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