The Value Distribution and Normality Criteria of a Class of Meromorphic Functions

YANG QI

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

The Value Distribution and Normality Criteria of a Class of Meromorphic Functions

YANG QI

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

Communicated by Ji You-qing

In this article,we use Zalcman Lemma to investigate the normal family of meromorphic functions concerning shared values,which improves some earlier related results.

meromorphic function,shared value,normal criterion

1 Introduction and Main Results

Let D be a domain of the open complex planeC,f(z)and g(z)be two nonconstant meromorphic functions defned in D,a be a fnite complex value.We say that f and g share a CM(or IM)in D provided that f−a and g−a have the same zeros counting(or ignoring) multiplicity in D.When a=∞,the zeros of f−a means the poles of f(see[1]).It is assumed that the reader is familiar with the standard notations and the basic results of Nevanlinna’s value-distribution theory(see[2]–[4]).

It is also interesting to fnd normality criteria from the point of view of shared values.In this area,Schwick[5]frst proved an interesting result that a family of meromorphic functions in a domain is normal if in which every function shares three distinct fnite complex numbers with its frst derivative.And later,more results about shared values’normality criteria related a Hayma conjecture of higher derivative have emerged(see[6]–[13]).

Lately,Chen[14]proved the following theorems.

Theorem 1.2Let D be a domain inCand let F be a family of meromorphic functions in D.Let k∈N+and a,b be two fnite complex numbers with a/=0.Suppose that every f∈F has all its zeros of multiplicity at least k+1 and all its poles of multiplicity at least k+2.If f(k)−af2and g(k)−ag2share the value b IM for every pair of functions(f,g)of F,then F is a normal family in D.

A natural problem arises:what can we say if f(k)−afnin Theorem 1.1 is replaced by the(f(k))m−afn?In this paper,we prove the following results.

Theorem 1.4Let D be a domain inCand let F be a family of meromorphic functions in D.Let k,m∈N+and a,b be two fnite complex numbers with a/=0.Suppose that every f∈F has all its zeros of multiplicity at least k+1 and all its poles of multiplicity at least mk+2.If(f(k))m−afm+1and(g(k))m−agm+1share the value b IM for every pair of functions(f,g)of F,then F is a normal family in D.

2 Some Lemmas

Lemma 2.1[15]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＜p.Then F is not normal at 0 if and only if there exist

a)a number 0＜r＜1;

b)points znwith|zn|＜r;

c)functions fn∈F;

d)positive numbers ρn→0

where

possibly outside a set with fnite linear measure.

Proof. Set

Since f(k)(z)/≡0,we have Φ(z)/≡0.Thus

Hence

So that

On the other hand,(2.2)gives

By(2.2),we have

From(2.3)–(2.6),we obtain

Since all zeros and poles of f are multiplicities at least k and d respectively,we get

So that

This completes the proof of Lemma 2.2.

Proof. Suppose to the contrary that(f(k))m−afnhas at most one zero.Since f/=0,we get f is a rational but not a polynomial.

Case 1.If(f(k))m−afnhas only zero z0with multiplicity l,then we set

where A is a nonzero constant and

For the sake of simplicity,we denote

From(2.7),we have

where g(z)is a polynomial such that deg(g(z))≤k(t−1).

From(2.7)and(2.8),we get

By the assumption that(f(k))m−afnhas exactly one zero z0with multiply l,we have

where C is a nonzero constant.Thus

Diferentiating(2.10),we obtain

For the sake of simplicity,we denote

Hence

Since(n−m)βi−mk−1＞0,we have

But g1(zi)/=0(i=1,2,···,t),a contradiction.

Case 2.If(f(k))m−afnhas no zeros,then l=0 for(2.9).We have

where C is a nonzero constant.Thus

This completes the proof of Lemma 2.3.

Proof. Suppose to the contrary that(f(k))m−afnhas at most one zero.

Case I.When f is a non-constant polynomial,noting that all zeros of f have multiplicity at least k+1,we know that(f(k))m−afnmust have zeros.We claim that f has exactly one zero.Otherwise,combing with the conditions of Lemma 2.4,we can get(f(k))m−afnhas at least two zeros,which contradicts with our assumption.

Set

where s≥k+1,B is a nonzero constant.Then

Since(s−k)m≥1,we obtain that sm(s−1)m···(s−k+1)m−aBn−m(z−z0)(n−m)s+mkhas least one zero which is not z0from(2.11).Therefore,(f(k))m−afnhas at least two distinct zeros,a contradiction.

Case II.When f is rational but not a polynomial,we consider two cases.

Case 1.Suppose that(f(k))m−afnhas only zero z0with multiplicity at least l.If f/=0, by Lemma 2.3,we get a contradiction.So f has zeros,and then we can deduce that z0is the only zero of f.Otherwise,(f(k))m−afnhas at least two distinct zeros,a contradiction.

We set

For the sake of simplicity,we denote

From(2.12),we have

where g(z)is a polynomial with deg(g)≤kt.

From(2.12)and(2.13),we get

By the assumption that(f(k))m−afnhas exactly one zero z0with multiply l,we have

where B is a nonzero constant.Thus

Case 1.2.If l=m(s−k),from(2.14),it follows that

Diferentiating(2.15),we have

For the sake of simplicity,we denote

Thus

Since(n−m)βi−mk−1＞0,we get

But g2(zi)/=0(i=1,2,···,t),a contradiction.

Case 2.If(f(k))m−afnhas no zeros,then f has no zeros.It is a contradiction with Lemma 2.3.

This completes the proof of Lemma 2.4.

Lemma 2.5Let f(z)be a transcendental meromorphic function,and let k,m∈N+and c∈C{0}.If all zeros of f are of multiplicity at least k+1 and all poles of f are of multiplicity at least mk+2,then(f(k))m−cfm+1has infnitely many zeros.

Proof. Suppose that(f(k))m−cfm+1has only fnitely many zeros.Then

Clearly,an arbitrary zero of f is a zero of(f(k))m−cfm+1.Since all zeros of f are of multiplicity at least k+1,we can deduce that f has only fnitely zeros,and so

Set

Similarly with the proof of Lemma 2.2,we can get

Since all poles of f are multiplicities at least mk+2,we obtain

So that

This contradicts with f is transcendental.

This completes the proof of Lemma 2.5.

Similarly to the proofs of Lemmas 2.3 and 2.4,we can get the following Lemmas.

f are of multiplicity at least d,then(f(k))m−afnhas at least two distinct zeros.

3 Proofs of Theorems

Proof Theorem 1.3Suppose that F is not normal in D.Then there exists at least one point z0such that F is not normal at the point z0.Without loss of generality we assume that z0=0.By Lemma 2.1,there exist points zj→0,positive numbers ρj→0 and functions fj∈F such that

locally uniformly with respect to the spherical metric,where g is a non-constant meromorphic function inCand whose poles and zeros are of multiplicity at least d and k+1,respectively. Moreover,the order of g is at most 2.

From(3.1)we know that

and

also locally uniformly with respect to the spherical metric.

If(g(k)(ξ))m−agn(ξ)≡0,since all poles of g have multiplicity at least d,we have

By Lemma 2.2,we have

Then

If(g(k)(ξ))m−agn(ξ)/=0,then(3.3)gives that g(ξ)is also a constant.Hence,(g(k)(ξ))m−agn(ξ)is a non-constant meromorphic function and has at least one zero.

Next we prove that(g(k)(ξ))m−agn(ξ)has just a unique zero.Suppose to the contrary, let ξ0and ξ∗0be two distinct zeros of(g(k)(ξ))m−agn(ξ),and choose δ(＞0)small enough such that

where

From(3.2)and by Hurwitz’s theorem,there exist points ξj∈D(ξ0,δ),∈D(,δ)such that for sufciently large j,

By the hypothesis that for each pair of functions f and g in F,(f(k)(ξ))m−afn(ξ)and (g(k)(ξ))m−agn(ξ)share b in D,we know that for any positive integer t,

Fix t,take j→∞,and note zj+ρjξj→0,zj+ρjξ∗j→0,then

Hence

Noting that g has poles and zeros of multiplicities at least d and k+1,respectively,(3.3) deduces that g(ξ)is a rational function with degree at most 2.By Lemmas 2.3 and 2.4,this is a contradiction.

This completes the proof of Theorem 1.3.

Proof Theorem 1.4Suppose that F is not normal in D.Then there exists at least one point z0such that F is not normal at the point z0.Without loss of generality we assume that z0=0.By Lemma 2.1,there exist points zj→0,positive numbers ρj→0 and functions fj∈F such that

locally uniformly with respect to the spherical metric,where g is a non-constant meromorphic function inCand whose poles and zeros are of multiplicity at least mk+2 and k+1, respectively.Moreover,the order of g is at most 2.

From(3.4)we know that

and

also locally uniformly with respect to the spherical metric.

If(g(k)(ξ))m−agm+1(ξ)≡0,since all poles of g have multiplicity at least mk+2,we can deduce that g(ξ)is an entire function easily.Thus

Therefore,g(ξ)is a constant,a contradiction.So

By Lemmas 2.5,2.6 and 2.7,(g(k)(ξ))m−agm+1(ξ)has at least two distinct zeros.Proceeding as in the later proof of Theorem 1.3,we will get a contradiction.The proof is completed.

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A

1674-5647(2017)01-0053-11

10.13447/j.1674-5647.2017.01.06

Received date:Oct.29,2015.

Foundation item:The NSF(2016D01A059)of Xinjiang.

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

2010 MR subject classifcation:30D35,30D45