\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 153, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/153\hfil Brezis-Nirenberg problems with fractional Laplacian]
{Multiple solutions for Brezis-Nirenberg problems with fractional Laplacian}

\author[H. Guo \hfil EJDE-2016/153\hfilneg]
{Hui Guo}

\address{Hui Guo \newline
College of Mathematics and Econometrics,
Hunan University,
Changsha 410082, China}
\email{huiguo\_math@163.com}

\thanks{Submitted April 9, 2016. Published June 20, 2016.}
\subjclass[2010]{35J20, 35J60, 35J67, 35R11}
\keywords{Fractional Laplacian; multiple solutions;
 Brezis-Nirenberg problem}

\begin{abstract}
 In this article, we prove the multiplicity of nontrivial solutions to
 the  critical problem with fractional Laplacian
 \begin{gather*}
 (-\Delta)^{\alpha/2}  u=|u|^{2^*_{\alpha}-2}u+\lambda u\quad\text{in } \Omega,\\
 u=0\quad \text{on } \partial \Omega,
 \end{gather*}
 where $0<\alpha<2$, $N>(1+\sqrt{2})\alpha$, $\Omega\subset \mathbb{R}^N$
 is a smooth bounded domain. More precisely,  for any $\lambda>0$,
 this problem has at least $[(N+1)/2]$ pairs of nontrivial weak solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks



\section{Introduction}

In recent decades, the study of nonlocal diffusion problems has attracted 
much attention from mathematicians, in particular, of equations involving 
fractional Laplace operator.
 As is known to us, the fractional Laplace operator appears in 
 anomalous diffusion phenomena in several fields such as physics, 
 biology and probability. Moreover, it can be viewed as the infinitesimal 
 generator of a stable L\'evy process. For more details and
 applications, one can  see \cite{Applebaum} and references therein.

In this article, we  focus  on the multiplicity of nontrivial solutions 
to the  Brezis-Nirenberg problem involving fractional Laplacian
\begin{equation}\label{1.1}
\begin{gathered}
(-\Delta)^{\alpha/2}  u=|u|^{2^*_{\alpha}-2}u+\lambda u\quad \text{in }\ \Omega,\\
u=0 \quad \text{on } \partial \Omega.
\end{gathered}
\end{equation}
where $\alpha\in(0,2)$, $N>(1+\sqrt{2})\alpha$, $\Omega$ is a bounded domain 
in $\mathbb{R}^N$, $(-\Delta)^{\alpha/2}$ stands for the fractional 
Laplacian operator, and $2^*_{\alpha}=\frac{2N}{N-\alpha}$ is the 
critical Sobolev exponent.

It is well known that \eqref{1.1} has been widely studied when $\alpha=2$.
In a pioneering work \cite{Brezis-Nirenberg}, Brezis and Nirenberg proved 
that \eqref{1.1} possesses a positive solution for $\lambda\in(0,\lambda_1)$, where
$\lambda_1$ denotes the eigenvalue  of $-\Delta$ in $\Omega$ with 
zero Dirichlet boundary data.
 Devillanova and Solimini \cite{Solimini} proved that there exist infinitely
many solutions of equation \eqref{1.1} when $N\geq 7$ and $\lambda>0$. 
For $N\geq 4$ and $\lambda>0$, Clapp and Weth \cite{Clapp-Weth} showed  
that \eqref{1.1} possesses  finitely many solutions with energy less than 
$\frac{2 }{N}{S}^{N/2}$. Later, based on these results, 
the authors in \cite{Chen-Shioji-Zou} showed that there are at least 
$[\frac{N+1}{2}]$ pairs of nontrivial solutions for $N\geq 5$ and 
$\lambda\geq \lambda_1$.Here $[a]$ is the least integer $n$ such that $n\geq a$.

When $\alpha\in (0,2)$, the operator $(-\Delta)^{\alpha/2}$ defined in a
 bounded domain $\Omega$ has several definitions, and these definitions 
are not necessarily equivalent to each other. 
In this article, we consider the fractional Laplace operator defined as 
in \cite{Brandle-Colorado-de Pablo,Cabre-Tan} by the spectral 
decomposition of the Laplacian,
\begin{equation}\label{definition1}
(-\Delta)^{\alpha/2}  u=\sum_{j=1}^{\infty} \lambda_j^{\alpha/2}a_j e_j,\quad
 \text{for } u=\sum_{j=1}^{\infty} a_j e_j\quad
 \text{with}\ \sum_{j=1}^{\infty} a_j^2 \lambda_j^{\alpha/2}<\infty.
\end{equation}
Here $(\lambda_j,e_j)$ denote the eigenvalues and eigenfunctions of
 $-\Delta$ in $\Omega$ with zero Dirichlet boundary data, and then 
$(\lambda_j^{\alpha/2},e_j)$ are the eigenvalues and eigenfunctions of 
$(-\Delta)^{\alpha/2}$ in $\Omega$ with zero Dirichlet boundary data. 
With this definition,  many results on the existence and nonexistence 
of nontrivial solutions of the fractional Brezis-Nirenberg problem \eqref{1.1} 
has been obtained by using the formulation of the fractional Laplacian 
through Dirichlet-Neumann maps introduced in \cite{Caffarelli-Silvestre}. 
When $\alpha=1$, Cabre and Tan \cite{Cabre-Tan} proved that there is no 
solution when $\lambda=0$ and $\Omega$ star-shaped domain.
 Later, Tan \cite{Tan} obtained a positive solution if 
$\lambda\in (0,\lambda_1^{1/2})$.For general $\alpha\in (0,2)$, 
the authors \cite{Barrios-Colorado-Pablo2012} showed that  problem \eqref{1.1} 
has no positive solution for $\lambda\geq \lambda_1^{\alpha/2}$, 
and has at least a positive solution for each 
$\lambda\in(0,\lambda_1^{\alpha/2})$.
By using the general Nehari manifold  method, Hua and Yu \cite{Hua-Yu} 
obtained a nontrivial ground state solution for any $\lambda>0$, 
provided $N>(1+\sqrt{2})\alpha$. For more related results, one may 
see \cite{Brandle-Colorado-de Pablo,Hua-Yu} and references therein. 
But to the best of our knowledge, there exist few results on the multiplicity 
of solutions for \eqref{1.1} with critical case.

Motivated by this, in this paper, we are devoted to  the multiplicity of 
nontrivial solutions of \eqref{1.1} with any $\alpha\in(0,2),N>(1+\sqrt{2})\alpha$ 
and $\lambda>0$.The first difficulty lies in that the fractional 
Laplacian operator $(-\Delta)^s$ is nonlocal. This nonlocal property makes 
some calculations difficult. To overcome this difficulty, we transform 
the nonlocal problem into a local problem by using the extension technique 
introduced by Caffarelli and Silvestre in \cite{Caffarelli-Silvestre}. 
More precisely, for any bounded domain $\Omega$, define cylinder 
$\mathcal{C}_{\Omega}:=\Omega\times(0,\infty)\subset \mathbb{R}^{N+1}_+$. 
If we  denote the points in $\mathcal{C}_{\Omega}$ by $(x,t)$,  
then for any $u\in H_0^{\alpha/2}(\Omega)$, the $\alpha$-harmornic
 extension $U=E_{\alpha}(u)$ can be defined as the solution of
\begin{equation}\label{2.1*}
\begin{gathered}
\operatorname{div}(t^{1-\alpha}\nabla U)=0\quad
\text{in } \mathcal{C}_{\Omega},\\
U=0\quad \text{on } \partial_{L}\mathcal{C}_{\Omega},\\
U(x,0)=u(x)\quad \text{on } \Omega\times\{t=0\},
\end{gathered}
\end{equation}
where $\partial_{L}\mathcal{C}_{\Omega}:=\partial\Omega\times[0,+\infty)$.
The relevance between $U$ and the fractional Laplacian of the original 
functions $u$ is through the formula
\begin{equation}\label{2.1}
-\lim_{t\to 0^+}t^{1-\alpha}\frac{\partial U}{\partial y}(x,t)
=\frac{1}{k_{\alpha}}(-\Delta)^{\alpha/2}u(x),
\end{equation}
where $k_{\alpha}$ is a normalization constant and  only depends on $N$ and $\alpha$.
Therefore, after this extension, problem \eqref{1.1} can be transformed into 
an equivalent form
\begin{equation}\label{1.2}
\begin{gathered}
L_{\alpha} U=0\quad \text{in }\ \mathcal{C}_{\Omega},\\
U=0\quad\text{on } \partial_{L}\mathcal{C}_{\Omega}\\
\frac{\partial U}{\partial \nu^{\alpha}}=|u|^{2^*_{\alpha}-2}u+\lambda u,
\quad\text{in } \Omega\times\{t=0\}.
\end{gathered}
\end{equation}
Here
$$
L_{\alpha} U:=-\operatorname{div}(t^{1-\alpha}\nabla U),\quad
\frac{\partial U}{\partial \nu^{\alpha}}:=-k_{\alpha} 
\lim_{t\to 0^+}t^a \frac{\partial U}{\partial t}.
$$

The second difficulty lies in that   \eqref{1.1} is a critical problem. 
Hence, the corresponding energy functional  does not satisfy the $(PS)$ condition. 
To  overcome this difficulty,  one has to use $(PS)_c$ condition instead of 
$(PS)$ condition. This idea has been widely used in the past decades, 
see \cite{Brezis-Nirenberg}.  For our  paper, we shall  use the global 
compactness results in fractional Sobolev space, see 
\cite[Proposition 2.1]{Yan-Yang-Yu}.

The third difficulty lies in that the $\alpha-harmornic$ extension function 
has no explicit expression. In order to find a critical value in some interval 
where the $(PS)_c$ condition holds, the usual way is to estimate some test 
functions. But for our problem, different from classical Laplace operator, 
our eigenfunctions in $\mathcal{C}_\Omega$ and test functions can not be 
written explicitly. To overcome this difficulty, we use the Poisson kernel, 
trace inequality and some asymptotic behavior of Bessel functions.

The fourth difficulty lies in how to find multiple solutions of problem \eqref{1.1}. 
Following the ideas in \cite{Chen-Shioji-Zou}, we can obtain the multiplicity 
of nontrivial solutions of \eqref{1.1} by  using the Krasnoselskii genus. 
This article extends the multiplicity results in \cite{Chen-Shioji-Zou} 
from  classical Laplace operator to the fractional case.
Now we are ready to state our main result.

\begin{theorem}\label{thm1}
Let $\alpha\in(0,2), N>(1+\sqrt{2})\alpha$ and $\Omega\subset \mathbb{R}^N$ 
be a smooth bounded domain. Then, for any $\lambda>0$, the problem \eqref{1.1}
 admits at least $[(N+1)/2]$ pairs of nontrivial solutions.
\end{theorem}

This article is organized as follows. In section 2, we introduce a 
variational setting for  problem  \eqref{1.1}, and present some preliminary
 results. In section 3, some useful estimates are obtained. 
In section 4, we are devoted to  the proof of Theorem \ref{thm1}.

\section{Preliminaries}

According to the definition of  \eqref{definition1},  the operator 
$(-\Delta)^{\alpha/2}$ is well defined on the space
$$
H^{\alpha/2}_0(\Omega)=\big\{u=\sum_{j=1}^{\infty} a_je_j\in L^2(\Omega):
 \|u\|_{H_0^{\alpha/2}(\Omega)}=(\sum_{j=1}^{\infty} 
a_j^2 \lambda_j^{\alpha/2})^{1/2}<+\infty\big\}.
$$
For each $u\in H_0^{\alpha/2}(\Omega)$, the corresponding  extension function 
$U:=E_{\alpha}(u)$ as a solution to \eqref{2.1*}, belongs to the space
\begin{align*}
X_0^{\alpha}(\mathcal{C}_{\Omega})
=\Big\{&U\in L^2(\mathcal{C}_{\Omega})
: U=0 \text{ on } \partial_L\mathcal{C}_{\Omega},\\
&\|U\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}
=\Big(k_{\alpha}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|\nabla U|^2\,dx\,dt\Big)^{1/2}
<\infty\Big\}
\end{align*}
with inner product
$$
(U,V)_{X_0^{\alpha}(\mathcal{C}_{\Omega})}
:=k_{\alpha}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}\nabla U \cdot \nabla V \,dx\,dt.
$$
Clearly, we have
\begin{equation}\label{2}
\|E_{\alpha}(u)\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}
=\|u\|_{H_0^{\alpha/2}(\Omega)},\quad \forall\ u\in H^{\alpha/2}_0(\Omega).
\end{equation}

Note that \eqref{1.2} is equivalent to \eqref{1.1} by extension technique 
(see \cite{Caffarelli-Silvestre}). Thus in this paper,  we shall focus our
 attention on looking for weak solutions of \eqref{1.2} in 
$X_0^{\alpha}(\mathcal{C}_{\Omega})$. First, consider the energy functional 
associated  to \eqref{1.2}
$$
I(U)=\frac{k_{\alpha}}{2}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}
|\nabla U(x,t)|^2\,dx\,dt
-\frac{\lambda}{2}\int_{\Omega}|U(x,0)|^2dx
-\frac{1}{2^*_{\alpha}}\int_{\Omega}|U(x,0)|^{2^*_{\alpha}}dx.
$$
It is well known that  for any critical point $U$ of $I$ in 
$X_0^{\alpha}(\mathcal{C}_{\Omega})$, the function $u:=U(\cdot,0)$ defined 
in the sense of traces, belongs to $H_0^{\alpha/2}(\Omega)$ and thus is a 
solution to problem \eqref{1.1}. The inverse is also true.

Next,  to use Krasnoselskii genus, we  consider a new functional 
as in \cite{Chen-Shioji-Zou},
\begin{equation}
\begin{aligned}
J(U)
:=&\frac{k_{\alpha}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|\nabla U(x,t)|^2\,dx\,dt
-\lambda\int_{\Omega}|u|^2dx}{\left(\int_{\Omega}|u|^{2^*_{\alpha}}
dx\right)^{2/2^*_{\alpha}}}\\
=&k_{\alpha}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|\nabla U(x,t)|^2\,dx\,dt
-\lambda\int_{\Omega}|u|^2dx
\end{aligned}
\end{equation}
defined on 
$$
M:=\{U\in X_0^{\alpha}(\mathcal{C}_{\Omega}):
 \|U(x,0)\|_{L^{2^*_{\alpha}}(\Omega)}=1\}.
$$
It is easy to check  $J\in C^1(M,\mathbb{R})$, and
 $U\in M$ is a critical point of $J$ with $J(U)=\beta>0$, if and only if 
$\tilde{U}=\beta^{\frac{1}{2^*_{\alpha}-2}}U$ is a critical point of $I$ 
with $I(\tilde{U})=\frac{\alpha}{2N}\beta^{N/\alpha}>0$.
Similarly, $(U_n)$ is a $(PS)_{\beta}$ sequence for $J$ if and only if 
the sequence $(\tilde{U}_n)$ is a $(PS)_{\tilde{\beta}}$ sequence for $I$ 
with $\tilde{\beta}=\frac{\alpha}{2N}\beta^{\frac{N}{\alpha}}$,
where $\tilde{U}_n:=\beta^{\frac{1}{2^*_{\alpha}-2}}U_n$.
Here we say a sequence $(U_n)$ in $M$ is a $(PS)_{\beta}$ sequence for $J$ if
$$
J(U_n)\to \beta\quad \text{and}\quad \|J'(U_n)\|\to 0,\quad \text{as}\ m\to\infty.
$$

Let
\begin{equation}\label{w1}
w_{\epsilon}(x)=(\frac{\epsilon}{\epsilon^2+|x|^2})^{\frac{N-\alpha}{2}},\quad
 \forall \epsilon>0,\; x\in \mathbb{R}^N,
\end{equation}
be the extremal function of Sobolev trace inequality
$$
\int_{\mathbb{R}^{N+1}_+}t^{1-\alpha}|\nabla U(x,t)|^2\,dx\,dt
\geq  {S_{\alpha,N}}\Big(\int_{\mathbb{R}^N}|U(x,0)|^{2^*_{\alpha}}dx
\Big)^{2/{2^*_{\alpha}}}.
$$
According to \cite{Chen-Li-Ou} and \cite{Li-Y}, after a translation, 
$w_{\epsilon}$ is the unique positive solution of
\begin{equation}\label{1.3}
(-\Delta)^{\alpha/2} u=|u|^{2^*_{\alpha}-2}u\quad \text{in }
 \dot{H}^{\alpha/2}(\mathbb{R}^N),
\end{equation}
and hence \begin{equation}\label{est3}
\frac{1}{k_{\alpha}}\|W_{\epsilon}\|_{X_0^{\alpha}(\mathbb{R}^{N+1}_+)}^2
=\|w_{\epsilon}\|_{L^{2^*_{\alpha}}(\mathbb{R}^N)}^{2^*_{\alpha}}
=(k_{\alpha}S_{\alpha,N})^{N/\alpha}.
\end{equation}
It is well known that when $\Omega=\mathbb{R}^N$, the $\alpha$-harmonic
 extension has an explicit expression in term of the Poisson kernel 
(see \cite{Caffarelli-Silvestre})
\begin{equation}\label{extension1}
U(x,t)=P_t^{\alpha}*u(x)=C_{N,\alpha}t^{\alpha}\int_{\mathbb{R}^N}\frac{u(y)}
{(|x-y|^2+t^2)^{\frac{N+\alpha}{2}}}dy,\quad
 \forall u\in H_0^{\alpha/2}(\mathbb{R}^N).
\end{equation}
where $C_{N,\alpha}$ is a constant.
So the $\alpha$-harmonic extension of $w_{\epsilon}$ can be written as
\begin{equation}\label{extension2}
W_{\epsilon}(x,t)=P_t^{\alpha}*w_{\epsilon}(x)
=C_{N,\alpha}t^{\alpha}\int_{\mathbb{R}^N}\frac{w_{\epsilon}(y)}
{(|x-y|^2+t^2)^{\frac{N+\alpha}{2}}}dy.
\end{equation}
One can see \cite[Remark 2.2]{Barrios-Colorado-Pablo2012} and references 
therein for more details.
Let
$$
\mathfrak{M}:=\{W_{\epsilon}(\cdot-(y,0)):  \epsilon>0, y\in \mathbb{R}^N\}.
$$
Then, we have the following compactness lemma.

\begin{lemma}\label{3.22}
Let $(U_n)$ be a $(PS)_{\beta_j}$ sequence for functional $J$.Up to a subsequence,
the following conclusions hold.

\noindent(a) If $\beta_j\in (0,{{k_{\alpha}S_{\alpha,N}}})$, then 
$(U_n)$ converges in $M$ and $\beta_j$ is a critical value of $J$.

\noindent(b) If $\beta_j\in ({{k_{\alpha}S_{\alpha,N}}},
 2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}})$, then one of the following cases follows:
\begin{itemize}
\item[(b.1)] $(U_n)$ converges in $M$ and $\beta_j$ is a critical value of $J$;

\item[(b.2)] There exists a critical point $u \in M$ of $J$ with
 $$
J(U)=\left(\beta_j^{N/\alpha}-{(k_{\alpha}S_{\alpha,N})}^{N/\alpha}
\right)^{\alpha/N}\in (0,{{k_{\alpha}S_{\alpha,N}}}).
$$
\end{itemize}

\noindent(c) If $\beta_j={{k_{\alpha}S_{\alpha,N}}}$, then one of the following 
cases holds:
\begin{itemize}
\item[(c.1)] $(U_n)$ converges in $M$ and $\beta_j$ is a critical value of $J$;
\item[(c.2)] $\operatorname{dist}(\beta_j^{\frac{1}{2^*_{\alpha}-2}}U_n, 
\mathfrak{M})\to 0$ or 
$\operatorname{dist}(\beta_j^{\frac{1}{2^*_{\alpha}-2}}U_n, -\mathfrak{M})\to 0$.
\end{itemize}
\end{lemma}

\begin{proof}
By using the standard argument, this lemma follows directly from the global 
compactness result in fractional Sobolev space 
\cite[Theorem 1.3]{Yan-Yang-Yu}.
\end{proof}

In the following, we write $\lambda_0=0$ for $k=0$.
It is easy to see that for each $\lambda>0$, there exists $k\geq 0$ 
such that $\lambda_k^{\alpha/2}\leq \lambda<\lambda_{k+1}^{\alpha/2}$.
Then, define
$$
H^-:=\operatorname{span}\{e_1,\dots,e_k\},\quad H^+
:=\overline{\operatorname{span}\{e_1,\dots,e_k\}^\perp}.
$$
Clearly, $H^-=\emptyset$ for $0<\lambda<\lambda_1^{\alpha/2}$.
Let $\mathcal{E}:=\{A\subset M: A \text{ is closed and symmetric} \}$.
For any integer $j\geq k+1$, we define
$\Sigma_{j}=\{A\in \mathcal{E}: \gamma(A)\geq j\}$, where $\gamma$ 
denotes the usual Krasnoselskii genus, and consider
$$
\beta_j:=\inf_{A\in \Sigma_j}\sup_{U\in A} J(U).
$$
Note that for each $A\in \Sigma_{j}$, $\gamma(A)>k$ and 
$A\cap\big\{U\in H^+: \|U(\cdot,0)\|_{L^{2^*_{\alpha}}(\Omega)}=1\big\}
\neq \emptyset$.Thus
\begin{equation}\label{beta1big0}
\beta_j>0\quad \text{for any integer } j\geq k+1.
\end{equation}
By using similar argument as in \cite[Lemma 2.2]{Chen-Shioji-Zou}, 
we can find a $(PS)_{\beta_j}$ sequence $(U_n)$ for  $J$.
 Moreover, we have the following lemma. Set 
$K^{\beta}:=\{U\in \mathfrak{M}: J'(U)=0\text{ and } J(U)=\beta\}$.

\begin{lemma}\label{2.2}
If $0<\beta_j=\beta_{j+1}<2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}}$, then 
$K^{\beta_j}$ is infinite.
\end{lemma}
\begin{proof}
By using standard arguments as in the proof of \cite[Lemma 2.4]{Chen-Shioji-Zou},
 we can obtain the result. So we omit the proof.
\end{proof}

\section{Some estimates}

In this section, we set
$\mathbb{B}_r^+(z_0):=\{z\in \overline{\mathbb{R}^{N+1}_+}: |z-z_0|<r\}$ and 
$B_r(x_0):=\{x\in \mathbb{R}^{N}: |x-x_0|<r\}$, and denote
$r_{xt}:=|(x,t)|=(|x|^2+|t|^2)^{1/2}$.For simplicity, we shall write 
$\mathbb{B}_r^+$ and $B_r$ instead of $\mathbb{B}_r^+(0)$ and $B_r(0)$, 
respectively.

Recall that $(\lambda_i^{\alpha/2},e_i)$ are the eigenvalues and eigenfunctions of 
$(-\Delta)^{\alpha/2}$ with zero Dirichlet boundary data.
Let $E_i$ denote the $\alpha$-harmonic extension of $e_i$, i.e., $E_i$ is 
the solution of
\begin{equation}\label{2.3}
\begin{gathered}
-\operatorname{div}(t^{1-\alpha}E_i)=0 \quad \text{in } \mathcal{C}_{\Omega},\\
E_i(x,t)=0 \quad \text{on } \partial_{L}\mathcal{C}_{\Omega},\\
-k_{\alpha}\lim_{t\to 0^+}t^{1-\alpha}\frac{\partial E_i}{\partial t}(x,t)
=\lambda_i^{\alpha/2} e_i(x)\quad\text{on } \Omega\times\{0\}.
\end{gathered}
\end{equation}
Then, we have the following Lemma.

\begin{lemma}\label{bounded}
There exists $C>0$ such that
$$
\sup_{(x,t)\in \mathcal{C}_{\Omega}}E_i(x,t)\leq C\quad \text{for all }
 i=1,\dots,k.
$$
\end{lemma}

\begin{proof}
For each $i=1,\dots,k$, it follows from 
\cite[Lemma 3.3]{Brandle-Colorado-de Pablo} that
$$
E_i=E_{\alpha}(e_i)=e_i(x)\psi(\lambda_i^{1/2}t),
$$
where $\psi$ is continuous and satisfies the following asymptotic behavior
\begin{gather*}
\psi(s)\sim 1-c_1 s^{\alpha}\quad \text{as } s\to 0,\\
\psi(s)\sim c_2 s^{\frac{\alpha-1}{2}}e^{-s}\ \text{as } s\to \infty,
\end{gather*}
where  $c_1=\frac{2^{1-\alpha}\Gamma(1-\alpha/2)}{\alpha\Gamma(\alpha/2)},
 c_2=\frac{2^{\frac{1-\alpha}{2}}\pi^{1/2}}{\Gamma(\alpha/2)}$, 
(see\cite{Brandle-Colorado-de Pablo} and \cite{Lebedev} for more details).
Clearly, $\psi$ is bounded. Since $e_i\in C^{\infty}(\Omega)$, we conclude 
that there exists $C>0$ such that
$$
\sup_{\mathcal{C}_{\Omega}}e_i(x)\psi(\lambda_i^{1/2}t)\leq C
\quad\text{uniformly for  } i=1,\dots,k.
$$
We completed the proof.
\end{proof}


Without loss of generality, we  assume $0\in \Omega$.
Then, we have
$\mathbb{B}_{2/m}^+\subset \mathcal{C}_{\Omega}$ for $m$ large enough.
Let
\begin{equation}\label{25*}
E_i^m(x,t):=\zeta_{2/m}(x,t) E_i(x,t),
\end{equation}
where $\zeta_{\eta}(x,t):=\bar{\zeta}(\frac{r_{xt}}{\eta})$ for any $\eta>0$,
and $\bar{\zeta}$ is defined by
\begin{equation}\label{2.33}
\bar{\zeta}(s)=\begin{cases}
0 &\text{if } s\in [0,\frac{1}{2}),\\
2s-1 &\text{if } s\in [\frac{1}{2},1),\\
1 &\text{if } s\in [1,+\infty).
\end{cases}
\end{equation}
Clearly,
\begin{equation}\label{xi}
|\nabla\zeta_{2/m}(x,t)|\leq m,\quad 
E_i^m(x,0)=\zeta_{2/m}(x,0)e_i,\quad 
 \operatorname{supp}E_i^m \subset \mathcal{C}_{\Omega}\backslash 
\overline{\mathbb{B}^+_{\frac{1}{m}}}.
\end{equation}
In the following, we denote $\zeta_{0}=1$ for $\eta=0$ and
 $A_m:=\{(x,t)\in \overline{\mathcal{C}_{\Omega}}: 
r_{xt}\in (\frac{1}{m},\frac{2}{m})\}$.

\begin{lemma}\label{lem1}
$\|E^m_i-E_i\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}\to 0$ as $m\to+\infty$.
\end{lemma}

\begin{proof}
Note that
\begin{equation}\label{2.4}
\int_{\Omega}e_i^2 dx=1\quad \text{and}\quad
 \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|\nabla E_i|^2 \,dx\,dt
=\lambda_i^{\alpha/2}.
\end{equation}
This, combined with Lemma \ref{bounded} and \eqref{xi}, implies that
\begin{equation}\label{3.1}
\begin{aligned}
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}
|\nabla \zeta_{2/m}|^2 |E_i|^2 \,dx\,dt
&= \int_{A_m}t^{1-\alpha}|\nabla \zeta_{2/m}|^2 |E_i|^2\,dx\,dt\\
&\leq Cm^2\int_{A_m}t^{1-\alpha}\,dx\,dt\\
&\leq Cm^{\alpha-N} \to 0\ \text{as}\ m\to\infty,
\end{aligned}
\end{equation}
and
\begin{equation}\label{3.2}
\begin{aligned}
&\big|\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}(\zeta_{2/m}-1)
E_i \nabla \zeta_{2/m}\cdot \nabla E_i \,dx\,dt\big| \\
&\leq \int_{A_m}t^{1-\alpha}|(\zeta_{2/m}-1)||E_i| |\nabla \zeta_{2/m}\cdot \nabla E_i|\,dx\,dt\\
&\leq Cm\int_{A_m}t^{1-\alpha}|\nabla E_i|\,dx\,dt\\
&\leq Cm(\int_{A_m}t^{1-\alpha}\,dx\,dt)^{1/2}
\Big(\int_{A_m}t^{1-\alpha}|\nabla E_i|^2\Big)^{1/2}\,dx\,dt\\
&\leq Cm^{\frac{\alpha-N}{2}} \to 0\ \text{as}\ m\to\infty.
\end{aligned}
\end{equation}
In addition, according to the absolutely continuity of the integral, we obtain
\begin{equation}\label{3.3}
\begin{aligned}
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}(\zeta_{2/m}-1)^2
|\nabla E_i|^2\,dx\,dt
&=\int_{\mathbb{B}_{2/m}^+}t^{1-\alpha}
(\zeta_{2/m}-1)^2|\nabla E_i|^2\,dx\,dt\\
&\leq \int_{\mathbb{B}_{2/m}^+}t^{1-\alpha}|\nabla E_i|^2\,dx\,dt
\to 0\quad\text{as } m\to\infty.
\end{aligned}
\end{equation}
Therefore,  from \eqref{3.1}, \eqref{3.2} and \eqref{3.3} it follows that
\begin{align*}
&\|E^m_i-E_i\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2\\
&=\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|\nabla 
 (\zeta_{2/m} E_i-E_i)|^2\,dx\,dt\\
&=\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}[|\nabla \zeta_{2/m}|^2
 |E_i|^2+2(\zeta_{2/m}-1)E_i \nabla \zeta_{2/m}\cdot 
\nabla E_i+(\zeta_{2/m}-1)^2|\nabla E_i|^2]\,dx\,dt\\
&\to  0\quad  \text{as } m\to\infty.
\end{align*}
The proof is complete.
\end{proof}

Define
$$
H_m^-:=\operatorname{span}\{E_1^m,\dots,E_k^m\}\quad \text{for }k\geq 1.
$$

\begin{lemma}\label{lem2}
Let $k\geq 1$. Then there exists $m_0>1$ such that for any $m\geq m_0$, it holds
\begin{equation}\label{26*}
\max_{\{U\in H_m^-,\|u\|_{L^2(\Omega)}=1\}}\|U\|_{X_0^{\alpha}
(\mathcal{C}_{\Omega})}^2\leq {\lambda_k^{\alpha/2}}+C_1 m^{\alpha-N},
\end{equation}
where $C_1$ is a positive constant independent of $m$.
\end{lemma}

\begin{proof}
First, we denote $e_i^m(x):=E_i^m(x,0)$, then according to \eqref{25*}, 
$e_i^m(x)=\zeta_{2/m}(x,0)e_i(x)$.
In what follows, we shall prove the following estimates:
\begin{gather}\label{2.10}
\|E_i^m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
\leq \lambda^{\alpha/2}_i+Cm^{\alpha-N},\quad i=1,2,\dots, \\
\label{2.11}
|( E_i^m, E_j^m)_{X_0^{\alpha}(\mathcal{C}_{\Omega})}|
\leq Cm^{\alpha-N},\quad i,j=1,2,\dots,i\neq j, \\
\label{2.12}
|( e_i^m, e_j^m)_{L^2(\Omega)}|\leq Cm^{-N},\quad i,j=1,2,\dots,i\neq j, \\
\label{2.13}
\|e_i^m\|_{L^2(\Omega)}^2\geq 1-Cm^{-N},\quad i=1,2,\dots.
\end{gather}
Indeed, by \eqref{25*}, we have
\begin{equation}\label{2.21}
\begin{aligned}
&\|E_i^m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2-\|E_i\|_{X_0^{\alpha}
(\mathcal{C}_{\Omega})}^2\\
&= \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}(|\nabla E_i^m|^2-|\nabla E_i|^2)\,dx\,dt\\
&=\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}
 [(\zeta_{2/m}^2-1)|\nabla E_i|^2+2\zeta_{2/m} 
E_i\nabla \zeta_{2/m}\cdot\nabla E_i
 +|E_i|^2|\nabla \zeta_{2/m}|^2]\,dx\,dt.
\end{aligned}
\end{equation}
On the other hand, multiplying \eqref{2.3} by $(\zeta_{2/m}^2-1)E_i$ 
and integrating by parts over $\mathcal{C}_{\Omega}$, we obtain
\begin{equation}\label{2.22}
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}[(\zeta_{2/m}^2-1)
|\nabla E_i|^2+2\zeta_{2/m} E_i\nabla 
\zeta_{2/m}\cdot\nabla E_i]\,dx\,dt
=\lambda \int_{\Omega}(\zeta_{2/m}^2(x,0)-1)e_i^2dx.
\end{equation}
Inserting \eqref{2.22} into \eqref{2.21}, we conclude from \eqref{2.4} 
and Lemma \ref{bounded} that
\begin{align*}
\|E_i^m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
&\leq \|E_i\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
+\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|E_i|^2|\nabla \zeta_{2/m}|^2\,dx\,dt
 +\lambda \int_{\Omega}(\zeta_{2/m}^2(x,0)-1)e_i^2dx\\
&\leq \|E_i\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
 +\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|E_i|^2
 |\nabla \zeta_{2/m}|^2\,dx\,dt\\
&\leq \lambda_i^{\alpha/2}+Cm^2 \int_{A_m}t^{1-\alpha}\,dx\,dt\\
&\leq \lambda_i^{\alpha/2}+Cm^{\alpha-N}.\\
\end{align*}
Hence \eqref{2.10} holds.

Observe that from \eqref{2.3},
$$
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}\nabla E_i\cdot \nabla E_j\,dx\,dt=0,
\quad\ i\neq j.
$$
Then we have
\begin{equation}\label{3.44}
\begin{aligned}
( E_i^m, E_j^m)_{X_0^{\alpha}(\mathcal{C}_{\Omega})}
&= \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}\nabla E_i^m\cdot \nabla E_j^m \,dx\,dt\\
&= \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}[\zeta_{2/m}^2\nabla E_i\cdot
 \nabla E_j+\zeta_{2/m} E_j\nabla E_i\cdot\nabla\zeta_{2/m}  \\
&\quad +\zeta_{2/m} E_i\nabla\zeta_{2/m}\cdot\nabla E_j
 +E_iE_j|\nabla \zeta_{2/m}|^2]\,dx\,dt\\
&= \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}[(\zeta_{2/m}^2-1)\nabla 
 E_i\cdot \nabla E_j+\zeta_{2/m} E_j\nabla\zeta_{2/m}\cdot\nabla E_i\\
&\quad +\zeta_{2/m} E_i\nabla\zeta_{2/m}\cdot\nabla E_j
 +|\nabla \zeta_{2/m}|^2E_iE_j]\,dx\,dt.
\end{aligned}
\end{equation}
Multiplying both sides of \eqref{2.3} by $(\zeta_{2/m}^2-1)E_j$ and 
integrating by parts, we obtain
$$
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}[(\zeta_{2/m}^2-1)\nabla E_i
\cdot \nabla E_j+2\zeta_{2/m} 
E_j\nabla\zeta_{2/m}\cdot\nabla E_i]\,dx\,dt
=\lambda_i^{\alpha/2}\int_{\Omega}(\zeta_{2/m}^2(x,0)-1)e_ie_jdx.
$$
Similarly, we obtain
$$
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}[(\zeta_{2/m}^2-1)\nabla E_i\cdot 
\nabla E_j+2\zeta_{2/m} E_i\nabla\zeta_{2/m}\cdot\nabla E_j]\,dx\,dt
=\lambda_j^{\alpha/2}\int_{\Omega}(\zeta_{2/m}^2(x,0)-1)e_ie_jdx.
$$
This, combined with \eqref{3.44}, implies 
\begin{equation}\label{3.5}
\begin{aligned}
&( E_i^m, E_j^m)_{X_0^{\alpha}(\mathcal{C}_{\Omega})} \\
&= \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|\nabla 
\zeta_{2/m}|^2E_iE_j\,dx\,dy
+\frac{\lambda_i^{\alpha/2}+\lambda_j^{\alpha/2}}{2}\int_{\Omega}
(\zeta_{2/m}^2(x,0)-1)e_ie_j\,dx.
\end{aligned}
\end{equation}
By \eqref{3.5} and Lemma \ref{bounded}, we have
\begin{align*}
&|( E_i^m, E_j^m)_{X_0^{\alpha}(\mathcal{C}_{\Omega})}|\\
&\leq Cm^2\int_{\mathbb{B}^+_{2/m}}t^{1-\alpha}|E_iE_j|\,dx\,dt
 +\frac{\lambda_i^{\alpha/2}
 +\lambda_j^{\alpha/2}}{2}
\int_{B_{2/m}}|(\zeta_{2/m}^2(x,0)-1)||e_ie_j|dx\\
&\leq  Cm^2\int_{\mathbb{B}^+_{2/m}}t^{1-\alpha}\,dx\,dt
 +C\int_{B_{2/m}}dx\\
&\leq  Cm^2 m^{\alpha-N-2}+Cm^{-N}\\
&\leq  Cm^{\alpha-N},
\end{align*}
which yields \eqref{2.11}.

Note that $\int_{\Omega}e_ie_jdx=0$ when $i\neq j$, then from 
Lemma \ref{bounded} it follows that
\begin{align*}
|( e_i^m, e_j^m)_{L^2(\Omega)}|
&= |\int_{\Omega}\zeta_{m}^2(x,0) e_i e_j dx|\\
&= |\int_{\Omega}(\zeta_{m}^2(x,0)-1) e_i e_j dx|\\
&\leq  |\int_{B_{2/m}}e_i e_j dx|
\leq  Cm^{-N}.
\end{align*}
So \eqref{2.12} holds.

In view of \eqref{2.4} and Lemma \ref{bounded}, we obtain
\begin{align*}
\|e_i^m\|^2_{L^2(\Omega)}
&= \int_{\Omega}e_i^2dx-\int_{\Omega}(1-\zeta_{2/m}^2(x,0))e_i^2dx\\
&\geq 1-\int_{B_{2/m}}e_i^2dx\\
&\geq 1-Cm^{-N},
\end{align*}
which implies \eqref{2.13}.

Now, by using the above estimates \eqref{2.10}--\eqref{2.13}, 
we are ready to prove \eqref{26*}.  Let $U_m\in H_m^-$ with the 
trace $\|u_m\|_{L^2(\Omega)}=1$ such that
\begin{equation}\label{extre}
\max_{\{U\in H^-_m,\|u\|_{L^2(\Omega)}=1\}}\|U\|^2_{X_0^{\alpha}(\mathcal{C}_{\Omega})}
=\|U_m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2.
\end{equation}
Then, there exist numbers $a_1^m,\dots,a_k^m$ such that 
$U_m=\sum_{i=1}^{k} a_i^m E_i^m$.
Thus, we have
\begin{gather*}
\|U_m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
=\sum_{i=1}^{k}(a_i^m)^2 \|E_i^m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
+2\sum_{1\leq i<j\leq k}a_i^ma_j^m(E_i^m,E_j^m)_{X_0^{\alpha}(\mathcal{C}_{\Omega})},
\\
1=\|u_m\|_{L^2(\Omega)}^2=\sum_{i=1}^{k}(a_i^m)^2 \|e_i^m\|_{L^2(\Omega)}^2
+2\sum_{1\leq i<j\leq k}a_i^ma_j^m(e_i^m,e_j^m)_{L^2(\Omega)}.
\end{gather*}
According to \eqref{2.12} and \eqref{2.13}, there exists $m_0>1$ such that 
for $m\geq m_0$,
$$
|( e_i^m, e_j^m)_{L^2({\Omega})}|\leq\frac{1}{4}\quad\text{when }
 i\neq j,\quad \text{and}\quad
\|e_i^m\|^2_{L^2({\Omega})}\geq\frac{3}{4}.
$$
Then, it holds
\begin{align*}
1&= \sum_{i=1}^{k}(a_i^m)^2 \|e_i^m\|_{L^2(\Omega)}^2
 +2\sum_{1\leq i<j\leq k}a_i^ma_j^m(e_i^m,e_j^m)_{L^2(\Omega)}\\
&\geq \sum_{i=1}^{k}(a_i^m)^2 \|e_i^m\|_{L^2(\Omega)}^2
 -2\sum_{1\leq i<j\leq k}|a_i^m||a_j^m||(e_i^m,e_j^m)_{L^2(\Omega)}|\\
&\geq \frac{3}{4}\sum_{i=1}^{k}(a_i^m)^2-\frac{1}{4}
\sum_{1\leq i<j\leq k}(|a_i^m|^2+|a_j^m|^2)\\
&\geq \frac{1}{4}\sum_{i=1}^{k}(a_i^m)^2,
\end{align*}
which implies
\begin{equation}\label{2.15}
|a_i^m|  \text{ are uniformly bounded for } m\geq m_0.
\end{equation}
By \eqref{2.12}, \eqref{2.13} and \eqref{2.15}, we conclude
\begin{align*}
1&\geq  \sum_{i=1}^{k}(a_i^m)^2 \|e_i^m\|_{L^2(\Omega)}^2-2\sum_{1\leq i<j\leq k}|a_i^m||a_j^m||(e_i^m,e_j^m)_{L^2(\Omega)}|\\
&\geq  \sum_{i=1}^{k}(a_i^m)^2 \|e_i^m\|_{L^2(\Omega)}^2-C\sum_{1\leq i<j\leq k}|(e_i^m,e_j^m)_{L^2(\Omega)}|\\
&\geq  \sum_{i=1}^{k}(a_i^m)^2 \|e_i^m\|_{L^2(\Omega)}^2-Cm^{-1-N}\\
&\geq  \sum_{i=1}^{k}(a_i^m)^2-C\sum_{i=1}^{k}(a_i^m)^2 m^{-1-N}-Cm^{-1-N}\\
&\geq  \sum_{i=1}^{k}(a_i^m)^2-Cm^{-1-N}.
\end{align*}
This, combined with \eqref{2.10},\eqref{2.11} and \eqref{2.15}, implies that
\begin{equation}\label{extr1}
\begin{aligned}
\|U_m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
&= \sum_{i=1}^{k}(a_i^m)^2 \|E_i^m\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
+2\sum_{1\leq i<j\leq k}a_i^ma_j^m(E_i^m,E_j^m)_{X_0^{\alpha}
(\mathcal{C}_{\Omega})}\\
&\leq  {\lambda_k^{\alpha/2}}\sum_{i=1}^{k}(a_i^m)^2+Cm^{\alpha-N}+Cm^{\alpha-N}\\
&\leq  {\lambda_k^{\alpha/2}}+Cm^{\alpha-N}+Cm^{\alpha-N}\\
&\leq  {\lambda_k^{\alpha/2}}+C_1 m^{\alpha-N}
\end{aligned}
\end{equation}
for some $C_1>0$.
Therefore,  \eqref{extre} and \eqref{extr1} yield the proof.
\end{proof}

Based on the estimate in Lemma \ref{lem2}, we have the following lemma.

\begin{lemma}\label{lem3}
Suppose $k\geq 1$ and $\lambda\geq {\lambda_k^{\alpha/2}}$.
Then for any $m\geq m_0$, it holds
$$
\sup_{U\in H_m^-} I(U)\leq C_2 m^{\frac{N(\alpha-N)}{\alpha}},
$$
where $C_2$ is a positive number independent of $m$.
\end{lemma}

\begin{proof}
In view of Lemmas \ref{lem1} and \ref{lem2}, there exists some constant $C_2>0$ 
such that for any $m\geq m_0$ and $U\in H_m^-$,
\begin{align*}
I(U)
&\leq  \frac{{\lambda_k^{\alpha/2}}-\lambda}{2}\int_{\Omega} |u|^2 dx
 +\frac{C_1 m^{\alpha-N}}{2} \int_{\Omega}|u|^2dx-\frac{1}{2^*_{\alpha}}
 \int_{\Omega}|u|^{2^*_{\alpha}} dx\\
&\leq  \frac{C_1 m^{\alpha-N}}{2} \int_{\Omega}|u|^2dx-\frac{1}{2^*_{\alpha}}
 \int_{\Omega}|u|^{2^*_{\alpha}} dx\\
&\leq  Cm^{\alpha-N}\|u\|_{L^{2^*_{\alpha}}(\Omega)}^2-\frac{1}{2^*_{\alpha}}
 \|u\|_{L^{2^*_{\alpha}}(\Omega)}^{2^*_{\alpha}}\\
&\leq  \max_{t\geq 0}(Cm^{\alpha-N}t^2-\frac{1}{2^*_{\alpha}}t^{2^*_{\alpha}})\\
&\leq  C_2 m^{\frac{N(\alpha-N)}{\alpha}}.
\end{align*}
Thus the proof is complete.
\end{proof}

In what follows, we shall introduce a lemma that describes the property of 
$W_1$ defined in \eqref{extension2}. This lemma plays a key role in our estimates 
in this section. Here, we write $W_1^{(\alpha)}$ instead of $W_1$ to emphasize 
the dependence on the parameter $\alpha$.

\begin{lemma}[{\cite[Lemma 3.7]{Barrios-Colorado-Pablo2012}}] \label{lemmaestimate} 
It holds
\begin{gather}\label{est1}
|\nabla W_1^{(\alpha)}(x,t)|\leq \frac{C}{t}W_1^{(\alpha)}(x,t),
\quad 0<\alpha<2, (x,t)\in \mathbb{R}^{N+1}_+, \\
\label{est2}
|\nabla W_1^{(\alpha)}(x,t)|\leq CW_1^{(\alpha-1)}(x,t),
\quad 1<\alpha<2, (x,t)\in \mathbb{R}^{N+1}_+.
\end{gather}
\end{lemma}

Now, we define a cut-off function $\bar{\phi}(s)\in C^{\infty}(\mathbb{R}^+)$ 
with $0\leq \bar{\phi}(s)\leq 1$, which is non-increasing and satisfies
\[
\bar{\phi}(s)=\begin{cases}
1 & \text{if } 0\leq s\leq \frac{1}{2},\\
0 & \text{if } s\geq 1,
\end{cases}
\]
and $|\nabla \bar{\phi}|$ is bounded.
For any $r>0$, set
$$
\phi_r(x,t)=\bar{\phi}(\frac{r_{xt}}{r})
$$
then $|\nabla \phi_r|\leq C/r$ for some positive constant $C$ independent of $r$.
Let $0<\epsilon<r<\frac{2}{m}$.According to \eqref{w1} and \eqref{extension2}, 
define
$$
W_{\epsilon}^r(x,t):=\phi_r(x,t) W_{\epsilon}(x,t)\quad \text{and}\quad
w_{\epsilon}^r(x):=W_{\epsilon}^r(x,0).
$$
Obviously, $W_{\epsilon}^r\in X_0^{\alpha}(\mathcal{C}_{\Omega})$ and 
$w_{\epsilon}^r(x)=\phi_r(x,0) w_{\epsilon}(x)$. Recalling that 
$\zeta_{\eta}(x,t)=\bar{\zeta}(\frac{r_{xt}}{\eta})$ defined in \eqref{2.33} 
and $\zeta_0=1$ for $\eta=0$, we have the following lemma.

\begin{lemma}\label{estimates}
Let $0\leq 2\eta<\epsilon<r$ and $\tilde{x}\in \Omega$.
Then the following estimates hold:
(a) 
\begin{align*}
&\|\zeta_{\eta}(x-\tilde{x},t)W_{\epsilon}^r\|_{X_0^{\alpha}
 (\mathcal{C}_{\Omega})}^2 \\
&\leq \begin{cases}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}+C(\frac{\epsilon}{r})^{N-\alpha}
+C(\frac{\eta}{\epsilon})^{N-\alpha}, &\text{if }\alpha\in(0,1),\\
(k_{\alpha}S_{\alpha,N})^{N/\alpha}
+C(\frac{\epsilon}{r})^{N-1}|\log{\frac{\epsilon}{r}}|
+C(\frac{\eta}{\epsilon})^{N-\alpha}, &\text{if }\alpha=1,\\
(k_{\alpha}S_{\alpha,N})^{N/\alpha}
+C(\frac{\epsilon}{r})^{N-\alpha}+C(\frac{\eta}{\epsilon})^{N-\alpha}, 
&\text{if }\alpha\in(1,2).
\end{cases}
\end{align*}
(b) 
\[
\int_{\Omega}|\zeta_{\eta}(x-\tilde{x},0)w_{\epsilon}^r(x)|^{2^*_{\alpha}}dx
\geq (k_{\alpha}S_{\alpha,N})^{N/\alpha}-C(\frac{\epsilon}{r})^{N}
-C(\frac{\eta}{\epsilon})^{N}.
\]
\end{lemma}

\begin{proof}
Let $r<1$ and $C_r:=\{(x,t)\in\mathbb{R}^{N+1}_+: r/2
\leq |(x-\tilde{x},t)|\leq r\}$.
According to \eqref{w1}, $w_1^{(\alpha)}(x)\leq |x|^{\alpha-N}$.
Then by\eqref{extension2}, for any $(x,t)\in C_{r/\epsilon}$, we have
\begin{equation}\label{3.4}
\begin{aligned}
W_1^{(\alpha)}(x,t)
&= \int_{|{y}|<\frac{r}{4\epsilon}}P^{\alpha}_t(x-{y})w_1({y})dy
 +\int_{|{y}|>\frac{r}{4\epsilon}}P^{\alpha}_t(x-{y})w_1({y})dy\\
&\leq  Ct^{\alpha}\int_{|{y}|<\frac{r}{4\epsilon}}\frac{w_1({y})}{(|x|^2
 +|t|^2-|{y}|^2)^{\frac{N+\alpha}{2}}}dy
 +C(\frac{\epsilon}{r})^{N-\alpha}\int_{\mathbb{R}^N}P^{\alpha}_t({y})dy\\
&\leq  Ct^{\alpha}\int_{|{y}|<\frac{r}{4\epsilon}}
 \frac{w_1({y})}{((\frac{r}{2\epsilon})^2-(\frac{r}{4\epsilon})^2)
 ^{\frac{N+\alpha}{2}}}dy
 +C(\frac{\epsilon}{r})^{N-\alpha}\int_{\mathbb{R}^N}P^{\alpha}_t({y})dy\\
&\leq  C(\frac{\epsilon}{r})^{N+\alpha}t^{\alpha}\int_{|{y}|
 <\frac{r}{4\epsilon}}w_1({y})dy
+C(\frac{\epsilon}{r})^{N-\alpha}\int_{\mathbb{R}^N}P^{\alpha}_t({y})dy\\
&\leq  C(\frac{\epsilon}{r})^{N+\alpha}t^{\alpha}\int_{|{y}|
 <\frac{r}{4\epsilon}}\frac{1}{|{y}|^{N-\alpha}}dy
+C(\frac{\epsilon}{r})^{N-\alpha}\\
&\leq  C(\frac{\epsilon}{r})^{N}t^{\alpha}+C(\frac{\epsilon}{r})^{N-\alpha}\\
&\leq  C(\frac{\epsilon}{r})^{N-\alpha}.
\end{aligned}
\end{equation}
Moreover, by Lemma \ref{lemmaestimate} and \eqref{3.4}, we obtain
\begin{equation}\label{422}
\begin{aligned}
&\int_{C_{\frac{r}{\epsilon}}}
t^{1-\alpha}|W_1^{(\alpha)}\nabla W_1^{(\alpha)}|\,dx\,dt \\
&\leq \begin{cases}
C(\frac{\epsilon}{r})^{2N-2\alpha}\int_{C_{r/\epsilon}}t^{-\alpha}\,dx\,dt\leq
C(\frac{\epsilon}{r})^{N-\alpha-1},& \text{if}\ \alpha\in(0,1),
\\
C(\frac{\epsilon}{r})^{2N-2}\int_{C_{r/\epsilon}}t^{-1}\,dx\,dt
 \leq C(\frac{\epsilon}{r})^{N-2}|\log{\frac{\epsilon}{r}}|, &\text{if}\ \alpha=1,
\\
C(\frac{\epsilon}{r})^{2N-\alpha+1}\int_{C_{r/\epsilon}}t^{1-\alpha}\,dx\,dt
 \leq C(\frac{\epsilon}{r})^{N-\alpha-1},&\text{if } \alpha\in(1,2).
\end{cases}
\end{aligned}
\end{equation}
Note that $W_{\epsilon}(x,t)
=\epsilon^{\frac{\alpha-N}{2}}W_1(\frac{x}{\epsilon},\frac{t}{\epsilon})$.
Then for the case $\eta=0$, we have
\begin{equation}\label{421}
\begin{aligned}
&\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}W_{\epsilon}
\phi_r \nabla\phi_r\cdot\nabla W_{\epsilon}\,dx\,dt\\
&\leq  Cr^{-1}\int_{C_r}t^{1-\alpha}|W_{\epsilon}||\nabla W_{\epsilon}|\,dx\,dt\\
&=  Cr^{-1}\epsilon\int_{C_{r/\epsilon}}t^{1-\alpha}|W_{1}
(x,t)||\nabla W_{1}(x,t)|\,dx\,dt\\
&\leq \begin{cases}
C(\frac{\epsilon}{r})^{N-\alpha}, &\text{if } \alpha\in(0,1),\\
C(\frac{\epsilon}{r})^{N-1}|\log{\frac{\epsilon}{r}}|,&\text{if } \alpha=1,\\
C(\frac{\epsilon}{r})^{N-\alpha},& \text{if } \alpha\in(1,2).
\end{cases}
\end{aligned}
\end{equation}
Since $0\leq w_{\epsilon}(x)\leq \epsilon^{\frac{N-\alpha}{2}}|x|^{\alpha-N}$
 and the $\alpha-$extension of $|x|^{\alpha-N}$ is $r_{xt}^{\alpha-N}$, 
we conclude that 
$W_{\epsilon}(x,t)\leq \epsilon^{\frac{N-\alpha}{2}}r_{xt}^{\alpha-N}$ and
\begin{equation}\label{4222}
\begin{aligned}
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|W_{\epsilon}\nabla \phi_r|^2 \,dx\,dt
&\leq  Cr^{-2}\int_{C_r}t^{1-\alpha}|W_{\epsilon}|^2\,dx\,dt\\
&\leq  C\frac{\epsilon^{N-\alpha}}{r^2}\int_{C_r}t^{1-\alpha}r_{xt}^{2(\alpha-N)}\,dx\,dt\\
&\leq  C\frac{\epsilon^{N-\alpha}}{r^{2N+2-2\alpha}}\int_{C_r}t^{1-\alpha}\,dx\,dt\\
&\leq  C\frac{\epsilon^{N-\alpha}}{r^{N-\alpha}}.
\end{aligned}
\end{equation}
From \eqref{421} and \eqref{4222} it follows that
\begin{equation}\label{42223}
\begin{aligned}
&\|W_{\epsilon}^r\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2\\
&= k_{\alpha}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}(|\phi_r\nabla W_{\epsilon}|^2+|W_{\epsilon}\nabla \phi_r|^2+2 W_{\epsilon}\phi_r \nabla\phi_r\cdot\nabla W_{\epsilon})\,dx\,dt\\
&\leq  \|W_{\epsilon}\|^2_{\mathbb{R}^{N+1}_+}
+k_{\alpha}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|W_{\epsilon}\nabla \phi_r|^2
+2 k_{\alpha} \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}W_{\epsilon}\phi_r \nabla\phi_r\cdot\nabla W_{\epsilon}\,dx\,dt\\
&\leq \begin{cases}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}+C(\frac{\epsilon}{r})^{N-\alpha},
&\text{if } \alpha\in(0,1),\\
(k_{\alpha}S_{\alpha,N})^{N/\alpha}+C(\frac{\epsilon}{r})^{N-1}
 +C(\frac{\epsilon}{r})^{N-1}|\log{\frac{\epsilon}{r}}|,&\text{if } \alpha=1,\\
(k_{\alpha}S_{\alpha,N})^{N/\alpha}+C(\frac{\epsilon}{r})^{N-\alpha},
&\text{if } \alpha\in(1,2).
\end{cases}
\end{aligned}
\end{equation}
In addition, since
\[
\int_{\mathbb{R}^N\backslash B_r}|W_{\epsilon}(x)|^{2^*_{\alpha}}dx
=C\int_{r}^{\infty}(\frac{\epsilon}{\epsilon^2+\rho^2})^N
\rho^{N-1}d\rho\leq C\epsilon^N r^{-N},
\]
from \eqref{est3} we conclude  that
\begin{equation}\label{422245}
\begin{aligned}
\int_{\Omega}|w_{\epsilon}^r|^{2^*_{\alpha}}dx
&\geq \int_{B(r/2)}|w_{\epsilon}|^{2^*_{\alpha}}dx\\
&= (k_{\alpha}S_{\alpha,N})^{N/\alpha}
 -\int_{\mathbb{R}^N\backslash B(r/2)}|w_{\epsilon}|^{2^*_{\alpha}}dx\\
&\geq   (k_{\alpha}S_{\alpha,N})^{N/\alpha}-C(\frac{\epsilon}{r})^{N}.
\end{aligned}
\end{equation}
Now, we turn to the case $\eta>0$.
Since  $w_{\epsilon}\leq C\epsilon^{(\alpha-N)/2}$ and 
$|\nabla w_{\epsilon}|\leq C\epsilon^{(\alpha-N-2)/2}$, 
from \eqref{extension2} we obtain 
\begin{equation}\label{ab1}
\begin{aligned}
W_{\epsilon}(x,t)
&\leq C\epsilon^{(\alpha-N)/2}\int_{\mathbb{R}^N}t^{\alpha}
\frac{1}{(|x-s|^2+t^2)^{\frac{N+\alpha}{2}}}ds\\
&= C\epsilon^{(\alpha-N)/2}\int_{\mathbb{R}^N}
\frac{1}{(|s|^2+1)^{\frac{N+\alpha}{2}}}ds\\
&\leq  C\epsilon^{(\alpha-N)/2}
\end{aligned}
\end{equation}
and
\begin{align*}
|\nabla W_{\epsilon}(x,t)|
&= \int_{\mathbb{R}^N}P_t^{\alpha}({y}) |\nabla w_{\epsilon}(x-{y})|dy\\
&\leq  C\epsilon^{(\alpha-N-2)/2}\int_{\mathbb{R}^N}P_t^{\alpha}({y})dy\\
&\leq  C\epsilon^{(\alpha-N-2)/2}.
\end{align*}
Furthermore,
\begin{equation}\label{ab2}
\begin{aligned}
|\nabla W_{\epsilon}^r(x,t)|
&\leq |W_{\epsilon}\nabla \phi_r|+\phi_r |\nabla W_{\epsilon}|\\
&\leq  Cr^{-1}W_{\epsilon}+|\nabla W_{\epsilon}|\\
&\leq  Cr^{-1}\epsilon^{(\alpha-N)/2}+\epsilon^{(\alpha-N)/2}\epsilon^{-1}\\
&\leq  C\epsilon^{(\alpha-N-2)/2}.
\end{aligned}
\end{equation}
Since $W_{\epsilon}^r\leq W_{\epsilon}$, by \eqref{ab1} and \eqref{ab2}, we have
\begin{equation}\label{50}
\begin{aligned}
\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|W_{\epsilon}^r(x,t)
\nabla \zeta_{\eta}(x-\tilde{x},t)|^2\,dx\,dt
&\leq  C\eta^{-2}\int_{C_{\eta}}t^{1-\alpha}|W_{\epsilon}^r|^2\,dx\,dt\\
&\leq  C\eta^{-2}\epsilon^{\alpha-N}\int_{C_{\eta}}t^{1-\alpha}\,dx\,dt\\
&\leq  C\frac{\eta^{N-\alpha}}{\epsilon^{N-\alpha}}
\end{aligned}
\end{equation}
and
\begin{equation}\label{51}
\begin{aligned}
&\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}\zeta_{\eta} 
\nabla\zeta_{\eta}(x-\tilde{x},t)\cdot W_{\epsilon}^r\nabla W_{\epsilon}^r(x,t) 
\,dx\,dt \\
&\leq  C\eta^{-1}\int_{C_{\eta}}t^{1-\alpha}|W_{\epsilon}^r|\,
 |\nabla W_{\epsilon}^r|\,dx\,dt\\
&\leq  C\eta^{-1}\epsilon^{\alpha-N-1}\int_{C_{\eta}}t^{1-\alpha}\,dx\,dt\\
&\leq  C(\frac{\eta}{\epsilon})^{N-\alpha+1}.
\end{aligned}
\end{equation}
From \eqref{est3}, \eqref{50} and \eqref{51} it follows that
\begin{equation}\label{42226}
\begin{aligned}
&\|\zeta_{\eta}(x-\tilde{x},t)W_{\epsilon}^r(x,t)\|_{X_0^{\alpha}
(\mathcal{C}_{\Omega})}^2 \\
&=k_{\alpha}\int_{\mathcal{C}_{\Omega}}
 t^{1-\alpha}(|\zeta_{\eta}\nabla W_{\epsilon}^r|^2
 +|W_{\epsilon}^r\nabla \zeta_{\eta}|^2+2 W_{\epsilon}^r\zeta_{\eta}
 \nabla\zeta_{\eta}\cdot\nabla W_{\epsilon}^r)\,dx\,dt\\
&\leq \|W_{\epsilon}^r\|^2_{\mathbb{R}^{N+1}_+}
+k_{\alpha}\int_{\mathcal{C}_{\Omega}}t^{1-\alpha}|W_{\epsilon}^r\nabla 
 \zeta_{\eta}|^2
+2 k_{\alpha} \int_{\mathcal{C}_{\Omega}}t^{1-\alpha}W_{\epsilon}^r\zeta_{\eta} 
 \nabla\zeta_{\eta}\cdot\nabla W_{\epsilon}^r\,dx\,dt\\
&\leq \|W_{\epsilon}^r\|^2_{\mathbb{R}^{N+1}_+}+C(\frac{\eta}{\epsilon})^{N-\alpha},
\end{aligned}
\end{equation}
which, together with \eqref{42223}, implies (a).

In addition, by \eqref{422245}, we have
\begin{equation}\label{42225}
\begin{aligned}
\int_{\Omega}|\zeta_{\eta}(x-\tilde{x},0)w_{\epsilon}^r|^{2^*_{\alpha}}dx
&= \int_{\Omega}|w_{\epsilon}^r|^{2^*_{\alpha}}dx
 -\int_{\Omega}(1-\zeta_{\eta}^{2^*_{\alpha}}(x,0))|w_{\epsilon}^r|^{2^*_{\alpha}}dx\\
&= \int_{\Omega}|w_{\epsilon}^r|^{2^*_{\alpha}}dx
-\int_{|(x-\tilde{x},t)|\leq \eta}|w_{\epsilon}|^{2^*_{\alpha}}dx\\
&\geq  (k_{\alpha}S_{\alpha,N})^{N/\alpha}-C(\frac{\epsilon}{r})^{N}
 -C\int_{|(x-\tilde{x},t)|\leq \eta}\epsilon^{-N} dx\\
&\geq  (k_{\alpha}S_{\alpha,N})^{N/\alpha}-C(\frac{\epsilon}{r})^{N}
 -C(\frac{\eta}{\epsilon})^{N},
\end{aligned}
\end{equation}
and then (b) follows.
The proof is complete.
\end{proof}


Note that $N>(1+\sqrt{2})\alpha$, then 
$\frac{N(N-2)-\alpha^2}{\alpha^2}>\frac{\alpha}{N-2\alpha}$.
Fix $\tilde{\theta}\in (\frac{\alpha}{N-2\alpha},\frac{N(N-2)-\alpha^2}{\alpha^2})$, 
then we can define $r_1:=\frac{1}{6m}$ and $\epsilon_r:=r^{\tilde{\theta}+1}$.
Set
\begin{equation}\label{45*}
\tilde{W}_r(x,t):=\phi_r(x,t) W_{\epsilon_r}(x,t)\quad \text{and}\quad 
\tilde{w}_r(x)=\tilde{W}_r(x,0).
\end{equation}
Obviously, $\tilde{W}_r$ and $\tilde{w}_r$ are continuous with respect to 
$r\in (0,r_1]$ in $X_0^{\alpha}(\mathcal{C}_{\Omega})$. In addition, for any 
$0<r\leq r_1$ and $\eta\in [0,r^{2\tilde{\theta}+1}]$,
the following result holds.

\begin{proposition}\label{lemma4.1}
There exist $C_3>0$ and $m_1>m_0$ such that for any $m\geq m_1$ and 
${\tilde{x}}\in \Omega$,
$$
\sup_{\tau\geq 0} I(\tau\zeta_{\eta}(x-{\tilde{x}},t)\tilde{W}_r(x,t))
\begin{cases}
<\frac{\alpha}{2N}(k_{\alpha}S_{\alpha,N})^{N/\alpha}
&\text{if } r\in (0,\frac{r_1}{2}],\\
\leq S_m &\text{if } r\in [\frac{r_1}{2},r_1],
\end{cases}
$$
where
\[
S_m=\frac{\alpha}{2N}(k_{\alpha}S_{\alpha,N})^{N/\alpha}
-C_3m^{-(\tilde{\theta}+1)\alpha}
\]
and 
$S_m+C_2 m^{\frac{N(\alpha-N)}{\alpha}}<\frac{\alpha}{2N}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}$. Here $m_0$ and $C_2$ are defined in Lemmas 
\ref{lem2} and \ref{lem3}, respectively.
\end{proposition}

\begin{proof}
By Lemma \ref{estimates} and \eqref{45*}, we have
\begin{align*}
&\|\zeta_{\eta}(x-{\tilde{x}},t)\tilde{W}_r(x,t)\|_{X_0^{\alpha}
 (\mathcal{C}_{\Omega})}^2\\
&\leq (k_{\alpha}S_{\alpha,N})^{N/\alpha}
 +C(\frac{{\epsilon_r}}{r})^{N-\alpha}
 + C(\frac{\eta}{{\epsilon_r}})^{N-\alpha}\\
&\leq \begin{cases}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}+Cr^{\tilde{\theta}(N-\alpha)}, 
 &\text{if } \alpha\in (0,1)\cup(1,2),\\
(k_{1}S_{1,N})^{N}+Cr^{\tilde{\theta}(N-1)}|\log{r}|,&\text{if }
 \alpha=1, 
\end{cases}\
\end{align*}
and
\begin{align*}
\int_{\Omega}|\zeta_{\eta}(x,0)\tilde{w}_r(x)|^{2^*_{\alpha}}dx
&\geq  (k_{\alpha}S_{\alpha,N})^{N/\alpha}
 -C(\frac{{\epsilon_r}}{r})^{N}-C(\frac{\eta}{{\epsilon_r}})^{N}\\
&\geq  (k_{\alpha}S_{\alpha,N})^{N/\alpha}-Cr^{\tilde{\theta} N}.
\end{align*}
Then for $r\in [\frac{r_1}{2},r_1]$, we have
\begin{equation}\label{54}
\begin{aligned}
&\|\zeta_{\eta}(x-{\tilde{x}},t)\tilde{W}_r(x,t)\|_{X_0^{\alpha}
(\mathcal{C}_{\Omega})}^2\\
&\leq \begin{cases}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}+Cm^{-\tilde{\theta}(N-\alpha)}
&\text{if } \alpha\in (0,1)\cup(1,2),\\
(k_{1}S_{1,N})^{N}+Cm^{-\tilde{\theta}(N-1)}\log m & \text{if } \alpha= 1,
\end{cases}
\end{aligned}
\end{equation}
and
\begin{equation}\label{55}
\begin{aligned}
\int_{\Omega}|\zeta_{\eta}(x,0)\tilde{W}_r(x,0)|^{2^*_{\alpha}}dx
&\geq  (k_{\alpha}S_{\alpha,N})^{N/\alpha}-C(\frac{{\epsilon_r}}{r})^{N}
 -C(\frac{\eta}{{\epsilon_r}})^{N}\\
&\geq  (k_{\alpha}S_{\alpha,N})^{N/\alpha}-Cm^{-\tilde{\theta} N}
\quad \text{if } r\in [\frac{r_1}{2},r_1].
\end{aligned}
\end{equation}

Note that 
$w_{\epsilon_r}(x)= (\frac{\epsilon_r}{\epsilon_r^2+|x|^2})^{\frac{N-\alpha}{2}}
\geq (\frac{1}{2\epsilon_r})^{\frac{N-\alpha}{2}}$ for $|x|\leq \epsilon_r$.
Then for $\eta=0$, we have
\begin{equation}\label{56}
\begin{aligned}
\int_{\Omega}|\tilde{w}_r(x)|^2dx
&\geq  \int_{B_{\epsilon_r}(0)}|w_{\epsilon_r}|^2dx\\
&\geq  C\int_{B_{\epsilon_r}(0)}\epsilon_r^{\alpha-N}dx\\
&\geq  C\epsilon_r^{\alpha}\\
&= Cr^{(\tilde{\theta}+1)\alpha}\quad
\big(\geq Cm^{-(\tilde{\theta}+1)\alpha} \text{ if } r\in [\frac{r_1}{2},r_1]\big).
\end{aligned}
\end{equation}
In addition, since $\tilde{w}_r(x)\leq C\epsilon_r^{\frac{\alpha-N}{2}}$, 
for any $\tilde{x}\in \Omega$ and $0<\eta\leq r^{2\tilde{\theta}+1}$, we obtain
\begin{equation}\label{57}
\begin{aligned}
\int_{\Omega}|\zeta_{\eta}(x-{\tilde{x}},0)\tilde{w}_r|^2dx
&= \int_{\Omega}|\tilde{w}_r|^2dx
-\int_{\Omega}(1-\zeta^2_{\eta}(x-{\tilde{x}},0))|\tilde{w}_r|^2dx\\
&\geq  C\epsilon_r^{\alpha}-\int_{|x-\tilde{x}|\leq \eta} \epsilon_r^{\alpha-N}dx\\
&\geq  C\epsilon_r^{\alpha}-\eta^{N} \epsilon_r^{\alpha-N}\\
&\geq  C r^{(\tilde{\theta}+1)\alpha}\ \left(\geq C m^{-(\tilde{\theta}+1)\alpha}\ \text{if}\ r\in [\frac{r_1}{2},r_1]\right).
\end{aligned}
\end{equation}
Since $\tilde{\theta}\in (\frac{\alpha}{N-2\alpha},
\frac{N(N-\alpha)-\alpha^2}{\alpha^2})$, we have 
$\tilde{\theta}(N-\alpha)>(\tilde{\theta}+1)\alpha>0$, and then there exists 
$r_0>0$ such that for any $0<r\leq r_0$, it holds
\begin{equation}\label{55*}
r^{\tilde{\theta}(N-\alpha)}
<r^{(\tilde{\theta}+1)\alpha}\quad \text{and}\quad 
r^{\tilde{\theta}(N-\alpha)}|\log r|<r^{(\tilde{\theta}+1)\alpha}.
\end{equation}
Thus for $\alpha\in(0,1)\cup (1,2)$ and $r\in[r_1/2,r_1]$ with $r_1<r_0$, 
by \eqref{54}-\eqref{55*}, we have
\begin{equation}\label{z1}
\begin{aligned}
&I(\tau\zeta_{\eta}(x-{\tilde{x}},t)\tilde{W}_r(x,t))\\
&= \frac{\tau^2}{2}[k_{\alpha}\int_{\mathcal{C}_{\Omega}}
t^{1-\alpha}|\zeta_{\eta}(x-{\tilde{x}},t)\tilde{W}_r(x,t)|^2\,dx\,dt\\
&\quad -\lambda\int_{\Omega}|\zeta_{\eta}(x-{\tilde{x}},0)\tilde{w}_r(x)|^2dx]
-\frac{\tau^{2^*_{\alpha}}}{2^*_{\alpha}}\int_{\Omega}|\tilde{w}_r(x)|^{2^*_{\alpha}}dx\\
&\leq  \max_{\tau\geq 0}\frac{\tau^2}{2} ((k_{\alpha}S_{\alpha,N})^{N/\alpha}+Cr^{\tilde{\theta}(N-\alpha)}
-Cr^{(\tilde{\theta}+1)\alpha})\\
&\quad -\frac{\tau^{2^*_{\alpha}}}{2^*_{\alpha}}((k_{\alpha}S_{\alpha,N})^{N/\alpha}
-Cr^{\tilde{\theta} N}) \\
&\leq  \frac{\alpha}{2N}
((k_{\alpha}S_{\alpha,N})^{N/\alpha}-Cr^{(\tilde{\theta}+1)\alpha})
\Big(\frac{(k_{\alpha}S_{\alpha,N})^{N/\alpha}-Cr^{(\tilde{\theta}+1)\alpha}}
{(k_{\alpha}S_{\alpha,N})^{N/\alpha}-Cr^{\tilde{\theta} N}}
\Big)^{\frac{N-\alpha}{2}}\\
&\leq \frac{\alpha}{2N}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}-C r^{(\tilde{\theta}+1)\alpha}\\
&= \frac{\alpha}{2N}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}-C m^{-(\tilde{\theta}+1)\alpha}.
\end{aligned}
\end{equation}
Similarly, for $\alpha=1$, by \eqref{54}--\eqref{55*}, we obtain
\begin{equation}\label{z2}
\begin{aligned}
&I(\tau\zeta_{\eta}(x-{\tilde{x}},t)\tilde{W}_r(x,t)) \\
&= \frac{\tau^2}{2}[k_{1}\int_{\mathcal{C}_{\Omega}}
|\zeta_{\eta}(x-{\tilde{x}},t)\tilde{W}_r(x,t)|^2\,dx\,dt\\
&\quad -\lambda\int_{\Omega}|\zeta_{\eta}(x-{\tilde{x}},0)\tilde{w}_r(x)|^2dx]
-\frac{\tau^{2^*_{1}}}{2^*_{1}}\int_{\Omega}|\tilde{w}_r(x)|^{2^*_{1}}dx\\
&\leq  \max_{\tau\geq 0}\frac{\tau^2}{2} ((k_{1}S_{1,N})^{N}
 +Cr^{\tilde{\theta}(N-1)}|\log r|-Cr^{(\tilde{\theta}+1)\alpha}) \\
&\quad  -\frac{\tau^{2^*_{1}}}{2^*_{1}}((k_{1}S_{1,N})^{N}
-Cr^{\tilde{\theta} N}) \\
&\leq  \frac{1}{2N}
((k_{1}S_{1,N})^{N}-Cr^{\tilde{\theta}+1})
(\frac{(k_{1}S_{1,N})^{N}-Cr^{\tilde{\theta}+1}}{(k_{1}S_{1,N})^{N}-Cr^{\tilde{\theta} N}})^{\frac{N-1}{2}}\\
&\leq \frac{1}{2N}
(k_{1}S_{1,N})^{N}-C r^{\tilde{\theta}+1}\\
&= \frac{1}{2N}
(k_{1}S_{1,N})^{N}-C m^{-(\tilde{\theta}+1)}.
\end{aligned}
\end{equation}
Therefore, by \eqref{z1} and \eqref{z2},  there exist $C_3>0$ and $m_1>m_0$ 
such that for any $\alpha\in(0,2)$ and $m\geq m_1$,
$$
\sup_{t\geq 0} I(t\zeta_{\eta}(x-{\tilde{x}},y)\tilde{W}_r(x,t))\leq \frac{\alpha}{2N}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}-C_3 m^{-(\tilde{\theta}+1)\alpha}=:S_m
$$
and
$C_3 m^{-(\tilde{\theta}+1)\alpha}>C_2 m^{\frac{N(\alpha-N)}{\alpha}}$
due to $0<(\tilde{\theta}+1)\alpha<\frac{N(N-\alpha)}{\alpha}$.
The lemma follows immediately.
\end{proof}

\section{Proof of Theorem \ref{thm1}}

In this section, we have all the tools to prove our main result.
Now, we fix $m\geq m_1$.Note that $r_1=\frac{1}{6m}$ and 
$\epsilon_r=r^{\tilde{\theta}+1}$.Let $\eta_r=r^{2\tilde{\theta}+1}$
 and $\tilde{x}\in \Omega$.Then for any $0<r\leq r_1$, we have
\begin{gather}\label{supp1}
\tilde{W}_{r}(x+\tilde{x},t)\in
X_0^{\alpha}\big(\overline{\mathbb{B}_r^+(-\tilde{x})}\big),\\
\label{supp}
\zeta_{\eta_r}(x,t)\tilde{W}_{r}(x+\tilde{x},t)\in
X_0^{\alpha}\Big(\overline{\mathbb{B}_r^+(-\tilde{x})
\backslash \mathbb{B}_{\frac{\eta_r}{2}}^+(0)}\Big).
\end{gather}
We write $B^j=\{x\in \mathbb{R}^j: |x|\leq 1\}$ and 
$S^j=\{x\in \mathbb{R}^{j+1}: |x|= 1\}$ for any integer $j\geq 1$.
Denote $u^{\pm}:=\max\{\pm u,0\}$. We have the following lemma.

\begin{lemma}\label{odd}
For any integer $k\geq 0$, there exists an odd continuous map
 $\bar{h}: \mathbb{R}^{k+N+2}\to X_0^{\alpha}(\mathcal{C}_{\Omega})$ 
such that $\lim_{|x|\to +\infty}I(\bar{h}(x))=-\infty$ and 
$\sup_{U\in \bar{h}(\mathbb{R}^{k+N+2})} I(U)
<\frac{\alpha}{N}(k_{\alpha}S_{\alpha,N})^{N/\alpha}$.
\end{lemma}

\begin{proof}
The proof follows the same idea as in \cite{Chen-Shioji-Zou}, 
so we only sketch the proof.
\smallskip
 
\noindent\textbf{Step 1:} First, we construct an odd continuous map
 $h_1: B^N\to X_0^{\alpha}(\mathbb{B}_{\frac{1}{2m}}(0))$ such that
\begin{equation}\label{construct1}
\operatorname{supp}h_1(y)^+\cap \operatorname{supp}h_1(y)^-
=\emptyset\text{ and } \sup_{\tau\geq 0} 
I(\tau h_1(y))<S_m+\frac{\alpha}{2N}(k_{\alpha} S_{\alpha,N})^{N/\alpha},
\end{equation}
for all $y\in B^N$.
For any $y\in B^N$, set $s=|y|,\theta=\frac{y}{|y|}$ and define 
$h_1: B^N\to X_0^{\alpha}(\mathbb{B}_{\frac{1}{2m}}(0))$ by
\begin{align*}
&h_1(y)(x,t)\\
&=\begin{cases}
\tilde{W}_{\frac{\eta_{r_1}}{2}}(x,t)-\xi_{\eta_{r_1}}(x,t)\tilde{W}_{r_1}
(x+4sr_1 \theta,t) \\
\quad\text{if } 0\leq s\leq \frac{1}{2},\\[4pt]
\tilde{W}_{s(2r_1-\eta_{r_1})-r_1+\eta_{r_1}}(x-2r_1 (2s\theta-\theta),t)
-\xi_{\eta_{r_1}}(x,t)\tilde{W}_{r_1}(x+2r_1 \theta,t) \\
\quad\text{if } \frac{1}{2}\leq s\leq 1.\\
\end{cases}
\end{align*}
Clearly, \eqref{construct1} follows from \eqref{supp1},\eqref{supp} 
and Proposition \ref{lemma4.1}.
\smallskip

\noindent\textbf{Step 2:} 
The map $h_1$ induces an odd continuous mapping 
$h_2: \mathbb{S}^{N}\to X_0^{\alpha}\big(\mathbb{B}_{\frac{1}{2m}}(0)\big)$ by
$$
h_2(x_1,\dots,x_{N+1})
=\begin{cases}
h_1(x_1,\dots,x_{N}) &\text{if } x_{N+1}\geq 0,\\
-h_1(-x_1,\dots,-x_{N}) &\text{if } x_{N+1}\leq 0.
\end{cases}
$$
Since $h_1$ is odd on $\mathbb{S}^{N-1}$, we have
\begin{equation}\label{construct2}
\operatorname{supp}h_2(\theta)^+\cap \operatorname{supp}h_2(\theta)^-
=\emptyset\ \text{and}\ \sup_{\tau\geq 0} I(\tau h_2(\theta))
<S_m+\frac{\alpha}{2N}S^{N/\alpha},\ \forall \theta\in \mathbb{S}^{N}.
\end{equation}

\noindent\textbf{Step 3:} There exists an odd continuous map 
$h_3: \mathbb{R}^{N+2}\to X_0^{\alpha}(\mathbb{B}_{\frac{1}{m}}(0))$ such that
\begin{equation}\label{33*}
\sup_{U\in h_3(\mathbb{R}^{N+2})} I(U)<S_m+\frac{\alpha}{2N}
(k_{\alpha}S_{\alpha,N})^{N/\alpha}.
\end{equation}
Indeed, define a cylindric surface in $\mathbb{R}^{N+2}$ by
$$
Z:=(\mathbb{S}^{N}\times[-1,1])\cup (B^{N+1}\times\{-1,1\})\subset \mathbb{R}^{N+2},
$$
and choose a positive function 
$v_0:=\xi_{\eta_{r_1}}(x,t)\tilde{W}_{\frac{1}{6m}}(x+y_0,t)
\in X_0^{\alpha}\big(\mathbb{B}_{\frac{1}{m}}(0)\backslash 
\mathbb{B}_{\frac{1}{2m}}(0)\big)$ with $y_0\in \Omega$
and $|y_0|=\frac{3}{4m}$.
Then, it follows from Proposition \ref{lemma4.1} that
\begin{equation}\label{supp2}
\sup_{t\geq 0}I(tv_0)\leq S_m.
\end{equation}
For $\theta\in \mathbb{S}^{N},{s_1}\in [0,1], {s_2}\in[-1,1]$, set
$$
\tilde{h}_2({s_1}\theta,{s_2}):=\begin{cases}
(1-{s_2})h_2(\theta)^-+(1+{s_2})h_2(\theta)^+ &\text{if } {s_1}=1,\\
2 s_1h_2(\theta)^++(1-{s_1})v_0 &\text{if } {s_2}=1,\\
2 s_1h_2(\theta)^-+(1-{s_1})v_0 &\text{if } {s_2}=-1.
\end{cases}
$$
It is easy to check that
$\operatorname{supp}\tilde{h}_2({s_1}\theta,{s_2})^+
\cap \operatorname{supp}\tilde{h}_2({s_1}\theta,{s_2})^-=\emptyset$.
Now, we extend $\tilde{h}_2$ to a map
$h_3: \mathbb{R}^{N+2}\to X_0^{\alpha}\left(\mathbb{B}_{\frac{1}{m}}(0)\right)$ 
by
$$
h_3(\tilde{t}z):=\tilde{t}\tilde{h}_2(z)\quad  \text{for } z\in Z,\ \tilde{t}\geq 0.
$$
Thus, \eqref{33*} follows from \eqref{supp1}, \eqref{supp}, \eqref{supp2} and 
Proposition \ref{lemma4.1} immediately.
\smallskip

\noindent\textbf{Step 4:} For $k\geq 1$, define an odd continuous map
$\bar{h}: \mathbb{R}^{k+N+2}\to X_0^{\alpha}(\mathcal{C}_{\Omega})$
by
$$
\bar{h}(y,z)=\tilde{h}_3(y)+h_3(z)\quad  \text{for all } y\in \mathbb{R}^k, 
z\in \mathbb{R}^{N+2},
$$
where $\tilde{h}_3: \mathbb{R}^k\to X_0^{\alpha}
\left(\Omega\backslash \mathbb{B}_{\frac{1}{m}}(0)\right)$ 
is an odd map defined by
$\tilde{h}_3(y_1,\dots,y_k):=\sum_{i=1}^{k} y_i E_i^m$.
It is easy to see that $\lim_{|(y,z)|\to +\infty}I(\bar{h}(y,z))=-\infty$.
Note that $\operatorname{supp}\tilde{h}_3(y)\cap \operatorname{supp}h_3(z)
=\emptyset$ for all $y\in \mathbb{R}^k, z\in \mathbb{R}^{N+2}$, then by 
Lemma \ref{lem3} and Proposition \ref{lemma4.1}, we have
\begin{align*}
\sup_{(y,z)\in \mathbb{R}^{k+N+2}}I(\bar{h}(y,z))
&\leq  \sup_{y\in \mathbb{R}^k} I(\tilde{h}_3(y))
 +\sup_{z\in \mathbb{R}^{N+2}}I(h_3(z))\\
&< C_2 m^{\frac{N(\alpha-N)}{\alpha}}+S_m
 +\frac{\alpha}{2N}(k_{\alpha}S_{\alpha,N})^{N/\alpha}\\
&< \frac{\alpha}{N}(k_{\alpha}S_{\alpha,N})^{N/\alpha}.
\end{align*}

\noindent\textbf{Step 5:} For $k=0$, we define
$\bar{h}: \mathbb{R}^{N+2}\to X_0^{\alpha}(\mathcal{C}_{\Omega})$ by $\bar{h}=h_3$.
Clearly, it follows from Proposition \ref{lemma4.1} that
$$
\sup_{z\in \mathbb{R}^{N+2}}I(\bar{h}(z))
<\frac{\alpha}{N}(k_{\alpha}S_{\alpha,N})^{N/\alpha}.
$$
Therefore, Step 4 and Step 5 yield our conclusion.
\end{proof}

Note that $\lambda_0=0$ and 
$\lambda_k^{\alpha/2}\leq \lambda< \lambda_{k+1}^{\alpha/2}$ for some 
$k\geq 0$. Then, we have the following lemma.

\begin{lemma}\label{beta1}
$0<\beta_{k+1}\leq \dots\leq \beta_{k+N+2}<2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}}$.
\end{lemma}

\begin{proof}
According to the definition of $\beta_j$, we obtain 
$\beta_{k+1}\leq \dots\leq \beta_{k+N+2}$. Moreover, by \eqref{beta1big0}, 
we have $\beta_{k+1}>0$. So we only need to verify 
$\beta_{k+N+2}<2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}}$.
By using the same idea as in \cite{Chen-Shioji-Zou}, we conclude that
$\gamma(\mathfrak{A})\geq k+N+2$, where
 $\mathfrak{A}:=\{U\in \bar{h}(\mathbb{R}^{k+N+2}):
 \|u\|_{L^{2^*_{\alpha}}(\Omega)}=1\}$.
Then, it follows from Lemma \ref{odd} that for any function $U\in \mathfrak{A}$,
$$
\frac{\alpha}{N}(k_{\alpha}S_{\alpha,N})^{N/\alpha}>\sup_{\tau\geq 0} I(\tau U)
\geq \frac{\alpha}{2N} \Big(\frac{\|U\|_{X_0^{\alpha}(\mathcal{C}_{\Omega})}^2
-\lambda\|u\|^2_{L^2(\Omega)}}{\|u\|_{L^{2^*_{\alpha}}(\Omega)}^2}\Big)^{N/\alpha}
=\frac{\alpha}{2N}J(U)^{N/\alpha}.
$$
This implies that 
$\sup_{U\in \mathfrak{A}}J(U)<2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}}$.
Therefore, by the definition of $\beta_{k+N+2}$, we conclude that
$\beta_{k+N+2}<2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}}$. We completed the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1}]
There are two cases to complete our proof.
If $K^{\beta}$ is infinite for some 
$\beta\in(0,2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}})$, then by Lemma \ref{2.2},  
$J$ has infinitely many critical points and hence we complete our proof. 
If $K^{\beta}$ is finite for all 
$\beta\in (0,2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}})$, then according to 
Lemmas \ref{2.2} and \ref{beta1}, we may assume 
$0<\beta_{k+1}<\dots<\beta_{k+N+2}<2^{\alpha/N}{{k_{\alpha}S_{\alpha,N}}}$.
Let $j_0\geq 1$ be an integer such that 
$\beta_{k+j_0}\geq {k_{\alpha}S_{\alpha,N}}$, Then Lemma \ref{3.22} 
implies that $J$ has at least $\max\{j_0-1, N+2-j_0\}\geq [\frac{N+1}{2}]$ 
pairs of nontrivial critical points, and so do $I$.
The proof is complete.
\end{proof}

\subsection*{Acknowledgments}
This research was supported by the
National Natural Science Foundation of  China
(Grant No.11371128 \& No.11171098).

\begin{thebibliography}{99}

\bibitem{Applebaum} Applebaum, D.;
\emph{L\'evy processes-from probability to finance and quantum groups}. 
Notices of the American Mathematical Society, 2004, 51(11).

\bibitem{Barrios-Colorado-Pablo2012}
Barrios, B; Colorado, E; De Pablo, A; et al;
\emph{On some critical problems for the fractional Laplacian operator}. 
Journal of Differential Equations, 2012, 252(11): 6133-6162.

\bibitem{Brandle-Colorado-de Pablo}
Brandle, C; Colorado, E; De Pablo, A; et al;
\emph{A concave-convex elliptic problem involving the fractional Laplacian}. 
Proceedings of the Royal Society of Edinburgh, 2013, 143(1):39-71.

\bibitem{Brezis-Nirenberg}
Brezis, H.; Nirenberg, L.;
\emph{Positive solutions of nonlinear elliptic equations involving critical 
sobolev exponents}. Communications on Pure and Applied Mathematics, 1983, 
36(4):437-477.

\bibitem{Cabre-Tan}
Cabre, X.; Tan, J.;
\emph{Positive solutions of nonlinear problems involving the square root of 
the Laplacian}. Advances in Mathematics, 2009, 42(1-2):2052-2093.

\bibitem{Caffarelli-Silvestre}
Caffarelli, L; Silvestre, L.;
\emph{An Extension Problem Related to the Fractional Laplacian}. 
Communications in Partial Differential Equations, 2006, 32(7-9):1245-1260.

\bibitem{Chang-Wang}
Chang, X.; Wang, Z. Q.;
\emph{Ground state of scalar field equations involving a fractional Laplacian 
with general nonlinearity}. Nonlinearity, 2013, 26(2):479-494(16).

\bibitem{Chen-Li-Ou}
Chen, W.;  Li, C.; Ou, B.;
\emph{Classification of solutions for an integral equation}. 
Communications on Pure and Applied Mathematics, 2006, 59(3): 330-343.

\bibitem{Chen-Shioji-Zou}
Chen, Z.; Shioji, N.; Zou, W.;
\emph{Ground state and multiple solutions for a critical exponent problem}. 
Nonlinear Differential Equations and Applications Nodea, 2012, 19(3):253-277.

\bibitem{Clapp-Weth}
Clapp, M.; Weth, T.;
\emph{Multiple solutions for the Brezis-Nirenberg problem}. 
Advances in Differential Equations, 2005, 10(10):463-480.

\bibitem{Solimini}
Devillanova, G.; Solimini, S.;
\emph{Concentration estimates and multiple solutions to
elliptic problems at critical growth}. Advances in Differential Equations, 
7, 1257-1280 (2002)

\bibitem{Hua-Yu}
Hua, Y.;  Yu, X.;
\emph{On the ground state solution for a critical fractional Laplacian equation}. 
Nonlinear Analysis, 2013, 87(87):116-125.

\bibitem{Lebedev}
Lebedev, N. N.;
\emph{Special functions and their applications}. Selected Russian Publications 
in the Mathematical Sciences, 1966, 20(93):70-72.

\bibitem{Li-Y}
Li, Y.;
\emph{Remark on some conformally invariant integral equations: 
the method of moving spheres}. Journal of the European Mathematical Society, 2003, 2(2):153-180.

\bibitem{Tan}
Tan, J.;
\emph{The Brezis-Nirenberg type problem involving the square root of the Laplacian}. 
Calculus of Variations and Partial Differential Equations, 2011, 42(1-2):21-41.

\bibitem{Yan-Yang-Yu}
Yan, S.;, Yang, J.; Yu, X.;
\emph{Equations involving fractional Laplacian operator: 
Compactness and application}. Journal of Functional Analysis, 2007, 269(1):47-79.
\end{thebibliography}





 \end{document}