Weakly Weighted Sharing and Uniqueness of Meromorphic Functions

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.

1. Introduction and Main Results

In this paper, a meromorphic function means meromorphic in the whole complex plane ℂ. 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.

Let 𝑓 be a nonconstant meromorphic function. A meromorphic function 𝑎 is called a small 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.

Let 𝑓 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]). Let 𝑓 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]). Let 𝑓(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.

Let 𝑓 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 𝑁1𝑟,−𝑓−𝑎𝑁𝑟,𝑓=𝑎,𝑔=𝑏=𝑆𝑟,𝑓,𝑁1𝑟,−𝑔−𝑏𝑁𝑟,𝑓=𝑎,𝑔=𝑏=𝑆𝑟,𝑔. (1)

As in Theorem A and throughout this paper, when 𝑓 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 by 𝑁(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 the weakly weighted IM sharing which is given by the following definition.

Definition 2 (see [5]). Let 𝑘 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 if 𝑓 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]). Let 𝑓 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)𝐼𝑀^∗, 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)3𝑇(𝑟,𝑓)=∑⁵_𝑖=1𝑁(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)≤3𝑁(𝑟,1/(𝑓−𝑎_𝑖))+𝑆(𝑟), 𝑖=1,…,5;(e)𝑇(𝑟,𝑓)=𝑁(𝑟,1/(𝑓−𝑎₆))+𝑆(𝑟);(f)𝑁(𝑟,𝑓=𝑎₆,𝑔=𝑏₆)≤(3/5)𝑇(𝑟,𝑓)+𝑆(𝑟);(g)𝑇(𝑟,𝑓)=𝑁(𝑟,1/(𝑓−𝑎₅))+𝑆(𝑟) and 𝑇(𝑟,𝑓)=2𝑁(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. Let 𝑓 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 𝑘₁,…,𝑘₅ be five positive integers or infinity with 𝐾≡∑⁵_𝑖=1(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 𝑁1𝑟,𝑓−𝑎₆+𝑁_𝑘5)𝑟,𝑓=𝑎₅,𝑔≠𝑏₅≤𝜆𝑇𝑟,𝑓+𝑆(𝑟),𝑁1𝑟,𝑔−𝑏₆+𝑁_𝑘5)𝑟,𝑔=𝑏₅,𝑓≠𝑎₅≤𝜆𝑇𝑟,𝑔+𝑆(𝑟), (2) 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)4𝑇(𝑟,𝑓)=∑⁶_𝑖=1𝑁(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)≤3𝑁(𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖)+𝑆(𝑟), 𝑖=1,…,5;(e)𝑇(𝑟,𝑓)=𝑁(𝑟,1/(𝑓−𝑎₆))+𝑁(𝑟,𝑓=𝑎₅,𝑔≠𝑏₅)+𝑆(𝑟);(f)𝑁(𝑟,𝑓=𝑎₆,𝑔=𝑏₆)≤𝑁(𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖)+𝑆(𝑟) and 𝑁(𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖)≤3𝑇(𝑟,𝑓)− 4𝑁(𝑟,𝑓=𝑎₆,𝑔=𝑏₆)+𝑆(𝑟), 𝑖=1,…,6;(g)𝑇(𝑟,𝑓)=𝑁(𝑟,𝑓=𝑎₅,𝑔=𝑏₅)+𝑆(𝑟) and 𝑇(𝑟,𝑓)=2𝑁(𝑟,1/(𝑓−𝑎_𝑖))+𝑆(𝑟) for 𝑖=1,…,4 if 𝑎_𝑖≡𝑏_𝑖, 𝑖=1,…,4;(h)𝑇(𝑟,𝑓)=2𝑁(𝑟,𝑓=𝑎₅,𝑔=𝑎₅)+𝑆(𝑟),𝑇(𝑟,𝑓)=𝑁(𝑟,1/(𝑓−𝑎₄))+𝑆(𝑟) and 𝑇(𝑟,𝑓)=2𝑁(𝑟,1/ (𝑓−𝑎_𝑖))+𝑆(𝑟) for 𝑖=1,2,3 if 𝑎₅≡𝑏₅ and 𝑎_𝑖≡𝑏_𝑖, 𝑖=1,2,3.The theorem is true, if 𝑓,𝑎₁,…,𝑎₆ are interchanged with 𝑔,𝑏₁,…,𝑏₆, respectively.

Remark 4. In Theorem 3, 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 Theorem 3, we can immediately obtain the following corollary.

Corollary 5. Let 𝑓 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)𝐼𝑀^∗, for 𝑖=1,…,4 and 𝑁1𝑟,𝑓−𝑎₆+𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅≤𝑇𝑟,𝑓+𝑆(𝑟),𝑁1𝑟,𝑔−𝑏₆+𝑁𝑟,𝑔=𝑏₅,𝑓≠𝑎₅≤𝑇𝑟,𝑔+𝑆(𝑟). (3) If there exists a number 𝜆∈[0,2/5) such that 𝑁𝑟,𝑓=𝑎₆,𝑔≠𝑏₆+𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅≤𝜆𝑇𝑟,𝑓+𝑆(𝑟), (4) 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=𝑎₆,𝑔≠𝑏₆)≤𝜆𝑇(𝑟,𝑓)+𝑆(𝑟). It is evident that Corollary 5 improves Theorem A and the result of Li and Yang [6, Corollary 2].

Obviously, Theorem 3 is a generalization of Theorem B, and Corollary 5 is a generalization of Theorem A and Corollary 2 in [6].

Example 7. Let 𝑓 and 𝑔 be defined as in Example 1, and let (𝑎₁,𝑏₁)=(0,0), (𝑎₂,𝑏₂)=(∞,∞), (𝑎₃,𝑏₃)=(1,1), (𝑎₄,𝑏₄)=(−1/8,−1/8), (𝑎₅,𝑏₅)=(−1/2,1/4) and 𝑎₆=𝑏₆=1/4+√3/4. It is easy to show that 𝑁(𝑟,𝑓=𝑎₆,𝑔≠𝑏₆)=𝑇(𝑟,𝑓), 𝑁(𝑟,𝑓=𝑎₅,𝑔≠𝑏₅)=𝑆(𝑟), 𝑁(𝑟,𝑔=𝑏₅,𝑓≠𝑎₅)=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. Let 𝑓 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 𝑘₁,…,𝑘₆ be six positive integers or infinity with 𝐾≡∑⁶_𝑖=1(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 𝜆₁ and 𝜆₂ inside the interval [0,1−𝐾] such that 36<8−5𝜆₁+𝐾8−5𝜆₂+𝐾 (5) and the following two inequalities hold 𝑁_𝑘6)𝑟,𝑓=𝑎₆,𝑔≠𝑏₆+𝑁_𝑘5)𝑟,𝑓=𝑎₅,𝑔≠𝑏₅≤𝜆₁𝑇𝑟,𝑓+𝑆(𝑟),𝑁_𝑘6)𝑟,𝑔=𝑏₆,𝑓≠𝑎₆+𝑁_𝑘5)𝑟,𝑔=𝑏₅,𝑓≠𝑎₅≤𝜆₂𝑇𝑟,𝑔+𝑆(𝑟), (6) then 𝑓 is a quasi-Möbius transformation of 𝑔.

We deduce, from Theorem 8, the following corollary which is an improvement to Theorem A.

Corollary 9. Let 𝑓 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)𝐼𝑀^∗, 1≤𝑖≤4. If there are two numbers 𝜆₁ and 𝜆₂ inside the interval [0,1] such that 36<(8−5𝜆₁)(8−5𝜆₂) and the following two inequalities hold 𝑁𝑟,𝑓=𝑎₆,𝑔≠𝑏₆+𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅≤𝜆₁𝑇𝑟,𝑓+𝑆(𝑟),𝑁𝑟,𝑔=𝑏₆,𝑓≠𝑎₆+𝑁𝑟,𝑔=𝑏₅,𝑓≠𝑎₅≤𝜆₂𝑇𝑟,𝑔+𝑆(𝑟), (7) 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. Let 𝑓 and 𝑔 be two nonconstant meromorphic functions such that 𝐹=𝑐₁𝑓²𝑔+𝑐₂𝑓𝑔+𝑐₃𝑓²+𝑐₄𝑓+𝑐₅𝑔+𝑐₆≡0, (8) where 𝑐₁,…,𝑐₆ 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)=2𝑇(𝑟,𝑓)+𝑆(𝑟).

Proof. Since 𝐹≡0, then 𝑔(𝑐₁𝑓²+𝑐₂𝑓+𝑐₅)=−(𝑐₃𝑓²+𝑐₄𝑓+𝑐₆). We note that 𝑐₁𝑓²+𝑐₂𝑓+𝑐₅≢0, because at least one of {𝑐₁,…,𝑐₆} is not zero. Hence, 𝑐𝑔=−³𝑓²+𝑐₄𝑓+𝑐₆𝑐₁𝑓²+𝑐₂𝑓+𝑐₅. (9) 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)=2𝑇(𝑟,𝑓)+𝑆(𝑟). This completes the proof of Lemma 10.

Lemma 11 (see [10]). Let 𝑓 be a nonconstant meromorphic function and let 𝑎₁,…,𝑎_𝑛 be 𝑛 distinct small functions of 𝑓. Then (𝑛−2−𝜖)𝑇𝑟,𝑓≤^𝑛_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖+𝑆𝑟,𝑓 (10) holds for any positive number 𝜖.

Lemma 12 (see [6]). Suppose that 𝑓 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)≤𝑝𝑇(𝑟,𝑓)+𝑞𝑇(𝑟,𝑔)+𝑆(𝑟).

3. Proofs of Theorems 3 and 8

If 𝑎₆=∞ (or 𝑏₆=∞), 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,𝑎₆=0, (or 𝑏₆=0, if 𝑏₆=∞), 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 Theorem 3

It is not difficult to find six small functions 𝑐_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 𝑐₁𝑎²_𝑖𝑏_𝑖+𝑐₂𝑎_𝑖𝑏_𝑖+𝑐₃𝑎²_𝑖+𝑐₄𝑎_𝑖+𝑐₅𝑏_𝑖+𝑐₆𝑑≡0,𝑖=1,2,3,4,5,₁𝑎_𝑖𝑏²_𝑖+𝑑₂𝑎_𝑖𝑏_𝑖+𝑑₃𝑏²_𝑖+𝑑₄𝑎_𝑖+𝑑₅𝑏_𝑖+𝑑₆≡0,𝑖=1,2,3,4,5. (11) Consequently, the following two functions: 𝐹∶=𝐹𝑓,𝑔=𝑐₁𝑓²𝑔+𝑐₂𝑓𝑔+𝑐₃𝑓²+𝑐₄𝑓+𝑐₅𝑔+𝑐₆,𝐺∶=𝐺𝑓,𝑔=𝑑₁𝑓𝑔²+𝑑₂𝑓𝑔+𝑑₃𝑔²+𝑑₄𝑓+𝑑₅𝑔+𝑑₆ (12)(13) satisfy 𝐹(𝑎_𝑖,𝑏_𝑖)≡0 and 𝐺(𝑎_𝑖,𝑏_𝑖)≡0, for (1≤𝑖≤5).

Suppose that 𝐹≢0. For all 1≤𝑖≤5, we denote by 𝑁_(𝑘𝑖+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 4𝑇𝑟,𝑓≤⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)≤+𝑆_5𝑖=1𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+⁵_𝑖=1𝑁_(𝑘𝑖+1𝑟,𝑓=𝑎_𝑖,𝑔≠𝑏_𝑖+𝑁_𝑘5)𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁1𝑟,𝑓−𝑎₆(𝑟)≤+𝑆_5𝑖=1𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+⁵_𝑖=1𝑁_(𝑘𝑖+1𝑟,𝑓=𝑎_𝑖,𝑔≠𝑏_𝑖(𝑟)≤+𝜆𝑇𝑟,𝑓+𝑆𝑁1𝑟,𝐹++𝜆𝑇𝑟,𝑓_5𝑖=11𝑘_𝑖𝑁1+1𝑟,𝑓−𝑎_𝑖(𝑟)+()𝑇(𝑟),+𝑆≤2𝑇𝑟,𝑓+𝑇𝑟,𝑔𝜆+𝐾𝑟,𝑓+𝑆 (14) and the inequality (14) gives us (2−𝐾−𝜆)𝑇𝑟,𝑓≤𝑇𝑟,𝑔+𝑆(𝑟). (15) In the same method as the above, it can be shown that 4𝑇𝑟,𝑔≤⁶_𝑖=1𝑁1𝑟,𝑔−𝑏_𝑖(𝑟)≤+𝑆_5𝑖=1𝑁𝑟,𝑔=𝑏_𝑖,𝑓=𝑎_𝑖+⁵_𝑖=1𝑁_(𝑘𝑖+1𝑟,𝑔=𝑏_𝑖,𝑓≠𝑎_𝑖(𝑟)≤+𝜆𝑇𝑟,𝑔+𝑆𝑁1𝑟,𝐹++𝜆𝑇𝑟,𝑔_5𝑖=11𝑘_𝑖𝑁1+1𝑟,𝑔−𝑏_𝑖(𝑟)+()𝑇(𝑟),+𝑆≤2𝑇𝑟,𝑓+𝑇𝑟,𝑔𝜆+𝐾𝑟,𝑔+𝑆 (16) and this gives us (3−𝐾−𝜆)𝑇𝑟,𝑔≤2𝑇𝑟,𝑓+𝑆(𝑟). (17) It follows from (15) and (17) that ()𝑇(𝑟)≤23−𝐾−𝜆𝑟,𝑔≤2𝑇𝑟,𝑓+𝑆𝑇(𝑟),2−𝐾−𝜆𝑟,𝑔+𝑆 (18) 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 𝐾+𝜆=1. (19) Consequently, we deduce (a) from (15), (17), and (19), and then from (14), we get (b). Also, (14) gives us 𝑁_𝑘5)𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁1𝑟,𝑓−𝑎₆+𝑆(𝑟)=𝜆𝑇𝑟,𝑓. (20) From (a), (14), and (19), we deduce the following relation: 𝑁_(𝑘𝑖+1𝑟,𝑓=𝑎_𝑖,𝑔≠𝑏_𝑖=1𝑘_𝑖𝑁1+1𝑟,𝑓−𝑎_𝑖(𝑟)=1+𝑆𝑘_𝑖𝑇(𝑟)+1𝑟,𝑓+𝑆,𝑖=1,…,5. (21)

Let us now prove 𝐾=0. We first assume that there exits 1≤𝑖≤5 such that 𝑘_𝑖<∞. Then from (21), we have 𝑁_𝑘𝑖)𝑟,𝑓=𝑎_𝑖,𝑔≠𝑏_𝑖+𝑁_𝑓𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+𝑁_𝑘𝑖𝑟,𝑓=𝑎_𝑖=𝑆(𝑟), (22) 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_𝑘𝑖(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 𝑇𝑟,𝑓=𝑘_𝑖+1𝑁𝑟,𝑓=𝑎_𝑖,𝑔≠𝑏_𝑖(𝑟)1+𝑆=𝑁𝑟,𝑓−𝑎_𝑖+𝑆(𝑟),if𝑘_𝑖<∞,1≤𝑖≤5. (23)Let us consider two cases.

Case 1. Suppose that there are two values of 𝑘_𝑖, (𝑖=1,…,5) that are finite. Suppose that 𝑘₁<∞ and 𝑘₂<∞. 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 ⁴_𝑖=3𝑁1𝑟,𝑓−𝑎_𝑖+𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅≤𝑁1𝑟,𝑔+𝑓−Ł⁴_𝑖=31𝑘_𝑖𝑁1+1𝑟,𝑓−𝑎_𝑖(𝑟)≤1+𝑆2+𝑘₃+1+1𝑘₄𝑇(𝑟).+1𝑟,𝑓+𝑆 (24) It follows from (19)–(23), (a), and (b) that 4𝑇𝑟,𝑓=⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖≤1𝑘₁+1+1𝑘₂𝑇++1𝑟,𝑓_6𝑖=3𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)≤+𝑆2+⁴_𝑖=11𝑘_𝑖𝑇++1𝑟,𝑓𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁1𝑟,𝑓−𝑎₆(𝑟)≤()𝑇(𝑟)(𝑟),+𝑆2+𝐾𝑟,𝑓+𝜆𝑇𝑟,𝑓+𝑆=3𝑇𝑟,𝑓+𝑆 (25) which is impossible.

By using the above method, it can be proved that there are no other two values of 𝑘_𝑖, (𝑖=1,…,5) that are simultaneously finite. Therefore, this case cannot occur.

Case 2. Suppose that there is 1≤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+𝑘₄). 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 ⁴_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖≤𝑁1𝑟,𝑔+1𝑓−Ł𝑘₄𝑁1+1𝑟,𝑓−𝑎₄(𝑟)≤1+𝑆2+𝑘₄𝑇(𝑟)=()𝑇(𝑟).+1𝑟,𝑓+𝑆2+𝐾𝑟,𝑓+𝑆 (26) From (19)–(23) and (26) we have 4𝑇𝑟,𝑓=⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)≤()𝑇++𝑆2+𝐾𝑟,𝑓𝑁1𝑟,𝑓−𝑎₅+𝑁1𝑟,𝑓−𝑎₆(𝑟)≤()𝑇++𝑆2+𝐾+𝜆𝑟,𝑓𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅(𝑟)(𝑟),+𝑆≤4𝑇𝑟,𝑓+𝑆 (27) which gives 𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅=𝑇𝑟,𝑓+𝑆(𝑟),𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁1𝑟,𝑓−𝑎₆(𝑟).+𝑆=𝜆𝑇𝑟,𝑓 (28) Thus, from (28), we deduce that 𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅=𝑆(𝑟). (29) That means 𝑁1𝑟,𝑓−𝑎₆+𝑆(𝑟)=𝜆𝑇𝑟,𝑓. (30) On the other hand, by using (23)–(29) we get _𝑖=1,2,5𝑁1𝑟,𝑓−𝑎_𝑖≤2𝑇𝑟,𝑓+𝑆(𝑟). (31)It follows from (19), (23), (30), (31) and by using (a) and (b), we see 4𝑇𝑟,𝑓=⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)≤()𝑇++𝑆2+𝐾+𝜆𝑟,𝑓𝑁1𝑟,𝑓−𝑎₃(𝑟)(𝑟);+𝑆≤4𝑇𝑟,𝑓+𝑆 (32) thus, the inequality (32) gives us 𝑁1𝑟,𝑓−𝑎₃=𝑇𝑟,𝑓+𝑆(𝑟). (33) Similarly, we have 𝑁1𝑟,𝑓−𝑎_𝑖=𝑇𝑟,𝑓+𝑆(𝑟),𝑖=1,2. (34) From (28), (29), (a), and (b), we get 4𝑇𝑟,𝑓=⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)++𝑆=4𝑇𝑟,𝑓𝑁1𝑟,𝑓−𝑎₄+𝑁1𝑟,𝑓−𝑎₆+𝑆(𝑟), (35) which yields 𝑁1𝑟,𝑓−𝑎₄+𝑁1𝑟,𝑓−𝑎₆=𝑆(𝑟). (36) From (21) and (36) (remembering 𝑘₄<∞), we have 𝑇(𝑟,𝑓)=𝑆(𝑟), which is impossible. That means that 𝑘₄=∞. 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 ³_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖+𝑆(𝑟)≤2𝑇𝑟,𝑓+𝑆(𝑟). (37) From (19)–(23), (37), (a), and (b), we get 4𝑇𝑟,𝑓=⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)++𝑆≤2𝑇𝑟,𝑓𝑁1𝑟,𝑓−𝑎₄(𝑟)(𝑟)(𝑟),+𝐾𝑇𝑟,𝑓+𝜆𝑇𝑟,𝑓+𝑆≤3𝑇𝑟,𝑓+𝐾𝑇𝑟,𝑓+𝜆𝑇𝑟,𝑓+𝑆=4𝑇𝑟,𝑓+𝑆 (38) and from (38), we deduce that 𝑁1𝑟,𝑓−𝑎₄=𝑇𝑟,𝑓+𝑆(𝑟),𝑁1𝑟,𝑓−𝑎₆=𝜆𝑇𝑟,𝑓+𝑆(𝑟). (39) Similarly, we have 𝑁1𝑟,𝑓−𝑎_𝑖=𝑇𝑟,𝑓+𝑆(𝑟),𝑖=1,2,3, (40) and from (40) and (b), it follows that 4𝑇𝑟,𝑓=⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)++𝑆=4𝑇𝑟,𝑓𝑁1𝑟,𝑓−𝑎₅+𝑁1𝑟,𝑓−𝑎₆+𝑆(𝑟), (41) which yields 𝑁1𝑟,𝑓−𝑎₅+𝑁1𝑟,𝑓−𝑎₆=𝑆(𝑟). (42) From (21) and (42) (remembering 𝑘₅<∞), we have 𝑇(𝑟,𝑓)=𝑆(𝑟), which is impossible. That means that 𝑘₅=∞. 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 3𝑇𝑟,𝑓=⁴_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖+𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅(𝑟)++𝑆≤2𝑇𝑟,𝑓𝑁1𝑟,𝑓−𝑎_𝑖+𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅+𝑆(𝑟). (43) In the same way, for all 1≤𝑖, 𝑗≤4 and 𝑖≠𝑗, we get 3𝑇𝑟,𝑓=⁴_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖+𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅(𝑟)++𝑆≤2𝑇𝑟,𝑓𝑁1𝑟,𝑓−𝑎_𝑖+𝑁1𝑟,𝑓−𝑎_𝑖. (44) It is obvious that (c) follows from (43) and (44). It follows from (c) that, for all 1≤𝑗≤4, 3𝑇𝑟,𝑓≤⁴_{𝑖=1𝑖≠𝑗}𝑁1𝑟,𝑓−𝑎_𝑖+3𝑁1𝑟,𝑓−𝑎_𝑗+𝑆(𝑟)≤2𝑇𝑟,𝑓+3𝑁1𝑟,𝑓−𝑎_𝑗(𝑟),≤+𝑆3𝑇𝑟,𝑓³_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖+3𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅+𝑆(𝑟)≤2𝑇𝑟,𝑓+3𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅+𝑆(𝑟). (45) From the last two inequalities, we deduce (d).

Let 1≤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 𝐻∶=𝐻𝑓,𝑔=𝐴₁𝑓²𝑔+𝐴₂𝑓𝑔+𝐴₃𝑓²+𝐴₄𝑓+𝐴₅𝑔+𝐴₆ (46) 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)=2𝑇(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 4𝑇𝑟,𝑓=⁶_𝑖=1𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁𝑟,𝑓=𝑎₆,𝑔≠𝑏₆(𝑟)≤+𝑆𝑁𝑟1𝐻+𝑁𝑟,𝑓=𝑎_𝑗,𝑔=𝑏_𝑗+𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁𝑟,𝑓=𝑎₆,𝑔≠𝑏₆(𝑟)++𝑆≤3𝑇𝑟,𝑓𝑁𝑟,𝑓=𝑎_𝑗,𝑔=𝑏_𝑗+𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁𝑟,𝑓=𝑎₆,𝑔≠𝑏₆+𝑆(𝑟), (47) which yields 𝑇𝑟,𝑓≤𝑁𝑟,𝑓=𝑎_𝑗,𝑔=𝑏_𝑗+𝑁𝑟,𝑓=𝑎₅,𝑔≠𝑏₅+𝑁𝑟,𝑓=𝑎₆,𝑔≠𝑏₆+𝑆(𝑟), (48) and from this and (e), we get 𝑁𝑟,𝑓=𝑎₆,𝑔=𝑏₆≤𝑁𝑟,𝑓=𝑎_𝑗,𝑔=𝑏_𝑗+𝑆(𝑟),𝑗=1,…,6, (49) and it follows from (49) and Lemma 11 that, for all 1≤𝑗≤6, we get 4𝑁𝑟,𝑓=𝑎₆,𝑔=𝑏₆+𝑁𝑟,𝑓=𝑎_𝑗,𝑔=𝑏_𝑗≤⁵_{𝑖=1𝑖≠𝑗}𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+𝑁𝑟,𝑓=𝑎_𝑗,𝑔=𝑏_𝑗(𝑟)(𝑟).+𝑆≤3𝑇𝑟,𝑓+𝑆 (50) From (49) and (50), we get (f).

Suppose that 𝑎_𝑖≡𝑏_𝑖, for 𝑖=1,…,4. By using (a), (b), and Lemma 11, we have 3𝑇𝑟,𝑓=⁵_𝑖=1𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖(𝑟)≤+𝑆𝑁1𝑟,+𝑓−𝑔𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅(𝑟)++𝑆≤2𝑇𝑟,𝑓𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅(𝑟)(𝑟),+𝑆≤3𝑇𝑟,𝑓+𝑆 (51) which gives 2𝑇𝑟,𝑓=⁴_𝑖=1𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖(𝑟),𝑇=+𝑆𝑟,𝑓𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅+𝑆(𝑟). (52) From (52) and (c), we obtain 𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+𝑁𝑟,𝑓=𝑎_𝑗,𝑔=𝑏_𝑗=𝑇𝑟,𝑓+𝑆(𝑟),𝑖,𝑗=1,…,4,𝑖≠𝑗, (53) which yields 𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖=12𝑇𝑟,𝑓+𝑆(𝑟),𝑖=1,…,4. (54) From (52) and (54), we get (g).

By using the same technique as in the proof of (g), we prove (h). If 𝑓,𝑎₁,…,𝑎₆ are interchanged with 𝑔,𝑏₁,…,𝑏₆, respectively, then from (14)–(19), we deduce that Theorem 3 is clear. This completes the proof of Theorem 3, when 𝐹≢0.

Now, assume that 𝐹≡0. Therefore, by Lemma 10, we deduce that 𝑇(𝑟,𝑔)=2𝑇(𝑟,𝑓)+𝑆(𝑟). If the possibility 𝐺≡0, we deduce that 𝑇(𝑟,𝑓)=2𝑇(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 Theorem 8

Assume that 𝑓 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 𝑁1𝑟,𝑓−𝑎₅+𝑁1𝑟,𝑓−𝑎₆≤𝑁𝑟,𝑓=𝑎₅,𝑔=𝑏₅+𝑁𝑟,𝑓=𝑎₆,𝑔=𝑏₆+𝜆₁+1𝑘₅+1+1𝑘₆𝑇(𝑟).+1𝑟,𝑓+𝑆 (55) Since 𝐹(𝑎_𝑖,𝑏_𝑖)≡0, for all 1≤𝑖≤5, it follows from (55), and by applying Lemmas 11 and 12, that 4𝑇𝑟,𝑓≤⁶_𝑖=1𝑁1𝑟,𝑓−𝑎_𝑖(𝑟)≤+𝑆_6𝑖=1𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+𝐾+𝜆₁𝑇(𝑟)≤𝑟,𝑓+𝑆𝑁1𝑟,𝐹+𝑁𝑟,𝑓=𝑎₆,𝑔=𝑏₆+𝐾+𝜆₁𝑇(𝑟)≤𝑟,𝑓+𝑆𝑁𝑟,𝑓=𝑎₆,𝑔=𝑏₆+2+𝐾+𝜆₁𝑇(𝑟),𝑟,𝑓+𝑇𝑟,𝑔+𝑆 (56) which implies that 2𝑇𝑟,𝑓≤𝑇𝑟,𝑔+𝑁𝑟,𝑓=𝑎₆,𝑔=𝑏₆+𝐾+𝜆₁𝑇𝑟,𝑓+𝑆(𝑟). (57) Of course, if 𝐹≡0, then according to Lemma 10, we get 𝑇(𝑟,𝑔)=2𝑇(𝑟,𝑓)+𝑆(𝑟), which means that (57) remains clear when 𝐹≡0. Similarly, we have 2𝑇𝑟,𝑓≤𝑇𝑟,𝑔+𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+𝐾+𝜆₁𝑇𝑟,𝑓+𝑆(𝑟),𝑖=1,…,5. (58) Symmetrically, by using (13) we have 2𝑇𝑟,𝑔≤𝑇𝑟,𝑓+𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+𝐾+𝜆₂𝑇𝑟,𝑔+𝑆(𝑟),𝑖=1,…,5. (59)

Suppose that 𝐹≡0. Then, by Lemma 10, we deduce that 𝑇(𝑟,𝑔)=2𝑇(𝑟,𝑓)+𝑆(𝑟), and 𝐺≢0 (otherwise, we have 𝑇(𝑟,𝑓)=2𝑇(𝑟,𝑔)+𝑆(𝑟)). It follows from (59) that 10𝑇𝑟,𝑔≤5𝑇𝑟,𝑓+⁵_𝑖=1𝑁𝑟,𝑓=𝑎_𝑖,𝑔=𝑏_𝑖+5𝐾+𝜆₂𝑇(𝑟)+𝑟,𝑔+𝑆≤5𝑇𝑟,𝑓𝑁1𝑟,𝐺+5𝐾+𝜆₂𝑇(𝑟)+5𝑟,𝑔+𝑆≤6𝑇𝑟,𝑓𝐾+𝜆₂𝑇(𝑟)≤5+2𝑟,𝑔+𝑆𝐾+𝜆₂+5𝑇𝑟,𝑔+𝑆(𝑟), (60) and the inequality (60) tells us that 1≤𝐾+𝜆₂; that is, 𝐾+𝜆₂=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 10𝑇𝑟,𝑓≤5𝑇𝑟,𝑔+𝑁1𝑟,𝐹+5𝑃+𝜆₁𝑇(𝑟)+5𝑟,𝑓+𝑆≤6𝑇𝑟,𝑔𝐾+𝜆₁+2𝑇𝑟,𝑓+𝑆(𝑟), (61) which yields 8−5𝐾+𝜆₁𝑇𝑟,𝑓≤6𝑇𝑟,𝑔+𝑆(𝑟). (62) In the same method as the above and by using (59), we can show that 8−5𝐾+𝜆₂𝑇𝑟,𝑔≤6𝑇𝑟,𝑓+𝑆(𝑟). (63) From (62) and (63), we obtain (8−5(𝜆₁+𝐾))(8−5(𝜆₂+𝐾))≤36, which is a contradiction with (5). That means that 𝑓 should be a quasi-Möbius transformation of 𝑔. This proves Theorem 8.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author thanks the referee for his/her helpful suggestions concerning the presentation of this paper.