1. Introduction and statement of the main results

Elliptic problems with singular nonlinearities appear in many nonlinear phenomena, for instance, in the study of chemical catalysts process, non-Newtonian fluids, and in the study of the temperature of electrical conductors whose resistance depends on the temperature (see e.g., ^[3,6,10,15] and the references therein). The seminal work ^[7] is the start point of a large literature concerning singular elliptic problems, see for instance, ^{[1,3,5,6,8,9,10,13,15,17,18,21,22,23,24]}, and ^[30]. For additional references and a systematic study of singular elliptic problems see also ^[26].

In ^[10], Diaz, Morel and Oswald considered problems of the form

$\left\{ \begin{array} [c]{c} -\Delta u = -u^{-\gamma}+\lambda h\left( x\right) \text{ in }\Omega, \\ u = 0\text{ on }\partial\Omega, \\ u>0\text{ in }\Omega \end{array} \right.$

(1.1)

where $\Omega$ is a bounded and regular enough domain, $0 < \gamma < 1,$ $\lambda>0$ and $h\in L^{\infty}\left(\Omega\right)$ is a nonnegative and nonidentically zero function. They proved (see ^[10], Theorem 1, Corollary 1, Lemma 2 and Theorem 3) that there exists $\lambda _{0}>0$ such that. for $\lambda>\lambda_{0},$ problem (1.1) has a unique maximal solution $u\in H_{0}^{1}\left(\Omega\right)$ and has no solution when $\lambda < \lambda_{0}.$

Concerning nonlocal singular problems, Barrios, De Bonis, Medina, and Peral proved in ^[2] that if $\Omega$ is a bounded and regular enough domain in $\mathbb{R}^{n},$ $0 < s < 1,$ $n>2s,$ $f$ is a nonnegative function in a suitable Lebesgue space, $\lambda>0,$ $M>0$ and $1 < p < \frac {n+2s}{n-2s},$ then the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = \lambda f\left( x\right) u^{-\gamma} +Mu^{p}\text{ in }\Omega, \\ u = 0\text{ on }\mathbb{R}^{n}\setminus\Omega, \\ u>0\text{ in }\Omega, \end{array} \right.$

(1.2)

has a solution, in a suitable weak sense whenever $\lambda>0$ and $M>0,$ and that, if $M = 1$ and $f = 1,$ then there exists $\Lambda>0$ such that (1.2) has at least two solutions when $\lambda < \Lambda$ and has no solution when $\lambda>\Lambda.$

A natural question is to ask if an analogous of the quoted result of ^[10] hold in the nonlocal case, i.e., when $-\Delta$ is replaced by the fractional laplacian $\left(-\Delta\right) ^{s},$ $s\in\left(0, 1\right),$ and with the boundary condition $u = 0$ on $\partial\Omega$ replaced by $u = 0$ on $\mathbb{R}^{n}\setminus\Omega.$ Our aim in this paper is to obtain such a result. Note that the approach of ^[10] need to be modified in order to be used in the fractional case. Indeed, a step in ^[10] was to observe that, if $\varphi_{1}$ denotes a positive principal eigenfunction for $-\Delta$ on $\Omega,$ with Dirichlet boundary condition, then

$-\Delta\varphi_{1}^{\frac{2}{1+\gamma}} = \frac{2}{1+\gamma}\lambda_{1} \varphi_{1}^{\frac{2}{1+\gamma}}-\frac{2\left( 1-\gamma\right) }{\left( \gamma+1\right) ^{2}}\left\vert \nabla\varphi_{1}\right\vert ^{2}\varphi _{1}^{-\frac{2\gamma}{1+\gamma}}\text{ in }\Omega, \label{D M O cuenta 1}$

(1.3)

where $\lambda_{1}$ is the corresponding principal eigenvalue. From this fact, and using the properties of a principal eigenfunction, Diaz, Morel and Oswald proved that, for $\varepsilon$ positive and small enough, $\varepsilon \varphi_{1}^{\frac{2}{1+\gamma}}$ is a subsolution of problem (1.1). Since formula (1.3), is not avalaible for the principal eigenfunction of $\left(-\Delta\right) ^{s},$ the arguments of ^[10] need to be modified in order to deal with the fractional case.

Let us state the functional setting for our problem. For $s\in\left(0, 1\right)$ and $n\in\mathbb{N},$ let

$H^{s}\left( \mathbb{R}^{n}\right) : = \left\{ u\in L^{2}\left( \mathbb{R}^{n}\right) :\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}} \frac{\left\vert u\left( x\right) -u\left( y\right) \right\vert ^{2} }{\left\vert x-y\right\vert ^{n+2s}}dxdy<\infty\right\} ,$

and for $u\in H^{s}\left(\mathbb{R}^{n}\right),$ let $\left\Vert u\right\Vert _{H^{s}\left(\mathbb{R}^{n}\right) }: = \left(\int _{\mathbb{R}^{n}}u^{2}+\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}} \frac{\left\vert u\left(x\right) -u\left(y\right) \right\vert ^{2} }{\left\vert x-y\right\vert ^{n+2s}}dxdy\right) ^{\frac{1}{2}}.$ Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with $C^{1, 1}$ boundary and let

$X_{0}^{s}\left( \Omega\right) : = \left\{ u\in H^{s}\left( \mathbb{R} ^{n}\right) :u = 0\text{ }a.e.\text{ in }\mathbb{R}^{n}\setminus\Omega\right\} ,$

and for $u\in X_{0}^{s}\left(\Omega\right),$ let $\left\Vert u\right\Vert _{X_{0}^{s}\left(\Omega\right) }: = \left(\int_{\mathbb{R}^{n} \times\mathbb{R}^{n}}\frac{\left\vert u\left(x\right) -u\left(y\right) \right\vert ^{2}}{\left\vert x-y\right\vert ^{n+2s}}dxdy\right) ^{\frac{1} {2}}.$

With these norms, $H^{s}\left(\mathbb{R}^{n}\right)$ and $X_{0}^{s}\left(\Omega\right)$ are Hilbert spaces (see e.g., ^[29], Lemma 7), $C_{c}^{\infty}\left(\Omega\right)$ is dense in $X_{0}^{s}\left(\Omega\right)$ (see ^[16], Theorem 6). Also, $X_{0}^{s}\left(\Omega\right)$ is a closed subspace of $H^{s}\left(\mathbb{R}^{n}\right)$, and from the fractional Poincaré inequality (as stated e.g., in ^[11], Theorem 6.5; see Remark 2.1 below), if $n>2s$ then $\left\Vert.\right\Vert _{X_{0}^{s}\left(\Omega\right) }$ and $\left\Vert.\right\Vert _{H^{s}\left(\mathbb{R}^{n}\right) }$ are equivalent norms on $X_{0} ^{s}\left(\Omega\right).$ For $f\in L_{loc}^{1}\left(\Omega\right)$ we say that $f\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}$ if there exists a positive constant $c$ such that $\left\vert \int_{\Omega }f\varphi\right\vert \leq c\left\Vert u\right\Vert _{X_{0}^{s}\left(\Omega\right) }$ for any $\varphi\in X_{0}^{s}\left(\Omega\right).$ For $f\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}$ we will write $\left(\left(-\Delta\right) ^{s}\right) ^{-1}f$ for the unique weak solution $u$ (given by the Riesz theorem) of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = f\text{ in }\Omega, \\ u = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(1.4)

Here and below, the notion of weak solution that we use is the given in the following definition:

Definition 1.1. Let $s\in\left(0, 1\right),$ let $f:\Omega\rightarrow\mathbb{R}$ be a Lebesgue measurable function such that $f\varphi\in L^{1}\left(\Omega\right)$ for any $\varphi\in X_{0} ^{s}\left(\Omega\right).$ We say that $u:\Omega\rightarrow\mathbb{R}$ is a weak solution to the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = f\text{ in }\Omega, \\ u = 0\text{ in }\mathbb{R}^{n}\setminus\Omega \end{array} \right.$

if $u\in X_{0}^{s}\left(\Omega\right),$ $u = 0$ in $\mathbb{R}^{n} \setminus\Omega$ and, for any $\varphi\in$ $X_{0}^{s}\left(\Omega\right),$

$\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( u\left( x\right) -u\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy = \int_{\Omega }f\varphi.$

For $u\in X_{0}^{s}\left(\Omega\right)$ and $f\in L_{loc}^{1}\left(\Omega\right),$ we will write $\left(-\Delta\right) ^{s}u\leq f$ in $\Omega$ (respectively $\left(-\Delta\right) ^{s}u\geq f$ in $\Omega$) to mean that, for any nonnegative $\varphi\in H_{0}^{s}\left(\Omega\right),$ it hold that $f\varphi\in L^{1}\left(\Omega\right)$ and

$\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( u\left( x\right) -u\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy\leq\int_{\Omega }f\varphi\text{ (resp. }\geq\int_{\Omega}f\varphi\text{).}$

For $u, v\in X_{0}^{s}\left(\Omega\right),$ we will write $\left(-\Delta\right) ^{s}u\leq\left(-\Delta\right) ^{s}v$ in $\Omega$ (respectively $\left(-\Delta\right) ^{s}u\geq\left(-\Delta\right) ^{s}v$ in $\Omega$)$,$ to mean that $\left(-\Delta\right) ^{s}\left(u-v\right) \leq0$ in $\Omega$ (resp. $\left(-\Delta\right) ^{s}\left(u-v\right) \geq0$ in $\Omega$).

Let

$\mathcal{E}: = \left\{ u\in X_{0}^{s}\left( \Omega\right) :cd_{\Omega} ^{s}\leq u\leq c^{\prime}d_{\Omega} ^{s}\text{ }a.e.\text{ in }\Omega, \text{ for some positive constants }c\text{ and }c^{\prime}\right\}$

where, for $x\in\Omega,$ $d_{\Omega}\left(x\right) : = dist\left(x, \partial\Omega\right).$ With these notations, our main results read as follows:

Theorem 1.2. Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with $C^{1, 1}$ boundary, let $s\in\left(0, 1\right),$ and assume $n>2s.$ Let $h\in L^{\infty}\left(\Omega\right)$ be such that $0\leq h\not \equiv 0$ in $\Omega$ (i.e., $\left\vert \left\{ x\in\Omega:h\left(x\right) >0\right\} \right\vert >0$) and let $g:\Omega\times\left(0, \infty\right) \rightarrow\left[0, \infty\right)$ be a function satisfying the following conditions g1)-g5)

g1) $g:\Omega\times\left(0, \infty\right) \rightarrow\left[0, \infty\right)$ is a Carathéodory function, $g\left(., s\right) \in L^{\infty}\left(\Omega\right)$ for any $s>0$ and $\lim_{\sigma\rightarrow\infty}\left\Vert \left(g\left(., \sigma\right) \right) \right\Vert _{\infty} = 0.$

g2) $\sigma\rightarrow g\left(x, \sigma\right)$ is non increasing on $\left(0, \infty\right)$ $a.e.$ $x\in\Omega.$

g3) $g\left(., \sigma d_{\Omega}^{s}\right) \in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}$ and $d_{\Omega} ^{-s}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(d_{\Omega} ^{s}g\left(., \sigma d_{\Omega}^{s}\right) \right) \in L^{\infty}\left(\Omega\right)$ for all $\sigma>0.$

g4) It hold that:

$\lim_{\sigma\rightarrow\infty}\left\Vert \left(\sigma d_{\Omega}^{s}\right) ^{-1}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(d_{\Omega} ^{s}g\left(., \sigma d_{\Omega}^{s}\right) \right) \right\Vert _{\infty } = 0$, and

$\lim_{\sigma\rightarrow\infty}\left\Vert d_{\Omega}^{-s}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(g\left(., \sigma\right) \right) \right\Vert _{L^{\infty}\left(\Omega\right) } = 0.$

g5) $d_{\Omega}^{s}g\left(., \sigma d_{\Omega }^{s}\right) \in L^{2}\left(\Omega\right)$ for any $\sigma>0.$

Consider the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = -g\left( ., u\right) +\lambda h\; in\; \Omega, \\ u = 0\; in \;\mathbb{R}^{n}\setminus\Omega, \\ u>0\; in \;\Omega \end{array} \right.$

(1.5)

Then there exists $\lambda^{\ast}\geq0$ such that:

$i)$ If $\lambda>\lambda^{\ast}$ then (1.5) has a weak solution $u^{\left(\lambda\right) }\in\mathcal{E}$, which is maximal in the following sense: If $v\in\mathcal{E}$ satisfies $\left(-\Delta\right) ^{s}v\leq-g\left(., v\right) +\lambda h$ in $\Omega,$ then $u^{\left(\lambda\right) }\geq v$ $a.e.$ in $\Omega$ .

$ii)$ If $\lambda < \lambda^{\ast},$ no weak solution exists in $\mathcal{E}.$

$iii)$ If, in addition, there exists $b\in L^{\infty}\left(\Omega\right)$ such that $0\leq b\not \equiv 0$ in $\Omega$ and $g\left(., s\right) \geq bs^{-\beta}$ $a.e.$ in $\Omega$ for any $s\in\left(0, \infty\right),$ then $\lambda^{\ast}>0.$

Theorem 1.2 allows $g\left(x, s\right)$ to be singular at $s = 0.$ In fact, in Lemma 3.2, using some estimates from ^[4] for the Green function of $\left(-\Delta\right) ^{s}$ in $\Omega$ (with homogeneous Dirichlet boundary condition on $\mathbb{R}^{n} \setminus\Omega$), we show that if $g\left(x, s\right) = as^{-\beta}$ with $a$ a nonnegative function in $L^{\infty}\left(\Omega\right)$ and $\beta\in\left[0, s\right),$ then $g$ satisfies the assumptions of Theorem 1.2. Thus, as a consequence of Theorem 1.2, we obtain the following:

Theorem 1.3. Let $\Omega,$ $s,$ and $h$ be as in the statement of Theorem 1.2, and let $g:\Omega\times\left(0, \infty\right) \rightarrow\left[0, \infty\right).$ Then the assertions $i)-iii)$ of Theorem 1.2 remain true if we assume, instead of the conditions g1)-g5), that the following conditions g6) and g7) hold:

g6) $g:\Omega\times\left(0, \infty\right) \rightarrow\left[0, \infty\right)$ is a Carathéodory function and $s\rightarrow g\left(x, s\right)$ is nonincreasing for $a.e.$ $x\in\Omega.$

g7) There exist positive constants $a$ and $\beta\in\left[ 0, s\right)$ such that $g\left(., s\right) \leq as^{-\beta}$ $a.e.$ in $\Omega$ for any $s\in\left(0, \infty\right).$

Let us sketch our approach: In Section 2 we consider, for $\varepsilon>0,$ the following approximated problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = -g\left( ., u+\varepsilon\right) +\lambda h\text{ in }\Omega, \\ u = 0\text{ in }\mathbb{R}^{n}\setminus\Omega, \\ u>0\text{ in }\Omega. \end{array} \right.$

(1.6)

Let us mention that, in order to deal with problems involving the $(p; q)$-Laplacian and a convection term, this type of approximation was considered in ^[14] (see problem $P_{\varepsilon}$ therein).

Lemma 2.5 gives a positive number $\lambda_{0},$ independent of $\varepsilon$ and such that, for $\lambda = \lambda_{0},$ problem (1.6) has a weak solution $w_{\varepsilon}$. From this result, and from some properties of the function $w_{\varepsilon},$ in Lemma 2.11 we show that, for $\lambda\geq\lambda_{0}$ and for any $\varepsilon>0,$ there exists a weak solution $u_{\varepsilon}$ of problem (1.6), with the following properties:

a) $cd_{\Omega}^{s}\leq u_{\varepsilon}\leq c^{\prime}d_{\Omega}^{s}$ for some positive constants $c$ and $c^{\prime}$ independent of $\varepsilon,$

b) $u_{\varepsilon} \leq\overline{u},$ where $\overline{u}$ is the solution of the problem $\left(-\Delta\right) ^{s}\overline{u} = \lambda h$ in $\Omega,$ $\overline{u} = 0$ in $\mathbb{R}^{n}\setminus\Omega,$

c) $u_{\varepsilon}\geq\psi$ for any $\psi\in X_{0}^{s}\left(\Omega\right)$ such that $\left(-\Delta\right) ^{s}\psi = -g\left(., \psi+\varepsilon \right) +\lambda h$ in $\Omega.$

In section 3 we prove Theorems 1.2 and 1.3. To prove Theorem 1.2, we consider a decreasing sequence $\left\{ \varepsilon_{j}\right\} _{j\in\mathbb{N}}$ such that $\lim_{j\rightarrow\infty}\varepsilon_{j} = 0,$ and we show that, for $\lambda\geq\lambda_{0},$ the sequence of functions $\left\{ u_{\varepsilon _{j}}\right\} _{j\in\mathbb{N}}$ given by Lemma 2.11 converges, in $X_{0}^{s}\left(\Omega\right),$ to a weak solution $u$ of problem (1.5) which has the properties required by the theorem. An adaptation of some of the arguments of ^[10] gives that, if problem (1.5) has a weak solution in $\mathcal{E}$, then it has a maximal (in the sense stated in the theorem) weak solution in $\mathcal{E}$ and that if for some $\lambda = \lambda^{\prime}$ (1.5) has a weak solution in $\mathcal{E}$, then it has a weak solution in $\mathcal{E}$ for any $\lambda\geq\lambda^{\prime}.$ Finally, the assertion $iii)$ of Theorem 1.2 is proved with the same argument given in ^[10].

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2. Preliminaries and auxiliary results

We fix, from now on, $h\in L^{\infty}\left(\Omega\right)$ such that $0\leq h\not \equiv 0$ in $\Omega.$ We assume also from now on (except in Lemma 3.2) that $g:\Omega\times\left(0, \infty\right) \rightarrow\left[0, \infty\right)$ satisfies the assumptions g1)-g5 of Theorem 1.2.

In the next remark we collect some general facts concerning the operator $\left(-\Delta\right) ^{s}$.

Remark 2.1. ⅰ) (see e.g., ^[27], Proposition 4.1 and Corollary 4.2) The following comparison principle holds: If $u, v\in X_{0} ^{s}\left(\Omega\right)$ and $\left(-\Delta\right) ^{s}u\geq\left(-\Delta\right) ^{s}v$ in $\Omega$ then $u\geq v$ in $\Omega.$ In particular, the following maximum principle holds: If $v\in X_{0}^{s}\left(\Omega\right),$ $\left(-\Delta\right) ^{s}v\geq0$ in $\Omega$ and $v\geq0$ in $\mathbb{R}^{n}\setminus\Omega,$ then $v\geq0$ in $\Omega.$

ⅱ) (see e.g., ^[27], Lemma 7.3) If $f:\Omega \rightarrow\mathbb{R}$ is a nonnegative and not identically zero measurable function in $f\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime},$ then the weak solution $u$ of problem (1.4) satisfies, for some positive constant $c,$

$u\geq cd_{\Omega}^{s}\text{ in }\Omega.$

(2.1)

ⅲ) (see e.g., ^[28], Proposition 1.1) If $f\in L^{\infty}\left(\Omega\right)$ then the weak solution $u$ of problem (1.4) belongs to $C^{s}\left(\mathbb{R}^{n}\right).$ In particular, there exists a positive constant $c$ such that

$\left\vert u\right\vert \leq cd_{\Omega}^{s}\text{ in }\Omega.$

(2.2)

ⅳ) (Poincaré inequality, see ^[11], Theorem 6.5) Let $s\in\left(0, 1\right)$ and let $2_{s}^{\ast}: = \frac{2n}{n-2s}.$ Then there exists a positive constant $C = C\left(n, s\right)$ such that, for any measurable and compactly supported function $f:\mathbb{R}^{n}\rightarrow \mathbb{R}$,

$\left\Vert f\right\Vert _{L^{2_{s}^{\ast}}\left( \mathbb{R}^{n}\right) }\leq C\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( f\left( x\right) -f\left( y\right) \right) ^{2}}{\left\vert x-y\right\vert ^{n+sp}}dxdy.$

ⅴ) If $v\in L^{\left(2_{s}^{\ast}\right) ^{\prime}}\left(\Omega\right)$ then $v\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime},$ and $\left\Vert v\right\Vert _{\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}}\leq C\left\Vert v\right\Vert _{\left(2_{s}^{\ast}\right) ^{\prime}},$ with $C$ as in i). Indeed, for $\varphi\in X_{0}^{s}\left(\Omega\right),$ from the Hölder inequality and iii), $\int_{\Omega}\left\vert v\varphi\right\vert \leq\left\Vert v\right\Vert _{\left(2_{s}^{\ast}\right) ^{\prime}}\left\Vert \varphi\right\Vert _{2_{s}^{\ast}}\leq C\left\Vert v\right\Vert _{\left(2_{s}^{\ast}\right) ^{\prime}}\left\Vert \varphi\right\Vert _{X_{0}^{s}\left(\Omega\right) }.$

ⅵ) (Hardy inequality, see ^[25], Theorem 2.1) There exists a positive constant $c$ such that, for any $\varphi\in X_{0}^{s}\left(\Omega\right),$

$\left\Vert d_{\Omega}^{-s}\varphi\right\Vert _{2}\leq c\left\Vert \varphi\right\Vert _{X_{0}^{s}\left( \Omega\right) }.$

(2.3)

Remark 2.2. Let $z^{\ast}\in H^{s}\left(\mathbb{R}^{n}\right)$ be the solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}z^{\ast} = \tau_{1}h\text{ in }\Omega\\ z^{\ast} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega, \end{array} \right.$

(2.4)

with $\tau_{1}$ chosen such that $\left\Vert z^{\ast}\right\Vert _{L^{\infty }\left(\mathbb{R}^{n}\right) } = 1.$ Since $h\in L^{\infty}\left(\mathbb{R}^{n}\right),$ Remark 2.1 ⅲ) gives $z^{\ast}\in C\left(\mathbb{R}^{n}\right)$ (see also ^[12], Theorem 1.2). Thus, since $supp\left(z^{\ast}\right) \subset\overline{\Omega}$ and $z^{\ast}\in C\left(\overline{\Omega}\right)$, we have $z^{\ast}\in L^{\infty}\left(\mathbb{R}^{n}\right).$ Moreover, by Remark 2.1 ⅱ), there exists a positive constant $c^{\ast}$ such that

$z^{\ast}\geq c^{\ast}d_{\Omega}^{s}\text{ in }\Omega.$

(2.5)

Remark 2.3. There exist positive numbers $M_{0}$ and $M_{1}$ such that

$\begin{align} \frac{1}{2}c^{\ast}M_{1} & \geq\left\Vert d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( g\left( ., \frac{1}{2}c^{\ast} M_{1}d_{\Omega}^{s}\right) \right) \right\Vert _{\infty} , \label{remark M0 M1 uno new}\\ M_{1} & <M_{0}, \nonumber\\ \frac{1}{2}c^{\ast}M_{1} & \geq\left\Vert d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( g\left( ., M_{0}\right) \right) \right\Vert _{L^{\infty}\left( \Omega\right) }.\nonumber \end{align}$

(2.6)

Indeed, by g4), $\lim_{\sigma\rightarrow\infty}\left\Vert \left(\sigma d_{\Omega}^{s}\right) ^{-1}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(d_{\Omega}^{s}g\left(., \sigma d_{\Omega}^{s}\right) \right) \right\Vert _{\infty} = 0$ and so the first one of the above inequalities hold for $M_{1}$ large enough. Fix such a $M_{1}.$ Since, from g4), $\lim _{\sigma\rightarrow\infty}\left\Vert d_{\Omega}^{-s}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(g\left(., \sigma\right) \right) \right\Vert _{L^{\infty}\left(\Omega\right) } = 0,$ the remaining inequalities of (2.6) hold for $M_{0}$ large enough.

Lemma 2.4. Let $\varepsilon>0$ and let $z^{\ast},$ $\tau_{1}$ and $c^{\ast}$ be as in Remark 2.2. Let $M_{0}$ and $M_{1}$ be as in Remark 2.3. Let $z: = M_{1}z^{\ast}$ and let $w^{0, \varepsilon}:\mathbb{R}^{n}\rightarrow\mathbb{R}$ be the constant function $w^{0, \varepsilon} = M_{0}.$ Then there exist sequences $\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ and $\left\{ \zeta ^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ in $X_{0}^{s}\left(\Omega\right) \cap L^{\infty}\left(\Omega\right)$ such that, for all $j\in\mathbb{N}$:

$i)$ $w^{j-1, \varepsilon}\geq w^{j, \varepsilon}\geq 0$ in $\mathbb{R}^{n},$

$ii)$ $w^{j, \varepsilon}\geq\frac{1}{2} c^{\ast}M_{1}d_{\Omega}^{s}$ in $\Omega,$

$iii)$ $w^{j, \varepsilon}$ is a weak solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}w^{j, \varepsilon} = -g\left( ., w^{j-1, \varepsilon }+\varepsilon\right) +\tau_{1}M_{1}h\; in \;\Omega, \\ w^{j, \varepsilon} = 0\; in \;\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.7)

$iv)$ $w^{j, \varepsilon} = z-\zeta^{j, \varepsilon}$ in $\mathbb{R}^{n}$ and $\zeta^{j, \varepsilon}$ is a weak solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\zeta^{j, \varepsilon} = g\left( ., w^{j-1, \varepsilon}+\varepsilon\right) \;in \;\Omega, \\ \zeta^{j, \varepsilon} = 0\; in \;\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.8)

$v)$ $\left\Vert w^{j, \varepsilon}\right\Vert _{X_{0}^{s}\left(\Omega\right) }\leq c$ for some positive constant $c$ independent of $j$ and $\varepsilon.$

Proof. The sequences $\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ and $\left\{ \zeta^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ with the properties $i)-v)$ will be constructed inductively. Let $\zeta ^{1, \varepsilon}\in X_{0}^{1}\left(\Omega\right)$ be the solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\zeta^{1, \varepsilon} = g\left( ., w^{0, \varepsilon }+\varepsilon\right) \text{ in }\Omega\\ \zeta^{1, \varepsilon} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega \end{array} \right.$

(thus $iv)$ holds for $j = 1$). From g1) and g2) we have $0\leq g\left(., w^{0, \varepsilon}+\varepsilon\right) \leq g\left(., \varepsilon\right) \in L^{\infty}\left(\Omega\right).$ Thus $g\left(., w^{0, \varepsilon}+\varepsilon\right) \in L^{\infty}\left(\Omega\right).$ Then, by Remark 2.1 $iii)$, $\zeta ^{1, \varepsilon}\in C\left(\mathbb{R}^{n}\right).$ Therefore, since $supp\left(\zeta^{1, \varepsilon}\right) \subset\overline{\Omega},$ we have $\zeta^{1, \varepsilon}\in L^{\infty}\left(\Omega\right).$ By g1), $g\left(., M_{0}\right) \in L^{\infty}\left(\Omega\right)$ and so $g\left(., M_{0}\right) \in\left(X_{0}^{1}\left(\Omega\right) \right) ^{\prime}.$ Let $u_{0}: = \left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(g\left(., M_{0}\right) \right).$ Then, by g1) and g3), $d_{\Omega}^{-s}u_{0}\in L^{\infty}\left(\Omega\right).$ We have, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\left( \zeta^{1, \varepsilon}-u_{0}\right) = g\left( ., w^{0, \varepsilon}+\varepsilon\right) -g\left( ., M_{0}\right) \leq0\text{ in }\Omega\\ \zeta^{1, \varepsilon}-u_{0} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Then, by the maximum principle of Remark 2.1 $i)$,

$0\leq\zeta^{1, \varepsilon}\leq u_{0}\leq\left\Vert d_{\Omega}^{-s} u_{0}\right\Vert _{L^{\infty}\left( \Omega\right) }d_{\Omega}^{s}\text{ in }\Omega.$

(2.9)

Let $z: = M_{1}z^{\ast}.$ By Remark 2.2, $z\in H^{s}\left(\mathbb{R}^{n}\right) \cap C\left(\overline{\Omega}\right)$ and

$z\geq c^{\ast}M_{1}d_{\Omega}^{s}\text{ in }\Omega.\label{zeta lower estimate}$

(2.10)

Also, $z\leq M_{1}$ in $\Omega,$ and $z$ is a weak solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}z = \tau_{1}M_{1}h\text{ in }\Omega, \\ z = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Let $w^{1, \varepsilon}: = z-\zeta^{1, \varepsilon}.$ Then $w^{1, \varepsilon}\in H^{s}\left(\mathbb{R}^{n}\right)$ and $w^{1, \varepsilon} = 0$ in $\mathbb{R}^{n}\setminus\Omega.$ Thus $w^{1, \varepsilon}\in X_{0}^{s}\left(\Omega\right).$ Also $w^{1, \varepsilon}\in L^{\infty}\left(\Omega\right).$ Since $\zeta^{1, \varepsilon}\geq0$ in $\Omega,$ we have

$w^{0, \varepsilon}-w^{1, \varepsilon} = M_{0}-z+\zeta^{1, \varepsilon}\geq M_{0}-z\geq M_{0}-M_{1}>0\text{ in }\Omega.$

Then $w^{1, \varepsilon}\leq w^{0, \varepsilon}$ in $\Omega.$ Thus $i)$ holds for $j = 1.$ Now, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}w^{1, \varepsilon} = \left( -\Delta\right) ^{s}\left( z-\zeta^{1, \varepsilon}\right) = \tau_{1}M_{1}h-\left( -\Delta\right) ^{s}\left( \zeta^{1, \varepsilon}\right) \\ = -g\left( ., w^{0, \varepsilon}+\varepsilon\right) +\tau_{1}M_{1}h\text{ in }\Omega, \\ w^{1, \varepsilon} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega, \end{array} \right.$

and so $iii)$ holds for $j = 1.$ Also, from (2.9), (2.10), and taking into account that (2.6),

$\begin{align*} w^{1, \varepsilon} & = z-\zeta^{1, \varepsilon}\geq c^{\ast}M_{1}d_{\Omega} ^{s}-\left\Vert d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( g\left( ., M_{0}\right) \right) \right\Vert _{L^{\infty}\left( \Omega\right) }d_{\Omega}^{s}\\ & \geq\frac{1}{2}c^{\ast}M_{1}d_{\Omega}^{s}\text{ in }\Omega. \end{align*}$

and then $w^{1, \varepsilon}\geq\frac{1}{2}c^{\ast}M_{1}d_{\Omega}^{s}$ in $\Omega.$ Thus $ii)$ holds for $j=1.$

Suppose constructed, for $k\geq1,$ functions $w^{1, \varepsilon}, ...,$ $w^{k, \varepsilon}$ and $\zeta^{1, \varepsilon}, ..., \zeta^{k, \varepsilon},$ belonging to $X_{0} ^{s}\left(\Omega\right) \cap L^{\infty}\left(\Omega\right),$ and with the properties $i)$-$iv)$. Let $\zeta^{k+1, \varepsilon}\in X_{0}^{s}\left(\Omega\right)$ be the solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\zeta^{k+1, \varepsilon} = g\left( ., w^{k, \varepsilon}+\varepsilon\right) \text{ in }\Omega, \\ \zeta^{k+1, \varepsilon} = 0\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.11)

(and so $iv)$ holds for $j = k+1$) and let $w^{k+1, \varepsilon} : = z-\zeta^{k+1, \varepsilon}.$ Then $w^{k+1, \varepsilon}\in H^{s}\left(\mathbb{R}^{n}\right)$ and $w^{k+1, \varepsilon} = 0$ in $\mathbb{R} ^{n}\setminus\Omega.$ Thus $w^{k+1, \varepsilon}\in X_{0}^{s}\left(\Omega\right).$ Also,

$w^{k, \varepsilon}-w^{k+1, \varepsilon} = \zeta^{k+1, \varepsilon}-\zeta ^{k, \varepsilon}\text{ in }\mathbb{R}^{n}$

(2.12)

and

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\left( \zeta^{k+1, \varepsilon}-\zeta ^{k, \varepsilon}\right) = g\left( ., w^{k, \varepsilon}+\varepsilon\right) -g\left( ., w^{k-1, \varepsilon}+\varepsilon\right) \geq0\text{ in }\Omega, \\ \zeta^{k+1, \varepsilon}-\zeta^{k, \varepsilon} = 0\text{ in }\mathbb{R} ^{n}\setminus\Omega, \end{array} \right.$

the last inequality because, by g1), $s\rightarrow g\left(., s\right)$ is nonincreasing and (by our inductive hypothesis) $w^{k, \varepsilon}\leq w^{k-1, \varepsilon}$ in $\Omega.$ Then, by the maximum principle, $\zeta^{k+1, \varepsilon}-\zeta^{k, \varepsilon}\geq0$ in $\mathbb{R}^{n}.$ Therefore, by (2.12), $w^{k, \varepsilon }\geq w^{k+1, \varepsilon}$ in $\mathbb{R}^{n},$and then $i)$ holds for $j = k+1.$ Also,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}w^{k+1, \varepsilon} = \left( -\Delta\right) ^{s}z-\left( -\Delta\right) ^{s}\zeta^{k+1, \varepsilon} = -g\left( ., w^{k, \varepsilon}+\varepsilon\right) +\tau_{1}M_{1}h\text{ in }\Omega, \\ w^{k+1, \varepsilon} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Then $iii)$ holds for $j = k+1.$ By g4), $g\left(., \frac{1} {2}c^{\ast}M_{1}d_{\Omega}^{s}\right) \in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}.$ Let $u_{1}: = \left(\left(-\Delta \right) ^{s}\right) ^{-1}\left(g\left(., \frac{1}{2}c^{\ast} M_{1}d_{\Omega}^{s}\right) \right) \in X_{0}^{s}\left(\Omega\right).$ By the inductive hypothesis we have $w^{k, \varepsilon}\geq\frac{1}{2}c^{\ast }M_{1}d_{\Omega}^{s}$ in $\Omega.$ Now,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\left( \zeta^{k+1, \varepsilon}-u_{1}\right) \\ = g\left( ., w^{k, \varepsilon}+\varepsilon\right) -g\left( ., \frac{1} {2}c^{\ast}M_{1}d_{\Omega}^{s}\right) \leq0\text{ in }\Omega, \\ \zeta^{k+1, \varepsilon}-u_{1} = 0\text{ on }\mathbb{R}^{n}\setminus\Omega, \end{array} \right.$

then the comparison principle gives $\zeta^{k+1, \varepsilon}\leq u_{1}.$ Thus, in $\Omega,$

$\begin{align*} w^{k+1, \varepsilon} & = z-\zeta^{k+1, \varepsilon}\geq c^{\ast}M_{1}d_{\Omega }^{s}-u_{1}\\ & = c^{\ast}M_{1}d_{\Omega}^{s}-\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( g\left( ., \frac{1}{2}c^{\ast}M_{1}d_{\Omega}^{s}\right) \right) \\ & \geq c^{\ast}M_{1}d_{\Omega}^{s}-\left\Vert d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( g\left( ., \frac{1}{2}c^{\ast} M_{1}d_{\Omega}^{s}\right) \right) \right\Vert _{\infty}d_{\Omega}^{s} \geq\frac{1}{2}c^{\ast}M_{1}d_{\Omega}^{s}, \end{align*}$

the last inequality by (2.6). Thus $ii)$ holds for $j = k+1.$ This complete the inductive construction of the sequences $\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ and $\left\{ \zeta^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ with the properties $i)-iv)$.

To see $v)$, we take $\zeta^{j, \varepsilon}$ as a test function in (2.8). Using $ii)$, the Hölder inequality, the Poincaré inequality of Remark 2.1 $iv)$, we get, for any $j\in\mathbb{N},$

$\begin{align*} \left\Vert \zeta^{j, \varepsilon}\right\Vert _{X_{0}^{s}\left( \Omega\right) }^{2} & = \int_{\Omega}g\left( ., w^{j-1, \varepsilon}+\varepsilon\right) \zeta^{j, \varepsilon}\leq\int_{\Omega}g\left( ., \frac{1}{2}c^{\ast} M_{1}d_{\Omega}^{s}\right) \zeta^{j, \varepsilon}\\ & = \int_{\Omega}d_{\Omega}^{s}g\left( ., \frac{1}{2}c^{\ast}M_{1}d_{\Omega }^{s}\right) d_{\Omega}^{-s}\zeta^{j, \varepsilon}\\ & \leq\left\Vert d_{\Omega}^{s}g\left( ., \frac{1}{2}c^{\ast}M_{1}d_{\Omega }^{s}\right) \right\Vert _{2}\left\Vert d_{\Omega}^{-s}\zeta^{j, \varepsilon }\right\Vert _{2}\leq c\left\Vert \zeta^{j, \varepsilon}\right\Vert _{X_{0} ^{s}\left( \Omega\right) }. \end{align*}$

where $c$ is a positive constant $c$ independent of $j$ and $\varepsilon,$ and where, in the last inequality, we have used g5). Then $\left\Vert \zeta^{j, \varepsilon}\right\Vert _{X_{0}^{s}\left(\Omega\right) }$ has an upper bound independent of $j$ and $\varepsilon.$ Since $w^{j, \varepsilon } = z-\zeta^{j, \varepsilon},$ the same assertion holds for $w^{j, \varepsilon}.$

Lemma 2.5. Let $\varepsilon>0$ and let $\tau _{1}$ and $c^{\ast}$ be as in Remark 2.2. Let $M_{0}$ and $M_{1}$ be as in Remark 2.3 and let $\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ and $\left\{ \zeta ^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ be as in Lemma 2.4. Let $w_{\varepsilon}: = \lim_{j\rightarrow\infty }w^{j, \varepsilon}$ and let $\zeta_{\varepsilon}: = \lim_{j\rightarrow\infty }\zeta^{j, \varepsilon}.$ Then

$i)$ $w_{\varepsilon}$ and $\zeta _{\varepsilon}$ belong to $H^{s}\left(\mathbb{R}^{n}\right) \cap C\left(\overline{\Omega}\right),$

$ii)$ $\frac{1}{2}c^{\ast}M_{1}d_{\Omega }^{s}\leq w_{\varepsilon}\leq M_{0}$ in $\Omega,$ and there exists a positive constant $c$ independent of $\varepsilon$ such that $w^{\varepsilon}\leq cd_{\Omega}^{s}$ in $\Omega.$

$iii)$ $w_{\varepsilon}$ satisfies, in weak sense,

$\left\{ \begin{array} [c]{c} -\Delta w_{\varepsilon} = -g\left( ., w_{\varepsilon}+\varepsilon\right) +\tau_{1}M_{1}h\; in \;\Omega, \\ w_{\varepsilon} = 0\; in \;\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.13)

$iv)$ $\zeta_{\varepsilon}$ satisfies, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\zeta_{\varepsilon} = g\left( ., w_{\varepsilon }+\varepsilon\right) \; in \;\Omega, \\ \zeta_{\varepsilon} = 0\; in \;\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.14)

Proof. Let $z^{\ast}$ be as in Remark 2.2, and let $z: = M_{1}z^{\ast}.$ Let $M_{0}$ and $M_{1}$ be as in Remark 2.3. By Lemma 2.4, $\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ is a nonincreasing sequence of nonnegative functions in $\mathbb{R}^{n},$ and so there exists $w_{\varepsilon} = \lim_{j\rightarrow\infty}w^{j, \varepsilon}.$ Since $\zeta^{j, \varepsilon} = z-w^{j-1, \varepsilon},$ there exists also $\zeta_{\varepsilon} = \lim_{j\rightarrow\infty}\zeta^{j, \varepsilon}.$ Again by Lemma 2.4 we have, for any $j\in\mathbb{N},$ $0\leq w^{j, \varepsilon} = z-\zeta^{j, \varepsilon}\leq z\in L^{\infty}\left(\Omega\right).$ Thus, by the Lebesgue dominated convergence theorem,

$\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}\text{ converges to }w_{\varepsilon}\text{ in }L^{p}\left( \Omega\right) \text{ for any } p\in\left[ 1, \infty\right) ,$

(2.15)

and so $\left\{ g\left(., w^{j, \varepsilon}+\varepsilon\right) \right\} _{j\in\mathbb{N}}$ converges to $g\left(., w_{\varepsilon}+\varepsilon \right)$ in $L^{p}\left(\Omega\right)$ for any $p\in\left[ 1, \infty\right).$ We claim that

$\zeta_{\varepsilon}\in X_{0}^{s}\left( \Omega\right) \text{ and }\left\{ \zeta^{j, \varepsilon}\right\} _{j\in\mathbb{N}}\text{ converges in }X_{0} ^{s}\left( \Omega\right) \text{ to }\zeta_{\varepsilon}.$

(2.16)

Indeed, for $j,$ $k\in\mathbb{N},$ from (2.8),

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\left( \zeta^{j, \varepsilon}-\zeta^{k, \varepsilon }\right) = g\left( ., w^{j-1, \varepsilon}+\varepsilon\right) -g\left( ., w^{k-1, \varepsilon}+\varepsilon\right) \text{ in }\Omega, \\ \zeta^{j, \varepsilon}-\zeta^{k, \varepsilon} = 0\text{ in }\mathbb{R} ^{n}\setminus\Omega. \end{array} \right.$

(2.17)

We take $\zeta^{j, \varepsilon}-\zeta^{k, \varepsilon}$ as a test function in (2.17) to obtain

$\begin{align*} \left\Vert \zeta^{j, \varepsilon}-\zeta^{k, \varepsilon}\right\Vert _{X_{0} ^{s}\left( \Omega\right) }^{2} & = \int_{\Omega}\left( \zeta ^{j, \varepsilon}-\zeta^{k, \varepsilon}\right) \left( g\left( ., w^{j-1, \varepsilon}+\varepsilon\right) -g\left( ., w^{k-1, \varepsilon }+\varepsilon\right) \right) \\ & \leq\left\Vert \zeta^{j, \varepsilon}-\zeta^{k, \varepsilon}\right\Vert _{2_{s}^{\ast}}\left\Vert g\left( ., w^{j-1, \varepsilon}+\varepsilon\right) -g\left( ., w^{k-1, \varepsilon}+\varepsilon\right) \right\Vert _{\left( 2_{s}^{\ast}\right) ^{\prime}} \end{align*}$

where $2_{s}^{\ast}: = \frac{2n}{n-2s}.$ Then

$\left\Vert \zeta^{j, \varepsilon}-\zeta^{k, \varepsilon}\right\Vert _{X_{0} ^{s}\left( \Omega\right) }\leq c\left\Vert g\left( ., w^{j-1, \varepsilon }+\varepsilon\right) -g\left( ., w^{k-1, \varepsilon}+\varepsilon\right) \right\Vert _{\left( 2_{s}^{\ast}\right) ^{\prime}}.$

where $c$ is a constant independent of $j$ and $k.$ Since $\left\{ g\left(., w^{j-1, \varepsilon}+\varepsilon\right) \right\} _{j\in\mathbb{N}}$ converges to $g\left(., w_{\varepsilon}+\varepsilon\right)$ in $L^{\left(2_{s}^{\ast}\right) ^{\prime}}\left(\Omega\right),$ we get

$\lim\limits_{j, k\rightarrow\infty}\left\Vert \zeta^{j, \varepsilon}-\zeta ^{k, \varepsilon}\right\Vert _{X_{0}^{s}\left( \Omega\right) } = 0,$

and thus $\left\{ \zeta^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ converges in $X_{0}^{s}\left(\Omega\right).$ Since $\left\{ \zeta^{j, \varepsilon }\right\} _{j\in\mathbb{N}}$ converges to $\zeta_{\varepsilon}$ in pointwise sense, (2.16) follows. Also, $w^{j, \varepsilon} = z-\zeta^{j, \varepsilon},$ and then $\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ converges to $w_{\varepsilon, \rho}$ in $X_{0}^{s}\left(\Omega\right).$ Thus

$w_{\varepsilon}\in X_{0}^{s}\left( \Omega\right) \text{ and }\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}\text{ converges in }X_{0} ^{s}\left( \Omega\right) \text{ to }w_{\varepsilon}.$

(2.18)

To prove (2.14) observe that, from (2.8), we have, for any $\varphi\in X_{0}^{s}\left(\Omega\right)$ and $j\in\mathbb{N},$

$\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( \zeta^{\varepsilon , j}\left( x\right) -\zeta^{\varepsilon, j}\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy = \int_{\Omega}g\left( ., w^{\varepsilon , j-1}+\varepsilon\right) \varphi.$

(2.19)

Taking $\lim_{j\rightarrow\infty}$ in (2.19) and using (2.16) and (2.15), we obtain (2.14). From (2.14) and since, by g1) and g2), $g\left(., w_{\varepsilon}+\varepsilon \right) \in L^{\infty}\left(\Omega\right),$ Remark 2.1 $iii)$ gives that, in addition, $\zeta_{\varepsilon}\in C\left(\overline{\Omega}\right)$ (and so, since $w_{\varepsilon} = z-\zeta_{\varepsilon},$ then also $w_{\varepsilon}\in C\left(\overline{\Omega}\right)$)$.$

Let us see that (2.13) holds. Let $\varphi\in X_{0}^{s}\left(\Omega\right).$ From (2.7), we have, for any $j\in\mathbb{N},$

$\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( w^{j, \varepsilon }\left( x\right) -w^{j, \varepsilon}\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy\label{todavia no limite}\\ = \int_{\Omega\setminus\overline{B_{\rho}\left( y\right) }}\left( -g\left( ., w^{j-1, \varepsilon}+\varepsilon\right) +\tau_{1}M_{1}h\right) \varphi.$

(2.20)

Since $\varphi\in X_{0}^{s}\left(\Omega\right)$ and $\left\{ w^{j, \varepsilon}\right\} _{j\in\mathbb{N}}$ converges to $w_{\varepsilon}$ in $X_{0}^{s}\left(\Omega\right)$ we have

$\lim\limits_{j\rightarrow\infty}\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}} \frac{\left( w^{j, \varepsilon}\left( x\right) -w^{j, \varepsilon}\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy\label{limite uno}\\ = \int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( w_{\varepsilon }\left( x\right) -w_{\varepsilon}\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy.$

(2.21)

Also, $w_{\varepsilon}\left(x\right) = \lim_{j\rightarrow\infty }w^{j, \varepsilon}\left(x\right)$ for any $x\in\Omega,$ and

$\left\vert g\left( ., w^{j-1, \varepsilon}+\varepsilon\right) \varphi \right\vert \leq g\left( ., \varepsilon\right) \left\vert \varphi\right\vert \in L^{1}\left( \Omega\right) ,$

and clearly $\left\vert \tau_{1}M_{1}h\varphi\right\vert \in L^{1}\left(\Omega\right).$ Then, by the Lebesgue dominated convergence theorem,

$\lim\limits_{j\rightarrow\infty}\int_{\Omega}\left( -g\left( ., w^{j-1, \varepsilon }+\varepsilon\right) +\tau_{1}M_{1}h\right) \varphi = \int_{\Omega}\left( -g\left( ., w_{\varepsilon}+\varepsilon\right) +\tau_{1}M_{1}h\right) \varphi.$

(2.22)

Now (2.13) follows from (2.20), (2.21) and (2.22). Finally, by Lemma 2.4 we have, for all $j\in\mathbb{N},$ $\frac{1}{2}c^{\ast }M_{1}d_{\Omega}^{s}\leq w^{j, \varepsilon}$ in $\Omega$ and so the same inequality hold with $w^{j, \varepsilon}$ replaced by $w_{\varepsilon}.$ Also, since $w^{j, \varepsilon}\leq z_{0}$ in $\Omega$ we have $w^{j, \varepsilon }\leq cd_{\Omega}^{s}$ with $c$ independent of $j$ and $\varepsilon.$

Remark 2.6. Let $G:\Omega\times\Omega\rightarrow\mathbb{R\cup}\left\{ \infty\right\}$ be the Green function for $\left(-\Delta\right) ^{s}$ in $\Omega,$ with homogeneous Dirichlet boundary condition on $\mathbb{R} ^{n}\setminus\Omega.$ Then, for $f\in C\left(\overline{\Omega}\right)$ the solution $u$ of problem (1.4) is given by $u\left(x\right) = \int_{\Omega}G\left(x, y\right) f\left(y\right) dy$ for $x\in\Omega$ and by $u\left(x\right) = 0$ for $x\in\mathbb{R}^{n} \setminus\Omega.$ Let us recall the following estimates from ^[4]:

ⅰ) (see ^[4], Theorems 1.1 and 1.2) There exist positive constants $c$ and $c^{\prime},$ depending only on $\Omega$ and $s,$ such that for $x; y\in\Omega,$

$G(x, y) \leq c\frac{d_{\Omega}\left( x\right) ^{s}}{\left\vert x-y\right\vert ^{n-s}},$

(2.23)

$G(x, y) \leq c\frac{d_{\Omega}\left( x\right) ^{s}}{d_{\Omega}\left( y\right) ^{s}}\frac{1}{\left\vert x-y\right\vert ^{n-2s}},$

(2.24)

$G(x, y) \leq c\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{s}}{\left\vert x-y\right\vert ^{n}}$

(2.25)

$G(x, y) \geq c^{\prime}\frac{1}{\left\vert x-y\right\vert ^{n-2s}}\text{ if }\left\vert x-y\right\vert \leq\max\left\{ \frac{d_{\Omega}\left( x\right) }{2}, \frac{d_{\Omega}\left( y\right) }{2}\right\}$

(2.26)

$G(x, y) \geq c^{\prime}\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega }\left( y\right) ^{s}}{\left\vert x-y\right\vert ^{n}}\text{ if }\left\vert x-y\right\vert >\max\left\{ \frac{d_{\Omega}\left( x\right) }{2} , \frac{d_{\Omega}\left( y\right) }{2}\right\}$

(2.27)

ⅱ) From ⅰ) it follows that there exists a positive constant $c^{\prime\prime },$ depending only on $\Omega$ and $s,$ such that for $x; y\in\Omega,$

$G(x;y)\geq c^{\prime\prime}d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{s}.$

(2.28)

Indeed:

If $\left\vert x-y\right\vert >\max\left\{ \frac{d_{\Omega }\left(x\right) }{2}, \frac{d_{\Omega}\left(y\right) }{2}\right\}$ then, from (2.27), $G(x; y)\geq c^{\prime}\frac{d_{\Omega}\left(x\right) ^{s}d_{\Omega}\left(y\right) ^{s}}{\left\vert x-y\right\vert ^{n}}$ and so $G(x; y)\geq c^{\prime}\frac{d_{\Omega}\left(x\right) ^{s}d_{\Omega}\left(y\right) ^{s}}{\left(diam\left(\Omega\right) \right) ^{n}}$.

If $\left\vert x-y\right\vert \leq\max\left\{ \frac{d_{\Omega}\left(x\right) }{2}, \frac{d_{\Omega}\left(y\right) } {2}\right\}$ then either $\left\vert x-y\right\vert \leq\frac{d_{\Omega }\left(x\right) }{2}$ or $\left\vert x-y\right\vert \leq\frac{d_{\Omega }\left(y\right) }{2}.$ If $\left\vert x-y\right\vert \leq\frac{d_{\Omega }\left(x\right) }{2}$ consider $z\in\partial\Omega$ such that $d_{\Omega }\left(y\right) = \left\vert z-y\right\vert.$ then $d_{\Omega}\left(y\right) = \left\vert z-y\right\vert \geq\left\vert x-z\right\vert -\left\vert x-y\right\vert \geq d_{\Omega}\left(x\right) -\left\vert x-y\right\vert \geq\frac{1}{2}d_{\Omega}\left(x\right).$ Then $d_{\Omega}\left(y\right) \geq\frac{1}{2}d_{\Omega}\left(x\right) \geq\left\vert x-y\right\vert.$ Thus, since also $\left\vert x-y\right\vert \leq \frac{d_{\Omega}\left(x\right) }{2},$ we have $\left\vert x-y\right\vert \leq\frac{1}{\sqrt{2}}\left(d_{\Omega}\left(x\right) d_{\Omega}\left(y\right) \right) ^{\frac{1}{2}},$ and so, from (2.26), $G(x, y)\geq c^{\prime}\frac{1}{\left\vert x-y\right\vert ^{n-2s}}\geq c^{\prime}\frac{1}{\left(\frac{1}{\sqrt{2}}\left(d_{\Omega}\left(x\right) d_{\Omega}\left(y\right) \right) ^{\frac{1}{2}}\right) ^{n-2s} } = 2^{\frac{n}{2}-s}c^{\prime}\frac{d_{\Omega}^{s}\left(x\right) d_{\Omega }^{s}\left(y\right) }{\left(d_{\Omega}\left(x\right) d_{\Omega}\left(y\right) \right) ^{\frac{n}{2}}}\geq\frac{2^{\frac{n}{2}-s}c^{\prime} }{\left(diam\left(\Omega\right) \right) ^{n}}d_{\Omega}^{s}\left(x\right) d_{\Omega}^{s}\left(y\right).$ If $\left\vert x-y\right\vert \leq\frac{d_{\Omega}\left(y\right) }{2},$ by interchanging the roles of $x$ and $y$ in the above argument, the same conclusion is reached.

ⅲ) If $0 < \beta < s,$ then

$G(x, y)\leq c\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{\beta}}{\left\vert x-y\right\vert ^{n-s+\beta}}.$

(2.29)

Indeed, If $d_{\Omega}\left(y\right) \geq\left\vert x-y\right\vert$ then, from (2.23),

$G(x, y)\leq c\frac{d_{\Omega}\left( x\right) ^{s}}{\left\vert x-y\right\vert ^{n-s}} = c\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{\beta}}{\left\vert x-y\right\vert ^{n-s}d_{\Omega}\left( y\right) ^{\beta }}\leq c\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{\beta}}{\left\vert x-y\right\vert ^{n-s+\beta}},$

and if $d_{\Omega}\left(y\right) \leq\left\vert x-y\right\vert$ then, from (2.27),

$G(x, y)\leq c\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{s}}{\left\vert x-y\right\vert ^{n}} = c\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{\beta}d_{\Omega}\left( y\right) ^{s-\beta }}{\left\vert x-y\right\vert ^{n-s+\beta}\left\vert x-y\right\vert ^{s-\beta} }\leq c\frac{d_{\Omega}\left( x\right) ^{s}d_{\Omega}\left( y\right) ^{\beta}}{\left\vert x-y\right\vert ^{n-s+\beta}},$

ⅳ) If $f\in C\left(\overline{\Omega}\right)$ then the unique solution $u\in X_{0}^{s}\left(\Omega\right)$ of problem (1.4) is given by $u\left(x\right) : = \int_{\Omega}G\left(x, y\right) f\left(y\right) dy$ for $x\in\Omega,$ and $u\left(x\right) : = 0$ for $x\in\mathbb{R}^{n}\setminus\Omega.$

Lemma 2.7. Let $a\in L^{\infty}\left(\Omega\right)$ and let $\beta\in\left[0, s\right).$ Then $ad_{\Omega}^{-\beta}\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}$ and the weak solution $u\in X_{0}^{s}\left(\Omega\right)$ of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = ad_{\Omega}^{-\beta}\; in \;\Omega, \\ u = 0\; in \;\mathbb{R}^{n}\setminus\Omega \end{array} \right.$

(2.30)

satisfies $d_{\Omega}^{-s}u\in L^{\infty}\left(\Omega\right).$

Proof. Let $\varphi\in X_{0}^{s}\left(\Omega\right).$ By the Hölder and Hardy inequalities we have $\int_{\Omega}\left\vert ad_{\Omega}^{-\beta} \varphi\right\vert = \int_{\Omega}\left\vert ad_{\Omega}^{s-\beta}d_{\Omega }^{-s}\varphi\right\vert \leq\left\Vert a\right\Vert _{\infty}\left\Vert d_{\Omega}^{s-\beta}\right\Vert _{2}\left\Vert d_{\Omega}^{-s}\varphi \right\Vert _{2}\leq c\left\Vert \varphi\right\Vert _{X_{0}^{s}\left(\Omega\right) }$ with $c$ a positive constant independent of $\varphi.$ Thus $ad_{\Omega}^{-\beta}\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}.$ Let $u\in X_{0}^{s}\left(\Omega\right)$ be the unique weak solution (given by the Riesz Theorem) of problem (2.30) and consider a decreasing sequence $\left\{ \varepsilon_{j}\right\} _{j\in N}$ in $\left(0, 1\right)$ such that $\lim_{j\rightarrow\infty}\varepsilon_{j} = 0.$ For $j\in\mathbb{N},$ let $u_{\varepsilon_{j}}\in X_{0}^{s}\left(\Omega\right)$ be the weak solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{\varepsilon_{j}} = a\left( d_{\Omega }+\varepsilon_{j}\right) ^{-\beta}\text{ in }\Omega, \\ u_{\varepsilon_{j}} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.31)

Thus $u_{\varepsilon_{j}} = \int_{\Omega}G\left(., y\right) a\left(y\right) \left(d_{\Omega}\left(y\right) +\varepsilon_{j}\right) ^{-\beta}dy$ in $\Omega,$ where $G$ is the Green function for $\left(-\Delta\right) ^{s}$ in $\Omega,$ with homogeneous Dirichlet boundary condition on $\mathbb{R} ^{n}\setminus\Omega.$ Since $\beta < s$ we have $\int_{\Omega}\frac {1}{\left\vert x-y\right\vert ^{n-s+\beta}}dy < \infty.$ Thus, recalling (2.29), there exists a positive constant $c$ such that, for any $j\in\mathbb{N}$ and $\left(x, y\right) \in\Omega\times\Omega,$

$\begin{align*} 0 & \leq G\left( x, y\right) a\left( y\right) \left( d_{\Omega}\left( y\right) +\varepsilon_{j}\right) ^{-\beta}\leq c\frac{d_{\Omega}^{s}\left( x\right) d_{\Omega}^{\beta}\left( y\right) }{\left\vert x-y\right\vert ^{n-s+\beta}}\left( d_{\Omega}\left( y\right) +\varepsilon_{j}\right) ^{-\beta}\\ & \leq cd_{\Omega}^{s}\left( x\right) \frac{1}{\left\vert x-y\right\vert ^{n-s+\beta}}\in L^{1}\left( \Omega, dy\right) . \end{align*}$

Since also $\lim_{j\rightarrow\infty}G\left(x, y\right) a\left(y\right) \left(d_{\Omega}\left(y\right) +\varepsilon_{j}\right) ^{-\beta } = G\left(x, y\right) a\left(y\right) d_{\Omega}^{-\beta}\left(y\right)$ for $a.e.$ $y\in\Omega,$ by the Lebesgue dominated convergence theorem, $\left\{ u_{\varepsilon_{j}}\left(x\right) \right\} _{j\in\mathbb{N}}$ converges to $\int_{\Omega}G\left(x, y\right) a\left(y\right) d_{\Omega }^{-\beta}\left(y\right) dy$ for any $x\in\Omega.$ Let $u\left(x\right) : = \lim_{j\rightarrow\infty}u_{\varepsilon_{j}}\left(x\right).$ Thus $u\left(x\right) = \int_{\Omega}G\left(x, y\right) a\left(y\right) d_{\Omega}^{-\beta}\left(y\right) dy.$ Again from (2.29), $u\leq cd_{\Omega}^{s}$ $a.e.$ in $\Omega,$ with $c$ constant $c$ independent of $x.$ Now we take $u_{\varepsilon_{j}}$ as a test function in (2.31) to obtain that

$\begin{align*} \int_{\Omega\times\Omega}\frac{\left( u_{\varepsilon_{j}}\left( x\right) -u_{\varepsilon_{j}}\left( y\right) \right) ^{2}}{\left\vert x-y\right\vert ^{n+2s}} & = \int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( u_{\varepsilon_{j}}\left( x\right) -u_{\varepsilon_{j}}\left( y\right) \right) ^{2}}{\left\vert x-y\right\vert ^{n+2s}}\\ & = \int_{\Omega}u_{\varepsilon_{j}}\left( y\right) \left( d_{\Omega }\left( y\right) +\varepsilon_{j}\right) ^{-\beta}dy\\ & \leq c\int_{\Omega}d_{\Omega}^{s}\left( y\right) \left( d_{\Omega }\left( y\right) +\varepsilon\right) j^{-\beta}dy\leq c^{\prime} \int_{\Omega}d_{\Omega}^{s-\beta}\left( y\right) dy = c^{\prime\prime}, \end{align*}$

with $c$ and $c^{\prime}$ constants independent of $j.$ For $j\in\mathbb{N},$ let $U_{\varepsilon_{j}}$ and $U$ be the functions, defined on $\mathbb{R} ^{n}\times\mathbb{R}^{n},$ by

$U_{\varepsilon_{j}}\left( x, y\right) : = u_{\varepsilon_{j}}\left( x\right) -u_{\varepsilon_{j}}\left( y\right) , \text{ }U\left( x, y\right) : = u\left( x\right) -u\left( y\right) .$

Then $\left\{ U_{\varepsilon_{j}}\right\} _{j\in\mathbb{N}}$ is bounded in $\mathcal{H} = L^{2}\left(\mathbb{R}^{n}\times\mathbb{R}^{n}, \frac {1}{\left\vert x-y\right\vert ^{n+2s}}dxdy\right).$ Thus, after pass to a subsequence if necessary, we can assume that $\left\{ U_{\varepsilon_{j} }\right\} _{j\in\mathbb{N}}$ is weakly convergent in $\mathcal{H}$ to some $V\in\mathcal{H}.$ Since $\left\{ U_{\varepsilon_{j}}\right\} _{j\in \mathbb{N}}$ converges pointwise to $U$ on $\mathbb{R}^{n}\times\mathbb{R} ^{n},$ we conclude that $U\in\mathcal{H}$ and that $\left\{ U_{\varepsilon _{j}}\right\} _{j\in\mathbb{N}}$ converges weakly to $U$ in $\mathcal{H}.$ Thus $u\in X_{0}^{s}\left(\Omega\right)$ and, for any $\varphi\in X_{0}^{s}\left(\Omega\right),$

$\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( u\left( x\right) -u\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy\\ = \lim\limits_{j\rightarrow\infty}\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}} \frac{\left( u_{\varepsilon_{j}}\left( x\right) -u_{\varepsilon_{j}}\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy\\ = \lim\limits_{j\rightarrow\infty}\int_{\Omega}a\left( d_{\Omega}+\varepsilon _{j}\right) ^{-\beta}\varphi = \int_{\Omega}ad_{\Omega}^{-\beta}\varphi,$

Then $u$ is the weak solution of (2.30). Finally, since for all $j,$ $u_{\varepsilon_{j}}\leq c^{\prime}d_{\Omega}^{s}$ $a.e.$ in $\Omega,$ we have $u\leq c^{\prime}d_{\Omega}^{s}$ $a.e.$ in $\Omega.$

Lemma 2.8. Let $\lambda>0$ and let $\varepsilon \geq0.$ Suppose that $\left\{ u_{j}\right\} _{j\in\mathbb{N}}\subset X_{0}^{s}\left(\Omega\right)$ is a nonincreasing sequence with the following properties $i)$ and $ii)$:

$i)$ There exist positive constants $c$ and $c^{\prime}$ such that $cd_{\Omega}^{s}\leq u_{j}\leq c^{\prime}d_{\Omega }^{s}$ $a.e.$ in $\Omega$ for any $j\in\mathbb{N}.$

$ii)$ for any $j\in\mathbb{N}$, $u_{j}$ is a weak solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{j} = -g\left( ., u_{j}+\varepsilon\right) +\lambda h\; in\;\Omega, \\ u_{j} = 0\; in \;\mathbb{R}^{n}\setminus\Omega, \\ u_{j}>.0\; in \;\Omega \end{array} \right.$

(2.32)

Then $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ converges in $X_{0} ^{s}\left(\Omega\right)$ to a weak solution $u$ of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = -g\left( ., u+\varepsilon\right) +\lambda h\; in \;\Omega, \\ u_{j} = 0\; in \;\mathbb{R}^{n}\setminus\Omega, \\ u>0\; in \;\Omega, \end{array} \right.$

(2.33)

which satisfies $cd_{\Omega}^{s}\leq u\leq c^{\prime}d_{\Omega}^{s}$ $a.e.$ in $\Omega.$ Moreover, the same conclusions holds if, instead of ii), we assume the following $ii')$:

$ii')$ for any $j\geq2,$ $u_{j}$ is a weak solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{j} = -g\left( ., u_{j-1}+\varepsilon\right) +\lambda h\; in\;\Omega, \\ u_{j} = 0\; in \;\mathbb{R}^{n}\setminus\Omega, \\ u_{j}>0\; in \;\Omega. \end{array} \right.$

Proof. Assume $i)$ and $ii)$. For $x\in\mathbb{R}^{n},$ let $u\left(x\right) : = \lim_{j\rightarrow\infty}u_{j}\left(x\right).$ For $j, k\in\mathbb{N}$ we have, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\left( u_{j}-u_{k}\right) = g\left( ., u_{k}+\varepsilon\right) -g\left( u_{j}+\varepsilon\right) \text{ in }\Omega, \\ u_{j}-u_{k} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.34)

We take $u_{j}-u_{k}$ as a test function in (2.34) to get

$\begin{align} \left\Vert u_{j}-u_{k}\right\Vert _{X_{0}^{s}\left( \Omega\right) }^{2} & = \int_{\Omega}\left( g\left( ., u_{k}+\varepsilon\right) -g\left( ., u_{j}+\varepsilon\right) \right) \left( u_{j}-u_{k}\right) \label{inecuacion domingo dos new}\\ & = \int_{\Omega}d_{\Omega}^{s}\left( g\left( ., u_{k}+\varepsilon\right) -g\left( ., u_{j}+\varepsilon\right) \right) d_{\Omega}^{-s}\left( u_{j}-u_{k}\right) \nonumber\\ & \leq\left\Vert d_{\Omega}^{-s}\left( \overline{u}_{j}-\overline{u} _{k}\right) \right\Vert _{2}\left\Vert d_{\Omega}^{s}\left( g\left( ., \overline{u}_{k}+\varepsilon\right) -g\left( ., \overline{u}_{j} +\varepsilon\right) \right) \right\Vert _{2}.\nonumber \end{align}$

(2.35)

By the Hardy inequality, $\left\Vert d_{\Omega}^{-s}\left(u_{j} -u_{k}\right) \right\Vert _{2}\leq c^{\prime\prime}\left\Vert u_{j} -u_{k}\right\Vert _{X_{0}^{s}\left(\Omega\right) }$ where $c^{\prime\prime }$ is a constant independent of $j$ and $k.$ Thus, from (2.35),

$\left\Vert u_{j}-u_{k}\right\Vert _{X_{0}^{s}\left( \Omega\right) }\leq c^{\prime\prime}\left\Vert d_{\Omega}^{s}\left( g\left( ., u_{k} +\varepsilon\right) -g\left( ., u_{j}+\varepsilon\right) \right) \right\Vert _{2}.$

(2.36)

Now, $\lim_{j, k\rightarrow\infty}\left\vert d_{\Omega}^{s}\left(g\left(., u_{k}+\varepsilon\right) -g\left(., u_{j}+\varepsilon\right) \right) \right\vert ^{2} = 0$ $a.e.$ in $\Omega.$ Also, since $u_{l}\geq cd_{\Omega} ^{s}$ $a.e.$ in $\Omega$ for any $l\in\mathbb{N},$ and taking into account g5) and g2),

$\left\vert d_{\Omega}^{s}\left( g\left( ., u_{k}+\varepsilon\right) -g\left( ., u_{j}+\varepsilon\right) \right) \right\vert ^{2}\leq c^{\prime }\left( d_{\Omega}^{s}g\left( ., cd_{\Omega}^{s}\right) \right) ^{2}\in L^{1}\left( \Omega\right) ,$

where $c^{\prime}$ is a constant independent of $j$ and $k.$ Then, by the Lebesgue dominated convergence theorem $\lim_{j, k\rightarrow\infty}\left\Vert d_{\Omega}^{s}\left(g\left(., u_{k}+\varepsilon\right) -g\left(., u_{j}+\varepsilon\right) \right) \right\Vert _{2} = 0.$ Therefore, from (2.36), $\lim_{j, k\rightarrow\infty}\left\Vert u_{j}-u_{k}\right\Vert _{X_{0}^{s}\left(\Omega\right) } = 0$ and so $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ converges in $X_{0}^{s}\left(\Omega\right)$ to some $u^{\ast}\in X_{0}^{s}\left(\Omega\right).$ Then, by the Poincaré inequality of Remark 2.1 $iv)$, $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ converges to $u^{\ast}$ in $L^{2_{s}^{\ast}}\left(\Omega\right),$ and thus there exists a subsequence $\left\{ u_{j_{k}}\right\} _{k\in\mathbb{N}}$ that converges to $u^{\ast}$ $a.e.$ in $\Omega$. Since $\left\{ u_{j_{k}}\right\} _{k\in\mathbb{N}}$ converges pointwise to $u_{\varepsilon},$ we conclude that $u^{\ast} = u.$ Then $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ converges to $u_{\varepsilon}$ in $X_{0}^{s}\left(\Omega\right).$ Now, for $\varphi\in X_{0}^{s}\left(\Omega\right)$ and $j\in\mathbb{N},$

$\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( u_{j}\left( x\right) -u_{j}\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s} }dxdy = \int_{\Omega}\left( -g\left( ., u_{j}+\varepsilon\right) +\lambda h\right) \varphi.$

(2.37)

Since $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ converges to $u$ in $X_{0}^{s}\left(\Omega\right),$ we have

$\lim\limits_{j\rightarrow\infty}\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}} \frac{\left( u_{j}\left( x\right) -u_{j}\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy\label{ecuacion domingo cinco}\\ = \int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( u\left( x\right) -u\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy.\nonumber$

(2.38)

On the other hand, $\left\vert \left(-g\left(., u_{j}+\varepsilon\right) +\lambda h\right) \varphi\right\vert \leq\left(g\left(., cd_{\Omega }\right) +\lambda\left\Vert h\right\Vert _{\infty}\right) \left\vert \varphi\right\vert \in L^{1}\left(\Omega\right)$ (with $c$ as in $i)$)$.$ Also, $\left\{ \left(-g\left(u_{j}+\varepsilon\right) +\lambda h\right) \varphi\right\} _{j\in\mathbb{N}}$ converges to $\left(-g\left(u+\varepsilon\right) +\lambda h\right) \varphi$ $a.e.$ in $\Omega.$ Then, by the Lebesgue dominated convergence theorem,

$\lim\limits_{j\rightarrow\infty}\int_{\Omega}\left( -g\left( ., u_{j}+\varepsilon \right) +\lambda h\right) \varphi = \int_{\Omega}\left( -g\left( ., u+\varepsilon\right) +\lambda h\right) \varphi.$

(2.39)

From (2.37), (2.38) and (2.39) we get, for any $\varphi\in X_{0}^{s}\left(\Omega\right),$

$\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\frac{\left( u\left( x\right) -u\left( y\right) \right) \left( \varphi\left( x\right) -\varphi\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy = \int_{\Omega }\left( -g\left( ., u+\varepsilon\right) +\lambda h\right) \varphi.$

and so $u$ is a weak solution of problem (2.33) which clearly satisfies $cd_{\Omega}^{s}\leq u\leq c^{\prime}d_{\Omega}^{s}$ $a.e.$ in $\Omega.$ If instead of $ii)$ we assume $ii')$, the proof is the same. Only replace, for $j\geq2,$ $k\geq2$ and in each appearance, $g\left(., u_{j}\right)$ and $g\left(., u_{k}\right)$ by $g\left(., u_{j-1}\right)$ and $g\left(., u_{k-1}\right)$ respectively.

Lemma 2.9. Let $\lambda>0,$ and let $\overline{u}$ be the weak solution of

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\overline{u} = \lambda h\; in \;\Omega, \\ \overline{u} = 0\; in \;\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.40)

Assume that, for each $\varepsilon>0,$ we have a function $\widetilde {v}_{\varepsilon}\in X_{0}^{s}\left(\Omega\right)$ satisfying, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\widetilde{v}_{\varepsilon}\leq-g\left( ., \widetilde{v}_{\varepsilon}+\varepsilon\right) +\lambda h\;in \; \Omega, \\ \widetilde{v}_{\varepsilon} = 0\; in \;\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

(2.41)

and such that $\widetilde{v}_{\varepsilon}\geq cd_{\Omega}^{s}$ $a.e.$ in $\Omega,$ where $c$ is a positive constant independent of $\varepsilon.$ Then for any $\varepsilon>0$ there exists a sequence $\left\{ u_{j}\right\} _{j\in\mathbb{N}}\subset X_{0}^{s}\left(\Omega\right)$ such that:

$i)$ $u_{1} = \overline{u}$ and $u_{j}\leq$ $u_{j-1}$ for any $j\geq2.$

$ii)$ $\widetilde{v}_{\varepsilon}\leq$ $u_{j}\leq\overline{u}$ for all $j\in N.$

$iii)$ For any $j\geq2,$ $u_{j}$ satisfies, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{j} = -g\left( ., u_{j-1}+\varepsilon\right) +\lambda h\; in \;\Omega, \\ u_{j} = 0\; in \;\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

$iv)$ There exist positive constants $c$ and $c^{\prime}$independent of $\varepsilon$ and $j$ such that, for all $j,$ $cd_{\Omega}^{s}\leq u_{j}\leq c^{\prime}d_{\Omega}^{s}$ $a.e.$ in $\Omega.$

Proof. By Remark 2.1 $iii)$, there exists a positive constant $c^{\prime}$ such that $\overline{u}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega.$ We construct inductively a sequence $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ satisfying the assertions $i)-iii)$ of the lemma: Let $u_{1}: = \overline{u}.$ Thus, in weak sense, $\left(-\Delta\right) ^{s}u_{1} = \lambda h\geq-g\left(., \widetilde{v}_{\varepsilon}+\varepsilon \right) +\lambda h$ in $\Omega.$ Thus $\left(-\Delta\right) ^{s}\left(u_{1}-\widetilde{v}_{\varepsilon}\right) \geq0$ in $\Omega.$ Then, by the maximum principle in Remark 2.1 $i)$, $u_{1} \geq\widetilde{v}_{\varepsilon}$ in $\Omega,$ and so $u_{1}\geq cd_{\Omega }^{s}$ in $\Omega.$ Then, for some positive constant $c^{\prime\prime},$ $\left\vert -g\left(., u_{1}+\varepsilon\right) +\lambda h\right\vert \leq c^{\prime\prime}\left(1+g\left(., cd_{\Omega}^{s}\right) \right)$ in $\Omega$ and, by g1), $g\left(., cd_{\Omega}^{s}\right) \in L^{\infty}\left(\Omega\right).$ Thus there exists a weak solution $u_{2}\in X_{0}^{s}\left(\Omega\right)$ to the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{2} = -g\left( ., u_{1}+\varepsilon\right) +\lambda h\text{ in }\Omega, \\ u_{2} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Since, in weak sense, $\left(-\Delta\right) ^{s}u_{2}\leq\lambda h = \left(-\Delta\right) ^{s}u_{1}$ in $\Omega,$ the maximum principle in Remark 2.1 $i)$ gives $u_{2}\leq u_{1}$ in $\Omega.$ Since $u_{1}\geq\widetilde{v}_{\varepsilon}$ in $\Omega$ we have, in weak sense, $\left(-\Delta\right) ^{s}u_{2} = -g\left(., u_{1}+\varepsilon \right) +\lambda h\geq-g\left(., \widetilde{v}_{\varepsilon}+\varepsilon \right) +\lambda h$ in $\Omega.$ Also, $\left(-\Delta\right) ^{s}\widetilde{v}_{\varepsilon}\leq-g\left(., \widetilde{v}_{\varepsilon }+\varepsilon\right) +\lambda h$ in $\Omega$ and so, by the maximum principle, $u_{2}\geq\widetilde{v}_{\varepsilon}$ in $\Omega.$ Then $i)-iii)$ hold for $j = 1.$

Supposed constructed $u_{1}, ..., u_{k}$ such that $i)-iii)$ hold for $1\leq j\leq k.$ Then, for some positive constant $c^{\prime\prime\prime},$ $\left\vert -g\left(., u_{k}+\varepsilon\right) +\lambda h\right\vert \leq c^{\prime\prime}\left(1+g\left(., cd_{\Omega}^{s}\right) \right)$ in $\Omega$ and so, as before, there exists a weak solution $u_{k+1}\in X_{0}^{s}\left(\Omega\right)$ to the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{k+1} = -g\left( ., u_{k}+\varepsilon\right) +\lambda h\text{ in }\Omega, \\ u_{k+1} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

By our inductive hypothesis, $u_{k}\geq\widetilde{v}_{\varepsilon}$ in $\Omega.$ Then, in weak sense, $\left(-\Delta\right) ^{s}u_{k+1} = -g\left(., u_{k}+\varepsilon\right) +\lambda h\geq-g\left(., \widetilde {v}_{\varepsilon}+\varepsilon\right) +\lambda h\geq\left(-\Delta\right) ^{s}\widetilde{v}_{\varepsilon}$ in $\Omega$ and thus, by the maximum principle, $u_{k+1}\geq\widetilde{v}_{\varepsilon}$ in $\Omega.$ If $k = 2$ we have $u_{k}\leq u_{k-1}$ in $\Omega.$ If $k>2,$ by the inductive hypothesis we have, in weak sense, $\left(-\Delta\right) ^{s}u_{k} = -g\left(., u_{k-1}+\varepsilon\right) +\lambda h\leq-g\left(., u_{k-2}+\varepsilon \right) +\lambda h$ in $\Omega.$ Also, $\left(-\Delta\right) ^{s} u_{k} = -g\left(., u_{k-1}+\varepsilon\right) +\lambda h$ in $\Omega.$ Thus, by the maximum principle, $u_{k+1}\leq u_{k}$ in $\Omega.$ Again by the inductive hypothesis $u_{k}\leq\overline{u}$ in $\Omega$ and then, since $u_{k+1}\leq u_{k}$ in $\Omega,$ we get $u_{k+1}\leq\overline{u}$ in $\Omega.$

Since for all $j,$ $v_{\varepsilon}\leq u_{j}\leq\overline{u}$ in $\Omega,$ $iv)$ follows from the facts that $\overline{u}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega,$ and that $\widetilde{v}_{\varepsilon }\geq cd_{\Omega}^{s}$ in $\Omega,$ with $c$ and $c^{\prime}$ positive constants independent of $\varepsilon$ and $j.$

Lemma 2.10. Let $\lambda>0.$ Assume that we have, for each $\varepsilon>0,$ a function $\widetilde{v}_{\varepsilon}\in X_{0}^{s}\left(\Omega\right)$ satisfying, in weak sense, (2.41), and such that $\widetilde{v}_{\varepsilon}\geq cd_{\Omega}^{s}$ $a.e.$ in $\Omega,$ with $c$ a positive constant independent of $\varepsilon.$ Then for any $\varepsilon>0$ there exists a weak solution $u_{\varepsilon}$ of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{\varepsilon} = -g\left( ., u_{\varepsilon }+\varepsilon\right) +\lambda h\;in \;\Omega, \\ u_{\varepsilon} = 0\; in \;\mathbb{R}^{n}\setminus\Omega, \\ u_{\varepsilon}>0\;in \;\Omega. \end{array} \right.$

(2.42)

such that:

$i)$ $u_{\varepsilon}\geq\widetilde{v}_{\varepsilon}$ and there exist positive constants $c$ and $c^{\prime}$ independent of $\varepsilon$ such that $cd_{\Omega}^{s}\leq u_{\varepsilon}\leq c^{\prime }d_{\Omega}^{s}$ in $\Omega,$

$ii)$ If $\underline{u}_{\varepsilon}\in X_{0}^{s}\left(\Omega\right)$ and $\left(-\Delta\right) ^{s} \underline{u}_{\varepsilon}\leq-g\left(., \underline{u}_{\varepsilon }+\varepsilon\right) +\lambda h$ in $\Omega,$ then $\underline{u} _{\varepsilon}\leq u_{\varepsilon}$ in $\Omega,$

$iii)$ If $0 < \varepsilon < \eta$ then $u_{\varepsilon}\leq u_{\eta}.$

Proof. Let $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ be as given by Lemma 2.9. For $x\in\Omega,$ let $u_{\varepsilon}\left(x\right) : = \lim_{j\rightarrow\infty}u_{j}\left(x\right).$ By Lemma 2.8, $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ converges to $u_{\varepsilon}$ in $X_{0} ^{s}\left(\Omega\right)$ and $u_{\varepsilon}$ is a weak solution to (2.42). From Lemma 2.9 $iv)$ we have $u_{\varepsilon}\geq\widetilde{v}_{\varepsilon}$ in $\Omega$ and $cd_{\Omega }^{s}\leq u_{\varepsilon}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega,$ for some positive constants $c$ and $c^{\prime}$ independent of $\varepsilon.$ Then $i)$ holds. If $\underline{u}_{\varepsilon}\in X_{0}^{s}\left(\Omega\right)$ and $\left(-\Delta\right) ^{s}\underline{u}_{\varepsilon }\leq-g\left(., \underline{u}_{\varepsilon}+\varepsilon\right) +\lambda h$ in $\Omega,$ then $\left(-\Delta\right) ^{s}\underline{u}_{\varepsilon} \leq\lambda h = \left(-\Delta\right) ^{s}u_{1}$ in $\Omega,$ and so $\underline{u}_{\varepsilon}\leq u_{1}.$ Thus $\left(-\Delta\right) ^{s}\underline{u}_{\varepsilon}\leq-g\left(., \underline{u}_{\varepsilon }+\varepsilon\right) +\lambda h\leq-g\left(., u_{1}+\varepsilon\right) +\lambda h = \left(-\Delta\right) ^{s}u_{2},$ then $\underline{u} _{\varepsilon}\leq u_{2}$ and, iterating this procedure, we obtain that $\underline{u}_{\varepsilon}\leq u_{j}$ for all $j.$ Then $\underline {u}_{\varepsilon}\leq u_{\varepsilon}.$ Thus $ii)$ holds. Finally, $iii)$ is immediate from $ii)$.

Lemma 2.11. Let $\varepsilon>0$ and let $\tau_{1}$ and $M_{1}$ be as in Remarks 2.2, and 2.3 respectively. Let $w_{\varepsilon}$ be as in Lemma 2.5. Then, for $\lambda\geq\tau_{1} M_{1},$ there exists a weak solution $u_{\varepsilon}\in X_{0}^{s}\left(\Omega\right)$ of problem (2.42) such that

$i)$ $u_{\varepsilon}\geq w_{\varepsilon}$ and there exist positive constants $c$ and $c^{\prime},$ both independent of $\varepsilon,$ such that $cd_{\Omega }^{s}\leq u_{\varepsilon}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega,$

$ii)$ If $\underline{u}_{\varepsilon}\in X_{0}^{s}\left(\Omega\right)$ and $\left(-\Delta\right) ^{s}\underline{u}_{\varepsilon}\leq-g\left(., \underline{u}_{\varepsilon}+\varepsilon\right) +\lambda h$ in $\Omega,$ then $\underline{u}_{\varepsilon}\leq u_{\varepsilon}$ in $\Omega,$

$iii)$ If $0 < \varepsilon_{1} < \varepsilon_{2}$ then $u_{\varepsilon_{1}}\leq u_{\varepsilon_{2}}.$

Proof. Let $\lambda\geq\tau_{1}M_{1}$ and let $w_{\varepsilon}$ be as in Lemma 2.5. We have, in weak sense,

$\left\{ \begin{array} [c]{c} -\Delta w_{\varepsilon} = -g\left( ., w_{\varepsilon}+\varepsilon\right) +\tau_{1}M_{1}h\text{ in }\Omega, \\ w_{\varepsilon} = 0\text{ on }\mathbb{R}^{n}\setminus\Omega, \end{array} \right.$

Also, $-g\left(., w_{\varepsilon}+\varepsilon\right) +\tau_{1}M_{1} h\leq-g\left(., w_{\varepsilon}+\varepsilon\right) +\lambda h$ in $\Omega,$ and $cd_{\Omega}^{s}\leq w_{\varepsilon}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega,$ with $c$ and $c^{\prime}$ positive constants independent of $\varepsilon.$ Then the lemma follows from Lemma 2.10.

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3. Proof of the main results

Lemma 3.1. Let $\lambda>0.$ If problem (1.5) has a weak solution in $\mathcal{E}$, then it has a weak solution $u\in\mathcal{E}$ satisfying $u\geq\psi$ $a.e.$ in $\Omega$ for any $\psi \in\mathcal{E}$ such that $-\Delta\psi\leq-g\left(., \psi\right) +\lambda h$ in $\Omega.$

Proof. Let $u^{\ast}\in\mathcal{E}$ be a weak solution of (1.5), and let $\overline{u}$ be as in (2.40). By the comparison principle $u^{\ast}\leq$ $\overline{u}$ in $\Omega.$ We construct inductively a sequence $\left\{ u_{j}\right\} _{j\in\mathbb{N}}\subset\mathcal{E}$ with the following properties: $u_{1} = \overline{u}$ and

$i)$ $u^{\ast}\leq u_{j}\leq\overline{u}$ for all $j\in\mathbb{N}$

$ii)$ $g\left(., u_{j}\right) \in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}$ for all $j\in\mathbb{N}$.

$iii)$ $u_{j}\leq u_{j-1}$ for all $j\geq2.$

$iv)$ For all $j\geq2$

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{j} = -g\left( ., u_{j-1}\right) +\lambda h\text{ in }\Omega, \\ u_{j} = 0\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

To do it, define $u_{1} = :\overline{u}.$ Thus $u_{1}\in\mathcal{E}$. By the comparison principle, $u^{\ast}\leq\overline{u},$ i.e., $u^{\ast}\leq u_{1}.$ By Remark 2.1there exist positive constants $c$ and $c^{\prime}$ such that $cd_{\Omega}^{s}\leq\overline{u}\leq c^{\prime }d_{\Omega}^{s}$ in $\Omega.$ Thus $\left\vert -g\left(., \overline {u}\right) +\lambda h\right\vert \leq g\left(., cd_{\Omega}^{s}\right) +\lambda\left\Vert h\right\Vert _{\infty}$ and so $-g\left(., u_{1}\right) +\lambda h\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}.$ Thus $i)$ and $ii)$ hold for $j = 1.$ Define $u_{2}$ as the weak solution of

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{2} = -g\left( ., u_{1}\right) +\lambda h\text{ in }\Omega, \\ u_{2} = 0\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Then, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{2}\leq\left( -\Delta\right) ^{s}u_{1}\text{ in }\Omega, \\ u_{2} = 0 = u_{1}\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

and so $u_{2}\leq u_{1} = \overline{u}$ $a.e.$ in $\Omega.$ Since $u_{1}\geq u^{\ast},$ we have, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{2} = -g\left( ., u_{1}\right) +\lambda h\geq-g\left( ., u^{\ast}\right) +\lambda h = \left( -\Delta\right) ^{s}u^{\ast}\text{ in }\Omega, \\ u_{2} = 0 = u^{\ast}\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

and then $u_{2}\geq u^{\ast}$ $a.e.$ in $\Omega.$ Thus $u^{\ast}\leq u_{2} \leq\overline{u}.$ In particular this gives $u_{2}\in\mathcal{E}.$ Let $c^{\prime\prime}>0$ such that $u^{\ast}\geq c^{\prime\prime}d_{\Omega}$ in $\Omega.$ Now, $\left\vert -g\left(., u_{2}\right) +\lambda h\right\vert \leq g\left(., u_{2}\right) +\lambda h\leq g\left(., u^{\ast}\right) +\lambda h\leq g\left(., c^{\prime\prime}d_{\Omega}^{s}\right) +\lambda\left\Vert h\right\Vert _{\infty}$ and so $-g\left(., u_{2}\right) +\lambda h\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}.$ Thus $i)-iv)$ hold for $j = 2.$ Suppose constructed, for $2\leq j\leq k,$ functions $u_{j}\in\mathcal{E}$ with the properties $i)-iv)$. Define $u_{k+1}$ by

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{k+1} = -g\left( ., u_{k}\right) +\lambda h\text{ in }\Omega, \\ u_{k+1} = 0\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Thus, by the comparison principle, $u_{k+1}\leq\overline{u}.$ Also, by the inductive hypothesis, $u_{k}\geq u^{\ast},$ then

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{k+1} = -g\left( ., u_{k}\right) +\lambda h\geq-g\left( ., u^{\ast}\right) +\lambda h\text{ in }\Omega, \\ u_{k+1} = 0 = u^{\ast}\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

and so $u_{k+1}\geq u^{\ast}.$ Then $u^{\ast}\leq u_{k+1}\leq\overline{u}.$ In particular $u_{k+1}\in\mathcal{E}.$ Again by the inductive hypothesis, $u_{k}\leq u_{k-1}.$ Then

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u_{k+1} = -g\left( ., u_{k}\right) +\lambda h\leq-g\left( ., u_{k-1}\right) +\lambda h = \left( -\Delta\right) ^{s} u_{k}\text{ in }\Omega, \\ u_{k+1} = 0 = u_{k}\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

and so $u_{k+1}\leq u_{k}.$ Also, $\left\vert -g\left(., u_{k+1}\right) +\lambda h\right\vert \leq g\left(., u_{k+1}\right) +\lambda h\leq g\left(., u^{\ast}\right) +\lambda h\leq g\left(., c^{\prime\prime}d_{\Omega} ^{s}\right) +\lambda\left\Vert h\right\Vert _{\infty}$ and so $-g\left(., u_{2}\right) +\lambda h\in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}.$ Thus $i)-iv)$ hold for $j = k+1,$ which completes the inductive construction of the sequence $\left\{ u_{j}\right\} _{j\in \mathbb{N}}.$ For $x\in\mathbb{R}^{n}$ let $u\left(x\right) : = \lim _{j\rightarrow\infty}u_{j}\left(x\right).$ By $i)$ we have $c^{\prime\prime}d_{\Omega}^{s}\leq u_{j}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega$ for all $j\in\mathbb{N}$, and so $c^{\prime\prime}d_{\Omega}^{s}\leq u\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega.$ By Lemma 2.8 $\left\{ u_{j}\right\} _{j\in \mathbb{N}}$ converges in $X_{0}^{s}\left(\Omega\right)$ to some weak solution $u^{\ast\ast}\in X_{0}^{s}\left(\Omega\right)$ of problem (1.5). Thus, by the Poincaré inequality, $\left\{ u_{j}\right\} _{j\in\mathbb{N}}$ converges to $u^{\ast\ast}$ in $L^{2_{s}^{\ast}}\left(\Omega\right),$ which implies $u = u^{\ast\ast}.$ Thus $u\in X_{0}^{s}\left(\Omega\right)$ and $u$ is a weak solution of problem (1.5). Since $c^{\prime\prime}d_{\Omega}^{s}\leq u_{j}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega$ for all $j,$ we have $c^{\prime\prime}d_{\Omega}^{s}\leq u\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega.$ Thus $u\in\mathcal{E}$. Let $\psi\in\mathcal{E}$ such that $-\Delta\psi\leq-g\left(., \psi\right) +\lambda h$ in $\Omega.$ By the comparison principle, $\psi\leq u$ $a.e.$ in $\Omega.$ An easy induction shows that $\psi\leq u_{j}$ for all $j.$ Indeed, by the comparison principle, $\psi\leq\overline{u} = u_{1}.$ Then

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\psi\leq-g\left( ., \psi\right) +\lambda h\leq-g\left( ., u_{1}\right) +\lambda h = \left( -\Delta\right) ^{s} u_{2}\text{ in }\Omega, \\ \psi = 0 = u_{2}\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Thus, again by the comparison principle, $\psi\leq u_{2}.$ Suppose that $k\geq2$ and $\psi\leq u_{k}.$ Then, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\psi\leq-g\left( ., \psi\right) +\lambda h\leq-g\left( ., u_{k}\right) +\lambda h = \left( -\Delta\right) ^{s} u_{k+1}\text{ in }\Omega, \\ \psi = 0 = u_{k+1}\text{ on }\mathbb{R}^{n}\setminus\Omega, \end{array} \right.$

which gives $\psi\leq u_{k+1}.$ Thus $\psi\leq u_{j}$ for all $j,$ and so $\psi\leq u.$

Proof of Theorem 1. Let $\left\{ \varepsilon_{j}\right\} _{j\in\mathbb{N}}\subset\left(0, \infty\right)$ be a decreasing sequence such that $\lim_{j\rightarrow\infty}\varepsilon_{j} = 0.$ For $\lambda\geq \tau_{1}M_{1}$ and $j\in\mathbb{N},$ let $u_{\varepsilon_{j}}$ be the weak solution of problem (2.42), given by Lemma 2.11, taking there $\varepsilon = \varepsilon_{j}$. Then $\left\{ u_{\varepsilon_{j}}\right\} _{j\in \mathbb{N}}$ is a nonincreasing sequence in $X_{0}^{s}\left(\Omega\right)$ and there exist positive constants $c$ and $c^{\prime}$ such that $cd_{\Omega }^{s}\leq u_{j}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega$ for all $j\in\mathbb{N}.$ Therefore, by Lemma 2.8, $\left\{ u_{\varepsilon_{j}}\right\} _{j\in\mathbb{N}}$ converges in $X_{0}^{s}\left(\Omega\right)$ to some weak solution $u\in X_{0} ^{s}\left(\Omega\right)$ of problem (1.5). Let

$\mathcal{T}: = \left\{ \lambda>0:\text{ problem (1.5) has a weak solution }u\in\mathcal{E}\right\} .$

Thus $\lambda\in\mathcal{T}$ whenever $\lambda\geq\tau_{1}M_{1}$. Consider now an arbitrary $\lambda\in\mathcal{T}$, and let $\lambda^{\prime}>\lambda.$ Let $u\in\mathcal{E}$ be a weak solution of the problem

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = -g\left( ., u\right) +\lambda h\text{ in }\Omega, \\ u = 0\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Let $\left\{ \varepsilon_{j}\right\} _{j\in\mathbb{N}}\subset\left(0, \infty\right)$ be a decreasing sequence such that $\lim_{j\rightarrow \infty}\varepsilon_{j} = 0.$ We have, in weak sense,

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}u = -g\left( ., u\right) +\lambda h\leq-g\left( ., u+\varepsilon_{j}\right) +\lambda^{\prime}h\text{ in }\Omega, \\ u = 0\text{ on }\mathbb{R}^{n}\setminus\Omega. \end{array} \right.$

Then, by Lemma 2.9, used with $\varepsilon = \varepsilon_{j},$ $\widetilde{v}_{\varepsilon_{j}} = u,$ and with $\lambda$ replaced by $\lambda^{\prime},$ there exists a nonincreasing sequence $\left\{ \widetilde{u}_{\varepsilon_{j}}\right\} _{j\in\mathbb{N} }\subset X_{0}^{s}\left(\Omega\right)$ such that

$\left\{ \begin{array} [c]{c} \left( -\Delta\right) ^{s}\widetilde{u}_{\varepsilon_{j}} = -g\left( ., \widetilde{u}_{\varepsilon_{j}}+\varepsilon_{j}\right) +\lambda^{\prime }h\text{ in }\Omega, \\ \widetilde{u}_{\varepsilon_{j}} = 0\text{ in }\mathbb{R}^{n}\setminus\Omega, \end{array} \right.$

satisfying that $\widetilde{u}_{\varepsilon_{j}}\geq u$ for all $j\in \mathbb{N},$ and $cd_{\Omega}^{s}\leq\widetilde{u}_{\varepsilon_{j}}\leq c^{\prime}d_{\Omega}^{s}$ in $\Omega$ for some positive constants $c$ and $c^{\prime}$ independent of $j.$ Let $\widetilde{u}: = \lim_{j\rightarrow\infty }\widetilde{u}_{\varepsilon_{j}}.$ Proceeding as in the first part of the proof, we get that $\widetilde{u}\in\mathcal{E}$ and that $\widetilde{u}$ is a weak solution of problem (1.5). Then $\lambda^{\prime} \in\mathcal{T}$ whenever $\lambda^{\prime}>\lambda$ for some $\lambda \in\mathcal{T}$. Thus there exists $\lambda^{\ast}\geq0$ such that $\left(\lambda^{\ast}, \infty\right) \subset\mathcal{T}\subset\left[\lambda^{\ast }, \infty\right).$

By Lemma 3.1, for any $\lambda\in\mathcal{T}$ there exists a weak solution $u\in\mathcal{E}$ of problem (1.5) such that $u\geq\psi$ $a.e.$ in $\Omega$ for any $\psi\in\mathcal{E}$ such that $\left(-\Delta\right) ^{s}\psi \leq-g\left(., \psi\right) +\lambda h$ in $\Omega.$

Suppose now that $g\left(., s\right) \geq bs^{-\beta}$ $a.e.$ in $\Omega$ for any $s\in\left(0, \infty\right)$ for some $b\in L^{\infty}\left(\Omega\right)$ such that $0\leq b\not \equiv 0$ in $\Omega.$ Then there exist a constant $\eta_{0}>0$ and a measurable set $\Omega_{0}\subset\Omega$ such that $\left\vert \Omega_{0}\right\vert >0$ and $b\geq\eta_{0}$ in $\Omega_{0}.$ Let $\lambda_{1}$ be the principal eigenvalue for $\left(-\Delta\right) ^{s}$ in $\Omega$ with Dirichlet boundary condition $\varphi_{1} = 0$ on $\mathbb{R}^{n}\setminus\Omega,$ and let $\varphi_{1}\in X_{0}^{s}\left(\Omega\right)$ be an associated positive principal eigenfunction. Then

$\lambda_{1}\int_{\Omega}\varphi\varphi_{1} = \int_{\mathbb{R}^{n}\times \mathbb{R}^{n}}\frac{\left( \varphi\left( x\right) -\varphi\left( y\right) \right) \left( \varphi_{1}\left( x\right) -\varphi_{1}\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy$

and $\varphi_{1}>0$ $a.e.$ in $\Omega$ (see e.g., ^[25], Theorem 3.1). Let $\lambda\in\mathcal{T}$ and let $u\in\mathcal{E}$ be a weak solution of (1.5). Thus

$\begin{align*} \lambda_{1}\int_{\Omega}u\varphi_{1} & = \int_{\mathbb{R}^{n}\times \mathbb{R}^{n}}\frac{\left( u\left( x\right) -u\left( y\right) \right) \left( \varphi_{1}\left( x\right) -\varphi_{1}\left( y\right) \right) }{\left\vert x-y\right\vert ^{n+2s}}dxdy\\ & = \int_{\Omega}\left( -\varphi_{1}g\left( ., u\right) +\lambda h\varphi_{1}\right) \leq\int_{\Omega}\left( -bu^{-\beta}\varphi_{1}+\lambda h\varphi_{1}\right) \end{align*}$

and so

$\lambda\int_{\Omega}h\varphi_{1}\geq\int_{\Omega_{0}}\left( \lambda _{1}u+bu^{-\beta}\right) \varphi_{1}\geq\inf\limits_{s>0}\left( \lambda_{1} s+\eta_{0}s^{-\beta}\right) \int_{\Omega_{0}}\varphi_{1}$

thus $\lambda\geq\inf_{s>0}\left(\lambda_{1}s+\eta_{0}s^{-\beta}\right) \left(\int_{\Omega}h\varphi_{1}\right) ^{-1}\int_{\Omega_{0}}\varphi_{1}$ for any $\lambda\in\mathcal{T}$. Then $\lambda^{\ast}>0.$

Lemma 3.2. Let $g:\Omega\times\left(0, \infty\right) \rightarrow\left[0, \infty\right)$ be a Carathéodory function. Assume that $s\rightarrow g\left(x, s\right)$ is nonincreasing for $a.e.$ $x\in\Omega,$ and that, for some $a\in L^{\infty}\left(\Omega\right)$ and $\beta\in\left[0, s\right),$ $g\left(., s\right) \leq as^{-\beta}$ $a.e.$ in $\Omega$ for any $s\in\left(0, \infty\right).$ Then $g$ satisfies the conditions g1)-g5) of Theorem 1.2.

Proof. Clearly $g$ satisfies g1) and g2). Let $\sigma>0.$ By Lemma 2.7, $0\leq g\left(., \sigma d_{\Omega} ^{s}\right) \leq a\sigma^{-\beta}d_{\Omega}^{-s\beta}\in\left(X_{0} ^{s}\left(\Omega\right) \right) ^{\prime}$ and so $g\left(., \sigma d_{\Omega}^{s}\right) \in\left(X_{0}^{s}\left(\Omega\right) \right) ^{\prime}.$ Also, from the comparison principle, $0\leq\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(d_{\Omega}^{s}g\left(., \sigma ad_{\Omega}^{s}\right) \right) \leq\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(\sigma^{-\beta}ad_{\Omega}^{s-\beta}\right)$ in $\Omega,$ and, since $ad_{\Omega}^{s-\beta}\in L^{\infty}\left(\Omega\right),$ by Remark 2.1 $iii)$, there exists a positive constant $c$ such that $\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(\sigma ad_{\Omega}^{s-\beta}\right) \leq cd_{\Omega}^{s}$ in $\Omega.$ Thus $d_{\Omega}^{-s}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(d_{\Omega}^{s}g\left(., \sigma d_{\Omega}^{s}\right) \right) \in L^{\infty}\left(\Omega\right).$ Then $g$ satisfies g3). In particular, $d_{\Omega}^{-s}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(d_{\Omega}^{s}g\left(., d_{\Omega }^{s}\right) \right) \in L^{\infty}\left(\Omega\right).$ Since, for $\sigma\geq1,$

$0\leq\left( \sigma d_{\Omega}^{s}\right) ^{-1}\left( \left( -\Delta \right) ^{s}\right) ^{-1}\left( d_{\Omega}^{s}g\left( ., \sigma d_{\Omega }^{s}\right) \right) \leq\sigma^{-1}d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( d_{\Omega}^{s}g\left( ., d_{\Omega }^{s}\right) \right) ,$

we get $\lim_{\sigma\rightarrow\infty}\left\Vert \left(\sigma d_{\Omega} ^{s}\right) ^{-1}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(d_{\Omega}^{s}g\left(., \sigma d_{\Omega}^{s}\right) \right) \right\Vert _{\infty} = 0.$ Also, by the comparison principle,

$0\leq d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( g\left( ., \sigma\right) \right) \leq\sigma^{-\beta}d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( a\right) ,$

and, by Remark 2.1 $iii)$, $d_{\Omega} ^{-s}\left(\left(-\Delta\right) ^{s}\right) ^{-1}\left(a\right) \in L^{\infty}\left(\Omega\right).$ Thus

$\lim\limits_{\sigma\rightarrow\infty}\left\Vert d_{\Omega}^{-s}\left( \left( -\Delta\right) ^{s}\right) ^{-1}\left( g\left( ., \sigma\right) \right) \right\Vert _{L^{\infty}\left( \Omega\right) } = 0.$

Then g4) holds. Finally, $0\leq d_{\Omega}^{s}g\left(., \sigma d_{\Omega}^{s}\right) \leq\sigma^{-\beta}d_{\Omega}^{s-\beta}a$ and so g5) holds.

Proof of Theorem 1.3. The theorem follows from Lemma 3.2 and Theorem 1.2.

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Conflict of interest

The author declare no conflicts of interest in this paper.

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