# Growth of Meromorphic Function Sharing Functions and Some Uniqueness Problems

Authors: Jianming Qi; Fanning Meng; Wenjun Yuan
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

Estimating the growth of meromorphic solutions has been an important topic of research in complex differential equations. In this paper, we devoted to considering uniqueness problems by estimating the growth of meromorphic functions. Further, some examples are given to show that the conclusions are meaningful.


---

## Body

## 1. Introduction and Main Results

Assuming that the reader is familiar with the notations and results on Nevanlinna theory [1] and the applications of normal family theory on estimating the growth of meromorphic functions (see [2–4]), it is an interesting attempt to consider the growth properties of meromorphic functions under the condition involved sharing value or some complex differential (or difference) equations (see [5–9]).For a meromorphic functionf, the order ρ
(
f
) and hyperorder σ
(
f
) of f are defined as follows [1]:(1)
ρ
f
≔
l
i
m
s
u
p
r
→
∞
log
⁡
T
r
,
f
log
⁡
r
,
σ
f
≔
l
i
m
s
u
p
r
→
∞
log
⁡
log
⁡
T
r
,
f
log
⁡
r
.Letf
(
z
) and g
(
z
) be two nonconstant meromorphic functions in the complex plane C, and let α
(
z
) be a meromorphic function or a finite complex number. If g
(
z
)
-
α
(
z
)
=
0 whenever f
(
z
)
-
α
(
z
)
=
0, we write f
(
z
)
=
α
(
z
)
⇒
g
(
z
)
=
α
(
z
). If f
(
z
)
=
α
(
z
)
⇒
g
(
z
)
=
α
(
z
) and g
(
z
)
=
α
(
z
)
⇒
f
(
z
)
=
α
(
z
), we write f
(
z
)
=
α
(
z
)
⇔
g
(
z
)
=
α
(
z
) and say that f
(
z
) and g
(
z
) share α
(
z
) IM (ignoring multiplicity). If g
(
z
)
-
α
(
z
)
=
0 whenever f
(
z
)
-
α
(
z
)
=
0 and the multiplicity of the zero z
0 of g
-
α is greater than or equal to that of the zero z
0 of f
-
α, then we denote this condition by f
(
z
)
-
α
(
z
)
=
0
↦
g
(
z
)
-
α
(
z
)
=
0. Let R be a rational function which behaves asymptotically c
r
β as r
→
∞, where c
≠
0
,
β are constants. The degree of R at infinity is defined as deg
⁡
R
≔
deg
∞
⁡
R
≔
max
⁡
{
0
,
β
}.In the following, for a linear differential polynomial off, we write(2)
L
f
≔
a
k
f
k
+
a
k
-
1
f
k
-
1
+
⋯
+
a
2
f
′
′
+
a
1
f
′
+
a
0
f
,where a
1
,
a
2
,
…
,
a
k
(
≠
0
) are constants and k
≥
2 is an integer.In 1986, Jank et al. [10] proved that, for an entire function f, if f and f
′ share a finite nonzero value a IM and if f
(
z
)
=
a
⇒
f
′
(
z
)
=
a, then f
≡
f
′. In 2006, Wang [11] replaced the value a by a polynomial Q
≢
0 and obtained the following result: let f be a nonconstant entire function, let Q be a polynomial of degree q
≥
1, and let k
>
q be an integer. If f and f
′ share Q CM and if f
(
z
)
=
Q
(
z
)
⇒
f
(
k
)
(
z
)
=
Q
(
z
), then f
≡
f
′.In 2010, Lü and Yi [12] obtained the following results.Theorem 1.
Letf be a nonconstant meromorphic function with finitely many poles, and let R be a nonzero rational function. If (3)
f
z
=
R
z
⟹
f
′
z
=
R
z
,
f
′
z
=
R
z
⟹
L
f
z
≤
M
+
o
1
R
z
,where M is a positive number, then f is of order at most 1.Theorem 2.
Letf be a nonconstant meromorphic function with finitely many poles, and let R be a nonzero rational function such that f and R have no common poles. If (4)
f
z
=
R
z
⟹
f
′
z
=
R
z
,
f
′
z
=
R
z
⟹
L
f
z
=
R
z
,
f
′
z
-
R
z
=
0
⟼
f
z
-
R
z
=
0
,then one of the following three cases holds:(i)
f
(
z
)
=
b
e
z and ∑
n
=
0
k
a
n
=
1, where b is a nonzero constant;
(ii)
R reduces to a constant, say a, and a
0
+
a
1
+
∑
n
=
2
k
‍
a
n
c
n
-
1
=
1 for some nonzero constant c such that (5)
f
z
=
b
e
c
z
-
a
1
-
c
c
,where b is a nonzero constant;
(iii)
R is a nonconstant polynomial with deg
⁡
R
=
l
≤
k
-
2, f
(
z
)
=
λ
e
c
z
+
Q
(
z
), and (6)
a
0
+
a
1
=
1
,
a
2
=
a
3
=
⋯
=
a
l
=
0
,
∑
n
=
l
+
1
k
a
n
c
n
-
l
+
1
=
0
,where λ
,
c are two nonzero constants and Q is a polynomial such that c
Q
-
Q
′
=
(
c
-
1
)
R
.Problem 3.
In Theorem1, we see that f, f
′, and L
(
f
) share one function with zero order. So it is natural to ask what will happen if they share a function of infinite order or positive finite order?Considering Problem3, we derive the following results.Theorem 4 (main theorem).
LetR be a nonzero rational function and let f
,
γ be two entire functions. Let k
(
≥
2
) be an integer and let L
(
f
) be defined as (2). If (7)
f
z
=
α
z
⟹
f
′
z
=
α
z
,
f
′
z
=
α
z
⟹
L
f
z
≤
M
+
o
1
α
z
,where α
=
R
e
γ
(
α
≢
α
′
), and if α
-
α
′ has at most finitely many zeros, then σ
(
f
)
≤
σ
(
α
)
=
ρ
(
γ
).The following examples show that our conclusionσ
(
f
)
≤
ρ
(
γ
) really exists and is sharp.Example 5.
Letf
(
z
)
=
A
e
z, where A is a nonzero constant. Let α
(
z
)
=
e
e
-
z
+
z. Noting that f
≡
f
′
≡
f
(
k
)
(
k
≥
2
), we have (8)
f
z
=
α
z
⟹
f
′
z
=
α
z
,
f
′
z
=
α
z
⟹
f
k
z
=
α
z
.Obviously, α
(
z
)
-
α
′
(
z
)
=
e
e
-
z has no zeros. Thus it satisfies the assumptions of Theorem 4 and σ
(
f
)
=
0
<
σ
(
α
)
=
1.Example 6.
Letf
(
z
)
=
2
e
z and α
(
z
)
=
(
4
z
2
-
z
+
2
)
e
z
2. Noting that f
≡
f
′
≡
f
(
k
), then (9)
f
z
=
α
z
⟹
f
′
z
=
α
z
,
f
′
z
=
α
z
⟹
f
k
z
=
α
z
. It satisfies the assumptions of Theorem 4 and σ
(
f
)
=
0.Example 7.
Letf
(
z
)
=
4
z
2
-
8
z
+
8, α
(
z
)
=
2
z
2. Noting that f
≢
f
′
≢
f
′
′, then f
(
z
)
-
α
(
z
)
=
2
(
z
-
2
)
2, f
′
(
z
)
-
α
(
z
)
=
-
2
(
z
-
2
)
2, and f
′
′
(
z
)
-
α
(
z
)
=
2
(
2
-
z
)
(
2
+
z
). It is easy to see f
(
z
)
=
α
(
z
)
⇒
f
′
(
z
)
=
α
(
z
) and f
′
(
z
)
=
α
(
z
)
⇒
f
′
′
(
z
)
=
α
(
z
) and σ
(
f
)
=
σ
(
α
). It satisfies the assumptions of Theorem 4 and σ
(
f
)
=
σ
(
α
)
=
0.Example 8.
Letf
(
z
)
=
z
4
A
e
z
+
z
4
+
8
z
3
+
24
z
2
+
48
z
+
48 and α
(
z
)
=
z
4
+
8
z
3
+
24
z
2
+
48
z
+
48, where A
=
e
4 is a constant. Differentiating f twice yields f
′
(
z
)
=
(
z
4
+
4
z
3
)
A
e
z
+
4
z
3
+
24
z
2
+
48
z
+
48 and f
′
′
(
z
)
=
(
z
4
+
8
z
3
+
12
z
2
)
A
e
z
+
12
z
2
+
48
z
+
48; then f
(
z
)
=
α
(
z
)
⇒
f
′
(
z
)
=
α
(
z
)
and
f
′
(
z
)
=
α
(
z
)
⇒
f
′
′
(
z
)
=
α
(
z
). Thus σ
(
f
)
≤
σ
(
α
), but f
≢
f
′.Theorem 9 (main theorem).
Letf be a meromorphic function with at most finitely many poles, and let α
=
R
e
Q
(
α
≢
α
′
), where R
(
≢
0
) is a rational function and Q is a nonconstant polynomial. Let k
(
≥
2
) be an integer and let L
(
f
) be defined as (2). Then ρ
(
f
)
≤
deg
⁡
Q, if (10)
f
z
=
α
z
⟹
f
′
z
=
α
z
,
f
′
z
=
α
z
⟹
L
f
z
≤
M
+
o
1
α
z
.Example 10.
Letf
(
z
)
=
4
e
z, α
(
z
)
=
z
e
z
+
1, so ρ
(
f
)
=
1, and deg
⁡
Q
=
1. Noting that (11)
f
z
=
α
z
⟹
f
′
z
=
α
z
,
f
′
z
=
α
z
⟹
f
k
z
=
α
z
,then ρ
(
f
)
=
1
≤
deg
⁡
Q
=
1.Remark 11.
Example10 illustrates that the conclusion of Theorem 9 really occurs.Problem 12.
If the rational functionR is replaced by a function α
=
R
e
Q (here Q is a polynomial) in Theorem 2, what will happen?Investigating Problem12, we obtain the following result.Theorem 13.
Letf be a nonconstant transcendental meromorphic function with finitely many poles. Let α
=
R
e
Q (here Q is a polynomial and α
≢
α
′) be a function and let R be a nonzero rational function such that f and R have no common poles, and let L
(
f
) be defined as (2). Let ρ
(
α
)
<
ρ
(
f
). If (12)
f
z
=
α
z
⟹
f
′
z
=
α
z
,
f
′
z
=
α
z
⟹
L
f
z
=
α
z
,and f
′
(
z
)
-
α
(
z
)
=
0
↦
f
(
z
)
-
α
(
z
)
=
0, then the conclusions of Theorem 2 still hold and Q
(
z
) must be a constant.


## 2. Some Lemmas

In order to prove our theorems, we need the following lemmas.Normal families, in particular, of holomorphic functions often appear in operator theory on spaces of analytic functions; for example, see in [13, Lemma 3] and in [14, Lemma 4]. Using the same method of the famous Pang-Zalcman Lemma [15, Lemma 2] and the result of Lü et al. [8, Lemma 2.1, page 595], it is easy to obtain the following lemma. It plays an important role in the proof of Theorems 4 and 9.Lemma 14 (see [8, 15]).
Let{
f
n
} be a family of meromorphic (analytic) functions in the unit disc Δ. If a
n
→
a, a
<
1, and f
n
♯
(
a
n
)
→
∞, and if there exists A
≥
1 such that f
n
′
z
≤
A whenever f
n
(
z
)
=
0, then there exist(i)
a subsequence off
n (which we still write as {
f
n
}),
(ii)
pointsz
n
→
z
0
z
0
<
1,
(iii)
positive numbersρ
n
→
0,such that ρ
n
-
1
f
n
(
z
n
+
ρ
n
ξ
)
=
g
n
(
ξ
)
→
g
(
ξ
) locally uniformly, where g is a nonconstant meromorphic (resp., entire) function on C, such that ρ
(
g
)
≤
2 (resp., ρ
(
g
)
≤
1), g
♯
(
ξ
)
≤
g
♯
(
0
)
=
A
+
1, and (13)
ρ
n
≤
M
f
n
♯
a
n
,where M is a constant which is independent of n.
Here, as usual,g
♯
(
ξ
)
=
g
′
ξ
/
1
+
g
ξ
2 is the spherical derivative.The next lemma is an extending result obtained by Lü and Qi in [16].Lemma 15 (see [16]).
Letf be a meromorphic function of hyperorder σ
(
f
)
>
0. Then, for any ϵ
>
0, there exists a sequence z
n
→
∞ such that (14)
f
♯
z
n
>
e
z
n
d
-
ϵ for large enough n, where d
=
σ
(
f
) if σ
(
f
)
<
∞ or d is an arbitrary positive number.Lemma 16 (see [5]).
Letf
(
z
) be an entire function with ρ
(
f
)
>
1; then for each 0
<
μ
<
ρ
(
f
)
-
1, there exist points a
n
→
∞
(
n
→
∞
), such that (15)
lim
n
→
∞
⁡
f
♯
a
n
a
n
μ
=
+
∞
.Lemma 17 (see [17]).
Letg be a nonconstant entire function with order ρ
(
g
)
≤
1, let k
≥
2 be an integer, and let a be a nonzero finite value. If g
(
z
)
=
0
⇒
g
′
(
z
)
=
a and g
′
(
z
)
=
a
⇒
g
(
k
)
(
z
)
=
0, then g
(
z
)
=
a
(
z
-
z
0
), where z
0 is a constant.Lemma 18 (see [1]).
Suppose thatf
(
z
) and g
(
z
) are two nonconstant meromorphic functions in the complex plane with ρ
(
f
) and ρ
(
g
) as their orders, respectively. Then (16)
ρ
f
g
≤
max
⁡
ρ
f
,
ρ
g
,
ρ
f
+
g
≤
max
⁡
ρ
f
,
ρ
g
.The following lemma is from the proof of Theorem  2 in [18] (see pages 493–495 in [18]). It plays an important role in the proof of Theorem 13.Lemma 19 (see [18]).
Letf and α be two meromorphic functions of finite order such that both f and α have finitely many poles, f and α have no common poles, and the order of α is less than the order of f. Let F
=
f
-
α and A
=
α
-
α
′, let P be a nonzero polynomial, and let Q be a polynomial. If F is a solution of differential equation (17)
F
′
F
-
1
P
e
Q
=
A
F
,then Q reduces to a constant and P reduces a constant.


## 3. Proof of Theorem4

In the proof, we use some ideas of [8, 19–21]. The proof of Theorem 4 is as follows.Noting thatα
=
R
e
γ, thus σ
(
α
)
=
ρ
(
γ
). So we just need to obtain σ
(
f
)
≤
ρ
(
γ
).On the contrary, suppose thatσ
(
f
)
>
ρ
(
γ
). Take d such that σ
(
f
)
>
d
>
c
≔
ρ
(
γ
), and set H
≔
f
-
α. Then(I)
H
=
0
⇒
H
′
=
α
-
α
′,
(II)
H
′
=
α
-
α
′
⇒
L
H
≤
M
+
o
1
α
+
L
α.Putβ
=
α
-
α
′
=
(
R
-
R
′
-
R
γ
′
)
e
γ and φ
=
M
+
o
1
α
+
L
α
=
A
P
R
,
γ
e
γ. Here P
R
,
γ
=
R
(
k
)
+
R
γ
(
k
)
+
R
(
k
-
1
)
γ
′
+
⋯
+
R
(
γ
′
)
k is a differential polynomial about R and γ and A is a positive number. Set F
=
H
/
β. Obviously, σ
(
F
)
=
σ
(
f
). By Lemma 15, then for 0
<
ϵ
<
d
-
c
/
2, there exists a sequence w
n
→
∞ as n
→
∞ such that (18)
F
♯
w
n
>
e
w
n
σ
F
-
ϵ
>
e
w
n
d
-
ϵ
.Noting thatβ
=
α
-
α
′ has at most finitely many zeros, then there exists a positive number r such that F has no poles in D
=
{
z
:
z
>
r
}.In view ofw
n
→
∞ as n
→
∞, without loss of generality, we may assume w
n
≥
r
+
1 for all n. Define D
1
=
{
z
:
z
<
1
} and (19)
F
n
z
=
F
w
n
+
z
=
H
w
n
+
z
β
w
n
+
z
,then every F
n is analytic in D
1. Now, fix z
∈
D
1. If F
n
(
z
)
=
0, then H
(
w
n
+
z
)
=
0. It is clear from (I) that H
′
(
w
n
+
z
)
=
β
(
w
n
+
z
). Hence (for n large enough) (20)
F
n
′
z
=
H
′
w
n
+
z
β
w
n
+
z
-
H
w
n
+
z
β
w
n
+
z
β
′
w
n
+
z
β
w
n
+
z
=
1
.Also F
n
♯
(
0
)
→
∞ as n
→
∞. It follows from Marty’s criterion that (
F
n
)
n is not normal at z
=
0.Therefore, we can apply Lemma14. Choosing an appropriate subsequence of (
F
n
)
n if necessary, we may assume that there exist sequences (
z
n
)
n and (
ρ
n
)
n with z
n
<
r
<
1 and ρ
n
→
0 such that sequence (
g
n
)
n is defined by(21)
g
n
ζ
=
ρ
n
-
1
F
n
z
n
+
ρ
n
ζ
=
ρ
n
-
1
H
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
ζlocally uniformly in C, where g is a nonconstant entire function of order at most 1. Moreover, g
♯
(
ξ
)
≤
g
♯
(
0
)
=
2 for all ξ
∈
C and(22)
ρ
n
≤
M
F
n
♯
0
=
M
F
♯
w
n
≤
M
e
-
w
n
d
-
ϵfor a positive number M.We claim that(23)
ρ
n
k
-
1
H
k
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
k
ζlocally uniformly in C.Using the mathematical induction, we prove the claim as follows.From (21), we have(24)
g
n
′
ζ
=
H
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
H
w
n
+
z
n
+
ρ
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
=
H
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
ρ
n
g
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
′
ζ
.Noting thatβ
=
α
-
α
′
=
(
R
-
R
′
-
R
γ
′
)
e
γ and φ
=
α
-
α
(
k
)
=
P
R
,
γ
e
γ, we have β
′
/
β
=
R
′
+
R
γ
′
-
R
′
′
-
2
R
′
γ
′
-
R
γ
′
′
-
R
γ
′
2
/
R
-
R
′
-
R
γ
′ and ρ
(
γ
′
′
)
=
ρ
(
γ
′
)
=
ρ
(
γ
′
2
)
=
ρ
(
γ
)
=
c. In view of the definition of order, we have(25)
β
′
β
z
=
w
n
+
z
n
+
ρ
n
ζ
=
R
′
+
R
γ
′
-
R
′
′
-
2
R
′
γ
′
-
R
γ
′
′
-
R
γ
′
2
R
-
R
′
-
R
γ
′
z
=
w
n
+
z
n
+
ρ
n
ζ
≤
w
n
q
M
w
n
+
z
n
+
ρ
n
ζ
,
γ
′
≤
w
n
q
M
2
w
n
,
γ
′
≤
w
n
q
e
A
w
n
c
+
ϵ
,where A is a positive constant and q is an integer. Noting that 0
<
ϵ
<
d
-
c
/
2, we have d
-
ϵ
>
c
+
ϵ. Then, combining (22) and (25) yields(26)
H
w
n
+
z
n
+
ρ
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
=
ρ
n
g
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
≤
M
g
n
ζ
w
n
q
e
A
w
n
c
+
ϵ
-
w
n
d
-
ϵ
⟶
0
,
as
n
⟶
∞
.From (24) and (26), we deduce that(27)
H
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
′
ζlocally uniformly in C, which implies that the claim is right when k
=
1.We assume that the claim is also right whenk
=
l; that is,(28)
ρ
n
l
-
1
H
l
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
l
ζlocally uniformly in C.DefineG
n
(
ζ
)
=
ρ
n
l
-
1
H
(
l
)
(
w
n
+
z
n
+
ρ
n
ζ
)
/
β
(
w
n
+
z
n
+
ρ
n
ζ
). Then(29)
G
n
′
ζ
=
ρ
n
l
H
l
+
1
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
ρ
n
l
H
l
w
n
+
z
n
+
ρ
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
=
ρ
n
l
H
l
+
1
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
ρ
n
G
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
l
+
1
ζlocally uniformly in C. Note the fact that(30)
ρ
n
G
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
0
.We immediately derive that(31)
ρ
n
l
H
l
+
1
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
l
+
1
ζlocally uniformly in C, which finishes the proof of the claim.Furthermore, we have(32)
ρ
n
k
-
1
L
H
w
n
+
z
n
+
ρ
n
ζ
a
k
β
w
n
+
z
n
+
ρ
n
ζ
=
ρ
n
k
-
1
H
k
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
+
⋯
+
ρ
n
k
-
1
a
0
H
w
n
+
z
n
+
ρ
n
ζ
a
k
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
k
ζlocally uniformly in C.We claim that(1)
g
(
ζ
)
=
0
⇒
g
′
(
ζ
)
=
1,
(2)
g
′
(
ζ
)
=
1
⇒
g
(
k
)
(
ζ
)
=
0.Suppose thatg
(
ζ
0
)
=
0; then by Hurwitz’s theorem there exist ζ
n, ζ
n
→
ζ
0, such that (for n sufficiently large) (33)
g
n
ζ
n
=
ρ
n
-
1
H
w
n
+
z
n
+
ρ
n
ζ
n
β
w
n
+
z
n
+
ρ
n
ζ
n
=
0
.Thus H
(
w
n
+
z
n
+
ρ
n
ζ
n
)
=
0; by (
I
) we have(34)
H
′
w
n
+
z
n
+
ρ
n
ζ
n
=
β
w
n
+
z
n
+
ρ
n
ζ
n
.By (27), we derive that (35)
g
′
ζ
0
=
lim
n
→
∞
⁡
H
′
w
n
+
z
n
+
ρ
n
ζ
n
β
w
n
+
z
n
+
ρ
n
ζ
n
=
1
, which implies that g
(
ζ
)
=
0
⇒
g
′
(
ζ
)
=
1.To prove (2), suppose thatg
′
(
η
0
)
=
1. We know g
′
≢
1; otherwise g
♯
(
0
)
≤
1
<
2 is a contradiction. Hence by (27) and Hurwitz’s theorem, there exist η
n
→
η
0 such that (for n sufficiently large) (36)
H
′
w
n
+
z
n
+
ρ
n
η
n
=
β
w
n
+
z
n
+
ρ
n
η
n
. It is obvious from (II) that L
H
w
n
+
z
n
+
ρ
n
η
n
≤
A
P
R
,
γ
e
γ
w
n
+
z
n
+
ρ
n
η
n. By (23), (32), and ρ
(
γ
k
)
=
ρ
(
γ
′
)
=
ρ
(
(
γ
′
)
k
)
=
ρ
(
γ
)
=
c, similarly with (25) and (26), we obtain(37)
g
k
η
0
=
lim
n
→
∞
⁡
ρ
n
k
-
1
L
H
w
n
+
z
n
+
ρ
n
η
n
a
k
β
w
n
+
z
n
+
ρ
n
η
n
=
0
≤
lim
n
→
∞
⁡
ρ
n
k
-
1
A
P
R
,
γ
a
k
R
-
R
′
-
R
γ
′
z
=
w
n
+
z
n
+
ρ
n
η
n
=
0which yields (2). By Lemma 17, it is easy to deduce that g
(
ζ
)
=
ζ
-
b
0, where b
0 is a constant; then g
♯
(
0
)
≤
1
<
2, which is also a contradiction.Thus, we complete the proof of Theorem4.


## 4. Proof of Theorem9

SetH
=
f
-
α. Then(I)
H
(
z
)
=
0
⇒
H
′
(
z
)
=
α
(
z
)
-
α
′
(
z
),
(II)
H
′
(
z
)
=
α
-
α
′
⇒
L
H
≤
M
+
o
1
α
+
L
α.Putβ
=
α
-
α
′
=
(
R
-
R
′
-
R
Q
′
)
e
Q
=
R
1
e
Q and φ
=
M
+
o
1
α
+
L
α
=
A
R
2
e
Q. Here R
2
=
R
(
k
)
+
R
Q
(
k
)
+
R
(
k
-
1
)
Q
′
+
⋯
+
R
(
Q
′
)
k, R
1
(
≢
0
) and R
2 are two rational functions, deg
⁡
R
2
=
deg
⁡
R
Q
′
k, A is a positive number. Set F
=
H
/
β.Noting thatβ
=
α
-
α
′ has at most finitely many zeros and f has finitely many poles, then there exists a positive number r such that F has no poles in D
=
{
z
:
z
>
r
}.Ifρ
(
F
)
>
deg
⁡
Q, by Lemma 16, for every 0
<
μ
<
ρ
(
F
)
-
1, there exist w
n
→
∞, such that, for n
→
∞,(38)
lim
n
→
∞
⁡
F
♯
w
n
w
n
μ
=
+
∞
.In view ofw
n
→
∞ as n
→
∞, without loss of generality, we may assume w
n
≥
r
+
1 for all n. Define D
1
=
{
z
:
z
<
1
} and (39)
F
n
z
=
F
w
n
+
z
=
H
w
n
+
z
β
w
n
+
z
; then every F
n is analytic in D
1. Now, fix z
∈
D
1. If F
n
(
z
)
=
0, then H
(
w
n
+
z
)
=
0. It is clear from (
I
) that H
′
(
w
n
+
z
)
=
β
(
w
n
+
z
). Hence (for n large enough) (40)
F
n
′
z
=
H
′
w
n
+
z
β
w
n
+
z
-
H
w
n
+
z
β
w
n
+
z
β
′
w
n
+
z
β
w
n
+
z
=
1
. Also F
n
♯
(
0
)
→
∞ as n
→
∞. It follows from Marty’s criterion that (
F
n
)
n is not normal at z
=
0.Therefore, we can apply Lemma14. Choosing an appropriate subsequence of (
F
n
)
n if necessary, we may assume that there exist sequences (
z
n
)
n and (
ρ
n
)
n with z
n
<
r
<
1 and ρ
n
→
0 such that sequences (
g
n
)
n is defined by(41)
g
n
ζ
=
ρ
n
-
1
F
n
z
n
+
ρ
n
ζ
⟶
g
ζlocally uniformly in C, where g is a nonconstant entire function of order at most 1. Moreover, g
♯
(
ξ
)
≤
g
♯
(
0
)
=
2 for all ξ
∈
C and(42)
ρ
n
≤
M
F
n
♯
0
=
M
F
♯
w
n
≤
M
w
n
-
μ
-
ϵfor a positive number M.We claim that(43)
ρ
n
k
-
1
H
k
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
k
ζlocally uniformly in C.Using the mathematical induction, we prove the claim as follows.From (41), we have(44)
g
n
′
ζ
=
H
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
H
w
n
+
z
n
+
ρ
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
=
H
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
ρ
n
g
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
′
ζ
.Noting thatβ
=
α
-
α
′
=
R
1
e
Q and φ
=
M
+
o
1
α
+
L
α
=
A
R
2
e
Q, we have(45)
β
′
β
z
=
w
n
+
z
n
+
ρ
n
ζ
=
R
1
′
+
R
1
Q
′
R
1
z
=
w
n
+
z
n
+
ρ
n
ζ
=
R
′
-
R
′
′
-
2
R
′
Q
′
-
R
Q
′
′
+
R
Q
′
-
R
Q
′
2
R
-
R
′
-
R
Q
′
z
=
w
n
+
z
n
+
ρ
n
ζ
=
O
w
n
l
1
.By (38) and (42), combining with ρ
(
F
)
>
deg
⁡
Q and 0
<
μ
<
ρ
(
F
)
-
1, we deduce that(46)
lim
n
→
∞
⁡
w
n
l
1
ρ
n
≤
lim
n
→
∞
⁡
M
w
n
l
1
-
μ
+
ϵ
=
0
,where l
1
=
deg
⁡
R
′
-
R
′
′
-
2
R
′
Q
′
-
R
Q
′
′
+
R
Q
′
-
R
Q
′
2
/
R
-
R
′
-
R
Q
′ = deg
⁡
R
′
/
R
-
R
′
′
/
R
-
2
R
′
/
R
Q
′
-
Q
′
′
+
Q
′
-
Q
′
2
/
1
-
R
′
/
R
-
Q
′ = deg
⁡
Q
′ is a fixed constant. Then, combining (38), (42), (45), and (46) yields(47)
H
w
n
+
z
n
+
ρ
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
=
ρ
n
g
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
≤
M
g
n
ζ
w
n
l
1
-
μ
-
ϵ
⟶
0
,
as
n
⟶
∞
.From (44) and (47), we deduce that(48)
H
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
′
ζlocally uniformly in C, which implies that the claim is right when k
=
1.We assume that the claim is also right whenk
=
l; that is,(49)
ρ
n
l
-
1
H
l
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
l
ζlocally uniformly in C.DefineG
n
(
ζ
)
=
ρ
n
l
-
1
H
(
l
)
(
w
n
+
z
n
+
ρ
n
ζ
)
/
β
(
w
n
+
z
n
+
ρ
n
ζ
). Then(50)
G
n
′
ζ
=
ρ
n
l
H
l
+
1
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
ρ
n
l
H
l
w
n
+
z
n
+
ρ
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
=
ρ
n
l
H
l
+
1
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
-
ρ
n
G
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
l
+
1
ζlocally uniformly in C. Note the fact that(51)
ρ
n
G
n
ζ
β
′
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
0
.We immediately derive that(52)
ρ
n
l
H
l
+
1
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
l
+
1
ζlocally uniformly in C, which finishes the proof of the claim.Furthermore, we have(53)
ρ
n
k
-
1
L
H
w
n
+
z
n
+
ρ
n
ζ
a
k
β
w
n
+
z
n
+
ρ
n
ζ
=
ρ
n
k
-
1
H
k
w
n
+
z
n
+
ρ
n
ζ
β
w
n
+
z
n
+
ρ
n
ζ
+
⋯
+
ρ
n
k
-
1
a
1
H
′
w
n
+
z
n
+
ρ
n
ζ
a
k
β
w
n
+
z
n
+
ρ
n
ζ
+
ρ
n
k
-
1
a
0
H
w
n
+
z
n
+
ρ
n
ζ
a
k
β
w
n
+
z
n
+
ρ
n
ζ
⟶
g
k
ζlocally uniformly in C.We claim that(1)g
(
ζ
)
=
0
⇒
g
′
(
ζ
)
=
1,(2)g
′
(
ζ
)
=
1
⇒
g
(
k
)
(
ζ
)
=
0.The proof of (1) is exactly the same as in the proof of Theorem4. To prove (2), just replace (53) in the previous proof by(54)
g
k
η
0
=
lim
n
→
∞
⁡
ρ
n
k
-
1
L
H
w
n
+
z
n
+
ρ
n
η
n
a
k
β
w
n
+
z
n
+
ρ
n
η
n
≤
lim
n
→
∞
⁡
ρ
n
k
-
1
φ
w
n
+
z
n
+
ρ
n
η
n
a
k
β
w
n
+
z
n
+
ρ
n
η
n
=
lim
n
→
∞
⁡
ρ
n
k
-
1
R
2
w
n
+
z
n
+
ρ
n
η
n
a
k
R
-
R
′
-
R
Q
′
w
n
+
z
n
+
ρ
n
η
n
=
lim
n
→
∞
⁡
ρ
n
k
-
1
O
w
n
l
2
.Here l
2
=
deg
⁡
R
2
/
R
-
R
′
-
R
Q
′
=
deg
⁡
R
2
/
R
/
1
-
R
′
/
R
-
Q
′
=
deg
⁡
(
Q
′
)
k
-
1 is also a fixed constant.By (38) and (42), combining with ρ
(
F
)
>
deg
⁡
Q and 0
<
μ
<
ρ
(
F
)
-
1, we deduce that(55)
lim
n
→
∞
⁡
w
n
l
2
ρ
n
k
-
1
≤
lim
n
→
∞
⁡
M
w
n
l
2
-
k
-
1
μ
+
ϵ
=
0
,which yields (2). From Lemma 17, it is easy to deduce that g
(
ζ
)
=
ζ
-
b
0, where b
0 is a constant; then g
♯
(
0
)
≤
1
<
2, which is also a contradiction.Soρ
(
F
)
≤
deg
⁡
Q. Next we will prove ρ
(
f
)
≤
ρ
(
F
)
≤
deg
⁡
Q.Ifρ
(
α
)
<
ρ
(
f
), noting that ρ
(
α
-
α
′
)
≤
ρ
(
α
), by Lemma 18, then ρ
(
F
(
α
-
α
′
)
)
≤
max
⁡
{
ρ
(
F
)
,
ρ
(
α
)
}. Due to F
=
f
-
α
/
α
-
α
′, we have f
=
α
+
F
(
α
-
α
′
). Thus, by Lemma 18, ρ
(
f
)
≤
max
⁡
ρ
α
,
ρ
F
α
-
α
′
≤
max
⁡
{
ρ
(
α
)
,
ρ
(
F
)
}. Then, it follows from ρ
(
α
)
<
ρ
(
f
) that ρ
(
f
)
≤
ρ
(
F
)
≤
deg
⁡
Q.Ifρ
(
α
)
=
ρ
(
f
), noting that ρ
(
α
)
=
deg
⁡
Q, we obtain ρ
(
f
)
=
ρ
(
α
)
=
deg
⁡
Q.Ifρ
(
α
)
>
ρ
(
f
), due to F
=
f
-
α
/
α
-
α
′
=
f
-
R
e
Q
/
R
1
e
Q
=
f
/
R
1
e
Q
-
R
/
R
1, hence f
=
(
F
+
R
/
R
1
)
R
1
e
Q. Because R
/
R
1 is a rational function, we get ρ
(
R
/
R
1
)
=
ρ
(
R
1
)
=
0. By Lemma 18, we may obtain deg
⁡
Q
=
ρ
(
R
e
Q
)
=
ρ
(
α
)
<
ρ
(
f
)
≤
max
⁡
ρ
F
+
R
/
R
1
,
ρ
R
1
e
Q
≤
max
⁡
{
ρ
(
F
)
,
ρ
(
R
1
e
Q
)
}
≤
deg
⁡
Q, a contradiction.At last, we obtainρ
(
f
)
≤
deg
⁡
Q.Thus, we complete the proof of Theorem9.


## 5. Proof of Theorem13

Now we distinguish two cases.Case 1 (ifQ is a nonconstant polynomial).
By the assumption of Theorem13 and the result of Theorem 9, we easily deduce that f is of order at most deg
⁡
Q. Define(56)
μ
=
f
′
-
α
f
-
α
.
The fact thatf is transcendental and ρ
(
α
)
<
ρ
(
f
) implies μ
≢
0.
Because off
′
(
z
)
-
α
(
z
)
=
0
↦
f
(
z
)
-
α
(
z
)
=
0, it is easy to obtain that μ has no zeros. With the assumption f
(
z
)
=
α
(
z
)
⇒
f
′
(
z
)
=
α
(
z
), we derive that f
-
α has finitely many multiple zeros. We know that the possible poles of μ are from the multiple zeros of f
-
α and the poles of f; thus μ has finitely many poles. Moreover, from (56), we have ρ
(
μ
)
≤
ρ
(
f
)
≤
deg
⁡
Q. Therefore, we can set(57)
μ
=
1
P
z
e
Q
∗
z
,where P is a nonzero polynomial and deg
⁡
Q
∗ is at most deg
⁡
Q.
SetF
=
f
-
α and A
=
α
-
α
′. We obtain(58)
F
′
F
-
1
P
e
Q
∗
=
A
F
.By Lemma 19, we get Q
∗ and P; both are constants. Thus μ is a nonzero constant. Set μ
=
c, rewriting (56) as(59)
f
′
=
c
f
+
1
-
c
α
=
c
f
+
H
,where H
=
(
1
-
c
)
α.
Ifc
=
1, then f
=
f
′ and f
(
z
)
=
b
e
z, where b is a nonzero constant. So ρ
(
f
)
=
1. Since Q is a nonconstant polynomial, we get ρ
(
α
)
=
ρ
(
R
e
Q
)
≥
1. It is a contradiction with the condition ρ
(
α
)
<
ρ
(
f
).
Next, we considerc
≠
1. Differentiating (59) k
-
1 times yields (60)
f
k
=
c
k
f
+
c
k
-
1
H
+
c
k
-
2
H
′
+
⋯
+
c
H
k
-
2
+
H
k
-
1
. Furthermore, we get(61)
L
f
=
∑
n
=
0
k
a
n
c
n
f
+
∑
n
=
1
k
a
n
c
n
-
1
H
+
∑
n
=
2
k
a
n
c
n
-
2
H
′
+
⋯
+
∑
n
=
p
+
1
k
a
n
c
n
-
p
+
1
H
p
+
⋯
+
a
k
H
k
-
1
.
Set(62)
ϕ
=
∑
n
=
0
k
a
n
c
n
α
+
∑
n
=
1
k
a
n
c
n
-
1
1
-
c
α
+
∑
n
=
2
k
a
n
c
n
-
2
1
-
c
α
′
+
⋯
+
∑
n
=
p
+
1
k
a
n
c
n
-
p
+
1
1
-
c
α
p
+
⋯
+
a
k
1
-
c
α
k
-
1
=
a
0
+
a
1
+
∑
n
=
2
k
a
n
c
n
-
1
α
+
∑
n
=
2
k
a
n
c
n
-
2
1
-
c
α
′
+
⋯
+
∑
n
=
p
+
1
k
a
n
c
n
-
p
+
1
1
-
c
α
p
+
⋯
+
a
k
1
-
c
α
k
-
1
=
R
1
e
Q
.Obviously, R
1 is a rational function.
Supposez
0 is a zero of f
-
α. Substituting H
=
(
1
-
c
)
α and z
0 into (61), we get α
(
z
0
)
=
ϕ
(
z
0
); this shows that z
0 is also a zero of α
-
ϕ.
Next, we will prove thatα
≡
ϕ. Otherwise, we assume that α
≢
ϕ. Combining the above discussion and the fact that f
-
α has finitely many multiple zeros yields(63)
N
r
,
1
f
-
α
=
N
1
)
r
,
1
f
-
α
+
N
(
2
r
,
1
f
-
α
≤
N
r
,
1
α
-
ϕ
+
O
log
⁡
r
=
N
r
,
1
R
-
R
1
e
Q
+
O
log
⁡
r
=
O
log
⁡
r
,and it implies that f
-
α has finitely many zeros. Thus we can set f
-
α
=
R
2
(
z
)
e
Q
1, where R
2 is a rational function and Q
1 is a polynomial and degree at most deg
⁡
Q. Differentiating the above equation leads to(64)
f
′
z
=
α
′
z
+
R
2
′
+
Q
1
′
R
2
e
Q
1
.
By the fact thatf
-
α has finitely many zeros and the assumption f
′
(
z
)
=
α
(
z
)
↦
f
(
z
)
=
α
(
z
), we deduce that f
′
-
α also has finitely many zeros. Noting that(65)
f
′
z
-
α
z
=
α
′
z
-
α
z
+
R
2
′
z
+
Q
1
′
z
R
2
z
e
Q
1
z
;thus α
-
α
′ or R
2
′
+
R
2
Q
1
′
≡
0. By the assumptions of Theorem 13, α
≢
α
′, this is impossible.
IfR
2
′
+
Q
1
′
R
2
=
0, then R
2
′
/
R
2
=
-
Q
1
′. By the definition, we get deg
⁡
R
2
′
/
R
2
=
0 and also deg
⁡
Q
1
′
=
0. So we may set Q
1
=
b
z
+
d or Q
1
=
a, where b is a nonzero constant and a
,
d are two constants.
IfQ
1
=
a, then f
=
α
+
a. So ρ
(
f
)
=
ρ
(
α
). It is a contradiction about the condition ρ
(
α
)
<
ρ
(
f
).
IfQ
1
=
b
z
+
d, then f
-
α
=
R
2
(
z
)
e
b
z
+
d. So (64) changes to(66)
f
′
=
α
′
+
R
2
′
+
b
R
2
e
b
z
+
d
,and (65) turns into(67)
f
′
-
α
=
α
′
-
α
+
R
2
′
+
b
R
2
e
b
z
+
d
.Thus α
-
α
′
=
0 or R
2
′
+
R
2
b
=
0. Also α
≠
α
′ is impossible. If R
2
′
+
R
2
b
=
0, then b
=
0, a contradiction.
At last, we complete the proof of(68)
α
≡
ϕ
.Next, we distinguish the following subcases.
Subcase  1.1 (R is not a polynomial). Suppose that a
0 is a pole of R with multiplicity m. By (62) and the fact that c
≠
1 and a
0 is a pole of ϕ with multiplicity m
+
k
-
1, it contradicts (68).
Subcase 1.2 (R is nonconstant polynomial). From (62), R
1 is a polynomial. By simple calculation, we get deg
⁡
R
1
=
deg
⁡
(
R
(
Q
′
)
k
-
1
). Combining with (68), we deduce deg
⁡
R
=
deg
⁡
R
1
=
deg
⁡
⁡
(
R
(
Q
′
)
k
-
1
)
=
deg
⁡
R
+
deg
⁡
(
Q
′
)
k
-
1. So Q
=
m
z
+
n and ρ
(
α
)
=
1. Here m is a nonzero constant, and n is a constant.By integrating (59), we have(69)
f
z
=
λ
e
c
z
+
A
z
e
m
z
,where λ is a nonzero constant and A is a polynomial.Soρ
(
f
)
≤
1
=
ρ
(
α
), a contradiction.Subcase  1.3 (R is a constant). Set R
=
a. So α
=
a
e
Q. As subcase  1.2, we also get Q
(
z
)
=
m
z
+
n. Here m is a nonzero constant, and n is a constant. Integrating (59) yields(70)
f
z
=
a
1
-
c
∫
e
Q
z
-
c
z
d
z
+
c
0
e
c
z
,where c
0 is a constant.Ifm
=
c, by calculation, f
(
z
)
=
(
a
b
(
1
-
c
)
z
+
c
0
∗
)
e
c
z (here b
=
e
n and c
0
∗ is an arbitrary constant), and ρ
(
f
)
=
ρ
(
α
)
=
1, a contradiction.Som
≠
c, by calculation; f
(
z
)
=
a
b
(
1
-
c
)
1
/
m
-
c
e
m
z
+
c
0
∗
e
c
z (here b
=
e
n and c
0
∗ is an arbitrary constant), and ρ
(
f
)
≤
1
=
ρ
(
α
), a contradiction.Case 2.
IfQ is a constant, because f is a transcendental meromorphic function, obviously the condition ρ
(
α
)
<
ρ
(
f
) and α
≢
α
′ hold.So Theorem13 turns into Theorem 2, and the results of Theorem 2 still hold.The proof is completed.



---

*Source: 6250867-2016-09-05.xml*