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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 94, pp. 1--12.\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 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/94\hfil Potential Landesman-Lazer type condition]
{Potential Landesman-Lazer type conditions and the
 Fu\v{c}\'{i}k spectrum}
\author[P. Tomiczek\hfil EJDE-2005/94\hfilneg]
{Petr Tomiczek}

\address{Petr Tomiczek \hfill\break
Department of Mathematics, University of West Bohemia\\
Univerzitn\'{\i} 22, 306 14 Plze\v{n}, Czech Republic}
\email{tomiczek@kma.zcu.cz}

\date{}
\thanks{Submitted November 26, 2004. Published August 29, 2005.}
\thanks{Partially supported by the Grant Agency of Czech Republic, MSM 4977751301.}
\subjclass[2000]{35J70, 58E05, 49B27}
\keywords{Resonance; eigenvalue; jumping nonlinearities; Fucik spectrum}

\begin{abstract}
 We prove the existence of solutions to the nonlinear problem
 $$\displaylines{
 u''(x)+\lambda_+ u^+(x)-\lambda_- u^-(x)+g(x,u(x))=f(x)\,,\quad
 x\in (0,\pi)\,,\cr
 u(0)=u(\pi)=0 }
 $$
 where the point $[\lambda_+,\lambda_-]$ is a point of the Fu\v{c}\'{\i}k
 spectrum  and the nonlinearity $g(x,u(x))$ satisfies a potential
 Landesman-Lazer type  condition.
 We use a variational method based on the generalization of the
 Saddle Point Theorem.
\end{abstract}

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

\section{Introduction}

We investigate the existence of solutions for the nonlinear boundary-value
problem
\begin{equation} \label{1.1}
\begin{gathered}
u''(x)+ \lambda_+ u^+(x)-\lambda_- u^-(x)+g(x,u(x))=f(x)
\,,\quad x\in (0,\pi)\,,\\
u(0)=u(\pi)=0\,.
\end{gathered}
\end{equation}
 Here $u^{\pm}=\max\{\pm u,0\}$, $\lambda_+,\lambda_- \in
 \mathbb{R}$, the nonlinearity $g\,\colon( 0,\pi )\times\mathbb{R} \mapsto
 \mathbb{R}$
 is a Caratheodory function and $f\in L^1(0,\pi)$.
For  $g\equiv 0$ and $f\equiv 0$ problem (\ref{1.1}) becomes
\begin{equation}\label{1.2}
\begin{gathered}
u''(x)+ \lambda_+ u^+(x)-\lambda_- u^-(x)=0\,,\quad x\in (0,\pi)\,,\\
u(0)=u(\pi)=0\,.
\end{gathered}
\end{equation}
We define $\Sigma=\{[\lambda_+,\lambda_-]\in\mathbb{R}^2 : \mbox{(\ref{1.2}) has a nontrivial
solution}\}$. This set is called the Fu\v{c}\'{\i}k spectrum
(see \cite{lit2}), and can be
expressed as $\Sigma=\bigcup_{j=1}^{\infty}\Sigma_j$  where
\begin{gather*}
 \Sigma_1  = \bigl\{[\lambda_+ ,\lambda_- ] \in \mathbb{R}^2: \;
(\lambda_+ -1)(\lambda_- -1) = 0 \bigr\}\,, \\
 \Sigma_{2i}  =  \bigl\{[\lambda_+,\lambda_-] \in \mathbb{R}^2: \;
 i\,
\Bigl({1\over {\sqrt {\lambda_+}}}+{1\over {\sqrt {\lambda_-}}} \Bigr)
= 1 \bigr\}\,, \\
 \Sigma_{2i+1}  = \Sigma_{2i+1,1} \cup  \Sigma_{2i+1,2} \quad \mbox{where} \\
 \Sigma_{2i+1,1}  =  \bigl\{[\lambda_+,\lambda_-] \in \mathbb{R}^2:
 \; i\,\Bigl({1\over {\sqrt {\lambda_+}}}+
{1\over {\sqrt {\lambda_-}}} \Bigr)+ {1\over {\sqrt {\lambda_+}}}=
1\bigr\}\,,\\
 \Sigma_{2i+1,2}  =  \bigl\{[\lambda_+,\lambda_-] \in \mathbb{R}^2:
 \; i\,\Bigl({1\over {\sqrt {\lambda_+}}}+
{1\over {\sqrt {\lambda_-}}}\Bigr)+ {1\over {\sqrt {\lambda_-}}}=
1 \bigr\}  \,.
\end{gather*}

We suppose that
\begin{equation}\label{1.3}
\begin{gathered}
\ [\lambda_+,\lambda_-]\in \Sigma_m\,,
\mbox{ if $m\in\mathbb{N}$ is even} \\
\ [\lambda_+,\lambda_-]\in \Sigma_{m2}\,,
\mbox{ if $m\in\mathbb{N}$ is odd}  \\
\mbox{and }\lambda_-<\lambda_+ <(m+1)^2\,.
\end{gathered}
\end{equation}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1}
\end{center}
%\includegraphics[width=16cm,height=21.5cm]{fucikclanek}
\caption{Fu\v{c}\'{i}k spectrum}
\end{figure}


\begin{remark} \label{rem1}  \rm
Assuming that $(m+1)^2>\lambda_+>\lambda_-$, if
 $[\lambda_+,\lambda_-]\in \Sigma_m$,
 $m\in\mathbb{N}$, then $\lambda_- > (m-1)^2$.
\end{remark}

 We define the potential of the nonlinearity $g$ as
$$ G(x,s)=\int_0^{s}g(x,t)\,dt $$
and
$$
G_+(x)=\liminf_{s\to +\infty}{{G(x,s)}\over s}\,,\quad
G_-(x)=\limsup_{s\to -\infty}{{G(x,s)}\over s}\,.
$$
  We denote by $\varphi_m$ a nontrivial
solution of (\ref{1.2}) corresponding to $[\lambda_+,\lambda_-]$ (see Remark \ref{rem2}). We assume
that for any $\varphi_m$ the following potential Landesman-Lazer type condition  holds:
\begin{equation} \label{1.4}
\int_0^{\pi}f(x)\varphi_m(x) \,dx < \int_0^{\pi}\bigl[
G_+(x)(\varphi_m(x))^+ -
 G_-(x)(\varphi_m(x))^-\bigr] \,dx\,.
\end{equation}
We  suppose that the nonlinearity $g$ is bounded, i.e.
there exists $p(x)\in L^1(0,\pi)$ such that
\begin{equation} \label{1.5}
 |g(x,s)|\le p(x)\quad \mbox{for
a.e.}\; x\in(0,\pi)\,,\; \forall \,s\in\mathbb{R}
\end{equation}
 and we prove the solvability of (\ref{1.1}) in Theorem (\ref{thm2}) below.

This article is inspired by a result in \cite{lit3}
 where the author studies the case when
$g(x,s)/s$ lies (in some sense) between $\Sigma_1$ and $\Sigma_2$ and
by a result in \cite{lit1} with the classical Landesman-Lazer type
condition \cite[Corollary 2]{lit1}.

\begin{remark} \label{rem2} \rm
First we note that if $m$ is even then two different functions
$\varphi_{m1},\varphi_{m2}$ of norm 1 correspond to the point
$[\lambda_+,\lambda_-]\in \Sigma_m$.
For example for $m=2$, $\lambda_+>\lambda_- $ we have
$$
\varphi_{21}(x)=\begin{cases}
k_1\sqrt{\lambda_-}\sin(\sqrt{\lambda_+}x),
&   x\in\langle 0,\pi/\sqrt{\lambda_+}\rangle ,\\
-k_1\sqrt{\lambda_+}\sin(\sqrt{\lambda_-}(x-\pi/\sqrt{\lambda_+})),
&     x\in\langle\pi/\sqrt{\lambda_+},\pi\rangle  ,
\end{cases}
$$
where $k_1>0$, and
$$
\varphi_{22}(x)=\begin{cases} -k_2\sqrt{\lambda_+}\sin(\sqrt{\lambda_-}x), &
  x\in\langle 0,\pi/\sqrt{\lambda_-}\rangle  ,\\
k_2\sqrt{\lambda_-}\sin(\sqrt{\lambda_+}(x-\pi/\sqrt{\lambda_-})),
&     x\in\langle\pi/\sqrt{\lambda_-},\pi\rangle  ,
\end{cases}
$$
where $k_2>0$.


 For $\lambda_+=\lambda_-=4$ we set $\varphi_{21}(x)=k_1\sin 2x$ and
$\varphi_{22}(x)=-k_2\sin 2x$, where $k_1,k_2>0$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig2}
\end{center}
%\includegraphics[width=17cm,height=19cm]{varphiclanek}
 \caption{Solutions corresponding to $\Sigma_2$}
\end{figure}

If $m$ is odd, then $\Sigma_m=\Sigma_{m1}\cup \Sigma_{m2}$ and
it corresponds only one function $\varphi_{m1}$ od norm 1 to the point
$[\lambda_+',\lambda_-'] \in \Sigma_{m1}$,\, one function
$\varphi_{m2}$ of norm 1 to the point $[\lambda_+,\lambda_-] \in
\Sigma_{m2}$, respectively.

 For $m=3$, $\lambda_+'>\lambda_-'$, $\lambda_+>\lambda_- $ we have
\begin{align*}
&\varphi_{31}(x)\\
&=\begin{cases}
k_1\sqrt{\lambda_-'}\sin(\sqrt{\lambda_+'}x), &
 x\in\langle 0,\pi/\sqrt{\lambda_+'}\rangle ,\\
-k_1\sqrt{\lambda_+'}\sin(\sqrt{\lambda_-'}(x-\pi/\sqrt{\lambda_+'})),
&
  x\in\langle\pi/\sqrt{\lambda_+'},\pi/\sqrt{\lambda_+'}+\pi/\sqrt{\lambda_-'}\rangle  ,\\
k_1\sqrt{\lambda_-'}\sin(\sqrt{\lambda_+'}(x-\pi/\sqrt{\lambda_+'}-\pi/\sqrt{\lambda_-'})),
&
x\in\langle\pi/\sqrt{\lambda_+'}+\pi/\sqrt{\lambda_-'},\pi\rangle ,
\end{cases}
\end{align*}
where $k_1>0$.
\begin{align*}
&\varphi_{32}(x)\\
&= \begin{cases}
-k_2\sqrt{\lambda_+}\sin(\sqrt{\lambda_-}x), &
 x\in\langle 0,\pi/\sqrt{\lambda_-}\rangle  ,\\
k_2\sqrt{\lambda_-}\sin(\sqrt{\lambda_+}(x-\pi/\sqrt{\lambda_-})),
& x\in\langle\pi/\sqrt{\lambda_-},\pi/\sqrt{\lambda_-}+\pi/\sqrt{\lambda_+}\rangle ,\\
-k_2\sqrt{\lambda_+}\sin(\sqrt{\lambda_-}(x-\pi/\sqrt{\lambda_-}-\pi/\sqrt{\lambda_+})),
& x\in\langle\pi/\sqrt{\lambda_-}+\pi/\sqrt{\lambda_+},\pi\rangle,
\end{cases}
\end{align*}
where $k_2>0$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig3}
\end{center}
%\includegraphics[width=17cm,height=19cm]{varphiclanek3}
 \caption{Solutions corresponding to $\Sigma_3$}
\end{figure}


For $\lambda_+=\lambda_-=m^2$ we set $\varphi_{m1}(x)=k_1\sin mx$, and
$\varphi_{m2}(x)=-k_2\sin mx$, where $k_1,k_2>0$ and from the condition
(\ref{1.4}) we obtain
$$
\int_0^{\pi}f(x)\sin mx
\,dx < \int_0^{\pi}\bigl[ G_+(x)(\sin mx)^+ - G_-(x)(\sin mx)^-\bigr] \,dx
$$
and
$$
\int_0^{\pi}f(x)(-\sin mx) \,dx < \int_0^{\pi}\bigl[ G_+(x)(-\sin mx)^+
- G_-(x)(-\sin mx)^-\bigr] \,dx\,.
$$
Hence it follows
\begin{equation}
\begin{aligned}
&\int_0^{\pi}\bigl[ G_-(x)(\sin mx)^+  -
G_+(x)(\sin mx)^-\bigr] \,dx   \\
&<\int_0^{\pi}f(x)\sin mx \,dx < \int_0^{\pi}\bigl[ G_+(x)(\sin mx)^+ -
 G_-(x)(\sin mx)^-\bigr] \,dx\,.
\end{aligned}
\end{equation}
We obtained the potential Landesman-Lazer type condition (see \cite{lit6}).
\end{remark}

\begin{remark} \label{rem3} \rm
We have
$$
\langle v, \sin mx\rangle=\int_0^{\pi}
v'(x)(\sin mx)'\,dx= m^2\int_0^{\pi} v(x)\sin mx\,dx \quad \forall\, v\in H
$$
($H$ is a Sobolev space defined below).
Since and from the definition of the functions $\varphi_{m1},\varphi_{m2}$
(see remark \ref{rem2})
it follows
\begin{equation}
\label{1.6} \langle \varphi_{m1}, \sin mx\rangle>0\,\quad \mbox{and}\quad \,
\langle \varphi_{m2}, \sin mx\rangle<0\,.
\end{equation}
\end{remark}

\section{Preliminaries}

\subsection*{Notation}
 We shall use the classical spaces
${C}(0,\pi)$, $L^p(0,\pi)$ of continuous and measurable
real-valued functions whose $p$-th power of the absolute value is
Lebesgue integrable, respectively. $\!H$ is the Sobolev space of
absolutely continuous functions $u \colon (0,\pi)\to \mathbb{R}$
such that $u'\!\in\!L^2(0,\pi)$ and $u(0)\!=\!u(\pi)\!=\!0.$ We
denote by the symbols $\| \cdot \|$, and $\| \cdot \|_2$ the norm
in $H$, and in $L^2(0,\pi)$, respectively. We denote
$\langle\,\cdot,\cdot\,\rangle$ the pairing in the space
$H$.

By a solution of (\ref{1.1}) we mean a function
$u\in {C}^1(0,\pi)$ such that $u'$ is absolutely
continuous, $u$
satisfies the boundary conditions and the equations
(\ref{1.1}) holds a.e. in $(0,\pi)$.

Let $I\,\colon\,H\to \mathbb{R}$ be a functional such that $I\in
{C}^1(H,\mathbb{R})$ (continuously differentiable). We say that
$u$ is a critical point of $I$, if
$$
\langle I'(u), v\rangle = 0 \quad \mbox{for all } v\in H\,.$$

We say that $\gamma$ is a critical value of $I$, if
there is $u_0\in H$ such that $I(u_0)=\gamma$ and
$I'(u_0)=0$.

We say that $I$ satisfies Palais-Smale condition (PS) if every
sequence $(u_n)$ for which $I(u_n)$ is bounded in $H$ and
$I'(u_n)\to 0$ (as $n\to \infty)$ possesses a~convergent
subsequence.

We study (\ref{1.1}) by the use of a variational method. More
precisely, we look for critical points of the functional $I:H\to
\mathbb{R}$, which is defined by
\begin{equation}
\label{2.1} I(u)={1\over 2}\int_0^{\pi
}\bigl[(u')^2-\lambda_+(u^+)^2- \lambda_-(u^-)^2 \bigr]\,dx -
\int_0^{\pi }\bigl[ G(x,u)-fu\bigr] \,dx\,.
\end{equation}
Every critical point $u\in H$ of the functional $I$ satisfies
$$
\int_0^{\pi}\bigl[ u'v'-(\lambda_+u^+ - \lambda_-u^-)v \bigr] \,dx
- \int_0^{\pi }\bigl[g(x,u)v-fv\bigr] \,dx=0\quad \mbox{for all
} v\in H\,.
$$
Then $u$ is also a weak solution of
(\ref{1.1}) and vice versa.

The usual regularity argument for ODE yields immediately
(see Fu\v{c}\'{\i}k \cite{lit2}) that any
weak solution of (\ref{1.1}) is also a solution in the sense mentioned above.

We will use the following variant of the Saddle Point Theorem (see \cite{lit4}) which is proved in
Struwe \cite[Theorem 8.4]{lit5}.

\begin{theorem} \label{thm1}
Let $S$ be a closed subset in $H$ and $Q$ a bounded subset in $H$
with boundary $\partial Q$. Set $\Gamma = \{h : \,h\in {\bf
C}(H,H), \,h(u)=u\,\mbox{ on } \partial Q\}$. Suppose $I\in
{C}^1(H,\mathbb{R})$ and
\begin{itemize}
\item[$(i)$] $S \cap \partial Q = \emptyset$,
\item[$(ii)$] $S \cap h(Q) \not= \emptyset$, for every $h\in\Gamma$,
\item[$(iii)$] there are constants $\mu,\nu$ such that\,
$\mu=\inf_{u\in S} I(u) > \sup_{u\in \partial Q}
I(u)=\nu$,
\item[$(iv)$] $I$ satisfies Palais-Smale condition.
\end{itemize}
Then the number
$$
\gamma=\inf_{h\in \Gamma}\sup_{u\in Q}  I(h(u))
$$
defines a critical value $\gamma>\nu$ of $I$.
\end{theorem}

We say that $S$ and $\partial Q$ {\it link} if they satisfy
conditions (i), (ii) of the theorem above.

We denote the first integral in the functional $I$ by
$$
J(u)=\int_0^\pi \bigl[(u')^2-\lambda_+(u^+)^2- \lambda_-(u^-)^2\bigr]
\,dx\,.
$$
Now we present a few results needed later.

\begin{lemma} \label{l1}
 Let  $\varphi$ be a solution of (\ref{1.2})
with $[\lambda_+,\lambda_-]\, \in \Sigma$,
$\lambda_+\ge\lambda_-$. We put $u=a\varphi+ w$, $ a\ge 0$,
$w\in H$. Then the following relation holds
\begin{equation}\label{2.2}
 \int_0^{\pi }\bigl[(w')^2-\lambda_+ w^2 \bigr] \,dx
 \le J(u)\le \int_0^{\pi }\bigl[(w')^2-\lambda_- w^2 \bigr] \,dx\,.
\end{equation}
\end{lemma}

\begin{proof}  We prove only the right inequality in
(\ref{2.2}), the proof of the left inequality is similar.
Since $\varphi$ is a solution of (\ref{1.2}) we have
\begin{equation}\label{2.3}
\int_0^\pi  \varphi'w' \,dx=\int_0^\pi \bigl[\lambda_+ \varphi^+
w -\lambda_-\varphi^- w \bigr] \,dx\quad \mbox{for } w\in H
\end{equation}
and
\begin{equation}\label{2.4}
\int_0^\pi  (\varphi')^2 \,dx=\int_0^\pi
\bigl[\lambda_+(\varphi^+)^2+\lambda_-(\varphi^-)^2\bigr] \,dx\,.
\end{equation}
By (\ref{2.3}) and (\ref{2.4}), we obtain
\begin{equation} \label{2.5}
\begin{aligned}
J(u) &= \int_0^{\pi} \Bigl[((a\varphi+w)')^2 -\lambda_+
((a\varphi+w)^+)^2-\lambda_-((a\varphi+w)^-)^2\Bigr]\,dx \\
&=  \int_0^{\pi} \Bigl[(a\varphi')^2 +2a\varphi'w'+(w')^2-
(\lambda_+-\lambda_-)((a\varphi+w)^+)^2   \\
&\quad -\lambda_-(a\varphi+w)^2\Bigr]\,dx \\
&= \int_0^{\pi} \Bigl[(\lambda_+-\lambda_-)(a\varphi^+)^2
+\lambda_-(a\varphi)^2+2a((\lambda_+-\lambda_-)\varphi^++\lambda_-\varphi)w
\\
&\quad +(w')^2
-(\lambda_+-\lambda_-)((a\varphi+w)^+)^2-\lambda_-((a\varphi)^2
+2a\varphi w+w^2)\Bigr]\,dx \\
&= \int_0^{\pi} \Bigl\{(\lambda_+-\lambda_-)\bigl[(a\varphi^+)^2
+2a\varphi^+w-((a\varphi+w)^+)^2\bigr]  \\
& \quad +(w')^2-\lambda_-w^2\Bigr\}\,dx\,.
\end{aligned}
\end{equation}
For the function $(a\varphi^+)^2+2a\varphi^+w-((a\varphi+w)^+)^2$
in the last integral in (\ref{2.5}) we have
\begin{align*}
&(a\varphi^+)^2+2a\varphi^+w-((a\varphi+w)^+)^2\\
&=\begin{cases}
-((a\varphi+w)^+)^2\le 0 & \varphi<0 \\
-w^2\le 0 & \varphi\ge0, a\varphi+w\ge 0 \\
a\varphi^+(a\varphi^++w+w)\le 0 &  \varphi\ge0, a\varphi+w< 0\,.
\end{cases}
\end{align*}
By the assumption $\lambda_+\ge\lambda_-$, we obtain the
assertion of the Lemma~\ref{l1}.
\end{proof}

\begin{remark}\label{rem4} \rm
It follows from the previous proof that we obtain the equality
$$J(u)=\int_0^\pi \bigl[(w')^2-\lambda_-w^2\bigr]
\,dx
$$
in (\ref{2.2}) if $a\varphi+w\le 0$ when $\varphi<0$, and
$w= 0$ when $\varphi\ge0$.
Consequently, if the equality holds and if
$w$ in span$\{\sin x, \dots, \sin kx\},\, k\in\mathbb{N}$, then $w=0$.
\end{remark}

\section{Main result}

\begin{theorem} \label{thm2}
 Under the assumptions (\ref{1.3}), (\ref{1.4}),
and (\ref{1.5}), Problem (\ref{1.1}) has at least one solution in
$H$.
\end{theorem}

\begin{proof}
First we suppose that $m$ is even.
We shall prove that the functional $I$ defined by (\ref{2.1})
satisfies the assumptions in Theorem~\ref{thm1}.
Let $\varphi_{m1}, \varphi_{m2}$ be the normalized solutions of
(\ref{1.2}) described above (see Remark \ref{rem2})\,.

Let $H^-$ be the subspace of $H$ spanned by functions $\sin x,
\dots, \sin (m-1)x$. We define $V\equiv V_1\cup V_2$ where
\begin{gather*}
V_1=\{u\in H : u=a_1\varphi_{m1}+w,\; 0\le a_1,
\; w\in H^-\}, \\
V_2=\{u\in H :  u=a_2\varphi_{m2}+w,\; 0\le a_2,
\; w\in H^-\}.
\end{gather*}
Let $K>0$, $L>0$ then we define $Q\equiv Q_1\cup Q_2$ where
\begin{gather*}
Q_1=\{u\in V_1 : 0\le a_1\le K,\; \|w\|\le L\},\\
Q_2=\{u\in V_2 : 0\le a_2\le K,\; \|w\|\le L\}.
\end{gather*}
Let $S$ be the subspace of $H$ spanned by functions
$\sin(m+1)x, \sin(m+2)x, \dots$.

 Next, we verify the assumptions of
Theorem \ref{thm1}. We see that $S$ is a closed subset in $H$
and $Q$ is a bounded subset in $H$.

\noindent(i) Firstly we note that for $z\in H^-\oplus S$ we have
$\langle z, \sin mx\rangle=0$.
We suppose for contradiction that there is $u\in \partial Q\cap S$. Then
$$
0\stackrel{u\in S}=\langle u, \sin mx\rangle \stackrel{u\in \partial Q}{=}
\langle a_i\varphi_{mi}+w,\sin mx\rangle \stackrel{w\in H^-}{=}
 a_i\langle \varphi_{mi},\sin mx\rangle\,
$$
$i=1,2$. From previous equalities and inequalities (\ref{1.6}) it follows
that $a_i=0$, $i=1,2$ and $u=w$.
For $u=w\in\partial Q$ we have $\|u\|=L>0$ and we obtain a contradiction
with $u\in H^-\cap S =\{o\}$.

\noindent(ii) We prove that $H=V\oplus S$. We can write a function $h\in H$
in the form
\begin{align*}
h&=\sum_{i=1}^{m-1} b_i \sin ix + b_m\sin mx
+\sum_{i=m+1}^{\infty}\! b_i \sin ix\\
&= \overline{h} + b_m\sin mx + \widetilde{h},\, b_i\in\mathbb{R},
\end{align*}
$i\in \mathbb{N}$. The inequalities (\ref{1.6}) yield that there are
constants $b_{m1}, b_{m2}>0$ such that
$\sin mx=b_{m1}(\varphi_{m1}-\overline{\varphi}_{m1}-\widetilde{\varphi}_{m1})$
 and
$-\sin mx=b_{m2}(\varphi_{m2}-\overline{\varphi}_{m2}-\widetilde{\varphi}_{m2})$.
Hence we have for $b_m\ge 0$,
\begin{align*}
h&= \overline{h} + b_mb_{m1}(\varphi_{m1}-\overline{\varphi}_{m1}
-\widetilde{\varphi}_{m1}) + \widetilde{h}\\
&= \underbrace{(\overbrace{\overline{h}
-b_mb_{m1}\overline{\varphi}_{m1}}^{\in H^-} + \overbrace{b_mb_{m1}}^{\ge 0}
\varphi_{m1})}_{\in V}
+\underbrace{(\widetilde{h}-b_mb_{m1}\widetilde{\varphi}_{m1})}_{\in S}\,.
\end{align*}
Similarly for $b_m\le 0$,
\begin{align*}
h&=\overline{h} + |b_m|b_{m2}(\varphi_{m2}-\overline{\varphi}_{m2}
-\widetilde{\varphi}_{m2}) + \widetilde{h}\\
&= \underbrace{(\overbrace{\overline{h}-|b_m|b_{m2}
\overline{\varphi}_{m2}}^{\in H^-}
+ \overbrace{|b_m|b_{m2}}^{\ge 0}\varphi_{m2})}_{\in V}
+\underbrace{(\widetilde{h}-|b_m|b_{m2}\widetilde{\varphi}_{m2})}_{\in S}\,.
\end{align*}
We proved that $H$ is spanned by $V$ and $S$.

The proof of the assumption  $S\cap h(Q)\not=\emptyset\quad \forall
h\in\Gamma$ is similar to the proof in \cite[example 8.2]{lit5}.
Let $\pi \colon H\to V$ be the continuous projection of $H$ onto $V$.
 We have to show that $0\in \pi(h(Q))$. For $t\in[0,1]$, $u\in Q$
we define
$$
h_t(u)=t \pi(h(u))+(1-t)u\,.
$$
The function $h_t$ defines a homotopy of $h_0= id$ with $h_1=\pi\circ h$.
Moreover, $h_t|\partial Q=\mathop{\rm id}$ for all $t\in[0,1]$.
Hence the topological degree
$\deg(h_t,Q,0)$ is well-defined and by homotopy invariance we have
$$
\deg(\pi\circ h,Q,0)=\deg(\mathrm{id},Q,0)=1\,.
$$
Hence $0\in \pi(h(Q))$, as needed.


\noindent (iii)  Firstly, we note that by assumption
(\ref{1.5}), one has
\begin{equation}
\label{3.1} \lim_{\|u\|\to \infty} \int_0^{\pi }
\frac{G(x,u) - fu}{\|u\|^2} \,dx  = 0\,.
\end{equation}
First we show that the infimum of functional $I$ on the set $S$ is
a real number. We prove for this that
\begin{equation}
\label{3.2} \lim_{\|u\|\to \infty} I(u)=
\infty\quad\mbox{for all } u\in S
\end{equation}
and $I$ is bounded on bounded sets.

 Because of the compact imbedding of $H$ into
${C}(0,\pi)$ $(\| u \|_{{C}(0,\pi)}\leq c_1\| u \|)$, and of
$H$ into L$^2(0,\pi)$ $(\| u \|_2\leq c_2\| u \|)$, and the
assumption (\ref{1.5}) one has
\begin{align*}
I(u)&=\frac{1}{2}\int_0^{\pi }\bigl[(u')^2-
\lambda_+(u^+)^2-\lambda_-(u^-)^2\bigr]\,dx
 -\int_0^{\pi } \bigl[G(x,u) - fu \bigr] \,dx \\
&\le \frac{1}{2}\left( \|u\|^2+\lambda_+ \|u^+\|_2^2+ \lambda_-
\|u^-\|_2^2 \right) + \int_0^{\pi } \bigl[\,(|p|+|f|) |u|\bigr]
\,dx  \\
&\le \frac{1}{2}\left( \|u\|^2+\lambda_+ c_2\|u^+\|^2+ \lambda_-
c_2 \|u^-\|^2 \right) + \left( \|p\|_1+\|f\|_1 \right) c_1\|u\|
\,. \nonumber
\end{align*}
Hence $I$ is bounded on bounded subsets of S.

To prove (\ref{3.2}), we argue by contradiction. We suppose that
there is a sequence $(u_n)\subset S$ such that $\|u_n\|\to \infty$
and a constant $c_3$ satisfying
\begin{equation}
\label{3.3} \liminf_{n\to \infty} I(u_n)\le c_3\,.
\end{equation}
For $u\in S$ the following relation holds
\begin{equation}\label{3.4}
\| u\|^2=\int_0^{\pi } (u')^2 \,dx \ge (m+1)^2\int_0^{\pi } u^2
\,dx= (m+1)^2 \| u\|_2^2\,.
\end{equation}
The definition of $I$, (\ref{3.1}), (\ref{3.3}) and (\ref{3.4})
yield
\begin{equation}
\label{3.5} 0\ge \liminf_{n\to \infty}
\frac{I(u_n)}{\|u_n\|^2}\ge \liminf_{n\to \infty}
\frac{((m+1)^2-\lambda_+)\|u_n^+\|_2^2+((m+1)^2-\lambda_-)\|u_n^-\|_2^2}{2\|u_n\|^2}\,.
\end{equation}
It follows from (\ref{3.5}) and (\ref{1.3}) that $\| u_n\|_2^2/ \|
u_n \|^2 \to 0$ and from the definition of $I$ and (\ref{3.1}) we
have
$$
\liminf_{\| u_n \|\to \infty} \frac{I(u_n)}{\| u_n \|^2} =
\frac{1}{2}
$$
a contradiction to (\ref{3.5}).
We proved that
there is $\mu\in\mathbb{R}$ such that $\inf_{u\in S} I(u) =\mu$.

 Second we estimate the value $I(u)$ for $u\in\partial Q$.
We remark that $u\in\partial Q$ can be either of the form $K\varphi_m+w$,
with $\|w\|\le L$ or of the form $a_i\varphi_{mi}$, with $0\le a_i\le K$,
$\|w\|=L$ ($i=1,2$).
We prove that
\begin{equation}
\label{3.6} \sup_{(K+L) \to \infty}I(K\varphi_m+w)
=\sup_{\| u \| \to \infty}I(u)=-\infty\,
\quad\mbox{for}\quad u\in\partial Q\,.
\end{equation}
For (\ref{3.6}), we argue by contradiction. Suppose that
(\ref{3.6}) is not true then there are a sequence $(u_n)\subset
\partial Q$ such that $\| u_n \| \to \infty$ and a constant $c_4$
satisfying
\begin{equation}
\label{3.7} \limsup_{n\to \infty} I(u_n)\ge c_4\,.
\end{equation}
Hence, it follows
\begin{equation}
\label{3.8} \limsup_{n\to \infty} \Bigl[ \frac{1}{2}\int_0^{\pi}
\frac{(u_n')^2-
\lambda_+(u_n^+)^2-\lambda_-(u_n^-)^2}{\|u_n\|^2}\,dx
 -\int_0^{\pi } \frac{G(x,u_n) - fu_n}{\|u_n\|^2} \,dx \Bigr]\ge 0\,.
\end{equation}
Set $v_n=u_n/\| u_n \|$. Since $\dim\partial Q<\infty $ there is
$v_0\in \partial Q$ such that $v_n\to v_0$ strongly in $H$
(also strongly in $L^2(0,\pi))$. Then (\ref{3.8}) and
(\ref{3.1}) yield
\begin{equation}
\label{3.9} \frac{1}{2}\int_0^{\pi} \bigl[ (v_0')^2-
\lambda_+(v_0^+)^2-\lambda_-(v_0^-)^2 \bigr] \,dx \ge 0\,.
\end{equation}
Let $v_0=a_0\varphi_m + w_0$,
$a_0\in\mathbb{R}_0^+, w_0\in H^-$.
It follows from Lemma \ref{l1} that
\begin{equation}
\label{3.10} \int_0^{\pi} \bigl[ (v_0')^2-
\lambda_+(v_0^+)^2-\lambda_-(v_0^-)^2 \bigr] \,dx \le
\int_0^{\pi} \bigl[ (w_0')^2- \lambda_-(w_0)^2 \bigr] \,dx\,.
\end{equation}
For $w_0\in H^-$  we have
\begin{equation}
\label{3.11}  \int_0^{\pi} \bigl[ (w_0')^2- \lambda_- w_0^2 \bigr]
\,dx\le \int_0^{\pi} \bigl[ ((m-1)^2- \lambda_-)\, w_0^2 \bigr]
\,dx \,.
\end{equation}
Since $(m-1)^2<\lambda_-$ (see Remark \ref{rem1}) then (\ref{3.9}), (\ref{3.10}) and (\ref{3.11})
yield
$$
\int_0^{\pi} \bigl[ (v_0')^2-
\lambda_+(v_0^+)^2-\lambda_-(v_0^-)^2\bigr] \,dx = ((m-1)^2-
\lambda_-) \|w_0\|_2^2 = 0\,.
$$
 Hence we obtain $w_0=0$ and $v_0=a_0 \varphi_m$,
$\|v_0 \|=1$.   Now we divide (\ref{3.7}) by $\|u_n\|$ then
\begin{equation}
\label{3.12} \limsup_{n\to \infty} \Bigl[ \frac{1}{2}\int_0^{\pi}
\frac{(u_n')^2- \lambda_+(u_n^+)^2-\lambda_-(u_n^-)^2}{\|u_n\|}\,dx
 -\int_0^{\pi } \frac{G(x,u_n) - fu_n}{\|u_n\|} \,dx \Bigr]\ge 0\,.
\end{equation}
By Lemma \ref{l1} the first integral in (\ref{3.12}) is less then or
equal to 0. Hence it follows
\begin{equation}
\label{3.13}
 \limsup_{n\to \infty}\int_0^{\pi } \frac{-G(x,u_n) + fu_n}{\|u_n\|} \,dx =
 \limsup_{n\to \infty}
 \int_0^{\pi } \Bigl[ \frac{-G(x,u_n)}{u_n} v_n + f v_n \Bigr] \,dx\ge 0\,.
\end{equation}


Because of the compact imbedding $H^-\subset {C}(0,\pi)$, we have
$v_n\to a_0 \varphi_m$ in ${C}(0,\pi)$ and we get
$$
\lim_{n\to \infty}u_n(x) =
\begin{cases}
+\infty &\mbox{for $x\in(0,\pi)$  such that
 $ \varphi_m(x)>0$}\,,\\
-\infty& \mbox{for $x\in(0,\pi)$  such that  $\varphi_m(x)<0$}\,.
\end{cases}
$$
We note that from (\ref{1.5}) it follows that $-|p(x)|\le G_+(x)$,
$G_-(x)\le |p(x)|$ for a.e. $x\in(0,\pi)$. We obtain from
Fatou's lemma and (\ref{3.13})
$$
\int_0^{\pi} f(x)\varphi_m(x) \,dx
\ge \int_0^{\pi} \bigl[ G_+(x) (\varphi_m(x))^+ -G_-(x)
(\varphi_m(x))^-\bigr] \,dx\,,
$$
a contradiction to (\ref{1.4}).
We proved that by choosing $K,L$ sufficiently large there is
$\nu\in\mathbb{R}$ such that
$\sup_{u\in \partial Q} I(u) =\nu<\mu$.
 Then Assumption (iii) of Theorem \ref{thm2} is verified.

\noindent (iv) Now we show
that $I$ satisfies the Palais-Smale condition. First, we suppose
that the sequence $(u_n)$ is unbounded and there exists a constant
$c_5$ such that
\begin{equation}   \label{3.14}
\Bigl| {1\over 2}\int_0^{\pi} \bigl[(u'_n)^2-\lambda_+
(u_n^+)^2-\lambda_- (u_n^-)^2 \bigr] \,dx -\int_0^{\pi} \bigl[
G(x,u_n)- f u_n \bigr] \,dx \Bigr| \le c_5
\end{equation}
and
\begin{equation}
\label{3.15} \lim_{n\to \infty}\| I'(u_n)\|=0\,.
\end{equation}
Let $(w_k)$ be an arbitrary sequence bounded in $H$. It follows
from (\ref{3.15}) and the Schwarz inequality that
\begin{equation}\label{3.16}
\begin{aligned}
&\Bigl| \lim_{{n\to \infty}\atop{k\to \infty}}
\int_0^{\pi}\bigl[ u_n'w_k'-(\lambda_+ u_n^+ -\lambda_-
u_n^-)\,w_k\bigr] \,dx -\int_0^{\pi} \bigl[ g(x,u_n)w_k - fw_k
\bigr] \,dx \,\Bigr|  \\
& = \bigl| \lim_{{n\to \infty}\atop{k\to \infty}}\langle I'(u_n),
w_k\rangle \bigr|\\
& \le \lim_{{n\to \infty}\atop{k\to \infty}} \| I'(u_n)\|\cdot\|
w_k\| =0\,.
\end{aligned}
\end{equation}
Put $v_n=u_n/ \| u_n \|$. Due to compact imbedding $H\subset
L^2(0,\pi)$ there is $v_0\in H$ such that (up to subsequence)
$v_n\rightharpoonup v_0$ weakly in H, $v_n\to v_0$ strongly in
$L^2(0,\pi)$. We divide (\ref{3.16}) by $\|u_n\|$ and we obtain
\begin{equation}
\label{3.17} \lim_{n,k\, \to \infty}
\int_0^{\pi}\bigl[ v_n'w_k'-(\lambda_+ v_n^+ -\lambda_-
v_n^-)\,w_k
 \bigr] \,dx=0
\end{equation}
and
\begin{equation}
\label{3.18} \lim_{i, k\,\to \infty}
 \int_0^{\pi}\bigl[ v_i'w_k'-(\lambda_+
v_i^+ -\lambda_- v_i^-)\, w_k \bigr] \,dx=0\,.
\end{equation}
We subtract equalities (\ref{3.17}) and (\ref{3.18}) we have
\begin{equation} \label{3.19}
\lim_{n,i,k\, \to \infty}
\int_0^{\pi}\bigl[ (v_n'-v_i')w_k'-(\lambda_+
(v_n^+-v_i^+) -\lambda_- (v_n^--v_i^-)\,)\, w_k \bigr] \,dx=0\,.
\end{equation}
Because $(w_k)$ is a arbitrary bounded sequence we can set $w_k=v_n-v_i$
in (\ref{3.19}) and we get
\begin{equation}
\label{3.20} \lim_{n,i\, \to \infty}
 \Bigl[ \|
v_n-v_i\|^2-\int_0^{\pi}\left[\,[\lambda_+ (v_n^+-v_i^+) -\lambda_-
(v_n^--v_i^-)](v_n-v_i)\,\right] \,dx\Bigr]=0\,.
\end{equation}
Since $v_n\to v_0$ strongly in $L^2(0,\pi)$ the integral in
(\ref{3.20}) converges to 0 and then $v_n$ is a Cauchy sequence
in $H$ and $v_n\to v_0$ strongly in $H$ and $\|v_0\|=1$.

It follows from (\ref{3.17}) and the usual regularity argument for ordinary differential equations
(see Fu\v{c}\'{\i}k \cite{lit2}) that $v_0$ is the solution of the equation
$$v_0''+\lambda_+ v_0^+ -\lambda_- v_0^-=0\,.$$ From the
assumption $[\lambda_+,\lambda_-]\in \Sigma_m$ it follows that
$v_0=a_0\varphi_{m}, a_0>0$.

We set $u_n=a_n\varphi_m+\widehat{u}_n$, where $a_n\ge 0,
\widehat{u}_n\in H^-\oplus S$. We remark that $u=u^+-u^-$  and using (\ref{2.3}) in the first
integral in (\ref{3.16}) we obtain
\begin{equation}
\label{3.21}
\begin{aligned}
I_1 &= \int_0^{\pi} \left[ (a_n \varphi_{m}+\widehat{u}_n)'w_k' - (\lambda_+ u_n^+ - \lambda_-u_n^-)
w_k \right] \,dx \\
 &= \int_0^{\pi} \left[ a_n \varphi_{m}'w_k'
+ (\widehat{u}_n)'w_k' - ((\lambda_+ -\lambda_-)u_n^+ + \lambda_-u_n)\,w_k \right] \,dx\\
 &= \int_0^{\pi} \left[ a_n(\lambda_+ \varphi_{m}^+-\lambda_- \varphi_{m}^-)w_k
+ (\widehat{u}_n)'w_k' - ((\lambda_+ -\lambda_-)u_n^+ + \lambda_-u_n)\,w_k \right] \,dx\\ &=
\int_0^{\pi} \big\{ a_n[(\lambda_+-\lambda_-)
\varphi_{m}^+ +\lambda_- \varphi_{m}]w_k + (\widehat{u}_n)'w_k'  \\
&\quad -  [(\lambda_+ -\lambda_-)(a_n\varphi_m+\widehat{u}_n)^+ +
\lambda_-(a_n\varphi_m+\widehat{u}_n)]\,w_k \big\} \,dx \\
&= \int_0^{\pi} \left[
(\lambda_+-\lambda_-) (a_n\varphi_{m}^+ - (a_n\varphi_m+\widehat{u}_n)^+)w_k
+ (\widehat{u}_n)'w_k'
- \lambda_- \widehat{u}_n w_k \right] \,dx\,.
\end{aligned}
\end{equation}
Similarly we obtain
\begin{equation}
\label{3.22}  I_1=  \int_0^{\pi} \left[ (\lambda_+-\lambda_-)
(a_n\varphi_{m}^- - (a_n\varphi_m+\widehat{u}_n)^-)w_k +
(\widehat{u}_n)'w_k' - \lambda_+ \widehat{u}_n w_k \right] \,dx\,.
\end{equation}
Adding (\ref{3.21}) and (\ref{3.22}) and  we have
\begin{equation}   \label{3.23}
2 I_1 = \int_0^{\pi} [ (\lambda_+ - \lambda_-)
(|a_n\varphi_{m}| - |a_n\varphi_m+\widehat{u}_n|) w_k
+ 2 (\widehat{u}_n)'w_k' - (\lambda_+ +
\lambda_-)\widehat{u}_n w_k ]\,dx\,.
\end{equation}
We set \, $\widehat{u}_n=\overline{u}_n+\widetilde{u}_n$ \,  where \,
 $\overline{u}_n\in H^- ,\, \widetilde{u}_n\in S  $ \,  and
 we put in (\ref{3.23}) \,
 $w_k=(\overline{u}_n-\widetilde{u}_n)/\|\widehat{u}_n\|$ then we have
\begin{equation} \label{3.24}
\begin{aligned}
2 I_1&=  \frac{1}{\|\widehat{u}_n\|}\int_0^{\pi} \big[
(\lambda_+-\lambda_-)(|a_n\varphi_{m}|
- |a_n\varphi_m+\overline{u}_n+\widetilde{u}_n|)
 (\overline{u}_n-\widetilde{u}_n) \\
&\quad + 2 \left.(\overline{u}_n')^2 -2(\widetilde{u}_n')^2 - (\lambda_+
+\lambda_-)(\overline{u}_n^2-\widetilde{u}_n^2) \right] \,dx\,.
\end{aligned}
\end{equation}
Hence
\begin{equation}\label{3.25}
\begin{aligned}
2 I_1 &\le  \frac{1}{\|\widehat{u}_n\|} \Bigl( \int_0^{\pi}
 [(\lambda_+-\lambda_-)\,|\overline{u}_n+\widetilde{u}_n| \,
 |\overline{u}_n-\widetilde{u}_n| ]\,dx \\
&\quad +  2 \|\overline{u}_n\|^2 -2\|\widetilde{u}_n\|^2 - (\lambda_+
 +\lambda_-)(\|\overline{u}_n\|_2^2-\|\widetilde{u}_n\|_2^2) \Bigr) \\
&= \frac{1}{\|\widehat{u}_n\|} \Bigl( \int_0^{\pi}
 [(\lambda_+-\lambda_-)\,|\overline{u}_n^2-\widetilde{u}_n^2| ]\,dx \\
& \quad+ 2 \|\overline{u}_n\|^2 -(\lambda_+ +\lambda_-) \|\overline{u}_n\|_2^2
-2\|\widetilde{u}_n\|^2 + (\lambda_+ +\lambda_-) \|\widetilde{u}_n\|_2^2
\Bigr) \,.
\end{aligned}
\end{equation}
The inequality $|a^2-b^2|\le \max\{a^2, b^2\}$,  (\ref{3.25}) and
(\ref{1.3}) yield
\begin{equation}
\label{3.26}
 I_1 \le  \max \bigl\{\|\overline{u}_n\|^2
- \lambda_- \|\overline{u}_n\|_2^2,
 -\|\widetilde{u}_n\|^2 + \lambda_+ \|\widetilde{u}_n\|_2^2\bigr\}\,
\frac{1}{\|\widehat{u}_n\|}
\,.
\end{equation}
We note that the following relations hold \, $\|\overline{u}_n\|^2 \le
(m-1)^2\|\overline{u}_n\|_2^2, $ \, $
 \|\widetilde{u}_n\|^2 \ge $ $(m+1)^2\|\widetilde{u}_n\|_2^2$. Hence from
 assumption (\ref{1.3}) and (\ref{3.26}) it follows that there is
 $\varepsilon>0$ such that
\begin{equation}
\label{3.27}
 I_1 \le  -\varepsilon \max \bigl\{\|\overline{u}_n\|^2 , \|\widetilde{u}_n\|^2 \bigr\}\frac{1}{\|\widehat{u}_n\|}
\,.
\end{equation}
 From (\ref{3.16}), (\ref{3.27}) it follows
\begin{equation}
\label{3.28}
 \lim_{n\to \infty} -\varepsilon \frac{\max
\bigl\{\|\overline{u}_n\|^2 , \|\widetilde{u}_n\|^2
\bigr\}}{\|\widehat{u}_n\|} -\int_0^{\pi} \big[ (g(x,u_n)-
f)\frac{\overline{u}_n-\widetilde{u}_n}{\|\widehat{u}_n\|} \big] \,dx \ge
0.
\end{equation}
Now we suppose that $\|\widehat{u}_n\|\to\infty$. We note that
$\|\widehat{u}_n\|^2=\|\overline{u}_n\|^2+\|\widetilde{u}_n\|^2$, we divide (\ref{3.28}) by
$\|\widehat{u}_n\|$ and using  (\ref{1.5}) we have
\begin{equation}
\label{3.29}
 -\frac{\varepsilon}{2}\ge\lim_{n\to \infty} -\varepsilon \frac{\max
\bigl\{\|\overline{u}_n\|^2 , \|\widetilde{u}_n\|^2 \bigr\}
}{\|\widehat{u}_n\|^2}-\int_0^{\pi}  \frac{g(x,u_n)-
f}{\|\widehat{u}_n\|}\,
\frac{\overline{u}_n-\widetilde{u}_n}{\|\widehat{u}_n\|} \,dx \ge 0
\end{equation}
a contradiction to $\varepsilon>0$.
This implies that the sequence $(\widehat{u}_n)$ is bounded. We
use (\ref{2.2}) from Lemma \ref{l1} with $w=\widehat{u}_n$ and we
obtain
$$
 \int_0^{\pi }\bigl[(\widehat{u}_n')^2-\lambda_+ \widehat{u}_n^2 \bigr] \,dx
 \le J(u_n)\le \int_0^{\pi }\bigl[(\widehat{u}_n')^2-\lambda_- \widehat{u}_n^2 \bigr] \,dx\,.
$$
 Hence
\begin{equation}\label{3.30}
\lim_{n\to\infty}
\frac{J(u_n)}{\|u_n\|}=\lim_{n\to\infty}\frac{ \int_0^{\pi
}\left[ (u_n')^2-\lambda_+ u_n^2-\lambda_- u_n^2 \right]
\,dx}{\|u_n\|}=0\,.
\end{equation}
We divide (\ref{3.14}) by $\|u_n\|$ and (\ref{3.30}) yield
\begin{equation}
\label{3.31}
 \lim_{n\to \infty}
 \int_0^{\pi } \bigl[ \frac{-G(x,u_n)+f u_n}{\|u_n\|} \bigr] \,dx= 0
\end{equation}
and using Fatou's lemma in (\ref{3.31}) we obtain a contradiction
to (\ref{1.4}).

This implies that the sequence $(u_n)$ is bounded. Then there
exists $u_0\in H$ such that $ u_n\rightharpoonup u_0$ in $H$,
$u_n\to u_0$ in $L^2(0,\pi)$ (up to subsequence). It follows from
the equality (\ref{3.16}) that
\begin{equation}
\label{3.32}
 \lim_{n,i,k\, \to \infty}
\int_0^{\pi}\left[(u_n-u_i)'w_k'- [\lambda_+ (u_n^+-u_i^+)
-\lambda_- (u_n^--u_i^-)]w_k\right] \,dx =0\,.
\end{equation}
We put $w_k=u_n-u_i$ in (\ref{3.32}) and the strong convergence
$u_n \to u_0$ in $L^2(0,\pi)$ and (\ref{3.32}) imply the strong
convergence $u_n \to u_0$ in $H$. This shows that the functional
$I$ satisfies Palais-Smale condition and the proof of
Theorem\,\ref{thm2} for $m$ even is complete.
\smallskip

Now we suppose that $m$ is odd.
We have $[\lambda_+,\lambda_-]\in\Sigma_{m2}$ and the nontrivial
solution $\varphi_{m2}$ of (\ref{1.2}) corresponding to
$[\lambda_+,\lambda_-]$. Then there is $k>0$ such that
$[\lambda_+-k,\lambda_--k]\in\Sigma_{m1}$ and solution
$\varphi_{m1}$ corresponding to $[\lambda_+-k,\lambda_--k]=[\lambda_+',\lambda_-']$ (see Remark \ref{rem2})\,.

We define the sets $Q$ and $S$ like for $m$ even and the proof of
the steps (i), (ii) of theorem \ref{thm2} is the same.
In the step (iii) we change inequality (\ref{3.10}) if
$v_0=a_0\varphi_{m1}$ as it follows
\begin{equation} \label{3.10p}
\begin{aligned}
&\int_0^{\pi} \left[ (v_0')^2-
\lambda_+(v_0^+)^2-\lambda_-(v_0^-)^2 \right] \,dx \\
&=\int_0^{\pi} \left[ (v_0')^2-
(\lambda_+-k)(v_0^+)^2-(\lambda_--k)(v_0^-)^2
\right] \,dx- k \int_0^{\pi}  v_0^2 \,dx  \\
&\le -k \int_0^{\pi}  v_0^2 \,dx + \int_0^{\pi} \left[ (w_0')^2-
\lambda_-(w_0)^2 \right] \,dx\,.
\end{aligned}
\end{equation}
Then by (\ref{3.9}), (\ref{3.10p}) and (\ref{3.11}) we obtain
$k \int_0^{\pi}  v_0^2 \,dx=0$,
a contradiction to $\|v_0\|=1$.
The proof of the step (iv) is similar to the prove for $m$ even. The
proof of the theorem \ref{thm2} is complete.
\end{proof}

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\end{document}