text:
\begin{document}

\title[Logarithmic Schr\"odinger equations]
{Semiclassical states for fractional logarithmic Schr\"{o}dinger equations}

 \author{Xiaoming An}

 \address{School of Mathematics and Statistics\, \&\, Guizhou University of Finance and Economics, Guiyang, 550025, P. R. China}
\email{xman@mail.gufe.edu.cn}

\begin{abstract}
In this paper, we consider the following fractional logarithmic Schr\"odinger equation
\begin{equation*}
\varepsilon^{2s}(-\Delta)^s u + V(x)u=u\log |u|^2\ \ \text{in}\ \R^N,
\end{equation*}
where $\varepsilon>0$, $N\ge 1$, $V(x)\in C(\R^N,[-1,+\infty))$. By introducing an interesting penalized function, we show that the problem has a positive solution $u_{\varepsilon}$ concentrating at a local minimum of $V$ as $\varepsilon\to 0$. There is no restriction on decay rates of $V$, especially it can be compactly supported.

{\bf Key words:} Fractional Logarithmic Schr\"odinger; penalized; concentration; compactly  supported.

{\bf AMS Subject Classifications:} 35J05, 35J20, 35J10.
\end{abstract}

\maketitle

\section{Introduction}\label{s1}
We study the following fractional Schr\"odinger equation with logarithmic nonlinear term:
\begin{equation}\label{eq1.1}
\varepsilon^{2s}(-\Delta)^s u + V(x)u=u\log |u|^2,\ \ x \in\ \R^N,
\end{equation}
where $\varepsilon>0$, $N\in \N$, $V(x)\in C(\R^N,\R)$ is a continuous potential. This type of problem comes from the study of standing waves $\psi(x,t) = e^{iEt/\varepsilon}u(x)$ of the following fractional nonlinear Schr\"odinger equation:
\begin{equation}\label{Adeq1.2}
i\frac{\partial \psi}{\partial t} = \varepsilon^{2s}(-\Delta)^s \psi + (V(x)+ E)\psi - f(\psi),
\end{equation}
where $f:\C\to \R$ is a function with $f(z) = g(|z|)z$ and $g:\R^+\to \R$ is a  real function(which is $\log|\cdot|^2$ in \eqref{eq1.1}). For power type nonlinearities, the fractional Schr\"odinger equation was introduced by Laskin (\cite{2}, \cite{3}) as an extension of the classical nonlinear Schr\"{o}dinger equations $s = 1$ in which the Brownian motion of the quantum paths is replaced by a L\`{e}vy flight.

Equation \eqref{eq1.1} is a generalization of the classical Nonlinear  Schr\"odinger Equation with logarithmic nonlinearity:

\begin{equation}\label{AAdeq1.4}
-\varepsilon^{2}\Delta u + V(x)u= u\log |u|^2,\ \ x \in\ \R^N
\end{equation}
which admits applications related to quantum mechanics, quantum optics, nuclear physics, transport and diffusion phenomena, open quantum systems, effective quantum gravity, theory of superfluidity and Bose-Einstein condensation(see \cite{K.G.Zlo-2010,Z.Q.-Wang-C.Zhang-ARMA-2019} and the references therein for more details). For this problem, one can check easily that there always exists $u\in H^1(\R^N)$ such that
$$
\int_{\R^N}|u|^2\log|u|^2=-\infty,
$$
which makes the natural functional corresponding to \eqref{AAdeq1.4} is not well defined in $H^1(\R^N)$. To overcome this difficulty, considering the case that $\varepsilon>0$ is fixed, M. Squassin et al. in \cite{M.Squassina-A.Szulkin-2015}, decompose the functional $I$ into the sum of a $C^1$ functional and a convex l.s.c(short for lower semicontinuous hereafter) functional and then use the Mountain Pass Theorem 3.2 in \cite{A.Szulkin-1986} to find a critical point.
W. Shuai in \cite{W.Shuai-JMP-2021} added some growth conditions such as $\liminf\limits_{|x|\to+\infty}V(x)|x|^{2\sigma}>0$(which makes $\int_{\R^N}|u|^2\log|u|^2$ is $C^1$ in $H^1_V(\R^N)=\{u\in H^1(\R^N):V(x)|u|^2\in L^1(\R^N)\}$) and then proved that the problem has a least energy sign-changing solution and a positive ground state. Considering that $\varepsilon>0$ is fixed and $V(x)\equiv\lambda>0$, Wiliam C. Troy in \cite{W.C.Troy-ARMA-2016} developed  a new comparison method to show that the positive solution of \eqref{AAdeq1.4} is unique up to translation when $1\le N\le 9$; Using the result of Serrin-Tang \cite{J.Serrin-M.Tang-IUMJ-2000}, D'Avenia et al. pointed out in \cite{P.D'Avenia-CCM-2014} that the positive solution of \eqref{AAdeq1.4} is also unique up to translation. Following, letting $u_p$ be the unique positive solution of \eqref{AAdeq1.4}, i.e.,
$$
-\Delta u + \lambda u =|u|^{p-2}{u},
$$
where $p>2$, Wang et al. in \cite{Z.Q.-Wang-C.Zhang-ARMA-2019} proved that $u_p$ will converge to the unique solution of \eqref{AAdeq1.4} in the sense of $C^{2,\alpha}(\R^N)$ if $p\to 2$. Considering the case that $\varepsilon\to 0$, by penalized idea, C. Zhang et al. in \cite{C.Zhang-X.Zhang-CV-2020} showed that \eqref{AAdeq1.4} has solutions $u_{\varepsilon}$ concentrating at various types of topological critical points of $V$ as $\varepsilon\to 0$ provided that $\lim\limits_{|x|\to\infty}V(x)|x|^{-2}>-\infty$($V$ can be unbounded below).

To our best knowledge, there are few results about \eqref{eq1.1} in the nonlocal case $0<s<1$. When $\varepsilon>0$ is fixed, D'Avenia et al. in \cite{P.D'Avenia-MMAS-2015} obtained existence of infinitely weak solutions. In \cite{A.H.Ardila-NA-2017}, it was proved by compactness method that \eqref{eq1.1} has ground states which are stable.

In this paper, we are interesting in semiclassical analysis of \eqref{eq1.1}. From a mathematical point of view, the transition from quantum to classical mechanics can be formally performed by letting $\varepsilon\to 0$. For small $\varepsilon > 0$, solutions $u_{\varepsilon}$ are usually referred to as semiclassical bound states.

In order to state our main result, we need to give some notations and assumptions.
For  $s\in(0,1)$, the fractional Sobolev space $H^s(\mathbb{R}^N)$ is defined as
$$
H^s(\mathbb{R}^N) = \Big\{u\in L^2(\mathbb{R}^N):\frac{u(x) - u(y)}{|x - y|^{N/2 + s}}\in L^2(\mathbb{R}^N\times\mathbb{R}^N)\Big\},
$$
endowed with the norm
$$
\|u\|_{H^s(\mathbb{R}^N)} = \Big(\int_{\mathbb{R}^N}|(-\Delta)^{s/2}u|^2 + u^2\,  {\mathrm{d}}x \Big)^{\frac{1}{2}},
$$
where
$$
\int_{\mathbb{R}^N}|(-\Delta)^{s/2}u|^2{\mathrm{d}}x  = \int_{\mathbb{R}^{2N}}\frac{|u(x) - u(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x \,  {\mathrm{d}}y.
$$
Like the classical case, we define the space $\dot{H}^s(\R^N)$ as the completion of $C^{\infty}_c(\R^N)$ under the norm
$$
\|u\|^2 = \int_{\mathbb{R}^N}|(-\Delta)^{s/2}u|^2{\mathrm{d}}x  = \int_{\mathbb{R}^{2N}}\frac{|u(x) - u(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x \,  {\mathrm{d}}y.
$$
We will use the following local fractional Sobolev space
$$
W^{s,2}(\Omega) = \Big\{u\in L^2(\Omega): \frac{|u(x) - u(y)|}{|x - y|^{\frac{N}{2} + s}}\in L^2(\Omega\times\Omega)\Big\}.
$$
It is easy to check that $W^{s,2}(\Omega)$ is a Hilbert space under the following inner product
$$
(u,v) = \int_{\Omega}\int_{\Omega}\frac{(u(x) - u(y))(v(x) - v(y))}{|x - y|^{N + 2s}}dxdy + \int_{\Omega}uvdx\ \ \forall u,v\in W^{s,2}(\Omega),
$$
see \cite{4} for more details.
Also from \cite{4},  the fractional Laplacian is defined as
\begin{align*}
(-\Delta)^s u(x) \, &= \, C  (N,s)P.V.\int_{\mathbb{R}^N}\frac{u(x) - u(y)}{|x - y|^{N + 2s}}\,{\mathrm{d}}y
\\[2mm]
\, & =\,   C(N,s)\lim_{\varepsilon\to 0}\int_{\mathbb{R}^N          \backslash        B_{\varepsilon}(x)      }\frac{u(x) - u(y)}{|x - y|^{N + 2s}}\,{\mathrm{d}}y.
\end{align*}
For the sake of simplicity, we define for every $u\in \dot{H}^s(\R^N)$ the fractional $(-\Delta)^s$ as
$$
(-\Delta)^s u(x) = 2\int_{\mathbb{R}^N}\frac{u(x) - u(y)}{|x - y|^{N + 2s}}\,{\mathrm{d}}y.
$$
Note that for every $\varphi\in C^{\infty}_c(\R^N)$, it holds
$$
\int_{\R^N}(-\Delta)^s u(x)\varphi(x) = \frac{d}{dt}\int_{\R^N}|(-\Delta)^{s/2}(u+t\varphi)|^2\large|_{t=0}.
$$
Our solutions will be found in the following weighted fractional Sobolev  space:
$$
\mathcal{D}^s_{V,\varepsilon}(\mathbb{R}^N) = \left\{u\in \dot{H}^s(\R^N):\,\,u\in L^2\big(\mathbb{R}^N,\big(V(x)+1\big)\,{\mathrm{d}}x\big)\right\},
$$
endowed with the norm
$$
\|u\|_{\mathcal{D}^s_{V,\varepsilon}(\mathbb{R}^N)} = \Big(\int_{\mathbb{R}^N}\varepsilon^{2s}|(-\Delta)^{s/2}u|^2 + (V(x) + 1)u^2\,{\mathrm{d}}x \Big)^{\frac{1}{2}}.
$$
For the potential term $V$, we assume that $V$ is continuous and

$(\mathcal{V}_1)$ $V(x)+1\ge 0$;

$(\mathcal{V}_2)$  there exist open bounded sets $\Lambda\subset\subset U$ with smooth boundaries $\partial\Lambda,\ \partial U$, such that
\begin{equation}\label{{Aeq1.3}}
0 < \lambda = \inf_{\Lambda}(V+1) < \inf_{U\backslash\Lambda}(V+1).
\end{equation}
Without loss of generality, we assume that $0\in\Lambda$.

Now, with the notations and assumptions above at hand, we are in a position to state our main result:

\begin{theorem}\label{th1.1}
Let $V$ satisfy $(\mathcal{V}_1)$ and $(\mathcal{V}_2)$. Then there exists an $\varepsilon_0>0$ such that \eqref{eq1.1} has a  positive solution $u_{\varepsilon}$ if $\varepsilon\in(0,\varepsilon_0)$. Moreover, $u_{\varepsilon}$ has a global maximum point $x_{\varepsilon}$ which satisfies $$\lim_{\varepsilon\to 0}V(x_{\varepsilon}) = \inf_{x\in\Lambda}V(x)$$ and
\begin{equation*}\label{eq6.4}
u_{\varepsilon}(x)\le\frac{C\varepsilon^{N+2s}}{\varepsilon^{N+2s}+|x - x_{\varepsilon}|^{N+2s}},\  x\in\R^N,
\end{equation*}
where $C$ is a positive constant.
\end{theorem}

The difficulties in the proof of Theorem \ref{th1.1} are stated as follows.
Firstly, there exists $u\in\mathcal{D}^s_{V,\varepsilon}(\R^N)$ such that
$$
\int_{\R^N}|u|^2\log|u|^2 = -\infty,
$$
which makes the natural Euler-Lagrange functional corresponding to \eqref{eq1.1}, i.e.,
\begin{equation*}
I_{\varepsilon}(u):=\frac{1}{2}\int_{\R^N}\big(\varepsilon^{2s}|(-\Delta)^{s/2}u|^2 + (V(x)+1)|u|^2\big)-\frac{1}{2}\int_{\R^N}u^2\log u^2 \mathrm{d}x,
\end{equation*}
is not well defined in $\mathcal{D}^s_{V,\varepsilon}(\R^N)$. Then all the methods developed for \eqref{eq1.1} in the case that the logarithmic term replaced by power type nonlinear term failed(see \cite{100,APX}). Moreover, the expectation that the concentration should occur at a local minimum of $V$ in $\Lambda$ makes us have to truncate the nonlinear term outside $\Lambda$.

Noting that since the ground states of limiting equation of \eqref{eq1.1} decay like $\frac{1}{|x|^{N+2s}}$, we can not use the penalized idea in \cite{C.Zhang-X.Zhang-CV-2020} which deal with the case $s=1$ to truncate the nonlinear term or the potential term.

Creatively, to overcome the two difficulties above, we use the characteristic function $4\chi_{\R^N\Lambda}$ to truncate the nonlinear term(see \eqref{AAAeq1} below), i.e., we firstly study the following penalized problem
\begin{equation}\label{Revisioneq2.1}
\varepsilon^{2s}(-\Delta)^su_{\varepsilon} + (V(x) + 1)u_{\varepsilon} = \chi_{\Lambda}(1 + \log|u|^2)u_{\varepsilon} - \chi_{\R^N\backslash\Lambda}\max\{2u,-u(1 + \log |u|^2)\}.
\end{equation}
The corresponding functional $J_{\varepsilon}$, all through  is  not well defined in $\mathcal{D}^s_{V,\varepsilon}$, is the sum of a $C^1$ functional and a convex l.s.c. functional(see \eqref{AAAeq1} and Remarks \ref{re2.3} below), whose critical points can be found  by the Mountain Pass Theorem 3.2 in \cite{A.Szulkin-1986}.
Then,  a penalized solution  $u$ of \eqref{Revisioneq2.1} is indeed a solution of the origin problem \eqref{eq1.1} if
$$
-(1+\log|u|^2)\ge 2\ \ \text{on}\ \R^N\backslash\Lambda.
$$
This,  after the proof of concentration of $u$, can be checked easily  by the well-known fact that a sub-solution of $(-\Delta)^su + u\le0$ decays like $\frac{1}{|x|^{N+2s}}$(see the last part of Section 3 for more details).

We need to emphasize that the nonlocal term $(-\Delta)^s$ makes the estimates more difficult than the classical case. In fact, for a smooth function $f\in C^{\infty}_c(\R^N)$, one can not compute $(-\Delta)^sf$ as precisely as $-\Delta f$(see Appendix A below for example).

The nonlocal effect makes us have to know the global $L^2$-norm information of penalized solution, which in \cite{APX} was given by the assumption $\liminf_{|x|\to \infty}(V(x)+1)|x|^{2s}>0$. But, in the present paper, the global $L^2$-norm information of penalized solution will be obtained by the fractional Hardy inequality: There exists a positive constant $C_{N,s}$ such that
\begin{equation}\label{Adeq1.4}
\int_{\R^N}\frac{|u(x)|^2}{|x|^{2s}}\,{\mathrm{d}}x\le C_{N,s}|(-\Delta)^{s/2}|^2_2
\end{equation}
for all $u\in \dot{H}^s(\R^N)$(see the proof of \eqref{Adeq3.3} in Appendix A for example).

\vspace{0.5cm}
\textbf{Plan of the paper.} In Section \ref{s2}, we obtain the penalized problem by truncating the logarithmic term in \eqref{eq1.1} outside by $4\chi_{\R^N\Lambda}u$, then we use the Mountain Pass Theorem 3.2 in \cite{A.Szulkin-1986} to obtain a penalized solution $u_{\varepsilon}$. In Section \ref{s3}, we study the concentration of $u_{\varepsilon}$ and then linearize the penalized equation in Section \ref{s2}. At the last part of Section \ref{s3},  we use the well-known decay estimates of fractional Schr\"oinger equations to prove  the asymptotic behaviour of $u_{\varepsilon}$, which implies that  $u_{\varepsilon}$ solves the origin problem.

\vspace{1cm}

\section{The penalized problem}\label{s2}
The following inequality exposes the relationship between $H^s(\R^N)$ and the Banach space $L^q(\R^N)$.
\begin{proposition}\label{wpr2.1} ({Fractional version of the Gagliardo$-$Nirenberg inequality \cite{4}})
For every $u\in H^s(\mathbb{R}^N)$,
\begin{align} \label{inequality2.1}
\|u\|_{q} \leq C \|(-\Delta)^{s/2}u\|^{\theta}_{2}\|u\|^{1 -\theta }_{2},
\end{align}
where $q\in [2,2^*_s]$ and $\theta $ satisfies $\frac{\theta}{2^*_s} + \frac{(1 -\theta)}{2} = \frac{1}{q}$.
\end{proposition}

By the proposition above, we know that $ H^s(\mathbb{R}^N)$ is continuously  embedded into $L^q(\mathbb{R}^N)$ for $\ q\in [2,2^*_s]$. Moreover, on bounded set, the embedding is compact (see \cite{4}), i.e.,
$$
H^s(\mathbb{R}^N)\subset\subset L^q_{loc}(\mathbb{R}^N)\ \text{compactly, if}\ q\in [1,2^*_s).
$$

Our proof will rely on the following fractional logarithmic Sobolev inequality(see \cite{A.Cotsiilis-N.Tavoularis-2005}):

\begin{proposition}\label{Adpr2.2} For any $u\in H^s(\R^N)$, it holds
$$
\int_{\R^N}|u|^2\log \Big(\frac{|u|^2}{\|u\|^2_2}\Big) + \Big(N+\frac{N}{s}\log a + \log \frac{s\Gamma(\frac{N}{2})}{\Gamma(\frac{N}{2s})}\Big)\|u\|^2_2\le \frac{a^2}{\pi^s}\|(-\Delta)^s\|^2_2,\ a>0.
$$
\end{proposition}

Now we are going to modify the origin problem \eqref{eq1.1}. In order to use the Mountain Pass Theorem 3.2 in \cite{A.Szulkin-1986}, considering the vanishing of $V$ and the concentration should occur in $\Lambda$, we modified the nonlinear term as follows. Define
\begin{align}\label{AAAeq1}
\nonumber G_1(x,s)&=\frac{1}{2}\chi_{\Lambda}(x)s^2_+\log s^2_+,\\  G_2(x,s)&=\frac{1}{2}\chi_{\R^N\backslash\Lambda}(x)\int_{0}^{s}\max\{4t_+,-2t_+(1 + \log t^2_+)\}dt.
\end{align}
\begin{remark}\label{re2.3}
\textbf{I}mportantly, since  $G'_2(x,s)$ is nondecreasing on $s$, $G_2(x,s)\ge 0$ is convex on $s$ for all $x\in\R^N\backslash\Lambda$ and the functional $\mathcal{G}^2:\mathcal{D}^s_{V,\varepsilon}(\R^N)\to \R$ given by
$$
\mathcal{G}^2(u)= \int_{\R^N}G_2(x,u)
$$
is convex and l.s.c by Fatou's Lemma.

Proposition \ref{wpr2.1} and the boundedness of $\Lambda$ imply that the functional $\mathcal{G}^1:\mathcal{D}^s_{V,\varepsilon}(\R^N)\to \R$ given by
$$
\mathcal{G}^1(u)= \int_{\R^N}G_1(x,u)
$$
is $C^1$. So the functional $$J_{\varepsilon}(u)=\Phi_{\varepsilon}(u) + \Psi(u)$$ with $$\Phi_{\varepsilon}(u) = \frac{1}{2}\|u\|^2_{V,\varepsilon}-\mathcal{G}^1(u)\ \ \text{and}\ \ \Psi=\mathcal{G}^2(u)$$
 has the form stated in \cite{A.Szulkin-1986}.
\end{remark}

 By the remark above, although $J_{\varepsilon}$ is not $C^1$, we can still use the Mountain Pass Theorem 3.2 in \cite{A.Szulkin-1986} to find a critical point for $J_{\varepsilon}$. We first state some necessary definitions corresponding to those functionals has the form of $J_{\varepsilon}$.
\begin{definition}\label{de2.1}
Let $E$ be a Banach space, $E'$ be the dual space of $E$ and $\langle \cdot,\cdot\rangle$ be the duality paring between $E'$ and $E$. Let $J:E\to \R$ be a functional of the form $J(u)=\Phi(u)+\Psi(u)$, where $\Phi\in C^1(E,\R)$ and  $\Psi$ is convex and l.s.c.. We have the following definitions:

$(i)$ A critical point of $J$ is a point $u\in E$ such that $J(u)<+\infty$ and $0\in\partial J(u)$, i.e.
$$
\langle\Phi'(u),v-u\rangle + \Psi(v)-\Psi(u)\ge 0,\ \forall v\in E.
$$

$(ii)$ A Palais-Smale sequence at level $c$ for $J$ is a sequence $(u_n)\subset E$ such that $J(u_n)\to c$ and there is a numerical sequence $\sigma_n\to 0^+$ with
$$
\langle\Phi'(u_n),v-u_n\rangle + \Psi(v)-\Psi(u_n)\ge -\sigma_n\|v-u_n\|,\ \forall v\in E.
$$

$(iii)$ The functional $J$ satisfies the Palais-Smale condition at level $c$ $(PS)_c$ condition if all Palais-Smale sequence at level $c$ has a convergent subsequence.

$(iv)$ The set $D(J):=\{u\in E:J(u)<+\infty\}$ is called the effective domain of $J$.

\end{definition}

According to the Corollary 2.6 in \cite{M.Squassina-A.Szulkin-2015}, we have
\begin{proposition}\label{Adpr2.4}
Let $\sup_nJ_{\varepsilon}(u_n)<+\infty$. Then $(u_n)$ is a Palais-Smale sequence if and only if $J'(u_n)\to 0$ in $\big(\mathcal{D}^s_{V,\varepsilon}(\R^N)\big)'$.
\end{proposition}

To use Theorem 3.2 in \cite{A.Szulkin-1986}, we need to prove that $J_{\varepsilon}$ satisfies the $(PS)_c$ condition $(iii)$ above.
\begin{proposition}\label{pr2.2} $J_{\varepsilon}$ satisfies $(PS)_c$ condition, i.e., each sequence $(u_n)\subset \mathcal{D}^s_{V,\varepsilon}(\R^N)$ with $\lim\limits_{n\to\infty}J_{\varepsilon}(u_n)\to c$

has a convergent subsequence in $\mathcal{D}^s_{V,\varepsilon}(\R^N)$.

\end{proposition}
\begin{proof}
We first show that $(u_n)$ is bounded in $\mathcal{D}^s_{V,\varepsilon}(\R^N)$. Observing firstly that $\Psi\ge 0\ \forall u\in\mathcal{D}^s_{V,\varepsilon}(\R^N)$, hence we have
\begin{equation}\label{SSEQ2.2}
J_{\varepsilon}(u_n)\ge \Phi_{\varepsilon}(u_n) = \frac{1}{2}\|u_n\|^2_{V,\varepsilon} - \frac{1}{2}\int_{\Lambda}|u_n|^2\log|u_n|^2.
\end{equation}
Moreover, by \eqref{AAAeq1}, Proposition \ref{Adpr2.4} and the fact that $\int\max\{f,g\}\ge \max\{\int f,\int g\}$, it holds
\begin{equation}\label{AAdeq2.4}
\int_{\R^N}|u_n|^2\mathrm{d}x\le 2J_{\varepsilon}(u_n) - \langle J'_{\varepsilon}(u_n),u_n\rangle\le C + o_n(1)\|u_n\|_{V,\varepsilon}.
\end{equation}

For a set $A\in\R^N$, we define $A^{d}:=\{x\in\R^N:dist(A,x)<d\}$.
Let $\eta\in C^{\infty}_c(\Lambda^{\delta})$ be a function satisfying $0\le \eta\le 1$ and $\eta\equiv 1$ on $\overline{\Lambda}$, where the $\delta$ is a small parameter such that $\Lambda^{2\delta}\subset\subset U$. Defining $v_n(x) = u_n(\varepsilon x)\eta(\varepsilon x)$, we have $v_n(x)\in H^s(\R^N)\backslash\{0\}$. Then, by Proposition \ref{Adpr2.2}, we have
\begin{align*}
\int_{\R^N}|v_n|^2\log|v_n|^2 \le \frac{a^s}{\pi}\|(-\Delta)^{s/2}v_n\|^2 + (C_{s,N} + \frac{N}{s}\log a)\|v_n\|^2_2 + \|v_n\|^2_2\log \|v_n\|^2_2.
\end{align*}
Resclaing back, we find

\begin{eqnarray}\label{Adeq2.4}
\begin{split}
&\quad\int_{\R^N}|\eta u_n|^2\log\|\eta u_n\|^2\\
&\le \frac{a\varepsilon^{2s}}{2}\|(-\Delta)^{s/2}(\eta u_n)\|^2_2 + (C_{s,N} + \frac{N}{s}\log a)\|\eta u_n\|_2^2+ \|\eta u_n\|^2_2 \log \Big(\frac{1}{\varepsilon^N}\|\eta u_n\|^2_2\Big).

\end{split}
\end{eqnarray}
By fractional Hardy inequality \eqref{Adeq1.4} and some delicate nonlocal estimates, we will prove in Appendix A that
\begin{equation}\label{Adeq2.5}
T_1(\eta):=\varepsilon^{2s}\|(-\Delta)^{s/2}(\eta u_n)\|^2_2\le C\|u_n\|^2_{V,\varepsilon},
\end{equation}
where $C$ is a positive constant. Hence, returning back to \eqref{Adeq2.4}, letting $a>0$ be small enough, by \eqref{AAdeq2.4}, we get
\begin{eqnarray}\label{Adeq2.6}
\begin{split}
&\quad\int_{\R^N}|\eta u_n|^2\log\|\eta u_n\|^2\le \frac{1}{2}\|u_n\|^2_{V,\varepsilon} + C(1 + \|\eta u_n\|_2^{2(1 + \delta)}) -N\|u_n\|^2_2\log\varepsilon\\
&\le \frac{1}{2}\|u_n\|^2_{V,\varepsilon} + C(1 + \|u_n\|^{1 + \delta}_{V,\varepsilon}) - N\big(C + \|u_n\|_{V,\varepsilon}\big)\log\varepsilon,

\end{split}
\end{eqnarray}
where $\delta\in(0,1)$ is a parameter.
Note that for every $f:\R^N\to \R$, it holds
$$
\int_{U\backslash\Lambda}|f|^2\log |f|^2\ge -\frac{1}{e}|U\backslash\Lambda|.
$$
Then by \eqref{Adeq2.6} and \eqref{SSEQ2.2}, we have
$$
\widetilde{C}\ge \frac{1}{2}\|u_n\|^2_{V,\varepsilon} - \|u_n\|^{1 + \delta}_{V,\varepsilon} + N(C + \|u_n\|_{V,\varepsilon})\log\varepsilon,
$$
where $\widetilde{C}$ is a positive constant. Consequently, since $0<\delta<1$, we can conclude that $(u_n)$ is bounded in $\mathcal{D}^s_{V,\varepsilon}(\R^N)$.
 Going if necessary to a subsequence, we assume that
 $$
 u_n\rightharpoonup u\ \text{weakly in}\ \mathcal{D}^s_{V,\varepsilon}(\R^N).
 $$

 Next we show that $(u_n)$ has a convergent subsequence. Since $u_n\rightharpoonup u\in \mathcal{D}^s_{V,\varepsilon}(\R^N)$, by the boundedness of $\Lambda$ and the fact that $|G_1(x,s)|\le C_1|t|^{\frac{3}{2}} + C_2|t|^{\frac{5}{2}}$, where $C_1,C_2>0$ are two positive constants, we have
 $$
 {G}_1'(x,u_n)u_n\to {G}_1'(x,u)u.
 $$
Noting that since $J'_{\varepsilon}(u_n)\varphi = o_n(1)\|\varphi\|_{V,\varepsilon}$ for all $\varphi\in C^{\infty}_c(\R^N)$, we deduce that $J'_{\varepsilon}(u)\varphi = 0$ for all $\varphi\in C^{\infty}_c(\R^N)$, and so, $J'_{\varepsilon}(u)u=0$. Combing with $|J'_{\varepsilon}(u_n)u_n|\le \|J'_{\varepsilon}(u_n)\|_{\mathcal{D}'_{V,\varepsilon}}\|u_n\|_{V,\varepsilon} =  o_n(1)\|u_n\|$, we have
\begin{equation}\label{Adeq2.8}
\|u_n\|^2_{V,\varepsilon} + \int_{\R^N}\chi_{\R^N\backslash\Lambda}G_2'(x,u_n)u_n = \|u\|^2_{V,\varepsilon} + \int_{\R^N}\chi_{\R^N\backslash\Lambda}G_2'(x,u)u + o_n(1).
\end{equation}
Finally, by $\|u\|\le \liminf\limits_{n\to\infty}\|u_n\|^2_{V,\varepsilon}$ and the l.s.c property(by Fatou's Lemma), we get that $u_n\to u$ strongly in $\mathcal{D}^s_{V,\varepsilon}(\R^N)$. This completes the proof.
\end{proof}

\begin{remark}
The proof of \eqref{Adeq2.5} is trivial if $s=1$, but delicate if $0<s<1$. It will need the global $L^2$ information of $u_n$, which is given by the nonlocal operator $(-\Delta)^s$ and the fractional Hardy inequality \eqref{Adeq1.4}(see \eqref{eqA.2} for example).
\end{remark}
Obviously, there exists $\rho>0$ such that
$$
J_{\varepsilon}(u)\ge \Phi_{\varepsilon}(u)\ge \frac{1}{2}\|u\|^2_{V,\varepsilon} - C\|u\|^p_{V,\varepsilon}>0\ \text{for all}\ u\ \text{with}\ \|u\|^2_{V,\varepsilon} = \rho
$$
and for each $u\in C^{\infty}_c(\Lambda)\backslash\{0\}$, it holds
$$
J_{\varepsilon}(su)\to-\infty\ \text{as}\ s\to+\infty,
$$
i.e., $J_{\varepsilon}$ owns mountain pass geometry. Thus by Proposition \ref{pr2.2} and Theorem 3.2 in \cite{A.Szulkin-1986}, we immediately have:
\begin{lemma}\label{le2.3}
The mountain pass value
$$
c_{\varepsilon}=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}J_{\varepsilon}(\gamma(t))
$$
is positive and can be achieved by a positive function $u_{\varepsilon}$ which is a critical point of $J_{\varepsilon}$ and solves the following penalized problem
\begin{equation}\label{eq2.1}
\varepsilon^{2s}(-\Delta)^s u_{\varepsilon} + (V(x) + 1)u_{\varepsilon} = G_1'(x,(u_{\varepsilon})_+) - G_2'(x,(u_{\varepsilon})_+).
\end{equation}
\end{lemma}

\begin{proof}
By Theorem 3.2 in \cite{A.Szulkin-1986}, $c_{\varepsilon}$ is a critical value, i.e., there exists $u_{\varepsilon}\in D(J_{\varepsilon})=\{v\in \mathcal{D}^s_{V,\varepsilon}:J_{\varepsilon}(v)<+\infty\}$ with $J_{\varepsilon}(u_{\varepsilon}) = c_{\varepsilon}$ such that
$$
\langle \Phi'_{\varepsilon}(u_{\varepsilon}),v-u_{\varepsilon}\rangle + \Psi(v) - \Psi(u_{\varepsilon})\ge 0\ \ \forall v\in \mathcal{D}^s_{V,\varepsilon}.
$$
In particular, letting $t>0$ and $v = u_{\varepsilon} + t\varphi$ with $\varphi\in C^{\infty}_c(\R^N)$, we have
$$
\langle \Phi'_{\varepsilon}(u_{\varepsilon}),\varphi\rangle + \frac{\Psi(u_{\varepsilon} + t\varphi) - \Psi(u_{\varepsilon})}{t}\ge 0\ \ \forall t>0
$$
and then
$$
\langle J'_{\varepsilon}(u_{\varepsilon}),\varphi\rangle = \langle \Phi'_{\varepsilon}(u_{\varepsilon}),\varphi\rangle + \int_{\R^N}G_2'(x,u_{\varepsilon})\varphi\ge 0.
$$
Rearranging $\varphi = -\psi$, we eventually have
$$
\langle J'_{\varepsilon}(u_{\varepsilon}),\varphi\rangle = 0\ \forall \varphi\in C^{\infty}_c(\R^N),
$$
which implies \eqref{eq2.1}.

Finally, letting $(u_{\varepsilon})_-$ be a test function to \eqref{eq2.1}, we find $u_{\varepsilon}\ge 0$. By the standard regularity assertion and maximum principle in \cite[Appendix D]{20}, we conclude that $u_{\varepsilon}$ is positive.
\end{proof}

\vspace{0.5cm}

\section{Concentration and the origin problem}\label{s3}

In this section we will prove the concentration phenomenon of $u_{\varepsilon}$ via energy comparison, by which we will prove at the last of this section that
\begin{equation}\label{Adeq3.1}
-(1 + \log|u_{\varepsilon}|^2)\ge 2\ \ \forall x\in\R^N\backslash\Lambda,
\end{equation}
which and \eqref{eq2.1} indicate that $u_{\varepsilon}$ solves the origin problem \eqref{eq1.1}.

\subsection{Concentration}
In the first subsection, we prove the concentration of $u_{\varepsilon}$ via comparing the energy $c_{\varepsilon}$ with the least energy of the limiting problem of \eqref{eq1.1}.

The limiting problem corresponding to \eqref{eq1.1} is
\begin{equation}\label{eq3.1}
(-\Delta)^s u + \lambda u = u\log |u|^2,
\end{equation}
where $\lambda>-1$. Its Euler-Lagrange functional is
$$
\mathcal{L}_{\lambda}(u) = \frac{1}{2}\int_{\R^N}|(-\Delta)^{s/2}u|^2 + (\lambda + 1)|u|^2 - \frac{1}{2}\int_{\R^N}u^2\log u^2.
$$
In \cite{A.H.Ardila-NA-2017}, it was proved that the limiting problem \eqref{eq3.1} has a least energy solution $U_{\lambda}$ with
$$
\mathcal{L}_{\lambda}(U_{\lambda}) = \mathcal{C}_{\lambda} := \inf_{\varphi\in H^s(\R^N)\backslash\{0\}}\max_{t>0}\mathcal{L}_{\lambda}(t\varphi)=\inf_{\varphi\in C^{\infty}_c(\R^N)\backslash\{0\},\varphi\ge0}\max_{t>0}\mathcal{L}_{\lambda}(t\varphi).
$$
For $\mathcal{C}_{\lambda}$, we have
\begin{proposition}\label{pr3.1}
the function $\mathcal{C}_{\cdot}:(-1,+\infty)\to (0,+\infty)$ is continuous and increasing.
\end{proposition}
\begin{proof}
Let $-1<\lambda<\lambda'<+\infty$ and $U_{\lambda'}$ be the ground solution of equation \eqref{eq3.1} with $\lambda=\lambda'$. An easy analysis shows that the following function $f:\R^+\to \R^+$
$$
f(t)=\mathcal{L}_{\lambda}(tU_{\lambda'}) = \mathcal{L}_{\lambda'}(tU_{\lambda'}) + \frac{t^2}{2}(\lambda - \lambda')\int_{\R^N}|U_{\lambda'}|^2
$$
has a unique maximum point $t'\in(0,+\infty)$, from which we have
\begin{align*}
\mathcal{C}_{\lambda}\le \max_{t>0}f(t) &= \mathcal{L}_{\lambda'}(t'U_{\lambda'}) + \frac{(t')^2}{2}(\lambda - \lambda')\int_{\R^N}|U_{\lambda'}|^2\\
&\le \mathcal{C}_{\lambda'} + \frac{(t')^2}{2}(\lambda - \lambda')\int_{\R^N}|U_{\lambda'}|^2\\
&<\mathcal{C}_{\lambda'}.
\end{align*}
Similarly, it holds
$$
\mathcal{C}_{\lambda'}\le \mathcal{C_{\lambda}} + \frac{\tilde{t}^2}{2}(\lambda' - \lambda)\int_{\R^N}|U_{\lambda}|^2
$$
for some unique $\tilde{t}\in(0,+\infty)$.
Then  $\mathcal{C}_{\lambda}$ is increasing and continuous.
\end{proof}

By the analysis above, we have the following upper bound of $c_{\varepsilon}$.
\begin{proposition}\label{pr3.2}
It holds
$$
\limsup_{\varepsilon\to 0}\frac{c_{\varepsilon}}{\varepsilon^N}\le\min_{x\in\Lambda}\mathcal{C}_{V(x)}.
$$
\end{proposition}
\begin{proof}
Let $\varphi\in C^{\infty}_c(\R^N)\backslash\{0\},\ \varphi\ge 0$ and define for each $x_0\in\Lambda$
$$
\varphi_{\varepsilon}(x) = \varphi\Big(\frac{x - x_0}{\varepsilon}\Big).
$$
Obviously, supp$\varphi_{\varepsilon}\subset \Lambda$ for small $\varepsilon$ and $\gamma_{\varepsilon}(t) = tT_0\varphi_{\varepsilon}\in \Gamma_{\varepsilon}$ for some $T_0$ large enough. Then we have
\begin{align*}
\frac{c_{\varepsilon}}{\varepsilon^N}\le\frac{\max_{t\in[0,1]}
J_{\varepsilon}(\gamma_{\varepsilon}(t))}{\varepsilon^N}\le \mathcal{L}_{V(x_0)}(t\varphi) + o_{\varepsilon}(1)
\end{align*}
and
$$
\limsup_{\varepsilon\to 0}\frac{c_{\varepsilon}}{\varepsilon^N}\le\inf_{{\varphi\in C^{\infty}_c(\R^N)\backslash\{0\}}\atop{\varphi\ge 0}}\mathcal{L}_{V(x_0)}(t\varphi)=\mathcal{C}_{V(x_0)},
$$
which completes the proof.
\end{proof}

Next, we give the lower bounds of solutions of \eqref{eq2.1}.

\begin{proposition}\label{pr3.3}
Let $(u_{\varepsilon_n})$ with $\varepsilon_n>0,\ \varepsilon_n\to 0\ \text{as}\ n\to\infty$ be a family of solutions of $\eqref{eq2.1}$. If for each $k\in \N$, there exists $k$ families of points $\{(x^i_{\varepsilon_n}):1\le i\le k\}$ with $\lim\limits_{n\to\infty}x^i_{\varepsilon_n}=x^i_*$ such that
$$
\liminf_{n\to\infty}\|u_{\varepsilon_n}\|_{L^{\infty}(B_{\varepsilon_n\rho})(x^i_{\varepsilon_n})}>0,\ V(x^i_*)+1>0,\ \ 1\le i\le k,
$$
$$
\liminf_{n\to\infty}\frac{|x^i_{\varepsilon_n} - x^j_{\varepsilon_n}|}{\varepsilon_n} =+\infty,\ \ 1\le i\ne j\le k
$$
and
$$
\limsup_{n\to\infty}\frac{J_{\varepsilon_n}(u_{\varepsilon_n})}{\varepsilon^N_n}<+\infty,
$$
then
$$
\liminf_{n\to\infty}\frac{J_{\varepsilon_n}(u_{\varepsilon_n})}{\varepsilon^N}\ge\sum_{i = 1}^k\mathcal{C}_{V(x^i_*)}.
$$
\end{proposition}

\begin{proof}
Fixing a $1\le i\le k$ and rescaling  the function $u_{\varepsilon_n}$ as $v^i_{n}(x) = u(\varepsilon_n x + x^i_{\varepsilon_n}),\ x\in\R^N$, we have by the estimate in Proposition \ref{pr2.2}  that
$$
\sup_{n}\int_{\R^N}\big(|(-\Delta)^{s/2} v^i_n|^2 + \big(V^i_n(x)+1\big)|v^i_n|^2\big)<+\infty,
$$
where $V^i_n(\cdot)=V(\varepsilon_n \cdot + x^i_{\varepsilon_n})$. Obviously, $v^i_n$ satisfies
\begin{equation}\label{Aeq3.2}
(-\Delta)^s v^i_n + \big(V^i_n(x)+1\big)v^i_n = {G}_1'(\varepsilon_n x + x^i_{\varepsilon_n},v^i_n) -{G}_2'(\varepsilon_n x + x^i_n,v^i_n)\ \ \text{in}\ \R^N.
\end{equation}
 Fixing $R>0$, we have by continuity that
\begin{align*}
&\quad\limsup_{n\to\infty}\|v^i_{n}\|^2_{H^s(B_R)}\\
&\le \limsup_{n\to\infty}(\inf_{\Lambda}(V(x) + 1))^{-1}\int_{\R^N}\big(|(-\Delta)^{s/2} v^i_n|^2 + \big( V^i_n(x)+1\big)|v^i_n|^2\big)<+\infty,
\end{align*}
which says that $(v^i_n)$ is bounded in $H^s_{loc}(\R^N)$ and then by diagonal argument, we can assume without loss of generality that
$
v^i_n\rightharpoonup v^i_*
$
weakly in $H^s_{loc}$ as $n\to \infty$. By
$$
\|v^i_*\|^2_{H^s(B_R)}\le \liminf_{n\to\infty}\|v^i_{n}\|^2_{H^s(B_R)}<+\infty,
$$
we have $v^i_*\in H^s(\R^N)$.

The smoothness of $\Lambda$ implies that the set $\Lambda^i_n=\{x:\varepsilon_n x + x^i_{\varepsilon_n}\in\Lambda\}$ converges to a set $\Lambda^i_*\in\{\emptyset,H,\R^N\}$ as $n\to \infty$, where $H$ is a half plane, by which we have
\begin{align*}
&\quad\int_{\R^N}\Big({G}_1'(\varepsilon x + x^i_{\varepsilon_n},v^i_n)\varphi - {G}_2'(\varepsilon x + x^i_{\varepsilon_n},v^i_n)\varphi\Big)\\
&\to \int_{\R^N}\chi_{\Lambda^i_*}(1 +\log |v^i_*|^2)v^i_*\varphi - \int_{\R^N}\chi_{\R^N\backslash\Lambda^i_*}\max\{2v^i_*,-(1 +\log |v^i_*|^2)v^i_*\}\varphi
\end{align*}
for all $\varphi\in C^{\infty}_c(\R^N)$ as $n\to \infty$.
Then we conclude that $v^i_*$ satisfies the following equation:
\begin{align}\label{eq3.2}
\nonumber&\
\quad(-\Delta)^s v^i_* + (V(x^i_*) + 1)v^i_*\\
&= \chi_{\Lambda^i_*}(x)v^i_*(1+\log|v^i_*|^2)-\chi_{\R^N\backslash\Lambda^i_*}\max\{2v^i_*,-(1 +\log |v^i_*|^2)v^i_*\}\ \ \text{in}\ R^N.
\end{align}
By the similar regularity argument in \cite[Appendix D]{20}, we have
$$
\|v^i_*\|_{L^{\infty}(B_{\rho}(x^i_*))} = \lim_{n\to\infty}|v^i_n\|_{L^{\infty}(B_{\rho}(x^i_n))} = \lim_{n\to\infty}\|u^i_n\|_{L^{\infty}(B_{\varepsilon_n\rho}(x^i_{\varepsilon_n}))}>0,
$$
which implies that $v^i_*$ is nontrivial.

The Euler-Lagrange functional corresponding to \eqref{eq3.2} is
\begin{align*}
J^i_*(u)&=\frac{1}{2}\int_{\R^N}(|(-\Delta)^{s/2}u|^2 + (V(x^i_*) + 1)|u|^2) - \frac{1}{2}\int_{\R^N}\chi_{\Lambda^i_*}|u|^2\log|u|^2\\
 &\quad+ \frac{1}{2}\int_{\R^N\backslash\Lambda^i_*}\mathrm{d}x\int_{0}^{|u(x)|}\max\{4s,-2s(1 + \log s^2)\}ds,
\end{align*}
which implies
\begin{align*}
J^i_*(v^i_*) = \max_{t>0}J^i_*(tv^i_*)\ge \max_{t>0}\mathcal{L}_{V(x^i_*)}(tv^i_*)\ge\mathcal{C}_{V(x^i_*)}.
\end{align*}
Then, after rescaling, we have
\begin{align*}
&\quad \liminf_{n\to\infty}\frac{1}{2\varepsilon^N_n}\int_{B_{\varepsilon_nR}(x^i_{\varepsilon_n})}\Big((
(\varepsilon^{2s}_n|(-\Delta)^{s/2}u_n|^2 + (V(x) + 1)|u_n|^2)\\
&\qquad - \chi_{\Lambda}|u_n|^2\log|u_n|^2+\chi_{\R^N\backslash\Lambda}{G}_2(x,u_n)\Big)\\
&\ge J^i_*(v^i_*) + o_R(1)\ge \mathcal{C}_{V(x^i_*)} + o_R(1).
\end{align*}

Now let us estimate the energy outside $\bigcup_{i=1}^kB_{\varepsilon_n}(x^i_{\varepsilon_n})$.

Choose $\eta$ as another cut-off  function with $\eta_R\equiv0$ in $B_{R}$ and $\eta_R\equiv1$ on $B^c_{2R}$ and define
$$
\eta_{n,R}(\cdot) = \prod_{i=1}^k\eta_R\Big(\frac{\cdot - x^i_{\varepsilon_n}}{\varepsilon_n}\Big).
$$
Testing  \eqref{eq2.1} against with $\eta_{n,R}u_{\varepsilon_n}$, by the definition of penalized function in \eqref{AAAeq1},
we find
\begin{align}\label{AAdeq3.5}
\nonumber&\quad\frac{1}{2}\int_{\Big(\bigcup_{i=1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c}(\varepsilon^{2s}_n|(-\Delta)^{s/2}u_n|^2 + (V(x) + 1)|u_n|^2)\\
\nonumber&- \frac{1}{2}\int_{\Big(\bigcup_{i=1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c\cap\Lambda}
{G}_1(x,u_n)+\frac{1}{2}\int_{\Big(\bigcup_{i=1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c\cap\R^N\backslash\Lambda}
{G}_2(x,u_n)\\
\nonumber& := T^2_{n,R}+\frac{1}{2}\int_{{\bigcup_{i=1}^k\Big(B_{2\varepsilon_nR}(x^i_{\varepsilon_n})\backslash B_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)}}(1-\eta_{n,R})

(\varepsilon^{2s}_n|(-\Delta)^{s/2}u_n|^2 + (V(x) + 1)|u_n|^2)\\
\nonumber&\quad + \frac{1}{2}\int_{\Lambda}\eta_{n,R}|u_n|^2(1+\log|u_n|^2) - \frac{1}{2}{\int_{\R^N}}\chi_{\Big(\bigcup_{i = 1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c\cap\Lambda}|u_n|^2\log|u_n|^2\\
\nonumber&\quad-\frac{1}{2}\int_{\R^N\backslash\Lambda}\eta_{n,R}(x)\max\{2|u_n|^2,-|u_n|^2(1+\log|u_n|^2)\}\\
&\quad + \frac{1}{2}\int_{\R^N\backslash\Lambda}\chi_{\Big(\bigcup_{i = 1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c}\mathrm{d}x\int_{0}^{|u_n|}\max\{4t_+,-2t_+(1+\log |t_+|^2)\}\\
\nonumber&\ge T^2_{n,R}
 + \frac{1}{2}\int_{\Lambda}\eta_{n,R}|u_n|^2(1+\log|u_n|^2) - \frac{1}{2}{\int_{\Lambda}}\chi_{\Big(\bigcup_{i = 1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c}|u_n|^2\log|u_n|^2\\
\nonumber&\quad-\frac{1}{2}\int_{\R^N\backslash\Lambda}\eta_{n,R}(x)\max\{2|u_n|^2,-|u_n|^2(1+\log|u_n|^2)\}\\
\nonumber&\quad + \frac{1}{2}\int_{\R^N\backslash\Lambda}\chi_{\Big(\bigcup_{i = 1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c}\max\{2|u_n|^2,-|u_n|^2(1+\log |u_n|^2)\}\mathrm{d}x\\
\nonumber& \ge T^2_{n,R}
+ \frac{1}{2}\int_{\bigcup_{i=1}^k\Big(B_{2\varepsilon_nR}(x^i_{\varepsilon_n})\backslash B_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)\cap\Lambda}(\eta_{n,R}-1)|u_n|^2\log|u_n|^2\\
\nonumber&\quad +\frac{1}{2}\int_{{\bigcup_{i=1}^k\Big(B_{2\varepsilon_nR}(x^i_{\varepsilon_n})\backslash B_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)}\cap\R^N\backslash\Lambda}(\eta_{n,R}-1)\max\{2|u_n|^2,-|u_n|^2(1 + \log|u_n|^2)\},

\end{align}
where
\begin{align*}
&T^2_{n,R}= \frac{\varepsilon^{2s}_n}{2}\int_{\R^N}\mathrm{d}x\int_{\R^N}\frac{(\eta_{n,R}(y)-\eta_{n,R}(x))u_n(y)(u_n(x)-u_n(y))}{|x-y|^{N+2s}}\mathrm{d}y.
\end{align*}

In Appendix A, we will prove by fractional Hardy inequality \eqref{Adeq1.4} that
\begin{equation}\label{Adeq3.5}
\limsup_{n\to\infty}\frac{T^2_{n,R}}{\varepsilon^N_n}\ge o_R(1).
\end{equation}
Hence, by the fact that $v^i_n\to v^i_*$ strongly in $L^q_{loc}(\R^N)$ with $1<q<2^*_s$, we conclude that
\begin{align*}
&\lim_{n\to\infty}\Big(\quad\frac{1}{2\varepsilon^N_n}\int_{\Big(\bigcup_{i=1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c}(\varepsilon^{2s}_n|(-\Delta)^{s/2}u_n|^2 + (V(x) + 1)|u_n|^2)\\
&- \frac{1}{2\varepsilon^N_n}\int_{\Big(\bigcup_{i=1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c}
{G}_1(x,u_n)+\frac{1}{2\varepsilon^N_n}\int_{\Big(\bigcup_{i=1}^kB_{\varepsilon_nR}(x^i_{\varepsilon_n})\Big)^c}
{G}_2(x,u_n)\Big)\\
&\ge o_R(1) - C\int_{B_{2R}\backslash B_{R}}(|v^i_*| + |v^i_*|^{q})\\
&=o_R(1).
\end{align*}

Finally, by the analysis above, we have
$$
\liminf_{n\to\infty}\frac{J_{\varepsilon_n}(u_{\varepsilon_n})}{\varepsilon^N_n}\ge \sum_{i = 1}^k\mathcal{C}_{V(x^i_*)} + o_R(1),
$$
the conclusion then follows by letting $R\to\infty$.

\end{proof}

\begin{remark}

It is easy to check that
$$
\int\max\{f,g\}\ge \max\{\int f,\int g\},
$$
which and the skillful choice of the truncated function in \eqref{AAAeq1} play a key role in the proof of \eqref{AAdeq3.5}. The estimates \eqref{Adeq3.5} is trivial in the classical case, but delicate under the nonlocal effect of $(-\Delta)^s(0<s<1)$, some skillful global estimates will be involved(e.g., the $L^2$ information of $u_{\varepsilon_n}$ outside $\Lambda$, see \eqref{Adeq3.3} in Appendix for example).
\end{remark}

Now we prove the concentration of $u_{\varepsilon}$.
\begin{lemma}\label{le3.4}
Let $\rho>0$ and $u_{\varepsilon}$ be the penalized solution given by Lemma \ref{le2.3}. There exists a family of points $(x_{\varepsilon})\subset\Lambda$ such that

$(i)$ $\liminf\limits_{\varepsilon\to 0}\|u_{\varepsilon}\|_{L^{\infty}(B_{\varepsilon\rho}(x_{\varepsilon}))}>0$.

$(ii)$ $\lim\limits_{\varepsilon\to 0}V(x_{\varepsilon}) = \inf_{\Lambda} V$.

$(iii)$ $\lim\limits_{{R\to\infty}\atop{\varepsilon\to0}}\|u_{\varepsilon}\|_{L^{\infty}(U\backslash B_{\varepsilon R}(x_{\varepsilon}))}=0$.

\end{lemma}

\begin{proof}
Easily, we have
\begin{align*}
0&<\int_{\R^N}(\varepsilon^{2s}|(-\Delta)^{s/2}u_{\varepsilon}|^2 + (V(x) + 1)|u_{\varepsilon}|^2)\le \int_{\Lambda\cap\{x:u_{\varepsilon}(x)> e^{-1/2}\}}|u_{\varepsilon}|^2(1 + \log|u_{\varepsilon}|^2)\\
&\qquad\le \|1 + \log|u_{\varepsilon}|^2\|_{L^{\infty}(\Lambda\cap\{x:u_{\varepsilon}(x)> e^{-1/2}\})}\int_{\Lambda}|u_{\varepsilon}|^2,
\end{align*}
which and the similar regularity assertion in \cite{20} imply that there exists $x_{\varepsilon}\in \overline{\Lambda}$ such that
$$
u_{\varepsilon}(x_{\varepsilon}) = \sup_{x\in\Lambda}u_{\varepsilon}(x)\ \ \text{and}\ \liminf_{\varepsilon\to 0}\|u_{\varepsilon}\|_{L^{\infty}(B_{\varepsilon\rho}(x_{\varepsilon}))}>0.
$$
This proves $(i)$.

For $(ii)$, assuming without loss of generality that $\lim\limits_{\varepsilon\to 0}x_{\varepsilon} = x_*$, by the lower and upper bounds of $u_{\varepsilon}$ in Propositions \ref{pr3.2} and \ref{pr3.3}, we have
$$
\min_{x\in \Lambda}\mathcal{C}_{V(x)}\ge \liminf_{\varepsilon\to 0}\frac{J_{\varepsilon}(u_{\varepsilon})}{\varepsilon^N}\ge\mathcal{C}_{V(x_*)},
$$
which implies $V(x_*) = \min_{x\in\Lambda}V(x)$.

For $(iii)$, if it is not true, then one will get a contradiction like
$$
\min_{x\in \Lambda}\mathcal{C}_{V(x)}\ge \liminf_{\varepsilon\to 0}\frac{J_{\varepsilon}(u_{\varepsilon})}{\varepsilon^N}\ge\mathcal{C}_{V(x_*)} + \mathcal{C}_{V(y_*)}
$$
for some $y_*\in \overline{U}$ by Proposition \ref{pr3.3}.
\end{proof}

\subsection{Back to the origin problem}
In this subsection, we use Lemma \ref{le3.4} to linearize equation \eqref{eq2.1} and then use the well-known decay estimates of positive solutions of fractional Schr\"odinger equations  to show that \eqref{Adeq3.1} is true.

Noting that by the regular assertion in Appendix D of \cite{20}, we can assume that
\begin{equation}\label{eq5.1}
\sup_{\Lambda}u_{\varepsilon}(x)\le C<\infty,
\end{equation}
where $C$ is a positive constant. Hence by Lemma \ref{le3.4},
we can linearize the penalized equation \eqref{eq2.1} as follows.
\begin{proposition}\label{pr5.1}
Let $\varepsilon>0$ be small enough, $x_{\varepsilon}$ be the point given by Lemma \ref{le3.4}. Then there exists $R>0$ such that
$$
\left\{
  \begin{array}{ll}
   \varepsilon^{2s}(-\Delta)^su_{\varepsilon} + \min\{\lambda,2\} u_{\varepsilon}\le 0, & \text{in}\ \R^N\backslash B_{\varepsilon R}(x_{\varepsilon}) \\
    u_{\varepsilon}\le C, &\text{in}\ B_{\varepsilon R}(x_{\varepsilon})
  \end{array}
\right.
$$
\end{proposition}

\begin{proof}
For $\varepsilon>0$ small enough, by Lemma \ref{le3.4}, there exists $R>0$ such that $
(1 + \log|u_{\varepsilon}|^2)\le 0
$
for all $x\in U\backslash B_{\varepsilon R}(x_{\varepsilon})$,
the conclusion then follows by the penalized function in \eqref{AAAeq1} and inserting \eqref{eq5.1} into \eqref{eq2.1}.
\end{proof}

At last, we prove Theorem \ref{th1.1}.

\textbf{Completes the proof of Theorem \ref{th1.1}}.

By Proposition \ref{pr5.1}, the rescaling function $v_{\varepsilon} = u_{\varepsilon}(\varepsilon x + x_{\varepsilon}$) satisfies
\begin{equation*}
\left\{
  \begin{array}{ll}
   (-\Delta)^s v_{\varepsilon} + \lambda v_{\varepsilon}\le 0, & \text{in}\ \R^N\backslash B_{R} \\
    v_{\varepsilon}\le C, &\text{in}\ B_{R}.
  \end{array}
\right.
\end{equation*}
Then by the well-known decay estimates of fractional Schr\"odinger equation(see \cite[Appendix D]{20} for example), we have
\begin{equation}\label{Adeq3.7}
v_{\varepsilon}(x)\le \frac{C}{1 + |x|^{N+2s}},\ x\in\R^N.
\end{equation}
Consequently, it holds
$$
u_{\varepsilon}(x)\le \frac{C\varepsilon^{N+2s}}{\varepsilon^{N+2s} + |x - x_{\varepsilon}|^{N+2s}},
$$
from which, we can conclude that
$$
\max\{2u_{\varepsilon},-\big(1 + \log(u_{\varepsilon})^2\big)\} = -\big(1 + \log(u_{\varepsilon})^2\big)\ \ \text{for all}\ x\in\R^N\backslash\Lambda
$$
if $\varepsilon>0$ is small. This proves that $u_{\varepsilon}$ solves the origin problem \eqref{eq1.1} and complete the proof of this paper.

\vspace{0.5cm}
\appendix
  \renewcommand{\appendixname}{Appendix~\Alph{section}}
  \section{}
In this section, we are going to verify \eqref{Adeq2.5} and \eqref{Adeq3.5}. The fractional Hardy inequality \eqref{Adeq1.4} will be involved in the proof. We first give the proof of \eqref{Adeq2.5}. By Cauchy inequality, we have
\begin{align*}
T_1(\eta)
&\le 2{\varepsilon^{2s}}\int_{\R^N}\mathrm{d}x\int_{\R^N} \frac{|u_n(x)|^2(\eta(x)-\eta(y))^2}{|x-y|^{N+2s}}\mathrm{d}y + C\|u_n\|^2_{V,\varepsilon}\\
&:= T_{11}(\eta) + C\|u_n\|^2_{V,\varepsilon}.
\end{align*}

By decomposition, denoting $f(x,y)=\frac{|u_n(x)|^2(\eta(x)-\eta(y))^2}{|x-y|^{N+2s}}$, we have
\begin{align*}
\frac{1}{2}T_{11}(\eta)
&=\varepsilon^{2s}\int_{\Lambda^{\delta}}\mathrm{d}x\int_{\R^N}f(x,y)\mathrm{d}y +  \varepsilon^{2s}\int_{\R^N\backslash\Lambda^{\delta}}\mathrm{d}x\int_{\Lambda^{\delta}}f(x,y)\mathrm{d}y\\
&:=\sum_{i=1}^2T_{11i}(\eta).
\end{align*}

Since
\begin{align*}
\int_{\R^N}\frac{(\eta(x) - \eta(y))^2}{|x-y|^{N+2s}}\mathrm{d}y
&=\int_{B_1(x)}\frac{(\eta(x) - \eta(y))^2}{|x-y|^{N+2s}}\mathrm{d}y+\int_{\R^N\backslash B_1(x)}\frac{(\eta(x) - \eta(y))^2}{|x-y|^{N+2s}}\mathrm{d}y\\
&\le C\Big(\int_{B_1(0)}\frac{1}{|z|^{N+2s-2}}\mathrm{d}y+\int_{\R^N\backslash B_1(0)}\frac{1}{|z|^{N+2s}}\mathrm{d}y\Big)\\
&\le C,
\end{align*}
we have
\begin{eqnarray}\label{eqA.1}
\begin{split}
T_{111}(\eta)
\le C\varepsilon^{2s}\int_{\Lambda^{\delta}}|u_n(x)|^2
\le C\|u_n\|^2_{V,\varepsilon}.
\end{split}
\end{eqnarray}

For $T_{112}(\eta)$, similar to the estimates above, by fractional Hardy inequality \eqref{Adeq1.4}, we have
\begin{eqnarray}\label{eqA.2}
\begin{split}
T_{112}(\eta)
&\le  C\varepsilon^{2s}\int_{\R^N\backslash\Lambda^{2\delta}}|u_n(x)|^2\mathrm{d}x\int_{\Lambda^{\delta}}\frac{1}{|x-y|^{N+2s}}\mathrm{d}y\\
&\le  C\varepsilon^{2s}\int_{\R^N\backslash\Lambda^{2\delta}}\frac{|u_n(x)|^2}{|x|^{2s}}\mathrm{d}x\int_{\Lambda^{\delta}}\frac{|x|^{2s}}{|x-y|^{N+2s}}\mathrm{d}y\\
&\le C\varepsilon^{2s}\int_{\R^N\backslash\Lambda^{2\delta}}\frac{|u_n(x)|^2}{|x|^{2s}}\mathrm{d}x\int_{\Lambda^{\delta}}\frac{|x|^{2s}}{\big(dist(x,\partial\Lambda^{2\delta}) + \delta + dist(y,\partial\Lambda^{\delta})\big)^{N+2s}}\mathrm{d}y\\
&\le C\|u_n\|^2_{V,\varepsilon}.
\end{split}
\end{eqnarray}

By \eqref{eqA.1} and \eqref{eqA.2}, we conclude that`
$$
T_1(\eta)\le C\|u_n\|^2_{V,\varepsilon}.
$$

Next, we give the proof of \eqref{Adeq3.5}. A change of variable tells us
\begin{eqnarray*}
\begin{split}
T^2_{n,R}& = \int_{\mathbb{R}^N}u_n(y)\,{\mathrm{d}}y\int_{\mathbb{R}^N}\frac{(u_n(x) - u_n(y))(\psi_{n,R}(x) - \psi_{n,R}(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\
 & = \varepsilon^{N - 2s}_n\sum_{l = 1}^k\int_{\mathbb{R}^N}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y\int_{\mathbb{R}^N}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))(\eta_R(x) - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x,
\end{split}
\end{eqnarray*}
where the functions $\beta^l_n$ and $\alpha^l_n$ are defined skillfully as
$$
\beta^l_n(y) = \prod^{l - 1}_{s = 0}\eta\Big(\frac{y}{R} + \frac{x^l_n - x^s_n}{\varepsilon_n R}\Big),\,\,\,\beta^0_n(y)\equiv 1
$$
and
$$
\alpha^l_n(x) = \prod^{k}_{s = l + 1}\eta\Big(\frac{x}{R} + \frac{x^l_n - x^s_n}{\varepsilon_n R}\Big),\,\,\, \alpha^k_n(x) \equiv 1.
$$
Following, we have
\begin{align*}
\varepsilon^{2s - N}_nT^2_{n,R}& = \sum_{l = 1}^k\int_{B_{2R}} v^l_n(y) \beta^l_n(y) \,{\mathrm{d}}y \int_{B^c_{2R}}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))(\eta_R(x) - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x
\\[1mm]
   &\quad + \sum_{l = 1}^k \int_{B^c_{2R}}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y  \int_{B_{2R}}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))(\eta_R(x) - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x
   \\[1mm]
   &\quad + \sum_{l = 1}^k\int_{B_{2R}}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y   \int_{B_{2R}}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))(\eta_R(x) - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
   &:=T^{21}_{n,R} +T^{22}_{n,R} + T^{23}_{n,R}.
\end{align*}
By the choice of $\eta_R$ and  $\lim\limits_{n\to\infty}\frac{|x^l_n - x^s_n|}{\varepsilon_n} = \infty$ if $l\neq s$, for $n$ large, we have
\begin{align*}
T^{21}_{n,R}&= \sum_{l = 1}^k\int_{B_{2R}\backslash B_R}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y\int_{B^c_{2R}}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))(1 - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
       &\quad + \sum_{l = 1}^k\int_{{B_R}}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y\int_{B^c_{2R}}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
       &= \sum_{l = 1}^k\int_{B_{2R}\backslash B_R}v^l_n(y)\,{\mathrm{d}}y\int_{B^c_{2R}}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))(1 - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
       &\quad + \sum_{l = 1}^k\int_{{B_R}}v^l_n(y)\,{\mathrm{d}}y\int_{B^c_{2R}}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
       :&=T^{211}_{n,R} +T^{212}_{n,R}
\end{align*}
and
\begin{align*}
        T^{22}_{n,R}
          =& \sum_{l = 1}^k\int_{B^c_{2R}}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y\int_{B_{2R}\backslash B_R}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))(\eta_R(x) - 1)}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
         & - \sum_{l = 1}^k\int_{B^c_{2R}}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y\int_{B_R}\frac{\alpha^l_n(x)(v^l_n(x) - v^l_n(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x \\[1mm]
         =& \sum_{l = 1}^k\int_{B^c_{2R}}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y\int_{B_{2R}\backslash B_R}\frac{(v^l_n(x) - v^l_n(y))(\eta_R(x) - 1)}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
         & - \sum_{l = 1}^k\int_{B^c_{2R}}v^l_n(y)\beta^l_n(y)\,{\mathrm{d}}y\int_{B_R}\frac{v^l_n(x) - v^l_n(y)}{|x - y|^{N + 2s}}\,{\mathrm{d}}x \\[1mm]
                  :=&T^{221}_{n,R} + T^{222}_{n,R}.
\end{align*}
Also, for large $n$,
\begin{align*}
T^{23}_{n,R} =&\sum_{l = 1}^k\int_{B_{2R}\backslash{B_R}}v^l_n(y)\,{\mathrm{d}}y\int_{B_{2R}}\frac{(v^l_n(x) - v^l_n(y))(\eta_R(x) - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
& + \sum_{l = 1}^k\int_{B_{2R}}v^l_n(y)\,{\mathrm{d}}y\int_{B_{2R}\backslash{B_R}}\frac{(v^l_n(x) - v^l_n(y))(\eta_R(x) - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
&+ \sum_{l = 1}^k\int_{B_{2R}\backslash{B_R}}v^l_n(y)\,{\mathrm{d}}y\int_{B_{2R}\backslash B_{R}}\frac{(v^l_n(x) - v^l_n(y))(\eta_R(x) - \eta_R(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
:=& T^{231}_{n,R} + T^{232}_{n,R} + T^{233}_{n,R}.
\end{align*}

For $|T^{2i2}_{n,R}|,\ i = 1,2$, it holds
\begin{align*}
\nonumber    &\limsup_{n\to\infty}  |T^{2i2}_{n,R}|  \leq CR^{-2s} + \limsup_{n\to\infty}2\sum_{l = 1}^k\int_{B^c_{2R}}\,{\mathrm{d}}y\int_{B_{R}}\frac{ (v^l_n(y))^2}{|x-y|^{N + 2s}}\,{\mathrm{d}}x.

\end{align*}
By the fractional Hardy inequality \eqref{Adeq1.4} and letting $\widetilde{R} = R^{\frac{N + 1}{N}}$, we find
\begin{align}\label{Adeq3.3}
\nonumber&\quad \limsup_{n\to\infty}\int_{B^c_{2R}}\big(v^l_n(y))^2\,{\mathrm{d}}y\int_{B_R}\frac{1}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\\[1mm]
\nonumber&\leq C\limsup_{n\to\infty}\int_{B^c_{2R}}\big(v^l_n(y))^2\frac{R^N}{|y|^{N + 2s}}\,{\mathrm{d}}y\\[1mm]
\nonumber& \le C\limsup_{n\to\infty}\int_{B_{\widetilde{R}}\backslash B_{2R}}\big(v^l_n(y))^2\frac{R^N}{|y|^{N + 2s}}\,{\mathrm{d}}y + C\limsup_{n\to\infty} \int_{B^c_{\widetilde{R}}}\big(v^l
_n(y))^2\frac{R^N}{|y|^{N + 2s}}\,{\mathrm{d}}y\\[1mm]
&\leq C\limsup_{n\to\infty}\int_{B_{\widetilde{R}}\backslash B_{2R}}\big(v^l_n(y))^2\,{\mathrm{d}}y + C \limsup_{n\to\infty}\int_{B^c_{\widetilde{R}}}\frac{(v^l_n(y))^2}{|y|^{2s}}\frac{R^N}{|y|^{N}}\,{\mathrm{d}}y\\[1mm]
\nonumber&\leq  C\int_{B_{\widetilde{R}}\backslash B_{2R}}\big(v^l_*(y))^2\,{\mathrm{d}}y + \frac{C}{R}\\[1mm]
\nonumber& = o_R(1).
\end{align}

Noting that for each $\Omega\subset\subset\R^N$ is smooth, it holds
$$
v^l_n\rightharpoonup v^l_*\ \text{weakly in}\ W^{s,2}(\Omega).
$$
Then, for $T^{211}_{n,R}$, by the estimates of  $T^{2i2}_{n,R}$($i=1,2$), we have

\begin{align*}
&\limsup_{n\to\infty}  |T^{211}_{n,R}| \nonumber
\\[1mm]
&\leq
  \limsup_{n\to\infty}  \sum_{l = 1}^k\int_{B_{2R}\backslash B_R}\,{\mathrm{d}}y\int_{B^c_{2R}}\frac{|v^l_n(x) - v^l_n(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x \nonumber
  \\[1mm]
 &
 + \limsup_{n\to\infty}\sum_{l = 1}^k \int_{B_{2R}\backslash B_R}(v^l_n(y))^2\,{\mathrm{d}}y \int_{B^c_{2R}}\frac{(1 - \eta_{R}(y))^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x \nonumber
  \\[1mm]
 & \leq  \limsup_{n\to\infty}  \sum_{l = 1}^k  \int_{B_{2R}\backslash B_R}\,{\mathrm{d}}y\int_{B^c_{2R}\cap B_{4R}}\frac{|v^l_n(x) - v^l_n(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x\nonumber
  \\[1mm]
 &+\limsup_{n\to\infty}\sum_{l = 1}^k \int_{B_{2R}\backslash B_R}\,{\mathrm{d}}y\int_{B^c_{4R}}\frac{|v^l_n(x) - v^l_n(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x + C\limsup_{n\to\infty}\sum_{l = 1}^k\int_{B_{2R}\backslash B_R}(v^l_n(y))^2\,{\mathrm{d}}y\nonumber
 \\[1mm]
 & \leq  \limsup_{n\to\infty}\sum_{l = 1}^k\int_{B_{4R}\backslash B_R}\,{\mathrm{d}}y\int_{B_{4R}\backslash B_R}\frac{|v^l_n(x) - v^l_n(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x + C\sum_{l = 1}^k\int_{\R^N}(v^*_l(y))^2\mathrm{d}y\nonumber
 \\[1mm]
& \leq C\int_{B_{4R}\backslash B_R}\,{\mathrm{d}}y\int_{R^N}\frac{|v^l_*(x) - v^l_*(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x + o_R(1)\\
&=o_R(1).
\end{align*}
Similarly, we get
\begin{align*}
    \limsup_{n\to\infty}  |T^{221}_{n,R}| &\leq o_R(1)
\end{align*}
and
\begin{align*}
  \limsup_{n\to\infty}|T^{231}_{n,R} + T^{232}_{n,R} + T^{233}_{n,R}|\leq o_R(1).
\end{align*}
Therefore
\begin{equation*}
\varepsilon^{2s - N}T^{2}_{n,R} \leq o_R(1).
\end{equation*}

\vspace{0.5cm}

\vspace{1cm}

config: arxiv


text:
\begin{document}
\maketitle
\abstract{In this paper we provide some sharp asymptotic results for
a stochastic model of species survival recently proposed by Guiol, Marchado, and  Schinazi.}
\section{Introduction and statement of results}
Recently, Guiol, Marchado, and  Schinazi \cite{GMS} proposed a new mathematical framework for modeling
species survival which is closely related to the discrete Bak-Sneppen evolution model. In the original  Bak-Sneppen model \cite{BSM}
a finite number  of species are arranged in a circle, each species being characterized by
its location and a parameter representing the  \emph{fitness} of the species and taking values between zero and one. The number
of species and their location on the circle are fixed and remain unchanged throughout the evolution of the system.
At discrete times $n=0,1,\ldots,$ the species with the
lowest fitness and its two immediate neighbors  update simultaneously  their fitness values at random. The Bak-Sneppen evolution model is
often referred to as an ``ecosystem" because of the local interaction between different species. The distinguishing feature of the
model is the emergence of self-organized criticality \cite{BAKB, evolution, GMW,jensen}  regardless the simplicity and
robustness of the underlying evolution mechanism. The Bak-Sneppen model has attracted significant attention over the past few decades,
but it has also been proven to be difficult for analytical study. See for instance \cite{GMW} for a relatively recent survey of the model.
\par
The asymptotic behavior of the Bak-Sneppen model, as the number of species gets arbitrarily large, was conjectured on the basis of computer
simulations in \cite{BSM}. It appears that the distribution of the fitness is asymptotically uniform on an interval $(f_{\mbox{crit}}, 1)$ for some
critical parameter $f_{\mbox{crit}}$, the value of which is close to $2/3$ \cite{BAKB, jensen}.
\par
Guiol, Marchado, and  Schinazi \cite{GMS} were able to prove a similar
result for a related model with a stochastically growing number of species.
Their analysis is based on a reduction to the study of a certain random walk,
which allows them to build a proof using well-known results
from the theory of random walks. The main result of \cite{GMS} is thus based on
general properties of Markov chains, and  suitable variations of the result can in
principle be carried out to other similar models.
\par
In this paper we  focus on the model introduced in \cite{GMS} as is. Our aim
is to elucidate the underlying mechanism responsible for the phenomenon described in
\cite{GMS} by sharpening the estimates that lead to the major qualitative statement therein.
We proceed with a description of the Guiol, Marchado, and  Schinazi (GMS) model.
\par
In contrast to the Bak-Sneppen model, the number of species in the GMS model is random and changes in time,
and only the species with the lowest fitness is randomly replaced.
The local interaction between species is not considered in the GMS model, and therefore the spatial structure of the population
is of no importance. Let $p>\frac 12$ be given and denote $q=1-p.$ Let $\Z_+$ denote the set of non-negative integers and let
$X=(X_n:n\in \Z_+)$ be a discrete-time birth and death process with the following transition probabilities: from each site, $X_n$ increases by $1$ with probability $p$; from each site different than $0$, $X_n$ decreases by $1$ with probability $q=(1-p)$; finally, at $0$, $X_n$ stays put with probability $q.$  Thus $X$ is a
nearest-neighbor transient random walk on the integer lattice $\Z_+$ with holding times and reflection barrier at zero. Throughout the paper we assume that
$X_0=0$ with probability one.
\par
The model describes the evolution of a population of species: newborn species  are given "equal opportunity" by assigning them a random fitness, and the least fit species are the first to die.  A jump to the right represents birth of a new species, whereas a jump to the left represents death of an existing species. Thus $X_n$ represents the number of {\it species alive} at time $n.$ When a new species is born,  it is assigned a {\it fitness}. The fitness is a uniform $[0,1]$ random variable independent on the fitness of all previously born species as well as of the path of the process $X.$ When $X$ jumps to the left, the species with the lowest fitness is eliminated. We remark
that, in a different context,  a similar model was considered by Liggett and Schinazi in \cite{LS09}.
\par
Fix $f \in (0,1)$. We examine the model by considering two coupled random processes, $L=(L_n:n\in \Z_+)$ (for {\it lower or left})
and $R=(R_n:n\in \Z_+)$ (respectively, for {\it right}),  where $L_n$ denotes the number of species alive at time $n$ whose fitness is less than $f$
while $R_n$ denotes the number of the remaining species alive at time $n$.
\par
Observe that  $L_n$ increases by $1$ if $X_n$ does and the newborn species has fitness less than $f,$ and $L_n$ decreases by $1$ whenever $X_n$ decreases by one and $L_n$ is not zero. The value of $L_n$ remains unchanged when either $X_n$ increases by $1$ and the newborn species has fitness at least $f$ or $X_n$
decreases by $1$ and $L_n=0.$
\par
Notice that when it is not at zero, the process $L$  evolves as a nearest-neighbor random walk with probability $pf$ of jumping to the right, probability $q$ of going
to the left, and probability $1-pf -q $ of staying put. When at zero, $L_n$ jumps to the right with probability  $pf$, and stays put with
the complementary probability $1-pf.$ Thus $L$ is a Markov chain. Since $p>q,$ it is positive recurrent if $pf < q,$ null-recurrent if $pf =q,$ and is otherwise transient. In what follows we will denote the critical value $q/p$ of the parameter $f$ by $f_c.$
\par
Consider the process $B=(B_n:n\in \Z_+),$ where $B_n$ is the total number of species born by time $n$ with fitness at least $f$. Observe that $B$ is a non-decreasing Markov chain (formed by sums of i.i.d. Bernulli variables), which jumps one step up with probabilities $p(1-f)=p-q$ and stays put with the complementary probability $2q.$ It is shown in \cite{GMS} that when $f=f_c,$ we have that for any $\veps >0$
\beqn
\label{icor}
0 \le \limsup\limits_{n\to\infty} \frac{B_n - R_n}{ n^{1/2+\veps} } \le \frac{2}{q}, \qquad \as,
\feqn
while each species with fitness less than the critical value disappears after a finite (random) time. It follows from \eqref{icor} that
the distribution of species with fitness higher than $f_c$ approaches a uniform law.
\par
We sharpen the above result by proving the following theorem. Recall that $f_c=q/p.$
\begin{theorem}
\label{th:limit}
Suppose that $f=f_c.$ Then
\begin{enumerate}
\item
$\displaystyle \limsup_{n\to\infty} \frac{B_n -R_n}{\sqrt{4q n \ln \ln n}} = 1,\qquad \mbox{\rm a.s.}$
\item
$\displaystyle  \frac{B_n -R_n}{\sqrt{2qn}} \Rightarrow|N(0,1)|,$ where $N(0,1)$ denotes
a mean-zero Gaussian random variable with variance one, and $ \Rightarrow$ stands for convergence in distribution.
\item $(L_n:n\in \N)$ is a recurrent Markov chain (visiting zero infinitely often) and hence each species with fitness less than the critical
value becomes eventually extinct with probability one.
\end{enumerate}
\end{theorem}
The proof of the theorem is included in Section~\ref{proofs}.  Notice that by the law of large numbers $B_n\sim n(1-f_c),$ $\as,$ as $n\to \infty.$ Moreover, the fitness of any randomly chosen species from $B_n$ is distributed uniformly on $(f_c,1).$ Hence the theorem implies that
this uniform law on $(f_c,1)$ is the limiting distribution for the fitness of a randomly chosen species being alive at time $n$ and having fitness larger than $f_c.$
\section{Proof of Theorem~\ref{th:limit}}
\label{proofs}
There is no loss of generality assuming that $X$ is obtained recursively from an i.i.d. sequence of Bernoulli random variables $J=(J_n:n\in \Z_+)$ with
$P(J_n=1)=q,$ $P(J_n=0)=p,$ as follows: $X_0=0$ and for $n\in \Z_+$ we have
\beqn
\label{eq:Xdefn}
X_{n+1} =X_n +  (1-J_n) - J_n (1-s_n),\mbox{ where } s_n:= \ch{X_n=0}.
\feqn
Let $G=(G_n:n\in\Z_+)$ be a Markov chain formed by the pairs $G_n = (L_n,J_n).$ Figure \ref{fig:Gprob} illustrates the transition mechanism of
$G.$
\begin{figure}[t]
\includegraphics[width=\textwidth]{surv_anal.pdf}
\caption{Transition probabilities for $G$}
\label{fig:Gprob}
\end{figure}
\subsection{Reduction from $B_n-R_n$ to an occupation time of $G$}
Let $\Delta=(\Delta_n:n\in\Z_+)$ be the process defined through  $\Delta_n= B_n -R_n.$
Notice that $\Delta_n$ increases by $1$ if  and only if  $R_n$ decreases by $1,$ and otherwise stays put. That is
\beq
\Delta_{n+1}-\Delta_n = \ch{X_{n+1}-X_n=-1}  \ch{L_n=0} .
\feq
Thus by \eqref{eq:Xdefn}, $\Delta_{n+1} -\Delta_n = (1-s_n) J_n  \ch{L_n=0} $, and we have
\beq
\Delta_n = \sum_{i=0}^{n-1} (1-s_i) J_i \ch{L_i=0} = \eta_n - \sum_{i=0}^{n-1} J_i s_i,\mbox{ where }  \eta_n:= \sum_{i=0}^{n-1} J_i \ch{L_i=0}.
\feq
Observe that $\eta_n$ is the occupation time (number of visits) of $G$ at $(0,1)$ up to time $n-1$ and that $\sum_{i=0}^{n-1} s_i J_i$
is bounded above by $\sum_{i=0}^\infty s_i.$ Since $X$ is transient and $X_0=0,$ the latter series can be represented
in the form $\sum_{i=0}^\infty s_i=\sum_{i=1}^g h_i,$
where $g$ is the total number of visits to zero (distributed geometrically, according to $P(g=k)=f_c^{k-1}(1-f_c),$ $k\in\N,$
see for instance \cite[p.~274]{durrett}) and $h_i$ are successive holding times of $X$
at zero (i.i.d. sequence of geometric random variables,
independent of $g,$ with $P(h_i=k)=q^{k-1}p,$ $k\in\N$). In particular,
\beq
E\Bigl(\sum_{i=0}^\infty s_i J_i\Bigr)\leq E(g)\cdot E(h_1)=\frac{1}{p(1-f_c)}=\frac{1}{2p-1}<\infty.
\feq
Since $L$ and consequently $G$ are recurrent (and thus
$\Delta_n$ is a non-decreasing sequence converging to $+\infty$ with probability one), it therefore suffices to show that Theorem~\ref{th:limit} holds with
$\Delta_n=B_n-R_n$ replaced by $\eta_n$ in its statement.
\paragraph{Excursion decomposition for the path of $L.$}
We decompose the path of $L$ into a sequence of successive excursions from $0,$
each one lasting until (but not including) the first time $L$ returns to $0$ from $1.$
Observe that once the Markov chain $L$ is at zero, it will stay put until $J_k=0$ and the fitness of the newborn particle
is less than $f_c.$ Therefore, the holding time of $L$ at zero during one excursion is a geometric random variable with parameter
$pf_c=q.$ Let $\tilde h_k$ be the i.i.d. sequence of holding times at zero during successive excursions of $L$ from zero.  That is,
$P(\tilde h_k=n)=p^{n-1}q,$ $n \in \N.$ In what follows we will use the notation $\mbox{Geom}(a)$ for the geometric distribution
with parameter $a$ (for instance, we could write $\tilde h_k \sim \mbox{Geom}(q)$). Let
$\sigma_k$ be the $k$-th excursion starting time, that is $\sigma_1=0$ and
\beq
\sigma_k=\inf\{n>\sigma_{k-1}: X_{n-1}=1, X_n=0\},\quad k=2,3,\ldots
\feq
Define
\beq
\mu_k=\sum_{i=\sigma_k}^{\sigma_{k+1}-1} J_i\ch{L_i=0}=\sum_{i=\sigma_k}^{\sigma_k+\tilde h_k-1} J_i\ch{L_i=0}.
\feq
Notice that $(\mu_k:k\in \N)$ is an i.i.d. sequence and
\beqn
\label{law-ln}
\sum_{k=1}^{N_n} \mu_k < \eta_n  \le \sum_{k=1}^{N_n+1} \mu_k .
\feqn
We now compute $\mu:=E(\mu_k)$ using the fact that the sequence of pairs $G_n=(J_n,L_n)$ forms
a Markov chain, the transition mechanism of which is illustrated in Fig.~\ref{fig:Gprob}.  The value of
$\mu$ is equal to the expected number of visits by this Markov chain to the state $(0,1)$ during the period of time
starting at the state $(0,1)$ with probability $q$ and at $(0,0)$ with probability $p,$
and lasting until $G$ leaves the set $\{(0,0), (0,1)\}.$ We thus have, using first step analysis,
\beq
\mu=q(1+\mu)+p(1-f_c)\mu+pf_c\cdot 0=q+\mu(1-pf_c)=q+\mu p.
\feq
Hence $\mu=1.$ Consequently, using \eqref{law-ln} and the law of large numbers, we obtain
\beqn
\label{eq:etan}
\eta_n \sim N_n ~\mbox{as}~n\to \infty,\quad \as
\feqn
\paragraph{Reduction to a simple random walk}
The duration of the $k$-th excursion of $L$ from zero is equal to the sum $\tilde h_k+\alpha_k$ of the holding time at zero $\tilde h_k$ distributed as
$\mbox{Geom}(q),$ and the time $\alpha_k$ that it takes for $L$ to return from 1 to zero. Notice that
\beq
\alpha_k=\sigma_k-(\sigma_{k-1}+\tilde h_k-1).
\feq
For $m\in\N$ let $V_m = \sum_{k=1}^m (\tilde h_k + \alpha_k)$ be the total duration of the first $m$ excursions of $L$ from $0.$
Then $N_k=\max\{m:V_m\le k\}.$ With each excursion we can associate a \emph{skeleton}, which is the path obtained from the excursion
by omitting all transitions from a state to itself, for all states. If we let $\tau_k$ denote the length of the skeleton, then due to the choice of $f_c$ it follows that $\tau_k$ has the same distribution as the time required for the simple symmetric random walk on $\Z$ to get back to $0$ starting from $0.$ Furthermore, the time spent by $L$ at each visit to a site is a geometric random variable, $\mbox{Geom}(2q)$ for sites different than $0$ and $\mbox{Geom}(q)$ for $0.$ Therefore, the length of a single excursion itself is a sum of one
$\mbox{Geom}(q)$ random variable plus a sum of $\tau_k-1$ independent
$\mbox{Geom}(2q)$ random variables. If we replace $\tilde h_k$ with a $\mbox{Geom}(2q)$, then the resulting modified ``excursion time" becomes
a sum of $\tau_k $ copies of a $\mbox{Geom}(2q)$ random variable. Let $V_m'$ denote the total length of the first $m$ excursions modified in this way.
Let $N_k'$ denote the number of such excursions occurred by time $k$, that is
$N_k'= \max\{m:V_m' \le k\}.$ Then $V_m$ stochastically dominates $V_m'$. By the law of large numbers,
\beqn
\label{eq:lln}
\lim_{m\to\infty}  \frac{ V_m - V_m'}{m}  = E (\tilde h_1) - E (h_1') = \frac{1}{2q},
\feqn
where $h_1'$ is a geometric random variable with parameter $2q.$  Letting $T_m = \sum_{k=1}^m \tau_k,$ we obtain
\beq
V_m' = \sum_{k=1}^{T_m} h_k'',
\feq
where $(h_k'':k\in\N)$ is an i.i.d. sequence of random variables, each one distributed as $\mbox{Geom}(2q).$ Thus, by the law of large numbers,
\beqn
\label{eq:Vl'}
V_m' \sim \frac{T_m}{2q}~\mbox{as}~m\to\infty, \quad \as
\feqn
Notice that $T_m$ is distributed the same as the total length of the first $m$ excursions from zero of a simple symmetric random walk.
\subsection{Completion of the proof: CLT and LIL for $\eta_n$}
\paragraph{LIL for $\eta_n$}
We need the following result. Although the claim is a ``folk fact", we give a short proof
for the sake of completeness.
\begin{lem}
\label{lem:T}
\beq
\liminf_{m\to\infty} \frac{T_m}{m^2 /(2 \ln \ln m)}= 1 \mbox{ \rm a.s. }
\feq
\end{lem}
\begin{proof}[Proof of Lemma \ref{lem:T}]
Let $S=(S_n:n\in\Z_+)$ denote the simple symmetric random walk on $\Z$. Let $\gamma_0=0$ and define inductively
$\gamma_{m+1}= \inf \{k> \gamma_m: S_k=m+1\}$. Let $Y_m= \gamma_{m+1}-\gamma_m$ with the usual convention that
the infimum over an empty set is $+\infty.$ Let $\phi(x) = \sqrt{2x \ln \ln x}$ for $x>0.$
By the law of iterated logarithm for $S,$
\beq
\limsup_{n\to\infty} \frac{S_n}{\phi(n) } = \limsup_{n\to\infty} \frac{S_{\gamma_n}}{\phi(\gamma_n)}=\limsup_{n\to\infty} \frac{n}{\phi(\gamma_n)}.
\feq
Since $\phi^{-1}(x)\sim x^2/(2 \ln \ln x)$ as $x \to\infty,$ we obtain
\beq
\liminf_{n\to\infty} \frac{\gamma_n}{n^2 / (2\ln \ln n)} =1.
\feq
Observe that $\gamma_n=\sum_{i=0}^{n-1} Y_i$. The distribution of $t_1$, the time to return to $0$ starting from $0$ for $S$, is equal to the distribution
of $t_1'$, the time to return to $0$ starting from $0$ for the reflected random walk $|S|.$ Since $t_1'$ is equal in distribution to $1+Y_1,$
\beq
\liminf_{m\to\infty} \frac{T_m}{m^2 /(2 \ln \ln m)}= \liminf_{n\to\infty}  \frac{\gamma_n+n}{n^2 / (2\ln \ln n)} =1,
\feq
completing the proof of the lemma.
\end{proof}
Using the lemma along with \eqref{eq:lln} and \eqref{eq:Vl'}, we obtain
\beq
\liminf_{m\to\infty} \frac{2q  V_m}{m^2 /(2\ln \ln m)}=\liminf_{m\to\infty} \frac{2q V_m'}{m^2/(2\ln \ln m)}=1, \quad \as
\feq
Consequently, since $N$ is the inverse of sequence of $V$, we obtain
\beq
\limsup_{k\to\infty} \frac{N_k}{\sqrt{4q k \ln \ln k}}=1, \quad \as
\feq
Combining this with \eqref{eq:etan} completes the proof of the law of iterated logarithm for $\eta_n.$
\paragraph{CLT for $\eta_n$}
We now turn to the proof of the central limit theorem. It is well known (see for instance \cite[p.~394]{durrett}) that
\beq
\lim_{m\to\infty} E \bigl(e^{-\theta T_m/m^2}\bigr) =E \bigl(e^{-\sqrt{2\theta}}\bigr),\qquad \theta \geq 0.
\feq
Therefore, it follows from \eqref{eq:lln} and \eqref{eq:Vl'} that
\beq
\lim_{m\to\infty} E \bigl(e^{-\theta V_m/m^2}\bigr) = \lim_{m\to\infty} E (e^{-\theta V_m'/m^2}\bigr) =
E \bigl(e^{-\sqrt{\theta/q}}\bigr).
\feq
The function $\theta \to e^{-c \sqrt{2  \theta}},$  $\theta \geq 0,$ is the Laplace transform of a positive stable law with index $1/2$
whose density function is given by (see for instance \cite[p.~395]{durrett})
\beq
\varphi_c(u) = \ch{u \geq 0} \frac{c e^{-c^2/2u} }{\sqrt{2\pi u^3}}.
\feq
We intend to use this formula with $c=\frac{1}{\sqrt{2q}}.$ Set $V_0=0$ and observe that for all $k\in \N,$
\beq
P(N_k \le u  ) = P_0 (V_{\lfloor u\rfloor} \ge  k).
\feq
Fix $s>0$ and let $u=\sqrt{k}s$. Then
\beq
P(N_k \le \sqrt{k}s) =P(V_{\lfloor \sqrt{k}s \rfloor}\ge k ) =
P\Bigl (\frac{V_{\lfloor \sqrt{k}s \rfloor}}{(\lfloor \sqrt{k} s\rfloor)^2} \ge \frac{1}{s^2}(1+o(1)) \Bigr) \underset{k\to\infty}{\to}
\int_{s^{-2}}^\infty \varphi_c (u) du.
\feq
Differentiating the right-hand side we obtain
\beq
\lim_{k\to\infty}  P(N_k \le \sqrt{k}s) =  \int_0^s \frac{2 e^{- u^2/(2c^{-2})}}{\sqrt{2\pi c^{-2}}}du.
\feq
Therefore $N_k /\sqrt{k}$ converges weakly to the absolute value of a centered normal random variable with variance equal to $c^{-2} = 2q$. Combining this with \eqref{eq:etan} completes the proof of the central limit theorem for $\eta_n.$ \qed


TITLE: Can the domain and co-domain be the same set? Is this a function?
QUESTION [0 upvotes]: Let $A$ denote the set of all real numbers. Let $B$ denote the same set as $A$. Let $f$ be the function that, to each number in $A$ assigns the cube of the number. Is $f$ a function?

REPLY [1 votes]: Yes, there is nothing at all against having a function with the same set as domain and codomain. There are lots of functions $\mathbb R\to\mathbb R$, among these $x\mapsto x^2$ and $x\mapsto x^3$.
(Note that there is no requirement either that every element of the codomain has to be hit by the function. So $x\mapsto x^2$ describes a perfectly good function $\mathbb R\to\mathbb R$. If if in fact every point in the codomain is hit -- such as for $x\mapsto x^3$, we say that the function is surjective. Some but not all functions are surjective).


TITLE: In a noncommutative ring, is there always a pair $x,y$ such that $xy-yx=1$?
QUESTION [4 upvotes]: Let $R$ be a non-commutative ring. Are there two element $x,y\in R$ such that $xy-yx=1_{R}$?
I have proved it is true for $R$ being an algebra with finite dimension.
Sorry, I made a mistake, should be not true for $R$ being a finite dimension  algebra over a field with char $0$.

REPLY [2 votes]: In unital Banach algebras, the unit is not a commutator.  This is mentioned without proof on the Wikipedia page for Banach algebras.  The proof I know relies on spectral theory: since $z\mapsto 1+z$ is entire, the spectral mapping theorem gives $\sigma(1+ba) = 1+\sigma(ba)$; but $\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}$, so if $ab=1+ba$ then it follows easily that $\sigma(ab)$ is unbounded, which is impossible.


TITLE: What is this strange 3D rotations expression?
QUESTION [1 upvotes]: I have found a 3D rotations expression that I don't know, and have to convert to a simple 3 x rotations sequence or matrix...
Here is a record of some rotations sequences ('Z30' means 'rotate 30° around Oz') :
// base
Ox: {x: 1, y: 0, z: 0}
Oy: {x: 0, y: 1, z: 0}
move: {x: 6.42995404746696, y: -500, z: -268.028464077285}

// base X45
Ox: {x: 1, y: 0, z: 0}
Oy: {x: 0, y: -0.707106781186548, z: 0.707106781186547}
move: {x: 6.42995404746696, y: -73.2233047033605, z: -444.805159373921}

// base Y60
Ox: {x: 0.500000000000001, y: 0, z: 0.866025403784438}
Oy: {x: 0, y: 1, z: 0}
move: {x: 6.42995404746766, y: -500, z: -268.028464077285}

// base Z30
Ox: {x: 0.866025403784439, y: 0.5, z: 0}
Oy: {x: 0.5, y: -0.866025403784439, z: 0}
move: {x: -118.570045952533, y: -33.4936490538881, z: -268.028464077284}

// base Z30 X45
Ox: {x: 0.866025403784439, y: 0.353553390593274, z: -0.353553390593273}
Oy: {x: 0.5, y: -0.612372435695795, z: 0.612372435695794}
move: {x: -118.570045952533, y: -96.9068910760492, z: -421.121573001233}

// base Z30 X45 Y60
Ox: {x: 0.739198919740117, y: 0.353553390593274, z: 0.573223304703363}
Oy: {x: 0.28033008588991, y: 0.612372435695795, z: -0.739198919740117}
move: {x: -63.6525674250102, y: -403.093108923947, z: -83.228734142255}

The names 'Ox' and 'Oy' are arbitrary, and the datas contain a third 'Oz' vector wich is always {0 0 1}.
Can someone tell what formalism it is and how it works ?

REPLY [1 votes]: The names $O_x$ and $O_y$ are not arbitrary, although "O" is an odd choice.
What's being given, in each case, is the result of transforming the standard basis vectors,
$$
e_1 = \pmatrix{1\\0\\0}, e_2 = \pmatrix{0\\1\\0}, e_3 = \pmatrix{0\\0\\1}
$$
by the specified rotation. For the fourth one, for instance, $Z30$ refers to rotating about $Z$ by 30 degrees. This transforms $e_1$ to
$$
\pmatrix{\cos 30\\ \sin 30 \\ 0 } \approx \pmatrix{0.866\\ 0.5 \\ 0}
$$
The thing labelled $O_y$ in each case is the result of transforming $e_2$ by the given transformation. What about transforming $e_3$? It's just the cross product $T(e_1) \times T(e_2)$, which the author seems to not have written down.
The good news is that the matrix for this transformation is easy to produce. It's
$$
\pmatrix{ 0.866 & -0.5 & * \\ 0.5 & 0.866 & * \\ 0 & 0 & *}
$$
where the first two columns are filled in with the values listed as $O_x$ and $O_y$, and the third column must be computed as the cross-product of these two; in this case, that's easy: you get
$$
\pmatrix{ 0.866 & -0.5 & 0 \\ 0.5 & 0.866 & 0 \\ 0 & 0 & 1}.
$$
Chances are good that the last item, "move", represents a translation. If you consider the point $(x, y, z)$ in 3-space to be represented by a four-tuple $(x, y, z, 1)$, then you can apply both translation and rotation in a single matrix form by writing a 4-by-4 matrix. The upper left 3-by-3 is the thing I just showed you; the last column contains the "move" vector, and the bottom right entry is a "1", so you get (for the "Z30" example), approximately
$$
\pmatrix{ 0.866 & -0.5 & 0 & -118 \\ 0.5 & 0.866 & 0 & -22 \\ 0 & 0 & 1 & -268 \\ 0 & 0 & 0 & 1}.
$$
To get the matrix for the entire sequence of xforms, you need to compute the matrix for each, and then multiple them all; whether you multiply them left-to-right or right-to-left isn't clear from the context you've given, but if I had to guess, I'll bet on right-to-left (i.e., you write down the first matrix; then the second one to the LEFT of it; then the third one to the LEFT of both, and so on, and then compute the entire product).
Given the actual rotation sequences, you might be better off writing them as quaternions and explicitly computing the product, and then converting to a matrix using something like Rodrigues's formula. I mean "better" here in the sense of "less chance for round-off error or typographical mistakes". But maybe that approach is better only for someone familiar with quaternions --- a beginner with those might be just as likely to make mistakes with THEM as with the cosines and sines here.


TITLE: $\mathbb Q$-metric spaces: how much is lost?
QUESTION [1 upvotes]: I want to know how weak (literally "punctured") the theory of metric spaces becomes if we impose the condition that the distance function can assume only rational (nonnegative) values.

REPLY [5 votes]: Clearly a $\Bbb Q$-metrizable space must be zero-dimensional. The space ${}^\omega\omega$ of irrationals is universal for separable zero-dimensional metrizable spaces and is $\Bbb Q$-metrizable, so every separable zero-dimensional metrizable space is $\Bbb Q$-metrizable. To define such a metric on ${}^\omega\omega$, for distinct $x=\langle x_n:n\in\omega\rangle$ and $y=\langle y_n:n\in\omega\rangle$ in ${}^\omega\omega$ let $\delta(x,y)=\min\{n\in\omega:x_n\ne y_n\}$ and $d(x,y)=2^{-\delta(x,y)}$.
More generally, let $D_\kappa$ be the discrete space of cardinality $\kappa$. Then ${}^\omega D_\kappa$ with the product topology is universal for strongly zero-dimensional metrizable spaces of weight at most $\kappa$, and a $\Bbb Q$-metric can be defined on it in the same way as for ${}^\omega\omega$. (These are in fact non-Archimedean metrics, also known as ultrametrics.)
There are (necessarily non-separable) zero-dimensional metric spaces that are not strongly zero-dimensional; I don’t know whether they admit $\Bbb Q$-metrics.


TITLE: characteristic functions - Probability
QUESTION [3 upvotes]: How to find characteristic function of real-valued random variable if I know that its distribution function is singular but isn't discrete? For example, how to find characteristic function of random variable with distribution function:
$F(x)=\left\{\begin{matrix}
0&,x<-3
\\ 1/3&,-3\leq x<2
\\ 1&,2\leq x
\end{matrix}\right.$

REPLY [1 votes]: The characteristic function is given by
$$
\varphi_X(t)=\operatorname Ee^{itX}.
$$
In this case,
$$
\varphi_X(t)=\frac13e^{-3it}+\frac23e^{2it}
$$
since the random variable $X$ corresponding to the distribution function $F$ takes two values $-3$ and $2$ with probabilities $\Pr\{X=-3\}=\frac13$ and $\Pr\{X=2\}=\frac23$.


TITLE: Prove that A₁D, B₁E, C₁F intersect at the same point
QUESTION [2 upvotes]: Let ABC be a triangle.AA₁, BB₁, CC₁ are the angle bisectors of the triangle. ω is circumcircle of ABC. ω∩AA₁=A₂, ω∩BB₁=B₂, ω∩CC₁=C₂. Circumcircles of AB₁B₂, BC₁C₂, CA₁A₂ intersect with AB,BC and CA at D,E,F respectively. Prove that A₁D, B₁E, C₁F intersect at the same point.
My try:
After some angle chasing I have found out that C₂E=BC₂, A₂C=A₂F, B₂A=B₂D. And after considering 3 equal triangles. I found out that AB=AF, BD=BC, CA=CE. But after that I haven't managed to get anything. I am suspecting we might use Pascal's theorem since there are 6 points on circle ω.

REPLY [1 votes]: Here is an other answer, based on my already given answer using barycentric coordinates to get the location of the intersection point. Well, we already have an accepted answer, so i will make it short.
Let $a,b,c$ be the sides of the given triangle.
From the OP we know that $D$ is placed on $BA$ so that $\Delta BCD$ is isosceles, i.e. $BD=BC=a$. We construct $A_3$ on $BC$ so that $C_1A_3\| AC$. And similarly $B_3$, $C_3$. Let $U=BB_3\cap CD$. Picture so far:

As constructed, $AA_3$, $BB_3$, $CC_3$ are concurrent in a point $X$, reciprocal of the theorem of Ceva:
$$
\frac{A_3B}{A_3C}\cdot
\frac{B_3C}{B_3A}\cdot
\frac{C_3A}{C_3B}
=
-
\frac ab\cdot
\frac bc\cdot
\frac ca\cdot
=-1\ .
$$
Let us show that $DA_1$ is also passing through $X$.

Menelaus for $\Delta ADC$ w.r.t. the transverse line $BB_3U$ gives
$$
1 =
\frac{BD}{BA}\cdot
\frac{B_3A}{B_3C}\cdot
\frac{UC}{UD}
=
-
\frac{a}{c}\cdot
\frac{c}{b}\cdot
\frac{UC}{UD}\ .
\qquad\text{ So }
\frac{UC}{UD} = -\frac ba\ .
$$
We need the position of $C_3$ on $BD$. From $\frac{C_3A}{C_3B}=\frac{B_1A}{B_1C}=\frac{BA}{BC}=\frac ca$ we have $C_3A=c^2/(a+c)$, $C_3B=ac/(a+c)$. This gives $C_3D=C_3A+AD=C_3A+(a-c)=a^2/(a+c)$.
We are now in position to apply the reciprocal of Ceva in $\Delta DBC$ for the points $A_1$, $U$, $C_3$, so we compute:
$$
\frac{A_1B}{A_1C}\cdot
\frac{UC}{UD}\cdot
\frac{C_3D}{C_3B}
=
-
\frac cb\cdot
\frac ba\cdot
\frac {a^2/(a+c)}{ac/(a+c)}
=-1
\ .
$$
So $A_1D$ passes through $BU\cap CC3=X$.
This shows the concurrence of the six lines $AA_3$, $BB_3$, $CC_3$; $A_1D$, $B_1E$, $C_1F$.
$\square$


TITLE: Expected value of two successive heads or tails (stuck on computation)
QUESTION [4 upvotes]: Problem:
This is problem 33 of section 2.6 (conditioning) in Bertsekas, Tsitsiklis intro to probability text.
We are given that a coin has probability of heads equal to p and tails equal to q and it is tossed successively and independently until a head comes twice in a row or a tail comes twice in a row. What isthe expected value of the number of tosses?
Solution:
Let $X$ be a random value that tracks the number of tosses. In addition, let $H_k, T_k$ be the event that a head or tail comes up on the kth flip respectively. The solution I am looking at involves using total expectation theorem.
I outline the following steps in the solution.
We can partition the sample space into $H_1$ and $T_1$, allowing us to invoke the total expectation theorem. Hence,
$E[X] = pE[X|H_1] + qE[X|T_1]$
Again by total expectation theorem, we can utilize another partition to obtain
$E[X|H_1] = pE[X|H_1,H_2] + qE[X|H_1,T_2] = 2p + q(1 + E[X|T_1])$
$E[X|T_1] = qE[X|T_1,T_2] + pE[X|T_1,H_2] = 2q + p(1+E[X|H_1])$
Stuck:
The solution follows up by saying that we can combine the above two relations and use the fact that p+q = 1 to obtain
$E[X|T_1] = \frac{2+p^2}{1-pq} , E[X|H_1] = \frac{2+q^2}{1-pq}$
I have been staring at this for a while now, but I cannot seem to see the steps of the computation. I'd be very grateful if someone could fill in the details regarding the computation. Thank you.

REPLY [0 votes]: An alternative solution to this problem that avoids using recursion and simultaneous equations is to split all possible terminating sequences into four blocks.
Block 1: Start Heads, ends 2 Heads
Block 2: Start Heads, ends 2 Tails
Block 3: Start Tails, ends 2 Tails
Block 4: Start Tails, end 2 Heads

pp
pqpp
pqpqpp

pqq
pqpqq
pqpqpqq

The other two blocks  are the same except it starts with q/tails so qq, qpqq ...
The expectation is the sum of the expectation of each block and so E[X] can the be given by the following sums
$$ E[X] = \sum_{x=1}^{\infty}2x\ p^{x+1}q^{x-1} + \sum_{x=1}^{\infty} (2x+1)p^{x}q^{x+1} +\sum_{x=1}^{\infty}2x\ q^{x+1}p^{x-1} + \sum_{x=1}^{\infty} (2x+1)q^{x}p^{x+1}
$$
The last two sums are basically the first two sums with p and q reversed. You can compute the infinity sums using the following two formulas.
$$\sum_{x=1}^\infty x^k=\frac{x}{x-1}$$
$$\sum_{x=1}^\infty k x^k=\frac{x}{(x-1)^2}$$
Honestly though I cheated and used mathematica to get the following formula. Given $p + q = 1$
$$E[X]=-1 + \frac{3}{1 + (-1 + p) p}$$


TITLE: Simplest Possible Closed Figure in an N-Dimensional Space?
QUESTION [4 upvotes]: The students in my physics class were playing with Rubik's Cubes this morning before class. This got us talking about solids. The traditional Rubik's Cube is a six-sided closed solid in a three dimensional space. Another student had a Rubik's Dodecahedron, which got us talking about the "simplest" possible Rubik's Solid. We decided that what we meant by simplest was "possessing the smallest number of sides". We settled on a tetrahedron (which apparently does exist in Rubik's form).
Then we got to talking about the simplest possible closed figure in spaces of varying dimension. It seems like the simplest possible closed figure composed of straight sides in a two dimensional space has three sides, a triangle. Similarly, the simplest possible closed figure composed of straight sides in a one dimensional space would be a line segment, with two sides (is this a stretch?).
Then class began.
Intuitively, it seems like the simplest possible closed figure whose sides are all straight in an $n$ dimensional space will have $n+1$ sides.
Is that intuition correct?
My formal training is in the Classics, so I'm not sure whether I've asked the question properly. What branch of mathematics thinks about questions like that, and is there a good introductory text for it?

REPLY [3 votes]: An entertaining introduction to thinking about shapes in different dimensions is Edwin A. Abbott's Flatland: A Romance of Many Dimensions (1884).
The generalization to higher dimensions of the triangle and the tetrahedron is the simplex.  Your intuition is correct; in $n$ dimensions, the $n$-simplex has $n+1$ "faces" that are $(n-1)$-simplexes themselves.  More properly, these are its $n$-faces.  Each $n$-face is the convex hull of $n$ of the $n+1$ vertices.  There are also $m$-faces for $2 \le m < n$, which are defined similarly, and which generalize the edges of the tetrahedron (i.e., they are faces of the faces, or faces of the faces of the faces, etc.)


TITLE: Prove that set of all lines in the plane is uncountable.
QUESTION [0 upvotes]: Let $L$ be the set of all lines in the plane. Prove that $L$ is uncountable, but only countably many of the lines in $L$ contain more than one rational point.

Attempt: Well, I was trying to define $L$ using linear combinations of points since a line is a linear combo of two points. So, I wanted to define $L=\{ax+yb:x+y+z=k,k∈Z_+\}$. But, this does not seem right. Anyway, once I define $L$ I would try to find a function from $L$ to some uncountable set D that is onto or a function from D to L that is 1-1. Help appreciated thank you.

REPLY [2 votes]: You can List all lines in the plane this way:
Use the fact that a line is described uniquely once you know its slope, and one of its
intercepts with the axes. The slope is indexed by the Reals, and so is the intercept, say the x-intercept. So we count all the possible pairs ( slope, x-intercept), and show
it is equal to $|\mathbb R|$
1) Consider all lines thru the origin $(0,0)$.  These are described uniquely by their
slope, and there are $|\mathbb R|$ of them, since the slope is parametrized by the Reals.
2) From 1) , we can cover all other cases of lines not going thru $(0,0)$ , by considering all possible ( say x-) intercepts of a line thru any point, with fixed slope$m$. For every line
in 1), there are $\mathbb R$ lines not going thru the origin, but with the same slope.
This means there are $|\mathbb R|\times |\mathbb R|=|\mathbb R|$ total lines in the plane.


TITLE: Intersection area of concentric ellipses
QUESTION [1 upvotes]: I need to find the area of intersection of two concentric ellipses. I imagine the concentricity should make this simpler than the general case, but the ellipses could be rotated.
Alternatively, if there is no simple solution to this. I could live with the area of parallelogram that is formed by the intersections points (when there are 4).
I don't have background in math or geometry, so any pointers is appreciated.
Intersecting ellipses:

REPLY [1 votes]: This is a partial answer.
I would say the hardest part is likely finding the points of intersection. In this post I discuss the generic approach, which involves solving a quadratic equation of at least degree $3$. However, since your ellipses are both centered around the origin, a quadratic equation will suffice.
Rewrite your ellipses in the following form (by expanding the numerator of the form you say you have):
$$a_1x^2+b_1y^2+c_1xy=d_1\qquad a_2x^2+b_2y^2+c_2cy=d_2$$
Then you can combine them: take $d_2$ times the first equation minus $d_1$ times the second and you get
$$(a_1d_2-a_2d_1)x^2+(b_1d_2-b_2d_1)y^2+(c_1d_2-c_2d_1)xy=0$$
This is the equation of a pair of lines passing through the origin and through the points of intersection. Next compute
$$t = \sqrt{(c_1d_2-c_2d_1)^2-4(a_1d_2-a_2d_1)(b_1d_2-b_2d_1)}$$
The lines are now described by one of these equations:
\begin{align*}
2(a_1d_2-a_2d_1)x + (c_1d_2-c_2d_1\pm t)y &= 0 \\
(c_1d_2-c_2d_1\pm t)x + 2(b_1d_2-b_2d_1)y &= 0
\end{align*}
Generally just picking one of these two equations is enough, since with the $\pm$ in it it will already give you both the lines. If you are unlucky, though, and one of the lines coincides with one of the coordinate axes, then either $(a_1d_2-a_2d_1)$ or $(b_1d_2-b_2d_1)$ will be zero. Using the equation with the zero there will cause one of the equations to become $0x+0y=0$ for one of the cases of the $\pm$. In those cases better use the equation where there is no zero.
Now you have the equation of a line, and you can intersect that with the ellipse. And then you have the points of intersection, and can use these for angle computation. I will not include all details for that in my answer, but I invite other answers to cover these parts.


TITLE: $a\cdot b =|a||b| \cos\gamma \quad \cos\gamma=\cos(\varphi - \varphi ') \sin\theta \sin \theta '+\cos\theta \cos \theta'$
QUESTION [0 upvotes]: see Spherical law of cosines
For two $a,b$ vectors in 3-dimensional  Euclidean space. if $\theta$,$\varphi$ and $\theta '$,$\varphi '$ are angular coordinates of $a$ and $b$ respectively.
$$a\cdot b =|a||b| \cos\gamma$$
Simplest way to show $\cos\gamma=\cos(\varphi - \varphi ') \sin\theta \sin \theta '+\cos\theta \cos \theta'$ $\quad$ without using tangents?

REPLY [1 votes]: $cos(\varphi-\varphi ')=cos\varphi cos\varphi'+sin\varphi sin\varphi'$
Use unit vectors $a,b$ i.e $|a|=1,|b|=1$
$a=(x,y,z)=(sin\theta cos\varphi,sin\theta sin\varphi,cos\theta)$
$b=(x',y',z')=(sin\theta' cos\varphi',sin\theta' sin\varphi',cos\theta')$
dot product over the $xy$ plane :
$cos(\varphi-\varphi ')=\frac {xx'+yy'}{\sqrt{x^2+y^2}\sqrt{x'^2+y'^2}}=\frac {xx'+yy'}{\sin\theta \sin\theta'}$
dot product $a.b$ : $\cos\gamma=xx'+yy'+zz'=xx'+yy'+cos\theta cos\theta'$
$\cos\gamma=cos(\varphi-\varphi ')\sin\theta \sin\theta'+cos\theta cos\theta$


TITLE: Improper integrals outside domains
QUESTION [1 upvotes]: What is the standard convention for computing something like
$$\int_0^3 \frac{1}{\sqrt{1-x}}\,\mathrm{d}x? $$
Is it equivalent, since the integrand is not defined on $(1,\infty)$, to
$$\int_0^1 \frac{1}{\sqrt{1-x}} \, \mathrm{d}x$$
where one can use the standard technique for an improper integral, or does it simply not exist?

REPLY [0 votes]: For $x>1$, the term $ \frac{1}{\sqrt{1-x}}$ is not defined. Hence $\int_0^3 \frac{1}{\sqrt{1-x}}\,\mathrm{d}x$ is nonsense.


TITLE: Uncoupled projections.
QUESTION [0 upvotes]: I have troubles understanding this sentence from book Linear Algebra and its Application by Prof.Gilbert Strang.

The hard part of least squares disappears when vectors are
  orthonormal. The projections onto the axes are uncoupled, and p is
  the sum $p=(q_1^T b)q_1+···+(q_n^T b)q_n$. We emphasize that those
  projections do...

In particular, i don't understand what word uncoupled mean in this case. I understand that in case of orthonormal basis projection simplifies to $QQ^T$ and it is not invertible procedure. As i can see it pieces of b lay down on corresponding parts of basis and in the same time scaled by this basis vectors, some pieces of b are lost because we reduce dimensionality (otherwise $P=I$). But what parts of b or P or Q coupled here? I don't understand. Thank you!

REPLY [1 votes]: Basically, it’s another way of saying that the orthogonal projections onto the axes are independent of each other when the axes are orthogonal: varying the length of the component of $b$ along one axis doesn’t affect its components along any other axis. If basis vectors aren’t orthogonal, there’s “cross talk” between them and the projection of $b$ onto one vector can pick up some of the components of $b$ in other directions. I emphasize orthogonal here because it’s possible to find uncoupled projections for any basis; it’s just that they won’t necessarily be orthogonal projections (as defined by the inner product $(a,b)=b^Ta$).
See this answer for a more detailed explanation, including diagrams, in $\mathbb R^2$.