The zeros of q-shift difference polynomials of meromorphic functions

Abstract

In this paper, we investigate the value distribution of difference polynomials $f{(z)}^{n}f(qz+c)$ and $f^n(z)+a[f(qz+c)-f(z)]$ related to two well-known differential polynomials, where $f(z)$ is a meromorphic function with finite logarithmic order.

MSC:30D35, 39B12.

1 Introduction

In this paper, we shall assume that the reader is familiar with the fundamental results and the standard notation of the Nevanlinna value distribution theory of meromorphic functions (see [1, 2]). The term ‘meromorphic function’ will mean meromorphic in the whole complex plane ℂ. In addition, we will use notations $\rho (f)$ to denote the order of growth of a meromorphic function $f(z)$, $\lambda (f)$ to denote the exponents of convergence of the zero-sequence of a meromorphic function $f(z)$, $\lambda (\frac{1}{f})$ to denote the exponents of convergence of the sequence of distinct poles of $f(z)$.

The non-autonomous Schröder q-difference equation

$$f(qz)=R(z,f(z)),$$

(1.1)

where the right-hand side is rational in both arguments, has been widely studied during the last decades (see, e.g., [3–7]). There is a variety of methods which can be used to study the value distribution of meromorphic solutions of (1.1).

Recently, the Nevanlinna theory involving q-difference has been developed to study q-difference equations and q-difference polynomials. Many papers have focused on complex difference, giving many difference analogues in value distribution theory of meromorphic functions (see [8–15]).

Hayman [16] posed the following famous conjecture.

Theorem A If f is a transcendental meromorphic function and $n\ge 1$, then $f^n{f}^{\prime }$ takes every finite nonzero value $b\in \mathbb{C}$ infinitely often.

This conjecture has been solved by Hayman [1] for $n\ge 3$, by Mues [17] for $n=2$, by Bergweiler and Eremenko [18] for $n=1$.

Hayman [16] also proved the following famous result.

Theorem B If $f(z)$ is a transcendental meromorphic function, $n\ge 5$ is an integer, and a (≠0) is a constant, then $f^{\prime }(z)-af{(z)}^n$ assumes all finite values $b\in \mathbb{C}$ infinitely often.

He also conjectured in [1] that the same result holds for $n=3$ and 4. However, Mues [17] proved that the conjecture is not true by providing a counterexample and proved that $f^{\prime }-a{f}^4$ has infinitely many zeros.

Liu and Qi proved two theorems which considered q-shift difference polynomials (see [13]), which can be seen as difference versions of the above classical results.

Theorem C Let f be a zero-order transcendental meromorphic function and q be a nonzero complex constant. Then, for $n\ge 6$, $f^n(z)f(qz+c)$ assumes every nonzero value $b\in \mathbb{C}$ infinitely often.

Theorem D Let f be a zero-order transcendental meromorphic function and a, q be nonzero complex constants. Then, for $n\ge 8$, $f^n(z)+a[f(qz+c)-f(z)]$ assumes every nonzero value $b\in \mathbb{C}$ infinitely often.

Remark 1 They also conjectured the numbers $n\ge 6$ and $n\ge 8$ can be reduced in Theorems C and D. But they could not deal with it. In fact, Zhang and Korhonen [19] also proved a result similar to Theorem C under the condition $n\ge 6$. Obviously, it is an interesting question to reduce the number n. In this paper, our results give some answers in some sense.

2 Main results

In order to express our results, we need to introduce some definitions (see [20, 21]).

A positive increasing function $S(r)$, defined for $r>0$, is said to be of finite logarithmic order λ if

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logS(r)}{loglogr}=\lambda .$$

(2.1)

$S(r)$ is said to be of infinite logarithmic order if the limit superior above is infinite.

Definition 1 If $f(z)$ is a function meromorphic in the complex plane ℂ, the logarithmic order of f is the logarithmic order of its characteristic function $T(r,f)$.

It is clear that the logarithmic order of a non-constant rational function is 1, but there exist infinitely many transcendental entire functions of logarithmic order 1 from Theorem 7.3 [20]. Hence, the transcendental meromorphic function is of the logarithmic order ≥1.

Let $f(z)$ be a meromorphic function of finite positive logarithmic order λ. A non-negative continuous function $\lambda (r)$ defined in $(0,+\mathrm{\infty })$ is said to be proximate logarithmic order of $T(r,f)$, if $\lambda (r)$ satisfies the following three conditions:

• (1) ${lim}_{r\to +\mathrm{\infty }}\lambda (r)=\lambda$.

• (2) ${\lambda }^{\prime }(r)$ exists everywhere in $(0,+\mathrm{\infty })$ except possibly in a countable set where ${\lambda }^{\prime }(r^{+})$ and ${\lambda }^{\prime }(r^{-})$ exist. Moreover, if we use the one-sided derivative ${\lambda }^{\prime }(r^{+})$ or ${\lambda }^{\prime }(r^{-})$ instead of ${\lambda }^{\prime }(r)$ of r in the exceptional set, then

  $$\underset{r\to +\mathrm{\infty }}{lim}r(logr){\lambda }^{\prime }(r)loglogr=0.$$

  (2.2)

• (3) Let $U(r,f)={(logr)}^{\lambda (r)}$, we have $T(r,f)\le U(r,f)$ for sufficiently large r and

  $$\frac{T(r,f)}{U(r,f)}=1.$$

  (2.3)

The above function $U(r,f)$ is called a logarithmic-type function of $T(r,f)$. If $f(z)$ is a meromorphic function of finite positive logarithmic order λ, then $T(r,f)$ has proximate logarithmic order $\lambda (r)$.

Let $f(z)$ be a meromorphic function, for each $a\in \hat{\mathbb{C}}=\mathbb{C}\cup \{\mathrm{\infty }\}$, an a-point of $f(z)$ means a root of the equation $f(z)=a$. Let $\{z_j(a)\}$ be the sequence of a-points of $f(z)$ with $r_j(a)\le r_{j+1}(a)$, where $r_j(a)=|z_j(a)|$. The logarithmic exponent of convergence of a-points of $f(z)$ is a number ${\rho }_{log}(a)$ which is defined by

$${\rho }_{log}(a)=inf\left\{\mu \right|\mu >0,\sum _{j}1/{|log{r}_j(a)|}^{\mu }<+\mathrm{\infty }\}.$$

(2.4)

This quantity plays an important role in measuring the value distribution of a-points of $f(z)$.

Throughout this paper, we denote the logarithmic order of $n(r,f=a)$ by ${\lambda }_{log}(a)$, where $n(r,f=a)$ is the number of roots of the equation $f(z)=a$ in $|z|\le r$. It is well known that if a meromorphic function $f(z)$ is of finite order, then the order of $n(r,f=a)$ equals the exponent of convergence of a-points of $f(z)$. The corresponding result for meromorphic functions of finite logarithmic order also holds. That is, if $f(z)$ is a non-constant meromorphic function and of finite logarithmic order, then for each $a\in \hat{\mathbb{C}}$, the logarithmic order of $n(r,f=a)$ equals the logarithmic exponent of convergence of a-points of $f(z)$.

Although for any given meromorphic function $f(z)$ with finite positive order and for any $a\in \hat{\mathbb{C}}$, the counting functions $N(r,f=a)$ and $n(r,f=a)$ both have the same order, the situation is different for functions of finite logarithmic order. That is, if $f(z)$ is a non-constant meromorphic function in ℂ, for each $a\in \hat{\mathbb{C}}$, $N(r,f=a)$ is of logarithmic order ${\lambda }_{log}(a)+1$ where ${\lambda }_{log}(a)$ is the logarithmic order of $n(r,f=a)$.

Theorem 2.1 If $f(z)$ is a transcendental meromorphic function of finite logarithmic order λ, with the logarithmic exponent of convergence of poles less than $\lambda -1$ and q, c are nonzero complex constants, then for $n\ge 2$, $f^n(z)f(qz+c)$ assumes every value $b\in \mathbb{C}$ infinitely often.

Remark 2 The following examples show that the hypothesis the logarithmic exponent of convergence of poles ${\lambda }_{log}(\mathrm{\infty })$ is less than $\lambda -1$ is sharp.

Example 1 Let $f(z)={\prod }_{j=0}^{\mathrm{\infty }}{(1-q^{j}z)}^{-1}$, $0<|p|<1$. Then $f(qz)=(1-z)f(z)$. But $f^n(z)f(qz)$ have only one zero. We know (see [22, 23])

$$T(r,f)=N(r,f)=O\left({(logr)}^2\right).$$

Thus, $\lambda =2$, and the logarithmic exponent of convergence of poles ${\lambda }_{log}(\mathrm{\infty })$ is equal to $\lambda -1=1$. Hence, the condition the logarithmic exponent of convergence of poles is less than $\lambda -1$ cannot be omitted.

Example 2 Let $f(z)={\prod }_{j=0}^{\mathrm{\infty }}{(1-z/q^j)}^{-1}$, $|p|>1$. Then $f(qz)=\frac{f(z)}{1-qz}$ and $f^n(z)f(qz)=\frac{f{(z)}^{n+1}}{1-qz}$ have no zeros. Note that $T(r,f)=N(r,f)=O({(logr)}^2)$ (see [22–24]), then $\lambda =2$, and the logarithmic exponent of convergence of poles ${\lambda }_{log}(\mathrm{\infty })$ is equal to $\lambda -1=1$. Hence, our condition the logarithmic exponent of convergence of poles is less than $\lambda -1$ cannot be omitted.

Remark 3 We note that the authors claimed b is nonzero in Theorem C. But b can be zero in Theorem 2.1.

Theorem 2.2 If $f(z)$ is a transcendental meromorphic function of finite logarithmic order λ, with the logarithmic exponent of convergence of poles less than $\lambda -1$, and a, q are nonzero complex constants, then for $n\ge 5$, $f^n(z)+a[f(qz+c)-f(z)]$ assumes every value $b\in \mathbb{C}$ infinitely often.

Remark 4 The authors also claimed b is nonzero in Theorem D. In fact, b can take the zeros in Theorem D from their proof.

In the following, we consider the difference polynomials similar to Theorem 2.2 and Theorem 1.5 in [25].

Theorem 2.3 If $f(z)$ is a transcendental meromorphic function of finite logarithmic order λ, with the logarithmic exponent of convergence of poles less than $\lambda -1$, and a, q are nonzero constants, then for $n\ge 3$, $f^n(z)-af(qz+c)$ assumes every value $b\in \mathbb{C}$ infinitely often.

3 Proof of Theorem 2.1

We need the following lemmas for the proof of Theorem 2.1.

For a transcendental meromorphic function $f(z)$, $T(r,f)$ is usually dominated by three integrated counting functions. However, when $f(z)$ is of finite logarithmic order, $T(r,f)$ can be dominated by two integrated counting functions as the following shows.

Lemma 3.1 ([20])

If $f(z)$ is a transcendental meromorphic function of finite logarithmic order λ, then for any two distinct extended complex values a and b, we have

$$T(r,f)\le N(r,f=a)+N(r,f=b)+o(U(r,f)),$$

(3.1)

where $U(r,f)={(logr)}^{\lambda (r)}$ is a logarithmic-type function of $T(r,f)$. Furthermore, if $T(r,f)$ has a finite lower logarithmic order

$$\mu =\underset{r\to +\mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{logT(r,f)}{loglogr},$$

(3.2)

with $\lambda -\mu <1$, then

$$T(r,f)\le N(r,f=a)+N(r,f=b)+o(T(r,f)).$$

(3.3)

Lemma 3.2 If $f(z)$ is a non-constant zero-order meromorphic function and $q\in \mathbb{C}\mathrm{\setminus }\{0\}$, then

$$T(r,f(qz))=(1+o(1))T(r,f(z))$$

(3.4)

on a set of lower logarithmic density 1.

Lemma 3.3 ([26])

Let $f(z)$ be a meromorphic function of finite order ρ, and let $c\in \mathbb{C}\mathrm{\setminus }\{0\}$. Then, for each $\epsilon >0$, one has

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

(3.5)

From the proof of Theorem 2.1 in [26], we know if $f(z)$ is of zero order, then (3.5) can be written into

$$T(r,f(z+c))=T(r,f)+O(logr).$$

(3.6)

From Lemma 3.2 and (3.6), we can obtain

Lemma 3.4 If $f(z)$ is a non-constant zero-order meromorphic function and $q\in \mathbb{C}\mathrm{\setminus }\{0\}$, then

$$T(r,f(qz+c))=(1+o(1))T(r,f(z))+O(logr)$$

(3.7)

on a set of lower logarithmic density 1.

Remark 5 If $f(z)$ is a transcendental meromorphic function of finite logarithmic order λ, then (3.7) can be rewritten into

$$T(r,f(qz+c))=(1+o(1))T(r,f(z)).$$

(3.8)

Lemma 3.5 ([27])

Let f be a non-constant meromorphic function, n be a positive integer. $P(f)=a_n{f}^n+a_{n-1}{f}^{n-1}+\cdots +a_{1}f$ where $a_i$ is a meromorphic function satisfying $T(r,a_i)=S(r,f)$ ($i=1,2,\dots ,n$). Then

$$T(r,P(f))=nT(r,f)+S(r,f).$$

Lemma 3.6 If $f(z)$ is a non-constant zero-order meromorphic function and $q\in \mathbb{C}\mathrm{\setminus }\{0\}$, then

$$N(r,f(qz+c))=(1+o(1))N(r,f(z))+O(logr)$$

(3.9)

on a set of lower logarithmic density 1.

From the proofs of Theorem 1.3 [19] and Theorem 2.2 [26], we can obtain the lemma easily. If $f(z)$ is a transcendental meromorphic function of finite logarithmic order λ, then (3.9) can be rewritten into

$$N(r,f(qz+c))=(1+o(1))N(r,f(z)).$$

(3.10)

Lemma 3.7 ([13, 28])

Let $f(z)$ be a non-constant meromorphic function of zero order, and let $q,c\in \mathbb{C}\mathrm{\setminus }\{0\}$. Then

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

on a set of logarithmic density 1.

Lemma 3.8 If $f(z)$ is a transcendental meromorphic function of finite logarithmic order λ and $c\in \mathbb{C}\mathrm{\setminus }\{0\}$, and $n\ge 2$ is an integer, set $G(z)=f^n(z)f(qz+c)$, then $T(r,G)=O(T(r,f))$.

Proof We can rewrite $G(z)$ in the form

$$G(z)=f{(z)}^{n+1}\frac{f(qz+c)}{f(z)}.$$

(3.11)

For each $\epsilon >0$, by Lemma 3.7 and (3.11), we get that

$$m(r,G)\le (n+1)m(r,f)+m(r,\frac{f(qz+c)}{f(z)})=(n+1)m(r,f)+o\{T(r,f)\}.$$

(3.12)

From Lemma 3.6, we have

$$N(r,G)\le nN(r,f)+N(r,f(qz+c))\le (n+1+o(1))N(r,f).$$

(3.13)

By (3.12) and (3.13), we have

$$T(r,G)\le O(T(r,f)).$$

(3.14)

By Lemma 3.5 and (3.8), we have

$$\begin{array}{rcl}nT(r,f(z))& =& T(r,f^n(z))=T(r,\frac{G(z)}{f(qz+c)})\le T(r,G)+T(r,f(qz+c))\\ =& T(r,G(z))+(1+o(1))T(r,f(z)).\end{array}$$

(3.15)

Thus, from (3.15) we have

$$(n-1-o(1))T(r,f)\le T(r,G).$$

(3.16)

That is, $T(r,f)\le O(T(r,G))$ from $n\ge 2$.

Thus, (3.14) and (3.16) give that $T(r,G)=O(T(r,f))$. □

Proof of Theorem 2.1 Denote $G(z)=f{(z)}^{n}f(qz+c)$.

We claim that $G(z)$ is transcendental if $n\ge 2$.

Suppose that $G(z)$ is a rational function $R(z)$. Then $f{(z)}^n=R(z)/f(qz+c)$. Therefore, by Lemma 3.5 and (3.8), $nT(r,f(z))=T(r,f{(z)}^n)=T(r,R(z)/f(qz+c))\le T(r,f(qz+c))+O(logr)=(1+o(1))T(r,f(z))$, which contradicts $n\ge 2$.

Hence, the claim holds.

By Lemma 3.1, Lemmas 3.5-3.6, Lemma 3.8, and (3.8), we have

$$\begin{array}{rcl}nT(r,f(z))& =& T(r,f{(z)}^n)=T(r,G(z)/f(qz+c))\\ \le & T(r,F(z))+T(r,f(qz+c))+O(1)\\ =& (1+o(1))T(r,f(z))+T(r,G)\\ =& (1+o(1))T(r,f(z))+N(r,G)+N(r,\frac{1}{G-b})+o(U(r,G))\\ \le & (1+o(1))T(r,f(z))+(n+1+o(1))N(r,f(z))\\ +N(r,\frac{1}{G-b})+o(U(r,f)),\end{array}$$

(3.17)

where $U(r,f)={(logr)}^{\lambda (r)}$ is a logarithmic-type function of $T(r,f)$.

Note that for the logarithmic exponent of convergence of poles less than $\lambda -1$, we have

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logN(r,f(z))}{loglogr}<\lambda .$$

(3.18)

Suppose that

$$N(r,\frac{1}{G-b})=o(T(r,G)).$$

(3.19)

By Lemma 3.8, we know (3.19) can be written into

$$N(r,\frac{1}{G-b})=o(T(r,f)).$$

(3.20)

From (3.17) and (3.20), we have

$$(n-1-o(1))T(r,f(z))\le (n+1+o(1))N(r,f(z))+o(U(r,f))$$

(3.21)

for sufficiently large r.

By (3.19), (3.21), and $n\ge 2$, we have

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logT(r,f)}{loglogr}<\lambda .$$

(3.22)

This contradicts the fact that $T(r,f)$ has logarithmic order λ. Hence,

$$N(r,\frac{1}{G-b})\ne o(T(r,G)).$$

That is, $G-b$ has infinitely many zeros, then $f^n(z)f(qz+c)-b$ has infinitely many zeros. This completes the proof of Theorem 2.1. □

4 Proof of Theorems 2.2 and 2.3

Let

$$\phi =\frac{b-af(z)-af(qz+c)}{f^n(z)}.$$

(4.1)

By Lemma 3.4 and (3.8), we obtain

$$\begin{array}{rcl}nT(r,f(z))& =& T(r,f{(z)}^n)=T(r,\phi (z)/(b-af(z)-af(qz+c)))+O(1)\\ \le & T(r,\phi (z))+T(r,f(qz+c))+T(r,f(z))+O(1)\\ =& T(r,\phi (z))+(2+o(1))T(r,f(z))+O(1),\end{array}$$

which implies that

$$T(r,\phi (z))\ge (n-2-o(1))T(r,f)+O(1).$$

(4.2)

From (4.1), we can easily get

$$T(r,\phi (z))\le (n+2+o(1))T(r,f)+O(1).$$

(4.3)

By (4.2)-(4.3) and $n\ge 3$, we have

$$T(r,\phi (z))=O(T(r,f)).$$

(4.4)

We claim that $\phi (z)$ is transcendental. Suppose that $\phi (z)$ is rational, it contradicts (4.2) if $n\ge 3$. The claim holds.

Suppose that $N(r,\frac{1}{\phi -1})=o\{T(r,\phi )\}$. In the following, we will get a contradiction. By (4.4), we have

$$N(r,\frac{1}{\phi -1})=o\{T(r,f)\}.$$

(4.5)

Note that for the logarithmic exponent of convergence of poles less than $\lambda -1$, we have

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logN(r,f(z))}{loglogr}<\lambda .$$

(4.6)

By Lemma 3.1 and (4.4)-(4.5), we obtain

$$\begin{array}{rcl}T(r,\phi (z))& \le & N(r,\frac{1}{\phi })+N(r,\frac{1}{\phi -1})+o(U(r,\phi ))\\ \le & N(r,\frac{1}{b-af(z)-af(qz+c)})+nN(r,f)+N(r,\frac{1}{\phi -1})+o(U(r,f))\\ \le & T(r,f(qz))+T(r,f(z))+nN(r,f)+N(r,\frac{1}{\phi -1})+o(U(r,f))\\ \le & (2+o(1))T(r,f(z))+nN(r,f)+N(r,\frac{1}{\phi -1})+o(U(r,f))\\ \le & (2+o(1))T(r,f(z))+nN(r,f)+o(T(r,f))+o(U(r,f)),\end{array}$$

(4.7)

where $U(r,f)={(logr)}^{\lambda (r)}$ is a logarithmic-type function of $T(r,f)$.

By (4.2) and (4.7), we have

$$(n-4-o(1))T(r,f)\le nN(r,f)+o(T(r,f))+o(U(r,f)),$$

for sufficiently large r. Hence, by (4.6) we have

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logT(r,f)}{loglogr}<\lambda .$$

(4.8)

This contradicts the fact that $T(r,f)$ has logarithmic order λ. Hence,

$$N(r,\frac{1}{\phi -1})\ne o(T(r,F)).$$

That is, $\phi -1$ has infinitely many zeros, then $f^n(z)+a[f(qz+c)-f(z)]-b$ has infinitely many zeros.

This completes the proof of Theorem 2.2.

The proof of Theorem 2.3 is similar to the proof of Theorem 2.2, we omit it here.