Received April 21st, 2010, Revised May 24th, 2010, Accepted for publication May 25th, 2010.
Integral Operator Defined by k−th Hadamard Product
Maslina Darus & Rabha W. Ibrahim School of Mathematical Sciences
Faculty of science and Technology, Universiti Kebangsaan Malaysia Bangi 43600, Selangor Darul Ehsan, Malaysia
Email: [email protected], [email protected]
Abstract. We introduce an integral operator on the class A of analytic functions in the unit disk involving
k
th Hadamard product (convolution) corresponding to the differential operator defined recently by Al-Shaqsi and Darus. New classes containing this operator are studied. Characterization and other properties of these classes are studied. Moreover, subordination and superordination results involving this operator are obtained.Keywords: Hadamard product; integral operator; subordination; superordination.
AMS Mathematics Subject Classification (2000): 30C45.
1 Introduction
Let H be the class of functions analytic in the unit disk
U
and H[a,n] be the subclass of H consisting of functions of the form. ...
= )
(z aanznan1zn1
f Let A be the subclass of H consisting of functions of the form
. ,
= ) (
2
=
U z z a z z
f n n
n
(1) The following differential operator is defined in [1] and studied in [2]
A A
D
k,:
by0, 0, {0}, ,
) , ( ] 1) ( [1
= ) (
2
=
,
N
D f z z n kC n anzn k
n k
(2) where
1). ( ) (
)
= (
= 1 ) ,
(
n n n
n C
Remark 1.1. When
=1,
=0 we get Sa
la
gean differential operator [3],
0
=
k
gives Ruscheweyh operator [4], = 0
implies Al-Oboudi differential operator of order (k) [5] and when = 1
operator (2) reduces to Al-shaqsi and Darus differential operator of order (k) [6].Given two functions n n
n
a z
z z f g
f , A , ( ) = =2 and
n n
b
nz z
z
g ( ) = =2 their convolution or Hadamard product f(z)g(z) is defined by
. ,
= ) ( ) (
2
=
U z z b a z z g z
f n n n
n
And for several functions f1(z),...,fm(z)A
. , , ) ...
(
= ) ( ...
)
( 1
2
=
1 z f z z a a z z U
f n mn n
n
m
Analogous to
D
k,f ( z ), z U
we define an integral operatorJ
k,: A A
as follows.Let
0.
1 , ) (1 := 1
)
(
2
z
z z
z z
z z
. , ) , ( ] 1) ( [1
=
) ] [(1 ) ( ...
) (
= ) (
0 2
=
1
N
k z n C n
z
z z z
z z
F
n k
n times k k
And let Fk(1) be defined such that
.
=
= 1 )
(
2
= 1)
(
n n k
k
z z
z F z
z F
Then
. 0 0, , ) ,
, ( ] 1) (
= [1
) ( ) ]
) (1 ( ...
) ( [
=
) (
= ) (
0 2
=
1) ( 1 1)
( ,
U z k
n z C n
z a
z z f
z z z
z f F z f
n k
n n
times k k k
N J
(3) Remark 1.2. When
=1,
=0 we get the integral operator [3], alsok = 0
gives Noor integral operator [7,8].Some of relations for this integral operator are discussed in the next lemma.
Lemma 1.1. Let fA. Then
) .
= ( ) ( ) (
), (
= ) ( ) (
0 1
1,0 0
,0
t dt t z f
f ii
z f z f i
zJ J
Proof.
), (
=
= ) ( ) (
2
= 0
,0f z z a z f z
i n n
n
J
).
(
=
=
] [1
) = ) (
(
1 1,0
2
=
1 2 0 = 0
z f
n z z a
dt t a t dt
t ii f
n n n
n n n z z
J
In the following definitions, we introduce new classes of analytic functions containing the integral operator (3):
Definition 1.1. Let f(z)A. Then
f ( z ) S
k,( )
if and only if. 1,
<
0 ,
>
) } (
] ) ( { [
,
,
z U
z f
z f z
k k
J J
Definition 1.2. Let f(z)A. Then
f ( z ) C
k,( )
if and only if. 1,
<
0 ,
>
) } ) ( (
] ) ) ( ( { [
,
,
z U
z f
z f z
k k
J J
Let F and
G
be analytic functions in the unit diskU .
The function F ise
subordinat
to G, written F G, ifG
is univalent, F(0)=G(0) and ).( )
(U GU
F In general, given two functions F(z) and G(z), which are analytic in U, the function F(z) is said to be subordination to G(z) in
U
if there exists a function h(z), analytic inU
withU z all for z
h and
h(0)=0 | ( )|<1 such that
. ))
( (
= )
(z G h z forallz U
F
Let
:C2 C and leth
be univalent inU .
If p is analytic inU
and satisfies the differential subordination
(p(z)),zp(z))h(z) then p is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, ifp q .
If p and
(p(z)),zp(z)) are univalent inU
and satisfy the differential superordination h(z)
(p(z)),zp(z)) then p is called a solution of the differential superordination. An analytic function q is called subordinant of the solution of the differential superordination ifq p .
Let
be an analytic function in a domain containing f(U), (0)=0 and (0)>0.The function fA is called
like if. 0,
>
)) } ( (
)
{ ( z U
z f
z f
z
This concept was introduced by Brickman [9] and established that a function
A
f is univalent if and only if f is
like for some .Definition 1.3. Let
be analytic function in a domain containing 1= (0) 0,
= (0) ),
(U
f and (
)0 for
f(U)0. Let q(z) be a fixed analytic function in U, q(0)=1. The function fA is called
like with respect to q if. ), )) (
( (
)
( q z z U
z f
z f
z
The paper is organized as follows: Section 2 discuses the characterization properties for functions belonging to the classes
S
k( ), C
k( )
and Section 3, gives the subordination and superordination results involving the integral operatorJ
k,f ( z ).
For this purpose we need to the following lemmas in the sequel.Definition 1.4. [10] Denote by Q the set of all functions f(z) that are analytic and injective on U E( f) where E(f):={
U:limz f(z)=} and are such that f(
)0 for
UE( f).Lemma 1.2. [11] Let q(z) be univalent in the unit disk
U
and
and
be analytic in a domain D containing q(U) with
(w)0 when wq(U). Set).
( )) ( ( :=
) ( )), ( ( ) ( :=
)
(z zq z q z h z q z Q z
Q
Suppose that1. Q(z) is starlike univalent in
U
, and2.
> 0
) (
) (
z Q
z h z
forz U .
If
)) ( ( ) ( )) ( ( )) ( ( ) ( )) (
(p z zp z
p z
q z zq z
q z
then
) ( ) (z q z
p
and q(z) is the best dominant.
Lemma 1.3. [12] Let q(z) be convex univalent in the unit disk
U
and
and
be analytic in a domain D containing q(U). Suppose that 1. zq(z)
(q(z)) is starlike univalent in U, and2.
} > 0
)) ( (
)) ( { (
z q
z q
forz U .
If p(z)H[q(0),1]Q, with p(U)D and
(p(z))zp(z)
(z) is univalent inU
and)) ( ( ) ( )) ( ( )) ( ( ) ( )) (
(q z zq z
q z
p z zp z
p z
then q(z) p(z) and q(z) is the best subordinant.
2 General Properties of
J
k,In this section we study the characterization properties for the function
A ) (z
f to belong to the classes
S
k,( )
andC
k,( )
by obtaining the coefficient bounds.Theorem 2.1. Let f(z)A. If
1,
<
0 , ) 1
, ( ] 1) ( [1
|
| ) (
2
=
nn kaCn nn
(4)
then
f ( z ) S
k,( ).
The result (4) is sharp.Proof. Suppose that (4) holds. Since
k n n
k n n
k n
n
n a n n
a n
n a
] 1) ( [1
|
| ]
1) ( [1
|
| ] 1) ( [1
|
||
1 |
2
= 2
= 2
=
then this implies that
,
>
] 1) ( [1
| 1 |
] 1) ( [1
| 1 |
2
= 2
=
k n n
k n n
n a n
a n
hence
.
>
) } (
] ) ( { [
,
,
z f
z f z
k k
J
J
We also note that the assertion (4) is sharp and the extremal function is given by
) . (
) )(1 , ( ] 1) (
= [1 ) (
2
=
n k
n
n z
n C z n
z
f
Corollary 2.1. Let the assumption of Theorem 2.1. Then
2.
) , (
) )(1 , ( ] 1) (
| [1
|
n
n
n C a n
k
n
Corollary 2.2. Let the assumption of Theorem 2.1. Then for
=
=0 and1
=
. 2, ,
|
|an nk1 n kN0
In the same way we can verify the following results:
Theorem 2.2. Let f(z)A. If
1,
<
0 , ] 1
1) ( )[1 , (
| 1
||
|
2
=
n kn C n n
n a
n (5)
then
f ( z ) C
k,( ).
The result (5) is sharp.Corollary 2.3. Let the assumption of Theorem 2.2. Then
2.
| , 1
|
) )(1 , ( ] 1) (
| [1
|
n
n n
n C a n
k
n
Also we have the following inclusion results
Theorem 2.3. Let
0
1
2< 1.
ThenS
k,(
1) S
k,(
2).
Proof. By Theorem 2.1.
Theorem 2.4. Let
0
1
2< 1.
ThenC
k,(
1) C
k,(
2).
Proof. By Theorem 2.2.
Theorem 2.5. Let
0
1
2.
Then ( ) , ( )., 2
1
k
k S
S
Proof. By Theorem 2.1.
Theorem 2.6. Let
0
1
2.
Then ( ) , ( ).2
1
k
k C
C
Proof. By Theorem 2.2.
Moreover, we introduce the following distortion theorems.
Theorem 2.7. Let fA and satisfies (4). Then for zU and 0
<12
,
| |
) (2
)
| (1
|
| ) (
|
kf z z z
J
and
.
| ) | (2
)
| (1
|
| ) (
|
k,f z z z
2
J
Proof. By using Theorem 2.1, one can verify that
)
[1( 1)| ]| ( , )
[1 (( 1))]| (| , ) 1(2
2
= 2
= n C n
a n
n C n
a
k n n
k n n
then
2 . 1 ) , ( ] 1) ( [1
|
|
2
=
n an kC nn
Thus we obtain
2
2 2
= 2
= ,
|
| 2 ] [ 1
|
|
| ] | 1) ( [1
|
| |
|
] | 1) (
=| [1
| ) (
|
z z
n z z a
n z z a
z f
k n n
n k n n
k
J
The other assertion can be proved as follows
.
|
| 2 ] [ 1
|
|
| ) | , ( ] 1) ( [1
|
| |
|
) | , ( ] 1) (
| [1
) | , ( ] 1) (
=| [1
| ) (
|
2
2 2
= 2
= 2
= ,
z z
n z C n
z a
n z C n
z a
n z C n
z a z
f
k n n
n k
n n
n k
n n
k
J
This complete the proof.
In the same way we can get the following results.
Theorem 2.8. Let f(z)A and satisfies (5). Then for zU and 0
<12
,
| |
) 2(2
)
| (1
|
| ) (
|
kf z z z
J
and
.
| ) | 2(2
)
| (1
|
| ) (
|
k,f z z z
2
J
Also, we have the following distortion results
Theorem 2.9. Let f(z)A and and satisfies (4). Then for U
z n C n
m[1( 1)
]k (
, ), and 0
<1|
2) | (2
)
| (1
|
| ) (
| m z
z z
f
and
.
| ) | (2
)
| (1
|
| ) (
| m z
2z z
f
Proof. By using Theorem 2.1, one can show that
) ) (1
, ( ] 1) ( [1
|
| )
| (
| ) (
|
| ) (2
2
= 2
= 2
=
a
n a m
nn kaCn n mn n n
n n
then
2 . )
| (1
|
2
=
an mn
Thus we obtain
2 2 2
= 2
=
| 2 |
)
| (1
|
|
||
|
|
|
|
=|
| ) (
|
m z z
z a z
z a z z
f
n n
n n n
The other assertion can be proved as follows
.
| 2 |
)
| (1
|
|
||
|
|
|
|
|
| ) (
|
2 2 2
= 2
=
m z z
z a z
z a z z
f
n n
n n n
This completes the proof.
In the same way we can get the following results.
Theorem 2.10. Let f(z)A and and satisfies (5). Then for zU and 1
<
0
|
2) | 2(2
)
| (1
|
| ) (
| m z
z z
f
and
.
| ) | 2(2
)
| (1
|
| ) (
| m z
2z z
f
3 Sandwich Result.
By making use of lemmas 1.2 and 1.3, we prove the following subordination and superordination results involving the integral operator (3).
Theorem 3.1. Let q0 be univalent in
U
such that) (
) (
z q
z q z
is starlike univalent in
U
and0.
, , 0,
>
) } (
) ) ( ) (
( ) ) ( (
{1
C
z q
z z q z
q z z q
(6) If fA satisfies the subordination
) , (
) ) ( ( )] } ( [
)]
( [ ] ) ( [
] ) ( 1 [
){
(
, , ,
,
z q
z z q z
f z f z
f z f
z z
kk k
k
J J J
J
then
) )] ( ( [
] ) ( [
,
,
q z
z f
z f z
k k
J J
(7) and q is the best dominant.
Proof. Our aim is to apply Lemma 1.2. Setting
)] .
( [
] ) ( := [
) (
, ,
z f
z f z z
p
kk
J J
By computation shows that
)]
( [
)]
( [ ]
) ( [
] ) ( 1 [
) = (
) (
, ,
,
f z
z f z
z f
z f J z z
p z p z
k k k
k
J J
J
which yields the following subordination
. , ) , (
) ) ( ) (
( ) ) (
( C
q z
z z q z
p z
z p
By setting
0, , :=
) ( :=
)
(
andit can be easily observed that
(
),
(
) are analytic in C\{0} and that 0) (
when
C\{0}. Also, by letting) (
)
= ( )) ( ( ) (
= )
( q z
z z q z q z q z z
Q
and
) , (
) ) ( ( ) = (
) ( )
( )
= ( ) ( )) ( (
= )
( q z
z z q z
q z z q z q
z z q
Q z q z
h
we find that Q(z) is starlike univalent in
U
and that0.
, , 0,
>
) } (
) ) ( ) (
( ) ) ( ( {1
= ) } (
)
{ (
C
z q
z z q z
q z z q z
Q z h z
Then the relation (7) follows by an application of Lemma 1.2.
Corollary 3.1. Let the assumptions of Theorem 3.1 hold. Then the subordination
) , (
) ( )]
( [
] ) ( [ ] ) ( [
] ) ( 1 [
, , ,
,
z q
z q z z f
z f z z f
z f z
k k k
k
J J J
J
implies
) )] (
( [
] ) ( [
,
,
q z
z f
z f z
k k
J
J
(8) and q is the best dominant.
Proof. By letting
=0,
=1,(
):=
.Corollary 3.2. If fA and assume that (7) holds then
) )(1 (1
) ( )]
( [
] ) ( [ ] ) ( [
] ) ( 1 [
, , ,
,
Bz Az
z B A z
f z f z z f
z f z
k k k
k
J J J
J
implies
1
<
1 1 ,
1 )]
( [
] ) ( [
,
,
A Bz B
Az z
f z f z
k k
J J
and Bz
Az
1
1 is the best dominant.
Proof. By setting (
):=
,
=1,
=0 andBz z Az
q
1 :=1 )
( where
1.
<
1
B A
Corollary 3.3. If fA and assume that (7) holds then
2 ,
, ,
,
1 2 )]
( [
] ) ( [ ] ) ( [
] ) ( 1 [
z z z
f z f z z f
z f z
k k k
k
J J J
J
implies
1 , 1 ) (
] ) ( [
, ,
z z z
f z f z
k k
J J
and z z
1
1 is the best dominant.
Proof. By setting (
):=
,
=0,
=1, and . 1 :=1 )( z
z z
q
Corollary 3.4. If fA and assume that (7) holds then
z Az f
z f z z f
z f z
k k k
k
)] ( [
] ) ( [ ] ) ( [
] ) ( 1 [
, , ,
,
J J J
J
implies
) , (
] ) ( [
,
, Az
k k
z e f
z f
z
J
J
and eAz is the best dominant.
Proof. By setting (
):=
,
=0,
=1, and q(z):=eAz,|A|<
.Theorem 3.2. Let q(z)0 be convex univalent in the unit disk
U .
Suppose thatU z z for
q z z q
q
C
} > 0, , , )
( ) ) (
( {
(9) and
) (
) (
z q
z q z
is starlike univalent in
U .
Ifq Q
z f
z f z
k k
[ (0),1]
)]
( [
] ) ( [
,
,
H
J J
where
,
A f
)] } ( [
)]
( [ ] ) ( [
] ) ( 1 [
){
(
, , ,
,
z f
z f z
f z f z z
k k k
k
J
J J
J
is univalent is
U
and the subordination)] }, ( [
)]
( [ ] ) ( [
] ) ( 1 [
){
) ( (
) ) ( (
, , ,
,
z f
z f z
f z f z z
z q
z
z q
kk k
k
J
J J
J
holds, then
)]
( [
] ) ( ) [
(
, ,
z f
z f z z
q
kk
J J
(10) and q is the best subordinant.
Proof. Our aim is to apply Lemma 1.3. Setting
)] . ( [
] ) ( := [
) (
, ,
z f
z f z z
p
kk
J J
By computation shows that
)]
( [
)]
( [ ]
) ( [
] ) ( 1 [
) = (
) (
, , ,
,
z f
z f z
z f
z f z z
p z p z
k k k
k
J J J
J
which yields the following subordination
. , ) , (
) ) ( ) (
( ) ) (
( C
p z
z z p z
q z
z q
By setting
0, , :=
) ( :=
)
(
andit can be easily observed that
(
),
(
) are analytic in C\{0} and that 0) (
when
C\{0}. Also, we obtain0.
>
) } (
) ) (
( {
= )) } ( (
)) ( { (
z q
z z q
z q q
z
q
Then (10) follows by an application of Lemma 1.3.
Combining Theorems 3.1 and 3.2 in order to get the following Sandwich theorems
Theorem 3.3 Let
q
1( z ) 0, q
2( z ) 0
be convex univalent in the unit diskU
satisfy (9) and (6) respectively. Suppose that and , =1,2) (
) (
' i
z q
z zq
i
i is starlike univalent in
U .
If fA andQ z q
f z f z
k k
[ (0),1]
)]
( [
] ) ( [
1 ,
,
H
J J
)] } ( [
)]
( [ ] ) ( [
] ) ( 1 [
){
(
, , ,
,
z f
z f z
f z f
z z
kk k
k
J
J J
J
is univalent in
U
and the subordination) (
) ( ) ' ( )] } ( [
)]
( [ ] ) ( [
] ) ( 1 [
){
) ( (
) ( ) ' (
2 2 ,
, ,
, 1
1
z q
z z q z
f z f z
f z f z z
z q
z
z q
kk k
k
J J J
J
holds, then
) )] (
( [
] ) ( ) [
(
2, ,
1
q z
z f
z f z z
q
kk
J J
and
q
1( z )
is the best subordinant andq
2( z )
is the best dominant.Acknowledgement
The work here was supported by UKM-ST-06-FRGS0107-2009, MOHE Malaysia. The authors also would like to thank the anonymous referee for the informative and creative comments given to the article.
References
[1] Al-Shaqsi, K. & Darus, M., Differential subordination with generalized derivative operator. (to appear in AJMMS)
[2] Darus, M. & Ibrahim, R., Generalization of differential operator, Journal of Mathematics and Statistics 4(3), pp. 138-144, 2008.
[3] S
a
la
gean, G.S., Subclasses of univalent functions, Lecture Notes in Math., 1013, Springer-Verlag, Berlin, pp. 362-372, 1983.
[4] Ruscheweyh, S., New criteria for univalent functions, Proc. Amer. Math.
Soc., 49, 109-115, 1975.
[5] Al-Oboudi, F.M., On univalent functions defined by a generalized S
a
l
a
gean operator, I.J.M.M.S, 27, pp. 1429-1436, 2004.
[6] Al-Shaqsi, Darus, M., An operator defined by convolusion involving polylogarthms functions, Journal of Mathematics and Statistics, 4(1), pp.
46-50, 2008.
[7] Noor, K.I., On new classes of integral operators, J.Nat.Geomet. 16, pp.
71-80, 1999.
[8] Noor, K.I. & Noor, M.A., On integral operators, J. of Math. Analy. and Appl. 238, pp. 341-352, 1999.
[9] Brickman, L.,
like analytic functions, I.Bull. Amer. Math. Soc. 79, pp. 555-558, 1973.[10] Bulboaca, T., Classes of first-order differential superordinations, Demonstr.Math. 35(2), pp. 287-292, 2002.
[11] Miller, S.S. & Mocanu, P.T., Subordinants of differential superordinations, Complex Variables, 48(10), pp.815-826, 2003.
[12] Miller, S.S. & Mocanu, P.T., Differential Subordinantions: Theory and Applications. Pure and Applied Mathematics, 225, Dekker, New York, 2000.
