# Some Basic Properties of Certain New Subclass of Meromorphic Functions

Authors: Lei Shi; Zhi-Gang Wang
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

We introduce and investigate a new subclassℳ
(
β
,
η
) of meromorphic functions. Some interesting properties such as inclusion relationship, coefficient estimates, neighborhoods, and partial sums are proved. Connections of the results with known results are also considered.


---

## Body

## 1. Introduction

LetΣ denote the class of functions f of the form
(1)
f
(
z
)
=
1
z
+
∑
k
=
1
∞
‍
a
k
z
k
,
which are analytic in the punctured open unit disk:
(2)
U
*
≔
{
z
:
z
∈
C
,
0
<
z
<
1
}
≕
U
∖
{
0
}
.LetP denote the class of functions of the form
(3)
p
(
z
)
=
1
+
∑
k
=
1
∞
‍
p
k
z
k
,
which are analytic U and satisfy the condition
(4)
R
p
(
z
)
>
0
(
z
∈
U
)
.A functionf
∈
Σ is said to be in the class M
S
*
(
α
) of meromorphic starlike functions of order α if it satisfies the inequality
(5)
R
z
f
′
(
z
)
f
(
z
)
<
-
α
(
z
∈
U
*
;
0
≤
α
<
1
)
.Forη
>
1, Wang et al. [1] (see also Nehari and Netanyahu [2]) introduced and studied a new subclass M
(
η
) of Σ consisting of functions f satisfying
(6)
R
z
f
′
(
z
)
f
(
z
)
>
-
η
(
z
∈
U
*
)
.We note that meromorphic starlike functions and related topics attract many authors’ attentions; see (for example) the earlier works [3–8] and the references cited therein.Let(7)
f
(
z
)
=
z
+
∑
k
=
m
+
1
∞
‍
a
k
z
k
m
∈
N
≔
{
1,2
,
3
…
}
be analytic in U. Assuming that α
∈
C and 0
≤
β
<
1, we say that a function f
∈
H
m
(
α
,
β
) if it satisfies the condition
(8)
R
z
f
′
(
z
)
f
(
z
)
+
α
z
f
′′
(
z
)
f
(
z
)
>
α
β
β
+
m
2
-
1
+
β
-
m
2
(
z
∈
U
)
.
The function class H
m
(
α
,
β
) was introduced and studied recently by Ravichandran et al. [9], Liu et al. [10], Singh and Gupta [11], and Wang et al. [12].In [13], Wang et al. introduced a subclass of meromorphic function H
~
(
β
,
λ
) which satisfies the condition
(9)
R
z
f
′
(
z
)
f
(
z
)
+
β
z
2
f
′′
(
z
)
f
(
z
)
<
β
λ
(
λ
+
1
2
)
+
β
2
-
λ
(
β
≥
0
;
1
2
≤
λ
<
1
;
z
∈
U
*
)
.
It was proved that the class H
~
(
β
,
λ
) is a subclass of the M
S
*
(
λ
) of meromorphically starlike functions of order λ.Motivated essentially by the above works, we introduce and investigate a new subclass ofΣ of meromorphic functions.Definition 1.
Suppose thatη
>
1. Let M
(
β
,
η
) denote a subclass of Σ consisting of functions satisfying the condition that
(10)
R
z
f
′
(
z
)
f
(
z
)
+
β
z
2
f
′′
(
z
)
f
(
z
)
>
1
2
β
(
2
η
2
+
η
+
1
)
-
η
(
z
∈
U
*
)
.We note that, forβ
=
0, the class M
(
0
,
η
) reduces to M
(
η
).In the present paper, we aim at proving some interesting properties such as inclusion relationship, coefficient estimates, neighborhoods, and partial sums for functions in the classM
(
β
,
η
).The following lemmas will be required in our investigation.Lemma 2 (see [14]).
LetΩ be a set in the complex plane C and suppose that Φ is mapping from C
2
×
U to C which satisfies Φ
(
i
x
,
y
;
z
)
∉
Ω for z
∈
U and for all real x, y such that y
≤
-
(
1
+
x
2
)
/
2. If the function ψ
(
z
)
=
1
+
c
1
z
+
c
2
z
2
+
⋯ is analytic in U and Φ
(
ψ
(
z
)
,
z
ψ
′
(
z
)
;
z
)
∈
Ω for all z
∈
U, then R
(
ψ
(
z
)
)
>
0.Lemma 3.
Letη
>
1, 0
≤
β
<
1, and -
1
+
2
β
+
γ
>
0. Suppose also that the sequence {
A
k
}
k
=
1
∞ is defined by
(11)
A
1
=
-
1
+
2
β
+
γ
1
-
β
,
A
k
=
2
(
-
1
+
2
β
+
γ
)
(
k
+
1
)
1
+
(
k
-
2
)
β
1
+
∑
m
=
1
k
-
1
‍
A
m
.
Then
(12)
A
k
=
-
1
+
2
β
+
γ
1
-
β
·
∏
m
=
1
k
-
1
‍
m
+
1
1
+
m
-
2
β
-
2
1
-
2
β
-
γ
m
+
2
1
+
m
-
1
β
(
k
≥
2
)
.Proof.
From (11), we have
(13)
k
+
1
1
+
k
-
2
β
A
k
=
2
-
1
+
2
β
+
γ
1
+
∑
m
=
1
k
-
1
‍
A
m
,
k
+
2
1
+
k
-
1
β
A
k
+
1

=
2
-
1
+
2
β
+
γ
1
+
∑
m
=
1
k
‍
A
m
.
Combining (13), we find that
(14)
A
k
+
1
A
k
=
(
k
+
1
)
1
+
(
k
-
2
)
β
+
2
(
-
1
+
2
β
+
γ
)
(
k
+
2
)
1
+
(
k
-
1
)
β
(
k
∈
N
)
.
Thus, for k
≥
2, we deduce from (14) that
(15)
A
k
=
A
k
A
k
-
1
×
A
k
-
1
A
k
-
2
×
⋯
×
A
2
A
1
×
A
1
=
k
1
+
(
k
-
3
)
β
+
2
(
-
1
+
2
β
+
γ
)
(
k
+
1
)
1
+
(
k
-
2
)
β
×
(
k
-
1
)
1
+
(
k
-
4
)
β
+
2
(
-
1
+
2
β
+
γ
)
k
1
+
(
k
-
3
)
β
×
⋯
×
2
(
1
-
β
)
+
2
(
-
1
+
2
β
+
γ
)
3
1
+
0
β
×
-
1
+
2
β
+
γ
1
-
β
=
-
1
+
2
β
+
γ
1
-
β
·
∏
m
=
1
k
-
1
‍
(
m
+
1
)
1
+
(
m
-
2
)
β
-
2
(
1
-
2
β
-
γ
)
(
m
+
2
)
1
+
(
m
-
1
)
β
.
This completes the proof of Lemma 3.Lemma 4.
Let(16)
η
>
1
,
0
≤
β
<
2
η
-
2
2
η
2
+
η
-
1
.
Suppose also that f
∈
Σ is given by (1) and
(17)
∑
k
=
1
∞
‍
k
+
β
k
(
k
-
1
)
+
γ
a
k
≤
γ
-
1
,
where (and throughout this paper unless otherwise mentioned) the parameter γ is defined as
(18)
γ
≔
η
-
1
2
β
2
η
2
+
η
+
1
.
Then f
∈
M
(
β
,
η
).The proof of Lemma4 is similar to that of Theorem 1 in Wang et al. [1] and so is omitted.


## 2. Main Results

We begin by proving the following result which shows thatM
(
β
,
η
) is a subclass of M
(
η
).Theorem 5.
Suppose thatη
>
1 and β
≥
0. Then
(19)
M
(
β
,
η
)
⊂
M
(
η
)
.Proof.
Define(20)
ρ
(
z
)
≔
z
f
′
(
z
)
/
f
(
z
)
+
η
η
-
1
(
η
>
1
;
z
∈
U
)
.
Then ρ is analytic in U. It follows from (20) that
(21)
-
z
f
′′
(
z
)
f
′
(
z
)
=
η
+
1
-
(
η
-
1
)
ρ
(
z
)
+
(
η
-
1
)
z
ρ
′
(
z
)
η
-
(
η
-
1
)
ρ
(
z
)
.
Combining (20) and (21), we obtain that
(22)
z
f
′
(
z
)
f
(
z
)
β
z
f
′′
(
z
)
f
′
(
z
)
+
1
=
β
(
η
-
1
)
z
ρ
′
(
z
)
+
β
η
-
1
2
ρ
2
(
z
)
-
(
η
-
1
)
(
2
β
η
+
β
-
1
)
ρ
(
z
)
+
η
(
β
η
+
β
-
1
)
=
Φ
(
ρ
(
z
)
,
z
ρ
′
(
z
)
;
z
)
,
where
(23)
Φ
r
,
s
;
t
=
β
(
η
-
1
)
s
+
β
η
-
1
2
r
2
-
(
η
-
1
)
(
2
β
η
+
β
-
1
)
r
+
η
(
β
η
+
β
-
1
)
.
For all real x and y satisfying y
≤
-
(
1
+
x
2
)
/
2, we have
(24)
R
ϕ
(
i
x
,
y
;
z
)
=
β
(
η
-
1
)
y
-
β
η
-
1
2
x
2
+
η
(
β
η
+
β
-
1
)
≤
-
β
(
η
-
1
)
1
+
x
2
2
-
β
η
-
1
2
x
2
+
η
(
β
η
+
β
-
1
)
≤
1
2
β
(
2
η
2
+
η
+
1
)
-
η
.
If we set
(25)
Ω
=
ξ
:
R
(
ξ
)
>
1
2
β
(
2
η
2
+
η
+
1
)
-
η
,
then Φ
(
i
x
,
y
;
z
)
∉
Ω for all real x
,
y such that y
≤
-
(
1
+
x
2
)
/
2. Moreover, from definition (10), we know that Φ
(
ρ
(
z
)
,
z
ρ
′
(
z
)
;
z
)
∈
Ω. Using Lemma 2, we conclude that R
ρ
(
z
)
>
0 for all z
∈
U, which implies that f
∈
M
(
η
). This completes the proof of Theorem 5.Now we consider the coefficient estimates for functions belonging to the classM
(
β
,
η
).Theorem 6.
Suppose that(26)
η
>
1
,
0
≤
β
<
2
η
-
2
2
η
2
+
η
-
1
.
If f
∈
M
(
β
,
η
), then
(27)
a
1
≤
-
1
+
2
β
+
γ
1
-
β
,
a
k
≤
-
1
+
2
β
+
γ
1
-
β
·
∏
m
=
1
k
-
1
‍
m
+
1
1
+
m
-
2
β
-
2
1
-
2
β
-
γ
m
+
2
1
+
m
-
1
β
(
k
∈
N
∖
{
1
}
)
.Proof.
Suppose thatf
∈
M
(
β
,
η
). Then there exists τ
∈
P such that
(28)
z
f
′
(
z
)
f
(
z
)
+
β
z
2
f
′′
(
z
)
f
(
z
)
+
γ
=
(
-
1
+
2
β
+
γ
)
τ
(
z
)
z
∈
U
*
.
It follows from (28) that
(29)
z
f
′
(
z
)
+
β
z
2
f
′′
(
z
)
=
(
-
1
+
2
β
+
γ
)
τ
(
z
)
-
γ
f
(
z
)
.
Combining (1) and (29), we have
(30)
-
1
z
+
∑
k
=
1
∞
‍
k
a
k
z
k
+
β
2
z
+
∑
k
=
1
∞
‍
k
(
k
-
1
)
a
k
z
k
=
(
-
1
+
2
β
+
γ
)
1
+
∑
k
=
1
∞
‍
τ
k
z
k
-
γ
·
1
z
+
∑
k
=
1
∞
‍
a
k
z
k
.
Evaluating the coefficient of z
n in both sides of (30) yields
(31)
2
1
-
β
a
1
=
-
1
+
2
β
+
γ
τ
2
,
(32)
(
k
+
1
)
1
+
(
k
-
2
)
β
a
k
=
(
-
1
+
2
β
+
γ
)
τ
k
+
1
+
∑
l
=
1
k
-
1
‍
τ
k
-
l
a
l
.
By observing the fact that τ
k
≤
2 for k
∈
N, we find from (31) and (32) that
(33)
a
1
≤
-
1
+
2
β
+
γ
1
-
β
,
a
k
≤
2
(
-
1
+
2
β
+
γ
)
(
k
+
1
)
1
+
(
k
-
2
)
β
1
+
∑
m
=
1
k
-
1
‍
a
m
(
k
≥
2
)
.
Now we define the sequence {
A
k
}
k
=
1
∞ as follows:
(34)
A
1
=
-
1
+
2
β
+
γ
1
-
β
,
A
k
=
2
(
-
1
+
2
β
+
γ
)
(
k
+
1
)
1
+
(
k
-
2
)
β
1
+
∑
m
=
1
k
-
1
‍
A
m
.
In order to prove that
(35)
a
k
≤
A
k
(
k
∈
N
)
,
we use the principle of mathematical induction by noting that
(36)
a
1
≤
A
1
=
-
1
+
2
β
+
γ
1
-
β
.
Therefore, we assume that
(37)
a
m
≤
A
m
(
m
=
1,2
,
…
,
k
;
k
∈
N
)
.
Combining (32) and (33), we get
(38)
a
k
+
1
≤
2
(
-
1
+
2
β
+
γ
)
(
k
+
1
)
1
+
(
k
-
2
)
β
1
+
∑
m
=
1
k
‍
a
m
≤
2
(
-
1
+
2
β
+
γ
)
(
k
+
1
)
1
+
(
k
-
2
)
β
1
+
∑
m
=
1
k
‍
A
m
=
A
k
+
1
.
Hence, by the principle of mathematical induction, we have
(39)
a
k
≤
A
k
(
k
∈
N
)
as desired. By means of Lemma 2 and (33), we know that (12) holds. Combining (39) and (12), we readily get the coefficient estimates asserted by Theorem 6.Using Lemma4, we introduce the δ-neighborhood of a function f
∈
Σ of the form (1) by means of the following definition:
(40)
N
δ
f
≔
g
∈
Σ
:
g
z
=
1
z
+
∑
k
=
1
∞
‍
b
k
z
k
,
∑
k
=
1
∞
‍
k
+
β
k
(
k
-
1
)
+
γ
γ
-
1
a
k
-
b
k
≤
δ
(
δ
≥
0
)
.By making use of definition (40), we obtain the following result.Theorem 7.
Iff
∈
Σ satisfies the condition
(41)
f
(
z
)
+
ε
z
-
1
1
+
ε
∈
M
(
β
,
η
)
(
ε
∈
C
;
ε
<
δ
;
δ
>
0
)
,
then
(42)
N
δ
(
f
)
⊂
M
(
β
,
η
)
.Proof.
It is easily seen from (10) that a function g
∈
M
(
β
,
η
) if and only if
(43)
z
g
′
(
z
)
+
β
z
2
g
′′
(
z
)
+
g
(
z
)
z
g
′
(
z
)
+
β
z
2
g
′′
(
z
)
+
(
2
γ
-
1
)
g
(
z
)
≠
σ
(
z
∈
U
;
σ
∈
C
;
σ
=
1
)
,
which is equivalent to
(44)
(
g
*
ϱ
)
(
z
)
z
-
1
≠
0
(
z
∈
U
)
,
where
(45)
ϱ
(
z
)
=
1
z
+
∑
k
=
1
∞
‍
c
k
z
k
c
k
≔
k
+
β
k
(
k
-
1
)
+
1
-
[
k
+
β
k
(
k
-
1
)
+
(
2
γ
-
1
)
]
σ
2
β
+
(
-
2
+
2
β
+
2
γ
)
σ
.
It follows from (45) that
(46)
c
k
=
k
+
β
k
(
k
-
1
)
+
1
-
[
k
+
β
k
(
k
-
1
)
+
(
2
γ
-
1
)
]
σ
2
β
+
(
-
2
+
2
β
+
2
γ
)
σ
≤
k
+
β
k
(
k
-
1
)
+
1
+
[
k
+
β
k
(
k
-
1
)
+
(
2
γ
-
1
)
]
σ
(
-
2
+
2
β
+
2
γ
)
σ
-
2
β
=
k
+
β
k
k
-
1
+
γ
γ
-
1
(
σ
=
1
)
.
Furthermore, under the hypotheses of Theorem 7, (44) yields the following inequality:
(47)
(
f
*
ϱ
)
(
z
)
z
-
1
≥
δ
(
z
∈
U
;
δ
>
0
)
.
Suppose that
(48)
χ
(
z
)
=
1
z
+
∑
k
=
1
∞
‍
d
k
z
k
∈
N
δ
(
f
)
.
It follows from (40) that
(49)
(
(
f
-
χ
)
*
ϱ
)
(
z
)
z
-
1
=
∑
k
=
1
∞
‍
a
k
-
d
k
c
k
z
k
+
1
≤
z
∑
k
=
1
∞
‍
k
+
β
k
(
k
-
1
)
+
γ
γ
-
1
a
k
-
d
k
<
δ
.
Combining (47) and (49), we have
(50)
(
χ
*
ϱ
)
(
z
)
z
-
1
=
(
[
f
+
(
χ
-
f
)
]
*
ϱ
)
(
z
)
z
-
1
≥
(
f
*
ϱ
)
(
z
)
z
-
1
-
(
(
χ
-
f
)
*
ϱ
)
(
z
)
z
-
1
>
0
,
which implies that
(51)
(
χ
*
ϱ
)
(
z
)
z
-
1
≠
0
(
z
∈
U
)
.
Thus, we have
(52)
χ
(
z
)
∈
N
δ
(
f
)
⊂
M
(
β
,
η
)
.
This completes the proof of Theorem 7.Finally, we derive the partial sums of functions in the classM
(
β
,
η
).Theorem 8.
Letf
∈
Σ be given by (1) and define the partial sums f
n
(
z
) of f by
(53)
f
n
(
z
)
=
1
z
+
∑
k
=
1
n
‍
a
k
z
k
(
n
∈
N
)
.
Suppose also that
(54)
∑
k
=
1
∞
‍
k
+
β
k
(
k
-
1
)
+
γ
γ
-
1
a
k
≤
1
.
Then (1)
f
∈
M
(
β
,
η
);
(2)
(55)
R
f
(
z
)
f
n
(
z
)
≥
n
+
β
n
(
n
+
1
)
+
2
n
+
β
n
(
n
+
1
)
+
1
+
γ
(
n
∈
N
;
z
∈
U
)
,
(56)
R
f
n
(
z
)
f
(
z
)
≥
n
+
β
n
(
n
+
1
)
+
1
+
γ
n
+
β
n
(
n
+
1
)
+
2
γ
(
n
∈
N
;
z
∈
U
)
.Each of the bounds in (55) and (56) is the best possible for each n
∈
N.Proof.
(1) It is easy to see that the result follows directly from Lemma4.
(2) Note that(57)
n
+
1
+
β
n
(
n
+
1
)
+
γ
γ
-
1
>
n
+
β
n
(
n
-
1
)
+
γ
γ
-
1
>
1
(
n
∈
N
)
.
Thus, we have
(58)
∑
k
=
1
n
‍
a
k
+
n
+
β
n
(
n
+
1
)
+
1
+
γ
γ
-
1
∑
k
=
n
+
1
∞
‍
a
k
≤
∑
k
=
1
∞
‍
k
+
β
k
(
k
-
1
)
+
γ
γ
-
1
a
k
≤
1
.
By setting
(59)
h
1
z
=
n
+
β
n
n
+
1
+
1
+
γ
γ
-
1
·
f
z
f
n
z
-
n
+
β
n
n
+
1
+
2
n
+
β
n
n
+
1
+
1
+
γ
=

1

+

(
(
n
+
β
n
(
n
+
1
)

+

1
+
γ
)
/

(
γ
-
1
)
)

∑
k
=
n
+
1
∞
a
k
z
k
+
1
1
+
∑
k
=
1
n
a
k
z
k
+
1
,
we find from (58) and (59) that
(60)
h
1
(
z
)
-
1
h
1
(
z
)
+
1
≤

(
(
n
+
β
n
n
+
1
+
1
+
γ
)

/

(
γ
-
1
)
)

∑
k
=
n
+
1
∞
a
k

2
-
2

∑
k
=
1
n

a
k

-

(
n
+

β
n
n
+
1

+

1
+
γ

/
γ
-
1
)

∑
k
=
n
+
1
∞

a
k
≤
1
(
z
∈
U
)
,
which implies inequality (55).
If we put(61)
f
(
z
)
=
1
z
-
γ
-
1
n
+
β
n
(
n
+
1
)
+
1
+
γ
z
n
+
1
,
then
(62)
f
z
f
n
z
=
1
-
γ
-
1
n
+
β
n
n
+
1
+
1
+
γ
z
n
+
2
⟶
n
+
β
n
n
+
1
+
2
n
+
β
n
n
+
1
+
1
+
γ
(
z
⟶
1
-
)
,
which shows that the bound in (55) is the best possible for each n
∈
N.
Now, we set(63)
h
2
z
=
n
+
β
n
n
+
1
+
2
γ
γ
-
1
·
f
n
z
f
z
-
n
+
β
n
n
+
1
+
1
+
γ
n
+
β
n
n
+
1
+
2
γ
=
1
-
(
(
n
+
β
n
(
n
+
1
)
+
2
γ
)
/
(
γ
-
1
)
)
∑
k
=
n
+
1
∞
a
k
z
k
+
1
1
+
∑
k
=
1
∞
a
k
z
k
+
1
.
In view of (58) and (63), we conclude that
(64)
h
2
(
z
)
-
1
h
2
(
z
)
+
1
≤
(
(
n
+
β
n
(
n
+
1
)
+
2
γ
)
/
(
γ
-
1
)
)
∑
k
=
n
+
1
∞
a
k
2

-

2
∑
k
=
1
n
a
k

-

(
(
n
+
β
n
(
n
+
1
)
+
2
)
/
(
γ
-
1
)
)
∑
k
=
n
+
1
∞

a
k
≤
1
(
z
∈
U
)
,
which leads to inequality (56) asserted in Theorem 8. The bound in (56) is sharp with the extremal function f given by (61). We thus complete the proof of Theorem 8.In what follows, we turn to quotients involving derivatives. The proof of Theorem9 is similar to that of Theorem 8 and so the details may be omitted.Theorem 9.
Letf
∈
Σ be given by (1) and define the partial sums f
n
(
z
) of f by (53). If the condition (54) holds, then
(65)
R
f
′
(
z
)
f
n
′
(
z
)
≥
β
n
(
n
+
1
)
-
n
γ
n
+
β
n
(
n
+
1
)
+
1
+
γ
(
n
∈
N
;
z
∈
U
)
,
R
f
n
′
(
z
)
f
′
(
z
)
≥
n
+
β
n
(
n
+
1
)
+
1
+
γ
β
n
(
n
+
1
)
+
(
n
+
2
)
γ
(
n
∈
N
;
z
∈
U
)
.
The bounds in (65) are sharp with the extremal function given by (61).



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*Source: 196060-2015-03-01.xml*