Normality Criteria of Meromorphic Functions Concerning Shared Analytic Function

Yang Qi

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

Communicated by Ji You-qing

Normality Criteria of Meromorphic Functions Concerning Shared Analytic Function

Yang Qi

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

Communicated by Ji You-qing

In this paper,we use Pang-Zalcman lemma to investigate the normal family of meromorphic functions concerning shared analytic function,which improves some earlier related results.

meromorphic function,entire function,shared function,normal family

2010 MR subject classification:30D35,30D45

Document code:A

Article ID:1674-5647(2016)01-0047-10

10.13447/j.1674-5647.2016.01.03

1　Introduction and Main Results

Let D be a domain in C,and F be a family of meromorphic functions defined in the domain D.F is said to be normal in D if any sequence{fn}⊂F contains a subsequence fnj,which converges spherically locally uniformly in D to a meromorphic function or∞(see[1]–[5]).

Let f(z)be a mermorphic function in a domain D and z0∈D.If f(z0)=z0,we say that z0is the fixed-point of f(z).Let f(z)and g(z)denote two meromorphic functions in D.If f(z)−ψ(z)and g(z)−ψ(z)have the same zeros(or ignoring multiplicity),then we say that f(z)and g(z)share ψ(z)CM(or IM).

In 1998,Wang and Fang[6]proved the following result:

Theorem 1.1Let k and n≥k+1 be two positive integers,and f be a transcendental merimorphic function.Then(fn)(k)assumes every finite nonzero value infinitely often.

Corresponding to Theorem 1.1,there are the following theorems about normal families.

Theorem 1.2[7]Let k and 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 2009,Li and Gu[8]proved:

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

Lately,many authors studied the functions of the form f(f(k))n.Hu and Meng[9]proved:

Theorem 1.4Take positive integers n and k with n,k≥2,and take a non-zero complex number a.Let F be a family of meromorphic functions in the plane domain D such that each f∈F has all its zeros of multiplicity at least k.For each pair(f,g)∈F,if f(f(k))nand g(g(k))nshare a IM,then F is normal in D.

Recently,Jiang and Gao[10]extended Theorem 1.4 as follows:

Theorem 1.5Let m≥0,n≥2m+2 and k≥2 be three positive integers and m be divisible by n+1.Suppose that a(z)(≢0)is a holomorphic function with zeros of multiplicity m in a domain D.Let F be a family of meromorphic functions in a domain D,and for each f∈F,f has all its zeros of multiplicity max{k+m,2m+2}at least.For each pair (f,g)∈F,if f(f(k))nand g(g(k))nshare a(z)IM,then F is normal in D.

A natural question is:What can be said if the function f(f(k))nin Theorem 1.5 is replaced by the function fd(f(k))n?In this paper,we answer this question by proving the following theorem:

Theorem 1.6Let F be a family of meromorphic functions defined in a domain D,and m≥0,n≥2m+2,k≥2,d≥1,p≥1 be five integers and m be divisible by n+d.Let ψ(z)≢0 be an analytic function with zeros of multiplicity m in a domain D.Suppose that every f∈F has all its zeros of multiplicity at leastFor each pair(f,g)∈F,if fd(f(k))nand gd(g(k))nshare ψ(z)IM,then F is normal in D.

Remark 1.1Obviously,from Theorem 1.6,we can get Theorem 1.5 when d=1.

2　Some Lemmas

In order to prove Theorem 1.6,we require the following results.

Lemma 2.1[11]Let F be a family of meromorphic functions on the unit disc satisfying 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)a number 0

b)points znwith|zn|

c)functions fn∈F;

d)positive numbers ρn→0

Lemma 2.2Let m≥0,k,n≥2,d≥1 be four integers,H(z)=amzm+am−1zm−1+ ···+a0be a polynomial,where am(≠0),am−1,···,a0are constants.If f is a non-constant polynomial,and the multiplicity of its all zeros is at least,then fd(z)(f(k)(z))n−H(z) has at least two distinct zeros,and fd(z)(f(k)(z))n−H(z)≢0.

Proof.Since f is a non-constant polynomial with zeros of multiplicityat least,we know that the degree of f isat least,and

Then fd(z)(f(k)(z))n−H(z)has at least one zero.

If fd(z)(f(k)(z))n−H(z)has only one zero,we may assume that

where λ is a non-zero constant,l is a positive integer.Compare the degrees of H(z)and f(z),we have

Then

Thus z0is the unique zero of(fd(z)(f(k)(z))n)(m+1).Since f is a non-constant polynomial with zeros of multiplicityat least,we know that z0is a zero of f.Thus

it contradicts with

Thus,fd(z)(f(k)(z))n−H(z)has at least two distinct zeros.

Lemma 2.3Let m ≥0,n≥2m+2,k,d≥1 be four integers,H(z)=amzm+ am−1zm−1+···+a0be a polynomial,where am(≠0),am−1,···,a0are constants.If f is a non-polynomial rational function,and the multiplicity of its all zeros is at least 2m+2, then fd(z)(f(k)(z))n−H(z)has at least two distinct zeros,and fd(z)(f(k)(z))n−H(z)≢0.

Proof.Since f is a non-polynomial rational function,it is obvious that

Let

where A is a non-zero constant,s,t≥1,mi≥2m+2(i=1,2,···,s),nj≥n(k+1)+d (j=1,2,···,t).For simplicity,we denote

By differentiating both sides of(2.1)step by step,we have

where g1(z)is a non-constant polynomial with deg(g1)≤(m+1)(s+t−1).

Now,we discuss two cases.

Case 1.If fd(z)(f(k)(z))n−H(z)has a unique zero z0,then we set

where B is a non-zero constant and l is a positive integer,P and Q are polynomials with degree M and N,also P and Q have no common factors.

Here we discuss two subcases.

Subcase 1.1.m≥l.

By differentiating both sides of(2.5),we have

where g2(z)is a polynomial with deg(g2)≤(m+1)t−(m−l+1).By(2.1)and(2.5),since m≥l,one has

From(2.4)and(2.6),

Then

it contradicts with m≥l.

Subcase 1.2.m

By differentiating both sides of(2.5),we have

where g3(z)is a polynomial with deg(g3)≤(m+1)t.

By differentiating both sides of(2.5)step by step for m times,we can get that z0is a zero of(fd(f(k))n)(m)=H(m).Since H(m)=am≠0,one has

Here we discuss in two subcases.

Subcase 1.2.1.l≠N+m.

From(2.1)and(2.5),we obtain deg(P)≥ deg(Q),that is,M ≥N.Since z0≠αi(i=1,2,···,s),(2.4)and(2.7)imply

So

By using(2.2)and(2.3),we obtain

which is a contradiction.

Subcase 1.2.2.l=N+m.

We further distinguish two subcases.

(i)M≥N.

By(2.4)and(2.7),we obtain

Similar to Subcase 1.2.1,we obtain a contradiction M

(ii)M

By using(2.4)and(2.7)again,we obtain

Hence

which is a contradiction.

Case 2.If fd(f(k))n−H(z)has no zero,then l=0 in(2.5).Proceeding as in the proof of Case 1,we get a contradiction.

Lemma 2.3 is proved.

Lemma 2.4[12]Suppose that f(z)is a transcendental meromorphic function,n,k,d are three positive integers.Then,when k≥1,n,d≥2,fd(f(k))n−φ(z)has infinitely many zeros,where φ(z)≢0,T(r,φ)=S(r,f).

3　Proof of Theorem 1.6

From Theorem 1.5,when d=1,Theorem 1.6 holds.

Next,we prove the case d≥2.

For any point z0∈D,either ψ(z0)=0 or ψ(z0)≠0.

Case 1.ψ(z0)=0.

We may assume z0=0 and ψ(z)=zm+am+1zm+1+···=zmh(z),where am+1,am+2, ···are constants,h(0)=1,and m can be divisible by n+d.

Let

If F1is not normal at 0,by Lemma 2.1,there exist a sequence{zj}of complex numbers with zj→z0and a sequence{ρj}of positive numbers with ρj→0 such that

locally uniformly on compact subsets of C,where g(ξ)is a non-constant meromorphic function in C,all of whose zeros have multiplicity at leastMoreover, g(ξ)has order at most 2.

Here we distinguish two cases.

locally uniformly on compact subsets of C disjoint from the poles of g,where H(ξ)is a non-constant meromorphic function in C,all of whose zeros have multiplicity at leastMoreover,H(ξ)has order at most 2.So

spherically locally uniformly in C disjoint from the poles of g.

If Hd(ξ)(H(k)(ξ))n≡ξm,since H has zeros with multiplicity at leastobviously there is a contradiction.Hence Hd(ξ)(H(k)(ξ))n≡ξm. Since the multiplicity of all zeros of H is at leastby Lemmas 2.2,2.3 and 2.4,Hd(ξ)(H(k)(ξ))n−ξmhas at least two distinct zeros.

Suppose that ξ0,ξ∗0are two distinct zeros of Hd(ξ)(H(k)(ξ))n−ξm.We choose a positive number δ small enough such that D1∩D2=∅and Hd(ξ)(H(k)(ξ))n−ξmhas no other zeros

By Hurwitz’s theorem,there exists a subsequence ofwe still denote it asthen there exist pointsand points ξj→ξ0such that when j is large enough,

Since,by the assumption in Theorem 1.6,share ψ(z),it follows that

Thus,when j is large enough,This contradicts with the factsThus F1is normal at 0.

where

Thus we have

Since

we have

On the other hand,for l=1,2,···,k,we have

Thus we have

spherically locally uniformly in C disjoint from the poles of g.

If gd(ξ)(g(k)(ξ))n≡1,then g has no zeros.Of course,g also has no poles.Since g is a non-constant Meromorphic function of order at most 2,there exist constants ci(i= 1,2),(c1,c2)≠(0,0),andObviously,this is contrary to the case gd(ξ)(g(k)(ξ))n≡1.Hence

Since the multiplicity of all zeros of g is at leastby Lemmas 2.2,2.3 and 2.4,gd(ξ)(g(k)(ξ))n−1 has at least two distinct zeros.

By Hurwitz’s theorem,there exists a subsequence ofwe still denote it asThen there exist pointsand pointssuch that when j is large enough,

Similar to the proof of Case 1.1,we get a contradiction.Then,F1is normal at 0.

From Cases 1.1 and 1.2,we know that F1is normal at 0,and there exist∆={z:|z|<ρ} and a subsequence of Fj,we still denote it as Fj,such that Fjconverges spherically locally uniformly to a meromorphic function F(z)or∞in∆.

Here we distinguish two cases.

Case(i).When j is large enough,fj(0)≠0.Then F(0)=∞.Thus,for each Fj(z)∈F1, there exists a δ>0 such that if F(z)∈F1,then|F(z)|>1 for all z∈∆δ={z:|z|<δ}. Thus,for sufficiently large j,is holomorphic in∆δ.Therefore,for all fj∈F, when|z|=δ/2,we have

By Maximum Principle and Montel’s Theorem,F is normal at z=0.

Case(ii).There exists a subsequence of fj,we still denote it as fj,such that fj(0)=0.

Since f∈F,the multiplicity of all zeros of f is at least p≥maxthen F(0)=0.Thus,there exists 0

By Cases(i)and(ii),F is normal at z=0.

Case 2.ψ(z0)≠0.

Suppose that F is not normal at z0.By Lemma 2.1 there exist a sequence{zj}of complex numbers with zj→z0,a sequence{ρn}of positive numbers with ρj→0 such that

locally uniformly on compact subsets of C,where g(ξ)is a non-constant meromorphic function in C,all of whose zeros have multiplicity at leastMoreover, g(ξ)has order at most 2.

Hence,by Lemmas 2.2,2.3 and 2.4,similar to the proof of Case 1.1,we get a contradiction.Thus F is normal at z0.

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date:May 27,2014.

The NSF(11461070)of China.

E-mail address:yangqi 8138@126.com(Yang Q).