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Mountain Pass Theorem, critical nonlinearities, best critical  
Sobolev constant, variational techniques, integrodifferential operators, 
fractional Laplacian.
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\title[The Brezis-Nirenberg result for the fractional Laplacian]{
The Brezis-Nirenberg result\\ for the fractional Laplacian}

\thanks{The first author was supported by
the MIUR National Research Project {\it Variational and Topological Methods
in the Study of Nonlinear Phenomena}, while the second one by the
ERC grant $\epsilon$ ({\it Elliptic Pde's and Symmetry of Interfaces and Layers
for Odd Nonlinearities}) and the FIRB project A\&B ({\it Analysis
and Beyond}).}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\author[R. Servadei]{Raffaella Servadei}
\address{Dipartimento di Matematica,
          Universit\`a della Calabria,
          Ponte Pietro Bucci 31 B, 87036 Arcavacata di Rende (Cosenza), Italy}
\email{\tt servadei@mat.unical.it}

\author[E. Valdinoci]{Enrico Valdinoci}
\address{Dipartimento di Matematica,
          Universit\`a di Roma `Tor Vergata',
          Via della Ricerca Scientifica, 00133 Roma, Italy}
\email{\tt valdinoci@mat.uniroma2.it}

\keywords{Mountain Pass Theorem, critical nonlinearities, best critical Sobolev constant, variational
techniques, integrodifferential
operators, fractional Laplacian.\\
\phantom{aa} 2010 AMS Subject Classification: Primary:
49J35, 35A15, 35S15;
Secondary: 47G20, 45G05.}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{abstract}
Aim of this paper is to deal with the non-local fractional counterpart of the Laplace equation involving critical nonlinearities studied in the famous paper of Brezis and Nirenberg \cite{bn}. Namely, our model is the following equation
$$ \left\{
\begin{array}{ll}
(-\Delta )^s u-\lambda u=|u|^{2^*-2}u & {\mbox{ in }} \Omega\\
u=0 & {\mbox{ in }} \RR^n\setminus \Omega\,,
\end{array} \right.
$$
where $(-\Delta )^s$ is the fractional Laplace operator, $s\in (0,1)$, $\Omega$ is an open bounded set of $\RR^n$, $n>2s$, with Lipschitz boundary, $\lambda>0$ is a real parameter and $2^*=2n/(n-2s)$ is a fractional critical Sobolev exponent.

In this paper we first study the problem in a general framework, indeed
we consider the equation
$$\left\{
\begin{array}{ll}
\mathcal L_K u+\lambda u+|u|^{2^*-2}u+f(x, u)=0 &  \mbox{in } \Omega\\
u=0 & \mbox{in } \RR^n\setminus \Omega\,,
\end{array}\right.$$
where $\mathcal L_K$ is a general non-local integrodifferential operator and $f$ is a lower order perturbation of the critical power $|u|^{2^*-2}u$\,. In this setting we prove an existence result through variational techniques.

Then, as a concrete example, we derive a Brezis-Nirenberg type
result for our model equation, that is we show that,
if~$\lambda_{1,s}$ is the first eigenvalue of the non-local operator~$(-\Delta)^s$ with homogeneous Dirichlet boundary datum, then for any~$\lambda\in (0, \lambda_{1,s})$
there exists a non-trivial solution of the above model equation, provided $n\geq 4s$\,. In this sense the present work may be seen as the extension
of the classical Brezis-Nirenberg result to the case of non-local fractional
operators.
\end{abstract}

\maketitle

\tableofcontents

\section{Introduction}\label{sec:introduzione}

\subsection{Fractional Laplace equations with critical nonlinearities}\label{subsec:intro1}
Recently, a great attention has been devoted to fractional and non-local
operators of elliptic type, both for their interesting theoretical
structure and in view of concrete applications (for several
motivations, an elementary
introduction to this topic and a -- still not exhaustive --
list of references see, e.g., \cite{valpal}).
Motivated by the interest shared by the mathematical community
in this topic, we consider here the non-local counterpart of the
following nonlinear elliptic partial differential critical equation
\begin{equation}\label{1} \left\{
\begin{array}{ll}
-\Delta u-\lambda u=|u|^{2_*-2}u & {\mbox{ in }} \Omega\\
u=0 & {\mbox{ on }} \partial \Omega\,, \quad 2_*=2n/(n-2)\,, n>2\,,
\end{array} \right.
\end{equation}
namely the non-local fractional equation
\begin{equation}\label{1s}
\left\{
\begin{array}{ll}
(-\Delta)^s u-\lambda u=|u|^{2^*-2}u & {\mbox{ in }} \Omega\\
u=0 & {\mbox{ in }} \RR^n\setminus \Omega\,,
\end{array} \right.
\end{equation}
where $s\in (0,1)$ is fixed and $(-\Delta)^s$ is the fractional Laplace operator,
which (up to normalization factors) may be defined as
\begin{equation} \label{2}
-(-\Delta)^s u(x)=
\frac12
\int_{\RR^n}\frac{u(x+y)+u(x-y)-2u(x)}{|y|^{n+2s}}\,dy\,,
\,\,\,\,\, x\in \RR^n\end{equation}
(see for instance~\cite{valpal} and references therein for further
details on the
fractional Laplacian), $\Omega\subset \RR^n$, $n>2s$, is open, bounded and with
Lipschitz boundary, $\lambda>0$ and
\begin{equation}\label{2star}
2^*=\frac{2n}{n-2s}\,,
\end{equation}
is the fractional critical Sobolev exponent.\footnote{We remark that
$$ 2_*=\frac{2n}{n-2}>\frac{2n}{n-2s}=2^*>2\,.$$
In our framework, the exponent $2^*$ plays the role of a fractional
critical Sobolev exponent (see, e.g. \cite[Theorem~6.5]{valpal}).}
The
homogeneous
Dirichlet datum in~\eqref{1s} is given
in $\RR^n\setminus \Omega$ and not simply on $\partial \Omega$, as it
happens in~\eqref{1},
consistently with the non-local character of the operator~$(-\Delta)^s$\,.

In the case when the nonlinear term in problems like~\eqref{1} has a
subcritical growth (i.e., for instance, is of the form $|u|^{q-2}u$,
with $q\in (2, 2_*)$) the existence of solutions was investigated by
many authors (see \cite{struwe, willem} and references therein), while
the fractional analogue of these results for problem~\eqref{1s} can be
found in \cite{svmountain, svlinking}.

Aim of this paper is to investigate the critical case. In the classical critical setting in the famous paper~\cite{bn} Brezis and Nirenberg proved that
\begin{itemize}
\item[$a)$] if $n\geq 4$, then for any $\lambda \in \big(0, \lambda_1(-\Delta)\big)$ problem~\eqref{1} has a positive solution;

\item[$b)$] if $n=3$ then there exists a constant $\lambda_*\in \big(0, \lambda_1(-\Delta)\big)$ such that for any $\lambda \in \big(\lambda_*, \lambda_1(-\Delta)\big)$ problem~\eqref{1} has a positive solution.
\end{itemize}
In the case when $\Omega$ is a ball in $b)$ they calculated explicitly $\lambda_*$ and proved that the following result (stronger than $b)$) holds true:
\begin{itemize}
\item[$c)$] when $n=3$ and $\Omega$ is a ball, problem~\eqref{1} has a positive solution if and only if $\lambda\in \big(\lambda_1(-\Delta)/4, \lambda_1(-\Delta)\big)$\,.
\end{itemize}
Here $\lambda_1(-\Delta)$ is the first eigenvalue of the Laplacian with homogeneous Dirichlet boundary conditions. In \cite{capfortpalm} Capozzi, Fortunato and Palmieri extended the existence result of \cite{bn}, proving that if $n\geq 4$, then problem~\eqref{1} has a solution for any $\lambda>0$\,.

Our aim is to show that the results of \cite{bn} can be extended to problem~\eqref{1s}. In order to do this, as in the classical case, we will use a variational technique: we can approach the problem at least in two different ways. The first one consists of noting that the solutions of~\eqref{1s} correspond to critical points of the `free' functional defined as follows
$$u \mapsto \frac 1 2 \int_{\RR^{2n}}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-\frac \lambda 2 \int_\Omega |u(x)|^2\,dx
-\frac{1}{2^*}\int_\Omega |u(x)|^{2^*}dx\,.$$
The second variational approach consists of looking for critical points of the functional
$$u \mapsto \int_{\RR^{2n}}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-\lambda \int_\Omega |u(x)|^2\,dx$$
on the sphere~$\big\{u : \|u\|_{L^{2^*}(\Omega)}=1\big\}$\,.
This second approach is  related to the existence of the best fractional critical Sobolev constant.

In the model case~\eqref{1s} these two approaches are equivalent. On the other hand, for more general nonlinearities with critical growth this is not true, since it is not always possible to reduce a boundary value problem to a constrained minimization problem: in this case, the use of the `free' functional becomes, therefore, somewhat unavoidable.

In the present paper we first study the following general integrodifferential equation
\begin{equation}\label{op}
\left\{
\begin{array}{ll}
\mathcal L_K u+\lambda u+|u|^{2^*-2}u+f(x, u)=0 &  \mbox{in } \Omega\\
u=0 & \mbox{in } \RR^n\setminus \Omega\,,
\end{array}\right.
\end{equation}
where $s\in (0,1)$ is fixed, $\Omega\subset \RR^n$ , $n>2s$, is open, bounded and with Lipschitz boundary, $\lambda>0$,
and $\mathcal L_K$ is the non-local operator defined as follows:
\begin{equation}\label{lk}
\mathcal L_Ku(x)=
\frac12
\int_{\RR^n}\Big(u(x+y)+u(x-y)-2u(x)\Big)K(y)\,dy\,,
\,\,\,\,\, x\in \RR^n\,.
\end{equation}
Here $K:\RR^n\setminus\{0\}\rightarrow(0,+\infty)$ is a function such that
\begin{equation}\label{kernel}
{\mbox{$m K\in L^1(\RR^n)$, where $m(x)=\min \{|x|^2,
1\}$\,;}}
\end{equation}
\begin{equation}\label{kernelfrac}
\mbox{there exists}\,\, \theta>0\,\,
\mbox{such that}\,\, K(x)\geq \theta |x|^{-(n+2s)}\,\,
\mbox{for any}\,\, x\in \RR^n \setminus\{0\}\,;\\
\end{equation}
\begin{equation}\label{evenkernel}
K(x)=K(-x)\,\, \mbox{for any}\,\, x\in \RR^n \setminus\{0\}\,.
\end{equation}
A model for $K$ is given by $K(x)=|x|^{-(n+2s)}$ and, in this case,
$\mathcal L_K$ is -- up to a normalization constant -- the fractional
Laplace
operator $-(-\Delta)^s$
defined in \eqref{2}.

The nonlinear term in
equation~\eqref{op} is a Carath\'eodory function
$f:\Omega\times \RR\to \RR$ verifying the following conditions:
\begin{equation}\label{supf}
\sup\Big\{|f(x,t)| : \mbox{a.e.}\,\, x\in \Omega\,,\,\, |t|\le M\Big\}<+\infty\,\,\,
\mbox{for any}\,\, M>0\,;
\end{equation}
\begin{equation}\label{cond0}
{\displaystyle \lim_{t\to 0}\frac{f(x,t)}{t}=0}\,\, \mbox{uniformly in}\,\, x\in \Omega\,;
\end{equation}
\begin{equation}\label{condinfinito}
{\displaystyle \lim_{|t|\to +\infty}\frac{f(x,t)}{|t|^{2^*-1}}=0}\,\, \mbox{uniformly in}\,\, x\in \Omega\,.
\end{equation}

The term $f$ represents a lower order perturbation of the critical nonlinearity~$|u|^{2^*-2}u$\,.
As a model for $f$ we can take the
nonlinearity~$f(x,t)=a(x)|t|^{q}$, with $a\in
L^\infty(\Omega)$ and $q\in (1, 2^*-1)$\,.

In particular we are interested in the weak formulation of~\eqref{op} given by the following problem (for this, it is convenient to assume \eqref{evenkernel})
\begin{equation}\label{problemaK0f}
\left\{\begin{array}{l} {\displaystyle \int_{\RR^{2n} }
(u(x)-u(y))(\varphi(x)-\varphi(y))K(x-y) dx\,dy-\lambda \int_\Omega
u(x)
\varphi(x)\,dx}\\
\\
\qquad \qquad \qquad
{\displaystyle = \int_\Omega |u(x)|^{2^*-2}u(x)\varphi(x)dx+\int_\Omega f(x,u(x))\varphi(x)dx}\,\,\,\,\,\,\, \forall\,\, \varphi \in X_0\\
\\
u\in X_0\,.
\end{array}\right.
\end{equation}

Here the functional space $X$ denotes the linear space of Lebesgue
measurable functions from $\RR^n$ to $\RR$ such that the restriction
to $\Omega$ of any function $g$ in $X$ belongs to $L^2(\Omega)$ and
$$\mbox{the map}\,\,\,
(x,y)\mapsto (g(x)-g(y))\sqrt{K(x-y)}\,\,\, \mbox{is in}\,\,\,
L^2\big(\RR^{2n} \setminus ({\mathcal C}\Omega\times
{\mathcal C}\Omega), dxdy\big)\,,$$
where ${\mathcal C}\Omega:=\RR^n \setminus\Omega$\,, while
$$X_0=\{g\in X : g=0\,\, \mbox{a.e. in}\,\,
\RR^n\setminus
\Omega\}\,.$$
We note that
\begin{equation}\label{C2}
C^2_0 (\Omega)\subseteq X_0,
\end{equation}
see, e.g., \cite[Lemma~11]{sv} (for this
we need condition~\eqref{kernel}), and so
$X$ and $X_0$ are non-empty.


\subsection{Main results of the paper}\label{susec:mainresults}
Assumption~\eqref{cond0} implies  that $f(x,0)=0$,
so that the
function~$u\equiv0$ is a (trivial) solution of~\eqref{op}.
The aim of this paper will be, then, to find non-trivial solutions for \eqref{op}\,.
For this, we will adapt the variational techniques of \cite{bn} to the non-local setting. More precisely, we will study the critical points of the functional $\mathcal J_{K,\,\lambda}:X_0\to \RR$ defined as follows
\begin{equation}\label{JKlambda}
\begin{aligned}
\mathcal J_{K,\,\lambda}(u)& =\frac 1 2 \int_{\RR^{2n}}|u(x)-u(y)|^2 K(x-y)\,dx\,dy-\frac \lambda 2 \int_\Omega |u(x)|^2\,dx
\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad
-\frac{1}{2^*}\int_\Omega |u(x)|^{2^*}\,dx
-\int_\Omega F(x, u(x))dx\,,
\end{aligned}
\end{equation}
where $F$ is given by
\begin{equation}\label{F}
{\displaystyle F(x,t)=\int_0^t f(x,\tau)d\tau}\,.
\end{equation}

As in the classical case of the Laplacian, in the non-local setting the
issue that needs to be overcome
is that the functional~$\mathcal J_{K,\,\lambda}$ does not satisfy the
Palais--Smale condition. This lack of compactness is due to the fact
that the embedding $X_0\hookrightarrow L^{2^*}(\RR^n)$ (or, in the case
of the fractional Laplacian, $H^s(\RR^n)\hookrightarrow L^{2^*}(\RR^n)$)
is not compact. For this the functional~$\mathcal J_{K,\,\lambda}$ does
not verify the Palais--Smale condition globally, but --
as we will see in the sequel -- only in a
suitable
range related to the best fractional critical Sobolev constant in the
embedding $X_0\hookrightarrow L^{2^*}(\RR^n)$. For this,
it is convenient to
define
\begin{equation}\label{SK}
{\displaystyle S_K:=\inf_{v\in X_0\setminus\{0\}}S_K(v)\,,}
\end{equation}
where for any $v\in X_0\setminus\{0\}$
\begin{equation}\label{SKv}
{\displaystyle  S_K(v):=\frac{{\displaystyle\int_{\RR^{2n}}|v(x)-v(y)|^2 K(x-y)\,dx\,dy}}{{\displaystyle\Big(\int_\Omega |v(x)|^{2^*}dx\Big)^{2/2^*}}}=\frac{{\displaystyle\int_{\RR^{2n}}|v(x)-v(y)|^2 K(x-y)\,dx\,dy}}{{\displaystyle\Big(\int_{\RR^n} |v(x)|^{2^*}dx\Big)^{2/2^*}}}\,.}
\end{equation}
Along the paper we also need the following constant
\begin{equation}\label{SKlambda}
{\displaystyle S_{K,\,\lambda}:=\inf_{v\in X_0\setminus\{0\}}S_{K,\,\lambda}(v)\,,}
\end{equation}
where for any $v\in X_0\setminus\{0\}$
\begin{equation}\label{SKlambdav}
\begin{aligned}
{\displaystyle  S_{K,\,\lambda}(v)}& {\displaystyle :=\frac{{\displaystyle\int_{\RR^{2n}}|v(x)-v(y)|^2 K(x-y)\,dx\,dy-\lambda\int_\Omega |v(x)|^2\,dx}}{{\displaystyle\Big(\int_\Omega |v(x)|^{2^*}dx\Big)^{2/2^*}}}}\\
& {\displaystyle =\frac{{\displaystyle\int_{\RR^{2n}}|v(x)-v(y)|^2 K(x-y)\,dx\,dy-\lambda\int_{\RR^n} |v(x)|^2\,dx}}{{\displaystyle\Big(\int_{\RR^n} |v(x)|^{2^*}dx\Big)^{2/2^*}}}\,.}
\end{aligned}
\end{equation}
Of course,~$S_{K,\,\lambda}(v)<S_{K}(v)$ for any~$v\in
X_0\setminus\{0\}$
and any~$\lambda>0$, so~$S_{K,\,\lambda}\le S_{K}$.
The topic of research of this paper will actually focus
on the case when the {\em strict} inequality occurs
(see~\eqref{SlambdaSK}
below).

In the usual fractional Sobolev space $H^s(\RR^n)$ the counterpart of $S_K$ and $S_{K,\,\lambda}$ are the constants $S_s$ and $S_{s,\,\lambda}$ defined as follows
\begin{equation}\label{Ss}
{\displaystyle S_s:=\inf_{v\in H^s(\RR^n)\setminus\{0\}}S_s(v)\,,}
\end{equation}
where for any $v\in H^s(\RR^n)\setminus\{0\}$
\begin{equation}\label{Ssv}
{\displaystyle S_s(v):=\frac{{\displaystyle\int_{\RR^{2n}}\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}} \,dx\,dy}}{{\displaystyle\Big(\int_{\RR^n} |v(x)|^{2^*}dx\Big)^{2/2^*}}}\,,}
\end{equation}
and
\begin{equation}\label{Sslambda}
{\displaystyle S_{s,\,\lambda}:=\inf_{v\in H^s(\RR^n)\setminus\{0\}}S_{s,\,\lambda}(v)\,,}
\end{equation}
where for any $v\in H^s(\RR^n)\setminus\{0\}$
\begin{equation}\label{Sslambdav}
{\displaystyle S_{s,\,\lambda}(v):=\frac{{\displaystyle\int_{\RR^{2n}}\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}} \,dx\,dy-\lambda \int_{\RR^n} |v(x)|^2\,dx}}{{\displaystyle\Big(\int_{\RR^n} |v(x)|^{2^*}dx\Big)^{2/2^*}}}\,.}
\end{equation}
We stress that~$S_{K,\lambda}$ and~$S_K$ depend on~$\Omega$ (since so
does~$X_0$), while~$S_{s,\lambda}$ and~$S_s$ do not
(because the minimization occurs on the whole of~$H^s(\RR^n)$).
The consistency of all these definitions will be discussed in next
results (see Lemmas~\ref{lemmaembeddings}, \ref{lemmaSs}, \ref{lemmaSKlambda} and \ref{lemmaSslambda} in Subsection~\ref{subsec:embeddings})\,.
Notice that~$S_{s,\,\lambda}(v)<S_{s}(v)$ for any~$v\in
H^s(\RR^n)\setminus\{0\}$
and any~$\lambda>0$, so~$S_{s,\,\lambda}\le S_{s}$: in the forthcoming
Theorems~\ref{lapfra0f} and~\ref{lapfra0} we will consider
the case in which the {\em strict} inequality occurs.

The first result of the present paper is the following one:
\begin{theorem}\label{generalkernel0f}
Let $s\in (0,1)$, $n>2s$ and $\Omega$ be an open bounded set of
$\RR^n$ with Lipschitz boundary and let $\lambda\in (0, \lambda_1)$\,, where
$\lambda_1$ is the first eigenvalue of the non-local operator~$-\mathcal
L_K$ with homogeneous
Dirichlet boundary conditions.

Let
$K:\RR^n\setminus\{0\}\rightarrow(0,+\infty)$ be a function satisfying conditions~\eqref{kernel}--\eqref{evenkernel} and let $f:\Omega\times \RR \to \RR$ be a Carath\'eodory function verifying \eqref{supf}--\eqref{condinfinito}\,. Finally, assume that
\begin{equation}\label{u0S}
\begin{aligned}
& \mbox{there exists $u_0\in X_0\setminus\{0\}$ with $u_0\geq 0$ a.e. in $\RR^n$\,, such that}\\
& \quad \qquad \qquad \sup_{\zeta\geq 0}\mathcal J_{K,\,\lambda}(\zeta u_0)<\frac s n\, S_K^{n/(2s)}\,.
\end{aligned}
\end{equation}
Then, problem~\eqref{problemaK0f} admits a
solution $u\in X_0$, which is not identically zero\,.
\end{theorem}

In the case when there is no
perturbation of the critical power nonlinearity, that is $f\equiv 0$ in $\Omega \times \RR$, problem~\eqref{problemaK0f} becomes the critical equation
\begin{equation}\label{problemaK0}
\left\{\begin{array}{l} {\displaystyle \int_{\RR^{2n} }
(u(x)-u(y))(\varphi(x)-\varphi(y))K(x-y) dx\,dy-\lambda \int_\Omega
u(x)
\varphi(x)\,dx}\\
\\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad
{\displaystyle = \int_\Omega |u(x)|^{2^*-2}u(x)\varphi(x)dx}\,\,\,\,\,\,\, \forall\,\, \varphi \in X_0\\
\\
u\in X_0\,,
\end{array}\right.
\end{equation}
for which Theorem~\ref{generalkernel0f}
yields the following existence result:

\begin{theorem}\label{generalkernel0}
Let $s\in (0,1)$, $n>2s$ and $\Omega$ be an open bounded set of
$\RR^n$ with Lipschitz boundary. Let
$K:\RR^n\setminus\{0\}\rightarrow(0,+\infty)$ be a function satisfying conditions~\eqref{kernel}--\eqref{evenkernel} and let $\lambda\in (0,\lambda_1)$\,, where $\lambda_1$ denotes the first eigenvalue of the non-local operator~$-\mathcal L_K$ with homogeneous Dirichlet boundary conditions\,. Assume that
\begin{equation}\label{SlambdaSK}
\begin{aligned}
{\displaystyle S_{K,\,\lambda}<S_K\,.}
\end{aligned}
\end{equation}

Then, problem~\eqref{problemaK0} admits a
solution $u\in X_0$, which is not identically zero\,.
\end{theorem}
In a sense, Theorem~\ref{generalkernel0}
clarifies condition~\eqref{u0S}: indeed, as we will see
in Section~\ref{subsec:mountainpassf0}, condition~\eqref{u0S},
which looks artificial at a first glance, boils down to
the more natural condition~\eqref{SlambdaSK}
when~$f\equiv0$.

In the non-local framework, the simplest example we can deal with is
given by the fractional Laplacian $(-\Delta)^s$, for which we have the following critical equation with a lower order perturbation:
\begin{equation}\label{problemalapfrac0f}
\left\{\begin{array}{l}
{\displaystyle \int_{\RR^{2n}}}
{\displaystyle \frac{(u(x)-u(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}\,dx\,dy -\lambda
\int_\Omega u(x) \varphi(x)\,dx}\\
\\
\qquad \qquad \qquad \qquad \qquad
{\displaystyle =\int_\Omega |u(x)|^{2^*-2}u(x)\varphi(x)dx+\int_\Omega
f(x,u(x))\,\varphi(x)\,dx}\\
\\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\,\,\,\,\,\,\, \forall\,\, \varphi \in H^s(\RR^n)\,\,\mbox{with}\,\,\, \varphi=0\,\,\, \mbox{a.e. in}\,\,\, \RR^n\setminus\Omega\\
\\
u\in H^s(\RR^n)\,.
\end{array}\right.
\end{equation}
This problem represents the Euler-Lagrange equation of the functional $\mathcal J_{s,\,\lambda}: H^s(\RR^n)\to \RR$ defined as
\begin{equation}\label{jsl}
\begin{aligned}
\mathcal J_{s,\,\lambda}(u) &=
\frac 1 2 \int_{\RR^{2n}}
\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-
\frac \lambda 2 \int_\Omega |u(x)|^2\,dx
-\frac{1}{2^*}\int_\Omega |u(x)|^{2^*}dx\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad-\int_\Omega F(x,
u(x))\,dx\,,
\end{aligned}
\end{equation}
with $F$ as in \eqref{F}\,. In this setting our existence result reads as follows:

\begin{theorem}\label{lapfra0f}
Let $s\in (0,1)$, $n>2s$ and $\Omega$ be an open bounded set of $\RR^n$
with
Lipschitz boundary and let $\lambda\in (0, \lambda_{1,s})$\,, where $\lambda_{1,s}$ is the first eigenvalue of the non-local operator~$(-\Delta)^s$ with homogeneous Dirichlet boundary datum.
Let $f:\Omega\times \RR \to \RR$ be a Carath\'eodory function
verifying \eqref{supf}--\eqref{condinfinito} and assume that
\begin{equation}\label{u0Sfrac}
\begin{aligned}
& \mbox{there exists $u_0\in H^s(\RR^n)\setminus\{0\}$ with $u_0\geq 0$ a.e. in $\Omega$}\\
& \qquad \qquad \mbox{and $u_0=0$ a.e. in $\RR^n\setminus \Omega$\,, such that}\\
& \qquad \qquad \qquad {\displaystyle \sup_{\zeta\geq 0} \mathcal J_{s,\,\lambda}(\zeta u_0)<\frac s n\, S_s^{n/(2s)}}\,.
\end{aligned}
\end{equation}

Then, problem~\eqref{problemalapfrac0f} admits a
solution $u\in H^s(\RR^n)$, which is not identically zero, such that $u=0$ a.e. in $\RR^n\setminus \Omega$\,.
\end{theorem}

When $f\equiv 0$ in $\Omega \times \RR^n$ problem~\eqref{problemalapfrac0f} becomes our critical model equation~\eqref{1s}, whose weak formulation is given by
\begin{equation}\label{problemalapfrac0}
\left\{\begin{array}{l}
{\displaystyle \int_{\RR^{2n}}}
{\displaystyle \frac{(u(x)-u(y))(\varphi(x)-\varphi(y))}{
|x-y|^{n+2s}}\,dx\,dy -\lambda
\int_\Omega u(x) \varphi(x)\,dx}\\
\\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
{\displaystyle =\int_\Omega |u(x)|^{2^*-2}u(x)\varphi(x)dx}\\
\\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad\,\,\,\,\,\,\, \forall\,\, \varphi \in H^s(\RR^n)\,\,\mbox{with}\,\,\, \varphi=0\,\,\, \mbox{a.e. in}\,\,\, \RR^n\setminus\Omega\\
\\
u\in H^s(\RR^n)\,.
\end{array}\right.
\end{equation}
In this framework, assumption~\eqref{u0Sfrac}
may be checked directly and
Theorem~\ref{lapfra0f} gives:

\begin{theorem}\label{lapfra0}
Let $s\in (0,1)$, $n\geq 4s$ and $\Omega$ be an open bounded set of $\RR^n$
with Lipschitz boundary.

Then, denoting by $\lambda_{1,s}$ the
first eigenvalue of the non-local operator~$(-\Delta)^s$
with homogeneous Dirichlet boundary datum,  for any
$\lambda\in (0, \lambda_{1,s})$ problem~\eqref{problemalapfrac0} admits a
solution $u\in H^s(\RR^n)$, which is not identically zero,
such that $u=0$ a.e. in $\RR^n\setminus \Omega$\,.
\end{theorem}

For the proof of Theorem~\ref{lapfra0} we use the fact the infimum in
\eqref{Ss} is attained by a suitable function $\widetilde u$, as it is
proved in \cite[Theorem~1.1]{cotsiolis}. Then, starting from $\widetilde
u$
we define a family $u_\varepsilon$ and we estimate $S_{s,\,\lambda}(u_\varepsilon)$\,, see Subsection~\ref{subsec:mountainpassf0lapfrac} for the details. With respect to the classical case of the Laplacian, here this estimate is more delicate, due to the non-local nature of the operator $(-\Delta)^s$\,. In particular, in the non-local framework, in order to control the Gagliardo seminorm of $u_\varepsilon$, pointwise estimates on $u_\varepsilon$ are not enough. Indeed, we also need some integral estimates which allow us to control the interactions between $B_\delta$ and $\RR^n\setminus B_\delta$ (for more details, see Proposition~\ref{stimagagliardo})\,.

As we said before, in this paper we adapt the variational approach used in \cite{bn}
to the non-local framework. For this we will work in
a functional analytical setting that is inspired by (but not equivalent
to) the fractional Sobolev spaces, in order to correctly encode
the Dirichlet boundary datum in the variational formulation.
Also the functional setting we will use allow
us to overcome the problems related to
the lack of compactness and to show that the Palais--Smale condition holds true in a suitable range related to the best fractional Sobolev constant in the embedding $X_0\hookrightarrow L^{2^*}(\RR^n)$ (or, in the case of the fractional Laplacian, $H^s(\RR^n)\hookrightarrow L^{2^*}(\RR^n)$), see Subsection~\ref{subsec:finalremarks} for more details.
In fact, this range of validity of the Palais--Smale condition
may be explicitly related to the strict sign of the inequality
obtained from the Sobolev constants, as we discuss in
Proposition~\ref{condequiv}.

Finally, note that when $s=1$ Theorem~\ref{lapfra0} reads as the results obtained in \cite{bn} for the Laplacian. For all these reasons we think that Theorem~\ref{lapfra0}
may be seen as the natural extension of the results proved in~\cite{bn} (see also \cite{struwe, willem}) to the non-local fractional framework.
With this respect, we recall the recent paper~\cite{tan}, where
a non-local version of the Brezis-Nirenberg result was also considered
(in the case~$s=1/2$ and for an operator conceptually different than the
one treated
here).

The paper is organized as follows. In Section~\ref{sec:preliminary}
we give some basic estimates on the nonlinearity $f$ and its primitive. Moreover, we introduce the functional setting we will work in and we discuss some embeddings of the spaces $X_0$ and $H^s(\RR^n)$ into the usual Lebesgue spaces. At the end of the section, we deal with an eigenvalue problem driven by the non-local integrodifferential operator $-\mathcal L_K$ and we recall some properties of its first eigenvalue.
In Section~\ref{sec:proofgeneral} we prove
Theorems~\ref{generalkernel0f} and \ref{generalkernel0} by
performing a
variational method (in fact, we
reduce
Theorem~\ref{generalkernel0}
to Theorem~\ref{generalkernel0f}
proving the equivalence of conditions~\eqref{u0S}
and~\eqref{SlambdaSK} when~$f\equiv0$). Finally, in
Section~\ref{sec:prooffraclap},
as
an application, we
consider the case of an equation driven by the fractional Laplacian
operator and we prove Theorems~\ref{lapfra0f} and \ref{lapfra0}
(for this, condition~\eqref{u0Sfrac}
will be reduced to~\eqref{u0S}
and so Theorem~\ref{generalkernel0f}
will be applied to prove Theorems~\ref{lapfra0f}, which, in turn,
will imply Theorem~\ref{lapfra0}).

The proofs are, in several points, different than the ones
of the classical case of the Laplacian:
these additional complications are not only
technical, but, sometimes, they somehow reflect
the different distribution of the energy density
due to the non-local interactions. For instance,
the far-away part of the energy in our case,
though it is small, gives a contribution to the functional
that cannot be neglected, since it is of order comparable with
the perturbation.

\section{Some preliminary results}\label{sec:preliminary}
In this section we prove some preliminary results which will be useful in the sequel.

\subsection{Estimates on the nonlinearity and its primitive}
Here we gather some elementary results which will be useful in the main
estimates of the paper.
We use the behavior of $f$ at zero and at infinity in order to deduce some estimates on
the nonlinear term and its primitive (with respect to its second variable) $F$ defined in \eqref{F}\,.
This part is quite standard and does not take into account the non-local nature of the
problem: the reader familiar with these topics may skip it
and go directly to Subsection~\ref{SN}.

\begin{lemma}\label{fFcritico}
Assume $f:\Omega\times \RR \to \RR$ is a Carath\'eodory function satisfying conditions~\eqref{supf}--\eqref{condinfinito}. Then, for any $\varepsilon>0$ there exists $M=M(\varepsilon)>0$ such that a.e. $x\in \Omega$ and for any $t\in \RR$
\begin{equation}\label{cond22critico}
|f(x,t)|\leq 2^*\varepsilon |t|^{2^*-1}+M(\varepsilon)\,
\end{equation}
and so, as a consequence,
\begin{equation}\label{cond22Fcritico}
|F(x,t)|\leq \varepsilon\,|t|^{2^*}+M(\varepsilon) |t|\,,
\end{equation}
where $F$ is defined as in \eqref{F}\,.
\end{lemma}
\begin{proof}
By assumption~\eqref{condinfinito} for any $\varepsilon>0$ there exists $\sigma=\sigma(\varepsilon)>0$ such that for any $t\in \RR$ with $|t|>\sigma$ and a.e. $x\in \Omega$ we get
\begin{equation}\label{cond00critico}
|f(x,t)|\leq 2^*\varepsilon |t|^{2^*-1}\,.
\end{equation}
Moreover, by \eqref{supf} there exists $M(\varepsilon)=M(\sigma(\varepsilon))>0$ such that a.e. $x\in \Omega$ and for any $t\in \RR$ with $|t|\leq \sigma$ we have
\begin{equation}\label{cond11critico}
|f(x,t)|\leq M(\varepsilon)\,.
\end{equation}
Combining \eqref{cond00critico} and \eqref{cond11critico} it easily follows
\eqref{cond22critico}\,. Using the definition of $F$ (see \eqref{F}), we
also get \eqref{cond22Fcritico}\,.
\end{proof}

\begin{lemma}\label{fF}
Assume $f:\Omega\times \RR \to \RR$ is a Carath\'eodory function satisfying conditions~\eqref{supf}--\eqref{condinfinito}. Then, for any $\varepsilon>0$ there exists $\delta=\delta(\varepsilon)$ such that a.e. $x\in \Omega$ and for any $t\in \RR$
\begin{equation}\label{cond22}
|f(x,t)|\leq 2\varepsilon |t|+2^*\delta(\varepsilon) |t|^{2^*-1}\,
\end{equation}
and so, as a consequence,
\begin{equation}\label{cond22F}
|F(x,t)|\leq \varepsilon\,|t|^2+\delta(\varepsilon)\, |t|^{2^*}\,,
\end{equation}
where $F$ is defined as in \eqref{F}\,.
\end{lemma}
\begin{proof}
By assumption~\eqref{cond0} for any $\varepsilon>0$ there exists $\sigma=\sigma(\varepsilon)>0$ such that for any $t\in \RR$ with $|t|<\sigma$ and a.e. $x\in \Omega$ we get
\begin{equation}\label{cond00}
|f(x,t)|\leq 2\varepsilon |t|\,.
\end{equation}
Moreover, by \eqref{cond22critico} with $\varepsilon=1$ we get that
$$|f(x,t)|\leq 2^*|t|^{2^*-1}+M(1)=|t|^{2^*-1}\Big(2^*+\frac{M(1)}{|t|^{2^*-1}}\Big)\,,$$
so that there exists $\delta(\varepsilon)=\delta(\sigma(\varepsilon))>0$ such that a.e. $x\in \Omega$ and for any $t\in \RR$ with $|t|\geq \sigma$ we have
\begin{equation}\label{cond11}
|f(x,t)|\leq 2^*\delta(\varepsilon) |t|^{2^*-1}\,.
\end{equation}
Combining \eqref{cond00} and \eqref{cond11} it easily follows
\eqref{cond22}\,. Using the definition of $F$ given in \eqref{F}, we
also deduce \eqref{cond22F}\,.
\end{proof}

\subsection{The functional setting}\label{SN}
Here, we give some basic results on the spaces
$X$ and $X_0$. The reader familiar
with the fractional Sobolev spaces and their extensions (as
in~\cite{svmountain, svlinking}) may go directly to
Section~\ref{sec:proofgeneral}.
In the
sequel we set
$Q=\RR^{2n}\setminus \mathcal
O$\,, where
\begin{equation}\label{O}
{\mathcal{O}}=
({\mathcal{C}}\Omega)\times({\mathcal{C}}\Omega)\subset\RR^{2n}\,,
\end{equation}
and $\mathcal C\Omega=\RR^n\setminus \Omega$\,.

The space $X$ is endowed with the norm defined as
\begin{equation}\label{norma}
\|g\|_X=\|g\|_{L^2(\Omega)}+\Big(\int_Q |g(x)-g(y)|^2K(x-y)dx\,dy\Big)^{1/2}\,.
\end{equation}
It is easily seen that $\|\cdot\|_X$ is a norm on $X$ (see, for instance, \cite{svmountain} for a proof).

In the following we denote by $H^s(\Omega)$ the usual fractional
Sobolev space endowed with the norm (the so-called \emph{Gagliardo norm})
\begin{equation}\label{gagliardonorm}
\|g\|_{H^s(\Omega)}=\|g\|_{L^2(\Omega)}+
\Big(\int_{\Omega\times
\Omega}\frac{\,\,\,|g(x)-g(y)|^2}{|x-y|^{n+2s}}\,dx\,dy\Big)^{1/2}\,.
\end{equation}
We remark that, even in the model case in which $K(x)=|x|^{-(n+2s)}$, the norms
in \eqref{norma} and \eqref{gagliardonorm}
are not the same, because $\Omega\times\Omega$ is strictly contained in $Q$
(this makes the classical fractional Sobolev space approach not
sufficient for studying the problem).

For further details on the fractional Sobolev spaces we refer to~\cite{valpal}
and to the references therein.

In the next result we give some connections between the space $X_0$ and the usual fractional Sobolev spaces $H^s(\RR^n)$\,, which will be useful along the paper\,.

\begin{lemma}\label{lemmaV}
The following assertions hold true:
\begin{itemize}
\item[$a)$] let $K:\RR^n\setminus\{0\}\rightarrow (0,+\infty)$
satisfy assumptions~\eqref{kernel}--\eqref{evenkernel}\,. Then $X_0 \subseteq H^s(\RR^n)$ and,
moreover
$$ \|v\|_{H^s(\Omega)}\leq
\|v\|_{H^s(\RR^n)}\leq c(\theta)\|v\|_X\,,$$
where $c(\theta)=\max\{1, \theta^{-1/2}\}$\,, with
$\theta$ given in \eqref{kernelfrac}\,;
\item[$b)$] let $K(x)=|x|^{-(n+2s)}$\,. Then
$$X_0=\big\{v\in H^s(\RR^n) : v=0 \,\,\, \mbox{a.e. in}\,\,\, \RR^n\setminus \Omega\big\}\,.$$
\end{itemize}
\end{lemma}

\begin{proof}
Part $a)$ was proved in \cite[Lemma~5-$b)$]{svmountain}.  Let us show part~$b)$\,.
By assertion~$a)$ it is easily seen that $X_0\subseteq \big\{v\in H^s(\RR^n) : v=0 \,\,\, \mbox{a.e. in}\,\,\, \RR^n\setminus \Omega\big\}\,.$

Conversely, let $v\in H^s(\RR^n)$ be such that $v=0$ a.e. in
$\RR^n\setminus \Omega$. Since $v\in H^s(\RR^n)$, we have that
$$\int_{\RR^n} |v(x)|^2\,dx<+\infty$$
and
$$\int_{\RR^{2n}}\frac{\,\,\,|v(x)-v(y)|^2}{|x-y|^{n+2s}}\,dx\,dy<+\infty\,,$$
from which, using the property that $v=0$ a.e. in $\RR^n\setminus \Omega$\,, we deduce
$$\int_\Omega |v(x)|^2\,dx=\int_{\RR^n} |v(x)|^2\,dx<+\infty$$
and
$$\int_Q\frac{\,\,\,|v(x)-v(y)|^2}{|x-y|^{n+2s}}\,dx\,dy=
\int_{\RR^{2n}}\frac{\,\,\,|v(x)-v(y)|^2}{|x-y|^{n+2s}}\,dx\,dy<+\infty\,.$$
 Hence, $v\in X_0$ and this ends the proof of Lemma~\ref{lemmaV}-$b)$\,.
\end{proof}

In the sequel we consider the function
\begin{equation}\label{normaX0}
X_0\ni v\mapsto \|v\|_{X_0}=\left(\int_Q|v(x)-v(y)|^2K(x-y)\,dx\,dy\right)^{1/2}
\end{equation}
and we take \eqref{normaX0} as norm on $X_0$. The following result, proved in \cite[Lemmas~6 and 7]{svmountain}, holds true:

\begin{lemma}\label{lemmanormaequiv}
Let $K:\RR^n\setminus\{0\}\rightarrow (0,+\infty)$
satisfy assumptions~\eqref{kernel}--\eqref{evenkernel}\,.
Then
\begin{itemize}
\item[$a)$] there exists a positive constant $c$, depending only on $n$ and $s$, such that for any $v\in X_0$
$$\|v\|_{L^{2^*}(\Omega)}^2=
\|v\|_{L^{2^*}(\RR^n)}^2\leq c
\int_{\RR^{2n}}\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}}\, dx\,dy\,,
$$
where $2^*$ is given in \eqref{2star}\,;

\item[$b)$] there exists a constant $C>1$, depending only on $n$, $s$, $\theta$ and $\Omega$, such that for any $v\in X_0$
$$\int_Q|v(x)-v(y)|^2K(x-y)\,dx\,dy\leq \|v\|_X^2\leq C \int_Q|v(x)-v(y)|^2K(x-y)\,dx\,dy\,,$$
that is formula~\eqref{normaX0} defines a norm on $X_0$ equivalent to the usual one given in \eqref{norma};
\item[$c)$] $\left(X_0, \|\cdot\|_{X_0}\right)$ is a Hilbert space,
with scalar product
$$\langle u,v\rangle_{X_0}=\int_Q
\big( u(x)-u(y)\big) \big( v(x)-v(y)\big)\,K(x-y)\,dx\,dy.$$
\end{itemize}
\end{lemma}

Note that in \eqref{normaX0} (and in the related scalar product, see Lemma~\ref{lemmanormaequiv}-$c)$) the integral can be extended to all $\RR^{2n}$, since $v\in X_0$ (and so $v=0$ a.e. in $\RR^n\setminus \Omega$).

\subsection{Some embeddings into the usual Lebesgue spaces}\label{subsec:embeddings}
In this subsection we discuss some results related to the embeddings of the spaces $X_0$ and $H^s(\RR^n)$ into the usual Lebesgue spaces, which allow us to define the best fractional critical Sobolev constants given in formulas~\eqref{SK}, \eqref{SKlambda}, \eqref{Ss} and \eqref{Sslambda}\,.

\begin{lemma}\label{lemmaembeddings}
Let $K:\RR^n\setminus\{0\}\rightarrow (0,+\infty)$
satisfy assumptions~\eqref{kernel}--\eqref{evenkernel}\,.
Then, the following assertions hold true:
\begin{itemize}
\item [$a)$] if $\Omega$ has a Lipschitz boundary, then the embedding $X_0\hookrightarrow L^\nu(\RR^n)$ is compact for any $\nu\in [1, 2^*)$;
\item[$b)$] the embedding $X_0\hookrightarrow L^{2^*}(\RR^n)$ is continuous.
\end{itemize}
\end{lemma}
\begin{proof}
Part~$a)$ is proved, for instance, in \cite[Lemma~8]{svmountain}. Let us
show $b)$\,.
For this, by Lemma~\ref{lemmanormaequiv}-$a)$ and assumption~\eqref{kernelfrac} we get that for any $v\in X_0$
$$\begin{aligned}
\|v\|_{L^{2^*}(\Omega)}^2 =\|v\|_{L^{2^*}(\RR^n)}^2 & \leq c
\int_{\RR^{2n}}\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}}\, dx\,dy\\
& \leq \frac c \theta \int_{\RR^{2n}}|v(x)-v(y)|^2 K(x-y)\,dx\,dy\\
&=\frac c \theta \|v\|_{X_0}^2\,,
\end{aligned}$$
for a suitable positive constant $c$, depending only on $n$ and $s$. Hence the assertion~$b)$ is proved.
\end{proof}
Thanks to Lemma~\ref{lemmaembeddings} we can define the constant~$S_K$ given in formula~\eqref{SK}
and see that~$S_K>0$. Moreover, note that, since in formula~\eqref{SKv} the integral
over $\Omega$
can be extended to all $\RR^n$ (being $v=0$ a.e. in $\RR^n\setminus \Omega$)\,, then the function $v\mapsto S_K(v)$ does not depend on the domain $\Omega$\,, while, in general, $S_K$ does (indeed, $X_0$ depends on $\Omega$).

The counterpart of
Lemma~\ref{lemmaembeddings} in the usual fractional Sobolev spaces is
given by the following result proved, for example, in
\cite[Theorem~6.5]{valpal}:
\begin{lemma}\label{lemmaSs}
The embedding $H^s(\RR^n)\hookrightarrow L^\nu(\RR^n)$ is continuous for any $\nu\in [2, 2^*]$\,.
\end{lemma}
Lemma~\ref{lemmaSs} allows us to define the constant~$S_s$
given in formula~\eqref{Ss} and see that~$S_s>0$\,.
\begin{remark}\label{remarkSKSs}
{\rm
In the particular case when $K(x)=|x|^{-(n+2s)}$, it is easily seen
that
$$\inf_{v\in H^s(\RR^n)\setminus\{0\}}S_s(v)\leq \inf_{v\in X_0\setminus\{0\}}S_K(v)\,,$$
since $X_0\subseteq H^s(\RR^n)$ by Lemma~\ref{lemmaV}-$a)$\,.
}\end{remark}

As a final observation, we note that, with respect to the embeddings in
the Lebesgue
spaces, the fractional Sobolev space $H^s(\RR^n)$ behaves like the usual
Sobolev space $H^1(\RR^n)$, while $X_0$ like~$H^1_0(\Omega)$ (this is
due
to the fact that the functions $v\in X_0$ are such that $v=0$ a.e. in
$\RR^n\setminus \Omega$ and so $X_0$ may be seen somehow as a space of
functions defined in the bounded set $\Omega$)\,.

\medskip

Due to the presence of the term $\lambda u$ in the equations we are interested in, it is convenient to define another norm in the spaces $X_0$ and $H^s(\RR^n)$, respectively, as follows:
\begin{equation}\label{normaX0lambda}
X_0\ni v\mapsto
\|v\|_{X_0,\,\lambda}:=\left(\int_{\RR^{2n}}|v(x)-v(y)|^2K(x-y)\,dx\,dy-\lambda
\int_\Omega |v(x)|^2\,dx\right)^{1/2}\,,
\end{equation}
and
\begin{equation}\label{normaHslambda}
H^s(\RR^n)\ni v\mapsto
\|v\|_{H^s(\RR^n),\,\lambda}:=\left(\int_{\RR^{2n}}\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-\lambda
\int_{\RR^n} |v(x)|^2\,dx\right)^{1/2}\,.
\end{equation}
In \cite[Lemma~10]{svlinking} we proved the following result:

\begin{lemma}\label{lemmanormalambda}
Let $K:\RR^n\setminus\{0\}\rightarrow (0,+\infty)$ satisfy
assumptions~\eqref{kernel}--\eqref{evenkernel} and let $\lambda\in (0, \lambda_1)$\,, where $\lambda_1$ is the first eigenvalue of the operator~$-\mathcal L_K$ with homogeneous Dirichlet boundary conditions.  Then, formula~\eqref{normaX0lambda} defines a norm on $X_0$ which is equivalent to the usual one given in~\eqref{normaX0} in the sense that
for any $v\in X_0$
\begin{equation}\label{mMlambda}
\left(1-\frac{\lambda}{\lambda_1}\right) \|v\|_{X_0}^2\leq
\|v\|_{X_0,\,\lambda}^2\leq \|v\|_{X_0}^2\,.
\end{equation}
\end{lemma}

As a consequence of Lemmas~\ref{lemmaembeddings}-$b)$ and \ref{lemmanormalambda} we have the following result:
\begin{lemma}\label{lemmaSKlambda}
Let $K:\RR^n\setminus\{0\}\rightarrow (0,+\infty)$
satisfy assumptions~\eqref{kernel}--\eqref{evenkernel} and let $\lambda\in (0, \lambda_1)$\,, where $\lambda_1$ is the first eigenvalue of the operator~$-\mathcal L_K$ with homogeneous Dirichlet boundary conditions.
Then the embedding $(X_0, \|\cdot\|_{X_0,\,\lambda})\hookrightarrow L^{2^*}(\RR^n)$ is continuous.
\end{lemma}
Thanks to Lemma~\ref{lemmaSKlambda} we can define the constant $S_{K,\,\lambda}$ as in formula~\eqref{SKlambda}
and get that~$S_{K,\,\lambda}>0$\,. Also in this case we note that,
since in
formula~\eqref{SKlambdav}
the integral over $\Omega$ can be extended to all $\RR^n$ (being $v=0$ a.e. in $\RR^n\setminus \Omega$)\,, then the function $v\mapsto S_{K,\,\lambda}(v)$ does not depend on the domain $\Omega$\,, while, in general, $S_{K,\,\lambda}$ does.

The counterpart of Lemmas~\ref{lemmanormalambda} and \ref{lemmaSKlambda} in the usual fractional Sobolev spaces is given by the following result:
\begin{lemma}\label{lemmaSslambda}
Let $\lambda\in (0, \lambda_{1,\,s})$\,, where $\lambda_{1,\,s}$ is the first eigenvalue of the operator~$(-\Delta)^s$ with homogeneous Dirichlet boundary conditions. Then formula~\eqref{normaHslambda} defines a norm on $H^s(\RR^n)$ equivalent to the usual one given in~\eqref{gagliardonorm}\,.

Moreover, the embedding $(H^s(\RR^n), \|\cdot\|_{H^s(\RR^n),\,\lambda})\hookrightarrow L^{2^*}(\RR^n)$ is continuous.
\end{lemma}
\begin{proof}
To see that formula~\eqref{normaHslambda} defines a norm on $H^s(\RR^n)$ equivalent to the usual one, it is enough to argue as in the proof of \cite[Lemma~10]{svlinking}\,.

The continuity of the embedding $(H^s(\RR^n), \|\cdot\|_{H^s(\RR^n),\,\lambda})\hookrightarrow L^{2^*}(\RR^n)$ comes from the fact that the norms~\eqref{gagliardonorm} and \eqref{normaHslambda} are equivalent and by Lemma~\ref{lemmaSs}\,.
\end{proof}

Thanks to Lemma~\ref{lemmaSslambda} we can define the constant
$S_{s,\,\lambda}$ given in formula~\eqref{Sslambda},
and conclude that~$S_{s,\,\lambda}>0$\,.

\begin{remark}\label{remarkSKSslambda1}
{\rm In the model case when $K(x)=|x|^{-(n+2s)}$, it is easy to prove
that
$$\inf_{v\in H^s(\RR^n)\setminus\{0\}}S_{s,\,\lambda}(v)\leq \inf_{v\in X_0\setminus\{0\}}S_{K,\,\lambda}(v)\,,$$
since $X_0\subseteq H^s(\RR^n)$ by Lemma~\ref{lemmaV}-$a)$\,.
}\end{remark}

\begin{remark}\label{remarkSKSslambda2}
{\rm In the case when $\lambda=0$, we get that
$$S_{K,\,\lambda}=S_K \qquad \mbox{and} \qquad S_{s,\,\lambda}=S_s\,,$$
while for $\lambda>0$ we have that
$$S_{K,\,\lambda}\leq S_K \qquad \mbox{and} \qquad S_{s,\,\lambda}\leq S_s\,.$$
}\end{remark}

\subsection{An eigenvalue problem}\label{sec:autovalori}
Here we consider the following eigenvalue problem driven by the non-local integrodifferential operator~$-\mathcal L_K$
\begin{equation}\label{problemaautovalori}
\left\{\begin{array}{ll}
-\mathcal L_K u=\lambda\, u & \mbox{in } \Omega\\
u=0 & \mbox{in } \RR^n\setminus \Omega\,,
\end{array}\right.
\end{equation}
where $s\in (0,1)$, $n>2s$, $\Omega$ is an open bounded set of
$\RR^n$ and
$K:\RR^n\setminus\{0\}\rightarrow(0,+\infty)$ is a function
satisfying \eqref{kernel}--\eqref{evenkernel}.

More precisely, we are interested in the weak formulation of~\eqref{problemaautovalori},
which consists in the following eigenvalue
problem
\begin{equation}\label{problemaautovaloriweak}
\left\{\begin{array}{l} {\displaystyle \int_{\RR^{2n} }
(u(x)-u(y))(\varphi(x)-\varphi(y))K(x-y) dx\,dy=\lambda \int_\Omega
u(x)
\varphi(x)\,dx}\,\,\,\,\,\,\, \forall\,\, \varphi \in X_0\\
\\
u\in X_0\,.
\end{array}\right.
\end{equation}

We recall that $\lambda\in \RR$ is an eigenvalue of $-\mathcal L_K$ provided there exists a non-trivial
solution $u\in X_0$ of problem~\eqref{problemaautovalori} -- in fact, of its weak formulation~\eqref{problemaautovaloriweak} --
and, in this case,
any solution will be called an eigenfunction
corresponding to the eigenvalue~$\lambda$.

For the purpose of this paper we need only the properties of the first
eigenvalue of $-\mathcal L_K$ and of its eigenfunction, that is the following result, whose proof can be found in \cite[Proposition~9 and Appendix~A]{svlinking}, where a complete study of the spectrum of $-\mathcal L_K$ is presented:

\begin{proposition}[First eigenvalue and first eigenfunction
of $-\mathcal L_K$]\label{propautovalori}
Let $s\in (0,1)$, $n>2s$, $\Omega$ be an open bounded set of
$\RR^n$ and let
$K:\RR^n\setminus\{0\}\rightarrow(0,+\infty)$ be a function
satisfying assumptions~\eqref{kernel}--\eqref{evenkernel}. Then,
\begin{itemize}
\item[$a)$] problem~\eqref{problemaautovaloriweak}
admits an eigenvalue
$\lambda_1$ which is positive and that can be characterized as follows
    \begin{equation}\label{caratterizzazionelambda1}
\lambda_1=\min_{{u\in X_0} \atop{\|u\|_{L^2(\Omega)}=1}}
\int_{\RR^{2n} }
|u(x)-u(y)|^2K(x-y) dx\,dy\,,
    \end{equation}
or, equivalently,
\begin{equation}\label{caratterizzazionelambda11}
\lambda_1=\min_{u\in X_0\setminus\{0\}}\frac{{\displaystyle\int_{\RR^{2n} }
|u(x)-u(y)|^2K(x-y) dx\,dy}}{{\displaystyle\int_\Omega |u(x)|^2\,dx}}\,;
    \end{equation}
\smallskip

\item[$b)$] there exists a non-negative
function $e_1\in X_0$, which is an eigenfunction corresponding to $\lambda_1$, attaining the minimum
in~\eqref{caratterizzazionelambda1}, that is
$\|e_1\|_{L^2(\Omega)}=1$ and
\begin{equation}\label{77}
    {\displaystyle \lambda_1=\int_{\RR^{2n} }
|e_1(x)-e_1(y)|^2K(x-y) dx\,dy}\,;
    \end{equation}
\smallskip

\item[$c)$] $\lambda_1$ is simple, that is if $u\in X_0$ is a solution of the following equation
\begin{equation}\label{problemalambda1weak}
{\displaystyle \int_{\RR^{2n} }
(u(x)-u(y))(\varphi(x)-\varphi(y))K(x-y) dx\,dy=\lambda_1 \int_\Omega
u(x)
\varphi(x)\,dx}\,\,\,\,\,\,\, \forall\,\, \varphi \in X_0\,,
\end{equation}
 then $u=\zeta e_1$, with $\zeta\in \RR$\,.
\end{itemize}
\end{proposition}

\section{A general critical integrodifferential
equation}\label{sec:proofgeneral}

In this section we study problem~\eqref{problemaK0f} which is the
Euler-Lagrange equation of the functional $\mathcal J_{K,\,\lambda}$ defined in \eqref{JKlambda}\,.
Notice that this functional is well defined thanks
to assumptions \eqref{supf} and \eqref{condinfinito}, to the fact that $\Omega$ is bounded, and to
Lemma~\ref{lemmaembeddings}-$b)$.
Moreover, $\mathcal J_{K,\,\lambda}$ is Fr\'echet
differentiable in $u\in X_0$ and for any $\varphi\in X_0$
$$\begin{aligned}
\langle \mathcal J_{K,\,\lambda}'(u), \varphi\rangle & = \int_{\RR^{2n}}
\big(u(x)-u(y)\big)\big(\varphi(x)-\varphi(y)\big)K(x-y)\,dx\,dy -\lambda \int_\Omega u(x) \varphi(x)\,dx \\
& \qquad \qquad \qquad \qquad\qquad \quad
 -\int_\Omega |u(x)|^{2^*-1}(x) \varphi(x)\,dx-\int_\Omega f(x, u(x))\varphi(x)\,dx\,.
\end{aligned}$$

Thus, critical points of $\mathcal J_{K,\,\lambda}$ are solutions to
problem~\eqref{problemaK0f}. In order to find these critical points,
we will apply a variant of the Mountain
Pass Theorem without the Palais-Smale condition,
as given in~\cite{ar} (see also \cite[Theorem~2.2]{bn})\,.

Indeed, here we can not apply the classical Mountain Pass Theorem, since, due to the lack of compactness in the embedding $X_0\hookrightarrow L^{2^*}(\RR^n)$ (see Lemma~\ref{lemmaembeddings}-$b)$), the functional $\mathcal J_{K,\,\lambda}$ does not verify the Palais-Smale condition globally, but only in an energy range determined by the best fractional critical Sobolev constant $S_K$ given in formula~\eqref{SK} (see also Lemma~\ref{lemmaembeddings}-$b)$)\,.

This variant of the Mountain Pass Theorem requires that the functional $\mathcal J_{K,\,\lambda}$
has a suitable {\em geometric structure}, as stated, e.g., in conditions~$(2.9)$ and $(2.10)$ of
\cite[Theorem~2.2]{bn}\,.

\subsection{The critical case with a lower order perturbation}\label{subsec:mountainpassf}
This subsection is devoted to problem~\eqref{problemaK0f} and to the proof of Theorem~\ref{generalkernel0f}\,.

First of all, we prove that the functional $\mathcal J_{K,\,\lambda}$ has the
geometric features required by \cite[Theorem~2.2]{bn}.

\begin{proposition}\label{lemmaJpositivoMP}
Let $\lambda\in (0, \lambda_1)$ and let $f$ be a Carath\'eodory function
satisfying conditions \eqref{supf}--\eqref{condinfinito}. Then,
there exist $\rho>0$ and $\beta>0$ such that for any $u\in X_0$ with
$\|u\|_{X_0}=\rho$ it results that $\mathcal J_{K,\,\lambda}(u)\geq
\beta$\,.
\end{proposition}
\begin{proof}
Let $u$ be a function in $X_0$. By \eqref{cond22F} we get that for any $\varepsilon>0$
\begin{equation}\label{infJfrac0}
\begin{aligned}
\mathcal J_{K,\,\lambda}(u)& \geq \frac 1 2\int_{\RR^{2n}}|u(x)-u(y)|^2 K(x-y)dx\,dy-\frac \lambda 2 \int_\Omega |u(x)|^2\,dx-\frac{1}{2^*}\int_\Omega |u(x)|^{2^*}dx\\
& \qquad \qquad \qquad \qquad \qquad \qquad -\varepsilon\int_\Omega |u(x)|^2dx - \delta(\varepsilon)\int_\Omega |u(x)|^{2^*}\,dx\\
& \geq \frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right) \int_{\RR^{2n}} |u(x)-u(y)|^2
K(x-y)dx\,dy-\varepsilon\|u\|_{L^2(\Omega)}^2\\
& \qquad \qquad \qquad \qquad
 -\Big(\frac{1}{2^*}+\delta(\varepsilon)\Big)\|u\|_{L^{2^*}(\Omega)}^{2^*} \\
& \geq \frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right) \int_{\RR^{2n}}|u(x)-u(y)|^2
K(x-y)\,dx\,dy-\varepsilon|\Omega|^{(2^*-2)/2^*}
\|u\|_{L^{2^*}(\Omega)}^2\\
& \qquad \qquad \qquad \qquad -\Big(\frac{1}{2^*}+\delta(\varepsilon)\Big)\|u\|_{L^{2^*}(\Omega)}^{2^*}\,,\\
\end{aligned}
\end{equation}
thanks to Lemma~\ref{lemmanormalambda} (here we need
$0<\lambda<\lambda_1$) and to the fact that $L^{2^*}(\Omega)\hookrightarrow
L^2(\Omega)$ continuously (being $\Omega$ bounded and $2<2^*$).

Using \eqref{kernelfrac} and Lemma~\ref{lemmanormaequiv}-$a)$-$b)$, we
deduce from \eqref{infJfrac0} that for any $\varepsilon>0$
\begin{equation}\label{infJfrac20}
\begin{aligned}
\mathcal J_{K,\,\lambda}(u) & \geq \frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right) \int_{\RR^{2n}} |u(x)-u(y)|^2 K(x-y)\,dx\,dy \\
& \qquad -\varepsilon c|\Omega|^{(2^*-2)/2^*}
 \int_{\RR^{2n}}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\, dx\,dy \\
& \qquad -\Big(\frac{1}{2^*}+\delta(\varepsilon)\Big)c^{2^*/2}\left( \int_{\RR^{2n}}\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\, dx\,dy\right)^{2^*/2}\\
& \ge \left[\frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right)  -\frac{\varepsilon
c|\Omega|^{(2^*-2)/2^*}}{\theta}\right]\int_{\RR^{2n}} |u(x)-u(y)|^2 K(x-y)\,dx\,dy\\
& \qquad
-\Big(\frac{1}{2^*}+\delta(\varepsilon)\Big)
\Big(\frac{c}{\theta}\Big)^{2^*/2}
\left( \int_{\RR^{2n}} |u(x)-u(y)|^2 K(x-y)\, dx\,dy\right)^{2^*/2}\\
& =\left[\frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right)  -\frac{\varepsilon
c|\Omega|^{(2^*-2)/2^*}}{\theta}\right]\|u\|_{X_0}^2-\Big(\frac{1}{2^*}+\delta(\varepsilon)\Big)
\Big(\frac{c}{\theta}\Big)^{2^*/2}
\|u\|_{X_0}^{2^*}\,.
\end{aligned}
\end{equation}

Choosing $\varepsilon>0$ such that $2\varepsilon c
|\Omega|^{(2^*-2)/2^*}<\theta(1-\lambda/\lambda_1)$, by \eqref{infJfrac20}
it easily follows that
$$\mathcal J_{K,\,\lambda}(u)\geq \alpha \|u\|_{X_0}^2\left(1-\kappa
\|u\|_{X_0}^{2^*-2}\right)\,,$$
for suitable positive constants $\alpha$ and $\kappa$\,.

Now, let $u\in X_0$ be such that $\|u\|_{X_0}=\rho>0$. Since $2^*>2$, we can choose $\rho$ sufficiently small (i.e. $\rho$ such that $1-\kappa\rho^{2^*-2}>0$), so that
$$\inf_{
{u\in X_0}\atop{\|u\|_{X_0}=\rho}
} \mathcal J_{K,\,\lambda}(u)\geq \alpha \rho^2(1-\kappa \rho^{2^*-2})=:\beta>0\,.$$
Hence, Proposition~\ref{lemmaJpositivoMP} is proved.
\end{proof}

\begin{proposition}\label{lemmaJnegativoMP}
Let $\lambda\in (0, \lambda_1)$ and let $f$ be a Carath\'eodory function
satisfying conditions \eqref{supf}--\eqref{condinfinito}. Then, there
exists $e\in X_0$ such that $e\ge 0$ a.e. in $\RR^n$,
$\|e\|_{X_0}>\rho$ and $\mathcal J_{K,\,\lambda}(e)<\beta$\,, where $\rho$
and $\beta$ are given in Proposition~\ref{lemmaJpositivoMP}.

In particular, if we assume condition~\eqref{u0S}, we can construct $e$ as follows
\begin{equation}\label{e}
e=\zeta_0u_0
\end{equation}
with $u_0$ as in~\eqref{u0S}
and $\zeta_0>0$ large enough.
\end{proposition}
\begin{proof} We fix $u\in X_0$ such that $\|u\|_{X_0}=1$ and $u\ge 0$ a.e. in $\RR^n$:
we remark that this choice is possible thanks to \eqref{C2}
(alternatively, one can replace any $u\in X_0$ with its positive part,
which belongs to $X_0$ too, thanks to \cite[Lemma~12]{sv}).

Now, let $\zeta>0$\,. By Lemma~\ref{lemmanormalambda}
(here we use again the fact that $\lambda<\lambda_1$) and Lemma~\ref{fFcritico} (see formula~\eqref{cond22Fcritico} used here with $\varepsilon=1/(2\cdot 2^*)$) we have
$$\begin{aligned}
\mathcal J_{K,\,\lambda}(\zeta u) & = \frac{\zeta^2}{2}\int_{\RR^{2n}} |u(x)- u(y)|^2
K(x-y)dx\,dy-\frac \lambda 2\, \zeta^2 \int_\Omega |u(x)|^2\,dx\\
& \qquad \qquad \qquad -\frac{\zeta^{2^*}}{\,2^*}\int_\Omega |u(x)|^{2^*}\,dx
-\int_\Omega F(x, \zeta u(x))\,dx\\
& \leq \frac{\zeta^2}{2}
-\Big(\frac{1}{\,2^*}-\frac{1}{2\cdot 2^*}\Big)\zeta^{2^*}\int_\Omega
|u(x)|^{2^*}\,dx + M\big(1/(2\cdot 2^*)\big)\int_\Omega|\zeta u(x)|\,dx
\\
& = \frac{\zeta^2}{2} -\frac{\zeta^{2^*}}{2\cdot
2^*}\int_\Omega |u(x)|^{2^*}\,dx +\zeta M\big(1/(2\cdot
2^*)\big)\int_\Omega|u(x)|\,dx
\end{aligned}$$
Since $2^*>2>1$, passing to the limit as $\zeta\to +\infty$, we get that
$\mathcal J_{K,\,\lambda}(\zeta u)\to -\infty$, so that the assertion follows
taking $e=\zeta u$, with $\zeta$ sufficiently large.

In particular, under condition~\eqref{u0S}, we can take
$u=u_0/\|u_0\|_{X_0}\in X_0$ (note that
$u_0\not \equiv 0$ by assumption) so that we can choose $e=\zeta_0u_0$ with $\zeta_0$ large enough\,.
\end{proof}

\subsubsection{End of the proof of Theorem~\ref{generalkernel0f}}
Propositions \ref{lemmaJpositivoMP} and~\ref{lemmaJnegativoMP} give
that the geometry of the variant of the Mountain Pass Theorem stated in \cite[Theorem~2.2]{bn}
is fulfilled by $\mathcal J_{K,\,\lambda}$. Moreover, since $F(0)=0$, we easily get
that $\mathcal J_{K,\,\lambda}(0)=0<\beta$, with $\beta$ given in Proposition~\ref{lemmaJpositivoMP}.
Now, set
\begin{equation}\label{cdef}
c=\inf_{P\in\mathcal P} \sup_{v\in P([0,1])} \mathcal J_{K,\,\lambda}(v)\,,
\end{equation}
where
$$\mathcal P=\Big\{P\in C([0,1]; X_0) : P(0)=0\,, \,\, P(1)=e\Big\}$$
with $e=\zeta_0u_0$ given in Proposition~\ref{lemmaJnegativoMP}\,.
In order to prove Theorem~\ref{generalkernel0f} we proceed by steps.

\medskip

\begin{claim}\label{claim1}
The constant $c$ given in \eqref{cdef} is such that
$$\beta\leq c<\frac s n\, S_K^{n/(2s)}\,,$$ where $\beta$ is given in Proposition~\ref{lemmaJpositivoMP} and $S_K$ is defined in formula~\eqref{SK} (see also Lemma~\ref{lemmaembeddings}-$b)$)\,.
\end{claim}
\begin{proof}
We observe that, for any $P\in \mathcal P$, the function $t\mapsto \|P(t)\|_{X_0}$ is continuous in $[0,1]$, that $\|P(0)\|_{X_0}=\|0\|_{X_0}=0<\rho$, and that $\|P(1)\|_{X_0}=\|e\|_{X_0}>\rho$, with $\rho$ as in Proposition~\ref{lemmaJpositivoMP}\,.
Accordingly, there exists $\overline t\in (0,1)$ such that $\|P(\overline t)\|_{X_0}=\rho$\,. Thus,
$$\sup_{v\in P([0,1])} \mathcal J_{K,\,\lambda}(v)\geq \mathcal J_{K,\,\lambda}(P(\overline t))\geq \inf_{
{v\in X_0}\atop{\|v\|_{X_0}=\rho}
} \mathcal J_{K,\,\lambda}(v)\,.$$
As a consequence $c\geq \beta$, where $\beta$ is given in Proposition~\ref{lemmaJpositivoMP}\,.

Now, we recall that $e=\zeta_0u_0$, thanks to~\eqref{e}, and so the map
$[0,1]\ni t\mapsto t\, \zeta_0u_0$ belongs
to $\mathcal P$. Hence,
by assumption~\eqref{u0S}, we conclude that
$$\inf_{P\in {\mathcal P}}\sup_{v\in P([0,1])} \mathcal J_{K,\,\lambda}(v)\leq \sup_{\zeta\geq 0} \mathcal J_{K,\,\lambda}(\zeta u_0)<\frac s n\, S_K^{n/(2s)}$$
so that
\begin{equation}\label{cS}
c<\frac s n\, S_K^{n/(2s)}\,.
\end{equation}
Hence, Claim~\ref{claim1} is proved.
\end{proof}

By \cite[Theorem~2.2]{bn} there exists a sequence~$u_j$ in $X_0$ such that
\begin{equation}\label{Jc0}
\mathcal J_{K,\,\lambda}(u_j)\to c
\end{equation}
and
\begin{equation}\label{J'00}
\sup\Big\{ \big|\langle\,\mathcal J_{K,\,\lambda}'(u_j),\varphi\,\rangle
\big|\,: \; \varphi\in X_0\,, \|\varphi\|_{X_0}=1\Big\}\to 0
\end{equation}
as $j\to +\infty$.
Now we prove some properties of the sequence~$u_j$\,.

\medskip

\begin{claim}\label{claim2}
The sequence $u_j$ is bounded in $X_0$\,.
\end{claim}
\begin{proof}
For any $j\in \NN$ by \eqref{Jc0} and \eqref{J'00} it easily follows
that there exists $\kappa>0$ such that
\begin{equation}\label{jlimitato0}
|\mathcal J_{K,\,\lambda}(u_j)|\leq \kappa\,,
\end{equation}
and
\begin{equation}\label{j'limitato0}
\Big|\langle \mathcal J_{K,\,\lambda}'(u_j),
\frac{u_j}{\|u_j\|_{X_0}}\rangle\Big| \leq \kappa\,.
\end{equation}
As a consequence of \eqref{jlimitato0} and \eqref{j'limitato0} we have
\begin{equation}\label{2k}
\mathcal J_{K,\,\lambda}(u_j)-\frac 1 2 \langle \mathcal J_{K,\,\lambda}'(u_j),
u_j\rangle\leq \kappa \left(1+ \|u_j\|_{X_0}\right)\,.
\end{equation}
By H\"older inequality, Lemma~\ref{lemmanormaequiv}-$a)$ and \eqref{kernelfrac}
\begin{equation}\label{H}
\int_\Omega |u_j(x)|\,dx\le
|\Omega|^{(2^*-1)/2^*}\left(\int_\Omega
|u_j(x)|^{2^*}\,dx\right)^{1/2^*}\le \widetilde \kappa
|\Omega|^{(2^*-1)/2^*}\|u_j\|_{X_0}\,,
\end{equation}
for some positive constant~$\widetilde \kappa$\,.
Accordingly, by Lemma~\ref{fFcritico} and \eqref{H} we obtain
\begin{equation}\label{jj'0}
\begin{aligned}
\mathcal J_{K,\,\lambda}(u_j)-\frac 1 2 \langle \mathcal J_{K,\,\lambda}'(u_j), u_j\rangle & =
-\left(\frac{1}{\,2^*} -\frac 1 2\right)\|u_j\|_{L^{2^*}(\Omega)}^{2^*} -\int_\Omega F(x, u_j(x))\,dx\\
& \qquad \qquad \qquad \qquad  +\frac 1 2 \int_\Omega f(x, u_j(x)) \,u_j(x)\,dx\\
& \geq \frac s n\, \|u_j\|_{L^{2^*}(\Omega)}^{2^*}-\varepsilon\Big(1+\frac{2^*}{2}\Big)\|u_j\|_{L^{2^*}(\Omega)}^{2^*}\\
& \qquad \qquad \qquad \qquad
-\frac{3M(\varepsilon)}{2}\int_\Omega |u_j(x)|\,dx\\
& \geq \frac s n\, \|u_j\|_{L^{2^*}(\Omega)}^{2^*}-\varepsilon\Big(1+\frac{2^*}{2}\Big)\|u_j\|_{L^{2^*}(\Omega)}^{2^*}\\
& \qquad \qquad \qquad \qquad
-C_\varepsilon\|u_j\|_{X_0}\,,
\end{aligned}
\end{equation}
for a suitable~$C_\varepsilon>0$, possibly depending on~$|\Omega|$.
Choosing $\varepsilon>0$ small enough (i.e. $\varepsilon$ such that $\frac s n>\varepsilon \Big(1+\frac{2^*}{2}\Big)$), \eqref{2k} and~\eqref{jj'0} imply that, for any $j\in \NN$
\begin{equation}\label{norma2star}
\|u_j\|_{L^{2^*}(\Omega)}^{2^*} \leq \kappa_*\left(1+\|u_j\|_{X_0}\right)
\end{equation}
for a suitable positive constant $\kappa_*$\,.

Finally, by Lemma~\ref{lemmanormalambda} (which holds true, since $0<\lambda<\lambda_1$) and Lemma~\ref{fFcritico} (used here with $\varepsilon=1$) it follows that
\begin{equation}\label{stimafin}
\begin{aligned}
\mathcal J_{K,\,\lambda}(u_j) & \geq \frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right) \|u_j\|_{X_0}^2-\frac{1}{2^*}\|u_j\|_{L^{2^*}(\Omega)}^{2^*} -\int_\Omega F(x, u_j(x))\,dx\\
& \geq
\frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right) \|u_j\|_{X_0}^2-\Big(\frac{1}{2^*}+1\Big)\|u_j\|_{L^{2^*}(\Omega)}^{2^*}
-M(1)\int_\Omega |u_j(x)|\,dx\,.
\end{aligned}
\end{equation}

Combining \eqref{jlimitato0}, \eqref{norma2star} and \eqref{stimafin} we conclude that
$$\begin{aligned}
\kappa\geq \mathcal J_{K,\,\lambda}(u_j) & \geq \frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right) \|u_j\|_{X_0}^2-\Big(\frac{1}{2^*}+1\Big)\|u_j\|_{L^{2^*}(\Omega)}^{2^*} -M(1)\int_\Omega |u_j(x)|\,dx
\\
& \geq
\frac{1}{2}\left(1-\frac{\lambda}{\lambda_1}\right) \|u_j\|_{X_0}^2-\kappa_*\Big(\frac{1}{2^*}+1\Big)\left(1+\|u_j\|_{X_0}\right)
-M(1)\int_\Omega |u_j(x)|\,dx\,,
\end{aligned}$$
so that, recalling~\eqref{H}, we see that, for any $j\in \NN$
$$\|u_j\|_{X_0}^2 \leq \kappa_{**}\left(1+\|u_j\|_{X_0}\right)$$
for a suitable positive constant $\kappa_{**}$\,. Hence, the proof of Claim~\ref{claim2} is complete\,.
\end{proof}

\begin{claim}\label{claim3}
Problem~\eqref{problemaK0f} admits a solution $u_\infty\in X_0$\,.
\end{claim}
\begin{proof}
Since $u_j$ is bounded in $X_0$ (thanks to Claim~\ref{claim2}) and $X_0$
is a reflexive space
(being a Hilbert space, by Lemma~\ref{lemmanormaequiv}-$c)$), up to a
subsequence, still denoted by $u_j$, there exists $u_\infty \in X_0$
such that~$u_j\to u_\infty$ weakly in~$X_0$, that is
\begin{equation}\label{convergenze0}
\begin{aligned}
 & \int_{\RR^{2n}}\big(u_j(x)-u_j(y)\big)\big(\varphi(x)-\varphi(y)\big) K(x-y)\,dx\,dy \to \\
& \qquad \qquad
\int_{\RR^{2n}}\big(u_\infty(x)-u_\infty(y)\big)\big(\varphi(x)-\varphi(y)\big)
K(x-y)\,dx\,dy  \quad \mbox{for any}\,\, \varphi\in X_0
\end{aligned}
\end{equation}
as $j\to +\infty$\,. Moreover, since $u_j$ is bounded in $X_0$,
we deduce from~\eqref{norma2star} that
\begin{equation}\label{DD}
{\mbox{$u_j$ is bounded in $L^{2^*}(\Omega)$\,.}}\end{equation}
Consequently, by Lemma~\ref{lemmaembeddings}-$b)$
and
the fact that
$L^{2^*}(\RR^n)$ is a reflexive space we have that, up to a subsequence
\begin{equation}\label{convergenzaweak}
u_j \to u_\infty \quad \mbox{weakly in}\,\, L^{2^*}(\RR^n)
\end{equation}
as $j\to +\infty$, while by Lemma~\ref{lemmaembeddings}-$a)$, up to a subsequence,
\begin{equation}\label{convergenze0bis}
u_j \to u_\infty \quad \mbox{in}\,\, L^\nu(\RR^n)
\end{equation}
\begin{equation}\label{convergenze0ter}
u_j \to u_\infty \quad \mbox{a.e. in}\,\, \RR^n
\end{equation}
as $j\to +\infty$ and there exists $\ell\in L^\nu(\RR^n)$ such that
\begin{equation}\label{dominata20}
|u_j(x)|\leq \ell(x) \quad \mbox{a.e. in}\,\, \RR^n\,\quad \mbox{for any}\,\,j\in \NN
\end{equation}
for any $\nu\in[1, 2^*)$ (see, for instance \cite[Theorem~IV.9]{brezis}).

By \eqref{convergenzaweak} and the fact that $|u_j|^{2^*-2}u_j$ is bounded in $L^{2^*/(2^*-1)}(\Omega)$ we see that
\begin{equation}\label{convergenzaweak1}
|u_j|^{2^*-2}u_j \to |u_\infty|^{2^*-2}u_\infty \quad \mbox{weakly in}\,\, L^{2^*/(2^*-1)}(\Omega)
\end{equation}
as $j\to +\infty$\,.

Furthermore,
by \eqref{DD} and Lemma~\ref{fFcritico} (see
formula~\eqref{cond22critico})
and the fact that $\Omega$ is bounded we get that
$$f(\cdot, u_j(\cdot)) {\mbox{ is bounded in }}
L^{2^*/(2^*-1)}(\Omega)\,.$$
More precisely, by \eqref{convergenze0ter}, \eqref{dominata20}, the
fact that the map $t\mapsto f(\cdot, t)$ is continuous in $t\in \RR$, \eqref{cond22critico}
and the Dominated Convergence Theorem applied in $L^{2^*/(2^*-1)}(\Omega)$ (note that $2^*/(2^*-1)<2^*$, being $2<2^*$) we get
\begin{equation}\label{convf0}
 f(\cdot, u_j(\cdot)) \to  f(\cdot, u_\infty(\cdot)) \quad \mbox{in} \quad L^{2^*/(2^*-1)}(\Omega)
\end{equation}
as $j\to +\infty$. Since the dual of~$L^{ 2^*/(2^*-1)}(\Omega)$
is $L^{2^*}(\Omega)$, by using
\eqref{convf0} it is easily seen that
$$\int_\Omega f(x, u_j(x))\varphi(x)\,dx \to  \int_\Omega f(x, u_\infty(x))\varphi(x)\,dx \quad \mbox{for any} \quad \varphi\in L^{2^*}(\Omega)\,,$$
and so, in particular,
\begin{equation}\label{convffi1}
\int_\Omega f(x, u_j(x))\varphi(x)\,dx \to  \int_\Omega f(x, u_\infty(x))\varphi(x)\,dx \quad \mbox{for any} \quad \varphi\in X_0
\end{equation}
as $j\to +\infty$ (here we use Lemma~\ref{lemmaembeddings}-$b)$).

Since \eqref{J'00} holds true, for any $\varphi \in X_0$
$$\begin{aligned}
0\leftarrow \langle \mathcal J_{K,\,\lambda}'(u_j), \varphi\rangle & =
\int_{\RR^{2n}}\big(u_j(x)-u_j(y)\big)\big(\varphi(x)-\varphi(y)\big) K(x-y)\,dx\,dy-\lambda \int_\Omega u_j(x)\varphi(x)\,dx\\
& \qquad \qquad \qquad \quad \quad  - \int_\Omega |u_j(x)|^{2^*-2}u_j(x)\varphi(x)\,dx -
\int_\Omega f(x,
u_j(x))\varphi(x)\,dx\,,
\end{aligned}$$
so that, passing to the limit in this expression as $j\to +\infty$ and taking into account \eqref{convergenze0}, \eqref{convergenze0bis}, \eqref{convergenzaweak1} and \eqref{convffi1} we get
$$\begin{aligned}
\int_{\RR^{2n}}\big(u_\infty(x)-u_\infty(y)\big)& \big(\varphi(x)-\varphi(y)\big) K(x-y)\,dx\,dy-\lambda \int_\Omega u_\infty(x)\varphi(x)\,dx\\
& \quad \quad  - \int_\Omega |u_\infty(x)|^{2^*-2}u_\infty(x)\varphi(x)\,dx -
\int_\Omega f(x,
u_\infty(x))\varphi(x)\,dx=0
\end{aligned}$$
for any $\varphi \in X_0$, that is $u_\infty$ is a solution of problem~\eqref{problemaK0} and Claim~\ref{claim3} follows.
\end{proof}

\begin{claim}\label{claim4}
The solution~$u_\infty$ is non-trivial, i.e. $u_\infty\not\equiv 0$ in
$\Omega$\,.
\end{claim}
\begin{proof}
Suppose, by contradiction, that $u_\infty \equiv 0$ in $\Omega$ (and so $u_\infty \equiv 0$ in $\RR^n$\,, being $u_\infty\in X_0$).

Using Lemma~\ref{fFcritico}
and~\eqref{DD}, we see that for any $\varepsilon>0$ and $j\in \NN$
\begin{equation}\label{fuj}
\begin{aligned}
\left|\int_\Omega f(x, u_j(x))u_j(x)\,dx\right| & \leq 2^*\varepsilon\int_\Omega |u_j(x)|^{2^*}\,dx+M(\varepsilon)\int_\Omega |u_j(x)|\,dx\\
& \leq 2^*\varepsilon C+ M(\varepsilon)\|u_j\|_{L^1(\Omega)}
\end{aligned}
\end{equation}
and
\begin{equation}\label{Fuj}
\begin{aligned}\left|
\int_\Omega F(x, u_j(x))\,dx\right| & \leq \varepsilon\int_\Omega |u_j(x)|^{2^*}\,dx+M(\varepsilon)\int_\Omega |u_j(x)|\,dx\\
& \leq \varepsilon C+ M(\varepsilon)\,\|u_j\|_{L^1(\Omega)}\,.
\end{aligned}
\end{equation}
Keeping $\varepsilon$ fixed
and taking
the limit in $j$, we obtain
\begin{eqnarray*}
&&\limsup_{j\rightarrow+\infty}
\left|\int_\Omega f(x, u_j(x))u_j(x)\,dx\right| \leq 2^*\varepsilon C\\
{\mbox{and }}&&\limsup_{j\rightarrow+\infty}\left|
\int_\Omega F(x, u_j(x))\,dx\right|\le \varepsilon C\,,
\end{eqnarray*} due to~\eqref{convergenze0bis}.
Hence, by sending $\varepsilon\rightarrow0^+$, we obtain
\begin{equation}\label{fujlim}
\begin{aligned}
\int_\Omega f(x, u_j(x))u_j(x)\,dx \to 0
\end{aligned}
\end{equation}
and
\begin{equation}\label{Fujlim}
\begin{aligned}
\int_\Omega F(x, u_j(x))\,dx \to 0
\end{aligned}
\end{equation}
as $j\to +\infty$\,.

Moreover, being $u_j$ bounded in $X_0$ by Claim~\ref{claim2}, from \eqref{J'00} it follows that
$$\begin{aligned}
0\leftarrow \langle \mathcal J'_{K,\,\lambda}(u_j), u_j\rangle & =
\int_{\RR^{2n}}|u_j(x)-u_j(y)|^2 K(x-y)\,dx\,dy-\lambda \int_\Omega |u_j(x)|^2\,dx\\
& \qquad \qquad \qquad \quad \quad  - \int_\Omega |u_j(x)|^{2^*}\,dx -
\int_\Omega f(x, u_j(x))u_j(x)\,dx\,,
\end{aligned}$$
which, thanks to \eqref{convergenze0bis} and \eqref{fujlim}, gives
\begin{equation}\label{nonso}
\int_{\RR^{2n}}|u_j(x)-u_j(y)|^2 K(x-y)\,dx\,dy- \int_\Omega |u_j(x)|^{2^*}\,dx\to 0
\end{equation}
as $j\to +\infty$\,.

Now, by Claim~\ref{claim2} the sequence $\|u_j\|_{X_0}$ is bounded in $\RR$. Hence, up to a subsequence, if necessary, we can assume that
\begin{equation}\label{L1}
\|u_j\|_{X_0}^2=\int_{\RR^{2n}}|u_j(x)-u_j(y)|^2 K(x-y)\,dx\,dy\to L
\end{equation}
and so, as a consequence of \eqref{nonso}
\begin{equation}\label{L2}
\int_\Omega |u_j(x)|^{2^*}\,dx\to L
\end{equation}
as $j\to +\infty$\,.
Of course $L\in [0, +\infty)$. Furthermore, by \eqref{Jc0} we have that, as $j\to +\infty$,
$$\begin{aligned}
\frac 1 2\int_{\RR^{2n}}&|u_j(x)-u_j(y)|^2 K(x-y)\,dx\,dy -\frac\lambda 2\int_\Omega |u_j(x)|^2\,dx\\
& \qquad \qquad \qquad \quad \quad  -\frac{1}{2^*} \int_\Omega |u_j(x)|^{2^*}\,dx -
\int_\Omega F(x, u_j(x))\,dx\to c\,,
\end{aligned}$$
so that, using \eqref{convergenze0bis} with $u_\infty\equiv 0$ in $\RR^n$, \eqref{Fujlim}, \eqref{L1} and \eqref{L2}, it follows
\begin{equation}\label{c}
c=\Big(\frac 1 2 -\frac{1}{2^*}\Big)L=\frac s n\, L\,.
\end{equation}
Since $c\geq \beta>0$ by Claim~\ref{claim1}, it is easily seen that $L>0$\,.

Moreover, by Lemma~\ref{lemmaembeddings}-$b)$ and \eqref{SK}
$$\int_{\RR^{2n}}|u_j(x)-u_j(y)|^2 K(x-y)\,dx\,dy\geq S_K\|u_j\|_{L^{2^*}(\Omega)}^2\,,$$
so that, passing to the limit as $j\to +\infty$, and taking into account \eqref{L1} and \eqref{L2}, we get
$$L\geq S_K L^{2/2^*}\,,$$
which, combined with \eqref{c} gives
$$c\geq \frac s n\, S_K^{2^*/(2^*-2)}=\frac s n\, S_K^{n/(2s)}\,.$$
This contradicts Claim~\ref{claim1}\,. Hence, $u_\infty\not\equiv 0$ in
$\Omega$ and this ends the proof of Claim~\ref{claim4}\,.

The proof of Theorem~\ref{generalkernel0f} is complete.
\end{proof}

\subsection{The critical case $\mathcal L_K u+\lambda u+|u|^{2^*-2}u=0$}
\label{subsec:mountainpassf0}
In this subsection we consider problem~\eqref{problemaK0f} with $f\equiv 0$, that is we study \eqref{problemaK0} and we prove Theorem~\ref{generalkernel0}\,.

In order to do this it is enough to show that, when $f\equiv 0$\,, conditions \eqref{u0S} and \eqref{SlambdaSK} are equivalent. For this we need the following proposition:
\begin{proposition}\label{propcondizioneSlambda}
For any~$u_0\in X_0\setminus\{0\}$\,,
\begin{equation}\label{supK}
\begin{aligned}
\sup_{\zeta\geq 0} & \left(\frac{\zeta^2}{2} \int_{\RR^{2n}}|u_0(x)-u_0(y)|^2 K(x-y)\,dx\,dy -\frac {\zeta^2\lambda}{2} \int_\Omega |u_0(x)|^2\,dx\right.
\\
& \qquad \qquad \qquad \qquad \qquad \qquad \left.-\frac{\zeta^{2^*}}{2^*}\int_\Omega |u_0(x)|^{2^*}\,dx
\right) = \frac s n\, S_{K,\,\lambda}^{n/(2s)}(u_0)\,,
\end{aligned}
\end{equation}
where the function $X_0\setminus\{0\}\ni v \mapsto S_{K,\,\lambda}(v)$ is defined in \eqref{SKlambdav}\,.

\end{proposition}

\begin{proof}
Let $g:[0, +\infty)\to \RR$ be the following function
$${\displaystyle g(\zeta) =\frac{\zeta^2}{2} \int_{\RR^{2n}}|u_0(x)-u_0(y)|^2 K(x-y)\,dx\,dy -\frac {\zeta^2\lambda}{2} \int_\Omega |u_0(x)|^2\,dx-\frac{\zeta^{2^*}}{2^*}\int_\Omega |u_0(x)|^{2^*}\,dx}\,.$$
Note that $g$ is differentiable in $(0, +\infty)$ and
$${\displaystyle g'(\zeta)=\zeta\int_{\RR^{2n}}|u_0(x)-u_0(y)|^2 K(x-y)\,dx\,dy - \zeta\lambda \int_\Omega |u_0(x)|^2\,dx-\zeta^{2^*-1}\int_\Omega |u_0(x)|^{2^*}\,dx}\,,$$
so that $g'(\zeta)\ge0$ if and only if
$$\zeta\le\bar
\zeta=\left(\frac{{\displaystyle\int_{\RR^{2n}}|u_0(x)-u_0(y)|^2 K(x-y)\,dx\,dy - \lambda \int_\Omega |u_0(x)|^2\,dx}}{{\displaystyle\int_\Omega |u_0(x)|^{2^*}\,dx}}\right)^{1/(2^*-2)}\,.$$
Therefore~$\bar \zeta$ is a maximum for $g$ and
$$\sup_{\zeta\geq 0}\,g(\zeta)=\max_{\zeta\geq 0}\,g(\zeta)=g(\bar \zeta)=\frac s n\, S_{K,\,\lambda}^{n/(2s)}(u_0)\,.$$
This concludes the proof of Proposition~\ref{propcondizioneSlambda}\,.
\end{proof}

\subsubsection{End of the proof of Theorem~\ref{generalkernel0}.}\label{subsec:dimtheorem2}
By Proposition~\ref{propcondizioneSlambda} we deduce that, when $f\equiv 0$ in $\Omega\times \RR$, condition~\eqref{u0S} reads as follows
\begin{equation}\label{u0Sequiv}
\begin{aligned}
&\mbox{there exists $u_0\in X_0\setminus\{0\}$ with $u_0\geq 0$  a.e. in $\RR^n$\,,}\\
& \qquad \qquad \mbox{such that } S_{K,\, \lambda}(u_0)<S_K\,.
\end{aligned}
\end{equation}

It is easily seen that conditions \eqref{u0Sequiv} and \eqref{SlambdaSK} are equivalent. Indeed, suppose that \eqref{u0Sequiv} holds true. Then,
\begin{equation}\label{3.31bis}
S_{K,\,\lambda}:=\inf_{v\in X_0\setminus\{0\}}\, S_{K,\,\lambda}(v)\leq
S_{K,\, \lambda}(u_0)<S_K\,.\end{equation}
While, if \eqref{SlambdaSK} is valid, then there exists $\widetilde v\in
X_0\setminus\{0\}$ such that
\begin{equation}\label{vtilde}
S_{K,\, \lambda}(\widetilde v)<S_K\,.
\end{equation}
Thus, condition~\eqref{u0Sequiv} follows by taking $u_0=|\widetilde
v|\in
X_0$: we remark that this choice is possible since for any $v\in X_0$ its positive part $v^+=\max\{v, 0\}$ and its negative part $v^-=\max\{-v, 0\}$ belong to $X_0$ too, by \cite[Lemma~12]{sv} (hence, $|v|=v^++v^-\in X_0$, being $X_0$ a linear space). Indeed, by triangle inequality, a.e. $x,y\in \RR^n$
$$ \Big| |\widetilde v(x)|-|\widetilde v(y)|\Big|\le
|\widetilde v(x)-\widetilde v(y)|.$$
This and~\eqref{vtilde} give
$$S_{K,\,\lambda}(u_0)=S_{K,\,\lambda}(|\widetilde v|)\leq
S_{K,\,\lambda}(\widetilde v)<S_K\,.$$

Hence, conditions \eqref{u0Sequiv} and \eqref{SlambdaSK} are equivalent and so Theorem~\ref{generalkernel0} comes from Theorem~\ref{generalkernel0f}\,.

\section{A critical fractional Laplace equation}\label{sec:prooffraclap}
Aim of this section is to consider the particular fractional kernel
$$K(x)=|x|^{-(n+2s)},$$ which gives rise to the fractional Laplace operator $-(-\Delta )^s$ defined is \eqref{2}\,.
Hence, here we study
$$\left\{
\begin{array}{ll}
(-\Delta)^s u-\lambda u=|u|^{2^*-2}u +f(x,u) & {\mbox{ in }} \Omega\\
u=0 & {\mbox{ in }} \RR^n\setminus \Omega\,,
\end{array} \right.$$
which reduces, when $f\equiv 0$ in $\Omega \times \RR$, to
problem~\eqref{1s}\,. More
precisely, we are interested in the weak formulation
of these problems, which is given by \eqref{problemalapfrac0f} and
\eqref{problemalapfrac0}, respectively.

\subsection{The critical case for $(-\Delta)^s$ with a lower order perturbation: proof of Theorem~\ref{lapfra0f}}\label{subsec:mountainpassflapfrac}

In order to prove Theorem~\ref{lapfra0f} it is enough to note that, in this setting, condition~\eqref{u0Sfrac} implies \eqref{u0S}, thanks to Lemma~\ref{lemmaV}-$b)$ and Remark~\ref{remarkSKSs}\,.

Hence, Theorem~\ref{lapfra0f} follows from Theorem~\ref{generalkernel0f} and from the fact that $X_0\subseteq H^s(\RR^n)$ by Lemma~\ref{lemmaV}-$a)$ (alternatively, we can use Lemma~\ref{lemmaV}-$b)$, since the solution $u$ provided by Theorem~\ref{generalkernel0f} is such that $u=0$ a.e. in $\RR^n\setminus\Omega$)\,.



\subsection{The critical case $(-\Delta)^s u-\lambda u=|u|^{2^*-2}u$: proof of Theorem~\ref{lapfra0}}
\label{subsec:mountainpassf0lapfrac}

The idea consists of applying Theorem~\ref{lapfra0f} with $f\equiv 0$ in $\Omega\times \RR$\,.

First of all, note that by Proposition~\ref{propcondizioneSlambda} with
$K(x)=|x|^{-(n+2s)}$ -- see also Lemma~\ref{lemmaV}-$b)$ --
we have that if $u_0\in H^s(\RR^n)\setminus\{0\}$, with~$u_0=0$ a.e.
outside~$\Omega$, then
\begin{equation}\label{supKfrac}
\begin{aligned}
\sup_{\zeta\geq 0} & \left(\frac{\zeta^2}{2} \int_{\RR^{2n}}\frac{|u_0(x)-u_0(y)|^2}{|x-y|^{+2s}}\,dx\,dy -\frac {\zeta^2\lambda}{2} \int_\Omega |u_0(x)|^2\,dx\right.
\\
& \qquad \qquad \qquad \qquad \qquad \qquad \left.-\frac{\zeta^{2^*}}{2^*}\int_\Omega |u_0(x)|^{2^*}\,dx
\right) = \frac s n\, S_{s,\,\lambda}^{n/(2s)}(u_0)\,,
\end{aligned}
\end{equation}
where the function $H^s(\RR^n)\setminus\{0\}\ni v \mapsto S_{s,\,\lambda}(v)$ is defined in \eqref{Sslambdav}\,.

Therefore, when $f\equiv 0$ in $\Omega\times \RR$, condition~\eqref{u0Sfrac} reads as follows
\begin{equation}\label{LAC}\begin{aligned}
&\mbox{there exists $u_0\in H^s(\RR^n)\setminus\{0\}$ with $u_0\geq 0$  a.e. in $\RR^n$ and $u_0=0$ a.e. in $\RR^n\setminus \Omega$\,,}\\
& \qquad \qquad \qquad \qquad \qquad \mbox{such that } S_{s,\,
\lambda}(u_0)<S_s\,.
\end{aligned}\end{equation}
Hence, in order to prove Theorem~\ref{lapfra0} it is enough to show that
condition~\eqref{LAC} is satisfied
for some appropriate function~$u_0$. For this, we proceed again by
steps.

\begin{claim}\label{claimubar}
The infimum in formula~\eqref{Ss} is attained, that is
$$S_s=S_s(\widetilde u)\,,$$
where
\begin{equation}\label{utilde}
\widetilde u(x)=\kappa{\Big(\mu^2+|x-x_0|^2\Big)^{-(n-2s)/2}}\,, \quad
x\in \RR^n\end{equation}
with $\kappa\in \RR\setminus\{0\}$, $\mu>0$ and $x_0\in \RR^n$ fixed constants.

Equivalently, the function $\bar u$ defined as
\begin{equation}\label{baru}
\bar u(x)=\frac{\widetilde u(x)}{\|\widetilde
u\|_{L^{2^*}(\RR^n)}}\end{equation}
is such that
\begin{equation}\label{utildeSs}
{\displaystyle S_s=\inf_{{v\in H^s(\RR^n)} \atop{\|v\|_{L^{2^*}(\RR^n)}=1}}{\displaystyle\int_{\RR^{2n}}\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}} \,dx\,dy}=\int_{\RR^{2n}}\frac{|\bar u(x)-\bar u(y)|^2}{|x-y|^{n+2s}} \,dx\,dy\,.}
\end{equation}
\end{claim}
\begin{proof}
This assertion comes from \cite[Theorem~1.1]{cotsiolis}\,.
\end{proof}

As in the classical case of the Laplacian, starting from $\bar u$ we can construct an explicit solution of the limiting problem
\begin{equation}\label{limitingproblem}
(-\Delta )^s u=|u|^{2^*-2}u \quad {\mbox{ in }} \RR^n
\end{equation}
as we show in the following result:
\begin{claim}\label{claimustar}
The function
\begin{equation}\label{ustar}
{\displaystyle u^*(x)=\bar u\left(\frac{x}{S_s^{1/(2s)}}\right)}\,, \qquad x\in \RR^n
\end{equation}
is a solution of problem~\eqref{limitingproblem} satisfying the property
\begin{equation}\label{normaSs}
\|u^*\|_{L^{2^*}(\RR^n)}^{2^*}=S_s^{n/(2s)}\,.
\end{equation}
\end{claim}
\begin{proof}
By Claim~\ref{claimubar} there exists a Lagrange multiplier $\gamma\in \RR$ such that
\begin{equation}\label{lagrange}
{\displaystyle \int_{\RR^{2n}}\frac{(\bar u(x)-\bar u(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}\,dx\,dy =\gamma\int_{\RR^n} |\bar u(x)|^{2^*-2}\bar u(x)\varphi(x)dx}
\end{equation}
for any $\varphi \in H^s(\RR^n)$\,.

Taking $\varphi=\bar u$ as a test function in \eqref{lagrange} and using again Claim~\ref{claimubar} we get
\begin{equation}\label{Ssclaim}
{\displaystyle S_s=\int_{\RR^{2n}}\frac{|\bar u(x)-\bar u(y)|^2}{|x-y|^{n+2s}} \,dx\,dy=\gamma\,,}
\end{equation}
since $\|\bar u\|_{L^{2^*}(\RR^n)}=1$ by construction (see \eqref{baru}).

Let $u^*$ be the function defined as in \eqref{ustar}\,.
Note that $u^*\in H^s(\RR^n)\setminus \{0\}$\,. Moreover, for any $\varphi \in H^s(\RR^n)$, by \eqref{lagrange} and \eqref{Ssclaim} we have
\begin{equation}\label{ustarvarphi}
\begin{aligned}
\int_{\RR^{2n}}& {\displaystyle \frac{(u^*(x)-u^*(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}\,dx\,dy}\\
& \qquad \qquad {\displaystyle =S_s^{(n-2s)/(2s)}\int_{\RR^{2n}}\frac{(\bar u(x)-\bar u(y))\big(\varphi(S_s^{1/(2s)}x)-\varphi(S_s^{1/(2s)}y)\big)}{|x-y|^{n+2s}}\,dx\,dy}\\
& \qquad \qquad {\displaystyle =S_s^{n/(2s)}\int_{\RR^n} |\bar u(x)|^{2^*-2}\bar u(x)\varphi(S_s^{1/(2s)}x)dx}\\
& \qquad \qquad {\displaystyle =\int_{\RR^n} |u^*(x)|^{2^*-2}u^*(x)\varphi(x)dx\,,}
\end{aligned}
\end{equation}
that is $u^*$ is a solution of problem~\eqref{limitingproblem}\,.

Finally, taking $\varphi=u^*$ as a test function in \eqref{ustarvarphi} and using \eqref{baru} and \eqref{ustar} we get
$${\displaystyle \int_{\RR^{2n}} \frac{|u^*(x)-u^*(y)|^2}{|x-y|^{n+2s}}\,dx\,dy=
\int_{\RR^n}|u^*(x)|^{2^*}\,dx=S_s^{n/(2s)}\int_{\RR^n}|\bar u(x)|^{2^*}\,dx=S_s^{n/(2s)}}\,,$$

 so that the proof of Claim~\ref{claimustar} is complete.
\end{proof}

Now, we consider the following family of functions $U_\varepsilon$ defined as
\begin{equation}\label{Uepsilon}
U_\varepsilon(x)=\varepsilon^{-(n-2s)/2}\,u^*(x/\varepsilon)\,, \qquad x\in \RR^n
\end{equation}
for any $\varepsilon>0$\,.
The scaling invariance of the problem is pointed out in the following
observation:

\begin{claim}\label{claimUepsilon}
The function $U_\varepsilon$ is a solution of problem~\eqref{limitingproblem} and verifies the following equalities
\begin{equation}\label{claim 7}
{\displaystyle \int_{\RR^{2n}} \frac{|U_\varepsilon(x)-U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,dx\,dy=
\int_{\RR^n}|U_\varepsilon(x)|^{2^*}\,dx=S_s^{n/(2s)}}\end{equation}
for any $\varepsilon>0$\,.
\end{claim}
\begin{proof}
By \eqref{Uepsilon} and Claim~\ref{claimustar} it easily follows that  $U_\varepsilon\in H^s(\RR^n)$ and
\begin{equation}\label{Uepsilonvarphi}
\begin{aligned}
\int_{\RR^{2n}}& {\displaystyle \frac{(U_\varepsilon(x)-U_\varepsilon(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}\,dx\,dy}\\
& \qquad \qquad {\displaystyle =\varepsilon^{(n-2s)/2}\int_{\RR^{2n}}\frac{(u^*(x)- u^*(y))\big(\varphi(\varepsilon x)-\varphi(\varepsilon y)\big)}{|x-y|^{n+2s}}\,dx\,dy}\\
& \qquad \qquad {\displaystyle =\varepsilon^{(n-2s)/2}\int_{\RR^n} |u^*(x)|^{2^*-2} u^*(x)\varphi(\varepsilon x)dx}\\
& \qquad \qquad {\displaystyle =\int_{\RR^n} |U_\varepsilon(x)|^{2^*-2}U_\varepsilon(x)\varphi(x)dx\,,}
\end{aligned}
\end{equation}
for any $\varphi \in H^s(\RR^n)$ and $\varepsilon>0$\,.
Hence, $U_\varepsilon$ is a solution of problem~\eqref{limitingproblem} for any $\varepsilon>0$\,.

Taking $\varphi=U_\varepsilon$ in \eqref{Uepsilonvarphi} (this choice is admissible, since $U_\varepsilon\in H^s(\RR^n)$), we get
$${\displaystyle \int_{\RR^{2n}} \frac{|U_\varepsilon(x)-U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,dx\,dy=
\int_{\RR^n}|U_\varepsilon(x)|^{2^*}\,dx}$$
for any $\varepsilon>0$\,.
The scale invariance of the norm in $L^{2^*}(\RR^n)$ under the scaling
$u^* \mapsto U_\varepsilon$ and \eqref{normaSs} give
$${\displaystyle \int_{\RR^n}|U_\varepsilon(x)|^{2^*}\,dx=\int_{\RR^n}|u^*(x)|^{2^*}\,dx=S_s^{n/(2s)}}$$
for any $\varepsilon>0$\,, and so Claim~\ref{claimUepsilon} is proved.
\end{proof}

In order to check~\eqref{LAC}, it is now
necessary to appropriately put~$U_\varepsilon$ to
zero outside~$\Omega$, according to the following procedure.
Let us fix $\delta>0$ such that
\begin{equation}\label{LAC2}
B_{4\delta}\subset\Omega\end{equation}
and let
$\eta \in C^\infty(\RR^n)$ be
such
that $0\leq \eta \leq 1$ in
$\RR^n$\,, $\eta\equiv 1$ in $B_\delta$ and $\eta\equiv
0$ in $\mathcal C B_{2\delta}$, where $B_{\delta}=B(0, \delta)$ and $\mathcal C B_\delta=\RR^n\setminus B_\delta$\,. For
every $\varepsilon>0$ we denote by $u_\varepsilon$ the following
function
\begin{equation}\label{uepsilon}
u_\varepsilon(x)=\eta(x)U_\varepsilon(x)\,, \quad x\in \RR^n\,,
\end{equation}
where $U_\varepsilon$ is given in \eqref{Uepsilon}\,.

In order to prove Theorem~\ref{lapfra0} we need to
check condition~\eqref{LAC} for~$u_\varepsilon$ (at least for some
suitably small~$\varepsilon$), and for this we will estimate
$S_{s,\,\lambda}(u_\varepsilon)$\,.
In the setting of the fractional Laplacian this estimate is more delicate than in the case of the Laplacian, due to the non-local nature of the operator~$(-\Delta)^s$\,.

For this we need some preliminary claims. In what follows we suppose that, up to a translation, $x_0=0$ in \eqref{utilde}\,.

\begin{claim}\label{claimstimauepsilon}
Let $\varrho>0$ and $\mu$ be as
in~\eqref{utilde}\,.
If $x\in{\mathcal C} B_\varrho$,
then
\begin{equation}\label{stima 0}
\big|u_\varepsilon(x)\big|\le\big|U_\varepsilon(x)\big|\le
C\varepsilon^{(n-2s)/2}
\end{equation}
for any $\varepsilon>0$ and for some positive constant
$C$, possibly depending on $\mu$, $\varrho$, $s$ and~$n$\,.
\end{claim}
\begin{proof} Using~\eqref{Uepsilon}, \eqref{ustar}, \eqref{baru}
and \eqref{utilde} (with~$x_0=0$), we have that
\begin{equation}\label{bal} U_\varepsilon(x)=\widetilde\kappa
\varepsilon^{-(n-2s)/2}
\left(\mu^2+\left|\frac{x}{\varepsilon
S_s^{1/(2s)}}\right|^2\right)^{-(n-2s)/2},\end{equation}
for a suitable~$\widetilde \kappa\in \RR\setminus\{0\}$\,.
If $x\in{\mathcal C} B_\varrho$,
we deduce that
$$\big|U_\varepsilon(x)\big|\le C\varepsilon^{-(n-2s)/2} \,\left(
\mu^2+\left(\frac{\varrho}{\varepsilon
S_s^{1/(2s)}}\right)^2\right)^{-(n-2s)/2}\le
C\varepsilon^{(n-2s)/2}\,,$$
and so Claim~\ref{claimstimauepsilon} follows, being $\eta\leq 1$ in $\RR^n$\,.
\end{proof}

\begin{claim}\label{claimstimagradienteuepsilon}
Let $\varrho>0$ and $\mu$ be as
in~\eqref{utilde}\,. If $x\in{\mathcal C} B_\varrho$,
then
\begin{equation}\label{stima gra}
\big|\nabla u_\varepsilon(x)\big|\le
C\varepsilon^{(n-2s)/2}
\end{equation}
for any $\varepsilon>0$ and for some positive constant $C$, possibly
depending on $\mu$, $\varrho$, $s$ and~$n$\,.
\end{claim}
\begin{proof}
First of all we observe that, for any $|\xi|\ge \varrho$, we have that
\begin{equation}\label{AUX}
\begin{split}
& \left( 1+\left|\frac{\xi}{\varepsilon}\right|^2\right)^{-(n-2s)/2}
+\frac{|\xi|}{\varepsilon^2}
\left( 1+\left|\frac{\xi}{\varepsilon}\right|^2\right)^{-1-(n-2s)/2}
\\&\qquad\le
\left( 1+\left|\frac{\xi}{\varepsilon}\right|^2\right)^{-(n-2s)/2}
+\frac1 \varrho \,\left|\frac{\xi}{\varepsilon}\right|^2
\left( 1+\left|\frac{\xi}{\varepsilon}\right|^2\right)^{-1-(n-2s)/2}\\ &\qquad\le
\left(1+\frac1 \varrho\right)\,
\left( 1+\left|\frac{\xi}{\varepsilon}\right|^2\right)^{-(n-2s)/2}
\\ &\qquad\le
\left(1+\frac1 \varrho\right)\,
\left( 1+\left|\frac{ \varrho}{\varepsilon}\right|^2\right)^{-(n-2s)/2}
\\ &\qquad\le
\left(1+\frac1 \varrho\right)\,\left( \frac{\varepsilon}{ \varrho}\right)^{n-2s}.
\end{split}
\end{equation}

As a consequence
of \eqref{uepsilon} and \eqref{AUX} we have that,
for any $x\in {\mathcal C} B_\varrho$,
$$\begin{aligned}
\big|\nabla u_\varepsilon(x)\big|
& \le C\varepsilon^{-(n-2s)/2} \left[
\left( \mu^2+\left|\frac{x}{\varepsilon S_s^{1/(2s)}}\right|^2\right)^{-(n-2s)/2}
\right. \\ &\qquad\quad\left.+\frac{1}{\varepsilon S_s^{1/(2s)}}
\left|\frac{x}{\varepsilon S_s^{1/(2s)}}\right|
\left( \mu^2+\left|\frac{x}{\varepsilon S_s^{1/(2s)}}\right|^2\right)^{-1-(n-2s)/2}\right]
\\ &\quad\le C\varepsilon^{-(n-2s)/2}\cdot \varepsilon^{n-2s}=
C\varepsilon^{(n-2s)/2}\,,
\end{aligned}$$
which proves Claim~\ref{claimstimagradienteuepsilon}\,.
\end{proof}

\begin{claim}\label{claimstimaminimo}
Let~$\delta$ be as in~\eqref{LAC2} and $\mu$ be as in \eqref{utilde}\,. The following assertions hold true
\begin{itemize}
\item [$a)$] for any $x\in\RR^n$ and $y\in {\mathcal{C}}B_\delta$,
with~$|x-y|\le\delta/2$,
\begin{equation}\label{stima 1}
\big| u_\varepsilon(x)-u_\varepsilon(y)\big|\le
C\varepsilon^{(n-2s)/2} |x-y|\,;
\end{equation}

\item [$b)$] for any $x, y\in\mathcal C B_\delta$
\begin{equation}\label{stima 11}
\big| u_\varepsilon(x)-u_\varepsilon(y)\big|\le
C\varepsilon^{(n-2s)/2} \min\{1, |x-y|\}
\end{equation}
\end{itemize}
for any $\varepsilon>0$ and for some positive constant $C$, possibly depending on $\mu$, $\delta$, $s$ and~$n$\,.
\end{claim}
\begin{proof}
Let us start by proving assertion~$a)$\,. For this let $x\in\RR^n$ and $y\in {\mathcal{C}}B_\delta$
with~$|x-y|\le\delta/2$ and
let $\xi$ be any point on the segment joining $x$ and $y$. Then
$$ \xi=tx+(1-t)y, \qquad{\mbox{ for some }}t\in[0,1]\,,$$
so that
$$ |\xi|=| y+t(x-y)|\ge |y|-t|x-y|\ge \delta -t(\delta/2)\ge \delta/2\,.$$
This and Claim~\ref{claimstimagradienteuepsilon} (here used with $\varrho=\delta/2$) imply that
$|\nabla u_\varepsilon(\xi)|\le
C\varepsilon^{(n-2s)/2}\,$, and so, by a first order Taylor expansion,
$$ \big| u_\varepsilon(x)-u_\varepsilon(y)\big|\le
C\varepsilon^{(n-2s)/2} |x-y|\,,$$
which proves \eqref{stima 1}.

Now, we show $b)$\,. For this let $x, y\in\mathcal C B_\delta$\,.
If $|x-y|\le \delta/2$ then $b)$ follows from $a)$\,, so we
may suppose~$|x-y|>\delta/2$\,: then
$$ \big| u_\varepsilon(x)-u_\varepsilon(y)\big|\le
\big| u_\varepsilon(x)\big|+\big|u_\varepsilon(y)\big|
\le
C\varepsilon^{(n-2s)/2}\,,$$ thanks to Claim~\ref{claimstimauepsilon} (here used with $\varrho=\delta$), and this
completes the proof of \eqref{stima 11}\,.
\end{proof}

Now we can estimate the Gagliardo seminorm of $u_\varepsilon$ according to the following result:

\begin{proposition}\label{stimagagliardo}
Let $s\in (0,1)$ and $n>2s$. Then, the  following estimate holds true
$${\displaystyle \int_{\RR^{2n}} \frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,dx\,dy
\le S_s^{n/(2s)}+\mathcal O(\varepsilon^{n-2s})}$$
as $\varepsilon\to 0$\,.
\end{proposition}
\begin{proof}
The proof makes use of the previous estimates
and it is a bit complicated -- definitely more
difficult than the one for similar results in the case
of the Laplacian. The additional complications
arise not only from technical obstacles (differentiating
is of course simpler than integrating), but also from
a different distribution of the energy density:
in particular, in the fractional case, the interaction
between the interior and the exterior of the cut-off
in~\eqref{uepsilon} is small but not anymore negligible
(indeed it will contribute to the total energy according
to the forthcoming estimate~\eqref{difficult5}, \eqref{difficult3} and \eqref{difficult2}).
To keep track of these non-local interactions, we introduce the notation
$$\mathbb D:=\big\{ (x,y)\in\RR^{2n} {\mbox{ : }}
x\in B_\delta,\ y\in{\mathcal{C}} B_\delta
{\mbox{ and }} |x-y|>\delta/2\big\}$$
and
$$\mathbb E:=\big\{ (x,y)\in\RR^{2n} {\mbox{ : }}
x\in B_\delta,\ y\in{\mathcal{C}} B_\delta
{\mbox{ and }} |x-y|\le \delta/2\big\}\,,$$
where $\delta$ is as in \eqref{LAC2}\,.

By \eqref{uepsilon} we have that
\begin{equation}\label{integraler2n}
\begin{aligned}
 \int_{\RR^{2n}}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy
&= \int_{B_\delta\times B_\delta}
\frac{|U_\varepsilon(x)-U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \qquad  +2 \int_{\mathbb D}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \qquad
+2\int_{\mathbb E}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \qquad +
\int_{(\mathcal C B_\delta )\times( \mathcal C B_\delta)}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\,.
\end{aligned}
\end{equation}
By Claim~\ref{claimstimaminimo} (here, in particular, we use \eqref{stima 11}) we have
\begin{equation}\label{stimacomplementare}
\int_{(\mathcal C B_\delta) \times (\mathcal C B_\delta)}
\!\!\!\!\!\!\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\leq C\varepsilon^{n-2s}
\int_{\RR^{2n}}\frac{\min\{1, |x-y|^2\}}{|x-y|^{n+2s}}\,
dx\,dy=\mathcal O(\varepsilon^{n-2s})\,,
\end{equation}
while, by \eqref{stima 1}
\begin{equation}\label{stimaE}
\begin{aligned}
\int_{\mathbb E}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy & \leq C\varepsilon^{n-2s}
\int_{{x\in
B_\delta, \,y\in \mathcal C B_\delta}\atop{|x-y|\leq \delta/2}}
\frac{|x-y|^2}{\,\,|x-y|^{n+2s}}\,dx\,dy\\
& \leq C\varepsilon^{n-2s} \int_{|x|\leq \delta}\, dx
\int_{|\xi|\leq \delta/2} \frac{1}{\,\,|\xi|^{n+2s-2}}\, d\,\xi\\
& = \mathcal O(\varepsilon^{n-2s})\,,
\end{aligned}
\end{equation}
as $\varepsilon \to 0$\,. In both these estimates we use that $s\in (0,1)$\,.

Now, in \eqref{integraler2n} it remains to estimate the integral on $\mathbb D$, that is
\begin{equation}\label{integraleD}
\int_{\mathbb D}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\,.
\end{equation}

For this, recalling that~$u_\varepsilon(x)=
U_\varepsilon(x)$ for any~$x\in B_\delta$ thanks to~\eqref{uepsilon},
we note that for any~$(x,y)\in \mathbb D$
$$\begin{aligned}
|u_\varepsilon(x)-u_\varepsilon(y)|^2 & =
|U_\varepsilon(x)-u_\varepsilon(y)|^2\\
& =\big|
\big(U_\varepsilon(x)-U_\varepsilon(y)\big)+
\big(U_\varepsilon(y)-u_\varepsilon(y)\big)\big|^2
\\
&\le
|U_\varepsilon(x)-U_\varepsilon(y)|^2+
|U_\varepsilon(y)-u_\varepsilon(y)|^2+2
|U_\varepsilon(x)-U_\varepsilon(y)|
\,|U_\varepsilon(y)-u_\varepsilon(y)|\,,
\end{aligned}$$
so that
\begin{equation}\label{difficult5}
\begin{aligned}
\int_{\mathbb D}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy &\le
\int_{\mathbb D}
\frac{|U_\varepsilon(x)-U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,dx\,dy\\
& \qquad
+ \int_{\mathbb D}
\frac{|U_\varepsilon(y)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,dx\,dy\\
& \qquad
+2 \int_{\mathbb D}\frac{
|U_\varepsilon(x)-U_\varepsilon(y)|
\,|U_\varepsilon(y)-u_\varepsilon(y)|}{|x-y|^{n+2s}}\,dx\,dy\,.
\end{aligned}
\end{equation}
Hence, in order to estimate~\eqref{integraleD},
we bound the last two terms in the right-hand side of
\eqref{difficult5}\,.

By exploiting Claim~\ref{claimstimauepsilon} (here used with $\varrho=\delta$)\,, we obtain
\begin{equation}\label{difficult3}
\begin{aligned}
\int_{\mathbb D}
\frac{|U_\varepsilon(y)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy & \le \int_{\mathbb D}
\frac{\big(|U_\varepsilon(y)|+|u_\varepsilon(y)|\big)^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \le 4\int_{\mathbb D}
\frac{|U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \le
C\varepsilon^{n-2s}
\int_{{x\in
B_\delta, \,y\in \mathcal C B_\delta }\atop{|x-y|>\delta/2}}\frac{1}{|x-y|^{n+2s}}\,
dx\,dy\\
& \leq C\varepsilon^{n-2s} \int_{|\zeta|\leq \delta}\, d\zeta \int_{|\xi|> \delta/2} \frac{1}{\,\,|\xi|^{n+2s}}\, d\,\xi\\
& =\mathcal O(\varepsilon^{n-2s})\,,
\end{aligned}\end{equation}
as $\varepsilon \to 0$\,.

In order to estimate the last term in the right-hand side of~\eqref{difficult5}, first of all we note that, once more  by~\eqref{bal} (which is valid for any $x\in \RR^n$) and Claim~\ref{claimstimauepsilon}\,,
$$ |U_\varepsilon(x)|\,|U_\varepsilon(y)|\le
C \left(\mu^2+\left|\frac{x}{\varepsilon
S_s^{1/(2s)}}\right|^2\right)^{-(n-2s)/2}\,$$
for any~$(x,y)\in \mathbb D$\,.

Therefore, by using the short-hand notation
$$\delta_\varepsilon:={\delta/ (\varepsilon S_s^{1/(2s)})}$$
and the change of variable~$\zeta:=x/
(\varepsilon
S_s^{1/(2s)})$ and~$\xi:=x-y$ (and up to renaming~$C$), we have that
\begin{equation*}
\begin{split}
\int_{\mathbb D} \frac{|U_\varepsilon(x)|\,|U_\varepsilon(y)|}{|x-y|^{n+2s}}\,
dx\,dy & \le
C \int_{\mathbb D} \left(\mu^2+\left|\frac{x}{\varepsilon
S_s^{1/(2s)}}\right|^2\right)^{-(n-2s)/2} |x-y|^{-(n+2s)}\,dx\,dy\\
&\qquad \le C \varepsilon^n
\int_{ { \zeta\in B_{\delta_\varepsilon}
}\atop{
|\xi|> \delta/2}}
\left(\mu^2+\left|\zeta\right|^2\right)^{-(n-2s)/2}|\xi|^{-(n+2s)}\,d\,\zeta\,d\,\xi\\
&\qquad \le C \varepsilon^n
\int_{ \zeta\in B_{ \delta_\varepsilon }}
\left(\mu^2+\left|\zeta\right|^2\right)^{-(n-2s)/2}\,d\,\zeta\\
&\qquad \le C \varepsilon^n\left[1+
\int_{\zeta\in B_{ \delta_\varepsilon} \setminus B_1 }
\left(\mu^2+\left|\zeta\right|^2\right)^{-(n-2s)/2}\,d\,\zeta
\right]\\
&\qquad \le C \varepsilon^n\left[1+
\int_{
\zeta\in B_{\delta_\varepsilon} \setminus B_1
}
\left|\zeta\right|^{-(n-2s)}\,d\,\zeta
\right]\\
&\qquad = C \varepsilon^n\left[1+
\int_1^{\delta/ (\varepsilon S_s^{1/(2s)})} \rho^{-(n-2s)+(n-1)}\,d\rho
\right]\\
&\qquad = C\varepsilon^n \big( 1+\varepsilon^{-2s}\big)
\\ &\qquad = \mathcal O(\varepsilon^{n-2s})\,,
\end{split}
\end{equation*}
as $\varepsilon\to 0$\,. Here we use again that $s\in (0,1)$\,.

As a consequence, recalling~\eqref{uepsilon} and Claim~\ref{claimstimauepsilon}, we get
\begin{equation}\label{difficult1}
\begin{split}
\int_{\mathbb D}
\frac{|U_\varepsilon(x)|\,|U_\varepsilon(y)-u_\varepsilon(y)|}{|x-y|^{n+2s}}\,
dx\,dy & \le
\int_{\mathbb D}
\frac{|U_\varepsilon(x)|\,\big(
|U_\varepsilon(y)|+|u_\varepsilon(y)|\big)}{|x-y|^{n+2s}}\,
dx\,dy
\\ & \le
2 \int_{\mathbb D}
\frac{|U_\varepsilon(x)|\,|U_\varepsilon(y)|}{|x-y|^{n+2s}}\,
dx\,dy
\\ & = \mathcal O(\varepsilon^{n-2s})\,,
\end{split}
\end{equation}
as $\varepsilon\to 0$\,.

On the other hand, again by~\eqref{uepsilon}
and Claim~\ref{claimstimauepsilon},
\begin{equation}\label{difficult mio}
\begin{aligned}
\int_{\mathbb D}
\frac{|U_\varepsilon(y)|\,|U_\varepsilon(y)-u_\varepsilon(y)|}{|x-y|^{n+2s}}\,
dx\,dy &\le
2 \int_{\mathbb D}
\frac{|U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \le C\varepsilon^{n-2s} \int_{{x\in
B_\delta, \, y\in \mathcal C B_\delta}\atop{|x-y|>\delta/2}}
\frac{1}{|x-y|^{n+2s}}\,
dx\,dy
\\
& = \mathcal O(\varepsilon^{n-2s})\,,
\end{aligned}
\end{equation}
as $\varepsilon\to 0$ (here we argue as in \eqref{difficult3}).
Putting together \eqref{difficult1} with \eqref{difficult mio}, we infer that
\begin{equation}\label{difficult2}
\begin{split}
\int_{\mathbb D} \frac{|U_\varepsilon(x)-U_\varepsilon(y)|
\,|U_\varepsilon(y)-u_\varepsilon(y)|}{|x-y|^{n+2s}}\,
dx\,dy & \le
\int_{\mathbb D}
\frac{|U_\varepsilon(x)|
\,|U_\varepsilon(y)-u_\varepsilon(y)|}{|x-y|^{n+2s}}\,
dx\,dy\\
& \qquad +
\int_{\mathbb D}
\frac{|U_\varepsilon(y)|
\,|U_\varepsilon(y)-u_\varepsilon(y)|}{|x-y|^{n+2s}}\,
dx\,dy\\
&  =\mathcal O(\varepsilon^{n-2s})\,,
\end{split}\end{equation}
as $\varepsilon\to 0$\,.

Finally, by \eqref{integraler2n}--\eqref{stimaE}, \eqref{difficult5},
\eqref{difficult3} and \eqref{difficult2} we get\footnote{It is
interesting to observe that the energy interaction outside
$B_\delta\times B_\delta$ is not negligible: indeed, while the
contributions in \eqref{stimacomplementare},
\eqref{stimaE}, \eqref{difficult3} and \eqref{difficult2} are
all~$\mathcal O(\varepsilon^{n-2s})$, the integral in~${\mathbb D}$
provides a relevant part that needs to be taken into account in
the full energy. }
$$\begin{aligned}
 \int_{\RR^{2n}}
\frac{|u_\varepsilon(x)-u_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy
&= \int_{B_\delta\times B_\delta}
\frac{|U_\varepsilon(x)-U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \qquad  +2 \int_{\mathbb D}
\frac{|U_\varepsilon(x)-U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,dx\,dy + \mathcal O(\varepsilon^{n-2s})\\
& \leq \int_{\RR^{2n}}
\frac{|U_\varepsilon(x)-U_\varepsilon(y)|^2}{|x-y|^{n+2s}}\,dx\,dy + \mathcal O(\varepsilon^{n-2s})\,,
\end{aligned}$$
as $\varepsilon \to 0$\,.
Then, the desired result now follows from~ Claim~\ref{claimUepsilon}.
\end{proof}

Now we consider the $L^2$ and the $L^{2^*}$-- norms of the
function~$u_\varepsilon$: in this case our estimates can be proved
exactly as in the case of the Laplacian, but we prefer to repeat them,
for reader's convenience.

\begin{proposition}\label{stima22star}
Let $s\in (0,1)$ and $n>2s$. Then, the  following estimates hold true
\begin{equation}\label{ug1}
\int_{\RR^n} |u_\varepsilon(x)|^2\,dx\geq \begin{cases}
C_s\varepsilon^{2s}+ \mathcal O(\varepsilon^{n-2s}) & \mbox{if} \quad n>4s\\
C_s\varepsilon^{2s}\big|\log\varepsilon\big|+ \mathcal O(\varepsilon^{2s})  & \mbox{if} \quad n=4s\\
C_s\varepsilon^{n-2s}+ \mathcal O(\varepsilon^{2s}) & \mbox{if} \quad n<4s\,,
\end{cases}\end{equation} and
\begin{equation}\label{ug2}\int_{\RR^n}
|u_\varepsilon(x)|^{2^*}\,dx=S_s^{n/(2s)}
+ \mathcal O(\varepsilon^n\,)\,,\end{equation} as $\varepsilon \to
0$\,, for some positive constant $C_s$ depending on $s$\,.
\end{proposition}
\begin{proof}
By definition of $u_\varepsilon$ and \eqref{Uepsilon} we get
$$\begin{aligned}
\int_{\RR^n} |u_\varepsilon(x)|^2\,dx & = \int_{B_\delta} |U_\varepsilon(x)|^2\,dx +\int_{B_{2\delta}\setminus B_\delta} |\eta(x)U_\varepsilon(x)|^2\,dx\\
& \geq \varepsilon^{-(n-2s)} \int_{B_\delta} |u^*(x/\varepsilon)|^2\,dx \\
& \geq \varepsilon^{2s} \int_R^{\delta/\varepsilon} |u^*(r)|^2\,r^{n-1}\,dr
\end{aligned}$$
for any $0<R<\delta/\varepsilon$\,.

Now, by \eqref{ustar} and Claim~\ref{claimubar} we get
$$\begin{aligned}
\int_{\RR^n} |u_\varepsilon(x)|^2\,dx & \geq \varepsilon^{2s} \int_R^{\delta/\varepsilon} |u^*(r)|^2\,r^{n-1}\,dr\\
& \cong \varepsilon^{2s} \int_R^{\delta/\varepsilon} r^{-n+4s-1}
dr\\
\\
& = \left\{\begin{array}{lll} -C_{1,s}\varepsilon^{n-2s}+C_{2,s}\varepsilon^{2s}
& \mbox{if} & n>4s\\\\
{\displaystyle C_{1,s}\varepsilon^{2s}\big|\log \varepsilon\big|+C_{2,s}\varepsilon^{2s}} & \mbox{if} &
n=4s\\\\
C_{1,s}\varepsilon^{n-2s}-C_{2,s}\varepsilon^{2s} & \mbox{if} & n<4s\,,
\end{array}\right.\\\\
& = \left\{\begin{array}{lll} C_s\varepsilon^{2s} + \mathcal O(\varepsilon^{n-2s})
& \mbox{if} & n>4s\\\\
{\displaystyle C_s\varepsilon^{2s}\big|\log \varepsilon\big| + \mathcal O(\varepsilon^{2s})} & \mbox{if} &
n=4s\\\\
C_s\varepsilon^{n-2s}+ \mathcal O(\varepsilon^{2s}) & \mbox{if} & n<4s\,,
\end{array}\right.\\\\
\end{aligned}$$
as $\varepsilon \to 0$\,, for some positive constant $C_{1,s}$, $C_{2,s}$ and $C_s$, depending on $s$, and for $\varepsilon$ sufficiently
small. This proves~\eqref{ug1} and we now prove~\eqref{ug2}.
Arguing in the same way and using Claim~\ref{claimUepsilon}, we have
\begin{equation}\label{ug3}\begin{aligned}
\int_{\RR^n} |u_\varepsilon(x)|^{2^*}\,dx &
=\int_{B_\delta} |U_\varepsilon(x)|^{2^*}\,dx+\int_{B_{2\delta}\setminus B_\delta} |\eta(x)U_\varepsilon(x)|^{2^*}\,dx\\
& \leq \int_{\RR^n} |U_\varepsilon(x)|^{2^*}\,dx+\int_{B_{2\delta}\setminus B_\delta} |U_\varepsilon(x)|^{2^*}\,dx\\
& = S_s^{n/(2s)} +\varepsilon^{-n}\int_{B_{2\delta}\setminus B_\delta} |u^*(x/\varepsilon)|^{2^*}\,dx\\
& \le S_s^{n/(2s)} +C\varepsilon^{n}\int_{B_{2\delta}\setminus B_\delta}
|x|^{-2n}\,dx\\
& = S_s^{n/(2s)}+\mathcal O(\varepsilon^n)
\end{aligned}\end{equation}
as $\varepsilon \to 0$\,. This ends the proof of Proposition~\ref{stima22star}\,.
\end{proof}

With Propositions~\ref{stimagagliardo} and \ref{stima22star} we can conclude the proof of Theorem~\ref{lapfra0}\,.

\subsubsection{End of the proof of Theorem~\ref{lapfra0}}\label{subsec:prooftheorem4}
We prove that~$u_\varepsilon$ satisfies condition~\eqref{LAC} if~$\varepsilon$
is suitably small (recall also that~$u_\epsilon$ vanishes
outside~$\Omega$ because of~\eqref{LAC2}). For
this, let
us consider two different cases.\\

\textbf{Case 1: $n>4s$\,.}
By the definition of the function $S_{s,\,\lambda}(\cdot)$ given in formula~\eqref{Sslambdav} and by
Propositions~\ref{stimagagliardo} and \ref{stima22star} we get
$$\begin{aligned}
{\displaystyle S_{s,\,\lambda}(u_\varepsilon)}& {\displaystyle \leq\frac{S_s^{n/(2s)}+\mathcal O(\varepsilon^{n-2s})-\lambda C_s\varepsilon^{2s}}{\Big(S_s^{n/(2s)}+\mathcal O(\varepsilon^{n}\Big)^{2/2^*}}}\\
& {\displaystyle \leq S_s +\mathcal O(\varepsilon^{n-2s})-\lambda \widetilde C_s\varepsilon^{2s}}\\
& {\displaystyle = S_s +\varepsilon^{2s}\Big(\mathcal O(\varepsilon^{n-4s})-\lambda \widetilde C_s\Big)<S_s\,,}
\end{aligned}$$
for any $\lambda>0$, if $\varepsilon>0$ is sufficiently small, for a suitable positive constant $\widetilde C_s$. Hence, condition~\eqref{LAC} is satisfied (with $u_0=u_\varepsilon$) for any $\lambda>0$ when $n>4s$\,. Thus, by Theorem~\ref{lapfra0f}, for any $\lambda\in (0, \lambda_{1,s})$ problem~\eqref{problemalapfrac0} admits a
solution $u\in H^s(\RR^n)$, which is not identically zero, such that $u=0$ a.e. in $\RR^n\setminus \Omega$\,.

\medskip

\textbf{Case 2: $n=4s$\,.}
Arguing as in Case~1 and taking into account that $n=4s$ for some positive constant $\widetilde C_s$ we get
$$\begin{aligned}
{\displaystyle S_{s,\,\lambda}(u_\varepsilon)}& {\displaystyle \leq\frac{S_s^{n/(2s)}+\mathcal O(\varepsilon^{n-2s})-\lambda C_s\varepsilon^{2s}\big|\log \varepsilon\big|+ \mathcal O(\varepsilon^{2s})}{\Big(S_s^{n/(2s)}+\mathcal O(\varepsilon^{n}\Big)^{2/2^*}}}\\
& {\displaystyle \leq S_s +\mathcal O(\varepsilon^{2s})-\lambda \widetilde C_s\varepsilon^{2s}\big|\log \varepsilon\big|}\\
& {\displaystyle = S_s +\varepsilon^{2s}\left(\mathcal O(1)-\lambda \widetilde C_s \big|\log \varepsilon\big|\right)<S_s
\,,}
\end{aligned}$$
for any $\lambda>0$, if $\varepsilon>0$ is sufficiently small. Hence, condition~\eqref{LAC} is satisfied with $u_0=u_\varepsilon$ for any $\lambda>0$ when $n=4s$\,. Therefore, thanks to Theorem~\ref{lapfra0f}\,, for any $\lambda\in (0, \lambda_{1,s})$ problem~\eqref{problemalapfrac0} admits a
solution $u\in H^s(\RR^n)$, which is not identically zero and such that $u=0$ a.e. in $\RR^n\setminus \Omega$\,.

\medskip

This completes the proof of Theorem~\ref{lapfra0}\,.

\begin{remark} {\rm We think that it is an interesting open problem
to check whether the same result
of Theorem~\ref{lapfra0} holds true when $2s<n<4s$\,.
In this last case, arguing as in Cases~1 and 2, we have
\begin{equation}\label{agg0}\begin{aligned}
{\displaystyle S_{s,\,\lambda}(u_\varepsilon)}& {\displaystyle \leq\frac{S_s^{n/(2s)}+\mathcal O(\varepsilon^{n-2s})-\lambda C_s\varepsilon^{n-2s}+ \mathcal O(\varepsilon^{2s})}{\Big(S_s^{n/(2s)}+\mathcal O(\varepsilon^{n}\Big)^{2/2^*}}}\\
& {\displaystyle \leq S_s +\varepsilon^{n-2s}\Big(\mathcal O(1)-\lambda \widetilde C_s\Big)+\mathcal O(\varepsilon^{2s})<S_s\,,}
\end{aligned}\end{equation}
if $\lambda$ is large enough, say $\lambda>\lambda_s>0$, when $\varepsilon>0$ is sufficiently small.
Hence, as a consequence of Theorem~\ref{lapfra0f}, when  $2s<n<4s$\,, if
\begin{equation} \label{agg}
\lambda_s<\lambda_{1,\,s}\end{equation} then
problem~\eqref{problemalapfrac0} admits a
solution $u\in H^s(\RR^n)$, which is not identically zero and such that $u=0$ a.e. in $\RR^n\setminus \Omega$\,, for any $\lambda\in (\lambda_s, \lambda_{1,s})$\,.
The difficulty here is to fix the competition in the above choices
of~$\lambda_s$, i.e. to
show that it is
possible to choose~$\lambda_s$ large enough, according to~\eqref{agg0},
but not too large, according to~\eqref{agg}.}\end{remark}

\subsubsection{Some remarks on condition~$S_{s,\,\lambda}<S_s$}\label{subsec:finalremarks}
As in the classical case (see the discussion in
\cite[pages 458--459]{bn}), also in the non-local setting
the Palais--Smale condition may globally fail for the functional $\mathcal J_{s,\,\lambda}$\,.
Therefore, it is necessary to restrict our attention only to a suitable range related to the best fractional critical Sobolev constant $S_s$ of the embedding $H^s(\RR^n)\hookrightarrow L^{2^*}(\RR^n)$\,. Indeed, this functional satisfies the Palais--Smale condition at any level $c$, with $c$ satisfying
\begin{equation}\label{cSs}
c<\frac s n\, S_s^{n/(2s)}\,.
\end{equation}
As discussed in the introduction,
another natural condition to look at is given by
\begin{equation}\label{SslambdaSs}
S_{s,\,\lambda}<S_s\,.
\end{equation}
It turns out that
conditions \eqref{cSs} and $\eqref{SslambdaSs}$ are related as follows:
\begin{proposition}\label{condequiv}
Let $0<\lambda<\lambda_{1,\,s}$\,, where $\lambda_{1,\,s}$ is the first eigenvalue of the operator~$(-\Delta)^s$with homogeneous Dirichlet boundary conditions and let
\begin{equation}\label{cdefs}
c=\inf_{P\in\mathcal P}\sup_{v\in P([0,1])} \mathcal J^0_{s,\,\lambda}(v)\,,
\end{equation}
where
$$\mathcal J^0_{s,\,\lambda}(u) =
\frac 1 2 \int_{\RR^{2n}}
\frac{|u(x)-u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-
\frac \lambda 2 \int_\Omega |u(x)|^2\,dx
-\frac{1}{2^*}\int_\Omega |u(x)|^{2^*}dx\,,$$
and
$$\mathcal P=\Big\{P\in C([0,1]; X_0) : P(0)=0\,,\,\, P(1)=e\Big\}$$
with $e\in H^s(\RR^n)$ such that $\mathcal J^0_{s,\,\lambda}(e)<0$\,.

Then, conditions \eqref{cSs} and $\eqref{SslambdaSs}$ are equivalent.
\end{proposition}

\begin{proof}
It is enough to show that
\begin{equation}\label{cfinale}
c=\frac s n\, S_{s,\,\lambda}^{n/(2s)}\,.
\end{equation}
For this, let $u\in H^s(\RR^n)\setminus\{0\}$ and let $P(\zeta)=\zeta u$ with $\zeta\in [0,1]$. By \eqref{supKfrac} and \eqref{cdefs} it is easily seen that
$$c\leq \sup_{\zeta\geq 0}\, \mathcal J_{s,\,\lambda}(\zeta u)=\frac s n\, S_{s,\,\lambda}^{n/(2s)}(u)\,,$$
so that, taking the infimum and using the definition~\eqref{Sslambda}, we obtain
\begin{equation}\label{cleq}
c\leq\frac s n\, S_{s,\,\lambda}^{n/(2s)}\,.
\end{equation}

Now, let us prove the other inequality.
In order to do this let us consider the continuous function
$$[0,1]\ni\zeta \mapsto  \langle (\mathcal
J^0_{s,\,\lambda})'(P(\zeta)),P(\zeta)\rangle\,,$$
for any fixed $P\in{\mathcal P}$\,.
Let
\begin{equation}\label{TZ}
\widetilde\zeta:=\sup\{ \zeta\in[0,1]\ :\
P(\zeta)=0\}\,.\end{equation}
Since~$P(0)=0$, $P(1)=e\ne0$ and $P$ is continuous, it follows
that
\begin{equation}\label{tilde zeta}
\widetilde\zeta \in[0,1)\,.
\end{equation}
Since $2<2^*$ and $\mathcal J^0_{s,\,\lambda}(e)<0$ by assumption, we have that
\begin{equation}\label{j'snegativo}
\begin{aligned}
\langle (\mathcal J^0_{s,\,\lambda})'(e),e\rangle &= \int_{\RR^{2n}}\frac{|e(x)-e(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-\,\lambda\int_\Omega |e(x)|^2\,dx
-\int_\Omega |e(x)|^{2^*}dx\\
& = 2\mathcal J^0_{s,\,\lambda}(e)+\Big(\frac{2}{\,2^*}-1\Big)\int_\Omega |e(x)|^{2^*}dx\\
& < 2\mathcal J^0_{s,\,\lambda}(e)<0\,.
\end{aligned}
\end{equation}
On the other hand, since~$0<\lambda<\lambda_{1,\,s}$\,, we get
\begin{equation*}
\begin{aligned}
\langle (\mathcal J^0_{s,\,\lambda})'(P(\zeta)),P(\zeta)\rangle & =
\int_{\RR^{2n}}\frac{|P(\zeta)(x)-P(\zeta)(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-
\lambda\int_\Omega
|P(\zeta)(x)|^2\,dx\\
& \qquad \qquad \qquad \qquad
-\int_\Omega |P(\zeta)(x)|^{2^*}dx\\
& \geq \Big(1-\frac{\lambda}{\lambda_{1,\,s}}\Big)
\int_{\RR^{2n}}\frac{|P(\zeta)(x)-P(\zeta)(y)|^2}{|x-y|^{n+2s}}\,dx\,dy
-\|P(\zeta)\|_{L^{2^*}(\Omega)}^{2^*}\,.
\end{aligned}
\end{equation*}
So, by Lemma~\ref{lemmanormaequiv}-$a)$
and the fact that~$2<2^*$,
$$ \langle (\mathcal J^0_{s,\,\lambda})'(P(\zeta)),P(\zeta)\rangle \ge
\frac1c \Big(1-\frac{\lambda}{\lambda_{1,\,s}}\Big)
\|P(\zeta)\|_{L^{2^*}(\Omega)}^{2}
-\|P(\zeta)\|_{L^{2^*}(\Omega)}^{2^*}>0\,,$$
provided that
$$\|P(\zeta)\|_{L^{2^*}(\Omega)}\in \left(0,
\left[ \frac1c \Big(1-\frac{\lambda}{\lambda_{1,\,s}}\Big)
\right]^{1/({2^*-2})} \right)\,.$$
In particular, since~$P$ is continuous,
recalling~\eqref{TZ} and~\eqref{tilde zeta}, we obtain that
there exists~$\widehat\zeta\in(\widetilde\zeta,1)$ such that
\begin{equation}\label{j'spositivo}
{\mbox{$\langle (\mathcal J^0_{s,\,\lambda})'(P(\zeta)),P(\zeta)\rangle
>0$ for any~$\zeta\in (\widetilde\zeta,\widehat\zeta)$\,.}}
\end{equation}
As a consequence of \eqref{j'snegativo} and \eqref{j'spositivo}
we get
that there exists $\bar \zeta\in (\widehat\zeta, 1)$ such that $\bar
u:=P(\bar \zeta)$ satisfies
\begin{equation}\label{ubar}
\langle (\mathcal J^0_{s,\,\lambda})'(\bar u),\bar u\rangle = \int_{\RR^{2n}}\frac{|\bar u(x)-\bar u(y)|^2}{|x-y|^{n+2s}}\,dx\,dy-\,\lambda\int_\Omega |\bar u(x)|^2\,dx
-\int_\Omega |\bar u(x)|^{2^*}dx=0\,.
\end{equation}
We stress that
\begin{equation}\label{bu}
\bar u\ne 0\,,\end{equation}
because~$\bar\zeta>\widehat\zeta>\widetilde\zeta$
by construction, and so~$P(\bar\zeta)\ne0$ by~\eqref{TZ}.

Also, using the definition of $\mathcal J^0_{s,\,\lambda}$ and
\eqref{ubar} we easily get
\begin{equation}\label{ubar2}
\mathcal J^0_{s,\,\lambda}(\bar u)=\Big(\frac 1 2 -\frac{1}{2^*}\Big)\|\bar u\|_{L^{2^*}}^{2^*}=\frac s n\,\|\bar u\|_{L^{2^*}}^{2^*}\,.
\end{equation}
Hence, by \eqref{Sslambda}, it holds true that
$$\begin{aligned}
S_{s,\,\lambda}& \leq S_{s,\,\lambda}(\bar u)\\
& =\frac{{\displaystyle\int_{\RR^{2n}}\frac{|\bar u(x)-\bar u(y)|^2}{|x-y|^{n+2s}} \,dx\,dy-\lambda \int_{\RR^n} |\bar u(x)|^2\,dx}}{{\displaystyle\Big(\int_{\RR^n} |\bar u(x)|^{2^*}dx\Big)^{2/2^*}}}\\
& \leq \frac{{\displaystyle\int_{\RR^{2n}}\frac{|\bar u(x)-\bar u(y)|^2}{|x-y|^{n+2s}} \,dx\,dy-\lambda \int_\Omega |\bar u(x)|^2\,dx}}{{\displaystyle\Big(\int_\Omega |\bar u(x)|^{2^*}dx\Big)^{2/2^*}}}\\
& = {\displaystyle\Big(\int_\Omega |\bar u(x)|^{2^*}dx\Big)^{(2^*-2)/2^*}}\\
& = {\displaystyle\Big(\frac n s\, \mathcal J^0_{s,\,\lambda}(\bar u)\Big)^{2s/n}}\\
& \leq {\displaystyle\Big(\frac n s\, \sup_{v\in P([0,1])}\,\mathcal J^0_{s,\,\lambda}(v)\Big)^{2s/n}}\,,
\end{aligned}$$
thanks to \eqref{ubar}--\eqref{ubar2}\,.
Taking the infimum in $P\in \mathcal P$\,, we conclude that
$${\displaystyle S_{s,\,\lambda}\leq \Big(\frac n s\, c\Big)^{2s/n}}\,,$$
that is
\begin{equation}\label{cgeq}
c\geq \frac s n\,S_{s,\,\lambda}^{n/(2s)}\,.
\end{equation}
By \eqref{cleq} and \eqref{cgeq} we deduce \eqref{cfinale}, and this ends the proof of Proposition~\ref{condequiv}\,.
\end{proof}

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