\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 226, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2012/226\hfil Growth of solutions]
{Growth of solutions to linear differential equations with entire coefficients}
\author[ H. Hu, X.-M. Zheng \hfil EJDE-2012/226\hfilneg]
{Hui Hu, Xiu-Min Zheng} % in alphabetical order
\address{Hui Hu \newline
Institute of Mathematics and Information Science, Jiangxi Normal University,
330022, China}
\email{h\_h87\_6@hotmail.com}
\address{Xiu-Min Zheng (Corresponding author)\newline
Institute of Mathematics and Information Science, Jiangxi Normal University,
330022, China}
\email{zhengxiumin2008@sina.com}
\thanks{Submitted May 8, 2012. Published December 17, 2012.}
\subjclass[2000]{30D35, 34M05}
\keywords{Iterated $p$-order; iterated $p$-lower order; iterated $p$-lower type;
\hfill\break\indent iterated exponent of convergence of distinct zeros}
\begin{abstract}
In this article, we study the growth of solutions of
linear differential equations with some dominant entire coefficients.
Especially, we obtain some results on the iterated $p$-lower order
of these solutions, which extend previous results.
Moreover, we investigate the iterated exponent of convergence of
distinct zeros of $f^{(j)}(z)-\varphi(z)$.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks
\def\R{\mathbb{R}}
\section{Introduction}
We shall assume that readers are familiar with the fundamental
results and the standard notations of Nevanlinna's theory; see e.g.
\cite{h1,l1,v1}. Let us define inductively for $r\in[0,+\infty)$,
$\exp_1 r=e^r$ and $\exp_{p+1}r=\exp(\exp_p r)$, $p\in \mathbb{N}$. For all
sufficiently large $r$, we define $\log_1 r=\log r$ and
$\log_{p+1}r=\log(\log_p r)$, $p\in \mathbb{N}$. We also denote
$\exp_0 r=r=\log_0 r$ and $\exp_{-1}r=\log_1 r$. We recall the
following definitions of finite iterated order; see e.g.
\cite{b2,c1,l1,k1,t1,t3}.
\begin{definition} \label{def1.1}\rm
The iterated $p$-order $\sigma_p(f)$ of a meromorphic function $f(z)$
is defined as
$$
\sigma_p(f)=\limsup_{r\to\infty} \frac{\log_pT(r,f)}{\log r}\quad
(p\in \mathbb{N}).
$$
\end{definition}
\begin{remark} \label{rmk1.2} \rm
If $f(z)$ is an entire function, then
$$
\sigma_p(f)=\limsup_{r\to\infty} \frac{\log_pT(r,f)}{\log r}
=\limsup_{r\to\infty} \frac{\log_{p+1}M(r,f)}{\log r}
=\limsup_{r\to\infty} \frac{\log_p{\nu_f(r)}}{\log r},
$$
where $p\in \mathbb{N}$, $\nu_f(r)$ is the central index of $f(z)$.
\end{remark}
\begin{definition} \label{def1.3} \rm
The iterated $p$-lower order $\mu_p(f)$ of a meromorphic function
$f(z)$ is defined by
$$
\mu_p(f)=\liminf_{r\to\infty} \frac{\log_pT(r,f)}{\log r}\quad (p\in \mathbb{N}).
$$
\end{definition}
\begin{remark} \label{rmk1.4} \rm
The iterated $p$-lower order $\mu_p(f)$ of an entire function $f(z)$
is defined by
$$
\mu_p(f)=\liminf_{r\to\infty} \frac{\log_pT(r,f)}{\log r}
=\liminf_{r\to\infty} \frac{\log_{p+1}M(r,f)}{\log r}
=\liminf_{r\to\infty} \frac{\log_p{\nu_f(r)}}{\log r}\quad (p\in \mathbb{N}).
$$
\end{remark}
\begin{definition} \label{def1.5} \rm
The finiteness degree of the order of a meromorphic function $f(z)$ is defined by
\begin{equation*}
{i}(f)=\begin{cases}
0, & \text{if $f$ is rational}; \\
\min\{j\in \mathbb{N}:\sigma _{j}(f)<\infty\}, &
\text{if $f$ is transcendental with}\\
&\sigma_{j}(f)<\infty \text{ for some }j\in\mathbb{N}; \\
\infty, & \text{if } \sigma _{j}(f)=\infty \text{ for all }j\in \mathbb{N}.
\end{cases}
\end{equation*}
\end{definition}
\begin{definition} \label{def1.6}\rm
The iterated convergence exponent of
the sequence of $a$-points of a meromorphic function $f(z)$ is
defined by
$$
\lambda_p(f-a)=\lambda_p(f,a)=\limsup_{r\to\infty}
\frac{\log_p N(r,\frac{1}{f-a})}{\log r}\quad (p\in \mathbb{N}),
$$
and the iterated convergence exponent of the sequence of distinct
$a$-points of a meromorphic function $f(z)$ is defined by
$$
\overline{\lambda}_p(f-a)=\overline{\lambda}_p(f,a)=\limsup_{r\to\infty}
\frac{\log_p \overline{N}(r,\frac{1}{f-a})}{\log r}\quad (p\in \mathbb{N}).
$$
If $a=0$, the iterated convergence exponent of the zeros or the iterated
convergence exponent of the distinct zeros is defined respectively by
$$
\lambda_p(f)=\lambda_p(f,0)=\limsup_{r\to\infty}
\frac{\log_p N(r,\frac{1}{f})}{\log r}~~(p\in \mathbb{N}),
$$
or
$$
\overline{\lambda}_p(f)=\overline{\lambda}_p(f,0)
=\limsup_{r\to\infty} \frac{\log_p \overline{N}(r,\frac{1}{f})}{\log r}\quad
(p\in \mathbb{N}).
$$
If $a=\infty$, the iterated convergence exponent of the poles or the
iterated convergence exponent of the distinct poles is defined respectively by
$$
\lambda_p(\frac{1}{f})=\limsup_{r\to\infty} \frac{\log_p N(r,f)}{\log r}\quad
(p\in \mathbb{N}),
$$
or
$$\overline{\lambda}_p(\frac{1}{f})
=\limsup_{r\to\infty} \frac{\log_p \overline{N}(r,f)}{\log r}\quad
(p\in \mathbb{N}).
$$
Furthermore, we can get the definitions of $\lambda_p(f-\varphi)$
and $\overline{\lambda}_p(f-\varphi)$, when $a$ is replaced by a
meromorphic function $\varphi$.
\end{definition}
\begin{definition} \label{def1.7}\rm
Let $f(z)$ be an entire function. Then the iterated $p$-type
of an entire function $f(z)$, with iterated
$p$-order $0<\sigma_p(f)<\infty$ is defined by
$$
\tau_p(f)=\limsup_{r\to\infty} \frac{\log_{p-1}T(r,f)}{r^{\sigma_p(f)}}
=\limsup_{r\to\infty} \frac{\log_{p}M(r,f)}{r^{\sigma_p(f)}}\quad
(p\in \mathbb{N}\backslash\{1\}).
$$
\end{definition}
We definite the iterated $p$-lower type of $f(z)$ as follows.
\begin{definition} \label{def1.8} \rm
Let $f(z)$ be an entire function. Then the iterated $p$-lower type
of an entire function $f(z)$, with iterated $p$-lower order
$0<\mu_p(f)<\infty$, is defined by
$$
\underline\tau_p(f)=\liminf_{r\to\infty}
\frac{\log_{p-1}T(r,f)}{r^{\mu_p(f)}}=\liminf_{r\to\infty}
\frac{\log_{p}M(r,f)}{r^{\mu_p(f)}}\quad(p\in \mathbb{N}\backslash\{1\}).
$$
\end{definition}
\begin{remark} \label{rmk1.9} \rm
If $p=1$, then the equalities
\begin{gather*}
\limsup_{r\to\infty} \frac{\log_{p-1}T(r,f)}{r^{\sigma_p(f)}}
=\limsup_{r\to\infty} \frac{\log_{p}M(r,f)}{r^{\sigma_p(f)}},\\
\liminf_{r\to\infty} \frac{\log_{p-1}T(r,f)}{r^{\mu_p(f)}}
=\liminf_{r\to\infty} \frac{\log_{p}M(r,f)}{r^{\mu_p(f)}}
\end{gather*}
in Definitions \ref{def1.7} and \ref{def1.8} respectively fail to hold.
For example, for the function $f(z)=e^z$, we have
$\lim_{r\to\infty}\frac{T(r,f)}{r}=\frac{1}{\pi}\neq1
=\lim_{r\to\infty}\frac{\log M(r,f)}{r}$.
Therefore, we assume $p\in\mathbb{N}\backslash\{1\}$ in the following.
\end{remark}
We denote the linear measure and the logarithmic measure of a
set $E\subset[0,+\infty)$ by $mE=\int_E dt$ and $m_l E=\int_E dt/t$
respectively (see e.g. \cite{h2}).
\section{Main Results}
In 1998, Kinnunen investigated complex oscillation properties of the
solutions of the higher order linear differential equations
\begin{equation}
f^{(n)}+A_{n-1}(z)f^{(n-1)}+\dots+A_1(z)f'+A_0(z)f=0 \label{e2.1}
\end{equation}
and
\begin{equation}
f^{(n)}+A_{n-1}(z)f^{(n-1)}+\dots+A_1(z)f'+A_0(z)f=F(z), \label{e2.2}
\end{equation}
with entire coefficients of finite iterated order and obtained the
following result in \cite{k1}.
\begin{theorem} \label{thmA}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be entire functions and
let $i(A_0)=p$, $0
0)$
and $\max\{\tau_p(A_j):\sigma_p(A_j)=\sigma_p(A_0)\}<\tau_p(A_0)
=\tau(0<\tau<\infty)$. Then every solution $f(z)\not\equiv 0$
of \eqref{e2.1} satisfies $i(f)=p+1$ and $\sigma_{p+1}(f)=\sigma_p(A_0)$.
\end{theorem}
Theorems \ref{thmA} and \ref{thmB} investigated the iterated order of solutions of
\eqref{e2.1}, when there is some dominating coefficient with iterated order.
Another question is: If there is some dominating coefficient with iterated
lower order, what can we say about the growth of solutions of \eqref{e2.1}.
For the special case $p=2$, Zhang-Tu in \cite{z1} discussed it and obtained
the following result.
\begin{theorem} \label{thmC}
Let $A_0(z), \dots,A_{n-1}(z)$ be entire functions satisfying
$\max\{\sigma(A_j)$,
$j=1,\dots,{n-1}\}<\mu(A_0)\leq\sigma(A_0)<\infty$, then every solution
$f(z)\not\equiv 0$ of \eqref{e2.1} satisfies
$$
\mu(A_0)=\mu_2(f)\leq\sigma_2(f)=\sigma(A_0).
$$
\end{theorem}
In this paper, we investigate the above problems.
Moreover, we investigate the iterated exponent of convergence of
distinct zeros of $f^{(j)}(z)-\varphi(z)$.
Firstly, we extend Theorem \ref{thmC} into a general case and obtain
the same result.
\begin{theorem} \label{thm2.1}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$
be entire functions of finite iterated order satisfying
$\max\{\sigma_p(A_j),j=1,\dots,n-1\}<\mu_p(A_0)\leq\sigma_p(A_0)<\infty$,
then every solution $f(z)\not\equiv 0$ of \eqref{e2.1}
satisfies
\begin{equation}
\mu_p(A_0)=\mu_{p+1}(f)\leq\sigma_{p+1}(f)=\sigma_p(A_0).\label{e2.3}
\end{equation}
\end{theorem}
Secondly, when there are some coefficients with iterated order equal
to $\mu_p(A_0)$, we obtain the following two results.
\begin{theorem} \label{thm2.2}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$
be entire functions, and let $i(A_0)=p$. Assume that
$\max\{\sigma_p(A_j):j\neq0\}\leq\mu_p(A_0)\leq\sigma_p(A_0)$ and
$\tau_1=\max\{\tau_p(A_j):\sigma_p(A_j)=\mu_p(A_0)\}<\underline\tau_p(A_0)
=\tau(0<\tau<\infty)$.
Then every solution $f(z)\not\equiv 0$ of \eqref{e2.1} satisfies
\begin{equation}
\mu_{p+1}(f)=\mu_p(A_0)\leq\sigma_p(A_0)=\sigma_{p+1}(f)
=\lambda_{p+1}(f-\varphi)=\overline{\lambda}_{p+1}(f-\varphi),\label{e2.4}
\end{equation}
where $\varphi(z)\not\equiv 0$ is an entire function satisfying
$\sigma_{p+1}(\varphi)<\mu_p(A_0)$.
\end{theorem}
\begin{theorem} \label{thm2.3}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$
be entire functions of finite iterated order satisfying
$\max\{\sigma_p(A_j),j\neq0\}\leq\mu_p(A_0)=\mu$ and
$$
\limsup_{r\to\infty} {\sum_{j=1}^{n-1}m(r,A_j)}/{m(r,A_0)}<1.
$$
Then every non-trivial solution $f(z)$ of \eqref{e2.1} satisfies
\begin{equation}
\mu_{p+1}(f)=\mu_p(A_0)\leq\sigma_p(A_0)=\sigma_{p+1}(f)
=\lambda_{p+1}(f-\varphi)=\overline{\lambda}_{p+1}(f-\varphi),\label{e2.5}
\end{equation}
where $\varphi(z)\not\equiv 0$ is an entire function satisfying
$\sigma_{p+1}(\varphi)<\mu_p(A_0)$.
\end{theorem}
\begin{remark} \label{rmk0}\rm
All solutions of \eqref{e2.1} in Theorems \ref{thm2.1}, \ref{thm2.2},
\ref{thm2.3} are of regular growth
$\mu_{p+1}(f)=\sigma_{p+1}(f)$, when the coefficient $A_0(z)$
is of regular growth $\mu_p(A_0)=\sigma_p(A_0)$.
\end{remark}
\begin{theorem} \label{thm2.4}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be meromorphic functions of finite
iterated order satisfying
$$
\max\{\lambda_p(\frac{1}{A_0}),\sigma_p(A_j),j=1,\dots,n-1\}
<\mu_p(A_0)\leq\sigma_p(A_0)<\infty,
$$
if $f(z)\not\equiv 0$ is a meromorphic solution of \eqref{e2.1}
satisfying $\frac{N(r,f)}{{\overline N}(r,f)}<\exp_{p-1}\{r^b\}$,
$(b<\mu_p(A_0))$, then we have
\begin{equation}
\sigma_p(A_0)=\sigma_{p+1}(f)={\lambda}_{p+1}(f^{(j)}-\varphi)
=\overline{\lambda}_{p+1}(f^{(j)}-\varphi),\ (j=0,1,\dots),\label{e2.6}
\end{equation}
where $\varphi(z)\not\equiv 0$ is a meromorphic function satisfying
$\sigma_{p+1}(\varphi)<\sigma_p(A_0)$.
\end{theorem}
\begin{corollary} \label{coro2.1}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ satisfying the hypotheses
of Theorem \ref{thm2.1}, then
every solution $f(z)\not\equiv 0$ of \eqref{e2.1}
satisfies
$$
\mu_{p+1}(f)=\mu_p(A_0)\leq\sigma_p(A_0)=\sigma_{p+1}(f)
=\lambda_{p+1}(f^{(j)}-\varphi)=\overline{\lambda}_{p+1}(f^{(j)}-\varphi),
$$
where $\varphi(z)\not\equiv 0$ is an entire function satisfying
$\sigma_{p+1}(\varphi)<\sigma_p(A_0)$.
\end{corollary}
\section{Lemmas for the proofs of main results}
\begin{lemma} \label{lem3.1}
Let $f(z)$ be a transcendental entire function. There exists a
set $E_1$ of $r$ of finite logarithmic measure, such that for all $z$ satisfying
$|z|=r\not\in E_1$ and $|f(z)|=M(r,f)$, we have
$$
\frac{f^{(k)}(z)}{f(z)}=\big(\frac{\nu_f(r)}{z}\big)^k(1+o(1)),\quad
(k\in \mathbb{N}, r\not\in E_1),
$$
where $\nu_f(r)$ is the central index of $f(z)$.
\end{lemma}
\begin{lemma}[\cite{h2,l1}] \label{lem3.2}
Let $g:[0,+\infty)\to\mathbb{R}$ and
$h:[0,+\infty)\to\mathbb{R}$ be monotone increasing
functions. If (i) $g(r)\leq h(r)$ outside of an exceptional set of
finite linear measure, or (ii) $g(r)\leq h(r)$, $r\not\in
E_2\cup(0,1]$, where $E_2\subset [1,\infty)$ is a set of finite
logarithmic measure, then for any constant $\alpha>1$, there exists
$r_0=r_0(\alpha)>0$ such that $g(r)\leq h(\alpha r)$ for all
$r>r_0$.
\end{lemma}
\begin{lemma}[\cite{t2}] \label{lem3.3}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be meromorphic functions
of finite iterated order satisfying
$\max\{\sigma_p(A_j),j=1,\dots,n-1\}<\mu_p(A_0)\leq\sigma_p(A_0)<\infty$,
if $f(z)\not\equiv 0$ is a meromorphic solution of \eqref{e2.1}
satisfying $\frac{N(r,f)}{{\overline N}(r,f)}<\exp_{p-1}\{r^b\}$,
$(b<\mu_p(A_0))$,
then $\sigma_{p+1}(f)=\sigma_p(A_0)$.
\end{lemma}
\begin{lemma} \label{lem3.4}
Let $f(z)$ be an entire function with $\mu_p(f)<\infty$, then for
any given $\varepsilon>0$, there exists a set $E_4\subset(1,+\infty)$
having infinite logarithmic measure such that for all $r\in E_4$,
we have
$$
\mu_p(f)=\lim_{r\to\infty,\, r\in E_4}\frac{\log_pT(r,f)}{\log r}
=\lim_{r\to\infty,\, r\in E_4} \frac{\log_{p+1}M(r,f)}{\log r}
=\lim_{r\to\infty,\, r\in E_4}\frac{\log_p{\nu_f(r)}}{\log r},
$$
and
$$
M(r,f)<\exp_p\{r^{\mu_p(f)+\varepsilon}\}.
$$
\end{lemma}
\begin{proof}
We use a similar proof as \cite[Lemma 3.8]{t2}.
By the definition of iterated $p$-lower order, there exists a
sequence $\{r_n\}_{n=1}^{\infty}$ tending to $\infty$ satisfying
$(1+\frac{1}{n})r_n0$, there exists an $n_1$ such that
for $n\geq n_1$ and any $r\in[r_n,(1+\frac{1}{n})r_n]$,
we have
$$
{\frac{\log_{p+1}M(r_n,f)}{\log(1+\frac{1}{n})r_n}}
\leq\frac{\log_{p+1}M(r,f)}{\log r}
\leq\frac{\log_{p+1}M((1+\frac{1}{n})r_n,f)}{\log r_n}.
$$
Let $E_4={\cup_{n=n_1}^{\infty}}[r_n,(1+\frac{1}{n})r_n]$,
then for any $r\in E_4$, we have
$$
\lim_{r\to\infty,\, r\in E_4} \frac{\log_{p+1}M(r,f)}{\log r}
=\lim_{r_n \to\infty}{\frac{\log_{p+1}M(r_n,f)}{\log r_n}}=\mu_p(f),
$$
and
$$
m_l E={\sum_{n=n_1}^{\infty}}{\int_{r_n}^{(1+\frac{1}{n})r_n}}{\frac{dt}{t}}
={\sum_{n=n_1}^{\infty}}\log(1+\frac{1}{n})=\infty.
$$
It is easy to see
$$
\lim_{r\to\infty,\, r\in E_4} \frac{\log_pT(r,f)}{\log r}
=\lim_{r\to\infty,\, r\in E_4}\frac{\log_{p+1}M(r,f)}{\log r}
=\lim_{r\to\infty,\, r\in E_4}\frac{\log_p{\nu_f(r)}}{\log r}.
$$
The proof is complete.
\end{proof}
\begin{lemma}[\cite{t2}] \label{lem3.5}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z)$ be entire functions
of finite iterated order satisfying $i(A_0)=p,~\sigma_p(A_0)=\sigma$
and
$$
\limsup_{r\to\infty} {\sum_{j=1}^{n-1}m(r,A_j)}/{m(r,A_0)}<1,
$$
then every non-trivial solution $f(z)$ of \eqref{e2.1} satisfies
$\sigma_{p+1}(f)=\sigma_p(A_0)=\sigma$.
\end{lemma}
\begin{lemma}[\cite{g1}] \label{lem3.6}
Let $f(z)$ be a transcendental meromorphic function.
Let $\alpha>1$ be a constant, and $k$ and $j$ be integers
satisfying $k>j\geq0$. Then the following two statements hold:
(a) There exists a set $E_6\subset[1,\infty)$
which has finite logarithmic measure, and a constant $C>0$,
such that for all $z$ satisfying $|z|=r\not\in E_6\cup[0,1]$, we have
\begin{equation}
\big|\frac{f^{(k)}(z)}{f^{(j)}(z)}\big|
\leq C\big[\frac{T(\alpha r,f)}{r}(\log r)^{\alpha}\log T(\alpha r,f)\big]^{k-j}.
\label{e3.1}
\end{equation}
(b) There exists a set $E_6'\subset[0,2\pi)$ which has
linear measure zero, such that if $\theta\in[0,2\pi)-E_6'$,
then there is a constant $R=R(\theta)>0$ such that \eqref{e3.1} holds for
all $z$ satisfying $\arg z=\theta$ and $|z|\geq R$.
\end{lemma}
\begin{lemma}[\cite{t1}] \label{lem3.7}
Let $A_0(z), A_1(z), \dots,A_{n-1}(z), F(z)\not\equiv 0$ be meromorphic
functions and
let $f(z)$ be a meromorphic solution of \eqref{e2.2} satisfying one of
the following two conditions
\begin{itemize}
\item[(i)] $\max\{i(F)=q, i(A_j), j=0,1,\dots,n-1\}0$ and sufficiently large $r$. By the
hypotheses of Lemma \ref{lem3.10}, there exists a set $E_9$ having infinite
logarithmic measure such that for all $|z|=r\in E_9$, we have
\begin{equation}
m(r,B_0)>\exp_{p-1}\{r^{\beta_0-\varepsilon}\}.\label{e3.15}
\end{equation}
By \eqref{e3.4},\eqref{e3.14} and \eqref{e3.15}, we have
\begin{equation}
\exp_{p-1}\{r^{\beta_0-\varepsilon}\}\leq
O\left(\log(rT(r,f))\right)+(n-1)\exp_{p-1}
\{r^{\beta_1+\varepsilon}\},\label{e3.16}
\end{equation}
for any given $\varepsilon(0<2\varepsilon<\beta_0-\beta_1)$, where
$r\in E_9\backslash E, r\to\infty$, and $E$ is a set of $r$
of finite linear measure. By \eqref{e3.16}, we have
$\sigma_{p+1}(f)\geq\beta_0$.
\end{proof}
\section{Proofs of main theorems}
\begin{proof}[Proof of Theorem \ref{thm2.1}]
By Theorem \ref{thmA}, we know that every solution $f(z)\not\equiv 0$ of
\eqref{e2.1} satisfies $\sigma_{p+1}(f)=\sigma_p(A_0)$.
Then we only need to prove that every solution $f(z)$
of \eqref{e2.1} satisfies $\mu_{p+1}(f)=\mu_p(A_0)$.
We rewrite \eqref{e2.1} as
\begin{equation}
|A_0(z)|\leq\big|\frac{f^{(n)}(z)}{f(z)}\big|+|A_{n-1}(z)|
\big|\frac{f^{(n-1)}(z)}{f(z)}\big|+\dots+|A_1(z)|\big|\frac{f'(z)}{f(z)}\big|.
\label{e4.1}
\end{equation}
Set $\max\{\sigma_p(A_j):j\neq0\}=c$, then for any given
$\varepsilon(0<2\varepsilon<\mu_p(A_0)-c)$ and for sufficiently large $r$,
we have
\begin{equation}
M(r,A_0)\geq\exp_{p}\{r^{\mu_p(A_0)-\varepsilon}\},\label{e4.2}
\end{equation}
and
\begin{equation}
M(r,A_j)\leq\exp_{p}\{r^{c+\varepsilon}\},\quad (j=1,2,\dots,n-1).\label{e4.3}
\end{equation}
By Lemma \ref{lem3.6}, there exists a set $E_6$ having finite logarithmic measure
and a constant $C>0$ such that for all $z$ satisfying
$|z|=r\not\in E_6\cup[0,1]$, we have
\begin{equation}
\big|\frac{f^{(k)}(z)}{f(z)}\big|\leq C(T(2r,f))^{k+1},\ (k\geq1).\label{e4.4}
\end{equation}
Substituting \eqref{e4.2}-\eqref{e4.4} into \eqref{e4.1}, for the above
$\varepsilon>0$, we have
\begin{equation}
\exp_{p}\{r^{\mu_p(A_0)-\varepsilon}\}\leq
Cn\exp_{p}\{r^{c+\varepsilon}\}\left(T(2r,f)\right)^{n+1},\label{e4.5}
\end{equation}
for all $z$ satisfying $|z|=r\not\in E_6\cup[0,1]$, $r\to\infty$ and
$|A_0(z)|=M(r,A_0)$. By Lemma \ref{lem3.2} and
\eqref{e4.5}, we have $\mu_{p+1}(f)\geq\mu_p(A_0)-\varepsilon$. Since
$\varepsilon>0$ is arbitrary, we obtain
\begin{equation}
\mu_{p+1}(f)\geq\mu_p(A_0).\label{e4.6}
\end{equation}
By \eqref{e2.1}, we have
\begin{equation}
\big|{\frac{f^{(n)}(z)}{f(z)}}\big|
\leq|A_{n-1}(z)|\big|{\frac{f^{(n-1)}(z)}{f(z)}}\big|+\dots
+|A_1(z)|\big|{\frac{f'(z)}{f(z)}}\big|+|A_0(z)|.\label{e4.7}
\end{equation}
By Lemma \ref{lem3.1}, there exists a set
$E_1\subset(1,+\infty)$ having finite logarithmic measure such that
for all $z$ satisfying $|z|=r\not\in E_1$, and $|f(z)|=M(r,f)$, we
have
\begin{equation}
\big|\frac{f^{(j)}(z)}{f(z)}\big|
=\big|\frac{\nu_f(r)}{z}\big|^j|1+o(1)|,\quad (j=1,\dots,n).\label{e4.8}
\end{equation}
By Lemma \ref{lem3.4}, there exists a set
$E_4\subset(1,+\infty)$ having infinite logarithmic measure such
that for all $|z|=r\in E_4\backslash E_1$, we have
\begin{equation}
|A_0(z)|\leq M(r,A_0)\leq\exp_p\{r^{\mu_p(A_0)+\varepsilon}\}. \label{e4.9}
\end{equation}
Hence, by \eqref{e4.3},\eqref{e4.7}-\eqref{e4.9}, we have
$$
|\nu_f(r)|^n|1+o(1)|\leq n\exp_p\{r^{\mu_p(A_0)+\varepsilon}\}
r^n|\nu_f(r)|^{n-1}|1+o(1)|,
$$
then we obtain
\begin{equation}
|\nu_f(r)||1+o(1)|\leq
nr^n\exp_p\{r^{\mu_p(A_0)+\varepsilon}\},\quad
(r\in E_4\backslash E_1).\label{e4.10}
\end{equation}
By the definition of iterated $p$-lower order
and \eqref{e4.10}, we have $\mu_{p+1}(f)\leq\mu_p(A_0)+\varepsilon$.
Since $\varepsilon>0$ is arbitrary, we have
\begin{equation}
\mu_{p+1}(f)\leq\mu_p(A_0).\label{e4.11}
\end{equation}
By \eqref{e4.6} and \eqref{e4.11}, we obtain $\mu_{p+1}(f)=\mu_p(A_0)$.
The proof is complete.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm2.2}]
By Theorem \ref{thmB}, we have $\sigma_{p+1}(f)=\sigma_p(A_0)$.
Now we need to prove
(1) $\mu_{p+1}(f)=\mu_p(A_0)$ and
(2) $\sigma_{p+1}(f)=\overline{\lambda}_{p+1}{(f-\varphi)}$.
(1) On the one hand, we set
$b=\max\{\sigma_p(A_j),\sigma_p(A_j)<\mu_p(A_0)\}$. If
$\sigma_p(A_j)<\mu_p(A_0)$, then for any given
$\varepsilon(0<2\varepsilon<\min\{\mu_p(A_0)-b,\tau-\tau_1\})$ and
for sufficiently large $r$, we have
\begin{equation}
M(r,A_j)\leq\exp_{p}\{r^{b+\varepsilon}\}\leq\exp_{p}
\{r^{\mu_p(A_0)-\varepsilon}\}.\label{e4.12}
\end{equation}
If $\sigma_p(A_j)=\mu_p(A_0),~\tau_p(A_j)\leq\tau_1<\tau
=\underline{\tau}_p(A_0)$,
then for sufficiently large $r$, we have
\begin{gather}
M(r,A_j)\leq\exp_{p}\{(\tau_1+\varepsilon)r^{\mu_p(A_0)}\},\label{e4.13}
\\
M(r,A_0)\geq\exp_{p}\{(\tau-\varepsilon)r^{\mu_p(A_0)}\}.\label{e4.14}
\end{gather}
By \eqref{e4.12}, \eqref{e4.13}, \eqref{e4.14}, \eqref{e4.1} and \eqref{e4.4},
we obtain
\begin{equation}
\exp_{p}\{(\tau-\varepsilon)r^{\mu_p(A_0)}\}\leq
n\exp_{p}\{(\tau_1+\varepsilon)r^{\mu_p(A_0)}\}CT(r,f)^{n+1},\label{e4.15}
\end{equation}
where $C>0$ is a constant, for all $z$ satisfying
$|z|=r\not\in E_6\cup[0,1]$, $r\to\infty$ and $|A_0(z)|=M(r,A_0)$.
By Lemma \ref{lem3.2} and \eqref{e4.15}, we have $\mu_{p+1}(f)\geq\mu_p(A_0)$.
On the other hand, by Lemma \ref{lem3.4}, there exists a set $E_4$ having
infinite logarithmic measure such that for all $r\in E_4$, we have
\begin{equation}
|A_0(z)|\leq M(r,A_0)\leq\exp_p\{(\tau+\varepsilon)r^{\mu_p(A_0)}\}.\label{e4.16}
\end{equation}
By \eqref{e4.7}, \eqref{e4.8}, \eqref{e4.12}, \eqref{e4.13} and \eqref{e4.16},
we have
\begin{equation}
|\nu_f(r)|^n|1+o(1)|\leq n\exp_p\{(\tau+\varepsilon)r^{\mu_p(A_0)}\}r^n|
\nu_f(r)|^{n-1}|1+o(1)|,\label{e4.17}
\end{equation}
where $r\in E_4\backslash E_1$, $r\to\infty$. By the
definition of iterated $p$-lower order and \eqref{e4.17}, we obtain
$\mu_{p+1}(f)\leq\mu_p(A_0)$. Thus, we have
$\mu_{p+1}(f)=\mu_p(A_0)$.
(2) We prove that
$\overline{\lambda}_{p+1}(f-\varphi)=\sigma_{p+1}(f)$. Assume that
$f(z)\not\equiv 0$ is a solution of \eqref{e2.1}, then
$\sigma_{p+1}(f)=\sigma_p(A_0)$. Set $g=f-\varphi$, since
$\sigma_{p+1}(\varphi)<\mu_p(A_0)\leq\sigma_p(A_0)$, then
$\sigma_{p+1}(g)=\sigma_{p+1}(f)=\sigma_p(A_0)$,
$\overline{\lambda}_{p+1}(g)=\overline{\lambda}_{p+1}(f-\varphi)$.
Substituting $f=g+\varphi, f'=g'+\varphi',\dots,
f^{(n)}=g^{(n)}+\varphi^{(n)}$, into \eqref{e2.1}, we obtain
\begin{equation}
g^{(n)}+A_{n-1}(z)g^{(n-1)}+\dots+A_0(z)g
=-[\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi].
\label{e4.18}
\end{equation}
If $F(z)=\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi\equiv
0$, then by Lemma \ref{lem3.8}, we have
$\sigma_{p+1}(\varphi)\geq\mu_p(A_0)$, which is a contradiction.
Since $F(z)\not\equiv 0$ and
$\sigma_{p+1}(F)<\sigma_{p+1}(f)=\sigma_{p+1}(g)$. By Lemma \ref{lem3.7} and
\eqref{e4.18}, we have
$\overline{\lambda}_{p+1}(g)=\lambda_{p+1}(g)=\sigma_{p+1}(g)=\sigma_p(A_0)$.
Therefore,
$\overline{\lambda}_{p+1}(f-\varphi)={\lambda}_{p+1}(f-\varphi)
=\sigma_{p+1}(f)=\sigma_p(A_0)$.
The proof is complete.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm2.3}]
By Lemma \ref{lem3.5}, we have
$\sigma_{p+1}(f)=\sigma_p(A_0)$. Now we need to prove
(1) $\mu_{p+1}(f)=\mu_p(A_0)$ and
(2) $\sigma_{p+1}(f)=\overline{\lambda}_{p+1}{(f-\varphi)}$.
(1) On the one hand, by \eqref{e4.1} and the logarithmic derivative
lemma, we have
\begin{equation}
m(r,A_0)\leq \sum_{j=1}^{n-1}m(r,A_j)+O(\log(rT(r,f))),\quad
(r\not\in E),\label{e4.19}
\end{equation}
where $E$ is a set of $r$ of finite linear measure.
Setting $\limsup_{r\to\infty} {\sum_{j=1}^{n-1}m(r,A_j)}/{m(r,A_0)}<\beta<1$,
for sufficiently large $r$, we have
\begin{equation}
\sum_{j=1}^{n-1}m(r,A_j)<\beta m(r,A_0).\label{e4.20}
\end{equation}
By \eqref{e4.19} and \eqref{e4.20}, we have
\begin{equation}
(1-\beta)m(r,A_0)\leq O(\log(rT(r,f))),\quad (r\not\in E).\label{e4.21}
\end{equation}
By $\mu_p(A_0)=\mu$, for any given $\varepsilon>0$ and sufficiently large $r$,
we have
\begin{equation}
m(r,A_0)\geq\exp_{p-1}\{r^{\mu-\varepsilon}\}.\label{e4.22}
\end{equation}
By \eqref{e4.21} and \eqref{e4.22}, for the above
$\varepsilon>0,r\not\in E,r\to\infty$, we have
\begin{equation}
(1-\beta)\exp_{p-1}\{r^{\mu-\varepsilon}\}\leq O\big(\log(rT(r,f))\big).
\label{e4.23}
\end{equation}
By Lemma \ref{lem3.2} and \eqref{e4.23}, we have
$\mu-\varepsilon\leq\mu_{p+1}(f)$. Since $\varepsilon>0$ is arbitrary,
we have $\mu_p(A_0)=\mu\leq\mu_{p+1}(f)$.
On the other hand, since
$\max\{\sigma_p(A_j),j\neq0\}\leq\mu_p(A_0)=\mu$, for any given
$\varepsilon>0$ and sufficiently large $r$, we have
\begin{equation}
|A_j(z)|\leq\exp_p\{r^{\mu+\varepsilon}\},\quad
(j=1,\dots,n-1).\label{e4.24}
\end{equation}
By Lemma \ref{lem3.4}, there exists a set of
$E_2$ having infinite logarithmic measure such that for all
$r\in E_2$, we have
\begin{equation}
|A_0(z)|\leq\exp_p\{r^{\mu+\varepsilon}\}.\label{e4.25}
\end{equation}
By \eqref{e4.7}, \eqref{e4.8}, \eqref{e4.24} and \eqref{e4.25}, we have
\begin{equation}
|\nu_f(r)|^n|1+o(1)|\leq n\exp_p\{r^{\mu+\varepsilon}\}r^n
|\nu_f(r)|^{n-1}|1+o(1)|.\label{e4.26}
\end{equation}
By \eqref{e4.26}, for the above $\varepsilon>0$, we obtain
\begin{equation}
|\nu_f(r)||1+o(1)|\leq nr^n\exp_p\{r^{\mu+\varepsilon}\},\label{e4.27}
\end{equation}
where $|z|=r\in E_2\backslash E_1,r\to\infty,|f(z)|=M(r,f)$.
By \eqref{e4.27}, we obtain $\mu_{p+1}(f)\leq\mu+\varepsilon$.
Since $\varepsilon>0$ is
arbitrary, we have $\mu_{p+1}(f)\leq\mu$. Thus, we have
$\mu_{p+1}(f)=\mu_p(A_0)$.
(2) We prove that
$\overline{\lambda}_{p+1}(f-\varphi)=\sigma_{p+1}(f)$. Setting
$g=f-\varphi$, since $\sigma_{p+1}(\varphi)<\mu_p(A_0)$, we have
$\sigma_{p+1}(g)=\sigma_{p+1}(f)=\sigma_p(A_0)$,
$\overline{\lambda}_{p+1}(g)=\overline{\lambda}_{p+1}(f-\varphi)$.
Substituting $f=g+\varphi,f'=g'+\varphi',\dots,
f^{(n)}=g^{(n)}+\varphi^{(n)}$ into \eqref{e2.1}, we obtain
\begin{equation}
g^{(n)}+A_{n-1}(z)g^{(n-1)}+\dots+A_0(z)g
=-[\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi].\label{e4.28}
\end{equation}
If $F(z)=\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi\equiv
0$, then by part (1), we have $\sigma_{p+1}(\varphi)\geq\mu_p(A_0)$,
which is a contradiction. Since $F(z)\not\equiv 0$ and
$\sigma_{p+1}(F)<\sigma_{p+1}(f)=\sigma_{p+1}(g)$. By Lemma \ref{lem3.7} and
\eqref{e4.28}, we have
$\overline{\lambda}_{p+1}(g)=\lambda_{p+1}(g)=\sigma_{p+1}(g)=\sigma_p(A_0)$.
Therefore,
$\mu_p(A_0)=\mu_{p+1}(f)\leq\sigma_{p+1}(f)=\sigma_p(A_0)
=\overline{\lambda}_{p+1}(f-\varphi)={\lambda}_{p+1}(f-\varphi)$.
The proof is complete.
\end{proof}
\section{Proof of Theorem \ref{thm2.4}}
By Lemma \ref{lem3.3}, we have $\sigma_{p+1}(f)=\sigma_p(A_0)$. Now we prove
that $\overline{\lambda}_{p+1}(f^{(j)}-\varphi)=\sigma_{p+1}(f)$.
(1) We prove the
$\overline{\lambda}_{p+1}(f-\varphi)=\sigma_{p+1}(f)$. Setting
$g=f-\varphi$, since $\sigma_{p+1}(\varphi)<\sigma_p(A_0)$, we have
$\sigma_{p+1}(g)=\sigma_{p+1}(f)=\sigma_p(A_0)$,
$\overline{\lambda}_{p+1}(g)=\overline{\lambda}_{p+1}(f-\varphi)$.
Substituting $f=g+\varphi, f'=g'+\varphi',\dots,
f^{(n)}=g^{(n)}+\varphi^{(n)}$ into \eqref{e2.1}, we obtain
\begin{equation}
g^{(n)}+A_{n-1}(z)g^{(n-1)}+\dots+A_0(z)g
=-[\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi].\label{e5.1}
\end{equation}
Since $\lambda_p(\frac{1}{A_0})<\mu_p(A_0)$, we have
$N(r,A_0)=o(T(r,A_0)),r\to\infty$. Therefore, by Lemma \ref{lem3.9},
we have
\begin{equation}
\begin{split}
\sigma_p(A_0)&=\limsup_{r\to\infty} \frac{\log_p T(r,A_0)}{\log r}
=\lim_{r\to\infty,\, r\in E_8}\frac{\log_p T(r,A_0)}{\log r}\\
&=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r},
\end{split}\label{e5.2}
\end{equation}
where $E_8$ is a subset of $r$ of infinite logarithmic measure.
Combining the assumption and \eqref{e5.2}, we have
\begin{equation}
\limsup_{r\to\infty} \frac{\log_p
m(r,A_j)}{\log r}<\lim_{r\to\infty,\, r\in E_8}\frac{\log_p m(r,A_0)}{\log r}=\sigma_p(A_0),\
j=1,\dots,n.\label{e5.3}
\end{equation}
If $F(z)=\varphi^{(n)}+A_{n-1}(z)\varphi^{(n-1)}+\dots+A_0(z)\varphi\equiv
0$, then by Lemma \ref{lem3.10}, we have
$\sigma_{p+1}(\varphi)\geq\sigma_p(A_0)$, which is a contradiction.
Since $F(z)\not\equiv 0$ and
$\sigma_{p+1}(F)<\sigma_{p+1}(f)=\sigma_{p+1}(g)$, by
Lemma \ref{lem3.7} and
\eqref{e5.1}, we have
$\overline{\lambda}_{p+1}(g)=\lambda_{p+1}(g)=\sigma_{p+1}(g)=\sigma_p(A_0)$.
Therefore,
$\overline{\lambda}_{p+1}(f-\varphi)={\lambda}_{p+1}(f-\varphi)
=\sigma_{p+1}(f)=\sigma_p(A_0)$.
(2) We prove that
$\overline{\lambda}_{p+1}(f'-\varphi)=\sigma_{p+1}(f)$.
Setting $g_1=f'-\varphi$, we have
$\sigma_{p+1}(g_1)=\sigma_{p+1}(f)=\sigma_p(A_0)$ and
\begin{equation}
f'=g_1+\varphi,\dots,f^{(n+1)}=g_1^{(n)}+\varphi^{(n)}.\label{e5.4}
\end{equation}
By \eqref{e2.1}, we have
\begin{equation}
f(z)=-\frac{1}{A_0(z)}\left(f^{(n)}+\dots+A_1(z)f'\right).\label{e5.5}
\end{equation}
The derivative of \eqref{e2.1} is
\begin{equation}
f^{(n+1)}+A_{n-1}f^{(n)}+(A_{n-1}'+A_{n-2})f^{(n-1)}+\dots
+(A_1'+A_0)f'+A_0'f=0.\label{e5.6}
\end{equation}
Substituting \eqref{e5.4} and \eqref{e5.5} into \eqref{e5.6}, we obtain
\begin{align*}
&g_1^{(n)}+(A_{n-1}-\frac{A_0'}{A_0})g_1^{(n-1)}+(A_{n-2}+A_{n-1}'
-\frac{A_{n-1}A_0'}{A_0})g_1^{(n-2)}
+\dots\\
& +(A_0+A_1'-\frac{A_1A_0'}{A_0})g_1\\
&=-\big[\varphi^{(n)}+(A_{n-1}-\frac{A_0'}{A_0})\varphi^{(n-1)}+\dots
+(A_0+A_1'-\frac{A_1A_0'}{A_0})\varphi\big].
\end{align*}
Setting
\begin{equation}
\begin{gathered}
B_{n-1}=A_{n-1}-\frac{A_0'}{A_0},\quad
B_{n-2}=A_{n-2}+A_{n-1}'-\frac{A_{n-1}A_0'}{A_0},\\
\dots,\quad
B_0=A_0+A_1'-\frac{A_1A_0'}{A_0},
\end{gathered}\label{e5.7}
\end{equation}
we have
\begin{equation}
g_1^{(n)}+B_{n-1}g_1^{(n-1)}+B_{n-2}g_1^{(n-2)}+\dots+B_0g_1
=-\big[\varphi^{(n)}+B_{n-1}\varphi^{(n-1)}+\dots+B_0\varphi\big].\label{e5.8}
\end{equation}
By \eqref{e5.3} and \eqref{e5.7}, we have
\begin{equation}
\limsup_{r\to\infty} \frac{\log_pm(r,B_j)}{\log
r}<\lim_{r\to\infty,\, r\in E_8}
\frac{\log_pm(r,A_0)}{\log r}=\sigma_p(A_0),\quad
(j\neq0),\label{e5.9}
\end{equation}
and
\begin{equation}
\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r}
=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,B_0)}{\log r},\label{e5.10}
\end{equation}
where $E_8$ is a subset of infinite logarithmic measure $r$. Let
$F_1(z)=\varphi^{(n)}+B_{n-1}\varphi^{(n-1)}+\dots+B_0\varphi$. We
affirm $F_1(z)\not\equiv 0$. If $F_1(z)\equiv 0$, then by \eqref{e5.9},
\eqref{e5.10} and Lemma \ref{lem3.10}, we obtain
$\sigma_{p+1}(\varphi)\geq\sigma_p(A_0)$, which is a contradiction.
Since $F_1(z)\not\equiv 0$, and
$\sigma_{p+1}(F_1)<\sigma_{p+1}(g_1)=\sigma_p(A_0)$. By Lemma \ref{lem3.7}
and \eqref{e5.8}, we obtain
$$
\overline{\lambda}_{p+1}(f'-\varphi)={\lambda}_{p+1}(f'-\varphi)=\sigma_{p+1}(f).
$$
(3) We prove that
$\overline{\lambda}_{p+1}(f''-\varphi)=\sigma_{p+1}(f)$.
Setting $g_2=f''-\varphi$, we have
$\sigma_{p+1}(g_2)=\sigma_{p+1}(f)=\sigma_p(A_0)$ and
\begin{equation}
f''=g_2+\varphi,\dots,f^{(n+2)}=g_2^{(n)}+\varphi^{(n)}.\label{e5.11}
\end{equation}
Substituting \eqref{e5.5} into \eqref{e5.6}, we have
\begin{equation}
\begin{split}
&f^{(n+1)}+(A_{n-1}-\frac{A_0'}{A_0})f^{(n)}+(A_{n-2}+A_{n-1}'
-\frac{A_{n-1}A_0'}{A_0})f^{(n-1)}+\dots\\
&+(A_0+A_1'-\frac{A_1A_0'}{A_0})f'=0.
\end{split}\label{e5.12}
\end{equation}
The derivative of \eqref{e5.12} is
\begin{equation}
\begin{split}
&f^{(n+2)}+(A_{n-1}-\frac{A_0'}{A_0})f^{(n+1)}
+\big[(A_{n-1}-\frac{A_0'}{A_0})'+(A_{n-2}+A_{n-1}'-\frac{A_{n-1}A_0'}{A_0})\big]
f^{(n)}\\
& +\dots+(A_0+A_1'-\frac{A_1A_0'}{A_0})'f'=0.
\end{split}\label{e5.13}
\end{equation}
By \eqref{e5.12}, we have
\begin{equation}
f'=-\big[\frac{1}{A_0+A_1'-\frac{A_1A_0'}{A_0}}f^{(n+1)}
+\frac{A_{n-1}-\frac{A_0'}{A_0}}{A_0+A_1'
-\frac{A_1A_0'}{A_0}}f^{(n)}+\dots+\frac{A_1+A_2'
-\frac{A_2A_0'}{A_0}}{A_0+A_1'-\frac{A_1A_0'}{A_0}}f''\big].\label{e5.14}
\end{equation}
Substituting \eqref{e5.14} into \eqref{e5.13}, we have
\begin{align*}
&f^{(n+2)}+\big[(A_{n-1}-\frac{A_0'}{A_0})-\frac{(A_0+A_1'
-\frac{A_1A_0'}{A_0})'}{A_0+A_1'-\frac{A_1A_0'}{A_0}}\big]f^{(n+1)}+\dots\\
& +\big[(A_0+A_1'-\frac{A_1A_0'}{A_0})+(A_1+A_2'-\frac{A_2A_0'}{A_0})'\\
&-\frac{(A_1+A_2'-\frac{A_2A_0'}{A_0})(A_0+A_1'-\frac{A_1A_0'}{A_0})'}{A_0+A_1'
-\frac{A_1A_0'}{A_0}}\big]f''
=0.
\end{align*}
Setting
\begin{equation}
\begin{gathered}
C_{n-1}=B_{n-1}-\frac{B_0'}{B_0},\quad
C_{n-2}=B_{n-2}+B_{n-1}'-\frac{B_{n-1}B_0'}{B_0},\\
\dots,\quad C_0=B_0+B_1'-\frac{B_1B_0'}{B_0},
\end{gathered}\label{e5.15}
\end{equation}
we obtain
\begin{equation}
f^{(n+2)}+C_{n-1}(z)f^{(n+1)}+\dots+C_0(z)f''=0.\label{e5.16}
\end{equation}
Substituting \eqref{e5.11} into \eqref{e5.16}, we obtain
\begin{equation}
g_2^{(n)}+C_{n-1}(z)g_2^{(n-1)}+\dots+C_0(z)g_2
=-\big[\varphi^{(n)}+C_{n-1}(z)\varphi^{(n-1)}+\dots+C_0(z)\varphi\big].
\label{e5.17}
\end{equation}
By \eqref{e5.2}, \eqref{e5.9}, \eqref{e5.10} and \eqref{e5.15}, we have
\begin{equation}
\limsup_{r\to\infty} \frac{\log_pm(r,C_j)}{\log
r}<\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log
r}=\sigma_p(A_0),(j\neq0),\label{e5.18}
\end{equation}
and
\begin{equation}
\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r}
=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,C_0)}{\log r},\label{e5.19}
\end{equation}
where $E_8$ is a subset of $r$ of infinite logarithmic measure. If
$F_2(z)\equiv\varphi^{(n)}+C_{n-1}(z)\varphi^{(n-1)}+\dots+C_0(z)\varphi\equiv
0$, then by \eqref{e5.18}, \eqref{e5.19} and Lemma \ref{lem3.10}, we have
$\sigma_{p+1}(\varphi)\geq\sigma_p(A_0)$, which is a contradiction.
Therefore, $F_2(z)\not\equiv 0$. Since
$\sigma_{p+1}(F_2)<\sigma_{p+1}(g_2)=\sigma_p(A_0)$, by Lemma \ref{lem3.7}
and \eqref{e5.17}, we have
$$
\overline{\lambda}_{p+1}(f''-\varphi)={\lambda}_{p+1}(f''-\varphi)
=\sigma_{p+1}(f).
$$
(4) We prove that
$\overline{\lambda}_{p+1}(f'''-\varphi)=\sigma_{p+1}(f)$.
Setting $g_3=f'''-\varphi$, then
$\sigma_{p+1}(g_3)=\sigma_{p+1}(f)=\sigma_p(A_0)$ and
\begin{equation}
f'''=g_3+\varphi,\quad \dots,\quad f^{(n+3)}=g_3^{(n)}+\varphi^{(n)}.\label{e5.20}
\end{equation}
The derivative of \eqref{e5.16} is
\begin{equation}
f^{(n+3)}+C_{n-1}f^{(n+2)}+(C_{n-1}'+C_{n-2})f^{(n+1)}+\dots
+(C_1'+C_0)f'''+C_0'f''=0.\label{e5.21}
\end{equation}
By \eqref{e5.16}, we have
\begin{equation}
f''=-\big[\frac{1}{C_0}f^{(n+2)}+\frac{C_{n-1}}{C_0}f^{(n+1)}
+\dots+\frac{C_1}{C_0}f'''\big].\label{e5.22}
\end{equation}
Substituting \eqref{e5.22} into \eqref{e5.21}, we have
\begin{equation}
\begin{aligned}
&f^{(n+3)}+\Big(C_{n-1}-\frac{C_0'}{C_0}\Big)f^{(n+2)}
+\Big(C_{n-2}+C_{n-1}'-\frac{C_{n-1}C_0'}{C_0}\Big)f^{(n+1)}\\
&+\dots+\Big(C_0+C_1'-\frac{C_1C_0'}{C_0}\Big)f'''=0.
\end{aligned} \label{e5.23}
\end{equation}
Setting
\begin{equation}
\begin{gathered}
D_{n-1}=C_{n-1}-\frac{C_0'}{C_0},\quad
D_{n-2}=C_{n-2}+C_{n-1}'-\frac{C_{n-1}C_0'}{C_0},\\
\dots,\quad
D_0=C_0+C_1'-\frac{C_1C_0'}{C_0},
\end{gathered}\label{e5.24}
\end{equation}
we have
\begin{equation}
f^{(n+3)}+D_{n-1}(z)f^{(n+2)}+\dots+D_0(z)f'''=0.\label{e5.25}
\end{equation}
Substituting \eqref{e5.20} into \eqref{e5.25}, we obtain
\begin{equation}
g_3^{(n)}+D_{n-1}(z)g_3^{(n-1)}+\dots+D_0(z)g_3
=-[\varphi^{(n)}+D_{n-1}(z)\varphi^{(n-1)}+\dots+D_0(z)\varphi].\label{e5.26}
\end{equation}
By \eqref{e5.18}, \eqref{e5.19} and \eqref{e5.24}, we have
\begin{equation}
\limsup_{r\to\infty} \frac{\log_pm(r,D_j)}{\log r}
<\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r}
=\sigma_p(A_0),\ (j\neq0),\label{e5.27}
\end{equation}
and
\begin{equation}
\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r}
=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,D_0)}{\log r},\label{e5.28}
\end{equation}
where $E_8$ is a subset of $r$ of infinite logarithmic measure. Let
$F_3(z)=\varphi^{(n)}+D_{n-1}(z)\varphi^{(n-1)}+\dots+D_0(z)\varphi\equiv
0$, by \eqref{e5.27}, \eqref{e5.28} and Lemma \ref{lem3.10}, we have
$F_3(z)\not\equiv 0$.
Since $\sigma_{p+1}(F_3)<\sigma_{p+1}(g_3)=\sigma_p(A_0)$, by
Lemma \ref{lem3.7} and \eqref{e5.26}, we have
$$
\overline{\lambda}_{p+1}(f'''-\varphi)={\lambda}_{p+1}(f'''-\varphi)
=\sigma_{p+1}(f).
$$
(5) We prove that
$\overline{\lambda}_{p+1}(f^{(j)}-\varphi)=\sigma_{p+1}(f)$, $(j>3)$.
Setting $g_j=f^{(j)}-\varphi$, $(j>3)$, then
$\sigma_{p+1}(g_j)=\sigma_{p+1}(f^{(j)})=\sigma_p(A_0)$ and
\begin{equation}
f^{(j+1)}=g_{j}'+\varphi',\quad \dots,f^{(n)}=g_{j}^{(n-j)}+\varphi^{(n-j)},\quad
(j>3).\label{e5.29}
\end{equation}
By successive derivation on \eqref{e5.25}, we also get an equation which
has similar form with \eqref{e5.23}. Furthermore, combining \eqref{e5.29}, we
can get
\begin{equation}
\begin{aligned}
&g_{j}^{(n)}+(H_{n-1}-\frac{H_0'}{H_0})g_{j}^{(n-1)}+\dots
+(H_0+H_1'-\frac{H_1H_0'}{H_0})g_j \\
&=-\big[\varphi^{(n)}+\dots+(H_0+H_1'-\frac{H_1H_0'}{H_0})\varphi\big],
\end{aligned} \label{e5.30}
\end{equation}
where $H_j(z)$, $(j=0,1,\dots,n-1)$ are meromorphic functions which
have the same form as $D_j(z),(j=1,\dots,n-1)$. Setting
$G_{n-1}=H_{n-1}-\frac{H_0'}{H_0}$, \dots,
$G_0=H_0+H_1'-\frac{H_1H_0'}{H_0}$, we have
$$
\limsup_{r\to\infty} \frac{\log_pm(r,G_j)}{\log r}
<\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,A_0)}{\log r}
=\sigma_p(A_0),\ (j\neq 0),
$$
and
$$
\sigma_p(A_0)=\lim_{r\to\infty,\, r\in E_8}
\frac{\log_pm(r,A_0)}{\log r}
=\lim_{r\to\infty,\, r\in E_8}\frac{\log_pm(r,G_0)}{\log r},
$$
where $E_8$ is a subset of $r$ of infinite logarithmic measure. By
Lemmas \ref{lem3.7} and \ref{lem3.10}, we can get
$\overline{\lambda}_{p+1}(g_j)={\lambda}_{p+1}(g_j)=\sigma_{p+1}(g_j)$;
i.e.,
$\overline{\lambda}_{p+1}(f^{(j)}-\varphi)
={\lambda}_{p+1}(f^{(j)}-\varphi)=\sigma_{p+1}(f)$.
the proof of Theorem \ref{thm2.4} is complete.
\subsection*{Acknowledgements}
The authors are grateful to the
referees and editors for their valuable comments which lead to the
improvement of this paper.
This project is supported by
the National Natural Science Foundation of China (11126145. 11171119),
and the Natural Science Foundation of Jiangxi Province in China
(20114BAB211003, 20122BAB211005).
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