\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small {\em Electronic Journal of
Differential Equations}, Vol. 2003(2003), No. 43, pp. 1--8.
\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/43\hfil Hardy inequalities with boundary terms]
{Hardy inequalities with boundary terms}

\author[Zhi-Qiang Wang \&  Meijun Zhu \hfil EJDE--2003/43\hfilneg]
{Zhi-Qiang Wang \&  Meijun Zhu}

\address{Zhi-Qiang Wang \hfill\break
Department of Mathematics and Statistics\\
Utah State University\\
Logan, Utah 84322-3900, USA}
\email{wang@math.usu.edu}

\address{Meijun Zhu \hfill\break
Department of Mathematics\\
University of Oklahoma\\
Norman, Oklahoma 73019, USA} 
\email{mzhu@math.ou.edu}

\date{}
\thanks{Submitted December 2, 2002. Published April 16, 2003.}
\subjclass[2000]{35J20, 35J25} 
\keywords{Hardy inequality with
boundary terms, weighted versions of \hfill\break\indent Hardy inequality}

\begin{abstract}
 In this  note, we present some Hardy type inequalities for
 functions which do not vanish on the boundary of a given domain.
 We establish these inequalities for both bounded and unbounded
 domains and also obtain the best embedding constants in these
 inequalities for special domains. Our results are motivated by and
 building upon some recent work in \cite{WangC0, WangC, LZgafa,
 Zhu}.
\end{abstract}
\maketitle

\newtheorem{thm}{Theorem}[section]
\newtheorem{rem}[thm]{Remark}
\numberwithin{equation}{section}

\section{Introduction}

The standard Hardy inequality states that for $N\geq 3$,
\begin{equation}
\frac {(N-2)^2}4  \int_{\mathbb{R}^N} \frac {u^2}{|x|^2}dx \le
\int_{\mathbb{R}^N} |\nabla u|^2 dx\label{1-1}
\end{equation}
for any $u\in C_0^{\infty}(\mathbb{R}^N)$. Here $(N-2)^2/4$ is the
best possible constant. This inequality can be extended to functions
in the space $\mathcal{D}^{1,2}(\mathbb{R}^N) $ which is the completion of
$C_0^{\infty}(\mathbb{R}^N)$ with respect to the norm
$$
\| u\|^2 =\int_{\mathbb{R}} |\nabla u|^2 dx.
$$
There are many generalizations of
this inequality, see for example, \cite{Br1,Br2,CKN,WangC,Chou,Da,Lie,Lin}
and references therein.
The weighted version of this inequality was given in
\cite{CKN}.
In this paper
we will consider another type of generalizations (motivated by
recent work of Li and Zhu \cite{LZgafa} and Zhu \cite{Zhu}). Let
$\Omega \subset \mathbb{R}^N$ be a bounded domain and $u \in
\mathcal{D}^{1,2}_0(\Omega)$. Since we can trivially extend $u$ to a new
function in $\mathcal{D}^{1,2}(\mathbb{R}^N)$ which vanishes outside $\Omega$,  we
 obtain  the following Hardy inequality on a bounded domain:
\begin{equation}
\frac {(N-2)^2}4  \int_{\Omega} \frac {u^2}{|x|^2}dx \le
\int_{\Omega} |\nabla u|^2dx. \label{1-2}
\end{equation}
Naturally, one may ask whether there are some analogous inequalities that
 hold for function $u \in H^1(\Omega) $ (Notice that $u(x)$ may not vanish on the boundary of $\Omega$).
 Since (\ref{1-2}) does not hold for any constant function, we shall expect,
like in the case of Sobolev
 inequality (see, for example, \cite{LZgafa}), the right hand side may include some lower order terms.


We shall consider a more general version of the Hardy inequality --
the weighted version (\cite{CKN}): for $a < \frac{N-2}{2}$, it holds for all
$u\in C^{\infty}_0(\mathbb{R}^N)$
$$
  \frac{(N-2-2a)^2}{4} \int_{\mathbb{R}^N} |x|^{-2(a+1)} u^2 dx
  \leq \int_{\mathbb{R}^N} |x|^{-2a} |\nabla u|^2 dx.
$$
Recently, in \cite{WangC0,WangC} a new formulation of this
inequality has been given by using a conformal transformation.
Based on this conformal transformation we first establish weighted
Hardy type inequalities with boundary terms in two specific
domains. Denote $B_1(0)= \{ x \in \mathbb{R}^N \ : \ |x|<1\}$, and
$B^c_1(0)=\mathbb{R}^N \setminus \overline B_1(0)$. We assume
below $N\geq 2$ when we treat the weighted version of the Hardy
inequality and $N\geq 3$ when we treat the classical Hardy
inequality. Let us define the weighted Sobolev space
$\mathcal{D}_a^{1,2}(\mathbb{R}^N)$ to be the completion of
$C^{\infty}_0(\mathbb{R}^N)$ with respect to the following norm
$$
\| u\|_a^2 = \int_{\mathbb{R}} |x|^{-2a}|\nabla u|^2 dx.
$$
In the following, for simplicity of notations we omit the
integration variables when the situation is clear.

\begin{thm}
Let $a< \frac{N-2}{2}$. Then for all $u\in \mathcal{D}_a^{1,2}(\mathbb{R}^N)$
\begin{equation}
\frac {(N-2-2a)^2}4 \int_{B_1(0)} \frac {u^2}{|x|^{2(a+1)}} <
\int_{B_1(0)} |x|^{-2a}|\nabla u|^2+\frac {N-2-2a}2 \int_{\partial B_1(0)} u^2,
\label{1-3}
\end{equation}
and
\begin{equation}
\frac {(N-2-2a)^2}4 \int_{B_1^c(0)} \frac {u^2}{|x|^{2(a+1)}} <
\int_{B_1^c(0)} |x|^{-2a}
|\nabla u|^2-\frac {N-2-2a}2 \int_{\partial B_1(0)} u^2.
\label{1-4}
\end{equation}
\label{thm1-1}
\end{thm}


\begin{rem} \rm
The strict inequalities are  due to the non-existence of extremal
functions in (\ref{1-3}) and (\ref{1-4}). The constants involved
in the above inequalities are sharp in the sense that $$ \frac
{(N-2-2a)^2}4 = \inf_{u\in \mathcal{D}_a^{1,2}(\mathbb{R}^N)\setminus \{0\}} \frac{
\int_{B_1(0)} |x|^{-2a}|\nabla u|^2+\frac {N-2-2a}2 \int_{\partial B_1(0)}
u^2}{ \int_{B_1(0)} \frac {u^2}{|x|^{2(a+1)}}} , $$ and a similar
statement holds for (\ref{1-4}).
\label{rem1-1}
\end{rem}

Using similar arguments
we obtain a Hardy inequality on any $C^1$ smooth domains with bounded
boundary and $0 \notin \partial \Omega$.

\begin{thm}
Let $a < \frac{N-2}{2}$. If $\Omega \subset \mathbb{R}^N$ is a
smooth domain with $\partial \Omega $ being bounded and $0 \notin
\partial \Omega$, then there is a constant $C_{h}$ (depending on
$\Omega$), such that for all $u\in
\mathcal{D}_a^{1,2}(\mathbb{R}^N)$
\begin{equation}
\frac {(N-2-2a)^2}4  \int_{\Omega} \frac {u^2}{|x|^{2(a+1)}}
\le \int_{\Omega}
|x|^{-2a} |\nabla u|^2 +C_{h}\int_{\partial \Omega} u^2.
\label{1-5}
\end{equation}
If in addition $\Omega \subset \mathbb{R}^N$ is bounded and
star-shaped with respect to the origin, then there is a constant $C'_{h}>0$
(depending on $\Omega$),
such that for all $u\in \mathcal{D}_a^{1,2}(\mathbb{R}^N)$
\begin{equation}
\frac {(N-2-2a)^2}4 \int_{\Omega^c} \frac {u^2}{|x|^{2(a+1)}} \le
\int_{\Omega^c} |x|^{-2a}|\nabla u|^2 - C'_{h}\int_{\partial \Omega} u^2.
\label{1-5-1}
\end{equation}
\label{thm1-2}
\end{thm}

A natural question following the theorem is
what may happen if $\Omega$ is convex
(thus star-shaped with respect to any interior point) but the origin lies
outside $\Omega$? Based on a new integral inequality established
in \cite{Zhu}, we have the following result.


\begin{thm}
Let $\Omega \subset \mathbb{R}^N$ be a bounded piecewise smooth
domain which  contains the origin. Assume that $\partial \Omega$ consists of two smooth hyper-surfaces $\Gamma_1$ and $\Gamma_2$. If $\Gamma_2$ is concave with respect to the domain $\Omega$ and is part of the boundary of a rotationally symmetric convex domain, then
\begin{equation}
2^{-2/N}  \frac {(N-2)^2}4  \int_{\Omega} \frac {u^2}{|x|^2}
\le \int_{\Omega} |\nabla u|^2
\label{1-5-2}
\end{equation}
holds for any $u \in \mathcal{D}^{1,2}(\mathbb{R}^N)$ with $u=0$ on $\Gamma_1$.
\label{thm1-3}
\end{thm}

\begin{rem}
Let $\Omega \subset \mathbb{R}^N$ be a smooth convex  revolution solid
which does not contain the origin.  As a simple corollary of
Theorem \ref{thm1-3}, we see that for any $u\in \mathcal{D}^{1,2}(\mathbb{R}^N)$,
\begin{equation}
2^{-2/N} \frac {(N-2)^2}4  \int_{\Omega^c} \frac {u^2}{|x|^2}
\le \int_{\Omega^c} |\nabla u|^2.
\label{1-5-2-2}
\end{equation}
\end{rem}

For any domain $\Omega$ and any  $u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}$, let
$$
I(u, \Omega):= \frac {\int_{\Omega} |\nabla u|^2}{\int_{\Omega}
\frac {u^2}{|x|^2}}\,.
$$
We will give an example of a domain that $\Omega$ satisfies
the conditions in Theorem \ref{thm1-3},
\begin{equation}
\inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N)
\setminus \{0\}, u=0 \ on \ \Gamma_1} I(u, \Omega) <  \frac {(N-2)^2}4.
\label{1-6}
\end{equation}
Quite similar to the case of Sobolev inequality, we have the following
theorem.

\begin{thm}
Let $\Omega \subset \mathbb{R}^N$ be a smooth domain such that
$\partial \Omega$ is bounded and
$0\notin \partial \Omega$.
If $ 0< \inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}} I(u, \Omega) <(N-2)^2/4 $,
then the infimum is achieved by a function $\bar u \in \mathcal{D}^{1,2}(\mathbb{R}^N)$.
\label{thm1-4}
\end{thm}

\begin{rem} \rm
Assume that $\Omega$ satisfies the conditions in Theorem
\ref{thm1-3}. Following the proof of Theorem \ref{thm1-4}, we
easily prove that if
$$
\inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) H^1(\Omega) \setminus \{0\},
\ u=0 \ on \ \Gamma_1} I(u, \Omega)<(N-2)^2/4 ,
$$
then $ \inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}, \ u=0 \ on \ \Gamma_1} I(u, \Omega)$
is achieved by some functions.
This indicates that if $\Omega \subset \mathbb{R}^N$ is a convex  domain which
does not contain the origin, then there might be no uniform lower bound
for $\inf_{u\in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}}I(u, \Omega^c)$.
\label{rem1-4}
\end{rem}

\section{Proofs of Theorems}

The proofs of Theorem \ref{thm1-1}--\ref{thm1-2} are based on the
following conformal transformation which was used in \cite{WangC0,
WangC} to give a new formulation of a family of weighted Sobolev
inequalities due to Caffarelli-Kohn-Nirenberg in \cite{CKN}. This
family of inequalities include the weighted version of the Hardy
inequalities.

We define $\varphi:\mathbb{R}^N  \to  \mathcal{C}:= \mathbb{R}
\times S^{N-1}$ as the conformal transformation
\begin{equation}
\varphi(x)=(-\ln|x|, \frac x{|x|}).
\label{2-1-1}
\end{equation}
Here we use $(t,\theta)\in \mathbb{R} \times S^{N-1}$. And we
define
\begin{equation}
u(x)=|x|^{-\frac {N-2-2a}2}v(-\ln|x|, \frac x{|x|}), \quad
 \forall x \in \mathbb{R}^N.
\label{2-1}
\end{equation}
Due to the density lemma in \cite[Lemma 2.1]{WangC}, we  need to prove
Theorem \ref{thm1-1}--\ref{thm1-2} only for functions in
$C_0^1(\mathbb{R}^N)$.


\begin{proof}[Proof of Theorem \ref{thm1-1}]
Let  $u \in C^1(B_1(0))$, and $v$ be given by (\ref{2-1}).
By \cite[Proposition 2.2]{WangC}, we know that
$v \in H^1(\mathcal{C}_+)$, where
$\mathcal{C}_+=\{(t, \theta) \in \mathbb{R} \times S^{N-1} : t >0\}$.
Denote $\mathcal{S}:= 0\times S^{N-1}$, we have
\begin{equation}
\begin{aligned}
\int_{B_1} |x|^{-2a}|\nabla u|^2
&=\int_{\mathcal{C}_+}(|\nabla_\theta v|^2 +(v_t+\frac {N-2-2a}{2} v)^2)
d \mu\\
&=\int_{\mathcal{C}_+}(|\nabla v|^2 +(N-2)v_t v+(\frac {N-2-2a}2)^2 v^2)
d \mu\\
&=\int_{\mathcal{C}_+}(|\nabla v|^2 +(\frac {N-2-2a}2)^2 v^2)d\mu+
\int_{\mathcal{C}_+}\frac {N-2-2a}2 (v^2)_td\mu,
\end{aligned}
\label{2-2}
\end{equation}
and
$$
\int_{\mathcal{C}_+}\frac {N-2-2a}2 (v^2)_td\mu = \int_0^\infty
\int_{\mathcal{S}_t } \frac {N-2-2a}2 (v^2)_t d\theta dt= -\frac
{N-2-2a}2  \int_\mathcal{S} v^2 d \theta,
$$
where $\mathcal{S}_t:=t  \times S^{N-1}$ and $d\mu =d\theta d t$.
Also, it is easy to check that
$$
\int_{\partial  B_1} u^2 d\theta = \int_{\partial
B_1}|x|^{-N+2+2a} v^2 d\theta  =\int_\mathcal{S} v^2 d\theta.
$$
Therefore, we have
$$
\int_{B_1}|x|^{-2a}|\nabla u|^2 +\frac {N-2-2a}2 \int_{\partial
B_1} u^2 d\theta  = \int_{\mathcal{C}_+}(|\nabla v|^2 +(\frac
{N-2-2a}2)^2 v^2)d\mu.
$$
On the other hand
$$
\int_{B_1} \frac {u^2}{|x|^{2(a+1)}} dx = \int_{\mathcal{C}_+}| v|^2 d \mu.
$$
It follows that
\begin{align*}
\frac {\int_{B_1}|x|^{-2a}|\nabla u|^2 dx +\frac {N-2-2a}2
\int_{\partial  B_1} u^2 d\theta }{\int_{B_1} \frac
{u^2}{|x|^{2(a+1)}} dx} &= \frac {\int_{\mathcal{C}_+}(|\nabla
v|^2 +(\frac {N-2-2a}2)^2 v^2)d\mu}
{\int_{\mathcal{C}_+}| v|^2 d \mu}\\
&> (\frac {N-2-2a}2)^2
\end{align*}
which yields (\ref{1-3}). The last inequality in the above
expression follows from $v$ being in $H^1(\mathcal{C}_+)$. The
inequality (\ref{1-4}) can be proved in the same spirit, and we
shall omit the details. \end{proof}


\begin{proof}[Proof of Theorem \ref{thm1-2}]
Without loss of generality, we can assume that $\partial
\Omega \subset B_1(0)$. For any $u(x)\in C^1(\Omega)$, let $\varphi$ be the transformation given by (\ref{2-1-1}), and $v(x)$ be given by (\ref{2-1}). Denote $\mathcal{C}_\omega= \varphi(\Omega)$. Thus $ \partial \mathcal{C}_\omega \subset \mathcal{C}_+$.
Similar to (\ref{2-2}), we have
\begin{align*}
\int_{\Omega}|x|^{-2a}|\nabla u|^2
&=\int_{\mathcal{C}_\omega}(|\nabla_\theta v|^2 +(v_t+\frac {N-2-2a}2 v)^2)d \mu
\\
&=\int_{\mathcal{C}_\omega}(|\nabla v|^2 +(\frac {N-2-2a}2)^2
v^2)d\mu+ \int_{\mathcal{C}_\omega } \frac {N-2-2a}2 (v^2)_t d\mu.
\end{align*}
But due to Green's formula
$$
\int_{\mathcal{C}_\omega}\frac {N-2-2a}2 (v^2)_td\mu =   \frac
{N-2-2a}2 \int_{\partial C_\omega} (v^2,0) \bar \eta d S_\omega,
$$
where $\bar \eta$ is the unit out norm vector of $\partial
C_\omega$ and $d S_\omega$ is the volume element on $\partial
C_\omega$. Then
\begin{align*}
 \big|\frac {N-2-2a}2 \int_{\partial C_\omega} (v^2,0) \bar \eta
dS_\omega\big|
&\leq  \frac {N-2-2a}2 \int_{\partial C_\omega} v^2 d S_\omega\\
&=  \frac {N-2-2a}2 \int_{\partial C_\omega}|x|^{(N-2-2a)}u^2(-\ln
|x|, x/|x|)
d S_\omega\\
&= \frac {N-2-2a}2 \int_{\partial \Omega} |x|^{-1 -2a} u^2 \\
&\leq  \frac {N-2-2a}2 C_0 \int_{\partial \Omega} u^2,
\end{align*}
where
$$
C_0= \begin{cases} (\max\{|x|: x\in \partial \Omega\})^{-1-2a}
&\mbox{if }-1-2a \geq 0,\\
 (\min\{|x|: x\in \partial \Omega\})^{-1-2a} & \mbox{if }-1-2a <0.
\end{cases}
$$
Therefore,
$$
\int_{\Omega}|x|^{-2a}|\nabla u|^2+\frac {N-2-2a}2 C_0
\int_{\partial \Omega} u^2  \ge \int_{\mathcal{C}_\omega}(|\nabla
v|^2 + (\frac {N-2-2a}2)^2 v^2)d\mu.
$$
On the other hand,
$$
\int_{\Omega} \frac {u^2}{|x|^{2(a+1)}} dx
= \int_{\mathcal{C}_\omega}| v|^2 d \mu.
$$
Above two inequalities yield (\ref{1-5}). (\ref{1-5-1}) can be proved
in the same spirit, and we shall omit details here.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1-3}]
We need to prove the inequality only for non-negative smooth
functions. Suppose that $u  \in C^1(\Omega)$ is a non-negative
function satisfying $u=0$ on $\Gamma_1$. Let $\Omega^*$ be the
ball centered at the origin which has the same volume as $\Omega$.
Let $u^*$ be the Schwartz symmetrization of $u$. Namely, we define
$$
u^*(x)= \sup \{ t: \mu(t) > \omega_N |x|^N\}, $$ where $\omega_N$
is the volume of the unit ball in $\mathbb{R}^N$, and $\mu(t)$ is
the Lebesgue measure of the set $\{x\in \Omega : u(x)>t\}$. Then,
it is well-known (see, e.g., Bandle \cite{Ba}) that
$$
\int_\Omega \frac {u^2}{|x|^2} \le \int_{\Omega^*} \frac {(u^*)^2}{|x|^2}.
$$
On the other hand,
from Zhu \cite{Zhu} (this is the place where we use the assumption on
$\Gamma_2$) we know that
$$
\int_{\Omega^*} |\nabla u^*|^2 \le 2^{2/N} \int_\Omega |\nabla u|^2.
$$
Since $u^*=0$ on $\partial \Omega^*$, we know from the standard Hardy
inequality that
$$
(\frac {N-2}2)^2 \int_{\Omega^*} \frac {(u^*)^2}{|x|^2} \le \int_{\Omega^*} |\nabla u^*|^2.
$$
These three inequalities yield Theorem \ref{thm1-3}.
\end{proof}

There might be a guess that for $\Omega$ satisfying the condition in Theorem
\ref{thm1-3},
$$
\inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}, u=0 \mbox{ on }\Gamma_1} I(u, \Omega)
$$
will  always be larger than or equal to $(N-2)^2/4$. However, we present
an example to show that this is not the case.

\subsection*{An example}
Let $\mathbb{R}^N_{-1}= \{(x', x_N) \in \mathbb{R}^N  :  x_N >-1\}$.
We are going to show that
$$
\inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}}
\frac {\int_{\mathbb{R}^N_{-1} } |\nabla u|^2}{\int_{\mathbb{R}^N_{-1}} \frac {u^2}{|x|^2}} <  \frac {(N-2)^2}4.
$$
Let $\varphi$ be the transformation given by (\ref{2-1-1}), and $v(x)$
be given by (\ref{2-1}) for $x \in \mathbb{R}^N_{-1} $.  Denote $\mathcal{C}_{-1}=
\varphi(\mathbb{R}^N_{-1})$. Then
$$
 \frac {\int_{\mathbb{R}^N_{-1} } |\nabla u|^2}{\int_{\mathbb{R}^N_{-1}}
 \frac {u^2}{|x|^2}}=  \frac {\int_{\mathcal{C}_{-1} } (|\nabla v|^2
 +(\frac {N-2}2)^2 v^2)d\mu + \frac {N-2}2
 \int_{\mathcal{C}_{-1} }(v^2)_t d\mu }{\int_{\mathcal{C}_{-1}} v^2 d\mu}.
$$
Now, we choose
$$
\tilde v(t, \theta)=
\begin{cases}
0, & t \le -R-R_0\\
(t+R+R_0)/R_0,   & -R-R_0 \le t \le -R\\
1, & -R \le t \le R\\
(R+R_0-t)/R_0,   &  R \le t \le R+R_0\\
0,  & t \ge R+R_0,
\end{cases}
$$
where $R_0 >4/(N-2)$, and $R$ will be chosen sufficiently large.
A simple calculation shows that
\begin{multline*}
\int_{\mathcal{C}_{-1} } (|\nabla \tilde v|^2+ (\frac {N-2}2)^2 \tilde v^2)d\mu
 + \frac {N-2}2 \int_{\mathcal{C}_{-1} }(\tilde v^2)_t d\mu \\
 =\int_{\mathcal{C}_{-1} }  (\frac {N-2}2)^2 \tilde v^2 d\mu+
(\frac 3 2+o(1)) \frac {|S^{N-1}|}{ R_0} -(\frac 12 +o(1)) \frac {N-2}2 |S^{N-1}|,
\end{multline*}
where $o(1) \to 0$ as $R\to \infty$. Let
$\tilde u(x)=|x|^{-\frac {N-2}2} \tilde v(-\ln|x|, \frac x{|x|})$.
It is easy to see that $\tilde u \in \mathcal{D}^{1,2}(\mathbb{R}^N) $. Thus for sufficiently large $R$
$$
\inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}} \frac {\int_{\mathbb{R}^N_{-1} } |\nabla u|^2}{\int_{\mathbb{R}^N_{-1}} \frac {u^2}{|x|^2}} \le \frac {\int_{\mathbb{R}^N_{-1} } |\nabla \tilde u|^2}{\int_{\mathbb{R}^N_{-1}} \frac {\tilde u^2}{|x|^2}}< (\frac {N-2}2)^2.
$$
When $\Omega={\mathbb{R}^N_{-1}} \cap (\mathop{\rm supp} \tilde u)^o $,
where $(\mathop{\rm supp} \tilde u)^o $ is the set of interior points of
$\mathop{\rm supp} \tilde u$, we easily see that
$$
\inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}, u=0 \ on \ \Gamma_1} I(u, \Omega)
<(N-2)^2/4.
$$
%\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1-4}]
Let $u_m \in \mathcal{D}^{1,2}(\mathbb{R}^N) $ be a minimizing sequence such that
$ \int_{\Omega} u_m^2/|x|^2 =1$. Then
$$
\int_\Omega |\nabla u_m|^2 \to \inf_{u \in \mathcal{D}^{1,2}(\mathbb{R}^N) \setminus \{0\}} I(u, \Omega)
:= \xi <(\frac {N-2}2)^2.
$$
If $\Omega$ is bounded,
$\int_{\Omega} u_m^2 \le C \int_{\Omega} u_m^2/|x|^2 =C$.
Thus $u_m$ is uniformly bounded in $H^1(\Omega)$. It follows that
$u_m \to \bar u$ weakly in $H^1(\Omega)$, thus
\begin{equation}
\xi +o_m(1)= \int_\Omega |\nabla u_m|^2
=\int_\Omega |\nabla u_m -\nabla \bar u|^2 +\int_\Omega
|\nabla \bar u|^2 +o_m(1),
\label{5-2}
\end{equation}
where $o_m(1) \to 0$ as $m \to \infty$.
If $\Omega $ is unbounded, since $\partial \Omega$ is bounded,
$\Omega$ contains the exterior of some ball domain. Then we may check
that $X= \{ u|_{\Omega}\;|\; u\in \mathcal{D}^{1,2}(\mathbb{R}^N)\}$ is a Hilbert space
with a norm $\|u\|^2_{X}=\int_{\Omega} |\nabla u|^2dx$, this is due to the
Sobolev inequality.
Then $u_m$ is bounded in $X$ and has a weak limit $\bar{u}$
and we again have (\ref{5-2}).
By the weak convergence of $ \frac{u_m}{|x|}$ to
$ \frac{\bar{u}}{|x|}$ in $L^2(\Omega)$, we also have
\begin{equation}
\int_\Omega \frac {\bar u^2}{|x|^2}
=\int_\Omega \frac {u_m^2}{|x|^2} - \int_\Omega \frac {(u_m-\bar u)^2}{|x|^2} +o_m(1).
\label{5-3}
\end{equation}
Therefore, we have
\begin{align*}
 \xi +o_m(1)
& =\int_\Omega |\nabla u_m -\nabla \bar u|^2 +\int_\Omega |\nabla \bar u|^2
+o_m(1) \quad\mbox{ by  (\ref{5-2})}\\
&\ge (\frac {N-2}2)^2 \int_\Omega \frac {|u_m-\bar u|^2}{|x|^2} -C_h
\int_{\partial \Omega } |u_m-\bar u|^2+\xi \int_\Omega \frac {\bar u^2}{|x|^2}
+o_m(1) \ \\
& \mbox{(by Theorem \ref{thm1-2}  and  the definition  of $\xi$)} \\
& \ge (\frac {N-2}2)^2 \int_\Omega \frac {|u_m-\bar u|^2}{|x|^2} +
\xi \int_\Omega \frac {\bar u^2}{|x|^2} +o_m(1)\\
&\mbox{(by  Sobolev  embedding)}\\
&=[(\frac {N-2}2)^2 - \xi] \int_\Omega \frac {|u_m-\bar u|^2}{|x|^2}
+\xi +o_m(1) \quad\mbox{(by  (\ref{5-3}))},
\end{align*}
which implies
$\int_\Omega \frac {|u_m-\bar u|^2}{|x|^2}  \to 0 \quad\mbox{as } m
\to \infty$.
It follows that $\int_\Omega |\bar u|^2/|x|^2 =1$, thus
 $\bar u$ is the minimizer of $I(u, \Omega)$.
\end{proof}

\noindent{\bf Notes Added in Proof.} After this paper was accepted
we found a paper by Adimurthi: Hardy-Sobolev inequality in
$H^1(\Omega)$ and its applications, Comm. Contem. Math., 4 (2002),
409-434,  which contains related results and uses different
methods.
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\end{document}