# Fixed Points of Difference Operator of Meromorphic Functions

Authors: Zhaojun Wu; Hongyan Xu
Journal: The Scientific World Journal
 (2014)

---

## Abstract

Letf be a transcendental meromorphic function of order less than one. The authors prove that the exact difference Δ
f
=
f
z
+
1
-
f
z has infinitely many fixed points, if a
∈
ℂ and ∞ are Borel exceptional values (or Nevanlinna deficiency values) of f. These results extend the related results obtained by Chen and Shon.


---

## Body

## 1. Introduction and Main Results

In this paper, we assume that the reader is familiar with the notations of frequency use in Nevanlinna theory (see [1–3]). Let f
(
z
) be a meromorphic function in the complex plane ℂ and a
∈
ℂ. We use the notations σ
(
f
) to denote the order of f
(
z
), λ
(
f
,
a
), and λ
(
1
/
f
), respectively, to denote the exponent of convergence of zeros of f
(
z
)
-
a and poles of f
(
z
). Especially, if a
=
0, we denote λ
(
f
,
0
)
=
λ
(
f
). A point z
∈
ℂ is called as a fixed point of f
(
z
) if f
(
z
)
=
z. There is a considerable number of results on the fixed points for meromorphic functions in the plane; we refer the reader to Chuang and Yang [4]. It follows Chen and Shon [5]; we use the notation τ
(
f
) to denote the exponent of convergence of fixed points of f that is defined as
(1)
τ
(
f
)
=
limsup
⁡
r
→
∞
log
⁡
⁡
N
(
r
,
1
/
(
f
-
z
)
)
log
⁡
⁡
r
.Letf be a transcendental meromorphic function in the complex plane ℂ. The exact differences Δ
f are defined by Δ
f
=
f
(
z
+
1
)
-
f
(
z
).Recently, there are a number of papers (including [6–16]) focusing on the differences analogues of Nevanlinna's theory and its application on the complex difference equations. For the fixed points of the difference operator Δ
f, Chen and Shon have proved the following.Theorem A (see [17]).
Letf be a transcendental entire function of order of growth σ
(
f
)
=
1 and have infinitely many zeros with the exponent of convergence of zeros λ
(
f
)
<
1. Then Δ
f has infinitely many zeros and infinitely many fixed points.When the order off is less than 1, Chen and Shon have proved the following.Theorem B (see [5]).
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
≤
1. Suppose that f satisfies λ
(
1
/
f
)
<
λ
(
f
)
<
1 or has infinitely many zeros (with λ
(
f
)
=
0) and finitely many poles. Then Δ
f has infinitely many fixed points and satisfies the exponent of convergence of fixed points τ
(
Δ
f
)
=
σ
(
f
).A natural question is, lettingf be a transcendental meromorphic function of order of growth σ
(
f
)
<
1, is there a similar result as that in Theorem B if λ
(
1
/
f
)
≥
λ
(
f
) or f has infinitely many zeros (with λ
(
f
)
=
0) and infinitely many poles?In this paper, we will prove the following theorem to answer the question.Theorem 1 (main).
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
<
1 and a
∈
ℂ. Suppose that f satisfies λ
(
1
/
f
)
<
σ
(
f
) and λ
(
f
,
a
)
<
σ
(
f
). Then Δ
f has infinitely many fixed points and satisfies the exponent of convergence of fixed points τ
(
Δ
f
)
=
σ
(
f
).From Theorem1, we can get the following corollary.Corollary 2.
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
<
1. Suppose that f satisfies λ
(
f
)
≤
λ
(
1
/
f
)
<
σ
(
f
). Then Δ
f has infinitely many fixed points and satisfies the exponent of convergence of fixed points τ
(
Δ
f
)
=
σ
(
f
).In Theorem1, we suppose that f satisfies λ
(
1
/
f
)
<
σ
(
f
) and λ
(
f
,
a
)
<
σ
(
f
). That is to say ∞ and a are Borel exceptional values of f. If we suppose that ∞ and a are Nevanlinna deficiency values of f, is there a similar result as that in Theorem B? In the following, we give Theorem 3 to answer this question.Letf
(
z
) be a meromorphic function in the complex plane ℂ and a
∈
ℂ
∞
=
ℂ
∪
{
∞
}. Nevanlinna’s deficiency of f with respect to a is defined by
(2)
δ
(
a
,
f
)
=
1
-
limsup
⁡
r
→
∞
N
(
r
,
1
/
(
f
-
a
)
)
T
(
r
,
f
)
.Ifa
=
∞, then one should replace N
(
r
,
1
/
(
f
-
a
)
) in the above formula by N
(
r
,
f
). If δ
(
a
,
f
)
>
0, then a is called a Nevanlinna deficiency value of f.Theorem 3 (main).
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
<
1 and a
∈
ℂ. Suppose that f satisfies δ
(
∞
,
f
)
=
1 and a is a Nevanlinna deficiency value of f. Then Δ
f has infinitely many fixed points.Corollary 4.
Letf be a transcendental entire function of order of growth σ
(
f
)
<
1 and a
∈
ℂ. Suppose that δ
(
a
,
f
)
>
0. Then Δ
f has infinitely many fixed points.


## 2.    Some Lemmas

Lemma 1 (lemma on the logarithmic derivative).
Letf
(
z
) be a meromorphic function. If the function f
(
z
) has finite order, then
(3)
m
(
r
,
f
(
k
)
f
)
=
O
(
log
⁡
r
)
holds for any positive integer k.Lemma 2 (see [18]).
Letf
(
z
) be a meromorphic function with the exponent of convergence of poles λ
(
1
/
f
)
=
λ
<
+
∞ and let c be a nonzero complex number. Then for each ɛ
>
0, we have
(4)
N
(
r
,
f
(
z
+
c
)
)
=
N
(
r
,
f
)
+
O
(
r
λ
-
1
+
ɛ
)
+
O
(
log
⁡
r
)
.Lemma 3.
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
<
1 and let c be a nonzero complex number. Then
(5)
N
(
r
,
f
(
z
+
c
)
)
=
N
(
r
,
f
)
+
O
(
log
⁡
r
)
.Proof.
Since the orderσ
(
f
)
≔
σ
<
1, then λ
(
1
/
f
)
=
λ
≤
σ
<
1. Therefore, for any 0
<
ɛ
<
1
-
σ, it follows from Lemma 2 that
(6)
N
(
r
,
f
(
z
+
c
)
)
=
N
(
r
,
f
)
+
O
(
r
λ
-
1
+
ɛ
)
+
O
(
log
⁡
⁡
r
)
=
N
(
r
,
f
)
+
O
(
1
)
+
O
(
log
⁡
⁡
r
)
.
That is,
(7)
N
(
r
,
f
(
z
+
c
)
)
=
N
(
r
,
f
)
+
O
(
log
⁡
r
)
.Lemma 4 (see [6]).
Letf be a function transcendental and meromorphic in the plane which satisfies
(8)
liminf
⁡
r
→
∞
T
(
r
,
f
)
r
=
0
.
Then Δ
f is transcendental.Lemma 5.
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
=
σ
<
1. Then Δ
f is transcendental.Proof.
Since the orderσ
(
f
)
≔
σ
<
1, then, for any positive ɛ
(
0
<
ɛ
<
1
-
σ
), there exists R
>
0 such that for any r
>
R we have
(9)
T
(
r
,
f
)
≤
r
σ
+
ɛ
.
Therefore,(10)
liminf
⁡
r
→
∞
T
(
r
,
f
)
r
=
0
.
Lemma 5 follows Lemma 4.Lemma 6 (see [7]).
Letf
(
z
) be a meromorphic function of finite order, then σ
(
Δ
f
)
≤
σ
(
f
).Lemma 7 (see [7]).
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
<
1. Then for any ɛ
>
0 and any positive integer k, there exists a set E
⊂
(
1
,
∞
) that depends on f and has finite logarithmic measure, such that for all z satisfying |
z
|
=
r
∉
E
∪
[
0,1
] we have
(11)
Δ
k
f

(
z
)
f

(
z
)
=
f
(
k
)

(
z
)
f

(
z
)
+
O
(
r
(
k
+
1
)
(
σ
-
1
)
+
ɛ
)
.It is easy to derive the following lemma from Lemma1 and Lemma 7.Lemma 8.
Letf be a transcendental meromorphic function of order of growth σ
(
f
)
<
1. Then for any positive integer k there exists a set E
⊂
(
1
,
∞
) that depends on f and has finite logarithmic measure, such that
(12)
m
(
r
,
Δ
k
f
(
z
)
f
(
z
)
)
=
O
(
log
⁡
⁡
r
)
,
r
∉
E
.


## 3.  Proof of Theorems

Proof.
Since(13)
1
f
=
Δ
f
z
f
-
z
Δ
2
f
-
Δ
f
z
f
Δ
f
-
z
z
Δ
2
f
-
Δ
f
,
then(14)
m
(
r
,
1
f
)
≤
m
(
r
,
Δ
f
z
f
)
+
m
(
r
,
z
Δ
2
f
-
Δ
f
z
f
)
+
m
(
r
,
Δ
f
-
z
z
Δ
2
f
-
Δ
f
)
+
O
(
1
)
≤
2
m
(
r
,
Δ
f
f
)
+
m
(
r
,
Δ
2
f
f
)
+
m
(
r
,
Δ
f
-
z
z
Δ
2
f
-
Δ
f
)
+
O
(
log
⁡
⁡
r
)
.
Applying the first fundamental theorem, we get(15)
m
(
r
,
1
f
)
=
T
(
r
,
f
)
-
N
(
r
,
1
f
)
+
O
(
1
)
,
m
(
r
,
Δ
f
-
z
z
Δ
2
f
-
Δ
f
)
=
m
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
+
N
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
-
N
(
r
,
Δ
f
-
z
z
Δ
2
f
-
Δ
f
)
+
O
(
1
)
≤
m
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
+
N
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
+
O
(
1
)
.
Combining (14)-(15) we have
(16)
T
(
r
,
f
)
≤
N
(
r
,
1
f
)
+
2
m
(
r
,
Δ
f
f
)
+
m
(
r
,
Δ
2
f
f
)
+
m
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
+
N
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
+
O
(
log
⁡
⁡
r
)
≤
N
(
r
,
1
f
)
+
N
(
r
,
1
Δ
f
-
z
)
+
N
(
r
,
z
Δ
2
f
-
Δ
f
)
+
2
m
(
r
,
Δ
f
f
)
+
m
(
r
,
Δ
2
f
f
)
+
m
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
+
O
(
log
⁡
⁡
r
)
.
Since(17)
Δ
2
f
=
Δ
(
f
(
z
+
1
)
-
f
(
z
)
)
=
f
(
z
+
2
)
-
2
f
(
z
+
1
)
+
f
(
z
)
,
Δ
(
Δ
f
-
z
)
=
Δ
(
f
(
z
+
1
)
-
f
(
z
)
-
z
)
=
f
(
z
+
2
)
-
2
f
(
z
+
1
)
+
f
(
z
)
-
1
,
then, we can get(18)
z
Δ
2
f
-
Δ
f
=
z
f
(
z
+
2
)
-
2
z
f
(
z
+
1
)
+
z
f
(
z
)
z
Δ
2
f
-
Δ
f
=
-
f
(
z
+
1
)
+
f
(
z
)
.
z
Δ
(
Δ
f
-
z
)
-
(
Δ
f
-
z
)
=
z
f
(
z
+
2
)
-
2
z
f
(
z
+
1
)
z
Δ
(
Δ
f
-
z
)
-
(
Δ
f
-
z
)
=
+
z
f
(
z
)
-
f
(
z
+
1
)
+
f
(
z
)
.
Therefore,(19)
z
Δ
2
f
-
Δ
f
Δ
f
-
z
=
z
Δ
(
Δ
f
-
z
)
-
(
Δ
f
-
z
)
Δ
f
-
z
=
z
Δ
(
Δ
f
-
z
)
Δ
f
-
z
-
1
,
(20)
N
(
r
,
z
Δ
2
f
-
Δ
f
)
≤
N
(
r
,
f
(
z
+
2
)
)
+
N
(
r
,
f
(
z
+
1
)
)
+
N
(
r
,
f
(
z
)
)
.
Thus from Lemma3 and (20), we deduce
(21)
N
(
r
,
z
Δ
2
f
-
Δ
f
)
≤
3
N
(
r
,
f
(
z
)
)
+
O
(
log
⁡
⁡
r
)
.
By Lemmas5 and 6, we know that Δ
f
-
z is a transcendental meromorphic function of order of growth σ
(
Δ
f
-
z
)
≤
σ
(
f
)
<
1. It follows from Lemma 8 and (19) that there exists a set E
⊂
(
1
,
∞
) that has finite logarithmic measure, such that for any r
∉
E we have
(22)
m
(
r
,
Δ
f
f
)
=
O
(
log
⁡
⁡
r
)
,
m
(
r
,
Δ
2
f
f
)
=
O
(
log
⁡
⁡
r
)
,
m
(
r
,
z
Δ
2
f
-
Δ
f
Δ
f
-
z
)
=
O
(
log
⁡
⁡
r
)
.
From (16) and (21)-(22), we have
(23)
T
(
r
,
f
)
≤
3
N
(
r
,
f
)
+
N
(
r
,
1
f
)
+
N
(
r
,
1
Δ
f
-
z
)
+
O
(
log
⁡
⁡
r
)
,
r
∉
E
.
Denotingg
≡
f
-
a by (23) we derive,
(24)
T
(
r
,
f
)
≤
T
(
r
,
g
)
+
O
(
1
)
≤
3
N
(
r
,
g
)
+
N
(
r
,
1
g
)
+
N
(
r
,
1
Δ
g
-
z
)
+
O
(
log
⁡
⁡
r
)
≤
3
N
(
r
,
f
)
+
N
(
r
,
1
f
-
a
)
+
N
(
r
,
1
Δ
f
-
z
)
+
O
(
log
⁡
⁡
r
)
,
r
∉
E
.


### 3.1.  The Rest of the Proof of Theorem1

By Lemma6, we know that τ
(
Δ
f
)
≤
σ
(
f
). If τ
(
Δ
f
)
<
σ
(
f
), by λ
(
1
/
f
)
<
σ
(
f
) and λ
(
f
,
a
)
<
σ
(
f
), there exists a number η
<
σ
(
f
), such that for any sufficient  r we have
(25)
N
(
r
,
f
)
<
r
η
,
N
(
r
,
1
f
-
a
)
<
r
η
,
N
(
r
,
1
Δ
f
-
z
)
<
r
η
.Combining (24) and (25), we can get a contradiction. Therefore, we have τ
(
Δ
f
)
=
σ
(
f
).


### 3.2.  The Rest of the Proof of Theorem3

Sinceδ
(
∞
,
f
)
=
1, then N
(
r
,
f
)
=
o
(
T
(
r
,
f
)
). By (24), we can get
(26)
(
1
-
o
(
1
)
)
T
(
r
,
f
)
≤
N
(
r
,
1
f
-
a
)
+
N
(
r
,
1
Δ
f
-
z
)
+
O
(
log
⁡
⁡
r
)
,
r
∉
E
.Sinceδ
(
a
,
f
)
>
0, then there is a positive number θ
<
1 such that
(27)
N
(
r
,
1
f
-
a
)
<
θ
T
(
r
,
f
)
.IfΔ
f has only a finite number of fixed points, then from (26) and (27) we would have
(28)
(
1
-
o
(
1
)
-
θ
)
T
(
r
,
f
)
≤
O
(
log
⁡
r
)
,
r
∉
E
.This contradictsf being transcendental. Therefore, Δ
f has infinitely many fixed points.


## 3.1.  The Rest of the Proof of Theorem1

By Lemma6, we know that τ
(
Δ
f
)
≤
σ
(
f
). If τ
(
Δ
f
)
<
σ
(
f
), by λ
(
1
/
f
)
<
σ
(
f
) and λ
(
f
,
a
)
<
σ
(
f
), there exists a number η
<
σ
(
f
), such that for any sufficient  r we have
(25)
N
(
r
,
f
)
<
r
η
,
N
(
r
,
1
f
-
a
)
<
r
η
,
N
(
r
,
1
Δ
f
-
z
)
<
r
η
.Combining (24) and (25), we can get a contradiction. Therefore, we have τ
(
Δ
f
)
=
σ
(
f
).


## 3.2.  The Rest of the Proof of Theorem3

Sinceδ
(
∞
,
f
)
=
1, then N
(
r
,
f
)
=
o
(
T
(
r
,
f
)
). By (24), we can get
(26)
(
1
-
o
(
1
)
)
T
(
r
,
f
)
≤
N
(
r
,
1
f
-
a
)
+
N
(
r
,
1
Δ
f
-
z
)
+
O
(
log
⁡
⁡
r
)
,
r
∉
E
.Sinceδ
(
a
,
f
)
>
0, then there is a positive number θ
<
1 such that
(27)
N
(
r
,
1
f
-
a
)
<
θ
T
(
r
,
f
)
.IfΔ
f has only a finite number of fixed points, then from (26) and (27) we would have
(28)
(
1
-
o
(
1
)
-
θ
)
T
(
r
,
f
)
≤
O
(
log
⁡
r
)
,
r
∉
E
.This contradictsf being transcendental. Therefore, Δ
f has infinitely many fixed points.



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*Source: 103249-2014-01-19.xml*