\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 233, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/233\hfil Value distribution of the $q$-difference product]
{Value distribution of the $q$-difference product of entire functions}

\author[N. Xu, T.-B. Cao, C.-P. Zhong \hfil EJDE-2014/233\hfilneg]
{Na Xu, Ting-Bin Cao, Chun-Ping Zhong}  % in alphabetical order

\address{Na Xu \newline
School of Mathematical Sciences, Xiamen University, Xiamen 361005, China }
\email{xuna406@163.com}

\address{Ting-Bin Cao (corresponding author)\newline
Department of Mathematics, Nanchang University, Nanchang,
Jiangxi 330031, China}
\email{tbcao@ncu.edu.cn}

\address{Chun-Ping Zhong \newline
School of Mathematical Sciences, Xiamen University,
Xiamen 361005,  China}
\email{zcp@xmu.edu.cn}


\thanks{Submitted April 7, 2014. Published November 3, 2014.}
\subjclass[2000]{30D35, 39A05}
\keywords{Nevanlinna theory; $q$-difference; entire functions}

\begin{abstract}
 For a complex value $q\neq 0, 1$, and a transcendental entire function
 $f(z)$ with order $0<\sigma(f)<\infty$, we study the value distribution
 of $q$-difference product $f(z)f(qz)$ and $f^{n}(z)(f(qz)-f(z))$.
 Properties of entire solution of a certain $q$-difference linear
 equation are also considered.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction and main results}

 A meromorphic function $f(z)$ means meromorphic in the complex plane
$\mathbb{C}$. If no poles occur, then $f(z)$ reduces to an entire function.
For every real number $x\geq 0$, we define
$\log^{+}x:=\max\{0, \log x\}$. Assume that $n(r, f)$ counts the number of
the poles of $f$ in $|z|\leq r$, each pole according to its multiplicity,
and that $\overline{n}(r,f)$ counts the number of the distinct poles of $f$
in $|z|\leq r$, ignoring the multiplicity.
The characteristic function of $f$ is defined by
$$
T(r,f):=m(r,f)+N(r,f),
$$
where
\begin{gather*}
N(r,f):=\int_{0}^{r}\frac{n(t,f)-n(0,f)}{t}dt+n(0,f)\log r, \\
m(r,f):=\frac{1}{2\pi}\int_{0}^{2\pi}\log^{+}|f(re^{i\theta})|d\theta.
\end{gather*}
The notation $\overline{N}(r,f)$ is similarly defined with
$\overline{n}(r,f)$ instead of $n(r,f)$. For more notations and
definitions of the Nevanlinna's value distribution theory of
meromorphic functions, refer to \cite{hayman-1, yl}.

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 $r\longrightarrow\infty$
outside a possible exceptional set $E$ of logarithmic density $0$.
The order and the exponent of convergence of zeros of meromorphic function
$f(z)$ is respectively defined as
\begin{gather*}
\sigma(f)=\limsup_{r\to\infty}\frac{\log {T(r,f)}}{\log r},\\
\lambda(f)=\limsup_{r\to\infty}\frac{\log {N(r,\frac{1}{f})}}{\log r}.
\end{gather*}
The reduced deficiency of $a$ with respect to $f(z)$ is defined by
 $$
\Theta(a,f)=1-\limsup_{r\to\infty}\frac{\overline{N}(r,\frac{1}{f-a})}{T(r,f)}.
$$
The difference operators for a meromorphic function $f$ are defined as
\begin{gather*}
\triangle_{c}f(z)=f(z+c)-f(z)\quad (c\neq 0),\\
\nabla_{q}f(z)=f(qz)-f(z) \quad (q\neq 0,1).
\end{gather*}
 A Borel exceptional value of $f(z)$ is any value $a$ satisfying
 $\lambda(f-a)<\sigma(f)$.

Recently, the difference variant of the Nevanlinna theory has been established
independently in \cite{Barnett, Chiang, Halburd 1, Halburd 2}.
Using these theories, value distributions of difference polynomials have been
studied by many papers. For example, Laine and Yang \cite{Laine} proved if
$f(z)$ is a transcendental entire function of finite order, $c$ is a
nonzero complex constant and $n\geq2$, then $f^{n}(z)f(z+c)$ takes every
nonzero value infinitely often.
Liu and Yang \cite{Liu 1} proved the following theorem.


\begin{theorem}[{\cite[Theorem 1.4]{Liu 1}}]  \label{thmA}
Let $f(z)$ be a transcendental entire function of finite order,
and $c$ be a nonzero complex constant, $\triangle_{c} f(z)=f(z+c)-f(z)\not\equiv0$.
Then for $n\geq2$, $f^{n}(z)\triangle_{c} f(z)-p(z)$ has infinitely many zeros,
where $p(z)\not\equiv0$ is a polynomial in $z$.
\end{theorem}

The following theorems  discussed the case $n\geq2$. For the case $n=1$,
 Chen \cite{Chen 2}, Chen-Huang-Zheng \cite{Chen 4} considered value
distributions of $f(z)f(z+c), f(z)\Delta_{c}f(z)$.


\begin{theorem}[{\cite[Corollary 1.3]{Chen 4}}] \label{thmB}
Let $f(z)$ be a transcendental entire function of finite order, and
$c$ 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.
\end{theorem}

\begin{theorem}[{\cite[Theorem 2]{Chen 2}}] \label{thmC}
Let $f(z)$ be a finite order transcendental entire function with a
finite Borel exceptional value $d$, and let $c\in\mathbb{C}\setminus\{0\}$
be a constant satisfying $f(z+c)\not\equiv f(z)$.
Set $H(z)=f(z)\Delta_{c}f(z)$ where $\triangle_{c} f(z)=f(z+c)-f(z)$.
Then the following statements hold:
\begin{itemize}
\item[(1)] $H(z)$ takes every nonzero value $a\in\mathbb{C}$ infinitely
often and satisfies $\lambda(H-a)=\sigma(f)$.

\item[(2)] If $d\neq0$, then $H(z)$ has no any finite Borel exceptional value.

\item[(3)] If $d=0$, then $0$ is also the Borel exceptional value of $H(z)$.
So that $H(z)$ has no nonzero finite Borel exceptional value.
\end{itemize}
\end{theorem}


\begin{theorem}[{\cite[Theorem 3]{Chen 2}}] \label{thmD}
Let $f(z)$ be a transcendental entire function of finite order and let
$c\in\mathbb{C}\setminus\{0\}$ be a constant satisfying $f(z+c)\not\equiv f(z)$.
Set $H(z)=f(z)\Delta_{c}f(z)$ where $\triangle_{c} f(z)=f(z+c)-f(z)$.
If $f(z)$ has infinitely many multi-order zeros, then $H(z)$ takes every
value $a\in\mathbb{C}$ infinitely often.
\end{theorem}

Chen \cite{Chen 2} also considered zeros of difference product
$H_{n}(z)=f^{n}(z)\Delta_{c}f(z)$ and gave some conditions guarantee
$H_{n}(z)$ has finitely many zeros or infinitely many zeros.


\begin{theorem}[{\cite[Theorem 1]{Chen 2}}] \label{thmE}
Let $f(z)$ be a transcendental entire function of finite order and
$c\in\mathbb{C}\setminus\{0\}$ be a constant satisfying $f(z+c)\not\equiv f(z)$.
Set $H_{n}(z)=f^{n}(z)\triangle_{c} f(z)$ where $\triangle_{c} f(z)=f(z+c)-f(z)$,
$n\geq 2$ is an integer. Then the following statements hold:
\begin{itemize}
\item[(1)] If $f(z)$ satisfies $\sigma(f)\neq 1$, or has infinitely
 many zeros, then $H_{n}(z)$ has infinitely many zeros.

\item[(2)] If $f(z)$ has only finitely many zeros and $\sigma(f)=1$,
then $H_{n}(z)$ has only finitely many zeros.
\end{itemize}
\end{theorem}

The Nevanlinna theory for the $q$-difference operator plays an important
part in considering value distributions of $q$-difference polynomials.
Since $q$-difference logarithmic derivative lemma is only use for meromorphic
functions of zero order, most papers only consider meromorphic functions of
zero order. For example, Zhang and Korhonen \cite{Zhang 2} proved that
for a transcendental entire function $f(z)$ of zero order and a nonzero complex
constant $q$ and $n\geq2$, $f^{n}(z)f(qz)$ assumes every nonzero value
$a\in\mathbb{C}$ infinitely often. Recently, Liu-Liu-Cao \cite{Liu-3}
extended this to consider zero distributions of $q$-difference products
$f^{n}(z)(f^{m}(z)-a)f(qz+c)$ and $f^{n}(z)(f^{m}(z)-a)[f(qz+c)-f(z)]$
for meromorphic function $f$ with order zero.

It is natural to ask how about value distribution of $q$-difference products
for functions with positive order? The main purpose of this paper is to
consider a transcendental entire function $f$ with positive and finite order,
and obtain some results on the value distributions of $q$-difference products
$f(z)f(qz)$ and $f^{n}(z)(f(qz)-f(z))$. However, in this case, we have to
add the condition that $f$ has finitely many zeros, or something like that.
The first main theorem is as follows.

\begin{theorem}\label{T-1}
Let $f(z)$ be a transcendental entire function of finite and positive order
 $\sigma(f)$, $q\in\mathbb{C}\setminus\{0\}$ be a constant satisfying
$q^{\sigma(f)}\neq-1$. Set $H(z)=f(z)f(qz)$. 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 $f(z)$.
\end{theorem}

If replacing by the condition that $f(z)$ has infinitely many multi-order
zeros and considering any value $a$ which can be zero, then we have another
theorem.

\begin{theorem}\label{T-2}
Let $f(z)$ be a transcendental entire function of finite and positive order
 $\sigma(f)$, $q\in\mathbb{C}\setminus\{0\}$ be a constant. Set $H(z)=f(z)f(qz)$.
 If $f(z)$ has infinitely many multi-order zeros, then $H(z)$ takes every value
$a\in\mathbb{C}$ infinitely often.
\end{theorem}

For the $q$-difference product $f(z)(f(qz)-f(z))$ we have the following main
theorem.

\begin{theorem}\label{T-3}
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 satisfying
$q^{\sigma(f)}\neq\pm1$ and $f(z)\not\equiv f(qz)$, set
$H(z)=f(z)\nabla_{q}f(z)$. If $f(z)$ has finitely many zeros,
then $H(z)-\alpha(z)$ has infinitely many zeros, where $\alpha(z)$ is
a small entire function with respect to $f(z)$.
\end{theorem}

By the definition of Borel exceptional value and the proof of Theorem
\ref{T-3}, the following result is immediately true.

\begin{corollary}\label{coro2}
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 satisfying
$q^{\sigma(f)}\neq\pm1$ and $f(z)\not\equiv f(qz)$, set
$H(z)=f(z)\nabla_{q}f(z)$. If $f(z)$ has finitely many zeros,
then $H(z)$ has no any finite Borel exceptional value.
\end{corollary}

If replacing by the condition that $f(z)$ has infinitely many multi-order
zeros, we also have the following theorem.

\begin{theorem}\label{T-4}
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 satisfying
$q^{\sigma(f)}\neq\pm1$ and $f(z)\not\equiv f(qz)$, set $H(z)=f(z)\nabla_{q}f(z)$.
If $f(z)$ has infinitely many multi-order zeros, then $H(z)$ takes every value
$a\in\mathbb{C}$ infinitely often.
\end{theorem}

If considering zero distribution of $q$-difference product
$f^{n}(z)\nabla_{q}f(z)$, we have the following result whether $f$ has
finitely many zeros or not.

\begin{theorem}\label{T-5}
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 satisfying
$q^{\sigma(f)}\neq1$ and $f(z)\not\equiv f(qz)$.
Set $H_{n}(z)=f^{n}(z)\nabla_{q}f(z)$, $n\geq1$ is an integer.
Then $H_{n}(z)$ has infinitely many zeros.
\end{theorem}

Chen \cite{Chen 3} considered complex linear difference equations and obtained
that the relation between $\lambda(g)$ and $\sigma(g)$ of entire solutions
to nonhomogeneous linear difference equations is better than that of homogeneous
equations. Next, we will consider a special $q$-difference linear equation and
obtain the following result.

\begin{theorem}\label{T-6}
 Let $F(z)$ and $h_{j}(z) (j=1, \ldots , n)$ be entire functions with orders
all less than one, such that at least one of $h_{j}(z)\not\equiv0$, and let
$q_{j}(j=1, \ldots ,n)\in\mathbb{C}\setminus\{0,1\}$
 be constants satisfying $(\frac{q_s}{q_t}) ^{\sigma(f)}\neq 1$ for any $s\neq t$.
Suppose that $f(z)$ is a finite and positive order transcendental entire solution
of linear $q$-difference equation

\begin{equation}\label{E-1.1}
h_{n}(z)f(q_{n}z)+ \ldots +h_{1}(z)f(q_{1}z)=F(z).
\end{equation}
Then $f(z)$ has infinitely many zeros.
\end{theorem}

There exist many solutions which satisfy the functional equation \eqref{E-1.1}.
For example:\par

\begin{example}\label{examp1} \rm
It is known that the transcendental entire function $f(z)=z+\cos z^{3}$
with order three has infinitely many zeros.
Let $h_{1}(z)=z^{5}=-h_{2}(z)$, and let $q_{2}=2$, $q_{1}=-2$. Obviously,
$(\frac{q_{2}}{q_{1}})^{\sigma(f)}=(-1)^{3}\neq 1$. Then the function $f(z)$
satisfies the non-homogeneous linear $q$-difference equation
$$
h_{2}(z)f(q_{2}z)+h_{1}(z)f(q_{1}z)=-4z^{6}.
$$
\end{example}

The following example shows that the condition
$(\frac{q_s}{q_t}) ^{\sigma(f)}\neq 1$ for any $s\neq t$ in Theorem \ref{T-6}
is necessary.

\begin{example} \label{examp2} \rm
Let $h_{1}(z)=-h_{2}(z)\not \equiv 0$, $q_{2}=q_{1}=q\neq 0, 1$.
Then the function $f(z)=e^{z}$ with order one satisfies the homogeneous linear
$q$-difference equation
$$
h_{2}(z)f(q_{2}z)+h_{1}(z)f(q_{1}z)=0.
$$
Here, $(\frac{q_2}{q_1}) ^{\sigma(f)}= 1$, and but $f(z)=e^{z}$ has no zeros.
\end{example}

 \section{Lemmas}

To prove our results, we need some lemmas. The first one is the well-known
Weierstrass factorization theorem and Hadamard factorization theorem.

\begin{lemma}[\cite{ash}] \label{L-0}
If an entire function $f$ has a finite
exponent of convergence $\lambda(f)$ for its zero-sequence, then $f$ has
a representation in the form $$f(z)=Q(z)e^{g(z)},$$ satisfying
$\lambda(Q)=\sigma(Q)=\lambda(f)$. Further, if $f$ is of finite order,
then $g$ in the above form is a polynomial of degree less or equal to the
order of $f$.
\end{lemma}

\begin{lemma}[\cite{Yi}] \label{L-1}
Suppose that $f_1(z), f_2(z), \ldots , f_n(z), (n\geq2)$ are meromorphic
functions and $g_1(z), g_2(z), \ldots , g_n(z)$ are entire functions
satisfying the following conditions
\begin{itemize}
\item[(1)] $\sum_{j=1}^{n}f_{j}(z)e^{g_{j}(z)}\equiv 0$;

\item[(2)] $g_j(z)-g_k(z)$ are not constants for $1\leq j<k\leq n$;

\item[(3)] For $1\leq j\leq n, 1\leq h < k\leq n$,
$T (r, f_{j}) = o(T (r, e^{g_{h}-g_{k}}))$ $(r\to\infty, r\not\in E)$.
\end{itemize}
Then $f_j(z)\equiv0 (j=1, 2, \ldots , n)$.
\end{lemma}

 \begin{lemma}\label{L-2}
Let $f(z)$ be a transcendental entire function of finite and positive
order $\sigma(f)$, $q\in\mathbb{C}\setminus\{0\}$ be a constant satisfying
$q^{\sigma(f)}\neq -1$. Set $H(z)=f(z)f(qz)$. If $f(z)$ has finitely many zeros,
then $H(z)$ is a transcendental entire function and $\sigma(H)=\sigma(f)$.
\end{lemma}

\begin{proof}
Since $f(z)$ is a transcendental entire function of finite order and has
finitely many zeros,  by Lemma \ref{L-0},  $f(z)$ can be written as
$$
f(z)=g(z)e^{h(z)},
$$
where $g(z)(\not\equiv 0)$, $h(z)$ are polynomials. Set
$$
h(z)=a_{k}z^{k}+ \ldots +a_{0},
$$
where $a_{k}, \ldots, a_{0}$ are constants, $a_{k}\neq0$. Since
$\sigma(f)\neq0$, it follows that  $\sigma(f)=\deg(h(z))=k\geq1$. So
$$
H(z)=f(z)f(qz)=g(z)g(qz)e^{(a_{k}+a_{k}q^{k})z^{k}+ \ldots +2a_{0}}.
$$
Since $g(z)(\not\equiv 0)$ is a polynomial, $q^{\sigma(f)}=q^{k}\neq -1$, it
follows that
$H(z)$ is a transcendental entire function and $\sigma(H)=\sigma(f)=k$.
\end{proof}

\section{Proofs of main results}

\subsection{Proof of Theorem \ref{T-1}}

 Since $f(z)$ is a transcendental entire function of finite order and has
finitely many zeros,  by Lemma \ref{L-0}, $f(z)$ can be written as
$$
f(z)=g(z)e^{h(z)},
$$
where $g(z)(\not\equiv 0)$, $h(z)$ are polynomials. Set
$$
h(z)=a_{k}z^{k}+ \ldots +a_{0},
$$
where $a_{k}, \ldots, a_{0}$ are constants, $a_{k}\neq0$. Since
$\sigma(f)\neq0$, it follows that $\sigma(f)=\deg(h(z))=k\geq1$. So
\begin{equation}\label{E-3.1}
H(z)=f(z)f(qz)=g(z)g(qz)p(z)e^{(1+q^{k})a_{k}z^{k}},
\end{equation}
where $p(z)=e^{(1+q^{k-1})a_{k-1}z^{k-1}+ \ldots +2a_{0}}$,
 $\sigma(p)\leq k-1<k$.

Suppose $H(z)-\alpha(z)$ has finitely many zeros, by Lemma \ref{L-2},
$\sigma(H-\alpha)=\sigma(H)=\sigma(f)$, then $H(z)-\alpha(z)$ can be written as
\begin{equation}\label{E-3.2}
H(z)-\alpha(z)=s(z)e^{tz^{k}},
 \end{equation}
where $s(z)$ is an entire function with $\sigma(s)<k$, $t\neq0$ is a constant.
By \eqref{E-3.1} and \eqref{E-3.2}, we obtain
\begin{equation}\label{E-3.3}
g(z)g(qz)p(z)e^{(1+q^{k})a_{k}z^{k}}-s(z)e^{tz^{k}}-\alpha(z)=0.
\end{equation}

 Case 1:  $(1+q^{k})a_{k}\neq t$. Since $\sigma(p)<k$, $\sigma(s)<k$,
$g(z)$ and $g(qz)$ are polynomials,  by Lemma \ref{L-1}, we obtain
$$
g(z)g(qz)p(z)\equiv 0, s(z)\equiv 0, \alpha(z)\equiv0.
$$
Which is a contradiction.

Case 2: $(1+q^{k})a_{k}=t$. Then, \eqref{E-3.3} can be written as
$$
g(z)g(qz)p(z)-s(z)=\alpha(z)e^{-tz^{k}}.
$$
Since $\sigma(g(z)g(qz)p(z)-s(z))<k$, while $\sigma(\alpha(z)e^{-tz^{k}})=k$,
which is a contradiction.

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


\subsection{Proof of Theorem \ref{T-2}}
If $a=0$, then $H(z)$ has obviously infinitely many zeros since $f(z)$
 has infinitely many zeros.

Now we consider $a\neq0$. Suppose $H(z)-a$ has finitely many zeros,
by Lemma \ref{L-0}, $H(z)-a$ can be rewritten as
 \begin{equation}\label{E-3.4}
 H(z)-a=f(z)f(qz)-a=g(z)e^{h(z)},
 \end{equation}
 where $g(z)(\not\equiv 0)$, $h(z)$ are polynomials, $\deg(h(z))\geq1$.
Differentiating \eqref{E-3.4} and eliminating $e^{h(z)}$, we obtain
\begin{equation}\label{E-3.5}
\frac{(f(z)f(qz))'}{f(z)f(qz)}=\frac{g'(z)+g(z)h'(z)}{g(z)}
-a\frac{g'(z)+g(z)h'(z)}{g(z)}\frac{1}{f(z)f(qz)}.
\end{equation}
Since $g(z)(\not\equiv 0)$, $h(z)$ are polynomials and
$ \deg(h(z))\geq1$, it follows that $g'(z)+g(z)h'(z)\not\equiv0$.
 As $f(z)$ has infinitely many multi-order zeros, there is a multi-order
zero $z_{0}$, such that $\mid z_{0}\mid$ is sufficiently large and
$g(z_{0})\neq0, g'(z_{0})+g(z_{0})h'(z_{0})\neq0$. Thus the right
side of \eqref{E-3.5} has a multi-order pole at $z_{0}$, but the left side
of \eqref{E-3.5} has only a simple pole at $z_{0}$, which is a contradiction.

Hence $H(z)$ takes every value $a\in\mathbb{C}$ infinitely often.

\subsection{Proof of Theorem \ref{T-3}}

Since $f(z)$ is a transcendental entire function of finite order and has
finitely many zeros,   by Lemma \ref{L-0},  $f(z)$ can be written as
$$
f(z)=g(z)e^{h(z)},
$$
where $g(z)(\not\equiv 0)$, $h(z)$ are polynomials, set
$$
h(z)=a_{k}z^{k}+ \ldots +a_{0},$$ where $a_{k}, \ldots, a_{0}$ are constants,
$a_{k}\neq0$. Since $\sigma(f)\neq0$, it follows that $\sigma(f)=\deg(h(z))=k\geq1$.
\begin{equation}\label{E-3.6}
H(z)=f(z)\nabla_{q}f(z)=g(z)g(qz)p_{1}(z)e^{(1+q^{k})a_{k}z^{k}}
-g^{2}(z)p_{2}(z)e^{2a_{k}z^{k}},
\end{equation}
where
\begin{gather*}
 p_{1}(z)=e^{(1+q^{k-1})a_{k-1}z^{k-1}+ \ldots +2a_{0}}, \quad
 \sigma(p_{1})\leq k-1<k;\\
 p_{2}(z)=e^{2a_{k-1}z^{k-1}+ \ldots +2a_{0}},\quad
 \sigma(p_{2})\leq k-1<k.
\end{gather*}
Since $g(z)(\not\equiv 0)$ is a polynomial, $q^{\sigma(f)}=q^{k}\neq \pm1$,
$\sigma(p_{1})<k$, $\sigma(p_{2})<k$, it follows $H(z)$ is a transcendental entire
function and $\sigma(H)=\sigma(f)=k$.

Suppose $H(z)-\alpha(z)$ has finitely many zeros, then
$\lambda(H-\alpha)<\sigma(H)=\sigma(f)$, $H(z)-\alpha(z)$ can be written as
\begin{equation}\label{E-3.7}
H(z)-\alpha(z)=s(z)e^{tz^{k}},
 \end{equation}
where $s(z)$ is an entire function with $\sigma(s)<k$, $t\neq0$ is a constant.
By \eqref{E-3.6} and \eqref{E-3.7}, we obtain
\begin{equation}\label{E-3.8}
g(z)g(qz)p_{1}(z)e^{(1+q^{k})a_{k}z^{k}}-g^{2}(z)p_{2}(z)e^{2a_{k}z^{k}}
-s(z)e^{tz^{k}}-\alpha(z)=0.
\end{equation}
Since $q^{\sigma(f)}=q^{k}\neq 1$, it follows that $(1+q^{k})a_{k}\neq 2a_{k}$.


Case 1: $(1+q^{k})a_{k}\neq t, 2a_{k}\neq t$. By Lemma \ref{L-1},
we obtain
$$
g(z)g(qz)p_{1}(z)\equiv0, g^{2}(z)p_{2}(z)\equiv0,
s(z)\equiv0, \alpha(z)\equiv0.
$$ This is a contradiction.

Case 2: $(1+q^{k})a_{k}=t$. Then \eqref{E-3.8} can be written as
$$
(g(z)g(qz)p_{1}(z)-s(z))e^{(1+q^{k})a_{k}z^{k}}-g^{2}(z)p_{2}(z)e^{2a_{k}z^{k}}
-\alpha(z)=0.
$$
By Lemma \ref{L-1}, we obtain
$$
g(z)g(qz)p_{1}(z)-s(z)\equiv0, g^{2}(z)p_{2}(z)\equiv0, \alpha(z)\equiv0,
$$
which is a contradiction.

Case 3: $2a_{k}=t$. Then using the same method as above, we also obtain
a contradiction.

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

\subsection{Proof of Theorem \ref{T-4}}

If $a=0$, then $H(z)$ has obviously infinitely many zeros as $f(z)$
has infinitely many zeros and $f(qz)-f(z)\not\equiv0$.

Now we consider $a\neq0$. Suppose $H(z)-a$ has finitely many zeros, from
the proof of Theorem \ref{T-3}, $H(z)$ is a transcendental entire function
and $\sigma(H)=\sigma(f)$. So  by Lemma \ref{L-0}, $H(z)-a$ can be rewritten as
 \begin{equation}\label{E-3.9}
 H(z)-a=f(z)f(qz)-f^{2}(z)-a=g(z)e^{h(z)},
\end{equation}
where $g(z)(\not\equiv 0)$, $h(z)$ are polynomials,
$\deg(h(z))\geq1$. Differentiating \eqref{E-3.9}and eliminating $e^{h(z)}$,
we obtain
\begin{equation}\label{E-3.10}
\frac{(f(z)f(qz))'}{f(z)f(qz)}-\frac{2f'(z)}{f(qz)}
=\frac{g'(z)+g(z)h'(z)}{g(z)}[1-\frac{f(z)}{f(qz)}-\frac{a}{f(z)f(qz)}].
\end{equation}
Since $g(z)(\not\equiv 0)$, $h(z)$ are polynomials and $\deg(h(z))\geq1$,
it follows that $g'(z)+g(z)h'(z)\not\equiv0$.
As $f(z)$ has infinitely many multi-order zeros, there is a
multi-order zero $z_{0}$ of multiplicity $k\geq 2$, such that
$|z_{0}|$ is sufficiently large and
$g(z_{0})\neq0, g'(z_{0})+g(z_{0})h'(z_{0})\neq0$.

If $f(qz)$ has zero at $z_{0}$ of multiplicity $s\geq 1$, then
$\frac{(f(z)f(qz))'}{f(z)f(qz)}$ has a simple pole at $z_{0}$;
$-\frac{2f'(z)}{f(qz)}$ has pole at $z_{0}$ of multiplicity $s-k+1$;
 $\frac{f(z)}{f(qz)}$ has pole at $z_{0}$ of multiplicity $s-k$; but
$\frac{a}{f(z)f(qz)}$ has pole at $z_{0}$ of multiplicity $s+k$.
This is a contradiction.

If $f(qz_{0})\neq0$, then $\frac{(f(z)f(qz))'}{f(z)f(qz)}$
has a simple pole at $z_{0}$; $\frac{f(z_{0})}{f(qz_{0})}=0$;
$-\frac{2f'(z_{0})}{f(qz_{0})}=0$; but $\frac{a}{f(z)f(qz)}$ has pole at $z_{0}$
of multiplicity $k\geq 2$. We also have a contradiction.

Hence $H(z)$ takes every value $a\in\mathbb{C}$ infinitely often.


\subsection{Proof of Theorem \ref{T-5}}

If $f(z)$ has infinitely many zeros, then $H_{n}(z)$ has infinitely
many zeros since $f(qz)-f(z)\not\equiv 0$.

Now we consider $f(z)$ has finitely many zeros, suppose $H_{n}(z)$
has only finitely many zeros. Since $f(z)$ is a transcendental entire
function of finite order and has finitely many zeros,   by
Lemma \ref{L-0}, $f(z)$ can be written as
$$
f(z)=g(z)e^{h(z)},
$$
 where $g(z)(\not\equiv 0)$, $h(z)$ are polynomials. Set
$$
h(z)=a_{k}z^{k}+ \ldots +a_{0},
$$ where $a_{k}, \ldots, a_{0}$ are constants, $a_{k}\neq0$.
Since $\sigma(f)\neq0$, it follows that $\sigma(f)=\deg(h(z))=k\geq1$.
So
\begin{equation}\label{E-3.11}
H_{n}(z)=f^{n}(z)\nabla_{q} f(z)=g^{n}(z)g(qz)e^{nh(z)+h(qz)}
-g^{n+1}(z)e^{(n+1)h(z)}
\end{equation}
and
$$
(n+1)h(z)-(nh(z)+h(qz))=(1-q^{k})a_{k}z^{k}+(1-q^{k-1})a_{k-1}z^{k-1}
+ \ldots +(1-q)a_{1}z.
$$
Since $q^{\sigma(f)}=q^{k}\neq 1$, it follows that
 $(n+1)h(z)-(nh(z)+h(qz))$ is not a constant.
So $H_{n}(z)$ is a transcendental entire function,  by Lemma \ref{L-0},
$H_{n}(z)$ can be written as
\begin{equation}\label{E-3.12}
H_{n}(z)=g_{1}(z)e^{h_{1}(z)},
\end{equation}
where $g_{1}(z)(\not\equiv 0)$, $h_{1}(z)$ are polynomials.
By \eqref{E-3.11} and \eqref{E-3.12}, we obtain
\begin{equation}\label{E-3.13}
g^{n}(z)g(qz)e^{nh(z)+h(qz)}-g^{n+1}(z)e^{(n+1)h(z)}-g_{1}(z)e^{h_{1}(z)}=0.
\end{equation}

Note that $(n+1)h(z)-(nh(z)+h(qz))$ is not a constant and $g(z)$
and $g_{1}(z)$ are polynomials. If $(nh(z)+h(qz))-h_{1}(z)$ and
$(n+1)h(z)-h_{1}(z)$ are not constants, then by \eqref{E-3.13}
and Lemma \ref{L-1}, we obtain
$$
g^{n}(z)g(qz)\equiv0,g^{n+1}(z)\equiv0,g_{1}(z)\equiv0.
$$
This is a contradiction.

If $(nh(z)+h(qz))-h_{1}(z)=c$, where $c$ is a constant, then
\eqref{E-3.13} can be written as
\begin{equation}\label{E-3.14}
(g^{n}(z)g(qz)-e^{-c}g_{1}(z))e^{nh(z)+h(qz)}-g^{n+1}(z)e^{(n+1)h(z)}=0.
\end{equation}

By \eqref{E-3.14} and Lemma \ref{L-1}, we obtain
$$
g^{n}(z)g(qz)-e^{-c}g_{1}(z)\equiv0, g^{n+1}(z)\equiv0,
$$
Which 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 $H_{n}(z)$ has infinitely many zeros.

\subsection{Proof of Theorem \ref{T-6}}

Suppose $f(z)$ has finitely many zeros, since $f(z)$ is a transcendental
entire function of finite and positive order, it follows
by Lemma \ref{L-0}, that $f(z)$ can be written as
\begin{equation}\label{E-3.15}
f(z)=g(z)e^{h(z)},
\end{equation}
where $g(z)(\not\equiv 0)$, $h(z)$ are polynomials. Set
$$
h(z)=a_{k}z^{k}+ \ldots +a_{0},
$$
where $a_{k}, \ldots, a_{0}$ are constants, $a_{k}\neq0$.
Since $\sigma(f)\neq0$, it follows that $\sigma(f)=\deg(h(z))=k\geq1$.
Substituting \eqref{E-3.15}into \eqref{E-1.1}, we obtain
\begin{equation}\label{E-3.16}
h_{n}(z)g(q_{n}z)e^{h(q_{n}z)}+ \ldots +h_{1}(z)g(q_{1}z)e^{h(q_{1}z)}=F(z)
\end{equation}
and
$$
h(q_{s}z)-h(q_{t}z)=a_{k}(q_{s}^{k}-q_{t}^{k})z^{k}+a_{k-1}(q_{s}^{k-1}
-q_{t}^{k-1})z^{k-1}+ \ldots +a_{1}(q_{s}-q_{t})z,
$$
where $s\neq t$. Since $q_{j}(j=1, \ldots, n)\in\mathbb{C}\setminus\{0,1\}$
 and $(\frac{q_s}{q_t})^{\sigma(f)}\neq 1$ for $s\neq t$, it follows that
$$
q_{s}^{k}\neq q_{t}^{k}, \deg(h(q_{s}z)-h(q_{t}z))=k,
\sigma(e^{h(q_{s}z)-h(q_{t}z)})=k.
$$
Since $\sigma(h_{j}(z)g(q_{j}z))<1<k$ for $j=1, \ldots, n$, $\sigma(F)<1<k$,
 by \eqref{E-3.16} and Lemma \ref{L-1}, we obtain
$$
h_{n}(z)g(q_{n}z)\equiv0, \ldots , h_{1}(z)g(q_{1}z)\equiv0, F(z)\equiv0.
$$
We get a contradiction. Hence $f(z)$ has infinitely many zeros.


\subsection*{Acknowledgements}
The authors would like to thank the anonymous
referee for making valuable suggestions and comments to
improve the present paper.

This research was partly supported by NSFC (no.11101201), 
CPSF(no. 2014M551865 ), CSC(no. 201308360070), PSF of Jiangxi, 
the NSF of Jiangxi (no.20122BAB211001), NSF of ED of Jiangxi (no. GJJ13077), 
the National Natural Science Foundation of China(11271304), 
the Natural Science Foundation of Fujian Province of China for Distinguished 
Young Scholars(2013J06001), and the Program for New Century Excellent 
Talents in University (NCET-13-0510).

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\end{document}