AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation84769010.1155/2009/847690847690Research ArticleUniqueness of Entire Functions Sharing Polynomials with Their DerivativesQiJianming1Feng2ChenAng1StevicStevo1School of Mathematics and System SciencesShandong UniversityJinan, Shandong 250100Chinasdu.edu.cn2Department of MathematicsChina University of PetroleumDongying, Shandong 257061Chinacup.edu.cn200917200920090112200810042009060520092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We use the theory of normal families to prove the following. Let Q1(z)=a1zp+a1,p1zp1++a1,0 and Q2(z)=a2zp+a2,p1zp1++a2,0 be two polynomials such that degQ1=degQ2=p (where p is a nonnegative integer) and a1,a2(a20) are two distinct complex numbers. Let f(z) be a transcendental entire function. If f(z) and f(z) share the polynomial Q1(z) CM and if f(z)=Q2(z) whenever f(z)=Q2(z), then ff. This result improves a result due to Li and Yi.

1. Introduction and Main Results

Let f(z) and g(z) be two nonconstant meromorphic functions in the complex plane ,and let P(z) be a polynomial or a finite complex number.  degP(z) denotes the degree of the polynomial P(z). To simplify the statement of our results in this paper, deviating from the common definition, we consider the zero polynomial as a polynomial of degree 0. If g(z)-P(z)=0 whenever f(z)-P(z)=0, we write f(z)=P(z)g(z)=P(z). If f(z)=P(z)g(z)=P(z) and g(z)=P(z)f(z)=P(z), we write f(z)=P(z)g(z)=P(z) and say that f(z) and g(z) share P(z) (IM ignoring multiplicity). If f(z)-P(z) and g(z)-P(z) have the same zeros with the same multiplicities, we write f(z)=P(z)ƒg(z)=P(z) and say that f(z) and g(z) share P(z) (CM counting multiplicity) (see, ). In addition, we use notations σ(f), σ2(f) to denote the order and the hyperorder of f(z), respectively, where σ(f)=lim suprlog+T(r,f)logr,σ2(f)=lim suprlog+log+T(r,f)logr. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory, as found in [1, 2].

In 1977, Rubel and Yang  proved the well-known theorem.

Theorem A.

Let a and b be two complex numbers such that  ba, and letf(z)  be a nonconstant entire function. If f(z)=aƒf(z)=a  andf(z)=bƒf(z)=b, then f(z)f(z).

This result has undergone various extensions and improvements (see, ).

In 1979, Mues and Steinmetz  proved the following result.

Theorem B.

Let a and b be two complex numbers such that ba, and let f(z) be a nonconstant entire function. If f(z)=af(z)=a and f(z)=bf(z)=b, then f(z)f(z).

In 2006, Li and Yi  proved the following related result.

Theorem C.

Let a  and  b  be two complex numbers such that  ba,0, and letf(z)  be a nonconstant entire function. If f(z)=af(z)=a  and f(z)=bf(z)=b, then  f(z)f(z).

Remark 1.1.

Meanwhile, Li and Yi  give an example to show that b0 cannot be omitted in Theorem C.

In recent years, there have been several papers dealing with entire functions that share a polynomial with their derivatives.

In 2006, Wang  proved the following result.

Theorem D.

Let  f(z)  be a nonconstant entire function, and let  Q(z)  be a polynomial of degree  q1.  Let  kq+1  be an integer. If  f(z)=Q(z)ƒf(z)=Q(z)  and if  f(k)(z)=Q(z)  for every  z  with  f(z)=Q(z),  then  f(z)f(z).

In 2007, Li and Yi  proved the following result.

Theorem E.

Let  f(z)  be a nonconstant entire function of hyperorder  σ2(f)<1/2,  and let  Q(z)  be a nonconstant polynomial. If   f(z)=Q(z)ƒf(z)=Q(z),   then f(z)-Q(z)f(z)-Q(z)c for some constant c0.

In 2008, Grahl and Meng  proved the following result.

Theorem F.

Let f(z)  be a nonconstant entire function, and let  Q(z)  be a nonconstant polynomial. Let  k2  be an integer. If  f(z)=Q(z)ƒf(z)=Q(z)  and if for some positive  M  we have  |f(k)(z)|M(1+|Q(z)|)  for every  z  with  f(z)=Q(z),  then f(z)-Q(z)f(z)-Q(z) is constant.

From the ideas of Theorem D to Theorem F, it is natural to ask whether the values a, b in Theorem C can be replaced by two polynomials Q1, Q2. The main purpose of this paper is to investigate this problem. We prove the following result.

Theorem 1.2.

Let   Q1(z)=a1zp+a1,p-1zp-1++a1,0   and   Q2(z)=a2zp+a2,p-1zp-1++a2,0   be two polynomials such that degQ1(z)=degQ2(z)=p  (where p is a nonnegative integer) and a1, a2(a20) are two distinct complex numbers. Let f(z) be a transcendental entire function. If f(z)=Q1(z)ƒf(z)=Q1(z) and f(z)=Q2f(z)=Q2(z), then f(z)f(z).

Remark 1.3.

The following example shows the hypothesis that f is transcendental cannot be omitted in Theorem 1.2.

Example 1.4.

Let f(z)=z3, Q1(z)=2z3-3z2 and Q2(z)=z3. Then f(z)-Q1(z)f(z)-Q1(z)=2,f(z)=Q2(z)f(z)=Q2(z). While f(z) does not satisfy the result of Theorem 1.2.

Remark 1.5.

The case p=0 of Theorem 1.2 yields Theorem C.

It seems that we cannot get the result by the methods used in [4, 5]. In order to prove our theorem, we need the following result which is interesting in its own right.

Theorem 1.6.

Let  Q1(z)=a1zp+a1,p-1zp-1++a1,0  and  Q2(z)=a2zp+a2,p-1zp-1++a2,0   be two polynomials such that degQ1(z)=degQ2(z)=p   (where  p is a nonnegative integer) and a1, a2(a20)  are two distinct complex numbers. Let f(z)  be a nonconstant entire function, and f(z)=Q1(z)f(z)=Q1(z) and f(z)=Q2(z)f(z)=Q2(z), then f(z)  is of finite order.

2. Some Lemmas

In order to prove our theorems, we need the following lemmas.

Let h be a meromorphic function in . h is called a normal function if there exists a positive M such that h#(z)M for all z, where h#(z)=|h(z)|1+|h(z)|2 denotes the spherical derivative of h.

Let be a family of meromorphic functions in a domain D. We say that is normal in D if every sequence {fn}n contains a subsequence which converges spherically and uniformly on compact subsets of D; see .

Normal families, in particular, of holomorphic functions often appear in operator theory on spaces of analytic functions; for example, see in [10, Lemma 3] and in [11, Lemma 4].

Lemma 2.1 (see [<xref ref-type="bibr" rid="B14">12</xref>]).

Let be a family of analytic functions in the unit disc Δ with the property that for each f(z), all zeros of f(z) have multiplicity at least k. Suppose that there exists a number A1 such that |f(k)(z)|A whenever f(z) and f(z)=0. If is not normal in Δ, then for 0αk, there exist

a number r(0,1),

a sequence of complex numbers zn, |zn|<r,

a sequence of functions fn, and

a sequence of positive numbers ρn0

such that gn(ξ)=ρn-αfn(zn+ρnξ) converges locally and uniformly (with respect to the spherical metric) to a nonconstant analytic function g(ξ) on , and moreover, the zeros of g(ξ) are of multiplicity at least k, g#(ξ)g#(0)=kA+1.

Lemma 2.2 (see [<xref ref-type="bibr" rid="B1">13</xref>]).

A normal meromorphic function has order at most two. A normal entire function is of exponential type and thus has order at most one.

Lemma 2.3 (see [<xref ref-type="bibr" rid="B10">9</xref>, Marty's criterion]).

A family of meromorphic functions on a domain D is normal if and only if, for each compact subset KD, there exists a constant M such that f#(z)M for each f and zK.

Lemma 2.4 (see [<xref ref-type="bibr" rid="B4">2</xref>]).

Let f(z) be a meromorphic function, and let a1(z), a2(z), a3(z) be three distinct meromorphic functions satisfying T(r,ai)=S(r,f),  i=1,2,3. Then T(r,f)N¯(r,1f-a1)+N¯(r,1f-a1)+N¯(r,1f-a3)+S(r,f).

Lemma 2.5 (see [<xref ref-type="bibr" rid="B6">5</xref>]).

Let be a family of functions holomorphic on a domain D, and let a and b be two finite complex numbers such that ba,0. If for each f, f(z)=af(z)=a and f(z)=bf(z)=b, then is normal in D.

3. Proof of Theorem <xref ref-type="statement" rid="thm1.2">1.6</xref>

If Q10, by degQ1=degQ2, we obtain p=0, a1=0, Q2a2(a20). From the conditions of Theorem 1.6, we obtain f(z)=0f(z)=0 and f(z)=a2f(z)=a2. By Lemmas 2.5 and 2.3 we obtain that f is a normal function in D. By Lemma 2.2 we obtain that f is a finite order function.

If Q10, by degQ1=degQ2 and a20, we obtain a10. Now we consider the function F=f/Q1-1, and we distinguish two cases.

Case 1.

If there exists a constant M such that F#(z)M, by Lemmas 2.3 and 2.2, then F is of finite order. Hence f=(F+1)Q1 is of finite order as well.

Case 2.

If there does not exist a constant M such that F#(z)M, then there exists a sequence (wn)n such that wn and F#(wn) for n.

Since Q1 is a polynomial, there exists an r1 such that |Q1(z)Q1(z)|2p|z|z  satisfying|z|r1. Obviously, if z, then 2p/|z|0. Let r>r1, and D={z:|z|r}, then F is analytic in D. Without loss of generality, we may assume |wn|r+1 for all n. We define D1={z:|z|<1} and Fn(z)=F(wn+z)=f(wn+z)Q1(wn+z)-1. Let zD1 be fixed; from the above equality, if F(wn+z)=0, then f(wn+z)=Q1(wn+z). Noting that f=Q1f=Q1, then we obtain the following: if n, then |Fn(z)|=|(f(wn+z)Q1(wn+z))|=|f(wn+z)Q1(wn+z)-f(wn+z)Q1(wn+z)Q1(wn+z)Q1(wn+z)||f(wn+z)Q1(wn+z)|+|f(wn+z)Q1(wn+z)||Q1(wn+z)Q1(wn+z)|<2.

Obviously, Fn(z) are analytic in D1 and Fn#(0)=F#(wn) as n. It follows from Lemma 2.3 that (Fn)n is not normal at z=0.

Therefore, we can apply Lemma 2.1, with (α=k=1andA=2). Choosing an appropriate subsequence of (Fn)n if necessary, we may assume that there exist sequences (zn)n and (ρn)n, such that zn0 and ρn0 and such that the sequence (gn)n defined by gn(ξ)=ρn-1Fn(zn+ρnξ)=ρn-1{f(wn+zn+ρnξ)Q1(wn+zn+ρnξ)-1}g(ξ) converges locally and uniformly in where g(ξ) is a nonconstant analytic function and g#(ξ)g#(0)=A+1=3. By lemma 2.2, the order of g(ξ) is at most 1.

First, we will prove that g=0g=1 on . Suppose that there exists a point ξ0 such that g(ξ0)=0. Then by Hurwitz's theorem, there exist ξn, ξnξ0 as n such that for n sufficiently large gn(ξn)=ρn-1{f(wn+zn+ρnξn)Q1(wn+zn+ρnξn)-1}=0. This implies f(wn+zn+ρnξn)=Q1(wn+zn+ρnξn). From the above, we obtain gn(ξ)=f(wn+zn+ρnξ)Q1(wn+zn+ρnξ)-f(wn+zn+ρnξ)Q1(wn+zn+ρnξ)Q1(wn+zn+ρnξ)Q1(wn+zn+ρnξ). Let Gn(ξ)=f(wn+zn+ρnξ)/Q1(wn+zn+ρnξ), by (3.1), (3.3) and (3.4), it is easy to obtain limnGn(ξ)=limngn(ξ)=g(ξ). Noting that f=Q1f=Q1, we have Gn(ξn)=f(wn+zn+ρnξn)Q1(wn+zn+ρnξn)=1(n) Thus g(ξ0)=limnGn(ξn)=1. This shows that g=0g=1.

Next we will prove that g(ξ)a2/a1 on . Suppose that there exists a point ξ0 such that g(ξ0)=a2/a1. If g(ξ)a2/a1, then g(ξ)=a2/a1ξ+c, where c is a constant, together with the fact that g=0g=1 gives a2/a1=1, which contradicts to the assumptions. Thus g(ξ)a2/a1. Since Gn(ξ)-Q2(wn+zn+ρnξ)/Q1(wn+zn+ρnξ)g(ξ)-a2/a1 as n and g(ξ0)=a2/a1, by Hurwitz's theorem, there exist ξnξ0 as n such that for n sufficiently large Gn(ξn)-Q2(wn+zn+ρnξn)Q1(wn+zn+ρnξn)=0f(wn+zn+ρnξn)=Q2(wn+zn+ρnξn). Noting that f=Q2f=Q2, from (3.4) and (3.9) (for n sufficiently large), we have gn(ξn)=ρn-1{f(wn+zn+ρnξn)Q1(wn+zn+ρnξn)-1}=ρn-1{Q2(wn+zn+ρnξn)Q1(wn+zn+ρnξn)-1}. Since a2a1(a10), degQ1=degQ2=p and ρn0, by (3.10), we get g(ξ0)=limngn(ξn)=, which contradicts g(ξ0)=a2/a1. This shows that g(ξ)a2/a1 on .

Since g is of order at most one, so is g, it follows that g(ξ)=a2a1+eb0+b1ξ, where b0,b1 are two finite constants. We divide this case into two subcases.Subcase 1.

If b1=0, from (3.12), we have g(ξ)=(a2a1+eb0)ξ+c0, where c0 is a constant. Since g=0g=1, from (3.13) we have a2/a1+eb0=1. By a simple calculation, we have g#(0)=1/(1+|c0|2), which contradicts g#(0)=3.

Subcase 2.

If b10, by g(ξ)=a2a1+eb0+b1ξ, we obtain g(ξ)=a2a1ξ+1b1eb0+b1ξ+B, where B is a constant. Obviously, g(ξ)=0 has infinitely many solutions. Suppose that there exists a point ξ0 such that g(ξ0)=0. By (3.14), (3.15), and g=0g=1, we get a unique ξ0=(a2-a1-b1Ba1)/b1a2. Which is a contradiction.

Thus f is of finite order. This completes the proof of the theorem.

4. Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.2</xref>

Now we distinguish two cases.

Case 1.

If p=0, by degQ1=degQ2=0, we deduce Q1a1 and Q2a2(a2a1,0). By Theorem C, we obtain ff.

Case 2.

If p1, by degQ1=degQ2=p and a20, we deduce a10. So Q1 is a nonconstant polynomial. By Theorem 1.6, we know that f is of finite order. Thus, the hyperorder σ2(f)=0. Then, by Theorem E, we have λ=f-Q1f-Q1, where λ is a nonzero constant. We rewrite it as f=λf+(1-λ)Q1. If λ=1, we obtain ff.

Now, we assume that λ1. Solving (4.2), we obtain f(z)=Aeλz+P(z), where A is a nonzero constant, and P(z) is a polynomial. Thus, we have f(z)=Aλeλz+P(z). Substituting (4.3) and (4.4) into (4.2), we get (λ-1)Q1-(λP-P)0. Next, we will prove that P(z)Q2(z). Suppose that P(z)Q2(z), by (4.4) we obtain

N¯(r,1f(z)-Q2(z))=N¯(r,1Aλeλz+P(z)-Q2(z)). Since f(z) is a transcendental entire function and P(z)-Q2(z) is a polynomial, we deduce T(r,P(z)-Q2(z))=S(r,f). It is well known that 0 and are the Picard values of eλz. By Lemma 2.4, we obtain

T(r,λAeλz)N¯(r,1Aλeλz+P(z)-Q2(z))+S(r,f). By the Nevanlinna First Fundamental Theorem, we immediately obtain N¯(r,1Aλeλz+P(z)-Q2(z))T(r,λAeλz)+S(r,f). If we combine (4.7) and (4.8), we obtain N¯(r,1Aλeλz+P(z)-Q2(z))=T(r,λAeλz)+S(r,f)S(r,f). Since P(z)Q2(z), we suppose z0 is a zero of f-Q2. By the assumption f(z)=Q2(z)f(z)=Q2(z), we have f(z0)=Q2(z0). Substituting z0 into (4.3) and (4.4), we have (λ-1)Q2(z0)=λP(z0)-P(z0). If (λ-1)Q2-(λP-P)0, noting that (λ-1)Q2-(λP-P) is a polynomial, we have N¯(r,1f-Q2)N¯(r,1(λ-1)Q2-(λP-P))T(r,(λ-1)Q2-(λP-P))=S(r,f), which contradicts with (4.9). Hence, (λ-1)Q2-(λP-P)0. Comparing the above equality to (4.5), we have Q1Q2, a contradiction. Thus, we prove P(z)Q2(z). It is easy to see degQ2=degP. By (4.5) we obtain degQ1=degP. Finally we deduce degQ1degQ2. This is a contradiction. So λ1 is impossible. This completes the proof of Theorem 1.2.

Acknowledgment

The authors are grateful to the referee for his(her) valuable suggestions and comments. The authors would like to express their hearty thanks to Professor Hongxun Yi for his valuable advice and helpful information. The work was supported by the NNSF of China (no. 10371065), the NSF of Shang-dong Province, China (no. Z2002A01), and the NSFC-RFBR.

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