# On the Uniqueness of Meromorphic Functions on Annuli in terms of Deficiencies

Authors: Dawei Meng; Nan Lu; Sanyang Liu
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The purpose of this article is to study the uniqueness of meromorphic functions on annuli. Under a certain condition about deficiencies, we prove some new uniqueness theorems of meromorphic functions on the annulusA=z:1/R0<z<R0, where 1<R0≤+∞.


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## Body

## 1. Introduction and Main Results

In this article, we assume that the readers are familiar with the classical notations and definitions of Nevanlinna theory (refer to [1, 2]). The main purpose of this article is to study the uniqueness of meromorphic functions on annuli. For the necessary concepts and notations of the Nevanlinna theory of meromorphic functions on annuli, such as T0r,f,m0r,f,andN0r,f, refer to the excellent summarizations [3–11].Leta be a value in the extended complex plane ℂ¯, and let fandg be two meromorphic functions on the annulus A=z:1/R0<z<R0, where 1<R0≤+∞. Then, we say that f and g share a IM (ignoring multiplicities) when f−a and g−a have the same zeros, and furthermore, we say that f and g share a CM (counting multiplicities) when f−a and g−a have the same zeros with the same multiplicities. As mentioned in [3, 4], the reduced counting function N¯0r,1/f−a is defined by(1)∫1/r1n¯1t,1/f−atdt+∫1rn¯2t,1/f−atdt,where n¯1t,1/f−a and n¯2t,1/f−a are the functions counting (counting only once) the zeros of f−a in z:t<z≤1 and z:1<z≤t, respectively. Similarly, we denote by N¯0Er,a N¯0Dr,a the reduced counting function of common zeros (different zeros) of f−a and g−a on A, where N¯0Dr,a=N¯0r,1/f−a+N¯0r,1/g−a−2N¯0Er,a. It is obvious that f and g share a IM on A if N¯0Dr,a=0. Following the definitions in [2], we say that f and g share a “IM” if N¯0Dr,a=oT0r,f+oT0r,g, and we say that f and g share a “CM” if N¯0r,1/f−a+N¯0r,1/g−a−2N¯0Cr,a=oT0r,f+oT0r,g, in which N¯0Cr,a denotes the reduced counting function of common zeros with the same multiplicities of f−a and g−a.For a nonconstant meromorphic functionf on the annulus A, it is named as a transcendental meromorphic function on A provided that(2)limsupr⟶∞T0r,flogr=∞,1<r<R0=+∞,orlimsupr⟶R0T0r,f−logR0−r=∞,1<r<R0<+∞,respectively. In fact, the transcendental meromorphic functions are also known as admissible meromorphic functions. If f is a transcendental meromorphic function on A, then we have Sr,f=oT0r,f for all 1<r<R0 except for a set Δr such that ∫Δrrλ−1dr<+∞ or a set Δr′ such that ∫Δr′dr/R0−rλ+1<+∞, respectively.There existed many famous results about the uniqueness theory of meromorphic functions sharing values. In 1926, Nevanlinna [12] proved the celebrated five-value theorem.Theorem 1.
Letf and g be two nonconstant meromorphic functions in ℂ, and let aii=1,2,3,4,5 be five distinct values in ℂ¯. If f and g share the values ai IM for i=1,2,3,4,5 in ℂ, then f≡g.Since that time, a series of results emerged in large numbers, which discussed and generalized the five-value theorem (Theroem1). For the main results about the generalizations of Theroem 1 in simply connected regions, we can refer to [7, 13–16]. For instance, Zheng [15, 16] and Fang [13] obtained the generalization of the five-value theorem in an angular domain and in the unit disc, respectively. For the generalizations on multiply-connected regions, we can refer to [5, 17, 18].Recently, Cao et al. [17, 18] proved the following five-value theorem on annuli (the case of R0=+∞ was derived from [5] by Kondratyuk and Laine).Theorem 2.
Letf and g be two transcendental meromorphic functions on the annulus A=z:1/R0<z<R0, where 1<R0≤+∞. Let ai i=1,2,3,4,5 be five distinct values in ℂ¯. If f and g share the values ai IM for i=1,2,3,4,5 on A, then f≡g.From the very point of sharing small functions, we studied above theorems in [19] and provided the following uniqueness theorem of meromorphic functions sharing four small functions on annuli.Theorem 3.
Letf and g be two transcendental meromorphic functions on the annulus A=z:1/R0<z<R0, where 1<R0≤+∞. Let ai≡aizi=1,2,3,4 be four distinct small functions with respect to f and g on A. If f and g share aii=1,2,3,4 IM and(3)∑i=14N˜0r,ai≠Sr,f,then f≡g, where N˜0r,ai is the reduced counting function which counts the multiple common zeros of f−ai and g−ai on A.In this article, we mainly investigate whether Theroem2 holds if f and g dissatisfy the condition of sharing values. From the very point of deficiencies, we deal with this question and propose the following uniqueness theorem without conditions of sharing values. This theorem generalizes Theroem 2.Theorem 4.
Letf and g be two transcendental meromorphic functions on the annulus A=z:1/R0<z<R0, where 1<R0≤+∞, and let aii=1,2,3,4,5 be five distinct complex numbers in ℂ¯. Then, we have f≡g provided that(4)∑i=15δ0Dai>143,where the deficiencies δ0Dai are defined as(5)1−limsupr⟶∞N¯0Dr,aiT0r,f+T0r,g,when R0=+∞, or(6)1−limsupr⟶R0N¯0Dr,aiT0r,f+T0r,g,when R0<+∞, respectively.In special, iff and g share aii=1,2,3,4,5 “IM,” then it is obvious that ∑i=15δ0Dai=5, which satisfies the condition(7)∑i=15δ0Dai>143,of Theroem 4. In view of the discussion above, we deduce a corollary as follows. This corollary partly improves Theroem 2 in the sense that IM is replaced with “IM.”Corollary 1.
Letf and g be two transcendental meromorphic functions on the annulus A=z:1/R0<z<R0, where 1<R0≤+∞, and let aii=1,2,3,4,5 be five distinct complex numbers in ℂ¯. If f and g share aii=1,2,3,4,5 “IM,” then f≡g.


## 2. Some Lemmas

In this section, we will give some necessary lemmas, where the third lemma is motivated by the ideas of [20–22].Lemma 1 (see [4], Theroem 1).
Letf be a nonconstant meromorphic function on the annulus A=z:1/R0<z<R0, where R0≤+∞, and let λ≥0. Then,(i)
IfR0=+∞, then m0r,f′/f=OlogrT0r,f for R∈1,∞ except for a set Δr such that ∫Δrrλ−1dr<+∞.(ii)
IfR0<+∞, then m0r,f′/f=OlogT0r,f/R0−r for r∈1,R0 except for a set Δr′ such that ∫Δr′dr/R0−rλ+1<+∞.Lemma 2 (see [18], Theorem 2.3).
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,(8)q−2T0r,f<∑j=1qN¯0r,1f−aj+Sr,f.Inspired by the ideas of [20–22], we propose the following lemma and give the proof.Lemma 3.
Letf and g be two transcendental meromorphic functions on A=z:1/R0<z<R0, where 1<R0≤+∞, and let aii=1,2,3,4,5 be five distinct complex numbers in ℂ∪∞. If f≡g, then(9)N¯0Er,ai≤∑j=1,j≠i5N¯0Dr,aj+Sr,f+Sr,g,where N¯0Er,ai N¯0Dr,ai is the reduced counting function of the common (different) zeros of f−ai and g−ai on A i=1,2,3,4,5.Proof.
Without loss of generality, we suppose thata1=0, a2=1, a3=∞, a4=a, and a5=b, in which a,b are two distinct complex numbers such that a,b≠0,1,∞. Otherwise, a Möbius transformation as(10)f−a1f−a3a2−a3a2−a1,will be done. Then, set(11)h=h1−h2,where(12)h1=f−gf′g′ff−1gg−1g−a,h2=f−gf′g′gg−1ff−1f−a.
Noting thath1 can be expressed by(13)f′f−1g′ag+g′aa−1g−a−g′a−1g−1+1a−1f′f−f′f−1g′g−a−g′g−1.
By Lemma1, we have m0r,h1=Sr,f+Sr,g. Similarly, we get m0r,h2=Sr,f+Sr,g, and thus,(14)m0r,h=Sr,f+Sr,g,holds.
Next, we will estimate the counting functionN0r,h. A simple computation yields(15)h=f−g2f′g′ff−1f−agg−1g−a.
Then, it is easy to see that the poles ofh only come from the zeros of f,g,f−1,g−1,f−a,andg−a and the poles of fandg on A. Now, let z0 be a common zero of f and g on A with multiplicity p and q, respectively. Without loss of generality, assume that p≥q. Then, it follows that z0 is a zero of f−g2 with multiplicity at least 2q and that z0 is a pole of(16)f′g′ff−1f−agg−1g−a,with multiplicity 2. We consequently know that z0 is not a pole of h, and hence the poles of h cannot occur at the common zeros of f and g. By similar methods, we can conclude that the poles of h cannot occur at the common zeros of f−1 and g−1, the common zeros of f−a and g−a, and the common poles of f and g, so the poles of h only come from the different zeros of f,g,f−1,g−1,f−a,andg−a and the different poles of fandg on A. In order to analyze these different zeros and different poles, we distinguish the following distinct cases.Case 1: letz1 be a zero of f which is not a zero of g. Then, by using the equation (15), we find that z1 is a pole of h with multiplicity at most 1 if gz1≠1,∞,a; otherwise, z1 is a pole of h with multiplicity at most 2.Case 2: letz2 be a zero of f−1, which is not a zero of g−1. It is clear that z2 is a pole of h with multiplicity at most 1 if gz2≠0,∞,a; otherwise, z2 is a pole of h with multiplicity at most 2.Case 3: letz3 be a pole of f, which is not a pole of g. It is clear that z2 is a pole of h with multiplicity at most 1 if gz3≠0,1,a; otherwise, z3 is a pole of h with multiplicity at most 2.Case 4: letz4 be a zero of f−a, which is not a zero of g−a. It is clear that z4 is a pole of h with multiplicity at most 1 if gz4≠0,1,∞; otherwise, z4 is a pole of h with multiplicity at most 2.Case 5: letz5 be a zero of g, which is not a zero of f. It is clear that z5 is a pole of h with multiplicity at most 1 if fz5≠1,∞,a; otherwise, z5 is a pole of h with multiplicity at most 2.Case 6: letz6 be a zero of g−1, which is not a zero of f−1. It is clear that z6 is a pole of h with multiplicity at most 1 if fz6≠0,∞,a; otherwise, z6 is a pole of h with multiplicity at most 2.Case 7: letz7 be a pole of g, which is not a pole of f. It is clear that z7 is a pole of h with multiplicity at most 1 if fz7≠0,1,a; otherwise, z7 is a pole of h with multiplicity at most 2.Case 8: letz8 be a zero of g−a, which is not a zero of f−a. It is clear that z8 is a pole of h with multiplicity at most 1 if fz8≠0,1,∞; otherwise, z8 is a pole of h with multiplicity at most 2.
In view of these cases, we obtain(17)N0r,h≤N¯0Dr,0+N¯0Dr,1+N¯0Dr,∞+N¯0Dr,a,which means(18)N0r,h≤∑i=14N¯0Dr,ai.
Combining (14) with (18), we get(19)T0r,h≤∑i=14N¯0Dr,ai+Sr,f+Sr,g.
Ifh≡0, then h1≡h2. This implies that f≡g, which is impossible, so h≡0 holds. Furthermore, it follows from (15) that the common zeros of f−b and g−b must be the zeros of h. This implies that(20)N¯0Er,b≤N0r,1h≤T0r,h,which further implies that(21)N¯0Er,a5≤∑i=14N¯0Dr,ai+Sr,f+Sr,g,combined with (19). Similarly, we can derive other inequations as(22)N¯0Er,ai≤∑j=1,j≠i5N¯0Dr,aj+Sr,f+Sr,g,i=1,2,3,4.
Therefore, we have proved Lemma3.


## 3. The Proof of Theorem 4

On the contrary, we suppose thatf≡g, and then it follows from Lemma 3 that(23)N¯0Er,ai≤∑j=1,j≠i5N¯0Dr,aj+Sr,f+Sr,g,for i=1,2,3,4,5. Thus, noting that(24)N¯0r,1f−ai+N¯0r,1g−ai=2N¯0Er,ai+N¯0Dr,ai,we know(25)N¯0r,1f−ai+N¯0r,1g−ai≤N¯0Dr,ai+∑j=1,j≠i52N¯0Dr,aj+Sr,f+Sr,g.This yields that(26)N¯0r,1f−ai+N¯0r,1g−ai≤∑j=152N¯0Dr,aj−N¯0Dr,ai+Sr,f+Sr,g,for i=1,2,3,4,5, which further yields that(27)∑i=15N¯0r,1f−ai+∑i=15N¯0r,1g−ai≤9∑i=15N¯0Dr,ai+Sr,f+Sr,g.On the other hand, by utilizing Lemma2, we find(28)3T0r,f<∑i=15N¯0r,1f−ai+Sr,f,3T0r,g<∑i=15N¯0r,1g−ai+Sr,f.Therefore, it follows from (27) that (29)3T0r,f+3T0r,g≤9∑i=15N¯0Dr,ai+Sr,f+Sr,g,which means(30)13≤∑i=15N¯0Dr,aiT0r,f+T0r,g+Sr,f+Sr,gT0r,f+T0r,g.Consequently, from (30), we have(31)∑i=15δ0Daj≤143,which is a contradiction. Hence, the proof is completed.



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*Source: 6482895-2020-01-28.xml*