# The Shared Set and Uniqueness of Meromorphic Functions on Annuli

Authors: Hong Yan Xu; Zhao Jun Wu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The purpose of this paper is to deal with the shared set and
uniqueness of meromorphic functions on annulus. The set of this paper is
different from the set of the paper by Cao and Deng, and our theorems are
improvement of the results given by Cao and Deng.


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## Body

## 1. Introduction

In 1929, Nevanlinna [1] first investigated the uniqueness of meromorphic functions in the whole complex plane and obtained the well-known result—5 IM theorem of two meromorphic functions sharing five distinct values.After his theorem, there are vast references on the uniqueness of meromorphic functions sharing values and sets in the whole complex plane, the unit disc; and angular domain (see [2–8]).The notations of the Nevanlinna theory such asT(r,f), m(r,f), and N(r,f) were usually used in those papers (see [5, 9, 10]). We use ℂ to denote the open complex plane, ℂ~:=ℂ∪{∞} to denote the extended complex plane, and 𝕏 to denote the subset of ℂ. Let S be a set of distinct elements in ℂ~ and 𝕏⊆ℂ. Define
(1)E𝕏(S,f)=⋃a∈S‍{z∈𝕏∣fa(z)=0,countingmultiplicities},E¯𝕏(S,f)=⋃a∈S{z∈𝕏∣fa(z)=0,ignoringmultiplicities}‍,
where fa(z)=f(z)-a if a∈ℂ and f∞(z)=1/f(z). We also define
(2)E¯1𝕏(S,f)=⋃a∈S‍{z∈𝕏:allthesimplezerosoffa(z)}.Fora∈ℂ~, we say that two meromorphic functions f and g share the value aCM(IM) in 𝕏 (or ℂ), if f(z)-a and g(z)-a have the same zeros with the same multiplicities (ignoring multiplicities) in 𝕏 (or ℂ).The whole complex planeℂ, the unit disc, and angular domain all can be regarded as simply connected regions those results of the uniqueness of shared values and sets in the above cases can also be regarded as the uniqueness of meromorphic functions in simply connected regions.Thus, it raises naturally an interesting subject on the uniqueness of the meromorphic functions in the multiply connected region.The main purpose of this paper is to study the uniqueness of meromorphic functions in doubly connected domains of complex planeℂ. From the doubly connected mapping theorem [11], we can get that each doubly connected domain is conformally equivalent to the annulus {z:r<|z|<R}, 0≤r<R≤+∞. There are two cases: (1) r=0 and R=+∞ and (2) 0<r<R<+∞; for case (2) 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. The basic notions of the Nevanlinna theory on annuli will be showed in the next section.Recently, there are some results on the Nevanlinna Theory of meromorphic functions on the annulus (see [12–19]). In 2005, Khrystiyanyn and Kondratyuk [13, 14] proposed the Nevanlinna theory for meromorphic functions on annuli (see also [20]). Lund and Ye [16] in 2009 studied meromorphic functions on the annuli with the form {z:R1<|z|<R2}, where R1≥0 and R2≤∞. However, there are few results about the uniqueness of meromorphic functions on the annulus. In 2009 and 2011, Cao et al. [21, 22] 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 ([22, Theorem 3.2] or [21, Corollary 3.3]).
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 ajIM for j=1,2,3,4,5, then f1(z)≡f2(z).Remark 2.
For the caseR0=+∞, the assertion was proved by Kondratyuk and Laine [20].In 2012, Cao and Deng [23] investigated the uniqueness of two meromorphic functions in 𝔸 sharing three or two finite sets; we obtain the following theorems which are an analog of results on ℂ according to Lin and Yi [24].Theorem 3.
Letf and g be two admissible meromorphic functions in the annulus 𝔸. Put S1={0}, S2={∞}, and S3={w:P(w)=0}, where
(3)P(w)=awn+n(n-1)w2+2n(n-2)bw-(n-1)(n-2)b2,
where n≥5 is an integer and a and b are two nonzero complex numbers satisfying abn-2≠1,2. If E¯(S2,f)=E¯(S2,g) and E(Sj,f)=E(Sj,g) (j=1,3), then f≡g.Theorem 4.
Letf and g be two admissible meromorphic functions in the annulus 𝔸. Put S1={∞} and S2={w:P(w)=0}, where P(w) is stated as in Theorem 3, and a and b are two nonzero complex numbers satisfying abn-2≠2, n≥8 is an integer. If  E¯(S1,f)=E¯(S1,g) and E(S2,f)=E(S2,g), then f≡g.In this paper, we will focus on the uniqueness problem of shared set of meromorphic functions on the annuli. In fact, we will study the uniqueness of meromorphic functions on the annuli sharing one setS={w∈𝔸:P1(w)=0}, where
(4)P1(w)=(n-1)(n-2)2wn-n(n-2)wn-1+n(n-1)2wn-2-c,
and c is a complex number satisfying c≠0,1 and we obtain the following results.Theorem 5.
Letf and g be two admissible meromorphic functions in the annulus 𝔸. If E𝔸(S,f)=E𝔸(S,g) and n is an integer ≥11, then f≡g.A setS is called a unique range set for meromorphic functions on annulus 𝔸, if, for any two nonconstant meromorphic functions f and g, the condition E𝔸(S,f)=E𝔸(S,g) implies f≡g. We denote by ♯S the cardinality of a set S. Thus, from Theorem 5, we can get the following corollary.Corollary 6.
There exists one finite setS with ♯S=7, such that any two admissible meromorphic functions f and g on 𝔸 must be identical if E𝔸(S,f)=E𝔸(S,g).Theorem 7.
Letf and g be two admissible meromorphic functions in the annulus 𝔸. If n is an integer ≥7, E𝔸(S,f)=E𝔸(S,g), and Θ0(∞,f)>3/4, Θ0(∞,g)>3/4, then f≡g.Corollary 8.
There exists one finite setS with ♯S=7, such that any two admissible analytic functions f and g on 𝔸 must be identical if E1𝔸(S,f)=E1𝔸(S,g).Theorem 9.
Letf and g be two admissible meromorphic functions in the annulus 𝔸. If E1𝔸(S,f)=E1𝔸(S,g) and n is an integer ≥15, then f≡g.A setS is called a unique range set, with weight 1 for meromorphic functions on annulus 𝔸, if for any two nonconstant meromorphic functions f and g, the condition E1𝔸(S,f)=E1𝔸(S,g) implies f≡g. Thus, from Theorem 9, we can get the following corollary.Corollary 10.
There exists one finite setS with ♯S=15, such that any two admissible meromorphic functions f and g on 𝔸 must be identical if E1𝔸(S,f)=E1𝔸(S,g).Theorem 11.
Letf and g be two admissible meromorphic functions in the annulus 𝔸. Let n be an integer ≥9 and S={w∈𝔸:P1(w)=0}, where P1(w) and c are stated as in Theorem 5. If E1𝔸(S,f)=E1𝔸(S,g) and Θ0(∞,f)>5/6, Θ0(∞,g)>5/6, then f≡g.From Theorem11, we can get the corollary as follows.Corollary 12.
There exists one finite setS with ♯S=9, such that any two admissible analytic functions f and g on 𝔸 must be identical if E1𝔸(S,f)=E1𝔸(S,g).


## 2. Preliminaries and Some Lemmas

Lettingf be a meromorphic function on whole plane ℂ, the classical notations of the Nevanlinna theory are denoted as follows:(5)N(R,f)=∫0R‍n(t,f)-n(0,f)tdt+n(0,f)logR,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}.Lettingf be a meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R<R0≤+∞, the notations of the Nevanlinna theory on annuli had been introduced in [13, 20], such as N0(R,f),m0(R,f),T0(R,f),…. In addition, we define
(6)Θ0(∞,f)=1-limsupR→∞N¯0(R,f)T0(R,f).
We also 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}, with each point being counted only once. Similarly, we have the notations N¯1k)(t,f), N¯1(k(t,f), N¯2k)(t,f), N¯2(k(t,f), N¯0k)(t,f), N¯0(k(t,f).For a nonconstant meromorphic functionf on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R<R0≤+∞, the following properties will be used in this paper (see [13]): (i)
(7)T0(R,f)=T0(R,1f),(ii)
(8)max{T0(R,f1·f2),T0(R,f1f2),T0(R,f1+f2)}≤T0(R,f1)+T0(R,f2)+O(1),(iii)
(9)T0(R,1f-a)=T0(R,f)+O(1),foreveryfixeda∈ℂ.Khrystiyanyn and Kondratyuk [14] also obtained the lemma on the logarithmic derivative on the annulus 𝔸.Lemma 13 (see [14], lemma on the logarithmic derivative).
Letf be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where R0≤+∞, and let λ>0. Then,(i)
in the caseR0=+∞,
(10)m0(R,f′f)=O(log(RT0(R,f))),forR∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞;(ii)
ifR0<+∞, then
(11)m0(R,f′f)=O(log(T0(R,f)R0-R)),forR∈(1,R0) except for the set ΔR′ such that ∫ΔR′(dR/(R0-R)λ-1)‍<+∞.Remark 14.
From [14, 20], the conclusions still hold if m0(R,f′/f) is replaced by m0(R,f(k)/f), k∈N+.In 2005, the second fundamental theorem on the annulus𝔸 was first obtained by Khrystiyanyn and Kondratyuk [14]. Later, Cao et al. [22] introduced other forms of the second fundamental theorem on annuli as follows.Lemma 15 ([22, 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~. Then,
(12)(q-2)T0(R,f)<∑j=1q‍N¯0(R,1f-aj)+S(R,f),
where (i) in the case R0=+∞,
(13)S(R,f)=O(log(RT0(R,f))),
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞.
(ii) IfR0<+∞, then
(14)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)<+∞.Remark 16.
In fact, from the proof of Theorem 2.3 in [22], under the assumptions of Lemma 15, we can get the following conclusion:
(15)(q-2)T0(R,f)<∑j=1q‍N¯0(R,1f-aj)-N00(R,1f′)+S(R,f),
where S(R,f) is stated as in Lemma 15 and N00(R,1/f′) is the counting function for the zeros of f′ in 𝔸, where f does not take one of the values aj (j=1,2,…,q).Definition 17.
Letf(z) be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0≤+∞. The function f is called an admissible meromorphic function on the annulus 𝔸 provided that
(16)limsupR→∞T0(R,f)logR=∞,1<R<R0=+∞,
or
(17)limsupR→R0T0(R,f)-log(R0-R)=∞,1<R<R0<+∞,
respectively.Thus, for an 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 Lemma 13, respectively.The following result can be derived from the proof of Frank-Reinders' theorem in [25].Lemma 18.
Letn≥6 and
(18)H(w)=(n-1)(n-2)2wn-n(n-2)wn-1+n(n-1)2wn-2.
Then, H(w) is a unique polynomial for admissible meromorphic functions; that is, for any two admissible meromorphic functions f and g in 𝔸, H(f)≡H(g) implies f≡g.By a similar discussion to the one in [26], one can obtain a stand and Valiron-Mohoko type result in 𝔸 as follows.Lemma 19 (see [23]).
Letf be a nonconstant meromorphic function in 𝔸, Q1(f) and let Q2(f) be two mutually prime polynomials in f with degree m and n, respectively. Then,
(19)T0(R,Q1(f)Q2(f))=max{m,n}T0(R,f)+S(R,f).Lemma 20.
Supposef is a nonconstant meromorphic function in 𝔸. Then,
(20)N0(R,1f′)≤N0(R,1f)+N¯0(R,f)+S(R,f)+O(1),
where S(R,f) is stated as in Lemma 15.Proof.
Since(21)m0(R,1f)≤m0(R,1f′)+m0(R,f′f)=m0(R,1f′)+S(R,f).
Then, from properties of T0(R,f), we have
(22)T0(R,f)-N0(R,1f)≤T0(R,f′)-N0(R,1f′)+S(r,f)+O(1);
that is,
(23)N0(R,1f′)≤T0(R,f′)-T0(R,f)+N0(R,1f)+S(r,f)+O(1).
Since
(24)T0(R,f′)=m0(R,f′)+N0(R,f′)-2m(1,f′)≤m0(R,f)+m0(R,f′f)+N0(R,f)+N¯0(R,f)-2m(1,f′)≤T0(R,f)+N¯0(R,f)+S(R,f)+O(1).
Then, from (23) and (24), we can get the conclusion of this lemma.Next, we will give the two main lemmas of this paper as follows.Lemma 21.
LetF and G be admissible meromorphic functions in A satisfying E𝔸(F,0)=E𝔸(G,0) and let c1,c2,…,cq be q (≥2) distinct nonzero complex numbers. If
(25)limsupR→∞,R∈I((3N¯0(R,F)+∑j=1q‍N¯0(2)(R,1F-cj)+N¯0(R,1F′))(3N¯0(R,F)+∑j=1q‍N¯0(2)(R,1F-cj)+N¯0(R,1F′))×(T0(R,F))-1∑j=1q‍N¯0(2)(R,1F-cj))<q,limsupR→∞,R∈I((3N¯0(R,G)+∑j=1q‍N¯0(2)(R,1G-cj)+N¯0(R,1G′))(3N¯0(R,F)+∑j=1q‍N¯0(2)(R,1F-cj)+N¯0(R,1F′))×(T0(R,G))-1∑j=1q‍N¯0(2)(R,1G-cj))<q,
where N¯0(2)(R,·)=N¯0(R,·)+N¯0(2(R,·), N¯0(2(r,·)=N¯0(r,·)-N¯01)(r,·), and I is some set of R of infinite linear measure, then
(26)F=aG+bcG+d,
where a,b,c,d∈ℂ are constants with ad-bc≠0.Proof.
Set(27)H≡F′′F′-2F′F-(G′′G′-2G′G).
Supposing thatH≢0, from Lemma 13 and Remark 14, we have
(28)m0(R,H)=S(R),
where S(R):=o{T0(R)}, T0(R)=max{T0(R,F),T0(R,G)}. Since E𝔸(F,0)=E𝔸(G,0), and by an elementary calculation, we can conclude that if z0 is a common simple zero of F and G in 𝔸, then H(z0)=0. Thus, we have
(29)N01)(R)≤N0(R,1H)≤T0(R,H)+O(1)≤N0(R,H)+S(R),
where N01)(R)=N01)(R,1/F)=N01)(R,1/G). The poles of H in 𝔸 can only occur at zeros of F′ and G′ in 𝔸 or poles of F and G in 𝔸. Moreover, H only has simple zeros in 𝔸. Hence, from (29), we have
(30)N01)(R)≤N¯0(R,F)+N¯0(R,G)+N¯00(R,1F′)+N¯00(R,1G′)+∑j=1q‍N¯0(2(R,1F-cj)+∑j=1q‍N¯0(2(R,1G-cj)+S(R),
where N¯00(R,1/F′) is the reduced counting function for the zeros of F′ in A, where F does not take one of the values 0,c1,c2,…,cq.
Since(31)N¯0(R,1F)+N¯0(R,1G)=2N01)(R)+N¯0(2(R,1F)+N¯0(2(R,1G).
Then, from (30) and (31), we have
(32)N¯0(R,1F)+N¯0(R,1G)≤2N¯0(R,F)+2N¯0(R,G)+2N¯00(R,1F′)+2N¯00(R,1G′)+N¯0(2(R,1F)+N¯0(2(R,1G)+2∑j=1q‍N¯0(2(R,1F-cj)+2∑j=1q‍N¯0(2(R,1G-cj)+S(R).
From Remark 16, we have
(33)qT0(R,F)≤N¯0(R,F)+N¯0(R,1F)+∑j=1q‍N¯0(R,1F-cj)-N00(R,1F′)+S(R),r∉E,qT0(R,G)≤N¯0(R,G)+N¯0(R,1G)+∑j=1q‍N¯0(R,1G-cj)-N00(R,1G′)+S(R),r∉E,
where E is a set of r of finite linear measure and it needs not to be the same at each occurrence. From (32)-(33), it follows that, for r∉E,
(34)q{T0(R,F)+T0(R,G)}≤3N¯0(R,F)+3N¯0(R,G)+∑j=1q‍N¯0(R,1F-cj)+∑j=1q‍N¯0(R,1G-cj)+2∑j=1q‍N¯0(2(R,1F-cj)+2∑j=1q‍N¯0(2(R,1G-cj)+N¯00(R,1F′)+N¯00(R,1G′)+N¯0(2(R,1F)+N¯0(2(R,1G)+S(R).
Since
(35)∑j=1q‍N¯0(2(R,1F-cj)+N¯0(2(R,1F)+N¯00(R,1F′)=N¯0(R,1F′),∑j=1q‍N¯0(2(R,1G-cj)+N¯0(2(R,1G)+N¯00(R,1G′)=N¯0(R,1G′).
From (34)-(35), we can get that, for R∉E,
(36)q{T0(R,F)+T0(R,G)}≤3N¯0(R,F)+3N¯0(R,G)+∑j=1q‍N¯0(2)(R,1F-cj)+∑j=1q‍N¯0(2)(R,1G-cj)+N¯0(R,1F′)+N¯0(R,1G′)+S(R).
From (25) and (36), since f, g are admissible functions in 𝔸, we can get that
(37)T0(R,F)+T0(R,G)≤o{T0(R,F)+T0(R,G)},R∉E,R∈I.
Thus, we can get a contradiction. Therefore, H(z)≡0; that is,
(38)F′′F′-2F′F≡G′′G′-2G′G.
For the above equality, by integration, we can get
(39)F≡aG+bcG+d,
where a,b,c,d∈ℂ and ad-bc≠0.Lemma 22.
LetF and G be admissble meromorphic functions in 𝔸 satisfying E1𝔸(F,0)=E1𝔸(G,0) and let c1,c2,…,cq be q (≥2) distinct nonzero complex numbers. If
(40)limsupR→∞,R∈I((3N¯0(R,F)+∑j=1q‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)+2N¯0(2(R,1F)∑j=1q‍N¯0(2)(R,1F-cj))×(T0(R,F))-1(3N¯0(R,F)+∑j=1q‍N¯0(2)(R,1F-cj))<q,limsupR→∞,R∈I((3N¯0(R,G)+∑j=1q‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)+2N¯0(2(R,1G)∑j=1q‍N¯0(2)(R,1G-cj))×(T0(R,G))-1(3N¯0(R,G)+∑j=1q‍N¯0(2)(R,1G-cj))<q,
where N¯0(2)(R,·), N¯0(2(r,·), and I are stated as in Lemma 21; then
(41)F=aG+bcG+d,
where a,b,c,d∈ℂ are constants with ad-bc≠0.Proof.
LetH be stated as in the proof of Lemma 21, since E1𝔸(F,0)=E1𝔸(G,0), we can get that
(42)N01)(R)≤N¯0(R,F)+N¯0(R,G)+N¯00(R,1F′)+N¯00(R,1G′)+N¯0(2(R,1F)+N¯0(2(R,1G)+∑j=1q‍N¯0(2(R,1F-cj)+∑j=1q‍N¯0(2(R,1G-cj).
Similar to the argument in Lemma 21, we can get that, for R∉E(43)q{T0(R,F)+T0(R,G)}≤3N¯0(R,F)+3N¯0(R,G)+∑j=1q‍N¯0(2)(R,1F-cj)+∑j=1q‍N¯0(2)(R,1G-cj)+2N¯0(2(R,1F)+2N¯0(2(R,1G)+N¯0(R,1F′)+N¯0(R,1G′)+S(R).
From (40) and (43), since f, g are admissible functions in 𝔸, we can get that
(44)T0(R,F)+T0(R,G)≤o{T0(R,F)+T0(R,G)},R∉E,R∈I.
Thus, we can get a contradiction. Therefore, H(z)≡0; that is,
(45)F′′F′-2F′F≡G′′G′-2G′G.
From the above equality, by integration, we can get
(46)F≡aG+bcG+d,
where a,b,c,d∈ℂ and ad-bc≠0.


## 3. Proofs of Theorems5 and 7

### 3.1. The Proof of Theorem5

From the definition ofP1(w), we can get that P1(1)=1-c:=c1≠0, P1(0)=-c:=c2≠0, and
(47)P1′(w)=n(n-1)(n-2)2(w-1)2wn-3,(48)P1(w)-c1=(w-1)3Q1(w),Q1(1)≠0,(49)P1(w)-c2=wn-2Q2(w),Q2(0)≠0,
where Q1, Q2 are polynomials of degrees n-3 and 2, respectively. We also see that Qi (i=1,2) and P1 have only simple zeros.LetF and G be defined as F=P1(f) and G=P1(g). Since E𝔸(f,S)=E𝔸(g,S), we have E𝔸(F,0)=E𝔸(G,0). From (48) and (49), we have
(50)N¯0(2)(R,1F-c1)=N¯0(R,1F-c1)+N¯0(2(R,1F-c1)≤2N¯0(R,1f-1)+∑i=1n-3‍N0(R,1f-ai)≤(n-1)T0(R,f)+S(R),N¯0(2)(R,1F-c2)=N¯0(R,1F-c2)+N¯0(2(R,1F-c2)≤2N¯0(R,1f)+∑j=12‍N0(R,1f-bj)≤4T0(R,f)+S(R),
where ai (i=1,…,n-3) and bj (j=1,2) are the zeros of Q1(w) and Q2(w) in 𝔸, respectively.From (47), we have
(51)N¯0(R,1F′)≤N¯0(R,1f)+N¯0(R,1f-1)+N¯(R,1f′).
From Lemma 19, we have T0(R,F)=nT0(R,f)+S(R). Thus, combining (50) and (51), by Lemmas 21 and 20 and n≥11, we have
(52)limsupR→∞,R∉E((3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)∑j=12‍N¯0(2)(R,1F-cj))×(T0(R,F))-1)≤limsupR→∞,R∉E4N¯0(R,f)+(n+6)T0(R,f)nT0(R,f)<2.
Similarly, we can obtain
(53)limsupR→∞,R∉E((3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1)≤limsupR→∞,R∉E4N¯0(R,g)+(n+6)T0(R,g)nT0(R,g)<2.
Thus, by Lemma 21, we have
(54)F′′F′-2F′F≡G′′G′-2G′G.
From the previous equality, by integration, we can get
(55)F≡aG+bcG+d,
where a,b,c,d∈ℂ and ad-bc≠0. Since E𝔸(f,S) is nonempty and E𝔸(f,S)=E𝔸(g,S), we have b=0, a≠0. Hence,
(56)F≡aGcG+d≡GAG+B,
where A=c/a, B=d/a≠0.Two cases will be considered as follows.Case 1 (A≠0).
From the definition ofP1(w) and (56), we can see that every zero of P1(g)+B/A in 𝔸 has a multiplicity of at least n. Here, three following subcases will be discussed.Subcase 1 (B/A=-c1).
From (48), we have
(57)P1(g)+BA=(g-1)3(g-a1)(g-a2)⋯(g-an-3),
where ai≠0,1 are distinct values. It follows that
(58)Θ0(ai,f)=1-limsupR→∞N¯0(R,a)T0(R,f)≥1-limsupR→∞N¯0(R,a)N0(R,f)≥12.
We can see that P1(g)+B/A has n-2 values satisfying the above inequality. Thus, from Lemma 15 and n≥11, we can get a contradiction.Subcase 2 (B/A=-c2).
From (48), we have
(59)P1(g)+BA=gn-2(g-b1)(g-b2),
where b1≠b2, bi≠0,1 (i=1,2). It follows that every zero of g in 𝔸 has a multiplicity of at least 2 and every zero of g-bi (i=1,2) in 𝔸 has a multiplicity of at least n. Then, by Lemma 15, we have
(60)T0(R,g)≤N¯0(R,1g)+N¯0(R,1g-b1)+N¯0(R,1g-b2)+S(R)≤12N0(R,1g)+1nN0(R,1g-b1)+1nN0(R,1g-b1)+S(R)≤(12+2n)T0(R,g)+S(R).
Since g is an admissible function in 𝔸 and n≥11, we can get a contradiction.Subcase 3 (B/A≠-c1,-c2).
By using the same argument as in Subcases1 or 2, we can get a contradiction.Case 2 (A=0).
IfB≠1, from (56); we have F=G/B; that is,
(61)P1(f)=1BP1(g).
From (49) and (61), we have
(62)P1(f)-c2B=1B(P1(g)-c2)=1Bgn-2(g-b1)(g-b2).
Since c2/B≠c2, from (47), it follows that P1(f)-c2/B has at least n-2 distinct zeros e1,e2,…,en-2. Then, by applying Lemma 15, we have
(63)(n-4)T0(R,f)≤∑i=1n-2‍N¯0(R,1f-ei)+S(R)≤N¯0(R,1g)+N¯0(R,1g-b1)+N¯0(R,1g-b2)+S(R)≤3T0(R,g)+S(R).
By applying Lemma21 to (61), and from (63), since n≥11 and f is an admissible function in 𝔸, we can get a contradiction.
Thus, we haveA=0 and B=1; that is, P1(f)≡P1(g). Noting the form of P1(w); we can get that P1(f)≡P1(g), that is,
(64)(n-1)(n-2)2fn-n(n-2)fn-1+n(n-1)2fn-2≡(n-1)(n-2)2gn-n(n-2)gn-1+n(n-1)2gn-2.
Since f, g are admissible functions in 𝔸, then it follows by Lemma 18 that f≡g.Therefore, the proof of Theorem5 is completed.


### 3.2. The Proof of Theorem7

SinceΘ0(∞,f)>3/4 and Θ0(∞,g)>3/4, it follows that
(65)limsupR→∞N¯0(R,f)T0(R,f)<14,limsupR→∞N¯0(R,g)T0(R,g)<14.
By applying (65), from (52) and (53), and since n≥7, we can get
(66)limsupR→∞,R∉E((3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)∑j=12‍N¯0(2)(R,1F-cj))×(T0(R,F))-1)≤limsupR→∞,R∉E4N¯0(R,f)+(n+6)T0(R,f)nT0(R,f)<2,limsupR→∞,R∉E((3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1)≤limsupR→∞,R∉E4N¯0(R,g)+(n+6)T0(R,g)nT0(R,g)<2.
Then, from Lemma 21, we have F≡(aG+b)/(cG+d), where a,b,c,d∈ℂ and ad-bc≠0. Thus, by using the same argument as that in Theorem 5, we can prove the conclusion of Theorem 7.


## 3.1. The Proof of Theorem5

From the definition ofP1(w), we can get that P1(1)=1-c:=c1≠0, P1(0)=-c:=c2≠0, and
(47)P1′(w)=n(n-1)(n-2)2(w-1)2wn-3,(48)P1(w)-c1=(w-1)3Q1(w),Q1(1)≠0,(49)P1(w)-c2=wn-2Q2(w),Q2(0)≠0,
where Q1, Q2 are polynomials of degrees n-3 and 2, respectively. We also see that Qi (i=1,2) and P1 have only simple zeros.LetF and G be defined as F=P1(f) and G=P1(g). Since E𝔸(f,S)=E𝔸(g,S), we have E𝔸(F,0)=E𝔸(G,0). From (48) and (49), we have
(50)N¯0(2)(R,1F-c1)=N¯0(R,1F-c1)+N¯0(2(R,1F-c1)≤2N¯0(R,1f-1)+∑i=1n-3‍N0(R,1f-ai)≤(n-1)T0(R,f)+S(R),N¯0(2)(R,1F-c2)=N¯0(R,1F-c2)+N¯0(2(R,1F-c2)≤2N¯0(R,1f)+∑j=12‍N0(R,1f-bj)≤4T0(R,f)+S(R),
where ai (i=1,…,n-3) and bj (j=1,2) are the zeros of Q1(w) and Q2(w) in 𝔸, respectively.From (47), we have
(51)N¯0(R,1F′)≤N¯0(R,1f)+N¯0(R,1f-1)+N¯(R,1f′).
From Lemma 19, we have T0(R,F)=nT0(R,f)+S(R). Thus, combining (50) and (51), by Lemmas 21 and 20 and n≥11, we have
(52)limsupR→∞,R∉E((3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)∑j=12‍N¯0(2)(R,1F-cj))×(T0(R,F))-1)≤limsupR→∞,R∉E4N¯0(R,f)+(n+6)T0(R,f)nT0(R,f)<2.
Similarly, we can obtain
(53)limsupR→∞,R∉E((3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1)≤limsupR→∞,R∉E4N¯0(R,g)+(n+6)T0(R,g)nT0(R,g)<2.
Thus, by Lemma 21, we have
(54)F′′F′-2F′F≡G′′G′-2G′G.
From the previous equality, by integration, we can get
(55)F≡aG+bcG+d,
where a,b,c,d∈ℂ and ad-bc≠0. Since E𝔸(f,S) is nonempty and E𝔸(f,S)=E𝔸(g,S), we have b=0, a≠0. Hence,
(56)F≡aGcG+d≡GAG+B,
where A=c/a, B=d/a≠0.Two cases will be considered as follows.Case 1 (A≠0).
From the definition ofP1(w) and (56), we can see that every zero of P1(g)+B/A in 𝔸 has a multiplicity of at least n. Here, three following subcases will be discussed.Subcase 1 (B/A=-c1).
From (48), we have
(57)P1(g)+BA=(g-1)3(g-a1)(g-a2)⋯(g-an-3),
where ai≠0,1 are distinct values. It follows that
(58)Θ0(ai,f)=1-limsupR→∞N¯0(R,a)T0(R,f)≥1-limsupR→∞N¯0(R,a)N0(R,f)≥12.
We can see that P1(g)+B/A has n-2 values satisfying the above inequality. Thus, from Lemma 15 and n≥11, we can get a contradiction.Subcase 2 (B/A=-c2).
From (48), we have
(59)P1(g)+BA=gn-2(g-b1)(g-b2),
where b1≠b2, bi≠0,1 (i=1,2). It follows that every zero of g in 𝔸 has a multiplicity of at least 2 and every zero of g-bi (i=1,2) in 𝔸 has a multiplicity of at least n. Then, by Lemma 15, we have
(60)T0(R,g)≤N¯0(R,1g)+N¯0(R,1g-b1)+N¯0(R,1g-b2)+S(R)≤12N0(R,1g)+1nN0(R,1g-b1)+1nN0(R,1g-b1)+S(R)≤(12+2n)T0(R,g)+S(R).
Since g is an admissible function in 𝔸 and n≥11, we can get a contradiction.Subcase 3 (B/A≠-c1,-c2).
By using the same argument as in Subcases1 or 2, we can get a contradiction.Case 2 (A=0).
IfB≠1, from (56); we have F=G/B; that is,
(61)P1(f)=1BP1(g).
From (49) and (61), we have
(62)P1(f)-c2B=1B(P1(g)-c2)=1Bgn-2(g-b1)(g-b2).
Since c2/B≠c2, from (47), it follows that P1(f)-c2/B has at least n-2 distinct zeros e1,e2,…,en-2. Then, by applying Lemma 15, we have
(63)(n-4)T0(R,f)≤∑i=1n-2‍N¯0(R,1f-ei)+S(R)≤N¯0(R,1g)+N¯0(R,1g-b1)+N¯0(R,1g-b2)+S(R)≤3T0(R,g)+S(R).
By applying Lemma21 to (61), and from (63), since n≥11 and f is an admissible function in 𝔸, we can get a contradiction.
Thus, we haveA=0 and B=1; that is, P1(f)≡P1(g). Noting the form of P1(w); we can get that P1(f)≡P1(g), that is,
(64)(n-1)(n-2)2fn-n(n-2)fn-1+n(n-1)2fn-2≡(n-1)(n-2)2gn-n(n-2)gn-1+n(n-1)2gn-2.
Since f, g are admissible functions in 𝔸, then it follows by Lemma 18 that f≡g.Therefore, the proof of Theorem5 is completed.


## 3.2. The Proof of Theorem7

SinceΘ0(∞,f)>3/4 and Θ0(∞,g)>3/4, it follows that
(65)limsupR→∞N¯0(R,f)T0(R,f)<14,limsupR→∞N¯0(R,g)T0(R,g)<14.
By applying (65), from (52) and (53), and since n≥7, we can get
(66)limsupR→∞,R∉E((3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)∑j=12‍N¯0(2)(R,1F-cj))×(T0(R,F))-1)≤limsupR→∞,R∉E4N¯0(R,f)+(n+6)T0(R,f)nT0(R,f)<2,limsupR→∞,R∉E((3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1)≤limsupR→∞,R∉E4N¯0(R,g)+(n+6)T0(R,g)nT0(R,g)<2.
Then, from Lemma 21, we have F≡(aG+b)/(cG+d), where a,b,c,d∈ℂ and ad-bc≠0. Thus, by using the same argument as that in Theorem 5, we can prove the conclusion of Theorem 7.


## 4. Proofs of Theorems9 and 11

### 4.1. The Proof of Theorem9

SinceE1𝔸(f,S)=E1𝔸(g,S), we have E1𝔸(F,0)=E1𝔸(G,0). From (47)–(49), we can get
(67)N¯0(2(R,1F)=∑i=1n‍N¯0(R,1f-di)≤N¯(R,1f′),
where di (i=1,…,n) are the distinct zeros of P1(w). And from (51) and (67), by Lemma 20, we have
(68)N¯0(R,1F′)+2N¯0(2(R,1F)≤N¯0(R,1f)+N¯0(R,1f-1)+3N0(R,1f)+3N¯0(R,f)≤5T0(R,f)+3N¯0(R,f)+S(R).
Then, from (50) and (68), since T0(R,F)=nT0(R,f)+S(R) and n≥15, we have
(69)limsupR→∞,R∉E((3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)+2N¯0(2(R,1F)∑j=12‍N¯0(2)(R,1F-cj))×(T0(R,F))-1∑j=12‍N¯0(2)(R,1F-cj))≤limsupR→∞,R∉E6N¯0(R,f)+(n+8)T0(R,f)nT0(R,f)<2.
Similarly, we can get
(70)limsupR→∞,R∉E((3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)+2N¯0(2(R,1G)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1∑j=12‍N¯0(2)(R,1G-cj))≤limsupR→∞,R∉E6N¯0(R,g)+(n+8)T0(R,g)nT0(R,g)<2.
Thus, by Lemma 22, we have
(71)F≡aG+bcG+d,
where a,b,c,d∈ℂ and ad-bc≠0. By using arguments similar to those in the proof of Theorem 5, we can get that f≡g.Therefore, this completes the proof of Theorem9.


### 4.2. The Proof of Theorem11

SinceΘ0(∞,f)>5/6 and Θ0(∞,g)>5/6, it follows that
(72)limsupR→∞N¯0(R,f)T0(R,f)<16,limsupR→∞N¯0(R,g)T0(R,g)<16.
By applying (72), from (69) and (70), since n≥7, we can get
(73)limsupR→∞,R∉E((∑j=12‍N¯0(2)(R,1G-cj)3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)+2N¯0(2(R,1F)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,F))-1∑j=12‍N¯0(2)(R,1F-cj))≤limsupR→∞,R∉E6N¯0(R,f)+(n+8)T0(R,f)nT0(R,f)<2,limsupR→∞,R∉E((∑j=12‍N¯0(2)(R,1G-cj)3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)+2N¯0(2(R,1G)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1∑j=12‍N¯0(2)(R,1G-cj))≤limsupR→∞,R∉E6N¯0(R,g)+(n+8)T0(R,g)nT0(R,g)<2.
Then, from Lemma 22, we have F≡(aG+b)/(cG+d), where a,b,c,d∈ℂ and ad-bc≠0. Thus, by using the same argument as in Theorem 5, we can prove the conclusion of Theorem 11.


## 4.1. The Proof of Theorem9

SinceE1𝔸(f,S)=E1𝔸(g,S), we have E1𝔸(F,0)=E1𝔸(G,0). From (47)–(49), we can get
(67)N¯0(2(R,1F)=∑i=1n‍N¯0(R,1f-di)≤N¯(R,1f′),
where di (i=1,…,n) are the distinct zeros of P1(w). And from (51) and (67), by Lemma 20, we have
(68)N¯0(R,1F′)+2N¯0(2(R,1F)≤N¯0(R,1f)+N¯0(R,1f-1)+3N0(R,1f)+3N¯0(R,f)≤5T0(R,f)+3N¯0(R,f)+S(R).
Then, from (50) and (68), since T0(R,F)=nT0(R,f)+S(R) and n≥15, we have
(69)limsupR→∞,R∉E((3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)+2N¯0(2(R,1F)∑j=12‍N¯0(2)(R,1F-cj))×(T0(R,F))-1∑j=12‍N¯0(2)(R,1F-cj))≤limsupR→∞,R∉E6N¯0(R,f)+(n+8)T0(R,f)nT0(R,f)<2.
Similarly, we can get
(70)limsupR→∞,R∉E((3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)+2N¯0(2(R,1G)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1∑j=12‍N¯0(2)(R,1G-cj))≤limsupR→∞,R∉E6N¯0(R,g)+(n+8)T0(R,g)nT0(R,g)<2.
Thus, by Lemma 22, we have
(71)F≡aG+bcG+d,
where a,b,c,d∈ℂ and ad-bc≠0. By using arguments similar to those in the proof of Theorem 5, we can get that f≡g.Therefore, this completes the proof of Theorem9.


## 4.2. The Proof of Theorem11

SinceΘ0(∞,f)>5/6 and Θ0(∞,g)>5/6, it follows that
(72)limsupR→∞N¯0(R,f)T0(R,f)<16,limsupR→∞N¯0(R,g)T0(R,g)<16.
By applying (72), from (69) and (70), since n≥7, we can get
(73)limsupR→∞,R∉E((∑j=12‍N¯0(2)(R,1G-cj)3N¯0(R,F)+∑j=12‍N¯0(2)(R,1F-cj)+N¯0(R,1F′)+2N¯0(2(R,1F)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,F))-1∑j=12‍N¯0(2)(R,1F-cj))≤limsupR→∞,R∉E6N¯0(R,f)+(n+8)T0(R,f)nT0(R,f)<2,limsupR→∞,R∉E((∑j=12‍N¯0(2)(R,1G-cj)3N¯0(R,G)+∑j=12‍N¯0(2)(R,1G-cj)+N¯0(R,1G′)+2N¯0(2(R,1G)∑j=12‍N¯0(2)(R,1G-cj))×(T0(R,G))-1∑j=12‍N¯0(2)(R,1G-cj))≤limsupR→∞,R∉E6N¯0(R,g)+(n+8)T0(R,g)nT0(R,g)<2.
Then, from Lemma 22, we have F≡(aG+b)/(cG+d), where a,b,c,d∈ℂ and ad-bc≠0. Thus, by using the same argument as in Theorem 5, we can prove the conclusion of Theorem 11.



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*Source: 758318-2013-11-07.xml*