# Weakly Weighted Sharing and Uniqueness of Meromorphic Functions

Authors: Thamir Alzahary
Journal: Journal of Complex Analysis
 (2014)

---

## Abstract

With the aid of the notion of weakly weighted sharing, we study the uniqueness of meromorphic
functions sharing four pairs of small functions. Our results improve and generalize some results
given by Czubiak and Gundersen, Li and Yang, and other authors.


---

## Body

## 1. Introduction and Main Results

In this paper, a meromorphic function means meromorphic in the whole complex planeC. We assume the reader is familiar with the standard notion used in the Nevanlinna value distribution theory such as T
(
r
,
f
), N
(
r
,
f
), m
(
r
,
f
), and N
¯
(
r
,
f
) (see [1, 2]). For any nonconstant meromorphic function f, the term S
(
r
,
f
) denotes any quantity that satisfies S
(
r
,
f
)
=
o
(
1
)
(
T
(
r
,
f
)
) as r
→
∞ outside a possible exceptional set of finite linear measure.Letf be a nonconstant meromorphic function. A meromorphic function a is called asmall function of f, if T
(
r
,
a
)
=
S
(
r
,
f
). If k is a positive integer, we denote by N
¯
k
)
(
r
,
f
) the reduced counting function of the poles of f whose multiplicities are less than or equal to k and denote by N
¯
(
k
(
r
,
f
) the reduced counting function of the poles of f whose multiplicities are greater than or equal to k.Letf and g be nonconstant meromorphic functions, and let a, b be two values in C. We say that f and g share the value a IM provided that f
(
z
)
-
a and g
(
z
)
-
a have the same zeros ignoring multiplicities. In addition, we say that f and g share the value ∞ IM, if 1
/
f and 1
/
g share 0 IM. We say that f and g share the pair of values (
a
,
b
) IM provided that f
(
z
)
-
a and g
(
z
)
-
b have the same zeros ignoring multiplicities.The following theorem is a well-known and significant result in the uniqueness theory of meromorphic functions and has been proved by Czubiak and Gundersen.Theorem A (see [3]).
Letf and g be two nonconstant meromorphic functions that share six pairs of values (
a
i
,
b
i
), 1
≤
i
≤
6 IM, where a
i
≠
a
j whenever i
≠
j and b
i
≠
b
j whenever i
≠
j. Then f is a Möbius transformation of g.The following example, found by Gundersen, shows that the number “six” in Theorem A cannot be replaced with “five.”Example 1 (see [4]).
Letf
(
z
)
=
(
e
z
+
1
)
/
(
e
z
-
1
)
2, g
(
z
)
=
(
e
z
+
1
)
2
/
8
(
e
z
-
1
). We see that f, g share (
0,0
), (
∞
,
∞
), (
1,1
), (
-
1
/
8
,
-
1
/
8
) and (
-
1
/
2,1
/
4
) IM, and f is not a Möbius transformation of g.Letf and g be nonconstant meromorphic functions and let a
,
b be two small meromorphic functions of f and g. We denote by N
¯
(
r
,
f
=
a
,
g
=
b
) the reduced counting function of the common zeros of f
-
a and g
-
b. We say that f and g share (
a
,
b
)
IM
*, if
(1)
N
¯
(
r
,
1
f
-
a
)
-
N
¯
(
r
,
f
=
a
,
g
=
b
)
=
S
(
r
,
f
)
,
N
¯
(
r
,
1
g
-
b
)
-
N
¯
(
r
,
f
=
a
,
g
=
b
)
=
S
(
r
,
g
)
.As in Theorem A and throughout this paper, whenf and g are nonconstant meromorphic functions, we let S
(
r
) denote the term which is both S
(
r
,
f
) and S
(
r
,
g
) simultaneously.We denote byN
¯
(
r
,
f
=
a
,
g
≠
b
) the reduced counting function of those a-points of f, which are not the b-points of g. We note that f and g share (
a
,
b
)
IM
* if and only if N
¯
(
r
,
f
=
a
,
g
≠
b
)
=
S
(
r
) and N
¯
(
r
,
g
=
b
,
f
≠
a
)
=
S
(
r
). According to this note, we generalize the definitions of IM and IM
* to theweakly weightedIMsharing which is given by the following definition.Definition 2 (see [5]).
Letk be a positive integer or infinity, and let a, b be two small functions of nonconstant meromorphic functions f and g. We denote by N
¯
k
)
(
r
,
f
=
a
,
g
≠
b
) the reduced counting function of those a-points of f whose multiplicities are less than or equal to k, that are not the b-points of g. If N
¯
k
)
(
r
,
f
=
a
,
g
≠
b
)
+
N
¯
k
)
(
r
,
g
=
b
,
f
≠
a
)
=
S
(
r
), we say that f and g share (
a
,
b
,
k
)
*IM.We note that iff and g share (
a
,
b
,
k
)
*IM, then f and g share (
a
,
b
,
p
)
*IM, for all integer 1
≤
p
≤
k. Also, we note that f and g share (
a
,
b
)
IM
* if and only if f and g share (
a
,
b
,
∞
)
*IM.Recently, Li and Yang have proved the following.Theorem B (see [6]).
Letf and g be two nonconstant meromorphic functions, and let a
i, b
i (i
=
1
,
…
,
6) be small functions of f and g, with a
i
≢
a
j, b
i
≢
b
j whenever i
≠
j. If f and g share the five pairs (
a
i
,
b
i
)
I
M
*, 1
≤
i
≤
5 and f is not a quasi-Möbius transformation of g, then the following identities or inequalities hold: (
a
)
T
(
r
,
f
)
=
T
(
r
,
g
)
+
S
(
r
);
(
b
)
3
T
(
r
,
f
)
=
∑
i
=
1
5
N
¯
(
r
,
1
/
(
f
-
a
i
)
)
+
S
(
r
);
(
c
)
T
(
r
,
f
)
≤
N
¯
(
r
,
1
/
(
f
-
a
i
)
)
+
N
¯
(
r
,
1
/
(
f
-
a
j
)
)
+
S
(
r
), i
≠
j, i
,
j
=
1
,
…
,
5;
(
d
)
T
(
r
,
f
)
≤
3
N
¯
(
r
,
1
/
(
f
-
a
i
)
)
+
S
(
r
), i
=
1
,
…
,
5;
(
e
)
T
(
r
,
f
)
=
N
¯
(
r
,
1
/
(
f
-
a
6
)
)
+
S
(
r
);
(
f
)
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
≤
(
3
/
5
)
T
(
r
,
f
)
+
S
(
r
);
(
g
)
T
(
r
,
f
)
=
N
¯
(
r
,
1
/
(
f
-
a
5
)
)
+
S
(
r
) and T
(
r
,
f
)
=
2
N
¯
(
r
,
1
/
(
f
-
a
i
)
)
+
S
(
r
) for i
=
1
,
…
,
4 if a
i
≡
b
i, i
=
1
,
…
,
4.One may ask the following question: is it possible to relax the condition “f and g share five pairs of small functions” in Theorem B to the condition “f and g share four pairs of small functions?”The goal of the present paper is to generalize and improve Theorems A and B by using the weakly weighted sharing. We now turn to state our results.Theorem 3.
Letf and g be two nonconstant meromorphic functions and let a
i, b
i (i
=
1
,
…
,
6) be small functions of f and g, with a
i
≢
a
j, b
i
≢
b
j whenever i
≠
j, and let k
1
,
…
,
k
5 be five positive integers or infinity with K
≡
∑
i
=
1
5
(
1
/
(
1
+
k
i
)
)
≤
1. Suppose that f and g share (
a
i
,
b
i
,
k
i
)
*IM,  i
=
1
,
…
,
4, and f is not a quasi-Möbius transformation of g. If there exists a number λ
∈
[
0,1
-
K
] such that
(2)
N
¯
(
r
,
1
f
-
a
6
)
+
N
¯
k
5
)
(
r
,
f
=
a
5
,
g
≠
b
5
)
≤
λ
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
(
r
,
1
g
-
b
6
)
+
N
¯
k
5
)
(
r
,
g
=
b
5
,
f
≠
a
5
)
≤
λ
T
(
r
,
g
)
+
S
(
r
)
,
then k
i
=
∞, for all 1
≤
i
≤
5 (that means f and g share (
a
i
,
b
i
,
∞
)
*IM,  i
=
1
,
…
,
4), and the following identities or inequalities hold:(a)
T
(
r
,
f
)
=
T
(
r
,
g
)
+
S
(
r
);
(b)
4
T
(
r
,
f
)
=
∑
i
=
1
6
N
¯
(
r
,
1
/
(
f
-
a
i
)
)
+
S
(
r
);
(c)
T
(
r
,
f
)
≤
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
S
(
r
), i
≠
j, i
,
j
=
1
,
…
,
5;
(d)
T
(
r
,
f
)
≤
3
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
S
(
r
), i
=
1
,
…
,
5;
(e)
T
(
r
,
f
)
=
N
¯
(
r
,
1
/
(
f
-
a
6
)
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
S
(
r
);
(f)
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
≤
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
S
(
r
) and N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
≤
3
T
(
r
,
f
)
-
4
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
S
(
r
), i
=
1
,
…
,
6;
(g)
T
(
r
,
f
)
=
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
) and T
(
r
,
f
)
=
2
N
¯
(
r
,
1
/
(
f
-
a
i
)
)
+
S
(
r
) for i
=
1
,
…
,
4 if a
i
≡
b
i, i
=
1
,
…
,
4;
(h)
T
(
r
,
f
)
=
2
N
¯
(
r
,
f
=
a
5
,
g
=
a
5
)
+
S
(
r
)
,
T
(
r
,
f
)
=
N
¯
(
r
,
1
/
(
f
-
a
4
)
)
+
S
(
r
) and T
(
r
,
f
)
=
2
N
¯
(
r
,
1
/
(
f
-
a
i
)
)
+
S
(
r
) for i
=
1,2
,
3 if a
5
≡
b
5 and a
i
≡
b
i, i
=
1,2
,
3.The theorem is true, if f
,
a
1
,
…
,
a
6 are interchanged with g
,
b
1
,
…
,
b
6, respectively.Remark 4.
In Theorem3, if f and g share the five pairs (
a
i
,
b
i
)
IM
*, 1
≤
i
≤
5, then from Definition 2 we deduce that condition (2) occurs, and then the properties “(a)–(h)” of Theorem 3 give us the properties “(a)–(g)” of Theorem B. We also see that if f and g share five distinct small functions IM
*, then, from (g) and (h) of Theorem 3, we deduce that f is a quasi-Möbius transformation of g, and hence, f
≡
g. This result was proved in [7] (entire case ) and [8] (meromorphic case).From (e) and (f) of Theorem3, we can immediately obtain the following corollary.Corollary 5.
Letf and g be two meromorphic functions, and let a
i, b
i (i
=
1
,
…
,
6) be small functions of f and g, with a
i
≢
a
j, b
i
≢
b
j whenever i
≠
j. Suppose that f and g share the four pairs (
a
i
,
b
i
)
I
M
*, for i
=
1
,
…
,
4 and
(3)
N
¯
(
r
,
1
f
-
a
6
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
≤
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
(
r
,
1
g
-
b
6
)
+
N
¯
(
r
,
g
=
b
5
,
f
≠
a
5
)
≤
T
(
r
,
g
)
+
S
(
r
)
.
If there exists a number λ
∈
[
0,2
/
5
) such that
(4)
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
≤
λ
T
(
r
,
f
)
+
S
(
r
)
,
then f must be a quasi-Möbius transformation of g.Remark 6.
Li and Yang [6, Corollary 2] proved Corollary 5, when f and g share the five pairs (
a
i
,
b
i
)
IM
*, 1
≤
i
≤
5, and if there exists a number λ
∈
[
0,2
/
5
) such that N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
≤
λ
T
(
r
,
f
)
+
S
(
r
). It is evident that Corollary 5 improves Theorem A and the result of Li and Yang [6, Corollary 2].Obviously, Theorem3 is a generalization of Theorem B, and Corollary 5 is a generalization of Theorem A and Corollary 2 in [6].Example 7.
Letf and g be defined as in Example 1, and let (
a
1
,
b
1
)
=
(
0,0
), (
a
2
,
b
2
)
=
(
∞
,
∞
), (
a
3
,
b
3
)
=
(
1,1
), (
a
4
,
b
4
)
=
(
-
1
/
8
,
-
1
/
8
), (
a
5
,
b
5
)
=
(
-
1
/
2,1
/
4
) and a
6
=
b
6
=
1
/
4
+
3
/
4. It is easy to show that N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
=
T
(
r
,
f
), N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
=
S
(
r
), N
¯
(
r
,
g
=
b
5
,
f
≠
a
5
)
=
S
(
r
) and f is not a quasi-Möbius transformation of g. This shows that the condition “λ
∈
[
0,2
/
5
) in Corollary 5” is necessary.We have the following more general result.Theorem 8.
Letf and g be two nonconstant meromorphic functions, and let a
i, b
i (i
=
1
,
…
,
6) be small functions of f and g, with a
i
≢
a
j, b
i
≢
b
j whenever i
≠
j, and let k
1
,
…
,
k
6 be six positive integers or infinity with K
≡
∑
i
=
1
6
‍
(
1
/
1
+
k
i
)
≤
1. Suppose that f and g share (
a
i
,
b
i
,
k
i
)
*IM, for i
=
1
,
…
,
4. If there are two numbers λ
1 and λ
2 inside the interval [
0,1
-
K
] such that
(5)
36
<
(
8
-
5
(
λ
1
+
K
)
)
(
8
-
5
(
λ
2
+
K
)
)
and the following two inequalities hold
(6)
N
¯
k
6
)
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
N
¯
k
5
)
(
r
,
f
=
a
5
,
g
≠
b
5
)
≤
λ
1
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
k
6
)
(
r
,
g
=
b
6
,
f
≠
a
6
)
+
N
¯
k
5
)
(
r
,
g
=
b
5
,
f
≠
a
5
)
≤
λ
2
T
(
r
,
g
)
+
S
(
r
)
,
then f is a quasi-Möbius transformation of g.We deduce, from Theorem8, the following corollary which is an improvement to Theorem A.Corollary 9.
Letf and g be two nonconstant meromorphic functions, and let a
i, b
i (i
=
1
,
…
,
6) be small functions of f and g, with a
i
≢
a
j, b
i
≢
b
j whenever i
≠
j. Suppose that f and g share the four pairs (
a
i
,
b
i
)
I
M
*, 1
≤
i
≤
4. If there are two numbers λ
1 and λ
2 inside the interval [
0,1
] such that 36
<
(
8
-
5
λ
1
)
(
8
-
5
λ
2
) and the following two inequalities hold
(7)
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
≤
λ
1
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
(
r
,
g
=
b
6
,
f
≠
a
6
)
+
N
¯
(
r
,
g
=
b
5
,
f
≠
a
5
)
≤
λ
2
T
(
r
,
g
)
+
S
(
r
)
,
then f is a quasi-Möbius transformation of g.


## 2. Lemmas

In this section, we introduce some lemmas that will be used to prove the main results in this paper.Lemma 10.
Letf and g be two nonconstant meromorphic functions such that
(8)
F
=
c
1
f
2
g
+
c
2
f
g
+
c
3
f
2
+
c
4
f
+
c
5
g
+
c
6
≡
0
,
where c
1
,
…
,
c
6 are small functions of f and g and at least one of them is not identically zero. If f is not a quasi-Möbius transformation of g, then T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
).Proof.
SinceF
≡
0, then g
(
c
1
f
2
+
c
2
f
+
c
5
)
=
-
(
c
3
f
2
+
c
4
f
+
c
6
). We note that c
1
f
2
+
c
2
f
+
c
5
≢
0, because at least one of {
c
1
,
…
,
c
6
} is not zero. Hence,
(9)
g
=
-
c
3
f
2
+
c
4
f
+
c
6
c
1
f
2
+
c
2
f
+
c
5
.
Since f is not a quasi-Möbius transformation of g, the right-hand side of the above equation is irreducible. Therefore, by applying Valiron-Mokhonko lemma [9] we get T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
). This completes the proof of Lemma 10.Lemma 11 (see [10]).
Letf be a nonconstant meromorphic function and let a
1
,
…
,
a
n be n distinct small functions of f. Then
(10)
(
n
-
2
-
ϵ
)
T
(
r
,
f
)
≤
∑
i
=
1
n
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
,
f
)
holds for any positive number ϵ.Lemma 12 (see [6]).
Suppose thatf and g are nonconstant meromorphic functions, and F
=
F
(
f
,
g
) is a polynomial in f and g with coefficients being small functions of f and g. The degree of F related to f is p, and the degree of F related to g is q. Then we have T
(
r
,
F
)
≤
p
T
(
r
,
f
)
+
q
T
(
r
,
g
)
+
S
(
r
).


## 3. Proofs of Theorems3 and 8

Ifa
6
=
∞ (or b
6
=
∞), then, for i
=
1,2
,
3,4
,
5, and we let F
=
1
/
(
f
-
a
), G
=
1
/
(
g
-
a
), a
¯
i
=
1
/
(
a
i
-
a
), and b
¯
i
=
1
/
(
b
i
-
a
), where a is any complex number a such that a
≢
a
i, b
i (i
=
1,2
,
3,4
,
5). We can take F
,
G
,
a
¯
i
,
a
¯
6
=
0, (or b
¯
6
=
0, if b
6
=
∞), and b
¯
i (i
=
1,2
,
3,4
,
5) in our proofs instead of f, g, a
i and b
i (i
=
1,2
,
3,4
,
6), respectively. Therefore, we assume that none of a
i and b
i (i
=
1
,
…
,
6) is infinity.


### 3.1. Proof of Theorem3

It is not difficult to find six small functionsc
i (1
≤
i
≤
6) of f and g that are not all zero and six other small functions d
i (1
≤
i
≤
6) of f and g that are not all zero, such that
(11)
c
1
a
i
2
b
i
+
c
2
a
i
b
i
+
c
3
a
i
2
+
c
4
a
i
+
c
5
b
i
+
c
6
≡
0
,
2222222222222222222222
i
=
1,2
,
3,4
,
5
,
d
1
a
i
b
i
2
+
d
2
a
i
b
i
+
d
3
b
i
2
+
d
4
a
i
+
d
5
b
i
+
d
6
≡
0
,
2222222222222222222222222
i
=
1,2
,
3,4
,
5
.
Consequently, the following two functions:
(12)
F
:
=
F
(
f
,
g
)
=
c
1
f
2
g
+
c
2
f
g
+
c
3
f
2
+
c
4
f
+
c
5
g
+
c
6
,
(13)
G
:
=
G
(
f
,
g
)
=
d
1
f
g
2
+
d
2
f
g
+
d
3
g
2
+
d
4
f
+
d
5
g
+
d
6
satisfy F
(
a
i
,
b
i
)
≡
0 and G
(
a
i
,
b
i
)
≡
0, for (1
≤
i
≤
5).Suppose thatF
≢
0. For all 1
≤
i
≤
5, we denote by N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
) the reduced counting function of those a
i-points of f whose multiplicities are greater than or equal to k
i
+
1, that are not the b
i-points of g in |
z
|
<
r. By using Lemmas 11 and 12, we have
(14)
4
T
(
r
,
f
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
∑
i
=
1
5
N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
N
¯
k
5
)
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
≤
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
∑
i
=
1
5
‍
N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
≤
N
¯
(
r
,
1
F
)
+
λ
T
(
r
,
f
)
+
∑
i
=
1
5
‍
1
k
i
+
1
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
T
(
r
,
g
)
+
(
λ
+
K
)
T
(
r
,
f
)
+
S
(
r
)
,
and the inequality (14) gives us
(15)
(
2
-
K
-
λ
)
T
(
r
,
f
)
≤
T
(
r
,
g
)
+
S
(
r
)
.
In the same method as the above, it can be shown that
(16)
4
T
(
r
,
g
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
1
g
-
b
i
)
+
S
(
r
)
≤
∑
i
=
1
5
‍
N
¯
(
r
,
g
=
b
i
,
f
=
a
i
)
+
∑
i
=
1
5
‍
N
¯
(
k
i
+
1
(
r
,
g
=
b
i
,
f
≠
a
i
)
+
λ
T
(
r
,
g
)
+
S
(
r
)
≤
N
¯
(
r
,
1
F
)
+
λ
T
(
r
,
g
)
+
∑
i
=
1
5
‍
1
k
i
+
1
N
(
r
,
1
g
-
b
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
T
(
r
,
g
)
+
(
λ
+
K
)
T
(
r
,
g
)
+
S
(
r
)
,
and this gives us
(17)
(
3
-
K
-
λ
)
T
(
r
,
g
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
.
It follows from (15) and (17) that
(18)
(
3
-
K
-
λ
)
T
(
r
,
g
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
≤
2
2
-
K
-
λ
T
(
r
,
g
)
+
S
(
r
)
,
which means that (
3
-
K
-
λ
)
(
2
-
K
-
λ
)
≤
2, and hence (
K
+
λ
-
4
)
(
K
+
λ
-
1
)
≤
0. Since K
+
λ
≤
1 (this follows from the assumption of Theorem 3), then, from the last inequality, we deduce that 0
≤
K
+
λ
-
1, which means that
(19)
K
+
λ
=
1
.
Consequently, we deduce (a) from (15), (17), and (19), and then from (14), we get (b). Also, (14) gives us
(20)
N
¯
k
5
)
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
=
λ
T
(
r
,
f
)
.
From (a), (14), and (19), we deduce the following relation:
(21)
N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
)
=
1
k
i
+
1
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
=
1
k
i
+
1
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1
,
…
,
5
.Let us now proveK
=
0. We first assume that there exits 1
≤
i
≤
5 such that k
i
<
∞. Then from (21), we have
(22)
N
k
i
)
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
N
f
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
k
i
(
r
,
f
=
a
i
)
=
S
(
r
)
,
where N
f
(
r
,
f
=
a
i
,
g
=
b
i
) is the counting function of the points f
-
a
i
=
0 that are zeros of g
-
b
i
=
0, and N
k
i
(
r
,
f
=
a
i
) is the counting function of those a
i-points of f whose multiplicities are not equal to k
i
+
1.It follows from the preceding equation and (21) that we have
(23)
T
(
r
,
f
)
=
(
k
i
+
1
)
N
¯
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
S
(
r
)
=
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
,
if
k
i
<
∞
,
1
≤
i
≤
5
.Let us consider two cases.Case 1.
Suppose that there are two values ofk
i, (i
=
1
,
…
,
5) that are finite. Suppose that k
1
<
∞ and k
2
<
∞. Let Ł be a quasi-Möbius transformation such that a
j
≡
Ł
(
b
j
), j
=
3,4
,
5. Since f is not a quasi-Möbius transformation of g, then f
≢
Ł
(
g
), and then by (a), we have
(24)
∑
i
=
3
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
≤
N
¯
(
r
,
1
f
-
Ł
(
g
)
)
+
∑
i
=
3
4
‍
1
k
i
+
1
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
1
k
3
+
1
+
1
k
4
+
1
)
T
(
r
,
f
)
+
S
(
r
)
.
It follows from (19)–(23), (a), and (b) that
(25)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
≤
(
1
k
1
+
1
+
1
k
2
+
1
)
T
(
r
,
f
)
+
∑
i
=
3
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
∑
i
=
1
4
‍
1
k
i
+
1
)
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
≤
(
2
+
K
)
T
(
r
,
f
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
=
3
T
(
r
,
f
)
+
S
(
r
)
,
which is impossible.By using the above method, it can be proved that there are no other two values ofk
i, (i
=
1
,
…
,
5) that are simultaneously finite. Therefore, this case cannot occur.Case 2.
Suppose that there is1
≤
s
≤
5 such that k
s
<
∞ and k
i
=
∞, for all 1
≤
i
≤
5 with i
≠
s.Case  2.1. Suppose that s
=
4. That is, k
i
=
∞, for all 1
≤
i
≤
5 with i
≠
4, and hence, K
=
1
/
(
1
+
k
4
). As in Case 1, let Ł be a quasi-Möbius transformation such that a
j
≡
Ł
(
b
j
), j
=
1,2
,
3. Since f is not a quasi-Möbius transformation of g, then f
≢
Ł
(
g
). From (23) and (a), we see
(26)
∑
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
≤
N
¯
(
r
,
1
f
-
Ł
(
g
)
)
+
1
k
4
+
1
N
(
r
,
1
f
-
a
4
)
+
S
(
r
)
≤
(
2
+
1
k
4
+
1
)
T
(
r
,
f
)
+
S
(
r
)
=
(
2
+
K
)
T
(
r
,
f
)
+
S
(
r
)
.
From (19)–(23) and (26) we have
(27)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
K
)
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
≤
(
2
+
K
+
λ
)
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
4
T
(
r
,
f
)
+
S
(
r
)
,
which gives
(28)
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
=
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
=
λ
T
(
r
,
f
)
.
Thus, from (28), we deduce that
(29)
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
=
S
(
r
)
.
That means
(30)
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
=
λ
T
(
r
,
f
)
.
On the other hand, by using (23)–(29) we get
(31)
∑
i
=
1,2
,
5
‍
N
¯
(
r
,
1
f
-
a
i
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
.It follows from (19), (23), (30), (31) and by using (a) and (b), we see
(32)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
K
+
λ
)
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
3
)
+
S
(
r
)
≤
4
T
(
r
,
f
)
+
S
(
r
)
;
thus, the inequality (32) gives us
(33)
N
¯
(
r
,
1
f
-
a
3
)
=
T
(
r
,
f
)
+
S
(
r
)
.
Similarly, we have
(34)
N
¯
(
r
,
1
f
-
a
i
)
=
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1,2
.
From (28), (29), (a), and (b), we get
(35)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
=
4
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
4
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
,
which yields
(36)
N
¯
(
r
,
1
f
-
a
4
)
+
N
¯
(
r
,
1
f
-
a
6
)
=
S
(
r
)
.
From (21) and (36) (remembering k
4
<
∞), we have T
(
r
,
f
)
=
S
(
r
), which is impossible. That means that k
4
=
∞. In the same method as the above, we can prove that k
i
=
∞, i
=
1,2
,
3. This proves that f and g share (
a
i
,
b
i
,
∞
)
*IM, for all 1
≤
i
≤
4.Case  2.2. Suppose that s
=
5. We do as in Case 2.1 to get
(37)
∑
i
=
1
3
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
.
From (19)–(23), (37), (a), and (b), we get
(38)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
4
)
+
K
T
(
r
,
f
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
K
T
(
r
,
f
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
=
4
T
(
r
,
f
)
+
S
(
r
)
,
and from (38), we deduce that
(39)
N
¯
(
r
,
1
f
-
a
4
)
=
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
(
r
,
1
f
-
a
6
)
=
λ
T
(
r
,
f
)
+
S
(
r
)
.
Similarly, we have
(40)
N
¯
(
r
,
1
f
-
a
i
)
=
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1,2
,
3
,
and from (40) and (b), it follows that
(41)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
=
4
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
,
which yields
(42)
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
=
S
(
r
)
.
From (21) and (42) (remembering k
5
<
∞), we have T
(
r
,
f
)
=
S
(
r
), which is impossible. That means that k
5
=
∞. Therefore, from (19) and (20), we get (e). By using the idea which is used in Case 1, then, from (a), (b), and (e) and for all 1
≤
i
≤
4, we observe
(43)
3
T
(
r
,
f
)
=
∑
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
.
In the same way, for all 1
≤
i, j
≤
4 and i
≠
j, we get
(44)
3
T
(
r
,
f
)
=
∑
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
1
f
-
a
i
)
.
It is obvious that (c) follows from (43) and (44). It follows from (c) that, for all 1
≤
j
≤
4,
(45)
3
T
(
r
,
f
)
≤
∑
i
≠
j
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
3
N
¯
(
r
,
1
f
-
a
j
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
3
N
¯
(
r
,
1
f
-
a
j
)
+
S
(
r
)
,
3
T
(
r
,
f
)
≤
∑
i
=
1
3
‍
N
¯
(
r
,
1
f
-
a
i
)
+
3
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
3
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
.
From the last two inequalities, we deduce (d).Let1
≤
j
≤
6. We select six small functions A
i (1
≤
i
≤
6) of f and g that are not all zero such that the following two functions
(46)
H
:
=
H
(
f
,
g
)
=
A
1
f
2
g
+
A
2
f
g
+
A
3
f
2
+
A
4
f
+
A
5
g
+
A
6
satisfy H
(
a
i
,
b
i
)
≡
0 for i
=
1
,
…
,
6 with i
≠
j. Since f is not a quasi-Möbius transformation of g, it follows from Lemma 10 that if H
≡
0, then T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
), and this contradicts (a). So, H
≢
0. It follows from (a) and (b), and by applying Lemma 12 to the function H, that
(47)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
≤
N
¯
(
r
1
H
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
,
which yields
(48)
T
(
r
,
f
)
≤
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
,
and from this and (e), we get
(49)
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
≤
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
S
(
r
)
,
j
=
1
,
…
,
6
,
and it follows from (49) and Lemma 11 that, for all 1
≤
j
≤
6, we get
(50)
4
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
≤
∑
i
≠
j
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
S
(
r
)
.
From (49) and (50), we get (f).Suppose thata
i
≡
b
i, for i
=
1
,
…
,
4. By using (a), (b), and Lemma 11, we have
(51)
3
T
(
r
,
f
)
=
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
S
(
r
)
≤
N
¯
(
r
,
1
f
-
g
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
S
(
r
)
,
which gives
(52)
2
T
(
r
,
f
)
=
∑
i
=
1
4
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
S
(
r
)
,
2
T
(
r
,
f
)
=
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
.
From (52) and (c), we obtain
(53)
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
=
T
(
r
,
f
)
+
S
(
r
)
,
i
,
j
=
1
,
…
,
4
,
i
≠
j
,
which yields
(54)
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
=
1
2
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1
,
…
,
4
.
From (52) and (54), we get (g).By using the same technique as in the proof of (g), we prove (h). Iff
,
a
1
,
…
,
a
6 are interchanged with g
,
b
1
,
…
,
b
6, respectively, then from (14)–(19), we deduce that Theorem 3 is clear. This completes the proof of Theorem 3, when F
≢
0.Now, assume thatF
≡
0. Therefore, by Lemma 10, we deduce that T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
). If the possibility G
≡
0, we deduce that T
(
r
,
f
)
=
2
T
(
r
,
g
)
+
S
(
r
), which is impossible. That means F and G are not simultaneously zero, and hence G
≢
0. We use the same way when F
≢
0 to show that Theorem 3 is clear. This proves Theorem 3.


### 3.2. Proof of Theorem8

Assume thatf is not a quasi-Möbius transformation of g. Let F and G be defined as in (12) and (13). Suppose that F
≢
0. Note that
(55)
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
≤
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
λ
1
+
1
k
5
+
1
+
1
k
6
+
1
)
T
(
r
,
f
)
+
S
(
r
)
.
Since F
(
a
i
,
b
i
)
≡
0, for all 1
≤
i
≤
5, it follows from (55), and by applying Lemmas 11 and 12, that
(56)
4
T
(
r
,
f
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
≤
N
¯
(
r
,
1
F
)
+
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
≤
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
2
+
K
+
λ
1
)
T
(
r
,
f
)
+
T
(
r
,
g
)
+
S
(
r
)
,
which implies that
(57)
2
T
(
r
,
f
)
≤
T
(
r
,
g
)
+
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
.
Of course, if F
≡
0, then according to Lemma 10, we get T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
), which means that (57) remains clear when F
≡
0. Similarly, we have
(58)
2
T
(
r
,
f
)
≤
T
(
r
,
g
)
+
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1
,
…
,
5
.
Symmetrically, by using (13) we have
(59)
2
T
(
r
,
g
)
≤
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
(
K
+
λ
2
)
T
(
r
,
g
)
+
S
(
r
)
,
i
=
1
,
…
,
5
.Suppose thatF
≡
0. Then, by Lemma 10, we deduce that T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
), and G
≢
0 (otherwise, we have T
(
r
,
f
)
=
2
T
(
r
,
g
)
+
S
(
r
)). It follows from (59) that
(60)
10
T
(
r
,
g
)
≤
5
T
(
r
,
f
)
+
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
5
(
K
+
λ
2
)
T
(
r
,
g
)
+
S
(
r
)
≤
5
T
(
r
,
f
)
+
N
¯
(
r
,
1
G
)
+
5
(
K
+
λ
2
)
T
(
r
,
g
)
+
S
(
r
)
≤
6
T
(
r
,
f
)
+
(
5
(
K
+
λ
2
)
+
2
)
T
(
r
,
g
)
+
S
(
r
)
≤
(
5
(
K
+
λ
2
)
+
5
)
T
(
r
,
g
)
+
S
(
r
)
,
and the inequality (60) tells us that 1
≤
K
+
λ
2; that is, K
+
λ
2
=
1; this together with the inequality (5), we get a contradiction. So, F
≢
0. In the same way, we prove that G
≢
0. From (58), we have
(61)
10
T
(
r
,
f
)
≤
5
T
(
r
,
g
)
+
N
¯
(
r
,
1
F
)
+
5
(
P
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
≤
6
T
(
r
,
g
)
+
(
5
(
K
+
λ
1
)
+
2
)
T
(
r
,
f
)
+
S
(
r
)
,
which yields
(62)
(
8
-
5
(
K
+
λ
1
)
)
T
(
r
,
f
)
≤
6
T
(
r
,
g
)
+
S
(
r
)
.
In the same method as the above and by using (59), we can show that
(63)
(
8
-
5
(
K
+
λ
2
)
)
T
(
r
,
g
)
≤
6
T
(
r
,
f
)
+
S
(
r
)
.
From (62) and (63), we obtain (
8
-
5
(
λ
1
+
K
)
)
(
8
-
5
(
λ
2
+
K
)
)
≤
36, which is a contradiction with (5). That means that f should be a quasi-Möbius transformation of g. This proves Theorem 8.


## 3.1. Proof of Theorem3

It is not difficult to find six small functionsc
i (1
≤
i
≤
6) of f and g that are not all zero and six other small functions d
i (1
≤
i
≤
6) of f and g that are not all zero, such that
(11)
c
1
a
i
2
b
i
+
c
2
a
i
b
i
+
c
3
a
i
2
+
c
4
a
i
+
c
5
b
i
+
c
6
≡
0
,
2222222222222222222222
i
=
1,2
,
3,4
,
5
,
d
1
a
i
b
i
2
+
d
2
a
i
b
i
+
d
3
b
i
2
+
d
4
a
i
+
d
5
b
i
+
d
6
≡
0
,
2222222222222222222222222
i
=
1,2
,
3,4
,
5
.
Consequently, the following two functions:
(12)
F
:
=
F
(
f
,
g
)
=
c
1
f
2
g
+
c
2
f
g
+
c
3
f
2
+
c
4
f
+
c
5
g
+
c
6
,
(13)
G
:
=
G
(
f
,
g
)
=
d
1
f
g
2
+
d
2
f
g
+
d
3
g
2
+
d
4
f
+
d
5
g
+
d
6
satisfy F
(
a
i
,
b
i
)
≡
0 and G
(
a
i
,
b
i
)
≡
0, for (1
≤
i
≤
5).Suppose thatF
≢
0. For all 1
≤
i
≤
5, we denote by N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
) the reduced counting function of those a
i-points of f whose multiplicities are greater than or equal to k
i
+
1, that are not the b
i-points of g in |
z
|
<
r. By using Lemmas 11 and 12, we have
(14)
4
T
(
r
,
f
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
∑
i
=
1
5
N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
N
¯
k
5
)
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
≤
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
∑
i
=
1
5
‍
N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
≤
N
¯
(
r
,
1
F
)
+
λ
T
(
r
,
f
)
+
∑
i
=
1
5
‍
1
k
i
+
1
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
T
(
r
,
g
)
+
(
λ
+
K
)
T
(
r
,
f
)
+
S
(
r
)
,
and the inequality (14) gives us
(15)
(
2
-
K
-
λ
)
T
(
r
,
f
)
≤
T
(
r
,
g
)
+
S
(
r
)
.
In the same method as the above, it can be shown that
(16)
4
T
(
r
,
g
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
1
g
-
b
i
)
+
S
(
r
)
≤
∑
i
=
1
5
‍
N
¯
(
r
,
g
=
b
i
,
f
=
a
i
)
+
∑
i
=
1
5
‍
N
¯
(
k
i
+
1
(
r
,
g
=
b
i
,
f
≠
a
i
)
+
λ
T
(
r
,
g
)
+
S
(
r
)
≤
N
¯
(
r
,
1
F
)
+
λ
T
(
r
,
g
)
+
∑
i
=
1
5
‍
1
k
i
+
1
N
(
r
,
1
g
-
b
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
T
(
r
,
g
)
+
(
λ
+
K
)
T
(
r
,
g
)
+
S
(
r
)
,
and this gives us
(17)
(
3
-
K
-
λ
)
T
(
r
,
g
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
.
It follows from (15) and (17) that
(18)
(
3
-
K
-
λ
)
T
(
r
,
g
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
≤
2
2
-
K
-
λ
T
(
r
,
g
)
+
S
(
r
)
,
which means that (
3
-
K
-
λ
)
(
2
-
K
-
λ
)
≤
2, and hence (
K
+
λ
-
4
)
(
K
+
λ
-
1
)
≤
0. Since K
+
λ
≤
1 (this follows from the assumption of Theorem 3), then, from the last inequality, we deduce that 0
≤
K
+
λ
-
1, which means that
(19)
K
+
λ
=
1
.
Consequently, we deduce (a) from (15), (17), and (19), and then from (14), we get (b). Also, (14) gives us
(20)
N
¯
k
5
)
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
=
λ
T
(
r
,
f
)
.
From (a), (14), and (19), we deduce the following relation:
(21)
N
¯
(
k
i
+
1
(
r
,
f
=
a
i
,
g
≠
b
i
)
=
1
k
i
+
1
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
=
1
k
i
+
1
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1
,
…
,
5
.Let us now proveK
=
0. We first assume that there exits 1
≤
i
≤
5 such that k
i
<
∞. Then from (21), we have
(22)
N
k
i
)
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
N
f
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
k
i
(
r
,
f
=
a
i
)
=
S
(
r
)
,
where N
f
(
r
,
f
=
a
i
,
g
=
b
i
) is the counting function of the points f
-
a
i
=
0 that are zeros of g
-
b
i
=
0, and N
k
i
(
r
,
f
=
a
i
) is the counting function of those a
i-points of f whose multiplicities are not equal to k
i
+
1.It follows from the preceding equation and (21) that we have
(23)
T
(
r
,
f
)
=
(
k
i
+
1
)
N
¯
(
r
,
f
=
a
i
,
g
≠
b
i
)
+
S
(
r
)
=
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
,
if
k
i
<
∞
,
1
≤
i
≤
5
.Let us consider two cases.Case 1.
Suppose that there are two values ofk
i, (i
=
1
,
…
,
5) that are finite. Suppose that k
1
<
∞ and k
2
<
∞. Let Ł be a quasi-Möbius transformation such that a
j
≡
Ł
(
b
j
), j
=
3,4
,
5. Since f is not a quasi-Möbius transformation of g, then f
≢
Ł
(
g
), and then by (a), we have
(24)
∑
i
=
3
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
≤
N
¯
(
r
,
1
f
-
Ł
(
g
)
)
+
∑
i
=
3
4
‍
1
k
i
+
1
N
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
1
k
3
+
1
+
1
k
4
+
1
)
T
(
r
,
f
)
+
S
(
r
)
.
It follows from (19)–(23), (a), and (b) that
(25)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
≤
(
1
k
1
+
1
+
1
k
2
+
1
)
T
(
r
,
f
)
+
∑
i
=
3
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
∑
i
=
1
4
‍
1
k
i
+
1
)
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
≤
(
2
+
K
)
T
(
r
,
f
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
=
3
T
(
r
,
f
)
+
S
(
r
)
,
which is impossible.By using the above method, it can be proved that there are no other two values ofk
i, (i
=
1
,
…
,
5) that are simultaneously finite. Therefore, this case cannot occur.Case 2.
Suppose that there is1
≤
s
≤
5 such that k
s
<
∞ and k
i
=
∞, for all 1
≤
i
≤
5 with i
≠
s.Case  2.1. Suppose that s
=
4. That is, k
i
=
∞, for all 1
≤
i
≤
5 with i
≠
4, and hence, K
=
1
/
(
1
+
k
4
). As in Case 1, let Ł be a quasi-Möbius transformation such that a
j
≡
Ł
(
b
j
), j
=
1,2
,
3. Since f is not a quasi-Möbius transformation of g, then f
≢
Ł
(
g
). From (23) and (a), we see
(26)
∑
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
≤
N
¯
(
r
,
1
f
-
Ł
(
g
)
)
+
1
k
4
+
1
N
(
r
,
1
f
-
a
4
)
+
S
(
r
)
≤
(
2
+
1
k
4
+
1
)
T
(
r
,
f
)
+
S
(
r
)
=
(
2
+
K
)
T
(
r
,
f
)
+
S
(
r
)
.
From (19)–(23) and (26) we have
(27)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
K
)
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
≤
(
2
+
K
+
λ
)
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
4
T
(
r
,
f
)
+
S
(
r
)
,
which gives
(28)
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
=
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
=
λ
T
(
r
,
f
)
.
Thus, from (28), we deduce that
(29)
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
=
S
(
r
)
.
That means
(30)
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
=
λ
T
(
r
,
f
)
.
On the other hand, by using (23)–(29) we get
(31)
∑
i
=
1,2
,
5
‍
N
¯
(
r
,
1
f
-
a
i
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
.It follows from (19), (23), (30), (31) and by using (a) and (b), we see
(32)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
(
2
+
K
+
λ
)
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
3
)
+
S
(
r
)
≤
4
T
(
r
,
f
)
+
S
(
r
)
;
thus, the inequality (32) gives us
(33)
N
¯
(
r
,
1
f
-
a
3
)
=
T
(
r
,
f
)
+
S
(
r
)
.
Similarly, we have
(34)
N
¯
(
r
,
1
f
-
a
i
)
=
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1,2
.
From (28), (29), (a), and (b), we get
(35)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
=
4
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
4
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
,
which yields
(36)
N
¯
(
r
,
1
f
-
a
4
)
+
N
¯
(
r
,
1
f
-
a
6
)
=
S
(
r
)
.
From (21) and (36) (remembering k
4
<
∞), we have T
(
r
,
f
)
=
S
(
r
), which is impossible. That means that k
4
=
∞. In the same method as the above, we can prove that k
i
=
∞, i
=
1,2
,
3. This proves that f and g share (
a
i
,
b
i
,
∞
)
*IM, for all 1
≤
i
≤
4.Case  2.2. Suppose that s
=
5. We do as in Case 2.1 to get
(37)
∑
i
=
1
3
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
S
(
r
)
.
From (19)–(23), (37), (a), and (b), we get
(38)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
4
)
+
K
T
(
r
,
f
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
K
T
(
r
,
f
)
+
λ
T
(
r
,
f
)
+
S
(
r
)
=
4
T
(
r
,
f
)
+
S
(
r
)
,
and from (38), we deduce that
(39)
N
¯
(
r
,
1
f
-
a
4
)
=
T
(
r
,
f
)
+
S
(
r
)
,
N
¯
(
r
,
1
f
-
a
6
)
=
λ
T
(
r
,
f
)
+
S
(
r
)
.
Similarly, we have
(40)
N
¯
(
r
,
1
f
-
a
i
)
=
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1,2
,
3
,
and from (40) and (b), it follows that
(41)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
=
4
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
+
S
(
r
)
,
which yields
(42)
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
=
S
(
r
)
.
From (21) and (42) (remembering k
5
<
∞), we have T
(
r
,
f
)
=
S
(
r
), which is impossible. That means that k
5
=
∞. Therefore, from (19) and (20), we get (e). By using the idea which is used in Case 1, then, from (a), (b), and (e) and for all 1
≤
i
≤
4, we observe
(43)
3
T
(
r
,
f
)
=
∑
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
.
In the same way, for all 1
≤
i, j
≤
4 and i
≠
j, we get
(44)
3
T
(
r
,
f
)
=
∑
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
1
f
-
a
i
)
+
N
¯
(
r
,
1
f
-
a
i
)
.
It is obvious that (c) follows from (43) and (44). It follows from (c) that, for all 1
≤
j
≤
4,
(45)
3
T
(
r
,
f
)
≤
∑
i
≠
j
i
=
1
4
‍
N
¯
(
r
,
1
f
-
a
i
)
+
3
N
¯
(
r
,
1
f
-
a
j
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
3
N
¯
(
r
,
1
f
-
a
j
)
+
S
(
r
)
,
3
T
(
r
,
f
)
≤
∑
i
=
1
3
‍
N
¯
(
r
,
1
f
-
a
i
)
+
3
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
3
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
.
From the last two inequalities, we deduce (d).Let1
≤
j
≤
6. We select six small functions A
i (1
≤
i
≤
6) of f and g that are not all zero such that the following two functions
(46)
H
:
=
H
(
f
,
g
)
=
A
1
f
2
g
+
A
2
f
g
+
A
3
f
2
+
A
4
f
+
A
5
g
+
A
6
satisfy H
(
a
i
,
b
i
)
≡
0 for i
=
1
,
…
,
6 with i
≠
j. Since f is not a quasi-Möbius transformation of g, it follows from Lemma 10 that if H
≡
0, then T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
), and this contradicts (a). So, H
≢
0. It follows from (a) and (b), and by applying Lemma 12 to the function H, that
(47)
4
T
(
r
,
f
)
=
∑
i
=
1
6
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
≤
N
¯
(
r
1
H
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
,
which yields
(48)
T
(
r
,
f
)
≤
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
N
¯
(
r
,
f
=
a
5
,
g
≠
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
≠
b
6
)
+
S
(
r
)
,
and from this and (e), we get
(49)
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
≤
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
S
(
r
)
,
j
=
1
,
…
,
6
,
and it follows from (49) and Lemma 11 that, for all 1
≤
j
≤
6, we get
(50)
4
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
≤
∑
i
≠
j
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
S
(
r
)
.
From (49) and (50), we get (f).Suppose thata
i
≡
b
i, for i
=
1
,
…
,
4. By using (a), (b), and Lemma 11, we have
(51)
3
T
(
r
,
f
)
=
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
S
(
r
)
≤
N
¯
(
r
,
1
f
-
g
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
2
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
≤
3
T
(
r
,
f
)
+
S
(
r
)
,
which gives
(52)
2
T
(
r
,
f
)
=
∑
i
=
1
4
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
S
(
r
)
,
2
T
(
r
,
f
)
=
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
S
(
r
)
.
From (52) and (c), we obtain
(53)
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
N
¯
(
r
,
f
=
a
j
,
g
=
b
j
)
=
T
(
r
,
f
)
+
S
(
r
)
,
i
,
j
=
1
,
…
,
4
,
i
≠
j
,
which yields
(54)
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
=
1
2
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1
,
…
,
4
.
From (52) and (54), we get (g).By using the same technique as in the proof of (g), we prove (h). Iff
,
a
1
,
…
,
a
6 are interchanged with g
,
b
1
,
…
,
b
6, respectively, then from (14)–(19), we deduce that Theorem 3 is clear. This completes the proof of Theorem 3, when F
≢
0.Now, assume thatF
≡
0. Therefore, by Lemma 10, we deduce that T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
). If the possibility G
≡
0, we deduce that T
(
r
,
f
)
=
2
T
(
r
,
g
)
+
S
(
r
), which is impossible. That means F and G are not simultaneously zero, and hence G
≢
0. We use the same way when F
≢
0 to show that Theorem 3 is clear. This proves Theorem 3.


## 3.2. Proof of Theorem8

Assume thatf is not a quasi-Möbius transformation of g. Let F and G be defined as in (12) and (13). Suppose that F
≢
0. Note that
(55)
N
¯
(
r
,
1
f
-
a
5
)
+
N
¯
(
r
,
1
f
-
a
6
)
≤
N
¯
(
r
,
f
=
a
5
,
g
=
b
5
)
+
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
λ
1
+
1
k
5
+
1
+
1
k
6
+
1
)
T
(
r
,
f
)
+
S
(
r
)
.
Since F
(
a
i
,
b
i
)
≡
0, for all 1
≤
i
≤
5, it follows from (55), and by applying Lemmas 11 and 12, that
(56)
4
T
(
r
,
f
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
1
f
-
a
i
)
+
S
(
r
)
≤
∑
i
=
1
6
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
≤
N
¯
(
r
,
1
F
)
+
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
≤
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
2
+
K
+
λ
1
)
T
(
r
,
f
)
+
T
(
r
,
g
)
+
S
(
r
)
,
which implies that
(57)
2
T
(
r
,
f
)
≤
T
(
r
,
g
)
+
N
¯
(
r
,
f
=
a
6
,
g
=
b
6
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
.
Of course, if F
≡
0, then according to Lemma 10, we get T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
), which means that (57) remains clear when F
≡
0. Similarly, we have
(58)
2
T
(
r
,
f
)
≤
T
(
r
,
g
)
+
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
(
K
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
,
i
=
1
,
…
,
5
.
Symmetrically, by using (13) we have
(59)
2
T
(
r
,
g
)
≤
T
(
r
,
f
)
+
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
(
K
+
λ
2
)
T
(
r
,
g
)
+
S
(
r
)
,
i
=
1
,
…
,
5
.Suppose thatF
≡
0. Then, by Lemma 10, we deduce that T
(
r
,
g
)
=
2
T
(
r
,
f
)
+
S
(
r
), and G
≢
0 (otherwise, we have T
(
r
,
f
)
=
2
T
(
r
,
g
)
+
S
(
r
)). It follows from (59) that
(60)
10
T
(
r
,
g
)
≤
5
T
(
r
,
f
)
+
∑
i
=
1
5
‍
N
¯
(
r
,
f
=
a
i
,
g
=
b
i
)
+
5
(
K
+
λ
2
)
T
(
r
,
g
)
+
S
(
r
)
≤
5
T
(
r
,
f
)
+
N
¯
(
r
,
1
G
)
+
5
(
K
+
λ
2
)
T
(
r
,
g
)
+
S
(
r
)
≤
6
T
(
r
,
f
)
+
(
5
(
K
+
λ
2
)
+
2
)
T
(
r
,
g
)
+
S
(
r
)
≤
(
5
(
K
+
λ
2
)
+
5
)
T
(
r
,
g
)
+
S
(
r
)
,
and the inequality (60) tells us that 1
≤
K
+
λ
2; that is, K
+
λ
2
=
1; this together with the inequality (5), we get a contradiction. So, F
≢
0. In the same way, we prove that G
≢
0. From (58), we have
(61)
10
T
(
r
,
f
)
≤
5
T
(
r
,
g
)
+
N
¯
(
r
,
1
F
)
+
5
(
P
+
λ
1
)
T
(
r
,
f
)
+
S
(
r
)
≤
6
T
(
r
,
g
)
+
(
5
(
K
+
λ
1
)
+
2
)
T
(
r
,
f
)
+
S
(
r
)
,
which yields
(62)
(
8
-
5
(
K
+
λ
1
)
)
T
(
r
,
f
)
≤
6
T
(
r
,
g
)
+
S
(
r
)
.
In the same method as the above and by using (59), we can show that
(63)
(
8
-
5
(
K
+
λ
2
)
)
T
(
r
,
g
)
≤
6
T
(
r
,
f
)
+
S
(
r
)
.
From (62) and (63), we obtain (
8
-
5
(
λ
1
+
K
)
)
(
8
-
5
(
λ
2
+
K
)
)
≤
36, which is a contradiction with (5). That means that f should be a quasi-Möbius transformation of g. This proves Theorem 8.



---

*Source: 378958-2014-05-22.xml*