\section{Introduction}
The main goal of this work is to study the evolution problem
	associated to the following nonlinear non-local operator 
	\begin{equation}\label{operador}
	\Lambda_1^s u(x)\coloneqq\inf_{\theta\in\mathbb{S}^{N-1}}\int_{\mathbb{R}}
\dfrac{{u}(x+\tau \theta)-{u}(x)}{|\tau|^{1+2s}} d\tau.
\end{equation}
	Here $s\in(0,1)$ and $\mathbb{S}^{N-1}$ denotes the $(N-1)-$sphere.
	Notice that in the previous definition the infimum among directions $z$ of the $1-$dimensional
	fractional Laplacian is computed in the line with direction $\theta$ that passes
	trough the point $x$.  
	This paper is a natural continuation of \cite{DPQR_FC} where the elliptic problem
	$\Lambda_1^s u(x)=0$ inside a convex domain with a Dirichlet exterior condition
	is studied. However, we want to highlight that in the present work we establish some relevant results also related with the elliptic problem like the regularity up to the boundary.
	
	
	
	
	The operator described in \eqref{operador} appears naturally when we extend the notion
	of convexity of a function to the non-local setting. Let us describe this 
	fact with more detail. 
	It is well
	known that a function $u\colon\Omega\to\mathbb{R}$ is said to be 
	convex in $\Omega$
	if for any two points $x,y\in\Omega$ such that the segment that goes from $x$ to $y$ is contained in $\Omega$, i.e.,  
	\[
		[x,y]\coloneqq\{tx+(1-t)y\colon t\in(0,1)\}\subseteq\Omega,
	\] 
	it holds that 
	\[
		u(tx+(1-t)y)\le t u(x)+(1-t) u(y), \quad t\in(0,1).
	\]
	See for instance, \cite{Vel}.
	It terms of a second order partial equations, a function
	$u$ is convex if and only if
	\begin{equation}\label{eq_intro_Du}
		\lambda_1(D^2u)(x)\ge0,
	\end{equation}
	in the viscosity sense. Here $\lambda_1(D^2u)$ denotes the 
	smallest eigenvalue of the Hessian, $D^2u,$ that is
	\begin{equation}\label{eq_intro_lamba1}
		\lambda_1(D^2u)(x)\coloneqq\inf
		\left\{	
			\langle D^2u(x) z,z\rangle\colon z\in\mathbb{S}^{N-1}
		\right\}.
	\end{equation} 
	We refer to \cite{BlancRossi,HL1,OS,Ober}, and references therein.
	
	In \cite{DPQR_FC}, inspired by the classical notion of convexity,
	 the authors propose the following natural extension 	
	of convexity for functions in the non-local setting: a function
	$u\colon \mathbb{R}^N\to \mathbb{R}$ is said $s-$convex in $\Omega$ if
	for any $x,y\in\Omega$ such that $[x,y]\subseteq\Omega$ it holds that
	\[
		u(tx+(1-t)y)\le v(tx+(1-t)y),\quad t\in(0,1),	
	\] 
	where now $v$ is the viscosity solution of the $1-$dimensional fractional 
	Laplacian of order $s$ in the line $t \mapsto tx + (1-t)y$ for $t \in (0,1)$ with
	exterior values $u(tx+(1-t)y)$ for $t \not\in (0,1)$, that is, $v$ is the solution to 
		\begin{align*}
		&\Delta_1^sv(tx+(1-t)y)
			\coloneqq \int_{\mathbb{R}}
			\dfrac{{v}(x+\tau z)-{v}(x)}{|\tau|^{1+2s}} d\tau=0 
			\quad t\in(0,1),\\
		&{v(tx+(1-t)y)}=u(tx+(1-t)y),\quad t\not \in(0,1).
	\end{align*}
	As usual, the integral is to be understood in the principal
	value sense. Finally, in \cite{DPQR_FC}, it is shown that $u$
	is $s-$convex in a bounded strictly convex $\mathcal{C}^{2}$ domain $\Omega$
	(from this point, $\Omega$ will be 
		a bounded domain with $\mathcal{C}^2$-boundary such that all the principal curvatures of the surface $\partial\Omega$ 
		are positive everywhere) if only if 
	it satisfies
	\[
		\Lambda_1^s u(x)\ge 0\text{ in } \Omega,
	\] 
	in the viscosity sense where $\Lambda_1^s$ is the operator given by \eqref{operador}.
	
We notice that this operator \eqref{operador} also appears as part of a family of operators
	called the truncated fractional Laplacians that where studied in \cite{TBG}.
	For the local counterpart of truncated Laplacians we refer to \cite{BGI,BGI2,BGL}.  
	As far as we know this is the first reference dealing with the associated evolution problem. 
	
	\medskip
	
	Coming back to the main objective of the work, we will study the following problem	
	\begin{equation}\label{ev_eq-1}
		\begin{cases}
			u_t(x,t) =\Lambda_1^s u(x,t) &\text{in }\Omega\times(0,\infty),\\
			u(x,t)=g(x) &\text{in }
				(\mathbb{R}^N\setminus\Omega)\times (0,\infty),\\
			u(x,0)=u_0(x) &\text{in }{ \mathbb{R}^N},
		\end{cases}
	\end{equation}
	where $\Omega$ is a bounded and strictly convex domain, 
	$g$ is a function defined in $\mathbb{R}^N\setminus\Omega$ and $u_0$ is a fixed initial condition defined in $\Omega$. 
	
	Since the operator \eqref{operador} is not in divergence form 
	the natural framework to treat this problem is to use the theory 
	of viscosity solutions (see Section \ref{sect-evolution} for more details). For this notion of solutions we refer to the celebrated work of \cite{CIL} for the classical local concept and, for instance, to \cite{RO} for a non-local counterpart. It is good to bear in mind that, in order to guarantee the attainability of the datum in the notion of the viscosity solution, we have to impose some reasonable conditions. Indeed in what follows we will assume that 
{	\begin{equation}\label{hip}	
			\Lambda_1^su_0(x)\in L^{\infty}(\Omega)  ,\, 
			g\in \mathcal{C}({\mathbb{R}^N\setminus\Omega})\cap 
		L^\infty({\mathbb{R} ^N\setminus\Omega})
		\mbox{ and } 
			u_0=g \mbox{ in }	\mathbb{R}^N\setminus\Omega.
			\end{equation}
}
	
	Our first result, presented in the following theorem, says the the evolution problem is well posed. 
	
	\begin{theorem} \label{teo-1.intro} If $\Omega$ is a bounded strictly convex domain,
	$u_0{\in\mathcal{C}(\mathbb{R}^N)}$ and $g$ verifying \eqref{hip}, then there exists a unique viscosity solution to 
	\begin{equation}\label{ev_eq}
		\begin{cases}
			u_t(x,t)=\Lambda_1^s u(x,t) &\text{in }\Omega\times(0,\infty),\\
			u(x,t)=g(x) &\text{in }
				(\mathbb{R}^N\setminus\Omega)\times(0,\infty),\\
			u(x,0)=u_0(x) &\text{in }{ \mathbb{R}^N}.
		\end{cases}
	\end{equation}
	Moreover, a comparison principle holds; that is, if $\overline{u}$ is a supersolution 
	 and $\underline{u}$ is a subsolution to \eqref{ev_eq} then it holds that
	 $$\overline{u} (x,t) \geq \underline{u} (x,t),$$ for every $(x,t) \in \Omega \times (0,\infty)$. 	
	\end{theorem}
	
	Our next objective is to study the asymptotic behavior of the solution as $t\to \infty$
	in the special case $g \equiv 0$. 
	When we deal with the homogeneous case, $g \equiv 0$, we can obtain an 
	exponential bound for the rate of convergence to zero. 
	In parabolic problems the exponential decay rate
	is associated with the first eigenvalue of the operator.
	In our case this eigenvalue problem reads as
		\begin{equation} \label{eigen.intro}
			\begin{cases}
				-\Lambda_1^s \varphi_1(x)=\mu_{1}\varphi_1 
				&\text{in }\Omega,\quad\\
			 \varphi_1=0&\text{in } \mathbb{R}^{N}\setminus\Omega.
				\end{cases}
		\end{equation}
At this point we mention that this eigenvalue problem and a regularity issue associated to the operator \eqref{operador} were also studied in the recent work \cite{biri} but we include the result here because our proof is different. To prove the existence of the eigenfunctions we need a regularity result that
we consider interesting in its own right and that reads as follows, 
\begin{theorem}\label{regularidad.intro} Assume $s>1/2$.
		Let $u$ be a viscosity solution
		of 
	
        $$\begin{cases}
		-\Lambda_1^s u(x)=f(x) & \mbox{in}\quad \Omega,\\
		u(x)=g(x) & \mbox{in}\quad \mathbb{R}^N\setminus\Omega.
		\end{cases}
   
    $$
Assume that $g$ satisfies a H\"older bound, so that there exist $M_g$ and $\gamma\in(0,2s-1)$ such that
\begin{equation}\label{holder-g}
|g(x)-g(y))| \leq M_g |x-y|^{\gamma},\, x,\, y\in \mathbb{R}^N\setminus\Omega.
\end{equation}
Then,  there exists $C>0$ such that
 		\[
 			\|u\|_{\mathcal{C}^\gamma({\bar\Omega)}}\leq C.
 		\]
	\end{theorem}
	

As we commented before, thanks to this regularity result we can show the existence of a first eigenvalue 
with a corresponding nontrivial 
eigenfunction. Then, we obtain the following result
for the asymptotic behavior of solutions to our evolution equation (first we show exponential convergence to the steady state and, in the
special case of $g \equiv 0$ and $u_0$ compactly supported inside
$\Omega$ we obtain more precise bounds that imply that the exponential rate is given by the first eigenvalue).




\begin{theorem} \label{teo.conver.intro} Let $s>1/2$.
If $u_0$ and $g$ satisfy \eqref{hip} 
		then there exist $C(u_0)>0$ and $\mu>0$ such that 
		\begin{equation}\label{c_uniforme}
			 \|  u(\cdot,t)-z(\cdot) \|_{L^\infty (\overline\Omega)} 
			\leq C e^{-\mu t},\quad t>0,
		\end{equation}
		where $u(x,t)$ is the unique viscosity solution of \eqref{ev_eq} 
		and $z$ is the unique viscosity solution of 
		\begin{equation}\label{eliptico0}
			\begin{cases}
				-\Lambda_1^s z(x)=0&\text{in }\Omega,\\
				z=g&\text{in } \mathbb{R}^{N}\setminus\Omega.\\
			\end{cases}
		\end{equation}

		
When $g \equiv 0$ and $u_0$ is compactly supported inside $\Omega$ 
we have that there exists $T$ such that
$$
C e^{-\mu_1 t} \varphi_1 (x) \leq u(x,t)  \leq 0, \qquad x \in \overline{\Omega}, \, t>T,
$$ 
here
$\mu = \mu_1$ is the first eigenvalue
		of our operator and $\varphi_1$ is a negative eigenfunction.
		
\end{theorem}


For similar results established in the local case for the evolution problem associated with 
	\eqref{eq_intro_lamba1} we refer to \cite{BlancRossi}. 
	
		 \begin{remark}
		{\rm The results obtained in the present work could be extended to other kind of operators like 
		\begin{equation}\label{operador.j}
	\Lambda_j^s u(x)\coloneqq \sup_{S } \inf_{\theta\in S \cap \mathbb{S}^{N-1}}\int_{\mathbb{R}}
\dfrac{{u}(x+\tau \theta)-{u}(x)}{|\tau|^{1+2s}} d\tau.
\end{equation}
Here $S$ is a subspace of dimension $j$ in $\mathbb{R}^{N}$.
This extension could be obtained as soon as one proves a comparison
principle for this kind of operators. }
	\end{remark} 

	
	\medskip
	
	The paper is organized as follows: in Section \ref{sect-regul}
	we deal with the regularity result and prove Theorem \ref{regularidad.intro}; 
	in Section \ref{sect-eigen} we prove the existence of the first eigenvalue (with
	a corresponding eigenfunction); finally, in Section \ref{sect-evolution} 
	is devote to the study of
	the evolution problem, that is, we prove Theorem \ref{teo-1.intro} and 
	Theorem \ref{teo.conver.intro}. 
	
	\medskip
	
	We remark here that along the work we will denote by $C$ a positive constant that 
	may change from line to line and that we use $B_R(x)$ to denote the ball of radius 
	$R>0$ centered at $x\in\mathbb{R}^{N}.$


\section{Regularity, up to the boundary, for the elliptic problem} \label{sect-regul}
    We will assume that $s>\nicefrac12$ in the whole section and the main objetive will be the study of the H\"older regularity, up to the boundary, of a viscosity solution to
    \begin{equation}\label{eq:regularidad}
        \begin{cases}
		-\Lambda_1^s u(x)=f(x) & \mbox{in}\quad \Omega,\, s>\nicefrac12\\
		u(x)=g(x) & \mbox{in}\quad \mathbb{R}^N\setminus\Omega,
		\end{cases}
    \end{equation}
	{where $f$ is a bounded function and $g\in\mathcal{C}^{\gamma}$ for some $0<\gamma<2s-1$. First of all, it is worth to bear in mind that the notion of viscosity solution that we will use for the elliptic problem can be found in \cite{DPQR_FC}. We start by introducing some notation. For every $x\in B_1(0)$ and $\theta\in \mathbb{S}^{N-1},$ we
    define
    \[
	  d(x)\coloneqq (1-|x|^2)_+, \quad \mu(x,\theta)\coloneqq\frac{\langle x,\theta \rangle}{|x|}\in[-1,1],
    \]
    and
    \[
         p_{x,\theta}(t)\coloneqq t^2+2\mu(x,\theta)|x|t-d(x).
    \]
    We denote by $t_0(x,\theta)$ and $t_1(x,\theta)$ the two roots of the $p_{x,\theta}(t),$ that is,
    \begin{align*}
	&t_0(x,\theta)\coloneqq-\langle x,\theta \rangle-\sqrt{d(x)+\langle x,\theta \rangle^2}< 0,\\
	&t_1(x,\theta)\coloneqq-\langle x,\theta \rangle+\sqrt{d(x)+\langle x,\theta \rangle^2}>0,
    \end{align*}
    that satisfy
    \begin{equation}\label{tt01}
        t_0(x,\theta)t_1(x,\theta) =-d(x).
    \end{equation}
    
    Let us define our first barrier by	
    \[
	w_\gamma(x)\coloneqq d(x)^\gamma, \quad \gamma\in\mathbb{R},
    \]
    and the integral
    \[
        \mathcal{I}_{\theta} (f)(x)\coloneqq\int_{\mathbb{R}}
			\dfrac{f(x+\tau \theta)-f(x)}{|\tau|^{1+2s}} d\tau,\quad \theta\in\mathbb{S}^{N-1}.    
    \]
    
    Taking into account the previous notations, we can formulate our first lemma.
    
    \begin{lemma} \label{barrera_2}
	If $0<\gamma<2s-1$ then there exits $c_1>0$ such that
	\[
	    \mathcal{I}_{\theta} (w_\gamma)(x)\leq w_\gamma(x)  \left[
	    	-\frac{t_1(x,\theta)^{-2s} + |t_0(x,\theta)|^{-2s}}{2s}
	    -c_1\left(\frac{2\mu(x,\theta)|x|}{d(x)}\right)^{2s}\right],
	\]
	for all $ x\in B_1(0)$ and $\theta\in \mathbb{S}^{N-1}.$ In particular, 
	there exists $c_2>0$ such that
	\[
	    \mathcal{I}_{\theta}(w_\gamma(x))\leq -c_2 d(x)^{\gamma-2s},\quad x\in B_1(0),\, \theta\in \mathbb{S}^{N-1}.\]
    \end{lemma}
    \begin{proof}
        Let $ x\in B_1(0)$ and $\theta\in \mathbb{S}^{N-1}.$
        To simplify the notation, let us now temporarily write  $w=w_\gamma,$
	$\mu=\mu(x,\theta),$ and $t_i=t_i(x,\theta)$ with $i=0,1.$ First of all we notice that
	\begin{equation}
            \label{naux1}
            \begin{aligned}
	        \mathcal{I}_{\theta} (w)(x)&=\int_{\mathbb{R}}\dfrac{(1-|x+\tau\theta|^2)^\gamma_+-(1-|x|^2)^\gamma}{|\tau|^{1+2s}}d\tau\\
		&=\int_{\mathbb{R}}\dfrac{(1-|x|^2-2\mu |x|\tau-\tau^2)^\gamma_+-(1-|x|^2)^\gamma}{|\tau|^{1+2s}} d\tau\\
		&=\int_{\mathbb{R}\setminus(t_0,t_1)}\dfrac{(1-|x|^2-2\mu |x|\tau-\tau^2)^\gamma_+-(1-|x|^2)^\gamma}{|\tau|^{1+2s}} d\tau
		+ J,
	    \end{aligned}
        \end{equation}
	where
	\[
	    J\coloneqq\int_{t_0}^{t_1}
			\dfrac{(1-|x|^2-2\mu |x|\tau-\tau^2)^\gamma-(1-|x|^2)^\gamma}{|\tau|^{1+2s}} d\tau.
	\]
	Since by definition of $t_i$, $i=0,1$, it is clear that
	\[
	    1-|x|^2-2\mu |x|\tau-\tau^2< 0, \quad\tau\in (-\infty,t_0)\cup(t_1,\infty),
	\]
        from \eqref{naux1} we get
	\[
	    \mathcal{I}_{\theta} (w)(x)\le -\frac{w(x)}{2s}\left( t_1^{-2s}  + |t_0|^{-2s}\right)+J.
	\]
	So, to conclude, we just need to estimate $J$. For that, by symmetry, we split the rest of the proof in the two cases $\mu>0$ and $\mu=0.$

        \medskip

	\noindent {\it Case 1: $\mu>0.$} Since it is clear that
		\begin{equation}\label{Rada1}
			J\leq\int_{t_0}^{t_1}
				\dfrac{(1-|x|^2-2\mu |x|\tau)^\gamma-(1-|x|^2)^\gamma}{|\tau|^{1+2s}} d\tau
				= w(x) \int_{t_0}^{t_1}
				\dfrac{\left(1-\tfrac{2\mu |x|}{d(x)}\tau\right)^\gamma-1}{|\tau|^{1+2s}} d\tau,
		\end{equation}
		by making the change of variable $\upsilon=\frac{2\mu |x|}{d(x)}\tau$ we get
		\[
			J\leq w(x) \left(\frac{2\mu|x|}{d(x)}\right)^{2s}
			\int_{\upsilon_0}^{\upsilon_1}\frac{(1-\upsilon)^\gamma-1}{|\upsilon|^{1+2s}}d\upsilon=:w(x) \left(\frac{2\mu|x|}{d(x)}\right)^{2s}
J_1,
		\]
		where $\upsilon_i=\tfrac{2\mu |x|}{d(x)}, t_i=1-\tfrac{t^2_i}{d(x)}$, $i=0,1.$ Proceeding now as in \cite[Lemma B.1]{BCGJ}, by making the change of variable  $e^\rho =1-\upsilon$, only in  $\rho \in [\rho_1(x), \rho_0(x)]$ (not in all the Euclidean space as in \cite{BCGJ}) where
		$\rho_i(x)=\ln \left(\tfrac{\upsilon_i^2}{d(x)}\right)$, $i=0,1$, we obtain that 
		\[
			J_1=2^{-2s}\int_{\rho_1}^{\rho_0}e^{(\gamma+1-2s)\frac{\rho}2}
				\sinh\left(\gamma \frac{\rho}2\right)|\sinh(\rho)|^{-(1+2s)}d\rho
				\begin{cases}
					<0 &\text{if } \gamma<2s-1,\\
					=0 &\text{if } \gamma=2s-1.\\
				\end{cases}
		\]
		Observe that, by \eqref{tt01}, in the previous estimate, we have used that $|\rho_0(x)|=|\rho_1(x)|$ for all $x\in B_1(0)$. Thus the conclusion follows in the case $\mu>0$.

		\medskip

		\noindent {\it Case 2: $\mu=0.$} As now $t_0=-\sqrt{d(x)},$ $t_1=\sqrt{d(x)}$, by symmetry of the integral it is follows that
		$$
		\begin{aligned}
		J&=2 w(x)\int_{0}^{\sqrt{d(x)}}\dfrac{(1-|x|^2-\tau^2)_{+}^\gamma-(1-|x|^2)^\gamma}{|\tau|^{1+2s}}d\tau\\
&=2w(x)\int_{0}^{\sqrt{d(x)}}\frac{\left(1-\frac{\tau^2}{d(x)}\right)^{\gamma}-1}{|\tau|^{1+2s}}\, d\tau\\
&=2\frac{w(x)}{d(x)^s}\int_{0}^{1}\frac{(1-z)^{\gamma}-1}{|z|^{1+s}}\, dz<-C\frac{w(x)}{d(x)^s},
\end{aligned}
$$
for some $C>0$ as wanted.
	\end{proof}

	Let us define the second barrier
		\[
			v_\gamma(x):=|x|^\gamma,  \quad \gamma\in\mathbb{R},
		\]
for which we have the next results.
	\begin{lemma}\label{barrera_1}
		For every
		$0<\gamma<2s-1$ there exists
		 $C>0$ such that
		\[
			\mathcal{I}_{\hat{x}} (v_\gamma)(x)\le -C|x|^{\gamma-2s},\, 
			x\in \mathbb{R}^{N}\setminus\{0\},
		\]
		where $\hat{x}=\tfrac{x}{|x|}\in \mathbb{S}^{N-1}.$
	\end{lemma}
	\begin{proof}
		Let $x\in \mathbb{R}^{N}\setminus\{0\},$ and $\gamma\in\left(0,2s-1\right)$. 
		By using the change of variable $t=\tfrac{\tau}{|x|},$ we have that
		\begin{equation}\label{b2_1}
			\begin{aligned}
				\mathcal{I}_{\hat{x}} (v_\gamma)(x)&=|x|^{\gamma-2s}
				\int_{\mathbb{R}}
				\dfrac{ |1+t|^\gamma-1}
				{|t|^{1+2s}}dt,\\
			
			
			
			
			
			
			
			
			\end{aligned}
		\end{equation}
		where the last integral is finite and negative by using, one more time, the ideas of \cite[Proof of Lemma B.1]{BCGJ}.
	\end{proof}


\begin{lemma} \label{cota1+} 
For every $0<\gamma<2s$ and $y_0\in\mathbb{R}^{N}$ there exists $C > 0$ such that
$$\Lambda_N^s (|x-y_0|)^\gamma\leq C (|x-y_0|^{\gamma-2s})\ ,\,x\in\mathbb{R}^{N},$$
here $\Lambda_N^s$ is a maximal operator define as
\begin{equation}\label{operador+}
	\Lambda_N^s (v)(x)\coloneqq\sup_{\theta\in\mathbb{S}^{N-1}}\int_{\mathbb{R}}
\dfrac{{v}(x+\tau \theta)-{v}(x)}{|\tau|^{1+2s}} d\tau.
\end{equation}
		 
	\end{lemma}
\begin{proof} Follows the same lines of the proof of \cite[Lemma 3.3]{DQT}.
      \end{proof}

      
	To prove our last auxiliary result we have to take into account the geometry properties of our domain. In particular we will use that, by \cite[Lemma 6.2]{BGI},
	\begin{equation}\label{geometria}
		\Omega= \bigcap\limits_{z\in \partial\Omega}B_R(z-R\nu(z)),
	\end{equation}
for some $R>0$ whose value it is related with the principal curvatures of $\partial\Omega$ and $\nu(z)$ denotes the outward normal unit vector of $\Omega$ in $z\in\partial\Omega$.
Now we can establish the following result.
	\begin{lemma} \label{cota11}
		Let $u$ be a {viscosity}  solution  of \eqref{eq:regularidad} with $g$ satisfying \eqref{hip} and \eqref{holder-g}. Then there exists $C>0$ such that
 		\[
 		u(x)-g(z)\geq -C \mbox{dist}(x,\partial\Omega)^\gamma, \quad x\in\overline{\Omega},
 		\]
with $z \in \partial \Omega$ verifying that $\mbox{dist}(x,\partial\Omega)=|x-z|$.
		
			
	\end{lemma}

	\begin{proof} Let us first consider the case $g=0$. For that let $\gamma\in(0,2s-1),$ $z\in \partial\Omega$ and $M>0$ a positive constant that will be chosen later. We define the nonpositive function
		\[
			g_{z,M}(x)\coloneqq -M\left(R^2-|x-y_z|^2\right)_{+}^\gamma,\quad\mbox{where}\quad y_z:=z-R\nu(z),\quad x\in\mathbb{R}^{N}.
		\] 
		Since, by Lemma \ref{barrera_2}, we know that there exists $C>0$ such that
		\[
			-\Lambda_1^s g_{z,M}(x)\le -CM\left(R^2-|x-y_z|^2\right)^{\gamma-2s},\quad x\in B_R(y_z),
		\]
		using the fact that $\gamma-2s\le0,$ we get
		\begin{equation}\label{desp}
			-\Lambda_1^s g_{z,M}(x)\le -CMR^{2(\gamma-2s)},\quad
			x\in B_R(y_z).
		\end{equation}
		Taking now $M\ge \tfrac{\|f\|_{\infty}}{CR^{2(\gamma-2s)}},$ (independent of $z$), it is clear that
		\[
			-\Lambda_1^s g_{z,M}(x)\le -\|f\|_{L^{\infty}(\Omega)}
			\le -\Lambda_1^s u(x), \quad 
			x \in B_R(y_z).
		\]
		
		
		On the other hand, since by \eqref{geometria} $\Omega\subseteq B_R(y_z)$, we also have that $g_{z,M}(x)=0 =u(x)$ for any $x\in\mathbb{R}^N\setminus B_R(y_z)\,$. Thus the comparison principle given in \cite{DPQR_FC} allows us to affirm
		\[
			u(x)\ge g_{z,M}(x),\quad x\in\mathbb{R}^N.
		\]
		
		Let us now consider a fixed but arbitrary $x\in \Omega$. Taking $z\in\partial\Omega$ such that
		$|z-x|=\mbox{dist} (x, \partial\Omega)$ by using the Cauchy–Schwarz inequality we conclude that
		\begin{align}
			g_{z,M}(x)&=-M\left(R^2-|x-z+R\nu(z)|^2\right)^\gamma\\
			&=-M\left(-|x-z|^2-2R(x-z)\cdot\nu(z)\right)^\gamma\\
			&=-M\left||x-z|^2+2R(x-z)\cdot\nu(z)\right|^\gamma\\
			&\ge-M\left(|x-z|^{2\gamma}+2R|x-z|^\gamma\right)\\
			&=-M|x-z|^\gamma\left(|x-z|^{\gamma}+2R\right)\\
			&\ge-M\mbox{dist} (x, \partial\Omega)^\gamma\left(\max_{y\in\Omega,\, z\in\partial\Omega}|y-z|^{\gamma}+2R\right),\label{despp}
		\end{align}
		as wanted.
		
		
		
		In the case $g\not=0$ we define, for every $x\in\mathbb{R}^{N}$ the auxiliary function,
		$$h(x)=g(z)-M_g|x-z|^{\gamma}+\mu g_{z,M}(x), z\in\partial\Omega,$$
		where $M_g$ is the H\"older constant given in \eqref{holder-g} and $\mu>0$ is a fixed, but arbitrary, parameter that will be taken later. Applying the standard extremal inequality
		$$-\Lambda_1^sv_1\leq- \Lambda_1^sv_2+ \Lambda_N^s(v_2-v_1),$$ 
		for $v_1=h(x)$ and $v_2=\mu g_{z,M}(x)$, Lemma \ref{cota1+} and \eqref{desp}, we get
		$$-\Lambda_1^sh(x)\leq M_g|x-z|^{\gamma-2s}-CM\mu R^{2(\gamma-2s)},$$
		with $x\in B_{R}(y_z),\, y_z=z-R\nu(z)$. Thus by using the fact that $\gamma<2s$ by choosing $\mu$ large enough, we obtain that $-\Lambda_1^sh(x)\leq -\|f\|_{L^{\infty}(\Omega)}$ in $B_R(y_z)$. Since, by \eqref{holder-g},  it is also true that $h(x)\leq g(x),\, x\in \mathbb{R}^{N}\setminus B_R(y_z)$ we obtain that the function $h$ is a subsolution of \eqref{eq:regularidad}. Thus, by comparison we have that $h(x)\leq g(x) $ for all $x \in  \mathbb{R}^N$. 
		Doing as we did in the case $g=0$ we consider now $z$ such that $|z-x|=\mbox{dist} (x,\partial\Omega)$. So that, by using \eqref{despp} we conclude 
		$$u(x)-g(z)\geq  h(x)-g(z)\geq -C\mbox{dist} (x, \partial\Omega)^\gamma.$$
		as wanted.		
	\end{proof}
	
	We are now able to conclude this section by proving the regularity result. 

	\begin{demoregularidad}

		We proceed again as in \cite{BGI}.
		Let us fix $\gamma\in(0,2s-1)$ and some points, fixed but arbitrary, $x_0,y_0\in\bar\Omega$. We have to prove that there exists $L>0$ such that
	\begin{equation}\label{puntito}
	u(x_0)\leq u(y_0)+L|x_0-y_0|^{\gamma}.
	\end{equation}	
For that, on one hand we notice that, by Lemma \ref{barrera_1}, there exists $C>0$, independent of
		$y_0,$ satisfying
$$-\Lambda_1^s H_{y_0}(x) \ge CL|x-y_0|^{\gamma-2s},\, x\neq y_0,$$ 
where $$H_{y_0}(x):=u(y_0)+L|x-y_0|^{\gamma},\quad x\in\mathbb{R}^{N}.$$ Therefore we can choose $L>0$ in order to have
\begin{equation}\label{ppa}
-\Lambda_1^s H_{y_0}(x) \ge CL\delta^{\gamma-2s}\geq\|f\|_{L^{\infty}(\Omega)}\geq -\Lambda_1^s u(x),\quad |x-y_0|<\delta,\, x\in\Omega,
\end{equation}
for a fixed but arbitrary $0<\delta<1$. Let us now prove that
\begin{equation}\label{ppa2}
u(x)\leq H_{y_0}(x)\quad\mbox{ in $\mathbb{R}^N\setminus (B_{\delta}(y_0)\cap\Omega)$.}
\end{equation}
Indeed if $x\in \Omega\cap (\mathbb{R}^N\setminus B_{\delta}(y_0))$ we can pick $L>0$ large enough such that \eqref{ppa2} is satisfied by using the fact that $u$ is a bounded function and $\gamma>0$. Moreover when $x\in\mathbb{R}^N\setminus \Omega$ applying Lemma \ref{cota11} we know that there exists $C>0$, independent of $y_0$, such that
$$u(x)-u(y_0)=-u(y_0)\leq C\mbox{dist}(y_0,\partial\Omega)^{\gamma}\leq C|x-y_0|^{\gamma}\leq L|x-y_0|^{\gamma},$$
by choosing $L\geq C$. Therefore, from \eqref{ppa} and \eqref{ppa2}, applying one more time the comparison principle given in \cite{DPQR_FC} we conclude that
$$u(x)\leq H_{y_0}(x),\quad x\in\mathbb{R}^{N},$$
that implies \eqref{puntito} by taking $x=x_0\in\bar{\Omega}.$
 

For the general case $g\not =0$ we borrow some ideas of \cite{DQT}. For that we take $y_0\in\overline{\Omega}$, and {a small parameter $\nicefrac12>{\delta}$.} For some $L>0$, fixed but arbitrary, we define the function 
$$\Phi (x):= u(x)-u(y_0)-L|x-y_0|^\gamma.$$
We claim that, for $L$ large enough, just in terms of the data and ${\delta}$, but not on $y_0$, $\Phi$ is nonpositive in $\mathbb{R}^{N}$. We notice that, from this, we get that
$$u(x)-u(y_0)\leq L|x-y_0|^\gamma \quad \mbox{for all}\quad \mathbb{R}^{N},$$
so, since $L$ does not depend on $y_0$, interchanging the role of $y_0$ and $x$ we conclude the $\gamma$-H\"older regularity of $u$ as wanted.
Suppose by contradiction that the previous claim is false, that is 
\begin{equation}\label{vv}
\sup_{\mathbb{R}^{N}} \Phi> 0.
\end{equation}
With an initial choice 
{$L \geq 2\nicefrac{\|u\|_\infty}{{\delta}^\gamma}$
it is follows that the supremum, that has to be positive, is achieved at some point $\bar x\neq y_0$, $\bar x\in B_{{\delta}}(y_0)$. Let us now prove that this would be impossible by the fact that $\bar x$ cannot belong to $\Omega$ nor to $\mathbb{R}^N\setminus\Omega$. Indeed, if $\bar x\in\Omega$} we consider $\phi(x)\coloneqq H_{y_0}(x) \coloneqq u(y_0){+}L|x-y_0|^\gamma$ as a test function  (regarded as a subsolution to the problem) at the maximum point $\bar x$ 
to obtain
$$-\Lambda_1^s H_{y_0}(\bar x) \le f,$$
{that contradicts \eqref{ppa} by taking $L>0$ large enough.
Let us now conclude by proving that $\bar x \in  (\mathbb{R}^{N}\setminus \Omega)$ is not possible too}. In fact, {if $\mbox{dist} (y_0,\partial\Omega):=|y_0-z|,\, z\in\partial\Omega$}, since by Lemma \ref{cota11}, we have
\[
u(\bar x)-u(y_0)\leq u(\bar x)-g(z)+g(z)-u(y_0) \leq g(\bar x)-g(z)+C \mbox{dist}({y_0},\partial\Omega)^\gamma,\, \bar x \in  (\mathbb{R}^{N}\setminus \Omega),
 \]
by the H\"older condition for $g$ we get
\begin{equation}\label{desppp}
u(\bar x)-u(y_0)\leq M_g |  \bar x-z |^\gamma+C |y_0-z|^\gamma \leq  \tilde C |\bar x -y_0|^\gamma.
\end{equation}
We notice that to obtain the last inequality we have used the simple fact that $|y_0-z| \leq |y_0-\bar x|$ and, therefore, $|\bar x -z|\leq 2 |\bar x -y_0|$.	We conclude by noticing that \eqref{desppp} is also not possible by taking $L \geq \tilde C$ since this implies $\Phi(\bar x)\leq 0$ against the hypothesis \eqref{vv}. So that the regularity follows.







	\end{demoregularidad}

\section{Eigenvalue problem} \label{sect-eigen}

	Before starting the study of the eigenvalue problem, we need to establish a maximum principle given in the next	lemma.
	\begin{lemma}[Strong Maximum Principle] \label{SMP}
		Let $u$ be a non-positive  viscosity solution of 
		$$
		 \begin{cases}
		-\Lambda_1^s u(x)\leq 0 & \mbox{in}\quad \Omega,\\
		u(x)=0 & \mbox{in}\quad \mathbb{R}^N\setminus\Omega.
		\end{cases}		
$$ Then $u =0$ or $u<0$.
	\end{lemma} 
	\begin{proof} 
		Let us suppose by contradiction that there exits $x_0 \in \Omega$ 
		such that $u(x_0)=0$ and $u<0$ in $B_r(x_1)\subset \Omega$ for some $r>0$, $x_1\in\Omega$.
 		We take now a non-positive function $\varphi \in \mathcal{C}^\infty(\bar\Omega)$  above $u$ 
 		(that is, $u \leq \varphi\leq 0$) such that 
 		$$spt (\varphi) \subset  B_r(x_1)\quad\mbox{ and }\quad  \varphi(x)<0,\quad x\in B_{\frac{r}2}(x_1).$$
		Thus, if we define $$\theta:=\frac{x_1-x_0}{|x_1-x_0|}\in\mathbb{S}^{N-1}$$ and we chose $\delta>0$ small enough such that
 		$B_\delta(x_0)\subset \Omega$, we get 
 		\[
 			\int_{-\delta}^{\delta}\frac{-{\varphi }(x_0+\tau \theta)}
			{|\tau|^{1+2s}} d\tau+ \int_{\mathbb{R}\setminus(-\delta,\delta)}
			\frac{-{u }(x_0+\tau \theta)}{|\tau|^{1+2s}} d\tau>0.		
		\]
  		Since $\varphi(x_0)=0$ the previous inequality implies
  		\begin{equation}\label{simpli}
  			\sup_{z\in\mathbb{S}^{N-1}}\int_{-\delta}^{\delta}\frac{\varphi(x_0)-{\varphi} (x_0+\tau z)}
			{|\tau|^{1+2s}} d\tau+ \int_{\mathbb{R}\setminus(-\delta,\delta)}
			\frac{\varphi(x_0)-{u }(x_0+\tau z)}{|\tau|^{1+2s}} d\tau
			>0.
		\end{equation}
		By using the fact that $x_0$ is a local maximum for 
		$u-\varphi$ and $-\Lambda_1^s u\leq 0$ in the viscosity sense, we also get 
		$$
			-\inf_{z\in\mathbb{S}^{N-1}}\int_{-\delta}^{\delta}\frac{-\varphi(x_0)+{\varphi} (x_0+\tau z)}
			{|\tau|^{1+2s}} d\tau+ \int_{(-\delta,\delta)^c}
			\frac{-\varphi(x_0)+{u }(x_0+\tau z)}{|\tau|^{1+2s}} d\tau
			\leq 0,
		$$
		that is in contradiction with \eqref{simpli}.
	\end{proof}
	
	We can now show the main result of this section.	
	\begin{theorem} \label{autof}
	    Assume $s>\nicefrac12.$
		There exist $0>\varphi_1\in\mathcal{C}(\mathbb{R}^{N})$ 	
			and $\mu_1>0$ solving the Dirichlet problem
		$$
			\begin{cases}
				-\Lambda_1^s \varphi_1(x)=\mu_{1}\varphi_1 
				&\text{in }\Omega,\quad\\
			 \varphi_1=0&\text{in } \mathbb{R}^{N}\setminus\Omega,\\
				\end{cases}
		$$
		in the viscosity sense.
	\end{theorem}

	\begin{proof}
		The idea is to apply the above Krein-Rutman Theorem in the real Banach of continuous functions  
		$X\coloneqq \mathcal{C}(\overline\Omega)$, with  the uniform norm. 
		The cone will be the set of nonpositive continuous functions 
		$K\coloneqq\mathcal{C}_{-}(\overline\Omega)$. 
		Notice that this produce a ``reverse inequality" since the standard is the nonnegative cone. 
		In fact, given $x,y \in X$  $ x\preceq y \iff y-x \in \mathcal{C}_{-}(\overline\Omega)$ 
		so  $y\leq x$ in $\overline\Omega$. For the  definition of 
		$\mathcal T$ we proceded to invert the operator $-\Lambda_1^s$. 
		For every $v\in\mathcal{C}(\overline{\Omega})$ we know that 
		there exists an unique viscosity solution $u\in\mathcal{C}(\overline{\Omega})$ of the problem
		$$
			\begin{cases}
			-\Lambda_1^s u=v &\text{in }\Omega,\quad\\
			u=0&\text{in } \mathbb{R}^{N}\setminus\Omega,\\
				\end{cases}
		$$
		{(see \cite[Theorem 1.1]{DPQR_FC})} then the operator 
		$\mathcal T\colon\mathcal{C}(\overline\Omega)\to 
		\mathcal{C}(\overline\Omega)$ given by 
		$\mathcal{T}(v)\coloneqq u$, is well defined.  
		Moreover, the operator $-\Lambda_1^s$ satisfies comparison principle 
		{(see \cite[Theorem 2.2]{DPQR_FC})}  
		thus $\mathcal{T}$ is increasing and also positively $1-$homogeneous.  
		With  the same arguments as in  \cite[Lemma 4.5]{CS}  (see also \cite[Corollary 4.7]{CS}) 
		$\mathcal{T}$ continuos with the uniforme norm.

		Moreover using the trivial function as a super solution and comparison we get that if 
			$v\in \mathcal{C}_{-}(\overline\Omega)$ then  $T(u_0)\in \mathcal{C}_{-}(\overline\Omega)$.
	So 
	$$
		\mathcal T:\mathcal{C}_{-}(\overline\Omega)
		\to \mathcal{C}_{-}(\overline\Omega),
	$$
	is weel define.
	
	 Now by Theorem \ref{regularidad.intro} we also know that if $v_n$ is a uniforme bound sequence then
	$$
		\|\mathcal T(v_n)\|_{\mathcal{C}^{\beta}
		(\overline\Omega)}\leq C,
	$$
	for some $C>0$ (independent of  $n$) and  
	for some $\beta\in (0,1)$, thus we get that the operator is 
	compact using Arzela-Ascoli Theorem.

	If we take $v_0 \in K$ non trivial with compact support in $\Omega$ then $\mathcal T(v_0)<0$ 
	by the strong maximum principle so there exists  $M$ large such 
	that  $M\mathcal  T(v_0)\leq v_0$. therefore we have check all the requirements to apply the Krein-Rutman 
	Theorem that allows us to conclude the existence of 
	$\varphi_1\in\mathcal{C}_{-}(\overline\Omega)$, 
	$\varphi_1\neq 0$ and $\mu_1>0$ such 
	that $\mathcal T(\varphi_1)=\mu_1\varphi_1$. 
	This implies the existence of an eigenvalue for 
	$-\Lambda_1^s$ associated to a negative eigenfunction 
	$\varphi_1$ as we wanted to show. Finally, notice that $\varphi_1<0$ by the strong maximum principle given in Lemma \ref{SMP}.
	\end{proof}
	
	\begin{remark}
		{\rm It is important to bear in mind that, as occurs in the local case, the simplicity of the eigenvalue found in the previous results is nowadays an open problem. }
	\end{remark}
\section{Evolution problem} \label{sect-evolution}
	This section is devoted to study the evolution problem given in 
	\eqref{ev_eq} under some conditions on the datum. We start by introducing the notion 
	of viscosity solution we will work with.
	
	\medskip
	
	Let $T>0.$ A bounded upper (resp. lower) semicontinuous function
	$u\colon\mathbb{R}^N\times[0,T]\to\mathbb{R}^N$
	is said to be a viscosity subsolution  (resp. supersolution) of 
	\begin{equation}\label{ev_eq_T}
		\begin{cases}
			u_t (x,t) =\Lambda_1^s u(x,t) &\text{in }\Omega\times(0,T),\\
			u(x,t)=g(x) &\text{in }
				(\mathbb{R}^N\setminus\Omega)\times(0,T),\\
			u(x,0)=u_0(x) &\text{in }\mathbb{R}^N,
		\end{cases}
	\end{equation}
	if
	\begin{itemize}
		\item $u(x,t)\leq g(x)$ in $(\mathbb{R}^N\setminus\Omega)\times(0,T);$
		\item $u(x,0)\leq u_0(x)$ in $\mathbb{R}^N;$
		\item For every smooth function 
			$\varphi\colon\mathbb{R}^N\times[0,\infty)\to \mathbb{R}$
			in every maximum (resp. minimum) point of $(x_0,t_0)\in\Omega\times (0,T)$ of
			$u-\varphi$ in $B_\delta(x_0)\times (t_0-\delta,t_0+\delta)
			\subseteq \Omega\times (0,T)$ for some $\delta>0$ we have that
			\[
				\bar{u}_t(x_0,t_0)-\Lambda_1^su(x_0,t_0)\le0\, (resp. \ge0)
			\]
			where
			\[
				\bar{u}(x,t)=
				\begin{cases}
					\varphi(x,t) &\text{ if } x\in
						B_\delta(x_0)\times (t_0-\delta,t_0+\delta),\\
					u(x,t) &\text{ if } x\not\in
						B_\delta(x_0)\times (t_0-\delta,t_0+\delta),
				\end{cases}
			\]
	\end{itemize}
	
	A bounded function $u\colon \mathbb{R}^N\times[0,T]\to\mathbb{R}^N$ is a
	viscosity solution of \eqref{ev_eq_T} if it is both subsolution and supersolution.
	
	\medskip
	
	Then, a bounded upper (resp. lower) semicontinuous function
	$u\colon\mathbb{R}^N\times[0,\infty) \to\mathbb{R}^N$
	is said to be a viscosity subsolution  (resp. supersolution) of \eqref{ev_eq} if
	$u$ is  a viscosity subsolution  (resp. supersolution) of \eqref{ev_eq_T} for any 
	$T>0.$ Consequently, a continuous function 
	$u\colon \mathbb{R}^N\times[0,\infty)\to\mathbb{R}^N$ is a
	viscosity solution of \eqref{ev_eq} if it is both subsolution and 
	supersolution.
	
	\medskip
	
	 	
	Our next aim is to show that, under reasonable hypothesis, there exists a unique viscosity solution of
	\eqref{ev_eq}, that is, to obtain the proof of Theorem \ref{teo-1.intro}. The proof is obtained as usual: first, we show a comparison 
	principle and then we conclude by the Perron's method using adequate sub and supersolutions.

	\begin{theorem} \label{comparacion}
		Let $T>0$, $u$ and $v$ be  viscosity subsolution and supersolution of 
		\eqref{ev_eq_T} respectively. If 
		$u_0$ and $g$ satisfy \eqref{hip} then
		$u\le v$ on $\Omega\times [0,T]$.
	\end{theorem}
	
	\begin{proof}
 		Following the  ideas given in \cite{CIL}, 
 		we {\it double variables} defining the function
 		$$
 			\widetilde{u}(x,t)\coloneqq u(x,t)-
 			\tfrac{\varepsilon}{T-t},\quad\varepsilon>0,
 		$$
  		that satisfies, in the viscosity sense,
  		$$
  			\begin{cases}
			\widetilde {u}_t(x,t)-\Lambda_1^s \widetilde{u}(x,t)\leq-
			\tfrac{\varepsilon}{(T-t)^2}<0 &\text{in }\Omega\times(0,T),\\
			\widetilde{u}\to -\infty, &\text{when }
				t\to T^{-}.\\
				\end{cases}
		$$
		It is clear that, if we prove 
		$\widetilde{u}(x,t)\leq v(x,t)$ for $x\in\Omega$ and $t\in (0,T)$ 
		then the conclusion follows by taking $\varepsilon\to 0$.  
		Therefore let us assume that 
  		$$
  		 	\begin{cases}
				u_t(x,t)-\Lambda_1^s u(x,t)<0 &\text{in }\Omega\times(0,T),\\
				u\to -\infty, &\text{when }
					t\to T^{-},\\
			\end{cases}
		$$
 		and we suppose, by contradiction, that
 		$$
 			\sup_{x\in\overline{\Omega},\, t\in [0,T)}{u(x,t)-v(x,t)}
 			\coloneqq\eta>0.
 		$$
 
 		Since $u$ and $-v$ are bounded from above, 
 		$\overline\Omega$ is compact and $u\to -\infty$ as $t\to T^{-}$ then,  
 		for some $\alpha$ big enough that will be chosen later, there exists 
 		$(x_{\alpha}, y_{\alpha}, t_{\alpha}, r_{\alpha})$ the maximum point of
 		$$
 			G_{\alpha}(x,y,t,r)\coloneqq u(x,t)-v(y,r)-
 			\alpha(|x-y|^{2}+|t-r|^2),\, (t,r)\in [0,T)^2,
 			 (x,y)\in \overline{\Omega}\times \overline{\Omega}.
 		$$
 		If we denote
 		$$
 			M_{\alpha}\coloneqq 
 			G_{\alpha}(x_{\alpha}, y_{\alpha}, t_{\alpha}, r_{\alpha})(\geq \eta>0),
 		$$
 		it is clear that 
 		$M_{\alpha}\searrow \widetilde{M}$ as $\alpha\to\infty,$ 
 		for some $\widetilde{M}\geq\eta>0$. 
 		Moreover, since
 		$$
 			M_{\frac{\alpha}{2}}
 			\geq G_{\frac{\alpha}{2}}(x_{\alpha}, y_{\alpha}, t_{\alpha}, 
 			r_{\alpha})
 			=M_{\alpha}+\frac{\alpha}{2}(|x_{\alpha}-y_{\alpha}|^{2}
 			+|t_{\alpha}-r_{\alpha}|^2),
 		$$
		it follows that 
		$$
			\lim_{\alpha\to\infty}\alpha |x_{\alpha}-y_{\alpha}|^{2}
			=\lim_{\alpha\to\infty}
			\alpha |t_{\alpha}-r_{\alpha}|^{2}= 0.
		$$
		We notice that the previos limits also implies that, if, by subsequence,
		\[
			\lim_{\alpha\to\infty}(x_\alpha,y_\alpha)\coloneqq
			(\widetilde x, \widetilde y)
			\in \overline{\Omega}\times \overline{\Omega},\text{ and } 
			\lim_{\alpha\to\infty}(t_\alpha,r_\alpha)=(\widetilde t, \widetilde r)
			\in [0,T)\times [0,T),
		\]
		then $\widetilde x=\widetilde y$ and $\widetilde t=\widetilde r.$
		We also observe that, since
		$$
			0<\eta\leq\widetilde{M}=\lim_{\alpha\to\infty}
			G_{\alpha}(x_{\alpha}, 
			y_{\alpha}, t_{\alpha}, r_{\alpha})
			\leq u(\widetilde x, \widetilde t)-v(\widetilde x, 
			\widetilde t)\leq \eta,
		$$
		clearly 
		\begin{equation}\label{mg}
		\eta=u(\widetilde x, \widetilde t)-v(\widetilde x, \widetilde t).
		\end{equation}
		By using the facts that $u(\cdot,t)\leq v(\cdot, t)$ in $\partial\Omega$ 
		and $u(x,0)=v(x,0),$ \eqref{mg} implies that 
		$\widetilde x=\widetilde y\notin \partial\Omega$ and 
		$\widetilde t=\widetilde r\neq 0$. Therefore, we can define
		\begin{equation}\label{interior}
			d_{\alpha}
			\coloneqq\min\{d(x_\alpha,\partial\Omega),\, 
			d(y_\alpha,\partial\Omega)\}>0,\mbox{ and }\delta_{\alpha}
			\coloneqq\min  \{t_{\alpha},\, r_{\alpha}\}>0.
		\end{equation}
		
		Without loss of generality, we can also assume that
		\begin{equation}\label{tecnico}
			\frac{d(\widetilde{x},\partial\Omega)}2<d_\alpha\text{ and }
				0<t_{\alpha}\leq r_{\alpha}.
		\end{equation} 
		
		We define now the regular function
		$$
			\psi_{\alpha}(z,t)\coloneqq v(y_{\alpha}, r_{\alpha})
			+\alpha(|t-r_{\alpha}|^2+|z-y_{\alpha}|^2),
		$$ 
		that clearly satisfies 
		\begin{equation}\label{entiempo}
			\partial_{t}\psi_{\alpha}(x_{\alpha}, t_{\alpha})=2\alpha(t_\alpha-r_\alpha).
		\end{equation} 
		Moreover $\psi_{\alpha}(z,t)$ touches $u$ from above in $(x_\alpha,t_{\alpha})$. 
		Indeed, by \eqref{interior}, if we consider $w\in\mathbb{R}^{N}$ and $\delta>0$
		such that $x_{\alpha}+w\in \Omega$ and $t_{\alpha}+\delta\in [0,T)$ since 			\[
			G_{\alpha}(x_{\alpha}, y_{\alpha}, t_{\alpha}, r_{\alpha})\geq 
			G_{\alpha}(x_{\alpha}+w, y_{\alpha}, t_{\alpha}+\delta, r_{\alpha}),
		\]
		 by \eqref{tecnico} we get that
		$$
			u(x_\alpha,t_{\alpha})-\alpha|x_{\alpha}-y_{\alpha}|^2
			-u(x_{\alpha}+w,t_{\alpha}+\delta)+\alpha|x_{\alpha}+w-y_{\alpha}|
			\geq -\alpha\delta^2.
		$$
		Thus
		$$
			u(x_{\alpha},t_{\alpha})-\psi_{\alpha}(x_{\alpha},t_{\alpha})
			\geq u(x_{\alpha}+w,t_{\alpha}+\delta)-\psi_{\alpha}(x_{\alpha}
			+w,t_{\alpha}+\delta),
		$$
		as wanted. Therefore, since $u$ is a subsolution, 
		by \eqref{entiempo} it follows that
		\begin{equation}\label{uno}
				0\geq 
				2\alpha(t_\alpha-r_\alpha)-\Lambda_1\psi_{\alpha}(x_{\alpha},t_{\alpha})
				=-\inf\{E_{z,\delta}(u,\psi_\alpha,x_\alpha,t_\alpha)\colon 
				z\in\mathbb{S}^{N-1}\}.
		\end{equation}
		for any $\delta<d_\alpha.$ Here
		$$
		\begin{array}{l}
			\displaystyle E_{z,\delta}(w,\phi,x,t)\coloneqq
			I^1_{z,\delta}(\phi,x,t)+
			I^2_{z,\delta}(w,x,t)+I^3_{z}(w,x,t)-2\alpha(t_\alpha-r_\alpha),\\[7pt]
			\displaystyle I^1_{z,\delta}(\phi,x,t)\coloneqq\int_{-\delta}^\delta
			\frac{\phi(x+\tau z,t)-\phi(x,t) }{|\tau|^{1+2s}} d\tau,\\[7pt]
			\displaystyle I^2_{z,\delta}(w,x,t)\coloneqq\int_{A_\delta^{z}(x)}
			\hspace{-.6cm}\frac{w(x+\tau z,t)-w(x,t) }{|\tau|^{1+2s}} d\tau
			\\[7pt] \qquad \text{ where } 
			A_\delta^{z}(x)\coloneqq \{t\in \mathbb{R}
			\setminus(-\delta,\delta)\colon x+tz\in\Omega\},\\[7pt]
			\displaystyle I^3_{z}(w,x,t)\coloneqq\int_{B^{z}(x)}
			\hspace{-.6cm}
			\frac{g(x+\tau z,t)-w(x,t) }{|\tau|^{1+2s}} d\tau \\[7pt]
			\qquad \text{ where } 
			B^{z}(x)\coloneqq \{t\in \mathbb{R}\colon x+tz\not\in\Omega\},
		\end{array}
		$$
		for $w$ and $\phi$ regular functions and $(x,t)\in\mathbb{R}^N\times(0,T).$
		Defining now 
		$$
			\varphi_{\alpha}(z,t)\coloneqq
			u(x_{\alpha}, t_{\alpha})-\alpha(|t-t_{\alpha}|^2+|z-x_{\alpha}|^2),
		$$
		reasoning in a similar way, we also get that $\varphi_{\alpha}(z,t)$ touches 
		$v$ from below in $(y_\alpha,r_{\alpha})$.
		Therefore, since it is also true that 
		$\partial_t\varphi_{\alpha}(y_{\alpha},r_{\alpha})
		=-2\alpha(r_\alpha-t_\alpha)$, 
		using now the fact that $v$ is supersolution, we obtain
		\begin{equation}\label{dos}
			0 \leq 2(t_\alpha-r_\alpha)-\Lambda_1\varphi_{\alpha}(y_{\alpha},r_{\alpha})
			=-\inf\{E_{z,\delta}(v,\varphi_\alpha,y_\alpha,r_\alpha)\colon 
				z\in\mathbb{S}^{N-1}\}.
		\end{equation}
		for any $\delta<d_{\alpha}.$
		
	
		Therefore, by \eqref{uno} and \eqref{dos}, for each $h>0$ 
		there exists $z_{\alpha,h}\in \mathbb{S}^{N-1} $ 
		such that
		\begin{equation}\label{EE}
			E_{(z_{\alpha,h}),\delta}(u,\psi_\alpha,x_\alpha,t_\alpha)\geq 0,
			\quad \mbox{and}
			\quad  
			E_{(z_{\alpha,h}),\delta}
			(v,\varphi_\alpha,y_\alpha,r_\alpha)
			\leq h,
		\end{equation}
		for any $\delta\in(0,d_\alpha),$ so
		%
		\begin{equation}\label{eq:estiinf}
			-h \le E_{(z_{\alpha,h}),\delta}(u,\psi_\alpha,x_\alpha,t_\alpha)- 
			E_{(z_{\alpha,h}),\delta}
			(v,\varphi_\alpha,y_\alpha,r_\alpha)
		\end{equation}
		for $0<\delta<\tfrac{d(\widetilde{x},\partial \Omega )}2.$ 
		We will get a contradiction from the previous estimate by finding an 
		upper bound and passing to the limit in a delicate way. 
		For that, let us first assume 
		\[
		    z_{\alpha,h} \to z_0,\quad\text{when } \alpha\to \infty, \text{ and } h \to 0,
		\]
		by a subsequence 
		if necessary. Let us now observe that there is a positive constant $C$ independent of $\delta,$ 
		$\alpha$ and $h$, such that
        \begin{equation}\label{eq:nueva1}
                  	\max 
			\left\{
				|I^1_{(z_{\alpha,h}),\delta}(\psi_{\alpha},x_\alpha,t_\alpha)|, 
				|I^1_{(z_{\alpha,h}),\delta}(\varphi_{\alpha},y_\alpha,r_\alpha)|, 
			\right\}
			\le C\alpha\delta^{2-2s}.
         \end{equation}		
		Let us now denote by $L_z(x)$ the line that passes trough $x$ with direction $z$, that is
		\[
			L_z (x) \coloneqq x+\tau z,\qquad \tau \in \mathbb{R}.
		\]
		To continue our argument it will be crucial to bear in mind that
		$z_{\alpha,h} \to z_0$ when $\alpha\to \infty,$ and  $h \to 0,$ and the fact that 
		$\Omega$ is a strictly convex domain. Indeed this allows us to affirm
		\[
			\mathbf{1}_{L_{z_{\alpha,h}}(y_\alpha)\cap \Omega}, 
			\mathbf{1}_{L_{z_{\alpha,h}}
			(x_\alpha)\cap \Omega} \to 
			\mathbf{1}_{L_{z_0}(\widetilde{x}) 
			\cap \Omega}\quad \mbox{ a.e.,}
		\]
		and
		\[
			\mathbf{1}_{L_{z_{\alpha,h}}(x_\alpha) \cap (\mathbb{R}^N \setminus \Omega)}, 
			\mathbf{1}_{L_{z_{\alpha,h}}(y_\alpha)\cap (\mathbb{R}^N \setminus \Omega)} 
			\to \mathbf{1}_{L_{z_0}(\widetilde{x})\cap (\mathbb{R}^N \setminus \Omega)} \quad\mbox{ a.e.},
		\]
		when $\alpha\to \infty, \text{ and } h \to 0.$ Thus since $(x_{\alpha}, y_{\alpha}, t_{\alpha}, r_{\alpha})$ is the maximum point of
 		$G_{\alpha}(x,y,t,r)$ and $u,v$ are bounded continuous functions, by using the previous two limits and the dominated convergence theorem, it follows that
            \begin{equation}\label{eq:nueva2}
                \limsup_{\alpha\to\infty\atop\delta,h\to0}
			I^2_{(z_{\alpha,h}),\delta}(u,x_\alpha,t_\alpha)
			-I^2_{(z_{\alpha,h}),\delta}(v,y_\alpha,r_\alpha)\le 0,
            \end{equation}		
		and
		\begin{equation}\label{eq:nueva3}
                  \lim_{\alpha\to\infty\atop h\to0}	
			I^3_{z_{\alpha,h}}(u,x_\alpha,t_\alpha)
			-I^3_{z_{\alpha,h}}(v,y_\alpha,r_\alpha)
			=-\widetilde{M} \int_{L_{z_0}(\bar x)\cap (\mathbb{R}^N \setminus \Omega)} 
			\frac{d\tau}{|\tau|^{1+2s}}.  
                \end{equation}	
		

		Therefore, 
		from \eqref{eq:nueva1}, \eqref{eq:nueva2}, and \eqref{eq:nueva3}, letting first 
		$\delta \to 0$, then $\alpha \to \infty $, 
		and $h \to 0,$ we get
		\begin{equation
			\begin{aligned}
				\limsup\limits_{\alpha\to\infty,\atop \delta,h\to 0}
				E_{(z_{\alpha,h}),\delta}(u, \phi_\alpha, x_\alpha,t_\alpha)&- 
				E_{(z_{\alpha,h}),\delta} 
				(v, \varphi_{\alpha}, y_\alpha,r_\alpha)	\\		
				&\le \displaystyle
						-\widetilde{M} \int_{L_{z_0}(\bar x)\cap (\mathbb{R}^N \setminus \Omega)}
						\frac{d\tau}{|\tau|^{1+2s}}.
			\end{aligned}
		\end{equation}	
		Thus by \eqref{eq:estiinf}, we conclude
		\[
			0\le -\widetilde{M} 
			\int_{L_{z_0}(\bar x)
			\cap (\mathbb{R}^N \setminus \Omega)} \frac{dt}{|t|^{1+2s}}<0,
		\]
		which implies the desirable contradiction.
	\end{proof}
	
	With the previous result the classical Perron's method can be used so we obtain the next result. Since we assumed that \eqref{hip} holds, by the results obtained in \cite{DPQR_FC}, we know that there exists a unique function $
	 \overline{w}\in\mathcal{C}(\overline\Omega)$ that satisfies, 
	\[
		\begin{cases}
			\Lambda_1^s \overline{w}(x)=-\|\Lambda_1^su_0\|_{L^{\infty}(\Omega)} &\text{in }\Omega,\\
			\overline{w}(x)=g(x) &\text{in }
				\mathbb{R}^N\setminus\Omega,
		\end{cases}
	\]
	in the viscosity sense, with $\overline{w}|_{\partial\Omega}=g|_{\partial\Omega}.$ 
	Since by the comparison principle proved in \cite[Theorem 2.2]{DPQR_FC} it is clear that 
	$\overline{w}\geq u_0$ in $\mathbb{R}^{N}$ we get that $\overline{u}(x,t)\coloneqq\overline{w}(x)$ 
	is a viscosity supersolution of \eqref{ev_eq} where the datum is attained in 
	a classical continuous way. Reasoning similarly we also get that 
	$\underline u(x,t)\coloneqq\underline{w}(x)$ with
	$$
		\begin{cases}
			\Lambda_1^s \underline{w}(x)=\|\Lambda_1^su_0\|_{L^{\infty}(\Omega)} &\text{in }\Omega,\\
			\underline{w}(x)=g(x) &\text{in }
				\mathbb{R}^N\setminus\Omega,\\
		\end{cases}
	$$
	is a subsolution of \eqref{ev_eq} where the datum is also attained in a classical continuous way. That is 
	by assuming \eqref{hip} we obtain a well orderer viscosity sub and supersolution continuous up to $
	\overline{\Omega}$, of problem \eqref{ev_eq}.
	Then, Perron's method gives us the existence of a viscosity solution to \eqref{ev_eq}. Since the uniqueness follows from the comparison principle, Theorem \ref{teo-1.intro} is proved.

	
	
	
	
	\bigskip 
	
	Once we have proven the existence and uniqueness of viscosity solutions of the parabolic problem, by using 
	the existence of the first eigenvalue and eigenfunction of $-\Lambda_1^s$ (see Theorem \ref{autof}) and 
	following some ideas from \cite{BlancRossi}, we can establish the asymptotic behavior of the solutions 
	of 	\eqref{ev_eq}. For that let us consider 
	\begin{equation}\label{auto1}
		\begin{cases}
			-\Lambda_1^s \varphi_R(x)=\mu_{R}\varphi_R &\text{in }\Omega_{R},\quad\\
			 \varphi_R=0&\text{in } \mathbb{R}^{N}\setminus\Omega_{R},\\
				\end{cases}
	\end{equation}
	and
	\begin{equation}\label{autoN}
		\begin{cases}
			-\Lambda_N^s \psi_R(x)=\mu_{R}\psi_R &\text{in }\Omega_{R},\\
			 \psi_R=0&\text{in } \mathbb{R}^{N}\setminus\Omega_{R},\\
				\end{cases}
		\end{equation}
	where $\Omega\subset\subset\Omega_R$ with $\Omega_R$ is a strictly convex domain of 
	$\mathbb{R}^{N}$, $\mu_R>0$, $0\leq \psi_R(x)=-\varphi_R(x)$, $x\in\Omega_R$. 
	
	
	Then we have the following result.
	\begin{theorem}\label{asymp}
		Assume $s>\nicefrac12.$ If $u_0$ and $g$ satisfy \eqref{hip} 
		then there exist $C_i=C_i(u_0)>0$, $i=1,\, 2$ such that 
		\begin{equation}\label{c_uniforme_2}
			C_1e^{-\mu_Rt}\varphi_R(x)\leq u(x,t)-z(x)\leq C_2 e^{-\mu_Rt}\psi_R(x),
			\quad x\in\overline{\Omega},\quad t>0,
		\end{equation}
		where $u(x,t)$ is the unique viscosity solution of \eqref{ev_eq} 
		given by Theorem \ref{teo-1.intro} and $z$ is the unique viscosity solution of 
		\begin{equation}\label{eliptico0}
			\begin{cases}
				-\Lambda_1^s z(x)=0&\text{in }\Omega,\\
				z=g&\text{in } \mathbb{R}^{N}\setminus\Omega.\\
			\end{cases}
		\end{equation}
	\end{theorem}
	
	
	
	\begin{proof}
		For every $x\in\overline{\Omega}$ and $t>0$ let us define, for some positive constants 
		that will be chosen later, the functions
		$$
			\underline u(x,t)\coloneqq z(x)+C_1e^{-\mu_Rt}\varphi_R(x)
			$$
			 and 
			 $$ 
			\overline u(x,t)\coloneqq z(x)+C_2e^{-\mu_Rt}\psi_R(x).
		$$
		On the first hand it is clear that 
		$\underline u(x,t)=g(x)=\overline u(x,t)$ in $\mathbb{R}^{N}\setminus\Omega$.  
		By choosing the constants in an appropriate way, depending on the datum $u_0$, since
		$$
			\mbox{$0< \psi_R(x) =-\varphi_R(x)$, (so that $0>\varphi_R(x)$), $x\in\Omega$,}
		$$
		it is also true that $\underline u(x,0)\leq u_0(x)\leq \overline u(x,0)$. 
		On the other hand, since for every functions $f,\, g$ we have that,
		$$
			\inf_{x\in\mathbb{R}^N} f(x)+\inf_{x\in\mathbb{R}^N} g(x)\leq \inf_{x\in\mathbb{R}^N} (f(x)+g(x))\leq \inf_{x\in\mathbb{R}^N} f(x)
			+\sup_{x\in\mathbb{R}^N} g(x),
		$$
		then
		$$
			\Lambda_1^{s}\underline u(x,t)\geq \Lambda_1^{s}z(x)+C_1e^{-\mu_Rt}\Lambda_1^{s}\varphi_R(x)
		$$
		and
		$$
			\Lambda_1^{s}\overline u(x,t)\leq \Lambda_1^{s}z(x)+C_2e^{-\mu_Rt}\Lambda_N^{s}\psi_R(x),
		$$
		for every $x\in\Omega_R$ and $t>0$. Therefore, by \eqref{auto1},\eqref{autoN} and \eqref{eliptico0} 
		we obtain that $\underline{u}(x,t)$ (resp. $\overline{u}(x,t)$) is a viscosity subsolution 
		(resp. supersolution) of \eqref{ev_eq}. 
		Thus, the conclusion follows by the comparison principle proved in Theorem \ref{comparacion}.
	\end{proof}
	
		\begin{remark} {\rm
		If we would like to obtain the same bound as the one given in \eqref{c_uniforme_2} 
		replacing $\varphi_R$ (resp. $\psi_R$) by $\varphi_1$ (resp. $\psi_1$), 
		the first eigenfunction in $\Omega$, not in $\Omega_R$, of $\Lambda_1^s$ (resp. $\Lambda_N^s$),  in the general case, 
		we would need to have a Hopf Lemma for these eigenfunction. That is, a bound like 
		$\varphi_1(x)\leq -C\delta(x)^{\beta}$ (resp. $\psi_1(x)\geq C\delta(x)^{\beta}$), 
		for some $\beta\in (0,1)$, as well as some additional hypothesis on the datum $g$, like H\"older regularity that would imply $z\in\mathcal{C}^{\beta}(\overline{\Omega})$, would be required 
		in order to guarantee $C\varphi_1(x)\leq u_0(x)-z(x)\leq C\psi_1(x),\, x\in\overline{\Omega}$. 
		
		\medskip
		
		Whether we have \eqref{c_uniforme_2} with $(\varphi_R, \psi_R)$ or with 
		$(\varphi_1, \psi_1)$ the conclusion of Theorem \ref{asymp} 
		implies the uniform convergence of the solution of the parabolic problem \eqref{ev_eq} 
		to the $s$-convex envelope of the datum $g$, $z$. That is, \eqref{c_uniforme} that we presented in Theorem \ref{teo.conver.intro}.}
	\end{remark}


	In some particular cases we can improve the exponential bound obtained in the previous result. 
	Indeed we have the following result.
	
	\begin{proposition}\label{particular}
		Asumme $s>\nicefrac12.$ 
		If $u_0$ and $g$ is an affine function in the whole $\mathbb{R}^N$ then 
		there exists $T=T(\Omega)$ such that 
		\begin{equation}\label{lonuestro}
			u(x,t)\leq z(x),\quad x\in\overline{\Omega},\quad t>T.
		\end{equation}
		with $u(x,t)$ the unique viscosity solution of \eqref{ev_eq} given 
		by Theorem \ref{teo-1.intro} and $z$ is the unique viscosity solution of \eqref{eliptico0}.
	\end{proposition}
	\begin{proof}
	        Notice that since $g$ is an affine function, we have that $\Lambda_1^s g(x)=0$ for any $x\in\mathbb{R}^N.$
	        Then $v(x,t)=u(x,t)-g(x)$ is a solution of 
	        \[
                    \begin{cases}
			v_t(x,t)=\Lambda_1^s v(x,t) &\text{in }\Omega\times(0,T),\\
			v(x,t)=0 &\text{in }
				(\mathbb{R}^N\setminus\Omega)\times(0,T),\\
			v(x,0)=u_0(x)-g(x) &\text{in }\mathbb{R}^N.
	    	    \end{cases}	        
	        \]
		Thus we could assume that $g\equiv0$ that in particular implies $z\equiv 0$. 
		Let us consider now $\Omega\subseteq B_{R}\coloneqq B_R(0)$, $R>0$ and we define
		$$
			0\leq w(x,t)\coloneqq e^{\mu_{R}\xi^{2s}(R-t)}\psi_{R}(\xi|x|)),\quad x\in B_{R},\, t>0,\, \xi>0,
		$$
		where $\mu_R>0$ and $\psi_R$ were given in \eqref{autoN} with $B_{R}$ instead of $\Omega_R$ 
		such that $\|\psi_{R}\|_{L^{\infty}(B_R)}=1$. Since clearly $\Lambda_1^s\leq \Lambda_N^s$ 
		it follows that
		$$
			\partial_t w(x,t)=-\mu_R\xi^{2s}w(x,t)=e^{\mu_{R}\xi^{2s}(R-t)}
			\Lambda_{N}^{s}\psi_{R}(\xi|x|))\geq \Lambda_1^s w(x,t),
		$$
		for every $(x,t)\in\Omega\times (0,\infty)$. 
		Moreover by using that $\psi_{R}>0$ in 
		$\Omega$ we get that $w(x,0)=e^{\mu_{R}\xi^{2s}R}\psi_{R}(\xi|x|))\geq u_0(x)$ if $x\in\Omega$. 
		That is, $w(x,t)$ is a supersolution of \eqref{ev_eq} so by Theorem \ref{comparacion} it follows that
		$$
			u(x,t)\leq e^{\mu_{R}\xi^{2s}(R-t)}\psi_{R}(\xi|x|)).
		$$
		Thus, by taking $\xi\to\infty$ there exists $T(=R)$ such that the conclusion follows.
	\end{proof}
	
	{
	In the special case that $g\equiv 0$ and $u_0$ is compactly supported 
	we have that the exponential rate is given by the first eigenvalue in $\Omega$.
	Notice that in this case 
	$z \equiv 0$ is the unique viscosity solution of \eqref{eliptico0}.
	
	\begin{theorem}\label{particular.fin}
	    Let $s>\nicefrac12.$
		If $g\equiv 0$ and $u_0$ is compactly supported inside
		$\Omega$ then  
		\begin{equation}\label{lonuestro.fin}
		C e^{-\mu_1 t}	\leq u(x,t)\leq 0,\quad x\in\overline{\Omega},\quad t>T.
		\end{equation}
		with $u(x,t)$ the unique viscosity solution of \eqref{ev_eq} given 
		by Theorem \ref{teo-1.intro}.
	\end{theorem}
	
	
	\begin{proof}
	
	Finally, in the special case that $u_0$ is compactly supported we have that
		$$u_0 (x) \geq C \varphi_1(x), \qquad x \in \mathbb{R}^N $$
		for a large constant $C$ (recall that the first eigenfunction is
		strictly negative in $\Omega$). Therefore, from a comparison argument, we get
		$$u(x,t)\geq C \varphi_1(x)e^{-\mu_1 t}.$$
		This bound together with \eqref{lonuestro} 
		give the result. 
		\end{proof}
		
		}

	
	
	\bigskip
	Of course the bound obtained in Proposition \ref{particular} cannot be expected in general, we mean, for instance $u(x,t)=e^{-\mu_1 t}
	\varphi_1(x)$, where $(\mu_1,\varphi_1)$ were given in Theorem \ref{autof}, is a solution of \eqref{ev_eq} 
	that it is not zero in finite time. That is, we could not expect, in general, that the solution of the 
	evolution problem lies bellow the $s$-convex enveloped in $\overline{\Omega}$ in finite time. Indeed we 
	have the following result.

	\begin{proposition}
		For every $x_0\in\Omega$ there exist $K_i>0$, $i=1,\,2$, 
		$u_0\in{\mathcal{C}(\mathbb{R}^N)}$ 
		and $g$ satisfy 
		\eqref{hip} such that  $u(x,t),$ the unique viscosity solution of \eqref{ev_eq}, satisfies
		$$
			u(x_0,t)\geq z(x_0)+K_1e^{-K_2t},\quad t>0,
		$$	
		with $z$ the unique viscosity solution of \eqref{eliptico0}.
	\end{proposition}
	\begin{proof}
	    Let us assume, without loss of generality that $x_0\in\Omega$ is the origin and  let us consider 
	    \begin{equation}\label{diff}
		B_r\coloneqq B_r(0)\subset\subset\Omega.
	    \end{equation} 
	    In what follows we  write $x=(x_1, x')\in\mathbb{R}^{N}$ and we define 
	    \[
	        B_r^{1}\coloneqq B_r\cap\{x'=0\}=[-r,r]\subseteq\mathbb{R}.
	    \]
		Therefore there exist (see \cite{SV}) $\mu_1^{1}>0$ a positive eigenvalue and $\varphi_1^{1}$ a 
		positive eigenfunction that is $\mathcal{C}^{s}$ up to the boundary such that
		$$	
			\begin{cases}
				-\Lambda_1^s\varphi_1^{1}(x)
				\left(=(-\Delta)^s_1 \varphi_1^{1}(x)\right)=\mu_{1}^1\varphi^{1}_1 &\text{in }[-r,r],\quad\\
					 \varphi_1^{1}=0&\text{in }(-\infty,-r)\cup(r,\infty).\\
			\end{cases}
		$$
		Since, by \eqref{diff}, if $(x_1,0)\in B_r$ clearly 
		$(x_1,0)\notin (\mathbb{R}^N\setminus\Omega)\cap\{x'=0\}$ 
		we can consider a nonnegative function 
		$g\in \mathcal{C}(\overline{\mathbb{R}^N\setminus\Omega})\cap 
		L^\infty(\overline{\mathbb{R} ^N\setminus\Omega})$ such that
		\begin{equation}\label {sobre_gg}
			g(x_1,x')\geq \varphi_1^{1}(x_1),\quad \{|x_1|\leq r,\, (x_1,x')\notin\Omega\
		\end{equation}
		\begin{equation}\label{sobre_g}
			g(x_1,0)=0,\quad (x_1,0)\in (\mathbb{R}^{N}\setminus\Omega)\cap\{x'=0\},
		\end{equation}
		as a datum of \eqref{eliptico0}. On one hand, by the comparison principle 
		(see \cite[Theorem 2.2]{DPQR_FC}) it is follows that $z\geq 0$. 
		On the other hand, taking into account the geometric interpretation 
		of the $s$-convex envelope (see \cite[Theorem 1.1]{DPQR_FC}) $z\leq z_{D}$ 
		with $z_D$ the convex envelope of $g$ in $\Omega\cap\{x'=0\}$. Since, 
		by \eqref{sobre_g}, $z_D\equiv 0$ it follows that, in particular,
		\begin{equation}\label{zetaen0}
			z(0)=0. 
		\end{equation}

		Let us now take $u_0\geq 0$ such that 
		$\Lambda_1^su_0(x)\in L^{\infty}(\Omega)$ and
		\begin{equation}\label{sobre_u0}
			u_0(x_1,x')\geq \varphi^1_1(x_1),\quad |x_1|\leq r. 
		
		\end{equation}
		Thus, by \eqref{sobre_gg}, \eqref{sobre_u0} and the fact that 
		$\varphi_1^1(x_1)=0$ if $|x_1|>r$, the function 
		$$
			v(x_1, x', t)\coloneqq\varphi_1^1(x_1)e^{-\mu_1^1t}\left(\leq \varphi_1^1(x_1)\right),
			\quad t>0,$$ satisfies
		$$
		\begin{cases}
					v_t(x,t)=\Lambda_1^s v(x,t) &\text{in }\widetilde{\Omega}_r\times(0,\infty),\\
					{v(x_1,x',t)\leq g(x)} &\text{in }
						(\mathbb{R}^N\setminus\widetilde{\Omega}_r)\times(0,\infty),\\
					v(x_1,x',0)\leq u_0(x) &\text{in }\mathbb{R}^N,
				\end{cases}
		$$
		where
		$$
		    \widetilde{\Omega}_r\coloneqq\{ (x_1,x')\in\Omega\colon |x_1|\leq r\}.
		   $$
		Therefore by Theorem \ref{comparacion} we get that $u(x_1,x',t)\geq v(x_1, x', t)$ so, 
		by \eqref{zetaen0}, we conclude
		$$u(0,t)\geq z(0)+\varphi_1^{1}(0)e^{-\mu_1^1t},\quad t>0,$$
		as wanted.
\end{proof}

\section*{Acknowledgements}
BB was partially supported by AEI MTM2016-80474-P grant and Ram\'on y Cajal fellowship RYC2018-026098-I (Spain).

    LMDP and JDR are partially supported by
    Agencia Nacional de Promoción de la Investigación, el Desarrollo
    Tecnológico y la Innovación PICT-2018-3183, and
    PICT-2019-00985 and UBACYT 20020190100367 (Argentina).

LMDP was also partially
    supported by the European Union’s Horizon 2020 research and innovation program
    under the Marie Sklodowska-Curie grant agreement No 777822,

