Value distribution of q-difference differential polynomials of entire functions

Abstract

For a complex value $q\ne 0,1$, and a transcendental entire function $f(z)$ with order, $0<\sigma (f)<\mathrm{\infty }$, we study the value distribution of q-difference differential polynomials ${[f^n(z)(f(qz)-f(z))]}^{(k)}$ and ${[f(z)f(qz)]}^{(k)}$.

MSC:30D35, 39A05.

1 Introduction and main results

A meromorphic function $f(z)$ means meromorphic in the complex plane ℂ. If no poles occur, then $f(z)$ reduces to an entire function. For every real number $x\ge 0$, we define ${log}^{+}x:=max\{0,logx\}$. Assume that $n(r,f)$ counts the number of poles of f in $|z|<r$, each pole according to its multiplicity, and that $\overline{n}(r,f)$ counts the number of the distinct poles of f in $|z|<r$, ignoring the multiplicity. The characteristic function of $f(z)$ is defined by

$$T(r,f):=m(r,f)+N(r,f),$$

where

$$N(r,f):={\int }_0^r\frac{n(t,f)-n(0,f)}{t}dt+n(0,f)logr$$

and

$$m(r,f):=\frac{1}{2\pi }{\int }_0^{2\pi }{log}^{+}\left|f\left(r{e}^{i\theta }\right)\right|d\theta .$$

The notation $\overline{N}(r,f)$ is similarly defined with $\overline{n}(r,f)$ instead of $n(r,f)$. More notations and definitions of the Nevanlinna value distribution theory of meromorphic functions can be found in [1, 2].

A meromorphic function $\alpha (z)$ is called a small function with respect to $f(z)$, if $T(r,\alpha )=S(r,f)$, where $S(r,f)$ denotes any quantity satisfying $S(r,f)=o(T(r,f))$ as $r\to \mathrm{\infty }$ outside a possible exceptional set E of finite logarithmic measure. The order and the exponent of convergence of zeros of meromorphic function $f(z)$ is, respectively, defined as

$$\sigma (f)=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logT(r,f)}{logr},\lambda (f)=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logN(r,\frac{1}{f})}{logr}.$$

The difference operators for a meromorphic function $f(z)$ are defined as

$$\begin{array}{c}{\mathrm{\Delta }}_{c}f(z)=f(z+c)-f(z)(c\ne 0),\hfill \\ {\mathrm{\nabla }}_{q}f(z)=f(qz)-f(z)(q\ne 0,1).\hfill \end{array}$$

A Borel exceptional value of $f(z)$ is any value $a\in \mathbb{C}\cup \{\mathrm{\infty }\}$ satisfying $\lambda (f-a)<\sigma (f)$, where $\lambda (f-\mathrm{\infty })$ means $\lambda (\frac{1}{f})$.

Recently, the difference variant of the Nevanlinna theory has been established independently in [3–6]. Using these theories, value distributions of difference polynomials have been studied by many papers. For example, Laine and Yang [7] proved that if $f(z)$ is a transcendental entire function of finite order, c is a nonzero complex constant and $n\ge 2$, then $f^n(z)f(z+c)$ takes every nonzero value infinitely often.

Chen [8] considered the value distribution of $f(z)f(z+c)$ and obtained the following theorem.

Theorem A [[8], Corollary 1.3]

Let $f(z)$ be a transcendental entire function of finite order, and let c be a nonzero complex constant. If $f(z)$ has the Borel exceptional value 0, then $H(z)=f(z)f(z+c)$ takes every nonzero value $a\in \mathbb{C}$ infinitely often.

Chen [9] considered zeros of difference product $H_n(z)=f^n(z){\mathrm{\Delta }}_{c}f(z)$ and gave some conditions that guarantee $H_n(z)$ has finitely many zeros or infinitely many zeros.

Theorem B [[9], Theorem 1]

Let $f(z)$ be a transcendental entire function of finite order and $c\in \mathbb{C}\setminus \{0\}$ be a constant such that $f(z+c)\not\equiv f(z)$. Set $H_n(z)=f^n(z){\mathrm{\Delta }}_{c}f(z)$ where ${\mathrm{\Delta }}_{c}f(z)=f(z+c)-f(z)$, $n\ge 2$ is an integer. Then the following statements hold.

• (1) If $f(z)$ satisfies $\sigma (f)\ne 1$, or has infinitely many zeros, then $H_n(z)$ has infinitely many zeros.

• (2) If $f(z)$ has only finitely many zeros and $\sigma (f)=1$, then $H_n(z)$ has only finitely many zeros.

The zero distribution of differential polynomials is a classical topic in the theory of meromorphic functions. Hayman [[10], Theorem 10] firstly considered the value distribution of $f^n{f}^{\prime }-1$, where f is a transcendental function.

Recently, Liu, Liu and Cao [11] investigated the zeros of ${[f{(z)}^{n}f(z+c)]}^{(k)}-\alpha (z)$ and ${[f{(z)}^n{\mathrm{\Delta }}_{c}f(z)]}^{(k)}-\alpha (z)$, where $\alpha (z)$ is a nonzero small function with respect to $f(z)$.

Theorem C [[11], Theorems 1.1 and 1.3]

Let $f(z)$ be a transcendental entire function of finite order and $\alpha (z)$ be a nonzero small function with respect to $f(z)$. If $n\ge k+2$, then ${[f{(z)}^{n}f(z+c)]}^{(k)}-\alpha (z)$ has infinitely many zeros. If $f(z)$ is not a periodic function with period c and $n\ge k+3$, then ${[f{(z)}^n{\mathrm{\Delta }}_{c}f(z)]}^{(k)}-\alpha (z)$ has infinitely many zeros.

The main purpose of this paper is to consider a transcendental entire function $f(z)$ with positive and finite order and obtain some results on the value distributions of the q-difference differential polynomials ${[f^n(z)(f(qz)-f(z))]}^{(k)}$ and ${[f(z)f(qz)]}^{(k)}$. The first theorem will consider what conditions guarantee that ${[f^n(z){\mathrm{\nabla }}_{q}f(z)]}^{(k)}$ has infinitely many zeros.

Theorem 1.1 Let $f(z)$ be a transcendental entire function of finite and positive order $\sigma (f)$, $q\in \mathbb{C}\setminus \{0,1\}$ be a constant such that $q^{\sigma (f)}\ne 1$ and $f(z)\not\equiv f(qz)$. Set $H(z)={[f^n(z){\mathrm{\nabla }}_{q}f(z)]}^{(k)}$, $n\ge 1$ is an integer. If $f(z)$ has finitely many zeros, then $H(z)-\alpha (z)$ has infinitely many zeros, where $\alpha (z)$ is a nonzero small entire function with respect to $f(z)$.

In the following, we will study the value distribution of ${[f(z)f(qz)]}^{(k)}$.

Theorem 1.2 Let $f(z)$ be a transcendental entire function of finite and positive order $\sigma (f)$, a be a finite Borel exceptional value of $f(z)$, $q\in \mathbb{C}\setminus \{0,1\}$ be a constant such that $q^{\sigma (f)}\ne \pm 1$. Set $H(z)={[f(z)f(qz)]}^{(k)}$, then the following statements hold.

• (1) If $a=0$, then 0 is also the Borel exceptional value of $H(z)$. So that $H(z)$ has no nonzero finite Borel exceptional value.

• (2) If $a\ne 0$, then $H(z)$ has no finite Borel exceptional value.

• (3) $H(z)$ takes every nonzero value $c\in \mathbb{C}$ infinitely often and satisfies $\lambda (H-c)=\sigma (f)$.

Using the similar method of the proof of Theorem 1.2(1), we get the following result immediately.

Corollary 1.1 Let $f(z)$ be a transcendental entire function of finite and positive order $\sigma (f)$, $q\in \mathbb{C}\setminus \{0,1\}$ be a constant such that $q^{\sigma (f)}\ne -n$. If 0 is a Borel exceptional value of $f(z)$, then 0 is also the Borel exceptional value of ${[f^n(z)f(qz)]}^{(k)}$.

2 Some lemmas

The following are the well-known Weierstrass factorization and Hadamard factorization theorems.

Lemma 2.1 [12]

If an entire function $f(z)$ has a finite exponent of convergence $\lambda (f)$ for its zero-sequence, then $f(z)$ has a representation in the form

$$f(z)=Q(z)e^{g(z)},$$

satisfying $\lambda (Q)=\sigma (Q)=\lambda (f)$. Further, if $f(z)$ is of finite order, then $g(z)$ in the above form is a polynomial of degree less or equal to the order of $f(z)$.

Lemma 2.2 [13]

Suppose that $f_1(z),f_2(z),\dots ,f_n(z)$ ($n\ge 2$) are meromorphic functions and $g_1(z),g_2(z),\dots ,g_n(z)$ are entire functions satisfying the following conditions:

• (1) ${\sum }_{j=1}^n{f}_j(z)e^{g_j(z)}\equiv 0$;

• (2) $g_j(z)-g_k(z)$ are not constants for $1\le j<k\le n$;

• (3) For $1\le j\le n$, $1\le h<k\le n$, $T(r,f_j)=o(T(r,e^{g_h-g_k}))$ ($r\to \mathrm{\infty }$, $r\notin E$).

Then $f_j(z)\equiv 0$ ($j=1,2,\dots ,n$).

3 The proofs

3.1 Proof of Theorem 1.1

Since $f(z)$ is a transcendental entire function of finite order and has finitely many zeros, then by Lemma 2.1, $f(z)$ can be written as

$$f(z)=g(z)e^{h(z)},$$

where $g(z)$ (≢0), $h(z)$ are polynomials. Set

$$h(z)=a_s{z}^s+\cdots +a_0,$$

where $a_s,a_{s-1},\dots ,a_0$ are constants, and $a_s\ne 0$. Since $\sigma (f)\ne 0$, then $\sigma (f)=deg(h(z))=s\ge 1$. So

$$\begin{array}{rcl}H(z)& =& {[f^n(z){\mathrm{\nabla }}_{q}f(z)]}^{(k)}\\ =& {[g^n(z)g(qz)e^{nh(z)+h(qz)}-g^{n+1}(z)e^{(n+1)h(z)}]}^{(k)}\\ =& g_1(z)e^{nh(z)+h(qz)}-g_2(z)e^{(n+1)h(z)},\end{array}$$

where $g_1(z)$, $g_2(z)$ are nonzero polynomials and

$$(n+1)h(z)-(nh(z)+h(qz))=(1-q^s)a_s{z}^s+(1-q^{s-1})a_{s-1}{z}^{s-1}+\cdots +(1-q)a_{1}z.$$

Since $q^{\sigma (f)}=q^s\ne 1$, then $(n+1)h(z)-(nh(z)+h(qz))$ is not a constant. So $H(z)$ is a transcendental entire function, suppose that $H(z)-\alpha (z)$ has finitely many zeros, then by Lemma 2.1, $H(z)-\alpha (z)$ can be written as

$$H(z)-\alpha (z)=g_3(z)e^{h_1(z)}.$$

Here $g_3(z)$ (≢0), $h_1(z)$ are polynomials. Combing the above equalities, we obtain

$$g_1(z)e^{nh(z)+h(qz)}-g_2(z)e^{(n+1)h(z)}-g_3(z)e^{h_1(z)}-\alpha (z)=0.$$

(3.1)

Note that $(n+1)h(z)-(nh(z)+h(qz))$ is not a constant and $g_1(z)$, $g_2(z)$, $g_3(z)$ are nonzero polynomials. If $(nh(z)+h(qz))-h_1(z)$ and $(n+1)h(z)-h_1(z)$ are not constants, then by (3.1) and Lemma 2.2, we obtain

$$g_1(z)\equiv 0,g_2(z)\equiv 0,g_3(z)\equiv 0,\alpha (z)\equiv 0.$$

This is a contradiction.

If $(nh(z)+h(qz))-h_1(z)=c$, where c is a constant, then (3.1) can be rewritten as

$$(g_1(z)-e^{-c}{g}_3(z))e^{nh(z)+h(qz)}-g_2(z)e^{(n+1)h(z)}-\alpha (z)=0.$$

(3.2)

By (3.2) and Lemma 2.2, we obtain

$$g_1(z)-e^{-c}{g}_3(z)\equiv 0,g_2(z)\equiv 0,\alpha (z)\equiv 0.$$

This is a contradiction.

If $(n+1)h(z)-h_1(z)=c$, where c is a constant, then using the same method as above, we also obtain a contradiction.

Hence $H(z)-\alpha (z)$ has infinitely many zeros.

3.2 Proof of Theorem 1.2

Since $f(z)$ is a transcendental entire function of finite and positive order with a Borel exceptional value a, then by Lemma 2.1, $f(z)$ can be written as

$$f(z)=a+g(z)e^{\alpha z^s}.$$

Here s is a positive integer, α is a nonzero constant, $g(z)$ is a nonzero entire function satisfying $\sigma (g)<\sigma (f)=s$, thus

$$\begin{array}{rcl}f(z)f(qz)& =& (a+g(z)e^{\alpha z^s})(a+g(qz)e^{\alpha q^s{z}^s})\\ =& a^2+ag(qz)e^{\alpha q^s{z}^s}+ag(z)e^{\alpha z^s}+g(z)g(qz)e^{\alpha (1+q^s)z^s}.\end{array}$$

Since $q^{\sigma (f)}=q^s\ne \pm 1$, $g(z)$ is a nonzero entire function satisfying $\sigma (g)<s$, $f(z)f(qz)$ is a transcendental entire function and $\sigma (f(z)f(qz))=s$, and by the classical result of Nevanlinna theory, we get $\sigma (H)=\sigma (f(z)f(qz))=s$. Then

$$\begin{array}{rcl}H(z)& =& {[f(z)f(qz)]}^{(k)}\\ =& {[a^2+ag(qz)e^{\alpha q^s{z}^s}+ag(z)e^{\alpha z^s}+g(z)g(qz)e^{\alpha (1+q^s)z^s}]}^{(k)}\\ =& g_1(z)e^{\alpha q^s{z}^s}+g_2(z)e^{\alpha z^s}+g_3(z)e^{\alpha (1+q^s)z^s}.\end{array}$$

Here $g_1(z)$, $g_2(z)$, $g_3(z)$ are differential polynomials of $g(z)$, $g(qz)$ and $\sigma (g_i(z))<s$, $i=1,2,3$.

Case 1. $a=0$, by the above equality, we get

$$H(z)=g_3(z)e^{\alpha (1+q^s)z^s}.$$

Since $q^{\sigma (f)}=q^s\ne -1$ and $\sigma (g_3(z))<s$, this implies that 0 is the Borel exceptional value of $H(z)$.

Case 2. $a\ne 0$, suppose $H(z)$ has a finite Borel exceptional value b, then by Lemma 2.1, $H(z)$ can be written as

$$H(z)=b+h(z)e^{\beta z^s}.$$

Here β is a nonzero constant, $h(z)$ is a nonzero entire function satisfying $\sigma (h)<\sigma (H)=s$. Combing the above equalities, we obtain

$$g_1(z)e^{\alpha q^s{z}^s}+g_2(z)e^{\alpha z^s}+g_3(z)e^{\alpha (1+q^s)z^s}-h(z)e^{\beta z^s}-b=0.$$

(3.3)

Since $q^{\sigma (f)}=q^s\ne \pm 1$, we have $\alpha q^s\ne \alpha \ne \alpha (1+q^s)$.

If $\alpha q^s\ne \beta$, $\alpha \ne \beta$, $\alpha (1+q^s)\ne \beta$. By (3.3) and Lemma 2.2, we obtain

$$g_1(z)\equiv 0,g_2(z)\equiv 0,g_3(z)\equiv 0,h(z)\equiv 0,b\equiv 0.$$

This is a contradiction.

If $\alpha q^s=\beta$, then (3.3) can be rewritten as

$$(g_1(z)-h(z))e^{\alpha q^s{z}^s}+g_2(z)e^{\alpha z^s}+g_3(z)e^{\alpha (1+q^s)z^s}-b=0.$$

(3.4)

By (3.4) and Lemma 2.2, we obtain

$$g_1(z)-h(z)\equiv 0,g_2(z)\equiv 0,g_3(z)\equiv 0,b\equiv 0.$$

Which is a contradiction.

If $\alpha =\beta$ or $\alpha (1+q^s)=\beta$, then using the same method as above, we also obtain a contradiction.

Case 3. From Case 1 and Case 2, we get that if $f(z)$ has a finite Borel exceptional value, then any nonzero finite value c must not be the Borel exceptional value of $H(z)$, so $H(z)$ takes every nonzero value $c\in \mathbb{C}$ infinitely often, since $\sigma (H)=s$, then $\lambda (H-c)=\sigma (H)=\sigma (f)$.

The proof of Theorem 1.2 is completed.