Value distribution of difference and q-difference polynomials

Abstract

In this paper, we investigate the value distribution of difference polynomial and obtain the following result, which improves a recent result of K. Liu and L.Z. Yang: Let f be a transcendental meromorphic function of finite order σ, c be a nonzero constant, and $\alpha (z)\not\equiv 0$ be a small function of f, and let

$$P(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0$$

be a polynomial with a multiple zero. If $\lambda (1/f)<\sigma$, then $P(f)f(z+c)-\alpha (z)$ has infinitely many zeros. We also obtain a result concerning the value distribution of q-difference polynomial.

MSC:30D35, 39A05.

1 Introduction and main results

Throughout the paper, we assume that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory as found in [1–3]. A function $f(z)$ is called the meromorphic function, if it is analytic in the complex plane except at isolated poles. For any non-constant meromorphic function f, we denote by $S(r,f)$ any quantity satisfying

$$\underset{r\to \mathrm{\infty }}{lim}\frac{S(r,f)}{T(r,f)}=0,$$

possibly outside of a set of finite linear measure in ${\mathbb{R}}^{+}$. A meromorphic function $a(z)$ is called a small function of $f(z)$ provided that $T(r,a)=S(r,f)$. As usual, we denote by $\sigma (f)$ the order of a meromorphic function $f(z)$, and denote by $\lambda (f)$ ($\lambda (1/f)$) the exponent of convergence of the zeros (poles) of $f(z)$.

Recently, a number of papers concerning the complex difference products and the differences analogues of Nevanlinna’s theory have been published (see [4–12] for example), and many excellent results have been obtained. In 2007, Laine and Yang [10] investigated the value distribution of difference products of entire functions, and obtained the following result.

Theorem A Let $f(z)$ be a transcendental entire function of finite order, and c be a non-zero complex constant. Then for $n\ge 2$, $f{(z)}^{n}f(z+c)$ assumes every non-zero value $a\in \mathbb{C}$ infinitely often.

Liu and Yang [11] improved Theorem A, and proved the next result.

Theorem B Let $f(z)$ be a transcendental entire function of finite order, and c be a non-zero complex constant. Then for $n\ge 2$, $f{(z)}^{n}f(z+c)-p(z)$ has infinitely many zeros, where $p(z)\not\equiv 0$ is a polynomial in z.

The purpose of this paper is to investigate the value distribution of difference polynomial $P(f)f(z+c)-\alpha (z)$ and q-difference polynomial $P(f)f(qz)-\alpha (z)$, where $P(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0$ with constant coefficients $a_n(\ne 0),a_{n-1},\dots ,a_0$, and $\alpha (z)$ is a mall function of $f(z)$.

For the sake of simplicity, we denote by $s(P)$ and $m(P)$ the number of the simple zeros and the number of multiple zeros of a polynomial

$$P(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0$$

respectively.

We obtain the following result which improves Theorem A and Theorem B.

Theorem 1.1 Let f be a transcendental meromorphic function of finite order $\sigma (f)=\sigma$, and c be a non-zero constant, and let

$$P(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0$$

be a polynomial with constant coefficients $a_n(\ne 0),a_{n-1},\dots ,a_0$ and $m(P)>0$. If $\lambda (\frac{1}{f})<\sigma$, then $P(f)f(z+c)-\alpha (z)$ has infinitely many zeros, where $\alpha (z)\not\equiv 0$ is a small function of f.

Remark 1 The result of Theorem 1.1 may be false if $\alpha (z)\equiv 0$, for example, $f(z)=\frac{e^{z^2}}{z}$, it is obvious that $f^{2}f(z+1)$ has no zeros. The following example shows that the assumption $\lambda (\frac{1}{f})<\sigma$ in Theorem 1.1 cannot be deleted. In fact, let $f(z)=\frac{1-e^z}{1+e^z}$, $c=\pi i$, $\alpha (z)=-1$, and $P(z)=z^2$. Then $\lambda (\frac{1}{f})=\sigma (f)=1$ and $P(f)f(z+c)-\alpha (z)=\frac{2}{1+e^z}$ has no zeros. Also, let $f(z)=i+e^z$, $c=\pi i$, $\alpha (z)=1$, and $P(z)=z(z-i+1)(z-i-1)$. Then $P(f)f(z+c)-\alpha (z)=-e^{4z}$ has no zeros. This shows that the restriction in Theorem 1.1 to the multiple zero case is essential.

Considering the value distribution of q-differences polynomials, we obtain the following result.

Theorem 1.2 Let $f(z)$ be a transcendental entire function of zero order, and $\alpha (z)\in S(r,f)$. Suppose that q is a non-zero complex constant and n is an integer. If $m(P)>0$, then $P(f)f(qz)-\alpha (z)$ has infinitely many zeros.

2 Some lemmas

Lemma 2.1 [6]

Given two distinct complex constants ${\eta }_1$, ${\eta }_2$, let f be a meromorphic function of finite order σ. Then, for each $\epsilon >0$, we have

$$m(r,\frac{f(z+{\eta }_1)}{f(z+{\eta }_2)})=O\left(r^{\sigma -1+\epsilon }\right).$$

Lemma 2.2 [6]

Let f be a transcendental meromorphic function of finite order σ, c be a complex number. Then, for each $\epsilon >0$, we have

$$T(r,f(z+c))=T(r,f(z))+O\left(r^{\sigma -1+\epsilon }\right)+O(logr).$$

The following lemma is a revised form of Lemma 2.4.2 in [2].

Lemma 2.3 Let $f(z)$ be a transcendental meromorphic solution of

$$f^{n}A(z,f)=B(z,f),$$

where $A(z,f)$, $B(z,f)$ are differential polynomials in f and its derivatives with meromorphic coefficients, say $\{a_{\lambda }\mid \lambda \in I\}$, n be a positive integer. If the total degree of $B(z,f)$ as a polynomial in f and its derivatives is less than or equal to n, then

$$m(r,A(z,f))\le \sum _{\lambda \in I}m(r,a_{\lambda })+S(r,f).$$

Lemma 2.4 [12]

Let $f(z)$ be a non-constant meromorphic function of finite order, $c\in \mathbb{C}$. Then

$$\begin{array}{c}N(r,\frac{1}{f(z+c)})\le N(r,\frac{1}{f(z)})+S(r,f),N(r,f(z+c))\le N(r,f)+S(r,f),\hfill \\ \overline{N}(r,\frac{1}{f(z+c)})\le \overline{N}(r,\frac{1}{f(z)})+S(r,f),\overline{N}(r,f(z+c))\le \overline{N}(r,f)+S(r,f),\hfill \end{array}$$

outside of a possible exceptional set E with finite logarithmic measure.

Lemma 2.5 [4]

Let f be a non-constant zero-order meromorphic function, and $q\in \mathbb{C}\setminus \{0\}$. Then

$$m(r,\frac{f(qz)}{f(z)})=o(T(r,f))$$

on a set of logarithmic density 1.

Remark 2 For the similar reason in Theorem 1.1 in [4], we can easily deduce that

$$m(r,\frac{f(z)}{f(qz)})=o(T(r,f))$$

also holds on a set of logarithmic density 1.

Proof

Using the identity

$$\frac{{\rho }^2-r^2}{{\rho }^2-2\rho rcos(\phi -\theta )+r^2}=Re\left(\frac{\rho e^{i\theta }+z}{\rho e^{i\theta }-z}\right),$$

and let Poisson-Jensen formula with $R=\rho$, we see

$$\begin{array}{rl}log\left|\frac{f(z)}{f(qz)}\right|=& {\int }_0^{2\pi }log\left|f\left(\rho e^{i\theta }\right)\right|Re(\frac{\rho e^{i\theta }+z}{\rho e^{i\theta }-z}-\frac{\rho e^{i\theta }+qz}{\rho e^{i\theta }-qz})\frac{d\theta }{2\pi }\\ +\sum _{|a_n|<\rho }log\left|\frac{(z-a_n)({\rho }^2-{\overline{a}}_{n}qz)}{(qz-a_n)({\rho }^2-{\overline{a}}_{n}z)}\right|\\ -\sum _{|b_m|<\rho }log\left|\frac{(z-b_m)({\rho }^2-{\overline{b}}_{m}qz)}{(qz-b_m)({\rho }^2-{\overline{b}}_{m}z)}\right|\\ =& S_1^{\prime }(z)+S_1^{\prime }(z)-S_3^{\prime }(z),\end{array}$$

where $\{a_n\}$ and $\{b_m\}$ are the zeros and poles of f, respectively. Integration on the set $E:=\{\phi \in [0,2\pi ]:|\frac{f(r{e}^{i\phi })}{f(qr{e}^{i\phi })}|\ge 1\}$ gives us the proximity function,

$$\begin{array}{rl}m(r,\frac{f(z)}{f(qz)})& ={\int }_{E}log\left|\frac{f(z)}{f(qz)}\right|\frac{d\psi }{2\pi }\\ ={\int }_E(S_1^{\prime }\left(r{e}^{i\psi }\right)+S_2^{\prime }\left(r{e}^{i\psi }\right)-S_3^{\prime }\left(r{e}^{i\psi }\right))\frac{d\psi }{2\pi }\\ \le {\int }_0^{2\pi }(\left|S_1^{\prime }\left(r{e}^{i\psi }\right)\right|+\left|S_2^{\prime }\left(r{e}^{i\psi }\right)\right|+\left|S_3^{\prime }\left(r{e}^{i\psi }\right)\right|)\frac{d\psi }{2\pi }.\end{array}$$

Since $S_i^{\prime }=-S_i$ ($i=1,2,3$) in [4], we get $|S_i^{\prime }|=|S_i|$ ($i=1,2,3$).

Following the similar method in the proof of Theorem 1.1 in [4], we get the result. □

Lemma 2.6 Let f be a non-constant zero-order entire function, and $q\in \mathbb{C}\setminus \{0\}$. Then

$$T(r,P(f)f(qz))=T(r,P(f)f(z))+S(r,f)$$

on a set of logarithmic density 1.

Proof Since f is an entire function of zero-order, we deduce from Lemma 2.5 that

$$\begin{array}{rl}T(r,P(f)f(qz))& =m(r,P(f)f(qz))\\ \le m(r,P(f)f(z))+m(r,\frac{f(qz)}{f(z)})\\ \le m(r,P(f)f(z))+S(r,f)\\ =T(r,P(f)f(z))+S(r,f),\end{array}$$

that is

$$T(r,P(f)f(qz))\le T(r,P(f)f(z))+S(r,f).$$

(2.1)

On the other hand, using Remark 2, we get

$$\begin{array}{rl}T(r,P(f)f(z))& =m(r,P(f)f(z))\\ \le m(r,P(f)f(qz))+m(r,\frac{f(z)}{f(qz)})\\ \le m(r,P(f)f(qz))+S(r,f)\\ =T(r,P(f)f(qz))+S(r,f),\end{array}$$

that is

$$T(r,P(f)f(z))\le T(r,P(f)f(qz))+S(r,f).$$

(2.2)

The assertion follows from (2.1) and (2.2). □

3 Proof of Theorem 1.1

Let $\beta (z)$ be the canonical products of the nonzero poles of $P(f)f(z+c)-\alpha (z)$. Since $\lambda (1/f)<\sigma$ and $\alpha (z)$ is a small function of $f(z)$, we know that $\sigma (\beta )=\lambda (\beta )<\sigma (f)$. Suppose on contrary to the assertion that $P(f)f(z+c)-\alpha (z)$ has finitely many zeros. Then we have

$$P(f)f(z+c)-\alpha (z)=R(z)e^{Q(z)}/\beta (z),$$

where $Q(z)$ is a polynomial, and $R(z)\not\equiv 0$ is a rational function. Set $H(z)=R(z)/\beta (z)$. Then

$$\sigma (H)<\sigma (f)=\sigma ,$$

(3.1)

and

$$P(f)f(z+c)-\alpha (z)=H(z)e^{Q(z)}.$$

(3.2)

Differentiating (3.2) and eliminating $e^{Q(z)}$, we obtain

$$\begin{array}{r}{P}^{\prime }(f)f^{\prime }(z)f(z+c)H(z)+P(f)f^{\prime }(z+c)H(z)-P(f)f(z+c)H^{\prime }(z)-P(f)f(z+c)Q^{\prime }(z)H(z)\\ ={\alpha }^{\prime }(z)H(z)-\alpha (z)H^{\prime }(z)-\alpha (z)Q^{\prime }(z)H(z).\end{array}$$

(3.3)

Let ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_t$ be the distinct zeros of $P(z)$. Then

$$P(f)=a_n{(f-{\alpha }_1)}^{n_1}{(f-{\alpha }_2)}^{n_2}\cdots {(f-{\alpha }_t)}^{n_t}.$$

Substituting this into (3.3), we have

$$\begin{array}{r}{a}_n\prod _{j=1}^t{(f-{\alpha }_j)}^{n_j-1}\{(n_1\prod _{j\ne 1}(f-{\alpha }_j)+n_2\prod _{j\ne 2}(f-{\alpha }_j)+\cdots +n_t\prod _{j\ne t}(f-{\alpha }_j))\\ \times f(z+c)H(z)f^{\prime }(z)+f^{\prime }(z+c)H(z)\\ \times \prod _{j=1}^t(f-{\alpha }_j)-f(z+c)(H^{\prime }(z)+Q^{\prime }(z)H(z))\prod _{j=1}^t(f-{\alpha }_j)\}\\ ={\alpha }^{\prime }(z)H(z)-\alpha (z)H^{\prime }(z)-\alpha (z)Q^{\prime }(z)H(z).\end{array}$$

Note that $P(z)$ has at least one multiple zero, we may assume that $n_1>1$ without loss of generality, and we have

$$a_n{(f-{\alpha }_1)}^{n_1-1}F(z,f)={\alpha }^{\prime }(z)H(z)-\alpha (z)H^{\prime }(z)-\alpha (z)Q^{\prime }(z)H(z),$$

(3.4)

where

$$\begin{array}{rl}F(z,f)=& \prod _{j=2}^t{(f-{\alpha }_j)}^{n_j-1}\{(n_1\prod _{j\ne 1}(f-{\alpha }_j)+n_2\prod _{j\ne 2}(f-{\alpha }_j)+\cdots +n_t\prod _{j\ne t}(f-{\alpha }_j))\\ \times f(z+c)H(z)f^{\prime }(z)+f^{\prime }(z+c)H(z)\prod _{j=1}^t(f-{\alpha }_j)\\ -f(z+c)(H^{\prime }(z)+Q^{\prime }(z)H(z))\prod _{j=1}^t(f-{\alpha }_j)\}.\end{array}$$

Now we distinguish two cases.

Case 1. $F(z,f)\equiv 0$. In this case, we obtain from (3.4) that

$${\alpha }^{\prime }(z)H(z)-\alpha (z)H^{\prime }(z)-\alpha (z)Q^{\prime }(z)H(z)\equiv 0.$$

Since $\alpha (z)\not\equiv 0$ and $H(z)\not\equiv 0$, by integrating, we have

$$\frac{\alpha (z)}{H(z)}=k{e}^{Q(z)},$$

(3.5)

where k is a non-zero constant. From (3.2) and (3.5), we have

$$P(f)f(z+c)=(\frac{1}{k}+1)\alpha (z).$$

By Lemma 2.2, we have

$$\begin{array}{rl}nT(r,f(z))& =T(r,P(f))+O(1)\\ \le T(r,f(z+c))+T(r,\alpha (z))+O(1)\\ =T(r,f(z))+O\left(r^{\sigma -1+\epsilon }\right)+S(r,f).\end{array}$$

Since $n\ge n_1\ge 2$, and $f(z)$ is a transcendental, this is impossible.

Case 2. $F(z,f)\not\equiv 0$. In this case, we set

$$\begin{array}{rl}{F}^{\ast }(z,f)=& \frac{F(z,f)}{f-{\alpha }_1}\\ =& \prod _{j=2}^t{(f-{\alpha }_j)}^{n_j-1}\{(n_1\prod _{j\ne 1}(f-{\alpha }_j)+n_2\prod _{j\ne 2}(f-{\alpha }_j)+\cdots +n_t\prod _{j\ne t}(f-{\alpha }_j))\\ \times \frac{f(z+c)}{f(z)}f(z)H(z)\frac{f^{\prime }(z)}{f-{\alpha }_1}+\frac{f^{\prime }(z+c)}{f(z+c)}\frac{f(z+c)}{f(z)}f(z)H(z)\prod _{j=2}^t(f-{\alpha }_j)\\ -\frac{f(z+c)}{f(z)}f(z)(H^{\prime }(z)+Q^{\prime }(z)H(z))\prod _{j=2}^t(f-{\alpha }_j)\}.\end{array}$$

Since $f(z)=(f(z)-{\alpha }_1)+{\alpha }_1$ and $f^{(k)}={(f-{\alpha }_1)}^{(k)}$, we know that $F^{\ast }(z,f)$ is a differential polynomial of $f(z)-{\alpha }_1$ with meromorphic coefficients, and

$$a_n{(f-{\alpha }_1)}^{n_1}{F}^{\ast }(z,f)={\alpha }^{\prime }(z)H(z)-\alpha (z)H^{\prime }(z)-\alpha (z)Q^{\prime }(z)H(z).$$

(3.6)

By Lemma 2.3, we have

$$\begin{array}{rl}m(r,{(f-{\alpha }_1)}^k{F}^{\ast }(z,f))\le & 3m(r,\frac{f(z+c)}{f(z)})+m(r,\frac{f^{\prime }(z+c)}{f(z+c)})+m(r,\frac{f^{\prime }(z)}{f-{\alpha }_1})\\ +5T(r,H)+S(r,f)\end{array}$$

(3.7)

for $k=0$ and $k=1$.

Now for any given ε ($0<\epsilon <1$), we obtain from Lemma 2.1, Lemma 2.2 and (3.1) that

$$m(r,\frac{f(z+c)}{f(z)})=O\left(r^{\sigma -\epsilon }\right),T(r,H)=O\left(r^{\sigma -\epsilon }\right),$$

(3.8)

$$m(r,\frac{f^{\prime }(z+c)}{f(z+c)})=O\left(r^{\sigma -\epsilon }\right)+S(r,f).$$

(3.9)

The lemma of logarithmic derivative implies that

$$m(r,\frac{f^{\prime }(z)}{f-{\alpha }_1})=S(r,f).$$

(3.10)

It follows from (3.7) to (3.10) that

$$m(r,F^{\ast }(z,f))=O\left(r^{\sigma -\epsilon }\right)+S(r,f),$$

(3.11)

$$m(r,(f-{\alpha }_1)F^{\ast }(z,f))=O\left(r^{\sigma -\epsilon }\right)+S(r,f).$$

(3.12)

Since $(f-{\alpha }_1)F^{\ast }(z,f)=F(z,f)$, we obtain from the definition of $F(z,f)$ that

$$N(r,F(z,f))=O(N(r,H(z))+N(r,f))=O\left(r^{\sigma -\epsilon }\right)+S(r,f).$$

Thus,

$$T(r,(f-{\alpha }_1)F^{\ast }(z,f))=O\left(r^{\sigma -\epsilon }\right)+S(r,f).$$

(3.13)

Note that, a zero of $f(z)-{\alpha }_1$ which is not a pole of $f(z+c)$ and $H(z)$, is a pole of $F^{\ast }(z,f)$ with the multiplicity at most 1, we know from (3.6), (3.1), Lemma 2.4 and $\lambda (1/f)<\sigma$ that

$$\begin{array}{rl}(n_1-1)N(r,\frac{1}{f(z)-{\alpha }_1})\le & N(r,\frac{1}{{\alpha }^{\prime }(z)H(z)-\alpha (z)H^{\prime }(z)-\alpha (z)Q^{\prime }(z)H(z)})\\ +O\left(N(r,f(z+c))\right)+O(N(r,H))\\ =& O\left(r^{\sigma -\epsilon }\right)\end{array}$$

(3.14)

for the positive ε sufficiently small. Hence (see the definition of $F^{\ast }(z,f)$),

$$\begin{array}{rl}N(r,F^{\ast }(z,f))& =O(N(r,\frac{1}{f-{\alpha }_1})+N(r,f)+N(r,H))\\ =O\left(r^{\sigma -\epsilon }\right)+S(r,f).\end{array}$$

(3.15)

It follows from (3.15) and (3.11) that

$$T(r,F^{\ast }(z,f))=O\left(r^{\sigma -\epsilon }\right)+S(r,f).$$

(3.16)

Thus, we deduce from (3.16) and (3.13) that

$$\begin{array}{rl}T(r,f(z))& =T(r,f(z)-{\alpha }_1)+O(1)=T(r,\frac{(f-{\alpha }_1)F^{\ast }(z,f)}{F^{\ast }(z,f)})\\ =O\left(r^{\sigma -\epsilon }\right)+S(r,f).\end{array}$$

This contradicts that f is of order σ. Theorem 1.1 is proved.

4 Proof of Theorem 1.2

Denote $F(z)=P(f)f(qz)$. From Lemma 2.6 and the standard Valiron-Mohon’ko theorem, we deduce

$$\begin{array}{rl}T(r,F(z))& =T(r,P(f)f(z))+S(r,f)\\ =(n+1)T(r,f(z))+S(r,f).\end{array}$$

Since f is a entire function, then by the second main theorem and Lemma 2.5, we have

$$\begin{array}{rcl}T(r,F(z))& \le & \overline{N}(r,F(z))+\overline{N}(r,\frac{1}{F(z)})+\overline{N}(r,\frac{1}{F(z)-\alpha (z)})+S(r,f)\\ \le & \overline{N}(r,\frac{1}{P(f)})+\overline{N}(r,\frac{1}{f(qz)})+\overline{N}(r,\frac{1}{F(z)-\alpha (z)})+S(r,f)\\ \le & (s(P)+m(P))T(r,f(z))+T(r,f(qz))\\ +\overline{N}(r,\frac{1}{F(z)-\alpha (z)})+S(r,f)\\ \le & (s(P)+m(P))T(r,f(z))+m(r,\frac{f(qz)}{f(z)})+m(r,f(z))\\ +\overline{N}(r,\frac{1}{F(z)-\alpha (z)})+S(r,f)\\ \le & (s(P)+m(P)+1)T(r,f(z))+\overline{N}(r,\frac{1}{F(z)-\alpha (z)})+S(r,f),\end{array}$$

that is,

$$\overline{N}(r,\frac{1}{F(z)-\alpha (z)})\ge (n-s(P)-m(P))T(r,f(z))+S(r,f(z)).$$

Since f is a transcendental entire function with $m(P)>0$, we deduce that $P(f)f(qz)-\alpha (z)$ has infinitely many zeros.