Asymptotic behavior of extremals for fractional Sobolev inequalities associated with singular problems

Abstract

Let $\Omega$ be a smooth, bounded domain of $\mathbb {R}^{N}$, $\omega$ be a positive, $L^{1}$-normalized function, and $0<s<1<p.$ We study the asymptotic behavior, as $p\rightarrow \infty ,$ of the pair $\left( \root p \of {\Lambda _{p} },u_{p}\right) ,$ where $\Lambda _{p}$ is the best constant C in the Sobolev-type inequality

$$\begin{aligned} C\exp \left( \int _{\Omega }(\log \left| u\right| ^{p})\omega \mathrm {d}x\right) \le \left[ u\right] _{s,p}^{p}\quad \forall \,u\in W_{0}^{s,p}(\Omega ) \end{aligned}$$

and $u_{p}$ is the positive, suitably normalized extremal function corresponding to $\Lambda _{p}$. We show that the limit pairs are closely related to the problem of minimizing the quotient $\left| u\right| _{s}/\exp \left( \int _{\Omega }(\log \left| u\right| )\omega \mathrm {d}x\right) ,$ where $\left| u\right| _{s}$ denotes the s-Hölder seminorm of a function $u\in C_{0}^{0,s}(\overline{\Omega }).$

1 Introduction

Let $\Omega$ be a smooth (at least Lipschitz) domain of $\mathbb {R}^{N}$, and consider the fractional Sobolev space

$$\begin{aligned} W_{0}^{s,p}(\Omega ):=\left\{ u\in L^{p}(\mathbb {R}^{N}):u=0\ \mathrm {in} \ \mathbb {R}^{N}{\setminus }\Omega \quad \mathrm {and}\quad \left[ u\right] _{s,p}<\infty \right\} , \quad 0<s<1<p, \end{aligned}$$

where

$$\begin{aligned} \left[ u\right] _{s,p}:=\left( \int _{\mathbb {R}^{N}}\int _{\mathbb {R}^{N} }\frac{\left| u(x)-u(y)\right| ^{p}}{\left| x-y\right| ^{N+sp}}\mathrm {d}x\mathrm {d}y\right) ^{\frac{1}{p}}. \end{aligned}$$

It is well known that the Gagliardo seminorm $\left[ \cdot \right] _{s,p}$ is a norm in $W_{0}^{s,p}(\Omega )$ and that this Banach space is uniformly convex. Actually,

$$\begin{aligned} W_{0}^{s,p}(\Omega )=\overline{C_{c}^{\infty }(\Omega )}^{\left[ \cdot \right] _{s,p}}. \end{aligned}$$

Let $\omega$ be a nonnegative function in $L^{1}(\Omega )$ satisfying $\left\| \omega \right\| _{L^{1}(\Omega )}=1$, and define

$$\begin{aligned} \mathcal {M}_{p}:=\left\{ u\in W_{0}^{s,p}(\Omega ):\int _{\Omega } (\log \left| u\right| )\omega \mathrm {d}x=0\right\} \end{aligned}$$

and

$$\begin{aligned} \Lambda _{p}:=\inf \left\{ \left[ u\right] _{s,p}^{p}:u\in \mathcal {M} _{p}\right\} . \end{aligned}$$

(1)

In the recent paper [9], it is proved that $\Lambda _{p}>0$ and that

$$\begin{aligned} \Lambda _{p}\exp \left( \int _{\Omega }(\log \left| u\right| ^{p} )\omega \mathrm {d}x\right) \le \left[ u\right] _{s,p}^{p} \quad \forall \,u\in W_{0}^{s,p}(\Omega ), \end{aligned}$$

(2)

provided that $\Lambda _{p}<\infty .$ Moreover, the equality in this Sobolev-type inequality holds if, and only if, u is a scalar multiple of the function $u_{p}\in \mathcal {M}_{p}$ which is the only weak solution of the problem

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \left( -\Delta _{p}\right) ^{s}u=\Lambda _{p}u^{-1}\omega &{} \mathrm {in} &{} \Omega \\ u>0 &{} \mathrm {in} &{} \Omega \\ u=0 &{} \mathrm {in} &{} \mathbb {R}^{N}{\setminus }\Omega . \end{array} \right. \end{aligned}$$

(3)

Here, $\left( -\Delta _{p}\right) ^{s}$ is the s-fractional p-Laplacian, formally defined by

$$\begin{aligned} \left( -\Delta _{p}\right) ^{s}u(x)=-2\int _{\mathbb {R}^{N}}\dfrac{\left| u(y)-u(x)\right| ^{p-2}(u(y)-u(x))}{\left| y-x\right| ^{N+sp} }\mathrm {d}y. \end{aligned}$$

We recall that a weak solution of the equation in (3) is a function $u\in W_{0}^{s,p}(\Omega )$ satisfying

$$\begin{aligned} \left\langle \left( -\Delta _{p}\right) ^{s}u,\varphi \right\rangle =\Lambda _{p}\int _{\Omega }u^{-1}\varphi \omega \mathrm {d}x\quad \forall \,\varphi \in W_{0}^{s,p}(\Omega ), \end{aligned}$$

where

$$\begin{aligned} \left\langle \left( -\Delta _{p}\right) ^{s}u,\varphi \right\rangle :=\int _{\mathbb {R}^{N}}\int _{\mathbb {R}^{N}}\dfrac{\left| u(x)-u(y)\right| ^{p-2}(u(x)-u(y))(\varphi (x)-\varphi (y))}{\left| x-y\right| ^{N+sp}}\mathrm {d}x\mathrm {d}y \end{aligned}$$

is the expression of $\left( -\Delta _{p}\right) ^{s}$ as an operator from $W_{0}^{s,p}(\Omega )$ into its dual.

The purpose of this paper is to determine both the asymptotic behavior of the pair $\left( \root p \of {\Lambda _{p}},u_{p}\right)$, as $p\rightarrow \infty$, and the corresponding limit problem of (3). In our study $s\in (0,1)$ is kept fixed.

After introducing, in Sect. 2, the notation used throughout the paper, we prove in Sect. 3 that $\Lambda _{p}<\infty$ by constructing a function $\xi \in C_{0}^{0,1}(\overline{\Omega })\cap \mathcal {M}_{p}.$ In the simplest case $\omega \equiv \left| \Omega \right| ^{-1}$ this was made in [10] where the inequality (2) corresponding to the standard Sobolev Space $W_{0}^{1,p} (\Omega )$ has been derived.

In Sect. 4, we show that the limit problem is closely related to the problem of minimizing the quotient

$$\begin{aligned} Q_{s}(u):=\frac{\left| u\right| _{s}}{\exp \left( \int _{\Omega } (\log \left| u\right| )\omega \mathrm {d}x\right) } \end{aligned}$$

on the Banach space $\left( C_{0}^{0,s}(\overline{\Omega }),\left| \cdot \right| _{s}\right)$ of the s-Hölder continuous functions in $\overline{\Omega }$ that are zero on the boundary $\partial \Omega .$ Here, $\left| u\right| _{s}$ denotes the s-Hölder seminorm of u (see (6)).

We prove that if $p_{n}\rightarrow \infty$ then (up to a subsequence)

$$\begin{aligned} u_{p_{n}}\rightarrow u_{\infty }\in C_{0}^{0,s}(\overline{\Omega } )\ \mathrm {uniformly\ in}\ \overline{\Omega },\quad \mathrm {and}\quad \root p_{n} \of {\Lambda _{p_{n}}}\rightarrow \left| u_{\infty }\right| _{s}. \end{aligned}$$

Moreover, the limit function $u_{\infty }$ satisfies

$$\begin{aligned} \int _{\Omega }(\log \left| u_{\infty }\right| )\omega \mathrm {d} x\ge 0\quad \mathrm {and}\quad Q_{s}(u_{\infty })\le Q_{s}(u)\quad \forall \ u\in C_{0}^{0,s}(\overline{\Omega }){\setminus }\left\{ 0\right\} \end{aligned}$$

and the only minimizers of the quotient $Q_{s}$ are the scalar multiples of $u_{\infty }.$

One of the difficulties we face in Sect. 4 is that $C_{c}^{\infty }(\Omega )$ is not dense in $\left( C_{0}^{0,s}(\Omega ),\left| \cdot \right| _{s}\right) .$ This makes it impossible to directly exploit the fact that $u_{p}$ is a weak solution of (3). We overcome this issue by using a convenient technical result proved in [18, Lemma 3.2] and employed in [2] to deal with a similar approximation matter.

In Sect. 5, motivated by [3, 13, 17], we derive the limit problem of (3). Assuming that $\omega$ is continuous and positive in $\Omega$, we prove that $u_{\infty }$ is a viscosity solution of

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \mathcal {L}_{\infty }^{-}u+\left| u\right| _{s}=0 &{} \mathrm {in} &{} \Omega \\ u=0 &{} \mathrm {in} &{}\quad \mathbb {R}^{N}{\setminus }\Omega \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} \left( \mathcal {L}_{\infty }^{-}u\right) (x):=\inf _{y\in \mathbb {R} ^{N}{\setminus }\left\{ x\right\} }\frac{u(y)-u(x)}{\left| y-x\right| ^{s}}. \end{aligned}$$

We also show $u_{\infty }$ is a viscosity supersolution of

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \mathcal {L}_{\infty }u=0 &{} \mathrm {in} &{} \Omega \\ u=0 &{} \mathrm {in} &{}\quad \mathbb {R}^{N}{\setminus }\Omega \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} \mathcal {L}_{\infty }:=\mathcal {L}_{\infty }^{+}+\mathcal {L}_{\infty }^{-} \end{aligned}$$

and

$$\begin{aligned} \left( \mathcal {L}_{\infty }^{+}u\right) (x):=\sup _{y\in \mathbb {R} ^{N}{\setminus }\left\{ x\right\} }\frac{u(y)-u(x)}{\left| y-x\right| ^{s}}. \end{aligned}$$

This fact guarantees that $u_{\infty }>0$ in $\Omega .$

The existing literature on the asymptotic behavior (as $p\rightarrow \infty$) of solutions of problems involving the p-Laplacian is most focused on the local version of the operator, that is, on the problem

$$\begin{aligned} \left\{ \begin{array} [c]{lll} -\Delta _{p}u=f(x,u) &{}\quad \mathrm {in} &{}\quad \Omega \\ u=0 &{}\quad \mathrm {on} &{}\quad \partial \Omega \end{array} \right. \end{aligned}$$

(4)

where $\Delta _{p}u={\text {div}}\left( \left| \nabla u\right| ^{p-2}\nabla u\right)$ is the standard p-Laplacian. This kind of asymptotic behavior has been studied for at least three decades (see [1, 14, 16]) and many new results, adding the dependence of p in the term f(x, u),  are still being produced (see [4,5,6, 8]). The solutions of (4) are obtained in the natural Sobolev space $W_{0}^{1,p}(\Omega )$, and an important property related to this space, crucial in the study of the asymptotic behavior of the corresponding family of solutions $\left\{ u_{p}\right\} ,$ is the inclusion

$$\begin{aligned} W_{0}^{1,p_{2}}(\Omega )\subset W_{0}^{1,p_{1}}(\Omega )\quad \mathrm {whenever} \quad 1<p_{1}<p_{2}. \end{aligned}$$

It allows us to show that any uniform limit function $u_{\infty }$ of the sequence $\left\{ u_{p_{n}}\right\}$ (with $p_{n}\rightarrow \infty$) is admissible as a test function in the weak formulation of (4), so that $u_{\infty }$ inherits certain properties of the functions of $\left\{ u_{p_{n}}\right\} .$

Since the inclusion $W_{0}^{s,p_{2}}(\Omega )\subset W_{0}^{s,p_{1}}(\Omega )$ does not hold when $0<s<1<p_{1}<p_{2}$ (see [19]), the asymptotic behavior, as $p\rightarrow \infty ,$ of the solutions of the problem

$$\begin{aligned} \left\{ \begin{array} [c]{lll} (-\Delta _{p})^{s}u=f(x,u) &{}\quad \mathrm {in} &{}\quad \Omega \\ u=0 &{}\quad \mathrm {in} &{}\quad \mathbb {R}^{N}{\setminus }\Omega \end{array} \right. \end{aligned}$$

(5)

is more difficult to be determined. For example, in the case considered in the present paper ($f(x,u)=\omega (x)/u$) we cannot ensure that the property

$$\begin{aligned} \int _{\Omega }(\log \left| u_{p_{n}}\right| )\omega \mathrm {d}x=0 \end{aligned}$$

is inherited by the limit function $u_{\infty }$ (see Remark 12). Actually, we are able to prove only that

$$\begin{aligned} \int _{\Omega }(\log u_{\infty })\omega \mathrm {d}x\ge 0. \end{aligned}$$

As a consequence, the limit functions of the family $\left\{ u_{p}\right\} _{p>1}$ might not be unique.

The study of the asymptotic behavior, as $p\rightarrow \infty ,$ of the solutions of (5) is quite recent and restricted to few works. In [17] the authors considered $f(x,u)=\lambda _{p}\left| u\right| ^{p-2}u$ where $\lambda _{p}$ is the first eigenvalue of the s-fractional p-Laplacian. Among other results, they proved that

$$\begin{aligned} \lim _{p\rightarrow \infty }\root p \of {\lambda _{p}}=R^{-s}, \end{aligned}$$

where R is the radius of the largest ball inscribed in $\Omega ,$ and that limit function $u_{\infty }$ of the family $\left\{ u_{p}\right\}$ is a positive viscosity solution of

$$\begin{aligned} \max \left\{ \mathcal {L}_{\infty }u\ ,\ \mathcal {L}_{\infty }^{-}u+R^{-s} u\right\} =0. \end{aligned}$$

The equation in (5) with $f=0$ and under the nonhomogeneous boundary condition $u=g$ in $\mathbb {R}^{N}{\setminus }\Omega$ was first studied in [3]. It is shown that the limit function is an optimal s-Hölder extension of $g\in C^{0,s}(\partial \Omega )$ and also a viscosity solution of the equation

$$\begin{aligned} \mathcal {L}_{\infty }u=0\quad \mathrm {in}\ \partial \Omega . \end{aligned}$$

Moreover, some tools for studying the behavior as $p\rightarrow \infty$ of the solutions of (5) are developed there.

In [13], also under the boundary condition $u=g$ in $\mathbb {R} ^{N}{\setminus }\Omega ,$ the cases $f=f(x)$ and $f=f(u)=\left| u\right| ^{\theta (p)-2}u$ with $\Theta :=\lim _{p\rightarrow \infty }\theta (p)/p<1$ are studied. In the first case, different limit equations involving the operators $\mathcal {L}_{\infty },$$\mathcal {L}_{\infty }^{+}$ and $\mathcal {L}_{\infty }^{-}$ are derived according to the sign of the function f(x),  what resembles the known results obtained in [1], where the standard p-Laplacian is considered. For example, the limit function $u_{\infty }$ is a viscosity solution of

$$\begin{aligned} -\mathcal {L}_{\infty }^{-}u=1\quad \mathrm {in}\ \left\{ f>0\right\} . \end{aligned}$$

As for the second case, the limit equation is

$$\begin{aligned} \min \left\{ -\mathcal {L}_{\infty }^{-}u-u^{\Theta },-\mathcal {L}_{\infty }u\right\} =0 \end{aligned}$$

which is consistent with the limit equation obtained in [4] for the standard p-Laplacian and $f(u)=\left| u\right| ^{\theta (p)-2}u$ satisfying $\Theta :=\lim _{p\rightarrow \infty }\theta (p)/p<1.$

2 Notation

The ball centered at $x\in \mathbb {R}^{N}$ with radius $\rho$ is denoted by $B(x,\rho )$, and $\delta$ stands for the distance function to the boundary $\partial \Omega ,$ defined by

$$\begin{aligned} \delta (x):=\min _{y\in \partial \Omega }\left| x-y\right| ,\quad x\in \overline{\Omega }. \end{aligned}$$

We recall that $\delta \in C_{0}^{0,1}(\overline{\Omega })$ and satisfies $\left| \nabla \delta \right| =1$ a.e. in $\Omega .$ Here,

$$\begin{aligned} C_{0}^{0,\beta }(\overline{\Omega }):=\left\{ u\in C^{0,\beta }(\overline{\Omega }):\,u=0\,\mathrm {on}\,\partial \Omega \right\} ,\quad 0<\beta \le 1, \end{aligned}$$

where $C^{0,\beta }(\overline{\Omega })$ is the well-known $\beta$-Hölder space endowed with the norm

$$\begin{aligned} \left\| u\right\| _{0,\beta }=\left\| u\right\| _{\infty }+\left| u\right| _{\beta } \end{aligned}$$

with $\left\| u\right\| _{\infty }$ denoting the sup norm of u and $\left| u\right| _{\beta }$ denoting the $\beta$-Hölder seminorm, that is,

$$\begin{aligned} \left| u\right| _{\beta }:=\sup _{x,y\in \overline{\Omega },x\not =y} \frac{\left| u(x)-u(y)\right| }{\left| x-y\right| ^{\beta }}. \end{aligned}$$

(6)

We recall that $\left( C_{0}^{0,\beta }(\overline{\Omega }),\left| \cdot \right| _{\beta }\right)$ is a Banach space. The fact that the $\beta$-Hölder seminorm $\left| \cdot \right| _{\beta }$ is a norm in $C_{0}^{0,\beta }(\overline{\Omega })$ equivalent to $\left\| u\right\| _{0,\beta }$ is a consequence of the estimate

$$\begin{aligned} \left\| u\right\| _{\infty }\le \left| u\right| _{\beta }\left\| \delta \right\| _{\infty }^{\beta }\quad \forall \,u\in C_{0}^{0,\beta }(\overline{\Omega }), \end{aligned}$$

which in turn follows from the following

$$\begin{aligned} \left| u(x)\right| =\left| u(x)-u(y_{x})\right| \le \left| u\right| _{\beta }\left| x-y_{x}\right| ^{\beta }=\left| u\right| _{\beta }\delta (x)^{\beta }\quad \forall \,x\in \Omega , \end{aligned}$$

(7)

where $y_{x}\in \partial \Omega$ is such that $\delta (x)=\left| x-y_{x}\right| .$

We also define

$$\begin{aligned} C_{c}^{\infty }(\Omega ):=\left\{ u\in C^{\infty }(\Omega ):{\text {*}}{supp} (f)\subset \subset \Omega \right\} \end{aligned}$$

where

$$\begin{aligned} {\text {*}}{supp}(u):=\left\{ x\in \Omega :u(x)\not =0\right\} \end{aligned}$$

is the support of u and $X\subset \subset Y$ means that $\overline{X}$ is a compact subset of Y. Analogously, we define $E_{c}$ if E is a space of functions (e.g., $C_{c}(\mathbb {R}^{N}),$$C_{c}(\mathbb {R}^{N};\mathbb {R} ^{N}),$$C_{c}^{0,\beta }(\overline{\Omega })$).

3 Finiteness of $\Lambda _{p}$

Let us recall the Federer’s co-area formula (see [12])

$$\begin{aligned} \int _{\Omega }g(x)\left| \nabla f(x)\right| \mathrm {d}x=\int _{-\infty }^{\infty }\left( \int _{f^{-1}\left\{ t\right\} }g(x)\mathrm {d} \mathcal {H}_{N-1}\right) \mathrm {d}t, \end{aligned}$$

which holds whenever $g\in L^{1}(\Omega )$ and $f\in C^{0,1}(\overline{\Omega })$. (In this formula $\mathcal {H}_{N-1}$ stands for the $(N-1)$-dimensional Hausdorff measure).

In the particular case $f=\delta$, the above formula becomes

$$\begin{aligned} \int _{\Omega }g(x)\mathrm {d}x=\int _{0}^{\left\| \delta \right\| _{\infty } }\left( \int _{\delta ^{-1}\left\{ t\right\} }g(x)\mathrm {d}\mathcal {H} _{N-1}\right) \mathrm {d}t. \end{aligned}$$

(8)

Proposition 1

Let $\omega \in L^{1}(\Omega )$ such that

$$\begin{aligned} \int _{\Omega }\omega \mathrm {d}x=1\quad \mathrm {and}\quad \omega \ge 0\quad \mathrm {a.e.}\,\mathrm {in}\,\Omega . \end{aligned}$$

(9)

There exists a nonnegative function $\xi \in C(\overline{\Omega })$ that vanishes on the boundary $\partial \Omega$ and satisfies

$$\begin{aligned} \int _{\Omega }(\log \left| \xi \right| )\omega \mathrm {d}x=0. \end{aligned}$$

If, in addition,

$$\begin{aligned} K_{\epsilon }:={\text {*}}{ess}_{0\le t\le \epsilon }\int _{\delta ^{-1}\left\{ t\right\} }\omega \mathrm {d}\mathcal {H}_{N-1}<\infty \end{aligned}$$

(10)

for some $\epsilon >0,$ then $\xi \in C_{0}^{0,1}(\overline{\Omega }).$

Proof

Let $\sigma :[0,\left\| \delta \right\| _{\infty }]\rightarrow [0,1]$ be the $\omega$-distribution associated with $\delta ,$ that is,

$$\begin{aligned} \sigma (t):=\int _{\Omega _{t}}\omega \mathrm {d}x,\quad t\in [0,\left\| \delta \right\| _{\infty }] \end{aligned}$$

where

$$\begin{aligned} \Omega _{t}:=\left\{ x\in \Omega :\delta (x)>t\right\} \end{aligned}$$

is the t-superlevel set of $\delta .$

We remark that $\sigma$ is continuous at each point $t\in [0,\left\| \delta \right\| _{\infty }]$ since the t-level set $\delta ^{-1}\left\{ t\right\}$ has Lebesgue measure zero. This follows, for example, from the Lebesgue density theorem (see [11], where the distance function to a general closed set in $\mathbb {R}^{N}$ is considered).

Thus, there exists a nonincreasing sequence $\left\{ t_{n}\right\} \subset [0,\left\| \delta \right\| _{\infty }]$ such that

$$\begin{aligned} \sigma (t_{n})=1-\frac{1}{2^{n}}. \end{aligned}$$

Now, choose a nondecreasing, piecewise linear function $\varphi \in C([0,\left\| \delta \right\| _{\infty }])$ satisfying

$$\begin{aligned} \varphi (0)=0\quad \mathrm {and}\quad \varphi (t_{n})=\frac{1}{2^{n}}, \end{aligned}$$

and take the function

$$\begin{aligned} \xi _{1}:=\varphi \circ \delta \in C_{0}(\overline{\Omega }). \end{aligned}$$

Taking into account that

$$\begin{aligned} t_{n+1}\le \delta (x)\le t_{n}\quad \mathrm {a.e.}\,x\in \Omega _{t_{n+1} }{\setminus }\Omega _{t_{n}}, \end{aligned}$$

one has

$$\begin{aligned} \frac{1}{2^{n+1}}=\varphi (t_{n+1})\le \xi _{1}(x)\le \varphi (t_{n})=\frac{1}{2^{n}}\quad \mathrm {a.e.}\,x\in \Omega _{t_{n+1}}{\setminus }\Omega _{t_{n}}. \end{aligned}$$

Consequently,

$$\begin{aligned} \int _{\Omega }\left| \xi _{1}\right| ^{\epsilon }\omega \mathrm {d}x&\ge \int _{\Omega _{t_{1}}}\left| \xi _{1}\right| ^{\epsilon } \omega \mathrm {d}x+\sum _{k=1}^{n}\int _{\Omega _{t_{k+1}}{\setminus }\Omega _{t_{k}} }\left| \xi _{1}\right| ^{\epsilon }\omega \mathrm {d}x\\&\ge \frac{1}{2^{\epsilon }}\int _{\Omega _{t_{1}}}\omega \mathrm {d}x+\sum _{k=1}^{n}\frac{1}{2^{\epsilon (k+1)}}\int _{\Omega _{t_{k+1}}{\setminus } \Omega _{t_{k}}}\omega \mathrm {d}x\\&=\frac{1}{2^{\epsilon }}\sigma (t_{1})+\sum _{k=1}^{n}\frac{1}{2^{\epsilon (k+1)}}\left( \sigma (t_{k+1})-\sigma (t_{k})\right) \\&=\frac{1}{2^{\epsilon }}\frac{1}{2}+\sum _{k=1}^{n}\frac{1}{2^{\epsilon (k+1)}}\frac{1}{2^{k+1}}=\sum _{k=1}^{n+1}\left( (1/2)^{\epsilon +1}\right) ^{k}. \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\left( \int _{\Omega }\left| \xi _{1}\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon }}\ge \lim _{\epsilon \rightarrow 0}\left( \sum _{k=1}^{\infty }\left( (1/2)^{\epsilon +1}\right) ^{k}\right) ^{\frac{1}{\epsilon }}=\lim _{\epsilon \rightarrow 0}\left( \frac{(1/2)^{\epsilon +1}}{1-(1/2)^{\epsilon +1}}\right) ^{\frac{1}{\epsilon }}=\frac{1}{4}. \end{aligned}$$

Taking $\xi :=k\xi _{1}$ with

$$\begin{aligned} k=\lim _{\epsilon \rightarrow 0}\left( \int _{\Omega }\left| \xi _{1}\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{-\frac{1}{\epsilon }} \end{aligned}$$

we obtain, by L’Hôpital’s rule,

$$\begin{aligned} 1=\lim _{\epsilon \rightarrow 0^{+}}\left( \int _{\Omega }\left| \xi \right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon }} =\exp \left( \int _{\Omega }(\log \left| \xi \right| )\omega \mathrm {d}x\right) . \end{aligned}$$

Hence,

$$\begin{aligned} \int _{\Omega }(\log \left| \xi \right| )\omega \mathrm {d}x=0. \end{aligned}$$

We now prove that $\xi _{1}\in C^{0,1}(\overline{\Omega })$ under the additional hypothesis (10). Since the nondecreasing function $\varphi$ can be chosen such that $\varphi ^{\prime }$ is bounded in any closed interval contained in $(0,\left\| \delta \right\| _{\infty }],$ we can assume that $\nabla \xi _{1}\in L_{{\text {loc}}}^{\infty }(\Omega )$ (note that $\left| \nabla \xi _{1}\right| =\left| \varphi ^{\prime } (\delta )\nabla \delta \right| =\left| \varphi ^{\prime }(\delta )\right|$ a.e. in $\Omega$).

Thus, it suffices to show that the quotient

$$\begin{aligned} Q(x,y):=\frac{\left| \xi _{1}(x)-\xi _{1}(y)\right| }{\left| x-y\right| } \end{aligned}$$

is bounded uniformly with respect to $y\in \partial \Omega$ and $x\in \Omega _{\epsilon }^{c}:=\left\{ x\in \overline{\Omega }:\delta (x)\le \epsilon \right\} ,$ where $\epsilon$ is given by (10).

Let $x\in \Omega _{\epsilon }^{c}$ and $y\in \partial \Omega$ be fixed and chose $n\in \mathbb {N}$ sufficiently large such that

$$\begin{aligned} t_{n+1}<\delta (x)\le t_{n}\le \epsilon . \end{aligned}$$

Since $\xi _{1}(y)=0$ and $\varphi$ is nondecreasing, one has

$$\begin{aligned} \left| \xi _{1}(x)-\xi _{1}(y)\right| =\xi _{1}(x)\le \varphi (t_{n})=\frac{1}{2^{n}}. \end{aligned}$$

Moreover,

$$\begin{aligned} t_{n+1}<\delta (x)\le \left| x-y\right| . \end{aligned}$$

Hence,

$$\begin{aligned} Q(x,y)\le \frac{1}{2^{n}t_{n+1}}\quad \mathrm {whenever}\,y\in \partial \Omega \,\mathrm {and}\,x\in \Omega _{\epsilon }^{c}. \end{aligned}$$

Applying the co-area formula (8) with $g=\omega$ and $\Omega =\Omega _{t_{n}+1}^{c}$, we find

$$\begin{aligned} \frac{1}{2^{n+1}}=\int _{\Omega _{t_{n}+1}^{c}}\omega \mathrm {d}x=\int _{0}^{t_{n+1}}\left( \int _{\delta ^{-1}\left\{ t\right\} }\omega \mathrm {d}\mathcal {H}_{N-1}\right) \mathrm {d}t\le K_{\epsilon }t_{n+1}. \end{aligned}$$

It follows that

$$\begin{aligned} Q(x,y)\le \frac{1}{2^{n}t_{n+1}}\le \frac{K_{\epsilon }2^{n+1}}{2^{n} }=2K_{\epsilon }\quad \mathrm {whenever}\,y\in \partial \Omega \,\mathrm {and} \,x\in \Omega _{\epsilon }^{c}, \end{aligned}$$

(11)

concluding thus the proof that $\xi _{1}\in C^{0,1}(\overline{\Omega }).$$\square$

Remark 2

The estimate (11) can also be obtained from the Weyl’s Formula (see [15]) provided that $\omega$ is bounded on an $\epsilon$-tubular neighborhood of $\partial \Omega .$

In the remaining of this section, $\xi$ denotes the function obtained in Proposition 1 extended as zero outside $\Omega .$ So,

$$\begin{aligned} \xi \in C_{0}^{0,1}(\overline{\Omega })\quad \mathrm {and}\quad \int _{\Omega } (\log \left| \xi \right| )\omega \mathrm {d}x=0. \end{aligned}$$

Since $C_{0}^{0,1}(\overline{\Omega })\subseteq W_{0}^{1,p}(\Omega )\subseteq W_{0}^{s,p}(\Omega )$, we have $\xi \in \mathcal {M}_{p}$ (for a proof of the second inclusion see [7]). Therefore,

$$\begin{aligned} \Lambda _{p}\le \left[ \xi \right] _{s,p}^{p}\quad \forall \,p>1. \end{aligned}$$

(12)

Combining (12) with the results proved in [9, Section 4] (which requires $\omega \in L^{r}(\Omega ),$ for some $r>1$), we have the following theorem.

Theorem 3

Let $\omega$ be a function in $L^{r}(\Omega ),$ for some $r>1,$ satisfying (9)–(10). For each $p>1,$ the infimum $\Lambda _{p}$ in (1) is attained by a function $u_{p} \in \mathcal {M}_{p}$ which is the only positive weak solution of

$$\begin{aligned} \left( -\Delta _{p}\right) ^{s}u=\Lambda _{p}u^{-1}\omega ,\quad u\in W_{0}^{s,p}(\Omega ). \end{aligned}$$

Summarizing,

$$\begin{aligned} \left[ u_{p}\right] _{s,p}^{p}=\Lambda _{p}:=\min \left\{ \left[ u\right] _{s,p}^{p}:u\in \mathcal {M}_{p}\right\} \le \left[ \xi \right] _{s,p} ^{p}\quad \forall \,p>1, \end{aligned}$$

(13)

and $u_{p}$ is the unique function in $W_{0}^{1,p}(\Omega )$ satisfying

$$\begin{aligned} u_{p}>0\quad \mathrm {in}\,\Omega \quad \mathrm {and}\quad \left\langle \left( -\Delta _{p}\right) ^{s}u_{p},\phi \right\rangle =\Lambda _{p}\int _{\Omega }\omega (u_{p})^{-1}\phi \mathrm {d}x\quad \forall \,\phi \in W_{0}^{s,p}(\Omega ). \end{aligned}$$

We also have

$$\begin{aligned} 0<\root p \of {\Lambda _{p}}\le \frac{\left[ u\right] _{s,p}}{\exp \left( \int _{\Omega }(\log \left| u\right| )\omega \mathrm {d}x\right) } \quad \forall \,u\in W_{0}^{s,p}(\Omega ), \end{aligned}$$

since the quotient is homogeneous.

Remark 4

It is worth pointing out that

$$\begin{aligned} \int _{\Omega }(\log \left| u\right| )\omega \mathrm {d}x=-\infty \end{aligned}$$

(14)

for any function $u\in L^{\infty }(\Omega )$ whose ${\text {*}}{supp}u$ is a proper subset of ${\text {*}}{supp}\omega .$ Indeed, in this case we have

$$\begin{aligned} 0\le \exp \left( \int _{\Omega }(\log \left| u\right| )\omega \mathrm {d}x\right) =\lim _{t\rightarrow 0^{+}}\left( \int _{\Omega }\left| u\right| ^{t}\omega \mathrm {d}x\right) ^{\frac{1}{t}}\le \left\| u\right\| _{\infty }\lim _{t\rightarrow 0^{+}}\left( \int _{{\text {*}}{supp}\left| u\right| }\omega \mathrm {d}x\right) ^{\frac{1}{t}}=0. \end{aligned}$$

Thus, if $\omega >0$ almost everywhere in $\Omega$ then (14) holds for every $u\in C_{c}^{\infty }(\Omega ){\setminus }\left\{ 0\right\} .$

4 The asymptotic behavior as $p\rightarrow \infty$

In this section, we assume that the weight $\omega$ satisfies the hypothesis of Theorem 3. Our goal is to relate the asymptotic behavior (as $p\rightarrow \infty$) of the pair $\left( \root p \of {\Lambda _{p}},u_{p}\right)$ with the problem of minimizing the homogeneous quotient $Q_{s}:C_{0} ^{0,s}(\overline{\Omega }){\setminus }\left\{ 0\right\} \rightarrow (0,\infty )$ defined by

$$\begin{aligned} Q_{s}(u):=\frac{\left| u\right| _{s}}{k(u)}\quad \mathrm {where}\quad k(u):=\exp \left( \int _{\Omega }(\log \left| u\right| )\omega \mathrm {d}x\right) . \end{aligned}$$

Note that $k(u)=0$ if, and only if, u satisfies (14). In particular, according to Remark 4,

$$\begin{aligned} \omega >0\quad \mathrm {a.e.}\,\mathrm {in}\,\Omega \Longrightarrow Q_{s} (u)=\infty \quad \forall \,u\in C_{c}^{\infty }(\Omega ){\setminus }\left\{ 0\right\} . \end{aligned}$$

We also observe that

$$\begin{aligned} 0\le k(u)\le \int _{\Omega }\left| u\right| \omega \mathrm {d} x<\infty \quad \forall \,u\in C_{0}^{0,s}(\overline{\Omega }){\setminus }\left\{ 0\right\} , \end{aligned}$$

(15)

where the second inequality is consequence of the Jensen’s inequality (since the logarithm is concave):

$$\begin{aligned} \int _{\Omega }(\log \left| u\right| )\omega \mathrm {d}x\le \log \left( \int _{\Omega }\left| u\right| \omega \mathrm {d}x\right) . \end{aligned}$$

(16)

Now, let us define

$$\begin{aligned} \mu _{s}:=\inf _{u\in C_{0}^{0,s}(\overline{\Omega }){\setminus }\left\{ 0\right\} }Q_{s}(u). \end{aligned}$$

Thanks to the homogeneity of $Q_{s}$, we have

$$\begin{aligned} \mu _{s}=\inf _{u\in \mathcal {M}_{s}}\left| u\right| _{s} \end{aligned}$$

where

$$\begin{aligned} \mathcal {M}_{s}:=\left\{ u\in C_{0}^{0,s}(\overline{\Omega }):k(u)=1\right\} . \end{aligned}$$

Combining (15) and (7), we obtain

$$\begin{aligned} 1\le \int _{\Omega }\left| u\right| \omega \mathrm {d}x\le \left| u\right| _{s}\int _{\Omega }\delta ^{s}\omega \mathrm {d}x\quad \forall \,u\in \mathcal {M}_{s}, \end{aligned}$$

what yields the following positive lower bound to $\mu _{s}$

$$\begin{aligned} \left( \int _{\Omega }\delta ^{s}\omega \mathrm {d}x\right) ^{-1}\le \mu _{s}. \end{aligned}$$

In the sequel we show that $\mu _{s}$ is in fact a minimum, attained at a unique nonnegative function. Before this, let us make an important remark.

Remark 5

If v minimizes $\left| \cdot \right| _{s}$ in $\mathcal {M}_{s}$ the same holds for $\left| v\right| ,$ since the function $w=\left| v\right|$ belongs to $\mathcal {M}_{s}$ and satisfies $\left| w\right| _{s}\le \left| v\right| _{s}.$

Proposition 6

There exists a unique nonnegative function $v\in \mathcal {M}_{s}$ such that

$$\begin{aligned} \mu _{s}=\left| v\right| _{s}. \end{aligned}$$

Proof

Let $\left\{ v_{n}\right\} _{n\in \mathbb {N}}\subset \mathcal {M}_{s}$ be such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\left| v_{n}\right| _{s}=\mu _{s}. \end{aligned}$$

(17)

Since the function $w_{n}=\left| v_{n}\right|$ belongs to $\mathcal {M}_{s}$ and satisfies $\left| w_{n}\right| _{s} \le \left| v_{n}\right| _{s}$, we can assume that $v_{n}\ge 0$ in $\Omega .$

It follows from (17) that $\left\{ v_{n}\right\} _{n\in \mathbb {N} }$ is bounded in $C_{0}^{0,s}(\overline{\Omega }).$ Hence, the compactness of the embedding $C_{0}^{0,s}(\overline{\Omega })\hookrightarrow C_{0} (\overline{\Omega })$ allows us to assume (by renaming a subsequence) that $\left\{ v_{n}\right\} _{n\in \mathbb {N}}$ converges uniformly to a function $v\in C_{0}(\overline{\Omega })$. Of course, $v\ge 0$ in $\Omega .$

Letting $n\rightarrow \infty$ in the inequality

$$\begin{aligned} \left| v_{n}(x)-v_{n}(y)\right| \le \left| v_{n}\right| _{s}\left| x-y\right| ^{s}\quad \forall \,x,y\in \overline{\Omega } \end{aligned}$$

and taking (17) into account, we obtain

$$\begin{aligned} \left| v(x)-v(y)\right| \le \mu _{s}\left| x-y\right| ^{s} \quad \forall \,x,y\in \overline{\Omega }. \end{aligned}$$

This implies that $v\in C_{0}^{0,s}(\overline{\Omega })$ and

$$\begin{aligned} \left| v\right| _{s}\le \mu _{s}. \end{aligned}$$

(18)

Thus, to prove that $\mu _{s}=\left| v\right| _{s}$ it suffices to verify that $v\in \mathcal {M}_{s}.$ Since

$$\begin{aligned} 1=k(v_{n})=\lim _{\epsilon \rightarrow 0^{+}}\left( \int _{\Omega }\left| v_{n}\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon } }\le \left( \int _{\Omega }\left| v_{n}\right| ^{t}\omega \mathrm {d}x\right) ^{\frac{1}{t}} \quad \forall \,t>0 \end{aligned}$$

the uniform convergence $v_{n}\rightarrow v$ yields

$$\begin{aligned} 1\le \left( \int _{\Omega }\left| v\right| ^{t}\omega \mathrm {d} x\right) ^{\frac{1}{t}} \quad \forall \,t>0. \end{aligned}$$

Hence,

$$\begin{aligned} 1\le \lim _{t\rightarrow 0^{+}}\left( \int _{\Omega }\left| v\right| ^{t}\mathrm {d}x\right) ^{\frac{1}{t}}=k(v). \end{aligned}$$

Thus, noticing that $(k(v))^{-1}v\in \mathcal {M}_{s}$ and taking (18) into account, we obtain

$$\begin{aligned} \mu _{s}\le \left| (k(v))^{-1}v\right| _{s}=(k(v))^{-1}\left| v\right| _{s}\le \left| v\right| _{s}\le \mu _{s}. \end{aligned}$$

Therefore, $k(v)=1,$$v\in \mathcal {M}_{s}$ and $\left| v\right| _{s}=\mu _{s}.$

Now, let $u\in \mathcal {M}_{s}$ be a nonnegative minimizer of $\left| \cdot \right| _{s}$ and consider the convex combination

$$\begin{aligned} w:=\theta u+(1-\theta )v\quad \mathrm {with}\quad 0<\theta <1. \end{aligned}$$

Since the logarithm is a concave function, we have

$$\begin{aligned} \int _{\Omega }(\log w)\omega \mathrm {d}x&\ge \int _{\Omega }(\theta \log (u)+(1-\theta )\log (v))\omega \mathrm {d}x\\&=\theta \int _{\Omega }(\log u)\omega \mathrm {d}x+(1-\theta )\int _{\Omega }(\log v)\omega \mathrm {d}x=0. \end{aligned}$$

This implies that $c^{-1}w\in \mathcal {M}_{s}$ where $c:=k(w)\ge 1.$

Hence,

$$\begin{aligned} \mu _{s}\le c^{-1}\left| w\right| _{s}\le \left| w\right| _{s}\le \theta \left| u\right| _{s}+(1-\theta )\left| v\right| _{s}=\theta \mu _{s}+(1-\theta )\mu _{s}=\mu _{s}. \end{aligned}$$

It follows that $c=1$ and the convex combination w minimizes $\left| \cdot \right| _{s}$ in $\mathcal {M}_{s}.$ Consequently,

$$\begin{aligned} 0=\int _{\Omega }\left[ \log (\theta u+(1-\theta )v)\right] \omega \mathrm {d}x\ge \int _{\Omega }\left[ \theta \log (u)+(1-\theta )\log (v)\right] \omega \mathrm {d}x=0. \end{aligned}$$

Since the concavity of the logarithm is strict, one must have $u=Cv$ for some positive constant C. Taking account that $1=k(u)=Ck(v)=C,$ we have $u=v.$$\square$

From now on, $v_{s}\in \mathcal {M}_{s}$ denotes the only nonnegative minimizer of $\left| \cdot \right| _{s}$ on $\mathcal {M}_{s},$ given by Proposition 6. The main result of this section, proved in the sequence, shows that if $p_{n}\rightarrow \infty$ then a subsequence of $\left\{ u_{p_{n}}\right\} _{n\in \mathbb {N}}$ converges uniformly to a scalar multiple of $v_{s,}$ say $u_{\infty }=k_{\infty }v_{s}$ where $k_{\infty }\ge 1.$

In the next section (see (37)), we show that $u_{\infty }$ is strictly positive in $\Omega ,$ implying thus that $-v_{s}$ and $v_{s}$ are the only minimizers of $\left| \cdot \right| _{s}$ on $\mathcal {M}_{s}.$ As consequence, the minimizers of $Q_{s}$ on $C_{0}^{0,s}(\overline{\Omega }){\setminus }\left\{ 0\right\}$ are precisely the scalar multiples of $v_{s}$ (or, equivalently, the scalar multiples of $u_{\infty }$). Further, we derive an equation satisfied by $v_{s}$ and $\mu _{s}$ in the viscosity sense (see Corollary 16).

Lemma 7

Let $u\in C_{0}^{0,s}(\overline{\Omega })$ be extended as zero outside $\Omega .$ If $u\in W^{s,q}(\Omega )$ for some $q>1,$ then $u\in W_{0}^{s,p}(\Omega )$ for all $p\ge q$ and

$$\begin{aligned} \lim _{p\rightarrow \infty }\left[ u\right] _{s,p}=\left| u\right| _{s}. \end{aligned}$$

(19)

Proof

First, note that the inequality

$$\begin{aligned} \left| u(x)-u(y)\right| \le \left| u\right| _{s}\left| x-y\right| ^{s} \end{aligned}$$

is valid for all $x,y\in \mathbb {R}^{N},$ not only for those $x,y\in \overline{\Omega }.$ In fact, this is obvious when $x,y\in \mathbb {R} ^{N}{\setminus }\overline{\Omega }.$ Now, if $x\in \Omega$ and $y\in \mathbb {R} ^{N}{\setminus }\overline{\Omega }$ then take $y_{1}\in \partial \Omega$ such that $\left| x-y_{1}\right| \le \left| x-y\right|$ (such $y_{1}$ can be taken on the straight line connecting x to y). Since $u(y)=u(y_{1} )=0,$ we have

$$\begin{aligned} \left| u(x)-u(y)\right| =\left| u(x)\right| =\left| u(x)-u(y_{1})\right| \le \left| u\right| _{s}\left| x-y_{1}\right| ^{s}\le \left| u\right| _{s}\left| x-y\right| ^{s}. \end{aligned}$$

For each $p>q$, we have

$$\begin{aligned} \left[ u\right] _{s,p}^{p}=\int _{\mathbb {R}^{N}}\int _{\mathbb {R}^{N}} \frac{\left| u(x)-u(y)\right| ^{p-q}}{\left| x-y\right| ^{s(p-q)}}\frac{\left| u(x)-u(y)\right| ^{q}}{\left| x-y\right| ^{N+sq}}\mathrm {d}x\mathrm {d}y\le (\left| u\right| _{s})^{(p-q)}\left[ u\right] _{s,q}^{q}. \end{aligned}$$

Thus, $u\in W_{0}^{s,p}(\Omega )$ and

$$\begin{aligned} \limsup _{p\rightarrow \infty }\left[ u\right] _{s,p}\le \lim _{p\rightarrow \infty }\left| u\right| _{s}^{(p-q)/p}\left[ u\right] _{s,q} ^{q/p}=\left| u\right| _{s}. \end{aligned}$$

(20)

Now, noticing that (by Fatou’s lemma)

$$\begin{aligned} \int _{\Omega }\int _{\Omega }\left( \frac{\left| u(x)-u(y)\right| }{\left| x-y\right| ^{s}}\right) ^{q}\mathrm {d}x\mathrm {d} y\le \liminf _{p\rightarrow \infty }\int _{\Omega }\int _{\Omega }\left( \frac{\left| u(x)-u(y)\right| }{\left| x-y\right| ^{\frac{N}{p}+s}}\right) ^{q}\mathrm {d}x\mathrm {d}y \end{aligned}$$

and (by Hölder’s inequality)

$$\begin{aligned} \int _{\Omega }\int _{\Omega }\left( \frac{\left| u(x)-u(y)\right| }{\left| x-y\right| ^{\frac{N}{p}+s}}\right) ^{q}\mathrm {d} x\mathrm {d}y&\le \left| \Omega \right| ^{2(1-\frac{q}{p})}\left( \int _{\Omega }\int _{\Omega }\left( \frac{\left| u(x)-u(y)\right| }{\left| x-y\right| ^{\frac{N}{p}+s}}\right) ^{p}\mathrm {d} x\mathrm {d}y\right) ^{\frac{q}{p}}\\&\le \left| \Omega \right| ^{2(1-\frac{q}{p})}\left[ u\right] _{s,p}^{q}, \end{aligned}$$

we obtain

$$\begin{aligned} \left( \int _{\Omega }\int _{\Omega }\left( \frac{\left| u(x)-u(y)\right| }{\left| x-y\right| ^{s}}\right) ^{q} \mathrm {d}x\mathrm {d}y\right) ^{\frac{1}{q}}\le \left| \Omega \right| ^{2/q}\liminf _{p\rightarrow \infty }\left[ u\right] _{s,p}. \end{aligned}$$

Hence, taking into account that

$$\begin{aligned} \left| u\right| _{s}=\lim _{q\rightarrow \infty }\left( \int _{\Omega }\int _{\Omega }\left( \frac{\left| u(x)-u(y)\right| }{\left| x-y\right| ^{s}}\right) ^{q}\mathrm {d}x\mathrm {d}y\right) ^{\frac{1}{q}} \end{aligned}$$

we arrive at

$$\begin{aligned} \left| u\right| _{s}\le \lim _{q\rightarrow \infty }\left| \Omega \right| ^{2/q}\left( \liminf _{p\rightarrow \infty }\left[ u\right] _{s,p}\right) =\liminf _{p\rightarrow \infty }\left[ u\right] _{s,p}. \end{aligned}$$

This estimate combined with (20) leads us to (19). $\square$

It is known (see [7, Theorem 8.2]) that if $p>\dfrac{N}{s}$ then there exists of a positive constant C such that

$$\begin{aligned} \left\| u\right\| _{C^{0,\beta }(\overline{\Omega })}\le C\left[ u\right] _{s,p}\quad \forall \,u\in W_{0}^{s,p}(\Omega ), \end{aligned}$$

(21)

where $\beta :=s-\dfrac{N}{p}\in (0,1).$ As pointed out in [13, Remark 2.2] the constant C in (21) can be chosen uniform with respect to p.

We remark that the family of positive numbers $\left\{ \root p \of {\Lambda _{p} }\right\} _{p>1}$ is bounded. Indeed, combining (12) with the previous lemma we obtain

$$\begin{aligned} \limsup _{p\rightarrow \infty }\root p \of {\Lambda _{p}}\le \left| \xi \right| _{s}. \end{aligned}$$

The next lemma, where ${\text {Id}}$ stands for the identity function, is extracted of the proof of [18, Lemma 3.2]. It helps us to overcome the fact that $C_{c}^{\infty }(\Omega )$ is not dense in $C_{0}^{0,s}(\overline{\Omega }).$

Lemma 8

[see [18, Lemma 3.2]]Let $\Omega \subset \mathbb {R}^{N}$ be a Lipschitz bounded domain. There exist $\phi \in C_{c}^{\infty } (\mathbb {R}^{N},\mathbb {R}^{N})$ and $0<\tau _{0}<(\left| \phi \right| _{1})^{-1}$ such that, for each $0\le \tau \le \tau _{0},$ the map

$$\begin{aligned} \Phi _{\tau }:={\text {Id}}+\tau \phi :\mathbb {R}^{N}\rightarrow \mathbb {R}^{N} \end{aligned}$$

is a diffeomorphism satisfying

• 1. $\Phi _{\tau }(\overline{\Omega })\subset \subset \Omega ,$

• 2. $\Phi _{\tau }\rightarrow {\text {Id}}$ and $(\Phi _{\tau } )^{-1}\rightarrow {\text {Id}}$ as $\tau \rightarrow 0^{+}$ uniformly on $\mathbb {R}^{N}$,

• 3. $\left| (\Phi _{\tau })^{-1}(x)-(\Phi _{\tau })^{-1}(y)\right| \le \dfrac{\left| x-y\right| }{1-\tau \left| \phi \right| _{1} }.$

Lemma 9

Let $u\in C_{0}^{0,s}(\overline{\Omega })$ be a nonnegative function extended as zero outside $\Omega .$ There exists a sequence of nonnegative functions $\left\{ u_{k}\right\} _{k\in \mathbf {N}}\subset C_{0}^{0,s}(\overline{\Omega })\cap W_{0}^{s,p}(\Omega ),$ for all $p>1,$ converging uniformly to u in $\overline{\Omega }$ and such that

$$\begin{aligned} \limsup _{k\rightarrow \infty }\left| u_{k}\right| _{s}\le \left| u\right| _{s}. \end{aligned}$$

Proof

For each $k\in \mathbb {N}$ let $\Psi _{k}$ denote the inverse of $\Phi _{1/k},$ given by Lemma 8, and set

$$\begin{aligned} \Omega _{k}:=\Phi _{1/k}(\overline{\Omega }). \end{aligned}$$

Since $\Omega _{k}\subset \subset \Omega$ there exists $U_{k},$ a subdomain of $\Omega ,$ such that

$$\begin{aligned} \overline{\Omega _{k}}\subset U_{k}\subset \overline{U_{k}}\subset \Omega . \end{aligned}$$

Let $\eta \in C^{\infty }(\mathbb {R}^{N})$ be a standard convolution kernel: $\eta (z)>0$ if $\left| z\right| <1,$$\eta (z)=0$ if $\left| z\right| \ge 1$ and $\int _{\left| z\right| \le 1}\phi (z)\mathrm {d}z=1.$

Define the function

$$\begin{aligned} u_{k}=(u\circ \Psi _{k})*\eta _{k}\in C^{\infty }(\mathbb {R}^{N}), \end{aligned}$$

where

$$\begin{aligned} \eta _{k}(x):=(\epsilon _{k})^{-N}\eta (\frac{x}{\epsilon _{k}}),\quad x\in \mathbb {R}^{N} \end{aligned}$$

and $\epsilon _{k}<{\text {dist}}(\Omega _{k},\partial U_{k}).$ Note that $\epsilon _{k}\rightarrow 0.$

Since

$$\begin{aligned} B(x,\epsilon _{k})\subset \mathbb {R}^{N}{\setminus }\Omega _{k}\quad \forall \,x\in \mathbb {R}^{N}{\setminus } U_{k}, \end{aligned}$$

we have

$$\begin{aligned} \Psi _{k}(B(x,\epsilon _{k}))\subset \mathbb {R}^{N}{\setminus }\Omega \quad \forall \,x\in \mathbb {R}^{N}{\setminus } U_{k}. \end{aligned}$$

Hence, observing that

$$\begin{aligned} u_{k}(x)=\int _{\mathbb {R}^{N}}\eta _{k}(x-z)u(\Psi _{k}(z))\mathrm {d} z=\int _{B(0,1)}\eta (z)u(\psi _{k}(x-\epsilon _{k}z))\mathrm {d}z\quad \forall \,x\in \mathbb {R}^{N} \end{aligned}$$

and that

$$\begin{aligned} \left| x-\epsilon _{k}z-x\right| \le \epsilon _{k}\quad \forall \,z\in B(0,1) \end{aligned}$$

we conclude that

$$\begin{aligned} u_{k}(x)=0\quad \forall \,x\in \mathbb {R}^{N}{\setminus } U_{k}. \end{aligned}$$

Therefore, $u_{k}\in C_{c}^{\infty }(\Omega )\subset W_{0}^{1,p}(\Omega )$ for all $p>1.$

Now, let $x,y\in \overline{\Omega }$ be fixed. According to item 3 of Lemma 8,

$$\begin{aligned} \left| u_{k}(x)-u_{k}(y)\right|&\le \int _{B(0,1)}\eta (z)\left| u(\Psi _{k}(x-\epsilon _{k}z))-u(\Psi _{k}(y-\epsilon _{k}z))\right| \mathrm {d}z\\&\le \left| u\right| _{s}\int _{B(0,1)}\eta (z)\left| \Psi _{k}(x-\epsilon _{k}z)-\Psi _{k}(y-\epsilon _{k}z))\right| ^{s}\mathrm {d}z\\&\le \dfrac{\left| u\right| _{s}}{(1-(1/k)\left| \phi \right| _{1})^{s}}\int _{B(0,1)}\eta (z)\left| x-y\right| ^{s}\mathrm {d}z\\&=\dfrac{\left| u\right| _{s}}{(1-(1/k)\left| \phi \right| _{1})^{s}}\left| x-y\right| ^{s}. \end{aligned}$$

It follows that $u_{k}\in C_{0}^{0,s}(\overline{\Omega })$ and

$$\begin{aligned} \limsup _{k\rightarrow \infty }\left| u_{k}\right| _{s}\le \lim _{k\rightarrow \infty }\dfrac{\left| u\right| _{s}}{(1-(1/k)\left| \phi \right| _{1})^{s}}=\left| u\right| _{s}. \end{aligned}$$

Consequently, up to a subsequence, $u_{k}\rightarrow \widetilde{u}\in C(\overline{\Omega })$ uniformly in $\overline{\Omega }.$ Hence, $\widetilde{u}=u$ since item 2 of Lemma 8 implies that

$$\begin{aligned} \lim _{k\rightarrow \infty }u_{k}(x)=\int _{B(0,1)}\eta (z)u(\lim _{k\rightarrow \infty }\Psi _{k}(x-\epsilon _{k}z))\mathrm {d}z=u(x)\int _{B(0,1)}\eta (z)\mathrm {d}z=u(x). \end{aligned}$$

$\square$

Theorem 10

Let $p_{n}\rightarrow \infty .$ Up to a subsequence, $\left\{ u_{p_{n}}\right\} _{n\in \mathbb {N}}$ converges uniformly to a nonnegative function $u_{\infty }\in C_{0}^{0,s}(\overline{\Omega })$ such that

$$\begin{aligned} \left| u_{\infty }\right| _{s}=\lim _{n\rightarrow \infty }\root p_{n} \of {\Lambda _{p_{n}}}. \end{aligned}$$

Furthermore,

$$\begin{aligned} v_{s}=(k_{\infty })^{-1}u_{\infty } \end{aligned}$$

(22)

where

$$\begin{aligned} k_{\infty }:=k(u_{\infty })=\exp \left( \int _{\Omega }(\log \left| u_{\infty }\right| )\omega \mathrm {d}x\right) \ge 1. \end{aligned}$$

(23)

Proof

Let $p_{0}>\dfrac{N}{s}$ be fixed and take $\beta _{0}=s-\frac{N}{p_{0}}.$ For each $(x,y)\in \Omega \times \Omega ,$ with $x\not =y,$ we obtain from (21)

$$\begin{aligned} \frac{\left| u_{p}(x)-u_{p}(y)\right| }{\left| x-y\right| ^{s-\frac{N}{p_{0}}}}&=\frac{\left| u_{p}(x)-u_{p}(y)\right| }{\left| x-y\right| ^{s-\frac{N}{p}}}\left| x-y\right| ^{N(\frac{1}{p_{0}}-\frac{1}{p})}\\&\le C\left[ u_{p}\right] _{s,p}{\text {diam}}(\Omega )^{N(\frac{1}{p_{0}}-\frac{1}{p})},\quad \forall \,p\ge p_{0}, \end{aligned}$$

where C is uniform with respect to p and ${\text {diam}}(\Omega )$ is the diameter of $\Omega .$ Hence, in view of (13) and (12) the family $\left\{ u_{p}\right\} _{p\ge p_{0}}$ is bounded in $C_{0} ^{0,\beta _{0}}(\overline{\Omega }),$ implying that, up to a subsequence, $u_{p_{n}}\rightarrow u_{\infty }\in C(\overline{\Omega })$ uniformly in $\overline{\Omega }.$ Of course, the limit function $u_{\infty }$ is nonnegative in $\Omega$ and vanishes on $\partial \Omega .$

Letting $n\rightarrow \infty$ in the inequality (which follows from (21))

$$\begin{aligned} \frac{\left| u_{p_{n}}(x)-u_{p_{n}}(y)\right| }{\left| x-y\right| ^{s-\frac{N}{p_{n}}}}\le C\left[ u_{p_{n}}\right] _{s,p_{n} }=C\root p_{n} \of {\Lambda _{p_{n}}} \end{aligned}$$

and taking (12) into account, we conclude that $u_{\infty }\in C_{0}^{0,s}(\overline{\Omega }).$

Up to another subsequence, we can assume that

$$\begin{aligned} \root p_{n} \of {\Lambda _{p_{n}}}\rightarrow L. \end{aligned}$$

Let $q>\dfrac{N}{s}$ be fixed. By Fatou’s Lemma and Hölder’s inequality,

$$\begin{aligned}&\int _{\Omega }\int _{\Omega }\left( \frac{\left| u_{\infty }(x)-u_{\infty }(y)\right| }{\left| x-y\right| ^{s}}\right) ^{q}\mathrm {d} x\mathrm {d}y \\&\quad \le \liminf _{n\rightarrow \infty }\int _{\Omega }\int _{\Omega }\left( \frac{\left| u_{p_{n}}(x)-u_{p_{n}}(y)\right| }{\left| x-y\right| ^{\frac{N}{p_{n}}+s}}\right) ^{q}\mathrm {d}x\mathrm {d}y\\&\quad \le \liminf _{n\rightarrow \infty }\left| \Omega \right| ^{2(1-\frac{q}{p_{n}})}\left( \int _{\Omega }\int _{\Omega }\left( \frac{\left| u_{p_{n}}(x)-u_{p_{n}}(y)\right| }{\left| x-y\right| ^{\frac{N}{p_{n}}+s}}\right) ^{p_{n}}\mathrm {d}x\mathrm {d}y\right) ^{\frac{q}{p_{n}}}\\&\quad \le \left| \Omega \right| ^{2}\liminf _{n\rightarrow \infty }\left[ u_{p_{n}}\right] _{s,p_{n}}^{q}=\left| \Omega \right| ^{2} \lim _{n\rightarrow \infty }(\root p_{n} \of {\Lambda _{p_{n}}})^{q}=\left| \Omega \right| ^{2}L^{q}. \end{aligned}$$

Therefore,

$$\begin{aligned} \left| u_{\infty }\right| _{s}=\lim _{q\rightarrow \infty }\left( \int _{\Omega }\int _{\Omega }\left( \frac{\left| u_{\infty }(x)-u_{\infty }(y)\right| }{\left| x-y\right| ^{s}}\right) ^{q}\mathrm {d} x\mathrm {d}y\right) ^{1/q}\le \lim _{q\rightarrow \infty }\left| \Omega \right| ^{\frac{2}{q}}L=L. \end{aligned}$$

(24)

To prove that $k_{\infty }\ge 1$, we first note that

$$\begin{aligned} \lim _{t\rightarrow 0^{+}}\left( {\displaystyle \int _{\Omega }} \left| u_{p_{n}}\right| ^{t}\omega \mathrm {d}x\right) ^{\frac{1}{t} }=\inf _{0<t<1}\left( {\displaystyle \int _{\Omega }} \left| u_{p_{n}}\right| ^{t}\omega \mathrm {d}x\right) ^{\frac{1}{t} }\le \left( {\displaystyle \int _{\Omega }} \left| u_{p_{n}}\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon }}\quad \forall \,\epsilon \in (0,1). \end{aligned}$$

Consequently,

$$\begin{aligned} 1=k(u_{p_{n}})=\lim _{t\rightarrow 0^{+}}\left( {\displaystyle \int _{\Omega }} \left| u_{p_{n}}\right| ^{t}\omega \mathrm {d}x\right) ^{\frac{1}{t} }\le \left( {\displaystyle \int _{\Omega }} \left| u_{p_{n}}\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon }}. \end{aligned}$$

The uniform convergence $u_{p_{n}}\rightarrow u_{\infty }$ then yields

$$\begin{aligned} 1\le \lim _{n\rightarrow \infty }\left( {\displaystyle \int _{\Omega }} \left| u_{p_{n}}\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon }}=\left( {\displaystyle \int _{\Omega }} \left| u_{\infty }\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon }}. \end{aligned}$$

Therefore,

$$\begin{aligned} k_{\infty }=k(u_{\infty })=\lim _{\epsilon \rightarrow 0^{+}}\left( \int _{\Omega }\left| u_{\infty }\right| ^{\epsilon }\omega \mathrm {d}x\right) ^{\frac{1}{\epsilon }}\ge 1. \end{aligned}$$

It follows that $(k_{\infty })^{-1}u_{\infty }\in \mathcal {M}_{s},$ so that

$$\begin{aligned} \mu _{s}\le \left| (k_{\infty })^{-1}u_{\infty }\right| _{s}=(k_{\infty })^{-1}\left| u_{\infty }\right| _{s}. \end{aligned}$$

(25)

In the next step, we prove that

$$\begin{aligned} \int _{\Omega }\frac{u}{u_{\infty }}\omega \mathrm {d}x\le \frac{\left| u\right| _{s}}{L}\quad \forall \,u\in C_{0}^{0,s}(\overline{\Omega }). \end{aligned}$$

(26)

According to Lemma 9, there exists a sequence of nonnegative functions $\left\{ u_{k}\right\} _{k\in \mathbf {N}}\subset C_{0}^{0,s}(\overline{\Omega })\cap W_{0}^{s,p}(\Omega ),$ for all $p>1,$ converging uniformly to u in $C(\overline{\Omega })$ and such that

$$\begin{aligned} \limsup _{k\rightarrow \infty }\left| u_{k}\right| _{s}\le \left| u\right| _{s}. \end{aligned}$$

Since $u_{p}$ is the weak solution of (3) and $\Lambda _{p}=\left[ u_{p}\right] _{s,p}^{p}$, we use Hölder’s inequality to get

$$\begin{aligned} \Lambda _{p}\int _{\Omega }\frac{u_{k}}{u_{p}}\omega \mathrm {d}x=\left\langle (-\Delta _{p})^{s}u_{p},u_{k}\right\rangle \le \left[ u_{p}\right] _{s,p}^{p-1}\left[ u_{k}\right] _{s,p}=(\Lambda _{p})^{\frac{p-1}{p}}\left[ u_{k}\right] _{s,p}. \end{aligned}$$

It follows that

$$\begin{aligned} \root p_{n} \of {\Lambda _{p_{n}}}\int _{\Omega }\frac{u_{k}}{u_{p_{n}}} \omega \mathrm {d}x\le \left[ u_{k}\right] _{s,p_{n}}. \end{aligned}$$

Combining Fatou’s lemma with the uniform convergence $u_{p_{n}}\rightarrow u_{\infty }$ and Lemma 7, we obtain

$$\begin{aligned} L\int _{\Omega }\frac{u_{k}}{u_{\infty }}\omega \mathrm {d}x\le L\liminf _{n\rightarrow \infty }\int _{\Omega }\frac{u_{k}}{u_{p_{n}}}\omega \mathrm {d} x\le \liminf _{n\rightarrow \infty }\left[ u_{k}\right] _{s,p_{n}}=\left| u_{k}\right| _{s}, \end{aligned}$$

that is,

$$\begin{aligned} L\int _{\Omega }\frac{u_{k}}{u_{\infty }}\omega \mathrm {d}x\le \left| u_{k}\right| _{s}. \end{aligned}$$

Letting $k\rightarrow \infty$ and applying Fatou’s lemma again, we arrive at (26):

$$\begin{aligned} L\int _{\Omega }\frac{u}{u_{\infty }}\omega \mathrm {d}x\le L\liminf _{k\rightarrow \infty }\int _{\Omega }\frac{u_{k}}{u_{\infty }}\omega \mathrm {d}x\le \liminf _{k\rightarrow \infty }\left| u_{k}\right| _{s}\le \left| u\right| _{s}. \end{aligned}$$

Taking $u=u_{\infty }$ in (26), we obtain

$$\begin{aligned} L\le \left| u_{\infty }\right| _{s} \end{aligned}$$

and combining this with (24) we conclude that

$$\begin{aligned} L=\left| u_{\infty }\right| _{s}. \end{aligned}$$

(27)

Now, let $0\le u\in \mathcal {M}_{s}$ be fixed. Then (16) yields

$$\begin{aligned} -\int _{\Omega }(\log u_{\infty })\omega \mathrm {d}x&=\int _{\Omega }(\log u)\omega \mathrm {d}x-\int _{\Omega }(\log u_{\infty })\omega \mathrm {d}x\\&=\int _{\Omega }(\log (\frac{u}{u_{\infty }}))\omega \mathrm {d}x\le \log \left( \int _{\Omega }\frac{u}{u_{\infty }}\omega \mathrm {d}x\right) . \end{aligned}$$

Hence, (26) and (27) imply that

$$\begin{aligned} (k_{\infty })^{-1}\le \int _{\Omega }\frac{u}{u_{\infty }}\omega \mathrm {d} x\le \frac{\left| u\right| _{s}}{\left| u_{\infty }\right| _{s}}\quad \mathrm {whenever}\quad 0\le u\in \mathcal {M}_{s}. \end{aligned}$$

(28)

Combining these estimates at $u=v_{s}$ with (25), we obtain

$$\begin{aligned} (k_{\infty })^{-1}\le \int _{\Omega }\frac{v_{s}}{u_{\infty }}\omega \mathrm {d}x\le \frac{\left| v_{s}\right| _{s}}{\left| u_{\infty }\right| _{s}}=\frac{\mu _{s}}{\left| u_{\infty }\right| _{s}} \le (k_{\infty })^{-1}, \end{aligned}$$

which leads us to conclude that

$$\begin{aligned} \mu _{s}=\left| (k_{\infty })^{-1}u_{\infty }\right| _{s}\quad \mathrm {and}\quad (k_{\infty })^{-1}=\int _{\Omega }\frac{v_{s}}{u_{\infty }} \omega \mathrm {d}x. \end{aligned}$$

Since $v_{s}$ is the only nonnegative minimizer of $\left| \cdot \right| _{s}$ on $\mathcal {M}_{s}$, we get (22). $\square$

Corollary 11

The following inequalities hold

$$\begin{aligned} k(u)\le \int _{\Omega }\frac{\left| u\right| }{v_{s}}\omega \mathrm {d}x\le \frac{\left| u\right| _{s}}{\mu _{s}}\quad \forall \,u\in C_{0}^{0,s}(\overline{\Omega }). \end{aligned}$$

(29)

Proof

Since we already know that $L=\left| u_{\infty }\right| _{s}$ and $u_{\infty }=k_{\infty }v_{s}$, the second inequality in (29) follows from (26), with u replaced with $w=\left| u\right|$ (note that $\left| w\right| _{s}\le \left| u\right| _{s}$). The first inequality in (29) is obvious when $k(u)=0$ and, when $k(u)>0,$ it follows from the first inequality in (28), with $w=(k(u))^{-1} \left| u\right| \in \mathcal {M}_{s}.$$\square$

Remark 12

In contrast with what happens in similar problems driven by the standard p-Laplacian, we are not able to prove that $u_{\infty }\in W_{0}^{s,q}(\Omega )$ for some $q>1.$ Such a property would guarantee that $u_{\infty }=v_{s}$ and, consequently,

$$\begin{aligned} \lim _{p\rightarrow \infty }u_{p}=v_{s} \end{aligned}$$

(that is, $v_{s}$ would be the only limit point of the family $\left\{ u_{p}\right\} _{p>1},$ as $p\rightarrow \infty$). Indeed, if $u_{\infty }\in W_{0}^{s,q}(\Omega )$ for some $q>1$ then, according to Lemma 7, $u_{\infty }\in W_{0}^{s,p_{n}}(\Omega )$ for all n sufficiently large (such that $p_{n}\ge q$) and

$$\begin{aligned} \lim _{n\rightarrow \infty }\left[ u_{\infty }\right] _{s,p_{n}}=\left| u_{\infty }\right| _{s}. \end{aligned}$$

Hence, proceeding as in the proof of Theorem 10, we would arrive at

$$\begin{aligned} 1\le k_{\infty }\le \int _{\Omega }\frac{u_{\infty }}{u_{p_{n}}}\omega \mathrm {d}x\le \frac{\left[ u_{\infty }\right] _{s,p_{n}}}{\root p_{n} \of {\Lambda _{p_{n}}}}. \end{aligned}$$

Since $\lim _{n\rightarrow \infty }\left[ u_{\infty }\right] _{s,p_{n}} =\lim _{n\rightarrow \infty }\root p_{n} \of {\Lambda _{p_{n}}}=\left| u_{\infty }\right| _{s}$ we would conclude that $k_{\infty }=1$ and $u_{\infty } =v_{s}.$

5 The limit problem

For a matter of compatibility with the viscosity approach, we add the hypotheses of continuity and strict positiveness to the weight $\omega$. So, we assume in this section that

$$\begin{aligned} \omega \in C(\Omega )\cap L^{r}(\Omega ),\,r>1,\quad \omega >0\quad \mathrm {in} \quad \Omega ,\quad \mathrm {and}\quad \int _{\Omega }\omega \mathrm {d}x=1. \end{aligned}$$

Note that such $\omega$ satisfies the hypotheses of Theorem 3.

For $1<p<\infty$ we write the s-fractional p-Laplacian, in its integral version, as $\left( -\Delta _{p}\right) ^{s}=-\mathcal {L}_{p}$ where

$$\begin{aligned} (\mathcal {L}_{p}u)(x):=2\int _{\mathbb {R}^{N}}\dfrac{\left| u(y)-u(x)\right| ^{p-2}(u(y)-u(x))}{\left| y-x\right| ^{N+sp} }\mathrm {d}y. \end{aligned}$$

(30)

Corresponding to the case $p=\infty$, we define operator $\mathcal {L}_{\infty }$ by

$$\begin{aligned} \mathcal {L}_{\infty }:=\mathcal {L}_{\infty }^{+}+\mathcal {L}_{\infty }^{-}, \end{aligned}$$

(31)

where

$$\begin{aligned} \left( \mathcal {L}_{\infty }^{+}u\right) (x):=\sup _{y\in \mathbb {R} ^{N}{\setminus }\left\{ x\right\} }\frac{u(y)-u(x)}{\left| y-x\right| ^{s}}\quad \mathrm {and}\quad \left( \mathcal {L}_{\infty }^{-}u\right) (x):=\inf _{y\in \mathbb {R}^{N}{\setminus }\left\{ x\right\} }\frac{u(y)-u(x)}{\left| y-x\right| ^{s}}. \end{aligned}$$

(32)

In the sequel, we consider, in the viscosity sense, the problem

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \mathcal {L}u=0 &{} \mathrm {in} &{} \Omega \\ u=0 &{} \mathrm {in} &{} \mathbb {R}^{N}{\setminus }\Omega , \end{array} \right. \end{aligned}$$

(33)

where either $\mathcal {L}u=\mathcal {L}_{p}u+\Lambda _{p}u^{-1}\omega ,$ with $1<p<\infty ,$ or

$$\begin{aligned} \mathcal {L}u=\mathcal {L}_{\infty }u\quad \mathrm {or}\quad \mathcal {L} u=\mathcal {L}_{\infty }^{-}u+\left| u_{\infty }\right| _{s}. \end{aligned}$$

We recall some definitions related to the viscosity approach for the problem (33).

Definition 13

Let $u\in C(\mathbb {R}^{N})$ such that $u>0$ in $\Omega$ and $u=0$ in $\mathbb {R}^{N}{\setminus }\Omega .$ We say that u is a viscosity supersolution of Eq. (33) if

$$\begin{aligned} (\mathcal {L}\varphi )(x_{0})\le 0 \end{aligned}$$

for all pair $\left( x_{0},\varphi \right) \in \Omega \times C_{0} ^{1}(\mathbb {R}^{N})$ satisfying

$$\begin{aligned} \varphi (x_{0})=u(x_{0})\quad \mathrm {and}\quad \varphi (x)\le u(x)\quad \forall \,x\in \mathbb {R}^{N}. \end{aligned}$$

Analogously, we say that u is a viscosity subsolution of (33) if

$$\begin{aligned} (\mathcal {L}\varphi )(x_{0})\ge 0 \end{aligned}$$

for all pair $\left( x_{0},\varphi \right) \in \Omega \times C_{0} ^{1}(\mathbb {R}^{N})$ satisfying

$$\begin{aligned} \varphi (x_{0})=u(x_{0})\quad \mathrm {and}\quad \varphi (x)\ge u(x)\quad \forall \,x\in \mathbb {R}^{N}. \end{aligned}$$

We say that u is a viscosity solution of (33) if it is simultaneously a subsolution and a supersolution of (33).

The next lemma can be proved by following, step by step, the proof of Proposition 11 of [17].

Lemma 14

Let $u\in W_{0}^{s,p}(\Omega )\cap C(\overline{\Omega })$ be a positive weak solution of (3). Then u is a viscosity solution of

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \mathcal {L}_{p}u+\Lambda _{p}u^{-1}\omega =0 &{}\quad \mathrm {in} &{}\quad \Omega \\ u=0 &{}\quad \mathrm {in} &{}\quad \mathbb {R}^{N}{\setminus }\Omega . \end{array} \right. \end{aligned}$$

(34)

Our main result in this section is the following, where $u_{\infty }\in C_{0}^{0,s}(\overline{\Omega })$ is the function given by Theorem 10.

Theorem 15

The function $u_{\infty }\in C_{0}^{0,s}(\overline{\Omega }),$ extended as zero outside $\Omega ,$ is both a viscosity supersolution of the problem

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \mathcal {L}_{\infty }u=0 &{}\quad \mathrm {in} &{}\quad \Omega \\ u=0 &{}\quad \mathrm {in} &{}\quad \mathbb {R}^{N}{\setminus }\Omega \end{array} \right. \end{aligned}$$

(35)

and a viscosity solution of the problem

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \mathcal {L}_{\infty }^{-}u+\left| u_{\infty }\right| _{s}=0 &{}\quad \mathrm {in} &{}\quad \Omega \\ u=0 &{}\quad \mathrm {in} &{}\quad \mathbb {R}^{N}{\setminus }\Omega . \end{array} \right. \end{aligned}$$

(36)

Moreover, $u_{\infty }$ is strictly positive in $\Omega$ and the only minimizers of $\left| \cdot \right| _{s}$ on $\mathcal {M}_{s}$ are

$$\begin{aligned} -v_{s}\quad \mathrm {and}\quad v_{s}. \end{aligned}$$

(37)

Proof

We begin by proving that $u_{\infty }$ is a viscosity supersolution of (36). For this, let us fix $\left( x_{0},\varphi \right) \in \Omega \times C_{0}^{1}(\mathbb {R}^{N})$ satisfying

$$\begin{aligned} \varphi (x_{0})=u_{\infty }(x_{0})\quad \mathrm {and}\quad \varphi (x)\le u_{\infty }(x)\quad \forall \,x\in \mathbb {R}^{N}. \end{aligned}$$

(38)

Without loss of generality, we can assume that

$$\begin{aligned} \varphi (x)<u_{\infty }(x)\quad \forall \,x\in \mathbb {R}^{N}, \end{aligned}$$

what allows us to assure that $u_{p_{n}}-\varphi$ assumes its minimum value at a point $x_{n},$ with $x_{n}\rightarrow x_{0}.$

Let $c_{n}:=u_{p_{n}}(x_{n})-\varphi (x_{n}).$ Of course, $c_{n}\rightarrow 0$ (due to the uniform convergence $u_{p_{n}}\rightarrow u_{\infty }$). By construction,

$$\begin{aligned} \varphi (x_{n})+c_{n}=u_{p_{n}}(x_{n})\quad \mathrm {and}\quad \varphi (x)+c_{n}\le u_{p_{n}}(x)\quad \forall \,x\in \mathbb {R}^{N}. \end{aligned}$$

According to the previous lemma, $u_{p}$ is a viscosity supersolution of (34) since it is a viscosity solution of the same problem. Therefore,

$$\begin{aligned} (\mathcal {L}_{p_{n}}\varphi )(x_{n})+\Lambda _{p_{n}}\frac{\omega (x_{n} )}{u_{p_{n}}(x_{n})}=(\mathcal {L}_{p_{n}}(\varphi +c_{n}))(x_{n})+\Lambda _{p_{n}}\frac{\omega (x_{n})}{\varphi (x_{n})+c_{n}}\le 0, \end{aligned}$$

an inequality that can be rewritten as

$$\begin{aligned} A_{n}^{p_{n}-1}+C_{n}^{p_{n}-1}\le B_{n}^{p_{n}-1} \end{aligned}$$

where

$$\begin{aligned} A_{n}^{p_{n}-1}=2\int _{\mathbb {R}^{N}}\dfrac{\left| \varphi (y)-\varphi (x_{n})\right| ^{p_{n}-2}(\varphi (y)-\varphi (x_{n}))^{+}}{\left| y-x\right| ^{N+sp_{n}}}\mathrm {d}y\ge 0, \end{aligned}$$

$$\begin{aligned} B_{n}^{p_{n}-1}=2\int _{\mathbb {R}^{N}}\dfrac{\left| \varphi (y)-\varphi (x_{n})\right| ^{p_{n}-2}(\varphi (y)-\varphi (x_{n}))^{-}}{\left| y-x\right| ^{N+sp_{n}}}\mathrm {d}y\ge 0, \end{aligned}$$

and

$$\begin{aligned} C_{n}^{p_{n}-1}=\Lambda _{p_{n}}\frac{\omega (x_{n})}{u_{p_{n}}(x_{n})}>0. \end{aligned}$$

(Here, $a^{+}:=\max \left\{ a,0\right\}$ and $a^{-}:=\max \left\{ -a,0\right\} ,$ so that $a=a^{+}-a^{-}.$)

According to Lemma 6.1 of [13], which was adapted from Lemma 6.5 of [3], we have

$$\begin{aligned} \lim _{n\rightarrow \infty }A_{n}=\left( \mathcal {L}_{\infty }^{+}\varphi \right) (x_{0})\quad \mathrm {and}\quad \lim _{n\rightarrow \infty }B_{n}=-\left( \mathcal {L}_{\infty }^{-}\varphi \right) (x_{0}). \end{aligned}$$

Hence, noticing that

$$\begin{aligned} A_{n}^{p_{n}-1}\le A_{n}^{p_{n}-1}+C_{n}^{p_{n}-1}\le B_{n}^{p_{n}-1} \end{aligned}$$

we conclude that

$$\begin{aligned} \left( \mathcal {L}_{\infty }\varphi \right) (x_{0})=\left( \mathcal {L} _{\infty }^{+}\varphi \right) (x_{0})+\left( \mathcal {L}_{\infty }^{-} \varphi \right) (x_{0})\le 0 \end{aligned}$$

since

$$\begin{aligned} \left( \mathcal {L}_{\infty }^{+}\varphi \right) (x_{0})=\lim _{n\rightarrow \infty }A_{n}\le \lim _{n\rightarrow \infty }B_{n}=-\left( \mathcal {L}_{\infty }^{-}\varphi \right) (x_{0}). \end{aligned}$$

We have proved that $u_{\infty }$ is a supersolution of (35). Therefore, by directly applying Lemma 22 of [17] we conclude $u_{\infty }>0$ in $\Omega .$

The strict positiveness of $u_{\infty }$ in $\Omega$ and the uniqueness of the nonnegative minimizers of $\left| \cdot \right| _{s}$ on $\mathcal {M} _{s}$ imply that if $w\in \mathcal {M}_{s}$ is such that

$$\begin{aligned} \left| w\right| _{s}=\min _{u\in \mathcal {M}_{s}}\left| u\right| _{s} \end{aligned}$$

then $\left| w\right| =v_{s}=(k_{\infty })^{-1}u_{\infty }>0$ in $\Omega$ (recall that $\left| w\right|$ is also a minimizer). The continuity of w then implies that either $w>0$ in $\Omega$ or $w<0$ in $\Omega .$ Consequently, $w=v_{s}$ or $w=-v_{s}.$

Now, recalling that

$$\begin{aligned} \lim _{n\rightarrow \infty }(\Lambda _{p_{n}})^{\frac{1}{p_{n}-1}}=\left| u_{\infty }\right| _{s} \end{aligned}$$

and using that $\omega (x_{0})>0$ and $u_{\infty }(x_{0})>0$ we have

$$\begin{aligned} \lim _{n\rightarrow \infty }C_{n}=\left| u_{\infty }\right| _{s} \end{aligned}$$

Hence, since

$$\begin{aligned} C_{n}^{p_{n}-1}\le A_{n}^{p_{n}-1}+C_{n}^{p_{n}-1}\le B_{n}^{p_{n}-1}, \end{aligned}$$

we obtain

$$\begin{aligned} \left| u_{\infty }\right| _{s}=\lim _{n\rightarrow \infty }C_{n}\le \lim _{n\rightarrow \infty }B_{n}=-\left( \mathcal {L}_{\infty }^{-} \varphi \right) (x_{0}). \end{aligned}$$

It follows that $u_{\infty }$ is a viscosity supersolution of (36).

Now, let us take a pair $\left( x_{0},\varphi \right) \in \Omega \times C_{0}^{1}(\mathbb {R}^{N})$ satisfying

$$\begin{aligned} \varphi (x_{0})=u_{\infty }(x_{0})\quad \mathrm {and}\quad \varphi (x)\ge u_{\infty }(x)\quad \forall \,x\in \mathbb {R}^{N}. \end{aligned}$$

(39)

Since

$$\begin{aligned} -\left| u_{\infty }\right| _{s}\le \frac{u_{\infty }(x)-u_{\infty } (x_{0})}{\left| x-x_{0}\right| ^{s}}\le \frac{\varphi (x)-\varphi (x_{0})}{\left| x-x_{0}\right| ^{s}}\quad \forall \,x\in \mathbb {R} ^{N}{\setminus }\left\{ x_{0}\right\} , \end{aligned}$$

we have

$$\begin{aligned} -\left| u_{\infty }\right| _{s}\le \inf _{x\in \mathbb {R}^{N} {\setminus }\left\{ x_{0}\right\} }\frac{\varphi (x)-\varphi (x_{0})}{\left| x-x_{0}\right| ^{s}}=\left( \mathcal {L}_{\infty }^{-}\varphi \right) (x_{0}). \end{aligned}$$

Therefore, $u_{\infty }$ is a viscosity subsolution of (36). $\square$

Since $v_{s}=(k_{\infty })^{-1}u_{\infty }$ is the only positive minimizer of $\left| \cdot \right| _{s}$ on $C_{0}^{0,s}(\overline{\Omega } ){\setminus }\left\{ 0\right\}$ and $\mathcal {L}_{\infty }^{-}(ku)=k\mathcal {L} _{\infty }^{-}u\$ for any positive constant k,  the following corollary is immediate.

Corollary 16

The minimizer $v_{s}$ is a viscosity solution of the problem

$$\begin{aligned} \left\{ \begin{array} [c]{lll} \mathcal {L}_{\infty }^{-}u+\mu _{s}=0 &{} \mathrm {in} &{} \Omega \\ u=0 &{} \mathrm {in} &{} \mathbb {R}^{N}{\setminus }\Omega . \end{array} \right. \end{aligned}$$