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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 126, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{7mm}}

\begin{document}
\title[\hfilneg EJDE-2018/126\hfil Perturbed fractional Schr\"odinger equation]
{Multiplicity of solutions for a perturbed fractional Schr\"odinger equation
involving oscillatory terms}

\author[C. Ji, F. Fang \hfil EJDE-2018/126\hfilneg]
{Chao Ji, Fei Fang}

\address{Chao Ji \newline
Department of Mathematics,
East China University of Science and Technology,
200237 Shanghai, China}
\email{jichao@ecust.edu.cn}

\address{Fei Fang \newline
Department of Mathematics, 
Beijing Technology and Business University,
100048 Beijing, China}
\email{fangfei68@163.com}


\dedicatory{Communicated by Binlin Zhang}

\thanks{Submitted January 7, 2018. Published June 18, 2018.}
\subjclass[2010]{35J60, 47J30}
\keywords{Fractional Schr\"odinger equation; multiple solutions; oscillatory terms}

\begin{abstract}
 In this article we study the perturbed fractional Schr\"odinger equation
 involving  oscillatory terms
 \begin{gather*}
 (-\Delta)^{\alpha} u+u =Q(x)\Big(f(u)+\epsilon g(u)\Big), \quad x\in \mathbb{R}^N\\
 u\geq 0,
 \end{gather*}
 where $\alpha\in (0, 1)$ and $N> 2\alpha$, $(-\Delta)^{\alpha}$ stands for
 the fractional Laplacian, $Q: \mathbb{R}^N\to \mathbb{R}^N$ is a radial,
 positive potential,  $f\in C([0, \infty), \mathbb{R})$ oscillates near the
 origin or at infinity and   $g\in C([0, \infty), \mathbb{R})$ with $g(0)=0$.
 By using the variational method  and the principle of symmetric criticality
 for non-smooth Szulkin-type functionals,  we establish that:
 (1) the unperturbed problem, i.e.\ with $\epsilon=0$  has infinitely many solutions;
 (2) the number of distinct solutions becomes greater and greater when
 $| \epsilon|$ is smaller and smaller. Moreover, various properties of the
 solutions are also described in terms of the $L^{\infty}$- and  
 $H^{\alpha}(\mathbb{R}^N)$-norms.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction} \label{intro}

In this paper we consider the multiplicity of positive solutions for the
 fractional Schr\"odinger equation
\begin{equation} \label{ePe}
\begin{gathered}
(-\Delta)^{\alpha} u+u =Q(x)\Big(f(u)+\epsilon g(u)\Big), \quad x\in\mathbb{R}^N,\\
 u\geq 0,
\end{gathered}
\end{equation}
where $\alpha\in (0, 1)$, $N> 2\alpha$,  $(-\Delta)^{\alpha}$ stands
for the fractional Laplacian, $Q:\mathbb{R}^N \to \mathbb{R}^N$ ia a radial,
 positive potential, $f:[0 +\infty)\to\mathbb{R}^N$ is a continuous nonlinearity
 which oscillates near the origin or at infinity and $g:[0, \infty) \to R$
is an continuous function with $g(0)=0$.

In the local case, that is, when $\alpha=1$, the arbitrarily many solutions
for the perturbed elliptic problem \eqref{ePe} involving oscillatory terms,
for the case $N\geq 2$, has been studied in \cite{rK}.
Krist\'aly  \cite{rK} first proved the unperturbed problem
\eqref{ePe} with $\epsilon=0$ in \eqref{ePe},
 has infinitely many distinct solutions. Then, he proved that the number
of distinct solutions for the perturbed problem \eqref{ePe}
 becomes greater and greater when
$| \epsilon|$ is smaller and smaller.

In the nonlocal case, that is, when $\alpha\in (0, 1)$, to the best of
our knowledge, there are no studies for the fractional nonlinear equation
\eqref{ePe}, maybe because technique developed for local case cannot
be adapted immediately, c.f.\ \cite{rS1}.
Motivated by  \cite{rK}, we establish the multiplicity of positive solutions
for \eqref{ePe}. Because of the nonlocal nature of the fractional Laplacian,
we would like to point out that some estimates in \cite{rK} cannot be
obtained directly when $\alpha\in (0, 1)$. In this paper, we will
overcome these difficulties by more careful estimates for the energy functional
associated with the auxiliary problem, see proof of Theorem \ref{thm2.1}.
Another novelty is the truncation function $\omega_{s}(x)$ in \cite{rK}
will be replaced by a more general function.

Throughout this paper,  we always assume
\begin{itemize}
\item[(A1)]   $Q:\mathbb{R}^N\to \mathbb{R}^N$ is a positive, continuous,
 radially symmetric potential such that $Q\in L^{p}(\mathbb{R}^N)$ for every $p\in [1, 2]$.
\end{itemize}
We recall that, for any $\alpha\in (0, 1)$, the fractional Laplacian
$(-\Delta)^{\alpha} u$ of a function $u:\mathbb{R}^N\to \mathbb{R}^N$,
 with sufficient decay, is defined by
\[  % \label{1}
\mathcal{F}((-\Delta)^{\alpha} u)(\xi)
=| \xi|^{2\alpha}\mathcal{F}(u)(\xi), \quad  \xi\in \mathbb{R}^N,
\]
where $\mathcal{F}$ denotes the Fourier transform,
\[
\mathcal{F}(\phi)(\xi)
=\frac{1}{(2\pi)^{N/2}}\int_{\mathbb{R}^N}  e^{-i\xi\cdot x}\phi(x)dx
\equiv \widehat{\phi}(\xi),
\]
for function $\phi$ in the Schwartz class.
 $(-\Delta)^{\alpha} u$ can also be computed by the following singular integral:
\[
(-\Delta)^{\alpha} u=c_{N, \alpha}\operatorname{P.V.}
\int_{\mathbb{R}^N} \frac{u(x)-u(y)}{| x-y|^{N+2\alpha}}dy,
\]
here P.V. is the principal value and $c_{N, \alpha}$ ia a normalization
constant.

The fractional Sobolev space $H^{\alpha}(\mathbb{R}^N)$ is defined by
\[  %\label{space}
 H^\alpha(\mathbb{R}^N)=\big\{u\in L^2(\mathbb{R}^N):
\frac{| u(x)-u(y)|}{| x-y|^{\frac{N}{2}+\alpha}}
\in L^2(\mathbb{R}^N\times \mathbb{R}^N)\big\},
\]
endowed with the norm
\[  %\label{space}
 \| u\|=\Big(\int_{\mathbb{R}^N} u^2dx
+\iint_{\mathbb{R}^N\times\mathbb{R}^N}\frac{| u(x)-u(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy
\Big)^{1/2},
\]
where the term
\[  %\label{space}
[u]_{H^\alpha(\mathbb{R}^N)}=\| (-\Delta)^{\alpha/2} u\|_{L^2(\mathbb{R}^N)}
:=\Big(\iint_{\mathbb{R}^N\times\mathbb{R}^N}\frac{| u(x)-u(y)|^2}
{| x-y|^{N+2\alpha}}\,dx\,dy\Big)^{1/2}
\]
is the so-called \emph{Gagliardo semi-norm} of $u$. 

For $N>2\alpha$, from \cite{rDPV} we know that there exists a constant
$C=C(N, \alpha)>0$ such that
\[
\| u\|_{L^{2_{\alpha}^{*}}(\mathbb{R}^N)}\leq C\| u\|_{H^{\alpha}(\mathbb{R}^N)}
\]
for every $u\in H^{\alpha}(\mathbb{R}^N)$, where $2_{\alpha}^{*}=\frac{2N}{N-2\alpha}$
is the fractional critical exponent. Moreover, the embedding
$H^{\alpha}(\mathbb{R}^N)\subset L^{q}(\mathbb{R}^N)$ is continuous for any 
$q\in [2, 2_{\alpha}^{*}]$,
and is locally compact whenever $q\in [2, 2_{\alpha}^{*})$.
For the basic properties of the fractional Sobolev space $H^{\alpha}(\mathbb{R}^N)$,
we refer to \cite{rDPV, rL, rMRS, rS1}. 

Let $f\in C([0, \infty), \mathbb{R})$ and $F(t)=\int_{0}^{t}f(\tau)d\tau$,
$t\geq 0$. We assume:
\begin{itemize}
\item[(A2)] $-\infty<\liminf_{t\to 0^{+}}\frac{F(t)}{t^2}
\leq \limsup_{t\to 0^{+}}\frac{F(t)}{t^2}=+\infty$.

\item[(A3)] There exists a sequence $(t_i)_i\subset (0, \infty)$
converging to 0 such that $f(t_i)< 0$ for every $i\in N$.
\end{itemize}

\begin{remark} \label{rmk1.1} \rm
(1) Assumptions (A2) and (A3) imply an oscillatory behavior of $f$ near the origin.

(2) Let $\alpha, \beta, \gamma\in \mathbb{R}^N$ such that $0<\alpha<1<\alpha+\beta$,
and $\gamma\in(0, 1)$. Then, the function  $f\in C([0, \infty), \mathbb{R})$
defined by $f(0)=0$ and $f(s)=s^{\alpha}(\gamma+\sin s^{-\beta})$ satisfies
 (A2) and (A3), respectively.
\end{remark}

For the unperturbed problem \eqref{ePe} with $\epsilon=0$, we have the
following result.

\begin{theorem} \label{thm1.1}
Assume {\rm (A1)} holds and $f\in C([0, \infty), \mathbb{R})$ satisfying
{\rm (A2)} and {\rm (A3)}. Then there exists a sequence
$\{u_i^{0}\}_i\subset H^{\alpha}(\mathbb{R}^N)$ of distinct, radially symmetric
weak solutions of \eqref{ePe} with $\epsilon=0$  such that
\begin{equation} \label{e1.1}
\lim_{i\to \infty} \| u_i^{0}\|_{L^{\infty}}
=\lim_{i\to \infty} \| u_i^{0}\|=0.
\end{equation}
\end{theorem}

For the perturbed problem \eqref{ePe}, one has the weaker result.

\begin{theorem} \label{thm1.2}
Assume {\rm (A1)} holds, $f\in C([0, \infty), \mathbb{R})$ satisfying
{\rm (A2)} and {\rm (A3)} and  $g\in C([0, \infty), \mathbb{R})$ with $g(0)=0$.
Then, for every $k\in N$, there exists $\epsilon_{k}^{0}>0$ such that
\eqref{ePe} has at least $k$ distinct, radially symmetric weak solutions
in $H^{\alpha}(\mathbb{R}^N)$ whenever
$\epsilon\in [-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$. Moreover, if this
$k$ solutions are denoted by $u_{i, \epsilon}^{0}\in H^{\alpha}(\mathbb{R}^N)$,
$i=1, \dots, k$, then
\begin{equation}  \label{e1.2}
\| u_{i, \epsilon}^{0}\|_{L^{\infty}}< \frac{1}{i}\quad \text{and}\quad
\| u_{i, \epsilon}^{0}\|< \frac{1}{i}\quad \text{for any } i=1, \dots, k.
\end{equation}
\end{theorem}

\begin{remark}  \label{rmk1.2}  \rm
Note \eqref{e1.1} and \eqref{e1.2} are in a perfect concordance.
Moreover, the perturbed and unperturbed ones are equivalent in the sense
that they are deducible from each other. Clearly, the perturbed problem
contains the unperturbed one by choosing $g=0$. Conversely,
exploiting the behavior of certain sequences which appear in the proof of
 Theorem \ref{thm1.1}, we can show that for every $k\in N$, there exists
$\epsilon_{k}^{0}>0$ such that the perturbed problem  has at least $k$
distinct solutions in $H^{\alpha}(\mathbb{R}^N)$ whenever
 $\epsilon\in [-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$.
\end{remark}

Next, we will state the counterparts of Theorems \ref{thm1.1} and \ref{thm1.2}
 when $f$ oscillates at infinity. We assume:
\begin{itemize}
\item[(A4)] $-\infty<\liminf_{t\to \infty}\frac{F(t)}{t^2}
\leq \limsup_{t\to \infty}\frac{F(t)}{t^2}=+\infty$.

\item[(A5)] There exists a sequence $(t_i)_i\subset (0, \infty)$ converging to $+\infty$ such that $f(t_i)< 0$ for every $i\in N$.
\end{itemize}

\begin{remark} \label{rmk1.3} \rm
(1) The assumptions (A4) and (A5) imply an oscillatory behavior of $f$ at infinity.

(2) Let $\alpha, \beta, \gamma\in \mathbb{R}^N$ such that $\alpha>1$,
$| \alpha-\beta|<1$,  and $\gamma\in(0, 1)$. Then, the function
$f\in C([0, \infty), \mathbb{R})$ defined by
$f(s)=s^{\alpha}(\gamma+\sin s^{-\beta})$ satisfies (A4) and (A5), respectively.
\end{remark}

For problem \eqref{ePe} with $\epsilon=0$, we have the counterpart of 
Theorem \ref{thm1.1}.

\begin{theorem} \label{thm1.3}
Assume {\rm (A1)} holds and $f\in C([0, \infty), \mathbb{R})$ satisfying
{\rm (A4), (A5)} and $f(0)=0$. Then there exists a sequence
$\{u_i^{\infty}\}_i\subset H^{\alpha}(\mathbb{R}^N)$ of radially symmetric
 weak solutions of \eqref{ePe} with $\epsilon=0$  such that
\begin{equation}  \label{e1.3}
\lim_{i\to \infty} \| u_i^{\infty}\|_{L^{\infty}}=\infty.
\end{equation}
\end{theorem}

\begin{remark} \label{rmk1.4} \rm
Beside of (A4) and (A5), no further growth condition is assumed on the nonlinear
 tern at infinity.
Actually, this is why we cannot give $H^{\alpha}(\mathbb{R}^N)$-norm estimates
for the solutions in Theorem \ref{thm1.3}. If we further assume that $f$
satisfies the following growth condition at infinity,
i.e., there exists $q\in (1, \frac{2_{\alpha}^{*}}{2})$ and $C>0$ such that
\begin{equation} \label{e1.4}
| f(t)|\leq C(1+t^{q-1} )\quad \text{for all } t\in [0, \infty).
\end{equation}
Then, we have
\begin{equation}  \label{e1.5}
\lim_{i\to \infty} \| u_i^{\infty}\|=\infty.
\end{equation}
It is easy to see that \eqref{e1.4} and the right side of (A4) imply $q>2$.
Thus, \eqref{e1.5} is possible for the lower dimensions $N=1, 2, 3$ and
adding some restriction for $\alpha$,  that is, when $4\alpha>N>2\alpha$,
 since $2<\frac{2_{\alpha}^{*}}{2}$. In fact, for \eqref{e1.5} holds,
we need to further assume that $\alpha\in (\frac{1}{4}, \frac{1}{2})$, if $N=1$;
 $\alpha\in (\frac{1}{2}, 1)$, if $N=2$; $\alpha\in (\frac{3}{4}, 1)$, if $N=3$.
Another way to guarantee \eqref{e1.5} is to complete assumption (A1) by
allowing for instance $Q\in L^{\infty}(\mathbb{R}^N)$ and \eqref{e1.4} with
$q\in(2, 2_{\alpha}^{*})$.
\end{remark}

For problem \eqref{ePe}, we also have the counterpart of Theorem \ref{thm1.2}.

\begin{theorem} \label{thm1.4}
Assume {\rm (A1)} holds, $f\in C([0, \infty), \mathbb{R})$ satisfying
{\rm (A4), (A5)} with $f(0)=0$, and  $g\in C([0, \infty), \mathbb{R})$ with $g(0)=0$.
Then, for every $k\in N$, there exists $\epsilon_{k}^{\infty}>0$ such that
\eqref{ePe} has at least $k$ distinct, radially symmetric weak solutions
in $H^{\alpha}(\mathbb{R}^N)$ whenever
$\epsilon\in [-\epsilon_{k}^{\infty}, \epsilon_{k}^{\infty}]$.
Moreover, for this $k$ solutions are denoted by
 $u_{i, \epsilon}^{\infty}\in H^{\alpha}(\mathbb{R}^N)$, $i=1, \dots, k$,
we have
\begin{equation}  \label{e1.6}
\| u_{i, \epsilon}^{\infty}\|_{L^{\infty}}> i-1\quad
\text{for } i=1, \dots, k.
\end{equation}
\end{theorem}

\begin{remark} \label{rmk1.5} \rm
Equations \eqref{e1.3} and \eqref{e1.6} are also in concordance.
Moreover, if both functions $f$ and $g$ verify \eqref{e1.4} with 
$q\in (2, \frac{2_{\alpha}^{*}}{2})$ and if $N=1$,
$\alpha\in (\frac{1}{4}, \frac{1}{2})$, if $N=2$, $\alpha\in (\frac{1}{2}, 1)$, 
if $N=3$, $\alpha\in (\frac{3}{4}, 1)$, then besides of \eqref{e1.6}, 
whenever $\epsilon\in [-\epsilon_{k}^{\infty}, \epsilon_{k}^{\infty}]$, we also have
\begin{equation} \label{e1.7}
\| u_{i, \epsilon}^{\infty}\|> i-1\quad  \text{for } i=1, \dots, k.
\end{equation}
\end{remark}

In recent years, the study of the various nonlinear equations or systems involving
fractional Laplacian has received considerable attention. These problems mainly
arise in fractional quantum mechanics \cite{rL1, rL2}, physics and
chemistry \cite{rMK}, obstacle problems \cite{rSi}, optimization and finance
\cite{rCT} and so on. The literature on non-local fractional Laplacian
operators and their application to differential
equations is quite large, we refer the interested reader to
\cite{rCS, rCW, rMR, rMRS, rPXZ1, rPXZ, rS1, rS2, rSi, rXzr1, rXzr2, rXzz} and the
references therein.

The rest of this article  is organized as follows.
In Section 2, we present an auxiliary result which is important for our problem.
In Section 3 we prove Theorems \ref{thm1.1} and \ref{thm1.2}.
In Section 4 we prove Theorems \ref{thm1.3} and  \ref{thm1.4}, 
Remarks \ref{rmk1.4} and  \ref{rmk1.5}.

\section{An auxiliary result}

In this section we consider the generic problem
\begin{equation} \label{e2.1}
\begin{gathered}
(-\Delta)^{\alpha} u+u =Q(x)h(u), \quad x\in\mathbb{R}^N\\
u\geq 0,
\end{gathered}
\end{equation}

Beside of the assumption (A1), we further assume that
\begin{itemize}
\item[(A6)] $h:[0, +\infty)\to \mathbb{R}^N$ is a continuous, bounded
function such that $h(0)=0$;

\item[(A7)] There are $0<a<b$ such that $h(s)\leq 0$ for all $s\in [a, b]$.
\end{itemize}
By  assumption (A6), we may put $h(s)=0$ for $s\leq 0$.
The energy functional $J_{h}$ on $H^{\alpha}(\mathbb{R}^N)$ associated with
problem \eqref{e2.1} is
\begin{equation}  \label{e2.2} % fcl
J_{h}(u):= \frac{1}{2}\int_{\mathbb{R}^N}  \Big(| (-\Delta)^{\alpha/2} u|^2+u^2\Big)dx
-\int_{\mathbb{R}^N} Q(x)H(u)\,dx,
\end{equation}
where $H(u)=\int_{0}^{u}h(s)ds$. By the mean value theorem and H\"older
inequality, for any $u\in H^{\alpha}(\mathbb{R}^N)$, we have
\[
\int_{\mathbb{R}^N} Q(x)H(u)\,dx\leq \int_{\mathbb{R}^N} Q(x)| H(u)|\,dx
\leq M_{h}\| Q\|_{L^2}\| u\|_{L^2}<\infty,
\]
where $M_{h}=\sup_{s\in \mathbb{R}}| h(s)|$, so the functional $J_{h}$ is well defined.
 Moreover, by the assumptions (A1), (A6) and Lebesgue dominated
convergence theorem, $J\in C^{1}(H^{\alpha}(\mathbb{R}^N), \mathbb{R})$
and its critical points are the solutions of problem \eqref{e2.1}.

Now, we denote by $H^{\alpha}_{\rm rad}(\mathbb{R}^N)$ radial functions in
$H^{\alpha}(\mathbb{R}^N)$, and let
\[
R_{h}=J_{h}\big|_{H^{\alpha}_{\rm rad}(\mathbb{R}^N)},
\]
i.e., the restriction of $J_{h}$ to $H^{\alpha}_{\rm rad}(\mathbb{R}^N)$. Moreover, for
$b\in \mathbb{R}^{+}$, we denote
\[
W^{b}=\{u\in H^{\alpha}(\mathbb{R}^N): \| u\|_{L^{\infty}}\leq
b\}\quad \text{and}\quad W^{b}_{\rm rad}=W^{b}\cap
H^{\alpha}_{\rm rad}(\mathbb{R}^N).
\]
Now we state the main result of this section.

\begin{theorem} \label{thm2.1}
Assume that {\rm (A1), (A6), (A7)} hold. Then
\begin{itemize}
\item[(i)] The functional $R_{h}$ is bounded from below on
$W^{b}_{\rm rad}$ and attains its infimum at $u_{h}\in
W^{b}_{\rm rad}$.

\item[(ii)] $u_{h}\in [0, a]$ for a.e. $x\in \mathbb{R}^N$.

\item[(iii)] $u_{h}$ is a radial weak solution of problem \eqref{e2.1}.
\end{itemize}
\end{theorem}

\begin{proof}
 (i) For any $u\in H^{\alpha}_{\rm rad}(\mathbb{R}^N)$, by (A1) and (A6), we have
\begin{align*}
R_{h}(u)
&=\frac{1}{2}\| u\|^2-\int_{\mathbb{R}^N}Q(x)H(u)dx\\
&\geq \frac{1}{2}\| u\|^2- M_{h}\| Q\|_{L^2}\| u\|_{L^2}\\
&\geq \frac{1}{2}\| u\|^2- M_{h}\| Q\|_{L^2}\| u\|\\
&\geq -\frac{1}{2}M_{h}^2\| Q\|_{L^2}^2,
\end{align*}
so the functional $R_{h}$ is bounded from below on
$W^{b}_{\rm rad}$. Now we prove that it attains infimum at
$u_{h}\in W^{b}_{\rm rad}$. Noting that $W_{\rm rad}^{b}$ is convex and
closed, so it is weakly closed. By the above inequality,
the functional $R_{h}$ is coercive, so we only need to show
that the functional $R_{h}$ is sequentially weakly lower
semicontinuous. Since $u\mapsto \| u\|$ is sequentially weakly lower
semicontinuous, it is enough to show that $u\mapsto \int_{\mathbb{R}^N}Q(x)H(u)$
is sequentially weakly continuous.  Arguing by contraction,
suppose that for a sequence $\{u_{n}\}_{n}\subset H^{\alpha}_{\rm rad}(\mathbb{R}^N)$
such that $u_{n}\rightharpoonup u\in H^{\alpha}_{\rm rad}(\mathbb{R}^N)$,
there exists a number $\epsilon_{0}>0$ such that
\[
0<\epsilon_{0}\leq \Big| \int_{\mathbb{R}^N}Q(x)H(u_{n})
- \int_{\mathbb{R}^N}Q(x)H(u)\Big|\quad\text{for all } n\in \mathbb{N}.
\]
By \cite{rL}, we can see that  $H^{\alpha}_{\rm rad}(\mathbb{R}^N)$ is compactly
embedded into $L^{q}(\mathbb{R}^N)$ for all $q\in (2, 2_{\alpha}^{*})$,
so $u_{n}\to u$ in $L^{q}(\mathbb{R}^N)$. By the mean value theorem and
H\"older inequality, we have
\begin{align*}
0&<\epsilon_{0}\leq \Big| \int_{\mathbb{R}^N}Q(x)H(u_{n})
 - \int_{\mathbb{R}^N}Q(x)H(u)\Big|\leq M_{h}\int_{\mathbb{R}^N}Q(x)| u_{n}-u| dx \\
&\leq  M_{h}\| Q\|_{L^{\frac{q}{q-1}} }\| u_{n}-u\|_{L^{q}},
\end{align*}
this is a contradiction and the proof part (i) is complete.

(ii) Let $A=\{x\in \mathbb{R}^N: u_{h}(x)\not\in [0, a]\}$ and suppose that
$| A|>0$, where $| A|$ denotes the Lebesgue measure of the set $A$.
Define the function $\gamma: \mathbb{R}\to \mathbb{R}^N$ by
$\gamma(s)=\min(s_{+}, a)$,
 where $s_{+}=\max(s, 0)$,  then $\gamma$ ia a Lipschitz function and
$\gamma(0)=0$. Set $\omega=\gamma\circ u_{h}$, it is clear that
 $\omega$ is radial, $0\leq \omega \leq a$ for a.e. $x\in \mathbb{R}^N$ and
 $\omega\in H^{\alpha}(\mathbb{R}^N)$.

Now we define the sets
\[
A_1=\{x\in A: u_{h}(x)<0\},\quad  A_{2}=\{x\in A: u_{h}(x)>a\}.
\]
Then $A=A_1\cup A_{2}$, and we have that $\omega(x)=u_{h}(x)$ for
all $x\in \mathbb{R}^N\setminus A$, $\omega(x)=0$ for all $x\in A_1$, and
$\omega(x)=a$ for all $x\in A_{2}$. Thus,
\begin{align*}
&R_{h}(\omega)-R_{h}(u_{h}) \\
&=\frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}\frac{| \omega(x)-\omega(y)|^2}
 {| x-y|^{N+2\alpha}}\,dx\,dy-\frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}
 \frac{| u_{h}(x)-u_{h}(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\frac{1}{2}\int_{\mathbb{R}^N} (\omega^2-u_{h}^2)dx+\int_{\mathbb{R}^N} Q(x)
 \big(H(\omega)-H(u_{h})\big)dx\\
&=\frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}\frac{| \omega(x)
 -\omega(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy
 -\frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}
 \frac{| u_{h}(x)-u_{h}(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\frac{1}{2}\int _{A}(\omega^2-u_{h}^2)dx
 +\int _{A}  Q(x)\Big(H(\omega)-H(u_{h})\Big)dx.
\end{align*}
If $x\in A_1$ and $y\in A_1$, then
\begin{equation} \label{e2.3}
| \omega(x)-\omega(y)|=0\leq | u_{h}(x)-u_{h}(y)|.
\end{equation}
If $x\in A_1$ and $y\in A_{2}$, then
\begin{equation} \label{e2.4}
| \omega(x)-\omega(y)|=a\leq | u_{h}(x)-u_{h}(y)|.
\end{equation}
If $x\in A_1$ and $y\in \mathbb{R}^N\setminus A$, then
\begin{equation} \label{e2.5}
| \omega(x)-\omega(y)|=u_{h}(y)\leq | u_{h}(x)-u_{h}(y)|.
\end{equation}
If $x\in A_{2}$, $y\in A_{2}$, then
\begin{equation} \label{e2.6}
| \omega(x)-\omega(y)|=0\leq | u_{h}(x)-u_{h}(y)|.
\end{equation}
If $x\in A_{2}$, $y\in \mathbb{R}^N\setminus A$, then
\begin{equation} \label{e2.7}
| \omega(x)-\omega(y)|=| a-u_{h}(y)|\leq | u_{h}(x)-u_{h}(y)|.
\end{equation}
If $x\in \mathbb{R}^N\setminus A$, $y\in \mathbb{R}^N\setminus A$, then
\begin{equation} \label{e2.8}
| \omega(x)-\omega(y)|=| u_{h}(x)-u_{h}(y)|.
\end{equation}
From \eqref{e2.3}-\eqref{e2.8}, for any $x, y\in\mathbb{R}^N$, one has
\[
| \omega(x)-\omega(y)|\leq | u_{h}(x)-u_{h}(y)|,
\]
so
\[
\frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}
\frac{| \omega(x)-\omega(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy
-\frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}
\frac{| u_{h}(x)-u_{h}(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy\leq 0.
\]
Note that
\[
\int _{A} (\omega^2-u_{h}^2)dx=-\int _{A_1}u_{h}^2dx
+\int _{A_{2}}(a^2-u_{h}^2)dx\leq 0.
\]
Since $h(s)=0$ for all $s\leq 0$, we have
\[
\int _{A_1}  Q(x)\Big(H(\omega)-H(u_{h})\Big)dx=0.
\]
By the mean value theorem, for a.e.\ $x\in A_{2}$, there exists
$\theta(x)\in [a, u_{h}(x)]\subseteq [a, b]$ such that
\[
H(\omega(x))-H(u_{h}(x))=H(a)-H(u_{h}(x))=h(\theta(x))(a-u_{h}(x)).
\]
By (A7), we have
\[
\int _{A_{2}}  Q(x)\Big(H(\omega)-H(u_{h})\Big)dx\leq 0.
\]
So $R_{h}(\omega)-R_{h}(u_{h})\leq 0$.
Moreover $R_{h}(\omega)-R_{h}(u_{h})\geq 0$
according to the definition of $u_{h}$. Thus
\begin{gather*}
 \frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}
 \frac{| \omega(x)-\omega(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy
 -\frac{1}{2}\iint_{\mathbb{R}^N\times\mathbb{R}^N}
 \frac{| u_{h}(x)-u_{h}(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy=0,\\
 \frac{1}{2}\int _{A}(\omega^2-u_{h}^2)dx=0,\quad
\int _{A}  Q(x)\Big(H(\omega)-H(u_{h})\Big)dx=0.
\end{gather*}
In particular,
\[
\int _{A_1}u_{h}^2dx=\int _{A_{2}}(a^2-u_{h}^2)dx=0,
\]
this implies that $\operatorname{meas}(A)$ should be zero and this
is a contradiction with the assumption.

(iii) By \cite{rSz},  we have
\begin{equation} \label{e2.9}
R_{h}'(u_{h})(\omega-u_{h})\geq 0 \quad \text{for every }\omega \in W^{b}.
\end{equation}
where we use a non-smooth symmetric critical principle for the
Szulkin-type functional.

Now we prove that $u_{h}$ is a weak solution of \eqref{e2.1}, that is, for all
$v\in H^{\alpha}(\mathbb{R}^N)$,
\[
\int_{\mathbb{R}^N} (-\Delta)^{\alpha/2} u_{h}(-\Delta)^{\alpha/2}v +u_{h}v\,dx
=\int_{\mathbb{R}^N} Q(x)h(u_{h})v\,dx.
\]
By \eqref{e2.9}, for all $\omega \in W^{b}$, it follows that
\begin{equation} \label{e2.10}
\begin{aligned}
&\int_{\mathbb{R}^N} (-\Delta)^{\alpha/2} u_{h}(-\Delta)^{\alpha/2}(\omega-u_{h})
 +u_{h}(\omega-u_{h})dx\\
&-\int_{\mathbb{R}^N} Q(x)h(u_{h})(\omega-u_{h})\,dx\geq 0.
\end{aligned}
\end{equation}
Define the function $\gamma(s)=\text{sgn}(s)\min(| s|,b)$, and fix
$\epsilon>0$ and $v\in H^{\alpha}(\mathbb{R}^N)$ arbitrarily. Since $\gamma$
is Lipschitz and $\gamma(0)=0$,
$\omega_{\gamma}=\gamma\circ (u_{h}+\epsilon v)\in H^{\alpha}(\mathbb{R}^N)$. The
explicit expression of the truncation function $\omega_{\gamma}$ is
\[
\omega_{\gamma}(x)=\begin{cases}
-b &\text{if }  x\in \{u_{h}+\epsilon v<-b\} ,\\
u_{h}(x)+\epsilon v(x) &\text{if } x\in \{-b\leq u_{h}+\epsilon v<b\},\\
b   &\text{if }  x\in \{u_{h}+\epsilon v\geq b\},
\end{cases}
\]
thus $\omega_{\gamma}\in W^{b}$.  Taking $\omega=\omega_{\gamma}$ as
a test function in  \eqref{e2.10}, we obtain
\begin{align*}
0&\leq  \int_{\mathbb{R}^N} (-\Delta)^{\alpha/2} u_{h}(-\Delta)^{\alpha/2}
 (\omega_{\gamma}-u_{h}) +u_{h}(\omega_{\gamma}-u_{h})dx \\
&\quad -\int_{\mathbb{R}^N} Q(x)h(u_{h})(\omega_{\gamma}-u_{h})\,dx\\
&= \iint_{\mathbb{R}^N\times\mathbb{R}^N}\frac{(u_{h}(x)-u_{h}(y))
 (\omega_{\gamma}(x)-u_{h}(x)-\omega_{\gamma}(y)+u_{h}(y))}{| x-y|^{N+2\alpha}}
 \,dx\,dy\\
&\quad+\int_{\mathbb{R}^N} u_{h}(\omega_{\gamma}-u_{h})dx
 -\int_{\mathbb{R}^N} Q(x)h(u_{h})(\omega_{\gamma}-u_{h})\,dx\\
&= \iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{-| u_{h}(x)-u_{h}(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v<-b\}\times \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(-b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad + \iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(-2b-u_{h}(x)+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(\epsilon v(x)+b+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\epsilon\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(\epsilon v(x)-b+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(u_{h}(y)-u_{h}(x))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\int_{\{u_{h}+\epsilon v<-b\}}[(b+u_{h})u_{h}-Q(x)h(u_{h})(b+u_{h})]dx\\
&\quad +\epsilon\int_{\{-b\leq u_{h}+\epsilon v<b\}}[u_{h}v-Q(x)h(u_{h})v]dx\\
&\quad -\int_{\{u_{h}+\epsilon v\geq b\}}[(u_{h}-b)u_{h}+Q(x)h(u_{h})(u_{h}-b)]dx.
\end{align*}
After a suitable rearrangement of the above terms, we obtain that
\begin{align*}
0&\leq  \epsilon\iint_{\mathbb{R}^N\times\mathbb{R}^N}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy
 +\epsilon\int_{\mathbb{R}^N}u_{h}v\,dx \\
&\quad -\epsilon\int_{\mathbb{R}^N}Q(x)h(u_{h})v\,dx\\
&\quad +\iint_{\{u_{h}+\epsilon v<-b\}\times 
 \{u_{h}+\epsilon v<-b\}}
 \frac{-| u_{h}(x)-u_{h}(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad-\epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times 
 \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy  \\
&\quad +\iint_{\{u_{h}+\epsilon v<-b\}\times
  \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(-b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(-2b-u_{h}(x)+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(\epsilon v(x)+b+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\epsilon\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(\epsilon v(x)-b+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\epsilon\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(u_{h}(y)-u_{h}(x))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\int_{\{u_{h}+\epsilon v<-b\}}[Q(x)h(u_{h})-u_{h}](b+u_{h}+\epsilon v)dx\\
&\quad +\int_{\{u_{h}+\epsilon v\geq b\}}[Q(x)h(u_{h})-u_{h}](-b+u_{h}+\epsilon v)dx.
\end{align*}
By the direct computation, one has
\begin{align*}
&\iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{-| u_{h}(x)-u_{h}(y)|^2}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq  \epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times 
 \{u_{h}+\epsilon v<-b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy, 
\end{align*}
\begin{align*}
&\iint_{\{u_{h}+\epsilon v<-b\}\times \{-b\leq u_{h}+\epsilon v<b\}}
\frac{(u_{h}(x)-u_{h}(y))(-b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq  \epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy,
\end{align*}
\begin{align*}
&\iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(-2b-u_{h}(x)+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq \epsilon\iint_{\{u_{h}+\epsilon v<-b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{| u_{h}(x)-u_{h}(y)|| v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy,
\end{align*}
\begin{align*}
&\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(\epsilon v(x)+b+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq  \epsilon\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{u_{h}+\epsilon v<-b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)| }{| x-y|^{N+2\alpha}}\,dx\,dy,
\end{align*}
\begin{align*}
&\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(\epsilon v(x)-b+u_{h}(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq  \epsilon\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times 
 \{u_{h}+\epsilon v\geq b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy,
\end{align*}
\begin{align*}
&\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq  \epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy,
\end{align*}
\begin{align*}
&\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(b-u_{h}(x)-\epsilon v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq  \epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times 
 \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy,
\end{align*}
and
\begin{align*}
&\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(u_{h}(y)-u_{h}(x))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&-\epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\leq  \epsilon\iint_{\{u_{h}+\epsilon v\geq b\}\times 
\{u_{h}+\epsilon v\geq b\}}
\frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy.
\end{align*}
Moreover, from $u_{h}\in[0, a]\subset [-b, b]$ for a.e. 
$x\in \mathbb{R}^N$, one has
\begin{align*}
&\int_{\{u_{h}+\epsilon v<-b\}}\Big(Q(x)h(u_{h})-u_{h}\Big)
(b+u_{h}+\epsilon v)\,dx \\
&\leq -\epsilon \int_{\{u_{h}+\epsilon v<-b\}}\Big(M_{h}Q(x)+u_{h}(x)\Big)v(x)dx
\end{align*}
and
\begin{align*}
&\int_{\{u_{h}+\epsilon v\geq b\}}\Big(Q(x)h(u_{h})-u_{h}\Big)(-b+u_{h}+\epsilon v)dx\\
&\leq \epsilon M_{h}\int_{\{u_{h}+\epsilon v\geq b\}}Q(x)v(x)dx.
\end{align*}
Using the above the estimates and dividing by $\epsilon>0$, we obtain
\begin{align*}
0&\leq  \iint_{\mathbb{R}^N\times\mathbb{R}^N}
 \frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy
 +\int_{\mathbb{R}^N}u_{h}v\,dx \\
&\quad -\int_{\mathbb{R}^N}Q(x)h(u_{h})v\,dx\\
&\quad +\iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v<-b\}\times \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v<-b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{| u_{h}(x)-u_{h}(y)|| v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)| }{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{-b\leq u_{h}+\epsilon v<b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v<-b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times \{-b\leq u_{h}+\epsilon v<b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad +\iint_{\{u_{h}+\epsilon v\geq b\}\times \{u_{h}+\epsilon v\geq b\}}
 \frac{| u_{h}(x)-u_{h}(y)| | v(x)-v(y)|}{| x-y|^{N+2\alpha}}\,dx\,dy\\
&\quad -\int_{\{u_{h}+\epsilon v<-b\}}\Big(M_{h}Q(x)+u_{h}(x)\Big)v(x)dx \\
&\quad +M_{h}\int_{\{u_{h}+\epsilon v\geq b\}}Q(x)v(x)dx.
\end{align*}
Letting $\epsilon\to 0^{+}$, we have
\[
\operatorname{meas}(\{u_{h}+\epsilon v<-b\})\to 0\quad \text{and}\quad 
\operatorname{meas}(\{u_{h}+\epsilon v\geq b\})\to 0,
\]
respectively. So, it follows that
\begin{align*}
0&\leq \iint_{\mathbb{R}^N\times\mathbb{R}^N}
\frac{(u_{h}(x)-u_{h}(y))(v(x)-v(y))}{| x-y|^{N+2\alpha}}\,dx\,dy \\
&\quad +\int_{\mathbb{R}^N}u_{h}v\,dx-\int_{\mathbb{R}^N}Q(x)h(u_{h})v\,dx.
\end{align*}
Using $-v$ instead of $v$, we also have the above inequality. 
So, $u_{h}$ is a weak solution of \eqref{e2.1}. The proof is complete.
\end{proof}

Fix $\sigma\in (0, 1)$ and $\rho>0$, for any $t>0$ we introduce the function
\begin{equation} \label{e2.11}
\omega_{\sigma}^{t}(x):=\begin{cases}
0 &\text{if } x\in \mathbb{R}^N\setminus B_{\rho},\\
\frac{t}{(1-\sigma)\rho}(\rho-| x|)&   \text{if } 
 x\in B_{\rho}\setminus B_{\sigma\rho},\\
t   &\text{if }  x\in  B_{\sigma\rho},
\end{cases}
\end{equation}
where $B_{r}$ denotes the N-dimensional ball with center 0 and radius $r>0$. 
It is clear that $\omega_{\sigma}^{t}(x)$ is radial. Later we will show
that $\omega_{\sigma}^{t}(x)\in H^{\alpha}(\mathbb{R}^N)$.
To prove the main theorems in this paper, we need a important estimate 
for the norm of $\omega_{\alpha}^{t}(x)$. For this, we set
\begin{equation} \label{e2.12}
\nu_{0}:=1+\frac{1}{\lambda_1}, \quad
\lambda_1=\inf_{u\in H_{0}^{1}(B_{\rho})\setminus\{0\}}
\frac{\| \nabla u\|_{L^2(B_{\rho})}^2}{\| u\|_{L^2(B_{\rho})}^2}.
\end{equation}

\begin{proposition} \label{prop2.1} % subdiff
Let $\sigma\in (0, 1)$, $\rho>0$ and $t>0$. Let $\omega_{\sigma}^{s}$ be 
the function given in \eqref{e2.11}, $S_{N-2}$ be the Lebesgue measure of 
the unit sphere in $\mathbb{R}^{N-1}$, and
\[
\Gamma(t):=\int_{0}^{+\infty}z^{t-1}e^{-z}dz,    \quad t>0,
\]
be the usual Gamma function. Then $\omega_{\sigma}^{t}\in H^{\alpha}(\mathbb{R}^N)$,
and one has
\begin{equation} \label{e2.13}
\begin{aligned}
\iint_{\mathbb{R}^N\times\mathbb{R}^N}\frac{| \omega_{\sigma}^{t}(x)
-\omega_{\sigma}^{t}(y)|^2 }{| x-y|^{N+2\alpha}}\,dx\,dy
&< \frac{t^2}{(1-\sigma)^2}\cdot\frac{\pi^{N/2}\rho^{N-2}
(1-\sigma^{N})}{\Gamma(1+\frac{N}{2})}\kappa_1\kappa_{2} \\
&=K(\rho, \sigma),
\end{aligned}
\end{equation}
where
\begin{gather*}
\kappa_1:=\begin{cases}
2\nu_{0} &\text{if }  N=1,\\
(\pi+\frac{4}{1+2\alpha})\nu_{0}&  \text{if } N=2,\\
S_{n-2}(\frac{\pi}{2}+\frac{2}{1+2\alpha})\nu_{0} & \text{if }  N\geq 3,
\end{cases},\\
\kappa_{2}:=\frac{1}{2(1-\alpha)}+\frac{2}{\alpha},
\end{gather*}
where $\nu_{0}$ is given in \eqref{e2.12}.
\end{proposition}

\begin{proof} 
The proof can be found in \cite{rMRS}, for the sake of completeness,
 we give it here. Computing the standard seminorm of the function
 $\omega_{\sigma}^{t}$ in $H^{1}(\mathbb{R}^N)$, one has
\begin{equation} \label{e2.14}
\begin{aligned}{}
[\omega_{\sigma}^{t}]_{H^{1}(\mathbb{R}^N)}^2
&= \int_{\mathbb{R}^N}| \nabla \omega_{\sigma}^{t}(x)|^2dx\\
&= \int_{B_{\rho}\setminus B_{\sigma\rho}}\frac{t^2}{(1-\sigma)^2\rho^2}dx
\\
&= \frac{t^2}{(1-\sigma)^2\rho^2}(| B_{\rho}|-| B_{\sigma\rho}|)\\
&= \frac{t^2}{(1-\sigma)^2}
\frac{\pi^{N/2}\rho^{N-2}(1-\sigma^{N})}{\Gamma(1+\frac{N}{2})}.
\end{aligned}
\end{equation}
Since $\omega_{\sigma}^{t}\in H_{0}^{1}(B_{\rho})$, by 
\cite[Proposition 1.1(b)]{rMRS}, it follows that 
$\omega_{\sigma}^{t}\in W^{\alpha, 2}(B_{\rho})$.
Moreover, the boundary $\partial B_{\rho}$ is Lipschitz, by 
\cite[Lemma 1.3]{rMRS}, we have that $\omega_{\sigma}^{t}\in H^{\alpha}(\mathbb{R}^N)$.

Hence, since $\alpha\in (0, 1)$, \cite[Corollary 1.15]{rMRS} yields
\begin{equation} \label{e2.15}
\begin{aligned}{}
[\omega_{\sigma}^{s}]_{H^{\alpha}(\mathbb{R}^N)}^2
&\leq 2\Big(\int_{\mathbb{R}^N}\frac{1-\cos x_1}{| x|^{N+2\alpha}}dx\Big)
 \int_{\mathbb{R}^N}| \xi|^{2\alpha}|| \mathcal{F}\omega_{\sigma}^{t}(\xi)|^2d\xi\\
&< 2\Big(\int_{\mathbb{R}^N}\frac{1-\cos x_1}{| x|^{N+2\alpha}}dx\Big)
 \int_{\mathbb{R}^N}(1+| \xi|^2|)| \mathcal{F}\omega_{\sigma}^{t}(\xi)|^2d\xi.
\end{aligned}
\end{equation}
Now, by standard arguments on the Fourier transform and Poincar\'e inequality,
 we have
\begin{equation} \label{e2.16}
\int_{\mathbb{R}^N}(1+| \xi|^2|)
| \mathcal{F}\omega_{\sigma}^{t}(\xi)|^2d\xi
\leq \nu_{0}[\omega_{\sigma}^{t}]_{H^{1}(\mathbb{R}^N)}^2.
\end{equation}
Moreover, by Parseval-Plancherel identity, it follows that
\[
\omega_{\sigma}^{s}\in L^2(\mathbb{R}^N)\quad \text{if and only if}\quad
\mathcal{F}\omega_{\sigma}^{t}\in L^2(\mathbb{R}^N)
\]
and
\begin{equation} \label{e2.17}
\|  \omega_{\sigma}^{t}\| _{L^2(\mathbb{R}^N)}^2
=\| \mathcal{F}\omega_{\sigma}^{t} \|_{L^2(\mathbb{R}^N)}^2.
\end{equation}
Moreover,
\[
| \omega_{\sigma}^{t}|\in L^2(\mathbb{R}^N)\quad \text{if and only if}\quad
| \xi|\mathcal{F}\omega_{\sigma}^{t}\in L^2(\mathbb{R}^N),
\]
and
\begin{equation} \label{e2.18}
\| \nabla \omega_{\sigma}^{t}\| _{L^2(\mathbb{R}^N)}^2
=\|\,| \xi| \mathcal{F}\omega_{\sigma}^{t} \|_{L^2(\mathbb{R}^N)}^2.
\end{equation}
By \eqref{e2.17} and \eqref{e2.18}, we have
\begin{align*}
\int_{\mathbb{R}^N}(1+| \xi|^2|)
| \mathcal{F}\omega_{\sigma}^{t}(\xi)|^2d\xi
=\|  \omega_{\sigma}^{t}\| _{L^2(\mathbb{R}^N)}^2
+\| \nabla \omega_{\sigma}^{t}\| _{L^2(\mathbb{R}^N)}^2.
\end{align*}
Hence, by \eqref{e2.12} and the definition of $\lambda_1$,
we have that the inequality \eqref{e2.16} is a direct consequence 
of above equality, taking into account that 
$\omega_{\sigma}^{t}\in H_{0}^{1}(B_{\rho})$.

Then, by \eqref{e2.15} and \eqref{e2.16}, it follows that
\[
\iint_{\mathbb{R}^N\times\mathbb{R}^N} \frac{| \omega_{\sigma}^{t}(x)
-\omega_{\sigma}^{t}(y)|^2 }{| x-y|^{N+2\alpha}}\,dx\,dy
< 2\Big(\int_{\mathbb{R}^N}\frac{1-\cos x_1}{| x|^{N+2\alpha}}dx\Big)
[\omega_{\sigma}^{t}]_{H^{1}(\mathbb{R}^N)}^2.
\]
Finally. from \eqref{e2.14} and \cite[Remark 1.13]{rMRS}, we can get the 
inequality \eqref{e2.13}.
\end{proof}

\section{Proofs of Theorems \ref{thm1.1} and \ref{thm1.2}} \label{pf1}

By (A3) and the continuity of $f$ and $g$, we may fix the positive sequences
 $\{a_i\}_i$, $\{b_i\}_i$ and $\{\epsilon_i\}_i$
such that $\lim_{i\to \infty}a_i=\lim_{i\to \infty} b_i=0$ and for all
$i\in N$,
\begin{gather} \label{e3.1}
b_{i+1}<a_i<t_i<b_i<1, \\
\label{e3.2}
f(t)+\epsilon g(t)\leq 0\quad\text{for all 
$ t\in[a_i, b_i]$ and $\epsilon\in [-\epsilon_i, \epsilon_i]$}.
\end{gather}
For every $i\in N$, we define the truncation function 
$f_i, g_i:[0, +\infty)\to \mathbb{R}^N$ by
\begin{equation} \label{e3.3}
f_i(t)=f(\min(t, b_i))\quad \text{and}\quad g_i(t)=g(\min(t, b_i)).
\end{equation}
By (A2) and (A3), it is easy to see that $f(0)=0$. Since $f_i(0)=g_i(0)=0$, 
we may extend continuously the function $f_i$ and $g_i$
to $\mathbb{R}^N$, taking 0 for negative arguments. For every $t\in \mathbb{R}^N$
 and $i\in N$, let $F_i(t)=\int_{0}^{t}f_i(\tau)d\tau$ and 
$G_i(t)=\int_{0}^{t}g_i(\tau)d\tau$.

For every $i\in N$ and $\epsilon\in [-\epsilon_i, \epsilon_i]$, 
the function $h_{i, \epsilon}^{0}:=f_i+\epsilon g_i$ is a continuous, 
bounded function with $h_{i, \epsilon}^{0}(0)=0$. 
By \eqref{e3.2} and \eqref{e3.3}, we have  $h_{i, \epsilon}^{0}(t)\leq 0$ 
for all $t\in [a_i, b_i]$. Thus, by Theorem \ref{thm2.1}, 
for every $i\in N$ and $\epsilon\in [-\epsilon_i, \epsilon_i]$, 
the  problem
\begin{equation}  \label{e3.4} % Ph
\begin{gathered}
(-\Delta)^{\alpha} u+u =Q(x)h_{i, \epsilon}^{0}(u), \quad x\in\mathbb{R}^N\\
u\geq 0,
\end{gathered}
\end{equation}
has a radially symmetric weak solution $u_{i, \epsilon}^{0}\in H^{\alpha}(\mathbb{R}^N)$
with
\begin{gather} \label{e3.5}
u_{i, \epsilon}^{0}\in[0, a_i]\quad\text{for a.e. } x\in \mathbb{R}^N, \\
 \label{e3.6}
u_{i, \epsilon}^{0}\text{ is the minimizer of the functional
$R_i^{\epsilon}$  on $W_{\rm rad}^{b_i}$},
\end{gather}
where $R_i^{\epsilon}$ is the functional associated with problem \eqref{e3.4},  and
\[
R_i^{\epsilon}(u)=\frac{1}{2}\| u\|^2
-\int_{\mathbb{R}^N}Q(x)\Big(F_i(u)+\epsilon G_i(u)\Big)dx,\quad 
u\in H^{\alpha}_{\rm rad}(\mathbb{R}^N).
\]
By \eqref{e3.3} and \eqref{e3.5}, $u_{i, \epsilon}^{0}$ is a weak solution 
not only for \eqref{e3.4} but also for problem \eqref{ePe}.

\begin{proof}[Proof of Theorem \ref{thm1.1}]
As an abbreviation, for every $i\in N$, write $u_i^{0}=u_{i, 0}^{0}$ and 
$R_i=R_i^{0}$. According the observation for problem \eqref{e3.4}, 
we only need to show that there are infinitely many distinct
elements in the sequence $\{u_{i, 0}^{0}\}_i$ verifying \eqref{e1.1}.

 We first prove that
\begin{gather} \label{e3.7}
R_i(u_i^{0})< 0\quad \text{for all } i\in N,\\
\label{e3.8} 
\lim_{i\to \infty} R_i(u_i^{0})= 0\,.
\end{gather}

The left side of (A2) implies that there exist $l_{0}>0$ and
$\delta\in (0, b_1)$ such that
\begin{equation} \label{e3.9}
F(t)\geq -l_{0}t^2\quad \text{for all } t\in (0, \delta).
\end{equation}
Let $L_{0}$ be large enough such that
\begin{equation} \label{e3.10}
\frac{1}{2}K(\rho, \sigma)
l_{0}\| Q\|_{L^{1}}< L_{0}(\sigma \rho)^{N}\omega_{N}\min{B_{\sigma\rho}} Q.
\end{equation}
where $\rho>0$ and $K(\rho, \sigma)$ come from \eqref{e2.13}.
The right side of (A2) implies there is a sequence $\{\widetilde{t}_i\}\subset(0,
\delta)$ such that $\widetilde{t}_i\leq a_i$ and $F(
\widetilde{t}_i)>L_{0}{\widetilde{t}_i}^2$ for all $i\in N$.
 Let $i\in N$ fixed and
$\omega_{\sigma}^{\widetilde{t}_i}\in H^{\alpha}_{\rm rad}(\mathbb{R}^N)$
 be the function from \eqref{e2.11} corresponding to
$\widetilde{t}_i>0$. Then $\omega_{\sigma}^{\widetilde{t}_i}\in
W_{\rm rad}^{b_i}$, and by \eqref{e2.13} and \eqref{e3.9}, we have
\begin{align*}
R_i(\omega_{\sigma}^{\widetilde{t}_i})
&= \frac{1}{2}\| \omega_{\sigma}^{\widetilde{t}_i}\|^2
-\int_{\mathbb{R}^N}Q(x)F_i(\omega_{\sigma}^{\widetilde{t}_i})dx\\
&= \frac{1}{2}\| \omega_{\sigma}^{\widetilde{t}_i}\|^2
-F(\widetilde{t}_i)\int_{B_{\sigma\rho}}Q(x)dx
 -\int_{B_{\rho\setminus B_{\sigma\rho}}}Q(x)
 F_i(\omega_{\sigma}^{\widetilde{t}_i})dx\\
&\leq \Big[\frac{1}{2}K(\rho, \sigma)
-L_{0}(\sigma \rho)^{N}\omega_{N}\min_{B_{\sigma\rho}} 
 Q+l_{0}\| Q\|_{L^{1}}\Big]\widetilde{t}_i^2
\end{align*}
So, by \eqref{e3.10}, one has
\begin{equation} \label{e3.11}
R_i(u_i^{0})=\min_{W_{rad}^{b_i}} R_i
\leq R_i(\omega_{\sigma}^{\widetilde{t}_i})< 0
\end{equation}
which proves \eqref{e3.7}.
Now we prove the limit \eqref{e3.8}.
 For every $i\in N$, by the mean value theorem, one has
\[
R_i(u_i^{0})\geq-\int_{\mathbb{R}^N}Q(x)F_i(u_i^{0})dx\geq -\| Q\|_{L^{1}}
\max_{t\in [0, 1]} | f(t)| a_i.
\]
Because of $\lim_{i\to \infty} a_i=0$,  the above inequality and 
\eqref{e3.11} leads to the limit \eqref{e3.8}.

By \eqref{e3.3} and \eqref{e3.5}, we know that
\[
R_i(u_i^{0})=R_1(u_i^{0}) \quad  \text{for all } i\in
N.
\]
Moreover, by \eqref{e3.7} and \eqref{e3.8},  we know that the sequence
$\{u_i^{0}\}$ are the infinitely many distinct radially symmetric weak
solutions of problem \eqref{ePe}.

Finally, we prove \eqref{e1.1}. It is clear that  
$\| u_i^{0}\|_{L^{\infty}}\leq  a_i$ for all $i\in N$ by \eqref{e3.5}.
 Since $\lim_{i\to \infty} a_i=0$, the first limit holds. For the second limit, 
by \eqref{e3.11},   \eqref{e3.1}, \eqref{e3.3} and \eqref{e3.5}, for  $i\in N$, 
one has
\[
\frac{1}{2}\| u_i^{0}\|^2\leq \| Q\|_{L^{1}}\max_{t\in [0, 1]} 
| f(t)| a_i\to 0,\quad \text{as } i\to\infty.
\]
The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.2}]
 By the observation for problem \eqref{e3.4},  to prove this theorem,
 we only need to prove that for every $k\in N$, there are at least 
$k$ distinct elements $u_{i, \epsilon}^{0}$ verifying \eqref{e1.2} 
when $\epsilon$ belongs to a certain interval around the origin.

Let $\{\theta_i\}_i$ be a sequence such that $\theta_i<0$ for all $i\in N$ and
$\lim_{i\to\infty} \theta_i=0$. By \eqref{e3.8} and \eqref{e3.11}, we have
$\lim_{i\to\infty} R_i(\omega_{\sigma}^{\widetilde{t}_i})=0$.
Thus, up to a subsequence, we may assume that the sequence
$\{(\theta_i, R_i(u_i^{0}), R_i(\omega_{\sigma}^{\widetilde{t}_i}),
a_i)\}_i\subset R^{4}$ which converges to $(0, 0, 0, 0)$ has the
property that for all $i\in N$,
\begin{gather} \label{e3.12}
\theta_i<R_i(u_i^{0})\leq
R_i(\omega_{\sigma}^{\widetilde{t}_i})<\theta_{i+1}, \\
\label{e3.13}
a_i<\min\Big\{\frac{1}{i}, \frac{1}{2i^2\| Q\|_{L^{1}}(\max_{[0, 1]}| f| 
+\max_{[0, 1]}| g|+1)}\Big\}.
\end{gather}
Denote
\[
\epsilon'_i=\frac{\theta_{i+1}-R_i(\omega_{\sigma}^{\widetilde{t}_i})}
{\| Q\|_{L^{1}}(\max_{[0, 1]}| g|+1)}\quad\text{and}\quad
\epsilon''_i=\frac{R_i(u_i^{0})-\theta_i}{\| Q\|_{L^{1}}(\max_{[0, 1]}| g|+1)} 
 \quad \forall i\in N.
\]
and fix $k\in N$.  By \eqref{e3.12},
\[
\epsilon_{k}^{0}=\min\{1, \epsilon_1, \dots , \epsilon_{k},
\epsilon'_1, \dots , \epsilon'_{k}, \epsilon''_1, \dots ,
\epsilon''_{k}\}>0.
\]
Then for every $i\in\{1, \dots , k\}$ and
$\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$,  it follows 
from \eqref{e3.1}, \eqref{e3.6} and the choice of $\epsilon'_i$ that
\begin{align*}
R_i^{\epsilon}(u_{i,\epsilon}^{0})
&\leq R_i^{\epsilon}(\omega_{\sigma}^{\widetilde{t}_i})\\
&= R_i(\omega_{\sigma}^{\widetilde{t}_i})-\epsilon\int_{\mathbb{R}^N}Q(x)
G_i(\omega_{\sigma}^{\widetilde{t}_i})dx
< \theta_{i+1}.
\end{align*}
Meanwhile, we also have from \eqref{e3.1}, \eqref{e3.6} for $\epsilon=0$ 
and the choice of $\epsilon''_i$,
\begin{align*}
R_i^{\epsilon}(u_{i,\epsilon}^{0})&=
R_i(u_{i,\epsilon}^{0})-\epsilon\int_{\mathbb{R}^N}Q(x)G_i(u_{i,\epsilon}^{0})dx\\
&\geq
R_i(u_i^{0})-\epsilon\int_{\mathbb{R}^N}Q(x)G_i(u_{i,\epsilon}^{0})dx
> \theta_i.
\end{align*}
Thus,  for every $i\in\{1, \dots , k\}$ and
$\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$,  one has
\[
\theta_i<R_i^{\epsilon}(u_{i,\epsilon}^{0})< \theta_{i+1},
\]
so
\[
R_1^{\epsilon}(u_{1,\epsilon}^{0})< \dots<
R_{k}^{\epsilon}(u_{k,\epsilon}^{0}).
\]
But $u_{i,\epsilon}^{0}\in W_{\rm rad}^{b_1}$ for every $i\in\{1,
\dots, k\}$, so
$R_i^{\epsilon}(u_{i,\epsilon}^{0})=R_1^{\epsilon}(u_{i,\epsilon}^{0})$ from \eqref{e3.3}.
Thus for every $\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$, we obtain
\[
R_1^{\epsilon}(u_{1,\epsilon}^{0})< \dots<
R_1^{\epsilon}(u_{k,\epsilon}^{0}),
\]
this shows that the elements $u_{1,\epsilon}^{0}$, $\dots$,
$u_{k,\epsilon}^{0}$ are distinct whenever
$\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$.

Now we show that \eqref{e1.2}. By \eqref{e3.5} and \eqref{e3.13}, the left
limit of \eqref{e1.2} holds. For the right limit of \eqref{e1.2}, it is easy to see that for every $i\in\{1, \dots , k\}$ and
$\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$,
\[
R_1^{\epsilon}(u_{i,\epsilon}^{0})=R_i^{\epsilon}(u_{i,\epsilon}^{0})<
\theta_{i+1}<0.
\]
Thus, for every $i\in\{1, \dots , k\}$ and
$\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$, by the mean value theorem,
 \eqref{e3.1}, \eqref{e3.5}, \eqref{e3.13} and $\epsilon_{k}^{0}\leq 1$,  one has
\begin{align*}
\frac{1}{2}\| u_{i,\epsilon}^{0}\|^2
&< \int_{\mathbb{R}^N}Q(x)\Big(F_i(u_{i,\epsilon}^{0})+\epsilon G_i(u_{i,\epsilon}^{0})\Big)\\
&\leq \| Q\|_{L^{1}}\Big(\max_{[0, 1]}| f|+\max_{[0, 1]}| g|\Big)a_i\\
&\leq \frac{1}{2i^2},
\end{align*}
which completes the proof.
\end{proof}


\section{Proofs of Theorems \ref{thm1.3} and  \ref{thm1.4}} \label{pf2}

The left-hand side of (A4) implies the existence of $l_{\infty}>0$ and
$\delta>0$ such that
\begin{equation} \label{e4.1}
F(t)\geq -l_{\infty}t^2\quad \text{for all } t>\delta.
\end{equation}
Let $L_{\infty}$ be large enough such that
\begin{equation} \label{e4.2}
\frac{1}{2}K(\rho, \sigma)+
l_{\infty}\| Q\|_{L^{1}}
< L_{\infty}(\sigma \rho)^{N}\omega_{N}\min_{B_{\sigma\rho}} Q,
\end{equation}
where $\rho>0$ and $K(\rho, \sigma)$ comes from \eqref{e2.13}.
 The right side of (A4)
implies there is a sequence $\{\widetilde{t}_i\}_i\subset(0,
\infty)$ such that $\lim_{i\to \infty}\widetilde{t}_i=\infty$ and
\begin{equation} \label{e4.3}
 F(\widetilde{t}_i)>L_{\infty}{\widetilde{t}_i}^2\quad \text{for all } i\in N.
\end{equation}
Since $\lim_{i\to \infty}\widetilde{t}_i=\infty$, we may fix a subsequence 
$\{\widetilde{t}_{m_i}\}_i$ of $\{\widetilde{t}_i\}_i$ such
that $\widetilde{t}_i\leq \widetilde{t}_{m_i}$ for all $i\in N$. Moreover,
since $f$ and $g$ are continuous, we may fix the positive sequences
 $\{a_i\}_i$, $\{b_i\}_i$ and $\{\epsilon_i\}_i$
such that $\lim_{i\to \infty} 
a_i=\lim_{i\to \infty} b_i=+\infty$ and for all $i\in N$,
\begin{gather}\label{e4.4}
a_i<\widetilde{t}_{m_i}<b_i<a_{i+1}, \\
\label{e4.5}
f(t)+\epsilon g(t)\leq 0\quad\text{for all $t\in[a_i, b_i]$ and
 $\epsilon\in [-\epsilon_i, \epsilon_i]$}.
\end{gather}
For every $i\in N$, we define the truncation function 
$f_i, g_i:[0, +\infty)\to \mathbb{R}^N$  as in \eqref{e3.3}.
Since $f_i(0)=g_i(0)=0$, so we may extend continuously the function $f_i$ and $g_i$
to $\mathbb{R}^N$, taking 0 for negative arguments. 
For every $t\in \mathbb{R}^N$ and $i\in N$, let 
$F_i(t)=\int_{0}^{t}f_i(\tau)d\tau$ and $G_i(t)=\int_{0}^{t}g_i(\tau)d\tau$.

For every $i\in N$ and $\epsilon\in [-\epsilon_i, \epsilon_i]$, 
the function $h_{i, \epsilon}^{\infty}:=f_i+\epsilon g_i$ is a continuous, 
bounded function with $h_{i, \epsilon}^{\infty}(0)=0$. 
By \eqref{e4.5} and \eqref{e3.3}, we have  $h_{i, \epsilon}^{\infty}(t)\leq 0$ 
for all $t\in [a_i, b_i]$. Thus, by Theorem \ref{thm2.1}, for every $i\in N$ and 
$\epsilon\in [-\epsilon_i, \epsilon_i]$, the  problem
\begin{equation} \label{e4.6} % Ph
\begin{gathered}
(-\Delta)^{\alpha} u+u =Q(x)h_{i, \epsilon}^{\infty}(u), \quad x\in\mathbb{R}^N\\
u\geq 0,
\end{gathered}
\end{equation}
has a radially symmetric weak solution $u_{i, \epsilon}^{\infty}\in H^{\alpha}(\mathbb{R}^N)$
with
\begin{gather} \label{e4.7}
u_{i, \epsilon}^{\infty}\in[0, a_i]\quad\text{for a.e. } x\in \mathbb{R}^N, \\
 \label{e4.8}
u_{i, \epsilon}^{\infty}\quad \text{is the minimizer of the functional 
$ R_i^{\epsilon} $ on $ W_{\rm rad}^{b_i}$,}
\end{gather}
where $R_i^{\epsilon}$ is defined as in section 4.
By \eqref{e3.3} and \eqref{e4.7}, $u_{i, \epsilon}^{\infty}$ is a weak 
solution not only for \eqref{e4.6} but also for problem \eqref{ePe}.


\begin{proof}[Proof of Theorem \ref{thm1.3}]
As in proof of Theorem \ref{thm1.1},  for every $i\in N$, write 
$u_i^{\infty}=u_{i, 0}^{\infty}$ and $R_i=R_i^{0}$. 
According the observation for problem \eqref{e4.6}, we only need to 
show that there are infinitely many distinct
elements in the sequence $\{u_{i, 0}^{\infty}\}_i$ verifying \eqref{e1.3}.
We  prove that
\begin{equation} \label{e4.9}
\lim_{i\to \infty} R_i(u_i^{\infty})= -\infty\,.
\end{equation}
 Let $i\in N$ be fixed and
$\omega_{\sigma}^{\widetilde{t}_i}\in H^{\alpha}_{\rm rad}(\mathbb{R}^N)$  be the
function from \eqref{e2.11} corresponding to
$\widetilde{t}_i>0$. Then $\omega_{\sigma}^{\widetilde{t}_i}\in
W_{\rm rad}^{b_i}$, and by \eqref{e2.13}, \eqref{e4.1} and \eqref{e4.3}, we have
\begin{align*}
R_i(\omega_{\sigma}^{\widetilde{t}_i})
&= \frac{1}{2}\| \omega_{\sigma}^{\widetilde{t}_i}\|^2
-\int_{\mathbb{R}^N}Q(x)F_i(\omega_{\sigma}^{\widetilde{t}_i})dx\\
 &= \frac{1}{2}\| \omega_{\sigma}^{\widetilde{s}_i}\|^2
-F(\widetilde{s}_i)\int_{B_{\sigma\rho}}Q(x)dx
 -\int_{(B_{\rho\setminus B_{\sigma\rho}})\cap
  \{\omega_{\sigma}^{\widetilde{t}_i}>\delta\}}
 Q(x)F_i(\omega_{\sigma}^{\widetilde{t}_i})dx\\
&\quad -\int_{(B_{\rho\setminus B_{\sigma\rho}})
 \cap \{\omega_{\sigma}^{\widetilde{t}_i}\leq \delta\}}
 Q(x)F_i(\omega_{\sigma}^{\widetilde{t}_i})dx\\
&\leq \Big[\frac{1}{2}K(\rho, \sigma)
 -L_{\infty}(\sigma \rho)^{N}\omega_{N}\min_{B_{\sigma\rho}} Q
 +l_{\infty}\| Q\|_{L^{1}}\Big]\widetilde{t}_i^2+\| Q\|_{L^{1}}
 \max_{t\in [0, \delta]} | F(t)|.
\end{align*}
Since $\lim_{i\to \infty}\widetilde{t}_i=\infty$  and \eqref{e4.2},
 one has $\lim_{i\to \infty}R_i(\omega_{\sigma}^{\widetilde{t}_i})=-\infty$.
But $R_i(u_i^{\infty})\leq R_i(\omega_{\sigma}^{\widetilde{t}_i})$ for all 
$i\in N$, which proves \eqref{e4.9}.

Suppose that in the sequence $\{u_i^{\infty}\}_i$ there are only finitely
 many distinct elements, denote $\{u_1^{\infty}, \dots, u_{i_{0}}^{\infty}\}$
for some $i_{0}\in N$. Thus the sequence $\{R_i(u_i^{\infty})\}_i$ 
reduces mostly to the finite set which contradicts \eqref{e4.9}.

Now we prove the limit \eqref{e1.3}. Argument by contradiction. 
Assume that there exists a subsequence $\{u_{k_i}^{\infty}\}_i$ 
of $\{u_i^{\infty}\}_i$ such that for all $i\in N$,  
$\| u_{k_i}^{\infty}\|_{L^{\infty}}\leq M$, for some $M>0$. 
In particular, $\{u_{k_i}^{\infty}\}_i\subset
W_{\rm rad}^{b_{l}}$ for some $l\in N$. Thus for every $k_i\geq l$, one has
\begin{align*}
R_{l}(u_{l}^{\infty})
&= \min_{W_{\rm rad}^{b_{l}}}
R_{l}=\underset{W_{\rm rad}^{b_{l}}}{\min}R_{k_i}\\
&\geq \min_{W_{\rm rad}^{b_{k_i}}} R_{k_i}=R_{k_i}(u_{k_i}^{\infty})\\
&\geq \min_{W_{\rm rad}^{b_{l}}} R_{k_i}\\
&= R_{l}(u_{l}^{\infty}).
\end{align*}
As a consequence,
\begin{equation} \label{e4.10}
R_{k_i}(u_{k_i}^{\infty})=R_{l}(u_{l}^{\infty}) \quad \text{for all } i\in N.
\end{equation}
Moreover the sequence $\{R_i(u_i^{\infty})\}_i$ is non-increasing.
By \eqref{e3.3} and \eqref{e4.8}, for all $i\in N$, we have
\[
R_{i+1}(u_{i+1}^{\infty})=\min_{W_{\rm rad}^{b_{i+1}}} R_{i+1}
\leq \min_ {W_{\rm rad}^{b_i}} R_{i+1}=\min_{W_{\rm rad}^{b_i}}
 R_i=R_i(u_i^{\infty}).
\]
Combining this latter fact with \eqref{e4.10}, one can find a number
$i_{0}\in N$ such that $R_i(u_i^{\infty})=R_{l}(u_{l}^{\infty})$
for all $i\geq i_{0}$, this is a contraction with \eqref{e4.9}. The
proof is complete.
\end{proof}

\begin{proof}[Proof of Remark \ref{rmk1.3}]
Assume that \eqref{e1.4} holds for $f$ with $q\in (2, \frac{2_{\alpha}^{*}}{2})$. 
Assume by contradiction that there exists a subsequence $\{u_{k_i}^{\infty}\}_i$ 
of $\{u_i^{\infty}\}_i$
such that for all $i\in N$, we have $\| u_{k_i}^{\infty}\|\leq \tilde{M}$ 
for some $\tilde{M}>0$. Fix $\beta\in [2q, 2_{\alpha}^{*})$.
By \eqref{e1.4} and the mean value theorem, one has
\begin{align*}
\big| \int_{\mathbb{R}^N}Q(x)F_{k_i}(u_{k_i}^{\infty})dx\big| 
&\leq  C\Big(\| Q\|_{L^2}\| u_{k_i}^{\infty}\|_{L^2}
 +\| Q\|_{L^{\frac{\beta}{\beta-q}}}\| u_{k_i}^{\infty}\|_{L^{\beta}}^{q}\Big)\\
&\leq  C_1\Big(\| Q\|_{L^2}\tilde{M}+\| Q\|_{L^{\frac{\beta}{\beta-q}}}
 \tilde{M}^{q}\Big)<\infty.
\end{align*}
Thus, the sequence $\{R_{k_i}(u_{k_i}^{\infty})\}_i$ is bounded. 
Since the sequence $\{R_{k_i}(u_{k_i}^{\infty})\}_i$ is non-increasing, it will be
bounded, this contradicts \eqref{e4.9}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.4}]
 By the observation for problem \eqref{e4.6}, in order to prove this theorem,
 we only need to prove that
for every $k\in N$, there are at least k distinct elements 
$u_{i, \epsilon}^{\infty}$ verifying \eqref{e1.6} when $\epsilon$ 
belongs to a certain interval around the origin.

Let $\{\theta_i\}_i$ be a sequence such that $\theta_i<0$ for all $i\in N$ and
$\lim_{i\to\infty} \theta_i=-\infty$. By the proof of Theorem \ref{thm1.3}, 
up to a subsequence, we may assume that the sequence
$\{(\theta_i, R_i(u_i^{\infty}), R_i(\omega_{\sigma}^{\widetilde{t}_i}),
a_i)\}_i\subset \mathbb{R}^{4}$ 
which converges to $(-\infty, -\infty, -\infty, \infty)$ has the
property that for all $i\in N$,
\begin{gather} \label{e4.11}
\theta_{i+1}<R_i(u_i^{\infty})\leq
R_i(\omega_{\sigma}^{\widetilde{t}_i})<\theta_i,\\
\label{e4.12}
a_i\geq i.
\end{gather}
Denote
\[
\epsilon'_i
=\frac{\theta_i-R_i(\omega_{\sigma}^{\widetilde{t}_i})}{\| Q\|_{L^{1}}
(\max_{[0, b_i]}| g|+1)b_i}\quad\text{and}\quad
\epsilon''_i=\frac{R_i(u_i^{\infty})-\theta_{i+1}}{\| Q\|_{L^{1}}
(\max_{[0, b_i]}| g|+1)b_i}  \quad  i\in N.
\]
and fix $k\in N$.  By \eqref{e4.11}, one has
\[
\epsilon_{k}^{\infty}=\min\{1, \epsilon_1, \dots , \epsilon_{k},
\epsilon'_1, \dots , \epsilon'_{k}, \epsilon''_1, \dots ,
\epsilon''_{k}\}>0.
\]
Then for every $i\in\{1, \dots , k\}$ and
$\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$,  
it follows from \eqref{e4.4}, \eqref{e4.8}, the choice of 
$\epsilon'_i$ and $\widetilde{t}_i\leq \widetilde{t}_{m_i}$ that
\begin{align*}
R_i^{\epsilon}(u_{i,\epsilon}^{\infty})&\leq
R_i^{\epsilon}(\omega_{\sigma}^{\widetilde{s}_i})\\
&= R_i(\omega_{\sigma}^{\widetilde{s}_i})
 -\epsilon\int_{\mathbb{R}^N}Q(x)G_i(\omega_{\sigma}^{\widetilde{s}_i})dx
< \theta_i.
\end{align*}
Meanwhile,  from \eqref{e4.8} for $\epsilon=0$, \eqref{e4.4},
 the choice of $\epsilon''_i$ and $\widetilde{t}_i\leq \widetilde{t}_{m_i}$ 
we have 
\begin{align*}
R_i^{\epsilon}(u_{i,\epsilon}^{\infty})
&= R_i(u_{i,\epsilon}^{\infty})-\epsilon\int_{\mathbb{R}^N}Q(x)G_i(u_{i,\epsilon}^{\infty})dx\\
&\geq
R_i(u_i^{\infty})-\epsilon\int_{\mathbb{R}^N}Q(x)G_i(u_{i,\epsilon}^{\infty})dx
> \theta_{i+1}.
\end{align*}
Thus,  for every $i\in\{1, \dots , k\}$ and
$\epsilon\in[-\epsilon_{k}^{\infty}, \epsilon_{k}^{\infty}]$,  one has
\begin{equation} \label{e4.13}
\theta_{i+1}<R_i^{\epsilon}(u_{i,\epsilon}^{\infty})< \theta_i,
\end{equation}
so
\begin{equation} \label{e4.14}
R_{k}^{\epsilon}(u_{k,\epsilon}^{\infty})< \dots<
R_1^{\epsilon}(u_{1,\epsilon}^{0})<0.
\end{equation}
But $u_{i,\epsilon}^{\infty}\in W_{\rm rad}^{b_{k}}$ for every $i\in\{1,
\dots, k\}$ by \eqref{e4.4}, so
$R_i^{\epsilon}(u_{i,\epsilon}^{\infty})=R_1^{\epsilon}(u_{i,\epsilon}^{\infty})$ by \eqref{e3.3}.
Thus for every $\epsilon\in[-\epsilon_{k}^{0}, \epsilon_{k}^{0}]$, 
it follows from \eqref{e4.14} that
\[
R_{k}^{\epsilon}(u_{k,\epsilon}^{\infty})< \dots<
R_{k}^{\epsilon}(u_{1,\epsilon}^{\infty})<0,
\]
this shows that the elements $u_{1,\epsilon}^{\infty}$, $\dots$,
$u_{k,\epsilon}^{\infty}$ are distinct whenever
$\epsilon\in[-\epsilon_{k}^{\infty}, \epsilon_{k}^{\infty}]$.

Now we prove  \eqref{e1.6}. Fix
$\epsilon\in[-\epsilon_{k}^{\infty}, \epsilon_{k}^{\infty}]$. First,
since
$R_1^{\epsilon}(u_{1,\epsilon}^{\infty})<0=R_1^{\epsilon}(0)$,
then $\| u_{1,\epsilon}^{\infty}\|_{L^{\infty}}>0$ which
proves \eqref{e1.6} for $i=1$. We further prove that
\begin{equation} \label{e4.15}
\| u_{i,\epsilon}^{\infty}\|_{L^{\infty}}>a_{i-1}\quad
\text{for all } i\in\{2, \dots, k\}.
\end{equation}
Argument by contradiction. Assume that there exists an element 
$i_{0}\in\{2, \dots, k\}$
such that $\| u_{i_{0},\epsilon}^{\infty}\|_{L^{\infty}}\leq
a_{i_{0}-1}$. Since $a_{i_{0}-1}< b_{i_{0}-1}$, then
$u_{i_{0},\epsilon}^{\infty}\in W_{\rm rad}^{b_{i_{0}-1}}$.
So, by \eqref{e3.3} and \eqref{e4.8}, it follows that
\[
R_{i_{0}-1}^{\epsilon}(u_{i_{0}-1,\epsilon}^{\infty})
=\min_{W_{\rm rad}^{b_{i_{0}-1}}} {R_{i_{0}-1}^{\epsilon}}\leq
R_{i_{0}-1}^{\epsilon}(u_{i_{0},\epsilon}^{\infty})
=R_{i_{0}}^{\epsilon}(u_{i_{0},\epsilon}^{\infty}),
\]
which contradicts \eqref{e4.14}. Thus, \eqref{e4.15} holds. 
By \eqref{e4.15} and \eqref{e4.12}, we can complete the proof.
\end{proof}

\begin{proof}[Proof of Remark \ref{rmk1.4}]
Assume that both functions $f$ and $g$ satisfy \eqref{e1.4} with 
$q\in (2, \frac{2_{\alpha}^{*}}{2})$. Fix $\beta\in (2q, 2_{\alpha}^{*})$. 
We may assume that the sequence $\{\theta_i\}_i$ from \eqref{e4.11} satisfies
\begin{align*}
\theta_i<-2C_1(1+| \epsilon|)\Big( \| Q\|_{L^2}(i_{0}-1)
+\| Q\|_{L^{\frac{\beta}{\beta-q}}}(i_{0}-1)^{q} \Big)\quad\text{for all } i\in N,
\end{align*}
where $C_1>0$. Fix $\epsilon\in [-\epsilon_{k}^{\infty}, \epsilon_{k}^{\infty}]$ 
and assume that $\| u_{i_{0}, \epsilon}^{\infty}\|\leq i_{0}-1$ for some 
$i_{0}\in\{1, \dots, k\}$. Then, we have
\begin{align*}
\frac{1}{2}\| u_{i_{0}, \epsilon}^{\infty}\|^2
&= R_{i_{0}}^{\epsilon}(u_{i_{0}, \epsilon}^{\infty})
 +\int_{\mathbb{R}^N}Q(x)\Big(F_{i_{0}}(u_{i_{0}, \epsilon}^{\infty})
 +G_{i_{0}}(u_{i_{0}, \epsilon}^{\infty})\Big)dx\\
&\leq \theta_{i_{0}}+C_1(1+| \epsilon|)
 \Big( \| Q\|_{L^2}\| u_{i_{0}, \epsilon}^{\infty}\|
 +\| Q\|_{L^{\frac{\beta}{\beta-q}}}\| u_{k_i}^{\infty}\|^{q} \Big)\\
&\leq \theta_{i_{0}}+2C_1\Big( \| Q\|_{L^2}(i_{0}-1)
 +\| Q\|_{L^{\frac{\beta}{\beta-q}}}(i_{0}-1)^{q} \Big)
<0.
\end{align*}
This is a contradiction. Therefore \eqref{e1.7} holds.
\end{proof}

\subsection*{Acknowledgments}
Chao Ji was supported by the Shanghai Natural Science Foundation Project
(18ZR1409100), by the  NSFC (grant No. 11301181), and by the
China Postdoctoral Science Foundation Project.
Fei Fang was supported by NSFC (grant No. 11626038).

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\end{document}