Malaysian Journal of Fundamental & Applied Sciences
available online at http://mjfas.ibnusina.utm.my
Partial Sums For Class Of Analytic Functions Defined By Integral Operator
Nagat. M. Mustafa* and Maslina DarusSchool of Mathematical Sciences, Faculty of Science and Technology KM Bang 43600 Selangor D. Ehsan, Malaysia Received 29 November 2011, Revised 10 June 2012, Accepted 15 August 2012, Available online 20 August 2012
ABSTRACT
In the present paper, we study the class of analytic functions involving generalized integral operator, which is defined by means of a general Hurwitz Lerch Zeta function denoted by
ℑ
s bα,f z ( )
with negative coefficients. The aim of the paper is to obtain the coefficient estimates and also partial sums of its sequenceℑ
αs b,f z ( )
. | Univalent functions | uniformly starlike functions | Hadamard product | partial sums | fractional derivatives and fractional integrals |
® 2012 Ibnu Sina Institute. All rights reserved.
http://dx.doi.org/10.11113/mjfas.v8n4.146 1. INTRODUCTION
Let
A
denote the class of all analytic functions in the open unit diskU {z ∈ = : z 1} <
given by the normalized power series of the form( )
2
(z)
n n, ( ) . 1
n
f z
∞a z z U
=
=
+ ∑ ∈
1 . 1 Definition: (Owa and. Srivastava [3]) Let the function
f
be analytic in a simply connected domain of the z-plane containing the origin. The fractional derivative off
of orderα
is defined by0
( )
1
( )( ) d t , (0 1),
(1 )
z
z f t
D f z d
dz z t
α α
α
= α ≤ <
Γ − ∫ −
where the multiplicityof
(z — t)
−αis removed by requiringlog(z — t)
to be real whenz — t 0 >
.Using Definition 1.1 and its known extensions involving fractional derivatives and fractional integrals, Owa and Srivastava [3] introduced the operator
Ω
α: A — A
as follows:f (z) (2 )
ZαD f z
zα( ), ( 2,3,4,...)
α
α α
Ω =Γ − ≠
*Corresponding author at:
n n
2
(n 1) (2 )
z a z ,( ).
(n 1 )
n
α z U α
∞
=
Γ + Γ −
= + ∈
Γ + −
∑
1.2 Definition:
(
Srivastava and Choi [2])
A general Hurwitz Lerch Zeta functionΦ ( , , ) z s b
definedby
( , , ) =
=0,
( )
n n s
z s b z
n b Φ
∞∑ +
where (
s ∈ ,b ∈ −
0when (|z| < 1),(b) 1
ℜ >
when (|z| = 1).We can write this function as :
( , , ) ( z ( , , )) Φ
*z s b = Φ b
sz s b ∗ f (z) ,
then*
=2
( , , ) = .
( 1)
s n n
n s
z s b z b a z
n b
Φ +
∞∑ + −
By using Definitions (1.1) and (1.2), the authors [1]
introduced the generalized integral operator
ℑ
s bα,:A A →
as the following:For
0 ≤ < α 1
ands ∈ ,b ∈ −
0,
( ) = ( ) (2 )
s *( , , ), ( 2,3,4,...)
s b z
α
f z b α z D
α αz s b α
ℑ Γ − Φ ≠
n n 2
(n 1) (2 ) b
z a z ,( ).
(n 1 ) (n b 1)
n
s
α z U α
∞
=
Γ + Γ −
= +
∑
Γ + − + − ∈Note that :
0
0,b
(z) f ≡ f z ( ) .
ℑ
Special cases of this operator includes:
0,b
f z ( ) f z ( )
α α
ℑ ≡ Ω
is Owa and Srivastava operator [3].
0
, 1
(z)
s,(z),
s b+
f ≡ J f
bℑ
is Srivastav and A. Attiyaintegral operator[5].
0
1,1
f (z) A( ) (z) , ≡ f
ℑ
is Alexander integral operators [6].0
1,1
(z) L( ) (z) ,
s+
f ≡ f
ℑ
is Libera integral operators [7].0
1,δ
(z) f ≡ L
δ( )(z); f
ℑ δ > − 1,
is Bernardi integral operator [8].0
,
(z) I (z);
sσ
f ≡
σf
ℑ
(σ > 0) is Jung- Kim-Srivastava integral operator [9].0
,1
(z)
n;
n
f S
−
≡
ℑ
(n ∈ N) is the Salagean derivative operator [13].0
, 1
(z)
s b
f
− +
ℑ
(s ∈ Z) multiplier transformations studied by Flett [14].Finally, for different choices of
s, b
andα
, several operators investigated earlier by other authors Cho and Kim [15], and Lin and Owa [16] are obtained.By using our integral operator we introduce the following class of
A
.A function
f ∈ A
is said to be in the class denoted by,
( , ),( 1 1), 0
SP
s bαβ γ − ≤ γ < β ≥
and satisfy, ,
, ,
( ( )) ( ( ))
( ) ( ) ( ) ( ) 1 ,
s b s b
s b s b
z f z z f z
f z f z
α α
α
γ β
α ℑ ′ ℑ ′
ℜ ℑ − ≥ ℑ −
where
s ∈ ,b ∈ − Z
0.
By suitably specializing the values of
α β γ , ,
ands
the classSP
s bα,( , ) β γ
reduces to the classes introduced and studied by various authors for example: For00,b
SP ( , ) SP( , ) β γ ≡ β γ
was introduced Rφnning [10], [2].2. RESULTS & DISCUSSION
2.1 Coefficient Estimates:
Before stating and proving our main results, we derive a sufficient condition giving the coefficient estimates for the function
f
to belong to the class,
( , )
SP
s bαβ γ
.The result is contained in the following:2.2 Theorem
A sufficient condition for a function
f
of the form (1) to be inSP
s bα,( , ) β γ
is that[ ]
=2
( 1) (2 )
(1 ) ( ) | | 1 ,
( 1 ) 1
s n n
n b
n a
n n b
β γ β α γ
α
∞ + − + Γ + Γ −Γ + − − + ≤ −
∑
where
s ∈ ,b ∈ −
0and( 1 − ≤ γ 1), < β ≥ 0
.Proof: It suffices to show that
, ,
, ,
( ( )) ( ( ))
( s bs b) ( ) ( s bs b) ( ) 1
z f z z f z
f z f z
α α
α
γ β
α ℑ ′ ℑ ′
ℜ ℑ − − ℑ −
, ,
( ( ))
(1 ) 1
( ) ( )
s b s b
z f z f z
α
β ℑ
α′
≤ + −
ℑ
n 2
n 2
(n 1) (2 ) b
(1 ) ( 1) a
(n 1 ) (n b 1)
(n 1) (2 ) b ,
1 ( 1) a
(n 1 ) (n b 1)
s
n
s n
n n β α
α α α
∞
=
∞
=
Γ + Γ −
+ − Γ + − + −
≤ − − ΓΓ + Γ −+ − + −
∑
∑
[ ]
=2
( 1) (2 )
(1 ) ( ) | | 1 ,
( 1 ) 1
s n n
n b
n a
n n b
β γ β α γ
α
∞ + − + Γ + Γ −Γ + − − + ≤ −
∑
the proof is complete.
2.3 partial sums
In this section we will examine the ratio of a function of the form (1) to its sequence of partial sums defined by
1
( ) =
f z z
andf z
k( ) = z + ∑
kn=2a z
n nwhen thecoefficients of
f
are sufficiently small to satisfy the condition (2). We will determine sharp lower boundsfor
( ) , ( ) , ( ) ( ) .
( )
k( ) ( )
k( )
k k
f z f z
f z f z and
f z f z f z f z
′ ′
ℜ ℜ ℜ ′ ℜ ′
2.4 Theorem
Let
f
be given by (1) satisfying (2), then1 , ( 2) (2 ) b
( 2 ) b
(1 )( 1) ( )
( ) ( ) 1
sk k
k k
k
f z f z
γ αβ γ β α
−
Γ + Γ −
Γ + − +
+ + − +
ℜ ≥ −
( z U ∈ ). (3) ( )
k
( ) f z f z
ℜ ≥
( 2) (2 ) b
[ ( 2 ) b
( 2) (2 ) b
] ( 2 ) b
(1 )( 1) ( )]
[(1 ) [(1 )( 1) ( ) ,
s s
k k k
k k k
k k
αα αα
β γ β
γ β γ β
Γ + Γ −
Γ + − +
Γ + Γ −
Γ + − +
+ + − +
− + + + − +
(z U∈ ). (4)
The results are sharp for every
k
with the function given by1 ,
( ) = (1 )
( 2) (2 ) b
(1 )( 1) ( )
( 2 ) b
k
s
f z z z
k k k k
γ
β γ β α
α
+
+ −
Γ + Γ −
+ + − +
Γ + − +
(z U k∈ , ≥0). (5) Proof:
We prove (3).Let
f
be given by (1) satisfying (2), by sitting1 ( 2) (2 )
( 2 )
( )
1 ( )
[(1 )( 1) ( ) ]
(
s
k
k k
b
k b f z f z
k
ααγ
β γ β
− Γ + Γ −
Γ + −
+
−
+ + − +
( 2) (2 )
( 2 )
1 (1 )( 1) ( )
s
)
k k b
k
αα k bγ
β γ β
−
Γ + Γ −
Γ + − +
−
+ + − +
1 ( ) 1 w z( )
.
w z +
=
−We find that ( ) w z =
[
n-1 n 1
nn-1 n n-1
2 1
( 2) (2 ) b
[ ( 2 ) b z
1 ,
( 2) (2 ) b a z( 2 ) b
2 2 a z 1
(1 )( 1) ( )]
(1 )( 1) ( )]
s
n k s
k
n n k
kk k
kk k
k k
αα γ
αα γ
β γ β
β γ β
∞
= +
∞
= = +
Γ + Γ −
Γ + − +
−
Γ + Γ −
Γ + − +
+ + −
+ + − +
+ + − +
∑
∑ ∑
( ) w z =
1
1 [
2 n
( n
[ n
( 2) (2 ) b 1 )( 1) ( )] ( 2 ) b
( 2) (2 ) b (1 )( 1) ( )] ( 2 ) b
2 2 a 1
a 1
a
.
n k
n k
s
s k
n
k kk k
k k k k
β γ β αα
β γ β αα
γ γ
∞
= +
∞
= +
+
=
Γ + Γ −
+ + − + Γ + − +
Γ + Γ −
+ + − + Γ + − +
− −
∑
−∑ ∑
Now
| w(z) | 1 ≤
if and only if1 n
( 2) (2 ) b
( 2 ) b
1
2((1 )( 1) ( )]
a
s
n k
k k k
k α α
γ
β γ β
∞
= +
Γ + Γ −
Γ + − +
−
+ + − +
∑
2 n
2 2
ka ,
n =
≤ − ∑
which is equivalent to
1 n
k bb ( 2) (2 ) [(1 )( 1) ( )] ( 2 )
1 n k a
k s
k k
α
β γ β α
γ
∞
= +
Γ + Γ − +
+ + − + Γ + −
−
∑
+2 n
k
a 1 . (6)
n =
∑ ≤
This will hold if we show that left-hand side of (6) is bounded above by
2
b n
n b 1
1 .
(n 1) (2 )
[(1 ) ( )] (n 1 ) a
n
n
sγ
β γ β α α
∞
=
+ −
−
Γ + Γ −
+ − +
Γ + −
∑
This is equivalent to showing that
2 1 an
(n 1) (2 ) b
[(1 ) ( )] (n 1 ) n b 1 1
n
n s γ
γ
β γ β αα
∞
=
− + Γ + Γ −
+ − + Γ + − + − − −
∑
1
(n 1) (2 )(n 1 ) n b 1b ( 2) (2 )( 2 ) k bb
1
(1 )( 1 ) ( ) s s
n k
kk
n k αα αα
γ
β γ β
∞
= +
Γ + Γ −Γ + − + − −Γ + Γ −Γ + − +
−
+ − − − +
∑
a
n≥ 0 , (7)
To see that
( )
(1 ) 1
( ) = ,
( 2) (2 ) b
( 2 ) b
(1 )( 1) ( )
k
s
f z z z
k
k k
k
γ
α
β γ β
α− +
+ Γ + Γ −
Γ + − +
+ + − +
( z U k ∈ , ≥ 0),
gives sharp result, we observe that for
i
z ren
π
that =
Re ( )
k( ) f z f z
≥
( 2) (2 ) b ,
( 2 ) b
[
1 (1 )( 1) ( )]
1
s zkk k k
k
ααγ
β γ β
Γ + Γ −
Γ + − +
−
+ + − +
+
→
1 1 ,
n b 1b (n 1) (2 )
[(1 )( 1) ( )] (n 1 )
k s
γ
β γ β αα
−
+ −
−
Γ + Γ −
+ + − +
Γ + − when z → 1−.
To prove the second part of this theorem, we write bb 1
( 2) (2 )
( 2 ) ( )
( )
(1 ) [(1 )( 1) ( ) ]
(
s
k k
k k f z
f z
k
ααγ
γ β γ β
+
Γ + Γ − Γ + −
− −
− + + + − +
[ ( 2) (2 ) b
( 2 ) b
( 2) (2 ) b
( 2 ) b
]
( ]
) 1 ) [(1 )( 1) ( )
(1 )( 1) ( )
ss
k k k
k k k
k
k
αααα
γ β γ β
β γ β
Γ + Γ −
Γ + − +
Γ + Γ −
Γ + − +
− + + + − +
+ + − +
1 ( ) 1 w z( )
.
w z +
=
−we find that
( ) w z =
1
n 1 2
. 1
( 2) (2 ) b
( ( 2 ) b n
( 2) (2 ) b
( ( 2 ) b n
2 2 a
1 ) [(1 )( 1) ( )]
1 1 ) (1 )( 1) ( )]
a
[ a
n k
n k
s
s k
n
kk k
kk k
k k
αα
αα
γ β γ β
γ
γ β γ β
γ
∞
= +
∞
= = +
Γ + Γ −
Γ + − +
Γ + Γ −
Γ + − +
− +
− + + + − +
−
− + + + − +
−
∑
∑ ∑
Now
| w(z) | 1 ≤
if and only ifn 1
( 2) (2 ) b
2 ( 2 ) b
1
(1 ) [(1 )( 1) ( )]
a
s
n k
kk k
k αα
γ
γ β γ β
∞
= +
Γ + Γ −
Γ + − +
−
− + + + − +
∑
n 2
2 2 k a ,
n =
≤ −
∑
which is equivalent to
1 n
( 2) (2 ) b
( 2 ) b
1 a
(1 ) [(1 )( 1) ( )]
n k
k s
k k
k αα
γ
γ β γ β
∞
= +
Γ + Γ −
Γ + − +
− +
− + + + − +
∑
2 n
k
a 1 .
n =
∑ ≤
Making use of (2) to get (7). Finally, equality holds in (4) for the extremal function
f
given by (5).This completes the proof.2.5 Theorem
Let
f
be given by (1) satisfying (2), then Re '( ) 1' ( )k f z f z
≥ −
( 1)(1 ) , (8)
( 2) (2 ) b
[(1 )( 1) ( )] ( 2 ) b
s
k
k k k k
γ
β γ β αα
+ −
Γ + Γ −
+ + − + Γ + − +
' ( ) Re
k'( )
f z f z
≥
1
( 2) (2 ) b
( 2 ) b
( 2) (2 ) b
([(1 )( 1) ( )] ( 2 ) b
.
( ) [(1 )( 1) ( )]
(9)
1 )( 1
n k
s s
k k k
k k k k
k
k ααβ γ β αα
β γ β
γ
∞
= +
Γ + Γ −
Γ + − +
Γ + Γ −
+ + − + Γ + − +
+ + + − +
− +
∑
The results are sharp for every
k
with the functiongive 1
(1 ) , ( ) =
( 2) (2 ) b
(1 )( 1) ( ) ( 2 ) b
k
z s
f z z
k k k k
γ
β γ β αα
+
+ −
Γ + Γ −
+ + − +
Γ + − +
( z U k ∈ , ≥ 0).
Proof: To prove the result (8), define thefunction w(z) by
( 2) (2 ) b
( 2 ) b
(
1 )
[ ]
f ' ( ) ( '( )
(1 )( 1) ( )
sk
k k k
z
f z
k
ααγ
β γ β
Γ + Γ −
Γ + − +
− −
+ + − +
( ) )
( 2) (2 ) b
( 2 ) b
1- )( 1 [(1 )( 1) ( )]
1
s,
k k k
k
k
ααγ
β γ β
Γ + Γ −
Γ + − +
+
+ + − +
−
where
w z = ( )
1
1
n n-1
( .
n n-1 n n-1
2 (
] a z
1 )( 1)
( 2) (2 ) b [(1 )( 1) ( )] ( 2 ) b a z 2 2 a z
( 2) (2 ) b
[(1 )( 1) ( ) ( 2 ) b
1 )( 1)
n k
n k
s k
n n s
k
n k k k k
n
k kk k
k γ
β γ β αα
β γ β αα
γ
∞
= +
∞
= + +
=
− +
Γ + Γ −
+ + − + Γ + − +
+
Γ + Γ −
+ + − + Γ + − +
− +
∑
∑
∑Now
| ( ) | 1 w z ≤
if and only ifn
) 1
( 2) (2 ) b
[(1 )( 1) ( )] ( 2 ) b
(1 )( 1 n k a
k s
k k k
k n
β γ β αα
γ
∞
= +
Γ + Γ −
+ + − + Γ + − +
− +
∑
n 2
k a 1 .
n
n
=
+
∑
≤Since the LHS of (8) is bounded above by
2 1 n
n b 1b
(n 1) (2 )
(1 ) ( )] (n 1 ) a .
s
n
n
γ
β γ β α α
∞
= −
Γ + Γ −
+ −+ − +
Γ + −
∑
The proof is complete.
To prove the second part of this theorem, we write ( 2) (2 ) b
( 2 ) b 1 (
(1 )( 1) [(1 )( 1) ( )]
f ' ( ) '( )
s
k
k k k
k k
f z z
ααγ
γ β γ β
Γ + Γ − Γ + − +
− −
− + + + + − +
( 2) (2 ) b
( 2 ) b
( 2) (2 ) b
( 2 ) b
[(1 )( 1) ( )]
1 ( ) , 1 ( ) (1 )( 1) [(1 )( 1) ( )]
s
s
kk k
kk k
k w z
k k w z
αα
αα
β γ β
γ β γ β
Γ + Γ −
Γ + − +
Γ + Γ −
Γ + − +
+ + − +
= +
− + + + + − + −
we find that
( ) w z =
1
1
n-1
1 (
n-1 1
n )
2
( 2) (2 ) b n ( 2 ) b 1 )( 1)
( 2) (2 ) b( 2 ) b n n-1
2 2 a z (1 )( 1
a z
[(1 )( 1) ( )]
[(1 )( 1) ( )]
a z .
n k
n k
s
s k
n
k k k
k
kk k
n k
k
k
ααγ
αα
γ
β γ β
β γ β
∞∞
= +
∞
= +
+
+
=
Γ + Γ − Γ + − +
− +
Γ + Γ − Γ + − +
+ + − +
+ + − +
+ + − +
∑
∑ ∑
Now
| ( ) | 1 w z ≤
if and only if1 n
( 2) (2 ) b
( 2 ) b
2 1 [(1 )( 1) ( )]
(1 )( 1)
a
s
n k
k k k
k
k
n
αα
β γ β
γ
∞
= +
Γ + Γ −
Γ + − +
+ + + − +
− +
∑
n 2
2 2 k a ,
n
n
=
≤ −
∑
which is equivalent to
( 2) (2 ) b
[ ( 2 ) b
(1 1 n
(1 )( 1) ( )]
)( 1)
1 a
k s
k k
n k
k
k n
αα
β γ β
γ
Γ + Γ − Γ + − + ∞
= +
+ + − +
− +
+ ∑
2 n
k a 1
n
n
=
+
∑
≤From the condition (2), it is sufficient to show that
1 ( ) n
( 2) (2 ) b
( 2 ) b
1 )( 1 ((1 )( 1) [(1 )( 1) ( )]( 1))
a
s n k
kk k
k
k k n k α
α γ
γ β γ β
∞
= +
Γ + Γ −
Γ + − +
− +
− + + + + − + − −
∑
(1 )( 1) n
( 2) (2 ) b
( 2 ) b
2 a 0 .
[(1 )( 1) ( )]
sk
k k
k k
n
k
γ
αα
β γ β
− +
Γ + Γ −
Γ + − +
+ = ≥
+ + − +
∑
which gives (9). The bound in ( 9) is sharp for all k ∈ N with the extremal function (5). Special cases of Theorems 2.4, 2.5 by setting
α = 0, s 0 and = β = 0
can be found in [11].2.5 Corollary
Let
f
be given by (1) satisfying2
( ) 1 ,
n
n γ γ
∞
=
− ≤ −
∑
Then
Re ( ) ( )
( ) 1
k
f z k z U
f z k γ
≥ ∈
+ −
,( ) 1 ( )
( ) 2 2
f z
kk z U
f z k
γ γ
+ −
ℜ ≥ + − ∈
,
'( ) ( )
' ( )
k1
f z k z U
f z k
γ γ
ℜ ≥ ∈
+ −
,and
'( ) 1
( )
' ( )
k( 1 ) ( 1)(1 )
f z k z U
f z k k
γ
γ γ
+ −
ℜ ≥ + − + + − ∈
. The results are sharp for every k with the function given by
( ) 1 ( ) .
f z z 1 z U
k γ
γ
= + − ∈
+ −
3. CONCLUSION
Our results certainly generalized several results obtained earlier. Therefore, partial sums of functions in the unit disk defined by a general integral operator are obtained The operator defined can be extended and can solve many new results and properties.
ACKNOWLEDGEMENT
The work here is support by the Universiti Kebangsaan Malaysia and the Libyan government
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