Content-Type: multipart/mixed; boundary="-------------1306111410369" This is a multi-part message in MIME format. ---------------1306111410369 Content-Type: text/plain; name="13-53.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="13-53.keywords" nonlocal equations, vibrating string, variational methods ---------------1306111410369 Content-Type: application/x-tex; name="Kirchoff_final.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="Kirchoff_final.tex" \documentclass{amsart} \parskip 4pt \parindent 4pt %\usepackage{epsfig} \usepackage{graphicx} \usepackage{color} \usepackage{epstopdf} %\usepackage[monochrome]{color} %\usepackage[notref,notcite]{showkeys} %%%%% THERE ARE SOME MACROS %%%%%%%%%%%%%%%%%%% \def\R{\mathbb R} \def\N{\mathbb N} \def\A{\mathbb A} \def\B{\mathbb B} \def\C{\mathbb C} \def\D{\mathbb D} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\const{\text{\upshape Const.}} \def\loc{{\text{\upshape loc}}} %%%%%%%%%%%% spazi negativi %%%%%%%%%%%%%%%%%%% \def\negquad{\!\!\!\!} \def\negqquad{\!\!\!\!\!\!\!\!} %%%%%%%%%%% il mio epsilon %%%%%%%%%%%%%%%%%%%% \def\eps{\varepsilon} \def\om{{\overline{m}}} \newcommand{\cal}[1]{{\mathcal #1}} %\numberwithin{equation}{section} \newtheorem{thm}{Theorem} \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \renewcommand{\proofname}{\textbf{Proof}} %\textwidth=16.4truecm \hoffset -1.4truecm \voffset -1.5truecm %\textheight=22.4truecm \pagestyle{plain} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \title[A critical Kirchhoff type problem involving a non--local operator] {A critical Kirchhoff type problem\\ involving a non--local operator} \author[A.~Fiscella]{Alessio Fiscella} \address{Dipartimento di Matematica ``Federigo Enriques'', Universit\`a di Milano\\ Via Cesare Saldini, 50 20133 Milano ITALY} \email{alessio.fiscella@unimi.it} \author[E. Valdinoci]{Enrico Valdinoci} \address{Weierstra{\ss} Institut f\"{u}r Angewandte Analysis und Stochastik, Hausvogteiplatz 11A, D-10117, Berlin, Germany} \email{\tt enrico.valdinoci@wias-berlin.de} \date{\today} %\subjclass{} \begin{abstract} In this paper we show the existence of non-negative solutions for a Kirchhoff type problem driven by a non--local integrodifferential operator, that is $$ -M(\left\|u\right\|^{2}_{Z})\mathcal L_Ku=\lambda f(x,u)+\left|u\right|^{2^* -2}u\quad \mbox{in }\,\,\Omega,\qquad u=0\quad\mbox{in}\,\,\mathbb{R}^{n}\setminus\Omega. $$ where $\mathcal L_K$ is an integrodifferential operator with kernel~$K$, $\Omega$ is a bounded subset of $\mathbb{R}^n$, $M$ and $f$ are continuous functions, $\left\|\cdot\right\|_Z$ is a functional norm and $2^*$ is a fractional Sobolev exponent. \end{abstract} \maketitle \section{Introduction} In this paper we deal with the following problem \begin{equation}\label{P} \mbox{ $\left\{\begin{array}{lll} $$-M\left(\displaystyle\int_{\mathbb{R}^{2n}} |u(x)-u(y)|^2K(x-y)dx\,dy\right)\;\mathcal L_Ku$$\\ $$=\lambda f(x,u)+\left|u\right|^{2^* -2}u & \mbox{in } \Omega,$$\\ $$u=0 & \mbox{in } \mathbb{R}^{n}\setminus\Omega$$ \end{array} \right.$} \end{equation} where $n> 2s$ with $s\in(0,1)$, $2^*=2n/(n-2s)$, $\lambda$ is a positive parameter, $\Omega\subset\mathbb{R}^{n}$ is an open bounded set, $M$ and $f$ are two continuous functions whose properties will be introduced later and $\mathcal L_K$ is a non-local operator defined as follows: \begin{equation*} \mathcal L_Ku(x)=\frac{1}{2}\int_{\mathbb{R}^{n}}(u(x+y)+u(x-y)-2u(x))K(y)dy, \end{equation*} for all $x\in\mathbb{R}^{n}$, where $K:\mathbb{R}^{n}\setminus \left\{0\right\}\rightarrow(0,+\infty)$ is a measurable function with the property that \begin{equation}\label{K2} \begin{alignedat}2 &\mbox{there exists}\,\, \theta >0 \,\,\mbox{and}\,\, s\in (0,1) \,\,\mbox{such that}\\ &\theta\left|x\right|^{-(n+2s)}\leq K(x)\leq\theta^{-1}\left|x\right|^{-(n+2s)} \mbox{for any}\,\, x\in\mathbb{R}^{n}\setminus\left\{0\right\}. \end{alignedat} \end{equation} It is immediate to observe that $mK\in L^{1}(\mathbb{R}^{n})$ by setting $m(x)=\min\left\{\left|x\right|^{2},1\right\}$. A typical example for $K$ is given by $K(x)=\left|x\right|^{-(n+2s)}$. In this case problem \eqref{P} becomes \begin{equation}\label{fraclapl} \mbox{ $\left\{\begin{array}{ll} $$M\left(\displaystyle\int_{\mathbb{R}^{2n}}\frac{|u(x)-u(y)|^2}{\left|x-y\right|^{n+2s}}dx\,dy\right)(-\Delta)^{s} u$$\\ $$=\lambda f(x,u)+\left|u\right|^{2^* -2}u & \mbox{in } \Omega,$$\\ $$u=0 & \mbox{in } \mathbb{R}^{n}\setminus \Omega,$$ \end{array} \right.$} \end{equation} where $-(-\Delta)^{s}$ is the fractional Laplace operator which (up to normalization factors) may be defined as \begin{equation*} -(-\Delta)^s u(x)= \frac{1}{2} \int_{\mathbb{R}^{n}}\frac{u(x+y)+u(x-y)-2u(x)}{|y|^{n+2s}}\,dy \end{equation*} for $x\in\mathbb{R}^{n}$ (see \cite{valpal} and references therein for further details on the fractional Laplacian and on the fractional Sobolev space $H^s(\mathbb{R}^n)$). Problems \eqref{P} and \eqref{fraclapl} have a variational nature and the natural space where finding solutions for them is the homogeneous fractional Sobolev space $H^s_0(\Omega)$ (see \cite{valpal}). In order to study \eqref{P} and \eqref{fraclapl} it is important to encode the `boundary condition' $u=0$ in $\mathbb{R}^n\setminus\Omega$ (which is different from the classical case of the Laplacian) in the weak formulation, by considering also that in the norm $\left\|u\right\|_{H^s(\mathbb{R}^n)}$ the interaction between $\Omega$ and $\mathbb{R}^n\setminus\Omega$ gives positive contribution. The functional space that takes into account these boundary condition will be denoted by $Z$ and it was introduced in \cite{fiscella} in the following way. First, we denote by $X$ the linear space of Lebesgue measurable functions $u:\mathbb{R}^n\rightarrow\mathbb{R}$ such that \[\mbox{the map}\,\,\, (x,y)\mapsto (u(x)-u(y))^2 K(x-y)\,\,\, \mbox{is in}\,\,\, L^1\big(Q,\,dxdy\big), \] where $Q:=\mathbb{R}^{2n}\setminus\left({\mathcal C}\Omega\times{\mathcal C}\Omega\right)$. The space $X$ is endowed with the norm \begin{equation}\label{norma} \left\|u\right\|_{X}=\Big(\|u\|_{L^2(\Omega)}+\int_Q |u(x)-u(y)|^2K(x-y)dx\,dy\Big)^{1/2}\,. \end{equation} It is immediate to observe that bounded and Lipschitz functions belong to $X$, thus $X$ is not reduced to $\left\{0\right\}$ (see \cite{sv2, sv3} for further details on space $X$). Now, the functional space $Z$ denotes the closure of $C^{\infty}_{0}(\Omega)$ in $X$. By \cite[Lemma 4]{fiscella}, the space $Z$ is an Hilbert space which can be endowed with the norm defined as \begin{equation}\label{normaz} \|u\|_{Z}=\Big(\int_Q |u(x)-u(y)|^2K(x-y)dx\,dy\Big)^{1/2}\,. \end{equation} Note that in \eqref{norma} and \eqref{normaz} the integrals can be extended to all $\mathbb{R}^{2n}$, since $u=0$ a.e. in $\mathbb{R}^n \setminus\Omega$. In view of our problem, we suppose that $M:\mathbb{R}^+ \rightarrow\mathbb{R}^+$ verifies the following conditions: \begin{equation}\label{m1} M\,\,\mbox{ is an increasing and continuous function}; \end{equation} \begin{equation}\label{m2} \mbox{there exists }\,\, m_0> 0\,\, \mbox{such that}\,\, M(t)\geq m_0= M(0)\,\,\mbox{for any}\,\,t\in\mathbb{R}^+\,. \end{equation} A typical example for $M$ is given by $M(t)=m_0 +tb$ with $b\geq 0$. Also, we assume that $f:\Omega\times\mathbb{R}\rightarrow\mathbb{R}$ is a continuous function that satisfies: \begin{equation}\label{f1} \lim_{\left|t\right|\rightarrow 0} \frac{f(x,t)}{t}=0,\,\,\mbox{uniformly in}\,\, x\in\Omega; \end{equation} \begin{equation}\label{f2} \mbox{there exists}\,\, q\in (2,2^*)\,\, \mbox{such that}\,\,\lim_{\left|t\right|\rightarrow \infty} \frac{f(x,t)}{t^{q-1}}=0\,\,\mbox{uniformly in}\,\,x\in\Omega; \end{equation} \begin{equation}\label{f3} \begin{aligned} &\mbox{there exists}\,\, \sigma\in (2,2^*)\,\, \mbox{such that for any}\,\,x\in\Omega\,\,\mbox{and}\,\, t>0\\ &0<\sigma F(x,t)=\sigma\int^{t}_{0}f(x,s)ds\leq tf(x,t)\,. \end{aligned} \end{equation} Moreover, since we intend to find non-negative solution, we assume this further condition for $f$ \begin{equation}\label{f4} f(x,t)=0\quad\mbox{for any}\,\,x\in\Omega\,\,\mbox{and}\,\,t\leq 0. \end{equation} An example of a function satisfying the conditions \eqref{f1}--\eqref{f4} is given by \[f(x,t)=\left\{\begin{array}{ll} $$0 & \mbox{if } t<0,$$\\ $$a(x)t^{q-1} & \mbox{if } 0< t < 1,$$\\ $$a(x)t^{q_1 -1} & \mbox{if } t\geq 1,$$ \end{array}\right. \] with $20$ for any $x\in\Omega$. The weak formulation of \eqref{P} is given by the following problem \begin{equation} \mbox{ $\left\{\begin{array}{lll} $$\displaystyle M(\left\|u\right\|^2_Z)\int_{\mathbb{R}^{2n}} (u(x)-u(y))(\varphi(x)-\varphi(y))K(x-y) dx\,dy\\ \qquad= \displaystyle\lambda\int_\Omega f(x, u(x))\varphi(x)\,dx+\int_\Omega \left|u(x)\right|^{2^*-2}u(x)\varphi(x)dx \,\,\,\,\,\,\,\forall\varphi \in Z$$\\ $$u\in Z$$. \end{array} \right.$}\label{wf} \end{equation} Thanks to our assumptions on $\Omega$, $M$, $f$ and $K$, all the integrals in \eqref{wf} are well defined if $u$, $\varphi\in Z$. We also point out that the odd part of function $K$ gives no contribution to the integral of the left-hand side of \eqref{wf}. Therefore, it would be not restrictive to assume that $K$ is even. Recently, some studies have been performed for critical problems in a non-local setting; we refer the interested readers to \cite{colorado, capella, sv1, sv2, sv3, tan}. Inspired by the above articles, in this paper we would like to investigate the existence of a nontrivial solution for problem \eqref{wf}, by extending the result in classical Laplacian case dealt with in \cite{fig}. \begin{thm} \label{Th1} Let $s\in(0,1)$, $n> 2s$ and $\Omega$ be a bounded open subset of $\mathbb{R}^{n}$. Assume that the functions $K$, $M$ and $f$ satisfy conditions \eqref{K2} and \eqref{m1}--\eqref{f3}. Then there exists $\lambda^* >0$ such that problem \eqref{wf} has a nontrivial solution $u_\lambda$ for all $\lambda\geq\lambda^*$. Such solution also verifies \[\lim_{\lambda\rightarrow\infty}\left\|u_\lambda\right\|_Z=0. \] \end{thm} The paper is organized as follows. In Section~\ref{sec auxiliary} we introduce a truncated problem whose weak solution will be a weak solution of the original problem \eqref{P}. In Section~\ref{sec variational} we prove some technical lemmas. In Section~\ref{sec result} we prove the existence of a solution for the truncated problem and our main result. Finally, in Section~\ref{sec nonnegative} we study the sign of the weak solutions of problem \eqref{P}. The paper ends with an appendix which presents some detailed motivation for our nonlocal equation, starting from some classical models for vibrating strings. \section{The auxiliary problem}\label{sec auxiliary} In order to prove Theorem \ref{Th1} we first study an auxiliary truncated problem, by assuming that $M$ is unbounded (otherwise the truncation on $M$ is not necessary). Given $\sigma$ as in \eqref{f3} and $a\in\mathbb{R}$ such that $m_00$ such that $M(t_0)=a$. Now, by setting \begin{equation*} M_a(t):= \left\{\begin{array}{ll} $$M(t) & \mbox{if } 0\leq t\leq t_0,$$\\ $$a & \mbox{if } t\geq t_0,$$ \end{array}\right. \end{equation*} we can introduce the following auxiliary problem \begin{equation} \mbox{ $\left\{\begin{array}{ll} $$-M_a(\left\|u\right\|^{2}_{Z})\mathcal L_Ku=\lambda f(x,u)+\left|u\right|^{2^* -2}u & \mbox{in } \Omega,$$\\ $$u=0 & \mbox{in } \mathbb{R}^{n}\setminus\Omega$$ \end{array} \right.$}\label{Pa} \end{equation} with $f$ and $\lambda$ defined as in Problem \eqref{P}. By \eqref{m1} we note also that \begin{equation}\label{ma} M_a(t)\leq a\quad\mbox{for any}\,\, t\geq 0. \end{equation} We obtain the following result. \begin{thm} \label{Th2} Let $s\in(0,1)$, $n> 2s$ and $\Omega$ be a bounded open subset of $\mathbb{R}^{n}$. Assume that conditions \eqref{K2} and \eqref{m1}--\eqref{f3} hold true. Then there exists $\lambda_0 >0$ such that problem \eqref{Pa} has a nontrivial weak solution, for all $\lambda\geq\lambda_0$ and for all $a\in(m_0,\displaystyle\frac{\sigma}{2}m_0)$. \end{thm} \section{Variational formulation and technical lemmas}\label{sec variational} For the proof of Theorem~\ref{Th2}, we observe that problem~\eqref{Pa} has a variational structure, indeed it is the Euler-Lagrange equation of the functional $\mathcal J_{a,\,\lambda}:Z\to \mathbb{R}$ defined as follows $$\mathcal J_{a,\,\lambda}(u)=\frac{1}{2}\widehat{M_a}(\left\|u\right\|^{2}_{Z})-\lambda\int_\Omega F(x, u(x))dx-\frac{1}{2^*}\int_\Omega \left|u(x)\right|^{2^*}dx\,.$$ where \[\widehat{M_a}(t)=\int^{t}_{0}M_a(s)ds. \] \noindent Note that the functional $\mathcal J_{a,\,\lambda}$ is Fr\'echet differentiable in $u\in Z$ and for any $\varphi\in Z$ \begin{equation}\label{derivata} \begin{alignedat}2 \mathcal J'_{a,\,\lambda}(u)(\varphi)&= M_a(\left\|u\right\|^{2}_{Z})\int_Q \big(u(x)-u(y)\big)\big(\varphi(x)-\varphi(y)\big)K(x-y)\,dx\,dy\\ &-\lambda\int_\Omega f(x, u(x))\varphi(x)\,dx-\int_\Omega \left|u(x)\right|^{2^*-2}u(x)\varphi(x)dx\,. \end{alignedat} \end{equation} Now we prove that the functional $\mathcal J_{a,\,\lambda}$ has the geometric features required by the Mountain Pass Theorem. \begin{lem}\label{mp1} Let $K$, $M$ and $f$ be three functions satisfying \eqref{K2} and \eqref{m1}--\eqref{f3}. Then there exist two positive constants $\rho$ and $\alpha$ such that \begin{equation} \mathcal J_{a,\,\lambda}(u)\geq\alpha>0, \end{equation} for any $u\in Z$ with $\left\|u\right\|_{Z}=\rho$. \end{lem} \begin{proof} By \eqref{f1} and \eqref{f2} it follows that, for any $\epsilon>0$ there exists $\delta=\delta(\epsilon)>0$ such that \begin{equation}\label{thebre} \left|F(x,t)\right|\leq \epsilon\left|t\right|^2 +\delta\left|t\right|^q\,. \end{equation} By \eqref{m2} and \eqref{thebre} we get \[\mathcal J_{a,\,\lambda}(u)\geq \frac{m_0}{2}\left\|u\right\|^{2}_{Z}-\epsilon\lambda\int_\Omega \left|u(x)\right|^2 dx-\delta\lambda\int_\Omega\left|u(x)\right|^q dx-\frac{1}{2^*}\int_\Omega\left|u(x)\right|^{2^*} dx. \] So, by using a fractional Sobolev inequality (see \cite[Theorem 6.5]{valpal}), there is a positive constant $C=C(\Omega)$ such that \[\mathcal J_{a,\,\lambda}(u)\geq \left(\frac{m_0}{2}-\epsilon\lambda C\right)\left\|u\right\|^{2}_{Z}-\delta\lambda C\left\|u\right\|^{q}_{Z}-C\left\|u\right\|^{2^*}_{Z}. \] Therefore, by fixing $\epsilon$ such that $k:=\displaystyle\frac{m_0}{2}-\epsilon\lambda C >0$, since $2\rho$. \end{lem} \begin{proof} We fix $u_0\in Z$ such that $\left\|u_0\right\|_{Z}=1$ and $u_0\geq 0$ a.e. in $\mathbb{R}^n$. Now, let $t>0$. By using \eqref{f3} and \eqref{ma}, we get \[\mathcal J_{a,\,\lambda}(tu_0)\leq a\frac{t^2}{2} -c_1t^\sigma \lambda\int_\Omega \left|u_0(x)\right|^\sigma dx+c_2 \left|\Omega\right|-\frac{t^{2^*}}{2^*}\int_\Omega \left|u_0(x)\right|^{2^*}dx. \] Since $\sigma>2$, passing to the limit as $t\rightarrow +\infty$, we get that $\mathcal J_{a,\,\lambda}(tu_0)\rightarrow -\infty$, so that the assertion follows taking $e=t_{*}u_0$, with $t_* >0$ large enough. \end{proof} Now, in order to prove the boundedness of the a Palais--Smale sequence we set \begin{equation*} c_{a,\,\lambda}:=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}\mathcal J_{a,\,\lambda}(\gamma(t))>0 \end{equation*} where \[\Gamma:=\left\{\gamma\in C([0,1],\,Z):\,\gamma(0)=0,\,\mathcal J_{a,\,\lambda}(\gamma(1))<0\right\}. \] \begin{lem}\label{limitatezza} Let $K$, $M$ and $f$ be three functions satisfying \eqref{K2} and \eqref{m1}--\eqref{f3}. Let $\left\{u_{j}\right\}_{j\in\mathbb{N}}$ be a sequence in $Z$ such that \begin{equation}\label{ps1} \mathcal J_{a,\,\lambda}(u_{j})\rightarrow c_{a,\,\lambda}, \end{equation} and \begin{equation}\label{ps2} \sup\left\{\left|\mathcal J'_{a,\,\lambda}(u_{j})(\phi)\right|:\,\phi\in Z,\,\left\|\phi\right\|_{Z}=1\right\}\rightarrow 0\quad\forall\phi\in Z, \end{equation} as $j\rightarrow +\infty$. Then $\left\{u_j\right\}_{j\in\mathbb{N}}$ is bounded in $Z$. \end{lem} \begin{proof} By \eqref{ps1} and \eqref{ps2} there exists $C>0$ such that \begin{equation}\label{4.1} \left|\mathcal J_{a,\,\lambda}(u_j)\right|\leq C\,\,\,\mbox{and}\,\,\, \left| \mathcal J'_{a,\,\lambda}(u_j)\left(\frac{u_j}{\left\|u_j\right\|_{Z}}\right)\right|\leq C, \end{equation} for any $j\in\mathbb{N}$. Moreover, by \eqref{m2}, \eqref{f3}, and \eqref{ma} it follows that \begin{equation}\label{4.2} \begin{alignedat}2 &\mathcal J_{a,\,\lambda}(u_j)-\frac{1}{\sigma}\mathcal J'_{a,\,\lambda}(u_j)(u_j)\\ &\qquad\quad\quad\geq\frac{1}{2}\widehat{M_a}(\left\|u_j\right\|^{2}_{Z})-\frac{1}{\sigma}M_a(\left\|u_j\right\|^{2}_{Z})\left\|u_j\right\|^{2}_{Z}\geq\left(\frac{1}{2}m_0-\frac{1}{\sigma}a\right)\left\|u_j\right\|^{2}_{Z} \end{alignedat} \end{equation} So, by combining \eqref{4.1} with \eqref{4.2} and by remembering that $m_00$ verifying $\mathcal J_{a,\,\lambda}(t_\lambda e)=\displaystyle\max_{t\geq 0}J_{a,\,\lambda}(te)$. Hence, $\mathcal J'_{a,\,\lambda}(t_\lambda e)(e)=0$ and by \eqref{derivata} we get \begin{equation}\label{3.1} t_\lambda \left\|e\right\|^2_Z M_a(t^{2}_{\lambda}\left\|e\right\|^2_Z)= \lambda\int_\Omega f(x, t_\lambda e(x))e(x)\,dx+t^{2^*-1}_{\lambda}\int_\Omega \left|e(x)\right|^{2^*}dx\,. \end{equation} Now, by construction $e\geq0$ a.e. in $\mathbb{R}^n$. So, by \eqref{f3}, \eqref{ma} and \eqref{3.1} it follows \[a\left\|e\right\|^2_Z\geq t^{2^*-2}_{\lambda}\int_\Omega \left|e(x)\right|^{2^*}dx, \] which implies that $t_\lambda$ is bounded for any $\lambda>0$. Thus, there exists $\beta\geq 0$ such that $t_{\lambda_j}\rightarrow \beta$ as $j\rightarrow+\infty$. So, by using also \eqref{ma} and \eqref{3.1} there exists $D>0$ such that \begin{equation}\label{D} \lambda_j\int_\Omega f(x, t_{\lambda_j} e(x))e(x)\,dx+t^{2^*-1}_{\lambda_j}\int_\Omega \left|e(x)\right|^{2^*}dx=t_{\lambda_j}M_a(t^{2}_{\lambda_j}\left\|e\right\|^2_Z)\leq D \end{equation} for any $j\in\mathbb{N}$. We claim that $\beta=0$. Indeed, if $\beta>0$ then by \eqref{f1}, \eqref{f2} for any $\epsilon>0$ there exists $\delta=\delta(\epsilon)>0$ such that \begin{equation*} \left|f(x,t)\right|\leq 2\epsilon\left|t\right| +q\delta\left|t\right|^{q-1}\,. \end{equation*} and so, by the Dominated Convergence Theorem, \[\int_\Omega f(x, t_{\lambda_j} e(x))e(x)\,dx\rightarrow\int_\Omega f(x, \beta e(x))e(x)\,dx\quad\mbox{as}\,\,j\rightarrow+\infty. \] By remembering that $\lambda_j\rightarrow+\infty$, we get \[\lim_{j\rightarrow+\infty}\lambda_j\int_\Omega f(x, t_{\lambda_j} e(x))e(x)\,dx+t^{2^*-1}_{\lambda_j}\int_\Omega \left|e(x)\right|^{2^*}dx=+\infty \] which contradicts \eqref{D}. Thus, we have that $\beta=0$. Now, we consider the following path $\gamma_*(t)=te$ for $t\in[0,1]$ which belongs to $\Gamma$. By using \eqref{f3} we get \begin{equation}\label{3.2} 0 \delta\}} |u(x)|^2 |x-y|^{-n-2s}\,dx\,dy=0. \end{equation} \end{prop} \begin{proof} We set $$ \xi_\delta:= \left( \int_{B_\delta(p)} |u(x)|^{2^*}\,dx\right)^{2/2^*}$$ and we remark that \begin{equation}\label{B796} \lim_{\delta\rightarrow0} \xi_\delta=0. \end{equation} Also we observe that, using the H\"older inequality with exponents $2^*/2=n/(n-2s)$ and $n/2s$, we have \begin{equation}\label{B7} \int_{B_\delta(p)} |u(x)|^2\,dx\le \left( \int_{B_\delta(p)} |u(x)|^{2^*}\,dx\right)^{2/2^*} \left( \int_{B_\delta(p)} 1\,dx\right)^{2s/n} \le C \xi_\delta \delta^{2s}, \end{equation} for some $C>0$ independent of $\delta$ (in what follows we will possibly change $C$ from line to line). Moreover \begin{equation}\label{eqB6} (U\times V)\cap \{|x-y|\le\delta\}\,\subseteq\, B_{2\delta}(p)\times B_{2\delta}(p). \end{equation} Indeed, if $(x,y)\in U\times V=B_\delta(p)\times \mathbb{R}^n$, with $|x-y|\le\delta$, we have that $$ |p-y|\le |p-x|+|x-y|\le \delta+\delta,$$ and so we get \eqref{eqB6}. On the other hand, if $(x,y)\in U\times V=\mathbb{R}^n\times B_\delta(p)$ with $|x-y|\le\delta$, we obtain $$ |p-x|\le |p-y|+|y-x|\le\delta+\delta,$$ and this completes the proof of \eqref{eqB6}. Now we use \eqref{eqB6}, we change variable $z:=x-y$ and we conclude that \begin{eqnarray*} && \int_{x\in U}\int_{y\in V\cap\{|x-y|\le \delta\}}|u(x)|^2 |x-y|^{2-n-2s}\,dx\,dy\\ &&\qquad \le \int_{x\in B_{2\delta}(p)}\int_{y\in B_{2\delta}(p)\cap\{|x-y|\le \delta\}} |u(x)|^2 |x-y|^{2-n-2s}\,dx\,dy \\ &&\qquad \le \int_{x\in B_{2\delta}(p)}\int_{z\in B_{\delta}} |u(x)|^2 |z|^{2-n-2s}\,dx\,dz\\ &&\qquad\le C\delta^{2-2s} \int_{x\in B_{2\delta}(p)}|u(x)|^2 \,dx. \end{eqnarray*} Using this and \eqref{B7} we obtain \begin{eqnarray*} && \delta^{-2}\int_U\int_{V\cap\{|x-y|\le \delta\}} |u(x)|^2 |x-y|^{2-n-2s}\,dx\,dy\\ &&\quad\le C\delta^{-2s} \int_{x\in B_{2\delta}(p)}|u(x)|^2 \,dx \le C\xi_\delta. \end{eqnarray*} This and \eqref{B796} imply \eqref{EX.1}. Now we prove \eqref{EX.2}. For this, we fix an auxiliary parameter $K>2$ (such parameter will be taken arbitrarily large at the end, after sending $\delta\rightarrow0$). We observe that \begin{equation}\label{BBeqB} U\times V\,\subseteq\, \big(B_{K\delta}(p)\times\mathbb{R}^n\big)\cup\big( (\mathbb{R}^n\setminus B_{K\delta}(p)) \times B_{\delta}(p) \big). \end{equation} Indeed, if $U\times V=B_\delta(p)\times \mathbb{R}^n$, then of course $U\times V\subseteq B_{K\delta}(p)\times\mathbb{R}^n$, hence \eqref{BBeqB} is obviuous. If instead $(x,y)\in U\times V=\mathbb{R}^n\times B_\delta(p)$, we distinguish two cases: if $x\in B_{K\delta}(p)$ then $(x,y)\in B_{K\delta}(p)\times \mathbb{R}^n$; if $x\in\mathbb{R}^n\setminus B_{K\delta}(p)$, then \begin{eqnarray*} &&(x,y)\in (\mathbb{R}^n\setminus B_{K\delta}(p))\times V= (\mathbb{R}^n\setminus B_{K\delta}(p))\times B_\delta(p).\end{eqnarray*} This completes the proof of \eqref{BBeqB}. Now we compute \begin{equation}\label{56.1} \begin{split} & \int_{x\in B_{K\delta}(p)}\int_{y\in\mathbb{R}^n\cap\{|x-y|> \delta\}} |u(x)|^2 |x-y|^{-n-2s}\,dx\,dy\\ &\qquad= \int_{x\in B_{K\delta}(p)}\int_{z\in\mathbb{R}^n\setminus B_\delta} |u(x)|^2 |z|^{-n-2s}\,dx\,dz \\ &\qquad= C\delta^{-2s} \int_{x\in B_{K\delta}(p)} |u(x)|^2 \,dx \\ &\qquad \le C\xi_{K\delta}, \end{split}\end{equation} where \eqref{B7} has been used again in the last step. Now we observe that if $x\in\mathbb{R}^n\setminus B_{K\delta}(p)$ and $y\in B_\delta(p)$ then \begin{eqnarray*} &&|x-y|\ge |x-p|-|y-p|=\frac{|x-p|}2+\frac{|x-p|}2-|y-p| \\ &&\qquad\ge \frac{|x-p|}2+\frac{K\delta}2-\delta\ge\frac{|x-p|}2.\end{eqnarray*} As a consequence, using the H\"older inequality with exponents $2^*/2=n/(n-2s)$ and $n/2s$, we infer that \begin{equation}\label{56.2.8} \begin{split} & \int_{x\in \mathbb{R}^n\setminus B_{K\delta}(p)} \int_{y\in B_\delta(p)} |u(x)|^2 |x-y|^{-n-2s}\,dx\,dy\\ \le\,& C \int_{x\in \mathbb{R}^n\setminus B_{K\delta}(p)} \int_{y\in B_\delta(p)} |u(x)|^2 |x-p|^{-n-2s}\,dx\,dy \\ =\,& C\delta^n \int_{x\in \mathbb{R}^n\setminus B_{K\delta}(p)}|u(x)|^2 |x-p|^{-n-2s}\,dx \\ \le\,& C\delta^n \left( \int_{x\in \mathbb{R}^n\setminus B_{K\delta}(p)}|u(x)|^{2^*}\,dx\right)^{2/2^*} \left(\int_{x\in \mathbb{R}^n\setminus B_{K\delta}(p)} |x-p|^{-(n+2s)n/2s}\,dx\right)^{2s/n} \\ \leq\,& C\delta^n \| u\|^2_{L^{2^*}(\mathbb{R}^n)} \left(\int_{K\delta}^{+\infty} \rho^{-((n+2s)n/2s)+(n-1)}d\rho \right)^{2s/n}\\ =\,& C\delta^n \| u\|^2_{L^{2^*}(\mathbb{R}^n)} \left( (K\delta)^{-n^2/2s} \right)^{2s/n}\\ =\,& CK^{-n}\| u\|^2_{L^{2^*}(\mathbb{R}^n)}. \end{split}\end{equation} By collecting the results in \eqref{BBeqB}, \eqref{56.1} and \eqref{56.2.8}, we obtain that \begin{eqnarray*} &&\int_U\int_{V\cap\{|x-y|> \delta\}} |u(x)|^2 |x-y|^{-n-2s}\,dx\,dy \\ &\le& \int_{x\in B_{K\delta}(p)}\int_{y\in\mathbb{R}^n\cap\{|x-y|> \delta\}} |u(x)|^2 |x-y|^{-n-2s}\,dx\,dy\\ &&\quad+ \int_{x\in \mathbb{R}^n\setminus B_{K\delta}(p)} \int_{y\in B_\delta(p)} |u(x)|^2 |x-y|^{-n-2s}\,dx\,dy\\ &\le& C\xi_{K\delta} +CK^{-n}\| u\|^2_{L^{2^*}(\mathbb{R}^n)}. \end{eqnarray*} {F}rom this, we send first $\delta\rightarrow0$ and then $K\rightarrow+\infty$ and we readily obtain \eqref{EX.2} (recall again \eqref{B796}). \end{proof} \section{Proof of Theorems \ref{Th1} and \ref{Th2}}\label{sec result} \begin{proof}[\bf Proof of Theorem \ref{Th2}] By Lemmas \ref{mp1} and \ref{mp2} the functional $\mathcal J_{a,\,\lambda}$ satisfies the geometric structure required by the Mountain Pass Theorem (see \cite[Theorem 2.2]{rabinowitz}) Now, it remains to check the validity of the Palais-Smale condition. Let $\left\{u_j\right\}_{j\in\mathbb{N}}$ be a sequence in $Z$ verifying \eqref{ps1} and \eqref{ps2}. Since by Lemma \ref{limitatezza} $\left\{u_j\right\}_{j\in\mathbb{N}}$ is bounded in $Z$, by applying also \cite[Lemma 4]{fiscella} and \cite[Theorem IV.9]{brezis}, up to a subsequence, there exists $u\in Z$ such that $u_j$ converges to $u$ weakly in $Z$, strongly in $L^q(\Omega)$ with $q\in[1,2^*)$ and a.e. in $\Omega$. Also, in particular there exists $h\in L^2(\Omega)$ such that \begin{equation}\label{h} \left|u_j(x)\right|\leq h(x)\quad\mbox{for any}\,\, j\in\mathbb{N}\,\,\mbox{and a.e.}\,\, x\in\Omega. \end{equation} We point out the above inequality and convergences are also verified in all $\mathbb{R}^n$, since $u_j=0=u$ a.e. in $\mathbb{R}^n\setminus\Omega$; in particular we shall assume that $h(x)=0$ for a.e. $x\in\mathbb{R}^n\setminus\Omega$. Moreover, up to a subsequence, there is $\alpha\geq 0$ such that $\left\|u_j\right\|_{Z}\rightarrow\alpha$, so by using \eqref{m1} it follows that $M_a(\left\|u_j\right\|^{2}_{Z})\rightarrow M_a(\alpha^2)$ as $j\rightarrow+\infty$. Now, we claim that \begin{equation}\label{claim} \left\|u_j\right\|^{2}_{Z}\rightarrow\left\|u\right\|^{2}_{Z}\qquad\mbox{as}\,\,j\rightarrow+\infty, \end{equation} which clearly implies that $u_j\rightarrow u$ in $Z$ as $j\rightarrow+\infty$. By \cite[Lemma 4]{fiscella} we know that $\left\{u_j\right\}_{j\in\mathbb{N}}$ is also bounded in $H^s_0(\Omega)$. So, by Phrokorov's Theorem we may suppose that there exist two positive measures $\mu$ and $\nu$ on $\mathbb{R}^n$ such that \begin{equation}\label{convergenza misure} \left|(-\Delta)^s u_j\right|^2 dx\stackrel{*}{\rightharpoonup}\mu\quad\mbox{and}\quad\left|u_j\right|^{2^*}\rightharpoonup\nu \end{equation} in the sense of measures. Moreover, by \cite[Theorem 2]{pal} we obtain an at most countable set of distinct points $\left\{x_i\right\}_{i\in J}$, positive numbers $\left\{\nu_i\right\}_{i\in J}$, $\left\{\mu_i\right\}_{i\in J}$ and a positive measure $\widetilde{\mu}$ with $Supp\, \widetilde{\mu}\subset\overline{\Omega}$ such that \begin{equation}\label{nu} \nu=\left|u\right|^{2^*}dx+\sum_{i\in J} \nu_i\delta_{x_i}, \end{equation} and \begin{equation}\label{mu} \mu=\left|(-\Delta)^s u\right|^2 dx+\widetilde{\mu}+\sum_{i\in J} \mu_i\delta_{x_i},\qquad\nu_i\leq S\mu^{2^*/2}_{i}, \end{equation} with $S$ the best constant of the Sobolev embedding. Our goal is to show that the set $J$ is empty. We argue by contradiction and suppose $J\neq\emptyset$. Then we fix $i\in J$ and for any $\delta>0$ we set $\psi_\delta(x):=\psi((x-x_i)/\delta)$ where $\psi\in C^{\infty}_{0}(\mathbb{R}^n,[0,1])$ is such that $\psi\equiv 1$ in $B(0,1)$ and $\psi\equiv0$ in $\mathbb{R}^n\setminus B(0,2)$. Since for a fixed $\delta>0$ $\left\{\psi_\delta u_j\right\}_{j\in\mathbb{N}}$ is bounded in $Z$ uniformly in $j$, by \eqref{ps2} it follows that $\mathcal J'_{a,\,\lambda}(u_j)(\psi_\delta u_j)\rightarrow 0$ as $j\rightarrow+\infty$, that is \begin{equation}\label{that is} \begin{alignedat}3 M_a&(\left\|u_j\right\|^{2}_{Z})\int_{\mathbb{R}^{2n}} u_j(x)\big(u_j(x)-u_j(y)\big)\big(\psi_\delta(x)-\psi_\delta(y)\big)K(x-y)\,dx\,dy\\ &=-M_a(\left\|u_j\right\|^{2}_{Z})\int_{\mathbb{R}^{2n}} \psi_\delta(y)\left|u_j(x)-u_j(y)\right|^2K(x-y)\,dx\,dy\\ &\quad+\lambda\int_\Omega f(x, u_j(x))\psi_\delta(x)u_j(x)dx+\int_\Omega \left|u_j(x)\right|^{2^*}\psi_\delta(x)dx\,+o_j(1), \end{alignedat} \end{equation} as $j\rightarrow+\infty$. By the Cauchy-Schwartz inequality we have \begin{equation}\label{4.3} \begin{alignedat}3 &\int_{\mathbb{R}^{2n}} u_j(x)\big(u_j(x)-u_j(y)\big)\big(\psi_\delta(x)-\psi_\delta(y)\big)K(x-y)\,dx\,dy\\ &\quad\quad\leq\left(\int_{\mathbb{R}^{2n}} \left|u_j(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy\right)^{1/2}\\ &\qquad\quad\left(\int_{\mathbb{R}^{2n}} \left|u_j(x)-u_j(y)\right|^2K(x-y)\,dx\,dy\right)^{1/2}, \end{alignedat} \end{equation} where the last term in the right-hand side is finite uniformly in $j$. Now, we claim that \begin{equation}\label{limite j e delta} \lim_{\delta\rightarrow0}\left[\lim_{j\rightarrow+\infty}\int_{\mathbb{R}^{2n}} \left|u_j(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy\right]=0. \end{equation} For this, we first fix $\delta>0$ and we observe that $u_j(x)\rightarrow u(x)$ a.e. $x\in\Omega$ as $j\rightarrow+\infty$. Since $u_j=0=u$ a.e. in $\mathbb{R}^n\setminus\Omega$, $u_j(x)\rightarrow u(x)$ a.e. $x\in\mathbb{R}^n$ as $j\rightarrow+\infty$. On the other hand, by \eqref{K2}, \eqref{h}, the boundedness and Lipschitz regularity of $\psi_\delta$ we get, for some $L>0$, \begin{equation*} \begin{alignedat}4 \int_{\mathbb{R}^{2n}}& \left|u_j(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy\\ &\leq\frac{1}{\theta}\int_{\mathbb{R}^{2n}} \left|u_j(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2\left|x-y\right|^{-n-2s}\,dx\,dy\\ &\leq \frac{L^2\delta^{-2}}{\theta}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n\cap\left\{\left|x-y\right|\leq \delta\right\}} \left|u_j(x)\right|^2\left|x-y\right|^{2-n-2s}\,dx\,dy\\ &\quad+\frac{4}{\theta}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n\cap\left\{\left|x-y\right|>\delta\right\}} \left|u_j(x)\right|^2\left|x-y\right|^{-n-2s}\,dx\,dy\\ &\leq C\frac{(L^2\delta^{-2}+4)}{\theta}\int_{\mathbb{R}^n}\left|u_j(x)\right|^2\,dx\,dy\leq C\frac{(L^2\delta^{-2}+4)}{\theta}\int_{\mathbb{R}^n}\left|h(x)\right|^2\,dx\,dy<+\infty \end{alignedat} \end{equation*} with $C=C(n,s,\delta)>0$. Thus, by the Dominated Convergence Theorem it follows that \begin{equation}\label{limitej} \begin{alignedat}2 \lim_{j\rightarrow+\infty}\int_{\mathbb{R}^{2n}} &\left|u_j(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy\\ &=\int_{\mathbb{R}^{2n}} \left|u(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy \end{alignedat} \end{equation} with $\delta>0$ fixed. By arguing as above we have \begin{equation}\label{4.4} \begin{alignedat}4 \int_{U\times V}& \left|u(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy\\ &\leq\frac{1}{\theta}\int_{\mathbb{R}^{2n}} \left|u(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2\left|x-y\right|^{-n-2s}\,dx\,dy\\ &\leq\frac{L^2}{\theta}\delta^{-2}\int_{U}\int_{V\cap\left\{\left|x-y\right|\leq \delta\right\}} \left|u(x)\right|^2 \left|x-y\right|^{2-n-2}\,dx\,dy\\ &\quad+\frac{4}{\theta}\int_{U}\int_{V\cap\left\{\left|x-y\right|>\delta\right\}} \left|u(x)\right|^2\left|x-y\right|^{-n-2}\,dx\,dy\\ \end{alignedat} \end{equation} where $U$ and $V$ are two generic subsets of $\mathbb{R}^n$. Now, we will prove that the term on the right-hand in \eqref{limitej} goes to 0 as $\delta\rightarrow0$, by using \eqref{4.4} case by case. First, we observe that when $U=V=\mathbb{R}^n\setminus B(x_i,\delta)$ all the integrals in \eqref{4.4} are equal to 0. When $U\times V=B(x_i,\delta)\times\mathbb{R}^n$ and $U\times V=\mathbb{R}^n\times B(x_i,\delta)$, we can use Proposition \ref{u per v} together with \eqref{4.4} to get \[\lim_{\delta\rightarrow0}\int_{U\times V} \left|u(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy=0. \] Thus, by using these last results we get \[\lim_{\delta\rightarrow0}\int_{\mathbb{R}^{2n}} \left|u(x)\right|^2\left|\psi_\delta(x)-\psi_\delta(y)\right|^2K(x-y)\,dx\,dy=0, \] and by combining this formula with \eqref{limitej} we prove \eqref{limite j e delta}; from this, by using also \eqref{4.3} it follows that \begin{equation}\label{zero} \lim_{\delta\rightarrow 0}\left[\lim_{j\rightarrow+\infty}M_a(\left\|u_j\right\|^{2}_{Z}) \int_{\mathbb{R}^{2n}} u_j(x)\big(u_j(x)-u_j(y)\big)\big(\psi_\delta(x)-\psi_\delta(y)\big)K(x-y)\,dx\,dy=0\right]. \end{equation} Now, by H\"{o}lder inequality and \eqref{K2} we observe that, for any $x\in\mathbb{R}^n$, \begin{equation}\label{4.5} \begin{alignedat}3 &\left|\int_{\mathbb{R}^n}\frac{u_j(x)-u_j(y)}{\left|x-y\right|^{n+2s}}dx\right|^2\\ \leq &\,2\left|u_j(y)\right|^2\left|\int_{\mathbb{R}^n\setminus\Omega}\frac{1}{\left|x-y\right|^{n+2s}}dx\right|^2+2\left|\int_{\Omega}\frac{u_j(x)-u_j(y)}{\left|x-y\right|^{n+2s}}dx\right|^2\\ \leq &\,C\left|u_j(y)\right|^2+2\frac{\left|\Omega\right|}{\theta}\int_{\Omega}\left|u_j(x)-u_j(y)\right|^2K(x-y)dx, \end{alignedat} \end{equation} with $C=C(\Omega)>0$. So, by \eqref{convergenza misure} and \eqref{4.5} we get \begin{equation}\label{term mu} \begin{alignedat}4 \liminf_{j\rightarrow+\infty} &\int_{\mathbb{R}^n} \psi_\delta(y)\int_\Omega\left|u_j(x)-u_j(y)\right|^2K(x-y)\,dx\,dy\\ &\geq \frac{\theta}{2\left|\Omega\right|}\frac{1}{c(n,s)}\liminf_{j\rightarrow+\infty}\int_{\mathbb{R}^n} \psi_\delta(y) \,c(n,s)\left|\int_{\mathbb{R}^n}\frac{\left|u_j(x)-u_j(y)\right|^2}{\left|x-y\right|^{n+2s}}\,dx\right|^2 dy\\ &\quad-C\liminf_{j\rightarrow+\infty}\int_{\mathbb{R}^n} \psi_\delta(y)\left|u_j(y)\right|^2 dy\\ &\geq\frac{\theta}{2\left|\Omega\right|}\frac{1}{c(n,s)}\int_{\mathbb{R}^n} \psi_\delta(y)d\mu-C\int_{B(x_i,\delta)} \left|u(y)\right|^2 dy. \end{alignedat} \end{equation} Moreover, by \eqref{f1}, \eqref{f2} for any $\epsilon>0$ there exists $\delta=\delta(\epsilon)>0$ such that \begin{equation}\label{f dominata} \left|f(x,t)\right|\leq 2\epsilon\left|t\right| +q\delta\left|t\right|^{q-1}\,. \end{equation} and so, by the Dominated Convergence Theorem we get \begin{equation}\label{term f} \int_{B(x_i,\delta)} f(x, u_j(x))u_j(x)\psi_\delta(x)dx\rightarrow\int_{B(x_i,\delta)} f(x, u(x))u(x)\psi_\delta(x)dx \end{equation} as $j\rightarrow+\infty$; we also observe that the resulting integral goes to 0 as $\delta\rightarrow0$. So, by \eqref{convergenza misure} it follows that \[\int_\Omega \left|u_j(x)\right|^{2^*}\psi_\delta(x)dx\rightarrow\int_\Omega \psi_\delta(x)d\nu\quad\mbox{as}\,\,j\rightarrow+\infty \] and by combining this last formula with \eqref{that is}, \eqref{zero}, \eqref{term mu} and \eqref{term f} we get \begin{equation*} \begin{alignedat}2 &\int_\Omega \psi_\delta(x)d\nu+\int_{B(x_i,\delta)} f(x, u(x))u(x)\psi_\delta(x)dx\\ &\qquad\qquad\quad\geq M_a(\alpha^2)C\left(\int_\Omega\psi_\delta(y)d\mu-\int_{B(x_i,\delta)}\left|u(y)\right|^2 dy\right)+o_\delta(1), \end{alignedat} \end{equation*} recalling that $M_a(\left\|u_j\right\|^{2}_{Z})\rightarrow M_a(\alpha^2)$ as $j\rightarrow+\infty$. By sending $\delta\rightarrow 0$ and by using \eqref{m2} we conclude that $\nu_i\geq M_a(\alpha^2)\mu_i\geq m_0 C\mu_i$ and by using also the inequality in \eqref{mu} we get \begin{equation}\label{4.6} \nu_i\geq\displaystyle\frac{(m_0 C)^{n/2s}}{S^{(n-2s)/2s}}, \end{equation} for any $i\in J$. Now we shall prove that the above expression cannot occur, and so the set $J$ is empty. By \eqref{ps1} and \eqref{ps2} we get \begin{equation}\label{4.7} \lim_{j\rightarrow+\infty}\left(\mathcal J_{a,\,\lambda}(u_j)-\frac{1}{\sigma} \mathcal J'_{a,\,\lambda}(u_j)(u_j)\right)=c_{a,\,\lambda}. \end{equation} Moreover, by \eqref{m2}, \eqref{f3} and remembering that $m_00=J_{a,\,\lambda}(0)$ we conclude that $u\not\equiv 0$. \end{proof} \begin{proof}[\bf Proof of Theorem \ref{Th1}] By Theorem \ref{Th2}, for any $\lambda\geq\lambda_0$ let $u_\lambda$ be a solution of problem \eqref{Pa}. Now, we claim that \begin{equation}\label{claim2} \mbox{there exists }\,\, \lambda^*\geq\lambda_0\,\, \mbox{such that}\,\, \left\|u_\lambda\right\|_{Z}\leq t_0\,\,\mbox{for any}\,\,\lambda\geq\lambda^*\,. \end{equation} where $t_0$ is given as at the beginning of Section \ref{sec auxiliary}. We argue by contradiction and suppose that there is a sequence $\left\{\lambda_j\right\}_{j\in\mathbb{N}}\subset\mathbb{R}$ such that $\left\|u_{\lambda_j}\right\|_{Z}\geq t_0$. Since $u_{\lambda_j}$ is a critical point of the functional $\mathcal J_{a,\,\lambda_j}$, by using also \eqref{m2} and \eqref{f3} it follows that \begin{align*} c_{a,\,\lambda_j}&\geq\frac{1}{2}\widehat{M_a}(\left\|u_{\lambda_j}\right\|^{2}_{Z})-\frac{1}{\sigma}M_a(\left\|u_{\lambda_j}\right\|^{2}_{Z})\left\|u_{\lambda_j}\right\|^{2}_{Z}\\ &\geq\left(\frac{1}{2}m_0-\frac{1}{\sigma}a\right)\left\|u_{\lambda_j}\right\|^{2}_{Z}\geq\left(\frac{1}{2}m_0-\frac{1}{\sigma}a\right)t^{2}_{0}, \end{align*} which contradicts Lemma \ref{infinito} since $m_00$. Since $u\in Z$, by costruction there exists $w\in C^\infty_0(\Omega)$ such that \begin{equation}\label{1} \left\|u-w\right\|_X<\frac{a}{2}. \end{equation} Now, for any $\epsilon>0$ and $x\in\mathbb{R}^n$, we set $v_\epsilon(x):=\left(\epsilon^2+w^2(x)\right)^{1/2}-\epsilon$. We observe that $v_\epsilon=0=w$ in $\mathbb{R}^n\setminus\Omega$ and it is a smooth function by construction. Hence, $v_\epsilon\in C^\infty_0(\Omega)$. Also, we have $v_\epsilon(x)\rightarrow\left|w(x)\right|$ a.e. $x\in\mathbb{R}^n$ as $\epsilon\rightarrow0$. Since $\left|v_\epsilon\right|\leq\left|w\right|$ for any $\epsilon>0$, by the Dominated Convergence Theorem, $v_\epsilon\rightarrow\left|w\right|$ in $L^2(\mathbb{R}^n)$ as $\epsilon\rightarrow0$. On the other hand, \[\left|\nabla v_\epsilon\right|=\frac{\left|w\right|\left|\nabla w\right|}{\left(\epsilon^2+w^2\right)^{1/2}}\leq\left|\nabla w\right|, \] uniformly in $\epsilon$. Therefore, by the boundedness and Lipschitz regularity of $w$ it follows that \begin{equation*} \begin{aligned} &\left|v_\epsilon(x)-\left|w(x)\right|-v_\epsilon(y)+\left|w(y)\right|\,\right|^2K(x-y)\\ &\qquad\qquad\qquad\leq 2\left(\left|v_\epsilon(x)-v_\epsilon(y)\right|^2+\left|\,\left|w(x)\right|-\left|w(y)\right|\,\right|^2\right)K(x-y) \\ &\qquad\qquad\qquad\leq C \,\min\left\{1,\,\left|x-y\right|^2\right\}K(x-y)\in L^1(\mathbb{R}^n\times\mathbb{R}^n). \end{aligned} \end{equation*} which is clearly finite thanks to \eqref{K2}. Thus, by the Dominated Convergence Theorem we get $v_\epsilon\rightarrow\left|w\right|$ in $X$ as $\epsilon\rightarrow0$, in particular \begin{equation}\label{2} \left\|v_\epsilon-\left|w\right|\right\|_X<\frac{a}{2} \end{equation} for $\epsilon$ sufficiently small, say $\epsilon\leq\bar\epsilon$, with $\bar\epsilon=\bar\epsilon(a)>0$. By \eqref{1} and \eqref{2} it is easy to see that \[\left\|\left|u\right|-v_{\bar\epsilon}\right\|_X\leq\left\|\left|u\right|-\left|w\right|\right\|_X+\left\|\left|w\right|-v_{\bar\epsilon}\right\|_X\leq\left\|u-w\right\|_X+\left\|\left|w\right|-v_{\bar\epsilon}\right\|_X 0$ is constant and~$\ell$ is the increment in the length of the string with respect to its rest position (in which the string has length~$2$), i.e. \begin{equation}\label{0099} \ell=\int_{-1}^1 \sqrt{1+u_x^2}\,dx-2. \end{equation} For small deformations, $\sqrt{1+u_x^2}= 1+\frac{u_x^2}{2}$ up to higher order terms, and so $$ \ell=\frac12 \int_{-1}^1 u_x^2\,dx.$$ By plugging this into~\eqref{hook19} we obtain $$ M=m_0+b\int_{-1}^1 u_x^2\,dx=m_0+b\int_{\mathbb{R}} u_x^2\,dx,$$ where we used the notation for which~$u$ is defined to vanish outside~$(-1,1)$. By inserting this into~\eqref{hook1}, one obtains the classical version of the Kirchhoff equation \begin{equation}\label{hook1.89} M\left( \int_{-\infty}^{+\infty} u_x^2\,dx\right)u_{xx}+f=0,\end{equation} with~$M(t)=m_0+bt$. As a historical remark, we mention that the equation was first introduced in~\cite{K1876, K1883} and then, probably independently, proposed in~\cite{carrier45, carrier49}; see also~\cite{oplinger} for a comparison between the theory and the experimental data. We observe that the first term in~\eqref{hook1.89} can be interpreted in a variational way, as arising from an energy of the form \begin{equation}\label{hook E} \frac12 \widehat{M}\left( \int_{-\infty}^{+\infty} u_x^2\,dx\right), \end{equation} where~$\widehat{M}$ is a primitive of~$M$. With this respect, the Kirchhoff equation of nonlocal type that we studied originates from the idea that the energy in~\eqref{hook E} does not depend on the~$H^1$ norm of the function that parameterizes the graph of the string, but rather on its~$H^s$ norm, namely we replaced~\eqref{hook E} with $$ \frac12 \widehat{M}\left( \int_{\mathbb{R}^2} \frac{(u(x)-u(y))^2}{|x-y|^{1+2s}}\,dx\,dy\right),$$ or even with more general kinds of fractional norms. In this sense, while the ``nonlocal'' feature of the tension in the classical Kirchhoff equation surfaces from the average of a ``local'' object (namely $u_x^2$), in the equation we took into account the ``nonlocal'' aspect of the tension arises from an object which is ``nonlocal'' as well. In general, we think it could be interesting to study even more general models in which the tension of the string is related to ``nonlocal'' measurments of the modification of the string from its rest position. Some of these models may be variational in nature (as the one considered here), some others may be not. Another way of obtaining the model we study from the classical Kirchhoff equation goes as follows. Following~\cite{CRS}, for~$\sigma\in(0,1)$, we consider the $\sigma$-length of the string as follows. Let~$E:=\{(x_1,x_2)\in\mathbb{R}^2{\mbox{ s.t. }} x_2