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Variational inequalities, integrodifferential operators, fractional Laplacian
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\begin{document}
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\title[Lewy-Stampacchia type estimates]{Lewy-Stampacchia type estimates for variational
inequalities driven by (non)local operators}
\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{Variational inequalities, integrodifferential
operators, fractional Laplacian.\\
\phantom{aa} 2010 AMS Subject Classification: Primary: 35J86, 35J87, 49J40;
Secondary: 35R11, 35R35.}

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

\begin{abstract}
The purpose of this paper is to derive some Lewy-Stampacchia
estimates in some cases of interest, such as the ones driven by
non-local operators. Since we will perform an abstract approach to
the problem, this will provide, as a byproduct, Lewy-Stampacchia
estimates in more classical cases as well. In particular, we can
recover the known estimates for the standard Laplacian, the
$p$-Laplacian, and the Laplacian in the Heisenberg group. In the
non-local framework we prove a Lewy-Stampacchia estimate for a
general integrodifferential operator and, as a particular case,
for the fractional Laplacian. As far as we
know, the results in the non-local setting are new.
\end{abstract}

\maketitle

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

\subsection{The classical obstacle problem and its modifications}

The simplest, classical example of obstacle problem
consists of an elastic membrane, with vertical displacement~$u$
on a domain~$\Omega$,
which is constrained at its boundary (say~$u=u_0$
along~$\partial\Omega$) and
it is forced to lie below some
obstacle (say, $u\le\psi$). Then, at the equilibrium,
whenever the membrane
does
not touch the obstacle, the elasticity provides
a balance of the tension of the membrane, that, geometrically,
reflects into a balance
of the principal curvatures of the surface described by~$u$.
On the other hand, when the membrane sticks to the obstacle,
its principal curvatures are expected to adapt to the ones of~$\psi$.
These physical considerations lead to the classical
variational inequality
\begin{equation}\label{E1}
\int_\Omega\nabla u(x)(\nabla v(x)-\nabla u(x))\,dx\ge0\end{equation} say,
for any test function~$v$, with~$v\le\psi$ and~$v=u_0$
along~$\partial\Omega$ (more formal details on the function spaces
will be discussed in the sequel).

If an external force~$-f$ is switched on,
the rest configuration of the membrane
will be such that the elastic tension of the membrane
equilibrates the force, so that~\eqref{E1}
becomes
\begin{equation}\label{E2}
\int_\Omega\nabla u(x)(\nabla v(x)-\nabla u(x))\,dx\ge\int_\Omega
f(x)\,(v(x)-u(x))\,dx\,.\end{equation} Of course, in the literature,
many extensions of this problem has been considered, particularly
for taking into account nonlinear elastic reactions of the membrane,
non commutative effects, and non-local interactions. For instance,
one may replace the linear elasticity (say, Hook's law) with a
power-like one: this would change~\eqref{E2} into the following
variational inequality of $p$-Laplace type:
\begin{equation}\label{E3}
\int_\Omega |\nabla u(x)|^{p-2}\nabla u(x)(\nabla v(x)-\nabla u(x))\,dx\ge\int_\Omega
f(x)\,(v(x)-u(x))\,dx\,,\end{equation}
for some~$p\in(1,\infty)$. These types of obstacle problems
have been recently considered in~\cite{Challal, Rod2}.

Similarly, one might replace the commutative Euclidean vector
fields with some non-commutative ones, such as the ones
of the Heisenberg group~${\HH^n}$
(see, e.g.,~\cite{VSCC92}). In this case,~\eqref{E2}
is replaced by
\begin{equation}\label{E4}
\int_\Omega\nabla_{\HH^n} u(\xi)(\nabla_{\HH^n} v(\xi)-\nabla_{\HH^n} u(\xi))\,d\xi\ge\int_\Omega
f(\xi)\,(v(\xi)-u(\xi))\,d\xi\,.\end{equation}
This type of variational inequalities
has been recently dealt with in~\cite{ValdinociPinamonti}.

Analogously, one might replace the local elastic reaction
in~\eqref{E2} with a non-local one, with the purpose of
taking into account the long-range interactions of particles.
For instance, one might replace the standard Laplacian~$\Delta$
with the so-called fractional Laplacian~$- (-\Delta)^s$,
with~$s\in(0,1)$.
In this case,~\eqref{E2} is replaced by the non-local
variational inequality
\begin{equation}\label{E5}
\begin{split}& \int_{\RR^{2n} \setminus ((\RR^n\setminus
\Omega)\times(\RR^n
\setminus \Omega))}\frac{(u(x)-u(y))(v(x)-v(y)-u(x)+u(y))}{|x-y|^{n+2s}}\,dx\,dy
\\ &\qquad\ge\int_\Omega
f(x)\,(v(x)-u(x))\,dx\,.\end{split}
\end{equation}
These kind of obstacle problems have been extensively studied
in~\cite{salsa, silvestre2, silvestre} (see also~\cite{land, stein,
sema} for the basic definitions and properties of the fractional
Laplacian). Obstacle problems for other integrodifferential kernels
have been also studied in~\cite{MasMat1, MasMat2}.

In Section~\ref{sec:abstract}, we will provide
a unified framework which will comprise
simultaneously, as particular cases,
all the variational inequalities in~\eqref{E2}, \eqref{E3},
\eqref{E4} and \eqref{E5}.

\subsection{Lewy-Stampacchia type estimates}

Solutions of the variational inequality~\eqref{E1}
have a bounded Laplacian. This fact may be euristically guessed
via the following argument. When~$u$ lies below the obstacle~$\psi$,
it is harmonic, hence~$\Delta u=0$. On the other hand,
at the points where~$u$ sticks to the obstacle,
one expects~$\Delta u$ to somewhat match with~$\Delta\psi$,
and the obstacle has to ``bend up'' at those contact points,
that is~$\Delta\psi\ge 0$. Therefore, though this argument is not rigorous
(since it does not really take into account possible singularities
that may occur at the free boundary~$\partial \{ u<\psi\}$,
which, in fact, could be in principle quite a wild set), one
can expect that solutions of the variational inequality~\eqref{E1}
have~$\Delta u$ comprised between~$0$ and the positive
part of~$\Delta\psi$, namely
\begin{equation}\label{LS1}
0\le \Delta u\le (\Delta \psi)^+\,.
\end{equation}
The content of the so-called Lewy-Stampacchia estimates
(named after~\cite{LewySt})
is exactly a rigorous derivation of~\eqref{LS1},
and possible
generalizations. Such kind of bounds are also called
``dual estimates'', since they are usually understood
as an integral bound in the dual of~$L^\infty(\Omega)$,
that is~\eqref{LS1} is derived in the distributional form
$$
0\le \int_\Omega \Delta u(x)\varphi(x) \,dx\le \int_\Omega (\Delta
\psi)^+(x)\varphi(x)\,dx
$$
for any nonnegative test function~$\varphi$. Also, when
an external force comes into play as in~\eqref{E2},
then~\eqref{LS1} gets modified as
$$ 0\le \Delta u+f\le (\Delta \psi+f)^+\,.$$

The purpose of this paper is to derive some Lewy-Stampacchia
estimates in some cases of interest, such as the ones driven by
non-local operators. Since we will perform an abstract approach to
the problem, this will provide, as a byproduct, Lewy-Stampacchia
estimates in more classical cases as well (in particular, we can
recover the known estimates for the standard Laplacian, the
$p$-Laplacian, and the Laplacian in the Heisenberg group).

In the non-local framework, the simplest example we
can deal with is given by the fractional\footnote{As customary,
we denoted by~$H^s(\Omega)$
the fractional Sobolev space, endowed by the so-called
Gagliardo norm, and by
$H^s_0(\Omega)$ the closure of $C^\infty_0(\Omega)$ with
respect to the norm~$\|\cdot \|_{H^s(\Omega)}$ (for further
details, see~\eqref{20}
in the sequel, and~\cite{valpal}
for a crash introduction
to fractional Sobolev spaces).}
Laplacian,
according to the following result:

\begin{theorem}[Lewy-Stampacchia type estimate for the 
fractional Laplacian]\label{lap fra}
Let $s\in (0,1)$, $n>2s$ and $\Omega$ be an open bounded set of $\RR^n$\,.
Let $f\in L^\infty(\Omega)$ and let $u_0$ and $\psi:\RR^n\to \RR$ be two
functions such that $u_0\in H^s(\Omega)\cap L^\infty(\RR^n\setminus
\Omega)$ and $\psi\in H^s(\Omega)$ with $u_0\leq \psi$ a.e. in
$\RR^n$ and $(-\Delta)^s\, \psi\in L^\infty(\Omega)$\,.

If $u:\RR^n\to \RR$ is a solution of the variational inequality
$$\left\{\begin{array}{l}
{\displaystyle \int_{\!\RR^{2n} \setminus \!((\RR^n\setminus
\Omega)\times(\RR^n \setminus \Omega))}
\!\!\!\!\!\!\!\!\!\frac{(u(x)-u(y))(v(x)-v(y)-u(x)+u(y))}{|x-y|^{n+2s}}\,dx\,dy\!\ge\!\!\int_\Omega
\!\!\!f(x)\,(v(x)-u(x))\,dx}\\
\\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \forall\,\, v\in
H^s(\Omega),\,\, v=u_0\,\, \mbox{a.e. in}\,\,
\RR^n\setminus\Omega,\,\, v\leq
\psi\,\, \mbox{a.e. in}\,\, \Omega\\
\\
u\in H^s(\Omega),\, u=u_0\,\, \mbox{a.e. in}\,\,
\RR^n\setminus\Omega,\, u\leq \psi\,\,
\mbox{a.e. in}\,\, \Omega\,,\\
\\
\end{array}\right.$$
then
\begin{equation}\label{Ex1}\begin{aligned}
& 0\leq-\int_{\RR^{2n} \setminus ((\RR^n\setminus
\Omega)\times(\RR^n \setminus \Omega))}
\frac{(u(x)-u(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}\,dx\,dy+\int_\Omega
f(x)\varphi(x)dx\\
& \qquad \qquad \qquad \qquad \qquad\leq\int_\Omega \Big(-(-\Delta)^s\,
\psi+f\Big)^+(x)\varphi(x) dx
\end{aligned}\end{equation}
for any $\varphi\in H^s_0(\Omega)$
with $\varphi\ge 0$ a.e. in $\Omega$ and $\varphi=0$ a.e. in
$\RR^n\setminus\Omega$.
\end{theorem}

Notice that~\eqref{Ex1} is a dual estimate, in the sense
that it can be interpreted, in the sense of distributions,
as a bound on the operator~$-(-\Delta)^s u$, by writing, concisely
$$ 0\leq -(-\Delta)^s u+f \le\big(
-(-\Delta)^s \psi+f\big)^+\,.$$
This interpretation in the sense of distribution is a general
property.

In fact, Theorem~\ref{lap fra} is a particular case of more general
result for integrodifferential operators of non-local type, as stated
in the next theorem, where we consider the integral operator $\mathcal L_K$ defined as follows:
\begin{equation}\label{lk}
\mathcal L_Ku(x)=\int_{\RR^n}\Big(u(x+y)+u(x-y)-2u(x)\Big)K(y)\,dy\,.
\end{equation}
Note that when $K(y)=|y|^{-(n+2s)}$, then $\mathcal L_K=-(-\Delta)^s$, $s\in (0,1)$\,.

\begin{theorem}[Lewy-Stampacchia type estimate for
general integrodifferential operators]\label{lap fra general kernel}
Let $K:\RR^n\setminus\{0\}\rightarrow(0,+\infty)$ be a function with the properties that
\begin{equation}\label{kernel}
{\mbox{$mK\in L^1(\RR^n)$, where $m(x)=\min \{|x|^2, 1\}$\,;}}
\end{equation}
\begin{equation}\label{kernelfrac}
\begin{aligned}
& \mbox{there exists}\,\, \lambda>0\,\, \mbox{and}\,\, s\in (0,1)\,\,
\mbox{such that for any}\,\, x\in \RR^n \setminus\{0\}\\
& \qquad \qquad \qquad \qquad \qquad K(x)\geq \lambda |x|^{-(n+2s)}\,;\\
\end{aligned}
\end{equation}
\begin{equation}\label{evenkernel}
K(x)=K(-x)\,\, \mbox{for any}\,\, x\in \RR^n \setminus\{0\}\,.
\end{equation}

Let $\Omega$ be an open bounded set of $\RR^n$, with $n>2s$.
Moreover, let $f\in L^\infty(\Omega)$ and let $u_0\in X\cap
L^\infty(\RR^n\setminus\Omega)$ and $\psi\in X$ with $u_0\leq \psi$
a.e. in $\RR^n$ and $\mathcal L_K\, \psi\in
L^\infty(\Omega)$\,, where $X$ is the linear space of Lebesgue measurable functions from $\RR^n$ to $\RR$ such that the restriction to $\Omega$ of any function in $X$ belongs to $L^2(\Omega)$ and
$$\mbox{the map}\,\,\,
(x,y)\mapsto (g(x)-g(y))\sqrt{K(x-y)}\,\,\, \mbox{is in}\,\,\,
L^2(Q, dxdy)\,,$$
with $Q=\RR^{2n} \setminus ((\RR^n\setminus
\Omega)\times(\RR^n \setminus \Omega))$\,.

If $u:\RR^n\to \RR$ is a solution of the variational inequality
\begin{equation}\label{varineqK}
{\small\left\{\begin{array}{l}
{\displaystyle \int_Q (u(x)-u(y))(v(x)-v(y)-u(x)+u(y))K(x-y) dx\,dy
\geq \int_\Omega f(x)(v(x)-u(x))dx}\\
\\
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \forall\,\, v\in
X,\,\, v=u_0\,\, \mbox{a.e. in}\,\, \RR^n\setminus\Omega,\,\, v\leq
\psi\,\, \mbox{a.e. in}\,\, \Omega\\
\\
u\in X,\, u=u_0\,\, \mbox{a.e. in}\,\, \RR^n\setminus\Omega,\, u\leq
\psi\,\,
\mbox{a.e. in}\,\, \Omega\,,\\
\\
\end{array}\right.}
\end{equation}
then
\begin{equation}\label{Ex2}
\begin{aligned}
& 0\leq -\int_Q
(u(x)-u(y))(\varphi(x)-\varphi(y))\,dx\,dy+\int_\Omega
f(x)\varphi(x)\,dx\\
& \qquad \qquad \qquad \qquad  \qquad \qquad \qquad\leq
\int_\Omega \Big(\mathcal L_K\, \psi+f\Big)^+(x)\varphi(x)
\,dx
\end{aligned}
\end{equation}
for any $\varphi\in C^\infty_0(\Omega)$ with $\varphi\ge 0$ in
$\Omega$ and $\varphi=0$ in $\RR^n\setminus\Omega$.
\end{theorem}

We observe that~\eqref{Ex2} may be stated,
in the distributional sense, as
$$ 0\leq \mathcal L_K u+f \le
(\mathcal L_K \psi+f)^+\,.$$
As far as we know, Theorems~\ref{lap fra}
and~\ref{lap fra general kernel} are new. On the other hand, as already
mentioned, since we
will perform a functional analytic approach, these results for us
will be the consequence of an abstract framework, which comprises
some known results as particular cases. For instance, we list some
classical and recent results that can be recovered by our approach:

\begin{theorem}[Lewy-Stampacchia type estimate for the 
Laplacian]\label{lap}
Let $\Omega \subset \RR^n$, $n\ge 1$, be an open bounded set of class $C^1$. Let $f\in L^\infty(\Omega)$ and let $u_0$, $\psi:\RR^n\to \RR$ be two functions
such that $u_0\in C(\overline \Omega)$ and $\psi \in C^2(\overline
\Omega)$ with $u_0\leq \psi$ a.e. in $\RR^n$\,.

If $u:\Omega\to \RR$ is a solution of the variational inequality
\begin{equation}\label{varineqlapl}
\left\{\begin{array}{l}
{\displaystyle \int_\Omega \nabla u(x)(\nabla v(x)-\nabla u(x)) dx
\geq \int_\Omega f(x)(v(x)-u(x))dx}\\
\\
\qquad \qquad \qquad \qquad \forall\,\, v\in H^1(\Omega),\,\,
v-u_0\in H^1_0(\Omega),\,\, v\leq
\psi\,\, \mbox{a.e. in}\,\, \Omega\\
\\
u\in H^1(\Omega),\, u-u_0\in H^1_0(\Omega),\, u\leq \psi\,\,
\mbox{a.e. in}\,\, \Omega\,,\\
\\
\end{array}\right.
\end{equation}
then
\begin{equation}\label{Ex3}
0\leq -\int_\Omega \nabla u(x)\nabla \varphi(x)
dx+\int_\Omega f(x)\varphi(x)dx \leq
\int_\Omega \Big(\Delta \psi+f\Big)^+(x)\varphi(x) dx\end{equation}
for any
$\varphi\in H^1_0(\Omega)$ with $\varphi\ge 0$ a.e. in $\Omega$\,.
\end{theorem}

\begin{theorem}[Lewy-Stampacchia type estimate for the 
$p$-Laplacian]\label{p lap}
Let $\Omega \subset \RR^n$, $n\ge 1$, be an open bounded set of class $C^{1, \alpha}$, with $\alpha \in (0, 1]$. Let $f\in L^\infty(\Omega)$ and let $u_0$, $\psi:\RR^n\to \RR$ be two functions such
that $u_0\in C^{1, \alpha}(\overline \Omega)$ and $\psi \in
C^2(\Omega)\cap C(\overline \Omega)$ with $u_0\leq \psi$ a.e. in
$\RR^n$ and $\Delta_p\, \psi\in L^\infty(\Omega)$, $p>1$\,.

If $u:\Omega\to \RR$ is a solution of the variational inequality
\begin{equation}\label{varineqplapl}
\left\{\begin{array}{l}
{\displaystyle \int_\Omega |\nabla u(x)|^{p-2}\nabla u(x)(\nabla
v(x)-\nabla u(x)) dx
\geq \int_\Omega f(x)(v(x)-u(x))dx}\\
\\
\qquad \qquad \qquad \qquad \forall\,\, v\in W^{1,p}(\Omega),\,\,
v-u_0\in W^{1,p}_0(\Omega),\,\, v\leq
\psi\,\, \mbox{a.e. in}\,\, \Omega\\
\\
u\in W^{1,p}(\Omega),\, u-u_0\in W^{1,p}_0(\Omega),\, u\leq \psi\,\,
\mbox{a.e. in}\,\, \Omega\,,\\
\\
\end{array}\right.
\end{equation}
then
\begin{equation}\label{Ex4}
0\leq -\int_\Omega |\nabla u(x)|^{p-2}\nabla
u(x)\nabla \varphi(x) dx+\int_\Omega f(x)\varphi(x)dx \leq
\int_\Omega \Big(\Delta_p\, \psi+f\Big)^+(x)\varphi(x)
dx\end{equation}
for any
$\varphi\in W^{1,p}_0(\Omega)$ with $\varphi\ge 0$ a.e. in
$\Omega$\,.
\end{theorem}

\begin{theorem}[Lewy-Stampacchia type estimate in the 
Heisenberg group]\label{hei}
Let $\Omega \subset \RR^n$, $n\ge 1$, be an open bounded set with
smooth boundary. Let $f\in L^\infty(\Omega)$ and let $u_0$, $\psi:\RR^n\to \RR$ be two functions
such that $u_0\in W^{1,p}_{\HH^n}(\Omega_*)\cap L^\infty(\Omega_*)$, with $\Omega_*$ smooth domain such that $\Omega \subset \subset \Omega_*$, and $\psi \in
C^2(\Omega)$ with $u_0\leq \psi$ a.e. in
$\RR^n$ and $\Delta_{\HH^n}\,\psi\in L^\infty(\Omega)$\,.

If $u:\Omega\to \RR$ is a solution of the variational inequality
$$\left\{\begin{array}{l}
{\displaystyle \int_\Omega \nabla_{\HH^n} u(x)(\nabla_{\HH^n}
v(x)-\nabla_{\HH^n} u(x)) dx
\geq \int_\Omega f(x)(v(x)-u(x))dx}\\
\\
\qquad \qquad \qquad \qquad \forall\,\, v\in
W^{1,2}_{\HH^n}(\Omega),\,\, v-u_0\in W^{1,2}_{\HH^n,0}(\Omega),\,\,
v\leq
\psi\,\, \mbox{a.e. in}\,\, \Omega\\
\\
u\in W^{1,2}_{\HH^n}(\Omega),\, u-u_0\in
W^{1,2}_{\HH^n,0}(\Omega),\, u\leq \psi\,\,
\mbox{a.e. in}\,\, \Omega\,,\\
\\
\end{array}\right.$$
then
\begin{equation}\label{Ex5}
0\leq -\int_\Omega \nabla_{\HH^n}
u(x)\nabla_{\HH^n} \varphi(x) dx+\int_\Omega f(x)\varphi(x)dx \leq
\int_\Omega \Big(\Delta_{\HH^n}\psi+f\Big)^+(x)\varphi(x) dx
\end{equation}
for
any $\varphi\in W^{1,2}_{\HH^n,0}(\Omega)$ with $\varphi\ge 0$ a.e.
in $\Omega$\,.
\end{theorem}

We observe that~\eqref{Ex3}, \eqref{Ex4} and~\eqref{Ex5}
may be interpreted in the sense of distributions, by concisely
writing
\begin{eqnarray*}
&& 0\leq \Delta u+f \le\big(\Delta \psi+f\big)^+\,,\\
&& 0\leq \Delta_p u+f \le\big(\Delta_p \psi+f\big)^+\\
{\mbox{and }}
&& 0\leq \Delta_{\HH^n} u+f \le\big(\Delta_{\HH^n}+ \psi+f\big)^+\,,
\end{eqnarray*}
respectively.

Theorem~\ref{lap} was proved in~\cite{LewySt} and the literature is
rich of many important extensions: see, among the
others,~\cite{BocGallouet90, DonatiF82, MoscoFre, Mok, Mok1, Mok2, MoscoTro} and
the references therein. A proof of Theorem~\ref{p lap} was recently
performed in~\cite{Challal}, and we found their approach very
inspiring for our setting
(see also~\cite{Rod2} for related
results). The case discussed in Theorem~\ref{hei}
was recently considered in~\cite{ValdinociPinamonti}
(see, e.g.,~\cite{VSCC92} for the basics of the Heisenberg
group and of the related Sobolev spaces).

The paper is organized as follows. In Sections~\ref{sec:abstract}
and~\ref{sec:stime} we state and
prove a Lewy-Stampacchia estimate in an abstract setting, while
Section~\ref{sec:laplacian} is devoted to the classical cases of the
Laplacian and the $p$-Laplacian operators,
and to the Laplacian in the
Heisenberg group. In Section~\ref{sec:kernel} we consider an
integrodifferential operator with slow decay and, in this
setting, we prove a dual estimate for the solutions of a variational
inequality driven by this operator. For us, the main
application of this is
a Lewy-Stampacchia type estimate for the fractional Laplacian
operator.

\section{The abstract setting}\label{sec:abstract}

In what follows let $Q\subset\mathbb R^m$, $\Omega$ be an open subset
of~$\RR^n$, $m,n\geq 1$, and let us denote by
$\mathcal C \Omega=\RR^n\setminus \Omega$.

Let $\mu$ be a measure on $Q$ (in particular,~$Q$
is~$\mu$-measurable), and let ${\displaystyle L^p(Q, d\mu)}$, with
$1\leq p\leq \infty$, be the standard Lebesgue space with respect to
the measure $\mu$\,. As usual, we also denote~$L^p(Q)=L^p(Q,dx)$
the standard Lebesgue space with respect to the Lebesgue measure~$dx$, while $\mathcal M(\RR^n)$ will be the linear space of Lebesgue measurable functions from $\RR^n$ to $\RR$.

We take $X$ to be a linear subspace of $\mathcal M(\RR^n)$,
with the property that
(the restriction to~$\Omega$
of) any function in~$X$ belongs to~$L^1(\Omega)$ (i.e., in simbols,
if~$u\in X$, then~$u_{|\Omega}\in L^1(\Omega)$).

Moreover, let $Y$ and $Z$ two sets such that it can be defined a product $$\begin{aligned}
 \mathcal P: & \,\,Y\times Z\to \mathbb R\\
& (y,z)\mapsto \mathcal P(y,z)=:yz
\end{aligned}$$
with the following property
\begin{equation}\label{integrabilita}
\forall\, y\in Y \,\, \mbox{and}\,\, \forall\, z\in Z\,\, \mbox{it holds}\,\, yz\in L^1(Q, d\mu)\,.
\end{equation}

We also consider two functions $u_0$, $\psi\in X$ with $u_0\leq\psi$
a.e. in $\RR^n$ and we define $\mathcal W$ as the set containing all
the nonnegative constants and the function $u_0-\psi$. We also
consider a linear subspace~$X_0$ and a set
$\widetilde X_0$ such that $\widetilde X_0\subseteq X_0\subseteq X$. We require that~$X_0$ satisfies the
following property:
\begin{equation}\label{x01}
\mbox{if $v\in X_0$\,\, \mbox{and}\,\, $w\in \mathcal W$}\,\,\, \mbox{then}\,\, (v+w)^+\in X_0\,,
\end{equation}
where $g^+$ denotes the positive part of
a function~$g$, that is $g^+(x)=\max\{g(x), 0\}$\,.

We also introduce the two functionals
$$a:X\to Y$$
and
$$b:X\to Z$$
and we define $\mathcal A:X\times X\to \RR$ by
$$\mathcal A[u, \varphi]:=-\int_Q a(u)b(\varphi) d\mu$$
for any $u, \varphi \in X$\,. We remark
that~$\mathcal A$ is well defined, thanks to~\eqref{integrabilita}.

If $u\in X$ and if there exists~$\Upsilon\in L^\infty(\Omega)$
such that
$$\mathcal A[u, \varphi]=\int_\Omega \Upsilon(x)\,\varphi(x)\, dx$$
for any $\varphi \in X_0$, we denote $\mathcal A(u):= \Upsilon$ and we say that
$\mathcal A(u)\in L^\infty(\Omega)$\,.

Throughout the paper we need the following assumptions on the functionals $a, b$ and $\mathcal A$:
\begin{equation}\label{conda}
a(v+\eta)=a(v) \quad \mbox{for any} \quad v
\in X \quad \mbox{and} \quad \eta\in \mathbb R \,;
\end{equation}
\begin{equation}\label{condb}
b(-v)=-~b(v) \quad \mbox{for any} \quad v\in X\,;
\end{equation}
\begin{equation}\begin{aligned}
& {\mbox{if $u,v\in X$, with $(u-v)^+\in X_0$ and
$\mathcal A[u,(u-v)^+]\ge
\mathcal A[v,(u-v)^+]$,}} \\&{\mbox{then $u\le v$ a.e. in
$\Omega$\,.}}\label{monotonia}
\end{aligned}\end{equation}

Next we prove our Lewy-Stampacchia type estimate in this general
framework.

\section{Lewy-Stampacchia type estimates}\label{sec:stime}

Let us fix $f\in L^\infty(\Omega)$ and suppose that~$\mathcal A(\psi) \in L^\infty(\Omega)$\,.

We consider the following variational inequality
\begin{equation}\label{varineq}
\left\{\begin{array}{l} {\displaystyle \int_Q a(u) b(v-u)d\mu \geq
\int_\Omega f(v-u)dx} \quad \forall\, v\in X,\,\, v-u_0\in X_0,\,\,
v\leq \psi\,\, \mbox{a.e. in}\,\, \Omega\\
\\
u\in X,\, u-u_0\in X_0,\, u\leq \psi\,\, \mbox{a.e. in}\,\, \Omega\,.
\end{array}\right.
\end{equation}

In what follows let $\eta\in (0,1)$ and let us denote by $h\in L^\infty(\Omega)$ and $H_\eta$ the functions
\begin{equation}\label{h}
h=(\mathcal A(\psi)+f)^+\end{equation}
and
\begin{equation}\label{Heta}
H_\eta(t)=\begin{cases}
0 & \mbox{if} \quad  t\leq 0\\
t/\eta & \mbox{if} \quad  0<t<\eta\\
1 & \mbox{if} \quad t\geq \eta\,.
\end{cases}
\end{equation}

We consider the
following approximated problem
\begin{equation}\label{problemauetaf}
\left\{\begin{array}{l}
\mathcal A(u_\eta)=h(1-H_\eta(\psi-u_\eta))-f \quad \mbox{in} \quad \Omega\\
\\
u_\eta \in X, \,\, u_\eta-u_0\in X_0\,.
\end{array}\right.
\end{equation}
More precisely, we say that $u_\eta$ is a solution of problem~\eqref{problemauetaf} if
\begin{equation}\label{problemaueta}
\left\{\begin{array}{l}
{\displaystyle \mathcal A[u_\eta, \varphi]=\int_\Omega \Big(h(1-H_\eta(\psi-u_\eta))-f\Big)\varphi\,dx \quad \forall\,\varphi\in X_0}\\
\\
u_\eta \in X, \,\, u_\eta-u_0\in X_0\,.
\end{array}\right.
\end{equation}

We assume that the following conditions hold true:
\begin{equation}\label{esistenzaueta}
\mbox{for any}\,\, \eta>0\,\, \mbox{there exists a solution}\,\,
u_\eta\,\, \mbox{of problem}\,\,~\eqref{problemaueta}\,;
\end{equation}
\begin{equation}\label{convA}
\begin{aligned}
& \mbox{if} \,\, u\,\, \mbox{is a solution of \eqref{varineq} and}\,\, u_\eta\,\, \mbox{is a sequence of solutions of \eqref{problemaueta}}\\
 & \mbox{such that} \,\, u_\eta\to u \,\, \mbox{uniformly in} \,\, \RR^n\,\, \mbox{as}\,\, \eta\to 0 \,\, \mbox{then, up to a subsequence,}  \\
& \qquad \qquad \mathcal A[u_\eta, \varphi]\to \mathcal A[u,
\varphi]\,\, \mbox{for any}\,\, \varphi\in \widetilde X_0\,.
\end{aligned}
\end{equation}

The main result of this section is the following:
\begin{theorem}\label{propLS}
Let $u_0, \psi \in X$, with $u_0\leq \psi$ a.e. in $\RR^n$ and $\mathcal A(\psi)\in L^\infty(\Omega)$, and let $f\in L^\infty(\Omega)$.
Assume conditions \eqref{integrabilita}--\eqref{monotonia},
\eqref{esistenzaueta} and \eqref{convA} hold true. If $u\in X$ is a
solution of the variational inequality \eqref{varineq}, then
\begin{equation}\label{EXA}
0\leq \mathcal A[u,\varphi]+\int_\Omega f\varphi \,dx\leq
\int_\Omega(\mathcal A(\psi)+f)^+\varphi\,dx \quad \forall \,\, \varphi
\in \widetilde X_0\,, \,\varphi\geq 0\,\, \mbox{a.e. in}\,\,
\Omega\,.\end{equation} \end{theorem}

\begin{proof}
Let $\eta\in (0,1)$ and $u_\eta$ be a solution of \eqref{problemaueta}.
First of all we prove that $u_\eta\leq \psi$ a.e. in $\Omega$\,.
For this, we notice that
$$ u_\eta-\psi=u_\eta-u_0+w\,,$$
with~$w=u_0-\psi\in \mathcal W$ by definition of $\mathcal W$. Hence
$$ (u_\eta-\psi)^+\in X_0\,,$$
thanks to~\eqref{x01} (applied here with~$v=u_\eta-u_0\in X_0)$.

Therefore, we can take~$\varphi=(u_\eta-\psi)^+$ in~\eqref{problemaueta}. We get
\begin{equation}\label{12}
\mathcal A [u_\eta, (u_\eta-\psi)^+]
=\int_\Omega \Big( h (1-H_\eta(\psi-u_\eta)) -f\Big)
(u_\eta-\psi)^+dx\,.
\end{equation}
Similarly,
\begin{equation}\label{13}
\mathcal A [\psi, (u_\eta-\psi)^+]
=
\int_\Omega \mathcal A(\psi)(u_\eta-\psi)^+dx\,,
\end{equation}
with~$\mathcal A(\psi)\in L^\infty(\Omega)$ by assumption.

Moreover, taking into account the definition of~$H_\eta$
and of the positive part,
we have
$$  h (1-H_\eta(\psi-u_\eta)) (u_\eta-\psi)^+ =
h(u_\eta-\psi)^+\quad \mbox{ a.e. in }\Omega\,.$$
So, \eqref{12} becomes
$$ \mathcal A [u_\eta, (u_\eta-\psi)^+]
=\int_\Omega ( h -f)
(u_\eta-\psi)^+dx\,.$$
{F}rom this and~\eqref{13}, we obtain that
$$\begin{aligned}
& \mathcal A[u_\eta,(u_\eta-\psi)^+]
-\mathcal A[\psi,
(u_\eta-\psi)^+] \\
& = \int_\Omega \Big(h-(\mathcal A(\psi)+f)\Big)(u_\eta-\psi)^+dx\geq
0\,.
\end{aligned}$$
Notice that
the last inequality is a consequence
of~\eqref{h}.
Then, the monotonicity condition~\eqref{monotonia}
implies that
\begin{equation}\label{uetapsi}
u_\eta\leq \psi \quad \mbox{a.e. in} \quad \Omega\,.
\end{equation}

Now we claim that $u\geq u_\eta$ a.e. in $\Omega$\,.
For this scope, let $\tau:=u+(u_\eta-u)^+$\,. It is easily seen that
$$\tau=\begin{cases}
u_\eta & \mbox{if} \quad u_\eta>u\\
u & \mbox{if}  \quad u_\eta \leq u\,,
\end{cases}$$
so that $\tau \leq \psi$ a.e. in $\Omega$\,. Moreover,
\begin{equation}\label{poi0}
u_\eta-u=(u_\eta-u_0)+(u_0-u)\in X_0\,,
\end{equation}
being~$X_0$ a linear space. As a consequence, applying~\eqref{x01}
with~$w=0\in \mathcal W$, we obtain that
\begin{equation}\label{poi}
(u_\eta-u)^+\in X_0\,.
\end{equation}
Therefore, by the definition of $\tau$
$$ \tau-u_0 = (u-u_0)+ (u_\eta-u)^+\in X_0\,.$$
Thus, we can use~\eqref{varineq}
with~$v=\tau$, getting
$$\int_Q a(u) b(\tau-u)d\mu \geq \int_\Omega f(\tau-u)dx\,,$$
i.e., using again the definition of $\tau$,
\begin{equation}\label{tautest}
\int_Q a(u) b((u_\eta-u)^+)d\mu \geq \int_\Omega f(u_\eta-u)^+dx\,.
\end{equation}
Moreover, by~\eqref{poi} we can take $\varphi=(u_\eta-u)^+$ in~\eqref{problemaueta}, so that we get
\begin{equation}\label{tautest2}
\begin{aligned}
\int_Q a(u_\eta)b((u_\eta-u)^+)d\mu & = -\int_\Omega
h(1-H_\eta(\psi-u_\eta))(u_\eta-u)^+dx+\int_\Omega f(u_\eta-u)^+dx\\
& \leq \int_\Omega f(u_\eta-u)^+dx\,,
\end{aligned}
\end{equation}
thanks to the definition of $H_\eta$.
Combining \eqref{tautest} and \eqref{tautest2} we get
$$\int_Q \left(a(u_\eta)- a(u)\right)b((u_\eta-u)^+)d\mu\leq 0\,.$$
{F}rom this relation and the monotonicity condition~\eqref{monotonia}
we conclude that
\begin{equation}\label{uetaubar}
u_\eta\leq u \quad \mbox{a.e. in} \quad \Omega\,,
\end{equation}as desired.

Now let us show that
\begin{equation}\label{uetaeta}
u\leq u_\eta+\eta \quad \mbox{a.e. in} \quad \Omega\,.
\end{equation}
To this goal, let $\theta:=u-(u-u_\eta-\eta)^+$\,. It is easy to see that
$$\theta\leq u\leq \psi \quad \mbox{a.e. in} \quad \Omega\,.$$
Moreover, recalling~\eqref{poi0} and~\eqref{x01} (used here
with~$w=-\eta\in \mathcal W$), we have
that
\begin{equation}\label{poi1}
(u-u_\eta-\eta)^+\in X_0\,.
\end{equation}
Hence
$$ \theta-u_0 = (u-u_0)-
(u-u_\eta-\eta)^+\in X_0\,.$$
Therefore, by taking~$v=\theta$ in~\eqref{varineq}, we have
$$\int_Q a(u) b(\theta-u)d\mu \geq \int_\Omega f(\theta-u)dx\,,$$
i.e., using the definition of $\theta$ and assumption \eqref{condb},
\begin{equation}\label{thetatest}
\int_Q a(u) b((u-u_\eta-\eta)^+)d\mu \leq \int_\Omega f(u-u_\eta-\eta)^+dx\,.
\end{equation}
Also, by~\eqref{poi1}, we can take~$\varphi=
(u-u_\eta-\eta)^+$ in~\eqref{problemaueta}. We have
\begin{equation}\label{testeta}
\begin{aligned}
-\!\!\int_Q\!\!\! a(u_\eta)b((u\!-\!u_\eta\!-\!\eta)^+)d\mu &\!=\!\!\int_\Omega\!\! h(1\!\!-\!\!H_\eta(\psi\!-\!u_\eta))(u\!-\!u_\eta\!-\!\eta)^+dx-\!\!\int_\Omega\!\!f(u\!-\!u_\eta\!-\!\eta)^+dx\\
& =-\int_\Omega f(u-u_\eta-\eta)^+\!dx\,,
\end{aligned}
\end{equation}
being
$$\{u-u_\eta-\eta>0\}\subseteq
\{\psi-u_\eta>\eta\}\subseteq\{1-H_\eta(\psi-u_\eta)=0\}\,.$$
Using assumption~\eqref{conda} and \eqref{testeta} we deduce
\begin{equation}\label{testeta2}
\int_Q a(u_\eta+\eta)b((u-u_\eta-\eta)^+)d\mu =\int_Q a(u_\eta)b((u-u_\eta-\eta)^+)d\mu =\int_\Omega f(u-u_\eta-\eta)^+dx\,.
\end{equation}
Combining \eqref{thetatest} and \eqref{testeta2} we have
$$\int_Q \left(a(u_\eta+\eta)- a(u) \right)b((u-u_\eta-\eta)^+)d\mu \geq 0\,.$$
Thus, using again the monotonicity assumption~\eqref{monotonia},
$$u\leq u_\eta+\eta \quad \mbox{a.e. in} \quad \Omega\,,$$
so that \eqref{uetaeta} is proved.

By \eqref{uetaubar} and \eqref{uetaeta} we have that
$$u-\eta \leq u_\eta \leq u \quad \mbox{a.e. in} \quad \Omega\,.$$
Then
\begin{equation}\label{convuniforme}
u_\eta \to u \quad \mbox{uniformly in} \quad \RR^n,
\quad \mbox{as} \quad \eta \to 0\,.
\end{equation}
Hence, condition \eqref{convA} implies that, up to subsequences,
\begin{equation}\label{Aueta}
\mathcal A[u_\eta,\varphi]\to \mathcal A[u,\varphi],
\quad \mbox{as} \quad \eta \to 0\,,
\end{equation}
for any~$\varphi\in \widetilde X_0$.

Taking into account that $u_\eta$
solves \eqref{problemaueta} and
the definition of $h$ and $H_\eta$ it is easy to check that
$$ 0\le \mathcal A[u_\eta,\varphi]+\int_\Omega f\varphi\,dx=\int_\Omega h\Big(1-H_\eta(\psi-u_\eta)\Big)\varphi\, dx\leq
\int_\Omega h\varphi\,dx$$
for any~$\varphi\in X_0$, with~$\varphi\ge0$ a.e. in $\Omega$.
So, passing to the limit as $\eta \to 0$ and using \eqref{Aueta} we get
$$ 0\le \mathcal A[u,\varphi]+\int_\Omega f\varphi\,dx\leq
\int_\Omega h\varphi\,dx$$ for any $\varphi\in \widetilde X_0$,
with~$\varphi\ge0$ a.e. in $\Omega$. Then, recalling the definition
of $h$ (see \eqref{h}), Theorem~\ref{propLS} is proved.
\end{proof}

\section{Some applications: the Laplacian, the
$p$-Laplacian, and the Laplacian in the Heisenberg
group}\label{sec:laplacian}
In this section we give some applications of Theorem~\ref{propLS} by
recovering some known results in
the case of the Laplacian, of the $p$-Laplacian and of the Laplacian
in the Heisenberg group. The
case of the
Laplacian is classical and it dates back to the original paper of \cite{LewySt}, which originated many important extensions and applications (see, among the others, \cite{BocGallouet90, DonatiF82, MoscoFre, Mok, Mok1, Mok2, MoscoTro}).
See \cite{Challal, Rod2} for the $p$-Laplacian and \cite{ValdinociPinamonti} for the Laplacian in the Heisenberg group.

Even if these cases have been already treated in the classical
or recent literature, we provide our arguments in full detail,
in order to clarify the abstract setting. Of course, the expert
reader, or the one interested only in the new non-local application,
may skip this part and go directly to Section~\ref{sec:kernel}.

\subsection{The Laplacian operator.}\label{subsec:laplacian}
Though the case of the Laplacian is known and the techniques
exploited are the standard (but tricky) Sobolev tools,
we provide full detail of the argument, both for the
readers's convenience and in order to make the abstract
setting concrete in a model case.

We start with some preliminary observations on the classical Sobolev
spaces which will be useful in the sequel.

\begin{lemma}\label{lemma0}
Let $\alpha, \beta: \RR^n\to \RR$, $n\ge 1$. Then for any $x,y\in
\RR^n$ the following inequality holds
$$|\alpha^+(x)-\beta^+(y)|\leq |\alpha(x)-\beta(y)|\,.$$
\end{lemma}
\begin{proof}
Suppose that $x,y\in \RR^n$ are such that $\alpha(x)>0$ and $\beta(y)>0$, then
$$|\alpha^+(x)-\beta^+(y)|=|\alpha(x)-\beta(y)|\,.$$

Now, let $x,y\in \RR^n$ be such that $\alpha(x)\leq 0$ and $\beta(y)\leq
0$. Then
$$|\alpha^+(x)-\beta^+(y)|=0\leq|\alpha(x)-\beta(y)|\,.$$

Finally, suppose that $x,y\in \RR^n$ are such that $\alpha(x)>0$ and
$\beta(y)\leq 0$. Then
$$|\alpha^+(x)-\beta^+(y)|=|\alpha(x)|=\alpha(x)\leq
\alpha(x)-\beta(y)\leq |\alpha(x)-\beta(y)|\,.$$
Hence the assertion is proved.
\end{proof}

\begin{lemma}\label{lemmaE}
Let $1\le p<\infty$ and let $v_j$ be a sequence of functions such that
$\|v_j-v_\infty\|_{L^p(\Omega)}\to 0$ as $j\to +\infty$\,. Then,
$$\|v_j^+-v_\infty^+\|_{L^p(\Omega)}\to 0$$ as $j\to +\infty$\,.
\end{lemma}
\begin{proof}
By Lemma~\ref{lemma0} we have that
$$\int_\Omega |v_j^+(x)-v_\infty^+(x)|^pdx \leq \int_\Omega |v_j(x)-v_\infty(x)|^pdx\to 0$$
as $j\to +\infty$\,.
\end{proof}

\begin{lemma}\label{lemmaF}
Let $1\le p<\infty$ and let $v_j$ be a sequence of functions such that
$\|v_j-v_\infty\|_{W^{1,p}(\Omega)}\to 0$ as $j\to +\infty$\,. Then,
up to a subsequence,
\begin{itemize}
\item[$a)$] $\nabla v_j^+\to \nabla v_\infty^+$ a.e. in $\Omega$
\item[$b)$] $\|v_j^+-v_\infty^+\|_{W^{1,^p}(\Omega)}\to 0$
\end{itemize}
as $j\to +\infty$\,.
\end{lemma}
\begin{proof}
By the assumption and by \cite[Theorem~IV.9]{brezis} we deduce that, up
to a subsequence, as
$j\to +\infty$
\begin{equation}\label{convlemmaF1}
 v_j\to v_\infty\,\, \mbox{in}\,\, L^p(\Omega)
\end{equation}
\begin{equation}\label{convlemmaF2}
 \nabla v_j\to \nabla v_\infty\,\, \mbox{in}\,\, L^p(\Omega)
\end{equation}
\begin{equation}\label{convlemmaF3}
 v_j\to v_\infty\,\,\,\, \mbox{and}\,\,\,\, \nabla v_j\to \nabla v_\infty\,\,
 \mbox{in}\,\,\, \Omega\setminus \mathcal N_0
\end{equation}
where $\mathcal N_0\subset \Omega$ is such that $|\mathcal
N_0|=0$\,.

By \cite[Lemma~7.7]{gilbargtrudinger} we also know that
\begin{equation}\label{nablavinfty}
\nabla v_\infty=0\,\,\, \mbox{for any}\,\, x\in \{v_\infty=0\}\setminus \mathcal N_1,\,\,
\mbox{with}\,\, |\mathcal N_1|=0
\end{equation}
and, by \cite[Lemma~7.6]{gilbargtrudinger} for any $j\in \NN\cup
\{\infty\}$
\begin{equation}\label{nabla1}
\nabla v_j^+(x)=\begin{cases}
\nabla v_j(x) & \mbox{if}\,\, x\in \{v_j>0\}\setminus \mathcal N_1^j\\
0 & \mbox{if}\,\, x\in \{v_j\leq 0\}\setminus \mathcal N_2^j,
\end{cases}
\end{equation}
 with $|\mathcal N_i^j|=0$\,, $i=1,2$\,.

First, let $\mathcal N=\mathcal N_0\cup \mathcal N_1\cup
\Big({\displaystyle \cup_{j=1}^\infty\mathcal N_1^j}\Big)\cup
\Big({\displaystyle \cup_{j=1}^\infty\mathcal N_2^j}\Big)$\,. Of course $|\mathcal N|=0$. Now,
we will show that $a)$ holds for any $x\in \Omega\setminus\mathcal
N$\,.

Let $x\in \{v_\infty>0\}\setminus \mathcal N$. Then, for $j$
sufficiently large, $v_j(x)>0$ by~\eqref{convlemmaF3},
so that $\nabla
v_j^+(x)=\nabla
v_j(x)$ by \eqref{nabla1}\,. Hence, again by \eqref{convlemmaF3} and \eqref{nabla1} we have
$$\lim_{j\to +\infty}\nabla v_j^+(x)=\lim_{j\to +\infty}\nabla v_j(x)=
\nabla v_\infty(x)=\nabla v_\infty^+(x)\,.$$

Now, let $x\in \{v_\infty<0\}\setminus \mathcal N$. Then, for $j$
sufficiently large, $v_j(x)<0$ by \eqref{convlemmaF3}, so that $\nabla v_j^+(x)=0=\nabla
v_\infty^+(x)$ by \eqref{nabla1}\,. Hence
$$\lim_{j\to +\infty}\nabla v_j^+(x)=\nabla v_\infty^+(x)\,.$$

Finally, let $x\in \{v_\infty=0\}\setminus \mathcal N$. Then,
$\nabla v_\infty(x)=0$ by \eqref{nablavinfty}. We also have that
$$|\nabla v_j^+(x)|\leq |\nabla v_j(x)|=|\nabla v_j(x)-\nabla v_\infty(x)|\to 0$$
as $j\to +\infty$, thanks to \eqref{nabla1}
and~\eqref{convlemmaF3}. Hence,
$$\lim_{j\to +\infty}\nabla v_j^+(x)=0=\nabla v_\infty^+(x)\,,$$
thanks to \eqref{nabla1}\,.
Then, assertion~$a)$ is proved.

Now, let us prove $b)$. For this note that by Lemma~\ref{lemmaE} and \eqref{convlemmaF1} we get
\begin{equation}\label{convFlp}
\|v_j^+ -v_\infty^+\|_{L^p(\Omega)}\to 0
\end{equation}
as $j\to +\infty$\,.
 Moreover, by \eqref{convlemmaF2} and \cite[Theorem~IV.9]{brezis} there exists $\ell\in
 L^p(\Omega)$ such that $|\nabla v_j(x)|\leq \ell(x)$ a.e. in
 $\Omega$ for any $j\in \NN$\,.
 Then, a.e. in $\Omega$
 $$\begin{aligned}
 |\nabla v_j^+(x)-\nabla v_\infty^+(x)|^p & \leq \Big(|\nabla v_j^+(x)|+|\nabla v_\infty^+(x)|\Big)^p\\
 & \leq
 \Big(|\nabla v_j(x)|+|\nabla v_\infty(x)|\Big)^p\\
 & \leq \Big(\ell(x)+|\nabla v_\infty(x)|\Big)^p\in L^1(\Omega)\,.
 \end{aligned}$$
Hence, the dominated convergence Theorem and $a)$ give
\begin{equation}\label{convFlp2}
\|\nabla v_j^+ -\nabla v_\infty^+\|_{L^p(\Omega)}\to 0
\end{equation}
as $j\to +\infty$\,. Assertion $b)$ follows by \eqref{convFlp} and
\eqref{convFlp2}.
\end{proof}

\begin{corollary}\label{corollario0}
Let $1\le p<\infty$ and $\Omega\subset \RR^n$, $n\ge 1$, be an
open set of class $C^1$\,. Let $v\in W^{1,p}_0(\Omega)$ and $w\in
W^{1,p}(\Omega)\cap C(\overline \Omega)$, with $w\leq 0$ in
$\overline\Omega$\,. Then $(v+w)^+\in W^{1,p}_0(\Omega)$\,.
\end{corollary}
\begin{proof}
Of course $v+w\in W^{1,p}(\Omega)$, so that, by
\cite[Lemma~7.6]{gilbargtrudinger} (see also \cite[Corollary~6.18]{lieb}) we have that $(v+w)^+\in W^{1,p}(\Omega)$\,.

Since $v\in W^{1,p}_0(\Omega)$, there exists a sequence $v_j\in
C^\infty_0(\Omega)$ such that $\|v_j-v\|_{W^{1,p}(\Omega)}\to 0$ as
$j\to +\infty$\,. Hence, $v_j+w\to v+w$ in $W^{1,p}(\Omega)$ and so,
by Lemma~\ref{lemmaF}
\begin{equation}\label{convh10}
(v_j+w)^+\to (v+w)^+\,\, \mbox{in}\,\, W^{1,p}(\Omega)\,.
\end{equation}
as $j\to +\infty$\,.

Moreover, $(v_j+w)^+\in C(\overline \Omega)$ and $v_j+w\leq v_j=0$
on $\partial \Omega$, being $w\leq 0$ in $\overline \Omega$ and $v_j\in C^\infty_0(\Omega)$\,, that
is $(v_j+w)^+=0$ on $\partial\Omega$ for any $j\in \NN$\,. Then, by \cite[Theorem~IX.17]{brezis}
\begin{equation}\label{h10}
(v_j+w)^+\in W^{1,p}_0(\Omega)
\end{equation}
for any $j\in\NN$\,.

Since $W^{1,p}_0(\Omega)$ is closed, \eqref{convh10} and \eqref{h10}
imply that $(v+w)^+\in W^{1,p}_0(\Omega)$\,.
\end{proof}

With respect to the abstract setting in this subsection we take
$Q=\Omega\subset \RR^n$, $n\ge 1$, $\Omega$ open, bounded set of class $C^1$, $d\mu=dx$ (i.e. the usual Lebesgue
measure),
\begin{eqnarray*}
&& X=\{g\in \mathcal M(\RR^n) : g_{|\Omega}\in
H^1(\Omega)\}\,,\\ && X_0=\{g\in \mathcal M(\RR^n) : g_{|\Omega}\in
H^1_0(\Omega)\}\,, \\ &&
\widetilde X_0=C^\infty_0(\Omega)=\{g:\RR^n\to \RR
:
g\in C^\infty(\RR^n)
\mbox{ and } g=0\,\, \mbox{in }
\mathcal C\Omega\}\,, \\{\mbox{and }}&&
Y=Z=\{g:\RR^n \to \RR^n:
g_{|\Omega}\in
(L^2(\Omega))^n\}\,,\end{eqnarray*}
while $\mathcal P$ is the scalar product
in
$\RR^n$\,. Here $H^1(\Omega)$ and $H^1_0(\Omega)$ are the usual
Sobolev spaces endowed with the norm
$\|u\|_{H^1(\Omega)}=\|u\|_{L^2(\Omega)}+\|\nabla
u\|_{L^2(\Omega)}$\,.

We also take $u_0\in \{g\in X: g_{|\Omega}\in
C(\overline\Omega)\}$ and $\psi\in \{g\in \mathcal M(\RR^n) : g_{|\Omega}\in
C^2(\overline \Omega)\}$ with $u_0\leq \psi$ a.e. in $\RR^n$\,.

Note that the restriction to~$\Omega$ of any function in~$X$ belongs
to~$L^1(\Omega)$, since $\Omega$ is bounded. Moreover, $u_0$,
$\psi\in X$ and $\widetilde X_0\subseteq X_0\subseteq X$. By
Corollary~\ref{corollario0} (applied here with $p=2$) we get that \eqref{x01} holds true. Here
we use that $u_0$, $\psi \in C(\overline \Omega)$\,.

Now, for any $u, \varphi \in X$ we define
$$\begin{aligned}
& a(u)=\nabla u\\
& b(\varphi)=\nabla \varphi\\
& \mathcal A[u,\varphi]=-\int_\Omega \nabla u(x)\nabla \varphi(x) dx\,.
\end{aligned}$$

Note that $\mathcal A$ is well defined, being $\nabla u, \nabla \varphi \in Y$, that is
\eqref{integrabilita} is satisfied. Moreover, it is easy to check that conditions
\eqref{conda} and \eqref{condb} are satisfied. In order to show \eqref{monotonia} it is enough to
note that for any $g\in X$
\begin{equation}\label{nabla+}
\nabla g^+(x)=\begin{cases}
\nabla g(x) & \mbox{a.e. if}\,\, g(x)> 0\\
0 & \mbox{a.e. if}\,\, g(x)\leq 0
\end{cases}
\end{equation}
(see, for instance, \cite[Lemma~7.6]{gilbargtrudinger}). Indeed, let $u,v\in X$ with $(u-v)^+\in X_0$ and
$\mathcal A[u,(u-v)^+]\ge\mathcal A[v,(u-v)^+]$\,. Then, we have
\begin{equation}\label{mon1}
\int_\Omega \nabla (u-v)(x)\nabla(u-v)^+(x) dx\leq 0\,.
\end{equation}
Using \eqref{nabla+} and \eqref{mon1} we have
$$0\geq\int_\Omega \nabla (u-v)(x)\nabla(u-v)^+(x) dx=\int_\Omega |\nabla(u-v)^+(x)|^2 dx\geq 0\,,$$
that is $\nabla(u-v)^+(x)=0$ a.e. in $\Omega$. Hence, $(u-v)^+$ is constant,
say~$u-v=c\in\RR$ a.e. in~$\Omega$. Since~$(u-v)^+\in X_0$,
we conclude that~$|c|\in H^1_0(\Omega)$, and so~$c=0$.
Accordingly,~$(u-v)^+=0$, and so~$u\leq v$ a.e. in $\Omega$, to wit
\eqref{monotonia} holds.

Now, let us fix $\eta\in (0,1)$ and let us consider the problem $\Delta u_\eta=g_\eta(x, u_\eta)$ in $\Omega$ with $u_\eta=u_0$ in $\mathcal C \Omega$, that is
\begin{equation}\label{problemalaplaciano}
\left\{\begin{array}{l}
{\displaystyle \int_\Omega \nabla u_\eta(x)\nabla \varphi(x)\, dx+\int_\Omega g_\eta(x, u_\eta(x))\varphi(x)\, dx=0 \quad \forall\, \varphi \in X_0}\\
\\
u_\eta\in X,\, u_\eta-u_0\in X_0\,,
\end{array}\right.
\end{equation}
where for a.e. $x\in \Omega$ and $t\in \RR$
\begin{equation}\label{geta}
g_\eta(x,t)=h(x)\big(1-H_\eta(\psi(x)-t)\big)-f(x)\,,
\end{equation}
with $h=(\Delta \psi +f)^+$, $f\in L^\infty(\Omega)$ and $H_\eta$ given in \eqref{Heta}\,. Note that $h\in L^\infty(\Omega)$, thanks to the regularity required on $\psi$, i.e. $\psi\in C^2(\overline \Omega)$.

We show that assumption~\eqref{esistenzaueta} holds true. For this we consider the space~$
X_{u_0}=\{u\in X : u-u_0\in X_0\}$ and the
functional $\mathcal J_\eta:
X_{u_0} \to \RR$ defined as follows
$$\mathcal J_\eta(u)=\frac 1 2\int_\Omega |\nabla u(x)|^2 dx+\int_\Omega G_\eta(x, u(x))dx\,,$$
where $G_\eta(x,t)=\int_0^t g_\eta(x,s) ds$\,.
First of all, we note that
\begin{equation}\label{inf}
{\displaystyle \inf_{u\in X_{u_0}}\mathcal J_\eta(u)>-\infty.}
\end{equation}
Indeed, using the definitions of $g_\eta$ and $H_\eta$ and the regularity
of $h$ and $f$ we get that $g_\eta(\cdot, u(\cdot))\in L^\infty(\Omega)$ for any $u\in X$ and
\begin{equation}\label{4.14bis}
\|g_\eta(\cdot,
u(\cdot))\|_{L^\infty(\Omega)}\leq
\|h\|_{L^\infty(\Omega)}+\|f\|_{L^\infty(\Omega)}=:\kappa\end{equation}
so that
\begin{equation}\label{crescitaG}
|G_\eta(x,u(x))|\leq \kappa |u(x)| \quad \mbox{a.e.}\,\, x\in \Omega\,.
\end{equation}
Hence, for any $\delta>0$ and $u\in X_{u_0}$
\begin{equation}\label{infJ}
\begin{aligned}
\mathcal J_\eta(u)& \geq \frac 1 2\int_\Omega |\nabla u(x)|^2 dx-\kappa\int_\Omega |u(x)|dx\\
& \geq \frac 1 2\int_\Omega |\nabla u(x)|^2 dx-\frac \kappa 2\left(\frac{|\Omega|}{\delta}+\int_\Omega \delta|u(x)|^2dx\right)\,,
\end{aligned}
\end{equation}
thanks to the Cauchy-Schwarz inequality.
Since $u-u_0\in X_0$, the Poincar\'e inequality (see, for instance, \cite[Corollary IX.19]{brezis}) gives
\begin{equation}\label{poincare}
\begin{aligned}
\|u-u_0\|_{L^2(\Omega)}& \leq C\|\nabla u-\nabla u_0\|_{L^2(\Omega)}\\
& \leq C\left(\|\nabla u\|_{L^2(\Omega)}+\|\nabla u_0\|_{L^2(\Omega)}\right)\,,
\end{aligned}
\end{equation}
where $C$ is a positive constant depending only on $\Omega$.
Using the fact that $\|u\|_{L^2(\Omega)}-\|u_0\|_{L^2(\Omega)}\leq \|u-u_0\|_{L^2(\Omega)}$,
by \eqref{poincare} we obtain
\begin{equation}\label{poincare2}
\|u\|_{L^2(\Omega)}\leq \tilde C \left(\|\nabla u\|_{L^2(\Omega)}+1\right)\,,
\end{equation}
where $\tilde C=\max\{C, C\|\nabla u_0\|_{L^2(\Omega)}+\| u_0\|_{L^2(\Omega)}\}$\,.
Then, by \eqref{infJ} and \eqref{poincare2} we get
\begin{align}\label{4.8bis}
\mathcal J_\eta(u)& \geq \frac 1 2\|\nabla u\|_{L^2(\Omega)}-\frac{\kappa|\Omega|}{2\delta}-\frac{\kappa\delta\tilde C}{2} \left(\|\nabla u\|_{L^2(\Omega)}+1\right)\,.
\end{align}
Choosing $\delta>0$ such that $\kappa\delta\tilde C<1$ it easily follows that
$$\mathcal J_\eta(u)\geq -\frac{\kappa}{2}\left(\frac{|\Omega|}{\delta}+\delta\tilde C\right)>-\infty\,,$$
for any $u\in X_{u_0}$\,, so that \eqref{inf} is proved.

Now, let us take a minimizing sequence $u_j$ for $\mathcal J_\eta$, i.e. a sequence $u_j$ in
$X_{u_0}$ such that
$${\displaystyle \mathcal J_\eta(u_j)\to \inf_{u\in X_{u_0}} \mathcal J_\eta(u)}>-\infty\,\, \mbox{as}\,\, j\to +\infty\,.$$
Then the sequence $\mathcal J_\eta(u_j)$ is bounded in $\RR$, hence,
by~\eqref{4.8bis} with $u=u_j$,
\begin{equation*}
\|\nabla u_j\|_{L^2(\Omega)}\leq \tilde \kappa \quad \mbox{for
any}\,\, j\in \NN
\end{equation*}
for some $\tilde \kappa>0$\,. This and
\eqref{poincare2} give that~$u_j$ is also bounded in~$L^2(\Omega)$.

Hence, the sequence $u_j-u_0$ is bounded in $H^1_0(\Omega)$\,.
Thus, up to a subsequence, we have that there exists $v_\infty \in H^1_0(\Omega)$ such that
\begin{equation}\label{convergenze}
\begin{aligned}
& \nabla (u_j-u_0) \to \nabla v_\infty \quad \mbox{weakly in}\,\, L^2(\Omega) \\
& u_j-u_0 \to v_\infty \quad \mbox{in}\,\, L^2(\Omega) \\
& u_j-u_0 \to v_\infty \quad \mbox{a.e. in}\,\, \Omega
\end{aligned}
\end{equation}
as $j\to +\infty$
and there exists $\ell\in L^2(\Omega)$ such that
\begin{equation}\label{dominata2}
|u_j(x)|\leq \ell(x) \quad \mbox{a.e. in}\,\, \Omega\,\quad \mbox{for any}\,\,j\in \NN
\end{equation}
(see, for instance \cite[Theorem~IV.9]{brezis}).
Now we define
$$u_\infty(x)=\begin{cases}
(v_\infty+u_0)(x) & \quad \mbox{if}\,\, x\in \Omega\\
u_0(x)  & \quad \mbox{if}\,\, x\in \mathcal C \Omega\,.
\end{cases}$$
Note that
\begin{equation}\label{uX0}
u_\infty\in X_{u_0}\,,
\end{equation}
since $(u_\infty-u_0)_{|\Omega}=v_\infty \in H^1_0(\Omega)$\,.

Using the weak lower semicontinuity of the norm in $L^2(\Omega)$, the fact
that the map $t\mapsto G_\eta(x, t)$ is continuous in $\RR$ a.e. $x\in
\Omega$, \eqref{crescitaG}, \eqref{convergenze}--\eqref{uX0} and the dominated convergence Theorem, we deduce that
$${\displaystyle \lim_{j\to +\infty}\mathcal J_\eta(u_j)\geq \frac 1 2\int_\Omega |\nabla u_\infty(x)|^2 dx+\int_\Omega G_\eta(x, u_\infty(x))dx=\mathcal J_\eta(u_\infty)\geq \inf_{u\in X_{u_0}}\mathcal J_\eta(u)}\,,$$
so that
$${\displaystyle \mathcal J_\eta(u_\infty)=\inf_{u\in X_{u_0}} \mathcal J_\eta(u)}\,.$$
Then problem~\eqref{problemalaplaciano} admits a solution in $X_{u_0}$ and so \eqref{esistenzaueta} follows.

Now we prove \eqref{convA}. Let $u_\eta$
be a sequence in $X$ converging
uniformly to some $u$ in $\RR^n$ as $\eta\to 0$.
For any $\varphi\in C^\infty_0(\Omega)$, integrating by parts we have
$$\mathcal A[u_\eta, \varphi]=-\int_\Omega \nabla u_\eta(x) \nabla
\varphi(x) dx=\int_\Omega u_\eta(x) \Delta \varphi(x) dx\,.$$
Hence,
$$\begin{aligned}
\Big|\mathcal A[u_\eta,\varphi]-\mathcal A[u, \varphi]\Big | & =
\Big|\int_\Omega (u_\eta(x)-u(x))\Delta \varphi(x) dx \Big|\\
& \leq \|\Delta \varphi\|_{L^\infty(\RR^n)}\int_\Omega |u_\eta(x)-u(x)|dx\\
& \leq \|\Delta \varphi\|_{L^\infty(\RR^n)}\|u_\eta-u\|_{L^\infty(\RR^n)} |\Omega|\to 0
\end{aligned}$$
as $\eta \to 0$. Thus, condition \eqref{convA} is proved.

Note that, in this proof,
we did not need to use that $u_\eta$ is
a solution of \eqref{problemalaplaciano} and that $u$ solves the variational inequality~\eqref{varineqlapl}\,.

\medskip

\begin{Prooflap}
Let $u$ be a solution of the variational inequality~\eqref{varineqlapl}.
In the setting of Theorem~\ref{lap} we can apply
Theorem~\ref{propLS} so that we get
$$0\leq -\int_\Omega \nabla u(x)\nabla \varphi(x) \,dx\leq \int_\Omega (\Delta \psi +f)^+(x)\varphi(x) \, dx\quad
\forall\, \varphi \in C^\infty_0(\Omega)\,,\, \varphi\geq
0\,\,\mbox{in}\,\, \Omega\,.$$
Note that, by density, the estimate holds true for any $\varphi\in H^1_0(\Omega)$ with
$\varphi\geq 0$ a.e. in $\Omega$\,.
\end{Prooflap}


\subsection{The $p$-Laplacian operator.}\label{subsec:plaplacian} With
respect to the abstract setting here we take $1<p<\infty$,
$Q=\Omega\subset
\RR^n$, $n\ge 1$, $\Omega$ open, bounded set with smooth boundary (say,
$\partial \Omega\in C^{1, \alpha}$ with $\alpha\in (0,1]$),
\begin{eqnarray*} &&
d\mu=dx\,, \\&& X=\{g\in \mathcal M(\RR^n) : g_{|\Omega}\in
W^{1,p}(\Omega)\}\,,\\&&
X_0=\{g\in \mathcal M(\RR^n) :
g_{|\Omega}\in W^{1,p}_0(\Omega)\}\,,\\ &&\widetilde
X_0=C^\infty_0(\Omega)\,, \\ && Y=\{g:\RR^n \to \RR^n: g_{|\Omega}\in
(L^{p'}(\Omega))^n\}\,, {\mbox{ with $1/p+1/p'=1$,}}\\
{\mbox{and }}&& Z=\{g:\RR^n \to
\RR^n:
g_{|\Omega}\in (L^{p}(\Omega))^n\}\,,\end{eqnarray*}
while $\mathcal P$ is the
scalar
product in $\RR^n$\,. Here $W^{1,p}(\Omega)$ and $W^{1,p}_0(\Omega)$
denote the usual Sobolev spaces endowed with the norm
$\|u\|_{W^{1,p}(\Omega)}=\|u\|_{L^p(\Omega)}+\|\nabla
u\|_{L^p(\Omega)}$\,.

Moreover we fix two functions $u_0$ and $\psi$ such that
\begin{equation}\label{regu0}
u_0\in \{g\in \mathcal M(\RR^n) : g_{|\overline \Omega}\in C^{1,
\alpha}(\overline \Omega)\}
\end{equation}
and
\begin{equation}\label{regupsi}
\psi\in \{g\in \mathcal M(\RR^n) : g_{|\overline \Omega}\in
C^{2}(\Omega)\cap C(\overline\Omega)\}
\end{equation}
with $\Delta_p\, \psi\in L^\infty(\Omega)$ and $u_0\leq \psi$ a.e.
in $\RR^n$\,. Here $\alpha$ is the one appearing in the regularity
of $\partial \Omega$\,. The choice of $u_0$ is admissible since
$C^{1, \alpha}(\overline \Omega)\subset W^{1,p}(\Omega)$.

Note that the restriction to~$\Omega$ of any function in~$X$ belongs
to~$L^1(\Omega)$, since $\Omega$ is bounded. Moreover, $\widetilde
X_0\subseteq X_0\subseteq X$\,.
In order to check assumption~\eqref{x01}
it
is enough to use Corollary~\ref{corollario0} and the fact that $u_0, \psi \in C(\overline \Omega)$\,.

For any $u, \varphi \in X$ we take
$$\begin{aligned}
& a(u)=|\nabla u|^{p-2}\nabla u\\
& b(\varphi)=\nabla \varphi\\
& \mathcal A[u,\varphi]=-\int_\Omega |\nabla u|^{p-2}(x)\nabla
u(x)\nabla \varphi(x) dx\,.
\end{aligned}$$

Note that $\mathcal A$ is well defined, thanks to the fact that $\nabla u\in Y$ and $\nabla \varphi \in Z$, so that \eqref{integrabilita} is satisfied. Furthermore, conditions \eqref{conda} and \eqref{condb} are trivially satisfied. In order to show \eqref{monotonia} we use the following inequality (see, for instance, \cite[formula (2.2), p. 210]{simon})
\begin{equation}\label{simon}
\langle |t|^{p-2}t-|t'|^{p-2}t', t-t'\rangle \geq \begin{cases}
C\frac{|t-t'|^2}{(|t|+|t'|)^{2-p}} & \mbox{if}\,\, 1<p<2\\
\\
C|t-t'|^p & \mbox{if}\,\, p\geq 2\,,
\end{cases}
\end{equation}
for any $t, t'\in \RR^n$\,, where $C$ is a positive constant.

Indeed, let $u,v\in X$ with $(u-v)^+\in X_0$ and
$\mathcal A[u,(u-v)^+]\ge\mathcal A[v,(u-v)^+]$. Then, we have
\begin{equation}\label{mon2}
\int_\Omega \Big(|\nabla u(x)|^{p-2}\nabla u(x) -|\nabla v(x)|^{p-2}\nabla v(x)\Big)\nabla(u-v)^+(x) dx\leq 0\,.
\end{equation}
By using \eqref{nabla+} (which still holds for any $g\in X$, with $X$ defined as in Subsection~\ref{subsec:plaplacian}, see,
for instance, \cite[Lemma~7.6]{gilbargtrudinger}), \eqref{simon} and \eqref{mon2} we get
$$\begin{aligned}
0\geq\int_\Omega \Big(|\nabla u(x)|^{p-2}\nabla u(x)& -|\nabla v(x)|^{p-2}\nabla v(x)\Big)\nabla(u-v)^+(x) dx\\
& = \int_{\{u>\,v\}} \Big(|\nabla u(x)|^{p-2}\nabla u(x) -|\nabla v(x)|^{p-2}\nabla v(x)\Big)\nabla(u-v)(x) dx\\
& \geq \begin{cases}
C{\displaystyle \int_{\{u>\,v\}} \frac{|\nabla (u-v)(x)|^2}{(|\nabla u(x)|+|\nabla v(x)|)^{2-p}}\,\,dx} & \mbox{if}\,\, 1<p<2\\
\\
 C{\displaystyle \int_{\{u>\,v\}} |\nabla (u-v)(x)|^p dx} & \mbox{if}\,\, p\geq 2
\end{cases}\\
& = \begin{cases}
 C{\displaystyle \int_\Omega \frac{|\nabla (u-v)^+(x)|^2}{(|\nabla u(x)|+|\nabla v(x)|)^{2-p}}\,\,dx\geq 0} & \mbox{if}\,\, 1<p<2\\
\\
 C{\displaystyle \int_\Omega |\nabla (u-v)(x)^+|^p dx \geq 0} & \mbox{if}\,\, p\geq 2\,,
\end{cases}\\
\end{aligned}$$
which implies that $\nabla(u-v)^+(x)=0$ a.e. in $\Omega$. Hence, as in the case of the Laplacian, we get that $u\leq v$ a.e. in
$\Omega$, and so \eqref{monotonia} holds true.

Now, let us fix $\eta \in (0,1)$ and let us consider the problem $\Delta_p u_\eta=g_\eta(x, u_\eta)$ in $\Omega$ with $u_\eta=u_0$ in $\mathcal C \Omega$, that is
\begin{equation}\label{problemaplaplaciano}
\left\{\begin{array}{l}
{\displaystyle \int_\Omega |\nabla u_\eta|^{p-2}(x)\nabla u_\eta(x)\nabla \varphi(x)\, dx+\int_\Omega g_\eta(x, u_\eta(x))\varphi(x)\, dx=0 \quad \forall\, \varphi \in X_0}\\
\\
u_\eta\in X,\, u_\eta-u_0\in X_0\,,
\end{array}\right.
\end{equation}
where $g_\eta(x,t)$ is given in \eqref{geta}, with $h=(\Delta_p\psi+f)^+\in L^\infty(\Omega)$\,, being $\Delta_p\,\psi, f\in L^\infty(\Omega)$\,.

As for condition~\eqref{esistenzaueta} it is enough to argue as in the case of the Laplacian, just substituting the Cauchy-Schwarz inequality with the Young inequality and use the Poincar\'e inequality in $L^p(\Omega)$.

Now we have to prove the validity of \eqref{convA}. The argument for this differs from the one of the Laplacian, since the operator $\Delta_p$ is nonlinear: in this case we will make use of the regularity theory for the $p$-Laplacian.

Let $u$ be a solution of \eqref{varineqplapl} and let $u_\eta$
be a sequence of solutions of \eqref{problemaplaplaciano} converging
uniformly to $u$ in $\RR^n$ as $\eta\to 0$\,, that is
\begin{equation}\label{convueta}
\|u_\eta -u\|_{L^\infty(\Omega)}\to 0 \quad \mbox{as}\,\, \eta \to 0\,.
\end{equation}
First of all, we show that $u_\eta\in L^\infty(\Omega)$ and $\|u_\eta\|_{L^\infty(\Omega)}$ may be bounded independently of $\eta$\,. Let $R>0$ be such that $\Omega\subset B_R$\,, where $B_R\subset \RR^n$ is the ball of radius $R$ centered in $0$\,. For $M>0$ we define
$$z(x)=e^{M(x_1+R)}-e^{2MR}-\|u_0\|_{L^\infty(\Omega)}\,, \quad x\in B_R\,.$$
We have that $z\in C^\infty(B_R)$ and
$$\Delta_p z(x)=M^p(p-1)e^{M(p-1)(x_1+R)}>M^p(p-1)$$
and so, recalling~\eqref{4.14bis}, we see that,
if $M$ is sufficiently
large,
$$\Delta_p z(x)>\|g_\eta(\cdot, u_\eta(\cdot))\|_{L^\infty(\Omega)}\geq -g_\eta(x, u_\eta(x))\,\,\,\, \mbox{a.e. in} \,\, \Omega\,,$$
that is
\begin{equation}\label{eqz}
\int_\Omega |\nabla z(x)|^{p-2}\nabla z(x)\nabla \varphi(x)\leq \int_\Omega g_\eta(x, u_\eta(x))\varphi(x)\,\, \mbox{for any} \, \varphi \in X_0,\, \varphi\ge 0\, \mbox{a.e. in}\,\,\Omega\,.
\end{equation}
Now, we extend $z$ to $\tilde z$ in all $\RR^n$ in such a way that
$\tilde z(x)=z_{|\Omega}(x)$ if $x\in \Omega$ and $\tilde z(x)\leq u_0(x)$ if $x\in \mathcal C \Omega$\,.
Of course $\tilde z\in X$, since $z_{|\Omega} \in C^\infty(\overline \Omega)\subset W^{1,p}(\Omega)$. Moreover, $\tilde z-u_0\in X$ and, using the definition of~$\tilde z$,
$$\tilde z(x)-u_0(x)\leq -\|u_0\|_{L^\infty(\Omega)}-u_0(x)\leq u_0(x)-u_0(x)= 0\,\,\,\, \mbox{if}\,\,\, x\in \Omega\,,$$ while
$$\tilde z(x)-u_0(x)\leq 0\,\,\, \mbox{if}\,\,\, x\in \mathcal C\Omega\,,$$
so that $\tilde z-u_0\leq 0$ in $\RR^n$\,.
We notice that
$$\tilde z-u_\eta=(\tilde z-u_0)+(u_0-u_\eta)\,.$$
Therefore, $(\tilde z-u_\eta)^+\in X_0$, thanks to Corollary~\ref{corollario0} applied here with $v=u_0-u_\eta\in X_0$ and $w=\tilde z-u_0\in X\cap C(\overline \Omega)$\,.

Hence, taking $\varphi=(\tilde z-u_\eta)^+$ in~\eqref{problemaplaplaciano}, we have
\begin{equation}\label{plapl1}
\int_\Omega |\nabla u_\eta(x)|^{p-2}\nabla u_\eta(x)\nabla(\tilde z -u_\eta)^+(x)dx
=\int_\Omega g_\eta(x, u_\eta(x))
(\tilde z -u_\eta)^+dx\,.
\end{equation}
Similarly, using \eqref{eqz}
\begin{equation}\label{plapl2}
\int_\Omega |\nabla z(x)|^{p-2}\nabla z(x)\nabla(\tilde z -u_\eta)^+(x)dx\leq\int_\Omega g_\eta(x, u_\eta(x))
(\tilde z -u_\eta)^+dx\,.
\end{equation}
By \eqref{plapl1} and \eqref{plapl2} and using the definition of $\tilde z$ we deduce
$$\int_\Omega \Big(|\nabla z(x)|^{p-2}\nabla z(x)-|\nabla u_\eta(x)|^{p-2}\nabla u_\eta(x)\Big)\nabla(z -u_\eta)^+(x)dx\leq 0\,.$$
By the monotonicity condition~\eqref{monotonia} we deduce that
$$z\leq u_\eta \,\,\,\, \mbox{a.e. in}\,\, \Omega\,.$$
Since $z$ is bounded in $\Omega$, we easily get
\begin{equation}\label{17.1}
u_\eta \geq z\geq -\|z\|_{L^\infty(\Omega)}\,\,\,\, \mbox{a.e. in}\,\, \Omega
\end{equation}
that is $u_\eta$ is bounded from below uniformly in $\eta$. Moreover,
since $u_\eta$ converges uniformly to $u$, which solves the variational inequality~\eqref{varineqplapl}, we have that, for $\eta$ sufficiently small
$$u_\eta \leq u+1\leq \psi +1 \,\,\,\, \mbox{a.e. in}\,\, \Omega\,.$$
{F}rom this and~\eqref{17.1}, since $\psi\in C(\overline \Omega)$, we get that 
$u_\eta\in L^\infty(\Omega)$ and
\begin{equation*} \|u_\eta\|_{L^\infty(\Omega)}\leq \hat 
\kappa\,,\end{equation*}
where $\hat \kappa$ is a suitable positive constant, independent of $\eta$\,.

With this result and using the choice of $\Omega$ and the regularity of $u_0$ given in \eqref{regu0}, we can apply \cite[Theorem~1]{liebermann} with $m=p-2$, $\kappa=0$, $\lambda=\Lambda=1$ and $M_0=\hat \kappa$. Then, we have that $u_\eta \in C^{1, \beta}(\overline \Omega)$ and
\begin{equation}\label{stimac1alpha}
\|u_\eta\|_{C^{1, \beta}(\overline \Omega)}\leq C \quad \mbox{for any $\eta>0$}\,,
\end{equation}
where $\beta\in (0,1)$ depends only on $\alpha$, $p$ and $n$, while $C$ is a positive constants depending only on $\alpha$, $p$, $n$, $\hat \kappa$, $\|u_0\|_{C^{1, \alpha}(\overline \Omega)}$ and $\Omega$\,.

Hence, by the Ascoli-Arzel\`a Theorem, up to a subsequence, we have that
\begin{equation}\label{unifconv}
\nabla u_\eta \to g \quad \mbox{uniformly in}\,\, \Omega \,\, \mbox{as
$\eta\to 0$}\,,
\end{equation}
for some~$g\in \big(C^{1,\beta}(\Omega)\big)^n$\,.
By \eqref{convueta} and \eqref{unifconv} it is easy to see that $g=\nabla u$ in $\Omega$ and so
$$\nabla u_\eta \to \nabla u \quad \mbox{uniformly in}\,\, \Omega \,\, \mbox{as $\eta\to 0$}\,.$$
Hence, in particular, we have that
\begin{equation}\label{unifconv1}
|\nabla u_\eta|^{p-2}\nabla u_\eta \to |\nabla u|^{p-2}\nabla u \quad \mbox{a.e. in}\,\, \Omega \,\, \mbox{as $\eta\to 0$}\,.
\end{equation}
Moreover, by \eqref{stimac1alpha} and the fact that $\Omega$ is bounded we get
$$|\nabla u_\eta(x)|\leq C\in L^p(\Omega) \quad \mbox{a.e. in }\, \Omega$$
and so
\begin{equation}\label{dominata}
\Big|\,|\nabla u_\eta(x)|^{p-2}\nabla u_\eta(x)\Big|\leq C^{p-1}\in L^{p'}(\Omega) \quad \mbox{a.e. in }\, \Omega
\end{equation}
for any $\eta>0$\,.
By \eqref{unifconv1}, \eqref{dominata} and the dominated convergence Theorem we get
\begin{equation}\label{convergenzalp'}
|\nabla u_\eta|^{p-2}\nabla u_\eta \to |\nabla u|^{p-2}\nabla u \quad \mbox{in }\,\, L^{p'}(\Omega) \,\, \mbox{as $\eta\to 0$}\,.
\end{equation}
Hence, for any $\varphi\in C^\infty_0(\Omega)$ we deduce that
$$\mathcal A[u_\eta, \varphi] = \int_\Omega |\nabla u_\eta(x)|^{p-2}\nabla u_\eta(x) \nabla \varphi(x) dx
\to \int_\Omega |\nabla u(x)|^{p-2}\nabla u(x) \nabla \varphi(x) dx = \mathcal A[u, \varphi]$$
as $\eta\to 0$, that is assumption~\eqref{convA} is proved.

It is interesting to remark that,
differently from the case of the
Laplacian, here, in order to
check assumption~\eqref{convA}, we took advantage of the fact
that~$u$ solves \eqref{varineqplapl} and of the fact that $u_\eta$ is a solution of an equation, so to use
the associated regularity theory.

\medskip

\begin{Proofplap}
Let $u$ be a solution of the variational inequality~\eqref{varineqplapl}.
In the setting of Theorem~\ref{p lap} we can apply
Theorem~\ref{propLS} so that we get
$$0\leq -\int_\Omega |\nabla u|^{p-2}(x)\nabla u(x)\nabla \varphi(x) \,dx\leq
\int_\Omega (\Delta_p \psi +f)^+(x)\varphi(x) \, dx$$
for any $\varphi
\in C^\infty_0(\Omega)$\,, $\varphi\geq 0$ in
$\Omega$\,. By density, it is easily seen that such estimate holds
for any $\varphi\in W^{1,p}_0(\Omega)$ with $\varphi\geq 0$ a.e. in
$\Omega$\,.
\end{Proofplap}

\subsection{The Laplacian operator in the Heisenberg group.}

Thanks to our abstract framework, the proof in the Heisenberg group
follows exactly the one for the standard Laplacian: it is enough to
replace~$\nabla$ with~$\nabla_{\HH^n}$, $\Delta$
with~$\Delta_{\HH^n}$ and the Sobolev space~$H^1$
with~$W^{1,2}_{\HH^n}$. We refer to~\cite{VSCC92} and the
introduction of~\cite{ValdinociPinamonti} for further details on the
Heisenberg group and on the above mentioned spaces.

In order to check condition~\eqref{esistenzaueta}
here we need some regularity assumptions on $u_0$ and $\psi$. For
instance, it is enough to
require that $u_0$, $\psi:\RR^n\to \RR$ are
such that $u_0\in W^{1,p}_{\HH^n}(\Omega_*)\cap L^\infty(\Omega_*)$, with $\Omega_*$ smooth domain such that $\Omega \subset \subset \Omega_*$, and $\psi \in
C^2(\Omega)$\,. For more details we refer to \cite{ValdinociPinamonti}.

\section{New applications: integral operators with even
kernel}\label{sec:kernel}

In this section we consider a
kernel~$K:\RR^n\setminus\{0\}\rightarrow (0,+\infty)$
with the properties \eqref{kernel}--\eqref{evenkernel}\,.
A typical example is given by the fractional Laplace
kernel~$K(x)=|x|^{-(n+2s)}$, $s\in (0,1)$.

With respect to the abstract setting here 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 $\Omega\subset \RR^n$, $n>2s$, is an open bounded set, $d\mu=dx\,dy$ (the standard Lebesgue measure
in $\mathbb R^{2n}$), $X$ is 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
$$\mbox{the map}\,\,\,
(x,y)\mapsto (g(x)-g(y))\sqrt{K(x-y)}\,\,\, \mbox{is in}\,\,\,
L^2(Q, dxdy)\,,$$
endowed with the norm defined as
$$\|g\|_X=\|g\|_{L^2(\Omega)}+\Big(\int_Q |g(x)-g(y)|^2K(x-y)dx\,dy\Big)^{1/2}\,.$$
It is easy to check that $\|\cdot\|_X$ is a norm on $X$. We only show that 
if $\|g\|_X=0$, then $g=0$ a.e. in~$\RR^n$. Indeed, by $\|g\|_X=0$ we get
$\|g\|_{L^2(\Omega)}=0$, which implies that
\begin{equation}\label{normag}
g=0 \quad \mbox{a.e. in} \,\, \Omega\,,
\end{equation}
and
\begin{equation}\label{normag1}
\int_Q |g(x)-g(y)|^2K(x-y)dx\,dy=0\,.
\end{equation}
By \eqref{normag1} we deduce that $g(x)=g(y)$ a.e. $(x,y)\in Q$, that is $g$ is constant a.e. in $\RR^n$, say $g=c\in \RR$ a.e. in $\RR^n$. By \eqref{normag} it easily follows that $c=0$, so that $g=0$ a.e. in $\RR^n$\,.\\

The following lemma is valid:
\begin{lemma}\label{lemmaCinfty0}
Let~$\varphi\in C^2_0(\Omega)$. Then
$${\mbox{the map }}
\mathbb R^{2n}\ni
(x,y)\mapsto |\varphi(x)-\varphi(y)|^2 K(x-y)
{\mbox{ belongs to }}
L^1(\mathbb R^{2n}).$$
\end{lemma}
\begin{proof}
Since~$\varphi$ vanishes outside~$\Omega$,
\begin{equation}\label{phi R}
\begin{aligned}
\int_{\RR^{2n}}\!\!\! |\varphi(x)-\varphi(y)|^2 K(x-y)\,dx\,dy &=
\!\!\int_{\Omega\times \Omega}\!\! |\varphi(x)-\varphi(y)|^2 K(x-y)\,dx\,dy\\
&\qquad+2 \int_{\Omega\times {\mathcal{C}} \Omega}
|\varphi(x)-\varphi(y)|^2 K(x-y)\,dx\,dy\\
& \le  2 \int_{\Omega\times \RR^n }\!\! |\varphi(x)-\varphi(y)|^2 K(x-y)\,dx\,dy.
\end{aligned}
\end{equation}
Now, we notice that
\begin{equation*}
|\varphi(x)-\varphi(y)|\le \|\nabla\varphi\|_{L^\infty(\RR^n)} |x-y|
\end{equation*}
and
$$|\varphi(x)-\varphi(y)|\le
2\|\varphi\|_{L^\infty(\RR^n)}.$$
Accordingly
\begin{equation*}
|\varphi(x)-\varphi(y)|\le 2\|\varphi\|_{C^1(\RR^n)} \min\{ |x-y|,
1\}
=2\|\varphi\|_{C^1(\RR^n)} \sqrt{m(x-y)},
\end{equation*}
where $m$ is defined in \eqref{kernel}.
Therefore,
from~\eqref{phi R} we deduce that
$$\begin{aligned}
\int_{\RR^{2n}} |\varphi(x)-\varphi(y)|^2 K(x-y)\,dx\,dy & \leq 8
\|\varphi\|_{C^1(\RR^n)}^2
\int_{\Omega\times \RR^n} m(x-y) K(x-y)\,dx\,dy\\
& = 8|\Omega| \|\varphi\|_{C^1(\RR^n)}^2 \int_{\RR^n}
m(\xi) K(\xi)\,d\xi\,.
\end{aligned}$$
Thus, Lemma~\ref{lemmaCinfty0} follows by~\eqref{kernel} and by the fact that $\Omega$ is bounded.
\end{proof}

As a trivial consequence of Lemma~\ref{lemmaCinfty0} we get that
$C^2_0(\Omega)\subseteq X$\,. We also note that, since $\Omega$ is
bounded, the restriction to $\Omega$ of any function in $X$ belongs to $L^1(\Omega)$, so that all the assumptions on $X$ are satisfied.

We also take \begin{eqnarray*}
&& X_0=\{g\in X : g=0\,\, \mbox{a.e. in}\,\,
\mathcal C
\Omega\}\,, \\ && \widetilde X_0=C^\infty_0(\Omega)\\
{\mbox{and
}} &&Y=Z=\{g:\RR^{2n}
\to\RR : g_{|Q} \in L^2(Q, dxdy)\}\,, \end{eqnarray*}
while $\mathcal P$ is
the usual
product between functions. Finally, we fix two functions $u_0\in
X\cap L^\infty(\mathcal C\Omega)$ and $\psi\in X$, with $u_0\leq
\psi$ a.e. in $\RR^n$ and $\mathcal L_K\,\psi\in
L^\infty(\Omega)$ (cf. \eqref{lk} for the definition of the operator $\mathcal L_K$).

Of course $\widetilde X_0\subseteq X_0\subseteq X$\,. In order to verify
condition~\eqref{x01} we need the following lemma.
\begin{lemma}\label{lemmaA}
Let $v$ be a function in $X$. Then $v^+\in X$\,.
\end{lemma}
\begin{proof}
By Lemma~\ref{lemma0} with $\beta\equiv 0$ we have
$$\int_\Omega |v^+(x)|^2dx \leq \int_\Omega |v(x)|^2dx<+\infty\,,$$
while, taking $\beta=\alpha$ we get
$$\int_Q |v^+(x)-v^+(y)|^2K(x-y)\,dx\,dy\leq \int_Q |v(x)-v(y)|^2K(x-y)\,dx\,dy<+\infty\,,$$
i.e. $v^+\in X$\,.
\end{proof}

With this result, we can prove \eqref{x01}. For this let $v\in X_0$ and $w\in X$
with $w\leq 0$ a.e. in $\RR^n$. Since $v+w\in X$, by
Lemma~\ref{lemmaA} we have that $(v+w)^+\in X$\,. Moreover, by assumption
$$(v+w)(x)\leq v(x)=0 \quad \mbox{a.e.}\,\, x\in \mathcal C\Omega\,,$$
so that $(v+w)^+=0$ in $\mathcal C \Omega$\,. Hence $(v+w)^+\in
X_0$\,, that is \eqref{x01} is verified.

\medskip

Finally, for any $u, \varphi\in X$ we define $a(u), b(\varphi):\mathbb R^{2n}\to \mathbb R$ as follows
$$\begin{aligned}
& a(u)(x,y)=(u(x)-u(y))\sqrt{K(x-y)}\\
& b(\varphi)(x,y)=(\varphi(x)-\varphi(y))\sqrt{K(x-y)}
\end{aligned}$$
and
$$\mathcal A[u,\varphi]=-\int_Q
(u(x)-u(y))(\varphi(x)-\varphi(y))K(x-y)dx\,dy\,.$$

Notice that $a(u)$ and $b(\varphi)$ belong to $Y$, since $u, \varphi \in X$. In the following lemma we prove that $\mathcal A$ is well defined.
\begin{lemma}\label{lemmaAwell}
Let~$v,\varphi\in X$\,. Then
\begin{equation*}
{\mbox{the map }}
Q\ni
(x,y)\mapsto (v(x)-v(y))(\varphi(x)-\varphi(y))K(x-y)
{\mbox{ belongs to }}
L^1(Q, dxdy).
\end{equation*}
\end{lemma}
\begin{proof}
By the Cauchy-Schwarz inequality we have
$$\begin{aligned}
& {2}\,|v(x)-v(y)|\,|\varphi(x)-\varphi(y)| \,\sqrt{K(x-y)}\,
\sqrt{K(x-y)}\\
& \qquad\le |v(x)-v(y)|^2 K(x-y)
+|\varphi(x)-\varphi(y)|^2 K(x-y).
\end{aligned}$$
This and the fact that $v, \varphi \in X$ give the assertion.
\end{proof}

Now, we have to check assumptions
\eqref{conda}--\eqref{monotonia}\,. Using the definitions of $a$ and
$b$ and \eqref{evenkernel} it is easily seen that \eqref{conda} and
\eqref{condb} holds true. Now, let us show \eqref{monotonia}. Let
$u,v \in X$ with $(u-v)^+\in X_0$ and $\mathcal A[u,(u-v)^+]\ge
\mathcal A[v,(u-v)^+]$\,. We set~$w:=u-v$ and we consider the
negative part~$w^-(x)=\min\{ w(x),0\}$. Notice that~$w^-\le0$,
$w=w^+ + w^-$ and~$w(x)w^+(x)=(w^+(x))^2$. Therefore
\begin{equation*}\begin{split}
(w(x)-w(y)) (w^+(x)-w^+(y)) \,&=
w(x)w^+(x) +w(y)w^+(y)-w(x)w^+(y)-w^+(x)w(y)\\
&=(w^+(x))^2+(w^+(y))^2-w(x)w^+(y)-w^+(x)w(y).
\end{split}\end{equation*}
Also, by \eqref{evenkernel}, we have that
$$ \int_Q w(x) w^+(y)K(x-y)\,dx\,dy=
\int_Q w^+(x) w(y) K(x-y)\,dx\,dy.$$
As a consequence,
\begin{equation*}\begin{split}
{0\geq \mathcal{A}}[v,(u-v)^+]&-{\mathcal{A}}[u,(u-v)^+]\,=
\int_Q a(w) b(w^+)\,d\mu\\
&=\int_Q (w(x)-w(y)) (w^+(x)-w^+(y)) \,K(x-y)\,dx\,dy\\
&=\int_Q \Big(
(w^+(x))^2+(w^+(y))^2-w(x)w^+(y)-w^+(x)w(y)
\Big) \,K(x-y)\,dx\,dy\\
&=\int_Q \Big(
(w^+(x))^2+(w^+(y))^2-2w^+(x)w(y)
\Big) \,K(x-y)\,dx\,dy \\
&=\int_Q \Big(
(w^+(x))^2+(w^+(y))^2-2w^+(x)w^+(y)
-2w^+(x)w^-(y)
\Big) \,K(x-y)\,dx\,dy \\
&=\int_Q \Big(
(w^+(x)-w^+(y))^2
-2w^+(x)w^-(y)
\Big) \,K(x-y)\,dx\,dy\geq 0\,,
\end{split}\end{equation*}
being $w^-\le0$\,.
Then, $(w^+(x)-w^+(y))^2
-2w^+(x)w^-(y)=0$ a.e. in $Q$, that is,
$$0\leq (w^+(x)-w^+(y))^2
=2w^+(x)w^-(y)\leq 0\,.$$
Hence, it follows that~$w^+(x)=w^+(y)$ a.e. $(x,y)\in Q$, which implies that $w^+$ is constant a.e. in $\RR^n$, say~$w^+(x)=c\ge 0$. Since $w^+\in X_0$ by assumption, then $c=0$, i.e.
$u\leq v$ a.e. in $\RR^n$\,.
Thus, condition~\eqref{monotonia} is verified.

Now, before going on, we need some preliminary results on $X$ and $X_0$.
In the following we denote by $H^s(\Omega)$ the usual fractional
Sobolev space endowed with the so-called Gagliardo norm
\begin{equation}\label{20} \|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}
while $H^s_0(\Omega)$ is the closure of $C^\infty_0(\Omega)$ with
respect to the norm $\|\cdot \|_{H^s(\Omega)}$\,. Of course,
$H^s_0(\RR^n)=H^s(\RR^n)$\,.
\begin{lemma}\label{lemmaV}
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 $v\in X$, then $v\in H^s(\Omega)$. Moreover
$$\|v\|_{H^s(\Omega)}\leq c(\lambda)\|v\|_X\,;$$
\item[$b)$] if $v\in X_0$, then $v\in H^s(\RR^n)$\,.
Moreover
$$\|v\|_{H^s(\RR^n)}\leq c(\lambda)\|v\|_X\,.$$
\end{itemize}
In both cases $c(\lambda)=\max\{1, 1/\lambda\}$\,.
\end{lemma}
\begin{proof}
Let us prove part $a)$. By \eqref{kernelfrac} we get
$$\begin{aligned}
\int_{\Omega\times
\Omega}\frac{\,\,\,|v(x)-v(y)|^2}{|x-y|^{n+2s}}\,dx\,dy & \leq
\frac 1\lambda\int_{\Omega\times \Omega}|v(x)-v(y)|^2K(x-y)\,dx\,dy \\
& \leq \frac 1\lambda\int_Q|v(x)-v(y)|^2K(x-y)\,dx\,dy<+\infty\,,
\end{aligned}$$
since $v\in X$\,. The first assertion is proved.

For part $b)$ note that $v\in X$ and $v=0$ a.e. in $\mathcal C\Omega$. As a consequence,
$$\|v\|_{L^2(\RR^n)}=\|v\|_{L^2(\Omega)}<+\infty\,,$$
and
$$\begin{aligned}
\int_{\RR^{2n}}\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}}\,dx\,dy &=\int_Q\frac{|v(x)-v(y)|^2}{|x-y|^{n+2s}}\,dx\,dy\\
& \leq \frac 1 \lambda \int_Q |v(x)-v(y)|^2K(x-y)\,dx\,dy<+\infty\,.
\end{aligned}$$
Hence $v\in H^s(\RR^n)$\,. The estimate on the norm easily follows.
\end{proof}

\begin{lemma}\label{lemmaZ}
Let $v_j$ be a sequence in $X$ such that
\begin{equation}\label{sup j}
\sup_{j\in \NN} \| v_j\|_X<+\infty
\end{equation}
and $v_j\to v_\infty$ a.e. in
$\RR^n$ as $j\to +\infty$. Then $v_\infty\in X$.

If, in addition, $v_j\in X_0$ for any $j\in\NN$, then $v_\infty\in
X_0$\,.
\end{lemma}
\begin{proof}
By \eqref{sup j} and the Fatou Lemma we have
$$+\infty>\liminf_{j\to +\infty}\int_\Omega|v_j(x)|^2dx\geq\int_\Omega|v_\infty(x)|^2dx\,,$$
i.e. $v_\infty\in L^2(\Omega)$, and
$$+\infty>\liminf_{j\to +\infty}\int_Q|v_j(x)-v_j(y)|^2K(x-y)\,dx\,dy
\geq\int_Q|v_\infty(x)-v_\infty(y)|^2K(x-y)\,dx\,dy\,.$$ Thus,
$v_\infty\in X$\,.

Now, suppose that $v_j=0$ a.e. in $\mathcal C \Omega$ for any
$j\in \NN$. Then, it is easy to see that $v_\infty=0$ a.e. in
$\mathcal C \Omega$\,. Hence, $v_\infty\in X_0$ and
the assertion is
proved.
\end{proof}

Now, let us fix $\eta\in (0,1)$. We consider the following problem $\mathcal L_K u_\eta=g_\eta(x, u_\eta)$ in $\Omega$ with $u_\eta=u_0$ in $\mathcal C\Omega$, that is
{\small\begin{equation}\label{problemakernel}
\left\{\begin{array}{l}
{\displaystyle \!\!\!\!\int_Q \!\!\left(u_\eta(x)-u_\eta(y)\right)\left(\varphi(x)-\varphi(y)\right)K(x-y)\,dx\,dy+\int_\Omega g_\eta(x, u_\eta(x))\varphi(x)\,dx=0 \quad \forall\, \varphi \in X_0}\\
\\
u_\eta\in X,\, u_\eta-u_0\in X_0\,,
\end{array}\right.
\end{equation}}
\noindent where $g_\eta$ is given in \eqref{geta}, with
$h=(\mathcal L_K\,\psi+f)^+\in L^\infty(\Omega)$, since $f, 
\mathcal L_K\psi\in L^\infty(\Omega)$\,. For the definition of $\mathcal L_K$ see 
\eqref{lk}.

In order to prove condition~\eqref{esistenzaueta}, we consider the
space $X_{u_0}=\{u\in X : u-u_0\in X_0\}$ and the functional
$\mathcal J_\eta: X_{u_0}\to \RR$ defined as follows
$$\mathcal J_\eta(u)=\frac 1 2 \int_Q|u(x)-u(y)|^2 K(x-y)dx\,dy+\int_\Omega G_\eta(x, u(x)) dx\,.$$
We recall that $G_\eta$ is the primitive of $g_\eta$ with respect to
its second variable.

First of all, we note that
\begin{equation}\label{inffrac}
{\displaystyle \inf_{u\in X_{u_0}}\mathcal J_\eta(u)>-\infty.}
\end{equation}
Indeed, as in the case of the Laplacian (cf.
\eqref{crescitaG}--\eqref{infJ}) we get that for any $\delta>0$ and
$u\in X_{u_0}$
\begin{equation}\label{infJfrac}
\begin{aligned}
\mathcal J_\eta(u)& \geq \frac 1 2\int_Q|u(x)-u(y)|^2 K(x-y)dx\,dy-\kappa\int_\Omega |u(x)|dx\\
& \geq \frac 1 2\int_Q|u(x)-u(y)|^2
K(x-y)dx\,dy-\frac{\kappa\delta}{2}\int_\Omega |u(x)|^2dx
-\frac{\kappa}{2\delta}|\Omega|\\
& \geq \frac 1 2\int_Q|u(x)-u(y)|^2
K(x-y)dx\,dy-\kappa\delta\|u-u_0\|_{L^2(\Omega)}^2\\
& \qquad \qquad \qquad \qquad -\kappa\delta\|u_0\|_{L^2(\Omega)}^2
-\frac{\kappa}{2\delta}|\Omega|\\
& \geq \frac 1 2\int_Q|u(x)-u(y)|^2
K(x-y)\,dx\,dy-\kappa\delta|\Omega|^{(2^*-2)/2^*}
\|u-u_0\|_{L^{2^*}(\Omega)}^2\\
& \qquad \qquad \qquad \qquad -\kappa\delta\|u_0\|_{L^2(\Omega)}^2-\frac{\kappa}{2\delta}|\Omega|\,,
\end{aligned}
\end{equation}
thanks to the Cauchy-Schwarz inequality, the Minkowski inequality and to the fact that
$L^{2^*}(\Omega)\hookrightarrow L^2(\Omega)$ continuously (being
$\Omega$ bounded and $2<2^*=2n/(n-2s)$). By Lemma~\ref{lemmaV}, we know that $u-u_0\in
H^s(\RR^n)$ and so, using \cite[Theorem~6.5]{valpal} (here with
$p=2$), we get
\begin{equation}\label{guida}
\|u-u_0\|_{L^{2^*}(\RR^n)}^2\leq C \int_{\RR^{2n}}\frac{|u(x)-u_0(x)-u(y)+u_0(y)|^2}{|x-y|^{n+2s}}\, dx\,dy\,,
\end{equation}
where $C$ is a positive constant depending only on $n$ and $s$\,.

Using \eqref{kernelfrac} and again the Minkowski inequality, by \eqref{infJfrac} and \eqref{guida} we
have that for any $\delta>0$ and
$u\in X_{u_0}$
\begin{equation}\label{infJfrac2}
\begin{aligned}
\mathcal J_\eta(u) & \geq \frac \lambda 2\int_Q\frac{|u(x)-u(y)|^2}
{|x-y|^{n+2s}}\,dx\,dy-\kappa\delta\|u_0\|_{L^2(\Omega)}^2-\frac{\kappa}{2\delta}|\Omega|\\
& \qquad -\kappa\delta C|\Omega|^{(2^*-2)/2^*}
 \int_{\RR^{2n}}\frac{|u(x)-u_0(x)-u(y)+u_0(y)|^2}{|x-y|^{n+2s}}\, dx\,dy\\
& \geq \lambda \int_Q\frac{|u(x)-u_0(x)-u(y)+u_0(y)|^2}
{|x-y|^{n+2s}}\,dx\,dy-\lambda \int_Q\frac{|u_0(x)-u_0(y)|^2}
{|x-y|^{n+2s}}\,dx\,dy\\
& \qquad -\kappa\delta C|\Omega|^{(2^*-2)/2^*}
 \int_{\RR^{2n}}\frac{|u(x)-u_0(x)-u(y)+u_0(y)|^2}{|x-y|^{n+2s}}\, dx\,dy\\
& \qquad  -\kappa\delta\|u_0\|_{L^2(\Omega)}^2-\frac{\kappa}{2\delta}|\Omega|\\
& = \Big( \lambda -\kappa\delta
C|\Omega|^{(2^*-2)/2^*}\Big)\int_Q\frac{|u(x)-u_0(x)-u(y)+u_0(y)|^2}{|x-y|^{n+2s}}\,dx\,dy\\
& \qquad
-\lambda \int_Q\frac{|u_0(x)-u_0(y)|^2}
{|x-y|^{n+2s}}\,dx\,dy-\kappa\delta\|u_0\|_{L^2(\Omega)}^2
-\frac{\kappa}{2\delta}|\Omega|\,.
\end{aligned}
\end{equation}
Note that, since $u_0\in X$ and \eqref{kernelfrac} holds true,
\begin{equation}\label{u0finito}
\int_Q\frac{|u_0(x)-u_0(y)|^2}
{|x-y|^{n+2s}}\,dx\,dy <+\infty \quad \mbox{and}\quad \|u_0\|_{L^2(\Omega)}<+\infty\,.
\end{equation}
Choosing $\delta>0$ such that $\kappa\delta C
|\Omega|^{(2^*-2)/2^*}<\lambda$, by \eqref{infJfrac2} and \eqref{u0finito} it easily
follows that
$$\mathcal J_\eta(u)\geq -\lambda \int_Q\frac{|u_0(x)-u_0(y)|^2}
{|x-y|^{n+2s}}\,dx\,dy-\kappa\delta\|u_0\|_{L^2(\Omega)}^2
-\frac{\kappa}{2\delta}|\Omega|>-\infty\,,$$
for any $u\in X_{u_0}$\,, so that \eqref{inffrac}
is proved.

Now, let us take a minimizing sequence $u_j$ for $\mathcal J_\eta$,
i.e. a sequence $u_j$ in $X_{u_0}$ such that
$${\displaystyle \mathcal J_\eta(u_j)\to \inf_{u\in X_{u_0}} \mathcal J_\eta(u)}>-\infty\,\, \mbox{as}\,\, j\to +\infty\,.$$
Then the sequence $\mathcal J_\eta(u_j)$ is bounded in $\RR$.  Hence,
using \eqref{infJfrac2} with $u=u_j$ and the fact that $u_j=u_0$ a.e. in $\mathcal C \Omega$, we have that for any $j\in \NN$
\begin{equation}\label{pre j}
\begin{aligned}
\int_{\RR^{2n}} &\frac{|u_j(x)-u_0(x)-u_j(y)+u_0(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\\
& \qquad \qquad \qquad = \int_Q \frac{|u_j(x)-u_0(x)-u_j(y)+u_0(y)|^2}{|x-y|^{n+2s}}\,
dx\,dy\leq \tilde \kappa\,,
\end{aligned}
\end{equation}
for some $\tilde \kappa>0$\,.  Moreover, since
$\|u_j-u_0\|_{L^2(\RR^n)}=\|u_j-u_0\|_{L^2(\Omega)}$, the
continuity of the embedding $L^{2^*}(\Omega)\hookrightarrow
L^2(\Omega)$ and \eqref{guida} give that~$u_j-u_0$ is bounded also in~$L^2(\RR^n)$\,.
Then, $u_j-u_0$ is bounded in $H^s(\RR^n)$\,.

Thus, by \cite[Theorem~6.5]{valpal}, up to a
subsequence, there exists $v_\infty \in L^q(\RR^n)$ with $q\in (2,
2^*)$ such that
$$u_j-u_0 \to v_\infty \quad \mbox{in}\,\, L^q(\RR^n)$$
and so, by \cite[Theorem~IV.9]{brezis}
\begin{equation}\label{convergenzefrac}
u_j-u_0 \to v_\infty \quad \mbox{a.e. in}\,\, \RR^n
\end{equation}
as $j\to +\infty$ and there exists $\ell\in L^q(\RR^n)$ such that
\begin{equation}\label{dominata22}
|u_j(x)|\leq \ell(x) \quad \mbox{a.e. in}\,\, \RR^n\,\quad \mbox{for any}\,\,j\in \NN\,.
\end{equation}
Also,~\eqref{sup j}
holds true here for~$v_j=u_j-u_0$, thanks to~\eqref{pre j}
and~\cite[Theorem~6.5]{valpal}:
as a consequence, by
Lemma~\ref{lemmaZ} we have that
$v_\infty\in X_0$\,.

Now, let $u_\infty=v_\infty+u_0$. Of course, $u_\infty\in X_{u_0}$
and, by \eqref{convergenzefrac}, \eqref{dominata22}, \eqref{crescitaG}, the fact that the map $t\mapsto
G_\eta(x, t)$ is continuous in $\RR$ a.e. $x\in \Omega$, the Fatou
Lemma and the dominated convergence theorem, we also get
$${\displaystyle \lim_{j\to +\infty}\mathcal J_\eta(u_j)\geq
\frac 1 2\int_Q |u_\infty(x)-u_\infty(y)|^2K(x-y)\,
dx\,dy+\int_\Omega G_\eta(x, u_\infty(x))dx=\mathcal
J_\eta(u_\infty)}\,,$$
so that
$${\displaystyle \mathcal J_\eta(u_\infty)=\inf_{u\in X_{u_0}} \mathcal J_\eta(u)}\,.$$
Hence, problem~\eqref{problemakernel} has a solution.

Now, in the following lemma we prove a sort of formula of integration by parts in $X$, which will be useful in the sequel.

\begin{lemma}\label{lemmaintparti}
Let~$\varphi\in C^2_0(\Omega)$ and~$v\in X\cap L^\infty(\mathcal C\Omega)$\,.
Then
\begin{equation*}
\begin{split}
&\int_{\RR^{2n}}
\big(v(x)-v(y)\big)\,\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy
\\ =\,&
\int_{ Q}
\big(v(x)-v(y)\big)\,\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy
\\ =\,&
\int_{\RR^{2n}}
v(x)\big(2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big)\,K(\xi)\,dx\,d\xi
.\end{split}\end{equation*}
\end{lemma}

\begin{proof} In what follows, for notation consistency,
it will be
convenient to denote
\begin{equation}\label{D0}
{\mathcal{D}}_0:=Q=\RR^{2n}\setminus {\mathcal{O}},
\end{equation}
with $\mathcal O$ as in \eqref{O}\,.
Also, given~$\epsilon\in[0,1)$, we define
$${{\mathcal{D}}_\epsilon} =
\Big\{
(x,y)\in {{\mathcal{D}}_0}
{\mbox{ s.t. }} |x-y|\ge\epsilon
\Big\}$$
and
$${\mathcal{D}}_\epsilon^\pm=
\Big\{
(x,\xi)\in \RR^{2n} {\mbox{ s.t.}}\,\, (x,x\pm\xi)\in{{\mathcal{D}}_0}
{\mbox{ and }} |\xi|\ge\epsilon
\Big\}.$$
We remark that the above notation for~${{\mathcal{D}}_\epsilon}$
is consistent with
the one in~\eqref{D0} for~$\epsilon=0$.
Recalling that~$\varphi$ vanishes
outside~$\Omega$ and \eqref{evenkernel}, we have
$$\begin{aligned}
\int_{{\mathcal{D}}_\epsilon}
|v(x)||\varphi(x)-&\varphi(y)|\,K(x-y)\,dx\,dy =
\int_{(\Omega\times \Omega)\cap \{|x-y|\ge\epsilon\}}
|v(x)||\varphi(x)-\varphi(y)|\,K(x-y)\,dx\,dy\\
&\qquad \qquad \qquad \qquad +
\int_{(\Omega\times {\mathcal{C}} \Omega)\cap \{|x-y|\ge\epsilon\}}
|v(x)||\varphi(x)-\varphi(y)|\,K(x-y)\,dx\,dy\\
&\qquad \qquad \qquad \qquad +
\int_{({\mathcal{C}} \Omega\times\Omega)\cap \{|x-y|\ge\epsilon\}}
|v(x)||\varphi(x)-\varphi(y)|\,K(x-y)\,dx\,dy\\
& = \int_{(\Omega\times \RR^n)\cap \{|x-y|\ge\epsilon\}}
|v(x)||\varphi(x)-\varphi(y)|\,K(x-y)\,dx\,dy\\
&\qquad \qquad \qquad \qquad +
\int_{(\Omega\times {\mathcal{C}} \Omega)\cap \{|x-y|\ge\epsilon\}}
|v(y)||\varphi(x)-\varphi(y)|\,K(x-y)\,dx\,dy\\
&\le 2\|\varphi\|_{L^\infty(\RR^n)}\left[
\int_{(\Omega\times \RR^n)\cap \{|x-y|\ge\epsilon\}}
|v(x)|K(x-y)\,dx\,dy
\right.\\
& \qquad \qquad \qquad \qquad \left. +
\int_{(\Omega\times {\mathcal{C}} \Omega)\cap \{|x-y|\ge\epsilon\}}
|v(y)|K(x-y)\,dx\,dy
\right]\\
&= \frac{2\|\varphi\|_{L^\infty(\RR^n)}}{\epsilon^2}
\left[\int_{(\Omega\times \RR^n)\cap \{|x-y|\ge\epsilon\}} |v(x)|m(x-y)
K(x-y)\,dx\,dy\right.\\
& \qquad \qquad \qquad \qquad \left. +
\int_{(\Omega\times {\mathcal{C}} \Omega)\cap \{|x-y|\ge\epsilon\}}
|v(y)|m(x-y)K(x-y)\,dx\,dy
\right]\\
&\le \frac{2\|\varphi\|_{L^\infty(\RR^n)}}{\epsilon^2}
\left[\int_{\Omega\times \RR^n} |v(x)|m(\xi)
K(\xi)\,dx\,d\xi\right.\\
& \qquad \qquad \qquad \qquad \left. +
\|v\|_{L^\infty(\mathcal C\Omega)}\int_{\Omega\times \RR^n}
m(\xi)K(\xi)\,dx\,d\xi
\right]\\
&= \frac{2\|\varphi\|_{L^\infty(\RR^n)}}{\epsilon^2}
\Big(\|v\|_{L^1(\Omega)}+|\Omega|\,\|v\|_{L^\infty(\mathcal C\Omega)}\Big)\int_{\RR^n} m(\xi)
K(\xi)\,d\xi\,,\\
\end{aligned}$$
which is finite, thanks to~\eqref{kernel}. As a consequence,
\begin{equation}\label{phi v}\begin{split}
&{\mbox{the maps }}
(x,y)\mapsto v(x)
\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\\
&\qquad{\mbox{ and }}
(x,y)\mapsto v(y)
\big(\varphi(x)-\varphi(y)\big)\,K(x-y)
{\mbox{ belong to }}
L^1({\mathcal{D}}_\epsilon, dxdy).
\end{split}
\end{equation}
Thanks to Lemma~\ref{lemmaAwell} and~\eqref{phi v} we can split the integrals
$$\begin{aligned}
\int_{ {\mathcal{D}}_\epsilon }
\big(v(x)-v(y)\big)\,\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy &
=\int_{ {\mathcal{D}}_\epsilon }
v(x)\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy\\
& \qquad +
\int_{ {\mathcal{D}}_\epsilon }
v(y)\big(\varphi(y)-\varphi(x)\big)\,K(x-y)\,dx\,dy\\
& = \int_{ {\mathcal{D}}_\epsilon }
v(x)\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy\\
&\qquad+ \int_{ {\mathcal{D}}_\epsilon }
v(x)\big(\varphi(x)-\varphi(y)\big)\,K(y-x)\,dx\,dy\,.
\end{aligned}$$
Now, with the change of variable $\xi=y-x$ in the first integral and $\xi=x-y$ in the second one, and taking into account the fact that $K$ is even (see \eqref{evenkernel}) and the definition of $\mathcal D_\epsilon^\pm$, we get
\begin{equation}\label{calculate}
\begin{aligned}
\int_{ {\mathcal{D}}_\epsilon }
\big(v(x)-v(y)\big)\,\big(\varphi(x)-\varphi(y)\big)&\,K(x-y)\,dx\,dy\\
&
= \int_{ {\mathcal{D}}_\epsilon }
v(x)\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy\\
&\qquad+ \int_{ {\mathcal{D}}_\epsilon }
v(x)\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy\\
& =\int_{ {\mathcal{D}}_\epsilon^+ }
v(x)\big(\varphi(x)-\varphi(x+\xi)\big)\,K(\xi)\,dx\,d\xi\\ &\qquad+
\int_{ {\mathcal{D}}_\epsilon^- }
v(x)\big(\varphi(x)-\varphi(x-\xi)\big)\,K(\xi)\,dx\,d\xi.
\end{aligned}
\end{equation}
Now, we claim that
\begin{equation}\label{delete}
\int_{ {\mathcal{D}}_\epsilon^- \setminus {\mathcal{D}}_\epsilon^+}
v(x)\big(\varphi(x)-\varphi(x+\xi)\big)\,K(\xi)\,dx\,d\xi=0.
\end{equation}
To check this, let~$(x,\xi)\in {\mathcal{D}}_\epsilon^- \setminus
{\mathcal{D}}_\epsilon^+$. By definitions of ${\mathcal{D}}_\epsilon^\pm$ it follows that
$(x, x-\xi)\in \mathcal D_0$\,, $(x, x+\xi)\not \in \mathcal D_0$ and $|\xi|\geq \epsilon$. Hence $(x, x+\xi)\in \mathcal O$, that is
\begin{equation}\label{where x}
x\in{\mathcal{C}}\Omega,
\end{equation}
and
\begin{equation}\label{where xi}
x+ \xi\in {\mathcal{C}}\Omega\,.
\end{equation}
Then, by~\eqref{where x} and~\eqref{where xi}, we have that~$\varphi(x)=0=\varphi(x+ \xi)$, for any $(x,\xi)\in {\mathcal{D}}_\epsilon^- \setminus
{\mathcal{D}}_\epsilon^+$\, which implies~\eqref{delete}.
Analogously, we get that
\begin{equation}\label{delete1}
\int_{ {\mathcal{D}}_\epsilon^+ \setminus {\mathcal{D}}_\epsilon^-}
v(x)\big(\varphi(x)-\varphi(x-\xi)\big)\,K(\xi)\,dx\,d\xi=0.
\end{equation}

In the light of~\eqref{calculate}, \eqref{delete} and~\eqref{delete1}, we obtain that
\begin{equation}\label{sum up}
\begin{aligned}
\int_{ {\mathcal{D}}_\epsilon }
\big(v(x)-v(y)\big)\,& \big(\varphi(x)-\varphi(y)\big) \,K(x-y)\,dx\,dy\\
&
= \int_{ {\mathcal{D}}_\epsilon^+ \cup {\mathcal{D}}_\epsilon^-}
v(x)\big(\varphi(x)-\varphi(x+\xi)\big)\,K(\xi)\,dx\,d\xi\\
&\qquad \qquad \qquad + \int_{ {\mathcal{D}}_\epsilon^- \cup {\mathcal{D}}_\epsilon^+ }
v(x)\big(\varphi(x)-\varphi(x-\xi)\big)\,K(\xi)\,dx\,d\xi
\\
& = \int_{ {\mathcal{D}}_\epsilon^+ \cup {\mathcal{D}}_\epsilon^-}
v(x)\big(2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big)\,K(\xi)\,dx\,d\xi\,.
\end{aligned}
\end{equation}
Now, let us define
\begin{equation}\label{notation}
\Phi(x,\xi)=\big(2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big)\,K(\xi).
\end{equation}
We claim that
\begin{equation}\label{psi good}
\Phi\in L^1(\RR^{2n}, dxdy).
\end{equation}
To establish this, we observe that
\begin{equation}\label{stimaPSI}
|\Phi(x,\xi)|\le 4\|\varphi\|_{L^\infty(\RR^n)} K(\xi)
\end{equation}
and, by a Taylor expansion, that
$$ |\Phi(x,\xi)|\le \|D^2\varphi\|_{L^\infty(\RR^n)} |\xi|^2
K(\xi).$$
Therefore
\begin{equation}\label{per psi}
|\Phi(x,\xi)|\le 4\|\varphi\|_{C^2(\RR^n)} m(\xi) K(\xi).
\end{equation}
Let
\begin{equation}\label{R}
R>1
\end{equation}
be such that $\Omega\subset B_R$\,.
If~$x\in {\mathcal{C}} B_{2R}$
and~$x\pm \xi\in\Omega\subset B_R$, then
\begin{equation}\label{normaxi>1}
|\xi|\ge |x|-|x\pm\xi|\ge 2R-R=R>1
\end{equation}
by \eqref{R}.
Now we define
$$\mathcal S=\Big\{(x,\xi)\in \RR^{2n} : x\in \mathcal C B_{2R}\,\, \mbox{and}\,\, \xi\in B_R(x)\cup B_R(-x)\Big\}$$
and
$$\mathcal S_*=\Big\{(x,\xi)\in \RR^{2n} : x\in B_R(\xi)\cup B_R(-\xi)\,\, \mbox{and}\,\, \xi\in \mathcal C B_1\Big\}\,.$$
Since \eqref{normaxi>1} holds true, then
\begin{equation}\label{SSstar}
\mathcal S\subset \mathcal S_*\,.
\end{equation}
Moreover, by definition of $m$ (see \eqref{kernel})
\begin{equation}\label{KS}
K(\xi)=m(\xi)K(\xi) \quad \mbox{if}\,\, (x, \xi)\in \mathcal S_*\,.
\end{equation}
Now, let $(x, \xi)\in \big(\mathcal C B_{2R}\times \RR^n\big)\setminus \mathcal S$. Then $x\in \mathcal C B_{2R}$ and $\xi \in \mathcal C\big(B_R(x)\cup B_R(-x)\big)$, that is $|x\pm \xi|>R$. As a consequence, since $\varphi$ vanishes outside $\Omega$
\begin{equation}\label{fi0}
\varphi(x)=0=\varphi(x\pm \xi) \quad \mbox{if}\quad (x, \xi)\in \big(\mathcal C B_{2R}\times \RR^n\big)\setminus \mathcal S\,.
\end{equation}
Thus, using \eqref{stimaPSI}, \eqref{SSstar}--\eqref{fi0} and \eqref{kernel} we have
\begin{equation}\label{integralefinito}
\begin{aligned}
\int_{({\mathcal{C}} B_{2R})\times\RR^n} |\Phi(x,\xi)|\,dx\,d\xi &
=\int_{\mathcal S} \big|2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big|\,K(\xi)\,dx\,d\xi \\
& \leq 4 \|\varphi\|_{L^\infty(\RR^n)}\int_{\mathcal S_*} K(\xi)\,dx\,d\xi \\
& = 4 \|\varphi\|_{L^\infty(\RR^n)}\int_{\mathcal S_*} m(\xi)K(\xi)\,dx\,d\xi \\
& \leq 8\omega_n R^n \|\varphi\|_{L^\infty(\RR^n)}\int_{\RR^n}
m(\xi)K(\xi)\,d\xi \\
& = C(n,R,K)<+\infty
\end{aligned}
\end{equation}
for a suitable positive constant $C(n,R,K)$, where $\omega_n$ is the volume of the unit ball in $\RR^n$, i.e. $\omega_n=2\pi^{n/2}/(n\Gamma(n/2))$.
By \eqref{per psi} and \eqref{integralefinito} we obtain
$$\begin{aligned}
\int_{\RR^{2n}} |\Phi(x,\xi)|\,dx\,d\xi & \leq C(n,R,K)+\int_{B_{2R}\times \RR^n } |\Phi(x,\xi)|\,dx\,d\xi\\
& \leq  C(n,R,K)+
4\|\varphi\|_{C^2(\RR^n)} \int_{B_{2R}\times \RR^n }
m(\xi) K(\xi)\,dx\,d\xi\\
& \leq  C(n,R,K)+
4\omega_n (2R)^n\|\varphi\|_{C^2(\RR^n)} \int_{\RR^n }
m(\xi) K(\xi)\,d\xi\,,
\end{aligned}$$
which, once more, is finite due to~\eqref{kernel}.
This establishes~\eqref{psi good}.

Owing to Lemma~\ref{lemmaAwell}
and~\eqref{psi good}, we can send~$\epsilon\rightarrow0^+$
in~\eqref{sum up}: we obtain
\begin{equation*}
\begin{split}
&\int_{ {\mathcal{D}}_0 }
\big(v(x)-v(y)\big)\,\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy
\\ =\,&
\int_{ {\mathcal{D}}_0^+ \cup {\mathcal{D}}_0^-}
v(x)\big(2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big)\,K(\xi)\,dx\,d\xi
.\end{split}\end{equation*}
By using once again
that~$\varphi$ vanishes outside~$\Omega$ we get the assertion of Lemma~\ref{lemmaintparti}.
\end{proof}

As a consequence of Lemma~\ref{lemmaintparti}, we can prove that $\mathcal A$ satisfies condition~\eqref{convA}. Indeed, the following corollary holds true.

\begin{corollary}\label{C2}
Let~$\varphi\in C^2_0(\Omega)$. Let~$v_j\in X$
be a sequence of functions converging
uniformly in~$\RR^n$ to $v_\infty \in X$ as $j\to +\infty$ and such that
$v_j-u_0\in X_0$ for any $j\in \NN\cup \{\infty\}$.
Then
\begin{equation*}
\begin{split}
&\lim_{j\rightarrow+\infty}\int_{ Q}
\big(v_j(x)-v_j(y)\big)\,\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy
\\ =\,&
\int_{ Q}
\big(v_\infty(x)-
v_\infty(y)\big)\,\big(\varphi(x)-\varphi(y)\big)\,K(x-y)\,dx\,dy
.\end{split}\end{equation*}
\end{corollary}

\begin{proof}
We define~$w_j:=v_j-v_\infty$\,. Notice that~$w_j\in X$,
being~$X$ a linear space, and $w_j=0$ a.e. in $\mathcal C\Omega$, since $v_j=u_0=v_\infty$ a.e. in $\mathcal C\Omega$. Hence, $w_j\in X\cap L^\infty(\mathcal C \Omega)$ for any $j\in \NN$.
Then, by Lemma~\ref{lemmaintparti} (applied to~$w_j$), we obtain that
\begin{equation}\label{Ri}
\begin{aligned}
\int_Q \big(
v_j(x)-& v_j(y)\big)\big(\varphi(x) -\varphi(y)\big)K(x-y)\,dx\,dy\\
&\qquad \qquad -
\int_Q \big(
v_\infty(x)-v_\infty(y)\big)\big(\varphi(x)-\varphi(y)\big)K(x-y)\,dx\,dy\\
&=
\int_Q \big(
w_j(x)-w_j(y)\big)\big(\varphi(x)-\varphi(y)\big)K(x-y)\,dx\,dy\\
& =
\int_Q w_j(x)\big(2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big)
K(\xi)\,dx\,d\xi
\\
& =
\int_Q \big(v_j(x)-v_\infty(x)\big)
\big(2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big) K(\xi)\,dx\,d\xi\,,
\end{aligned}
\end{equation}
for any $\varphi\in C^2_0(\Omega)$ and any $j\in \NN$\,.

Moreover, exploiting the notation in~\eqref{notation}
and~\eqref{psi good}, we have
\begin{equation*}
\begin{split}
&\lim_{j\rightarrow+\infty}\left|\int_{\RR^{2n}}
\big(v_j(x)-v_\infty(x)\big)
\big(2\varphi(x)-\varphi(x+\xi)-\varphi(x-\xi)\big)\,K(\xi)\,dx\,d\xi
\right|
\\ &\qquad =\lim_{j\rightarrow+\infty}
\left|\int_{\RR^{2n}}
\big(v_j(x)-v_\infty(x)\big)
\Phi(x,\xi)\,dx\,d\xi
\right|
\\ &\qquad\le
\lim_{j\rightarrow+\infty}
\|v_j-v_\infty\|_{L^\infty(\RR^n)} \|\Phi\|_{L^1(\RR^{2n})}=0\,,
\end{split}
\end{equation*}
since $v_j\to v_\infty$ uniformly in $\RR^n$ as $j\to +\infty$\,.

This and~\eqref{Ri} imply the assertion.
\end{proof}

Notice that here, in order to check assumption~\eqref{convA}
(which follows from Corollary~\ref{C2}), we do not take advantage from the
fact
that $u$ is a solution of the variational inequality \eqref{varineqK} and that $u_\eta$ solves
the approximated equation~\eqref{problemakernel}\,.

\medskip

\begin{Proofgeneral}
Let $u$ be a solution of the variational inequality~\eqref{varineqK}.
In the setting of Theorem~\ref{lap fra general kernel} we can apply
Theorem~\ref{propLS}, so that we get
\begin{equation}\label{pippo}
\begin{aligned}
& 0\leq -\int_Q (u(x)-u(y))(\varphi(x)-\varphi(y))K(x-y)\,dx\,dy+\int_\Omega f(x)\varphi(x)dx\\
& \qquad \qquad \qquad \qquad \qquad  \leq
\int_\Omega \Big(\mathcal L_K\, \psi+f\Big)^+(x)\varphi(x)
dx
\end{aligned}
\end{equation}
for any $\varphi\in C^\infty_0(\Omega)$ with
$\varphi\ge 0$ in $\Omega$ and $\varphi=0$ in $\mathcal C \Omega$.
\end{Proofgeneral}

\subsection{The fractional Laplacian operator.}
As an application of Theorem~\ref{lap fra general
kernel}, now we consider the case of the fractional Laplace kernel,
i.e. the case when
\begin{equation}\label{5.34bis}
K(x)=|x|^{-(n+2s)}\quad x\in \RR^n\setminus\{0\}\,,
\end{equation}
with~$s\in(0,1)$, $n\in\NN$, $n>2s$.

\begin{Prooflapfra}
By~\eqref{pippo} and~\eqref{5.34bis},
$$\begin{aligned}
0\leq-\int_Q & \frac{(u(x)-u(y))(\varphi(x)-\varphi(y))}{|x-y|^{n+2s}}\,dx\,dy+\int_\Omega
f(x)\varphi(x)dx\\
&  \qquad \qquad \qquad \qquad \leq\int_\Omega \Big(-(-\Delta)^s\,
\psi+f\Big)^+(x)\varphi(x) dx
\end{aligned}$$
for any $\varphi\in
C^\infty_0(\Omega)$ with $\varphi\ge 0$ in $\Omega$ and $\varphi=0$
in $\mathcal C \Omega$. Note that, using the definition of $H^s_0(\Omega)$ and
a density argument, we get that the estimate holds true for any
$\varphi\in H^s_0(\Omega)$ with $\varphi\ge 0$ a.e. in $\Omega$ and
$\varphi=0$ a.e. in $\mathcal C \Omega$.
\end{Prooflapfra}

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\end{thebibliography}



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