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 subclassT 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
LetAdenote the class of functions of the form
f(z) =z+ ∞ X
k=2
akzk (1)
which are analytic in the unit disk U ={z: |z| <1}. In [4], Al-oboudi defined a differential operator as follows, for a functionf ∈A,
D0f(z) =f(z),
Df(z) =D1f(z) = (1−λ)f(z) +λzf′(z) =Dλf(z), λ≥0, (2)
in general
Dnf(z) =Dλ(Dn−1f(z)). (3)
Iff(z) is given by (1), then from (2) and (3) we observe that
Dnf(z) =z+ ∞ X
k=2
[1 + (k−1)λ]nakzk (4)
Received 20-06-2008, Accepted 18-09-2009.
2000 Mathematics Subject Classification: 30C45
Key words and Phrases: Univalent function, Distortion theorem, Neighbourhood, Hadamard product
when λ = 1, get Salagean differential operator [7]. Further, let T denote the subclass ofAwhich consists of functions of the form
Since|Re(z)|<|z|for allz, we obtain
Re
½ −P∞
k=2[1 + (k−1)λ]nkakzk−1 2ξ(1−α)−(2ξ−1)P∞
k=2[1 + (k−1)λ]nkakzk−1 ¾
< β.
Lettingz→1− through real values, we get (8). The result is sharp for the function
f(z) =z− 2βξ(1−α)
[1 + (k−1)λ]nk[1 +β(2ξ−1)]z
k, k≥2.
Corollary 1. Letf ∈T belong to the class T Dλ(α, β, ξ;n)then
ak ≤
2βξ(1−α)
[1 + (k−1)λ]nk[1 +β(2ξ−1)], k≥2. (9)
Theorem 2. Let f ∈T belong to the class T Dλ(α, β, ξ;n), then for |z| ≤r <1, we have
r−r2 βξ(1−α) 1 +β(2ξ−1) ≤ |D
nf(z)| ≤r+r2 βξ(1−α)
1 +β(2ξ−1) (10)
and
1−r 2βξ(1−α)
1 +β(2ξ−1) ≤ |(D
nf(z))′| ≤1 +r 2βξ(1−α)
1 +β(2ξ−1). (11)
bounds given by (10) and (11) are sharp.
Proof. By Theorem 1, we have
∞ X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)]ak≤2βξ(1−α)
then, we have
2(1 +λ)n[1 +β(2ξ−1)]a k≤
∞ X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)]a
k≤2βξ(1−α),
then,
∞ X
k=2 ak≤
Hence
|Dnf(z)| ≤ |z|+ ∞ X
k=2 ¯
¯[1 + (k−1)λ]nakzk ¯ ¯
≤ |z|+|z|2(1 +λ)n ∞ X
k=2 ak
≤r+r2(1 +λ)n ∞ X
k=2 ak
≤r+r2 βξ(1−α) 1 +β(2ξ−1),
and
|Dnf(z)| ≥ |z| − ∞ X
k=2 ¯
¯[1 + (k−1)λ]na kzk
¯ ¯
≥ |z| − |z|2(1 +λ)n ∞ X
k=2 ak
≥r−r2(1 +λ)n ∞ X
k=2 ak
≥r−r2 βξ(1−α) 1 +β(2ξ−1),
thus (10) is true. Further,
|(Dnf(z))′| ≤1 + 2r(1 +λ)n ∞ X
k=2 ak
≤1 +r 2βξ(1−α) 1 +β(2ξ−1),
and
|(Dnf(z))′| ≥1−2r(1 +λ)n ∞ X
k=2 ak
≥1−r 2βξ(1−α) 1 +β(2ξ−1).
The result is sharp for the functionf(z) defined by
f(z) =z− 2βξ(1−α)
1 +β(2ξ−1)z
Theorem 3. Let n∈N∪ {0},λ≥0,0≤α1≤α2< 12ξ,0< β ≤1,1/2≤ξ≤1. ThenT Dλ(α2, β, ξ;n)⊂T Dλ(α1, β, ξ;n).
Proof. By assumption we have
2βξ(1−α2)
[1 + (k−1)λ]nk[1 +β(2ξ−1)] ≤
2βξ(1−α1)
[1 + (k−1)λ]nk[1 +β(2ξ−1)]
Thus,f(z)∈T Dλ(α2, β, ξ;n) implies that
∞ X
k=2
[1 + (k−1)λ]na k ≤
2βξ(1−α2) k[1 +β(2ξ−1)] ≤
2βξ(1−α1) k[1 +β(2ξ−1)]
thenf(z)∈T Dλ(α1, β, ξ;n).
Theorem 4. The set T Dλ(α, β, ξ;n)is the convex set.
Proof. Let fi(z) = z−P∞
k=2ak,izk (i = 1,2) belong to T Dλ(α, β, ξ;n) and let g(z) =ζ1f1(z) +ζ2f1(z) withζ1andζ2 non negative andζ1+ζ2= 1, we can write
g(z) =z−
∞ X
k=2
(ζ1ak,1+ζ2ak,2)zk.
It is sufficient to show thatg(z) =T Dλ(α, β, ξ;n) that means
∞
X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)](ζ1ak,1+ζ2ak,2)
=ζ1
∞
X
k=2
[1 + (k−1)λ]n
k[1 +β(2ξ−1)]ak,1+ζ2
∞
X
k=2
[1 + (k−1)λ]n
k[1 +β(2ξ−1)]ak,2
≤ζ1(2βξ(1−α)) +ζ2(2βξ(1−α))
= (ζ1+ζ2)(2βξ(1−α))
= 2βξ(1−α)
Thusg(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)
for all k≥2, n∈N ∪ {0}, λ≥0, 0 ≤α < 12ξ, 0< β ≤1,1/2 ≤ξ ≤1. Then
Theorem 5. Let
γ= 2βξ(1−α) (1 +λ)n[1 +β(2ξ−1)].
ThenT Dλ(α, β, ξ;n)⊂N(k,γ)(e).
Proof. Letf ∈T Dλ(α, β, ξ;n) then we have
2(1 +λ)n[1 +β(2ξ−1)] ∞ X
k=2 ak
≤
∞ X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)]ak
≤2βξ(1−α),
therefore
∞ X
k=2 ak≤
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
k=2 kak.
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)
(1 +λ)n[1 +β(2ξ−1)] =γ,
thereforef ∈N(k,γ)(e).
Definition 2. The function f(z) defined by (5) is said to be a member of the subclassT Dλ(α, β, ξ, ζ;n) if there exits a functiong∈T Dλ(α, β, ξ;n)such that
¯ ¯ ¯ ¯
f(z) g(z) −1
¯ ¯ ¯ ¯
Theorem 6. Let g∈T Dλ(α, β, ξ;n)and
ζ= 1−γ
2d(α, β, ξ;n). (15)
Then N(k,γ)(g) ⊂ T Dλ(α, β, ξ, ζ;n) where n ∈ N ∪ {0}, λ ≥ 0, 0 ≤ α < 12ξ, 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. Letf ∈N(k,γ)(g) then by (14) we haveP∞
k=2k|ak−bk| ≤γ, thenP∞k=2|ak− bk| ≤ γ2. Sinceg∈T Dλ(α, β, ξ;n) we have
∞ X
k=2 bk≤
βξ(1−α) (1 +λ)n[1 +β(2ξ−1)],
therefore,
¯ ¯ ¯ ¯
f(z) g(z)−1
¯ ¯ ¯ ¯
< P∞
k=2|ak−bk| 1−P∞
k=2bk
≤ γ
2
µ (1 +λ)n[1 +β(2ξ−1)]
(1 +λ)n[1 +β(2ξ−1)]−βξ(1−α) ¶
= γ
2d(α, β, ξ;n) = 1−ζ.
Then by Definition 2, we getf ∈T Dλ(α, β, ξ, ζ;n).
Theorem 7. Let f(z) and g(z) ∈ T Dλ(α1, β, ξ;n) be of the form (5) such that f(z) = z−P∞
k=2akzk and g(z) = z−P∞k=2bkzk where ak, bk ≥ 0. Then the Hadamard product h(z) defined by h(z) = z −P∞
k=2akbkzk is in the subclass T Dλ(α2, β, ξ;n)where
α2≤
[1 + (k−1)λ]nk[1 +β(2ξ−1)]−2βξ(1−α1)2 [1 + (k−1)λ]nk[1 +β(2ξ−1)] .
Proof. By Theorem 1, we have
∞ X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)]
2βξ(1−α1) ak ≤1 (16)
and ∞
X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)]
We have only to find the largestα2 such that
∞ X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)]
2βξ(1−α2) akbk ≤1.
Now, by Cauchy-Schwarz inequality, we obtain
∞ X
k=2
[1 + (k−1)λ]nk[1 +β(2ξ−1)] 2βξ(1−α1)
p
akbk≤1,
we need only to show that
[1 + (k−1)λ]nk[1 +β(2ξ−1)] 2βξ(1−α2) akbk
≤ [1 + (k−1)λ]
nk[1 +β(2ξ−1)]
2βξ(1−α1)
p
akbk, (18)
equivalently,
p akbk≤
[1 + (k−1)λ]nk[1 +β(2ξ−1)] 2βξ(1−α1) ×
2βξ(1−α2)
[1 + (k−1)λ]nk[1 +β(2ξ−1)]
≤ 1−α2
1−α1 .
But from (18), we have
p akbk≤
2βξ(1−α1)
[1 + (k−1)λ]nk[1 +β(2ξ−1)]
Consequently, we need to prove that
2βξ(1−α1)
[1 + (k−1)λ]nk[1 +β(2ξ−1)] ≤ 1−α2 1−α1.
or equivalently, that
α2≤
[1 + (k−1)λ]nk[1 +β(2ξ−1)]−2βξ(1−α1)2 [1 + (k−1)λ]nk[1 +β(2ξ−1)] .
Theorem 8. Letf ∈T Dλ(α, β, ξ;n)be defined by (5) andc any real number with c >−1 than the functionG(z)defined as G(z) = c+1zc
Rz 0 s
Proof. By virtue ofG(z) it follows from (5) that
≤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
Proof. Letf defined by (5) then
Fµ(z) = (1−µ)z+µ
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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.
