# Exceptional Values of Meromorphic Function on Annulus

Authors: Hong-Yan Xu
Journal: The Scientific World Journal
 (2013)

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## Abstract

The main purpose of this paper is to study the exceptional values of meromorphic function and its derivative on annulus. We also give some theorems and corollaries about exceptional values of meromorphic function on the annulus, which are the improvement of the previous results given by Chen and Wu.


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## Body

## 1. Introduction

It is a very interesting problem on the exceptional values of meromorphic functions in the value distribution theory and argument distributed theory such as the Picard exceptional value and the Borel exceptional value. It is well known that every class of exceptional value is always responding to a singular direction, such as, the Picard exceptional value relating with Julia’s direction and the Borel exceptional value relating with Borel’s direction. In the 1970s, Gopalakrishna and Bhoosnurmath [1, 2] and Singh and Gopalakrishna [3] investigated Borel exceptional values of meromorphic function and its derivative on the whole complex plane. In 2011, Peng and Sun [4] gave some examples on T exceptional value which is an exceptional value relating with T direction.In fact, Peng, Sun, Singh, Gopalakrishna, and so forth only studied exceptional values of meromorphic functions in the whole complex plane—single-connected region. For meromorphic function on the double-connected domain and several-connected region, there were few papers about its exceptional values. In 2005, Khrystiyanyn and Kondratyuk [5, 6] proposed the Nevanlinna theory for meromorphic functions on annuli (see also [7]). In 2010, Fernández [8] further investigated the value distribution of meromorphic functions on annuli. In 2012, Xu and Xuan [9] studied the uniqueness of meromorphic functions sharing some values on annuli. At the same year, Chen and Wu [10] firstly studied the Borel exceptional values of meromorphic function and its derivative on annulus. In this paper, we will further investigate the exceptional value of meromorphic function and its derivative on annulus and obtain a series of results which are improvement of previous theorems given by Chen and Wu [10].The structure of this paper is as follows. In Section2, we introduce the basic notations and fundamental theorems of meromorphic functions on annulus. Sections 3 and 4 are devoted to study the Borel exceptional values of meromorphic function on annulus and give some consequences of our theorems. Section 5 are devoted to study the Borel exceptional values of meromorphic function and its derivative on annulus.


## 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≤+∞. The notations of the Nevanlinna theory on annuli will be introduced as follows. Let
(1)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)T0(R,f)=m0(R,f)-2m(1,f)+N0(R,f).
Similarly, for a∈ℂ¯, we have
(3)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 have the notations N¯1k)(t,f), N¯1(k(t,f), N¯2k)(t,f), N¯2(k(t,f), and 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 [5]): (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),(iii)
T0(R,1/(f-a))=T0(R,f)+O(1), for every fixed a∈ℂ.Khrystiyanyn and Kondratyuk [6] also obtained the lemma on the logarithmic derivative on the annulus 𝔸.Theorem 1 (see [6] 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=+∞,
(4)m0(R,f′f)=O(log(RT0(R,f)))for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞,(ii)
ifR0<+∞, then
(5)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)<+∞.In 2005, The second fundamental theorem on the annulus𝔸 was first obtained by Khrystiyanyn and Kondratyuk [6]. Cao et al. [11] introduce other forms of the second fundamental theorem on annuli as follows.Theorem 2 ([11, 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 k1,k2,…,kq be q positive integers, and let λ≥0. Then
(6)(q-2)T0(R,f)<∑j=1q‍N¯0(R,1f-aj)+S(R,f),
where (i) in the case R0=+∞,
(7)S(R,f)=O(log(RT0(R,f)))
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞,(ii)
ifR0<+∞, then
(8)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 3.
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  which provided that
(9)limsupR→∞T0(R,f)logR=∞,1<R<R0=+∞
or
(10)limsupR→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 1, respectively.Definition 4 (see [10], Definition 2.1).
Letf(z) be a nonconstant meromorphic function on the annulus 𝔸={z:1/R0<|z|<R0}, where R0=+∞. Then the order of f(z) is defined by
(11)ρ(f)=limsupR→+∞T0(R,f)logR.Definition 5 (see [10], Definition 3.1).
Letf(z) be a nonconstant meromorphic function of order ρ(0<ρ<∞) on the annulus 𝔸={z:1/R0<|z|<R0}, where 1<R0≤+∞, and let a∈ℂ^. Then we say that a is(i)
an evB (exceptional value in the sense of Borel) forf on 𝔸 for distinct zeros of order ≤k if ρ¯k(a,f)<ρ;(ii)
an evB (exceptional value in the sense of Borel) forf on 𝔸 for distinct zeros if ρ¯(a,f)<ρ;(iii)
an evB (the Borel exceptional value) forf on 𝔸 if ρ(a,f)<ρ, where
(12)ρ¯k(a,f)=limsupR→+∞N¯0k)(R,1/(f-a))logR,ρ¯(a,f)=limsupR→+∞N¯0(R,1/(f-a))logR,ρ(a,f)=limsupR→+∞N0(R,1/(f-a))logR.In particular, we say thata is an evB for f on 𝔸 for simple zeros if k=1 and  a is an evB for f on 𝔸 for simple and double zeros if k=2.


## 3. The Main Results

Now, the main theorems of this paper are listed as follows.Theorem 6.
Letf be transcendental meromorphic function of order ρ(0<ρ<∞) on the annulus 𝔸:={z:1/R0<|z|<R0}, where R0=+∞. If there exist a11,…,ap11,a12,…,ap22,…,a1ν,…,apνν∈ℂ^ such that a1i,a2i,…,apii are evBs for f on 𝔸 for distinct zeros of order ≤ki, i=1,2,…,ν, where p1,…,pν,ν∈ℕ+, and k1,k2,…,kν are positive integers or infinity. Then
(13)Ξ:=∑i=1ν‍piki1+ki+∑i=1ν‍11+ki∑j=1pi‍δ0ki(aji,f)≤2.Remark 7.
Under the assumptions of Theorem6, if ν=3, p1=r,p2=t,p3=s and k1=k, k2=l, k3=m, since δ0ki(aji,f)≥0, then we can get
(14)krk+1+tll+1+smm+1≤∑i=1ν‍piki1+ki+∑i=1ν‍11+ki∑j=1pi‍δ0ki(aji,f)≤2.
This shows that Theorem 6 is an improvement of results given by Chen and Wu [10].Definition 8.
For positive integersk,m, we define that
(15)δ0k(a,f)=1-limsupR→+∞N0k(R,1/(f-a))T0(R,f),Θ0(a,f)=1-limsupR→+∞N¯0(R,1/(f-a))T0(R,f),
where N0k(R,1/(f-a)) is the counting function of a-points of f on 𝔸 where an a-point of multiplicity m is counted m times if m≤k and 1+k times if m>k. In particular, if k=∞, then
(16)δ0∞(a,f)=δ0(a,f)=liminfR→+∞m0(R,1/(f-a))T0(R,f)=1-limsupR→+∞N0(R,1/(f-a))T0(R,f).Theorem 9.
Letf be transcendental meromorphic function of order ρ(0<ρ<∞) on the annulus 𝔸:={z:1/R0<|z|<R0}, where R0=+∞. If there exist a∈ℂ^ and two positive integers k and p such that
(17)(1+k)Θ0(a,f)+∑b≠a‍δ0(b,f)>2-k(p-1),
then there exist at most p elements ℂ^∖{a} which are evBs for f on 𝔸 for distinct zeros of order not exceeding k.


### 3.1. Proof of Theorem6

Proof.
For any positive integerk or ∞ and a∈ℂ^, we have
(18)N¯0(R,1f-a)≤k1+kN¯0k)(R,1f-a)+11+kN0k(R,1f-a),
where k/(k+1)=1 and 1/(k+1)=0 if k=∞. Then, from (18) and Theorem 2, we have
(19)(∑i=1ν‍pi-2)T0(R,f)≤∑i=1ν‍ki1+ki∑j=1pi‍N¯0ki)(R,1f-aji)+∑i=1ν‍11+ki∑j=1pi‍N0ki(R,1f-aji)+S(R,f).
From Definition 5 and the assumptions of Theorem 6, there exists a constant η(0<η<ρ) such that for sufficiently large R,
(20)N¯0ki)(R,1f-aji)<Rη,j=1,2,…,pi,i=1,2,…,ν.
Hence, from (19) and (20) and for sufficiently large R, we have
(21)(∑i=1ν‍pi-2)T0(R,f)≤∑i=1ν‍11+ki∑j=1pi‍N0ki(R,1f-aji)+O(Rη)+S(R,f).
Thus, for sufficiently large R and taking arbitrary ε(>0), we can get from (21) and the definition of δk(a,f) that
(22)(∑i=1ν‍pi-2)T0(R,f)≤∑i=1ν‍11+ki∑j=1pi‍(1-δkiθ(aji,f)+ε)T0(R,f)+O(Rη)+S(R,f).
Since R=+∞, it follows that
(23)(Ξ-2-ε)T0(R,f)≤O(Rη)+O(logRT0(R,f)),
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞. If I=[R,R′)∈ΔR, we have
(24)R′λ-Rλ=λ∫RR′‍xλ-1dx≤λ∫ΔR‍Rλ-1dR<M,
where M is a constant. Thus,
(25)(Ξ-2-ε)T0(R,f)≤O(Rη)+O(logRT0(R,f))
holds for all R. If Ξ>2, we can choose an arbitrary ε(0<ε<Ξ-2) satisfying η+ε<ρ. Thus, from (25) and for sufficiently large R, we can get a contradiction to (13) easily.
Therefore, we can get the conclusion of Theorem6.


### 3.2. The Proof of Theorem9

Lemma 10.
Letf be transcendental meromorphic function of order ρ(0<ρ<∞) on the annulus 𝔸:={z:1/R0<|z|<R0}, where 1<R0≤+∞. Then
(26)limsupR→+∞m0(R,1/f′)T0(R,f)≤2-Θ0(∞,f)-∑b∈ℂ‍δ0(b,f).Proof.
Suppose thatb1,b2,…,bt∈ℂ are t distinct complex constants. From the definitions of m0(R,f),T0(R,f), we have
(27)∑i=1t‍m0(r,1f-bi)≤m0(r,1f′)+S(R,f)≤T0(R,f′)-N0(R,1f′)+S(R,f).
And since T0(R,f′)≤T0(R,f)+N¯0(R,f)+S(R,f), then we have
(28)∑i=1t‍m0(R,1f-bi)+N0(R,1f′)≤T0(R,f)+N¯0(R,f)+S(R,f).
It follows from the definitions of δ0(a,f),Θ0(a,f) that
(29)∑i=1t‍δ0(bi,f)+limsupR→+∞N0(R,1/f′)T0(R,f)≤2-Θ0(∞,f).
From (29) and t is arbitrary, the proof of Lemma 10 is completed easily.Proof of Theorem9.
Without loss of generality, we assume thata=∞. Next, we will employ reductio ad absurdum to prove the conclusion of Theorem 9. Suppose that there exist p+1 elements a1,a2,…,ap+1∈ℂ which are evBs for f on 𝔸 for distinct zeros of multiplicity ≤k. We know that if z0 is a zero of f-b on 𝔸 of multiplicity s(>1) for b∈ℂ, then z0 is a zero of f′ on 𝔸 of multiplicity s-1. It follows that
(30)∑i=1p+1‍N¯0(R,1f-ai)≤∑i=1p+1‍N¯0k)(R,1f-ai)+1kN0(R,1f′).
Therefore, by Theorem 2 we have
(31)pT0(R,f)≤∑i=1p+1‍N¯0(R,1f-ai)+N¯0(R,f)+S(R,f)≤∑i=1p+1‍N¯0k)(R,1f-ai)+1kN0(R,1f′)+N¯0(R,f)+S(R,f).
From the definitions of order and evB of f on 𝔸, there exists a constant η(0<η<ρ) such that for sufficiently large R, we have
(32)N¯0k)(R,1f-ai)<Rη,i=1,2,…,p+1.
From (31), (32), and Lemma 10, for sufficiently large R, it follows that
(33)(p-1-1k(2-Θ0(∞,f)-∑b≠∞‍δ0(b,f))+Θ0(∞,f))×T0(R,f)≤Rη+S(R,f),
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞. Since R0=+∞, similar to the argument as in Theorem 6, from (33) we find that
(34)(p-1-2-(k+1)Θ0(∞,f)-∑b≠∞δ0(b,f)k)T0(R,f)≤Rη+O(log(RT0(R,f)))
holds for all R. Since η<ρ, from (34) and for sufficiently large R, we can get
(35)(1+k)Θ0(∞,f)+∑b≠∞‍δ0(b,f)>2-k(p-1),
which is contradiction with the assumption of Theorem 9.
Thus, this completes the proof of Theorem9.


## 3.1. Proof of Theorem6

Proof.
For any positive integerk or ∞ and a∈ℂ^, we have
(18)N¯0(R,1f-a)≤k1+kN¯0k)(R,1f-a)+11+kN0k(R,1f-a),
where k/(k+1)=1 and 1/(k+1)=0 if k=∞. Then, from (18) and Theorem 2, we have
(19)(∑i=1ν‍pi-2)T0(R,f)≤∑i=1ν‍ki1+ki∑j=1pi‍N¯0ki)(R,1f-aji)+∑i=1ν‍11+ki∑j=1pi‍N0ki(R,1f-aji)+S(R,f).
From Definition 5 and the assumptions of Theorem 6, there exists a constant η(0<η<ρ) such that for sufficiently large R,
(20)N¯0ki)(R,1f-aji)<Rη,j=1,2,…,pi,i=1,2,…,ν.
Hence, from (19) and (20) and for sufficiently large R, we have
(21)(∑i=1ν‍pi-2)T0(R,f)≤∑i=1ν‍11+ki∑j=1pi‍N0ki(R,1f-aji)+O(Rη)+S(R,f).
Thus, for sufficiently large R and taking arbitrary ε(>0), we can get from (21) and the definition of δk(a,f) that
(22)(∑i=1ν‍pi-2)T0(R,f)≤∑i=1ν‍11+ki∑j=1pi‍(1-δkiθ(aji,f)+ε)T0(R,f)+O(Rη)+S(R,f).
Since R=+∞, it follows that
(23)(Ξ-2-ε)T0(R,f)≤O(Rη)+O(logRT0(R,f)),
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞. If I=[R,R′)∈ΔR, we have
(24)R′λ-Rλ=λ∫RR′‍xλ-1dx≤λ∫ΔR‍Rλ-1dR<M,
where M is a constant. Thus,
(25)(Ξ-2-ε)T0(R,f)≤O(Rη)+O(logRT0(R,f))
holds for all R. If Ξ>2, we can choose an arbitrary ε(0<ε<Ξ-2) satisfying η+ε<ρ. Thus, from (25) and for sufficiently large R, we can get a contradiction to (13) easily.
Therefore, we can get the conclusion of Theorem6.


## 3.2. The Proof of Theorem9

Lemma 10.
Letf be transcendental meromorphic function of order ρ(0<ρ<∞) on the annulus 𝔸:={z:1/R0<|z|<R0}, where 1<R0≤+∞. Then
(26)limsupR→+∞m0(R,1/f′)T0(R,f)≤2-Θ0(∞,f)-∑b∈ℂ‍δ0(b,f).Proof.
Suppose thatb1,b2,…,bt∈ℂ are t distinct complex constants. From the definitions of m0(R,f),T0(R,f), we have
(27)∑i=1t‍m0(r,1f-bi)≤m0(r,1f′)+S(R,f)≤T0(R,f′)-N0(R,1f′)+S(R,f).
And since T0(R,f′)≤T0(R,f)+N¯0(R,f)+S(R,f), then we have
(28)∑i=1t‍m0(R,1f-bi)+N0(R,1f′)≤T0(R,f)+N¯0(R,f)+S(R,f).
It follows from the definitions of δ0(a,f),Θ0(a,f) that
(29)∑i=1t‍δ0(bi,f)+limsupR→+∞N0(R,1/f′)T0(R,f)≤2-Θ0(∞,f).
From (29) and t is arbitrary, the proof of Lemma 10 is completed easily.Proof of Theorem9.
Without loss of generality, we assume thata=∞. Next, we will employ reductio ad absurdum to prove the conclusion of Theorem 9. Suppose that there exist p+1 elements a1,a2,…,ap+1∈ℂ which are evBs for f on 𝔸 for distinct zeros of multiplicity ≤k. We know that if z0 is a zero of f-b on 𝔸 of multiplicity s(>1) for b∈ℂ, then z0 is a zero of f′ on 𝔸 of multiplicity s-1. It follows that
(30)∑i=1p+1‍N¯0(R,1f-ai)≤∑i=1p+1‍N¯0k)(R,1f-ai)+1kN0(R,1f′).
Therefore, by Theorem 2 we have
(31)pT0(R,f)≤∑i=1p+1‍N¯0(R,1f-ai)+N¯0(R,f)+S(R,f)≤∑i=1p+1‍N¯0k)(R,1f-ai)+1kN0(R,1f′)+N¯0(R,f)+S(R,f).
From the definitions of order and evB of f on 𝔸, there exists a constant η(0<η<ρ) such that for sufficiently large R, we have
(32)N¯0k)(R,1f-ai)<Rη,i=1,2,…,p+1.
From (31), (32), and Lemma 10, for sufficiently large R, it follows that
(33)(p-1-1k(2-Θ0(∞,f)-∑b≠∞‍δ0(b,f))+Θ0(∞,f))×T0(R,f)≤Rη+S(R,f),
for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞. Since R0=+∞, similar to the argument as in Theorem 6, from (33) we find that
(34)(p-1-2-(k+1)Θ0(∞,f)-∑b≠∞δ0(b,f)k)T0(R,f)≤Rη+O(log(RT0(R,f)))
holds for all R. Since η<ρ, from (34) and for sufficiently large R, we can get
(35)(1+k)Θ0(∞,f)+∑b≠∞‍δ0(b,f)>2-k(p-1),
which is contradiction with the assumption of Theorem 9.
Thus, this completes the proof of Theorem9.


## 4. Some Consequences of Theorems6 and 9

In this section, we will give some consequences of Theorems6 and 9. Before we give these results, some definitions will be introduced below.Definition 11.
Letf be meromorphic function on the annulus 𝔸:={z:1/R0<|z|<R0}, where R0=+∞. For a∈ℂ^, then we say that(i)
a is called an exceptional value in the sense of Nevanlinna (evN for short) for f on 𝔸, if δ0(a,f)>0;(ii)
a is called a normal value in the sense of Nevanlinna (nvN for short) for f on 𝔸, if δ0(a,f)=0.
In addition, similar to the Picard exceptional value in the whole complex plane, we define thata is called an exceptional value in the sense of Picard (evP for short) for f on 𝔸, if f has at most a finite number of a-points on 𝔸.Consequence 1.
Under the assumptions of Theorem6, if k1=1, from Theorem 6, we get
(36)p1+2∑i=2ν‍piki1+ki+∑j=1p1‍δ01(aj1,f)+∑i=2ν‍21+ki∑j=1pi‍δ01(aji,f)≤4.
Since pi≥0 and δ01(aji,f)≥0 for i=1,…,ν, it follows that(i)
iff has an evB for simple zeros which is also an evN for f on 𝔸, then f has at most three evBs for simple zeros on 𝔸;(ii)
ifa1 and a2 are two evPs for f on 𝔸 then no other element is an evB for f for simple zeros on 𝔸;(iii)
there exist at most four elements which are evBs forf simple zeros on 𝔸 since δ1(aj1,f)≥0. Moreover, all these four values are nvNs for f on 𝔸.Consequence 2.
Under the assumptions of Theorem6, let k1=1,p1=1, then(a)
ifk2=2, we have
(37)2p23+∑i=3ν‍piki1+ki+12δ01(a11,f)+13∑j=1p2‍δ02(aj2,f)+∑i=3ν‍11+ki∑j=1pi‍δ0ki(aji,f)≤32.Since pi≥0 and δ0ki(aji,f)≥0, it follows from (37) that p2<3. Thus, if a11 is an evB for f on 𝔸 for simple zeros, that is, δ01(a11,f)>0, then there exist at most two other elements which are evBs for f on 𝔸 for distinct simple zeros and double zeros. Furthermore,(i)
ifδ01(a11,f)=1/3, then two other elements evBs are also evNs for f on 𝔸;(ii)
if any   of the two other elementsa12 and a22, say a12, satisfies δ01(a12,f)=1/2, then a11, a22 are also evNs for f on 𝔸;(b)
ifk2=3, we have
(38)3p24+∑i=3ν‍piki1+ki+12δ01(a1,f)+14∑j=1p2‍δ03(aj2,f)+∑i=3ν‍11+ki∑j=1pi‍δ0ki(aji,f)≤32.From the above inequality and δ0(a,f)≥0 for a∈ℂ^, we get that p1=1, p2≤2 and p1=1, p2=2, pi=0(i=3,…,ν). Thus, if f has an evB for simple zeros on 𝔸, then there exist at most two other elements which are evBs for f distinct zeros of multiplicity ≤3 on 𝔸. Moreover, all these exceptional values are nvNs for f on 𝔸.Consequence 3.
Under the assumptions of Theorem6, if k1=2, then we have
(39)2p13+∑i=2ν‍piki1+ki+13∑j=1p1‍δ02(aj1,f)+∑i=2ν‍11+ki∑j=1pi‍δ0ki(aji,f)≤2.
Since ∑i=2ν‍(piki/(1+ki))+∑i=2ν‍(1/(1+ki))∑j=1pi‍δ0ki(aji,f)≥0, then from (39) we have
(40)2p13+13∑j=1p1‍δ02(aj1,f)≤2.
Thus, it follows that p1≤3 and p1=3,pi=0, i=2,…,ν. Hence, we have that f has at most three evBs for distinct simple and double zeros on 𝔸. Moreover, all three evBs for distinct simple and double zeros are nvNs for f on 𝔸.Consequence 4.
Under the assumptions of Theorem6, let k1=2,p1=1.(a)
Ifk2=1, then we have
(41)p2+2∑i=3ν‍piki1+ki+23δ02(a11,f)+∑j=1p2‍δ01(aj2,f)+∑i=3ν‍21+ki∑j=1pi‍δ0ki(aji,f)≤83.Thus, it follows that p2≤2. So, if there exists an evB for f on 𝔸 for distinct and double zeros, say a11, then there exist at most two other evBs for f on 𝔸 for simple zeros, say a12, a22. Furthermore, if p1=1, p2=2, it follows that
(42)2δ02(a11,f)+3δ01(a12,f)+δ01(a22,f)≤2.
Thus, we can see that any one of a12, a22 may not be an evP for f on 𝔸. Furthermore, if a11 is an evP for f on 𝔸. Then a12, a22 are nvNs for f on 𝔸;(b)
ifp2=1, k2=3,k3=1, then we have
(43)p3+∑i=4ν‍2piki1+ki+23δ02(a11,f)+12δ03(a12,f)+∑j=1p3‍δ01(aj3,f)+∑i=4ν‍21+ki∑j=1pi‍δ0ki(aji,f)≤76.Thus, we can get that p3≤1. That is, if f have two evBs on 𝔸, say a11 and a12  and a11 is for distinct simple and double zeros, a12 for distinct zeros of order ≤3, then f has at most one evB for simple zeros on 𝔸. Furthermore, if p3=1, then
(44)23δ02(a11,f)+12δ03(a12,f)+δ01(a13,f)≤16.It follows thatδ02(a11,f)≤1/4,δ03(a12,f)≤1/3 and δ01(a13,f)≤1/6. Thus, if any equality of these three inequalities holds, then the other two are nvNs for f on 𝔸.Now, we will give some consequences of Theorem9 as follows.Consequence 5.
Under the assumptions of Theorem9, if k=1 and
(45)2Θ0(a,f)+∑b≠a‍δ0(b,f)>3-p,
we have the following:(i)
ifp=1 and 2Θ0(a,f)+∑b≠a‍δ0(b,f)>2, then there exists at most one element b≠a which is an evB for f on 𝔸 for simple zeros; in particular, this holds if there exists an a∈ℂ^ satisfying Θ0(a,f)=1;(ii)
ifp=2 and 2Θ0(a,f)+∑b≠a‍δ0(b,f)>1, then there exists at most two elements b1,b2≠a which are evBs for f on 𝔸 for simple zeros; in particular, this holds if there exists an a∈ℂ^ satisfying Θ0(a,f)>1/2;(iii)
ifp=3 and 2Θ0(a,f)+∑b≠a‍δ0(b,f)>0, then there exists at most three elements b1,b2,b3≠a which are evBs for f on 𝔸 for simple zeros; in particular, this holds if there exists an a∈ℂ^ satisfying Θ0(a,f)>0.Remark 12.
Under the assumptions of Theorem9, from Consequence 5, we have that if there exist four distinct elements b1,b2,b3,b4∈ℂ^ which are evBs for f on 𝔸 for simple zeros, then Θ0(bi,f)≤1/2 and Θ0(a,f)=0 for a≠bi and i=1,2,3,4.Consequence 6.
Under the assumptions of Theorem9, if k=2 and
(46)3Θ0(a,f)+∑b≠a‍δ0(b,f)>4-2p,
we have the following:(i)
ifp=1 and 3Θ0(a,f)+∑b≠a‍δ0(b,f)>2, then there exists at most one element b≠a which is an evB for f on 𝔸 for distinct simple and double zeros; in particular, this holds if there exists an a∈ℂ^ satisfying Θ0(a,f)>2/3;(ii)
ifp=2 and 3Θ0(a,f)+∑b≠a‍δ0(b,f)>0, then there exists at most two elements b1,b2≠a which are evBs for f on 𝔸 for distinct simple and double zeros; in particular, this holds if there exists an a∈ℂ^ satisfying Θ0(a,f)>0.Remark 13.
Under the assumptions of Theorem9, from Consequence 6, we have that if there exist four distinct elements b1,b2,b3∈ℂ^ which are evBs for f on 𝔸 for distinct simple and double zeros, then Θ0(bi,f)≤2/3 and Θ0(a,f)=0 for a≠bi and i=1,2,3.


## 5. Exceptional Value of Meromorphic Function and Its Derivative

In this section, we will study the exceptional value of meromorphic functionf on 𝔸 and its differential polynomial of the form
(47)M[f]:=M[f,f′,…,f(k)]=af(z)i0(f′)i1⋯(f(k))ik,
where i0,i1,…,ik∈ℕ, k∈ℕ+, and a(≠0)∈ℂ is a complex constant.Theorem 14.
Letf be transcendental meromorphic function of order ρ(0<ρ<∞) on the annulus 𝔸:={z:1/R0<|z|<R0}, where R0=+∞. If 0 and ∞ are evBs for f on 𝔸 for distinct zeros, then M[f] has no evB for simple zeros in ℂ^∖{0,∞}.Proof.
From (21), set σ=i0+i1+⋯+ik. Since f is a transcendental meromorphic function on 𝔸, similar to the method of [12, Theorem 1.21], we can get that f and M[f] have the same order ρ on 𝔸. By applying Theorem 1 for M[f], for b(≠0)∈ℂ we have
(48)T0(R,M[f])≤N¯0(R,1M[f])+N¯0(R,M[f])+N¯0(R,1M[f]-b)+S(R,M[f])≤N¯0(R,1M[f])+N¯0(R,f)+12N¯01)(R,1M[f]-b)+12N¯01(R,1M[f]-b)+S(R,f)≤N¯0(R,1M[f])+N¯0(R,f)+12N¯01)(R,1M[f]-b)+12T0(R,f)+S(R,f).
Since
(49)N¯0(R,1M[f])≤N0(R,fσM[f])+N¯0(R,1fσ)≤T0(R,M[f]fσ)+N¯0(R,1f)+S(R,f)≤N0(R,M[f]fσ)+N¯0(R,1f)+S(R,f)≤O(N¯0(R,1f)+N¯0(R,f))+S(R,f).
and 0 and ∞ are evBs for f on 𝔸 for distinct zeros, then there exists a constant η(0<η<ρ) such that for sufficiently large R, we have
(50)N¯0(R,1f)<Rη,N¯0(R,f)<Rη.
Suppose thatb is an evB for M[f] on 𝔸 for simple zeros, since M[f] is of order ρ. Then for sufficiently large R, there exists a constant β<ρ such that
(51)N¯01)(R,1M[f]-b)<Rβ.
Thus, from (48)–(50), it follows that
(52)12T0(R,M[f])≤O(rβ′)+O(logRT0(R,f)),
where β′=max{η,β} and for R∈(1,+∞) except for the set ΔR such that ∫ΔR‍Rλ-1dR<+∞. Since M[f] is of order ρ, then there exists a sequence {Rn} tending to ∞ such that
(53)T0(Rn,M[f])>Rnα,
where α(β′<α<ρ) is a constant. Thus, we can get a contradiction by using the same argument as in Theorem 6.
Therefore,M[f] has no evB for simple zeros in ℂ^∖{0,∞}.



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*Source: 937584-2013-08-05.xml*