\documentclass[12pt]{article}
\usepackage{epsfig,latexsym,amsfonts,amsmath,amssymb}
%\catcode`\@=11
%\@addtoreset{equation}{section}
\def\thesection{\arabic{section}}
\def\thesubsection{\thesection.\arabic{subsection}}
\def\theequation{\thesection.\arabic{equation}}
%\catcode`\@=12
\usepackage{colortbl}%
\usepackage[margin=3cm]{geometry}


\usepackage[notref,notcite]{showkeys}
%\usepackage{showkeys}



\def\R{\mathbb{R}}
\def\Rbuco{{R^n\!\setminus\!\{0\}}}

\def\C{\mathcal{C}}
\def\S{\mathbb{S}}
\def\f{\varphi}
\def\ep{\varepsilon}
\def\ep{\varepsilon}
\def\sstartar{{\theta^{*\!*}}}
\def\starstar{{2^{*\!*}}}
\def\thetastar{{\theta^{*\!*}}}
\def\uu{\underline u}
\def\div{{\rm div}}
\def\weak{{\,\rightharpoonup\,}}
\def\irn{\int\limits_{\R^n}}


\def\sstar{{2_s^*}}
\def\sstars{{2_s^*}}

\def\red{\textcolor{red}}



%%%%%%%%%% FRACTIONAL SOBOLEV SPACES


%\def\Ds{\left(-\Delta\right)^{\!s}\!_{\!\!\sharp}}
\def\DrD{\left(-\Delta\right)^{\!m}_{rD}}
\def\DshalfD{\left(-\Delta\right)^{\!\frac{s}{2}}}


\def\DsD{\left(-\Delta\right)^{\!m}_{sD}}
\def\DshalfN{\left(-\Delta_\Omega\right)^{\!\frac{s}{2}}}

\def\DsNeps{\left(-\Delta_{\Omega_h}\right)^{\!s}\!}
\def\DshalfNeps{\left(-\Delta_{\Omega_h}\right)^{\!\frac{s}{2}}}
\def\DsNepsh{(-\Delta_{\Omega_h})^{s}}

\def\DsNtilde{\left(-\Delta_{\widetilde\Omega}\right)^{\!s}\!}

\def\eps{\varepsilon}


%%%%% SPACES
\def\SoOm{{H^s({\Omega})}}
\def\SoOmNull{H^s_0({\Omega})}
\def\tildeSobolev{\widetilde{H}^s({\Omega})}
\def\SoOmsi{{H^\sigma({\Omega})}}
\def\SoOmNullsi{H^\sigma_0({\Omega})}
\def\tildeSobolevsi{\widetilde{H}^\sigma({\Omega})}
%\def\tildeSobolevN{\widetilde{H}^s_N({\Omega})}
\def\SoOmmin{{H^{-\sigma}({\Omega})}}
\def\tildeSobolevmin{\widetilde{H}^{-\sigma}({\Omega})}
\def\SoRn{H^s(\mathbb R^n)}
%
\def\what{{\cal F}}



%%%%%%%%%%

\def\proof{\noindent{\textbf{Proof. }}}
\def\QED{\hfill {$\square$}\goodbreak \medskip}

\newtheorem{Theorem}{Theorem}[section]
\newtheorem{Lemma}[Theorem]{Lemma}
\newtheorem{Proposition}[Theorem]{Proposition}
\newtheorem{Corollary}[Theorem]{Corollary}
\newtheorem{Remark}[Theorem]{Remark}
\newtheorem{Definition}[Theorem]{Definition}
\newtheorem{Example}{Example}


%\font\sc=cmcsc9
\linespread{1.2}
%\textwidth=14truecm
\hoffset=-.5truecm




\begin{document}


\title %{\vspace{-10mm}
{Remark on fractional Laplacians}


\author%{Roberta Musina\footnote{Dipartimento di Matematica ed Informatica, Universit\`a di Udine,
%via delle Scienze, 206 -- 33100 Udine, Italy. Email: {roberta.musina@uniud.it}. 
%{Partially supported by Miur-PRIN 201274FYK7\_004.}}~ and
{Alexander I. Nazarov\footnote{
St.Petersburg Department of Steklov Institute, Fontanka 27, St.Petersburg, 191023, Russia, 
and St.Petersburg State University, 
Universitetskii pr. 28, St.Petersburg, 198504, Russia. E-mail: al.il.nazarov@gmail.com.
Supported by RFBR grant 14-01-00534.}
}
%\date{}
%


\maketitle

\begin{abstract}
\footnotesize
We compare norms of $u$ and $|u|$ generated by fractional Lapalcian
(restricted or spectral) of order $s\in(1,3/2)$, in a bounded domain $\Omega\subset\R^n$.
\medskip



%\noindent
%\textbf{Keywords:} {Sobolev inequality,.}
%\medskip
%
%\noindent
%\textit{2010 Mathematics Subject Classification:} {.}
\end{abstract}

\normalsize

Let $\Omega\subset\R^n$ be a bounded Lipschitz domain. 
Recall that the spectral fractional Dirichlet Laplacian (or the ``Navier'' fractional Laplacian) of order $m\in\mathbb{R}_+$ on $\Omega$ is given by the formula
$$
\DsD\!u=\sum_{j=1}^\infty \lambda_j^m~\Big(\int\limits_\Omega u\f_j\Big)\f_j~\!,
$$
where $\lambda_j$ and $\f_j$, $j\ge 1$, are, respectively, the eigenvalues 
and (orthonormal in $L_2(\Omega)$) eigenfunctions of the conventional Dirichlet Laplacian $-\Delta_D$ in $\Omega$. Here the series converges in the sence of 
distributions. 

 It is well known (see, e.g., classical monograph \cite{Tr}), that for $0<m<3/2$ the domain of the corresponding quadratic form 
$Q_m^{sD}[u]=\langle\DsD u,u\rangle$ is the space
$$
{\widetilde H^m(\Omega)}:=\{u\in {H^m(\R^n)}~|~u\equiv 0~~\text{on}~~\R^n\setminus\overline\Omega\}.
$$

Next, the restricted fractional Dirichlet Laplacian of order $m\in\mathbb{R}_+$ is defined via the Fourier transform by
$$
\mathcal F[\DrD u](\xi)=|\xi|^{2m}\mathcal F[u](\xi)= \frac{|\xi|^{2m}}{(2\pi)^{n/2}}\irn e^{-i~\!\!\xi\cdot x}u(x)~\!dx~\!.
$$
The domain of its quadratic form $Q_m^{rD}[u]=\langle\DrD u,u\rangle$ is also the space $\widetilde H^m(\Omega)$ (for all $m\in\mathbb{R}_+$).

From a general result of \cite{BM} it follows that for $0<m<3/2$ the nonlinear operator $u\mapsto |u|$ maps $\widetilde H^m(\Omega)$ into
itself. So, we can try to compare $Q_m^{sD}[|u|]$ and $Q_m^{sD}[u]$ ($Q_m^{rD}[|u|]$ and $Q_m^{rD}[u]$) for $u\in\widetilde H^m(\Omega)$.

Obviously, for $m=1$ we have
$$
Q_1^{sD}[|u|]\equiv Q_1^{sD}[u]\equiv Q_1^{rD}[|u|]\equiv Q_1^{rD}[u]\equiv\int\limits_\Omega |\nabla u|^2~,\qquad u\in\widetilde H^1(\Omega)=H^1_0(\Omega).
$$
In contrast, %for $0<m<1$ 
the following result was established in \cite[Theorem 3]{MN-Mit}:\medskip

{\it Let $0<m<1$. For any $u\in\widetilde H^m(\Omega)$ such that $u\ne|u|$ (i.e. $u$ changes sign) we have 
$Q_m^{sD}[|u|]<Q_m^{sD}[u]$ and $Q_m^{rD}[|u|]<Q_m^{rD}[u]$.}\medskip

Here we show that for $1<m<3/2$ the reverse inequalities hold.


\section{Restricted operators}

{\bf Lemma 1}. {\it Let $u, v\in {\cal C}^\infty_0$ be nonnegative functions, and ${\rm supp}(u)\cap{\rm supp}(v)=\emptyset$. Then for $m>0$, $m\notin\mathbb{N}$ 
we have}
\begin{eqnarray*}
\langle\DrD u,v\rangle<0, & {\rm if} & \lfloor m\rfloor \mbox{\rm\quad is\quad even};
%\label{corol1}
\\
\langle\DrD u,v\rangle>0, & {\rm if} & \lfloor m\rfloor \mbox{\rm\quad is\quad odd}.
%\label{corol2}
\end{eqnarray*}

\proof
We write
\begin{eqnarray*}
&&\langle\DrD u,v\rangle=\langle\left(-\Delta\right)^{\!m-\lfloor m\rfloor}_{rD}u,\left(-\Delta_D\right)^{\!\lfloor m\rfloor}v\rangle\\
&&=C_{n,m} \cdot \int\limits_{\Omega} V.P.\int\limits_{\R^n}\frac{u(x)-u(y)}{|x-y|^{n+2(m-\lfloor m\rfloor)}}~\!dy\,
\left(-\Delta_D\right)^{\!\lfloor m\rfloor}v(x)~\!dx,
\end{eqnarray*}
Here $V.P.$ stands for the principal value of the integral
while $C_{n,m}=2^{2m+\frac n2}\,\frac {\Gamma(m+\frac{n}2)}{\Gamma(-m)}$.

Since the supports of $u$ and $v$ are separated, we have
$$
\langle\DrD u,v\rangle=-C_{n,m} \cdot \int\limits_{\Omega}\int\limits_{\Omega}\frac{u(y)\,\left(-\Delta_D\right)^{\!\lfloor m\rfloor}v(x)}
{|x-y|^{n+2(m-\lfloor m\rfloor)}}~\!dydx.
$$
We integrate by parts and use the relation $-\Delta_D |x|^{-(n+a)}=-(n+a)(a+2)|x|^{-(n+a+2)}$, and the statement follows.
\QED

{\bf Theorem 1}. {\it Let $m>0$, $m\notin\mathbb{N}$. Suppose that $u$ is such that $u_+, u_- \in\widetilde H^m(\Omega)$, $u_{\pm}\ne0$. Then}
\begin{eqnarray*}
Q_m^{rD}[|u|]<Q_m^{rD}[u], & {\rm if} & \lfloor m\rfloor \mbox{\rm\quad is\quad even};
%\label{corol1}
\\
Q_m^{rD}[|u|]>Q_m^{rD}[u], & {\rm if} & \lfloor m\rfloor \mbox{\rm\quad is\quad odd}.
%\label{corol2}
\end{eqnarray*}

\proof
 For $u_+, u_-\in {\cal C}^\infty_0$ with separated supports, the statement follows from Lemma and relations $u=u_+-u_-$, $|u|=u_++u_-$. In general case we proceed 
 by approximation.
 \QED

 {\bf Corollary}. {\it Let $1<m<3/2$. Then $Q_m^{rD}[|u|]>Q_m^{rD}[u]$ for all $u\in\widetilde H^m(\Omega)$ such that $u\ne|u|$.}\medskip

\proof
 This statement  follows from \cite{BM} and Theorem 1.
 \QED

 
\section{Spectral operators}


{\bf Lemma 2}. {\it Let $m<0$ or $m\in(1,3/2)$. Then $Q_m^{sD}[|\f_k|]>Q_m^{sD}[\f_k]$ for any $k\ge2$.
}
\medskip

\proof
Denote by $a_j=\int\limits_{\Omega}|\f_k|\f_j$ the Fourier coefficients of the function $|\f_k|$. Then for arbitrary $m<3/2$
 $$
 Q_m^{sD}[|\f_k|]=\sum\limits_{j=1}^{\infty}\lambda_j^m\,|a_j|^2;\qquad  Q_m^{sD}[\f_k]=\lambda_k^m.
 $$
It is well known that $\f_k$ changes sign. Hence $a_j\ne0$ for some $j\ne k$, and therefore 
 $$
 \frac {d^2}{dm^2}\left(\frac { Q_m^{sD}[|\f_k|]}{ Q_m^{sD}[\f_k]}\right)=
 \sum\limits_{j=0}^{\infty}\left(\frac {\lambda_j}{\lambda_k}\right)^m|a_j|^2\ln^2\left(\frac {\lambda_j}{\lambda_k}\right)>0,
 $$
 Thus, the quotient in the left-hand side is strictly convex in $m$. Since 
 $Q_0^{sD}[|\f_k|]=Q_0^{sD}[\f_k]$ and $Q_1^{sD}[|\f_k|]=Q_1^{sD}[\f_k]$, the statement follows.
\QED


{\bf Theorem 2}. {\it Let $1<m<3/2$. Then $Q_m^{sD}[|u|]>Q_m^{sD}[u]$ for all $u\in\widetilde H^m(\Omega)$ such that $u\ne|u|$.}\medskip

\proof
First, let $u=b_1\f_1+b_i\f_i+b_k\f_k$. Denote by $a_j$ the Fourier coefficients of the function $|u|$. Then
 $$
 Q_m^{sD}[|u|]=\sum\limits_{j=1}^{\infty}\lambda_j^m\,|a_j|^2;\qquad  Q_m^{sD}[u]=\lambda_1^m\,|b_1|^2+\lambda_i^m\,|b_i|^2+\lambda_k^m\,|b_k|^2.
 $$
 Note that $a_1>|b_1|$ since $u\ne|u|$. Suppose that $|b_i|> |a_i|$ and $|b_k|> |a_k|$ (other possibilities only simplify the proof).
 Then we can choose $c'_j, c''_j\ge0$ such that
 $$
 c'_1+c''_1=|a_1|^2-|b_1|^2;\qquad c'_i=c''_i=c'_k=c''_k=0;\qquad  c'_j+c''_j=|a_j|^2,\quad j\ne 1,i,k; 
 $$
 $$
 \sum\limits_{j=1}^{\infty}c'_j=|b_i|^2-|a_i|^2;\qquad  \sum\limits_{j=1}^{\infty}c''_j=|b_k|^2-|a_k|^2;
 $$
 $$
 \sum\limits_{j=1}^{\infty}\lambda_jc'_j=\lambda_i(|b_i|^2-|a_i|^2);\qquad  \sum\limits_{j=1}^{\infty}c''_j=\lambda_k(|b_k|^2-|a_k|^2).
 $$
 By the same argument as in previous Lemma, we have for $1<m<3/2$
 $$
 \sum\limits_{j=1}^{\infty}\lambda_j^mc'_j>\lambda_i^m(|b_i|^2-|a_i|^2);\qquad  \sum\limits_{j=1}^{\infty}\lambda_j^mc''_j>\lambda_k^m(|b_k|^2-|a_k|^2),
 $$
 and thus  $Q_m^{sD}[|u|]>Q_m^{sD}[u]$.
 
 For arbitrary finite linear combination of eigenfunctions the arguments are the same. In general case we proceed 
 by approximation.
\QED


\begin{thebibliography}{AB}


\bibitem{BM} 
G. Bourdaud and Y. Meyer, {Fonctions qui op\`erent sur les espaces de Sobolev}, J. Funct.
Anal. {\bf 97} (1991), 351--360.

\bibitem{MN-Mit} R. Musina\ and\ A. I. Nazarov, 
On the Sobolev and Hardy constants for the fractional Navier Laplacian, Nonlinear Analysis -- TMA,
{\bf 121} (2015), 123--129.

\bibitem{Tr}
H. Triebel, {Interpolation theory, function spaces, differential operators}, Deutscher Verlag Wissensch., Berlin, 1978. 

\end{thebibliography}











\end{document}
\section{Introduction}
%\red{Maximum principle in Cappella, Davila, Dupaigne, Sire, lemma 2.4}\\
Let $\Omega$ be a bounded and Lipschitz  domain in $\R^n$, $n\ge 1$. 
Given $s\in(0,1)$, a measurable function $\psi$ on $\Omega$ and $f\in \widetilde H^s(\Omega)'$, 
we consider the 
variational inequality
\begin{equation*}\tag{$\mathcal P_\Omega(\psi,f)$}
\label{eq:vi}
u\in K^s_\psi~,\qquad
\langle \DsD  u-f, v-u\rangle\ge 0\quad\forall v\in K^s_\psi~\!,
\end{equation*}
where $\DsD$ is the {\em Navier}~\!
Laplacian, that is the $s$-th power of the standard Laplacian in the sense of spectral theory,
and
$$
%\begin{equation*}\tag{A}
K^s_\psi=\left\{v\in{\widetilde H^s(\Omega)}~|~v\ge \psi ~\textit{a.e. on ${{\Omega}}$}~\right\}
%\end{equation*}
$$
(see below for notations and main definitions). 
We will always assume that the closed and convex set $K^s_\psi$
is not empty, also when not explicitly stated. 


Problem \ref{eq:vi} admits a unique solution $u$.
More precisely, by standard arguments $u$ is the unique solution to the equivalent minimization problem
\begin{equation}
\label{eq:minimization}
\inf_{v\in K^s_\psi}~\frac{1}{2}\langle \DsD  v,v\rangle-\langle f, v\rangle~\!.
\end{equation} 
In addition,  \ref{eq:vi} is  related to the free boundary problem
\begin{equation}
{\label{eq:problem}}
\begin{cases}
u\ge \psi&\text{in ${\Omega}$}\\
\DsD u\ge f&\text{in ${\Omega}$}\\
\DsD u= f&\text{in $\{u>\psi\}$}\\
%u\in{H^s(\R^n)}~,~~
u=  0&\text{in  $\R^n\setminus\overline\Omega$,}
\end{cases}
\end{equation}





\red{blabla}


\bigskip


We remark that 
our results
plainly cover non-homogeneous Dirichlet's problems of the form
\begin{equation}
\label{eq:g}
\begin{cases}
u\ge \psi&\text{in ${\Omega}$}\\
\DsD u\ge f&\text{in ${\Omega}$}\\
\DsD u= f&\text{in $\{u>\psi\}$}\\
%u\in{H^s(\R^n)}~,~~
u=  g&\text{in  $\R^n\setminus\overline\Omega$,}
\end{cases}
\end{equation}
for suitable boundary datum $g\in H^s(\R^n)$.  Notice indeed that $u$ solves (\ref{eq:g}) if and only if
$u-g$ solves $\mathcal P_\Omega(\psi-g,f+\DsD g)$.

Free boundary problems for the operator $\DsD u+u$
can be considered as well, with minor modifications in the statements and in the proofs.

\section{Preliminaries}
\label{S:preliminaries}
For a bounded and Lipschitz domain  $\Omega\subset\R^n$ we denote by $-\Delta_\Omega$ the Laplace operator
acting on $L^2(\Omega)$, that is
$$
\langle-\Delta_\Omega u,\f\rangle=\int\limits_\Omega u(-\Delta\f)~\!dx\quad u\in L^2(\Omega), ~\f\in C^\infty_0(\Omega).
$$
The domain of the corresponding quadratic
form $u\mapsto \langle-\Delta_\Omega u,u\rangle$ is the space
$H^1_0(\Omega)$. It turns out that $H^1_0(\Omega)$ is the completion of $C^\infty_0(\Omega)$ with respect to the Hilbertian norm
$\|\cdot\|^2=\|\nabla \cdot\|^2_2=\langle-\Delta_\Omega~\!\cdot~\!,\cdot\rangle$.

The eigenvalues of $-\Delta_\Omega$ on $H^1_0(\Omega)$ can be arranged in a non-decreasing
and unbounded sequence $\lambda_j$, $j\ge 1$, where each eigenvalue is repeated accordingly to its
multiplicity. We fix an orthonormal basis $\f_j$, $j\ge 1$, satisfying
$$
\f_j\in H^1_0(\Omega)~,\quad -\Delta\f_j=\lambda_j\f_j~,\quad\|\f_j\|_2=1.
$$
For $u\in H^1_0(\Omega)$ we have
\begin{equation}
\label{eq:H_basis}
u=\sum_{j=1}^\infty \Big(\int\limits_\Omega u\f_j\Big)\f_j~,\quad -\Delta_\Omega u=\sum_{j=1}^\infty \lambda_j~\!\Big(\int\limits_\Omega u\f_j\Big)\f_j,
\end{equation}
where the first limit is taken in $H^1_0(\Omega)$, while the second one has to be intended on the
sense of distributions. Thus
$$
\|u\|_2^2=\sum_{j=1}^\infty \Big(\int\limits_\Omega u\f_j\Big)^2~,\quad \langle-\Delta_\Omega u,u\rangle= 
\sum_{j=1}^\infty \lambda_j\Big(\int\limits_\Omega u\f_j\Big)^2~\!.
$$


Next, take $s\in(0,1)$. The ``Navier'' (or spectral) fractional Laplacian of order $s$ on $\Omega$ is the distribution 
$$
\DsD\!u=\sum_{j=1}^\infty \lambda_j^s~\Big(\int\limits_\Omega u\f_j\Big)\f_j~\!.
$$
 It is known that
$$
{\widetilde H^s(\Omega)}:=\{u\in {H^s(\R^n)}~|~u\equiv 0~~\text{on}~~\R^n\setminus\overline\Omega\}
$$
is the domain of the corresponding quadratic form $u\mapsto\langle\DsD u,u\rangle$, see for instance \cite[Lemma 1]{FL}.
The standard reference for the Sobolev space $H^s(\R^n)$ is the monograph
\cite{Tr} by Triebel. We endow $\widetilde H^s(\Omega)$ with
the Hilbertian norm
$$
\|u\|_{\widetilde H^s(\Omega)}^2=\sum_{j=1}^\infty\lambda_j^s\Big(\int\limits_\Omega u\f_j\Big)^2_2= \langle\DsD u,u\rangle=
\|\DshalfN u\|_2^2~\!.
$$
Notice that $\lambda_j^s$ is the $j$-th eigenvalue of $\DsD$. That is, $\DsD$ is the $s$-th power of $-\Delta_\Omega$ 
in the sense of spectral theory.

\red{scrivere qualcosa del Dirichlet nell'introduzione; la definizione nella sezione in cui si paragonano i 2
problemi}\\
For any function $u\in H^s(\R^n)\supset \widetilde H^s(\Omega)$ we can compute the ''Dirichlet'' (or restricted) Laplacian $\DrD u$, that is defined via the
Fourier transform by
$$
\mathcal F[\DrD u](\xi)=|\xi|^{2s}\mathcal F[u](\xi)= \frac{|\xi|^{2s}}{(2\pi)^{n/2}}\irn e^{-i~\!\!\xi\cdot x}u(x)~\!dx~\!.
$$
By \cite[Corollary 1]{FL}, an equivalent norm in $\widetilde H^s(\Omega)$ is given by 
$$
\|u\|^2=
\int\limits_{\R^n}|\DshalfD u|^2~\!dx=\int\limits_{\R^n}|\xi|^{2s}|\mathcal F[u]|^2~\!d\xi~\!.
$$



\subsection{The extension argument}
We recall here some basic facts from \cite{ST}. For $u\in\widetilde H^s(\Omega)$ we put
\begin{gather*}
{\mathcal E}(w)=\!\!\int\limits_0^\infty\!\int\limits_{\R^n} y^{1-2s}|\nabla w|^2\,dxdy~,\\
{\cal W}^\Omega_u=\Big\{w(x,y)~\!|~\!
{\mathcal E}(w)<\infty~,\ \ w\big|_{y=0}=u~,~w(\cdot, y)\equiv 0~~\text{on $\R^n\setminus\overline\Omega$}\Big\}.
\end{gather*}
The minimization problem
\begin{equation*}\tag{$\mathcal M^\Omega_u$}
\label{eq:CSmini}
\inf_{w\in{\cal W}^\Omega_u} {\mathcal E}(w)
\end{equation*}
has a unique solution $w^\Omega_u:\R^n\times\R_+~\longrightarrow~\R$, that solves the Dirichlet problem
\begin{equation}
\label{eq:ST}
\begin{cases}
-\div (y^{1-2s}\nabla w)=0\qquad\text{in $\Omega\times\mathbb R_+$}\\
w(\cdot,y)\equiv 0\quad\text{on $(\R^n\setminus\overline\Omega)\times\R_+$},~
w\big|_{y=0}=u.
\end{cases}
\end{equation}
The  results  in   \cite[Theorem 1.1]{ST} (see also Section 2 there-in), and integration by parts imply that
\begin{equation}
\label{eq:ST_energy}
%\int\limits_{\Omega}|\DshalfN u|^2~\!dx 
\|\DshalfN\!u\|_2^2=  {c_s}~\! {\mathcal E}(w^\Omega_u)~\!,
\end{equation}
for an explicitly known constant $c_s>0$. In addition, we have that
\begin{equation}
\label{eq:neumann}
\DsD u(x)=-c_s~\!\lim\limits_{y\to0^+} y^{1-2s}\partial_yw_u^\Omega(x,y)=
-(2sc_s)\lim\limits_{y\to0^+} \frac{w_u^\Omega(x,y)-u(x)}{y^{2s}},
\end{equation}
where the limits have to be intended in the sense of distributions.   %\red{is the proof below needed?}

Next, notice that trivially $u$ achieves the infimum
$$
\inf_{v\in  \widetilde H^s(\Omega)} %\int\limits_\Omega|\DshalfN v|^2~\!dx
\|\DshalfN\!v\|_2^2-2\langle\DsD u, v\rangle=- %\int\limits_\Omega|\DshalfN u|^2~\!,
\|\DshalfN\!u\|_2^2
$$
so that $w^\Omega_u$ achieves
$$
\inf_{{\mathcal E}(w)<\infty\atop\scriptstyle w(\cdot,y)\equiv 0~ \text{on}~ \R^n\setminus\overline\Omega}
\frac{c_s}2{\mathcal E}(w)-\langle\DsD u, w\big|_{y=0}\rangle.
$$
In particular for any function $w$ on $\R^n\times\R_+$, such that $w\big|_{y=0}\in\widetilde H^s(\Omega)$ and $\mathcal E(w)<\infty$,
one has that
\begin{equation}
\label{eq:w_variational}
c_s \int\limits_0^\infty\!\int\limits_{\R^n} y^{1-2s}\nabla w^\Omega_u\cdot\nabla w\,dxdy=
\langle\DsD u, w\big|_{y=0}\rangle~\!.
\end{equation}


\subsection{Dependence of the Navier Laplacian on the domain}
\label{SS:ST}

%  For the next lemma, see \red{Theorem 1 in \cite{FL}, see also [FL2]}.\\  I DISAGREE

Let $\Omega,\tilde\Omega $ be bounded and Lipschitz domains in $\R^n$, with $\Omega\subseteq\widetilde\Omega$. 
Since ${\cal W}^\Omega_u\subseteq {\cal W}^{\widetilde\Omega}_u$, the Stinga-Torrea extension argument readly gives
\begin{equation}
\label{eq:monotonicity}
\langle\DsNtilde u, u\rangle\le\langle\DsD u, u\rangle\quad\text{for any $u\in\widetilde H^s(\Omega)\hookrightarrow \widetilde H^s(\widetilde\Omega)$}.
\end{equation}
A related result is stated in the next lemma.

\begin{Lemma}
\label{L:lemma2}
Let $\Omega,\tilde\Omega $ be bounded and Lipschitz domains in $\R^n$, with $\Omega\subseteq\widetilde\Omega$.
Let $u\in \widetilde H^s(\Omega)$, $u\ge 0$. Then
$\DsNtilde u\le \DsD u$
in the distributional sense on $\Omega$.
\end{Lemma}

\proof
We follow the proof of Theorem 1 in \cite{FL}.
{Consider the Stinga-Torrea extensions $w^\Omega_u, w^{\widetilde\Omega}_u$ of $u$.
Hence $w^\Omega_u$ solves the minimization problem $\mathcal M^\Omega_u$ and the 
Euler-Lagrange equations (\ref{eq:ST}); the function $w^{\widetilde\Omega}_u$ solves $\mathcal M^{\widetilde\Omega}_u$
and appropriate Euler-Lagrange equations on $\widetilde\Omega\times \R_+$. Since $u$ is nonnegative
then $w^{\widetilde\Omega}_u$
is nonnegative by the maximum principle. Put
$W=w^{\widetilde\Omega}_u-w^{\Omega}_u$. Then $W$ solves
$$
\begin{cases}
-\div (y^{1-2s}\nabla W)=0\qquad\text{in $\Omega\times\mathbb R_+$}\\
W(\cdot,y)\ge 0\quad\text{on $(\R^n\setminus\overline\Omega)\times\R_+$}~,\quad
W\big|_{y=0}=0.
\end{cases}
$$
Thus the maximum principle gives $W\ge 0$ on $\Omega\times\mathbb R_+$ and hence
$$
0\le 2s\lim_{y\to 0^+}\frac{W(x,y)-W(x,0)}{y^{2s}}=%\lim_{y\to 0^+}y^{1-2s}\partial_y W=
\lim_{y\to 0^+}y^{1-2s}\big(\partial_yw^{\widetilde\Omega}_u-
\partial_yw^{\Omega}_u\big)~,\quad x\in\Omega.
$$
The conclusion readily follows from (\ref{eq:neumann}).
\QED}

\begin{Remark}
If $u\neq 0$ and $\overline\Omega\subset \widetilde\Omega$, the Hopf boundary point 
\red{Lemma in \cite{ABMMZ}} gives the strict inequality.
\end{Remark}

Now, for any integer $h\ge 1$, let $\Omega_h$ be a Lipschitz domain, such that 
$$
\overline\Omega\subset \Omega_h\subseteq\Big\{ x\in\R^n~|~d(x,\Omega)<\frac1h\Big\}. %WHY COMPACTLY CONTAINED?
$$
It is convenient to regard at $\widetilde H^s(\Omega)$, $s\in(0,1]$, 
as a subspace of $\widetilde H^s(B_R)$, where $B_R$ is an open  ball containing $\overline\Omega$, so that we 
have continuous embeddings
$\widetilde H^s_0(\Omega)\hookrightarrow \widetilde H^s_0(\Omega_h)\hookrightarrow \widetilde H^s_0(B_R)$. In particular,
 $H^1_0(\Omega)\hookrightarrow H^1_0(\Omega_h)\hookrightarrow H^1_0(B_R)$.

Now we recall some facts from \cite{BDM}, see also \cite{BB} and \cite[Theorem 2.3.2]{He}. 
It is easy to show that the domains $\Omega_h$ $\gamma-$converge to $\Omega$ as $h\to\infty$. That is,
for any $f\in L^2(B_R)$, it turns out that $v_{h}\to v$ in $H^1_0(B_R)$,
where the functions $v_{h}\in H^1_0({\Omega_h})$ and $v\in H^1_0(\Omega)$ are defined via
$$
-\Delta_{\Omega_h}\!v_{h}=f\quad\text{in ${\Omega_h}$}~,\quad -\Delta_{\Omega}v=f\quad\text{in $\Omega$.}
$$
Arrange the eigenvalues $\lambda_j^h$ and the eigenvectors $\f_j^h$ of $-\Delta_{\Omega_h}$ on $H^1_0(\Omega_h)$
as described at the beginning of the present section. The $\gamma$-convergence of the domains implies that
\begin{equation}
\label{eq:i)}
\lambda_j^h\to \lambda_j~,\qquad\f_j^h\to \f_j\quad\text{in $H^1_0(B_R)$}
\end{equation}
as $h\to\infty$, where $\lambda_j$ is an eigenvalue of $-\Delta_\Omega$ on $H^1_0(\Omega)$
and $\f_j$ is an eigenfunction of $-\Delta_\Omega$ on $H^1_0(\Omega)$
relative to the eigenvalue $\lambda_j$, see \cite[Example 2.1]{BDM}.
%The $\gamma$-convergence of the domains implies 
%the convergence of the corresponding resolvent operators $R_h:H^{-1}(\Omega_h)\to H^1_0(\Omega_h)$ to the 
%limit operator $R:H^{-1}(\Omega)\to H^1_0(\Omega)$ in the uniform operator topology of $\mathcal L(L^2(B_R))$, 
%see \cite[Example 2.1]{BDM}. 
Finally, the sequence 
$\lambda_j$ fills the whole spectrum of $-\Delta_\Omega$ on $H^1_0(\Omega)$ and  $\{\f_j\}$ is a complete
orthonormal system in $L^2(\Omega)$. In particular, for $u\in \widetilde H^s(\Omega)\hookrightarrow \widetilde H^s(\Omega_h)$ we can write
$$
\DsD u=\sum_{j=1}^\infty (\lambda_j)^s\Big(\int\limits_\Omega u\f_j\Big)\f_j~,\quad 
\DsNeps u=\sum_{j=1}^\infty  (\lambda^h_j)^s\Big(\int\limits_{\Omega_h} u\f^h_j\Big)\f_j^h~\!.
$$
We start to study the behavior of the fractional Laplacian $\DsNeps $ as $h\to \infty$.
 %\begin{gather*}
%u=\sum_{j=1}^\infty \Big(\int\limits_\Omega u\f_j\Big)\f_j=\sum_{j=1}^\infty \Big(\int\limits_\Omega u\f^h_j\Big)\f^h_j~,\\
%\langle\DsD u,u\rangle=\sum_{j=1}^\infty (\lambda_j)^s\sum_{j=1}^\infty \Big(\int\limits_\Omega u\f_j\Big)^2~,\quad 
%\langle\DsNeps u,u\rangle=\sum_{j=1}^\infty  (\lambda^h_j)^s\Big(\int\limits_\Omega u\f^h_j\Big)^2~\!.
%\end{gather*}

%\begin{Lemma}
%\label{L:lemma3}
%Let $\eta\in C^\infty_0(\Omega), \red{\eta\ge 0 ??}$. Then $\DsDeps\eta\to\DsN\eta$ in $L^2(B_R)$.
%\end{Lemma}
%
%\proof
%Using Lemma \ref{L:lemma2}, we can estimate
%$|\DsDeps\eta|\le |\DsN\eta|+|(-\Delta_{B_R}~\!)^{s}\eta|$ on the smallest domain $\Omega$.
%Since $\DsN\eta, (-\Delta_{B_R}~\!)^{s}\eta\in L^2(B_R)$, we only have to prove pointwise convergence. We write 
%$$
%\DsN \eta-\DsNeps\eta=\sum_{j=1}^\infty \Big(~\!(\lambda_j)^s a_j\f_j-(\lambda^h_j)^s a_j^h\f^h_j~\!\Big)\quad\text{in $\Omega$}~\!,
%$$
%where $\displaystyle{a_j=\displaystyle\int_\Omega u\f_j}$, $\displaystyle{a_j^h=\displaystyle\int_{\Omega_h} u\f_j^h}$.
%For $x\in \Omega$ and $\eps>0$ we can find an integer $m\ge 1$ such that
%$$
%(\DsD\eta-\DsNeps\eta)(x)\le\sum_{j=1}^m \Big(~\!(\lambda_j)^s a_j\f_j(x)-(\lambda^h_j)^s a_j^h\f^h_j(x)~\!\Big)+\eps~\!.
%$$
%Thus 
%$$
%\limsup_{h\to\infty}\left((\DsD\eta-\DsNeps\eta)(x)\right)\le\eps~\!,
%$$
%as $\lambda_j^h\to \lambda_j$, $\f_j^h\to \f_j$ in $L^2(\Omega)$ and $a_j^h\to a_j$ for any index $j\ge 1$.
%Using also Lemma \ref{L:lemma2}, 
%we readily infer that 
%$$
%0\le \liminf_{h\to \infty} (\DsD\eta-\DsNeps\eta)\le \limsup_{h\to \infty} (\DsD\eta-\DsNeps\eta)\le 0
%$$
%a.e. in $\Omega$. Thus $\DsD\eta-\DsNeps\eta\to 0$ a.e., and the proof is complete.
%\QED

\begin{Lemma}
\label{L:lemma3}
Let $\eta\in C^\infty_0(\Omega)$. Then $\DsNeps\eta\to\DsD\eta$ in $L^2(B_R)$.
\end{Lemma}

\proof
To simplify notation we put
$$
\psi_j^h= (\lambda_j)^s \Big(\int\limits_\Omega \eta\f_j\Big)\f_j-(\lambda^h_j)^s \Big(\int\limits_\Omega \eta\f^h_j\Big)\f^h_j~\!.
$$
Notice that $\psi_j^h\to 0$ in $L^2(\Omega)$ by (\ref{eq:i)}). In addition,
$\DsD \eta-\DsNeps\eta=\displaystyle\sum_{j=1}^\infty \psi^h_j$, with convergence in $L^2(\Omega)$. In particular, for any $\eps>0$ we can find an integer $m\ge 1$ such that
$$\|\DsD\eta-\DsNeps\eta\|_2^2\le\displaystyle\sum_{j=1}^m \|\psi^h_j\|_2^2+\eps=o(1)+\eps.$$ The conclusion is immediate.
\QED

The next lemma, of independent interest, will be crucially used in the proof of our regularity results

\begin{Lemma}\label{L:eige_ueps}
Let $u_{h}\in \widetilde H^s(\Omega_h)$ be a bounded sequence
in $\widetilde H^s(\Omega_h)$ such that $u_{h}\to u$  in $L^2(B_R)$. Then
$u\in \widetilde H^s(\Omega)$ and 
$$
i)\quad\displaystyle{\langle\DsD u,u\rangle\le \liminf_{h\to \infty}\langle\DsNepsh u_{h},u_{h}\rangle}~;\qquad
ii)\quad \displaystyle{\lim_{h\to\infty}\langle\DsNeps u,u\rangle=\langle\DsD u,u\rangle}~\!.
$$
\end{Lemma}

\proof
Clearly, $u$ is the weak limit of the sequence $u_h$ in $\widetilde H^s(B_R)$
and $u_h\to u$ almost everywhere. Hence $u\in\widetilde H^s(\Omega)$.  %Indeed, the proof does not need this information.
Next, for any  integer $m\ge 1$ we have that
$$
\langle\DsNepsh u_h,u_h\rangle\ge  \sum_{j=1}^m (\lambda^h_j)^s\Big(\int\limits_\Omega u\f^h_j\Big)^2=
\sum_{j=1}^m \lambda_j^s\Big(\int\limits_\Omega u\f_j\Big)^2+o(1)
$$
by (\ref{eq:i)}). Thus 
$$
\liminf_{h\to\infty}\langle\DsNepsh u_h,u_h\rangle\ge \sum_{j=1}^m \lambda_j^s \Big(\int\limits_\Omega u\f_j\Big)^2~\!.
$$
Taking the limit as $m\to \infty$ we infer
$$
\liminf_{h\to \infty}\langle\DsNeps u,u\rangle\ge \sum_{j=1}^\infty \lambda_j^s\Big(\int\limits_\Omega u\f_j\Big)^2=
\langle\DsD u,u\rangle,
$$
that ends the proof of $i)$. 
The last claim is an immediate consequence of $i)$ and of (\ref{eq:monotonicity}).
%Now we claim that
%\begin{equation}
%\label{eq:u_claim}
%P^\eps_j u\to P_j u\quad\text{in $L^2(B_R)$}
%\end{equation}
%for any index $j\ge 1$ (actually, it is possible to infer the $H^1_0$ convergence). We 
%use $iii)$ in Lemma \ref{L:eige} to get the following strong
%convergences in $H^1_0(B_R)$:
%\begin{equation}
%\label{eq:Pepsu}
%P^\eps_j(P_ju)\to  P_ju~,\quad P^\eps_j(P_ku)\to 0\quad\text{if $k\neq j$}.
%\end{equation}
%Now, for any fixed integer $m>j$ we let $u_m=\displaystyle\sum_{k=1}^m P_k u$ and
%we notice that 
%$$
%P^\eps_ju_m=\sum_{k=1}^m P^\eps_j(P_k u)= P_j u+o(1),
%$$
%with convergence in $L^2(B_R)$. Thus
%$$
%\|P^\eps_j u-P_j u\|_2\le\|P^\eps_j u-P^\eps_ju_m\|_2+\|P^\eps_j u_m-P^\eps_ju\|_2\le
%\|u-u_m\|_2+o(1),
%$$
%that implies
%$$
%\limsup_{h\to\infty}\|P^\eps_j u-P_j u\|_2\le
%\|u-u_m\|_2
%$$
%for any index $m>j$. Claim (\ref{eq:u_claim}) follows by recalling that $\|u-u_m\|_2\to 0$
%as $m\to \infty$.
%
%Now, for any large integer $m\ge 1$ we have that
%$$
%\langle\DsNeps u,u\rangle\ge  \sum_{j=1}^m (\lambda^h_j)^s \|P_j u\|^2_2~\to~
%\sum_{j=1}^m \lambda_j^s \|P_j u\|^2_2
%$$
%by $i)$ in Lemma \ref{L:eige} and  (\ref{eq:u_claim}). That implies
%$$
%\liminf_{h\to\infty}\langle\DsNeps u,u\rangle\ge \sum_{j=1}^m \lambda_j^s \|P_j u\|^2_2~\!,
%$$
%and taking the limit as $m\to \infty$ we infer
%$$
%\liminf_{h\to\infty}\langle\DsNeps u,u\rangle\ge \sum_{j=1}^\infty \lambda_j^s\|P_j u\|^2_2=
%\langle\DsD u,u\rangle,
%$$
%that ends the proof.
\QED


\begin{Remark}
Let us introduce the functionals $L^2(B_R)\to \R\cup\{\infty\}$,
$$
Q_s(u)=\begin{cases}
\langle\DsD u,u\rangle&\text{if $u\in \widetilde H^s(\Omega)$}\\
\infty&\text{otherwise;}
\end{cases}
~~
Q^h_s(u)=
\begin{cases}
\langle\DsNeps u,u\rangle&\text{if $u\in \widetilde H^s(\Omega_h)$}\\
\infty&\text{otherwise.}
\end{cases}
$$
Lemma \ref{L:eige_ueps} says that $Q_s=\Gamma$-$\displaystyle{\lim_{h\to\infty} Q_s^h}$.
For $s=1$ the result holds for any sequence of perturbating domains $\Omega_h$ that 
$\gamma$-converges to $\Omega$, see \cite[Theorem 13.12]{DM}. 
Differently from the fractional case $s\in(0,1)$, for $s=1$ the quadratic forms $Q^h_1, Q_1$ coincide on the intersection of
their domains.
\end{Remark}


\subsection{Truncations}
Truncation operators play an important role in studying  obstacle problems. For measurable functions $v,w$ we put
$$
v\vee w=\max\{v,w\}~,\quad v\wedge w=\min\{v,w\}~,\quad v^+=v\vee 0~,\quad v_-=-(v\wedge 0),
$$
so that $v=v^+-v^-$ and $|v|=v^++v^-$.  It is well known that 
$v\vee w\in {H^s(\R^n)}$ and $v\wedge w\in {H^s(\R^n)}$
if $v,w\in {H^s(\R^n)}$. Moreover, if $v\in {\widetilde H^s(\Omega)}$ and  $m\ge 0$ is a constant,
then 
$$(v+m)^-~,\quad  (v-m)^+~,\quad  v\wedge m\in {\widetilde H^s(\Omega)}~\!,
$$
see \cite[Lemma 2.4]{MNS}. 

\begin{Lemma}
\label{L:m_new}
Let $v\in {\widetilde H^s(\Omega)}$ and  $m\ge 0$. 
\begin{itemize}
\item[$i)$]~ If $(v\!+\!m)$ changes sign, \!then
\qquad$\displaystyle{\langle {\DsD} v,(v+m)^-\rangle+%\int\limits_\Omega|\DshalfN (v+m)^-|^2~\!dx< 0}
\|\DshalfN (v+m)^-\|_2^2< 0};$
\item[$ii)$]~ If $(v\!-\!m)$ changes sign, \!then 
\qquad$\displaystyle{\langle {\DsD} v,(v-m)^+\rangle- %\int\limits_\Omega|\DshalfN (v-m)^+|^2~\!dx
\|\DshalfN (v-m)^+\|_2^2> 0};$
\item[$iii)$]~ If~ $(v~\!-~\!m)$  changes sign, then 
\qquad
$
\displaystyle{\|\DshalfN (v\wedge m)\|_2^2<
\|\DshalfN v\|_2^2- \|\DshalfN (v- m)^+\|_2^2}.$
%$$
%\displaystyle{\int\limits_\Omega|\DshalfN (v\wedge m)|^2~\!dx\le 
%\int\limits_\Omega|\DshalfN v|^2~\!dx- \int\limits_\Omega|\DshalfN (v- m)^+|^2~\!dx}.$$
\end{itemize}
\end{Lemma}

\proof
We use again we the Stinga-Torrea extension argument in Subsection \ref{SS:ST}.
%For $u\in\widetilde H^s(\Omega)\subset \widetilde H^s(\Omega_h)$, let $w^\Omega_u$, $w^{\Omega_h}_u$ be the
%solution to the minimum problem $\mathcal M^\Omega_u$, $\mathcal M^{\widetilde\Omega}_u$, respectively. 
Let $w^\Omega_v$ be the solution to 
$\mathcal M^\Omega_v$ and let $w^\Omega_{(v+m)^-}$ be the 
solution to $\mathcal M^\Omega_{(v+m)^-}$. Since ~$(w^\Omega_v+m)^-\in {\cal W}^\Omega_{(v+m)^-}$, we have   
\begin{equation}
\label{eq:CSstrict}
{\mathcal E}((w^\Omega_v+m)^-)\ge{\mathcal E}(w^\Omega_{(v+m)^-})~\!.
%\int\limits_0^\infty\!\int\limits_{\mathbb{R}^n} y^{1-2s}|\nabla (w^\Omega_v+m)^-|^2\,dxdy>
%\int\limits_0^\infty\!\int\limits_{\mathbb{R}^n} y^{1-2s}|\nabla w^\Omega_{(v+m)^-}|^2\,dxdy.
\end{equation}
%%QUI
In addition, from (\ref{eq:w_variational}) we get  
\begin{equation}
\label{eq:WW}
c_s \int\limits_0^\infty\!\int\limits_{\R^n} y^{1-2s}\nabla w^\Omega_v\cdot\nabla (w^\Omega_v+m)^-\,dxdy=
\langle\DsD v, (v+m)^-\Big|_{y=0}\rangle.
\end{equation}
It is well known that $\nabla w^\Omega_v\cdot\nabla (w^\Omega_v+m)^-=-|\nabla (w^\Omega_v+m)^-|^2$ a.e. in $\Omega\times\R_+$.
Thus, (\ref{eq:WW}), (\ref{eq:CSstrict}) and (\ref{eq:ST_energy}) give
\begin{equation}
 \label{eq:lemma}
\langle\DsD v, (v+m)^-\rangle= -c_s{\mathcal E}((w^\Omega_v+m)^-)\le
 -c_s{\mathcal E}(w^\Omega_{(v+m)^-}) \le -\|\DshalfN (v+m)^-\|_2^2~\!,
 \end{equation}
and $i)$ with a large inequality follows. 

Now, assume that $(v+m)^-\neq 0$ and that equality holds in $i)$. We have to show that $v+m$ is nonnegative.
Since equality holds everywhere in (\ref{eq:lemma}), then ${\mathcal E}((w^\Omega_v+m)^-)=
{\mathcal E}(w^\Omega_{(v+m)^-})$. 
In particular $(w^\Omega_v+m)^-=w^\Omega_{(v+m)^-}$, because the minimization problem $\mathcal M^\Omega_{(v+m)^-}$ 
admits a unique solution. Since $w^\Omega_{(v+m)^-}$ solves (\ref{eq:ST}) (with  $u=(v+m)^-\ge 0$),
then the analytic function  $(w^\Omega_v+m)^-$ is positive in $\Omega\times\R_+$ by the maximum principle. That is,
$w^\Omega_v+m<0$ in $\Omega\times\R_+$, hence $w^\Omega_v+m\big|_{\{y=0\}}=v+m\le 0$ on $\Omega$. %(v+m)^+\equiv 0$.
 
To check $ii)$ notice that $(v-m)^+=((-v)+m)^-$ and then use $i)$ with 
$(-v)$ instead of $v$.

Next, we use $v\wedge m=v-(v-m)^+$ and $ii)$ to estimate
\begin{eqnarray*}
\|\DshalfN (v\wedge m)\|_2^2
&=&\|\DshalfN v\|_2^2-2
\langle\DsD v,(v-m)^+\rangle
+ \|\DshalfN (v-m)^+\|_2^2\\
&<&
\|\DshalfN v\|_2^2-
\|\DshalfN (v-m)^+\|_2^2~\!.
\end{eqnarray*}
%\begin{multline*}
%\int\limits_\Omega|\DshalfN (v\wedge m)|^2~\!dx\\
%=\int\limits_\Omega|\DshalfN v|^2~\!dx
%-2
%\langle\DsD v,(v-m)^+\rangle
%+ \int\limits_\Omega|\DshalfN (v-m)^+|^2~\!dx\\
%<
%\int\limits_\Omega|\DshalfN v|^2~\!dx-
%\int\limits_\Omega|\DshalfN (v-m)^+|^2~\!dx~\!.
%\end{multline*}
Thus $ii)$ holds true, and the lemma is completely proved.
\QED

\begin{Remark}
\red{One can improve \cite[Lemma 2.4]{MNS} by using the Caffarelli-Silvestre \cite{CaSi} 
extension argument instead of \cite{ST}, and arguing as in the above proof.}
\end{Remark}


\begin{Remark}
\label{R:truncation}
The ''Navier'' counterpart of  Lemma 2.1 in \cite{MNS} is easily obtained by taking $m=0$ in Lemma \ref{L:m_new}.
That is, for $v\in {\widetilde H^s(\Omega)}$ with $v^+, v^-\neq 0$ the following facts hold:
\begin{description}
\item$~~i)$\quad $\displaystyle{
\langle \DsD  v,v^-\rangle+ \|\DshalfN v^-\|_2^2}< 0$~\!;
%\int\limits_\Omega\DshalfN\!v^-~\!\DshalfN\!v^+~\!dx< 0}$~\!;
\item$~ii)$\quad $\displaystyle{
\langle \DsD  v,v^+\rangle- \|\DshalfN v^+\|_2^2}>0$~\!;
%\int\limits_\Omega\DshalfN\!v~\!\DshalfN\!v^-~\!dx+\int\limits_\Omega|\DshalfN v^-|^2~\!dx<0}$~\!;
\item$iii)$\quad $\displaystyle{
\langle \DsD  v^+,v^-\rangle=\langle \DsD v^-,v^+\rangle< 0}$~\!.
%\int\limits_\Omega\DshalfN\!v~\!\DshalfN\!v^+~\!dx-\int\limits_\Omega|\DshalfN v^+|^2~\!dx>0}$.
\end{description}
\end{Remark}

\begin{Remark}
\label{R:MP}
One can use $ii)$ in Remark \ref{R:truncation} to get the well known
weak maximum principle for $\DsD$. A strong maximum principle was proved
in \cite[Lemma 2.4]{CDDS}. Namely, if $u\in \widetilde H^s(\Omega)\setminus\{0\}$
and $\DsD u\ge 0$ in $\Omega$ then $u$ is positive on every compact set $K\subset\Omega$.
\end{Remark}

\begin{Remark}
\label{R:continuity}
\red{Assume  $v_h\to v$ in $\widetilde H^s(\Omega)$ and $v\le 0$. 
Then $\|\DshalfN v_h^+\|_2\to 0$. Indeed, 
$\|\DshalfN\cdot\|_2$ is a norm on $\widetilde H^s(\Omega)$ that is equivalent to
the norm $\|\DshalfD\cdot\|$ induced by $H^s(\R^n)$,
see \cite[Corollary 1]{FL}. Conclude using \cite[Corollary 2.3]{MNS}. }
%Another proof goes as follows. By $ii)$ in Remark \ref{R:truncation} we have
%$$
%\|\DshalfN v_h^+\|_2^2\le \langle \DsD  v_h,v_h^+\rangle.
%$$
%First, we infer that $v_h^+\to 0$ 
%weakly in $\widetilde H^s(\Omega)$. Thus
%$\langle \DsD  v_h,v_h^+\rangle\to 0$, as $\DsD v_h$ converges  in $\widetilde H^s(\Omega)'$,
%hence  $\|\DshalfN v_h^+\|_2\to 0$.}
\end{Remark}

\section{Equivalent formulations\\ and continuous dependence results}
We start by recalling the notion of (distributional) supersolution.
 
\begin{Definition}
A function ${\cal U}\in\widetilde H^s(\Omega)$ is a supersolution for $\DsD v=f$ if
$$
\langle \DsD{\cal U}-f,\f\rangle\ge  0\qquad\text{for any $\f\in \widetilde H^s(\Omega)$, $\f\ge 0$.}
$$
\end{Definition}
Thanks to the results in the previous section, the arguments in Section 3 of \cite{MNS} 
can be easily adapted to cover problem \ref{eq:vi}. %We limit ourselves to sketch their proofs.
We start by pointing out some equivalent formulations for \ref{eq:vi}. For the proof, argue
as for Theorem 3.2 in \cite{MNS}.

\begin{Theorem} %[Comparison principle]
\label{T:sup}
Let $u\in K^s_\psi$. The following sentences are equivalent.
\begin{itemize}
\item[$a)$] $u$ is the solution to problem \ref{eq:vi};
\item[$b)$] $u$ is the smallest supersolution for $\DsD v=f$ in the convex set $K^s_\psi$. That is, 
${\cal U}\ge u$ almost everywhere in ${\Omega}$,  for any supersolution ${\cal U}\in K^s_\psi$;
\item[$c)$]  $u$ is a supersolution for $\DsD v=f$ and 
$$\displaystyle{\langle\DsD u-f,(v-u)^-\rangle= 0}\qquad\text{for any $v\in K^s_\psi$.}
$$
\item[$d)$] $\displaystyle{\langle\DsD v-f,v-u\rangle\ge 0}$ for any $v\in K^s_\psi$.
\end{itemize}
\end{Theorem}

%\proof
%With no restriction, we take $f=0$ to simplify notation.
%
%\medskip
%
%{$a) \Longleftrightarrow b)$}.
%Assume that $u$ solves \ref{eq:vi}. Let $\f\in \widetilde H^s({\Omega})$ be any nonnegative test function. 
%Testing \ref{eq:vi} with $u+\f\in K^s_\psi$ one easily gets that $u$ is a supersolution. Next,
%let ${\cal U}\in K^s_\psi$ be a supersolution. Since $u-(u-{\cal U})^+={\cal U}\wedge u \in K^s_\psi$,
%from \ref{eq:vi} we obtain%Since $({(u-{\cal U})^+\ge 0$, we have %Add the inequalities
%\begin{eqnarray*}
%0&\ge& \langle \DsD  u,(u-{\cal U})^+\rangle\\ &=&\langle \DsD  (u-{\cal U}),(u-{\cal U})^+\rangle
%+\langle \DsD  {\cal U}),(u-{\cal U})^+\rangle\\
%&\ge& \langle \DsD  (u-{\cal U}),(u-{\cal U})^+\rangle
%\end{eqnarray*}
%by $ii)$ in Remark \ref{R:truncation}. Thus $u-{\cal U})^+=0$, and $b)$ holds.
%The opposite implication is immediate, as there exist at most one ''smallest'' supersolution in $K^s_\psi$.
%\\
%
%\noindent
%$a) \Longleftrightarrow c)$.
%If $u$ solves \ref{eq:vi}, then $u$ is  supersolution since $a) \Rightarrow b)$. Take $v\in K^s_\psi$
%and test \ref{eq:vi} with $u\pm(v-u)^-\in K^s_\psi$ to check that $u$ satisfies $b)$. 
%To prove the opposite implication, call $\tilde u \in K^s_\psi$  the solution to \ref{eq:vi}. 
%Since $\tilde u$ is the smallest supersolution in $K^s_\psi$, then $\tilde u\le u$ and
%$$
%\langle\DsD u,u-\tilde u\rangle=\langle\DsD u,(\tilde u-u)^-\rangle= 0
%$$
%because of $c)$ on $u$. Since $\tilde u$ solves \ref{eq:vi}, we also get
%$\langle\DsD \tilde u,u-\tilde u\rangle\ge  0$. 
%Substracting, we easily infer that $u=\tilde u$. 
%\\
%
%\noindent
%$a) \Longleftrightarrow d)$. Clearly $a)$ implies $d)$ because
%\begin{multline*}
%\langle\DsD v,v-u\rangle\\=\langle\DsD u,v-u\rangle+\langle\DsD (v-u),v-u\rangle\ge
%\langle\DsD u,v-u\rangle~\!.
%\end{multline*}
%To prove the converse, notice that for $u,v\in K^s_\psi$ we have. 
%From $\frac{v+u}{2}\in K^s_\psi$ and $d)$ we obtain
%$$
%4 \langle\DsD \Big(\frac{v+u}{2}\Big),\frac{v+u}{2}-u\rangle=
%\langle\DsD u,u\rangle-\langle f,u\rangle.
%$$
%Then clearly $d)$ implies $a)$ by the equivalence between \ref{eq:vi} and the minimization problem
%(\ref{eq:minimization}).
%\QED

The next corollary is an immediate consequence of $a)\Rightarrow b)$ in Theorem \ref{T:sup}.

\begin{Corollary}\label{compare_f}
Let $f_1, f_2\in \widetilde H^s(\Omega)'$ and let
 $u_i$ be  the solution to $\mathcal P_\Omega(\psi,f_i)$, $i=1,2$.
%\begin{equation*}\tag{$\mathcal P_\Omega_i$}
%u_i\in K^s_{\psi}~,\qquad
%\langle \DsD  u_i- f_i, v-u\rangle\ge 0\qquad\forall v\in K^s_{\psi}~\!, ~~i=1,2,
%\end{equation*}
%respectively. 
If  $f_1\ge f_2$ in the sense of distributions, then $u_1\ge u_2$
a.e. in $\Omega$.
\end{Corollary}

\begin{Remark}
\label{R:uniqueness}
Let  $u\in K^s_\psi$ be such that $\DsD u\ge f$ in $\Omega$. Then
$\DsD u-f$ can be identified with a nonnegative Radon measure on $\Omega$.
Assume that the support of this measure is contained in the coincidence set $\{u=\psi\}$, 
so that $u$ solves the free boundary problem
(\ref{eq:problem}). Let $v\in K^s_\psi$. Since $(v-u)^-$ vanishes on $\{u=\psi\}$, we have
$\langle \DsD u-f,(v-u)^-\rangle=0$. Hence $u$ solves \ref{eq:vi} by Theorem \ref{T:sup}.
\end{Remark}

Now we can state our continuous dependance results. The proof of the
next theorem is totally similar to the proof of Theorem 4.1 in \cite{MNS}, and we omit it.


\begin{Theorem}
\label{T:bounded1}
Let $\psi_1,\psi_2$ be  given  obstacles, $f\in \widetilde H^s(\Omega)'$ and let
 $u_i$ be the  solution to $\mathcal P_\Omega(\psi_i,f)$, $i=1,2$.
%\begin{equation*}\tag{$\mathcal P_\Omega_i$}
%u_i\in K^s_{\psi_i}~,\qquad
%\langle \DsD  u_i- f, v-u_i\rangle\ge 0\qquad\forall v\in K^s_{\psi_i}~\!, ~~i=1,2,
%\end{equation*}
%respectively.
%
If $\psi_1-\psi_2\in L^\infty({\Omega})$, then the difference $u_1-u_2$ is bounded, and
$$
i)~~\|(u_1-u_2)^+\|_\infty\le\|(\psi_1-\psi_2)^+\|_\infty~,\quad
ii)~~\|(u_1-u_2)^-\|_\infty\le\|(\psi_1-\psi_2)^-\|_\infty.
$$
In particular, $\|u_1-u_1\|_\infty\le\|\psi_1-\psi_2\|_\infty$.
\end{Theorem}

%\proof
%Put $m:=\|(\psi_1-\psi_2)^+\|_\infty$. Use $(u_2-u_1+m)^-\in K^s_\psi$ to test $\mathcal P(\psi_1,f)$
%and $u_2+(u_2-u_1+m)^-$ to test $\mathcal P_\Omega(\psi_1,f)$. Then add the two resulting inequalities
%and to get $\langle \DsD  (u_2-u_1),(u_2-u_1+m)^-\rangle \ge 0$. Hence,  $(u_2-u_1+m)^-=0$
%by Lemma \ref{L:m_new}. 
%We  proved that $(u_1-u_2)^+\le m$ a.e. in $\Omega$, hence $i)$ holds. 
%Inequality $ii)$ can be proved in the same way.
%\QED


\begin{Corollary}
\label{C:infty}
Let $\psi\in  L^\infty({\Omega})$ and $f\in L^p({\Omega})$, with $p\in(1,\infty)$, $p>n/2s$.
Let $u\in{\widetilde H^s(\Omega)}$ 
be the unique solution to \ref{eq:vi}.
Then $u\in L^\infty({\Omega})$ and
\begin{equation}
\label{eq:Linf}
\psi\vee\omega_f \le u\le \|\psi^+\|_\infty+c\|f^+\|_p\quad\text{a.e. in $\Omega$,}
\end{equation}
where $\omega_f$ solves (\ref{eq:e}) 
and $c$ depends only on $n,s,p$ and  ${\Omega}$.
In particular, if $f=0$ then 
$$
\psi^+\le u\le \|\psi^+\|_\infty~\!.
$$
\end{Corollary}

\proof 
Notice that $f\in \widetilde H^s(\Omega)'$ by Sobolev embedding theorem.
Since $u$ is supersolution of (\ref{eq:e}), the first inequality in (\ref{eq:Linf}) follows
by the maximum principle in Remark \ref{R:MP}. To prove the right hand side inequality 
in (\ref{eq:Linf}) we introduce the functions 
$\omega^N_{f^+}, \omega^D_{f^+}\in \widetilde H^s(\Omega)$ via
$$
\DsD \omega^N_{f^+}=\DrD \omega^D_{f^+}=f^+\qquad\text{in $\Omega$}.
$$
It has been proved in \cite{MNS}, proof of Corollary 4.2, that $\omega^D_{f^+}\le c\|f^+\|_p$, where
the constant $c>0$  does not depend on $f$. 

Next, 
since $\omega^D_{f^+}\ge 0$ by the maximum principle, then $\DsD \omega^D_{f^+}
\ge \DrD \omega^D_{f^+}$ in $\Omega$ by \cite[Theorem 1]{FL}. Thus $\DsD(\omega^D_{f^+}-\omega^N_{f^+})\ge 0$
in $\Omega$, that implies $\omega^D_{f^+}\ge \omega^N_{f^+}$ by the weak maximum principle in Remark 
\ref{R:MP}. In particular we have that $\omega^N_{f^+}\le c\|f^+\|_p$ a.e. in $\Omega$.

Now let $u_1$ be the unique solution of $\mathcal P_\Omega(\psi,f^+)$. 
We can consider $\omega^N_{f^+}$ as the solution of the problem $\mathcal P_\Omega(\omega_{f^+},f^+)$,
so that 
Theorem \ref{T:bounded1} gives
$$
u\le (u_1-\omega^N_{f^+})^++\omega^N_{f^+}\le \|(\psi-\omega^N_{f^+})^+\|_\infty+\omega^N_{f^+},
$$
and the last inequality in (\ref{eq:Linf}) readily follows.
\QED

To prove the next continuous dependence results, argue as  in \cite{MNS},
proofs of Theorems 4.3 and 4.4, respectively.

\begin{Theorem}
\label{T:Linfty}
Let $\psi_h\in L^\infty(\Omega)$ be  a sequence of obstacles and let $f\in \widetilde H^s(\Omega)'$ be given.
Assume that there exists $v_0\in \widetilde H^s(\Omega)$, such that $v_0\ge \psi_h$ for any $h$.

Denote by  $u_h$ the solution to the obstacle problem $\mathcal P_\Omega(\psi_h,f)$.
%\begin{equation*}\tag{$\mathcal P_\Omega(\psi_h,f)$}
%\label{eq:Ph}
%u_h\in K^s_{\psi_h}~,\qquad
%\langle \DsD  u_h- f, v-u_h\rangle\ge 0\qquad\forall v\in K^s_{\psi_h}~\!.
%\end{equation*}
If $\psi_h\to \psi$ in $L^\infty(\Omega)$, then 
$u_h\to u$ in $\widetilde H^s(\Omega)$, where $u$ is the solution to the limiting problem
\ref{eq:vi}.
\end{Theorem}

%\proof
%We can assume that $f=0$. Using Theorem \ref{T:bounded1}, we have that $u_h\to u$ a.e. in $\Omega$.
%Test $\mathcal P_\Omega(\psi_h,f)$ with $v_0$ to obtain that the sequence $u_h$ is bounded in $\widetilde H^s(\Omega)$,
%hence $u_h\to u$ weakly in $\widetilde H^s(\Omega)$. Next, if $\eps>0$ is small enough, then
%$v_{\eps}:=u+(v_0-u)\wedge {{\eps}}\in K^s_\psi$. 
%Since we can take $v_{\eps}$ as test function in $\mathcal P_\Omega(\psi_h,f)$,  we can estimate 
%\begin{multline*}
%\limsup_{h\to \infty} \|\DshalfN u_h\|^2_2\le
%\limsup_{h\to \infty}\left(
%\langle \DsD  u_h, u_h-v_\eps\rangle+ \langle \DsD  u_h, v_\eps\rangle\right)\\
%\le
%\langle \DsD  u, v_\eps\rangle=
%\|\DshalfN u\|_2^2+ \langle \DsD  u, (v_0-u)\wedge\eps\rangle.
%\end{multline*}
%Notice that $(v_0-u)\wedge\eps\to -(v_0-u)^-$ weakly in $\widetilde H^s(\Omega)$,
%use $iii)$ in Lemma \ref{L:m_new}. Therefore, we get
%$$
%\limsup_{h\to \infty} \|\DshalfN u_h\|^2_2\le
%\|\DshalfN u\|_2^2- \langle \DsD  u, (v_0-u)^-\rangle\le \|\DshalfN u\|_2^2,
%$$
%that is sufficient to conclude that $u_h\to u$ in $\widetilde H^s(\Omega)$.
%\QED


 \begin{Theorem}
 \label{T:Hs2}
Let $\psi_h\in {H^s(\R^n)}$ be  a sequence of obstacles such that $\psi_h^+\in \widetilde H^s(\Omega)$,
and let $f_h$ be a sequence in 
$\widetilde H^s(\Omega)'$.
Assume that
$$
\psi_h\to \psi\quad\text{in $H^s(\R^n)$, and}\quad f_h\to f\quad\text{in $H^s(\R^n)'$}.
$$
Denote by  $u_h$ the solution to the obstacle problem
$\mathcal P_\Omega(\psi_h,f_h)$. 
Then $u_h\to u$ in ${\widetilde H^s(\Omega)}$, where $u$ is the solution of the limiting obstacle problem \ref{eq:vi}.
\end{Theorem}

%\proof
%We can assume that $f_h, f=0$. Otherwise, replace $\psi_h$, $\psi$ with
%$\psi_h-\omega_{f_h}$ and $\psi-\omega_f$, respectively, see (\ref{eq:e}). 
%
%Clearly $u\wedge \psi\in K^s_\psi$ and
%$u\vee \psi_h=u+(\psi_h-u)^+\to u$ in $\widetilde H^s(\Omega)$,
%see Remark \ref{R:continuity}. Thus, from $\mathcal P_\Omega(\psi_h,0)$ we get
%\begin{eqnarray*}
%\langle \DsD  u_h, u_h\rangle&=&
%\langle \DsD  u_h, u_h-u\vee\psi_h\rangle+\langle \DsD  u_h, u\vee\psi_h\rangle\\
%&\le& \langle \DsD  u_h, u\rangle+o(1).
%\end{eqnarray*}
%First, we infer that the sequence $u_h$ is bounded, hence we can assume 
%$u_h\to \tilde u$ in $\widetilde H^s(\Omega)$. We get also
%$\displaystyle\limsup_{h\to \infty}
%\langle \DsD  u_h, u_h\rangle\le \langle \DsD  \tilde u, u\rangle$,
%and in particular $\|\DshalfN\tilde u\|_2\le \|\DshalfN u\|_2$. Since $\tilde u\in K^s_\psi$
%and since $u$ is the unique minimizer for (\ref{eq:minimization}) (with $f=0$), we can
%conclude that $\tilde u=u$. The strong convergence $u_h\to u$
%is readily proved via a semicontinuity argument.
%\QED





\section{Regularity results}
\label{S:regularity1}

Let $u$ be the solution to \ref{eq:vi}. We already noticed that  $\DsD u-f\ge 0$
as a distribution on $\Omega$. Thus, $\DsD u-f$ can be identified with a Radon measure on $\Omega$.
In this Section we obtain estimates on the measure $\DsD u-f$ and
a regularity result. We essentially use the same arguments as in \cite[Theorem 1.1]{MNS}. 
However, the proofs need more care, because of the dependence on the domain of the Navier quadratic form.
The preliminary results in Subsection \ref{SS:ST} will be largely used.

To state the main result in this section we define $\omega_f$ to
be the unique solution to
\begin{equation}
\label{eq:e}
\DsD\omega_f=f\quad\text{in $\Omega$}~,\qquad \omega_f\in \widetilde H^s(\Omega).
\end{equation}



\begin{Theorem}
\label{T:measure}
Assume that
$\psi$, $f\in\widetilde H^s(\Omega)'$ satisfy the following conditions.
\begin{itemize}
\item[$A1)$] $(\psi-\omega_f)^+\in \widetilde H^s(\Omega)$;
\item[$A2)$] $\DsD(\psi-\omega_f)^+-f$ is a  locally finite signed measure on $\Omega$.
\end{itemize}
Let $u\in{\widetilde H^s(\Omega)}$ be the unique solution to \ref{eq:vi}.
Then 
$$
0\le \DsD u-f\le (\DsD(\psi-\omega_f)^+-f)^+\quad\text{in the distributional sense on $\Omega$.}
$$
\end{Theorem}

\proof
We only indicate the main changes with respect to the proof of Theorem 1.1 in \cite{MNS}.
An important tool is the penalty method by Lewy-Stampacchia \cite{LS1}.

Notice that we can assume  $f=0$, 
$\psi\in \widetilde H^s(\Omega)$ and $\psi\ge 0$ in $\Omega$. In fact,
since
$u-\omega_f\in \widetilde H^s(\Omega)$ and $\DsD(u-\omega_f)\ge 0$,
then $u-\omega_f\ge 0$ in $\Omega$ (use the maximum principle in Remark \ref{R:MP}). 
Thus $u-\omega_f\ge \psi\vee\omega_f-\omega_f=(\psi-\omega_f)^+$ and
$u-\omega_f$ solves the obstacle problem
$\mathcal P_\Omega((\psi-\omega_f)^+,0)$. 

In conclusion, we only have to show that 
\begin{equation}
\label{eq:ineq}
0\le \DsD u\le (\DsD\psi)^+\quad\text{in the distributional sense on $\Omega$,}
\end{equation}
where $u$ solves $\mathcal P_\Omega(\psi,0)$ and $\psi$ is a nonnegative obstacle in $\widetilde H^s(\Omega)$, 
such that $\DsD\psi$
is a measure on $\Omega$. The right hand side inequality holds by Theorem 
\ref{T:sup}. Now we prove the following claim:
\begin{equation}
\label{eq:step1}
\text{\em Assume ${\DsD\psi}\in L^p(\Omega)$ for any $p>1$. Then (\ref{eq:ineq}) holds.}
\end{equation}
We take $p\ge \frac{2n}{n+2s}$, that is needed only if $n>2s$. Then
$\widetilde H^s(\Omega)\hookrightarrow L^{p'}(\Omega)$ and
$L^{p}(\Omega)\subset \widetilde H^s(\Omega)'$ by Sobolev embeddings. In
particular $({\DsD\psi})^+\in \widetilde H^s(\Omega)'$. 
Take a function $\theta_\eps\in C^\infty(\R)$ such that $0\le \theta_\eps\le 1$, and
$\theta_\eps(t)=1$ for $t\le 0$, $\theta_\eps(t)=0$ for $t\ge \eps$. Let $u_\eps\in \widetilde H^s(\Omega)$
be the unique solution to 
\begin{equation}
\label{eq:penalty}
\DsD u_\eps=\theta_\eps(u_\eps-\psi)~\!({\DsD\psi})^+\quad\text{in $\Omega$.}
\end{equation}
Notice that $(1-\theta_\eps(u_\eps-\psi))(\psi-u_\eps)^+=0$ a.e. in $\Omega$. Therefore,
using $ii)$ in Remark \ref{R:truncation} with $v=\psi-u_\eps$ one gets $(\psi-u_\eps)^+\equiv 0$. 
In particular we infer that $u_\eps\in K^s_\psi$. On the other hand, $\DsD u_\eps\ge 0$; thus $u_\eps\ge u$
by $b)$ in Theorem \ref{T:sup}. 
Next, notice that $\DsD(u_\eps-u)\le \DsD u_\eps$ and that $\theta_\eps(u_\eps-\psi)(u_\eps-u-\eps)^+=0$ a.e. in $\Omega$.
Then again $ii)$ in Remark \ref{R:truncation} plainly implies that $(u_\eps-u-\eps)^+\equiv 0$. In conclusion, we 
have that $u\le u_\eps\le u+\eps$, hence 
$\|u_\eps-u\|_\infty\to 0$ as $h\to\infty$. Therefore, for any nonnegative test
function $\f\in C^\infty_0(\Omega)$ we have that
\begin{eqnarray*}
\langle \DsD u,\f\rangle&=&\int\limits_\Omega u\DsD\f~\!dx=
\int\limits_\Omega u_\eps\DsD\f ~\!dx+o(1)\\
&=&\langle \DsD u_\eps,\f\rangle+o(1)
\le \langle (\DsD \psi)^+,\f\rangle+o(1).
\end{eqnarray*}
Hence, $\DsD u\le (\DsD \psi)^+$ in the distributional sense in $\Omega$, and the claim is proved.

\medskip
Now we use an approximation argument that needs to enlarge the domain $\Omega$.
%Take a large balla $B_R\supset\overline\Omega$. 
For any integer $h\ge 1$ let $\Omega_h\supset\overline\Omega$ be a bounded Lipschitz domain   
such that 
$\overline\Omega\subset \Omega_h\subseteq\Big\{ x\in\R^n~|~d(x,\Omega)<\frac1h\Big\}$,
as in Subsection \ref{SS:ST}.
The convex set
$$
K_h(\psi)=\{ v\in \widetilde H^s(\Omega_h)~|~v\ge \psi~~\text{a.e. on $\R^n$}~\}
$$
contains $K^s_\psi$, hence it is not empty. Let $u_h\in \widetilde H^s(\Omega_h)$ be the unique solution
to
\begin{equation*}\tag{$\mathcal P_{\Omega_h}$}
\label{eq:eps}
u_h\in K_h(\psi)~,\qquad
\langle \DsNeps  u_h, v-u_h\rangle\ge 0\qquad\forall v\in K_h(\psi)~\!.
\end{equation*}
%Trivially, $u_h$ is nonnegative. 
Now we prove that 
\begin{equation}
\label{eq:claim1}
%0\le 
\DsNeps  u_h\le (\DsD \psi)^+\quad \text{in the distributional sense on $\Omega$.}
\end{equation}
Fix $h$, and approximate $\psi$ via convolution with a sequence of smooth obstacles
$\psi^{k}=\psi*\rho_{k}$, where supp$(\rho_k)\subset B_{\frac{1}{k}}$. For $k$ large enough
we have that $\psi^{k}\in C^\infty_0(\Omega_h)$. In addition
$\psi^{k}\to \psi$ in $\widetilde H^s(B_R)$ as $k\to\infty$, where $B_R$ is any ball containing $\Omega_h$. 
Now, let $u_{h}^k\in \widetilde H^s(\Omega_h)$ be the unique solution to
\begin{equation*}\tag{$\mathcal P_{\Omega_h}^k$}
\label{eq:vih}
u_{h}^k\in K_{h}(\psi^{k})~,\qquad
\langle \DsNeps  u_{h}^k, v-u_{h}^k\rangle\ge 0 \qquad\forall v\in K_{h}(\psi^{k})~\!,
\end{equation*}
where $K_{h}(\psi^{k}):=\{v\in \widetilde H^s(\Omega_h)~|~u\ge \psi^{k}\}$. Then
$u_{h}^k\to u_h$  in $\widetilde H^s(\Omega_h)$ as $k\to \infty$ by Theorem \ref{T:Hs2}, and (\ref{eq:step1}) gives
\begin{equation}
\label{eq:eps_h}
%0\le  
\DsNeps  u_{h}^k\le (\DsNeps \psi^{k})^+\quad\text{in the distributional sense on $\Omega$.}
\end{equation}
Next, $(\DsNeps \psi)^+*\rho_k$ is a nonnegative smooth function, and
$$(\DsNeps \psi)^+*\rho_k\ge (\DsNeps \psi)*\rho_k=\DsNeps\psi^{k}~\!.$$
Thus
$(\DsNeps \psi)^+*\rho_h\ge (\DsNeps\psi^{k})^+$, and (\ref{eq:eps_h}) implies
$$
\DsNeps  u_h^k\le (\DsNeps \psi)^+*\rho_k\quad\text{in the distributional sense on $\Omega$.}
$$
Now, as $k\to \infty$ we have that
 $(\DsNeps \psi)^+*\rho_k\to (\DsNeps \psi)^+$ in the sense of measures,
and $\DsNeps u^k_h\to \DsNeps u_h$ in the sense of distributions. Thus
$$
%0\le  
\DsNeps  u_h\le (\DsNeps \psi)^+\quad\text{in the distributional sense on $\Omega$.}
$$
Since $\psi\in\widetilde H^s(\Omega)$ is nonnegative, Lemma \ref{L:lemma2}
gives {$(\DsNeps \psi)^+\le (\DsD \psi)^+$}, and  (\ref{eq:claim1}) follows.

The main difference with respect to the Dirichlet case is in taking the limit in (\ref{eq:claim1})
as the domains $\Omega_h$ shrink to $\Omega$.
First, we notice that
$u\in \widetilde H^s(\Omega_h)$ and in particular $u\in K_h(\psi)$. 
Therefore, using the variational characterization of the unique solution $u_h$ to $(\mathcal P_{\Omega_h})$ 
and (\ref{eq:monotonicity}) we find
\begin{equation}
\label{eq:uff}
\langle  \DsNeps  u_h,u_h\rangle\le 
\langle  \DsNeps  u,u\rangle\le \langle  \DsD  u,u\rangle~\!.
\end{equation}
Inequality (\ref{eq:monotonicity}) gives also 
$\langle (-\Delta_{B_R}\!)^{s}u_h,u_h\rangle \le \langle  \DsNeps  u_h,u_h\rangle$.
Thus
(\ref{eq:uff}) implies that the sequence $u_h$ is bounded in $\widetilde H^s(B_R)$,
and therefore 
we can assume that $u_h\to \tilde u$ weakly in $\widetilde H^s(B_R)$. Using 
Lemma \ref{L:eige_ueps} and  (\ref{eq:uff}) we readily get that  $u\in \widetilde H^s(\Omega)$ and
\begin{eqnarray}
\langle\DsD \tilde u,\tilde u\rangle\le \liminf_{h\to\infty}\langle\DsNepsh u_h,u_h\rangle
\le  \limsup_{h\to\infty}\langle\DsNepsh u_h,u_h\rangle
\le\langle\DsD u,u\rangle~\!,
\label{eq:party}
\end{eqnarray}
that is, $\langle  \DsD  \tilde u,\tilde u\rangle\le \langle  \DsD  u,u\rangle$.
On the other hand, $u_h\to \tilde u$ almost everywhere and 
$ u_h\ge \psi$ on ${\Omega}$. Thus $\tilde u\in K^s_\psi$.
Using the characterization of $u$ as the unique solution to the minimization problem (\ref{eq:minimization})
(with $f\equiv 0$), we first get that $\tilde u=u$. Then we use (\ref{eq:party})
to infer that $u_h\to u$ in $\widetilde H^s(B_R)$. 

To conclude take any nonnegative function $\eta\in C^\infty_0(\Omega)$. We use (\ref{eq:claim1})
and we recall that $u_h\to u$, $\DsNeps\eta\to\DsD\eta$ in $L^2(B_R)$,
see Lemma \ref{L:lemma3}, to estimate
$$
\langle(\DsD \psi)^+,\eta\rangle\ge
\langle\DsNeps u_h,\eta\rangle
=\langle\DsNeps \eta,u_h\rangle
=
\langle\DsD \eta,u\rangle+o(1)= \langle\DsD u, \eta\rangle+o(1)~\!.
$$
The proof  is complete.
\QED

\medskip

In the next results we adopt ''pointwise'' definitions of the contact set and on the non-contact set:
\begin{equation}
\label{eq:set}
\{u=\psi\}:=\{x\in\Omega~|~u(x)=\psi(x)\}~,\quad
\{u>\psi\}:=\{x\in\Omega~|~u_D(x)=\psi(x)\}~\!.
\end{equation}
Clearly, $\{u=\psi\}$ and $\{u>\psi\}$ are determined up to negligible sets if $u\in \widetilde H^s(\Omega)$.


\begin{Theorem}
\label{T:regularity}
Let $\psi$ and $f\in\widetilde H^s(\Omega)'$ satisfying 
assumptions $A1)$ and $A2)$ in Theorem \ref{T:measure} and
\begin{itemize}
\item[$A3)$] $(\DsD(\psi-\omega_f)^+-f)^+\in L^p_{\rm loc}(\Omega)$ for some $p\in[1,\infty]$.
\end{itemize}
Let $u\in{\widetilde H^s(\Omega)}$ be the unique solution to \ref{eq:vi}.
Then the following facts hold.
\begin{itemize}
\item[$i)$] $\DsD u-f\in L^p_{\rm loc}({\Omega})$;
\item[$ii)$] $0\le \DsD u-f\le (\DsD(\psi-\omega_f)^+-f)^+$ a.e. on $\Omega$;
\item[$iii)$] $\DsD  u=f$ a.e. on $\{u>\psi\}$.
\end{itemize}
In particular, $u$ solves the free boundary problem (\ref{eq:problem}).
%If in addition $p> n/2s$ \textcolor{red}{and $n>2s$???}, then $u$ is (locally) H\"older continuous in $\Omega$ and
\end{Theorem}

\proof 
We follow the proof of Theorem 1.1 in \cite{MNS}.
Statements $i)$ and $ii)$ hold by 
Theorem \ref{T:measure}. Let us prove the last claim. As before, we  assume that 
$f\equiv 0$. Then, $\DsD u\ge 0$ and the maximum principle in Remark \ref{R:MP} gives $u\ge 0$.
In particular, $u\ge \psi^+$ and $\{u>\psi\}=\{u>\psi^+\}$.

Use $c)$ in Theorem \ref{T:sup} with $v= \psi^+\in \widetilde H^s(\Omega)$, to get
$\langle\DsD u,u-\psi^+\rangle =0$. 
Let $\f\in C^\infty_0(\Omega)$ be any nonnegative cut off function; for $m\ge 1$ put
$g_m=(u-\psi^+)\wedge m$. 
Since $\DsD u\ge 0$, $\DsD u\in L^1_{\rm loc}(\Omega)$, $u-\psi^+\ge \f g_m$ and $\f g_m\in L^\infty(\Omega)$
has compact support in $\Omega$, we have that
$$
0=\langle \DsD u,u-\psi^+\rangle\ge\langle \DsD u,\f g_m\rangle=\int\limits_{\Omega} {\DsD u}\cdot(\f g_m)dx~\!.
$$
Next, use the monotone convergence theorem to get 
$$
0\ge \lim_{m\to\infty}\int\limits_{\Omega} {\DsD u}~\!\!\cdot \f g_m~\!dx=
\int\limits_{\Omega} ({\DsD u}~\!\!\cdot (u-\psi^+))\f~\!dx.
$$
Since $({\DsD u}~\!\!\cdot (u-\psi^+))\f\ge 0$ a.e. on $\Omega$, this implies that 
$({\DsD u}~\!\!\cdot (u-\psi^+))\f=0$ a.e. in $\Omega$. Since $\f$ was arbitrarily chosen, we 
can conclude that $\DsD u~\!\!\cdot (u-\psi^+)=0$  a.e. in $\Omega$, and $iii)$ is proved.
\QED




\begin{Remark}
\label{R:regularity}
\red{check!}
To obtain better regularity results for $u$, one can apply the regularity
theory for
$$
\DsD u=g\in L^p(\Omega)\quad\text{in $\Omega$}~,\qquad u\in \widetilde H^s(\Omega).
$$
In particular, using \cite{Se1, Se2} (see also \cite{Gr}) we obtain that $u\in H^{2s}(\Omega)$. In particular, if 
$p>\frac{n}{2s}$, then $u$ is H\"older continuous in $\Omega$.
\end{Remark}


We conclude this section by giving a sufficient condition for the continuity of $u$.

\begin{Theorem}
\label{T:2}
\red{check}
%Assume that $\Omega$ is a bounded \red{Lipschitz domain satisfying the exterior ball condition}.
Let $\psi\in C^0(\overline\Omega)$ be a given obstacle, such that $\psi\le 0$ on $\partial\Omega$.
Let $f\in L^p(\Omega)$, with $p>n/2s$.
Then $u$ is continuous on $\R^n$.
\end{Theorem}
\bigskip

\proof
The argument is the same as in \cite{MNS}, proof of Theorem 4.8.
Fix a small $\eps>0$. We can assume that $\psi-\eps\in C^0_0(\R^n)$ and 
$\psi-\eps\le 0$ outside $\Omega$. Let $\psi^\eps_{h}$ be a sequence in $C^\infty_0(\R^n)$ such that
$\psi^\eps_h\to \psi-\eps$ uniformly on $\R^n$, as $h\to\infty$.

By Theorem \ref{T:regularity}, the solution $u_h\in \widetilde H^s(\Omega)$ to 
$\mathcal P_\Omega(\psi^\eps_h,f)$ satisfies $\DsD u^\eps_h\in L^p(\Omega)$ and therefore 
$u^\eps_h$ is H\"older continuous, see Remark \ref{R:regularity}.
Moreover, the estimates in Theorem \ref{T:bounded1} imply
that $u^\eps_h \to u^\eps$ uniformly on  $\Omega$, where $u^\eps$ solves
$\mathcal P_\Omega(\psi-\eps,f)$. In particular, $u^\eps\in C^0(\overline\Omega)$.
Finally, use again Theorem \ref{T:bounded1} to get that $u^\eps\to u$ uniformly, 
and conclude the proof.
\QED



\section{Comparing the Navier and the Dirichlet problems}

In this section we compare the unique solutions $u_N, u_D$ to
the variational inequalities 
\begin{equation*}\tag{$\mathcal P_N(\psi)$}
\label{eq:N}
u_N\in K^s_\psi~,\qquad
\langle \DsD  u_N, v-u_N\rangle\ge 0\quad\forall v\in K^s_\psi
\end{equation*}
\begin{equation*}\tag{$\mathcal P_D(\psi)$}
\label{eq:D}
u_D\in K^s_\psi~,\qquad
~~\!\langle \DrD  u_D, ~\!v-u_D~\!\rangle~\ge 0\quad\forall v\in K^s_\psi~\!,
\end{equation*}
respectively. Problem \ref{eq:D}
has been investigated in \cite{MNS}. Recall that
\begin{equation}
\label{eq:L_positive}
\DsD u_N\ge 0~,\quad\DrD u_D\ge 0\qquad\text{on $\Omega$}
\end{equation}
in the sense of distributions, see Theorem \ref{T:sup} and \cite[Theorem 3.2]{MNS}, so that
$u_N, u_D$ are nonnegative by the maximum principle, and $u_N, u_D\ge \psi^+$. 
It is not restrictive to assume that
\begin{equation}
\label{eq:psi_positive}
\psi\ge 0\quad\text{almost everywhere in $\Omega$, and $\psi\neq 0$}
\end{equation}
In fact, it is easy to show that $u_N, u_D$ solve problems 
$\mathcal P_N(\psi^+)$, $\mathcal P_D(\psi^+)$, so that, if needed, we can replace the obstacle $\psi$
by $\psi^+$. If $\psi\equiv 0$, then clearly $u_N=u_D=0$.

If (\ref{eq:psi_positive}) holds, then
\begin{equation}
\label{eq:u_positive}
u_N> 0\quad\text{and}\quad u_D> 0\qquad\text{on $\Omega$},
\end{equation}
in the sense that $u_N, u_D$ are essentially bounded away from $0$ on every
compact set $K\subset\Omega$, use \cite[Lemma 2.4]{CDDS} and 
\cite[Theorem 2.5]{IMS}.

In this section we need to refine the notion of contact set that has been introduced
in (\ref{eq:set}). We essentially use an idea 
due to Lewy and Stampacchia \cite{LS1, LS2}, see also \cite[Section 6]{KS}. 
We start with some preliminaries.

\begin{Definition}
Let $v$ be a nonnegative measurable function on the open set $\Omega$, and let $x\in \Omega$.
We say that $v(x)>0$ if
there exist $\rho,\eps>0$ such that $B_\rho(x)\subseteq\Omega$ and
$$
v-\eps\ge 0~~\text{almost everywhere in $B_\rho(x)$}~\!.
$$
In addition, we put
$$
P[v]=\{~\!x\in \Omega~|~v(x)>0~\!\}
$$
\end{Definition}
and $I[v]=\Omega\setminus P[v]$. The set $P[v]$ is clearly open; thus $I[v]$ is closed in $\Omega$. 
\begin{Lemma}
\label{L:meas1}
Let $v\ge 0$ be a  measurable function on $\Omega$. Then $v>0$ a.e. in $P[v]$.
\end{Lemma}

\proof
We denote by $|E|$  the Lebesgue measure of a measurable set $E\subseteq\R^n$.

Put $N=\{x\in P[v]~|~ v(x)=0\}$ and let $K$ be any compact set with $K\subset N$.
Since $K\subset P[v]\subseteq\Omega$, for $x\in K$ we can find a ball $B_x$ about x such that
${B}_x\subset\Omega$ and $v\ge \eps_x>0$ on $B_x$. Since $K$ is compact,
we can find a finite number of points $x_i\in K$ such that $K$ is covered by the finite family $B_{x_i}$. Let $\eps_0=\min_i\eps_{x_i}$.
Then $v\ge \eps_0$ a.e. on $K$. Since $K\subset N$, this shows that $|K|=0$.
Hence $|N|=0$, and the lemma is proved.
\QED
Now we go back to the obstacle problems \ref{eq:N}, \ref{eq:D}.
By Lemma \ref{L:meas1} we have  the inclusions
$$
P[u_N-\psi]\subseteq\{u_N>\psi\}~,\quad P[u_D-\psi]\subseteq\{u_D>\psi\}.
$$
%and thus
%$\{u_N=\psi\}\subseteq I[u_N-\psi]$, $ \{u_D=\psi\}\subseteq I[u_D-\psi]$.
It might happen that 
$\{u_N>\psi\}$ and $\{u_D>\psi\}$ have positive measure but $P[u_N-\psi]$ and
$P[u_D-\psi]$ are empty, see Remark \ref{R:example} below.

\begin{Theorem}
\label{T:comparing1}
If (\ref{eq:psi_positive}) holds, then the following facts hold true.
\begin{itemize}
\item[$i)$] $\DsD u_N=0$ on $P[u_N-\psi]$ and $\DrD u_D=0$ on $P[u_D-\psi]$;
\item[$ii)$] $\DsD u_D\ge 0$ in the distributional sense on $\Omega$;
\item[$iii)$] $u_N\le u_D$;
\item[$iv)$] $u_D$ is the solution to the obstacle problem $\mathcal P_D(u_N)$;
\item[$v)$] $\|\DshalfD u_D\|_2\le \|\DshalfD u_N\|_2<\|\DshalfN u_N\|_2\le \|\DshalfN u_D\|_2$.
\end{itemize}
\end{Theorem}

\proof
Take $x\in P[u_N-\psi]$ and find $\rho,\eps>0$ such that $u_N-\eps\ge \psi$ a.e. on $B_\rho(x)\subset\Omega$.
Let $\f\in C^\infty_0(B_\rho(x))$. For $t\in(0,\eps\|\f\|^{-1}_\infty)$, we have
$u_N\mp t\f\in K_\psi^s$. Thus
$\pm t\langle \DsD  u_N, \f\rangle\ge 0$,
that proves $i)$ for $u_N$. The argument for $u_D$ is the same.

Now we use \cite[Theorem 1]{FL}, that gives $\DsD u_D\ge \DrD u_D$, 
as $u$ is nonnegative. Then $ii)$  follows from (\ref{eq:L_positive}). 
By $b)$ in Theorem \ref{T:sup}, we know that $u_N$ is the smallest supersolution to $\DsD v=0$ 
in the set ${\cal U}\in K^s_\psi$. Thus $u_D\ge u_N$ by $ii)$, and $iii)$ is proved.


To check $iv)$, we let $\tilde u$ to be the solution to $\mathcal P_D(u_N)$. In particular,
$\tilde u$ is the smallest supersolution ${\cal U}$ to $\DrD v=0$ such that
${\cal U}\in \widetilde H^s(\Omega)$, ${\cal U}\ge u_N$, see \cite[Theorem 3.2]{MNS}.
Thus $u_D\ge \tilde u$ by $iii)$ and since $\DrD u_D\ge 0$. On the other hand,
we also have that $\tilde u\ge u_N\ge \psi$, and $u_D$ is the smallest supersolution to $\DrD v=0$
in $K^s_\psi$, use again  \cite[Theorem 3.2]{MNS}. Thus $\tilde u\ge u_D$, and $iv)$ is proved.

Theorem 2 in \cite{FL} gives $\|\DshalfD u_N\|_2<\|\DshalfN u_N\|_2$.
The remaining inequalities on $v)$ follow from the variational formulation of the obstacle problems 
\ref{eq:N}, and \ref{eq:D},
see (\ref{eq:minimization}) (with $f=0$) and the corresponding formula in \cite{MNS} for $u_D$.
\QED

\begin{Theorem}
\label{T:comparing2}
Assume $\psi\ge 0$. If $P[u_N-\psi]$ is not empty, then % $u_D\neq u_N$.
$u_N, u_D$ are smooth on $P[u_N-\psi]$, and
\begin{equation}
\label{eq:comp9}
\DrD u_N<0~,\qquad u_D>u_N>0\quad\text{everywhere on $P[u_N-\psi]$.}
\end{equation}
\end{Theorem}

\proof
Since $\psi$ can not be the null function, then (\ref{eq:u_positive}) holds. 
Let $A=P[u_N-\psi]$. Clearly $A\subseteq P[u_D-\psi]$ since $u_D\ge u_N$,
and therefore $\DsD u_N=\DrD u_D= 0$ on the open set $A$ by Theorem \ref{T:comparing1}. 
In particular, $u_N, u_D$ are smooth positive functions on $A$.
We divide the proof in three steps.

\medskip

\noindent{\bf Step 1:} {\em $u_D\neq u_N$ on a set of positive measure.}\\ 
We argue by contradiction. Assume that $u_D=u_N$ a.e. and put $u=u_N$.
Then $u\ge 0$ and  $\DsD u=\DrD u= 0$ in $A$. To get a contradiction we follow the argument
in \cite{FL}, proof of Theorem 1. Denote by $w^N_u$ the Stinga-Torrea extension of $u$, see Subsection
\ref{SS:ST}. Thus $w_u^N$ solves
\begin{equation}
\label{eq:bpST}
-\div (y^{1-2s}\nabla w^N)=0\quad \mbox{in}\quad \Omega\times\mathbb R_+;\qquad w^N\big|_{y=0}=u;
\qquad w^N\big|_{x\in\partial\Omega}=0.
\end{equation}
Since $u$ is smooth in $A$, then 
\begin{equation}
0=\DsD u(x)=-c_s\cdot\lim\limits_{y\to0^+} 
\frac{w^N_u(x,y)-u(x)}{y^{2s}}
\label{extension_N}
\end{equation}
for any $x\in A$.
Next, let $w^D_u$ be the Caffarelli-Silvestre \cite{CaSi} extension of $u$. Then $w^D_u$ solves the boundary value problem
\begin{equation}
\label{eq:bpCS}
-\div (y^{1-2s}\nabla w^D)=0\quad \mbox{in}\quad \mathbb R^n\times\mathbb R_+;\qquad w^D\big|_{y=0}=u,
\end{equation}
and for $x\in A$ it results that
\begin{equation}
0=\DrD u(x)=-c_u\cdot\lim\limits_{y\to0^+} 
\frac{w_u^D(x,y)-u(x)}{y^{2s}}.
\label{extension_D}
\end{equation}
By the maximum principle we have that $w_u^D\ge0$ in 
$\mathbb R^n\times\mathbb R_+$. Thus, $w_u^D\ge w_u^N$ at 
$\partial\Omega\times\mathbb R_+$ and, again by the maximum principle, 
$W:=w^D_u-w^N_u\ge 0$ in $\Omega\times\mathbb R_+$. In addition, $W$ solves
$$
-\div (y^{1-2s}\nabla W)=0\quad \mbox{in}\quad A\times\mathbb R_+;\qquad 
W\big|_{A\times\{y=0\}}=0~\!.
%;\qquad w\big|_{x\in\partial\Omega}=0.
$$
From (\ref{extension_N}), (\ref{extension_D}) we also have that 
\begin{equation}
\label{eq:Maz}
\lim\limits_{y\to0^+} 
\frac{W(x,y)}{y^{2s}}=0\quad\text{on $A\times\{y=0\}$}.
\end{equation}
After changing of the variable $t=y^{2s}$ the function $W(x,t)$ solves 
\begin{equation}
\Delta_xW+4s^2t^{\frac{2s-1}s}W_{tt}=0\quad\mbox{in}\quad \Omega\times\mathbb R_+;\qquad 
W\big|_{A\times\{t=0\}}=0.
\label{Maz}
\end{equation}
The differential operator in (\ref{Maz}) satisfies the assumptions of \cite[Theorem 1.4]{ABMMZ} (the boundary point lemma)
at any point $(x_0,0)\in A\times\{0\}$ (see \cite{FL} for details). Thus, 
we have for any $x\in A$
$$\liminf\limits_{y\to 0^+}\frac{W(x,y)}{y^{2s}}=\liminf\limits_{t\to 0^+}\frac{W(x,t)}{t}>0.
$$
We reached a contradiction with (\ref{eq:Maz}). Hence $u_D$ and $u_N$ do not collapse.


\medskip

\noindent{\bf Step 2:} {\em $\DsD u_N>\DrD u_N$ in $A$.}\\ 
The argument is essentially the same as in \cite[Theorem 1]{FL} or in Step 1 above.
We sketch few details for the convenience
of the reader. 

Use as before \cite{ST},  \cite{CaSi} to find the solutions
$w^N, w^D:\R^n\times\R_+\to \R$ to (\ref{eq:bpST}) and (\ref{eq:bpCS}), respectively,
with boundary datum $u=u_N\ge 0$. Notice that $W:=w^D- w^N$ is nonnegative and solves
$$
-\div (y^{1-2s}\nabla W)=0\quad \mbox{in}\quad A\times\mathbb R_+;\qquad 
W\big|_{A\times\{y=0\}}=0~\!.
%;\qquad w\big|_{x\in\partial\Omega}=0.
$$
Then \cite[Theorem 1.4]{ABMMZ} (the boundary point lemma)
gives that
\begin{eqnarray*}
0&<&c_s\liminf\limits_{y\to 0^+}\frac{W(x,y)}{y^{2s}}=c_s\lim\limits_{y\to 0^+}\frac{(w^D(x,y)-u_N(x))-(w^N(x,y)-u_N(x))}{y^{2s}}\\
&=&-\left(\DrD u_N(x)-\DsD u_N(x)\right)
\end{eqnarray*}
for $x\in A$. Thus Step 2 is done.


\medskip

\noindent{\bf Step 2:} {\em Conclusion of the proof}\\
Since $\DsD u_N=0$ on $A$, 
then $\DrD u_N<0$ on $A$ by Step 2. Next, we have $u^D\ge u^N$, $u^D\neq u^N$ and $\DrD(u_D-u_N)=-\DrD u_N> 0$ on $A$. 
Hence $u_D-u_N>0$ in $A$ by the strong maximum principle, use for instance \cite[Theorem 2.5]{IMS}.
\QED

\begin{Remark}
\label{R:example}
Take a smooth function $\eta\in \widetilde H^s(\Omega)$ satisfying
$\DrD \eta\ge 0$ and $\eta>0$ in $\Omega$. Let ${\kappa}\subset\Omega$ be a compact set, having positive measure but empty interior. Consider
the obstacle $\psi=\eta\chi_{\Omega\setminus\kappa}$ and the solutions $u_N, u_D$ to \ref{eq:N}, \ref{eq:D}, respectively.
Clearly $\eta\in K^s_{\psi}$. Thus $\eta\ge u_D$ because
$u_D$ is the smallest supersolution for $\DrD v=0$ in $K^s_{\psi}$. But then
we have
$\eta\ge u_D\ge u_N\ge \psi=\eta\chi_{\Omega\setminus\kappa}$. In particular, $u_D= u_N={\psi}$ on $\Omega\setminus {\kappa}$.
Actually $\{u_D={\psi}\}=\{u_N={\psi}\}=\Omega\setminus {\kappa}$, because $u_N, u_D$ are positive in $\Omega$.
In particular the sets $\{u_D>{\psi}\},\{u_N>{\psi}\}$ have positive measure as
they coincide with ${\kappa}$, and $P[u_N-{\psi}]=P[u_D-{\psi}]=\emptyset$.
\end{Remark}

\begin{Remark}
\label{R:true}
Sufficient conditions in order to have that $P[u_N-\psi]$ is not empty can be easily obtained. For instance,
if $\psi$ vanishes on an open set $B\subset\Omega$, then $B\subseteq P[u_N-\psi]$, since $u_N$ is positive and 
the strong maximum principle in \cite{CDDS} holds. If $\Omega$ is regular enough and $\psi$ is continuous on
$\overline\Omega$, then $u_N$ is continuous as well. Thus $u_N\equiv \psi$, or $P[u_N-\psi]=\{u_N>\psi\}$ is
not empty.
\end{Remark}





\begin{thebibliography}{CS}

\small

\bibitem{ABMMZ}
R. Alvarado, D. Brigham, V. Maz'ya, M. Mitrea\ and\ E. Ziad\'e, {\it On the regularity of domains 
satisfying a uniform hour-glass condition and a sharp version of the Hopf--Oleinik boundary point principle},
Probl. Mat. Anal., {\bf 57} (2011), 3--68 (Russian); English transl.: J. Math. Sci.,
{\bf 176} (2011), 281--360.

%\bibitem{BrSt} H. R. Brezis\ and\ G. Stampacchia, Sur la r\'egularit\'e de la solution d'in\'equations elliptiques, Bull. Soc. Math. France {\bf 96} (1968), 153--180.
%
\bibitem{BB}
D. Bucur\ and\ G. Buttazzo, {\it Variational methods in shape optimization problems}, Progress in Nonlinear Differential Equations and their Applications, 65, Birkh\"auser Boston, Boston, MA, 2005. 

\bibitem{BDM}
G. Buttazzo\ and\ G. Dal Maso, An existence result for a class of shape optimization problems, Arch. Rational Mech. Anal. {\bf 122} (1993), no.~2, 183--195.

\bibitem{CaFi}
L. Caffarelli\ and\ A. Figalli, Regularity of solutions to the parabolic fractional obstacle problem, J. Reine Angew. Math. {\bf 680} (2013), 191--233.

\bibitem{CaSi}
L. Caffarelli\ and\ L. Silvestre, {An extension problem related to the fractional Laplacian}, Comm. Part. Diff. Eqs. {\bf 32} (2007), no.~7-9, 1245--1260.

%\bibitem{CSS}
%L. A. Caffarelli, S. Salsa\ and\ L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for the fractional Laplacian, Invent. Math. {\bf 171} (2008), no.~2, 425--461.

\bibitem{CDDS}
A. Capella, J. D\'avila, L. Dupaigne\ and Y. Sire, Regularity of radial extremal solutions for some non-local semilinear equations, Comm. Partial Differential Equations {\bf 36} (2011), no.~8, 1353--1384.
%
%\bibitem{Co}
%M. Cozzi, Interior regularity of solutions of non-local equations in Sobolev and Nikol'skii spaces, preprint
%(2015).

\bibitem{DM}
G. Dal Maso, {\it An introduction to $\Gamma$-convergence}, Progress in Nonlinear Differential Equations and their Applications, 8, Birkh\"auser Boston, Boston, MA, 1993.
%
%\bibitem{Da}
%E. B. Davies, Sharp boundary estimates for elliptic operators, Math. Proc. Cambridge Philos. Soc. {\bf 129} (2000), no.~1, 165--178. 

\bibitem{Gr}
G. Grubb, Spectral results for mixed problems and fractional elliptic operators, J. Math. Anal. Appl. {\bf 421} (2015), no.~2, 1616--1634.

\bibitem{He}
A. Henrot, {\it Extremum problems for eigenvalues of elliptic operators}, 
Frontiers in Mathematics, Birkh\"auser, Basel, 2006. 

\bibitem{IMS}
A. Iannizzotto, S. Mosconi\ and\ M. Squassina, $H\sp s$ versus $C\sp 0$-weighted minimizers, NoDEA Nonlinear Differential Equations Appl. {\bf 22} (2015), no.~3, 477--497.

\bibitem{KS} D. Kinderlehrer\ and\ G. Stampacchia, {An introduction to variational inequalities and their applications}, reprint of the 1980 original, Classics in Applied Mathematics, 31, SIAM, Philadelphia, PA, 2000.

\bibitem{LS1}
H. Lewy\ and\ G. Stampacchia, On the regularity of the solution of a variational inequality, Comm. Pure Appl. Math. {\bf 22} (1969), 153--188.

\bibitem{LS2}
H. Lewy\ and\ G. Stampacchia, On the smoothness of superharmonics which solve a minimum problem, J. Analyse Math. {\bf 23} (1970), 227--236.

\bibitem{PG}
N. Garofalo\ and\ A. Petrosyan, Some new monotonicity formulas and the singular set in the lower dimensional obstacle problem, Invent. Math. {\bf 177} (2009), no.~2, 415--461.

\bibitem{JLL}
J. L. Lions, Partial differential inequalities, Uspehi Mat. Nauk {\bf 26} (1971), no.~2(158), 205--263.
%
%\bibitem{Mi}
%G. J. Minty, Monotone (nonlinear) operators in Hilbert space, Duke Math. J. {\bf 29} (1962), 341--346.

\bibitem{FL} R. Musina\ and\ A. I. Nazarov, On fractional Laplacians, 
Comm. Part. Diff. Eqs. {\bf 39} (2014), no.~9, 1780--1790.
%
%\bibitem{MNHS} 
%R. Musina\ and\ A. I. Nazarov, On the Sobolev and Hardy constants for the fractional Navier Laplacian, 
%Nonlinear Anal. {\bf 121} (2015), 123--129.

\bibitem{MNS} 
R. Musina\ and\ A. I. Nazarov, K. Sreenadh, Variational inequalities for the fractional Laplacian (2015),
preprint.

\bibitem{PP}
A. Petrosyan\ and\ C. A. Pop, Optimal regularity of solutions to the obstacle problem for the fractional Laplacian with drift, 
J. Funct. Anal. {\bf 268} (2015), no.~2, 417--472.
%
%\bibitem{RoS} X. Ros-Oton\ and\ J. Serra.,
%The extremal solution for the fractional Laplacian, preprint arXiv
%
%\bibitem{ROS2}
%X. Ros-Oton\ and\ J. Serra, The Dirichlet problem for the fractional Laplacian: regularity up to the boundary, J. Math. Pures Appl. (9) {\bf 101} (2014), no.~3, 275--302.

\bibitem{Se1}
R. Seeley, Norms and domains of the complex powers $A\sb{B}z$, Amer. J. Math. {\bf 93} (1971), 299--309.

\bibitem{Se2}
R. Seeley, Interpolation in $L\sp{p}$ with boundary conditions, Studia Math. {\bf 44} (1972), 47--60.

\bibitem{SV.obs}
R. Servadei\ and\ E. Valdinoci, 
Lewy-Stampacchia type estimates for variational inequalities driven by (non) local operators, Rev. Mat. Iberoamericana,
to appear (2013).

\bibitem{Sil}
L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, 
Comm. Pure Appl. Math. {\bf 60} (2007), no.~1, 67--112.

\bibitem{ST}
P. R. Stinga\ and\ J. L. Torrea, {Extension problem and Harnack's inequality for some fractional operators}, 
Comm. Part. Diff. Eqs. {\bf 35} (2010), no.~11, 2092--2122.

\bibitem{Tr}
H. Triebel, {Interpolation theory, function spaces, differential operators}, Deutscher Verlag Wissensch., Berlin, 1978. 

%\bibitem{U0}
%N.N. Ural'tzeva The regularity of the solutions of variational inequalities, 
%Zap. Nauchn. Semin. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI), 27 (1972), 211-219 [Russian]; 
%English transl. in J. Soviet Math. 3, no. 4 (1975), 565-574.
%
%
%\bibitem{U}
%N. N. Ural'tseva, {On the regularity of solutions of variational inequalities}, 
%Uspekhi Mat. Nauk {\bf 42} (1987), no.~6(258), 151--174, 248.
%English transl. in Russian Math. Surveys, 42, no. 6 (1987), 191-219.

%
%\bibitem{U2}
%N.N. Ural'tseva\\
%Strong solutions of the generalized Signorini problem, Uspekhi Mat. Nauk, 33, no. 4 (1978) [Russian].\\
%Strong solutions of the generalized Signorini problem, Sibirsk. Mat. Zh., 19, no. 5 (1978), 1204-1212 [Russian].\\
%A problem with one-sided constraints on the interface of two domains, Vestnik Leningrad. Univ. Mat. Mekh. Astronom. 8, vyp. 2 (1985), 36-42 [Russian]\\
%Estimation on the boundary of the domain of derivatives of solutions of variational inequalities, Probl. Mat. Anal., no. 10 (1986), 92-105 [Russian]; English transl. in J. Soviet Math. 45, no. 3 (1989), 11811191.\\
%Regularity of the solution of a problem with a two-sided constraint on the boundary, Vestnik Leningrad. Univ. Mat. Mekh. Astronom. 1, vyp. 1 (1986), 3-10 [Russian]\\
%Estimates of derivatives of solutions of elliptic and parabolic inequalities, Proc. Int. Congr. Math., AMS (1987), pp. 1143-1149.\\
%On the existence of smooth solutions, for parabolic systems, of problems with convex constraints on the boundary of the domain, Zap. Nauchn. Semin. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI), 171 (1989), 5-11 [Russian]; English transl. in J. Soviet Math. 56, no. 2 (1990), 2281-2285


\end{thebibliography}

\end{document}
