# On the Tumura-Clunie Theorem and Its Application

Authors: Gaixian Xue; Jinjin Huang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We cast aside the restriction of the simple pole in the Tumura-Clunie type theorems for meromorphic functions and obtain a better result which improves the earlier results of Y. D. Ren. Furthermore, as an application, we improve a theorem given by B. Y. Su.


---

## Body

## 1. Introduction and Main Results

A meromorphic function will always mean meromorphic in the complex planeℂ. We adopt the standard notation in the Nevanlinna value distribution theory of meromorphic functions such as T
(
r
,
f
), m
(
r
,
f
), N
(
r
,
f
), and N
¯
(
r
,
f
) as explained in [1, 2]. For any nonconstant meromorphic function f, we denote by S
(
r
,
f
) any quantity satisfying S
(
r
,
f
)
=
o
(
T
(
r
,
f
)
) as r
→
∞ possibly outside a set of finite linear measures that is not necessarily the same at each occurrence.Definition 1 (see [1]).
A meromorphic function “a
(
z
)” is said to be a small function of f if T
(
r
,
a
(
z
)
)
=
S
(
r
,
f
).Definition 2.
Throughout this paper one denotes bya
j
(
z
) meromorphic functions satisfying (
r
,
a
j
(
z
)
)
=
S
(
r
,
f
)
(
j
=
0,1
,
…
,
n
). If a
n
≢
0, we call P
[
f
]
=
a
n
f
n
+
a
n
-
1
f
n
-
1
+
⋯
+
a
1
f
+
a
0 a polynomial in f with degree n. If n
0
,
n
1
,
…
,
n
k are nonnegative integers, we call M
[
f
]
=
f
n
0
(
f
′
)
n
1
⋯
(
f
(
k
)
)
n
k a differential monomial in f of degree Υ
M
=
n
0
+
n
1
+
⋯
+
n
k and of weight Γ
M
=
n
0
+
2
n
1
+
⋯
+
(
k
+
1
)
n
k. If M
1
,
M
2
,
…
,
M
n are differential monomials in f, we call Q
[
f
]
=
∑
j
=
1
n
a
j
(
z
)
M
j
[
f
] a differential polynomial in f and define the degree Υ
Q and the weight Γ
Q by Υ
Q
=
max
⁡
j
=
1
n
Υ
M
j and Γ
Q
=
max
⁡
j
=
1
n
Γ
M
j, respectively.
AlsoQ
[
f
] is called a quasi-differential polynomial generated by f if, instead of assuming T
(
r
,
a
j
(
z
)
)
=
S
(
r
,
f
), we just assume that m
(
r
,
a
j
(
z
)
)
=
S
(
r
,
f
) for the coefficients a
j
(
z
)
(
j
=
1,2
,
…
,
n
).Definition 3.
Letk be a positive integer; for any a in the complex plane, one denotes by N
k
)
(
r
,
1
/
(
f
-
a
)
) the counting function of a-points of f with multiplicity less than or equal to k, by N
(
k
(
r
,
1
/
(
f
-
a
)
) the counting function of a-points of f with multiplicity more than or equal to k, and by N
k
(
r
,
1
/
(
f
-
a
)
) the counting function of a-points of f with multiplicity of k. Denote the reduced counting function by N
¯
k
)
(
r
,
1
/
(
f
-
a
)
), N
¯
(
k
(
r
,
1
/
(
f
-
a
)
), and N
¯
k
(
r
,
(
1
/
f
-
a
)
), respectively.
Letf be a nonconstant meromorphic function and let
(1)
Ϝ
=
f
n
+
Q
[
f
]
be a differential polynomial, where Q
[
f
] is also a differential polynomial and Υ
Q
≤
n
-
1.
Hua (see [3, page 69]) proved the following result.Theorem A.
Letf be a nonconstant meromorphic function and let Ϝ be given by (1) with Υ
Q
≤
n
-
1. If
(2)
N
(
r
,
f
)
+
N
(
r
,
1
Ϝ
)
=
S
(
r
,
f
)
,
then
(3)
Ϝ
=
(
f
+
a
(
z
)
n
)
n
,
where a
(
z
) is a small function of f.
ThenϜ
=
g
n, g
=
f
+
(
a
(
z
)
/
n
), and a
(
z
)
g
n
-
1 is obtained by substituting g for f, g
′ for f
′, and so forth in the terms of degree n
-
1 in Q
[
f
].Remark 4.
The conclusion still holds good if condition (2) is replaced with
(4)
N
(
r
,
f
)
+
N
(
r
,
1
Ϝ
)
=
S
o
(
r
,
f
)
,
where S
o
(
r
,
f
) denotes any quantity which satisfies S
o
(
r
,
f
)
=
o
(
T
(
r
,
f
)
) as r
→
+
∞ through a set of r of infinite measure.
Hua (see [3]) improved Theorem A and obtained the following result.Theorem B.
Letf be a nonconstant meromorphic function and let Ϝ be given by (1) with Υ
Q
≤
n
-
1. If
(5)
N
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
=
S
(
r
,
f
)
,
then
(6)
Ϝ
=
(
f
+
a
(
z
)
n
)
n
,
where a
(
z
) is a small function of f.Another theorem is due to Zhang and Li (see [4]), which can be stated as follows.Theorem C.
Letf be a nonconstant meromorphic function and let Ϝ be given by (1), where n
(
≥
Υ
Q
+
1
) is an integer. Then one of the following occurs.
(
i
) If Γ
Q
>
n
-
1, then
(7)
T
(
r
,
f
)
≤
{
1
+
2
(
Γ
Q
-
n
+
1
)
}
N
¯
(
r
,
f
)
+
(
Γ
Q
-
n
+
2
)
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
f
)
.
Or there exists a small proximity function a
(
z
) of f such that
(8)
Ϝ
=
(
f
+
a
(
z
)
n
)
n
,
and N
(
r
,
a
(
z
)
)
≤
(
Γ
Q
-
n
+
1
)
{
N
¯
(
r
,
f
)
+
N
¯
(
r
,
1
/
Ϝ
)
}
+
S
(
r
,
f
).
(
ii
) If Γ
Q
≤
n
-
1, then
(9)
T
(
r
,
f
)
≤
2
N
¯
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
f
)
,
or
(10)
Ϝ
=
(
f
+
a
(
z
)
n
)
n
,
where a
(
z
) is a small function of f.
(
iii
) In the special case, if Q
[
f
]
=
a
n
-
1
f
n
-
1
+
P
[
f
], where Γ
P
≤
n
-
2, then
(11)
T
(
r
,
f
)
≤
N
¯
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
f
)
,
or
(12)
Ϝ
=
(
f
+
a
(
z
)
n
)
n
,
where a
(
z
) is a small function of f.Corollary 5.
From Theorem C we know that if condition (2) is replaced with “N
¯
(
r
,
f
)
+
N
¯
(
r
,
1
/
Ϝ
)
=
S
(
r
,
f
)” in Theorem A, then the conclusion remains valid.
In this direction Ren (see [5]) also generalized Tumura-Clunie’s theorem concerning differential polynomials.
Combining the methods used in their proofs we show the following theorem.Theorem 6.
Letf be a nonconstant meromorphic function and let Ϝ be given by (1), where n
(
≥
Υ
Q
+
1
) is an integer and Γ
Ϝ
(
≠
2
) is the weight of Ϝ. If
(13)
N
¯
(
2
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
=
S
(
r
,
f
)
,
then
(14)
Ϝ
=
(
f
+
a
(
z
)
n
)
n
,
where a
(
z
) is a small function of f.It is easily seen from the following example thatΓ
Ϝ
≠
2 in Theorem 6 is necessary.Example 7.
Letf
=
tan
z and Ϝ
=
f
2
+
1. Obviously, (13) is obtained but (14) does not hold.


## 2. Some Lemmas

To prove our results, we need some lemmas.Lemma 8 (see [1]).
Letf
1 and f
2 be two nonzero meromorphic functions in the complex plane; then
(15)
N
(
r
,
f
1
f
2
)
-
N
(
r
,
1
f
1
f
2
)
=
N
(
r
,
f
1
)
+
N
(
r
,
f
2
)
-
N
(
r
,
1
f
1
)
-
N
(
r
,
1
f
2
)
.Lemma 9.
IfN
(
r
,
0
;
f
(
k
)
∣
f
≠
0
) denotes the counting function of those zeros of f
(
k
) which are not the zeros of f, where a zero of f
(
k
) is counted according to its multiplicity, then
(16)
N
(
r
,
0
;
f
(
k
)
∣
f
≠
0
)
≤
k
N
¯
(
r
,
f
)
+
N
(
r
,
0
;
f
∣
<
k
)
+
k
N
¯
(
r
,
0
;
f
∣
≥
k
)
+
S
(
r
,
f
)
.Lemma 10.
Suppose thatQ
[
f
] is given in Definition 2. Let z
0 be a pole of f of order p and neither a zero nor a pole of coefficients of Q
[
f
]. Then z
0 is a pole of Q
[
f
] of order at most p
Υ
Q
+
(
Γ
Q
-
Υ
Q
).Lemma 11 (see [6]).
Letf be a nonconstant meromorphic function and let Q
[
f
] be given in Definition 2. Then
(17)
m
(
r
,
Q
[
f
]
)
≤
Υ
Q
m
(
r
,
f
)
+
∑
j
=
1
n
‍
m
(
r
,
a
j
)
+
S
(
r
,
f
)
,
N
(
r
,
Q
[
f
]
)
≤
Γ
Q
N
(
r
,
f
)
+
∑
j
=
1
n
‍
N
(
r
,
a
j
)
+
S
(
r
,
f
)
.Lemma 12.
Suppose thatf is a nonconstant meromorphic function and Q
[
f
] is given in Definition 2. Then S
(
r
,
Q
)
=
S
(
r
,
f
).Proof.
It is straightforward by Lemma11.Lemma 13 (see [7]).
Letf be a nonconstant meromorphic function in the complex plane and let Q
1
[
f
] and Q
2
[
f
] be quasi-differential polynomials in f. If Υ
Q
2
≤
n and f
n
Q
1
[
f
]
=
Q
2
[
f
], then m
(
r
,
Q
1
[
f
]
)
=
S
(
r
,
f
).Lemma 14.
Letf be a nonconstant meromorphic function and let Ϝ be given by (1). Then
(18)
(
Γ
Ϝ
-
2
)
N
1
(
r
,
f
)
≤
2
N
¯
(
2
(
r
,
f
)
+
2
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
f
)
.Proof.
IfΓ
Ϝ
≤
2, the conclusion of Lemma 14 holds obviously.
In the following we suppose thatΓ
Ϝ
>
2.
WithϜ
=
f
n
+
Q
[
f
], we set
(19)
g
(
z
)
=
{
Ϝ
′
}
Γ
Ϝ
{
Ϝ
}
Γ
Ϝ
+
1
.
Let z
0 be a simple pole of f and not a zero of coefficients of Q
[
f
]; then
(20)
f
(
z
)
=
a
z
-
z
0
+
O
(
1
)
,
a
≠
0
as
z
⟶
z
0
.
From Lemma 10 we know that z
0 is a pole of Ϝ of order at most Γ
Ϝ; then we have
(21)
Ϝ
(
z
)
=
b
(
z
-
z
0
)
Γ
Ϝ
+
O
(
1
)
,
Ϝ
′
(
z
)
=
-
b
Γ
Ϝ
(
z
-
z
0
)
Γ
Ϝ
+
1
+
O
(
1
)
,
where b
≠
0.
Then(22)
Ϝ
(
z
)
=
b
(
z
-
z
0
)
Γ
Ϝ
{
1
+
O
(
z
-
z
0
)
Γ
Ϝ
}
,
Ϝ
′
(
z
)
=
-
b
Γ
Ϝ
(
z
-
z
0
)
Γ
Ϝ
+
1
{
1
+
O
(
z
-
z
0
)
Γ
Ϝ
+
1
}
,
g
(
z
)
=
(
-
1
)
Γ
Ϝ
Γ
Ϝ
Γ
Ϝ
b
{
1
+
O
(
z
-
z
0
)
Γ
Ϝ
}
.
Sog
(
z
0
)
≠
0
,
∞. But z
0 is a zero of g
′
(
z
) of order at least Γ
Ϝ
-
1. Then
(23)
(
Γ
Ϝ
-
1
)
N
1
(
r
,
f
)
≤
N
0
(
r
,
1
g
′
)
,
where N
0
(
r
,
1
/
g
′
) denotes the counting function of the zeros of g
′, not of g.
By Lemma8 and Nevanlinna first fundamental theorem, we get
(24)
N
(
r
,
g
g
′
)
-
N
(
r
,
g
′
g
)
=
N
(
r
,
1
g
′
)
+
N
(
r
,
g
)
-
N
(
r
,
g
′
)
-
N
(
r
,
1
g
)
=
N
0
(
r
,
1
g
′
)
-
N
¯
(
r
,
g
)
-
N
¯
(
r
,
1
g
)
,
N
(
r
,
g
g
′
)
-
N
(
r
,
g
′
g
)
=
m
(
r
,
g
′
g
)
-
m
(
r
,
g
g
′
)
+
O
(
1
)
.
From (24), we have
(25)
N
0
(
r
,
1
g
′
)
≤
N
¯
(
r
,
1
g
)
+
N
¯
(
r
,
g
)
+
m
(
r
,
g
′
g
)
+
O
(
1
)
≤
N
¯
(
r
,
1
g
)
+
N
¯
(
r
,
g
)
+
S
(
r
,
f
)
.
From (19), we know that the poles and zeros of g
(
z
) can only occur at the multiple zeros of f
(
z
), the zeros of Ϝ, and the zeros of Ϝ
′. Hence
(26)
N
¯
(
r
,
g
)
+
N
¯
(
r
,
1
g
)
≤
N
¯
(
2
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
+
N
0
(
r
,
1
Ϝ
′
)
+
S
(
r
,
f
)
,
where N
0
(
r
,
1
/
Ϝ
′
) denotes the counting function of the zeros of Ϝ
′, not of Ϝ.
By Lemmas9 and 12, we obtain
(27)
N
0
(
r
,
1
Ϝ
′
)
≤
N
¯
(
r
,
Ϝ
)
+
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
Ϝ
)
≤
N
¯
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
f
)
,
N
¯
(
r
,
f
)
=
N
1
(
r
,
f
)
+
N
¯
(
2
(
r
,
f
)
.
Combining (23), (25), (26), and (27), we obtain (18).
This completes the proof of Lemma14.Proof of Theorem6.
We consider two cases.
Case
1. If Γ
Ϝ
=
1, (14) holds obviously.
Case
2. If Γ
Ϝ
>
2, by Lemma 14 and (13) we have
(28)
N
¯
(
r
,
f
)
=
N
1
(
r
,
f
)
+
N
¯
(
2
(
r
,
f
)
≤
Γ
Ϝ
Γ
Ϝ
-
2
N
¯
(
2
(
r
,
f
)
+
2
Γ
Ϝ
-
2
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
f
)
≤
S
(
r
,
f
)
.
This shows that
(29)
N
¯
(
r
,
f
)
=
S
(
r
,
f
)
.
Suppose that Ϝ
≡
0.
So we havef
n
=
-
Q
[
f
] and Q
[
f
]
≢
0; moreover T
(
r
,
Q
[
f
]
)
=
n
T
(
r
,
f
)
+
S
(
r
,
f
).
By Lemma11 we get m
(
r
,
Q
[
f
]
)
≤
Υ
Q
m
(
r
,
f
)
+
S
(
r
,
f
).
On the other hand, we have(30)
n
m
(
r
,
f
)
=
m
(
r
,
f
n
)
=
m
(
r
,
Ϝ
-
Q
[
f
]
)
≤
m
(
r
,
Ϝ
)
+
m
(
r
,
Q
[
f
]
)
+
S
(
r
,
f
)
≤
Υ
Q
m
(
r
,
f
)
+
S
(
r
,
f
)
.
It follows that m
(
r
,
f
)
=
S
(
r
,
f
), which is impossible.
Therefore,Ϝ
≢
0.
Then(31)
T
(
r
,
Ϝ
′
Ϝ
)
≤
N
¯
(
r
,
Ϝ
)
+
N
¯
(
r
,
1
Ϝ
)
+
m
(
r
,
Ϝ
′
Ϝ
)
+
S
(
r
,
f
)
≤
N
¯
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
+
S
(
r
,
f
)
.
From (29) and the condition of the theorem, we know T
(
r
,
Ϝ
′
/
Ϝ
)
=
S
(
r
,
f
).
ByϜ
=
f
n
+
Q
[
f
], we have
(32)
Ϝ
′
=
Ϝ
′
Ϝ
f
n
+
Ϝ
′
Ϝ
Q
[
f
]
,
Ϝ
′
=
n
f
n
-
1
f
′
+
Q
′
[
f
]
.
And hence
(33)
f
n
-
1
(
f
Ϝ
′
Ϝ
-
n
f
′
)
=
Q
[
f
]
(
Q
′
[
f
]
Q
[
f
]
-
Ϝ
′
Ϝ
)
.
Let
(34)
Ω
1
[
f
]
=
f
Ϝ
′
Ϝ
-
n
f
′
,
Ω
2
[
f
]
=
Q
[
f
]
(
Q
′
[
f
]
Q
[
f
]
-
Ϝ
′
Ϝ
)
.
Then(35)
f
n
-
1
Ω
1
[
f
]
=
Ω
2
[
f
]
,
where Ω
1
[
f
] and Ω
2
[
f
] are quasi-differential polynomials.
By Lemma13 we have
(36)
m
(
r
,
Ω
1
[
f
]
)
=
S
(
r
,
f
)
.
By Lemma10 and (35) we obtain
(37)
N
(
r
,
Ω
1
[
f
]
)
=
N
(
r
,
Ω
2
[
f
]
)
-
(
n
-
1
)
N
(
r
,
f
)
+
S
(
r
,
f
)
≤
Υ
Q
N
(
r
,
f
)
+
(
Γ
Q
-
Υ
Q
+
1
)
N
¯
(
r
,
f
)
-
(
n
-
1
)
N
(
r
,
f
)
+
S
(
r
,
f
)
≤
(
Γ
Q
-
Υ
Q
+
1
)
N
¯
(
r
,
f
)
+
S
(
r
,
f
)
.
Note thatN
¯
(
r
,
f
)
=
S
(
r
,
f
).
SoT
(
r
,
Ω
1
[
f
]
)
=
S
(
r
,
f
).
From (34) we know that Q
[
f
] is a polynomial and Υ
Q
≤
n
-
1.
Set(38)
Q
[
f
]
=
b
(
z
)
f
n
-
1
+
P
[
f
]
,
where P
[
f
] is a polynomial and b
(
z
) is a small function of f; moreover Υ
P
≤
n
-
2.
Setg
=
f
+
(
b
(
z
)
/
n
); we have
(39)
Ϝ
=
g
n
+
R
[
g
]
,
where R
[
g
] is a polynomial and Υ
R
≤
n
-
2.
Now proceeding as the above proof, we get(40)
g
n
-
1
(
g
Ϝ
′
Ϝ
-
n
g
′
)
=
R
[
g
]
(
R
′
[
g
]
R
[
g
]
-
Ϝ
′
Ϝ
)
.
By Lemma13 we obtain
(41)
m
(
r
,
(
g
Ϝ
′
Ϝ
-
n
g
′
)
g
)
=
S
(
r
,
f
)
,
m
(
r
,
g
Ϝ
′
Ϝ
-
n
g
′
)
=
S
(
r
,
f
)
.
Therefore we have(42)
T
(
r
,
(
g
Ϝ
′
Ϝ
-
n
g
′
)
g
)
=
S
(
r
,
f
)
,
T
(
r
,
g
Ϝ
′
Ϝ
-
n
g
′
)
=
S
(
r
,
f
)
.
Notice thatT
(
r
,
g
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
≠
S
(
r
,
f
).
We can getg
(
Ϝ
′
/
Ϝ
)
-
n
g
′
≡
0.
SoϜ
≡
c
g
n, where c is a constant. Obviously c
=
1.
This proves Theorem6.


## 3. Application

Very recently, Yi (see [8, 9]) proved the following result.Theorem D.
Letf be a transcendental meromorphic function and let p
(
z
) be a polynomial, p
(
z
)
≢
0. If f and f
′ share 0 in ℂ, then f
′
-
p
(
z
) has infinitely many zeros.Remark 15.
From the hypothesis of Theorem E, it can be easily seen that all zeros off have multiplicity at least two.
Ren and Yang 2013 (see [10]) obtained the following result.Theorem E.
Letf be a transcendental meromorphic function and let R be a rational function, R
≢
0. Suppose that, with the exception of possibly finitely many, all zeros and poles of f are multiple. Then f
′
-
R has infinitely many zeros.It is natural to ask the following question: what can we say iff
′ is replaced by f
(
k
) and p
(
z
) and R are replaced by a small function relative to f in Theorems D and E?Later, Yang (see [11]) answered the above question and obtained the following result.Theorem F.
Letf be a transcendental meromorphic function satisfying
(43)
N
(
r
,
1
f
)
=
S
(
r
,
f
)
.
Then, for any k
≥
1 and any small function a
(
z
)
(
≢
0
,
∞
) of f,
(44)
N
(
r
,
1
f
(
k
)
-
a
(
z
)
)
≠
S
(
r
,
f
)
.We supplement Theorems D and E, improve Theorem F, and obtain the following result.Theorem 16.
Leth be a transcendental meromorphic function satisfying
(45)
N
¯
(
2
(
r
,
1
h
)
=
S
(
r
,
h
)
.
Then, for any n
≥
2 and any small function a
(
z
)
(
≢
0
,
∞
) of h,
(46)
N
(
r
,
1
h
(
n
)
-
a
(
z
)
)
≠
S
(
r
,
h
)
.The method of our proof essentially belongs to Yang. For the completeness, we give the proof here.Proof.
Set(47)
h
=
1
f
.
Then(48)
T
(
r
,
f
)
=
T
(
r
,
h
)
+
O
(
1
)
,
N
¯
(
2
(
r
,
1
h
)
=
N
¯
(
2
(
r
,
f
)
.
Obviously(49)
S
(
r
,
f
)
=
S
(
r
,
h
)
.
Now(50)
h
′′
=
-
f
f
′
+
2
(
f
′
)
2
f
3
,
h
′′′
=
-
6
(
f
′
)
3
-
f
2
f
′′
+
2
f
(
f
′
)
2
+
4
f
f
′
f
′′
f
4
⋯
.
Thus, in general,(51)
h
(
n
)
=
Q
n
(
f
)
f
n
+
1
,
where Q
n
(
f
) denotes a homogeneous differential polynomial in f of degree n. So
(52)
h
(
n
)
-
a
(
z
)
=
Q
n
(
f
)
-
a
(
z
)
f
n
+
1
f
n
+
1
.
If the assertion of the theorem was false, that is,(53)
N
(
r
,
1
h
(
n
)
-
a
(
z
)
)
=
S
(
r
,
f
)
,
then from (52) we have
(54)
Ϝ
=
f
n
+
1
-
Q
n
(
f
)
a
(
z
)
.
Thus from (48), (53), and (54), we obtain
(55)
N
¯
(
2
(
r
,
f
)
+
N
¯
(
r
,
1
Ϝ
)
=
S
(
r
,
f
)
.
Combining Theorem6, (55) gives
(56)
Ϝ
=
(
f
+
c
n
+
1
)
n
+
1
,
where c (a small function of f) is determined by the two equations: g
=
f
+
(
c
/
(
n
+
1
)
) and c
g
n
=
-
(
Q
n
(
g
)
/
a
(
z
)
).
We may claim that(i)
S
(
r
,
f
)
=
S
(
r
,
g
);
(ii)
N
¯
(
r
,
g
)
=
S
(
r
,
g
);
(iii)
T
(
r
,
g
(
k
)
/
g
)
=
S
(
r
,
g
) for all k
∈
ℕ.
In fact, from the definition ofg we know that the claim (i) above holds.
By (54) we have Γ
Ϝ
>
2.
Fromg
=
f
+
(
c
/
(
n
+
1
)
), Γ
Ϝ
>
2, and (29) we get
(57)
N
¯
(
r
,
g
)
=
N
¯
(
r
,
f
)
+
N
¯
(
r
,
c
)
=
S
(
r
,
f
)
=
S
(
r
,
g
)
.
That is, the claim (ii) above holds.
Combining (53) and the claims (i) and (ii), we may deduce
(58)
T
(
r
,
g
(
k
)
g
)
=
N
(
r
,
g
(
k
)
g
)
+
m
(
r
,
g
(
k
)
g
)
≤
k
N
¯
(
r
,
g
)
+
N
(
r
,
1
g
)
+
S
(
r
,
g
)
≤
k
N
¯
(
r
,
g
)
+
N
(
r
,
1
Ϝ
)
+
S
(
r
,
g
)
≤
S
(
r
,
g
)
.
Then the claim (iii) is true also.
Thus, by (54) and (56), we obtain
(59)
(
f
+
c
n
+
1
)
n
+
1
=
f
n
+
1
+
c
f
n
+
∑
k
=
2
n
+
1
‍
C
n
+
1
k
(
c
n
+
1
)
k
f
n
+
1
-
k
=
f
n
+
1
-
Q
n
(
f
)
a
(
z
)
.
Sincec
f
n
≡
-
(
Q
n
(
f
)
/
a
(
z
)
), it follows that
(60)
∑
k
=
2
n
+
1
‍
C
n
+
1
k
(
c
n
+
1
)
k
f
n
+
1
-
k
≡
0
,
which is impossible unless c
≡
0.
But then, from (59), -
(
Q
n
(
f
)
/
a
(
z
)
)
≡
0 and we have h
(
n
)
≡
0 which contradicts the fact that h is a transcendental meromorphic function.
This completes the proof of Theorem16.Remark 17.
Forn
=
1, from the proof of Theorem 16 and Corollary 5, we know that if the condition “N
¯
(
2
(
r
,
1
/
h
)
=
S
(
r
,
h
)” is replaced with “N
¯
(
r
,
1
/
h
)
=
S
(
r
,
h
)” in Theorem 16, then the conclusion still holds.



---

*Source: 615351-2014-04-22.xml*