2. Auxiliary results

Let us recall some important results that are used in this paper.

Let $T(\Omega ):=\mathbb {R}^{2n}\setminus (\mathbb {R}^{n}\setminus \Omega )^{2}$ be a cross-shaped set on a bounded domain $\Omega \subset \mathbb {R}^n$ . Define

(2-1) $$ \begin{align} H_{\Omega}^{s}:=\{ u: \mathbb{R}^{n}\longrightarrow \mathbb{R} \text{ measurable}: {\Vert u\Vert}_{H_{\Omega}^{s}}< \infty \}, \end{align} $$

which is equipped with the norm

$$ \begin{align*} \Vert u\Vert_{H_{\Omega}^{s}}:=\bigg(\Vert u\Vert_{L^{2} (\Omega)}^{2}+\int_{T(\Omega)}\frac{{\lvert u(x)-u(y)\rvert}^{2}}{{\lvert x-y \rvert}^{n+2s}}\,dx\,dy \bigg)^{{1}/{2}}. \end{align*} $$

Let us define the following set:

$$ \begin{align*}\mathcal{L}_{s}:= \bigg\{u: \mathbb{R}^{n}\longrightarrow \mathbb{R} \text{ measurable}: \int_{\mathbb{R}^n} \frac{\vert u(x)\vert}{1+\vert x\vert^{n+2s}}\,dx < \infty \bigg\}.\end{align*} $$

The condition $u \in \mathcal {L}_{s}$ is useful to give a sense to the pointwise definition of fractional Laplacians (1-2).

Next, we recall a few known results about the fractional Schrödinger equation (1-5).

Definition 2.4. A measurable function $u: \mathbb {R}^{n} \longrightarrow \mathbb {R}$ is called a weak solution of (1-5) if it satisfies the following equation:

$$ \begin{align*} & \frac{c_{n,s}}{2} \int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}} \frac{(u(x)-u(y))(\psi(x)-\psi(y))}{\vert x-y\vert^{n+2s}}\,dx\,dy\\ &\quad + \int_{\mathbb{R}^{n}}u(x)\psi(x)\,dx = \int_{\mathbb{R}^{n}} \vert u(x)\vert^{p-1}u(x)\psi(x)\,dx \end{align*} $$

for all $\psi \in C_{0}^{1}(\mathbb {R}^{n}).$

We define the corresponding energy functional $F: H^{s}(\mathbb {R}^n) \longrightarrow \mathbb {R} $ as follows:

(2-2) $$ \begin{align} F(u):=\frac{1}{2} \bigg[ \frac{c_{n,s}}{2} \int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}} \frac{\vert u(x)-u(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\mathbb{R}^{n}}u^{2}\,dx \bigg] -\frac{1}{p+1}\int_{\mathbb{R}^{n}}|u|^{p+1}\,dx. \end{align} $$

The weak solutions of (1-5) correspond to the critical points of $F.$

The next result gives us a positive, radially symmetric solution of (1-5), which decays at infinity.

Theorem 2.6 [Reference Felmer and Quaas20, Theorem 3.4]

Let u be the weak solution of (1-5). Then, $u \in L^{q}(\mathbb {R}^{n}) \cap C^{\alpha }(\mathbb {R}^{n})$ for some $q \in [2, \infty ) $ and $\alpha \in (0,1).$ Moreover,

$$ \begin{align*} \lim_{\vert x\vert \rightarrow \infty}u(x)=0. \end{align*} $$

The following theorem shows that the solutions of (1-5) have a power type of decay at infinity.

Theorem 2.8 [Reference Felmer and Quaas20, Theorem 1.5]

Let u be a positive classical solution of (1-5) such that

$$ \begin{align*}\lim_{\vert x\vert \rightarrow \infty}u(x)=0.\end{align*} $$

Then, there exist constants $0< C_{1} \leq C_{2}$ such that

$$ \begin{align*} \frac{C_{1}}{\vert x\vert^{n+2s}}\leq u(x) \leq \frac{C_{2}}{\vert x\vert^{n+2s}} \quad \text{for all } \vert x\vert\geq 1. \end{align*} $$

One can see that there exist some $m>0$ and $s_{0}>0$ such that for $f(u)=u^{p}-u,$

$$ \begin{align*} \frac{f(v)-f(u)}{v-u}\leq \frac{v^{p}-u^{p}}{v-u}\leq C (v+u)^{m} \quad\text{for all } 0 < u< v < s_{0}, \end{align*} $$

where $C>0$ is some constant. Also, it is simple to see that $f: [0,\infty ) \to \mathbb {R}$ is locally Lipschitz. Consequently, we have the following result on the radial symmetry and monotonicity property of positive solutions of (1-5).

Theorem 2.9 [Reference Felmer and Wang21, Theorem 1.2]

Let u be a positive classical solution of (1-5) such that

$$ \begin{align*}\lim_{\vert x\vert \rightarrow \infty}u(x)=0.\end{align*} $$

Further, assume that there exists

$$ \begin{align*}t> \max \bigg\{\frac{2s}{m}, \frac{n}{m+2} \bigg\}\end{align*} $$

such that u satisfies $u(x)=O({1}/{\vert x\vert ^{t}})$ as $ \vert x\vert \rightarrow \infty .$ Then, u is radially symmetric and strictly decreasing about some point in $\mathbb {R}^{n}.$

Remark 2.10. Since

$$ \begin{align*} \frac{C_{1}}{\vert x\vert^{n+2s}}\leq u(x) \leq \frac{C_{2}}{\vert x\vert^{n+2s}} \quad \text{for all } \vert x\vert\geq 1, \end{align*} $$

we can take $t=n+2s$ in the above theorem.

Now, [Reference Secchi37, Proposition 4.1] ascertains that if $u \in \mathbb {R}^n$ is a weak solution of (1-5), then u satisfies the following Pohozaev identity:

$$ \begin{align*} \mathcal{P}(u):= \frac{(n-2s)c_{n,s}}{4}\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \frac{\vert u(x)-u(y)\vert^2}{\vert x-y\vert^{n+2s}}\,dx\,dy + \frac{n}{2}\int_{\mathbb{R}^n} u^2 \,dx-\frac{n}{p+1} \int_{\mathbb{R}^n} u^{p+1}=0. \end{align*} $$

Let us define

$$ \begin{align*} \mathcal{G}:=\{ u \in H^s(\mathbb{R}^n) \setminus \{0\} \mid \mathcal{P}(u)=0 \}. \end{align*} $$

In [Reference Chen and Gao8], the authors have obtained a weak solution $w \in H^{s}(\mathbb {R}^{n})$ of (1-5) with the least energy among all other solutions. In particular, they have proved the following result.

Theorem 2.11 [Reference Chen and Gao8, Theorem 1.2]

Equation (1-5) has a weak solution $w \in H^s(\mathbb {R}^n)$ such that

$$ \begin{align*}0<F(w)=\inf_{u \in \mathcal{G}}F(u) .\end{align*} $$

Combining Theorems 2.7, 2.8, 2.9 and 2.11, we have the following result.

3. Regularity and bounds for the least energy solution $u_{d}$

Let $s\in (0,1)$ and $\Omega \subset \mathbb {R}^{n}$ be a bounded domain of class $C^{1,1}.$

Definition 3.1. A measurable function $u: \mathbb {R}^{n} \longrightarrow \mathbb {R}$ is said to be a weak solution of (1-1) if it satisfies the following equation:

(3-1) $$ \begin{align} &\frac{dc_{n,s}}{2} \int_{T(\Omega)} \frac{(u(x)-u(y))(\psi (x)-\psi(y))}{\vert x-y\vert^{n+2s}}\,dx\,dy\nonumber\\ &\quad+ \int_{\Omega}u(x)\psi(x)\,dx=\int_{\Omega} \vert u(x)\vert^{p-1}u(x)\psi(x)\,dx \end{align} $$

for all $\psi \in H_{\Omega }^{s}.$

We have the following result on the existence of a weak solution of (1-1).

Theorem 3.2 ([Reference Barrios, Montoro, Peral and Soria4, Theorem 6.1], [Reference Chen7, Theorem 1.1])

There exists a nonnegative weak solution $u_{d}$ of (1-1) with critical value $c_{d}$ , provided d is sufficiently small. Moreover, $u_{d}$ satisfies

$$ \begin{align*}0< J_{d}(u_{d}) \leq C d^{{n}/{2s}},\end{align*} $$

where the constant C is independent of $d.$ Consequently, $u_{d}$ is nonconstant.

Define

$$ \begin{align*}M[v]:=\sup_{t \geq 0}J_{d}(tv), \quad v\in H_{\Omega}^{s}.\end{align*} $$

In the next lemma, we indicate a useful characterisation of the critical value $c_{d}.$ We follow similar lines of proof to [Reference Ni and Takagi35, Lemma 3.1].

Lemma 3.3. The critical value $c_{d}$ is independent of the choice of $u \in H_{\Omega }^{s}$ such that $u \geq 0,~u \not \equiv 0$ and $J_{d}(u)=0.$ In fact, $c_{d}$ is the least positive critical value of $J_{d},$ which is given by

(3-2) $$ \begin{align} c_{d}=\inf\{M[v] \mid v\in H_{\Omega}^{s},~v\not \equiv 0, v\geq 0 \text{ in } \Omega \}. \end{align} $$

Proof. For $v\in H_{\Omega }^{s},$ let

$$ \begin{align*}\Omega^{+}=\{x \in \Omega \mid v(x)>0\}.\end{align*} $$

Now, for all those v satisfying $\vert \Omega ^{+}\vert>0,$ define

$$ \begin{align*}g_{d}(t):=J_{d}(tv) \quad \text{for } t \geq 0. \end{align*} $$

First, we show that $g_{d}(t)$ has a unique maximum. For this,

$$ \begin{align*} g^{\prime}_{d}(t)= t\bigg[ \frac{dc_{n,s}}{2} \int_{T(\Omega)} \frac{\vert v(x)-v(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}v^{2}\,dx \bigg] -t^{p}\int_{\Omega}{v }^{p+1}\,dx. \end{align*} $$

Therefore, $g^{\prime }_{d}(t_{0})=0$ for some $t_{0}>0$ if and only if

$$ \begin{align*}\frac{dc_{n,s}}{2} \int_{T(\Omega)} \frac{\vert v(x)-v(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}v^{2}\,dx = t_{0}^{p-1}\int_{\Omega} v^{p+1}\,dx.\end{align*} $$

Note that the right-hand side is strictly increasing in $t_{0}.$ And hence there exists a unique $t_{0}>0$ such that $g^{\prime }_{d}(t_{0})=0.$ Since $g_{d}(t)>0$ for $t>0$ small and $g_{d}(t) \rightarrow -\infty $ as $t \rightarrow +\infty ,$ one easily find that $g_{d}(t)$ has a unique maximum.

Let us fix a function $u \not \equiv 0, u\geq 0$ in $H_{\Omega }^{s} $ with $J_{d}(u)=0.$ Let $u_{d}$ be a positive solution of (1-1) obtained by applying the Mountain-Pass lemma and $c_{d}$ the corresponding critical value. We have $J_{d}(u_{d})=c_{d}$ and $J_{d}^{'}(u_{d})=0.$ Since $u_{d}>0$ and $J_{d}^{'}(u_{d})=0,$

(3-3) $$ \begin{align} M[u_{d}]=c_{d}, \end{align} $$

and hence

(3-4) $$ \begin{align} c_{d}\geq \inf\{M[v] \mid v\in H_{\Omega}^{s},~v\not \equiv 0, v\geq 0 \text{ in } \Omega \}. \end{align} $$

In contrast, assume that strict inequality occurs in (3-4). Then,

$$ \begin{align*}M[v_{0}]<c_{d},\end{align*} $$

for some $v_{0} \geq 0,\, v_{0} \not \equiv 0 $ in $H_{\Omega }^{s}. $ Therefore, there exists some $t_{1}>0$ such that $t_{1}v_{0}=u_{0}$ satisfies $J_{d}(u_{0})=0.$ Denote by U the subspace of $H_{\Omega }^{s}$ spanned by u and $u_{0}.$ Consider the subset of U defined as follows:

$$ \begin{align*}U^{+}:=\{ \alpha u + \beta u_{0}\mid \alpha,~\beta \geq 0 \}.\end{align*} $$

Suppose S is a circle on U of radius R so large that $R> \max \{ \Vert u\Vert , \Vert u_{0}\Vert \}$ and $J_{d}\leq 0$ on $S \cap U^{+}.$ Assume that $\gamma $ is the path made up of the line segment with endpoints $0$ and ${Ru_{0}}/{\Vert u_{0}\Vert },$ the circular arc $ S\cap U^{+}$ and the line segment with endpoints ${Ru}/{\Vert u\Vert }$ and $u.$ One can easily see that, along $\gamma $ , $J_{d}$ is positive only on the line segment joining $0$ and $u_{0}.$ Hence,

$$ \begin{align*} \max_{v \in \gamma}J_{d}(v)=M[v_{0}] < c_{d},\end{align*} $$

which is a contradiction to (1-4). Thus, we have equality in (3-4), that is,

$$ \begin{align*} c_{d}= \inf\{M[v] \mid v\in H_{\Omega}^{s},~v\not \equiv 0, v\geq 0 \text{ in } \Omega \}. \end{align*} $$

Note that $J_{d}(v)=J_{d}(-v)$ for any $v \in H^{s}_{\Omega }$ . Since any nontrivial critical point of $J_{d}$ is either positive or negative a.e. in $\Omega ,$ from the above discussion, one can see that $c_{d}$ is the least positive critical value of $J_{d}$ . This completes the proof.

The following lemma gives us the regularity estimate. A similar result is already proved in [Reference Cinti and Colasuonno10, Lemma 3.6] and [Reference Cinti and Colasuonno11, Remark 4.9].

Now, we prove that the least energy solution $u_{d}$ is bounded by some constant independent of $d.$

Proof of Theorem 1.2

The proof of the first inequality of Theorem 1.2 is fairly standard and simple, and can be seen in the literature; for instance, see [Reference Chen and Gao8, Theorem 1.1]. Since it is short, for the sake of completeness, we include it here. For this,

$$ \begin{align*} J_{d}(u_{d}):=\frac{1}{2} \bigg[ \frac{c_{n,s}d}{2} \int_{T(\Omega)} \frac{\vert u_{d}(x)-u_{d}(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u^{2}\,dx \bigg] -\frac{1}{p+1}\int_{\Omega}u_{d}^{p+1}\,dx. \end{align*} $$

Since $u_{d}$ is a critical point of $J_{d}$ ,

$$ \begin{align*} J_{d}^{\prime}(u_{d})=0 \quad\text{on } H_{\Omega}^{s}. \end{align*} $$

This implies that

(3-5) $$ \begin{align} d\frac{c_{n,s}}{2} \int_{T(\Omega)} \frac{\vert u_{d}(x)-u_{d}(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u_{d}^{2}\,dx =\int_{\Omega}u_{d}^{p+1}\,dx. \end{align} $$

Hence, from the above equations,

(3-6) $$ \begin{align} J_{d}(u_{d})&= \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{\Omega}u_{d}^{p+1}\,dx \\&= \frac{(p-1)}{2(p+1)} \int_{\Omega}u_{d}^{p+1}\,dx.\nonumber \end{align} $$

Now, by Theorem 3.2, we have $J_{d}(u_{d})\leq C d^{{n}/{2s}}$ , where the constant C depends only on $p.$ Using this inequality in the above equation,

$$ \begin{align*} \int_{\Omega}u_{d}^{p+1}\,dx \leq \frac{2(p+1)}{p-1}Cd^{{n}/{2s}}. \end{align*} $$

Taking $C_{0}={2(p+1)}/({p-1})C$ proves the first inequality of Theorem 1.2. The proof of the second inequality of Theorem 1.2 is a little constructive. We claim that

$$ \begin{align*}\displaystyle \sup_{\Omega}u_{d}(x) \leq C_{1}\end{align*} $$

for some constant $C_{1}>0$ depending on p and $\Omega $ only. Multiplying (1-1) by $u_{d}^{2t-1}$ and integrating over $\Omega ,$

(3-7) $$ \begin{align} \frac{c_{n,s}d}{2}\int_{T(\Omega)} \frac{(u_{d}(x)-u_{d}(y))(u_{d}^{2t-1}(x)-u_{d}^{2t-1}(y))}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u_{d}^{2t}\,dx = \int_{\Omega}u_{d}^{p+2t-1}\,dx. \end{align} $$

Now, we use the following inequality. We give the proof of this inequality in the Appendix. Let $x, y \geq 0$ be real numbers and $k\geq 1$ , then

(3-8) $$ \begin{align} \frac{1}{k}(x^{k}-y^{k})^{2}\leq (x-y)(x^{2k-1}-y^{2k-1}). \end{align} $$

Consequently,

(3-9) $$ \begin{align} \frac{1}{t}\int_{T(\Omega)} \frac{(u_{d}^{t}(x)-u_{d}^{t}(y))^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy \leq \int_{T(\Omega)} \frac{(u_{d}(x)-u_{d}(y))(u_{d}^{2t-1}(x)-u_{d}^{2t-1}(y))}{\vert x-y\vert^{n+2s}}\,dx\,dy. \end{align} $$

From (3-7) and (3-9),

(3-10) $$ \begin{align} \frac{dc_{n,s}}{2t}\int_{T(\Omega)} \frac{(u_{d}^{t}(x)-u_{d}^{t}(y))^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u_{d}^{2t}\,dx \leq \int_{\Omega}u_{d}^{p+2t-1}\,dx. \end{align} $$

Further, by the fractional Sobolev embedding (Theorem 2.1),

(3-11) $$ \begin{align} \bigg(\int_{\Omega}{\vert v\vert^{2_{s}^{*}}} \bigg)^{2/2_{s}^{*}} \leq \frac{A}{d}\bigg(d\frac{c_{n,s}}{2}\int_{\Omega}\int_{\Omega} \frac{\vert v(x)-v(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}\vert v\vert^{2}\,dx \bigg), \end{align} $$

where $d\in (0,d_{0})$ for some $d_{0}>0$ , $A>0$ some constant, $v\in H^{s}(\Omega )$ and $2_{s}^{*}={2n}/({n-2s}).$ The embedding constant A depends only on n, s, $d_{0}$ and $\Omega .$ To see this, let us define

$$ \begin{align*} \Omega_{d}:=\bigg \{ y : \frac{y}{d^{1/2s}} \in \Omega \bigg \} \quad\text{and}\quad w(y):=v \bigg (\frac{y}{d^{1/2s}} \bigg ),\quad\text{where } y \in \Omega_{d}. \end{align*} $$

Now,

$$ \begin{align*} &d\int_{\Omega}\int_{\Omega} \frac{\vert v(x)-v(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}{v}^{2}\,dx\\&\quad= \frac{1}{d^{{n}/{2s}}} \bigg [ \int_{\Omega_{d}}\int_{\Omega_{d}} \frac{\vert v (\tfrac{x'}{d^{{1}/{2s}}} )-v(\tfrac{y'}{d^{{1}/{2s}}} )\vert^{2}}{\vert x'-y'\vert^{n+2s}}\,dx'\,dy' + \int_{\Omega_{d}}{v\bigg (\frac{x'}{d^{{1}/{2s}}} \bigg )}^{2}\,dx' \bigg ] \\&\quad= \frac{1}{d^{{n}/{2s}}} \bigg [ \int_{\Omega_{d}}\int_{\Omega_{d}} \frac{\vert w(x')-w(y')\vert^{2}}{\vert x'-y'\vert^{n+2s}}\,dx'\,dy' + \int_{\Omega_{d}}{w(x')}^{2}\,dx' \bigg ] \\&\quad\geq \frac{A}{d^{{n}/{2s}}} \bigg(\int_{\Omega_{d}}{\vert w\vert^{2_{s}^{*}}} \,dx' \bigg)^{{2}/{2_{s}^{*}}} \\&\quad= Ad^{({2}/{2_{s}^{*}}-1 ){n}/{2s}} \bigg(\int_{\Omega}{\vert v\vert^{2_{s}^{*}}} \,dx \bigg)^{{2}/{2_{s}^{*}}}. \end{align*} $$

Therefore, we observe that A is uniform for $d \in (0, d_{0}).$

It is easy to see that $\Omega \times \Omega \subset T(\Omega ).$ Then, by virtue of (3-10) and (3-11),

(3-12) $$ \begin{align} \bigg(\int_{\Omega}{\vert u_{d}\vert^{t2_{s}^{*}}} \bigg)^{{2}/{2_{s}^{*}}} \leq \frac{tA}{d}\int_{\Omega}u_{d}^{p+2t-1}\,dx. \end{align} $$

Now, we define two sequences $\{L_{j} \}$ and $\{M_{j} \}$ by the following recurrence relations:

(3-13) $$ \begin{align} p-1+2L_{0}= & \, 2_{s}^{*}, \nonumber \\ p-1+2L_{j+1}= & \, 2_{s}^{*}L_{j}, \quad j=0,1,2,\ldots \end{align} $$

(3-14) $$ \begin{align} M_{0}=& \, (AC_{0})^{{2_{s}^{*}}/{2}}, \nonumber \\ M_{j+1}=& \, (AL_{j}M_{j})^{{2_{s}^{*}}/{2}}, \quad j=0,1,2,\ldots \end{align} $$

We note that $L_{j}$ is explicitly given by

(3-15) $$ \begin{align} L_{j}=\frac{1}{(2_{s}^{*}-2)}\bigg( \bigg(\frac{2_{s}^{*}}{2}\bigg)^{j+1}(2_{s}^{*}-p-1)+p-1 \bigg). \end{align} $$

Since $1<p<2_{s}^{*}-1$ , it follows that $L_{j}\geq 1$ for all $j \geq 0$ and $L_{j} \rightarrow \infty $ as $j \rightarrow \infty .$ We show that

(3-16) $$ \begin{align} \int_{\Omega}u_{d}^{p-1+2L_{j}}\,dx \leq & \, M_{j}d^{{n}/{2s}} \quad\text{for all } j\geq 0, \end{align} $$

and

(3-17) $$ \begin{align} M_{j}\leq e^{mL_{j-1}} \end{align} $$

for some constant $m>0.$ Then,

$$ \begin{align*} \sup_{\Omega}u_{d}(x)\leq C_{1},\end{align*} $$

where $C_{1}>0$ depends only on $C_{0}$ and $\Omega .$ In fact, (3-15) and (3-16) give

$$ \begin{align*} \Vert u\Vert_{L^{2_{s}^{*}L_{j-1}}(\Omega)} &\leq ( e^{mL_{j-1}}d^{{n}/{2s}} )^{{1}/{(2_{s}^{*}L_{j-1})}} \nonumber \\ &= e^{{m}/{2_{s}^{*}} d^{{(n-2s)}/{4L_{j-1}}}} \end{align*} $$

and hence letting $j \rightarrow \infty ,$

$$ \begin{align*}\Vert u\Vert_{L^{\infty}(\Omega)} \leq e^{{m}/{2_{s}^{*}}}.\end{align*} $$

First, we verify (3-16). By virtue of (1-6) and (3-11),

$$ \begin{align*} \bigg(\int_{\Omega}{\vert u_{d}\vert^{2_{s}^{*}}} \bigg)^{{2}/{2_{s}^{*}}} &\leq \frac{A}{d}\bigg( \frac{c_{n,s}d}{2}\int_{T(\Omega)} \frac{\vert u_{d}(x)-u_{d}(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}\vert u_{d}\vert^{2}\,dx \bigg) \nonumber \\ &\leq \frac{A}{d}C_{0}d^{{n}/{2s}} \nonumber \\ &= AC_{0}d^{{n}/{s2_{s}^{*}}}. \end{align*} $$

Hence, (3-16) holds for $j=0.$ Suppose that we have proved (3-16) for $j \geq 0.$ Then, by (3-12),

$$ \begin{align*} \int_{\Omega}{\vert u_{d}\vert^{p-1+2L_{j+1}}}\,dx &\leq \bigg (\frac{L_{j}A}{d}\int_{\Omega}u_{d}^{p+2L_{j}-1}\,dx \bigg )^{{2_{s}^{*}}/{2}} \nonumber \\ &\leq ( AL_{j}d^{-1}M_{j}d^{{n}/{2s}} )^{{2_{s}^{*}}/{2}} \nonumber \\ &= (AL_{j}M_{j} )^{{2_{s}^{*}}/{2}}d^{{n}/{2s}}. \end{align*} $$

This implies that (3-16) is also true for $j+1.$ Therefore, it remains to show (3-17). Put

(3-18) $$ \begin{align} \lambda_{j}= & \, \frac{2_{s}^{*}}{2} \cdot \log(AL_{j}) \quad\text{and}\quad \eta_{j}= \log(M_{j}). \end{align} $$

Hence,

$$ \begin{align*} \eta_{j+1}=& \, \frac{2_{s}^{*}}{2}\cdot \eta_{j} +\lambda_{j}. \end{align*} $$

The explicit value of $L_{j}$ is given by

(3-19) $$ \begin{align} L_{j}= (2_{s}^{*}-2)^{-1} ( (2^{-1}2_{s}^{*})^{j+1}(2_{s}^{*}-p-1)+p-1 ). \end{align} $$

Now,

$$ \begin{align*} \lambda_{j}=& \, \frac{2_{s}^{*}}{2}\log\bigg[ \frac{A}{(2_{s}^{*}-2)}((2^{-1}2_{s}^{*})^{j+1}(2_{s}^{*}-p-1)+p-1 ) \bigg ]\\ =& \, \frac{2_{s}^{*}}{2} \bigg [ \log(A(2_{s}^{*}-2))+ \log ((2^{-1}2_{s}^{*})^{j+1}(2_{s}^{*}-p-1)+p-1 ) \bigg ].\nonumber \end{align*} $$

Therefore, we can find some $C^{*}$ such that

$$ \begin{align*} \lambda_{j} \leq C^{*}(j+1). \end{align*} $$

We now define a sequence $ \{\gamma _{j} \}$ by

(3-20) $$ \begin{align} \gamma_{0}= \eta_{0} \quad\text{and}\quad \gamma_{j+1}= \frac{2_{s}^{*}}{2}\gamma_{j} + C^{*}(j+1) \end{align} $$

for $j \geq 1.$ Clearly, $\eta _{j} \leq \gamma _{j}$ for all $j \geq 0.$ Moreover, since

$$ \begin{align*} \gamma_{j}= \bigg (\frac{2_{s}^{*}}{2} \bigg )^{j}( \eta_{0}+2C^{*}2_{s}^{*}(2_{s}^{*}-2)^{-2} )-2C^{*}(2_{s}^{*}-2)^{-1} (j+ 2_{s}^{*}(2_{s}^{*}-2)), \end{align*} $$

in view of (3-19), there exists $m>0$ such that $\gamma _{j} \leq mL_{j-1}.$ Hence, $\log (M_{j}) \leq m L_{j-1}$ and we obtain (3-17). Observe that m depends only on $\eta _{0}$ , $2_{s}^{*}$ and $C^{*},$ whereas $C^{*}$ depends only on $2_{s}^{*}$ , p and $A.$ This completes the proof.

Remark 3.5. It is known that if $u \in \mathcal {L}_{s}(\mathbb {R}^{n})\cap C^{2s+\epsilon }(\Omega ), $ when $0<s<\tfrac 12, 2s+\epsilon <1$ or $u \in \mathcal {L}_{s}(\mathbb {R}^{n})\cap C^{1,2s+\epsilon -1}(\Omega ), $ when $\tfrac 12\leq s<1, 2s+\epsilon -1<1$ , one can compute $(-\Delta )^{s}u(x)$ pointwise for all x in $\Omega .$ In fact, one can write

$$ \begin{align*} (-\Delta)^{s}u(x)=c_{n,s}\ \mathrm{PV} \int_{\mathbb{R}^{n}} \frac{u(x)-u(y)}{\vert x-y\vert^{n+2s}}\,dy. \end{align*} $$

We make similar remarks as in [Reference Biagi, Dipierro, Valdinoci and Vecchi6], which offers a relation between the weak and classical solutions of (1-1).

Remark 3.7. Let $u_{d}$ be the least energy solution of (1-1) in $H^{s}_{\Omega }.$ Then, by Lemma 2.3, Theorem 1.2 and Lemma 3.4:

1. (1) for $0<s<\tfrac 12$ , $u_{d} \in \mathcal {L}_{s}(\mathbb {R}^n)\cap C^{2}(\Omega )$ if $p>3-2s$ and $u_{d} \in \mathcal {L}_{s}(\mathbb {R}^n) \cap C^{1, p-2+2s}(\Omega )$ if $2<p\leq 3-2s$ ;

2. (2) for $\tfrac 12 \leq s < 1$ , $u_d \in \mathcal {L}_{s}(\mathbb {R}^n) \cap C^{2}(\Omega ).$

Now, using the nonlocal integration by parts formulae given in [Reference Dipierro, Ros-Oton and Valdinoci14], one can easily check that

$$ \begin{align*} d(-\Delta)^{s}u_{d}(x) + u_{d}(x)=\vert u_{d}(x)\vert^{p-1}u_{d}(x) \end{align*} $$

holds pointwise in $\Omega .$ This implies that $u_{d}$ is a classical solution of (1-1). Conversely, if $u_{d}$ is a classical solution of (1-1) satisfying $u_{d} \in H^{s}_{\Omega },$ then $u_{d}$ is a weak solution of (1-1).

The following lemma shows that the maximum of the least energy solution is always greater than unity.

Lemma 3.8. Let $u_d$ be the least energy solution of (1-1). Let

$$ \begin{align*} M_{d}=\sup_{x\in \overline{\Omega}} u_{d}(x). \end{align*} $$

Then, $M_{d}>1.$

Proof. Since $u_{d}$ is a weak solution of (1-1),

$$ \begin{align*} &d\frac{c_{n,s}}{2}\int_{T(\Omega)}\frac{(u_{d}(x)-u_{d}(y))(w(x)-w(y))}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u_{d}w \,dx =\int_{\Omega} u_{d}^{p}w \,dx \end{align*} $$

holds for all $w \in H_{\Omega}^{s}$ . Taking $w=1$ in the above equation,

$$ \begin{align*} \int_{\Omega}u_{d}(x)\,dx=& \,\int_{\Omega}u_{d}^{p}(x)\,dx. \end{align*} $$

This implies that

$$ \begin{align*} \int_{\Omega}u_{d}(x)(1-u_{d}^{p-1}(x))\,dx=& \,0. \end{align*} $$

Now, if $u_{d}(x)\leq 1$ for all $x \in \overline {\Omega },$ then

$$ \begin{align*}1-u_{d}(x)\geq 0 \quad\text{for all } x \in {\overline{\Omega}} .\end{align*} $$

Thus, from the above equation, we get that $u_{d}(x)= 1~a.e.$ in $\overline {\Omega }.$ Now, by Lemma 3.4, we can assume that $u_{d}$ is continuous and hence $u_{d}\equiv 1$ in $\overline {\Omega },$ which is a contradiction to our assumption that $u_{d}$ is a nonconstant solution. Therefore, there exists $x_{0}$ in $\overline {\Omega }$ such that $u_{d}(x_{0})>1.$ Thus, $M_{d}>1.$

4. $L^{r}$ -estimates on $u_{d}$

Here, we derive an $L^{r}$ -estimate for $u_{d}.$ The following results are generalisations of [Reference Lin, Ni and Takagi31, Proposition 2.2 and Lemma 2.3] to the nonlocal case.

Proposition 4.1. For $d_{0}>0$ fixed, there is a constant $K_{0}$ such that

(4-1) $$ \begin{align} d\frac{c_{n,s}}{2}\int_{T(\Omega)}\frac{(u_{d}(x)-u_{d}(y))^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u_{d}^{2} \,dx \geq K_{0}d^{{n}/{2s}}, \end{align} $$

where $u_{d}$ is the least energy solution of (1-1) with $0<d<d_{0}.$

Proof. In contrast, suppose that there is a sequence $\{d_{k}\}$ contained in the interval $(0,d_{0})$ and a sequence of positive solutions $\{u_{k}\}$ to (1-1) with $d=d_{k}$ such that

$$ \begin{align*} \zeta_{k}:= \frac{1}{d^{{n}/{2s}}}\bigg(d \frac{c_{n,s}}{2} \int_{T(\Omega)}\frac{(u_{k}(x)-u_{k}(y))^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u_{k}^{2} \,dx \bigg) \rightarrow 0 \quad\text{as } k\rightarrow \infty. \end{align*} $$

We are going to follow the same arguments as used in the proof of Lemma 1.2 to prove this proposition. Once again, define the sequences $\{L_{k}\}$ and $\{M_{j}\}$ as defined earlier in (3-13) and (3-14), respectively. Instead of $C_{0},$ we write $\zeta _{k}$ in the definition of $\{M_{j}\}$ :

$$ \begin{align*} p-1+2L_{0}= & \, 2_{s}^{*}, \nonumber \\ p-1+2L_{j+1}= & \, 2_{s}^{*}L_{j}, \quad j=0,1,2,\ldots \end{align*} $$

and

$$ \begin{align*} M_{0}=& \, (A\zeta_{k})^{{2_{s}^{*}}/{2}}, \nonumber \\ M_{j+1}=& \, (AL_{j}M_{j})^{{2_{s}^{*}}/{2}}, \quad j=0,1,2,\ldots. \end{align*} $$

Further, define the sequences $\{\lambda _{j}\}$ , $\{\eta _{j}\}$ and $\{\gamma _{j}\}$ as defined earlier in (3-18) and (3-20). From (3-16),

(4-2) $$ \begin{align} \bigg(\int_{\Omega}u_{k}^{2_{s}^{*}L_{j-1}}\,dx\bigg)^{(2_{s}^{*}L_{j-1})} \leq & \, (M_{j}d_{k}^{n/2s} )^{1/(2_{s}^{*}L_{j-1})}. \end{align} $$

Since

$$ \begin{align*}\log(M_{j})=\eta_{j} \leq \gamma_{j},\end{align*} $$

we have

$$ \begin{align*} \frac{\log(M_{j})}{2_{s}^{*}L_{j-1}} \leq \frac{\eta_{j}}{2_{s}^{*}L_{j-1}}. \end{align*} $$

Now,

$$ \begin{align*} \lim_{j \rightarrow \infty} \frac{\eta_{j}}{2_{s}^{*}L_{j-1}}&= \lim_{j \rightarrow \infty}\frac{\big(\tfrac{2_{s}^{*}}{2}\big)^{j}[ \eta_{0}+2C^{*}2_{s}^{*}(2_{s}^{*}-2)^{-2} ]-2C^{*}(2_{s}^{*}-2)^{-1} [j+ 2_{s}^{*}(2_{s}^{*}-2)]}{\tfrac{2_{s}^{*}}{(2_{s}^{*}-2)}\Big [ \big(\tfrac{2_{s}^{*}}{2}\big)^{j}(2_{s}^{*}-p-1)+p-1 \Big ]}\\[3pt] &= \frac{(2_{s}^{*}-2)(\eta_{0}+2C^{*}2_{s}^{*}(2_{s}^{*}-2)^{-2})}{2_{s}^{*}(2_{s}^{*}-p-1)}. \end{align*} $$

Letting $j \rightarrow \infty $ in (4-2),

(4-3) $$ \begin{align} \Vert u_{k}\Vert_{L^{\infty}(\Omega)} \leq e^{a_{1}(\eta_{0}+a_{2})}, \end{align} $$

with $a_{1}$ and $a_{2}$ depending only on $2_{s}^{*},~p$ and $C^{*}.$ Since

$$ \begin{align*} \eta_{0}= \log (M_{0})= \frac{2_{s}^{*}}{2}\log(A\zeta_{k}), \end{align*} $$

as $k \rightarrow \infty ,~\eta _{0}\rightarrow -\infty .$ Thus, in view of (4-3),

$$ \begin{align*} \Vert u_{k}\Vert_{L^{\infty}(\Omega)} \rightarrow 0, \end{align*} $$

which leads to a contradiction to Lemma 3.8.

Proof of Theorem 1.3

First, we show the second part of (1-7).

Case I. $r \geq 2_{s}^{*}={2n}/({n-2s}).$ Let $\{L_{j} \}$ be the sequence defined in (3-13). If $r \in \{2_{s}^{*}L_{j} \}$ , then the second inequality of (1-7) follows from (3-16). So assume that $2_{s}^{*}L_{j}<r< 2_{s}^{*}L_{j+1}$ for some $j \geq 0.$ We have

$$ \begin{align*}r=t2_{s}^{*}L_{j}+(1-t)2_{s}^{*}L_{j+1} \text{ for some } t \in (0,1).\end{align*} $$

Using the Hölder inequality and (3-16),

$$ \begin{align*} \int_{\Omega}u_{d}^{r}\,dx = & \, \int_{\Omega}u_{d}^{t2_{s}^{*}L_{j}+(1-t)2_{s}^{*}L_{j+1}}\,dx, \\ \leq & \, \bigg( \int_{\Omega}u_{d}^{2_{s}^{*}L_{j}}\,dx \bigg)^{t}\bigg( \int_{\Omega}u_{d}^{2_{s}^{*}L_{j+1}}\,dx \bigg)^{1-t} \\ \leq & \, (M_{j-1}d^{n/2s} )^{t}(M_{j}d^{n/2s})^{1-t}\\ = & \, M_{j-1}^{t}M_{j}^{1-t}d^{{n}/{2s}}. \end{align*} $$

Case II. $2 \leq r \leq 2_{s}^{*}$ . We write

$$ \begin{align*}r=2t+(1-t)2_{s}^{*}, \end{align*} $$

for some $t \in [0,1].$ Then, using the Hölder inequality, from (1-6) and (3-16) with $j=0$ ,

$$ \begin{align*} \int_{\Omega}u_{d}^{r}\,dx \leq & \, \bigg( \int_{\Omega}u_{d}^{2}\,dx \bigg)^{t} \bigg( \int_{\Omega}u_{d}^{2_{s}^{*}}\,dx \bigg)^{1-t}\\ \leq & \, C_{0}^{t}M_{0}^{(1-t)}d^{{n}/{2s}}, \end{align*} $$

where the constant $C_{0}$ is independent of $d.$

Case III. $1 \leq r < p+1.$ Integrating both sides of (1-1) and using the condition $\mathcal {N}_{s}u(x)=0$ for $x\in \mathcal {C}\Omega ,$

(4-4) $$ \begin{align} \int_{\Omega}u_{d}\,dx= \int_{\Omega}u_{d}^{p}\,dx. \end{align} $$

It is easy to see that

$$ \begin{align*} p=t+(1-t)(p+1) \quad\text{with } t=\frac{1}{p} \in (0,1). \end{align*} $$

Notice that $p+1 \in (2, 2_{s}^{*}).$ Therefore, using the Hölder inequality and (4-4),

$$ \begin{align*} \int_{\Omega}u_{d}^{p}\,dx \leq & \, \bigg( \int_{\Omega}u_{d}\,dx \bigg)^{t}\bigg( \int_{\Omega}u_{d}^{p+1}\,dx \bigg)^{(1-t)}, \\ \int_{\Omega}u_{d}^{p}\,dx \leq & \, \int_{\Omega}u_{d}^{p+1}\,dx \leq C_{0}d^{{n}/{2s}} \quad (\text{by}\ ({1\text{-}6})), \end{align*} $$

where the constant $C_{0}$ depends only upon $p+1.$

Also, in view of (4-4) and (1-6), we observe that the second inequality of (1-7) holds for $r=1.$ Now, repeating the interpolation between $1$ and $p+1$ , we see that the second inequality of (1-7) holds for all r $\geq 1.$

Case IV. Let $0<r \leq 1.$ Taking $F=u_{d}^{r},\, G=1, \, p=\frac {1}{r}, \, q=\frac {1}{1-r}$ and using the Hölder inequality,

$$ \begin{align*} \int_{\Omega}u_{d}^{r}\,dx \leq \Vert F\Vert_{p}\Vert G\Vert_{q}=\vert\Omega\vert^{1-r} \bigg(\int_{\Omega}u_{d}\,dx \bigg)^{r} \leq \vert\Omega\vert^{1-r} B(1)^{r}d^{{nr}/{2s}}. \end{align*} $$

This proves the second inequality of (1-8).

Now, let us prove the first inequality of (1-7) and (1-8). In view of (3-5) and (4-1),

$$ \begin{align*} \int_{\Omega}u_{d}^{p+1} \geq K_{0}d^{{n}/{2s}}. \end{align*} $$

Since

$$ \begin{align*}\displaystyle \sup_{\Omega}u_{d}(x) \leq C_{1}\quad \text{for some constant } C_{1}>0,\end{align*} $$

we have

$$ \begin{align*} K_{0}d^{{n}/{2s}}\leq \int_{\Omega}u_{d}^{p+1} &= \, \int_{\Omega}(u_{d}^{p+1-r})(u_{d}^{r})\,dx \\ &\leq C_{1}^{p+1-r}\int_{\Omega}u_{d}^{r}\,dx. \end{align*} $$

This implies that

$$ \begin{align*}\int_{\Omega}u_{d}^{r}\,dx \geq K_{0}C_{1}^{r-p-1}d^{{n}/{2s}}, ~r<p+1.\end{align*} $$

For $r>p+1$ , we write $p+1=1+(1-t)r.$ Therefore,

$$ \begin{align*} K_{0}d^{{n}/{2s}}\leq & \, \int_{\Omega}u_{d}^{p+1}\,dx \\ =& \, \int_{\Omega}u_{d}^{1+(1-t)r}\,dx \\ \leq & \, (u_{d}\,dx )^{t}(u_{d}^{r}\,dx )^{1-t} \\ \leq & \, ( B(1)d^{{n}/{2s}} )^{t} ( u_{d}^{r}\,dx )^{1-t}. \end{align*} $$

This yields that

$$ \begin{align*}\int_{\Omega}u_{d}^{r}\,dx \geq (K_{0}B(1)^{-t})^{{1}/({1-t})}d^{{n}/{2s}}.\\[-41pt] \end{align*} $$

5. Proof of Theorem 1.4

In this section, we prove Theorem 1.4. Its proof is intricate and requires some scaling and compactness arguments. We prove the statements of the theorem one by one. Let $z_{d} \in \overline {\Omega } $ be a point of maximum of $u_{d}.$ We approximate $u_{d}$ around $z_{d}$ by a scaled positive radial solution of (1-5). It gives us an upper bound on $c_{d},$ which is closely related to the location of point $z_{d}.$

Step I. We prove that there exists a positive constant $K_{*}$ such that

(5-1) $$ \begin{align} \rho (z_{d}, \partial \Omega) \leq K_{*}d^{{1}/{2s}} .\end{align} $$

If the inequality in (5-1) is not true, then there is a decreasing sequence $d_{j}\downarrow 0$ such that

(5-2) $$ \begin{align} \rho_{j}:=\frac{\rho(z_{j}, \partial \Omega)}{d_{j}^{{1}/{2s}}} \rightarrow +\infty \quad\text{as } j \rightarrow \infty, \end{align} $$

where $z_{j}:=z_{d_{j}}$ is a point of maximum of $u_{d_{j}}$ on $\overline {\Omega }.$ Define

$$ \begin{align*}\phi_{j}(y):=u_{d_{j}}(yd_{j}^{{1}/{2s}}+z_{j}) \quad \text{for } y \in \mathbb{R}^{n}.\end{align*} $$

Since $u_{d}$ is a classical solution of (1-1),

(5-3) $$ \begin{align} (-\Delta)^{s}\phi_{j}+\phi_{j}=\phi_{j}^{p} \quad \text{in } B_{\rho_{j}}, \end{align} $$

and:

1. (1) $ \phi _{j} \in C^{0,2s+\epsilon }(B_{\rho _{j}}),$ when $0<s<\tfrac 12$ , $ 2s+\epsilon <1; $

2. (2) $ \phi _{j} \in C^{1,2s+\epsilon -1}(B_{\rho _{j}}),$ when $\tfrac 12 \leq s <1$ , $2s+\epsilon -1<1.$

First, we claim that the sequence $\{\phi _{j}\}$ contains a convergent subsequence. Let $\{R_{k}\}$ be a monotone increasing sequence of positive numbers with $R_{k} \rightarrow +\infty $ as $k \rightarrow \infty .$ Therefore, we have for each $k,$ there is a number $j_{k}$ such that $4R_{k}< \rho _{j}$ whenever $j \geq j_{k}.$ Since $u_{d} \in L^{\infty }(\mathbb {R}^n) \cap \mathcal {L}_{s}(\mathbb {R}^n),$ we have $ \phi _{{j}} \in L^{\infty }(\mathbb {R}^n) \cap \mathcal {L}_{s}(\mathbb {R}^n)$ for each $j \geq 1.$ Now, we can use [Reference Fall19, Theorem 1.4] to get the following estimates.

For $0<s<\tfrac 12$ , $ 2s+\epsilon <1$ :

1. (i) let $4s+\epsilon <1,$ then

   $$ \begin{align*} \Vert\phi_{{j}}\Vert_{C^{0,4s+\epsilon} (B_{2R_{k}})} \leq C (\Vert\phi_j\Vert_{L^{\infty} (\mathbb{R}^{n})} + \Vert\phi_{j}^{p}-\phi_j\Vert_{C^{0,2s+\epsilon}(B_{4R_{k}})} ); \end{align*} $$

2. (ii) let $1<4s+\epsilon <2,$ then

   $$ \begin{align*} \Vert\phi_{{j}}\Vert_{C^{1,4s+\epsilon-1} (B_{2R_{k}})} \leq C (\Vert\phi_j\Vert_{L^{\infty} (\mathbb{R}^{n})} + \Vert\phi_{j}^{p}-\phi_j\Vert_{C^{0,2s+\epsilon}(B_{4R_{k}})} ); \end{align*} $$

and for $\tfrac 12 \leq s <1$ , $2s+\epsilon -1<1$ :

1. (iii) let $4s+\epsilon -1<1,$ then

   $$ \begin{align*} \Vert\phi_{{j}}\Vert_{C^{1,4s+\epsilon-1} (B_{2R_{k}})} \leq C (\Vert\phi_j\Vert_{L^{\infty} (\mathbb{R}^{n})} + \Vert\phi_{j}^{p}-\phi_j\Vert_{C^{1,2s+\epsilon-1}(B_{4R_{k}})} ); \end{align*} $$

2. (iv) let $1<4s+\epsilon -1<2,$ then

   $$ \begin{align*} \Vert\phi_{{j}}\Vert_{C^{2,4s+\epsilon-1} (B_{2R_{k}})} \leq C (\Vert\phi_j\Vert_{L^{\infty} (\mathbb{R}^{n})} + \Vert\phi_{j}^{p}-\phi_j\Vert_{C^{1,2s+\epsilon-1}(B_{4R_{k}})} ), \end{align*} $$

where the constant $C>0$ is independent of $j.$

Let us recall the inequality (1-6) here:

$$ \begin{align*} d \frac{c_{n,s}}{2} \int_{T(\Omega)} \frac{\vert u_{d}(x)-u_{d}(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}u^{2}\,dx=\int_{\Omega} u_{d}^{p+1} \leq C_{0}d^{{n}/{2s}}, \end{align*} $$

where $C_{0}$ is independent of $d.$ This yields

$$ \begin{align*} \int_{B_{\rho_{j}}} \phi_{j}^{p+1} \leq C_{0}, \end{align*} $$

and

(5-4) $$ \begin{align} \Vert\phi_{j}\Vert_{H^{s}(B_{\rho_{j}})} \leq C_{0} \quad\text{for all } j\geq 1. \end{align} $$

Also, by Theorem 1.3,

$$ \begin{align*} \int_{\Omega} u_{d}^{r} \leq B(r)d^{{n}/{2s}} \quad \text{for all } r \geq 1, \end{align*} $$

which implies that

(5-5) $$ \begin{align} \int_{B_{\rho_{j}}} \phi_{j}^r \leq B(r)\quad\text{for all } j\geq 1 \text{ and } r \geq 1. \end{align} $$

By Lemma 3.4 and Theorem 1.2,

(5-6) $$ \begin{align} \Vert u_{d}\Vert_{L^{\infty}(\mathbb{R}^{n})} \leq C_{1}, \end{align} $$

where the constant $C_{1}$ is independent of the diffusion constant $d.$ So, (5-5), (5-6) and [Reference Fall19, Theorem 1.3] imply that

$$ \begin{align*} \Vert\phi_{j}\Vert_{X_{s}(\overline{B}_{R_{k}})} < C_2 \quad \text{for all } j \geq j_{k}, \end{align*} $$

where the constant $C_2>0$ is independent of j and the space $X_{s}(\overline {B}_{R_{k}})$ is identified with one of the spaces $C^{0,4s+\epsilon }(\overline {B}_{R_{k}})$ , $C^{1,4s+\epsilon -1}(\overline {B}_{R_{k}})$ or $C^{2,4s+\epsilon -1}(\overline {B}_{R_{k}})$ with the same assumptions on s and $\epsilon $ as above. Therefore, $\{\phi _{j}\}$ is a relatively compact set in $X_{s}(\overline {B}_{R_{k}}),$ and hence by the standard diagonal process, we can extract a convergent subsequence of $\{\phi _{j}\},$ which we continue to denote by $\{\phi _{j}\}$ itself such that

$$ \begin{align*} \phi_{j} \rightarrow v \quad \text{in } C^{0, 2s+\epsilon}_{loc}(\mathbb{R}^{n}) \quad\text{when } 0<s<\tfrac{1}{2}, 2s+\epsilon<1 \end{align*} $$

or

$$ \begin{align*} \phi_{j} \rightarrow v \quad \text{in } C^{1, 2s+\epsilon-1}_{loc}(\mathbb{R}^{n}) \quad\text{when } \tfrac{1}{2}<s<1, 2s+\epsilon-1<1 \end{align*} $$

for some $v.$ The limit $v \in C^{0,2s+\epsilon }(\mathbb {R}^{n}) \cap H^{s}(\mathbb {R}^{n})$ when $0<s<\tfrac 12, 2s+\epsilon <1$ or $v \in C^{1,2s+\epsilon -1}(\mathbb {R}^{n}) \cap H^{s}(\mathbb {R}^{n})$ when $\tfrac 12<s<1, 2s+\epsilon -1<1$ follows from (5-4). Consequently,

$$ \begin{align*} \lim_{\vert x\vert\rightarrow \infty}v(x)=0. \end{align*} $$

Using [Reference Du, Jin, Xiong and Yang17, Theorem 1.1], we have $(-\Delta )^{s}\phi _{j}(x) $ converges to $ (-\Delta )^{s}v(x)$ point-wise in $\mathbb {R}^{n}.$ Consequently, we see that the limit v satisfies the equation

$$ \begin{align*} (-\Delta)^{s}v+v=v^{p} \quad \text{in } \mathbb{R}^{n}. \end{align*} $$

Clearly, $v \geq 0$ because each $\phi _{j} \geq 0.$ Since by Lemma 3.8 we have $\phi _{j}(0)=u_{d_{j}}(z_{j})>1$ for each $j \geq 1,$ one can see that $v \not \equiv 0.$

Using Theorem 2.9, one can see that v is radially symmetric and decreasing about some point in $\mathbb {R}^{n}.$ Since

$$ \begin{align*} \nabla v(0)= \lim_{j \rightarrow \infty} \nabla \phi_{j}(0)=0, \end{align*} $$

v is radially symmetric about the origin. Additionally, by Theorem 2.8, v has a power type of decay at infinity, that is,

$$ \begin{align*} v(r)\leq \frac{C_{2}}{r^{n+2s}}, \quad r \geq 1. \end{align*} $$

Now we derive a lower bound on the critical value $c_{d_{j}}.$ Let us define

(5-7) $$ \begin{align} \delta_{R}:= \frac{C_{2}}{R^{n+2s}}, \end{align} $$

where $R>0$ is an arbitrarily large real number. Then, there exists a positive integer $j_{R}$ such that if $j \geq j_{R}$ , then $\rho _{j}\geq 2R$ and

(5-8) $$ \begin{align} \Vert\phi_{j}-v\Vert_{C^{2}(\overline{B}_{2R})} \leq \delta_{R}. \end{align} $$

By Lemma 3.3,

$$ \begin{align*}c_{d_{j}}= M[u_{d_{j}}]=J_{d_{j}}(u_{d_{j}}).\end{align*} $$

Using this fact and (3-6),

$$ \begin{align*} c_{d_{j}}&= \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{\Omega}u_{d_{j}}^{p+1}\,dx \\ &\geq \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg) \int_{\vert x-z_{j}\vert<d_{j}^{{1}/{2s}}R}u_{d_{j}}^{p+1}\,dx \nonumber \\ &= d_{j}^{{n}/{2s}} \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg) \int_{\vert y\vert < R} \phi_{j}^{p+1}\,dy. \nonumber \end{align*} $$

Now,

(5-9) $$ \begin{align} c_{d_{j}}=& \, d_{j}^{{n}/{2s}} \bigg(\bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{B_{R}} v^{p+1}\,dy+ F_{j} \bigg), \end{align} $$

where

$$ \begin{align*}F_{j}:= \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg) \int_{B_{R}}\bigg( \phi_{{j}}^{p+1}- v^{p+1} \bigg)\,dy. \end{align*} $$

By Equation (5-8), we have for all $y \in B_{R},~ j \geq j_{R}$ ,

$$ \begin{align*} \vert \phi_{{j}}^{p+1}- v^{p+1}\vert \leq C \vert\phi_{{j}}-v\vert \leq \delta_{R}, \end{align*} $$

where $C>0$ is some constant. This implies that

$$ \begin{align*} \vert F_{j}\vert \leq \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)C \vert B_{R}\vert \delta_{R}=C_{3}R^{n}\delta_{R}, \end{align*} $$

where

$$ \begin{align*}C_{3}= \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\frac{w_{n}}{n} C\end{align*} $$

and $w_{n}$ denotes the surface area of the unit sphere in $\mathbb {R}^{n}.$ Consequently, (5-9) becomes

(5-10) $$ \begin{align} c_{d_{j}}\geq d_{j}^{{n}/{2s}} \bigg[\bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{B_{R}} v^{p+1}\,dy-C_{3}R^{n}\delta_{R} \bigg]. \end{align} $$

Now, it is easy to see that

$$ \begin{align*} \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{B_{R}} v^{p+1}\,dy= F(v)-\bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{\vert y\vert>R} v^{p+1}\,dy, \end{align*} $$

where $F(v)$ is defined earlier in (2-2). Simplifying the second term on right-hand side,

$$ \begin{align*} \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{\vert y\vert>R} v^{p+1}\,dy&= \, \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg) \int_{R}^{\infty} \frac{r^{n-1}w_{n}}{r^{(n+2s)(p+1)}}\,dr \\ &= \, \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg) \frac{w_{n}}{(n+2s)p+2s}\frac{1}{R^{(n+2s)p+2s}}=\frac{C_{4}}{R^{(n+2s)p+2s}}. \end{align*} $$

Therefore, one can write

(5-11) $$ \begin{align} \bigg(\frac{1}{2}-\frac{1}{p+1}\bigg)\int_{B_{R}} v^{p+1}\,dy = F(v)-\frac{C_{4}}{R^{(n+2s)p+2s}}. \end{align} $$

On combining (5-7), (5-10) and (5-11), we get for $j \geq j_{R},$

(5-12) $$ \begin{align} c_{d_{j}} &\geq d_{j}^{{n}/{2s}}\bigg( F(v)- \frac{C_{4}}{R^{(n+2s)p+2s}}-\frac{C_{2}C_{3}}{R^{2s}} \bigg) \nonumber \\ &\geq d_{j}^{{n}/{2s}}\bigg( F(v)- \frac{C_{5}}{R^{2s}} \bigg) , \end{align} $$

where $C_{5}$ is independent of j and $R.$

Now, we derive an upper bound on the critical value $c_{d_{j}}.$ Without loss of generality, we may assume that the domain $\Omega $ is a subset of $\mathbb {R}^{n}_{+}$ and $0 \in \partial \Omega .$ Given Definition 2.13, let w be the ground state solution of (1-5). Define

$$ \begin{align*} \Omega_{d}&:=\bigg\{\frac{x}{d^{{1}/{2s}}} \mid x\in \Omega \bigg\},\\ w_{d}(x)&:=w\bigg( \frac{x}{d^{{1}/{2s}}}\bigg), \quad\text{for } x\in \mathbb{R}^{n}. \end{align*} $$

Since $w \geq 0$ , this implies that $w_{d} \geq 0$ . Define

$$ \begin{align*}g_{d}(t):=J_{d}(tw_{d}),\quad t \geq 0.\end{align*} $$

Then, by Lemma 3.3, there exists a unique $t_{0}=t_{0}(d)>0$ at which $g_{d}$ attains a maximum. It is easy to see that $t_{0}(d) \rightarrow 1$ as $d \downarrow 0.$ Hence,

$$ \begin{align*} M[w_{d}]&= J_{d}\bigg(t_{0}w_{d}\bigg) \\&= \frac{t_{0}^{2}}{2} \bigg[\frac{dc_{n,s}}{2}\int_{T(\Omega)}\frac{\vert w_{d}(x)-w_{d}(y)\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy+ \int_{\Omega}w_{d}^{2}\,dx\bigg]-\frac{t_{0}^{p+1}}{p+1}\int_{\Omega}w_{d}^{p+1}\,dx \\&= \frac{t_{0}^{2}}{2} \bigg[\frac{dc_{n,s}}{2}\int_{T(\Omega)}\frac{\bigg\vert w\bigg(\dfrac{x}{d^{{1}/{2s}}}\bigg)-w\bigg(\dfrac{y}{d^{{1}/{2s}}}\bigg)\bigg\vert^{2}}{\vert x-y\vert^{n+2s}}\,dx\,dy + \int_{\Omega}w^{2}\bigg(\frac{x}{d^{{1}/{2s}}}\bigg)\,dx\bigg]\\&\quad -\frac{t_{0}^{p+1}}{p+1}\int_{\Omega}w^{p+1}\bigg(\frac{x}{d^{{1}/{2s}}}\bigg)\,dx. \end{align*} $$

The change of variables

$$ \begin{align*} \frac{x}{d^{{1}/{2s}}}=a,\, \frac{y}{d^{{1}/{2s}}}=b, \end{align*} $$

gives us

$$ \begin{align*} M[w_{d}]&= d^{{n}/{2s}} \bigg(\frac{t_{0}^{2}}{2} \bigg[\frac{c_{n,s}}{2}\int_{T(\Omega_{d})}\frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db+ \int_{\Omega_{d}}w^{2}\,da\bigg]-\frac{t_{0}^{p+1}}{p+1}\int_{\Omega_{d}}w^{p+1}\,da \bigg) \\&= d^{{n}/{2s}} I_d \end{align*} $$

where $I_{d}$ is the expression

$$ \begin{align*} &\frac{t_{0}^{2}}{2} \bigg[\frac{c_{n,s}}{2}\!\int_{\Omega_{d}}\! \int_{\Omega_{d}}\! \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db+ 2c_{n,s}\int_{\mathcal{C}\Omega_{d}}\! \int_{\Omega_{d}}\! \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db +\! \int_{\Omega_{d}}\!w^{2}\,da\bigg]\\&\quad -\frac{t_{0}^{p+1}}{p+1}\int_{\Omega_{d}}w^{p+1}\,da. \end{align*} $$

Since

$$ \begin{align*} t_{0}(d) \rightarrow 1 \quad\text{as }d\downarrow 0, \end{align*} $$

we get for $I_d$

$$ \begin{align*} &\frac{1}{2} \bigg[\frac{c_{n,s}}{2}\!\int_{\mathbb{R}^{n}_{+}} \!\int_{\mathbb{R}^{n}_{+}}\! \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db+ 2c_{n,s}\!\int_{\mathcal{C} \mathbb{R}^{n}_{+}}\! \int_{\mathbb{R}^{n}_{+}}\! \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db + \!\int_{\mathbb{R}^{n}_{+}}\! w^{2}\,da\bigg]\\&\quad -\frac{1}{p+1}\int_{\mathbb{R}^{n}_{+}}w^{p+1}\,da + o(1)\end{align*} $$

as $d \downarrow 0.$ Further, w being nonnegative and radially symmetric implies that

$$ \begin{align*}\int_{\mathbb{R}^{n}_{+}}w^{2}\,da=\frac{1}{2}\int_{\mathbb{R}^{n}}w^{2}\,da,\,\, \int_{\mathbb{R}^{n}_{+}}w^{p+1}\,da=\frac{1}{2}\int_{\mathbb{R}^{n}}w^{p+1}\,da,\end{align*} $$

$$ \begin{align*}\int_{\mathbb{R}^{n}_{+}} \int_{\mathbb{R}^{n}_{+}} \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db = \frac{1}{4} \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db,\end{align*} $$

$$ \begin{align*} \int_{\mathcal{C}\mathbb{R}^{n}_{+}} \int_{\mathbb{R}^{n}_{+}} \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db = \frac{1}{4} \int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db.\end{align*} $$

Using these estimates,

$$ \begin{align*} I_{d} &< \frac{1}{2}\bigg( \frac{1}{2} \bigg[ \frac{c_{n,s}}{2}\int_{\mathbb{R}^{n}} \int_{\mathbb{R}^{n}} \frac{\vert w(a)-w(b)\vert^{2}}{\vert a-b\vert^{n+2s}}\,da\,db + \int_{\mathbb{R}^{n}}w^{2}\,da \bigg]- \frac{1}{p+1} \int_{\mathbb{R}^{n}}w^{p+1}\,da\bigg)+ o(1)\\& = \frac{1}{2}F(w)+o(1),\end{align*} $$

as $d \downarrow 0.$ Thus,

$$ \begin{align*} M[w_{d}]=d^{{n}/{2s}}I_{d}< \frac{d^{{n}/{2s}}}{2}F(w)+o(1), \end{align*} $$

as $d\downarrow 0.$ Using part (c) of Theorem 2.12, we have $0<F(w) \leq F(v)$ for any nonnegative nonzero classical solution v of (1-5) and by Lemma 3.3,

$$ \begin{align*}c_{d_{j}} \leq M[w_{d_{j}}]< \frac{d_{j}^{{n}/{2s}}}{2}F(v)\end{align*} $$

for $d_{j}$ sufficiently small. By letting R be sufficiently large in (5-12), we reach a contradiction. This proves (5-1).

Step II. Now, we claim that $z_{d} \in \partial \Omega .$ Suppose that there is a decreasing sequence $d_{j} \downarrow 0$ such that $z_{d_{j}}:=z_{j} \in \Omega .$ We have from Lemma 1.4 that the sequence $\{z_{j}\}$ converges to some $z \in \partial \Omega .$ Without loss of generality, let us assume that $z=0.$ Define

$$ \begin{align*} \widehat{u}_{j}(x):= \begin{cases} u_{d_{j} }(x)& \, \text{in } \mathbb{R}^{n}_{+}, \\ u_{d_{j} }(x', -x_{n}) & \, \text{in } \mathbb{R}^{n}_{-}, \end{cases} \end{align*} $$

where

$$ \begin{align*}x'=(x_{1},x_{2},\ldots, x_{n-1}),\quad \mathbb{R}^{n}_{+}=\{(x',x_{n}) \mid x_{n} \geq 0 \},\,\,\mathbb{R}^{n}_{-}=\{(x',x_{n}) \mid x_{n} \leq 0 \}. \end{align*} $$

Also, define a scaled function

(5-13) $$ \begin{align} \psi_{j}(y):=\widehat{u}_{j}(yd_{j}^{{1}/{2s}}+z_{j})\quad\text{for } y \in \mathbb{R}^{n}. \end{align} $$

Now, for $z_{j}=(z_{j}^{\prime },z_{jn}),$ we can write $z_{jn}=\alpha _{j}d_{j}^{{1}/{2s}}$ for some $\alpha _{j}>0.$ The sequence $\{\alpha _{j} \} $ is bounded, which follows from Lemma 1.4. Let

$$ \begin{align*} \rho_{j}:=\frac{\rho(z_{j}, \partial \Omega)}{d_{j}^{{1}/{2s}}} , \end{align*} $$

where $\rho (z_{j}, \partial \Omega )$ denotes the distance between $z_{j}$ and $ \partial \Omega .$ One can see easily that the function $\psi _{j}$ satisfies the equation

$$ \begin{align*} (-\Delta)^{s}\psi_{j}(y)+ \psi_{j}(y)= \psi_{j}(y)^{p}+ d_{j}{h}(y) \quad \text{in } B_{\rho_{j}} \end{align*} $$

for some function ${h}$ of $y.$ To see this, let $y \in B_{\rho _{j}}$ , so

(5-14) $$ \begin{align} (-\Delta)^{s}\psi_{j}(y)&=c_{n,s}\ \mathrm{PV} \int_{\mathbb{R}^{n}}\frac{\psi_{j}(y)-\psi_{j}(x)}{\vert y-x\vert^{n+2s}}\,dx= c_{n,s} \lim_{\epsilon \rightarrow 0} \int_{\mathcal{C} B_{\epsilon}(y)}\frac{\psi_{j}(y)-\psi_{j}(x)}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\ &= c_{n,s} \lim_{\epsilon \rightarrow 0} \bigg[ \int_{\mathcal{C} B_{\epsilon}(y)}\frac{\widehat{u}_{j}(yd_{j}^{{1}/{2s}}+z_{j})-\widehat{u}_{j}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx \bigg] \nonumber \\ &= c_{n,s} \lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \geq -\alpha_{j}\}\bigcap \mathcal{C} B_{\epsilon}(y)}\frac{\widehat{u}_{j}(yd_{j}^{{1}/{2s}}+z_{j})-\widehat{u}_{j}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\ & \quad+ c_{n,s} \lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \leq-\alpha_{j}\}\bigcap \mathcal{C}B_{\epsilon}(y)}\frac{\widehat{u}_{j}(yd_{j}^{{1}/{2s}}+z_{j})-\widehat{u}_{j}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx. \end{align} $$

For $y_{n} \geq -\alpha _{j},$

$$ \begin{align*} &(-\Delta)^{s}\psi_{j}(y)\nonumber\\&= c_{n,s}\lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \geq -\alpha_{j}\}\bigcap \mathcal{C} B_{\epsilon}(y)}\frac{{u}_{{j}}(yd_{j}^{{1}/{2s}}+z_{j})-{u}_{{j}}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\& \quad+ c_{n,s}\lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \leq-\alpha_{j}\}\bigcap \mathcal{C} B_{\epsilon}(y)}\frac{{u}_{{j}}(yd_{j}^{{1}/{2s}}+z_{j})-{u}_{{j}}(x' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(x_{n}+\alpha_{j}d_{j}^{{1}/{2s}}))}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\&= c_{n,s} \lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \geq -\alpha_{j}\} \bigcap \mathcal{C} B_{\epsilon}(y)}\frac{{u}_{{j}}(yd_{j}^{{1}/{2s}}+z_{j})-{u}_{{j}}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\&\quad + c_{n,s}\lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \leq-\alpha_{j}\}\bigcap \mathcal{C} B_{\epsilon}(y)}\frac{{u}_{{j}}(yd_{j}^{{1}/{2s}}+z_{j})-{u}_{{j}}(x'd_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(x_{n}+\alpha_{j}d_{j}^{{1}/{2s}}))}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\&= c_{n,s} \lim_{\epsilon \rightarrow 0} \int_{\mathcal{C} B_{\epsilon}(y)}\frac{{u}_{{j}}(yd_{j}^{{1}/{2s}}+z_{j})-{u}_{{j}}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\&\quad + c_{n,s}\lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \leq-\alpha_{j}\} \bigcap \mathcal{C}B_{\epsilon}(y)}\frac{{u}_{{j}}(xd_{j}^{{1}/{2s}}+z_{j}) -{u}_{{j}}(x'd_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(x_{n}+\alpha_{j}d_{j}^{{1}/{2s}}))}{\vert y-x\vert^{n+2s}}\,dx \nonumber \\&= c_{n,s} \lim_{\epsilon \rightarrow 0} \int_{\mathcal{C} B_{\epsilon}(y)}\frac{{u}_{{j}}(yd_{j}^{{1}/{2s}}+z_{j})-{u}_{{j}}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx\nonumber\\&\quad + c_{n,s}\lim_{\epsilon \rightarrow 0} \int_{\{x_{n} \leq-\alpha_{j}\}\bigcap \mathcal{C} B_{\epsilon}(y)}\frac{{u}_{{j}}(xd_{j}^{{1}/{2s}}+z_{j}) -\widehat{u}_{{j}}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx . \end{align*} $$

Making the change of variables

$$ \begin{align*}yd_{j}^{{1}/{2s}}+z_{j}=a \text{ and } xd_{j}^{{1}/{2s}}+z_{j}=b,\end{align*} $$

we get

(5-15) $$ \begin{align} (-\Delta)^{s}\psi_{j}(y)&= d_{j}(-\Delta)^{s}u_{j}(a)+ d_{j} c_{n,s} \lim_{\eta \rightarrow 0} \int_{\{b_{n} \leq 0\} \bigcap \mathcal{C} B_{\eta }(a)}\frac{{u}_{{j}}(b) -\widehat{u}_{{j}}(b)}{\vert a-b\vert^{n+2s}}db \nonumber\\ &= d_{j}(-\Delta)^{s}u_{j}(a)+ d_{j}h(a), \end{align} $$

where

$$ \begin{align*}\eta = \epsilon d_{j}^{{1}/{2s}} \end{align*} $$

and

$$ \begin{align*}h(a)= c_{n,s} \lim_{\eta \rightarrow 0} \int_{\{b_{n} \leq 0\} \bigcap \mathcal{C} B_{\eta }(a)}\frac{{u}_{{j}}(b) -\widehat{u}_{{j}}(b)}{\vert a-b\vert^{n+2s}}\,db.\end{align*} $$

Note that $a \in \Omega .$

Now, consider the case $y_{n} \leq -\alpha _{j}.$ Equation (5-14) becomes

$$ \begin{align*} &(-\Delta)^{s}\psi_{j}(y)= I_1 + I_2 \end{align*} $$

where

$$ \begin{align*}&I_1 = \int_{\{x_{n} \geq -\alpha_{j}\}\bigcap \mathcal{C} B_{\epsilon}(y)}\frac{{u}_{j}(y' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(y_{n}d_{j}^{{1}/{2s}}+\alpha_{j}d_{j}^{{1}/{2s}}))-{u}_{j}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx \end{align*} $$

and

(5-16) $$ \begin{align} &I_2 = \int_{\{x_{n} \leq-\alpha_{j}\}\bigcap \mathcal{C} B_{\epsilon}(y)}\frac{{u}_{j}(y' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(y_{n}d_{j}^{{1}/{2s}}+\alpha_{j}d_{j}^{{1}/{2s}}))- {u}_{j}(x' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(x_{n}d_{j}^{{1}/{2s}}+\alpha_{j}d_{j}^{{1}/{2s}}))}{\vert y-x\vert^{n+2s}}\,dx \end{align} $$

Let us introduce some notation. We write $\widehat {x}= (x',-x_{n})$ , $\widetilde {x}=(\widehat {x}',\widehat {x}_{n})$ and $\widehat {x}_{n}=-x_{n}$ for $x=(x',x_{n}) \in \mathbb {R}^{n}, n>1.$ Using these, let us compute

$$ \begin{align*} I_{2}&= c_{n,s} \lim_{\epsilon \rightarrow 0} \bigg[\int_{\{\widehat{x}_{n} \geq \alpha_{j}\} \bigcap \mathcal{C} B_{\epsilon}(\widehat{y})}\frac{{u}_{j}(\widehat{y}' d_{j}^{{1}/{2s}}+\hat{z}_{j}^{\prime}, \widehat{y}_{n}d_{j}^{{1}/{2s}}+\widehat{\alpha}_{j}d_{j}^{{1}/{2s}})- {u}_{j}(\widehat{x}' d_{j}^{{1}/{2s}}+\widehat{z}_{j}^{\prime}, \widehat{x}_{n}d_{j}^{{1}/{2s}}+\widehat{\alpha}_{j}d_{j}^{{1}/{2s}})}{\vert\widehat{y}-\widehat{x}\vert^{n+2s}}d\widetilde{x} \bigg] \nonumber \\ &= c_{n,s} \lim_{\epsilon \rightarrow 0} \bigg[\int_{\{\widehat{x}_{n} \geq \alpha_{j}\} \bigcap \mathcal{C} B_{\epsilon}(\widehat{y})}\frac{{u}_{j}(\widetilde{y}d_{j}^{{1}/{2s}}+\widetilde{z}_{j})- {u}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j} )}{\vert\widehat{y}-\widehat{x}\vert^{n+2s}}d\widetilde{x} \bigg] \nonumber\\ &= c_{n,s} \lim_{\epsilon \rightarrow 0} \bigg[\int_{\{\widehat{x}_{n} \geq \alpha_{j}\} \bigcap \mathcal{C} B_{\epsilon}(\widetilde{y})}\frac{{u}_{j}(\widetilde{y}d_{j}^{{1}/{2s}}+\widetilde{z}_{j})- {u}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j} )}{\vert\widetilde{y}-\widetilde{x}\vert^{n+2s}}d\widetilde{x} \bigg]. \end{align*} $$

Now, we simplify $I_{1}$ :

$$ \begin{align*} I_{1}&= c_{n,s} \lim_{\epsilon \rightarrow 0} \bigg[ \int_{\{x_{n} \geq -\alpha_{j}\} \bigcap \mathcal{C} B_{\epsilon}({y})}\!\!\frac{{u}_{j}(y' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(y_{n}d_{j}^{{1}/{2s}}+\alpha_{j}d_{j}^{{1}/{2s}}))- {u}_{j}(x' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(x_{n}d_{j}^{{1}/{2s}}+\alpha_{j}d_{j}^{{1}/{2s}}))}{\vert y-x\vert^{n+2s}} \nonumber \\ & \quad+ \int_{\{x_{n} \geq -\alpha_{j}\} \cap \mathcal{C} B_{\epsilon}({y})} \frac{ {u}_{j}(w'd_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(x_{n}d_{j}^{{1}/{2s}}+\alpha_{j}d_{j}^{{1}/{2s}}))-{u}_{j}(xd_{j}^{{1}/{2s}}+z_{j})}{\vert y-x\vert^{n+2s}}\,dx \bigg] \nonumber \\ &= c_{n,s}\lim_{\epsilon \rightarrow 0} \bigg[ \int_{\{\widehat{x}_{n} \leq \alpha_{j}\} \bigcap \mathcal{C}B_{\epsilon}(\widetilde{y})}\frac{{u}_{j}(\widetilde{y}d_{j}^{{1}/{2s}}+\widetilde{z}_{j})- {u}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j} )}{\vert\widetilde{y}-\widetilde{x}\vert^{n+2s}}d\widetilde{x}\nonumber\\ &\quad+ \int_{\{{x}_{n} \geq -\alpha_{j}\} \cap \mathcal{C}B_{\epsilon}(\widetilde{y})}\frac{{u}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j})- \widehat{{u}}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j} )}{\vert\widetilde{y}-\widetilde{x}\vert^{n+2s}}d\widetilde{x} \bigg]. \end{align*} $$

Using these estimates for $I_{1}$ and $I_{2}$ in (5-16),

$$ \begin{align*} (-\Delta)^{s}\psi_{j}(y)&= c_{n,s}\ \mathrm{PV} \int_{\mathbb{R}^{n}}\frac{{u}_{j}(\widetilde{y}d_{j}^{{1}/{2s}}+\widetilde{z}_{j})- {u}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j} )}{\vert\widetilde{y}-\widetilde{x}\vert^{n+2s}}d\widetilde{x}\nonumber\\ &\quad+ c_{n,s} \lim_{\epsilon \rightarrow 0} \int_{\{{x}_{n} \geq -\alpha_{j}\} \bigcap \mathcal{C}B_{\epsilon}(\widetilde{y})}\frac{{u}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j})- \widehat{{u}}_{j}(\widetilde{x}d_{j}^{{1}/{2s}}+\widetilde{z}_{j} )}{\vert\widetilde{y}-\widetilde{x}\vert^{n+2s}}d\widetilde{x}. \end{align*} $$

By the change of variables

$$ \begin{align*}\widetilde{y}d_{j}^{{1}/{2s}}+\widetilde{z}_{j}=e \quad\text{and}\quad \widetilde{w}d_{j}^{{1}/{2s}}+\widetilde{z}_{j}=f,\end{align*} $$

we get

(5-17) $$ \begin{align} &(-\Delta)^{s}\psi_{j}(y)= d_{j}(-\Delta)^{s}u_{j}(e)\nonumber\\ &\qquad+ d_{j} c_{n,s} \lim_{\eta \rightarrow 0} \int_{\{f_{n} \leq 0\}\bigcap \mathcal{C}B_{\eta}(e)}\frac{{u}_{{j}}(f) -\widehat{u}_{{j}}(f)}{\vert e-f\vert^{n+2s}}\,df, \,\mbox{where} f_n \mbox{ is the } n\,\mbox{th coordinate of } f \nonumber \\ &\quad= d_{j}(-\Delta)^{s}u_{j}(e) + d_{j}h(e). \end{align} $$

Note that $e \in \Omega .$ Further, for $y \in B_{\rho _{j}},$

$$ \begin{align*} \psi_{j}(y)= \widehat{u}_{j}(yd_{j}^{{1}/{2s}}+ z_{j})= \begin{cases} u_{d_{j} }(yd_{j}^{{1}/{2s}}+z_{j})& \text{if } y_{n} \geq - \alpha_{j}, \\ u_{d_{j} }(y' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(y_{n}d_{j}^{{1}/{2s}}+ \alpha_{j}d_{j}^{{1}/{2s}})) & \text{if } y_{n} \leq -\alpha_{j}. \end{cases} \end{align*} $$

We can write

$$ \begin{align*}(y' d_{j}^{{1}/{2s}}+z_{j}^{\prime}, -(y_{n}d_{j}^{{1}/{2s}}+ \alpha_{j}d_{j}^{{1}/{2s}}))=\widetilde{y}d_{j}^{{1}/{2s}}+\widetilde{z}_{j}.\end{align*} $$

Again re-naming the variables $yd_{j}^{{1}/{2s}}+ z_{j}$ and $\widetilde {y}d_{j}^{{1}/{2s}}+\widetilde {z}_{j}$ by a and $e,$ respectively,

$$ \begin{align*} \psi_{j}(y)= \begin{cases} u_{d_{j} }(a)& \text{if } y_{n} \geq -\alpha_{j}, \\ u_{d_{j} }(e) & \text{if } y_{n} \leq -\alpha_{j}. \end{cases} \end{align*} $$

We know that $u_{j}$ satisfies (1-1) in the point-wise sense as well. Therefore, combining above equation with (5-15), (5-17), we have for $y \in B_{\rho _{j}}, $

$$ \begin{align*} (-\Delta)^{s}\psi_{j}(y)+\psi_{j}(y)=\psi_{j}(y)^{p}+ d_{j}{h}(y). \end{align*} $$

Now, arguing as in the proof of Step I with minor modifications, one can obtain a convergent subsequence of $\{\psi _{j} \},$ which we denote again by $\{\psi _{j} \}$ such that $\psi _{j} \rightarrow v$ in $C_{loc}^{2}(\mathbb {R}^{n}).$ Therefore, as $d_{j}\downarrow 0,$

$$ \begin{align*} (-\Delta)^{s}v+ v= v^{p}\quad \text{in } \mathbb{R}^{n}. \end{align*} $$

Since $v \in H^{s}(\mathbb {R}^{n})$ and v is radially decreasing, v is spherically symmetric to $y=0.$ Moreover, v has power-type decay at infinity, which follows from Theorem 2.8, that is,

$$ \begin{align*} v(r) \leq \frac{C_{2}}{r^{n+2s}}, \quad r \geq 1, \end{align*} $$

for some constant $C_{2}>0.$ Let us define $\delta _{R}$ as in (5-7), that is,

$$ \begin{align*}\delta_{R}:= \frac{C_{2}}{R^{n+2s}}\end{align*} $$

for R sufficiently large to be defined later. Then, there exists an integer $j_{R}$ such that for $j \geq j_{R},$

(5-18) $$ \begin{align} \Vert\psi_{j}-v\Vert_{C^{2}(\overline{B_{4R}})} \leq \delta_{R}. \end{align} $$

We choose R sufficiently large that $R> \alpha _{j}$ for all $j,$ where the $\alpha _{j}$ terms are the same as defined earlier right after (5-13). We can choose such an R because $\{\alpha _{j}\}$ is a bounded sequence. The following lemma is very useful to prove our claim that $z_{d} \in \partial \Omega .$

Now, we use this lemma to show that $\psi _{j}$ has only one local maximum point in $B_{R}.$ For this, we choose two numbers $k,l$ $(0<k<l)$ such that $v"(r)<0 $ for $0 \leq r \leq k.$ Further, we see that $v"(0)<0$ and $ v(k)<1.$ Let us define

$$ \begin{align*} \theta = \min \{ \vert v'(r)\vert \mid k \leq r \leq l \}. \end{align*} $$

It is easy to observe that $\theta>0$ because $v'<0$ for $r>0.$ Then for $\delta _{R}<\theta ,$ we have by (5-18) that

$$ \begin{align*} 0< \theta - \delta_{R} \leq \vert\nabla v(y)\vert -\vert\nabla \psi_{j}(y)-\nabla v(y)\vert \leq \vert\nabla \psi_{j}(y)\vert \text{ for } k \leq \vert y\vert \leq l. \end{align*} $$

Applying Lemma 5.2 in the ball $\overline {B}_{k},$ we conclude that $y=0$ is the only local maximum point of $\psi _{j}$ in $B_{l}.$ If $y_{j}$ is a maximum point of $\psi _{j}$ in $B_{R},$ then by Lemma 3.8, we have $\psi _{j} \geq 1.$ Choose $R>0$ sufficiently large so that $\delta _{R}< 1-v(l).$ Therefore,

$$ \begin{align*} \psi_{j}(y) \leq v(y) + \delta_{R} \leq v(l)+ \delta_{R}<1. \end{align*} $$

Hence, $y_{j} \in B_{l}$ and therefore $y_{j}=0.$

If $\alpha _{j}>0,$ then by the definition of $\widehat {u}_{j}$ , $z_{R}^{*}=(z_{j}^{\prime }, -\alpha _{j}d_{j}^{{1}/{2s}})$ is also a maximum point of $\widehat {u}_{j}.$ This implies that $(0,-\alpha _{j})$ is another maximum point of $\psi _{j}$ in $B_{R},$ which is a contradiction. This proves our claim and hence completes the proof of Theorem 1.4.