\section{Introduction  and main results}



For $n \geq 3$  Hardy inequality states,  that for any $u \in 
C^{\infty}_{c}({\mathbb{R}}^n)$ there holds
\[
\int_{{\mathbb{R}}^n} |\nabla u|^2 dx 
\geq  \Big( \frac{n-2}{2}\Big)^2  \int_{{\mathbb{R}}^n} \frac{u^2}{|x|^2} dx \ ,
\]
where $\frac{(n-2)^2}{4}$ is the best constant. On the other hand Sobolev inequality reads as follows
\[
\int_{\Omega} |\nabla u|^2 dx \geq S_{n} \bigg( \int_{\Omega}  |u|^{\frac{2n}{n-2}}     dx \bigg)^{\frac{n-2}{n}}, \qquad u\in C^{\infty}_c(\Omega)
\]
where $S_n=
 \pi n(n-2) \left( \Gamma(\frac{n}{2})/ \Gamma(n) \right)^{\frac2n}$ is the best Sobolev constant for any domain $\Omega\subset{\mathbb{R}}^n$.
 
 
There are 
various improved versions of either  Hardy or Sobolev inequalities   in the case  of a bounded domain $\Omega$ containing the origin see e.g
\cite{BV,VZ,Te,ACR,FT2002,BFT1,BFT2,FMoT,BN,BL}. We mention in particular the
following sharp Hardy--Sobolev inequality  from \cite{FT2002,AFT} that combines both inequalities
\[
\int_{\Omega} |\nabla u|^2 dx 
\geq   \Big( \frac{n-2}{2}\Big)^2
 \int_{\Omega} \frac{u^2}{|x|^2} dx \\
+ (n-2)^{-\frac{2(n-1)}{n}}  \, S_{n} \bigg( \int_{\Omega}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \bigg)^{\frac{n-2}{n}}
\]
for all $u \in C^{\infty}_{c}(\Omega)$. Here $X_1=X_1(|x|/ D)$, with 
\[
X_1(t) = \frac{1}{1- \ln t}, \;\;\; t \in (0,1), \quad\quad D:=\sup_{x \in \Omega}|x| \ .
\]

A natural question is what are the analogues of Hardy and Hardy--Sobolev inequalities in case the origin is on the boundary of $\Omega$ instead of being in the interior. As we shall see, contrary to the previous case, the geometry of $\Omega$ plays an important role. In the simplest case of the  half space ${\mathbb{R}}^n_{+}=\{(x',x_n):~x_n>0\}$, Hardy 
inequality with best constant reads (cf. \cite{N, Fil3})
\begin{equation}\label{in10}
\int_{{\mathbb{R}}^n_{+}} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{{\mathbb{R}}^n_{+}} \frac{u^2}{|x|^2} dx, ~~~~~~~~~~\forall  u \in 
C^{\infty}_c({\mathbb{R}}^n_+)\ .
\end{equation}
In the more general case where the domain is a cone ${\mathscr C}$ with its vertex at the origin  the sharp Hardy inequality reads
(cf. \cite{N})
\begin{equation}
\label{cone}
\int_{{\mathscr C} } |\nabla u|^2 dx \geq  \left( \Big( \frac{n-2}{4}\Big)^2  + \mu_1(\Sigma) \right) \int_{{\mathscr C} } \frac{u^2}{|x|^2} dx,  \qquad \forall  u \in C_c^{\infty}({\mathscr C} ) \ ,
\end{equation}
where $ \Sigma = {\mathscr C} \cap S^{n-1}$ and $\mu_1(\Sigma)$ is the first Dirichlet eigenvalue of the Dirichlet Laplace-Beltrami operator on $\Sigma$.
 
If on the other hand, the origin is on the boundary of a smooth near zero domain, then, related types of problems have been studied in \cite{GK, GR1, CL, GR2}. More precisely the following minimization problem has been considered for $0<s<2$ and $n \geq 4$,
 \begin{equation}\label{13}
 \mu_{s}(\Omega) = \inf_{u \in H^1_{0}(\Omega)} \frac{\int_{\Omega} |\nabla u|^2 dx }{\bigg(\int_{ \Omega}  \frac{|u|^{\frac{2(n-s)}{n-2}}}{|x|^s}     dx \bigg)^{\frac{n-2}{n-s}}}  \ ,
 \end{equation}
and it was established that the geometry of $\Omega$ around zero plays an important role. In particular if the mean curvature at zero is negative then $ \mu_{s}(\Omega) < \mu_{s}({\mathbb{R}}^n_{+})$
 and there exists a minimizer for (\ref{13}). In the limit case $s=2$ the infimum  $ \mu_{2}(\Omega)$ is the best Hardy  constant and under certain geometric assumptions on $\Omega$ has been studied in \cite{Ccras1,Ccras2,Cfa,Csiam,FM1}.

 
  In \cite{F1} it was realized  that    the geometry  plays no role for the local best Hardy constant. That is,  
  for $r>0$ small enough if  we denote by $B_r$ the ball of radius $r$ centered at the origin, then for a smooth near zero domain $\Omega$ one has
 \begin{equation}\label{fa}
\int_{\Omega \cap B_r} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega \cap B_r} \frac{u^2}{|x|^2} dx, ~~~~~~~ u \in C^{\infty}_{c}(\Omega \cap B_r) \ ,
\end{equation}
which in particular implies the existence of a constant $\lambda \geq 0$
such that
\begin{eqnarray*}
\lambda  \int_{\Omega} u^2 dx  + \int_{\Omega} |\nabla u|^2 dx 
\geq  \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx , \quad u \in C^{\infty}_{c}(\Omega) \ .
\end{eqnarray*}
 
The first question we raise in this work  is to find a more quantitative result
 that connects the local inequality (\ref{fa}) to the global inequality in the half space (\ref{in10}). To state our first result we denote by $B_{\rho}(x_0)$ the ball of radius $\rho$ centered at $x_0$ and simply by  $B_{\rho}$ in case the ball is centered at the  origin; we also denote by ${\cal C} A$ the complement of a set $A\subset{\mathbb{R}}^n$.

Throughout this work $\Omega \subset {\mathbb{R}}^n$, $n \geq 2$, is a bounded domain with $0 \in \partial \Omega$  satisfying an exterior ball condition
at zero, that is there exists a ball
\[B_{\rho}(- \rho e_n) \subset {\cal C} \overline{\Omega} \ .
\]
 We also denote 
\[
D:=\sup_{\Omega}|x|.
\]


\begin{theorem}
\label{b}
 There exists a positive  constant $\tau_n$ depending only on $n$ such that
if  the radius of the exterior ball satisfies $\rho\geq D/ \tau_n$ then
\begin{equation}
\int_{\Omega} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx \ ,
\label{eq:ioa}
\end{equation}
for all $u \in C^{\infty}_{c}(\Omega)$. 
If in addition $\Omega$ satisfies an interior ball condition at $0$ then the constant $n^2/4$ is sharp. 
\end{theorem}






Thus in the case of a smooth (near zero) domain $\Omega$, if the exterior ball at zero is large enough compared to the size of $\Omega$ then the Hardy constant is $n^2/4$. If however the (largest) exterior ball is not large enough, at the end of Section \ref{sec3} we present an Example 
where the Hardy constant is smaller than $n^2/4$.




We next improve Hardy inequality by adding a Sobolev term:

\begin{theorem}
\label{c}
Let   $n \geq 3$. There exist positive constants $\sigma_n $ and $C_n$  that depend only on $n$ such that, if the radius of the exterior ball satisfies  $\rho\geq D/\sigma_n$ the following holds true: 
\[
\int_{\Omega} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx \\
+  C_n \bigg( \int_{\Omega}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \bigg)^{\frac{n-2}{n}},
\]
for all $u \in C^{\infty}_{c}(\Omega)$. Here $X_1=X_1(|x|/ 3D)$.
If in addition $\Omega$ satisfies an interior ball condition at $0$ then 
the exponent $(2n-2)/(n-2)$ of $X_1$ is sharp.
\end{theorem}

If  the radius of the exterior ball is small then there exists  a non negative constant $\lambda$ (that depends on $\Omega$) so that we have 
\[
\lambda \int_{\Omega} u^2 dx + \int_{\Omega} |\nabla u|^2 dx 
\geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx \\
+  C_n \Big( \int_{\Omega}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \Big)^{\frac{n-2}{n}}, ~~~~u \in C^{\infty}_{c}(\Omega) \ ;
\]
for the precise statement see Theorem \ref{4c}.

Under the assumptions of Theorem \ref{c}, a simple application of Holder's inequality yields that for any $\alpha>2$  there exists a positive constant $c(\alpha, \Omega)$ such  that
\[
\int_{\Omega} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx \\
+  c(\alpha, \Omega)  \int_{\Omega} \frac{X_1^{\alpha}}{|x|^2}u^{2} dx , ~~~~~u \in C^{\infty}_{c}(\Omega) \ .
\]
If $c(\alpha, \Omega)$ is the best constant then this inequality cannot be further improved, see Theorem \ref{d}. On the other hand, as we shall see,  in the limiting case  $\alpha=2$ the inequality  is also true, that is
\[
\int_{\Omega} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx \\
+  \frac14  \int_{\Omega} \frac{X_1^{2}}{|x|^2}u^{2} dx , ~~~~~u \in C^{\infty}_{c}(\Omega) \ ,
\]
and  the constant $1/4$ is sharp. In contrast with the case $\alpha>2$ this inequality can be further improved.

This is a particular case of a more general situation where one has a non negative potential $V$  that for some $\lambda$ non negative  and some sharp  positive constant $C$ the following inequality is true:
\begin{equation}
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx +C \int_{\Omega} Vu^2dx, \quad u \in  C^{\infty}_{c}(\Omega) \ .
\label{in301}
\end{equation}


In Section \ref{section:impr} we characterize maximal potentials, that is potentials $V$ such that (\ref{in301}) cannot be improved, with $C$ being the best constant for (\ref{in301}); such examples are the subcritical potentials, see Definition \ref{def:2}.
The main result of Section 5 is Theorem \ref{d}. We note that this description of maximal potentials is analogues to the description 
in \cite{Te,FT2002} for the interior point singularity case. 

In Section \ref{sub:log:impr} we consider the problem of successively improving Hardy inequality by critical potentials.
Before stating our result we first define the iterated logarithms (cf. \cite{FT2002})
\[
X_{k+1}(t) = X_k(X_1(t)), \qquad t\in (0,1],~~~~k=1,2,\ldots
\]
One can check that for $t \in (0,1)$ the series $ \sum_{i=1}^{\infty} X_1(t)X_{2}(t)\ldots X_i(t)$ converges
(see the proof of Lemma 6.3 in \cite{FT2002} or the Appendix in \cite{De}) and
that it is a strictly increasing function of $t$. We denote by $\kappa$ the unique $\kappa >1$ for which
\begin{equation}
\sum_{i=1}^{\infty}X_1(1/\kappa)\ldots X_i(1/\kappa)=\frac{1}{4}.
\label{kappa}
\end{equation}
We then have 
\begin{theorem}
\label{f}
 There exists $\sigma_n >0$ that depends only on $n$ such that if the radius of the exterior ball satisfies  $\rho\geq D /\sigma_n$ the following holds true: 
\begin{eqnarray*}
 \int_{\Omega} |\nabla u|^2 dx
&\geq &   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx + \frac{1}{4} \sum_{i=1}^{\infty} \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx   ,
\end{eqnarray*}
for all $u \in C^{\infty}_{c}(\Omega)$; here $X_i=X_i(|x|/ (3\kappa D))$.
If in addition $\Omega$ satisfies an interior ball condition at $0$ then the constants $1/4$ are sharp  at each step, that is
\[
\inf_{u \in C^{\infty}_{c}(\Omega)} \frac{\int_{\Omega} |\nabla u|^2 dx -
\frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx }
{ \int_{\Omega} \frac{u^2}{|x|^2}X_1^2 dx} =  \frac{1}{4} \ ,
\]
and for each $m=2,3,\ldots $
\[
\inf_{u \in C^{\infty}_{c}(\Omega)} \frac{\int_{\Omega} |\nabla u|^2 dx -
\frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx - \frac{1}{4} \sum_{i=1}^{m-1} \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx}
{ \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_m^2 dx} =  \frac{1}{4} \ .
\]
\end{theorem}
We also have the Hardy-Sobolev analogue:
\begin{theorem}
\label{g}
Let  $n \geq 3$. There exist positive constants $\sigma_n $ and $C_n$  that depend only on $n$ such that, if the radius of the exterior ball satisfies  $\rho\geq D/\sigma_n$ then   for any $m\in{\mathbb{N}}$ the following holds true:
\[
 \int_{\Omega} |\nabla u|^2 dx  \geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx +
 \frac{1}{4} \sum_{i=1}^m \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx 
+  C_n \left( \int_{\Omega}  (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}} ,
\]
for all $u \in C^{\infty}_{c}(\Omega)$; here $X_i=X_i( |x| / (3\kappa D))$. 
If in addition $\Omega$ satisfies an interior ball condition at $0$ then the exponents $(2n-2)/(n-2)$ of $X_i$ are also sharp.
\end{theorem}

We then proceed to obtain a characterization for maximal potentials in the context of logarithmic improvements; see Theorem \ref{h}.


Analogues of these theorems hold true if the domain $\Omega$ is a cone with vertex at zero and Section \ref{section:cone} is entirely
devoted to this. What is interesting in this case is that the Sobolev constant depends on the cone. As a typical result we mention here the following theorem that
refers to a bounded cone ${\mathscr C}_1 := {\mathscr C} \cap B_1$, the intersection of an infinite cone ${\mathscr C}$ with vertex at the origin with the unit ball $B_1$.
\begin{theorem}
Let $n\geq 3$. There exists a positive constant $C$ that depends only on $\Sigma$ such that
\begin{equation}
\int_{{\mathscr C}_{1}}|\nabla u|^2dx \geq \Big( \frac{(n-2)^2}{4} +\mu_1(\Sigma)\Big) \int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}dx    + C \left( \int_{{\mathscr C}_{1}}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}},
\label{in99}
\end{equation}
for all $u\in C^{\infty}_c({\mathscr C}_1)$; here $X_1=X_1(|x|)$. The exponent $(2n-2)/(n-2)$ of $X_1$ is the best possible. Moreover the best constant $C$ for inequality (\ref{in99}) satisfies the estimate
\begin{equation}
C \leq   C_n|\Sigma|^{\frac{2}{n}}
\label{in101}
\end{equation}
for some positive constant $C_n$ that depends only on $n$.
In particular the best constant $C$ of inequality (\ref{in99})
cannot be taken to be independent of $\Sigma$.
\label{i}
\end{theorem}


Finally, a similar analysis goes through if one has potentials with multiple singularities on the boundary, see  Theorem \ref{th72} for one such result.


Our results  about point singularities on the boundary, are analogous to the case of interior point singularities 
see \cite{FT2002, Te, AFT}. We note however that whereas in the interior singularity case the geometry of $\Omega$ is irrelevant, in this work  the curvature of the boundary introduces several technical difficulties even in the case of the plain Hardy inequality (\ref{eq:ioa}) as already noted in several recent  works see e.g. \cite{Ccras1, Ccras2, Cfa, Csiam, F1, FM1, GR1, GR2}.
 To overcome these difficulties we produce new improved inequalities in the flat case, see Lemmas
\ref{gera:lem:1.12im_ex}, \ref{gera:lem:123}, \ref{gera:lem_mlogs}
and then we use suitable conformal transformations thus obtaining sharp inequalities under the exterior ball assumption.




\section{Distance from  the vertex of a cone}
\label{section:cone}

In this section we consider the case of a finite cone and we obtain both homogeneous and nonhomogeneous improvements of the Hardy inequality (\ref{cone}). We pay particular attention to the special case where the cone is the half ball $B^{+}_{R}$. In this case the  estimates we obtain are stronger than in the case of a general cone and play a crucial role in our subsequent  analysis.

Let $\Sigma  \subset S^{n-1}$ be a domain in $S^{n-1}$ (that is a set that is open and connected in the relative topology)
with Lipschitz boundary. Let $\mu_k=\mu_k(\Sigma)$ be the $k$th Dirichlet eigenvalue of the Laplace-Beltrami operator on $\Sigma$
and let $\phi_k$ be a corresponding eigenfunction that is,
\[
\darr{- \Delta_{S^{n-1}} \phi_k(\omega)  =  \mu_k \phi_k(\omega),\;\; }{ \omega \in \Sigma \ ,} 
{\phi_k \big|_{\partial  \Sigma}   = 0  .}{}
\]
We may assume that $\{\phi_k\}$ is a complete orthonormal system in $L^2(\Sigma)$. We note that $\mu_1$ is a simple eigenvalue and we take $\phi_1$ to be positive.

We define
\[
{\mathscr C} = \{ x \in {\mathbb{R}}^n \smallsetminus \{0\}: ~~\frac{x}{|x|} \in \Sigma \} \, , \qquad
{\mathscr C}_{1} = {\mathscr C}  \cap B_1 = \{ x \in {\mathbb{R}}^n \smallsetminus \{0\}: ~~\frac{x}{|x|} \in \Sigma \, ,
|x|<1  \} \ .
\]
{\bf \em Proof of Theorem \ref{i}.} Let $u\in C^{\infty}_c({\mathscr C}_1)$ be given and let
\[
u(x)=\sum_{k=1}^{\infty}u_{k}(r)\phi _{k}(\omega)
\]
be its decomposition into the spherical harmonics of $\Sigma$. We then have
\[
u_k(r)=\int_{\Sigma} u(x)\phi_k(\omega)dS(\omega).
\]
Let $\omega_{n-1}$ denote the surface measure of the unit sphere $S^{n-1}$. Throughout this proof for any radial function $G$ (which sometimes shall be written as $G(x)$ and sometimes as $G(r)$) we shall use the notation
\[
\int_{B_1}G(x)dx =\omega_{n-1}\int_0^1 G(r)r^{n-1}dr. 
\]
It then easily follows that
\begin{eqnarray*}
\int_{{\mathscr C}_1}|\nabla u|^2dx &=& \frac{1}{\omega_{n-1}} \int_{B_1} \Big( |\nabla u_1|^2 +\mu_1\frac{u_1^2}{|x|^2} \Big) dx 
 + \frac{1}{\omega_{n-1}}\sum_{k=2}^{\infty}\int_{B_1} \Big( |\nabla u_k|^2 +\mu_k\frac{u_k^2}{|x|^2}\Big)dx \\
&=& \frac{1}{\omega_{n-1}}\int_{B_1} \Big( |\nabla u_1|^2 +\mu_1\frac{u_1^2}{|x|^2} \Big) dx 
+ \int_{{\mathscr C}_1}|\nabla (u-u_1 \phi_1)|^2dx \, .
\end{eqnarray*}
Moreover for any bounded radial function $G$ we have
\[
\int_{{\mathscr C}_1}Gu^2 dx  =	\frac{|\Sigma|}{\omega_{n-1}}\int_{B_1}Gu_1^2 dx  
+ \int_{{\mathscr C}_1}G(u-u_1 \phi_1)^2 dx \, .
\]
Therefore
\begin{eqnarray*}
&& \hspace{-1.5cm}\int_{{\mathscr C}_1}|\nabla u|^2dx - 
\Big( \frac{(n-2)^2}{4} +\mu_1\Big) \int_{{\mathscr C}_1}\frac{u^2}{|x|^2}dx \\
&=& \frac{1}{\omega_{n-1}} \bigg[ 
\int_{B_1} |\nabla u_1|^2dx   - \Big( \frac{n-2}{2} \Big)^2\int_{B_1}\frac{u_1^2}{|x|^2}dx \bigg]  \\
&& + \frac{1}{\omega_{n-1}}\sum_{k=2}^{\infty}\int_{B_1} \Big( |\nabla u_k|^2  +\mu_k\frac{u_k^2}{|x|^2}
-\Big( \frac{(n-2)^2}{4} +\mu_1\Big) \frac{u_k^2}{|x|^2}\Big)dx \\
&\geq&  \frac{1}{\omega_{n-1}} \bigg[ 
\int_{B_1} |\nabla u_1|^2dx   - \Big( \frac{n-2}{2} \Big)^2\int_{B_1}\frac{u_1^2}{|x|^2}dx \bigg]  \\
&& + \frac{1}{\omega_{n-1}} \cdot \frac{\mu_2-\mu_1}{ (\frac{n-2}{2})^2+ \mu_2}  
\sum_{k=2}^{\infty}\int_{B_1} \Big( |\nabla u_k|^2  +\mu_k\frac{u_k^2}{|x|^2}\Big) dx \\
&\geq& C_n \Big(\int_{B_1}X_1^{\frac{2(n-1)}{n-2}}|u_1|^{\frac{2n}{n-2}}dx\Big)^{\frac{n-2}{2}} + 
 \frac{\mu_2-\mu_1}{ (\frac{n-2}{2})^2+ \mu_2} \int_{{\mathscr C}_1} |\nabla (u-u_1\phi_1)|^2 dx \\
&\geq& \frac{C_n}{1+\mu_1}  \Big(\int_{{\mathscr C}_1}
X_1^{\frac{2(n-1)}{n-2}}|u_1\phi_1|^{\frac{2n}{n-2}}dx\Big)^{\frac{n-2}{2}} + 
 \frac{(\mu_2-\mu_1)S_n}{ (\frac{n-2}{2})^2+ \mu_2}
\Big(\int_{{\mathscr C}_1} |u-u_1\phi_1|^{\frac{2n}{n-2}} dx \Big)^{\frac{n-2}{2}}\\
&\geq& C(n,\Sigma)\Big(\int_{{\mathscr C}_1} X_1^{\frac{2(n-1)}{n-2}}|u|^{\frac{2n}{n-2}}dx\Big)^{\frac{n-2}{2}}.
\end{eqnarray*}
For the optimality of the exponent, suppose to the contrary that there exists $p<(2n-2)/(n-2)$ such that
\[
\int_{{\mathscr C}_1}|\nabla u|^2dx \geq \Big( \frac{(n-2)^2}{4} +\mu_1(\Sigma)\Big) \int_{{\mathscr C}_1}\frac{u^2}{|x|^2}dx    + C \left( \int_{{\mathscr C}_1}  X_1^{p}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}},
\]
for all $u\in C^{\infty}_c({\mathscr C}_1)$. Considering functions $u$ of the form $u(x)=v(r)\phi_1(\omega)$ with
$v(1)=0$ we obtain that any such $v$ satisfies
\begin{equation}
\int_{B_1}|\nabla v|^2 dx \geq \Big( \frac{n-2}{2}\Big)^2  \int_{B_1}\frac{v^2}{|x|^2}dx 
+    C \left( \int_{B_1}  X_1^{p}|v|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}}.
\label{123}
\end{equation}
This is a contradiction since the best exponent of $X_1$ in (\ref{123}) is $2(n-1)/(n-2)$; see \cite{FT2002}.

To prove estimate (\ref{in101})  we test inequality (\ref{in99}) with a function of the form
$u(x)=v(r)\phi_1(\omega)$. Then an easy calculation gives
\begin{eqnarray*}
C&\leq& \frac{\int_{{\mathscr C}_1}|\nabla u|^2dx - \Big( \frac{(n-2)^2}{4} +\mu_1\Big) \int_{{\mathscr C}_1}\frac{u^2}{|x|^2}dx}
{\Big(\int_{{\mathscr C}_1}|u|^{\frac{2n}{n-2}} X_1^{\frac{2n-2}{n-2}}dx \Big)^{\frac{n-2}{n}}} \\
&=&   \frac{\frac{1}{\omega_{n-1}} \int_{B_1} \Big[ |\nabla v|^2dx - \Big( \frac{n-2}{2} \Big)^2 \frac{v^2}{|x|^2}\Big] dx}
{\Big(\frac{1}{\omega_{n-1}}\int_{B_1}|v|^{\frac{2n}{n-2}} X_1^{\frac{2n-2}{n-2}}dx \Big)^{\frac{n-2}{n}}
\Big( \int_{\Sigma}|\phi_1|^{\frac{2n}{n-2}} dS \Big)^{\frac{n-2}{n}}}
\end{eqnarray*}
Minimizing with respect to $v$ (see \cite[Theorem B]{AFT}) we conclude that
\[
C\leq \frac{\omega_{n-1}^{-\frac{2}{n}}(n-2)^{-\frac{2}{n}}S_n }{\Big(\int_{\Sigma}|\phi_1|^{\frac{2n}{n-2}}dS\Big)^{\frac{n-2}{n}}}.
\]
By the normalization of $\phi_1$ and H\"{o}lder inequality we conclude that
\[
C\leq \omega_{n-1}^{-\frac{2(n-1)}{n}}(n-2)^{-\frac{2}{n}}S_n |\Sigma|^{\frac{2}{n}},
\]
which concludes the proof. $\hfill\Box$


In a similar fashion we obtain
\begin{theorem}
Let $n\geq 3$. There exists a constant $C$ that depends only on $\Sigma$ such that for any $m\in{\mathbb{N}}$
\begin{eqnarray}
\int_{{\mathscr C}_{1}}|\nabla u|^2dx &\geq&
\Big( \frac{(n-2)^2}{4} +\mu_1(\Sigma)\Big)\int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}dx   + \frac{1}{4}\sum_{i=1}^m
\int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx  \nonumber \\
&&+ C\left( \int_{{\mathscr C}_{1}}  (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}},
\label{gera:eq:xxxx_cone}
\end{eqnarray}
for all $u\in C^{\infty}_c({\mathscr C}_1)$; here $X_i=X_i(|x|)$. Each constant $1/4$ is the best possible, that is,
\[
\inf_{u \in C^{\infty}_{c}({\mathscr C}_1) } \frac{\int_{{\mathscr C}_1} |\nabla u|^2 dx - \big( (\frac{n-2}{2})^2 +\mu_1(\Sigma) \big) \int_{{\mathscr C}_1} \frac{u^2}{|x|^2} dx }
{ \int_{{\mathscr C}_1} \frac{u^2}{|x|^2}X_1^2 dx} =  \frac{1}{4} \ ,
\]
and for each $m=2,3,\ldots $
\[
\inf_{u \in C^{\infty}_{c}({\mathscr C}_1)} \frac{\int_{{\mathscr C}_1} |\nabla u|^2 dx -
\big( (\frac{n-2}{2})^2 +\mu_1(\Sigma) \big)  \int_{{\mathscr C}_1} \frac{u^2}{|x|^2} dx - \frac{1}{4} \sum_{i=1}^{m-1} \int_{{\mathscr C}_1} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx}
{ \int_{{\mathscr C}_1} \frac{u^2}{|x|^2}X_1^2\ldots X_m^2 dx} =  \frac{1}{4}  \, .
\]
The exponent $(2n-2)/(n-2)$
is the best possible. Moreover the best constant $C$ for inequality (\ref{gera:eq:xxxx_cone}) satisfies the estimate
\begin{equation}
C \leq   C_n|\Sigma|^{\frac{2}{n}}
\label{indep_log}
\end{equation}
for some positive constant $C_n$ that depends only on $n$.
In particular the best constant $C$ of inequality (\ref{gera:eq:xxxx_cone}) cannot be taken to be independent of $\Sigma$.
\label{thm_hardy_sob_BR_cone}
\end{theorem}
{\bf \em Proof.} Arguing as in the proof of Theorem~\ref{i} we arrive at

\begin{eqnarray*}
&& \int_{{\mathscr C}_{1}}|\nabla u|^2dx - 
\Big( \frac{(n-2)^2}{4} +\mu_1\Big) \int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}dx
-\frac{1}{4}\sum_{i=1}^m\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx \\
&=& \frac{1}{\omega_{n-1}} \bigg[ 
\int_{B_1} |\nabla u_1|^2dx   - \Big( \frac{n-2}{2} \Big)^2\int_{B_1}\frac{u_1^2}{|x|^2}dx
-\frac{1}{4}\sum_{i=1}^m\int_{B_1}\frac{u_1^2}{|x|^2}X_1^2\ldots X_i^2 dx
 \bigg]  \\
&& + \frac{1}{\omega_{n-1}}\sum_{k=2}^{\infty}\int_{B_1} \Big( |\nabla u_k|^2  +\mu_k\frac{u_k^2}{|x|^2}
-\Big( \frac{(n-2)^2}{4} +\mu_1\Big) \frac{u_k^2}{|x|^2} -\frac{1}{4}\sum_{i=1}^m 
\frac{u_k^2}{|x|^2} X_1^2\ldots X_i^2 \Big)dx \\
&\geq&  \frac{1}{\omega_{n-1}} \bigg[ 
\int_{B_1} |\nabla u_1|^2dx   - \Big( \frac{n-2}{2} \Big)^2\int_{B_1}\frac{u_1^2}{|x|^2}dx
- \frac{1}{4}\sum_{i=1}^m\int_{B_1}\frac{u_1^2}{|x|^2}X_1^2\ldots X_i^2 dx \bigg]  \\
&& + \frac{\mu_2-\mu_1}{2\big[(\frac{n-2}{2})^2+ \mu_2\big]} \frac{1}{\omega_{n-1}} 
\sum_{k=2}^{\infty}\int_{B_1} \Big( |\nabla u_k|^2  +\mu_k\frac{u_k^2}{|x|^2}\Big) dx \\
&\geq& C_n  \Big(\int_{B_1}(X_1\ldots X_{m+1})^{\frac{2(n-1)}{n-2}}|u_1|^{\frac{2n}{n-2}}dx\Big)^{\frac{n-2}{2}} + 
 \frac{\mu_2-\mu_1}{2\big[ (\frac{n-2}{2})^2+ \mu_2\big]} \int_{{\mathscr C}_1} |\nabla (u-u_1\phi_1)|^2 dx \\
&\geq& \frac{C_n}{1+\mu_1}  \Big(\int_{{\mathscr C}_1}
(X_1\ldots X_{m+1})^{\frac{2(n-1)}{n-2}}|u_1\phi_1|^{\frac{2n}{n-2}}dx\Big)^{\frac{n-2}{2}} + 
 \frac{(\mu_2-\mu_1)S_n}{2\big[ (\frac{n-2}{2})^2+ \mu_2\big]}
\Big(\int_{{\mathscr C}_1} |u-u_1\phi_1|^{\frac{2n}{n-2}} dx \Big)^{\frac{n-2}{2}}\\
&\geq& C(n,\Sigma)  \Big(\int_{{\mathscr C}_1} (X_1\ldots X_{m+1})^{\frac{2(n-1)}{n-2}}|u|^{\frac{2n}{n-2}}dx\Big)^{\frac{n-2}{2}}.
\end{eqnarray*}
For the optimality of the constants $1/4$ we make once again the choice $u(x)=v(r)\phi_1(\omega)$ to conclude that
\begin{eqnarray*}
&& \inf_{ C^{\infty}_c({\mathscr C}_1)}
\frac{  \int_{{\mathscr C}_1}|\nabla u|^2dx - \big( (\frac{n-2}{2})^2 +\mu_1(\Sigma) \big)
\int_{{\mathscr C}_1}\frac{u^2}{|x|^2}dx   - \frac{1}{4}\sum_{i=1}^{m-1}
\int_{{\mathscr C}_1}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx }{ \int_{{\mathscr C}_1}\frac{u^2}{|x|^2}X_1^2\ldots X_{m}^2
dx} \\
&\leq&  \inf_{ C^{\infty}_c(B_1)}
\frac{  \int_{B_1}|\nabla v|^2dx - \big( \frac{n-2}{2} \big)^2
\int_{B_1}\frac{v^2}{|x|^2}dx   - \frac{1}{4}\sum_{i=1}^{m-1}
\int_{B_1}\frac{v^2}{|x|^2}X_1^2\ldots X_i^2 dx }{ \int_{B_1}\frac{v^2}{|x|^2}X_1^2\ldots X_{m}^2
dx}\\
&=&\frac{1}{4},
\end{eqnarray*}
by \cite[Theorem 6.1]{FT2002}. The optimality of the exponent in the Sobolev term follows as before from the optimality of the corresponding exponent of the Hardy-Sobolev inequality for an interior point, \cite[Theorem A']{FT2002}. 

Finally to prove estimate (\ref{indep_log}) we once again test inequality (\ref{gera:eq:xxxx_cone}) with a function of the form
$u(x)=v(r)\phi_1(\omega)$. We then obtain
\begin{eqnarray*}
C&\leq& \frac{\int_{{\mathscr C}_1}|\nabla u|^2dx - \big( \frac{(n-2)^2}{4} +\mu_1\big) \int_{{\mathscr C}_1}\frac{u^2}{|x|^2}dx
- \frac{1}{4}\sum_{i=1}^{m}
\int_{{\mathscr C}_1}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx }{\Big(\int_{{\mathscr C}_1}|u|^{\frac{2n}{n-2}} (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}dx \Big)^{\frac{n-2}{n}}} \\
&=&   \frac{\frac{1}{\omega_{n-1}} \int_{B_1} \Big[ |\nabla v|^2dx - \big( \frac{n-2}{2} \big)^2 \frac{v^2}{|x|^2}
-\frac{1}{4}\sum_{i=1}^{m}\int_{B_1}\frac{v^2}{|x|^2}X_1^2\ldots X_i^2 \Big] dx}
{\Big(\frac{1}{\omega_{n-1}}\int_{B_1}|v|^{\frac{2n}{n-2}} (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}dx \Big)^{\frac{n-2}{n}}
\Big( \int_{\Sigma}|\phi_1|^{\frac{2n}{n-2}} dS \Big)^{\frac{n-2}{n}}}
\end{eqnarray*}
Minimizing with respect to $v$ (see \cite[Theorem B]{AFT}) and using H\"{o}lder inequality we conclude that
\[
C\leq \frac{\omega_{n-1}^{-\frac{2}{n}}(n-2)^{-\frac{2(n-1)}{n}}S_n }{\Big(\int_{\Sigma}|\phi_1|^{\frac{2n}{n-2}}dS\Big)^{\frac{n-2}{n}}}
\leq \omega_{n-1}^{-\frac{2}{n}}(n-2)^{-\frac{2}{n}}S_n |\Sigma|^{\frac{2}{n}},
\]
which concludes the proof. $\hfill\Box$




\begin{theorem}
Let $n\geq 2$. There holds
\[
\int_{{\mathscr C}_{1}}|\nabla u|^2dx \geq \bigg(   \Big( \frac{n-2}{2}\Big)^2 +\mu_1(\Sigma)\bigg)
\int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}dx   +\frac{1}{4}\sum_{i=1}^{\infty}\int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2dx 
\]
for all $u\in C^{\infty}_c({\mathscr C}_{1})$; here $X_i=X_i(|x|)$. Each constant $1/4$ is sharp.
\end{theorem}
{\bf \em Proof.} This follows from Theorem \ref{thm_hardy_sob_BR_cone} by letting $m\to +\infty$. The optimality of the constants $1/4$ has been established in Theorem~\ref{thm_hardy_sob_BR_cone}.
$\hfill\Box$







\subsection{Improved Hardy inequalities in half balls}


The case of half ball where  $\Sigma=S^{n-1}_+$ is of particular importance for our approach. In this case the Hardy constant becomes
\[
 \Big( \frac{n-2}{2} \Big)^2 +n-1 =\frac{n^2}{4},
\]
and the Sobolev constants of Theorems \ref{i} and \ref{thm_hardy_sob_BR_cone}
depend only on $n$. As a special case of the previous results
we have the following sharp inequalities for all functions $u\in C^{\infty}_c(B_R^+)$:
\begin{equation}
\int_{B_{R}^+}|\nabla u|^2dx \geq\frac{n^2}{4}\int_{B_{R}^+}\frac{u^2}{|x|^2}dx    + C_n \left( \int_{B_R^+}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}},
\label{gera:h_sob_br}
\end{equation}
\begin{eqnarray}
\int_{B_{R}^+}|\nabla u|^2dx &\geq&\frac{n^2}{4}\int_{B_{R}^+}\frac{u^2}{|x|^2}dx   + \frac{1}{4}\sum_{i=1}^m
\int_{B_R^+}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx  \nonumber \\
&&+ C_n \left( \int_{B_R^{+}}  (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}},
\label{gera:eq:xxxx}
\end{eqnarray}
\begin{equation}
\int_{B_{R}^+}|\nabla u|^2dx \geq\frac{n^2}{4}\int_{B_{R}^+}\frac{u^2}{|x|^2}dx   +\frac{1}{4}\sum_{i=1}^{\infty}\int_{B_{R}^+}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2dx  \ ,
\label{gera:eq:1.12im}
\end{equation}
where $X_i=X_i(|x|/ R)$.

In these inequalities the singularity lies on a flat part of the boundary. However if the boundary is not flat near the singularity, then curvature plays a role.  To overcome these difficulties, in the next three lemmas we establish stronger versions of (\ref{gera:h_sob_br}), (\ref{gera:eq:xxxx}) and (\ref{gera:eq:1.12im}) that will be used to prove Theorems \ref{c}, \ref{f} and \ref{g}.


We recall (cf.(\ref{kappa})) that $\kappa$ is the unique $\kappa >1$ for which $\sum_{i=1}^{\infty}X_1(1/\kappa)\ldots X_i(1/\kappa)=\frac{1}{4}$.
We also denote for $t \in (0,1)$,
\[
\eta(t):=  \sum_{i=1}^{\infty} X_1(t)X_2(t)\ldots  X_i(t), \quad 
B(t):=  \sum_{i=1}^{\infty} X_1^2(t)X_2^2(t)\ldots  X_i^2(t) \ .
\]
Using the identity
\[
 \frac{d}{dt}X_k (t) =\frac{1}{t}X_1(t)\ldots X_{k-1}(t)X_k^2(t)
\]
we easily obtain cf \cite{BFT2}
\[
\frac{d}{dt}\eta(t) =\frac{1}{2t} \big(\eta(t)^2 +B(t) \big)  \ , \quad t \in (0,1) \ .
\]
We next have.
\begin{lemma}
For any $R>0$ there holds
\begin{equation}
\int_{B_{R}^+}|\nabla u|^2dx \geq
\frac{n^2}{4}\int_{B_{R}^+}\frac{u^2}{|x|^2}dx   +\frac{1}{4}\sum_{i=1}^{\infty}\int_{B_{R}^+}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2dx  + \frac{1}{8R^{1/2}}\int_{B_{R}^+}\frac{u^2}{|x|^{3/2}}dx  ,
\label{gera:eq:1.12im_ex}
\end{equation}
for all $u\in C^{\infty}_c(B_{R}^+)$; here $X_i=X_i(|x|/\kappa R)$.
\label{gera:lem:1.12im_ex}
\end{lemma}
\noindent{\bf \em Proof.} Let ${\bf T }$ be a $C^1$ vector field in $B_R^+$ and $u\in C^{\infty}_c(B_R^+)$. We have
\[
\int_{B_R^+} {\rm div}{\bf T}\,  u^2 dx =- 2\int_{B_R^+} u\nabla u\cdot {\bf T}dx \leq \int_{B_R^+} \big( |\nabla u|^2 + |{\bf T}|^2 u^2 \big) dx 
\]
and therefore
\[
\int_{B_R^+}|\nabla u|^2dx \geq \int_{B_R^+} \Big( {\rm div} {\bf T} -|{\bf T}|^2 \Big)u^2 dx \; .
\]
We shall apply this for the vector field
\[
{\bf T} =\frac{n}{2}\frac{x}{|x|^2}  -\frac{e_n}{x_n} +\frac{\eta}{2}\frac{x}{|x|^2} +\frac{1}{2(R^{1/2}- |x|^{1/2})}\frac{x}{|x|^{3/2}} \, , \qquad (\eta=\eta(|x|/\kappa R)).
\]
We  have
\begin{eqnarray*}
{\rm div}{\bf T}&  =& \frac{n(n-2)}{2|x|^2} +\frac{1}{x_n^2} +\frac{n-2}{2|x|^2}\eta  +\frac{\eta^2 +B}{4|x|^2} \\
&& + \frac{n-\frac{3}{2}}{2(R^{1/2} -|x|^{1/2})|x|^{3/2}} + \frac{1}{4(R^{1/2} -|x|^{1/2})^2|x|} ,
\end{eqnarray*}
hence
\begin{eqnarray*}
{\rm div}{\bf T} -|{\bf T}|^2 &=&\frac{n^2}{4|x|^2} +\frac{1}{4|x|^2}\sum_{i=1}^{\infty}X_1^2\ldots X_i^2  
+\frac{\frac{1}{2}-\eta}{2(R^{1/2} -|x|^{1/2})|x|^{3/2}} \\
&\geq& \frac{n^2}{4|x|^2} +\frac{1}{4|x|^2}\sum_{i=1}^{\infty}X_1^2\ldots X_i^2  
+\frac{1}{8R^{1/2}|x|^{3/2}},
\end{eqnarray*}
where in the last inequality we used that $\eta \leq \frac14$, because of the choice of $\kappa$,
and the result follows. $\hfill\Box$



\begin{lemma}
Let $n\geq 3$. There exists a constant $C_n$ that depends only on $n$ such that for any $R>0$ there holds
\[
\int_{B_{R}^+}|\nabla u|^2dx \geq\frac{n^2}{4}\int_{B_{R}^+}\frac{u^2}{|x|^2}dx    + 
\frac{1}{16R^{1/2}}\int_{B_R^+}\frac{u^2}{|x|^{3/2}}dx 
 + C_n \left( \int_{B_R^+}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}} dx \right)^{\frac{n-2}{n}},
\]
for all $u\in C^{\infty}_c(B_R^+)$; here $X_1=X_1(|x|/R)$.
\label{gera:lem:123}
\end{lemma}
{\bf \em Proof.} The result follows by taking a convex combination of (\ref{gera:h_sob_br}) and (\ref{gera:eq:1.12im_ex}) and discarding the logarithmic terms that do not come with the sharp constant; see also the next lemma.
$\hfill\Box$


\begin{lemma}
Let $n\geq 3$ and $m\in{\mathbb{N}}$. There exists a constant $C_n$ that depends only on $n$ such that for all $R>0$ there holds
\begin{eqnarray*}
\int_{B_{R}^+}|\nabla u|^2dx &\geq&\frac{n^2}{4}\int_{B_{R}^+}\frac{u^2}{|x|^2}dx   + \frac{1}{4}\sum_{i=1}^m
\int_{B_R^+}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx  \\
&& \hspace{-2cm} + \frac{1}{16R^{1/2}}\int_{B_{R}^+}\frac{u^2}{|x|^{3/2}}dx 
+ C_n \left( \int_{B_R^+}  (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}},
\end{eqnarray*}
for all $u\in C^{\infty}_c(B_R^+)$; here $X_i=X_i(|x|/\kappa R)$.
\label{gera:lem_mlogs}
\end{lemma}
{\bf \em Proof.} This follows by taking a convex combination of (\ref{gera:eq:xxxx}) and (\ref{gera:eq:1.12im_ex}).
$\hfill\Box$







\section{Hardy inequality in  bounded  domains}
\label{sec3}



In this section  we provide the proof of Theorem \ref{b} and use an example to establish the necessity of a relatively large exterior ball assumption.
We also analyse the Hardy constant in the case of annuli (see Theorem \ref{a}).


We initially establish that $n^2/4$ is an upper bound for the Hardy constant under an interior ball condition.
\begin{lemma}
If $\Omega$  satisfies an interior ball condition at 0 then
for any $r>0$ we have
\[
\inf_{u\in C^{\infty}_c(\Omega \cap B_r)} 
\frac{\int_{\Omega \cap B_r} |\nabla u|^2 dx}{\int_{\Omega \cap B_r} \frac{u^2}{|x|^2} dx } \leq \frac{n^2}{4}.
\]
\label{lem:sharpness}
\end{lemma}
\noindent{\bf \em Proof.} Without loss of generality we may assume that the interior ball is $B_{\rho}(\rho e_n)$ and satisfies
$B_{\rho}(\rho e_n)  \subset \Omega \cap B_r$,
therefore it is enough to establish that
\[
\inf_{u\in C^{\infty}_c(B_{\rho}(\rho e_n))} 
\frac{\int_{B_{\rho}(\rho e_n)} |\nabla u|^2 dx}{\int_{B_{\rho}(\rho e_n)} \frac{u^2}{|x|^2} dx } \leq \frac{n^2}{4}.
\]
Using a scaling argument we find that this infimum is equal to
\[
\inf_{u\in C^{\infty}_c({\mathbb{R}}^n_+)} 
\frac{\int_{{\mathbb{R}}^n_+} |\nabla u|^2 dx}{\int_{{\mathbb{R}}^n_+} \frac{u^2}{|x|^2} dx } ,
\]
which is equal to $n^2/4$. $\hfill\Box$


We shall next prove a result about annuli. We use the notation 
 \[
{\mathscr D}( x_0 ; r_1,r_2) :=\{ x\in{\mathbb{R}}^n \; : \; r_1< |x-x_0|<r_2\}.
\]
or simply ${\mathscr D}( r_1,r_2)$ in case $x_0=0$. Also, $e_n$ shall denote the unit vector in the $x_n$ direction.


\begin{theorem}
\label{a}
Let $n\geq 2$ and let $\lambda_{\tau}$ denote the best constant for the Hardy inequality
\begin{equation}
\label{in22}
\hspace{-1.1cm}  \int_{{\mathscr D}( \rho ,\rho(1+\tau))} |\nabla u|^2 dx     \geq  
 \lambda_{\tau}  \int_{{\mathscr D}( \rho ,\rho(1+\tau))} \frac{u^2}{ |x- \rho e_n|^2} dx  \, , \qquad u\in C^{\infty}_c({\mathscr D}( \rho ,\rho(1+\tau))).
\end{equation}
There exists a constant $\tau_n>0$ which depends only on $n$ such that 

$\quad({\rm i})$ For all $0<\tau\leq \tau_n$ there holds $\lambda_{\tau}=n^2/4$

$\quad({\rm ii})$ For all $\tau >\tau_n$ there holds $\lambda_{\tau}<n^2/4$.

Moreover $\lambda_{\tau}$ is strictly decreasing in $(\tau_n,+\infty)$ and $\lim_{\tau\to+\infty}\lambda_{\tau}=
(n-2)^2/4$.
\end{theorem}

\noindent {\bf\em Proof.} It is enough to establish the result for $\rho=1$, the general case then follows by scaling.
To prove $({\rm i})$ it is enough to establish that for small enough $\tau>0$ we have inequality (\ref{in22}).
We apply (\ref{gera:eq:1.12im}) with $R=2$ where we place the singularity at $e_n$ and we obtain the inequality
\begin{equation}\label{hi:59}
\int_{B_1} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{B_1 } \frac{u^2}{|x-e_n|^2} dx  + \frac{1}{4}  \int_{B_1 } \frac{u^2}{|x-e_n|^2}X_1^2 dx,  
\quad\forall  u \in C^{\infty}_{0}(B_1)  \ , 
\end{equation}
where $X_1=X_1(|x-e_n|/2)$. Next we apply the Kelvin transform
\[
u(x)= |y|^{n-2} v(y), \qquad y = \frac{x}{|x|^2}   \ .
\]
Then by standard  calculations using the conformality of the Kelvin transform we have
\[
\int_{B_1} |\nabla u(x)|^2 dx = \int_{{\cal C} B_1} |\nabla v(y)|^2 dy \ ,
\]
and since 
\[
|x -e_n| = \frac{|y -e_n|}{|y|}  \ ,
\]
inequality (\ref{hi:59}) takes the equivalent form
\begin{equation}
\label{hi:63}
\int_{{\cal C} B_1} |\nabla v|^2 dy  \geq     \frac{n^2}{4}  \int_{{\cal C} B_1 } \frac{v^2}{|y|^2 |y-e_n|^2} dy 
   + \frac{1}{4}  \int_{{\cal C} B_1 } \frac{X_1^2}{|y|^2 |y-e_n|^2}v^2 dy  
\end{equation}
for all $v \in C^{\infty}_c( {\cal C} B_1)$, where $X_1=X_1( |y-e_n| / 2|y| )$.

It follows from (\ref{hi:63}) that for any $\tau>0$ and any $v \in C^{\infty}_c(B_{1+\tau} \smallsetminus B_{1})$ there holds
\[
 \int_{B_{1+\tau} \smallsetminus B_{1}} |\nabla v|^2 dy    \geq 
  \frac{n^2}{4}  \int_{B_{1+\tau} \smallsetminus B_{1}} \frac{v^2}{ |y-  e_n|^2} dy   
   +   \frac{1}{4} \int_{B_{1+\tau} \smallsetminus B_{1}} \frac{  n^2  -n^2 |y|^2 + X_1^2\left( \frac{|y- e_n|}{2|y|}  \right) }{|y|^2 |y-  e_n|^2}  v^2   dy  \ .
\]
To conclude the proof  it suffices  to show that the last term above is nonnegative for  small enough $\tau>0$. For this it is enough to have the inequality
\[
 X_1^2\Big( \frac{|y- e_n|}{2|y|}\Big) \geq n^2 \left(|y|^2-1 \right) \ ,  \qquad 1 < |y| < 1+ \tau \ .
\]
Writing $|y|=1+t$,  $0 < t < \tau$, we have that  $|y-e_n| \geq t$. Hence
\[
 X_1^2 \Big( \frac{|y- e_n|}{2|y|}\Big) \geq  X_1^2 \Big( \frac{t}{2(t+1)}\Big) \ ,
\]
and therefore it is enough to have
\[
X_1^2 \Big( \frac{t}{2(t+1)}\Big) \geq n^2 t (t+2),  \qquad 0<t<\tau \ .
\]
Since 
$\lim_{t \to 0^{+}} X_1^2(t)/t = +\infty$, the result follows.



We shall next establish that the set of all $\tau >0$ for which inequality (\ref{in22}) holds true 
is bounded and therefore we may define
\[
\tau_n= \sup\{ \tau >0 : \mbox{ inequality (\ref{in22}) holds true} \}.
\]
For this we first note that for $\tau>2$ we have the inclusion
\[
B_{\tau}\setminus B_2 \subset B_{1+\tau}(-e_n) \setminus B_1(-e_n).
\]
and therefore
\[
\inf_{C^{\infty}_c( {\mathscr D}( -e_n ; 1, 1+\tau ))} 
\frac{ \int_{{\mathscr D}( -e_n ; 1, 1+\tau )} |\nabla u|^2dx}{\int_{{\mathscr D}( -e_n ; 1, 1+\tau )}
\frac{u^2}{|x|^2}dx}
\leq
\inf_{C^{\infty}_c(B_{\tau}\setminus B_2 ) }
\frac{ \int_{B_{\tau}\setminus B_2 } |\nabla u|^2dx}{\int_{B_{\tau}\setminus B_2 } \frac{u^2}{|x|^2}dx}.
\]
Using the radial function
\[
 u(r) =r^{-\frac{n-2}{2}} \sin \Big( \frac{\ln (r/2) \, \pi}{ \ln(\tau/2)} \Big) , \quad\quad  2<r<\tau ,
\]
we easily see that the last infimum is equal to
$
\big( \frac{n-2}{2}  \big)^2 + \big( \frac{\pi}{\ln\frac{\tau}{2}}  \big)^2
$
and in particular it is smaller than $n^2/4$ if
\[
\tau >2e^{\frac{\pi}{\sqrt{n-1}}}.
\]
This implies  the existence of an $H^1_0$ minimizer (see e.g. \cite{GR1}, Theorem 4.2)  and therefore the strict monotonicity of $\lambda_{\tau}$ for $\tau>\tau_n$. The above computation also gives that
$\lim_{\tau\to +\infty}\lambda_{\tau}\leq (n-2)^2/4$; this combined with the standard Hardy inequality gives
$\lim_{\tau\to +\infty}\lambda_{\tau}= (n-2)^2/4$ thus concluding the proof of the theorem. $\hfill\Box$

We next have

{\bf\em Proof of Theorem \ref{b}:} As we shall see, the constant $\tau_n$ of Theorem \ref{b} is the same as that of Theorem \ref{a} above.
Since $\Omega \cap B_{\rho\tau_n} \subset {\mathscr D}(-\rho e_n ; \rho , \rho(1+\tau_n))$, it follows from Theorem \ref{a} that
\begin{equation}
\int_{\Omega \cap B_{\rho\tau_n}}  |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega \cap B_{\rho\tau_n}}  \frac{u^2}{|x|^2} dx  , \quad\quad
u\in C^{\infty}_c(\Omega \cap B_{\rho\tau_n}).
\label{555b}
\end{equation}
The assumption $\rho\tau_n \geq D$ implies $\Omega\subset B_{\rho\tau_n}$ and therefore (\ref{eq:ioa}) follows from (\ref{555b}).
The sharpness of the constant $n^2/4$ follows directly from Lemma \ref{lem:sharpness}. $\hfill\Box$




It is natural to ask whether the assumption of having a large exterior ball at zero is necessary in order to have the Hardy inequality with constant $n^2/4$. In the following example we will see that for small exterior balls inequality (\ref{eq:ioa}) fails.






{\bf Example.} Given $\rho \in (0,1/2)$ and $\theta\in (0,\pi/2)$ we define the domain
\[
{\mathscr A}_{\rho,\theta} =\{ x=(x',x_n)\in B_1 \; : \; x_n<\cot\theta |x'| \mbox{ and } |x-\rho e_n| >\rho \}.
\]
Let $\Omega$ be a domain containing ${\mathscr A}_{\rho,\theta}$ and having the same largest exterior ball at zero, namely $B(\rho e_n,\rho)$.


We denote by $\lambda_1(n,\theta)$ the first Dirichlet eigenvalue of the Laplace operator on the spherical cap
\[
\Sigma_{\theta}= \{ (x',x_n) \in S^{n-1} : x_n< \cot\theta \; |x'| \}.
\]
By monotonicity it follows that for $\theta <\pi/2$ we have $\lambda_1(n,\theta) < \lambda_1(n,\pi/2) =n-1$.
We shall prove that if
\begin{equation}
\rho < \frac{1}{2\cos \theta} e^{- \frac{\pi}{\sqrt{n-1-\lambda_1(n,\theta)}}} \ ,
\label{888}
\end{equation}
then
\begin{equation}\label{3.35}
\inf_{C^{\infty}_c(\Omega)} \frac{\int_{\Omega}|\nabla u|^2dx}{\int_{\Omega}
\frac{u^2}{|x|^2}dx} \leq
\inf_{C^{\infty}_c({\mathscr A}_{\rho,\theta})} \frac{\int_{{\mathscr A}_{\rho,\theta}}|\nabla u|^2dx}{\int_{{\mathscr A}_{\rho,\theta}}
\frac{u^2}{|x|^2}dx} <\frac{n^2}{4},
\end{equation}
that is the Hardy inequality with constant $n^2/4$ fails in $\Omega$  if the exterior ball at zero is small enough.



{\bf \em Proof of (\ref{3.35}).} We first note that
\[
 (B_1 \setminus B_{2\rho\cos\theta}) \cap  \{ (x',x_n) : x_n<\cot\theta \; |x'| \} \subset {\mathscr A}_{\rho,\theta}.
\]
Separating variables we then conclude that
\begin{eqnarray*}
\inf_{C^{\infty}_c({\mathscr A}_{\rho,\theta})} \frac{\int_{{\mathscr A}_{\rho,\theta}}|\nabla u|^2dx}{\int_{{\mathscr A}_{\rho,\theta}}
\frac{u^2}{|x|^2}dx} &\leq &
\inf_{f(2\rho\cos\theta)=f(1)=0}\frac{ \int_{2\rho\cos\theta}^1   f'(r)^2 r^{n-1} dr }{\int_{2\rho\cos\theta}^1   f(r)^2 r^{n-3} dr } +
\inf_{g(\theta)=g'(\pi)=0}\frac{ \int_{\theta}^{\pi}  \sin^{n-2}t  g'(t)^2 dt }{\int_{\theta}^{\pi}  \sin^{n-2}t  g(t)^2 dt }\\
&=& \Big( \frac{n-2}{2}\Big)^2 + \Big( \frac{\pi}{\ln (2\rho\cos\theta)}\Big)^2 +\lambda_1(n,\theta) 
< \frac{n^2}{4},
\end{eqnarray*}
by assumption (\ref{888}). 


\section{Improved Hardy-Sobolev inequalities for bounded domains}



In this section we shall establish improved Hardy and Hardy-Sobolev inequalities and in particular we will provide the proof of Theorem \ref{c}.
We start with the following lemma.
\begin{lemma}
Let $n\geq 3$. There exist $\sigma_n \in (0,1)$  and a constant $C_n>0$, both depending only on $n$, such that for all $\rho>0$
and all $r\leq \sigma_n\rho$ we have
\begin{equation}
\int_{{\cal C} B(\rho) \cap B(\rho e_n,r  )}|\nabla u|^2dx \geq \frac{n^2}{4}\int_{{\cal C} B(\rho) \cap B(\rho e_n, r)}
\frac{u^2}{|x-\rho e_n|^2}dx 
+ C_n \left( \int_{{\cal C} B(\rho) \cap B(\rho e_n, r)}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}} 
dx \right)^{\frac{n-2}{n}} , 
\label{gera:eq:1.33aaa} 
\end{equation}
for all $u\in C^{\infty}_c({\cal C} B(\rho) \cap B(\rho e_n, r ))$; here $X_1=X_1(|x-\rho e_n|/3r)$.
\label{gera:thm:general_sobolev}
\end{lemma}
{\bf \em Proof.} We establish (\ref{gera:eq:1.33aaa}) for $\rho=1$, the general case will then follow by scaling.
The map
\begin{equation}
S(v)= \frac{1}{|v +e_n|^2} (2v',1-|v|^2)
\label{gera:map:s}
\end{equation}
maps conformally ${\mathbb{R}}^n_+$ onto the unit ball $B_1$. We note that
\begin{equation}
|S(v)| =\frac{|v-e_n|}{|v+e_n|}.
\label{gera:S1}
\end{equation}
Composing $S$ with the Kelvin transform $K$ we obtain that the map
\begin{equation}
T(v) =(KS)(v)=  \frac{1}{ |v-e_n|^2}(2v',1-|v|^2) 
\label{gera:def_T}
\end{equation}
maps conformally ${\mathbb{R}}^n_+$ onto ${\cal C} B_1$.
The Jacobian determinant $JS(v)$ of $S$ can be computed explicitly and one finds
\[
|JS(v)| =\frac{2^n}{ |v +e_n|^{2n}} .
\] 
The Jacobian of the Kelvin map $K(y)$ is $|y|^{-2n}$ hence, using also (\ref{gera:S1}), the Jacobian of $T$ is
\begin{equation}
|JT(v)| = | JK(Sv)|  \,\, |JS(v)| = \frac{2^n |Sv|^{-2n}}{  |v+e_n|^{2n}} = \frac{2^n}{|v-e_n|^{2n}}.
\label{gera:jt1}
\end{equation}
Now,  simple computations give that $S^{-1}=S$ and therefore $T^{-1}=S^{-1}K^{-1}=SK$. From this we find
\[
T^{-1}(x)=\frac{1}{|x'|^2+(x_n+1)^2}(2x',|x|^2 -1)
\]  
and therefore
\begin{equation}
|T^{-1}(x)| =\frac{|x-e_n|}{|x+e_n|}.
\label{gera:T1112}
\end{equation}
Now let $r<1$ be fixed (this will be chosen later on)
and let $F\in C^{\infty}_c(T(B_r^+))$ be given. We define the function $G$ on $B_r^+$ by
\[
G(v) =F(T(v))  | (JT)(v)|^{\frac{n-2}{2n}}  = F(T(v)) \Big( \frac{2}{|v-e_n|^2} \Big) ^{\frac{n-2}{2}}.
\]
We then have by Lemma \ref{gera:lem:123},
\begin{equation}
\int_{B_r^+}|\nabla G|^2  dv \geq \frac{n^2}{4} \int_{B_r^+}\frac{G^2}{|v|^2} dv  +
 \frac{1}{16r^{1/2}}\int_{B_{r}^+}\frac{G^2}{|v|^{3/2}}dv  + C_n \left( \int_{B_r+}  X_1^{\frac{2n-2}{n-2}}|G|^{\frac{2n}{n-2}}  dv \right)^{\frac{n-2}{n}}
 ,  \label{gera:josbir1}
\end{equation}
where $X_1=X_1(|v|/r)$. We next change variables in (\ref{gera:josbir1}). 


We have
\[
G(v) = 2^{\frac{n-2}{2}} F(T(v))  |v-e_n|^{2-n}
\]
and therefore
\begin{eqnarray*}
|\nabla_v G(v)|^2 & =& 2^{n-2} \Big(  |\nabla F(T(v))|^2 |v-e_n|^{2(2-n)}
+ 2 |v-e_n|^{2-n}F(T(v)) \nabla (F(T(v)) \cdot \nabla |v-e_n|^{2-n} \\
&& +F(T(v))^2 | \nabla |v-e_n|^{2-n} |^2  \Big). 
\end{eqnarray*}
After integration over $B_r^+$ and a change of variables the first term turns out to be equal to $\int_{T(B_r^+)}|\nabla F|^2  dx$.
Integrating the other two terms yields
\begin{eqnarray*}
&&\hspace{-1.5cm}\int_{B_r^+}\Big(  
2 |v-e_n|^{2-n}F(T(v)) \nabla (F(T(v)) \cdot \nabla |v-e_n|^{2-n}
+F(T(v))^2 | \nabla |v-e_n|^{2-n} |^2  \Big)dv \\
&=&  \int_{B_r^+}\Big(  
 |v-e_n|^{2-n} \nabla (F(T(v))^2  \cdot \nabla |v-e_n|^{2-n}
+F(T(v))^2 | \nabla |v-e_n|^{2-n} |^2  \Big)dv \\
&=&  \int_{B_r^+} (F(T(v)))^2  \Big(  - {\rm div} \big(  |v-e_n|^{2-n}  \nabla |v-e_n|^{2-n} \big)
+ \big| \nabla |v-e_n|^{2-n} \big|^2  \Big)dv \\
&=&0.
\end{eqnarray*}
We thus conclude that
\[
\int_{B_r^+}|\nabla G|^2  dv  =  \int_{T(B_r^+)}|\nabla F|^2  dx \, .
\]
Using  (\ref{gera:jt1}) and (\ref{gera:T1112}) we also find that
\[
\int_{B_r^+}\frac{G^2}{|v|^2} dv = \int_{T(B_r^+)}\frac{4F^2}{|x-e_n|^2|x+e_n|^2} dx \, .
\]
The other two integrals in (\ref{gera:josbir1}) can similarly be transformed and we conclude that (\ref{gera:josbir1})
takes the form
\begin{eqnarray*}
 \int_{T(B_r^+)}|\nabla F|^2dx &\geq& \frac{n^2}{4} \int_{T(B_r^+)}\frac{4F^2}{|x-e_n|^2|x+e_n|^2}dx   + \frac{1}{16r^{1/2}} \int_{T(B_r^+)}\frac{4F^2}{|x-e_n|^{3/2} |x+e_n|^{5/2}}dx  \\
&& +C_n \left( \int_{T(B_r+)}  X_1^{\frac{2n-2}{n-2}}|F|^{\frac{2n}{n-2}}  dv \right)^{\frac{n-2}{n}},
\end{eqnarray*}
where $X_1=X_1( |x-e_n| /r|x+e_n|)$.
Now, it follows from (\ref{gera:T1112}) and some simple geometry that for any $r<1$
\[
T(B(r))=\{ x\in{\mathbb{R}}^n \, : \, |x-e_n| <  r|x+e_n|\} =B\big(\frac{1+r^2}{1-r^2} e_n , \frac{2r}{1-r^2}\big) \supset B(e_n, r),
\]
therefore
\begin{equation}
T(B_r^+)  \supset B_1^{c}  \cap B(e_n, r)  .
\label{inclusion}
\end{equation}
We will choose $\sigma_n \in (0,1)$ such that for all $r\leq\sigma_n$ and for all $x\in B_1^{c}\cap B(e_n, r)
\subset T(B_r^+)$ there holds
\[
\frac{n^2}{4}\frac{4}{|x-e_n|^2|x+e_n|^2} +  \frac{1}{16r^{1/2}}\frac{4}{|x-e_n|^{3/2} |x+e_n|^{5/2}}  
\geq \frac{n^2}{4|x-e_n|^2} ,
\]
or equivalently,
\begin{equation}
|x-e_n|^{1/2} \geq  n^2r^{1/2}|x+e_n|^{5/2} \Big( 1-\frac{4}{|x+e_n|^2}\Big).
\label{gef}
\end{equation}
Indeed, this is immediate for $|x+e_n| \leq 2$.  Assuming that $|x+e_n| > 2$ we 
set $t=|x-e_n|$. We then have $|x+e_n|\leq t+2$ and therefore (\ref{gef}) will follow provided
\[
n^2 r^{1/2} t^{1/2}(t+4)(t+2)^{1/2} \leq 1,
\]
for all $t\leq r$. Simple computations give that the last inequality holds true provided $ t\leq 1 / (75n^4r)$.
This will be true for all $t\leq r$ and all $r\leq\sigma_n$ if $\sigma_n$ is chosen as
\[
\sigma_n =\frac{1}{\sqrt{75}n^2}.
\]
Finally, the inequality $|x+e_n| \leq 3$ implies $X_1( |x-e_n| /r|x+e_n|)\geq X_1(|x-e_n|/3r)$. This completes the proof.
$\hfill\Box$








\noindent
{\bf \em Proof of Theorem \ref{c}.} We shall actually prove that the constant $\sigma_n$ in the statement of the theorem is the same as the constant $\sigma_n$ in the statement of Lemma~\ref{gera:thm:general_sobolev}. Without loss of generality we may assume that $\rho=1$, the general case then follows by scaling.

We first note that from Lemma \ref{gera:thm:general_sobolev} and the inclusions
\[
\Omega \cap B_{\rho} \subset {\cal C} B(-e_n, 1) 
\cap B_{\rho}   \subset  {\cal C} B(-e_n, 1) \cap B_{\sigma_n} .
\]
we obtain that for all $r \leq\sigma_n\rho $ there holds
\begin{equation}
\int_{\Omega \cap B_r} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega \cap B_r} \frac{u^2}{|x|^2} dx 
+  C_n \bigg( \int_{\Omega \cap B_r}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \bigg)^{\frac{n-2}{n}},
\label{wasa}
\end{equation}
for all $u \in C^{\infty}_{c}(\Omega \cap B_r)$, where $X_1=X_1(|x|/3r)$.


We apply (\ref{wasa}) for $r=D$ (which is allowed since $D\leq\sigma_n\rho$) and the result follows immediately from the inclusion $\Omega\subset B_D$.
To establish the optimality of  the exponent $(2n-2)/(n-2)$,  it is enough to show the following

{\bf Claim.} If $p<(2n-2)/(n-2)$ then there is no $\sigma>1$ such that the inequality
\begin{equation}
\int_{ B_1\cap B_{\rho}(e_n)} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{ B_1\cap B_{\rho}(e_n)} \frac{u^2}{|x-e_n|^2} dx 
+  C \left( \int_{ B_1\cap B_{\rho}(e_n)}  X_1^{p}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}}  \label{gera:dbp:opt}
\end{equation}
with $X_1=X_1(|x-e_n|/\sigma\rho)$ holds
true for some small $\rho>0$ and some $C>0$ and all $u \in C^{\infty}_{c}(\Omega \cap B_{\rho})$.

Suppose to the contrary that (\ref{gera:dbp:opt}) is true for all $u\in C^{\infty}_c(B_1\cap B_{\rho}(e_n))$.
We use the conformal map $S$ defined by (\ref{gera:map:s}) to pull-back (\ref{gera:dbp:opt}) to $S^{-1}(B_1\cap B_{\rho}(e_n) )$. We write $x=Sv$ and define
\[
w(v) =u(S(v)) \Big(  \frac{2}{|v+e_n|^2}\Big)^{\frac{n-2}{2}} .
\]
Noting that $|x-e_n| =2|v|/|v+e_n|$ we obtain that
there exists $R>0$ such that the following inequality holds true for all $w \in C^{\infty}_c(B^+_R)$
\[
\int_{B^+_R} |\nabla w|^2 dv \geq   \frac{n^2}{4}  \int_{ B^+_R} \frac{w^2}{|v|^2|v+e_n|^2} dv 
+  C \left( \int_{ B^+_R}  X_1^{p}|w|^{\frac{2n}{n-2}}     dv \right)^{\frac{n-2}{n}},  
\]
or equivalently
\begin{equation}
\frac{n^2}{4}\int_{B^+_R} \frac{|v|^2 +2v_n}{|v|^2|v+e_n|^2}w^2dv  +
\int_{B^+_R} |\nabla w|^2 dv \geq   \frac{n^2}{4}  \int_{ B^+_R} \frac{w^2}{|v|^2} dv 
+  C \left( \int_{ B^+_R}  X_1^{p}|w|^{\frac{2n}{n-2}}     dv \right)^{\frac{n-2}{n}},  
\label{444}
\end{equation}
where $X_1=X_1(2|v|/ \sigma \rho |v+e_n|)$.
Now, from Lemma \ref{gera:lem:1.12im_ex} we have the inequality
\begin{equation}
\int_{B_{R}^+}\frac{w^2}{|x|^{3/2}}dx \leq  4R^{1/2} \Big( 
\int_{B_{R}^+}|\nabla w|^2dx - \frac{n^2}{4}\int_{B_{R}^+}\frac{w^2}{|x|^2}dx  \Big).
\label{444a}
\end{equation}
By taking $R$ small enough we obtain from (\ref{444}) and (\ref{444a}) that
\[
\int_{B^+_R} |\nabla w|^2 dv \geq   \frac{n^2}{4}  \int_{ B^+_R} \frac{w^2}{|v|^2} dv 
+  C \left( \int_{ B^+_R}  X_1^{p}|w|^{\frac{2n}{n-2}}     dv \right)^{\frac{n-2}{n}} , \qquad w\in C^{\infty}_c(B^+_R).
\]
This violates the optimality of the exponent $2(n-1)/(n-2)$ of Theorem~\ref{i}, concluding the proof.
$\hfill\Box$

If the  radius of the  exterior ball is small we then have

\begin{theorem}
\label{4c}
Let $n \geq 3$. There exist positive constants $\lambda_n $ and $C_n$  that depend only on $n$ such that, if the radius of the exterior ball satisfies  $\rho < D / \sigma_n$ the following holds true: 
\[
\int_{\Omega} |\nabla u|^2 dx + \frac{\lambda_{n}}{\rho^2} \int_{\Omega} u^2 dx
\geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx \\
+  C_n \bigg( \int_{\Omega}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \bigg)^{\frac{n-2}{n}},
\]
for all $u \in C^{\infty}_{c}(\Omega)$; here $X_1=X_1(|x|/ 3D)$.
If in addition $\Omega$ satisfies an interior ball condition at $0$ then the constant $n^2/4$ and
the exponent $(2n-2)/(n-2)$ of $X_1$ are sharp in both inequalities.
\end{theorem}








{\bf \em Proof.} Without loss of generality we may assume that $\rho =1$, the general case following by scaling.
We consider a  $C^{\infty}$ cutoff function $\phi(r)$ 
such that $\phi(r) = 1$ for $0\leq r \leq \sigma_n/2$ and $\phi(r) = 0$ for $ r \geq \sigma_n$, and  $\phi(|x|)$ we have $0 \leq \phi \leq 1$, $|\nabla \phi| \leq C_1$, $|\Delta \phi|< C_2$ for some constants
 depending only on $n$. We then  compute
\begin{eqnarray*}
\int_{\Omega}|\nabla u|^2 dx  &=&  \int_{\Omega}|\nabla (\phi u) +\nabla ((1-\phi)u)|^2 dx\\
&=&  \int_{\Omega}|\nabla (\phi u)|^2 dx  +\int_{\Omega} |\nabla ((1-\phi)u)|^2 dx  + 2 \int_{\Omega}\phi(1-\phi)|\nabla u|^2dx \\
&& + \int_{\Omega}(2\phi-1)\Delta\phi \,  u^2 dx \\
&\geq&  \int_{\Omega}|\nabla (\phi u)|^2 dx  +\int_{\Omega} |\nabla ((1-\phi)u)|^2 dx  - c_n\int_{\Omega}u^2 dx \\
\mbox{ (by (\ref{wasa}))} \;\; &\geq&  \frac{n^2}{4} \int_{\Omega \cap B(\sigma_n)} \frac{ \phi^2 u^2}{|x|^2}dx  + 
  C_n \Big( \int_{\Omega \cap B(\sigma_n)}  X_1^{\frac{2n-2}{n-2}} (|x|/3\sigma_n) |\phi u|^{\frac{2n}{n-2}}  dx \Big)^{\frac{n-2}{n}} \\
	&& + S_n \Big( \int_{\Omega}  |(1-\phi) u|^{\frac{2n}{n-2}}  dx \Big)^{\frac{n-2}{n}} - c_n\int_{\Omega}u^2 dx \\
&\geq& \frac{n^2}{4} \int_{\Omega} \frac{ u^2}{|x|^2}dx  + 
  C_n \Big( \int_{\Omega }  X_1^{\frac{2n-2}{n-2}} (|x|/3\sigma_n) |\phi u|^{\frac{2n}{n-2}}  dx \Big)^{\frac{n-2}{n}} \\
	&& + S_n \Big( \int_{\Omega} X_1^{\frac{2n-2}{n-2}} (|x|/3\sigma_n) |(1-\phi) u|^{\frac{2n}{n-2}}dx \Big)^{\frac{n-2}{n}}-c_n'\int_{\Omega}u^2 dx \\
&\geq& \frac{n^2}{4} \int_{\Omega} \frac{ u^2}{|x|^2}dx  + 
  C_n'\Big( \int_{\Omega}   X_1^{\frac{2n-2}{n-2}}(|x|/3D)|u|^{\frac{2n}{n-2}}  dx \Big)^{\frac{n-2}{n}} -c_n'\int_{\Omega}u^2 dx ,
\end{eqnarray*}
where for the last inequality we used the fact that $D>\sigma_n$.

The sharpness of the constant $n^2/4$ and of the  exponent $(2n-2)/(n-2)$ follow as in the  proof of Theorems \ref{b} and \ref{c}.
{\hfill $\Box$}








\section{Characterizing maximal potentials}
\label{section:impr}


Throughout this section we assume  that $\Omega$ satisfies both an interior and exterior ball condition at $0$.
Without loss of generality we may  assume that the exterior ball at 0 is $B(-2\rho e_n,2\rho )$ for some
$\rho>0$.

Our starting point is the following improved Hardy inequality contained in Theorem \ref{4c},
\begin{equation}
 \lambda \int_{\Omega} u^2 dx  + \int_{\Omega} |\nabla u|^2 dx  \geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx 
\label{spe}
\end{equation}
for all $u \in C^{\infty}_{c}(\Omega)$. We shall be interested in the problem of improvements of (\ref{spe}) and whether corresponding best constants are attained.
In connection with this we make the following definition 
\begin{definition}
\label{def:1}
A non-negative potential $V\in L^{n/2}_{\rm loc}(\overline{\Omega}\setminus\{0\})$ is called {\em admissible} if there exist
$\lambda\geq 0$ and $C>0$ such that 
\begin{equation}
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx +C \int_{\Omega} Vu^2dx, \quad u \in  C^{\infty}_{c}(\Omega),
\label{in301a}
\end{equation}
The  class of all admissible potentials for the domain $\Omega$ is denoted by ${\cal A}(\Omega)$.
\end{definition}
For a given $V \in {\cal A}(\Omega)$ we denote by $b(\lambda)>0$ the best constant $C$ of inequality (\ref{in301a}). We next address the question whether
there exists non-negative potentials $W \in {\cal A}(\Omega)$ and a positive constant $C$ such that
\[
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx +b(\lambda)\int_{\Omega} Vu^2dx   +C \int_{\Omega} Wu^2dx,          \quad u\in C^{\infty}_{c}(\Omega) .
\]
In case there does not exist such a potential $W$  we say that the potential
\[
\frac{n^2}{4} \frac{1}{|x|^2} +b(\lambda)V(x) \ ,
\]
is a {\it maximal potential}. Our next goal is to characterize maximal potentials. In this direction 
for $V\in{\cal A}(\Omega)$  and small $r>0$ we define
\begin{equation}
\label{crv}
C_r(V) =\inf_{u\in C^{\infty}_{c}(\Omega)}
\frac{\lambda\int_{\Omega\cap B_r}u^2dx +
\int_{\Omega\cap B_r}|\nabla u|^2dx - \frac{n^2}{4} 
\int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx}{\int_{\Omega\cap B_r}Vu^2dx}.
\end{equation}
Since $C_r(V)$ is a non-increasing function we can define
\[
{\cal C}^0(V)=\lim_{r\to 0+}C_r(V),
\]
which may also be equal to $+\infty$.  
This definition gives the impression that ${\cal C}^0(V)$ might depend on the choice of $\lambda$. We will now establish that ${\cal C}^0(V)$ is independent of $\lambda$. Let us denote at the moment the infimum in   (\ref{crv}) by    ${\cal C}^0(V, \lambda)$ to express the dependence on $\lambda$.
We have seen (cf. (\ref{wasa}))  that for small $r>0$ there exists a positive constant $C_n$ that depends only on $n$ so that 
 \[
\int_{\Omega \cap B_r} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega \cap B_r} \frac{u^2}{|x|^2} dx 
+  C_n \bigg( \int_{\Omega \cap B_r}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \bigg)^{\frac{n-2}{n}}, \qquad 
u\in C^{\infty}_c(\Omega \cap B_r).
\]
Using Holder's inequality we conclude the existence of a positive constant $c$ independent of $r$, so that for small $r$ we have
\[
\int_{\Omega\cap B_r}|\nabla u|^2dx \geq \frac{n^2}{4} \int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx + \frac{c}{r^2}
\int_{\Omega\cap B_r}u^2 dx \; , \qquad u \in C^{\infty}_c(\Omega\cap B_r).
\]
This implies
\begin{eqnarray*}
0&\leq &\frac{ \int_{\Omega\cap B_r}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx}{\int_{\Omega\cap B_r}Vu^2dx} \leq   \frac{ \lambda\int_{\Omega\cap B_r}u^2dx +
\int_{\Omega\cap B_r}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx}{\int_{\Omega\cap B_r}Vu^2dx} \\
&\leq& (1+\lambda cr^2)\frac{ \int_{\Omega\cap B_r}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx}{\int_{\Omega\cap B_r}Vu^2dx}.
\end{eqnarray*}
Hence $C_r(V,0)\leq C_r(V, \lambda)\leq  (1+\lambda cr^2) C_r(V,0)$. Letting $r\to 0$ we conclude that ${\cal C}^0(V)$ is indeed
independent of the choice of $\lambda \geq 0$.

\begin{definition}
We say that the potential $V\in {\cal A}(\Omega)$ is subcritical if ${\cal C}^0(V) =+\infty$.
\label{def:2}
\end{definition}


\begin{lemma}
\label{lem:77}
Let $V$ be a non-negative potential satisfying
\begin{equation}
\int_{\Omega} V^{n/2}X_1^{1-n}dx <+\infty,
\label{77}
\end{equation}
where $X_1=X_1(|x|/D)$. Then $V$ is a subcritical potential.
\end{lemma}
{\bf \em Proof.} Applying  Theorem \ref{c} we obtain that for $r>0$ small enough we have
\[
\int_{ \Omega \cap B_r} |\nabla u|^2 dx \geq   \frac{n^2}{4}  \int_{\Omega \cap B_{r}} \frac{u^2}{|x|^2} dx 
+  C_n \bigg( \int_{\Omega \cap B_{r}}  X_1^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \bigg)^{\frac{n-2}{n}}
\]
for all $u\in C^{\infty}_c(\Omega \cap B_{r})$, where $X_1=X_1(|x|/D)$. Applying H\"{o}lder inequality we then easily obtain that
\[
\frac{\int_{\Omega\cap B_r}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx}
{\int_{\Omega\cap B_r}Vu^2dx} \geq \frac{1}{ C_n} \Big(   \int_{\Omega\cap B_r} V^{n/2}X_1^{1-n}dx \Big)^{-\frac{2}{n}}.
\]
Letting $r\to 0+$ we conclude that ${\cal C}^0(V) =+\infty$.
 $\hfill\Box$


We shall also consider the following more general situation. Assume that $V,W_1,W_2$ are non-negative potentials in ${\cal A}(\Omega)$ and assume that there exist
$c>0$ and a radius $R>0$ so that
\begin{equation}
\int_{\Omega\cap B_{R}}W_1 u^2dx +
\int_{\Omega\cap B_{R}}|\nabla u|^2dx \geq \frac{n^2}{4} \int_{\Omega\cap B_{R}}\frac{u^2}{|x|^2}dx +
\int_{\Omega\cap B_{R}}W_2 u^2dx + c\int_{\Omega\cap B_{R}}V u^2dx 
\label{w1w2}
\end{equation}
for all $u\in C^{\infty}_c(\Omega\cap B_{R})$. For $0<r<R$ we define
\[
C_r(W_1,W_2;V) =\inf_{C^{\infty}_c(\Omega\cap B_{r})}\frac{\int_{\Omega\cap B_{r}}W_1 u^2dx +\int_{\Omega\cap B_{r}}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_{r}}\frac{u^2}{|x|^2}dx -
\int_{\Omega\cap B_{r}}W_2 u^2dx}{\int_{\Omega\cap B_{r}}V u^2dx}
\]
and we denote
\[
{\cal C}^0(W_1,W_2;V) =\lim_{r\to 0} C_r(W_1,W_2;V).
\]

We next show that subcritical potentials do not affect the concentration level ${\cal C}^0(V)$. More precisely we have
\begin{lemma}
Let $V,W_1,W_2$ be non-negative potentials in ${\cal A}(\Omega)$ and assume that for some $R>0$ there exists $c>0$ such that (\ref{w1w2}) holds true.
If in addition $W_1,W_2$ are subcritical then ${\cal C}^0(W_1,W_2;V) ={\cal C}^0(V)$.
\label{lem:w1w2}
\end{lemma}
{\bf \em Proof.} The subcriticality of $W_i$ implies that for small $r>0$ there holds
\begin{equation}
\label{ww1}
\int_{\Omega\cap B_{r}}|\nabla u|^2dx \geq \frac{n^2}{4} \int_{\Omega\cap B_{r}}\frac{u^2}{|x|^2}dx +
c_r(W_i)\int_{\Omega\cap B_{r}}W_i u^2dx \; , \quad u \in C^{\infty}_c(\Omega\cap B_{r}) \ ,
\end{equation}
with $\lim_{r\to 0} c_r(W_i) =+\infty$. From inequalities (\ref{w1w2}) and (\ref{ww1}) follows that
for $r>0$ small enough we have
\begin{eqnarray*}
&&\hspace{-4cm}\Big(1 -\frac{1}{c_r(W_2)}\Big)\frac{\int_{\Omega\cap B_{r}}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_{r}}\frac{u^2}{|x|^2}dx}{ \int_{\Omega\cap B_{r}}V u^2dx}\\
&\leq&
\frac{\int_{\Omega\cap B_{r}}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_{r}}\frac{u^2}{|x|^2}dx +
 \int_{\Omega\cap B_{r}}(W_1-W_2) u^2dx}{ \int_{\Omega\cap B_{r}}V u^2dx} \\
&\leq&\Big(1 +\frac{1}{c_r(W_1)}\Big)
\frac{\int_{\Omega\cap B_{r}}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_{r}}\frac{u^2}{|x|^2}dx }{ \int_{\Omega\cap B_{r}}V u^2dx} \ .
\end{eqnarray*}
This implies that 
\[
\Big(1 -\frac{1}{c_r(W_2)}\Big) c_r(V) \leq   C_r(W_1,W_2;V) \leq \Big(1 +\frac{1}{c_r(W_1)}\Big) c_r(V),
\]
and the result follows by letting $r\to 0+$.

 $\hfill\Box$









Given $u\in C^{\infty}_c(\Omega)$ we define the function $w$ by
\begin{equation}
u(x) =  \frac{|x+\rho e_n|^2- \rho^2}{|x|^{\frac{n}{2}}   |x+2\rho e_n|^{\frac{n}{2}}} w(x).
\label{uw}
\end{equation}
Then $w\in C^{\infty}_c(\Omega)$ by our assumption that the exterior ball at zero is $B(-2\rho e_n,2\rho)$.
After some computations and using integration by parts we arrive at
\[
\int_{\Omega}|\nabla u|^2dx =\int_{\Omega}  \frac{\big(  |x+\rho e_n|^2- \rho^2 \big)^2 }{|x|^{n}   |x+2\rho e_n|^{n}}|\nabla w|^2dx +
 n^2 \rho^2 \int_{\Omega}  \frac{\big(  |x+\rho e_n|^2- \rho^2 \big)^2 }{|x|^{n+2}   |x+2\rho e_n|^{n+2}} w^2dx ,
\]
so inequality (\ref{in301a}) is written
 \begin{eqnarray*}
\lambda\int_{\Omega} \frac{\big( |x+ \rho e_n|^2- \rho^2 \big)^2 }{|x|^{n}   |x+2\rho e_n|^{n}}   w^2dx &+&
\int_{\Omega}  \frac{\big(  |x+\rho e_n|^2- \rho^2 \big)^2 }{|x|^{n}   |x+2\rho e_n|^{n}}|\nabla w|^2dx
+  n^2 \rho^2 \int_{\Omega}  \frac{\big(  |x+\rho e_n|^2-\rho^2 \big)^2 }{|x|^{n+2}   |x+2\rho e_n|^{n+2}} w^2dx \\
& \geq & \frac{n^2}{4}\int_{\Omega}  \frac{\big(  |x+\rho e_n|^2-\rho^2 \big)^2 }{|x|^{n+2}   |x+2\rho e_n|^{n}} w^2dx 
+ c \int_{\Omega} \frac{\big( |x+\rho e_n|^2-\rho^2 \big)^2 }{|x|^{n}   |x+2\rho e_n|^{n}}  V w^2 dx
\end{eqnarray*}
which can also take the equivalent form
 \begin{eqnarray}
\lambda\int_{\Omega} \frac{\big( |x+\rho e_n|^2- \rho^2 \big)^2 }{|x|^{n}   |x+2\rho e_n|^{n}}   w^2dx &+&
\int_{\Omega}  \frac{\big(  |x+\rho e_n|^2-\rho^2 \big)^2 }{|x|^{n}   |x+2\rho e_n|^{n}}|\nabla w|^2dx \nonumber \\
&  &\hspace{-3cm}\geq \frac{n^2}{4}\int_{\Omega}  \frac{\big(  |x+ \rho e_n|^2-\rho^2 \big)^2 (|x|^2+4\rho x_n)}{|x|^{n+2}   |x+2\rho e_n|^{n+2}} w^2dx 
+ c \int_{\Omega} \frac{\big( |x+\rho e_n|^2-\rho^2 \big)^2 }{|x|^{n}   |x+2\rho e_n|^{n}}  V w^2 dx.
\label{w}
\end{eqnarray}
It is clear from the above that (\ref{w}) is valid for all functions $w\in C^{\infty}_c(\Omega)$ and moreover,
for a fixed $\lambda \geq 0$ the best constants $c$ for inequalities (\ref{in301a}) and (\ref{w}) coincide.
That common best constant shall be denoted by $b=b(\lambda)$.


Denoting
\[
\phi(x)=\frac{|x+ \rho e_n|^2-\rho^2 }{|x|^{\frac{n}{2}}   |x+2\rho e_n|^{\frac{n}{2}}} 
\]
it then follows that (cf. (\ref{crv}))
\[
 C_r(V) = 
\inf_{ W^{1,2}_0( \Omega\cap B_r ; \phi^2)}
\hspace{-0.3cm}\frac{\lambda \int_{\Omega\cap B_r}\phi^2   w^2dx +
\int_{\Omega\cap B_r} \phi^2|\nabla w|^2dx -
 \frac{n^2}{4}\int_{\Omega\cap B_r} \frac{|x|^2+4\rho x_n}{|x|^{2}   |x+2\rho e_n|^{2}} \phi^2 w^2dx}
{  \int_{\Omega\cap B_r} V \phi^2  w^2 dx}
\]
where $W^{1,2}_0( \Omega\cap B_r ; \phi^2)$ denotes the closure of $C^{\infty}_c(\Omega\cap B_r)$ under the norm
\[
\int_{\Omega\cap B_r}\phi^2|\nabla w|^2dx +
 \int_{\Omega\cap B_r} \phi^2  w^2dx .
\]


We shall now see a simpler way for expressing ${\cal C}^0(V)=\lim_{r\to 0+}C_r(V)$ in terms of the weight $\phi^2$. For this we define
\[
C_r(V;\phi^2) =\inf_{W^{1,2}_0( \Omega\cap B_r ; \phi^2)}
\frac{ \int_{\Omega\cap B_r}  \phi^2|\nabla w|^2dx}{ \int_{\Omega\cap B_r} V  \phi^2  w^2dx}.
\]
and
\[
{\cal C}^0(V;\phi^2) =\lim_{r\to 0+} C_r(V;\phi^2).
\]


\begin{lemma}
Let $V$ be a non-negative potential in ${\cal A}(\Omega)$. Then
${\cal C}^0(V) ={\cal C}^0(V;\phi^2)$.
\label{lem:fr_ir}
\end{lemma}
{\bf \em Proof.} For the sake of simplicity we assume that $\rho=1$; the general case then follows by scaling.
On the one hand we have for small $r>0$ that
\[
cr^{-2}\int_{\Omega\cap B_r}\phi^2w^2 dx \leq  \int_{\Omega\cap B_r} \phi^2|\nabla w|^2dx -
 \frac{n^2}{4}\int_{\Omega\cap B_r} \frac{|x|^2+4x_n}{|x|^{2}   |x+2e_n|^{2}} \phi^2 w^2dx \; , \qquad
u\in C^{\infty}_c(\Omega\cap B_r)
\]
for some universal constant $c>0$, which implies the inequality
\begin{equation}
\frac{\lambda \int_{\Omega\cap B_r}\phi^2   w^2dx +
\int_{\Omega\cap B_r} \phi^2|\nabla w|^2dx -
 \frac{n^2}{4}\int_{\Omega\cap B_r} \frac{|x|^2+4x_n}{|x|^{2}   |x+2e_n|^{2}} \phi^2 w^2dx}
{  \int_{\Omega\cap B_r} \phi^2 V w^2 dx}
\leq
\Big( 1+\frac{\lambda}{c}r^2\Big)\frac{\int_{\Omega\cap B_r} \phi^2|\nabla w|^2dx }{  \int_{\Omega\cap B_r} \phi^2 V w^2 dx}.
\label{ineq1}
\end{equation}
On the other hand, since
\[
0\leq \frac{|x|^2 +4x_n}{|x|^2 |x+2e_n|^2} \leq \frac{c}{|x|},
\]
we have the inequality
\[
cr^{-1}\int_{\Omega\cap B_r} \frac{|x|^2+4x_n}{|x|^{2}   |x+2e_n|^{2}} \phi^2 w^2dx \leq \int_{\Omega\cap B_r} \phi^2|\nabla w|^2dx \, , \qquad  w\in C^{\infty}_c(\Omega\cap B_r) 
\]
which in turn implies
\begin{equation}
\frac{\int_{\Omega\cap B_r} \phi^2|\nabla w|^2dx }{  \int_{\Omega\cap B_r} \phi^2 V w^2 dx}
(1-cr)
\leq 
\frac{\lambda \int_{\Omega\cap B_r}\phi^2   w^2dx +
\int_{\Omega\cap B_r} \phi^2|\nabla w|^2dx -
 \frac{n^2}{4}\int_{\Omega\cap B_r} \frac{|x|^2+4x_n}{|x|^{2}   |x+2e_n|^{2}} \phi^2 w^2dx}
{  \int_{\Omega\cap B_r} \phi^2 V w^2 dx} 
\label{ineq2}
\end{equation}
The result follows by combining inequalities (\ref{ineq1}) and (\ref{ineq2}) 
and letting $r\to 0+$. $\hfill\Box$



One important consequence of subcriticality is the following compactness property.
\begin{lemma}
Assume that the positive potential $V\in{\cal A}(\Omega)$ is subcritical. Then for any sequence $(w_k)$ which is bounded in $W^{1,2}_0(\Omega ;\phi^2)$ there exists a subsequence, also denoted by $(w_k)$, and a function $w_0\in W^{1,2}_0(\Omega ;\phi^2)$ so that
\begin{eqnarray*}
({\rm i}) && w_k \rightharpoonup w_0 \mbox{ in } W^{1,2}_0(\Omega ;\phi^2)\\
({\rm ii}) && \int_{\Omega}\phi^2V(w_k-w_0)^2dx \longrightarrow 0 .
\end{eqnarray*}
\label{compactness}
\end{lemma}
{\bf \em Proof.} Part (i) is standard. To prove (ii) we may assume without loss of generality that $w_0=0$.
We consider a small $r>0$ and a smooth cut-off function $\psi$ such that $\psi=1$ in $B_{r/2}$ and $\psi=0$ outside $B_r$.
We then have
\begin{eqnarray*}
 \int_{\Omega}\phi^2|\nabla w_k|^2 dx &=& \int_{\Omega}\phi^2|\nabla (\psi w_k)  +  \nabla ((1-\psi) w_k)  |^2 dx \\
&=&
 \int_{\Omega}\phi^2 |\nabla (\psi w_k)|^2dx +  \int_{\Omega}\phi^2 |\nabla \big( (1-\psi) w_k\big)|^2dx  +
2 \int_{\Omega}\phi^2 \psi(1-\psi)|\nabla w_k|^2dx \\
&&  + \int_{\Omega}\phi^2 (1-2\psi)w_k\nabla\psi\cdot \nabla w_k dx - \int_{\Omega}\phi^2  |\nabla \psi|^2 w_k^2dx \\
&\geq& \int_{\Omega}\phi^2 |\nabla (\psi w_k)|^2dx +o(1) \\
&\geq& C_r(V;\phi^2) \int_{\Omega}\phi^2 V \psi^2w_k^2dx +o(1)\\
&=&  C_r(V;\phi^2) \int_{\Omega}\phi^2 V w_k^2dx +o(1) , 
\end{eqnarray*}
that is
\begin{equation}
\int_{\Omega} V\phi^2  w_k^2dx \leq \frac{1}{C_r(V;\phi^2)} \int_{\Omega}\phi^2|\nabla w_k|^2 dx +o(1) \ .
\label{pas}
\end{equation}
The result follows by noting that the RHS of (\ref{pas}) can be made arbitrarily small by choosing $r>0$ small enough.
$\hfill\Box$

We can now state and prove the main result of this section.
\begin{theorem}
Let $V\in{\cal A}(\Omega)$ and $\lambda\geq 0$ be given and let $b(\lambda)$ be the best constant for the inequality
\begin{equation}
\lambda\int_{\Omega} \phi^2 w^2dx +
\int_{\Omega} \phi^2|\nabla w|^2dx \geq  \frac{n^2}{4} \int_{\Omega}   \frac{|x|^2+4\rho x_n}{|x|^{2} |x+2\rho e_n|^{2}}  \phi^2 w^2
dx +b(\lambda)\int_{\Omega} V \phi^2 w^2dx , \qquad w\in W^{1,2}_0(\Omega;\phi^2).
\label{eq:w:no1}
\end{equation}
If in addition
\[
b(\lambda)<{\cal C}^0(V)
\]
then the best constant $b(\lambda)$ in (\ref{eq:w:no1}) is realized by a function $w_0\in W^{1,2}_0(\Omega;\phi^2)$. In particular the best constant $b(\lambda)$ is realized if the potential $V$ is subcritical.
\label{gera:thm:exmin}
\end{theorem}
{\bf \em Proof.} We denote
\[
Q(x)=\frac{n^2}{4}  \frac{|x|^2+4\rho x_n}{|x|^{2} |x+2\rho e_n|^{2}}
\]
Then it is easily seen that
\[
0\leq Q\leq \frac{c}{|x|}, \qquad x\in\Omega,
\]
which implies that $Q$ is a subcritical potential by Lemma \ref{lem:77}.
We consider a minimizing sequence $(w_k)$ for (\ref{eq:w:no1}) and without loss of generality we assume that
\begin{equation}
\int_{\Omega} Q \phi^2  w_k^2dx  +  b(\lambda) \int_{\Omega} V \phi^2  w_k^2 dx
=1, \qquad   \lambda\int_{\Omega} \phi^2  w_k^2dx + \int_{\Omega}  \phi^2 |\nabla w_k|^2dx  \longrightarrow 1, \mbox{ as }k\to \infty.
\label{g11}
\end{equation}
Since $(w_k)$ is bounded in $W^{1,2}_0(\Omega ; \phi^2)$, it has a subsequence, which we assume is $(w_k)$ itself, which converges weakly
to some $w_0\in W^{1,2}_0(\Omega ; \phi^2)$. We define $v_k=w_k-w_0$. 

We consider a small enough $r>0$ so that $C_r(V;\phi^2)>b(\lambda)$ and a smooth cut-off function $\psi$
such that $\psi=1$ in $B_{r/2}$ and $\psi=0$ outside $B_r$. Arguing as in the proof of Lemma \ref{compactness} we have
\begin{eqnarray}
\int_{\Omega}\phi^2 |\nabla v_k|^2 dx &=&  \int_{\Omega}\phi^2 |\nabla [\psi v_k +(1-\psi)v_k] |^2dx \nonumber \\
&\geq&  \int_{\Omega}\phi^2 |\nabla (\psi v_k)|^2dx  +o(1) \nonumber \\
&\geq&  C_r(V;\phi^2) \int_{\Omega}\phi^2 V \psi^2 v_k^2 dx +o(1) \nonumber  \\
&=&  C_r(V;\phi^2) \int_{\Omega}\phi^2 V v_k^2 dx +o(1) \ .  \label{equ}
\end{eqnarray}
Now, substituting $w_k=v_k+w_0$ in the normalization relations (\ref{g11}) and using Lemma \ref{compactness} we obtain
\begin{equation}
\int_{\Omega} \phi^2 Q w_0^2dx  +  b(\lambda) \int_{\Omega} \phi^2 V w_0^2 dx + b(\lambda) \int_{\Omega} \phi^2 V v_k^2 dx  =1 +o(1)
\label{norm1}
\end{equation}
and
\begin{equation}
\lambda\int_{\Omega} \phi^2  w_0^2dx + \int_{\Omega}  \phi^2 |\nabla w_0|^2dx + \int_{\Omega}  \phi^2 |\nabla v_k|^2dx = 1+o(1).
\label{norm2}
\end{equation}
From (\ref{equ}) and (\ref{norm1}) we obtain
\begin{equation}
\int_{\Omega}\phi^2 |\nabla v_k|^2 dx \geq \frac{ C_r(V;\phi^2)}{b(\lambda)}\Big(  1-    \int_{\Omega} \phi^2 Q w_0^2dx  +  b(\lambda) \int_{\Omega} V \phi^2  w_0^2 dx        \Big) +o(1).
\label{norm7}
\end{equation}
Moreover using (\ref{eq:w:no1}) for $w=w_0$ we obtain from (\ref{norm2}) that
\begin{equation}
\int_{\Omega}  \phi^2 |\nabla v_k|^2dx \leq 1 -\int_{\Omega}\phi^2Qw_0^2dx -b(\lambda)\int_{\Omega} \phi^2Vw_0^2dx +o(1) .
\label{norm3}
\end{equation}
From (\ref{norm7}) and (\ref{norm3}) we conclude that
\[
\Big(  1- \frac{ C_r(V;\phi^2)}{b(\lambda)}\Big) \bigg(      1-    \int_{\Omega} \phi^2 Q w_0^2dx  +  b(\lambda) \int_{\Omega} V \phi^2  w_0^2 dx       \bigg) \geq 0.
\]
Since $C_r(V;\phi^2)>b(\lambda)$, this implies that
\[
 \int_{\Omega} \phi^2 Q w_0^2dx  +  b(\lambda) \int_{\Omega} \phi^2 V w_0^2 dx \geq 1.
\]
But by lower semicontinuity,
\[
 \int_{\Omega} \phi^2 Q w_0^2dx  +  b(\lambda) \int_{\Omega} \phi^2 V w_0^2 dx \leq 1.
\]
Hence $w_0$ is a minimizer. $\hfill\Box$


The next theorem is an immediate consequence of Theorem \ref{gera:thm:exmin}
\begin{theorem}
\label{d}
Let $V$ be a non-negative potential in ${\cal A}(\Omega)$. (a) Let $\lambda\geq 0$ and $b(\lambda)>0$ be such that
\begin{equation}
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx +b(\lambda)\int_{\Omega} Vu^2dx, \quad u\in C_{c}^{\infty}(\Omega),
\label{in30}
\end{equation}
where $b(\lambda)$ is the best constant. If in addition
\[
b(\lambda)<{\cal C}^0(V) \ ,
\]
then the potential $n^2/4|x|^2 +b(\lambda)V(x)$ is a maximal potential, that is inequality
(\ref{in30}) cannot be improved by adding a non-negative potential $W$ in the RHS. \newline
(b) If $V$ is a subcritical potential then there exist $\lambda\geq 0$ and a best constant $b(\lambda)>0$ such that (\ref{in30}) is true. Moreover the potential $n^2/4|x|^2 +b(\lambda)V(x)$ is a maximal potential.
\end{theorem}
{\bf \em Proof.} $(a)$ Suppose that  $n^2/4|x|^2 +b(\lambda)V(x)$ is not a maximal potential, that is there exists a non-trivial potential
$W\geq 0$ in ${\cal A}(\Omega)$ such that
\[
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx +b(\lambda)\int_{\Omega} Vu^2dx + \int_{\Omega} Wu^2dx
\]
holds true for all $u\in H^1_0(\Omega)$. Using the transformation (\ref{uw}) this is equivalently written as
\[
\lambda\int_{\Omega} \phi^2 w^2dx +
\int_{\Omega} \phi^2|\nabla w|^2dx \geq   \int_{\Omega} Q\phi^2  w^2
dx +b(\lambda)\int_{\Omega} \phi^2Vw^2dx  +\int_{\Omega} W\phi^2w^2 dx , \qquad w\in W^{1,2}_0(\Omega;\phi^2).
\]
Using $w=w_0$ where $w_0$ is the minimizer from Theorem \ref{gera:thm:exmin} we conclude that $\int_{\Omega} W\phi^2w_0^2 dx\leq 0$, which is a contradiction. \newline
Part $(b)$ is an immediate consequence of part $(a)$ since any subcritical potential $V$ is in ${\cal A}(\Omega)$ and satisfies
${\cal C}^0(V) =+\infty$.
$\hfill\Box$




\section{Logarithmic improvements and maximal potentials}
\label{sub:log:impr}

Throughout this section we continue to assume  that $\Omega$ satisfies both an interior and exterior ball condition at $0$.
We also continue to assume that the exterior ball at 0 is $B(-2\rho e_n,2\rho)$ for some $\rho>0$. 



In this section we will provide the proofs of Theorems \ref{f}, \ref{g} and also study maximal potentials in the context of logarithmic improvements of Hardy
inequality.



\subsection{Logarithmic improvements}


To prove Theorems  \ref{f} and \ref{g} we first establish the following lemmas:
\begin{lemma}
Let $n\geq 2$. There exists a positive constant  $\sigma_n$ depending only on $n$ such that for all $\rho>0$ and all $r\leq\sigma_n \rho$ we have
\begin{eqnarray}
({\rm i}) && \int_{{\cal C} B(\rho) \cap B(\rho e_n, r)} |\nabla u|^2 dx \geq
\frac{n^2}{4} \int_{{\cal C} B(\rho) \cap B(\rho e_n,r)} \frac{u^2}{|x-\rho e_n|^2}dx  \nonumber\\[0.2cm]
&& +
\frac{1}{4}\sum_{i=1}^{\infty} \int_{{\cal C} B(\rho) \cap B(\rho e_n, r)} \frac{u^2}{|x-\rho e_n|^2} X_1^2\ldots X_i^2 dx \nonumber\\[0.2cm]
({\rm ii}) && \mbox{If in addition $n\geq 3$ there exists a constant $C_n$ depending only on $n$ such that for all $m \in {\mathbb{N}}$} \nonumber \\[0.2cm]
&& \int_{{\cal C} B(\rho) \cap B(\rho e_n,  r)}|\nabla u|^2dx  \geq\frac{n^2}{4}\int_{{\cal C} B(\rho) \cap B(\rho e_n, r)}
\frac{u^2}{|x-\rho e_n|^2}dx  \nonumber \\
&& \hspace{3cm} 
+ \frac{1}{4}\sum_{i=1}^m \int_{{\cal C} B(\rho) \cap B(\rho e_n, r)}\frac{u^2}{|x-\rho e_n|^2}X_1^2\ldots X_i^2 dx  \nonumber \\  
&& \hspace{3cm} + C_n \left( \int_{{\cal C} B(\rho) \cap B(\rho e_n, r)} 
 (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}}.
\label{gera:eq:1.33}
\end{eqnarray}
Both inequalities are valid for all $u\in C^{\infty}_c({\cal C} B(\rho) \cap B(\rho e_n, r))$ and
in both cases $X_i=X_i(|x-\rho e_n|/ (3\kappa r))$. 
\label{gera:general_k_sobolev}
\end{lemma}
{\bf \em Proof.} To prove (i) it is enough to consider the case $\rho=1$.
 We fix $r<1$ and we apply Lemma \ref{gera:lem:1.12im_ex}. Changing variables via $T$ (cf. (\ref{gera:def_T})) we obtain
\begin{eqnarray}
&&\int_{T(B_r^+)}|\nabla u|^2dx \geq \frac{n^2}{4} \int_{T(B_r^+)}\frac{4u^2}{|x-e_n|^2|x+e_n|^2}dx  \nonumber \\
&&+\frac{1}{4}\sum_{i=1}^{\infty} \int_{T(B_r^+)} \frac{4u^2\, X_1^2 \ldots X_i^2}{|x-e_n|^2|x+e_n|^2}   dx 
 + \frac{1}{8r^{1/2}} \int_{T(B_r^+)}\frac{4u^2}{|x-e_n|^{3/2} |x+e_n|^{5/2}}dx
\label{33333}
\end{eqnarray}
for all $u\in C^{\infty}_c(T(B_r^+))$; here $X_k=X_k( |x-e_n| / (\kappa r|x+e_n|)$. As already noted (cf. (\ref{inclusion})) we have 
\[
{\cal C} B_1  \cap B(e_n, r) \subset T(B_r^+) .
\]
so integrals in (\ref{33333}) can be taken over ${\cal C} B_1  \cap B(e_n, r)$.

Once again it is enough to find $\sigma_n<1$ such that for all $r\leq\sigma_n$ 
and all $x\in {\cal C} B_1\cap B(e_n, r)$ there holds
\begin{eqnarray*}
&&\frac{n^2}{4}\frac{4}{|x-e_n|^2|x+e_n|^2} +\frac{1}{4}\sum_{i=1}^{\infty} \frac{4}{|x-e_n|^2|x+e_n|^2}  X_1^2 \ldots X_i^2  \\
&& +  \frac{4}{8r^{1/2}|x-e_n|^{3/2} |x+e_n|^{5/2}}  
\geq \frac{n^2}{4|x-e_n|^2} +\frac{1}{4|x-e_n|^2}\sum_{i=1}^{\infty} X_1^2 \ldots X_i^2. 
\end{eqnarray*}
or equivalently,
\[
|x-e_n|^{1/2} \geq \frac{1}{2} r^{1/2} |x+e_n|^{5/2} \Big( 1-\frac{4}{|x+e_n|^2} \Big)
\Big(n^2 +\sum_{i=1}^{\infty} X_1^2 \ldots X_i^2 \Big).
\]
This is immediate if  $|x+e_n| \leq 2$. Assuming $|x+e_n| \geq 2$ we will actually establish the stronger inequality
(recalling that $\sum_k X_1^2\ldots X_k^2 \leq 1/4$)
\begin{equation}
|x-e_n|^{1/2} \geq \frac{n^2 +\frac{1}{4}}{2} r^{1/2} |x+e_n|^{5/2} \Big( 1-\frac{4}{|x+e_n|^2} \Big).
\label{gera:1212}
\end{equation}
But this is almost the same as inequality (\ref{gef}), the only difference being that in the place of $n^2$ we now have $(n^2 +1/4)/2$; we omit further details.  

The proof of (ii) is analogous, but we now use Lemma \ref{gera:lem_mlogs}
instead of Lemma \ref{gera:lem:1.12im_ex}. Again, we may take $\rho=1$. We then fix $r<1$ and changing variables via $T$ we obtain
\begin{eqnarray}
&& \int_{T(B_r^+)}|\nabla u|^2dx \geq \frac{n^2}{4} \int_{T(B_r^+)}\frac{4u^2}{|x-e_n|^2|x+e_n|^2}dx  
+\frac{1}{4}\sum_{i=1}^{m} \int_{T(B_r^+)} \frac{4u^2\, X_1^2 \ldots X_i^2}{|x-e_n|^2|x+e_n|^2}   dx  \nonumber  \\
&& + \frac{1}{16r^{1/2}} \int_{T(B_r^+)}\frac{4u^2}{|x-e_n|^{3/2} |x+e_n|^{5/2}}dx  +  C_n \left( \int_{T(B_r^+)} 
 (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}   dx \right)^{\frac{n-2}{n}}
\label{gera:cy33}
\end{eqnarray}
for all $u\in C^{\infty}_c(T(B_r^+))$; here $X_k=X_k( |x-e_n| / (\kappa r|x+e_n|)$. As in the proof of pert (i), this is also true if the integrals are taken over ${\cal C} B_1  \cap B(e_n, r)$ and $u\in C^{\infty}_c({\cal C} B_1  \cap B(e_n, r)) $

Hence the result will follow once we establish for all $x\in {\cal C} B_1\cap B(e_n, r)$the inequality
\begin{eqnarray*}
&&\frac{n^2}{4}\frac{4}{|x-e_n|^2|x+e_n|^2} +\frac{1}{4}\sum_{i=1}^{m} \frac{4}{|x-e_n|^2|x+e_n|^2}  X_1^2 \ldots X_i^2  \\
&& +  \frac{4}{16r^{1/2}|x-e_n|^{3/2} |x+e_n|^{5/2}}  
\geq \frac{n^2}{4|x-e_n|^2} +\frac{1}{4|x-e_n|^2}\sum_{i=1}^{m} X_1^2 \ldots X_i^2. 
\end{eqnarray*}
This is equivalent to
\[
|x-e_n|^{1/2} \geq  r^{1/2} |x+e_n|^{5/2} \Big( 1-\frac{4}{|x+e_n|^2} \Big)
\Big(n^2 +\sum_{i=1}^{m} X_1^2 \ldots X_i^2 \Big).
\]
The argument now goes as in part (i); we omit further details. $\hfill\Box$


\noindent
{\bf\em Proof of Theorem \ref{f}}.  Without loss of generality we assume that $\rho=1$.
We use  part (i) of Lemma \ref{gera:general_k_sobolev} for $r=\sigma_n$ making a translation of
(\ref{gera:eq:1.33}) by $- e_n$. We obtain
\[
\int_{{\cal C} B(- e_n,1) \cap B(\sigma_n)} |\nabla u|^2 dx \geq
\frac{n^2}{4} \int_{{\cal C} B(- e_n,1) \cap B(\sigma_n)} \frac{u^2}{|x|^2}dx  
 + \frac{1}{4}\sum_{i=1}^{\infty} \int_{{\cal C} B(- e_n, 1) \cap B(\sigma_n)} \frac{u^2}{|x|^2} X_1^2\ldots X_i^2 dx ,
\]
for all $u\in C^{\infty}_c({\cal C} B(- e_n, 1) \cap B(\sigma_n))$, where $X_i=X_i(|x|/(3\kappa \sigma_n))$.
Since $\Omega \subset {\cal C} B(- e_n, 1) \cap B(\sigma_n)$, the result follows. 
{\hfill $\Box$}


\begin{theorem}
\label{6f}

Let $n \geq 2$. There exists a positive constant $\lambda_n $  that depends only on $n$ such that, if the radius of the exterior ball satisfies  $\rho < D / \sigma_n$, with $\sigma_n$ as in Theorem \ref{g}, the following holds true: 
\begin{eqnarray*}
 \int_{\Omega} |\nabla u|^2 dx + \frac{\lambda_{n}}{\rho^2} \int_{\Omega} u^2 dx 
&\geq&   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx + \frac{1}{4} \sum_{i=1}^{\infty} \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx  
\end{eqnarray*}
for all $u \in C^{\infty}_{c}(\Omega)$; here $X_i=X_i(|x|/ (3 \kappa D))$.
If in addition $\Omega$ satisfies an interior ball condition at $0$ then the constants $1/4$ are sharp  at each step.

\end{theorem}

{\bf\em Proof.}
We argue as in the proof of Theorem \ref{c}. This time however we also  use the global
estimate $\sum_i{X_1^2\ldots X_i^2}\leq 1/4$ in order to estimate uniformly the constant in front of the $L^2$ term. We omit further details.
The sharpness of the constants $1/4$ has already been proved. $\hfill\Box$


\noindent
{\bf\em Proof of Theorem \ref{g}} We argue as in the proof of Theorem \ref{f}, using now part (ii) of Lemma \ref{gera:general_k_sobolev}.
To prove the sharpness of the constants $1/4$ and the exponent $(2n-2)/(n-2)$ we argue as in the proof of Theorem~\ref{c};
we omit the details. $\hfill\Box$


In  case the exterior ball is small, working as in Theorem \ref{4c} we have the following

\begin{theorem}
\label{6g}
Let $n \geq 3$. There exist positive constants $\lambda_n $ and $C_n$  that depend only on $n$ such that, if the radius of the exterior ball satisfies  $\rho < D / \sigma_n$, with $\sigma_n$ as in Theorem \ref{g},  then  for any $m\in{\mathbb{N}}$  the following holds true: 
\[
 \int_{\Omega} |\nabla u|^2 dx + \frac{\lambda_{n}}{\rho^2} \int_{\Omega} u^2 dx
\geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx + \frac{1}{4} \sum_{i=1}^m \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx 
+  C_n \left( \int_{\Omega}  (X_1\ldots X_{m+1})^{\frac{2n-2}{n-2}}|u|^{\frac{2n}{n-2}}     dx \right)^{\frac{n-2}{n}} ,
\]
for all $u \in C^{\infty}_{c}(\Omega)$; here  $X_i=X_i(|x|/ (3 \kappa D))$.
If in addition $\Omega$ satisfies an interior ball condition at $0$ then the exponents $(2n-2)/(n-2)$ of $X_i$ are also sharp.
\end{theorem}



\subsection{Maximal logarithmic potentials}



Here we characterize maximal potentials in the context of logarithmically improved Hardy inequalities.

Our starting point in this subsection is the improved Hardy inequality contained in Theorem \ref{g},
\begin{equation}\label{62.13}
 \lambda \int_{\Omega} u^2 dx  + \int_{\Omega} |\nabla u|^2 dx 
\geq   \frac{n^2}{4}  \int_{\Omega} \frac{u^2}{|x|^2} dx + \frac{1}{4} \sum_{i=1}^m \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx 
\end{equation}
for all $u \in C^{\infty}_{c}(\Omega)$; here $X_i=X_i(|x|/3 \kappa \tilde D)$ where $\tilde{D} \geq D$. 
We shall be interested in the problem of improvements of (\ref{62.13}) and whether the corresponding best constants are attained.

The analysis that will follow is analogous to that of Section \ref{section:impr}; for this reason we shall avoid the details in cases where the arguments are quite similar.


\begin{definition}
A non-negative potential $V\in L^{n/2}_{\rm loc}(\overline{\Omega}\setminus\{0\})$ is called $m$-{\em admissible} if there exist
$\lambda\geq 0$, $\tilde D\geq D$ and $C>0$ such that
\begin{equation}
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx 
+ \frac{1}{4} \sum_{i=1}^m \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx+
C\int_{\Omega} Vu^2dx, \quad u \in C^{\infty}_{c}(\Omega) \ ,
\label{inma}
\end{equation}
where $X_i=X_i(|x|/3 \kappa \tilde{D})$. The class of all $m$-admissible potentials for the domain $\Omega$ is denoted  by ${\cal A}_m(\Omega)$.
\end{definition}
We note that there is a big variety of $m$-admissible potentials. For example if $V$ satisfies 
\begin{equation}
\int_{\Omega} V^{\frac{n}{2}}(X_1 \ldots X_{m+1})^{1-n}dx <+\infty,
\label{inexa}
\end{equation}
where $X_i=X_i(|x|/3 \kappa D)$, $i=1,\ldots, m+1$, then $V$ is $m$-admissible by
Theorem~\ref{g}.


For a given $V \in {\cal A}_m(\Omega)$ we denote by $b_m(\lambda)>0$ the best constant $C$ of inequality (\ref{inma}). We next address the question whether inequality (\ref{inma}) with best constant $b_m(\lambda)$ can be further improved. That is, whether there exists
potential $W \in {\cal A}_m(\Omega)$ and a positive constant $C$ such that
the following inequality holds true as well
\[
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx +
\sum_{i=1}^m \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx
 +b_m(\lambda)\int_{\Omega} Vu^2dx   +C \int_{\Omega} Wu^2dx           \ ,
\]
for $u\in C^{\infty}_{c}(\Omega)$.
In case there does not exist  such a potential $W$,  we say that the potential
\[
\frac{n^2}{4} \frac{1}{|x|^2} +\sum_{i=1}^m  \frac{X_1^2\ldots X_i^2 }{|x|^2} +b_m(\lambda)V(x) \ ,
\]
is an   {\it  m--maximal potential}. Our next goal is to characterize $m$--maximal potentials. In this direction 
for $V\in{\cal A}_m(\Omega)$  and small $r>0$ we define
\[
C_{m,r}(V) =\inf_{u\in C_{c}^{\infty}(\Omega\cap B_r)}
\frac{\lambda\int_{\Omega\cap B_r}u^2dx +
\int_{\Omega\cap B_r}|\nabla u|^2dx - \frac{n^2}{4} 
\int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx
+\frac{1}{4} \sum_{i=1}^m \int_{\Omega \cap B_r } \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx}{\int_{\Omega\cap B_r}Vu^2dx},
\]
where $X_i=X_i(|x|/3 \kappa \tilde{D})$ and $\tilde{D}\geq D$.
We also define
\[
{\cal C}^0_m(V)=\lim_{r\to 0+}C_{m,r}(V).
\]
Arguing as in Section \ref{section:impr} we can see that ${\cal C}^0_m(V)$ is independent of the specific choice of $\lambda\geq 0$ and
$\tilde{D}\geq D$.
\begin{definition}
The potential $V\in {\cal A}_m(\Omega)$ is $m$-subcritical if ${\cal C}_m^0 =+\infty$.
\end{definition}
\begin{lemma}
Let $V$ be a non-negative potential satisfying
\[
\int_{\Omega} V^{n/2}(X_1\ldots X_{m+1})^{1-n}dx <+\infty,
\]
where $X_i=X_i(|x|/3 \kappa D)$, $i=1,\ldots, m+1$. Then $V$ is an $m$-subcritical potential.
\label{lem:24}
\end{lemma}
{\bf \em Proof.} The proof is quite similar to the proof of Lemma \ref{lem:77} and makes use of Theorem \ref{g}
to establish the inequality 
\begin{eqnarray*}
&&\hspace{-3cm} \frac{\int_{\Omega\cap B_r}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_r}\frac{u^2}{|x|^2}dx
-\frac{1}{4}\sum_{k=1}^m \int_{\Omega\cap B_r}\frac{u^2}{|x|^2}X_1^2\ldots X_k^2 dx}
{\int_{\Omega\cap B_r}Vu^2dx} \\
&& \hspace{3cm} \geq \frac{1}{ C_n} \Big(   \int_{\Omega\cap B_r} V^{n/2}(X_1 \ldots X_{m+1})^{1-n}dx \Big)^{-\frac{2}{n}}.
\end{eqnarray*}
The result then follows. $\hfill\Box$






As in Section \ref{section:impr} we shall also consider the following more general situation. We consider non-negative potentials
$V,W_1,W_2\in {\cal A}_m(\Omega)$ and assume that there exist
$c>0$ and a radius $R>0$ so that
\begin{eqnarray}
 \int_{\Omega\cap B_{R}}W_1 u^2dx +
\int_{\Omega\cap B_{R}}|\nabla u|^2dx &\geq& \frac{n^2}{4} \int_{\Omega\cap B_{R}}\frac{u^2}{|x|^2}dx  
+ \frac{1}{4} \sum_{i=1}^m \int_{\Omega \cap B_r } \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx \nonumber \\
&& + \int_{\Omega\cap B_{R}}W_2 u^2dx + c\int_{\Omega\cap B_{R}}V u^2dx 
\label{w1w2m}
\end{eqnarray}
for all $u\in C^{\infty}_c(\Omega\cap B_{R})$. For $0<r\leq R$ we define
\begin{eqnarray*}
& &\hspace{-6mm} C_{m,r}(W_1,W_2;V) =\\ 
&&  \hspace{-6mm} \inf_{C^{\infty}_c(\Omega\cap B_{r})}\hspace{-14pt}\frac{\int_{\Omega\cap B_{r}}\! W_1 u^2dx +\int_{\Omega\cap B_{r}}|\nabla u|^2dx - \frac{n^2}{4} \int_{\Omega\cap B_{r}}\frac{u^2}{|x|^2}dx - 
\frac{1}{4} \sum_{i=1}^m \int_{\Omega \cap B_r } \frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx -
\int_{\Omega\cap B_{r}}\! W_2 u^2dx}{\int_{\Omega\cap B_{r}}V u^2dx}
\end{eqnarray*}
and we denote
\[
{\cal C}_m^0(W_1,W_2;V) =\lim_{r\to 0+} C_{m,r}(W_1,W_2;V).
\]
The proof of the following lemma is similar to the proof of Lemma \ref{lem:w1w2} and is omitted.
\begin{lemma}
Let $V,W_1,W_2$ be non-negative potentials in ${\cal A}_m(\Omega)$ and assume that  there exist $R>0$ and
$c>0$ such that (\ref{w1w2m}) holds true. If in addition $W_1,W_2$ are subcritical then
${\cal C}_m^0(W_1,W_2;V) ={\cal C}_m^0(V)$.
\end{lemma}

Let $\tilde{D}\geq D$ be fixed. Given $u\in C^{\infty}_c(\Omega)$ we define the function $w$ by
\begin{eqnarray*}
u(x) &=& 
 \frac{|x+\rho  e_n|^2-\rho^2}{|x|^{\frac{n}{2}}   |x+2\rho e_n|^{\frac{n}{2}}} X_1^{-\frac{1}{2}} \ldots X_m^{-\frac{1}{2}} w(x)  \\
&=:& \phi_m(x)w(x) ,  
\end{eqnarray*}
where $X_i=X_i(|x|/3 \kappa \tilde{D})$, $i=1,\ldots,m$.
Then $w\in C^{\infty}_c(\Omega)$ by our assumption that $B(-2\rho e_n,2\rho)$ is an exterior ball. 
After some computations we arrive at 
\begin{eqnarray*}
 \int_{\Omega} |\nabla u|^2dx &=&\int_{\Omega}  \phi_m^2  |\nabla w|^2dx +  n^2 \rho^2 \int_{\Omega}  \frac{  \phi_m^2w^2   }{|x|^{2}   |x+2\rho e_n|^{2}} dx  \\
& +& \int_{\Omega} \frac{\phi_m^2 w^2}{|x|^2} \bigg\{ \frac{1}{4}\sum_{k=1}^mX_1^2 \ldots X_k^2   -\frac{n}{2} \frac{ |x|^2+2\rho x_n}{|x+2\rho e_n|^2}  
\sum_{k=1}^m X_1 \ldots X_k +\frac{|x|^2}{|x|^2+2\rho x_n} \sum_{k=1}^mX_1 \ldots X_k   \bigg\}dx,
\end{eqnarray*}
so inequality (\ref{inma}) is written
 \begin{eqnarray}
&& \lambda\int_{\Omega} \phi_m^2  w^2dx + \int_{\Omega}  \phi_m^2 |\nabla w|^2dx +\int_{\Omega} \frac{\phi_m^2 w^2}{|x|^2}
\bigg\{  -\frac{n}{2} \frac{ |x|^2+2\rho x_n}{|x+2\rho e_n|^2}  
 +\frac{|x|^2}{|x|^2+2\rho x_n}  \bigg\} \Big(\sum_{k=1}^m X_1 \ldots X_k \Big) dx \nonumber \\
 & \geq & \frac{n^2}{4}\int_{\Omega}  \frac{ |x|^2+4\rho x_n }{|x|^2   |x+2\rho e_n|^2}   \phi_m^2 w^2dx 
+ c \int_{\Omega}   V \phi_m^2 w^2 dx,
\label{w1}
\end{eqnarray}
It is clear from the above that (\ref{w1}) is valid for all functions $w\in C^{\infty}_c(\Omega)$ and moreover,
for a fixed $\lambda \geq 0$ the best constants $c$ for inequalities (\ref{inma}) and (\ref{w1}) coincide.
That common best constant shall be denoted by $b_m=b_m(\lambda)$.

Defining
\[
Q_m(x)=    \frac{n^2}{4} \frac{ |x|^2+4\rho x_n }{|x|^2   |x+2\rho e_n|^2}    +
\frac{1}{|x|^2}
\bigg\{  \frac{n}{2} \frac{ |x|^2+2\rho x_n}{|x+2\rho e_n|^2}  
 -\frac{|x|^2}{|x|^2+2\rho x_n}  \bigg\} \Big(\sum_{k=1}^m X_1 \ldots X_k \Big) 
\]
it then follows that
\[
 C_{m,r}(V) = 
\inf_{ W^{1,2}_0( \Omega\cap B_r ; \phi_m^2)}
\hspace{-0.3cm}\frac{\lambda \int_{\Omega\cap B_r}\phi_m^2   w^2dx +
\int_{\Omega\cap B_r} \phi_m^2|\nabla w|^2dx -
 \int_{\Omega\cap B_r} Q_m  \phi_m^2 w^2dx}
{  \int_{\Omega\cap B_r} \phi_m^2 V w^2 dx}
\]
where $W^{1,2}_0( \Omega\cap B_r ; \phi_m^2)$ denotes the closure of $C^{\infty}_c(\Omega\cap B_r)$ under the norm
\[
 \int_{\Omega\cap B_r}\phi_m ^2|\nabla w|^2dx + \int_{\Omega\cap B_r} \phi_m^2 w^2dx .
\]

Similarly to Section \ref{section:impr} we shall use a simpler way for expressing ${\cal C}^0_m(V)=\lim_{r\to 0+}C_{m,r}(V)$. For this we define
\[
C_{m,r}(V;\phi_m^2) =\inf_{W^{1,2}_0( \Omega\cap B_r ; \phi_m^2)}
\frac{ \int_{\Omega\cap B_r}  \phi_m^2|\nabla w|^2dx}{ \int_{\Omega\cap B_r}  V \phi_m^2  w^2dx}.
\]
and
\[
{\cal C}^0_m(V;\phi_m^2) =\lim_{r\to 0+} C_{m,r}(V;\phi_m^2).
\]


\begin{lemma}
Let $V$ be a non-negative potential in ${\cal A}_m(\Omega)$. Then
${\cal C}^0_m(V) ={\cal C}^0_m(V;\phi_m^2)$.
\end{lemma}
{\bf \em Proof.} The proof is quite similar to the proof of Lemma \ref{lem:fr_ir}. In particular we make use of the fact that
\[
\int_{\Omega} \Big(\frac{X_1}{|x|^2+2x_n}\Big)^{n/2} (X_1\ldots X_{m+1})^{1-n} < +\infty ,
\]
from which it easily follows that $|Q_m|$ is an $m$-subcritical potential. We omit further details. $\hfill\Box$



One important consequence of $m$-subcriticality is the following compactness property whose proof is similar to 
that of Lemma \ref{compactness} and is therefore omitted.
\begin{lemma}
Assume that the positive potential $V\in{\cal A}_m(\Omega)$ is $m$-subcritical. Then for any sequence $(w_k)$ which is bounded in $W^{1,2}_0(\Omega ;\phi_m^2)$ there exists a subsequence, also denoted by $(w_k)$, and a function $w_0\in W^{1,2}_0(\Omega ;\phi_m^2)$ so that
\begin{eqnarray*}
({\rm i}) && w_k \rightharpoonup w_0 \mbox{ in } W^{1,2}_0(\Omega ;\phi_m^2)\\
({\rm ii}) && \int_{\Omega} V \phi_m^2(w_k-w_0)^2dx \longrightarrow 0 .
\end{eqnarray*}
\end{lemma}


We can now state and prove the main theorems of this section.
\begin{theorem}
Let $V\in{\cal A}_m(\Omega)$ and $\lambda\geq 0$ be given and let $b_m(\lambda)$ be the best constant for the inequality
\begin{equation}
 \lambda\int_{\Omega} \phi_m^2  w^2dx + \int_{\Omega} \phi_m^2 |\nabla w|^2dx \geq  \int_{\Omega}  Q_m  \phi_m^2 w^2dx  + c \int_{\Omega}   V \phi_m^2 w^2 dx,
 \qquad w\in  W^{1,2}_0(\Omega;\phi_m^2),
\label{eq:insta}
\end{equation}
where
\[
 \phi_m(x)=
\frac{|x+\rho  e_n|^2-\rho^2}{|x|^{\frac{n}{2}}   |x+2\rho e_n|^{\frac{n}{2}}} X_1^{-\frac{1}{2}} \ldots X_m^{-\frac{1}{2}}
\]
and
\[
Q_m(x)=    \frac{n^2}{4} \frac{ |x|^2+4\rho x_n }{|x|^2   |x+2\rho e_n|^2}    +
\frac{1}{|x|^2}
\bigg\{  \frac{n}{2} \frac{ |x|^2+2\rho x_n}{|x+2\rho e_n|^2}  
 -\frac{|x|^2}{|x|^2+2\rho x_n}  \bigg\} \Big(\sum_{k=1}^m X_1 \ldots X_k \Big) ,
\]
with $X_i=X_i(|x|/3  \kappa \tilde{D})$, $\tilde{D}\geq D$.
If in addition
\[
b_m(\lambda)<{\cal C}_m^0
\]
then the best constant $b_m(\lambda)$ in (\ref{eq:insta}) is realized by a function $w_0\in W^{1,2}_0(\Omega;\phi_m^2)$. In particular the best constant $b_m(\lambda)$ is realized if $V$ is an $m$-subcritical potential.
\label{gera:thm:exmin1}
\end{theorem}
{\bf \em Proof.} The proof is similar to the proof of Theorem \ref{gera:thm:exmin} so we shall only give a sketch of the proof. What is important for our argument is that the potential $Q_m$ is a subcritical potential.
We consider a minimizing sequence $(w_k)$ for (\ref{eq:insta}) and without loss of generality we assume that
\[
\int_{\Omega}  Q_m \phi_m^2  w_k^2dx  +  b_m(\lambda) \int_{\Omega} V \phi_m^2  w_k^2 dx
=1, \qquad   \lambda\int_{\Omega} \phi_m^2  w_k^2dx + \int_{\Omega}  \phi_m^2 |\nabla w_k|^2dx  \longrightarrow 1, \; \mbox{ as }k\to \infty.
\]
Since $(w_k)$ is bounded in $W^{1,2}_0(\Omega ; \phi_m^2)$, it has a subsequence, which we assume is $(w_k)$ itself, which converges weakly to some $w_0\in W^{1,2}_0(\Omega ; \phi_m^2)$. We define $v_k=w_k-w_0$.

We consider a small enough $r>0$ so that $C_{m,r}(V;\phi_m^2)>b_m(\lambda)$ and a smooth cut-off function $\psi$
such that $\psi=1$ in $B_{r/2}$ and $\psi=0$ outside $B_r$. Arguing as in the proof of Theorem \ref{gera:thm:exmin} we obtain
\begin{equation}
\int_{\Omega}\phi_m^2 |\nabla v_k|^2 dx \geq  C_{m,r}(V;\phi_m^2) \int_{\Omega}V \phi_m^2  v_k^2 dx +o(1) 
\label{equm}
\end{equation}
and also 
\begin{equation}
\int_{\Omega} Q_m\phi_m^2  w_0^2dx  +  b_m(\lambda) \int_{\Omega} V \phi_m^2  w_0^2 dx + b_m(\lambda) \int_{\Omega} V \phi_m^2  v_k^2 dx  =1 +o(1)
\label{norm1a}
\end{equation}
and
\begin{equation}
\lambda \int_{\Omega} \phi_m^2  w_0^2dx + \int_{\Omega}  \phi_m^2 |\nabla w_0|^2dx + \int_{\Omega}  \phi_m^2 |\nabla v_k|^2dx = 1+o(1).
\label{norm2a}
\end{equation}
From (\ref{equm}) and (\ref{norm1a}) we obtain
\[
\int_{\Omega}\phi_m^2 |\nabla v_k|^2 dx \geq \frac{ C_{m,r}(V;\phi_m^2)}{b_m(\lambda)}\Big(  1-    \int_{\Omega} Q_m \phi_m^2  w_0^2dx  +  b_m(\lambda) \int_{\Omega} V \phi_m^2  w_0^2 dx        \Big) +o(1).
\]
Writing (\ref{eq:insta}) for $w=w_0$ we obtain from (\ref{norm2a}) that
\[
\int_{\Omega}  \phi_m^2 |\nabla v_k|^2dx \leq 1 -\int_{\Omega}Q_m \phi_m^2w_0^2dx -b_m(\lambda)\int_{\Omega} V\phi_m^2w_0^2dx +o(1) 
\]
and arguing as before we conclude that
\[
 \int_{\Omega} Q_m \phi_m^2  w_0^2dx  +  b_m(\lambda) \int_{\Omega} V \phi_m^2 V w_0^2 dx =1,
\]
that is $w_0$ is a minimizer for (\ref{eq:insta}). 
$\hfill\Box$


Finally, the next is a direct consequence of Theorem \ref{gera:thm:exmin1}.
\begin{theorem}
\label{h}
Let $n \geq 3$.  $V$ is a non-negative potential in ${\cal A}_m(\Omega)$ and $\tilde{D}\geq D$.
(a) Let $\lambda\geq 0$ and $b_m(\lambda)>0$ be the best constant for the following inequality
\begin{equation}
\lambda\int_{\Omega} u^2dx +
\int_{\Omega}|\nabla u|^2dx \geq  \frac{n^2}{4} \int_{\Omega}\frac{u^2}{|x|^2}dx + \frac{1}{4} \sum_{k=1}^m \int_{\Omega} \frac{u^2}{|x|^2}X_1^2\ldots X_k^2 dx
+ b_m(\lambda)\int_{\Omega} Vu^2dx, \quad u\in C^{\infty}_{c}(\Omega),
\label{in60}
\end{equation}
where $X_i=X_i(|x|/3 \kappa \tilde{D})$. If in addition
\[
b_m(\lambda)<{\cal C}_m^0 \ ,
\]
then the potential $n^2/4|x|^2 + 1/4|x|^2 \sum_{k=1}^m X_1^2\ldots X_k^2  +  b_m(\lambda)V(x)$ is a maximal potential, that is, inequality
(\ref{in60}) cannot be improved by adding a non-negative potential $W$ in the RHS. \newline
(b) If $V$ is an $m$-subcritical potential then there exist $\lambda\geq 0$ and a best constant $b_m(\lambda)>0$ such that (\ref{in60}) is true. Moreover the potential $n^2/4|x|^2 +  1/4|x|^2 \sum_{k=1}^m X_1^2\ldots X_k^2 + b_m(\lambda)V(x)$ is a maximal potential.
\end{theorem}




\section{Maximal potentials in finite cones}

In the previous two sections we characterized maximal potentials in bounded domains satisfying the exterior and interior ball condition. Analogous results  also hold true in the case of finite cones.

Let ${\mathscr C}_1$  be the cone  determined by the domain $\Sigma\subset S^{n-1}$ as defined in Section \ref{section:cone}; more
generally  we set ${\mathscr C}_r:={\mathscr C} \cap B_r$.
In this subsection we shall initially  be interested in the question of characterizing maximal potentials
for improved versions of inequality (\ref{cone}).
\begin{definition}
A non-negative potential $V\in L^{n/2}_{\rm loc}(\overline{{\mathscr C}_1}\setminus\{0\})$ is called {\em admissible} if there exists
$c>0$ such that
\begin{equation}
\int_{{\mathscr C}_1}|\nabla u|^2dx \geq  \left(  \Big(\frac{n-2}{2}\Big)^2  + \mu_1(\Sigma) \right) \int_{{\mathscr C}_1} \frac{u^2}{|x|^2} dx
+c  \int_{{\mathscr C}_1} Vu^2dx, \quad u\in C^{\infty}_{c}({\mathscr C}_1).
\label{1234c}
\end{equation}
We denote by ${\cal A}({\mathscr C}_1)$ the class of all admissible potentials. 
\end{definition}
 
Once again there is a big variety of admissible potentials. For example if $V$ satisfies 
$\int_{{\mathscr C}_1} V^{n/2}X_1^{1-n}dx <+\infty$
where $X_1=X_1(|x|)$, then $V$ is admissible by Theorem \ref{i}.


Given $V\in{\cal A}({\mathscr C}_1)$ and $r\in (0,1)$ we define
\[
C_r(V) =\inf_{u\in C^{\infty}_c({\mathscr C}_r)}
\frac{\int_{{\mathscr C}_r}|\nabla u|^2dx - \left(  \Big(\frac{n-2}{2}\Big)^2  + \mu_1(\Sigma) \right)   
\int_{{\mathscr C}_r}\frac{u^2}{|x|^2}dx}{\int_{{\mathscr C}_r}Vu^2dx}
\]
and
\[
{\cal C}^0(V)=\lim_{r\to 0+}C_r(V).
\]



\begin{definition}
We say that the potential $V\in {\cal A}({\mathscr C}_1)$ is subcritical if ${\cal C}^0(V) =+\infty$.
\end{definition}

The analogue of Lemma \ref{lem:77} is the following
\begin{lemma}
Let $V$ be a non-negative potential satisfying
\[
\int_{{\mathscr C}_1} V^{n/2}X_1^{1-n}dx <+\infty,
\]
where $X_1=X_1(|x|)$. Then $V$ is a subcritical potential.
\end{lemma}


Given $u\in C^{\infty}_c({\mathscr C}_1)$ we define the function $w$ by
\begin{equation}
u(x) = |x|^{-\frac{n-2}{2}}\phi_1(\omega) w(x) =:\psi(x) w(x),
\label{uwc}
\end{equation}
where $\omega:=\frac{x}{|x|}  \in \Sigma$  and  $\phi_1(\omega)$, is the first eigenfunction of the Dirichlet Laplacian in $\Sigma$. 
After some computations we obtain
\[
\int_{{\mathscr C}_1}|\nabla u|^2dx =\int_{{\mathscr C}_1}  \psi^2  |\nabla w|^2dx + \left(  \Big(\frac{n-2}{2}\Big)^2  + \mu_1(\Sigma) \right)   \int_{{\mathscr C}_1}  \psi^2 w^2dx .
\]
As usual we denote by 
$W^{1,2}_0({\mathscr C}_1  ; \psi^2)$ the closure of $C^{\infty}_c({\mathscr C}_1)$ under the norm
\[
\int_{{\mathscr C}_1}\psi^2|\nabla w|^2dx +
 \int_{{\mathscr C}_1} \psi^2  w^2dx .
\]
It is easily seen that inequality (\ref{1234c}) 
 under the change of variables (\ref{uwc}) is equivalent to
\[
\int_{{\mathscr C}_1} \psi ^2|\nabla w|^2dx \geq 
  c \int_{{\mathscr C}_1} V \psi^2 w^2dx , \qquad w\in W^{1,2}_0({\mathscr C}_1;\psi^2).
\]
The analogues of Theorems \ref{gera:thm:exmin} and \ref{d} read as follows: 

\begin{theorem}
Let $V\in{\cal A}({\mathscr C}_1)$  and suppose that $b$ is the best constant for the inequality
\begin{equation}
\int_{{\mathscr C}_1} \psi ^2|\nabla w|^2dx \geq 
  b \int_{{\mathscr C}_1} V \psi^2 w^2dx , \qquad w\in W^{1,2}_0({\mathscr C}_1;\psi^2).
\label{eq:w:no1c}
\end{equation}
If in addition $b<{\cal C}^0(V)$ 
then the best constant $b$ in (\ref{eq:w:no1c}) is realized by a function $w_0\in W^{1,2}_0({\mathscr C}_1;\psi^2)$. In particular the best constant $b$ is realized if the potential $V$ is subcritical.
\end{theorem}


\begin{theorem}
Let $V$ be a non-negative potential in ${\cal A}({\mathscr C}_1)$. \newline
 (a) Let $b>0$ be the best constant in the following inequality
\begin{equation}
\int_{{\mathscr C}_1}|\nabla u|^2dx \geq  \bigg(  \Big(\frac{n-2}{2}\Big)^2  + \mu_1(\Sigma) \bigg) 
\int_{{\mathscr C}_1}\frac{u^2}{|x|^2}dx +b \int_{{\mathscr C}_1} Vu^2dx,
 \quad u\in  C^{\infty}_{c}({\mathscr C}_1) \ .
\label{eq:thm:123:c}
\end{equation}
 If in addition $b<{\cal C}^0(V)$
then the potential
$\big[  \big(\frac{n-2}{2}\big)^2  + \mu_1(\Sigma) \big]   |x|^{-2} +b V(x)$
is a maximal potential, that is inequality
(\ref{eq:thm:123:c}) cannot be improved by adding a non-negative potential $W$ in the RHS. \newline
\noindent (b) If $V$ is a subcritical potential then there exists a best constant $b>0$ such that (\ref{eq:thm:123:c}) is true. Moreover the potential
$\big[  \big(\frac{n-2}{2}\big)^2  + \mu_1(\Sigma) \big]   |x|^{-2}   +bV(x)$ is a maximal potential.
\end{theorem}

The proofs of these theorems are quite similar and slightly simpler to the proofs of of Theorems \ref{gera:thm:exmin} and \ref{d}.


In analogy to the results of Section \ref{sub:log:impr}  we have  similar theorems for the improved Hardy inequality involving logarithmic corrections. In particular we have


\begin{definition}
A non-negative potential $V\in L^{n/2}_{\rm loc}(\overline{{\mathscr C}_1}\setminus\{0\})$ is called {\em m--admissible} if there exists
$c>0$ such that for $ u\in C^{\infty}_{c}({\mathscr C}_1)$, there holds
\begin{equation}
\int_{{\mathscr C}_1}|\nabla u|^2dx \geq  \left(  \Big(\frac{n-2}{2}\Big)^2  + \mu_1(\Sigma) \right) \int_{{\mathscr C}_1} \frac{u^2}{|x|^2} dx
+\frac{1}{4}\sum_{i=1}^m
\int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx  
+c  \int_{{\mathscr C}_1} Vu^2dx \ .
\label{1234cc}
\end{equation}
\end{definition}
We denote by ${\cal A}_m({\mathscr C}_1)$ the class of all $m$--admissible potentials. 


Given $V\in{\cal A}_m({\mathscr C}_1)$ and $r\in (0,1)$ we define
\[
C_{m,r}(V) =\inf_{u\in C^{\infty}_c({\mathscr C}_r)}
\frac{\int_{{\mathscr C}_r}|\nabla u|^2dx - \bigg(  \Big(\frac{n-2}{2}\Big)^2  + \mu_1(\Sigma) \bigg)   
\int_{{\mathscr C}_r}\frac{u^2}{|x|^2}dx- \frac{1}{4}\sum_{i=1}^m
\int_{{\mathscr C}_{r}}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx  }{\int_{{\mathscr C}_r}Vu^2dx}
\]
and
\[
{\cal C}^0_m(V)=\lim_{r\to 0+}C_{m,r}(V).
\]



\begin{definition}
We say that the potential $V\in {\cal A}_m({\mathscr C}_1)$ is $m$-subcritical if ${\cal C}^0_m(V) =+\infty$.
\end{definition}

Changing variables by 
\[
u(x) =|x|^{-\frac{n-2}{2}}\phi_1(\omega)X_1^{-\frac{1}{2}}\ldots  X_m^{-\frac{1}{2}} w(x)=:\psi_m(x) w(x) \ ,
\]
inequality (\ref{1234cc}) is equivalent to
\[
\int_{{\mathscr C}_1} \psi_m^2|\nabla w|^2dx \geq 
  c \int_{{\mathscr C}_1} V \psi^2_m w^2dx , \qquad w\in W^{1,2}_0({\mathscr C}_1;\psi^2_m).
\]

Now the analogues of Theorems \ref{gera:thm:exmin1} and \ref{h} are as follows

\begin{theorem}
Let $V\in{\cal A}_m({\mathscr C}_1)$ and let $b_m$ be the best constant for the inequality
\begin{equation}
  \int_{{\mathscr C}_1}  \psi_m^2 |\nabla w|^2dx \geq  b_m \int_{{\mathscr C}_1}  V \psi_m^2 w^2 dx,
 \qquad w\in  W^{1,2}_0(\Omega;\psi_m^2),
\label{eq:instac}
\end{equation}
If in addition $b_m<{\cal C}^0_m(V)$ 
then the best constant $b_m$ in (\ref{eq:instac})  is realized   in  $W^{1,2}_0({\mathscr C}_1;\psi^2_m)$. In particular the best constant $b_m$ is realized if the potential $V$ is $m$--subcritical.
\end{theorem}

\begin{theorem}
Let $V$ be a non-negative potential in ${\cal A}_m({\mathscr C}_1)$. \newline
 (a) Let $b_m>0$ be the best constant in the following inequality
\begin{equation}
\int_{{\mathscr C}_1}|\nabla u|^2dx \geq  \bigg(  \Big(\frac{n-2}{2}\Big)^2  + \mu_1(\Sigma) \bigg)\int_{{\mathscr C}_1}\frac{u^2}{|x|^2}dx +\frac{1}{4}\sum_{i=1}^m   \int_{{\mathscr C}_{1}}\frac{u^2}{|x|^2}X_1^2\ldots X_i^2 dx +
+b_m \int_{{\mathscr C}_1} Vu^2dx, 
\label{eq:thm:123:cc}
\end{equation}
where $u\in  C^{\infty}_{c}({\mathscr C}_1)$.  \newline
 If in addition $b_m<{\cal C}^0_m(V)$
then the potential
$\big[  \big(\frac{n-2}{2}\big)^2  + \mu_1(\Sigma)    
 + \frac{1}{4}\sum_{i=1}^m
X_1^2\ldots X_i^2  \big] |x|^{-2} +b_m V(x)$
is a maximal potential, that is inequality
(\ref{eq:thm:123:cc}) cannot be improved by adding a non-negative potential $W$ in the RHS. \newline
\noindent (b) If $V$ is a subcritical potential then there exists a best constant $b_m>0$ such that (\ref{eq:thm:123:cc}) is true. Moreover the potential
$\big[  \big(\frac{n-2}{2}\big)^2  + \mu_1(\Sigma) + \frac{1}{4}\sum_{i=1}^m
X_1^2\ldots X_i^2   \big]   |x|^{-2}   +b_mV(x)$ is a maximal potential.
\end{theorem}






{\bf Remark.} All the above results involve a single point singularity on the boundary. Similar results however can be obtained when there
are multiple singularities. For instance we have the following result


\begin{theorem}
\label{th72}
Assume that $\Omega \subset {\mathbb{R}}^n$, $n \geq 3$, is a bounded domain that satisfies an exterior ball condition at each of the points $a_1,\ldots,a_m\in \partial\Omega$.
Then there exist a positive constant $c=c(n,m)$ depending only on $n$ and $m$ and a positive constant $\lambda$ such that
\[
\lambda \int_{\Omega}u^2 dx + \int_{\Omega}|  \nabla u|^2 dx \geq \frac{n^2}{4}\sum_{k=1}^m \int_{\Omega}\frac{u^2}{|x-a_k|^2}dx
+c \Big( \int_{\Omega} |u|^{\frac{2n}{n-2}}W dx\Big)^{\frac{n-2}{n}} \ , ~~~~~u\in C^{\infty}_c(\Omega) \ ,
\]
where 
\[ 
 W=W(x):=\prod_{k=1}^{m} X_1^{\frac{2n-2}{n-2}}(\frac{|x-a_k|}{D}),~~~~~{\rm and }~~~~ D:=  \max_{k=1,\ldots, m}~ \sup_{x \in \Omega} |x-a_k|  \ .
 \]
\end{theorem}
The proof uses ideas that we have used so far in connection with standard partition of unity arguments; we omit further details.






\medskip

{ \bf Acknowledgement} We would like to thank the referees for their comments which led to a considerable improvement of the presentation
of this work. AT acknowledges partial support by ELKE grant, University of Crete.















