Normal Families of Holomorphic Functions Concerning Zero Numbers

YANG QI

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

Communicated by Ji You-qing

1 Introduction and Main Results

Let F be a meromorphic function in C,and D be a domain in C.F is said to be normal in D if any sequence{fn}⊂F contains a subsequence fnjsuch that fnjconverges spherically locally uniformly in D,to a meromorphic function or ∞ (see[1]–[3]).

In 1959,Hayman[4]proved the following result.

Theorem 1.1Letfbe a meromorphic function inC,n≥5be a positive integer,anda(0),bbe two finite constants.Iff′− afnb,thenfis a constant.

The following normality criterion corresponding to Hayman’s result was proved by Drasin[5]and Ye[6].

Theorem 1.2Letn≥2be a positive integer,a(0),bbe two finite constants,andFbe a family of Holomorphic functions in a domainD.If for eachf ∈ F,f′−afnb,thenFis normal inD.

Recently,by the idea of concerning zero numbers,Denget al.[7]proved the following result.

Theorem 1.3Letm,n,kbe three positive integers satisfyingn≥m+1,a(0),bbe two finite constants,andFbe a family of Holomorphic functions in a domainD,all of whosezeros have multiplicity at leastk.If for each functionf∈F,f(k)−afn−bhas at mostmkdistinct zeros inD,thenFis normal inD.

A natural problem arises:what can we say if f(k)in Theorem 1.3 is replaced by the(f(k))d?In this paper,we prove the following result.

Theorem 1.4Letm,n,k,dbe four positive integers satisfyingn≥(m+1)d,a(0),bbe two finite constants,andFbe a family of holomorphic functions in a domainD,all of whose zeros have multiplicity at leastk.If for each functionf∈F,(f(k))d−afn−bhas at mostmdkdistinct zeros inD,thenFis normal inD.

Example 1.1Let n,k,d be three positive integers,a be a nonzero finite constant,and F={fj=jzk−1:j=1,2,3,···},D={z:|z|＜ 1}.Then,for each f ∈ F,(f(k))d−afn−0 has just one distinct zero in D,but F is not normal in D.This shows that the zeros of function f∈F have multiplicity at least k is necessary in Theorem 1.4.

Example 1.2Let n,k,d be three positive integers,a be a nonzero finite constant,and F={fj=jzk:j=1,2,3,···},D={z:|z|＜ 1}.Then,for each f ∈ F,(f(k))d−af(m+1)d−1−0 has exactly[(m+1)d−1]k ≥ mdk distinct zero in D,and(f(k))d−af(m+1)d−0 has exactly(m+1)dk≥mdk+1,but F is not normal in D.This shows that both n≥(m+1)d and(f(k))d−afn−b have at most mdk distinct zeros in Theorem 1.4 are best possible.

2Some Lemmas

In order to prove our theorems,we require the following results.

Lemma 2.1[8]LetFbe a family of meromorphic functions on the unit disc Δ satisfying all zeros of functions inFhave multiplicity≥pand all poles of functions inFhave multiplicity≥ q.Letαbe a real number satisfying−p＜ α＜ q.ThenFis not normal at a pointz0if and only if there exist

(i)pointszn∈Δ,zn→z0;

(ii)positive numbersρn,ρn→ 0;

(iii)functionsfn∈F

such thatfn(zn+ ρnζ)→ g(ζ)spherically uniformly on each compact subset ofC,whereg(ζ)is a nonconstant meromorphic function satisfying the zeros ofgare of multiplicities≥pand the poles ofgare of multiplicities≥q.Moreover,the order ofgis at most2.Ifgis holomorphic,thengis of exponential type and the order ofgis at most1.

Lemma 2.2[9]Letfbe a nonconstant meromorphic(entire)function in the complex plane,a(0)be a finite constant,andnbe a positive integer withn≥4(n≥2).Thenf′−afnhas at least two distinct zeros.

Lemma 2.3Takefbe an entire function,leta(/0)be a finite constant,andn,m,d,kbe four positive integers satisfyingn≥(m+1)dandmdk≥2.If all zeros offhave multiplicity at leastk,then

Proof.Set

Since all zeros of f have multiplicity at least k,it follows f(k)(z)0.Otherwise,f is a polynomial with degf ≤ k − 1,thus f is a constant,a contradiction.HenceΨ(z)0.

By(2.1),we have

Thus,we get

it follows that

On the other hand,by(2.2),we have

whereis the counting function of zeros of bothΨand f(k).It follows that

By(2.2),we have

andΨ(z)1.Otherwise,we get

Because f is an entire function,we have

Then

Thus

a contradiction.HenceΨ(z)1.

Thus by(2.3)–(2.5)and Nevanlinna’s first and second fundamental theorems,we have

Since

it follows from(2.6)–(2.8)that

Lemma 2.3 is proved.

3 Proof of Theorem 1.4

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

spherically uniformly on compact subsets C,where g is a nonconstant holomorphic function on C and whose zeros have multiplicity at least k.Moreover,the order of g is at most 1.Obviously,(g(k)(ξ))d− agn(ξ)0.Suppose not,since g is an entire function,we have

Then

So g is a constant,a contradiction.

We claim that(g(k)(ξ))d−agn(ξ)has at most mdk distinct zeros.Suppose that(g(k)(ξ))d−agn(ξ)has mdk+1 distinct zeros ξi(i=1,2,···,mdk+1).We have

uniformly on compact subsets of C.By Hurwitz’s theorem,for j sufficiently large,there exist points ξj,i(i=1,2,···,mdk+1)such that ξj,i→ ξiand

However,− b have at most mdk distinct zeros in D,and zj+ ρjξj,i→ z0,which is a contradiction.So we can get(g(k)(ξ))d− agn(ξ)has at most mdk distinct zeros.If mdk=1,we get m=1,d=1,k=1,and this is a contradiction with Lemma 2.2.

Next,we consider the case mdk≥2.

Suppose that(g(k))d−agnhas l(≤ mdk)distinct zeros.Since g is an entire function all zeros of g have multiplicity at least k,by Lemma 2.3,we can obtain

Considering n≥(m+1)d,by(3.1),we get

Thus,we deduce that g is a nonconstant rational function satisfying

Next we consider two cases.

Case 1md≥2 and k≥1.

We consider two subcases again.

Case 1.1md≥2 and k=1.

By(3.2),we get

Next we consider two subcases again.

Case 1.1.1md=2.

In this case we have

We deduce that degg≤2.Next we consider two subcases again.

Case 1.1.1.1degg=1.

We can write g(ξ)=Aξ+B,where A0.So,

Obviously,(g′)d−agnhas at least n ≥ (m+1)d=mdk+d≥ mdk+1 distinct zeros,this is a contradiction.

Case 1.1.1.2degg=2.

We can write g(ξ)=B1(ξ− α1)(ξ− α2),where B10.

When α1= α2,we have

By(3.1)and noticing that md=2,k=1,we have

So,we know

this is a contradiction.

If α1α2,then we are supposed to notice that n ≥ (m+1)d=2+d.If n ≥ 3+d,then by(3.1)and noticing that md=2,k=1,we get

which is a contradiction.

If n=2+d,then g(ξ)=B(ξ− α)(ξ− α),where B0 and αα.Set φ =

112112noticing that md=2,k=1,next we consider two subcases again.

Case 1.1.1.2.1m=2,d=1,k=1.

In this case,we get

Obviously,g′− ag3and φ′φ +a have the same zeros.

SetThen we get ψ′= φ′φ,and we obtain

whereis a non-zero constant.So we have

where q1(ξ)is a polynomial with degq1=1.

By Lemma 2.2,we get that g′− ag3has at least two distinct zeros,then φ′φ +a has at least two distinct zeros.Suppose that φ′φ +a has only two distinct zeros b1,b2,and set

where C is a non-zero constant,and l1+l2=6.So

where q2(ξ)is a polynomial with degq2≤ 2.

From(3.3),we also have

where q3(ξ)is a polynomial with degq3≤ 2.Noticing that α1,α2and b1,b2are distinct,from(3.4)and(3.5),we get l1+l2−2≤degq3≤2.Then l1+l2≤4,this contradicts with l1+l2=6.Then ψ′+a has at least three distinct zeros.So g′−ag3has at least 3=mdk+1 distinct zeros,this is a contradiction.

Case 1.1.1.2.2m=1,d=2,k=1.

In this case,we have

Obviously,(g′)2− ag4and(φ′)2− a have the same zeros.By a simple calculation,we get

where q4(ξ)=[2ξ− (α1+ α2)]2.Obviously,(φ′)2− a has at least two distinct zeros.Suppose that(φ′)2− a has only two distinct zeros b3,b4,and set

where C is a non-zero constant,and l3+l4=8.Then

where q5(ξ)is a polynomial with degq5≤ 2.From(3.6),we also have

where q6(ξ)is a polynomial with degq6≤ 3.Noticing that α1,α2and b3,b4are distinct,from(3.7)and(3.8),we get

i.e.,l3+l4≤ 5,this contradicts with l3+l4=8.Then(φ′)2− a has at least three distinct zeros,so(g′)2− ag4has at least 3=mdk+1 distinct zeros,this is a contradiction.

Case 1.1.2md≥3.

By(3.2),we obtain

So we get degg=1.As discussed in case 1.1.1.1,we get a contradiction.

Case 1.2md≥2,k≥2.

By(3.2),we get

Then we know that g has at most one zero since all zeros of g have multiplicity at least k.Assume that g(ξ)=A(ξ−c)l,where k ≤ l≤ k+1.Then,

Obviously,(g(k))d−agnhas nl−(l−k)d≥mdk+dk distinct zeros,this is a contradiction.

Case 2md=1,k≥2.

By(3.2),we get

Next,we consider two subcases.

Case 2.1k=2.

By(3.9),we get

So degg≤4.We divide into two subcases again.

Case 2.1.1degg=4.

We deduce that g has two distinct zeros at most since all zeros of g have multiplicity at least k=2.If g has only one zero,as discussed in Case 1.2,we get a contradiction.Then g has two distinct zeros.In the condition of n≥(m+1)d≥2,we discuss it in two conditions.If n≥3,then by(3.1)and noticing that md=1,k=2,we obtain

which contradicts with degg=4.If n=2,we can write g(ξ)=A(ξ− α1)2(ξ−α2)2,where A0 is a constant and α1α2,and set

where β2=A,then

Obviously,g′′− ag2and −2ψ′′ψ +6(ψ′)2− a have the same zeros.From(3.10),we have

where q7(ξ)= −B[2ξ−(α1+α2)],B=and

where q8(ξ)is a polynomial with degq8≤ 2.From(3.10)–(3.12),we get

where h(ξ)is a polynomial with degh ≤ 2.Obviously,ϕ−a has at least two distinct zeros.Suppose that ϕ−a has only two distinct zeros b1,b2,and set

where C is a non-zero constant,and l1+l2=8.So

where q9(ξ)is a polynomial with degq9≤ 2.From(3.13),we also have

where q10(ξ)is a polynomial with degq10≤ 3.Noticing that α1,α2and b1,b2are distinct,from(3.14)and(3.15),we have

i.e.,l1+l2≤ 5,this contradicts with l1+l2=8.Then ϕ−a has at least three distinct zeros,so g′′− ag2has at least 3=mdk+1 distinct zeros,this is a contradiction.

Case 2.1.2degg≤3.

We get that g has only one zero.As discussed in Case 1.2,we get a contradiction.

Case 2.2k＞2.

By(3.9),we get degg≤k+2.So we get that g has only one zero.As discussed in Case 1.2,we get a contradiction.Thus,we deduce that(g(k)(ξ))d− agn(ξ)has at least mdk+1 distinct zeros,this is a contradiction.This shows that F is normal at z0.Thus F is normal in D.

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