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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 164, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/164\hfil Growth of solutions]
{Growth of solutions to higher order linear homogeneous
differential equations in angular domains}

\author[N. Wu\hfil EJDE-2010/164\hfilneg]
{Nan Wu}

\address{Nan Wu \newline
Department of Mathematical Sciences,
Tsinghua University, Beijing, 100084, China}
\email{wunan07@gmail.com}

\thanks{Submitted February 23, 2010. Published November 17, 2010.}
\thanks{Supported by grant 10871108 from the NSF of China}
\subjclass[2000]{30D10, 30D20, 30B10, 34M05}
\keywords{Meromorphic solutions; Nevanlinna theory; order}

\begin{abstract}
 In this article, we discuss the growth of meromorphic solutions
 to higher order homogeneous differential equations in some angular
 domains, instead of the whole complex plane.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction and statement of main results}

By a transcendental meromorphic function, we mean a function that
is meromorphic on the whole complex plane, and is not a rational
function; in other words, $\infty$ is an essential singular point.
We assume the reader is familiar with the Nevanlinna theory of
meromorphic functions and basic notation such as: Nevanlinna
characteristic $T(r,f)$, integrated counting function $N(r,f)$,
and proximity function $m(r,f)$, and  the deficiency $\delta(a,f)$
of $f(z)$. For the details, see \cite{Hayman,Yang}. The order
$\lambda$ and the lower order $\mu$ are defined as follows:
$$
\lambda(f)=\limsup_{r\to\infty}\frac{\log T(r,f)}{\log r},\quad
\mu(f)=\liminf_{r\to\infty}\frac{\log T(r,f)}{\log r}.
$$
It is known the growth of meromorphic solutions of differential
equations with meromorphic coefficients in the complex plane
$\mathbb{C}$ attracted a lot research. In this article, we
discuss the growth of meromorphic solutions of differential
equations with transcendental meromorphic coefficients in a proper
subset of $\mathbb{C}$. Let $f(z)$ be a meromorphic function in an
angular region $\overline{\Omega}(\alpha,\beta)=\{z:\alpha\leq\operatorname{arg}
z\leq\beta\}$. Recall the definition of Ahlfors-Shimizu
characteristic in an angle (see \cite{Tsuji}). Set
$$
\Omega(r)=\Omega(\alpha,\beta)\cap\{z:1<|z|<r\}=\{z:\alpha<\operatorname{arg}
z<\beta, 1<|z|<r \}.
$$
 Define
$$
\mathcal{S}(r,\Omega,f)=\frac{1}{\pi}\iint_{\Omega(r)}
\Big(\frac{|f'(z)|}{1+|f(z)|^2}\Big)^2 d\sigma, \quad
\mathcal{T}(r,\Omega,f)=\int_1^r
\frac{\mathcal{S}(t,\Omega,f)}{t}dt.
$$
The order and lower order of $f$ on $\Omega$ are defined as
follows
$$
\sigma_{\alpha,\beta}(f)=\limsup_{r\to\infty}
\frac{\log \mathcal {T}(r,\Omega,f)}{\log r},\quad
\mu_{\alpha,\beta}(f)=\liminf_{r\to\infty}
\frac{\log \mathcal {T}(r,\Omega,f)}{\log r}.
$$
\begin{remark} \label{rmk1} \rm
The order $\sigma_{\alpha,\beta}(f)$ of a
meormorphic function $f$ on an angular region here we give is
reasonable, because $\mathcal {T}(r, \mathbb{C}, f)=T(r, f)+O(1)$.
\end{remark}

Nevanlinna theory on the angular domain plays an important role in
value distribution of meromrorphic functions. Let us recall the
following terms \cite{Goldberg}:
\begin{gather*}
A_{\alpha,\beta}(r,f)=\frac{\omega}{\pi}\int_1^r\Big(\frac{1}
{t^\omega}-\frac{t^\omega}{r^{2\omega}}\Big)\{\log^+|f(te^{i\alpha})|
+\log^+|f(te^{i\beta})|\}\frac{dt}{t},\\
B_{\alpha,\beta}(r,f)=\frac{2\omega}{\pi
r^\omega}\int_{\alpha}^\beta\log^+|f(re^{i\theta})|\sin\omega(\theta-\alpha)d\theta,\\
C_{\alpha,\beta}(r,f)=2\sum_{1<|b_n|<r}\Big(\frac{1}{|b_n|^\omega}-
\frac{|b_n|^\omega}{r^{2\omega}}\Big)\sin\omega(\theta_n-\alpha),
\end{gather*}
where $\omega=\pi/(\beta-\alpha)$, and $b_n=|b_n|e^{i\theta_n}$ is a
pole of $f(z)$ in the angular domain $\Omega(\alpha,\beta)$, appears
according to its multiplicity. The Nevanlinna's angular
characteristic is defined as follows:
\[
S_{\alpha,\beta}(r,f)=A_{\alpha,\beta}(r,f)+B_{\alpha,\beta}(r,f)
+C_{\alpha,\beta}(r,f).
\]
Some articles define the order and lower order of $f$ on $\Omega$
as:
$$
\overline{\sigma}_{\alpha,\beta}(f)=\limsup_{r\to\infty}
\frac{\log S_{\alpha,\beta}(r,f)}{\log r}, \quad
\overline{\mu}_{\alpha,\beta}(f)
=\liminf_{r\to\infty}\frac{\log S_{\alpha,\beta}(r,f)}{\log r}.
$$
According to the inequality
$$
S_{\alpha,\beta}(r,f)\leq2\omega^2
\frac{\mathcal {T}(r,\Omega,f)}{r^\omega}
+\omega^3\int_1^r\frac{\mathcal {T}(t,\Omega,f)}{t^{\omega+1}}dt+O(1),
$$
showed by Zheng \cite[Theorem 2.4.7]{JH01}, if
$\sigma_{\alpha,\beta}(r,f)<\infty$, then
$\overline{\sigma}_{\alpha,\beta}(r,f)<\infty$.

We consider $q$ pairs of real numbers $\{\alpha_j,\beta_j\}$ such
that
\begin{equation}\label{1.1}
-\pi\leq\alpha_1<\beta_1\leq\alpha_2<\beta_2\leq\dots
\alpha_q<\beta_q\leq\pi
\end{equation}
and the angular domains $X=\cup_{j=1}^q\{z:\alpha_j\leq\operatorname{arg}
z\leq\beta_j\}$. For a function $f$ meromorphic in the complex plane
$\mathbb{C}$, we define the order of $f$ on $X$ as
$$
\sigma_X(f)=\limsup_{r\to\infty}
\frac{\log \mathcal {T}(r, X, f)}{\log r}.
$$
It is obvious that
$\sigma_{\alpha_j,\beta_j}(f)\leq\sigma_X(f)
\leq\sum_{j=1}^q\sigma_{\alpha_j,\beta_j}(f)$.
$j=1,2,\dots,q$. And $\sigma_X(f)=+\infty$ if and only if there
exists at least one $1\leq j_0\leq q$ such that
$\sigma_{\alpha_{j_0},\beta_{j_0}}(f)=+\infty$.  We will establish
the following results.

\begin{theorem} \label{thm1.1}
Let $A_0(z)$ be a meromorphic function in $\mathbb{C}$ with finite
lower order $\mu<\infty$ and nonzero order $0<\lambda\leq\infty$ and
$\delta=\delta(\infty,A_0)>0$. For $q$ pair of real numbers
$\{\alpha_j,\beta_j\}$ satisfying \eqref{1.1} and
\begin{equation}\label{thmeq1.1}
\sum_{j=1}^q(\alpha_{j+1}-\beta_j)<\frac{4}{\sigma}
\arcsin\sqrt{\delta/2}
\end{equation}
where $\sigma>0$ with $\mu\leq\sigma\leq\lambda$. If
$A_j(z)(j=1,2,\dots,n)$ are meromorphic functions in $\mathbb{C}$
with $T(r,A_j)=o(T(r,A_0))$, then every solution $f\not\equiv0$ to
the equation
$$
A_nf^{(n)}+A_{n-1}f^{(n-1)}+\dots+A_0f=0
$$
has the order $\sigma_X(f)=+\infty$ in
$X=\cup_{j=1}^q\{z: \alpha_j\leq\operatorname{arg} z\leq\beta_j\}$.
\end{theorem}

If we remove the condition $\mu(A_0)<\infty$ in
Theorem \ref{thm1.1}, we can establish the following result.

\begin{theorem} \label{thm1.2}
Let $A_0(z)$ be a meromorphic function in $\mathbb{C}$ with nonzero
order $0<\lambda\leq\infty$ and $\delta(\infty,A_0)>0$. Suppose that
for $q$ directions $\operatorname{arg} z=\alpha_j(1\leq j\leq q)$, satisfying
$$
-\pi\leq\alpha_1<\alpha_2<\dots<\alpha_q<\pi, \alpha_{q+1}
=\alpha_1+2\pi,
$$
$A_j(z)$, $j=1,2,\dots,n$, are meromorphic functions in $\mathbb{C}$
with finite lower order and $T(r, A_j)=o(T(r, A_0))$. Then every
solution $f\not\equiv0$ to the equation
$$
A_nf^{(n)}+A_{n-1}f^{(n-1)}+\dots+A_0f=0
$$
has order $\sigma_X(f)=+\infty$ in
$X=\mathbb{C}\backslash\cup_{j=1}^q\{z:\operatorname{arg} z=\alpha_j\}$.
\end{theorem}

The method in this paper was firstly used by Zheng \cite{Zheng} to
investigate the growth of transcendental meromorphic functions with
radially distributed values.

\section{Some auxiliary results}

To prove the theorems, we give some lemmas. The following
result is from \cite{Yang01,JH01,Zheng}.

\begin{lemma}\label{lem1.1}
Let $f(z)$ be a transcendental meromorphic function with lower order
$\mu<\infty$ and order $0<\lambda\leq\infty$, then for any positive
number $\mu\leq\sigma\leq\lambda$ and any set $E$ with finite
measure, there exist a sequence $\{r_n\}$, such that
\begin{itemize}
\item[(1)] $r_n\notin E$, $\lim_{n\to\infty}\frac{r_n}{n}=\infty$;

\item[(2)] $\liminf_{n\to\infty}\frac{\log T(r_n,f)}{\log
r_n}\geq\sigma$;

\item[(3)] $T(t,f)<(1+o(1))(\frac{2t}{r_n})^\sigma
T(r_n/2,f)$, $t\in[r_n/n,nr_n]$;

\item[(4)] $T(t,f)/t^{\sigma-\varepsilon_n}
\leq2^{\sigma+1}T(r_n,f)/r_n^{\sigma-\varepsilon_n}$,
$1\leq t\leq nr_n$, $\varepsilon_n=[\log n]^{-2}$.
\end{itemize}
\end{lemma}

We recall that $\{r_n\}$ is called the P\'{o}lya peaks of order
$\sigma$ outside $E$. Given a positive function $\Lambda(r)$
satisfying $\lim_{r\to\infty}\Lambda(r)=0$. For $r>0$ and
$a\in\mathbb{C}$, define
$$
D_\Lambda(r,a)=\{\theta\in[-\pi,\pi):\log^+
\frac{1}{|f(re^{i\theta})-a|}>\Lambda(r)T(r,f)\},
$$
and
$$
D_\Lambda(r,\infty)=\{\theta\in[-\pi,\pi):\log^+
|f(re^{i\theta})|>\Lambda(r)T(r,f)\}.
$$
The following result is called the spread relation, which was
conjectured by Edrei \cite{Edrei} and proved by Baernstein
\cite{Baernstein}.

\begin{lemma}\label{lem1.2}
Let $f(z)$ be transcendental and meromorphic in $\mathbb{C}$ with
the finite lower order $\mu<\infty$ and the positive order
$0<\lambda\leq\infty$ and has one deficient values
$a\in\widehat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$. Then for any
sequence of P\'{o}lya peaks $\{r_n\}$ of order
$\sigma>0,\mu\leq\sigma\leq\lambda$ and any positive function
$\Lambda(r)\to0$ as $r\to+\infty$, we have
$$
\liminf_{n\to\infty} \operatorname{meas} D_\Lambda(r_n,a) \geq\min\{2\pi,
\frac{4}{\sigma}\arcsin\sqrt{\delta(a,f)/2}\}.
$$
\end{lemma}

To make it clearly, we give the definition of $\mathbb{R}$-set on
the complex plane $\mathbb{C}$.

\begin{definition} \label{def2.2} \rm
Let $B(z_n, r_n)=\{z: |z-z_n|<r_n\}$ be an open disk on the complex
plane. If $\sum_{n=1}^\infty r_n<\infty$,
$\cup_{n=1}^\infty B(z_n, r_n)$ is called an $\mathbb{R}$-set.
\end{definition}

\begin{lemma}[\cite{Wu}]\label{lem1.4}
Let $f$ be a meromorphic function on the angular region
$\overline{\Omega}(\alpha,\beta)$ with finite order $\rho$, let
$\Gamma=\{(n_1,m_1),(n_2,m_2),\dots,(n_j,m_j)\}$ denote a finite
set of distinct pair of integers which satisfying $n_i>m_i\geq0$ for
$i=1,2,\dots,j$, and let $\varepsilon>0$ and $\delta>0$ be given
constants. Then there exists $K>0$ depending only on
$f,\varepsilon,\delta$ such that
\begin{equation}\label{2.1a}
\Big|\frac{f^{(n)}(z)}{f^{(m)}(z)}\Big|
<K|z|^{(n-m)(k_\delta+2\rho+1+\varepsilon)}(\sin
k_\delta(\varphi-\alpha-\delta))^{-2^{n-m}},
\end{equation}
for all $(n,m)\in\Gamma$ and all $z=re^{i\varphi}\in
\Omega(\alpha+\delta,\beta-\delta)$ except for a $\mathbb{R}$-set,
that is, a countable union of discs whose radii have finite sum,
where $k_\delta=\frac{\pi}{\beta-\alpha-2\delta}$.
\end{lemma}

To prove Theorem \ref{thm1.2}, we need a result from Edrei \cite{Edrei}.

\begin{lemma}\label{lem1.5}
Let $f(z)$ be a meromorphic function with
$\delta=\delta(\infty,f)>0$. Then given $\varepsilon>0$, we have
$$
\operatorname{meas} E(r,f)>\frac{1}{T^\varepsilon(r,f)[\log r]^{1+\varepsilon}},
r\notin F,
$$
where
$$
E(r,f)=\{\theta\in[-\pi,\pi):
\log^+|f(re^{i\theta})|>\frac{\delta}{4}T(r,f)\}
$$
and $F$ is a set of positive real numbers with finite logarithmic
 measure depending on $\varepsilon$.
\end{lemma}

\section{Proof of the Theorems}

\begin{proof}[Proof of Theorem \ref{thm1.1}]

We suppose that there exists a nontrival meromorphic solution $f$
such that $\sigma_{\alpha_j,\beta_j}(f)<+\infty$,  $j=1,2,\dots, q$.
In view of Lemma \ref{lem1.4}, there exists a constant $M>0$ not
depending on $z$ such that
\[
\left|\frac{f^{(j)}(z)}{f(z)}\right|<|z|^M,\quad  j=1,2,\dots,n\,.
\]
for all $z\in\Omega(\alpha_j+\varepsilon,\beta_j-\varepsilon)$,
$j=1,2,\dots,q$, except for a $\mathbb{R}$-set $E$. For $E$, we can
define a set $F=\{r>0|
 \exists z\in E, s.t. |z|=r\}$ thus
$$\operatorname{meas} F<\infty.$$

(I) $\lambda(A_0)>\mu(A_0)$. Then $\lambda(A_0)>\sigma\geq\mu(A_0)$.
By the inequality \eqref{thmeq1.1}, we can take a real number
$\varepsilon>0$ such that
\begin{equation}\label{2.1}
\sum_{j=1}^q(\alpha_{j+1}-\beta_j+2\varepsilon)+2\varepsilon
<\frac{4}{\sigma+2\varepsilon}\arcsin\sqrt{\delta/2},
\end{equation}
where $\alpha_{q+1}=2\pi+\alpha_1$, and
$$
\lambda(A_0)>\sigma+2\varepsilon>\mu(A_0).
$$
Applying Lemma \ref{lem1.1} to $A(z)$ gives the existence of the
P\'{o}lya peak $\{r_n\}$ of order $\sigma+2\varepsilon$ of $A(z)$
such that $r_n\notin F$, and then from Lemma \ref{lem1.2} for
sufficiently large $n$ we have
\begin{equation}\label{2.2}
\operatorname{meas} D(r_n,\infty)>\frac{4}{\sigma+2\varepsilon}\arcsin
\sqrt{\delta/2}-\varepsilon.
\end{equation}
We can assume for all the $n$, above holds. Set
$$
K:=\operatorname{meas} (D(r_n,\infty)\cap\cup_{j=1}^q(\alpha_j+\varepsilon,
\beta_j-\varepsilon)).
$$
Then from \eqref{2.1} and \eqref{2.2} it follows that
\begin{align*}
K&\geq \operatorname{meas}(D(r_n,\infty))-\operatorname{meas}([0,2\pi)\backslash
 \cup_{j=1}^q(\alpha_j+\varepsilon,\beta_j-\varepsilon))\\
&=\operatorname{meas}(D(r_n,\infty))-\operatorname{meas}(\cup_{j=1}^q(\beta_j
 -\varepsilon,\alpha_{j+1}+\varepsilon))\\
&=\operatorname{meas}(D(r_n,\infty))-\sum_{j=1}^q(\alpha_{j+1}-\beta_j
 +2\varepsilon)>\varepsilon>0.
\end{align*}
It is easy to see that there exists a $j_0$ such that for infinitely
many $n$, we have
\begin{equation}\label{2.3}
\operatorname{meas}(D(r_n,\infty)\cap(\alpha_{j_0}+\varepsilon,\beta_{j_0}
-\varepsilon))>\frac{K}{q}.
\end{equation}
We can assume for all the $n$, \eqref{2.3} holds. We define a real
function by
\[
\Lambda(r)^2=\max\left\{\frac{T(r_n,A_j)}{T(r_n,A_0)}, \frac{\log
r_n}{T(r_n,A_0)}; j=1,2,\dots,n\right\},
\]
for $ r_n\leq r<r_{n+1}$. Obviously $\lim_{r\to\infty}\Lambda(r)=0$
and
$$
\operatorname{meas} D_\Lambda'(r_n)=\operatorname{meas}\{\theta:
r_ne^{i\theta}\in E\}=0.
$$
Set
$$
D_n=(\alpha_{j_0}+\varepsilon,\beta_{j_0}-\varepsilon)\backslash
D_\Lambda'(r_n),\quad
E_n=D(r_n,\infty)\cap(\alpha_{j_0}+\varepsilon,
\beta_{j_0}-\varepsilon).
$$
Thus from the definition of $D(r,\infty)$ it follows that
\begin{equation}\label{2.4}
\begin{aligned}
\int_{\alpha_{j_0}+\varepsilon}^{\beta_{j_0}-\varepsilon}
\log^+|A_0(r_ne^{i\theta})|d\theta&\geq\int_{E_n}
\log^+|A_0(r_ne^{i\theta})|d\theta\\
&\geq \operatorname{meas}(E_n)\Lambda(r_n)T(r_n, A_0)\\
&>\frac{K}{q}\Lambda(r_n)T(r_n, A_0).
\end{aligned}
\end{equation}
Thus, we have
\begin{equation}\label{2.5}
\begin{split}
&\int_{\alpha_{j_0}+\varepsilon}^{\beta_{j_0}-\varepsilon}
\log^+|A_0(r_ne^{i\theta})|d\theta\\
&\leq\int_{\alpha_{j_0}+\varepsilon}^{\beta_{j_0}-\varepsilon}
\sum_{j=1}^n\left(\log^+\left|\frac{f^{(j)}(r_ne^{i\theta})}
{f(r_ne^{i\theta})}\right|+\log^+|A_j(r_ne^{i\theta})|\right) d\theta\\
&=\Big(\int_{D_\Lambda'(r_n)}+\int_{D_n}\Big)
\sum_{j=1}^n\left(\log^+\left|\frac{f^{(j)}(r_ne^{i\theta})}{f(r_n
e^{i\theta})}\right|+\log^+|A_j(r_ne^{i\theta})|\right)d\theta\\
&\leq\int_{\alpha_{j_0}+\varepsilon}^{\beta_{j_0}-\varepsilon}
\sum_{j=1}^n\log^+|A_j(r_ne^{i\theta})|d\theta+O(\log r_n)\\
&\leq \sum_{j=1}^nT(r_n, A_j)+O(\log r_n)\\
&\leq \Lambda^2(r_n)T(r_n, A_0).
\end{split}
\end{equation}
Therefore,
$$
\frac{K}{q}\Lambda(r_n)<\Lambda^2(r_n).
$$
This contradicts that $\Lambda(r)\to 0$.

(II) $\lambda(A_0)=\mu(A_0)$. Then $\lambda(A_0)=\sigma=\mu(A_0)$. By
the same argument as in (I) with all the $\sigma+2\varepsilon$
replaced by $\sigma$, we can derive a contradiction. The proof is
complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
Applying Lemma \ref{lem1.1} to $A_0(z)$ confirms the existence
of a sequence $\{r_n\}$ of positive numbers such that
$r_n\notin E$ and
\begin{equation}\label{3.6}
\operatorname{meas} E(r_n, A_0)>\frac{1}{T^\varepsilon(r_n,A_0)[\log
r_n]^{1+\varepsilon}},
\end{equation}
where $E(r_n, A_0)$ is defined as in Lemma \ref{lem1.5}. Set
$$
\varepsilon_n=\frac{1}{2q+1}\frac{1}{T^\varepsilon(r_n,A_0)[\log
r_n]^{1+\varepsilon}}.
$$
Then for \eqref{3.6} it follows that
\begin{align*}
&\operatorname{meas} (E(r_n, A_0)\cap\cup_{j=1}^q(\alpha_j+\varepsilon_n,
\alpha_{j+1}-\varepsilon_n))\\
&\geq \operatorname{meas} E(r_n, A_0)
-\operatorname{meas}(\cup_{j=1}^q(\alpha_j+\varepsilon_n,
\alpha_{j+1}-\varepsilon_n))\\
&\geq(2q+1)\varepsilon_n-2q\varepsilon_n=\varepsilon_n>0.
\end{align*}
so that there exists a $j$ such that for infinitely many $n$, we
have
\begin{equation}\label{3.7}
\operatorname{meas} E_n>\frac{\varepsilon_n}{q},
\end{equation}
where $E_n=E(r_n, A_0)\cap(\alpha_j+\varepsilon_n,
\alpha_{j+1}-\varepsilon_n)$. We can assume that \eqref{3.7} holds
for all the $n$. Thus
\begin{equation}\label{3.8}
\begin{split}
\int_{\alpha_j+\varepsilon_n}^{\alpha_{j+1}-\varepsilon_n}\log^+
|A_0(r_ne^{i\theta})|d\theta&\geq\int_{E_n}\log^+|A_0(r_ne^{i\theta})|d\theta\\
&\geq \operatorname{meas}(E_n)\frac{\delta}{4}T(r_n, A_0)\\
&\geq \frac{\delta\varepsilon_n}{4q}T(r_n, A_0).
\end{split}
\end{equation}
On the other hand,
\begin{equation}\label{3.9}
\int_{\alpha_j+\varepsilon_n}^{\alpha_{j+1}-\varepsilon_n}\log^+
|A_0(r_ne^{i\theta})|d\theta<\sum_{j=1}^n
T(r_n, A_j)+O(\log r_n)
\end{equation}
Combining \eqref{3.8} and \eqref{3.9} gives
$$
\varepsilon_n T(r_n, A_0)\leq \frac{4q}{\delta}\sum_{j=1}^n
T(r_n, A_j)+O(\log r_n),
$$
so that
$$
T^{1-\varepsilon}(r_n, A_0)\leq\frac{4q(2q+1)}{\delta}
[\log r_n]^{1+\varepsilon}\sum_{j=1}^n
T(r_n, A_j)+O(\log^{2+\varepsilon} r_n),
$$
we have $\mu(A_0)\leq\max_{1\leq j\leq q}(\mu(A_j))/(1-\varepsilon)$.
By the same method as in Theorem \ref{thm1.1}, we obtain a contradiction,
which completes the proof.
\end{proof}


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\end{document}