\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 65, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/65\hfil  Finite order solutions]
{Finite order solutions of \\
complex linear differential equations}

\author[I. Laine \& R. Yang\hfil EJDE-2004/65\hfilneg]
{Ilpo Laine \& Ronghua Yang } % in alphabetical order

\address{Ilpo Laine \hfill\break
University of Joensuu, Department of Mathematics \\
P.O. Box 111, FIN-80101 Joensuu, Finland}
\email{ilpo.laine@joensuu.fi}

\address{Ronghua Yang \hfill\break
University of Joensuu, Department of Mathematics \\
P.O. Box 111, FIN-80101 Joensuu, Finland}
\email{ronghua.yang@joensuu.fi}

\date{}
\thanks{Submitted December 10, 2003. Published April 28, 2004.}
\thanks{Partially supported by grant 50981 from the Academy of Finland.}
\subjclass[2000]{30D35, 34M10}
\keywords{Linear differential equations, growth of solutions, iterated order}

\begin{abstract}
 We shall consider the growth of solutions of complex linear
 homogeneous differential equations
 $$
 f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_1(z)f'+A_0(z)f=0
 $$
 with entire coefficients. If one of the intermediate coefficients
 in exponentially dominating in a sector and $f$ is of finite
 order, then a derivative $f^{(j)}$ is asymptotically constant in a
 slightly smaller sector. We also find conditions on the
 coefficients to ensure that all transcendental solutions are of
 infinite order. This paper  extends previous results due to
 Gundersen and to Bela\"{\i}di and Hamani.
\end{abstract}

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

\section{Introduction}\label{intro}

It is well known that all solutions of the linear differential
equation
\begin{equation}\label{0lin}
 f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_{1}(z)f'+A_{0}(z)f=0
\end{equation}
are entire functions, provided the coefficients
$A_{0}(z)\not\equiv 0$, $A_{1}(z),\ldots ,A_{k-1}(z)$ are entire. A
classical result, due to Wittich, tells that all solutions of
(\ref{0lin}) are of finite order of growth if and only if all
coefficients $A_{j}(z)$, $j=0,\ldots ,k-1,$ are polynomials. For a
complete analysis of possible orders in the polynomial case, see
\cite{GSW}. If some (or all) of the coefficients are
transcendental, a natural question is to ask when and how many
solutions of finite order may appear. Partial results have been
available since a paper of Frei \cite{Fr}. In its all generality,
however, the problem remains open. Our starting point for this
paper is a result due to Gundersen in \cite{Gary1}:

\begin{theorem}\label{thma}
Let $A_{0}(z)\not\equiv 0, A_{1}(z)$ be entire functions such that
for some real constants $\alpha >0$, $\beta >0$, $\theta_{1}<\theta_{2}$
we have
\begin{gather*}
|A_{1}(z)|\geq \exp ((1+o(1))\alpha |z|^{\beta}),\\
|A_{0}(z)|\leq \exp (o(1)|z|^{\beta})
\end{gather*}
 as $z\to\infty$ in the sector $S(0):\theta_{1}\leq \arg z\leq \theta_{2}$. Given
$\varepsilon >0$ small enough, let $S(\varepsilon )$ denote the
sector $\theta_{1}+\varepsilon \leq \arg z \leq
\theta_{2}-\varepsilon$. If $f$ is a nontrivial solution of
\begin{equation}\label{2lin}
 f''+A_{1}(z)f'+A_{0}(z)f=0
\end{equation}
of finite order, then the following conditions hold:
\begin{itemize}
\item[(i)] There exists a constant $b\neq 0$ such that $f(z)\to
b$ as $z\to\infty$ in $S(\varepsilon)$. Indeed,
$$|f(z)-b|\leq \exp (-(1+o(1))\alpha |z|^{\beta}).
$$
\item[(ii)] For each integer $k\geq 1$,
$$|f^{(k)}(z)|\leq \exp (-(1+o(1))\alpha|z|^{\beta})
$$
as $z\to\infty$ in $S(\varepsilon)$.
\end{itemize}
\end{theorem}

This result has been recently generalized to the higher order case
(\ref{0lin}) by Bela\"{\i}di and Hamani, see \cite{BEL}, as
follows:

\begin{theorem}\label{thmb}
Let $A_{0}(z)\not\equiv 0,A_{1}(z),\ldots ,A_{k-1}(z)$ be entire
functions such that for some real constants $\alpha >0,\beta >0,
\theta_{1}<\theta_{2}$, we have
\begin{gather*}
|A_{1}(z)|\geq \exp((1+o(1))\alpha |z|^{\beta}),\\
|A_{j}(z)|\leq \exp (o(1)|z|^{\beta}),\quad j=0,2,3,\ldots ,k-1,
\end{gather*} as
$z\to\infty$ in $S(0)$. If $f$ is a nontrivial solution of
(\ref{0lin}) of finite order, then the following conditions hold,
provided $\varepsilon >0$ is small enough:
\begin{itemize}
\item[(i)] There exists a constant $b\neq 0$ such that $f(z)\to
b$ as $z\to\infty$ in $S(\varepsilon )$. Indeed,
$$|f(z)-b|\leq \exp (-(1+o(1))\alpha |z|^{\beta}).$$

\item[(ii)] For each integer $k\geq 1$, $$|f^{(k)}(z)|\leq \exp (-(1+o(1))\alpha
|z|^{\beta})$$ as $z\to\infty$ in $S(\varepsilon )$.
\end{itemize}
\end{theorem}

A natural question is now to ask about a counterpart of Theorem \ref{thmb}
in the case that the coefficient in the same sense as in Theorem \ref{thmb}
is $A_{s}(z)$ instead of $A_{1}(z)$. We are going to present such
a counterpart in this paper, see Theorem~\ref{Gary} below. As the
proof now is more complicated than the corresponding proof of
Theorem \ref{thmb}, see \cite{BEL}, we express the growth conditions for
the coefficients more explicitly. In fact, making use of $o(1)$
only as in Theorem \ref{thmb} might leave some doubts on the necessary
uniformity in the course of the proof.

\section{Notation and results}\label{notats}

Given $\varepsilon >0$, and $\theta_{1},\theta_{2}\in [0,2\pi ),
\theta_{1}<\theta_{2}$, we denote by
$S(\theta_{1},\theta_{2},\varepsilon )$, resp.
$S(R,\theta_{1},\theta_{2},\varepsilon )$, the sector $\{
z|\theta_{1}+\varepsilon \leq \arg z\leq \theta_{2}-\varepsilon\}$,
resp. the truncated sector
$S(\theta_{1},\theta_{2},\varepsilon )\cap \{ |z|\geq R\}$. If the
sector boundaries are clear, and there is no possibility of
confusion, we apply the shorter notations $S(\varepsilon)$, resp.
$S(R,\varepsilon )$. In the proof of Theorem~1 below, we agree
that whenever stating that $r\geq r_{j}$ to indicate that $|z|=r$
has to be large enough, we always assume that $r_{j}\geq r_{j-1}$.
Hence, in such a situation, all corresponding previous conditions
for $r\geq r_{j-1}$ remain valid, without saying this explicitly
in what follows.

\begin{theorem}\label{Gary}
Let $\theta_{1}<\theta_{2}$ be given to fix a sector $S(0)$, let
$k\geq 2$ be a natural number, and let $\delta >0$ be any real
number such that $k\delta < 1$. Suppose that $A_{0}(z),\ldots
A_{k-1}(z)$ with $A_{0}(z)\not\equiv 0$ are entire functions such
that for real constants $\alpha >0$, $\beta
>0$, we have, for some $s=1,\ldots ,k-1$,
\begin{gather}\label{large}
 |A_{s}(z)|\geq \exp ((1+\delta )\alpha |z|^{\beta}),\\
\label{small}
 |A_{j}(z)|\leq \exp (\delta \alpha |z|^{\beta})
\end{gather}
for all $j=0,\ldots ,s-1,s+1,\ldots ,k-1$ whenever $|z|=r\geq
r_{\delta}$ in the sector $S(0)$. Given $\varepsilon
>0$ small enough, if $f$ is a transcendental solution of
finite order $\rho <\infty$ of the linear differential equation
\begin{equation}\label{lin}
 f^{(k)}+A_{k-1}(z)f^{(k-1)}+\dots +A_{1}(z)f'+A_{0}(z)f=0,
\end{equation}
then the following conditions hold:

\begin{itemize}
\item[(i)] There exists $j\in \{ 0,\ldots ,s-1\}$ and a complex
constant $b_{j}\neq 0$ such that $f^{(j)}(z)\to b_{j}$ as
$z\to\infty$ in the sector $S(\varepsilon)$. More
precisely,
\begin{equation}\label{almostb}
 |f^{(j)}(z)-b_{j}|\leq \exp (-(1-k\delta )\alpha |z|^{\beta})
\end{equation}
in $S(\varepsilon )$, provided $|z|$ is large enough.

\item[(ii)] For each integer $m\geq j+1$,
\begin{equation}\label{dersmall}
 |f^{(m)}(z)|\leq \exp (-(1-k\delta )\alpha |z|^{\beta})
\end{equation}
in $S(3\varepsilon )$ for all $|z|$ large enough.
\end{itemize}
\end{theorem}

\begin{remark} \rm In Theorem \ref{Gary}, it may happen that $j<s-1$.
Indeed, $f(z)=e^{z}+1$ satisfies
\begin{equation}\label{esim}
 f'''+2e^{-z}f''-e^{z}f'+(-2+e^{z})f=0.
\end{equation}
Obviously, (\ref{esim}) fulfills the assumptions of Theorem
\ref{Gary} in the sector $\frac{2\pi}{3}<\theta <\frac{3\pi}{4}$.
In this example, $A_{2}(z)=2e^{-z}$ is the dominating coefficient,
while we have $j=0$.
\end{remark}

\begin{remark} \rm
The following two theorems are natural counterparts to
\cite[Theorem 5]{Gary1}, and \cite[Theorem 1.6]{BEL}, respectively
to \cite[Theorem 1.7]{BEL}.
\end{remark}

\begin{theorem}\label{Gary2}
Let $A_{0}(z)\not\equiv 0$, $A_{1}(z),\ldots,A_{k-1}(z)$, $k\geq 2,$
be entire functions, let $\alpha >0$, $\beta>0$ be given constants,
let $\delta >0$ be a real number such that $k\delta <1$
and let $s$ be an integer such that $1\leq s\leq k-1$.
Suppose that (i) $\rho (A_{j})<\beta$ for $j\neq s$
and (ii) for any given $\varepsilon>0$, there exists two finite
collections of real numbers
$(\phi_{m})$ and $(\theta_{m})$ that satisfy
$\phi_{1}<\theta_{1}<\phi_{2}<\theta_{2}<\dots
<\phi_{n}<\theta_{n}<\phi_{n+1}=\phi_{1}+2\pi$, such that
\begin{gather}\label{epsilon}
 \sum_{m=1}^{n}(\phi_{m+1}-\theta_{m})<\varepsilon,\\
\label{suuri}
 |A_{s}(z)|\geq \exp ((1+\delta )\alpha |z|^{\beta})
\end{gather}
as $z\to\infty$ in $\phi_{m}\leq \arg z \leq \theta_{m}$,
$m=1,\ldots ,n$. Then every transcendental solution $f$ of
(\ref{lin}) is of infinite order.
\end{theorem}

Invoking the iterated order
$\rho_{p}(f):=\limsup_{r\to\infty}(\log_{p}T(r,f))/\log r$
for entire functions, we add our final theorem which is a simple
extension of  \cite[Theorem 1.7]{BEL}.

\begin{theorem}\label{triv}
Let $A_{0}(z),\ldots ,A_{k-1}(z)$ be
entire functions such that for some integer $s$, $1\leq s \leq
k-1$, we have $\rho_{p} (A_{j})\leq \alpha <\beta =\rho_{p}
(A_{s})\leq +\infty$ for all $j\neq s$. Then every transcendental
solution $f$ of (\ref{lin}) satisfies $\rho_{p} (f)\geq \rho_{p}
(A_{s})$.
\end{theorem}

\section{Preparations for proofs}\label{preps}

To prove Theorem \ref{Gary}, we need two preparatory lemmas. The
first one is a simple extension of \cite[Lemma 4]{Gary1}. See also
\cite[Lemma 3]{GS}.

\begin{lemma}\label{logderline} Let $f(z)$ be an entire function,
and suppose that $|f^{(k)}(z)|$ is unbounded on a ray $\arg
z=\theta$. Then there exists a sequence $z_{n}=r_{n}e^{i\theta}$
tending to infinity such that $f^{(k)}(z_{n})\to\infty$
and that
\begin{equation}\label{quotient}
\left|\frac{f^{(j)}(z_{n})}{f^{(k)}(z_{n})}\right|\leq
 \frac{1}{(k-j)!}(1+o(1))|z_{n}|^{k-j},
\end{equation}
provided $j<k$.
\end{lemma}

\begin{proof} Let $M(r,\theta ,f^{(k)})$ denote the maximum modulus
of $f^{(k)}$ on the line segment $[0,re^{i\theta}]$. Clearly, we
may construct a sequence of points $z_{n}=r_{n}e^{i\theta}$,
$r_{n}\to\infty$, such that $M(r_{n},\theta
,f^{(k)})=|f^{(k)}(r_{n}e^{i\theta})|\to\infty$ as
$n\to\infty$. For each $n$, we obtain by $(k-j)$-fold
iterated integration along the line segment $[0,z_{n}]$,
\begin{equation}\label{int1}
\begin{split}
f^{(j)}(z_{n})=&f^{(j)}(0)+f^{(j+1)}(0)z_{n}+\dots
+\frac{1}{(k-j-1)!}f^{(k-1)}(0)z_{n}^{k-j-1}\\
&+\int_{0}^{z_{n}}\dots \int_{0}^{z_{n}}f^{(k)}(t)dt\dots dt.
\end{split}
\end{equation}
Therefore, by an elementary triangle inequality estimate,
\begin{equation}
\begin{split}
|f^{(j)}(z_{n})|\leq& |f^{(j)}(0)|+|f^{(j+1)}(0)|r_{n}+\dots\\
&+\frac{1}{(k-j-1)!}|f^{(k-1)}(0)|r_{n}^{k-j-1}
+\frac{1}{(k-j)!}|f^{(k)}(z_{n})|r_{n}^{k-j}.
\end{split}\label{int2}
\end{equation}
The assertion immediately follows.
\end{proof}

\begin{lemma}\label{integral}
Given $\alpha >0$, $\beta >0$, $K\geq
\frac{1}{2}$ and $0<\eta <\frac{1}{2}$, the integral
$I(r):=\int_{r}^{+\infty}\exp (-K\alpha t^{\beta})dt$ converges.
More precisely, if $\beta >1$, then $I(r)\leq \exp (-K\alpha
r^{\beta})$, whenever $r^{\beta -1}\geq \frac{1}{K\alpha\beta}$
and if $\beta \leq 1$, then $I(r)\leq \exp (-(K-\eta )\alpha
r^{\beta})$ for any given $\eta \in (0,1/2)$, provided $\eta\alpha
r^{\beta}\geq (1-\beta )\log r + \log \frac{2}{\alpha\beta}$.
\end{lemma}


\begin{remark} \rm
Observe that the lower bound obtained for $r$ above
is independent of $K$, in both cases. Moreover, if we take $r$
large enough, say $r\geq r_{0}\geq r_{\delta }$, then $I(r)\leq
\exp (-(K-\eta )\alpha r^{\beta})$, in both cases again.
\end{remark}


\begin{proof} Clearly,
\begin{equation}\label{arvio}
\begin{split}
I(r)&=\int_{r}^{+\infty}\exp (-K\alpha t^{\beta})dt =
\frac{K\alpha\beta r^{\beta -1}}{K\alpha\beta r^{\beta
-1}}\int_{r}^{+\infty}\exp (-K\alpha t^{\beta})dt
\\
&\leq -\frac{1}{K\alpha\beta r^{\beta
-1}}\int_{r}^{+\infty}-K\alpha\beta t^{\beta -1}\exp (-K\alpha
t^{\beta})dt
\\
&= \frac{1}{K\alpha\beta r^{\beta -1}}\exp (-K\alpha r^{\beta}).
\end{split}
\end{equation}
If now $\beta >1$, it suffices to have $K\alpha\beta r^{\beta
-1}\geq 1$, as required. If $\beta \leq 1$, we may write the above
estimate for $I(r)$ as
\begin{equation}\label{next}
 I(r)\leq \frac{2}{\alpha\beta}r^{1-\beta}\exp (-\eta\alpha
 r^{\beta})\exp(-(K-\eta )\alpha r^{\beta}).
\end{equation}
Provided $\frac{2}{\alpha\beta}r^{1-\beta}\exp (-\eta\alpha
r^{\beta}) \leq 1$, we again have the required estimate.
\end{proof}

\section{Proofs of theorems}\label{proofs}

\begin{proof}[Proof of Theorem $\ref{Gary}$]
\textbf{Boundedness of $f^{(s)}$ in $S(\varepsilon )$.}
Recall first that by \cite{Gary2}, Corollary 1, there exists a set
$E\subset [0,2\pi )$ of linear measure zero such that for all
$k\geq s\geq 0$, all $j=s+1,\ldots ,k$, and all $r\geq r_{1}$,
\begin{equation}\label{dersmall2}
\Big|\frac{f^{(j)}(z)}{f^{(s)}(z)}\Big|\leq |z|^{(j-s)(\rho
-1+\varepsilon)} \leq |z|^{k\rho}
\end{equation}
along any ray $\arg z = \psi$ such that $\psi \in
[0,2\pi)\setminus E$, provided $0<\varepsilon <1$.

Suppose now, for a while, that $|f^{(s)}(z)|$ is unbounded on some
ray $\arg z = \phi \in S(0)\setminus E$. By Lemma
\ref{logderline}, there exists a sequence of points
$z_{n}=r_{n}e^{i\phi}$, $r_{n}\to\infty$ such that
$f^{(s)}(z_{n})\to\infty$ and so
\begin{equation}\label{dersmall3}
\Big|\frac{f^{(j)}(z_{n})}{f^{(s)}(z_{n})}\Big|\leq
\frac{1}{(s-j)!}(1+o(1))|z_{n}|^{s-j}\leq 2|z_{n}|^{k}
\end{equation}
for all $j=0,\ldots ,s-1$ and all $n$ large enough, say
$|z_{n}|\geq r_{2}$. From (\ref{lin}), we next conclude that
\begin{equation}\label{estAs}
\begin{split}
|A_{s}|\leq& \Big|\frac{f^{(k)}}{f^{(s)}}\Big|
+|A_{k-1}|\Big|\frac{f^{(k-1)}}{f^{(s)}}\Big|+\dots
+|A_{s+1}|\Big|\frac{f^{(s+1)}}{f^{(s)}}\Big|
\\
&+|A_{s-1}|\Big|\frac{f^{(s-1)}}{f^{(s)}}\Big|+\dots
+|A_{1}|\Big|\frac{f'}{f^{(s)}}\Big|
+|A_{0}|\Big|\frac{f}{f^{(s)}}\Big|.
\end{split}
\end{equation}
Combining now (\ref{small}), (\ref{dersmall2}) and
(\ref{dersmall3}) with the above estimate (\ref{estAs}) for
$A_{s}$, it is straightforward to see that $|A_{s}(z_{n})|\leq
\exp (3\delta\alpha r_{n}^{\beta})$ for all $n$ large enough in
the sequence $z_{n}$ on the ray $\arg z=\phi$, contradicting
(\ref{large}). Therefore, $|f^{(s)}(z)|$ remains bounded on all
rays $\arg z =\phi \in S(0)\setminus E$. By a standard application
of the Phragm\'{e}n--Lindel\"{o}f principle, we conclude that
$f^{(s)}(z)$ is bounded, say $|f^{(s)}(z)|\leq M$, in the whole
sector $S(\varepsilon)$. \smallskip

\noindent \textbf{Preliminary estimate for $|f^{(m)}(z)|$, $m\leq s$.}
We now proceed to show that $|f^{(m)}(z)|=O(|z|^{s-m})$ on any ray
$\arg z = \phi \in S(0)\setminus E$, for all $m\leq s$. Of course,
it suffices to consider $m<s$. By $(s-m)$-fold iterated
integration along the ray under consideration, see (\ref{int2}),
\begin{equation}\label{estder}
\begin{split}
|f^{(m)}(z)|\leq &|f^{(m)}(0)|+|f^{(m+1)}(0)||z|+\dots
+\frac{1}{(s-m-1)!}|f^{(s-1)}(0)||z|^{s-m-1}
\\
&+M\int_{0}^{|z|}\dots \int_{0}^{|z|}dt\dots dt = O(|z|^{s-m}).
\end{split}
\end{equation}

\noindent\textbf{Proof of the second assertion for $m=s$.}
Writing now (\ref{lin}) in the form
\begin{equation}\label{linmod1}
 f^{(s)}=\frac{1}{A_{s}}(f^{(k)}+A_{k-1}f^{(k-1)}+\dots
 +A_{s+1}f^{(s+1)}+A_{s-1}f^{(s-1)}+\dots
 A_{1}f'+A_{0}f),
\end{equation}
and recalling (\ref{dersmall2}), (\ref{estder}), the boundedness
of $f^{(s)}$ and the assumptions (\ref{large}) and (\ref{small}),
we conclude that whenever $r\geq r_{3}$, we get
\begin{equation}\label{estders}
 |f^{(s)}(z)|\leq \exp (-(1-\delta)\alpha |z|^{\beta})
\end{equation}
along any ray $\arg z =\phi \in S(\varepsilon)\setminus E$. By the
Phragm\'{e}n--Lindel\"{o}f principle again, (\ref{estders})
remains true in the sector $S(2\varepsilon )$, proving the second
assertion in the case of $m=s$. \smallskip


\noindent\textbf{Proof of the second assertion for $m>s$.}
We may now restrict ourselves to the sector $S(3\varepsilon )$. We
assume that $r\geq r_{4}$ is large enough to satisfy that for an
arbitrary $z=re^{i\theta}\in S(3\varepsilon )$, the disk $\Gamma
(z)$ of radius at most $\rho = \max_{s<m\leq
k}((m-s)!)^{1/(m-s)}$, centered at $z$, is contained in
$S(2\varepsilon )$, i.e. we must take $r_{4}\geq \rho /\sin
\varepsilon $. Given now $m>s$, we may use (\ref{estders}) in the
Cauchy formula to see that
\begin{equation}\label{Cauchy}
 |f^{(m)}(z)|\leq \frac{(m-s)!}{2\pi}\int_{\Gamma (z)}\frac{|f^{(s)}(\zeta )|}{|z-\zeta
 |^{m-s+1}}d\zeta .
\end{equation}
By the selection of $\rho$ above, we may combine (\ref{estders})
and (\ref{Cauchy}) to conclude that
\begin{equation}\label{fm}
|f^{(m)}(z)|\leq \exp (-(1-\delta )\alpha |z|^{\beta}).
\end{equation}


\noindent\textbf{Proof of the first assertion for $j=s-1$.}
Fix now $\theta \in S(2\varepsilon)$, and define
\begin{equation}\label{as}
a_{s}:=\int_{0}^{+\infty}f^{(s)}(te^{i\theta})e^{i\theta}dt
= \lim_{R\to\infty}\int_{0}^{R}f^{(s)}(te^{i\theta
})e^{i\theta }dt.
\end{equation}
By (\ref{estders}), it is not difficult to see that
$a_{s}\in\mathbb{C}$. Moreover, the definition of $a_{s}$ is
independent of $\theta$. Indeed, integrating $f^{(s)}(\zeta )$
along the sector boundary $0\to Re^{i\phi }\to
Re^{i\theta }\to 0$, and using (\ref{estders}) to conclude
that the integral of $f^{(s)}(\zeta )$ over the arc $[Re^{i\phi
},Re^{i\theta }]$ tends to zero as $R\to\infty$, the
independence from $\theta$ immediately follows. Define now
$b_{s-1}:=f^{(s)}(0)+a_{s}$, and suppose that $b_{s-1}\neq 0$. Let
$z=re^{i\phi}$ be an arbitrary point in $S(2\varepsilon )$ such
that $r\geq r_{4}$. Then, since
\begin{equation}\label{erotus}
 f^{(s-1)}(z)-b_{s-1}=\int_{0}^{z}f^{(s)}(\zeta )d\zeta -
 \int_{0}^{+\infty}f^{(s)}(te^{i\phi })e^{i\phi }dt,
\end{equation}
we may apply (\ref{estders}) and Lemma \ref{integral} to conclude
that
\begin{equation}\label{estb}
\begin{split}
 |f^{(s-1)}(z)-b_{s-1}|
& = \Bigl|\int_{0}^{z}f^{(s)}(\zeta )d\zeta
-\int_{0}^{\infty}f^{(s)}(te^{i\phi })e^{i\phi }dt\Bigr|
\\
& = \Bigl|\int_{\infty }^{z}f^{(s)}(te^{i\phi })e^{i\phi
}dt\Bigr|\leq \int_{|z|}^{\infty}|f^{(s)}(te^{i\phi })|dt
\\
& \leq \int_{r}^{\infty}\exp (-(1-\delta )\alpha t^{\beta})dt \\
&\leq \exp (-(1-2\delta )\alpha r^{\beta}),
\end{split}
\end{equation}
provided that $r\geq r_{4}$. Since we assumed that $b_{s-1}\neq
0$, we have completed the proof of the first assertion in this
case.
\smallskip

\noindent\textbf{Proof of the first assertion for $j<s-1$, the first part.}
We now have $b_{s-1}=0$. To continue, we define $a_{s-1}$
replacing $f^{(s)}$ by $f^{(s-1)}$ in (\ref{as}), and
$b_{s-2}:=f^{(s-2)}(0)+a_{s-1}$. To estimate
$f^{(s-2)}(z)-b_{s-2}$, we apply Lemma \ref{integral} and
$|f^{(s-1)}(z)|\leq \exp (-(1-2\delta )\alpha r^{\beta})$ in place
of (\ref{estders}) exactly as in (\ref{estb}) to obtain
\begin{equation}\label{as-1}
 |f^{(s-2)}(z)-b_{s-2}|\leq \exp (-(1-3\delta )\alpha r^{\beta}).
\end{equation}
for $r\geq r_{4}$.

We may now continue inductively. If $b_{j}\neq 0$ for some
$j=s-t$, $t = 2,\ldots ,s-1$, we obtain
\begin{equation}\label{as-k}
 |f^{(s-t)}(z)-b_{s-t}|\leq \exp (-(1-(t+1)\delta )\alpha
 r^{\beta}).
\end{equation}
Otherwise, we have $b_{s-1}=b_{s-2}=\dots =b_{1}=0$, and we have
the estimate
\begin{equation}\label{est0}
 |f(z)-b_{0}|\leq \exp (-(1-(s+1)\delta )\alpha |z|^{\beta}).
\end{equation}
If now $b_{0}\neq 0$, we have proved the first assertion. It
remains to show that the case $b_{0}=0$ is not possible.
\smallskip

\noindent\textbf{Proof of the first assertion for $j<s-1$, the second part
(impossibility of $b_{s-1}=\dots =b_{0}=0$).}
(a) First step. Writing now (\ref{lin}) in the form
\begin{equation}\label{linmod2}
\begin{split}
-\frac{f^{(s)}}{f}=&\frac{A_{0}}{A_{s}}+\frac{A_{1}}{A_{s}}
\frac{f'}{f}+\dots+\frac{A_{s-1}}{A_{s}}\frac{f^{(s-1)}}{f}
\\
&+\frac{A_{s+1}}{A_{s}}\frac{f^{(s+1)}}{f}+\dots
+\frac{A_{k-1}}{A_{s}}\frac{f^{(k-1)}}{f}
+\frac{1}{A_{s}}\frac{f^{(k)}}{f},
\end{split}
\end{equation}
we may use (\ref{large}), (\ref{small}) and (\ref{dersmall2}) to
conclude that
\begin{equation}\label{ders0}
 \Big|\frac{f^{(s)}(z)}{f(z)}\Big|\leq \exp (-(1-\delta )\alpha |z|^{\beta})
\end{equation}
in $S(2\varepsilon)\setminus E$. Therefore, by (\ref{ders0}) and
(\ref{est0}) with $b_{0}=0$, we infer that
\begin{equation}\label{estders2}
 |f^{(s)}(z)|\leq \exp (-(2-(s+2)\delta )\alpha r^{\beta})
\end{equation}
in $S(2\varepsilon)\setminus E$, hence in $S(2\varepsilon
+\varepsilon /2)$ by the Phragm\'{e}n--Lindel\"{o}f principle.

\noindent (b) Inductive step. Suppose now that we have been able to prove
that the estimate
\begin{equation}\label{estderT1}
 |f^{(s)}(z)|\leq \exp (-(T-((T-1)s+T)\delta )\alpha |z|^{\beta})
\end{equation}
holds good in the sector $S(2\varepsilon
+\sum_{j=1}^{T-1}\frac{\varepsilon}{2^{j}})$. Combining now Lemma
\ref{integral} with (\ref{estderT1}), we may repeat the reasoning
in (\ref{estb}) to obtain
\begin{equation}\label{estbT}
 |f^{(s-1)}(z)|\leq \exp (-(T-((T-1)s+T)\delta -\delta )\alpha
 r^{\beta}).
\end{equation}
Since $b_{s-1}=\dots =b_{0}=0$, we apply a parallel reasoning as
in 4.6. of this proof above to get
\begin{equation}\label{fT}
 |f(z)|\leq \exp (-(T-((T-1)s+T)\delta -s\delta )\alpha r^{\beta}),
\end{equation}
valid in $S(r_{4},2\varepsilon +\sum_{j=1}^{T-1}\frac{\varepsilon
}{2^{j}})$. Combining now (\ref{fT}) with (\ref{ders0}), we obtain
$$|f^{(s)}(z)|\leq \exp (-(T+1-(Ts+T+1)\delta )\alpha |z|^{\beta})$$
in $S(2\varepsilon
+\sum_{j=1}^{T-1}\frac{\varepsilon}{2^{j}})\setminus E$, provided
$r\geq r_{4}$. By the Phragm\'{e}n--Lindel\"{o}f principle, this
inequality remains valid in the whole sector $S(2\varepsilon
+\sum_{j=1}^{\infty}\frac{\varepsilon}{2^{j}})=S(3\varepsilon )$,
completing the inductive step.

\noindent
(c) Final conclusion. We have proved that, in this special case of
$b_{s-1}=\dots = b_{0}=0$, the inequality (\ref{estderT1}) is
valid in $S(3\varepsilon )$ for all $T\in \mathbb{N}$, provided
$r\geq r_{4}$. Fix now a finite line segment in
$S(r_{4},3\varepsilon)$. Since $k\delta <1$, and $s+1\leq k$, it
follows that $T-((T-1)s+T)\delta \to\infty$ as
$T\to\infty$. Hence, $f^{(s)}$ vanishes identically on
such a line segment. Therefore, by the standard uniqueness theorem
of entire functions, $f$ has to be a polynomial, a contradiction.
\end{proof}


\begin{proof}[Proof of Theorem \ref{Gary2}]
Suppose that $f$ is a
transcendental solution of (\ref{lin}) of finite order of growth.
Given $\varepsilon >0$, let $(\phi_{m})$ and $(\theta_{m})$ be as
in the assumptions. From (\ref{suuri}) and the supposition that
$\rho (A_{j})<\beta$ whenever $j\neq s$, we conclude by using
Theorem \ref{Gary}(ii) that $|f^{(s)}(z)|$ is bounded in each of
the sectors $\phi_{m}+3\varepsilon\leq\arg
z\leq\theta_{m}-3\varepsilon, m=1,\ldots ,n$. As $\varepsilon$ is
arbitrarily small, we infer from (\ref{epsilon}) and the
Phragm\'{e}n--Lindel\"{o}f principle that $|f^{(s)}(z)|$ must be
bounded in the whole complex plane. By the Liouville theorem, $f$
has to be a polynomial, a contradiction.
\end{proof}

\begin{proof}[Proof of Theorem \ref{triv}]
 Suppose $f$ is a transcendental
solution of (\ref{lin}) such that $\rho_{p}(f)<\rho_{p}(A_{s})$.
Writing (\ref{lin}) in the form
\begin{equation}\label{mod}
 A_{s}(z)f^{(s)}=-\sum_{j=0,j\neq s}^{k}A_{j}(z)f^{(j)},
\end{equation}
and making use of the elementary iterated order (in)equalities and
the invariance of the iterated order under differentiation, we
immediately observe that the left hand side of (\ref{mod}) is of
iterated order $\rho_{p}(A_{s})$, while the right hand side must
be of iterated order $\leq \max_{j\neq
s}(\rho_{p}(f),\rho_{p}(A_{j}))<\rho_{p}(A_{s})$, a contradiction.
\end{proof}

\noindent\textbf{Acknowledgement.} The authors would like to thank the
anonymous referee for his/her careful reading of the manuscript.

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\end{thebibliography}


\end{document}