Directory UMM :Journals:Journal_of_mathematics:OTHER:
ACTA UNIVERSITATIS APULENSIS
No 12/2006
FRACTIONAL CALCULUS AND SOME PROPERTIES OF
SALAGEAN-TYPE K-STARLIKE FUNCTIONS
WITH NEGATIVE COEFFICIENTS
¨
¨
H. Ozlem
GUNEY
Abstract.We define and investigate a new class of Salagean-type k-starlike
functions with negative coefficients. We obtain some properties of this subclass. Furthermore, we give Hadamard product of several functions and some
distortion theorems for fractional calculus of this generalized class.
Keywords. analytic function, univalent function, k-starlike, S˘al˘agean operator, Hadamard product.
2000 Mathematics Subject Classification : 30C45.
1. Introduction
Let A denote the class of functions of the form
f (z) = z +
∞
X
ak z k
k=2
which are analytic in the open unit disk U = {z ∈ C : |z| < 1} . For f (z)
belongs to A, Salagean [12] has introduced the following operator called the
Salagean operator :
D0 f (z) = f (z)
D1 f (z) = Df (z) = zf ′ (z)
..
.
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Dn f (z) = D(Dn−1 f (z)) , n ∈ N0 = {0} ∪ {1, 2, · · ·}.
Let Sp (p ≥ 1) denote the class of functions of the form f(z) = z p +
p+n
that are holomorphic and p-valent in the unit disk U .
n=1 ap+n z
Also, let Ap (n) denote the subclass of Sp (p ≥ 1) consisting of functions
that can be expressed in the form
P∞
f (z) = z p −
∞
X
am z m
;
am ≥ 0.
(1)
m=p+n
We can write the following equalities for the functions f (z) belonging to the
class Ap (n):
D0 f (z) = f (z)
∞
X
D1 f (z) = Df (z) = zf ′ (z) = pz p −
mam z m
m=p+n
..
.
Dλ f (z) = D(Dλ−1 f (z)) = pλ z p −
∞
X
mλ am z m , λ ∈ N0 = {0} ∪ N.
m=p+n
Denote by S the class of functions of the form f (z) = z + a2 z 2 + ..., analytic
and univalent, by ST (α) subclass consisting of starlike and univalent of order
α and by k − ST (0 ≤ k < ∞) a class of k−starlike univalent functions in U ,
introduced and investigated Lecko and Wisniowska in [1].
It is known that every f ∈ k − ST has a continuous extension to U , f (U )
is bounded and f (∂U ) is a rectifiable curve [2].
In the present paper, a subclass (k, n, p, α, λ) − ST of starlike functions in
the open unit disk U is introduced. A functions f (z) ∈ Ap (n) is said to be in
the class (k, n, p, α, λ) − ST if it is satisfies
ζ (z − ζ)(Dλ f (z))′
Re
+
z
Dλ f (z)
(
94
)
≥α
(2)
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for some α (0 ≤ α < p) and z ∈ U .
We note that
i)(0, 1, 1, α, 0) − ST ≡ T ∗ (α) was introduced by Silverman [3],
ii)(k, 1, 1, 0, 0) − ST ≡ k − ST was introduced by Lecko and Wisniowska
[1],
iii)(k, 1, 1, α, 0) − ST ≡ k − ST (α) was studied by Baharati et.al. [5],
iv)(k, n, 1, α, 0) − ST ≡ (k, n, α) − ST was studied by Guney et.al [4],
v)(k, 1, 1, α, 1) − ST ≡ (k, 1, −1, α) − UCV ≡ k − UCV(α) was investigated
by Baharati et.al [5]. Also, the class (k, A, B, α) − U CV was introduced by
Guney et al.[6].
2.Some results of the class (k, n, p, α, λ) − ST
Theorem 2.1 A function f ∈ Ap (n) is in the class (k, n, p, α, λ) − ST iff
∞
X
mλ [k(m − 1) + m − α]am ≤ pλ (p − α) + pλ k(p − 1).
(3)
m=p+n
Proof. Let (k, n, p, α, λ) − ST . Then we have from (2)
ζ (z − ζ)(Dλ f (z))′
Re
+
z
Dλ f (z)
(
∞
P
λ p−1
λ+1 p
λ+1 p−1
ζp z +p z −ζp z +ζ
(mλ+1 −mλ )am z m−1 −
pλ z p −
P
∞
P
m=p+n
m=p+n
∞
=Re
)
mλ a
m
zm
m=p+n
mλ+1 am z m
If we choose z and ζ real and z → 1− and ζ → −k + , we get,
−kpλ + pλ+1 + kpλ+1 + k
∞
P
(mλ − mλ+1 )am −
m=p+n
∞
P
pλ −
m λ am
m=p+n
or
∞
X
∞
P
m=p+n
mλ+1 am
mλ [k(m − 1) + m − α]am ≤ pλ (p − α) + pλ k(p − 1)
m=p+n
95
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which is equivalent to (3).
Conversely, assume that
(3) is true.
Then
n
ζ
z
+
(z−ζ)(Dλ f (z))′
Dλ f (z)
o
(z−ζ) pλ+1 z p−1 −
= zζ +
pλ z p −
m=p+n
∞
P
mλ am z m
∞
P
(mλ+1 −mλ )am z m−1 −
m=p+n
m=p+n
∞
pλ z p −
mλ+1 am z m−1
P
mλ am z m
m=p+n
∞
P
λ p−1
λ+1 p
λ+1 p−1
ζp z +p z −ζp z +ζ
=
∞
P
mλ+1 am z m
≥ α.
m=p+n
for |z| < 1. If we choose z → 1− and ζ → −k + through real values, we obtain
λ
(
ζ (z − ζ)(D f (z))
Re
+
z
Dλ f (z)
′
)
pλ k(p − 1) + pλ+1 −
=
pλ −
∞
P
m=p+n
∞
P
m=p+n
mλ [k(m − 1) + m]am
.
m λ am
(4)
If (3) is rewritten as
∞
X
mλ [k(m − 1) + m]am ≤ pλ+1 − pλ α + pλ k(p − 1) + α
∞
X
m λ am ,
m=p+n
m=p+n
and (4) is used, then we obtain
λ
ζ (z − ζ)(Dλ f (z))′
+
Re
z
Dλ f (z)
(
)
α p −
≥
pλ
−
∞
P
λ
m=p+n
∞
P
m=p+n
m am
mλ a
!
= α.
m
Thus f ∈ (k, n, p, α, λ) − ST .
Theorem 2.2 If f ∈ (k, n, p, α, λ) − ST , then for |z| = r < 1 we obtain
pλ (p − α) + pλ k(p − 1)
rp+n ≤ |f (z)| ≤
r −
λ
(p + n) [(1 + k)(p + n) − k − α]
p
≤ rp +
pλ (p − α) + pλ k(p − 1)
rp+n
(p + n)λ [(1 + k)(p + n) − k − α]
and
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pr
p−1
pλ (p − α) + pλ k(p − 1)
−
rp+n−1 ≤ |f ′ (z)| ≤
λ−1
(p + n) [(1 + k)(p + n) − k − α]
≤ prp−1 +
pλ (p − α) + pλ k(p − 1)
rp+n−1 .
(p + n)λ−1 [(1 + k)(p + n) − k − α]
(6)
All inequalities are sharp.
Proof From (3), we have
pλ (p − α) + pλ k(p − 1)
am ≤
(p + n)λ [(1 + k)(p + n) − k − α]
m=p+n
∞
X
(7)
and
∞
X
mam ≤
m=p+n
Thus
|f (z)| ≤ rp +
pλ (p − α) + pλ k(p − 1)
.
(p + n)λ−1 [(1 + k)(p + n) − k − α]
∞
X
am rm ≤ rp + rp+n
m=p+n
≤ rp +
∞
X
(8)
am ≤
m=p+n
pλ (p − α) + pλ k(p − 1)
rp+n
(p + n)λ [(1 + k)(p + n) − k − α]
and
|f (z)| ≥ rp −
∞
X
am rm ≥ rp − rp+n
m=p+n
≥ rp −
∞
X
am ≥
m=p+n
pλ (p − α) + pλ k(p − 1)
rp+n .
(p + n)λ [(1 + k)(p + n) − k − α]
which prove that the assertion (5) of Theorem 2.2.
Furthermore, for |z| = r < 1 and (6), we have
|f ′ (z)| ≤ prp−1 +
∞
X
mam rm−1 ≤ prp−1 + rp+n−1
m=p+n
∞
X
m=p+n
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≤ prp−1 +
pλ (p − α) + pλ k(p − 1)
rp+n−1
(p + n)λ−1 [(1 + k)(p + n) − k − α]
and
′
|f (z)| ≥ pr
p−1
−
∞
X
mam r
m−1
≥ pr
p−1
−r
p+n−1
m=p+n
≥ prp−1 −
∞
X
mam
m=p+n
pλ (p − α) + pλ k(p − 1)
rp+n−1
(p + n)λ−1 [(1 + k)(p + n) − k − α]
which prove that the assertion (6) of Theorem 2.2.
The bounds in (5) and (6) are attained for the function f given by
f (z) = z p −
pλ (p − α) + pλ k(p − 1)
z p+n
(p + n)λ [(1 + k)(p + n) − k − α]
Theorem 2.3 Let the functions
∞
X
f (z) = z p −
am z m ; am ≥ 0.
m=p+n
and
p
g (z) = z −
∞
X
bm z m ; bm ≥ 0.
m=p+n
be in the class (k, n, p, α, λ) − ST . Then for 0 ≤ ρ ≤ 1,
h (z) = (1 − ρ)f (z) + ρ g(z) = z p −
∞
X
cm z m ; cm ≥ 0.
m=p+n
is in the class (k, n, p, α, λ) − ST .
Proof. Assume that f, g ∈ (k, n, p, α, λ) − ST .
Then we have from Theorem 2.1
∞
X
mλ [k(m − 1) + m − α]am ≤ pλ (p − α) + pλ k(p − 1)
m=p+n
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and
∞
X
mλ [k(m − 1) + m − α]bm ≤ pλ (p − α) + pλ k(p − 1).
m=p+n
Therefore we can see that
∞
X
mλ [k(m − 1) + m − α]cm =
m=p+n
∞
X
=
mλ [k(m − 1) + m − α][(1 − ρ)am + ρbm ] ≤ pλ (p − α) + pλ k(p − 1)
m=p+n
which completes the proof of Theorem 2.3.
Definition 2.1 The Modified Hadamard Product f ∗ g of two functions
f (z) = z p −
∞
X
a m z m ; am ≥ 0
m=p+n
and
g (z) = z p −
∞
X
bm z m ; bm ≥ 0
m=p+n
are denoted by
(f ∗ g) (z) = z p −
∞
X
am b m z m .
m=p+n
Theorem 2.4 If f, g ∈ (k, n, p, α, λ)−ST then f ∗g ∈ (k, n, p, β, λ)−ST ,
where
β = β(k, n, p, α, λ) =
=
(pλ+1 + pλ k(p − 1))(p + n)λ [(1 + k)(p + n) − k − α]2
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
−
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)(p + n) − k]
.
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
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The result is sharp for f (z) and g(z) given by
f (z) = g(z) = z p −
pλ (p − α) + pλ k(p − 1)
z p+n
(p + n)λ [(1 + k)(p + n) − k − α]
where 0 ≤ α < p and 0 ≤ k < ∞.
Proof. From Theorem 2.1, we have
mλ [k(m − 1) + m − α]
am ≤ 1
λ
λ
m=p+n p (p − α) + p k(p − 1)
(10)
mλ [k(m − 1) + m − α]
bm ≤ 1.
λ
λ
m=p+n p (p − α) + p k(p − 1)
(11)
∞
X
and
∞
X
We have to find the largest β such that
mλ [k(m − 1) + m − β]
am bm ≤ 1.
λ
λ
m=p+n p (p − β) + p k(p − 1)
∞
X
(12)
From (10) and (11) we obtain, by means of Cauchy-Schwarz inequality,that
mλ [k(m − 1) + m − α] q
am bm ≤ 1.
λ
λ
m=p+n p (p − α) + p k(p − 1)
∞
X
Therefore, (12) is true if
mλ [k(m − 1) + m − β]
mλ [k(m − 1) + m − α] q
a
b
≤
am b m
m m
pλ (p − β) + pλ k(p − 1)
pλ (p − α) + pλ k(p − 1)
or
[pλ (p − β) + pλ k(p − 1)][k(m − 1) + m − α]
.
am b m ≤ λ
[p (p − α) + pλ k(p − 1)][k(m − 1) + m − β]
q
Note that from (13)
q
am b m ≤
[pλ (p − α) + pλ k(p − 1)]
.
mλ [k(m − 1) + m − α]
Thus if
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[pλ (p − α) + pλ k(p − 1)]
[pλ (p − β) + pλ k(p − 1)][k(m − 1) + m − α]
≤
mλ [k(m − 1) + m − α]
[pλ (p − α) + pλ k(p − 1)][k(m − 1) + m − β]
or, equivalently, if
(pλ+1 + pλ k(p − 1))mλ [(1 + k)m − k − α]2
β≤ λ λ
p m [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
−
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)m − k]
pλ mλ [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
then (12) satisfied. Defining the function Φ(m) by
Φ(m) =
−
(pλ+1 + pλ k(p − 1))mλ [(1 + k)m − k − α]2
pλ mλ [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)m − k]
.
pλ mλ [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
We can see that Φ(m) is increasing function of m. Therefore,
β ≤ Φ(p+n) =
−
(pλ+1 + pλ k(p − 1))(p + n)λ [(1 + k)(p + n) − k − α]2
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)(p + n) − k]
.
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
which completes the assertion of theorem.
3. Extreme Points for (k, n, p, α, λ) − ST
λ
λ
k(p−1) m
z , m=
Theorem 3.1 Let fp+n−1 (z) = z p and fm (z) = z p − pmλ(p−α)+p
[k(m−1)+m−α]
p + n, p + n + 1, .... Then If f ∈ (k, n, p, α, λ) − ST iff it can be expressed in
the form
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f (z) =
∞
X
ξm fm (z),
∞
X
ξm fm (z).
m=p+n+1
where ξm ≥ 0 and
∞
P
m=p+n−1
ξm = 1.
Proof. Assume that
f (z) =
m=p+n−1
Then
f (z) = ξp+n−1 fp+n−1 (z) +
∞
X
ξm fm (z) =
m=p+n
p
= ξp+n−1 z +
∞
X
ξm
m=p+n
=
∞
X
m=p+n−1
ξm z p −
"
pλ (p − α) + pλ k(p − 1) m
z
z − λ
m [k(m − 1) + m − α]
p
∞
X
ξm
m=p+n
#
pλ (p − α) + pλ k(p − 1) m
z
mλ [k(m − 1) + m − α]
pλ (p − α) + pλ k(p − 1) m
z .
ξm λ
=z −
m [k(m − 1) + m − α]
m=p+n
p
∞
X
Thus
∞
X
m=p+n
ξm
pλ (p − α) + pλ k(p − 1)
mλ [k(m − 1) + m − α]
=
∞
X
!
mλ [k(m − 1) + m − α]
pλ (p − α) + pλ k(p − 1)
ξm − ξp+n−1 = 1 − ξm ≤ 1.
m=p+n−1
Hence f ∈ (k, n, p, α, λ) − ST .
102
!
=
∞
X
m=p+n
ξm
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Conversely, suppose that f ∈ (k, n, p, α, λ) − ST . Since
|am | ≤
pλ (p − α) + pλ k(p − 1)
mλ [k(m − 1) + m − α]
m = p + n, p + n + 1, ...,
we can set
ξm =
mλ [k(m − 1) + m − α]
am ,
pλ (p − α) + pλ k(p − 1)
and
ξp+n−1 = 1 −
m = p + n, p + n + 1, ...
∞
X
ξm .
m=p+n
Then
f (z) = z p −
∞
X
am z m = z p −
m=p+n
= zp −
∞
X
m=p+n
pλ (p − α) + pλ k(p − 1)
ξ zm
λ [k(m − 1) + m − α] m
m
m=p+n
∞
X
ξm (z p − fm (z)) = 1 −
= ξn+p−1 z p +
∞
X
∞
X
m=p+n
ξm z p +
ξm fm (z) = ξn+p−1 fn+p−1 (z) +
m=p+n
∞
X
ξm fm (z)
m=p+n
∞
X
ξm fm (z)
m=p+n
=
∞
X
ξm fm (z).
m=p+n−1
This completes the assertion of theorem.
Corollary 3.1 The extreme points of (k, n, p, α, λ) − ST are given by
fp+n−1 (z) = z p
and
fm (z) = z p −
pλ (p − α) + pλ k(p − 1) m
z ,
mλ [k(m − 1) + m − α]
103
m = p + n, p + n + 1, ....
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4. Definitions and Applications of the Fractional Calculus
In this section, we shall prove several distortion theorems for functions to
general class (k, n, p, α, λ)−ST . Each of these theorems would involve certain
operators of fractional calculus we find it to be convenient to recall here the
following definition which were used recently by Owa [7] (and more recently,
by Owa and Srivastava [3], and Srivastava and Owa [8] and Srivastava and
Owa [9] ; see also Srivastava et all. [10] )
Definition 4.1. The fractional integral of order λ is defined, for a function f , by
Dz−µ
z
1 Z
f (ζ)
f (z) =
dζ
Γ (µ )
( z − ζ )1−µ
(µ > 0 )
(14)
0
where f is an analytic function in a simply connected region of the z -plane
containing the origin, and the multiplicity of ( z −ζ )µ−1 is removed by requiring
log ( z − ζ ) to be real when z − ζ > 0.
Definition 4.2. The fractional derivative of order µ is defined, for a
function f , by
Dzµ
z
f (ζ)
d Z
1
dζ(0 ≤ µ < 1 )
f (z) =
Γ (1 − µ ) d z
( z − ζ )µ
(15)
0
where f is constrained, and the multiplicity of (z − ζ )−µ is removed, as in
Definition 1.
Definition 4.3. Under the hypotheses of Definition 2, the fractional
derivative of order ( n + µ ) is defined by
dn µ
D f (z)
(0 ≤ µ < 1 )
f (z) =
d zn z
S
where 0 ≤ µ < 1 and n ∈ IN0 = IN { 0 }.
From Definition 4.2, we have
Dzn+µ
Dz0 f (z) = f (z)
which, in view of Definition 4.3 yields,
104
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dn 0
D f (z) = f n (z).
d zn z
Thus, it follows from (17) and (18) that
Dzn+0 f (z) =
(18)
lim Dz−µ f (z) = f (z)
µ→0
and
lim Dz1−µ f (z) = f ′ (z).
µ→0
Theorem 4.1. Let the function f defined by
f (z) = z p −
∞
X
am z m
;
am ≥ 0
m=p+n
be in the class (k, n, p, α, λ) − ST . Then
−µ
Dz (D i f (z)) ≥ |z|p+µ ·
pλ (p − α) + pλ k(p − 1)
Γ(p + n + 1)
Γ(p + 1)
·
−
|z|n
Γ(p + µ + 1) Γ(p + n + µ + 1) (p + n)λ−i [(1 + k)(p + n) − k − α]
(19)
and
)
(
−µ
Dz (D i f (z)) ≤ |z|p+µ ·
Γ(p + 1)
pλ (p − α) + pλ k(p − 1)
Γ(p + n + 1)
·
+
|z|n
Γ(p + µ + 1) Γ(p + n + µ + 1) (p + n)λ−i [(1 + k)(p + n) − k − α]
(20)
for µ > 0, 0 ≤ i ≤ λ and z ∈ U. The equalities in (19) and (20) are attained
for the function f (z) given by
(
)
f (z) = z p −
pλ (p − α) + pλ k(p − 1)
z p+n .
[(1 + k)(p + n) − k − α](p + n)λ
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Proof. We note that
∞
X
Γ(p + µ + 1) −µ −µ i
Γ(m + 1)Γ(p + µ + 1) i
z Dz (D f (z)) = z p −
m am z m .
Γ(p + 1)
Γ(p
+
1)Γ(m
+
µ
+
1)
m=p+n
Defining the function ϕ(m)
ϕ(m) =
Γ(m + 1)Γ(p + µ + 1)
; µ > 0.
Γ(p + 1)Γ(m + µ + 1)
We can see that ϕ(m) is decreasing in m, that is
0 < ϕ(m) ≤ ϕ(p + n) =
Γ(p + n + 1)Γ(p + µ + 1)
.
Γ(p + 1)Γ(p + n + µ + 1)
On other hand, from [11]
∞
X
m i am ≤
m=p+n
pλ (p − α) + pλ k(p − 1)
(p + n)−(λ−i) ;
[(1 + k)(p + n) − k − α]
0 ≤ i ≤ λ.
Therefore,
Γ(p + µ + 1)
z −µ Dz−µ (Di f (z)) ≥ |z|p − ϕ(p + n)|z|p+n
Γ(p + 1)
≥ |z|p −
∞
X
m i am
m=p+n
Γ(p + n + 1)Γ(p + µ + 1) pλ (p − α) + pλ k(p − 1)
(p + n)−(λ−i) |z|p+n
Γ(p + 1)Γ(p + n + µ + 1) mλ [k(m − 1) + m − α]
and
Γ(p + µ + 1)
−µ −µ
i
≤ |z|p + ϕ(p + n)|z|p+n
z
D
(D
f
(z))
z
Γ(p + 1)
≤ |z|p +
∞
X
m i am
m=p+n
Γ(p + n + 1)Γ(p + µ + 1) pλ (p − α) + pλ k(p − 1)
(p + n)−(λ−i) |z|p+n .
Γ(p + 1)Γ(p + n + µ + 1) mλ [k(m − 1) + m − α]
which completes the proof of theorem.
Next, we prove
106
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Theorem 4.2. Let the function f defined by
f (z) = z p −
∞
X
am z m ;
am ≥ 0
m=p+n
be in the class (k, n, p, α, λ) − ST . Then
µ
Dz (D i f (z)) ≥ |z|p−µ ·
pλ (p − α) + pλ k(p − 1)
Γ(p + n)
Γ(p + 1)
−
|z|n
·
λ−i−1
Γ(p − µ + 1) Γ(p + n − µ + 1) (p + n)
[(1 + k)(p + n) − k − α]
(22)
and
)
(
µ
Dz (D i f (z)) ≤ |z|p−µ ·
pλ (p − α) + pλ k(p − 1)
Γ(p + n)
Γ(p + 1)
+
|z|n
·
λ−i−1
Γ(p − µ + 1) Γ(p + n − µ + 1) (p + n)
[(1 + k)(p + n) − k − α]
(23)
for 0 ≤ µ < 1, 0 ≤ i ≤ λ − 1 and z ∈ U. The equalities (22) and (23) are
attained for the function f (z) given by (21).
)
(
Proof. Using similar argument as given by Theorem 4.1, we can get result.
Remark. Letting µ → 0 and i = 0 in Theorem 4.1 and taking µ → 1
and i = 0 in Theorem 4.2, we obtain the inequalities (5) and (6) in Theorem
2.2, respectively.
References
[1] A.Lecko and A.Wisniowska, Geometric properties of subclasses of starlike functions, J.Comp. and Appl. Math. 155(2003),383-387.
[2] S.Kanas and A.Wisniowska, Conic regions and k-uniform convexity,
J.Comp. and Appl. Math. 105(1999),327-336.
107
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H. Ozlem
G¨
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[3] H.Silverman, Univalent functions with negative coefficients, Proc.Amer.
Math. Soc. 51(1975),109-116.
¨
[4] H.Ozlem
Guney and S.S.Eker , On a subclass of certain k-starlike functions with negative coefficients, Bull.Brazillian Math. Soc. 37(1)(2006), 19-28.
[5] R.Baharati,R.Parvathom and A.Swaminathan, On a subclasses of uniformly convex functions and corresponding class of starlike functions, tamkang
J. Math 28(1997),17-32.
¨
[6] H.Ozlem
Guney, S.S.Eker and S.Owa, Fractional calculus and some
properties of k-uniform convex functions with negative coefficients, Taiwanese
J.Math Vol.10, No.6 (2006), 1671-1683.
[7] S.Owa, On the distortion theorems, Kyunpook Math J. 18 (1978 ), 53-59.
[8] S.Owa and H.M.Srivastava, Univalent and Starlike generalized hypergeometric functions, Canad. J. Math. 39 (1987 ),1057-1077.
[9] H.M.Srivastava and S.Owa, Some Characterization and distortion theorems involving fractional calculus, linear operators and certain subclasses of
analytic functions, Nagoya Math.J. 106 (1987),1-28.
[10] H.M.Srivastava, M.Saigo and S.Owa, A class of distortion theorems
involving certain operators of fractional calculus, J.Math.Anal.Appl. 131(1988
), 412-420.
[11] S.Owa and S.K.Lee, Certain generalized class of analytic functions with
negative coefficients, to appear.
[12] G.S.S˘al˘agean, Subclasses of univalent function, Complex Analysis-Fifth
Romanian Finish Seminar, Bucharest, 1(1981 ), 362-372.
Author:
¨
H. Ozlem
G¨
uney
Department of Mathematics
Faculty of Science and Art
University of Dicle
21280 Diyarbakır
Turkey
e-mail:[email protected]
108
No 12/2006
FRACTIONAL CALCULUS AND SOME PROPERTIES OF
SALAGEAN-TYPE K-STARLIKE FUNCTIONS
WITH NEGATIVE COEFFICIENTS
¨
¨
H. Ozlem
GUNEY
Abstract.We define and investigate a new class of Salagean-type k-starlike
functions with negative coefficients. We obtain some properties of this subclass. Furthermore, we give Hadamard product of several functions and some
distortion theorems for fractional calculus of this generalized class.
Keywords. analytic function, univalent function, k-starlike, S˘al˘agean operator, Hadamard product.
2000 Mathematics Subject Classification : 30C45.
1. Introduction
Let A denote the class of functions of the form
f (z) = z +
∞
X
ak z k
k=2
which are analytic in the open unit disk U = {z ∈ C : |z| < 1} . For f (z)
belongs to A, Salagean [12] has introduced the following operator called the
Salagean operator :
D0 f (z) = f (z)
D1 f (z) = Df (z) = zf ′ (z)
..
.
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Dn f (z) = D(Dn−1 f (z)) , n ∈ N0 = {0} ∪ {1, 2, · · ·}.
Let Sp (p ≥ 1) denote the class of functions of the form f(z) = z p +
p+n
that are holomorphic and p-valent in the unit disk U .
n=1 ap+n z
Also, let Ap (n) denote the subclass of Sp (p ≥ 1) consisting of functions
that can be expressed in the form
P∞
f (z) = z p −
∞
X
am z m
;
am ≥ 0.
(1)
m=p+n
We can write the following equalities for the functions f (z) belonging to the
class Ap (n):
D0 f (z) = f (z)
∞
X
D1 f (z) = Df (z) = zf ′ (z) = pz p −
mam z m
m=p+n
..
.
Dλ f (z) = D(Dλ−1 f (z)) = pλ z p −
∞
X
mλ am z m , λ ∈ N0 = {0} ∪ N.
m=p+n
Denote by S the class of functions of the form f (z) = z + a2 z 2 + ..., analytic
and univalent, by ST (α) subclass consisting of starlike and univalent of order
α and by k − ST (0 ≤ k < ∞) a class of k−starlike univalent functions in U ,
introduced and investigated Lecko and Wisniowska in [1].
It is known that every f ∈ k − ST has a continuous extension to U , f (U )
is bounded and f (∂U ) is a rectifiable curve [2].
In the present paper, a subclass (k, n, p, α, λ) − ST of starlike functions in
the open unit disk U is introduced. A functions f (z) ∈ Ap (n) is said to be in
the class (k, n, p, α, λ) − ST if it is satisfies
ζ (z − ζ)(Dλ f (z))′
Re
+
z
Dλ f (z)
(
94
)
≥α
(2)
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for some α (0 ≤ α < p) and z ∈ U .
We note that
i)(0, 1, 1, α, 0) − ST ≡ T ∗ (α) was introduced by Silverman [3],
ii)(k, 1, 1, 0, 0) − ST ≡ k − ST was introduced by Lecko and Wisniowska
[1],
iii)(k, 1, 1, α, 0) − ST ≡ k − ST (α) was studied by Baharati et.al. [5],
iv)(k, n, 1, α, 0) − ST ≡ (k, n, α) − ST was studied by Guney et.al [4],
v)(k, 1, 1, α, 1) − ST ≡ (k, 1, −1, α) − UCV ≡ k − UCV(α) was investigated
by Baharati et.al [5]. Also, the class (k, A, B, α) − U CV was introduced by
Guney et al.[6].
2.Some results of the class (k, n, p, α, λ) − ST
Theorem 2.1 A function f ∈ Ap (n) is in the class (k, n, p, α, λ) − ST iff
∞
X
mλ [k(m − 1) + m − α]am ≤ pλ (p − α) + pλ k(p − 1).
(3)
m=p+n
Proof. Let (k, n, p, α, λ) − ST . Then we have from (2)
ζ (z − ζ)(Dλ f (z))′
Re
+
z
Dλ f (z)
(
∞
P
λ p−1
λ+1 p
λ+1 p−1
ζp z +p z −ζp z +ζ
(mλ+1 −mλ )am z m−1 −
pλ z p −
P
∞
P
m=p+n
m=p+n
∞
=Re
)
mλ a
m
zm
m=p+n
mλ+1 am z m
If we choose z and ζ real and z → 1− and ζ → −k + , we get,
−kpλ + pλ+1 + kpλ+1 + k
∞
P
(mλ − mλ+1 )am −
m=p+n
∞
P
pλ −
m λ am
m=p+n
or
∞
X
∞
P
m=p+n
mλ+1 am
mλ [k(m − 1) + m − α]am ≤ pλ (p − α) + pλ k(p − 1)
m=p+n
95
≥ α.
≥α
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which is equivalent to (3).
Conversely, assume that
(3) is true.
Then
n
ζ
z
+
(z−ζ)(Dλ f (z))′
Dλ f (z)
o
(z−ζ) pλ+1 z p−1 −
= zζ +
pλ z p −
m=p+n
∞
P
mλ am z m
∞
P
(mλ+1 −mλ )am z m−1 −
m=p+n
m=p+n
∞
pλ z p −
mλ+1 am z m−1
P
mλ am z m
m=p+n
∞
P
λ p−1
λ+1 p
λ+1 p−1
ζp z +p z −ζp z +ζ
=
∞
P
mλ+1 am z m
≥ α.
m=p+n
for |z| < 1. If we choose z → 1− and ζ → −k + through real values, we obtain
λ
(
ζ (z − ζ)(D f (z))
Re
+
z
Dλ f (z)
′
)
pλ k(p − 1) + pλ+1 −
=
pλ −
∞
P
m=p+n
∞
P
m=p+n
mλ [k(m − 1) + m]am
.
m λ am
(4)
If (3) is rewritten as
∞
X
mλ [k(m − 1) + m]am ≤ pλ+1 − pλ α + pλ k(p − 1) + α
∞
X
m λ am ,
m=p+n
m=p+n
and (4) is used, then we obtain
λ
ζ (z − ζ)(Dλ f (z))′
+
Re
z
Dλ f (z)
(
)
α p −
≥
pλ
−
∞
P
λ
m=p+n
∞
P
m=p+n
m am
mλ a
!
= α.
m
Thus f ∈ (k, n, p, α, λ) − ST .
Theorem 2.2 If f ∈ (k, n, p, α, λ) − ST , then for |z| = r < 1 we obtain
pλ (p − α) + pλ k(p − 1)
rp+n ≤ |f (z)| ≤
r −
λ
(p + n) [(1 + k)(p + n) − k − α]
p
≤ rp +
pλ (p − α) + pλ k(p − 1)
rp+n
(p + n)λ [(1 + k)(p + n) − k − α]
and
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pr
p−1
pλ (p − α) + pλ k(p − 1)
−
rp+n−1 ≤ |f ′ (z)| ≤
λ−1
(p + n) [(1 + k)(p + n) − k − α]
≤ prp−1 +
pλ (p − α) + pλ k(p − 1)
rp+n−1 .
(p + n)λ−1 [(1 + k)(p + n) − k − α]
(6)
All inequalities are sharp.
Proof From (3), we have
pλ (p − α) + pλ k(p − 1)
am ≤
(p + n)λ [(1 + k)(p + n) − k − α]
m=p+n
∞
X
(7)
and
∞
X
mam ≤
m=p+n
Thus
|f (z)| ≤ rp +
pλ (p − α) + pλ k(p − 1)
.
(p + n)λ−1 [(1 + k)(p + n) − k − α]
∞
X
am rm ≤ rp + rp+n
m=p+n
≤ rp +
∞
X
(8)
am ≤
m=p+n
pλ (p − α) + pλ k(p − 1)
rp+n
(p + n)λ [(1 + k)(p + n) − k − α]
and
|f (z)| ≥ rp −
∞
X
am rm ≥ rp − rp+n
m=p+n
≥ rp −
∞
X
am ≥
m=p+n
pλ (p − α) + pλ k(p − 1)
rp+n .
(p + n)λ [(1 + k)(p + n) − k − α]
which prove that the assertion (5) of Theorem 2.2.
Furthermore, for |z| = r < 1 and (6), we have
|f ′ (z)| ≤ prp−1 +
∞
X
mam rm−1 ≤ prp−1 + rp+n−1
m=p+n
∞
X
m=p+n
97
mam
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≤ prp−1 +
pλ (p − α) + pλ k(p − 1)
rp+n−1
(p + n)λ−1 [(1 + k)(p + n) − k − α]
and
′
|f (z)| ≥ pr
p−1
−
∞
X
mam r
m−1
≥ pr
p−1
−r
p+n−1
m=p+n
≥ prp−1 −
∞
X
mam
m=p+n
pλ (p − α) + pλ k(p − 1)
rp+n−1
(p + n)λ−1 [(1 + k)(p + n) − k − α]
which prove that the assertion (6) of Theorem 2.2.
The bounds in (5) and (6) are attained for the function f given by
f (z) = z p −
pλ (p − α) + pλ k(p − 1)
z p+n
(p + n)λ [(1 + k)(p + n) − k − α]
Theorem 2.3 Let the functions
∞
X
f (z) = z p −
am z m ; am ≥ 0.
m=p+n
and
p
g (z) = z −
∞
X
bm z m ; bm ≥ 0.
m=p+n
be in the class (k, n, p, α, λ) − ST . Then for 0 ≤ ρ ≤ 1,
h (z) = (1 − ρ)f (z) + ρ g(z) = z p −
∞
X
cm z m ; cm ≥ 0.
m=p+n
is in the class (k, n, p, α, λ) − ST .
Proof. Assume that f, g ∈ (k, n, p, α, λ) − ST .
Then we have from Theorem 2.1
∞
X
mλ [k(m − 1) + m − α]am ≤ pλ (p − α) + pλ k(p − 1)
m=p+n
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and
∞
X
mλ [k(m − 1) + m − α]bm ≤ pλ (p − α) + pλ k(p − 1).
m=p+n
Therefore we can see that
∞
X
mλ [k(m − 1) + m − α]cm =
m=p+n
∞
X
=
mλ [k(m − 1) + m − α][(1 − ρ)am + ρbm ] ≤ pλ (p − α) + pλ k(p − 1)
m=p+n
which completes the proof of Theorem 2.3.
Definition 2.1 The Modified Hadamard Product f ∗ g of two functions
f (z) = z p −
∞
X
a m z m ; am ≥ 0
m=p+n
and
g (z) = z p −
∞
X
bm z m ; bm ≥ 0
m=p+n
are denoted by
(f ∗ g) (z) = z p −
∞
X
am b m z m .
m=p+n
Theorem 2.4 If f, g ∈ (k, n, p, α, λ)−ST then f ∗g ∈ (k, n, p, β, λ)−ST ,
where
β = β(k, n, p, α, λ) =
=
(pλ+1 + pλ k(p − 1))(p + n)λ [(1 + k)(p + n) − k − α]2
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
−
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)(p + n) − k]
.
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
99
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The result is sharp for f (z) and g(z) given by
f (z) = g(z) = z p −
pλ (p − α) + pλ k(p − 1)
z p+n
(p + n)λ [(1 + k)(p + n) − k − α]
where 0 ≤ α < p and 0 ≤ k < ∞.
Proof. From Theorem 2.1, we have
mλ [k(m − 1) + m − α]
am ≤ 1
λ
λ
m=p+n p (p − α) + p k(p − 1)
(10)
mλ [k(m − 1) + m − α]
bm ≤ 1.
λ
λ
m=p+n p (p − α) + p k(p − 1)
(11)
∞
X
and
∞
X
We have to find the largest β such that
mλ [k(m − 1) + m − β]
am bm ≤ 1.
λ
λ
m=p+n p (p − β) + p k(p − 1)
∞
X
(12)
From (10) and (11) we obtain, by means of Cauchy-Schwarz inequality,that
mλ [k(m − 1) + m − α] q
am bm ≤ 1.
λ
λ
m=p+n p (p − α) + p k(p − 1)
∞
X
Therefore, (12) is true if
mλ [k(m − 1) + m − β]
mλ [k(m − 1) + m − α] q
a
b
≤
am b m
m m
pλ (p − β) + pλ k(p − 1)
pλ (p − α) + pλ k(p − 1)
or
[pλ (p − β) + pλ k(p − 1)][k(m − 1) + m − α]
.
am b m ≤ λ
[p (p − α) + pλ k(p − 1)][k(m − 1) + m − β]
q
Note that from (13)
q
am b m ≤
[pλ (p − α) + pλ k(p − 1)]
.
mλ [k(m − 1) + m − α]
Thus if
100
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[pλ (p − α) + pλ k(p − 1)]
[pλ (p − β) + pλ k(p − 1)][k(m − 1) + m − α]
≤
mλ [k(m − 1) + m − α]
[pλ (p − α) + pλ k(p − 1)][k(m − 1) + m − β]
or, equivalently, if
(pλ+1 + pλ k(p − 1))mλ [(1 + k)m − k − α]2
β≤ λ λ
p m [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
−
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)m − k]
pλ mλ [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
then (12) satisfied. Defining the function Φ(m) by
Φ(m) =
−
(pλ+1 + pλ k(p − 1))mλ [(1 + k)m − k − α]2
pλ mλ [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)m − k]
.
pλ mλ [(1 + k)m − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
We can see that Φ(m) is increasing function of m. Therefore,
β ≤ Φ(p+n) =
−
(pλ+1 + pλ k(p − 1))(p + n)λ [(1 + k)(p + n) − k − α]2
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
(pλ (p − α) + pλ k(p − 1))2 [(1 + k)(p + n) − k]
.
pλ (p + n)λ [(1 + k)(p + n) − k − α]2 − (pλ (p − α) + pλ k(p − 1))2
which completes the assertion of theorem.
3. Extreme Points for (k, n, p, α, λ) − ST
λ
λ
k(p−1) m
z , m=
Theorem 3.1 Let fp+n−1 (z) = z p and fm (z) = z p − pmλ(p−α)+p
[k(m−1)+m−α]
p + n, p + n + 1, .... Then If f ∈ (k, n, p, α, λ) − ST iff it can be expressed in
the form
101
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f (z) =
∞
X
ξm fm (z),
∞
X
ξm fm (z).
m=p+n+1
where ξm ≥ 0 and
∞
P
m=p+n−1
ξm = 1.
Proof. Assume that
f (z) =
m=p+n−1
Then
f (z) = ξp+n−1 fp+n−1 (z) +
∞
X
ξm fm (z) =
m=p+n
p
= ξp+n−1 z +
∞
X
ξm
m=p+n
=
∞
X
m=p+n−1
ξm z p −
"
pλ (p − α) + pλ k(p − 1) m
z
z − λ
m [k(m − 1) + m − α]
p
∞
X
ξm
m=p+n
#
pλ (p − α) + pλ k(p − 1) m
z
mλ [k(m − 1) + m − α]
pλ (p − α) + pλ k(p − 1) m
z .
ξm λ
=z −
m [k(m − 1) + m − α]
m=p+n
p
∞
X
Thus
∞
X
m=p+n
ξm
pλ (p − α) + pλ k(p − 1)
mλ [k(m − 1) + m − α]
=
∞
X
!
mλ [k(m − 1) + m − α]
pλ (p − α) + pλ k(p − 1)
ξm − ξp+n−1 = 1 − ξm ≤ 1.
m=p+n−1
Hence f ∈ (k, n, p, α, λ) − ST .
102
!
=
∞
X
m=p+n
ξm
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Conversely, suppose that f ∈ (k, n, p, α, λ) − ST . Since
|am | ≤
pλ (p − α) + pλ k(p − 1)
mλ [k(m − 1) + m − α]
m = p + n, p + n + 1, ...,
we can set
ξm =
mλ [k(m − 1) + m − α]
am ,
pλ (p − α) + pλ k(p − 1)
and
ξp+n−1 = 1 −
m = p + n, p + n + 1, ...
∞
X
ξm .
m=p+n
Then
f (z) = z p −
∞
X
am z m = z p −
m=p+n
= zp −
∞
X
m=p+n
pλ (p − α) + pλ k(p − 1)
ξ zm
λ [k(m − 1) + m − α] m
m
m=p+n
∞
X
ξm (z p − fm (z)) = 1 −
= ξn+p−1 z p +
∞
X
∞
X
m=p+n
ξm z p +
ξm fm (z) = ξn+p−1 fn+p−1 (z) +
m=p+n
∞
X
ξm fm (z)
m=p+n
∞
X
ξm fm (z)
m=p+n
=
∞
X
ξm fm (z).
m=p+n−1
This completes the assertion of theorem.
Corollary 3.1 The extreme points of (k, n, p, α, λ) − ST are given by
fp+n−1 (z) = z p
and
fm (z) = z p −
pλ (p − α) + pλ k(p − 1) m
z ,
mλ [k(m − 1) + m − α]
103
m = p + n, p + n + 1, ....
¨
H. Ozlem
G¨
uney - Fractional calculus and some properties of ...
4. Definitions and Applications of the Fractional Calculus
In this section, we shall prove several distortion theorems for functions to
general class (k, n, p, α, λ)−ST . Each of these theorems would involve certain
operators of fractional calculus we find it to be convenient to recall here the
following definition which were used recently by Owa [7] (and more recently,
by Owa and Srivastava [3], and Srivastava and Owa [8] and Srivastava and
Owa [9] ; see also Srivastava et all. [10] )
Definition 4.1. The fractional integral of order λ is defined, for a function f , by
Dz−µ
z
1 Z
f (ζ)
f (z) =
dζ
Γ (µ )
( z − ζ )1−µ
(µ > 0 )
(14)
0
where f is an analytic function in a simply connected region of the z -plane
containing the origin, and the multiplicity of ( z −ζ )µ−1 is removed by requiring
log ( z − ζ ) to be real when z − ζ > 0.
Definition 4.2. The fractional derivative of order µ is defined, for a
function f , by
Dzµ
z
f (ζ)
d Z
1
dζ(0 ≤ µ < 1 )
f (z) =
Γ (1 − µ ) d z
( z − ζ )µ
(15)
0
where f is constrained, and the multiplicity of (z − ζ )−µ is removed, as in
Definition 1.
Definition 4.3. Under the hypotheses of Definition 2, the fractional
derivative of order ( n + µ ) is defined by
dn µ
D f (z)
(0 ≤ µ < 1 )
f (z) =
d zn z
S
where 0 ≤ µ < 1 and n ∈ IN0 = IN { 0 }.
From Definition 4.2, we have
Dzn+µ
Dz0 f (z) = f (z)
which, in view of Definition 4.3 yields,
104
(16)
(17)
¨
H. Ozlem
G¨
uney - Fractional calculus and some properties of ...
dn 0
D f (z) = f n (z).
d zn z
Thus, it follows from (17) and (18) that
Dzn+0 f (z) =
(18)
lim Dz−µ f (z) = f (z)
µ→0
and
lim Dz1−µ f (z) = f ′ (z).
µ→0
Theorem 4.1. Let the function f defined by
f (z) = z p −
∞
X
am z m
;
am ≥ 0
m=p+n
be in the class (k, n, p, α, λ) − ST . Then
−µ
Dz (D i f (z)) ≥ |z|p+µ ·
pλ (p − α) + pλ k(p − 1)
Γ(p + n + 1)
Γ(p + 1)
·
−
|z|n
Γ(p + µ + 1) Γ(p + n + µ + 1) (p + n)λ−i [(1 + k)(p + n) − k − α]
(19)
and
)
(
−µ
Dz (D i f (z)) ≤ |z|p+µ ·
Γ(p + 1)
pλ (p − α) + pλ k(p − 1)
Γ(p + n + 1)
·
+
|z|n
Γ(p + µ + 1) Γ(p + n + µ + 1) (p + n)λ−i [(1 + k)(p + n) − k − α]
(20)
for µ > 0, 0 ≤ i ≤ λ and z ∈ U. The equalities in (19) and (20) are attained
for the function f (z) given by
(
)
f (z) = z p −
pλ (p − α) + pλ k(p − 1)
z p+n .
[(1 + k)(p + n) − k − α](p + n)λ
105
(21)
¨
H. Ozlem
G¨
uney - Fractional calculus and some properties of ...
Proof. We note that
∞
X
Γ(p + µ + 1) −µ −µ i
Γ(m + 1)Γ(p + µ + 1) i
z Dz (D f (z)) = z p −
m am z m .
Γ(p + 1)
Γ(p
+
1)Γ(m
+
µ
+
1)
m=p+n
Defining the function ϕ(m)
ϕ(m) =
Γ(m + 1)Γ(p + µ + 1)
; µ > 0.
Γ(p + 1)Γ(m + µ + 1)
We can see that ϕ(m) is decreasing in m, that is
0 < ϕ(m) ≤ ϕ(p + n) =
Γ(p + n + 1)Γ(p + µ + 1)
.
Γ(p + 1)Γ(p + n + µ + 1)
On other hand, from [11]
∞
X
m i am ≤
m=p+n
pλ (p − α) + pλ k(p − 1)
(p + n)−(λ−i) ;
[(1 + k)(p + n) − k − α]
0 ≤ i ≤ λ.
Therefore,
Γ(p + µ + 1)
z −µ Dz−µ (Di f (z)) ≥ |z|p − ϕ(p + n)|z|p+n
Γ(p + 1)
≥ |z|p −
∞
X
m i am
m=p+n
Γ(p + n + 1)Γ(p + µ + 1) pλ (p − α) + pλ k(p − 1)
(p + n)−(λ−i) |z|p+n
Γ(p + 1)Γ(p + n + µ + 1) mλ [k(m − 1) + m − α]
and
Γ(p + µ + 1)
−µ −µ
i
≤ |z|p + ϕ(p + n)|z|p+n
z
D
(D
f
(z))
z
Γ(p + 1)
≤ |z|p +
∞
X
m i am
m=p+n
Γ(p + n + 1)Γ(p + µ + 1) pλ (p − α) + pλ k(p − 1)
(p + n)−(λ−i) |z|p+n .
Γ(p + 1)Γ(p + n + µ + 1) mλ [k(m − 1) + m − α]
which completes the proof of theorem.
Next, we prove
106
¨
H. Ozlem
G¨
uney - Fractional calculus and some properties of ...
Theorem 4.2. Let the function f defined by
f (z) = z p −
∞
X
am z m ;
am ≥ 0
m=p+n
be in the class (k, n, p, α, λ) − ST . Then
µ
Dz (D i f (z)) ≥ |z|p−µ ·
pλ (p − α) + pλ k(p − 1)
Γ(p + n)
Γ(p + 1)
−
|z|n
·
λ−i−1
Γ(p − µ + 1) Γ(p + n − µ + 1) (p + n)
[(1 + k)(p + n) − k − α]
(22)
and
)
(
µ
Dz (D i f (z)) ≤ |z|p−µ ·
pλ (p − α) + pλ k(p − 1)
Γ(p + n)
Γ(p + 1)
+
|z|n
·
λ−i−1
Γ(p − µ + 1) Γ(p + n − µ + 1) (p + n)
[(1 + k)(p + n) − k − α]
(23)
for 0 ≤ µ < 1, 0 ≤ i ≤ λ − 1 and z ∈ U. The equalities (22) and (23) are
attained for the function f (z) given by (21).
)
(
Proof. Using similar argument as given by Theorem 4.1, we can get result.
Remark. Letting µ → 0 and i = 0 in Theorem 4.1 and taking µ → 1
and i = 0 in Theorem 4.2, we obtain the inequalities (5) and (6) in Theorem
2.2, respectively.
References
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J.Comp. and Appl. Math. 105(1999),327-336.
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¨
H. Ozlem
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uney - Fractional calculus and some properties of ...
[3] H.Silverman, Univalent functions with negative coefficients, Proc.Amer.
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¨
[4] H.Ozlem
Guney and S.S.Eker , On a subclass of certain k-starlike functions with negative coefficients, Bull.Brazillian Math. Soc. 37(1)(2006), 19-28.
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[6] H.Ozlem
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[8] S.Owa and H.M.Srivastava, Univalent and Starlike generalized hypergeometric functions, Canad. J. Math. 39 (1987 ),1057-1077.
[9] H.M.Srivastava and S.Owa, Some Characterization and distortion theorems involving fractional calculus, linear operators and certain subclasses of
analytic functions, Nagoya Math.J. 106 (1987),1-28.
[10] H.M.Srivastava, M.Saigo and S.Owa, A class of distortion theorems
involving certain operators of fractional calculus, J.Math.Anal.Appl. 131(1988
), 412-420.
[11] S.Owa and S.K.Lee, Certain generalized class of analytic functions with
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Author:
¨
H. Ozlem
G¨
uney
Department of Mathematics
Faculty of Science and Art
University of Dicle
21280 Diyarbakır
Turkey
e-mail:[email protected]
108