General Mathematics Vol. 15, No. 4 (2007), 69-82
Certain Subclasses of Analytic Functions
with Negative Coefficients Defined by
Generalized Sălăgean Operator1
H. E. Darwish
Abstract
We introduce the subclass Tj (n, m, γ, α, λ) of analytic functions
with negative coefficients defined by generalized Sălăgean operator
Dλn . Coefficient estimates, some important properties of the class
Tj (n, m, γ, α, λ) and distortion theorems are determined. Further,
extremal properties and radii of close-to-convexity, starlikeness and
convexity of the class Tj (n, m, γ, α, λ) are obtained.
2000 Mathematical Subject Classification:30C45
Key words:Analytic functions, negative coefficients, distortion theorems.
1
Introduction
Let Aj denote the class of functions of the form
(1.1)
f (z) = z +
∞
X
ak z k (j ∈ N = {1, 2, ....})
k=j+1
1
Received 11 July, 2007
Accepted for publication (in revised form) 25 December, 2007
69
70
H.E. Darwish
which are analytic in the unit disc U = {z : |z| < 1}. In [1] AL-Oboudi
defined the generalized Sălăgean operator as following :
(1.2)
Dλ0 f (z) = f (z),
(1.3)
Dλ1 f (z) = (1 − λ)f (z) + λzf ′ (z)
= Dλ f (z)
,λ ≥ 0
and
Dλn f (z) = Dλ (Dn−1 f (z)).
(1.4)
If f (z) is given by (1.1), then from (1.3) and (1.4) we see that
Dλn f (z)
(1.5)
=z+
∞
X
[1 + (k − 1)λ]n ak z k .
k=j+1
When λ = 1, j = 1 we get Sălăgean’s differential operator [9].
Let Tj denote the subclass of Aj consisting of functions of the form
(1.6)
f (z) = z −
∞
X
ak z k (ak ≥ 0 , j ∈ N ).
k=j+1
With the above operator Dλn , we say that a function f (z) belonging to
Aj is in the class Aj (n, m, γ, α, λ) if and only if
(1.7)
Re
½
Dλn+m f (z)/Dλn f (z)
γ (Dλn+m f (z)/Dλn f (z)) + 1 − γ
¾
> α, z ∈ U,
for some α, γ ∈ [0, 1), λ ≥ 0, j ∈ N, n, m ∈ N0∗ = N ∪ {0} .
Furthermore, by specializing the parameters n, m, γ, α and λ, we obtain
the following subclasses studied by various other authors:
(i) T1 (0, 1, 0, α, 1) = T ∗ (α) and T1 (1, 1, 0, α, 1) = C(α)(Silverman [11]);
71
Certain Subclasses of ...
(ii) Tj (0, 1, 0, α, 1) = Tα (j) and Tj (1, 1, 0, α, 1) = Cα (j)(Chatterjea [5]
and Srivastava et al. [12]);
(iii) T1 (n, 1, 0, α, 1) = T (n, α) (Hur and Oh [8]);
(iv)T1 (0, 1, γ, α, 1) = T (γ, α) and T1 (1, 1, γ, α, 1) = C(γ, α)
(Altintas and Owa [2]);
(iv) T1 (n, 1, γ, α, 1) = Tn (γ, α) (Aouf and Cho [3] and Cho and Aouf [6]);
(v) Tj (n, m, 0, α, 1) = Tj (n, m, α) (Sekine [10] and Hossen, Sălăgean and
Aouf [7]);
(vi) Tj (n, m, γ, α, 1) = Tj (n, m, γ, α)(Aouf, Darwish and Attiya[4]).
2
Coefficient estimates
Lemma 2.1. Let the function f (z) be defined by (1.6) with j = 1.Then
f (z) ∈ Tj (n, m, γ, α, λ) if and only if
∞
X
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}ak ≤ 1 − α,
(2.1)
k=2
for n ∈ N0 , m ∈ N0 , 0 ≤ α < 1, λ ≥ 0 and 0 ≤ γ < 1.The result is sharp .
Assume the inequality (2.1) holds and let |z| = 1.Then we have
(2.2)
(1 − γ)
≤
1−γ
∞
X
k=2
∞
X
¯
¯
¯
¯
Dλn+m f (z)
¯
¯
n
¯
¯
D
f
(z)
λ
¯≤
¯
−
1
¯
¯ Dn+m f (z)
¯
¯γ λ
+
1
−
γ
¯
¯
Dλn f (z)
[1 + (k − 1)λ]
n+m
ak |z|
k−1
− (1 − γ)
[1 + (k − 1)λ]n+m ak |z|k−1 − (1 − γ)
k=2
∞
X
k=2
∞
X
[1 + (k − 1)]n ak |z|k−1
[1 + (k − 1)λ]n ak |z|k−1
k=2
72
H.E. Darwish
(1 − γ)
½∞
P
[1 + (k − 1)λ]
n+m
ak −
k=2
≤
1−γ
∞
P
∞
P
n
[1 + (k − 1)λ] ak
k=2
[1 + (k − 1)λ]n+m ak − (1 − γ)
k=2
∞
P
¾
≤1−α
[1 + (k − 1)λ]n ak
k=2
which implies (1.7). Thus it follows from this fact that f (z) ∈ T1 (n, m, γ, α, λ).
Conversely, assume that the function f (z)is in the class T1 (n, m, γ, α, λ).Then

Dλn+m f (z)



Dλn f (z)
=
Re


Dλn+m f (z)


γ
+ 1 − γ
Dλn f (z)




(2.3)
= Re






1−
∞
X
[1 + (k − 1)λ]
n+m






ak z k−1
k=2




 1− γ
∞
X
[1 + (k − 1)λ]
n+m
ak z
k−1
− (1 − γ)
∞
X
k=2
k=2


[1 + (k − 1)λ]n ak z k−1 


> α(z ∈ U )
for z ∈ U.Choose values of z on the real axis so that Dλn+m f (z)/Dλn+m f (z)
©
ª
/ γ Dλn+m f (z)/Dλn f (z) + 1 − γ is real. Upon clearing the denominator in
(2.3) and letting z → 1− through real values, we obtain
(
∞
∞
X
X
[1 + (k − 1)λ]n+m ak
[1 + (k − 1)λ]n+m ak ≥ α 1 − γ
1−
k=2
(2.4)
−(1 − γ)
k=2
∞
X
k=2
[1 + (k − 1)]n ak
)
which gives (2.1). Finally the result is sharp with the extremal function
given by
(2.5) f (z) = z −
1−α
zk
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}
(k ≥ 2).
With the aid of Lemma 2.1, we prove
73
Certain Subclasses of ...
Theorem 2.1. Let the function f (z) be defined by (1.6). Then f (z) ∈
Tj (n, m, γ, α, λ)if and only if
(2.6)
∞
X
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − γα) − α(1 − γ)}ak ≤ 1 − α
k=j+1
for n ∈ N0 , m ∈ N0 , 0 ≤ α < 1, 0 ≤ γ < 1 and λ ≥ 0. The result is sharp
for the function
(2.7) f (z) = z −
[1 + (k −
1)λ]n {[1
(1 − α)
zk
m
+ (k − 1)λ] (1 − γα) − α(1 − γ)}
(k ≥ j + 1).
Putting ak = 0(k = 2, 3, ..., j) in Lemma 2.1, we can prove the assertion
of Theorem1.
Corollary 2.1. Let the function f (z) defined by (1.6) be in the class
Tj (n, m, γ, α, λ). Then
(2.8)
1−α
ak ≤
(k ≥ j + 1).
n
[1 + (k − 1)λ] {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}
The equality in (2.8) is attained for the function f (z) given by (2.7).
3
Some Properties of the class Tj (n, m, γ, α, λ)
Theorem 3.1. Let 0 ≤ α < 1, 0 ≤ γ1 ≤ γ2 < 1, λ ≥ 0, n ∈ N0 , and m ∈
N0 .Then
Tj (n, m, γ1 , α, λ) ⊂ Tj (n, m, γ2 , α, λ).
It follows from Theorem 2.1., that
74
H.E. Darwish
∞
X
≤
k=j+1
∞
X
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ2 ) − α(1 − γ2 )}ak
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ1 ) − α(1 − γ1 )}ak
k=j+1
≤ 1−α
for f (z) ∈ Tj (n, m, γ1 , α, λ). Hence f (z) ∈ Tj (n, m, γ2 , α, λ).
Theorem 3.2. Let 0 ≤ α < 1, 0 ≤ γ < 1, λ ≥ 0, and n, m ∈ N0 .Then
(i) Tj (n + 1, m, γ, α, λ) ⊂ Tj (n, m, γ, α, λ),
(ii) Tj (n, m + 1, γ, α, λ) ⊂ Tj (n, m, γ, α, λ),
and
(iii) Tj (n + 1, m + 1, γ, α, λ) ⊂ Tj (n, m, γ, α, λ).
The proof follows immediatly from Theorem 2.1.
4
Distortion Theorems
Theorem 4.1. Let the function f (z) defined by (1.6) be in the class
Tj (n, m, γ, α, λ). Then we have
¯
¯
(4.1) ¯Dλi f (z)¯ ≥ |z| −
[1 +
jλ]n−i {[1
(1 − α)
|z|j+1
m
+ jλ] (1 − αγ) − α(1 − γ)}
[1 +
λj]n−i {[1
(1 − α)
|z|j+1
m
+ jλ] (1 − αγ) − α(1 − γ)}
and
¯
¯
(4.2) ¯Dλi f (z)¯ ≤ |z| +
for z ∈ U, where 0 ≤ i ≤ n.Then equalities in (4.1) and (4.2) are attained
for the function f (z) given by
(4.3)
f (z) = z −
[1 +
jλ]n {[1
(1 − α)
z j+1 .
+ λj]m (1 − αγ) − α(1 − γ)}
75
Certain Subclasses of ...
Proof. Note that f (z) ∈ Tj (n, m, γ, α, λ) if and only if Di f (z) ∈ Tj (n−
i, m, γ, α, λ) and that
Dλi f (z)
(4.4)
=z−
∞
X
[1 + (k − 1)λ]i ak z k .
k=j+1
Using Theorem 1, we note that
(4.5)
[1 + λj]
n−i
m
{[1 + λj] (1 − γα) − α(1 − γ)}
∞
X
[1 + (k − 1)λ]i ak ≤ 1 − α),
k=j+1
that is, that
(4.6)
∞
X
[1 + (k − 1)λ]i ak ≤
k=j+1
(1 − α)
.
[1 + λj]n−i {[1 + λj ]m (1 − αγ) − α(1 − γ)}
It follows from (4.4) and (4.6) that
¯
¯
(4.7) ¯Dλi f (z)¯ ≥ |z| −
[1 +
λj]n−i {[1
(1 − α)
|z|j+1
+ λj]m (1 − αγ) − α(1 − γ)}
and
¯
¯
(4.8) ¯Dλi f (z)¯ ≤ |z| +
[1 +
λj]n−i {[1
(1 − α)
|z|j+1 .
+ λj]m (1 − αγ) − α(1 − γ)}
Finally, we note that the equalities in (4.1) and (4.2) are attained
for the function f (z) defined by
(4.9)
Dλi f (z) = z −
(1 − α)
|z|j+1 .
n−i
m
[1 + λj ] {[1 + λj ] (1 − αγ) − α(1 − γ)}
This complete the proof of Theorem 4.1.
Corollary 4.1. Let the function f (z) defined by (1.6) be in the class
Tj (n, m, γ, α, λ). Then we have
(4.10)
|f (z)| ≥ |z| −
[1 +
jλ]n {[1
(1 − α)
|z|j+1
m
+ jλ] (1 − αγ) − α(1 − γ)}
76
H.E. Darwish
and
(4.11)
|f (z)| ≤ |z| +
[1 +
λj]n {[1
(1 − α)
|z|j+1
+ λj]m (1 − αγ) − α(1 − γ)}
for z ∈ U. The equalities in (4.10) and (4.11) are attained for the function
f (z) given by (4.3).
Proof. Taking i = 0 in Theorem 4.1., we can easily show (4.10) and (4.11).
5
Extremal properties of the class
Tj (n,m,γ,α,λ)
Theorem 5.1. The class Tj (n, m, γ, α, λ) is convex.
Proof. Let the functions
(5.1)
fν (z) = z −
∞
X
aν,k z k (aν.k ≥ 0, ν = 1, 2)
k=j+1
be in the class Tj (n, m, γ, α, λ) . It is sufficient to show that the function
h(z) defined by
(5.2)
h(z) = tf1 (z) + (1 − t) f2 (z) (0 ≤ t ≤ 1)
is in the class Tj (n, m, γ, α, λ) . Since
(5.3)
h(z) = z −
∞
X
[ta1,k + (1 − t)a2,k ] z k ,
k=j+1
with the aid of Theorem 2.1. , we have
∞
X
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}.
k=j+1
(5.4)
{ta1,k + (1 − t)a2,k } ≤ 1 − α
77
Certain Subclasses of ...
which implies that h(z) ∈ Tj (n, m, γ, α, λ). Hence Tj (n, m, γ, α, λ) is convex.
As a consequence of Theorem 5.1. we can obtain the extreme points of
the class Tj (n, m, γ, α, λ).
Theorem 5.2. Let fj (z) = z and
(5.5) fk (z) = z −
1−α
zk
n
m
[1 + (k − 1)λ] {[1 + (k − 1)λ] (1 − αγ) − α(1 − γ)}
(n, m ∈ N o, k ≥ j + 1)
for 0 ≤ α < 1, and 0 ≤ γ < 1, λ ≥ 0. Then f (z) is in the class
Tj (n, m, γ, α, λ) if and only if it can be expressed in the form
(5.6)
f (z) =
∞
X
µk fk (z),
k=j
where µk ≥ 0 for k ≥ j and
∞
X
µk = 1.
k=j
Proof. Suppose that
(5.7)
f (z) =
∞
X
µk fk (z)
k=j
= µj fj (z) +
∞
X
µk fk (z)
k=j+1
= (1 −
∞
X
µk )+
k=j+1
+
∞
X
k=j+1
½
µk z −
1−α
n
[1 + (k − 1)λ] {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}
¾
78
H.E. Darwish
=z−
∞
X
k=j+1
[1 + (k −
1)λ]n {[1
1−α
zk .
+ (k − 1)λ]m (1 − αγ) − α(1 − γ)}
Then it follows that
∞
X
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}(1 − α)µk
(1 − α)[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}
k=j+1
(5.8)
=
∞
X
µk = 1 − µj ≤ 1.
k=j+1
So by Theorem 2.1., f (z) ∈ Tj (n, m, γ, α, λ).
Conversely, assume that the function f (z) defined by (1.6) belongs to
the class Tj (n, m, γ, α, λ).Then
(5.9)
ak ≤
[1 + (k −
1)λ]n {[1
(1 − α)
(k ≥ j + 1).
+ (k − 1)λ]m (1 − αγ) − α(1 − γ)}
Setting
(5.10)
µk =
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}
ak (k ≥ j + 1)
(1 − α)
and
µj = 1 −
∞
X
µk
k=j+1
we notice that f (z) can be expressed in the form (5.6). This completes the
proof of Theorem 5.2.
Corollary 5.1. The extreme points of the class Tj (n, m, γ, α, λ) are the
functions fk (z)(k ≥ j) given by Theorem 5.2.
79
Certain Subclasses of ...
6
Radii of close-to-convexity, starlikeness and
convexity
Theorem 6.1. Let the function f (z) defined by (1.6) be in the class
Tj (n, m, γ, α, λ).Then f (z) is close - to- convex of order ρ(0 ≤ ρ < 1) in
|z| < r1 (n, m, γ, α, λ, ρ), where
(6.1)
= inf k
r1 (n, m, γ, α, λ, ρ) =
½
(1 − ρ)[1 + (k − 1)λ]
n+1
m
{[1 + (k − 1)λ] (1 − αγ) − α(1 − γ)}
k(1 − α)
(k ≥ j + 1).
The result is sharp, with the extremal function f (z) given by (2.7).
1
¾ k−1
It is sufficient to show that
′
|f (z) − 1| = 1 − ρ(0 ≤ ρ < 1)
for |z| < r1 (n, m, γ, α, λ, ρ).
We have
¯
¯
∞
∞
¯
¯ X
X
′
¯
k−1 ¯
kak |z|k−1 .
kak z ¯ ≤
|f (z) − 1| = ¯−
¯
¯
k=j+1
k=j+1
′
Thus | f (z) − 1| ≤ 1 − ρ if
∞
X
(6.2)
(
k=j+1
k
) ak |z|k−1 ≤ 1.
1−ρ
Hence, by Theorem 2.1., (6.2) will be true if
(6.3)
k |z|k−1
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}
≤
(1 − ρ)
(1 − α)
or if
(1 − ρ)[1 + (k + 1)λ]n {[1 + (k − 1)λ]m (1 − αγ) − α(1 − γ)}
|z| ≤
k(1 − α)
·
1
¸ k−1
80
H.E. Darwish
(k ≥ j + 1).
The theorem follows easily from (6.3).
Theorem 6.2. Let the function f (z) defined by (1.6) be in the class
Tj (n, m, γ, α, λ); then f (z) is starlike of order ρ(0 ≤ ρ < 1) in |z| <
r2 (n, m, γ, α, λ, ρ),where
(6.4)
r2 (n, m, γ, α, λ, ρ) =
= inf
k
½
(1 − ρ)[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − γα) − α(1 − γ)}
(k − ρ)(1 − α)
1
¾ k−1
(k ≥ j + 1).
The result is sharp, with extremal function f (z) given by (2.7).
¯ ′
¯
¯ zf (z)
¯
¯
Proof. We must show that ¯
− 1¯¯ ≤ 1−ρ for |z| < r2 (n, m, γ, α, λ, ρ).
f (z)
We have
¯
¯ ′
¯
¯ zf (z)
¯
¯
¯ f (z) − 1¯ ≤
∞
P
(k − 1) ak |z|k−1
k=j+1
1−
.
ak |z|
k−1
k=j+1
¯
¯
¯
¯ zf ′ (z)
− 1¯¯ ≤ 1 − ρ if
Thus ¯¯
f (z)
(6.5)
∞
P
∞
X
(k − ρ) ak |z|k−1
≤ 1.
(1 − ρ)
k=j+1
Hence, by Theorem 2.1., (6.5) will be true if
[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − γα) − α(1 − γ)}
(k − ρ)ak |z|k−1
≤
(1 − ρ)
(1 − α)
or if
(6.6)
|z| ≤
81
Certain Subclasses of ...
≤
½
(1 − ρ)[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − γα) − α(1 − γ)}
(k − ρ)(1 − α)
1
¾ k−1
,
(k ≥ j + 1).
The theorem follows easily from (6.6).
Corollary 6.1. Let the function f (z) defined by (1.6) be in the class
Tj (n, m, γ, α, λ); then f (z) is convex of order ρ(0 ≤ ρ < 1) in
|z| < r3 (n, m, γ, α, λ, ρ),where
r3 (n, m, γ, α, λ, ρ)
(6.7)
= inf
k
½
(1 − ρ)[1 + (k − 1)λ]n {[1 + (k − 1)λ]m (1 − γα) − α(1 − γ)}
k(k − ρ)(1 − α)
1
¾ k−1
(k ≥ j + 1).
The result is sharp, with extremal function f (z) given by (2.7).
Acknowledgment. The author wishes to thank Prof. M. K. Aouf for
his kind encouragement and help in the preparation of this paper.
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Department of Mathematics
Faculty of Science
University of Mansoura
Mansoura, Egypt
E-mail: Darwish333@yahoo.com