# Some New Results on Fixed Points of Meromorphic Functions Defined in Annuli

Authors: Zhaojun Wu; Zuxing Xuan; Yuxian Chen
Journal: Journal of Function Spaces
 (2015)

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## Abstract

The purpose of this paper is to investigate the fixed points of meromorphic functions in annuli. Some well-known facts of fixed points for meromorphic functions in the plane will be considered in annuli.


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## Body

## 1. Introduction and Main Results

Letf(z) be a meromorphic function. A point z is called a fixed point if f(z)=z. There are a considerable number of results on the fixed points for meromorphic functions in the plane; we refer the reader to Chuang and Yang [1]. In 1988, Zhu [2] has proved the following.Theorem A.
Letf be a transcendental meromorphic function in the plane. Then either f or f′ has infinitely many fixed points.
In the 1960s, Baker [3] proved that if f(z) is a transcendental entire function in the plane such that there exists a∈C with δ(a,f)>0, then f has fixed points of order one. In 1993, Lahiri [4] gave an extension of Baker’s results and proved the following theorem.Theorem B.
Letf be a transcendental meromorphic function in the plane. Suppose that there exists a∈C with δ(a,f)>0 and N(r,f)=S(r,f). Then f has infinitely many fixed points.
In recent ten years some well-known facts of the value distribution theory for meromorphic function in the plane were extended for the meromorphic function in the annuli:(1)AR0=z:1R0<z<R0,1<R0≤+∞.In 2005, Khrystiyanyn and Kondratyuk [5, 6] gave an extension of the Nevanlinna value distribution theory for meromorphic functions in annuli. In their extension the main characteristics of meromorphic functions are one-parameter and possess the same properties as in the classical case of a simply connected domain. In [5, 6], we can get the analogues of Jensen’s formula, the first fundamental theorem, the lemma on logarithmic derivative, and the second fundamental theorem of the Nevanlinna theory for meromorphic functions in annuli. Lund and Ye [7] investigated logarithmic derivatives in the annuli {z:R1<|z|<R2}. After [5, 6], Cao et al. [8–11] study the uniqueness of the meromorphic functions in annuli, Chen and Wu [12] study the exceptional values for meromorphic function and its derivatives in annuli, and Fernández [13] studies the value distribution of meromorphic functions in the punctured plane A(∞). The main purpose of this paper is to study the fixed points of meromorphic functions in annuli. We will prove the following theorems by using the similar method as in [2, 4].Theorem 1.
Letf be an admissible meromorphic function in A(R0), 1<R0≤+∞. Then either f or f′ has infinitely many fixed points.Theorem 2.
Letf be an admissible meromorphic function in A(R0), 1<R0≤+∞. Suppose that there exists a∈C with δ0(a,f)>0 and N0(R,f)=S(R,f). Then both f and f′ have infinitely many fixed points.


## 2. Nevanlinna Theory in Annuli

In the following, we introduce the definitions, notations, and results of [5, 6, 8, 14] which will be used in this paper. Let f(z) be a meromorphic function in A(R0). Denote (2)mR,1f-a=12π∫02πlog+⁡1fReiθ-adθ,mR,f=12π∫02πlog+⁡fReiθdθ,where a∈C and 1/R0<R<R0. Let (3)m0R,1f-a=mR,1f-a+m1R,1f-a-2m1,1f-a,1<R<R0,m0R,f=mR,f+m1R,f-2m1,f,1<R<R0.Put (4)N1R,1f-a=∫1/R1n1t,1/f-atdt,N2R,1f-a=∫1Rn2t,1/f-atdt,where 1<R<R0, n1(t,1/f-a) is the counting function of poles of the function 1/f-a in {z:t<|z|≤1}, and n2(t,1/f-a) is the counting function of poles of the function 1/f-a in {z:1<|z|≤t}. Denote also (5)N1R,f=∫1/R1n1t,ftdt,N2R,f=∫1Rn2t,ftdt,where 1<R<R0, n1(t,f) is the counting function of poles of the function f in {z:t<|z|≤1}, and n2(t,f) is the counting function of poles of the function f in {z:1<|z|≤t}. Let (6)N0R,1f-a=N1R,1f-a+N2R,1f-a,N0R,f=N1R,f+N2R,f.Finally, we define the Nevanlinna characteristic of f in A(R0) by (7)T0R,f=m0R,f+N0R,f,1<R<R0,where R0≤+∞. Suppose that f, g are two meromorphic functions in A(R0), where 1<R0≤+∞. Then(8)T0R,f=T0R,1f,(9)max⁡m0R,fg,m0R,fg,m0R,f±g≤m0R,f+m0R,g+O1,(10)max⁡T0R,fg,T0R,fg,T0R,f±g≤T0R,f+T0R,g+O1.Definition 3.
Letf be a nonconstant meromorphic function in A(R0). One calls f admissible if (11)limsupR→+∞⁡T0R,flog⁡R=∞,1≤R<R0=+∞,or (12)limsupR→R0⁡T0R,f-log⁡R0-R=∞,1≤R<R0=+∞.Definition 4.
Letf be a nonconstant meromorphic function in A(R0), 1<R0≤+∞, and a∈C. Then the value (13)δ0a,f=1-limsupR→R0⁡N0R,1/f-aT0R,fis called the deficiency of the function f for the value a. For a=∞, one sets (14)δ0∞,f=1-limsupR→R0⁡N0R,fT0R,f.Throughout, we denote byS(R,∗) quantities that satisfy the following:(1)
for the caseR0=+∞, (15)SR,∗=Olog⁡RT0R,f,forR∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞;(2)
for the caseR0<+∞, (16)SR,∗=Olog⁡T0R,fR0-R,forR∈(1,R0) except for the set ΔR′ such that ∫ΔR′‍dR/(R0-R)λ-1<+∞.Thus, for an admissible meromorphic function in A(R0), 1<R0≤+∞, S(R,f)=O(T0(R,f)) holds for all 1≤R<R0≤+∞ except for the set ΔR or the set ΔR′ mentioned above, respectively. Under the above notations, we give the following theorems which will be used in the proof of Theorems 1 and 2.Theorem C (the first fundamental theorem).
Letf be a nonconstant meromorphic function in A(R0), where 1<R0≤+∞. Then (17)T0R,1f-a=T0R,f+O1,1<R<R0,for every fixed a∈C.Theorem D (lemma on the logarithmic derivative).
Letf be an admissible meromorphic function in A(R0), 1<R0≤+∞. Then m0R,f(k)/f=S(R,f) holds for every positive integer k.


## 3. Proof of Theorem1

Lemma 5.
Letf be an admissible meromorphic function in A(R0), 1<R0≤+∞. Then(18)T0R,f≤N¯0R,f+N0R,1f+N0R,1f′-z+1+SR,f.Proof.
Consider(19)1f=f′φf-φf′′-f′φff′-φφf′′-f′,where φ=φ(z)=z-1. This leads to(20)m0R,1f≤m0R,f′φf+m0R,φf′′-f′φf+m0R,f′-φφf′′-f′.Applying the first fundamental theorem, we get(21)m0R,1f=T0R,f-N0R,1f+O1,m0R,f′-φφf′′-f′=m0R,φf′′-f′f′-φ+N0R,φf′′-f′f′-φ-N0R,f′-φφf′′-f′+O1=m0R,φf′′-f′f′-φ+N0R,1f′-φ+N0R,φf′′-f′-N0R,f′-φ-N0R,1φf′′-f′+O1≤m0R,φf′′-f′f′-φ+N0R,1f′-φ+N¯0R,f-N0R,1φf′′-f′+O1.Substituting (21) into (20), noting that m0(R,φ)=S(R,f), S(R,f′-φ)=S(R,f), we have (22)T0R,f≤N0R,1f+2m0R,f′φf+m0R,f′f+m0R,φf′′-f′f′-φ+N0R,1f′-φ+N¯0R,f-N0R,1φf′′-f′+O1≤N0R,1f+2m0R,f′φf+m0R,f′f+m0R,f′-φ′-f′-φf′-φ+N0R,1f′-φ+N¯0R,f-N0R,1φf′′-f′+O1=N¯0R,f+N0R,1f+N0R,1f′-φ-N0R,1φf′′-f′+SR,f.Lemma 6.
Suppose thatf is an admissible meromorphic function in A(R0), 1<R0≤+∞. Then(23)N01R,f≤9N0R,1f+9N0R,1f′-z+1+SR,f,where N01)(R,f) is the N0 function of n1)(t,f) which denote the number of simple poles of f.Proof.
We define(24)Pz=ff′-φφf′′-f′,Gz=Pφ′′+Pφ′′+φ=3P′+φP′′+φ,where φ=φ(z)=z-1. Let z0∈A(R0) be a simple pole of f(z). Then as in [2] we have G(z0)=0. Therefore, when G(z)≢0, by the first fundamental theorem and (10), noting that T0(R,φ)=S(R,f), S(R,P)=S(R,f), we have(25)N01R,f≤N0R,1G≤T0R,1G=T0R,G+O1=T0R,P3P′P+φP′′P+φ+O1≤T0R,P+T0R,P′+T0R,P′′+2T0R,1P+SR,f≤7T0R,P+2T0R,1P+SR,f+SR,P≤9T0R,1P+SR,f.Now we estimate T0(R,1/P). Firstly, we have(26)m0R,1P=m0R,φf′′-f′ff′-φ≤m0R,f′-φ/f′′f′-φ/f′+m0R,f′f=SR,f.In [2], Zhu indicated that poles of 1/P can occur only at zeros of f(z) or f′-φ. But every zero with multiplicity k of f′-φ gives a pole of 1/P once at most, for it must be the zero of φf′′-f′ with multiplicity being at least k-1. So we have(27)N0R,1P≤N0R,1f+N0R,1f′-φ.From (25), (26), and (27) we derive (28)N01R,f≤9N0R,1f+9N0R,1f′-φ+SR,f.By a similar discussion as in [2], one can prove Lemma 6 in the case of G(z)≡0.We are now in the position to prove Theorem1.Proof.
NoteN¯0(R,f)=N01)(R,f)+N¯0(2(R,f), where N¯0(2 is the N0 function of n¯(2(t,f) which denote the number of multiple poles of f, ignoring multiplicity. By (18) we have (29)2N¯02R,f+N01R,f≤N0R,f≤T0R,f≤N01R,f+N¯02R,f+N0R,1f+N0R,1f′-z+1+SR,f.Thus,(30)N¯02R,f≤N0R,1f+N0R,1f′-z+1+SR,f.By (18), (23), and (30) we have(31)T0R,f≤N01R,f+N¯02R,f+N0R,1f+N0R,1f′-z+1+SR,f≤11N0R,1f+11N0R,1f′-z+1+SR,f.Denoting g≡f-z, by (31) we derive(32)T0R,f≤T0R,g+O1≤11N0R,1g+11N0R,1g′-z+1+SR,g+O1≤11N0R,1f-z+11N0R,1f′-z+SR,f.If both f and f′ have only a finite number of fixed points, then from (32) we would have (33)T0R,f≤SR,f.This leads to a contradiction and Theorem 1 is proved.


## 4. Proof of Theorem2

Proof.
Consider(34)1f=f′zf-zf′′-f′zff′-zzf′′-f′.This leads to(35)m0R,1f≤m0R,f′zf+m0R,zf′′-f′zf+m0R,f′-zzf′′-f′.Applying the first fundamental theorem, we get(36)m0R,1f=T0R,f-N0R,1f+O1,m0R,f′-zzf′′-f′=m0R,zf′′-f′f′-z+N0R,zf′′-f′f′-z-N0R,f′-zzf′′-f′+O1=m0R,zf′′-f′f′-z+N0R,1f′-z+N0R,zf′′-f′-N0R,f′-z-N0R,1zf′′-f′+O1≤m0R,zf′′-f′f′-z+N0R,1f′-z+N¯0R,f-N0R,1zf′′-f′+O1.Substituting (36) into (35), we have(37)T0R,f≤N0R,1f+2m0R,f′zf+m0R,f′f+m0R,zf′′-f′f′-z+N0R,1f′-z+N¯0R,f-N0R,1zf′′-f′+O1≤N0R,1f+2m0R,f′zf+m0R,f′f+m0R,f′-z′-f′-zf′-z+N0R,1f′-z+N¯0R,f-N0R,1zf′′-f′+O1≤N¯0R,f+N0R,1f+N0R,1f′-z+SR,f.Denoting g≡f-a, by (37) we derive(38)T0R,f≤T0R,g+O1≤N¯0R,g+N0R,1g+N0R,1g′-z+SR,g≤N0R,f+N0R,1f-a+N0R,1f′-z+SR,f.Since N0(R,f)=S(R,f), by (38), we can get(39)T0R,f≤N0R,1f-a+N0R,1f′-z+SR,f.Since δ0(a,f)>0, then there is a positive number θ<1 such that(40)N0R,1f-a<θT0R,f.If f′ have only a finite number of fixed points, then from (39) and (40) we would have (41)1-θT0R,f≤SR,f.This leads to a contradiction and f′ has infinitely many fixed points. In the following, we will prove that f also has infinitely many fixed points under the condition of Theorem 2.
Ifa[1],a[2],…,a[p] are distinct finite complex numbers, put (42)Fz=∑i=1p1fz-ai.Then as in [15] we have(43)mR,F≥∑i=1pmR,1fz-ai+O1,m1R,F≥∑i=1pm1R,1fz-ai+O1.Since (44)mR,F≤mR,f′F+mR,1f′≤∑i=1pmR,f′f-ai+mR,1f′,m1R,F≤m1R,f′F+m1R,1f′≤∑i=1pm1R,f′f-ai+m1R,1f′,therefore,(45)m0R,F≤∑i=1pm0R,f′f-ai+m0R,1f′+O1.It follows from (43)–(45) and Theorem D that(46)∑i=1pm0R,1fz-ai≤m0R,1f′+SR,f.In particular, we have(47)m0R,1fz-a≤m0R,1f′+SR,f.Applying (46) to the function f-z we obtain(48)m0R,1fz-z≤m0R,1f′-1+SR,f.Further application of (46) gives(49)m0R,1f′+m0R,1f′-1≤m0R,1f′′+SR,f′≤T0R,f′′+SR,f.On the other hand, in view of Theorems C and D, one has(50)T0R,f′′=m0R,f′′+N0R,f′′≤m0R,f+m0R,f′′f+N0R,f′′=m0R,f+N0R,f′′+SR,f.If z0 is a pole of order k, it is a pole of f′′ or order k+2≤3k. So N0(R,f′′)≤3N0(R,f) and by the given condition N0(R,f′′)≤S(R,f). Therefore, we get from (50) that(51)T0R,f′′≤T0R,f+SR,f.Combining (47), (48), and (49) with (51), we have(52)m0R,1fz-a+m0R,1fz-z≤T0R,f+SR,f.By Theorem C and (10), we see that(53)T0R,f≤N0R,1fz-a+N0R,1fz-z+SR,f.If f has only a finite number of fixed points, then from (40) and (53) we would have (54)1-θT0R,f≤SR,f.This leads to a contradiction and f has infinitely many fixed points.



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*Source: 426576-2015-12-17.xml*