Existence results for nonlocal problems governed by the regional fractional Laplacian

Abstract

The aim of the present paper is to study existence results of minimizers of the critical fractional Sobolev constant on bounded domains. Under some values of the fractional parameter we show that the best constant is achieved. If moreover the underlying domain is a ball, we obtain positive radial minimizers for all possible values of the fractional parameter in higher dimension, while we impose a positive mass condition in low dimension.

1 introduction and main results

Let $\Omega$ be a Lipschitz open set of $\mathbb {R}^N$, $s\in (1/2,1)$ and $N>2s$. The purpose of this paper is to study the existence of minimizers to the best Sobolev critical constant

$$\begin{aligned} S_{N,s}(\Omega )=\inf _{\begin{array}{c} u\in H^s_0(\Omega )\\ u\ne 0 \end{array}}\frac{Q_{N,s,\Omega }(u)}{\Vert u\Vert ^2_{L^{2^*_s}(\Omega )}}, \end{aligned}$$

(1.1)

where $H^s_0(\Omega )$ is the completion of $C^{\infty }_c(\Omega )$ with respect to the $H^s(\Omega )$-norm, $2^*_s:=\frac{2N}{N-2s}$ is the so-called fractional critical Sobolev exponent and $Q_{N,s,\Omega }(\cdot )$ is a nonnegative quadratic form defined on $H^s_0(\Omega )$ by

$$\begin{aligned} Q_{N,s,\Omega }(u):=\frac{c_{N,s}}{2}\int _{\Omega }\int _{\Omega }\frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}\ dxdy. \end{aligned}$$

We notice that for $s\in (0,1/2]$ and $\Omega$ bounded, the constant function 1 belongs to $H^s_0(\Omega )$, and thus, the above Sobolev constant is zero in this case. We refer the reader to Appendix A below for more details and the definition of Lipschitz domains in this paper.

We recall that nonnegative minimizers of the constant $S_{N,s}(\Omega )$ are weak solutions to nonlinear Dirichlet problem

$$\begin{aligned} \left\{ \begin{aligned} (-\Delta )^s_{\Omega }u&=u^{2^*_s-1}\quad \text {in}\quad \Omega \\ u&=0\quad \quad \quad \text {on}\quad \partial \Omega , \end{aligned} \right. \end{aligned}$$

(1.2)

where $(-\Delta )^s_{\Omega }$ is the regional fractional Laplacian defined as

$$\begin{aligned} (-\Delta )^s_{\Omega }u(x)=c_{N,s}P.V.\int _{\Omega }\frac{u(x)-u(y)}{|x-y|^{N+2s}}\ dy,~~x\in \Omega . \end{aligned}$$

Here, $c_{N,s}$ is the usual positive normalization constant of $(-\Delta )^s$ and P.V. stands for the principal value of the integral.

In the theory of partial differential equations, the existence of solutions of nonlinear equations appears as a natural question. This strongly depends on the type of nonlinearities that are considered. For instance, nonlinear equations involving subcritical power nonlinearities, say $f(t)=|t|^{p-1}$ with $p<2^*_s$, are quite well-understood and due to compactness, the existence of solutions can be easily established by using for example the Mountain Pass theorem. One can also study the corresponding minimization problem and prove that a minimizer exists. Besides, at the critical exponent $p=2^*_s$ we lose compactness and therefore standard argument of calculus of variation cannot be applied to derive the existence of solutions. As a typical example, when $\Omega$ is a star-shaped bounded domain, it has been proved that the Dirichlet problem

$$\begin{aligned} (-\Delta )^su=u^{2^*_s-1},\quad \quad u>0\quad \text {in}~~\Omega ,\ \ \ \ \ u=0\quad \text {in}~~\mathbb {R}^N\setminus \Omega \end{aligned}$$

(1.3)

does not admit a solution. Such a nonexistrence result was first proved in [11] and later in [17, 18] by means of a fractional Pohozaev type identity. However, (1.2) can have a solution even if $\Omega$ is star-shaped and smooth. It is therefore interesting to understand the type of domains and exponents for which (1.2) does not admit a solution.

In the case where $\Omega =\mathbb {R}^N$ or $\Omega =\mathbb {R}^N_+$, the infinimum $S_{N,s}(\Omega )>0$ for all $s\in (0,1)$. Moreover, see e.g. [2, 16] all minimizers of $S_{N,s}(\mathbb {R}^N)$ are of the form

$$\begin{aligned} u(x)=a\Big (\frac{1}{b^2+|x-x_0|^2}\Big )^{\frac{N-2s}{2}},\quad \quad x\in \mathbb {R}^N \end{aligned}$$

(1.4)

where a, b are positive constants and $x_0\in \mathbb {R}^N$.

Problem of type (1.2) is less understood in contrast with (1.3). The only paper investigating it is [12]. Precisely, the authors in [12] considered the equivalent minimization problem and obtain existence of minimizers under some assumptions on $\Omega$ and the range of the parameter s. In particular, it is proved in [12] that if a portion of $\partial \Omega$ lies on a hyperplane and $N\ge 4s$, then $S_{N,s}(\Omega )$ is achieved.

Our first main result removes this assumption on $\Omega$ provided s is close to 1/2.

Theorem 1.1

Let $N\ge 2$ and $\Omega \subset \mathbb {R}^N$ be a bounded $C^1$ open set. Then there exists $s_0\in (1/2,1)$ such that for all $s\in (1/2,s_0)$, the infimum $S_{N,s}(\Omega )$ is achieved by a positive function $u\in H^s_0(\Omega )$ satisfying (1.2).

The main ingredient to prove Theorem 1.1 is to show that $S_{N,s}(\Omega )<S_{N,s}(\mathbb {R}^N_+)$ for s closed to 1/2. In fact, this strict inequality allows for a sort of compactness. We achieve this by showing that $S_{N,1/2}(\Omega )=0$ provided $\Omega$ is a bounded Lipschitz open set. We notice here that our notion of Lipschitz open set is that $\partial \Omega$ is locally given by the restriction of a bi-Lipschitz map. This is strictly weaker than the strongly Lipschitz property, meaning that $\partial \Omega$ is locally given by a graph of a Lipschitz function, see Definiton A.2 and Remark A.3 below.

Next, let ${\mathcal B}$ denote the unit centered ball in $\mathbb {R}^N$. We consider the minimization problem (1.1) on the space $H^s_{0,rad}({\mathcal B})$, the completion of the space of radial functions belonging to $C^\infty _c({\mathcal B})$ with respect to the norm $H^s_0({\mathcal B})$. More precisely, we consider the infinimum problem, for $h\in L^\infty ({\mathcal B})$ being radial,

$$\begin{aligned} S_{N,s,rad}({\mathcal B},h)=\inf _{\begin{array}{c} u\in H^s_{0,rad}({\mathcal B})\\ u\ne 0 \end{array}}\frac{Q_{N,s,{\mathcal B}}(u)+\int _{{\mathcal B}}hu^2 dx}{\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}}. \end{aligned}$$

(1.5)

Our next result is related to the existence of minimizers for the infimum $S_{N,s,rad}({\mathcal B},0)$ in high dimension $N\ge 4s$. Our second main result is the following.

Theorem 1.2

Let $s\in (1/2,1)$ and $N\ge 4s$. Then the infinimum

$$\begin{aligned} S_{N,s,rad}({\mathcal B},0)=\inf _{\begin{array}{c} u\in H^s_{0,rad}({\mathcal B})\\ u\ne 0 \end{array}}\frac{Q_{N,s,{\mathcal B}}(u) }{\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}} \end{aligned}$$

(1.6)

is achieved by a positive function $u\in H^s_{0,rad}({\mathcal B})$, satisfying

$$\begin{aligned} \begin{aligned} (-\Delta )^s_{{\mathcal B}}u&=u^{2^*_s-1}\quad \text {in}\quad {\mathcal B},\qquad u=0\quad \text {on}\quad \partial {\mathcal B}. \end{aligned} \end{aligned}$$

We now turn our attention to the minimization problem $S_{N,s,rad}({\mathcal B},h)$ in low dimension $N<4s$. This Sobolev constant is related to the Schrödinger operator $(-\Delta )^s_{{\mathcal B}}+h$. As a necessary condition for the existence of positive minimizers, it is important to assume that $(-\Delta )^s_{{\mathcal B}}+h$ defines a coercive bilinear form on $H^s_{0,rad}({\mathcal B})$.

Before stating our next result, we need to introduce the mass of ${\mathcal B}$ at 0 associated to the Schrödinger operator $(-\Delta )^s+h$, where $(-\Delta )^s$ is the standard fractional Laplacian. Indeed, let G(x, y) be the Green function of the operator $(-\Delta )^s+h$ on ${\mathcal B}$ and ${\mathcal R}$ be the fundamental solution of $(-\Delta )^s$ on $\mathbb {R}^N$. Then the function $x\mapsto \textbf{k}(x)=G(x,0)-{\mathcal R}(x)$ is continuous in ${\mathcal B}$. The mass of the operator $(-\Delta )^s+h$ at 0 is given by $\textbf{k}(0)$. Our next existence result is a consequence of the fact that the mass is positive, see [13, 19].

Theorem 1.3

Let $s\in (1/2,1)$, $2\le N<4s$, $h\in L^\infty _{rad}({\mathcal B})$ and suppose that $S_{N,s,rad}({\mathcal B},h)>0$. Assume that $\textbf{k}(0)>0$. Then $S_{N,s,rad}({\mathcal B},h)$ is achieved by a positive function $u\in H^s_{0,rad}({\mathcal B})$, satisfying

$$\begin{aligned} \begin{aligned} (-\Delta )^s_{{\mathcal B}}u+ h u&=u^{2^*_s-1}\quad \text {in}\quad {\mathcal B},\qquad u=0\quad \text {on}\quad \partial {\mathcal B}. \end{aligned} \end{aligned}$$

The role of the mass in proving the existence of minimizers (for Sobolev constant) in low dimensions is very crucial. As we will see later, it helps us to restore the compactness. Indeed, the strict positivity $\textbf{k}(0)>0$ implies that the Sobolev constant in ${\mathcal B}$ is strictly less than that of $\mathbb {R}^N$, and thereby produces the existence of minimizers.

An interesting question that arises is whether symmetry breaking occurs? More generally, for $p\ge 1$, is every positive solution to $u\in H^s_0({\mathcal B})$ to

$$\begin{aligned} \begin{aligned} (-\Delta )^s_{{\mathcal B}}u&=u^{p}\quad \text {in}\quad {\mathcal B}, \qquad u=0 \quad \text {on}\quad \partial {\mathcal B}, \end{aligned} \end{aligned}$$

is radial? We conjecture that the answer to this question is no.

In Proposition 2.3 we obtain a priori $L^{\infty }$-bounds of minimizers. Hence, by the ineterior regularity theory and standard boostrap arguments, they belong to $C^\infty (\Omega )$, provided $h\in C^\infty (\Omega )$. In addition, the boundary regularity result in [4, 10] implies that minimizers are actually $C^{2s-1}(\overline{\Omega })$.

The rest of the paper is organized as follows. in Sect. 2 we give some preliminaries that will be useful throughout this paper. In Sect. 3 we prove Theorems 1.1. In Sect. 4 we collect some useful results needed to prove Theorems 1.2 and 1.3 whereas in Sect. 5 we establish Theorems 1.2 and 1.3. Finally in the Appendix A we prove that the constant function 1 belongs to $H^s_0(\Omega )$ for $s\in (0,1/2]$.

2 Preliminary

In this section, we introduce some preliminary properties which will be useful in this work. For all $s\in (0,1)$, the fractional Sobolev space $H^s(\Omega )$ is defined as the set of all measurable functions u such that

$$\begin{aligned}{}[u]^2_{H^s(\Omega )}:=\frac{c_{N,s}}{2}\int _{\Omega }\int _{\Omega }\frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}\ dxdy \end{aligned}$$

is finite. It is a Hilbert space endowed with the norm

$$\begin{aligned} \Vert u\Vert ^2_{H^s(\Omega )}=\Vert u\Vert ^2_{L^2(\Omega )}+[u]^2_{H^s(\Omega )}. \end{aligned}$$

We refer to [7] for more details on this fractional Sobolev spaces. Next, we denote by $H^s_0(\Omega )$ the completion of $C^{\infty }_c(\Omega )$ under the norm $\Vert \cdot \Vert _{H^s(\Omega )}$. Moreover, for $s\in (1/2,1)$, $H^s_0(\Omega )$ is a Hilbert space equipped with the norm

$$\begin{aligned} \Vert u\Vert ^2_{H^s_0(\Omega )}=\frac{c_{N,s}}{2}\int _{\Omega }\int _{\Omega }\frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}\ dxdy \end{aligned}$$

which is equivalent to the usual one in $H^s(\Omega )$ thanks to Poincaré inequality. We define the Hilbert space

$$\begin{aligned} {\mathcal H}^s_0(\Omega )=\{u\in H^s(\mathbb {R}^N):u=0~\text {in}~\mathbb {R}^N\setminus \Omega \} \end{aligned}$$

endowed with the norm $\Vert \cdot \Vert _{H^s(\mathbb {R}^N)}$, which is the completion of $C^{\infty }_c(\Omega )$ with respect to the norm $\Vert \cdot \Vert _{H^s(\mathbb {R}^N)}$. In the sequel, $H^s_{0,rad}(\Omega )$ and ${\mathcal H}^s_{0,rad}(\Omega )$ are respectively the space of radially symmetric functions of $H^s_0(\Omega )$ and ${\mathcal H}^s_0(\Omega )$. We denote by $L^{\infty }_{rad}(\Omega )$ the space of radial functions u belonging to $L^{\infty }(\Omega )$.

Given $x\in \Omega$ and $r>0$, we denote by $B_r(x)$ the open ball centered at x with radius r. When the center is not specified, we will understand that it’s the origin, e.g. $B_2(0)=B_2$. The upper half-ball centered at x with radius r is denoted by $B^+_r(x)$. We will always use $\delta _{\Omega }(x)=\text {dist}(x,\partial \Omega )$ for the distance from x to the boundary. For every set $A\subset \mathbb {R}^N$, we denote by $\mathbbm {1}_{A}$ its characteristic function.

Proposition 2.1

(see [5, 7]) The embedding $H^s_0(\Omega )\hookrightarrow L^p(\Omega )$ is continuous for any $p\in [2,2^*_s]$, and compact for any $p\in [2,2^*_s)$.

The next proposition gives an elementary result regarding the role of convex functions applied to $(-\Delta )^s_{\Omega }$.

Proposition 2.2

Assume that $\phi :\mathbb {R}\rightarrow \mathbb {R}$ is a Lipschitz convex function such that $\phi (0)=0$. Then if $u\in H^s_0(\Omega )$ we have

$$\begin{aligned} (-\Delta )^s_{\Omega }\phi (u)\le \phi '(u)(-\Delta )^s_{\Omega }u\quad \text {weakly in}\quad \Omega . \end{aligned}$$

(2.1)

Proof

The proof of the above lemma is standard. In fact, using that every convex $\phi$ satisfies $\phi (a)-\phi (b)\le \phi '(a)(a-b)$ for all $a,b\in \mathbb {R}$, the proof follows.

$\square$

We conclude this section showing in proposition below, the boundedness of any nonnegative solution of (1.2). The argument uses Moser’s iteration method. A similar result has been established in [1] for the case of fractional Laplacian.

Proposition 2.3

Let $u\in H^s_0(\Omega )$ be a nonnegative solution to problem (1.2). Then $u\in L^{\infty }(\Omega )$.

Proof

For $\beta \ge 1$ and $T>0$ large, we define the following convex function

$$\begin{aligned} \phi _{T,\beta }(t)=\left\{ \begin{aligned}&0,\quad \quad \quad \quad \quad \quad \quad \quad \quad \text {if}\quad t\le 0\\&t^{\beta },\quad \quad \quad \quad \quad \quad \quad \quad ~~\text {if}\quad 0<t<T\\&\beta T^{\beta -1}(t-T)+T^{\beta },\quad \text {if}\quad t\ge T. \end{aligned} \right. \end{aligned}$$

Throughout the proof, we will use $\phi _{T,\beta }=:\phi$ for the sake of simplicity. Since $\phi$ is Lipschitz, with constant $\Lambda _{\phi }=\beta T^{\beta -1}$, and $\phi (0)=0$, then $\phi (u)\in H^s_0(\Omega )$ and by the convexity of $\phi$, we have, according to Proposition 2.2 that

$$\begin{aligned} (-\Delta )^s_{\Omega }\phi (u)\le \phi '(u)(-\Delta )^s_{\Omega }u. \end{aligned}$$

(2.2)

By Proposition 2.1 and inequality (2.2) we have that

$$\begin{aligned} \Vert \phi (u)\Vert ^2_{L^{2^*_s}(\Omega )}&\le C\Vert \phi (u)\Vert ^2_{H^s_0(\Omega )}=C\int _{\Omega }\phi (u)(-\Delta )^s_{\Omega }\phi (u)\ dx\\&\le C\int _{\Omega }\phi (u)\phi '(u)(-\Delta )^s_{\Omega }u\ dx\\&=C\int _{\Omega }\phi (u)\phi '(u)u^{2^*_s-1}\ dx. \end{aligned}$$

Moreover, since $u\phi '(u)\le \beta \phi (u)$, we have that

$$\begin{aligned} \Vert \phi (u)\Vert ^2_{L^{2^*_s}(\Omega )}\le C\beta \int _{\Omega }(\phi (u))^2u^{2^*_s-2}\ dx. \end{aligned}$$

(2.3)

We point out that the integral on the right-hand side of the above inequality is finite. Indeed, using that $\beta \ge 1$ and $\phi (u)$ is linear when $u\ge T$, we have from a quick computation that

$$\begin{aligned} \int _{\Omega }(\phi (u))^2u^{2^*_s-2}\ dx&=\int _{\{u\le T\}}(\phi (u))^2u^{2^*_s-2}\ dx+\int _{\{u>T\}}(\phi (u))^2u^{2^*_s-2}\ dx\\ {}&\le T^{2\beta -2}\int _{\Omega }u^{2^*_s}\ dx+C\int _{\Omega }u^{2^*_s}\ dx<\infty . \end{aligned}$$

We now choose $\beta$ in (2.3) so that $2\beta -1 = 2^*_s$. Denoting by $\beta _1$ such a value, then we can equivalently write

$$\begin{aligned} \beta _1:=\frac{2^*_s+1}{2}. \end{aligned}$$

(2.4)

Let $K>0$ be a positive number whose value will be fixed later on. Then applying H$\ddot{\text {o}}$lder’s inequality with exponents $q:=2^*_s/2$ and $q':=2^*_s/(2^*_s-2)$ in the integral on the right-hand side of inequality (2.3), we find that

$$\begin{aligned}&\int _{\Omega }(\phi (u))^2u^{2^*_s-2}\ dx=\int _{\{u\le K\}}(\phi (u))^2u^{2^*_s-2}\ dx+\int _{\{u> K\}}(\phi (u))^2u^{2^*_s-2}\ dx \nonumber \\&\le \int _{\{u\le K\}}\frac{(\phi (u))^2}{u}K^{2^*_s-1}\ dx+\Bigg (\int _{\Omega }(\phi (u))^{2^*_s}\ dx\Bigg )^{2/2^*_s}\Bigg (\int _{\{u>K\}}u^{2^*_s}\ dx\Bigg )^{\frac{2^*_s-2}{2^*_s}}. \end{aligned}$$

(2.5)

Now, thanks to Monotone Convergence Theorem, we can choose K as big as we wish so that

$$\begin{aligned} \Bigg (\int _{\{u>K\}}u^{2^*_s}\ dx\Bigg )^{\frac{2^*_s-2}{2^*_s}}\le \frac{1}{2C\beta _1}, \end{aligned}$$

(2.6)

where C is the positive constant appearing in (2.3). Therefore, by taking into account (2.6) in (2.5) and by using also (2.4), we deduce from (2.3) that

$$\begin{aligned} \Vert \phi (u)\Vert ^2_{L^{2^*_s}(\Omega )}\le 2C\beta _1\Bigg (K^{2^*_s-1}\int _{\Omega }\frac{(\phi (u))^2}{u}\ dx\Bigg ). \end{aligned}$$

Since $\phi (u)\le u^{\beta _1}$ and recalling (2.4), and by letting $T\rightarrow \infty$, we get that

$$\begin{aligned} \Bigg (\int _{\Omega }u^{2^*_s\beta _1}\ dx\Bigg )^{2/2^*_s}\le 2C\beta _1\Bigg (K^{2^*_s-1}\int _{\Omega }u^{2^*_s}\ dx\Bigg )<\infty , \end{aligned}$$

and therefore

$$\begin{aligned} u\in L^{2^*_s\beta _1}(\Omega ). \end{aligned}$$

(2.7)

Suppose now that $\beta >\beta _1$. Thus, using that $\phi (u)\le u^{\beta }$ in the right hand side of (2.3) and letting $T\rightarrow \infty$ we get

$$\begin{aligned} \Bigg (\int _{\Omega }u^{2^*_s\beta }\ dx\Bigg )^{2/2^*_s}\le C\beta \Bigg (\int _{\Omega }u^{2\beta +2^*_s-2}\ dx\Bigg ). \end{aligned}$$

(2.8)

Therefore,

$$\begin{aligned} \Bigg (\int _{\Omega }u^{2^*_s\beta }\ dx\Bigg )^{\frac{1}{2^*_s(\beta -1)}}\le (C\beta )^{\frac{1}{2(\beta -1)}}\Bigg (\int _{\Omega }u^{2\beta +2^*_s-2}\ dx\Bigg )^{\frac{1}{2(\beta -1)}}. \end{aligned}$$

(2.9)

We are now in position to use an iterative argument as in [1, Proposition 2.2]. For that, we define inductively the sequence $\beta _{m+1},~m\ge 1$ by

$$\begin{aligned} 2\beta _{m+1}+2^*_s-2=2^*_s\beta _m, \end{aligned}$$

from which we deduce that,

$$\begin{aligned} \beta _{m+1}-1=\Big (\frac{2^*_s}{2}\Big )^m(\beta _1-1). \end{aligned}$$

Now by using $\beta _{m+1}$ in place of $\beta$, in (2.9), it follows that

$$\begin{aligned} \Bigg (\int _{\Omega }u^{2^*_s\beta _{m+1}}\ dx\Bigg )^{\frac{1}{2^*_s(\beta _{m+1}-1)}}\le (C\beta _{m+1})^{\frac{1}{2(\beta _{m+1}-1)}}\Bigg (\int _{\Omega }u^{2^*_s\beta _m}\ dx\Bigg )^{\frac{1}{2^*_s(\beta _m-1)}}. \end{aligned}$$

For the sake of clarity, we set

$$\begin{aligned} C_{m+1}:=(C\beta _{m+1})^{\frac{1}{2(\beta _{m+1}-1)}}\quad \text {and}\quad A_m:=\Bigg (\int _{\Omega }u^{2^*_s\beta _m}\ dx\Bigg )^{\frac{1}{2^*_s(\beta _m-1)}} \end{aligned}$$

so that

$$\begin{aligned} A_{m+1}\le C_{m+1}A_m,~~m\ge 1. \end{aligned}$$

(2.10)

Then iterating the above inequality, we find that

$$\begin{aligned} A_{m+1}\le \prod ^{m+1}_{i=2}C_iA_1, \end{aligned}$$

which implies that

$$\begin{aligned} \log A_{m+1}&\le \sum _{i=2}^{m+1}\log C_i+\log A_1\\&\le \sum _{i=2}^{\infty }\log C_i+\log A_1. \end{aligned}$$

Since $\beta _{m+1}=(\beta _1-1/2)^m(\beta _1-1)+1$ then the serie $\sum _{i=2}^{\infty }\log C_i$ converges. Also, since $u\in L^{2^*_s\beta _1}(\Omega )$ (see (2.7)), then $A_1\le C$. From this, we find that

$$\begin{aligned} \log A_{m+1}\le C_0 \end{aligned}$$

(2.11)

with being $C_0>0$ a positive constant independent of m. By letting $m\rightarrow \infty$, it follows that

$$\begin{aligned} \Vert u\Vert _{L^{\infty }(\Omega )}\le C_0'<\infty . \end{aligned}$$

This completes the proof. $\square$

3 Existence of minimizers for s close to 1/2

We aim to study the existence of nontrivial solutions of (1.2). As pointed point out in the introduction the embedding $H^s_0(\Omega )\hookrightarrow L^{2^*_s}(\Omega )$ fails to be compact and due to this, the functional energy associated to (1.2) does not satisfy the Palais-Smale compactness condition. Hence finding the critical points by standard variational methods become a very tough task. Therefore, a natural question arises:

$$\begin{aligned} {({\mathcal Q})}\, Does\, problem\, (1.2)\, admits\, a\, nontrivial\, solution? \end{aligned}$$

In other words, we are looking at whether the quantity

$$\begin{aligned} S_{N,s}(\Omega )=\inf _{\begin{array}{c} u\in H^s_0(\Omega )\\ u\ne 0 \end{array}}\frac{Q_{N,s,\Omega }(u)}{\Vert u\Vert ^2_{L^{2^*_s}(\Omega )}} \end{aligned}$$

(3.1)

is attained or not. Here $Q_{N,s,\Omega }(\cdot )$ is a nonnegative quadratic form define on $H^s_0(\Omega )$ by

$$\begin{aligned} Q_{N,s,\Omega }(u):=\frac{c_{N,s}}{2}\int _{\Omega }\int _{\Omega }\frac{(u(x)-u(y))^2}{|x-y|^{N+2s}}\ dxdy. \end{aligned}$$

As a quick comment on the above question, Frank et al. [12, Theorem 4] gave a positive answer in the special case of a class of $C^1$ open sets whose boundary has a flat part, that is $C^1$ domains $\Omega$ with the shape $B^+_r(z)\subset \Omega \subset \mathbb {R}^N_+$ for some $r>0$ and $z\in \partial \mathbb {R}^N_+$, and such that $\mathbb {R}^N_+\setminus \Omega$ has nonempty interior. This flatness assumption on the boundary of $\Omega$ allows the authors in [12] to obtain the strict inequality $S_{N,s}(\Omega )<S_{N,s}(\mathbb {R}^N_+)$, which is the crucial ingredient for the proof of Theorem 4 in there. Notice that in [12], the question remains open for a larger class of sets.

In the sequel, we give a positive affirmation to the above question in the case of arbitrary open sets with $C^1$ boundary, provided that s is close to 1/2. As a consequence, one has in contrast with the fractional Laplacian that the above question has a positive answer even if $\Omega$ is convex and of class $C^\infty$.

For the reader’s convenience, we restate our main result in the following.

Theorem 3.1

Let $N\ge 2$ and $\Omega \subset \mathbb {R}^N$ be a bounded Lipschitz open set. There exists $s_0\in (1/2,1)$ such that for all $s\in (1/2,s_0)$, any minimizing sequence for $S_{N,s}(\Omega )$, normalized in $H^s_0(\Omega )$ is relatively compact in $H^s_0(\Omega )$. In particular, the infimum is achieved.

The proof of the above main theorem is a direct consequence of the key proposition below (see Proposition 3.2), in which we examine the asymptotic behavior of the Sobolev critical constant $S_{N,s}(\Omega )$ as s tends to $1/2^{+}$, by showing that the latter goes to zero. The proof of this only requires the domain to be Lipschitz. Our key proposition is stated as follows.

Proposition 3.2

Let $\Omega \subset \mathbb {R}^N$ be a bounded Lipschitz open set. Then

$$\begin{aligned} \lim \limits _{s\searrow 1/2}S_{N,s}(\Omega )=0. \end{aligned}$$

(3.2)

We now collect some interesting results that are needed to complete the proof of Proposition 3.2 above. Let us start with the following upper semicontinuous lemma.

Lemma 3.3

Let $\Omega \subset \mathbb {R}^N$ be a bounded Lipschitz open set. Fix $s_0\in [1/2,1)$. Then

$$\begin{aligned} \limsup _{s\searrow s_0}S_{N,s}(\Omega )\le S_{N,s_0}(\Omega ). \end{aligned}$$

(3.3)

Proof

For $t\in \mathbb {R}$, we recall the elementary inequality

$$\begin{aligned} |e^t-1|\le \sum _{k=1}^{+\infty }\frac{|t|^k}{k!}\le \sum _{k=1}^{+\infty }\frac{|t|^k}{(k-1)!}\le |t|e^{|t|}. \end{aligned}$$

(3.4)

For all $r,\gamma >0$, we also recall the following growth regarding the logarithmic function:

$$\begin{aligned} |\log |z||\le \frac{1}{e\gamma }|z|^{-\gamma }~~\text {if}~~ |z|\le r~~~~\text {and}~~~~|\log |z||\le \frac{1}{e\gamma }|z|^{\gamma }~~ \text {if}~~|z|\ge r. \end{aligned}$$

(3.5)

Let $\varepsilon >0$ and let $u_{\varepsilon }\in C^{\infty }_c(\Omega )$ such that $\Vert u_{\varepsilon }\Vert _{L^{2^*_s}(\Omega )}=1$ and $Q_{N,s_0,\Omega }(u_{\varepsilon })\le S_{N,s_0}(\Omega )+\varepsilon$. Then $S_{N,s}(\Omega )\le Q_{N,s,\Omega }(u_{\varepsilon })$. From this, we obtain that

$$\begin{aligned} S_{N,s}(\Omega )-S_{N,s_0}(\Omega )\le Q_{N,s,\Omega }(u_{\varepsilon })- Q_{N,s_0,\Omega }(u_{\varepsilon })+\varepsilon . \end{aligned}$$

(3.6)

On the other hand,

$$\begin{aligned}&|Q_{N,s,\Omega }(u_{\varepsilon })- Q_{N,s_0,\Omega }(u_{\varepsilon })|\\&\quad \le \frac{1}{2}|c_{N,s}-c_{N,s_0}|\int _{\Omega }\int _{\Omega }\frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0}}\ dxdy\\&\qquad +\frac{c_{N,s}}{2}\int _{\Omega }\int _{\Omega }\frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0}}||x-y|^{2(s_0-s)}-1|\ dxdy\\&\quad \le \frac{1}{c_{N,s_0}}(S_{N,s_0}(\Omega )+\varepsilon )|c_{N,s}-c_{N,s_0}|\\&\qquad +\frac{c_{N,s}}{2}\int _{\Omega }\int _{\Omega }\frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0}}||x-y|^{2(s_0-s)}-1|\ dxdy. \end{aligned}$$

Next, from (3.4) we have that

$$\begin{aligned}&||x-y|^{2(s_0-s)}-1|=|e^{2(s_0-s)\log |x-y|}-1|\\&\quad \le 2|s_0-s||\log |x-y||e^{2|s_0-s||\log |x-y||}\\&\quad =2|s_0-s||\log |x-y|||x-y|^{2|s_0-s|}. \end{aligned}$$

Taking this into account and using the regularity of $u_{\varepsilon }$ and the property (3.5) (with $\gamma <2(1-s_0)$), we have, with

$$\begin{aligned}{} & {} A_{\Omega }:=\{(x,y)\in \Omega \times \Omega :|x-y|\le 1\}\quad \text {and}\quad \\{} & {} B_{\Omega }:=\{(x,y)\in \Omega \times \Omega :|x-y|>1\}, \end{aligned}$$

the estimate

$$\begin{aligned}&\int _{\Omega }\int _{\Omega }\frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0}}||x-y|^{2(s_0-s)}-1|\ dxdy\\&\quad =2|s_0-s|\int _{\Omega }\int _{\Omega }\frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0}}|\log |x-y|||x-y|^{2|s_0-s|}\ dxdy\\&\quad \le 2|s_0-s|\text {diam}(\Omega )^{2|s_0-s|}\int _{\Omega }\int _{\Omega }\frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0}}|\log |x-y||\ dxdy\\&\quad =2|s_0-s|\text {diam}(\Omega )^{2|s_0-s|}\Bigg (\iint _{A_{\Omega }}\cdots +\iint _{B_{\Omega }}\cdots \Bigg )\\&\quad \frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0}}|\log |x-y||\ dxdy\\&\quad \le 2(e\gamma )^{-1}|s_0-s|\text {diam}(\Omega )^{2|s_0-s|}\Bigg (\iint _{A_{\Omega }}\frac{(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2}{|x-y|^{N+2s_0+\gamma }}\\&\qquad +\text {diam}(\Omega )^{\gamma }\iint _{B_{\Omega }}(u_{\varepsilon }(x)-u_{\varepsilon }(y))^2\Bigg )\ dxdy\\&\quad \le 2(e\gamma )^{-1}|s_0-s|\text {diam}(\Omega )^{2|s_0-s|}\Bigg (\Vert u_{\varepsilon }\Vert ^2_{C^1(\Omega )}\iint _{A_{\Omega }}|x-y|^{2-N-2s_0-\gamma }\ dxdy\\&\qquad +4|\Omega |\text {diam}(\Omega )^{\gamma }\Bigg )\\&\quad =C\text {diam}(\Omega )^{2|s_0-s|} |s_0-s| \end{aligned}$$

where $\text {diam}(\Omega )=\sup \{|x-y|:x,y\in \Omega \}$ is the diameter of $\Omega$ and $C=C(N,s_0,\gamma ,\Omega ,u_{\varepsilon })>0$ is a positive constant.

From the above estimate, we find that

$$\begin{aligned}&|Q_{N,s,\Omega }(u_{\varepsilon })- Q_{N,s_0,\Omega }(u_{\varepsilon })|\nonumber \\&\quad \le \frac{1}{c_{N,s_0}}(S_{N,s_0}(\Omega )+\varepsilon )|c_{N,s}-c_{N,s_0}|+\frac{Cc_{N,s}}{2}\text {diam}(\Omega )^{2|s_0-s|}|s_0-s| \end{aligned}$$

(3.7)

and from this, we deduce from (3.6) that

$$\begin{aligned} \limsup _{s\searrow s_0}S_{N,s}(\Omega )\le S_{N,s_0}(\Omega )+\varepsilon . \end{aligned}$$

(3.8)

Since $\varepsilon$ can be chosen arbitrarily small, (3.3) follows. This finishes the proof.

$\square$

We have the following proposition. While this result is known (see e.g. [14]) and since we could not find a detailed proof, we include its proof in Appendix A. The idea of proof is to construct a sequence of functions with compact support in $\Omega$ and approximate the constant function 1. This allows us to deduce that $1\in H^{1/2}_0(\Omega )$ and thus $S_{N,1/2}(\Omega )=0$.

Proposition 3.4

Let $\Omega$ be a bounded Lipschitz open set of $\mathbb {R}^N$. Then

$$\begin{aligned} S_{N,1/2}(\Omega )=0. \end{aligned}$$

(3.9)

We can now give the proof of our key proposition.

Proof of Proposition 3.2

Since $S_{N,s}(\Omega )>0$ then if follows that

$$\begin{aligned} \liminf _{s\searrow 1/2} S_{N,s}(\Omega )\ge 0. \end{aligned}$$

(3.10)

On the other hand, applying Lemma 3.3 together with Proposition 3.4, we have that

$$\begin{aligned} \limsup _{s\searrow 1/2}S_{N,s}(\Omega )\le S_{N,1/2}(\Omega )=0, \end{aligned}$$

(3.11)

Now, from (3.10) and (3.11) we deduce (3.2), and this ends the proof of Proposition 3.2. $\square$

Having the above key tools in mind, we can now give the proof of Theorem 3.1.

Proof of Theorem 3.1

Let $s\in (1/2,1)$ with s close to 1/2. Then by Proposition 3.2, we have that $S_{N,s}(\Omega )\rightarrow 0$ as $s\searrow 1/2$. Consequently, for s close to 1/2, and since $S_{N,s}(\mathbb {R}^N_+)>0$ for all $s\in (0,1)$ (see e.g. [9, Lemma 2.1]), we deduce that

$$\begin{aligned} 0<S_{N,s}(\Omega )<S_{N,s}(\mathbb {R}^N_+)\quad \text {for all}\quad s\in (1/2,s_0) \end{aligned}$$

(3.12)

for some $s_0\in (1/2,1)$. With the above key inequality, we complete the proof by following closely the argument developed by Frank et al. [12] for the proof of Theorem 4 in there. $\square$

4 The radial problem

In the present section, we consider the existence of minimizers to quotient

$$\begin{aligned} S_{N,s,rad}({\mathcal B},h):=\inf _{u\in C^\infty _{c,rad}({\mathcal B})} \frac{[u]_{H^s({\mathcal B})}^2+\int _{{\mathcal B}} hu^2dx}{\Vert u\Vert _{L^{2^*_s}({\mathcal B})}^2}. \end{aligned}$$

(4.1)

Here and in the following, we consider the class of radial potentials $h\in L^\infty ({\mathcal B})$ such that

$$\begin{aligned} S_{N,s,rad}({\mathcal B},h)>0. \end{aligned}$$

(4.2)

We observe that if $h(x)\equiv -\lambda$ with $\lambda <\lambda _1({\mathcal B})$, the first eigenvalue of $(-\Delta )^s_{{\mathcal B}}$, then (4.2) holds. The aim of this section is to provide situations in which $S_{N,s,rad}({\mathcal B},h)<S_{N,s}(\mathbb {R}^N) .$

Remark 4.1

We observe that if h satisfies (4.2), then if $u\in H^s_0({\mathcal B})$ satisfies, weakly, $(-\Delta )^s_{{\mathcal B}} u+hu=f$ in ${\mathcal B}$ with $f \in L^p({\mathcal B})$, for some $p>\frac{N}{2s}$, then $u\in C({\mathcal B})\cap L^\infty ({\mathcal B})$. This follows from the argument of Proposition 2.3 and the interior regularity.

We start recalling the following result from [12].

Proposition 4.2

([12, Proposition 7]) Let $s\in (1/2,1)$ and $N\ge 4s$. Then

$$\begin{aligned} S_{N,s,rad}({\mathcal B}, 0)<S_{N,s}(\mathbb {R}^N) . \end{aligned}$$

(4.3)

The following result plays a crucial role for the existence theorems.

Proposition 4.3

Let $1/2<s<1$ and $N\ge 2$. Then there is a constant $C=C(N,s)>0$ such that for all $u\in H^s_{0,rad}({\mathcal B})$,

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u)\ge S_{N,s}(\mathbb {R}^N)\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}-C_{{\mathcal B}}\Vert u\Vert ^2_{L^2({\mathcal B})}. \end{aligned}$$

(4.4)

For this, we need the following two lemmas.

Lemma 4.4

For every $\rho \in (0,1)$, there exists $K_{\rho }>0$ with the property that

$$\begin{aligned}{} & {} Q_{N,s,{\mathcal B}}(u)\ge S_{N,s}(\mathbb {R}^N)\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}-K_{\rho }\Vert u\Vert ^2_{L^2({\mathcal B})}\\{} & {} \quad \text {for every}~u\in H^s_{0,rad}({\mathcal B})~\text {with}~\text {supp}~u\subset B_{\rho }. \end{aligned}$$

Proof

Let $u\in H^s_{0,rad}({\mathcal B})$ with $\text {supp}~u\subset B_{\rho }$. We have

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u)= & {} Q_{N,s,\mathbb {R}^N}(u)-\int _{{\mathcal B}}\kappa _{{\mathcal B}}(x)u(x)^2\ dx\ge S_{N,s}(\mathbb {R}^N)\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}\\{} & {} -\int _{{\mathcal B}}\kappa _{{\mathcal B}}(x)u(x)^2\ dx, \end{aligned}$$

with being $\kappa _{{\mathcal B}}$ the killing measure for ${\mathcal B}$ defined as $\kappa _{{\mathcal B}}(x)=c_{N,s}\int _{\mathbb {R}^N\setminus {\mathcal B}}\frac{1}{|x-y|^{N+2s}}\ dy,~x\in {\mathcal B}$. On the other hand, since $\text {supp}u\subset B_{\rho }$, then

$$\begin{aligned} \int _{{\mathcal B}}\kappa _{{\mathcal B}}(x)u(x)^2\ dx= \int _{B_{\rho }}\kappa _{{\mathcal B}}(x)u(x)^2\ dx \end{aligned}$$

and for every $x\in B_\rho$,

$$\begin{aligned} \kappa _{{\mathcal B}}(x)=c_{N,s}\int _{\mathbb {R}^N\setminus {\mathcal B}}\frac{dy}{|x-y|^{N+2s}}\le c_{N,s}\int _{|z|\ge 1-\rho }|z|^{-N-2s}\ dz=a_{N,s}(1-\rho )^{-2s}. \end{aligned}$$

Taking this into account, we find that

$$\begin{aligned} \int _{{\mathcal B}}\kappa _{{\mathcal B}}(x)u(x)^2\ dx \le a_{N,s}(1-\rho )^{-2s}\int _{B_{\rho }}u(x)^2\ dx \le K_{\rho }\Vert u\Vert ^2_{L^2(B_\rho )} \le K_{\rho }\Vert u\Vert ^2_{L^2({\mathcal B})}, \end{aligned}$$

with $K_{\rho }=a_{N,s}(1-\rho )^{-2s}$. From this, we get that

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u)\ge S_{N,s}(\mathbb {R}^N)\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}-K_{\rho }\Vert u\Vert ^2_{L^2({\mathcal B})}, \end{aligned}$$

concluding the proof. $\square$

Lemma 4.5

For every $M, \rho >0$ there exists $C_{\rho ,M}>0$ with

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u)\ge & {} M\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}-C_{\rho ,M}\Vert u\Vert ^2_{L^2({\mathcal B})}\quad \text {for every}~u\in H^s_{0,rad}({\mathcal B})\\{} & {} \quad \text {with}~u\equiv 0~\text {in}~B_\rho . \end{aligned}$$

Proof

We first recall that for $s\in (1/2,1)$, $H^s_0({\mathcal B})={\mathcal H}^s_0({\mathcal B})$. Therefore, for every $u\in H^s_{0,rad}({\mathcal B})\subset H^s_0({\mathcal B})={\mathcal H}^s_0({\mathcal B})$, we have $u\in {\mathcal H}^s_{0,rad}({\mathcal B})$. Thus, combining the fractional version of the Strauss radial lemma (see [6, Lemma 2.5]) and the Hardy inequality (see [8]) we get that

$$\begin{aligned} |u(x)|^2&\le \gamma _{N,s}|x|^{-(N-2s)}Q_{N,s,\mathbb {R}^N}(u) \nonumber \\&=\gamma _{N,s}|x|^{-(N-2s)}\Bigg (Q_{N,s,{\mathcal B}}(u)+\int _{{\mathcal B}}\kappa _{{\mathcal B}}(x)u(x)^2\ dx\Bigg )\nonumber \\&\le \gamma _{N,s}|x|^{-(N-2s)}\Bigg (Q_{N,s,{\mathcal B}}(u)+\gamma _{N,s,{\mathcal B}}\int _{{\mathcal B}}\delta _{{\mathcal B}}(x)^{-2s}u(x)^2\ dx\Bigg )\nonumber \\&\le d_{N,s,{\mathcal B}}|x|^{-(N-2s)}Q_{N,s,{\mathcal B}}(u), \end{aligned}$$

(4.5)

which implies that

$$\begin{aligned} \Vert u\Vert ^2_{L^{\infty }({\mathcal B}\setminus B_{\rho })}\le & {} d_{N,s,{\mathcal B}}\rho ^{-(N-2s)}Q_{N,s,{\mathcal B}}(u)\nonumber \\{} & {} \text {for every}~u\in H^s_{0,rad}({\mathcal B})~\text {with}~u\equiv 0~\text {in}~B_\rho . \end{aligned}$$

(4.6)

Consequently, using interpolation and Young’s inequality with exponents $p=2/\alpha$ and $p'=2/(2-\alpha )$, we find that, for all $M>0$,

$$\begin{aligned} \Vert u\Vert ^2_{L^{2^*_s}({\mathcal B}\setminus B_{\rho })}&\le C\Vert u\Vert ^{\alpha }_{L^2({\mathcal B}\setminus B_{\rho })}\Vert u\Vert ^{2-\alpha }_{L^{\infty }({\mathcal B}\setminus B_{\rho })}\\&\le \frac{1}{Md_{N,s,{\mathcal B}}\rho ^{-(N-2s)}}\Vert u\Vert ^2_{L^{\infty }({\mathcal B}\setminus B_{\rho })}+\frac{C_{\rho ,M}}{M}\Vert u\Vert ^2_{L^{2}({\mathcal B}\setminus B_{\rho })} \end{aligned}$$

with suitable constants $\alpha \in (0,2)$ and $C_{\rho ,M}>0$, and hence

$$\begin{aligned} M\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B}\setminus B_{\rho })}&\le \frac{1}{d_{N,s,{\mathcal B}}\rho ^{-(N-2s)}}\Vert u\Vert ^2_{L^{\infty }({\mathcal B}\setminus B_{\rho })}+C_{\rho ,M}\Vert u\Vert ^2_{L^{2}({\mathcal B}\setminus B_{\rho })}\\&\le Q_{N,s,{\mathcal B}}(u)+C_{\rho ,M}\Vert u\Vert ^2_{L^2({\mathcal B})} \end{aligned}$$

for every $u\in H^s_{0,rad}({\mathcal B})$ with $u\equiv 0$ in $B_\rho$. The claim follows. $\square$

In the following, we give the

Proof of Proposition 4.3

We choose $0<\rho _2<\rho _1<1$. Moreover, let $\chi _1, \chi _2\in C^{\infty }_c(\mathbb {R}^N)$ with $0\le \chi _i\le 1$, $\chi ^2_1+\chi ^2_2\equiv 1$ in ${\mathcal B}$ and $\text {supp}~\chi _1\subset B_{\rho _1}$, $\text {supp}~\chi _2\subset \mathbb {R}^N\setminus \overline{B_{\rho _2}}$. Then we can write $u=\chi ^2_1u+\chi ^2_2u$ in ${\mathcal B}$.

Applying $Q_{N,s,{\mathcal B}}(\cdot )$ to $u=\sum _{i=1}^{2}\chi ^2_iu$, we easily find that

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u)= & {} \sum _{i=1}^{2}Q_{N,s,{\mathcal B}}(\chi _iu)\nonumber \\{} & {} -\frac{c_{N,s}}{2}\sum _{i=1}^{2}\int _{{\mathcal B}}\int _{{\mathcal B}}\frac{(\chi _i(x)-\chi _i(y))^2}{|x-y|^{N+2s}}u(x)u(y)\ dxdy. \end{aligned}$$

(4.7)

By the regularity of $\chi _i$, we observe that there is no singularity in the double integral and therefore it follows from the Schur test that there exists a positive constant $C>0$ such that

$$\begin{aligned} \sum _{i=1}^{2}\int _{{\mathcal B}}\int _{{\mathcal B}}\frac{(\chi _i(x)-\chi _i(y))^2}{|x-y|^{N+2s}}u(x)u(y)\ dxdy\le C\int _{{\mathcal B}}u^2\ dx. \end{aligned}$$

(4.8)

In fact, we can write

$$\begin{aligned} \int _{{\mathcal B}}\int _{{\mathcal B}}\frac{(\chi _i(x)-\chi _i(y))^2}{|x-y|^{N+2s}}u(x)u(y)\ dxdy&\le C\int _{{\mathcal B}}\int _{{\mathcal B}}K(x,y)u(x)u(y)\ dxdy \end{aligned}$$

(4.9)

$$\begin{aligned}&=C\int _{{\mathcal B}}Tu(x)u(x)\ dx \end{aligned}$$

(4.10)

where

$$\begin{aligned} Tu(x)=\int _{{\mathcal B}}K(x,y)u(y)\ dy\quad \quad \text {with}\quad \quad K(x,y)=|x-y|^{2-N-2s}. \end{aligned}$$

Moreover, by H$\ddot{\text {o}}$lder inequality,

$$\begin{aligned} \int _{{\mathcal B}}Tu(x)u(x)\ dx\le \Vert Tu\Vert _{L^2({\mathcal B})}\Vert u\Vert _{L^2({\mathcal B})}. \end{aligned}$$

(4.11)

Now, the Schur test implies that there is $C>0$ such that

$$\begin{aligned} \Vert Tu\Vert _{L^2({\mathcal B})}\le C\Vert u\Vert _{L^2({\mathcal B})}. \end{aligned}$$

(4.12)

Therefore, inequality (4.8) follows by combining (4.9), (4.11) and (4.12).

On the other hand, by Lemmas 4.4 and 4.5, there exists a positive constant $C>0$, depending on $\rho _1$ and $\rho _2$ with the property that

$$\begin{aligned} Q_{N,s,{\mathcal B}}(\chi _iu)\ge S_{N,s}(\mathbb {R}^N)\Vert \chi _iu\Vert ^2_{L^{2^*_s}({\mathcal B})}-C\Vert \chi _iu\Vert ^2_{L^2({\mathcal B})}. \end{aligned}$$

(4.13)

Plugging (4.8) and (4.13) into (4.7), we find that

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u)\ge S_{N,s}(\mathbb {R}^N)\sum _{i=1}^{2}\Vert \chi _iu\Vert ^2_{L^{2^*_s}({\mathcal B})}-C\sum _{i=1}^{2}\Vert \chi _iu\Vert ^2_{L^2({\mathcal B})}. \end{aligned}$$

(4.14)

Next, since $\sum _{i=1}^{2}\chi ^2_i=1$, we have

$$\begin{aligned} \sum _{i=1}^{2}\Vert \chi _iu\Vert ^2_{L^{2^*_s}({\mathcal B})}&=\sum _{i=1}^{2}\Big \Vert \chi ^2_iu^2\Big \Vert _{L^{\frac{N}{N-2s}}({\mathcal B})}\ge \Bigg \Vert \sum _{i=1}^{2}\chi ^2_iu^2\Bigg \Vert _{L^{\frac{N}{N-2s}}({\mathcal B})}\\ {}&=\Vert u^2\Vert _{L^{\frac{N}{N-2s}}({\mathcal B})}=\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}. \end{aligned}$$

Using this in (4.13), it follows that

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u)\ge S_{N,s}(\mathbb {R}^N)\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}-C\Vert u\Vert ^2_{L^2({\mathcal B})}, \end{aligned}$$

completing the proof. $\square$

4.1 The case $2s<N<4s$

We now let G(x, y) be the Green function of $(-\Delta )^s+h$, with zero exterior Dirichlet boundary data. Letting $G(x)=G(x,0)$, we have that

$$\begin{aligned} {\left\{ \begin{array}{ll} (-\Delta )^sG(x)+h(x)G(x)=\delta _0(x)&{}\hbox { in}\ {\mathcal B}\\ ~~~~~~~~~~~~~~~~~~~~~~~~G(x)=0&{} \hbox { in}\ \mathbb {R}^N\setminus {\mathcal B}, \end{array}\right. } \end{aligned}$$

(4.15)

where $\delta _0$ is the Dirac mass at 0 and $h\in L^{\infty }({\mathcal B})$ a radial function. We recall that G is a radial function. In fact this follows from the construction and uniqueness of Green function. We let ${\mathcal R}(x)=t_{N,s} |x|^{2s-N}$ be the fundamental solution of $(-\Delta )^s$ on $\mathbb {R}^N$. It satisfies

$$\begin{aligned} (-\Delta )^s{\mathcal R}(x)=\delta _0(x), \end{aligned}$$

(4.16)

where $t_{N,s}:=\pi ^{-\frac{N}{2}}2^{-s}\frac{\Gamma ((N-s)/2)}{\Gamma (s/2)}$. We now define $\overline{\textbf{k}}\in L^1({\mathcal B})$, by

$$\begin{aligned} \overline{\textbf{k}}(x):=G(x)- {\mathcal R}(x). \end{aligned}$$

(4.17)

It then follows, from (4.15), that

$$\begin{aligned} (-\Delta )^s\overline{\textbf{k}}(x)+h(x)\overline{\textbf{k}}(x)=-h(x){\mathcal R}(x). \end{aligned}$$

(4.18)

Since $N<4s$, we have that $\overline{\textbf{k}}\in L^2({\mathcal B})$ and $h{\mathcal R}\in L^p({\mathcal B})\cap L^2({\mathcal B})$, for some $p>\frac{N}{2s}$. Therefore, by regularity theory, $\overline{\textbf{k}}\in C(\overline{{\mathcal B}}) .$ Recall that $\overline{\textbf{k}}(y)$ is the mass of ${\mathcal B}$ associated to the operator $\mathcal {L}_{\mathbb {R}^N}:=(-\Delta )^s+ h(x)$. We remark that if $\chi \in C^\infty _c({\mathcal B})$, with $\chi =1$ in a neighborhood of 0, then letting

$$\begin{aligned} \textbf{k}(x):=G(x)- \chi (x){\mathcal R}(x), \end{aligned}$$

then, by continuity, $\textbf{k}(y)=\overline{\textbf{k}}(y)$, for all $y\in {\mathcal B}$. This follows from the fact that $(-\Delta )^s\textbf{k}+h \textbf{k}\in L^p({\mathcal B})$, for some $p>\frac{N}{2s}$ and thus $\textbf{k}\in C({\mathcal B})$.

Remark 4.6

It would be interesting to find potential h for which $\textbf{k}(0)>0$.

First, for $\varepsilon >0$ we set

$$\begin{aligned} u_{\varepsilon }(x)=\gamma _0\Big (\frac{\varepsilon }{\varepsilon ^2+ |x|^2}\Big )^{\frac{N-2s}{2}}, \end{aligned}$$

where $\gamma _0$ is a positive constant (independent of $\varepsilon$) such that $\Vert u_{\varepsilon }\Vert _{L^{2^*_s}(\mathbb {R}^N)}=1$. It is known that $u_{\varepsilon }$ satisfies the Euler-Lagrange equation

$$\begin{aligned} (-\Delta )^su_{\varepsilon }=S_{N,s} u^{2^*_s-1}_{\varepsilon }\quad \text {in}\quad \mathbb {R}^N. \end{aligned}$$

(4.19)

Our next result shows that in low dimension $N<4s$, the positive mass implies existence of minimizers.

Lemma 4.7

Suppose that $2s<N<4s$. Suppose that $\textbf{k}(0)>0$. Then

$$\begin{aligned} S_{N,s,rad}({\mathcal B},h)< S_{N,s} :=S_{N,s}(\mathbb {R}^N) . \end{aligned}$$

(4.20)

Proof

For $r\in (0,1/4)$, we let $\eta \in C^\infty _c(B_{2r})$ be radial, with $\eta =1$ on $B_{r}$. We define the test function $v_\varepsilon \in H^s_{0,rad}({\mathcal B})$ given by

$$\begin{aligned} v_\varepsilon (x)&=\eta (x) u_\varepsilon (x)+\varepsilon ^{\frac{N-2s}{2}}\frac{\gamma _0}{t_{N,s}} \left( G(x)- \eta (x){\mathcal R}(x) \right) \nonumber \\&=\eta (x) u_\varepsilon (x)+\varepsilon ^{\frac{N-2s}{2}}\frac{\gamma _0}{t_{N,s}} \textbf{k}(x) . \end{aligned}$$

(4.21)

We define $W_\varepsilon :=\eta u_\varepsilon -\varepsilon ^{\frac{N-2s}{2}}\frac{\gamma _0}{t_{N,s}} \eta {\mathcal R}$ and $a_s:=\frac{\gamma _0}{t_{N,s}}$.

Note that $\varepsilon ^{-\frac{N-2s}{2}} W_\varepsilon \rightarrow 0 \in C_{loc}(\mathbb {R}^N\setminus \{0\})\cap L^1({\mathcal B})$ and $|\varepsilon ^{-\frac{N-2s}{2}} u_\varepsilon (x)|\le \gamma _0 |x|^{2s-N}$. Hence, since $N<4s$, we deduce that $|x|^{2(2s-N)}\in L^1_{loc}(\mathbb {R}^N)$ and thus by the dominated convergence theorem,

$$\begin{aligned} \int _{{\mathcal B}} u_\varepsilon (x) h(x)W_\varepsilon (x)\, dx=o(\varepsilon ^{N-2s}). \end{aligned}$$

(4.22)

We then have

$$\begin{aligned}&{[}v_\varepsilon ]^{2}_{H^s({\mathcal B})}+ \int _{{\mathcal B}}hv_\varepsilon ^2\, dx\le [v_\varepsilon ]^{2}_{H^s(\mathbb {R}^N)}+ \int _{{\mathcal B}}hv_\varepsilon ^2\, dx= \int _{{\mathcal B}} v_\varepsilon (x) \mathcal {L}_{\mathbb {R}^N} v_\varepsilon (x)\, dx\\&\quad \le \varepsilon ^{\frac{N-2s}{2}}a_s \int _{{\mathcal B}} v_\varepsilon (x) \mathcal {L}_{\mathbb {R}^N} G(x) \, dx + \int _{{\mathcal B}} v_\varepsilon (x) \mathcal {L}_{\mathbb {R}^N} W_\varepsilon (x)\, dx \\&\quad \le \varepsilon ^{\frac{N-2s}{2}}a_s u_\varepsilon (0)+ \varepsilon ^{{N-2s}}a_s^2 \textbf{k}(0) + \int _{{\mathcal B}}\eta u_\varepsilon (x) (-\Delta )^sW_\varepsilon (x)\, dx\\&\qquad + \varepsilon ^{\frac{N-2s}{2}} a_s\int _{{\mathcal B}}\textbf{k}(x) \mathcal {L}_{\mathbb {R}^N} W_\varepsilon (x)\, dx+ o(\varepsilon ^{N-2s})\\&\quad \le \varepsilon ^{\frac{N-2s}{2}} a_s u_\varepsilon (0)+ \varepsilon ^{{N-2s}}a_s^2 \textbf{k}(0) + \int _{{\mathcal B}}\eta u_\varepsilon (x) (-\Delta )^s(\eta u_\varepsilon )(x) \, dx\\&\qquad - \varepsilon ^{\frac{N-2s}{2}} a_s \int _{{\mathcal B}}\eta u_\varepsilon (x) (-\Delta )^s( \eta {\mathcal R})(x)\, dx \\&\qquad + \varepsilon ^{\frac{N-2s}{2}} a_s\int _{{\mathcal B}}\textbf{k}(x) \mathcal {L}_{\mathbb {R}^N} W_\varepsilon (x)\, dx + o(\varepsilon ^{N-2s}) \\&\quad \le \varepsilon ^{\frac{N-2s}{2}} a_s u_\varepsilon (0)+ \varepsilon ^{{N-2s}}a_s^2 \textbf{k}(0) \\&\qquad + \int _{\mathbb {R}^N}\eta u_\varepsilon (x) (-\Delta )^s(\eta u_\varepsilon )(x) \, dx - \varepsilon ^{\frac{N-2s}{2}} a_s \int _{\mathbb {R}^N}\eta u_\varepsilon (x) (-\Delta )^s( \eta {\mathcal R})(x)\, dx \\&\qquad + \varepsilon ^{\frac{N-2s}{2}} a_s\int _{\mathbb {R}^N}\textbf{k}(x) \mathcal {L}_{\mathbb {R}^N} W_\varepsilon (x)\, dx+ o(\varepsilon ^{N-2s}). \end{aligned}$$

Letting $\overline{W}_\varepsilon =u_\varepsilon - \varepsilon ^{\frac{N-2s}{2}} a_s {\mathcal R}(x)$, since $N<4s$, we have that

$$\begin{aligned} \varepsilon ^{-\frac{N-2s}{2}} \overline{W}_\varepsilon \rightarrow 0\qquad \text { in } C^1_{loc}(\mathbb {R}^N\setminus \{0\})\cap {\mathcal {L}}^1_s\cap L^2_{loc}(\mathbb {R}^N). \end{aligned}$$

(4.23)

Therefore, using that $(-\Delta )^s{\mathcal R}=\delta _0$ and $(-\Delta )^su_\varepsilon =S_{N,s} u_\varepsilon ^{2^*_s-1}$, we get

$$\begin{aligned}&\varepsilon ^{\frac{N-2s}{2}} a_s u_\varepsilon (0)+ \int _{\mathbb {R}^N}\eta u_\varepsilon (x) (-\Delta )^s(\eta u_\varepsilon )(x) \, dx\\&\qquad - \varepsilon ^{\frac{N-2s}{2}} a_s \int _{\mathbb {R}^N}\eta u_\varepsilon (x) (-\Delta )^s( \eta {\mathcal R})(x)\, dx\\&\quad =\varepsilon ^{\frac{N-2s}{2}} a_s u_\varepsilon (0) + \int _{\mathbb {R}^N}\eta ^2 u_\varepsilon (x) (-\Delta )^su_\varepsilon (x) \, dx \\&\qquad - \varepsilon ^{\frac{N-2s}{2}} a_s \int _{\mathbb {R}^N}\eta u_\varepsilon (x) (-\Delta )^s{\mathcal R}(x) \, dx\\&\qquad + \int _{\mathbb {R}^N}\eta u_\varepsilon (x) \overline{W}_\varepsilon (x) (-\Delta )^s\eta (x) \, dx - \int _{B_{2r}}\eta u_\varepsilon (x) J_\varepsilon (x)dx\\&\quad =S_{N,s}\int _{\mathbb {R}^N}\eta ^2 u_\varepsilon ^{2^*_s}+ \int _{\mathbb {R}^N}\eta u_\varepsilon (x) \overline{W}_\varepsilon (x) (-\Delta )^s\eta (x) \, dx - \int _{B_{2r}}\eta u_\varepsilon (x)J_\varepsilon (x) dx\\&\quad =S_{N,s}\int _{\mathbb {R}^N}\eta ^2 u_\varepsilon ^{2^*_s}+ o( \varepsilon ^{N-2s}) -\int _{B_{2r}}\eta u_\varepsilon (x) J_\varepsilon (x) dx, \end{aligned}$$

where $J_\varepsilon (x):={c_{N,s}} \int _{\mathbb {R}^{N}}\frac{({\overline{W}}_\varepsilon (x)-{\overline{W}}_\varepsilon (y))( \eta (x)-\eta (y))}{|x-y|^{N+2s}}\, dy$. To estimate $J_\varepsilon$, we consider first $x\in B_{r/2}$ and thus

$$\begin{aligned} J_\varepsilon (x)= c_{N,s}\int _{|y|>r}\frac{({\overline{W}}_\varepsilon (x)-{\overline{W}}_\varepsilon (y))( \eta (x)-\eta (y))}{|x-y|^{N+2s}}\, dy=o( \varepsilon ^{\frac{N-2s}{2}}) O( |x|^{\frac{N-2s}{2}}) . \end{aligned}$$

If now $|x|\ge r/2$, we estimate

$$\begin{aligned} | J_\varepsilon (x)|&\le c_{N,s}\int _{|y|<{r/4}}\frac{|({\overline{W}}_\varepsilon (x)-{\overline{W}}_\varepsilon (y))( \eta (x)-\eta (y))|}{|x-y|^{N+2s}}\, dy\\&\quad +c_{N,s} \int _{|y|>{r/4}}\frac{|({\overline{W}}_\varepsilon (x)-{\overline{W}}_\varepsilon (y))( \eta (x)-\eta (y))|}{|x-y|^{N+2s}}\, dy\\&\le o( \varepsilon ^{\frac{N-2s}{2}})+ \Vert \nabla \eta \Vert _{L^\infty (\mathbb {R}^N)} \int _{4r>|y|>{r/4}}\frac{\sup _{t\in [0,1]}|\nabla {\overline{W}}_\varepsilon (\gamma _{x,y}(t))| |\gamma '_{x,y}(t)| }{|x-y|^{N+2s-1}}\, dy\\&=o( \varepsilon ^{\frac{N-2s}{2}}) \end{aligned}$$

where $\gamma _{x,y}:[0,1]\rightarrow B_{r/2}\setminus B_{r/4}$ is the $C^1$ shortest curve satisfying $\gamma _{x,y}(0)=x$, $\gamma _{x,y}(1)=y$ and $\sup _{t\in [0,1]} |\gamma '_{x,y}(t)|\le C |x-y|$.

Since $N<4s$, by (4.18) and (4.23), we have

$$\begin{aligned}&\left| \int _{\mathbb {R}^N}\textbf{k}(x) \mathcal {L}_{\mathbb {R}^N} W_\varepsilon (x)\, dx \right| \le \left| \int _{B_{2r}}|\mathcal {L}_{\mathbb {R}^N}\textbf{k}(x)| |W_\varepsilon (x)|\, dx \right| =o( \varepsilon ^{\frac{N-2s}{2}} ) . \end{aligned}$$

We thus conclude that

$$\begin{aligned}{}[v_\varepsilon ]^{2}_{H^s({\mathcal B})}+ \int _{{\mathcal B}} h v_\varepsilon ^2\, dx&\le S_{N,s}\int _{\mathbb {R}^N}\eta ^2 u_\varepsilon ^{2^*_s}\nonumber \\&\qquad + \varepsilon ^{{N-2s}}a_s^2 \textbf{k}(0)+o( \varepsilon ^{{N-2s}})+ O( \varepsilon ^{{N-2s}})o_r(1) \nonumber \\&\le S_{N,s}+ \varepsilon ^{{N-2s}}a_s^2 \textbf{k}(0)+o( \varepsilon ^{{N-2s}})+ O(r^{4s-N} \varepsilon ^{{N-2s}}). \end{aligned}$$

(4.24)

Since $2^*_s>2$, there exists a positive constant C(N, s) such that

$$\begin{aligned} ||a+b|^{2^*_s}-|a|^{2^*_s}-2^*_s ab |a|^{2^*_s-2}| \le C(N,s) \left( |a|^{2^*_s-2} b^2+|b|^{2^*_s}\right) \quad \text {for all } a,b \in \mathbb {R}. \end{aligned}$$

As a consequence, with $a=\eta (x) u_\varepsilon (x)$ and $b=\varepsilon ^{\frac{N-2s}{2}} a_s\textbf{k}(x)$, we obtain

$$\begin{aligned}&\int _{{\mathcal B}}v_\varepsilon ^{2^*_s}- \int _{\mathbb {R}^N}(\eta u_\varepsilon )^{2^*_s}=2^*_s \varepsilon ^{\frac{N-2s}{2}} a_s \int _{{\mathcal B}} (\eta u_\varepsilon )^{2^*_s-1}\textbf{k}(x)\, dx\\&\quad + o( \varepsilon ^{{N-2s}})+ O\left( \varepsilon ^{N-2s} \int _{\mathbb {R}^N}| \eta (x) u_\varepsilon (x)|^{2^*_s-2} \textbf{k}^2(x)dx \right) \\&\quad =2^*_s \varepsilon ^{\frac{N-2s}{2}} \frac{a_s}{S_{N,s}} \int _{{\mathcal B}} \eta ^{2^*_s-1}\textbf{k}(x) (-\Delta )^su_\varepsilon \, dx+ o( \varepsilon ^{{N-2s}})\\&\qquad +\varepsilon ^{{N-2s}} O\left( \Vert \eta u_\varepsilon \Vert _{L^{2^*_s}(B_{2r})}^{{2^*_s-2}} \Vert \textbf{k}\Vert _{L^{2^*_s}(B_{2r})}^{{2}} \right) . \\&\quad =2^*_s \varepsilon ^{\frac{N-2s}{2}} \frac{a_s}{S_{N,s}} \int _{{\mathcal B}} \textbf{k}(x) (-\Delta )^s{\overline{W}}_\varepsilon \, dx\\&\qquad + 2^*_s \varepsilon ^{\frac{N-2s}{2}} \frac{a_s}{S_{N,s}} \int _{{\mathcal B}}(\eta ^{2^*_s-1}-1)\textbf{k}(x) (-\Delta )^s{\overline{W}}_\varepsilon \, dx\\&\qquad + 2^*_s \varepsilon ^{{N-2s}} \frac{a_s^2}{S_{N,s}}\textbf{k}(0)+ o( \varepsilon ^{{N-2s}})+ O(\varepsilon ^{{N-2s}} r^{N-2s} ) \\&\quad =2^*_s \varepsilon ^{\frac{N-2s}{2}} \frac{a_s}{S_{N,s}} \int _{{\mathcal B}} {\overline{W}}_\varepsilon (x) \mathcal {L}_{\mathbb {R}^N}\textbf{k}(x) \, dx\\&\qquad + 2^*_s \varepsilon ^{\frac{N-2s}{2}} \frac{a_s}{S_{N,s}} \int _{{\mathcal B}}(\eta ^{2^*_s-1}-1)\textbf{k}(x) (-\Delta )^s{\overline{W}}_\varepsilon \, dx\\&\qquad + 2^*_s \varepsilon ^{{N-2s}} \frac{a_s^2}{S_{N,s}}\textbf{k}(0)+ o( \varepsilon ^{{N-2s}}) + O(\varepsilon ^{{N-2s}} r^{N-2s} )\\&\quad =2^*_s \varepsilon ^{{N-2s}} \frac{a_s^2}{S_{N,s}}O\left( \int _{|x|<2r} |x|^{2s-N}\left( \frac{1}{(\varepsilon ^2+|x|^2)^{\frac{N-2s}{2}}} -\frac{1}{|x|^{N-2s}} \right) \, dx\right) \\&\qquad + 2^*_s \varepsilon ^{\frac{N-2s}{2}} \frac{a_s}{S_{N,s}} \int _{{\mathcal B}}(\eta ^{2^*_s-1}-1)\textbf{k}(x) (-\Delta )^s\overline{W}_\varepsilon \, dx \\&\qquad + 2^*_s \varepsilon ^{{N-2s}} \frac{a_s^2}{S_{N,s}}\textbf{k}(0)+ o( \varepsilon ^{{N-2s}})+O(\varepsilon ^{{N-2s}} ) o_r(1). \end{aligned}$$

We estimate

$$\begin{aligned}&\int _{{\mathcal B}}(\eta ^{2^*_s-1}-1)\textbf{k}(x) (-\Delta )^s\overline{W}_\varepsilon \, dx= \int _{{\mathcal B}}(\eta ^{2^*_s-1}-1)\textbf{k}(x) (-\Delta )^s(\eta _{r/4} \overline{W}_\varepsilon )\, dx+o( \varepsilon ^{\frac{N-2s}{2}} ) \\&\quad =c_{N,s}\int _{|x|\ge r}(1-\eta ^{2^*_s-1}(x))\textbf{k}(x)\int _{ |y|<r/2}\frac{\eta _{r/4} (y){\overline{W}}_\varepsilon (y)\, dy}{|x-y|^{N+2s}}\, dy +o( \varepsilon ^{\frac{N-2s}{2}} )=o( \varepsilon ^{\frac{N-2s}{2}} ). \end{aligned}$$

Here, from the definition of $\eta$, we define $\eta _{r/4}\in C^{\infty }_c(B_{r/2})$ with $\eta _{r/4}=1$ on $B_{r/4}$. From the above estimates, we then obtain

$$\begin{aligned} \int _{{\mathcal B}}v_\varepsilon ^{2^*_s}&= \int _{\mathbb {R}^N}(\eta u_\varepsilon )^{2^*_s} + 2^*_s \varepsilon ^{{N-2s}} \frac{a_s^2}{S_{N,s}}\textbf{k}(0)+ o( \varepsilon ^{{N-2s}})+O(\varepsilon ^{{N-2s}} ) o_r(1)\\&=1+ 2^*_s \varepsilon ^{{N-2s}} \frac{a_s^2}{S_{N,s}}\textbf{k}(0)+ o( \varepsilon ^{{N-2s}})+O(\varepsilon ^{{N-2s}} ) o_r(1). \end{aligned}$$

Combining this with (4.24), we finally get

$$\begin{aligned} \frac{ [v_\varepsilon ]_{H^s({\mathcal B})}^2+ \int _{{\mathcal B}}hv_\varepsilon ^2\, dx}{\Vert v_\varepsilon \Vert _{L^{2^*_s}({\mathcal B})}^2}\le S_{N,s}- \varepsilon ^{{N-2s}} {a_s^2} \textbf{k}(0)+ o( \varepsilon ^{{N-2s}})+O(\varepsilon ^{{N-2s}} ) o_r(1). \end{aligned}$$

This finishes the proof. $\square$

5 Existence of radial minimizers

The goal of this section is to investigate the existence of a radial solution of problem (1.2) in the case when $\Omega ={\mathcal B}$ is the unit ball of $\mathbb {R}^N$, $N>2s$. More precisely, we aim to analyze the attainability of the following radial critical level

$$\begin{aligned} S_{N,s,rad}({\mathcal B},h)=\inf _{\begin{array}{c} u\in H^s_{0,rad}({\mathcal B})\\ u\ne 0 \end{array}}\frac{Q_{N,s,{\mathcal B}}(u)+\int _{{\mathcal B}}hu^2\, dx}{\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}}. \end{aligned}$$

(5.1)

To this end, we make use of the method of missing mass as in [12]. The idea is to prove that a minimizing sequence for $S_{N,s,rad}({\mathcal B},h)$ does not concentrate at the origin. For that, we will exploit inequalities (4.3) and (4.20) respectively for high $(N\ge 4s)$ and low $(2s<N<4s)$ dimensions.

For the reader’s convenience, we restate the main result of this subsection in the following.

Theorem 5.1

Let $s\in (1/2,1)$, $N>2s$ and $h\in L^\infty ({\mathcal B})$ be a radial function. Suppose that $0<S_{N,s,rad}({\mathcal B},h)<S_{N,s}(\mathbb {R}^N).$ Then any minimizing sequence for $S_{N,s,rad}({\mathcal B},h)$, normalized in $H^s_{0,rad}({\mathcal B})$ is relatively compact in $H^s_{0,rad}({\mathcal B})$. In particular, the infimum is achieved.

To prove the above theorem, we first collect some useful results. Let’s introduce

$$\begin{aligned} S^*_{N,s,rad}({\mathcal B}):= & {} \inf \Big \{\liminf _{k\rightarrow \infty }\Vert u_k\Vert ^{-2}_{L^{2^*_s}({\mathcal B})}:Q_{N,s,{\mathcal B}}(u_k)\nonumber \\= & {} 1,~u_k\rightharpoonup 0~\text {in}~H^s_{0,rad}({\mathcal B})\Big \}. \end{aligned}$$

(5.2)

As we will see in the sequel, the infimum $S^*_{N,s,rad}({\mathcal B})$ is crucial in showing that normalized minimizing sequences that weakly converge to zero in $H^s_{0,rad}({\mathcal B})$ move away from the origin in such a way that the concentration at the origin is excluded.

We have the following interesting one-sided inequality.

Proposition 5.2

Let $1/2<s<1$ and $N\ge 2$. Then

$$\begin{aligned} S^*_{N,s,rad}({\mathcal B})\ge S_{N,s}(\mathbb {R}^N). \end{aligned}$$

(5.3)

Proof

Let $(u_k)\subset H^s_{0,rad}({\mathcal B})$ with $Q_{N,s,{\mathcal B}}(u_k)=1$ and $u_k\rightharpoonup 0$ in $H^s_{0,rad}({\mathcal B})$. Then by Proposition 4.3 there is $C_{{\mathcal B}}>0$ such that

$$\begin{aligned} Q_{N,s,{\mathcal B}}(u_k)\ge S_{N,s}(\mathbb {R}^N)\Vert u_k\Vert ^2_{L^{2^*_s}({\mathcal B})}-C_{{\mathcal B}}\Vert u_k\Vert ^2_{L^2({\mathcal B})}. \end{aligned}$$

By the compact embedding $H^s_{0,rad}({\mathcal B})\hookrightarrow L^2({\mathcal B})$, we have $u_k\rightarrow 0$ in $L^2({\mathcal B})$. Using this and by passing to the limit in the above inequality, we find that

$$\begin{aligned} 1\ge S_{N,s}(\mathbb {R}^N)\limsup _{k\rightarrow \infty }\Vert u_k\Vert ^2_{L^{2^*_s}({\mathcal B})}, \end{aligned}$$

that is,

$$\begin{aligned} \liminf _{k\rightarrow \infty }\Vert u_k\Vert ^{-2}_{L^{2^*_s}({\mathcal B})}\ge S_{N,s}(\mathbb {R}^N). \end{aligned}$$

From the above inequality, we conclude the proof. $\square$

Having collected the above results, we are ready to prove our main result.

Proof of Theorem 5.1

Let $(u_k)$ be a minimizing sequence for $S_{N,s,rad}({\mathcal B},h)$, which is normalized in $H^s_{0,rad}({\mathcal B})$. Then after passing to a subsequence, there is $u\in H^s_{0,rad}({\mathcal B})$ such that

$$\begin{aligned} \begin{aligned}&u_k\rightharpoonup u\quad \text {weakly in}~~H^s_{0,rad}({\mathcal B})\\&u_k\rightarrow u\quad \text {strongly in}~~L^2({\mathcal B})\\&u_k\rightarrow u\quad \text {a.e. in}~~{\mathcal B}. \end{aligned} \end{aligned}$$

(5.4)

Now, by setting $w_k=u_k-u$, it follows that $w_k\rightharpoonup 0$ weakly in $H^s_{0,rad}({\mathcal B})$. Using this, we have that

$$\begin{aligned} 1= & {} Q_{N,s,{\mathcal B},h}(u_k):=Q_{N,s,{\mathcal B}}(u_k)+\int _{{\mathcal B}}hu_k^2\,dx\nonumber \\= & {} Q_{N,s,{\mathcal B},h}(u)+Q_{N,s,{\mathcal B}}(w_k)+o(1), \end{aligned}$$

(5.5)

where $Q_{N,s,{\mathcal B},h}(u):=Q_{N,s,{\mathcal B}}(u)+\int _{{\mathcal B}}hu^2\,dx$. From the above identities, we see that $Q_{N,s,{\mathcal B}}(w_k)$ converges, say, to $R_1$, which satisfies according to the above equality,

$$\begin{aligned} 1=Q_{N,s,{\mathcal B},h}(u)+R_1. \end{aligned}$$

(5.6)

Moreover, using that $u_k\rightarrow u$ a.e. in ${\mathcal B}$ and the Brezis-Lieb lemma [3], we get that

$$\begin{aligned} S_{N,s,rad}({\mathcal B},h)^{-\frac{N}{N-2s}}+o(1)= & {} \Vert u_k\Vert ^{\frac{2N}{N-2s}}_{L^{2^*_s}({\mathcal B})}\nonumber \\= & {} \Vert u\Vert ^{\frac{2N}{N-2s}}_{L^{2^*_s}({\mathcal B})}+\Vert w_k\Vert ^{\frac{2N}{N-2s}}_{L^{2^*_s}({\mathcal B})}+o(1), \end{aligned}$$

(5.7)

from which we deduce that $\int _{{\mathcal B}}|w_k|^{\frac{2N}{N-2s}}\ dx$ converges, say, to $R_2$ satisfying

$$\begin{aligned} S_{N,s,rad}({\mathcal B},h)^{-\frac{N}{N-2s}}=\Vert u\Vert ^{\frac{2N}{N-2s}}_{L^{2^*_s}({\mathcal B})}+R_2. \end{aligned}$$

(5.8)

Now by Proposition 5.2 we easily see that

$$\begin{aligned} R_1\ge S_{N,s}(\mathbb {R}^N)R_2^{\frac{N-2s}{N}}. \end{aligned}$$

(5.9)

Indeed, (5.9) follows immediately if $R_2=0$. Otherwise, if $R_2>0$, then it suffices to use ${\tilde{w}}_k:=w_k/Q_{N,s,{\mathcal B}}(w_k)^{1/2}$ in the definition of $S^*_{N,s,rad}({\mathcal B})$ since ${\tilde{w}}_k\rightharpoonup 0$ weakly in $H^s_{0,rad}({\mathcal B})$ and $Q_{N,s,{\mathcal B}}({\tilde{w}}_k)=1$ as well.

From (5.6), (5.8), (5.9) and by using the elementary inequality¹

$$\begin{aligned} (a-b)^{\alpha }\ge a^{\alpha }-b^{\alpha }\quad \text {for}~0\le \alpha \le 1,~a\ge b\ge 0 \end{aligned}$$

(5.10)

with $\alpha =(N-2s)/N$, we find that

$$\begin{aligned} 1&=Q_{N,s,{\mathcal B},h}(u)+R_1\\&\ge Q_{N,s,{\mathcal B},h}(u)+S_{N,s}(\mathbb {R}^N)R_2^{\frac{N-2s}{N}}\\&=Q_{N,s,{\mathcal B},h}(u)+(S_{N,s}(\mathbb {R}^N)-S_{N,s,rad}({\mathcal B},h))R^{\frac{N-2s}{N}}_2\\&\quad +S_{N,s,rad}({\mathcal B})\Big (S_{N,s,rad}({\mathcal B},h)^{-\frac{N}{N-2s}}-\Vert u\Vert ^{\frac{2N}{N-2s}}_{L^{2^*_s}({\mathcal B})}\Big )^{\frac{N-2s}{N}}\\&\ge Q_{N,s,{\mathcal B},h}(u)+(S_{N,s}(\mathbb {R}^N)-S_{N,s,rad}({\mathcal B},h))R^{\frac{N-2s}{N}}_2\\&\quad +S_{N,s,rad}({\mathcal B},h)\Big (S_{N,s,rad}({\mathcal B},h)^{-1}-\Vert u\Vert ^{2}_{L^{2^*_s}({\mathcal B})}\Big )\\&=Q_{N,s,{\mathcal B},h}(u)+(S_{N,s}(\mathbb {R}^N)-S_{N,s,rad}({\mathcal B},h))R^{\frac{N-2s}{N}}_2\\&\quad +1-S_{N,s,rad}({\mathcal B},h)\Vert u\Vert ^{2}_{L^{2^*_s}({\mathcal B})}. \end{aligned}$$

Thus,

$$\begin{aligned}&Q_{N,s,{\mathcal B},h}(u)-S_{N,s,rad}({\mathcal B},h)\Vert u\Vert ^{2}_{L^{2^*_s}({\mathcal B})}+(S_{N,s}(\mathbb {R}^N)\nonumber \\&\quad -S_{N,s,rad}({\mathcal B},h))R^{\frac{N-2s}{N}}_2\le 0. \end{aligned}$$

(5.11)

Since $Q_{N,s,{\mathcal B},h}(u)\ge S_{N,s,rad}({\mathcal B},h)\Vert u\Vert ^{2}_{L^{2^*_s}({\mathcal B})}$ and $S_{N,s}(\mathbb {R}^N)>S_{N,s,rad}({\mathcal B},h)$ by assumption, it follows from (5.11) that $R_2=0$ which implies that $u\not \equiv 0$ thanks to (5.8). Therefore,

$$\begin{aligned} Q_{N,s,{\mathcal B},h}(u)\le S_{N,s,rad}({\mathcal B},h)\Vert u\Vert ^2_{L^{2^*_s}({\mathcal B})}, \end{aligned}$$

which implies that u is an optimizer. Therefore, instead of the inequality (5.9), we have equality, yielding $R_1=0$. This implies that $Q_{N,s,{\mathcal B},h}(u)=1$ and from this, we conclude that $(u_k)$ converges strongly in $H^s_{0,rad}({\mathcal B})$. The proof is therefore finished. $\square$

Proof of Theorem 1.2 and Theorem 1.3 (completed)

The proof of Theorem 1.2 and Theorem 1.3 are immediate consequences of Theorem 5.1, Lemma 4.7 and Proposition 4.2. $\square$

Appendix

In this section, we prove that the constant function 1 belongs to $H^s_0(\Omega )$ for $s\in (0,1/2]$. By Sobolev embedding, it is enough to treat the case $s=1/2$.

For every $k\in \mathbb {N}$, we define $\chi _k\in C^{0,1}(\mathbb {R}_+)$ by

$$\begin{aligned} \chi _k(t)=\left\{ \begin{aligned}&0\quad \quad \quad \quad \quad \text {if}\quad t\le \frac{1}{k^2}, \\&\frac{\log k^2t}{|\log 1/k|}\quad ~~\text {if}\quad \frac{1}{k^2}\le t\le \frac{1}{k}, \\&1\quad \quad \quad \quad \quad \text {if}\quad t\ge \frac{1}{k}. \end{aligned} \right. \end{aligned}$$

(A.1)

We wish now to approximate the constant function 1 with respect to the $H^{1/2}(\Omega )$-norm. The general strategy is to build an approximation sequence with $\chi _k$ together with a partition of unity. Before going further in our analysis, we need first of all a one-dimensional approximation argument.

Lemma A.1

We have

$$\begin{aligned} \chi _k\rightarrow 1\quad \quad \text {in}~H^{1/2}(\mathbb {R}_+)~~\text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.2)

Proof

Clearly, by definition $\chi _k\rightarrow 1$ a.e. in $\mathbb {R}_+$. The goal is to show that

$$\begin{aligned} \Vert \chi _k-1\Vert _{H^{1/2}(\mathbb {R}_+)}\rightarrow 0\quad \quad \text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.3)

We start by proving that

$$\begin{aligned} \Vert \chi _k-1\Vert _{L^2(\mathbb {R}_+)}\rightarrow 0\quad \quad \text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.4)

We have

$$\begin{aligned} \Vert \chi _k-1\Vert ^2_{L^2(\mathbb {R}_+)}&=\int _{0}^{\infty }(\chi _k-1)^2\ dt=\int _{0}^{1/k^2}(\chi _k-1)^2\ dt+\int _{1/k^2}^{1/k}(\chi _k-1)^2\ dt\\&=\frac{1}{k^2}+\int _{1/k^2}^{1/k}\Big (\frac{\log k^2t}{\log k}-1\Big )^2\ dt=\frac{1}{k^2}+\frac{1}{k^2}\int _{1}^{k}\Big (\frac{\log t}{\log k}-1\Big )^2\ dt\\&=\frac{1}{k^2}+\frac{1}{k\log ^2k}\int _{1/k}^{1}\log ^2t\ dt\\&=\frac{1}{k^2}+\frac{1}{k^2\log ^2k}\Big (2-\frac{\log ^2k}{k}-\frac{2\log k}{k}-\frac{2}{k}\Big ). \end{aligned}$$

From the estimate above, (A.4) follows.

Next, we also prove that

$$\begin{aligned}{}[\chi _k-1]_{H^{1/2}(\mathbb {R}_+)}\rightarrow 0\quad \quad \text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.5)

We have

$$\begin{aligned}&[\chi _k-1]^2_{H^{1/2}(\mathbb {R}_+)}=\frac{c_{1,1/2}}{2}\int _{0}^{\infty }\int _{0}^{\infty }\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\\&\quad =c\Bigg (\int _{0}^{1/k}\int _{0}^{1/k}\cdots +2\int _{0}^{1/k}\int _{1/k}^{\infty }\cdots +\int _{1/k}^{\infty }\int _{1/k}^{\infty }\cdots \Bigg )\\&\qquad \times \frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy. \end{aligned}$$

Since $\chi _k(x)=\chi _k(y)=1$ for $(x,y)\in (1/k,\infty )\times (1/k,\infty )$ then the third integral in the above equality vanishes. Therefore,

$$\begin{aligned}{}[\chi _k-1]^2_{H^{1/2}(\mathbb {R}_+)}=c\int _{0}^{\infty }\int _{0}^{\infty }\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy=c(I_k+J_k) \end{aligned}$$

where

$$\begin{aligned} I_k:= & {} \int _{0}^{1/k}\int _{0}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\quad \text {and}\quad J_k\\:= & {} 2\int _{0}^{1/k}\int _{1/k}^{\infty }\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy. \end{aligned}$$

Estimate of $J_k$. We have

$$\begin{aligned}&\int _{0}^{1/k}\int _{1/k}^{\infty }\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\\&\quad =\Bigg (\int _{0}^{1/k^2}\int _{1/k}^{\infty }\cdots +\int _{1/k^2}^{1/k}\int _{1/k}^{\infty }\cdots \Bigg )\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\\&\quad =J^1_k+J^2_k \end{aligned}$$

where

$$\begin{aligned} J^1_k:= & {} \int _{0}^{1/k^2}\int _{1/k}^{\infty }\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\quad \text {and}\\ J^2_k:= & {} \int _{1/k^2}^{1/k}\int _{1/k}^{\infty }\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy. \end{aligned}$$

Regarding $J^1_k$, we have from the definition of $\chi _k$ that

$$\begin{aligned} J^1_k&=\int _{0}^{1/k^2}\int _{1/k}^{\infty }\frac{1}{(x-y)^2}\ dxdy{\mathop {=}\limits ^{\tau =\frac{x}{y}}}\int _{0}^{1/k^2}\frac{1}{y}\int _{1/ky}^{\infty }\frac{1}{(\tau -1)^2}\ d\tau dy\nonumber \\&=\int _{0}^{1/k^2}\frac{k}{1-ky}\ dy=-\log \Big (1-\frac{1}{k}\Big ). \end{aligned}$$

(A.6)

For $J^2_k$, we also use the definition of $\chi _k$ to see that

$$\begin{aligned} J^2_k&=\int _{1/k^2}^{1/k}\int _{1/k}^{\infty }\frac{\Big (1-\frac{\log k^2x}{\log k}\Big )^2}{(x-y)^2}\ dxdy=\frac{1}{\log ^2k}\int _{1/k^2}^{1/k}\int _{1/k}^{\infty }\frac{(\log k-\log k^2x)^2}{(x-y)}\ dxdy\nonumber \\&=\frac{1}{\log ^2k}\int _{1/k^2}^{1/k}\int _{1/k}^{\infty }\frac{(\log kx)^2}{(x-y)^2}\ dxdy{\mathop {=}\limits ^{\begin{array}{c} \tau =kx\\ t=ky \end{array}}}\frac{1}{\log ^2k}\int _{1/k}^{1}\int _{1}^{\infty }\frac{\log ^2\tau }{(\tau -t)^2}\ d\tau dt\nonumber \\&=\frac{1}{\log ^2k}\int _{1}^{\infty }\Big (\frac{1}{(\tau -\frac{1}{k})}-\frac{1}{(\tau -1)}\Big )\log ^2\tau \ d\tau \nonumber \\&=\frac{1}{\log ^2k}\int _{1}^{\infty }\frac{\frac{1}{k}-1}{(\tau -\frac{1}{k})(\tau -1)}\log ^2\tau \ d\tau . \end{aligned}$$

(A.7)

Using that $\log \tau \sim \tau -1$ as $\tau \rightarrow 1$ and $\frac{\log ^2\tau }{(\tau -\frac{1}{k})(\tau -1)}\sim \frac{\log ^2\tau }{\tau ^2}\le \frac{c}{\tau ^{2-\varepsilon }}$ as $\tau \rightarrow \infty$, for every $\varepsilon >0$, then the above integral is convergence for k sufficiently large. This implies that

$$\begin{aligned} J^2_k=o(1)\quad \quad \text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.8)

Combining (A.6) and (A.7), and by using (A.8), we find that

$$\begin{aligned} J_k&=2\Bigg (-\log \Big (1-\frac{1}{k}\Big )+\frac{1}{\log ^2k}\int _{1}^{\infty }\frac{\frac{1}{k}-1}{(\tau -\frac{1}{k})(\tau -1)}\log ^2\tau \ d\tau \Bigg )\nonumber \\&\rightarrow 0\quad \quad \text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.9)

Estimate of $I_k$. We have

$$\begin{aligned} I_k&=\Bigg (\int _{0}^{1/k^2}\int _{0}^{2/k^2}\cdots +\int _{0}^{1/k^2}\int _{2/k^2}^{1/k}\cdots \\&\quad +\int _{1/k^2}^{1/k}\int _{0}^{2/k^2}\cdots +\int _{1/k^2}^{1/k}\int _{2/k^2}^{1/k}\cdots \Bigg )\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\\&=I^1_k+I^2_k+I^3_k \end{aligned}$$

where

$$\begin{aligned}{} & {} I^1_k:=\int _{0}^{1/k^2}\int _{0}^{2/k^2}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy,\\{} & {} I^2_k:=\int _{1/k^2}^{1/k}\int _{2/k^2}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy \end{aligned}$$

and

$$\begin{aligned} I^3_k:=\Bigg (\int _{0}^{1/k^2}\int _{2/k^2}^{1/k}\cdots +\int _{1/k^2}^{1/k}\int _{0}^{2/k^2}\cdots \Bigg )\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy. \end{aligned}$$

It now suffices to estimate $I^1_k, I^2_k$ and $I^3_k$.

Concerning $I^1_k$, we have

$$\begin{aligned} I^1_k&=\int _{0}^{1/k^2}\int _{1/k^2}^{2/k^2}\frac{\chi _k(x)^2}{(x-y)^2}\ dxdy=\frac{1}{\log ^2k}\int _{0}^{1/k^2}\int _{1/k^2}^{2/k^2}\frac{(\log k^2x)^2}{(x-y)^2}\ dxdy\nonumber \\&{\mathop {=}\limits ^{\begin{array}{c} \tau =k^2x\\ t=k^2y \end{array}}}\frac{1}{\log ^2k}\int _{0}^{1}\int _{1}^{2}\frac{\log ^2 \tau }{(\tau -t)^2}\ d\tau dt=\frac{1}{\log ^2k}\int _{0}^{1}\int _{1}^{2}\frac{(\log \tau -\log 1)^2}{(\tau -t)^2}\ d\tau dt\nonumber \\&\quad \le \frac{c}{\log ^2k}\int _{0}^{1}\int _{1}^{2}\frac{(\tau -1)^2}{(\tau -t)^2}\ d\tau dt=\frac{c}{\log ^2k}\int _{1}^{2}\int _{0}^{1}\frac{(\tau -1)^2}{(\tau -t)^2}\ dtd\tau \nonumber \\&\quad =\frac{c}{\log ^2k}\int _{1}^{2}(\tau -1)^2\Big (\frac{1}{\tau -1}-\frac{1}{\tau }\Big )=\frac{c'}{\log ^2k}. \end{aligned}$$

(A.10)

Next, as regards $I^2_k$, the change of variables $\tau =k^2x$ and $t=k^2y$ gives

$$\begin{aligned} I^2_k&=\int _{1/k^2}^{1/k}\int _{2/k^2}^{1/k}\frac{(\log k^2x-\log k^2y)^2}{(x-y)^2}\ dxdy=\frac{1}{\log ^2k}\int _{1}^{k}\int _{2}^{k}\frac{(\log \tau -\log t)^2}{(\tau -t)^2}\ d\tau dt\nonumber \\&=\frac{1}{\log ^2k}\int _{1}^{k}\int _{2}^{k}\frac{(\log (\tau /t))^2}{(\tau -t)^2}\ d\tau dt{\mathop {=}\limits ^{r=\tau /t}}\frac{1}{\log ^2k}\int _{1}^{k}\frac{1}{t}\int _{2/t}^{k/t}\frac{\log ^2r}{(r-1)^2}\ drdt\nonumber \\&\le \frac{1}{\log ^2k}\int _{1}^{k}\frac{dt}{t}\int _{0}^{\infty }\frac{\log ^2r}{(r-1)^2}\ dr=\frac{c}{\log k}. \end{aligned}$$

(A.11)

For $I^3_k$, we have

$$\begin{aligned} I^3_k&\le 2\int _{0}^{2/k^2}\int _{1/k^2}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\\&=2\int _{0}^{2/k^2}\int _{1/k^2}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\\&=2\int _{0}^{1/k^2}\int _{1/k^2}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy\\&\quad +2\int _{1/k^2}^{2/k^2}\int _{1/k^2}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy. \end{aligned}$$

Now,

$$\begin{aligned}&\int _{0}^{1/k^2}\int _{1/k^2}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy=\frac{1}{\log ^2k}\int _{0}^{1/k^2}\int _{1/k^2}^{1/k}\frac{(\log k^2x)^2}{(x-y)^2}\ dxdy\nonumber \\&{\mathop {=}\limits ^{\begin{array}{c} \tau =k^2x\\ t=k^2y \end{array}}}\frac{1}{\log ^2k}\int _{0}^{1}\int _{1}^{k}\frac{\log ^2\tau }{(\tau -t)^2}\ d\tau dt=\frac{1}{\log ^2k}\int _{1}^{k}\Big (\frac{1}{(\tau -1)^2}-\frac{1}{\tau ^2}\Big )\log ^2\tau \ d\tau \nonumber \\&\le \frac{1}{\log ^2k}\int _{1}^{\infty }\Big (\frac{1}{(\tau -1)^2}-\frac{1}{\tau ^2}\Big )\log ^2\tau \ d\tau =\frac{c}{\log ^2k}. \end{aligned}$$

(A.12)

Arguing as in the case of $I^2_k$, we have that

$$\begin{aligned}&\int _{1/k^2}^{2/k^2}\int _{1/k^2}^{1/k}\frac{(\chi _k(x)-\chi _k(y))^2}{(x-y)^2}\ dxdy=\frac{1}{\log ^2k}\int _{1}^{k}\int _{1}^{2}\frac{(\log t-\log \tau )^2}{(t-\tau )^2}\ dtd\tau \nonumber \\&{\mathop {=}\limits ^{r=t/\tau }}\frac{1}{\log ^2k}\int _{1}^{k}\frac{d\tau }{\tau }\int _{1/\tau }^{2/\tau }\frac{\log ^2r}{(r-1)^2}\ dr\le \frac{1}{\log ^2k}\int _{1}^{k}\frac{d\tau }{\tau }\int _{1}^{\infty }\frac{\log ^2r}{(r-1)^2}\ dr\nonumber \\&~~=\frac{c}{\log k}. \end{aligned}$$

(A.13)

Putting together (A.10), (A.11), (A.12) and (A.13), we find that

$$\begin{aligned} I_k\le \frac{c}{\log ^2k}+\frac{c}{\log k}\rightarrow 0\quad \quad \text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.14)

From (A.9) and (A.14), we conclude that

$$\begin{aligned}{}[\chi _k-1]_{H^{1/2}(\mathbb {R}_+)}\rightarrow 0\quad \quad \text {as}~~ k\rightarrow \infty . \end{aligned}$$

(A.15)

Now, (A.3) follows by combining (A.4) and (A.15). As wanted. $\square$

Definition A.2

We say that an open subset $\Omega$ of $\mathbb {R}^N$ is Lipschitz if for each $q\in \partial \Omega$, there exist a tangent hyperplane $H_q$, a normal $N_q$ of $H_q$, $r_q>0$, open $r_q$-balls $B_{r_q}\subset H_q$ and a function $\Phi _q: B_{r_q}\times I\rightarrow \mathbb {R}^N$ such that

• (i) $\Phi _q(B_{r_q}\cap H^+_q)\subset \Omega$

• (ii) $\Phi _q(B_{r_q}\cap \partial H^+_q)\subset \partial \Omega$

• (iii) $C^{-1}|x-y|\le |\Phi _q(x)-\Phi _q(y)|\le C|x-y|,~~~C>1,~~x,y\in B_{r_q}\times I,~~I\subset \mathbb {R}$.

Here, $H^+_q$ is the upper half-tangent hyperplane containing $N_q$. Put $Q_q:=B_{r_q}\times (-r_q,r_q)$ and we recall that $B_{r_q}$ is a $(N-1)$-ball.

Remark A.3

We would like to make the following observation. It is well-known that a domain $\Omega$ is said to be strongly Lipschitz if its boundary can be seen as a local graph of a Lipschitz function $\phi :\mathbb {R}^{N-1}\rightarrow \mathbb {R}$. Moreover, by mean of a vectorfield $\eta$ (with $|\eta |=1$ on $\partial \Omega$) which is globally transversal² to $\partial \Omega$, one can construct a bi-Lipschitz mapping via $\phi$. In particular, $\Omega$ fulfills properties (i)-(iii). However, every Lipschitz domain in the sense of definition (i)-(iii) is not necessarily a local graph of a Lipschitz function, see [15] for a counterexample.

Clearly, there exists $\beta >0$ such that

$$\begin{aligned} \overline{\Omega _{\beta }}:=\{0\le \delta _{\Omega }(x)\le \beta \}\subset \cup _{q\in \partial \Omega }\Phi _q(Q_{q}). \end{aligned}$$

(A.16)

We recall that $\Omega _{\beta }$ is the so-called inner tubular neighbourhood of $\Omega$. By compactness, there exists $m\in \mathbb {N}$ such that

$$\begin{aligned} \overline{\Omega _{\beta }}:=\{0\le \delta _{\Omega }(x)\le \beta \}\subset \cup ^{m}_{j=1}\Phi _{q_j}(Q_{q_j}). \end{aligned}$$

(A.17)

We will write j in the place of $q_j$ provided there is no ambiguity. For $j=1,\dots ,m$, let $u_k^j$ be a sequence define by

$$\begin{aligned} u_k^j(\Phi _{j}(x))=\chi _k(x_N),~~~\forall x\in Q_j, \end{aligned}$$

where $\chi _k$ is defined in (A.1). Equivalently, $u_k^j$ can be defined as

$$\begin{aligned} u_k^j(x)=\chi _k(\Phi ^{-1}_{j}(x)\cdot N_{j}),~~~\forall x\in \Omega . \end{aligned}$$

(A.18)

Define ${\mathcal O}_j:=\Phi _j(Q_j)$ and ${\mathcal O}_{m+1}=\Omega \setminus \overline{\Omega _{\beta }}$. We also write $Q_j^+:=B_{r_j}\times (0,r_j)$.

We have the following.

Lemma A.4

For all $j=1,\dots ,m$ there exists a positive constant $C>0$ depending only on $j, m, \Omega$ and N such that

$$\begin{aligned} \Vert u^j_k-\mathbbm {1}_{\Omega }\Vert _{H^{1/2}({\mathcal O}_j\cap \Omega )}\le C\Vert \chi _k-1\Vert _{H^{1/2}(0,r_j)}. \end{aligned}$$

(A.19)

Proof

For $j=1,\dots ,m$, by using the change of variables $x=\Phi _j(z)$ and $y=\Phi _j({\overline{z}})$, we get

$$\begin{aligned}&\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{(u_k(x)-u_k(y))^2}{|x-y|^{N+1}}\ dxdy\nonumber \\&\quad =\int _{Q_j^+}\int _{Q_j^+}\frac{(u_k(\Phi _j(z))-u_k(\Phi _j({\overline{z}})))^2}{|\Phi _j(z)-\Phi _j({\overline{z}})|^{N+1}}\ dzd{\overline{z}} \nonumber \\&\quad =\int _{Q_j^+}\int _{Q_j^+}\frac{(\chi _k(z_N)-\chi _k({\overline{z}}_N))^2}{|\Phi _j(z)-\Phi _j({\overline{z}})|^{N+1}}\ dzd{\overline{z}}\nonumber \\&\quad \le C\int _{Q_j^+}\int _{Q_j^+}\frac{(\chi _k(z_N)-\chi _k({\overline{z}}_N))^2}{|z-{\overline{z}}|^{N+1}}\ dzd{\overline{z}} \nonumber \\&\quad \le C\int _{B_{r_j}}\int _{B_{r_j}}\int _{0}^{r_j}\int _{0}^{r_j}\frac{(\chi _k(z_N)-\chi _k({\overline{z}}_N))^2}{|z-{\overline{z}}|^{N+1}}\ dzd{\overline{z}} \nonumber \\&\quad \le C\int _{B_{r_j}}dz'\int _{H_j}d\overline{z}'\int _{0}^{r_j}\int _{0}^{r_j}\frac{(\chi _k(z_N)-\chi _k({\overline{z}}_N))^2}{(|z'-{\overline{z}}'|^2+|z_N-{\overline{z}}_N|^2)^{\frac{N+1}{2}}}\ dz_Nd{\overline{z}}_N . \end{aligned}$$

(A.20)

By translation and rotation, we have

$$\begin{aligned}&\int _{B_{r_j}}dz'\int _{H_j}d\overline{z}'\int _{0}^{r_j}\int _{0}^{r_j}\frac{(\chi _k(z_N)-\chi _k({\overline{z}}_N))^2}{(|z'-{\overline{z}}'|^2+|z_N-{\overline{z}}_N|^2)^{\frac{N+1}{2}}}\ dz_Nd{\overline{z}}_N \\&\quad = \int _{ B_{r_j}} dz'\int _{\mathbb {R}^{N-1}} d\overline{z}'\int _{0}^{r_j}\int _{0}^{r_j}\frac{(\chi _k(z_N)-\chi _k({\overline{z}}_N))^2}{(|z'-{\overline{z}}'|^2+|z_N-{\overline{z}}_N|^2)^{\frac{N+1}{2}}}\ dz_Nd{\overline{z}}_N \\&\quad \le CA\int _{0}^{r_j}\int _{0}^{r_j}\frac{(\chi _k(z_N)-\chi _k({\overline{z}}_N))^2}{ |z_N-{\overline{z}}_N|^2 }\ dz_Nd{\overline{z}}_N , \end{aligned}$$

where $A= \int _{\mathbb {R}^{N-1}}\frac{dl}{(1+|l|^2)^{(N+1)/2}}\le C$ and $B_{r_j}$ is a bounded open subset of $\mathbb {R}^{N-1}$. Therefore, since the estimate of the $L^2$ norm follows easily, this and (A.20) give (A.19), concluding the proof. $\square$

Consider $0\le \psi _j\in C^\infty _c({\mathcal O}_j)$ a partitioning of unity subordinated to $\{{\mathcal O}_j\}_{j=1,\dots ,m+1}$. Define

$$\begin{aligned} u_k:=\sum _{j=1}^{m+1}\psi _ju_k^j \in C^{0,1}_c(\Omega ), \end{aligned}$$

(A.21)

where $u^{m+1}_k\equiv 1$ on $\Omega$. We have the following approximation.

Lemma A.5

There holds

$$\begin{aligned} \Vert u_k-\mathbbm {1}_{\Omega }\Vert _{H^{1/2}(\Omega )}\rightarrow 0\quad \quad \text {as}~~k\rightarrow \infty . \end{aligned}$$

(A.22)

Proof

We estimate

$$\begin{aligned}{}[u_k-\mathbbm {1}_\Omega ]_{H^{1/2}(\Omega )}^2&\le \left( \sum _{j=1}^{m+1} [\psi _j u_k^j-\psi _j]_{H^{1/2}(\Omega )}\right) ^2 \le m \sum _{j=1}^m [\psi _j u_k^j-\psi _j]_{H^{1/2}(\Omega )}^2\\&\le C \sum _{j=1}^{m} \int _{{\mathcal O}_j\cap \Omega \times {\mathcal O}_j\cap \Omega }\dots \,dxdy+ C\sum _{j=1}^{m} \int _{ \Omega \setminus {\mathcal O}_j\times \Omega \cap {\mathcal O}_j }\dots \, dxdy\\&=: C I_1(k)+ C I_2(k). \end{aligned}$$

We now estimate $I_1(k)$ and $I_2(k)$. Let us start with $I_2(k)$.

We have

$$\begin{aligned} I_2(k)&= \sum _{j=1}^{m} \int _{\Omega \setminus {\mathcal O}_j\times \Omega \cap {\mathcal O}_j}\frac{[(\psi _j u_k^j-\psi _j)(x)-(\psi _j u_k^j-\psi _j)(y)]^2}{|x-y|^{N+1}}dxdy\nonumber \\&= \sum _{j=1}^{m} \int _{ \Omega \setminus {\mathcal O}_j}\frac{dx}{|x-y|^{N+1}} \int _{ \Omega \cap \text {Supp}\psi _j}{ (\psi _j u_k^j-\psi _j)(y)^2}dy\nonumber \\&\le C \sum _{j=1}^{m} \text {dist}( \text {Supp}\psi _j, \partial {\mathcal O}_j)^{-N-1} \int _{ \Omega \cap {\mathcal O}_j }\psi _j^2 | u_k^j(y)-1|^2dy\nonumber \\&\le C(N)\max _{1\le j\le {m}} \text {dist}( \text {Supp}\psi _j, \partial {\mathcal O}_j)^{-N-1} \sum _{j=1}^{m} \Vert u_k^j-\mathbbm {1}_{\Omega }\Vert _{L^{2}( \Omega \cap {\mathcal O}_j )}^2. \end{aligned}$$

(A.23)

Now regarding $I_1(k)$, we have

$$\begin{aligned} I_1(k)&=\sum _{j=1}^{m}\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{[\psi _j(x)(u^j_k(x)-1)-\psi _j(y)(u^j_k(y)-1)]^2}{|x-y|^{N+1}}\ dxdy\\&=\sum _{j=1}^{m}\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\\&\frac{[\psi _j(x)((u^j_k(x)-1)-(u^j_k(y)-1))+(\psi _j(x)-\psi _j(y))(u^j_k(y)-1)]^2}{|x-y|^{N+1}}\ dxdy\\&\le 2\sum _{j=1}^{m}\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{\psi _j(x)^2[(u^j_k(x)-1)-(u^j_k(y)-1)]^2}{|x-y|^{N+1}}\ dxdy\\&\quad +2\sum _{j=1}^{m}\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{(\psi _j(x)-\psi _j(y))^2(u^j_k(y)-1)^2}{|x-y|^{N+1}}\ dxdy\\&=I^1_1(k)+I^2_1(k), \end{aligned}$$

where

$$\begin{aligned} I^1_1(k)&=2\sum _{j=1}^{m}\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{\psi _j(x)^2[(u^j_k(x)-1)-(u^j_k(y)-1)]^2}{|x-y|^{N+1}}\ dxdy\nonumber \\&\le 2\sum _{j=1}^{m}\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{[(u^j_k(x)-1)-(u^j_k(y)-1)]^2}{|x-y|^{N+1}}dxdy\nonumber \\&\quad \qquad \quad (\text {since}~0\le \psi _j\le 1)\nonumber \\&=c\sum _{j=1}^{m}[u^j-\mathbbm {1}_{\Omega }]^2_{H^{1/2}({\mathcal O}_j\cap \Omega )} \end{aligned}$$

(A.24)

and

$$\begin{aligned} I^2_1(k)=2\sum _{j=1}^{m}\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{(\psi _j(x)-\psi _j(y))^2(u^j_k(y)-1)^2}{|x-y|^{N+1}}\ dxdy. \end{aligned}$$

Using that $\psi _j$ is Lipschitz, we get

$$\begin{aligned}&2\int _{{\mathcal O}_j\cap \Omega }\int _{{\mathcal O}_j\cap \Omega }\frac{(\psi _j(x)-\psi _j(y))^2(u^j_k(y)-1)^2}{|x-y|^{N+1}}\ dxdy\\&\quad \le c(j)^2\iint _{|x-y|<1}\frac{(u^j_k(y)-1)^2|x-y|^2}{|x-y|^{N+1}}\ dxdy+8\iint _{|x-y|\ge 1}\frac{(u^j_k(y)-1)^2}{|x-y|^{N+1}}\ dxdy\\&\quad \le {\tilde{c}}(j)\Vert u^j_k-\mathbbm {1}_{\Omega }\Vert ^2_{L^2({\mathcal O}_j\cap \Omega )} \end{aligned}$$

which implies that

$$\begin{aligned} I^2_1(k)\le \max _{1\le j\le m}{\tilde{c}}(j)\sum _{j=1}^{m}\Vert u^j_k-\mathbbm {1}_{\Omega }\Vert ^2_{L^2({\mathcal O}_j\cap \Omega )}. \end{aligned}$$

(A.25)

Finally, (A.23), (A.24) and (A.25) yield

$$\begin{aligned} \Vert u_k-\mathbbm {1}_\Omega \Vert _{H^{1/2}(\Omega )}^2&=\Vert u_k-\mathbbm {1}_\Omega \Vert ^2_{L^2(\Omega )}+[u_k-\mathbbm {1}_{\Omega }]^2_{H^{1/2}(\Omega )}\nonumber \\&\le c\sum _{j=1}^{m}\Vert u^j_k-\mathbbm {1}_{\Omega }\Vert ^2_{L^2({\mathcal O}_j\cap \Omega )}+C I_1(k)+C I_2(k)\nonumber \\&={\tilde{c}}\sum _{j=1}^{m}\Vert u^j_k-\mathbbm {1}_{\Omega }\Vert ^2_{H^{1/2}({\mathcal O}_j\cap \Omega )}\nonumber \\&\le C(N,m) \sum _{j=1}^{m} \Vert \chi _k-1\Vert _{H^{1/2}(0,r_j)}^2. \end{aligned}$$

(A.26)

In the latter inequality, we used Lemma A.4. Now, since from Lemma A.1 there holds $\Vert \chi _k-1\Vert _{H^{1/2}(0,r_j)}^2\rightarrow 0$ as $k\rightarrow \infty$, we complete the proof by letting $k\rightarrow \infty$ in the inequality (A.26). $\square$

As a direct consequence of the above approximation results, we have the following.

Proposition A.6

Let $N\ge 2,~s\in (0,1/2]$ and let $\Omega \subset \mathbb {R}^N$ be a bounded Lipschitz domain. Then

$$\begin{aligned} S_{N,s}(\Omega )=0. \end{aligned}$$

(A.27)

Before proving the proposition above, we mention that our result extends to $s=1/2$ the one obtained in [12, Lemma 16]. Below, we give the

Proof of Proposition A.6

By definition

$$\begin{aligned} S_{N,s}(\Omega )=\inf _{\begin{array}{c} u\in H^{s}_0(\Omega )\\ u\ne 0 \end{array}}\frac{Q_{N,s,\Omega }(u)}{\Vert u\Vert ^2_{L^{2^*_s}(\Omega )}}=\inf _{\begin{array}{c} u\in C^{0,1}_c(\Omega )\\ u\ne 0 \end{array}}\frac{Q_{N,s,\Omega }(u)}{\Vert u\Vert ^2_{L^{2^*_s}(\Omega )}}, \end{aligned}$$

(A.28)

where $C^{0,1}_c(\Omega )$ is the space of Lipschitz functions with compact support. Now by Lemma A.5, we get

$$\begin{aligned}{} & {} 0\le S_{N,s}(\Omega )\le \frac{Q_{N,s,\Omega }(u_k)}{\Vert u_k\Vert ^2_{L^{2^*_s}(\Omega )}}\le C(N,s) \frac{Q_{N,1/2,\Omega }(u_k)}{\Vert u_k\Vert ^2_{L^{2^*_s}(\Omega )}}\nonumber \\{} & {} \quad =C(N,s)\frac{[u_k-\mathbbm {1}_{\Omega }]_{H^{1/2}(\Omega )}}{\Vert u_k\Vert ^2_{L^{2^*_s}(\Omega )}}\rightarrow 0~~\text {as}~~k\rightarrow \infty , \end{aligned}$$

(A.29)

where $u_k$ is defined by (A.21), which satisfies $\liminf _{k\rightarrow \infty }\Vert u_k\Vert ^2_{L^{2^*_s}(\Omega )}>0$. $\square$