# Unicity of Meromorphic Functions Sharing Sets with Their Linear Difference Polynomials

Authors: Sheng Li; BaoQin Chen
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We mainly investigate the unicity of meromorphic functions sharing two or three sets with their linear difference polynomials and prove some results.


---

## Body

## 1. Introduction and Main Results

In this paper, we assume the reader is familiar with the fundamental results and the basic notations of the Nevanlinna theory of meromorphic functions (see, e.g., [1–3]). Let f
(
z
) be meromorphic in the whole plane. We use the notation ρ
(
f
) to denote the order of growth of the meromorphic function f
(
z
). In addition, we denote by S
(
r
,
f
) any quantity satisfying S
(
r
,
f
)
=
o
(
T
(
r
,
f
)
), as r
→
∞ outside of a possible exceptional set of finite logarithmic measure. We say that a meromorphic function a
(
z
) is a small function of f
(
z
) provided that T
(
r
,
a
)
=
S
(
r
,
f
). Let S
(
f
) be the set of all small functions of f
(
z
).For a setS
⊂
S
(
f
), we define the following:
(1)
E
f
(
S
)
=
⋃
a
∈
S
{
z
∣
f
(
z
)
-
a
(
z
)
=
0
,
counting
multiplicities
}
,
E
¯
f
(
S
)
=
⋃
a
∈
S
{
z
∣
f
(
z
)
-
a
(
z
)
=
0
,
ignoring
multiplicities
}
.Letf and g be meromorphic functions. If E
f
(
S
)
=
E
g
(
S
) and E
¯
f
(
S
)
=
E
¯
g
(
S
), respectively, then we say that f and g share a set S CM and IM, respectively.Furthermore, letc be a nonzero complex constant. We define the shift of f
(
z
) by f
(
z
+
c
), and define the difference operators of f
(
z
) by
(2)
Δ
c
f
(
z
)
=
f
(
z
+
c
)
-
f
(
z
)
,
Δ
c
n
f
(
z
)
=
Δ
c
n
-
1
(
Δ
c
f
(
z
)
)
,
n
∈
ℕ
,
n
≥
2
.The unicity theory of meromorphic functions sharing sets is an important topic of the uniqueness theory. First of all, we recall the following theorem given by Li and Yang in [4].Theorem A (see [4]).
Letm
≥
2 and let n
>
2
m
+
6 with n and n
-
m having no common factors. Let a and b be two nonzero constants such that the equation ω
n
+
a
ω
n
-
m
+
b
=
0 has no multiple roots. Let S
=
{
ω
∣
ω
n
+
a
ω
n
-
m
+
b
=
0
}. Then, for any two nonconstant meromorphic functions f and g, the conditions E
f
(
S
)
=
E
g
(
S
) and E
f
(
{
∞
}
)
=
E
g
(
{
∞
}
) imply f
=
g.Yi and Lin considered the casem
=
1 with the condition that two meromorphic functions share three sets and got the result as follows.Theorem B (see [5]).
LetS
1
=
{
ω
:
ω
n
+
a
ω
n
-
1
+
b
=
0
}, S
2
=
{
0
}, and S
3
=
{
∞
}, where a, b are nonzero constants such that ω
n
+
a
ω
n
-
1
+
b
=
0 has no repeated root and n
(
≥
4
) is an integer. If, for two nonconstant meromorphic functions f and g, E
f
(
S
j
)
=
E
g
(
S
j
) for j
=
1,2
,
3, and Θ
(
∞
;
f
)
>
0, then f
≡
g.Recently, a number of papers have focused on difference analogues of the Nevanlinna theory (see, e.g., [6–9]). In particular, there has been an increasing interest in studying the uniqueness problems related to meromorphic functions and their shifts or their difference operators (see, e.g., [10–16]).In 2010, Zhang considered a meromorphic functionf
(
z
) sharing sets with its shift f
(
z
+
c
) and proved the following result.Theorem C (see [16]).
Letm
≥
2 and let n
≥
2
m
+
4 with n and n
-
m having no common factors. Let a and b be two nonzero constants such that the equation ω
n
+
a
ω
n
-
m
+
b
=
0 has no multiple roots. Let S
=
{
ω
∣
ω
n
+
a
ω
n
-
m
+
b
=
0
}. Suppose that f
(
z
) is a nonconstant meromorphic function of finite order. Then E
f
(
z
)
(
S
)
=
E
f
(
z
+
c
)
(
S
) and E
f
(
z
)
(
{
∞
}
)
=
E
f
(
z
+
c
)
(
{
∞
}
) imply f
(
z
)
≡
f
(
z
+
c
).For an analogue result in difference operator, B. Chen and Z. Chen proved the following theorem in [10].Theorem D (see [10]).
Letm
≥
2 and let n
≥
2
m
+
4 with n and n
-
m having no common factors. Let a and b be two nonzero constants such that the equation ω
n
+
a
ω
n
-
m
+
b
=
0 has no multiple roots. Let S
=
{
ω
∣
ω
n
+
a
ω
n
-
m
+
b
=
0
}. Suppose that f
(
z
) is a nonconstant meromorphic function of finite order satisfying E
f
(
z
)
(
S
)
=
E
Δ
c
f
(
S
) and E
f
(
z
)
(
{
∞
}
)
=
E
Δ
c
f
(
{
∞
}
). If
(3)
N
(
r
,
1
Δ
c
f
)
=
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
,
then Δ
c
f
≡
f
(
z
).It is natural to ask what happens if the shiftf
(
z
+
c
) or difference operator Δ
c
f
(
z
) is replaced by a general expression of f
(
z
), such as a linear difference polynomial of f
(
z
).Here, a linear difference polynomial off
(
z
) is an expression of the form
(4)
L
(
z
,
f
)
=
b
k
(
z
)
f
(
z
+
c
k
)
+
⋯
+
b
0
(
z
)
f
(
z
+
c
0
)
,
where b
k
(
z
)
≢
0, b
0
(
z
)
,
…
,
b
k
(
z
) are small functions of f
(
z
), c
0
,
…
,
c
k are complex constants, and k is a nonnegative integer.In this paper, our aim is to investigate the uniqueness problems of linear difference polynomials off
(
z
). In particular, we primarily consider the linear difference polynomial L
(
z
,
f
) which satisfies one of the following conditions:
(5)
(
i
)
b
0
(
z
)
+
⋯
+
b
k
(
z
)
≡
1
,
(
ii
)
b
0
(
z
)
+
⋯
+
b
k
(
z
)
≡
0
,
N
(
r
,
1
L
(
z
,
f
)
)
=
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
.Corresponding to the above question, we obtain the following results.Theorem 1.
Letm
≥
2 and let n
≥
2
m
+
4 with n and n
-
m having no common factors. Let a and b be two nonzero constants such that the equation ω
n
+
a
ω
n
-
m
+
b
=
0 has no multiple roots. Let S
=
{
ω
∣
ω
n
+
a
ω
n
-
m
+
b
=
0
}. Suppose that f
(
z
) is a nonconstant meromorphic function of finite order and L
(
z
,
f
) is of the form (4) satisfying the condition in (5). If E
f
(
z
)
(
S
)
=
E
L
(
z
,
f
)
(
S
) and E
f
(
z
)
(
{
∞
}
)
=
E
L
(
z
,
f
)
(
{
∞
}
), then L
(
z
,
f
)
≡
f
(
z
).Corollary 2.
Letn, m, and S be given as in Theorem 1. Suppose that f
(
z
) is a nonconstant meromorphic function of finite order satisfying the following:
(6)
N
(
r
,
1
Δ
c
k
f
)
=
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
.
If E
f
(
z
)
(
S
)
=
E
Δ
c
k
f
(
z
)
(
S
) and E
f
(
z
)
(
{
∞
}
)
=
E
Δ
c
k
f
(
z
)
(
{
∞
}
), then Δ
c
k
f
≡
f
(
z
).With an additional restriction on the order of growth off
(
z
), we prove the following fact.Theorem 3.
Letn, m, and S be given as in Theorem 1. Suppose that f
(
z
) is a nonconstant meromorphic function of finite order such that ρ
(
f
)
∉
ℕ. If E
f
(
z
)
(
S
)
=
E
L
(
z
,
f
)
(
S
) and E
f
(
z
)
(
{
∞
}
)
=
E
L
(
z
,
f
)
(
{
∞
}
), then L
(
z
,
f
)
≡
f
(
z
).Remark 4.
Note that, in Theorem3, we do not assume that the linear polynomial L
(
z
,
f
) satisfies the condition in (5). In fact, since ρ
(
f
)
∉
ℕ, by (19), we can easily get ρ
(
e
h
(
z
)
)
=
deg
⁡
(
h
(
z
)
)
<
ρ
(
f
), which implies T
(
r
,
e
h
(
z
)
)
=
S
(
r
,
f
). Then using a similar method as in the proof of Theorem 1, we can complete the proof of Theorem 3.Now we may ask what happens if the conditionm
≥
2 in Theorem 1 is replaced by a weaker condition containing the case m
=
1 or even m
=
0. By considering three sets, we get the following theorem.Theorem 5.
Letn, m be nonnegative integers such that n
>
m. Let a and b be nonzero constants such that ω
n
+
a
ω
n
-
m
+
b
=
0 has no multiple roots. Let S
1
=
{
ω
:
ω
n
+
a
ω
n
-
m
+
b
=
0
}
≠
⌀, S
2
=
{
∞
}, and S
3
=
{
0
}. Suppose that f
(
z
) is a nonconstant meromorphic function of finite order, L
(
z
,
f
) is of the form (4) satisfying the condition in (5), and E
f
(
z
)
(
S
j
)
=
E
L
(
z
,
f
)
(
S
j
) for j
=
1,2
,
3. Then one has the following.(i)
Ifm
=
0, then L
(
z
,
f
)
≡
t
f
(
z
), where t
n
=
1.
(ii)
Ifn and m are coprime, then L
(
z
,
f
)
≡
f
(
z
).Remark 6.
Takingm
=
1 in Theorem 5, we can obtain an analogue result of Theorem B related to linear difference polynomials.Furthermore, the following result is a corollary of Theorem5 related to difference operators.Corollary 7.
Letn, m, and S
j, j
=
1,2
,
3, be given as in Theorem 5. Suppose that f
(
z
) is a nonconstant meromorphic function of finite order satisfying
(7)
N
(
r
,
1
f
(
z
)
)
=
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
,
and E
f
(
z
)
(
S
j
)
=
E
Δ
c
k
f
(
S
j
) for j
=
1,2
,
3. Then one has the following. (i)
Ifm
=
0, then Δ
c
k
f
≡
t
f
(
z
), where t
n
=
1.
(ii)
Ifn and m are coprime, then Δ
c
k
f
≡
f
(
z
).Finally, we give some examples for our results.Examples.
In the following, letg
(
z
) be an entire function with period 1 such that ρ
(
g
)
∈
(
1
,
∞
)
∖
ℕ (see [17]).(1)
For the case (i) of condition (5), let f
1
(
z
)
=
e
2
π
i
z, f
2
(
z
)
=
g
(
z
)
e
2
π
i
z, f
3
(
z
)
=
e
2
π
i
z
/
g
(
z
) and let L
(
z
,
f
j
)
=
2
f
j
(
z
)
-
f
j
(
z
+
1
). Then for j
=
1,2
,
3, L
(
z
,
f
j
)
=
f
j
(
z
) and the sum of the coefficients of L
(
z
,
f
j
) is equal to 1. These examples satisfy Theorems 1 and 5 but do not satisfy Theorem D.
(2)
For the case (ii) of condition (5), let f
(
z
)
=
e
z
log
⁡
2
g
(
z
) and let L
(
z
,
f
)
=
Δ
f
(
z
)
=
f
(
z
+
1
)
-
f
(
z
). Then L
(
z
,
f
)
=
Δ
f
(
z
)
=
f
(
z
), the sum of the coefficients of L
(
z
,
f
) equals 0, and
(8)
N
(
r
,
1
Δ
f
)
=
N
(
r
,
1
f
)
=
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
.
This example satisfies Theorems1 and 5 and Corollaries 2 and 7.
(3)
For Theorem3, let f
(
z
)
=
e
z
log
⁡
3
/
g
(
z
) and let L
(
z
,
f
)
=
f
(
z
+
1
)
-
2
f
(
z
). Then L
(
z
,
f
)
=
f
(
z
) and the sum of the coefficients of L
(
z
,
f
) equals -
1. This example satisfies Theorem 3 but does not satisfy Theorem D and Theorems 1 and 5.


## 2. Proof of Theorem1

We need the following lemmas for the proof of Theorem1.The difference analogue of the logarithmic derivative lemma was given by Halburd-Korhonen [7] and Chiang-Feng [6] independently. We recall the following lemmas.Lemma 8 (see [7]).
Letf
(
z
) be a nonconstant meromorphic function of finite order, c
∈
ℂ and δ
<
1. Then
(9)
m
(
r
,
f
(
z
+
c
)
f
(
z
)
)
=
o
(
T
(
r
+
|
c
|
,
f
)
r
δ
)
,
for all r outside of a possible exceptional set with finite logarithmic measure.Lemma 9 (see [8]).
Letc
∈
ℂ, let n
∈
ℕ, and let f
(
z
) be a meromorphic function of finite order. Then for any small periodic function a
(
z
)
∈
S
(
f
) with period c, consider the following:
(10)
m
(
r
,
Δ
c
n
f
f
(
z
)
-
a
(
z
)
)
=
S
(
r
,
f
)
,
where the exceptional set associated with S
(
r
,
f
) is of at most finite logarithmic measure.Letf
(
z
) be a meromorphic function of finite order. Notice that if L
(
z
,
f
) (≢0) is of the form (4) such that b
0
(
z
)
+
⋯
+
b
k
(
z
)
≡
0, then, for any given complex constant a, L
(
z
,
a
)
=
0. This indicates that L
(
z
,
f
)
=
L
(
z
,
f
-
a
) and hence
(11)
m
(
r
,
L
(
z
,
f
)
f
-
a
)
=
m
(
r
,
L
(
z
,
f
-
a
)
f
-
a
)
≤
∑
j
=
0
k
m
(
r
,
b
j
(
z
)
(
f
(
z
+
c
j
)
-
a
)
f
-
a
)
+
S
(
r
,
f
)
=
S
(
r
,
f
)
.
With this, one can easily prove Lemma 10 below by a similar reasoning as in the proof of the difference analogue of the second main theorem of the Nevanlinna theory in [8] by Halburd and Korhonen. We omit those details.Lemma 10.
Letc
∈
ℂ, let f
(
z
) be a meromorphic function of finite order, and let L
(
z
,
f
)
≢
0 be of the form (4) such that b
0
(
z
)
+
⋯
+
b
k
(
z
)
≡
0. Let q
≥
2 and let a
1
,
…
,
a
q be distinct complex constants. Then
(12)
m
(
r
,
f
)
+
∑
i
=
1
q
m
(
r
,
1
f
-
a
i
)
≤
2
T
(
r
,
f
)
-
N
*
(
r
,
f
)
+
S
(
r
,
f
)
,
where
(13)
N
*
(
r
,
f
)
:
=
2
N
(
r
,
f
)
-
N
(
r
,
L
(
z
,
f
)
)
+
N
(
r
,
1
L
(
z
,
f
)
)
and the exceptional set associated with S
(
r
,
f
) is of at most finite logarithmic measure.Remark 11.
If the linear difference polynomialL
(
z
,
f
) is replaced by
(14)
L
*
(
z
,
f
)
=
b
k
(
z
)
f
(
z
+
k
c
)
+
⋯
+
b
1
(
z
)
f
(
z
+
c
)
+
b
0
(
z
)
f
(
z
)
,
Lemma 10 also holds even if the distinct complex constants a
1
,
…
,
a
q are replaced by a
1
(
z
)
,
…
,
a
q
(
z
) which are distinct meromorphic periodic functions with period c such that a
i
∈
S
(
f
) for all i
=
1
,
…
,
q.The following is the standard Valiron-Mohon’ko theorem; (see Theorem 2.2.5 in the book of Laine [2]).Lemma 12 (see [2]).
Letf
(
z
) be a meromorphic function. Then, for all irreducible rational functions in f,
(15)
R
(
z
,
f
)
=
P
(
z
,
f
)
Q
(
z
,
f
)
=
∑
i
=
0
p
‍
a
i
(
z
)
f
i
∑
j
=
0
q
‍
b
j
(
z
)
f
j
,
with meromorphic coefficients a
i
(
z
), b
j
(
z
) such that
(16)
T
(
r
,
a
i
)
=
S
(
r
,
f
)
,
i
=
0
,
…
,
p
,
T
(
r
,
b
j
)
=
S
(
r
,
f
)
,
j
=
0
,
…
,
q
.
The characteristic function of R
(
z
,
f
) satisfies
(17)
T
(
r
,
R
(
z
,
f
)
)
=
d
T
(
r
,
f
)
+
S
(
r
,
f
)
,
where d
=
max
⁡
{
p
,
q
}.Proof of Theorem1.
Sincef
(
z
) and L
(
z
,
f
) share ∞ CM, we see that L
(
z
,
f
)
≢
0 and N
(
r
,
L
(
z
,
f
)
)
=
N
(
r
,
f
(
z
)
). Then by Lemma 8, we have
(18)
T
(
r
,
L
(
z
,
f
)
)
=
m
(
r
,
L
(
z
,
f
)
)
+
N
(
r
,
L
(
z
,
f
)
)
≤
m
(
r
,
L
(
z
,
f
)
f
(
z
)
)
+
m
(
r
,
f
(
z
)
)
+
N
(
r
,
f
(
z
)
)
≤
∑
i
=
0
k
m
(
r
,
f
(
z
+
c
i
)
f
(
z
)
)
+
∑
i
=
0
k
m
(
r
,
b
i
(
z
)
)
+
T
(
r
,
f
(
z
)
)
≤
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
.
SinceE
f
(
z
)
(
S
)
=
E
L
(
z
,
f
)
(
S
), where S
=
{
ω
∣
ω
n
+
a
ω
n
-
m
+
b
=
0
} and the equation ω
n
+
a
ω
n
-
m
+
b
=
0 has no multiple roots, we know that (
L
(
z
,
f
)
)
n
+
a
(
L
(
z
,
f
)
)
n
-
m
+
b and f
(
z
)
n
+
a
f
(
z
)
n
-
m
+
b share 0 CM. Then from this and the condition E
f
(
z
)
(
{
∞
}
)
=
E
L
(
z
,
f
)
(
{
∞
}
), there exists a polynomial h
(
z
) such that
(19)
(
L
(
z
,
f
)
)
n
+
a
(
L
(
z
,
f
)
)
n
-
m
+
b
f
(
z
)
n
+
a
f
(
z
)
n
-
m
+
b
=
e
h
(
z
)
.
Suppose thate
h
(
z
)
≢
1. Note that S
=
{
ω
∣
ω
n
+
a
ω
n
-
m
+
b
=
0
} and the equation ω
n
+
a
ω
n
-
m
+
b
=
0 has no multiple roots. Let ω
1
,
…
,
ω
n denote all different roots of the equation ω
n
+
a
ω
n
-
m
+
b
=
0.
Next we prove thatT
(
r
,
e
h
(
z
)
)
=
S
(
r
,
f
). We know that
(20)
L
(
z
,
f
)
-
ω
i
=
b
k
(
z
)
(
f
(
z
+
c
k
)
-
f
(
z
)
)
+
⋯
+
b
0
(
z
)
(
f
(
z
+
c
0
)
-
f
(
z
)
)
+
(
b
k
(
z
)
+
⋯
+
b
0
(
z
)
)
f
(
z
)
-
ω
i
,
=
b
k
(
z
)
Δ
c
k
f
+
⋯
+
b
0
(
z
)
Δ
c
0
f
+
(
b
k
(
z
)
+
⋯
+
b
0
(
z
)
)
f
(
z
)
-
ω
i
.
(i)
Ifb
0
(
z
)
+
⋯
+
b
k
(
z
)
≡
1, we see that
(21)
L
(
z
,
f
)
-
ω
i
=
b
k
(
z
)
Δ
c
k
f
+
⋯
+
b
0
(
z
)
Δ
c
0
f
+
(
f
(
z
)
-
ω
i
)
.Then we deduce from this, (19), and Lemma 9 that
(22)
T
(
r
,
e
h
(
z
)
)
=
m
(
r
,
e
h
(
z
)
)
=
m
(
r
,
(
L
(
z
,
f
)
)
n
+
a
(
L
(
z
,
f
)
)
n
-
m
+
b
f
(
z
)
n
+
a
f
(
z
)
n
-
m
+
b
)
=
m
(
r
,
(
L
(
z
,
f
)
-
ω
1
)
⋯
(
L
(
z
,
f
)
-
ω
n
)
(
f
(
z
)
-
ω
1
)
⋯
(
f
(
z
)
-
ω
n
)
)
≤
∑
i
=
1
n
m
(
r
,
L
(
z
,
f
)
-
ω
i
f
(
z
)
-
ω
i
)
+
S
(
r
,
f
)
≤
∑
i
=
1
n
∑
j
=
0
k
m
(
r
,
Δ
c
j
f
f
(
z
)
-
ω
i
)
+
∑
i
=
1
n
∑
j
=
0
k
m
(
r
,
b
j
(
z
)
)
+
S
(
r
,
f
)
=
S
(
r
,
f
)
.
(ii)
Ifb
0
(
z
)
+
⋯
+
b
k
(
z
)
≡
0, we have
(23)
L
(
z
,
f
)
-
ω
i
=
b
k
(
z
)
Δ
c
k
f
+
⋯
+
b
0
(
z
)
Δ
c
0
f
-
ω
i
.From this, (19), and Lemma 9, we get
(24)
T
(
r
,
e
h
(
z
)
)
=
m
(
r
,
e
h
(
z
)
)
≤
∑
i
=
1
n
m
(
r
,
L
(
z
,
f
)
-
ω
i
f
(
z
)
-
ω
i
)
+
S
(
r
,
f
)
≤
∑
i
=
1
n
∑
j
=
0
k
m
(
r
,
Δ
c
j
f
f
(
z
)
-
ω
i
)
+
∑
i
=
1
n
m
(
r
,
1
f
(
z
)
-
ω
i
)
+
S
(
r
,
f
)
=
∑
i
=
1
n
m
(
r
,
1
f
(
z
)
-
ω
i
)
+
S
(
r
,
f
)
.
Applying Lemma10 to f
(
z
), we get
(25)
∑
i
=
1
n
m
(
r
,
1
f
(
z
)
-
ω
i
)
≤
2
T
(
r
,
f
(
z
)
)
-
m
(
r
,
f
(
z
)
)
-
2
N
(
r
,
f
(
z
)
)
+
N
(
r
,
L
(
z
,
f
)
)
-
N
(
r
,
1
L
(
z
,
f
)
)
+
S
(
r
,
f
)
=
T
(
r
,
f
(
z
)
)
-
N
(
r
,
1
L
(
z
,
f
)
)
+
S
(
r
,
f
)
.
Then the assumptions in (5), (24), and (25) yield the following:
(26)
T
(
r
,
e
h
(
z
)
)
≤
T
(
r
,
f
(
z
)
)
-
N
(
r
,
1
L
(
z
,
f
)
)
+
S
(
r
,
f
)
=
S
(
r
,
f
)
.
To sum up, we now prove that T
(
r
,
e
h
(
z
)
)
=
S
(
r
,
f
). Rewriting (19), we get
(27)
(
L
(
z
,
f
)
)
n
-
m
[
(
L
(
z
,
f
)
)
m
+
a
]
=
[
f
(
z
)
n
+
a
f
(
z
)
n
-
m
+
b
-
b
e
-
h
(
z
)
]
e
h
(
z
)
.
DenoteF
(
z
)
=
f
(
z
)
n
+
a
f
(
z
)
n
-
m. It follows from Lemma 12 and m
>
0 that
(28)
T
(
r
,
F
(
z
)
)
=
n
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
.
Hence, S
(
r
,
F
)
=
S
(
r
,
f
).
By (18) and (27) and applying the second main theorem for three small target functions, we deduce the following:
(29)
T
(
r
,
F
(
z
)
)
≤
N
¯
(
r
,
F
(
z
)
)
+
N
¯
(
r
,
1
F
(
z
)
)
+
N
¯
(
r
,
1
F
(
z
)
+
b
-
b
e
-
p
(
z
)
)
+
S
(
r
,
F
)
≤
N
¯
(
r
,
f
(
z
)
)
+
N
¯
(
r
,
1
f
(
z
)
n
-
m
[
f
(
z
)
m
+
a
]
)
+
N
¯
(
r
,
1
(
L
(
z
,
f
)
)
n
-
m
)
+
N
¯
(
r
,
1
(
L
(
z
,
f
)
)
m
+
a
)
+
S
(
r
,
f
)
≤
N
¯
(
r
,
f
(
z
)
)
+
N
¯
(
r
,
1
f
(
z
)
)
+
N
¯
(
r
,
1
f
(
z
)
m
+
a
)
+
N
¯
(
r
,
1
L
(
z
,
f
)
)
+
T
(
r
,
1
(
L
(
z
,
f
)
)
m
+
a
)
+
S
(
r
,
f
)
≤
T
(
r
,
f
(
z
)
)
+
T
(
r
,
1
f
(
z
)
)
+
T
(
r
,
1
f
(
z
)
m
+
a
)
+
T
(
r
,
1
L
(
z
,
f
)
)
+
m
T
(
r
,
L
(
z
,
f
)
)
+
S
(
r
,
f
)
≤
(
m
+
2
)
T
(
r
,
f
(
z
)
)
+
(
m
+
1
)
T
(
r
,
L
(
z
,
f
)
)
+
S
(
r
,
f
)
≤
(
2
m
+
3
)
T
(
r
,
f
(
z
)
)
+
S
(
r
,
f
)
.
By combining (28) and (29), we have
(30)
(
n
-
2
m
-
3
)
T
(
r
,
f
(
z
)
)
≤
S
(
r
,
f
)
,
which contradicts with n
≥
2
m
+
4.
Now we turn to consider the casee
h
(
z
)
≡
1. Equation (19) yields the following:
(31)
(
L
(
z
,
f
)
)
n
+
a
(
L
(
z
,
f
)
)
n
-
m
≡
f
(
z
)
n
+
a
f
(
z
)
n
-
m
.
Set φ
(
z
)
=
L
(
z
,
f
)
/
f
(
z
), and we have
(32)
f
(
z
)
m
(
φ
(
z
)
n
-
1
)
=
-
a
(
φ
(
z
)
n
-
m
-
1
)
.
Ifφ
(
z
) is not a constant, (32) can be rewritten as
(33)
f
(
z
)
m
(
φ
(
z
)
-
1
)
(
φ
(
z
)
-
μ
)
⋯
(
φ
(
z
)
-
μ
n
-
1
)
=
-
a
(
φ
(
z
)
-
1
)
(
φ
(
z
)
-
ν
)
⋯
(
φ
(
z
)
-
ν
n
-
m
-
1
)
,
where μ
=
cos
⁡
(
2
π
/
n
)
+
i
sin
(
2
π
/
n
) and ν
=
cos
⁡
(
2
π
/
(
n
-
m
)
)
+
i
sin
(
2
π
/
(
n
-
m
)
).
By the assumption thatn and n
-
m have no common factors, we see that μ
,
…
,
μ
n
-
1, ν
,
…
,
ν
n
-
m
-
1 are different. Assume that z
0 is a μ
j-point of φ
(
z
) of multiplicity u
j
>
0, where 1
≤
j
≤
n
-
1. Notice that
(34)
-
a
(
φ
(
z
0
)
-
1
)
(
φ
(
z
0
)
-
ν
)
⋯
(
φ
(
z
0
)
-
ν
n
-
m
-
1
)
is a constant. Then (33) implies that z
0 is a pole of f
(
z
)
m. Thus, u
j
≥
m. This yields the following, for 1
≤
j
≤
n
-
1:
(35)
m
N
¯
(
r
,
1
φ
(
z
)
-
μ
j
)
≤
N
(
r
,
1
φ
(
z
)
-
μ
j
)
≤
T
(
r
,
φ
(
z
)
)
+
S
(
r
,
h
)
.
Then by (35), we get
(36)
2
≥
∑
j
=
1
n
-
1
Θ
(
μ
j
,
φ
(
z
)
)
=
∑
j
=
1
n
-
1
{
1
-
lim
⁡
r
→
∞
¯
N
¯
(
r
,
1
/
(
φ
(
z
)
-
μ
j
)
)
T
(
r
,
φ
(
z
)
)
}
≥
∑
j
=
1
n
-
1
(
1
-
1
m
)
=
(
n
-
1
)
(
1
-
1
m
)
,
which is impossible with m
≥
2 and n
≥
2
m
+
4.
Hence,φ
(
z
) is a constant. Since f
(
z
) is a nonconstant meromorphic function, we deduce from (32) that φ
(
z
)
≡
1. This yields L
(
z
,
f
)
≡
f
(
z
), which completes the proof of Theorem 1.


## 3. Proof of Theorem5

Sincef
(
z
) is a nonconstant meromorphic function of finite order, E
f
(
z
)
(
S
j
)
=
E
L
(
z
,
f
)
(
S
j
) for j
=
1,2
,
3, S
1
=
{
ω
:
ω
n
+
a
ω
n
-
m
+
b
=
0
}, S
2
=
{
∞
}, and S
3
=
{
0
}, we have L
(
z
,
f
)
≢
0, N
(
r
,
L
(
z
,
f
)
)
=
N
(
r
,
f
(
z
)
), and N
(
r
,
1
/
L
(
z
,
f
)
)
=
N
(
r
,
1
/
f
(
z
)
), and we also get (18) and (19).Sincef
(
z
) and L
(
z
,
f
) share 0, ∞ CM, there exists a polynomial h
*
(
z
) such that
(37)
L
(
z
,
f
)
f
(
z
)
=
e
h
*
(
z
)
.By Lemma8, we see that
(38)
T
(
r
,
e
h
*
(
z
)
)
=
m
(
r
,
e
h
*
(
z
)
)
=
m
(
r
,
L
(
z
,
f
)
f
(
z
)
)
≤
∑
j
=
0
k
m
(
r
,
f
(
z
+
c
j
)
f
(
z
)
)
+
∑
j
=
0
k
m
(
r
,
b
j
(
z
)
)
+
S
(
r
,
f
)
=
S
(
r
,
f
)
.As in the proof of Theorem1, we see that T
(
r
,
e
h
(
z
)
)
=
S
(
r
,
f
) still holds in both cases (i) and (ii).Rewriting (19), we have
(39)
(
L
(
z
,
f
)
)
n
+
a
(
L
(
z
,
f
)
)
n
-
m
-
e
h
(
z
)
f
(
z
)
n
-
a
e
h
(
z
)
f
(
z
)
n
-
m
=
b
(
e
h
(
z
)
-
1
)
.
Combining this and (37), we get
(40)
(
e
n
h
*
(
z
)
-
e
h
(
z
)
)
f
(
z
)
n
+
a
(
e
(
n
-
m
)
h
*
(
z
)
-
e
h
(
z
)
)
f
(
z
)
n
-
m
=
b
(
e
h
(
z
)
-
1
)
.Suppose thate
n
h
*
(
z
)
-
e
h
(
z
)
≢
0. If m
=
0, (40) becomes
(41)
(
a
+
1
)
(
e
n
h
*
(
z
)
-
e
h
(
z
)
)
f
(
z
)
n
=
b
(
e
h
(
z
)
-
1
)
.
By the condition that b
≠
0, S
1
=
{
ω
:
(
a
+
1
)
ω
n
+
b
=
0
}
≠
⌀ implies a
≠
-
1.It follows from (38), (41), and T
(
r
,
e
h
(
z
)
)
=
S
(
r
,
f
) that
(42)
n
T
(
r
,
f
)
+
S
(
r
,
f
)
=
T
(
r
,
(
e
n
h
*
(
z
)
-
e
h
(
z
)
)
f
(
z
)
n
)
=
T
(
r
,
b
(
e
h
(
z
)
-
1
)
)
=
S
(
r
,
f
)
,
which is a contradiction, since n
≥
1.Ifm
≥
1, it follows from (38), (41), and T
(
r
,
e
h
(
z
)
)
=
S
(
r
,
f
) that
(43)
n
T
(
r
,
f
)
+
S
(
r
,
f
)
=
T
(
r
,
(
e
n
h
*
(
z
)
-
e
h
(
z
)
)
f
(
z
)
n
)
=
T
(
r
,
-
a
(
e
(
n
-
m
)
h
*
(
z
)
-
e
h
(
z
)
)
f
(
z
)
n
-
m
hh
h
+
b
(
e
h
(
z
)
-
1
)
(
e
(
n
-
m
)
h
*
(
z
)
-
e
h
(
z
)
)
)
≤
(
n
-
m
)
T
(
r
,
f
)
+
S
(
r
,
f
)
.
That is impossible.Therefore,e
n
h
*
(
z
)
-
e
h
(
z
)
≡
0. Notice that a
,
b
≠
0. Using a similar method, we can prove that e
(
n
-
m
)
h
*
(
z
)
-
e
h
(
z
)
≡
0. Then (40) implies that e
h
(
z
)
≡
1.Ifm
=
0, we have e
n
h
*
(
z
)
≡
1. Obviously, e
h
*
(
z
) is a constant. Set t
=
e
h
*
(
z
). Thus, by (37), we get L
(
z
,
f
)
≡
t
f
(
z
), where t
n
=
1.Ifn and m are coprime, e
n
h
*
(
z
)
≡
1 and e
m
h
*
(
z
)
≡
1 imply that e
h
*
(
z
)
≡
1. Thus, by (37), we get L
(
z
,
f
)
≡
f
(
z
). Thus, Theorem 5 is proved.



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*Source: 894968-2014-04-13.xml*