NEW SUBCLASS OF UNIVALENT FUNCTIONS DEFINED BY USING GENERALIZED SALAGEAN OPERATOR | Joshi | Journal of the Indonesian Mathematical Society 46 144 1 PB

J. Indones. Math. Soc. (MIHMI)
Vol. 15, No. 2 (2009), pp. 79–89.

NEW SUBCLASS OF UNIVALENT
FUNCTIONS DEFINED BY USING
GENERALIZED SALAGEAN OPERATOR
S. B. Joshi and N. D. Sangle
Abstract. In this paper, we have introduced and studied a new subclass T Dλ (α, β, ξ; n)
of univalent functions defined by using generalized Salagean operator in the unit disk
U = {z : |z| < 1} We have obtained among others results like, coefficient inequalities,
distortion theorem, extreme points, neighbourhood and Hadamard product properties.

1. INTRODUCTION
Let A denote the class of functions of the form
f (z) = z +

∞
X

ak z k

(1)

k=2

which are analytic in the unit disk U = {z : |z| < 1}. In [4], Al-oboudi defined a
differential operator as follows, for a function f ∈ A,
D0 f (z) = f (z),
Df (z) = D1 f (z) = (1 − λ)f (z) + λzf ′ (z) = Dλ f (z), λ ≥ 0,

(2)

Dn f (z) = Dλ (Dn−1 f (z)).

(3)

in general
If f (z) is given by (1), then from (2) and (3) we observe that
Dn f (z) = z +

∞

X

[1 + (k − 1)λ]n ak z k

(4)

k=2
Received 20-06-2008, Accepted 18-09-2009.
2000 Mathematics Subject Classification: 30C45
Key words and Phrases: Univalent function, Distortion theorem, Neighbourhood, Hadamard product

79

80

S.B. Joshi and N.D. Sangle

when λ = 1, get Salagean differential operator [7]. Further, let T denote the
subclass of A which consists of functions of the form
f (z) = z −

∞
X

ak z k , ak ≥ 0.

(5)

k=2

A function f (z) belonging to A is in the class Dλ (α, β, ξ; n), if and only if
¯
¯
¯
¯
(Dn f (z))′ − 1
¯
¯
¯ 2ξ[(Dn f (z))′ − α] − [(Dn f (z))′ − 1] ¯ < β

(6)

where 0 ≤ α < 12 ξ, 0 < β ≤ 1, 1/2 ≤ ξ ≤ 1, n ∈ N ∪ {0}, z ∈ U . Let
T Dλ (α, β, ξ; n) = T ∩ Dλ (α, β, ξ; n).

(7)

2. MAIN RESULTS
Theorem 1. Let f be defined by (5). Then f ∈ T Dλ (α, β, ξ; n) if and only if
∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]ak ≤ 2βξ(1 − α)

(8)

k=2

0 ≤ α < 12 ξ, 0 < β ≤ 1, 1/2 ≤ ξ ≤ 1, n ∈ N ∪ {0}, λ ≥ 0.
Proof. For |z| = 1, we get

|(Dn f (z))′ − 1| − β|2ξ[(Dn f (z))′ − α] − [(Dn f (z))′ − 1]|
¯ ∞
¯
¯ X
¯
¯
¯
[1 + (k − 1)λ]n kak z k−1 ¯
= ¯−
¯
¯
k=2
¯
¯
∞
∞
¯
¯
X
X

¯
n
k−1
n
k−1 ¯
[1 + (k − 1)λ] kak z
+
− β ¯2ξ(1 − α) − 2ξ
[1 + (k − 1)λ] kak z
¯
¯
¯
k=2

k=2

≤

∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]ak − 2βξ(1 − α)

k=2

≤ 0,

by hypothesis. Thus by maximum modulus theorem, we have f ∈ T Dλ (α, β, ξ; n).
Conversely, suppose that f ∈ T Dλ (α, β, ξ; n) hence the condition (6) gives us
¯
¯
¯
¯
(Dn f (z))′ − 1
¯
¯
¯ 2ξ[(Dn f (z))′ − α] − [(Dn f (z))′ − 1] ¯
¯
¯
P∞

¯
¯
− k=2 [1 + (k − 1)λ]n kak z k−1
¯ < β.
P∞
= ¯¯
¯
n
k−1
2ξ(1 − α) − (2ξ − 1)
[1 + (k − 1)λ] kak z
k=2

New Subclass of Univalent Functions

81

Since |Re(z)| < |z| for all z, we obtain
Re

½

P∞
¾
− k=2 [1 + (k − 1)λ]n kak z k−1
P∞
< β.
2ξ(1 − α) − (2ξ − 1) k=2 [1 + (k − 1)λ]n kak z k−1

Letting z → 1− through real values, we get (8). The result is sharp for the
function
2βξ(1 − α)
f (z) = z −
z k , k ≥ 2.
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

Corollary 1. Let f ∈ T belong to the class T Dλ (α, β, ξ; n) then
ak ≤

2βξ(1 − α)

, k ≥ 2.
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

(9)

Theorem 2. Let f ∈ T belong to the class T Dλ (α, β, ξ; n), then for |z| ≤ r < 1,
we have
βξ(1 − α)
βξ(1 − α)
≤ |Dn f (z)| ≤ r + r2
(10)
r − r2
1 + β(2ξ − 1)
1 + β(2ξ − 1)
and
1−r

2βξ(1 − α)
2βξ(1 − α)
≤ |(Dn f (z))′ | ≤ 1 + r

.
1 + β(2ξ − 1)
1 + β(2ξ − 1)

(11)

bounds given by (10) and (11) are sharp.
Proof. By Theorem 1, we have
∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]ak ≤ 2βξ(1 − α)

k=2

then, we have
n

2(1 + λ) [1 + β(2ξ − 1)]ak ≤

∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]ak ≤ 2βξ(1 − α),

k=2

then,
∞
X

k=2

ak ≤

2βξ(1 − α)
2(1 + λ)n [1 + β(2ξ − 1)]

82

S.B. Joshi and N.D. Sangle

Hence
|Dn f (z)| ≤ |z| +

∞
X
¯
¯
¯[1 + (k − 1)λ]n ak z k ¯

k=2

≤ |z| + |z|2 (1 + λ)n

∞
X

ak

k=2

≤ r + r2 (1 + λ)n

∞
X

ak

k=2

≤ r + r2

βξ(1 − α)
,
1 + β(2ξ − 1)

and
|Dn f (z)| ≥ |z| −

∞
X
¯
¯
¯[1 + (k − 1)λ]n ak z k ¯

k=2

≥ |z| − |z|2 (1 + λ)n

∞
X

ak

k=2

≥ r − r2 (1 + λ)n

∞
X

ak

k=2

≥ r − r2

βξ(1 − α)
,
1 + β(2ξ − 1)

thus (10) is true. Further,
|(Dn f (z))′ | ≤ 1 + 2r(1 + λ)n

∞
X

ak

∞
X

ak

k=2

≤1+r

2βξ(1 − α)
,
1 + β(2ξ − 1)

and
|(Dn f (z))′ | ≥ 1 − 2r(1 + λ)n

k=2

≥1−r

2βξ(1 − α)
.
1 + β(2ξ − 1)

The result is sharp for the function f (z) defined by
f (z) = z −

2βξ(1 − α) 2
z , z = ±r.
1 + β(2ξ − 1)

New Subclass of Univalent Functions

83

Theorem 3. Let n ∈ N ∪ {0}, λ ≥ 0, 0 ≤ α1 ≤ α2 < 12 ξ, 0 < β ≤ 1, 1/2 ≤ ξ ≤ 1.
Then T Dλ (α2 , β, ξ; n) ⊂ T Dλ (α1 , β, ξ; n).
Proof. By assumption we have
2βξ(1 − α2 )
2βξ(1 − α1 )
≤
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
Thus, f (z) ∈ T Dλ (α2 , β, ξ; n) implies that
∞
X

[1 + (k − 1)λ]n ak ≤

k=2

2βξ(1 − α1 )
2βξ(1 − α2 )
≤
k[1 + β(2ξ − 1)]
k[1 + β(2ξ − 1)]

then f (z) ∈ T Dλ (α1 , β, ξ; n).

Theorem 4. The set T Dλ (α, β, ξ; n) is the convex set.
P∞
Proof. Let fi (z) = z − k=2 ak,i z k (i = 1, 2) belong to T Dλ (α, β, ξ; n) and let
g(z) = ζ1 f1 (z) + ζ2 f1 (z) with ζ1 and ζ2 non negative and ζ1 + ζ2 = 1, we can write
g(z) = z −

∞
X

(ζ1 ak,1 + ζ2 ak,2 )z k .

k=2

It is sufficient to show that g(z) = T Dλ (α, β, ξ; n) that means
∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)](ζ1 ak,1 + ζ2 ak,2 )

k=2

= ζ1

∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]ak,1 + ζ2

k=2

∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]ak,2

k=2

≤ ζ1 (2βξ(1 − α)) + ζ2 (2βξ(1 − α))
= (ζ1 + ζ2 )(2βξ(1 − α))
= 2βξ(1 − α)

Thus g(z) ∈ T Dλ (α, β, ξ; n).

We shall now present a result on extreme points in the following theorem
Theorem 5. Let f1 (z) = z and
f (z) = z −

2βξ(1 − α)
zk
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

84

S.B. Joshi and N.D. Sangle

for all k ≥ 2, n ∈ N ∪ {0}, λ ≥ 0, 0 ≤ α < 12 ξ, 0 < β ≤ 1, 1/2 ≤ ξ ≤ 1. Then
f (z) is P
in the subclass T Dλ (α, β, ξ; n)P
if and only if it can be expressed
in the form
P∞
∞
∞
f (z) = k=2 γk z k where γk ≥ 0 and k=2 γk = 1 or 1 = γ1 + k=2 γk .
Proof. Let f (z) =

P∞

k=2

γk z k where γk ≥ 0 and

f (z) = z −

∞
X

k=2

P∞

k=2

γk = 1. Thus

2βξ(1 − α)
γk z k
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

and we obtain
¶
∞ µ
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
2βξ(1 − α)
γk ×
2βξ(1 − α)
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
k=2

=

∞
X

γk = 1 − γ1 ≤ 1.

k=2

In view of Theorem 1, this show that f (z) ∈ T Dλ (α, β, ξ; n).
Conversely, suppose that of the form (5) belong to T Dλ (α, β, ξ; n) then
ak ≤

2βξ(1 − α)
, k ≥ 2,
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

Putting

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
2βξ(1 − α)
P∞
P∞
and γ1 = 1 − k=2 γk , then we have f (z) = γ1 f1 (z) + k=2 γk fk (z). This completes the proof.
γk =

3. NEIGHBOURHOOD AND HADAMARD PRODUCT
PROPERTIES
Definition 1. [6] Let γk ≥ 0 and f (z) ∈ T of the form (5). The (k, γ)-neighbourhood
of a function f (z) defined by
(
)
∞
∞
X
X
N(k,γ) (f ) = g ∈ T : g(z) = z −
bk z k and
k|ak − bk | ≤ γ
(12)
k=2

For the identity function e(z) = z we have
(
N(k,γ) (e) =

g ∈ T : g(z) = z −

k=2

∞
X

k=2

k

bk z and

∞
X

k=2

k|bk | ≤ γ

)

.

(13)

85

New Subclass of Univalent Functions

Theorem 5. Let
γ=

2βξ(1 − α)
.
(1 + λ)n [1 + β(2ξ − 1)]

Then T Dλ (α, β, ξ; n) ⊂ N(k,γ) (e).
Proof. Let f ∈ T Dλ (α, β, ξ; n) then we have
2(1 + λ)n [1 + β(2ξ − 1)]
≤

∞
X

ak

k=2
∞
X

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]ak

k=2

≤ 2βξ(1 − α),
therefore

∞
X

ak ≤

k=2

βξ(1 − α)
,
(1 + λ)n [1 + β(2ξ − 1)]

(14)

also we have for |z| < r
|f ′ (z)| ≤ 1 + |z|

∞
X

k=2

kak ≤ 1 + r

∞
X

kak .

k=2

In view (14), we have
|f ′ (z)| ≤ 1 + r

2βξ(1 − α)
.
(1 + λ)n [1 + β(2ξ − 1)]

From above inequalities we get
∞
X

k=2

kak ≤

2βξ(1 − α)
= γ,
(1 + λ)n [1 + β(2ξ − 1)]

therefore f ∈ N(k,γ) (e).

Definition 2. The function f (z) defined by (5) is said to be a member of the
subclass T Dλ (α, β, ξ, ζ; n) if there exits a function g ∈ T Dλ (α, β, ξ; n) such that
¯
¯
¯ f (z)
¯
¯
¯ ≤ 1 − ζ, z ∈ U, 0 ≤ ζ < 1.
−
1
¯ g(z)
¯

86

S.B. Joshi and N.D. Sangle

Theorem 6. Let g ∈ T Dλ (α, β, ξ; n) and
ζ =1−

γ
d(α, β, ξ; n).
2

(15)

Then N(k,γ) (g) ⊂ T Dλ (α, β, ξ, ζ; n) where n ∈ N ∪ {0}, λ ≥ 0, 0 ≤ α <
0 < β ≤ 1, 1/2 ≤ ξ ≤ 1, 0 ≤ ζ < 1 and
d(α, β, ξ; n) =

(1 + λ)n [1 + β(2ξ − 1)]
.
(1 + λ)n [1 + β(2ξ − 1)] − βξ(1 − α)

Proof. Let f ∈ N(k,γ) (g) then by (14) we have
bk | ≤ γ2 . Since g ∈ T Dλ (α, β, ξ; n) we have
∞
X

1
2 ξ,

bk ≤

k=2

P∞

k=2

k|ak −bk | ≤ γ, then

P∞

k=2

|ak −

βξ(1 − α)
,
(1 + λ)n [1 + β(2ξ − 1)]

therefore,
¯ P∞
¯
¯
¯ f (z)
|ak − bk |
¯ < k=2P
¯
−
1
∞
¯
¯ g(z)
1 − k=2 bk
µ
¶
γ
(1 + λ)n [1 + β(2ξ − 1)]
≤
2 (1 + λ)n [1 + β(2ξ − 1)] − βξ(1 − α)
γ
= d(α, β, ξ; n) = 1 − ζ.
2
Then by Definition 2, we get f ∈ T Dλ (α, β, ξ, ζ; n).

Theorem 7.PLet f (z) and g(z) ∈ T Dλ (α
, β, ξ; n) be of the form (5) such that
P1∞
∞
k
where ak , bk ≥ 0. Then the
f (z) = z − k=2 ak z k and g(z) = z − k=2 bk zP
∞
Hadamard product h(z) defined by h(z) = z − k=2 ak bk z k is in the subclass
T Dλ (α2 , β, ξ; n) where
α2 ≤

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)] − 2βξ(1 − α1 )2
.
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

Proof. By Theorem 1, we have
∞
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

ak ≤ 1

(16)

∞
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

bk ≤ 1.

(17)

k=2

and

k=2

2βξ(1 − α1 )

2βξ(1 − α1 )

New Subclass of Univalent Functions

87

We have only to find the largest α2 such that
∞
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

k=2

2βξ(1 − α2 )

ak bk ≤ 1.

Now, by Cauchy-Schwarz inequality, we obtain
∞
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)] p

k=2

2βξ(1 − α1 )

ak bk ≤ 1,

we need only to show that
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
ak bk
2βξ(1 − α2 )
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)] p
≤
ak bk , (18)
2βξ(1 − α1 )
equivalently,
p
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
2βξ(1 − α2 )
ak bk ≤
×
2βξ(1 − α1 )
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
1 − α2
.
≤
1 − α1
But from (18), we have
p
ak bk ≤

2βξ(1 − α1 )
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

Consequently, we need to prove that
1 − α2
2βξ(1 − α1 )
.
≤
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]
1 − α1
or equivalently, that
α2 ≤

[1 + (k − 1)λ]n k[1 + β(2ξ − 1)] − 2βξ(1 − α1 )2
.
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)]

Theorem 8. Let f ∈ T Dλ (α, β, ξ; n) be defined by (5)Rand c any real number with
z c−1
s f (s)ds, c > −1, also
c > −1 than the function G(z) defined as G(z) = c+1
zc
0
belongs to T Dλ (α, β, ξ; n).

88

S.B. Joshi and N.D. Sangle

Proof. By virtue of G(z) it follows from (5) that
!
Z Ã
∞
X
c+1 z
c
k+c−1
s −
ak s
ds
G(z) =
zc
0
k=2
¶
∞ µ
X
c+1
=z−
ak z k .
c+k
k=2

But

µ
¶
∞
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)] c + 1
2βξ(1 − α)

k=2

since

³

c+1
c+k

´

c+k

ak ≤ 1,

≤ 1 and by Theorem 1, so the proof is complete.

Theorem 8. Let f ∈ T Dλ (α, β, ξ; n) be defined by (5) and
Z z
f (s)
Fµ (z) = (1 − µ)z + µ
ds (µ ≥ 0, z ∈ U ).
s
0
Then Fµ (z) is also in T Dλ (α, β, ξ; n) if 0 ≤ µ ≤ 2.
Proof. Let f defined by (5) then
Fµ (z) = (1 − µ)z + µ

Z

z

0

=z−

∞
X
µ

k=2

By Theorem 1 and since

¡µ
k

k

s−

P∞

k=2

ak sk

s

¶

ds

ak z k .

¢
≤ 1 we have

∞
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)] ³ µ ´

k=2

µ

2βξ(1 − α)

k

≤

ak

∞
X
[1 + (k − 1)λ]n k[1 + β(2ξ − 1)] ³ µ ´

k=2

2βξ(1 − α)

2

ak ≤ 1,

then Fµ (z) is in T Dλ (α, β, ξ; n).

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New Subclass of Univalent Functions

89

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S. B. Joshi: Department of Mathematics, Walchand College of Engineering, Sangli (M.S),
India 416 415.
E-mail: joshisb@hotmail.com.
N. D. Sangle: Department of Mathematics, Annasaheb Dange College of Engineering,
Ashta, Sangli (M.S), India 416 301.
E-mail: navneet sangle@rediffmail.com.