ISSN1842-6298 (electronic), 1843-7265 (print) Volume5(2010), 35 – 47
SOME APPLICATIONS OF GENERALIZED
RUSCHEWEYH DERIVATIVES INVOLVING A
GENERAL FRACTIONAL DERIVATIVE
OPERATOR TO A CLASS OF ANALYTIC
FUNCTIONS WITH NEGATIVE COEFFICIENTS I
Waggas Galib Atshan and S. R. Kulkarni
Abstract. For certain univalent functionf, we study a class of functionsfas defined by making use of the generalized Ruscheweyh derivatives involving a general fractional derivative operator, satisfying
Re
✭
z(J1λ,µf(z))′ (1−γ)J1λ,µf(z) +γz2(Jλ,µ
1 f(z)) ′′
✮
> β.
A necessary and sufficient condition for a function to be in the class Aλ,µ,νγ (n, β) is obtained. In addition, our paper includes distortion theorem, radii of starlikeness, convexity and close-to-convexity, extreme points. Also, we get some results in this paper.
1
Introduction
Let Ω denote the class of functions which are analytic in the unit diskU ={z∈C :
|z|<1} and letA(n) denote the subclass of Ω consisting of functions of the form:
f(z) =z−
∞ X
k=n+1
akzk, (ak≥0, n∈IN), (1.1)
where f(z) is analytic and univalent in the unit diskU. Then the function f(z)∈
A(n) is said to be in the class S(n, α), if and only if
Re
zf′
(z) f(z)
> α , (z∈U,0≤α <1). (1.2)
A function f(z) ∈S(n, α) is called starlike function of order α. A function f(z) ∈
A(n) is said to be in the class C(n, α) if and only if
Re
1 +zf
′′
(z) f′(z)
> α, (z∈U,0≤α <1). (1.3)
2000 Mathematics Subject Classification: 30C45.
A function f(z)∈C(n, α) is called convex function of order α. It is observed that
f(z)∈C(n, α) if and only ifzf′(z)∈S(n, α) ∀ n∈IN [2]. (1.4)
A functionf(z)∈A(n) is said to be in the classK(n, α) if there is a convex function g(z) such that
Re
f′
(z) g′(z)
> α, (∀z∈U,0≤α <1). (1.5)
A function f(z)∈K(n, α) is called close-to-convex of orderα.
We shall need the fractional derivative operator ([10], [11]) in this paper. Let a, b, c ∈C withC 6= 0,−1,−2,· · ·. The Gaussian hypergeometric function
2F1 is defined by
2F1(z)≡ 2F1(a, b;c;z) =
∞ X
n=0
(a)n(b)n
(c)n
zn
n!, (1.6)
where (λ)nis the pochhammer symbol defined, in terms of the Gamma function, by
(λ)n=
Γ(λ+n) Γ(λ) =
1 (n= 0)
λ(λ+ 1)· · ·(λ+n−1) (n∈IN)
Definition 1. Let 0≤λ < 1 and µ, ν ∈ IR. Then, in terms of familiar (Gauss’s) hypergeometric function 2F1, the generalized fractional derivative operatorJ0λ,µ,ν,z of a functionf(z) is defined by:
J0λ,µ,ν,z f(z) =
1 Γ(1−λ)
d dz
zλ−µRz
0(z− E)
−λf(E)·
2F1(µ−λ,1−ν; 1−λ; 1−Ez)dE (0≤λ <1)
dn
dznJ
λ−n,µ,ν
0,z f(z), (n≤λ < n+ 1, n∈IN)
(1.7) where the functionf(z) is analytic in a simply-connected region of thez-plane con-taining the origin, with the order
f(z) =O(|z|ǫ),(z→0), (1.8)
for ǫ >max{0, µ−ν} −1, and the multiplicity of(z− E)−λ is removed by requiring
log(z− E) to be real, when z− E >0.
The fractional derivative of order λ of a function f(z) is defined by
Dzλ{f(z)}= 1 Γ(1−λ)
d dz
Z z
0
f(E)
(z− E)λdE, 0≤λ <1, (1.9)
where f(z) it is chosen as in (1.7), and the multiplicity of (z− E)−λ is removed by
By comparing (e1.7) with (1.9), we find
J0λ,λ,ν,z f(z) =Dλz{f(z)}, (0≤λ <1). (1.10)
In terms of gamma function, we have
J0λ,µ,ν,z zk= Γ(k+ 1)Γ(1−µ+ν+k) Γ(1−µ+k)Γ(1−λ+ν+k)z
k−µ, (1.11)
(0≤λ <1, µ, ν ∈IR and k >max{0, µ−ν} −1).
Definition 2. Let f(z) ∈ A(n) be given by (1.1). Then the class Aλ,µ,νγ (n, β) is
defined by
Aλ,µ,νγ (n, β) =
(
f ∈A(n) :Re
(
z(J1λ,µf(z))′ (1−γ)J1λ,µf(z) +γz2(Jλ,µ
1 f(z))′′
)
> β,
(z∈U,0≤γ <1, n∈IN; 0≤β <1;λ >−1)}, (1.12)
where J1λ,µf is a generalized Ruscheweyh derivative defined by Goyal and Goyal [3, p. 442] as
J1λ,µf(z) = Γ(µ−λ+ν+ 2) Γ(ν+ 2)Γ(µ+ 1)zJ
λ,µ,ν
0,z (z
µ−1f(z)) (1.13)
= z−
∞ X
k=n+1
akC1λ,µ(k)zk, (1.14)
where
C1λ,µ(k) = Γ(k+µ)Γ(ν+ 2 +µ−λ)Γ(k+ν+ 1)
Γ(k)Γ(k+ν+ 1 +µ−λ)Γ(ν+ 2)Γ(1 +µ). (1.15) For µ = λ= α, ν = 1, the generalized Ruscheweyh derivatives reduces to ordinary Ruscheweyh derivatives of f(z) of order α [7]:
Dαf(z) = z Γ(α+ 1)D
α(zα−1f(z)) =z−
∞ X
k=n+1
akCk(α)zk, (1.16)
where
Ck(α) =
(α+ 1)(α+ 2)· · ·(α+k−1)
(k−1)! . (1.17)
The class Aλ,µ,νγ (n, β) contains many well-known classes of analytic functions, for
example:
(i) Ifµ=λ=α, ν= 1, n= 1, we get the classAλ,λ,γ 1(1, β)was studied by Tehranchi
(ii) If µ=λ= 0, ν = 1, α=β, γ = 0, we get the class of starlike function of order said to be properly subordinate to f.
2
Main Results
or equivalently
and so by the mean value theorem, we have
Re and so the proof is complete.
Corollary 6. Let f ∈Aλ,µ,νγ (n, β), then
ak<
1−β(1−γ)
Theorem 7. The class Aλ,µ,νγ (n, β) is convex set.
Proof. Letf(z), g(z) be the arbitrary elements ofAλ,µ,νγ (n, β), then for everyt (0<
t <1), we show that (1−t)f(z) +tg(z)∈Aλ,µ,νγ (n, β) thus, we have
(1−t)f(z) +tg(z) =z−
∞ X
k=n+1
[(1−t)ak+tbk]zk
and
∞ X
k=n+1
γβ(1 +k−k2) +k−β 1−β(1−γ)
[(1−t)ak+tbk]C1λ,µ(k)
= (1−t)
∞ X
k=n+1
γβ(1 +k−k2) +k−β 1−β(1−γ) akC
λ,µ
1 (k)
+t
∞ X
k=n+1
γβ(1 +k−k2) +k−β 1−β(1−γ) bkC
λ,µ
1 (k)<1.
This completes the proof.
Corollary 8. Assume that f(z) and g(z) belong to Aλ,µ,νγ (n, β), then the function
y(z) defined by y(z) = 12(f(z) +g(z))also belong to Aλ,µ,νγ (n, β).
3
Distortion Theorem
In the next theorem, we will find distortion bound forJ1λ,µf(z).
Theorem 9. Let f(z)∈Aλ,µ,νγ (n, β), then
|z| − 1−β(1−γ)
γβ(2 +n−(1 +n)2) + (n+ 1)−β|z|
n+1≤ |Jλ,µ
1 f(z)|
≤ |z|+ 1−β(1−γ)
γβ(2 +n−(1 +n)2) + (n+ 1)−β|z|
n+1.
Proof. Letf(z)∈Aλ,µ,νγ (n, β). By Theorem5, we have
∞ X
k=n+1
C1λ,µ(k)ak<
1−β(1−γ)
γβ(2 +n−(1 +n)2) + (n+ 1)−β, n∈IN ={1,2,· · · }.
Therefore
|J1λ,µf(z)| ≤ |z|+|z|n+1
∞ X
k=n+1
C1λ,µ(k)ak<|z|+
1−β(1−γ)
γβ(2 +n−(1 +n)2) + (n+ 1)−β|z|
and
This completes the proof.
4
Radii of Starlikeness, Convexity and Close-to-convexity
By using (1.4) (Alexander’s theorem [2] : fis convex if and only ifzf′ the proof of Theorem 11.
Proof. Letf(z)∈Aλ,µ,νγ (n, β). Then by Theorem 5
≤1 but, by Theorem5above inequality
holds true if
|z|k−1≤ (1−δ)[βγ(1 +k−k2) +k−β] k(1−β(1−γ)) C
λ,µ
1 (k).
This completes the proof.
5
Extreme Points
Proof. Firstly, let us expressf as in the above theorem, therefore we can write
where
gk=
1−β(1−γ) [γβ(1 +k−k2) +k−β]Cλ,µ
1 (k) σk.
Therefore, f ∈Aλ,µ,νγ (n, β), since
∞ X
k=n+1
gk[γβ(1 +k−k2) +k−β]C1λ,µ(k)
1−β(1−γ) =
∞ X
k=n+1
σk= 1−σn<1.
Conversely, assume thatf ∈Aλ,µ,νγ (n, β), then by (2.1) we may set
σk=
[γβ(1 +k−k2) +k−β]Cλ,µ
1 (k)
1−β(1−γ) ak, k≥n+ 1 and 1−
∞ X
k=n+1
σk=σn.
Then
f(z) = z−
∞ X
k=n+1
akzk=z−
∞ X
k=n+1
(1−β(1−γ))σk
[γβ(1 +k−k2) +k−β]Cλ,µ
1 (k) zk
= z−
∞ X
k=n+1
σk(z−fk(z)) =z 1−
∞ X
k=n+1 σk
!
+
∞ X
k=n+1
σkfk(z)
= σnz+
∞ X
k=n+1
σkfk(z) =
∞ X
k=n
σkfk(z).
This completes the proof.
6
Subordination
Theorem 14. For n= 1, let f(z)∈Aλ,µ,νγ (1, β) and h(z) be an arbitrary element
of A(1) such that h≺f, defined in Definition 4, and if
hk=
1 k!
dk(f(w(z)))
dzk
z=0
(6.1)
also if
∞ P
k=2
[β(1−γ) +βγk(k−1)−k]|hk|
|h1|
<(1−β(1−γ)). (6.2)
Proof. Since h ≺ f by definition of subordination there is analytic function w(z) such that |w(z)| ≤ |z| and h(z) = f(w(z)). But h(z) is the composition of two analytic functions in the unit disk, therefore we can expand this function in terms of Taylor series at origin as below
h(z) =
Therefore, we can write
h(z) =h1z+
, and using the mean value theorem, we have to prove
Re
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Waggas Galib Atshan S. R. Kulkarni
Department of Mathematics, Department of Mathematics, College of Computer Science and Mathematics, Fergusson College, Pune - 411004, University of AL-Qadisiya, Diwaniya, Iraq. India.
