text:
\begin{document}
\title{\bf A concave-convex elliptic problem involving the fractional
Laplacian}

\author{
{\bf C. Br\"andle, E. Colorado, A. de Pablo}\\
\small Departamento de Matem{\'a}ticas, Universidad Carlos III de
Madrid, \\ 28911 Legan{\'e}s, Spain.}

\maketitle

\begin{abstract}
We study a nonlinear elliptic problem defined in a bounded domain involving
fractional powers of the Laplacian operator together with a concave-convex
term. We characterize completely the range of parameters for which solutions of
the problem exist and prove a multiplicity result.

\

\

\noindent{\it Keywords:} semilinear elliptic equations, fractional Laplacian, Sobolev
trace inequality.

\noindent{\it 2000 Mathematics Subject classification:}
35R11,
35J20,
46E35,

\end{abstract}

\section{Introduction}\label{sect-intro}

In the past decades the problem
$$
\left\{
    \begin{array}
      {rcl@{\qquad}l}
        -\Delta u&=&f(u)&\mbox{in }\Omega\subset\mathbb{R}^N,\\
    u&=&0&\mbox{on }\partial \Omega,
    \end{array}
    \right.
$$
has been widely investigated. See~\cite{A} for a survey, and for
example the list (far from complete)~\cite{ABC,Brezis-Nirenberg2,Lions85b} for more specific
problems, where different nonlinearities and different classes of domains,
bounded or not, are considered. Other different diffusion operators, like the $p$--Laplacian, fully nonlinear operators, etc, have been also treated,  see for
example~\cite{BEP, Caffarelli-Cabre,GP} and the references there in. We deal
here with a nonlocal version of the above problem, for a particular type of
nonlinearities, i.e., we study a concave-convex problem involving the fractional
Laplacian operator
\begin{equation}
  \label{prob:main.nonlocal}
\left\{
    \begin{array}
      {rcl@{\qquad}l}
        (-\Delta)^{\alpha/2}u&=&\lambda u^q+u^p,\qquad u>0&\mbox{ in } \Omega,\\
    u&=&0&\mbox{ on } \partial \Omega,
    \end{array}
    \right.
\end{equation}
with $0<\alpha<2$, $0<q<1<p< \frac{N+\alpha}{N-\alpha}$, $N>\alpha$,
$\lambda>0$ and $\Omega\subset\mathbb{R}^N$ a smooth bounded domain.

The nonlocal operator $(-\Delta)^{\alpha/2}$ in $\mathbb{R}^N$ is defined on
the Schwartz class of functions $g\in
\mathcal{S}$ through the Fourier transform,
$$
    [(-\Delta)^{\alpha/2}  g]^{\wedge}\,(\xi)=(2\pi|\xi|)^\alpha \,\widehat{g}(\xi),
$$
or via the Riesz potential, see for example~\cite{Landkof,Stein}. Observe that
$\alpha=2$ corresponds to the standard local Laplacian.

This type of diffusion operators arises in several areas such as physics,
probability and finance, see for
instance~\cite{Applebaum,Bertoin,Cont-Tankov,VIKH}. In particular, the
fractional Laplacian can be understood as the infinitesimal generator of a stable L{\'e}vy
process,~\cite{Bertoin}.

There is another way of defining this operator. In fact, in the case $\alpha=1$
there is an explicit form of calculating the half-Laplacian acting on a
function $u$ in the whole space $\mathbb{R}^N$, as the normal derivative on the
boundary of its harmonic extension to the upper half-space
$\mathbb{R}^{N+1}_+$, the so-called Dirichlet to Neumann operator. The
``$\alpha$ derivative" $(-\Delta)^{\alpha/2}$ can be characterized in a similar
way, defining the $\alpha$-harmonic extension to the upper half-space,
see~\cite{Caffarelli-Silvestre} and Section~\ref{sect:whole-space} for details.
This extension is commonly used in the recent literature since it allows to
write nonlocal problems in a local way and this permits to use the variational
techniques for these kind of problems.

In the case of the operator defined in bounded domains $\Omega$, the above
characterization has to be adapted. The fractional powers of a linear positive
operator in $\Omega$ are defined by means of the spectral decomposition.
In~\cite{Cabre-Tan},  the authors consider the fractional operator
$(-\Delta)^{1/2}$ defined using the mentioned Dirichlet to Neumann operator,
but restricted to the cylinder
$\Omega\times\mathbb{R}_+\subset\mathbb{R}^{N+1}_+$, and show that this
definition is coherent with the spectral one, see also~\cite{stinga-torrea} for
the case $\alpha\neq1$. We recall that this is not the unique possibility of
defining a nonlocal operator related to the fractional Laplacian in a bounded
domain. See for instance the definition of the so called {\it regional
fractional Laplacian} in~\cite{BBC,Guan-Ma}, where the authors consider the
Riesz integral restricted to the domain $\Omega$. This leads to a different
operator related to a Neumann problem.

As to the concave-convex nonlinearity, there is a huge amount of results
involving different (local) operators, see for
instance~\cite{AbCP,ABC,BEP,CCP,CP,GP}. We quoted the work~\cite{ABC} from where
some ideas are used in the present paper. In most of the problems considered in
those papers a critical exponent appears, which generically separates the range
where compactness results can be applied or can not (in the fully nonlinear
case the situation is slightly different, but still a critical exponent
appears,
\cite{CCP}). In our case, the critical exponent with respect to the corresponding Sobolev
embedding is given by $2^*_\alpha=\frac{2N}{N-\alpha}$. This is a reason why
problem \eqref{prob:main.nonlocal} is studied in the subcritical case
$p<2^*_\alpha-1=\frac{N+\alpha}{N-\alpha}$; see also the nonexistence result
for supercritical nonlinearities in Corollary~\ref{supercritical}.

The main results we prove characterize the existence of solutions
of~\eqref{prob:main.nonlocal} in terms of the parameter $\lambda$. A
competition between the sublinear and superlinear terms plays a role, which
leads to different results concerning existence and multiplicity of solutions,
among others. By a solution we mean an energy solution, see the precise
definition in Section~\ref{sect-f(u)}.
\begin{thm}\label{thm:main}
  There exists $\Lambda>0$ such that for Problem~\eqref{prob:main.nonlocal}
  there holds:
\begin{itemize}
\item[{\rm 1.}] If $0<\lambda <\Lambda$ there is a minimal solution.
Moreover, the family of minimal solutions is increasing with respect to
$\lambda$.
 \item[{\rm 2.}] If $\lambda =\Lambda$ there is at least one solution.
 \item[{\rm 3.}] If $\lambda >\Lambda$ there is no solution.
 \item[{\rm 4.}] For any $0<\lambda <\Lambda$ there exist at least two
 solutions.
\end{itemize}
\end{thm}

For $\alpha\in[1,2)$ and $p$ subcritical, we also prove that there exists a
universal $L^\infty$-bound for every solution independently of $\lambda$.
\begin{thm}\label{th:universalbound}
  Let $\alpha\ge1$. Then there exists a constant $C>0$ such that, for any
  $0<\lambda\le\Lambda$,
  every solution to Problem~\eqref{prob:main.nonlocal} satisfies
  $$
\|u\|_{\infty}\le C.
  $$
\end{thm}
The proof of this last result relies on the classical argument of rescaling
introduced in \cite{Gidas-Spruck} which yields to problems on unbounded domains, which require some Liouville-type results, which can be seen in~\cite{dps}. This is the point where the
restriction $\alpha\ge1$ appears.

The paper is organized as follows: we devote Section~\ref{sect:bounded} to
settle the preliminaries, such as define properly the fractional Laplacian in a
bounded domain, by means of the use of the $\alpha$-harmonic extension, and
study an associated linear equation in the local version. The main section,
Section~\ref{sect-f(u)}, contains the results related to the nonlocal nonlinear
problem~\eqref{prob:main.nonlocal}, where we prove
Theorems~\ref{thm:main}--\ref{th:universalbound}. We include an appendix where
some trace inequality, needed in the proofs,   is obtained, with sharp
constant.

\section{The fractional Laplacian in a bounded domain}
\label{sect:bounded}
\setcounter{equation}{0}

\subsection{The $\alpha$-harmonic extension}

To define the fractional Laplacian in a bounded domain we
follow~\cite{Cabre-Tan}, see also~\cite{stinga-torrea}. The idea is to use the
$\alpha$-harmonic extension introduced in \cite{Caffarelli-Silvestre} to define
the same operator in the whole space, see also the Appendix, but restricted to
our bounded domain. To this aim we consider the cylinder
$$
  {C}_\Omega=\{(x,y) : x\in\Omega,\  y\in
  \mathbb{R}_+\}\subset\mathbb{R}^{N+1}_+\,,
$$
and denote by $\partial_L{C}_\Omega$ its lateral boundary.

We first define the extension operator and fractional Laplacian for smooth
functions.

 \begin{defi}
Given a regular function $u$, we define its
$\alpha$-harmonic extension $w=\ext_\alpha(u)$ to the cylinder
${C}_\Omega$ as the solution to the problem
\begin{equation}
  \left\{
    \begin{array}
      {rcl@{\qquad}l}
        -\diver(y^{1-\alpha}\nabla w)&=&0&\mbox{ in }\ {C}_\Omega,\\ [5pt]
        w&=&0&\mbox{ on }\  \partial_L{C}_\Omega,\\ [5pt]
        w&=&u&\mbox{ on }\ \Omega.
    \end{array}
    \right.
\label{eq:local.prob.CS.bdd}\end{equation}
\end{defi}
As in the whole space, there is also a Poisson formula for the extension
operator in a bounded domain, defined through the Laplace transform and the
heat semigroup generator $e^{t\Delta}$, see~\cite{stinga-torrea} for details.

\begin{defi}
The fractional operator $(-\Delta)^{\alpha/2}$ in $\Omega$, acting
on a regular function $u$, is defined by
\begin{equation}
    \label{normalder2}
(-\Delta)^{\alpha/2}u(x)=-\frac{1}{\kappa_\alpha}\lim\limits_{y\to0^+}y^{1-\alpha}\dfrac{\partial
w}{\partial y}(x,y),
\end{equation}
where $w=\ext_\alpha(u)$ and $\kappa_\alpha$ is given as in~\eqref{normalder}
in the Appendix.
\end{defi}

This operator can be extended by density to a fractional Sobolev space.

\subsection{Spectral decomposition}

It is classical that the powers of a positive operator in a bounded domain (or
in an unbounded domain provided the spectrum is discrete) are defined through
the spectral decomposition using the powers of the eigenvalues of the original
operator. We show next that in this case this is coherent with the
Dirichlet-Neumann operator defined above. Let $(\varphi_j,\lambda_j)$ be the
eigenfunctions and eigenvectors of $-\Delta$ in $\Omega$ with  Dirichlet
boundary data. Define the space of functions defined in our domain $\Omega$,
$$
H^{\alpha/2}_0(\Omega)=\{u=\sum a_j
  \varphi_j\in L^2(\Omega)\  :\  \|u\|_{H^{\alpha/2}_0}=
  \left(\sum a_j^2\lambda_j^{\alpha/2}\right)^{1/2}<\infty\},
$$
its topological dual $H^{-\alpha/2}(\Omega)$, and also the energy space of
functions defined in the cylinder ${C}_\Omega$,
$$
X_0^\alpha({C}_\Omega)=\{z\in L^{2}({C}_\Omega)\,:\, z=0 \mbox{ on }
\partial_L{C}_\Omega,\  \|z\|_{X_0^\alpha}^2=
\left(\int_{{C}_\Omega} y^{1-\alpha}|\nabla
z|^2\right)^{1/2}<\infty\}.
$$

\begin{lemma}
\begin{enumerate}
  \item[{\rm 1.}] The eigenfunctions and eigenvectors of
$(-\Delta)^{\alpha/2}$ in $\Omega$  with  Dirichlet boundary data are given by
$(\varphi_j,\lambda_j^{\alpha/2})$.
\item[{\rm 2.}] If $u=\sum a_j
  \varphi_j\in H^{\alpha/2}_0(\Omega)$ then  $\ext_\alpha(u)\in X_0^\alpha({C}_\Omega)$ and
$$
\ext_\alpha(u)(x,y)=\sum a_j\varphi_j(x)\psi(\lambda_j^{1/2} y),
  $$
where $\psi(s)$ solves the problem
$$
 \left\{\begin{array}
      {rcl@{\qquad}l}
      \psi''+\dfrac{(1-\alpha)}s\psi'&=&\psi,&s>0,\\ [3mm]
      -\lim\limits_{s\to0^+}s^{1-\alpha}\psi'(s)&=&\kappa_\alpha,\\ [3mm]
      \psi(0)&=&1.
    \end{array}\right.
$$
\item[{\rm 3.}] In the same hypotheses, $(-\Delta)^{\alpha/2}u\in H^{-\alpha/2}(\Omega)$,
and $(-\Delta)^{\alpha/2}u=\sum
  a_j\lambda_j^{\alpha/2}\varphi_j$.
\item[{\rm 4.}] It holds
$$
\|(-\Delta)^{\alpha/2} u\|_{H^{-\alpha/2}}=
\|(-\Delta)^{\alpha/4} u\|_{L^2}=\|u\|_{H^{\alpha/2}_0}=
\kappa_\alpha^{-1/2}\|\ext_\alpha(u)\|_{X^\alpha_0}. $$
\end{enumerate}
\label{lem:spectral}\end{lemma}

The proof of this result is straightforward. The function $\psi$ coincides with the
solution $\phi_\alpha$ in problem~\eqref{prob:phialpha}. The calculation of the
norms is also straightforward: Using the orthogonality of the family
$\{\varphi_j\}$, together with $\int_\Omega\varphi_j^2=1$,
$\int_\Omega|\nabla\varphi_j|^2=\lambda_j$, and~\eqref{bessel2}, we have
$$
\begin{array}{l}
\displaystyle\int_{\mathbb{R}^{N+1}_+}y^{1-\alpha}|\nabla
\ext_\alpha(u)(x,y)|^2\,dxdy\\ [4mm] = \displaystyle\int_0^\infty
y^{1-\alpha}\int_\Omega\Big( \sum
a_j^2|\nabla\varphi_j(x)|^2\psi(\lambda_j^{1/2}y)^2+
a_j^2\lambda_j\varphi_j(x)^2(\psi'(\lambda_j^{1/2}y))^2\Big)\,dxdy\\
[4mm] = \displaystyle\int_0^\infty y^{1-\alpha}\sum a_j^2\lambda_j\Big(
\psi(\lambda_j^{1/2}y)^2+ (\psi'(\lambda_j^{1/2}y))^2\Big)\,dy\\ [4mm] =
\displaystyle\sum a_j^2\lambda_j^{\alpha/2}\int_0^\infty s^{1-\alpha}\Big(
\psi(s)^2+ (\psi'(s))^2\Big)\,ds= \kappa_\alpha\displaystyle\sum
a_j^2\lambda_j^{\alpha/2}\\ [4mm] =
\displaystyle\kappa_\alpha\displaystyle\sum
(a_j\lambda_j^{\alpha/4})^2.
\end{array}
$$

\subsection{The linear problem}

We now use the extension problem~\eqref{eq:local.prob.CS.bdd} and the
expression~\eqref{normalder2} to reformulate the nonlocal problems in a local
way. Let $g$ be a regular function and consider the following problems, the
nonlocal problem
 \begin{equation}
  \label{prob:general}
\left\{
    \begin{array}{rcl@{\quad}l}
    (-\Delta)^{\alpha/2}u & = & g(x)&\mbox{ in }\  \Omega,\\
    u & = & 0&\mbox{ on }\  \partial \Omega,
    \end{array}
    \right.
    \end{equation}
and the corresponding local one
 \begin{equation}
  \left\{
    \begin{array}{rcl@{\quad}l}
        -\diver(y^{1-\alpha}\nabla w) & = & 0&\mbox{ in }\  {C}_\Omega,\\[5pt]
         w&=&0&\mbox{ on }\  \partial_L{C}_\Omega,\\ [5pt]
         -\dfrac1\kappa_\alpha\lim\limits_{y\to0^+}y^{1-\alpha}\dfrac{\partial w}{\partial
y}& = & g(x)&\mbox{ on }\  \Omega.
    \end{array}
    \right.
\label{prob:local.general}\end{equation}

We want to define the concept of solution to~\eqref{prob:general}, which is
done in terms of the solution to problem~\eqref{prob:local.general}.

\begin{defi}\label{def:weak.trace}
  We say that $w\in X^\alpha_0({C}_\Omega)$ is an energy solution
to problem~\eqref{prob:local.general},  if for every
function $\varphi\in \mathcal{C}_0^1({C}_\Omega)$ it holds
\begin{equation}
    \int_{{C}_\Omega} y^{1-\alpha}\big\langle\nabla w(x,y),\nabla\varphi(x,y)\big\rangle\,dxdy=
    \int_\Omega \kappa_\alpha g(x)\varphi(x,0)\,dx.
    \label{energysol}
\end{equation}
\end{defi}
In fact more general test functions can be used in the above formula, whenever
the integrals make sense. A supersolution (subsolution) is a function that
verifies~\eqref{energysol} with equality replaced by $\ge$ ($\le$) for every
nonnegative test function.
\begin{defi}\label{def:weak}
  We say that $u\in H_0^{\alpha/2}(\Omega)$ is an energy solution to
problem~\eqref{prob:general} if it is the trace on $\Omega$ of a function $w$
which is an energy solution to problem~\eqref{prob:local.general}.
 \end{defi}

It is clear that a solution exists for instance for every $g\in
H^{-\alpha/2}(\Omega)$. In order to deal with
problem~\eqref{prob:local.general} we will assume, without loss of generality,
$\kappa_\alpha=1$, by changing the function $g$.

In~\cite{cabre-sire} this linear problem is also mentioned. There some results
are obtained using the theory of degenerate elliptic equations developed
in~\cite{fabes-kenig-serapioni}, in particular a regularity result for bounded
solutions to this problem is obtained in~\cite{cabre-sire}. We prove here that
the solutions are in fact bounded if $g$ satisfies a minimal integrability
condition.

\begin{thm}\label{th:linear}
Let $w$ be a solution to problem~\eqref{prob:local.general}.  If $g\in
L^{r}(\Omega)$, $r>\frac{N}{\alpha}$, then $w\in L^\infty({C}_\Omega)$.
\end{thm}
\demo The proof follows from the well-known Moser's iterative technique, that we take
from~\cite[Theorem 8.15]{GT}, and uses the trace inequality
\eqref{eq:trace-bdd}. Without loss of generality we may assume $w\ge0$, and
this simplifies notation. The general case is obtained in a similar way.

We define for $\beta\geq 1$ and $K\geq k$ ($k$ to be chosen later) a ${\cal
C}^1([k,\infty))$ function $H$, as follows:
$$
    H(z)=\left\{
    \begin{array}
      {l@{\qquad}l}
      z^\beta-k^\beta,&z\in[k,K],\\
      \mbox{linear},&z>K.
    \end{array}
    \right.
$$
Let us also define $v=w+k$, $\nu=\tr(v)$, and choose as test function
$\varphi$,
$$
  \varphi=G(v)=\int_k^v |H'(s)|^2\,ds,\qquad \nabla\varphi=|H'(v)|^2\nabla v.
$$
Observe that it is an admissible test function, though it is not $\mathcal{C}^1$.
Replacing this test function into the definition of energy solution we obtain
on one hand:
 \begin{equation}
   \label{GT1}
     \begin{array}
   {rcl}
        \displaystyle\int_{{C}_\Omega} y^{1-\alpha}\big\langle\nabla w,\nabla\varphi\big\rangle\,dxdy &=&
    \displaystyle\int_{{C}_\Omega} y^{1-\alpha}\big|\nabla v|^2|H'(v)|^2\,dxdy
    \\ [12pt]
     &=&\displaystyle \int_{{C}_\Omega} y^{1-\alpha}\big|\nabla H(v)|^2\,dxdy
     \\ [12pt]
     &\geq& \displaystyle\bigg(\int_\Omega |H(\nu)|^{\frac{2N}{N-\alpha}} \,dx\bigg)^{\frac{N-\alpha}N}=
     \|H(\nu)\|^2_{\frac{2N}{N-\alpha}}\, ,
 \end{array}
  \end{equation}
 where the last inequality follows by \eqref{eq:trace-bdd}.

 On the other hand
 \begin{equation}
   \label{GT2}
  \begin{aligned}
   \int_\Omega g(x)\varphi(x,0)\,dx&=\int_\Omega g(x)G(\nu)\,dx\leq \int_\Omega g(x)\nu G'(\nu)\,dx\\
   &\leq \frac{1}{k}\int_\Omega g(x)\nu^2|H'(\nu)|^2\,dx= \frac{1}{k}\int_\Omega
   g(x)|\nu H'(\nu)|^2\,dx.
 \end{aligned}
 \end{equation}
Inequality~\eqref{GT1} together with~\eqref{GT2}, leads to
\begin{equation}\label{eq:ineq}
  \|H(\nu)\|_{\frac{2N}{N-\alpha}}\leq \left(\frac{1}{k}\|g\|_{r}\right)^{1/2}
  \|(\nu H'(\nu))^2\|^{1/2}_{\frac{r}{r-1}}=\|\nu H'(\nu)\|_{\frac{2r}{r-1}},
\end{equation}
by choosing $k=\|g\|_{r}$. Letting $K\to \infty$ in the definition
of $H$, the inequality~\eqref{eq:ineq} becomes
$$
  \|\nu\|_{\frac{2N\beta}{N-\alpha}}\leq \|\nu\|_{\frac{2 r\beta}{r-1}}.
$$
Hence for all $\beta\geq 1$ the inclusion $\nu\in L^{\frac{2
r\beta}{r-1}}(\Omega)$ implies the stronger inclusion $\nu\in
L^{\frac{2N\beta}{N-\alpha}}(\Omega)$, since $\frac{2N\beta}{N-\alpha}>\frac{2
r\beta}{r-1}$ provided $r>\frac N\alpha$. The result follows now, as
in~\cite{GT}, by an iteration argument, starting with
$\beta=\frac{N(r-1)}{r(N-\alpha)}>1$ and $\nu\in
L^{\frac{2N}{N-\alpha}}(\Omega)$. This gives $\nu\in L^\infty(\Omega)$, and
then $w\in L^\infty({C}_\Omega)$. In fact we get the estimate
$$
\|w\|_\infty\le c(\|w\|_{X^\alpha}+\|g\|_r).
$$
\hfill$\Box$

\begin{corol}
\label{cor:GT}
  Let $w$ be a solution to problem~\eqref{prob:local.general}.
  If $g\in L^\infty(\Omega)$, then $w\in \mathcal{C}^\gamma(\overline{{C}_\Omega})$ for some $\gamma\in(0,1)$.
\end{corol}

\demo
    Using Theorem~\ref{th:linear}, the result follows directly from~\cite[Lemma~4.4]{cabre-sire},
    where it is proved that any bounded solution to problem~\eqref{prob:local.general}
    with a bounded $g$ is $\mathcal{C}^\gamma$.~\hfill$\Box$

\section{The nonlinear nonlocal problem}\label{sect-f(u)}
\setcounter{equation}{0}

\subsection{The local realization}

We deal now with the core of the paper; i.e. the study of the nonlocal
problem~\eqref{prob:main.nonlocal}. We write that problem in local version in
the following way: a solution to problem~\eqref{prob:main.nonlocal} is a
function $u=\tr(w)$, the trace of $w$ on $\Omega\times\{y=0\}$, where $w$
solves the local problem

 \begin{equation}
  \left\{
    \begin{array}
      {rcl@{\qquad}l}
        -\diver(y^{1-\alpha}\nabla w)&=&0&\mbox{ in }\  {C}_\Omega,\\ [5pt]
        w&=&0&\mbox{ on } \partial_L{C}_\Omega,\\[5pt]
       \dfrac{\partial w}{\partial\nu^\alpha}&=&f(w)&\mbox{ in }\  \Omega,\\
        \end{array}
    \right.
\label{mainprob}\end{equation}
where
\begin{equation}\label{nualpha}
\frac{\partial
w}{\partial\nu^\alpha}(x)=-\lim\limits_{y\to0^+}y^{1-\alpha}\dfrac{\partial
w}{\partial y}(x,y).
\end{equation}

\noindent{\bf Note.\ } In order to simplify notation in what follows we will denote, when no confusio arises,
$w$ for the function defined in the cylinder ${C}_\Omega$ as well as for
its trace $\tr(w)$ on $\Omega\times\{y=0\}$.

As we have said, we will focus on the particular nonlinearity
\begin{equation}\label{eq:cc}
  f(s)=f_\lambda(s)=\lambda s^q+s^p.
\end{equation}
However many auxiliary results will be proved for more general reactions $f$
satisfying the growth condition
\begin{equation}
  \label{eq:hyp.crecimiento.f}
  0\le f(s)\leq c(1+|s|^p), \qquad \mbox{for some\ } p>0.
\end{equation}

\begin{obse}
  \label{rem:kappa}
    In the definition \eqref{nualpha} we have neglected the constant $\kappa_\alpha$
appearing in~\eqref{normalder2} by a simple rescaling. Therefore,
the results on the coefficient $\lambda$ for the local
problem~\eqref{mainprob}--\eqref{eq:cc} in this section are
translated into problem~\eqref{prob:main.nonlocal} with $\lambda$
multiplied by $\kappa_\alpha^{p(q-1)-1}$.
\end{obse}

Following Definition~\ref{def:weak.trace}, we say that $w\in X_0^\alpha({\cal
C}_\Omega)$ is an energy solution of~\eqref{mainprob} if the following identity
holds
$$ \int_{{C}_\Omega}y^{1-\alpha}\big\langle\nabla w,\nabla\varphi\big\rangle\,dxdy=
  \int_\Omega f(w)\varphi\,dx
$$
 for every regular test function $\varphi$. In the analogous way we define sub-
and supersolution.

We consider now the functional
$$ J(w)=\frac12\int_{{C}_\Omega} y^{1-\alpha}|\nabla w|^2\,dxdy-\int_{\Omega} F(w)\,
  dx,
$$
 where $F(s)=\int_0^s f(\tau)\,d\tau$. For simplicity of notation, we
define $f(s)=0$ for $s\le0$. Recall that the trace satisfies $w\in
L^{r}(\Omega)$, (again this means $\tr(w)\in L^r(\Omega)$), for every $1\le
r\le
\frac{2N}{N-\alpha}$ if $N>\alpha$, $1<r\le\infty$ if $N\le\alpha$. In
particular if $p\leq
\frac{N+\alpha}{N-\alpha}$, and $f$
verifies~\eqref{eq:hyp.crecimiento.f}  then $F(w)\in L^1(\Omega)$,
and the functional is well defined and bounded from below.

It is well known that critical points of $J$ are solutions to~\eqref{mainprob}
with a general reaction $f$. We consider also the minimization problem
$$
  I=\inf\Big\{\int_{{C}_\Omega} y^{1-\alpha}|\nabla w|^2\,dxdy\,:\,w\in X_0^\alpha({C}_\Omega),\,\int_{\Omega}F (w)\,
  dx=1\Big\} ,
$$
 for which, by classical variational techniques, one has that below the critical
exponent the infimum $I$ is achieved. This gives a nonnegative solution. Later
on we will see that this infimum is positive provided $\lambda>0$ is small
enough. On the contrary, for $\lambda$ large enough the infimum is the trivial
solution.

We now establish two preliminary results. The first one is a classical
procedure of sub- and supersolutions to obtain a solution. We omit its proof.

\begin{lemma}\label{lem:iteration}
Assume there exist a subsolution $w_1$ and a supersolution $w_2$ to
problem~\eqref{mainprob} verifying $w_1\le w_2$. Then there also exists a solution $w$
satisfying $w_1\le w\le w_2$ in~${C}_\Omega$.
\end{lemma}

The second one is a comparison result for concave nonlinearities. The proof
follows the lines of the corresponding one for the Laplacian performed
in~\cite{Brezis-Kamin}.

\begin{lemma}\label{lem:comparison}
  Assume the function $f(t)/t$ is decreasing for $t>0$ and consider $w_1,w_2\in  X^{\alpha}_0(C_\Omega)$
  positive subsolution and supersolution, respectively,
  to problem~\eqref{mainprob}. Then $w_1\leq w_2$ in  $\overline{{C}_\Omega}$.
\end{lemma}

\demo By definition we have, for the nonnegative test functions
$\varphi_1$ and $\varphi_2$ to be chosen in an appropriate way,
$$
\begin{array}{l}
\displaystyle\int_{{C}_\Omega}y^{1-\alpha}\big\langle\nabla
w_1,\nabla\varphi_1\big\rangle\,dxdy\le
  \int_\Omega f(w_1)\varphi_1\,dx,  \\ [3mm]
\displaystyle\int_{{C}_\Omega}y^{1-\alpha}\big\langle\nabla
w_2,\nabla\varphi_2\big\rangle\,dxdy\ge
  \int_\Omega f(w_2)\varphi_2\,dx.
  \end{array}
$$
Now let $\theta(t)$ be a smooth nondecreasing function such that
$\theta(t)=0$ for $t\le0$, $\theta(t)=1$ for $t\ge1$, and set
$\theta_\varepsilon(t)=\theta(\frac{t}{\varepsilon})$. If we put, in the
above inequalities
$$
\varphi_1=w_2\,\theta_\varepsilon(w_1-
w_2),\qquad\varphi_2=w_1\,\theta_\varepsilon(w_1-w_2),
$$
we get
$$ I_1\ge \displaystyle
  \int_\Omega w_1w_2\Big(\frac{f(w_2)}{w_2}-
  \frac{f(w_1)}{w_1}\Big)\theta_\varepsilon(w_1-w_2)\,dx,
   $$ where
$$
  I_1:= \displaystyle\int_{{C}_\Omega}y^{1-\alpha}\big\langle w_1\,\nabla
w_2-w_2\,\nabla w_1,\nabla(w_1-w_2)\big\rangle\,\theta_\varepsilon'(w_1-w_2)\,dxdy.
$$
Now we estimate $I_1$ as follows:
$$
\begin{array}{rl}
\displaystyle I_1&\displaystyle \le\int_{{\cal
C}_\Omega}y^{1-\alpha}\big\langle \nabla w_1,(w_1-w_2)
\nabla(w_1-w_2)\big\rangle\,\theta_\varepsilon'(w_1-w_2)\,dxdy\\
[3mm]
  &\displaystyle=\int_{{\cal
C}_\Omega}y^{1-\alpha}\big\langle \nabla
w_1,\nabla\gamma_\varepsilon(w_1-w_2)\big\rangle\,dxdy \end{array}$$
where $\gamma_\epsilon'(t)=t\theta_\varepsilon'(t)$. Therefore,
since $0\le \gamma_\varepsilon\le\varepsilon$, we have
$$
I_1\le\int_{\Omega}f(w_1)\gamma_\varepsilon(w_1-w_2)\,dx\le
c\varepsilon.
$$
We end as in~\cite{ABC}. Letting $\varepsilon$ tend to zero, we
obtain
$$
\int_{\Omega\cap\{w_1>w_2\}} w_1w_2\left(\frac{f(w_2)}{w_2}-
  \frac{f(w_1)}{w_1}\right)\,dx\le0,
$$
which together with the hypothesis on $f$ gives $w_1\le w_2$ in $\Omega$.
Comparison in ${C}_\Omega$ follows easily by the maximum principle.~
\hfill$\Box$

Now we show that the solutions to problem
\eqref{mainprob}--\eqref{eq:hyp.crecimiento.f} are bounded and H\"older
continuous. Later on, in Section~\ref{sec-further}, we will obtain a uniform
$L^\infty$-estimate in the case where $f$ is given by \eqref{eq:cc} and the
convex power is subcritical.

\begin{prop}
  Let $f$ satisfy~\eqref{eq:hyp.crecimiento.f} with $p< \frac{N+\alpha}{N-\alpha}$, and let
  $w\in X_0^\alpha({C}_\Omega)$ be an energy solution to problem~\eqref{mainprob}.
  Then $w\in L^\infty({C}_\Omega)\cap \mathcal{C}^{\gamma}(\overline{{C}_\Omega})$ for some $0<\gamma<1$.
\label{pro:c-gamma}\end{prop}
\demo
The proof follows closely the technique of~\cite{Brezis-Kato}. As in the proof
of Theorem~\ref{th:linear}, we assume $w\ge0$. We consider, formally, the test
function $\varphi=w^{\beta-p}$, for some $\beta>p+1$. The justification of the
following calculations can be made substituting $\varphi$ by some approximated
truncature. We therefore proceed with the formal analysis. We get, using the
trace immersion, the inequality
$$
\left(\int_\Omega w^{\frac{(\beta-p+1)N}{N-\alpha}}\right)^{\frac{N-\alpha}N}\le
C(\beta,\alpha,N,\Omega)\int_\Omega w^{\beta}.
$$
This estimate allows us to obtain the following iterative process
$$
\|w\|_{\beta_{j+1}}\le C\|w\|_{\beta_j}^{\frac{\beta_j}{\beta_j-p+1}},
$$
with $\beta_{j+1}=\frac{N}{N-\alpha}(\beta_j+1-p)$. To have
$\beta_{j+1}>\beta_j$ we need $\beta_j>\frac{(p-1)N}{\alpha}$. Since $w\in
L^{2^*_\alpha}(\Omega)$, starting with $\beta_0=\frac{2N}{N-\alpha}$, we get
the above restriction provided $p<\frac{N+\alpha}{N-\alpha}$. It is clear that
in a finite number of steps we get, for $g(x)=f(w(x,0))$, the regularity $g\in
L^r$ for some $r>\frac N\alpha$. As a consequence, we obtain the conclusion
applying Theorem~\ref{th:linear} and Corollary~\ref{cor:GT}.~\hfill$\Box$

\subsection{A nonexistence result}
The following result relies on a the use of a classical Pohozaev type
multiplier.

\begin{prop}
Assume $f$ is a $\mathcal{C}^1$ function with primitive $F$, and $w$ is an energy
solution to problem~\eqref{mainprob}. Then  the
following Pohozaev-type identity holds
$$
\frac12\int_{\partial_L{C}_\Omega}y^{1-\alpha}\big\langle (x,y),\nu\big\rangle|\nabla
w|^2\,d\sigma-N\int_\Omega F(w)\,dx+\frac{N-\alpha}2\int_\Omega wf(w)\,dx=0.
$$
\end{prop}
\demo
Just use the identity
$$
\begin{array}{l}
\displaystyle\big\langle (x,y),\nu\big\rangle y^{\alpha-1}\diver(y^{1-\alpha}\nabla w)
+\diver\Big[y^{1-\alpha}\Big(\big\langle(x,y),\nabla w\big\rangle-\frac12(x,y)|\nabla
w|^2\Big)\Big] \\ [4mm]
\displaystyle +\Big(\frac{N+2-\alpha}2-1\Big)|\nabla w|^2=0,
\end{array}
$$
where $\nu$ is the (exterior) normal vector  to $\partial\Omega$. It is
calculus matter to check this equality.
\hfill$\Box$

As a consequence we obtain a nonexistence result in the supercritical case for
domains with particular geometry.
\begin{thm}
If $\Omega$ is starshaped and the nonlinearity $f$ satisfies the inequality
$((N-\alpha)sf(s)-2N F(s))\ge0 $, then  problem~\eqref{mainprob}  has no
solution. In particular, in the case $f(s)=s^p$ this means that there is no
solution for any $p\ge \frac{N+\alpha}{N-\alpha}$.
\label{supercritical}\end{thm}

The case $\alpha=1$ has been proved in~\cite{Cabre-Tan}. The
corresponding result for the Laplacian (problem
\eqref{prob:main.nonlocal} with $\alpha=2$) comes from
\cite{Pohozaev}.

\subsection{Proof of Theorem~\ref{thm:main}}

We prove here Theorem~\ref{thm:main} in terms of the solution of the local
version~\eqref{mainprob}. For the sake of readability we split the proof of
into several lemmas. From now on we will denote
$$
(P_\lambda)\equiv\left\{\begin{array} {rcl@{\qquad}l}
    -\diver(y^{1-\alpha}\nabla w)&=&0,&\mbox{ in }\  {C}_\Omega,\\ [5pt]
    w&=&0,&\mbox{ on }\  \partial_L{C}_\Omega,\\[5pt]
    \dfrac{\partial
w}{\partial\nu^\alpha} &=&\lambda w^q+w^p, \quad u>0 &\mbox{ in }\  \Omega,\\
\end{array}\right.
$$
and consider the associated energy functional
$$
J_\lambda (w) =
\displaystyle\frac12\int_{{C}_\Omega} y^{1-\alpha}|\nabla w|^2\,dxdy-
  \int_{\Omega} F_\lambda(w)\,dx,
$$
where
$$
F_\lambda(s)=\frac\lambda{q+1}s^{q+1}+\frac1{p+1}s^{p+1}.
$$

\begin{lemma}\label{lem:>}
Let $\Lambda$ be defined by
$$
\Lambda =\sup\{ \lambda >0\ :\ \mbox{Problem
$({P}_{\lambda })$ has solution}\} .
$$
Then $0<\Lambda <\infty$.
\end{lemma}
\demo
Consider the eigenvalue problem associated to the first eigenvalue $\lambda_1$,
and let $\varphi_1 >0$ be an associated eigenfunction, see
Lemma~\ref{lem:spectral}. Then using $\varphi_1$ as a test function in
$(P_\lambda)$ we have that
\begin{equation}
  \label{eq:lem.>}
  \int_\Omega (\lambda w^q+w^p)\varphi_1\,dx=\lambda_1\int_\Omega w\varphi_1\, dx.
\end{equation}
Since there exist positive constants $c,\delta$ such that $\lambda
t^q+t^p>c\lambda^\delta t$, for any $t>0$ we obtain from~\eqref{eq:lem.>}
(recall that $u>0$) that $c\lambda^\delta<\lambda_1$ which implies
$\Lambda<\infty$.

To prove $\Lambda>0$ we use the sub- and supersolution technique to construct a
solution for any small $\lambda$. In fact a subsolution is obtained as
$\underline w=\varepsilon\varphi_1$, $\varepsilon>0$ small. A supersolution is
a suitable multiple of the function $e$ solution to
$$
  \left\{
    \begin{array}
      {rcl@{\qquad}l}
        -\diver(y^{1-\alpha}\nabla e)&=&0&\mbox{ in }\  {C}_\Omega,\\ [5pt]
        e&=&0&\mbox{ on } \partial_L{C}_\Omega,\\[5pt]
       \dfrac{\partial e}{\partial\nu^\alpha}&=&1&\mbox{ in }\  \Omega.\\
        \end{array}
    \right.
$$
\hfill$\Box$

 This proves the third statement in Theorem~\ref{thm:main}.

\begin{lemma}\label{lem:<}
Problem $(P_\lambda)$ has a positive solution for every $0<\lambda <\Lambda$.
Moreover, the family $\{w_\lambda\}$ of minimal solutions is increasing with
respect to~$\lambda$.
 \end{lemma}

\begin{obse}
  \label{rem:Lambda}
  Although this $\Lambda$ is not exactly the same as that of~Theorem~\ref{thm:main},
  see Remark~\ref{rem:kappa}, we have not changed the notation for the sake of simplicity.
\end{obse}
\demo
We already proved in the previous lemma that Problem~$(P_\lambda)$ has a
solution for every $\lambda>0$ small. Another way of proving this result is to
look at the associated functional $J_\lambda$. Using
Theorem~\ref{cor:trace-bdd}, we have that this functional verifies
$$
\begin{array}{rcl}
  J_\lambda (w) & = & \displaystyle\frac12\int_{{C}_\Omega} y^{1-\alpha}|\nabla w|^2\,dxdy-
  \int_{\Omega} F_\lambda(w)\,
  dx \\ [3mm] & \ge &\displaystyle
\dfrac{1}{2}\int_{{C}_\Omega}y^{1-\alpha}|\nabla w|^2\,dxdy- \lambda
C_1\Big(\int_{{C}_\Omega}y^{1-\alpha}|\nabla
w|^2\,dxdy\Big)^{\frac{q+1}{2}}
\\ [3mm] & &\displaystyle -C_2\Big(\int_{{C}_\Omega}
y^{1-\alpha}|\nabla w|^2\,dxdy\Big)^{\frac{p+1}{2}},
\end{array}
$$
for some positive constants $C_1$ and $C_2$. Then for $\lambda$ small enough
there exist two solutions of problem  $(P_{\lambda})$, one given by
minimization and another one given by the Mountain-Pass Theorem,
\cite{AR}. The proof is standard, based on the geometry of the
function $g(t)=\frac12t^2-\lambda C_1t^{q+1}-C_2t^{p+1}$, see for
instance~\cite{GP} for more details.

We now show that there exists a solution for every $\lambda\in(0,\Lambda)$. By
definition of $\Lambda$, we know that there exists a solution corresponding to
any value of $\lambda$ close to $\Lambda$. Let us denote it by $\mu$, and let
$w_\mu$ be the associated solution. Now $w_\mu$ is a supersolution for all
problems $(P_\lambda)$ with $\lambda<\mu$. Take $v_\lambda$ the unique solution
to problem
\eqref{mainprob} with $f(s)=\lambda s^q$. Obviously $v_\lambda$ is a
subsolution to problem $(P_\lambda)$. By Lemma~\ref{lem:comparison}
$v_\lambda\le w_\mu$. Therefore by Lemma~\ref{lem:iteration} we conclude that
there is a solution for all $\lambda\in (0,\mu)$, and as a consequence, for the
whole open interval $(0,\Lambda)$. Moreover, this solution is the minimal one.
The monotonicity follows directly from the comparison lemma.
\hfill$\Box$

 This proves the first statement in Theorem~\ref{thm:main}.

\begin{lemma}\label{lem:=}
  Problem~$(P_\lambda)$ has at least one solution if $\lambda=\Lambda$.
\end{lemma}
\demo
Let $\{\lambda_n\}$ be a sequence such that $\lambda_n\nearrow
\Lambda $. We denote by $w_n=w_{\lambda_n}$ the minimal solution to
problem $(P_{\lambda_n})$. As in~\cite{ABC}, we can prove that the linearized
equation at the minimal solution has nonnegative eigenvalues. Then  it follows,
as in~\cite{ABC} again, $J_{\lambda_n}(w_n)<0$. Since $J'(w_{\lambda})=0$, one
easily gets the bound $\|w_n\|_{X_0^\alpha({C}_\Omega)}\le k$. Hence,
there exists a weakly convergent subsequence in $X_0^\alpha({C}_\Omega)$
and as a consequence a weak solution of $(P_{\lambda})$ for $\lambda =\Lambda$.
\hfill$\Box$

This proves the second statement in Theorem~\ref{thm:main}. To conclude the
proof of that theorem, we show next the existence of a  second solution for
every $0<\lambda<\Lambda$. It is essential to have that the first solution is
given as a local minimum of the associated functional, $J_{\lambda}$. To prove
this last assertion we follow some ideas developed in
\cite{Alama}.

Let $\lambda_0\in (0,\Lambda)$ be fixed and consider
$\lambda_0<\bar{\lambda}_1<\Lambda$. Take $\phi_0=w_{\lambda_0}$,
$\phi_1=w_{\bar\lambda_1}$ the two minimal solutions to problem $(P_\lambda)$
with $\lambda=\lambda_0$ and $\lambda=\bar{\lambda}_1$ respectively, then by
comparison, $\phi_0<\phi_1$. We define
$$
M=\{w\in X^\alpha_0({C}_\Omega):\,0\le w\le \phi_1\}.
$$
Notice that $M$ is a convex closed set of $X^\alpha_0({C}_\Omega)$. Since $J_{\lambda_0}$ is
bounded from bellow in $M$ and it is semicontinuous on $M$, we get the
existence of $\underline{\omega}\in M$ such that $J_{\lambda_0}(\underline{\omega})=\inf_{w\in
M}J_{\lambda_0}(w)$.
Let $v_0$ be the unique positive solution to  problem
\begin{equation}\label{eq:concave}
\left\{
\begin{array}{rcl@{\qquad}l}
-\diver(y^{1-\alpha}\nabla v_0)&=&0,&\mbox{ in }\  {C}_\Omega,\\ [5pt] v_0
&=& 0,&\mbox{ on }\  \partial_L{C}_\Omega,\\ [5pt]
\dfrac{\partial
v_0}{\partial\nu^\alpha} &=&  v_{0}^q,&\mbox{ in }\  \Omega.
\end{array}\right.
\end{equation}
(The existence and uniqueness of this solution is clear). Since for
$0<\varepsilon<<\lambda_0$, and $J_{\lambda_0}(\varepsilon v_0)<0$, we have
$\varepsilon v_0\in M$, then $\underline{\omega}\neq 0$. Therefore
$J_{\lambda_0}(\underline{\omega})<0$. By arguments similar to those
in~\cite[Theorem 2.4]{Stru}, we obtain that $\underline{\omega}$ is a solution
to problem~$(P_{\lambda_0})$. There are two possibilities:
\begin{itemize}
\item
If $\underline{\omega}\not \equiv w_{\lambda_0}$, then the result follows.

\item If  $\underline{\omega}\equiv w_{\lambda_0}$, we have
just to prove that  $\underline{\omega}$ is a local minimum of $J_{\lambda_0}$.
Assuming that this is true, the conclusion in part 4 of Theorem~\ref{thm:main}
follows by using a classical argument: The second solution is given by  the
Mountain Pass Theorem, see for instance \cite{AR}.
\end{itemize}

We prove now that the minimal solution $w_{\lambda_0}$ is in fact a local
minimum of $J_{\lambda_0}$. We argue by contradiction.

Suppose that $\underline{\omega}$
is not a local minimum of $J_{\lambda_0}$ in $X_0^\alpha ({C}_\Omega )$, then there exists a sequence
$\{v_n\}\subset X_0^\alpha ({C}_\Omega )$ such that $\|v_n-\underline{\omega}\|_{X_0^\alpha}\to 0$ and
$J_{\lambda_0}(v_n)<J_{\lambda_0}(\underline{\omega})$.

\noindent Let $w_n=(v_n-\phi_1)^{+}$ and
$z_n=\max\{0,\min\{v_n,\phi_1\}\}$.  It is clear that $z_n\in M$ and
$$
z_n(x,y)= \left\{
\begin{array}{l@{\quad}l}
 0 & \mbox{  if   }v_n(x,y)\le 0,\\[6pt]
  v_n(x,y) & \mbox{ if }0\le v_n(x,y)\le
\phi_1(x,y),\\[6pt]
  \phi_1(x,y) & \mbox{ if }\phi_1(x,y)\le v_n(x,y).
\end{array}
\right.
$$
We set
$$
\begin{array}{ll}
  T_n\equiv \{(x,y)\in {C}_\Omega :\,z_n(x,y)=v_n(x,y)\},\qquad &S_n\equiv
\text{supp}(w_n),\\[6pt]
\widetilde{T}_n=\overline{T_n}\cap\Omega, \qquad &\widetilde{S}_n=S_n\cap\Omega.
\end{array}
$$
Notice that $\text{supp}(v_n^+)=T_n\cup S_n$.
We claim that
\begin{equation}
  \label{claim}
 |\widetilde{S}_n|_{\Omega}\to 0\quad\mbox{  as } n\to \infty,
\end{equation}
where $|A|_\Omega\equiv
\int_{\Omega}\chi_A (x)\, dx$.

\noindent By the definition of $F_\lambda$, we set
$F_{\lambda_0}(s)=\dfrac{\l_{0}}{q+1}s_+^{q+1}+\dfrac{1}{p+1}s_{+}^{p+1}$, for $s\in\mathbb{R}$,
and get
$$
\begin{aligned}
\displaystyle J_{\l_0}(v_n) =&
     \displaystyle  \frac12 \int_{C_\Omega} y^{1-\alpha}|\nabla v_n|^2\, dxdy -\int_\Omega F_{\lambda_0}(v_n)\,dx
\\
=& \displaystyle  \frac12 \int_{T_n} y^{1-\alpha}|\nabla z_n|^2\, dx
dy-\int_{\widetilde{T}_n}F_{\lambda_0}(z_n)\,dx +\frac12 \int_{S_n}y^{1-\alpha}  |\nabla v_n|^2\,dxdy\\
 &-\displaystyle\int_{\widetilde{S}_n}F_{\lambda_0}(v_n)\,dx+\frac12 \int_{C_\Omega} y^{1-\alpha} |\nabla v^-_n|^2\,dxdy\\
=&
\frac12 \displaystyle\int_{T_n} y^{1-\alpha}|\nabla z_n|^2\, dxdy -\int_{\widetilde{T}_n}F_{\lambda_0}(z_n)\,dx\\
 & \displaystyle+\frac12 \int_{S_n} y^{1-\alpha} |\nabla (w_n+\phi_1)|^2\,dxdy-\int_{\widetilde{S}_n}F_{\lambda_0}(w_n+\phi_1)\,dx\\
 &\displaystyle+\frac12 \int_{C_\Omega}y^{1-\alpha} |\nabla v^-_n|^2\, dxdy. \end{aligned}
$$
Since
$$
\int_{C_\Omega} y^{1-\alpha}|\nabla z_n |^2\, dxdy=\int_{T_n}y^{1-\alpha}  |\nabla v_n |^2\,dxdy+\int_{S_n}y^{1-\alpha}|\nabla\phi_1|^2\, dxdy
$$
and
$$
\int_\Omega F_{\lambda_0}(z_n)\,dx=\int_{\widetilde{T}_n}F_{\lambda_0}(v_n)\,dx+\int_{\widetilde{S}_n}
F_{\lambda_0}(\phi_1)\,dx,
$$
by using the fact that $\phi_1$ is a supersolution to $(P_\lambda)$ with
$\l=\l_0$, we conclude that
\begin{eqnarray*}
J_{\l_0}(v_n)& = & J_{\l_0}(z_n)+\frac12 \int_{S_n}y^{1-\alpha}  (|\nabla
(w_n+\phi_1)|^2-|\nabla \phi_1|^2)\,dxdy\\
&&-\int_{\widetilde{S}_n}(F_{\lambda_0}(w_n+\phi_1)-F_{\lambda_0}(\phi_1))\,dx+\frac12
\int_{C_\Omega}y^{1-\alpha}|\nabla v^-_n|^2\,dxdy\\ & \ge &
J_{\l_0}(z_n)+\frac{1}{2}\|w_n\|_{X^\alpha_0}^2+\frac{1}{2}
\|v_n^-\|_{X^\alpha_0}^2\\
&&- \int_\Omega
\left\{F_{\lambda_0}(w_n+\phi_1)-F_{\lambda_0}(\phi_1)-(F_{\lambda_0})_{u}(\phi_1)w_n\right\}\,dx\\
&\ge&
J_{\l_0}(\underline{\omega})+\frac{1}{2}\|w_n\|_{X^\alpha_0}^2+\frac{1}{2}
\|v_n^-\|_{X^\alpha_0}^2\\
&&- \int_\Omega
\left\{F_{\lambda_0}(w_n+\phi_1)-F_{\lambda_0}(\phi_1)-(F_{\lambda_0})_{u}(\phi_1)w_n\right\}\,dx.
\end{eqnarray*}
On one hand, taking into account that $0<q+1<2$, one obtains that
$$0\le \frac{1}{q+1}(w_n+\phi_1)^{q+1}-\frac{1}{q+1}\phi_1^{q+1}-\phi_1^{q}w_n\le
\frac{q}{2}\frac{w_n^2}{\phi_1^{1-q}}.$$
The well known Picone's inequality (see \cite{Picone}) establish:
$$
|\nabla u|^2- \nabla \left(\dfrac{u^2}{v}\right)\cdot\nabla v\ge 0,
$$
for differentiable functions $v>0$, $u\ge 0$. In our case, by an approximation argument we get
$$
\l_0\int_\Omega \frac{w_n^2}{\phi_1^{1-q}}
dx \le \|w_n\|_{X_0^\alpha}^2.
$$
On the other hand, since $p+1>2$,
$$
\begin{array}{rcl}
0 & \le &\displaystyle
\frac{1}{p+1}(w_n+\phi_1)^{p+1}-\frac{1}{p+1}\phi_1^{p+1}-\phi_1^{p}w_n \le
\frac{r}{2}w_n^2(w_n+\phi_1)^{p-1}\\ & & \\
 & \le &
C(p)(\phi_1^{p-1}w_n^2+w_n^{p+1}).
\end{array}
$$
Hence using that $p+1<2^*_\alpha$ and the claim~\eqref{claim}
$$
\int_\Omega\left\{ \frac{1}{p+1}(w_n+\phi_1)^{p+1}-\frac{1}{p+1}\phi_1^{p+1}-\phi_1^{p}w_n\right\}dx
\le o(1)\|w_n\|_{X_0^\alpha}^2. $$ As a consequence we obtain that
\begin{eqnarray*}
J_{\l_0}(v_n)&\ge&
J_{\l_0}(\underline{\omega})+\frac{1}{2}\|w_n\|_{X_0^\alpha}^2(1-q-o(1))+
\frac{1}{2}\|v_n^-\|_{X_0^\alpha}^2\\
&\equiv&
J_{\l_0}(\underline{\omega})+\frac{1}{2}\|w_n\|_{X_0^\alpha}^2(1-q-o(1))+o(1).
\end{eqnarray*}
Since $q<1$, there results that
$J_{\l_0}(\underline{\omega})>J_{\l_0}(v_n)\ge J_{\l_0}(\underline{\omega})$ for
$n>n_0$, a contradiction with the main hypothesis. Hence $\underline\omega $ is a minimum.

 To finish the proof we have to prove the claim~\eqref{claim}. For $\varepsilon>0$ small, and $\delta>0$ ($\delta$ to be chosen later), we consider
$$
\begin{array}{rcl}
E_n & = & \{x\in
\Omega:\,v_n(x)\ge\phi_1(x)\quad \wedge\quad \phi_1(x)>\underline{\omega}(x)+\delta\},\\
 F_n & = & \{x\in
\Omega:\,v_n(x)\ge\phi_1(x)\quad\wedge\quad\phi_1(x)\le
\underline{\omega}(x)+\delta\}.
\end{array}
$$
Using the fact that
\begin{eqnarray*} 0 & = & |\{x\in \Omega:\,
\phi_1(x)< \underline{\omega}(x)\}|=\left|\bigcap_{j=1}^\infty \left\{x\in \Omega:\,
\phi_1(x)\le \underline{\omega}(x)+\frac{1}{j}\right\}\right|\\ & = &
\lim_{j\to \infty}\left|\left\{x\in \Omega:\, \phi_1(x)\le
\underline{\omega}(x)+\frac{1}{j}\right\}\right|,
\end{eqnarray*}
we get for $j_0$ large enough, that if $\delta<
\frac{1}{j_0}$ then  $$|\{x\in \Omega:\,
\phi_1(x)\le \underline{\omega}(x)+\delta\}|\le \frac{\varepsilon}{2}.$$ Hence
we conclude that $|F_n|_\Omega\le \frac{\varepsilon}{2}$.

Since $\| v_n-\underline{\omega}\|_{X_0^\alpha}\to 0$ as $n\to \infty$, in
particular by the trace embedding, $\|
v_n-\underline{\omega}\|_{L^2(\Omega)}\to 0$. We obtain that, for $n\ge n_0$
large,
$$
\frac{\delta^2\varepsilon}{2}\ge\int_{C_\Omega}|v_n-\underline{\omega}|^2 dx\ge \int_{E_n}
|v_n-\underline{\omega}|^2 dx \ge \delta^2|E_n|_{\Omega}.
$$
Therefore $|E_n|_{\Omega}\le \dfrac{\varepsilon}{2}$. Since
$\widetilde{S}_n\subset F_n\cup E_n$ we conclude that
$|\widetilde{S}_n|_\Omega\le
\varepsilon$ for $n\le n_0$. Hence $|\widetilde{S}_n|_\Omega\to 0$ as $n\to \infty$ and
the claim follows.~\hfill$\Box$

\subsection{Proof of Theorem~\ref{th:universalbound} and further results}\label{sec-further}
We start with the uniform $L^{\infty}$-estimates for solutions to
problem~\eqref{prob:main.nonlocal} in its local version given by $(P_\lambda)$.
\begin{thm}\label{thm:uniform}
  Assume $\alpha\ge1$, $p<\frac{N+\alpha}{N-\alpha}$ and $N\geq 2$. Then there exists a constant $C=C(p,\Omega)>0$
  such that every solution to problem~$(P_\lambda)$ satisfies
  $$
\|w\|_{\infty}\le C,
  $$
  for every $0\leq \lambda\leq\Lambda$.
\end{thm}

The proof is based in a rescaling method and the following two nonexistence results, proved in~\cite{dps}:

\begin{thm}
\label{thm:L1}
  Let $\alpha\ge1$. Then the problem in the half-space $\mathbb{R}^{N+1}_{+}$,
$$
  \left\{
    \begin{array}{rcl@{\quad}l}
        -\diver(y^{1-\alpha}\nabla w)&=&0&\mbox{ in }\mathbb{R}^{N+1}_{+}\\
       \dfrac{\partial w}{\partial\nu^\alpha}(x)&=&w^p(x,0)& \mbox{ on }\partial\mathbb{R}^{N+1}_{+}=\mathbb{R}^N
    \end{array}
    \right.
$$
  has no solution provided $p<\frac{N+\alpha}{N-\alpha}$, if $N\ge2$, or for every $p$ if $N=1$.
\end{thm}

\begin{thm}
    \label{thm:L2}
  The problem in the first quarter
  $$\mathbb{R}^{N+1}_{++}=
  \{z=(x',x_N,y)\,|\,x'\in\mathbb{R}^{N-1},\,x_N>0,\,y>0\},$$
$$
  \left\{
    \begin{array}{rcl@{\quad}l}
        -\diver(y^{1-\alpha}\nabla w)&=&0,&x_N>0,\,y>0,\\
       \dfrac{\partial w}{\partial\nu^\alpha}(x',x_N)&=&w^p(x',x_N,0),&\\
        w(x',0,y)&=&0,&
    \end{array}
    \right.
$$
  has no positive bounded solution provided $p<\frac{N+\alpha}{N-\alpha}$.
\end{thm}

\demodel{Theorem~{\rm \ref{thm:uniform}}}Assume by contradiction that there
exists a sequence $\{w_n\}\subset X^{\alpha}_0({{C}}_\Omega)$ of solutions to
$(P_\lambda)$ verifying that $M_n=\|w_n\|_\infty\to\infty$, as $n\to \infty$.
By the Maximum Principle, which holds for our problem, see
\cite{fabes-kenig-serapioni}, the maximum of $w_n$ is attained at a point
$(x_n,0)$ where $x_n\in\Omega$. We define
$\Omega_n=\frac{1}{\mu_n}(\Omega-x_n)$, with $\mu_n=M_n^{(1-p)/\alpha}$, i.e.,
we center at $x_n$ and dilate by $\frac{1}{\mu_n}\to\infty$ as $n\to \infty$.

We consider the scaled functions
$$
v_n(x,y)=\dfrac{w_n(x_n+\mu_n x,\mu_n y)}{M_n}, \quad\mbox{for }
x\in\Omega_n,\: y\ge 0.
$$
It is clear that $\| v_n\|\le 1$, $v_n(0,0)=1$ and moreover
\begin{equation}
  \label{eq:vn}
  \left\{\begin{array}
{rcl@{\qquad}l} -\diver(y^{1-\alpha}\nabla v_n)&=&0&\mbox{ in }\  {
C}_{\Omega_n},\\ [5pt]
    v_n&=&0&\mbox{ on }\  \partial_L{C}_{\Omega_n},\\ [5pt]
\dfrac{\partial v_n}{\partial\nu^\alpha} &=&\lambda M_n^{q-p}v_n^q+v_n^p  &\mbox{ in }\  \Omega_n\times \{ 0\} .
\end{array}\right.
\end{equation}
By Arzel\`a-Ascoli Theorem (the solution is $\mathcal{C}^\gamma$, see
Proposition~\ref{pro:c-gamma}), there exists a subsequence, which we denote
again by $v_n$, which converges to some function $v$ as $n\to\infty$. In order
to see the problem satisfied by $v$ we pass to the limit in the weak
formulation of~\eqref{eq:vn}. We observe that $\| v_n\|_\infty\le 1$ implies
$\|v_n\|_{X_0^\alpha(C_\Omega)}\leq C$.

We define $d_n=\mbox{dist} (x_n,\partial\Omega)$, then there are two
possibilities as $n\to\infty$ according the behaviour of the ratio
$\frac{d_n}{\mu_n}$:
\begin{enumerate}
\item $\left\{\dfrac{d_n}{\mu_n}\right\}_n$ is not bounded.
\item $\left\{\dfrac{d_n}{\mu_n}\right\}_n$ remains bounded.
\end{enumerate}

In the first case, since $B_{d_n/\mu_n}(0)\subset \Omega_n$, and $\Omega_n$ is
smooth, it is clear that $\Omega_n$ tends to $\mathbb{R}^N$ and $v$ is a
solution to
$$
\left\{\begin{array}
{rcl@{\qquad}l}
-\diver(y^{1-\alpha}\nabla v)&=&0\quad&\mbox{ in } \mathbb{R}_+^{N+1},\\[5pt]
\dfrac{\partial
v}{\partial\nu^\alpha}&=&v^p &\mbox{ on } \partial\mathbb{R}_+^{N+1}.
\end{array}\right.
$$
Moreover, $v(0,0)=1$ and $v>0$ which is a contradiction with Theorem~\ref{thm:L1}.

In the second case, we may assume that $\frac{d_n}{\mu_n}\to s\ge 0$ as
$n\to\infty.$ As a consequence, passing to the limit, the domains $\Omega_n$
converge (up to a rotation) to some half-space $H_s=\{x\in\mathbb{R}^N :
x_N>-s\}$.    We obtain here that $v$ is a solution to
$$
\left\{\begin{array}
{rcl@{\qquad}l}
-\diver(y^{1-\alpha}\nabla v)&=&0\quad&\mbox{ in } H_s\times(0,\infty),\\[5pt]
\dfrac{\partial
v}{\partial\nu^\alpha}&=&v^p &\mbox{ on } H_s\times\{0\},
\end{array}\right.
$$
with $\|v\|_\infty=1$, $v(0,0)=1$. In the case $s=0$ this is a contradiction
with the continuity of $v$. If $s>0$, the contradiction comes from
Theorem~\ref{thm:L2}.~\hfill$\Box$

We next prove a uniqueness result for solutions with small norm.
\begin{thm}\label{lem:smallnorm}
 There exists at most one solution to problem $(P_\lambda)$
with small norm.
\end{thm}
 We follow closely the arguments in \cite{ABC}, so we establish the following previous result:
\begin{lemma}
Let $z$ be the unique solution to problem \eqref{eq:concave}. There exists a
constant $\beta>0$ such that
\begin{equation}\label{eq:eigenpositive}
\|\phi\|^2_{X_0^\alpha ({{C}}_\Omega)}-q\int_{\Omega}z^{q-1}\phi^2\, dx\ge \beta\|\phi\|_{L^2(\Omega)}^2,\quad \forall \phi\in X_0^\alpha({{C}}_\Omega).
\end{equation}
\end{lemma}
\demo
We recall that $z$ can be obtained by minimization
$$
\min \left\{ \frac 12 \| \omega\|^2_{X_0^\alpha ({{C}}_\Omega)}-\frac{1}{q+1}\| w\|^{q+1}_{L^{q+1}(\Omega)}:\quad \omega \in X_0^\alpha ({{C}}_\Omega)\right\}.
$$
As a consequence,
$$
 \| \phi \|^2_{X_0^\alpha ({{C}}_\Omega)}-q\int_\Omega z^{q-1}\phi^2\, dx\ge 0,\quad \forall \phi\in X_0^\alpha({{C}}_\Omega).
$$
This implies that the first eigenvalue $a_1$ of the linearized problem
$$
\left\{\begin{array}
{rcl@{\qquad}l}
-\diver(y^{1-\alpha}\nabla \phi)&=&0,\quad&\mbox{ in }\  {C}_\Omega,\\[5pt]
\phi&=&0,\quad&\mbox{ on }\  \partial_L{C}_\Omega,\\[5pt]
\dfrac{\partial
\phi}{\partial\nu^\alpha}-qz^{q-1}\phi&=&a\phi, &\mbox{ on }\  \Omega\times\{0\},
\end{array}\right.
$$
is nonnegative.

Assume first that $a_1=0$ and let $\varphi$ be a corresponding eigenfunction.
Taking into account that $z$ is the solution to~\eqref{eq:concave} we obtain that
$$q\int_\Omega z^q\varphi\, dx=\int_\Omega z^q\varphi\, dx$$ which is a contradiction.

Hence $a_1>0$, which proves \eqref{eq:eigenpositive}.
\hfill$\Box$

\demodel{Theorem {\rm \ref{lem:smallnorm}}}
Consider $A>0$ such that $pA^{p-1}<\beta$, where $\beta$ is given in
\eqref{eq:eigenpositive}.  Now we prove that problem $(P_\lambda)$ has at most one
solution with $L^\infty$-norm less than $A$.

Assume by contradiction that  $(P_\lambda)$ has a second solution
$w=w_\lambda+v$ verifying  $\| w\|_\infty<A$. Since $w_\lambda$ is the minimal
solution, it follows that $v>0$ in $\Omega\times[0,\infty)$. We define now
$\eta=\lambda^{\frac{1}{1-q}}z$, where $z$ is the solution to
\eqref{eq:concave}. Then it verifies $-\div(y^{1-\alpha}\nabla \eta)=0$, with boundary
condition $\lambda \eta^q$. Moreover, $w_\lambda$ is a supersolution to the
problem that $\eta$ verifies. Then by comparison, Lemma~\ref{lem:comparison},
applied with $f(t)=\lambda t^q$, $v=\eta$ and $w=w_\lambda$, we get
\begin{equation}\label{eq:ineq-concave}
w_\lambda\ge \lambda^{\frac{1}{1-q}}z \qquad \mbox{on } \Omega\times\{0\}.
\end{equation}
Since $w=w_\lambda+v$ is solution to $(P_\lambda)$ we have, on
$\Omega\times\{0\}$,
$$
\dfrac{\partial (w_\lambda+v)}{\partial\nu^\alpha}=\lambda (w_\lambda+v)^q+(w_\lambda+v)^p\le
\lambda w_\lambda^q+\lambda q w_\lambda^{q-1}v+(w_\lambda+v)^p,
$$
where the inequality is a consequence of the concavity, hence
$$
\dfrac{\partial v}{\partial\nu^\alpha}\le\lambda q w_\lambda^{q-1} v +(w_\lambda +v)^p-w_\lambda^p.
$$
Moreover, \eqref{eq:ineq-concave} implies $w_\lambda^{q-1}\ge
\lambda^{-1}z^{q-1}.$ From the previous two inequalities we get
$$
\dfrac{\partial v}{\partial\nu^\alpha}\le q z^{q-1} v +(w_\lambda +v)^p-w_\lambda^p.
$$
Using that $\|w_\lambda +v\|_\infty\le A$, we obtain
$(w_\lambda+v)^p-w_\lambda^p\le pA^{p-1}v$. As a consequence,
$$
\dfrac{\partial v}{\partial\nu^\alpha}- q z^{q-1} v \le pA^{p-1}v.
$$
Taking $v$ as a test function and $\phi=v$ in \eqref{eq:eigenpositive} we arrive to
$$
\beta\int_\Omega v^2 \, dx\le p A^{p-1}\int_\Omega v^2 \, dx.
$$
Since $pA^{p-1}<\beta$ we conclude that $v\equiv 0$, which gives the desired
contradiction.~\hfill$\Box$
\begin{obse}
This proof also provides the asymptotic behavior of $w_\lambda$ near $\lambda=
0$, namely $w_\lambda \approx\lambda^{ \frac{1}{1-q}}z$, where $z$ is the
unique solution to problem~\eqref{eq:concave}.
\end{obse}

\vskip1cm

\setcounter{section}{0}
\renewcommand{\thesection}{\Alph{section}}
\renewcommand{\theequation}{\thesection.\arabic{equation}}

\section{Appendix: A trace inequality}
\label{sect:whole-space}
\setcounter{equation}{0}

\subsection{Preliminaries. The problem in the whole space}
Here we recall the definition of the fractional Laplacian in the whole space
and its local representation by means of the extension to one more variable, as
it is done in~\cite{Caffarelli-Silvestre}.

Let $u$ be a regular function in $\mathbb{R}^N$. We say that $w=\ext_\alpha(u)$
is its $\alpha$-harmonic extension  to the upper half-space,
$\mathbb{R}^{N+1}_+$, if $w$ is a solution to the problem
$$
  \left\{
    \begin{array}
      {rcl@{\qquad}l}
        -\diver(y^{1-\alpha}\nabla w)&=&0&\mbox{ in } \mathbb{R}^{N+1}_+,\\ [5pt]
        w&=&u&\mbox{ on } \mathbb{R}^N\times\{y=0\}.
    \end{array}
    \right.
$$
In~\cite{Caffarelli-Silvestre} it is proved that
 \begin{equation}
\lim\limits_{y\to0^+}y^{1-\alpha}\dfrac{\partial w}{\partial
y}(x,y)=-\kappa_\alpha(-\Delta)^{\alpha/2}u(x),
\label{normalder}\end{equation} where
$\kappa_\alpha=\frac{2^{1-\alpha}\Gamma(1-\alpha/2)}{\Gamma(\alpha/2)}$.

The appropriate functional spaces to work with are
$X^\alpha(\mathbb{R}^{N+1}_+)$ and $\dot{H}^{\alpha/2}(\mathbb{R}^N)$, defined
as the completion of $\mathcal{C}_0^\infty(\overline{\mathbb{R}^{N+1}_+})$ and
$\mathcal{C}_0^\infty(\mathbb{R}^N)$, respectively, under the norms
$$
 \begin{array}{l}
    \displaystyle\|\phi\|^2_{X^\alpha}=\int_{\mathbb{R}^{N+1}_+} y^{1-\alpha}|\nabla
    \phi(x,y)|^2\,dx dy,\\[12pt]
    \displaystyle\|\psi\|^2_{\dot{H}^{\alpha/2}}=\int_{\mathbb{R}^N}|2\pi\xi|^\alpha|\widehat{\psi}(\xi)|^2\,d\xi
    =\int_{\mathbb{R}^N}|(-\Delta)^{\alpha/4}\psi(x)|^2\, dx.\\
  \end{array}
$$

The extension operator is well defined for smooth functions through a Poisson
kernel, whose explicit expression is given in~\cite{Caffarelli-Silvestre}.
 It  can also be defined in the space $\dot H^{\alpha/2}(\mathbb{R}^N)$, and in fact
\begin{equation}
  \label{eq:ext}
    \|\ext_\alpha(\psi)\|_{X^\alpha}=c_\alpha\|\psi\|_{\dot{H}^{\alpha/2}}\, , \quad \forall \, \psi\in \dot H^{\alpha/2}(\mathbb{R}^N),
\end{equation}
where $c_\alpha=\sqrt\kappa_\alpha$ , see Lemma~\ref{lemma:well-ext}. On the
other hand, for a function $\phi\in X^\alpha(\mathbb{R}^{N+1}_+)$, we will
denote its trace on $\mathbb{R}^N\times\{y=0\}$ as $\tr(\phi)$. This  trace
operator is also well defined and it satisfies
\begin{equation}
  \label{eq:well-trace}
  \|\tr(\phi)\|_{\dot{H}^{\alpha/2}}\le
c_\alpha^{-1}\|\phi\|_{X^\alpha}.
\end{equation}

\subsection{The trace immersion}

From \eqref{eq:well-trace}, the Sobolev embedding yields then that the trace
also belongs to $L^{2^*_\alpha}(\mathbb{R}^N)$, where
$2^*_\alpha=\frac{2N}{N-\alpha}$. Even the best constant associated to this
inclusion is attained and can be characterized. Although most of the  results
used in order to prove the following theorem are known, we have collected them
for the readers convenience.

\begin{thm}\label{th:trace}
For every $z\in X^\alpha(\mathbb{R}^{N+1}_+)$ it holds
\begin{equation}
  \label{eq:trace}
  \left(\int_{\mathbb{R}^{N}} |v(x)|^{\frac{2N}{N-\alpha}}\,dx\right)^{\frac{N-\alpha}N}\le S(\alpha,N)
  \int_{\mathbb{R}^{N+1}_+} y^{1-\alpha}|\nabla z(x,y)|^2\,dxdy,
\end{equation}
where $v=\tr(z)$. The best constant takes the exact value
\begin{equation}
S(\alpha,N)=\frac{\Gamma(\frac\alpha2)\Gamma(\frac{N-\alpha}2)(\Gamma(N))^{\frac\alpha
N}} {2\pi^{\frac\alpha2}\Gamma(\frac{2-\alpha}2)
\Gamma(\frac{N+\alpha}2)(\Gamma(\frac N2))^{\frac\alpha
N}}
\label{constant-s}\end{equation} and it is achieved when $v$ takes
the form
\begin{equation}
v(x)=\tau^{\frac{N-\alpha}2}\big(|x-x_0|^2+\tau^2\big)^{-\frac{N-\alpha}2},
\label{func-min}
\end{equation} for some $x_0\in\mathbb{R}^N$, $\tau>0$, and $z=\ext_\alpha(v)$.
\end{thm}

The analogous results for the classical  Laplace operator can be found
in~\cite{Escobar,Lions85b}.

\begin{lemma}
\label{lemma:well-ext}
    Let $v\in\dot H^{\alpha/2}(\mathbb{R}^N)$ and let $z=\ext_\beta(v)$ be its $\beta$-harmonic extension, $\beta\in(\alpha/2,2)$. Then
      $z\in X^{\alpha}(\mathbb{R}^{N+1}_+)$ and moreover there exists a
      positive universal constant $c(\alpha,\beta)$ such that
    \begin{equation}
      \|v\|_{\dot H^{\alpha/2}}=c(\alpha,\beta)\|z\|_{X^{\alpha}}.
    \label{eq:x2}\end{equation}
In particular if $\beta=\alpha$ we have
$c(\alpha,\alpha)=1/\sqrt{\kappa_\alpha}$.\end{lemma}
Inequality~\eqref{eq:trace} needs only the case $\beta=\alpha$, which is
deduced directly from the proof of  the local characterization of
$(-\Delta)^{\alpha/2}$ in~\cite{Caffarelli-Silvestre}. The calculations
performed in~\cite{Caffarelli-Silvestre} can be extended to cover the  range
$\alpha/2<\beta<2$ and in particular includes the case $\beta=1$ proved
in~\cite{Xiao}.

\demo Since $z=\ext_\beta(v)$, by definition $z$ solves
$\diver(y^{1-\beta}\nabla z)=0$,  which is equivalent to
$$
\Delta_x z+\frac{1-\beta}y\,\frac{\partial z}{\partial y}+\frac{\partial^2 z}{\partial
y^2}=0.
$$
Taking Fourier transform in $x\in\mathbb{R}^N$ for $y>0$ fixed, we have
$$
-4\pi^2|\xi|^2\hat z+\frac{1-\beta}y\,\frac{\partial \hat z}{\partial
y}+\frac{\partial^2
\hat z}{\partial y^2}=0.
$$
and $\hat z(\xi,0)=\hat v(\xi)$. Therefore $\hat z(\xi,y)=\hat
v(\xi)\phi_\beta(2\pi|\xi|y)$, where $\phi_\beta$ solves the problem
\begin{equation}
  -\phi+\frac{1-\beta}s
\phi'+\phi''=0,\qquad\phi(0)=1,\quad\lim\limits_{s\to\infty}\phi(s)=0.
\label{prob:phialpha}
\end{equation}
In fact, $\phi_\beta$ minimizes the functional
$$
H_\beta(\phi)=\int_0^\infty(|\phi(s)|^2+|\phi'(s)|^2)s^{1-\beta}\,ds.
$$
and it can be shown that it is a combination of Bessel functions,
see~\cite{Lebedev}. More precisely, $\phi_\beta$ satisfies the following
asymptotic behaviour
\begin{equation}
  \begin{array}{ll}
\phi_\beta(s)\sim 1-c_1 s^{\beta},&\quad\mbox{for }\
s\to0,\\ [4mm] \phi_\beta(s)\sim c_2s^{\frac{\beta-1}2}e^{-s},&\quad\mbox{for
}\ s\to\infty, \end{array}
\label{bessel}\end{equation}
where
$$
c_1(\beta)=\frac{2^{1-\beta}\Gamma(1-\frac\beta2)}{\beta\Gamma(\frac\beta2)},\quad
c_2(\beta)=\frac{2^{\frac{1-\beta}2}\pi^{1/2}}{\Gamma(\frac\beta2)}.
$$

Now we observe that
$$
\begin{array}{rl}
\displaystyle\int_{\mathbb{R}^N} |\nabla z(x,y)|^2\,dx&=\displaystyle\int_{\mathbb{R}^N}
\left(|\nabla_x z(x,y)|^2+\Big|\frac{\partial z}{\partial
y}(x,y)\Big|^2\right)\,dx\\ [4mm]
&=\displaystyle\int_{\mathbb{R}^N}\left(4\pi^2|\xi|^2|\hat
z(\xi,y)|^2+\Big|\frac{\partial
\hat z}{\partial y}(\xi,y)\Big|^2\right)\,d\xi.
\end{array}$$
Then, multiplying by $y^{1-\alpha}$ and integrating in $y$,
$$
\begin{array}{l}
\displaystyle\int_0^\infty\int_{\mathbb{R}^N} y^{1-\alpha}|\nabla
z(x,y)|^2\,dxdy\\ [4mm] =\displaystyle
\int_0^\infty\int_{\mathbb{R}^N} 4\pi^2|\xi|^2|\hat
v(\xi)|^2\big(|\phi_\beta(2\pi|\xi|y)|^2+
|\phi_\beta'(2\pi|\xi|y)|^2\big)y^{1-\alpha}\,d\xi dy \\ [4mm]
=\displaystyle\int_0^\infty(|\phi_\beta(s)|^2+
|\phi_\beta'(s)|^2)s^{1-\alpha}\,ds\,\int_{\mathbb{R}^N}|2\pi\xi|^{\alpha}|\hat
v(\xi)|^2\,d\xi.
\end{array}$$
Using \eqref{bessel} we see that the integral
$\int_0^\infty(|\phi_\beta|^2+|\phi_\beta'|^2)s^{1-\alpha}\,ds$ is convergent
provided $\beta>\alpha/2$. This proves~\eqref{eq:x2} with
$c(\alpha,\beta)=(H_\alpha(\phi_\beta))^{-1/2}$.~\hfill$\Box$

\begin{obse}
If $\beta=1$ we have $\phi_1(s)=e^{-s}$, and
$H_\alpha(\phi_1)=2^{\alpha-1}\Gamma(2-\alpha)$, see \cite{Xiao}. Moreover,
when $\beta=\alpha$, integrating by parts and using  the equation in
\eqref{prob:phialpha}, and~\eqref{bessel}, we obtain
\begin{equation}
\label{bessel2}
H_\alpha(\phi_\alpha)=\int_0^\infty
[\phi_\alpha^2(s)+(\phi_\alpha')^2(s)]s^{1-\alpha}\,ds=
-\lim_{s\to0}s^{1-\alpha}\phi_\alpha'(s)=\alpha c_1(\alpha)=\kappa_\alpha.
\end{equation}
\end{obse}

\begin{lemma}
Let $z\in X^\alpha(\mathbb{R}^{N+1}_+)$ and let $w=\ext_\alpha(\tr(z))$ be its
$\alpha$-harmonic associated function (the extension of the trace). Then
  $$
    \|z\|_{X^\alpha}^2= \|w\|_{X^\alpha}^2+\|z-w\|_{X^\alpha}^2.
  $$
\label{lemma:minimum}\end{lemma}
  \demo Observe that, for $h=z-w$, we have
$$
\|z\|_{X^\alpha}^2=\int_{\mathbb{R}^{N+1}_+} y^{1-\alpha}(|\nabla
w|^2+|\nabla h|^2+2 \big\langle\nabla w,\nabla h\big\rangle).
$$
But, since $\tr(h)=0$, we have $
\displaystyle\int_{\mathbb{R}^{N+1}_+} y^{1-\alpha}\big\langle\nabla
w,\nabla h\big\rangle\,dxdy =0 $.~\hfill$\Box$

\begin{lemma}
\label{lem:HL1}
 If $g\in L^{\frac{2N}{N+\alpha}}(\mathbb{R}^N)$, and $f\in \dot
H^{\alpha/2}(\mathbb{R}^N)$, then there exists a constant $\ell(\alpha,N)>0$
such that
\begin{equation}
\left|\int f(x)\,g(x)\,dx\right|\le
\ell(\alpha, N)\|f\|_{\dot{H}^{\alpha/2}}\|g\|_{\frac{2N}{N+\alpha}}.
\label{fg-alpha}\end{equation}
Moreover, the equality in \eqref{fg-alpha} with the best constant holds when
$f$ and $g$ take the form~\eqref{func-min}.
\end{lemma}

The proof follows by an standard argument that can be found, for instance
in~\cite{cotsiolis, Xiao}.
\demo
By Par\c{c}eval's identity and Cauchy-Schwarz's inequality, we have
$$
\begin{aligned}
\displaystyle\left(\int_{\mathbb{R}^N} f(x)\,g(x)\,dx\right)^2&\displaystyle=
\left(\int_{\mathbb{R}^N} \widehat f(\xi)\,\widehat g(\xi)\,d\xi\right)^2 \\  &\displaystyle\le
\left(\int_{\mathbb{R}^N} |2\pi\xi|^\alpha\,|\widehat f(\xi)|^2\,d\xi\right)\,
\left(\int_{\mathbb{R}^N} |2\pi\xi|^{-\alpha}\,|\widehat g(\xi)|^2\,d\xi\right).
\end{aligned}$$
The second term can be written using~\cite{Lieb-Loss} as
$$
\int_{\mathbb{R}^N} |2\pi\xi|^{-\alpha}\,|\widehat g(\xi)|^2\,d\xi=b(\alpha,N)
\int_{\mathbb{R}^{2N}}\frac{g(x)g(x')}{|x-x'|^{N-\alpha}}\,dxdx',
$$
where
$$
b(\alpha,N)=\frac{\Gamma(\frac{N-\alpha}2)}{2^\alpha\pi^{N/2}\Gamma(\frac\alpha2)}\,,
$$
We now use the following Hardy-Littlewood-Sobolev inequality, see
again~\cite{Lieb-Loss},
$$
\int_{\mathbb{R}^{2N}}\frac{g(x)g(x')}{|x-x'|^{N-\alpha}}\,dxdx'\le
d(\alpha,N)\|g\|^2_{\frac{2N}{N+\alpha}},
$$
where
$$
d(\alpha,N)=\frac{\pi^{\frac{N-\alpha}2}\Gamma(\frac\alpha2)(\Gamma(N))^{\frac\alpha
N}}{\Gamma(\frac{N+\alpha}2)(\Gamma(\frac{N}2))^{\frac\alpha N}},
$$
with equality if $g$ takes the form~\eqref{func-min}. From this we obtain the
desired estimate \eqref{fg-alpha} with the constant $
    \ell(\alpha,N)=\sqrt{ b(\alpha,N)d(\alpha,N)}
$.

When applying Cauchy-Schwarz's inequality, we obtain an identity provided the
functions $|\xi|^{\alpha/2}\widehat f(\xi)$ and $|\xi|^{-\alpha/2}\widehat
g(\xi)$ are proportional. This means
$$
\widehat g(\xi)=c|\xi|^{\alpha}\widehat f(\xi)=c[(-\Delta)^{\alpha/2}
f]^{\wedge}\,(\xi).
$$
We end by observing that if $g$ takes the form~\eqref{func-min} and
$g=c(-\Delta)^{\alpha/2}f$ then $f$ also takes the
form~\eqref{func-min}.~\hfill$\Box$

\demodel{Theorem~{\rm\ref{th:trace}}}We apply Lemma~\ref{lem:HL1}
with $g=|f|^{\frac{N+\alpha}{N-\alpha}-1}f$, then use
Lemma~\ref{lemma:well-ext} and conclude using Lemma~\ref{lemma:minimum}. The
best constant is $S(\alpha,N)=\ell^2(\alpha,N)/\kappa_\alpha$.~\hfill$\Box$

Related to this result we quote the work \cite{Chen-Li-Ou}, where it is proved
that the only positive regular solutions to
$(-\Delta)^{\alpha/2}f=cf^{\frac{N+\alpha}{N-\alpha}}$ take the
form~\eqref{func-min}.

\begin{obse}
If we let $\alpha$ tend to 2, when $N>2$, we recover the classical Sobolev
inequality for a function in $H^1(\mathbb{R}^N)$, with the same constant,
see~\cite{talenti}. In order to pass to the limit in the right-hand side of
\eqref{eq:trace}, at least formally, we observe that
$(2-\alpha)y^{1-\alpha}\,dy$ is a measure on compact sets of $\mathbb{R}_+$
converging (in the weak-* sense) to a Dirac delta. Hence
$$
\lim_{\alpha\to2^-}\int_0^1\left(\int_{\mathbb{R}^{N}}
|\nabla z(x,y)|^2\,dx\right)(2-\alpha)y^{1-\alpha}\,dy=\int_{\mathbb{R}^{N}}
|\nabla v(x)|^2\,dx.
$$
We then obtain
$$ \left(\int_{\mathbb{R}^{N}} |v(x)|^{\frac{2N}{N-2}}\,dx\right)^{\frac{N-2}N}\le S(N)
  \int_{\mathbb{R}^{N}} |\nabla v(x)|^2\,dx,
$$
with the best constant
$S(N)=\lim\limits_{\alpha\to2^-}\frac{S(\alpha,N)}{2-\alpha}=\frac{1}{\pi
N(N-2)}\Big(\frac{\Gamma(N)}{\Gamma(\frac{N}2)}\Big)^{\frac 2N}$. It is achieved when
$v$ takes the form
\eqref{func-min} with $\alpha$ replaced by 2.
\end{obse}

Let now $v\in X_0^\alpha({C}_\Omega)$. Its extension by zero outside the
cylinder ${C}_\Omega$ can be approximated by functions with compact
support in $\overline{\mathbb{R}^{N+1}_+}$. Thus, the trace
inequality~\eqref{eq:trace}, together with H\"older's inequality, gives a trace
inequality for bounded domains.

\begin{thm}\label{cor:trace-bdd}
For any $1\le r\le \frac{2N}{N-\alpha}$, and every $z\in X_0^\alpha({
C}_\Omega)$, it holds
\begin{equation}
  \label{eq:trace-bdd}
  \left(\int_{\Omega} |v(x)|^{r}\,dx\right)^{2/r}\le C(r,\alpha,N,|\Omega|)
  \int_{{C}_\Omega} y^{1-\alpha}|\nabla z(x,y)|^2\,dxdy,
\end{equation}
where $v=\tr(z)$.
\end{thm}

\vskip .5cm

{\small
\noindent \textsc{Acknowledgments.}\\
 CB and AdP partially supported by
Spanish Project MTM2008-06326-C02-02; EC partially supported by
Spanish Project MTM2009-10878.}


TITLE: In a lottery game with differing weighted probabilities among players, how to calculate the weights if I know previous winners & player distribution?
QUESTION [0 upvotes]: I apologize if this turns out to be a simple question with an overly long premise.
I'm playing a game of chance where players are distributed into 5 categories. Call them A, B, C, D, E. I know exactly who is in each category.
The game essentially works like this - each day, every player is given a ticket corresponding to their category. And each day, there are 100 empty buckets on the game board. Every player independently chooses which bucket to place his ticket in. At the end of the day, a ticket is drawn randomly from each bucket, and those drawn are winners for the day.
However, the probability of a player's ticket being drawn is weighted according to category. One category A ticket has a much higher chance of being drawn than a category E ticket, but that exact weight is unknown.
At the end of each day, I know everything about what happened that day except for these weights - I know exactly who placed tickets in each bucket, the categories of each player, and the winner of each bucket.
I do not have the weights assigned to each category of players, and this is what I am trying to calculate.
In other words, I have multiple days of data that looks like this:

Bucket | # of A | # of B | # of C | # of D | # of E | Winner Category
   I   |   2    |   6    |   12   |   36   |   90   |      *B*
  II   |   0    |   1    |   5    |   10   |   25   |      *D*
  ...

Given this, how do I calculate the weighted probabilities for the different categories?

REPLY [0 votes]: This is really more of a problem of statistics (rather than probability).
Let $W_A, W_B,\dots,W_E$ be the corresponding weights and $D$ be your observed data.
One approach would be to fix a value for, say, $W_A$ $($for instance, $W_A = 1)$ and let the other weights be unknown.
Then, for each value of $\mathbf w = (1, w_b,w_c,w_d,w_e)$ calculate
$$f(\mathbf w) = \mathbb P \left.\Big(D\,\,\middle|\,(W_A,W_B,W_C,W_D,W_E) = \mathbf w\Big)\right.,$$
and pick some value of $\mathbf w$ that maximizes $f$.
This way, we're taking a maximum likelihood approach. Does this make sense to you?

For each bucket $\beta$ and class $l \in \{a,b,c,d,e\}$, let $n_{l,\beta}$ denote the number of tickets of class $l$ in bucket $\beta$.
Then, the probability that the winning ticket is from class $j \in \{a,b,c,d,e\}$ is
$$p_{j,\beta}(\mathbf w) = \mathbb P(\text{$j$ = class of winning ticket from bucket $\beta$}\,|\,\mathbf w)
=
\frac{n_{j,\beta}\,w_j}{\sum_{l \in \{a,b,c,d,e\}}\,n_{l,\beta}\,w_l}.$$
If we denote by $k(\beta)$ the observed class of the winning ticket for bucket $\beta$, then independence gives us that
\begin{align}
f(\mathbf w)
&=
\prod_{\text{bucket $\beta$}}\,\left(
\frac{n_{k(\beta),\beta}\,w_{k(\beta)}}
{n_{a,\beta}+n_{b,\beta}\,w_b+n_{c,\beta}\,w_c+n_{d,\beta}\,w_d+n_{e,\beta}\,w_e}
\right)
\\&=
\prod_{\text{bucket $\beta$}}\,{\left(
1+\frac{\sum_{l\neq k(\beta)}\,n_l\,w_l}{n_{k(\beta),\beta}\,w_{k(\beta)}}
\right)}^{-1}
.\end{align}
Write
$$g(\beta; w_a,w_b,\dots,w_e) = {\left(
1+\frac{\sum_{l\neq k(\beta)}\,n_l\,w_l}{n_{k(\beta),\beta}\,w_{k(\beta)}}
\right)}^{-1} = p_{k(\beta),\beta}(\mathbf w),$$
so that $f(\mathbf w) = \prod_{\text{bucket $\beta$}}\,g(\beta;\mathbf w)$.
Then
\begin{align}
\frac{\partial f}{\partial w_j}(\mathbf w)
=
\sum_{\text{bucket $\beta$}}\,\left(
\left(\prod_{\text{bucket $\gamma \neq \beta$}}\,g(\gamma;\mathbf w)\right)
\cdot
\frac{\partial g}{\partial w_j}(\beta;\mathbf w)
\right).
\end{align}
On the other hand, a direct calculation shows that
$$
\frac{\partial g}{\partial w_j}(\beta;\mathbf w)
=
\left\{
\begin{array}{ccc}
\displaystyle g(\beta;\mathbf w)^2 \cdot \frac{\sum_{l\neq k(\beta)}\,n_{l,\beta}\,w_l}
{n_{k(\beta),\beta}\,{w_{k(\beta)}}^2}&;&j = k(\beta)\\
\\
\displaystyle -g(\beta;\mathbf w)^2 \cdot \frac{n_{j,\beta}}
{n_{k(\beta),\beta}\,w_{k(\beta)}}&;&j\neq k(\beta)
\end{array}
\right.
$$
Thus, we have that
\begin{align}
\frac{\partial f}{\partial w_j}(\mathbf w)
=
f(\mathbf w)\cdot\left[
\sum_{\substack{\text{bucket $\beta$}\\j=k(\beta)}}\left(
g(\beta;\mathbf w) \cdot \frac{\sum_{l\neq j}\,n_{l,\beta}\,w_l}
{n_{j,\beta}\,{w_j}^2}
\right)
-
\sum_{\substack{\text{bucket $\beta$}\\j\neq k(\beta)}}\left(
g(\beta;\mathbf w) \cdot \frac{n_{j,\beta}}
{n_{k(\beta),\beta}\,w_{k(\beta)}}
\right)
\right]
.\end{align}
Letting
$$
h_j(\beta;\mathbf w)
=
\left\{
\begin{array}{ccc}
\displaystyle \frac{\sum_{l\neq j}\,n_{l,\beta}\,w_l}
{n_{j,\beta}\,{w_j}^2}&;&j = k(\beta)\\
\\
\displaystyle -\frac{n_{j,\beta}}
{n_{k(\beta),\beta}\,w_{k(\beta)}}&;&j \neq k(\beta)
\end{array}
\right.
$$
simple manipulations show that
$$h_j(\beta, \mathbf w) = \frac{\delta_{j,k(\beta)} - p_{j,\beta}(\mathbf w)}
{w_j\cdot p_{k(\beta),\beta}(\mathbf w)},$$
where $\delta_{ij}$ is the Kronecker delta.
We can hence simplify the derivative with
\begin{align}
\frac{\partial f}{\partial w_j}(\mathbf w)
&=
f(\mathbf w)\cdot\left(
\sum_{\text{bucket $\beta$}}
g(\beta;\mathbf w) \cdot h_j(\beta;\mathbf w)
\right)
\\&=
\frac{f(\mathbf w)}{w_j}\cdot\left(
\sum_{\text{bucket $\beta$}}
\delta_{j,k(\beta)} - p_{j,\beta}(\mathbf w)
\right)
.\end{align}
Assuming all weights are positive, $f>0$.
This implies that
\begin{align}
\nabla f(\mathbf w) = 0
&\iff
\Big(\forall j \in \{b,c,d,e\}\Big)\,\sum_{\text{bucket $\beta$}}
\delta_{j,k(\beta)} - p_{j,\beta}(\mathbf w) = 0
\\&\iff
\Big(\forall j \in \{b,c,d,e\}\Big)\,
\sum_{\text{bucket $\beta$}}
p_{j,\beta}(\mathbf w)
=
\#\{\,\text{buckets $\beta$ with $j=k(\beta)$}\,\}
.\end{align}


TITLE: Is a perfect Matching always a Maximum Matching?
QUESTION [0 upvotes]: Is a perfect Matching always a Maximum Matching ?

REPLY [1 votes]: Yes of course, if you're using "maximum" in the sense of "maximum cardinality".  However, "maximum matching" might sometimes mean "maximum weighted matching", and  a perfect matching might not have maximum weight.


TITLE: Showing that $T\in End(\mathbb{R}^3)$
QUESTION [0 upvotes]: Consider the map:
$$T:(x_1,x_2,x_3)\mapsto(2x_1-x_2,x_2+x_3,4x_3)\qquad\mathbb{R}^3\mapsto\mathbb{R}^3$$
I want to show that $T\in End(\mathbb{R})$, which, if I'm not mistaken, means that I must show that it is well-defined and linear. I know how to show that it is linear, but I do not know how to show that it is also well-defined. I need to show that there exists only one element in the codomain for each element in the domain, but how can I do that?

REPLY [1 votes]: This question is not really about linear algebra at all, I think, but rather the notations we use when defining functions in general.
First of all, you have a slight typo in your post: the definition of $T$ should read
$$T : (x_1, x_2, x_3) \mapsto (2x_1 - x_2, x_2 + x_3, 4x_3) \quad \mathbb{R}^3 \to \mathbb{R}^3$$
or (even better in my opinion)
$$T = (x_1, x_2, x_3) \mapsto (2x_1 - x_2, x_2 + x_3, 4x_3) : \mathbb{R}^3 \to \mathbb{R}^3$$
Either of these two options would be fine, but the way you wrote it (with $\mathbb{R}^3 \mapsto \mathbb{R}^3$ instead of $\mathbb{R}^3 \to \mathbb{R}^3$) is definitely not correct (w.r.t. any convention I've seen).
The symbol $\mapsto$ (read as "maps to") is used to describe what a function actually does – more precisely, $[\text{something}] \mapsto [\text{other thing}]$ means "the function which, when given the input $[\text{something}]$, returns $[\text{other thing}]$". On the other hand, the symbol $\to$ (read as "to") is used to described the domain and codomain of a function – $f : A \to B$ means "$f$ is a function whose domain is $A$ and whose codomain is $B$". These two arrows convey very different kinds of information, so it's important to use the right symbols in the right places, at least if you want to convince others you know what's going on.
With that out of the way, let's talk about the actual definition of $T$ here. As previously mentioned, $(x_1, x_2, x_3) \mapsto (2x_1 - x_2, x_2 + x_3, 4x_3)$ is telling you that $T$ is meant to be the unique function such that $T(x_1, x_2, x_3) = (2x_1 - x_2, x_2 + x_3, 4x_3)$. Such a function clearly exists (formally, it is the set of ordered pairs $$\{((x_1, x_2, x_3), (2x_1 - x_2, x_2 + x_3, 4x_3)) : (x_1, x_2, x_3) \in \mathbb{R}^3\}$$), and this is all that's meant by $T$ being "well-defined". The reason we sometimes actually need to check well-definedness is that occasionally the inputs to the purported function (when written as $[\text{input}] \mapsto [\text{output}]$) aren't precisely the elements of the domain, or sometimes the outputs aren't clearly elements of the codomain. In your case, these were non-issues: the input was precisely the form of an arbitrary element of $\mathbb{R}^3$, and the output was clearly an element of $\mathbb{R}^3$.
As others have mentioned in the comments, though, these concerns can often be serious. For example, when trying to define functions from sets of equivalence classes (e.g. defining addition in modular arithmetic), it's important to check that the output of the function doesn't depend on the way you chose to represent the input – that is, while an arbitrary equivalence class may be represented by $[a]$, this representation is not unique unless each equivalence class has exactly $1$ element. As such, a function whose definition looks like $[a] \mapsto 2a$ might fail to be well-defined, because $[a]$ and $[a']$ might be exactly the same without $2a$ and $2a'$ being the same (then we'd have one element of the domain which is being sent to two different elements of the codomain, violating the definition of "function").
Another different example would be (for example) the function
$$n = z \mapsto z \overline{z} : \mathbb{C} \to \mathbb{R}$$
Here, the input to the function is just an arbitrary element of $\mathbb{C}$, so there's no issue there. However, the output of the function, $z \overline{z}$, is not obviously a real number, so to check that $n$ is well-defined, you would need to prove that $z \overline{z}$ is a real number for each complex number $z$ (indeed this is true, as I'm sure you're well aware ).


text:
{\bf Problem.} Let $f(n)$ be the integer closest to $\sqrt[4]{n}.$ Find $\sum_{k=1}^{1995}\frac 1{f(k)}.$
{\bf Level.} Level 4
{\bf Type.} Intermediate Algebra
{\bf Solution.} We have $f(n) = m$ if and only if \[m - \frac{1}{2} < \sqrt[4]{n} < m + \frac{1}{2},\]or \[\left(m - \frac{1}{2}\right)^4 < n < \left(m + \frac{1}{2}\right)^4.\]Expanding the fourth powers, we get \[m^4 - 2m^3 + \frac{3}{2}m^2 - \frac{1}{2}m + \frac{1}{16} < n < m^4+ 2m^3 + \frac{3}{2}m^2 + \frac{1}{2}m + \frac{1}{16}.\]The leftmost and rightmost expressions are both non-integers, and their difference is $4m^3 + m$. Therefore, there are exactly $4m^3 + m$ values of $n$ that satisfy this inequality.

For each $m$, there are $4m^3 + m$ terms of the form $\frac{1}{m}$ in the sum, so those terms contribute $(4m^3+m) \cdot \frac{1}{m} = 4m^2 + 1$ to the sum. Thus, from $m=1$ to $m=6$, we get $4(1+4+9+16+25+36) + 6 = 370$.

The remaining terms have $m=7$. Since $6.5^4 = 1785 \frac{1}{16}$, these are the terms from $n=1786$ to $n=1995$, inclusive. There are $1995 - 1786 + 1 = 210$ such terms, so they contribute $210 \cdot \frac{1}{7} = 30$ to the sum. Therefore, the final answer is $370 + 30 = \boxed{400}$.

set_name: MATH


TITLE: What is the height of liquid risen when a solid cylinder is inserted into a tub of water?
QUESTION [0 upvotes]: Problem: In a cylindrical tub of area $A$ and height $H$, water is upto a level of $H/2$. A solid cylinder of length $L$ and area $A/5$ is inserted into the tub and it is floating vertically with $L/4$ portion immersed. What is the increase in water level?
My Solution:
Let us assume the new level to be $H'$ and conserved the volume of water.
$H/2.A = L/4.(A-A/5) + (H'-L/4).A$
from which I got $H' = H/2 + L/20$. Which is even matching with answer given in the book.
But in the solution given in the book author has taken a different approach which is what I want to understand.
Author's Solution:
Let $x$ be the height of liquid risen above previous level (H/2).
Volume of water displaced by the cylinder ($= L/4.A/5)$ = $x.A$
Therefore, $x=L/20$ and new level of water is $H/2 + L/20$
I didn't understand how he equated volume of water displaced by the cylinder to $x.A$.

REPLY [1 votes]: To appreciate that xA is indeed the volume of water displaced, redraw the figure with the immersed part of the cylinder moved to the bottom (just for ease of calculation of the volume of the displaced water). In this new picture the top layer of thickness(height) x contains the displaced water and has a volume of xA.
By the way, your method is correct too. It has a typo on the LHS of the first equation (H should be replaced by H/2).