The partially shared values and small functions for meromorphic functions in a k punctured complex plane

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R E S E A R C H

Open Access

The partially shared values and small

functions for meromorphic functions in a

k-punctured complex plane

Hong Yan Xu

1*

_{, Yong Ming Li}

1

_{and Shan Liu}

2

*_{Correspondence:}

[email protected] 1_{School of Mathematics and} Computer Science, Shangrao Normal University, Shangrao, China Full list of author information is available at the end of the article

Abstract

The main aim of this article is to discuss the uniqueness of meromorphic functions partially sharing some values and small functions in ak-punctured complex plane

Ω

. We proved the following: Letf1,f2be two admissible meromorphic functions in

Ω

and

α

j(j= 1, 2,. . .,l) bel(≥5) distinct small functions with respect tofandg. If E(

α

j,

Ω

,f1)⊆E(

α

j,

Ω

,f2) (j= 1, 2,. . .,l) and

lim inf

r→+∞ l

j=1N0(r,_f₁1_–αj)

l

j=1N0(r,_f₂1_–_α_j)

> 5 2l– 5,

thenf1≡f2. Our results are some improvements and extension of previous theorems

given by Cao–Yi and Ge–Wu.

MSC: Primary 30D30; secondary 30D35

Keywords: Meromorphic function; Partially sharing; Small function;k-punctured

1 Introduction

This article is devoted to the study of uniqueness of functions which are meromorphic in a multiply-connected domain–k-punctured complex planeΩ. In 1920s, Nevanlinna gave the definition of characterized functionT(r,f) of meromorphic function and established the famous first and second main theorem, lemma on the logarithmic derivatives etc. of Nevalinna theory (see Hayman [1], Yang [2] and Yi and Yang [3]). Nowadays, Nevanlinna theory is a powerful tool in studying the properties of meromorphic functions in the fields of complex analysis. By applying this theory, the following well-known five-value theorem was given by Nevanlinna [4].

Theorem A(see [4]) If f and g are two nonconstant meromorphic functions that share ﬁve distinct values a1,a2,a3,a4,a5IM in X=C,then f(z)≡g(z).

Nevanlinna [4] also pointed out the following question.

Question A(see [4]) Does TheoremAstill hold if the ﬁve distinct values a1,a2,a3,a4,a5

are replaced by ﬁve distinct small functionsαj(j= 1, 2, . . . , 5)?

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Around TheoremA and QuestionA, the value distribution theory of meromorphic functions occupies one of the central places in complex analysis. Moreover,it is always an interesting topic how to extend and improve some important uniqueness theorems in the complex plane to the subsetX(including the unit disc, the angular domain, the annu-lus, etc.). Many scholars have paid significant attention to this topic and obtained lots of meaningful and important results (see [3,5–9]). For example, Fang [10] in 1999 proved the five-value theorem for meromorphic functions in the unit disc; Zheng [11] in 2003 obtained the five-value theorem for meromorphic functions in an angular domain; Cao, Yi, and Xu [12] in 2009 gave the five-value theorem for meromorphic functions in the an-nuli with the help of the Nevanlinna theory for meromorphic functions on anan-nuli given by Khrystiyanyn and Kondratyuk [13,14], or [15] in 2005, or [16] in 2004 (see [12]), etc. including [3,10,11,17–22]; Yi and Yang, Lahiri, and Xu improved a series of uniqueness theorems about weight-shared and partially shared (see [3,23–26]); there are a series of beautiful and important results related to QuestionA (see [27–32]). Especially, Yi [31] gave a positive answer to QuestionAand extended the five-value theorem to the case of sharing five distinct small functions.

Theorem B([31] The ﬁve small functions theorem) Let f and g be two nonconstant mero-morphic functions in a complex planeCand aj(j= 1, 2, 3, 4, 5)be ﬁve distinct small func-tions with respect to f and g.If f and g share aj(j= 1, 2, 3, 4, 5)IM inC,then f ≡g.

In 2016, the authors investigated the uniqueness of meromorphic functions sharing some finite sets in a special multiply-connected region—k-punctured complex plane— and obtained an analog of Nevanlinna’s famous five-value theorem for meromorphic func-tionsfandgin ak-punctured complex plane [33,34]. To state the result, some basic nota-tions and a definition aboutk-punctured complex plane should be introduced as follows, which can be found in [35].

Forkdistinct pointscj∈C,j∈ {1, 2, . . . ,k},Ω=C\

k

j=1{cj}can be called ak-punctured

complex plane. Of course, the annulus is a specialk-punctured plane ask= 1. Letk≥2,

d=1

2min{|cs–cj|:j=s}, andr0= 1

d+max{|cj|:j∈ {1, 2, . . . ,k}}, thus it yields that

1

r0 <d,

D1/r0(cj)∩D1/r0(cs) =∅ forj=s

and

D1/r0(cj)⊂Dr0(0) forj∈ {1, 2, . . . ,k},

whereDδ(c) ={z:|z–c|<δ}andDδ(c) ={z:|z–c| ≤δ}. Deﬁne

Ωr=Dr(0)\ k

j=1

D1/r(cj) for anyr≥r0.

Thus, it follows thatΩr⊃Ωr0 forr0<r≤+∞. Obviously,Ωris a multiple connected and

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For a meromorphic functionf in thek-punctured planeΩandr0≤r< +∞, letn0(r,f)

denote the counting function of its poles inΩr, and

N0(r,f) = r

r0

n0(t,f)

t dt, m0(r,f) =

1 2π

2π

0

log+freiθ _d_θ₊ 1

2π

m

j=1 2π

0

log+f

cj+

1

re iθ_d_θ

– 1 2π

2π

0

log+fr0eiθ dθ–

1 2π

m

j=1 2π

0

log+f

cj+

1

r0

eiθdθ,

wherelog+x=max{logx, 0}, then

T0(r,f) =m0(r,f) +N0(r,f)

is called the Nevanlinna characteristic offin thek-punctured complex plane. Besides, we useS(r,f) to denote any quantity satisfyingS(r,f) =o(T0(r,f)) for allroutside a possible

exceptional setEof ﬁnite linear measure.

Deﬁnition 1.1(see [33]) Letf be a nonconstant meromorphic function in ak-punctured planeΩ. The functionf is called admissible in ak-punctured planeΩprovided that

lim sup

r→+∞

T0(r,f)

logr = +∞, r0≤r< +∞.

Remark 1.1 (see [33]) From Theorem 5 in [35], a meromorphic function f in a k -punctured plane is rational iff satisﬁes

lim sup

r→+∞

T0(r,f)

logr < +∞, r0≤r< +∞.

Theorem C(see [33, Theorem 3.1]) Let f and g be two admissible meromorphic functions inΩ;if f,g share ﬁve distinct values a1,a2,a3,a4,a5IM inΩ,then f(z)≡g(z).

2 Results

The purpose of this article is to extend and improve some uniqueness results (including TheoremsA–C) to a special multiply-connected region—k-punctured complex plane.

By relaxing the form of sharing values IMto the partially sharing in Theorem C, we obtain the ﬁrst result of this article, which is an improvement of TheoremC.

Theorem 2.1 Let f1,f2be two admissible meromorphic functions inΩ,a1,a2, . . . ,albe l(≥

5)distinct values.IfE(aj,Ω,f1)⊆E(aj,Ω,f2)for all1≤j≤l and

lim inf

r→+∞

l

j=1N0(r,_f₁_–1_a_j) l

j=1N0(r,_f₂_–1_a

j)

> 1

l– 3,

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From Theorem2.1, we can obtain the following corollary immediately.

Corollary 2.1 Let f1,f2 be two admissible meromorphic functions inΩ, a1,a2, . . . ,al be l(≥5)distinct values.IfE(aj,Ω,f1)⊆E(aj,Ω,f2)for all1≤j≤l and f1≡f2,then

lim inf

r→+∞

l

j=1N0(r,_f₁_–1_a

j)

l

j=1N0(r,_f₂_–1_a_j)

≤ 1 l– 3.

Remark2.1 Whenl= 5 andE(aj,Ω,f1) =E(aj,Ω,f2) for all 1≤j≤l, in this case,

lim inf

r→+∞

5

j=1N0(r,_f₁_–1_a_j) 5

j=1N0(r,_f₂_–1_a_j)

= 1 >1 2.

Then it followsf1≡f2by Theorem2.1. Thus, this shows that Theorem2.1is an

improve-ment of TheoremC.

Inspired by QuestionA, TheoremB, and Theorem2.1,the second purpose of this paper is to investigate the uniqueness of meromorphic functions concerning small functions, and we obtain an analog of Nevanlinna’s ﬁve-value theorem for meromorphic functions in a k-punctured complex plane.

Theorem 2.2 Let f1,f2be two admissible meromorphic functions inΩ,α1,α2, . . . ,αlbe l(≥

5)distinct small functions with respect to f1,f2.IfE(αj,Ω,f1)⊆E(αj,Ω,f2)for all1≤j≤l

and

lim inf

r→+∞

l

j=1N0(r,_f₁_–1_α_j) l

j=1N0(r,_f₂_–1αj)

> 5 2l– 5,

whereE(α,Ω,h) ={z|h(z) –α(z) = 0,z∈Ω}for a meromorphic function h(z)inΩ,where each zero is counted only once,then f1≡f2.

Remark2.2 Let f be a nonconstant meromorphic function in ak-punctured planeΩ, we denote byS(f) a set of meromorphic functiona(z) in ak-punctured planeΩsatisfying

T0(r,a) =S(r,f), and such a meromorphic functiona(z) in ak-punctured planeΩis called

a small function with respect tof.

From Theorem2.2, the following results can be obtained immediately.

Corollary 2.2 Let f1,f2be two admissible meromorphic functions inΩ,and letα1,α2, . . . ,αl be l(≥5)distinct small functions with respect to f1,f2.IfE(αj,Ω,f1)⊆E(αj,Ω,f2)for all

1≤j≤l and f1≡f2,then

lim inf

r→+∞

l

j=1N0(r,_f₁_–1_α

j)

l

j=1N0(r,_f₂_–1αj)

≤ 5

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Corollary 2.3 Let f1,f2be two admissible meromorphic functions inΩ,and letα1,α2, . . . ,α6

be six distinct small functions with respect to f1,f2. IfE(αj,Ω,f1) =E(αj,Ω,f2) for all

1≤j≤6,then f1≡f2.

3 The proof of Theorem2.1

To prove Theorem2.1, we require the following lemmas.

Lemma 3.1(see [35, Theorem 3]) Let f,f1,f2be meromorphic functions in a k-punctured

planeΩ.Then:

(i) the functionT0(r,f)is nonnegative,continuous,nondecreasing,and convex with respect tologron[r0, +∞),T0(r0,f) = 0;

(ii) iff identically equals a constant,thenT0(r,f)vanishes identically; (iii) iff is not identically equal to zero,thenT0(r,f) =T0(r, 1/f),r0≤r< +∞;

(iv) T0(r,f1f2)≤T0(r,f1) +T0(r,f2) +O(1)andT0(r,f1+f2)≤T0(r,f1) +T0(r,f2) +O(1) forr0≤r< +∞;

(v) T0(r,_f_–1_a) =T0(r,f) +O(1)for any ﬁxeda∈C.

By using Lemma 6 in [35], we can get the following lemma easily.

Lemma 3.2 Let f be a nonconstant meromorphic function in a k-punctured planeΩand p be a positive integer,then

m0

r,f

(p)

f

=OlogT0(r,f) +O

log+r :=S(r,f), r→+∞,

outside a set E of ﬁnite linear measure.

Remark3.1 Obviously, iff is admissible in ak-punctured planeΩ, then

m0

r,f

(p)

f

=S(r,f) =oT0(r,f) .

Lemma 3.3([34, Theorem 2.5]) Let f be a nonconstant meromorphic function in a k-punctured planeΩ,and let a1,a2, . . . ,aq(q≥3)be distinct complex numbers in the ex-tended complex planeC:=C∪ {∞}.Then,for r0≤r< +∞,

(q– 2)T0(r,f)≤

q

ν=1

N0

r, 1

f–aν

+S(r,f),

wheren0(r,_f_–1_a)is the counting function of zeros of f–a inΩrwith the multiplicities reduced by1,

N0

r, 1

f–aν

=

r

r0

n0(t,_f_–1_a_ν)

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Proof of Theorem2.1 Without loss of generality, assume thataj(j= 1, 2, . . . ,l) are ﬁnite. In

view of3.3, it follows

(l– 2)T0(r,f1)≤

l

j=1

N0

r, 1

f1–aj

+S(r,f1)

and

(l– 2)T0(r,f2)≤

l

j=1

N0

r, 1

f2–aj

+S(r,f2).

Suppose thatf1≡f2. In view ofE(aj,Ω,f1)⊆E(aj,Ω,f2) for all 1≤j≤l, then it yields

l

j=1

N0

r, 1

f1–aj

≤N0

r, 1

f1–f2

≤T0(r,f1) +T0(r,f2) +O(1).

Thus, we have

l

j=1

N0

r, 1

f1–aj

≤

1

l– 2+o(1)

l

j=1

N0

r, 1

f1–aj

+

1

l– 2+o(1)

l

j=1

N0

r, 1

f2–aj

for allr∈/E, which implies

l– 3

l– 2+o(1)

l

j=1

N0

r, 1

f1–aj

≤

1

l– 2+o(1)

l

j=1

N0

r, 1

f2–aj

for allr∈/E. Hence, it follows

lim inf

r→+∞ l

j=1N0(r,_f₁_–1αj)

l

j=1N0(r,_f₂_–1αj)

≤ 1 l– 3,

which is a contradiction. Thus,f1≡f2.

Therefore, this completes the proof of Theorem2.1.

4 The proof of Theorem2.2

To prove Theorem2.2, we will require the following lemmas.

Lemma 4.1 Let f1(z)and f2(z)be two admissible meromorphic functions in a k-punctured

planeΩ,at(z)(≡0, 1)∈S(r) :=S(f1)∩S(f2),t= 1, 2,be not equal to constants

simultane-ously,and let

Fs(z) =

fsfs fs fs2–fs a1a1 a1 a21–a1

a2a2 a2 a22–a2

(7)

Then:

(i) Fs(z)≡0fors= 1, 2.

(ii)

2T0(r,fs) <N0

r, 1

fs– 1

+N0

r,1

fs

+N0(r,fs) +N0

r, 1

fs–a1

+N0

r, 1

fs–a2

+S(r,f1) +S(r,f2) fors= 1, 2.

Proof (i) Assume thatF1(z)≡0. Thus, we can rewrite (4.1) as the following form:

a1

a1 –a 2 a2 f1

f1– 1

– a

2

a2– 1

–

a1

a1– 1

– a

2

a2– 1 f1 f1 –a 2 a2

≡0. (4.2)

Next, we will divide the proof into four cases as follows.

Case1. If a1

a1 ≡

a2

a2, then it followsa1=η1a2, whereη1= 1 is a constant. From the

as-sumptions of this lemma, it yields a1

a1–1 ≡

a₂

a2–1, which implies

f₁ f1 ≡

a₂

a2. Thus, we have

f1(z) =η2a2(z), whereη2is a constant. Therefore, we get a contradiction.

Case2. If a1

a1–1≡

a₂

a2–1. By using the same argument as in Case 1, we also get a

contradic-tion.

Case3. If a1

a1 ≡

a2

a2,

a1

a1–1≡

a2

a2–1and

a1

a1 –

a2

a2 ≡

a1

a1–1–

a2

a2–1. Thus it follows from (4.1) that

f₁ f1– 1

–f

1

f1 ≡

a₂ a2– 1

–a

2

a2

.

By a simple integral, we have_f1

1 = 1 –η3(1 – 1

a2), whereη3is a constant, a contradiction.

Case4. Ifa1

a1 ≡

a₂ a2,

a₁ a1–1≡

a₂ a2–1 and

a₁ a1 –

a₂ a2 ≡

a₁ a1–1–

a₂

a2–1. Thus, we can rewrite (4.2) as

the following form:

a₁ a1 –a 2 a2 f₁ f1– 1

–

a₁ a1– 1

– a

2

a2– 1

f₁ f1

≡ a1

a1

a₂ a2– 1

–a

2

a2

a₁ a1– 1

. (4.3)

By observing (4.3), the zeros off1– 1 inΩcan only occur at the zeros, 1-points and poles

ofa1(z) anda2(z), and the zeros of

a₁ a1 –

a₂

a2 inΩ. Thus, we have

N0

r, 1

f1– 1 ≤ 2 j=1

N0(r,aj) +N0

r, 1

aj

+N0

r, 1

aj– 1

+N0

r, 1

a₁ a1 –

a₂ a2

=S(r,f1) +S(r,f2). (4.4)

Similarly, we have

N0

r,1

f1

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Further, the poles off1inΩcan only occur at the zeros, 1-points and poles ofa1(z) and

a2(z), and the zeros of (a

1

a1 –

a2

a2) – (

a1

a1–1–

a2

a2–1) inΩ. By a simple calculation, we have

N0(r,f1) =S(r,f1) +S(r,f2). (4.6)

By Lemma3.3and from (4.4)–(4.6), it follows

T0(r,f1) <N0

r, 1

f1– 1

+N0

r, 1

f1

+N0(r,f1) +S(r,f1)

=S(r,f1) +S(r,f2),

a contradiction.

IfF2(z)≡0, by using the same argument as above, we also get a contradiction. Then we

prove (i). (ii) Let

δ(z) = 1 3min

1,a1(z),a2(z),a1(z) – 1,a2(z) – 1,a1(z) –a2(z),z∈Ω

,

θt(r) =

θ:f1

reiθ –at

riθ ≤δriθ (t= 1, 2),

θ3(r) =

θ:f1

reiθ ≤δriθ ,

θ4(r) =

θ:f1

reiθ – 1≤δriθ . Then it follows

1 2π

2π

0

log 1

δ(reiθ₎dθ

≤ 1

2π 2π

0

log max

1, 1

|a1(z)|

, 1

|a2(z)|

, 1

|a1(z) – 1|

, 1

|a2(z) – 1|

, 1

|a1(z) –a2(z)|

,z∈Ω

dθ+log3

≤m

r, 1

a1

+m

r, 1

a2

+m

r, 1

a1– 1

+m

r, 1

a2– 1

+m

r, 1

a1–a2

+ 4log2.

Similarly, for anycj,j= 1, 2, . . . ,k, we have

1 2π

2π

0

log 1

δ(cj+1_reiθ) dθ

≤ 1

2π 2π

0

log max

1, 1

|a1(cj+1_reiθ)|

, 1

|a2(cj+1_reiθ)|

, 1

|a1(cj+1_reiθ) – 1|

, 1

|a2(cj+1_reiθ) – 1|

, 1

|a1(cj+1_reiθ) –a2(cj+1_reiθ)|

,z∈Ω

dθ+log3

≤m

1

r,

1

a1(cj+z)

+m

r, 1

a2(cj+z)

+m

r, 1

a1(cj+z) – 1

(9)

+m

r, 1

a2(cj+z) – 1

+m

r, 1

a1(cj+z) –a2(cj+z)

+ 4log2.

Further, forj= 1, 2, . . . ,k, 1

2π 2π

0

log 1

δ(r0eiθ)

dθ=O(1), 1 2π

2π

0

log 1

δ(cj+_r1₀eiθ)

dθ=O(1).

Thus, it follows

1 2π

2π

0

log 1

δ(reiθ₎dθ+

k

j=1

1 2π

2π

0

log 1

δ(cj+1_reiθ) dθ

– 1 2π

2π

0

log 1

δ(r0eiθ)

dθ–

k

j=1

1 2π

2π

0

log 1

δ(cj+_r1₀eiθ) dθ

≤m0

r, 1

a1

+m0

r, 1

a2

+m0

r, 1

a1– 1

+m0

r, 1

a2– 1

+m

r, 1

a1–a2

+O(1)

≤S(r,f1) +S(r,f2). (4.7)

On the other hand, taking

f1f1= (f1–a1)

f₁–a₁ +a₁(f1–a1) +a1

f₁–a₁ +a1a2:=F1,

f₁=f₁–a₁ +a₁:=F2,

f₁2–f1= (f1–a1)2+ (2a1– 1)(f1–a1) +a21–a1:=F3,

and substituting these into (4.22), by a simple calculation, we have

F=

F1–a1a2 F2–a1 F3–a21+a1

a1a1 a1 a21–a1

a2a2 a2 a22–a2

. (4.8)

From the deﬁnition ofθ1(r) andδ(z), we have f1

reiθ –a1

reiθ ≤δreiθ ≤1 +a1

reiθ asθ∈θ1(r) (4.9)

and 1 2π

θ1(r)

log+ F

f1–a1 dθ

≤m

r,f

1–a1

f1–a1

+Om(r,a1) +m(r,a2) +m

r,a₁ +mr,a₂

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On the other hand, we have|f1(reiθ) –a1(reiθ)| ≥δ(reiθ) asθ∈/θ1(r), that is,

1

|f1(reiθ) –a1(reiθ)|≤

1

δ(reiθ₎ asθ∈/θ1(r). (4.11)

By combining (4.10) and (4.11), we have

m

r, 1

f1–a1

≤ 1

2π

θ1(r)

log+ F

f1–a1 dθ+ 1

2π

θ1(r) log+1

F

dθ

+ 1 2π

[0,2π]–θ1(r) log1

δ dθ

≤ 1

2π

θ1(r)

log+ 1

F(reiθ₎

dθ+ 1

2π 2π

0

log 1 δ(reiθ₎

dθ

+S(r,f1) +S(r,f2). (4.12)

Similarly, we have

m

1

r,

1

f1(cj+1_reiθ) –a1(cj+1_reiθ)

≤ 1

2π

θ1(r)

log+ F(cj+ 1

re iθ₎

f1(cj+1_reiθ) –a1(cj+1_reiθ)

dθ

+ 1 2π

θ1(r)

log+ 1

F(cj+1_reiθ)

dθ+ 1

2π

[0,2π]–θ1(r)

log 1

δ(cj+1_reiθ)

dθ

≤ 1

2π

θ1(r)

log+ 1

F(cj+1_reiθ)

dθ+ 1

2π 2π

0

log 1

δ(cj+1_reiθ)

dθ

+S(r,f1) +S(r,f2). (4.13)

Since

m

r0,

1

f1–a1

=O(1), 1 2π

θ1(r0)

log+ 1

F(r0eiθ)

dθ=O(1),

m

1

r0

, 1

f1(cj+_r1₀eiθ) –a1(cj+_r1₀eiθ)

=O(1),

and 1 2π

θ1(r0)

log+ 1

F(cj+_r1₀eiθ)

dθ=O(1), j= 1, 2, . . . ,k,

by combining (4.7), (4.12), and (4.13), it follows

m0

r, 1

(11)

≤ 1

2π

θ1(r)

log+ 1

F(reiθ₎

dθ+

k j=1 1 2π

θ1(r)

log+ 1

F(cj+1_reiθ)

dθ

– 1 2π

θ1(r0)

log+ 1

F(r0eiθ) dθ–

k j=1 1 2π

θ1(r0)

log+ 1

F(cj+_r1₀eiθ)

dθ

+S(r,f1) +S(r,f2). (4.14)

By using the same argument as above, we have

m0

r, 1

f1–a2

≤ 1

2π

θ2(r)

log+ 1

F(reiθ₎

dθ+

k j=1 1 2π

θ2(r)

log+ 1

F(cj+1_reiθ)

dθ

– 1 2π

θ2(r0)

log+ 1

F(r0eiθ) dθ–

k j=1 1 2π

θ2(r0)

log+ 1

F(cj+_r1₀eiθ)

dθ

+S(r,f1) +S(r,f2); (4.15)

m0

r,1

f1

≤ 1

2π

θ3(r)

log+ 1

F(reiθ₎

dθ+

k j=1 1 2π

θ3(r)

log+ 1

F(cj+1_reiθ)

dθ

– 1 2π

θ3(r0)

log+ 1

F(r0eiθ) dθ–

k j=1 1 2π

θ3(r0)

log+ 1

F(cj+_r1₀eiθ)

dθ

+S(r,f1) +S(r,f2); (4.16)

and

m0

r, 1

f1– 1

≤ 1

2π

θ4(r)

log+ 1

F(reiθ₎

dθ+

k j=1 1 2π

θ4(r)

log+ 1

F(cj+1_reiθ)

dθ

– 1 2π

θ4(r0)

log+ 1

F(r0eiθ) dθ–

k j=1 1 2π

θ4(r0)

log+ 1

F(cj+_r1₀eiθ)

dθ

+S(r,f1) +S(r,f2). (4.17)

Sinceθ(r)∈[0, 2π), then from (4.14)–(4.17) it yields

m0

r, 1

f1–a1

+m0

r, 1

f1–a2

+m0

r, 1

f1

+m0

r, 1

f1– 1

<m0

r,1

F

(12)

Ifz0is a zero off1orf1– 1 orf1–a1orf1–a2inΩof multipliesp> 1 and not a pole ofa1

ora2inΩ, thenz0must be a zero ofF1(z) inΩof multipliesp– 1. Thus, it follows

4T0(r,f1) <N0

r, 1

f1

+N0

r, 1

f1– 1

+N0

r, 1

f1–a1

+N0

r, 1

f1–a2

–N0

r, 1

F1

+T0(r,F1) +O(1) +S(r,f1) +S(r,f2)

<N0

r, 1

f1

+N0

r, 1

f1– 1

+N0

r, 1

f1–a1

+N0

r, 1

f1–a2

+T0(r,F1) +O(1) +S(r,f1) +S(r,f2). (4.19)

In addition, from the deﬁnition ofF1(z), we can get

m0(r,F1) < 2m0(r,f1) +S(r,f1) +S(r,f2), (4.20)

N0(r,F1) < 2N0(r,f1) +N0(r,f1) +S(r,f1) +S(r,f2). (4.21)

Hence, from (4.19)–(4.21), we can get Lemma4.1(ii).

Therefore, this completes the proof of Lemma4.1.

Lemma 4.2 Let f1(z)and f2(z)be two admissible meromorphic functions in a k-punctured

planeΩ,αj(z)(≡0, 1)∈S(f1)∩S(f2),j= 1, 2, . . . , 5,be ﬁve distinct meromorphic functions

in a k-punctured planeΩ,then

2T0(r,fs) <

5

j=1

N0

r, 1

fs–αj

+S(r,f1) +S(r,f2), s= 1, 2. (4.22)

Proof Set

gs= fs–α4

fs–α5 α3–α5 α3–α4

(s= 1, 2),

aj=

αj–α4 αj–α5

α3–α5 α3–α4

(j= 1, 2).

Then it yields

T0(r,gs) –T0(r,fs)<S(r,f1) +S(r,f2), fors= 1, 2, (4.23)

S(r,f1) +S(r,f2) =S(r,g1) +S(r,g2). (4.24)

Here we will consider three cases as follows.

Case1. Ifg1andg2are admissible, then by applying Lemma4.1forg1,g2,a1,a2, we have

2T0(r,gs) <N0(r,gs) +N0

r, 1

gs

+N0

r, 1

gs– 1

+N0

r, 1

gs–a1

+N0

r, 1

gs–a2

(13)

fors= 1, 2. From the notations ofg1andg2andαj∈S(f1)∩S(f2), we have

N0(r,gs) <N0

r, 1

fs–α5

+O

₅

j=1

T0(r,αj)

, (4.26)

N0

r, 1

gs

<N0

r, 1

fs–α4

+O

₅

j=1

T0(r,αj)

, (4.27)

and

N0

r, 1

gs–aj

<N0

r, 1

fs–αj

+O

₅

j=1

T0(r,αj)

(4.28)

fors= 1, 2;j= 1, 2. Substituting (4.26)–(4.28) into (4.25), we can get (4.22) easily.

Case 2. Ifg1 andg2 are rational, then from Remark2.1we haveT0(r,gs) =O(logr) = S(r,f1) +S(r,f2) fors= 1, 2. Thus, combining (4.23) and (4.24), it yieldsT0(r,fs) =S(r,f1) +

S(r,f2) fors= 1, 2. Hence the conclusions hold.

Case3. If one ofg1,g2is rational, without loss of generality assume thatg1is rational and

g2is admissible. From Case 2 and Case 1, we haveT0(r,f1) =S(r,f1) +S(r,f2) and

T0(r,f2) < 5

j=1

N0

r, 1

f2–αj

+S(r,f1) +S(r,f2).

Then the conclusion holds.

From Cases 1–3, this completes the proof of Lemma4.2.

Proof of Theorem2.2 Take any distincts1, . . . ,s5∈ {1, 2, . . . ,l}, and in view of Lemma4.2,

it follows

2T0(r,fi) <

5

j=1

N0

r, 1

fi–αsj

+S(r), i= 1, 2. (4.29)

Thus, we can conclude

2

l

5

T0(r,fi)≤

5

l

5

_l

j=1

N0

r, 1

fi–αj

+S(r), i= 1, 2,

that is,

T0(r,fi)≤

5 2l

l

j=1

N0

r, 1

fi–αj

+S(r), i= 1, 2. (4.30)

Suppose thatf1≡f2. In view ofE(αj,Ω,f1)⊆E(αj,Ω,f2) for all 1≤j≤l, we have

l

j=1

N0

r, 1

f1–αj

≤N0

r, 1

f1–f2

(14)

In view of (4.30) and (4.31), we can deduce

l

j=1

N0

r, 1

f1–αj

≤

5 2l+o(1)

l

j=1

N0

r, 1

f1–αj

+

5 2l+o(1)

l

j=1

N0

r, 1

f2–αj

forr∈/E, which implies

2l– 5 2l +o(1)

l

j=1

N0

r, 1

f1–αj

≤

5 2l+o(1)

l

j=1

N0

r, 1

f2–αj

forr∈/E. This leads to

lim inf

r→+∞ l

j=1N0(r,_f₁_–1αj)

l

j=1N0(r,_f₂_–1αj)

≤ 5

2l– 5,

which is a contradiction with the assumption of Theorem2.2. Thus,f1≡f2.

Therefore, this completes the proof of Theorem2.2.

Acknowledgements

We thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation.

Funding

This work was supported by the National Natural Science Foundation of China (11561033, 11761035), the Natural Science Foundation of Jiangxi Province in China (20181BAB201001, 20171BAB201002, 20151BAB201008), and the Foundation of Education Department of Jiangxi (GJJ180734) of China. The third author was supported by CSC 2019.

Availability of data and materials No data were used to support this study.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The main idea of this paper was proposed by HYX. HYX prepared the manuscript initially and performed all the steps of the proofs in this research. All authors read and approved the ﬁnal manuscript.

Author details

1_{School of Mathematics and Computer Science, Shangrao Normal University, Shangrao, China.}2_{Department of}

Mathematics & Computer Science, Jiangxi Science & Technology Normal College, Nanchang, China.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 28 February 2019 Accepted: 1 November 2019 References

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