CHEN JuN-FAN AND CAI XIAO-HuA

(Department of Mathematics,Fujian Normal University,Fuzhou,350117)

Communicated by Ji You-qing

1 Introduction and Main Results

First of all we recall that a family F of functions meromorphic in a plane domain D is called to be normal in D,in the sense of Montel,if every sequence{fn}?F contains a subsequence{fnj}which converges spherically locally uniformly in D,to a meromorphic function or the constant ∞ (see[1]–[3]).

Let f and g be meromorphic in a domain D,b∈ C∪{∞}.If f(z)?b and g(z)?b assume the same zeros ignoring multiplicity,we say that f and g share b in D.

Inspired by heuristic Bloch’s principle(see[4]–[5])that there is an analogue in normal family theory corresponding to every Liouville-Picard type theorem,Gu[6]proved the following famous normality criterion related to the well-known Hayman’s alternative(see[7]).

Theorem A[6]LetFbe a family of meromorphic functions in a domainD,kbe a positive integer,andbbe a nonzero finite complex number.If for eachf∈F,f0andf(k)binD,thenFis normal inD.

Recently,by the idea of shared values,Fang and Zalcman[8],[9]extended Theorem A as follows.

Theorem B[8],[9]Letkbe a positive integer andFbe a family of meromorphic functions in a domainDsuch that for eachf∈F,all zeros offare of multiplicity at leastk+2.Letaandb0be two finite complex numbers.If for each pair of functionsf,g∈F,fandgsharea,f(k)andg(k)sharebinD,thenFis normal inD.

In 1989,Schwick[10]obtained the following theorem.

Theorem C[10]LetFbe a family of meromorphic functions in a domainD,n,kbe positive integers withn≥k+3,andbbe a nonzero finite complex number.If for eachf∈F,(fn)(k)binD,thenFis normal inD.

In 2009,Li and Gu[11]improved Theorem C and proved the following result with the idea of shared values.

Theorem D[11]LetFbe a family of meromorphic functions in a domainD,n,kbe positive integers withn≥k+2,andbbe a nonzero finite complex number.If for each pair of functionsf,g∈F,(fn)(k)and(gn)(k)sharebinD,thenFis normal inD.

In 1998,Wang and Fang[12]proved the following theorem.

Theorem E[12]Letkbe a positive integer andFbe a family of meromorphic functions in a domainDsuch that for eachf∈F,all poles offare of multiplicity at least2,and all zeros offare of multiplicity at leastk+1.Letbbe a nonzero finite complex number.If for eachf∈F,f(k)binD,thenFis normal inD.

It is natural to ask whether Theorem E can be extended in the same way that Theorem B extends Theorem A or Theorem D extends Theorem C.In this paper,we offer such an extension.

Theorem 1.1Letkbe a positive integer andFbe a family of meromorphic functions in a domainDsuch that for eachf∈F,all poles offare of multiplicity at least2,and all zeros offare of multiplicity at leastk+1.Letaandbbe two distinct finite complex numbers.If for eachf∈F,all zeros off(k)?aare of multiplicity at least2,and for each pair of functionsf,g∈F,f(k)andg(k)sharebinD,thenFis normal inD.

Corollary 1.1Letkbe a positive integer andFbe a family of holomorphic functions in a domainDsuch that for eachf∈F,all zeros offare of multiplicity at leastk+1.Letaandbbe two distinct finite complex numbers.If for eachf∈F,all zeros off(k)?aare of multiplicity at least2,and for each pair of functionsf,g∈F,f(k)andg(k)sharebinD,thenFis normal inD.

Moreover,we can prove the following result by restricting the numbers of the zeros of f(k)?b.

Theorem 1.2Letkbe a positive integer andFbe a family of meromorphic functions in a domainDsuch that for eachf∈F,all poles offare of multiplicity at least2,and all zeros offare of multiplicity at leastk+1.Letaandbbe two distinct finite complex numbers.If for eachf∈F,all zeros off(k)?aare of multiplicity at least2,f(k)?bhas at most1zero inDignoring multiplicity,thenFis normal inD.

Corollary 1.2Letkbe a positive integer andFbe a family of holomorphic functions in a domainDsuch that for eachf∈F,all zeros offare of multiplicity at leastk+1.Letaandbbe two distinct finite complex numbers.If for eachf∈F,all zeros off(k)?aare of multiplicity at least2,f(k)?bhas at most1zero inDignoring multiplicity,thenFis normal inD.

Example 1.1Let D={z:|z| ＜ 1}and F={fm:m=1,2,3,···},where fm(z)=mzk+2.Then,for each fm∈F,all zeros of fmare of multiplicity at least k+1 in D,and

Thus we see that for each pair of functions fm,fn∈F,andshare 0 in D,and for each fm∈F,has only one zero z=0 of exact multiplicity 2 in D.But F fails to be normal in D.This shows that the condition in Theorems 1.1 and 1.2 that a and b be two distinct finite complex numbers is necessary.

2Some Lemmas

Lemma 2.1[13]Letkbe a positive integer andFbe a family of meromorphic functions in a domainDonC,all of whose zeros have multiplicity at leastk,and suppose that there existsA≥1such that|f(k)(z)|≤Awheneverf(z)=0,f∈F.Then,ifFis not normal at some pointz0∈D,there exist,for each0≤α≤k,

(a)pointszn∈D,zn→z0;

(b)functionsfn∈F,and

(c)positive numbersρn→ 0

such that

locally uniformly with respect to the spherical metric,wheregis a nonconstant meromorphic function onC,all of whose zeros have multiplicity at leastk,such thatg#(ζ) ≤ g#(0)=kA+1.In particular,gis of order at most two.

Lemma 2.2[12]Letkbe a positive integer andfbe a transcendental meromorphic function inCsuch that all poles offare of multiplicity at least2,and all zeros offare of multiplicity at leastk+1.Thenf(k)assumes every nonzero finite complex number in finitely often.

Lemma 2.3Letkbe a positive integer andfbe a nonconstant rational function such that all poles offare of multiplicity at least2,and all zeros offare of multiplicity at leastk+1.Letaandbbe two distinct finite complex numbers.If all zeros off(k)?aare of multiplicity at least2,thenf(k)?bhas at least two distinct zeros.Proof.We consider two cases.

Case 1. f(k)?b has exactly one zero z0.

For this case we consider two subcases.

Subcase 1.1. f is a nonconstant polynomial.

Clearly,f(k)(z)a and f(k)(z)b,for otherwise f would be a polynomial of degree at most k,which contradicts the fact that all zeros of f are of multiplicity at least k+1.Since f(k)?b has exactly one zero z0and all zeros of f(k)?a are of multiplicity at least 2,then we can put

where A is a nonzero constant,n ≥ 2,and lp≥ 2,β1αp(p=1,2,···,q).Thus,by(2.1)and(2.2),we get

Differentiating(2.1)and(2.2)it follows that

where ? is a polynomial.Now from above two equalities and the fact that β1αp(p=1,2,···,q),we have

i.e.,

This contradicts with(2.3).

Subcase 1.2. f is a nonpolynomial rational function.

From the conditions we can set

where B is a nonzero constant,P and Q are two polynomials having no common factors,mi≥ 2(i=1,2,···,s),and nj≥ k+2(j=1,2,···,t).For the sake of convenience,we put

Since f(k)?b has exactly one zero z0,from(2.4)it follows that

where B1is a nonzero constant,z0ξi(i=1,2,···,s)since ab.Differentiating(2.4)we have

where degg ≤ s+t?1.Differentiating(2.7)we get

where

with some constants d0,d1,···,dt?1.

Next we distinguish two subcases again.

Subcase 1.2.1. lN.

By(2.4)–(2.7)we see that degP ≥ degQ and so M ≥ N.Since z0ξi(i=1,2,···,s),it follows from(2.8)and(2.9)that M?s≤degh=t and so M≤s+t.This together with(2.5)and(2.6)gives

a contradiction.

Subcase 1.2.2.l=N.

If M≥N,then proceeding as in the argument in Subcase 1.2.1,from(2.8)and(2.9)we also have M＜M,a contradiction.

If M ＜N,then(2.8)and(2.9)means that l?1≤s+t?1 and so by(2.5)and(2.6)it follows that

a contradiction.

Case 2. f(k)?b has no zeros,i.e.,f(k)b.

Since all poles of f are of multiplicity at least 2,all zeros of f are of multiplicity at least k+1,and all zeros of f(k)? a are of multiplicity at least 2,it follows by the Nevanlinna’s first and second fundamental theorem that

which implies that T(r,f(k))=S(r,f(k)),a contradiction.

This completes the proof of Lemma 2.3.

Example 2.1LetThen by a simple calculation it follows that

and

Clearly,not all zeros of f′(z)are of multiplicity at least 2,and f′(z)? 1 has only one zeroThis shows the condition in Lemma 2.3 that all zeros of f(k)?a are of multiplicity at least 2 cannot be omitted.

3 Proof of Theorem 1.1

Let z0∈D.We show that F is normal at z0.Let f∈F.We consider two cases.

Case 1. f(k)(z0)b.

There exists a δ＞ 0 such that f(k)(z)b in Dδ={z:|z?z0|＜ δ}? D.Thus by the assumptions,for each h∈F,all poles of h are of multiplicity at least 2,all zeros of h are of multiplicity at least k+1,and h(k)(z)b in Dδ.By Theorem E,F is normal in Dδ.Hence,F is normal at z0.

Case 2. f(k)(z0)=b.

There exists a δ＞0 such that f(k)(z)b in D0δ={z:0＜|z?z0|＜δ}?D.Thus by the assumptions,for each h∈F,all poles of h are of multiplicity at least 2,all zeros of h are of multiplicity at least k+1,and h(k)(z)b in D0δ.By Theorem E,F is normal in D0δ.

Next we prove that F is normal at z0.Without loss of generality,we may assume that z0=0.Suppose,on the contrary,that F is not normal at z0.Then from Lemma 2.1 there exist

(i)points zj∈D,zj→z0;

(ii)functions fj∈F,and

(iii)positive numbers ρj→ 0 such that

locally uniformly with respect to the spherical metric,where g is a nonconstant meromorphic function on C.Moreover,all poles of g are of multiplicity at least 2,and all zeros of g are of multiplicity at least k+1.

Now,from(3.1)we have

uniformly on compact subsets of C disjoint from the poles of g.

We claim that all zeros of g(k)(ζ)? a are of multiplicity at least 2.

Suppose that g(k)(ζ0)=a.Clearly,g(k)(ζ)a,for otherwise g would be a polynomial of degree at most k,which contradicts the fact that all zeros of g are of multiplicity at least k+1.Then by(3.2)and Hurwitz’s theorem there exist ζj, ζj→ ζ0,such that,for j sufficiently large,

Thus

since all zeros of f(k)(z)?a are of multiplicity at least 2.It now follows that

which implies that all zeros of g(k)(ζ)?a are of multiplicity at least 2.The claim is proved.

Obviously,g(k)(ζ)b.Next we consider further two subcases.

Subcase 2.1.

By Hurwitz’s theorem we see that g(k)(ζ)b.This contradicts with Lemmas 2.2 and 2.3.

Subcase 2.2.a finite complex number.

By Hurwitz’s theorem we deduce that g(k)(ζ)b for ζα,and g(k)(α)=b.This implies that g(k)(ζ)? b has only one zero ζ= α,contradicting with Lemmas 2.2 and 2.3 again.

Hence F is normal at z0and so F is normal in D.The proof of Theorem 1.1 is completed.

4 Proof of Theorem 1.2

Suppose that F is not normal in D.Then there exists at least one z0∈D such that F is not normal at the point z0.Without loss of generality,we can assume that z0=0.Thus from Lemma 2.1 we can find

(i)points zj∈D,zj→z0;

(ii)functions fj∈F,and

(iii) positive numbers ρj→ 0 such that

locally uniformly with respect to the spherical metric,where g is a nonconstant meromorphic function on C.Moreover,all poles of g are of multiplicity at least 2,and all zeros of g are of multiplicity at least k+1.

Now,from(4.1)we have

uniformly on compact subsets of C disjoint from the poles of g.

Proceeding as in the proof in Theorem 1.1,we can prove that all zeros of g(k)(ζ)?a are of multiplicity at least 2.

We claim that g(k)(ζ)? b has at most 1 zero ignoring multiplicity.

Suppose that g(k)(ζ)? b has at least 2 distinct zero ζ1and ζ2.Clearly,g(k)(ζ)b,for otherwise g would be a polynomial of degree at most k,which contradicts the fact that all zeros of g are of multiplicity at least k+1.Then by(4.2)and Hurwitz’s theorem there exist ζj,1,ζj,2,ζj,1→ ζ1,ζj,2→ ζ2,such that,for j sufficiently large,

Since f(k)j(z)? b has at most 1 zero in D ignoring multiplicity and zj+ ρjζj,1→ z0,zj+ρjζj,2→ z0,it follows that

Then ζj,1= ζj,2and so ζ1= ζ2.This is a contradiction and the claim thus is proved.

However,from Lemmas 2.2 and 2.3,we see that there do not exist nonconstant meromorphic functions that have the above properties.This contradiction shows that F is normal in D,and therefore the proof of Theorem 1.2 is completed.

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