text:
NYGHoops All State Team 2015
Co- Players of the year
Brianna Fraser 6"03" P (South Shore HS, Brooklyn, NY 2015)
The Maryland bound Fraser lead her team to its first ever Class AA City Championship while averaging 16 points, 10 rebounds and 3 blocks per game. In her final game in a Viking uniform she broke the NY State federation single game scoring record by scoring 46 points in a semifinal lost.
Lauren "Boogi" Brozoski 5'06" PG (Long Island Lutheran HS, NY 2015)
Michigan bound Brozoski did something that most players envision or talk about every year. Playing on the last day for a chip, Boogie's fairytale career ended with 4 State Federation Championships at the B, A and two at the AA level in her four years. As a senior she led the Crusaders to 19-4 record while averaging 18 points, 7 assist 3 rebounds and 2 steals.
Jalay Knowles 6'00' F (Ossining HS, Ossining, NY 2015)
The rugged five-year varsity starter led her team deep into a hunt for a federation championship. As a senior she averaged a double- double while playing a very competitive schedule (23.1 points and 10.4 Rebounds). She will continue her playing career at Hartford University.
Dominiques Toussain 5'09" G (Christ the King HS, Queens, NY 2016)
Toussain became the leader of a young Lady Royals team that would eventually go on to win the CHSAA State championship. She averaged 21.5 points, 6.4 rebounds and 3.2 assists per game, while also being as Christ the King's best perimeter defender.
Arella Guirantes 5'10" G (Bellport HS, Bellport, NY 2015)
At press time Guirantes has announced that she will reclassify into the class of 2016. It's hard to argue with her productivity averaging 35 points per game in her final season at Bellport. Guirantes currently hold multiple offers from BCS Schools.
Brianna Jones 5'09" G (North Babylon HS, North Babylon, NY 2015)
The Louisville bound Jones faced all types of junk defenses this past season. She still managed to put up impressive numbers averaging 28 points, 14 rebounds, 3 assists and 3 steals per game as a senior.
Gabby Giacone 6'00" F (Bethlehem HS, Troy, NY 2015)
The programs second all- time scorer, was a matchup night mare for most teams she faced in her senior campaign. Giacone averaged 21.8 points and 11.2 rebound, while shooting 58 percent from the field. Gabby will play her collegiate ball at Northeastern University.
Jordan Augustus 5'11" W (St Mary's HS, Long Island 15') St Johns
Shanique Edwards 6'01' F (Grand Street HS, Brooklyn 17')
Alex Smith 6'04" P (Murry Bergtraum HS, New York 15') Middle Tennessee
Mei-Lyn Beautista 5'06" PG (St Mary's HS, Long Island 15') GWU
McKayla Roberts 5'03" PG (Syracuse CNA, Syracuse NY 2016) LeMoyne
Third Team
Dana Watts 6'01" F (Rush Henrietta HS, NY 15') AIC
Carly Boland 6'00' W/F (Shenendehowa HS, Clifton Park 16')
Ashanae McLaughlin 5'07" P (Murry Bergtraum HS, NY 15') Seton Hall
Sydney Zambrotta 5'08" G (Christ the King HS, Queens NY 16')
Alexia Douglass 5'08" G (Brentwood HS, Brentwood NY 15') Fordham


The latest figures show the Irish grocery market grew by 3.5% in the 12 weeks.
And not a set menu in sight.
'He asked me if I wanted to go to a wedding with him the next month'.
This is what Valentine's Day is all about.
Your stomach's in knots, your heart flutters – but here's what happens to your brain when you're in love.
Wanna skip it this year?
A donation fund has been launched to help search efforts to find another frog of his kind.
It's both Pancake Tuesday and Valentine's Day this week so we asked a master crêpier asked for something sweet and savory using aphrodisiac ingredients of his choosing.
Make your love life miles better.
The new movie is Elba's directorial debut.
Poll: Would you be into a public wedding proposal?
Dream come true or worst nightmare?
This was the strangest episode any of us have seen in a while.
Have you ever seen such a well-oiled audience on The Late Late?
Irish people are renowned the world over for being romantic. We just love showering affection on people!
If you're in a relationship Valentine's Day can be really fun and smug. But if you're not, here is Alice from First Dates on how to find your boo.
Romantic Ireland is NOT dead and gone.
Of all the things that companies are trying to flog this Valentine's Day, the best gift is your heart. Your actual pound-of-flesh beating heart.
The Revolutionary Alliance of Men that Women find Unattractive is spearheading the campaign.
For when you need to say it with a little more sarcasm.


Today I am very pleased to showcase my yet another post of short blue bridesmaid dresses with lace! We give you a wide range of fashion Dresses that suits every occation. Shop our clothing sale for women's dresses and skirts and enjoy the same quality and style for less. A dressy bag with a chain helps you hold crudites and a cocktail-plus, it lends a cool edge to evening wear, says stylist Cher Coulter. Wear the chain diagonally across your body for the best effect. On the other hand, ladies who love to wear open back attire should prefer gold mini dress with long sleeves. In short, both green pink and gold mini dress with long sleeves are amazing choice for formal events. Scroll down and choose an awesome outfit for you. Cheers!


Mega churches have mega-sized rooms and have mega-sized needs for acoustical portable walls. Such is the case for Lake Point Church. Their fellowship hall is approximately 8000 sq. ft. in size and needed to both be kept open for certain programs and divided into smaller rooms for others. By using several Screenflex Portable Room Dividers, the church administrators easily set up 7 rooms capable of holding 48 students each, and one which holds 96 students. The 7 rooms average approximately 24 wide by 29 long while the larger Sunday School room is approximately 20 by 50.


text:
\section{Introduction and main results}

In this paper, we consider the fractional Schr\"{o}dinger equation
\begin{equation}\label{eq1.1}
\epsilon^{2s}(-\Delta)^s u + V(x)u = |u|^{p - 2}u ,\,\,x\in\,\,\mathbb{R}^N,
\end{equation}
where $N > 2s$, $s\in (0,1)$, $\epsilon > 0$ is a small parameter, $p\in (2, 2^*_s)$, $2^*_s=2N/(N-2s)$, $V(x)\in C(\mathbb R^N,[0,+\infty))$ is a continuous nonnegative potential. Problem~\eqref{eq1.1}  arise from the study of time-independent waves $\psi(x,t) = e^{-iEt/\epsilon}u(x)$ of the following nonlinear fractional Schr\"{o}dinger equation
\begin{equation}\label{eqNLFS}
i\epsilon\frac{\partial{\psi}}{{\partial t}} = \epsilon^{2s}(-\Delta)^s{\psi} + U(x)\psi - f(\psi)\ x\in \mathbb{R}^N.
\end{equation}
Equation  \eqref{eq1.1} has a wide application in Physics, for example, the Einstein's theory of relativity, phase transition, conservation laws and fractional quantum mechanic, see \cite{2,3,4} and the references therein for more details. The parameter $\epsilon$ is the adimensionalised Planck constant; when $\epsilon$ goes to $0$, the quantum physics will shift to the classical physics, in which case the analysis for \eqref{eq1.1} is called semiclassical analysis. In the local case $s=1$, the existence and concentration of solutions for equation \eqref{eq1.1} have been studied extensively, see  \cite{{5},{6},{7},{8},{9},{10},{12},{13},{14},{15}} and their references therein for example.

In the nonlocal case $0<s<1$, to our best knowledge, there are few results on existence and concentration phenomena for \eqref{eq1.1}. Actually,  due to  the nonlocal effect of $(-\Delta)^s$,  the techniques created for local case can not be applied easily, see \cite{19,20} for example. Considering
the non-vanishing case, i.e., $\inf_{x\in \mathbb{R}^N}V(x) > 0$, Alves et al. in \cite{100} used the penalized method developed by Del Pino et al  in \cite{10} and extension technical  developed by Caffarelli et al. in \cite{24} to construct a family of positive solutions that concentrate at a local minimum of $V$ as $\epsilon\to 0$. Also in the non-vanishing case, references \cite{{25},{26},{27}} proved that the concentration phenomena occurs at the nondegenerate critical points of $V$ by using a reduction method. For the other results of \eqref{eq1.1} with non-vanishing potentials, we would like to refer the readers to \cite{19,20,J.Davila-M.del Pino-J.Wei-2014,V.Ambrosio-2019}.

Throughout all the works before, an assumption that $\inf_{\mathbb R^N}V>0$ was required. The study of the case that $V>0$ but $\inf_{\mathbb R^N}V=0\, ( \liminf_{|x|\to\infty}V(x)=0$) was initiated by X.  An et al. in \cite{APX}, where it was proved by constructing a new penalized function  that equation \eqref{eq1.1} has a family of positive solutions concentrating at a positive local minimum of $V$ if $\liminf_{|x|\to\infty}V(x)|x|^{2s}>0$.

In this paper, we try to solve the question of existence and concentration for more general potentials including fast decaying potentials, i.e. potentials for which
$$
\liminf_{|x|\to\infty}V(x)|x|^{2s}=0.
$$
A special case of a fast decaying potential is a potential $V$ that vanishes identically, i.e., $V$ is compactly supported

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 $\dot{H}^s(\mathbb R^N)$ as the completion of $C^{\infty}_c(\mathbb R^N)$ under the norm
$$
\|u\|^2 = \int_{\mathbb{R}^N}|(-\Delta)^{s/2}u|^2{\mathrm{d}}x  = \int_{\mathbb{R}^{2N}}\frac{|u(x) - u(y)|^2}{|x - y|^{N + 2s}}\,{\mathrm{d}}x \,  {\mathrm{d}}y.
$$

Generally, the fractional Laplacian  (see \cite{4} for example) 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_{\epsilon\to 0}\int_{\mathbb{R}^N          \backslash        B_{\epsilon}(x)      }\frac{u(x) - u(y)}{|x - y|^{N + 2s}}\,{\mathrm{d}}y.
\end{align*}
For the sake of simplicity, we define for every $u\in \dot{H}^s(\mathbb R^N)$ the fractional $(-\Delta)^s$ as
$$
(-\Delta)^s u(x) = \int_{\mathbb{R}^N}\frac{u(x) - u(y)}{|x - y|^{N + 2s}}\,{\mathrm{d}}y.
$$
Our solutions will be found in the following weighted fractional Sobolev  space:
$$
\mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N) = \left\{u\in \dot{H}^s(\mathbb R^N):\,\,u\in L^2(\mathbb{R}^N,V(x)\,{\mathrm{d}}x)\right\},
$$
endowed with the norm
$$
\|u\|_{\mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N)} = \Big(\int_{\mathbb{R}^N}\epsilon^{2s}|(-\Delta)^{s/2}u|^2 + Vu^2\,{\mathrm{d}}x \Big)^{\frac{1}{2}}.
$$
For the potential term $V$, we assume that $V\in C\big(\mathbb{R}^N,[0,\infty)\big)$ and

$(\mathcal{A})$  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 < \inf_{U\backslash\Lambda}V.
\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 $N > 2s$, $s\in(0,1)$, $p\in (2 + \frac{2s}{N - 2s}, 2^*_s)$ and $V\in C\big(\mathbb{R}^N,[0,\infty)\big)$ satisfy the assumption $(\mathcal{A})$.
 Then there exists an $\epsilon_0>0$ such that problem \eqref{eq1.1} has a  positive solution $u_{\epsilon}\in \mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N)$ if $\epsilon\in(0,\epsilon_0)$. Moreover, there exist  an $\alpha\in (2s/(p - 2),N - 2s)$ and a family of points $\{x_{\epsilon}\in \overline{\Lambda}:\epsilon\in(0,\epsilon_0)\}$ such that $u_{\epsilon}(x_{\epsilon}) = \sup_{x\in \Lambda}u_{\epsilon}(x)$,
 $$
 \lim_{\epsilon\to 0}V(x_{\epsilon}) = \min_{\Lambda}V(x)
 $$
 and
\begin{align*}
u_{\epsilon}(x)\le \frac{C\epsilon^{\alpha}}{\epsilon^{\alpha} + |x - x_{\epsilon}|^{\alpha}},
\end{align*}
where $C$ is positive constant.
\end{theorem}
Particularly, we solve the fractional version  Ambrosetti-Malchiodi question(\cite{AA}) completely, which in the local case $s=1$ have been settled recently (see \cite{BaNa,Byeon-Wang,Cao-Peng,MV}).

Our approach in this work follows the variational penalization scheme in \cite{APX}, which was first introduced in \cite{10}. Naturally, equation \eqref{eq1.1} is the Euler-Lagrange equation of the functional
$$
I_{\epsilon}(u):=\frac{1}{2}\int_{\mathbb R^N}(\epsilon^{2s}|(-\Delta)^{s/2}u|^2 + V|u|^2) - \frac{1}{p+1}\int_{\mathbb R^N}|u|^p.
$$
However, an elementary (but tedious) calculation shows that when $2<p<2^*_s$ and $V$ vanishes faster than $|x|^{-2s}$, the second integral is not finite in $\mathcal{D}^s_{V,\epsilon}(\mathbb R^N)$. del Pino and Felmer provided an effective way to overcome such difficulty: to modify the nonlinear term. According to the idea,  assuming that $V$ satisfies $\liminf_{|x|\to\infty}V(x)|x|^{2s}>0$, in \cite{APX} the penalized problem
$$
\epsilon^{2s}(-\Delta)^su_{\epsilon} + V(x)u_{\epsilon}=\chi_{\Lambda}u^{p-1}_{\epsilon} + \chi_{\mathbb R^N\backslash\Lambda}\min\{P_{\epsilon}(x)u_{\epsilon},u^{p-1}_{\epsilon}\}
$$
was considered, where $P_{\epsilon}(x)$ satisfies $P_{\epsilon}(x) = 0$ on $\Lambda$, $P_{\epsilon}(x)>0$ outside $\Lambda$ and $P_{\epsilon}(x)\le \epsilon^{\nu}V(x)$ outside $\Lambda$ with $\nu>0$ is a small parameter. A penalized solution then follows by the fact that the Euler-Lagrange functional corresponding to the penalized problem owns Mountain Pass geometry and is a class of $C^1$. One has then to show that $u^{p-2}_{\epsilon}\le P_{\epsilon}$ outside $\Lambda$. However, when $V$ is compactly supported this approach fails, because one should then have that the solution $u_{\epsilon}$ are compactly supported. In brief, the potential $V$ can not dominated the nonlinear term if it is compactly supported. Our key observation in this paper is that, in order to overcome this difficulty, we can modify the problem by the following fractional Hardy inequality: There exists a positive constant $C_{N,s}$ such that
\begin{equation}\label{eq1.4}
\int_{\mathbb R^N}\frac{|u(x)|^2}{|x|^{2s}}\,{\mathrm{d}}x\le C_{N,s}|(-\Delta)^{s/2}|^2_2
\end{equation}
for all $u\in \dot{H}^s(\mathbb R^N)$(see \cite{FS})

One difference between equations in local case $s=1$ and nonlocal case $0<s<1$ is the construction of penalized function. In fact, for a smooth function $f\in C^{\infty}_c(\mathbb R^N)$, one can not compute $(-\Delta)^sf$ as precisely as $-\Delta f$. Another difference is the energy estimates. 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 $\liminf_{|x|\to 0}V(x)|x|^{2s}>0$. But, in the present paper,  the global $L^2$-norm information of penalized solution will be obtained by the Hardy inequality above (see \eqref{eq3.3} in the proof of Lemma \ref{le3.1} for more details).

\vspace{0.5cm}

  The paper is organized as follows: in Section \ref{s2}, we establish the penalized scheme and obtain a penalized solution $u_{\epsilon}$. In Section \ref{s3} we study the concentration phenomenon of $u_{\epsilon}$. In Section \ref{s4}, we prove the penalized solution $u_{\epsilon}$ solves the origin problem by constructing a special penalized function. In Section \ref{s5}, we give a short proof for the existence and concentration for equation \eqref{eq1.1} with more general nonlinear term.

\vspace{0.5cm}
\section{The penalized problem}\label{s2}

The following inequality exposes the relationship between $H^s(\mathbb R^N)$ and the Banach space $L^q(\mathbb R^N)$.
\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.,
$$
H^s(\mathbb{R}^N)\subset\subset L^q_{loc}(\mathbb{R}^N)\ \text{compactly, if}\ q\in [1,2^*_s).
$$

Now we are going to modify the origin problem \eqref{eq1.1}.
According to the fractional Hardy inequality \eqref{eq1.4}, we choose a family of  penalized potentials $\mathcal{P}_{\epsilon}\in L^{\infty}(\mathbb{R}^N,[0,\infty))$ for $\epsilon > 0$ small in such a way that
\begin{eqnarray}\label{eq2.1}
\begin{split}
\mathcal{P}_{\epsilon}(x) = 0\ \text{for all}\ x\in\Lambda
\end{split}
\end{eqnarray}
{and}
\begin{align}
\ \lim\limits_{\epsilon \to 0}\sup_{\mathbb{R}^N\backslash \Lambda} \mathcal{P}_{\epsilon}(x)\epsilon^{-(2s + 3\kappa/2)}|x|^{2s + \kappa} = 0,
\end{align}
where $\kappa>0$ is a small parameter. Noting that by \eqref{eq1.4}, when $\epsilon>0$ is small enough, it holds
\begin{align*}
\int_{A}\mathcal{P}_{\epsilon}(x)|u|^2\le C_{N,s}\frac{\epsilon^{2s +\frac{3\kappa}{2}}}{\inf_{x\in(\mathbb R^N\backslash\Lambda)\cap A}{|x|^{\kappa}}}\int_{\mathbb R^N}|(-\Delta)^{s/2}u|^2
\end{align*}
where $C_{N,s}$ is the constant in \eqref{eq1.4}. This type of estimate plays a key role in the paper. Now we give the penalized problem according to the choice of $\mathcal{P}_{\epsilon}$:
\begin{equation}\label{eq2.2}
\epsilon^{2s}(-\Delta)^su + Vu = \chi_{\Lambda}(x)u^{p - 1}_{+} + \chi_{\mathbb{R}^N\backslash \Lambda}(x)\min(u^{p - 1}_{+}, \mathcal{P}_{\epsilon}(x)u_+).
\end{equation}
It is easy to check that if a solution $u_{\epsilon}$ of \eqref{eq2.2} satisfies
$$
u_{\epsilon}\le \mathcal{P}_{\epsilon}\ \ \text{on}\ \mathbb R^N\backslash\Lambda,
$$
then $u_{\epsilon}$ is a solution of \eqref{eq1.1}.

For simplicity, we define
$$
g_{\epsilon}(x,s): = \chi_{\Lambda}(x)s^{p - 1}_{+} + \chi_{\mathbb{R}^N\backslash \Lambda}(x)\min(s^{p - 1}_{+}, \mathcal{P}_{\epsilon}(x)s_+),
$$
$$G_{\epsilon}(x,t) = \int_{0}^{t}g_{\epsilon}(x,s) {\mathrm{d}}s,$$
$$
\mathbf{g}_{\epsilon}(u)(x) = g_{\epsilon}(x,u(x))
$$
and
$$
 \mathbf{G}_{\epsilon}(u)(x) = G_{\epsilon}(x,u(x)).
$$
Then the Euler-Lagrange functional of \eqref{eq2.2}
is
$$
\mathbf{J}_{\epsilon}(u) = \frac{1}{2}\int_{\mathbb{R}^N}\big(\epsilon^{2s}|(-\Delta)^{s/2} u|^2 + V(x)|u|^2\big) - \int_{\mathbb{R}^N}\mathbf{G}_{\epsilon}(u).
$$

By  the Hardy inequality \eqref{eq1.4},  we know that $\mathbf{J}_{\epsilon}$ is well-defined on $\mathcal{D}^s_{V,\epsilon}(\mathbb R^N)$. In order to obtain a critical point to $\mathbf{J}_{\epsilon}$ via variational method, we need to verify that $\mathbf{J}_{\epsilon}$ is $C^1$ and satisfies the (P.S.) condition.
\begin{lemma}\label{le2.2}
$\mathbf{J}_{\epsilon}:\mathcal{D}^s_{V,\epsilon}(\mathbb R^N)\to \mathbb R$ is $C^1$ and satisfies (P.S.) condition.
\end{lemma}

\begin{proof}
We only need to show that the nonlinear term
$$
\mathcal{I}_{\epsilon}(u) = \int_{\mathbb R^N}\mathbf{G}_{\epsilon}(u)
$$
is $C^1$ since the other terms are obvious. For every $\varphi\in C^{\infty}_c(\mathbb R^N)$ and $0<|t|<1$, we have
$$
|\mathbf{G}_{\epsilon}(u + t\varphi) - \mathbf{G}_{\epsilon}(u)|/t\leq C\Big((|u|^p + |\varphi|^p)\chi_{\Lambda} + \mathcal{P}_{\epsilon}(|u|^2 + |\varphi|^2)\Big)\in L^1(\mathbb{R}^N).
$$
The existence of first order Gateaux derivative then follows  by Dominated Convergence Theorems.

Let $(u_n),\ u\in \mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N)$ satisfy  $u_n\to u\in \mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N)$. By Proposition \ref{wpr2.1}, the construction of $\mathcal{P}_{\epsilon}$, \eqref{eq1.4} and Dominated Convergence Theorem, we deduce that for every $\varphi\in H^{s}_{V,\epsilon}(\mathbb{\mathbb{R}^N})$ with $\|\varphi\|_{H^{s}_{V,\epsilon}(\mathbb{R}^N)} \leq 1$,
\begin{align*}
 &\quad\quad|\langle\mathcal{I}'_{\epsilon}(u_n) - \mathcal{I}'_{\epsilon}(u),\varphi\rangle|&\\
&\leq C\|u_n - u\|^p_{H^{s}_{V,\epsilon}(\mathbb{R}^N)} + \int_{\mathbb{R}^N\backslash{\Lambda}}\big|\min\{\mathcal{P}_{\epsilon},(u_n)^{p - 2}_+\}u_n\varphi - \min\{\mathcal{P}_{\epsilon},u^{p - 2}_+\}u\varphi \big| \,{\mathrm{d}}x
\\[2mm]
 & \leq  o_n(1) + \int_{\mathbb{R}^N\backslash{\Lambda}}\big|\min\{\mathcal{P}_{\epsilon},(u_n)^{p - 2}_+\}(u_n - u)\varphi\big|
 \\[2mm]
 &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \qquad + \big|\min\{\mathcal{P}_{\epsilon},(u_n)^{p - 2}_+\} - \min\{\mathcal{P}_{\epsilon},u^{p - 2}_+\}\big|u\varphi \,{\mathrm{d}}x
 \\[2mm]
 & \leq o_n(1) + \int_{\mathbb{R}^N\backslash{\Lambda}}\big|\min\{\mathcal{P}_{\epsilon},(u_n)^{p - 2}_+\} - \min\{\mathcal{P}_{\epsilon},u^{p - 2}_+\}\big||u|^2\,{\mathrm{d}}x
 \\[2mm]
 & = o_n(1),
\end{align*}
which says that $\mathbf{J}_{\epsilon}$ is a class of $C^1$.

 Next we show that $\mathbf{J}_{\epsilon}$ satisfies (P.S.) condition, i.e., to prove that any sequence $\{u_n\}\subset  \mathcal{D}^s_{V,\epsilon}(\mathbb R^N)$ satisfying
$$
\mathbf{J}_{\epsilon}(u_n)\to c,\ \ \text{and}\ \ J'_{\epsilon}(u_n)\to 0
$$
is relatively compact.

It is standard to verify using the fact $p>2$ and the construction of $\mathcal{P}_{\epsilon}$ that $\{u_n\}$ is bounded in $\mathcal{D}^s_{V,\epsilon}(\mathbb R^N)$. By Proposition \ref{wpr2.1}, one has thus $u_n\to u$ in $L^{p}_{loc}(\mathbb R^N)$.

On the other hand, one has, for every $\sigma>0$, by the fractional Hardy inequality \eqref{eq1.4}, for $R>0$ large enough,
\begin{align*}
\int_{\mathbb R^N\backslash B_R(0)}\mathcal{P}_{\epsilon}(x)|u_n|^2 \,{\mathrm{d}}x  &= \sup_{x\in\mathbb R^N\backslash B_R(0)}(C^{-1}_{N,s}\mathcal{P}_{\epsilon}(x)|x|^{2s})\, C_{N,s}\, \int_{\mathbb R^N\backslash B_R(0)}\frac{|u_n|^2}{|x|^{2s}}\,{\mathrm{d}}x
\\[1mm]
&\le \sup_{x\in\mathbb R^N\backslash B_R(0)}(\mathcal{P}_{\epsilon}(x)|x|^{2s})|(-\Delta)^{s/2} u_n|^2_2\\[1mm]
&\le \frac{C}{R^{\kappa}}
< \sigma.
\end{align*}
Hence
\begin{align*}
  \limsup_{n\to\infty}\|u_n - u\|^2_{\mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N)}
 &= \limsup\Big(\langle J'_{\epsilon}(u_n) - J'_{\epsilon}(u),u_n - u \rangle\\
 &\quad + \int_{\mathbb{R}^N}(\mathbf{g}_{\epsilon}(u_n) - \mathbf{g}_{\epsilon}(u))(u
_n(x) - u(x))\,{\mathrm{d}}x\Big)\\
  &\le \limsup_{n\to\infty}\Big(\int_{\mathbb{R}^N\backslash B_R(0)}\mathcal{P}_{\epsilon}(|u_n| + |u|)^2\,{\mathrm{d}}x\Big)^\frac{1}{2}\\
  &\le\sigma.
\end{align*}
Since $\sigma>0$ is arbitrary, this proves the lemma.

\end{proof}

It is standard to show that $\mathbf{J}_{\epsilon}$ owns Mountain Pass geometry. Hence, from Lemma \ref{le2.2}, we can use the min-max procedure in \cite{MW} to find a critical point for the functional $\mathbf{J}_{\epsilon}$.

\begin{lemma}\label{le2.3}
The Mountain Pass value
$$
c_{\epsilon}: = \inf_{\gamma\in\Gamma_{\epsilon}}\max_{t\in[0,1]}\mathbf{J}_{\epsilon}(\gamma(t))
$$
can be achieved by a positive function $u_{\epsilon}$ which is $C^{1,\gamma}$ for some $\gamma\in(0,1)$ and is a solution to the penalized problem \eqref{eq2.2},
where
$$
\Gamma_{\epsilon}:=\big\{\gamma\in C\big(\, [0,1],\, \mathcal{D}^s_{V,\epsilon}(\mathbb R^N)\, \big)|\gamma(0) = 0, \mathbf{J}_{\epsilon}(\gamma(1))<0\big\}.
$$
\end{lemma}

\begin{proof}
The existence of $u_{\epsilon}$ can be checked by the Mountain Pass Lemma in \cite{MW}. By the regularity argument in \cite[Appendix D]{20}, $u_{\epsilon}$ is $C^{1,\gamma}$ for some $\gamma\in(0,1)$. {Testing the penalized equation \eqref{eq2.2} with $(u_{\epsilon})_{-}$ and integrating, we can see that $u_{\epsilon}\geq 0$.}

{Supposing to the contrary that there exists  $x_0\in \mathbb{R}^N$ such that $u_{\epsilon}(x_0) = 0$, we then have
$$
0 = (-\Delta)^s{u_{\epsilon}}(x_0) + V(x_0)u_{\epsilon}(x_0) < 0,
$$
which is a contradiction. Therefore, $u_{\epsilon}>0$.}

\end{proof}

\section{Concentration}\label{s3}

In this section, we prove that the solution  $u_{\epsilon}$ obtained by Lemma \ref{le2.3} will concentrate at a local minimum of $\Lambda$ as $\epsilon\to 0$ {via comparing the energy $c_{\epsilon}$ in Lemma \ref{le2.3} with the least energy of the limiting problem corresponding to \eqref{eq1.1}.} This concentration can help to linearize the penalized equation \eqref{eq2.2}\,(see \eqref{AAeq4.2} below).

\subsection{The limiting problem}

\noindent  {For $a > 0$, the limiting problem associated to  \eqref{eq1.1} is
$$
(-\Delta)^s v + a v = |v|^{p - 2}v,
\eqno(\mathcal{P}_{a})
$$
whose Lagrange-Euler  functional $J_{a}: H^s(\mathbb{R}^N)\to \mathbb{R}$ is defined  as
$$
J_{a}(v) = \frac{1}{2}\int_{\mathbb{R}^N}(|(-\Delta)^{s/2} v|^2 + a |v|^2)dx - \frac{1}{p}\int_{\mathbb{R}^N}|v|^p dx.
$$}

{From \cite{20,FQT}, we know that the limiting problem has a positive ground state. We define the limiting energy by
\begin{equation}\label{eq3.1}
\mathcal{C}(a) = \inf_{v\in H^s(\mathbb{R}^N)\backslash\{0\}}\max_{t > 0}J_{a}(tv).
\end{equation}
Since for every $v\in H^s(\mathbb{R}^N),\ J_{a}(|v|) \leq J_{a}(v)$, by the  density of $C_0^\infty(\mathbb{R}^N)$ in   $H^s(\mathbb{R}^N)$, we have
\begin{equation}\label{eq3.2}
  \mathcal{C}(a) = \inf_{{C^{\infty}_c(\mathbb{R}^N)\backslash\{0\}}\atop{v\geq 0}}\max_{t > 0}J_{a}(tv).
\end{equation}
The following proposition is obvious.
\begin{proposition}
Let $a > 0$ and $v\in H^s(\mathbb{R}^N)$. Define
$$
v_{a}(y) = a^{\frac{1}{p - 2}}v(a^{1/2s}y).
$$
Then
$$
J_{a}(v_{a}) = a^{\frac{p}{p - 2} - \frac{N}{2s}}J_1(v_1).
$$
In particular, $v$ is a solution of $(\mathcal{P}_1)$  if and only if $v_a$ is a solution of $(\mathcal{P}_{a})$ and
$$
\mathcal{C}({a}) = \mathcal{C}({1})a^{\frac{p}{p - 2} - \frac{N}{2s}}.
$$
Moreover, $\mathcal{C}({a})$ is continuous and increasing in $(0,\infty)$.
\end{proposition}}

{Now we are going to compare the energy $c_{\epsilon}$ with $\mathcal{C}(V(x)),x\in\Lambda$. Firstly,
we give the upper bound of the energy $c_{\epsilon}$:
\begin{lemma}\label{GGle3.2}(Upper bound of the energy) we have
\begin{equation}\label{eq3.4}
 \limsup_{\epsilon\to 0}\frac{c_{\epsilon}}{\epsilon^N}\leq \inf_{x\in\Lambda}\mathcal{C}(V(x)).
\end{equation}
\end{lemma}}

\begin{proof}
For any given $x_0 \in \Lambda$ and nonnegative function $v \in C^{\infty}_c (\mathbb{R}^N)\backslash\{0\}$, we define
$$
v_{\epsilon}(x) = v\Big(\frac{x - x_0}{\epsilon}\Big).
$$
Obviously, $supp \,\,v \subset \Lambda$ when $\epsilon$ is small enough. It is easy  to check that $\gamma_{\epsilon}(t) = tT_0v_{\epsilon}\in \Gamma_{\epsilon}$ for some $T_0$ large enough. So
\begin{align*}
  \frac{c_{\epsilon}}{\epsilon^N}&\leq \max_{t\in [0,1]}J_{\epsilon}(\gamma_{\epsilon}(t))\leq \max_{t>0}J_{V(x_0)}(tv) + o_\epsilon(1),
\end{align*}
and
$$
\limsup_{\epsilon\to 0}\frac{c_{\epsilon}}{\epsilon^N}\leq \inf_{{C^{\infty}_c(\mathbb{R}^N)\backslash\{0\}}\atop{v\geq 0}}\max_{t > 0}J_{V(x_0)}(tv) \leq \mathcal{C}(V(x_0)),
$$
which implies our conclusion.
\end{proof}

Secondly, we  give the lower estimate on energy $c_{\epsilon}$.

\begin{lemma}\label{le3.1}
 Let $\{ \epsilon_n \}_{n\in \mathbb{N}}$ be a sequence of positive numbers converging to $0$,  $\{ u_n\} $ be a sequence of critical points given by Lemma \ref{le2.3},
  and for $j\in {1,2,\cdots,k}$,  $\{ x^j_n\} $ be a sequence in $\mathbb{R}^N$  converging  to $x^j_{*}\in \mathbb{R}^N$. If
$$
\limsup_{n\to\infty}\frac{1}{\epsilon^N_n}\int_{\mathbb{R}^N} \big\{ \epsilon^{2s}_n|(-\Delta)^{s/2}u_n|^2 + V|u_n|^2 \big\} \,{\mathrm{d}}x  < \infty,
$$

$$
V(x^j_{*}) > 0\ \text{and}\ \lim\limits_{n\to \infty}\frac{|x^i_n - x^j_n|}{\epsilon_n} = \infty\ \ \text{if}\ i\neq j \quad \text{for all}\ i,j = 1,2,\cdots,k,
$$
and for some $\rho > 0$,
$$
\liminf_{n\to \infty}\|u_n\|_{L^{\infty}(B_{\epsilon_n\rho}(x^j_n))} > 0 \quad     \text{for all}\,  j = 1,2,\cdots k,
$$
then $x^j_{*}\in \bar{\Lambda}$ and
$$
\liminf_{n\to \infty}\frac{J_{\epsilon_n}(u_n)}{{\epsilon^N_n}}\geq\sum_{j = 1}^{k}\mathcal{C}(V(x^j_*)).
$$
\end{lemma}

Some details in the proof of this lemma will be omitted since they are similar to Proposition 3.4 in \cite{APX}, and more attentions will be paid on the difference caused by the fast decaying potential $V$ (see \eqref{eq3.3} below for more details).

\begin{proof}
 For $j\in \{1,\ldots,k\}$, the rescaling  function $v^j_n\in H^s_{loc}(\mathbb{R}^N)$ defined by
$$
v^j_n(y) = u_n(x^j_n + \epsilon_ny)
$$
satisfies weakly the rescaled equation
$$
(-\Delta)^{s} v^j_n + V^j_n v^j_n = \textsf{g}^j_n(v^j_n)\ \text{in}\ \mathbb{R}^N,
$$
where  $V^j_n(y) = V(x^j_n + \epsilon_n y),\ \textsf{g}^j_n(v^j_n) = g_{\epsilon_n}(x^j_n + \epsilon_ny, v^j_n).
$ By the assumption, $$\sup_{n\in \mathbb N}\|u_n\|_{\mathcal{D}^s_{V,\epsilon}(\mathbb R^N)}<+\infty, $$
 the sequence $\{ v^j_n\}_{n\ge 1}$ is bounded in $H^s_{loc}(\mathbb R^N)$. By the regularity assertion in Appendix D of \cite{20}, the Liouville-type Lemma in Lemma 3.3 of \cite{APX}, we can conclude that there exist $x^j_*\in\bar{\Lambda}$ and a nonnegative $v^j_*\in H^s(\mathbb R^N)\backslash\{0\}$ such that $v^j_n\to v^j_*$ in $H^s_{loc}(\mathbb R^N)\cap C^1_{loc}(\mathbb R^N)$, $x^j_n\to x^j_*$ and
$$
(-\Delta)^s v^j_* + V(x^j_*) v^j_* = (v^j_*)^{p - 1}.
$$
Since $v^j_* \geq 0$, we see
\begin{align*}
  &\liminf_{n\to\infty}\frac{1}{{\epsilon^N_n}}\int_{B_{\epsilon_nR}(x^j_n)}  \Big\{ \, \frac{1}{2}(\epsilon^{2s}_n|(-\Delta)^{s/2}u_n|^2 + V|u_n|^2) - \mathbf{G}_{\epsilon_n}(u_n)   \, \Big\}\,{\mathrm{d}}x \\

 \geq &\mathcal{C}(V(x^j_*)) - C\int_{\mathbb{R}^N\backslash B_R} \Big\{ \,|  (-\Delta)^{s/2}v^j_*|^2 + V(x^j_*)|v^j_*|^2 \, \Big\}\,{\mathrm{d}}x,
\end{align*}
where $C > 0$ is a universal constant.

In order to study the integral outside $B_{\epsilon_nR}(x^j_n)$, we choose cut-off function  $\eta(x) \in C^{\infty}(\mathbb{R}^N)$ such that $0\leq \eta\leq 1,\ \eta = 0$ if  $ 0 \leq |x| \leq 1$ and $\eta = 1$ when  $     |x| \geq  2$. Define
$$
\psi_{n,R}(x) = \prod^{k}_{j = 1}\eta\Big(\frac{x - x^j_n}{\epsilon_n R}\Big).
$$
Since $u_n$ is a solution to the penalized problem $\eqref{eq2.2}$, we have by taking $\psi_{n,R}u_n$ as a test function in the penalized problem $\eqref{eq2.2}$ that
\begin{align}\label{teq}
 \nonumber &\int_{\mathbb{R}^N\backslash\cup^{k}_{j = 1}B_{\epsilon_nR}(x^j_n)}  \Big\{ \epsilon^{2s}_n   \psi_{n,R}|(-\Delta)^{s/2}u_n|^2 + V\psi_{n,R}|u_n|^2  \Big\} \,{\mathrm{d}}x
 \\[1mm]
   = &\int_{\mathbb{R}^N\backslash\cup^{k}_{j = 1}B_{\epsilon_nR}(x^j_n)}\mathbf{g}_{\epsilon_n}(u_n)u_n\psi_{n,R} \,{\mathrm{d}}x
   \\[1mm]
\nonumber  &-\int_{\mathbb{R}^N\backslash\cup^{k}_{j = 1}B_{\epsilon_nR}(x^j_n)}\int_{\mathbb{R}^N}\frac{u_n(y)(\psi_{n,R}(x) - \psi_{n,R}(y))(u_n(x) - u_n(y))}{|x - y|^{N + 2s}}\,{\mathrm{d}}y\,{\mathrm{d}}x
\\[1mm]
\nonumber  :=& \int_{\mathbb{R}^N\backslash\cup^{k}_{j = 1}B_{\epsilon_nR}(x^j_n)}\mathbf{g}_{\epsilon_n}(u_n)u_n\psi_{n,R}     \,{\mathrm{d}}x  + R_n.
\end{align}
Hence
\begin{align*}
  &\displaystyle\int_{\mathbb{R}^N\backslash\cup^{k}_{j = 1}B_{\epsilon_nR}(x^j_n)}   \Big\{\,  \frac{1}{2}(\epsilon^{2s}_n|(-\Delta)^{s/2}u_n|^2 + V|u_n|^2) - \mathbf{G}_{\epsilon_n}  \,  \Big\}  \,{\mathrm{d}}x
  \\
  \geq & \frac{1}{2}\int_{\mathbb{R}^N\backslash\cup^{k}_{j = 1}B_{\epsilon_nR}(x^j_n)} \Big\{\,  \epsilon^{2s}_n\psi_{n,R}|(-\Delta)^{s/2}u_n|^2 + V\psi_{n,R}|u_n|^2 - \mathbf{g}_{\epsilon_n}(u_n)u_n \,  \Big\}  \,{\mathrm{d}}x
  \\
   =& -\frac{\epsilon^{2s}_n}{2}R_n + \int_{\mathbb{R}^N\backslash\cup^{k}_{j = 1}B_{\epsilon_nR}(x^j_n)}\mathbf{g}_{\epsilon_n}(u_n)(\psi_{n,R} - 1)u_n  \,{\mathrm{d}}x.
\end{align*}
By scaling, $B_{\epsilon_nR}(x^j_n)\cap B_{\epsilon_nR}(x^l_n) = \emptyset$ if $n$ is large enough. Hence, by the fact that $v^j_n\to v^j_*$ in $L^{p}_{loc}(\mathbb{R}^N)$, we have
$$
\lim\sup_{n\to\infty}\Big|\frac{1}{2}\sum_{j = 1}^{k}\int_{B_{2R}\backslash B_{R}}\textsf{g}^j_n(v^j_n)v^j_n   \,{\mathrm{d}}x \Big|      = \Big|\frac{1}{2}\sum_{j = 1}^{k}\int_{B_{2R}\backslash B_{R}}(v^j_*)^p    \,{\mathrm{d}}x \Big| = o_R(1).
$$

Now we estimate $R_n$. A change of variable tells us
\begin{align}\label{eqi1}
R_n & = \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\\
\nonumber    & = \epsilon^{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{align}
where the functions $\beta^l_n$ and $\alpha^l_n$ are defined skillfully as
$$
\beta^l_n(y) = \prod^{l - 1}_{s = 0}\eta\Big(\frac{y}{R} + \frac{x^l_n - x^s_n}{\epsilon_n R}\Big),\,\,\,\beta^0_n(y)\equiv 1
$$
and
$$
\alpha^l_n(x) = \prod^{k}_{s = l + 1}\eta\Big(\frac{x}{R} + \frac{x^l_n - x^s_n}{\epsilon_n R}\Big),\,\,\, \alpha^k_n(x) \equiv 1.
$$
Following, we have
\begin{align*}
\epsilon^{2s - N}_nR_n & = \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]
   &:=R^{(1)}_n + R^{(2)}_n + R^{(3)}_n.
\end{align*}
By the choice of $\eta$ and  $\lim\limits_{n\to\infty}\frac{|x^l_n - x^s_n|}{\epsilon_n} = \infty$ if $l\neq s$, for $n$ large, we have
\begin{align*}
R^{(1)}_n&= \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]
       :&=R^{(11)}_n + R^{(12)}_n
\end{align*}
and
\begin{align*}
        R^{(2)}_n
          =& \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]
                  :=&R^{(21)}_n + R^{(22)}_n.
\end{align*}
Also, for large $n$,
\begin{align*}
R^{(3)}_n =&\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]
:=& R^{(31)}_n + R^{(32)}_n + R^{(33)}_n.
\end{align*}

For $|R^{(i2)}_n|,\ i = 1,2$, it holds
\begin{align*}
\nonumber    &\limsup_{n\to\infty}  |R^{(i2)}_n|  \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}{(|y| - R)^{N + 2s}}\,{\mathrm{d}}x.

\end{align*}
By the fractional Hardy inequality \eqref{eq1.4} and letting $\widetilde{R} = R^{\frac{N + 1}{N}}$, we find
\begin{align}\label{eq3.3}
\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}

For $R^{(11)}_n$, by the estimates of  $R^{(i2)}_n$ above and the similar estimates of Proposition 3.4 in \cite{APX}, we have

\begin{align}\label{eq3.8}
&\limsup_{n\to\infty}  |R^{(11)}_n| \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_{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 + o_R(1)\nonumber
 \\[1mm]
& \leq C\int_{B_{2R}\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).
\end{align}
Similarly, we get
\begin{align*}
    \limsup_{n\to\infty}  |R^{(21)}_n| &\leq o_R(1)
\end{align*}
and
\begin{align*}
  \limsup_{n\to\infty}|R^{(31)}_n| + |R^{(32)}_n| + |R^{(33)}_n|\leq o_R(1).
\end{align*}
Therefore
\begin{equation}\label{impotrant}
\epsilon^{2s - N}|R_n| \leq o_R(1).
\end{equation}
Finally, letting $R\to\infty$, we  have
$$
\liminf_{n\to \infty}\frac{J_{\epsilon_n}(u_n)}{{\epsilon^N_n}}\geq\sum_{j = 1}^{k}\mathcal{C}(V(x^j_*)).
$$
\end{proof}

At last, as a result of Lemma \ref{le3.1}, we have the  following concentration for the solution $u_{\epsilon}$.

\begin{lemma}\label{le2.5}
 Let $\rho > 0$. There exists a family of points $\{x_{\epsilon}\}\subset\Lambda$ such that
\begin{eqnarray*}
  &&(i)\ \liminf_{\epsilon\to 0}\|u_{\epsilon}\|_{L^{\infty}(B_{\epsilon\rho}(x_{\epsilon}))}> 0,\\
  &&(ii)\ \lim\limits_{\epsilon\to 0}V(x_{\epsilon}) = \inf_{\Lambda}V(x),\\
  &&(iii)\ \lim\limits_{{R\to \infty}\atop{{\epsilon\to 0}}}\|u_{\epsilon}\|_{L^{\infty}(U\backslash B_{\epsilon R}(x_{\epsilon}))} =  0.
\end{eqnarray*}
\end{lemma}

\begin{proof}
It is easy to verify using the construction of $\mathcal{P}_{\epsilon}$ that
$$
\liminf_{\epsilon\to 0}\|u_{\epsilon}\|_{L^{\infty}(\Lambda)} > 0.
$$
Then by the regularity assertion in Appendix D of \cite{20}, we get the existence of $x_{\epsilon}\in\bar{\Lambda}$. We assume that $x_{\epsilon}\to x_*$.

By Lemma \ref{le3.1}, it holds
$$
\liminf_{\epsilon\to 0}\frac{\mathbf{J}_{\epsilon}(u_{\epsilon})}{\epsilon^N}\ge {\mathcal{C}}(V(x_*)),
$$
But,
$$
\inf_{\Lambda}\mathcal{C}(V(x))\ge \limsup_{\epsilon\to 0}\frac{\mathbf{J}_{\epsilon}(u_{\epsilon})}{\epsilon^N}.
$$
Hence by the monotonicity of $\mathcal{C}(\cdot)$, $V(x_*) = \inf_{\Lambda}V(x).$

Arguing by contradiction, if $(iii)$ is  not hold, we can infer from Lemma \ref{le3.1} that
$$
\inf_{\Lambda}\mathcal{C}(V(x))\ge\liminf_{\epsilon\to 0}\frac{\mathbf{J}_{\epsilon}(u_{\epsilon})}{\epsilon^N}\ge 2\inf_{\Lambda}\mathcal{C}(V(x)),
$$
which is a contradiction. The proof is then completed.

\end{proof}

\vspace{0.5cm}

\section{Back to the original problem}\label{s4}

\noindent In this section, we show that $u_{\epsilon}^{p - 2}\leq \mathcal{P}_{\epsilon}$ by comparison principle.

{By Lemma \ref{le2.5} and the $L^{\infty}$ estimate in \cite[Appendix D]{20}, we can assume that there exists a positive constant $C_{\infty}$ such that
\begin{equation}\label{eq4.0}
\sup_{x\in \Lambda}|u_{\epsilon}(x)|\leq C_{\infty}
\end{equation}
if $\epsilon$ is small enough,
from which we can linearize the penalized problem \eqref{eq2.2} outside small balls as follows.}

\begin{proposition}\label{pr4.1} For $\epsilon > 0$ small enough and $\delta\in (0,1)$, there exist $R > 0$ and $x_{\epsilon}\in \Lambda$  such that
\begin{equation}\label{AAeq4.2}
  \left\{
    \begin{array}{ll}
      \epsilon^{2s}(-\Delta)^{s} u_{\epsilon} + (1 - \delta)Vu_{\epsilon}\leq \mathcal{P}_{\epsilon}(x)u_{\epsilon}(x), & \text{in}\ \mathbb{R}^N\backslash  B_{R\epsilon}(x_{\epsilon}), \vspace{2mm}\\
      u_{\epsilon}\leq C_{\infty} & \text{in}\ \Lambda.
    \end{array}
  \right.
\end{equation}
\end{proposition}
\noindent Letting $v_{\epsilon}(x) = u_{\epsilon}(\epsilon x + x_{\epsilon})$, it is easy to check that
\begin{equation}\label{AAeq4.3}
  \left\{
    \begin{array}{ll}
      (-\Delta)^{s} v_{\epsilon} + (1 - \delta)V_{\epsilon}(x)v_{\epsilon}\leq \widetilde{P}_{\epsilon}(x)u_{\epsilon}(x), & \text{in}\ \mathbb{R}^N\backslash  B_{R}(0), \vspace{2mm}\\
      v_{\epsilon}\leq C_{\infty} & \text{in}\ B_{R}(0),
    \end{array}
  \right.
\end{equation}
where $V_{\epsilon}(\cdot) = u_{\epsilon}(\epsilon \cdot + x_{\epsilon})$ and $\widetilde{P}_{\epsilon}(\cdot) = \mathcal{P}_{\epsilon}(\epsilon \cdot + x_{\epsilon})$.

 Now we construct a suitable sup-solution to equation \eqref{AAeq4.3}. Let ${ \tilde{\eta}_\beta }(|x|)  :\mathbb R^N\to [0,1]$ be a smooth non-increasing function with ${ \tilde{\eta}_{\beta}  }(s) \equiv 1$         when  $ s \in [-1,1]$ and $\tilde{\eta}_{\beta}\equiv 0$ when  $  s\in (-\infty,-1 - \beta)\bigcup(1 + \beta,+\infty)$, where $\beta>0$ is a small parameter.
Define $\eta_{\beta}(|x|) = \tilde{\eta}_{\beta}(R|x|)$. It is worth mentioning  that the non-increasing of $\eta$ is important, see \eqref{Ieq1} below.
Setting $0<\alpha< N -2s$ and denoting  $$f^{\beta}_{\alpha}(x) = \eta_{\beta}(x)\frac{1}{R^{\alpha}} + (1 - \eta_{\beta}(x))\frac{1}{|x|^{\alpha}},$$  we have:
\begin{proposition}\label{pr3.2}
Let $\epsilon>0$ be small enough. Then for every $x\in \mathbb{R}^N\backslash B_R(0)$, it holds
\begin{equation}\label{AAeq4.5}
      (-\Delta)^{s} f^{\beta}_{\alpha} + V_{\epsilon}(x)f^{\beta}_{\alpha} - \widetilde{P}_{\epsilon}(x)f^{\beta}_{\alpha}\geq 0.
\end{equation}
\end{proposition}

\begin{proof}
For $x\in \mathbb R^N\backslash B_{2R}(0)$, letting $\beta>0$ be small enough, it holds
\begin{align*}
&(-\Delta)^sf^{\beta}_{\alpha}(x)\\
                  & = \int_{{ B_R(0)}}\frac{|x|^{-\alpha} - {R^{-\alpha}}}{|x - y|^{N + 2s}}\,{\mathrm{d}}y + \int_{\mathbb{R}^N\backslash{{ B_R(0)}}}\frac{|x|^{-\alpha} - |y|^{-\alpha}}{|x - y|^{N + 2s}}\,{\mathrm{d}}y + \int_{\mathbb{R}^N\backslash{{ B_R(0)}}}\frac{\eta(y)|y|^{-\alpha} - \eta(y){R^{-\alpha}}}{|x - y|^{N + 2s}}\,{\mathrm{d}}y
                  \\[1mm]
                  & = (-\Delta)^s(|\cdot|^{\alpha})(x) +  \int_{{{ B_R(0)}}}\frac{|y|^{-\alpha} - {R^{-\alpha}}}{|x - y|^{N + 2s}}\,{\mathrm{d}}y + \int_{\mathbb{R}^N\backslash{{ B_R(0)}}}\frac{\eta(y)|y|^{-\alpha} - \eta(y){R^{-\alpha}}}{|x - y|^{N + 2s}}\,{\mathrm{d}}y
                  \\[1mm]
                  & \ge \frac{f^{\beta}_{\alpha}(x)}{|x|^{2s}}\Big(\int_{(B_1(0))^c}\frac{1- |z|^{\alpha}}{|z|^{N - 2s}|z - \vec{e}_1|^{N + 2s}}dz + \int_{(B_1(0))^c}\frac{|z|^{\alpha} - 1 }{|z|^{\alpha}|z - \vec{e}_1|^{N + 2s}}dz\Big)
                  \\[1mm]
                  &\quad +  R^{N - \alpha}\Big(\int_{B_1(0)}\frac{|y|^{-\alpha} - 1}{|x - yR|^{N + 2s}}\,{\mathrm{d}}y + \int_{B_{{1 + \beta}}(0)\backslash{{ B_1(0)}}}\frac{|y|^{-\alpha} - {1}}{|x - yR|^{N + 2s}}\,{\mathrm{d}}y\Big)
                  \\[1mm]
                  &\ge C_{\alpha}\frac{f^{\beta}_{\alpha}(x)}{|x|^{2s}}
\end{align*}

When $x\in B_{2R}(0)\backslash B_R(0)$, by the construction of $\eta_{\beta}$ and the computation above, it holds
\begin{align*}
&\quad(-\Delta)^sf^{\beta}_{\alpha}(x)\\
                  & = \int_{\mathbb R^N}\frac{(|x|^{-\alpha} - |y|^{-\alpha}) + \eta_{\beta}(y)(|y|^{-\alpha} - |x|^{-\alpha})}{|x - y|^{N + 2s}}\,{\mathrm{d}}y
                   \\[1mm]
                  & \quad + \int_{\mathbb R^N}\frac{(\eta_{\beta}(x) - \eta_{\beta}(y))R^{-\alpha} + (\eta_{\beta}(y) - \eta_{\beta}(x))|x|^{-\alpha}}{|x - y|^{N + 2s}}\,{\mathrm{d}}y
                  \\[1mm]
                  &= (R^{-\alpha} - |x|^{-\alpha})\int_{\mathbb R^N}\frac{\eta_{\beta}(x)- \eta_{\beta}(y)}{|x - y|^{N + 2s}}\,{\mathrm{d}}y + \int_{\mathbb R^N}\frac{\eta_{\beta}(y)(|y|^{-\alpha} - |x|^{-\alpha})}{|x - y|^{N + 2s}}\,{\mathrm{d}}y
                  \\[1mm]
                  &= (R^{-\alpha} - |x|^{-\alpha})\frac{(-\Delta)^s\tilde{\eta}_{\beta}(R/x)}{R^{2s}} + \int_{\mathbb R^N}\frac{\eta_{\beta}(y)(|y|^{-\alpha} - |x|^{-\alpha})}{|x - y|^{N + 2s}}\,{\mathrm{d}}y
                  \\[1mm]
                  & :\ge {-C_{\beta}}{R^{-(\alpha + 2s)}} + K^{\beta}_{\alpha},
\end{align*}
where $C_{\beta}$ is a positive constant depending only on $\beta$.

For $K^{\beta}_{\alpha}$, by Change-Of-Variable Theorem and the non-increasing of $\tilde{\eta}_{\beta}$, it holds
\begin{align}\label{Ieq1}
|x|^{\alpha + 2s}K^{\beta}_{\alpha}
\nonumber& = \int_{B_{1}(0)}\frac{\eta_{\beta}(y|x|)(|y|^{-\alpha} - 1)}{|y - \vec{e}_1|^{N + 2s}}\,{\mathrm{d}}y + \int_{B^c_{1}(0)\cap B_{1 + \beta}(0)}\frac{\eta_{\beta}(y|x|)(|y|^{-\alpha} - 1)}{|y - \vec{e}_1|^{N + 2s}}\,{\mathrm{d}}y
\\[1mm]
\nonumber &\ge \int_{B_{1}(0)}\frac{\eta_{\beta}(|x|)(|y|^{-\alpha} - 1)}{|y - \vec{e}_1|^{N + 2s}}\,{\mathrm{d}}y + \int_{B^c_{1}(0)\cap B_{1 + \beta}(0)}\frac{\eta_{\beta}(|x|)(|y|^{-\alpha} - 1)}{|y - \vec{e}_1|^{N + 2s}}\,{\mathrm{d}}y
\\[1mm]
& \ge \eta_{\beta}(|x|)\int_{B_{3}(\vec{e}_1)}\frac{(|y|^{-\alpha} - 1)}{|y - \vec{e}_1|^{N + 2s}}\,{\mathrm{d}}y\\[1mm]
\nonumber&\ge \widetilde{C}_{\alpha},
\end{align}
where $\widetilde{C}_{\alpha}$ is a constant depending only on $\alpha$(it will hold $\widetilde{C}_{N - 2s} > 0$ if $\alpha = N - 2s$).

Finally, by the computation above, we conclude that if $R$ is large enough and $\epsilon$ is small enough, it holds
\begin{align}\label{AAeq4.5}
     \nonumber &\quad(-\Delta)^{s} f^{\beta}_{\alpha} + V_{\epsilon}(x)f^{\beta}_{\alpha} - \widetilde{P}_{\epsilon}(x)f^{\beta}_{\alpha}\\[2mm]
&\ge \left\{
       \begin{array}{ll}
         -(C_{\beta} + \widetilde{C}_{\alpha}) R^{-\alpha - 2s} + \big(\inf_{x\in\Lambda_{\epsilon}}V(x)\big)f^{\beta}_{\alpha}(x), & x\in B_{2R}(0)\backslash B_R(0), \\[5mm]
         C_{\alpha}\frac{f^{\beta}_{\alpha}}{|x|^{2s}} + V_{\epsilon}(x)f^{\beta}_{\alpha} - \widetilde{P}_{\epsilon}(x)f^{\beta}_{\alpha}, & x\in\mathbb R^N\backslash B_{2R}(0),
       \end{array}
     \right.\\[2mm]
\nonumber&\ge 0
\end{align}
for every $x\in \mathbb R^N\backslash B_R(0)$, where $\Lambda_{\epsilon} = \{x:\epsilon x + x_{\epsilon}\in\Lambda\}$. This completes the proof.

\end{proof}

\noindent At the last of this section, we give the proof of  Theorem \ref{th1.1}

\textbf{Proof of Theorem \ref{th1.1}}.
Let
\begin{align}\label{eq4.5}
\left\{
  \begin{array}{ll}
    \alpha \in\Big(\frac{2s}{p - 2},N - 2s\Big),\ \kappa = \frac{\alpha(p - 2) - 2s}{4},
\mathcal{P}_{\epsilon}(x) = \frac{\epsilon^{2s + 2\kappa}}{|x|^{2s + \kappa}}\chi_{\mathbb R^N\backslash\Lambda}(x), & \\[5mm]
    \overline{U}(x) = CR^{\alpha}f^{\beta}_{\alpha}(x), &
  \end{array}
\right.
\end{align}
where the function $f^{\beta}_{\alpha}$ is that in Proposition \ref{pr3.2}.

 It is easy to check that $\mathcal{P}_{\epsilon}$ satisfies the assumption \eqref{eq2.1}.

 By Proposition \ref{pr3.2}, letting the constant  $C>0$ above be large enough and $\tilde{v}_{\epsilon}(x) = \overline{U}(x) - v_{\epsilon}(x)$, we have
\begin{equation*}
  \left\{
    \begin{array}{ll}
      (-\Delta)^{s} \tilde{v}_{\epsilon}(x) + V_{\epsilon}(x)\tilde{v}_{\epsilon}(x) - \widetilde{P}_{\epsilon}(x)\tilde{v}_{\epsilon}(x)\geq 0, & \text{in}\ \mathbb{R}^N\backslash  B_{R}(0), \vspace{2mm}\\[5mm]
      \tilde{v}_{\epsilon}(x)\ge 0 & \text{in}\ B_R(0).
    \end{array}
  \right.
\end{equation*}
Then, since $\tilde{v}_{\epsilon}\in \dot{H}^s(\mathbb R^N)$   (when $\alpha$ is closed to $N - 2s$), testing the equation above against with $\tilde{v}^-_{\epsilon}(x)$, by the fractional Hardy inequality in \eqref{eq1.4}, we find $\tilde{v}^-_{\epsilon}(x) = 0,\ x\in\mathbb R^N$. Hence $\tilde{v}_{\epsilon}(x)\ge 0,\ x\in\mathbb R^N$. Especially, we have
$$
u_{\epsilon}(x)\le \frac{C\epsilon^{\alpha}}{\epsilon^{\alpha} + |x - x_{\epsilon}|^{\alpha}},\ \forall \,  x\in\mathbb R^N.
$$
Moreover, it holds
$$
\big(u_{\epsilon}(x)\big)^{p - 2}\le \mathcal{P}_{\epsilon}(x),\ \forall  \, x\in\mathbb R^N\backslash\Lambda.
$$

As a result, $u_{\epsilon}$ solves the origin problem. This completes the proof.

\vspace{0.5cm}
\section{Further results}\label{s5}

In this section, we will consider \eqref{eq1.1} with general nonlinearity, i.e.,
\begin{equation}\label{eq5.1}
  \epsilon^{2s}(-\Delta)^su + V(x)u = f(u),
\end{equation}
where the potential $V(x)$ is the same as before, the nonlinearity $f:\mathbb{R}\to \mathbb{R}$ is assumed to satisfy the following properties:

$(\textbf{f}_1)$ $f$ is an odd function and $f(t) = o(t^{1 + \tilde{\kappa}})$ as $t\to 0^+$, where $\tilde{\kappa} = \frac{2s + 2\kappa}{\alpha -\nu}>0$ with $\nu>0$ is a small parameter and $\kappa$ is the parameter in \eqref{eq4.5}.

$(\bf{f}_2)$  $\lim_{t\to\infty}\frac{f(t)}{t^p} = 0$ for some $1<p<2^*_{s} - 1$.

$({ \bf{f} }_3)$ There exists $2<\theta \leq p + 1$ such that
$$
0\leq\theta F(t) < f(t)t\ \text{for all}\ t>0,
$$
\ \ \ \ \ \ \ \ \ where $F(t) = \displaystyle\int_{0}^tf(\alpha)\,{\mathrm{d}} \alpha$.

$(\textbf{f}_4)$ The map $t\mapsto \frac{f(t)}{t}$ is increasing on $(0,+\infty)$.

\vspace{0.2cm}

\noindent We have the following results  which are  same as Theorem \ref{th1.1}.

\begin{theorem}\label{th5.1}
Let $N > 2s$, $s\in(0,1)$, $p\in\Big(2+\frac{2s}{N - 2s},2^*_s\Big)$, $f$ satisfy $(\bf{f}_1)$- $(\bf{f}_4)$ and $V$ be the same as before. Then problem \eqref{eq1.1} has a positive solution $\hat{u}_{\epsilon}\in \mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N)$ if $\epsilon>0$ is small enough. Moreover, there exist a $\hat{x}_{\epsilon}\in\Lambda$ and an $\alpha\in(2s/(p - 2),2^*_s)$ such that
$$
 \lim_{\epsilon\to 0}V(\hat{x}_{\epsilon}) = \min_{\Lambda}V(x)
 $$
 and
\begin{align*}
u_{\epsilon}(x)\le \frac{C\epsilon^{\alpha}}{\epsilon^{\alpha} + |x - \hat{x}_{\epsilon}|^{\alpha}},
\end{align*}
where $C$ is positive constant.

\end{theorem}

\begin{proof}

We define the penalized nonlinearity   $\hat{g}_{\epsilon}:\mathbb{R}^N\times\mathbb{R}$ as
$$
\hat{g}_{\epsilon}(x,s):=\chi_{\Lambda}f(s_+) + \chi_{\mathbb{R}^N\backslash\Lambda}\min\{f(s_+),\mathcal{P}_{\epsilon}(x)s_+\}.
$$
In the sequel, we denote $\widehat{G}_{\epsilon}(x,t) = \displaystyle\int_{0}^{t}\hat{g}_{\epsilon}(x,s) \,{\mathrm{d}}s $ and define the penalized superposition operators $\hat{\mathbf{g}}_{\epsilon}$ and $\widehat{\mathbf{G}}_{\epsilon}$ as
$$
\hat{\mathbf{g}}_{\epsilon}(u)(x) = \hat{g}_{\epsilon}(x,u(x))\ \text{and}\ \widehat{\mathbf{G}}_{\epsilon}(u)(x) = \widehat{G}_{\epsilon}(x,u(x)).
$$
Accordingly, the penalized functional $\hat{J}_{\epsilon}:\mathcal{D}^s_{V,\epsilon}(\mathbb{R}^N)\to \mathbb{R}$ is given by
$$
\hat{J}_{\epsilon}(u) = \frac{1}{2}\int_{\mathbb{R}^N}(\epsilon^{2s}|(-\Delta)^{s/2} u|^2 + V(x)|u|^2) \,{\mathrm{d}}y   - \int_{\mathbb{R}^N}\widehat{\mathbf{G}}_{\epsilon}(u) \,{\mathrm{d}}y.
$$
By conditions $(f_2)$ and $(f_3)$, we can verify using the same argument in the proof of Lemma \ref{le2.2} that $\hat{J}_{\epsilon}$ is $C^1$ and satisfies (P.S.) condition. Then similar to Lemma \ref{le2.3}, one can show that the Mountain Pass value
$$
\hat{c}_{\epsilon} := \inf_{\gamma\in\widehat{\Gamma}_{\epsilon}}\max_{t\in [0,1]}\hat{J}_{\epsilon}(\gamma(t))
$$
can be achieved by a positive function $\hat{u}_{\epsilon}\in \mathcal{D}^s_{V,\epsilon}$, where
\begin{equation*}
\widehat{\Gamma}_{\epsilon} : = \{\gamma\in (C[0,1],\mathcal{D}^s_{V,\epsilon}(\mathbb R^N)):\gamma(0) = 0,\ \hat{J}_{\epsilon}(\gamma(1)) < 0\}.
\end{equation*}
Furthermore, it holds
\begin{equation}\label{eq5.2}
\epsilon^{2s}(-\Delta)^s\hat{u}_{\epsilon} + V(x)\hat{u}_{\epsilon} = \hat{\mathbf{g}}_{\epsilon}(\hat{u}_{\epsilon}).
\end{equation}

From \cite{S,FQT} and the condition on $f$, the following limiting problem
$$
(-\Delta)^su + au = f(u)
$$
has a positive ground state $\hat{u}$. Moreover, the least energy
$$
\hat{\mathcal{C}}(a) = \frac{1}{2}\Big(\int_{\mathbb R^N}|(-\Delta)^{s/2}\hat{u}|^2 + a|\hat{u}|^2\Big) \,{\mathrm{d}}x   - \int_{\mathbb R^N}F(\hat{u}) \,{\mathrm{d}}x
$$
is continuous and increasing.  Then, by the same argument in Lemma \ref{le2.5}, there exists a $\hat{x}_{\epsilon}\in\Lambda$ such that
\begin{eqnarray}\label{eq5.9}
\begin{split}
  &(1')\ \liminf_{\epsilon\to 0}\|\hat{u}_{\epsilon}\|_{L^{\infty}(B_{\epsilon\rho}(\hat{x}_{\epsilon}))}> 0,\\
  &(2')\ \lim_{\epsilon\to 0}V(\hat{x}_{\epsilon}) = \inf_{\Lambda}V,\\
  &(3')\ \lim\limits_{{R\to \infty}\atop{{\epsilon\to 0}}}\|\hat{u}_{\epsilon}\|_{L^{\infty}(U\backslash B_{\epsilon R}(\hat{x}_{\epsilon}))} =  0.
\end{split}
\end{eqnarray}
As a result, using $(f_1)$ to linearize \eqref{eq5.2}, we get
\begin{equation}\label{eq5.10}
  \left\{
    \begin{array}{ll}
      \epsilon^{2s}(-\Delta)^{s} \hat{u}_{\epsilon} + (1 - \delta)V\hat{u}_{\epsilon}\leq {P}_{\epsilon}\hat{u}_{\epsilon}, & \text{in}\ \mathbb{R}^N\backslash  B_{R\epsilon}(x_{\epsilon}), \vspace{2mm}\\
      \hat{u}_{\epsilon}\leq \widehat{C}_{\infty} & \text{in}\ \Lambda.
    \end{array}
  \right.
\end{equation}
Then, by the same argument in Section \ref{s4}, it holds
$$
\hat{u}_{\epsilon}(x)\le \frac{C\epsilon^{\alpha}}{\epsilon^{\alpha} + |x - \hat{x}_{\epsilon}|^{\alpha}},\ \forall x\in\mathbb R^N,
$$
where $\alpha$ is the same as that in \eqref{eq4.5}. Following, for every $x\in\mathbb R^N\backslash\Lambda$, since $\alpha\tilde{\kappa}>2s + 2\kappa$, letting $\epsilon>0$ be small enough, it holds
\begin{align*}
 \frac{f(\hat{u}_{\epsilon})}{\hat{u}_{\epsilon}}&\le (\hat{u}_{\epsilon})^{\tilde{\kappa}}\le \frac{C\epsilon^{\alpha\tilde{\kappa}}}{|x|^{\alpha\tilde{\kappa}}}\le \frac{\epsilon^{2s + 2\kappa}}{|x|^{2s + \kappa}} = \mathcal{P}_{\epsilon}(x).
\end{align*}

As a result, $\hat{u}_{\epsilon}$ is a solution to the origin problem \eqref{eq5.1}. This completes the paper.

\end{proof}

\bigskip
\bigskip
{\bf Acknowledgements: }   L.  Duan  is partially  supported by China Scholarship Council
(No.201906770024).
Y.  Peng  is supported by  the grant NSFC (No. 11501143).

redpajama_set_name: RedPajamaArXiv


text:
#NAMA Has Been a Dream Come True for Many US Vulture Funds...
Dáil Diary no 48 - 10 July 2015
The controversy over NAMA's sale of the Nothern Ireland loan portfolio, Project Eagle, has not gone away. It suited the Government just fine when the PAC said they would bring NAMA before the committee. - It also suited NAMA, whose polished performance was always going to give them some breathing space, given that the PAC, - no disrespect to any of it's members - just doesn't have the wherewithal to hold them to account. Here is a part of my piece on the workings of NAMA from last Tuesday, when we were debating a Housing Bill . -
"Aside from what seem to be a lot of spurious deals where it appears the taxpayer could have fared a great deal better, NAMA is actually boasting that it will make approximately €1 billion in profit over the six to eight year period. That still leaves the taxpayer about €40 billion short. Given that the market has been rising for a while, it is outrageous that more was not realised.
The point the Deputy made about the social dividend and NAMA's failure to provide it is a frightening one. What is really uncomfortable is the fact that we have a serious level of inconsistency in how NAMA has done its business. Sometimes it applies its rules rigidly and sometimes it does not. A simple example is that of the Dublin GAA trying to get hold of the Spawell but failing to do so. It beggars belief that a sporting organisation in Ireland would not have been facilitated and given a little bit of an extra chance. It is not like the GAA was going to make money on it. It intended to provide a social outlet for young people and should have been helped by NAMA. NAMA will argue that it is under an obligation to maximise the potential of the asset for the taxpayer. When it makes that argument, I would like NAMA to explain its approach when a bidder went to buy not the loans but the debt of the Chicago Spire, which was at $78 million plus costs which brought it to approximately $93 million. An investor sought to buy the debt, and this was every penny that was owed to the bank. This was not the reduced value, but the par value.
In other words, this investor was prepared to pay the debt in full but NAMA gave it to Jones Lang LaSalle in New York to sell. This was a site in Chicago. Even if NAMA thought it could get more for it, it was not in New York that it would have got it. It would have been interesting if it had marketed it in Chicago. Why could NAMA not accept the debt being bought out? It is estimated that it was sold for $35 million. NAMA refused $78 million, plus the cost, and it accepted a figure in the region of €35 million. That was claimed to be in the interests of the taxpayer in the same way as NAMA not accepting the bid from Dublin GAA for the Spawell because it claimed it was maximising the potential of the asset.
It is horrendous that NAMA could not deliver housing units for Dublin, in particular where there is such a serious housing crisis. It beggars belief. The argument was spun that certain property was not really suitable. The funny thing about that is that NAMA did not want to give much of the property that was suitable to the State because it was attractive to investors. What has happened is that vulture funds, mostly from the US, were allowed to cherry pick the best of it because NAMA sought to sell the best of it to them. It then considered some of the other property, which was not quite as attractive to the vulture funds, for social or affordable housing.
NAMA is a State body over which governments - this one and the previous one - do not seem to have much control. Issues are raised about it. The idea that the Comptroller and Auditor General has it all in hand beggars belief. I have seen too many things which I do not believe the Comptroller and Auditor General has seen, and I am going to give him the benefit of the doubt that he is honest. It would be an incredible challenge for the Comptroller and Auditor General and the few people working with him to keep tabs on everything that is happening within NAMA. It would be a massive challenge, and it is not only about auditing. Does the Comptroller and Auditor General have expertise in the area of markets, construction and the future potential of assets? Does he have all those people working for him in order to be able to test the waters and ensure NAMA is doing what is best for the taxpayer at every turn? It is asking way to much of the Comptroller and Auditor General to think that he could come up with all those answers. He would be an amazing man, irrespective of how many people he had working with him.
Sadly, there is now the idea that the Committee of Public Accounts can examine this, behind which the Government can hide, as if it is an Oireachtas investigation and the committee will see if everything is all right. That is not what will happen. If those in government want to know the truth and if they really want to know if the taxpayer was best served in these deals, they will eventually have to initiate an independent inquiry into the workings of NAMA. It is extremely important because not only are we dealing with things that have happened - people must be held responsible for what they have done - but there are so many assets still to be sold. Will that be done right? At present, we do not know and given that NAMA remains a secret society, we do not have a real opportunity to see how it operates. An independent investigation might bring us that. There is a great deal of money at stake. The figures are astronomical so it is extremely important the State addresses this.
The Government should not put this off and kick the can down the road. We need an independent inquiry established right now. How it would be structured is an argument for another day. The Government will probably have to bring in some people from outside the country. I refer to many of the more senior players in the services the Government might use. There is such an incestuous nature to much of what goes on this country that the Government will probably have to bring in some external expertise."
Good heartiness is a result of proper supply and hygiene. How can medicaments hels up? Circumstances that can influence your choice when you are buying medications are different. Below are basic reasons about cialis vs levitra vs viagra which one is better. Surely there are also other momentous questions. Choosing the perfect treatment option for a racy disease can get really confusing considering the advantages and disadvantages of the existing treatment methodologies. When you buy remedies like Cialis you have to have in mind about levitra vs cialis vs viagra. The most significant thing you must look for is which works better viagra or cialis or levitra. A long list of prescription drugs can lead to erectile disfunction, including many blood stress medicines, pain remedies, and most of antidepressants. Sometimes the treatment options may switch on erectile dysfunction remedies or hormone treatments.

redpajama_set_name: RedPajamaCommonCrawl


text:
The Deep North
One more home in Cyberspace
Image Composer Undo
Hack to have multiple Undo Levels in Image Composer:
Set a registry entry thus:
[HKEY_CURRENT_USER\Software\Microsoft\Microsoft Image Composer\Settings]
"MaxUndoLevels"=dword:00000010
I had to actually create the entry, but YMMV.
I was amazed that this worked. Why this is not a default or at least a configuration option is beyond me.
Sadly, this no longer works (posted July2022). About 2 Windows 10 updates back, the multiple undo's just stopped working. Whilst you can undo multiple "text" creations, it only allows ONE undo for any sprite created, effectively rendering the program virtually unusable.
I did that mod many many years back and have been using MIC as my default graphic editor since it came out on the old FrontPage CD. I do graphics professionally, so losing the multiple undo fuinctionality is a massive blow.
I will have to setup a PC running Windows 7 and get rid of this latest "bloat/spyware" called Windows 10
DeepNorth said…
Yikes. I tested and undo was dead on my system as well. Mercifully, it is just because the value was deleted from the registry. Putting it back in fixed the problem. Microsoft is psychopathic. Thanks for making the comment. It would have driven me nuts if I found it first!
The system cannot execute the specified program
It always annoys me no end when I get messages like the following: "The system cannot execute the specified program." I got the above error from Windows XP when I tried to execute a program I use all the time. The message is hugely aggravating because it says the obvious without giving any actionable information. If you have such a problem and you are executing from a deep directory structure that may be your problem. It was in my case. Looking on the web with that phrase brought up a bunch of arcane stuff that did not apply to me. It mostly brought up long threads (as these things tend to do) which follow this pattern: 'Q' is the guy with the problem asking for help 'A' can be any number of people who jump in to 'help'. Q: I got this error "The system cannot execute the specified program." when I tried to ... [long list of things tried] A: What program were you running, what operating system, where is the program? What type of
The JWST did not take resources away from the poor.
The JWST project employed many people for a long time. It helped them and their communities. Like other NASA projects, it has funded technology breakthroughs that will pay back the investment. Right now, mathematical physicists are working on a new theory that combines quantum physics and relativity and explains how the two emerge from a common root structure. This will be aided by knowledge of the early universe we get from JWST. A breakthrough there could lead to essentially limitless low-cost energy. It could possibly lead to crazy stuff like anti-gravity and magnetic monopoles. It could extend the periodic table, and allow us to create exotic substances, it could allow us to create room temperature superconductors, and it could conceivably allow us to figure out how to explore the galaxy. Our understanding of electromagnetism since the 19th century has allowed us to do miraculous things that people in earlier centuries would simply think of as pure magic. Reaching for the stars is
Coming Soon: General Artificial Intelligence
The closer you get to experts who understand the nuts and bolts and history of AI, the more you find them saying that what we have is not nearly General Artificial Intelligence (GAI), and that GAI seems far away. I think we already have the roots in place with Neural Networks (NN), Deep Learning (DL), Machine Learning (ML), and primitive domain limited Artificial Intelligence (AI). Things like computer vision, voice recognition, and language translation are already in production. These are tough problems, but in some ways, machines are already better than humans are. I expect GAI to be an emergent property as systems mature, join, and augment one another. I was around during the 70s AI winter, and was involved in the 80s AI winter as one of the naysayers. I built a demonstration system with a Sperry voice recognition card in 1984. I could demonstrate it in a quiet room, but as a practical matter, it was not production ready at all. Around 1988 we built demonstration expert systems usin
DeepNorth
Brockville, Ontario, Canada
The web's worst kept secret identity.
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Is it possible to use the Handle Animal skill to calm down a hostile animal?
We've recently encountered a pack of wolves that were trained by goblins, but their masters weren't anywhere nearby. Is it possible to use Handle animal to make them less hostile?
Should readying an action to set a brace weapon dissuade an intelligent enemy from charging?
Can you hire an hireling in order to produce things for you?
Do hybrid classes get everything from both classes during advancement?
How do you roll for a business when you are out of town?
Does Aegis Augment Weapon suit option affect ranged weapons they use?


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text:
Dominican National Charged with Identity Theft, Government Theft
From the U.S. Attorney's Office, District of Massachusetts:
BOSTON – A Dominican national was charged yesterday in federal court in Boston in connection with making false statements, stealing public money, and aggravated identity theft.
Daniel Polonia Morillo, 57, a Dominican national residing in Lawrence, was indicted on one count of false statement on an application for Social Security benefits, one count of false statement in an application and use of a passport, two counts of theft of public money, and one count of aggravated identity theft. Polonia Morillo was arrested and charged by criminal complaint during an October 2018 federal law enforcement sweep in and around the City of Lawrence.
According to the charging documents, Polonia Morillo used the name, Social Security number, and date of birth of another individual to apply for Social Security benefits in December 2016. At the time he applied, Polonia Morillo produced a United States passport in the other person's name as proof of his identity. In addition, Polonia Morillo is charged with stealing over $40,000 in Medicaid benefits from March 2012 through October 2018, and over $4,000 in Food Stamps from April 2015 through October 2018.
Aggravated identity theft carries a mandatory sentence of two-years in prison that must run consecutively to any other sentence imposed, one year of supervised release, and a fine of $250,000. The charge of theft of public money provides for a sentence of no greater than 10 years in prison, three years of supervised release, and a fine of $250,000 or twice the gross gain or loss, whichever is greater. The charge of false statement for Social Security benefits provides for a sentence of no greater than five years in prison, three years of supervised release, and a fine of $250,000. The charge of false statement in an application of a passport provides for a sentence of no greater than 10 years in prison, three years of supervised release, and a fine of $25,000. Polonia Morillo will be subject to deportation proceedings. Sentences are imposed by a federal district court judge based upon the U.S. Sentencing Guidelines and other statutory factors.
United States Attorney Andrew E. Lelling; Scott Antolik, Special Agent in Charge of the Social Security Administration, Office of Inspector General, Boston Field Office; Peter C. Fitzhugh, Special Agent in Charge of Homeland Security Investigations in Boston; and Massachusetts State Auditor Suzanne M. Bump made the announcement. Special Assistant U.S. Attorney Karen Burzycki of Lelling's Major Crimes Unit is prosecuting the case.
The details contained in the charging documents are allegations. The defendant is presumed innocent unless and until proven guilty beyond a reasonable doubt in a court of law.


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text:
China detains rights lawyer critical of President Xi Jinping
Chinese human rights lawyer Yu Wensheng poses for a photograph in Beijing in 2016. | AFP-JIJI
BEIJING – Beijing authorities on Friday took outspoken rights lawyer Yu Wensheng into custody as he was taking his child to school, his wife said, days after Yu said he was stripped of his legal license for criticizing President Xi Jinping.
Yu, a Beijing-based lawyer and activist, has repeatedly criticized the ruling Communist Party over a sweeping crackdown on rights lawyers and activists, which has seen hundreds detained and dozens arrested in recent years.
The authorities revoked Yu's license on Monday, citing rules requiring he be employed at a registered firm, but Yu told Reuters he believed the decision was in part reprisal for an open letter he wrote criticizing Xi's "totalitarian" rule.
Early on Friday morning, as Yu was taking his child to school, he was stopped by a squad of four police vehicles and over 10 officers, and taken into custody, Yu's wife Xu Yan told Reuters.
"I have not received any legal document about his detention and don't even know what crime he is suspected of committing," she said.
Xu added that she is in the process of searching for a lawyer to defend Yu if necessary, but many of those lawyers she knows in Beijing also have problems with their licenses.
It is unclear where Yu is being held and if he has been charged with any crime. China's ministry of public security did not immediately respond to a request for comment.
On Thursday, Yu released another open letter calling for changes to China's state constitution to make elections fairer, reduce military power and delete a preamble that gives the Communist Party primacy in national leadership.
Yu had been involved in a number of high-profile and politically sensitive cases, including defending fellow veteran human rights lawyer Wang Yu, following her detention in the summer of 2016.
Yu was detained briefly in October 2017 during a meeting of China's top leadership after he wrote a letter in which he said Xi was unsuited to rule China and calling for his replacement.
Most recently, Yu had been attempting to defend another detained rights lawyer, Wang Quanzhang, but the authorities have barred him from meeting Wang.
Wang was detained 920 days ago and his case has yet to come to trial. Neither his wife or any of the lawyers that she has asked to defend Wang, including Yu, have been able to meet with him since.
China, rights, Xi Jinping, Yu Wensheng


The use of mobile devices in small and enterprise businesses has exploded in the last 2 years, and with increased demand for use of non-corporate owned devices in the workplace, and a range of mixed mobile platforms such as iOS, Android, Windows Phone and Blackberry has only increased the challenge for businesses to both manage and safely enable employees . Thankfully, there is a multitude of approaches to providing a risk-balanced means of use for these devices through policy, procedures, guidelines and technology enforcement, and the experts at JAW Consulting UK can help to create an ideal security architecture for your business.
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Pierre Bodin 1721–1824 – webtrees Nos Trois Branches - Le G.A.C.
Pierre Bodin, son of Pierre Bodin and Marie Perrineau, was born on April 25, 1721 in Saint-André-sur-Sèvre, Deux-Sèvres, Poitou-Charentes and died on November 9, 1824 in Saint-Marsault, La Forêt-sur-Sèvre, Deux-Sèvres at the age of 103. He married Perrine Maupillier, daughter of Pierre Maupillier and Marie Bichon, before 1765.
Perrine Bodin, bordière, daughter of Pierre Bodin and Perrine Maupillier, was born in 1766 and died on July 15, 1840 in Saint-Marsault, Deux-Sèvres, Poitou-Charentes at the age of 74. She married 2 times. The first time she married Jean Bruneau. The second time she married Pierre Blochas.
Jean Baptiste Bruneau, bordier (on July 15, 1840), son of Jean Bruneau and Perrine Bodin, was born calculated 1807.
Louis Blochas, metayer (on July 15, 1840), son of Pierre Blochas and Perrine Bodin, was born calculated 1783.


There's a lesson I'm trying to subtly telegraph: all reporters need to be busy pointing the O-D's spotlight at the community, illuminating issues for public discussion.
"Shining the light" is one of the most important jobs on any newspaper's to-do list. Sunshine, after all, is the best disinfectant – and it's especially effective with government, which often grows mold when no one is looking.
I like to think of the O-D as the aerosol can containing the sunshine. We need to be spritzing it everywhere.
This duty takes on a heightened importance at budget time, when our elected officials figure out how to spend our tax dollars. Those discussions usually include a decision whether to take more out of our billfolds in the form of tax increases to help pay government's bills.
I was thinking about this as two of our reporters – Jennifer Fusco, the O-D's newly assigned watchdog reporter, and Bryon Ackerman, who covers Utica's city beat – took on the task of examining the city of Utica's 2009-10 budget proposal.
Their special report, which appears on today's front page, shines a Klieg light on the city's latest financial plan while taking a historical look at previous city budgets.
Fusco and Ackerman sorted though hundreds of pages of documents and online databases in an effort to provide readers with the clearest possible picture of the city's planned expenditures.
The challenge for any reporter is to take this avalanche of information and make it understandable. It's hard work that requires patience and attention to detail. It can be maddening. It also can be great fun.
It's worth noting that it's not just the city that gets this scrutiny.
In late 2008, we sprayed our sunshine on Oneida County's 2009 budget plan, which proposed a staggering 8.9 percent property tax increase at a time when many of us are suffering financially.
An O-D investigation found that the proposed budget contained raises of 10 percent or more for nearly 20 percent of county employees. The county legislature, responding to a citizen outcry, agreed to do away with most of the raises and cut the tax hike to 5.1 percent.
Today's look at the city's budget won't be the last; it's one of many budget reports we're preparing on local governments and school districts to help you, our readers, better understand how the public sector is spending the hard-earned cash it took from you through taxation.
We're arming you, our readers, with information you need to participate in government. It's a look at the facts, plain and simple.
What you think about how local governments spend your money is up to you to decide.
We'll keep the light on – I promise.
Joe Kieta is editor of the Observer-Dispatch. E-mail him at jkieta@uticaod.com.


For the better part of a century aerial photography has been one of the archaeologist's most important instruments for the detection of archaeological features and the representation of the resulting evidence through cartography.
In particular, 'historical' photographs from the 1930s, 1940s and 1950s show a landscape profoundly different from that of today. Building development, infrastructure projects, land-use change and mechanised agriculture have radically changed the landscape, completely destroying or partially concealing significant parts of the archaeological resource.
Only through the detailed analysis of this historical air photo coverage will it be possible to recover, in part at least, the presence, siting and inter-relationship of settlements and other man-made features and natural phenomena. Further reasons for interest in this source lie in the breadth of coverage represented in vertical aerial photography (extending in most cases to at least the provincial or regional scale) and in the huge amount of photographic material available, often without charge, in provincial, regional and military archives and in private collections. More recent vertical images offer an up-to-date record of the landscape for comparison with the earlier coverage, thereby defining in detail the transformations that have taken place in any chosen study area in recent years.
and in Great Britain by a variety of bodies.
It has been demonstrated that vertical coverage which coincides with a window of high aerial visibility can document many hundreds of archaeological sites and features in a few tens of minutes. Any kind of cost-benefit comparison with other investigative techniques would see this as an extraordinarily favourable return.


Oxford University's Centre for Statistics in Medicine has released independent validation of fragility fracture risk assessment tool QFracture, supporting its inclusion in the recent draft NICE guidelines on preventing osteoporotic fracture.
QFracture uses a simple algorithm to identify individuals at risk of developing fracture of the hip, spine or wrist.
Developed using data from the QResearch database www.qresearch.org, a joint not-for-profit project run by the University of Nottingham and leading healthcare IT provider EMIS, the Oxford validation shows that QFracture has the potential to significantly reduce the current £2bn bill for treating osteoporotic fractures in the UK.
The independent assessment saw Oxford researchers test the tool against 2.2 million patients drawn from the THIN database.
The findings come at an important time for QFracture which has also just been included in the draft scope of new NICE guidelines on the prevention of osteoporotic fracture. The guidelines will set national standards on how best to assess individuals patients risk of fracture.
The QFracture algorithm, which features in the development plan for EMIS Web, is available online for patients to use by answering a series of simple questions on their medical health (www.qfracture.org). Alternatively software which implements QFracture is available for use with NHS clinical computer systems via ClinRisk Ltd.


text:
Maxomys rajah är en däggdjursart som först beskrevs av Thomas 1894.  Maxomys rajah ingår i släktet taggråttor och familjen råttdjur. IUCN kategoriserar arten globalt som sårbar. Inga underarter finns listade i Catalogue of Life.

Denna gnagare förekommer på södra Malackahalvön, Borneo, Sumatra och på flera mindre öar i regionen. Arten lever i låglandet och i bergstrakter upp till 1100 meter över havet. Den vistas i olika slags skogar. I sällsynta fall vandrar den genom oljepalmodlingar för att nå andra skogsområden.

Källor

Externa länkar

Taggråttor
rajah
Däggdjur i orientaliska regionen

redpajama_set_name: RedPajamaWikipedia


peterzx necesitaba un nuevo diseño de papelería y lanzó un concurso de diseño en 99designs.
Black and White. My logo is quotation marks (see website) because I am an authority in the field and also a news publisher. I don't have artwork for this image.
BUSINESS CARD: On the front of the business card the legal practice intelligence logo should not be shown. Back of card is ok. Peter Frankl logo can be front and or back.


text:
First Supercars star locked in for Bathurst 12H
By Stefan Bartholomaeus
Tim Slade is the first Supercars star secured for the 2018 Bathurst 12 Hour, confirming his place aboard Objective Racing's McLaren 650S.
Slade returns to the team for the third straight year, teamed with Mobil 1 HSV Racing co-driver Warren Luff, car owner Tony Walls and new recruit Jaxon Evans.
A host of Supercars stars are again expected to be part of the growing GT race, with 12 of the championship's full-timers having been on the grid this year.
"It's great to be heading back to Bathurst with Tony and the Objective Racing crew," said Slade, whose entry is expected to run in the Pro-Am category for the February 2-4 event.
"Things haven't gone to plan the last couple of years and we haven't been able to get the kind of results we know we are capable of - but the car is fast and if things roll our way we know we will be competitive.
"Thanks to Tony for inviting me back. I'm really looking forward to getting back behind the wheel of those cars."
Slade will be hoping for a smoother run at the 12 Hour than he endured in the recent Supercheap Auto Bathurst 1000, where cooling issues resulted in him being dragged from his car.
This year has seen Slade contest all three of Bathurst's major events, having also campaigned a Mitsubishi Lancer Evo in Easter's 6 Hour production car contest.
Mostert wins on international GT debut
Fast Facts - ITM Auckland SuperSprint
Flashback: When rookie McLaughlin won at home
Saturday Sleuthing: The Bowe Falcon living in NZ
Race for Richie Stanaway heats up
Wood to step down from Supercars drive
Warren Luff	›
Tim Slade	›


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text:
La Estación Ecológica de Taim está situada en el estado de Río Grande del Sur, Brasil, a 370 km de Porto Alegre, encargada de preservar uno de los ecosistemas más vulnerables del estado. En el área hay unas 230 especies de aves, 70 de mamíferos, y 60 especies de peces.

La región de Taim está próxima a la frontera con Uruguay. Con la firma del tratado de San Ildefonso el 1 de octubre de 1777 culminó un largo pleito de las dos potencias coloniales, España y Portugal, por la demarcación de sus dominios en Sudamérica. De acuerdo a dicho tratado, Colonia del Sacramento volvió a ser española y Portugal afianzó su expansión hacia el oeste de la línea de Tordesillas. El territorio entre Chuy y Taim pasó a denominarse "Campo Neutral". En ese suelo ni portugueses ni españoles podían localizar tropas o campamentos ni permitir afincarse a sus súbditos. Con el tiempo, España perdería ascendencia sobre estos territorios.

Áreas protegidas de Brasil
Ecología de Brasil
Áreas protegidas del estado de Río Grande del Sur
Sitios relevantes para la observación de aves en Brasil
Río Grande (Brasil)
Áreas protegidas establecidas en 1988
Brasil en 1988
Bienes inscritos en la Lista Indicativa de Brasil


A: Your username is your email address for the APGA Community. If you have forgotten your APGA Community password or need assistance with your login information, please click here.
During the beta launch of the APGA Community, you will have two sets of login credentials - one for the current www.apga.org website and one for the APGA Community tied to your email address. When we switch over to our new APGA website at the end of July, your APGA Community login will become your only APGA login.
*During the Beta APGA Community Launch, this feature will be turned off.
A: Communities allow you to participate in discussions and share resources with other members. All APGA members have access to the APGA Community Open Forum, and there are separate communities for each of the APGA Committees.


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text:
Library of Congress > Performing Arts Databases > About
All Performing Arts Databases
Chasanoff/Elozua "Amazing Grace" Collection
It's Showtime! Sheet Music from Stage and Screen
Jazz on the Screen Filmography
Show Music on Record
Silent Feature Film
Tap Dance in America
Traditional Music & Spoken Word
U.S. ISMN Public Archive
Jeru: In the Words of Gerry Mulligan
An Oral Autobiography
Home | About | Introduction | Growing Up | Gil's Place | Miles Davis | Charlie Parker
"Young Blood" | Pianoless Quartet | Chet Baker | On Counterpoint | Addictions
Thelonious Monk | Live vs. Studio | Gene Krupa | Astor & Franca | Editor's Note
About the Oral Autobiography
[Gerry Mulligan, Harpers Bazaar
Cover], [1974]. Franca R. Mulligan,
photographer. The Gerry Mulligan
Collection, Performing Arts Reading
Room, Library of Congress.
This oral autobiography by Gerry Mulligan with Ken Poston was conducted in 1995. The autobiography was later edited by Jon and Iris Newsom. Mulligan's wife, Franca R. Mulligan, has kindly granted permission for the Library of Congress to make portions of this autobiography available on the The Library of Congress Presents: Music, Theater and Dance Web site. It is hoped that more from this autobiography will be added to this site in the future. Photographs were supplied by Franca R. Mulligan and were chosen to supplement the autobiography. Additional photographs from other collections at the Library of Congress were also used to illustrate different periods of Mulligan's life.
Digitizing the Collection
The photographs were scanned as jpeg images and for the most part were taken from Franca R. Mulligan's personal collection. Derivatives were made by the Network Development team at the Library of Congress.
WAV files of Mulligan's interviews were made from original source tape by Jon Newsom. MP3 and RealMedia streaming files were created by the IHAS staff. The EQ was adjusted with Sound Forge 5.0. The files were then de-clicked, amplified, and beginnings and ends faded using Cool-Edit Pro 1.0. The derivative files were generated using Sound Forge 5.0., the MP3s at 96K, the RealMedia files at 56K.
Copyright and Other Restrictions
The Library of Congress is providing access to these materials for educational and research purposes and makes no warranty with regard to their use for other purposes. The written permission of the copyright owners and/or other rights holders (such as holders of publicity and/or privacy rights) is required for distribution, reproduction, or other use of protected items beyond that allowed by fair use or other statutory exemptions. There may be content that is protected as "works for hire" (copyright may be held by the party that commissioned the original work) and/or under the copyright or neighboring-rights laws of other nations.
Responsibility for making an independent legal assessment of an item and securing any necessary permissions ultimately rests with persons desiring to use the item. Users should consult the bibliographic information that accompanies each item for specific information. This catalog data provides the details known to the Library of Congress regarding the corresponding items and may assist users in making independent assessments of the legal status of these items as related to their desired uses.
Items included here with the permission of the rights holders are indicated as such in the bibliographic record for each item.
While every effort has been made to obtain permission to use these songs or portions of these songs, in some cases the rights' owner may have only granted permission to use a portion of the material online. In those cases, only 30-second excerpts of sound recordings are used, and only one or two pages of print or manuscript materials are used.
In some cases, the Library was unable to identify a possible rights holder and has elected to place some of those items online as an exercise of fair use for strictly non-commercial educational uses. The Library of Congress would like to learn more about these materials and would like to hear from individuals or institutions that have any additional information or know of their history. Please contact: Performing Arts Reading Room.
Suggested credit line: Library of Congress, Music Division.
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