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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 16, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/16\hfil Meromorphic solutions]
{Meromorphic solutions to complex difference and $q$-difference
equations of Malmquist type}

\author[J. Zhang, J. Zhang \hfil EJDE-2014/16\hfilneg]
{Jie Zhang, Jianjun Zhang}  % in alphabetical order

\address{Jie Zhang  \newline
 College of Science,
China University of Mining and Technology, Xuzhou 221116, China}
\email{zhangjie1981@cumt.edu.cn}

\address{Jianjun Zhang (corresponding author)\newline
Mathematics and Information Technology School,
Jiangsu Second Normal University, Nanjing 210013, China}
\email{zhangjianjun1982@163.com}


\thanks{Submitted September 13, 2013. Published January 10, 2014.}
\subjclass[2000]{30D35, 34M10}
\keywords{Meromorphic; differential-difference equation;
fixed point; order}

\begin{abstract}
 In this article, we study the zeros, poles and fixed points
 of finite order transcendental meromorphic solutions of complex difference
 and $q$-difference equations of Malmquist type respectively.
 Some examples are structured to show that our results are sharp.
\end{abstract}

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


\section{Introduction}

 In this article, a meromorphic function always means it is
meromorphic in the whole complex plane $\mathbb{C}$. We assume that
the reader is familiar with the standard notations in the Nevanlinna
theory. We use the following standard notations in value
distribution theory (see \cite {Hayman-book, Yang-Yi-book, L.Yang-book}):
$$
T(r,f), m(r,f), N(r,f), \overline {N}(r,f), \dots.
$$
And we denote by $S(r,f)$ any quantity satisfying
$S(r,f)=o\{T(r,f)\}$,  as $r\to\infty$, 
possibly outside of a set $E$ with finite linear or logarithmic
measure, not necessarily the same at each occurrence. We also use
the notation $\tau(f)$ to denote the exponent of convergence of
fixed points of $f$, namely
$$
\tau(f)=\limsup_{r\to\infty}\frac{\log N(r,\frac{1}{f-z})}{\log r}.
$$
The deficiency of $a$ with respect to $f(z)$ is defined by
$$
\delta(a,f)=1-\limsup_{r\to\infty}\frac{N(r,\frac{1}{f-a})}{T(r,f)}.
$$
We use $\lambda(f)$ and $\bar{\lambda}(f)$ to denote the exponent of 
convergence of zeros of $f$ counting multiplicities
and ignoring multiplicities respectively, namely
$$
\lambda(f)=\limsup_{r\to\infty}\frac{\log N(r,\frac{1}{f})}{\log r},\quad
\bar{\lambda}(f)=\limsup_{r\to\infty}
\frac{\log \overline{N}(r,\frac{1}{f})}{\log r}.
$$
A polynomial $Q(z,f)$ is
called a differential-difference polynomial in $f$ if $Q$ is a
polynomial in $f$, its derivatives and shifts with small meromorphic
coefficients, say $\{a_\lambda| \lambda\in I\}$, such that
$T(r,a_\lambda)=S(r,f)$ for all $\lambda\in I$. We define the
difference operator $\Delta f=f(z+1)-f(z)$.

Recently, a large number of researches focusing on complex difference
and q-difference equation emerged. For example, Gundersen et al \cite {GG} 
considered the complex q-difference equation of Malmquist type and obtained the
following result.

\begin{theorem} \label{thmA}
 Let $f$ be a transcendental meromorphic solution of
the $q$-difference equation 
\begin{equation}\label{1}
f(qz)=R(z,f)=\frac{a_0(z)+a_1(z)f(z)+\dots +a_p(z)f^p(z)}{b_0(z)+b_1(z)f(z)
+\dots +b_t(z)f^t(z)},
\end{equation}
where $q\in \mathbb{C}$, $|q|\geq1$, $a_p(z)\not\equiv0$, $b_t(z)\equiv1$, 
and meromorphic coefficients
$a_i(z)$ $(i=0,1, \dots ,p)$ and $b_j(z)$ $(j=0,1, \dots ,t-1)$ are of
growth $S(r,f)$. If
$$
\overline N(r,f)+ \overline{N}(r,\frac{1}{f})=S(r,f),
$$
then  \eqref{1} is either the form
$$f(qz)=a_p(z)f^p(z)   \quad  \text{or}\quad
f(qz)=\frac{a_0(z)}{f^t(z)}.
$$
\end{theorem}

Here we consider a $q$-difference equation whose form is more general 
than in Equation \eqref{1}
under a condition similar to Theorem \ref{thmA} and obtain some results as follows.

\begin{theorem} \label{thm1}
 Let $f$ be a transcendental meromorphic solution of
 a $q$-difference equation of the following form
\begin{equation}\label{2}
\prod_{i=1}^nf(q_iz)=R(z,f)=\frac{a_0(z)+a_1(z)f(z)+\dots 
+a_p(z)f^p(z)}{b_0(z)+b_1(z)f(z)+\dots +b_t(z)f^t(z)},
\end{equation}
 where $q_i\neq0, 1$, $i=1,2\dots n$, and $R(z,f)$ is an
irreducible rational function in $f$ with meromorphic coefficients
$a_i(z)$ $(i=0, 1,\dots p)$ and $b_j(z)$ $(j=0, 1,\dots t)$ of
growth $S(r,f)$ such that $a_p(z)\not\equiv0, b_t(z)\equiv1$. If
$$
\max\{\overline{\lambda}(f), \overline{\lambda}(\frac{1}{f})\}<\sigma(f)
=\sigma\leq\infty,
$$
then  \eqref{2} is reduced to the form
$$
\prod_{i=1}^nf(q_iz)=a_p(z)f^p(z)   \quad \text{or}\quad
\prod_{i=1}^nf(q_iz)=\frac{a_0(z)}{f^t(z)}.
$$
\end{theorem}

The author in \cite {ZJ} considered a special complex difference 
equation of Malmquist type and obtained the following result.

\begin{theorem} \label{thmB}
 Let $R(z)$ be a non-constant rational function. For the 
difference equation 
$$
f(z+1)=R\circ  f(z),
$$
 (1) suppose it admits a non-constant rational solution $f(z)$, then 
both $R(z)$ and $f(z)$ are fractional linear functions;
\\
 (2) suppose it admits a transcendental meromorphic function $f(z)$ of 
finite order $\sigma(f)$,  then $R(z)$ is a fractional linear function, 
and suppose that it is denoted by $R(z)=\frac{az+b}{cz+d}$, 
where $ad-bc\neq0$, furthermore: 
\\
(2.1) if $bc\neq0$, then $\lambda(f)=\lambda(\frac{1}{f})=\tau(f)=\sigma(f)$; \\
(2.2) if $R\neq id$ and $\sigma(f)>0$, then \\ 
(2.2.1) $f(z)$ has at most one finite Borel exceptional value provided 
 $(d-a)^2+4b=0$ when $c\neq0$;\\
(2.2.2) if $f(z)$ has Borel exceptional value $\infty$, then
$f(z)$ has at most one finite Borel exceptional value
$\frac{b}{1-a}$ .
\end{theorem}

 Here we consider a type of difference equation more
general than in Theorem \ref{thmB} and obtain some results as follows.

\begin{theorem} \label{thm2}
Suppose that $c_1, c_2,\dots, c_n$ are distinct nonzero constants. 
If complex difference equation of Malmquist type
\begin{equation}\label{3}
\sum_{j=1}^{n}f(z + c_j)=R(f(z)) =\frac{P(f(z))}{ Q(f(z))}
=\frac{a_pf^p(z) + a_{p-1}f^{p-1}(z) +\dots + a_0}{ b_qf^q(z) 
+ b_{q-1}f^{q-1}(z) + \dots + b_0}
\end{equation}
admits a transcendental meromorphic solution $f(z)$ of finite order,
where $P(f)$ and $Q(f)$ are relatively prime polynomials in $f$ with constant 
coefficients $a_s$ $(s = 0,1,\dots,p)$ and $b_t$ $(t = 0,1,\dots,q)$
such that $a_0a_pb_q \neq 0$, and 
$$
d=\deg R(z)=\max{\{\deg P(z), \deg Q(z)\}}\geq 2,
$$ 
then\\
 (1) $f(z)$ has infinitely many zeros and satisfies $\delta(0,f)=0$;\\
 (2) $f(z)$ has infinitely many fixed points and satisfies $\tau(f)=\sigma(f)$;\\
 (3) $f(z)$ has infinitely many poles and satisfies 
$\lambda (\frac{1}{f})= \sigma(f)$;\\
 (4) $f(z)$ has no deficiency value $b$ except that the value $b$ satisfies 
$$
a_pb^p + a_{p-1}b^{p-1} +\dots + a_0=nb(b_qb^q + b_{q-1}b^{q-1} + \dots + b_0).
$$
\end{theorem}

\begin{example} \label{examp1} \rm
Let $f(z)=1/(e^{\pi iz}-1)$, then we see the following
identical equation holds. 
$$
(f(z+1)+f(z+2))(2f(z)+1)=2{f}^2(z),
$$
which means $f(z)$ satisfies the complex difference equation of
Malmquist type
$$
\sum_{j=1}^{2}f(z + c_j)=R(f(z)),
$$ 
where $c_1=1, c_2=2$ and $R(z)=2z^2/(2z+1)$. But $f(z)\neq0$.
\end{example}

This example shows that the assumption $a_0\neq0$ is necessary for our 
result (1) in Theorem \ref{thm2}.

\begin{example} \label{examp2} \rm
Let $f(z)=e^{\pi iz}+z$, then we see the following identical
equation holds. $$f(z+2)+f(z+4)=2f(z)+6,$$ which means $f(z)$
satisfies the complex difference equation of Malmquist type
$$
\sum_{j=1}^{2}f(z + c_j)=R(f(z)),
$$ 
where $c_1=2, c_2=4$ and $R(z)=2z+6$. But $f(z)\neq z, \infty$.
\end{example}

This example shows that the assumption $\deg R(z)\geq 2$ is necessary 
for our results (2)-(3) in Theorem \ref{thm2}.

\begin{example} \label{exap3} \rm
Let $f(z)=e^{\pi iz}+1$, then we see the following identical equation
holds.
$$
f(z-\frac{\log2}{i\pi})+f(z-\frac{\log2}{i\pi}+2)=f(z)+1,
$$
which means $f(z)$ satisfies the complex difference equation of
Malmquist type
$$
\sum_{j=1}^{2}f(z + c_j)=R(f(z)),
$$ 
where $c_1=-\frac{\log2}{i\pi}$, $c_2=2+c_1$ and $R(z)=z+1$. 
But $f(z)\neq 1, \infty$.
\end{example}

This example shows that the assumption $\deg R(z)\geq 2$ is necessary 
for our result (3) and the assumption $a_pb^p + a_{p-1}b^{p-1} +\dots +
a_0=nb(b_qb^q + b_{q-1}b^{q-1} + \dots + b_0)$ is necessary for our
result (4) in Theorem \ref{thm2}.

\begin{example} \label{examp4} \rm
Let $f(z)=2+\frac{1}{e^{\pi iz}-1}$, then by a simple calculation,
we see the following identical equation holds.
$$
(f(z+2)+f(z+1))(f(z)-\frac{3}{2})={f^2(z)-2},
$$ 
which means $f(z)$ satisfies the complex difference equation of Malmquist 
type
$$
\sum_{j=1}^{2}f(z + c_j)=R(f(z))=\frac{f^2(z)-2}{f(z)-\frac{3}{2}},
$$ 
where $c_1=1, c_2=2$ and $R(z)=\frac{z^2-2}{z-\frac{3}{2}}$. But $f(z)\neq 2$.
\end{example}

This example shows that the assumption
$a_pb^p + a_{p-1}b^{p-1} +\dots + a_0=nb(b_qb^q + b_{q-1}b^{q-1} + \dots + b_0)$
is necessary for our result (4) in Theorem \ref{thm2}
 even when $\deg R(z)\geq 2$.

 In 2007, Laine and Yang \cite{L-CCY} considered zeros of
one certain type of difference polynomials and obtained the
following classic theorem.

\begin{theorem} \label{thmC}
Let $f$ be a transcendental entire function of finite order and $c$
be a nonzero complex constant. If $n\geq2$, then $f^{n}(z)f(z+c)-a$
has infinitely many zeros, where $a\in \mathbb{C}\setminus\{0\}$.
\end{theorem}

 At last, we also consider one special difference polynomial 
$f^n{(\Delta f)}^s-\alpha(z)$ corresponding to Theorem \ref{thmC} as follows.

\begin{theorem} \label{thm3}
Let $f$ be a transcendental entire function of finite order, 
$\alpha(z)(\not\equiv0)$ be a small function of $f$ and $\Delta f\not\equiv0$. 
Then $f^n{(\Delta f)}^s-\alpha(z) (n\geq2)$ has infinitely many zeros.
\end{theorem}

\section{Some lemmas} 

To prove our results, we need some lemmas as follows.

\begin{lemma}[\cite{Yi-a}]\label{m2}
 Let $f$ be a meromorphic function with a finite order $\sigma$,
and $\eta$ be a nonzero constant. For any $\varepsilon>0$, we have
$$
m(r,\frac{f(z+\eta)}{f(z)})=O(r^{\sigma-1+\varepsilon}).
$$
\end{lemma}

\begin{lemma}[\cite{Yi-a}]\label{mt}
 Let $f(z)$ be a transcendental meromorphic function with finite
order $\sigma$ and $\eta$ be a nonzero complex number. Then for each
$\varepsilon>0$, we have
\[
T(r,f(z+\eta))=T(r,f)+O(r^{\sigma-1+\varepsilon})+O(\log r);
\]
i.e., $T(r,f(z+\eta))=T(r,f)+S(r,f)$.
\end{lemma}

\begin{lemma}[\cite{Yi-a}]\label{t}
Let $f(z)$ be a meromorphic function with finite exponent of convergence 
of poles $\lambda=\lambda(\frac{1}{f})<\infty$, and $\eta$ be a fixed number.
Then for each $\varepsilon>0$, we have
$$
N(r,f(z+\eta))=N(r,f)+O(r^{\lambda-1+\varepsilon})+O(\log r).
$$
\end{lemma}

\begin{lemma}[\cite{L-Y}]\label{cl}
Let $w(z)$ be a finite order transcendental meromorphic solution
of the equation
$$
P(z,w)=0,$$ where $P(z,w)$ is a differential-difference polynomial 
in $w$ and its shifts. If
$P(z,a)\not\equiv0$ for a meromorphic function $a\in S(r,w)$, then
$$
m(r,\frac{1}{w-a})=S(r,w).
$$
\end{lemma}

\begin{lemma}[\cite{Yang-Yi-book}] \label{T}
Let $f(z)$ be a non-constant meromorphic function in the complex
plane and 
$$
R(f)=\frac{p(f)}{q(f)},
$$ 
where $p(f)=\sum_{k=0}^{p}a_{k}f^{k}$ and
$q(f)=\sum_{j=0}^{q}b_{j}f^{j}$
are two mutually prime polynomials in $f(z)$. If the coefficients
$a_{k}, b_{j}$ are small functions of $f(z)$ and
$a_{p}(z)\not\equiv0, b_{q}(z)\not\equiv0$, then
$$
T(r,R(f))=\max \{p, q\}T(r,f)+S(r,f).
$$
\end{lemma}

\begin{lemma}[\cite{GW}] \label{CT}
Let $f$ be a transcendental meromorphic function and
$$
F=a_nf^n+a_{n-1}f^{n-1}+\dots +a_0\quad (a_n\not\equiv0)
$$
be a polynomial in $f$ with coefficients being small functions of
$f$. Then either
$$
F=a_n(f+\frac{a_{n-1}}{na_n})^n\quad \text{or}\quad
T(r,f)\leq \overline{N}(r,\frac{1}{F})+\overline{N}(r,f)+S(r,f).
$$
\end{lemma}

\begin{remark}\label{rmk1} \rm
From the definition, we have (see \cite {BW})
$$
m(r,f(cz))=m(|c|r,f(z))\quad \text{and}\quad
N(|c|r,f(z))=N(r,f(cz))+n(0,f(cz))\log|c|,
$$
i.e.,
$$
N(|c|r,f(z))=N(r,f(cz))+O(1)\quad \text{and}\quad
T(|c|r,f(z))=T(r,f(cz))+O(1).
$$
\end{remark}

\section{Proofs of theorems}
\begin{proof}[Proof of Theorem \ref{thm1}] 
 First of all, we suppose that both $p$ and $t$ are
positive integers, and rewrite Equation \eqref{2} as 
\begin{equation}\label{1.4}
H(z):=\prod_{i=1}^nf(q_iz)=A(z)\frac{P(z,f)}{T(z,f)},
\end{equation}
where
$$
A(z)=\frac{a_p(z)}{b_t(z)},\quad
P(z,f)=\frac {a_0(z)}{a_p(z)}+\frac {a_1(z)}{a_p(z)}f(z)+\dots +f^p(z)
$$
and
$$
T(z,f)=\frac{b_0(z)}{b_t(z)}+\frac{b_1(z)}{b_t(z)}f(z)+\dots +f^t(z).
$$
Then from the definitions of $H, P$ and $T$ in Equation \eqref{1.4}, we obtain
\begin{gather}\label{1.5}
\frac{H'}{H}=\sum_{i=1}^n\frac{q_if'(q_iz)}{f(q_iz)},\\
\label{1.6} (\frac{H'}{H}-\frac{A'}{A})PT=P'T-PT'.
\end{gather}
Fixing constants $\beta, \gamma$ such that
$$
\max\{\overline{\lambda}(f),
\overline{\lambda}(\frac{1}{f})\}<\beta<\gamma<\sigma,
$$
then we obtain
\begin{equation}\label{1.7}
T(r,\frac{f'}{f})=m(r,\frac{f'}{f})+\overline{N}(r,f)+\overline{N}(r,\frac{1}{f})
=O(r^\beta)+S(r,f).
\end{equation}
By choosing subsequence $r_n$ such that $T(r_n,f)>{r_n}^{\gamma}$, and 
noticing Equation \eqref{1.7} simultaneously, we obtain
\begin{equation}\label{1.8}
T(r_n,\frac{f'}{f})=S(r_n,f).
\end{equation}
Using the same method which leads to  \eqref{1.7}, and taking note of 
Remark \ref{rmk1},
 we can get
\begin{align*}
T(r,\frac{H'}{H})
&= m(r,\frac{H'}{H})+\overline{N}(r,H) +\overline{N}(r,\frac{1}{H})\\
&\leq \sum_{i=1}^n\overline{N}(r,f(q_iz))
 +\sum_{i=1}^n\overline{N}(r,\frac{1}{f(q_iz)})+S(r,f)\\
&=\sum_{i=1}^n\overline{N}(|q_i|r,f(z))+\sum_{i=1}^n\overline{N}
 (|q_i|r,\frac{1}{f(z)})+S(r,f)\\
&=O(r^\beta)+S(r,f),
\end{align*}
which shows
\begin{equation}\label{1.9}
T(r_n,\frac{H'}{H})=S(r_n,f).
\end{equation}
By substituting $P,T$ which are defined in \eqref{1.4} into \eqref{1.6}, we obtain
$$
(\frac{H'}{H}-\frac{A'}{A})(f^{p+t}+Q_{p+t-1})=P'T-PT'
=(p-t)\frac{f'}{f}f^{p+t}+\tilde{Q}_{p+t-1}
$$
where $Q_{p+t-1}$ and $\tilde{Q}_{p+t-1}$ are differential polynomials in $f$
with coefficients being small functions with degree at most $p+t-1$.
Thus from the equation above, we obtain
\begin{equation}\label{1.10}
[\frac{H'}{H}-\frac{A'}{A}-(p-t)\frac{f'}{f}]f^{p+t}
=\tilde{Q}_{p+t-1}-(\frac{H'}{H}-\frac{A'}{A})Q_{p+t-1}.
\end{equation}
From  \eqref{1.8} and \eqref{1.9}, we see 
$\frac{H'}{H}-\frac{A'}{A}-(p-t)\frac{f'}{f}$ is
 small function of $f(z)$ (for $r_n$). Thus if
$\frac{H'}{H}-\frac{A'}{A}-(p-t)\frac{f'}{f}\not\equiv0$, then by
applying Lemma \ref{T} to  \eqref{1.10}, we obtain
$$
(p+t)T(r_n,f)\leq(p+t-1)T(r_n,f)+S(r_n,f),
$$
which is impossible. Thus
$$
\frac{H'}{H}-\frac{A'}{A}-(p-t)\frac{f'}{f}\equiv0.
$$ 
Then we solve the equation above and get
\begin{equation}\label{1.11}
f^{p-t}=k\frac{H}{A}=k\frac{P}{T},
\end{equation}
 where $k$ is a nonzero constant. By  \eqref{1.11} and
Lemma \ref{T}, we obtain
$|p-t|=\max\{p,t\}$,
which is impossible since $p$ and $t$ are positive integers. 
Thus $p=0$ or $t=0$, and we distinguish two cases as follows. 

\noindent\textbf{Case 1.} $t=0$, then  \eqref{2} becomes
$$
F:=\prod_{i=1}^nf(q_iz)={a_0(z)+a_1(z)f(z)+\dots
+a_p(z)f^p(z)}.
$$ 
By Lemma \ref{CT} and the equation above, we obtain
\begin{equation}\label{1.12}
T(r,f)\leq \overline{N}(r,\frac{1}{F})+\overline{N}(r,f)+S(r,f)
\end{equation}
or
\begin{equation}\label{1.13}
F=a_p(f+\frac{a_{p-1}}{pa_p})^p.
\end{equation}
If Equation \eqref{1.12} holds, then 
$$
T(r,f)\leq \sum_{i=1}^n\overline{N}(r,\frac{1}{f(q_iz)})
 +\overline{N}(r,f)+S(r,f)=O(r^\beta)+S(r,f).
$$
Thus by the same discussion above, we see $T(r_n,f)\leq S(r_n,f)$,
which is impossible.

If Equation \eqref{1.13} holds, and $a_{p-1}\not\equiv0$, then the
second main theorem related to small functions implies
\begin{align*}
T(r,f)&\leq \overline{N}(r,\frac{1}{f})+\overline{N}(r,f)
 +\overline{N}(r,\frac{1}{f+\frac{a_{p-1}}{pa_p}})+\varepsilon
 T(r,f)+S(r,f)\\
&\leq O(r^\beta)+\overline{N}(r,\frac{1}{F})+\varepsilon T(r,f)+S(r,f)\\
&\leq \sum_{i=1}^n\overline{N}(r,\frac{1}{f(q_iz)})+O(r^\beta)+\varepsilon
T(r,f)+S(r,f)\\
&= O(r^\beta)+\varepsilon T(r,f)+S(r,f).
\end{align*}
That is,
$$ 
T(r_n,f)\leq \varepsilon T(r_n,f)+S(r_n,f),
$$
which is impossible since can $\varepsilon$  be set small enough. 
Thus $a_{p-1}\equiv0$, and then Equation \eqref{2} becomes
$\prod_{i=1}^nf(q_iz)=a_pf^p$.

\noindent\textbf{Case 2.} 
$p=0$. then Equation \eqref{2} becomes
$$
\tilde{F}:=(\prod_{i=1}^nf(q_iz))^{-1}
=\frac{1}{a_0}(b_0(z)+b_1(z)f(z)+\dots +b_t(z)f^t(z)).
$$
Using the similar method in case 1, we obtain
$\prod_{i=1}^nf(q_iz)=\frac{a_0(z)}{f^t(z)}$.
The proof of Theorem \ref{thm1} is complete. 
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm2}]
First of all, we suppose that $f$ is a finite order
transcendental meromorphic solution of  \eqref{3}. Then by Lemma \ref{mt},  
Lemma \ref{T} and Equation \eqref{3}, we can deduce 
$$
dT(r,f)+S(r,f)=T(r,\sum_{j=1}^{n}f(z + c_j))\leq nT(r,f)+S(r,f),
$$ 
which leads to $n\geq d\geq2$. 

(1) From \eqref{3}, it is easy to see
\begin{align*}
P(z,f)&:=\Big[\sum_{j=1}^{n}f(z + c_j)][ b_q{f(z)}^q + b_{q-1}{f(z)}^{q-1} 
 + \dots + b_0\Big]\\
&\quad -[a_p{f(z)}^p + a_{p-1}{f(z)}^{p-1} +\dots + a_0]\equiv0.
\end{align*}
Then we see $P(z,0)=-a_0\not\equiv0$. It follows from
Lemma \ref{cl} that
$$
m(r,\frac{1}{f})=S(r,f)
$$
possibly out a set with logarithmic measure, which leads to
$$
N(r,\frac{1}{f})=T(r,f)+S(r,f).
$$
Thus $\delta(0,f)=0$, i.e., $\lambda(f)=\sigma(f)$.

(2) Let $f(z)=g(z)+z$ and substitute it into  \eqref{3}, we see
\begin{align*}
\tilde{P}(z,g)&:=[\sum_{j=1}^{n}g(z + c_j)+\sum_{j=1}^{n}c_j+nz]
 [ b_q{(g(z)+z)}^q + b_{q-1}{(g(z)+z)}^{q-1} +\dots \\
&\quad + b_0]-[a_p{(g(z)+z)}^p + a_{p-1}{(g(z)+z)}^{p-1} +\dots 
+ a_0]\equiv0.
\end{align*}
Thus 
\[
\tilde{P}(z,0)=(nz+\sum_{j=1}^{n}c_j)Q(z)-P(z).
\]
If
$(nz+\sum_{j=1}^{n}c_j)Q(z)-P(z)\equiv0$, then $Q(z)|P(z)$. But
$Q(z), P(z)$ are relatively prime polynomials, so $Q(z)$ is a
nonzero constant and then $P(z)$ is a polynomial with degree 1. This
is impossible since $\deg R(z)\geq2$. Thus
$(nz+\sum_{j=1}^{n}c_j)Q(z)-P(z)\not\equiv0$, that is
$\tilde{P}(z,0)\not\equiv0$. It follows from Lemma \ref{cl} once
again that
$$
m(r,\frac{1}{f-z})=m(r,\frac{1}{g})=S(r,f)
$$
possibly out a set with logarithmic measure, which leads to
$$
N(r,\frac{1}{f-z})=T(r,f)+S(r,f).
$$
Thus $\tau(f)=\sigma(f)$.

(3) By applying Lemmas \ref{m2}, \ref{t}, \ref{T} to  \eqref{3}, we have
\begin{align*}
dT(r,f)+S(r,f)
&=T(r,\sum_{j=1}^{n}f(z + c_j))\\
&=m\big(r,\frac{\sum_{j=1}^{n}f(z + c_j)}{f}f\big)
 +N(r,\sum_{j=1}^{n}f(z + c_j))\\
&\leq m(r,f)+\sum_{j=1}^{n}N(r,f(z + c_j))+S(r,f)\\
&=T(r,f)-N(r,f)+nN(r,f)+S(r,f).
\end{align*}
That is,
$$
T(r,f)\leq(d-1)T(r,f)\leq (n-1)N(r,f)+S(r,f),
$$
which leads to $\lambda(\frac{1}{f})=\sigma(f)$.

(4) Suppose that $b$ is a deficiency value with respect to $f$, 
from result (3), we see $b\neq \infty$. Set $f(z)=g(z)+b$ and substitute 
it into  \eqref{3}, we have
$$
nb+\sum_{j=1}^{n}g(z + c_j)=R(z)\circ(g(z)+b)=R(z)\circ(z+b)\circ g(z).
$$
That is,
$$
\sum_{j=1}^{n}g(z + c_j)=\tilde{R}(z)\circ g(z),
$$
where $\tilde{R}(z)=R(z+b)-nb$.
If
$$
\tilde{a}_0=a_pb^p +
a_{p-1}b^{p-1} +\dots + a_0-nb(b_qb^q + b_{q-1}b^{q-1} + \dots + b_0)\neq0,
$$
then from result (1), we see
$$
N(r,\frac{1}{f-b})=N(r,\frac{1}{g})=T(r,f)+S(r,f),
$$
which implies $b$ is not a deficiency value of $f$.
The proof of Theorem \ref{thm2} is complete. 
\end{proof} 

\begin{proof}[Proof of Theorem \ref{thm3}]
If $n\geq3$, then by Lemma \ref{m2}, it is easy to see that
\begin{gather*}
T(r,\frac{\Delta f}{f})=N(r,\frac{\Delta f}{f})+m(r,\frac{\Delta f}{f})\leq
T(r,f)+S(r,f),\\
T(r,\Delta f)=m(r,\Delta f)\leq m(r,\frac{\Delta f}{f})+m(r,f)\leq
T(r,f)+S(r,f).
\end{gather*}
Thus
\begin{align*}
&nT(r,f)+S(r,f)\\
& \leq(n+s)T(r,f)-sT(r,\frac{\Delta f}{f})\\
&\leq T(r,f^{n+s}{(\frac{\Delta f}{f})}^s)\\
&=T(r,f^n{(\Delta f)}^s)\\
&\leq \overline{N}(r,\frac{1}{f})+\overline{N}(r,\frac{1}{\Delta
f})+\overline{N}(r,\frac{1}{f^n{(\Delta
f)}^s-\alpha(z)})+\varepsilon T(r,f)+S(r,f)\\
&\leq2T(r,f) +\overline{N}(r,\frac{1}{f^n{(\Delta f)}^s-\alpha(z)})
 +\varepsilon T(r,f)+S(r,f).
\end{align*}
That is,
$$
(n-2-\varepsilon)T(r,f)\leq\overline{N}(r,\frac{1}{f^n{(\Delta f)}^s-\alpha})
+S(r,f).
$$
Thus $f^n{(\Delta f)}^s-\alpha(z)$ has infinitely many zeros since
$\varepsilon$ can be fixed small enough. Now we just need to
consider the case $n=2$. On the contrary, we suppose that
$f^n{(\Delta f)}^s-\alpha(z)$ has just only finitely many zeros, then
we obtain that there exists two polynomials said $p$ and $Q$ such
that
\begin{equation}\label{3.1}
f^2{(\Delta f)}^s-\alpha=pe^Q.
\end{equation}
 By differentiating  \eqref{3.1} and eliminating
$e^{Q(z)}$, we obtain
\begin{equation}\label{3.2}
f[2pf'{(\Delta f)}^s+sp{(\Delta f)}^{s-1}{(\Delta f)}'f-(p'+pQ'){(\Delta f)}^sf]
=p{\alpha}'-\alpha(p'+pQ').
\end{equation}
 If $p{\alpha}'-\alpha(p'+pQ')\not\equiv0$, then from  \eqref{3.2}, we see
$$
N(r,\frac{1}{f})\leq N(r,
\frac{1}{p{\alpha}'-\alpha(p'+pQ')})=S(r,f).
$$
Hence
\begin{align*}
2T(r,f)
&\leq T(r,f^2{(\Delta f)}^s)+S(r,f)\\
&\leq \overline{N}(r,\frac{1}{f})+\overline{N}(r,\frac{1}{\Delta
f})+\overline{N}(r,\frac{1}{f^2{(\Delta
f)}^s-\alpha(z)})+\varepsilon T(r,f)+S(r,f)\\
&\leq T(r,f)+\varepsilon T(r,f)+S(r,f),
\end{align*}
which is impossible.

If $p{\alpha}'-\alpha(p'+pQ')\equiv0$, we see $pe^Q=k\alpha$, where
$k$ is a constant. Thus we substitute it into  \eqref{3.1},
 and obtain
 $$
f^2{(\Delta f)}^s=(k+1)\alpha.
$$
Thus
 $$
2T(r,f)\leq T(r,f^2{(\Delta f)}^s)+S(r,f)=T(r,(k+1)\alpha)+S(r,f)=S(r,f),
$$
which is a contradiction.
The proof of Theorem \ref{thm3} is complete.
\end{proof} 


\subsection*{Acknowledgments}
This research was supported by the Tianyuan Funds for Mathematics
(No. 11326085) and the Fundamental Research Funds for the
Central Universities (No. 2011QNA25).
The authors would like to thank the anonymous
referees for their comments and suggestions.

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