# The Uniqueness of Analytic Functions on Annuli Sharing Some Values

Authors: Hong-Yan Xu; Zu-Xing Xuan
Journal: Abstract and Applied Analysis
 (2012)

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## 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.


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## Body

## 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 useℂ 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]).
Iff 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]).
Letf1 and f2 be two transcendental or admissible meromorphic functions on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0≤+∞. Let aj(j=1,2,3,4,5) be five distinct complex numbers in ℂ¯. If f1,f2 share aj  IM for j=1,2,3,4,5, then f1(z)≡f2(z).Remark 1.3.
For the caseR0=+∞, the assertion was proved by Kondratyuk and Laine [16].From Theorem1.2, we can get the following results easily.Theorem 1.4.
Under the assumptions of Theorem1.2, if f1,f2 are two transcendental or admissible analytic functions on annulus 𝔸 and f1,f2 share aj  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 Section3), and these theorems improve Theorem 1.4.


## 2. Basic Notions in the Nevanlinna Theory on Annuli

Letf 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:
(2.1)N(R,f)=∫0R‍n(t,f)-n(0,f)tdt+n(0,f)log⁡R,m(R,f)=12π∫02π‍log⁡+|f(Reiθ)|dθ,T(R,f)=N(R,f)+m(R,f),
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
(2.2)N1(R,f)=∫1/R1‍n1(t,f)tdt,N2(R,f)=∫1R‍n2(t,f)tdt,m0(R,f)=m(R,f)+m(1R,f),N0(R,f)=N1(R,f)+N2(R,f),
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
(2.3)T0(R,f)=m0(R,f)-2m(1,f)+N0(R,f)
and has the following properties.Proposition 2.1 (see [14]).
Letf 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 Proposition2.1, 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 annulus𝔸={z:1/R0<|z|<R0}, where 1<R<R0≤+∞. Then
(2.4)T0(R,1f-a)=T0(R,f)+O(1)
for every fixed a∈ℂ.Khrystiyanyn and Kondratyuk also obtained the lemma on the logarithmic derivative on the annulus𝔸.Theorem 2.3 (see [15] (lemma on the logarithmic derivative)).
Let f be a nonconstant meromorphic function on the annulus𝔸={z:1/R0<|z|<R0}, where R0≤+∞, and let λ>0. Then,(i)
in the caseR0=+∞,
(2.5)m0(R,f′f)=O(log⁡(RT0(R,f)))
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞;(ii)
ifR0<+∞, then
(2.6)m0(R,f′f)=O(log⁡(T0(R,f)R0-R))
for R∈(1,R0) except for the set ΔR′ such that ∫ΔR′‍dR/(R0-R)λ-1<+∞.We denote the deficiency ofa∈ℂ¯=ℂ∪{∞} with respect to a meromorphic function f on the annulus 𝔸 by
(2.7)δ0(a,f)=δ0(0,f-a)=liminf⁡r→R0m0(r,1/(f-a))T0(r,f)=1-limsup⁡r→R0N0(r,1/(f-a))T0(r,f)
and denote the reduced deficiency by
(2.8)Θ0(a,f)=Θ0(0,f-a)=1-limsup⁡r→R0N¯0(r,1/(f-a))T0(r,f),
where
(2.9)N¯0(r,1f-a)=N¯1(R,1f-a)+N¯2(R,1f-a)=∫1/R1‍n¯1(t,1/(f-a))tdt+∫1R‍n¯2(t,1/(f-a))tdt
in which each zero of the function f-a is counted only once. In addition, we use n¯1k)(t,1/(f-a)) (or n¯1(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 N¯1k)(t,f), N¯1(k(t,f), N¯2k)(t,f), N¯2(k(t,f), N¯0k)(t,f), and N¯0(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)).
Letf be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0≤+∞. Let a1,a2,…,aq be q distinct complex numbers in the extended complex plane ℂ¯. Let λ≥0. Then, (i)
(q-2)T0(R,f)<∑j=1qN0(R,1/(f-aj))-N0(1)(R,f)+S(R,f),(ii)
(q-2)T0(R,f)<∑j=1qN¯0(R,1/(f-aj))+S(R,f), where
(2.10)N0(1)(R,f)=N0(R,1f′)+2N0(R,f)-N0(R,f′),
and (i) in the case R0=+∞,
(2.11)S(R,f)=O(log⁡(RT0(R,f)))
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞; (ii) if R0<+∞, then
(2.12)S(R,f)=O(log⁡(T0(R,f)R0-R))
for R∈(1,R0) except for the set ΔR′ such that ∫ΔR′‍dR/(R0-R)λ-1<+∞.Definition 2.5.
Letf(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
(2.13)limsup⁡R→∞T0(R,f)log⁡R=∞,1<R<R0=+∞,
or
(2.14)limsup⁡R→R0T0(R,f)-log⁡(R0-R)=∞,1<R<R0<+∞,
respectively.Thus, for a transcendental or admissible meromorphic function on the annulus𝔸, S(R,f)=o(T0(R,f)) holds for all 1<R<R0 except for the set ΔR or the set ΔR′ mentioned in Theorem 2.3, respectively.


## 3. The Main Theorems and Some Lemmas

Now we show our main results, which improve Theorem1.4.Theorem 3.1.
Letf,g be two analytic functions on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0≤+∞, and let aj∈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 Theorem3.1, with CM replaced by IM, we have either f(z)≡g(z) or
(3.1)f≡a3g-a1a2g-a4,
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 Theorems3.1 and 3.2 are improvement of Theorem 1.4.To prove the above theorems, we need some lemmas as follows.Lemma 3.4.
Letf,g be two distinct analytic functions on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0≤+∞, and let aj∈ℂ(j=1,2,3,4) be four distinct complex numbers. If f=aj⇒g=aj 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 Lemma3.4 and applying Theorem 2.4(ii), we can get(3.2)3T0(R,f)≤∑j=14‍N¯0(R,1f-aj)+S(R,f)≤∑j=14‍N¯0(R,1g-aj)+S(R,f)≤4T0(R,g)+S(R,f).
Therefore(3.3)T0(R,f)≤4T0(R,g)+o(T0(R,f))
holds for all 1<R<R0 except for the set ΔR or the set ΔR′ 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 thatf is a transcendental or admissible meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0≤+∞. Let P(f)=a0fp+a1fp-1+⋯+ap(a0≠0) be a polynomial of f with degree p, where the coefficients aj(j=0,1,…,p) are constants, and let bj(j=1,2,…,q) be q(q≥p+1) distinct finite complex numbers. Then,
(3.4)m0(R,P(f)⋅f′(f-b1)(f-b2)⋯(f-bq))=S(R,f).Proof.
From Theorem2.3 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.
Letf,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 aj∈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)=N¯0(R,1/(f-a3))+N¯0(R,1/(f-a4))+S(R),(iv)
T0(R,f)=N¯0(R,1/(f-aj))+S(R),j=1,2,(v)
T0(R,g)=N¯0(R,1/(g-aj))+S(R),j=3,4,(vi)
T0(R,f′)=T0(R,f)+S(R),T0(R,g′)=T0(R,g)+S(R),
whereS(R):=S(R,f)=S(R,g).Proof.
By the assumption of this lemma and by Theorem2.4(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(3.5)η:=f′g′(f-g)(f-a3)(f-a4)(g-a1)(g-a2).
From the conditions of this lemma, we can get thatη 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).
Sincef,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
(3.6)3T0(R,f)≤∑j=14‍N¯0(R,1f-aj)+S(R,f),(3.7)≤N¯0(r,1f-g)+S(R,f)=T0(R,f-g)+S(R,f),(3.8)≤T0(R,f)+T0(R,g)+S(R),(3.9)T0(R,g)≤N¯0(R,1g-a1)+N¯0(R,1g-a2)+S(R,g),(3.10)=N¯0(R,1f-a1)+N¯0(R,1f-a2)+S(R),(3.11)≤2T0(R,f)+S(R).
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
(3.12)f=f′g′η(g-a1)(g-a2)+f′g′(a3f+a4f-a3a4-fg)η(f-a3)(f-a4)(g-a1)(g-a2).
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
(3.13)T0(R,f)≤T0(R,f′)+2m(1,f′)-2m(1,f)+S(R,f).
From the fact that f is transcendental or admissible, we have
(3.14)T0(R,f)≤T0(R,f′)+S(R,f)+O(1)=T0(R,f′)+S(R,f).
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
(3.15)T0(R,f′)≤T0(R,f)+S(R,f)+O(1)=T0(R,f)+S(R,f).
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 Theorem3.1

Supposef≢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
(4.1)ψ1∶=f′(f-a3)(f-a1)(f-a2)-g′(g-a3)(g-a1)(g-a2),ψ2∶=f′(f-a4)(f-a1)(f-a2)-g′(g-a4)(g-a1)(g-a2).By Lemma3.4, we can get
(4.2)m0(R,ψi)=S(R,f)+S(R,g)=S(R),i=1,2.
Moreover, we can prove N0(R,ψi)=O(1)(i=1,2). In fact, the poles of ψi in 𝔸 only can occur at the zeros of f-aj and g-aj(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-aj with multiplicity m(≥1), then z0∈𝔸 is a zero of g-aj(j=1,2) with multiplicity m(≥1). Suppose that
(4.3)f-aj=(z-z0)mαj(z),g-aj=(z-z0)mβj(z),
where αj(z),βj(z) are analytic functions in 𝔸 and αj(z0)≠0,βj(z0)≠0(j=1,2); by a simple calculation, we have
(4.4)ψi(z0)=K(αj′(z0)αj(z0)-βj′(z0)βj(z0))(i,j=1,2),
where K is a constant. Therefore, we can get that ψi(i=1,2) are analytic in 𝔸. Thus, from (4.2), we can get T0(R,ψi)=m0(R,ψi)+O(1)=S(R)(i=1,2).Ifψi≢0,i=1,2, then we have
(4.5)N¯0(R,1f-a3)≤N0(R,1ψ1)≤T0(R,ψ1)+S(R,f)+O(1)=S(R),N¯0(R,1f-a4)≤N0(R,1ψ2)≤T0(R,ψ2)+S(R,f)+O(1)=S(R).
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 ofψ1 and ψ2 is identically zero, say ψ1≡0; then we have
(4.6)N¯0(2(R,1g-a4)=N¯0(2(R,1f-a4).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
(4.7)N¯01)(R,1g-a4)=N¯01)(R,1f-a4)+S(R).From (4.6) and (4.7), we can get
(4.8)N¯0(R,1g-a4)=N¯0(R,1f-a4)+S(R).Similarly, whenψ2≡0, we can get
(4.9)N¯0(R,1g-a3)=N¯0(R,1f-a3)+S(R).From (4.8), (4.9), and Lemma 3.6(i), (v), we can get
(4.10)2T0(R,f)=N¯0(R,1f-a3)+S(R)
or
(4.11)2T0(R,f)=N¯0(R,1f-a4)+S(R).
Since f,g are transcendental or admissible functions on the annulus 𝔸, we can get a contradiction again.Thus, we complete the proof of Theorem3.1.


## 5. The Proof of Theorem3.2

Suppose thatf≢g. By Theorem 2.4(ii) and the fact that f is transcendental or admissible on 𝔸, we have
(5.1)2T0(R,f)+N¯0(R,1g-a4)≤N¯0(R,1f-a1)+N¯0(R,1f-a2)+N¯0(R,1f-a3)+N¯0(R,1g-a4)+S(R,f)≤N¯0(R,1f-g)+S(R,f)≤T0(R,f)+T0(R,g)+S(R,f)+S(R,g).
Therefore, we have
(5.2)T0(R,f)+N¯0(R,1g-a4)≤T0(R,g)+S(R,f)+S(R,g).
Similarly, we have
(5.3)T0(R,g)+N¯0(R,1f-a3)≤T0(R,f)+S(R,g)+S(R,f).From (5.2) and (5.3), we can see that T0(R,f)=T0(R,g)+S(R,f)+S(R,g), and
(5.4)N¯0(R,1f-a3)=S(R,f)+S(R,g),N¯0(R,1g-a4)=S(R,f)+S(R,g).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
(5.5)2T0(R,f)=N¯0(R,1f-a1)+N¯0(R,1f-a2)+S(R).From (5.5), we can see that “almost all” of zeros of f-ai(i=1,2) in 𝔸 are simple. Similarly, “almost all” of zeros of g-ai(i=1,2) in 𝔸 are simple, too. Let
(5.6)φ1∶=(a1-a3)f′(f-a2)(f-a1)(f-a3)-(a1-a4)g′(g-a2)(g-a1)(g-a4),φ2∶=(a2-a3)f′(f-a1)(f-a2)(f-a3)-(a2-a4)g′(g-a1)(g-a2)(g-a4).
By Lemma 3.5, we can get that m0(R,φi)=S(R)(i=1,2). Since f,g share a1,a2  IM in 𝔸 and from (5.2), we have N0(R,φi)=S(R)(i=1,2). Therefore, we can get T0(R,φi)=S(R),(i=1,2).Ifφ1≢0, then we have N¯0(R,1/(f-a2))≤N¯0(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,
(5.7)f′f-a3-g′g-a4-f′f-a1+g′g-a1-f′f-a2+g′g-a2≡0,
which implies that
(5.8)f-a3g-a4⋅(g-a1)(g-a2)(f-a1)(f-a2)≡c,
where c is a nonzero constant. Rewrite (5.8) as
(5.9)g2-(a1+a2-cγ(f)f-a3)g+a1a2+ca4γ(f)f-a3≡0,
where γ(f):=(f-a1)(f-a2). The discriminant of (5.9) is
(5.10)Δ(f)=(a1+a2-cγ(f)f-a3)2-4(a1a2+ca4γ(f)f-a3)∶=℘(f)(f-a3)2,
where
(5.11)℘(z)∶=((a1+a2)(z-a3)-cγ(z))2-4a1a2(z-a3)2-4ca4γ(z)(z-a3)
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
(5.12)g(z)=12(a1+a2-cγ(a)a-a3):=b.
Set
(5.13)ϕ1:=f′g′(f-g)(f-a1)(g-a2)(f-a3)(g-a4),ϕ2:=f′g′(f-g)(f-a2)(g-a1)(f-a3)(g-a4),ϕ=ϕ2ϕ1=(f-a1)(g-a2)(f-a2)(g-a1).By Lemma3.5, we can get m0(R,ϕi)=S(R)(i=1,2). And by a simple calculation, we can get N0(R,ϕi)=S(R)(i=1,2). Then we have T0(R,ϕi)=S(R)(i=1,2), thus we have T0(R,ϕ)=S(R).Assume thatf is not a Möbius transformation of g; then ϕ is a nonconstant function. Since
(5.14)℘(a1)=((a1+a2)(a1-a3))2-4a1a2(a1-a3)2=(a1-a3)2(a1-a2)2≠0,℘(a2)=((a1+a3)(a2-a3))2-4a1a2(a2-a3)2=(a2-a3)2(a1-a2)2≠0.
From a≠ai(i=1,2) and (5.8), we can get
(5.15)N¯0(R,1f-a)≤N¯0(R,1ϕ-ξ)≤T0(R,ϕ)=S(R),
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
(5.16)T0(R,f)≤N¯0(R,1f-a3)+N¯0(R,1f-a)+S(R)=S(R).Sincef,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
(5.17)f≡a3g-a1a2g-a4.
Furthermore, a3,a4 are Picard exceptional values of f and g in 𝔸, respectively.Thus, we complete the proof of Theorem3.2.



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*Source: 896596-2012-08-13.xml*