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The zeros of complex differentialdifference polynomials
Advances in Difference Equations volume 2014, Article number: 157 (2014)
Abstract
This paper is devoted to considering the zeros of complex differentialdifference polynomials of different types. Our results can be seen as the differentialdifference analogues of Hayman conjecture (Ann. Math. 70:942, 1959).
MSC:30D35, 39A05.
1 Introduction and main results
Let $f(z)$ be a meromorphic function in the complex domain. Assume that the reader is familiar with standard symbols and fundamental results of Nevanlinna theory [1, 2]. Recall that $a(z)$ is a small function with respect to $f(z)$, if $T(r,a)=S(r,f)$, where $S(r,f)$ is used to denote any quantity satisfying $S(r,f)=o(T(r,f))$ as $r\to \mathrm{\infty}$ outside of a possible exceptional set of finite logarithmic measure. Denote by $\rho (f)$ and ${\rho}_{2}(f)$ the order and the hyperorder of f. In this paper, c is a nonzero complex constant, n, k are positive integers, unless otherwise specified.
Hayman [3] conjectured that if f is a transcendental meromorphic function, then ${f}^{n}{f}^{\prime}$ takes every finite nonzero value infinitely often. In fact, Hayman [3] proved that if f is a transcendental meromorphic function and $n\ge 3$, then ${f}^{n}{f}^{\prime}$ takes every finite nonzero value infinitely often. Later, the case $n=2$ was settled by Mues [4]. Bergweiler and Eremenko [5], Chen and Fang [[6], Theorem 1] proved the case of $n=1$, respectively. In the past years, the topic on the zeros of differential polynomials has always been an important research problem in value distribution of meromorphic functions. With the development of the difference analogues of Nevanlinna theory, some authors paid their attention to the zeros of difference polynomials. Laine and Yang [[7], Theorem 2] firstly considered the zeros distribution of $f{(z)}^{n}f(z+c)a$, where a is a nonzero constant, which can be seen as a difference analogue of Hayman conjecture. Recently, many authors were interested in the zeros distribution of difference polynomials of different types, such as [8–13].
A polynomial $Q(z,f)$ can be called a differentialdifference polynomial in f whenever $Q(z,f)$ is a polynomial in $f(z)$, its shifts $f(z+c)$ and their derivatives, with small functions of $f(z)$ as the coefficients. It is interesting to consider the zeros of differentialdifference polynomials. The aim of the paper is to explore the differences or analogues among the zeros of differential polynomials, difference polynomials, differentialdifference polynomials. Liu et al. [[14], Theorems 1.1 and 1.3] considered this problem and obtained the following result, where ${\mathrm{\Delta}}_{c}f=f(z+c)f(z)$.
Theorem A Let f be a transcendental entire function of finite order and $a(z)$ be a nonzero small function with respect to $f(z)$. If $n\ge k+2$, then ${[f{(z)}^{n}f(z+c)]}^{(k)}a(z)$ has infinitely many zeros. If f is not a periodic function with period c and $n\ge k+3$, then ${[f{(z)}^{n}{\mathrm{\Delta}}_{c}f]}^{(k)}a(z)$ has infinitely many zeros.
If $a(z)\equiv 0$ in Theorem A, some results can be found in [15]. In this paper, we will consider the zeros of differentialdifference polynomials of $f{(z)}^{n}{f}^{(k)}(z+c)a(z)$ and $f{(z)}^{n}{({\mathrm{\Delta}}_{c}f)}^{(k)}a(z)$.
Theorem 1.1 Let f be a transcendental entire function of hyperorder ${\rho}_{2}(f)<1$. If $n\ge 3$, then $f{(z)}^{n}{f}^{(k)}(z+c)a(z)$ has infinitely many zeros, where $a(z)$ is a nonzero small function with respect to $f(z)$.
Remark 1 (1) The condition that $a(z)$ is a nonzero small function cannot be removed, which can be seen by $f(z)={e}^{z}$ and ${e}^{c}=2$. Thus we get $f{(z)}^{n}{f}^{(k)}(z+c)=2{e}^{(n+1)z}$ has no zeros.

(2)
The condition ${\rho}_{2}(f)<1$ cannot be deleted, which can be seen by $f(z)={e}^{{e}^{z}}$ of ${\rho}_{2}(f)=1$, thus $f{(z)}^{n}{f}^{\prime}(z+c)+n{e}^{z}+P(z)=P(z)$ has finitely many zeros, where ${e}^{c}=n$ and $P(z)$ is a nonzero polynomial. In fact, for any integer k, we can choose appropriate $\alpha (z)$ to make $f{(z)}^{n}{f}^{(k)}(z+c)+\alpha (z)+P(z)=P(z)$, $\alpha (z)$ is a polynomial in ${e}^{z}$.
If f is a finite order transcendental entire function, we prove the following result.
Theorem 1.2 Let f be a finite order transcendental entire function. If $n\ge 2$, then $f{(z)}^{n}{f}^{(k)}(z+c)a(z)$ has infinitely many zeros, where $a(z)$ is an entire function with $\rho (a)<\rho (f)$.
Definition 1 Define that a polynomial $p(z)$ is a Borel exceptional polynomial of $f(z)$ when
where $\lambda (f(z)p(z))$ is the exponent of convergence of zeros of $f(z)p(z)$.
Theorem 1.3 Let f be a finite order transcendental entire function with a Borel exceptional polynomial $d(z)$. If $n\ge 1$, then $f{(z)}^{n}{f}^{(k)}(z+c)b$ has infinitely many zeros, where b is a nonzero constant.
Remark 2 (1) The condition that b is a nonzero constant cannot be removed, which can be seen by $f(z)={e}^{z}$ which has a Borel exceptional value 0. Thus, we get $f{(z)}^{n}{f}^{(k)}(z+c)={e}^{c}{e}^{(n+1)z}$ has no zeros.

(2)
From the above three theorems, we can reduce the value of n with additional conditions. However, we hope that the condition $n\ge 3$ can be reduced to $n\ge 1$ in Theorem 1.1. Unfortunately, we have not succeeded in doing that.
If $f(z)$ is a transcendental meromorphic function, we obtain the next result.
Theorem 1.4 Let f be a transcendental meromorphic function of hyperorder ${\rho}_{2}(f)<1$. If $n\ge 2k+6$, then $f{(z)}^{n}{f}^{(k)}(z+c)a(z)$ has infinitely many zeros, where $a(z)$ is a nonzero small function with respect to $f(z)$.
Using the similar method of proofs of Theorems 1.1 and 1.4 below, we can get the following result.
Theorem 1.5 Let f be a transcendental meromorphic (entire) function of hyperorder ${\rho}_{2}(f)<1$. If $n\ge 4k+9$ ($n\ge 4$), then $f{(z)}^{n}{({\mathrm{\Delta}}_{c}f)}^{(k)}a(z)$ has infinitely many zeros, where $a(z)$ is a nonzero small function with respect to $f(z)$.
Finally, we recall the classical results due to Hayman [[3], Theorems 8 and 9], which can be combined as follows.
Theorem B Let f be a transcendental meromorphic function and $a\ne 0$, b be a finite complex constant. Then ${f}^{n}+a{f}^{\prime}b$ has infinitely many zeros for $n\ge 5$. If f is transcendental entire, this holds for $n\ge 3$, resp. $n\ge 2$, if $b=0$.
We then proceed to consider the zeros of $f{(z)}^{n}+a(z){f}^{(k)}(z+c)b(z)$, which can be seen as the differentialdifference analogues of Theorem B.
Theorem 1.6 Let f be a transcendental entire function with finite order, let $a(z)$, $b(z)$ be small functions with respect to f. Then $f{(z)}^{n}+a(z){f}^{(k)}(z+c)+b(z)$ has infinitely many zeros for $n\ge 3$, resp. $n\ge 2$, if $b(z)\equiv 0$.
Remark 3 The condition $n\ge 3$ cannot be improved if $b(z)\not\equiv 0$, which can be seen by the function $f(z)={e}^{z}+1$ and ${e}^{c}=2$, thus $f{(z)}^{2}{f}^{\prime}(z+c)1={e}^{2z}$ has no zeros. The condition $n\ge 2$ cannot be improved if $b(z)\equiv 0$, which can be seen by the function $f(z)={e}^{z}$ and ${e}^{c}=2$, thus $f(z){f}^{\prime}(z+c)={e}^{z}$ has no zeros.
2 Some lemmas
The difference analogue of logarithmic derivative lemma, given by Chiang and Feng [[16], Corollary 2.5], Halburd and Korhonen [[17], Theorem 2.1], plays an important part in considering the difference analogues of Nevanlinna theory. Afterwards, Halburd, Korhonen and Tohge improved the condition of growth from $\rho <\mathrm{\infty}$ to ${\rho}_{2}(f)<1$ as follows.
Lemma 2.1 [[18], Theorem 5.1]
Let f be a transcendental meromorphic function of ${\rho}_{2}(f)<1$, $\varsigma <1$, ε is a number small enough. Then
for all r outside of a set of finite logarithmic measure.
Lemma 2.2 [[18], Lemma 8.3]
Let $T:[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ be a nondecreasing continuous function and let $s\in (0,\mathrm{\infty})$. If the hyperorder of T is strictly less than one, i.e.,
and $\delta \in (0,1\varsigma )$, then
for all r runs to infinity outside of a set of finite logarithmic measure.
From Lemma 2.2, then we get the following lemma.
Lemma 2.3 Let $f(z)$ be a transcendental meromorphic function of ${\rho}_{2}(f)<1$. Then
and
Lemma 2.4 Let f be a transcendental meromorphic function of ${\rho}_{2}(f)<1$. Then
for all r outside of a set of finite logarithmic measure.
Proof Combining the lemma of logarithmic derivative with Lemma 2.1, Lemma 2.3, we get
□
Lemma 2.5 [[2], Theorems 1.22 and 1.24]
Let $f(z)$ be a transcendental meromorphic function. Then
Lemma 2.6 Let $f(z)$ be a transcendental meromorphic function of ${\rho}_{2}(f)<1$, and let $F(z)=f{(z)}^{n}{f}^{(k)}(z+c)$. Then
If $f(z)$ is a transcendental entire function of ${\rho}_{2}(f)<1$, then
Proof Remark that
From Lemma 2.1, Lemma 2.3 and the standard ValironMohon’ko theorem [2, 19] and f is a transcendental meromorphic function, then we obtain
On the other hand, using Lemma 2.3, we have
Thus, (2.9) follows from (2.11) and (2.12). If f is a transcendental entire function with ${\rho}_{2}(f)<1$, then
Thus, (2.10) follows from (2.12) and (2.13). □
Using the similar method as the proof of Lemma 2.6, we get the following result, which is important in the proof of Theorem 1.5.
Lemma 2.7 Let $f(z)$ be a transcendental meromorphic function of ${\rho}_{2}(f)<1$, and let $G(z)=f{(z)}^{n}{({\mathrm{\Delta}}_{c}f)}^{(k)}$. Then
If $f(z)$ is a transcendental entire function of ${\rho}_{2}(f)<1$, then
Remark (1) The right inequality of (2.14) cannot be improved, which can be seen by $f(z)=\frac{1}{1+{e}^{z}}$, ${e}^{c}=1$, thus $f{(z)}^{2}{({\mathrm{\Delta}}_{c}f)}^{\prime}=\frac{2{e}^{3z}+2{e}^{z}}{{(1+{e}^{z})}^{4}{(1{e}^{z})}^{2}}$, which implies that $T(r,f{(z)}^{2}{({\mathrm{\Delta}}_{c}f)}^{\prime})=6T(r,f)+S(r,f)$.

(2)
Inequality (2.15) cannot be improved. If $f(z)={e}^{z}+{z}^{2}$, ${e}^{c}=1$, thus $f{(z)}^{n}{({\mathrm{\Delta}}_{c}f)}^{\prime}=2c{({e}^{z}+{z}^{2})}^{n}$, which implies that $T(r,f{(z)}^{n}{({\mathrm{\Delta}}_{c}f)}^{\prime})=nT(r,f)+S(r,f)$. If $f(z)={e}^{z}$, ${e}^{c}=2$, thus $f{(z)}^{n}{({\mathrm{\Delta}}_{c}f)}^{\prime}={e}^{(n+1)z}$, which implies that $T(r,f{(z)}^{n}{({\mathrm{\Delta}}_{c}f)}^{\prime})=(n+1)T(r,f)+S(r,f)$.
The following two results are due to Yang and Yi, see [2].
Lemma 2.8 [[2], Theorem 1.56]
Let ${f}_{1}$, ${f}_{2}$, ${f}_{3}$ be meromorphic functions such that ${f}_{1}$ is not a constant. If ${f}_{1}+{f}_{2}+{f}_{3}=1$ and if
where $\lambda <1$ and $T(r):={max}_{1\le j\le 3}T(r,{f}_{j})$, then either ${f}_{2}=1$ or ${f}_{3}=1$.
Lemma 2.9 [[2], Theorem 1.52]
If ${f}_{j}(z)$ ($j=1,2,\dots ,n$) ($n\ge 2$), ${g}_{j}(z)$ ($j=1,2,\dots ,n$) are entire functions satisfying

(i)
${\sum}_{j=1}^{n}{f}_{j}(z){e}^{{g}_{j}(z)}\equiv 0$,

(ii)
the order of ${f}_{j}$ is less than that of ${e}^{{g}_{h}(z){g}_{k}(z)}$ for $1\le j\le n$, $1\le h<k\le n$, then ${f}_{j}(z)\equiv 0$ ($j=1,2,\dots ,n$).
3 Proofs of Theorem 1.1 and Theorem 1.4
Denote $F(z)=f{(z)}^{n}{f}^{(k)}(z+c)$. From Lemma 2.6, then $F(z)$ is not a constant. Assume that $F(z)a(z)$ has only finitely many zeros, from the second main theorem for three small functions [[1], Theorem 2.5] and Lemma 2.5, then we get
From Lemma 2.6, if $f(z)$ is a transcendental meromorphic function, we get
which is a contradiction with $n\ge 2k+6$. If $f(z)$ is a transcendental entire function, we get
which is a contradiction with $n\ge 3$.
4 Proof of Theorem 1.2
From Theorem 1.1, we just need to prove the case that $n=2$. Suppose contrary to the assertion that $f{(z)}^{2}{f}^{(k)}(z+c)a(z)$ has finitely many zeros, where $a(z)$ is an entire function with $\rho (a)<\rho (f)$. Then from the Hadamard factorization theorem, we have
where $H(z)$, $Q(z)$ are nonzero polynomials and $degQ\le \rho (f)$. Differentiating (4.1) and eliminating ${e}^{Q(z)}$, we obtain
where
and
We affirm that $F(z,f)$ cannot vanish identically. Indeed, if $F(z,f)\equiv 0$, then
which implies that
By integrating the above equation, we have
where A is a nonzero constant. Since $a(z)$ is an entire function with $\rho (a)<\rho (f)$ and $H(z)$ is a nonzero polynomial. Thus, we get $degQ<\rho (f)$. Therefore $f{(z)}^{2}{f}^{(k)}(z+c)=(A+1)H(z){e}^{Q(z)}$, from Lemma 2.1 and Lemma 2.5 we get
which contradicts the assumption that $f(z)$ is transcendental of finite order $\rho (f)$. From (4.2), we get
From the Clunie lemma [[19], Theorem 2.4.2], we get
obviously, $F(z,f)$ is an entire function. Thus, from (4.4) and (4.5), we get $T(r,f)=S(r,f)$, which is a contradiction.
5 Proof of Theorem 1.3
If $d(z)\not\equiv 0$ is a Borel exceptional polynomial of $f(z)$, thus the value 1 is a Borel exceptional value of $\frac{f(z)}{d(z)}$, if $d(z)\equiv 0$, then the value 0 is a Borel exceptional value of $f(z)$, then $f(z)$ must have positive integer order [[2], p.106, Corollary]. Without loss of generality, assume that $\rho (f)=s$, s is a positive integer, then the transcendental entire function $f(z)$ can be written as $f(z)=d(z)+h(z){e}^{\alpha {z}^{s}}$, where α is a nonzero constant and $h(z)$ is a nonzero entire function with $\lambda (h)\le \rho (h)<\rho (f)=s$. Hence,
where
From Theorem 1.2, we need to prove the case of $n=1$ only. Suppose that $f(z){f}^{(k)}(z+c)b$ has finitely many zeros, from the Hadamard factorization theorem, then we assume that
where $A(z)$ is an entire function with order $\rho (A)<s$ and has finitely many zeros, β is a nonzero constant. Thus, we get
Let ${B}_{1}(z)=s\alpha {h}_{1}(z){z}^{s1}+{h}_{1}^{\prime}(z)$, ${B}_{2}(z)=s\alpha {B}_{1}(z){z}^{s1}+{B}_{1}^{\prime}(z)$, …, ${B}_{k}(z)=s\alpha {B}_{k1}(z){z}^{s1}+{B}_{k1}^{\prime}(z)$. Thus, we have
Since $d(z)$ is a nonzero polynomial, then we get $d(z){d}^{(k)}(z+c)b\not\equiv 0$. Let ${f}_{1}=\frac{[h(z){d}^{(k)}(z+c)+d(z){B}_{k}(z)]{e}^{\alpha {z}^{s}}}{bd(z){d}^{(k)}(z+c)}$, ${f}_{2}=\frac{h(z){B}_{k}(z){e}^{2\alpha {z}^{s}}}{bd(z){d}^{(k)}(z+c)}$, ${f}_{3}=\frac{A(z){e}^{\beta {z}^{s}}}{bd(z){d}^{(k)}(z+c)}$. Thus, we get ${f}_{1}+{f}_{2}+{f}_{3}=1$. Since $\rho (h(z))<s$, $\rho (A(z))<s$, which implies that $\rho ({B}_{k}(z))<s$, we get ${f}_{1}$, ${f}_{2}$, ${f}_{3}$ are not constants, which is a contradiction with Lemma 2.8. The proof of Theorem 1.3 is completed.
6 Proof of Theorem 1.6
Let $\psi :=\frac{a(z){f}^{(k)}(z+c)+b(z)}{f{(z)}^{n}}$. We proceed to proving that $\psi +1$ has infinitely many zeros, thus $f{(z)}^{n}+a(z){f}^{(k)}(z+c)+b(z)$ has infinitely many zeros. If f is a transcendental entire function with finite order, we will prove
Applying the first main theorem and Lemma 2.3, Lemma 2.5, we observe that
From (6.2), we easily obtain (6.1). We will estimate the zeros and poles of ψ,
and
Using the second main theorem, Lemma 2.3, Lemma 2.5, we get
Since $n\ge 4$, then (6.5) implies that $\psi +1$ has infinitely many zeros. In what follows, we will prove that if f is a transcendental entire function with finite order and $b(z)\ne 0$, then n can be reduced to $n\ge 3$.
We suppose that $f{(z)}^{n}+a(z){f}^{(k)}(z+c)+b(z)$ has finitely many zeros, from the Hadamard factorization theorem, then there exist two polynomials $r(z)$ and $p(z)$ such that
Differentiating (6.6) and eliminating ${e}^{p(z)}$, we obtain
If $n{f}^{\prime}({p}^{\prime}+{r}^{\prime}/r)f\equiv 0$, then $f{(z)}^{n}=Cr(z){e}^{p(z)}$. Thus, from (6.6), we get
Thus $C=1$, otherwise, $nT(r,f)=T(r,a(z){f}^{(k)}(z+c)+b(z))\le T(r,f)+S(r,f)$, which is a contradiction with $n\ge 3$. Hence $a(z){f}^{(k)}(z+c)+b(z)=0$, which is also a contradiction. Thus $n{f}^{\prime}({p}^{\prime}+{r}^{\prime}/r)f\not\equiv 0$. Since $n\ge 3$, we may apply the Clunie lemma [[19], Lemma 2.4.2], Lemma 2.4 and (6.7) to conclude that
and
Combining the above two estimates, we obtain $T(r,f)=S(r,f)$, a contradiction.
It remains to prove the case $n=2$ and $b(z)=0$. Thus (6.7) now takes the form
Similarly as the case $n\ge 3$, we also conclude that $\varphi :=2{f}^{\prime}({p}^{\prime}+{r}^{\prime}/r)f\ne 0$. We have
Differentiating $\varphi (z)$, we obtain
and so
This can be written as
We proceed to show that $N(r,1/f)=S(r,f)$. Suppose that ${z}_{0}$ is a zero of f with multiplicity k. If $k\ge 2$, then ${z}_{0}$ is a zero of ϕ, the contribution to $N(r,1/f)$ is $S(r,f)$. If the zero of f is simple and we must have that ${p}^{\prime}+\frac{{r}^{\prime}}{r}+2\frac{{\varphi}^{\prime}}{\varphi}$ vanishes at ${z}_{0}$, which implies $N(r,\frac{1}{f})=S(r,f)$. Therefore, we can assume that $f(z)$ takes the form $f(z)=\phi (z){e}^{\alpha {z}^{s}}$, where $q(z)$ is a polynomial and $N(r,\frac{1}{\phi})=S(r,f)$ and $\rho (\phi )<s$. Substituting this expression into (6.6), we obtain
where $g(z)$ is an entire function with $\rho (g)<s$, β is a constant. If $\beta =2\alpha $, which implies that ${[\phi (z+c){e}^{\alpha {(z+c)}^{s}}]}^{(k)}\equiv 0$, which is impossible. If $\beta =\alpha $, then $\phi (z)\equiv 0$, which also is impossible. Thus, $\beta \ne 2\alpha $ and $\beta \ne \alpha $, from Lemma 2.9, we get $\phi {(z)}^{2}{e}^{2\alpha {z}^{s}}\equiv 0$, $a(z){[\phi (z+c){e}^{\alpha {(z+c)}^{s}}]}^{(k)}\equiv 0$ and $g(z){e}^{\beta {z}^{s}}\equiv 0$, which are impossible. Thus, we have completed the proof of Theorem 1.6.
Remark 4 Inequality (6.1) is not valid for $f(z)$ is a transcendental meromorphic function, which can be seen by $f(z)=tanz$, thus $\psi :=\frac{{tan}^{\prime}(z+\pi )1}{{tan}^{n}z}=\frac{1}{{tan}^{n2}z}$. Thus, $T(r,\psi )=(n2)T(r,f)+S(r,f)$.
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Acknowledgements
The authors would like to thank the referees for valuable suggestions for improving our paper. This work was partially supported by the NSFC (No. 11301260, 11101201), the NSF of Jiangxi (No. 20132BAB211003) and the YFED of Jiangxi (No. GJJ13078) of China.
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Liu, X., Liu, K. & Zhou, L. The zeros of complex differentialdifference polynomials. Adv Differ Equ 2014, 157 (2014). https://doi.org/10.1186/168718472014157
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Keywords
 differentialdifference polynomial
 zeros
 Borel exceptional polynomial