General Mathematics Vol. 18, No. 2 (2010), 121–132
On a subclass of analytic functions with negative
coefficient associated with convolution structure 1
V.B.L. Chaurasia, Anil Sharma
Abstract
The main object of this paper is to study the subclass SC(γ, λ, β)
of analytic univalent functions with negative coefficients in unit disc U
= {z : |z| < 1}. Further coefficient estimates, distortion theorem and
integral operators for this class are also obtained. We also discuss radii
of convexity and closure properties for functions belonging to the class
SC(γ, λ, β).
2000 Mathematics Subject Classification: 30C45, 33C20
Key words and phrases: Coefficient estimates, Hadamard product,
Distortion theorem, Closure properties.
1
Introduction
Let A denote the class of the functions
(1)
f (z) = z +
∞
X
ak z k
k=2
1
Received 26 November, 2008
Accepted for publication (in revised form) 9 February, 2009
121
122
V.B.L. Chaurasia, Anil Sharma
which are analytic in the unit disk U = {z : |z| < 1}.
A function f ∈ A is said to belong to the class A of Starlike functions of
order α (0 ≤ α < 1), if it is satisfies, for z ∈ U , the conditions
½ 0 ¾
zf (z)
(2)
Re
> α.
f (z)
We denote this class by S ∗ (α). Further, f ∈ A is said to be convex function
of order α in U, if it satisfies
½
¾
zf 00 (z)
(3)
Re 1 + 0
> α,
f (z)
z ∈ U,
for some α (0 ≤ α < 1). We denote this class K(α).
Let T denote subclass of A, consisting functions of the form
(4)
f (z) = z −
∞
X
ak z k , ak ≥ 0.
k=2
The function
(5)
Sα (z) = z(1 − z)−2(1−α) , α(0 ≤ α ≤ 1)
is the familiar extremal function for the class S ∗ (α), setting
k
Π (i − 2α)
(6)
C (α, k) =
i=2
(k − 1)!
, k ≥ 2,
using (5) and (6) we can write
(7)
Sα (z) = z +
∞
X
C(α, k)z k .
k=2
Clearly, C(α, k) is a decreasing function in α, and that



∞, α < 1/2


(8)
lim C (α, k) =
1, α = 1/2

k→∞


 0, α > 1/ .
2
On a subclass of analytic functions with negative coefficient...
123
If we now define g(z) as
(9)
g(z) = z +
∞
X
bk z k
k=2
then the Hadamard product (or convolution) of two analytic functions f(z)
and g(z), where f(z) , g(z) is given by equations (1) and (9) respectively, is
defined as
(10)
(f ∗ g)(z) = z +
∞
X
ak bk z k
k=2
For a function f(z) in A, we can define the differential operator Dn , introduced by Salagean [9] as
D0 f (z) = f (z)
D1 f (z) = Df (z) = zf 0 (z) = z +
∞
X
kak z k
k=2
D2 f (z) = D(Df (z)) = z +
∞
X
k2 ak z k
k=2
(11)
n
n−1
D f (z) = D(D)
f (z) = z +
∞
X
kn ak z k
k=2
We also define a subclass of A consisting of functions f(z), denoted by SC(γ, λ, β)
which satisfy the following condition
·
µ
¶¸
1 z [λz(Dn f ∗ Sα )0 (z) + (1 − λ)(Dn f ∗ Sα )0 (z)]
(12) Re 1 +
−1
> β,
γ
λz(Dn f ∗ Sα )0 (z) + (1 − λ)(Dn f ∗ Sα )(z)
(0 ≤ λ ≤ 1, 0 ≤ β< 1; γ ∈ C, z ∈ U ).
Special case of class.
(a) When λ = 0, and α = 1/2 then our class reduces in class of starlike
124
V.B.L. Chaurasia, Anil Sharma
functions of order β.
(b) When λ = 0, then our class reduces in class of starlike functions of complex
order γ.
(c) When α = 1/2 then this class reduces in class defined and studied by
Altintas, Irmak, Owa and Srivastava [5].
2
Coefficient estimates.
Theorem 1 Let the function f (z)εA is in the class SC(γ, λ, β), if and only
if
(13)
∞
X
k n [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)ak ≤ γ (1−β)
k=2
Proof. Assume that the inequality (13) holds true, then
¯ µ
¶¯
¯ 1 z [λz(Dn f ∗ Sα )0 (z) + (1 − λ)(Dn f ∗ Sα )0 (z)]
¯
¯
¯
−
1
¯γ
¯
λz(Dn f ∗ Sα )0 (z) + (1 − λ)(Dn f ∗ Sα )(z)
¯ P
¯
∞
¯
¯
¯
kn [λk + 1 − λ] (1 − k)C(α, k)ak zk−1 ¯
¯
¯ 1  k=2
¯
= ¯¯ 
∞

¯
P
¯
¯γ
n
k−1
1−
k [λk + 1 − λ] C(α, k)ak z
¯
¯
k=2
≤ (1 − β).
Hence, by using the maximum modulus principle, f (z) ∈ SC(γ, λ, β). Conversely, assume that the function f(z) defined by (1) is in the class SC(γ, λ, β).
Then we will have
µ
¶¾
½
1 z [λ z(Dn f ∗ Sα )0 (z) + (1 − λ)(Dn f ∗ Sα )0 (z)]
−1
> β,
Re 1 +
γ
λz(Dn f ∗ Sα )0 (z) + (1 − λ)(Dn f ∗ Sα )(z)

P

∞
n
k



k [λk + 1 − λ] (1 − k)C(α, k)ak z 



1
k=2

 > β,
Re 1 +
∞

P

γ


n
k
 z−
k [λk + 1 − λ] C(α, k)ak z 
k=2
On a subclass of analytic functions with negative coefficient...
125

P
∞
n
k−1




k [λk + 1 − λ] (1 − k)C(α, k)ak z



1
k=2
 > β,
Re 
1
+
∞


P

γ


n
k−1
 1−

k [λk + 1 − λ] C(α, k)ak z

k=2
and now when z → 1− , we obtain
∞
P
k=2
kn [λk + 1 − λ] (1 − k)C(α, k)ak
1−
∞
P
≥ γ(β − 1)
kn [λk
k=2
+ 1 − λ] C(α, k)ak
and finally,
∞
X
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)ak ≤ γ(1 − β).
k=2
Corollary 1 Let the function f(z) defined by (1) be in the class SC(γ, λ, β).
Then
(14)
ak ≤
kn [λk
γ(1 − β)
,
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
(k ≥ 2)
and the equality is attained for the function f(z) given by
(15)
3
f (z) = z −
kn [λk
γ(1 − β)
zk .
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
Distortion Theorem.
Theorem 2 Let the function f(z) be in class SC(γ, λ, β) then for 0 ≤| z |= r
(16)
(17)
r−
kn [λk
≤r+
γ(1 − β)
rk ≤ |f(z)|
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
kn [λk
γ(1 − β)
rk .
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
126
V.B.L. Chaurasia, Anil Sharma
Proof. From equation (15), easy to find that
|z| −
kn [λk
≤ |z| +
γ(1 − β)
|z|k ≤ |f(z)|
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
γ(1 − β)
|z|k
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
Now using the fact that |z| = r < 1, we obtain the desired result.
Corollary 2 If the function f(z) is in the class SC(γ, λ, β) then f(z) is included in a disc with centre at the origin and radius r, where
(18)
r=1+
kn [λk
γ(1 − β)
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
Theorem 3 Let the function f(z) be in the class SC(γ, λ, β) , then
1−
kn−1 [λk
≤1+
γ(1 − β)
rk−1 ≤ |f(z)|
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
kn−1 [λk
γ(1 − β)
rk−1
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
Where equality holds for the function f(z) given by (15).
1−
kγ(1 − β)
|z|k−1 ≤ |f(z)|
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
≤1+
kn [λk
kγ(1 − β)
|z|k−1
+ 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
Again using the fact that |z| = r, we obtain the desired result.
4
Integral Operators
Theorem 4 Let the function f(z) defined by (1) be in the class SC(γ, λ, β)
and let c be a real number such that c > −1. Then F(z), defined by
Z z
c+1
tc−1 f (t)dt
(19)
F(z) = c
z
0
also belongs to the class SC(γ, λ, β).
On a subclass of analytic functions with negative coefficient...
127
Proof. From the representation of F(z), it is obtained that
(20)
F(z) = z −
∞
X
bk z k ,
k=2
³
where bk =
c+1
k+c
Therefore
´
ak
∞
X
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)bk
k=2
=
µ
∞
X
n
k [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
k=2
≤
∞
X
c+1
k+c
¶
ak
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)ak
k=2
≤ γ(β − 1),
since f(z) belongs to SC(γ, λ, β) so by virtue of Theorem 1, F(z) is also element
of SC(γ, λ, β).
Theorem 5 Let the function
F (z) = z −
∞
X
ak z k , ak ≥ 0
k=2
be in the class SC(γ, λ, β) and is defined by equation (19). Now if c > −1,
then F(z) is univalent in |z| < R∗ , where
½
(21)
∗
R = inf
kn−1 [λk+1−λ] [k−1−γ(β −1)] C(α, k)(c+1)
(c+k)γ(1−β)
The result is sharp.
Proof. From (19) we have
∞
X
z 1−c (z c F (z))0
f (z) =
=z−
c+1
k=2
µ
¶
c+k
ak z k .
c+1
¾ k−11
, k≥2
128
V.B.L. Chaurasia, Anil Sharma
In order to obtain the required result, it is sufficient to prove that
¯ 0
¯
¯f (z) − 1¯ < 1 f or |z| < R∗ .
Now since
¯ ∞ µ
¯
¶
¯
¯0
¯ ¯¯ X
c
+
k
¯
¯f (z) − 1¯ = ¯−
k
ak z k−1 ¯
¯
¯
c+1
(22)
k=2
≤
¶
µ
∞
X
c+k
ak |z|k−1
k
c+1
k=2
µ
¶
∞
X
c+k
≤
k
ak |z|k−1
c+1
k=2
¶
µ
∞
X
c+k
ak |z|k−1 ≤ 1
≤
k
c+1
k=2
But from Theorem 1, we know that
∞
X
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)ak
<1
γ(1 − β)
(23)
k=2
From equation (22) and (23) we have
µ
k
c+k
c+1
¶
|z|k−1 ≤
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
γ(1 − β)
or
½
(24)
kn−1 [λk+1−λ] [k−1−γ(β −1)] C(α, k)(c+1)
|z| ≤
(c + k)γ(1−β)
1
¾ k−1
, (k ≥ 2),
we obtain the desired result. The result is sharp for the function
f(z) = z −
kn [λk
(c + k)γ(1 − β)
zk , (k ≥ 2).
+ 1 − λ] [k − 1 − γ(β − 1)] (c + k)C(α, k)
On a subclass of analytic functions with negative coefficient...
5
129
Radius of Convexity
Theorem 6 If f(z) given by (1) is in class SC(γ, λ, β) then f(z) is convex in
|z| < Rp, where
½
(25)
Rp = inf
kn−2 [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)ak
γ(1 − β)
1
¾ (k−1)
The result is sharp.
Proof. In order to establish the required result it is sufficient to show that
¯ 0 ¯
¯ zf (z) ¯
¯
¯
¯ f 0 (z) ¯ < 1, |z| < Rp
in view of (1), we have
¯ 00 ¯
¯ zf (z) ¯
¯
¯
¯ f 0 (z) ¯ ≤
∞
P
k=2
k(k − 1)ak |z|k−1
1−
∞
P
k=2
kak |z|k−1
Hence, we obtain
∞
X
(26)
k 2 ak |z|k−1 ≤ 1
k=2
but from Theorem 1, we know that
(27)
∞
X
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)ak
<1
γ(1 − β)
k=2
and thus from (26) and (27) we have
k 2 |z|k−1 ≤
or
½
|z| ≤
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
γ(1 − β)
kn−2 [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)
γ(1 − β)
1
¾ (k−1)
Hence, f(z) is convex in |z| < Rp . The result is sharp and is given by (25).
130
6
V.B.L. Chaurasia, Anil Sharma
Closure Theorem
Theorem 7 Let the function fj (z), (j = 1, 2, , m) be defined by
(28)
fj (z) = z −
∞
X
akj zk
(akj > 0)
k=2
for z ∈ U , be in the class SC(γ, λ, β) then the function h(z) defined by
h(z) = z −
∞
X
bk zk
k=2
also belongs to the class SC(γ, λ, β), where
m
1 X
bk =
akj
m
j=1
Proof. Since fj (z) SC(γ, λ, β), it follows from Theorem 1, that
∞
X
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)akj < γ(1 − β),
(j = 1, 2, , m).
k=2
Therefore,
∞
X
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)bk
k=2
=
∞
X
k=2


m
X
1
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k) 
akj 
m
j=1
m
1 X
=
m
j=1
Ã
∞
X
!
kn [λk + 1 − λ] [k − 1 − γ(β − 1)] C(α, k)akj
k=2
< γ(1 − β).
Hence by Theorem 1, h(z) ∈ SC(γ, λ, β) also.
Theorem 8 The class SC(γ, λ, β) is closed under linear combination.
Proof. Employing same techniques used by Silverman [14] with the aid of
Theorem 8, it can be easily proved.
On a subclass of analytic functions with negative coefficient...
131
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V.B.L. Chaurasia, Anil Sharma
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V.B.L. Chaurasia
University of Rajasthan
Department of Mathematics
Jaipur-302055, India.
Anil Sharma
Suresh Gyanvihar University
Department of Mathematics
Jaipur-302011, India.