# New Applications of Nunokawa’s Lemmas

Authors: Yi-Hui Xu; Jin-Lin Liu
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We derive several univalent andp-valent conditions on certain meromorphic functions by applying well-known Nunokawa’s lemmas.


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## Body

## 1. A Sufficient Condition onp-Valent Analytic Functions

Let a functionf be univalent in the unit disk U
=
{
z
∈
C
:
|
z
|
<
1
} normalized by the conditions f
(
0
)
=
0 and f
′
(
0
)
=
1. Then, f maps U onto a starlike domain with respect to the origin if and only if(1)
R
e
z
f
′
z
f
z
>
0
,
z
∈
U
.If a function f is p-valent in U and satisfies (1), then f is called p-valent starlike in U.In [1], Ozaki proved that if (2)
f
z
=
z
p
+
∑
n
=
p
+
1
∞
a
n
z
n
p
∈
N
=
1,2
,
3
,
…is analytic in U and for some real α, (3)
R
e
exp
⁡
i
α
f
p
z
>
0
,
z
∈
U
,then f is at most p-valent in U. Ozaki’s p-valent condition is a generalization of the well-known Noshiro-Warschawski univalence condition. Very recently, Nunokawa, Cho, Kwon, and Sokól (an extension of Ozaki’s p-valent conditions, preprint) considered similar problems in the unit disc U. In this section, we will extend these p-valent conditions.To prove our results, we need the following well-known Nunokawa lemmas.Lemma 1 (see [2, 3]).
Letp
(
z
) be analytic function in U of the form (4)
p
z
=
1
+
∑
n
=
m
∞
c
n
z
n
,
c
m
≠
0
,
m
≥
1
,with p
(
z
)
≠
0 in U. If there exists a point z
0, |
z
0
|
<
1, such that (5)
arg
⁡
p
z
<
π
φ
2for |
z
|
<
|
z
0
| and (6)
arg
⁡
p
z
0
=
π
φ
2for some φ
>
0, then (7)
z
0
p
′
z
0
z
0
=
i
l
φ
,where (8)
l
≥
m
2
a
+
1
a
≥
m
when
arg
⁡
p
z
0
=
π
φ
2
,
l
≤
-
m
2
a
+
1
a
≤
-
m
when
arg
⁡
p
z
0
=
-
π
φ
2
,
p
z
0
1
/
φ
=
±
i
a
,
a
>
0
.Lemma 2 (see [4]).
Letp
≥
2 and 1
≤
k
≤
p. Also let f
(
z
)
=
z
p
+
∑
n
=
p
+
1
∞
‍
a
n
z
n be analytic in U. If there exists (
p
-
k
+
1
)-valent starlike function g
(
z
)
=
z
p
-
k
+
1
+
∑
n
=
p
-
k
+
2
∞
‍
b
n
z
n that satisfies(9)
R
e
z
f
k
z
g
z
>
0
,
z
∈
U
,then f is p-valent in U.Theorem 3.
Letp
≥
3 and also let f
(
z
)
=
z
p
+
∑
n
=
p
+
1
∞
‍
a
n
z
n be analytic in U. Suppose that, for any positive integer k with 3
≤
k
≤
p,(10)
arg
⁡
f
k
z
z
p
-
k
<
π
2
α
+
2
π
tan
-
1
⁡
α
p
-
k
+
1
,
z
∈
U
,where (11)
α
=
1
+
2
π
tan
-
1
⁡
1
p
-
k
+
2
.Then, f is p-valent in U.Proof.
First we prove that(12)
arg
⁡
f
k
-
1
z
z
p
-
k
+
1
<
π
α
2
,
z
∈
U
.For 3
≤
k
≤
p, we define(13)
p
1
z
=
f
k
-
1
z
p
p
-
1
⋯
p
-
k
+
2
z
p
-
k
+
1
,
p
1
0
=
1
.If there exists a point z
0
∈
U such that (14)
arg
⁡
p
1
z
<
π
α
2
for
z
<
z
0
,
arg
⁡
p
1
z
0
=
π
α
2
,then, from Lemma 1, we have(15)
z
0
p
1
′
z
0
p
1
z
0
=
i
l
α
.For the case arg
⁡
{
p
1
(
z
0
)
}
=
π
α
/
2, we have (16)
a
r
g
f
k
z
0
z
0
p
-
k
=
arg
⁡
p
-
k
+
1
p
1
z
0
+
z
0
p
1
′
z
0
=
arg
⁡
p
1
z
0
1
+
z
0
p
1
′
z
0
p
-
k
+
1
p
1
z
0
=
arg
⁡
p
1
z
0
+
arg
⁡
1
+
i
l
α
p
-
k
+
1
≥
π
2
α
+
2
π
tan
-
1
⁡
α
p
-
k
+
1
.This contradicts (10). For the case arg
⁡
p
1
(
z
0
)
=
-
(
π
α
/
2
), applying the same method as the above, we have (17)
a
r
g
f
k
z
0
z
0
p
-
k
=
arg
⁡
p
1
z
0
1
+
z
0
p
1
′
z
0
p
-
k
+
1
p
1
z
0
=
arg
⁡
p
1
z
0
+
arg
⁡
1
-
i
l
α
p
-
k
+
1
≤
-
π
2
α
+
2
π
tan
-
1
⁡
α
p
-
k
+
1
.This also contradicts (10). Therefore,(18)
arg
⁡
f
k
-
1
z
z
p
-
k
+
1
<
π
α
2
,
z
∈
U
.Because α
=
1
+
(
2
/
π
)
tan
-
1
⁡
1
/
p
-
k
+
2
>
1, we cannot apply Lemma 2 at this moment. We need the following step.
Put(19)
p
2
z
=
f
k
-
2
z
p
p
-
1
⋯
p
-
k
+
3
z
p
-
k
+
2
,
p
2
0
=
1
.Then, it follows that(20)
f
k
-
1
z
p
p
-
1
⋯
p
-
k
+
2
z
p
-
k
+
1
=
p
2
z
1
+
z
p
2
′
z
p
-
k
+
2
p
2
z
.We want to show R
e
p
2
(
z
)
>
0
(
z
∈
U
). If there exists a point z
1
∈
U such that (21)
arg
⁡
p
2
z
<
π
2
for
z
<
z
1
,
a
r
g
p
2
z
=
π
2
,then, from Lemma 1, we have(22)
z
1
p
2
′
z
1
p
2
z
1
=
i
l
.For the case a
r
g
p
2
(
z
1
)
=
π
/
2, by (20), we have (23)
arg
⁡
f
k
-
1
z
1
z
1
p
-
k
+
1
=
arg
⁡
p
2
z
1
1
+
z
1
p
2
′
z
1
p
-
k
+
2
p
2
z
1
=
arg
⁡
p
2
z
1
+
arg
⁡
1
+
i
l
p
-
k
+
2
≥
π
2
1
+
2
π
tan
-
1
⁡
1
p
-
k
+
2
=
π
α
2
.It contradicts (18). For the case arg p
2
(
z
1
)
=
-
(
π
/
2
), applying the same method as the above, we have (24)
arg
⁡
f
k
-
1
z
1
z
1
p
-
k
+
1
≤
-
π
2
1
+
2
π
tan
-
1
⁡
1
p
-
k
+
2
=
-
π
α
2
.It also contradicts (18). This shows that (25)
arg
⁡
f
k
-
2
z
z
p
-
k
+
2
<
π
2
,
z
∈
U
,or (26)
R
e
f
k
-
2
z
z
p
-
k
+
2
=
R
e
z
f
k
-
2
z
z
p
-
k
+
3
>
0
,
z
∈
U
.Therefore, from Lemma 2, we obtain that f is p-valent in U. Now the proof is complete.Takingk
=
p in Theorem 3, we have the following corollary.Corollary 4.
Letf
(
z
)
=
z
p
+
∑
n
=
p
+
1
∞
‍
a
n
z
n be analytic in U. Suppose that, for positive integer p
≥
3, (27)
arg
⁡
f
p
z
<
π
2
α
+
2
π
tan
-
1
⁡
α
,
z
∈
U
,where (28)
α
=
1
+
2
π
tan
-
1
⁡
1
2
=
1.2951672353
…
,
α
+
2
π
tan
-
1
⁡
α
=
1.876592028
…
.Then, f is p-valent in U.Example 5.
The functionf
(
z
)
=
z
p
(
p
≥
3
) satisfies condition (10) in Theorem 3 and is p-valent in U.


## 2. A Sufficient Condition on Univalency of Meromorphic Functions

LetA be the class of functions of the form(29)
f
z
=
z
+
∑
n
=
2
∞
a
n
z
n
,which are analytic in U. For two functions f and g analytic in U, the function f is said to be subordinate to g, written as f
(
z
)
≺
g
(
z
)
(
z
∈
U
), if there exists a function w analytic in U with w
(
0
)
=
0 and |
w
(
z
)
|
<
1 such that f
(
z
)
=
g
(
w
(
z
)
). In particular, if the function g is univalent in U, then f
(
z
)
≺
g
(
z
) is equivalent to f
(
0
)
=
g
(
0
) and f
(
U
)
⊂
g
(
U
).Also letΣ denote the class of functions of the form(30)
f
z
=
1
z
+
a
0
+
a
1
z
+
⋯which are analytic in the punctured unit disc U
∗
=
U
∖
{
0
}.In [5], Nunokawa et al. proved the following theorem.Theorem A.
Letf
∈
A with f
(
z
)
/
z
≠
0 for 0
<
|
z
|
<
1 and let(31)
z
f
z
′
′
≤
1
,
z
∈
U
.Then, f is univalent in U.In [6], Yang and Liu generalized the result of Theorem A and obtained sharp bound 2 of (31). Many interesting properties of analytic functions were considered by several authors in some recent works (e.g., [7–9]). In this section, we will derive a sufficient condition on univalency of meromorphic functions in Σ.For our purpose, we need the following lemmas.Lemma 6 (see [10]).
Letf and g be analytic in U with f
(
0
)
=
g
(
0
). If h
(
z
)
=
z
g
′
(
z
) is starlike in U and z
f
′
(
z
)
≺
h
(
z
), then (32)
f
z
≺
g
z
=
g
0
+
∫
0
z
h
t
t
d
t
.Lemma 7 (see [11]).
Letg be analytic and convex univalent in U and let f
(
z
)
=
g
(
0
)
+
∑
k
=
n
∞
‍
a
k
z
k
(
n
∈
N
=
{
1,2
,
3
,
…
}
) be analytic in U. If f
(
z
)
≺
g
(
z
), then (33)
z
-
c
∫
0
z
t
c
-
1
f
t
d
t
≺
1
n
z
-
c
/
n
∫
0
z
t
c
/
n
-
1
g
t
d
t
,where R
e
(
c
)
≥
0 and c
≠
0.Theorem 8.
Letf
(
z
)
=
(
1
/
z
)
+
a
0
+
a
1
z
+
⋯
∈
Σ with z
f
(
z
)
≠
0 for 0
<
|
z
|
<
1. If(34)
1
z
f
z
′
′
≤
α
,
z
∈
U
,where 0
<
α
<
M
=
(
1
-
2
|
a
0
|
)
/
2 and |
a
0
|
<
1
/
2, then f is meromorphic univalent in U
∗.Proof.
Let(35)
p
z
=
1
z
f
z
′
′
.Then, the function p is analytic in U. By integration from 0 to z, we get(36)
1
z
f
z
=
1
-
a
0
z
+
∫
0
z
d
v
∫
0
v
p
u
d
u
.Thus, we have(37)
f
z
=
1
z
-
a
0
z
2
+
z
φ
z
,where φ
(
z
)
=
∫
0
z
‍
d
v
∫
0
v
‍
p
(
u
)
d
u.
One can see easily from (34) and (35) that |
φ
(
z
)
|
≤
α and |
z
φ
′
(
z
)
|
≤
α, which shows that |
z
φ
z
′
|
≤
2
α for z
∈
U. Therefore,(38)
z
2
φ
z
2
-
z
1
φ
z
1
=
∫
z
1
z
2
z
φ
z
′
d
z
≤
2
α
z
2
-
z
1for z
1
,
z
2
∈
U
∗ and z
1
≠
z
2.
Ifz
1
≠
z
2, it follows from (37) and (38) that(39)
f
z
1
-
f
z
2
=
z
2
-
z
1
-
a
0
z
2
2
-
z
1
2
+
z
2
φ
z
2
-
z
1
φ
z
1
z
1
-
a
0
z
1
2
+
z
1
φ
z
1
z
2
-
a
0
z
2
2
+
z
2
φ
z
2
≥
z
2
-
z
1
-
2
a
0
z
2
-
z
1
-
2
α
z
2
-
z
1
z
1
-
a
0
z
1
2
+
z
1
φ
z
1
z
2
-
a
0
z
2
2
+
z
2
φ
z
2
>
0
.Hence, f is meromorphic univalent in U
∗. The proof is complete.Takinga
0
=
0 in Theorem 8, we obtain the following.Corollary 9.
Letf
(
z
)
=
1
/
z
+
a
1
z
+
⋯
∈
Σ with z
f
(
z
)
≠
0 for 0
<
|
z
|
<
1. If(40)
1
z
f
z
′
′
<
1
2
,
z
∈
U
,then f is meromorphic univalent in U
∗.Theorem 10.
Letf
(
z
)
=
1
/
z
+
a
0
+
a
1
z
+
⋯
∈
Σ with z
f
(
z
)
≠
0 for 0
<
|
z
|
<
1. Also let f satisfy condition (34). Then, for z
∈
U
∗,(41)
1
z
f
z
-
1
≤
z
a
0
+
α
2
z
,
1
-
z
a
0
+
α
2
z
≤
R
e
1
z
f
z
≤
1
+
z
a
0
+
α
2
z
,
f
z
≥
1
z
1
+
a
0
z
+
α
/
2
z
2
.Equalities in (41) are attained if we take(42)
f
z
=
1
z
1
±
a
0
z
+
α
/
2
z
2
.Proof.
In view of (34), we have(43)
z
1
z
f
z
′
′
≺
α
z
.Applying Lemma 6 to (43), we find that(44)
1
z
f
z
′
+
a
0
≺
α
z
.By using Lemma 7, (44) gives that(45)
1
z
∫
0
z
1
t
f
t
′
+
a
0
d
t
≺
α
2
z
;that is,(46)
1
z
f
z
=
1
-
a
0
z
+
α
2
z
w
z
,where w is analytic in U and |
w
(
z
)
|
≤
|
z
|
(
z
∈
U
) by Schwarz lemma.
Now, from (46), we can easily derive inequalities (41). The proof is complete.Example 11.
The functionf
(
z
)
=
1
/
z satisfies the conditions in Theorems 8 and 10.An Open Problem. The maximum value of M in Theorem 8 for which (34) implies meromorphic univalence is unknown.



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