text:
\section{Introduction}
\label{sec:int}
Hardy inequalities
 are of paramount importance in harmonic analysis, functional analysis, partial differential equations, potential theory and probability.
They are applied to
embedding theorems, Gagliardo--Nirenberg interpolation inequalities and in real interpolation theory, see Chua \cite{MR2138501},
 Ka{\l}amajska and Pietruska-Pa{\l}uba \cite{MR2852869}.
They
yield contractivity of operator semigroups,
a priori estimates, existence and regularity results  for solutions of PDEs,
plus their asymptotics
and qualitative properties, see, e.g.,  Maz'ya \cite{MR2777530}, Arendt, Goldstein and Goldstein \cite{MR2259099},  Barras and Goldstein \cite{MR742415}, and Vazquez and Zuazua \cite{MR1760280}.
The important connection betweeen Hardy-type inequalities and superharmonic functions was exploited, e.g., by Ancona \cite{MR856511}, Fitzsimmons \cite{MR1786080}, Bogdan and Dyda \cite{MR2663757}, Dyda \cite{Dyda12},
Devyver, Fraas and Pinchover \cite{MR3170212},
Bogdan, Dyda and Kim \cite{MR3460023}, Bogdan, Jakubowski, Grzywny and Pilarczyk \cite{MR3933622}.
In particular, the
inequalities are connected to sharp estimates of the heat kernel of $\Delta^{\alpha/2} + \kappa|x|^{-\alpha}$  \cite{MR3933622}, see also Calvaruso, Metafune, Negro and Spina \cite{MR4043021}.

For an account of
the history of Hardy-type inequalities we refer to Opic and Kufner \cite{MR1069756}.
The subject was initiated in 1920, when Hardy \cite{MR1544414} discovered that
\begin{equation}\label{e:hardycls}
\int_{0}^{\infty} \left[u'(x)\right]^2 dx \geq \frac{1}{4} \int_0^{\infty} \frac{u(x)^2}{x^2} dx,\end{equation}
for absolutely continuous functions $u$ such that $u(0)=0$ and $u' \in L^2(0,\infty)$.
The classical Hardy inequality in ${\mathbb{R}^d}$ for $d \geq 2$
is
\begin{equation}\label{e:hardycl}
\int_{{\mathbb{R}^d}} |\nabla u(x)|^2 \ge \frac{(d-2)^2}{4}\int_{{\mathbb{R}^d}} \frac{u(x)^2}{|x|^2} dx, \quad u\in L^2({\mathbb{R}^d}).
\end{equation}
Here the left-hand side of \eqref{e:hardycl} is considered infinite if the distributional gradient of $u$ is not a square-integrable function,  see, e.g., \cite[(30) and (32)]{MR3460023} for this formulation.

In 2000 Fitzsimmons \cite{MR1786080}
proved an abstract analogue of \eqref{e:hardycl}, in which the Dirichlet integral appearing on the left-hand side of \eqref{e:hardycl} is replaced
by a general symmetric Dirichlet form $\mathcal{E}$ in the sense of  Fukushima, Oshima, Takeda \cite{MR2778606}.
The rule stemming from \cite{MR1786080} is the following: If $\mathcal{L}$ is the generator of the
form
and function $h$ is superharmonic, i.e., $h\ge 0$ and $\mathcal{L}h\le 0$, then
$\mathcal{E}(u,u)\ge -\int u^2 \mathcal{L}h/h$.

The paper \cite{MR3460023} gives similar results in the setting of symmetric transition densities, with
explicit construction of the
function $h$, Riesz' charge
$-\mathcal{L}h$,
and  the Hardy weight, or Fitzsimmons' ratio,
$-\mathcal{L}h/h$.
The resulting
Hardy inequalities, in fact, Hardy identities, in some cases are optimal in the sense of large weight and large functional domain, desirably the whole of $L^2$.

The present work extends part of the results of \cite{MR3460023} to the setting of $L^p$ spaces with arbitrary $p\in (1,\infty)$.
We focus on integral forms
related to the semigroup of fractional Laplacian and give optimal inequalities in this case. We also show that the inequalities lead to optimal contractivity results for related operator semigroups on $L^p$.
From our presentation it should also be evident that the approach applies to more general sub-Markovian semigroups.

\subsection{Sobolev-Bregman forms}
Let  $d \in {\mathbb{N}}$ and $0<\alpha<2$.
We consider the fractional Laplacian,
$$
\Delta^{\alpha/2}u(x) := -(-\Delta)^{\alpha/2}u(x):=\underset{\epsilon \to 0+}{\lim} \int_{|y-x|>\epsilon} \left(u(y)-u(x)\right) \nu(x,y) \,dy,\quad x\in \mathbb R^d.
$$
Here, say, $u\in C_c^{2}(\mathbb R^d)$,
\[
\nu(x,y)=\mathcal{A}_{d,-\alpha}|y-x|^{-d-\alpha}, \quad \quad x, y \in {\mathbb{R}^d},
\]
and $\mathcal{A}_{d,-\alpha}=
2^{\alpha}\Gamma\big((d+\alpha)/2\big)\pi^{-d/2}/|\Gamma(-\alpha/2)|$.
We further consider
\begin{equation}\label{e:def-quad-form}
\mathcal{E}
{[u]}:= \mathcal{E}(u,u):= \, \frac{1}{2} \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} (u(x)-u(y))^2 \nu(x,y) \,dy \,dx,
\end{equation}
 defined for every (Borel measurable) $u: {\mathbb{R}^d} \to \mathbb{R}$.
The natural domain $\mathcal D(\mathcal E)$ of the form
consists of those functions $u \in L^2({\mathbb{R}^d})$
 for which $\mathcal{E}[u]<\infty$.
By \cite[Proposition 5]{MR3460023},
for all $0<\alpha<d \wedge 2$, $0 \leq \beta \leq d-\alpha$,
and
$u \in L^2({\mathbb{R}^d})$, we have the Hardy-type identity
\begin{align}\label{e:p2}
&\mathcal{E}{[u]}=
\kappa_{\beta}
\int_{\mathbb{R}^d} \frac{u(x)^2}{|x|^\alpha}\,dx
 + \frac{1}{2} \int_{\mathbb{R}^d}\!\int_{\mathbb{R}^d}
\left[\frac{u(x)}{h(x)}-\frac{u(y)}{h(y)}\right]^2
h(x)h(y) \nu(x,y)
\,dy\,dx,
\end{align}
where $h(x):=|x|^{-\beta}$, and
\begin{equation}\label{e.dkb}
\kappa_{\beta}=\frac{2^\alpha\Gamma\big(\frac{\beta+\alpha}{2}\big) \Gamma\big(\frac{d-\beta}{2}\big)}{\Gamma\big(\frac{\beta}{2}\big)\Gamma\big(\frac{d-\beta-\alpha}{2}\big)}.
\end{equation}
Before \cite{MR3460023}, the identity \eqref{e:p2} was given by Frank, Lieb and Seiringer in \cite[(4.3)]{MR2425175} for functions $u\in C^\infty_0({\mathbb{R}^d}\setminus\{0\})$ and $\beta \in [0, (d-\alpha)/2]$. We note that the
 coefficient $\kappa_\beta$
is increasing on $[0,(d-\alpha)/2]$, symmetric with respect to $(d-\alpha)/2$ and takes on the maximal value at
$\beta=(d-\alpha)/2$, which is
$$
\kappa_{(d-\alpha)/2}=2^\alpha\Gamma\left(\frac{d+\alpha}{4}\right)^2
\Gamma\left(\frac{d-\alpha}{4}\right)^{-2},
$$
see \cite[Lemma 3.2]{MR2425175} or \cite[p.~237]{MR3460023}.
Correspondingly, the following optimal fractional Hardy inequality holds
\begin{equation}\label{e:hardy-quad}
 \mathcal{E}{[u]}\geq
\kappa_{(d-\alpha)/2}
\int_{\mathbb{R}^d} \frac{u(x)^2}{|x|^\alpha}\,dx\quad \text{ for all } \quad  u \in L^2({\mathbb{R}^d}).
 \end{equation}
The inequality is also known as Hardy-Rellich inequality and  was proved by Herbst \cite[(2.6)]{MR0436854}, Beckner \cite[Theorem 2]{MR1254832} and Yafaev \cite{MR1717839}.
We will propose an optimal analogue of the inequality appropriate for $L^p({\mathbb{R}^d})$. To this end
for  $p \in (1, \infty)$ and (Borel measurable) $u: {\mathbb{R}^d} \to \mathbb{R}$ we define the Sobolev-Bregman form, or the $p$-form,
\begin{equation}\label{e:dEp}
\mathcal{E}_p[u] :=
\frac{1}{2}
\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} (u(x)-u(y)) (u(x)^{\langle p - 1 \rangle}  -u(y)^{\langle p - 1 \rangle} ) \nu(x,y) \,dy \,dx.
\end{equation}
Here and below we use the notation $$a^{\langle k \rangle} := \left|a \right|^k \operatorname{sgn} a,\quad a, k \in \mathbb{R},$$
where $0^k=0$.
The $p$-form
is well defined since  the integrand in \eqref{e:dEp} is nonnegative for every $u$.
In fact, to compare $\mathcal{E}_p$ with $\mathcal{E}=\mathcal{E}_2$, we recall that
\begin{equation}\label{e.cHk}
4(p-1)p^{-2}(b^{\langle p/2\rangle}-a^{\langle p/2\rangle})^2\leq (b-a)(b^{\langle p-1\rangle}-a^{\langle p-1\rangle})\leq 2(b^{\langle p/2\rangle}-a^{\langle p/2\rangle})^2.
\end{equation}
The inequality
holds true for all $p\in (1,\infty)$ and $a,b\in \mathbb{R}$, see, e.g., Liskevich, Perelmuter and Semenov \cite[Lemma 2.1]{MR1407327}.
By \eqref{e:hardy-quad}
and \eqref{e.cHk},
\begin{equation}\label{e:subgoal}
\mathcal{E}_p[u] \geq 4(p-1)p^{-2} \kappa_{(d-\alpha)/2}
\int_{{\mathbb{R}^d}} \frac{\left| u(x) \right|^p} {|x|^{\alpha}} \,dx, \quad u \in L^p({\mathbb{R}^d}).
\end{equation}
We remark that the inequality \eqref{e:subgoal} is given in Cialdea and Maz'ya \cite[p. 231]{MR3235527}.
Our goal is to improve the constant.
To this end, inspired by Bogdan, Dyda and Luks \cite[(9)]{MR3251822},
we consider the \textit{Bregman divergence}:
\begin{equation}
F_p(a,b) := |b|^p - |a|^p - pa^{\langle p - 1 \rangle} (b-a), \quad a, b \in \mathbb{R}.
\end{equation}
It is the second-order Taylor remainder of the convex function $\mathbb{R}\ni x\mapsto |x|^p$, so $F_p(a,b) \geq 0$. For instance, $F_2(a,b)= (b-a)^2$.
The function $F_p$  may be used to
quantify the regularity of functions in integral forms generalizing
\eqref{e:def-quad-form}. Indeed,
the \textit{symmetrization} of $F_p$ is
\begin{equation}\label{e:fp}
\frac12 (F_p(a,b) + F_p(b,a)) =\frac{p}{2} (b-a)(b^{\langle p - 1 \rangle} -a^{\langle p - 1 \rangle} ),
\end{equation}
which is the expression in the definition  of $\mathcal E_p$, up to the factor of $p$.
In passing we refer the reader to Bogdan, Grzywny, Pietruska-Pa{\l}uba and Rutkowski \cite{2020arXiv200601932B} for references to applications of Bregman divergence in analysis, statistical learning and optimization, and to Bogdan and Wi\c{e}cek \cite{bogdan2021Burkholder} for a martingale connection.

\subsection{Main results}
For $\beta\in \mathbb{R}$ we denote $h_\beta(x) := \left|x \right|^{-\beta}$, $x\in {\mathbb{R}^d}$.
Of course, for $a\in \mathbb{R}$,
\begin{equation}\label{e:hbeta}
h_{\beta}(x)^a = h_{a\beta}(x), \quad x\in {\mathbb{R}^d}.
\end{equation}
We propose the following Hardy-type identity for the Sobolev-Bregman forms.
\begin{thm}\label{thm:Theorem1}
If $0<\alpha<d \wedge 2$, $0 \leq \beta \leq (d-\alpha) \wedge (d-\alpha)/(p-1)$, $h=h_\beta$ and   $u\in L^p(\mathbb R^d)$, then
\begin{align}\label{e:main}
\mathcal{E}_p{[u]} &= \frac{\kappa_{(p-1)\beta}+(p-1)\kappa_{\beta}}{p} \int_{\mathbb{R}^d} \frac{\left|u(x)\right|^p}{\left|x\right|^{\alpha}} dx \\[2mm]
&\quad  +\frac{1}{p}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}}  F_p\left(\frac{u(x)}{h(x)}, \frac{u(y)}{h(y)}\right) h(x)^{p-1}h(y)  \nu(x,y) \,dy \,dx. \nonumber
\end{align}
\end{thm}
\noindent
In particular, for $\beta=(d-\alpha)/p$ we obtain
\begin{align}\label{e:ip}
\mathcal{E}_p{[u]} &= \kappa_{(d-\alpha)/p} \int_{{\mathbb{R}^d}} \frac{|u(x)|^p}{|x|^{\alpha}} dx \\
&\quad+ \frac{1}{p}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} F_p\left(\frac{u(x)}{h(x)}, \frac{u(y)}{h(y)}\right) h(x)^{p-1}h(y) \nu(x,y) \,dy \,dx,\nonumber
\end{align}
and, of course,
\begin{align}
\mathcal{E}_p{[u]}&\geq \kappa_{(d-\alpha)/p} \int_{{\mathbb{R}^d}} \frac{|u(x)|^p}{|x|^{\alpha}} \,dx
\quad \text{ for all } \quad u \in L^p({\mathbb{R}^d}).
\label{e.Hp}
\end{align}
The results extend the Hardy identities from \cite{MR3460023} and the ground-state representations of
Frank, Lieb and Seiringer in \cite[Proposition 4.1]{MR2425175}, see also  Frank and Seiringer \cite{MR2469027} and Beckner \cite{MR2984215}.
The proofs of \eqref{e:main} and \eqref{e:ip} are given in Section~\ref{sec:Hi}.
In Lemma~\ref{thm:lem7} of Section~\ref{sec:Hi}
we also show that \eqref{e.Hp} improves \eqref{e:subgoal}, namely for $p\neq 2$ we have
\begin{equation}\label{e.i}
\kappa_{(d-\alpha)/p}>\frac{4(p-1)}{p^2} \,
\frac{2^\alpha\Gamma\left(\frac{d+\alpha}{4}\right)^2}
{\Gamma\left(\frac{d-\alpha}{4}\right)^{2}}.
\end{equation}
\begin{figure}
\includegraphics[width=12cm]{rys2.png}
\caption{Illustration of \eqref{e.i} for $d=3, \alpha=1$}\label{fig:comp}
\end{figure}
Figure \ref{fig:comp} compares both sides of \eqref{e.i}, by showing $\kappa_{(d-\alpha)/p}$ of \eqref{e.Hp} (solid line) and $4(p-1)p^{-2}\kappa_{(d-\alpha)/2}$ of \eqref{e:subgoal} (dashed line) as functions of $1/p\in (0,1]$ for $d=3$ and $\alpha=1$.
Here is a statement deeper than \eqref{e.i}, which we prove in Section~\ref{s:opt}.
\begin{thm}\label{thm:Theorem2}
The constant in \eqref{e.Hp} is sharp.
\end{thm}
The result was previously known only for $p=2$; we again refer to \cite[(2.6)]{MR0436854}, \cite[proof of Theorem 2]{MR1254832} or \cite{MR1717839}.

For the sake of comparison we recall
the fractional Hardy-Sobolev inequality from
\cite[Theorem 1.1]{MR2469027}:
\begin{equation}\label{e:frank}
 \int_{{\mathbb{R}^d}}\int_{{\mathbb{R}^d}} \frac{|u(x)-u(y)|^p}{|x-y|^{d+ps} } \,dx\,dy
\geq \mathcal C_{d,s,p} \int_{\mathbb{R}^d} \frac{|u(x)|^p}{|x|^{ps}}\,dx.
\end{equation}
The constant
$C_{d,s,p}$ is known and sharp, too.
Of course,
the forms in \eqref{e:dEp} and  \eqref{e:frank}
are different generalizations of the quadratic form in \eqref{e:def-quad-form}. Correspondingly, the inequalities \eqref{e.Hp} and \eqref{e:frank} are different optimal generalizations of the fractional Hardy inequality \eqref{e:hardy-quad}.
In passing we refer the reader to \cite[Section 6]{2020arXiv200601932B} for an additional insight into the comparison of \eqref{e.Hp} and \eqref{e:frank} and to Frank and Seiringer \cite{MR2723817} for a version of \eqref{e:frank} for half-spaces.

Not unexpectedly, our main application of
Theorem~\ref{thm:Theorem1} and \ref{thm:Theorem2} is to the $L^p$ contractivity of the Feynman-Kac semigroup $\tilde P_t$ generated by $\Delta^{\alpha/2}+\kappa_\delta |x|^{-\alpha}$ with $\kappa_\delta$ given by \eqref{e.dkb}.
\begin{thm}\label{t.znHp}
Let $0<\alpha<2\wedge d$,
$\delta\in[0,(d-\alpha)/2]$,
$1<p<\infty$ and $0<t<\infty$. The operator $\tilde P_t$ is a contraction on $L^p({\mathbb{R}^d})$ if and only if $\kappa_\delta \le \kappa_{(d-\alpha)/p}$.
\end{thm}
The proof of this characterization is given in Section~\ref{s.ap}, see also Remark~\ref{r.n} there.
The result is analogous to the classical case ($\alpha=2$), where the operator $\Delta + \kappa|x|^{-2}$ generates a contraction semigroup on $L^p(\mathbb{R}^d)$ if and only if  $\kappa\le \kappa_{(d-2)/p}= (d-2)^2 (p-1)p^{-2}$,
see Kovalenko, Perelmuter and Semenov \cite{MR622855}, see also Liskevich and Semenov \cite[Theorems 1 and 2]{MR1409835} and \cite[Corollary 1.2]{MR2259099}.
Theorem~\ref{t.znHp}
is illustrated by Figure~\ref{fig:thm3}.
Notably,   if $\alpha=2$  then $\kappa_\beta=\beta(d-2-\beta)$ and \eqref{e.i} becomes equality, so we have improved Hardy inequality
for the \textit{nonlocal} operator $\Delta^{\alpha/2}$ but not for the \textit{local} operator $\Delta$.
\begin{figure}
\begin{tikzpicture}[
    scale=5,
    axis/.style={thick, ->},
    important line/.style={thick},
    dashed line/.style={dashed, thin},
    ]
    \draw[axis] (-0.05,0)  -- (1.65,0) node(xline)[right]{$\delta$};
    \draw[axis] (0,-0.05) -- (0,1.15) node(yline)[above] {$\frac{1}{p}$};
		\draw[dashed line] (0,1) -- (1.5,1);
		\draw[dashed line] (1.5,1) -- (1.5,0);
		\draw[important line] (0,1) -- (0.6,0.5);
		\draw[important line] (0,0) -- (0.6,0.5);
		\draw[dashed line] (0.6,1) -- (0.6,0);
		\draw [pattern=north west lines] (0,0) -- (0.6,0.5) -- (0,1);
		\draw[-] (0.75,0.0) -- (0.75,0.02);
		\draw[-] (0.6,0.0) -- (0.6,0.02);
		\draw[-] (1.2,0.0) -- (1.2,0.02);
		\draw[-] (0.0,0.5) -- (0.02,0.5);
		\draw[-] (0.0,1) -- (0.02,1);
		\draw[-] (0,0) -- (0.6,0) coordinate[label = {below:$\frac{d-\alpha}{2}$}];
		\draw[-] (0,0) -- (0.75,0) coordinate[label = {below:$\frac{d}{2}$}];
		\draw[-] (0,0) -- (1.5,0) coordinate[label = {below:$d$}];
    \draw[-] (0,0) -- (1.2,0) coordinate[label = {below:$d\!-\!\alpha$}];
		\draw[-] (0,0) -- (0,0.5) coordinate[label = {left:$\frac{1}{2}$}];
		\draw[-] (0,0) -- (0,1) coordinate[label = {left:$1$}];
    \draw[-] (0,0) -- (0.05,0) coordinate[label = {below:$0$}];
    \end{tikzpicture}
		\caption{The shaded area shows the values of $1/p$, for which, given $\delta\in [0,(d-\alpha)/2]$ and $\Delta^{\alpha/2}+\kappa_{\delta}|x|^{-\alpha}$,
$\tilde P_t$ are contractive on $L^p({\mathbb{R}^d})$.}\label{fig:thm3}
		\end{figure}

Here is the last theorem of the paper.
\begin{thm}\label{t.bd}
Let $0<\alpha<2\wedge d$, $\delta\in[0,(d-\alpha)/2]$, $1<p<\infty$ and $0<t<\infty$. The operator $\tilde P_t$ is bounded on $L^p({\mathbb{R}^d})$ if and only if $\delta < {d/p^*}$, where $p^*=\max\{p,p/(p-1)\}$.
\end{thm}
The result is proved in Section~\ref{s.ap}.
As we shall see, the result is a consequence of the estimates of $\tilde P_t$ given in \cite{MR3933622}.
An illustration of Theorem~\ref{t.bd} is given in Figure~\ref{fig:thm4}.
\begin{figure}
			\centering
				\begin{tikzpicture}[
    scale=5,
    axis/.style={thick, ->},
    important line/.style={thick},
    dashed line/.style={dashed, thin},
		dashed important line/.style={dashed, thick},
    ]
    \draw[axis] (-0.05,0)  -- (1.65,0) node(xline)[right]{$\delta$};
    \draw[axis] (0,-0.05) -- (0,1.15) node(yline)[above] {$\frac{1}{p}$};
    \draw[dashed line] (0.6,0.4) -- (1.5,1);
    \draw[dashed line] (0.6,0.6) -- (1.5,0);
		\draw[dashed line] (0,1) -- (1.5,1);
		\draw[dashed line] (1.5,1) -- (1.5,0);
		\draw[important line] (0.6,0.4) -- (0.6,0.6);
		\draw[dashed line] (0.6,1) -- (0.6,0);
		\draw[-] (0.75,0.0) -- (0.75,0.02);
		\draw[-] (0.6,0.0) -- (0.6,0.02);
		\draw[-] (1.2,0.0) -- (1.2,0.02);
		\draw[-] (0.0,0.5) -- (0.02,0.5);
		\draw[-] (0.0,1) -- (0.02,1);
		\draw[-] (0,0) -- (0.6,0) coordinate[label = {below:$\frac{d-\alpha}{2}$}];
		\draw[-] (0,0) -- (0.75,0) coordinate[label = {below:$\frac{d}{2}$}];
		\draw[-] (0,0) -- (1.5,0) coordinate[label = {below:$d$}];
    \draw[-] (0,0) -- (1.2,0) coordinate[label = {below:$d\!-\!\alpha$}];
		\draw[-] (0,0) -- (0,0.5) coordinate[label = {left:$\frac{1}{2}$}];
		\draw[-] (0,0) -- (0,1) coordinate[label = {left:$1$}];
    \draw[-] (0,0) -- (0.05,0) coordinate[label = {below:$0$}];
		\draw[dashed important line] (0,1) -- (0.6,0.6);
		\draw[dashed important line] (0,0) -- (0.6,0.4);
		\draw [pattern=north west lines] (0.00,0.005) -- (0.6,0.405) -- (0.6,0.595) -- (0,0.995);

    \end{tikzpicture}
		\caption{The shaded area shows the values of $1/p$, for which, given  $\delta\in [0,(d-\alpha)/2]$ and $\Delta^{\alpha/2}+\kappa_{\delta}|x|^{-\alpha}$,
$\tilde P_t$ are bounded on $L^p({\mathbb{R}^d})$.}\label{fig:thm4}
	\end{figure}

The structure of the paper is as follows. In Section~\ref{sec:Pre} we give preliminaries and ponder  definitions. The proof of Theorem~\ref{thm:Theorem1} is given in Section~\ref{sec:Hi}. The  proof of Theorem~\ref{thm:Theorem2} is in Section~\ref{s:opt}.
In Section~\ref{s.ap} we prove Theorem~\ref{t.znHp} and Theorem~\ref{t.bd}.
In Section~\ref{s.d}
we point out to related results in the literature and broader perspectives.

\textbf{Acknowledgements.} We thank Rupert Frank, Damir Kinzebulatov, Bart\l{}omiej Dyda, Jerome Goldstein, Kamil Kaleta, Mateusz Kwa\'{s}nicki, Adam Nowak, Sergey Pirogov, R\'ene Schilling and Karol Szczypkowski for helpful comments, discussions and references. We thank Yuli Semenov for an inspiring lecture in September 2018 in Wroc\l{}aw.

\subsection{Further discussion}\label{s.d}

The main results of the paper are the optimal Hardy identity \eqref{e:ip} and inequality \eqref{e.Hp}. In this section, however, we try to give a broader perspective for the considered Sobolev-Bregman forms and point out to connections and inspirations. It is well known that
the ramification of the potential theory of \textit{symmetric} Markovian processes and semigroups by means of the theory of Dirichlet forms on $L^2$ spaces turned out extremely succesful \cite{MR2778606}, Ma and R\"{o}ckner \cite{MR1214375}. Therefore many authors worked to extend it to the case of the $L^p$ spaces. The development is related to the fact that
$\langle -A f, f^{p-1}\rangle\ge 0$ for $f$ in the domain of the generator,
see \eqref{e.pfag}
below. We recall that for $p\in (1,\infty)$ the dual space of $L^p$ is, of course,  $L^{p/(p-1)}$ and for $u\in L^p$ we have $u^{\langle p-1\rangle}\in L^{p/(p-1)}$, and $\|u\|_p^{p}=\int |u|^p=\int |u^{\langle p-1\rangle}|^{p/(p-1)}=\int u^{\langle p-1\rangle}u$. Therefore $\|u\|_p^{2-p} u^{\langle p-1\rangle}$ yields a linear functional on $L^p$ appropriate for testing the dissipativity of generators in the Lumer--Phillips theorem,
see, e.g., Pazy \cite[Section 1.4]{MR710486} or \cite[p. 3]{MR3235527}. For the semigroups generated by local operators we refer to Langer and Maz'ya \cite{MR1694522}, Sobol and Vogt \cite[Theorem 1.1]{MR1923627} and the monograph
\cite{MR3235527},
see also
the historical comments in Section~4.7 therein.

As we shall experience first-hand in Section~\ref{s.ap}, the forms $\mathcal E_p$ capture the evolution of $L^p$ norms of functions upon the action of operator semigroups. The connection is well known, since at least
the paper of
Varopoulos \cite[(1,1)]{MR803094}.
It was then used, e.g., to study
perturbations of semigroups on $L^p$
in \cite[Theorem 3.2]{MR1407327}
and \cite{MR1409835}.
For nonlocal operators we refer the reader to Farkas, Jacob and Schilling \cite[(2.4)]{MR1808433} and to the monograph of Jacob \cite[(4.294)]{MR1873235}.
As a rule the authors use the intermediary $L^2$ setting, relying on inequalities similar to \eqref{e.cHk} and to
\eqref{e.cHkf} below. Therefore the resulting inequalities
do not allow for optimal constants, certainly in the case of nonlocal generators, cf. \eqref{e:subgoal} and \eqref{e.Hp}; see also \cite[p. 231]{MR3235527}.
Suboptimal constants may lead to suboptimal qualitative results,
as shall be evident from the proof of Theorem~\ref{t.znHp}.
We avoid the inaccuracy of \eqref{e.cHk} by systematically using the Bregman divergence -- we work \emph{intrinsically} in $L^p$ and therefore obtain the optimal constants and the optimal range of exponents in Theorems \ref{thm:Theorem2} and \ref{t.znHp}.

We remark that \eqref{e.cHkf} and \eqref{e.pfag} can be traced back to the papers \cite{MR803094} and Liskevich, Semenov \cite{MR1160303}. In our notation it reads there as $\langle -A f, f^{\langle p-1\rangle}\rangle\approx \langle -A^{1/2}f^{\langle p/2\rangle},A^{1/2}f^{\langle p/2\rangle}\rangle$, see \cite[Theorem~1]{MR1160303}, see also Kinzebulatov and Semenov \cite[Proposition 8]{kinzebulatov2020fractional}
for recent developments
and \cite{kinzebulatov2020admissible} for applications to
symmetric stable processes with drift, understood as
solutions of stochastic differential equations.

Finally, it may be hard to point out the first occurrence of \eqref{e.cHk}. Our best guess is \cite[Lemma 1]{MR1160303} and  \cite[Lemma 2.1]{MR1407327}, see also  \cite[p. 246]{MR803094}, Bakry \cite[p. 39]{MR1307412}, Stroock \cite[Lemma 9.9 and p. 134]{MR755154} and Carlen, Kusuoka and Stroock \cite[p. 269]{MR898496} for formulations with nonnegative arguments or one-sided comparison. We also point out to the calculations with forms and powers in Davies \cite[Chapter 2 and 3]{MR1103113}, to the calculations in
\cite[Section 7.6]{MR3235527} and
\cite[(2.19)]{2020arXiv200601932B}, and to Lemma~\ref{thm:incrineq} below.

\section{{Preliminaries}}\label{sec:Pre}

We use ``$:=$" to indicate definitions, e.g.,
$a \wedge b := \min \{ a, b\}$, $a \vee b := \max \{ a, b\}$, and $a_+:=a\vee 0$.
All the functions considered below are Borel measurable either by construction or assumptions. If $\mathcal F$ is a family of functions, then we let $\mathcal F_+=\{f\in \mathcal F: f\ge 0\}$.
For nonnegative functions $f$ and $g$ we write $f(x)\approx g(x)$ to indicate that there are numbers $0<c<C<\infty$ such that $cf(x)\le  g(x)\le C f(x)$ for all the considered arguments $x$.
We call such comparisons two-sided or \textit{sharp}.
For an open subset $D$ of the $d$-dimensional Euclidean space $\mathbb{R}^d$, we let $C^\infty_c (D)$ be the space of smooth functions with compact support in $D$.

Note that $(|x|^p)'=p x^{\langle p - 1 \rangle}$ on $\mathbb{R}$ for $p\in (1,\infty)$. We shall focus on related inequalities.
\begin{lem}
\label{thm:incrineq}
For $p \in (1,\infty)$ there are constants $C,C'>0$ such that for $a,b\in \mathbb{R}$,
\begin{align}\label{eq:2d}
0\leq|b|^p - |a|^p - pa^{\langle p - 1 \rangle} (b-a) &\leq C |b-a|^{\lambda} (|b|+|a|)^{p-\lambda}, \quad \lambda \in [0,2],\\
\label{eq:2c}
||b|^p - |a|^p|&\le (p+C) |b-a|(|b|+|a|)^{p-1},\\
\label{eq:2dd}
| b^{\langle p - 1 \rangle} - a^{\langle p - 1 \rangle}|&\leq C' |b-a|^{\lambda} (|b|+|a|)^{p-1-\lambda}, \quad \lambda \in [0,1].
\end{align}
\end{lem}
\begin{proof}
The inequality \eqref{eq:2d} with $\lambda=2$ follows from \cite[Lemma 6]{MR3251822}. In fact, we have
\begin{equation}\label{eq:2e}
|b|^p - |a|^p - pa^{\langle p - 1 \rangle} (b-a) \approx (b-a)^2 (|b|+|a|)^{p-2}, \quad a,b\in \mathbb{R},
\end{equation}
see \cite[(11)]{MR3251822}.
Here is, however, another proof:
if $a \neq 0$, then we let $x=b/a$ and arrive at the following equivalent statement of \eqref{eq:2d}:
$$
0\le |x|^p-1 - p(x-1) \leq C |x-1|^\lambda (|x|+1)^{p-\lambda}, \quad x \in \mathbb{R}.
$$
Since $x\mapsto |x|^p$ is strictly convex, both sides are continuous and strictly positive on $\mathbb{R}\setminus\{1\}$. Let first $\lambda=2.$
By a compactness argument it is enough to notice that
$$
\lim_{x \to \pm \infty} \frac{|x|^p  -1-p(x-1)}{(x-1)^2(|x|+1)^{p-2}} = 1,
$$
and, by L'H\^{o}pital's rule,
\begin{equation*}
\lim_{x \to 1} \frac{|x|^p  -1 -p(x-1)}{(x-1)^2}
= \frac{p(p-1)}{2}.
\end{equation*}
 When $\lambda\in [0,2]$, we have that  $|b-a|^2(|b|+|a|)^{p-2}\leq |b-a|^\lambda (|b|+|a|)^{p-\lambda}$, which gives
 \eqref{eq:2d}.
\eqref{eq:2c} follows from   \eqref{eq:2d} with $\lambda=1$, and
\eqref{eq:2dd} can be proved like \eqref{eq:2d}.
\end{proof}

\subsection{Functions and kernels}
Let $g$ denote the Gaussian kernel
\begin{equation}\label{e:Gk}
g_t(x) = (4\pi t)^{-d/2} e^{-|x|^2/(4t)}\,, \quad t>0,\quad x\in {\mathbb{R}^d}\,.
\end{equation}
Let $\eta_t(s)\ge 0$ be the density function of the $\alpha/2$-stable subordinator at time~$t$, see, e.g., Bliedtner and Hansen \cite[Section V.3]{MR850715}, Schilling, Song and Vondra{\v{c}}ek \cite[Chapter 13]{MR2978140} or Sato \cite[Section 6.30]{MR1739520}.
In particular,
$\eta_t(s) = 0$ for $s\le 0$, and we have the following formula for the Laplace transform of $\eta_t$:
\[
 \int_0^\infty e^{-us} \eta_t(s)\,ds = e^{-tu^{\alpha/2}}, \quad u\geq 0, \quad t>0.
\]
We define, by \textit{subordination}, the convolution semigroup of functions (the $\alpha$-stable semigroup):
\begin{equation}\label{e:stable-pt}
 p_t(x) = \int_0^\infty g_s(x) \eta_t(s)\,ds \,, \quad t>0,\quad x\in {\mathbb{R}^d}.
\end{equation}
It is well known that the functions $p_t$ satisfy the following scaling property:
\begin{equation}
  \label{e:sca}
  p_t(x)=t^{-\frac{d}{\alpha}}p_1(t^{-\frac{1}{\alpha}}x)\, \quad t>0,\quad x\in {\mathbb{R}^d}\,.
\end{equation}
We consider the following transition probability density
\begin{equation}\label{e.dp}
p_t(x,y) = p_t(y-x), \quad t>0, \, x,y \in {\mathbb{R}^d}.
\end{equation}
Clearly, $p_t(x,y)$ is symmetric in the space variables: $p_t(x,y)=p_t(y,x)$.
It is well known that
\begin{equation}\label{e.mpa}
p_t(x,y) \leq c \min \left(t^{-d/\alpha}, t|x-y|^{-d-\alpha}\right), \quad t>0,\quad x, y \in {\mathbb{R}^d},
\end{equation}
and so we can record the following convenient estimate
\begin{equation}\label{e.mp}
p_t(x,y)/t \leq c \nu(x,y), \quad t>0,\quad x, y \in {\mathbb{R}^d},
\end{equation}
see, e.g., Bogdan, Grzywny and Ryznar \cite[Theorem 21]{MR3165234}. Also,
\begin{equation}\label{e.zp}
p_t(x,y)/t \rightarrow \nu(x,y) \mbox{ as } t \rightarrow 0^+,
\end{equation}
see, e.g., Cygan, Grzywny and Trojan \cite[Proof of Theorem 6]{MR3646773}. We will use \eqref{e.mp} and \eqref{e.zp} to support arguments based on the following general fact.
 \begin{lem}\label{l:ru}
If nonnegative functions satisfy
$f_n\le c f$ for $n\in {\mathbb{N}}$, and $f=\lim f_n$, then
$\lim \int f_n d\mu= \int f d\mu$ for any measure $\mu$.
\end{lem}
\begin{proof}
If the integral $\int f d\mu$ is finite, then the dominated convergence theorem applies. Otherwise, by Fatou's lemma, $ \int f d\mu=\infty=\liminf \int f_n d\mu=\lim \int f_n d\mu$, too.
\end{proof}

For $\alpha < d$ and $\beta \in (0, d)$, we let
$$f_{\beta}(t) = c t^{(d-\alpha -\beta)/\alpha}_+ ,\quad t\in \mathbb{R}.$$
Here $c\in (0, \infty)$ is a normalizing constant so chosen that
\begin{equation}\label{e.dfb}
 \int_0^\infty f_{\beta}(t) p_t(x) dt =|x|^{-\beta}=h_\beta(x),  \quad x\in {\mathbb{R}^d}.
\end{equation}
For $\beta \in (0, d-\alpha)$ we let
\begin{equation}\label{e.dqb}
     q_\beta(x): = \frac{1}{h_\beta(x)}\int_0^{\infty }f_{\beta}'(t) p_t(x) dt,\quad x\in {\mathbb{R}^d}.
\end{equation}
We note that the choice of $c$ does not affect $q_\beta$.
By \cite[Section~4]{MR3460023},
\begin{equation}\label{e:qbeta}
  q_\beta(x) = \kappa_\beta |x|^{-\alpha }.
\end{equation}
By \cite[the proof of Proposition~5]{MR3460023}, the function $\beta \mapsto \kappa_\beta$ is increasing on $(0,(d-\alpha)/2]$ and decreasing on $[(d-\alpha)/2,d-\alpha)$. Furthermore, $\kappa_\beta = \kappa_{d-\alpha-\beta}$.

We denote, as usual, $P_t u(x)=\int_{{\mathbb{R}^d}} u(y) p_t(x,y) \, dy$ for $u:{\mathbb{R}^d} \to \mathbb{R}$ and $x \in {\mathbb{R}^d}$, if the integral is well defined.

Since $\int_{\mathbb{R}^d} p_t(x,y)dy=\int_{\mathbb{R}^d} p_t(x,y)dx=1$, by Schur's test, for every $p\in [1,\infty]$,
\begin{equation}\label{e.cLp}
\|P_tf\|_p\le \|f\|_p, \quad f\in L^p({\mathbb{R}^d}).
\end{equation}
In fact, since $p_t(x,y)\le p_t(0,0)=t^{-d/\alpha}p_1(0,0)<\infty$, by Young inequality,
\begin{equation}\label{e.cLi}
\|P_tf\|_\infty \le c_t \|f\|_p, \quad f\in L^p({\mathbb{R}^d}).
\end{equation}
By \cite[Eq. 7]{MR3460023}, for $0\leq \beta \leq d-\alpha$,
\begin{equation}\label{e:supermedian}
P_th_\beta \leq h_\beta.
\end{equation}
In this sense, $h_\beta$ is supermedian when $0\leq \beta \leq d-\alpha$.

\subsection{The domain of the form $\mathcal E_p$}
Recall that $p \in (1,\infty)$. As usual, $L^p({\mathbb{R}^d})$ is the collection of all  the real-valued functions on ${\mathbb{R}^d}$ such that $\int_{{\mathbb{R}^d}} |f(x)|^p dx < \infty$;  we identify functions $u,v \in L^p({\mathbb{R}^d})$ if $u=v$ a.e. on ${\mathbb{R}^d}$.  The dual space of $L^p({\mathbb{R}^d})$ is, of course, $L^q({\mathbb{R}^d})$, where $q=p/(p-1)$.  We also consider the canonical pairing $\langle u, v\rangle= \langle v, u \rangle = \int_{{\mathbb{R}^d}} u(x)v(x)dx$, $u \in L^p({\mathbb{R}^d}), v \in L^q({\mathbb{R}^d})$.

Since for any $a,b\in \mathbb{R}$ we have $(b-a)(b^{\langle p-1\rangle}-a^{\langle p-1 \rangle})\geq 0$, the nonlinear form
\[\mathcal{E}_p{[u]} = \frac{1}{2} \int_{\mathbb R^d}\int_{\mathbb R^d} (u(x)-u(y))(u^{\langle p-1\rangle}(x)-u^{\langle p-1 \rangle}(y))\nu(x,y) \,dy \,dx\] is well-defined, perhaps infinite, for every $u:{\mathbb{R}^d}\to\mathbb{R}.$ We define its natural domain:
\begin{equation}\label{def:DE}
\mathcal D(\mathcal E_p)= \{u\in L^p({\mathbb{R}^d}) : \mathcal E_p{[u]}<\infty\}.
\end{equation}
For $p=2$, as usual, we let
\begin{equation*}
\mathcal E{[u]}=\lim_{t\to 0}\frac{1}{t}\langle u-P_tu, u\rangle,
\end{equation*}
where $u\in L^2({\mathbb{R}^d})$ is arbitrary and the expression under the limit is decreasing in $t$ (see Hille and Phillips \cite[Section 22.3]{MR0423094} and \cite[Lemma 1.3.4]{MR2778606}).
We also let
\begin{equation*}
\mathcal D(\mathcal E)= \{u\in L^2({\mathbb{R}^d}): {\mathcal{E}[u]<\infty}\}.
\end{equation*}
Note that the quantity  suggested by the definition of
$\mathcal{E}[u]$, \[\mathcal{E}^{(t)}(u,v): = \frac{1}{t} \langle u - P_t u, v \rangle, \quad t>0,\]
makes sense for all
 $u \in L^p({\mathbb{R}^d})$ and $v \in L^q({\mathbb{R}^d})$ because $P_tL^p({\mathbb{R}^d})\subset L^p({\mathbb{R}^d})$. In fact, $(P_t)_{t\ge 0}$ form a strongly continuous semigroup of contractions on $L^p({\mathbb{R}^d})$, see, e.g., Kwa\'{s}nicki \cite{MR3613319}, and we denote by $\mathcal D_p(\Delta^{\alpha/2})$ the domain of $\Delta^{\alpha/2}$ considered as the generator of this semigroup.
We can now characterize $\mathcal D(\mathcal{E}_p)$. We note that \eqref{e.cHkf} is  a variant of  \cite[Theorem 3.1]{MR1407327}, see also \cite[Remark (3)]{MR1407327}.
\begin{lem}
Let $p>1$. Then for every $u\in L^p(\mathbb R^d)$ we have
\begin{equation}\label{e:wpf}
\mathcal{E}_p[u]= \lim_{t\to 0} \mathcal{E}^{(t)}(u,u^{\langle p-1 \rangle}).
\end{equation}
Furthermore,
\begin{align}\label{e:dom_def-1}
 \mathcal D(\mathcal{E}_p) & = \{u\in L^p({\mathbb{R}^d}): \sup_{t>0}\mathcal E^{(t)}(u,u^{\langle p-1 \rangle})<\infty\}\\
 \label{e:dom_def-2}
 &=\{u\in L^p({\mathbb{R}^d}): \text{ finite } \lim_{t\to 0}\mathcal E^{(t)}(u,u^{\langle p-1 \rangle}) \text{ exists}\}.
 \end{align}
For arbitrary Borel measurable $u: {\mathbb{R}^d} \to \mathbb{R}$ we also have
\begin{equation}\label{e.cHkf}
\frac{4(p-1)}{p^2}\mathcal{E}[u^{\langle p/2\rangle}]\le \mathcal{E}_p[u]\le 2 \mathcal{E}[u^{\langle p/2\rangle}]
\end{equation}
and $\mathcal D(\mathcal{E}_p)=\mathcal D(\mathcal{E})^{\langle 2/p\rangle}:=\{v^{\langle 2/p\rangle}: v\in \mathcal D(\mathcal{E})\}$.
Finally, $\mathcal D_p(\Delta^{\alpha/2})\subset  \mathcal D(\mathcal{E}_p)$ and
\begin{equation}\label{e.pfag}
\mathcal{E}_p[u]=-\langle\Delta^{\alpha/2}u, u^{\langle p-1\rangle}\rangle, \quad u\in \mathcal D_p(\Delta^{\alpha/2}).
\end{equation}
\end{lem}

\begin{proof}
Since $\int_{{\mathbb{R}^d}} p_t(x,y) dy = 1$ for all $t>0, \ x \in {\mathbb{R}^d}$, the symmetry of $p_t(x,y)$
yields
\begin{align*}
\mathcal{E}^{(t)}(u, u^{\langle p - 1 \rangle} ) &= \frac{1}{t} \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} p_t(x,y) (u(x)-u(y)) dy \, u(x)^{\langle p - 1 \rangle}  \, dx \\
&= \frac{1}{t} \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} p_t(y,x) (u(y)-u(x)) dx \, u(y)^{\langle p - 1 \rangle}  \,dy \\
&= \frac{1}{2t} \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} p_t(x,y) (u(x)-u(y)) (u(x)^{\langle p - 1 \rangle}  - u(y)^{\langle p - 1 \rangle} ) \,dy \,dx
\end{align*}
for every $u\in L^p(\mathbb R^d)$.
By Lemma~\ref{l:ru}, \eqref{e.mp} and \eqref{e.zp}
we get \eqref{e:wpf}. Since the limit  in \eqref{e:wpf} exists for all $u\in L^p(\mathbb R^d)$, we obtain \eqref{e:dom_def-1} and \eqref{e:dom_def-2}, using dominated convergence.
From \eqref{e.cHk} we obtain \eqref{e.cHkf}. Of course, if $v\in \mathcal D(\mathcal{E})$ and $u=v^{\langle 2/p\rangle}$, then by \eqref{e.cHkf}, $\mathcal{E}_p[u]\le 2 \mathcal{E}[(v^{\langle 2/p\rangle})^{\langle p/2\rangle}]=2\mathcal{E}[v]<\infty$, so $u\in \mathcal D(\mathcal{E}_p)$. Conversely, if $u\in \mathcal D(\mathcal{E}_p)$ and we define $v=u^{\langle p/2\rangle}$, then by \eqref{e.cHkf}, $\mathcal{E}[v]\le p^2(4(p-1))^{-1} \mathcal{E}_p[u]<\infty$, so $v\in \mathcal D(\mathcal{E})$ -- and clearly $u=v^{\langle 2/p\rangle}$.

Let $u \in \mathcal D_p(\Delta^{\alpha/2})$. Consider again $\frac{1}{t} \langle u - P_t u,u^{\langle p-1\rangle} \rangle$ as $t\to 0^+$. By the definition of the semigroup generator \cite{MR3613319}, we see that
$(P_tu-u)/t$ converges to $\Delta^{\alpha/2}u$ in $L^p({\mathbb{R}^d})$. Since $u^{\langle p-1\rangle}\in L^{p/(p-1)}({\mathbb{R}^d})$, the expression $\mathcal E^{(t)}(u,u^{\langle p-1\rangle})$ tends to (finite) $-\langle\Delta^{\alpha/2}u, u^{\langle p-1\rangle}\rangle$. On the other hand, it converges to $\mathcal{E}_p[u]$ by \eqref{e:wpf}.
\end{proof}

\section{Hardy identity and inequality}
\label{sec:Hi}
\begin{proof}[Proof of Theorem~\ref{thm:Theorem1}] For $\beta$ as in the assumptions of the theorem, we write $h=h_\beta$, $f=f_\beta$, etc. Take $u\in L^p({\mathbb{R}^d})$ and
let $v=u/h$, so that $vh=u$ a.e. The factorization $vh=u$ is the essence of Doob's conditioning, which inspired the calculations in \cite[Theorem 2]{MR3460023} and in what follows. Let $t>0$. Of course,
$vh \in L^p({\mathbb{R}^d})$, and by \eqref{e:supermedian}, $vP_th \in L^p({\mathbb{R}^d})$. Consider \eqref{e:wpf}. We have
\begin{align*}
\mathcal{E}^{(t)}(vh, (vh)^{\langle p - 1 \rangle} ) &= \frac{1}{t} \langle vh, (vh)^{\langle p - 1 \rangle}  \rangle
 - \frac{1}{t}\langle P_t(vh), (vh)^{\langle p - 1 \rangle}  \rangle
\\
& =  \frac{p-1}{p} \langle v \frac{h-P_th}{t}, (vh)^{\langle p - 1 \rangle}  \rangle  + \frac{p-1}{p} \langle v\frac{P_th}{t} , (vh)^{\langle p - 1 \rangle}  \rangle\\
&\quad+ \frac{1}{p} \langle v^{\langle p-1\rangle} \frac{h^{p-1}-P_th^{p-1}}{t}, vh \rangle+ \frac{1}{p} \langle v^{\langle p - 1 \rangle} \frac{ P_th^{p-1}}{t}, vh\rangle \\
&\quad- \langle \frac{ P_t(vh)}{t}, (vh)^{\langle p - 1 \rangle}  \rangle = I_t^{(1)} + I_t^{(2)} + J_t,
\end{align*}
where
\begin{align*}
I_t^{(1)}&:=  \frac{p-1}{p} \langle v \frac{h-P_th}{t}, (vh)^{\langle p - 1 \rangle}  \rangle=\frac{p-1}{p} \int_{{\mathbb{R}^d}}  |v(x)|^p h(x)^{p-1} \frac{(h-P_th)(x)}{t}\,dx,\\
I_t^{(2)}&:=  \frac{1}{p} \langle v^{\langle p-1\rangle} \frac{h^{p-1}-P_th^{p-1}}{t}, vh \rangle
=\frac{1}{p} \int_{{\mathbb{R}^d}}  |v(x)|^p h(x) \frac{(h^{p-1}-P_th^{p-1})(x)}{t}\,dx, \\
J_t&:=\frac{1}{p} \langle v^{\langle p - 1 \rangle} \frac{ P_th^{p-1}}{t}, vh \rangle + \frac{p-1}{p} \langle v\frac{P_th}{t} , (vh)^{\langle p - 1 \rangle}  \rangle - \langle \frac{ P_t(vh)}{t}, (vh)^{\langle p - 1 \rangle}  \rangle.
\end{align*}
By the symmetry of $p_t(x,y)$ and the definition of $F_p$,
\begin{align*}
J_t &= \frac{1}{p} \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} |v(x)|^p h(y)^{p-1}h(x) \, \frac{p_t(x,y)}{t} \,dy \,dx \\
&\quad + \frac{p-1}{p} \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} |v(x)|^p h(x)^{p-1} h(y)  \, \frac{p_t(x,y)}{t} \,dy \,dx \\
&\quad - \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} v(x)^{\langle p - 1 \rangle}  v(y) h(x)^{p-1} h(y)  \, \frac{p_t(x,y)}{t} \,dy \,dx \\
&= \frac{1}{p}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} \left[|v(y)|^p - |v(x)|^p -p v(x)^{\langle p - 1 \rangle}  \left(v(y) - v(x) \right) \right] h(x)^{p-1}h(y) \, \frac{p_t(x,y)}{t} \, dy \,dx \\
&=\frac{1}{p}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} F_p\left({v(x)}, {v(y)} \right) h(x)^{p-1}h(y) \frac{p_t(x,y)}{t} \, dy \, dx.
\end{align*}
By Lemma~\ref{l:ru}, \eqref{e.mp} and \eqref{e.zp},
\begin{align}\label{e.f1}
\lim_{t\to 0}J_t &=\frac{1}{p}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} F_p\left({v(x)}, {v(y)} \right) h(x)^{p-1}h(y) \nu(x,y) \, dy \, dx \nonumber \\
&=\frac{1}{p}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} F_p\left(\frac{u(x)}{h(x)}, \frac{u(y)}{h(y)} \right) h(x)^{p-1}h(y) \nu(x,y) \, dy \, dx.
\end{align}
We next consider $I_t^{(1)}$ and
recall that $h(x)=\int_0^\infty f(s) p_s(x)\,ds$, so for $x\neq 0$,
\begin{align*}
 (h-P_th)(x) &= \int_0^\infty f(s) p_s(x) \,ds - \int_0^\infty f(s) p_{s+t}(x) \,ds \\
&=\int_0^\infty [f(s) - f(s-t)] p_s(x) \,ds.
\end{align*}
Thus,
\begin{align*}
I_t^{(1)}&=\frac{p-1}{p} \int_{{\mathbb{R}^d}} |v(x)|^p h(x)^{p-1} \int_0^\infty \frac{1}{t}[f(s) - f(s-t)]\ p_s(x) \,ds\, dx.
\end{align*}
We have $0\leq f(s)-f(s-t)\leq C t f'(s)$ for $s,t>0$.
By Lemma~\ref{l:ru} and \eqref{e:qbeta} we get
\begin{align}
\underset{t \to 0} {\lim} \, I_t^{(1)}&= \frac{p-1}{p}  \int_{{\mathbb{R}^d}}  \, |v(x)|^p h(x)^{p-1} \int_0^{\infty} f'(s)p_s(x)\, ds\, dx\nonumber\\
&= \frac{p-1}{p}  \int_{{\mathbb{R}^d}} |v(x)|^p h(x)^{p}q(x) \, dx \label{e.f2}
=
\frac{(p-1)\kappa_{\beta}}{p} \int_{{\mathbb{R}^d}} \frac{|u(x)|^p}{|x|^{\alpha}}\, dx.
\end{align}
We can treat $I_t^{(2)}$ analogously. The analogy has several layers. We let $p'=p/(p-1)$. We have $p=p'/(p'-1)$ and $(p'-1)(p-1)=1$.
By considering
\eqref{e:hbeta} and \eqref{e.dfb} we get
\begin{align*}
I_t^{(2)} &=\frac{p'-1}{p'} \int_{{\mathbb{R}^d}}  |v(x)^{\langle 1/(p'-1) \rangle}|^{p'} h^{p'-1}_{(p-1)\beta}(x) \frac{h_{(p-1)\beta}(x)-P_th_{(p-1)\beta}(x)}{t} \, dx.
\end{align*}
We note that $0\le (p-1)\beta \leq d-\alpha$ and by the case of $I^{(1)}_t$ we get
\begin{align}\label{e.f3}
\underset{t \to 0} {\lim} \,I_t^{(2)}&=\frac{\kappa_{(p-1)\beta}}{p} \int_{{\mathbb{R}^d}} \frac{|u(x)|^p}{|x|^{\alpha}}\, dx.
\end{align}
By \eqref{e:wpf}, \eqref{e.f1}, \eqref{e.f2} and \eqref{e.f3} we get
\eqref{e:main}.
\end{proof}
For $\beta = (d-\alpha)/p$, we have $\kappa_{(p-1)\beta} = \kappa_{d-\alpha-\beta} = \kappa_{\beta}$, which yields \eqref{e:ip}. As aforementioned in Introduction, our constant is better than that in \eqref{e:subgoal}.
\begin{lem}\label{thm:lem7}
If $p\neq 2$, then
$
\kappa_{(d-\alpha)/p} > 4(p-1)p^{-2}\kappa_{(d-\alpha)/2}
$.
\end{lem}
\begin{proof}
Let $p>2$.
Denote
$$
r(t)=\kappa_{2t}=\frac{{2^\alpha}\Gamma(\frac{\alpha}{2}+t) \Gamma(\frac{d}{2}-t)}{\Gamma(t)\Gamma((d-\alpha)/2 -t)}, \quad t\in(0,\tfrac{d-\alpha}{4}).
$$
Put $t=\frac{d-\alpha}{2p}$. Then $p=\frac{d-\alpha}{2t}$ and
$$
\frac{4(p-1)}{p^2} = \frac{16t^2}{(d-\alpha)^2} \left(\frac{d-\alpha}{2t}-1\right) = \frac{16}{(d-\alpha)^2} t \left((d-\alpha)/2 -t \right) =: s(t).
$$
We only need to verify that
$$
r(t) > \kappa_{(d-\alpha)/2} s(t), \quad t \in (0,\tfrac{d-\alpha}{4}).
$$
Notice that $r(0^+) =0$ and  $s(0^+)=0$ and $r(\frac{d-\alpha}{4})=\kappa_{(d-\alpha)/2} = \kappa_{(d-\alpha)/2} s(\frac{d-\alpha}{4})$.
Let
$$
F(t) = \ln r(t) - \ln \kappa_{(d-\alpha)/2} s(t).
$$
It suffices to show that $F(t)>0$ for $t \in (0,\frac{d-\alpha}{4})$.
By \cite[Proof of Proposition 5]{MR3460023},
\begin{align*}
F'(t) &= \sum_{k=0}^{\infty} \left(\frac{1}{\frac{d}{2}+k-t} - \frac{1}{t+\frac{\alpha}{2}+k} - \frac{1}{(d-\alpha)/2+k-t} + \frac{1}{t+k}\right) - \left(\frac{1}{t} - \frac{1}{(d-\alpha)/2 -t} \right),
\end{align*}
hence
\begin{align*}
F'(t) &= \sum_{k=0}^{\infty} \left(\frac{1}{\frac{d}{2}+k-t} - \frac{1}{t+\frac{\alpha}{2}+k} - \frac{1}{(d-\alpha)/2+1+k-t} + \frac{1}{t+1+k}\right)\\
&= \sum_{k=0}^{\infty} \left(\frac{\frac{\alpha}{2}-1}{(t+1+k)(t+\frac{\alpha}{2}+k)} - \frac{\frac{\alpha}{2}-1}{(\frac{d}{2}+k-t)(\frac{d-\alpha+2}{2}+k-t)} \right) \\
&= \frac{\alpha-2}{2} \sum_{k=0}^{\infty} \frac{k(d-\alpha)+(d-\alpha)(d+2)/4 - t(d+2+4k)}{(t+1+k)(t+\frac{\alpha}{2} +k)(\frac{d}{2}+k-t) (\frac{d-\alpha+2}{2}+k-t)}.
\end{align*}
The numerator of every term in the last series is decreasing in $t$ and equal to $0$ for $t=\frac{d-\alpha}{4}$, so
$F'(t) <0$ for $t\in(0,\frac{d-\alpha}{4})$. Since $F(\frac{d-\alpha}{4})=0$,  $F(t)>0$ for $t\in(0,\frac{d-\alpha}{4})$.
We now consider the case $p \in(1,2)$. Let $q= \frac{p}{p-1}$. Of course, $q \in (2, \infty)$ and $\frac{1}{p}+\frac{1}{q}=1$.
We have
$ \kappa_{(d-\alpha)/q}=  \kappa_{(d-\alpha)/p}$, since $(d-\alpha)/q+(d-\alpha)/p=d-\alpha$ and
$\kappa_\beta$ is symmetric with respect to $(d-\alpha)/2$.
Furthermore, $4(p-1)p^{-2}=4/(pq)=4(q-1)q^{-2}$.
\end{proof}

\section{Optimality}\label{s:opt}

\noindent
In this section we prove Theorem~\ref{thm:Theorem2}.
The argument is rather technical due to integrability issues with the intended test function
for \eqref{e.Hp}.
We start with three auxiliary lemmas.
The following decomposition of $\mathcal{E}_p{[u]}$ is different and simpler than \eqref{e:main} but we should note that the second term needs not be nonnegative or finite.
\begin{lem}\label{thm:lem4}
Under the assumptions of Theorem~\ref{thm:Theorem1} we have
\begin{align*}
&\mathcal{E}_p{[u]} = \kappa_{\beta} \int_{{\mathbb{R}^d}} \frac{\left|u(x)\right|^p}{\left|x\right|^{\alpha}} dx \\
&+ \underset{t\to 0}\lim \ \frac{1}{2}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} \left( \frac{u(x)}{h(x)}-\frac{u(y)}{h(y)} \right) \left( \frac{u(x)^{\langle p - 1 \rangle} }{h(x)}-\frac{u(y)^{\langle p - 1 \rangle} }{h(y)} \right) h(x) h(y) \frac{p_t(x,y)}{t} \,dy \,dx,
\end{align*}
if additionally $u \in L^p({\mathbb{R}^d}, |x|^{-\alpha} dx)$.
\end{lem}
\begin{proof}
Thus, $u \in L^p({\mathbb{R}^d}, (1+ |x|^{-\alpha}) dx)$. Let $v=u/h$. By \eqref{e:wpf},
$$\mathcal{E}_p{[u]} = \lim_{t\to 0^+} \mathcal{E}^{(t)}(vh,(vh)^{\langle p - 1 \rangle} ).$$
By the proof of Theorem~\ref{thm:Theorem1}, $vP_th \in L^p({\mathbb{R}^d})$. We have
\begin{align*}
\mathcal{E}^{(t)}(vh, (vh)^{\langle p - 1 \rangle} ) &= \langle v \frac{h-P_th}{t}, (vh)^{\langle p - 1 \rangle}  \rangle +
   \langle  \frac{vP_th-P_t(vh)}{t} , (vh)^{\langle p - 1 \rangle}  \rangle =: I_t + J_t.
\end{align*}
By \eqref{e.f2} and the assumption $u \in L^p({\mathbb{R}^d}, |x|^{-\alpha} dx)$
\begin{align*}
\underset{t \to 0}{\lim} \, I_t = \kappa_{\beta} \int_{{\mathbb{R}^d}} \frac{|v(x)h(x)|^p}{|x|^{\alpha}} \,dx<\infty.
\end{align*}
Symmetrizing the integrand in $J_t$ we get
\begin{align*}
& J_t  =
 \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} \left(v(x)-v(y)\right) h(y) v(x)^{\langle p - 1 \rangle}  h(x)^{p-1} \, \frac{p_t(x,y)}{t}\, dy \, dx \\
&=\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} \!\!\!\left(v(x)-v(y)\right) \left( v(x)^{\langle p - 1 \rangle}  h(x)^{p-1}h(y) - v(y)^{\langle p - 1 \rangle} h(y)^{p-1}h(x) \right) \frac{p_t(x,y)}{2t} dy dx.
\end{align*}
Since $u=vh$, the result follows. \end{proof}

Here is an informal idea of the proof of Theorem~\ref{thm:Theorem2}.
Let
\begin{equation}\label{e.mi2}
u(x)=\left|x \right|^{-\beta/(p-1)} \wedge \left|x \right|^{-\beta}, \quad x\in {\mathbb{R}^d}.
\end{equation}
Since $h=h_\beta$, the integrand under the limit in Lemma~\ref{thm:lem4} is zero if $|x|,|y|\le 1$ and if $|x|,|y|\ge 1$, and it has negative sign otherwise.
This reverses the Hardy inequality, but, unfortunately we need to face
the aforementioned integrability issues.
The next result gives a preparation.
For $r>0$ we let $B(0,r) = \{x \in {\mathbb{R}^d}: |x|<r\}$. Also, we let $B_1=B(0,1)$ and $B_1^c={\mathbb{R}^d}\setminus B_1$.
\begin{lem}\label{thm:lem5}
If $p>2$ and $\mu>\beta\vee (d-\alpha)/p$, then
\begin{align*}
\int_{B_1^c}\int_{B_1^c}\left| |x|^{\beta-\mu} - |y|^{\beta-\mu} \right| \left||x|^{\beta-(p-1)\mu}-|y|^{\beta-(p-1)\mu}\right|&|x|^{-\beta}|y|^{-\beta} \nu(x,y) \, dy \, dx
<\infty.
\end{align*}
\end{lem}
\begin{proof}
For $x \in B_1^c$ we let
$D_1(x) = B_1^c \cap B(0, |x|/2)=\{y:1\leq |y|< |x|/2\}$, $D_2(x) = B_1^c \cap B(0, 2|x|)^c = B(0,2|x|)^c=\{y:2|x|\leq |y|\}$, and $D_3(x) = B_1^c \setminus (D_1(x) \cup D_2(x))=\{y:|x/2|\vee 1\leq |y|<2|x|\}$. The sets form a partition of $B_1^c$.
Note that $p\mu+\alpha > (p\beta)\vee(d-\alpha)+\alpha\ge d$.
With the above integrand, $\int_{B_1^c}\int_{D_1(x)}=\int_{B_1^c}\int_{D_2(x)}$.
Indeed, the integrand is symmetric in $x,y$, on the left we integrate over $\{x,y\in B_1^c: 2|y|<|x|\}$, and on the right -- over $\{x,y\in B_1^c: 2|x|<|y|\}$.
 Then for $y \in D_1(x)$ we have $|y| \le |x|$ and $|x-y| \ge |x|/2$, hence
\begin{align*}
\int_{B_1^c}\int_{D_1(x)}
&\le 2^{d+\alpha} \int_{B_1^c} \int_{D_1(x)} |x|^{-d-\alpha} |y|^{-p\mu}  dy \, dx\\
&= 2^{d+\alpha}\int_{B_1^c}\int_{B(0,2|y|)^c} |x|^{-d-\alpha} |y|^{-p\mu}   dx \, dy
= c  \int_{B_1^c} |y|^{-p\mu-\alpha} \, dy<\infty.
\end{align*}
For $y \in D_2(x)$ we have $|y| \ge |x|$ and $|x-y| \ge |y|/2$, hence
\begin{align*}
\int_{B_1^c}\int_{D_2(x)}
&\le 2^{d+\alpha} \int_{B_1^c} \int_{D_2(x)} |x|^{\beta-p\mu} |y|^{-\beta-d-\alpha} dy \, dx = c \int_{B_1^c} |x|^{-p\mu-\alpha} dx<\infty.
\end{align*}
We next note that for $a,b>0$ and $\gamma>0$,
\begin{equation}\label{eq:lemlem}
|a^{-\gamma} -b^{-\gamma}| \leq \gamma |a-b| (a\wedge b)^{-\gamma-1}.
\end{equation}
Indeed, if, say, $a\leq b$, then
$$
a^{-\gamma} - b^{-\gamma} = \gamma \int_a^b s^{-\gamma-1} ds \leq \gamma (b-a)a^{-\gamma-1} .
$$
For $y \in D_3(x)$ we have $|x|/2 <|y|<2|x|$, therefore,
\begin{align*}
\int_{B_1^c} \int_{D_3(x)}
&\leq  c_1 \int_{B_1^c} \int_{D_3(x)} |x|^{-p\mu -2} |x-y|^{-d-\alpha+2} dy \, dx \\
&\leq  c_{1} \int_{B_1^c} \int_{B(0,3|x|)} |x|^{-p\mu -2} |z|^{-d-\alpha+2} dz \, dx
= c_{2} \int_{B_1^c} |x|^{-p\mu-\alpha} dx<\infty.
\end{align*}
\end{proof}
Here is another auxiliary estimate.
\begin{lem}\label{thm:lem6}
If $p>2$, $0<\beta<\mu$, $R>1$ and $A_R = \{x \in {\mathbb{R}^d}: 1/R<|x|<1\}$, then
\begin{align}
\nonumber
&\int_{A_R}\int_{B_1^c} \left| 1-|y|^{\beta-\mu}\right| \left||x|^{-(p-2)\beta}-|y|^{\beta-(p-1)\mu} \right||x|^{-\beta}|y|^{-\beta} \nu(x,y) \, dy \,dx \\
& \le c (1 \vee R^{(p-1)\beta-d} \vee \operatorname{log} R).
\end{align}
\end{lem}
\begin{proof}
We split the integral as follows:
$$
\int_{A_R}\int_{B_1^c}=\int_{D_1}\int_{B_1^c} + \int_{D_2}\int_{D_3} + \int_{D_2}\int_{D_4} =: I_1 + I_2 + I_3,
$$
where $D_1 = \{ 1/R < |x| \leq 1/2 \}$, $D_2 = \{ 1/2 \le |x|<1\}$, $D_3 = \{1<|y| < 2\}$ and $D_4 = \{|y| \geq 2\}$.
If $x \in A_R$ and $y \in B_1^c$, then $|y|^{\beta-\mu} \le 1$ and $|x|^{-(p-2)\beta} \ge |y|^{\beta-(p-1)\mu}$.
Furthermore, for $x \in D_1$ and $y \in B_1^c$, we have $|x-y|>|y|/2$, hence
\begin{align*}
I_1 &\leq c_1  \int_{D_1} \int_{B_1^c} |x|^{-(p-1)\beta} |y|^{-d-\alpha-\beta} dy \, dx
=c_2 \int_{D_1} |x|^{-(p-1)\beta} dx \\
& \le c(1 \vee R^{(p-1)\beta  - d} \vee \operatorname{log} R).
\end{align*}
If $x \in D_2$, $y\in D_3$, then $1-|y|^{\beta-\mu} < |x|^{\beta-\mu} - |y|^{\beta-\mu}$ and $|x|^{-(p-2)\beta}-|y|^{\beta-(p-1)\mu} < |x|^{\beta-(p-1)\mu}-|y|^{\beta-(p-1)\mu}$, so by \eqref{eq:lemlem},
\begin{align*}
|I_2| &\leq c \int_{D_2} \int_{D_3} |x|^{\beta-p\mu-2} |y|^{-\beta} |x-y|^{-d-\alpha+2} dy \, dx \leq c_1 \int_{D_2}\int_{D_3} |x-y|^{-d-\alpha+2} dy \, dx \\
&\leq c_1 \int_{D_2}\int_{B(0,3)} |z|^{-d-\alpha+2} dz \, dx < \infty.
\end{align*}
To estimate $I_3$, we note that for $x\in D_2$, $y\in D_4$  we have  $|x-y| \geq \frac{1}{2}|y|$, so
\begin{align*}
I_3 &\leq c \int_{D_2} \int_{D_4}  |x-y|^{-d-\alpha} |y|^{-\beta}dy \, dx \leq c_2\int_{D_4}  |y|^{-d-\beta-\alpha} dy \, dx  <\infty.
\end{align*}
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:Theorem2}]
By Theorem \ref{thm:Theorem1} and \eqref{e:ip}, $\mathcal{E}_p[u] \geq \kappa_{(d-\alpha)/p} \int_{{\mathbb{R}^d}} |u(x)|^p|x|^{-\alpha} dx$  for $u\in L^p(dx)$. Let $\kappa>\kappa_{(d-\alpha)/p}$. It suffices to verify that
\begin{equation}\label{e.onH}
\mathcal{E}_p[u] < \kappa \int_{{\mathbb{R}^d}} \frac{|u(x)|^p}{|x|^{\alpha}} dx<\infty
\end{equation}
for some $u \in L^p({\mathbb{R}^d}, (1 \vee |x|^{-\alpha}) dx)$.

We first consider the case $p>2$.
Let $(d-\alpha)/p<\beta<(p-1)(d-\alpha)/p$.
We note that $\kappa_{\beta} > \kappa_{(d-\alpha)/p}$.
As usual, let $h(x)=h_\beta(x)= |x|^{-\beta}.$
Let $\mu>\beta \vee d/p$, $R>1$ and
\begin{equation}\label{e.mi}
u_R(x) =  \left|x \right|^{-\beta/(p-1)} \wedge \left|x \right|^{-\beta} \wedge R^{\mu -\beta} |x|^{-\mu},\quad x\in {\mathbb{R}^d},
\end{equation}
comp. \eqref{e.mi2}. This will be our test function for the Hardy inequality \eqref{e.Hp}.
 We have
$u_R \in L^p({\mathbb{R}^d}, (1+ |x|^{-\alpha}) dx)$. Therefore not only the Hardy inequality holds true for $u_R$, but also the representation from Lemma \ref{thm:lem4} is valid:
\begin{align*}
&\mathcal{E}_p[u_R] = \kappa_{\beta} \int_{{\mathbb{R}^d}} \frac{\left|u_R(x)\right|^p}{\left|x\right|^{\alpha}} dx \\
 &+ \underset{t\to 0}\lim \ \frac{1}{2}\int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} \left( \frac{u_R(x)}{h(x)}-\frac{u_R(y)}{h(y)} \right) \left( \frac{u_R(x)^{\langle p - 1 \rangle} }{h(x)}-\frac{u_R(y)^{\langle p - 1 \rangle} }{h(y)} \right) h(x) h(y) \frac{p_t(x,y)}{t} \,dy \,dx,
\end{align*}
and
the first integral on the right-hand side is finite. We next show that the limit  of the above double integral can be made negative by choosing a sufficiently large $R.$
Let
$B_1=B(0,1)$,
$B_2=B_2^R=B(0,R)\setminus B(0,1)$, $B_3 =B_3^R= B(0,R)^c$. By symmetry, we can split the integral as follows:
\begin{align*}
\int_{{\mathbb{R}^d}}\int_{{\mathbb{R}^d}} &= \int_{B_1}\int_{B_1} + \int_{B_2}\int_{B_2} + \ 2 \int_{B_1}\int_{B_2} + \int_{B_3}\int_{B_3} + \ 2 \int_{B_1}\int_{B_3} + \ 2\int_{B_2}\int_{B_3} \\
&=: I_1 + I_2 + 2I_3^R + I_4^R + 2I_5^R + 2I_6^R.
\end{align*}
Note that $u_R(x)=|x|^{-\beta/(p-1)}$ on $B_1$, $u_R(x)=|x|^{-\beta}$ on $B_2$ and $u_R(x)=R^{\mu-\beta}|x|^{-\mu}$ on $B_3$.
  We see that $I_1=I_2=0$  and the the integrand in $I_3^R$  is negative.
Moreover, observe that $I_3^R$ decreases when $R$ increases, so
there is a constant $A_0>0$ such that
\begin{align*}
&\limsup_{t\to 0} \int\limits_{B_1}\int\limits_{B_2^R} \left( \frac{u_R(x)}{h(x)}-\frac{u_R(y)}{h(y)} \right) \left( \frac{u_R(x)^{\langle p - 1 \rangle} }{h(x)}-\frac{u_R(y)^{\langle p - 1 \rangle} }{h(y)} \right) h(x) h(y) \frac{p_t(x,y)}{t} \,dy \,dx\\
& <-A_0
\end{align*}
for all  $R>1$. We also have
\begin{align*}
|I_4^R| &\leq  C\int_{B_3} \int_{B_3} R^{p(\mu-\beta)} \left| |x|^{\beta-\mu} - |y|^{\beta-\mu} \right| \left||x|^{\beta-(p-1)\mu}-|y|^{\beta-(p-1)\mu}\right| |x|^{-\beta} |y|^{-\beta} \nu(x,y) \, dy \, dx \\
&=  C R^{d-\alpha-p\beta} \int_{B_1^c}\int_{B_1^c}\left| |x|^{\beta-\mu} - |y|^{\beta-\mu} \right| \left||x|^{\beta-(p-1)\mu}-|y|^{\beta-(p-1)\mu}\right| |x|^{-\beta}|y|^{-\beta} \nu(x,y) \, dy \, dx \\
&\to 0 \text{ as } R\to \infty,
\end{align*}
see Lemma \ref{thm:lem5}.
Furthermore,
\begin{align*}
|I_5^R| &\leq C\int_{B_1}\int_{B_3} \left| |x|^{(p-2)\beta/(p-1)} - R^{\mu-\beta} |y|^{\beta-\mu} \right| \left|1-R^{(p-1)(\mu-\beta)}|y|^{\beta-(p-1)\mu}\right| |x|^{-\beta}|y|^{-\beta}\nu(x,y) \, dy \, dx \\
& \leq C\int_{B_1}\int_{B_3} |x|^{-\beta/(p-1)} |y|^{-\beta} \nu(x,y) \, dy \, dx +C \int_{B_1} \int_{B_3} |x|^{-\beta}|y|^{\beta-p \mu} R^{p(\mu-\beta)} \nu(x,y) \, dy \, dx \\
&\leq c\int_{B_3} |y|^{-\beta-d-\alpha} dy + cR^{p(\mu-\beta)} \int_{B_3} |y|^{\beta-p\mu-d-\alpha} dy \\
& = c_1 \left(R^{-\beta-\alpha} + R^{-(p-1)\beta-\alpha}\right)\to 0 \text{ as } R\to \infty.
\end{align*}
Finally,
\begin{align*}
|I_6^R| &\leq C\int_{B_2}\int_{B_3} \left| 1-R^{\mu-\beta}|y|^{\beta-\mu}\right| \left||x|^{-(p-2)\beta} - R^{(p-1)(\mu-\beta)}|y|^{\beta-(p-1)\mu}\right| |x|^{-\beta} |y|^{-\beta} \nu(x,y) \, dy \, dx  \\
&= CR^{d-\alpha-p\beta}\int_{A_R}\int_{B_1^c} \left| 1-|y|^{\beta-\mu}\right| \left||x|^{-(p-2)\beta}-|y|^{\beta-(p-1)\mu} \right| |x|^{-\beta}|y|^{-\beta} \nu(x,y) \, dy \,dx \\
&\to 0 \text{ as } R\to \infty,
\end{align*}
see Lemma \ref{thm:lem6}.
Hence, for $R$ sufficiently large,
$\mathcal{E}_p[u_R] < \kappa_{\beta} \int_{{\mathbb{R}^d}} |u_R(x)|^p|x|^{-\alpha} dx$.

Since $\beta\mapsto\kappa_{\beta}$ is continuous, symmetric about $(d-\alpha)/2$ and increasing on $\left(0,(d-\alpha)/2\right)$,
there is $\beta \in \left((d-\alpha)/p, (p-1)(d-\alpha)/p \right)$ such that $\kappa_{(d-\alpha)/p} < \kappa_{\beta} < \kappa$. Then $\mathcal{E}_p[u_R] < \kappa \int_{{\mathbb{R}^d}} |u_R(x)|^p |x|^{-\alpha} dx,$ which proves \eqref{e.onH}, and the theorem, for $p>2$.

We now consider $p \in(1,2)$. Let $p'= \frac{p}{p-1}$ be its H\"{o}lder conjugate. Of course, $p' \in (2, \infty)$ and $(p-1)(p'-1)=1$, or $(p-1)p'=p$.
For $u \in L^p({\mathbb{R}^d})$ we let $v=u^{\langle p-1 \rangle}$, i.e., $v^{\langle p'-1\rangle}=u$. We have $|u|^p=|v|^{p'}$,
so $\int_{\mathbb{R}^d} |v|^{p'} dx=\int_{\mathbb{R}^d} |u|^p dx$ and $v \in L^{p'}({\mathbb{R}^d})$. Also,
\begin{align}
\mathcal{E}_{p'}[v] &=\frac{1}{2} \int_{{\mathbb{R}^d}} \int_{{\mathbb{R}^d}} (v(x)^{\langle p'- 1\rangle} - v(y)^{\langle p' - 1 \rangle})(v(x) - v(y))  \nu(x,y) \, dy \, dx
= \mathcal{E}_p[u]
\end{align}
and
\begin{align}
\int_{{\mathbb{R}^d}} \frac{|v(x)|^{p'}}{|x|^{\alpha}} dx=\int_{{\mathbb{R}^d}} \frac{|u(x)|^p}{|x|^{\alpha}} dx.
\end{align}
It follows from the first part of the proof that the Hardy inequality holds for the exponents $p$ and $p'$ with the same constants. In particular, the constant $\kappa_{(d-\alpha)/p}$ is optimal on $L^{p'}({\mathbb{R}^d})$. This conforms with our claim, since
$ \kappa_{(d-\alpha)/p'}=  \kappa_{(d-\alpha)/p}$, as noted in the proof of Lemma~\ref{thm:lem7}.
\end{proof}
The following technical result will be useful in Section~\ref{s.ap}. Its proof may be of independent interest, since it sheds some light on the structure of $\mathcal D(\mathcal{E}_p)$.
\begin{lem}\label{l.tm}
If $p>2$ then the inequality \eqref{e.onH} holds for some  function in $C^\infty_c({\mathbb{R}^d})_+$.
\end{lem}
\begin{proof}
Assume that
$\kappa >\kappa_{(d-\alpha)/p}$.
Let $u$ be the function defined by \eqref{e.mi} with suitable $R$
and such that  \eqref{e.onH} holds.
The function is radially decreasing, meaning that $u(x)\le u(y)$ if $|x| \ge |y|$.
For $\epsilon>0$ we let $\phi_\epsilon(a)=(a-\epsilon)\vee 0$, $a\in\mathbb R$.
Of course,
\begin{equation}\label{e.ctr}
|\phi_\epsilon(a)| \leq |a| \qquad \mbox{and}\qquad |\phi_\epsilon(b)-\phi_\epsilon(a)|\leq |b-a|,\quad a,b\in\mathbb R.
\end{equation}
Consequently, by \eqref{eq:2e},
\begin{eqnarray*}
F_p(\phi_\epsilon(u(x)),\phi_\epsilon(u(y))) &\approx& (\phi_\epsilon(u(y))-\phi_\epsilon(u(x)))^2(|\phi_\epsilon(u(y))|+|\phi_\epsilon(u(x))|)^{p-2}\\
&\leq &(u(y)-u(x))^2(|u(y)|+|u(x)|)^{p-2} \\
&\approx & F_p(u(x),u(y)).
\end{eqnarray*}
Since $\phi_\epsilon(u)\to u$ as $\epsilon\to 0$, from the dominated convergence theorem we get
\[\mathcal E_p[\phi_\epsilon(u)]\to \mathcal E_p[u].\]
By Fatou's lemma
\[\liminf_{\epsilon\to 0}\int\frac{|\phi_\epsilon(u)|^p}{|x|^\alpha}\,dx \ge \int\frac{|u|^p}{|x|^\alpha}\,dx,\]
therefore taking a sufficiently small $\epsilon$ we get that \eqref{e.onH} holds with $\phi_\epsilon(u)$.
Slightly abusing the notation, below we write $u$ for the latter function, so $u:=\phi_\epsilon(u)$.
The function is radially decreasing and compactly supported and
we have $u^{\langle p/2\rangle}\in \mathcal D(\mathcal E),$ which follows from \eqref{e.cHk}.
Let $\varphi:\mathbb R^d\to\mathbb R_+$ be smooth, compactly supported, radially decreasing and such that $\int \varphi=1$.
For $\eta>0$ denote $\varphi_\eta(x)=\eta^{-d}\varphi(x/\eta)$
and let
\[
v_\eta=u^{\langle p/2\rangle}\ast \varphi_\eta.
\]
Each $v_\eta$ is smooth, nonnegative and of compact support. It is also radially decreasing, as a convolution of two such functions, see Beckner \cite[page 171]{MR385456}.
It is evident that $v_\eta\to u^{\langle p/2\rangle}$ (pointwise) as $\eta\to 0$.
By the arguments from \cite[the proof Lemma A.5]{MR4088505} we see that $\mathcal E[v_\eta-u^{\langle p/2\rangle}]\to 0$, hence by
Vitali's theorem, the functions
\[\mathbb R^d\times\mathbb R^d\ni(x,y)\mapsto (v_\eta(x)-v_\eta(y))^2\]
are uniformly integrable with respect to $\nu(x,y)dxdy$,
see Schilling \cite[Theorem 22.7]{MR3644418}.
Let
\begin{equation}\label{eq:approxi}
u_{\eta} = \left(v_\eta\right)^{\langle 2/p\rangle}=\left(u^{\langle p/2\rangle}\ast \varphi_\eta\right)^{\langle 2/p\rangle}.
\end{equation}
Since $v_\eta\ge 0$ is smooth and radially decreasing, $u_{\eta}$ is smooth, despite the fractional exponent in its definition.
It is clear that $u_{\eta}\to u$ pointwise. Since
\[F_p(u_{\eta}(x),u_{\eta}(y))\approx (v_\eta(x)-v_\eta(y))^2,\]
the left-hand side is uniformly integrable with respect to $\nu(x,y)dxdy$.
By Vitali's theorem again, $\mathcal E_p[u_{\eta}]\to\mathcal E_p[u]$ as $\eta\to 0$.
By Fatou's lemma, taking sufficiently small $\eta$ we see that \eqref{e.onH} holds with $u:=u_\eta.$
The proof is complete.
\end{proof}

\section{Application to parabolic equation}\label{s.ap}
In this section we prove for $p\in (1,\infty)$ that the Feynman-Kac semigroup generated by the Schr\"{o}dinger operator $\Delta^{\alpha/2}+\kappa_{(d-\alpha)/p}|x|^{-\alpha}$ is a contraction on $L^p({\mathbb{R}^d})$, and the threshold $\kappa_{(d-\alpha)/p}$ is sharp.
We shall also see that the semigroup generated by
$\Delta^{\alpha/2}+
\kappa |x|^{-\alpha}$ is bounded on $L^p({\mathbb{R}^d})$
if and only if either $d/p^*\ge (d-\alpha)/2$ and $\kappa \le \kappa_{(d-\alpha)/2}$ or
$d/p^*< (d-\alpha)/2$ and $\kappa \le \kappa_{d/p^*}$
Here $p^*=\max\{p,p/(p-1)\}$.

As usual, let $\alpha\in (0,2)$ and $\alpha<d\in {\mathbb{N}}$. For $\delta\in [0, (d-\alpha)/2]$, $\kappa=\kappa_\delta$ and  $q=q_\delta$ as in \eqref{e:qbeta},  we define (cf. \cite{MR3933622}) the Schr\"odinger perturbation of $p_t$ from \eqref{e.dp} by $q=q_\delta:$
\begin{equation}\label{def_p_tilde}
   \tilde p_t = \sum_{n=0}^{\infty }p_{t}^{(n)}.
\end{equation}
Here for $t>0$ and $x,y\in {\mathbb{R}^d}$ we let $p_{t}^{(0)}(x, y) = p_t(x, y)$ and
\begin{align}\label{pn}
     p_{t}^{(n)}(x, y) &= \int_0^t \int_{{\mathbb{R}^d}} p_s(x, z) q(z)p_{t-s}^{(n-1)}(z, y) \, dz ds\\
     &=\int_0^t \int_{{\mathbb{R}^d}} p_{s}^{(n-1)}(x, z) q(z)p_{t-s}(z, y) \, dz ds
     , \quad n \geq 1.\notag
\end{align}
From the general theory \cite{MR3460023}, $\tilde p_t$ is a symmetric transition density, i.e., the following Chapman-Kolmogorov equation holds:
\begin{align}\label{eq:ChK}
\int_{\mathbb{R}^d} \tilde{p}_s(x,z) \tilde{p}_t(z,y)\, dz = \tilde{p}_{t+s}(x,y)\,.
\end{align}
The scaling of $\tilde{p}_t(x,y)$ is the same as that of $p_t(x,y)$ \cite[Lemma 2.2]{MR3933622}:
\begin{equation}\label{ss form}
  \tilde p_t(x, y)=t^{-\frac{d}{\alpha }}\tilde p_1\big(xt^{-\frac{1}{\alpha }}, yt^{-\frac{1}{\alpha }}\big), \qquad  t>0,  \ x,y \in {\mathbb{R}^d}.
\end{equation}
For $t>0$ and $x\neq 0$ we define the Feynman-Kac semigroup
$$\tilde P_t f(x) = \int_{\mathbb{R}^d} \tilde p_t(x, y) f(y) dy.$$
The integral certainly makes sense if $f$ is nonnegative, but the following result in fact shows that $\tilde P_t$ may be contractive on $L^p({\mathbb{R}^d})$.
Before we proceed, we note that for nonnegative functions $f,g$ on ${\mathbb{R}^d}$ we have
\begin{equation}\label{e.duality}
\int_{\mathbb{R}^d}\int_{\mathbb{R}^d} \tilde p_t(x,y)f(y)dy\ g(x)dx=
\int_{\mathbb{R}^d}\int_{\mathbb{R}^d} \tilde p_t(x,y)g(y)dy\ f(x)dx.
\end{equation}
By H\"{o}lder inequality, for each $p\in (1,\infty)$ the operator norm of $\tilde P_t$ on $L^p$ is the same as on $L^{p/(p-1)}$ -- below this fact will be referred to as the duality argument.
Furthermore, by \eqref{ss form} we have
\begin{equation}\label{e.es}
\|\tilde P_t\|_{L^p({\mathbb{R}^d})\to L^p({\mathbb{R}^d})}=\|\tilde P_1\|_{L^p({\mathbb{R}^d})\to L^p({\mathbb{R}^d})}, \quad t>0.
\end{equation}
\begin{rem}\label{r.n}
For clarity we note that the heat kernel $\tilde p_t$ of $\Delta^{\alpha/2}+\kappa |x|^{-\alpha}$  can be defined also for
$\kappa\in (-\infty, 0)$, see Jakubowski and Wang \cite{MR4140086}, see also Cho, Kim, Song and Vondra\v{c}ek \cite{MR4163128}. Then
$0\le \tilde p_t\le p_t$, so $\|\tilde P_t\|_{L^p({\mathbb{R}^d})\to L^p({\mathbb{R}^d})}\le \|P_t\|_{L^p({\mathbb{R}^d})\to L^p({\mathbb{R}^d})}=1$ for every $p\ge 1$.
Further, for $\kappa>\kappa_{(d-\alpha)/2}$ we have $\tilde p_t(x,y)\equiv \infty$ by \cite[Corollary~4.11]{MR3933622}, so $\|\tilde P_t\|_{L^p({\mathbb{R}^d})\to L^p({\mathbb{R}^d})}=\infty$ in this case. The remaining cases are resolved in
Theorem ~\ref{t.znHp} and \ref{t.bd}, by considering $\kappa=\kappa_\delta$  with $\delta\in [0, (d-\alpha)/2]$.
\end{rem}
We recall that analogous perturbations of local elliptic operators have been widely investigated.
We refer, e.g., to Kovalenko, Perelmuter and Semenov \cite{MR622855}. In particular, the approach covers Hardy-type perturbations of the classical Laplacian. See also
\cite{MR4043021}.
 The relation of the Hardy inequality to the contractivity of the corresponding Feynman-Kac semigroups is also the subject of \cite{MR2259099}. In
\cite[Corollary 1.2]{MR2259099} the authors use the Hardy inequality to prove that
 the semigroup corresponding to the operator $-\Delta+\kappa|x|^{-2}$
is contractive on $L^p({\mathbb{R}^d})$ for $d\ge 2$ if and only if $\kappa\le (d-2)^2(p-1)/p^2$. This accords well with Theorem~\ref{t.znHp} because
$(d-2)^2(p-1)/p^2=\kappa_{(d-\alpha)/p}$, if we let $\alpha=2$.

Let us first present an informal idea of the proof of Theorem~\ref{t.znHp}.
Consider
\begin{align}\label{eq:CP}
\begin{cases}
u_t = \Delta^{\alpha/2} u + \kappa |x|^{-\alpha} u, \qquad t>0, x \in \mathbb{R}^d,\\
u(0,x) = f(x), \qquad x \in \mathbb{R}^d.
\end{cases}
\end{align}
The semigroup solution of this Cauchy problem is $u(t,x)=\tilde P_t f(x)$, at least for $f$ in the domain of the generator.
We multiply both sides of \eqref{eq:CP} by $u^{\langle p-1\rangle}$ and integrate
\begin{align*}
\int_{\mathbb{R}^d} u_t u^{\langle p-1\rangle} dx = \int_{\mathbb{R}^d} u^{\langle p-1\rangle} \Delta^{\alpha/2}u dx + \kappa \int_{\mathbb{R}^d} |u|^p |x|^{-\alpha} dx.
\end{align*}
By calculus and \eqref{e.Hp},
\begin{align}\label{e.znH}
\frac{\partial}{\partial t}\int_{\mathbb{R}^d} \frac{|u|^p}{p}  dx \le  \left(- \kappa_{(d-\alpha)/p}+\kappa\right) \int_{\mathbb{R}^d} |u|^p |x|^{-\alpha} dx \le 0,
\end{align}
hence $\|u(t,\cdot)\|_p^p$ is decreasing, and so $\|\tilde P_t f\|_p \le \|f\|_p$. On the other hand, for $\kappa>\kappa_{(d-\alpha)/p}$ we have the opposite inequality for the function from Lemma~\ref{l.tm}.
We now present rigorous arguments (some technical results used in the proof are still deferred to Section~\ref{s.A}).
\begin{proof}[Proof of Theorem~\ref{t.znHp}]
In the case of $p=2$, the condition $\delta\le (d-\alpha)/2$ is clearly met, while the contractivity of $\tilde P_t$ is proved in \cite[Proposition 2.4]{MR3933622}, using Schur's test.
Let $p\in (2,\infty)$.
 For $M\in (0,\infty)$ let $q^{(M)}(x)=q(x)\wedge M$, where, recall, $q=q_\delta$.
 In a similar manner as above, we define the Schr\"odinger perturbation of $p_t$ by $q^{(M)}$, which we denote $\tilde p^{(M)}_t$. Since $q^{(M)}$ is bounded, by Phillips' perturbation theorem the domain of the generator $\Delta^{\alpha/2}$ of the strongly continuous semigroup $P_t$ on $L^p({\mathbb{R}^d})$ is the same as the domain of the generator, $\Delta^{\alpha/2}+q^{(M)}$, of the strongly continuous semigroup
$$\tilde P^{(M)}_t f(x) := \int_{\mathbb{R}^d} \tilde p^{(M)}_t(x, y) f(y) dy.$$
We refer to Phillips \cite[Theorem~3.2]{MR54167} and \cite[Lemma 4.2]{MR3613319} for details.
Let $f$ be in the domain of $\Delta^{\alpha/2}$ on $L^p({\mathbb{R}^d})$.
Let $u^{(M)}(t,x)=\tilde P^{(M)}_t f(x)$.
Denote $u^{(M)}(t)=u^{(M)}(t,\cdot)$.
From the general semigroup theory, the mapping $[0,\infty)\ni t\mapsto u^{(M)}(t)\in L^p({\mathbb{R}^d})$ is continuously differentiable.
We then verify that $t \mapsto |u^{(M)}(t)|^p$ and $t \mapsto u^{(M)}(t)^{\langle p - 1 \rangle}$ are continuous in $L^1({\mathbb{R}^d})$ and $L^{\frac{p}{p-1}}({\mathbb{R}^d})$, respectively, see Lemma \ref{thm:difpow0}.
Then it follows from Lemma \ref{lem:pdif} that $|u^{(M)}(t)|^p$ is continuously differentiable in $L^1({\mathbb{R}^d})$ and
\begin{align}\label{eq:diff}
\frac{d}{dt}\int_{\mathbb{R}^d} |u^{(M)}(t)|^p dx&=
\int_{\mathbb{R}^d} \frac{d}{dt}|u^{(M)}(t)|^p dx=\int_{\mathbb{R}^d} p u^{(M)}(t)^{\langle p-1 \rangle}\frac{d}{dt}u^{(M)}(t) dx \nonumber \\
&=
p\int_{\mathbb{R}^d} u^{(M)}(t)^{\langle p-1 \rangle}(\Delta^{\alpha/2} u^{(M)}(t)+q^{(M)}u^{(M)}(t)) dx\nonumber \\
&= p\left(-\mathcal E_p[u^{(M)}(t)] +\int_{\mathbb R^d}q^{(M)}|u^{(M)}(t)|^p dx\right)
\le 0,
\end{align}
provided $\kappa\in [0,\kappa_{(d-\alpha)/p}]$, see \eqref{e.znH}. We then get $\int_{\mathbb{R}^d} |\tilde P_t^{(M)} f|^pdx\le \int_{\mathbb{R}^d} |f|^p dx$. This extends to all $f\in L^p({\mathbb{R}^d})$ by the density of the domain of the generator.
We then let $M\to \infty$. By monotone convergence,  $\tilde p_t^{(M)}\uparrow \tilde p_t$ and for every nonnegative $f\in L^p({\mathbb{R}^d})$,
 \[\int_{\mathbb{R}^d} (\tilde P_t f)^p dx\leftarrow \int_{\mathbb{R}^d} (\tilde P^{(M)}_t f)^p dx\le \int_{\mathbb{R}^d} f^p dx.\]
The estimate for general $f\in L^p({\mathbb{R}^d})$ follows by considering $f=f_+-f_-$, see \eqref{eq:fgen}.

The case of $p\in (1,2)$ results from duality with the same range of $\kappa$ as for the H\"{o}lder conjugate of $p$, that is
the contractivity of $\tilde P_t$ holds for $\kappa\in[0,\kappa_{(d-\alpha)/p'}]$ with $p'=p/(p-1)$. But $\kappa_{(d-\alpha)/p}=\kappa_{(d-\alpha)-(d-\alpha)/p} = \kappa_{(d-\alpha)/p'}$. Thus, the contractivity of $\tilde P_t$ on $L^p({\mathbb{R}^d})$ is proved if $p\in (1,\infty)$ and $\kappa\le \kappa_{(d-\alpha)/p}$.

We next assume $\kappa >\kappa_{(d-\alpha)/p}$ and prove that the contractivity fails. To this end
we first consider $p\in (2,\infty)$.
By Lemma~\ref{l.tm}
there is a nonnegative $f\in C_c^\infty(\mathbb R^d)$ such that
\begin{equation}\label{e.onHf}
\mathcal{E}_p[f] < \kappa \int_{{\mathbb{R}^d}} \frac{|f(x)|^p}{|x|^{\alpha}} dx<\infty.
\end{equation}
It is well known that $C_c^\infty(\mathbb R^d)$ is a subset of the domain of the generator of the semigroup $\{P_t,t\ge 0\}$ acting on $L^p$, see, e.g., Jacob \cite[Theorem 3.3.11]{MR1917230} or Farkas, Jacob and Schilling \cite[Theorem 1.4.2 or Proposition 2.1.1]{MR1840499}; or use \cite[Theorem~31.5]{MR1739520}, \eqref{e.mp} and \eqref{e.zp}. Therefore $f$ is in the domain of $\Delta^{\alpha/2}$ on $L^p({\mathbb{R}^d})$.
For $M\in (0,\infty)$ we write $u^{(M)}(t,x)=\tilde P_t^{(M)} f(x)$ or $u^{(M)}(t)=\tilde P_t^{(M)} f$.
By \eqref{eq:diff},
\begin{eqnarray*}
\frac{d}{dt}\int_{\mathbb R^d} |u^{(M)}(t)|^p dx &=&p\int_{\mathbb R^d} u^{(M)}(t)^{\langle p-1\rangle}\left(\Delta^{\alpha/2}u^{(M)}(t) +q^{(M)} u^{(M)}(t)\right)\,dx\\
&=& p\left(-\mathcal E_p[u^{(M)}(t)] +\int_{\mathbb R^d}q^{(M)}|u^{(M)}(t)|^p dx\right).
\end{eqnarray*}
In particular,
\begin{eqnarray}\label{eq:contr1}
&&\left.\frac{d}{dt}\int_{\mathbb R^d} |u^{(M)}(t)|^p dx \right|_{t=0}
=p\left(-\mathcal E_p[f] +\int_{\mathbb R^d} q^{(M)}|f|^pdx\right)\nonumber
\\
&=& p\left(\int_{\mathbb R^d}\left(q^{(M)}(x)-\frac{\kappa}{|x|^\alpha}\right)
|f|^pdx\right)
+p\left(\kappa\int_{\mathbb R^d} \frac{|f|^p}{|x|^\alpha}dx - \mathcal E_p[f]\right).
\end{eqnarray}
The last term is strictly positive by \eqref{e.onHf}. Since $q^{(M)}(x) \to\kappa|x|^{-\alpha}$, for sufficiently large $M$ the expression in \eqref{eq:contr1} is strictly positive, and so for such $M$ we get,
\[\frac{d}{dt}\|u_\kappa^{(M_0)}(t)\|_p^p\bigg|_{t=0} >0.\]
Therefore for small $t>0$ we have
\[ \|f\|_p^p= \|u_\kappa^{(M)}(0)\|_p^p < \|u_\kappa^{(M)}(t)\|_p^p\le \|\tilde P_t f\|_p^p.\]
The case $p\in (1,2)$ follows by the duality argument.
\end{proof}

To complement Theorem~\ref{t.znHp} we note that for $x,y\in {\mathbb{R}^d}$ and $t>0$,
\begin{equation}
\label{eq:mainThmEst}
\tilde{p}_t(x,y)\approx \left(1+t^{\delta/\alpha}|x|^{-\delta}  \right)\left(1+t^{\delta/\alpha}|y|^{-\delta}  \right) \left( t^{-d/\alpha}\wedge \frac{t}{|x-y|^{d+\alpha}} \right).
\end{equation}
The result is given in \cite[Theorem~1]{MR3933622} (for $x=0$ or $y=0$ we have $p^{(1)}_t(x,y) = \infty$; see \cite[Lemma 3.7 and the comment before Corollary 3.9]{MR4140086}, so $\tilde{p}_t(x,y) = \infty$).
Denote $H(x)=|x|^{-\delta}\vee 1$, $x\in {\mathbb{R}^d}$.
Clearly,
\begin{equation}\label{e.oHH}
\tilde p_1(x,y)\approx H(x)H(y)p_1(x,y), \quad x,y\in {\mathbb{R}^d}.
\end{equation}
\begin{proof}[Proof of Theorem~\ref{t.bd}]
By \eqref{e.es} it is enough to consider $t=1$.
Let $B=B(0,1)\subset {\mathbb{R}^d}$. Denote $q=p/(p-1)$. If $\delta\ge d/q$, then $\int_B H^q=\infty$, so there is $f\in L^p_+$ such that $\int_B H(y)f(y)=\infty$, see, e.g., \cite[Corollary~4.4.5]{MR2267655}. By \eqref{e.oHH}, $\tilde P_1 f=\infty$ on ${\mathbb{R}^d}$.
Also, if $\delta\ge d/p$, then $\tilde P_t \mathbf{1}_B\ge c H\mathbf{1}_B$, so $f:=\mathbf{1}_B\in L^p$ but $\int |\tilde P_t f|^p\ge c \int_B H^p=\infty$.
If $0\le \delta<d/p^*$, then we let $f\in L^p_+$, $g\in L^q_+$ and consider
\begin{align*}
I(f,g)&:=\int_{\mathbb{R}^d}\int_{\mathbb{R}^d} g(x)H(x)H(y)p_1(x,y)f(y)dydx\\
&=I(f\mathbf{1}_B,g\mathbf{1}_{B})+I(f\mathbf{1}_B,g\mathbf{1}_{B^c})+I(f\mathbf{1}_{B^c},g\mathbf{1}_{B})+I(f\mathbf{1}_{B^c},g\mathbf{1}_{B^c}).
\end{align*}
By \eqref{e.cLp} and H\"{o}lder inequality,
$$
I(f\mathbf{1}_{B^c},g\mathbf{1}_{B^c})\le \|f\|_p\|g\|_q.
$$
By H\"{o}lder inequality, $\int_B gH\le c\|g\|_q$, where $c=\big(\int_B |x|^{-\delta p}\big)^{1/p}<\infty$. Similarly,
$\int_B fH\le c\|f\|_p$, where $c=\big(\int_B |x|^{-\delta q}\big)^{1/q}<\infty$.
Furthermore, since $p_1$ is bounded, $I(f\mathbf{1}_B,g\mathbf{1}_{B})\le c \int_B gH \int_B f H\le c\|g\|_q\|f\|_p$.
By \eqref{e.cLi},
$I(f\mathbf{1}_{B^c},g\mathbf{1}_{B})\le \int_B gH \|p_1f\|_\infty\le c\|g\|_q\|f\|_p$. Similarly, $I(f\mathbf{1}_B,g\mathbf{1}_{B^c})\le c\|g\|_q\|f\|_p$.
By \eqref{e.oHH}, $\tilde P_1$ is bounded on $L^p$.
\end{proof}

\section{Appendix}\label{s.A}
For the convenience of the reader, we add some details to the arguments used in the proof of Theorem \ref{t.znHp}.
In what follows, we consider a measure space $(X,\mu)$ and the spaces $L^p(X,d\mu)$ for $p\geq 1$.
For simplicity, we write $L^p$ for $L^p(X,d\mu).$
Recall that $p/(p-1)$ is the H\"{o}lder conjugate exponent of $p\in (1,\infty)$ (i.e. we have $p^{-1} + (p/(p-1))^{-1} = 1$).
\begin{lem}[Continuity]\label{thm:continuity}
Let $p \in (1, \infty)$ and $r \in \big[\frac{p}{p-1}, \infty \big)$.
If $f\in L^p$, $g\in L^r$, then $\| fg\|_{\frac{pr}{p+r}} \leq \|f\|_p \|g\|_r$. If $f_n \to f$ in $L^p$ and $g_n \to g$ in $L^r$, then $f_ng_n \to fg$ in $L^{\frac{pr}{p+r}}$.
\end{lem}
\begin{proof}
Of course, $r>1$. Also, $\frac{p+r}{pr} = \frac{1}{r} + \frac{1}{p} \leq \frac{p-1}{p} + \frac{1}{p} = 1$, thus $\frac{pr}{p+r} \in [1, \infty)$. For $f \in L^p$, $g \in L^r$, by H\"older inequality with exponents $\frac{p+r}{r}$ and $ \frac{p+r}{p}$,
$$
\int |fg|^{\frac{pr}{p+r}} \leq \left(\int |f|^p \right)^{\frac{r}{p+r}} \left(\int |g|^r \right)^{\frac{p}{p+r}} < \infty,
$$
and we get the first statement.
The second statement is verified as follows,
\begin{align*}
\|f_ng_n - fg\|_{\frac{pr}{p+r}} &= \|f_ng_n - f_ng +f_ng -fg\|_{\frac{pr}{p+r}} \\
&\leq \|f_n\|_p \|g_n-g\|_r + \|f_n - f \|_p \|g\|_r \to 0,
\end{align*}
as the sequence $f_n$ is bounded in $L^p.$
\end{proof}
Let $p \in (1,\infty)$. Assume that
$[0,\infty) \ni t \mapsto u(t) \in L^p$.
We will relate the continuity and differentiability properties of $u$ in $L^p$ to those of $|u|^p$ in $L^1$.
We denote
$$\Delta_hu(t) = u(t+h)-u(t), \quad \textnormal{if} \,\, t, t+h\ge 0.$$

We say that $u$ is continuous in $L^p$ at $t\ge 0,$ if $\Delta_hu(t) \to 0$  in $L^p$ as $h\to 0$,
and we say $u$ is continuously differentiable at $t\ge 0$ if  $u'(t) := \lim_{h \to 0} \frac{1}{h} \Delta_hu(t)$ in $L^p$ with
continuous $u'$. Of course, $u'(0)$ is the right-hand side derivative in the above setting.

\begin{lem}\label{thm:difpow0}
If $u$ is continuous in $L^p$, then $|u|^p$  and $u^{\langle p - 1 \rangle}$ are continuous in $L^1$ and $L^{\frac{p}{p-1}}$, respectively.
\end{lem}
\begin{proof}
By \eqref{eq:2c},
$|\Delta_h |u(t)|^p|\le (p+C)|\Delta_h u(t)|(|u(t+h)|+|u(t)|)^{p-1}$.
Similarly, by \eqref{eq:2dd}, $|\Delta_h u(t)^{\langle p - 1 \rangle}|\le c'|\Delta_h u(t)|^\lambda (|u(t+h)|+|u(t)|)^{p-1-\lambda}$, and we can pick $\lambda>0$ such that $p-1-\lambda>0$.
The results follow from Lemma \ref{thm:continuity}.
\end{proof}

\begin{lem}[Differentiability]\label{thm:difpow}
If $[0,\infty)\ni t \mapsto u(t)$ is continuously differentiable in $L^p$,
then $|u|^p$ is continuously differentiable in $L^1$ and $(|u|^p)' = pu^{\langle p - 1 \rangle} u'$.
\end{lem}
For comparison, Marinelli, R\"ockner \cite[p. 4]{MR3463679} assert that for $p \geq 2$ the function
$\phi: H \ni x \mapsto \|x\|^p$ is weekly differentiable for every Hilbert space $H$, with the Fr\'echet derivative
$D\phi(x): y \mapsto p \|x\|^{p-2} (x,y)$.
\begin{proof}[Proof of Lemma \ref{thm:difpow}]
By Lemma~\ref{thm:continuity},
$u^{\langle p - 1 \rangle} u'$ is continuous in $L^1$.
By \eqref{eq:2d}, for
$\lambda\in [0,2]$,
\begin{align*}
\left|\frac{1}{h} \left(\Delta_h |u|^p (t) - p u(t)^{\langle p - 1 \rangle} \Delta_h u(t)\right)\right|
&\leq Ch^{\lambda-1} |\frac{1}{h} \Delta_hu(t)|^{\lambda} (|u(t+h)| +|u(t)|)^{p-\lambda}.
\end{align*}
We pick
$\lambda>1$ such that $p-\lambda>0$ and the result follows.
\end{proof}

Recall that $(P_t, t \geq 0)$ is a strongly continuous operator semigroup on $L^p$. Let $f$ be in the domain of its generator $A$. Let $u(t) = P_tf$. Then $u'(t) = P_tAf = AP_tf = Au(t)$.
By Lemma \ref{thm:difpow} we obtain the following result.
\begin{lem}\label{lem:pdif}
$|u(t)|^p$ is differentiable in $L^1$ with the derivative
\begin{equation}\label{e.ppp}
(|u(t)|^p)' =pu(t)^{\langle p - 1 \rangle} u'(t)= pu(t)^{\langle p - 1 \rangle} P_t Af, \quad t \geq 0.
\end{equation}
\end{lem}
Finally, recall that in the proof of Theorem \ref{t.znHp} we only proved the contractivity of the semigroup for $\kappa\le \kappa_{(d-\alpha)/p}$ and \textit{nonnegative} $f \in L^p$. This suffices because for general $f \in L^p$ we may write $f=f_{+} - f_{-}$ and
we have
$
(\tilde{P}_t f)_+ \leq \tilde{P}_t f_+,
$
and
$
(\tilde{P}_t f)_- \leq \tilde{P}_t f_-,
$
hence
$
|\tilde{P}_t f|^p \leq (\tilde{P}_tf_{+})^{p} + (\tilde{P}_tf_{-})^{p}.
$
Therefore,
\begin{align}
\int |\tilde{P}_tf(x)|^p dx &\leq \int \tilde{P}_tf_+(x)^p dx + \int \tilde{P}_tf_-(x)^p dx \label{eq:fgen}\\
&\leq \int f_+(x)^p dx + \int f_-(x)^p dx = \int |f(x)|^p dx. \nonumber
\end{align}

redpajama_set_name: RedPajamaArXiv


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text:
import { Name } from '@serenity-js/core/lib/model';

import { SerenityBDDReportContext } from '../SerenityBDDReportContext';

/**
 * @package
 */
export function testRunnerCalled<Context extends SerenityBDDReportContext>(name: Name): (context: Context) => Context {
    return (context: Context): Context => {
        context.report.testSource = name.value;

        return context;
    }
}

redpajama_set_name: RedPajamaGithub


text:
Bowie Bears Keep Their First Day of School Tradition Alive
Emily Slape Published: July 26, 2021
Francisco J. Ordaz via Twitter
The first day of school can be a little scary not only for students BUT for new staff.
As a way to make the first day exciting and fun, the staff over at Bowie High School welcomed their new staff into the new school year by singing the school fight song.
Sadly, I do not even remember my high school fight song, but I kind of know UTEP's. I am going to have to go back to the Fabens website and reteach myself the song. I am incredibly jealous of just how much school pride is oozing from the osos!
Oso is Spanish for bear.
I have always admired the Bowie bear community for their deep love of school pride! El Pasoans are very proud of not only the city but from where they graduated.
Bowie wasn't the only school showing their pride for a new school year, many El Paso Independent School District schools did something special for the first day of school.
As the first day of school for 2021 - 2022 will come and go. The memories will last forever. It may be a scary year with some students returning to a classroom after being homeschooled for so long, but we will all get through it together.
We here at KISSFM wish the best of luck to all the students and staff returning to school!
Categories: Celebrities

redpajama_set_name: RedPajamaCommonCrawl


The Claret Partners Limited Founder Ana Sainz talks about how her firm acts as a kind of 'hired gun' for companies when they need additional resources to tackle knotty tax issues.
Being a smaller firm in an industry of giants has advantages.
"Unlike the 'Big 4' consulting firms, we don't charge a fortune for our services," says Ana Sainz, Founder of The Claret Partners Limited.
Large consulting firms are burdened with layers of expensive in-house talent and overhead costs, so they must charge premium fees for their services. As a small firm, Claret Partners is able to deliver the same kind of specialized tax advice and services for its clients, on-demand and often in-house. The Toronto-based firm works primarily with banks, insurance companies, and the private equity industry to develop tax strategies and to identify tax risk in compliance, accounting, governance, and resources.
It's a valuable role to be able to fill-in for companies that don't have that kind of in-house specialization.
For highly specialized tax work, Sainz can easily bring in one of her trusted contractors to get the work done.
Claret clients like this model because they're not stuck paying big-firm fees.
"Some industries, like the private equity industry, really need very specific tax assistance. But it's not easy to find – you'd have to go to a 'Big 4' firm – and the trade-off is they're going to charge you an enormous amount of money to have someone seconded to your office for that kind of help," says Sainz.
"I'm kind of a hired gun for companies when they either need additional resources at peak periods or, I function as an in-house tax department," she explains.
Within The Claret Partners Tax Resource Roadmap, we will help you create a plan to develop your people into better resources and link this plan to a performance measurement matrix.
No matter what your business or the size, tax departments have a significant responsibility – to deliver efficient, credible tax provisions, reporting, and compliance. Achieving this goal requires your data, technology, people, and process be effectively integrated.


text:
Q: Modifying a variable in enclosing scope There is a common problem in languages that assume variable declarations are local. How do you get at variables in enclosing scopes.
Is there a way in Opa?
For example:
start() =
    name = Random.string(5)

    set_name(new_name) =
        old_name = name
        name = new_name
        log("User {old_name} changed name to {new_name}")

This doesn't work. We get a warning that name is unused in set_name, and the value of name in start is never changed.
In a language like Lua, Javascript or Scheme, there is explicit marking of locals, so variables not marked in that way can be found in the scope stack. In Python there is no such marking, and so this would be impossible. In Python you can get at global (toplevel) variables, and I've found the @toplevel directive in Opa too. But I'm interested in intermediate points in the scope chain.
There are workarounds, of course, by using records, but is there a direct route?

A: One solution is to use Reference module :
Reference.create, Reference.get, Reference.set
http://opalang.org/resources/doc/index.html#stdlib.core.reference.opa.html/!/value_stdlib.core.Reference

A: Reference module mentioned in Fred's answer is indeed one solution to this and it's closest to what you are asking for.
But it's also good to know that when programming the "right way" most of the state in your Opa program will be captured in Sessions, with a change to the session's state triggered by a message send to it.

redpajama_set_name: RedPajamaStackExchange


Hi, I made a page, inside it I have iFrame. I want the scroll bar to be on the left side, but if I put the dir=rtl in the body the text becomes wrong. I've tried to put the direction inside the table but it does't work. I've been looking a lot in the forums I didn't find an answer.
well i added dir="rtl" to the <body> tag and it looks fine... the only thing i noticed is that it puts a bottom scroll bar which can be fixed simply by changeing the table size to 425 rather than 430.
unless there is something i am missing?
here I add also, and it's not looks good. Maybe I don't know where to add it.
and about the scroll bar at the bottom I must check it too.
ummm i dont understand what that last one means? could you please spruce up your english?
sorry about my english - it's not my mother language.
anyway someone helped me allready in anouther forum and it's fine. Now I just want to get red of the scroll bar at the bottom. the width of the table in the page with the text is 430 and the width of the table in the page that call the text is 455 and still I have scroll bar at the bottom.


This novel is written a bit like a play itself, a stylistic choice that lends to sparse prose and descriptions that read more like dictations of stage directions. Again, much like the original play, every sentence feels like it only exists for the sake of ushering the plot along.
Meanwhile, the plot itself is often difficult to buy into, largely hindered by "young-people" slang-y dialogues and "young-people" Shakespeare rap lyrics made inevitably awkward by the fact that they were very obviously written by a not-young person.


text:
package org.apache.oodt.commons.activity;

import java.io.OutputStreamWriter;

/**
 * Storage that sends activity reports to the standard output as plain text.
 *
 * @author Kelly
 * @version $Revision: 1.1 $
 */
public class StandardOutputStorage extends WriterStorage {
	/**
	 * Creates a new {@link StandardOutputStorage} instance.
	 */
	public StandardOutputStorage() {
		super(new OutputStreamWriter(System.out));
	}
}


(561) 367-8777The Catholic University of AmericaSouthern Methodist UniversityU.S. District Court, Middle District of Florida and FloridaMartindale-Hubbell, Martindale-Hubbell, Martindale-Hubbell and Martindale-HubbellHernando County Bar Association, The Academy of Florida Trial Lawyers..."The legal perspective on car insurance"
(561) 266-9191University of DelawareFlorida, New Jersey and District of ColumbiaForum Publishing Group, Forum Publishing Group, Forum Publishing Group and Forum Publishing GroupPhiladelphia Bar Association, Washington, DC Bar...What to do in case of Car Accident?
(866) 353-5020University of North Carolina School of Law and Harvard Law SchoolUniversity of North Carolina - Chapel Hill and Phillips Exeter AcademyRhode IslandSuperlawyers , Providence Business News and Providence Business NewsRhode Island Bar Association and Thurgood Marshall Law Society"The Best Insurance Policy in Business is a No-Risk Culture" and Home-birth Blogging Battle Leads to Copyright Suit - "Fair Use"
The LII Lawyer Directory contains lawyers who have claimed their profiles and are actively seeking clients. Find more Palm Beach County, Florida Lawyers in the Justia Legal Services and Lawyers Directory which includes profiles of more than one million lawyers licensed to practice in the United States, in addition to profiles of legal aid, pro bono and legal service organizations.


REFRIGERANT TO THE OUTSIDE OF THE IMPELLER.
The important thing to remember is that a compressor performs the same function no matter what type it is.
forced into the top of the condenser coil by the compressor.
and, again, cannot have a temperature/pressure relationship.
Figure 3.31 An illustration of the internal working mechanism of a screw compressor.


The 7 grandchildren with their parents.
getting more on the outside than on the inside.
Garth is trying to teach Greysen how to freeze brand.
Greysen does not seem to be paying much attention.
Amanda passed that dominant gene for smiling on to Greysen.
When the grandkids get together it can get chaotic.
at Brenda's Slavick House. You can see Debbie and the photographer in the wall mirror.
Greysen on his birthday 7/09/01.
but the little one obviously had benefitted by 2 generations of improved genetics.
Greysen does not seem very excited about his first KSU football game.
Mark, Eva, Cole, Ransom, and Quanah.
My face may be a little messy but it still is a good day!
Nanny and Greysen seem to enjoy each other.
Quanah and Sierra hold Cole down.
for the April 6 sale. After all they got to work with Debbie Norton.
Sierra has her 5th birthday.
4th of July Turtle Race. They didn't win.
reached the age that Grant is now, Grant will then be 20 years old. Time flies!
This young fellow looks quite handsome in his new cowboy hat.


text:
Horizon thanks shareholders for their overwhelming support for its new independent and highly qualified board nominees; Calls on all shareholders to join in supporting the Horizon Nominees and voting against the proposed transaction with Tecpetrol before the July 27 deadline
Horizon Capital Management Inc.
Jul 21, 2015, 09:34 ET
PANAMA CITY, Republic of Panama, July 21, 2015 /CNW/ - Horizon Capital Management Inc. ("Horizon") today thanked its fellow Americas Petrogas Inc. ("Americas Petrogas" or the "Company") (TSXV: BOE) shareholders for their overwhelming support for Horizon's three independent and highly qualified nominees (the "Horizon Nominees") to the board of the Company (the "Board") at the upcoming shareholders meeting currently scheduled for July 29, 2015 (the "Meeting"). Horizon also called on all Americas Petrogas shareholders to join in supporting the Horizon Nominees by voting before the deadline of 9:00 a.m. (Mountain Time) on Monday July 27, 2015 (or any earlier deadline indicated by your broker). Information about the Horizon Nominees is provided below under the heading "The Horizon Nominees".
As a shareholder, you can VOTE FOR the Horizon Nominees using the form of proxy or voting instruction form that you received with your Meeting materials from the Company. Please carefully read and follow the instructions below under the heading "How to Vote for the Horizon Nominees" for more information about how to support the Horizon Nominees.
Horizon also once again urged shareholders to VOTE AGAINST the proposed sale by the Company of all of the outstanding common shares of Americas Petrogas Argentina S.A. ("APASA") to Tecpetrol International S.A. and its affiliates ("Tecpetrol") and related transactions (the "Proposed Transaction"). However, even if you vote for the Proposed Transaction you can still VOTE FOR the Horizon Nominees to protect the value of your investment if the Proposed Transaction is completed.
Even if you have already voted for management's director nominees or for the Proposed Transaction, you can change your vote by submitting another form of proxy or voting instruction form bearing a later date prior to the deadline above (or, if you are a registered shareholder, by attending the Meeting and voting in person).
THE HORIZON NOMINEES
Juan Argento. Mr. Argento is a Senior Advisor to Horizon Capital. Prior to joining Horizon, Mr. Argento held senior positions at Millennium Global, a London-based USD 13 billion investment firm, Rubikon Partners, a mid-market private equity firm, focused on the European market and chaired by Dr. Henry Kissinger, Texas Pacific Group and Salomon Brothers Inc. He has been Chairman and/or a board member of several private companies in Latin America and the CEO of a technology company in Latin America. He holds an A.B. in Economics with honors from Harvard University.
Aris Tsikouras. Aris is a consultant with Horizon Capital. Previously, Aris was a Senior Partner with McKinsey & Company over the past 16 years. While at McKinsey, he was based out of New York, Athens, São Paulo, and most recently Rio de Janeiro. Aris led more than 80 strategic, operational, and organizational projects with electricity, oil, and gas clients in emerging markets. He led McKinsey's Oil & Gas Hub in Rio de Janeiro as well as the firm's Global Emerging Markets Power & Gas Practice. Aris holds a M.S. in Engineering-Economic Systems from Stanford University and a M.Eng. in Engineering, Economics, and Management from Oxford University.
Alex Zyngier. Mr. Zyngier is a shareholder of the Company, a restructuring and turnaround expert, a graduate of the University of Chicago's MBA program and has held important positions in first-rate firms such as Goldman Sachs, McKinsey & Company and Deutsche Bank.
HOW TO VOTE FOR THE HORIZON NOMINEES
You can vote for the Horizon Nominees using the form of proxy or voting instruction form that you received with your Meeting materials by doing the following prior to the deadline of 9:00 a.m., Mountain time, on Monday July 27, 2015 (or any earlier deadline indicated by your broker):
Insert the name "Christopher Sunstrum" (the "Horizon Representative") as your proxy nominee in the blank space provided on your form of proxy or voting instruction form (Mr. Sunstrum is a partner with Goodmans LLP, Horizon's legal counsel). If you hold your Americas Petrogas shares through a broker, you may need to contact your broker immediately in order to appoint the Horizon Representative as your proxy nominee at the Meeting.
Do not provide any voting instructions for the election of directors (i.e., do not check any of the "for" or "withhold" boxes for any of management's director nominees); and
Otherwise properly complete, sign, date and return your form of proxy or voting instruction form (either by mail or via the Internet) by carefully following the instructions provided on your form of proxy or voting instruction form. You cannot vote by telephone if you want to support the Horizon Nominees.
Your form of proxy or voting instruction form must be submitted prior to 9:00 a.m., Mountain Time, on Monday, July 27, 2015 (or any earlier deadline indicated by your broker). Horizon encourages shareholders to submit their proxies and voting instructions (in accordance with the instructions provided above) via the Internet to ensure they are submitted prior to the deadline.
Please also send a copy of your form of proxy or voting instruction form (or a written confirmation if you vote via the Internet) to the attention of Juan Argento at E-mail: [email protected].
If you appoint the Horizon Representative as your proxy nominee with discretionary authority for the election of directors, your common shares will be voted FOR the Horizon Nominees and for the following management nominees: Ken Geren, Easton Wren and Ross C. McCutcheon. This will guarantee some necessary continuity, the necessary number of Canadian directors, and technically qualified members in the Reserves and Governance committees.
If you have misplaced your form of proxy or voting instruction form, or if you have any questions or need assistance in completing and submitting your proxy or voting instruction form or changing your vote, please contact Juan Argento at Tel: +1 347 759 6074 E-mail: [email protected].
INFORMATION IN SUPPORT OF PUBLIC BROADCAST SOLICITATION
The following information is provided in accordance with Canadian corporate and securities laws applicable to public broadcast solicitations. Horizon is relying on the exemption under section 9.2(4) of National Instrument 51-102 – Continuous Disclosure Obligations ("NI 51-102") to make this public broadcast solicitation. This solicitation is being made by Horizon and not by or on behalf of the management of Americas Petrogas. The registered office address of Americas Petrogas is 3911 Trasimene Crescent S.W., Calgary, Alberta T3E 7J6.
Horizon has filed this press release containing the information required by section 9.2(4)(c) of NI 51-102 and has filed a separate document containing the information required by Form 51-102F5 - Information Circular in respect of the Horizon Nominees for election to the board of directors of the Company at the Meeting, including any and all adjournments or postponements thereof, on the Company's profile on SEDAR at www.sedar.com.
Horizon may solicit proxies in reliance upon the public broadcast exemption to the solicitation requirements under applicable Canadian corporate and securities laws, conveyed by way of public broadcast, including through press releases, speeches or publications, and by any other manner permitted under applicable Canadian laws. All costs incurred for the solicitation will be borne by Horizon.
A registered holder of common shares of Americas Petrogas that gives a proxy may revoke it: (a) by completing and signing a valid proxy bearing a later date and returning it in accordance with the instructions contained in the form of proxy, or as otherwise provided in Americas Petrogas' management information circular; (b) by depositing an instrument in writing executed by the shareholder or by the shareholder's attorney authorized in writing, as the case may be: (i) with Computershare Trust Company of Canada, at any time, not less than 48 hours, excluding Saturdays, Sundays and statutory holidays in the Province of Alberta, preceding the Meeting or an adjournment or postponement thereof; (ii) at the registered office of Americas Petrogas at any time up to and including the last business day preceding the Meeting or any adjournment or postponement thereof, or (iii) with the chairman of the Meeting prior to its commencement on the day of the Meeting or any adjournment or postponement thereof; or (c) in any other manner permitted by law.
A non-registered holder of common shares of Americas Petrogas will be entitled to revoke a form of proxy or voting instruction form given to an intermediary at any time by written notice to the intermediary in accordance with the instructions given to the non-registered holder by its intermediary. It should be noted that revocation of proxies or voting instructions by a non-registered holder can take several days or even longer to complete and, accordingly, any such revocation should be completed well in advance of the deadline prescribed in the form of proxy or voting instruction form to ensure it is given effect in respect of the Meeting.
Horizon Capital Global Investments Inc., an affiliate of Horizon, has beneficial ownership over 12,757,000 common shares of Americas Petrogas, representing approximately 5.72% of the issued and outstanding common shares of Americas Petrogas.
With the exception of the foregoing, to the knowledge of Horizon, none of Horizon or any of the Horizon Nominees or any of their respective associates or affiliates has any material interest, direct or indirect, by way of beneficial ownership of securities or otherwise, in any matter currently known to be acted upon at the Meeting other than the election of directors. In addition, none of Horizon or any of the Horizon Nominees or any of their respective associates or affiliates has any material interest, direct or indirect, in any transaction since the beginning of the Company's most recently completed financial year or in any proposed transaction that has materially affected or would materially affect the Company or any of its subsidiaries.
ABOUT HORIZON CAPITAL MANAGEMENT INC.
Founded in 2004, Horizon Capital Management Inc. is a merchant banking firm with advisory and principal investment activities and with substantial focus on the oil and gas industry in Latin America.
All statements, other than statements of historical fact, included in this news release constitute "forward-looking information" as such term is defined in applicable Canadian securities legislation. Forward-looking information can generally be identified by the use of forward-looking language such as "will", "expect", "intend", "plan", "estimate", "anticipate", "believe" or "continue" (and grammatical variations and the negatives thereof) Such forward-looking information is based on certain understandings, assumptions, beliefs, opinions and expectations of Horizon. Shareholders should not place undue reliance on such forward-looking information, which is not a guarantee that any particular outcome, event, result, performance or other achievement will occur. Many risks, uncertainties and other factors could cause the actual outcomes, events, results, performance or achievements expressed or implied by such forward-looking information to vary materially from those described herein should any of those risks, uncertainties or other factors materialize. Accordingly, readers of this news release are cautioned not to place undue reliance on any forward-looking information contained in this herein. All forward-looking information contained herein is expressly qualified in its entirety by this cautionary statement. All forward-looking information contained herein is made as of July 21, 2015 and Horizon undertakes no obligation to publicly update or revise any such forward-looking information, except as required by law.
SOURCE Horizon Capital Management Inc.
For further information: Juan Argento, Horizon Capital Management Inc., Calle 53E, Urbanización Marbella, MMG Tower, Piso 16, Panamá, República de Panamá, Tel: +1 347 759 6074, E-mail: [email protected]


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Contact Boltz to schedule an appointment to discuss which hearth product is best for your needs or stop by office to visit our showroom and experience the warmth and ambiance in person.