The Uniqueness of Analytic Functions on Annuli Sharing Some Values

Abstract

The purpose of this paper is to deal with the shared values and uniqueness of analytic functions on annulus. Two theorems about analytic functions on annulus sharing four distinct values are obtained, and these theorems are improvement of the results given by Cao and Yi.

1. Introduction

In this paper, we will study the uniqueness problem of analytic functions in the field of complex analysis and adopt the standard notations of the Nevanlinna theory of meromorphic functions as explained (see [1–3]).

We use ℂ to denote the open complex plane, ℂ to denote the extended complex plane, and 𝕏 to denote the subset of ℂ. For 𝑎∈ℂ, we say that 𝑓(𝑧)−𝑎 and 𝑔(𝑧)−𝑎 have the same zeros with the same multiplicities (ignoring multiplicities) in 𝕏 (or ℂ) if two meromorphic functions 𝑓 and 𝑔 share the value 𝑎  CM  (IM) in 𝕏 (or ℂ). In addition, we also use 𝑓=𝑎⇌𝑔=𝑎 in 𝕏 (or ℂ) to express that 𝑓 and 𝑔 share the value 𝑎  CM in 𝕏 (or ℂ), 𝑓=𝑎⇔𝑔=𝑎 in 𝕏 (or ℂ) to express that 𝑓 and 𝑔 share the value 𝑎  IM in 𝕏 (or ℂ), and 𝑓=𝑎⇒𝑔=𝑎 in 𝕏 (or ℂ) to express that 𝑓=𝑎 implies 𝑔=𝑎 in 𝕏 (or ℂ).

In 1929, Nevanlinna (see [4]) proved the following well-known theorem.

Theorem 1.1 (see [4]). If 𝑓 and 𝑔 are two nonconstant meromorphic functions that share five distinct values 𝑎₁,𝑎₂,𝑎₃,𝑎₄, and 𝑎₅  IM in ℂ, then 𝑓(𝑧)≡𝑔(𝑧).

After his theorem, the uniqueness theory of meromorphic functions sharing values in the whole complex plane attracted many investigations (see [2]). In 2003, Zheng [5] studied the uniqueness problem under the condition that five values are shared in some angular domain in ℂ. There were many results in the field of the uniqueness with shared values in the complex plane and angular domain, see ([5–12]). The whole complex plane ℂ and angular domain all can be regarded as simply connected region. Thus, it is interesting to consider the uniqueness theory of meromorphic functions in the multiply connected region. Here, we will mainly study the uniqueness of meromorphic functions in doubly connected domains of complex plane ℂ. By the doubly connected mapping theorem [13] each doubly connected domain is conformally equivalent to the annulus {𝑧∶𝑟<|𝑧|<𝑅}, 0≤𝑟<𝑅≤+∞. We consider only two cases: 𝑟=0, 𝑅=+∞ simultaneously and 0<𝑟<𝑅<+∞. In the latter case the homothety 𝑧↦𝑧/√𝑟𝑅 reduces the given domain to the annulus {𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 𝑅₀=𝑅/𝑟. Thus, every annulus is invariant with respect to the inversion 𝑧↦1/𝑧 in two cases.

In 2005, Khrystiyanyn and Kondratyuk [14, 15] proposed the Nevanlinna theory for meromorphic functions on annuli (see also [16]). We will show the basic notions of the Nevanlinna theory on annuli in the next section. In 2009, Cao et al. [17, 18] investigated the uniqueness of meromorphic functions on annuli sharing some values and some sets and obtained an analog of Nevanlinna’s famous five-value theorem as follows.

Theorem 1.2 (see [18, Theorem  3.2]). Let 𝑓₁ and 𝑓₂ be two transcendental or admissible meromorphic functions on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅₀≤+∞. Let 𝑎_𝑗(𝑗=1,2,3,4,5) be five distinct complex numbers in ℂ. If 𝑓₁,𝑓₂ share 𝑎_𝑗  𝐼𝑀 for 𝑗=1,2,3,4,5, then 𝑓₁(𝑧)≡𝑓₂(𝑧).

Remark 1.3. For the case 𝑅₀=+∞, the assertion was proved by Kondratyuk and Laine [16].

From Theorem 1.2, we can get the following results easily.

Theorem 1.4. Under the assumptions of Theorem 1.2, if 𝑓₁,𝑓₂ are two transcendental or admissible analytic functions on annulus 𝔸 and 𝑓₁,𝑓₂ share 𝑎_𝑗  𝐼𝑀 for 𝑗=1,2,3,4, then 𝑓₁(𝑧)≡𝑓₂(𝑧).

In fact, we will prove some general theorems on the uniqueness of analytic functions on the annuli sharing four values in this paper (see Section 3), and these theorems improve Theorem 1.4.

2. Basic Notions in the Nevanlinna Theory on Annuli

Let 𝑓 be a meromorphic function on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅<𝑅₀≤+∞. We recall the classical notations of the Nevanlinna theory as follows: 𝑁𝑅,𝑓=^𝑅₀𝑛𝑡,𝑓−𝑛0,𝑓𝑡𝑚=1𝑑𝑡+𝑛0,𝑓log𝑅,𝑅,𝑓2𝜋₀^2𝜋log⁺𝑓Re^𝑖𝜃𝑑𝜃,𝑇𝑅,𝑓=𝑁𝑅,𝑓+𝑚𝑅,𝑓, (2.1) where log⁺𝑥=max{log𝑥,0} and 𝑛(𝑡,𝑓) is the counting function of poles of the function 𝑓 in {𝑧∶|𝑧|≤𝑡}. We here show the notations of the Nevanlinna theory on annuli. Let 𝑁₁𝑅,𝑓=¹_1/𝑅𝑛₁𝑡,𝑓𝑡𝑑𝑡,𝑁₂𝑅,𝑓=^𝑅₁𝑛₂𝑡,𝑓𝑡𝑚𝑑𝑡,₀1𝑅,𝑓=𝑚𝑅,𝑓+𝑚𝑅,𝑓,𝑁₀𝑅,𝑓=𝑁₁𝑅,𝑓+𝑁₂𝑅,𝑓, (2.2) where 𝑛₁(𝑡,𝑓) and 𝑛₂(𝑡,𝑓) are the counting functions of poles of the function 𝑓 in {𝑧∶𝑡<|𝑧|≤1} and {𝑧∶1<|𝑧|≤𝑡}, respectively. The Nevanlinna characteristic of 𝑓 on the annulus 𝔸 is defined by 𝑇₀𝑅,𝑓=𝑚₀𝑅,𝑓−2𝑚1,𝑓+𝑁₀𝑅,𝑓 (2.3) and has the following properties.

Proposition 2.1 (see [14]). Let 𝑓 be a nonconstant meromorphic function on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅<𝑅₀≤+∞. Then,(i)𝑇₀(𝑅,𝑓)=𝑇₀(𝑅,1/𝑓), (ii)max{𝑇₀(𝑅,𝑓₁⋅𝑓₂),𝑇₀(𝑅,𝑓₁/𝑓₂),𝑇₀(𝑅,𝑓₁+𝑓₂)}≤𝑇₀(𝑅,𝑓₁)+𝑇₀(𝑅,𝑓₂)+𝑂(1).

By Proposition 2.1, the first fundamental theorem on the annulus 𝔸 is immediately obtained.

Theorem 2.2 (see [14] (the first fundamental theorem)). Let f be a nonconstant meromorphic function on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅<𝑅₀≤+∞. Then 𝑇₀1𝑅,𝑓−𝑎=𝑇₀𝑅,𝑓+𝑂(1) (2.4) for every fixed 𝑎∈ℂ.

Khrystiyanyn and Kondratyuk also obtained the lemma on the logarithmic derivative on the annulus 𝔸.

Theorem 2.3 (see [15] (lemma on the logarithmic derivative)). Let f be a nonconstant meromorphic function on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 𝑅₀≤+∞, and let 𝜆>0. Then,(i)in the case 𝑅₀=+∞, 𝑚₀𝑓𝑅,^′𝑓=𝑂log𝑅𝑇₀𝑅,𝑓 (2.5) for 𝑅∈(1,+∞) except for the set Δ_𝑅 such that ∫_Δ𝑅𝑅^𝜆−1𝑑𝑅<+∞;(ii)if 𝑅₀<+∞, then 𝑚₀𝑓𝑅,^′𝑓𝑇=𝑂log⁰𝑅,𝑓𝑅₀−𝑅 (2.6) for 𝑅∈(1,𝑅₀) except for the set Δ^′_𝑅 such that ∫_Δ′𝑅𝑑𝑅/(𝑅₀−𝑅)^𝜆−1<+∞.

We denote the deficiency of 𝑎∈ℂ=ℂ∪{∞} with respect to a meromorphic function 𝑓 on the annulus 𝔸 by 𝛿₀𝑎,𝑓=𝛿₀0,𝑓−𝑎=liminf_𝑟→𝑅0𝑚₀𝑟,1/𝑓−𝑎𝑇₀𝑟,𝑓=1−limsup_𝑟→𝑅0𝑁₀𝑟,1/𝑓−𝑎𝑇₀𝑟,𝑓 (2.7) and denote the reduced deficiency by Θ₀𝑎,𝑓=Θ₀0,𝑓−𝑎=1−limsup_𝑟→𝑅0𝑁₀𝑟,1/𝑓−𝑎𝑇₀,𝑟,𝑓 (2.8) where 𝑁₀1𝑟,=𝑓−𝑎𝑁₁1𝑅,+𝑓−𝑎𝑁₂1𝑅,=𝑓−𝑎_11/𝑅𝑛₁𝑡,1/𝑓−𝑎𝑡𝑑𝑡+^𝑅₁𝑛₂𝑡,1/𝑓−𝑎𝑡𝑑𝑡 (2.9) in which each zero of the function 𝑓−𝑎 is counted only once. In addition, we use 𝑛₁^𝑘)(𝑡,1/(𝑓−𝑎)) (or 𝑛₁^(𝑘(𝑡,1/(𝑓−𝑎))) to denote the counting function of poles of the function 1/(𝑓−𝑎) with multiplicities ≤𝑘 (or >𝑘) in {𝑧∶𝑡<|𝑧|≤1}, each point counted only once. Similarly, we can give the notations 𝑁₁^𝑘)(𝑡,𝑓), 𝑁₁^(𝑘(𝑡,𝑓), 𝑁₂^𝑘)(𝑡,𝑓), 𝑁₂^(𝑘(𝑡,𝑓), 𝑁₀^𝑘)(𝑡,𝑓), and 𝑁₀^(𝑘(𝑡,𝑓).

Khrystiyanyn and Kondratyuk [15] first obtained the second fundamental theorem on the the annulus 𝔸. Later, Cao et al. [18] introduced other forms of the second fundamental theorem on annuli as follows.

Theorem 2.4 (see [18, Theorem  2.3] (the second fundamental theorem)). Let 𝑓 be a nonconstant meromorphic function on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅₀≤+∞. Let 𝑎₁,𝑎₂,…,𝑎_𝑞 be 𝑞 distinct complex numbers in the extended complex plane ℂ. Let 𝜆≥0. Then, (i)(𝑞−2)𝑇₀(𝑅,𝑓)<∑^𝑞_𝑗=1𝑁₀(𝑅,1/(𝑓−𝑎_𝑗))−𝑁₀⁽¹⁾(𝑅,𝑓)+𝑆(𝑅,𝑓), (ii)(𝑞−2)𝑇₀(𝑅,𝑓)<∑^𝑞_𝑗=1𝑁₀(𝑅,1/(𝑓−𝑎_𝑗))+𝑆(𝑅,𝑓), where 𝑁₀⁽¹⁾𝑅,𝑓=𝑁₀1𝑅,𝑓^′+2𝑁₀𝑅,𝑓−𝑁₀𝑅,𝑓^′, (2.10) and (i) in the case 𝑅₀=+∞, 𝑆𝑅,𝑓=𝑂log𝑅𝑇₀𝑅,𝑓 (2.11) for 𝑅∈(1,+∞) except for the set Δ_𝑅 such that ∫_Δ𝑅𝑅^𝜆−1𝑑𝑅<+∞; (ii) if 𝑅₀<+∞, then 𝑆𝑇𝑅,𝑓=𝑂log⁰𝑅,𝑓𝑅₀−𝑅 (2.12) for 𝑅∈(1,𝑅₀) except for the set Δ^′_𝑅 such that ∫_Δ′𝑅𝑑𝑅/(𝑅₀−𝑅)^𝜆−1<+∞.

Definition 2.5. Let 𝑓(𝑧) be a nonconstant meromorphic function on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅₀≤+∞. The function 𝑓 is called a transcendental or admissible meromorphic function on the annulus 𝔸 provided that limsup_𝑅→∞𝑇₀𝑅,𝑓log𝑅=∞,1<𝑅<𝑅₀=+∞, (2.13) or limsup_𝑅→𝑅0𝑇₀𝑅,𝑓−log𝑅₀−𝑅=∞,1<𝑅<𝑅₀<+∞, (2.14) respectively.

Thus, for a transcendental or admissible meromorphic function on the annulus 𝔸, 𝑆(𝑅,𝑓)=𝑜(𝑇₀(𝑅,𝑓)) holds for all 1<𝑅<𝑅₀ except for the set Δ_𝑅 or the set Δ^′_𝑅 mentioned in Theorem 2.3, respectively.

3. The Main Theorems and Some Lemmas

Now we show our main results, which improve Theorem 1.4.

Theorem 3.1. Let 𝑓,𝑔 be two analytic functions on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅₀≤+∞, and let 𝑎_𝑗∈ℂ(𝑗=1,2,3,4) be four distinct values. If 𝑓 and 𝑔 share the two distinct values 𝑎₁,𝑎₂  𝐶𝑀 in 𝔸 and 𝑓=𝑎₃⇒𝑔=𝑎₃ in 𝔸 and 𝑓=𝑎₄⇒𝑔=𝑎₄ in 𝔸, and 𝑓 is transcendental or admissible on 𝔸, then 𝑓(𝑧)≡𝑔(𝑧).

Theorem 3.2. Under the assumptions of Theorem 3.1, with 𝐶𝑀 replaced by IM, we have either 𝑓(𝑧)≡𝑔(𝑧) or 𝑎𝑓≡³𝑔−𝑎₁𝑎₂𝑔−𝑎₄, (3.1) and 𝑎₁+𝑎₂=𝑎₃+𝑎₄, 𝑎₃, and 𝑎₄ are exceptional values of 𝑓 and 𝑔 in 𝔸, respectively.

Remark 3.3. It is easily seen that Theorems 3.1 and 3.2 are improvement of Theorem 1.4.

To prove the above theorems, we need some lemmas as follows.

Lemma 3.4. Let 𝑓,𝑔 be two distinct analytic functions on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅₀≤+∞, and let 𝑎_𝑗∈ℂ(𝑗=1,2,3,4) be four distinct complex numbers. If 𝑓=𝑎_𝑗⇒𝑔=𝑎_𝑗 in 𝔸 for 𝑗=1,2,3,4 and if 𝑓 is transcendental or admissible on 𝔸, then 𝑔 is also transcendental or admissible.

Proof. By the assumption of Lemma 3.4 and applying Theorem 2.4(ii), we can get3𝑇₀𝑅,𝑓≤⁴_𝑗=1𝑁₀1𝑅,𝑓−𝑎_𝑗≤+𝑆𝑅,𝑓_4𝑗=1𝑁₀1𝑅,𝑔−𝑎_𝑗+𝑆𝑅,𝑓≤4𝑇₀𝑅,𝑔+𝑆𝑅,𝑓. (3.2)
Therefore 𝑇₀𝑅,𝑓≤4𝑇₀𝑅,𝑔+𝑜𝑇₀𝑅,𝑓 (3.3) holds for all 1<𝑅<𝑅₀ except for the set Δ_𝑅 or the set Δ^′_𝑅 mentioned in Theorem 2.3, respectively. Then, from Definition 2.5, we get that 𝑔 is transcendental or admissible on 𝔸.

Lemma 3.5. Suppose that 𝑓 is a transcendental or admissible meromorphic function on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅₀≤+∞. Let 𝑃(𝑓)=𝑎₀𝑓^𝑝+𝑎₁𝑓^𝑝−1+⋯+𝑎_𝑝(𝑎₀≠0) be a polynomial of 𝑓 with degree 𝑝, where the coefficients 𝑎_𝑗(𝑗=0,1,…,𝑝) are constants, and let 𝑏_𝑗(𝑗=1,2,…,𝑞) be 𝑞(𝑞≥𝑝+1) distinct finite complex numbers. Then, 𝑚₀𝑃𝑓𝑅,⋅𝑓^′𝑓−𝑏₁𝑓−𝑏₂⋯𝑓−𝑏_𝑞=𝑆𝑅,𝑓. (3.4)

Proof. From Theorem 2.3 and the definition of 𝑚₀(𝑅,𝑓), transcendental and admissible function, we can get this lemma by using the same argument as in Lemma  4.3 in [2].

Lemma 3.6. Let 𝑓,𝑔 be two distinct analytic functions on the annulus 𝔸={𝑧∶1/𝑅₀<|𝑧|<𝑅₀}, where 1<𝑅₀≤+∞. Suppose that 𝑓 and 𝑔 share 𝑎₁,𝑎₂  𝐼𝑀 in 𝔸, and 𝑓=𝑎₃⇒𝑔=𝑎₃ in 𝔸 and 𝑓=𝑎₄⇒𝑔=𝑎₄ in 𝔸, and 𝑎_𝑗∈ℂ(𝑗=1,2,3,4) are four distinct finite complex numbers. If 𝑓 is a transcendental or admissible function on 𝔸, then 𝑔 is also transcendental or admissible, and(i)𝑇₀(𝑅,𝑔)=2𝑇₀(𝑅,𝑓)+𝑆(𝑅), (ii)𝑇₀(𝑅,𝑓−𝑔)=3𝑇₀(𝑅,𝑓)+𝑆(𝑅); (iii)𝑇₀(𝑅,𝑓)=𝑁₀(𝑅,1/(𝑓−𝑎₃))+𝑁₀(𝑅,1/(𝑓−𝑎₄))+𝑆(𝑅), (iv)𝑇₀(𝑅,𝑓)=𝑁₀(𝑅,1/(𝑓−𝑎_𝑗))+𝑆(𝑅),𝑗=1,2, (v)𝑇₀(𝑅,𝑔)=𝑁₀(𝑅,1/(𝑔−𝑎_𝑗))+𝑆(𝑅),𝑗=3,4, (vi)𝑇₀(𝑅,𝑓^′)=𝑇₀(𝑅,𝑓)+𝑆(𝑅),𝑇₀(𝑅,𝑔^′)=𝑇₀(𝑅,𝑔)+𝑆(𝑅),
where 𝑆(𝑅)∶=𝑆(𝑅,𝑓)=𝑆(𝑅,𝑔).

Proof. By the assumption of this lemma and by Theorem 2.4(ii), we have 𝑇₀(𝑅,𝑓)≤3𝑇₀(𝑅,𝑔)+𝑆(𝑅,𝑓) and 𝑇₀(𝑅,𝑔)≤3𝑇₀(𝑅,𝑓)+𝑆(𝑅,𝑔). Thus, we can get 𝑆(𝑅,𝑓)=𝑆(𝑅,𝑔).
Let 𝑓𝜂∶=^′𝑔^′𝑓−𝑔𝑓−𝑎₃𝑓−𝑎₄𝑔−𝑎₁𝑔−𝑎₂. (3.5)
From the conditions of this lemma, we can get that 𝜂 is analytic on 𝔸 and 𝜂≢0 unless 𝑓≡𝑔. By Lemma 3.5, we have 𝑚₀(𝑅,𝜂)=𝑆(𝑅,𝑓)+𝑆(𝑅,𝑔)=𝑆(𝑅). Thus, we can get 𝑆(𝑅,𝜂)=𝑆(𝑅).
Since 𝑓,𝑔 are two nonconstant analytic functions on annulus 𝔸 and share 𝑎₁,𝑎₂  IM in 𝔸 and 𝑓=𝑎₃⇒𝑔=𝑎₃ and 𝑓=𝑎₄⇒𝑔=𝑎₄ in 𝔸, again by Theorem 2.4, we have 3𝑇₀𝑅,𝑓≤⁴_𝑗=1𝑁₀1𝑅,𝑓−𝑎_𝑗,≤+𝑆𝑅,𝑓𝑁₀1𝑟,𝑓−𝑔+𝑆𝑅,𝑓=𝑇₀𝑅,𝑓−𝑔+𝑆𝑅,𝑓,≤𝑇₀𝑅,𝑓+𝑇₀(𝑅),𝑇𝑅,𝑔+𝑆₀𝑅,𝑔≤𝑁₀1𝑅,𝑔−𝑎₁+𝑁₀1𝑅,𝑔−𝑎₂,=+𝑆𝑅,𝑔𝑁₀1𝑅,𝑓−𝑎₁+𝑁₀1𝑅,𝑓−𝑎₂+𝑆(𝑅),≤2𝑇₀𝑅,𝑓+𝑆(𝑅). (3.6)(3.7)(3.8)(3.9)(3.10)(3.11)
From (3.8) and (3.11), we can get (i), and from (3.7), (3.8), and (i), we can get (ii), and from (3.6), (3.8), (3.10), (3.11), and (i), we can get (iii). Thus, we can deduce that (iv) and (v) hold easily from (3.6)–(3.11) and (i)–(iii). Now, we will prove that (vi) holds as follows.
First, we can rewrite (3.5) as 𝑔𝑓=𝑓′^′𝜂𝑔−𝑎₁𝑔−𝑎₂+𝑓^′𝑔^′𝑎₃𝑓+𝑎₄𝑓−𝑎₃𝑎₄−𝑓𝑔𝜂𝑓−𝑎₃𝑓−𝑎₄𝑔−𝑎₁𝑔−𝑎₂. (3.12)
From (3.12) and Lemma 3.5, we can get 𝑚₀(𝑅,𝑓)≤𝑚₀(𝑅,𝑓′)+𝑆(𝑅,𝑓). Since 𝑓 is analytic on 𝔸, we have 𝑇₀𝑅,𝑓≤𝑇₀𝑅,𝑓^′+2𝑚1,𝑓^′−2𝑚1,𝑓+𝑆𝑅,𝑓. (3.13) From the fact that 𝑓 is transcendental or admissible, we have 𝑇₀𝑅,𝑓≤𝑇₀𝑅,𝑓^′+𝑆𝑅,𝑓+𝑂(1)=𝑇₀𝑅,𝑓^′+𝑆𝑅,𝑓. (3.14) On the other hand, since 𝑚₀(𝑅,𝑓^′)=𝑚₀(𝑅,𝑓(𝑓^′/𝑓))≤𝑚₀(𝑅,𝑓)+𝑚₀(𝑅,𝑓^′/𝑓)+𝑂(1), from Theorem 2.3, we have 𝑚₀(𝑅,𝑓^′)≤𝑚₀(𝑅,𝑓)+𝑆(𝑅,𝑓). Thus, we can get 𝑇₀𝑅,𝑓^′≤𝑇₀𝑅,𝑓+𝑆𝑅,𝑓+𝑂(1)=𝑇₀𝑅,𝑓+𝑆𝑅,𝑓. (3.15)
From (3.14), (3.15) and the fact that 𝑓 is transcendental or admissible, we can get 𝑇₀(𝑅,𝑓^′)=𝑇₀(𝑅,𝑓)+𝑆(𝑅,𝑓). Similarly, we can get 𝑇₀(𝑅,𝑔′)=𝑇₀(𝑅,𝑔)+𝑆(𝑅,𝑔).
Thus, we complete the proof of this lemma.

4. The Proof of Theorem 3.1

Suppose 𝑓≢𝑔. By the assumption of Theorem 3.1, we can get the conclusions (i)–(vi) of Lemma 3.6 and that 𝑔 is transcendental or admissible on 𝔸. Set 𝜓₁𝑓∶=^′𝑓−𝑎₃𝑓−𝑎₁𝑓−𝑎₂−𝑔^′𝑔−𝑎₃𝑔−𝑎₁𝑔−𝑎₂,𝜓₂𝑓∶=^′𝑓−𝑎₄𝑓−𝑎₁𝑓−𝑎₂−𝑔^′𝑔−𝑎₄𝑔−𝑎₁𝑔−𝑎₂. (4.1)

By Lemma 3.4, we can get 𝑚₀𝑅,𝜓_𝑖=𝑆𝑅,𝑓+𝑆𝑅,𝑔=𝑆(𝑅),𝑖=1,2. (4.2) Moreover, we can prove 𝑁₀(𝑅,𝜓_𝑖)=𝑂(1)(𝑖=1,2). In fact, the poles of 𝜓_𝑖 in 𝔸 only can occur at the zeros of 𝑓−𝑎_𝑗 and 𝑔−𝑎_𝑗(𝑖,𝑗=1,2) in 𝔸. Since 𝑓,𝑔 share 𝑎₁,𝑎₂  CM in 𝔸, we can see that if 𝑧₀∈𝔸 is a zero of 𝑓−𝑎_𝑗 with multiplicity 𝑚(≥1), then 𝑧₀∈𝔸 is a zero of 𝑔−𝑎_𝑗(𝑗=1,2) with multiplicity 𝑚(≥1). Suppose that 𝑓−𝑎_𝑗=𝑧−𝑧₀^𝑚𝛼_𝑗(𝑧),𝑔−𝑎_𝑗=𝑧−𝑧₀^𝑚𝛽_𝑗(𝑧), (4.3) where 𝛼_𝑗(𝑧),𝛽_𝑗(𝑧) are analytic functions in 𝔸 and 𝛼_𝑗(𝑧₀)≠0,𝛽_𝑗(𝑧₀)≠0(𝑗=1,2); by a simple calculation, we have 𝜓_𝑖𝑧₀𝛼=𝐾^′𝑗𝑧₀𝛼_𝑗𝑧₀−𝛽^′_𝑗𝑧₀𝛽_𝑗𝑧₀𝑖,𝑗=1,2, (4.4) where 𝐾 is a constant. Therefore, we can get that 𝜓_𝑖(𝑖=1,2) are analytic in 𝔸. Thus, from (4.2), we can get 𝑇₀(𝑅,𝜓_𝑖)=𝑚₀(𝑅,𝜓_𝑖)+𝑂(1)=𝑆(𝑅)(𝑖=1,2).

If 𝜓_𝑖≢0,𝑖=1,2, then we have 𝑁₀1𝑅,𝑓−𝑎₃≤𝑁₀1𝑅,𝜓₁≤𝑇₀𝑅,𝜓₁+𝑆𝑅,𝑓+𝑂(1)=𝑆(𝑅),𝑁₀1𝑅,𝑓−𝑎₄≤𝑁₀1𝑅,𝜓₂≤𝑇₀𝑅,𝜓₂+𝑆𝑅,𝑓+𝑂(1)=𝑆(𝑅). (4.5) From (4.5) and Lemma 3.6(iv), we have 𝑇₀(𝑅,𝑓)≤𝑆(𝑅). Thus, since 𝑓,𝑔 are transcendental or admissible functions on 𝔸, that is, 𝑓 and 𝑔 are of unbounded characteristic, and from the definition of 𝑆(𝑅), we can get a contradiction.

Assume that one of 𝜓₁ and 𝜓₂ is identically zero, say 𝜓₁≡0; then we have 𝑁₀⁽²1𝑅,𝑔−𝑎₄=𝑁₀⁽²1𝑅,𝑓−𝑎₄. (4.6)

From (3.5), we can see that 𝑔(𝑧₁)=𝑎₄ implies that 𝑓(𝑧₁)=𝑎₄ for such 𝑧₁∈𝔸 satisfying 𝜂(𝑧₁)≠0. Since 𝑇₀(𝑅,𝜂)=𝑆(𝑅), we have 𝑁₀¹⁾1𝑅,𝑔−𝑎₄=𝑁₀¹⁾1𝑅,𝑓−𝑎₄+𝑆(𝑅). (4.7)

From (4.6) and (4.7), we can get 𝑁₀1𝑅,𝑔−𝑎₄=𝑁₀1𝑅,𝑓−𝑎₄+𝑆(𝑅). (4.8)

Similarly, when 𝜓₂≡0, we can get 𝑁₀1𝑅,𝑔−𝑎₃=𝑁₀1𝑅,𝑓−𝑎₃+𝑆(𝑅). (4.9)

From (4.8), (4.9), and Lemma 3.6(i), (v), we can get 2𝑇₀𝑅,𝑓=𝑁₀1𝑅,𝑓−𝑎₃+𝑆(𝑅) (4.10) or 2𝑇₀𝑅,𝑓=𝑁₀1𝑅,𝑓−𝑎₄+𝑆(𝑅). (4.11) Since 𝑓,𝑔 are transcendental or admissible functions on the annulus 𝔸, we can get a contradiction again.

Thus, we complete the proof of Theorem 3.1.

5. The Proof of Theorem 3.2

Suppose that 𝑓≢𝑔. By Theorem 2.4(ii) and the fact that 𝑓 is transcendental or admissible on 𝔸, we have 2𝑇₀𝑅,𝑓+𝑁₀1𝑅,𝑔−𝑎₄≤𝑁₀1𝑅,𝑓−𝑎₁+𝑁₀1𝑅,𝑓−𝑎₂+𝑁₀1𝑅,𝑓−𝑎₃+𝑁₀1𝑅,𝑔−𝑎₄≤+𝑆𝑅,𝑓𝑁₀1𝑅,𝑓−𝑔+𝑆𝑅,𝑓≤𝑇₀𝑅,𝑓+𝑇₀𝑅,𝑔+𝑆𝑅,𝑓+𝑆𝑅,𝑔. (5.1) Therefore, we have 𝑇₀𝑅,𝑓+𝑁₀1𝑅,𝑔−𝑎₄≤𝑇₀𝑅,𝑔+𝑆𝑅,𝑓+𝑆𝑅,𝑔. (5.2) Similarly, we have 𝑇₀𝑅,𝑔+𝑁₀1𝑅,𝑓−𝑎₃≤𝑇₀𝑅,𝑓+𝑆𝑅,𝑔+𝑆𝑅,𝑓. (5.3)

From (5.2) and (5.3), we can see that 𝑇₀(𝑅,𝑓)=𝑇₀(𝑅,𝑔)+𝑆(𝑅,𝑓)+𝑆(𝑅,𝑔), and 𝑁₀1𝑅,𝑓−𝑎₃=𝑆𝑅,𝑓+𝑆𝑅,𝑔,𝑁₀1𝑅,𝑔−𝑎₄=𝑆𝑅,𝑓+𝑆𝑅,𝑔. (5.4)

Thus, from (5.2), (5.3), and the definition of 𝑆(𝑅), we can get that 𝑔 is also transcendental or admissible on 𝔸 when 𝑓 is transcendental or admissible on 𝔸.

From (5.1)–(5.4), we can also get 2𝑇₀𝑅,𝑓=𝑁₀1𝑅,𝑓−𝑎₁+𝑁₀1𝑅,𝑓−𝑎₂+𝑆(𝑅). (5.5)

From (5.5), we can see that “almost all” of zeros of 𝑓−𝑎_𝑖(𝑖=1,2) in 𝔸 are simple. Similarly, “almost all” of zeros of 𝑔−𝑎_𝑖(𝑖=1,2) in 𝔸 are simple, too. Let 𝜑₁𝑎∶=¹−𝑎₃𝑓^′𝑓−𝑎₂𝑓−𝑎₁𝑓−𝑎₃−𝑎₁−𝑎₄𝑔^′𝑔−𝑎₂𝑔−𝑎₁𝑔−𝑎₄,𝜑₂𝑎∶=²−𝑎₃𝑓^′𝑓−𝑎₁𝑓−𝑎₂𝑓−𝑎₃−𝑎₂−𝑎₄𝑔^′𝑔−𝑎₁𝑔−𝑎₂𝑔−𝑎₄. (5.6) By Lemma 3.5, we can get that 𝑚₀(𝑅,𝜑_𝑖)=𝑆(𝑅)(𝑖=1,2). Since 𝑓,𝑔 share 𝑎₁,𝑎₂  IM in 𝔸 and from (5.2), we have 𝑁₀(𝑅,𝜑_𝑖)=𝑆(𝑅)(𝑖=1,2). Therefore, we can get 𝑇₀(𝑅,𝜑_𝑖)=𝑆(𝑅),(𝑖=1,2).

If 𝜑₁≢0, then we have 𝑁₀(𝑅,1/(𝑓−𝑎₂))≤𝑁₀(𝑅,1/𝜑₁)=𝑆(𝑅). Thus, from (5.5), we can get a contradiction easily. Similarly, when 𝜑₂≢0, we can get a contradiction, too. Hence, 𝜑₁,𝜑₂ are identically equal to 0. Then, we have (𝜑₁−𝜑₂)/(𝑎₁−𝑎₂)≡0, that is, 𝑓^′𝑓−𝑎₃−𝑔^′𝑔−𝑎₄−𝑓^′𝑓−𝑎₁+𝑔^′𝑔−𝑎₁−𝑓^′𝑓−𝑎₂+𝑔^′𝑔−𝑎₂≡0, (5.7) which implies that 𝑓−𝑎₃𝑔−𝑎₄⋅𝑔−𝑎₁𝑔−𝑎₂𝑓−𝑎₁𝑓−𝑎₂≡𝑐, (5.8) where 𝑐 is a nonzero constant. Rewrite (5.8) as 𝑔²−𝑎₁+𝑎₂−𝑐𝛾𝑓𝑓−𝑎₃𝑔+𝑎₁𝑎₂+𝑐𝑎₄𝛾𝑓𝑓−𝑎₃≡0, (5.9) where 𝛾(𝑓)∶=(𝑓−𝑎₁)(𝑓−𝑎₂). The discriminant of (5.9) is Δ𝑓=𝑎₁+𝑎₂−𝑐𝛾(𝑓)𝑓−𝑎₃²−4𝑎₁𝑎₂+𝑐𝑎₄𝛾𝑓𝑓−𝑎₃℘𝑓∶=𝑓−𝑎₃², (5.10) where ℘(𝑧)∶=𝑎₁+𝑎₂𝑧−𝑎₃−𝑐𝛾(𝑧)²−4𝑎₁𝑎₂𝑧−𝑎₃²−4𝑐𝑎₄𝛾(𝑧)𝑧−𝑎₃ (5.11) is a polynomial of degree 4 in 𝑧. If 𝑎 is a zero of ℘(𝑧) in 𝔸, obviously 𝑎≠𝑎₃. Then, from (5.9), 𝑓(𝑧)=𝑎 implies that 𝑔(𝑧)=12𝑎₁+𝑎₂−𝑐𝛾(𝑎)𝑎−𝑎₃∶=𝑏. (5.12) Set 𝜙₁𝑓∶=^′𝑔^′𝑓−𝑔𝑓−𝑎₁𝑔−𝑎₂𝑓−𝑎₃𝑔−𝑎₄,𝜙₂𝑓∶=^′𝑔^′𝑓−𝑔𝑓−𝑎₂𝑔−𝑎₁𝑓−𝑎₃𝑔−𝑎₄,𝜙𝜙=²𝜙₁=𝑓−𝑎₁𝑔−𝑎₂𝑓−𝑎₂𝑔−𝑎₁. (5.13)

By Lemma 3.5, we can get 𝑚₀(𝑅,𝜙_𝑖)=𝑆(𝑅)(𝑖=1,2). And by a simple calculation, we can get 𝑁₀(𝑅,𝜙_𝑖)=𝑆(𝑅)(𝑖=1,2). Then we have 𝑇₀(𝑅,𝜙_𝑖)=𝑆(𝑅)(𝑖=1,2), thus we have 𝑇₀(𝑅,𝜙)=𝑆(𝑅).

Assume that 𝑓 is not a Möbius transformation of 𝑔; then 𝜙 is a nonconstant function. Since ℘𝑎₁=𝑎₁+𝑎₂𝑎₁−𝑎₃²−4𝑎₁𝑎₂𝑎₁−𝑎₃²=𝑎₁−𝑎₃²𝑎₁−𝑎₂²℘𝑎≠0,₂=𝑎₁+𝑎₃𝑎₂−𝑎₃²−4𝑎₁𝑎₂𝑎₂−𝑎₃²=𝑎₂−𝑎₃²𝑎₁−𝑎₂²≠0. (5.14) From 𝑎≠𝑎_𝑖(𝑖=1,2) and (5.8), we can get 𝑁₀1𝑅,≤𝑓−𝑎𝑁₀1𝑅,𝜙−𝜉≤𝑇₀𝑅,𝜙=𝑆(𝑅), (5.15) where 𝜉=(𝑎−𝑎₁)(𝑏−𝑎₂)/(𝑎−𝑎₂)(𝑏−𝑎₁). Since 𝑓 is transcendental or admissible analytic in 𝔸, by Theorem 2.4(ii) and (5.4), we can get 𝑇₀𝑅,𝑓≤𝑁₀1𝑅,𝑓−𝑎₃+𝑁₀1𝑅,(𝑅)(𝑅).𝑓−𝑎+𝑆=𝑆 (5.16)

Since 𝑓,𝑔 are transcendental or admissible functions on 𝔸, from the above inequality, we can get a contradiction. Therefore, we can get that 𝑓 is a Möbius transformation of 𝑔 on 𝔸. Since 𝑓,𝑔 are transcendental or admissible functions on 𝔸, by a simple calculation, we can get easily that 𝑎₁+𝑎₂=𝑎₃+𝑎₄ and 𝑎𝑓≡³𝑔−𝑎₁𝑎₂𝑔−𝑎₄. (5.17) Furthermore, 𝑎₃,𝑎₄ are Picard exceptional values of 𝑓 and 𝑔 in 𝔸, respectively.

Thus, we complete the proof of Theorem 3.2.

Acknowledgments

Z.-X. Xuan is supported in part by Beijing Municipal Research Foundation for The Excellent Scholars Ministry (2011D005022000009); Science and Technology Research Program of Beijng Municipal Commission of Education (KM201211417011); Funding Project for Academic Human Resources Development in Beijing Union University. This work was supported by the NSF of Jiangxi of China (Grant no. 2010GQS0119).