\documentstyle[twoside]{article}
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\markboth{\hfil Existence of Positive Solutions\hfil EJDE--1995/10}%
{EJDE--1995/10\hfil  M. Garc\'\i a-Huidobro, R. Man\'asevich \&
P. Ubilla  \hfil}
\begin{document}
\ifx\Box\undefined \newcommand{\Box}{\diamondsuit}\fi
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 1995}(1995), No. 10, pp. 1--22. \newline
ISSN 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113 }
 \vspace{\bigskipamount} \\
Existence of Positive  Solutions for some Dirichlet Problems with 
an Asymptotically Homogeneous Operator
\thanks{ {\em 1991 Mathematics Subject Classifications:}
 35J65.\newline\indent
{\em Key words and phrases:} Dirichlet Problem, Positive Solution, Blow up. 
\newline\indent
\copyright 1995 Southwest Texas State University  and University of
North Texas.\newline\indent
Submitted February 12, 1995. Published August 11, 1995.
\newline\indent
Supported by grants CI $1^*$-CT93-0323 from  EC, and 1940409-94 
from Fondecyt \newline\indent (MG-H and RM).
} }
\date{}
\author{Marta Garc\'\i a-Huidobro, Raul Man\'asevich \&
Pedro Ubilla} 
\maketitle

\begin{abstract}
Existence of positive radially symmetric solutions to a 
Dirichlet problem of the form
\begin{eqnarray*}&-{\rm div} (A(|Du|)Du)= f(u)& \mbox{ in } \Omega \\
   &u = 0& \mbox{ on } \partial\Omega \end{eqnarray*}
is studied by using blow-up techniques. It is proven here that by 
choosing the functions $sA(s)$ and $f(s)$ among a certain class called 
{\em asymptotically homogeneous}, 
the blow-up method still provides the a-priori bounds for positive
solutions. Existence is proved then by using degree theory.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{coro}{Corollary}[section]
\newcommand{\dis}{\displaystyle}

\section{Introduction}
In this paper we consider the existence of positive  radially
symmetric solutions for the problem
$$(D)\quad\cases{-{\rm div} (A(|Du|)Du)=f(u) &in $\Omega$\cr
 \cr \qquad\qquad\qquad\quad\ u = 0 &on $\partial\Omega$}$$
where $\Omega=B(0,R),$ $ R>0$, is the ball of radius $R$ in ${\Bbb R}^N$
and the function $f:{\Bbb R}\to{\Bbb R}$ is continuous.
For some functions $A:{\Bbb R}\to{\Bbb R}$, the radial solutions of $(D)$ 
satisfy the 
nonlinear boundary value problem
$$(D_r)\quad\cases{-(r^{N-1}\phi(u'))' 
= r^{N-1} f(u) &in $(0,R)$\cr \cr
\qquad\quad\ u'(0)= 0= u(R) & }$$
where $r=|x|,\ x\in {\Bbb R}^N$ and $\phi:{\Bbb R}\to{\Bbb R} $
is an odd increasing homeomorphism of ${\Bbb R}$, that is, an
odd increasing homeomorphism from $\Bbb R$ onto $\Bbb R$, given by $\phi(s)=sA(s).$
In $(D_r)$,\  $'$ denotes  derivative with respect 
to $r.$ 

In the rest of the paper we will deal with problem $(D_r)$ in the 
\lq\lq superlinear\rq\rq\  case,
that is, when
\begin{equation}
\lim_{s\to+\infty}\frac{f(s)}{\phi(s)}=+\infty.\label{superlinear}
\end{equation}
and $\phi,f$ belong to a class of functions to be described later.

By a solution to 
this problem   we will understand a function
$u\in C^1[0,R]$ with $\phi(u')\in C^1[0,R]$ and such that  $(D_r)$ is satisfied.
  

It is well known that for the homogeneous 
case, that is when $\phi(s)=|s|^{p-2}s$, $p>1,$  
 the use of blow up 
techniques allows to transform 
the question of a-priori bounds for 
positive  solutions to some superlinear problems  into 
a problem of non-existence of positive 
solutions in ${\Bbb R}^N$ for a certain limiting 
equation. This  limiting equation having the same left hand side 
nonlinear operator as the original equation, due to the homogeneity.
See \cite{GS} for the case of a scalar 
equation and $p=2,$ and  \cite{CMM} for the case of a system of 
$p,q$-Laplacians. 
See also  
\cite{PvV} for related results.

The natural question of whether this method can be extended to cover 
the radial situation posed by problem  
$(D_r),$ when the function $\phi$ is no 
longer homogeneous arises.

We give here a positive answer to this question by restricting
 the functions $\phi, f$ to a special class.
 This,  being strongly motivated by some 
 previous works done
for the one  dimensional case, see \cite{GMZ1}, \cite{GMZ2},  
\cite{GMZ3}, and \cite{U}. 

We now describe the class of functions $\phi,f$ we will consider in order to 
formulate 
our model problem. Throughout 
this paper we will assume that $\phi$ is an odd homeomorphism of ${\Bbb R}$
satisfying
\begin{eqnarray}
\lim_{|s|\to\infty}\frac{\phi(\sigma s)}{\phi(s)}=\sigma^{p-1}\quad\mbox{for 
all}\quad
\sigma\in{\Bbb R}_+
\label{phicond}
\end{eqnarray}
for some $p>1$ and $f$ satisfies
\begin{eqnarray}
\lim\limits_{s\to +\infty}\frac{f(\sigma s)}{f(s)}
=\sigma^{\delta}
\mbox{\qquad for all\qquad} \sigma\in{\Bbb R}_+\label{fcond}
\end{eqnarray}
for some $\delta>0$, where ${\Bbb R}_+:=[0,+\infty)$.

Conditions of this type, even without the monotonicity assumption on $\phi,$
have been very much used in Applied Probability in a different context 
than the one 
we will do here, see for instance
\cite{R}, \cite{S} and the references therein.
Indeed from \cite{R} or  \cite{S} we have the following general definition.
Let $h:{\Bbb R}_+\to {\Bbb R}_+$ be a measurable function that satisfies
\begin{eqnarray}
\lim\limits_{s\to +\infty}\frac{h(\sigma s)}{h(s)}
=\sigma^{q}
\mbox{\qquad for all\qquad} \sigma\in{\Bbb R}_+.\label{aho}
\end{eqnarray}
We will say then  that $h$ is {\it asymptotically homogeneous 
with index $q$},  for short AH or $q$-AH. We point out that
in \cite{R}, \cite{S}, functions $h$ satisfying (\ref{aho}) are called 
{\em regularly varying of index $q$}, nevertheless from our point of view it 
will be 
more illustrative to call them  asymptotically homogeneous.

In this sense if the function $\phi(s)=sA(s)$
in $(D_r)$ is AH of index $p-1,$ we will say that the corresponding 
operator in $(D_r)$ or $(D)$ is an {\em asymptotically homogeneous operator}.
 

Thus, with this notation, we will require that $\phi$ be a $(p-1)$-AH odd
homeomorphism of $\Bbb R$ for some $p>1$ and that the continuous function $f$ 
be $\delta$-AH for some $\delta>0$. 
 

Condition (\ref{superlinear}) implies that $\delta\ge p-1$, see Section 2. 
We observe that the case $\delta=p-1$ is indeed allowed as the 
following example shows:
\begin{eqnarray*}
\phi(s)=|s|^{p-2}s\frac{|s|}{\sqrt{1+s^2}}\ ,\quad
f(s)=|s|^{p-2}s\log(1+|s|).
\end{eqnarray*}
 

For later use we define the following functions
\begin{eqnarray}
\Phi (s) = \int^s_0\phi(\tau)d\tau\ ,\qquad\qquad
\Phi_*(s) = \int^s_0\phi^{-1}(\tau)d\tau.\label{1.2}
\end{eqnarray}
and following \cite{FP} or \cite{PS} we define the Legendre 
transform of $\Phi$ by
\begin{eqnarray}
H(s)= s\phi(s) - \Phi(s).\label{1.3}
\end{eqnarray}
Also we will set 
\begin{eqnarray*}
\phi_p(s)={|s|}^{p-2}s \mbox{\quad for all\quad} s\in{\Bbb R} 
\mbox{\quad and\quad}p>1,
\end{eqnarray*}
and  $\dis {p^*=\frac{p}{p-1}}.$ 
 

We end this section by establishing the organization of this paper. 
In Section~2 we establish and prove our main result for existence of 
positive  solutions. In Section 3
we show some examples that illustrate our results, and finally in the Appendix,
we prove some properties of the class of
asymptotically homogeneous functions that we use  throughout this paper. 



\section{Existence of positive solutions.}

In this section we will show that problem $(D_r),$ that we recall next
$$(D_r)\quad\cases{-(r^{N-1}\phi(u'))' = r^{N-1}f(u) &in $(0,R)$\cr\cr
\qquad\quad\ u'(0)=0=u(R),&}$$
has positive solutions
assuming that the homeomorphism $\phi$  satisfies (\ref{phicond}) and 
the nonlinearity $f$ satisfies (\ref{fcond}), $sf(s)\ge 0$ for $s\ge 0$ and 
it is superlinear with respect to 
 $\phi$.
 
Let $\Phi, \Phi_*$ and $H$ as
in  (\ref{1.2}) and (\ref{1.3}) respectively. We
have that
\begin{eqnarray}
H(s) = \Phi_* (\phi(s)),\label{3.0}
\end{eqnarray}
and that  $H$ is an even function of $s$. Since $\Phi_*$ is $p^*$-AH and $\phi$ 
is $(p-1)$-AH, it is easy to see,
by using Proposition \ref{phi-1lim} in the Appendix that
\begin{equation}
\lim\limits_{s\to+\infty}\frac{H(\sigma s)}{H(s)}=\sigma^p 
\mbox{\quad for all }\sigma\in {\Bbb R}_+,\label{Hconv}
\end{equation}
and hence $H$ is p-AH.
 

Let $F(s)=\int_0^sf(\tau)d\tau.$ We note that from Karamata's theorem, 
see \cite[Theorem 0.6]{R}, by using (\ref{phicond}) and (\ref{fcond}) it 
follows  that
\begin{eqnarray} 
\lim\limits_{s\to +\infty}\frac{\Phi(s)}{s\phi(s)}=\frac{1}{p}
\quad\mbox{and}\quad
\lim\limits_{s\to +\infty}\frac{F(s)}{sf(s)}=\frac{1}{\delta+1}.
\label{quotient}
\end{eqnarray}
Now, from (\ref{quotient}), given $\varepsilon>0$ there exists 
$s_0>0$ such that for all $s\ge s_0 $
$$ \delta+1-\varepsilon<\frac{sF'(s)}{F(s)}<\delta+1+\varepsilon.$$
Hence, solving this differential inequality we obtain
\begin{eqnarray}
A_1s^{\delta-\varepsilon}\le f(s)\le A_2s^{\delta+\varepsilon}\quad\mbox
{for all $s\ge s_0.$}\label{hof}
\end{eqnarray} 
Similarly, it can be shown that
\begin{eqnarray}
A_3s^{p-1-\varepsilon}\le \phi(s)\le A_4s^{p-1+\varepsilon}\quad\mbox
{for all $s\ge s_0$}\label{hophi}
\end{eqnarray}
where $A_i:=A_i(\varepsilon)>0\ \ i=1,\ldots,4.$ Hence (\ref{hof}) and  
(\ref{hophi})  yield that 
\begin{eqnarray}
 \frac{f(s)}{\phi(s)}
\le \frac{A_2}{A_3}s^{\delta-(p-1)+2\varepsilon}
\label{foverg}
\end{eqnarray}
and thus we see that (\ref{superlinear}) 
implies that $\delta\ge p-1$.
 

Next, and for later purposes we consider the equation
\begin{eqnarray}
H(z)= F(s).\label{3.1}
\end{eqnarray}

Since $F$ is strictly increasing for $s$ greater than some $s_0>0,$ and 
$F(s)\to +\infty$ as $s\to +\infty$ it is clear that  for each $s> s_0,$
equation (\ref{3.1}) will have a unique solution which we denote by
  $z(s).$  
Define now
$g:(s_0, +\infty)\to {\Bbb R}^+$, where ${\Bbb R}^+:=(0,+\infty)$ by
\begin{eqnarray}
g(s)= \frac{z(s)}{s}.\label{3.2}
\end{eqnarray}
We have the following proposition.

\begin{prop}
If (\ref{superlinear}) holds, then $g(s)\to+\infty$ as $s\to+\infty$.
\label{prop3.1}
\end{prop}

\paragraph{Proof.} We have that
$$F(s)=H(sg(s)) $$
thus, if  $g(s_n)\le M$ for some sequence $\{s_n\}$,  $s_n\to+\infty$, then 
\begin{eqnarray}
\frac{F(s_n)}{H(s_n)}\le \frac{H(s_nM)}{H(s_n)}\to M^p \quad{\rm as}\quad n\to 
\infty. 
\label{3.4}\end{eqnarray}
On the other hand, 
$$\frac{F(s)}{H(s)}=\frac{F(s)}{\Phi(s)}\cdot \frac{\Phi(s)}{\Phi_*(\phi(s))}$$
and
$$\frac{\Phi(s)}{\Phi_*(\phi(s))}=\frac{\Phi(s)}{s\phi(s)-\Phi(s)}=
\frac{1}{\frac{s\phi(s)}{\Phi(s)}-1},$$
hence, from (\ref{quotient}),
$$\lim_{s\to+\infty}\frac{\Phi(s)}{\Phi_*(\phi(s))}=\frac{1}{p-1}.$$
Thus, (\ref{3.4})  contradicts (\ref{superlinear}) 
by L'H\^opital's rule.
 

Now we establish our main existence result.

\begin{theorem}
Suppose that $\phi$ is an increasing odd homeomorphism of ${\Bbb R},$  
$f:{\Bbb R}\to {\Bbb R}$ is continuous , satisfies $sf(s)\ge 0,$ 
and is ultimately increasing.
Assume  also that $\phi$ and $f$ satisfy the superlinear condition
(\ref{superlinear}) and that there exist $p$, with $1<p<N,$ and $\delta>0$
 such that
\begin{itemize}
\item[\rm{(i)}] $\dis\lim\limits_{s\to+\infty}\frac{\phi(\sigma s)}{\phi(s)} =
\sigma^{p-1}$ 
{\rm\quad for all}\quad $\sigma\in {\Bbb R}_+.$
\item[\rm{(ii)}] 
 $\dis\lim_{s\to +\infty}\frac{f(s\sigma)}{f(s)}
=\sigma^{\delta},$ \quad{\rm for all\quad} 
 $\sigma\in {\Bbb R}_+.$  
\item[\rm{(iii)}] $\dis\lim_{t\to 0}\frac{\phi(t)}{f(t)}=+\infty$ 
{\rm\quad and \quad}  $\dis\liminf\limits_{t\to 0}
\frac{\phi(t\sigma)}{\phi(t)}>0,$ 
\quad{\rm for every\quad}  $\sigma\in {\Bbb R}^+,$ 
\item[\rm{(iv)}] $\dis\delta<\frac{N(p-1)+p}{N-p}.$
\end{itemize} 

Then problem $(D_r)$ has a positive solution.
\label{main}
\end{theorem}
 


The proof of this theorem will be done in three steps. 
We note that finding positive solutions of problem $(D_r)$ 
is equivalent to finding nontrivial solutions to the problem


$$(A)\qquad\cases{-(r^{N-1}\phi(u'))'= r^{N-1}f(|u|) & in $ (0,R)$\cr
\qquad\quad u'(0)=0=u(R) & }$$
Indeed, if $u(r)$ is a nontrivial solution of $(A)$, then $u'(R)<0$
for all $r\in(0,R)$ and since $u(R)=0$ we find that 
$u(r)>0$, for all $r\in(0,R)$. 
This shows that $u(r)$ is a positive solution of problem $(D_r)$.
 

\paragraph{ Step 1.  Abstract formulation of problem $(A)$.} 
 Let $C_\#$ denote the closed subspace of $C[0,R]$ defined by
$$C_\# = \{u\in C[0,R]\ :\ u(R)=0\},$$
then $C_\#$ is a Banach space for the norm $\|\ \|:=\|\ \|_\infty.$

By direct verification it can be seen that $u$ is a solution to $(A)$ if and 
only if $u$ is a fixed point of the operator 
 $T_0:C_\#\to C_\# $ defined by
\begin{eqnarray}
T_0(u)(r)=\int^R_r \phi^{-1}\Bigl[\frac{1}{s^{N-1}}\int^s_0
\xi^{N-1}f(|u(\xi)|) d\xi\Bigr]ds.\label{3.12}
\end{eqnarray}
  Define now the operator
$T: C_\#\times {\Bbb R}_+\to C_\#,$
by
\begin{eqnarray}
T(u,\tau)(r)=\int^R_r\phi^{-1}\Bigl[
\frac{1}{s^{N-1}}\int^s_0 \xi^{N-1}(f(|u(\xi)|)+\tau) d\xi\Bigr]ds\label{3.13}
\end{eqnarray}

We have that $T$ sends bounded sets of $C_\#\times {\Bbb R}_+$ into bounded
sets of $C_\#$ and that $T(u,0)=T_0(u).$ We prove now the following.

\begin{prop} 
The operator $T$ is completely  continuous.\label{prop3.2}
\end{prop}

\paragraph{Proof.} Let $\{(u_n,\tau_n)\}$ be a bounded sequence in
$C_\#\times{\Bbb R}_+$\ , say
\begin{eqnarray}
\|u_n\|_\infty+\tau_n\le C, \mbox{\quad  for all } n\in{\Bbb N},\label{3.7+}
\end{eqnarray}
and set
$$v_n=T(u_n,\tau_n),\quad n\in{\Bbb N}.$$
We want to show that $\{ v_n\}$ has a convergent subsequence.
By (\ref{3.7+}) and (\ref{3.13}),  $v_n\in 
C^1[0,R],$ for all $ n\in{\Bbb N}$ and satisfies
\begin{eqnarray}
|\phi(v'_n(r))|&=& 
\frac{1}{r^{N-1}}\int^r_0\xi^{N-1}
(f(|u_n(\xi)|) +\tau_n) d\xi\nonumber\\
 &\le& \frac{\tilde C R}{N},\nonumber
\end{eqnarray}
where $\tilde C$ is a positive constant. Thus the sequence $\{v'_n\}$ is  
bounded and since  the
sequence $\{v_n\}$ is  bounded also, the existence of  a
convergent subsequence follows from the Ascoli Arz\`ela's Theorem.

To show now that $T$ is continuous, let $\{(u_n,\tau_n)\}$ be a sequence
in $C_\#\times{\Bbb R}_+$ converging to $(u,\tau)\in C_\#\times{\Bbb R}_+$ and
 set
\begin{eqnarray*}
v_n(r)=\int^R_r h_n(s)ds\quad n\in{\Bbb N},
\end{eqnarray*}
where
\begin{eqnarray*}
h_n(s) =\phi^{-1}\Bigl[\frac{1}{s^{N-1}}\int^s_0 \xi^{N-1}
(f(|u_n(\xi)|)+ \tau_n) d\xi\Bigr].
\end{eqnarray*}
Clearly we have that $h_n(s)\to h(s)$ for each
$s\in[0,R],$  where
\begin{eqnarray*}
h(s):=\phi^{-1}\Bigl[\frac{1}{s^{N-1}}\int^s_0 \xi^{N-1}
(f(|u(\xi)|)+ \tau) d\xi\Bigr].
\end{eqnarray*}
Thus, since  $\{h_n\}$ is  bounded, it follows 
from Lebesgue's dominated convergence theorem that
$$\|h_n-h\|_{L^1(0,R)}\to 0\mbox{ as } n\to \infty.$$
If
$$v(r):= \int^R_r h(s)ds,$$
then
$$\|v_n-v\|\le \|h_n-h\|_{L^1(0,R)}$$
and hence
$$T(u_n,\tau_n) = v_n\to v= T(u,\tau)\quad \mbox{ as } n\to\infty.$$
This concludes the proof of proposition \ref{prop3.2}. $\Box$

\paragraph{Step 2. A-priori bounds.} 
We will show here that solutions $(u,\tau)\in C_\#\times {\Bbb R}_+$ 
of the equation
\begin{eqnarray}
u=T(u,\tau)\label{3.17}
\end{eqnarray}
are a priori bounded. This will be done by using {\it blow-up }techniques.
 

We first prove the following.

\begin{prop}
Suppose that there exists a sequence $\{(u_n,\tau_n)\}$ of solutions of 
(\ref{3.17})
such that
\begin{eqnarray}
\|u_n\|+\tau_n\to\infty\mbox{ as } n\to\infty,\label{3.18}
\end{eqnarray}
 then
\begin{itemize}
\item[\rm{(i)}] $\|u_n\|\to\infty$ as $n\to\infty$; and
\item[\rm{(ii)}]  $\dis \frac{\tau_n}{f(\|u_n\|)}\to 0.$
\end{itemize}\label{prop3.3}
\end{prop}

\paragraph{Proof.} We have that for each $n\in{\Bbb N}$ the pair 
$ (u_n,\tau_n)$ satisfies
\begin{eqnarray}
u_n(r)=\int^R_r\phi^{-1}\Bigl[\frac{1}{s^{N-1}}\int^s_0\xi^{N-1} (f(|u_n|)+
\tau_n)d\xi\Bigr]\ ds\nonumber
\end{eqnarray}
and thus
\begin{eqnarray}
||u_n||=u_n(0)\ge \int^R_{R/2}\phi^{-1}
(\frac{s\tau_n}{N})\ ds\ge \frac{R}{2}\phi^{-1}(\frac{R\tau_n}{2N})
\label{3.18'}
\end{eqnarray}
from which (i) follows. To show (ii) we have that by (\ref{3.18'})
and large $n,$

\begin{eqnarray}
\frac{R}{2N}\frac{\tau_n}{f(\|u_n\|)}\le
\frac{\phi(\frac{2}{R}\|u_n\|)}{f(\|u_n\|)}=
\frac{\phi(\frac{2}{R}\|u_n\|)}{\phi(\|u_n\|)}
\frac{\phi(\|u_n\|)}{f(\|u_n\|)}
.\label{3.20}
\end{eqnarray}
Thus by (\ref{superlinear}), and the fact that $\phi$ is (p-1)-AH,
 we obtain that 

$$\frac{\tau_n}{f(\|u_n\|)}\to 0\mbox{ as }n\to\infty.\quad \Box$$.
 

We will now prove that solutions to (\ref{3.17}) are {\em a-priori} bounded.
 

\begin{lemma} Suppose $(u,\tau)\in C_\#\times{\Bbb R}_+$ is 
a solution
of (\ref{3.17}), then there is a constant $C$, independent of 
$u$ and $\tau$, such that
\begin{eqnarray}
\|u\|+\tau\le C\label{3.22}
\end{eqnarray}\label{aprioribound}
\end{lemma}

\paragraph{Proof.} We argue by contradiction and thus we assume there
is a sequence $\{(u_n,\tau_n)\}$ in $C_\#\times {\Bbb R}_+$ such that
$(u_n,\tau_n)$ satisfies (\ref{3.17}) and such that
$$\|u_n\|+\tau_n\to\infty\mbox{ as } n\to\infty.$$
In order to simplify the writing, we set $t_n:=||u_n||$ and $z_n:=z(t_n)$ 
for each
$n\in{\Bbb N},$  where the function $z(\cdot)$ is defined in (\ref{3.1}).  Let 
us 
consider the change of variables
$$\cases{\dis \qquad y&$\dis=\quad\frac{z_n}{t_n}r$\cr
\dis w_n(y)&$\dis=\quad\frac{u_n(r)}{t_n}$.\cr}$$
Then from (\ref{3.17}), we find that $w_n$ satisfies

\begin{eqnarray}
-\frac{d}{dy}(y^{N-1}\phi(z_n\dot w_n(y)))z_n = t_ny^{N-1} 
(f(t_n w_n(y))+\tau_n)  \label{3.23}\\
w_n(0) =1, \ \dot w_n(0)=0,\ w_n(R\frac{z_n}{t_n})=0,\qquad\quad\label{3.23'}
 \end{eqnarray}
where here and henceforth $\dis \dot {(\ )}:=\frac{d}{dy}(\ ).$ We note that
$\dis \frac{z_n}{t_n}= g(t_n),$ with
$g$ as defined in (\ref{3.2}), and thus from propositions \ref{prop3.1} and 
 \ref{prop3.3}, 
$\dis\frac{z_n}{t_n}\to+\infty$
as $n\to\infty$.

Let $M_0>0$ be a constant. In the next argument we will suppose that 
$Rg(t_n)>M_0,$ for all $n,$ by passing to a subsequence if necessary.

Dividing both sides of (\ref{3.23}) by $y^{N-1}$, we may re-write it as
\begin{eqnarray*}
\frac{d}{dy}[\phi(z_n\dot w_n(y))]z_n + 
(f(t_n w_n(y)) +\tau_n)t_n
= -\frac{N-1}{y}\phi(z_n\dot w_n(y))z_n,
\end{eqnarray*}
and on multiplying both sides of this equation by $\dot w_n$ we 
obtain that
\begin{eqnarray}
\frac{d}{dy}[H(z_n\dot w_n(y))+F(t_nw_n(y)) +
\tau_nt_n w_n(y)]\le 0.\label{3.24}
\end{eqnarray}
Hence, by integrating (\ref{3.24}) on $(0,y)$, we find 
that
\begin{eqnarray*}
H(z_n\dot w_n(y)) +F(t_nw_n(y))+\tau_nt_n w_n(y) 
\le F(t_n)+\tau_nt_n,
\end{eqnarray*}
and thus
\begin{eqnarray*}
H(z_n\dot w_n(y))&\le & F(t_n)\Bigl[1+\frac{\tau_nt_n}{ F(t_n)}\Bigr]\\
&=&H(z_n)\Bigl[1+\frac{\tau_n}{f(t_n)}\frac{t_nf(t_n)}{ F(t_n)}\Bigr].
\end{eqnarray*}
But $\dis {\tau_n\over f(t_n)}\to 0$  and 
$\dis {\frac{t_n f(t_n)}{F(t_n)}}
\to\delta+1$
as $n\to\infty$, 
hence  there exists a constant $C>0$ such that
$$H(z_n \dot w_n(y))\le C H(z_n).$$
This implies that 
\begin{eqnarray}
|\dot w_n(y)|\le\frac{H^{-1}(CH(z_n))}{H^{-1}(H(z_n))},
\label{3.27}
\end{eqnarray}
and since from (\ref{Hconv}) and Proposition \ref{phi-1lim} in the Appendix
$$\frac{H^{-1}(Cs)}{H^{-1}(s)}\to C^{\frac{1}{p}}
\mbox{\quad as\quad}s\to+\infty,$$
there exists a constant $C_1>0$ such that
$$|\dot w_n(y)|\le C_1\quad\mbox{ for all }\quad n\in{\Bbb N} 
\quad\mbox{ and all }\quad y\in [0, M_0].$$
Thus  the sequence $\{w_n\}$ is equicontinuous. Since it is also 
uniformly bounded, an application of Ascoli 
Arz\`ela's theorem  yields that $\{w_n\}$ contains a convergent 
subsequence, 
which we denote again by $\{w_n\}$, say $w_n\to w $  in 
$C[0, M_0]$ as $n\to\infty$. Integrating (\ref{3.23})
on $[0,y]\subset [0,M_0]$, we find that

\begin{eqnarray}
-\phi(z_n\dot w_n(y)) = \frac{t_nf(t_n)}{z_n}h_n(y)\label{3.28}
\end{eqnarray}
where
\begin{eqnarray}
h_n(y) = \frac{1}{y^{N-1}} \int^y_0 s^{N-1} 
\Bigl(\frac{f(t_n w_n(s))}{f(t_n)}+\frac{\tau_n}{f(t_n)}\Bigr) \ ds.
\label{ma5}
\end{eqnarray}
We show now that 
$\{ h_n(y)\}$ is a convergent sequence for each 
$y\in [0,M_0]$ by an application of
 Lebesgue's dominated convergence theorem.
 

Using that $f$ is ultimately increasing, say for $x\ge x_1>0,$
we have that  there is a $x_0>0$ such that 
\begin{eqnarray}
\frac{f(\sigma x)}{f(x)} \le 1 \label{ma1}
 \end{eqnarray}
for all $x\ge x_0$ and for all $\sigma\in [0,1].$
Indeed 
there is a unique $x_0\ge x_1$ such that
\begin{eqnarray*}
f(x_0)=\max\limits_{s\in [0,x_1]}f(s):=M.
\end{eqnarray*}
Let $\sigma\in (0,1)$ and consider the term $\dis \frac{f(\sigma x)}{f(x)}$
for $x\ge x_0.$ If $\sigma x\ge x_0,$ then $f(\sigma x)\le f(x)$
and thus (\ref{ma1}) holds.
If now $\sigma x< x_0,$ then 
\begin{eqnarray*}
f(\sigma x)\le M =f(x_0)\le f(x)
\end{eqnarray*}
and again (\ref{ma1}) holds.
 

Thus from (\ref{ma1}) we have that 
\begin{eqnarray}
\frac{f(t_n w_n(s))}{f(t_n)} \le 1\label{ma6}
\end{eqnarray}
for all $s\in [0, M_0]$ and large $n.$
In particular this implies that $\{ h_n(y)\}$ is a bounded sequence.
We will show next that for each $s_0\in [0,M_0]$
\begin{eqnarray}
\lim\limits_{n\to \infty}\frac{f(t_n w_n(s_0))}{f(t_n)}=(w(s_0))^\delta.
\label{ma2}
\end{eqnarray}
We know that $w_n(s_0)\to w(s_0)$ as $n\to\infty.$  If $w(s_0)>0$,
then for large $n,$ $t_nw(s_0)>x_0$ and (\ref{ma2}) follows from the 
fact that  $f$ is ultimately increasing and $(ii)$. 
If $ w(s_0)=0,$ we have to distinguish the two cases corresponding to the 
 sequence $\{t_n w_n(s_0)\}$ being bounded or not. We only show 
that 
\begin{eqnarray}
\lim\limits_{n\to \infty}\frac{f(t_n w_n(s_0))}{f(t_n)}=0\label{ma4}
    \end{eqnarray}
for the latter situation. We argue by contradiction and thus we suppose
that 
\begin{eqnarray*}
\limsup\limits_{n\to \infty}\frac{f(t_n w_n(s_0))}{f(t_n)}=\mu_0>0.
 \end{eqnarray*}
Note that by (\ref{ma6}), $\mu_0\le 1.$ We have that  there is 
subsequence $\{n_k \}$ of positive integers such that 
\begin{eqnarray*}
\lim\limits_{k\to \infty}\frac{f(t_{n_k} w_{n_k}(s_0))}{f(t_{n_k})}=
\mu_0
\end{eqnarray*}
with $\{t_{n_k}w_{n_k}(s_0)\}$ an  unbounded sequence. Thus, 
passing to a subsequence if necessary, we can suppose that 
$t_{n_k}w_{n_k}(s_0)>x_0.$
 

Now, since $w_{n_k}(s_0)\to 0,$ given $\varepsilon>0 $ there is a
$k_0:=k_0(\varepsilon,\mu_0)$ such that 
\begin{eqnarray*}
 t_{n_k}w_{n_k}(s_0)\le \varepsilon t_{n_k}\mu_0,\mbox{\quad for 
all\quad}k>k_0,
\end{eqnarray*}
then
\begin{eqnarray}
\frac{f(t_{n_k} w_{n_k}(s_0))}{f(t_{n_k})}\le 
\frac{f(t_{n_k}\varepsilon\mu_0 )}{f(t_{n_k})}.\label{ma3}
\end{eqnarray}
But 
\begin{eqnarray*}
\lim\limits_{k\to \infty}
\frac{f(t_{n_k}\varepsilon\mu_0 )}{f(t_{n_k})}=
{(\varepsilon\mu_0 )}^\delta\le {\varepsilon}^\delta,
\end{eqnarray*}
since $f$ is $\delta-$AH, thus letting $k\to \infty$ in (\ref{ma3})
we find that
\begin{eqnarray*}
\mu_0\le {\varepsilon}^\delta,
\end{eqnarray*}
which is a contradiction, and then (\ref{ma4}) holds.
 

Applying Lebesgue's dominated convergence theorem 
 to the right hand side of (\ref{ma5}) 
we conclude  that $\{h_n(y)\}$ converges to 
\begin{eqnarray*}
h(y)= \frac{1}{y^{N-1}} \int^y_0 s^{N-1}w^\delta(s)ds,
\end{eqnarray*}
for each $y\in [0,M_0]$. 

 Solving for $\dot w_n(y)$ in 
(\ref{3.28}) we find
\begin{eqnarray}
-\dot w_n(y) 
= \frac{\phi^{-1}(\alpha_n(y)\phi(z_n))}{\phi^{-1}
(\phi(z_n))} \mbox{\qquad for\qquad } y\in (0,M_0]\label{3.30}
\end{eqnarray}
where
\begin{eqnarray*}
\alpha_n(y)= \frac{t_n f(t_n)}{z_n\phi(z_n)}h_n(y) =
\frac{t_n f(t_n)}{F(t_n)}\cdot \frac{H(z_n)}{z_n\phi(z_n)} h_n(y).
\end{eqnarray*}
From (\ref{1.3}) and (\ref{quotient})  it follows that
$$\frac{t_n f(t_n)}{F(t_n)}\cdot\frac{H(z_n)}{z_n\phi(z_n)}
\to (\delta+1)\Bigl(1-\frac{1}{p}\Bigr)=:\beta\quad\mbox{ as } n\to\infty,$$
Thus for each $y\in [0,M_0]$
\begin{eqnarray}
\lim\limits_{n\to\infty} \alpha_n(y)=\beta h(y).\label{3.31}
\end{eqnarray}
Integrating (\ref{3.30}) from $0$ to $y\in(0,M_0]$,
we obtain
\begin{eqnarray}
1-w_n(y)=\int^y_0 \frac{\phi^{-1}(\alpha_n(s)\phi(z_n))}
{\phi^{-1}(\phi(z_n))}ds,\label{3.32}
\end{eqnarray}
and by using (\ref{3.31}), Proposition \ref{phi-1lim}, 
and  Lebesgue's dominated convergence theorem,  
and by letting $n\to\infty$ in (\ref{3.32}), we find that
\begin{eqnarray}
1-w(y)= \int^y_0 \beta^{p^*-1}(h(s))^{p^*-1}ds.\label{3.33}
\end{eqnarray}
Then, differentiating (\ref{3.33}) we obtain
$$-w'(y) = \beta^{p^*-1}(h(y))^{p^*-1},$$
which yields
\begin{eqnarray*}
-\phi_p(w'(y))= \frac{\beta}{y^{N-1}}\int^y_0 s^{N-1}w^\delta(s)ds.
\end{eqnarray*}
Thus $w$ is a nonnegative nontrivial solution in $[0,M_0]$ 
to the initial value problem
 \begin{eqnarray}
-[y^{N-1}\phi_p(w'(y))]' = \beta y^{N-1}w^\delta(y)\label{3.35}\\
 w'(0)=0,\quad w(0)=1. \label{3.36}
\end{eqnarray}
By using next a diagonal iterative scheme, see for example the last part
of the proof of \cite[Proposition 4.1]{CMM}, 
$w$ can be
extended to all ${\Bbb R}_+$, as a nonnegative solution of 
(\ref{3.35})-(\ref{3.36}). 
Furthermore, and  arguing like in \cite{CMM}, it can be shown 
that $w$ is indeed a positive solution of class $C^2(0,+\infty)$ of 
 (\ref{3.35})-(\ref{3.36}). Since $\dis\delta<\frac{N(p-1)+p}{N-p}$, this is 
a contradiction in the case 
that $\delta>p-1$, see \cite{NS} or \cite{CMM}. In case that $\delta=p-1$,
it is well known (see for example \cite[Lemma 5.3]{DM}), that every solution of (\ref{3.35})-(\ref{3.36}) with $\beta>0$ 
is oscillatory in $(0,+\infty)$ and hence the contradiction.
Thus, lemma \ref{aprioribound} is proved. $\Box$

\paragraph{Step 3. Proof of Theorem \ref{main}.} 
From Lemma \ref{aprioribound}, if $(u,\tau)$ is a solution of (\ref{3.17}), 
i.e.,
\begin{eqnarray*}
u = T(u,\tau)
\end{eqnarray*}
then $\|u\|\le C$ and $0\le\tau\le C$, where $C$ is a positive constant.
Thus if $B(0,R_1)$ denotes the ball centered at $0$ in $C_\#$ with radius
$R_1>C$, we have that
\begin{eqnarray*}
u\ne T(u,\tau)
\end{eqnarray*}
for any $(u,\tau)\in \partial B(0,R_1)\times [0,R_1]$. Hence if $I$ 
denotes the identity
in $C_\#$ we have that the Leray-Schauder degree of the operator 
$$I-T(\cdot,\tau):\overline {B(0,R)}\to C_\#$$ 
is well defined for every
$\tau\in [0,R_1].$ Then, by the properties of the Leray-Schauder
 degree, we have that
\begin{eqnarray}
{\rm deg}_{LS}(I-T(\cdot,\tau),B(0,R_1),0) &=& {\rm deg}_{LS}(I-T(\cdot,R_1), 
B(0,R_1),0)\nonumber \\
&=& 0, \label{3.37}
\end{eqnarray}
since (\ref{3.17}) does not have solutions on $\overline{B(0,R_1))}\times 
\{R_1\}$. Thus from (\ref{3.37}) and the fact that $T(u,0)= T_0(u)$ 
\begin{eqnarray}
{\rm deg}_{LS}(I-T_0, B(0,R_1),0)=0. \label{3.38}
\end{eqnarray}
Next, let us define the operator $S:[0,1]\times C_\#\to C_\#$,
\begin{eqnarray}
S(\lambda,u)=\int^R_r\phi^{-1}[\frac{\lambda}{s^{N-1}}\int^s_0
\xi^{N-1}f(|u(\xi)|)d\xi]ds.\label{(3.39)}
\end{eqnarray}
Then as in step 1, it can be proved that $S$ is a completely continuous 
operator.  We note that $S(1,\cdot)=T_0$.

\paragraph{Claim.} There exist an $\varepsilon>0$ such that the equation
\begin{eqnarray}
u=S(\lambda,u)\label{3.40}
\end{eqnarray}
has no solution $(u,\lambda)$ with $u\in\partial B(0,\varepsilon)$ and
$\lambda\in [0,1]$.

\paragraph{Proof of the claim.} We argue by contradiction and thus we 
assume that there are sequences $\{u_n\}$ and $\{\lambda_n\}$ with $\|u_n\|=
\varepsilon_n\to 0$ as $n\to\infty$ and $\lambda_n\in[0,1]$ 
such that $(u_n,\lambda_n)$ satisfies (\ref{3.40}) for each $n\in{\Bbb N}$. 
We have that $(u_n,\lambda_n)$ satisfies
\begin{eqnarray*}
u_n(r)= \int^R_r\phi^{-1}\Bigl[\frac{\lambda_n}{s^{N-1}}\int^s_0
\xi^{N-1}f(|u_n(\xi)|)d\xi\Bigr]ds
\end{eqnarray*}
which, by the first in (iii), implies that for $n$ large
\begin{eqnarray*}
\varepsilon_n\le \phi^{-1}\Bigl(\phi(\varepsilon_n)\frac{\mu R}{N}\Bigr)R,
\end{eqnarray*}
where $\mu$ is a positive arbitrarily small number. Thus
\begin{equation}
\phi\Bigl(\frac{\varepsilon_n}{R}\Bigr)\le \frac{\mu R}{N} \phi(\varepsilon_n).
\label{3.31'}
\end{equation}
If $R\le 1,$ we immediately reach a contradiction. If  now $R>1, $ 
let us set  $\sigma=1/R,$
then 
$$\frac{\phi\Bigl({\sigma\varepsilon_n}\Bigr)}
{\phi\Bigl({\varepsilon_n}\Bigr)}
\le \frac{\mu R}{N},
$$
and we reach a contradiction by the second of (iii) and the fact that $\mu$
is arbitrary. Thus the claim holds.

It follows from this claim and the properties of the Leray Schauder 
degree  that for $\varepsilon>0$ small,
\begin{eqnarray*}
d_{LS}(I-S(\lambda,\cdot), B(0,\varepsilon),0)= 
\mbox{ constant \quad for all\quad} \lambda\in [0,1].
\end{eqnarray*}
Thus
\begin{eqnarray*}
d_{LS}(I-T_0,B(0,\varepsilon),0)= d_{LS} (I,B(0,\varepsilon),0)=1,\label{(3.41)}
\end{eqnarray*}
and then by (\ref{3.38}) and (\ref{3.40}) and the excision property of the 
Leray-Schauder degree we obtain that there must be a solution of the equation
$$u=T_0(u)$$
in $B(0,R_1)\setminus \overline{B(0,\varepsilon)}$. This concludes the proof of
the theorem. $\Box$

\section{Examples.} 
In this section we wish to show by mean of 
simple examples the applicability of our main theorem.

\paragraph{Example 3.1.} Let $\phi$ be defined by
\begin{eqnarray*}
\phi(s)=\sum\limits_{i=1}^n\alpha_i\phi_{p_i}(s)
\end{eqnarray*}
where for simplicity we assume $\alpha_i>0$ for $i=1,n.$ Also 
 $p_{i+1}>p_i>1,$ for $i=1,...,n-1.$



In a similar form let $f$ be defined by
 \begin{eqnarray*}
f(s)=\sum\limits_{j=1}^m\beta_j|s|^{\delta_j-1}s,
\end{eqnarray*}
where again for simplicity we assume $\beta_j>0$ for $j=1,\ldots,m,$ with
 $\delta_{j+1}>\delta_j>1,$ for $j=1,...,m-1.$

Then it is clear that 
$$\lim\limits_{s\to +\infty}\frac{\phi(\sigma s)}{\phi(s)}
=\sigma^{p_n-1}
\mbox{\quad and \quad} 
\lim\limits_{s\to 0}\frac{\phi(\sigma s)}{\phi(s)}
=\sigma^{p_1-1}
\mbox{\quad for all\quad} \sigma\in{\Bbb R}_+,$$

and that
$$\lim\limits_{s\to +\infty}\frac{f(\sigma s)}{f(s)}
=\sigma^{\delta_m},\mbox{\quad for all\quad} \sigma\in{\Bbb R}_+.$$

It can also be verified that if $\delta_1 >p_1-1, $ then
$$\lim\limits_{s\to 0}\frac{\phi(s)}{f(s)}=+\infty.$$

Thus,  if $N>p_n,$ and
$$p_n-1<\delta_m<\dis {\frac{N(p_n-1)+p_n}{N-p_n}}$$
all the conditions of Theorem \ref{main} are satisfied and hence for $\phi$ and 
$f$
as defined above, problem $(D_r)$ has a positive solution, and therefore  
the problem
$$\quad\cases{-{\rm div} (\sum\limits_{i=1}^{n}\alpha_i|Du|^{p_i-2}Du)=
\sum\limits_{j=1}^{m}\beta_j|u|^{\delta_j-1}u &in $\Omega$\cr
 \cr\qquad\qquad\quad\qquad\quad\quad\quad u  = 0 &on $\partial\Omega$}$$
has a positive radial solution of class $C^1.$
\medskip

In the first example  the operator $\phi $ 
and the nonlinearity $f,$ are asymptotic to powers at $+\infty,$ in 
the sense that, $f(s)$ or $\phi(s)$ divided by a suitable power of $s$
tends to a constant as $s\to +\infty,$
as it can be directly checked. We will give next  an example where both 
$\phi $ and $f$ are not asymptotic to a power.
 

\paragraph{Example 3.2.} Let us define the increasing 
homeomorphism $\psi$ of ${\Bbb R}$ and the 
function $g$ by 
$$\psi(s)=\phi_q(s)\log(1+|s|) \quad{\rm and}\quad 
g(s)=|s|^{\mu-1}\log(1+|s|)s,$$
with $q>1$ and $\mu>0$. Then it can be checked  that 
$$\lim\limits_{s\to +\infty}\frac{\psi(\sigma s)}{\psi(s)}
=\sigma^{q-1}
\mbox{\quad and \quad} 
\lim\limits_{s\to 0}\frac{\psi(\sigma s)}{\psi(s)}
=\sigma^{q}
\mbox{\quad for all\quad} \sigma\in{\Bbb R}_+,$$


$$\lim\limits_{s\to +\infty}\frac{g(\sigma s)}{g(s)}
=\sigma^\mu
\mbox{\quad for all\quad} \sigma\in{\Bbb R}_+,$$
where  neither  $\psi $ nor $g$ are asymptotic to a power at $+\infty.$
Also, it can be directly checked that if $\mu >q-1,$ then
$$\lim\limits_{s\to +\infty}\frac{g(s)}{\phi(s)}=+\infty
\mbox{\quad and\quad} 
\lim\limits_{s\to 0}\frac{\phi(s)}{g(s)}=+\infty.$$
Thus,  if $N>q,$ and
$$q-1<\mu<\dis {\frac{N(q-1)+q}{N-q}},$$ we have that
all the conditions of Theorem \ref{main} are satisfied  with  $\psi$ and $g$ 
in the place of $\phi$ and $f$ and hence 
$(D_r)$ has a positive solution, and therefore  
the problem
$$\quad\cases{-{\rm div} (|Du|^{q-2}\log(1+|Du|)Du)=
|u|^{\mu-1}\log(1+|u|)u &in $\Omega$\cr
 \cr\qquad\qquad\qquad\qquad\qquad\quad\qquad u = 0 &on $\partial\Omega$}$$
has a positive radial solution of class $C^1.$


It follows from  Proposition \ref{prop2.1} $(ii)$ in the Appendix that
the composition of two asymptotically homogeneous operators is also
asymptotically homogeneous. We use this fact to
obtain a third example as a combination of the previous 
two.


\paragraph{Example 3.3.} Define the functions $\theta$ and $h$ as follows
$$\theta(s)=(\phi\circ\psi)(s) \quad\mbox{ and}\quad h=f\circ g$$
where $\phi,\ f$ are as in example 3.1 and $\psi,\ g$ as in example 3.2, 
then in particular $\theta$ is an odd increasing homeomorphism of ${\Bbb R}.$

By Proposition \ref{prop2.1}  $(ii),$ we have immediately that $\theta$ is 
$(r-1)$-AH
and $h$ is $\rho$-AH, with
$$r=(p_n-1)(q-1)+1 \mbox{\quad and\quad} \rho=\delta_m\mu.$$

 It can be directly verified that 
 
$$\lim\limits_{s\to +0}\frac{\theta(\sigma s)}{\theta(s)}
=\sigma^{(q-1)(p_1-1)}
\mbox{\quad for all\quad} \sigma\in{\Bbb R}_+,$$
and if   
$$\mu\delta_1-(q-1)(p_1-1)\ge 0 \mbox{\quad and\quad}
\delta_1\ge (p_1-1)$$
with at least one of the inequalities strict, then
$$\lim\limits_{s\to 0}\frac{\theta(s)}{h(s)}=+\infty.$$

Finally, if

$$(q-1)(p_n-1)<\delta_m\mu<\frac {N(q-1)(p_n-1)+(q-1)(p_n-1)+
1}{N-(q-1)(p_n-1)-1},$$
with $\delta_m\ge (p_n-1),$
we have that all the  conditions of Theorem \ref{main} are fulfilled 
with  $\theta$ and $h$ 
in the place of $\phi$ and $f$ in that theorem. Hence 
$(D_r)$ has a positive solution and therefore  
the problem

$$
\quad\cases{-{\rm div} (\sum\limits_{i=1}^{n}
\alpha_i|Du|^{(p_i-1)(q-1)-1}{(\log(1+|Du|)}^{p_i-1}Du) \cr \qquad\quad=
\sum\limits_{j=1}^{m}\beta_j|u|^{\mu\delta_j-1}
{(\log(1+|u|)}^{\delta_j}u \mbox{\quad in\quad} \Omega\cr\cr
 \qquad\qquad u = 0 \mbox{\quad on\quad} \partial\Omega}
$$
has a positive radial solution of class $C^1.$

{\bf Remark. }For simplicity the right hand side functions in the above examples
have been chosen as  increasing odd homeomorphisms of ${\Bbb R}.$ It is 
nevertheless
clear how to modify these functions so that they satisfy the asymptotic 
conditions required by Theorem \ref{main}.


\section{Appendix}


Here, and for the sake of completeness, we briefly state and prove   
some of the properties of  AH functions we have used. 
For other properties 
of AH functions we have used, we refer to \cite{R} or \cite{S}. 
 

Our first proposition shows that if $\phi$ is AH
then so is $\phi^{-1}$, and that the composition of  AH functions is also AH.
 

\begin{prop}{ }

(i) Suppose that $\phi$  is an increasing odd homeomorphism of ${\Bbb R}$ 
that  is (p-1)-AH, then
\begin{eqnarray}
\lim\limits_{x\to+\infty} \frac{\phi^{-1}(\sigma x)}{\phi^{-1}(x)}=
\phi_{p^*}(\sigma) \mbox{\quad {\rm for all  }}\sigma\in {\Bbb R}.\label{2.16}
\end{eqnarray}
 

(ii)  Suppose  $\chi,\ \psi:{\Bbb R}_+\to{\Bbb R}_+$ are (p-1)-AH 
and (q-1)-AH
respectively, with $\chi$ ultimately increasing, and $\chi(s),\ 
\psi(s)\to+\infty$ as $s\to+\infty$.
Then
$\chi\circ\psi$ is (r-1)-AH,
with $r=(p-1)(q-1)+1$.\label{phi-1lim}
\label{prop2.1}
\end{prop}

\paragraph{Proof}. $(i)$ It suffices to prove the result for 
$\sigma\in(0,1).$ 
Hence, let $\sigma\in (0,1)$ be a fixed number 
and $\{x_n\}$ a sequence such that $x_n\to +\infty $ as 
$n\to \infty.$ The sequence 
$\dis \{\frac{\phi^{-1}(\sigma x_n)}{\phi^{-1}(x_n)}\}$ is a bounded sequence
and thus it contains a convergent subsequence, that we called the same, 
  say
$$\lim\limits_{n\to \infty}\frac{\phi^{-1}(\sigma x_n)}{\phi^{-1}(x_n)}
=L\in [0,1].$$
Then,  
given $\varepsilon>0$, we can find an $n_0\in\Bbb N$ such that
$$L-\varepsilon<\frac{\phi^{-1}(\sigma x_n)}{\phi^{-1}(x_n)}<
L+\varepsilon\qquad\mbox{for all}\quad n\ge n_0.$$
Setting  $t_n=\phi^{-1}(x_n), n\in{\Bbb N}$, and using that  
$\phi$ is increasing, we find that 
$$\phi_p(L-\varepsilon)\le\sigma\le\phi_p(L+\varepsilon)$$
from where $(i)$ follows.

To show $(ii)$, let $\sigma\in (0,1)$ and $\varepsilon>0$ be given. 
Then, there is $s_0=s_0(\sigma)$ such that
\begin{eqnarray}
-\varepsilon+\phi_q(\sigma)<\frac{\psi(\sigma s)}
{\psi(s)}< \varepsilon +\phi_q(\sigma),\label{2.26}
\end{eqnarray}
for all $s>s_0$  where we take $\varepsilon<\phi_q(\sigma)$. 
Since $\chi$ is ultimately increasing, we have from (\ref{2.26})
that
\begin{eqnarray}
\frac{\chi((\phi_q(\sigma)-\varepsilon)\psi(s))}{\chi(\psi(s))}
\le\frac{\chi(\psi(\sigma s))}{\chi(\psi(s))}
\le\frac{\chi((\phi_q(\sigma)+\varepsilon)\psi(s))}{\chi(\psi(s))}\label{2.26'}
\end{eqnarray}
for large $s$. Thus, letting $s$ go to $+\infty$ in (\ref{2.26'}) we find that
\begin{eqnarray}
\phi_p(\phi_q(\sigma)-\varepsilon) &\le& \liminf\limits_{s\to+\infty}
\frac{\chi(\psi(\sigma s))}{\chi(\psi(s))}\nonumber\\
&\le& \limsup\limits_{s\to+\infty}\frac{\chi(\psi(\sigma s))}
{\chi(\psi(s))}\le \phi_p (\phi_q(\sigma)+\varepsilon).\nonumber
\end{eqnarray}
 Then $(ii)$ follows,
since $0<\varepsilon<\phi_q(\sigma)$ is arbitrary. $\Box$
 
The final part of this section is dedicated to the following natural question,
which is interesting in its own.
 What happens if instead of (\ref{phicond}) we assume that 
$$\dis \lim\limits_{s\to +\infty}\frac{\phi(\sigma s)}{\phi(s)}
=e(\sigma)
\mbox{\quad for each\quad} \sigma\in{\Bbb R}_+,\leqno\mbox{(He)}$$ 
where $e$ is not supposed a-priori to be  a positive power. 
If (He) is assumed, and since $\displaystyle\frac{\phi(\sigma s)}{\phi(s)}$ is
an increasing function of $\sigma$, then $e$ is nondecreasing and hence 
locally integrable.  We will show 
in our next proposition that a simple condition on the average of $e$ 
in $[0,1]$  implies that  $e$  is a positive power. Let us set
$$\overline e=\int_0^1e(\sigma)\ d\sigma,$$
then we have

\begin{prop}
Suppose that $\psi:{\Bbb R}_+\to{\Bbb R}_+$ is a continuous function such that 
$\psi(s)>0$
 for $ s>0$
with $\psi(s)\to +\infty$ as $s\to +\infty,$  and satisfying $(He)$  with $\psi$ 
in the place of $\phi$.
Then
\begin{eqnarray}
\overline e\in (0,1)\qquad\mbox{if and only if\qquad}
e(\sigma)=\sigma^{p-1}\mbox{\quad for all\quad}\sigma\in{\Bbb 
R}_+,\label{average} 
\end{eqnarray}
 where $p>1.$ In this case 
$\dis \overline e=\frac{1}{p}.$
\label{philim}%prop2.3
\end{prop}

\paragraph{Proof.} We only have to show that  
$\overline e\in (0,1)$  implies that 
$e(\sigma)=\sigma^{p-1}.$ We will first prove the proposition in the case
that $\psi$ is increasing. Let $\sigma\in (0,1].$ Then for $x>0$,
\begin{eqnarray}
\frac{\Psi(\sigma x)}{x\psi(x)}=\int^{\sigma x}_0 \frac{\psi
(\tau)d\tau}{x\psi(x)},\label{2.6}
\end{eqnarray}
where $\Psi(x):=\int_0^x\psi(\tau)d\tau.$ Making the change of variables 
$\tau=xs$ in the integral of 
(\ref{2.6}), we find that
\begin{eqnarray}
\frac{\Psi(\sigma x)}{x\psi(x)}=\int^\sigma_0 
\frac{\psi(xs)}{\psi(x)}ds.\label{2.7}
\end{eqnarray}
 Since $\dis \frac{\psi(xs)}{\psi(x)}\le 1$, for each  $s\in [0,1],$ 
 from   Lebesgue's dominated convergence 
theorem it follows that
\begin{eqnarray}
\lim\limits_{x\to+\infty}\frac{\Psi(\sigma x)}{x\psi(x)}=
\int^\sigma_0 e(s)ds:= E(\sigma).\label{2.8}
\end{eqnarray}
In particular, for $\sigma=1,$
\begin{eqnarray}
\lim\limits_{x\to+\infty} \frac{\Psi(x)}{x\psi(x)}=E(1)=
\bar e.\label{2.9}
\end{eqnarray}
On the other hand, since $\Psi$ is of class $C^1$, by L'H\^opital's rule
it follows that
\begin{eqnarray}
\lim\limits_{x\to+\infty} \frac{\Psi(\sigma x)}{\Psi(x)}=\sigma\ 
e(\sigma).\label{2.10}
\end{eqnarray}
Letting $x\to+\infty$ in 
\begin{eqnarray}
\frac{\Psi(\sigma x)}{x\psi(x)}= \frac{\Psi(\sigma x)}{\Psi(x)}
\frac{\Psi(x)}{x\psi(x)}\ ,
\end{eqnarray}
and using  (\ref{2.8}), (\ref{2.9}), (\ref{2.10}), 
 it follows that
\begin{eqnarray*}
E(\sigma) = \bar e\sigma e(\sigma),
%\label{2.11}
\end{eqnarray*}
thus
\begin{eqnarray*}
e(\sigma)=\frac{E(\sigma)}{\bar e\sigma}\ ,
\end{eqnarray*}
and hence $e$ is continuous in $(0,1].$ Then
$E$  satisfies the differential equation
\begin{eqnarray} 
\frac{E'(\sigma)}{E(\sigma)}= \frac{1}{\bar e\sigma}. \label{2.12}
\end{eqnarray}
Now,  $\bar e=E(1)\in (0,1),$ and  the fact that $E$ is  
nondecreasing in
$[0,1]$ imply  the existence of a $\sigma_*\in [0,1]$ such that 
$$\sigma_*= \inf \{\sigma\in (0,1]\ |\  E(\sigma)>0\}.$$
Integrating (\ref{2.12})
on $[\sigma,1]$, with $\sigma\in[\sigma_*,1]$, we find that
\begin{eqnarray}
E(\sigma)= \frac{1}{p}\sigma^p,\label{2.13}
\end{eqnarray}
with $\dis p=\frac{1}{\bar e}.$
Clearly then, $\sigma_*$ must be zero and  (\ref{2.13}) holds
for all $\sigma\in [0,1].$ Thus
$$e(\sigma)=E'(\sigma)=\sigma^{p-1},$$
for all $\sigma\in [0,1],$ and $\dis p=\frac{1}{\bar e}>1.$

That $e(\sigma)=\sigma^{p-1}$ holds  for 
all $\sigma\in (1,+\infty)$ follows by setting
$\mu=1/\sigma$ in (He). 
 

We prove now the result for the general case. To this end we consider 
 $\Psi:{\Bbb R}_+\to {\Bbb R}_+$  defined by 
$\Psi (s) = \int^s_0\psi(\tau)d\tau.$ 
By (He) and an application of L'H\^opital's rule we have that 
\begin{eqnarray*}
\dis \lim\limits_{s\to +\infty}\frac{\Psi(\sigma s)}{\Psi(s)}
=\sigma e(\sigma):=E(\sigma)
\mbox{\quad for all\quad} \sigma\in{\Bbb R}_+.
\end{eqnarray*}

Let now 
$\overline E:=\int_0^1\sigma e(\sigma)\ d\sigma,$ then 
$\overline e\in (0,1)$ implies that 
 $\overline E\in (0,1).$ Thus by the previous argument,
$E(\sigma)=\sigma^{q-1}$ with  $q={\dis \frac{1}{\overline E}>1}.$
It follows that 
$$e(\sigma)=\frac{E(\sigma)}{\sigma}=\sigma^{q-2},
\mbox{\quad for all\quad} \sigma\in{\Bbb R}^+.$$
Using then that 
$$\overline E=\overline e-\int^1_0(\int^t_0e(\sigma)\ d\sigma\ )dt$$
we find that $q=p+1$ for some $p>1$. This concludes the proof of the 
proposition. $\Box$

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\vbox{\sc Marta Garc\'\i a-Huidobro \newline
 Departamento de  Matem\'aticas\newline
  Facultad de  Matem\'aticas,\newline
Universidad Cat\'olica de Chile \newline
Casilla 306, Correo 22\newline 
Santiago, Chile \newline
{\rm E-mail address: mgarcia@poincare.mat.puc.cl}} \medskip

\vbox{\sc  Raul Man\'asevich \newline
 Departamento de Ingenier\'\i a Matem\'atica, F.C.F.M. \newline
Universidad de Chile \newline
Casilla 170, Correo 3\newline
Santiago, Chile \newline
{\rm E-mail address:  manasevi@dim.uchile.cl}} 
\bigskip

\vbox{\sc Pedro Ubilla  \newline
Departamento de Matem\'aticas \newline
Universidad de Santiago de Chile \newline
Casilla 307, Correo 2\newline
Santiago, Chile \newline
{\rm E-mail address: pubilla@fermat.usach.cl} }


 
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