text:
\begin{document}

\title[Finsler Hardy inequalities]{Finsler Hardy inequalities}
\author[A. Mercaldo]{A. Mercaldo$^1$}
\author[M. Sano]{M. Sano$^2$}
\author[F. Takahashi]{F. Takahashi$^3$}

\date{\today}
\setcounter{footnote}{1}
\footnotetext{
Universit\`a di Napoli Federico II, Dipartimento di Matematica e Applicazioni
`R. Caccioppoli'',
Complesso Monte S. Angelo, via Cintia, 80126 Napoli, Italy;\\
e-mail: {\tt mercaldo@unina.it}}

\setcounter{footnote}{2}
\footnotetext{
Department of Mathematics, Tokyo Institute of Technology,
2-12-1 Ookayama, Meguro-ku, Tokyo 152-8551, Japan. \\
e-mail:{\tt megumisano0609@gmail.com}}

\setcounter{footnote}{3}
\footnotetext{
Department of Mathematics, Osaka City University,
3-3-138 Sugimoto, Sumiyoshi-ku, Osaka 558-8585, Japan. \\
e-mail:{\tt futoshi@sci.osaka-cu.ac.jp}}

\begin{abstract}
In this paper we present a unified simple approach to anisotropic Hardy inequalities in various settings.
We consider Hardy inequalities which involve a Finsler distance from a point or from the boundary of a domain.
The sharpness and the non-attainability of the constants in the inequalities are also proved.

\medskip
\noindent
{\sl Key words: Hardy inequality, Finsler norm, Best constant}
\rm
\\[0.1cm]
{\sl 2010 Mathematics Subject Classification: 26D10, 26D15}

\rm
\end{abstract}
\maketitle

\bigskip

\section{Introduction}
\label{section:Introduction}
The interest in the so-called anisotropic problems arose from G. Wulff's work on crystal shapes and minimization of anisotropic surface tensions in 1901
and it is becoming increasingly important in different contexts, as in the field of phase changes and phase of separation in multiphase materials (cf. \cite{BCS(book)}, \cite{Bellettini-Paolini}).
This justifies the necessity to extend to anisotropic case many of the classical tools, which are useful in classical variational problems.
In this paper we are interested in sharp anisotropic Hardy-type inequalities.
The basic idea is to endow the space $\re^N$ with the distance obtained by a Finsler metric and to extend several Hardy-type inequalities in such a new geometrical context.

The classical Hardy inequality asserts that for any $p \ge 1$, $p \ne N$, if $\Omega$ is a domain of $\re^N (N \ge 2)$ containing the origin,
then
\begin{equation}
\label{classical Hardy}
	\left | \frac{N-p}{p} \right|^p \int_{\Omega} \frac{|u|^p}{|x|^p} dx \le \int_{\Omega} \left| \nabla u \right|^p dx
\end{equation}
holds for $ u \in C_0^{\infty}(\Omega)$ if $1 \le p < N$ and for $u \in C_0^{\infty}(\Omega \setminus \{ 0 \})$ if $p > N$.
Here the constant $\left| \frac{N-p}{p} \right|^p$ is sharp and never attained when $p > 1$.
The critical Hardy inequality corresponding to the case $p=N$ has also been studied (cf. \cite{BFT1}, \cite{BFT2}, \cite{Ioku-Ishiwata}, \cite{Sano-TF(EJDE)}, \cite{TF});
in this case, for example, if $\Omega$ is a ball having center at the origin and radius $R$,
then $|x|^N$ appearing in \eqref{classical Hardy} is replaced by the Hardy potential of the type $|x|^N(\log \frac{R}{|x|})^N$.

Several variants of the Hardy inequalities \eqref{classical Hardy} have been known.
Among these we recall the {\it geometric type Hardy inequality} which asserts that, if $1 < p < \infty$ and $\Omega$ is a convex, possibly unbounded domain in $\re^N$,
then
\begin{equation}
\label{geo_Hardy}
	\(\frac{p-1}{p}\)^p \intO \frac{|u|^p}{(d(x))^p} dx \le \intO |\nabla u|^p dx\,, \qquad u \in C_0^{\infty}(\Omega)
\end{equation}
where $d(x)={\rm dist}(x, \partial \Omega)$ denotes the usual distance function from the boundary of $\Omega$ and the constant $\left( \frac{p-1}{p} \right)^p$ is sharp.
An improved version of \eqref{geo_Hardy} has been proved in \cite{BFT1}
where the best constant is given for a larger class of domains which verify the geometric assumption that $d$ is $p$-superharmonic in $\Omega$, i.e.,
\begin{equation}
\label{delta d}
	-\Delta_p d \ge 0
\end{equation}
in the distribution sense.
Here $\Delta_p$ is the $p$-Laplace operator $\Delta_p u = {\rm div}(|\nabla u|^{p-2} \nabla u)$.

Anisotropic Hardy inequalities are also known.
For example \eqref{classical Hardy} and \eqref{geo_Hardy} have been extended to the case where the Euclidean norm is replaced by a Finsler norm in \cite{Van Schaftingen}, \cite{Della Pietra-Blasio-Gavitone}
when $p=2$, and \cite{Bal}, \cite{BGM}, \cite{Brasco-Franzina} when $p \ne 2$, respectively.
The method in \cite{Brasco-Franzina} and \cite{Bal} is to use Picone type identities in Finsler setting.

In this paper further anisotropic sharp Hardy inequalities will be proved.
We consider a Finsler norm $H$ and its polar function $H^0$, whose definitions are given in \S \ref{section:Notation}.
Our first main result gives the following {\sl sharp anisotropic subcritical Hardy inequality} which is a consequence of Theorem \ref{Thm:H_p} in \S \ref{section:Hardy}
and Theorem \ref{Thm:H_p attainability} in \S \ref{section:attainability}.

\begin{theorem}{\rm (Sharp anisotropic subcritical Hardy inequality)}
Assume  $1 \le p < N$ or  $ p> N$. Let $\Omega$ be a domain in $\re^N$. Then the following inequality
\begin{equation}
\label{FH_p_intro}
	\left| \frac{N-p}{p} \right|^p \intO \frac{|u|^p}{\( H^0(x) \)^p} dx \le \intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p dx
\end{equation}
holds true for any $u \in C_0^{\infty}(\Omega)$ if $1 \le p < N$,
and for any $u \in C_0^{\infty}(\Omega \setminus \{ 0 \}$ if $p > N$.
Moreover if $0 \in \Omega$,
the constant $\( \frac{N-p}{p} \)^p$ is sharp and not attained if $1 < p < N$ and the constant $N-1$ is attained for any nonnegative $H^0$-radially decreasing, compactly supported function when $p=1$.
\end{theorem}
For the notion of $H^0$-radially decreasing function, see \S \ref{section:attainability}.

The critical case $p=N$ is also studied and the following result is a consequence of Theorem \ref{Thm:H_N} and Theorem \ref{Thm:H_N attainability}.
\begin{theorem}{\rm (Sharp anisotropic critical Hardy inequality)}
Let $\Omega \subset \re^N$, $N \ge 2$, be a bounded domain containing the origin and put $R = \sup_{x \in \Omega} H^0(x)$.
Then the inequality
\begin{equation}\label{FH_N_intro}
	\( \dfrac{N-1}{N} \)^N \intO \frac{|u|^N}{(H^0(x))^N (\log \frac{R}{H^0(x)})^N} dx
	\le \intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^N dx
\end{equation}
holds for any $u \in C_0^{\infty}(\Omega)$.
Moreover the constant $\( \frac{N-1}{N} \)^N$ is sharp and not attained.
\end{theorem}

In \S \ref{section:geo Hardy} an {\sl anisotropic Hardy inequality of geometric type} is proved,
while the attainability of the best constant is also studied in \S \ref{section:attainability}.
\begin{theorem}{\rm (Anisotropic Hardy inequality of geometric type)}
Let $1 < p < \infty$ and suppose $\delta = \delta(x)$ is a nonnegative, $\Delta_{H,p}$-superharmonic function on $\Omega$, i.e.,
\begin{equation}
\label{delta_p d}
	-\Delta_{H,p} \delta \ge 0
\end{equation}
in the distributional sense, where
\[
	\Delta_{H,p} \delta(x) = {\rm div} \( H^{p-1}(\nabla \delta(x))(\nabla H)(\nabla \delta(x)) \)
\]
denotes the Finsler $p$-Laplacian of $\delta$.
Then the inequality
\begin{equation}
\label{eq:geometric Finsler Hardy}
	\( \frac{p-1}{p} \)^p \intO \frac{|u|^p}{\delta^p} H^p(\nabla \delta) dx \le \intO |\nabla u \cdot (\nabla H)(\nabla \delta)|^p dx
\end{equation}
holds true for any $u \in C_0^{\infty}(\Omega)$.
\end{theorem}

Condition \eqref{delta_p d} will be discussed in \S \ref{section:geo Hardy}.
Here we just remark that \eqref{delta_p d} coincides with \eqref{delta d} when we choose the Euclidean norm as the Finsler norm
and $d$ as $\delta$ in \eqref{delta_p d}.

In Section \S \ref{section:Weighted} a weighted Finsler-Hardy-Poincar\'e inequality have been proved with respect to a weight $\rho$ which satisfies suitable assumptions.

Our Hardy inequalities will be proved by using a simple unified approach valid for any choice of Hardy potential.
A related approach has been adopted in \cite{BFT1}, \cite{BFT2}.

Finally we fix our attention on two anisotropic Hardy inequalities (\ref{FH_p_intro}) and (\ref{FH_N_intro}) which are quite different from each other in view of their forms, scaling structures and optimal constants.
However, according to \cite{Sano-Takahashi}, we can reveal an unexpected relation between the critical and the subcritical anisotropic Hardy inequalities
and show that the critical anisotropic Hardy inequality on a ball is embedded into a family of the subcritical anisotropic Hardy inequalities on the whole space
by using a transformation which connects both inequalities.
In \S \ref{section:Relation} we show that the transformation conserves not only the best constants but also the scale invariance structures of both inequalities,
at least in the $H^0$-radial setting.

\vspace{1em}\noindent
{\it Note added to Proof.}

After completing this work, the authors of this paper are informed by Professor M. Ruzhansky of his recent seminal works on the Hardy, Rellich, and other functional inequalities on homogeneous groups
with arbitrary quasi-norms \cite{ORS(arXiv2016)}, \cite{RS(arXiv2016)}, \cite{RS(Adv.Math)}, \cite{RS(CCM)}, \cite{RSY(TAMS)}, \cite{RSY(IEOT)}.
In \cite{RS(Adv.Math)}, for example, the following $L^p$-Hardy inequality
\[
	\left\| \frac{f}{|x|} \right\|_{L^p(\mathbb{G})} \le \frac{p}{Q-p} \| \mathcal{R} f \|_{L^p(\mathbb{G})}, \quad 1 < p < Q
\]
is proved on a homogeneous group $\mathbb{G}$ with the homogeneous dimension $Q$ and a homogeneous quasi-norm $| \cdot |$.
Here the operator $\mathcal{R} = \mathcal{R}_{| \cdot |} = \frac{d}{d|x|}$ is called a radial operator.
Other problems such as the optimality of constants and the existence of remainder terms are also studied in the above and subsequent papers.
If $\mathbb{G}$ is chosen as an abelian group $(\re^N, +)$ and $| \cdot |$ as $H^0(\cdot)$, our Hardy inequality \eqref{FH_p_intro} is nothing but the above inequality
since $\mathcal{R} f = \frac{x}{H^0(x)} \cdot \nabla f$ in this situation.
Their proof is based on the polar coordinate decomposition
\[
	\int_{\mathbb{G}} f(x) dx = \int_0^{\infty} \int_{\mathcal{S}} f(r\omega) r^{Q-1} d\sigma(\omega) dr
\]
for $f \in L^1(\mathbb{G})$ where $\mathcal{S} = \{ x \in \mathbb{G} \, : \, |x| = 1 \}$, and is different from ours in this paper, which depends basically on the use of divergence theorem.
In this sense, many results in the present paper can be seen as special cases of the results above with different proofs.
We stick to the Finsler setting since we want to apply our inequalities to the nonlinear problems involving the Finsler Laplacian.
Also we believe that our method of proof will be useful in such possible applications.

\section{Notation and basic properties}
\label{section:Notation}

Let $H: \re^N \to \re$ be a nonnegative, convex function of class $C^2(\re^N \setminus \{ 0 \})$,
which is even and positively homogeneous of degree 1:
\begin{equation}
\label{pos_H}
	H(t \xi ) = |t| H(\xi), \quad \forall \xi \in \re^N, \, \forall t \in \re.
\end{equation}
The above assumptions give the existence of positive constants $\alpha$ and $\beta$ such that
\[
	\alpha |\xi| \le H(\xi) \le \beta |\xi|, \quad \xi \in \re^N.
\]
Let $K$ denote the convex closed set
\[
	K = \{ \xi \in \re^N \,: \, H(\xi) \le 1 \}.
\]
Sometimes we will say that $H$ is  {\it the gauge} of $K$.
The {\it polar function} of $H$  is the function $H^0: \re^N \to \re$
 defined by
\[
	H^0(x) = \sup_{\xi \in \re^N \setminus \{ 0 \}} \frac{\xi \cdot x}{H(\xi)} = \sup_{\xi \in K} (\xi \cdot x)\,, \qquad x \in \re^N\,.
\]
Throughout this paper $\xi \cdot x = \sum_{j=1}^N \xi_j x_j$ denotes the usual inner product of $\re^N$.

\noindent Note that, by definition of $H^0$, the {\sl Schwarz inequality} holds true, i.e.,
\begin{equation}
\label{Schwarz}
	|\xi \cdot x| \le H(\xi) H^0(x), \quad \forall \xi, x \in \re^N.
\end{equation}
It is well-known that $H^0$ is a convex, positively homogeneous of degree 1, continuous function on $\re^N$,
and the following inequality is satisfied
\[
	\frac{1}{\beta} |x| \le H^0(x) \le\frac{1}{\alpha} |x|, \quad \forall x \in \re^N.
\]
Also the following equality
\[
	H(\xi) = (H^0)^0(\xi) = \sup_{x \in \re^N \setminus \{ 0 \}} \frac{x \cdot \xi}{H^0(x)}\,, \qquad \xi \in \re^N,
\]
holds and $H^0$ itself is the gauge of the closed convex set
\[
	K^0 = \{ x \in \re^N \,: \, H^0(x) \le 1 \}.
\]
We say that $K$ and $K_0$ are polar to each other.
The interior set of $K^0$, i.e.,
\[
	\mathcal{W} = \{ x \in \re^N \,: \, H^0(x) < 1 \}
\]
is called the {\it Wulff ball}, or $H^0$-unit ball, and we denote $\kappa_N = \mathcal{H}^N(\mathcal{W})$.
In this case, the {\it anisotropic $H$-perimeter} of $\mathcal{W}$, denoted by $P_H(\mathcal{W}; \re^N)$, is
$P_H(\mathcal{W}, \re^N) = N \kappa_N$.
For more explanation about the anisotropic perimeter, see \cite{AFTL} and {\cite{Belloni-Ferone-Kawohl}.
Throughout the paper, we denote
\[
	\w_{N-1} = P_H(\mathcal{W}; \re^N) = N \kappa_N.
\]
Denote
$$
	\mathcal{W}_R = \{ x \in \re^N \,|\, H^0(x) <R \}
$$
for any $R>0$ and we identify $\mathcal{W}_\infty$ with $\re^N$.
A function  $H \in C^2 \(\re^N\setminus \{0\} \)$ is a {\it Finsler  norm} if it satisfies properties \eqref{pos_H},
and moreover $H$ is strongly convex in the sense that the Hesse matrix of $H^2$, ${\rm Hess}(H^2)$ is positive definite.
For references about Finsler norms (or, more in general, for Finsler metrics) see \cite{BCS(book)}, \cite{Bellettini-Paolini}.

Here we just recall further properties, whose proofs are contained in \cite{Bellettini-Paolini} Lemma 2.1, 2.2, or \cite{Van Schaftingen} Proposition 6.2.
\begin{proposition}
\label{Prop:identities} Let $H$ be a Finsler norm on $\re^N$.
Then the following properties hold true:
\begin{enumerate}
\item[\rm(1)] $\nabla_{\xi} H(\xi) \cdot \xi = H(\xi)$, \qquad $\xi \ne 0$.
\item[\rm(2)] $\(\nabla_{\xi} H \)(t \xi) = \frac{t}{|t|} \( \nabla_{\xi} H \)(\xi)$, \qquad $\xi \ne 0, t \ne 0$.
\item[\rm(3)] $\(\nabla^2_{\xi} H \)(t \xi) = \frac{1}{|t|} \( \nabla_{\xi} H \)(\xi)$, \qquad $\xi \ne 0, t \ne 0$.
\item[\rm(4)] $H\( \nabla H^0(x) \) = 1$.
\item[\rm(5)] $H^0(x) \( \nabla_{\xi} H \) \( \nabla_x H^0(x) \) = x$.
\end{enumerate}
Similarly, following properties also hold true:
\begin{enumerate}
\item[\rm(1')] $\nabla_{x} H^0(x) \cdot x = H^0(x)$, \qquad $x \ne 0$.
\item[\rm(2')] $\(\nabla_{x} H^0 \)(t x) = \frac{t}{|t|} \( \nabla_{x} H^0 \)(x)$, \qquad $x \ne 0, t \ne 0$.
\item[\rm(3')] $\(\nabla^2_{x} H^0 \)(t x) = \frac{1}{|t|} \( \nabla_{x} H^0 \)(x)$, \qquad $x \ne 0, t \ne 0$.
\item[\rm(4')] $H^0 \( \nabla_{\xi} H(\xi) \) = 1$.
\item[\rm(5')] $H(\xi) \( \nabla_{x} H^0 \) \( \nabla_{\xi} H(\xi) \) = \xi$.
\end{enumerate}
\end{proposition}

Finally, given a smooth function $u$ on $\re^N$, {\it the Finsler Laplace operator} of $u$ (associated with $H$) is defined by
\begin{align*}
	\Delta_H u(x) &= {\rm div} \( H(\nabla u(x)) \( \nabla_{\xi} H \)(\nabla u(x)) \) \\
	&= \sum_{j=1}^N \frac{\pd}{\pd x_j} \( H(\xi) H_{\xi_j}(\xi) \Big|_{\xi = \nabla u(x)} \)
\end{align*}
and, more generally, for any $1<p<\infty$, {\it the Finsler $p$-Laplace operator} $\Delta_{H,p}$ by
\[
	\Delta_{H,p} u(x) = \mbox{div} \( H^{p-1}(\nabla u(x))(\nabla_{\xi} H)(\nabla u(x)) \).
\]
\vspace{1em}
Note that though the Finsler gradient vector
\[
	\nabla_H u(x) = H(\nabla u(x)) \( \nabla_{\xi} H \)(\nabla u(x)) = \nabla_{\xi} \( \frac{1}{2} H^2(\xi) \) \Big|_{\xi = \nabla u(x)}
\]
is a nonlinear operator, thanks to the strict convexity oh $H$, $\Delta_H$ and $\Delta_{H,p}$ is a uniformly elliptic operator locally.
The Finsler Laplacian has been widely investigated in literature and its notion goes back to the work of G. Wulff, who considered it to describe the theory of crystals.
Many other authors developed the related theory in several settings, considering both analytic and geometric points of view,
see (\cite{AIS}, \cite{Bellettini-Paolini}, \cite{Belloni-Ferone-Kawohl}, \cite{BCS}, \cite{Cianci-Salani}, \cite{Della Pietra-Blasio}, \cite{Ferone-Kawohl} and references therein).

\section{Hardy type inequalities}
\label{section:Hardy}

In this section, we prove several Finsler Hardy type inequalities in a unified method.
This simple approach is motivated by \cite{BFT1}, \cite{BFT2}, and \cite{TF}.

\begin{theorem}{\rm (Sharp anisotropic subcritical Hardy inequality)}
\label{Thm:H_p}
Assume  $1 \le p < N$ or $ p> N$.
Let $\Omega$ be a domain in $\re^N$. Then the following inequality holds true
\begin{equation}
\label{FH_p}
	\left| \frac{N-p}{p} \right|^p \intO \frac{|u|^p}{\( H^0(x) \)^p} dx \le \intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p dx
\end{equation}
for any $u \in C_0^{\infty}(\Omega)$ if $1 \le p < N$,
and for any $u \in C_0^{\infty}(\Omega \setminus \{ 0 \})$ if $p > N$.
\end{theorem}

\begin{remark}\rm
The anisotropic subcritical Hardy inequality (\ref{FH_p}) is invariant under the scaling $u_\la (x) = \la^{\frac{N-p}{p}} u(\la x)$, $(\la >0)$ when $\Omega =\re^N$.
Indeed, by (\ref{pos_H}) we can easily check the following equalities
\begin{align*}
\intRN \frac{|u_\la (x) |^p}{\( H^0(x) \)^p} dx &= \intRN \frac{|u (y) |^p}{\( H^0(y) \)^p} dy, \\
\intRN \left| \frac{x}{H^0(x)} \cdot \nabla u_\la (x) \right|^p dx &= \intRN \left| \frac{y}{H^0(y)} \cdot \nabla u (y) \right|^p dy.
\end{align*}
\end{remark}

\begin{proof}
We just prove the assertion when $1 \le p < N$, since the proof of the case where $p > N$ is similar.
Define
\[
	F(x) = \frac{x}{\( H^0(x) \)^{\la}}, \quad x \in \Omega.
\]
Then we have
\begin{align*}
	{\rm div} F(x) &= \frac{N}{\( H^0(x) \)^{\la}} + (-\la) \(H^0(x) \)^{-\la-1} \nabla_x H^0(x) \cdot x \\

	&= \frac{N-\la}{\( H^0(x) \)^{\la}},
\end{align*}
since $\nabla H^0(x) \cdot x = H^0(x)$ by Proposition \ref{Prop:identities} (1').
Take $\la = p$.
Then for any $u \in C_0^{\infty}(\Omega)$,
we compute
\begin{align*}
	&\intO \frac{|u(x)|^p}{\( H^0(x) \)^p} dx
	= \left| \frac{1}{N-p} \intO |u(x)|^p \mbox{div} F(x) dx \right| \\
	&= \left| - \frac{1}{N-p} \intO \nabla (|u|^p ) \cdot \frac{x}{\( H^0(x) \)^p} dx \right| \\
	&= \left| - \frac{p}{N-p} \intO |u|^{p-2} u \nabla u  \cdot \frac{x}{\( H^0(x) \)^p} dx \right| \\
	&\le \left| \frac{p}{N-p} \right| \( \intO \frac{|u|^p}{\( H^0(x) \)^p} dx \)^{(p-1)/p} \( \intO \left| \frac{x}{\( H^0(x) \)} \cdot \nabla u \right|^p dx \)^{1/p}.
\end{align*}
This yields the conclusion.
\end{proof}

\begin{remark}\rm
Note that by \eqref{Schwarz} and the positively 1-homogeniety of $H^0$,
the right-hand side of the inequality \eqref{FH_p} is estimated as
\[
	\intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p dx  \le \intO \( H( \nabla u(x)) \)^p dx.
\]
Thus Theorem \ref{Thm:H_p} improves the following inequality by Van Schaftingen (\cite{Van Schaftingen} Proposition 7.5),
\[
	\left| \frac{N-p}{p} \right|^p \intO \frac{|u|^p}{\( H^0(x) \)^p} dx\le \intO \( H( \nabla u(x)) \)^p dx,
\]
which is obtained by the use of symmetrization.
\end{remark}
\bigskip
Next result concerns the critical case $p=N$.

\begin{theorem}{\rm (Sharp anisotropic critical Hardy inequality)}
\label{Thm:H_N}
Let $\Omega \subset \re^N$, $N \ge 2$, be a bounded domain and put $R = \sup_{x \in \Omega} H^0(x)$.
Then the inequality
\begin{equation}\label{FH_N}
	\( \dfrac{N-1}{N} \)^N \intO \frac{|u|^N}{(H^0(x))^N (\log \frac{R}{H^0(x)})^N} dx
	\le \intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^N dx
\end{equation}
holds for any $u \in C_0^{\infty}(\Omega)$.
\end{theorem}

\begin{remark}\rm
The anisotropic critical Hardy inequality (\ref{FH_N}) is invariant under the scaling $u_\la (x) = \la^{-\frac{N-1}{N}} u\( \( \frac{H^0(x)}{R}\)^{\la -1} x \)$, $(\la >0)$
when $\Omega =\mathcal{W}_R = \{ x \in \re^N \,: \, H^0(x) < R \}$.
For the proof, see \S \ref{section:scale FH_N}.
\end{remark}

\begin{proof}
Define
\[
	G(x) = \frac{x}{\( H^0(x) \)^N \( \log \frac{R}{H^0(x)} \)^{\la}}, \quad x \in \Omega.
\]
Note that
\[
	{\rm div} \( \frac{x}{(H^0(x))^N} \) = 0
\]
by the former calculation in the proof of Theorem \ref{Thm:H_p}.
Then we have
\begin{align*}
	\mbox{div}\, G(x) &=  \frac{x}{(H^0(x))^N} \cdot (-\la) \( \log \frac{R}{H^0(x)} \)^{-\la-1} \( -\frac{1}{H^0(x)} \) \nabla H^0(x) \\
	&= \la \frac{x \cdot \nabla H^0(x)}{(H^0(x))^{N+1}} \( \log \frac{R}{H^0(x)} \)^{-\la-1} \\
	&= \frac{\la}{(H^0(x))^N \( \log \frac{R}{H^0(x)} \)^{\la +1}}.
\end{align*}

In particular, by choosing $\la = N-1$, we have
\begin{align*}
	&\int_{\Omega} \frac{|u(x)|^N}{(H^0(x))^N \( \log \frac{R}{H^0(x)} \)^N} dx \\

	&= \left| \frac{1}{N-1} \int_{\Omega} |u|^N \mbox{div} F(x) dx \right| \\
	&= \left| - \frac{1}{N-1} \int_{\Omega} \nabla (|u|^N ) \cdot \frac{x}{(H^0(x))^N \( \log \frac{R}{H^0(x)} \)^{N-1}} dx \right| \\
	&= \left| - \frac{N}{N-1} \int_{\Omega} |u|^{N-2} u \nabla u  \cdot \frac{\frac{x}{H^0(x)}}{(H^0(x))^{N-1} \( \log \frac{R}{H^0(x)} \)^{N-1}} dx \right| \\
	&\le  \frac{N}{N-1} \( \int_{\Omega} \frac{|u|^N}{(H^0(x))^N \( \log \frac{R}{H^0(x)} \)^N} dx \)^{\frac{N-1}{N}}
	\( \int_{\Omega} \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^N dx \)^{\frac{1}{N}}
\end{align*}
for $u \in C_0^{\infty}(\Omega)$. This yields the conclusion.

\end{proof}

\section{Hardy inequality of geometric type}
\label{section:geo Hardy}

Let $\Omega$ be a domain in $\re^N$ with Lipschitz boundary and let
\begin{equation}
\label{d_H}
	d_H(x) = \inf_{y \in \pd\Omega} H^0(x-y)
\end{equation}
be {\it the anisotropic distance} of $x \in \ol{\Omega}$ to the boundary of $\Omega \subset \re^N$.
Then we have
\[
	H(\nabla d_H(x)) = 1 \quad a.e. \ \text{in} \ \Omega
\]
and
\[
	\frac{1}{\beta} d(x) \le d_H(x) \le \frac{1}{\alpha} d(x)
\]
where $d(x) = \inf_{y \in \pd\Omega} |x-y|$ is the Euclidean distance from the boundary $\pd\Omega$.
In \cite{Della Pietra-Blasio-Gavitone}, the authors studied the {\it anisotropic Hardy inequality of geometric type} as follows:
\begin{theorem}{\rm (\cite{Della Pietra-Blasio-Gavitone})}
\label{Thm:geometric H_2}
Suppose $d_H$ is a $\Delta_H$-superharmonic in $\Omega$, i.e.,
\[
	-\Delta_H d_H \ge 0
\]
in the distribution sense.
Then the inequality
\begin{equation}
\label{eq:DPBG}
	\frac{1}{4} \intO \frac{|u|^2}{(d_H(x))^2} dx \le \intO (H(\nabla u))^2 dx
\end{equation}
holds true for any $u \in C_0^{\infty}(\Omega)$.
\end{theorem}
Note that if $\Omega$ is convex, the assumption $-\Delta_H d_H \ge 0$ holds true.
In \cite{Della Pietra-Blasio-Gavitone}, it is shown that there exists a non-convex domain $\Omega$ such that $d_H$ is $\Delta_H$-superharmonic on $\Omega$.
For the Euclidean geometric type Hardy inequalities, see \cite{Brezis-Marcus}, \cite{FTT(JEMS)}, \cite{MMP}, \cite{Tidblom(PAMS)}, and references there in.

In the next theorem, we improve their result in the following form:
\begin{theorem}{\rm (Anisotropic $L^p$-Hardy inequality of geometric type)}
\label{Thm:geometric H_p}
Let $1 < p < \infty$ and suppose $\delta = \delta(x)$ is a nonnegative, $\Delta_{H,p}$-superharmonic function on $\Omega$, i.e.,
\[
	-\Delta_{H,p} \delta \ge 0
\]
in the distributional sense.

Then the inequality
\begin{equation}
\label{eq:geometric Finsler Hardy}
	\( \frac{p-1}{p} \)^p \intO \frac{|u|^p}{\delta^p} H^p(\nabla \delta) dx \le \intO |\nabla u \cdot (\nabla_{\xi} H)(\nabla \delta)|^p dx
\end{equation}
holds true for any $u \in C_0^{\infty}(\Omega)$.
\end{theorem}

\begin{remark}\label{rem:geom}\rm
Note that by (\ref{Schwarz}) and Proposition \ref{Prop:identities} (4'), we have
\[
	|\nabla u \cdot (\nabla_{\xi} H)(\nabla \delta)| \le H(\nabla u) H^0((\nabla_{\xi} H)(\nabla \delta)) = H(\nabla u).
\]
Thus, by \eqref{eq:geometric Finsler Hardy}, we have the following inequality
\begin{equation}
\label{eq:geometric Finsler Hardy_d}
	\( \frac{p-1}{p} \)^p \intO \frac{|u|^p}{(d_H(x))^p} dx \le \intO (H(\nabla u))^p dx.
\end{equation}
Moreover if the domain satisfies the assumption $-\Delta_H d_H \ge 0$ in the distribution sense
(this is the case if $\Omega$ is convex),
then taking $\delta = d_H$ and using $H(\nabla d_H) = 1$ a.e. in $\Omega$,
we have $-\Delta_{H,p} d_H = -\Delta_{H} d_H \ge 0$.
Thus  Theorem \ref{Thm:geometric H_p} is an improvement of the result proved in \cite{Della Pietra-Blasio-Gavitone},
which gives the inequality \eqref{eq:geometric Finsler Hardy_d} when $p = 2$ under the assumption $-\Delta_{H} d_H \ge 0$ .
\end{remark}
\bigskip

\begin{proof}
For $x \in \Omega$, define
\[
	F(x) = \frac{(H(\nabla \delta(x)))^{p-2}}{\delta(x)^{p-1}} \nabla \delta(x).
\]
Then we have
\begin{align*}
	H(F)(\nabla_{\xi} H)(F) =  \frac{(H(\nabla \delta))^{p-1}}{\delta^{p-1}} (\nabla_{\xi} H)(\nabla \delta)
\end{align*}
and
\begin{align*}
	&\mbox{div} (H(F)(\nabla_{\xi} H)(F)) \\
	&= \delta^{1-p} \mbox{div} \( (H(\nabla \delta))^{p-1} (\nabla_{\xi} H)(\nabla \delta) \)
	+ (1-p) \delta^{-p} \nabla \delta \cdot (H(\nabla \delta))^{p-1} (\nabla_{\xi} H)(\nabla \delta) \\
	&= \frac{\Delta_{H,p} \delta}{\delta^{p-1}} - (p-1) \frac{(H(\nabla \delta))^p}{\delta^p}.
\end{align*}
Then for $u \in C_0^{\infty}(\Omega)$, we have
\begin{align*}
	&\int_{\Omega} |u|^p \( \frac{\Delta_{H,p} \delta}{\delta^{p-1}} - (p-1) \frac{(H(\nabla \delta))^p}{\delta^p} \) dx \\
	&= \int_{\Omega} |u|^p \mbox{div} (H(F)(\nabla_{\xi} H)(F)) dx \\
	&= -p \int_{\Omega} |u|^{p-1} (sgn(u)) \nabla u \cdot  (H(F) (\nabla_{\xi} H)(F) dx \\
	&= -p \int_{\Omega} |u|^{p-1} (sgn(u)) \nabla u \cdot \frac{(H(\nabla \delta))^{p-1}}{\delta^{p-1}} (\nabla_{\xi} H)(\nabla \delta) dx.
\end{align*}
Now, by the assumption $\Delta_{H,p} \delta \le 0$, we see
\begin{align*}
	0 &\ge \int_{\Omega} |u|^p \( 0 - (p-1) \frac{(H(\nabla \delta))^p}{\delta^p} \) dx \\
	&\ge -p \int_{\Omega} |u|^{p-1} (sgn(u)) \nabla u \cdot  \frac{(H(\nabla \delta))^{p-1}}{\delta^{p-1}} (\nabla_{\xi} H)(\nabla \delta) dx,
\end{align*}
and, by the H\"older inequality, this leads to
\begin{align*}
	(p-1) \int_{\Omega} |u|^p \frac{(H(\nabla \delta))^p}{\delta^p} dx &\le p \int_{\Omega} |u|^{p-1} \Big| \nabla u \cdot (\nabla_{\xi} H)(\nabla \delta) \Big| \frac{(H(\nabla \delta))^{p-1}}{\delta^{p-1}} dx\\
	&\le p \left (\int_{\Omega} |u|^{p}\frac{(H(\nabla \delta))^p}{\delta^p} dx\right)^\frac{p-1}{p}
	\left (\int_{\Omega}\Big| \nabla u \cdot (\nabla_{\xi} H)(\nabla \delta) \Big|^p	\right)^\frac{1}{p}.
\end{align*}
This gives us the result.
\end{proof}

Now some consequences of Theorem \ref{Thm:geometric H_p} are proved.
The first one is an anisotropic Hardy inequality when $\Omega$ is the half-space
\[
	\re^N_{+} = \{ x = (x_1,x_2, \cdots, x_N) \, : \, x_N > 0 \}
\]
with $N \ge 2$.
Denote $d_H(x) = \inf_{y \in \pd \re^N_{+}} H^0(x-y)$ the anisotropic distance from the boundary for $x \in \re^N_{+}$.
Note that $H(\nabla d_H(x)) = 1$ a.e. and,  by the convexity of $\re^N_{+}$,  $-\Delta_{H,p} d_H \ge 0$ holds for $1 < p < \infty$ and $N \ge 2$.
By Theorem \ref{Thm:geometric H_p} and the Remark \ref{rem:geom}, we have {\it the anisotropic Hardy inequality on the half-space}:

\begin{corollary}{\rm (Anisotropic Hardy inequality on the half-space)}
\label{cor:geometric H_p}
Assume $1 < p < \infty$ and $N \ge 2$.
Then the inequality
\begin{equation}
\label{H_p half}
	\(\frac{p-1}{p}\)^p \int_{\re^N_{+}} \frac{|u|^p}{\( d_H(x) \)^p} dx \le \int_{\re^N_{+}} |\nabla u \cdot (\nabla_{\xi} H)(\nabla d_H)|^p dx
\end{equation}
for any $u \in C_0^{\infty}(\re^N_{+})$.
\end{corollary}

The second  consequence of Theorem \ref{Thm:geometric H_p} is a lower bound for $\la_{1,p}(\Omega)$, the first eigenvalue of the Finsler $p$-Laplacian $\Delta_{H,p}$, by means of anisotropic inradius.
Define
\[
	\la_{1,p}(\Omega) = \min_{u \in W^{1,p}_0(\Omega), u \ne 0} \frac{\intO (H(\nabla u))^p dx}{\intO |u|^p dx}
\]
and assume that $\tau_H$, {\it the anisotropic inradius} of $\Omega$, is finite, i.e.,
\[
	\tau_H = \sup_{x \in \Omega} d_H(x) < \infty.
\]
We prove the following result:
\begin{corollary}
\label{Cor:lower bound}
Let $\Omega$ be a bounded domain in $\re^N$ satisfying $-\Delta_{H} d_H \ge 0$ in the distribution sense and $\tau_H < \infty$.
Let $\la_{1,p}(\Omega)$ be the first eigenvalue of the Finsler $p$-Laplacian $\Delta_{H,p}$.
Then it holds that
\[
	\la_{1,p}(\Omega) \ge \(\frac{p-1}{p}\)^p \(\frac{1}{\tau_H}\)^p.
\]
\end{corollary}

\begin{proof}

Applying \eqref{eq:geometric Finsler Hardy_d} to the first eigenfunction $\phi$ of $\la_{1,p}(\Omega)$, normalized as $\| \phi \|_{L^p(\Omega)} = 1$, we obtain
\[
	\la_{1,p}(\Omega) = \intO (H(\nabla \phi))^p dx \ge \( \frac{p-1}{p} \)^p \intO \frac{|\phi|^p}{(d_H(x))^p} dx \ge \( \frac{p-1}{p} \)^p \frac{1}{(\tau_H)^p}.
\]
\end{proof}

\section{Weighted Finsler-Hardy-Poincar\'e inequalities}
\label{section:Weighted}

In this section, we prove weighted version of Finsler Hardy-Poincar\'e type inequalities on $\re^N$, following arguments in \cite{Kombe-Ozaydin(2009)} and \cite{Kombe-Ozaydin(2013)}.

\begin{theorem}{\rm (Weighted anisotropic Hardy-Poincar\'e inequality)}
\label{Thm:WFH_p}
Let $1 \le p < \infty$.
Assume there exists a nonnegative function $\rho$ on $\re^N \setminus \{0\}$ such that $H(\nabla \rho) = 1$ and $\Delta_H \rho \ge \frac{C}{\rho}$ in the sense of distributions
where $C > 0$.
Then for $\alpha \in \re$ such that $C + \alpha > -1$, the following inequality holds true
\begin{equation}
\label{WFH_p}
	\( \frac{C + \alpha + 1}{p} \)^p \int_{\re^N} \rho^{\alpha} |u|^p dx \le \int_{\re^N} \rho^{\alpha+p} \left| (\nabla_{\xi} H)(\nabla \rho) \cdot \nabla u \right|^p dx
\end{equation}
for any $u \in C_0^{\infty}(\re^N \setminus \{ 0 \}$.
\end{theorem}

\begin{proof}
From the assumptions, we see $\nabla \rho \cdot (\nabla_{\xi} H)(\nabla \rho) = H(\nabla \rho) = 1$ and
\[
	\rho \Delta_H \rho = \rho \, {\rm div} ((\nabla_{\xi} H)(\nabla \rho)) \ge C.
\]
Thus
\begin{align*}
	{\rm div}(\rho (\nabla_{\xi} H(\nabla \rho)) = \rho \, {\rm div}((\nabla_{\xi} H)(\nabla \rho)) + \nabla \rho \cdot (\nabla_{\xi} H)(\nabla \rho) \ge C + 1.
\end{align*}
Multiplying this inequality by $\rho^{\alpha} |u|^p$ and integrating over $\re^N$, we have
\[
	(C + 1) \intRN \rho^{\alpha} |u|^p dx \le \intRN {\rm div}(\rho (\nabla_{\xi} H(\nabla \rho)) \rho^{\alpha} |u|^p dx.
\]
The divergence theorem and the H\"older inequality implies that
\begin{align*}
	(C + \alpha + 1) \intRN \rho^{\alpha} |u|^p dx &\le \left| -p \intRN \rho^{\alpha+1} |u|^{p-2} u \nabla u \cdot (\nabla_{\xi} H(\nabla \rho)) dx \right| \\
	&\le p \( \intRN \rho^{\alpha} |u|^p dx \)^{\frac{p-1}{p}} \( \intRN \rho^{\alpha+p} |\nabla u \cdot (\nabla_{\xi} H(\nabla \rho))|^p dx \)^{\frac{1}{p}}.
\end{align*}
After some manipulations, we have \eqref{WFH_p}.
\end{proof}

\begin{remark} \rm
\label{remark:H^0}
Since by Proposition \ref{Prop:identities} (4),  $\rho(x) = H^0(x)$ satisfies that $H(\nabla \rho) = 1$ and $\Delta_H \rho = \frac{N-1}{\rho}$, we have from Theorem \ref{Thm:WFH_p} that
\begin{align*}
	\( \frac{N + \alpha}{p} \)^p \int_{\re^N} (H^0(x))^{\alpha} |u|^p dx \le \int_{\re^N} (H^0(x))^{\alpha+p} \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p dx
\end{align*}
for any $u \in C_0^{\infty}(\re^N \setminus \{ 0 \})$.
This improves Theorem 5.4 by Brasco-Franzina \cite{Brasco-Franzina}, because the right-hand side of the above inequality is less than or equal to
$\int_{\re^N} (H^0(x))^{\alpha+p} H^p(\nabla u) dx$.

\noindent Finally we recall that in the Euclidean case, i.e. $\rho(x)=H_0(x)=|x|$,  weighted  Hardy-Poincar\'e inequalities are well-known (see, for example,  \cite{ACP}, \cite{ABCMP1}, \cite{Sano-TF(DIE)} and references therein)
\end{remark}

{\it The Uncertainty Principle} in quantum mechanics, sometimes called Heisenberg-Pauli-Weyl inequality, is well known in Euclidean context and it claims that
\[
	\frac{N^2}{4} \( \intRN |f(x)|^2 dx \)^2 \le \( \intRN |x|^2 |f(x)|^2 dx \) \( \intRN |\nabla f(x)|^2 dx \)
\]
for any $f \in C_0^{\infty}(\re^N)$.
In Finsler context, we obtain the following:

\begin{theorem}{\rm (Anisotropic uncertainty principle inequality)}
\label{Thm:FUP}
Let $1 \le p < \infty$ and $N \ge 2$.
Assume there exists a nonnegative function $\rho$ on $\re^N \setminus \{0\}$ such that $H(\nabla \rho) = 1$ and $\Delta_H \rho \ge \frac{C}{\rho}$ in the sense of distributions
where $C > 0$.
Then the following inequality holds true:
\begin{equation}
\label{FUP}
	\( \frac{C + 1}{2} \)^2 \( \int_{\re^N} |u|^2 dx \)^2 \le \( \int_{\re^N} \rho^2 |u|^2 dx \) \(\int_{\re^N} \left| (\nabla_{\xi} H)(\nabla \rho) \cdot \nabla u \right|^2 dx \)
\end{equation}
for any $u \in C_0^{\infty}(\re^N)$.
Especially, we have
\[
	\( \frac{N}{2} \)^2 \( \int_{\re^N} |u|^2 dx \)^2 \le \( \int_{\re^N} (H^0(x))^2 |u|^2 dx \) \(\int_{\re^N} \left| \frac{x}{\nabla (H^0(x))} \cdot \nabla u \right|^2 dx \)
\]
for any $u \in C_0^{\infty}(\re^N)$.
\end{theorem}

\begin{proof}
Since $H(\nabla \rho^2) = 2\rho H(\nabla \rho)$ and $(\nabla_{\xi} H)(\nabla \rho^2) = (\nabla_{\xi} H)(\nabla \rho)$ by Proposition \ref{Prop:identities},
by using assumptions, we have
\[
	\Delta_H (\rho^2) = {\rm div} (2 \rho H(\nabla \rho) (\nabla_{\xi} H)(\nabla \rho)) = 2H^2(\nabla \rho) + 2\rho \Delta_H \rho \ge 2 + 2C.
\]
Multiplying this inequality by $|u|^2$ and integrating over $\re^N$, we have
\[
	\intRN \Delta_H (\rho^2) |u|^2 dx \ge 2(1 + C) \intRN |u|^2 dx.
\]
On the other hand, integration by parts and Schwarz inequality implies
\begin{align*}
	\intRN \Delta_H (\rho^2) |u|^2 dx &= \intRN {\rm div} (2 \rho (\nabla_{\xi} H)(\nabla \rho)) |u|^2 dx \\
	&= - \intRN 2 \rho (\nabla_{\xi} H)(\nabla \rho) \cdot 2 u \nabla u dx \\
	&\le 4 \( \intRN \rho^2 |u|^2 dx \)^{\frac{1}{2}} \(\intRN |(\nabla_{\xi} H)(\nabla \rho) \cdot \nabla u |^2 dx \)^{\frac{1}{2}}.
\end{align*}
After some computations, we have \eqref{FUP}.
The last claim follows from Remark \ref{remark:H^0} and \eqref{FUP}.
\end{proof}

\section{The best constant and its attainability on Hardy inequalities}
\label{section:attainability}

In this section we investigate the sharpness and the attainability of the constants in anisotropic Hardy inequalities in the previous sections.
We call a function $f$ defined on $\re^N$ is {\it $H^0$-radial} if there exists a function $F = F(r)$ defined on $\re_{+}$ such that $f(x) = F(H^0(x))$.
If $F$ is decreasing on $\re_{+}$, then $f$ is called $H^0$-radially decreasing.
For a function space $X$, we define $X_{H^0rad} = \{ u \in X \,:\, \text{$u$ is $H^0$-radial} \}$.
Admitting some ambiguity, we sometimes write $f(x) = f(r)$ with $r = H^0(x)$ for $H^0$-radial function $f$.
Let us begin with two preliminary results.

\begin{proposition}
\label{Prop:spherical}
Let $R \in (0, +\infty]$.
For any $u \in W_0^{1,p}(\mathcal{W}_R)$,

there exists a $H^0$-radial function $U \in W_0^{1,p}(\mathcal{W}_R)$ such that the followings hold true:
\begin{align}
\label{simm_1}
	\int_{\mathcal{W}_R} V(H^0(x)) |U|^p \,dx &= \int_{\mathcal{W}_R} V(H^0(x)) |u |^p dx, \\
\label{simm_2}
	\int_{\mathcal{W}_R}  |\nabla U|^p \,dx &\le \int_{\mathcal{W}_R} \left| \nabla u \cdot \frac{x}{H^0(x)} \right|^p dx
\end{align}
where $V = V(r)$ is any function on $[0, R]$.
\end{proposition}

\begin{proof}
For $x \in \re^N \setminus \{0\}$, let us write $x=r\w$ where $r=H^0(x)$, $\w \in \pd \mathcal{W}$ and $\w_{N-1}= P_H(\mathcal{W}; \re^N) = N \kappa_N$.
It is enough to show Proposition \ref{Prop:spherical} for $u \in C_0^1(\mathcal{W}_R)$ by the density argument.
For any $u \in C_0^1(\mathcal{W}_R)$, we set
\begin{align*}
	\ol{U}(r) = \(\w_{N-1}^{-1} \int_{\pd \mathcal{W}} |u(r \w)|^p dS_{\w} \)^{\frac{1}{p}},
\end{align*}
where $dS_{\w}$ denotes a measure on $\pd \mathcal{W}$ such that
$\int_{\pd \mathcal{W}} dS_{\w} = P_H(\mathcal{W}; \re^N) = \w_{N-1}$ holds true.
Then by the H\"older inequality, we have
\begin{align*}
	\left| \ol{U}'(r) \right| &= \w_{N-1}^{-\frac{1}{p}} \( \int_{\pd \mathcal{W}} |u (r \w)|^p dS_{\w} \)^{\frac{1}{p}-1} \left| \int_{\pd \mathcal{W}} |u(r \w)|^{p-2} u(r\w) \frac{\pd u}{\pd r}(r\w) dS_{\w} \right| \\
	&\le \(\w_{N-1}^{-1} \int_{\pd \mathcal{W}} \left| \frac{\pd u}{\pd r}(r \w) \right|^p dS_{\w} \)^{\frac{1}{p}}.
\end{align*}
Therefore for $U(x) = \ol{U}(H^0(x))$, we obtain
\begin{align*}
	\int_{\mathcal{W}_R} |\nabla U|^p dx &= \w_{N-1} \int_0^R |\ol{U}'(r)|^p r^{N-1} dr \\
	&\le \int_0^R \int_{\pd \mathcal{W}} \left| \frac{\pd u}{\pd r}(r \w) \right|^p r^{N-1} dr dS_{\w} \\
	&= \int_{\mathcal{W}_R} \left |\nabla u \cdot \frac{x}{H^0(x)} \right|^p dx < \infty.
\end{align*}
Thus $U \in W^{1,p}_0(\mathcal{W}_R)$ and \eqref{simm_2} is proved.
Moreover we obtain
\begin{align*}
\int_{\mathcal{W}_R} V(H^0(x)) |U|^p \,dx &= \w_{N-1} \int_0^R V(r) |U(r)|^p r^{N-1} \,dr \\
&= \int_0^R \int_{\pd \mathcal{W}} V(r) |u (r\w )|^p r^{N-1} dr dS_{\w} \\
&= \int_{\mathcal{W}_R} V(H^0(x)) |u |^p dx.
\end{align*}
Hence \eqref{simm_1} is proved.
\end{proof}

\begin{proposition}
\label{Prop:radial lemma}
For $R \in (0, +\infty]$, let $U \in C^1(0, R)$ with $U(R) := \lim_{r \to R} U(r) = 0$.
Then the following pointwise estimates hold for any $r \in (0,R)$.
\begin{align}
\label{subcritical}
	|U(r)| &\le \( \frac{N-p}{p-1} \)^{\frac{p-1}{p}} \( \int_r^R |U'(s)|^p s^{N-1}\,ds \)^{\frac{1}{p}} r^{-\frac{N-p}{p}}, \quad (1< p < N), \\
\label{critical}
	|U(r)| &\le \( \int_r^R |U'(s)|^N s^{N-1}\,ds \)^{\frac{1}{N}} \( \log \frac{R}{r} \)^{\frac{N-1}{N}}.
\end{align}
\end{proposition}

\begin{proof}
Since
\begin{align*}
	|U(r)| &= \left| - \int_r^R U'(s) \,ds \right| \\
	&\le \( \int_r^R |U'(s)|^p s^{N-1}\,ds \)^{\frac{1}{p}} \( \int_r^R s^{-\frac{N-1}{p-1}}\,ds \)^{\frac{p-1}{p}},
\end{align*}
we obtain (\ref{subcritical}) and (\ref{critical}).
\end{proof}

For $\Omega \subseteq \re^N$ and $1 \le p < N$,
let us define the best constant of the anisotropic subcritical $L^p$-Hardy inequality on $\Omega$ as
\begin{equation}
\label{H_p}
	H_p(\Omega) = \inf_{0 \not\equiv u \in W_0^{1,p}(\Omega )} \frac{\disp \intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p dx}{\disp \intO \frac{|u|^p}{H^0(x)^p} dx}.
\end{equation}
In order to prove Theorem \ref{Thm:H_p attainability} below, we need the following result.

\begin{lemma}
\label{lem:regularity}
If $u(x) = U(H^0(x)) \in W_0^{1,p}(\re^N)$ is a $H^0$-radial minimizer of \eqref{H_p} with $\Omega = \re^N$, then $U \in C^1(0, \infty)$.
\end{lemma}

\begin{proof}
The proof is similar to that of Lemma 2.4 in \cite{Ioku-Ishiwata}.

Take any $0< a < b < \infty$.  Note that $U \in L^p(a,b)$.
By the characterization of one dimensional Sobolev space $W^{1,p}(a,b)$, $r \mapsto U(r)$ is locally absolutely continuous.
Particularly, $U(r)$ is differentiable for almost all $r \in (a,b)$.
Put $f(r)=r^{N-1}| \pd_r U(r)|^{p-2} \pd_r U(r)$, $g(r)= H_p(\re^N) r^{-p+N-1} |U(r)|^{p-2} U(r)$.
From the weak form of the Euler Lagrange equation of $H_p(\re^N)$:
\begin{align*}
	\int_0^\infty f(r) \pd_r v \,dr = \int_0^\infty g(r) v \,dr \quad ({}^{\forall}v \in W_{0, H^0rad}^{1,p}(\re^N )),
\end{align*}
we see that $f$ is weakly differentiable and its weak derivative is $\pd_r f(r) = -g(r)$ a.e. $r \in (a,b)$.
Moreover we have
\begin{align*}
	\int_a^b |\pd_r f(r)| \,dr &= H_p(\re^N) \int_a^b r^{N-p-1} |U(r)|^{p-1}\,dr < \infty.
\end{align*}
Therefore $f \in W^{1,1}(a,b)$ which yields that $f$ is absolutely continuous on $(a,b)$.
Since $a,b >0$ are arbitrary, we see that $U \in C^1(0, \infty)$.
\end{proof}

The first main result of this section is Theorem \ref{Thm:H_p attainability} below which concerns the sharpness of the constant in the anisotropic subcritical Hardy inequality given in Theorem \ref{Thm:H_p}
\begin{theorem}
\label{Thm:H_p attainability}
Let $\Omega$ be a domain in $\re^N$ with $0 \in \Omega$ and $1 \le p < N$. Then $H_p(\Omega)$ in \eqref{H_p} is $H_p(\Omega) = \( \frac{N-p}{p} \)^p$.
Moreover $H_p(\Omega)$ is not attained if $1 < p < N$.
On the other hand, $H_1(\Omega) = N-1$ is attained by any nonnegative function which is $H^0$-radially decreasing on its support.
\end{theorem}

\begin{proof}
Let $\delta >0$ satisfy $\mathcal{W}_{2\delta} \subset \Omega$ and $\alpha < \frac{N-p}{p}$.
Set
\begin{align*}
	\varphi_\alpha (x) =
	\begin{cases}
	(H^0(x))^{-\alpha} \quad &\text{if} \,\,H^0(x) \le \delta, \\
	\delta^{-\alpha -1} (2\delta -H^0(x))  &\text{if} \,\, \delta < H^0(x) < 2 \delta, \\
	0 &\text{if} \,\,2 \delta \le H^0(x).
\end{cases}
\end{align*}
For $x$ with $H^0(x) \le \delta$, by Proposition \ref{Prop:identities}. (1'), we have
\begin{align*}
	\frac{x}{H^0(x)} \cdot \nabla \varphi_\alpha (x)
	= -\alpha (H^0(x))^{-\alpha -1} \frac{x}{H^0(x)} \cdot \nabla H^0(x)
	= -\alpha (H^0(x))^{-\alpha -1}
\end{align*}
and
\begin{align*}
	\intO \left| \frac{x}{H^0(x)} \cdot \nabla \varphi_\alpha \right|^p dx
	&=\alpha^p \int_{\mathcal{W}_\delta} (H^0(x))^{-\alpha p -p} dx + C(\delta ) \\
	&=\alpha^p \w_{N-1} \int_0^\delta r^{-\alpha p -p +N -1} dr + C(\delta ) \\
	&=\alpha^p \w_{N-1} (N-p-\alpha p)^{-1} \delta^{N-\alpha p -p} + C(\delta).
\end{align*}
This yields that
\begin{align}
\label{nume}
	&\intO \left| \frac{x}{H^0(x)} \cdot \nabla \varphi_\alpha \right|^p dx \notag \\
	&= \alpha^p \w_{N-1} p^{-1} \( \frac{N-p}{p} - \alpha \)^{-1} \delta^{N-\alpha p -p} + o\( \( \frac{N-p}{p} - \alpha \)^{-1} \)
\end{align}
as $\alpha \nearrow \frac{N-p}{p}$.
On the other hand, we have
\begin{align}
\label{deno}
	&\intO \frac{|\varphi_\alpha|^p}{(H^0(x))^p} dx = \int_{\mathcal{W}_\delta} (H^0(x))^{-\alpha p -p} dx + C(\delta ) \notag \\
	&=\w_{N-1} p^{-1} \( \frac{N-p}{p} - \alpha \)^{-1} \delta^{N-\alpha p -p} + o\( \( \frac{N-p}{p} - \alpha \)^{-1} \).
\end{align}
From (\ref{nume}), (\ref{deno}) and Theorem \ref{Thm:H_p}, we see that
\begin{align*}
\( \frac{N-p}{p} \)^p \le H_p(\Omega) \le \frac{\intO \left| \frac{x}{H^0(x)} \cdot \nabla \varphi_\delta \right|^p dx}{\intO \frac{|\varphi_\alpha |^p}{(H^0(x))^p} dx} = \alpha^p + o(1) = \( \frac{N-p}{p} \)^p + o(1).
\end{align*}
Hence $H_p(\Omega)=(\frac{N-p}{p} )^p$.

Next we shall show the attainability of $H_1(\Omega)$.

For any $u \in C_0^\infty (\Omega)$, there exists $R > 0$ such that ${\rm supp} (u) \subset \mathcal{W}_R$.
Since $u$ is nonnegative $H^0$-radially decreasing function on its support, we obtain
\begin{align*}
	&\intO \frac{|u|}{H^0(x)} dx = \int_{\mathcal{W}_R} \frac{u(x)}{H^0(x)} dx
	=\w_{N-1} \int_0^R u(r) r^{N-2} dr \\
	&= -\frac{\w_{N-1}}{N-1} \int_0^R u'(r) r^{N-1} dr =\frac{1}{N-1} \intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right| dx.
\end{align*}
Therefore we see that $H_1(\Omega) = N-1$ is attained by $u$.

Finally we show the non-attainability of $H_p(\Omega)$ when $1<p<N$.
Assume by contradiction that $\tilde{u} \in W_0^{1,p}(\Omega)$ is a minimizer of $H_p(\Omega) = (\frac{N-p}{p})^p = H_p(\re^N)$.
Then by zero-extension there exists a minimizer $\overline{u} \in W_0^{1,p}(\re^N)$ of $H_p(\re^N)$.
By Proposition \ref{Prop:spherical}, there also exists a $H^0$-radial minimizer $u \in W_0^{1,p}(\re^N)$ of $H_p(\re^N)$.
Write $u(x) = U(H^0(x))$.
From Lemma \ref{lem:regularity} we see that $u \in C^1(\re^N \setminus \{ 0\})$.
Now we set
\[
	J(u) = \intRN \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p dx - \( \frac{N-p}{p} \)^p \intRN \frac{|u|^p}{(H^0(x))^p} dx
\]
and consider $v(x) = V(H^0(x))= (H^0(x))^{\frac{N-p}{p}} U(H^0(x))$.
Note that $\lim_{r \to \infty} |V(r)| =0$ from Proposition \ref{Prop:radial lemma}.
Indeed
\begin{align*}
	\lim_{r \to \infty} |V(r)| = \lim_{r \to \infty} r^{\frac{N-p}{p}}|U(r)| \le C \lim_{r \to \infty} \| H( \nabla u)\|_{L^p(\re^N \setminus \mathcal{W}_r)} =0.
\end{align*}
Since
\[
	\nabla u(x) = -\frac{N-p}{p} (H^0(x))^{-\frac{N}{p}} \nabla H^0(x) v(x) + (H^0(x))^{-\frac{N-p}{p}} \nabla v(x),
\]
we have
\begin{equation}
\label{radial H_p}
	\left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p
	= \left| -\frac{N-p}{p} (H^0(x))^{-\frac{N}{p}} v(x) + (H^0(x))^{-\frac{N-p}{p}} \frac{x}{H^0(x)} \cdot \nabla v(x) \right|^p.
\end{equation}
By recalling the inequality $|a+b|^p \ge |a|^p + p|a|^{p-2}a b$, $(p>1, a,b \in \re)$ and that the equality holds iff $b=0$,
we see
\begin{align*}
	&\left| \frac{x}{H^0(x)} \cdot \nabla u \right|^p \\
	&\ge \( \frac{N-p}{p} \)^p (H^0(x))^{-N} |v|^p - p \( \frac{N-p}{p} \)^{p-1} |v|^{p-2} v \nabla v \cdot \frac{x}{H^0(x)} (H^0(x))^{-N+1}.
\end{align*}
Therefore we have
\begin{align*}
	J(u) &\ge \( \frac{N-p}{p} \)^p \intRN \frac{|v|^p}{(H^0(x))^N} dx - \( \frac{N-p}{p} \)^{p-1} \intRN \nabla ( |v|^p ) \cdot \frac{x}{(H^0(x))^N} dx \\
	&\hspace{10em}-\( \frac{N-p}{p} \)^p \intRN \frac{|u|^p}{(H^0(x))^p} dx. \\
	&= -\lim_{R \to \infty}\( \frac{N-p}{p} \)^{p-1} \int_{\mathcal{W}_R} \nabla ( |v|^p ) \cdot \frac{x}{(H^0(x))^N} dx \\
	&= -\lim_{R \to \infty}\( \frac{N-p}{p} \)^{p-1} \frac{|V(R)|^p}{R^N} \int_{\pd \mathcal{W}_R} x \cdot \nu dS_x=0
\end{align*}
since $\lim_{R \to \infty} |V(R)| =0$, where $\nu$ is an outer normal vector and we have used the fact ${\rm div} \( \frac{x}{(H^0(x))^N} \) = 0$.
Since $u$ is a $H^0$-radial minimizer, $J(u)=0$, which implies
\[
	(H^0(x))^{-\frac{N-p}{p}} \frac{x}{H^0(x)} \cdot \nabla v(x) = 0
\]
by \eqref{radial H_p}.
This yields that $v(x)$ is a constant and $u(x) = c (H^0(x))^{-\frac{N-p}{p}}$ for some $c \in \re$ for $x \in \re^N \setminus \{ 0\}$.
However $(H^0(x))^{-\frac{N-p}{p}} \not\in W_0^{1,p}(\re^N)$. This is a contradiction.
Hence $H_p(\Omega)$ is not attained if $1 < p < N$.
\end{proof}

\begin{theorem}
\label{Thm:H_N attainability}
Let $\Omega$ be a bounded domain in $\re^N \,(N\ge 2)$, $0 \in \Omega$ and $R=\sup_{x \in \Omega} H^0(x)$.
Then
\begin{equation*}
	H_N(\Omega) := \inf_{0 \not\equiv u \in W_0^{1,N}(\Omega)}
	\frac{\displaystyle\intO \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^N dx}{\displaystyle\intO \frac{|u|^N}{H^0(x)^N (\log \frac{R}{H^0(x)})^N} dx}
	= \( \frac{N-1}{N} \)^N.
\end{equation*}
Moreover $H_N(\Omega)$ is not attained.
\end{theorem}

\begin{proof}
Let $\delta >0$ satisfy $\mathcal{W}_{2\delta} \subset \Omega$ and $\alpha < \frac{N-1}{N}$. Set
\begin{align*}
\varphi_\alpha (x) =
\begin{cases}
\( \log \frac{R}{H^0(x)} \)^{\alpha} \quad &\text{if} \,\,H^0(x) \le \delta, \\
\( \log \frac{R}{\delta} \)^{\alpha -1} (2\delta -H^0(x))  &\text{if} \,\, \delta < H^0(x) < 2 \delta, \\
0 &\text{if} \,\,2 \delta \le H^0(x).
\end{cases}
\end{align*}
Since for $x$ such that $H^0(x) \le \delta$, by Proposition \ref{Prop:identities}. (1') we have
\begin{align*}
\frac{x}{H^0(x)} \cdot \nabla \varphi_\alpha (x)

	&= -\alpha \( \log \frac{R}{H^0(x)} \)^{\alpha -1} \frac{1}{H^0(x)},
\end{align*}
and
\begin{align}\label{nume H_N}
	&\intO \left| \frac{x}{H^0(x)} \cdot \nabla \varphi_\alpha \right|^N dx \notag \\
	&=\alpha^N \int_{\mathcal{W}_\delta} \( \log \frac{R}{H^0(x)} \)^{\alpha N -N} \frac{1}{(H^0(x))^N} dx + C(\delta ) \notag \\
	&=\alpha^N \w_{N-1} \int_0^\delta \( \log \frac{R}{r} \)^{\alpha N -N} \frac{dr}{r} + C(\delta ) \notag \\
	&= \alpha^N \w_{N-1} N^{-1} \( \frac{N-1}{N} - \alpha \)^{-1} \( \log \frac{R}{\delta} \)^{\alpha N -N+1} + o\( \( \frac{N-1}{N} - \alpha \)^{-1} \)
\end{align}
as $\alpha \nearrow \frac{N-1}{N}$. On the other hand, we have
\begin{align}\label{deno H_N}
	&\intO \frac{|\varphi_\alpha |^N}{(H^0(x))^N (\log \frac{R}{H^0(x)})^N} dx
	= \int_{\mathcal{W}_\delta} \( \log \frac{R}{H^0(x)} \)^{\alpha N -N} \frac{1}{(H^0(x))^N} dx + C(\delta ) \notag \\
	&=\w_{N-1} N^{-1} \( \frac{N-1}{N} - \alpha \)^{-1} \( \log \frac{R}{\delta} \)^{\alpha N -N+1} + o\( \( \frac{N-1}{N} - \alpha \)^{-1} \).
\end{align}
From (\ref{nume H_N}), (\ref{deno H_N}) and Theorem \ref{Thm:H_N}, we see that
\begin{align*}
	\( \frac{N-1}{N} \)^N \le H_N(\Omega)
	\le \disp{ \frac{\intO \left| \frac{x}{H^0(x)} \cdot \nabla \varphi_\alpha \right|^N dx}{\intO \frac{|\varphi_\alpha |^N}{(H^0(x))^N (\log \frac{R}{H^0(x)})^N} dx} }
	= \alpha^N + o(1)
	= \( \frac{N-1}{N} \)^N + o(1).
	\end{align*}
Hence $H_N=(\frac{N-1}{N})^N$.

Next we shall show the non-attainability of $H_N(\Omega)$ by a contradiction.
Assume there exists a minimizer $\tilde{u} \in W_0^{1,N}(\Omega)$ of $H_N(\Omega)$.
Then by zero-extension there exists a minimizer $\overline{u} \in W_0^{1,N}(\mathcal{W}_R)$ of $H_N(\mathcal{W}_R)$, where $R=\sup_{x \in \Omega} H^0(x)$.
By Proposition \ref{Prop:spherical}, there also exists a $H^0$-radial minimizer $u \in W_0^{1,N}(\mathcal{W}_R)$ of $H_N(\mathcal{W}_R)$.
Write $u(x) = U(H^0(x))$.
We see that $u \in C^1(\mathcal{W}_R \setminus \{ 0\})$ in the same way as Lemma 2.4 in \cite{Ioku-Ishiwata}.
Now we set
\[
	J(u) = \int_{\mathcal{W}_R} \left| \frac{x}{H^0(x)} \cdot \nabla u \right|^N dx - \( \frac{N-1}{N} \)^N \int_{\mathcal{W}_R} \frac{|u|^N}{H^0(x)^N (\log \frac{R}{H^0(x)})^N} dx
\]
and consider $v(x) = V(H^0(x)) = (\log \frac{R}{H^0(x)})^{-\frac{N-1}{N}} U(H^0(x))$.
Note that $|V(R)| =\lim_{r \to R} |V(r)| =0$ from Proposition \ref{Prop:radial lemma}.
Indeed
\begin{align*}
	\lim_{r \to R} |V(r)| = \lim_{r \to R} \( \log \frac{R}{r} \)^{-\frac{N-1}{N}}|U(r)| \le C \lim_{r \to R} \| H( \nabla u)\|_{L^N(\mathcal{W}_R \setminus \mathcal{W}_r)} =0.
\end{align*}
Since
\[
	\nabla u(x) = \frac{N-1}{N} \( \log \frac{R}{H^0(x)} \)^{-\frac{1}{N}} \frac{\nabla H^0(x)}{H^0(x)} v(x) + \( \log \frac{R}{H^0(x)} \)^{-\frac{N-1}{N}} \nabla v(x),
\]
we have
\begin{align}
\label{radial H_N}
	&\left| \frac{x}{H^0(x)} \cdot \nabla u \right|^N \\
	&= \left| \frac{N-1}{N} \( \log \frac{R}{H^0(x)} \)^{-\frac{1}{N}} \frac{v(x)}{H^0(x)} + \( \log \frac{R}{H^0(x)} \)^{\frac{N-1}{N}} \frac{x}{H^0(x)} \cdot \nabla v(x) \right|^N. \notag
\end{align}
By recalling the inequality $|a+b|^N \ge |a|^N + N|a|^{N-2}a b$, $(N>1, a,b \in \re)$ and that the equality holds iff $b=0$,
we see
\begin{align*}
	\left| \frac{x}{H^0(x)} \cdot \nabla u \right|^N
	&\ge \( \frac{N-1}{N} \)^N \frac{|v|^N}{(H^0(x))^N} \( \log \frac{R}{H^0(x)} \)^{-1} \\
	&- N \( \frac{N-1}{N} \)^{N-1} |v|^{N-2} v \nabla v \cdot \frac{x}{H^0(x)} (H^0(x))^{-N+1}.
\end{align*}
Therefore we have
\begin{align*}
	J(u) &\ge \( \frac{N-1}{N} \)^N \int_{\mathcal{W}_R} \frac{|v|^N}{(H^0(x))^N} \( \log \frac{R}{H^0(x)} \)^{-1} dx \\
	&- \( \frac{N-1}{N} \)^{N-1} \int_{\mathcal{W}_R} \nabla (|v|^N ) \cdot \frac{x}{(H^0(x))^N} dx \\
	&- \( \frac{N-1}{N} \)^N \int_{\mathcal{W}_R} \frac{|u|^N}{H^0(x)^N (\log \frac{R}{(H^0(x))})^N} dx \\
	&=-  \( \frac{N-1}{N} \)^{N-1} \frac{|V(R)|^N}{R^N} \int_{\pd \mathcal{W}_R} x \cdot \nu dS_x = 0
\end{align*}
by $|V(R)| = 0$.
Since $u$ is a $H^0$-radial minimizer, $J(u)=0$, which implies
\[
	 \( \log \frac{R}{H^0(x)} \)^{\frac{N-1}{N}} \frac{x}{H^0(x)} \cdot \nabla v(x)=0
\]
by \eqref{radial H_N}.
This in turn yields that $v(x)$ is a constant and  $u(x)=c \( \log \frac{R}{H^0(x)} \)^{\frac{N-1}{N}}$ for some $c \in \re$ for $x \in \mathcal{W}_R \setminus \{ 0\}$.
However $\( \log \frac{R}{H^0(x)} \)^{\frac{N-1}{N}} \not\in W_0^{1,N}(\mathcal{W}_R)$.
This is a contradiction and $H_N(\Omega)$ is not attained.
\end{proof}

\begin{theorem}
\label{Thm:H_p-half best constant}
Let $N \ge 2$ and $1 < p < \infty$. Define $d_H$ as in \eqref{d_H}.
Then
\begin{equation*}
	C_p(\re^N_{+}) := \inf_{0 \not\equiv u \in W_0^{1,p}(\re^N_{+})} \frac{\displaystyle\int_{\re^N_{+}} |\nabla u \cdot (\nabla_{\xi} H)(\nabla d_H)|^p dx}{\displaystyle\int_{\re^N_{+}} \frac{|u|^p}{(d_H(x))^p} dx}
	= \( \frac{p-1}{p} \)^p.
\end{equation*}
\end{theorem}

\begin{proof}
The inequality (\ref{H_p half}) implies that $C_p(\re^N_{+}) \ge \( \frac{p-1}{p} \)^p$.

For $R > 0$, let
\begin{equation}
\label{QR}
	Q_R = \{ x = (x^{\prime}, x_N) \in \re^N_{+} \, | \, |x^{\prime}| < R,  0 < x_N < R \}
\end{equation}
be an open cube and
let $\eta$ be a smooth cut-off function with $0 \le \eta \le 1$, $\eta \equiv 1$ on $Q_R$, $\eta \equiv 0$ on $(Q_{2R})^c$.
For $\alpha > \frac{p-1}{p}$, put $u_{\alpha}(x) = \eta(x) (d_H(x))^\alpha$.
Then $u_{\alpha} \in W^{1,p}_0(\re^N_{+} \cap Q_{2R})$ and
\begin{align*}
	\nabla u_{\alpha} = \alpha \eta (d_H)^{\alpha-1} \nabla d_H + (\nabla \eta) (d_H)^{\alpha}.
\end{align*}
Thus
\begin{align*}
	&\nabla u_{\alpha} = \alpha (d_H)^{\alpha-1} \nabla d_H \quad \text{on} \quad \re^N_{+} \cap Q_R, \\
	&\int_{\re^N_{+} \cap Q_R} H^p(\nabla u_{\alpha}) dx = \alpha^p \int_{\re^N_{+} \cap Q_R} (d_H(x))^{p(\alpha-1)} dx, \\
	&\int_{\re^N_{+} \cap Q_R} \frac{|u_{\alpha}|^p}{(d_H(x))^p} dx = \int_{\re^N_{+} \cap Q_R} (d_H(x))^{p(\alpha-1)} dx.
\end{align*}
Now, since the inequality $\frac{1}{\alpha_2} |x| \le H^0(x) \le \frac{1}{\alpha_1} |x|$ holds, we have
$\frac{1}{\alpha_2} d_E(x) \le d_H(x) \le \frac{1}{\alpha_1} d_E(x)$,
where $d_E(x) = x_N$ denotes the Euclidean distance of $x \in \re^N_{+}$ from the boundary: $d_E(x) = \inf_{y \in \pd \re^N_{+}} |x-y|$.
Since
\[
	\int_{\re^N_{+} \cap Q_R} x_N^{p(\alpha-1)} dx = \int_{|x^{\prime}| < R} \int_0^R  x_N^{p(\alpha-1)} dx_N dx^{\prime} = C(R) \frac{R^{p(\alpha-1) + 1}}{p(\alpha-1) + 1},
\]
where $C(R) = \int_{|x^{\prime}| < R} dx^{\prime}$,
we have
\[
	\int_{\re^N_{+} \cap Q_R} (d_H(x))^{p(\alpha-1)} dx = O\( \frac{1}{p(\alpha-1) + 1} \) \quad \text{as} \quad \alpha \searrow \frac{p-1}{p}
\]
for any fixed $R > 0$.
On the other hand, by the convexity of $H$ and the fact $H(\nabla d_H) = 1$, we have
\begin{align*}
	H(\nabla u_{\alpha}) &= H\( \alpha \eta (d_H)^{\alpha-1} \nabla d_H + (\nabla \eta) (d_H)^{\alpha} \) \\
	&\le H(\alpha \eta (d_H)^{\alpha-1} \nabla d_H) + H((d_H)^{\alpha} (\nabla \eta)) \\
	&\le \alpha (d_H)^{\alpha-1} + (d_H)^{\alpha} H(\nabla \eta).

\end{align*}
Since $(2R)^{p(\alpha-1) +1} - R^{p(\alpha-1) +1} \to 0$ as $\alpha \searrow \frac{p-1}{p}$,
we have
\begin{align*}
	&\int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} x_N^{p(\alpha-1)} dx = \int_{R < |x^{\prime}| < 2R} \int_{R}^{2R}  x_N^{p(\alpha-1)} dx_N dx^{\prime} \\
	&= D(R) \frac{(2R)^{p(\alpha-1) +1} - R^{p(\alpha-1) +1}}{p(\alpha-1) + 1} = o(\frac{1}{p(\alpha-1)+1})
\end{align*}
as $\alpha \searrow \frac{p-1}{p}$ for any fixed $R>0$, where $D(R) = \int_{R < |x^{\prime}| < 2R} dx^{\prime}$.
Also we have
\begin{align*}

	&\int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} x_N^{p\alpha} dx = \int_{R < |x^{\prime}| < 2R} \int_{R}^{2R}  x_N^{p\alpha} dx_N dx^{\prime} = O(1).
\end{align*}
Then
\begin{align*}
	&\int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} (d_H(x))^{p(\alpha-1)} dx =  o(\frac{1}{p(\alpha-1)+1}), \\

	&\int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} (d_H(x))^{p\alpha} dx =  O(1)
\end{align*}
as $\alpha \searrow \frac{p-1}{p}$ for any fixed $R>0$ and thus
\begin{align*}
	&\int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} H(\nabla u_{\alpha})^p dx \\
	&\le 2^{p-1} \alpha^p \int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} (d_H)^{p(\alpha-1)} dx + 2^{p-1}  \sup_{x \in Q_{2R}} H^p(\nabla \eta(x)) \int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} (d_H)^{p\alpha} dx \\
	&= o(\frac{1}{p(\alpha-1)+1}) + O(1).
\end{align*}
Therefore, we see
\begin{align*}
	C_p(\re^N_{+} \cap Q_{2R}) &\le \frac{\disp\int_{\re^N_{+} \cap Q_{2R}} |\nabla u_{\alpha} \cdot (\nabla_{\xi} H)(\nabla d_H)|^p dx}{\disp\int_{\re^N_{+} \cap Q_{2R}} \frac{|u_{\alpha}|^p}{(d_H(x))^p} dx} \\
	&\le \frac{\displaystyle\int_{\re^N_{+} \cap Q_R} H^p(\nabla u_{\alpha}) dx
	+ \disp \int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} H^p(\nabla u_{\alpha}) dx}{\disp\int_{\re^N_{+} \cap Q_R} \frac{|u_{\alpha}|^p}{(d_H(x))^p} dx} \\
	&= \frac{\alpha^p \disp\int_{\re^N_{+} \cap Q_R} (d_H(x))^{p(\alpha-1)} dx + \disp\int_{\re^N_{+} \cap (Q_{2R} \setminus Q_R)} H^p(\nabla u_{\alpha}) dx}{\disp\int_{\re^N_{+} \cap Q_R} (d_H(x))^{p(\alpha-1)} dx} \\
	&= \alpha^p + \frac{o(\frac{1}{p(\alpha-1)+1}) + O(1)}{O(\frac{1}{p(\alpha-1)+1})}
\end{align*}
as $\alpha \searrow \frac{p-1}{p}$.
Then taking the limit $\alpha \searrow \frac{p-1}{p}$, we have $C_p(\re^N_{+}) \le (\frac{p-1}{p})^p$.
Thus we have proven the result.
\end{proof}

\begin{remark}\rm
Let $\Omega \subset \re^N_{+}$ be a domain with a flat boundary portion on $\pd\re^N_{+}$,
that is,
\[
	\text{$Q_{4R} \subset \Omega$ for some $R > 0$}
\]
where $Q_{4R}$ be an open cube as in (\ref{QR}).
Then we have $C_p(\Omega) = \( \frac{p-1}{p} \)^p$.
Because for such domain,
$d_E(x) = \inf_{y \in \pd\Omega} |x-y| = x_N$ for $x \in Q_{2R} \cap \Omega$
and the same proof as Theorem \ref{Thm:H_p-half best constant} works well.
\end{remark}

\begin{remark}\rm
Let $\Omega$ be a domain in $\re^N$ satisfying that $d_H$ is weakly twice differentiable and $-\Delta d_H \ge 0$ a.e.in $\Omega$,
where $d_H(x) = \inf_{y \in \pd\Omega} H^0(x-y)$.
Concerning the attainability of the best constant of (\ref{eq:geometric Finsler Hardy_d}), i.e.,
\begin{equation}
\label{C_p}
	C_p(\Omega) := \inf_{0 \not\equiv u \in C_0^{\infty}(\Omega)} \frac{\displaystyle\intO |\nabla u \cdot (\nabla_{\xi} H)(\nabla d_H)|^p dx}{\displaystyle\intO \frac{|u|^p}{(d_H(x))^p} dx},
\end{equation}
we will have the following observation.

First, for $u \in C_0^{\infty}(\Omega)$, define $v(x) = u(x) d_H^{-(\frac{p-1}{p})}$, $v$ is a Lipschitz function and $v = 0$ on $\pd\Omega$.
We compute
\begin{align*}
	&\nabla u = \(\frac{p-1}{p}\) d_H^{-\frac{1}{p}} v(x) \nabla d_H + d_H^{\frac{p-1}{p}} \nabla v, \\
	&\nabla u \cdot (\nabla_{\xi} H)(\nabla d_H) = \(\frac{p-1}{p}\) d_H^{-\frac{1}{p}} v(x) \nabla d_H \cdot (\nabla_{\xi} H)(\nabla d_H) \\
	&\hspace{7em} + d_H^{\frac{p-1}{p}} \nabla v \cdot (\nabla_{\xi} H)(\nabla d_H).
\end{align*}
Thus
\begin{align*}
	&|\nabla u \cdot (\nabla_{\xi} H)(\nabla d_H)|^p = \Big|\(\frac{p-1}{p}\) d_H^{-\frac{1}{p}} v(x) + d_H^{\frac{p-1}{p}} \nabla v \cdot (\nabla_{\xi} H)(\nabla d_H) \Big|^p \\
	&\ge \(\frac{p-1}{p}\)^p d_H^{-1} |v|^p + p \(\frac{p-1}{p}\)^{p-1} |v|^{p-2} v \nabla v \cdot (\nabla_{\xi} H)(\nabla d_H) \\

	&= \(\frac{p-1}{p}\)^p \frac{|u|^p}{d_H^p} + \(\frac{p-1}{p}\)^{p-1} \nabla (|v|^p) \cdot (\nabla_{\xi} H)(\nabla d_H), \\

\end{align*}
where, as before, we have used the fact that $|a + b|^p \ge |a|^p + p|a|^{p-2}ab$ for $p > 1$ and $a, b \in \re$.
Note that the equality holds true if and only if $b = 0$.
Thus we have
\begin{align*}
	J(u) &:= \intO |\nabla u \cdot (\nabla_{\xi} H)(\nabla d_H)|^p dx - \( \frac{p-1}{p} \)^p \intO \frac{|u|^p}{(d_H)^p} dx \\
	&= \(\frac{p-1}{p}\)^{p-1} \intO \nabla (|v|^p) \cdot (\nabla_{\xi} H)(\nabla d_H) dx \\

	&= -\(\frac{p-1}{p}\)^{p-1} \intO |v|^p (\Delta_H d_H) dx,
\end{align*}
since $H(\nabla d_H) = 1$ a.e.
Therefore, if $J(u) = 0$ for some $u \in C_0^{\infty}(\Omega)$, then we must have
\begin{align*}
	\begin{cases}
	&\Delta_H d_H = 0 \quad a.e. \ \text{in} \, \Omega, \\
	&|\nabla v \cdot (\nabla_{\xi} H)(\nabla d_H)| = 0 \quad a.e. \ \text{in} \, \Omega,
	\end{cases}
\end{align*}
since we assume that $\Delta_H d_H \le 0$ a.e.in $\Omega$.
In particular, we can claim that
{\it if $C_p(\Omega) = \( \frac{p-1}{p} \)^p$ and $-\Delta_H d_H > 0$ on a positive measure, then $C_p(\Omega)$ is not attained}.
\end{remark}

Let $1 < p < \infty$.
In the Euclidean case (i.e., $H(\xi) = |\xi|$, $H^0(x) = |x|$ for $\xi, x \in \re^N$),
the following facts are known \cite{MMP} \cite{MS(NA)}:
\begin{itemize}
\item For any convex domain $\Omega$, $C_p(\Omega) = (\frac{p-1}{p})^p$.
\item For any domain $\Omega$ such that $\pd\Omega$ has a tangent hyperplane at least one point in $\pd\Omega$, $C_p(\Omega) \le (\frac{p-1}{p})^p$.
\item For any bounded $C^2$-domain $\Omega$, if $C_p(\Omega) < (\frac{p-1}{p})^p$ then $C_p(\Omega)$ is attained.
\item For any bounded $C^2$-domain $\Omega$, $C_2(\Omega) < \frac{1}{4}$ if and only if $C_2(\Omega)$ is attained.
\end{itemize}
It could be interesting to study corresponding results for the best constant of the geometric Finsler Hardy inequality (\ref{C_p}).

\section{The scale invariance of the anisotropic critical Hardy inequality}
\label{section:scale FH_N}

In this section,
we shall show that the anisotropic critical Hardy inequality (\ref{FH_N}) is invariant under the scaling
\[
	u_\la (x) = \la^{-\frac{N-1}{N}} u\( \( \frac{H^0(x)}{R}\)^{\la -1} x \), \quad (\la >0)
\]
when $\Omega =\mathcal{W}_R$.
In order to show that, we need the following lemma.
\begin{lemma}\label{Lem:scale}
Let $c > 0$ and $a \in \re$.
For $y \in \re^N$, let $x = c H^0(y)^a y$. Then the Jacobian of the transformation $y \mapsto x$ is
\[
	\left| {\rm det} \( \frac{\pd (x_1, \cdots , x_N)}{\pd (y_1, \cdots , y_N)} \) \right| = c^N (1+a) (H^0(y))^{aN}.
\]
\end{lemma}
In the special case $H(\xi)=|\xi|$, Lemma \ref{Lem:scale} is shown by \cite{Ioku-Ishiwata}.

\begin{proof}

Let us assume $y \ne 0$. Then $x \ne 0$ and we may employ ``polar coordinate" $x = r \w$, $y = \rho \w$, where $r = H^0(x)$, $\rho = H^0(y)$ and $\w \in \pd \mathcal{W}$.
By homogeneity, we see $r = H^0(x) = c (H^0(y))^{a+1} = c \rho^{a+1}$, which implies $dr = c (a+1) \rho^a d\rho$.
Also we see $dx = r^{N-1} dr dS_{\w}$, $dy = \rho^{N-1} d\rho dS_{\w}$, where $dS_{\w}$ is an $(N-1)$-dimensional measure such that
\[
	\int_{\pd \mathcal{W}} dS_{\w} = P_H(\mathcal{W}; \re^N) = \omega_{N-1} = N \kappa_N.
\]
When $\pd \mathcal{W}$ is Lipschitz,
$dS_{\w}$ can be written $dS_{\w} = H(\nu(\w)) d\mathcal{H}^{N-1}$ where $\nu(\w) = \frac{\nabla H^0(\w)}{|\nabla H^0(\w)|}$ is an unit normal vector of $\pd \mathcal{W}$.
Now,
\begin{align*}
	dx &= r^{N-1} dr dS_{\w} = (c \rho^{a+1})^{N-1} \frac{dr}{d\rho} d\rho dS_{\w} \\
	&= c^N (a+1) (\rho^{a+1})^{N-1} \rho^a d\rho dS_{\w} = c^N (a+1) \rho^{aN} \rho^{N-1} d\rho dS_{\w} \\
	&=  c^N (a+1) \rho^{aN} dy.
\end{align*}
On the other hand, $dx = \left| {\rm det} \( \frac{\pd (x_1, \cdots , x_N)}{\pd (y_1, \cdots , y_N)} \) \right| dy$ by definition.
Thus we have the conclusion.
\end{proof}

\begin{remark}\rm
By a direct calculation, we see that the Jacobi matrix
\begin{align*}
	A &= \( \frac{\pd x_i}{\pd y_j} \)_{1 \le i,j \le N} = c (H^0(y))^a \left[ {\rm Id.} +  \frac{a}{H^0(y)} B \right], \\
	B &= ( H^0_{y_j}(y) y_i )_{1 \le i,j \le N}
\end{align*}
has eigenvalues
\begin{itemize}
\item $c (H^0(y))^a$ with multiplicity $N-1$, whose eigenspace is the orthogonal space of the vector $\nabla H^0(y)$, $y \ne 0$.
\item $c (1 + a) (H^0(y))^a$ with multiplicity $1$, whose eigenspace is $\re y$, $y \ne 0$.
\end{itemize}
Thus actually
\[
	{\rm det} A = {\rm det} \( \frac{\pd (x_1, \cdots , x_N)}{\pd (y_1, \cdots , y_N)} \) = c^N (1+a) (H^0(y))^{aN}.
\]
\end{remark}

Set $y= \( \frac{H^0(x)}{R}\)^{\la -1} x$, that is $x=R^{1-\frac{1}{\la}} (H^0(y))^{\frac{1}{\la}-1} y$.
Since
\begin{align*}
&\frac{\pd u(y)}{\pd x_i} \frac{x_i}{H^0(x)} = \sum_{j=1}^N \frac{\pd u(y)}{\pd y_j} \frac{\pd y_j}{\pd x_i} \frac{x_i}{H^0(x)} \\
&=\sum_{j=1}^N \frac{\pd u(y)}{\pd y_j} \frac{x_i}{H^0(x)} R^{1-\la} \left[ (\la -1) (H^0(x))^{\la -2} H^0_{x_i}(x) x_j + (H^0(x))^{\la -1} \delta_{ij} \right] \\
&=R^{1-\la} (\la -1) (H^0(x))^{\la -2} H^0_{x_i}(x) \frac{x_i}{H^0(x)} ( \nabla_y u(y) \cdot x ) + R^{1-\la} (H^0(x))^{\la -1} \frac{\pd u(y)}{\pd y_i} \frac{x_i}{H^0(x)},
\end{align*}
we obtain
\begin{align*}
&\nabla_x u(y) \cdot \frac{x}{H^0(x)} \\
&=R^{1-\la} (\la -1) (H^0(x))^{\la -1} \( \nabla_y u(y) \cdot \frac{x}{H^0(x)} \) + R^{1-\la} (H^0(x))^{\la -1} \( \nabla_y u(y) \cdot \frac{x}{H^0(x)} \) \\
&=\la R^{1-\la} (H^0(x))^{\la -1} \( \nabla_y u(y) \cdot \frac{x}{H^0(x)} \) =\la R^{\frac{1}{\la}-1} (H^0(y))^{1-\frac{1}{\la}} \( \nabla_y u(y) \cdot \frac{y}{H^0(y)} \).
\end{align*}
Therefore we see that
\begin{align*}
&\int_{\mathcal{W}_R} \left| \frac{x}{H^0(x)} \cdot \nabla u_\la (x) \right|^N dx \\
&=\la^{-N+1} \int_{\mathcal{W}_R} \la^N R^{\frac{N}{\la}-N} (H^0(y))^{N-\frac{N}{\la}} \left| \frac{y}{H^0(y)} \cdot \nabla u(y) \right|^N {\rm det} \( \frac{\pd (x_1, \cdots , x_N)}{\pd (y_1, \cdots , y_N)} \) dy \\
&=\la R^{\frac{N}{\la}-N} \int_{\mathcal{W}_R} (H^0(y))^{N-\frac{N}{\la}} \left| \frac{y}{H^0(y)} \cdot \nabla u(y) \right|^N R^{N-\frac{N}{\la}} \frac{1}{\la} (H^0(y))^{\frac{N}{\la}-N} dy \\
&=\int_{\mathcal{W}_R} \left| \frac{y}{H^0(y)} \cdot \nabla u (y) \right|^N dy,
\end{align*}
where the second equality comes from Lemma \ref{Lem:scale}, on taking $c=R^{1-\frac{1}{\la}}$ and $a=\frac{1}{\la} -1$. In the same way as above, we see that
\begin{align*}
&\int_{\mathcal{W}_R} \frac{|u_\la (x)|^N}{(H^0(x))^N (\log \frac{R}{H^0(x)})^N} dx \\
&= \la^{-N+1} \int_{\mathcal{W}_R} \frac{|u (y)|^N}{R^{N-\frac{N}{\la}} ((H^0(y)))^{\frac{N}{\la}} (\frac{1}{\la} \log \frac{R}{H^0(y)})^N} R^{N-\frac{N}{\la}} \frac{1}{\la} (H^0(y))^{\frac{N}{\la}-N} dy \\
&=\int_{\mathcal{W}_R} \frac{|u (y)|^N}{(H^0(y))^N (\log \frac{R}{H^0(y)})^N} dy.
\end{align*}
Hence the inequality (\ref{FH_N}) is invariant.

\section{Relation between the subcritical and the critical anisotropic Hardy inequalities}
\label{section:Relation}

In this section,
according to \cite{Sano-Takahashi},
a relation between the critical and the subcritical anisotropic Hardy inequalities (\ref{FH_p}), (\ref{FH_N}) is presented.
It will be shown  that the critical anisotropic Hardy inequality on a ball is embedded into a family of the subcritical anisotropic Hardy inequalities on the whole space
by using a transformation which connects both inequalities.

\begin{theorem}
\label{Thm:H_p to H_N eq}
Let $m,N \in \N$ satisfy $N \geq 2$ and $m > N$, and $\mathcal{W}^N_R:=\{ y \in \re^N \,|\, H^0(y) <R \}$. Set
\begin{align*}
	I(u)&=\int_{\re^m} \left| \nabla u \cdot \frac{x}{H^0(x)} \right|^N \,dx - \( \frac{m-N}{N} \)^N \int_{\re^m} \frac{|u|^N}{H^0(x)^N} \,dx, \\
	J(w)&=\int_{\mathcal{W}^N_R} \left| \nabla w \cdot \frac{y}{H^0(y)} \right|^N \,dy \\
	&\hspace{5em}- \( \frac{N-1}{N} \)^N \int_{\mathcal{W}^N_R} \dfrac{|w|^N}{H^0(y)^N \( \log \frac{R}{H^0(y)} \)^N} \,dy.
\end{align*}
Then for any $w \in C^1_{H^0rad}(\mathcal{W}^N_R \setminus \{ 0\})$ (resp. $u \in C_{H^0rad}^1(\re^m \setminus \{ 0\})$),
there exists $u \in C_{H^0rad}^1(\re^m \setminus \{ 0\})$ (resp. $w \in C^1_{H^0rad}(\mathcal{W}^N_R \setminus \{ 0\})$) such that the equality
\begin{align}
\label{H_p to H_N eq}
	I(u)= \frac{\w_{m-1}}{\w_{N-1}} \( \frac{m-N}{N-1} \)^{N-1} J(u)
\end{align}
holds true where $\w_{N-1}= N \kappa_N$ and $\w_{m-1}= m \kappa_m$.
\end{theorem}

Before the proof, we define a transformation which connects the critical and the subcritical anisotropic Hardy inequalities according to \cite{Sano-Takahashi}.
Let $m, N$ be integers such that $m > N$ and let $R > 0$ be fixed.
For a given $r \in [0,+\infty)$ (resp. $s \in [0,R)$ ), define a new variable
$s \in [0, R)$ (resp. $r \in [0, +\infty)$ ) by the relation
\begin{equation}
\label{s:r}
	\( \log \frac{R}{s} \)^{\frac{N-1}{N}} = r^{-\frac{m-N}{N}},
\end{equation}
that is,
\begin{equation}
\label{s(r):r(s)}
	s = s(r) = R \exp (-r^{-\alpha}), \quad (\text{resp.} \; r = r(s) =  \( \log \frac{R}{s} \)^{-1/\alpha} )
\end{equation}
where
\begin{equation}
\label{alpha}
	\alpha = \frac{m-N}{N-1}.
\end{equation}
Note that
the left-hand side of (\ref{s:r}) is the virtual extremal for (\ref{FH_N}) on $\mathcal{W}^N_R$
and the right-hand side of (\ref{s:r}) is the virtual extremal for (\ref{FH_p}) on the whole space $\re^m$ when $p = N < m$.
Easy computation shows that
\begin{equation}
\label{dsdr}
	\frac{ds}{dr} = \alpha s r^{-\alpha -1} > 0,
\end{equation}
so when $r$ varies from $0$ to $+\infty$ then $s$ varies from $0$ to $R$, and vice versa.

Let $r=H^0(x)$, $x \in \re^m$ and $s=H^0(y)$, $y \in \mathcal{W}^N_R$.
Now, for a given $u = u(r) \in C_{H^0rad}^1(\re^m \setminus \{ 0\})$ (resp. $w  = w(s) \in C_{H^0rad}^1(\mathcal{W}^N_R \setminus \{ 0\})$),
define a new function
$w  = w(s) \in C_{H^0rad}^1(\mathcal{W}^N_R \setminus \{ 0\})$ (resp.  $u = u(r) \in C_{H^0rad}^1(\re^m \setminus \{ 0\})$)
by
\begin{equation}
\label{w:u}
	w(s) = u(r),
\end{equation}
where variables $s$ and $r$ are related as in (\ref{s:r}).
Note that $\lim_{s \to R} w(s) =0$ is equivalent to $\lim_{r \to \infty} u(r) =0$.
Namely, under the transformation (\ref{w:u}), the boundary $\pd \mathcal{W}^N_R$ corresponds to the infinity point $\infty$ in $\re^m$.

\begin{proof}[{\bf Proof of Theorem \ref{Thm:H_p to H_N eq}}]

Define $u$ and $w$ as in \eqref{w:u}.
Then we obtain
\begin{align*}
	\int_{\re^m} \left| \nabla u \cdot \frac{x}{H^0(x)} \right|^N dx &= \w_{m-1} \int_{0}^{\infty} |u^{\prime}(r)|^{N} r^{m-1} dr \\
	&= \w_{m-1} \int_{0}^{R} \left| w^{\prime}(s) \frac{ds}{dr} \right|^{N} r(s)^{m-1} \frac{dr}{ds} \,ds \\
	&=\w_{m-1} \int_{0}^{R} |w^{\prime}(s)|^N \( \alpha s r(s)^{-\alpha-1} \)^{N-1} r(s)^{m-1} \,ds \\
	&=\w_{m-1} \alpha^{N-1} \int_{0}^{R} |w^{\prime}(s)|^N s^{N-1} r(s)^{m-1 - (\alpha+1)(N-1)} \,ds \\
	&= \frac{\w_{m-1}}{\w_{N-1}} \alpha^{N-1} \int_{0}^{R} |w^{\prime}(s)|^N s^{N-1} \,ds \\
	&= \frac{\w_{m-1}}{\w_{N-1}} \alpha^{N-1} \int_{\mathcal{W}^N_R} \left| \nabla w \cdot \frac{y}{H^0(y)} \right|^N \,dy,
\end{align*}
here we have used (\ref{dsdr}) and $m-1 - (\alpha +1)(N-1) = 0$ by (\ref{alpha}).

On the other hand, we have
\begin{align*}
	\int_{\re^m} \dfrac{|u(x)|^N}{H^0(x)^N} dx &= \w_{m-1} \int_{0}^{\infty} |u(r)|^N r^{m-N-1} \,dr \\
	&= \w_{m-1} \int_{0}^{R} |w(s)|^N r(s)^{m-N-1} \,\frac{dr}{ds} \,ds \\
	&= \w_{m-1} \int_{0}^{R} |w(s)|^N r(s)^{m-N-1} \alpha^{-1} s^{-1} r(s)^{\alpha+1}  \,ds \\
	&= \frac{\w_{m-1}}{\alpha} \int_{0}^{R} \frac{|w(s)|^N}{s} r(s)^{m-N+\alpha} \,ds \\
	&= \frac{\w_{m-1}}{\alpha} \int_{0}^{R} \frac{|w(s)|^N}{s (\log \frac{R}{s} )^{N}} \,ds \\
	&= \frac{\w_{m-1}}{\alpha \w_{N-1}} \int_{\mathcal{W}^N_R} \dfrac{|w(y)|^N}{H^0(y)^N \( \log \frac{R}{H^0(y)} \)^N} \,dy,
\end{align*}
since $r(s)^{m-N+\alpha} = (\log \frac{R}{s} )^{-N}$ by (\ref{s:r}) and (\ref{alpha}).

By combining these identities, we obtain Theorem \ref{Thm:H_p to H_N eq}.
\end{proof}

Lastly we show that the transformation preserves the scale invariance structures of the subcritical and the critical anisotropic Hardy inequalities.

\begin{proposition}
\label{Prop:scale}
Let $m, N$ be integers such that $m > N$.
For functions $u = u(r)$, $r \in [0, +\infty)$ and $w = w(s)$, $s \in [0,R)$,
define the scaled functions
\begin{align*}
	&u_{\mu} (r) = \mu^{\frac{m-N}{N}} u ( \mu r ), \\
	&w^{\la} (s) = \la^{-\frac{N-1}{N}} w (R^{1-\la} s^{\la})
\end{align*}
for $\mu, \la >0$.
Then we have
\begin{align*}
	&w^{\la}(s(r)) = u_{\mu}(r), \quad \text{where} \; \mu = \la^{-1/\alpha}, \\
	&u_{\mu}(r(s)) = w^{\la}(s), \quad \text{where} \; \la = \mu^{-\alpha},
\end{align*}
where $s = s(r)$ and $r = r(s)$ are as in (\ref{s(r):r(s)}) and  $\alpha$ is defined in (\ref{alpha}).
\end{proposition}

\begin{proof}
By direct calculation,
\begin{align*}
	R^{1-\la} s(r)^{\la} = R^{1-\la} \( R \exp (-r^{-\alpha} ) \)^{\la} &= R \exp \( -\la r^{-\alpha} \) \\
	&= R \exp \( \( - (\la^{-1/\alpha} r) \)^{-\alpha} \) = s( \mu r ),
\end{align*}
where $\mu = \la^{-1/\alpha}$.
Therefore we obtain
\begin{align*}
	w^{\la}(s(r)) &= \la^{-\frac{N-1}{N}} w(R^{1-\la} s(r)^{\la}) = \( \la^{-1/\alpha} \)^{\frac{m-N}{N}} w(s(\mu r)) \\
	&= \mu^{\frac{m-N}{N}}  u (\mu r) = u_{\mu}(r).
\end{align*}
The proof of $u_{\mu}(r(s)) = w^{\la}(s)$ for $\la = \mu^{-\alpha}$, is similar.
\end{proof}

\begin{comment}
In the $H^0$-radial case, the critical Hardy inequality is equivalent to the one of the subcritical anisotropic Hardy inequalities as described above.
However, in the $H^0$-non-radial case, since $m$ (dimension of the subcritical anisotropic Hardy inequality) is larger than $N$ (dimension of the critical one),
some information concerning the $H^0$-spherical variables may be lost.
Nevertheless we can deduce the critical anisotropic Hardy inequality from the subcritical one as follows.

\begin{theorem}(Non-radial case.)\label{Thm:H_p to H_N eq(non-rad)}
Let $m,N \in \N$ satisfy $N \geq 2$ and $m > N$.
Then for any $w \in C^1(B_R(0) \setminus \{ 0\})$
there exists $u \in C^1(\re^m \setminus \{ 0\})$ such that
\begin{align}
\label{H_p to H_N eq(non-rad)}
	&\int_{\re^m} \left| \nabla u \cdot \frac{x}{|x|} \right|^N \,dx - \( \frac{m-N}{N} \)^N \int_{\re^m} \frac{|u|^N}{|x|^N} \,dx \\
	&= \frac{\w_m}{\w_N} \( \frac{m-N}{N-1} \)^{N-1} \notag \\
&\( \int_{B_R^N(0)} \left| \nabla w \cdot \frac{y}{|y|} \right|^N \,dy - \( \frac{N-1}{N} \)^N \int_{B_R^N(0)} \dfrac{|w|^N}{|y|^N \( \log \frac{R}{|y|} \)^N} \,dy \) \notag
\end{align}
holds true.
\end{theorem}
\end{comment}

\vspace{1em}\noindent
{\bf Acknowledgments.}

The authors of this paper thank to Prof. M. Ruzhansky for his kind comments and useful information.

A part of this work was done while the second (M.S.) and the third author (F.T.)  visited Universit\`a di Napoli Federico II, Dipartimento di Matematica e Applicazioni `R. Caccioppoli''in September, 2017.
They thank the warm hospitality of the department.

The third author (F.T.) was supported by JSPS Grant-in-Aid for Scientific Research (B), No.15H03631.


text:
\begin{document}

\title{Quasi-optimal grouping for broadcast systems with hierarchical modulation}
\author{Hugo M{\'e}ric, Jos{\'e} Miguel Piquer and J{\'e}r{\^o}me Lacan}

\abstract{Recently, we proposed to combine time sharing with hierarchical modulation to increase the transmission rate of broadcast systems. Our proposal involves to group the receivers in pairs in order to transmit with hierarchical modulation. We introduced several grouping strategies but the optimal matching remained an open question. In this letter, we show that the optimal grouping is the solution of an assignment problem, for which efficient algorithms exist such as the Hungarian method. Based on this algorithm, we study the performance of the optimal grouping in terms of spectrum efficiency for a DVB-S2 system.}

\maketitle

\section{Introduction}
Modern broadcasting systems such as DVB-S2\footnote{Digital Video Broadcasting - Satellite - Second Generation} or DVB-SH\footnote{Digital Video Broadcasting - Satellite to Handheld} rely mainly on a time sharing strategy to optimise the transmission rate. Recently, we showed that combining time sharing with hierarchical modulation, a technique that merges several data streams in a same symbol, can provide significant gains (in terms of spectrum efficiency) compared to the best time sharing strategy \cite{1,2}.

In this letter, we consider one source communicating with $n$ receivers. The objective is to offer the \emph{same average spectrum efficiency to all the receivers}. We assume that the transmitter has knowledge of the signal-to-noise ratio (SNR) at the receivers. A concrete example is a DVB-S2 system that implements adaptive coding and modulation (ACM). Moreover, the system also implements \emph{hierarchical modulation with two layers}, i.e., two data streams are merged in a same symbol.

Receiver $i$ ($1 \leqslant i \leqslant n$) has a spectrum efficiency $R_i$ which corresponds to the best spectrum efficiency it can manage. The value of $R_i$ depends on $SNR_i$, the SNR of receiver $i$, and the transmission parameters (code rate and modulation). For instance, if the source transmits a QPSK modulated signal and the code rate is 1/3, then $R_i = 2\times 1/3$ bit/symbol. However, if receiver $i$ is paired with receiver $j$ ($1 \leqslant j \leqslant n$, $j \neq i$) and hierarchical modulation is used, the spectrum efficiency of receivers $i$ and $j$ is $R_{ij}^{hm}$ which depends on $SNR_i$, $SNR_j$ and the transmission parameters. The computation of $R_{ij}^{hm}$ is detailed in \cite{1}.

During the transmission, the source can either communicate directly with a receiver (called a single receiver) or group it with another receiver and use hierarchical modulation (called paired receivers). This process is the grouping strategy or matching. Once the strategy is decided, the average spectrum efficiency offered to all the receivers is
\begin{equation}
R = \left( \sum_{k} \frac{1}{R_{k}} + \sum_{(i,j)} \frac{1}{R^{hm}_{ij}} \right)^{-1},
\label{rate}
\end{equation}
where the sum over $k$ takes into account the single receivers and the sum over $(i,j)$ the paired receivers. Equation (\ref{rate}) is a direct extension of (9) in \cite{2}. As an example, \figurename~\ref{hm} illustrates a system with 8 receivers and a given grouping strategy. In that case, the average spectrum efficiency is
\begin{equation}
R = \left( \frac{1}{R^{hm}_{1,2}} + \frac{1}{R^{hm}_{3,4}} + \frac{1}{R_5} + \frac{1}{R_6} + \frac{1}{R^{hm}_{7,8}} \right)^{-1}.
\end{equation}

Compared to our previous works \cite{1,2}, the matching considered in this letter allows to communicate directly with a receiver without pairing it with another receiver. Thus the framework is more general.

\begin{figure}[h]
\centering
\includegraphics[width=0.7\columnwidth]{images/broad}
\caption{Broadcast system with 8 receivers and a given grouping strategy}
\label{hm}
\end{figure}

\section{Optimal matching}
In \cite{1}, we introduced several grouping strategies. Among the proposed strategies, the one that achieves the best performance consists in grouping the two receivers with the largest SNR difference and repeating this operation until each receiver is in a pair. However, the optimal matching, i.e., the one that maximises the average spectrum efficiency in (\ref{rate}), remained an open question.

For a system with $n$ receivers, we note $s_n$ the number of possible strategies consisting in single or paired receivers. The sequence $(s_n)_{n \geqslant 1}$ verifies
\begin{equation}
s_n = s_{n-1} + (n-1) s_{n-2},
\label{sn}
\end{equation}
for $n \geqslant 3$, with $s_1=1$ and $s_2=2$. To obtain (\ref{sn}), we consider a system with $n$ receivers. Then receiver $n$ can either be a single receiver and in that case we group the remaining $n-1$ receivers; or  receiver $n$ can be paired with another receiver (there are exactly $n-1$ possibilities) and we group the remaining $n-2$ receivers. From (\ref{sn}), we can show by recursion that $s_k \geqslant 2^k$ for $k \geqslant 5$. For large broadcast systems, it is thus impossible to test all the possible matchings to determine the optimal one.

To obtain the optimal strategy, we write the problem in matrix form. A grouping strategy can be represented by a $n \times n$ \emph{assignment matrix} $X$ where
\begin{equation}
X_{i,j} =  \begin{cases} 1 , & \text{if receiver } i \text{ is paired with receiver } j \\ 0 , & \text{otherwise.}\end{cases}
\end{equation}
By definition, $X$ is symmetric, contains exactly $n$ non-zero entries and there is exactly one non-zero entry in each row and column. In other words, $X$ is a symmetric permutation matrix. Note that the ones in the diagonal correspond to the single receivers. Then, we define the $n \times n$ \emph{cost matrix} $C$ by
\begin{equation}
C_{i,j} =  \begin{cases} 1/R_i , & \text{if } i=j \\ 1/\left(2R_{ij}^{hm}\right) , & \text{if } i \neq j \end{cases}
\label{cost}
\end{equation}
where $1 \leqslant i,j \leqslant n$. The cost matrix is also symmetric.

In combinatorial optimisation, the \emph{assignment problem} can be formulated as follow: given a cost matrix $C$, find an assignment (i.e., a set of $n$ entry positions so no two of which lie in the same row or column) such that the sum of the $n$ entries is the smallest possible. The sum of the $n$ entries is the \emph{assignment cost}. By noting that maximising the average spectrum efficiency in (\ref{rate}) amounts to minimising the term
\begin{equation}
\sum_{k} \frac{1}{R_{k}} + \sum_{(i,j)} \frac{1}{R^{hm}_{ij}},
\label{term_to_minimise}
\end{equation}
the optimal grouping strategy is (almost) equivalent to an assignment problem with the cost matrix $C$ defined in (\ref{cost}). Indeed, for a given grouping with assignment matrix $X$, the assignment cost of $X$ is equal to (\ref{term_to_minimise}). The only difference with the classical problem is that $X$ requires to be symmetric.

Several efficient algorithms solve the general assignment problem. For instance, the Munkres' assignment algorithm or Hungarian method solves it in polynomial time \cite{3,4}. For a $n \times n$ cost matrix, the algorithm can achieve a $\mathcal{O} (n^3)$ time complexity. The implementation requires easy operations on the cost matrix. We use this method with the cost matrix defined previously to obtain a quasi-optimal grouping.

\section{Implementation issue}
Henceforth, we have an heuristic to compute the optimal matching. Given the cost matrix defined in (\ref{cost}), the Hungarian method will always find a solution. However the assignment matrix provided by the algorithm may not be symmetric, while our problem requires a symmetric $X$. We will now discuss the following two questions: (a) knowing that the cost matrix $C$ is symmetric, is there always a symmetric solution? (b) Can the Hungarian method find an optimal symmetric $X$ solution to our problem?

To answer (a), we consider the following cost matrix
\begin{equation}
C =
\begin{pmatrix}
3             & \underline{4} & \mathbf{1}\\
\mathbf{4}    & 7             & \underline{3}\\
\underline{1} & \mathbf{3}    & 2
\end{pmatrix},
\label{counterexample}
\end{equation}
where the bold and underlined numbers correspond to the two optimal assignments. We conclude that even if $C$ is symmetric, there is not always a symmetric assignment matrix that solves the problem. Thus \emph{the Hungarian method only provides an upper bound for the spectral efficiency}, where the bound equals the inverse of the assignment cost returned by the algorithm.

Concerning (b), we will present how to use the Munkres' algorithm to obtain a quasi-optimal matching. Our simulations with the cost matrix defined previously show that the assignment matrix returned by the Hungarian method is generally not symmetric. This may be due to the fact that many coefficients in the matrix are the same. However, we remarked that for random matrices, the probability to obtain a symmetric assignment matrix is larger. Thus our idea is to add a small perturbation to the cost matrix $C$ and run the Hungarian method on the perturbed matrix $C'$. In practice, we compute $C'$ as follows: $C' = C + \epsilon$, where $\epsilon$ is a symmetric matrix whose coefficients are drawn according to $\mathcal{N}(0,\sigma^2)$ with $\sigma=10^{-3}$.

Using this technique, we obtained a symmetric assignment most of the time (more details in the next section). We then compared the average spectrum efficiency of this grouping with the upper bound computed by the Hungarian method. As the results will point out, the performance of the matching obtained with the perturbed cost matrix is very close to the upper bound, justifying the term quasi-optimal of our method.

\section{Performance evaluation}
To evaluate the performance of the quasi-optimal matching, we use the simulation framework proposed in \cite{1,2}. We quickly give the main characteristics. More details can be found in the previous papers. First, the physical layer is based on the DVB-S2 standard \cite{5}. The code rates and modulations are resumed in Table~1. The hierarchical 16-APSK, which is not in the DVB-S2 standard, was introduced in \cite{1}. Then, the transmission is subject to additive white Gaussian noise (AWGN). Finally, the channel model to estimate the SNR distribution of the receivers in a spot beam takes into account two sources of attenuation: the relative location of the terminal with respect to the center of (beam) coverage and the weather. For a given simulation, the only parameter to set is the SNR at the center of the spot beam $SNR_{max}$.
\begin{table}[h]
\processtable{Transmission parameters}
{\renewcommand{\arraystretch}{1.05}
\begin{tabular}{l|c}\hline
Code rate  & 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6, 8/9, 9/10 \\\hline
Modulation & QPSK, 8-PSK, 16-APSK, 32-APSK\\
           & hierarchical 8-PSK, hierarchical 16-APSK \\\hline
\end{tabular}}{}
\end{table}

\figurename~\ref{results} presents the gains (in terms of spectrum efficiency) when combining hierarchical modulation with time sharing for a broadcasting area with 500 receivers. For one system configuration (i.e., the parameter $SNR_{max}$ is set), we present the average, minimum and maximum gains over 100 simulations for the quasi-optimal matching obtained with the Hungarian method and for the best strategy in \cite{1} (which was described previously). The results point out that the quasi-optimal strategy provides some gains compared to the strategy proposed in \cite{1}. However the margin is small: in the best case ($SNR_{max}=9$ dB), we increase the performance from 6\% to 8\%. Even if we only notice a slight improvement, the maximum gain is now known for our framework.
\begin{figure}[h]
\centering
\includegraphics[width=0.975\columnwidth]{images/perf_randn}
\caption{Average spectrum efficiency gains (in comparison to time sharing without hierarchical modulation) with 500 receivers}
\label{results}
\end{figure}

We mentioned before that the Hungarian algorithm does not generally return a symmetric assignment. To tackle this problem, we proposed to add a small perturbation to the cost matrix. During our simulations, our technique did not succeed to find a symmetric assignment for only a few cases: 8 simulations over 108 for $SNR_{max}=9$ dB and 4 over 104 for $SNR_{max}=10$ dB. Our method is thus quite reliable to find symmetric assignment. Moreover, we compared the performance of the symmetric assignment and the optimal assignment (provided by the Hungarian method and not necessarily symmetric). In all the simulations, the difference between the two is less than 1\%. In conclusion, we may not find the optimal symmetric assignment (as we use a perturbed cost matrix) but the performance is very close to the upper bound.

Finally, we are interested to study the structure of the quasi-optimal groupings. Are the matchings random or is there a pattern (e.g., group the receivers with the largest SNR difference as in \cite{1})? To that end, we use the simulations to compute the probability that a coefficient in the assignment matrix $X$ is equal to one, i.e., $\Pr(X_{i,j}=1)$.

In order to see if there exists a pattern in the grouping process, we sort the receivers by increasing SNR. With that representation, the best strategy proposed in \cite{1} corresponds to the anti-diagonal matrix. Indeed, this strategy consists in grouping the two receivers with the largest SNR difference and repeating this operation. After sorting the receiver by increasing SNR, receiver $1$ which has the lowest SNR is matched to receiver $n$ which has the largest SNR, so $X_{1,n}=X_{n,1}=1$. For $1 \leqslant k \leqslant n$, receiver $k$ is matched to receiver $n+1-k$ and the corresponding assignment matrix is the anti-diagonal matrix. Note that time sharing, where there is no grouping, always corresponds to the identity matrix.

We present the structure of the assignment matrix in \figurename~\ref{pattern} for several values of $SNR_{max}$. We see clearly in \figurename~\ref{stats2} and \figurename~\ref{stats3} that the quasi-optimal grouping usually matches low SNR receivers with large SNR receivers. This explains why the matching proposed in \cite{1} performs well. Moreover, when the gains compared to time sharing are low (see \figurename~\ref{stats1} and \figurename~\ref{stats4}), the optimal matching exhibits many single receivers as time sharing.
\begin{figure}[!ht]
\centering
\subfigure[$SNR_{max} = 7$ dB]{
\includegraphics[width=0.475\columnwidth]{./images/results_7}
\label{stats1}
}
\subfigure[$SNR_{max} = 10$ dB]{
\includegraphics[width=0.475\columnwidth]{./images/results_10}
\label{stats2}
}
\subfigure[$SNR_{max} = 13$ dB]{
\includegraphics[width=0.475\columnwidth]{./images/results_13}
\label{stats3}
}
\subfigure[$SNR_{max} = 16$ dB]{
\includegraphics[width=0.475\columnwidth]{./images/results_16}
\label{stats4}
}
\caption{Structure of the quasi-optimal assignment matrix ($\Pr(X_{i,j}=1)$)}
\label{pattern}
\end{figure}

\section{Conclusion}
We showed how to obtain a close-to-optimal (in terms of spectrum efficiency) grouping strategy in a broadcast system that relies on time sharing and hierarchical modulation with two layers. The matching is the solution of an assignment problem that we solve using the Hungarian method. We compared the performance of the quasi-optimal strategy with a previously proposed matching and also studied its behaviour. In a future work, we will investigate how to obtain the optimal symmetric matching.
\vskip5pt

\noindent Hugo M{\'e}ric (\textit{INRIA Chile, Santiago, Chile})\\
\noindent E-mail: hugo.meric@inria.cl
\vskip3pt
\noindent Jos{\'e} Miguel Piquer (\textit{NIC Chile Research Lab, Santiago, Chile})

\vskip3pt
\noindent J{\'e}r{\^o}me Lacan (\textit{Universit{\'e} de Toulouse, Toulouse, France})

\vskip3pt

config: arxiv


TITLE: Probability distribution similar to log-normal.
QUESTION [0 upvotes]: I just got some moments of a random variable that have the form as follows.
$E(X^n)=\frac{n!}{(2n-1)!!}e^{n(n+1)x}$
It looks like a log-normal distribution, but not a log-normal distribution.
Is there a well-known probability distribution in which moments have the mentioned form?

REPLY [0 votes]: Too long for a comment:
When a lognormal variable is defined as
$$
Y=e^{\mu+\sigma Z}\,,\quad Z\sim N(0,1)
$$
then the moments of $Y$  are -as we know-
$$
\mathbb E[Y^n]=e^{\mu n +\frac{1}{2}n^2\sigma^2}\,.
$$
Therefore, with $\mu=x$ and $\sigma^2=2x$ we get
$$
\mathbb E[Y^n]=e^{n(n+n)x}\,.
$$
Writing
$$
(2n-1)!!=\frac{(2n)!}{2^nn!}
$$
we get
$$
\frac{n!}{(2n-1)!!}=\frac{2^n(n!)^2}{(2n)!}=\frac{2^n}{2n\choose{n}}\approx\frac{n!}{n^n}
$$
where we used that, for large $n$,
$$
{2n\choose n}\approx\frac{(2n)^n}{n!}
$$
holds.


TITLE: How to prove that two Groups Isomorphic?
QUESTION [0 upvotes]: We know that when two groups are cyclic and have the same order then these two groups are isomorphic.
If we know that one Group is cyclic and the other is not and they have the same order could we say that they are not isomorphic to each other?

REPLY [1 votes]: "If we know that one Group is cyclic, say $C_4$, and the other is not, say $C_2\times C_2$, and they have the same order, namely $4$, could we say that they are not isomorphic to each other?" Yes, we could.
The property of being cyclic, or abelian (or being nilpotent, solvable etc.),is preserved under isomorphism. The idea of an "isomorphism" is, that all "algebraic" properties are preserved.


TITLE: Indeterminate form : $ 0 \cdot \infty$
QUESTION [1 upvotes]: I have a question about the limit of :
$$\lim_{t \to \infty}t\cdot e^{-t}$$
Why is that equal to $0$ ? Shoudln't $ 0 \cdot \infty$ be an inderminate form ?
It's true that with the graphic we can determine that this limit is equal to  $0$ :

But all I know is that  : $ 0 \cdot \infty$ is supposed to be an inderminate form .

REPLY [1 votes]: A nice way to show this is:
$$f(t)=te^{-t}=\frac{t}{e^t}$$
and we know that:
$$e^t=\sum_{n=0}^\infty\frac{t^n}{n!}=1+t+\frac{t^2}{2}+\frac{t^3}{6}+...$$
and so:
$$f(t)=\frac{t}{1+t+t^2/2+t^3/3+...}=\frac{1}{1/t+1+t/2+t^2/3+...}$$
now if we look at this we can see that:
$$\lim_{t\to\infty}\frac1t=0$$
and so on the bottom we have something of the form:
$$\frac 1{1+\infty+\infty+...}$$
which will obviously tend to $0$


text:
\begin{document}

\maketitle

\begin{abstract}
Adaptive regularization methods come in diagonal and full-matrix variants. However, only the former have enjoyed widespread adoption in training large-scale deep models. This is due to the computational overhead of manipulating a full matrix in high dimension. In this paper, we show how to make full-matrix adaptive regularization practical and useful. We present GGT, a truly scalable full-matrix adaptive optimizer. At the heart of our algorithm is an efficient method for computing the inverse square root of a low-rank matrix. We show that GGT converges to first-order local minima, providing the first rigorous theoretical analysis of adaptive regularization in non-convex optimization. In preliminary experiments, GGT trains faster across a variety of synthetic tasks and standard deep learning benchmarks.
\end{abstract}

\input{1-intro}
\input{2-algorithm}
\input{3-experiments}
\input{4-theory-sketch}

\section{Conclusion}
This work investigates full-matrix adaptive regularization: our main contribution is to make this technique viable for large-scale optimization, by a method for efficient multiplication by the inverse square root of a full second-moment matrix over a short window of gradients. This leads to a new algorithm, GGT, a truly scalable optimization algorithm with full-matrix adaptive preconditioning.

Through synthetic experiments, we have shown that GGT accelerates optimization in ill-conditioned loss landscapes; this is supported by accompanying adaptive convergence guarantees. Preliminary experiments show accelerated convergence on standard deep learning benchmarks, with very different training dynamics from existing diagonal adaptive methods. We accompany our algorithm and experiments with the first theoretical guarantees for adaptive regularization in the non-convex setting, giving examples of provably faster convergence to first-order critical points. We hope that GGT will be the first of a new class of algorithms for the modern large-scale optimization toolbox, and to foster new discussion towards an ever-elusive understanding of loss landscapes in deep learning.

\section*{Acknowledgments}
We are grateful to Yoram Singer, Tomer Koren, Nadav Cohen, and Sanjeev Arora
for helpful discussions.

\bibliography{main}
\bibliographystyle{alpha}

\appendix
\input{A-theory}

\end{document}


TITLE: Polynomial Transformation
QUESTION [0 upvotes]: I'm currently at Desmos & GeoGebra tinkering with my project. I wanted to form a two letter A's with a domain restriction cutting its legs beneath it. How do I enlarge this polynomial equation in every possible direction? (like double or triple its original size)?
$$y=-4.6x^4-2.6x^3+9.9x^2+2.9x+4.2$$
It seems that changing the variables present within, only increases its height and that's not what I wanted to happen. Moreover, what variable should I add/remove within the expression to come up with that endeavor of mine?

REPLY [0 votes]: For any function $y(x)$, use $2y(x/2)$ to double it. In general $\lambda y(x/\lambda)$ scales it by $\lambda$ in both $x$ and $y$ directions.
For your specific example, $$Y(x)=2(-4.6 (x/2)^4 - 2.6 (x/2)^3 + 9.9 (x/2)^2 + 2.9 (x/2) + 4.2)$$ does the trick.