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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 59, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/59\hfil Properties of meromorphic solutions]
{Properties of meromorphic solutions of $q$-difference equations}

\author[X. Qi, L. Yang \hfil EJDE-2015/59\hfilneg]
{Xiaoguang Qi, Lianzhong Yang}

\address{Xiao-Guang Qi \newline
Jinan University, School of Mathematics,
Jinan, Shandong 250022, China}
\email{xiaogqi@mail.sdu.edu.cn, sms\_qixg@ujn.edu.cn}

\address{Lian-Zhong Yang \newline
Shandong University, School of Mathematics,
Jinan, Shandong 250100, China}
\email{lzyang@sdu.edu.cn}

\thanks{Submitted August 19, 2014. Published March 10, 2015.}
\subjclass[2000]{39A05, 30D35}
\keywords{Meromorphic functions; q-difference equation; growth; zero order}

\begin{abstract}
 In this article, we  utilize Nevanlinna value distribution theory
 to study the solvability and the growth of meromorphic function
 $f(z)$ that satisfies some $q$-difference equations, which can be
 seen the $q$-difference analogues of Painlev\'e I and II equations.
 This article extends earlier results by Chen et al \cite{C1,C2}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

In this article, we use the basic notions in Nevanlinna theory of
meromorphic functions, as found in \cite{yang and yi}. In addition,
we use $\delta(f)$ and $\lambda(\frac{1}{f})$ to
denote the order, the exponent of convergence of poles of
a meromorphic function $f(z)$, respectively.

A century ago, Painlev\'e \cite{PP1, PP2}, Fuchs \cite{LF} and 
Gambier \cite{BG} classified a large class of second order differential
equations in terms of a characteristic which is now known as the Painlev\'e
property.  Painlev\'e and his colleagues looked at the class
$$
w''(z) = F(z,w,w'),
$$
where $F$ is rational in $w$ and $w'$ and (locally) analytic in $z$, 
rejecting those equations which did not have
the Painlev\'e property. They singled out a list of 50 equations, 
six of which could not be integrated in terms of known functions. 
These equations are now known as the Painlev\'e equations. 
The first two of these equations are $P_I$ and $P_{II}$:
$$
w''=6w^2 + z, \quad  w''=2w^2 + zw + \alpha,
$$
where $\alpha$ is a constant. On the other hand, the equation that
are now traditionally called the discrete Painlev\'e
equations are special cases of QRT (Quispel-Robert-Thompson)
difference equations. The QRT difference equations was a starting
point in the discovery of the discrete Painlev\'e equations.
As we all known, the discrete $P_I$ and $P_{II}$ can be expressed in
the form
$$
y_{n+1}+ y_{n-1} = \frac{a n +b}{y_n}+c, \quad
 y_{n+1}+ y_{n-1}=\frac{(an+b)y_n+c}{1-y^{2}_{n}}.
$$
Recently, Halburd and Korhonen \cite{RG} used Nevanlinna theory to single
 out difference equations in the  form
\begin{equation}\label{0.1}
w(z+1)+w(z-1)=R(z, w),
\end{equation}
where $R(z, w)$ is rational in $w$ and meromorphic in $z$. 
They obtained that if \eqref{0.1} has an admissible meromorphic
solution of finite order, then either $w$ satisfies a difference Riccati equation, 
or \eqref{0.1} can be transformed by a linear change in $w$ to some difference 
equations, which include the difference $P_I$ and $P_{II}$ equations
\begin{equation}\label{0.2}
\begin{gathered}
f(z+1)+f(z-1)=\frac{az+b}{f}+c,\\
f(z+1)+f(z-1)=\frac{(az+b)f+c}{1-f^2}.
\end{gathered}
\end{equation}
Moreover, Chen et al \cite{C1, C2} investigated properties of finite-order 
transcendental meromorphic solutions of \eqref{0.2}.
Closely related to difference expressions are $q$-difference
expressions, where the usual shift $f(z+c)$ of a meromorphic
function will be replaced by the $q$-difference $f(qz)$, $q \in
\mathbb{C}\setminus\{0\}$. The Nevanlinna theory of $q$-difference
expressions and its applications to $q$-difference equations have
recently been considered, see \cite{B}. In addition, some
results about solutions of zero order for complex $q$-difference
equations, can be found in the introduction in \cite{B}.

A natural question is: what is the result if we give $q$-difference
 analogues of \eqref{0.2}. Corresponding to this question, we consider 
the following two equations:
\begin{gather}\label{1.1}
f(qz)+f(\frac{z}{q})=\frac{az+b}{f}+c,\\
\label{1.2}
f(qz)+f(\frac{z}{q})=\frac{(az+b)f+c}{1-f^2}.
\end{gather}

\begin{theorem}\label{T1}
Let $f(z)$ be a transcendental meromorphic solution with zero order
of equation \eqref{1.1}, and $a$, $b$, $c$ be three constants such
that $a$, $b$ cannot vanish simultaneously. Then,
\begin{itemize}
\item[(i)] $f(z)$ has infinitely many poles.
\item[(ii)] If $a\neq 0$, then $f(z)$ has infinitely many finite values.
\item[(iii)] If $a=0$ and $f(z)$ takes a finite
value $A$ finitely often, then $A$ is a solution of $2z^2-cz-b=0$.
\end{itemize}
\end{theorem}

\begin{theorem}\label{T2}
Let $a$, $b$, $c$ and $|q|\neq 1$ be four constants,
\begin{itemize}
\item[(i)] if $a\neq 0$, then equation \eqref{1.1} has no rational
solution;
\item[(ii)] if $a=0$, then the rational solutions of the
equation \eqref{1.1} must satisfy $f(z)=B+\frac{P(z)}{Q(z)}$, where
$P(z)$ and $Q(z)$ are relatively prime polynomials and satisfy $\deg
P < \deg Q$ and $2B^2-cB-b=0$.
\end{itemize}
\end{theorem}

\begin{theorem}\label{T3}
Let $a, b, c$ be constants with $ac\neq0$, and let $f(z)$ be a transcendental 
meromorphic solution with zero order of equation \eqref{1.2}. 
Then $f(z)$ has infinitely many poles and infinitely many finite values.
\end{theorem}

Using similar methods as in the proof of Theorem \ref{T1}, we can prove 
Theorem \ref{T3}. Here, we omit the details.

\begin{theorem}\label{T4}
Let $a, b, c$ be constants with $ac\neq0$ and $|q|\neq 1$. Suppose that 
a rational function
$$
f(z)=\frac{P(z)}{Q(z)}
=\frac{sz^m+p_{m-1}z^{m-1}+\dots+p_0z}{tz^n+q_{n-1}z^{m-1}+\dots+q_0z}
$$
is a solution of \eqref{1.2}, where $P(z)$ and $Q(z)$ are relatively prime 
polynomials, $s\neq 0, p_{m-1},\dots , p_0$ and
$t\neq 0, q_{n-1},\dots q_0$ are constants. Then
$n=m+1$ and $s=-\frac{c}{a}t$.
\end{theorem}

As for the next result from
this point of view, see \cite{AR}. Several papers have appeared in which 
the solutions of the following equation are studied
\begin{equation}\label{1.4}
f(z+1) + f(z-1) = a(z)f(z)^2 + b(z)f(z) + c(z).
\end{equation}
The reader is invited to see \cite{IY}.  The following result can be
seen as a $q$-difference counterpart to  \eqref{1.4}.

\begin{theorem}\label{T6}
Let $|q|\neq 1$ and $n\geq 2$, let $f(z)$ be a meromorphic solution of
\begin{equation}\label{1.3}
f(qz) + f(\frac{z}{q}) = a(z)f(z)^n + b(z)f(z) + c(z)
\end{equation}
with meromorphic coefficients satisfying $T(r, a) = S(r, f)$, $T(r, b) = S(r, f)$
 and $T(r, c) = S(r, f)$. Then $f(z)$ is of positive order of growth.
\end{theorem}

We remark that Theorem \ref{T6} is not true, when $n=1$.
This can be seen by considering $f(z)=z+1$,
$f(\frac{1}{2}z)=\frac{1}{2}z+1$ and $f(2z)=2z+1$. Then $f(z)$ is a solution of
$f(2z)+f(\frac{z}{2})=\frac{5}{2}f(z)-\frac{1}{2}$.


Also we remark that the right hand side of the above equations are essentially 
like the function $A(z, f)$, where $A(z, f)$ is a rational function of $f(z)$. 
Reversing the order of composition on the right hand side results in a functional 
$q$-difference equation.
The following theorem gives an example.

\begin{theorem}\label{T5}
Let $q\neq 0$ be a complex constant, and suppose $f(z)$ be a
transcendental meromorphic function of the equation
       \begin{equation}\label{5.1}
        A(qz, f(qz))=f(p(z)),
       \end{equation}
where $p(z)$ is a polynomial of degree $k\geq 2$, $A(z, y)$ is a
rational with meromorphic coefficients of growth $S(r, f)$ such that
$A(z, y)$ is irreducible in $y$. If $\deg_{f}A=n\geq k$, then for any 
$\varepsilon>0$
$$
T(r, f)= O((\log r)^{\alpha+\varepsilon}),
$$
where $\alpha=\frac{\log n}{\log k}$.
\end{theorem}

 Some ideas in this paper come from \cite{C1,IY}.

\section{Some Lemmas}

\begin{lemma}[{\cite[Theorem 2.1]{B}}] \label{lem2.1}
Let $f(z)$ be a non-constant zero order meromorphic solution of
$$
f(z)^nP(z, f) = Q(z, f),
 $$
where $P(z, f)$ and $Q(z, f)$ are q-difference polynomials in
$f(z)$. If the degree of $Q(z, f)$ as a polynomial in $f(z)$ and its
$q$-shifts is at most $n$, then 
$$
m(r, P(z, f)) = o(T(r, f))
$$ 
on a set of logarithmic density 1.
\end{lemma}

\begin{lemma}[{\cite[Theorem 2.2]{B}}]  \label{lem2.2}
Let $f(z)$ be a non-constant zero order meromorphic solution of
$$
H(z, f) = 0,
$$
where $H(z, f)$ is a $q$-difference polynomial in $f(z)$. 
If $H(z, a)\not\equiv 0$ for slowly moving target $a(z)$, then
$$
m(\frac{1}{f-a})= o(T(r, f))
$$ 
on a set of logarithmic density $1$.
\end{lemma}

\begin{lemma}[{\cite[Theorems 1.1 and 1.3]{zhang}}] \label{lem2.3}
Let $f(z)$ be a zero order meromorphic function, and
$q\in\mathbb{C}\setminus\{0\}$. Then
\begin{gather}\label{1}
T(r, f(qz))=(1+o(1))T(r, f(z)) \\
\label{2}
N(r, f(qz))=(1+o(1))N(r, f(z))
\end{gather}
on a set of lower logarithmic density $1$.
\end{lemma}

We recall some notation and a lemma from \cite{RG}. $\omega(z)$ has more than
$S(r, \omega)$ poles of a certain type, which means that the integrated counting 
function of these poles is not of type $S(r, \omega)$. 
We use the notation $D(z_0, r)$ to
denote an open disc of radius $r$ centered at $z_0\in C$. Also, $\infty^k$ 
denotes a pole of $\omega$ with multiplicity $k$. We now recall the following lemma
from \cite[Lemma 3.1]{RG}.

\begin{lemma}\label{lem2.4}
Let $\omega(z)$ be a meromorphic function having more than $S(r, \omega)$
poles, and let $a_s$ $(s =1,2, \dots, n)$ be small meromorphic functions
with respect to $\omega$. Denote by $m_j$ the maximum order of zeros and poles 
of the functions as at $z_j$ . Then for any $\varepsilon>0$, there are at most 
$S(r, \omega)$  points $z_j$ such that
$$
\omega(z_j) = \infty^{k_j},
$$
where $m_j \geq \varepsilon k_j$.
\end{lemma}

\begin{lemma}[\cite{R.Gg}] \label{L1}
Let $p(z)=a_kz^{k}+a_{k-1}z^{k-1}+\dots+a_1z+a_0$, $a_k\neq 0$, be
a non-constant polynomial of degree $k$ and let $f(z)$ be a
transcendental meromorphic function. Given $0< \varrho <|a_k|$,
denote $\zeta= |a_k|+\varrho$ and $\eta= |a_k|-\varrho$. Then, given
$\varepsilon>0$, we have
$$
(1-\varepsilon)T(\eta r^{k}, f)\leq T(r, f\circ p)\leq
(1+\varepsilon)T(\zeta r^{k}, f)
$$
for all $r$ large enough.
\end{lemma}

\begin{lemma}[\cite{R.Gg2}] \label{L1qb}
Let $\psi: [r_{0}, \infty)\to(0,\infty)$ be positive and
bounded in every finite interval, and suppose that 
$\psi(\mu r^m)\leq A\psi(r)+B$ holds for all $r$ large enough, where
$\mu>0$, $m>1$, $A>1$ and $B$ are real constants. Then
$$
\psi(r)=O((\log r)^{\alpha}),
$$
where $\alpha=\frac{\log A}{\log m}$.
\end{lemma}

\section{Proof of Theorem \ref{T1}}

(i) Suppose that $f(z)$ is a zero order transcendental meromorphic
solution of \eqref{1.1}. From \eqref{1.1}, we obtain
$$
fP(z, f)=Q(z, f),
$$
where
\begin{equation}\label{3.1}
P(z, f)=f(qz)+f(\frac{z}{q}),\quad Q(z, f)=az+b+cf(z).
\end{equation}
Applying Lemma \ref{lem2.1}, we obtain
\begin{equation}\label{3.2}
m(r, P(z, f))=o(T(r, f))
\end{equation}
on a set of logarithmic density $1$. Combining the Valiron-Mohon'ko theorem 
and \eqref{1.1},
\begin{equation}\label{3.3}
T(r, P(z, f))=T(r, f)+S(r, f)
\end{equation}
follows. From Lemma \ref{lem2.3}, we obtain
$$
N(r, P(z, f))\leq N(r, f(qz))+N(r,
f(\frac{z}{q}))=2(1+o(1))N(r, f)
$$
on a set of lower logarithmic density 1. This, together with
\eqref{3.2} and \eqref{3.3}, yields
$$
2(1+o(1))N(r, f)\geq T(r, f)+S(r, f)
$$
on a set of logarithmic density $1$. Hence, $f(z)$ has infinitely many poles.


(ii) For any finite value $A$, and set
$$
g(z)=f(z)-A.
$$
This, combining with \eqref{1.1}, it follows that
$$
g(qz)+g(\frac{z}{q})+2A=\frac{az+b}{g(z)+A}+c.
$$
Rewrite above equation in the form
\begin{equation}\label{3.4}
H(z, g)=(g(qz)+g(\frac{z}{q})+2A)(g(z)+A)-c(g(z)+A)-az-b=0.
\end{equation}
By the assumption that $a\neq 0$ and \eqref{3.4}, we have 
$$
H(z, 0)=2A^2-cA-b-az\not\equiv0.
$$
Hence, applying Lemma \ref{lem2.2}, we obtain that
$$
m(r, \frac{1}{g})= o(T(r, g))
$$ 
on a set of logarithmic density $1$.
That is,
\begin{equation}\label{3.5}
N(r, \frac{1}{f-A})=N(r, \frac{1}{g})=T(r, g)(1+o(1))=T(r, f)(1+o(1))
\end{equation}
on a set of logarithmic density $1$. The conclusion holds.


(iii) If $a=0$ and $A$ is not a solution of $2z^2-cz-b=0$; that is,
$$
H(z, 0)=2A^2-cA-b\neq 0.
$$ 
Then, we obtain $N(r, \frac{1}{f-A})=T(r,f)(1+o(1))$
as well, using the same way as the above argument. 
This contradicts the assumption, and the conclusion follows.

\section{Proof of Theorem \ref{T2}}

Suppose that $f(z)$ is a rational solution of \eqref{1.1}, and has
poles $z_1, \dots z_k$. Thus, $f(z)$ can be represented in the
form
\begin{equation}\label{3.5b}
f(z)=\sum_{i=1}^{k}\Big(\frac{d_{is_i}}{(z-z_i)^{s_i}}+\dots
+\frac{d_{is_1}}{(z-z_i)}\Big)+B+B_1z+\dots+B_mz^m,
\end{equation}
where $d_{is_i}\neq0,\dots d_{is_1}$, $B$, $B_1\dots B_m$ are
constants.


In the following, we prove that $B_1=\dots = B_m=0$. Suppose,
contrary to the assertion, that $B_m\neq0$, $(m\geq 1)$. For
sufficiently large $z$, from \eqref{3.5b}, we know
\begin{equation}\label{3.6}
\begin{gathered}
f(z)=B_mz^m(1+o(1)),\\
f(qz)=B_mz^mq^m(1+o(1)),\\
f(qz)=B_mz^m\frac{1}{q^m}(1+o(1)).
\end{gathered}
\end{equation}
From  \eqref{1.1}, we have
\begin{equation}\label{3.6*}
f(z)(f(qz)+f(\frac{z}{q}))=az+b+cf(z).
\end{equation}
This and \eqref{3.6} imply that
$$
(q^m+\frac{1}{q^m})B_{m}^{2}z^{2m}(1+o(1))=az+b+cB_mz^m(1+o(1)),
$$
Since $B_m\neq0$ and $|q|\neq 1$, we see the above equation is a 
contradiction for sufficiently large $z$. Therefore,
$B_1=\dots = B_m=0$.


Case 1. $a\neq 0$. If $B\neq 0$, then from \eqref{3.5}, we have
$$
f(qz)=f(z)=f(\frac{1}{q}z)=B+o(1),
$$
for sufficiently large $z$. Substituting the above equation into \eqref{3.6*}, 
we conclude that
$$
(B+o(1))(2B+o(1))=az+b+c(B+o(1)),
$$
which is a contradiction to the assumption that $a\neq 0$. 
Thus, $B=B_1=\dots = B_m=0$. Hence, we obtain
\begin{equation}\label{3.7}
f(z)=\frac{P(z)}{Q(z)},
\end{equation} 
where $P(z)$, $Q(z)$ are polynomials such that $\deg P<\deg Q$. 
Combining \eqref{3.7} and \eqref{1.1},
\begin{equation}\label{3.8}
\begin{split}
&P(qz)P(z)Q(\frac{z}{q})+P(z)P(\frac{z}{q})Q(qz)\\
&=(az+b)Q(qz)Q(z)Q(\frac{z}{q})+cP(z)Q(qz)Q(\frac{z}{q}).
\end{split}
\end{equation}
Observing the above equation, we see that the degree of both sides of 
\eqref{3.8} are not equal. This is impossible. Hence,
if $a\neq 0$, then \eqref{1.1} has no rational solution.


Case 2. $a=0$. From the above argument, we know that
$$
f(z)=\frac{P_1(z)}{Q_1(z)}+B,
$$
where $P_1(z)$, $Q_1(z)$ are polynomials such that $\deg P_1<\deg Q_2$.
Moreover, we have
$$
f(qz)=f(z)=f(\frac{1}{q}z)=B+o(1),
$$
for sufficiently large $z$. Combining above equation and \eqref{1.1}, 
we obtain that $2B^2-cB-b=0$. The conclusion holds.

\section{Proof of Theorem \ref{T4}}

Assume that $f(z)$ is a rational solution of \eqref{1.2}, and has
poles $z_1, \dots z_k$. Hence, $f(z)$ can be expressed as the
following the form:
\begin{equation}\label{4.1}
f(z)=\sum_{i=1}^{k}\Big(\frac{b_{is_i}}{(z-z_i)^{s_i}}+\dots
+\frac{b_{is_1}}{(z-z_i)}\Big)+D+D_1z+\dots+D_jz^j,
\end{equation}
where $b_{is_i}\neq0,\dots b_{is_1}$, $D$, $D_1\dots D_j$ are
constants. Using a similar method as in the proof of Theorem \ref{T2},
 we have $D_1=\dots = D_j=0$ as well.
If $D\neq 0$, then from \eqref{4.1}, we have
$$
f(qz)=f(z)=f(\frac{1}{q}z)=D+o(1),
$$
for sufficiently large $z$. Combining the above equation and  \eqref{1.2},
$$
(az+b)(D+o(1))=-(D^2+o(1))(2D+o(1))+(2D+o(1))-c
$$ 
follows. From the assumption that $a\neq0$ and $D\neq 0$, we obtain a contradiction.
Therefore, $D=0$. Furthermore, $f(z)$ can be expressed as
\begin{equation}\label{4.2}
f(z)=\frac{P(z)}{Q(z)},
\end{equation}
where
\begin{gather*}
P(z)=sz^m+p_{m-1}z^{m-1}+\dots+p_0z,\\
Q(z)=tz^n+q_{n-1}z^{m-1}+\dots+q_0z.
\end{gather*}
Substituting \eqref{4.2} into \eqref{1.2}, we obtain
\begin{equation}\label{4.3}
\begin{split}
&cQ (qz)Q (\frac{z}{q})Q(z)^2 + (az+b)Q(qz)Q(\frac{z}{q})P(z)Q(z)\\
&=P(qz)Q(\frac{z}{q})Q(z)^2 -P(qz)Q(\frac{z}{q})P(z)^2 + P(\frac{z}{q})Q(qz)Q(z)^2\\
&\quad -P(\frac{z}{q})Q(qz)P(z)^2.
\end{split}
\end{equation}
As $ac\neq 0$, by comparing the degrees of all terms of \eqref{4.3}, we obtain 
that $n=m +1$. From \eqref{4.2} and \eqref{1.2},
it follows that
\begin{equation}\label{4.4}
\frac{P(qz)}{Q(qz)}+\frac{P(\frac{z}{q})}{Q(\frac{z}{q})}
=\frac{(az+b)P(z)Q(z)+cQ(z)^2}{Q(z)^2-P(z)^2}.
\end{equation}
From this and $n=m+1$, we conclude that
$$
\frac{P(qz)}{Q(qz)}+\frac{P(\frac{z}{q})}{Q(\frac{z}{q})}\to 0,
$$
and
$$
\frac{(az+b)P(z)Q(z)+cQ(z)^2}{Q(z)^2-P(z)^2}=\frac{(ast+ct^2)z^{2n}
 (1+o(1))}{s^2z^{2n}(1+o(1))}
$$
for sufficiently large $z$. Comparing the two above equations, we 
know $ast+ct^2=0$, that is, $s=-\frac{c}{a}t$.

\section{Proof of Theorem \ref{T6}}

Let $f(z)$ be a meromorphic solution of \eqref{1.3}.
Suppose, contrary to the assertion, that $\sigma(f)=0$.
If $n>2$, then the conclusion holds by Lemma \ref{lem2.3}, clearly. 
 Now we consider the case $n=2$. Writing \eqref{1.3} in the form
$$
af^2=f(qz)+f(\frac{z}{q})-bf-c.
$$
By Lemma \ref{lem2.1}, we obtain that
$$
m(r, f) = o(T(r, f))
$$ 
on a set of logarithmic density 1. Hence, $f(z)$ has more than $S(r, f)$ poles 
counting multiplicities. We use $z_j$ to denote points in the pole sequence. 
Lemma \ref{lem2.4} implies that there are more
than $S(r, f)$ points such that the multiplicity of $f(z_j) = \infty$ is $k_j$, 
where $\varepsilon k_j > m_j$. Here $m_j$ refers to $a, b, c$. 
Denoting the sequence of such poles by $z_{1,j}$ , we take this sequence
as our starting point. For $\varepsilon<1/2$, by \eqref{1.3}, 
we have at least one of the points $qz_{1,j}$, $z_{1,j}/q$ is 
a pole of $f$ of multiplicity $k_{2,j}> (2-\varepsilon) k_{1,j}$. 
By Lemma \ref{lem2.4}, we obtain that $f(z)$ has
more than $S(r, f)$ such points $z_{2,j}$ such that the multiplicity of
$f(z_{2,j}) = \infty$ is $k_{2,j}$, where $\varepsilon k_{2,j} > m_{2,j}$. 
Then we only choose one of these points and denote
it by $z_{2,j}$. Continuing to the next phase. By \eqref{1.3}, we have 
that $z_{3,j}= qz_{2,j}$ or $z_{3,j}= \frac{z_{2,j}}{q}$
is a pole of $f$ of multiplicity $k_{3,j}$, where
$$
k_{3,j}\geq (2-\varepsilon)k_{2,j}> (2-\varepsilon)^2 k_{1,j}.
$$
 By induction, we can choose a sequence $z_n$ of poles of $f(z)$, the multiplicity
of which is $k_n$, and 
$k_n \geq (2-\varepsilon)^{n-1} k_{1,j}\geq (2-\varepsilon)^{n-1}$.


Case a. If $|q|>1$, then by a simple geometric observation, we obtain
$$
z_n\in B(0, |q|^{n}|z_1|)=B(0, |q|^{n}r_1)=B(0, r_n).
$$
Therefore,
$$
n(r_n, f)\geq (\frac{3}{2})^{n-1}
=(\frac{3}{2})^{\frac{\log r_n-\log r_1-\log |q|}{\log |q|}}.
$$
So, we obtain $\lambda(\frac{1}{f})=\frac{\log \frac{3}{2}}{\log|q|}>0$,
 contradicting our hypothesis $\sigma(f)=0$. Hence $\sigma(f)>0$.


Case b. If $|q|<1$, then by a simple geometric observation, it follows
$$
z_n\in B(0, |\frac{1}{q}|^{n}|z_1|)=B(0, |\frac{1}{q}|^{n}r_1)=B(0, r_n).
$$ 
Using the same argument as Case a, we can get the conclusion as well.



\section{Proof of Theorem \ref{T5}}

We first replace $z$ by $z/q$ in equation \eqref{5.1}, then applying the
Valiron-Mohon'ko theorem to the left hand side of \eqref{5.1}, and
combining Lemma \ref {L1}, we obtain that
$$
nT(r, f)+S(r, f)=T(r, A)=T(r, f(p(\frac{z}{q})))\geq
(1-\varepsilon)T(\mu r^{k}, f).
$$
Since we may assume $r$ to be large enough to satisfy
$$
n(1+\varepsilon)T(r, f)\geq (1-\varepsilon)T(\mu r^{k}, f)
$$
outside of a possible exceptional set of finite linear measure.
 We conclude that, for every $\lambda>1$, there
exists an $r_0>0$ such that
     \begin{equation}\label{5.2}
       n(1+\varepsilon)T(\lambda r, f)\geq (1-\varepsilon)T(\mu r^{k}, f)
     \end{equation}
holds for all $r\geq r_0$. Denoting $t=\lambda r$, then \eqref{5.2}
can be written as
$$
T(\frac{\mu}{\lambda^{k}}t^{k},
f)\leq\frac{n(1+\varepsilon)}{1-\varepsilon}T(t, f).
$$
Thus, we apply Lemma \ref{L1qb} to conclude that
$$
\alpha=\frac{\log n(1+\varepsilon)/(1-\varepsilon)}{\log
k}=\frac{\log n}{\log k}+o(1).
$$

\subsection*{Acknowledgments}
This work was supported by the National Natural Science Foundation
of China (No. 11301220 and No. 11371225), the NSF of Shandong
Province, China (No. ZR2012AQ020) and the Fund of Doctoral Program
Research of University of Jinan (XBS1211).

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\end{document}