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Normality Criteria of Meromorphic Functions

2016-10-12WangQiongYuanWenjunChenWeiandTianHonggen

Wang Qiong,Yuan Wen-jun,Chen Weiand Tian Hong-gen,*

(1.School of Mathematics Science,Xinjiang Normal University,Urumqi,830054)

(2.School of Mathematics and Information Sciences,Guangzhou University, Guangzhou,510006)

(3.School of Mathematics Sciences,Shandong University,Jinan,250000)

Communicated by Ji You-qing

Normality Criteria of Meromorphic Functions

Wang Qiong1,Yuan Wen-jun2,Chen Wei3and Tian Hong-gen1,*

(1.School of Mathematics Science,Xinjiang Normal University,Urumqi,830054)

(2.School of Mathematics and Information Sciences,Guangzhou University, Guangzhou,510006)

(3.School of Mathematics Sciences,Shandong University,Jinan,250000)

Communicated by Ji You-qing

In this paper,we consider normality criteria for a family of meromorphic functions concerning shared values.Let F be a family of meromorphic functions defined in a domain D,m,n,k and d be four positive integers satisfying m≥n+2 andand a(≠0),b be two finite constants.Suppose that every f∈F has all its zeros and poles of multiplicity at least k and d,respectively.If(fn)(k)−afmand(gn)(k)−agmshare the value b for every pair of functions(f,g)of F,then F is normal in D.Our results improve the related theorems of Schwick(Schwick W. Normality criteria for families of meromorphic function.J.Anal.Math.,1989,52: 241–289),Li and Gu(Li Y T,Gu Y X.On normal families of meromorphic functions. J.Math.Anal.Appl.,2009,354:421–425).

meromorphic function,shared value,normal criterion

2010 MR subject classification:30D30,30D45

Document code:A

Article ID:1674-5647(2016)01-0088-09

10.13447/j.1674-5647.2016.01.07

1 Introduction and Main Results

Let C be the set of complex numbers,D be a domain in C,which means that D is a connected nonempty open subset of C.Let F be a family of meromorphic functions defined in D.For{f,g}⊂F,{a,b}⊂P1=C∪{∞},we write f=a⇒g=b(f=a⇔g=b) if f−1(a)⊂g−1(b)(f−1(a)=g−1(b)),and say that f and g share a ignoring multiplicities(IM,for short)if f−1(a)=g−1(a)(see[1]).Here,the family F is said to be normal in D if any sequence of F must contain a subsequence that locally uniformly spherically converges to a meromorphic function or∞in D(see[2]).

In 1989,Schwick[3]proved a normality criterion:

Theorem 1.1Let k,n(≥k+3)be two positive integers,and F be a family of meromorphic functions defined in a domain D.If(fn)(k)≠1 for every function f∈F,then F is normal in D.

In 1998,Wang and Fang[4]proved:

Theorem 1.2Let k,n(≥ k+1)be two positive integers,and f be a transcendental meromorphic function.Then(fn)(k)assumes every finite non-zero value infinitely often.

For families of meromorphic functions,the connection between normality and shared values has been studied frequently.

By the ideas of shared values,Li and Gu[5]proved the following results:

Theorem 1.3Let F be a family of meromorphic functions defined in a domain D,k, n(≥k+2)be two positive integers,and a≠0 be a finite complex number.If(fn)(k)and (gn)(k)share a in D for every pair of functions f,g∈F,then F is normal in D.

In 2011,Liu and Li[6]studied Theorem 1.3,in which the value a was replaced by the fix-point z,and got the following result:

Theorem 1.4Let F be a family of meromorphic functions defined in a domain D,k, n(≥k+1)be two positive integers.If(fn)(k)and(gn)(k)share z in D for every pair of functions f,g∈F,then F is normal in D.

Lately,some theorems in this area appear.Hu and Meng[7],Jiang and Gao[8]studied the functions of the form f(f(k))n.Ding et al.[9]studied the functions of the form fm(f(k))nand Sun[10]studied the form P(f)(f(k))m.

Naturally,we pose the following question:

QuestionWhether the form(fn)(k)−afmin above Theorems can have similar results?

In this paper,we prove the following theorems and deal with this question.

Theorem 1.5Let F be a family of meromorphic functions defined in a domain D,m, n,k be three positive integers satisfying m≥n+k+3,and a(≠0),b be two finite complex constants.If(fn)(k)−afm≠b for every functions f of F,then F is normal in D.

Whether the condition m≥n+k+3 in Theorem 1.5 can be improved?We get the following results:

Theorem 1.6Let F be a family of meromorphic functions defined in a domain D,m, n,k(≥2)and d be four positive integers satisfying m≥n+k+1 and d≥2,and a(≠0),b be two finite complex constants.Suppose that every f∈F has all its poles of multiplicity at least d and(fn)(k)−afm≠b,then F is normal in D.

By the ideas of shared values,we can get the following results:

Theorem 1.7Let F be a family of meromorphic functions defined in a domain D,m, n,k and d be four positive integers satisfyingand a(≠0),b be two finite constants.Suppose that every f∈F has all its zeros and poles of multiplicity at least k and d,respectively.If(fn)(k)−afmand(gn)(k)−agmshare the value b IM for every pair of functions(f,g)of F,then F is normal in D.

2 Some Lemmas

In order to improve our theorems,we require the following Lemmas.

Lemma 2.1[11]Let F be a family of meromorphic functions on the unit disc∆ such that all zeros of functions in F have multiplicity≥p,and all poles of functions in F have multiplicity≥q.Let α be a real number satisfying−q<α

(a)points zj∈∆,zj→z0;

(b)functions fj∈F,and

(c)positive numbers ρj→0,

Lemma 2.2Let f(z)be meromorphic functions such that(fn)(k)(z)≢0,a(≠0)be a finite constant,and m,n,k and d be four positive integers satisfying m≥n+k+1.Then

Proof.Set

Since(fn)(k)(z)≢0,we know that Ψ(z)≢0.

By(2.1),we have

Thus,we get

which implies that

We see that a zero Ψ is attained at pole of f and zeros of(fn)(k)which is not zero of f, and a pole of f must be zero of Ψ by the condition m≥n+k+1.The pole of f cannot be zero of Ψ−1.Hence,if we denote¯N0(r)by the counting function of zeros of both Ψ and (fn)(k),we see that

On the other hand,we have

So we have

Therefore,by(2.3)–(2.7)and Nevanlinna’s first and second fundamental theorems,we have

Then,we have the inequality

Lemma 2.3Let F be a family of meromorphic functions defined in a domain D,m,n, k and d be four positive integers satisfyingb be two finite complex constants.Suppose that every f∈F has all its zeros and poles of multiplicity at least k and d,respectively,then

Proof.By the argument as Lemma 2.2,since the condition that all zeros and poles of f are multiplicities at least k and d,respectively,we get

Hence,by(2.9),(2.10)and the inequality(2.8),we get

3 Proof of Theorems

Proof of Theorem 1.5Suppose that F is not normal at z0.Then by Lemma 2.1,there exist fj∈F,zj→z0and ρj→0+such that

spherically uniformly on compact subsets of C,where g(ξ)is a nonconstant meromorphic function on C.We have

spherically uniformly on compact subsets of C outside poles of g.By hypothesis,(fn)(k)−afm≠b for every functions f of F.Applying Hurwitz theorem,we obtain that

or

If(gn)(k)−agm≡0,then g has not poles.By the logarithmic derivative lemma,we get

Hence

and this contradicts with g is nonconstant meromorphic function.Thus,

Then(gn)(k)≡0.

Indeed,if(gn)(k)≡0,then gnis polynomial with degree at most k−1,which is a contradiction with(gn)(k)−agm≠0.Applying Lemma 2.2 with meromorphic function g, we get

This implies

By m≥n+k+3,we conclude that g is constant function.This is a contradiction.Hence, F is normal in D.

Proof of Theorem 1.6By the argument as Theorem 1.5,we can assume that F is not normal at z0.Then by Lemma 2.1,there exist fj∈F,zj→z0and ρj→0+such that

spherically uniformly on compact subsets of C,where g(ξ)is a nonconstant meromorphic function on C and all its poles has multiplicity at least 2.Hence,from the inequality(3.1), we get

By hypothesis,

we get that g is a constant function.This is a impossible.Hence,F is normal in D.

Proof of Theorem 1.7Suppose that F is not normal at z0.Then by Lemma 2.1,there exist fj∈F,zj→z0and ρj→0+such that

spherically uniformly on compact subsets of C,where g(ξ)is a nonconstant meromorphic function on C and whose zeros and poles has multiplicity at least k and d,respectively. Moreover,the order of g is at most 2.We have

spherically uniformly on compact subsets of C outside poles of g.Hence,we apply Hurwitz theorem and obtain that

or

If(gn)(k)−agm≡0,since all poles of g have multiplicity at least d,we have

Therefore,g is a constant,a contradiction.So

By Lemma 2.3,we have

Then

If(gn)(k)−agm≠0,then(3.2)gives that g is a constant.Hence,(gn)(k)−agmis a meromorphic function and has at least one zero.

Next,we prove that(gn)(k)−agmhas just a unique zero.

Suppose to the contrary,let ξ0,be two distinct zeros of(gn)(k)(ξ)−agm(ξ),and choose δ>0 small enough such that

where

By Hurwitz theorem,there exists a sequence of points ξj∈D(ξ0,δ)andsuch that for large enough j,

By the assumption that for each pair of functions f,g∈F,(fn)(k)−afmand(gn)(k)−agmshare b in D,we know that for any positive integer m,

Hence,

Noting that g has zeros and poles of multiplicities at least k and d respectively,then (3.2)deduces that g is a rational function with degree at most 2.

If g is a polynomial and noting that degg≤2 and the multiplicities of zeros are at least k,we distinguish two cases.

Case 1 degg=1.

We can write g=A(ξ−ξ1),where A is a nonzero constant.So

Obviously,(gn)′+agmhas at least two distinct zeros,a contradiction.

Case 2 degg=2.

We distinguish two cases again.

Case 2.1 k=1.

We can write g=A(ξ−ξ1)(ξ−ξ2),where A is a nonzero constant.Then

Obviously,(gn)′+agmhas at least three distinct zeros,a contradiction.

Case 2.2 k=2.

We can write g=A(ξ−ξ1)2,where A is a nonzero constant.Then

Obviously,(gn)′+agmhas at least two distinct zeros,a contradiction.

Suppose that g is a rational function with degg≤2 and noting the multiplicities of poles are at leastwe also distinguish two subcases.

Subcase 1 degg=1.

We can write

where A,C are nonzero constants and AD+BC≠0.Then

Noting m≥n+2,(gn)′+agmhas at least two distinct zeros,a contradiction.

Subcase 2 degg=2.

We distinguish three cases.

Subcase 2.1 g=0,and g has only one zero.

In this case we have

where A,A1are two nonzero constants.We conclude that k=2.It follows thata contradiction.

Subcase 2.2 g=0,and g has two distinct zeros.

In this case we have

where A,A1are two nonzero constants.We conclude that k=1 and

Furthermore,

where A is a nonzero constant.So

where degg1<2.Obviously,(gn)′+agmhas at least three distinct zeros,a contradiction.

Subcase 2.3 g≠=0.

From Lemma 2.3,we get

which gives that g is a constant.This is a contradiction.

The proof is completed.

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date:Jan.6.2015.

The NSF(11461070,11271090)of China,the NSF(S2012010010121,2015A030313346)of Guangdong Province,and the Graduate Research,and Innovation Projects(XJGRI2015106)of Xinjiang Province.

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E-mail address:wq1298592600@163.com(Wang Q),tianhg@xjnu.edu.cn(Tian H G).