Gen. Math. Notes, Vol. 26, No. 2, February 2015, pp.23-33
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2015
www.i-csrs.org
Available free online at http://www.geman.in
On a New Class of Multivalent Functions
With Missing Coefficients
Xiaoli Liu1 and Liangpeng Xiong2
1,2
The Engineering and Technical College of
ChengDu University of Technology
614000, Leshan, Sichuan, P.R. China
1
E-mail: xiaoliliu2006@126.com
2
E-mail: xlpwxf@163.com
(Received: 19-11-14 / Accepted: 7-1-15)
Abstract
In this paper, we investigate a new class Θp,λ
ξ1 ,ξ2 of analytic functions in
the open unit disk. By using the geometry function theory, we discuss the
radius problems between the Θp,λ
ξ1 ,ξ2 and the convex functions or close-to-convex
functions. Several properties as the sufficient and necessary conditions and
modified-Hadamard product are given.
Keywords: Multivalent function, Convex function, Cauchy-schwarz inequality, Modified-Hadamard product.
1
Introduction
Let Ap be the class of functions of the form
p
f (z) = z +
∞
X
an z n ,
p ∈ Z+ = {1, 2, 3, ...},
(1)
n=p+1
that are p-valently analytic in the open unit disk U = {z ∈ C : |z| < 1}. If
two functions f1 (z) ∈ Ap , f2 (z) ∈ Ap and
fi (z) = z p +
∞
X
n=p+1
an,i z n , i = 1, 2, z ∈ U,
24
Xiaoli Liu et al.
then we define the f1 ⊕ f2 (z) as
f1 ⊕ f2 (z) = z p +
∞
X
(an,1 + an,2 )z n , z ∈ U.
n=p+1
Also, let Kp (α) denote the subclass of Ap consisting of f (z) which satisfy
!
zf 00 (z)
> α, (z ∈ U)
(2)
< 1+ 0
f (z)
for some real α(0 ≤ α < p). A function f (z) ∈ Kp (α) is said to be pvalently convex of order α in U. We note that K1 (α) ≡ K is usual convex class.
Moreover, a function f (z) ∈ Ap is in the class Cp (α) if
!
f 0 (z)
<
> α, z ∈ U
(3)
pz p−1
for some real α(0 ≤ α < 1). C1 (0) ≡ C is the close-to-convex class. These are
many results on the classes Kp (α) and Cp (α)(See [1, 2, 8, 9, 10, 13]).
Let Ap (θ) denote the subclass of Ap consisting of functions f (z) with the
coefficients an = |an |ei((n−p)θ+π) (n ≥ p + 1). Here, we introduce the subclasses
Cp (θ, α) and Kp (θ, α) as follows: Cp (θ, α) = Ap (θ)∩Cp (α), Kp (θ, α) = Ap (θ)∩
Kp (α). In fact, The C1 (θ, α) was introduced by Uyanik, Owa [12] and the
K1 (θ, α) ≡ K(θ, α) was introduced by Frasin [7].
In some earlier investigations, various interesting subclasses of the class Ap
and Ap (θ) have been studied with different view points(see [3, 4]). Motivated
by the aforementioned works done by Uyanik et al.[11, 12] and Frasin et al.[5,
6, 7], we now introduce the following subclass Θp,λ
ξ1 ,ξ2 ,ξ3 of analytic functions:
Definition 1.1 For the functions f (z) ∈ Ap given by (1), we say that
∞
P
p
f (z) ∈ Θp,λ
,
if
there
exists
a
function
g(z)
=
z
+
bn z n ∈ G such
ξ1 ,ξ2
n=p+1
that
!0
!00 f
(z)
⊕
g(z)
f
(z)
⊕
g(z)
2
+
ξ
z
ξ
z
≤ λ, z ∈ U,
1
2
zp
zp
where ξ1 , ξ2 ∈ C, λ > 0, p ∈ Z+ and
n
1
G = g(z) ∈ Ap : bp+1 = 0, bp+2 = − ap+2 ,
2
o
2
1
bp+3 = − ap+3 , ..., bn =
− 1 an , ... .
3
n−p
(4)
(5)
In the present paper, some properties for Θp,λ
ξ1 ,ξ2 are given. We discuss the
radius problems for f (z) belonging to Cp (θ, α) or Kp (θ, α) to be in the class
Θp,λ
ξ1 ,ξ2 , and obtain the modified-Hadamard product results.
25
On a New Class of Multivalent Functions...
2
Sufficient and Necessary Conditions
Theorem 2.1 If the function f (z) given by (1) satisfies the condition
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an | ≤ λ,
(6)
n=p+1
then f (z) ∈ Θp,λ
ξ1 ,ξ2 with a function
∞
X
p
g(z) = z +
bn z n ∈ G,
n=p+1
where ξ1 , ξ2 ∈ C, λ > 0 and p∈ Z+ = {1, 2, 3, ...}.
Proof For f (z) ∈ Ap and g(z) ∈ G, using the (5), then we have
!0
!00 f (z) ⊕ g(z)
f
(z)
⊕
g(z)
2
+ ξ2 z
ξ1 z
p
p
z
z
X
∞
n−p
= [ξ1 (n − p) + ξ2 (n − p)(n − p − 1)](an + bn )z ≤
=
(7)
n=p+1
∞
X
[|ξ1 |(n − p) + |ξ2 |(n − p)(n − p − 1)]|an + bn |
n=p+1
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an |.
n=p+1
It follows from(4), (6) and (7), then f (z) ∈ Θp,λ
ξ1 ,ξ2 . The proof of the theorem
is complete.
Theorem 2.2 If f (z) = z p +
∞
P
n=p+1
p
g(z) = z +
an z n ∈ Θp,λ
ξ1 ,ξ2 with a function
∞
X
bn z n ∈ G,
n=p+1
and arg ξ1 = arg ξ2 = γ and an = |an |ei((n−p)θ)−γ) , then we have
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an | ≤ λ.
n=p+1
26
Xiaoli Liu et al.
i((n−p)θ)−γ)
,
Proof If f (z) ∈ Θp,λ
ξ1 ,ξ2 with arg ξ1 = arg ξ2 = γ and an = |an |e
applying the (5), then we get
!0
!00 f (z) ⊕ g(z)
2 f (z) ⊕ g(z)
+ ξ2 z
ξ1 z
=
p
p
z
z
∞
X
n−p [ξ1 (n − p) + ξ2 (n − p)(n − p − 1)](an + bn )z =
(8)
n=p+1
X
∞
n−p [ξ1 + ξ2 (n − p − 1)]an z =
n=p+1
X
∞
iγ
i((n−p)θ−γ) n−p =
[|ξ1 | + |ξ2 |(n − p − 1)]e |an |e
z n=p+1
∞
X
i(n−p)θ n−p [|ξ1 | + |ξ2 |(n − p − 1)]|an |e
z ≤λ
=
n=p+1
for all z ∈ U. Letting z ∈ U such that z = |z|e−iθ , then we have that
∞
X
i(n−p)θ n−p [|ξ1 | + |ξ2 |(n − p − 1)]|an |e
z n=p+1
∞
X
=
(9)
[|ξ1 | + |ξ2 |(n − p − 1)]|an ||z|n−p
n=p+1
Now, taking |z| → 1− , form (8) and (9), it gives the required result. The proof
of the theorem is complete.
3
Radius Problems with Convex and Close-toConvex Functions
Working in a similar way as in Uyanı̀k, Owa [11, Lemma 3.1] and Frasin [6,
Lemma 4.1], we give the following Lemma 3.1 and Lemma 3.2:
Lemma 3.1 Suppose f (z) = z p +
∞
P
an z n ∈ Cp (θ, α), then we have
n=p+1
∞
X
n=p+1
n|an | ≤ p(1 − α), (0 ≤ α < 1).
27
On a New Class of Multivalent Functions...
∞
P
Lemma 3.2 Suppose f (z) = z p +
an z n ∈ Kp (θ, α), then we have
n=p+1
∞
X
n
(n − α)|an | 6 p − α, (0 ≤ α < p).
p
n=p+1
∞
P
Theorem 3.3 Let f (z) = z p +
an z n ∈ Cp (θ, α) and δ(0 < |δ| < 1)
n=p+1
is a complex number, then δ1p f (δz) ∈ Θp,λ
ξ1 ,ξ2 with a function g(z) ∈ G for
0 < |δ| ≤ |δ0 (λ)|, where |δ0 (λ)| is the smallest positive root of the equation
p
|ξ1 ||δ| p(1 − α)(1 − |δ|2 )
q
p
3
+ |ξ2 | 1 + |δ|2 |δ|2 p(1 − α) − |ap+1 |2 − λ(1 − |δ|2 ) 2 = 0.
Proof If f (z) ∈ Cp (θ, α), then we have that
∞
X
1
p
an δ n−p z n .
f (δz) = z +
δp
n=p+1
Applying Theorem 2.1, we need to show that
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an ||δ|n−p ≤ λ.
n=p+1
By using the Cauchy–Schwarz inequality, we can obtain
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an ||δ|n−p
(10)
n=p+1
|ξ1 |
≤ p
|δ|
|ξ2 |
+ p
|δ|
! 21
∞
X
|δ|2n
n=p+1
∞
X
∞
X
! 21
|an |2
n=p+1
! 21
(n − p − 1)2 |δ|2n
n=p+2
∞
X
! 12
|an |2
n=p+2
In fact, Lemma 3.1 implies that
∞
X
2
|an | ≤
n=p+1
≤
∞
X
n=p+1
∞
X
n=p+1
|an |
n|an | ≤ p(1 − α),
(11)
28
Xiaoli Liu et al.
So we also have
∞
X
|an |2 ≤ p(1 − α) − |an+1 |2 .
(12)
n=p+1
Moreover, putting x = |δ|2 , then we have
∞
X
2n
|δ|
∞
X
=
n=p+1
xn =
n=p+1
xp+1
1−x
(13)
and
∞
X
(n − p − 1)2 |δ|2n
n=p+2
∞
X
=
(14)
1 + x p+2
x .
(1 − x)3
(n − p − 1)2 xn =
n=p+2
Following (10)-(14), we can obtain that
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an ||δ|n−p
(15)
n=p+1
≤
|ξ1 |
|δ|p
|ξ2 |
+ p
|δ|
≤
|ξ1 |
|δ|p
! 12
∞
X
|δ|2n
n=p+1
∞
X
∞
X
! 21
|an |2
n=p+1
! 12
(n − p − 1)2 |δ|2n
n=p+2
xp+1
1−x
∞
X
! 21
|an |2
n=p+2
! 12
! 21
p(1 − α)
! 12
! 12
1 + x p+2
x
p(1 − α) − |ap+1 |2
(1 − x)3
! 12
! 21
|ξ1 | xp+1
≤ p
p(1 − α)
|δ| 1 − x
! 12
! 12
|ξ2 |
1 + x p+2
+ p
x
p(1 − α) − |ap+1 |2
|δ| (1 − x)3
p
p
p
1 + |δ|2 |δ|2 p(1 − α) − |ap+1 |2
|δ| p(1 − α)
= |ξ1 |
+ |ξ2 |
.
1
3
(1 − |δ|2 ) 2
(1 − |δ|2 ) 2
|ξ2 |
+ p
|δ|
29
On a New Class of Multivalent Functions...
We need to consider the complex number δ(0 < |δ| < 1) such that
|ξ1 |
p
|δ| p(1 − α)
1
(1 − |δ|2 ) 2
+ |ξ2 |
p
p
1 + |δ|2 |δ|2 p(1 − α) − |ap+1 |2
3
(1 − |δ|2 ) 2
= λ.
Hence, we definite the following function with |δ(λ)| by
p
F (|δ(λ)|) = |ξ1 ||δ| p(1 − α)(1 − |δ|2 )
q
p
3
+ |ξ2 | 1 + |δ|2 |δ|2 p(1 − α) − |ap+1 |2 − λ(1 − |δ|2 ) 2 .
It is easily to know that F (0) = −λ < 0 and
q
√
F (1) = 2|ξ2 | p(1 − α) − |ap+1 |2 > 0,
which implies that there exists some δ0 (λ) such that F (|δ0 (λ)|) = 0(0 <
|δ0 (λ)| < 1). The proof of the theorem is complete.
Theorem 3.4 Let f (z) = z p +
∞
P
an z n ∈ Kp (θ, α) and δ(0 < |δ| < 1)
n=p+1
is a complex number. Then δ1p f (δz) ∈ Θp,λ
ξ1 ,ξ2 with a function g(z) ∈ G for
0 < |δ| ≤ |δ0 (λ)|, where |δ0 (λ)| is the smallest positive root of the equation
q
p
√
3
2
2
2
|ξ1 ||δ| p − α(1 − |δ| ) + |ξ2 | 1 + |δ| |δ| p − α − |ap+1 |2 − λ(1 − |δ|2 ) 2 = 0.
Proof Since f (z) ∈ Kp (θ, α), using Lemma 3.2, we have that
∞
X
n
(n − α)|an | 6 p − α,
p
n=p+1
which leads to
∞
X
∞
X
n
|an | 6
(n − p)|an | 6
(n − α)|an |2
p
n=p+1
n=p+1
n=p+1
2
∞
X
2
∞
X
n
6
(n − α)|an | 6 p − α.
p
n=p+1
(16)
30
Xiaoli Liu et al.
Hence, from (15), we can also note that
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an ||δ|n−p
(17)
n=p+1
+
∞
X
|δ|2n
n=p+1
∞
X
|ξ2 |
|δ|p
≤
! 21
∞
X
|ξ1 |
≤ p
|δ|
! 21
|an |2
n=p+1
! 21
(n − p − 1)2 |δ|2n
xp+1
1−x
! 21
|an |2
n=p+2
n=p+2
|ξ1 |
|δ|p
∞
X
! 21
! 12
p − α)
! 12
! 21
1 + x p+2
x
p − α − |ap+1 |2
(1 − x)3
p
p
√
1 + |δ|2 |δ|2 p − α − |ap+1 |2
|δ| p − α
= |ξ1 |
1 + |ξ2 |
3
(1 − |δ|2 ) 2
(1 − |δ|2 ) 2
|ξ2 |
+ p
|δ|
Using the same technique as in the proof of Theorem 3.3, we derive the result.
The proof of the theorem is complete.
4
Modified-Hadamard Product
Let f (z) = z p +
∞
P
|an |ei((n−p)θ)−γ) z n , g(z) = z p +
n=p+1
∞
P
|bn |ei((n−p)θ)−γ) z n .
n=p+1
We define modified Hadamard product for the functions f , g as follows:
∞
X
p
(f ∗ g)(z) = z +
|an ||bn |ei((n−p)θ)−γ) z n , z ∈ U.
n=p+1
∞
P
Theorem 4.1 If f1 (z) = z p +
n=p+1
G, f2 (z) = z p +
∞
P
n=p+1
1
|an,1 |ei((n−p)θ)−γ) z n ∈ Θp,λ
ξ1 ,ξ2 with g1 (z) ∈
2
|an,2 |ei((n−p)θ)−γ) z n ∈ Θp,λ
ξ1 ,ξ2 with a function g2 (z) ∈ G and
arg ξ1 = arg ξ2 = γ , then we have
∗
(f1 ∗ f2 )(z) ∈ Θp,λ
ξ1 ,ξ2
with a function g(z) ∈ G, where
λ∗ =
1
λ1 λ2 .
|ξ1 |
31
On a New Class of Multivalent Functions...
∞
P
Proof Suppose f1 (z) = z p +
n=p+1
zp +
∞
P
n=p+1
1
|an,1 |ei((n−p)θ)−γ) z n ∈ Θp,λ
ξ1 ,ξ2 , f2 (z) =
2
|an,2 |ei((n−p)θ)−γ) z n ∈ Θp,λ
ξ1 ,ξ2 and arg ξ1 = arg ξ2 = γ, then from
Theorem 2.2, we have
and
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an,1 |
≤1
λ1
n=p+1
(18)
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an,2 |
≤ 1.
λ
2
n=p+1
(19)
Moreover, (18) and (19) imply that
and
∞
n X
[|ξ1 | + |ξ2 |(n − p − 1)]|an,1 | o 12
≤1
λ1
n=p+1
(20)
∞
n X
[|ξ1 | + |ξ2 |(n − p − 1)]|an,2 | o 12
≤ 1.
λ
2
n=p+1
(21)
By using the Holder inequality with (20) and (21), we get
∞ n
q
X
[|ξ1 | + |ξ2 |(n − p − 1)] o 12 n [|ξ1 | + |ξ2 |(n − p − 1)] o 21
|an,1 ||an,2 | ≤ 1,
λ
λ
1
2
n=p+1
so
∞
X
n 1 o 12 n 1 o 12 q
|an,1 ||bn,2 | ≤ 1.
|ξ1 | + |ξ2 |(n − p − 1)]
λ
λ
1
2
n=p+1
(22)
∗
In order to obtain the (f ∗ g)(z) ∈ Θp,λ
ξ1 ,ξ2 with a function g(z) ∈ G, we have to
∗
find the corresponding λ such that
∞
X
[|ξ1 | + |ξ2 |(n − p − 1)]|an,1 ||bn,2 |
≤ 1.
λ∗
n=p+1
(23)
Following (22), then (23) hold true if for any n ≥ p + 1,
1
1 1 1 1
1
≤ ( )2 ( )2 p
∗
λ
λ1
λ2
|an,1 ||bn,2 |
or
q
λ ≥ (λ1 ) (λ2 )
|an,1 ||bn,2 |.
∗
1
2
1
2
(24)
32
Xiaoli Liu et al.
In fact, (24) implies that
1
1
λ∗ = max{L (n)|L (n) = (λ1 ) 2 (λ2 ) 2
q
|an,1 ||bn,1 |, ∀n ≥ 1 + p}.
Furthermore, from (22), it is easy to know that
q
1
1
|an,1 ||bn,1 | ≤
(λ1 λ2 ) 2 ,
|ξ1 | + |ξ2 |(n − p − 1)
(25)
since |ξ1 | + |ξ2 |(n − p − 1) is increasing in n, following (25), then we can see
that
q
1
1
1
2
2
|an,1 ||bn,1 | ≤
λ1 λ2
L (n) = (λ1 ) (λ2 )
|ξ1 | + |ξ2 |(n − p − 1)
1
1
λ1 λ2 =
≤
λ1 λ2 .
[|ξ1 | + |ξ2 |(n − p − 1)]|n=p+1
|ξ1 |
The proof of the theorem is complete.
Acknowledgements: This research was supported by Supported by Scientific Research Fund of Sichuan Provincial Education Department of China(Grant
no.14ZB0364).
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