Uniqueness of Entire Functions Concerning Differences

(Institute of Applied Mathematics,South China Agricultural University,Guangzhou 510642,China)

Abstract:In this paper,we study the uniqueness of entire functions and prove the following theorem.Let f be a transcendental entire function of finite order.Then there exists at most one positive integer k,such that f(z)Δkcf(z)-R(z)has finitely many zeros,where R(z) is a non-vanishing rational function and c is a nonzero complex number.Our result is an improvement of the theorem given by Andasmas and Latreuch [1].

Keywords:Nevanlinna theory;Uniqueness;Entire functions;Difference operators

§1.Introduction and main results

Let C denote the complex plane andfbe a meromorphic function on C.In this paper,we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna value distribution theory such asT(r,f),m(r,f) andN(r,f) (see [9,15,16]).In addition,S(r,f)=o(T(r,f)),asr →∞outside of a possible exceptional set of finite linear measure.A meromorphic functionais called small function with respect tof,provided thatT(r,a)=S(r,f),S(f) denotes the set of all small functions offandρ(f) denotes the order of growth off.

For a meromorphic functionf(z),we define its difference operators by

In 1959,W.K.Hayman [8] proved that iffis a transcendental meromorphic function(entire function),nis an integer satisfyingn≥3 (n≥2) andcis a nonzero complex number,thenfnf′=chas infinitely many solutions.In 1967,J.Clunie [6] proved that the above result is also true whenn=1 for a transcendental entire function.Later on,Hayman [10] conjectured that the theorem remains to be valid whenn=2 andn=1 for a transcendental meromorphic function.Then,the casen=2 was settled by E.Mues [12] in 1979 and casen=1 was settled by Bergweiler-Eremenko [2] and Chen-Fang [3] in 1995.

In conclusion,we have the following result.

Theorem 1.1.Let f be a transcendental meromorphic function.If n is an integer satisfying n≥1and c is a nonzero complex number.Then fnf′=c has infinitely many solutions.

Up to now,this problem about the form offnf′has been resolved.Naturally,a similar question was raised:whether the form offf(k) have the same conclusion?

In 1996,Yang and Hu proved the following theorem.

Theorem 1.2.[14] Let f be a transcendental entire function of finite order.Then there exists at most one integer k ≥2,such that ff(k) has a nonzero and finite exceptional value.

In recent years,the value distribution of meromorphic functions with respect to difference analogue has become a subject of some interests,Andasmas and Latreuch proved the following theorem in 2020.

Theorem 1.3.[1] Let f(z)be a transcendental entire function of finite order,let c be a nonzero complex number and n,m(m>n)be two positive integers,such that,whereare two non-vanishing difference opeartors of f.Then,for all non-vanishing polynomial q(z),at least one ofhas infinitely many zeros.

In this paper,we study these problems further and prove the following result.

Theorem 1.4.Let f(z)be a transcendental entire function of finite order.Then there exists at most one positive integer k,such thathas finitely many zeros,where R(z)is a non-vanishing rational function and c is a nonzero complex number.

By Theorem 1.4,we have the following corollary,which is an improvement of Theorem 1.3.

Corollary 1.1.Let f(z)be a transcendental entire function of finite order,let c be a nonzero complex number and n,m(m>n)be two positive integers,such that,whereare two non-vanishing difference opeartors of f.Then,for all non-vanishing rational function R(z),at least one ofhas infinitely many zeros.

§2.Some lemmas

Lemma 2.1.[13] Let f be a meromorphic function,an(/=0),an−1,···,a0be constants,then

Lemma 2.2.[4,5] Let η1,η2be two arbitrary complex numbers such that η1/=η2and let f(z)be a finite order meromorphic function of order ρ.Then for each ε>0,we have

Lemma 2.3.[11] Suppose fj(j=1,2,···,n)are meromorphic functions and gk(k=1,2,···,n)(n≥2)are entire functions satisfying the following conditions:

(i),

(ii) gj −gk are not constants for1≤j

(iii) There exists a set E ⊂(1,+∞)that has linear measure or logarithmic measure.For1≤j ≤n,1≤h

Lemma 2.4.[7] Let f(z)be a transcendental meromorphic function of finite order ρ,and let ε>0be a given constant.Then,there exists a set E0⊂(0,+∞)that has finite logarithmic measure,such that for all z satisfying |z|/∈E0∪[0,1],and for all k,j(0≤j

Lemma 2.5.[9] Let f be a meromorphic function and let k ∈Z+.Then

wherepossibly outside a set E1⊂[0,+∞)of a finite linear measure.If f is a meromorphic function of finite order,then

Lemma 2.6.[4] Let f be a non-constant,finite order meromorphic solution of

where P(z,f),Q(z,f)are difference polynomials in f with meromorphic coefficients aj(j=1,2,...,s),and let δ<1.If the degree of Q(z,f)as a polynomial in f(z)and its shifts is at most n,then

Lemma 2.7.Let α(z),β(z)be two polynomial functions and R1(z),R2(z),T1(z),T2(z)be non-vanishing rational functions.Let f(z)be a finite order transcendental entire solution of the system

where c is a nonzero complex number,are two non-vanishing difference operators of f and n,m(n

Furthermore,for all rational functions di(i=1,2),which are not all vanishing,we have

Proof.Firstly,we prove thatdegα=ρ(f)>0.It is clear thatdegα≤ρ(f).Assume thatdegα<ρ(f).Then by (2.1),we have

By (2.2),we know that each zero offis the zero ofR1+T1eα.Hence,

By the First Fundamental Theorem,the inequality (2.4) and Lemma 2.1,Lemma 2.2,for eachε1>0,we have

which leads to a contradiction thatρ(f)≤max{ρ(f)−1+ε1,degα}.Similarly,we can deduce thatdegβ=ρ(f).Sincefis transcendental,we knowρ(f)>0.Therefore,

Next,we will provedeg(α+β)=ρ(f).By (2.5),we getdeg(α+β)≤ρ(f).Assume thatdeg(α+β)<ρ(f),which leads toeα+β ∈S(f).Combining (2.1) and (2.2),we have

Now we prove that.Assume the contrary,we get

which meansis an entire function.By Lemma 2.2,we have

By Lemma 2.6 to (2.7),we get

Therefore,we have

which is a contradiction.Applying Lemma 2.6 to (2.6),we get

Combining the above discussion,we have

which is a contradiction.Hence,we havedeg(α+β)=ρ(f).

Then we prove thatdeg(α−β)=ρ(f).By (2.5),we getdeg(α−β)≤ρ(f).Assume thatdeg(α−β)<ρ(f).Combining (2.1) and (2.2),we have

It is obvious thatR1T2−R2T1eα−β/≡0.Otherwise,assume that

which leads to

By (2.10) and (2.11),we have

which is a contradiction,with (2.1) and (2.2) is a equation system.By (2.9),we know that each zero offis the zero ofR1T2−R2T1eα−β.Hence,

By the First Fundamental Theorem,the inequality (2.12) and Lemma 2.1,Lemma 2.2,for eachε2>0,we have

which leads to a contradiction thatρ(f)≤max{ρ(f)−1+ε2,deg(α−β)}.Hencedeg(α−β)=ρ(f).

Finally,in order to prove that,whered1andd2are not all vanishing rational functions.Assume that.By (2.1) and (2.2),we have

Sincedegα=degβ=deg(α−β)=ρ(f)>0 and by Lemma 2.3,we obtaind1≡d2≡0,which is a contradiction.Thus,the Lemma is proved.

§3.Proof of Theorem 1.4

Proof.We will prove this theorem by contradiction.

Firstly,assume the contrary to our assertion that there exist two positive integersn,m,such that bothhave finitely many zeros,whereR1(z),R2(z) are two non-vanishing rational functions.Without loss of generality,we assume thatn

By differentiating (3.1) and eliminatingeα,

where.By Lemma 2.7,

Now,we prove thatA1/≡0.Assume thatA1≡0,then

The above result means that there exists a constantC1such thatT1eα=C1,which leads to the contradictionρ(f)=degα=0.By the same method,we can prove thatB1/≡0.As same as the above,we get

where.It is clear from (3.3) and (3.4) that each multiple zero offis the zero ofBi(i=1,2).Hence,

wheredenotes the counting function of zeros offwith multiplicity no less than 2.Letz0be the simple zero off.Substitutingz0into (3.3) and (3.4),we have

Combining (3.5) with (3.6),we obtain

Then,by Lemma 2.7,.By (3.7),we know the function

has no simple pole.By the above discussion,we have

On the other hand,by Lemma 2.2,for eachε3>0,we have

Hence,T(r,h)=N(r,h)+m(r,h)=S(r,f).Moreover,by (3.8),we have

Differentiating (3.9),we get

Substituting (3.9) and (3.10) into (3.3),we have

Combining the above equation to (3.11),we get

In the following,we will prove that

By Lemma 2.7,we get,and it is clear thatThen we assume to the contrary that.By the definition ofA1and simple integration,we get

whereC2is a nonzero constant.Thus,ρ(f)=degα≤ρ(h)≤ρ−1,which is a contradiction.

whereC3is a nonzero constant,which leads to the contradictionρ(f)=deg(α−β)=0.

By (3.12),we have

Substituting (3.13) into (3.9),we get

Differentiating (3.13),we have

Substituting (3.13) and (3.16) into (3.4),we get

Differentiating (3.17),we get

Letz0be the simple zero off.Substitutingz0into (3.17) and (3.18),z0is the zero ofHence,the function

has no simple pole.Using the same method as previous,we getT(r,H)=S(r,f).Then,

Substituting (3.19) into (3.17),we get

Then,we provev2/≡0.Assume the contrary and by (3.14),we get that

By simple integration the above equation,we get

whereC4is a nonzero constant,which leads to the contradictionρ(f)=deg(α+β)≤ρ(h).Sov2/≡0.Differentiating (3.20),we obtain

Letz0be the simple zero off.Substitutingz0into (3.20) and (3.22),we get thatz0is the zero ofHence,the function

has no simple pole.Using the same method as previous,we getT(r,U)=S(r,f).Then,

Substituting (3.23) into (3.22),we get

Multiplying (3.20) byand combining it with (3.24),we get

By (3.25) andh,H,U∈S(f),we deduce that

EliminatingUfrom (3.26) and (3.27),we obtain

Two cases will be discussed in the following.

Case 1..Dividing both sides of the equation (3.28) byv3(4v1v3−v22),we have

On the other hand,according to the definition ofv2andv3,we have

Combining (3.29) and (3.30),

According to the definition ofAi(i=1,2)and by simple integration,we deduce thatdeg(α+β)<ρ(f),which is a contradiction.

Case 2..By (3.26),we get,so the equation (3.23) can be rewritten as

Combining (3.31) and (3.19),we get

On the other hand,according to the definition ofv1,v3,we have

Combining (3.32) and (3.33),we obtain

Substitutinginto (3.34),we have

According to the definition ofv2,v3,we have

By simple calculation,we get

whereci(i=1,2,···,18) is a constant.By Lemma 2.4,for each 0<ε4<1 andk=1,2,we have

for allzsatisfying|z|/∈E0∪[0,1],whereE0⊂(0,+∞) is a set of finite logarithmic measure.

Dividing both sides of the equation (3.37) byNoting thatand,whereS(z) is a arbitrary nonzero rational function,we obtain

By the definition ofA1,A2,we have

Notingα=aρzρ+aρ−1zρ−1+···+a0andβ=bρzρ+bρ−1zρ−1+···+b0,whereai(aρ/=0),bi(bρ/=0) (i=0,1,···,ρ) are complex numbers,we get

Two cases will be considered in the following.

Firstly,we consider the caseCombining (3.1) and (3.15),we get

Combining (3.2) and (3.13),we get

whereCombining(3.40)with (3.41),we get

Hence,

By (3.15),we know thatφ1f+ψ1f′is an entire function.Therefore,

whereE1={θ:|f(reiθ)|≤1}andE2={θ:|f(reiθ)|>1}.By Lemma 2.5,we have

By (3.42) and Lemma 2.5,

For the case,by the same argument as the above,

Combining (3.1) with (3.2),we get that

By(3.43)and(3.44),we getai(z)∈S(f)(i=1,2,3).Sincedeg(α−β)=degα=degβ=ρ(f)>0,we getT1≡T2≡0 by Lemma 2.3,which is a contradiction.

Therefore,there exists at most one positive integer k,such thatf(z)Δkcf(z)−R(z) has finitely many zeros.