
V a c g z a c g z
zh z h z
a h
zh z k h z
a h
zh z k h z
a h
By applying Lemma 3, for hi( = 0,1, 2)i and = < 1
z r , we have
0
2
2 Re
1
2 1
Re 1
1 1 2 1
2
Re 1
1 1 1 2
1 2 2 1
Re .
1 1 1 2
i i
i
i
i
zh z h z
a h
r r
h z
a r r
r h z
a r r
a r a r a
h z
a r r
(31) Hence, the hand right side of (31) is positive for
= <
z r R, where is given by (30). This completes the proof of Theorem 7.
R
Theorem 8. Let fp
k a c, , , , , ,0
. Then
, , , , , ,
f p k a c for z <R , where is given by
R
2 2
1= 1
2 1
R p p
p p
.
(32)
Proof. Let
=
, ,p p
a c f z H z
a c g z
and
0
, =
1,
p p
a c g z H z
a c g z
and since f p
k a c, , , ,0, ,0
, it follows that
k
H and H0
. Then, by using samemethods as in the proof of Theorem 7, we obtain the required result.
3.Acknowledgements
This work was supported by Daegu National University of Education Research Grant in 2010.
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