\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 $00$, $m\notin\mathbb{N}$ 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 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 $1Q_m^{rD}[u]$ 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 $1Q_m^{sD}[u]$ 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\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 $\Omega\setminus {\kappa}$. 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