\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 199, pp. 1--22.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2018/199\hfil On the second eigenvalue] {On the second eigenvalue of nonlinear eigenvalue problems} \author[M. Degiovanni, M. Marzocchi \hfil EJDE-2018/199\hfilneg] {Marco Degiovanni, Marco Marzocchi} \address{Marco Degiovanni \newline Dipartimento di Matematica e Fisica\\ Universit\`a Cattolica del Sacro Cuore\\ Via dei Musei 41\\ 25121 Brescia, Italy} \email{marco.degiovanni@unicatt.it} \address{Marco Marzocchi \newline Dipartimento di Matematica e Fisica\\ Universit\`a Cattolica del Sacro Cuore\\ Via dei Musei 41\\ 25121 Brescia, Italy} \email{marco.marzocchi@unicatt.it} \dedicatory{Communicated by Marco Squassina} \thanks{Submitted December 7, 2018. Published December 14, 2018.} \subjclass[2010]{58E05, 35J66} \keywords{Nonlinear eigenvalue problems; variational methods; \hfill\break\indent quasilinear elliptic equations} \begin{abstract} This article is devoted to the characterization of the second eigenvalue of nonlinear eigenvalue problems. We propose an abstract approach which allows to treat nonsmooth quasilinear problems and also to recover, in a unified way, previous results concerning the $p$-Laplacian. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} Consider the nonlinear eigenvalue problem \begin{equation}\label{eq:eig} \begin{gathered} - \Delta_p u = \lambda V |u|^{p-2}u\quad\text{in $\Omega$}\,,\\ u = 0 \quad\text{on $\partial\Omega$}\,, \end{gathered} \end{equation} where $\Omega$ is an open subset of $\mathbb{R}^n$, $\Delta_p u:=\operatorname{div}(|\nabla u|^{p-2}\nabla u)$ denotes the $p$-Laplacian and $V$ is a possibly sign-changing weight. A real number $\lambda$ is said to be an \emph{eigenvalue} if~\eqref{eq:eig} admits a nontrivial solution~$u$. The existence of a diverging sequence $(\lambda_k)$ of eigenvalues has been proved, under quite general assumptions, in \cite{szulkin_willem1999}. In the case $V=1$, a different characterization of $\lambda_2$ has been provided in \cite{drabek_robinson1999}, in connection with the introduction of a possibly different sequence of eigenvalues. Further characterizations of $\lambda_2$, under various sets of assumptions, have been provided in \cite{anane_tsouli1996} and in particular in \cite{cuesta_defigueiredo_gossez1999,arias_campos_cuesta_gossez2002} by a mountain pass description. More recently, the mountain pass characterization of the second eigenvalue has been proved also for the fractional $p$-Laplacian in \cite{brasco_parini2016}. In all these papers the main techniques involved concern regularity theory for the solutions $u$ of \eqref{eq:eig} and variational methods, as the eigenvalues $\lambda$ can be characterized as the critical values of the functional \[ f(u) = \int_\Omega |\nabla u|^p\,dx \] on the manifold \[ M = \big\{u:\int_\Omega V|u|^p\,dx = 1\big\} \cup \big\{u:\int_\Omega V|u|^p\,dx = -1\big\}\,. \] More precisely, eigenvalues $\lambda$ with $\lambda>0$ are characterized by means of the manifold \[ M = \{u:\int_\Omega V|u|^p\,dx = 1\} \] and those with $\lambda<0$ by means of the manifold \[ M = \big\{u:\int_\Omega V|u|^p\,dx = -1\big\} \,. \] In the recent paper \cite{fusco_mukherjee_zhang2018} the case in which $\Omega$ is a $p$-quasi open set is considered and the mountain pass characterization of the second eigenvalue is proved also in that setting. In this last paper, some typical techniques of critical point theory are replaced by the use of the minimizing movements. \par On the other hand, the existence of a diverging sequence $(\lambda_k)$ of eigenvalues has also been proved in \cite{lucia_schuricht2013} when, more generally, $f$ is a convex functional of the form \[ f(u) = \int_\Omega a(x,\nabla u)\,dx \,. \] Since $a(x,\cdot)$ is not supposed to be of class $C^1$, the metric critical point theory developed independently in \cite{corvellec_degiovanni_marzocchi1993, degiovanni_marzocchi1994} and in \cite{ioffe_schwartzman1996, katriel1994} is applied in this case. \par The main purpose of this paper is to extend the characterizations of the second eigenvalue to the case treated in \cite{lucia_schuricht2013} by an abstract approach, based on techniques of metric critical point theory, which allows to recover in a unified way also the previous results on the second eigenvalue of the $p$-Laplacian. \par After recalling the main tools of metric critical point theory in Section~\ref{sect:recalls}, we will propose in Section~\ref{sect:general} our abstract setting and prove in Section~\ref{sect:main} the main results. They will be applied in Section~\ref{sect:nq} to the setting of \cite{lucia_schuricht2013}, while Section~\ref{sect:qopen} is devoted to problems on $p$-quasi open sets as in \cite{fusco_mukherjee_zhang2018}, but without the use of minimizing movements, and Section~\ref{sect:frac} to the fractional $p$-Laplacian considered in \cite{brasco_parini2016}. \section{Metric critical point theory} \label{sect:recalls} Let $M$ be a metric space endowed with the distance $d$ and $f:M \to \mathbb{R}$ a continuous function. We will denote by $B_{\delta}(u)$ the open ball of center $u$ and radius $\delta$. \subsection{First basic facts} The next notion has been independently introduced in \cite{corvellec_degiovanni_marzocchi1993, degiovanni_marzocchi1994} and in \cite{katriel1994}, while a variant has been considered in \cite{ioffe_schwartzman1996}. \begin{definition}\label{defn:ws} \rm For every $u\in M$, we denote by $|df|(u)$ the supremum of the $\sigma$'s in $[0, +\infty[$ such that there exist $\delta > 0$ and a continuous map \[ \mathcal{H}:B_{\delta}(u) \times [0,\delta] \to M \] satisfying \[ d(\mathcal{H}(v,t), t) \leq t \,,\quad f(\mathcal{H}(v,t))\leq f(v) - \sigma t\,, \] for every $v \in B_{\delta}(u)$ and $t \in [0,\delta]$. The extended real number $|df|(u)$ is called the \emph{weak slope} of $f$ at $u$. \end{definition} \begin{remark}\label{rem:wsc1} \rm Let $M$ be an open subset of a normed space and let $f:M \to \mathbb{R}$ be of class~$C^1$. Then $|df|(u)=\|f'(u)\|$ for any $u\in M$. \end{remark} \begin{remark} \label{rem:minws} \rm Let $u\in M$ be a local minimum of $f$. Then $|df|(u)=0$. \end{remark} \begin{remark} \label{rem:complip} \rm Let $\widehat{M}$ be another metric space and $\Psi:\widehat{M}\to M$ a homeomorphism which is Lipschitz continuous of constant $L$. Then, for every $u\in\widehat{M}$, we have \[ |d(f\circ\Psi)|(u) \leq L\,|df|(\Psi(u))\,. \] \end{remark} \begin{example}\rm Let $M=\mathbb{R}$, $\widehat{M}=[0,+\infty[$ and $f:M\to\mathbb{R}$ defined as $f(u)=-|u|$. Then $|df|(0)=0$, while $\big|d(f\bigl|_{[0,+\infty[})\big|(0)=1$. On the other hand, the inclusion map $[0,+\infty[\subseteq\mathbb{R}$ is Lipschitz continuous, but it is not a homeomorphism. \end{example} \begin{definition} \label{defn:crit} \rm We say that $u \in M$ is a \emph{(lower) critical point} of $f$ if $|df|(u)=0$. We say that $c \in \mathbb{R}$ is a {\emph{(lower) critical value}} of~$f$ if there exists $u \in M$ such that $f(u)=c$ and $|df|(u)=0$. \end{definition} \begin{definition} \label{defn:ps} \rm Given $c \in \mathbb{R}$, we say that $f$ satisfies the \emph{ Palais-Smale condition at level~$c$ } (${(PS)}_c$, for short), if every sequence $(u_k)$ in $M$, with $f(u_k) \to c$ and $|df|(u_k) \to 0$, admits a convergent subsequence in $M$. \end{definition} The next concept was first introduced in \cite{cerami1978}. \begin{definition} \label{defn:cps} \rm Let $\hat{u}\in M$. Given $c \in \mathbb{R}$, we say that $f$ satisfies the \emph{ Cerami-Palais-Smale condition at level~$c$ } (${(CPS)}_c$, for short), if every sequence $(u_k)$ in $M$, with $f(u_k) \to c$ and $(1+d(u_k,\hat{u}))|df|(u_k) \to 0$, admits a convergent subsequence in $M$. \end{definition} Since \[ (1+d(u_k,\hat{u}))|df|(u_k) \leq (1+d(u_k,\check{u}))|df|(u_k) + d(\check{u},\hat{u})|df|(u_k) \,, \] it is easily seen that $(CPS)_c$ is independent of the choice of $\hat{u}$. It is also clear that $(PS)_c$ implies $(CPS)_c$. When $f$ is smooth, the next result is contained in \cite[Theorem~1]{pucci_serrin1985}, which in turn developed some variants of the celebrated Mountain Pass Theorem (see \cite{ambrosetti_rabinowitz1973, rabinowitz1986}). \begin{theorem} \label{thm:mp} Let $v\in M$ be a local minimum of $f$, let $w\in M$ with $w\neq v$ and $f(w)\leq f(v)$ and set \[ \Phi = \{\varphi\in C([-1,1];M): \varphi(-1)=v\,,\,\,\varphi(1)=w\}\,. \] Assume that $M$ is complete, $\Phi\neq\emptyset$ and that $f$ satisfies $(CPS)_c$ at the level \[ c = \inf_{\varphi\in\Phi} \max_{-1\leq t\leq 1} f(\varphi(t))\,. \] Then there exists a critical point $u$ of $f$ with $u\neq v$, $u\neq w$ and $f(u)=c$. \end{theorem} \begin{proof} According to \cite[Theorem~4.1]{corvellec1999}, the $(CPS)_c$ condition is just the $(PS)_c$ condition with respect to an auxiliary distance which keeps the completeness of $M$ and does not change the critical points of $f$ and the topology of $M$. Therefore, we may assume without loss of generality that $f$ satisfies $(PS)_c$. Let $r>0$ be such that $d(w,v) > r$ and \[ f(z)\geq f(v)\quad\text{whenever $d(z,v) \leq r$}\,. \] If we set \[ A = \{z\in M:d(z,v) = r\}\,, \] we infer from \cite[Theorem~3.7]{corvellec_degiovanni_marzocchi1993} that there exists a critical point $u$ of $f$ with $f(u)=c$ and, moreover, that $u\in A$ if $c=\inf\limits_A f$. In both cases $c>\inf\limits_A f$ and $c=\inf\limits_A f$ we have that $u\neq v$, $u\neq w$ and the assertion follows. \end{proof} \begin{theorem} \label{thm:min} Let \[ b > a := \inf_M f > -\infty\,. \] Assume that $M$ is complete and that $f$ has no critical value in $]a,b[$ and satisfies $(CPS)_c$ for every $c\in[a,b[$. Suppose also that either the set $f^{-1}(a)$ is finite or each $v\in f^{-1}(a)$ admits a path connected neighborhood in $\{w\in M:f(w)\leq b\}$. Then, for every $u\in M$ with $f(u)\leq b$ and $|df|(u)\neq 0$, there exists a continuous map $\varphi:[-1,1]\to M$ such that $\varphi(-1)=u$, $f(\varphi(1))=a$ and $f(\varphi(t))\leq b$ for any $t\in[-1,1]$. \end{theorem} \begin{proof} As before, we may assume without loss of generality that $f$ satisfies $(PS)_c$ for every $c\in[a,b[$. Let $u\in M$ with $f(u)\leq b$ and $|df|(u)\neq 0$. By Definition~\ref{defn:ws} there exists a continuous map $\psi:[-1,0]\to M$ with $\psi(-1)=u$, $f(\psi(0))0$. On the other hand, there is no path connecting points $(x,y)$ with $f(x,y)>0$ and a minimum of $f$. \end{remark} \subsection{A case with symmetry} Let now $\Psi:M\to M$ be an isometry such that $\Psi\circ\Psi=\mathrm{Id}$. We assume that $f$ is also \emph{$\Psi$-invariant}, namely that $f(\Psi(u))=f(u)$ for any $u\in M$, and we set \[ \operatorname{Fix}(M) = \{u\in M:\Psi(u)=u\}\,. \] \begin{definition} \rm A subset $A$ of $M$ is said to be \emph{$\Psi$-invariant} if $\Psi(A)\subseteq A$. A map $\varphi:A\to \mathbb{R}^k$, where $A$ is a $\Psi$-invariant subset of $M$, is said to be \emph{$\Psi$-equivariant} if $\varphi(\Psi(u))=-\varphi(u)$ for any $u\in A$. Finally, a map $\varphi:S\to M$, where $S\subseteq \mathbb{R}^k$ is symmetric with respect to the origin, is said to be \emph{$\Psi$-equivariant} if $\varphi(-u)=\Psi(\varphi(u))$ for any $u\in S$. \end{definition} For every nonempty $\Psi$-invariant subset $A$ of $M$, we set \begin{align*} \gamma(A) = \min \big\{&k\geq 1: \text{there exists a $\Psi$-equivariant and continuous map} \\ &\varphi:A\to\mathbb{R}^k\setminus\{0\}\big\}\,. \end{align*} We agree that $\gamma(A)=\infty$ if there is no such $k$ and we set $\gamma(\emptyset)=0$. We also set \begin{align*} \overline{\gamma}(A) = \sup \big\{&k\geq 1: \text{there exists a $\Psi$-equivariant and continuous map}\\ &\varphi:\mathbb{R}^k\setminus\{0\}\to A \big\}\,. \end{align*} Again, we set $\overline{\gamma}(\emptyset)=0$. It is well known (see e.g.\ \cite{chang1993, krasnoselskii1964}) that \[ \overline{\gamma}(A) \leq \gamma(A) \quad\text{for every $\Psi$-invariant subset $A$ of $M$} \] and it is clear that $\overline{\gamma}(A) = \gamma(A) = \infty$ whenever $A\cap \operatorname{Fix}(M)\neq\emptyset$. Then, for every $k\geq 1$, we set \begin{gather*} \begin{aligned} \underline{c}_k = \inf\big\{&\max_A f: A\text{ is a compact and $\Psi$-invariant subset of $M$}\\ & \text{with } \gamma(A)\geq k \big\}\,, \end{aligned} \\ \begin{aligned} \overline{c}_k = \inf\big\{&\max_A f: A \text{ is a compact and $\Psi$-invariant subset of $M$}\\ &\text{with }\overline{\gamma}(A)\geq k \big\}\,, \end{aligned} \end{gather*} where we agree that $\underline{c}_k=+\infty$ (resp. $\overline{c}_k=+\infty$) if there is no $A$ with $\gamma(A)\geq k$ (resp. $\overline{\gamma}(A)\geq k$). It is easily seen that \begin{gather*} \underline{c}_k \leq \underline{c}_{k+1}\,,\quad \overline{c}_k \leq \overline{c}_{k+1}\,,\quad \underline{c}_k \leq \overline{c}_k\,,\quad \text{for every $k\geq 1$}\,,\\ \underline{c}_1 = \overline{c}_1 = \inf_M f\,. \end{gather*} \begin{theorem} \label{thm:general0} Assume that $M$ is complete. Then the following facts hold: \begin{itemize} \item[(a)] if \[ -\infty < \underline{c}_k < \inf\{f(u):u\in \operatorname{Fix}(M)\} \] and $f$ satisfies $(CPS)_{\underline{c}_k}$, then $\underline{c}_k$ is a critical value of $f$ (we agree that $\inf\emptyset = +\infty$); \item[(b)] if \[ -\infty < \overline{c}_k < \inf\{f(u):u\in \operatorname{Fix}(M)\} \] and $f$ satisfies $(CPS)_{\overline{c}_k}$, then $\overline{c}_k$ is a critical value of $f$; \item[(c)] if \[ -\infty < \underline{c}_k = \cdots = \underline{c}_{k+m-1} < \inf\{f(u):u\in \operatorname{Fix}(M)\} \] and $f$ satisfies $(CPS)_{\underline{c}_k}$, then \[ \gamma(\{u\in M: f(u)=\underline{c}_k\,,\,\,|df|(u)=0\}) \geq m\,; \] \item[(d)]if $f$ is bounded from below, \[ b<\inf\{f(u):u\in \operatorname{Fix}(M)\} \] and $f$ satisfies $(CPS)_c$ for every $c\leq b$, then we have \[ \gamma(\{u\in M:f(u) \leq b\}) < \infty\,. \] \end{itemize} \end{theorem} \begin{proof} Again, the proof of \cite[Theorem~4.1]{corvellec1999} is compatible with the symmetry structure. Therefore one can assume $(PS)_c$ instead of $(CPS)_c$. Then the argument is the same as in the proof of \cite[Theorem~2.5]{degiovanni_schuricht1998}. \end{proof} \subsection{Constrained problems} Let now $X$ be a real Banach space. In the following, $\partial f(u)$ will denote Clarke's subdifferential and $f^0(u;v)$ the associated generalized directional derivative (see \cite{clarke1983}). If $f$ is locally Lipschitz, we have \begin{gather*} f^0(u;v) := \limsup_{z\to u\,,\,\,t\to 0^+}\, \frac{f(z+tv)-f(z)}{t} = \limsup_{z\to u\,,\,\,w\to v\,,\,\,t\to 0^+}\, \frac{f(z+tw)-f(z)}{t} \,,\\ \partial f(u) = \{\alpha\in X':\text{$\langle\alpha,v\rangle\leq f^0(u;v)$ for any $v\in X$}\}\,. \end{gather*} If $f$ is locally Lipschitz and convex, we also have that \[ f^0(u;v) = \lim_{t\to 0^+}\, \frac{f(u+tv)-f(u)}{t} = \lim_{w\to v\,,\,\,t\to 0^+}\, \frac{f(u+tw)-f(u)}{t} \,, \] and $\partial f(u)$ agrees with the subdifferential of convex analysis. \begin{theorem} \label{thm:laglip} Let $U$ be an open subset of $X$, $f,g:U \to \mathbb{R}$ two locally Lipschitz functions, \[ M = \{v\in U:g(v) = 0\} \] and $u\in M$ with $0\not\in\partial g(u)$. Then we have \[ \big|d(f\bigr|_{M})\big|(u) \geq \min\{\|\alpha - \lambda \beta\|: \alpha\in\partial f(u)\,,\,\,\beta\in\partial g(u)\,,\,\, \lambda\in\mathbb{R}\}\,. \] \end{theorem} \begin{proof} Since $0\not\in\partial g(u)$, there exists $v\in X$ such that $g^0(u;v)<0$, namely \[ g^0(u;u_- -u) < 0\,,\quad g^0(u;u-u_+) < 0\,, \] if we set $u_- = u + v$ and $u_+ = u - v$. Then the assertion follows from \cite[Theorem~3.5]{degiovanni_schuricht1998}. \end{proof} \begin{theorem} \label{thm:lagc1} Let $U$ be a convex and open subset of $X$, $f:U \to \mathbb{R}$ a lower semicontinuous and convex function, $g:U \to \mathbb{R}$ a function of class $C^1$, \[ M = \{v\in U:g(v) = 0\} \] and $u\in M$ with $g'(u)\neq 0$. Then $f$ is locally Lipschitz and we have \[ \big|d(f\bigr|_{M})\big|(u) = \min\{\|\alpha - \lambda g'(u)\|: \alpha\in\partial f(u)\,,\,\,\lambda\in\mathbb{R}\}\,. \] \end{theorem} \begin{proof} By \cite[Corollaries~2.5 and~2.4]{ekeland_temam1974}, $f$ is locally Lipschitz. Then, from Theorem~\ref{thm:laglip} we infer that \[ \big|d(f\bigr|_{M})\big|(u) \geq \min\{\|\alpha - \lambda g'(u)\|: \alpha\in\partial f(u)\,,\,\,\lambda\in\mathbb{R}\}\,. \] Let now $\alpha\in\partial f(u)$, $\lambda\in\mathbb{R}$ and let \[ \mathcal{H}:(B_{\delta}(u)\cap M) \times [0,\delta] \to M \] be as in Definition~\ref{defn:ws}. Let \[ g(v) = g(u) + \langle g'(u),v-u\rangle + \|v-u\|\omega(v)\,, \] where $\omega$ is continuous with $\omega(u)=0$. Then we have \[ (f-\lambda g)(\mathcal{H}(u,t)) - (f-\lambda g)(u) = f(\mathcal{H}(u,t)) - f(u) \leq - \sigma t \] and \begin{align*} &(f-\lambda g)(\mathcal{H}(u,t)) - (f-\lambda g)(u)\\ &\geq \langle\alpha - \lambda g'(u),\mathcal{H}(u,t) - u\rangle - \lambda \|\mathcal{H}(u,t) - u\|\,\omega(\mathcal{H}(u,t)) \\ &\geq - (\|\alpha - \lambda g'(u)\| + |\lambda|\,|\omega(\mathcal{H}(u,t))|)\, \|\mathcal{H}(u,t) - u\| \\ &\geq - (\|\alpha - \lambda g'(u)\| + |\lambda| \,|\omega(\mathcal{H}(u,t))|)\,t \,. \end{align*} We infer that \[ \sigma \leq \|\alpha - \lambda g'(u)\| + |\lambda| \,|\omega(\mathcal{H}(u,t))| \quad\text{for every $t\in]0,\delta]$}\,. \] Going to the limit as $t\to 0$, we conclude that \[ \sigma \leq \|\alpha - \lambda g'(u)\| \] and the assertion follows. \end{proof} \begin{proposition}\label{prop:lag} Let $U$ be an open subset of $X$, $f,g:U \to \mathbb{R}$ two functions of class $C^1$, \[ M = \{v\in U:g(v) = 0\} \] and $u\in M$ with $g'(u)\neq 0$. Then we have \[ \big|d(f\bigr|_{M})\big|(u) = \min\{\|f'(u) - \lambda g'(u)\|:\lambda\in\mathbb{R}\}\,. \] \end{proposition} For a proof of the above proposition see \cite[Corollary~2.12]{degiovanni_marzocchi1994}. \begin{proposition} \label{prop:homog} Let $p\in\mathbb{R}$ and let $g:X\setminus\{0\}\to\mathbb{R}$ be a locally Lipschitz function which is positively homogeneous of degree $p$. Then we have \begin{gather*} g^0(u;u) = p \,g(u)\,,\quad g^0(u;-u) = - p \,g(u) \quad\text{for any $u\neq 0$}\,,\\ \langle\alpha,u\rangle = p\,g(u) \quad\text{for any $u\neq 0$ and $\alpha\in\partial g(u)$}\,. \end{gather*} \end{proposition} \begin{proof} If $L$ is a Lipschitz constant in a neighborhood of $u$, for $v$ close to $u$ and $t$ close to~$0$ we have \begin{align*} \frac{g(v+tu)-g(v)}{t} &= \frac{g(v+tv)-g(v)}{t} + \frac{g(v+tu)-g(v+tv)}{t} \\ &= \frac{(1+t)^p- 1}{t}\,g(v) + \frac{g(v+tu)-g(v+tv)}{t} \,, \end{align*} whence \[ \big\|\frac{g(v+tu)-g(v)}{t} - \frac{(1+t)^p- 1}{t}\,g(v)\big\| \leq L\|v-u\|\,. \] It follows \[ g^0(u;u) = \limsup_{v\to u\,,\,\,t\to 0^+}\, \frac{g(v+tu)-g(v)}{t} = p\,g(u)\,. \] The generalized directional derivative $g^0(u;-u)$ can be treated in a similar way. Then we also have \[ p\,g(u) = - g^0(u;-u) \leq \langle\alpha,u\rangle \leq g^0(u;u) = p\,g(u) \] and the proof is complete. \end{proof} \begin{corollary}\label{cor:homoglip} Let $f,g:X\setminus\{0\} \to \mathbb{R}$ be two locally Lipschitz functions which are positively homogeneous of the same degree $p\neq 0$. Then the following facts hold: \begin{itemize} \item[(a)] for every $u\in X\setminus\{0\}$ with $g(u)\neq 0$, we have $0\not\in\partial g(u)$; \item[(b)] if $u\in X\setminus\{0\}$ and $\alpha\in\partial f(u)$, $\beta\in\partial g(u)$, $\lambda\in\mathbb{R}$ satisfy $\alpha=\lambda\beta$, then \[ f(u)= \lambda \,g(u)\,. \] \end{itemize} \end{corollary} \begin{proof} (a) If $p>0$ and $g(u)>0$, we have $g^0(u;-u)=-p\,g(u)<0$ by Proposition~\ref{prop:homog}, whence $0\not\in\partial g(u)$. The other cases can be treated in a similar way. By Proposition~\ref{prop:homog} we have \[ p\,f(u) = \langle\alpha,u\rangle = \lambda\langle\beta,u\rangle = \lambda\,p\,g(u)\,, \] whence the assertion. \end{proof} % page 8 \section{General facts on nonlinear eigenvalue problems} \label{sect:general} Let $X$ be a real Banach space with $X\neq\{0\}$ and let $f,g:X\to\mathbb{R}$ be two functions such that: \begin{itemize} \item[(i)] $f$ and $g$ are even, continuous and positively homogeneous of the same degree $p > 0$. \end{itemize} \begin{definition} \label{defn:eigenvector} \rm We say that $u\in X$ is an \emph{eigenvector} if $g(u)\neq 0$ and $u$ is a critical point of $f\bigl|_{M_u}$, where \[ M_u = \{v\in X:g(v)=g(u)\}\,. \] In such a case we say that \[ \lambda = \frac{f(u)}{g(u)} \] is the \emph{eigenvalue} associated with the eigenvector $u$. \end{definition} \begin{proposition} \label{prop:cone} If $u$ is an eigenvector with eigenvalue $\lambda$ then, for every $t\neq 0$, we have that $tu$ is an eigenvector with the same eigenvalue. \end{proposition} \begin{proof} Since $\Psi(u)=tu$ is a homeomorphism such that $\Psi$ and $\Psi^{-1}$ are both Lipschitz continuous, it follows from Remark~\ref{rem:complip} that $u$ is a critical point of $f$ restricted to $M_u$ if and only if $tu$ is a critical point of $f$ restricted to \[ \{v\in X:g(t^{-1}v) = g(u)\}\,. \] Then the assertion easily follows. \end{proof} \begin{definition} \rm An eigenvalue $\lambda$ is said to be \emph{simple}, if it is not associated with two linearly independent eigenvectors. \end{definition} In the following, we will only consider eigenvectors with $g(u)>0$. Observe that~$\lambda$ is an eigenvalue associated with an eigenvector $u$ with $g(u)>0$ if and only if $\lambda$ is a critical value of $f$ restricted to \[ M :=\{v\in X:g(v)=1\}\,. \] \indent We also assume that \begin{itemize} \item[(ii)] the set $M$ is not empty and the function $f\bigl|_M$ is bounded from below and satisfies $(CPS)_c$ for any $c\in\mathbb{R}$. \end{itemize} This section is devoted to the consequences of (i) and (ii). We consider the isometry $\Psi:X\to X$ defined as $\Psi(u)=-u$ and define $\gamma(A)$ and $\overline{\gamma}(A)$, for every symmetric subset $A$ of $X$, according to Section~\ref{sect:recalls}. Then, for every $k\geq 1$, we set \begin{gather*} \begin{aligned} \underline{\lambda}_k = \inf\big\{&\max_A f: A \text{ is a compact and symmetric subset of } M \\ &\text{with }\gamma(A)\geq k\big\}\,, \end{aligned} \\ \begin{aligned} \overline{\lambda}_k = \inf\big\{&\max_A f: A \text{ is a compact and symmetric subset of } M \\ &\text{with } \overline{\gamma}(A)\geq k \big\}\,, \end{aligned} \end{gather*} where we agree that $\underline{\lambda}_k=+\infty$ (resp. $\overline{\lambda}_k=+\infty$) if there is no $A$ with $\gamma(A)\geq k$ (resp. $\overline{\gamma}(A)\geq k$). According to Section~\ref{sect:recalls}, we have that \begin{gather*} \underline{\lambda}_k \leq \underline{\lambda}_{k+1}\,,\quad \overline{\lambda}_k \leq \overline{\lambda}_{k+1}\,,\quad \underline{\lambda}_k \leq \overline{\lambda}_k\,,\quad \text{for every $k\geq 1$}\,,\\ \underline{\lambda}_1 = \overline{\lambda}_1 = \inf_M f\,. \end{gather*} Since $\underline{\lambda}_1 = \overline{\lambda}_1$, in the following we will simply write $\lambda_1$. \begin{theorem} \label{thm:general1} The following facts hold: \begin{itemize} \item[(a)] if $\underline{\lambda}_k < +\infty$, then $\underline{\lambda}_k$ is an eigenvalue; \item[(b)] if $\overline{\lambda}_k < +\infty$, then $\overline{\lambda}_k$ is an eigenvalue; \item[(c)] $\inf_M f$ is achieved, so that $\lambda_1 = \min_M f$; \item[(d)] if $\lambda_1$ is simple, then $\lambda_1 < \underline{\lambda}_2$; \item[(e)] for every $b\in\mathbb{R}$, we have \[ \gamma(\{u\in M:f(u) \leq b\}) < \infty\,, \] whence $\lim_k \underline{\lambda}_k = +\infty$. \end{itemize} \end{theorem} \begin{proof} Assertions (a), (b) and (e) follow from (a), (b) and (d) of Theorem~\ref{thm:general0}, while assertion (c) is a particular case of (a) or (b). (d) If $\lambda_1 = \underline{\lambda}_2$, we have \[ \gamma(\{u\in M:f(u)=\lambda_1\}) \geq 2 \] by (c) of Theorem~\ref{thm:general0}, while $\gamma(\{u,-u\})=1$ for every $u\in M$. \end{proof} \begin{definition} \rm Let $u\in M$ be an eigenvector with eigenvalue $\lambda_1$ and let $\Phi$ be the set of the continuous maps $\varphi:[-1,1]\to M$ such that $\varphi(-1)=-u$ and $\varphi(1)=u$. If $\Phi\neq\emptyset$, we define the ``mountain pass eigenvalue'' associated with $u$ \[ \lambda_{mp}(u) = \inf_{\varphi\in\Phi}\, \max_{-1\leq t\leq 1}\, f(\varphi(t))\,, \] otherwise we set $\lambda_{mp}(u)=+\infty$. \end{definition} \begin{theorem} \label{thm:general2} For every eigenvector $u\in M$ with eigenvalue $\lambda_1$, the following facts hold: \begin{itemize} \item[(a)] if $\lambda_{mp}(u) < +\infty$, then $\lambda_{mp}(u)$ is an eigenvalue; \item[(b)] we have $\lambda_1 \leq \underline{\lambda}_2 \leq \overline{\lambda}_2 \leq \lambda_{mp}(u)$; \item[(c)] if $b\in\mathbb{R}$ and there exists an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$ such that $u\in \varphi(\mathbb{R}^2\setminus\{0\})$ and $f(\varphi(t_1,t_2))\leq b$ for any $(t_1,t_2)\in \mathbb{R}^2\setminus\{0\}$, then \[ \lambda_{mp}(u) \leq b\,. \] \end{itemize} \end{theorem} \begin{proof} Assertion (a) follows from Theorem~\ref{thm:mp}. (b) We already know that $\lambda_1 \leq \underline{\lambda}_2 \leq\overline{\lambda}_2$. If $\varphi:[-1,1]\to M$ is a continuous map such that $\varphi(-1)=-u$ and $\varphi(1)=u$, then $\psi:\mathbb{R}^2\setminus\{0\}\to M$ defined as \[ \psi(t_1,t_2) = \begin{cases} \varphi\Big(\frac{t_1}{\sqrt{t_1^2+t_2^2}}\Big) &\text{if } t_2\geq 0\,,\\[4pt] - \varphi\Big( -\frac{t_1}{\sqrt{t_1^2+t_2^2}}\Big) &\text{if } t_2\leq 0\,, \end{cases} \] is continuous and odd, whence $\overline{\lambda}_2 \leq \lambda_{mp}(u)$. (c) Let $u=\varphi(\tau_1,\tau_2)$ with $(\tau_1,\tau_2)\in \mathbb{R}^2\setminus\{0\}$, whence $-u=\varphi(-\tau_1,-\tau_2)$. There exists a continuous map $\psi:[-1,1]\to\mathbb{R}^2\setminus\{0\}$ such that $\psi(-1)=(-\tau_1,-\tau_2)$ and $\psi(1)=(\tau_1,\tau_2)$. Then $(\varphi\circ\psi):[-1,1]\to M$ is continuous and satisfies $(\varphi\circ\psi)(-1)=-u$, $(\varphi\circ\psi)(1)=u$ and $f((\varphi\circ\psi)(t))\leq b$ for any $t\in[-1,1]$, whence $\lambda_{mp}(u)\leq b$. \end{proof} \begin{example} \rm Let $f,g:\mathbb{R}^3\to\mathbb{R}$ be defined as \begin{gather*} f(x,y,z) = 8z^6 - 15(x^2+y^2+z^2)z^4 +6 (x^2+y^2+z^2)^2 z^2 + 2(x^2+y^2+z^2)^3\,,\\ g(x,y,z) = (x^2+y^2+z^2)^3\,. \end{gather*} Then we have \[ \lambda_1 = 1\,,\quad \underline{\lambda}_2 = \overline{\lambda}_2 = 2\,, \] while $\pm u$ with $u = (0,0,1)$ are the eigenvectors in $M$ with eigenvalue $\lambda_1$. On the other hand, $\lambda_{mp}(u) = 43/16$ so that \[ \lambda_1 < \underline{\lambda}_2 = \overline{\lambda}_2 < \lambda_{mp}(u)\,. \] \end{example} The proof of \cite[Proposition~4.2]{brasco_parini2016} has suggested us the next concept. \begin{definition} \label{defn:decomposable} \rm Let $u\in X$ with $g(u)>0$. We say that $(u_1,u_2)$ is a \emph{decomposition of $u$}, if $u_1, u_2\in X$, $u=u_1+u_2$, $g(u_j)>0$ for $j=1,2$ and \begin{gather*} g(t_1u_1 + t_2u_2) \geq g(t_1u_1) + g(t_2u_2)\,,\\ f(t_1u_1 + t_2u_2) \leq \frac{f(u)}{g(u)}\,g(t_1u_1 + t_2u_2)\,, \end{gather*} for every $t_1,t_2\in \mathbb{R}$. An element $u\in X$ with $g(u)>0$ is said to be \emph{decomposable}, if it admits a decomposition~$(u_1,u_2)$. \end{definition} \begin{proposition} \label{prop:decomposable} Let $b\in\mathbb{R}$ and let $u_1, u_2\in X$ with $g(u_j)>0$ for $j=1,2$ and \begin{gather*} g(t_1u_1 + t_2u_2) \geq g(t_1u_1) + g(t_2u_2)\,,\\ f(t_1u_1 + t_2u_2) \leq b\,g(t_1u_1 + t_2u_2)\,, \end{gather*} for every $t_1,t_2\in \mathbb{R}$. Then there exists an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$, such that \[ \frac{u_1}{g(u_1)^{1/p}}\,,\,\, \frac{u_2}{g(u_2)^{1/p}}\,,\,\, \frac{u_1+u_2}{g(u_1+u_2)^{1/p}}\in \varphi(\mathbb{R}^2\setminus\{0\}) \] and \[ f(\varphi(t_1,t_2)) \leq b \quad\text{for every } (t_1,t_2)\in \mathbb{R}^2\setminus\{0\}\,. \] \end{proposition} \begin{proof} Since \[ g(t_1u_1 + t_2u_2) \geq g(t_1u_1) + g(t_2u_2) = |t_1|^p g(u_1) + |t_2|^p g(u_2) \,, \] we can define an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$ as \[ \varphi(t_1,t_2) = \frac{t_1u_1+t_2u_2}{g(t_1u_1 + t_2u_2)^{1/p}}\,. \] Of course, \[ \frac{u_1}{g(u_1)^{1/p}}=\varphi(1,0)\,,\,\, \frac{u_2}{g(u_2)^{1/p}}=\varphi(0,1)\,,\,\, \frac{u_1+u_2}{g(u_1+u_2)^{1/p}}=\varphi(1,1) \] and \[ f(\varphi(t_1,t_2)) = \frac{f(t_1u_1+t_2u_2)}{g(t_1u_1+t_2u_2)}\leq b\,, \] whence the assertion. \end{proof} \begin{theorem}\label{thm:decomposable} If $\lambda$ is an eigenvalue which admits a decomposable eigenvector, then $\lambda \geq \overline{\lambda}_2$. \end{theorem} \begin{proof} Let $u$ be a decomposable eigenvector with eigenvalue $\lambda$ and let $(u_1,u_2)$ be a decomposition of $u$. By Proposition~\ref{prop:decomposable}, there exists an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$ such that \[ f(\varphi(t_1,t_2)) \leq \frac{f(u)}{g(u)} = \lambda \quad\text{for every $(t_1,t_2)\in \mathbb{R}^2\setminus\{0\}$}\,, \] whence $\overline{\lambda}_2 \leq \lambda$. \end{proof} \section{Main results} \label{sect:main} Let again $X$ be a real Banach space and $f,g:X\to\mathbb{R}$ be two functions satisfying~(i) and~(ii). As before, we will consider only eigenvectors $u$ with $g(u)>0$. Throughout this section, we also assume that: \begin{itemize} \item[(iii)] if $u$ is an eigenvector with eigenvalue $\lambda$ and $v$ is an eigenvector with eigenvalue~$\mu$ (possibly with $\lambda=\mu$), such that $u$ and $v$ are linearly independent and $v$ is not decomposable, then one at least of the following facts holds: \begin{itemize} \item[(a)] we have \begin{gather*} g(t_1u + t_2v) \geq g(t_1u) + g(t_2v)\,,\\ f(t_1u + t_2v) \leq \max\{\lambda,\mu\}\, g(t_1u + t_2v)\,, \end{gather*} for every $t_1,t_2\in \mathbb{R}$. \item[(b)] $u$ is decomposable and admits a decomposition $(u_1,u_2)$ such that \begin{gather*} g(t_1u_1 + t_2v) \geq g(t_1u_1) + g(t_2v)\,,\\ f(t_1u_1 + t_2v) \leq \max\{\lambda,\mu\}\, g(t_1u_1 + t_2v)\,, \end{gather*} for every $t_1, t_2 \in \mathbb{R}$; \item[(c)] $u$ is decomposable and admits a decomposition $(u_1,u_2)$ such that $u_1$ is not an eigenvector. \end{itemize} \end{itemize} This section is devoted to study the consequences of (i), (ii) and (iii). \begin{theorem} \label{thm:simple} The following facts are equivalent: \begin{itemize} \item[(a)] $\lambda_1$ is simple; \item[(b)] we have $\lambda_1 < \underline{\lambda}_2$; \item[(c)] each eigenvector with eigenvalue $\lambda_1$ is not decomposable. \end{itemize} \end{theorem} \begin{proof} By (d) of Theorem~\ref{thm:general1} and Theorem~\ref{thm:decomposable}, it is enough to prove that (c) $\Rightarrow$ (a). Assume, for a contradiction, that $u, v$ are two linearly independent eigenvectors with eigenvalue $\lambda_1$. We know that $u$ and $v$ are not decomposable. Then assertion (a) of assumption (iii) holds. It easily follows that $g(u+v)>0$. Moreover, if $w\in X$ with $g(w)=g(u+v)$, we have \[ f(w)\geq \lambda_1 g(w) = \lambda_1 g(u+v) \geq f(u+v)\,. \] By Remark \ref{rem:minws}, we infer that $(u+v)$ is an eigenvector. Of course $\lambda_1$ is the associated eigenvalue and $u+v$ admits the decomposition $(u,v)$, whence a contradiction. \end{proof} \begin{theorem} \label{thm:empty} There is no eigenvalue $\lambda$ satisfying $\lambda_1 < \lambda < \overline{\lambda}_2$. Moreover, we have $\underline{\lambda}_2 = \overline{\lambda}_2$. \end{theorem} \begin{proof} Assume, for a contradiction, that $\lambda$ is an eigenvalue such that $\lambda_1 < \lambda < \overline{\lambda}_2$ and let $u$ be an eigenvector with eigenvalue $\lambda$ and $v$ an eigenvector with eigenvalue~$\lambda_1$. By Proposition~\ref{prop:cone} we have that $u$ and $v$ are linearly independent. From Theorem \ref{thm:decomposable} we infer that $u$ and $v$ are not decomposable, so that assertion (a) of assumption (iii) holds. By Proposition \ref{prop:decomposable} there exists an odd and continuous map \[ \varphi:\mathbb{R}^2\setminus\{0\}\to M \] such that \[ f(\varphi(t_1,t_2)) \leq \max\{\lambda_1,\lambda\} = \lambda \quad\text{for every } (t_1,t_2)\in \mathbb{R}^2\setminus\{0\}\,, \] whence $\overline{\lambda}_2 \leq \lambda$ and a contradiction follows. If $\lambda_1=\overline{\lambda}_2$, it is obvious that $\lambda_1=\underline{\lambda}_2=\overline{\lambda}_2$. If $\lambda_1<\overline{\lambda}_2$, it follows from Theorems~\ref{thm:decomposable} and~\ref{thm:simple} that $\lambda_1<\underline{\lambda}_2$, whence $\underline{\lambda}_2=\overline{\lambda}_2$. \end{proof} \begin{theorem} If $\lambda_1$ is simple, then $\lambda_1$ is isolated in the set of the eigenvalues. \end{theorem} The above theorem follows from Theorems~\ref{thm:simple} and~\ref{thm:empty}. Now we can prove the main result of the paper. \begin{theorem} \label{thm:main} For every eigenvector $u\in M$ with eigenvalue $\lambda_1$, we have \[ \lambda_{mp}(u) =\underline{\lambda}_2 = \overline{\lambda}_2 \,. \] In particular, $\lambda_{mp}(u)$ is independent of $u$. \end{theorem} \begin{proof} Let $u\in M$ be an eigenvector with eigenvalue $\lambda_1$. By (b) of Theorem~\ref{thm:general2}, it is sufficient to prove that $\lambda_{mp}(u) \leq \underline\lambda_2$. We deal with several possible scenarios. \smallskip \noindent\textbf{Case 1:} $\lambda_1$ is not simple. \smallskip \noindent\textbf{Subcase 1.1:} $u$ is decomposable. Let $(u_1,u_2)$ be a decomposition of $u$. By Proposition~\ref{prop:decomposable}, there exists an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$ such that \[ \frac{u_1}{g(u_1)^{1/p}}\,,\,\, \frac{u_2}{g(u_2)^{1/p}}\,,\,\, u\in \varphi(\mathbb{R}^2\setminus\{0\}) \] and \[ f(\varphi(t_1,t_2)) \leq \lambda_1 \quad\text{for every $(t_1,t_2)\in \mathbb{R}^2\setminus\{0\}$}\,. \] Actually, in this case equality holds and $u_1, u_2$ also are eigenvectors with eigenvalue~$\lambda_1$. Taking into account assertion~(c) of Theorem~\ref{thm:general2}, we conclude that \[ \lambda_{mp}(u) \leq \lambda_1 \leq \underline\lambda_2\,. \] \smallskip \noindent\textbf{Subcase 1.2:} $u$ is not decomposable. Then, by Theorem \ref{thm:simple}, the eigenvalue $\lambda_1$ admits another eigenvector $v$ which is decomposable. Clearly $u$ and $v$ are linearly independent and we take into account assumption~(iii). As in the previous case, if $(v_1,v_2)$ is a decomposition of $v$, then~$v_1$ and $v_2$ also are eigenvectors with eigenvalue $\lambda_1$. Therefore assertion~(c) of assumption~(iii) cannot hold. If~(a) holds, by Proposition~\ref{prop:decomposable} there exists an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$ such that $u\in \varphi(\mathbb{R}^2\setminus\{0\})$ and \[ f(\varphi(t_1,t_2)) = \lambda_1 \quad\text{for every } (t_1,t_2)\in \mathbb{R}^2\setminus\{0\}\,. \] Again, taking into account assertion~(c) of Theorem~\ref{thm:general2}, we conclude that \[ \lambda_{mp}(u) \leq \lambda_1 \leq \underline\lambda_2\,. \] If $(v_1,v_2)$ is a decomposition of $v$ as in~(b), again by Proposition~\ref{prop:decomposable} there exists an odd and continuous map \[ \varphi:\mathbb{R}^2\setminus\{0\}\to M \] with the same properties as in the previous case, whence \[ \lambda_{mp}(u) \leq \lambda_1 \leq \underline\lambda_2\,. \] \smallskip \noindent\textbf{Case 2:} $\lambda_1$ is simple. Now, by Theorem \ref{thm:simple}, it is $\lambda_1 < \underline\lambda_2$ and $u$ is not decomposable. If $\underline\lambda_2 = +\infty$ it is obvious that $\lambda_{mp}(u) \leq \underline\lambda_2$. Otherwise, let $\underline\lambda_2 < +\infty$ and let $v$ be an eigenvector associated with $\underline\lambda_2$. Since $\lambda_1\neq\underline\lambda_2$, from Proposition~\ref{prop:cone} it follows that $u$ and $v$ are linearly independent. This time, all the three scenarios~(a), (b) and (c) of assumption~(iii) are possible. In the cases (a) and (b) we find again, by Proposition~\ref{prop:decomposable}, an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$ such that $u\in \varphi(\mathbb{R}^2\setminus\{0\})$ and \[ f(\varphi(t_1,t_2)) \leq \max\{\lambda_1,\underline\lambda_2\} = \underline\lambda_2 \quad\text{for every $(t_1,t_2)\in \mathbb{R}^2\setminus\{0\}$}\,. \] By assertion~(c) of Theorem~\ref{thm:general2} we conclude that \[ \lambda_{mp}(u) \leq \underline\lambda_2\,. \] Finally, let $(v_1,v_2)$ be a decomposition of $v$ as in~(c) of assumption~(iii). Without loss of generality, we may assume that $g(v_1)=1$. By Proposition~\ref{prop:decomposable} there exists an odd and continuous map $\varphi:\mathbb{R}^2\setminus\{0\}\to M$ such that $v_1\in \varphi(\mathbb{R}^2\setminus\{0\})$ and \[ f(\varphi(t_1,t_2)) \leq \underline\lambda_2 \quad\text{for every } (t_1,t_2)\in \mathbb{R}^2\setminus\{0\}\,. \] If $v_1=\varphi(\tau_1,\tau_2)$, then $-v_1=\varphi(-\tau_1,-\tau_2)$ and there exists a continuous map $\psi:[-1,1]\to \mathbb{R}^2\setminus\{0\}$ such that $\psi(-1)=(-\tau_1,-\tau_2)$ and $\psi(1)=(\tau_1,\tau_2)$. Then $(\varphi\circ\psi):[-1,1]\to M$ satisfies $(\varphi\circ\psi)(-1)=-v_1$, $(\varphi\circ\psi)(1)=v_1$ and $f((\varphi\circ\psi)(t))\leq \underline\lambda_2$ for any $t\in[-1,1]$. On the other hand, it follows from Theorem~\ref{thm:empty} that $f\bigl|_M$ has no critical value in $]\lambda_1,\underline\lambda_2[$. Furthermore, it is $f^{-1}(\lambda_1) = \{ u, -u \}$ and $f(v_1) \leq \underline\lambda_2$ with $\big|d(f\bigr|_{M})\big|(v_1) \neq 0$, as $v_1$ is not an eigenvector. From Theorem~\ref{thm:min} with $a=\lambda_1$ and $b=\underline\lambda_2$, we infer that there exists a continuous map $\eta:[-1,1]\to M$ such that $\eta(-1)=v_1$, $f(\eta(1)) = \lambda_1$ and $f(\eta(t))\leq \underline\lambda_2$ for any $t\in[-1,1]$. It follows that $\eta(1)$ is either $u$ or $-u$. If we define $\zeta:[-1,1]\to M$ by \[ \zeta(t) = \begin{cases} -\eta(-3-4t) & \text{if $-1\leq t\leq -1/2$}\,,\\ (\varphi\circ\psi)(2t) & \text{if $-1/2\leq t\leq 1/2$}\,,\\ \eta(4t-3) & \text{if $1/2\leq t\leq 1$}\,, \end{cases} \] then it is easily seen that $\zeta$ is a continuous map connecting $-u$ and $u$ with $f(\zeta(t)) \leq \underline\lambda_2$ for any $t\in[-1,1]$, whence $\lambda_{mp}(u) \leq \underline\lambda_2$ and the proof is complete. \end{proof} \section{Nonsmooth quasilinear elliptic problems} \label{sect:nq} This section is devoted to the setting of \cite{lucia_schuricht2013}. In the following, we set \[ s^+ = \max\{s,0\}\,,\quad s^- = \max\{-s,0\}\,. \] Let $\Omega$ be an open subset of $\mathbb{R}^n$ and let~$10$ such that \[ \nu|\xi|^p \leq a(x,\xi) \leq b |\xi|^p \] for a.e. $x\in \Omega$ and every $\xi\in\mathbb{R}^n$; \item[(h3)] we have $a(x,t\xi) = |t|^p a(x,\xi)$ for a.e. $x\in\Omega$ and every $t\in\mathbb{R}$ and $\xi\in\mathbb{R}^n$; \item[(h4)] we have $V > 0$ on a set of positive measure and \begin{itemize} \item if $p1$; \item if $p>n$, then $\Omega$ is bounded and $V^+\in L^1(\Omega)$. \end{itemize} \end{itemize} By \cite[Corollaries~2.3 and~2.4]{ekeland_temam1974}, the function $a(x,\cdot)$ is locally Lipschitz for a.e.\ $x\in\Omega$. According to Section~\ref{sect:recalls}, we set \[ a^0(x,\xi;\eta) = \lim_{t \to 0^+}\,\frac{a(x,\xi+t\eta)-a(x,\xi)}{t} = \lim_{k}\,k\left[a(x,\xi+(1/k)\eta)-a(x,\xi)\right]\,. \] It follows that \[ \{x\mapsto a^0(x,U(x);W(x))\} \] is measurable, whenever $U, W:\Omega\to\mathbb{R}^n$ are measurable. We denote by $D^{1,p}_0(\Omega)$ the completion of $C^\infty_c(\Omega)$ with respect to the norm \[ \|\nabla u\|_p = \Big(\int_\Omega |\nabla u|^p\,dx\Big)^{1/p}\,. \] Then we consider \[ X = \big\{u\in D^{1,p}_0(\Omega): V^-|u|^p\in L^1(\Omega)\big\} \] endowed with the norm \[ \|u\| = \Big(\int_\Omega |\nabla u|^p\,dx + \int_\Omega V^- |u|^p\,dx\Big)^{1/p} \] and define $f,g:X\to\mathbb{R}$ by \[ f(u) = \int_\Omega a(x,\nabla u)\,dx\,,\quad g(u) = \frac{1}{p}\,\int_\Omega V |u|^p\,dx\,. \] We also denote by $L^\infty_c(\Omega)$ the set of functions $u\in L^\infty(\Omega)$ vanishing a.e.\ outside some compact subset of $\Omega$. \begin{theorem} \label{thm:ls} The following facts hold: \begin{itemize} \item[(a)] $X$ is a Banach space naturally embedded in $W^{1,p}_{loc}(\Omega)$; \item[(b)] $f$ and $g$ satisfy assumptions~(i) and~(ii); moreover, $f$ is convex and locally Lipschitz, while $g$ is of class $C^1$; \item[(c)] for every $u\in X$, we have that $u$ is an eigenvector in the sense of Definition~\ref{defn:eigenvector} if and only if $u\neq 0$ and there exist $\lambda\in\mathbb{R}$ and $\alpha\in L^{p'}(\Omega;\mathbb{R}^n)$ such that $\alpha\in\partial_{\xi}a(x,\nabla u)$ a.e. in $\Omega$ and \[ \int_\Omega \alpha\cdot\nabla w\,dx = \lambda\, \int_\Omega V |u|^{p-2}u w\,dx \quad\text{for any $w\in X$}\,. \] Moreover, $\lambda$ is the associated eigenvalue in the sense of Definition~\ref{defn:eigenvector}. \end{itemize} \end{theorem} \begin{proof} Assertions (a) and (b) are proved in \cite{lucia_schuricht2013}. Since $f(u)=0$ only if $u=0$, assertion~(c) follows from Corollary~\ref{cor:homoglip}, Theorem~\ref{thm:lagc1} and \cite[Lemma~3.1]{lucia_schuricht2013}. \end{proof} According to Sections~\ref{sect:general} and~\ref{sect:main}, we will consider only eigenvectors $u$ with \[ \int_\Omega V |u|^{p}\,dx > 0\,. \] Several basic properties of eigenvalues and eigenvectors, such as the simplicity of the first eigenvalue and a Strong Maximum Principle, are already proved in \cite{lucia_schuricht2013}. Moreover, it is shown that $\underline{\lambda}_k<+\infty$ for any $k\geq 1$, so that $(\underline{\lambda}_k)$ is a diverging sequence of eigenvalues. We aim first to prove also the extension of a well known property (see \cite{anane1987, lindqvist1990, cuesta2001, kawohl_lindqvist2006, lucia_prashanth2006, kawohl_lucia_prashanth2007, lindqvist2008, brasco_franzina2012}), namely that only the first eigenvalue admits an eigenvector with constant sign, if~$\Omega$ is connected. \begin{lemma} \label{lem:ineq} Let $\underline{a}:\mathbb{R}^n\to\mathbb{R}$ be a convex function which is positively homogeneous of degree $p$. Then $\underline{a}$ is locally Lipschitz and we have \[ \underline{a}(\xi_1) \geq \underline{a}(\xi_0) + \frac{1}{p}\, \underline{a}^0(\xi_0;ps^{p-1}\xi_1 - (p-1)s^p\xi_0 - \xi_0) \] for every $\xi_0, \xi_1\in \mathbb{R}^n$ and $s\in\mathbb{R}$ such that either $s>0$ or $s=0$ and $\xi_1=0$. \end{lemma} \begin{proof} As before, $\underline{a}$ is locally Lipschitz. Assume first that $s>0$. As in the proof of \cite[Lemma~2.1]{brasco_franzina2012}, for every $t\in[0,1]$ we have \begin{align*} \underline{a}(\frac{(1-t)\xi_0 + t s^{p-1}\xi_1}{ ((1-t) + ts^p)^{\frac{p-1}{p}}}) &=((1-t) + ts^p) \underline{a}(\frac{1-t}{(1-t) + ts^p}\,\xi_0 + \frac{t s^p}{(1-t) + ts^p}\,\frac{\xi_1}{s}) \\ &\leq (1-t)\underline{a}(\xi_0) + t s^p\underline{a}\big(\frac{\xi_1}{s}\big) \\ &= (1-t)\underline{a}(\xi_0) + t \,\underline{a}(\xi_1) \,. \end{align*} On the other hand, if we set \[ \eta(t) = \frac{(1-t)\xi_0 + t s^{p-1}\xi_1}{ ((1-t) + ts^p)^{\frac{p-1}{p}}}\,, \] it is easily seen that \[ \eta'(0) = s^{p-1}\xi_1 - \frac{p-1}{p}\,s^p\xi_0 - \frac{1}{p}\,\xi_0\,, \] whence \[ \underline{a}(\xi_1) - \underline{a}(\xi_0) \geq \lim_{t\to 0^+}\, \frac{\underline{a}(\eta(t)) - \underline{a}(\xi_0)}{t} = \frac{1}{p}\, \underline{a}^0(\xi_0;ps^{p-1}\xi_1 - (p-1)s^p\xi_0 - \xi_0) \,. \] In the case $s=0$ and $\xi_1=0$, by Proposition~\ref{prop:homog} we have \[ \underline{a}^0(\xi_0;-\xi_0) = - p \,\underline{a}(\xi_0)\,, \] whence \[ \underline{a}(\xi_1) = 0 = \underline{a}(\xi_0) + \frac{1}{p}\,\underline{a}^0(\xi_0; -\xi_0) \] and the proof is complete. \end{proof} \begin{theorem} \label{thm:poseig} If $u$ is an eigenvector, then the following facts hold: \begin{itemize} \item[(a)] if $u>0$ a.e. in $\Omega$, then the associated eigenvalue is $\lambda_1$; \item[(b)] if $u\geq 0$ a.e. in $\Omega$ and $\Omega$ is connected, then the associated eigenvalue is $\lambda_1$. \end{itemize} \end{theorem} \begin{proof} Let $u$ be an eigenvector with eigenvalue $\lambda$ such that $u\geq 0$ a.e. in $\Omega$. For every $w\in W^{1,p}_{loc}(\Omega)\cap L^\infty(\Omega)$ with $\nabla w \in L^p(\Omega;\mathbb{R}^N)$ and $w\geq 0$ a.e. in $\Omega$ and for every $\varepsilon>0$, it is easily seen that \[ \frac{w^{p}}{(u+\varepsilon)^{p-1}} \in W^{1,p}_{loc}(\Omega)\cap L^\infty(\Omega) \] with \[ \nabla\frac{w^{p}}{(u+\varepsilon)^{p-1}} = p\,\frac{w^{p-1}}{(u+\varepsilon)^{p-1}}\,\nabla w - (p-1)\,\frac{w^p}{(u+\varepsilon)^p}\,\nabla u \in L^p(\Omega;\mathbb{R}^n)\,. \] From Lemma~\ref{lem:ineq} we infer that \[ \int_\Omega a(x,\nabla w)\,dx - \int_\Omega a(x,\nabla u)\,dx \geq \frac{1}{p}\int_{\Omega} a^0\Big(x,\nabla u; \nabla \frac{w^{p}}{(u+\varepsilon)^{p-1}} - \nabla u\Big)\,dx \,. \] Let now $w\in D^{1,p}_0(\Omega)\cap L^\infty_c(\Omega)$ with $w\geq 0$ a.e.\ in $\Omega$, so that \[ \frac{w^{p}}{(u+\varepsilon)^{p-1}} \in X\,. \] Taking into account~(c) of Theorem~\ref{thm:ls}, it follows that \begin{align*} \int_\Omega a(x,\nabla w)\,dx - \int_\Omega a(x,\nabla u)\,dx &\geq \frac{1}{p}\,\int_{\Omega} a^0(x,\nabla u; \nabla\,\frac{w^{p}}{(u+\varepsilon)^{p-1}} - \nabla u)\,dx \\ &\geq \frac{1}{p}\, \int_\Omega \alpha\cdot ( \nabla\frac{w^{p}}{(u+\varepsilon)^{p-1}} - \nabla u)\,dx\\ &= \frac{\lambda}{p}\,\int_\Omega V u ^{p-1} (\frac{w^{p}}{(u+\varepsilon)^{p-1}} - u)\,dx \\ &=\frac{\lambda}{p}\,\int_\Omega V \frac{u^{p-1}}{(u+\varepsilon)^{p-1}}\,w^p\,dx - \frac{\lambda}{p}\,\int_\Omega V u ^p\,dx \\ &= \frac{\lambda}{p}\,\int_\Omega V \frac{u^{p-1}}{(u+\varepsilon)^{p-1}}\,w^p\,dx - \int_\Omega a(x,\nabla u)\,dx \,, \end{align*} whence \[ \int_\Omega a(x,\nabla w)\,dx \geq \frac{\lambda}{p}\,\int_\Omega V \frac{u^{p-1}}{(u+\varepsilon)^{p-1}}\,w^p\,dx\,. \] Now let $w\in X$ with $w\geq 0$ a.e. in $\Omega$, let $(\hat{w}_k)$ be a sequence in $C^\infty_c(\Omega)$ converging to $w$ in $D^{1,p}_0(\Omega)$ and let \[ w_k = \min\{\hat{w}_k^+,w\}\,. \] Then $w_k \in D^{1,p}_0(\Omega)\cap L^\infty_c(\Omega)$ with $0\leq w_k\leq w$ a.e.\ in $\Omega$, whence \[ \int_\Omega a(x,\nabla w_k)\,dx \geq \frac{\lambda}{p}\,\int_\Omega V \frac{u^{p-1}}{(u+\varepsilon)^{p-1}}\,w_k^p\,dx\,. \] Going to the limit as $k\to\infty$ and $\varepsilon\to 0$, we obtain \[ \int_\Omega a(x,\nabla w)\,dx \geq \frac{\lambda}{p}\,\int_{\{u>0\}} V \,w^p\,dx \quad\text{for every $w\in X$ with $w\geq 0$ a.e. in $\Omega$}\,. \] Now, if $u>0$ a.e. in $\Omega$ we actually have \[ \int_\Omega a(x,\nabla w)\,dx \geq \frac{\lambda}{p}\,\int_\Omega V \,w^p\,dx \quad\text{for every $w\in X$ with $w\geq 0$ a.e.\ in $\Omega$}\,. \] By \cite[Proposition~6.1]{lucia_schuricht2013}, the eigenvalue $\lambda_1$ admits an eigenvector $v$ with $v\geq 0$ a.e. in $\Omega$. Without loss of generality, we may assume that $g(v)=1$. Then we have \[ \lambda_1 = \int_\Omega a(x,\nabla v)\,dx \geq \lambda \] and assertion (a) follows. If $u\geq 0$ a.e.\ in $\Omega$ and $\Omega$ is connected, from the Strong Maximum Principle (see \cite[Proposition~5.1]{lucia_schuricht2013}) we infer that $u>0$ a.e. in $\Omega$ and assertion~(b) follows from assertion~(a). \end{proof} \begin{lemma} \label{lem:declip} The following facts hold: \begin{itemize} \item[(a)] if $u\in X$ is an eigenvector and $u$ is sign-changing, then $u$ is decomposable and $(u^+,-u^-)$ is a decomposition of $u$; \item[(b)] if $u\in X$ is an eigenvector with eigenvalue $\lambda$ and there exists a connected component~$\omega$ of $\Omega$ such that $u$ is not a.e. vanishing on $\omega$ and on $\Omega\setminus\omega$, then $u$ is decomposable and $(u_1,u_2)$ given by \[ u_1=u\chi_\omega\,,\,\,u_2=u\chi_{\Omega\setminus\omega} \] is a decomposition of $u$ satisfying \[ f(u_j) = \lambda\,g(u_j) \quad\text{for $j=1,2$}\,; \] \item[(c)] if $u,v \in X$ are two linearly independent eigenvectors and there exists a connected component $\omega$ of $\Omega$ such that \[ u\chi_{\Omega\setminus\omega}=v\chi_{\Omega\setminus\omega}=0\,, \] then one at least, say $u$, is sign-changing and $u^+$, $-u^-$ are not eigenvectors. \end{itemize} \end{lemma} \begin{proof} Since $u^{\pm}\in X$ whenever $u\in X$, as in the proof of \cite[Proposition~6.1]{lucia_schuricht2013} we infer that \[ f(u^{\pm}) = \lambda\,g(u^{\pm})\,. \] Then assertion (a) easily follows. If $\omega$ is a connected component of $\Omega$ and $u_1=u\chi_\omega$, $u_2=u\chi_{\Omega\setminus\omega}$, then $u_1, u_2\in X$ and in a similar way it turns out that \[ f(u_j) = \lambda\,g(u_j)\,. \] Then assertion (b) also follows. Finally, let $u,v$ be two eigenvectors as in assertion (c). Without loss of generality, we may assume that $\omega=\Omega$ with $\Omega$ connected. First of all we claim that one at least is sign-changing. Assume, for a contradiction, that $u$ and $v$ are both of constant sign. From Theorem~\ref{thm:poseig} we infer they are both with eigenvalue $\lambda_1$. But this fact contradicts the simplicity of the first eigenvalue (see \cite[Proposition~6.4]{lucia_schuricht2013}). Assume that $u$ is sign-changing. By assertion(a) we have that $(u^+,-u^-)$ is a decomposition of $u$ and from the Strong Maximum Principle (see \cite[Proposition~5.1]{lucia_schuricht2013}) we infer that $u^+$, $-u^-$ are not eigenvectors. \end{proof} \begin{theorem} \label{thm:lsiii} The functions $f$ and $g$ satisfy also assumption (iii). \end{theorem} \begin{proof} Let $u$ be an eigenvector with eigenvalue $\lambda$ and $v$ an eigenvector with eigenvalue $\mu$ such that $u$ and $v$ are linearly independent and $v$ is not decomposable. By (a) and (b) of Lemma~\ref{lem:declip}, we have that $v$ has constant sign and there exists a connected component $\omega$ of $\Omega$ such that $v\chi_{\Omega\setminus\omega}=0$. If $u\chi_\omega=0$, then assertion~(a) of assumption~(iii) holds. If $u\chi_\omega$ and $u\chi_{\Omega\setminus\omega}$ are both different from $0$, then by~(b) of Lemma~\ref{lem:declip} \[ u_1=u\chi_{\Omega\setminus\omega}\,,\quad u_2=u\chi_\omega \] provide a decomposition of $u$ satisfying assertion~(b) of assumption~(iii). Finally, assume that $u\chi_{\Omega\setminus\omega}=0$. By~(c) of Lemma~\ref{lem:declip} we have that $(u^+,-u^-)$ is a decomposition of $u$ and $u^+$, $-u^-$ are not eigenvectors. Therefore, assertion~(c) of assumption~(iii) holds. \end{proof} \begin{remark} \rm If $\Omega$ is connected, then assertion~(c) of assumption~(iii) always holds. \end{remark} Now all the results of Section~\ref{sect:main} can be applied. Let us point out that $\Omega$ is not assumed to be connected. In particular, let us summarize the results concerning the second eigenvalue. \begin{theorem} For every eigenvector $u\in M$ with eigenvalue $\lambda_1$, we have \[ \lambda_{mp}(u) = \underline{\lambda}_2 = \overline{\lambda}_2 = \min\{\lambda\in\mathbb{R}: \text{$\lambda$ is an eigenvalue with $\lambda\neq \lambda_1$}\}\,. \] \end{theorem} \begin{proof} Since $(\underline{\lambda}_k)$ is a diverging sequence of eigenvalues, the set \[ \{\lambda\in\mathbb{R}: \lambda \text{ is an eigenvalue with } \lambda\neq \lambda_1\} \] is not empty. Then the assertion follows from Theorems~\ref{thm:empty} and~\ref{thm:main}. \end{proof} \section{A problem on quasi open sets} \label{sect:qopen} In this section we show that the abstract setting of Section~\ref{sect:main} can be applied also to the $p$-Laplacian on $p$-quasi open sets. In this way we provide a different proof, without the use of minimizing movements, of the main result of \cite{fusco_mukherjee_zhang2018}. \par Let $1