SUBCLASS OF MEROMORPHIC FUNCTIONS WITH POSITIVE COEFFICIENTS DEFINED BY RUSCHEWEYH DERIVATIVE II

advertisement
Surveys in Mathematics and its Applications
ISSN 1842-6298 (electronic), 1843 - 7265 (print)
Volume 3 (2008), 67 – 77
SUBCLASS OF MEROMORPHIC FUNCTIONS
WITH POSITIVE COEFFICIENTS DEFINED BY
RUSCHEWEYH DERIVATIVE II
Waggas Galib Atshan
Abstract. New class Σλ (α, β, γ) of univalent meromorphic functions defined by Ruscheweyh
derivative in the punctured unit disk U ∗ is introduced. We study several Hadamard product properties. Some results connected to inclusion relations, neighborhoods of the elements of class Σλ (α, β, γ)
and integral operators are obtained.
1
Introduction
Let Σ denote the class of meromorphic functions in the punctured unit disk
U ∗ = {z ∈ C : 0 < |z| < 1} of the form
f (z) = z −1 +
∞
X
an z n ,
(an ≥ 0, n ∈ IN = {1, 2, 3, · · · }).
(1)
n=1
A function f ∈ Σ is meromorphic starlike function of order β (0 ≤ β < 1) if
0 zf (z)
− Re
> β, (z ∈ U = U ∗ ∪ {0}).
(2)
f (z)
The class of all such functions is denoted by Σ∗ (β). A function f ∈ Σ is meromorphic convex function of order β (0 ≤ β < 1) if
zf 00 (z)
− Re 1 + 0
> β, (z ∈ U = U ∗ ∪ {0}).
(3)
f (z)
Definition 1. The Ruscheweyh derivative of order λ is denoted by Dλ f and is
defined as following: If
∞
X
f (z) = z −1 +
an z n ,
n=1
2000 Mathematics Subject Classification: 30C45.
Keywords: Meromorphic functions; Ruscheweyh derivative; Hadamard product; Coefficients
inequalities; Integral operator; Neighborhood.
******************************************************************************
http://www.utgjiu.ro/math/sma
68
W. G. Atshan
then
∞
X
1
−1
D f (z) =
∗
f
(z)
=
z
+
Dn (λ)an z n , λ > −1, z ∈ U ∗ ,
z(1 − z)λ+1
λ
(4)
n=1
where
Dn (λ) =
(λ + 1)(λ + 2) · · · (λ + n + 1)
.
(n + 1)!
(5)
Definition 2. The Hadamard product (or convolution) of two functions f (z) given
by (1) and
∞
X
g(z) = z −1 +
bn z n
(6)
n=1
is defined by
(f ∗ g)(z) = z
−1
+
∞
X
an bn z n .
n=1
Definition 3. Let f (z) ∈ Σ be given by (1). The new class Σλ (α, β, γ) is defined by
z(Dλ f (z))0 + γz 2 (Dλ f (z))00
f ∈ Σ : −Re
(7)
(1 − γ)Dλ f (z) + γz(Dλ f (z))0
z(Dλ f (z))0 + γz 2 (Dλ f (z))00
≥ α + 1 + β, z ∈ U, 0 ≤ β < 1,
λ
λ
0
(1 − γ)D f (z) + γz(D f (z))
1
λ > −1, 0 ≤ γ < , α ≥ 0 .
2
Σλ (α, β, γ) =
2
Main Results
We will need the following lemma whose details can be found in [1].
Lemma 4. The function f (z) defined by (1) is in the class Σλ (α, β, γ) if and only
if
∞
X
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)an ≤ (1 − β)(1 − 2γ)
(8)
n=1
where 0 ≤ β < 1, α ≥ 0, 0 ≤ γ < 12 , λ > −1, Dn (λ) given by (5).
Theorem 5. Let the function f (z) defined by (1) and the function g(z) = z −1 +
∞
P
bn z n (bn ≥ 0, n ∈ IN ) be in the class Σλ (α, β, γ). Then the function k(z) defined
n=1
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
69
Subclass of meromorphic functions with positive coefficients
by k(z) = z −1 +
∞
P
n=1
an bn z n is in the class Σλ (α, σ, γ), where (α ≥ 0, 0 ≤ γ < 21 ,
0 ≤ β < 1, 0 ≤ σ < 1, n ∈ IN and n ≥ 1) and σ is given by
σ =1−
(1 − β)2 (1 − 2γ)(n + 1)(α + 1)
.
(1 − β)2 (1 − 2γ) + (nγ − γ + 1)[n(α + 1) + (α + β)]2
Proof. We must find the largest σ such that
∞
X
(nγ − γ + 1)[n(α + 1) + (α + σ)]Dn (λ)
(1 − σ)(1 − 2γ)
n=1
an bn ≤ 1, Dn (λ) is given by (5).
Since f (z) and g(z) are in Σλ (α, β, γ), then
∞
X
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
(1 − β)(1 − 2γ)
n=1
and
∞
X
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
(1 − β)(1 − 2γ)
n=1
an ≤ 1
bn ≤ 1.
By Cauchy-Schwarz inequality, we get
∞
X
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ) p
(1 − β)(1 − 2γ)
n=1
an bn ≤ 1.
We want only to show that
(nγ − γ + 1)[n(α + 1) + (α + σ)]Dn (λ)
an bn
(1 − σ)(1 − 2γ)
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ) p
≤
an bn .
(1 − β)(1 − 2γ)
This equivalently to
p
(1 − σ)[n(α + 1) + (α + β)]
an bn ≤
.
(1 − β)[n(α + 1) + (α + σ)]
Then
σ ≤1−
(1 − β)2 (1 − 2γ)(n + 1)(α + 1)
.
(1 − β)2 (1 − 2γ) + (nγ − γ + 1)[n(α + 1) + (α + β)]2
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
70
W. G. Atshan
Theorem 6. Let the function f (z) defined by (1) and g(z) given by g(z) = z −1 +
∞
P
bn z n be in the class Σλ (α, β, γ), then the function k(z) defined by k(z) = z −1 +
n=1
∞
P
n=1
(a2n + b2n )z n is in the class Σλ (α, σ, γ), where 0 ≤ γ < 12 , λ > −1, 0 ≤ σ < 1,
0 ≤ β < 1, α ≥ 0 and
σ =1−
8(1 − β)2 (1 − 2γ)(1 + α)
.
(λ + 1)(λ + 2)(2α + β + 1)2 + 4(1 − β)2 (1 − 2γ)
Proof. We must find the largest σ such that
∞
X
(nγ − γ + 1)[n(α + 1) + (α + σ)]Dn (λ)
n=1
(1 − σ)(1 − 2γ)
(a2n + b2n ) ≤ 1.
Since f (z) and g(z) are in Σλ (α, β, γ), we get
2
∞ X
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
an
≤1
(1 − β)(1 − 2γ)
(9)
(10)
n=1
and
∞ X
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
n=1
(1 − β)(1 − 2γ)
2
bn
≤ 1.
(11)
Combining the inequalities (10) and (11), gives
∞
X
1 (nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ) 2 2
(an + b2n ) ≤ 1.
2
(1 − β)(1 − 2γ)
n=1
But, k(z) ∈ Σλ (α, β, γ) if and only if
∞ X
(nγ − γ + 1)[n(α + 1) + (α + σ)]Dn (λ)
(a2n + b2n ) ≤ 1.
(1 − σ)(1 − 2γ)
(12)
n=1
The inequality (12) would obviously imply (9) if
(nγ − γ + 1)[n(1 + α) + (α + σ)]Dn (λ)
(1 − σ)(1 − 2γ)
[(nγ − γ + 1)(n(α + 1) + (α + β))Dn (λ)]2
≤
2[(1 − β)(1 − 2γ)]2
q2
= .
2
Hence
q2
(nγ − γ + 1)[n(α + 1) + (α + σ)]Dn (λ)
≤
(1 − σ)(1 − 2γ)
2
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
71
Subclass of meromorphic functions with positive coefficients
or
(1 − σ)
2(n + 1)(nγ − γ + 1)Dn (λ)
≥
.
(1 + α)
(1 − 2γ)q 2 + 2(nγ − γ + 1)Dn (λ)
The right hand is decreasing function of n and its maximum if n = 1.
Now
(1 − σ)
(13)
(1 + α)
2(n + 1)(nγ − γ + 1)(1 − β)2 (1 − 2γ)Dn (λ)
≥
[(nγ − γ + 1)(n(α + 1) + (α + β))Dn (λ)]2 + 2(nγ − γ + 1)(1 − β)2 (1 − 2γ)Dn (λ)
Simplify (13) we obtain
(1 − σ)
8(1 − β)2 (1 − 2γ)
≥
(1 + α)
(λ + 1)(λ + 2)(2α + β + 1)2 + 4(1 − β)2 (1 − 2γ)
or
σ ≤1−
8(1 − β)2 (1 − 2γ)(1 + α)
.
(λ + 1)(λ + 2)(2α + β + 1)2 + 4(1 − β)2 (1 − 2γ)
This completes the proof of theorem.
We get the inclusive properties of the class Σλ (α, β, γ).
Theorem 7. Let α ≥ 0, 0 ≤ β < 1, 0 ≤ γ < 21 , σ ≥ 0, then
Σλ (α, β, γ) ⊆ Σλ (α, σ, 0)
where
σ ≤1−
(n + 1)(α + 1)(1 − β)(1 − 2γ)
, n ∈ IN, n ≥ 1.
(nγ − γ + 1)[n(1 + α) + (α + β)] + (1 − β)(1 − 2γ)
Proof. Let f (z) ∈ Σλ (α, β, γ), then from Lemma 4, we have
∞
X
(nγ − γ + 1)[n(1 + α) + (α + β)]Dn (λ)
(1 − β)(1 − 2γ)
n=1
an ≤ 1.
(14)
We want to find the value σ such that
∞
X
[n(1 + α) + (α + σ)]Dn (λ)
n=1
(1 − σ)
an ≤ 1.
(15)
The inequality (14) would obviously imply (15) if
[n(1 + α) + (α + σ)]Dn (λ)
(nγ − γ + 1)[n(1 + α) + (α + β)]Dn (λ)
≤
= q.
(1 − σ)
(1 − β)(1 − 2γ)
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
72
W. G. Atshan
Therefore,
[n(1 + α) + (α + σ)]Dn (λ)
≤ q.
(1 − σ)
(16)
(1 − σ)
(n + 1)Dn (λ)
≥
(n ≥ 1, n ∈ IN ).
(1 + α)
(q + Dn (λ))
(17)
Hence
The right hand side of (17) decreases as n increases and so is maximum for n = 1.
So (17) is satisfied provided
(1 − σ)
(n + 1)(1 − β)(1 − 2γ)
≥
= y.
(1 + α)
(nγ − γ + 1)[n(1 + α) + (α + β)] + (1 − β)(1 − 2γ)
Obviously y < 1 and
σ ≤1−
(n + 1)(α + 1)(1 − β)(1 − 2γ)
.
(nγ − γ + 1)[n(1 + α) + (α + β)] + (1 − β)(1 − 2γ)
This completes the proof of the theorem.
3
(n, δ)-Neighborhoods on Σσλ (α, β, γ)
The next, we determine the inclusion relation involving (n, δ) - neighborhoods. Following the earlier works on neighborhoods of analytic functions by Goodman [2] and
Ruscheweyh [4], but for meromorphic function studied by Liu and Srivastava [3], we
define the (n, δ)-neighborhood of a function f (z) ∈ Σ by
(
Nn,δ (f ) =
g ∈ Σ : g(z) = z −1 +
∞
X
bn z n and
n=1
∞
X
)
n|an − bn | ≤ δ
.
(18)
n=1
For the identity function e(z) = z, we have
(
Nn,δ (e) =
g ∈ Σ : g(z) = z
−1
+
∞
X
n=1
n
bn z and
∞
X
)
n|bn | ≤ δ
.
n=1
Definition 8. A function f ∈ Σ is said to be in the class Σσλ (α, β, γ) if there exists
a function g ∈ Σλ (α, β, γ) such that
f (z)
g(z) − 1 < 1 − σ, (z ∈ U, 0 ≤ σ < 1).
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
Subclass of meromorphic functions with positive coefficients
73
Theorem 9. If g ∈ Σλ (α, β, γ) and
σ =1−
δ(2α + β + 1)(λ + 1)(λ + 2)
(2α + β + 1)(λ + 1)(λ + 2) − 2(1 − β)(1 − 2γ)
(19)
then Nn,δ (g) ⊂ Σσλ (α, β, γ).
Proof. Let f ∈ Nn,δ (g), then we get from (18) that
∞
X
n|an − bn | ≤ δ
n=1
which implies the coefficient inequality
∞
X
|an − bn | ≤ δ, (n ∈ IN ).
n=1
Since g ∈ Σλ (α, β, γ), we have from Lemma 4
∞
X
bn ≤
n=1
2(1 − β)(1 − 2γ)
(2α + β + 1)(λ + 1)(λ + 2)
so that
∞
P
f (z)
<
−
1
g(z)
|an − bn |
n=1
1−
∞
P
≤
bn
δ(2α + β + 1)(λ + 1)(λ + 2)
= 1−σ.
(2α + β + 1)(λ + 1)(λ + 2) − 2(1 − β)(1 − 2γ)
n=1
Thus, by Definition 8, f ∈ Σσλ (α, β, γ) for σ given by (19).
4
Integral Operators
Next, we consider integral transforms of functions in the class Σλ (α, β, γ).
Theorem 10. Let the function f (z) given by (1) be in Σλ (α, β, γ). Then the
integral operator
Z 1
F (z) = c
uc f (uz)du, (0 < u ≤ 1, 0 < c < ∞)
(20)
0
is in Σλ (α, δ, γ), where
δ=
(c + 2)(2α + β + 1) − c(1 − β)(2α + 1)
.
(c + 2)(2α + β + 1) + c(1 − β)
The result is sharp for the function
f (z) =
1
2(1 − β)(1 − 2γ)
+
z.
z (2α + β + 1)(λ + 1)(λ + 2)
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
74
W. G. Atshan
∞
P
Proof. Let f (z) = z −1 +
an z n in Σλ (α, β, γ). Then
n=1
Z
1
uc f (uz)du
F (z) = c
0
Z
∞
1
uc−1 X
+
an un+c z n
z
= c
0
!
du
n=1
∞
=
1 X
c
+
an z n .
z
c+n+1
n=1
It is sufficient to show that
∞
X
c(nγ − γ + 1)[n(1 + α) + (α + δ)]Dn (λ)
n=1
(c + n + 1)(1 − δ)(1 − 2γ)
an ≤ 1.
(21)
Since f ∈ Σλ (α, β, γ), we have
∞
X
(nγ − γ + 1)[n(1 + α) + (α + β)]Dn (λ)
n=1
(1 − β)(1 − 2γ)
an ≤ 1.
Note that (21) is satisfied if
(nγ − γ + 1)[n(1 + α) + (α + β)]Dn (λ)
c(nγ − γ + 1)[n(1 + α) + (α + δ)]Dn (λ)
≤
.
(c + n + 1)(1 − δ)(1 − 2γ)
(1 − β)(1 − 2γ)
Rewriting the inequality, we have
c(nγ − γ + 1)[n(1 + α) + (α + δ)]Dn (λ)(1 − β)(1 − 2γ)
≤ (c + n + 1)(1 − δ)(1 − 2γ)(nγ − γ + 1)[n(1 + α) + (α + β)]Dn (λ).
Solving for δ, we have
δ≤
(c + n + 1)[n(1 + α) + (α + β)] − c(1 − β)[n(1 + α) + α]
= F (n).
(c + n + 1)[n(1 + α) + (α + β)] + c(1 − β)
A computation shows that
F (n + 1) − F (n) =
= [((c + n + 2)((n + 1)(1 + α) + (α + β)) − c(1 − β)((n + 1)(1 + α) + α)) ×
((c + n + 1)(n(1 + α) + (α + β)) + c(1 − β)) − ((c + n + 1)(n(1 + α) + (α + β))
−c(1 − β)(n(1 + α) + α))((c + n + 2)((n + 1)(1 + α) + (α + β)) + c(1 − β))]/
[((c + n + 1)(n(1 + α) + (α + β)) + c(1 − β)) ×
((c + n + 2)((n + 1)(1 + α) + (α + β) + c(1 − β))] > 0
for all n. This means that F (n) is increasing and F (n) ≥ F (1). Using this, the
result follows.
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
75
Subclass of meromorphic functions with positive coefficients
Theorem 11. If f ∈ Σλ (α, β, γ), then the integral operator
Z 1
F (z) = c
uc f (uz)du, (0 < u ≤ 1, 0 < c < ∞)
0
is in Σλ (α,
1+βc
2+c , γ).
The result is sharp for
fn (z) =
1
(1 − β)(1 − 2γ)
+
zn.
z (nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
Proof. By Definition of F , we get
Z 1
∞
1 X
c
c
F (z) = c
u f (uz)du = +
an z n .
z
c+n+1
0
n=1
By Lemma 4, it is sufficient to show that
∞
X
c(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
n=1
(1 −
1+βc
2+c )(1
− 2γ)(c + n + 1)
an ≤ 1.
(22)
Since if f ∈ Σλ (α, β, γ), then (22) satisfies if
c
(1 −
1+βc
2+c )(1
− 2γ)(c + n + 1)
1
(1 − β)(1 − 2γ)
≤
or equivalently, when
Y (n, c, β) =
c(1 − β)
(1 −
1+βc
2+c )(c
+ n + 1)
≤1
since Y (n, c, β) is a decreasing function of n (n ≥ 1), then the proof is completed.
The result is sharp for
fn (z) =
(1 − β)(1 − 2γ)
1
+
zn.
z (nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
Theorem 12. Let the function f (z) given by (1) be in Σλ (α, β, γ),
∞
Xc+n+1
1
F (z) = [(c + 1)f (z) + zf 0 (z)] = z −1 +
an z n , c > 0.
c
c
(23)
n=1
Then F (z) is in Σλ (α, β, γ) for |z| ≤ r(α, β, c, δ), where
r(α, β, c, δ) = inf
n
c(1 − δ)[n(α + 1) + (α + β)]
(1 − β)(c + n + 1)[n(α + 1) + (α + δ)]
1
n+1
,
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
76
W. G. Atshan
n = 1, 2, 3, · · · .
The result is sharp for the function
fn (z) =
1
(1 − β)(1 − 2γ)
+
z n , n = 1, 2, 3, · · · .
z (nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
Proof. Let
w=−
z(Dλ f (z))0 + γz 2 (Dλ f (z))00
.
(1 − γ)Dλ f (z) + z(Dλ f (z))0
Then it is sufficient to show that
|w − 1| < |w + 1 − 2δ|.
A computation shows that this is satisfied if
∞
X
(nγ − γ + 1)[n(α + 1) + (α + δ)]Dn (λ)(c + n + 1)
c(1 − δ)(1 − 2γ)
n=1
an |z|n+1 ≤ 1
(24)
since f ∈ Σλ (α, β, γ), then by Lemma 4, we have
∞
X
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)an ≤ (1 − β)(1 − 2γ).
n=1
The equation (24) is satisfied if
≤
(nγ − γ + 1)[n(α + 1) + (α + δ)]Dn (λ)(c + n + 1)
an |z|n+1
c(1 − δ)(1 − 2γ)
(nγ − γ + 1)[n(α + 1) + (α + β)]Dn (λ)
an .
(1 − β)(1 − 2γ)
Solving for |z|, we obtain the result.
Corollary 13. Let the function f (z) given by (1) be in Σλ (0, β, γ) and F (z) given
by (23). Then F (z) is in Σλ (0, β, γ) for |z| < r(0, β, γ, c, δ), where
r(0, β, γ, c, δ) = inf
n
c(1 − δ)(n + β)
(1 − β)(c + n + 1)(n + δ)
1
n+1
, n = 1, 2, 3, · · · .
The result is sharp for the function
fn (z) =
1
(1 − β)(1 − 2γ)
+
z n , n = 1, 2, 3, · · · .
z (nγ − γ + 1)(n + β)Dn (λ)
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
Subclass of meromorphic functions with positive coefficients
77
References
[1] W. G. Atshan and S. R. Kulkarni, Subclass of meromorphic functions with positive coefficients defined by Ruscheweyh derivative (I), J. Rajasthan Acad. Phy.
Sci., 6(2) (2007), 129-140. MR2332617(2008d:30013).
[2] A. W. Goodman, Univalent functions and nonanalytic curve, Proc. Amer. Math.
Soc., 8 (1957), 598-601. MR0086879(19,260a). Zbl 0166.33002.
[3] Jin-Lin Liu and H. M. Srivastava, A linear operator and associated families of
meromorphically multivalent functions, J. Math. Anal. Appl. 259 (2001), 566581. MR1842079(2002i:30014). Zbl 0997.30009.
[4] S. Ruscheweyh, Neighborhoods of univalent functions, Proc. Amer. Math. Soc.,
81 (1981), 521-527. MR0601721(82c:30016). Zbl 0458.30008.
[5] B. A. Uralegaddi and M. G. Ganigi, A certain class of meromorphically starlike
functions with positive coefficients, Pure Appl. Math. Sci., 26 (1987), 75-81.
MR0935517(89h:30027). Zbl 0638.30014.
Waggas Galib Atshan
Department of Mathematics,
College of Computer Science and Mathematics,
Al-Qadisiya University,
Iraq.
e-mail: waggashnd@yahoo.com
******************************************************************************
Surveys in Mathematics and its Applications 3 (2008), 67 – 77
http://www.utgjiu.ro/math/sma
Download