Blow-up behaviour of a fractional Adams-Moser-Trudinger type inequality in odd dimension
Ali Maalaoui
Luca Martinazzi\thanks{L.M.
A.M. supported by Swiss National Science Foundation.
Given a smoothly bounded domain $\Omega\Subset\R^n$ with $n\ge 1$ odd, we study the blow-up of bounded sequences $(u_k)\subset H^\frac{n{2_{00(\Omega)$ of solutions to the non-local equation $$(-\Delta)^\frac n2 u_k=\lambda_k u_ke^{\frac n2 u_k^2\quad \text{in \Omega,$$ where $\lambda_k\to\lambda_\infty \in [0,\infty)$, and % critical points to the fractional Moser-Trudinger-type inequality %$$\sup_{u\in H^{n/2_{00(\Omega), \, \|(-\Delta)^{n/4 u\|_{L^2(\R^n)^2\le \Lambda_1 \int_{\Omega e^{\frac{n{2u^2dx <\infty,\quad \Lambda_1:=(n-1)! \mathrm{vol(S^n)$$ $H^{\frac n2_{00(\Omega)$ denotes the Lions-Magenes spaces of functions $u\in L^2(\R^n)$ which are supported in $\Omega$ and with $(-\Delta)^\frac{n{4u\in L^2(\R^n)$. Extending previous works of Druet, Robert-Struwe and the second author, we show that if the sequence $(u_k)$ is not bounded in $L^\infty(\Omega)$, a suitably rescaled subsequence $\eta_k$ converges to the function $\eta_0(x)=\log\left(\frac{2{1+|x|^2\right)$, which solves the prescribed non-local $Q$-curvature equation $$(-\Delta)^\frac n2 \eta =(n-1)!e^{n\eta\quad \text{in \R^n$$ recently studied by Da Lio-Martinazzi-Rivi\`ere when $n=1$, Jin-Maalaoui-Martinazzi-Xiong when $n=3$, and Hyder when $n\ge 5$ is odd. We infer that blow-up can occur only if $\Lambda:=\limsup_{k\to \infty\|(-\Delta)^\frac n4 u_k\|_{L^2^2\ge \Lambda_1:= (n-1)!|S^n|$.% Moreover if $\Lambda=\Lambda_1$, and blow up occurs, then $u_k\rightharpoonup 0$ weakly in $H^\frac{n{2_{00(\Omega)$.
1504.00254.tex
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\begin{document}
\title{Blow-up behaviour of a fractional Adams-Moser-Trudinger type inequality in odd dimension}
\author{Ali Maalaoui, Luca Martinazzi\thanks{L.M., A.M. supported by Swiss National Science Foundation.}, Armin Schikorra\thanks{A.S.' research leading to these results has received funding from the European Research Council under the European Union's Seventh Framework Programme (FP7/2007-2013) / ERC grant agreement n� 267087. A.S. is partially supported by Swiss National Science Foundation.}}
\maketitle


\begin{abstract}
Given a smoothly bounded domain $\Omega\Subset\R^n$ with $n\ge 1$ odd, we study the blow-up of bounded sequences $(u_k)\subset H^\frac{n}{2}_{00}(\Omega)$ of
solutions to 
the non-local equation
$$(-\Delta)^\frac n2 u_k=\lambda_k u_ke^{\frac n2 u_k^2}\quad \text{in }\Omega,$$
where $\lambda_k\to\lambda_\infty \in [0,\infty)$, and
% critical points to the fractional Moser-Trudinger-type inequality
%$$\sup_{u\in H^{n/2}_{00}(\Omega), \, \|(-\Delta)^{n/4} u\|_{L^2(\R^n)}^2\le \Lambda_1} \int_{\Omega} e^{\frac{n}{2}u^2}dx <\infty,\quad \Lambda_1:=(n-1)! \mathrm{vol}(S^n)$$
$H^{\frac n2}_{00}(\Omega)$ denotes the Lions-Magenes spaces of functions $u\in L^2(\R^n)$ which are supported in $\Omega$ and with $(-\Delta)^\frac{n}{4}u\in L^2(\R^n)$.
Extending previous works of Druet, Robert-Struwe and the second author, we show that if the sequence $(u_k)$ is not bounded in $L^\infty(\Omega)$, a suitably rescaled subsequence $\eta_k$ converges to the function $\eta_0(x)=\log\left(\frac{2}{1+|x|^2}\right)$, which solves the prescribed non-local $Q$-curvature equation
$$(-\Delta)^\frac n2 \eta =(n-1)!e^{n\eta}\quad \text{in }\R^n$$
recently studied by Da Lio-Martinazzi-Rivi\`ere when $n=1$, Jin-Maalaoui-Martinazzi-Xiong when $n=3$, and Hyder when $n\ge 5$ is odd.
We infer that blow-up can occur only if $\Lambda:=\limsup_{k\to \infty}\|(-\Delta)^\frac n4 u_k\|_{L^2}^2\ge \Lambda_1:= (n-1)!|S^n|$.% Moreover if $\Lambda=\Lambda_1$, and blow up occurs, then $u_k\rightharpoonup 0$ weakly in $H^\frac{n}{2}_{00}(\Omega)$.
\end{abstract}
\section{Introduction}

In this paper we study some compactness properties of the embedding of $H^{\frac n2}_{00}(\Omega)$ into Orlicz spaces, where $\Omega$ is a smoothly bounded domain in $\R^n$. In order to introduce the relevant function spaces we start by recalling various definitions of fractional Laplacians.

Let $\mathcal{S}(\R^n)$ denote Schwarz space of smooth and rapidly decreasing functions on $\R^n$. For a function $u\in\mathcal{S}(\R^n)$ and for $s \in (0,\infty)$,
we define
\[
 \laps{s}u := (|\cdot|^{2s} u^\wedge)^\vee.
\]
Here the Fourier transform is defined via
\[
  u^\wedge(\xi) \equiv \mathcal{F}u(\xi) :=\frac{1}{ (2\pi)^{n/2}} \int_{\R^n} e^{-i x\cdot \xi}\ u(x) dx.
\]
and $u^\vee$ is its inverse.

For $s \in (0,2)$ one can also prove (see e.g. \cite{DNPV}) that for a certain constant $c_{n,s} \in \R$
\[
 \laps{s}u(x) = c_{n,s} P.V. \int_{\R^n} \frac{u(x+h)-u(x)}{|h|^{n+s}}\, dh. 
\]
In order to define the operator $(-\Delta)^s$ on a space larger than the Schwarz space, set for $s>0$
\begin{equation}\label{defLs} 
L_s(\R^n):=\left\{u\in L^1_{\loc}(\R^n): \int_{\R^n}\frac{|u(x)|}{1+|x|^{n+s}}dx<\infty\right\}.
\end{equation}
Then for $u\in L_s(\R^n)$ we can define
$(-\Delta)^s u$ as a tempered distribution as follows:
$$\langle (-\Delta)^\frac{s}{2} u,\varphi\rangle:=\int_{\R^n} u(-\Delta)^\frac{s}{2} \varphi dx,\quad \text{for }\varphi\in\mathcal{S}(\R^n).$$
This is due to the fact that for $\varphi\in \mathcal{S}(\R^n)$ one has $(1+|x|^{n+s})|(-\Delta)^\frac{s}{2} \varphi(x)|\le C$ for a constant depending on $\varphi$ but not on $x$, see \cite[Proposition 2.2]{Ali} and \cite{Sil}. % \armin{shouldn't we for politeness cite Silvestre's work here?}.

We can now define the space
$$H^s(\R^{n}):=\{u\in L^2(\R^n):(-\Delta)^\frac s2 u\in L^2(\R^n)\},$$
endowed with the norm  
$$\|u\|_{H^s(\R^n)}^{2}:=\|u\|_{L^2(\R^n)}^2+\|(-\Delta)^\frac s2 u\|_{L^2(\R^n)}^2,$$
where with the expression $(-\Delta)^\frac s2 u\in L^2(\R^n)$ we mean that the tempered distribution $(-\Delta)^\frac s2 u$ can be represented by a square-summable function. 

%Other definitions of $H^s(\R^n)$ are possible. For instance we mention that when $s\in (0,1)$ then a norm equivalent to $\|\cdot \|_{H^s(\R^n)}$ is
%\[
%\|u\|^2_{W^{s,2}(\R^n)} :=\|u\|_{L^2(\R^n)}^2+\int_{\R^n} \int_{\R^n} \frac{|u(x+h)+u(x)|^2}{|h|^{n+2s}}\ dh\ dx,
%\]
%see e.g. \cite{DNPV}.

Given a bounded set $\Omega\Subset \R^n$ we also define its subspace
$$H_{00}^{s}(\Omega):=\{u\in H^s(\mathbb{R}^{n}) : u\equiv0 \text{ on } \Omega^{c}\}.$$
In particular we will consider the space $X(\Omega):= H^{\frac n2}_{00}(\Omega)$ for $n$ is odd,
endowed with the norm
$$\|u\|_{X}^{2}:=\|(-\lap)^\frac n4 u\|_{L^2(\R^n)}^2=\int_{\mathbb{R}^{n}}|\xi|^{n}|\hat{u}(\xi)|^{2}d\xi.$$
The norms $\|\cdot \|_X$ and $\|\cdot\|_{H^\frac n2(\R^n)}$ are equivalent on $H_{00}^\frac n2(\Omega)$ by a Poincar\'e-type inequality. % This follows for instance from Theorem 7.1 in \cite{gru}. ( \armin{with the constant depending on $|\Omega|$, prove via Poincare or cite?}. \luca{We can discuss it.})
The space $H_{00}^s(\Omega)$ is also known as Lions-Magenes space, and is sometimes denoted by
$\tilde H^s(\Omega)$, or even $L^{s,2}_0(\Omega)$.

\medskip

We recall the following fractional version of the Adams-Moser-Trudinger inequality, see \cite[Theorem 1]{MS}:
\begin{theorem}\label{trmMT}
For any integer $n>0$ there exists a constant $C_n>0$ such that for every open set $\Omega\subset\R^n$ with finite volume $|\Omega|$ one has
\begin{equation}\label{probsup}
\sup_{u\in X(\Omega),\,\|u\|_{X}^2\le \Lambda_1}\int_{\Omega} e^{\frac n2 u^{2}} dx\le C_n|\Omega|,
\end{equation}
where $\Lambda_1:=(n-1)!|S^n|$.
\end{theorem}

When $n=2$ the above theorem is a special case of the Moser-Trudinger inequality \cite{tru}, and when $n>2$ is even it is a special case of Adams' inequality \cite{ada}.


\medskip

In this paper we want  to study the blow-up behavior of extremals of \eqref{probsup}, i.e.
weak solutions $u\in X(\Omega)$ of the  Euler-Lagrange equation  
\begin{equation}\label{equaprob}
(-\Delta)^{\frac{n}{2}}u=\lambda ue^{\frac n2 u^{2}},\quad \text{for some }\lambda\in \R,
\end{equation}
which can be intended in the following sense:

\begin{definition}
Given $f\in L^{2}(\Omega)$, a function with $u\in X(\Omega)+\R$ (i.e. $u+c\in X(\Omega)$ for some $c\in\R$) is a weak solution to 
\begin{equation}\label{eq}
(-\Delta)^{\frac{n}{2}}u=f\quad \text{in $\Omega$}
\end{equation}
if
\begin{equation}\label{eqweak}
\int_{\R^n} (-\lap)^\frac n4 u\, (-\lap)^\frac n4 \varphi dx = \int_{\R^n} f \varphi dx, \quad \forall \varphi\in X(\Omega).
\end{equation}
\end{definition}

\begin{remark}\label{rmkinteg} It follows from Theorem \ref{trmMT} that for $u\in X(\Omega)$ one has $e^{u^2}\in L^p(\Omega)$ for every $p\in [1,\infty)$ (see also \cite[Theorem 9.1]{pee}). In particular the right-hand side of \eqref{equaprob} belongs to $L^p(\Omega)$ for $p\in [1,\infty)$.
\end{remark}



The Lagrange multiplier $\lambda$ in \eqref{equaprob} can be computed by testing the equation with $\varphi=u$ (in the spirit of \eqref{eqweak}). This leads to 
\begin{equation}\label{energy}
\|(-\lap)^{\frac{n}{4}} u\|_{L^2(\R^n)}^2=\lambda\int_\Omega u^2e^{\frac n2 u^2}dx,
\end{equation}
whence $\lambda>0$, unless $u\equiv 0$.
% We also define the space 
% $$H^{\frac{1}{2}}_{00}(\Omega):=\{u\in H^{1/2}(\mathbb{R}^{3}); u=0 \text{ on } \Omega^{c}\},$$
% endowed with the norm 
% $$\|u\|_{H^{1/2}}^{2}:=\int_{\mathbb{R}^{3}}|\xi||\hat{u}|^{2}d\xi =\int_{\mathbb{R}^{3}}|(-\Delta)^{\frac{1}{4}}u|^{2}dx.$$

We are interested in the study of the blowing-up behavior of a sequence of continuous solution to the following problem :
\begin{equation}\label{eq0}
\left\{\begin{array}{ll}
(-\Delta)^{\frac{n}{2}}u_{k}=\lambda_{k}u_{k}e^{\frac{n}{2}u_{k}^{2}} &\text{in $\Omega$}\\
%u_k > 0 &\text{in $\Omega$}\\
u_{k} \in X(\Omega) &
\end{array}
\right.
\end{equation}
where $\lambda_{k}\geq 0$. 

\begin{remark}\label{rmkreg} It follows from Remark \ref{rmkinteg}, from the estimates in \cite{gru} and bootstrapping, that every solution $u$ to \eqref{equaprob} belongs to $C^{\frac{n-1}{2},\alpha}(\bar\Omega)\cap C^\infty(\Omega)$ for some $\alpha\in (0,1)$, and in fact  the function $d^{-\frac n2}u:\Omega\to\R$, where $d$ is the distance function from $\partial\Omega$, can be extended to a function in $C^\infty(\bar\Omega)$. In particular, $\sup_{\Omega} u_k \in \R$.
\end{remark}

The main result of this paper can be stated as follows :

\begin{theorem}\label{main}
Consider a \emph{bounded} sequence $(u_{k})_{k\in \mathbb{N}} \subset X(\Omega)$ of solutions to (\ref{eq0}). Set $m_{k}:=\sup_{\Omega} |u_{k}|$ and 
$$\Lambda:=\limsup_{k\to\infty}\|u_k\|_{X}^2<\infty.$$ Up to possibly replacing $u_k$ with $-u_k$ we can assume that $m_k=\sup_{\Omega} u_k$ for every $k$. Assume also that $0<\lambda_k\le \bar \lambda$ for some $\bar \lambda<\infty$ and that $\lambda_k\to\lambda_\infty$ as $k\to\infty$. Then up to extracting a subsequence one of the following holds:\\
\begin{itemize}
\item[(i)] $\lim_{k\to\infty} m_{k}<\infty$ and $u_{k}$ converges to $u_{\infty}$ in $C^{\ell}_{\loc}( \Omega)\cap C^\frac{n-1}{2}(\bar\Omega)$  for any $\ell\in\mathbb{N}$ and $u_{\infty}\in C^\ell_{\loc}(\Omega)\cap C^\frac{n-1}{2}(\bar\Omega)\cap X(\Omega)$ solves
$$(-\Delta)^{\frac{n}{2}}u_\infty=\lambda_\infty u_\infty e^{\frac{n}{2}u_\infty^{2}} \quad \text{in }\Omega.$$

\item[(ii)]  $\lim_{k\to\infty} m_{k}=\infty$, $\Lambda\ge \Lambda_1$, with $\Lambda_1$ as in Theorem \ref{trmMT}, % \luca{$u_k\rightharpoonup u_\infty$ in $X(\Omega)$ and
%\begin{equation}\label{energylim}
%\lim_{k\to\infty} \|u_{k}\|_{X}^2\geq \|u_\infty\|_X^2+ \Lambda_{1}:=(n-1)!|S^n|.
%\end{equation}
%}
and setting $r_{k}$ such that
\begin{equation}\label{eq:rkdef}
\lambda_{k}m_{k}^{2}e^{\frac{n}{2} m_{k}^{2}}r_{k}^{n}=2^n(n-1)!,
\end{equation} 
and
\begin{equation}\label{defetak}
\eta_{k}(x):=m_{k}(u_{k}(x_{k}+r_{k}x)-m_{k}),\quad \eta_0(x):=\log\left(\frac{2}{1+|x|^{2}}\right),
\end{equation}
 one has $\eta_k +\log 2 \to\eta_0$ in $C^{\ell}_{\loc}(\R^n)$ for every $\ell\ge 0$.
\end{itemize}
%In particular in $0\le \Lambda<\Lambda_1$, then only case (i) can occur.
\end{theorem}

Since Theorem \ref{main} was proven in \cite{AS}, \cite{dru}, \cite{RS} and \cite{mar4} when $n$ is even, we shall only consider the remaining case $n$ odd. In some proofs we will focus on the case $n\ge 3$, but simple modifications make every argument work for the case $n=1$. In fact the case $n=1$ is a slightly simpler, since comparison principles and in particular the Harnack inequality are available. 

The general strategy of the proof is similar to the one in the even-dimensional case, but some new difficulties arise due to the nonlocal nature of the operator $(-\Delta)^\frac{n}{2}$, as we shall now describe.

One would like to shows that in case of blow-up (Case (ii) in Theorem \ref{main}) the functions $\eta_k$ converge to a function $\eta_0\in L_n(\R^n)$ solving
\begin{equation}\label{eqeta0}
(-\Delta)^\frac n2 \eta_0=(n-1)! e^{n\eta_0}\quad \text{in }\R^n,\quad V:=\int_{\R^n} e^{n\eta_0}dx<\infty,
\end{equation}
and then prove that, among all solutions to \eqref{eqeta0}, $\eta_0$ has the special form given by \eqref{defetak}.

The first problem is that the local convergence of $\eta_k$ to a function $\eta_0$ rests on local gradient bounds for $\eta_k$ not depending on $k$ (when $n=1,2$ such bound are not necessary, thanks to the Harnack inequality). This is the content of Propositions \ref{la:ulapuest} and \ref{gradest}, one of the crucial parts of the paper.  In particular we will show that for $s < n$,
\begin{equation}\label{stimagu}
\int_{B_\rho(x_0)}|u_k(-\Delta)^\frac s2 u_k|\, dx\le C \rho^{n-s},\quad \text{for }B_{10\rho}(x_0)\subset\Omega.
\end{equation}
In the previous work \cite{mar4} an analogous estimate was obtained by noticing that $(-\Delta)^\frac{n}{2}(u_k^2)$ is uniformly bounded in $L^1(\Omega)$ when $n$ is even. Unfortunately this was based on an explicit expansion of $(-\Delta)^\frac{n}{2} (u_k^2)$ as sum of partial derivatives of $u_k$, which is of course not possible when $n$ is odd. Here instead we reduce \eqref{stimagu} to a the bound
$$\|u_k(-\Delta)^\frac s2 u_k\|_{L^{(\frac ns,\infty)}(B_\rho(x_0))}\le C,$$
which will be proven writing $u_k(-\Delta)^\frac s2 u_k$ in terms of the Riesz potential. The formal heuristic argument goes as follows. Write formally
\begin{equation}\label{formalid}
\begin{split}
u_k(-\Delta)^\frac s2 u_k &=u_k I_{n-s}(-\Delta)^\frac{n}{2} u_k\\
&=: (I_{\frac n2}(-\Delta)^\frac{n}{4}u_k)I_{n-s}(\theta(-\Delta)^\frac{n}{2} u_k) + I_{n-s}(u_k(-\Delta)^\frac{n}{2} u_k)+E\\
&=: A+B+E,
\end{split}
\end{equation}
where $\theta\in C^\infty_0(B_{2\rho}(x_0))$ is a cut-off functions, $I_t$ denotes the Riesz potential, and $E$ is an error term, which can be bounded using a commutator-type estimate.
%\armin{I don't think this describes it correctly. What I inherently used is \eqref{eq:estIII} (this is why I chose this argument). So it should say:
%\[
% u_k(-\Delta)^\frac s2 u_k = I_{n-s}(u_k(-\Delta)^\frac{n}{2} u_k) + ...
%\]
%The first term is bounedd by the PDE, and the ... are also bounded by other arguments. 
%}
Then one has to bound the term $A$ in $L^{(\frac{n}{s},\infty)}(B_\rho(x_0))$ using that $(-\Delta)^\frac{n}{4} u_k$ is bounded in $L^2(\R^n)$, while $(-\Delta)^\frac{n}{2} u_k$ is bounded in $L\log^\frac12 L(\Omega)$. These are borderline estimates, for instance because $I_{\frac{n}{2}}$ fails to send $L^2$ into $L^\infty$. Using elementary tricks we are able to circumvent this problem, obtaining Propositions \ref{borderline1}  and \ref{borderline2}. In order to bound $B$ one uses the PDE, and in particular that $u_k(-\Delta)^\frac n2 u_k$ is bounded in $L^1(\Omega)$.
Finally, to move from the formal argument to a rigorous one, and in particular to replace the first identity in \eqref{formalid} with a correct identity, we have to approximate $u_k$ with functions in $C^\infty_c(\R^n)$. The necessary technical results are contained in Section \ref{sec:borderline} and the appendix. 

The second problem, still related to the non-local nature of $(-\Delta)^\frac n2$, is that uniform estimates on the derivatives of the blown-up functions $\eta_k$ do indeed guarantee that $\eta_k\to\eta_0$ in $C^\ell_{\loc}(\R^n)$ (up to the additive constant $\log 2$ which we shall now ignore) for $\ell\le n-1$, but why should the convergence
\begin{equation}\label{convetak}
(-\Delta)^\frac{n}{2} \eta_k \to (-\Delta)^\frac{n}{2} \eta_0 \quad\text{in }\mathcal{S}'(\R^n) \text{ as }k\to\infty
\end{equation}
hold? Indeed \eqref{convetak} means that
\begin{equation}\label{convetak2}
\lim_{k\to\infty}\int_{\R^n} \eta_k \,(-\Delta)^\frac{n}{2} \varphi dx \to \int_{\R^n} \eta_0 \,(-\Delta)^\frac{n}{2} \varphi dx \quad \text{for every }\varphi \in \mathcal{S}(\R^n),
\end{equation}
and since even for $\varphi\in C^\infty_c(\R^n)$ we have that $(-\Delta)^\frac{n}{2}\varphi$ is not compactly supported, the local convergence of $\eta_k$ is not sufficient to guarantee \eqref{convetak2}. A priori it is not to be ruled out that while $\eta_k\to \eta_0$ very nicely in a compact set, ``at infinity'' $\eta_k$ has a wild behaviour. To rule this out we shall prove uniform bounds of $\eta_k$ in $L_s(\R^n)$ for any $s>0$, which is the content of Proposition \ref{boundLe}. Here we critically use that $\eta_k$ is uniformly upper bounded by construction, and the local bounds on the derivatives of $\eta_k$.

At this point it will be easy to conclude that $\eta_k$ locally converges to a function $\eta_0\in L_n(\R^n)$ solving \eqref{eqeta0}. Now we are faced with the problem of determining $\eta_0$. Indeed, similarly to what was shown in \cite{CC}, also in odd dimension $3$ or higher, Problem \eqref{eqeta0} has many solutions, as shown in \cite{JMMX} (when $n=3$) and \cite{Ali0} (for any $n\ge 3$ odd). Here we are able to use the following recent result of Ali Hyder, together with the previous bounds to show that among all solutions of \eqref{eqeta0} actually $\eta_0$ is a special one, precisely the one given in \eqref{defetak}.

\begin{theorem}[A. Hyder \cite{Ali}]\label{thmali}
Let $\eta_0\in L_n(\R^n)$ solve \eqref{eqeta0}. Then $\eta_0$ can be decomposed as $\eta_0= v+P$, where $P$ is a polynomial of degree at most $n-1$, and $v(x)=-\alpha \log(|x|)+o(\log|x|)$ as $|x|\to \infty$. Moreover $P$ is constant if and only if
\begin{equation}\label{spherical}
\eta_0(x)=\log\frac{2\lambda}{1+\lambda^2|x-x_0|^2},\quad \text{for some }\lambda>0,\quad x_0\in\R^n.
\end{equation}
%If $P$ is not constant, then there exists $j$ with $1\le j\le \frac{n-1}{2}$ and a constant $c\ne 0$ such that
%\begin{equation}\label{conda}
%\lim_{|x|\to\infty}\Delta^j \eta_0= c.
%\end{equation}
\end{theorem}

Indeed, if $\eta_0$ is not of the form \eqref{spherical}, then $\eta_0$ at infinity behaves like a logarithm plus a polynomial, only the former belonging to $L_s(\R^n)$ for $s$ small. This is in contradiction to the fact that $\eta_0\in L_s(\R^n)$ for every $s>0$. This argument is different from the one used in the even dimensional case, first introduced in \cite{RS} and then also applied in \cite{mar4} and other works.

In the case $n=1$ Theorem \ref{thmali} is not necessary because Da~Lio-Martinazzi-Rivi\`ere \cite{DLMR} proved that every function  $\eta_0\in L_1(\R)$ solving \eqref{eqeta0} for $n=1$ has necessarily the form \eqref{spherical}.

\medskip

It has to be mentioned that in even dimension the analog of Theorem \ref{main} was complemented in \cite{dru}, \cite{MS2} and \cite{str2} by a quantization result, saying that in case of blow-up
$$\Lambda=\int_\Omega |\nabla^\frac n2 u_\infty|^2 dx + L\Lambda_1\quad \text{for an integer }L>0.$$
In other words the energy loss in the weak limit is an integer multiple of the fixed quantity $\Lambda_1$. Although it is natural to expect this to hold true also in our non-local case, we remark that in the local case the proofs make abundant use of ODE techniques, which are not available when dealing with fractional Laplacians. On the other hand in the case of half-harmonic maps, precise energy quantization was obtained in \cite{DaLioBubbles}.


\subsection*{Notation}
The space $C^\alpha(\Omega) \equiv C^{\alpha_0,\tilde{\alpha}}(\Omega)$,  for $\alpha = \alpha_0 + \tilde{\alpha}$ with $\tilde{\alpha} \in (0,1]$, $\alpha_0 \in \N_0$, is the space of $\alpha_0$-times differentiable functions with $\alpha_0$th derivative H\"older continous of order $\tilde{\alpha}$.
We define the semi-norm
\[
 [f]_{C^{\alpha}(\Omega)}= \sup_{x\neq y \in \Omega} \frac{|\nabla^{\alpha_0} f(x) - \nabla^{\alpha_0}f(y)|}{|x-y|^{\tilde{\alpha}}},
\]
and the norm
\[
 \|f\|_{C^{\alpha}(\Omega)} := \sum_{k=0}^{\alpha_0} \|\nabla^{k}  f\|_{L^{\infty}(\Omega)} + [f]_{C^{\alpha}(\Omega)}.
\]


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

%\luca{
%\begin{lemma}\label{lambdak} For every $k$ we have $\lambda_k\le \lambda_1(\Omega)$, where $\lambda_1(\Omega)$ denotes the first eigenvalue of $(-\Delta)^\frac{n}{2}$ in $\Omega$,

%\end{lemma}

%}


\begin{proposition} If $\sup_k m_k\le C$ then up to a subsequence $u_k\to u_\infty$ in $C^{\ell}_{\loc}(\Omega)\cap C^{\frac{n-1}2}(\bar \Omega)$ for every $\ell>0$, where $u_\infty$ solves \eqref{eq0}.
\end{proposition}
\begin{proof}
This follows from Lemma~\ref{la:LinftyRHSest}, from the estimates in \cite{gru} (compare also to \cite{RS}), and the theorem of Arzel\`a-Ascoli.%\footnote{\armin{Do we have $\sup_k \lambda_k < \infty$?} \luca{Good that you noticed that. We don't have this in general, we have to assume it. I changed the statement of the theorem.}}
\end{proof}



We shall now assume that, up to a subsequence, $m_k\to \infty$ as $k\to\infty$  and we consider  $x_{k} \in {\Omega}$ so that 
\begin{equation}\label{eq:xkdef}
m_k \equiv \sup_{ \Omega}u_{k}=u_{k}(x_{k}) \to\infty \quad \text{as }k\to\infty.
\end{equation}
%Take $v_{k}=m_{k}^{-1}u(r_{k}x+x_{k})$ where $r_{k}$ is chosen as in \eqref{eq:rkdef}.
%We will assume from now on that $m_{k}\to +\infty$ and $\|u_{k}\|_{X}\leq C$.


\subsection{Rescaling and Convergence}

\begin{lemma}\label{rk}
Let $r_k$ and $x_k$ be defined by \eqref{eq:rkdef} and \eqref{eq:xkdef} respectively. Then we have
$$\lim_{k\to \infty} \frac{\dist(x_{k},\partial \Omega)}{r_{k}}=+\infty$$
\end{lemma}
\begin{proof}
For the sake of contradiction, we assume that $$\lim_{k\to \infty} \frac{\dist(x_{k},\partial \Omega)}{r_{k}}<\infty.$$
Let us assume that 
\begin{equation}\label{limrk}
0 < \lim_{k\to \infty} \frac{\dist(x_{k},\partial \Omega)}{r_{k}} < \infty.
\end{equation}
If the above limit vanishes then the argument is similar.
We set $\Omega_{k}=\{r_{k}^{-1}(x-x_{k});x\in \Omega\}$. Then
$$v_{k}(x):=\frac{u_k(r_k x+x_k)}{m_k}$$
satisfies
\begin{equation}
\begin{cases}
(-\Delta)^{\frac{n}{2}}v_{k}= \frac{2^n(n-1)!}{m_{k}^{2}}v_{k}e^{\frac{n}{2}m_{k}^{2}(v_{k}^{2}-1)} \quad &\text{ in $\Omega_{k}$}\\
v_{k}\in X(\Omega_k). \quad &
\end{cases}
\end{equation}
Notice that,
\[
 \vrac{(-\Delta)^{\frac{n}{4}}v_{k}}_{L^2(\R^n)} = (m_k)^{-1} \vrac{(-\Delta)^{\frac{n}{4}}u_{k}}_{L^2(\R^n)} \xrightarrow{k \to \infty} 0.
\]
Then by the Sobolev embedding, Proposition~\ref{pr:Sobolev} using also \eqref{Rieszprop}, the boundedness of the Riesz transform, and that $\laps{1} = \lapms{\frac{n}{2}-1} (-\lap)^{\frac{n}{4}}$,
\begin{equation}\label{gradvk}
\begin{split}
 &\vrac{\nabla v_{k}}_{L^n(\R^n)}= c\vrac{\Rz \laps{1} v_{k}}_{L^n(\R^n)} \aleq \vrac{ \laps{1} v_{k}}_{L^n(\R^n)}\\
 \aleq & \vrac{\lapms{\frac{n}{2}-1} (-\lap)^{\frac{n}{4}} v_{k}}_{L^n(\R^n)} \leq  C\vrac{(-\Delta)^{\frac{n}{4}}v_{k}}_{L^2(\R^n)} \xrightarrow{k \to \infty} 0.
\end{split}
 \end{equation}
On the other hand, by \eqref{limrk} there exists some $R > 0$ so that $B_{4R}(0) \subset \Omega_k$ for all $k \in \N$. Then
\[
\vrac{(-\Delta)^{\frac{n}{2}}v_{k}}_{L^\infty(B_{3R}(0))} \xrightarrow{k \to \infty} 0.
\]
This implies that for any $\alpha \in (0,n)$, Lemma~\ref{la:LinftyRHSest},
\[
 [v_{k}]_{C^{\alpha}(B_{2R}(0))} \le  C.
\]
So recalling that $|v_k|\le 1$, by Arzel\`a-Ascoli (up to a subsequence) we have that $v_k\to v$ in $C^{n-1}(B_R)$ for some $v$. 
Since at the same time $\nabla v_k \to 0$ in $L^n(\R^n)$ and $v_k (0) = 1$, we know that $v \equiv 1$ in $B_R$.

On the other hand, take $R_1 > R$ so that $B_{\frac{R_1}{2}}(0) \cap \partial \Omega_k \neq 0$ for all but possibly finitely many $k \in \N$. Using \eqref{gradvk}, and noticing that $v_k \equiv 0$ on a fixed part of positive measure of $B_{R_1}$, we know that $v_k \to 0$ in $L^n(B_{R_1}(0))$, hence $v \equiv 0$ in $B_R$. This contradicts $v \equiv 1$.
% 
% Moreover,
% \[
%  v_k(0) = 1\quad \forall k
% \]
% 
% 
% Then up to a rotation and after passing to a subsequence, $\Omega_{k}\to \mathbb{R}^{3}_{+}$. On the other hand, we have $$\|(-\Delta)^{\frac{3}{2}}v_{k}\|_{L^{\infty}(\Omega_{k})}\leq \frac{C}{m_{k}^{2}}.$$
% Now using the fact that $\|v_{k}\|_{L^{\infty}}=1$. Again we split $v_{k}=f_{k}+h_{k}$, where $(-\Delta)^{\frac{3}{2}}f_{k}=(-\Delta)^{\frac{3}{2}}v_{k}$ in $B_R$ and $f_k\in H^\frac32_{00}(B_R)$, while $h_{k}$ is $\frac{3}{2}$-harmonic in $B_R$.
% Then we have that $f_{k}\to 0$ in $C^{2,\alpha}$ (see appendix \ref{har2}). On the other hand $h_{k}$ is uniformly bounded, since $v_{k}$ is bounded. Therefore, from the estimate \ref{har1}(appendix), we have that up to a subsequence $v_{k} \to v$ in $C^{2,\alpha}_{loc}(\mathbb{R}^{3})$ and $v$ satisfies 
% $$
% \left\{\begin{array}{ll}
% (-\Delta)^{\frac{3}{2}}v=0 \text{ in $\mathbb{R}^{3}_{+}$}\\
% v\in H^\frac32_{00}( \mathbb{R}^{3}_{+}).
% \end{array}
% \right.
% $$
% Notice that $\nabla u_{k}=0$ on $\Omega^{c}$ since $u_{k}=0$ on the same set. From the Sobolev embedding of $H^{\frac{1}{2}}_{00}(\Omega) \hookrightarrow L^{3}(\Omega)$, we have that 
% $$\int_{\Omega_{k}} |\nabla v_{k}|^{3}dx =\frac{1}{m_{k}^{3}} \int_{\Omega} |\nabla u_{k}|^{3}dx \leq \frac{C}{m_{k}^{3}}.$$
% Indeed,
% $$\|\nabla u_{k}\|^{2}_{H^{1/2}}=\int_{\mathbb{R}^{3}}|\xi||\widehat{\nabla u}_{k}|^{2}d\xi=\int_{\mathbb{R}^{3}}|\xi|^3|\hat{u}_{k}|^{2}d\xi =\|u_{k}\|_{H}^{2}\leq C$$
% 
% 
% 
% Thus $\nabla v =0$ hence $v$ is constant but $v=0$ on $\partial \mathbb{R}^{3}_{+}$ thus $v=0$ which contradicts the fact that $v(0)=1$.
\end{proof}



%\begin{lemma}\label{diff} We have
%\begin{equation}
%u_{k}(x_{k}+r_{k}x)-m_{k}\to 0 \text{ in } C^{2}_{loc}(\mathbb{R}^{3}) \text{ as }k\to \infty.
%\end{equation}
%\end{lemma}



\begin{lemma}\label{diff} Let $m_k$ be as in \eqref{eq:xkdef}. Then we have
\begin{equation}
u_{k}(x_{k}+r_{k}x)-m_{k}\to 0 \text{ in } C^{n-1}_{\loc}(\mathbb{R}^{n}) \text{ as }k\to \infty.
\end{equation}
\end{lemma}

\begin{proof}
Let $\tilde{u}_k:=u_{k}(x_{k}+r_{k}x)$. Then $\tilde{u}_k \in C^0_c(\R^n)\cap X(\Omega_k)$ and
\[
 \sup_{x \in \R^n} |\tilde{u}_k(x)| = \tilde{u}_k(0) = m_k \in [0,\infty).
\]
As above by Sobolev embedding, $\tilde{u}_k \in W^{1,n}_0(\Omega_k)$
\[\begin{split}
 \limsup_{k\to\infty}\|\nabla  \tilde{u}_k\|_{L^n(\R^n)} &\le C \limsup_{k\to\infty} \|(-\lap)^\frac n4 \tilde{u}_k\|_{L^2(\R^n)} \\
&= C\limsup_{k\to\infty}\|(-\lap)^\frac n4 u_k\|_{L^2(\R^n)}\\
& \leq C(\Lambda).
\end{split}
\]
and from \eqref{Rieszprop-1}, \eqref{Rieszprop0} and \eqref{Rieszprop} below for $k$ large enough we get
\begin{equation}\label{gkbound}
\begin{split}
\|\laps{1} \tilde{u}_k\|_{L^n(\R^n)} &=  \sum_{i=1}^n\|\Rz_i\Rz_i \laps{1} \tilde{u}_k\|_{L^n(\R^n)}\\
&\le C \sum_{i=1}^n\|\Rz_i \laps{1} \tilde{u}_k\|_{L^n(\R^n)}  \\
&\le C \|\nabla  \tilde{u}_k\|_{L^n(\R^n)} \\
& \leq C(\Lambda).
\end{split}
\end{equation}
Notice that 
$$|(-\Delta)^\frac n2 \tilde{u}_k|\le \frac{C}{m_k}\quad \text{in }\Omega_k.$$
Finally, by Lemma~\ref{rk} for any $\varphi \in C_c^\infty(\R^3)$, for all sufficiently large $k$ depending on the size of the support of $\varphi$,
\begin{equation}\label{eq:pdeforwk}
 \bigg|\int_{\R^n} (-\lap)^{\frac{n}{4}} \tilde{u}_k\, (-\lap)^{\frac{n}{4}} \varphi\, dx\bigg| \leq C\ \frac{1}{m_k} \int_{\R^3} |\varphi|\,dx.
\end{equation}
Let $g_k := \laps{1} \tilde{u}_k$, bounded in $L^n(\R^n)$, according to \eqref{gkbound}.
There is a weakly convergent subsequence $g_k \rightharpoonup g$ in $L^n(\R^n)$. Moreover, we have for any $\varphi \in C_c^\infty(\R^n)$, by \eqref{eq:pdeforwk}
\[
  \int_{\R^n} g\, (-\lap)^\frac{n-1}{2} \varphi \, dx= \lim_{k \to \infty} \int_{\R^n} g_k\, (-\lap)^\frac{n-1}{2} \varphi \, dx=  \lim_{k \to \infty} \int_{\R^n} (-\lap)^\frac n4 \tilde{u}_k\, (-\lap)^\frac n4 \varphi\, dx \xrightarrow{k \to \infty} 0.
\]
Consequently, $g \in C^\infty(\R^n)\cap L^n(\R^n)$, and pointwise $(-\lap)^\frac{n-1}{2}g \equiv 0$. This implies that $g\equiv 0$. Indeed by elliptic estimates (see e.g. \cite[Proposition 4]{mar1}) and H\"older's inequality it follows that 
$$\|g\|_{L^\infty(B_1)}\le C\|g\|_{L^1(B_2)}\le \tilde C \|g\|_{L^n(B_2)},$$
which scaled gives
$$\|g\|_{L^\infty(B_R)} \le\tilde C R^{-1}\|g\|_{L^n(B_{2R})}\to 0\quad \text{as }R\to\infty.$$ 


So we have obtained, that $\laps{1} \tilde{u}_k \rightharpoonup 0$ in $L^n(\R^n)$. Then, using \eqref{Rieszprop} and \eqref{eq:RiezPI} we also have
$$\nabla \tilde{u}_k = \Rz \laps{1} \tilde{u}_k \rightharpoonup 0\quad \text{in }L^n(\R^n).$$
Since $\tilde{u}_k$ is uniformly bounded in $H^\frac n2(\R^n)$, since $n\geq 3$, we also have strong convergence in $W^{1,2}_{\loc}(\R^n)$. In particular up to choosing a subsequence, for any $R > 1$, 
\begin{equation}\label{grad0}
\nabla \tilde{u}_k \to 0 \quad \text{in }L^2(B_R).
\end{equation}
On the other hand, observe the following: For any $R > 1$, for all large $k \in \N$,  we have  $B_{2R}\subset \Omega_k$. From \eqref{eq:pdeforwk},  Lemma~\ref{la:LinftyRHSest} we obtain
\[
 \|\nabla \tilde{u}_k\|_{C^{n-2,\alpha}(B_R)} \leq C
\]
for a uniform constant $C$ and $\alpha\in (0,1)$.

Since $\tilde{u}_k(0) = m_k$, we have
\[
\|\tilde{u}_k-m_k\|_{L^\infty(B_R)} \leq  \|\nabla \tilde{u}_k\|_{L^\infty(B_R)} \leq C,
\]
and consequently we have shown that
\[
\|\tilde{u}_k-m_k\|_{C^{n-1,\alpha}(B_R)} \leq C
\]
Now Arzel\`a-Ascoli gives (up to a further subsequence) $C^{n-1}(B_R)$-convergence of $\tilde{u}_k-m_k$, and using \eqref{grad0} we have that $\tilde{u}_k - m_k \to 0$  in $C^{n-1}(B_R)$.  Since $R$ is arbitrary the proof is complete.
\end{proof}


% 
% \begin{lemma}\label{la:Konstante}
% Let $w \in C^{1}_{loc}(\R^3)$ so that
% \begin{equation}\label{eq:lap12winl3}
%  (-\lap)^{1/2} w, \nabla w \in L^3(\R^3),
% \end{equation}
% \[
% (- \lap)^{3/4} w \in L^2(\R^3),
% \]
% and
% \[
% \int_{\R^3} (-\lap)^{3/4} w\ (-\lap)^{3/4} \varphi dx = 0, 
% \]
% for all $\varphi \in C_c^\infty(\R^n)$.
% Then
% \[
%  w \equiv const.
% \]
% \end{lemma}
% \begin{proof}
% Since $\lap^{1/2}$ and $(-\lap)^{\frac{3}{4}}$ are well-defined (??????), we can write the PDE as
% \[
% \int_{\R^3} \lap^{1/2} w\ \lap \varphi = 0. 
% \]
% In particular, $\lap^{1/2} w \in C^\infty$ and \eqref{eq:lap12winl3} implies that
% \[
%  \lim_{|x| \to \infty} \lap^{1/2} w(x)= 0 (???????).
% \]
% Since $\lap^{1/2} w$ is a weakly harmonic function with boundary value $0$, we obtain that
% \[
%  \lap^{1/2} w \equiv 0.
% \]
% In particular,
% \[
%  \nabla w \equiv 0,
% \]
% which implies that $w$ is a constant.
% \end{proof}
% \begin{proof}
% {\color{magenta} rewrite this in two Lemmas: 1. using higher-order Poincar\'e estimates to get that the only limit of $w_k$ is zero, 2. using local estimates to show that we have $C^{2,\alpha}_{loc}$-convergence)}
% The function $w_{k}(x):=u_{k}(x_{k}+r_{k}x)-m_{k}$ in $\Omega_{k}$ solves
% $$(-\Delta)^{\frac{3}{2}}w_{k}=16\frac{v_{k}}{m_{k}}e^{\frac{3}{2}m_{k}^{2}(v_{k}^{2}-1)},$$
% and 
% $$\|\nabla w_{k}\|_{L^{3}(\Omega_{k})}=\|\nabla u_{k}\|_{L^{3}(\Omega)}\leq C.$$
% We write then $w_{k}=h_{1,k}+h_{2,k}$, where $h_{2,k}(x)=\int_{\mathbb{R}^{3}}\frac{1}{|x-y|}g_{k}$ and $g_{k}$ satisfies 
% 
% \begin{equation}
% \left\{ \begin{array}{ll}
% (-\Delta)^{\frac{1}{2}}g_{k}=16\frac{v_{k}}{m_{k}}e^{\frac{3}{2}m_{k}^{2}(v_{k}^{2}-1)} & \text{On } B_{R} \\
%  g_{k} = 0 & \text{On } \mathbb{R}\setminus B_{R} 
% \end{array}
% \right.
% \end{equation}
% 
% Where $B_{R}\subset \Omega_{k}$ is a fixed ball. Hence, using the elliptic estimates as in \ref{eq}, we have that $h_{2,k}\to 0$ in $C^{2,\alpha}(\overline{B_{R}})$ {\color{blue}(I don't understand. More details)}{\color{red}(I added more details)}.
% 
% Notice that this way $h_{1,k}$ and $h_{2,k}$ are globally defined, also we have for $f_{k}=-\Delta h_{1,k}$, $(-\Delta)^{\frac{1}{2}}f_{k}=0$ on $B_{R}$ {\color{blue} (How do we define $(-\Delta)^{\frac{1}{2}}g_{k}$ if $g_k$ has been defined only in $B_R$?)} hence, 
% $$\|f_{k}\|_{C^{l,\alpha}(B_{R/4})}\leq C||f_{k}||_{L^{1}(B_{R/2})}$$
% But on $B_{R/2}$,  $f_{k}=\gamma_{\lambda}*(-\Delta h_{1,k})=\nabla \gamma_{\lambda}*\nabla h_{1,k}$, hence 
% $$\|f_{k}\|_{L^{1}(B_{R/2})}\leq C \|\nabla h_{1,k}\|_{L^{1}(B_{R/2})}\leq C'$$
% Thus
% $$\|-\Delta h_{1,k}\|_{C^{l,\alpha}(B_{R/4})}\leq C.$$
% By Pizzetti's formula, using the fact that $|h_{1,k}(0)|=|h_{2,k}(0)|\leq C$ and $h_{1,k}\leq -h_{2,k} \leq C$, we have that $\|h_{1,k}\|_{L^{1}(B_{R/4})}<C$ and therefore $\|h_{1,k}\|_{C^{k,\alpha}(B_{\frac{R}{4}})}$ is uniformly bounded hence we have the convergence of $w_{k}$ in $C^{2,\alpha}_{loc}$ to a certain $w$ satisfying $(-\Delta)^{\frac{3}{2}}w=0$, this in particular means that $-\Delta w $ is constant. On the other hand, we have that $$   \|\nabla w\|_{L^{3}(\mathbb{R}^{3})}\leq C$$
% Then $w$ is a constant with $w(0)=0$ hence $w=0$.
% \end{proof}

%The equation
%$$(-\Delta)^\frac32 u=f\in H^{-\frac32}(\mathbb R^{3})$$
%means
%$$\langle (-\Delta)^\frac34 u,(-\Delta)^\frac34 \varphi\rangle:=\int_{\mathbb{R}^3}\hat u(\xi) |\xi|^3 \hat\varphi(\xi)d\xi = \langle f,\varphi\rangle\quad\text{for every }\varphi\in H^\frac32_{00}(\Omega).$$

%{\color{yellow} {\color{magenta} Again, I don't think this is possible:} Notice that if $f\in L^{2}(\Omega)$ and $u\in H^{2}(\mathbb{R}^{3})$ then 
%$$\langle (-\Delta)^\frac34 u,(-\Delta)^\frac34 \varphi\rangle:=\int_{\mathbb{R}^3}\hat u(\xi) |\xi|^3 \hat\varphi(\xi)d\xi = \int_{\mathbb{R}^3}\hat{(-\Delta u)}(\xi) |\xi| \hat\varphi(\xi)d\xi= \langle f,\varphi\rangle\quad\text{for every }\varphi\in H^\frac32_{00}(\Omega).$$
%Hence, $(-\Delta)^{\frac{3}{2}}u=(-\Delta)^{\frac{1}{2}}(-\Delta u)$.\\}



% \begin{proof} 
% We write $u_{k}=h_{k}+\overline{v}_{k}$, where $v_{k}=\int_{\mathbb{R}^{3}}\frac{1}{|x-y|}g_{k}(y)dy$ and $g_{k}$ satisfies 
% \begin{equation}\label{eq}
% \left\{\begin{array}{ll}
% (-\Delta)^{\frac 12}g_{k}=\lambda_{k}u_{k}e^{\frac 32 u^{2}_{k}}& \text{on } \Omega  \\
%  g_{k}=0 & \text{on } \mathbb{R}^{3}\setminus \Omega 
% \end{array}
% \right.
% \end{equation}
%  we have first that $g_{k}$ converges in $C^{0,\alpha}(\mathbb{R}^{3})$ for all $\alpha<\frac{1}{2}$, since $u_{k}e^{\frac{3}{2}u_{k}^{2}}$ is uniformly bounded in $L^{\infty}$, but again, $\overline{v}_{k}$ satisfies
% 
% \begin{equation}
% -\Delta \overline{v}_{k}=g_{k} \quad \text{ on } \mathbb{R}^{3}  \\
% \end{equation}
% 
% Therefore $\overline{v}_{k}\to \overline{v}_{\infty}$ in $C^{2,\alpha}(\overline{\Omega})$.\\
% Notice now that $h_{k}=u_{k}-\overline{v}_{k}$ and $-\Delta h_{k} \in L^{2}(\mathbb{R}^{3}$. Moreover we have that 
% $$(-\Delta) ^{\frac{1}{2}}(-\Delta h_{k}) = 0 \text{ On  } \Omega$$ 
% and $-\Delta h_{k} =0$ on $\Omega^{c}$ hence $-\Delta h_{k} = 0$ with $h_{k}=\overline{v}_{k}$ on $\Omega^{c}$. Thus we have the convergence of $h_{k}$ in $C^{2,\alpha}_{loc}(\Omega)$. 
% \end{proof}


% 
% \begin{lemma}
% The sequence $(-\Delta)^{\frac{3}{2}}u_{k}$ is bounded in $L(\log L)^{\frac{1}{2}}(\Omega)$. Moreover, $||\nabla^{l} u_{k}||_{\frac{3}{l},2}$ is bounded.
% \end{lemma}
% \begin{proof}
% Since $(-\Delta)^{\frac{3}{2}}u_{k}=\lambda_{k}u_{k}e^{\frac{3}{2} u_{k}^{2}}$ we have that 
% $$ \log(2+|(-\Delta)^{\frac{3}{2}}u_{k}|) = \log ( 2 +\lambda_{k}u_{k}e^{\frac{3}{2} u_{k}^{2}}) \leq C(1+u_{k})^{2},$$
% hence,
% $$\int_{\Omega} |(-\Delta)^{\frac{3}{2}}u_{k}|\log(2+|(-\Delta)^{\frac{3}{2}}u_{k}|)^{\frac{1}{2}} dx \leq C\left(2 \int_{\Omega} \lambda_{k}u_{k}^{2}e^{\frac{3}{2}u_{k}^{2}}dx + \int_{u_{k}\leq 1} \lambda_{k} u_{k}e^{\frac{3}{2}u_{k}^{2}}dx \right) $$
% $$\leq C\left(||u_{k}||_{H^{\frac{3}{2}}}^{2}+C_{2}|\Omega|\right).$$
% Which is clearly bounded independently of $k$.\\
% Now notice that ({\color{magenta} Check reference!})
% $$|\nabla^{l}u_{k}(x)|\leq C \int_{\Omega}\frac{1}{|x-y|^{l}}|(-\Delta)^{\frac{3}{2}}u_{k}| dx$$
% $$\leq C I_{3-l}(|(-\Delta)^{\frac{3}{2}}u_{k}| \chi_{\Omega} ).$$
% But the fractional integral $I_{3-l}$ is of strong type from $L(LogL)^{\frac{1}{2}}$ to $L^{\frac{3}{l},2}$, hence,
% $$||\nabla^{l} u_{k}||_{\frac{3}{l},2}\leq C ||(-\Delta)^{\frac{3}{2}}u_{k}||_{L(LogL)^{\frac{1}{2}}} <\tilde{C}.$$
% \end{proof}
% 

% We define now $\eta_{k}=m_{k} (u_k(x_k + r_k \cdot) - m_k) +\log (2)$. Then we have the following corollary
% 
% \armin{Rephrase this corollary! $v$?}
% \begin{corollary}\label{eta}
% The function $\eta_{k}$ satisfies \armin{what does this mean? we did not define $\laps{s}$ of constants}
% $$(-\Delta)^{\frac{3}{2}}\eta_{k}=V_{k}e^{3a_{k}\eta_{k}},$$
% where $V_{k}=2^{\frac{3}{2}(1-v_{k})+1}\to 2$ and $a_{k}=\frac{1}{2}(v_{k}+1)\to 1$ in $C^0_{\loc}(\R^3)$.
% \end{corollary}


\subsection{Gradient-type estimates}
Note that from \eqref{energy}
\[
 \limsup_{k \to \infty} \|u_k \laps{n} u_k\|_{L^1(\Omega)} \leq \Lambda.
\]
Moreover, as in \cite[Proof of Lemma 5]{mar4}, we know that for the Orlicz space $L \log^{\frac{1}{2}} L(\Omega)$,
\[
 \limsup_{k \to \infty} \|\laps{n} u_k \|_{L\log^{\frac{1}{2}} L(\Omega)} \leq C(\Lambda,\Omega).
\]
We will now need the following crucial estimate applied to $u=u_k$ and $\rho=Rr_k$ for a given $R>0$ and $k$ so large that $B_{10\rho} (x_k) \subset \Omega$ (compare to Lemma \ref{rk}).

\begin{proposition}\label{la:ulapuest}
Let $\Omega$ be a smoothly bounded domain, and consider $u \in X(\Omega)$ such that $\laps{n} u = f$ weakly in $\Omega$ for some $f \in L\log^{\frac{1}{2}} L(\Omega)\cap L^\infty(\Omega)$. Assume moreover that
\begin{equation}\label{eq:uniformbound}
 \vrac{u\, f}_{L^1(\Omega)} +  \|f\|_{L\log^{\frac{1}{2}} L(\Omega)} + \vrac{(-\lap)^{\frac{n}{4}} u}_{L^2(\R^n)} \leq C_1.
\end{equation}
Then for a constant depending $C_2$ depending only on $C_1$ and $s \in (0,n)$ we have %for any $B_{10\rho}(x) \subset \Omega$,
\[
 \sup_{B_{4\rho}(x_0) \subset \Omega} \rho^{s-n} \int_{B_{\rho}(x_0)}|u\laps{s} u|\, dx \leq C_2
\]
\end{proposition}
\begin{proof}
We will use the Lorentz spaces $L^{(p,q)}$, for which we refer the reader to the appendix. Using the H\"older-type inequality (see \cite{one})
$$\|gh\|_{L^1(\Omega)}\le \|g\|_{L^{(\frac{n}{n-s},1)}(\Omega)}\, \|h\|_{L^{(\frac{n}{s},\infty)}(\Omega)},$$
we get (for $B_\rho=B_\rho(x_0)$, to simplify the notation)
\[\begin{split}
\rho^{s-n} \int_{B_{\rho}}|u\laps{s} u| dx &\le \rho^{s-n} \|\chi_{B_\rho}\|_{L^{(\frac{n}{n-s},1)}(B_\rho)}  \vrac{u\laps{s} u}_{L^{(\frac{n}{s},\infty)}(B_\rho)}\\
&= C  \vrac{u\laps{s} u}_{L^{(\frac{n}{s},\infty)}(B_\rho)},
\end{split}\]
so that it remains to show the bound
\[
\sup_{B_{4\rho} \subset \Omega} \vrac{u\laps{s} u}_{L^{(\frac{n}{s},\infty)}(B_\rho)} \leq C_2.
\]
For $\eps > 0$ we denote with $u^\eps \in C_c^\infty(\R^n)$ the usual mollification.

Consider now a cut-off function $\theta_{B_1}\in C^\infty(B_2)$, $\theta_{B_1}\equiv 1$ in $B_1$ and $0\le \theta_{B_1}\le 1$ everywhere. Set $\theta_{B_{2\rho}}:=\theta_{B_1}(\cdot/2\rho)\in C^\infty_c(B_{4\rho})$.
Then since $u^\eps \in C^\infty_c(\R^n)$ we have for $s \in (0,n)$ pointwise in $B_\rho$:
%\begin{align*}
% |u\laps{s} u^\eps| = \lapms{t} (\laps{t} u)\, \laps{s} u^\eps
%\end{align*}
%Duality $g \in L^{\frac{n}{s}}$,
%\[
%\int g\, \lapms{t} (\laps{t} u)\, \laps{s} u^\eps\, dx = \int (\laps{t} u)\, \lapms{t} (g \laps{s} u^\eps) dx
%\]
\begin{align*}
 |u\laps{s} u^\eps| = &| u\lapms{n-s}\laps{n} u^\eps| \\
 \le & |u\lapms{n-s}(\theta_{B_{2\rho}}\laps{n} u^\eps)|+|u\lapms{n-s}((1-\theta_{B_{2\rho}})\laps{n} u^\eps)|\\ 
 \le&|u\lapms{n-s}(\theta_{B_{2\rho}}\laps{n} u^\eps)|\\
&+|u\lapms{n-s}((1-\theta_{B_{2\rho}})\laps{n} u^\eps)-\lapms{n-s}(u (1-\theta_{B_{2\rho}})\laps{n} u^\eps)|\\
&+|\lapms{n-s}(u \laps{n} u^\eps)|\\
=:& I + II + III.
\end{align*}
By Proposition~\ref{borderline1} and Proposition~\ref{borderline2}, using that $u = \lapms{\frac{n}{2}}(-\lap)^{\frac{n}{4}} u$ we infer
\begin{equation}\label{eq:estI}
 \vrac{I}_{L^{(\frac {n}{s},\infty)}(\R^n)} \aleq \vrac{(-\lap)^\frac n4 u}_{L^2(\R^n)} \vrac{\laps{n} u^\eps}_{L\log^{\frac{1}{2}} L(B_{4\rho)}} .
\end{equation}
From the disjoint-support commutator estimate, see Proposition~\ref{la:disjcommutator}, we have
\begin{equation}\label{eq:estII}
\vrac{II}_{L^{(\frac {n}{s},\infty)}(B_\rho)} \aleq \| (-\lap)^{\frac{n}{4}} u \|_{L^2(\R^n)}^2.
\end{equation}
Since the support of $u$ is contained in $\Omega$, by the Sobolev inequality
\begin{equation}\label{eq:estIII}
 \vrac{III}_{L^{(\frac{n}{s},\infty)}(\R^n)} \aleq \vrac{u \laps{n} u^\eps}_{L^1(\Omega)}
\end{equation}


Combining the estimates \eqref{eq:estIII}, \eqref{eq:estI}, \eqref{eq:estII} we arrive at
\[
\begin{split}
 \|u\laps{s} u^\eps\|_{L^{(\frac{n}{2},\infty)}(B_\rho)} &\aleq \|u \laps{n} u^\eps \|_{L^1(\Omega)} + \|(-\lap)^{\frac{n}{4}} u\|_{L^2(\R^n)}\ \|\laps{n} u^\eps\|_{L\log^{\frac{1}{2}} L(B_{4\rho})} \\
 &\quad + \| (-\lap)^{\frac{n}{4}} u\|_{L^2(\R^n)}^2\\
 &\overset{\eqref{eq:uniformbound}}{\leq} 
\|u \laps{n} u^\eps \|_{L^1(\Omega)} + C_1\ \|\laps{n} u^\eps\|_{L\log^{\frac{1}{2}} L(B_{4\rho})} + (C_1)^2.
 \end{split}
 \]
It remains to take $\eps \to 0$, but some care is needed, since $\laps{n} u$ is in general not a function, but a distribution.
 
Firstly, since $B_{4\rho} \subset \Omega$, for $\eps < \rho$ we have that
\[
 \laps{n} u^\eps = (\laps{n} u)^\eps \text{\quad in } B_{4\rho}.
\]
In particular, for $\eps < \rho$
\[
 \|\laps{n} u^\eps\|_{L\log^{\frac{1}{2}} L(B_{4\rho})} \aleq \|\laps{n} u\|_{L\log^{\frac{1}{2}} L(B_{4\rho})} \leq \|\laps{n} u\|_{L\log^{\frac{1}{2}} L(\Omega)} \leq C_1.
\]
For the remaining term $\|u \laps{n} u^\eps \|_{L^1(\Omega)}$, we need to argue as follows. Firstly, since $u^\eps$ is the usual mollification, we have
\begin{equation}\label{eqeps}
 \|\laps{n} u^\eps\|_{L^2(\R^n)} \leq \eps^{-\frac{n}{2}} \| (-\lap)^{\frac{n}{4}} u\|_{L^2(\R^n)}
\end{equation}
Moreover, since $\Omega$ is smooth and bounded and $u \in X(\Omega)$, the results by \cite{gru}, see also \cite[Theorem~1.2]{XR2}, using that 
\[
\|\laps{n} u\|_{L^\infty(\Omega)} =: C_3 < \infty,
\]
then if we set $\Omega_{-\eps} := \{x \in \Omega: \dist(x,\partial \Omega) > \eps\}$
\[
 \|u\|_{L^\infty(\Omega \backslash \Omega_{-\eps})} \aleq \eps^{\frac{n}{2}}\, C_3.
\]
In particular with \eqref{eqeps} we have 
\[\begin{split}
\|u\laps{n} u^\eps\|_{L^1(\Omega)} &\leq \|u\laps{n} u^\eps\|_{L^1(\Omega_{-\eps})} +  |\Omega \backslash \Omega_{-\eps}|^{\frac{1}{2}} \| (-\lap)^{\frac{n}{4}} u\|_{L^2(\R^n)}\\
&=  \|u\laps{n} u^\eps\|_{L^1(\Omega_{-\eps})} +  o(1) \text{\quad as $\eps \to 0$.}
\end{split}
\]
Now note again that
\[
 \laps{n} u^\eps = (\laps{n} u)^\eps \quad \mbox{pointwise in $\Omega_{-\eps}$}.
\]
Consequently, 
\[
 \|u\laps{n} u^\eps\|_{L^1(\Omega)} \leq \|u(\laps{n} u)^\eps\|_{L^1(\Omega_{-\eps})} +o(1)
\xrightarrow{\eps \to 0} \|u\ \laps{n} u\|_{L^1(\Omega)}.
\]
This concludes the proof of Lemma~\ref{la:ulapuest}.
\end{proof}


\subsection{Convergence of \texorpdfstring{$\eta_k$}{etak}}
Let $\eta_k$ be as in \eqref{defetak}.
%\[
% \eta_{k}(x):=m_{k}(u_{k}(x_{k}+r_{k}x)-m_{k}).
%\]
From Proposition~\ref{la:ulapuest} we now infer:

\begin{proposition}\label{gradest}
For every $s\in (0,n)$ there exists $C>0$ such that for every $R>0$ and $k$ large enough (depending on $R$ and $s$) we have
\begin{equation}\label{eq:lapsest}\int_{B_{R}}|(-\Delta)^\frac{s}{2} \eta_{k}| dx \leq C R^{n-s}.\end{equation}
\end{proposition}
\begin{proof}
According to Lemma \ref{diff} we have
$$m_k\le 2u_k\quad \text{on } B_{Rr_k}(x_k)\text{ for }k\text{ large enough,}$$
hence with Proposition~\ref{la:ulapuest} applied with $u=u_k$ and $\rho=Rr_k$ we obtain (note that $\laps{s} (m_k^2) = 0$)
\[
\begin{split}
\int_{B_{R}}|(-\Delta)^\frac{s}{2} \eta_{k}| dx&= \frac{m_k}{r_{k}^{n-s}} \int_{B_{Rr_{k}}(x_k)}|(-\Delta)^\frac s2 u_{k}| dx \\
&\le \frac{2}{r_{k}^{n-s}} \int_{B_{Rr_{k}}(x_k)}|u_k (-\Delta)^\frac s2 u_{k}| dx\\
& \leq C R^{n-s},
\end{split}
\]
as claimed.
\end{proof}

% % 
% % \begin{proof}
% % 
% % We fix $R>0$. Then using Propositions \ref{est1} and \ref{est2}, we have,
% % 
% % $$\int_{B_{Rr_{k}}(x_{k})}|\nabla^{l}(u_{k}^{2})(x)|dx\leq C (Rr_{k})^{3-l}.$$
% % On the other hand, from (\ref{diff}), we have that for $x\in B_{Rr_{k}}(x_{k})$,
% % $$m_{k}|\nabla u_{k}|(x)\leq C u_{k}(x) |\nabla u_{k}|$$
% % Hence  $$u_{k}(x_{k})\int_{B_{Rr_{k}}}|\nabla u_{k}(x)| dx \leq C (Rr_{k})^{2}.$$
% % A similar inequality hold for the second derivative.
% % \end{proof}
% 
% Let
% \[
%  \eta_{k}(x):=m_{k}(u_{k}(x_{k}+r_{k}x)-m_{k})
% \]
% 


\begin{proposition}\label{pr:boundedness}
For every $B_R \subset \R^n$ and any $\alpha \in [0,1)$ there exists a constant $C_{R,\alpha}$ so that
\[
 \| \eta_k \|_{C^{n-1+\alpha}(B_R)} \leq C_{R,\alpha}.
\]
for $k$ large enough.
\end{proposition}

\begin{proof} We have that $|(-\Delta)^\frac{n}{2} \eta_k|\le C(R)$ in $B_R$, in the sense that
\[\bigg|\int_{\R^n}(-\Delta)^\frac{n}{4} \eta_k (-\Delta)^\frac{n}{4} \varphi dx\bigg| \le C\|\varphi\|_{L^1(B_R)},\quad \text{for }\varphi\in C^\infty_c(B_R).\]
This can be rewritten as
\begin{equation}\label{eqinfty}
\bigg|\int_{\R^n}(-\Delta)^\frac{1}{2} \eta_k (-\Delta)^\frac{n-1}{2} \varphi dx\bigg| \le C\|\varphi\|_{L^1(B_R)},\quad \text{for }\varphi\in C^\infty_c(B_R),
\end{equation}
which means that the function $\psi_k:=(-\Delta)^\frac12 \eta_k$ satisfies
$$|(-\Delta)^\frac{n-1}{2}\psi_k| \le C_R\quad \text{in }B_R$$
in the sense of distributions (notice that $(-\Delta)^\frac{n-1}{2}$ is an integer power of $-\Delta$ since $n$ is odd).
This, together with the estimate
$$\|\psi_k\|_{L^1(B_R)}\le C R^{n-1}$$
given by Proposition \ref{gradest}, and standard elliptic estimates (see e.g. Proposition 4 and Lemma 20 in \cite{mar1}) implies that
$$\|\psi_k\|_{C^{n-2,\alpha}(B_{R/2})}\le C_{R,\alpha}\quad \text{for }0\le \alpha<1,$$
as claimed. Together with Harnack's inequality (see \cite{IMM}) we get
$$\|\eta_k\|_{C^{n-1,\alpha}(B_{R/4})}\le C_{R,\alpha}\quad \text{for }0\le \alpha<1,$$
and replacing $R$ with $4R$ we conclude.
% Split
%$\eta_k= \eta_{1,k}+\eta_{2,k}$ with $\eta_{1,k}\in H^\frac{1}{2}_{00}(B_{R/2})$ solving
%$$(-\Delta)^\frac12 \eta_{1,k}=(-\Delta)^\frac12 \eta_{k}\quad \text{in }B_{R/2}$$
%and $\eta_{2,k}\in C^\infty(B_R)$ satisfying
%$$(-\Delta)^\frac12 \eta_{2,k}=0\quad \text{in }B_{R/2},\quad \eta_{2,k}=\eta_k\quad \text{in }\R^n\setminus B_{R/2}.$$
%By (nonlocal) elliptic estimates one gets
\end{proof}

% \begin{proof}
% Set $\tilde{u}_k(x) := u_k(x_k + r_k x)$ and $w_k :=m_k \tilde{u}_k \equiv \eta_k+m_k^2 \in X(\Omega_k)$.
% % Note, however, that $w_k$ is not uniformly bounded in $X(\Omega_k)$. 
% Then $w_k$ satisfies weakly
% \[
% (-\Delta)^{\frac{n}{2}}w_k= 2^n(n-1)!\ \frac{\tilde{u}_{k}}{m_k} e^{\frac{n}{2}((\tilde{u}_k)^2-(m_k)^2)} \quad \text{ in $\Omega_{k}$}.
% \]
% In particular, since $m_k = \sup_{\Omega} |u_k|$,
% \begin{equation}\label{eq:lap32unofbd}
%  \sup_{k \in \N} \| \laps{n} w_k \|_{\infty,\Omega_k} < \infty.
% \end{equation}
% Moreover, we may assume, by Proposition~\ref{rk} for large $k \in \N$, that $\Omega_k$ contains $B_{10R}$.
% 
% % In particular we already know by Proposition~\ref{gradest}, 
% % \[
% %  \limsup_{k \to \infty} \|\lap w_k \|_{L^1(B_{2R})} \leq C_R.
% % \]
% From \eqref{eq:lap32unofbd} we are going to deduce that for any $j \in \{1,\ldots,n-1\}$ and any $p \in (1,\infty)$,
% \begin{equation}\label{eq:boundednessgoal}
%  \limsup_{k \to \infty} \| \nabla^j w_k \|_{L^p(B_R)} \leq C_{j,p}\ \brac{R^{\frac{n}{p}+n-j}+R^{\frac{n}{p}-j}}.
% \end{equation}
% Note that for $p=1$ this is a worse estimate than the one from Proposition~\ref{la:ulapuest} as $R \to \infty$ (\ToDo).
% % By the non-locality of $\laps{n}$ usual interpolation implies
% % \[
% %  \|\lap w_k \|_{\infty,B_R} \aleq C_R \|\laps{n} w_k \|_{\infty,B_{2R}} + C_R \vrac{\lap w_k}_{L^1(\R^3)},
% % \]
% % which is insufficient for our purposes because the second term may be unbounded.
% 
% We follow a similar strategy as in Lemma~\ref{la:ulapuest}. Since $\operatorname{supp} w_k \subset \subset \R^n$ and $w_k \in H^{n/2}(\R^n)$ we can write the following - up to the same approximation argument as in Lemma~\ref{la:ulapuest} (\ToDo check). Let $\theta_{B_{2R}} \in C_c^\infty(B_{3R})$ be a cutoff function $\theta_{B_{2R}} \equiv 1$ on $B_{2R}$. Let $T^j := \nabla^j \circ \lapms{j} :\, L^p(\R^n) \to L^p(\R^n,\R^{n^j})$. Set also $\tlapms{n-j} := \lapms{n-j} T^j$. Note that $T^j$ can be represented as a matrix of combinations of the Riesz transforms, for example, $(T^2)_{\alpha,\beta} = \Rz_\alpha \Rz_\beta$. In particular, $T^j$ plays no crucial role in the following computations, and the scared reader may think of $\tlapms{n-j}$ as $\lapms{n-j}$.
% For $0 < j < n$,
% \begin{equation}\label{eq:lapwk}
%  \nabla^j w_k = \tlapms{n-j} \laps{n} w_k = \tlapms{n-j}(\theta_{B_{2R}}\laps{n} w_k) + \tlapms{n-j}((1-\theta_{B_{2R}})\laps{n} w_k).
% \end{equation}
% Firstly, since $0 < j < n$, the boundedness of Riesz potential and Riesz transforms implies $\tlapms{n-j} :L^{(\frac{n}{n-j},1)}(\R^n) \to L^\infty(\R^n)$, see Proposition \ref{pr:Sobolev}, and then a H\"older type inequality on Lorentz spaces:
% \[
%  \vrac{\tlapms{n-j}(\theta_{B_{2R}}\laps{n} w_k))}_{\infty} \aleq \vrac{\laps{n} w_k}_{(\frac{n}{n-j},1),B_{3R}} \aleq R^{n-j}\  \vrac{\laps{n} w_k}_{\infty,B_{3R}}.
% \]
% This is uniformly bounded whenever $B_{3R} \subset \subset \Omega_k$, by \eqref{eq:lap32unofbd}. 
% 
% % Secondly by the disjoint support, Lemma~\ref{la:nonloc}, Lemma~\ref{disjointoperator}, \armin{write this for $\lapms{s}$},
% % \[
% % % marked
% %  \vrac{\lapms{n-j}(\chi_{\R^3 \backslash B_{2R}}\laps{n} w_k)}_{\infty,B_R} \aleq \vrac{\lap w_k}_{L^1(\R^3)}.
% % \]
% For the remaining part we decompose (recall $w_k = m_k \tilde{u}_k$),
% \[
% \begin{split}
% & \tlapms{n-j}((1-\theta_{B_{2R}})\laps{n} w_k)\\
% =& m_k\ \tlapms{n-j}((1-\theta_{B_{2R}})\laps{n} \tilde{u}_k)\\ 
% =& (m_k-\tilde{u}_k)\ \tlapms{n-j}((1-\theta_{B_{2R}})\laps{n} \tilde{u}_k)\\ 
% &+ \tilde{u}_k\ \tlapms{n-j}((1-\theta_{B_{2R}})\laps{n} \tilde{u}_k)\\ 
% =& (m_k-\tilde{u}_k) \tlapms{n-j}((1-\theta_{B_{2R}})\laps{n} \tilde{u}_k)\\ 
% &+ \tlapms{n-j}((1-\theta_{B_{2R}})\tilde{u}_k \laps{n} \tilde{u}_k)\\ 
% &+ \brac{\tilde{u}_k\ \tlapms{n-j}((1-\theta_{B_{2R}})\laps{n} \tilde{u}_k) -\tlapms{n-j}((1-\theta_{B_{2R}})\tilde{u}_k \laps{n} \tilde{u}_k)} \\
% =:& I_k + II_k + III_k.
% \end{split}
%  \]
% Firstly, by Lemma~\ref{diff} 
% \[
%  (\tilde{u}_k-m_k) \xrightarrow{k \to \infty} 0 \quad \text{in $C^{n-1}(\overline{B_R})$}.
% \]
% Moreover, by Lemma~\ref{la:freakshowestimate1}
% \[
% \begin{split}
% &\|\tlapms{n-j}((1-\theta_{B_{2R}})\ \laps{n} \tilde{u}_k)\|_{\infty,B_R}  \\
% \aleq & \sum_{\ell=1}^\infty \|\tlapms{n-j}(\theta_{A_{2^\ell R}}\ \laps{n} \tilde{u}_k)\|_{\infty,B_R}\\ 
% \aleq & \sum_{\ell=1}^\infty (2^\ell R)^{-j}  \|(-\lap)^{n/4} \tilde{u}_k\|_{2,\R^n}\\ 
% \aleq & R^{-j}  \|(-\lap)^{n/4} \tilde{u}_k\|_{2,\R^n}.
% \end{split}
% \]
% The right-hand side is uniformly bounded for $k \to \infty$. In particular, we thus have established
% \[
%  \limsup_{k \to \infty} \|I_k\|_{\infty,B_R} = 0.
% \]
% As for $II_k$, and by the disjoint support lemma, Lemma~\ref{la:nonloc},
% \[
%  \|II_k\|_{\infty,B_R} \aleq R^{-j} ||\tilde{u}_k \laps{n} \tilde{u}_k ||_{1,\R^n} =  R^{-j} ||\tilde{u}_k \laps{n} \tilde{u}_k ||_{1,\Omega_k} =  R^{-j} ||u_k \laps{n} u_k ||_{1,\Omega}
% \]
% Here we used again that $\operatorname{supp} \tilde{u}_k \subset \overline{\Omega_k}$. The right-hand side is again uniformly bounded (\ToDo check/write somewhere!).
% 
% It remains to treat $III_k$, and we use the same strategy as for the expression $II$ in the proof of Lemma~\ref{la:ulapuest}. Using the commutator structure, on $B_R$ we rewrite
% \[
% \begin{split}
% III_k =& \sum_{\ell = 2}^\infty \brac{\tilde{u}_k\ \tlapms{n-j}(\theta_{A_{2^\ell R}}\laps{n} \tilde{u}_k) -\tlapms{n-j}(\theta_{A_{2^\ell R}}\tilde{u}_k \laps{n} \tilde{u}_k)}\\
% =& \sum_{\ell = 2}^\infty \brac{v_k^\ell\ \tlapms{n-j}(\theta_{A_{2^\ell R}}\laps{n} \tilde{u}_k) -\tlapms{n-j}(\theta_{A_{2^\ell R}}v^\ell_k \laps{n} \tilde{u}_k)}\\
% \end{split}
% \]
% where
% \[
%  v^\ell_k := \theta_{B_{2^{l+1}R}}(\tilde{u}_k-(\tilde{u}_k)_{B^{2^{l+1}R}}).
% \]
% Then, just like in Lemma~\ref{la:ulapuest} we arrive for any $p \in (1,\infty)$, at the estimate
% \[
%  \|III_k\|_{p,B_R} \leq C_{p,j}\ R^{\frac{n}{p}-j}\ \|(-\lap)^{\frac{n}{4}} u_k\|_{2}^2.
% \]
% This implies our \eqref{eq:boundednessgoal}. That is for any $p \in (1,\infty)$,
% \[
%  \limsup_{k \to \infty} \|\nabla \eta_k\|_{W^{n-2,p}(B_R)} \leq C_{p,R}.
% \]
% In particular, for any $\beta \in (0,1)$ taking $p > \frac{n}{1-\beta}$, we have
% \[
%  \limsup_{k \to \infty} \| \nabla \eta_k \|_{C^{n-3+\beta}(B_R)} \aleq C_{\beta,R}.
% \]
% Using that $\eta_k(0) = 0$, this implies the claim.
% \end{proof}
% 


\begin{proposition}\label{boundLe}
The sequence $(\eta_k)$ is uniformly bounded in $L_s(\R^n)$ for any $s > 0$.
\end{proposition}
\begin{proof}
Since by Proposition \ref{pr:boundedness} the sequence $(\eta_k)$ is bounded in $L^\infty(B_1)$, it is easy to see that boundedness of $(\eta_k)$ in $L_s(\R^n)$ for some $s>0$ implies boundedness in $L_{s'}(\R^n)$ for every $s'>s$. Therefore without loss of generality we can assume that $s<1$. We then have
\begin{equation}\label{lapshwk}
\laps{s} \eta_k(x) = C_{n,s} \int_{\R^n} \frac{\eta_k(y)+\eta_k(x)}{|x-y|^{n+s}}\, dy.
\end{equation}
Consequently,  for an arbitrary $\varphi\in C^\infty_c(B_1)$ using \eqref{eq:lapsest}
\[
\begin{split}
C\|\varphi\|_{L^\infty(B_1)}&\overset{\eqref{eq:lapsest}}{\ge}%\bigg|\int_{\R^n} (-\Delta)^\frac n4 \eta_k\ (-\Delta)^\frac n4 \varphi\, dx\bigg|\\&=
\bigg|\int_{B_1} \laps{s} \eta_k\, \varphi\, dx\bigg|\\
&\overset{\eqref{lapshwk}}{\geq}
\int_{B_1} \int_{B_{2}}\frac{\eta_k(x)-\eta_k(y)}{|x-y|^{n+s}}dy\, \varphi(x)\, dx\\
&\quad +\int_{B_1} \int_{B_{2}^c}\frac{1}{|x-y|^{n+s}}dy\, \eta_k(x)\,  \varphi(x)\, dx \\
&\quad +\int_{B_1} \int_{B_{2}^c}\frac{-\eta_k(y)}{|x-y|^{n+s}}dy\, \varphi(x)\, dx\\
% 
% -c\int_{B_1} \int_{|h| < 2}\frac{|\eta_k(x+h)+\eta_k(x-h)-2\eta_k(x)|}{|h|^{n+1}}\, dh\ (-\Delta)^\frac{n-1}2 \varphi(x)\, dx\\
% &\quad -c\int_{B_1} \int_{|h| > 2}\frac{1}{|h|^{n+1}}\, dh\ 2|\eta_k(x)| |(-\Delta)^\frac{n-1}2 \varphi(x)|\, dx \\
% &\quad +c|\int_{B_1} \int_{|h| > 2} \frac{\eta_k(x+h)+\eta_k(x-h)}{|h|^{n+1}}\, dh\ (-\Delta)^\frac{n-1}2 \varphi(x)\, dx|\\
&=:I+II +III.
\end{split}
\]
Since by Proposition~\ref{pr:boundedness},
\[
 |\eta_k(x) - \eta_k(y)| \aleq |x-y| \quad \forall x,y \in B_2,
\]
we have that
\[
 |I| \aleq \int_{B_1} |\varphi(x)| \int_{|x-y| \leq 3} |x-y|^{-n+1-s}\ dy\ dx \aleq \|\varphi\|_{1}. 
\]
Since we also have $|\eta_k(x)| \leq C$ for all $x \in B_2$,
\[
 |II| \aleq \int_{B_1} |\varphi(x)| \int_{|x-y| > 1} |x-y|^{-n-s}\ dy\ dx \aleq \|\varphi\|_{1}. 
\]
Finally, since $-\eta_k(y) = |\eta_k(y)|$, we arrive at
\[
 \int_{B_1} \int_{B_{2}^c}\frac{|\eta_k(y)|}{|x-y|^{n+s}}dy\ \varphi(x)\, dx \leq C,
\]
for a constant depending on $\varphi$ and $s$, but independent of $k$. Taking $\varphi(x)$ to be non-negative and so that $\varphi \equiv 1$ on $B_{1/2}$, we arrive at
\[
 \int_{|y|> 2}\frac{|\eta_k(y)|}{1+|y|^{n+s}}dy \leq C.
\]
Since again by Proposition~\ref{pr:boundedness} for a $C$ uniform in $k$,
\[
 \int_{|y|< 2}\frac{|\eta_k(y)|}{1+|y|^{n+s}}dy \leq C.
\]
we have shown that
\[
 \sup_{k \in \N} \|\eta_k\|_{L_s(\R^n)} \leq C.
\]
\end{proof}

\begin{proposition}
Up to a subsequence,  $\eta_{k} +\log 2\to \eta_0 =\log(\frac{2}{1+|\cdot |^{2}})$ in $C^{\ell}_{\loc}(\mathbb{R}^n)$ for every $\ell\ge 0$, and
\begin{equation}\label{eneretak}
\lim_{R\to\infty}\lim_{k\to\infty} \int_{B_{Rr_k}(x_k)}\lambda_k u_k^2 e^{\frac n2 u_k^2}dx =(n-1)!\int_{\R^n} e^{n\eta_0}dx =\Lambda_1.
\end{equation}
\end{proposition}

\begin{proof}
Let $\eta_0$ be the pointwise limit of $\eta_k +\log 2$, which exists up to a subsequence, by Proposition \ref{pr:boundedness} and Arzel\`a-Ascoli's theorem. In fact the limit is in $C^\ell_{\loc}(\R^n)$ for every $\ell\ge 0$ since with Proposition \ref{boundLe} one can bootstrap regularity for the operator $(-\Delta)^\frac n2$, see e.g. \cite[Corollary 24]{JMMX}. It follows from Proposition \ref{boundLe} that 
\begin{equation}\label{eta0s}
\eta_0 \in L_s(\R^n)\quad \text{for every }s>0.
\end{equation}
We then have 
$$\lim_{k\to\infty}\int_{\R^n}e^{n(\eta_k+\log 2)} \varphi dx =\int_{\R^n}e^{n\eta_0}\varphi dx \quad  \text{for every }\varphi \in {C^\infty_c(\R^n)}.$$
We will show that moreover
\begin{equation}\label{eq:convergencePDEgoal}
  \lim_{k \to \infty} \int_{\R^n} \eta_k\ \laps{n} \varphi dx =  \int_{\R^n} \eta_0 \laps{n} \varphi dx\quad \text{for every }\varphi \in {C^\infty_c(\R^n)}.
\end{equation}
Then $\eta_0$ satisfies $(-\Delta)^\frac n2 \eta_0=(n-1)!e^{n\eta_0}$ as a distribution, and in fact also as tempered distribution. Then from to Theorem \ref{thmali} we infer that $\eta_0=v+P$ where $|v|\le C(1+\log(1+|\cdot|))\in L_{s}(\R^n)$ for every $s>0$, and $P$ is polynomial bounded from above. It is easy to see that if $P$ is not constant, then $P\not\in L_s(\R^n)$ for any $s>0$, which contadicts \eqref{eta0s}. Therefore $P$ is constant and $\eta_0$ is as in \eqref{spherical}. It remains to determine $\lambda$ and $x_0$ in \eqref{spherical}, but this is easy since $\eta_k(0)=0=\max_{\R^n}\eta_k$, so that $\eta_0(0)=\log 2=\max_{\R^n}\eta_0$, i.e. $x_0=0$, $\lambda=1$ and
%Also, considering that in Proposition \ref{gradest} we have (\ToDo for what needed ?, doesnt $L_\eps$ already rule out freaky polynomials?)
%YES
%$$\int_{B_R}|\Delta\eta_0|dx\le C R^{n-2},$$
$\eta_0(x)=\log\frac{2}{1+|x|^2}$.



In order to obtain \eqref{eq:convergencePDEgoal}, Assume that for some $R > 0$, $\operatorname{supp} \varphi \subset B_R$ and let $\theta > 1$. Then,
\[
\begin{split}
\int \eta_0 \laps{n} \varphi\, dx  - \int \eta_k \laps{n} \varphi\, dx
=&\int_{B_{\theta R}} (\eta_0 \laps{n} \varphi  - \eta_k\ \laps{n} \varphi)\, dx\\
&+\int_{\R^n \backslash B_{\theta R}} (\eta_0 \laps{n} \varphi  - \eta_k\ \laps{n} \varphi)\, dx\\
=:& I + II.
\end{split}
 \]
Notice that by the disjoint support $\varphi$, see Lemma~\ref{la:nonloc},
\[
\begin{split}
|II| &\aleq \|\varphi\|_{L^1(\R^n)}\, \int_{\R^n \backslash B_{\theta R}} \frac{|\eta_0(x)|+|\eta_k(x)|}{|x|^{2n}}\, dx\\
&
\aleq \|\varphi\|_{L^1(\R^n)}\, \theta^{s-n} \int_{\R^n \backslash B_{\theta R}} \frac{|\eta_0(x)|+|\eta_k(x)|}{1+|x|^{n+s}}\, dx
\end{split}
\]
and the uniform bound of $\eta_0$ and $\eta_k$ in $L_{s}(\R^n)$ implies that
\[
 |II| \aleq \theta^{s -n},
\]
for a constant independent of $k$. On the other hand, $\eta_0 - \eta_k \to 0$ uniformly in $B_{\theta R}$, which implies that $\lim_{k \to \infty} I = 0$. Consequently,
\[
 \lim_{k\to \infty}\bigg|\int \eta_0 \laps{n} \varphi\, dx - \int \eta_k\ \laps{n} \varphi\, dx\bigg| \aleq  \theta^{s -n},
\]
for any $\theta > 1$, and letting $\theta \to \infty$ we conclude the proof of \eqref{eq:convergencePDEgoal}.

Finally, using Lemma \ref{diff} and the definition of $r_k$, we obtain
\[\begin{split}
\int_{B_{Rr_k}(x_k)}\lambda_k u_k^2 e^{\frac n2 u_k^2}dx&=\int_{B_R}r_k^n \lambda_k u_k^2(r_k\cdot)e^{\frac n2 m_k^2}e^{\frac n2(u_k(r_k\cdot)-m_k)^2}e^{n\eta_k}dx\\
&=\int_{B_R}r_k^n \lambda_k m_k^2(1+o(1))e^{\frac n2 m_k^2}e^{o(1)}e^{n\eta_k}dx\\
&=2^n(n-1)!\int_{B_R}(1+o(1))e^{n\eta_k}dx\\
&=(n-1)!\int_{B_R}e^{n\eta_0}dx+o(1)
\end{split}\]
with $o(1)\to 0$ as $k\to\infty$. Now letting also $R\to\infty$ and noticing that
$$\int_{\R^n}e^{n\eta_0}dx=|S^n|,$$
we infer \eqref{eneretak}.
\end{proof}



% \ToDo old proof
% \begin{proof}
% We will show firstly that $\eta_{k}$ is uniformly bounded in $L_{\frac n2}(\R^n)$, see \eqref{defLs}.
% %\begin{equation}\label{stimaetak}
% %\|(-\Delta)^\frac n2 w_k\|_{\infty,B_1} \le C\quad \text{for all }k\ge k_0.
% %\end{equation}
% %Recall also that since for $w_k \in X(\Omega_k)$,
% Since $\eta_k$ is smooth in $B_R$ ($\eta_k\in C^{1,\alpha}(B_R)$ for some $\alpha>0$ would suffice) we can write%have the following principal value representation for the $\laps{1} _k$,
% \begin{equation}\label{lapshwk}
% \laps{1} \eta_k(x) = C_n P.V.\int_{\R^n} \frac{\eta_k(y)+\eta_k(x)}{|x-y|^{n+1}}\ dy.
% \end{equation}
% Consequently,  for an arbitrary $\varphi\in C^\infty_c(B_1)$ using \eqref{eqinfty} and in the second line the fractional integration by parts-Lemma~\ref{la:pi}
% \[
% \begin{split}
% C\|\varphi\|_{L^1(B_1)}&\overset{\eqref{eqinfty}}{\ge}%\bigg|\int_{\R^n} (-\Delta)^\frac n4 \eta_k\ (-\Delta)^\frac n4 \varphi\, dx\bigg|\\&=
% \bigg|\int_{B_1} \laps{1} \eta_k\ (-\Delta)^\frac{n-1}2 \varphi\, dx\bigg|\\
% &\overset{\eqref{lapshwk}}{\geq}
% \int_{B_1}P.V. \int_{B_{2}}\frac{\eta_k(x)-\eta_k(y)}{|x-y|^{n+1}}dy (-\Delta)^\frac{n-1}2 \varphi(x)\, dx\\
% &\quad +\int_{B_1} \int_{B_{2}^c}\frac{1}{|x-y|^{n+1}}dy\ \eta_k(x)\ (-\Delta)^\frac{n-1}2 \varphi(x)\, dx \\
% &\quad +\int_{B_1} \int_{B_{2}^c}\frac{-\eta_k(y)}{|x-y|^{n+1}}dy (-\Delta)^\frac{n-1}2 \varphi(x)\, dx\\
% % 
% % -c\int_{B_1} \int_{|h| < 2}\frac{|\eta_k(x+h)+\eta_k(x-h)-2\eta_k(x)|}{|h|^{n+1}}\, dh\ (-\Delta)^\frac{n-1}2 \varphi(x)\, dx\\
% % &\quad -c\int_{B_1} \int_{|h| > 2}\frac{1}{|h|^{n+1}}\, dh\ 2|\eta_k(x)| |(-\Delta)^\frac{n-1}2 \varphi(x)|\, dx \\
% % &\quad +c|\int_{B_1} \int_{|h| > 2} \frac{\eta_k(x+h)+\eta_k(x-h)}{|h|^{n+1}}\, dh\ (-\Delta)^\frac{n-1}2 \varphi(x)\, dx|\\
% &=:-I-II +III.
% \end{split}
% \]
% Using Proposition~\ref{pr:boundedness}, we in particular have for any $x \in B_1$ and for any $y \in B_2$
% \[
%  |\eta_k(x)+\nabla \eta_k (x)(x-y)-\eta_k(y)| \aleq |x-y|^{3/2}.
% \]
% Since on the other hand
% \[
%  \int_{B_2} \frac{x-y}{|x-y|^{n+1}} dy = \Rz \chi_{B_2}(x)
% \]
% is bounded in $L^p(\R^n)$ for any $p \in (1,\infty)$, together with the estimate $|\nabla u(x)| \leq C$ in $B_1$ we have that $I\le C$ with a constant $C$ depending on $\varphi$. 
% On the other hand
% \[
%  |\eta_k(x)| \aleq 1,
% \]
% hence $II\le C$ with a constant $C$ depending on $\varphi$.
% 
% Then, we have shown that
% \begin{equation}\label{estIII}
% |\int_{B_1}\int_{B_2} \frac{-\eta_k(y)}{|x-y|^{n+1}}\, dh\ (-\Delta)^\frac{n-1}2 \varphi(x)\, dx| \leq C_\varphi.
% \end{equation}
% Now, integrating by parts and recalling that
% $$(-\Delta)|x|^{-k}=k(n-k+2)|x|^{-k-2}\quad \text{for } k<n-2,$$
% 
%  obtains
% $$III=C_n\int_{B_1}\int_{B_{2}^c} \frac{-\eta_k(y)}{|x-y|^{2n}}dy\varphi(x)dx.$$
% Upon choosing $\varphi\equiv 1$ in $B_{1/2}$ and $\varphi\ge 0$ in $B_1$, and recalling that $\eta_k\le 0$ we obtain
% $$III\ge C(n,R) \int_{B_{2}^c}\frac{|\eta_k(y)|}{|y|^{2n}}dy,$$
% which, together with Proposition~\ref{pr:boundedness} and \eqref{estIII} implies
% $$\sup_{k \in \N} \|\eta_k\|_{L_\frac n2(\R^n)}\le C.$$
% Now  Proposition~\ref{pr:boundedness} implies that (up to a subsequence) $\eta_k\to \eta_0$ in $C^0_{\loc}(\R^n)$ for some function $\eta_0$, and by Fatou's lemma we easily get
% $$\|\eta_0\|_{L_\frac n2(\R^n)}=\lim_{k\to\infty}\|\eta_k\|_{L_\frac n2(\R^n)},\quad \lim_{k\to\infty} \|\eta_k-\eta_0\|_{L_\frac n2(\R^n)}=0.$$
% In particular $\eta_0$ satisfies
% $$(-\Delta)^\frac n2 \eta_0=(n-1)! e^{n\eta_0}\quad \text{in }\R^n$$
% in the weak sense, i.e.
% $$\int_{\R^n}\eta_0(-\Delta)^\frac n2 \varphi dx= (n-1)!\int_{\R^n}e^{n\eta_0}\varphi dx\quad \text{for }\varphi\in\mathcal{S}.$$
% Also, considering Proposition \ref{gradest} we have
% $$\int_{B_R}|\Delta\eta_0|dx\le C R^{n-2},$$
% and using Ali Hyder's result we infer that $\eta_0(x)+\log 2=\log\frac{2}{1+|x|^2}$.
% % 
% % 
% % 
% % First we claim that $\eta_{k}$ converges locally in $C^{2,\alpha}$ to $\overline{\eta}$ a solution to 
% % $$(-\Delta)^{\frac{3}{2}}\overline{\eta}=2e^{3\overline{\eta}}.$$
% % 
% % Since we have that $\eta_{k}\leq \log(2)$, $V_{k}e^{3a_{k} \eta_{k}}$ is locally uniformly bounded in $\R^3$,  $\eta_k$ and $(-\lap)^{3/2} \eta_k$ are uniformly bounded in $L^\infty_{\loc}(\R^3)$. By elliptic estimates this implies uniform estimates in $C^{2,\alpha}_{\loc}(\R^3)$, and Arzela-Ascoli implies convergence up to a subsequence in $C^{2,\alpha}$.
% % 
% % 
% % Now note that we have by Proposition~\ref{gradest} for any fixed $B_R$
% % \[
% %  \int_{B_R} |\lap \eta_k| = R\ (Rr_k)^{-1} m_k \int_{B_{Rr_k}} |\lap u_k| \xrightarrow{k \to \infty} C\ R < \infty.
% % \]
% 
% % We first write $\eta_{k}=f_{k}+h_{k}$, where $f_{k}$ satisfies $$(-\Delta)^{\frac{3}{2}}f_{k}=V_{k}e^{3a_{k}\eta_{k}}$$
% % and $f_{k}=0$ in $B_{R}^{c}$.\\
% % 
% % Step 1: We claim that $f_{k}$ is convergent in $C^{2,\alpha}_{loc}$. , hence $V_{k}e^{3a_{k} \eta_{k}} \in L^{\infty}_{loc}(B_{R})$ hence, from elliptic estimates (see []) we have that $f_{k}$ converges in $C^{2,\alpha}_{loc}(B_{R})$ to a function $f$.\\
% % 
% %  We consider now the sequence $h_{k}$. If $h_{k}$ is bounded in $L^{1}(B_{R})$ then $h_{k}$ is bounded in $C^{2,\alpha}_{loc}(B_{R})$. Thus we have that
% % we have the convergence of $\eta_{k}$, or else, $$||h_{k}||_{L^{1}}=\beta_{k}\to \infty.$$ So we define $\varphi_{k}=\frac{h_{k}}{\beta_{k}}$ then clearly $\varphi_{k}$ converges to $\varphi$ in $C^{2,\alpha}_{loc}$ and $(-\Delta)^{\frac{3}{2}}\varphi=0$. Notice that $-\Delta \varphi \not \equiv 0$, since $||\varphi||_{L^{1}}=1$, we have that $\varphi=0$ which is impossible. So there exists $R>0$ so that $\int_{B_{R}}|\Delta \varphi| dx >0$. Therefore,
% % 
% % $$\lim_{k\to \infty} \int_{B_{R}} |\Delta \eta_{k}| dx =\lim_{k\to \infty} \beta_{k} \int_{B_{R}} |\Delta \varphi| dx = + \infty.$$
% % On the other hand, from Proposition \ref{gradest}. we have 
% % $$\int_{B_{R}} |\nabla^{l}\eta_{k}| dx = m_{k}r_{k}^{l-3}\int_{B_{Rr_{k}}}|\nabla^{l} u_{k}| dx \leq C R^{3-l}.$$
% % Which is a contradiction, therefore, $h_{k}$ is bounded $L^{1}$ and indeed we have convergence of $\eta_{k}\to \eta$ in $C^{2,\alpha}$.\\
% 
% % Now according to \armin{???}, $\eta = v + p$, where $p$ is a polynomial of degree $0$ or $2$. But, notice that if $p$ is not of degree zero, then $\Delta \eta_{k} \to \Delta \eta =-a<0$, where $a$ is a constant. Now again, we have that
% %  $$\lim_{k\to \infty}\int_{B_{R}} |\nabla^{l}\eta_{k}| dx = +\infty$$
% %  And this is ruled out again by Proposition \ref{gradest}. So the degree of $p$ is zero and $\eta$ is a spherical solution. That is, there exists $x_{0}$ and $\lambda>0$ so that $$\eta(x)=\ln(\frac{2\lambda}{1+\lambda^{2}|x-x_{0}|^{2}}).$$
% %  Since $\eta_{k}(x)\leq \eta_{k}(0)=\ln(2)$, we have that $x_{0}=0$ and $\lambda=1$. Moreover, It follows easily that $$\lim_{R\to +\infty}\lim_{k\to +\infty} \int_{B_{Rr_{k}}(x_{k})} \lambda_{k}u_{k}^{2}e^{\frac{3}{2}u_{k}^{2}}dx=Q_{S^{3}}|S^{3}|.$$
% \end{proof}
% 


%\section{Case of multiple Blow-up points}
%Given $\ell \in \mathbb{N}$, we define, for a sequence of points $(x_{i,k})_{k\geq 1}$ such that $x_{i,k}\to x^{(i)}$ as $k\to \infty$ and $1\leq i\leq \ell$, the function $R_{\ell,k}(x)$ as follows
%$$R_{\ell,k}(x)=\inf_{1\leq i\leq \ell} |x-x_{i,k}|.$$
%We say that $(H_{\ell})$ holds if there are $\ell$ sequences $(x_{i,k})_{k\geq 1}$ as above, such that
%$$\sup_{x\in \Omega} \lambda_{k}R_{\ell,k}^{3}(x)u_{k}^{2}(x)e^{\frac{3}{2}u_{k}^{2}(x)}\leq C.$$
%We say that $(E_{\ell})$ holds, if there exists $\ell$ sequences of points $(x_{i,k})_{k\geq 1}$ such that $x_{i,k}\to x^{(i)}$ as $k\to \infty$ and $1\leq i\leq \ell$ such that for $r_{i,k}$ satisfying 
%$$\lambda_{k}r_{i,k}^{3}u_{k}^{2}(x_{i,k})e^{\frac{3}{2}u_{k}^{2}(x_{i,k})}=16,$$
%the following holds true :\\

%$(E_{\ell}^{1})$ : for all $1\leq i\not = j \leq \ell$, we have 
%$$\lim_{k\to \infty}\frac{dist(x_{i,k},\partial \Omega)}{r_{i,k}}=+\infty \text{ and }\lim_{k\to \infty}\frac{|x_{i,k}-x_{j,k}|}{r_{i,k}}=+\infty .$$
%$(E_{\ell}^{2})$ : $\eta_{i,k}(x)\to \eta_{0}(x)=\ln(\frac{2}{1+|x|^{2}})$ in $C^{2}_{loc}(\mathbb{R}^{3})$ for $1\leq i \leq \ell$.\\
%$(E_{\ell}^{3})$ : $$\lim_{R\infty} \lim_{k\to \infty} \int_{\bigcup_{1\leq i \leq \ell}B_{Rr_{i,k}}(x_{i,k})}\lambda_{k}u_{k}^{2}e^{\frac{3}{2}u_{k}^{2}} dx =\ell \Lambda_{1}.$$

%Notice that $(E_{1})$ holds from the previous section. Now for the sake of contradiction, we assume that $(E_{\ell})$ holds for some $\ell$ but not $(H_{\ell})$. Then, we choose $x_{\ell+1,k}\in \Omega$ such that
%\begin{equation}\label{ind}
%\lambda_{k}R_{\ell,k}(x_{\ell+1,k})u_{k}^{2}(x_{\ell+1,k})e^{\frac{3}{2}u_{k}^{2}(x_{\ell+1,k})}=\lambda_{k}\sup_{\Omega} R_{\ell,k}u_{k}^{2}e^{\frac{3}{2}u_{k}^{2}}\to +\infty, \text{ as } k\to \infty
%\end{equation}
%and choose $r_{\ell+1,k}$ as in (). It is easy to deduce from (\ref{ind}), that for $1\leq i \leq \ell$,
%$$lim_{k\to \infty}\frac{|x_{\ell+1,k}-x_{i,k}|}{r_{\ell+1,k}}= \infty.$$
%From $(E_{\ell}^{2})$ and (\ref{ind}) we have for $1\leq i \leq \ell$,
%$$lim_{k\to \infty}\frac{|x_{\ell+1,k}-x_{i,k}|}{r_{i,k}}= \infty.$$
%Now we prove the following lemma 
%\begin{lemma}
%Under the previous assumptions, we have
%\begin{equation}\label{ind2}
%\lim_{k\to \infty} \frac{dist(x_{\ell+1,k},\partial \Omega)}{r_{\ell+1,k}}=+\infty
%\end{equation}
%and
%\begin{equation} \label{ind3}
 %\lim_{k\to \infty} u_{k}(r_{\ell+1,k}x+x_{\ell+1,k})-u_{k}(x_{\ell+1,k})=0 \text{ in $C^{2}_{loc}(\mathbb{R}^{3})$ as } k\to \infty. 
%\end{equation}
%\end{lemma}
%\begin{proof}
%In this proof, we will set $y_{k}=x_{\ell+1,k}$ and $\varrho_{k}=r_{\ell+1,k}$. Now notice that from (\ref{ind}), if we test with the variable $y_{k}+\varrho_{k}x$, have that 
%$$\inf_{1\leq i\leq \ell}\left(|y_{k}-x_{i,k}+\varrho_{k}x|^{3}\right)u_{k}^{2}(y_{k}+\varrho_{k} x)e^{\frac{3}{2}u_{k}^{2}(y_{k}+\varrho_{k} x)}\leq \inf_{1\leq i \leq \ell}|y_{k}-x_{i,k}|^{3}u_{k}^{2}(y_{k})e^{\frac{3}{2}u_{k}^{2}(y_{k})}.$$

%Therefore, if we set $v_{\ell+1,k}=\frac{u_{k}(y_{k}+\varrho_{k} x)}{u_{k}(y_{k})}$ we have that
%$$v_{\ell+1,k}^{2}(x)e^{\frac{3}{2}u_{k}^{2}(y_{k})(v_{\ell+1,k}^{2}-1)}\leq \frac{\inf_{1\leq i \leq \ell}|y_{k}-x_{i,k}|^{3}}{\inf_{1\leq i\leq \ell}\left(|y_{k}-x_{i,k}+\varrho_{k}x|^{3}\right)}=1+o(1),$$
%where $o(1)\to 0$ as $k\to \infty$ locally uniformly. Therefore, we have that $v_{\ell+1,k}$ satisfies
%$$(-\Delta)^{\frac{3}{2}}v_{\ell+1,k}=\frac{16}{u_{k}^{2}(y_{k})}v_{\ell+1,k}^{2}e^{\frac{3}{2}u_{k}^{2}(y_{k})(v_{\ell+1,k}^{2}-1)}=o(1)$$
%Then using the same argument as in lemma (), one has the proof of (\ref{ind2}).\\
%Now we define the function $w_{\ell+1,k}(x)=u_{k}(y_{k}+\varrho_{k}x)-u_{k}(y_{k})$. Notice that $u_{k}(y_{k}+\varrho_{k}x)\to +\infty$ locally uniformly in $\mathbb{R}^{n}$.  

%\end{proof}



\section{Borderline and commutator estimates}\label{sec:borderline}
We have the following two borderline estimates:
\begin{proposition}\label{borderline1}
Let $g \in L^2(\R^n)$, $f \in L\log^{1/2}L(\R^n)$, $s \in (0,n)$. Then 
\[
\| \lapms{n-s} (f\, \lapms{\frac{n}{2}}g)\|_{(\frac{n}{s},\infty)} \aleq  \vrac{g}_{L^2} \vrac{f}_{L\log^{1/2}L} 
\]
\end{proposition}
\begin{proof}
By Fubini's theorem and Proposition~\ref{pr:orliczsobolev},
\[\begin{split}
\|f\lapms{\frac{n}{2}} g\|_{L^1}&\aleq  \int_{\R^n}\int_{\R^n}\frac{|g(z)|}{|z-y|^{n/2}}dz\ |f(y)|dy\\
&= \int_{\R^n}\int_{\R^n} \frac{|f(y)|}{|z-y|^{n/2}}dy\ |g(z)| dz\\
&=\||g|\ I_\frac{n}{2}|f|\|_{L^1}\\
&\le \|g\|_{L^2}\|I_\frac n2|f|\|_{L^2}\\
&\aleq\|g\|_{L^2}\|f\|_{L(\log^{1/2} L)},
\end{split}
\]
so that by Proposition~\ref{pr:Sobolev},
$$\|I_{n-s}(f\ I_{\frac n2}(g))\|_{(\frac ns,\infty)}\aleq \|f\ \lapms{\frac{n}{2}} g\|_{L^1}\aleq  \|g\|_{L^2}\|f\|_{L(\log^{1/2} L)}.$$
\end{proof}


\begin{proposition}\label{borderline2}
Let $g \in L^2(\R^n)$, $f \in L\log^{1/2}L(\R^n)$, $s \in (0,n)$. Then 
\[
\| \lapms{\frac{n}{2}} g\, \lapms{n-s}f\|_{(\frac{n}{s},\infty)} \aleq  \vrac{g}_{L^2} \vrac{f}_{L\log^{1/2}L} 
\]
\end{proposition}

\begin{proof} We write
$$\lapms{\frac{n}{2}} g\, \lapms{n-s}f(x)=\int_{\R^n}\int_{\R^n} k(x,y,z) g(z) f(y) dz dy,\quad k(x,y,z):=\frac{1}{|x-z|^\frac n2 |x-y|^s}.$$
For $\eps\in (0,s)$ we can now bound (cf. \cite{Sfracenergy})
$$k(x,y,z) \le |x-z|^{-\frac n2-\eps}|x-y|^{\eps-s}+ C|y-z|^{-\frac n2}|x-y|^{-s}=:II+III.$$
Indeed if $|y-z|\ge 2|x-z|$, then $|x-z|\le |x-y|$ (by the triangular inequality), hence $I\le II$. If $|y-z|\le 2|x-z|$, then $I\le C\,III.$
Therefore we have
$$|\lapms{\frac{n}{2}} g\, \lapms{n-s}f|\le I_{\frac n2-\eps} (|g|)\, I_{n-s+\eps}(|f|)+ CI_{n-s}(|f|\,I_{\frac n2}(|g|)).$$
The first term on the right-hand side can be bounded as
\[
\|  \lapms{\frac{n}{2}-\eps} g\, \lapms{n+\eps - s} f
\|_{(\frac{n}{s},\infty)} \aleq \|\lapms{\frac{n}{2}-\eps} (|g|)\|_{L^\frac{n}{\eps}}\, \|\lapms{n+\eps - s} (|f|)\|_{(\frac{n}{s-\eps},\infty)}
 \aleq \|g\|_{L^2}\,\|f\|_{L^1},
 \]
while the second term can be bounded by Proposition~\ref{borderline1}.
\end{proof}

% \armin{\begin{remark} Note that Proposition~\ref{borderline1} and Proposition~\ref{borderline2} imply
% \[
% \vrac{\lapms{\frac{n}{2}}bf)-b\lapms{\alpha} f}_{(\frac{np}{n-\alpha p},q)}
% 
% \| \lapms{\frac{n}{2}} g\, \lapms{n-s}f\|_{(\frac{n}{s},\infty)} \aleq \|I_{n-s}(|f|\, \lapms{\frac{n}{2}} (|g|))\|_{(\frac{n}{s},\infty)} \aleq \vrac{g}_{L^2} \vrac{f}_{L\log^{1/2}L}. 
% \]
% Chanillo's commutator estimates on the Riesz potential \cite{cha}
% \[
% \vrac{\lapms{\alpha}(bf)-b\lapms{\alpha} f}_{(\frac{np}{n-\alpha p},q)} \aleq [b]_{BMO}\ \vrac{f}_{(p,q)}
% \]
% cannot be extended to $(p,q) = (1,1)$, see \cite{DLZ}. What does hold (cf. \cite{YC}) is
% \[
% \vrac{\lapms{\alpha}(bf)-b\lapms{\alpha} f}_{(\frac{n}{n-\alpha},\infty)} \aleq [b]_{BMO}\ \vrac{f}_{L \log L}.
% \]
% This is not enough for our case, but as observed in \cite{Seps}, allowing instead of $[b]_{BMO}$ a term $\vrac{\laps{s} b}_{\frac{n}{s}}$ on the right-hand side allows for $f$ to be controlled on a larger space.  For $L^1$-Orlicz space $L \log^{\frac{1}{2}} L$ the arguments in \cite{Sfracenergy,Seps} can be extended with the help of \cite[Corollary 6.16]{BS}, see Proposition~\ref{pr:orliczsobolev}.
% \end{remark}
% }
% 
% To cite Armin'
% \begin{theorem}\label{la:commutator}
% Let $g \in L^2(\R^n)$, $f \in L\log^{1/2}L(\R^n)$, $s \in (0,n)$
% \[
% \| \lapms{n-s}(\lapms{\frac{n}{2}} g\, f) - \lapms{\frac{n}{2}} g\, \lapms{n-s}f\|_{(\frac{n}{s},\infty),\R^n} \aleq \vrac{g}_{2,\R^n}\ \vrac{f}_{L\log^{1/2}L} 
% \]
% In particular,
% \[
% \vrac{\lapms{n-s}(bf)-b\lapms{n-s} f}_{(\frac{n}{s},\infty),\R^n} \aleq \vrac{(-\lap)^{n/4}b}_{L^2}\ \vrac{f}_{L \log^{1/2} L}.
% \]
% \end{theorem}
% \begin{proof}
% Without loss of generality we can assume that $f$ and $g$ are non-negative, since otherwise one can write $g = g_+ - g_-$ and $f = f_+ + f_-$, and consider each of the four resulting commutators. 
% % \luca{[Really? I am not sure. The difference on the left-hand side might get small when replacing $f$ and $g$ with $|f|$ and $|g|$. On the other hand, that does not matter, considering the comments below]}. 
% Let
% \[
%  k(x,y,z) := \luca{-}\brac{ |x-z|^{-\frac{n}{2}}  - |y-z|^{-\frac{n}{2}}} |x-y|^{-s},
% \]
% then
% \[
%   \brac{\lapms{n-s}(\lapms{\frac{n}{2}} g\, f) - \lapms{\frac{n}{2}} g\, \lapms{n-s}f}(x) = \int_{\R^n}\int_{\R^n} k(x,y,z)\, g(z)\, f(y)\, dz\, dy
% \]
% then for $\eps \in (0,s)$,
% \[
% \begin{split}
%  k(x,y,z) \leq 
%  & |x-z|^{-\frac{n}{2}-\eps}  \ |x-y|^{\eps  -s}
%  + \chi_{|y-z| \ll |x-y|} \chi_{|y-z| \ll |x-z|}  |y-z|^{-\frac{n}{2}}  \ |x-y|^{-s}\\
% \aleq & |x-z|^{-\frac{n}{2}-\eps}  \ |x-y|^{\eps  -s}
%  + |y-z|^{-\frac{n}{2}}  \ |x-z|^{-s}
% \end{split}
%  \]
% 
% \armin{But how can you bound the III after integrating?, i.e. how do you get 
% \[
%  \| \lapms{n-s}(\lapms{\frac{n}{2}} g\, f)\|_{(\frac{n}{s},\infty),\R^n} \aleq \vrac{g}_{2,\R^n}\ \vrac{f}_{L\log^{1/2}L} 
% \]
% ? We have $\lapms{\frac{n}{2}} g \in BMO$ and $f \in L\log^{\frac{1}{2}} L$, so I guess in general $\lapms{\frac{n}{2}} g\, f \not \in L^1$.
 
 
%If instead of $III$ you have $|y-z|^{\frac{n}{2}-n}  \ |x-z|^{-s}$, i.e. $y \leftrightarrow z$ switched, and integrating you get the situation where $f$ and $g$ are switched:
%\[
%  \| \lapms{n-s}(\lapms{\frac{n}{2}} f\, g)\|_{(\frac{n}{s},\infty),\R^n}. 
% \]
%Now $\lapms{\frac{n}{2}} f \in L^2$ and $g \in L^2$
% }
%  
%  Thus, 
% \[
% \begin{split}
% & \int_{\R^n} \int_{\R^n} \brac{ |x-z|^{\frac{n}{2}-n}  - |y-z|^{\frac{n}{2}-n}}\, |x-y|^{s-n}\, g(z)\, f(y)\, dz\, dy\\
% \aleq & \lapms{\frac{n}{2}-\eps} g(x)\, \lapms{n+\eps - s} f(x) + \lapms{n-s}(g\lapms{\frac{n}{2}}f)(x).
% \end{split}
% \]
% The first term is fine, by H\"older's inequality and boundedness of Riesz potentials
% \[
% \|  \lapms{\frac{n}{2}-\eps} g\, \lapms{n+\eps - s} f
% \|_{(\frac{n}{s},\infty),\R^n} \aleq \|\lapms{\frac{n}{2}-\eps} g\|_{\frac{n}{\eps}}\, \|\lapms{n+\eps - s} f\|_{(\frac{n}{s-\eps},\infty)}
%  \aleq \|g\|_{2}\,\|f\|_{1}.
%  \]
% 
% 
% Now note that Proposition~\ref{pr:orliczsobolev}
% \[
%  \vrac{\lapms{\frac{n}{2}} f}_{2} \aleq \vrac{f}_{L\log^{1/2} L},
% \]
% and thus
% \[
%  \|g\lapms{\frac{n}{2}} f\|_{1} \aleq \vrac{f}_{L\log^{1/2}L}\ \vrac{g}_{2}
% \]
% and consequently, by the boundedness of $\lapms{n-s} : L^{1}(\R^n) \to L^{(\luca{\frac{n}{s}},\infty)}(\R^n)$,
% \[
%   \vrac{\lapms{n-s}(g\luca{\lapms{\frac n2} } f)}_{(\luca{\frac{n}{s}},\infty)} \aleq \vrac{f}_{L\log^{1/2}L}\ \vrac{g}_{2}.
% \]
% 
% \end{proof}
% % % % % % % \armin{
% % % % % % % Proposition~\ref{borderline1} and \ref{borderline2} in particular imply that
% % % % % % % \[
% % % % % % % \| \lapms{n-s}(\lapms{\frac{n}{p}} g\, f) - \lapms{\frac{n}{p}} g\, \lapms{n-s}f\|_{L^{(\frac{n}{s},\infty)}} \aleq \vrac{g}_{L^{2}}\ \vrac{f}_{L\log^{1/2}L}
% % % % % % % \]
% % % % % % % \armin{For no reason whatsoever}, we shall remark this commutator estimate actually can be improved with the techniques from \cite{Sfracenergy,Seps}.
% % % % % % % }
% % % % % % % \begin{theorem}\label{la:commutator2}
% % % % % % % Let $g \in L^{(p,q)}(\R^n)$, $f \in L\log^{1/r} L(\R^n)$, $s \in (0,n)$
% % % % % % % \[
% % % % % % % \| \lapms{n-s}(\lapms{\frac{n}{p}} g\, f) - \lapms{\frac{n}{p}} g\, \lapms{n-s}f\|_{L^{(\frac{n}{s},\tau)}} \aleq \vrac{g}_{L^{(p,q)}}\ \vrac{f}_{L\log^{1/r}L},\quad  \frac{1}{\tau}=\frac{1}{q}+\frac{1}{r}.
% % % % % % % \]
% % % % % % % \end{theorem}
% % % % % % % \begin{proof}
% % % % % % % As before we assume $f,g\ge 0$ and we write
% % % % % % % $$\lapms{n-s}(\lapms{\frac{n}{p}} g\, f)(x) - \lapms{\frac{n}{p}} g\, \lapms{n-s}f(x)=\int_{\R^n}\int_{\R^n}k(x,y,z)f(y)g(z)dy dz,$$
% % % % % % % with
% % % % % % % $$k(x,y,z)=-\left(\frac{1}{|x-z|^{n-\frac np}} - \frac{1}{|y-z|^{n-\frac np}} \right)\frac{1}{|x-y|^s}.$$
% % % % % % % By Lemma \ref{xyz} for $\eps>0$ sufficiently small we have
% % % % % % % $$|k(x,y,z)|\le C\left(\frac{1}{|x-z|^{n-\frac np+\eps}}\frac{1}{|x-y|^{s-\eps}}+\frac{1}{|y-z|^{n-\frac np+\eps}}\frac{1}{|x-z|^{s-\eps}}\right),$$
% % % % % % % hence
% % % % % % % $$|\lapms{n-s}(\lapms{\frac{n}{p}} g\, f) - \lapms{\frac{n}{p}} g\, \lapms{n-s}f|\aleq I_{\frac np-\eps}g\,I_{n-s+\eps}f + I_{n-s+\eps}(g\,I_{\frac np-\eps}f).$$
% % % % % % % For the first term on the right-hand side we have $I_{\frac np-\eps}g\in L^{(\frac n \eps,q)}(\R^n)$ and $I_{n-s+\eps}f\in L^{(\frac{n}{s-\eps},r)}(\R^n)$, and H\"older's inequality (for Lorentz spaces) and Proposition~\ref{pr:orliczsobolev} give
% % % % % % % \[
% % % % % % % \begin{split}
% % % % % % % \|I_{\frac np-\eps}g\,I_{n-s+\eps}f\|_{L^{(\frac{n}{s},\tau)}}&\aleq
% % % % % % % \|I_{\frac np-\eps}g \|_{L^{(\frac n \eps,q)}} \|I_{n-s+\eps}f\|_{L^{(\frac{n}{s-\eps},r)}}\\
% % % % % % % &\aleq \|g\|_{L^{(p,q)}}\|f\|_{L\log^{1/r}L}.
% % % % % % % \end{split}
% % % % % % % \]
% % % % % % % As for the second term we get
% % % % % % % \[\begin{split}
% % % % % % % \|I_{n-s+\eps}(g\,I_{\frac np-\eps}f) \|_{L^{(\frac ns,\tau)}}
% % % % % % % &\aleq \|g\,I_{\frac np-\eps}f\|_{L^{(\frac{n}{n+\eps}, \tau)}}\\
% % % % % % % &\aleq \|g\|_{L^{(p,q)}}\|I_{\frac np-\eps}f\|_{L^{(\frac{np}{np-n+\eps p},r)}}\\
% % % % % % % &\aleq \|g\|_{L^{(p,q)}}\|f\|_{L\log^{1/r}L}.
% % % % % % % \end{split}\]
% % % % % % % \end{proof}
% % % % % % % 
% % % % % % % 
% % % % % % % \armin{We could just cite this Lemma?, e.g. look my Transactions-Paper with Simon/Reiter Lemma A.4.}
% % % % % % % \luca{
% % % % % % % \begin{lemma}\label{xyz} Given $a,s>0$ and $\eps\in (0,1]$ with $\eps<\min\{a,s\}$, and $x,y,z\in \R^n$ distinct, we have
% % % % % % % \begin{equation}\label{stimaxyz}
% % % % % % % \left|\frac{1}{|x-z|^a}-\frac{1}{|y-z|^a}\right|\frac{1}{|x-y|^s}\le C \left(\frac{1}{|x-z|^{a+\eps}}\frac{1}{|x-y|^{s-\eps}}+\frac{1}{|y-z|^{a+\eps}}\frac{1}{|x-z|^{s-\eps}}\right),
% % % % % % % \end{equation}
% % % % % % % for a constant $C$ depending on $n$ and $s$.
% % % % % % % \end{lemma}
% % % % % % % \begin{proof}
% % % % % % % Up to a translation it suffices to consider the case $z=0$, and multiplying by $|x-y|^s$ we reduce to proving
% % % % % % % $$|I-II|:=\left|\frac{1}{|x|^a}-\frac{1}{|y|^a}\right|\aleq  \left(\frac{|x-y|^\eps}{|x|^{a+\eps}} +\frac{|x-y|^s}{|y|^{a+\eps}|x|^{s-\eps}}\right)=:III+IV.$$
% % % % % % % This follows easily by considering 3 cases.
% % % % % % % 
% % % % % % % Case 1: $|y|\ge 2|x|$. Then $|x|\le |x-y|$, hence $I\le III$ and $II< I$, whence $|I-II|\le I\le III$.
% % % % % % % 
% % % % % % % Case 2: $|x|\ge 2|y|$. Now we have $|y|\le |x-y|$ and $|x|\le 2|x-y|$, so that
% % % % % % % $$|I-II|\le II=\frac{1}{|y|^a}\le \frac{|x-y|^\eps}{|y|^{a+\eps}}\le 2^{s-\eps}\frac{|x-y|^s}{|y|^{a+\eps}|x|^{s-\eps}}\le 2^s IV.$$
% % % % % % % 
% % % % % % % Case 3: $\frac{y}{2}\le |x|\le 2|y|$. Then $|x-y|\le 3\min\{|x|,|y|\}$. Integrating the derivative of the function $f(x)=|x|^{-a}$ along a suitable path joining $x$ and $y$ contained in the annulus $A:=B_{2|y|}(0)\setminus B_{|y|}(0)$ and of length comparable to $|x-y|$ one gets
% % % % % % % $$|I-II|\aleq |x-y|\sup_{A}|\nabla f|\aleq \frac{|x-y|}{|x|^{a+1}}\aleq\frac{|x-y|^\eps}{|x|^{a+\eps}}\le III.$$
% % % % % % % \end{proof}
% % % % % % % }
% % % % % % % 


\subsection{Disjoint-support estimates}

When $\operatorname{supp} \varphi \subset K$ for a compact set $K$ then in general we have no information on the support of $\laps{s} \varphi$, since $\laps{s}$ is a non-local operator. In particular $\laps{s} \varphi(x) \neq 0$ also for $x$ far away from $K$. However, there is a decay of $|\laps{s} \varphi(x)|$ as $\dist(x,K) \to \infty$. We shall call this pseudo-local behavior of $\laps{s}$.
It has been used allover the literature, for statements in the following form see \cite{BRS}.

\begin{definition}[Cut-off functions]\label{def:cutoff} With $\theta_{B_1}$ we will denote a fixed smooth function $\theta_{B_1}\in C^\infty_c(B_2)$ with $\theta_{B_1}\equiv 1$ in $B_1$ and $0\le \theta_{B_1}\le 1$ everywhere. Define
$$\theta_{B_\rho}(x):= \theta_{B_1}(x/\rho)\in C^\infty_c(B_{2\rho}),\quad \theta_{A_\rho}:=\theta_{B_\rho}-\theta_{B_\frac\rho2}\in C^\infty_c(B_{2\rho}\setminus B_{\frac \rho2}).$$
%$\theta_{A_1}:=\theta_{B_1}-\theta_{B_\frac12}$.
\end{definition}

In the proof of Proposition \ref{la:ulapuest} we used the following ``disjoint-support'' commutator estimate. 
Compared to the usual commutator estimates \cite{CRW,Seps} the estimates here are simpler, due to the disjoint support. Note that going through the proof, one may obtain a BMO-estimate, which is false for the commutator without disjoint support, see \cite{DLZ}.

\begin{proposition}\label{la:disjcommutator}
Define the commutator $[u,\lapms{t}](v) := u\lapms{t} v - \lapms{t} (uv)$.
Then for any $u, v \in C_c^\infty(\R^n)$,
\[
\vrac{ [u,\lapms{n-s}]((1-\theta_{B_{2\rho}})(-\Delta)^\frac n4 f)}_{L^{(\frac {n}{s},\infty)}(B_\rho)} \aleq \| (-\lap)^{\frac{n}{4}} u \|_{L^2(\R^n)}  \| f\|_{L^2(\R^n)}.
\]
\end{proposition}
\begin{proof}
By scaling we can assume that $\rho = 1$. With $\theta_\ell:=\theta_{A_{2^\ell}}\in C^\infty_c (B_{2^{\ell+1}} \backslash B_{2^{\ell-1}})$ as in Definition \ref{def:cutoff} we have pointwise in $\R^n$
\[
 1-\theta_{B_2} = \sum_{l=2}^\infty \theta_{\ell}.
\]
Moreover for $t \in (0,\infty)$ and $p \in [1,\infty]$ we have
\[
 \| \laps{t} \theta_{\ell} \|_{L^p(\R^n)} \leq C_{t,p}\ 2^{\ell (\frac{n}{p}-t)}.
\]
Since $u, v \in C_c^\infty(\R^n)$, we have then
\begin{equation}\label{eqcom}
 |[u,\lapms{n-s}]((1-\theta_{B_2})(-\Delta)^\frac n4 f)| \leq \sum_{\ell=2}^\infty |u\lapms{n-s}(\theta_{\ell}(-\Delta)^\frac n4 f)-\lapms{n-s}(u \theta_{\ell}(-\Delta)^\frac n4 f)|.
\end{equation}
Now set
\[
 u_{\ell} := \theta_{B_{2^{\ell+2}}} (u-(u)_{B_{2^{\ell+2}}}).
\]
Since $\theta_{B_{2^{\ell+2}}} \equiv 1$ in $B_{2^{\ell+2}}\supset B_1 \cup\operatorname{supp} \theta_{\ell}$, and the constant $(u)_{B_{2^{\ell+2}}}$ commutes with $\lapms{n-s}$, multiplying each term on the right-hand side of \eqref{eqcom} by $\theta_{B_{2^{\ell+2}}}$ and summing and subtracting the term
$$\theta_{B_{2^{\ell+1}}}(u)_{B_{2^{\ell+1}}} I_{n-s}(\theta_\ell (-\Delta)^\frac n4 f),$$ we find
\[\begin{split}
 |[u,\lapms{n-s}]((1-\theta_{B_2})(-\Delta)^\frac n4 f)|& \leq \sum_{\ell=2}^\infty |u_\ell\lapms{n-s}(\theta_{\ell}(-\Delta)^\frac n4 f)| + \sum_{\ell=2}^\infty |\lapms{n-s}(u_\ell \theta_{\ell}(-\Delta)^\frac n4 f)| \\
 &=: \sum_{\ell=2}^\infty ((I)_{\ell} + (II)_{\ell})\quad \text{in }B_1.
 \end{split}
\]
Now, by H\"older inequality and Lemma~\ref{la:freakshowestimate1}
\[
\begin{split}
 \vrac{u_\ell\lapms{n-s}(\theta_{\ell}(-\Delta)^\frac n4 f)}_{L^{(\frac{n}{s},\infty)}(B_1)} & \aleq \vrac{u_\ell}_{L^\frac{n}{s}(B_1)} \|\lapms{n-s}(\theta_{\ell}(-\Delta)^\frac n4 f)\|_{L^\infty(B_1)} \\
 &\aleq \vrac{u_\ell}_{L^\frac{n}{s}(B_1)}\ 2^{-\ell s} \vrac{f}_{L^2(\R^n)}\\
 \end{split}
\]
Note that for any $p < \infty$,
\[
 \vrac{u_\ell}_{p,\R^n} \aleq C_p 2^{\ell\frac{n}{p}} [u]_{BMO} \aleq 2^{\ell \frac{n}{p}} \vrac{(-\lap)^\frac n4 u}_{2,\R^n}.
\]
Taking $p>\frac{n}{s}$ and $\delta=\frac {n}{p}$
\[
 \vrac{u_\ell}_{\frac{n}{s},B_1}\aleq \vrac{u_\ell}_{p,B_1}\aleq 2^{\ell \delta}\ \vrac{(-\lap)^\frac n4 u}_{2,\R^n}.
\]
Together, we arrive at
\[
 \vrac{u_\ell\lapms{n-s}(\theta_{\ell}(-\Delta)^\frac n4 f)}_{(\frac{n}{s},\infty),B_1}
%  \aleq  2^{\ell \delta}\ 2^{\ell ((n-s)-n)}\ \vrac{(-\lap)^{n/4} u}_{2,\R^n}\ ^2 = 
\aleq 2^{\ell (\delta-s)}\ \vrac{(-\lap)^\frac n4 u}_{2,\R^n}\ \vrac{f}_{2,\R^n},
\]
and for $\delta < s$, this ensures
\begin{equation}\label{eq:sumII1l}
 \sum_{\ell =2}^\infty (I)_{\ell} \aleq \vrac{(-\lap)^\frac n4 u}_{2,\R^n}\ \vrac{f}_{2,\R^n}.
\end{equation}
It remains to treat $(II)_{\ell}$, and we do that with Lemma~\ref{la:freakshowestimate2}:
\begin{align*}
 \vrac{\lapms{n-s}(u_\ell\ \theta_{\ell}(-\Delta)^\frac n4 f)}_{(\frac{n}{s},\infty),B_1}
 \aleq &\vrac{\lapms{n-s}(u_\ell\ \theta_{\ell}(-\Delta)^\frac n4 f)}_{\frac{n}{s},B_1}\\
 \aleq &\max_{t \in [0,\frac{n}{2}]} \vrac{\laps{t}u_\ell}_2\ (2^\ell )^{t-\frac{n}{2} -s}\ \vrac{f}_{2,\R^n}%\\
% \aleq &\max_{t \in [0,\frac{n}{2}]} \vrac{\laps{t}u_\ell}_2\ (2^\ell )^{t-\frac{n}{2} -s}\ \vrac{(-\lap)^\frac n4 u}_{2,\R^n}. 
\end{align*}
Now for any $t \in [0,\frac{n}{2}]$, by the construction of $u_\ell$ and Poincar\'e and Sobolev-embeddings,
\[
 \vrac{\laps{t}u_\ell}_2 \aleq (2^\ell )^{\frac{n}{2}-t} \vrac{(-\lap)^\frac n4 u}_{L^2(\R^n)}.
\]
This leads to
\[
 \vrac{\lapms{n-s}(u_\ell \theta_{\ell}(-\Delta)^\frac n4 f)}_{(\frac{n}{2},\infty),B_1}
 %\aleq&\max_{t \in [0,\frac{n}{2}]} (2^\ell )^{\frac{n}{2}-t}\ (2^\ell )^{t-\frac{n}{2} -s}\ \vrac{(-\lap)^\frac n4 u}^2_{2,\R^n}\\
 \aleq 2^{-s\ell}\ \vrac{(-\lap)^\frac n4 u}_{2,\R^n}\ \vrac{f}_{2,\R^n}. 
\]
Again, this ensures
\begin{equation}\label{eq:estsumII2}
 \sum_{\ell =2}^\infty (II)_{\ell} \aleq \vrac{(-\lap)^\frac n4 u}_{2,\R^n}\ \vrac{f}_{2,\R^n}.
\end{equation}
\end{proof}





\begin{lemma}\label{la:nonloc}
Let $\varphi \in C_c^\infty(K)$ for some compact set $K$ and let $\Omega \subset \R^n$ be an open set containing $K$ with $\dist(\partial\Omega,K) \ge d$ for some $d > 0$. Then for any $p \in [1,\infty]$ and $s \in (0,\infty)$ we have
\[
 \vrac{\laps{s} \varphi}_{L^p(\R^n \backslash \Omega)} \le C_{n,s,p}d^{n-(n+s)p} \vrac{\varphi}_{L^1(K)},
\]
and for any $s \in (0,n)$ and $p > \frac{n}{n-s}$ we have
\[
 \vrac{\lapms{s} \varphi}_{L^p(\R^n\setminus\Omega)} \le C_{n,s,p} d^{-(\frac{n}{p'}-s)}\ \vrac{\varphi}_{L^1(K)}.
\]
\end{lemma}

\begin{proof}
Since convolution and multiplication are transformed into each other under Fourier transform and $(|\cdot|^s)^\wedge = c |\cdot|^{-s-n}$, for $x$ away from the support of $\varphi$ we have
$$\laps{s} \varphi(x) = c_{n,s}|\cdot|^{-n-s}\ast \varphi (x).$$
In particular
\[
 |\chi_{\R^n \backslash \Omega} \laps{s} \varphi| \leq \big(|\cdot|^{-n-s}\chi_{|\cdot| \geq \frac{d}{2}} \big)\ast \varphi.
\]
Now the first claim follows by Young's inequality:
\[
  \vrac{\chi_{\R^n \backslash \Omega} \laps{s} \varphi}_{L^p(\R^n)} \aleq \vrac{|\cdot|^{-n-s}\chi_{|\cdot| \geq \frac{d}{2}}}_{L^p(\R^n)}\ \vrac{\varphi}_{L^1(\R^n)}.
\]
The proof of the second claim is very similar.
\end{proof}



\begin{lemma}\label{disjointoperator}
Consider two functions $\theta_1,\theta_2 \in C_c^\infty(\R^n)$. Suppose that $\theta_1$ and $\theta_2$ have disjoint support, i.e. for some $d > 0$,
\begin{equation}\label{eq:do:disjointK}
 \dist (\operatorname{supp} \theta_1, \operatorname{supp} \theta_2 ) \ge d.
\end{equation}
For $s \in (0,n)$ let the operator $T$ be defined via 
\[
 Tf := \theta_1\, \lapms{s} (\theta_2 f), \quad f \in \Sw(\R^n).
\]
Then for any $t > 0$ the operator $ T \laps{t}$, originally defined on $\mathcal{S}(\R^n)$, extends to a linear bounded operator from $L^p(\R^n)$ to $L^q(\R^n)$ whenever $1+\frac{1}{q}- \frac{1}{p} \in [0,1]$,  with the estimate
\[
 \|T \laps{t}f \|_{L^q(\R^n)} \le C_{\theta_1,\theta_2,p,q,t} \|f\|_{L^p(\R^n)}
\]
\end{lemma}
\begin{proof}
First set
\[
 k(x,y) := |x-y|^{s-n}\, \theta_1(x)\, \theta_2(y).
\]
Notice that based on $\theta_1$ and $\theta_2$ and the disjoint support of the two functions \eqref{eq:do:disjointK} we can find $\theta_3 \in C^\infty_c(\R^n)$, $\theta_3 \equiv 0$ in the ball $B_{d/2}$ so that
\[
 k(x,y) = \theta_3(x-y) |x-y|^{s-n}\ \theta_1(x)\ \theta_2(y)\ 
\]
Note that 
\[
 \theta_3(\cdot) |\cdot|^{s-n} \in C^\infty(\R^n)
\]
In particular, $k(\cdot,y) \in C_c^\infty(\R^n)$ for any $y \in \R^n$ and $k(x,\cdot) \in C_c^\infty(\R^n)$ for any $x \in \R^n$. Morever, $\laps{t}_x k(x,\cdot) \in C_c^\infty(\R^n)$ for any $x \in \R^n$. Then for $f \in \Sw(\R^n)$
\[
\begin{split}
 T \laps{t} f(x) =&  \int_{\R^n} k(x,y) \laps{t}_y f(y)\, dy\\
=&\int_{\R^n} \laps{t}_y k(x,y)\,  f(y)\, dy,
\end{split}
\]
where we integrated by parts.

Setting
\[
 \tilde{k}(x,y) := \laps{t}_y k(x,y),
\]
and using that $(-\Delta)^\frac t2 \varphi(y)$ decays like $|y|^{-n-t}$ at infinity if $\varphi$ is compactly supported, we bound
$$\sup_{x \in \R^n} \|\tilde{k}(x,\cdot)\|_{L^{r}(\R^n)}<\infty,\quad \sup_{y \in \R^n}\|\tilde{k}(\cdot,y)\|_{L^{r}(\R^n)}<\infty$$
for every $r\in [1,\infty]$.
Then, by a straightforward adaption of Young's convolution inequality, if $1+\frac{1}{q} = \frac{1}{p} + \frac{1}{r}$ we get
\[\begin{split}
 \vrac{T \laps{t} f}_{L^q(\R^n)} &\leq \left(\sup_{x \in \R^n} \|\tilde{k}(x,\cdot)\|_{L^{r}(\R^n)}+\sup_{y \in \R^n}\|\tilde{k}(\cdot,y)\|_{L^{r}(\R^n)}\right)\, \vrac{f}_{L^p(\R^n)}\\
 & = C_{\theta_1,\theta_2,t,p,q}\, \vrac{f}_{L^p(\R^n)}. 
\end{split}
\]
\end{proof}


In some special cases we need to compute the constant in the Lemma above.
\begin{lemma}\label{la:freakshowestimate1}
For any $p \in [1,\infty]$, $q \in [1,\infty)$, any $\rho > 0$, $k \geq 2$, $s \in (0,n)$, we have the following estimate for any $f \in \Sw(\R^n)$,
\begin{equation}\label{eq:freakshowestimate1est1}
 \vrac{\lapms{s} (\theta_{A_{2^k\rho}} \laps{t} f)}_{L^p(B_\rho)} \aleq (2^k \rho)^{s-t-\frac{n}{q}}\ \rho^{\frac{n}{p}}\, \vrac{f}_{L^q(\R^n)}.
\end{equation}
Similarly, for any $g \in C_c^\infty(B_\rho)$,
\[
 \vrac{\laps{t} (\theta_{A_{2^k\rho}} \lapms{s} g)}_{L^{q'}(\R^n)}  \aleq (2^k \rho)^{s-t-\frac{n}{q}}\ \rho^{\frac{n}{p}} \vrac{g}_{L^{p'}(B_\rho)}.
\]

\end{lemma}
\begin{proof}
The second estimate follows from the first one by duality. Indeed
\[
 \begin{split}
\vrac{\laps{t} (\theta_{A_{2^k\rho}} \lapms{s} g)}_{L^{q'}(\R^n)}
& \aleq \sup_{f\in \Sw(\R^n),\|f\|_{L^q(\R^n)\le 1}}  \int_{\R^n} f\, \laps{t} (\theta_{A_{2^k\rho}} \lapms{s} g)\, dx \\
& =  \sup_{f\in \Sw(\R^n),\|f\|_{L^q(\R^n)}\le 1} \int_{\R^n} \lapms{s}(\theta_{A_{2^k\rho}} \laps{t}f)\, g\, dx\\
&\aleq  (2^k \rho)^{s-t-\frac{n}{q}}\ \rho^{\frac{n}{p}}\ \vrac{g}_{L^{p'}(\R^n)},
 \end{split}
\]
where we used integration by parts twice, cf. \eqref{eq:fracPI}, and \eqref{eq:freakshowestimate1est1}.
%, we thus have
%\[
%\begin{split}
%&\vrac{\laps{t} (\theta_{A_{2^k\rho}} \lapms{s} g)}_{q',\R^n} \aleq \|g\|_{p'}\ \vrac{\lapms{s} (\theta_{A_{2^k\rho}} \laps{t} f)}_{p,B_\rho} \\
%\overset{\eqref{eq:freakshowestimate1est1}}{\aleq} &\|g\|_{p'}\ (2^k \rho)^{s-t-\frac{n}{q}}\ \rho^{\frac{n}{p}}\ \|f\|_{q} \leq (2^k \rho)^{s-t-\frac{n}{q}}\ \rho^{\frac{n}{p}} \|g\|_{p'}
%\end{split}
%\]

The estimate \eqref{eq:freakshowestimate1est1} for $p\in [1,\infty)$ follows via H\"older's inequality from the case $p=\infty$ which we shall now prove. Up to scaling we can take $\rho=1$ and then \eqref{eq:freakshowestimate1est1} reduces to 
\begin{equation}\label{eq:freakshowestimate1est1b}
\vrac{\lapms{s} (\theta_{A_{2^k}} \laps{t} f)}_{L^\infty(B_1)} \aleq (2^k)^{s-t-\frac{n}{q}} \vrac{f}_{L^q(\R^n)}.
\end{equation}
For $k=2$ \eqref{eq:freakshowestimate1est1b} follows from Lemma~\ref{disjointoperator}:
\[
\vrac{\lapms{s} (\theta_{A_4} \laps{t} f)}_{L^\infty(B_1)} \le C_1 \vrac{f}_{L^q(\R^n)},
\]
with $C_1$ depending on $s,t,n,q$ and the chosen cut-off function $\theta_{B_1}$ (fixed in Definition \ref{def:cutoff}).
The case $k> 2$ follows from the case $k=2$ by scaling:
\[\begin{split}
\vrac{\lapms{s} (\theta_{A_{2^{k+2}}} \laps{t} f)}_{L^\infty(B_1)}&\le \vrac{\lapms{s} (\theta_{A_{2^{k+2}}} \laps{t} f)}_{L^\infty(B_{2^{k}})}\\
 &=(2^k)^{s-t}\vrac{\lapms{s} (\theta_{A_4} \laps{t} f(2^k\cdot))}_{L^\infty(B_1)}\\
 &\le C_1 (2^k)^{s-t}\|f(2^k\cdot)\|_{L^q(\R^n)}\\
 &= C_1 (2^k)^{s-t-\frac nq}\|f\|_{L^q(\R^n)}.
\end{split}\]
\end{proof}


Considering above $\theta_{A_{2^k \rho}} g$ instead of $\theta_{A_{2^k \rho}}$ we also have the following:
\begin{lemma}\label{la:freakshowestimate2}
For any $\rho > 0$, $p \in (1,\infty)$
\[
 \vrac{\lapms{s} (\theta_{A_{2^k\rho}} g (-\lap)^\frac n4 f)}_{L^p(B_\rho)} \aleq \max_{t \in [0,\frac{n}{2}]} \vrac{\laps{t}g}_{L^2(\R^n)} (2^k \rho)^{t-\frac{n}{2} + s - n}\rho^{\frac{n}{p}}\ \vrac{f}_{L^2(\R^n)},
\]
for any $f,g\in \Sw(\R^n)$.
\end{lemma}
\begin{proof}
By duality, the claim follows if we show for any $\varphi \in C_c^\infty(B_\rho)$
\begin{equation}\label{eq:freak2claim2}
  \vrac{(-\lap)^\frac{n}{4} (\theta_{A_{2^k\rho}} g(\lapms{s} \varphi))}_{L^2(\R^n)} \aleq \max_{t \in [0,\frac{n}{2}]} \vrac{\laps{t}g}_{L^2(\R^n)} (2^k \rho)^{t-\frac{n}{2} + s - m} \rho^{\frac{n}{p}}\ \vrac{\varphi}_{L^{p'}(\R^n)}.
\end{equation}
By the definition of the three-term-commutator $H_{\frac{n}{2}}$, H\"older inequality for a small $t > 0$, and the related estimates, see Theorem~\ref{th:bicomest}, 
\[
\begin{split}
\vrac{(-\lap)^\frac n4 (\theta_{A_{2^k\rho}} g(\lapms{s} \varphi))}_{L^2(\R^n)} 
 \aleq &\vrac{(-\lap)^\frac n4 g}_{L^2(\R^n)}\, \vrac{\theta_{A_{2^k\rho}} \lapms{s} \varphi}_{L^\infty(\R^n)} \\
 &+ \vrac{g}_{L^\frac{2n}{n-2t}(\R^n)}\, \vrac{(-\lap)^\frac n4(\theta_{A_{2^k\rho}} \lapms{s} \varphi)}_{L^\frac{n}{t}(\R^n)}\\ 
 &+ \| H_{\frac{n}{2}} (g,\theta_{A_{2^k\rho}} (\lapms{s} \varphi)) \|_{L^2(\R^n)}\\
%  
\aleq &\vrac{(-\lap)^\frac n4 g}_{L^2(\R^n)}\ \vrac{\theta_{A_{2^k\rho}} \lapms{s} \varphi}_{L^\infty(\R^n)} \\
&+ \vrac{g}_{L^\frac{2n}{n-2t}(\R^n)}\, \vrac{(-\lap)^\frac n4(\theta_{A_{2^k\rho}} \lapms{s} \varphi)}_{L^\frac nt(\R^n)}\\ 
 &+ \| (-\lap)^\frac n4 g\|_{L^2(\R^n)}\ \vrac{(-\lap)^\frac n4(\theta_{A_{2^k\rho}} \lapms{s} \varphi)}_{L^2(\R^n)}\\
  \end{split}
 \]
By the Sobolev inequality,
\[
 \vrac{g}_{L^\frac{2n}{n-2t}(\R^n)} \aleq \vrac{\laps{t} g}_{L^2(\R^n)},
\]
and from Lemma~\ref{la:nonloc} with $\operatorname{supp} \varphi \subset B_\rho$
\[
 \vrac{\theta_{A_{2^k\rho}} \lapms{s} \varphi}_{L^\infty(\R^n)}  \aleq (2^k \rho)^{s-n} \vrac{\varphi}_{L^1(\R^n)} \aleq (2^k \rho)^{s-n} \rho^{\frac{n}{p'}}\ \| \varphi\|_{L^{p'}(\R^n)}.
\]
The remaining terms can be estimated with Lemma~\ref{la:freakshowestimate1}, and \eqref{eq:freak2claim2} follows.
\end{proof}






Let for $t > 0$ the three term commutator given as
\[
 H_t(a,b) := \laps{t}(ab) - b\laps{t}a -a\laps{t}b.
\]
A version similar to $H$ was first was introduced in \cite{DRSphere}. For subsequent similar results and extended arguments see also \cite{DRManifold,Sfracenergy},\cite[Lemma A.5]{BRS},\cite{DSpalphSphere}.
\begin{theorem}\label{th:bicomest}
Given $p \in (1,\infty)$, $t\ge 0$, $p_1, p_2 \in (1,\frac{n}{t}]$ satisfying
\[
 \frac{1}{p} = \frac{1}{p_1} + \frac{1}{p_2} - \frac{t}{n},
\]
it holds
%For any small $\eps \geq 0$, 
\[
 \vrac{H_t(a,b)}_{L^p(\R^n)} \aleq \vrac{\laps{t} a}_{L^{p_1}(\R^n)}\ \vrac{\laps{t} b}_{L^{p_2}(\R^n)},\quad \text{for }a,b\in \mathcal{S}(\R^n).
\]
\end{theorem}





\appendix
\section{Appendix}
% 
% \section{Estimates}
% %\begin{lemma}
% %Let $u\in H^{3/2}_{00}(\Omega)$ such that 
% %$$(-\Delta)^{\frac{3}{2}}u=f \text{ On } \Omega$$
% %in the sens of (). Then $u\in H^{3}(\mathbb{R}^{3})$. In particular $-\Delta u \in L^2$.
% %\end{lemma}
% 
% %\begin{proof}
% %Indeed from the equation we have that for all $\varphi \in H^{3/2}$ 
% %$$\int_{\mathbb{R}^{3}}\hat{u}|\xi|^{3}\hat{\varphi}d\xi =\int_{\mathbb{R}^{3}} \varphi f dx$$
% 
% %In particular, one can define the operator
% %$$T:\varphi \mapsto \int_{\mathbb{R}^{3}}\hat{u}|\xi|^{3}\varphi d\xi$$
% %for all $\varphi \in \mathcal{S}(\mathbb{R}^{3})$. Thus,
% %$$|T\varphi|=|\int f \mathcal{F}^{-1}\varphi dx| \leq ||f||_{L^{2}}||\varphi||_{L^{2}}$$
% %Hence since $\mathcal{S}(\mathbb{R}^{3})$ is dense in $L^{2}$ it extends to an operator on $L^{2}$, therefore by duality $\hat{u}|\xi|^{3} \in L^{2}$
% 
% 
% 
% 
% %\end{proof}
% 
% We consider in this section a function $h$ such that $h\in W^{2,p}(\Omega)$, here $\Omega$ is a bounded set of $\mathbb{R}^{3}$. 
% 
% {\color{blue}[This proposition is probably false]}
% 
% \begin{proposition}\label{har1}
% Assume that $(-\Delta)^{\frac{3}{2}}h=0$ in $B_{R}$, then $h$ is smooth and 
% $$||h||_{C^{k,\alpha}(B_{\frac{R}{2}})}\leq C ||h||_{L^{1}(B_{R})}.$$
% \end{proposition}
% 
% \begin{proof}
% Recall first that if $u\in L_{\frac{1}{2}}$ and $(-\Delta)^{\frac{1}{2}}u=0$ in $\Omega$, then $u(x)=\gamma_{\lambda}*u(x)$ for $\lambda\leq dist(x,\partial \Omega)$. It follows in particular from this that
% $$||u||_{C^{k}(B_{\frac{R}{2}})}\leq C ||u||_{L_{\frac{1}{2}}}.$$
% In our case, if we apply this inequality to $u=-\Delta h$ we find 
% $$\|\Delta h\|_{C^{k}(\frac{R}{2})} \leq C \|\Delta h \|_{L_{\frac{1}{2}}}\leq \frac{C}{R^{3}} \|\Delta h \|_{L^{1}},$$
% 
% but, notice that using the same representation for $-\Delta h=-\Delta \gamma_{\lambda}*h(x)$ we have that 
% $$||-\Delta h ||_{L^{1}}\leq C ||h||_{L^{1}}$$
% Hence, the desired inequality follows.
% \end{proof}
% 
% Now we consider the Dirichlet problem for the fractional Laplacian, that is 
% \begin{equation}\label{dir}
% \left\{ \begin{array}{ll}
% (-\Delta)^{s}u=g \text{ in $\Omega$}\\
% u=0 \text{ in $\mathbb{R}^{n}\setminus\Omega$ }\\
% u\in H^s(\R^n)
% \end{array}
% \right.
% \end{equation}
% Then the following result hold
% \begin{theorem}[\cite{XR}]\label{har2}
% Let $u\in H^s(\R^n)$ be a weak solution to (\ref{dir}), for some $g\in L^{\infty}(\Omega)$. Then $u\in C^{s}(\mathbb{R}^{n})$ and there exists $C>0$ independent of $u$ and $g$, such that 
% $$\|u\|_{C^{s}(\mathbb{R}^{n})}\leq C \|g\|_{L^{\infty}(\R^n)}.$$
% \end{theorem}
\subsection{The Riesz transform and Riesz potential}
We define the Riesz potential of $u$ for $s \in (0,n)$ and $u \in \Sw(\R^n)$ 
\[
\lapms{s} u  := |\cdot|^{s-n} \ast u,
\]
By the density of the Schwartz class $\Sw(\R^n)$ in $L^p(\R^n)$, the Riesz potential $\lapms{s}$ can be extended to an operator mapping $L^p(\R^n)$ into $L^{\frac{np}{n-s}}(\R^n)$ whenever $p, \frac{np}{n-s} \in (1,\infty)$.
Up to a constant, the Riesz potential $\lapms{s}$ is the inverse of the fractional laplacian $\laps{s}$, in the sense that for a constant $c_{n,s} \in \R$
\[
\laps{s} \lapms{s} f =\lapms{s} \laps{s} f = c_{n,s} f \quad \forall f \in \Sw(\R^n).
\]
The Riesz transform $\Rz = (\Rz_1,\dots ,\Rz_n)$ is defined as 
\[
  \Rz u(x) := \int_{\R^n} \frac{x-y}{|x-y|^{n+1}}\, u(y)\, dy,\quad u\in \mathcal{S}(\R^n),
\]
and by density can be extended to a continuous operator from $L^p(\R^n)$ into itself:
\begin{equation}\label{Rieszprop-1}
\|\Rz u\|_{L^p(\R^n)}\le c_{p,n}\|u\|_{L^p(\R^n)}\quad \text{for }u\in L^p(\R^n).
\end{equation}
One crucial properties of the Riesz transform is the that
\begin{equation}\label{Rieszprop0}
\sum_{i=1}^n \Rz_i \Rz_i = c_n\, Id,
\end{equation}
and
\begin{equation}\label{Rieszprop}
 \Rz\laps{1} f = c_n\, \nabla f,\quad u\in \mathcal{S}(\R^n).
\end{equation}

We also recall the following property:

\begin{lemma}[``Integration by parts'']\label{la:pi}
For any $f \in L^p(\R^n)$, $g \in L^{p'}(\R^n)$, $p \in [1,\infty]$ so that $\Rz f \in L^p(\R^n)$, $\Rz g \in L^{p'}(\R^n)$ it holds
\begin{equation}\label{eq:RiezPI}
 \int_{\R^n} \Rz f\, g\, dx = -\int_{\R^n} f\, \Rz g\, dx. 
\end{equation}
For any $f \in L^p(\R^n)$, $g \in L^{p'}(\R^n)$, $p \in (1,\infty)$ so that $\laps{s} f \in L^{p}(\R^n)$, $\laps{s} g \in L^{p'}(\R^n)$,
\begin{equation}\label{eq:fracPI}
 \int_{\R^n} \laps{s}f\, g\, dx = \int_{\R^n} f\, \laps{s} g\, dx. 
\end{equation}
 \end{lemma}

Note that together \eqref{eq:RiezPI} and \eqref{eq:fracPI} imply the usual integration by parts formula
\[
 \int_{\R^n} \nabla f\, g\, dx = c\int_{\R^n} \Rz \laps{1} f\, g\, dx = -c\int_{\R^n}  f\, \Rz \laps{1}g\, dx = -\int_{\R^n}  f\, \nabla g
 \,dx.
\]
%This justifies (for one of the authors) the term ``integration by parts''.


\subsection{Lorentz spaces and Sobolev inequality}
\begin{definition}
For $1< p <\infty$ and $1\leq q \leq \infty$, we define the Lorentz space $L^{(p,q)}(\mathbb{R}^{n})$ as the space of measurable functions $f$ for which
$$\|f\|_{L^{(p,q)}}:=p^{1/q} \|\lambda |\{|f|>\lambda\}|^{1/p}\|_{L^{q}(\frac{d\lambda}{\lambda})}<\infty$$
\end{definition}
It is important to notice that $L^{(p,p)}=L^{p}$ and $L^{(p,q)}\subset L^{(p,r)}$ if $r\geq q$.

\begin{proposition}[Sobolev inequality]\label{pr:Sobolev}
Let $1<p<\frac{n}{\alpha}$ and $1\leq r \leq\infty$. If $f\in L^{(p,q)}(\mathbb{R}^{n})$, then $I_{\alpha}f \in L^{(q,r)}(\mathbb{R}^{n})$ for $q=\frac{np}{n-\alpha p}$. Moreover, there exists $C> $ such that
$$\|\lapms{\alpha}f\|_{L^{(p,r)}(\R^n)}\leq C\|f\|_{L^{(q,r)}(\R^n)}.$$
For $p=1$, $I_{\alpha}$ maps $L^1(\R^n)$ into $L^{(q,\infty)}(\R^n)$ for $q=\frac{n}{n-\alpha}$. 
For $p = \frac{n}{\alpha}$, $I_{\alpha}$ is bounded from $L^{(p,1)}$ into $L^\infty(\R^n)$.
\end{proposition}

From \cite[Corollary 6.16]{BS} we have
\begin{proposition}\label{pr:orliczsobolev}
$I_{\alpha}$ is a bounded linear operator from $L \log^r L(\R^n)$ to $L^{(\frac{n}{n-\alpha},\frac{1}{r})}(\R^n)$ whenever $r \leq 1$, $\alpha \in (0,n)$.
\end{proposition}





%\section{Things that should be true, but need checking}
%\begin{lemma}\label{la:Lp-theory}
%Let $\Omega \subset \R^n$ be a domain with smooth boundary (or the half-space). Then for any $u: \R^n \to \R$ such that (in a weak sense)
%\[
 %\laps{t} u \in L^p(\Omega)
%\]
%there exists $v : \R^n \to \R$, $v \equiv u$ in $\Omega$ and
%\[
 %\vrac{\laps{t}v}_{L^p(\R^n)} \leq C\ \vrac{\laps{t} u}_{L^p(\Omega)},
%\]
%for a uniform constant $C$ depending only on $\Omega$ and $p$.
%\end{lemma}



% \section{Some other stuff}
% \begin{lemma}
% Assume that
% \[
%  \vrac{\laps{m-\eps} (u_k)^2|}_{(\frac{m}{m-\eps},\infty), \Omega} \leq C_1.
% \]
% Then for any $B_{2R}(x_0) \subset \Omega$,
% \[
%  \int_{\Omega} \int_{B_R(x_0)} |x-y|^{-l-\eps} |\laps{m-\eps} (u_k)^2(y)| dx dy \leq C\ R^{m-l}.
% \]
% \end{lemma}
% \begin{proof}
% We have
% \begin{align*}
%   &\int_{\Omega} \int_{B_R(x_0)} |x-y|^{-l-\eps} |\laps{m-\eps} (u_k)^2(y)| dx dy\\
%   =&\int_{B_{2R}(x_0)} \int_{B_R(x_0)} |x-y|^{-l-\eps} |\laps{m-\eps} (u_k)^2(y)| dx dy\\
%   &+\sum_{k=1}^\infty \int_{|y-x_0|\approx 2^kR } \int_{B_R(x_0)} |x-y|^{-l-\eps} |\laps{m-\eps} (u_k)^2(y)| dx dy\\
%   =:& I +\sum_{k=0}^\infty II_k.
% \end{align*}
% For the second guys,
% \begin{align*}
%  |II_k| &\leq (2^k R)^{-l-\eps} R^m \int_{|y-x_0|\approx 2^kR } |\laps{m-\eps} (u_k)^2(y)| dy\\
%  &\leq (2^k R)^{-l-\eps} R^m \vrac{\laps{m-\eps} (u_k)^2|}_{(\frac{m}{m-\eps},\infty), \{|\cdot-x_0|\approx 2^kR \} \cap \Omega}\ (2^kR)^{\eps}\\
%  &\leq C_1\ (2^k R)^{-l} R^m \\
% \end{align*}
% In particular, since $l \geq 1$,
% \[
%  \sum_{k \geq 1} |II_k| \leq C\ R^{m-l}. 
% \]
% For the remaining term $I$, (recall that $l \leq m-1$, so $|x-y|^{-l-\eps}$ is locally integrable for $\eps < 1$)
% \begin{align*}
%  |I| &\leq \int_{B_{2R}(x_0)}  |\laps{m-\eps} (u_k)^2(y)| dy\ \sup_{y \in B_{2R}} \int_{B_R(x_0)} |x-y|^{-l-\eps} dx\\
%  &\leq  R^{m-l-\eps} \int_{B_{2R}(x_0)}  |\laps{m-\eps} (u_k)^2(y)| dy\\
%  &\leq  R^{m-l-\eps} R^\eps \ C_1.
% \end{align*}
% 
% \end{proof}




\subsection{Interior estimates}
The following are a few estimates which could be seen as $L^p$-theory for the fractional Laplacian. Since we only need interior estimates, the proofs are long, but elementary -- just relying on the definitions of Riesz potential, Riesz transform and fractional Laplacian. 
\begin{lemma}\label{la:sharmonicest}
Let $\Omega\subset\R^n$ be open. Then for any $h \in H^\frac{n}{2}(\R^n)$ satisfying 
\begin{equation}\label{eq:lan2heq0}
 \int_{\R^n} (-\lap)^\frac{n}{4} h\, (-\lap)^\frac{n}{4} \varphi\, dx = 0 \quad \forall \varphi \in C_c^\infty(\Omega),
\end{equation}
we have $h \in C^\infty_{\loc}(\Omega)$, and for any compact set $K\Subset\Omega$ and any $\ell \in \N_0$, $\alpha \in (0,1]$ we have
\[
 [\nabla^\ell h ]_{C^{0,\alpha}(K)} \leq C_{\ell,\alpha, K,\Omega}\ \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)}.
\]
\end{lemma}
\begin{proof}
The smoothness $h$, i.e. $h \in C^\infty_{\loc}(\Omega)$, follows via an approximation argument from the a priori estimates below.
% To realize that $h$ is smooth in $K$, approximate by convolution $h$ in $H^{3/2}(\R^3)$ by smooth $h^\eps = h \ast \eta_\eps$. Given any $K\subset \Omega$, for any $\eps < \dist (K,\partial \Omega)$, we have that any $\varphi \in C_c^\infty(K)$ also satisfies $\varphi^\eps \in C_c^\infty(\Omega)$, and thus Fubini's theorem and 
% $$(-\Delta)^\frac34(h*\eta_\eps)=((-\Delta)^\frac34h)*\eta_\eps$$
% imply
% \[
%   \int (-\lap)^{3/4} h^\eps\ (-\lap)^{3/4} (\lap^i \varphi) = \int (-\lap)^{3/4} h\ (-\lap)^{3/4} (\lap^i \varphi)^\eps = 0 \quad \forall \varphi \in C_c^\infty(K), \forall i \in \N_0.
% \]
% Now consider $\tilde{h}^\eps := \theta_{B_{1/\eps}} h^\eps \in C_c^\infty(\R^3)$. Using this and H\"older's inequality, as well as ``fractional integration by parts`` (cf. Lemma~\ref{la:pi}), for any $\varphi \in C_c^\infty(K)$ 
% \begin{equation}\label{eqhtilde}
% \begin{split}
%   &\int (-\lap)^{3/4} \tilde{h}^\eps\ (-\lap)^{3/4} (\lap^i \varphi)\\
%   = &\int h^\eps\ (-\lap)^{3/2} (\lap^i \varphi) - \int h^\eps\ (1-\theta_{B_{1/\eps}})(-\lap)^{3/2} (\lap^i \varphi)\\
%   \leq &\int (-\lap)^{3/4} h^\eps\ (-\lap)^{3/4} (\lap^i \varphi) + \vrac{h^\eps}_{2}\ \vrac{(1-\theta_{B_{1/\eps}})(-\lap)^{3/2+i} \varphi}_{2}\\
%   =&0 + \vrac{h}_{H^{3/2}(\R^3)}\, \vrac{(1-\theta_{B_{1/\eps}})(-\lap)^{3/2+i} \varphi}_{2}
%  \end{split}
%  \end{equation}
% The disjoint support estimates Lemma~\ref{la:nonloc}, 
% \[
%  \vrac{(1-\theta_{B_{1/\eps}})(-\lap)^{3/2+i} \varphi}_{2} \le C_i \eps^{6+i} \vrac{\varphi}_{1}.
% \]
% In particular, since $\tilde{h}^\eps \in C_c^\infty(\R^3)$ we have ``integrating by parts`` (cf. Lemma~\ref{la:pi}) the first integral in \eqref{eqhtilde} that for any $\varphi \in C_c^\infty(K)$ 
% \[
% \sup_{\varphi \in C_c^\infty(K), \|\varphi\|_{1} \leq 1} \int (-\lap)^{3/2+i} \tilde{h}^\eps\  \varphi \leq C_{i}\vrac{h}_{H^{3/2}(\R^3)}.
% \]
% In particular,
% \begin{equation}\label{eq:lap32piest}
%  \vrac{(-\lap)^{3/2+i} \tilde{h}^\eps}_{p,K} \leq C_{p,K,h,i}
% \end{equation}
% Now, we use the Riesz transform,
% \[
%  \partial_\ell (-\lap)^{1+i} \tilde{h}^\eps = c \Rz_\ell [(-\lap)^{3/2+i} \tilde{h}^\eps ]
% \]
% Thus for any $K_1 \subset K_2 \subset K$, with $\dist(K_i,\partial K_{i+1}) > 0$,
% \begin{align*}
% & \chi_{K_1} \Rz_l [(-\lap)^{3/2+i} \tilde{h}^\eps]\\
% =& \chi_{K_1} \Rz_l [\eta_{K}(-\lap)^{3/2+i} (\tilde{h}^\eps)]\\
% &+\chi_{K_1} \Rz_l [(1-\eta_{K})(-\lap)^{3/2+i} \tilde{h}^\eps].
% \end{align*}
% By the boundedness of the Riesz transform on $L^p$-spaces for $p \in (1,\infty)$, and \eqref{eq:lap32piest},
% \[
%  \|\chi_{K_1} \Rz_l [\eta_{K}(-\lap)^{3/2+i} \tilde{h}^\eps]\|_{p,K}
%  \aleq C_{p,K,h,i}.
% \]
% On the other hand, by Lemma~\ref{disjointoperator},
% \[
% \| \chi_{K_1} \Rz_l [(1-\eta_{K})(-\lap)^{3/2+i} \tilde{h}^\eps]\|_{L^2} \aleq C_{K,K_1,i} \vrac{\tilde{h}^\eps}_2.
% \]
% So together we have shown that
% \[
%  \vrac{\partial_l (-\lap)^{1+i} \tilde{h}^\eps}_{2,K_1} \aleq C_{i,K,K_1} \vrac{h}_{H^{3/2}(\R^3)},
% \]
% which is to say that $\tilde{h}^\eps$ is uniformly bounded in $W^{3+2i,2}_{loc}(K)$. Letting $\eps$ go to zero, we obtain that $h \in W^{3+2i,2}_{loc}$ for any $i \in \N$, in particular $h \in C^\infty(K)$.
% 
Notice that \eqref{eq:lan2heq0} can be rewritten as

\begin{equation}\label{eq:lan2heq0bis}
 \int_{\R^n} \nabla h\cdot \nabla \laps{n-2} \varphi\,dx= 0, \quad \forall \varphi \in C_c^\infty(\Omega).
\end{equation}
Fix now $K \subset \subset K_1 \subset \subset K_2 \subset \subset \Omega$. For arbitary $\psi \in C_c^\infty(K_1)$ we have for $k \in \N_0$, 
\[
 \lap^k \psi = \laps{n-2} \lapms{n-2} \lap^k\psi.
\]
Thus, taking a cutoff-function $\eta_{K_2} \in C_c^\infty(\Omega)$, $\eta_{K_2} \equiv 1$ on $K_2$, 
\[
  \lap^k\psi = \laps{n-2} (\eta_{K_1}\lapms{n-2} \lap^k\psi) + \laps{n-2} ((1-\eta_{K_1}) \lapms{n-2} \lap^k\psi).
\]
Thus for any $\psi \in C_c^\infty(K_1)$, using \eqref{eq:lan2heq0bis} with $\varphi:=\eta_{K_2} I_{n-2}\Delta^k\psi$, we get
 \begin{align*}
\int_{\R^n}\nabla h \cdot \nabla  \lap^k \psi\, dx
= &\int_{\R^n} \nabla  h\cdot  \nabla \laps{n-2} ((1-\eta_{K_2}) \lapms{n-2} \lap^k\psi)\,dx \\
\leq & \vrac{\nabla h}_{L^n(\R^n)}  \vrac{\nabla \laps{n-2} ((1-\eta_{K_2}) \lapms{n-2} \lap^k\psi)}_{L^{n'}(\R^n)}\\
\leq & \vrac{\nabla h}_{L^n(\R^n)}  \vrac{\laps{n-1} ((1-\eta_{K_2}) \lapms{n-2} \lap^k\psi)}_{L^{n'}(\R^n)}\\
\leq & C_{K_1,K_2}\, \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)}\, \vrac{\psi}_{L^1(\R^n)}.
\end{align*}
The second-to-last step follows again from $\nabla = \Rz \laps{1}$ and because the Riesz transform $\Rz$ is bounded on $L^{n'}$. In the last step we used that the support of $1-\eta_{K_2}$ and $\psi$ are disjoint to apply Lemma~\ref{disjointoperator}, and Sobolev inequality. Classical regularity theory of elliptic PDE ensures that $h$ belongs to any Sobolev space $W^{\ell,p}_{\loc}(K_1)$ for any $\ell \in \N$, $p \in (1,\infty)$ together with the estimates
\[
 \vrac{h}_{W^{\ell,p}(K)} \leq C_{\ell,p,K,\Omega}\ (\vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)} + \vrac{h}_{L^2(\R^n)}) \aleq  \|h\|_{H^{\frac{n}{2}}(\R^n)},
\]
and
\[
 \vrac{\nabla h}_{W^{\ell+1,p}(K)} \leq C_{\ell,p,K,\Omega}\ \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)}
\]
The latter implies via the Morrey-Sobolev imbedding that for any $\alpha \in (0,1)$, $l \in \N_0$
\[
 [\nabla^\ell h]_{C^{0,\alpha}(K)} \leq C_{\ell,\alpha, K,\Omega}\, \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)} .
 \]
\end{proof}

\begin{lemma}\label{la:LinftyRHSest}
Let $\Omega$ be an open set of $\R^n$. Then for any $h \in H^\frac n2(\R^n)$ satisfying 
\[
 \int_{\R^n} (-\lap)^\frac{n}{4} h\, (-\lap)^\frac n4 \varphi\, dx = \int_{\R^n} f \varphi\, dx \quad \forall \varphi \in C_c^\infty(\Omega),
\]
and for any $\ell \in \{0,1,\ldots,n-1\}$, $\alpha \in (0,1)$ we have on any compact $K \subset\subset \Omega$,
\[
[\nabla^\ell h]_{C^{0,\alpha}(K)} \leq C_{\ell,\alpha,\Omega,K}\, \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)} + C_{\ell,\Omega,K}\, \vrac{f}_{L^\infty(\Omega)}.
\]
\end{lemma}

\begin{proof}
The proof is very similar to the one of Lemma~\ref{la:sharmonicest}. Fix again $K \subset \subset K_1 \subset \subset K_2  \subset \subset \Omega$. 
% For arbitary $\psi \in C_c^\infty(K_2 )$ we now have
%  \begin{align*}
% \int \partial^\alpha h\ \partial^\alpha \lap^k \psi
% = &\int \partial^\alpha h\  \partial^\alpha \laps{n-2} ((1-\eta_{K_2 }) \lapms{n-2} \lap^k\psi) + \int f\ \eta_{K_2 } \lapms{n-2} \lap^k\psi\\
% \leq & C_{K_2,K_2 }\ \vrac{(-\lap)^{n/4} h}_{L^2}\ \vrac{\psi}_{L^1} + \|f\|_{\infty} \vrac{\lapms{n-2} \lap^k \psi}_{1,K_2 }\\
% \end{align*}
% For $0 \leq 2k \leq n-2$, (note that this implies that $2k \leq n-3$ for odd $n$), we have
% \[
%  \vrac{\lapms{n-2} \lap^k \psi}_{1,K_2 } \leq C_{K_2 } \vrac{\psi}_{L^1(\R^n)}\
% \]
% and thus elliptic regularity theory gives us that
% \[
%  h \in W^{n-1,p}_{loc}(K_2 ) \quad \mbox{for any $p \in (0,\infty)$},
% \]
% with the estimate
% \[
%  \vrac{\nabla h}_{W^{n-2,p}(K_2)} \leq \vrac{(-\lap)^{n/4} h}_{L^2(\R^n)} + \vrac{f}_{L^\infty}.
% \]
% In particular by Sobolev-Morrey embedding we have the claim for $l \in \{0,\ldots,n-3\}$.
% 
We use that the following equation (note that $n-1$ is even and thus $\laps{n-1}$ is the classical $(n-1)$-Laplacian),
\[
 \int_{\R^n} \laps{1} h\, \lap^{\frac{n-1}{2}} \varphi\, dx = \int_{\R^n} f \varphi\, dx \quad \forall \varphi \in C_c^\infty(\Omega).
\]
Elliptic theory implies $\laps{1} h \in W^{n-1,p}_{loc}(\Omega)$ for any $p \in (1,\infty)$, with the estimate
\begin{equation}\label{eq:lapshhwnm1p}
 \vrac{\laps{1}h}_{W^{n-1,p}(K_2 )}  \aleq \vrac{f}_{L^\infty(\Omega)} + \vrac{\laps{1} h}_{L^n(\R^n)} \aleq \vrac{f}_{L^\infty(\Omega)} + \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)}.
\end{equation}
Here again we used that $\laps{1} h \in L^n(\R^n)$ by Sobolev embedding. With this in mind, we can write $\nabla h$ in terms of the Riesz transform $\Rz$ and $(-\Delta)^\frac12$,
\begin{equation}\label{eq:partialalphahdecomp}
\nabla h = \Rz (\eta_{K_1} \laps{1} h) + \Rz ((1-\eta_{K_1}) \laps{1} h),
\end{equation}
where we have a cutoff function $\eta_{K_1} \in C_c^\infty(K_2 )$ and $\eta_{K_1} \equiv 1$ in $K_1$.
The first term on the right-hand side belongs to $W^{n-1,p}(\R^n)$ by \eqref{eq:lapshhwnm1p} and the boundedness of the Riesz transform, and we have
\[
 \vrac{\Rz (\eta_{K_1} \laps{1} h)}_{W^{n-1,p}(\R^n)} \aleq  \vrac{f}_{\infty} + \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)}.
\]
The second term on the right-hand side of \eqref{eq:partialalphahdecomp} is smooth in $K$, by the disjoint support of $\chi_K$ and $(1-\eta_{K_1})$. Indeed, by Lemma~\ref{la:nonloc} for any $\ell \geq 0$,
\begin{align*}
\vrac{\nabla^\ell \Rz ((1-\eta_{K_1}) \laps{1} h)}_{L^\infty(K)} \aleq C_{K,K_1}\ \vrac{\laps{1} h}_{L^n(\R^n)} \aleq C_{K,K_1}\ \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)}.
\end{align*}
Together, we have shown that for any $0 \leq \ell \leq n-1$, $p \in (1,\infty)$,
\[
 \vrac{\nabla h}_{W^{\ell,p}(K)} \aleq \vrac{f}_{L^\infty(\Omega)} + \vrac{(-\lap)^\frac n4 h}_{L^2(\R^n)}.
\]
Now the Sobolev-Morrey embedding gives the claim.
\end{proof}

% 
% Write $h_k := w_k - v_k$, where $v_k \in H_{00}^{3/2}(B_{10R})$ is the unique weak solution of
% $$(-\Delta)^\frac32 v_k = (-\Delta)^\frac32 w_k\quad \text{in }B_{10R}.$$
% Grubb's paper {\color{blue}[I'm not sure that Grubb's paper contains this estimate]}, otherwise do it by hand, then should imply that in the smaller ball $B_{9R}$,
% \[
%  \|v_k\|_{C^{2,\alpha}(B_{9R})} \xrightarrow{k \to \infty} 0\quad \text{for every }\alpha\in [0,1). 
% \]
% On the other hand, by definition
% \[
%  \int \lap^{{3}/{4}} h_k \ \lap^{{3}/{4}} \varphi = 0  \quad \forall \varphi \in C_c^\infty(B_{10R}).
% \]
% Now note that since $h_k = 0$ in $(K\cup \mathrm{supp} (w_k))^c$,  by approximation of $h$ with smooth, compactly supported functions, using also that
% \[
%  ||h||_{H^{3/2}} \leq ||v||_{H^{3/2}} + ||w||_{H^{3/2}} < \infty,
% \]
% the following is true
% \[
%  h = \lap^{-3/2} (\theta_{B_{9R}} \lap^{3/2} h) + \lap^{-3/2} ((1-\theta_{B_{9R}}) \lap^{3/2} h)
% \]
% [Define $(-\Delta)^{-3/2}$]
% As seen before, the operator
% \[
%  f \mapsto \theta_{B_R} \lap^{-3/2} ((1-\theta_{B_{9R}}) (-\lap)^{\frac{3}{4}} f) 
% \]
% is a mollifying operator, and we have for any $\beta > 0$ (but it blows up for $\beta = 0$), CHECK!
% \[
%  ||\laps{\beta}(\theta_{B_R} \lap^{-3/2} ((1-\theta_{B_{9R}}) (-\lap)^{\frac{3}{4}} f)) ||_{L^\infty} \aleq C_{R,\beta} ||f||_{L^2}.
% \]
% thus for any $\alpha > 0$,
% \[
%  [\lap^{-3/2} ((1-\theta_{B_{9R}}) \lap^{3/2} h)]_{\dot{C}^{\alpha}(B_R)} \aleq C_\alpha \vrac{(-\lap)^{\frac{3}{4}} h}_{L^2(\R^n)}, 
% \]
% which is bounded, depending on $R$ but uniformly in $k$. On the other hand, by the PDE, if $\alpha < 3/2$,
% \[
%  ||\lap^{-3/2} (\theta_{B_{9R}} \lap^{3/2} h)||_{C^{\alpha}(B_R)} \aleq C_R.
% \]
% Consequently we have for any $\alpha \in (0,3/2)$,
% \[
%  [h]_{\dot{C}^{\alpha}(B_R)} \aleq C_\alpha \vrac{(-\lap)^{\frac{3}{4}} w}_{L^2(\R^n)}.
% \]
% Together with the estimates for $v$ for any $\alpha \in (0,3/2)$ we have shown that
% \[
%  [w_k]_{\dot{C}^{\alpha}(B_R)} \aleq C_\alpha \vrac{(-\lap)^{\frac{3}{4}} w}_{L^2(\R^n)}.
% \]









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\end{document}



1504.00254.extracted_bib
{Aba \textsc{N. Abatangelo, Large $s$-harmonic functions and boundary blow-up solutions for the fractional Laplacian. Preprint.
1504.00254.extracted_bib
{ada \textsc{D. Adams, A sharp inequality of J. Moser for higher order derivatives, Ann. of Math. 128 (1988), 385-398.
1504.00254.extracted_bib
{AS \textsc{Adimurthi, M. Struwe, Global compactness properties of semilinear elliptic equations with critical exponential growth, J. Functional Analysis 175 (2000), 125-167. %
1504.00254.extracted_bib
{Beck \textsc{W. Beckner, Sharp Sobolev inequalities on the sphere and the Moser-Trudinger inequality, Ann. of Math. (2) 138 (1993), 213-242.
1504.00254.extracted_bib
{BS \textsc{C. Bennet, R. Sharpley, Interpolation of operators, Pure and Applied Mathematics 129 (1988). %
1504.00254.extracted_bib
{BLW \textsc{M. Birkner, J. A. L\'opez-Mimbela, A. Wakolbinger, Comparison results and steady states for the Fujita equation with fractional Laplacian, Ann. I. H. Poincaré – AN 22 (2005) 83–97.
1504.00254.extracted_bib
{BRS \textsc{S. Blatt, Ph. Reiter, A. Schikorra, Harmonic analysis meets critical knots (Stationary points of the Moebius energy are smooth), Trans. Amer. Math. Soc. (to appear) (2014).
1504.00254.extracted_bib
{CC \textsc{S-Y. A. Chang, W. Chen: A note on a class of higher order conformally covariant equations, Discrete Contin. Dynam. Systems 63 (2001), 275-281. %
1504.00254.extracted_bib
{B\textsc{H. Br\'ezis, Functional analysis, Sobolev spaces and partial differential equations. Universitext. Springer, New York, 2011. %
1504.00254.extracted_bib
{BM \textsc{H. Br\'ezis, F. Merle, Uniform estimates and blow-up behavior for solutions of $-\Delta u=V(x)e^u$ in two dimensions, Comm. Partial Differential Equations %
1504.00254.extracted_bib
{CS \textsc{X. Cabr\'{e, Y. Sire Nonlinear equations for fractional Laplacians I: Regularity, maximum principles, and Hamiltonian estimates, Preprint.
1504.00254.extracted_bib
{CRW \textsc{R.~R. Coifman, R.~Rochberg, and G.~Weiss: Factorization theorems for {Hardy spaces in several variables, Ann. of Math., 103(3) (1976), 611--635. %
1504.00254.extracted_bib
{cha \textsc{S. Chanillo, A Note on Commutators, Indiana Univ. Math. J. 31 (1982), 7�16. %
1504.00254.extracted_bib
{DM \textsc{F. Da Lio, L. Martinazzi, T. Rivi\`ere, Non-local Liouville equation, preprint.
1504.00254.extracted_bib
{DaLioBubbles \textsc{F.~Da~Lio, Compactness and bubbles analysis for half-harmonic maps into spheres, \newblock {Ann Inst. Henri Poincar\'e, Analyse non lin\`{eaire, 3 (2015), 201-224
1504.00254.extracted_bib
{DRManifold \textsc{F.~Da~Lio, T.~Rivi{\`ere, Sub-criticality of non-local Schr\"odinger systems with antisymmetric potentials and applications to half-harmonic maps. \newblock {Adv. in Math. 227(3) (2011), 1300-1348.
1504.00254.extracted_bib
{DRSphere \textsc{F.~Da~Lio, T.~Rivi{\`ere, Three-term commutator estimates and the regularity of 1/2-harmonic maps into spheres, \newblock {Analysis and PDE, 4(1) (2011), 149-190.
1504.00254.extracted_bib
{DLMR \textsc{F. Da Lio, L. Martinazzi, T. Rivi\`ere, Blow-up analysis of a nonlocal Liouville-type equation, preprint (2015).
1504.00254.extracted_bib
{DSpalphSphere \textsc{F.~Da~Lio, A.~Schikorra, n/p-harmonic maps: regularity for the sphere case, Adv. Calc. Var. 7 (2014), 1-26. %
1504.00254.extracted_bib
{DV \textsc{G. Di Blasio, B. Volzone, Comparison and regularity results for the fractional Laplacian via symmetrization methods, J. Differential Equations 253 (2012), no. 9, 2593-2615.
1504.00254.extracted_bib
{DNPV\textsc{E. Di Nezza, G. Palatucci, E. Valdinoci, Hitchhiker's guide to the fractional Sobolev spaces, Bull. Sci. math., Vol. 136 (2012), No. 5, 521--573.
1504.00254.extracted_bib
{DLZ \textsc{Y. Ding, S. Z. Lu, P. Zhang, Weak estimates for commutators of fractional integral operators, Sci. China, Ser. A 44(7) (2001), 877-888.
1504.00254.extracted_bib
{dru \textsc{O. Druet, Multibumps analysis in dimension $2$: quantification of blow-up levels, Duke Math. J. 132 (2006), 217-269. %
1504.00254.extracted_bib
{fon \textsc{L. Fontana, Sharp borderline Sobolev inequalities on compact Riemannian manifolds, Comment. Math. Helv. 68 (1993), 415-454. %
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