# Uniqueness of Entire Functions Sharing Polynomials with Their Derivatives

Authors: Jianming Qi; Feng Lü; Ang Chen
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We use the theory of normal families to prove the following. LetQ1(z)=a1zp+a1,p−1zp−1+⋯+a1,0 and Q2(z)=a2zp+a2,p−1zp−1+⋯+a2,0 be two polynomials such that degQ1=degQ2=p (where p is a nonnegative integer) and a1,a2(a2≠0) 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 f≡f′. This result improves a result due to Li and Yi.


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## Body

## 1. Introduction and Main Results

Letf(z) and g(z) be two nonconstant meromorphic functions in the complex plane ℂ ,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)⇌g(z)=P(z) and say that f(z) and g(z) share P(z) (CM counting multiplicity) (see, [1]). In addition, we use notations σ(f), σ2(f) to denote the order and the hyperorder of f(z), respectively, where
(1.1)σ(f)=lim supr→∞log+T(r,f)logr,σ2(f)=lim supr→∞log+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 [3] proved the well-known theorem.Theorem A.
Leta and b be two complex numbers such thatb≠a,  and letf(z)be a nonconstant entire function. If f(z)=a⇌f′(z)=aandf(z)=b⇌f′(z)=b,  then f(z)≡f′(z).This result has undergone various extensions and improvements (see, [1]).In 1979, Mues and Steinmetz [4] proved the following result.Theorem B.
Leta and b be two complex numbers such that b≠a, and let f(z) be a nonconstant entire function. If f(z)=a⇔f′(z)=a and f(z)=b⇔f′(z)=b, then f(z)≡f′(z).In 2006, Li and Yi [5] proved the following related result.Theorem C.
Letaandbbe two complex numbers such thatb≠a,0,  and letf(z)be a nonconstant entire function. If f(z)=a⇌f′(z)=aand f′(z)=b⇒f(z)=b,  thenf(z)≡f′(z).Remark 1.1.
Meanwhile, Li and Yi [5] give an example to show that b≠0 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 [6] proved the following result.Theorem D.
Letf(z)be a nonconstant entire function, and letQ(z)be a polynomial of degreeq≥1.Letk≥q+1be an integer. Iff(z)=Q(z)⇌f′(z)=Q(z)and iff(k)(z)=Q(z)for everyz∈ℂwithf(z)=Q(z),thenf(z)≡f′(z).In 2007, Li and Yi [7] proved the following result.Theorem E.
Letf(z)be a nonconstant entire function of hyperorderσ2(f)<1/2,and letQ(z)be a nonconstant polynomial. If f(z)=Q(z)⇌f′(z)=Q(z), then
(1.2)f′(z)-Q(z)f(z)-Q(z)≡c
for some constant c≠0.In 2008, Grahl and Meng [8] proved the following result.Theorem F.
Letf(z)be a nonconstant entire function, and letQ(z)be a nonconstant polynomial. Letk≥2be an integer. Iff(z)=Q(z)⇌f′(z)=Q(z)and if for some positiveMwe have|f(k)(z)|≤M(1+|Q(z)|)for everyz∈ℂwithf(z)=Q(z),then
(1.3)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 valuesa, 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.
LetQ1(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(a2≠0) are two distinct complex numbers. Let f(z) be a transcendental entire function. If f(z)=Q1(z)⇌f′(z)=Q1(z) and f′(z)=Q2⇒f(z)=Q2(z), then f(z)≡f′(z).Remark 1.3.
The following example shows the hypothesis thatf is transcendental cannot be omitted in Theorem 1.2.Example 1.4.
Letf(z)=z3, Q1(z)=2z3-3z2 and Q2(z)=z3. Then
(1.4)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 casep=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.
LetQ1(z)=a1zp+a1,p-1zp-1+⋯+a1,0andQ2(z)=a2zp+a2,p-1zp-1+⋯+a2,0 be two polynomials such that degQ1(z)=degQ2(z)=p (wherep is a nonnegative integer) and a1,  a2(a2≠0)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.Leth be a meromorphic function in ℂ. h is called a normal function if there exists a positive M such that h#(z)≤M for all z∈ℂ, where
(2.1)h#(z)=|h′(z)|1+|h(z)|2
denotes the spherical derivative of h.Letℱ be a family of meromorphic functions in a domain D⊂ℂ. We say that ℱ is normal in D if every sequence {fn}n⊆ℱ contains a subsequence which converges spherically and uniformly on compact subsets of D; see [9].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 [12]).
Letℱ be a family of analytic functions in the unit disc Δ with the property that for each f(z)∈ℱ, all zeros of f(z) have multiplicity at least k. Suppose that there exists a number A≥1 such that |f(k)(z)|≤A whenever f(z)∈ℱ and f(z)=0. If ℱ is not normal in Δ, then for 0≤α≤k, there exist(1)
a numberr∈(0,1),(2)
a sequence of complex numberszn, |zn|<r,(3)
a sequence of functionsfn∈ℱ, and(4)
a sequence of positive numbersρn→0such 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 ℂ, and moreover, the zeros of g(ξ) are of multiplicity at least k, g#(ξ)≤g#(0)=kA+1.Lemma 2.2 (see [13]).
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 [9, Marty's criterion]).
A familyℱ of meromorphic functions on a domain D is normal if and only if, for each compact subset K⊆D, there exists a constant M such that f#(z)≤M for each f∈ℱ and z∈K.Lemma 2.4 (see [2]).
Letf(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
(2.2)T(r,f)≤N¯(r,1f-a1)+N¯(r,1f-a1)+N¯(r,1f-a3)+S(r,f).Lemma 2.5 (see [5]).
Letℱ be a family of functions holomorphic on a domain D, and let a and b be two finite complex numbers such that b≠a,0. If for each f∈ℱ, f(z)=a⇒f′(z)=a and f′(z)=b⇒f(z)=b, then ℱ is normal in D.


## 3. Proof of Theorem1.6

IfQ1≡0, by degQ1=degQ2, we obtain p=0, a1=0, Q2≡a2(a2≠0). From the conditions of Theorem 1.6, we obtain f(z)=0⇒f′(z)=0 and f′(z)=a2⇒f(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.IfQ1≢0, by degQ1=degQ2 and a2≠0, we obtain a1≠0. Now we consider the function F=f/Q1-1, and we distinguish two cases.Case 1.
If there exists a constantM 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 constantM such that F#(z)≤M, then there exists a sequence (wn)n such that wn→∞ and F#(wn)→∞ for n→∞.SinceQ1 is a polynomial, there exists an r1 such that
(3.1)|Q1′(z)Q1(z)|≤2p|z|∀z∈ℂsatisfying|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
(3.2)Fn(z)=F(wn+z)=f(wn+z)Q1(wn+z)-1.
Let z∈D1 be fixed; from the above equality, if F(wn+z)=0, then f(wn+z)=Q1(wn+z). Noting that f=Q1⇒f′=Q1, then we obtain the following: if n→∞, then
(3.3)|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 Lemma2.1, 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 zn→0 and ρn→0 and such that the sequence (gn)n defined by
(3.4)gn(ξ)=ρn-1Fn(zn+ρnξ)=ρn-1{f(wn+zn+ρnξ)Q1(wn+zn+ρnξ)-1}→g(ξ)
converges locally and uniformly in ℂ 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 thatg=0⇒g′=1 on ℂ. Suppose that there exists a point ξ0 such that g(ξ0)=0. Then by Hurwitz's theorem, there exist ξn, ξn→ξ0 as n→∞ such that for n sufficiently large
(3.5)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
(3.6)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 limn→∞Gn(ξ)=limn→∞gn′(ξ)=g′(ξ). Noting that f=Q1⇒f′=Q1, we have
(3.7)Gn(ξn)=f′(wn+zn+ρnξn)Q1(wn+zn+ρnξn)=1(n→∞)
Thus
(3.8)g′(ξ0)=limn→∞Gn(ξn)=1.
This shows that g=0⇒g′=1.Next we will prove thatg′(ξ)≠a2/a1 on ℂ. 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=0⇒g′=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 ξn→ξ0 as n→∞ such that for n sufficiently large
(3.9)Gn(ξn)-Q2(wn+zn+ρnξn)Q1(wn+zn+ρnξn)=0⇒f′(wn+zn+ρnξn)=Q2(wn+zn+ρnξn).
Noting that f′=Q2⇒f=Q2, from (3.4) and (3.9) (for n sufficiently large), we have
(3.10)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 a2≠a1(a1≠0), degQ1=degQ2=p and ρn→0, by (3.10), we get
(3.11)g(ξ0)=limn→∞gn(ξn)=∞,
which contradicts g′(ξ0)=a2/a1. This shows that g′(ξ)≠a2/a1 on ℂ.Sinceg is of order at most one, so is g′, it follows that
(3.12)g′(ξ)=a2a1+eb0+b1ξ,
where b0,b1 are two finite constants. We divide this case into two subcases.Subcase 1.
Ifb1=0, from (3.12), we have
(3.13)g(ξ)=(a2a1+eb0)ξ+c0,
where c0 is a constant. Since g=0⇒g′=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.
Ifb1≠0, by
(3.14)g′(ξ)=a2a1+eb0+b1ξ,
we obtain
(3.15)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=0⇒g′=1, we get a unique ξ0=(a2-a1-b1Ba1)/b1a2. Which is a contradiction.Thusf is of finite order. This completes the proof of the theorem.


## 4. Proof of Theorem1.2

Now we distinguish two cases.Case 1.
Ifp=0, by degQ1=degQ2=0, we deduce Q1≡a1 and Q2≡a2(a2≠a1,0). By Theorem C, we obtain f≡f′.Case 2.
Ifp≥1, by degQ1=degQ2=p and a2≠0, we deduce a1≠0. 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
(4.1)λ=f′-Q1f-Q1,
where λ is a nonzero constant. We rewrite it as
(4.2)f′=λf+(1-λ)Q1.
If λ=1, we obtain f≡f′.Now, we assume thatλ≠1. Solving (4.2), we obtain
(4.3)f(z)=Aeλz+P(z),
where A is a nonzero constant, and P(z) is a polynomial. Thus, we have
(4.4)f′(z)=Aλeλz+P′(z).
Substituting (4.3) and (4.4) into (4.2), we get
(4.5)(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(4.6)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(4.7)T(r,λAeλz)≤N¯(r,1Aλeλz+P′(z)-Q2(z))+S(r,f).
By the Nevanlinna First Fundamental Theorem, we immediately obtain
(4.8)N¯(r,1Aλeλz+P′(z)-Q2(z))≤T(r,λAeλz)+S(r,f).
If we combine (4.7) and (4.8), we obtain
(4.9)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
(4.10)(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
(4.11)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,
(4.12)(λ-1)Q2-(λP-P′)≡0.
Comparing the above equality to (4.5), we have Q1≡Q2, 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 degQ1≠degQ2. This is a contradiction. So λ≠1 is impossible. This completes the proof of Theorem 1.2.



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