# Unicity of Entire Functions concerning Shifts and Difference Operators

Authors: Dan Liu; Degui Yang; Mingliang Fang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We prove a unicity theorem of entire functions that share two distinct small functions with their shifts. The corollary of the theorem confirms the conjecture posed by Li and Gao (2011).


---

## Body

## 1. Introduction

Letf be a nonconstant meromorphic function in the complex plane ℂ. We will use the standard notations in Nevanlinna theory of meromorphic functions such as T
(
r
,
f
), N
(
r
,
f
), and m
(
r
,
f
) (see [1, 2]). The notation S
(
r
,
f
) is defined to be any quantity satisfying S
(
r
,
f
)
=
o
(
T
(
r
,
f
)
) as r
→
∞ possibly outside a set of finite linear measures. A meromorphic function a is called a small function related to f provided that T
(
r
,
a
)
=
S
(
r
,
f
).Letf and g be two nonconstant meromorphic functions, and let a be a small function related to both f and g. We say that f and g share a CM if f
-
a and g
-
a have the same zeros with the same multiplicities. f and g are said to share a IM if f
-
a and g
-
a have the same zeros ignoring multiplicities.LetN
(
r
,
a
) be the counting functions of all common zeros with the same multiplicities of f
-
a and g
-
a. If
(1)
N
(
r
,
1
f
-
a
)
+
N
(
r
,
1
g
-
a
)
-
2
N
(
r
,
a
)
=
S
(
r
,
f
)
+
S
(
r
,
g
)
,
then we say that f and g share a CM almost.For a nonzero complex constantc
∈
ℂ, we define difference operators as Δ
c
f
(
z
)
=
f
(
z
+
c
)
-
f
(
z
) and Δ
c
n
f
(
z
)
=
Δ
c
(
Δ
c
n
-
1
f
(
z
)
), n
∈
ℕ, n
>
2.In 1977, Rubel and Yang [3] proved the following result.Theorem A.
Letf be a nonconstant entire function. If f
(
z
) and f
′
(
z
) share two distinct finite values CM, then f
(
z
)
≡
f
′
(
z
).In fact, the conclusion still holds if the two CM values are replaced by two IM values (see Gundersen [4, 5], Mues and Steinmetz [6]).Recently, a number of articles focused on value distribution in shifts or difference operators of meromorphic functions (see [7–11]). In particular, some papers studied the unicity of meromorphic functions sharing values with their shifts or difference operators (see [12–14]). In 2009, Heittokangas et al. [12] proved the following result concerning shifts.Theorem B.
Letf be a nonconstant entire function of finite order, c
∈
ℂ. If f
(
z
) and f
(
z
+
c
) share two distinct finite values CM, then f
(
z
)
≡
f
(
z
+
c
).In 2011, Li and Gao [14] proved the following result concerning difference operators.Theorem C.
Letf be a nonconstant entire function of finite order, c
∈
ℂ, and let n be a positive integer. Suppose that f
(
z
) and Δ
c
n
f
(
z
) share two distinct finite values a, b CM and one of the following cases is satisfied:(i)
a
b
=
0
;
(ii)
a
b
≠
0
and
ρ
(
f
)
∉
N
.
Thenf
(
z
)
≡
Δ
c
n
f
(
z
).In [14], Li and Gao conjectured that the restriction ρ
(
f
)
∉
ℕ for the case a
b
≠
0 can be removed. In this paper, we confirm their conjecture. In fact, we prove the following more general results.Theorem 1.
Letf be a nonconstant entire function of finite order, let n be a positive integer, let a
(
z
), b
(
z
) be two distinct small functions related to f
(
z
), let m
1
,
m
2
,
…
,
m
n be nonzero complex numbers and c
1
,
c
2
,
…
,
c
n distinct finite values, and let
(2)
F
(
z
)
=
m
1
f
(
z
+
c
1
)
+
m
2
f
(
z
+
c
2
)
+
⋯
+
m
n
f
(
z
+
c
n
)
.
If f
(
z
) and F
(
z
) share a
(
z
), b
(
z
) CM, then f
(
z
)
≡
F
(
z
).Corollary 2.
Letf be a nonconstant entire function of finite order, let c be a nonzero finite complex number, let n be a positive integer, and let a, b be two distinct finite values. If f
(
z
) and Δ
c
n
f
(
z
) share a, b CM, then f
(
z
)
≡
Δ
c
n
f
(
z
).Remark 3.
Corollary2 confirms the conjecture of Li and Gao in [14].Corollary 4.
Letf be a nonconstant entire function of finite order, let c be a nonzero finite complex number, and let a
(
z
), b
(
z
) be two distinct small functions related to f. If f
(
z
) and f
(
z
+
c
) share a
(
z
), b
(
z
) CM, then f
(
z
)
≡
f
(
z
+
c
).


## 2. Some Lemmas

For the proof of Theorem1, we require the following results.Lemma 5 (see [15]).
Letf and g be two nonconstant meromorphic functions satisfying
(3)
N
(
r
,
1
f
)
+
N
(
r
,
f
)
=
S
(
r
,
f
)
,
N
(
r
,
1
g
)
+
N
(
r
,
g
)
=
S
(
r
,
g
)
.
If f
(
z
) and g
(
z
) share 1 CM almost, then either f
(
z
)
≡
g
(
z
) or f
(
z
)
g
(
z
)
≡
1.Lemma 6 (see [15]).
Letf and g be two nonconstant meromorphic functions satisfying
(4)
N
(
r
,
f
)
=
S
(
r
,
f
)
,
N
(
r
,
g
)
=
S
(
r
,
g
)
.
If f
(
z
) and g
(
z
) share 0 and 1 CM almost, and
(5)
lim
-
r
→
∞
r
∈
I
N
(
r
,
0
)
+
N
(
r
,
1
)
T
(
r
,
f
)
+
T
(
r
,
g
)
<
2
3
,
where I
⊂
[
0
,
∞
) is a set of infinitely linear measure, then
(6)
f
(
z
)
=
a
g
(
z
)
+
b
c
g
(
z
)
+
d
,
where a, b, c, and d are constants satisfying a
d
-
b
c
≠
0.Lemma 7 (see [10]).
Letf be a nonconstant meromorphic function of finite order, c
∈
ℂ. Then
(7)
m
(
r
,
f
(
z
+
c
)
f
(
z
)
)
=
S
(
r
,
f
)
,
for all r outside a possible exceptional set E with finite logarithmic measure ∫
E
‍
d
r
/
r
<
∞.In the following,S
(
r
,
f
) denotes any function satisfying S
(
r
,
f
)
=
o
(
T
(
r
,
f
)
) as r
→
∞, possibly outside a set with finite logarithmic measure.


## 3. Proof of Theorem1

We prove Theorem1 by contradiction. Suppose that f
(
z
)
≢
F
(
z
). Then it follows from f
(
z
) and F
(
z
) being two distinct entire functions that f
(
z
) and F
(
z
) share a
(
z
), b
(
z
), and ∞ CM. By the Nevanlinna second fundamental theorem for three small functions, we have
(8)
T
(
r
,
f
)
≤
N
¯
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
)
+
N
¯
(
r
,
1
f
-
b
)
+
S
(
r
,
f
)
≤
N
¯
(
r
,
1
F
-
a
)
+
N
¯
(
r
,
1
F
-
b
)
+
S
(
r
,
f
)
≤
2
T
(
r
,
F
)
+
S
(
r
,
f
)
.
Similarly, we have T
(
r
,
F
)
≤
2
T
(
r
,
f
)
+
S
(
r
,
F
). Therefore, S
(
r
,
f
)
=
S
(
r
,
F
).Set(9)
f
1
(
z
)
=
f
(
z
)
-
a
(
z
)
b
(
z
)
-
a
(
z
)
,
F
1
(
z
)
=
F
(
z
)
-
a
(
z
)
b
(
z
)
-
a
(
z
)
.
Thus f
1
(
z
), F
1
(
z
) share 0, 1, and ∞ CM almost.Obviously, we have(10)
T
(
r
,
f
1
)
=
T
(
r
,
f
)
+
S
(
r
,
f
)
,
T
(
r
,
F
1
)
=
T
(
r
,
F
)
+
S
(
r
,
f
)
,
S
(
r
,
F
)
=
S
(
r
,
F
1
)
=
S
(
r
,
f
1
)
=
S
(
r
,
f
)
.
By Nevanlinna’s second fundamental theorem, we have
(11)
T
(
r
,
f
1
)
≤
N
(
r
,
1
f
1
)
+
N
(
r
,
1
f
1
-
1
)
+
N
(
r
,
f
1
)
+
S
(
r
,
f
1
)
≤
N
(
r
,
0
)
+
N
(
r
,
1
)
+
S
(
r
,
f
)
≤
N
(
r
,
1
F
1
-
f
1
)
+
S
(
r
,
f
)
≤
T
(
r
,
F
1
-
f
1
)
+
S
(
r
,
f
)
≤
T
(
r
,
F
-
f
)
+
S
(
r
,
f
)
≤
m
(
r
,
F
-
f
)
+
S
(
r
,
f
)
.SinceF
-
f
=
m
1
f
(
z
+
c
1
)
+
m
2
f
(
z
+
c
2
)
+
⋯
+
m
n
f
(
z
+
c
n
)
-
f
(
z
)
=
f
(
z
)
[
m
1
(
f
(
z
+
c
1
)
/
f
(
z
)
)
+  m
2
(
f
(
z
+
c
2
)
/
f
(
z
)
)
+
⋯
+
m
n
(
f
(
z
+
c
n
)
/
f
(
z
)
)
-
1
], thus
(12)
m
(
r
,
F
-
f
)
≤
m
(
r
,
f
)
+
m
(
r
,
m
1
f
(
z
+
c
1
)
f
(
z
)
+
⋯
+
m
n
f
(
z
+
c
n
)
f
(
z
)
-
1
)
≤
m
(
r
,
f
)
+
S
(
r
,
f
)
.
By (11), we have
(13)
T
(
r
,
f
1
)
≤
N
(
r
,
0
)
+
N
(
r
,
1
)
+
S
(
r
,
f
)
≤
m
(
r
,
f
)
+
S
(
r
,
f
)
≤
T
(
r
,
f
)
+
S
(
r
,
f
)
=
T
(
r
,
f
1
)
+
S
(
r
,
f
)
.
It follows that
(14)
N
(
r
,
0
)
+
N
(
r
,
1
)
=
T
(
r
,
f
1
)
+
S
(
r
,
f
)
.
On the other hand, by Nevanlinna first fundamental theorem, we have
(15)
2
T
(
r
,
f
1
)
=
T
(
r
,
1
f
1
)
+
T
(
r
,
1
f
1
-
1
)
+
S
(
r
,
f
)
≤
N
(
r
,
0
)
+
N
(
r
,
1
)
+
m
(
r
,
1
f
1
)
+
m
(
r
,
1
f
1
-
1
)
+
S
(
r
,
f
)
≤
T
(
r
,
f
1
)
+
m
(
r
,
1
f
1
)
+
m
(
r
,
1
f
1
-
1
)
+
S
(
r
,
f
)
.
So we get
(16)
T
(
r
,
f
1
)
≤
m
(
r
,
1
f
1
)
+
m
(
r
,
1
f
1
-
1
)
+
S
(
r
,
f
)
≤
m
(
r
,
1
f
-
a
)
+
m
(
r
,
1
f
-
b
)
+
S
(
r
,
f
)
.Set(17)
a
1
(
z
)
=
m
1
a
(
z
+
c
1
)
+
m
2
a
(
z
+
c
2
)
+
⋯
+
m
n
a
(
z
+
c
n
)
,
b
1
(
z
)
=
m
1
b
(
z
+
c
1
)
+
m
2
b
(
z
+
c
2
)
+
⋯
+
m
n
b
(
z
+
c
n
)
.Ifa
1
(
z
)
≡
b
1
(
z
), we can deduce by (16) that
(18)
T
(
r
,
f
1
)
≤
m
(
r
,
1
f
-
a
+
1
f
-
b
)
+
S
(
r
,
f
)
≤
m
(
r
,
F
-
a
1
f
-
a
+
F
-
b
1
f
-
b
)
+
m
(
r
,
1
F
-
a
1
)
+
S
(
r
,
f
)
≤
T
(
r
,
F
)
+
S
(
r
,
f
)
≤
T
(
r
,
m
1
f
(
z
+
c
1
)
+
m
2
f
(
z
+
c
2
)
(
z
+
c
n
)
z
+
c
1
+
m
2
+
⋯
+
m
n
f
(
z
+
c
n
)
)
+
S
(
r
,
f
)
=
m
(
r
,
m
1
f
(
z
+
c
1
)
+
m
2
f
(
z
+
c
2
)
+
⋯
+
m
n
f
(
z
+
c
n
)
f
(
z
+
c
2
)
+
⋯
+
m
n
f
(
z
+
c
n
)
)
+
S
(
r
,
f
)
≤
m
(
r
,
f
)
+
S
(
r
,
f
)
≤
T
(
r
,
f
)
+
S
(
r
,
f
)
=
T
(
r
,
f
1
)
+
S
(
r
,
f
)
.Ifa
1
(
z
)
≢
b
1
(
z
), set
(19)
L
(
F
)
=
|
F
a
1
b
1
F
′
a
1
′
b
1
′
F
′′
a
1
′′
b
1
′′
|
.Then we have(20)
m
(
r
,
F
-
a
1
f
-
a
)
=
m
(
r
,
F
-
b
1
f
-
b
)
=
S
(
r
,
f
)
,
m
(
r
,
L
(
F
)
F
-
a
1
)
=
m
(
r
,
L
(
F
)
F
-
b
1
)
=
S
(
r
,
f
)
.It followed from (16) that
(21)
T
(
r
,
f
1
)
≤
m
(
r
,
F
-
a
1
f
-
a
)
+
m
(
r
,
1
F
-
a
1
)
+
m
(
r
,
F
-
b
1
f
-
b
)
+
m
(
r
,
1
F
-
b
1
)
+
S
(
r
,
f
)
≤
m
(
r
,
1
F
-
a
1
)
+
m
(
r
,
1
F
-
b
1
)
+
S
(
r
,
f
)
≤
m
(
r
,
1
F
-
a
1
+
1
F
-
b
1
)
+
S
(
r
,
f
)
≤
m
(
r
,
1
L
(
F
)
)
+
S
(
r
,
f
)
≤
T
(
r
,
L
(
F
)
)
+
S
(
r
,
f
)
≤
T
(
r
,
F
)
+
S
(
r
,
f
)
≤
T
(
r
,
m
1
f
(
z
+
c
1
)
+
m
2
f
(
z
+
c
2
)
(
z
+
c
n
)
+
⋯
+
m
n
f
(
z
+
c
n
)
)
+
S
(
r
,
f
)
=
m
(
r
,
m
1
f
(
z
+
c
1
)
+
m
2
f
(
z
+
c
2
)
+
⋯
+
m
n
f
(
z
+
c
n
)
)
+
S
(
r
,
f
)
≤
m
(
r
,
f
)
+
S
(
r
,
f
)
≤
T
(
r
,
f
)
+
S
(
r
,
f
)
=
T
(
r
,
f
1
)
+
S
(
r
,
f
)
.By (18) and (21), we can deduce that
(22)
T
(
r
,
f
1
)
=
T
(
r
,
F
)
+
S
(
r
,
f
)
=
T
(
r
,
F
1
)
+
S
(
r
,
f
)
.It follows from (14) and (22) that
(23)
lim
-
r
→
∞
r
∈
I
N
(
r
,
0
)
+
N
(
r
,
1
)
T
(
r
,
f
1
)
+
T
(
r
,
F
1
)
=
1
2
<
2
3
.By Lemma6, we have
(24)
f
1
(
z
)
=
A
F
1
(
z
)
+
B
C
F
1
(
z
)
+
D
,
where A, B, C, and D are complex numbers satisfying A
D
-
B
C
≠
0.Now, we consider three cases.Case 1.
ConsiderN
(
r
,
0
)
=
S
(
r
,
f
1
). Thus
(25)
N
(
r
,
1
f
1
)
+
N
(
r
,
f
1
)
=
S
(
r
,
f
1
)
=
S
(
r
,
f
)
.Similarly, we have(26)
N
(
r
,
1
F
1
)
+
N
(
r
,
F
1
)
=
S
(
r
,
F
1
)
=
S
(
r
,
f
)
.By Lemma5, we get that either f
1
≡
F
1 or f
1
F
1
≡
1.Iff
1
≡
F
1, we can easily deduce that f
≡
F, which is a contradiction with our assumption.Iff
1
F
1
≡
1, that is
(27)
(
f
(
z
)
-
a
)
(
F
(
z
)
-
a
)
≡
(
b
-
a
)
2
,
then we have
(28)
(
f
-
a
)
2
=
(
b
-
a
)
2
(
F
-
a
)
/
(
f
-
a
)
.From (28), we have
(29)
2
T
(
r
,
f
)
≤
T
(
r
,
(
f
-
a
)
2
)
+
S
(
r
,
f
)
=
T
(
r
,
1
(
b
-
a
)
2
/
(
(
F
-
a
)
/
(
f
-
a
)
)
)
+
S
(
r
,
f
)
≤
T
(
r
,
F
-
a
f
-
a
)
+
S
(
r
,
f
)
=
N
(
r
,
F
-
a
f
-
a
)
+
m
(
r
,
F
-
a
f
-
a
)
+
S
(
r
,
f
)
≤
m
(
r
,
(
F
-
a
1
)
+
(
a
1
-
a
)
f
-
a
)
+
S
(
r
,
f
)
≤
m
(
r
,
a
1
-
a
f
-
a
)
+
S
(
r
,
f
)
≤
T
(
r
,
f
)
+
S
(
r
,
f
)
.It follows thatT
(
r
,
f
)
≤
S
(
r
,
f
), a contradiction.Case 2.
ConsiderN
(
r
,
1
)
=
S
(
r
,
f
1
). Using the same argument as used in Case 1, we deduce that T
(
r
,
f
)
≤
S
(
r
,
f
), a contradiction.Case 3.
ConsiderN
(
r
,
0
)
≠
S
(
r
,
f
1
), N
(
r
,
1
)
≠
S
(
r
,
f
1
). Since f
1 and F
1 share 0, 1 CM almost, we deduce from (24) that
(30)
f
1
(
z
)
=
(
C
+
D
)
F
1
(
z
)
C
F
1
(
z
)
+
D
.IfC
=
0, then f
1
≡
F
1; that is, f
≡
F, a contradiction.HenceC
≠
0. Thus we have
(31)
N
(
r
,
1
F
1
+
(
D
/
C
)
)
=
N
(
r
,
f
1
)
=
S
(
r
,
f
1
)
=
S
(
r
,
f
)
.Obviously,D
/
C
≠
0, D
/
C
≠
-
1. Thus by Nevanlinna second fundamental theorem and (14), we get
(32)
2
T
(
r
,
f
1
)
=
2
T
(
r
,
F
1
)
+
S
(
r
,
f
1
)
≤
N
¯
(
r
,
1
F
1
)
+
N
¯
(
r
,
1
F
1
-
1
)
+
N
¯
(
r
,
1
F
1
+
(
D
/
C
)
)
+
S
(
r
,
f
)
≤
N
(
r
,
0
)
+
N
(
r
,
1
)
+
S
(
r
,
f
)
≤
T
(
r
,
f
1
)
+
S
(
r
,
f
)
.It follows thatT
(
r
,
f
1
)
≤
S
(
r
,
f
1
), a contradiction. Thus we prove that f
(
z
)
≡
F
(
z
). This completes the proof of Theorem 1.



---

*Source: 380910-2014-02-03.xml*