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R E S E A R C H

Open Access

On zeros and deficiencies of differences of

meromorphic functions

Jianren Long

1,2*

, Pengcheng Wu

2

and Jun Zhu

3

*Correspondence:

[email protected] 1School of Mathematical Sciences,

Xiamen University, Xiamen, 361005, P.R. China

2School of Mathematics and Computer Science, Guizhou Normal University, Guiyang, 550001, P.R. China

Full list of author information is available at the end of the article

Abstract

For a transcendental entire functionf(z) in the complex plane, we study its divided differencesGn(z). We partially prove a conjecture posed by Bergweiler and Langley under the additional condition that the lower order off(z) is smaller than12. Furthermore, we prove that if zero is a deficient value off(z), then

δ

(0,G) < 1, where

G(z) = (f(z+c) –f(z))/f(z).

MSC: 30D35

Keywords: complex difference; zero; deficiency

1 Introduction and main results

In this paper, we assume that the reader is familiar with the standard notations of Nevan-linna theory of meromorphic functions (see [, ] or []). In particular, for a meromor-phic functionf(z) in the complex planeC, we useρ(f) andμ(f) to denote its order and lower order respectively, andλ(f) to denote the exponent of the convergence of the zero-sequences.

Letf(z) be a transcendental meromorphic function inC. The forward differencesnf(z)

are defined in the standard way by

f(z) =f(z+ ) –f(z), n+f(z) =nf(z+ ) –nf(z), n= , , . . . . (.) The divided differences are defined by

G(z) =

f(z)

f(z) , Gn(z) =

nf(z)

f(z) , n= , , . . . . (.)

Recently, a number of papers including [–] have focused on the complex difference equations and differences. In [] Bergweiler and Langley investigated the existence of zeros off(z) andG(z). Their result may be viewed as discrete analogs of the following theorem on the zeros off(z).

Theorem A([–]) Let f(z)be transcendental and meromorphic in the complex plane with

lim inf

r→∞ T(r,f)

r = . (.)

Then fhas infinitely many zeros.

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Theorem A is sharp, as shown byez,tanzand examples of arbitrary order greater than 

constructed in []. Forf(z) as in the hypotheses of Theorem A, it follows from Hurwitz’s theorem that, ifzis a zero off(z) thenf(z+c) –f(z) has a zero nearzfor all sufficiently smallc∈C\{}. Thus it is natural to ask, for such functionsf(z), whetherf(z+c) –f(z) must always have infinitely many zeros or not. In [], Bergweiler and Langley answered this problem and obtained Theorem B and Theorem C.

Theorem B([]) Let f(z)be a transcendental entire function of orderρ(f) <.If Gn(z) defined by(.)is transcendental,then Gn(z)has infinitely many zeros.In particular, if

ρ(f) <min{n,},then Gn(z)is transcendental and has infinitely many zeros.

Remark  Recently, Theorem B was extended by Langley to the caseρ(f) = in []. We only state the following situation which was proved in [].

Theorem C([]) There existsδ∈(,)with the following property.Let f(z)be a

tran-scendental entire function with orderρ(f)≤ρ<+δ< .Then G(z)defined by(.)has

infinitely many zeros.

In [], Bergweiler and Langley conjecture that the conclusion of Theorem C holds for

ρ(f) < . In this paper, we will prove this conjecture under the additional condition that

μ(f) <.

Theorem  Let nNand let f be a transcendental entire function of orderρ(f) < .If

μ(f) < andμ(f)=n,n, . . . ,[ n

]

n ,then Gn(z)defined by(.)is transcendental and has in-finitely many zeros.

Using Theorem , we easily obtain the following corollary.

Corollary  Let nNand let f be a transcendental entire function of orderρ(f) < .If

μ(f) <min{,n},then Gn(z)is transcendental and has infinitely many zeros.

In [], Bergweiler and Langley also proved that, for a transcendental meromorphicf(z) of orderρ(f) < , iff(z) has finitely many poleszj,zksuch thatzjzk=c, theng(z) =f(z+ c) –f(z) has infinitely many zeros (see [], Theorem .). Furthermore, for a transcendental entire functionf of orderρ(f) < , Chen and Shon proved thatλ(g) =ρ(g) =ρ(f), and if

f(z) has finitely many zeroszj,zksuch thatzjzk=c, thenG(z) =g(z)/f(z) has infinitely

many zeros andλ(G) =ρ(G) =ρ(f) (see []). This result implies that zero is not the Borel exceptional value ofG(z).

In [], Langley investigated the deficiency of divided differenceG(z) defined by (.). He obtained that iff(z) is a transcendental entire function of orderρ(f) <  andμ(G) <, thenδ(,G) < . In particular, ifρ(f) < , thenδ(,G) <  (see [], Theorem .). The proof of his result depends oncosπρtheorem which is invalid forμ(G)≥.

In this paper, we consider some more general cases. Forc∈C\{}, we define

g(z) =f(z+c) –f(z), G(z) =g(z)

f(z). (.)

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Theorem  Let f(z)be a transcendental entire function of orderρ(f) < .Suppose that f(z)

has at most finitely many zeros zj,zk such that zjzk=c.If G(z)is defined by(.),then the following two statements hold:

(i) Ifδ(,G) = ,then there exists a setE⊂(,∞)of positive upper logarithmic density such thatm(r,f) =o(logM(r,f)),asr→ ∞,rE,whereM(r,f) =max|z|=r|f(z)|.

(ii) If zero is a deficient value off(z),thenδ(,G) < .

It is clear that, for a given transcendental entire functionf(z), all but countably many

c∈Csuch thatf(z) has at most finitely many zeroszj,zksuch thatzjzk=c. Furthermore,

we know that, for an entire functionf(z), iff(z) has a finite deficient value thenμ(f) >. Hence, Theorem  implies that, for some particular functionsf(z) of orderρ(f) > , we obtain a similar conclusion.

Example There is an example for Theorem . Let <μ< . Set

f(z) =

k=

 + z

.

Then μ(f) =ρ(f) =μandδ(,f) =  –sinμπ>  (see [, p.]). If we letc= , then it follows from Theorem  thatδ(,G) < .

The paper is organized as follows. In Section , we shall collect some notations and give some lemmas which will be used later. In Section , we shall prove Theorem . In Section , we shall prove Theorem .

2 Preliminaries and lemmas

Letf(z) =∞k=akzkbe an entire function in the complex plane. Then (see [, p.]) we

have

ρ(f) =lim sup

r→∞

log logM(r)

logr =lim supr→∞

log logμ(r)

logr =lim supr→∞

logν(r)

logr

and

μ(f) =lim inf

r→∞

log logM(r)

logr =lim infr→∞

log logμ(r)

logr =lim infr→∞

logν(r)

logr ,

whereM(r) =max{|f(z)|:|z|=r},μ(r) =max≤n<∞|an|rnis the maximum term andν(r) =

max{m:|am|rm=μ(r)}is the central index. It is well known thatν(r) is a nondecreasing

and right continuous function. Furthermore, iff(z) is transcendental entire, thenν(r)→

∞asr→ ∞.

For a setE⊂[,∞), we define its Lebesgue measure bym(E) and its logarithmic measure byml(E) =

E dt

t .

We also define the upper and lower logarithmic density ofE⊂[,∞), respectively, by

logdensE= lim

r→∞

ml(E∩[,r])

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and

logdensE= lim

r→∞

ml(E∩[,r])

logr .

Following Hayman [, pp.-], we say that a setEis anε-set ifEis a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. IfEis anε-set, then the set ofr≥ for which the circleS(,r) meets has finite logarithmic measure and hence zero upper logarithmic density. Moreover, for almost all realθ, the intersection ofEwith the rayargz=θis a bounded set.

The following lemma contains a basic property of meromorphic functions of finite order.

Lemma .([]) Let f(z)be a meromorphic function withρ(f) <∞.Then,for given real constants c> and H (>ρ(f)),there exists a set E⊂(,∞)such thatlogdensE≥ –ρH(f),

where E={t|T(tec,f)ekT(r,f)}and k=cH.

The following lemma is a version of the celebratedcosπρtheorem of [].

Lemma . ([]) Let f(z) be a transcendental entire function with lower order ≤

μ(f) < .Then,for eachα∈(μ(f), ),there exists a set E⊂[,∞)such thatlogdensE

 – μα(f), where E={r∈[,∞) :A(r) >B(r)cosπ α}, A(r) =inf|z|=rlog|f(z)|, and B(r) =

sup|z|=rlog|f(z)|.

We collect some important properties of the differences of meromorphic functions in the following lemmas.

Lemma .([]) Let f(z)be a transcendental meromorphic function inCwhich satisfies

(.).Then,with the notation(.)and(.),f(z)and G(z)are both transcendental.

Lemma .([]) Let f(z)be a meromorphic function of finite orderρ(f) =ρand let c be a non-zero finite complex number.Then,for eachε> ,we have

m

r,f(z+c)

f(z)

+m

r, f(z)

f(z+c)

=Orρ–+ε (.)

and

Tr,f(z+c)=T(r,f) +Orρ–+ε+O(logr). (.)

Lemma .([]) Let nNand let f(z)be a transcendental meromorphic function of order smaller thanin the complex planeC.Then there exists anε-set Ensuch that

nf(z)∼f(n)(z), z→ ∞,z∈C\En. (.)

3 Proof of Theorem 1

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Lemma . Let T(r) (> )be a nonconstant increasing function of finite lower orderμin r∈(,∞),i.e.

lim inf

r→∞

logT(r)

logr =μ<∞. (.)

For any givenμ> such thatμ<μ,define

E(μ) = r≥ :T(r) <

.

ThenlogdensE(μ) > .

Proof of Theorem  Since f is a transcendental entire function of order ρ(f) < , by Lemma ., we know that there exists anε-setEn, such that

nf(z)∼f(n)(z), z→ ∞,z∈C\En. (.)

From the Wiman-Valiron theory ([] or []), there is a subsetF⊂[,∞) withml(F) <∞

such that for allzsatisfying|z|=r∈/Fand|f(z)|=M(r,f),

f(n)(z)

f(z) =

ν(r)

z

n

 +o(), |z|=r→ ∞. (.)

SetH={|z|=r:zEn}. We know thatml(H) <∞. Therefore, by (.) and (.), for allz

satisfying|z|=r∈/(FH) and|f(z)|=M(r,f), we get

Gn(z) =

ν(r)

z

n

 +o(), |z|=r→ ∞. (.)

Now we divide the proof into two steps. Firstly, we prove thatGn(z) is transcendental. To

do this, we assume contrarily thatGn(z) is a rational function and seek a contradiction. By

using Lemma . toν(r) with lower orderμ(f), we see that, for any givenε,  <ε<  –μ(f), there exists a sequence{tj} ∈E(ε)\(FH) such that the following inequalities:

tj(μ(f)––ε)n< ν

(tj) tj

n

<tj(μ(f)–+ε)n (.)

hold for all sufficiently largej, whereE(ε) ={r≥ :ν(r) <(f)+ε}. Since (μ(f) –  +ε)n< 

andGn(z) is a rational function, we deduce from (.) and (.) that

Gn(z) =βzk

 +o(), z→ ∞, (.)

whereβ(= ) is a constant andkis a positive integer. By using (.), (.) and (.), we get

tj((f)+kn)<|β| +o()<tj((f)+kn+), tj→ ∞.

Sinceεcan be arbitrary small, we must have(f) +kn= ,i.e.,

μ(f) =  –k

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This contradicts the assumption thatμ(f)=n,n, . . . ,[ n

]

n . ThusGn(z) must be

transcen-dental.

In the second step, we prove that Gn(z) has infinitely many zeros. If this is not true,

assume thatGn(z) has only finitely many zeros. Then /Gn(z) is also transcendental having

finitely many poles. By using (.) and the Jensen formula, for  <ε<  –ρ(f), we have

T(r,Gn)≤(n+ )T(r,f) +o() +O(logr), r→ ∞. (.)

Thus, It follows from (.) thatμ(Gn) <

. Using Lemma . to 

Gn, we deduce that there exists a subsetF⊂[,∞) withlogdensF>  such that

log|Gn|< –cT(r,Gn) < –nlogr, |z|=rF, (.)

wherecis a positive real number. Hence, by (.) and (.), we get

Gn(z)|z|n=ν(r)n +o()< , |z|=rF

\(FH). (.)

Sinceν(r)→ ∞asr→ ∞, (.) gives a contradiction. Therefore,Gn(z) must have

in-finitely many zeros and the proof of Theorem  is completed.

4 Proof of Theorem 2

To prove Theorem , we first prove the following lemma.

Lemma . Let f(z)be a transcendental entire function of orderρ(f) < .Suppose that f(z)has at most finitely many zeros zk,zjsatisfying zkzj=c and the deficiencyδ(,G) = . Then,for any given <ε<  –ρ(f),there exists a constant r(ε) (> )such that the following

inequalities:

( –ε)T(r,G)≤m(r,f) –m

r,

f

≤( +ε)T(r,G) (.)

hold for all rr(ε).

Proof Letε>  such thatρ(f) +ε< . Using (.), we get

m

r,f(z+c)

f(z)

=O(), r→ ∞. (.)

Hence, it follows from (.), (.) and the Jensen formula that

m(r,G) =O(), T(r,G) =N(r,G) +O(), r→ ∞. (.)

So, by (.) and the Jensen formula, we get

T(r,G)≤N

r,

f

+O(), T(r,G)≤m(r,f) –m

r,

f

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It follows from Theorem  in [] thatG(z) is transcendental. Hence, we obtain

On the other hand, from our hypothesis, we can easily deduce that

n

Hence, by using (.), (.) and the Jensen formula, we see that the following inequalities

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By using the standard estimates for entire functions (see [] or []), we have

logm(r,g)≤logm(r,g)≤logM(r,g)≤logM(r,g), (.)

logm(r,f) +logM(r,f)≤logm(r,f) +logM(r,f), (.)

and

logM(r,g) =r

N(t,g)

(r+t)dt, logM(r,f) =r

N(t,f)

(r+t)dt. (.)

Hence, by (.), (.) and (.), we see that

logM(r,g) =ologM(r,f)

, r→ ∞. (.)

To finish the proof of (i), we need to consider the following two cases.

Case .μ(f) <. By Lemma ., there exists a setE⊂(,∞) withlogdensE≥ –μα(f),

α=

μ(f)+

 , such that

cosπ αlogM(r,f) <logm(r,f) (.)

holds for allrE. SetE=E. It follows from (.) and (.) that

logM(r,f)≤logM(r,f), rE. (.)

Case . μ(f) < . By Lemma ., there exists a setE⊂(,∞) withlogdensE≥  –μ(f)

α such that

–logm(r,f) <clogM(r,f), rE, (.)

where  <c= –cosπ α<  andα=μ(f)+. SetE=E. It follows from (.) and (.) that

logM(r,f)≤ 

 –clogM(r,f), rE. (.)

Hence, by using (.), (.) and (.), we deduce that there exists a setE⊂(,∞) with

logdensE> , such that

logM(r,g) =ologM(r,f), r→ ∞,rE. (.)

Using (.) and noting thatT(r,f) –T(r,g) –O()≤T(r,G), we see that the following inequalities

m

r,

f

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hold for allrr(ε). Hence, there exists a constantr(ε) >r(ε) satisfyingM(r,f) >  and

M(r,g) > . By using (.) and (.), we deduce that

m

r,

f

– ( –ε)O()≤εlogM(r,f) + ( –ε)ologM(r,f) (.)

hold forrEandrr(ε). Asεcan be arbitrarily small, we get

m

r,

f

=ologM(r,f), r→ ∞,rE. (.)

This gives (i).

In order to prove (ii), we suppose contrarily thatδ(,G) =  and seek a contradiction. From (i), we know that there exists a setE⊂(,∞) withlogdensE>  satisfying (.). Assume that the zero is a deficient value off(z) with deficiencyδ(,f) =δ> . It follows from definition of deficiency, we easily find that, for all sufficiently larger

m

r,

f

>δ

m(r,f). (.)

By (.) and (.), we have

T(r,f) =m(r,f) =ologM(r,f), r→ ∞,rE. (.)

Obviously, for an entire functionf(z), the following inequalities (see [] or [])

T(r,f)≤logM(r,f)≤T(r,f) (.)

hold for all sufficiently larger. Set

E= t|T(t,f)≤HT(r,f)

. (.)

By using Lemma ., we have

logdensE≥ –

ρ(f)

H ,

whereH= (logdensE)–ρ(f) +  >ρ(f). SetE=EE. By a simple computation, we can get

logdensE≥logdensE

ρ(f)

H > .

Hence, by (.), (.) and (.), we obtain

≤Ho(), r→ ∞,rE

. (.)

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Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the manuscript. All authors read and approved the final manuscript.

Author details

1School of Mathematical Sciences, Xiamen University, Xiamen, 361005, P.R. China.2School of Mathematics and Computer

Science, Guizhou Normal University, Guiyang, 550001, P.R. China. 3School of Mathematics Science, Peking University, Beijing, 100871, P.R. China.

Acknowledgements

The authors would like to thank the referee for his/her valuable suggestions, which greatly improved the present article. The authors wish to thank professor Wu Shengjian in Peking University for valuable suggestions and numerous important helps. The work was supported by the United Technology Foundation of Science and Technology Department of Guizhou Province and Guizhou Normal University (Grant No. LKS[2012]12), and National Natural Science Foundation of China (Grant No. 11171080).

Received: 22 January 2014 Accepted: 31 March 2014 Published:06 May 2014 References

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