TITLE: What is the condition for n by n symmetric matrix with 1 on diagonal and x everywhere else to be positive definite
QUESTION [0 upvotes]: What is the condition on x for n by n symmetric matrix with 1 on diagonal and x everywhere else to be positive definite
Let the matrix be
\begin{bmatrix}
1 & x & x & ... &x\\
x & 1 & x & ... &x\\
x & x & 1 & ... &x\\
.. & .. & ..& ...& ..\\
x & x & x & ... &1
\end{bmatrix}

REPLY [0 votes]: My long answer would be based on the fact that all the pivots of Positive-Definite matrix are always greater than 0
Now when $n=1$, matrix is $ 1 X 1 $, and pivot is $1$
When $n=2$, row reduced echelon form is \begin{bmatrix}
1 & x\\
0 & 1-x^2
\end{bmatrix}
and hence its pivots are $1, (1-x^2)$
When $n=3$, row reduced echelon form is,
\begin{bmatrix}
1 & x & x \\
0 & 1-x^2 & x(1-x) \\
0 & 0 & \frac{(1-x)(1+2x)}{(1+x)}
\end{bmatrix}
and hence its pivots are $1, (1-x^2), \frac{(1-x)(1+2x)}{(1+x)}$
When $n=4$, row reduced echelon form is,
\begin{bmatrix}
1 & x & x & x\\
0 & 1-x^2 & x(1-x) & x(1-x)\\
0 & 0 & \frac{(1-x)(1+2x)}{(1+x)} & \frac{(1-x)x}{(1+x)}\\
0 & 0 & 0 & \frac{(1-x)(1+3x)}{(1+2x)}
\end{bmatrix}
and hence its pivots are $1, (1-x^2),\frac{(1-x)(1+2x)}{(1+x)}, \frac{(1-x)(1+3x)}{(1+2x)}$
Generalizing the results for $n$, pivots are $1, \frac{(1-x)(1+x)}{(1+0x)},\frac{(1-x)(1+2x)}{(1+x)}, \frac{(1-x)(1+3x)}{(1+2x)}, ... ,\frac{(1-x)(1+(n-1)x)}{(1+(n-2)x)}$
For matrix to be positive-definite, all the pivots should be greater than $0$
Clearly first pivot, $(1)$ is $>0$
for second pivot, $(1-X^2) > 0 \implies -1<x<1$
for third pivot, $\frac{(1-x)(1+2x)}{(1+x)} > 0 \implies \frac{-1}{2}<x<1$
for fourth pivot, $\frac{(1-x)(1+3x)}{(1+2x)} > 0 \implies \frac{-1}{3}<x<1$
Generalizing this way, for n pivots, $\frac{-1}{(n-1)}<x<1$


text:
\begin{document}
\begin{frontmatter}
\title{A Characterization of the Two-weight Inequality for Riesz Potentials on Cones of Radially Decreasing Functions}
\author[add1,add2]{Alexander Meskhi}
\ead{alex72meskhi@yahoo.com; meskhi@rmi.ge}
\author[add3]{Ghulam Murtaza}
\ead{gmnizami@googlemail.com}
\author[add4]{Muhammad Sarwar}
\ead{sarwar@uom.edu.pk}

\address[add1]{Department of Mathematical Analysis, A. Razmadze Mathematical Institute, I. Javakhishvili Tbilisi State University, 2. University Str., 0186 Tbilisi, Georgia.}
\address[add2]{Department of Mathematics,  Faculty of Informatics and Control Systems, Georgian Technical University, 77, Kostava St., Tbilisi, Georgia.}
\address[add3]{Department of Mathematics,  GC University, Faisalabad, Pakistan.}
\address[add4]{ Department of Mathematics University of Malakand, Chakdara, Dir(L), Khyber Pakhtunkhwa, Pakistan}

\begin{abstract}
We establish necessary and sufficient conditions on a weight pair $(v,w)$ governing the boundedness of the Riesz potential operator $I_{\alpha}$ defined on a homogeneous group $G$ from $L^p_{dec,r}(w, G)$ to $L^q(v, G)$, where $L^p_{dec,r}(w, G)$ is the Lebesgue space defined for non-negative radially decreasing functions on $G$. The same problem is also studied for the potential operator with product kernels $I_{\alpha_1, \alpha_2}$ defined on a product of two homogeneous groups $G_1\times G_2$. In the latter case weights, in general, are not of product type. The derived results  are new even for Euclidean spaces. To get the main results we use Sawyer type duality theorems (which are also discussed in this paper) and two--weight Hardy type inequalities on $G$ and $G_1\times G_2$ respectively.
\end{abstract}

\begin{keyword}
Riesz potential, multiple Riesz potential, homogeneous group, cone of decreasing functions, two-weight inequality, Sawyer's duality theorem
  \MSC[2010] 42B20, 42B25.
  \end{keyword}
\end{frontmatter}

\newpage

\section{Introduction}
A homogeneous group is a simply connected nilpotent Lie
group G on a Lie algebra g with the one-parameter group of
transformations $\delta_{t}=exp(A\ log\ t)$, $t>0$, where A is a
diagonalized linear operator in $G$ with positive eigenvalues. In
the homogeneous group $G$ the mappings $exp\ o \ \delta_t\ o\
exp^{-1}$, $t>0$, are automorphisms in G, which will be again
denoted by $\delta_{t} $. The number $Q=tr\ A $ is the homogeneous
dimension of $G$. The symbol $e$ will stand for the neutral element in
$G$.
\par
It is possible to equip $G$ with a homogeneous norm $ r:G\rightarrow [
\ 0,\infty)$ which is continuous on $G$, smooth on $G\backslash\{e\}$
and satisfies the conditions:

(i) \ \ \  $r(x)=r(x^{-1})$  for every $x\in \ G$;

(ii)\ \ \ $r(\delta_t x)= tr(x) $ for every $x\in G$ and $t>0$;

(iii)\ \ $ r(x) = 0$ if and only if  $x=e$ ;

(iv) \  There exists $ c_o >0 $ such that
$$  r(xy) \leq c_o (r(x)+r(y)),\;\; x,y  \in  G. $$

In the sequel we denote by $B(a,t)$ an open
ball with the center $a$ and radius
$t>0$, \emph{i.e.}\\
$$ B(a,t ) : = \{ y \in \ G; \ r(ay^{-1}) < t  \}.$$

It can be observed that $ \delta _ t B(e,1) =B(e, t).$

Let us fix a Haar measure $|\cdot|$ in G such that $|B(e,1)|=1$. Then
$|\delta_{t} E|=t^Q|E|$. In particular, $|B(x,t)|=t^Q \hbox { for }x
\ \in \ G , \ t>0$.

Examples of homogeneous groups are: the Euclidean n-dimensional
space $\mathbb{R}^{n}$,  the Heisenberg   group, upper triangular
groups, etc. For the definition and basic properties of the
homogeneous  group we refer to  \cite{FS}, p. 12.

\vskip+0.2cm

\indent An everywhere positive function $\rho$ on $G$ will be called a weight. Denote by $L^p(\rho, G) \ (1<p<\infty)$ the
weighted Lebesgue space, which is the space of all measurable
functions $f:G\rightarrow \mathbb{C}$ defined by the norm
$$\|f\|_{L^p(\rho, G)}=\Big(\int\limits_G
|f(x)|^p \rho(x)dx\Big)^{\frac{1}{p}}<\infty. \nonumber $$ If $\rho\equiv
1$, then    we  we use the notation  $L^p(G).$

Denote by ${\mathcal{DR}}(G)$ the class of all radially decreasing functions on $G$ with values in ${\Bbb{R}}_+$, i.e. the fact that $\phi \in {\mathcal{DR}}(G)$ means that there is decreasing $\bar{\phi}: {\Bbb{R}}_+ \mapsto {\Bbb{R}}_+$ such that $\varphi(x)= \bar{\phi}(r(x))$. In the sequel we will use the symbol $\phi$ itself for $\bar{\phi}$; the fact that $\phi \in {\mathcal{DR}}(G)$ will be written also by the symbol $\varphi \downarrow r$. Let $G_1$ and $G_2$ be homogeneous groups. We say that a function $\psi: G_1\times G_2\mapsto {\Bbb{R}}_+$ is radially decreasing if it is such in each variable separately uniformly to another one. The fact that $\psi$ is radially decreasing on $G_1\times G_2$  will be denoted as $\psi \in {\mathcal{DR}}(G_1\times G_2)$.

Let $$(I_{\alpha}f)(x) = \int\limits_G f(y) \big(r(xy^{-1})\big)^{\alpha-Q} dy, \;\;\;\; 0<\alpha <Q, $$
be the Riesz potential defined on $G$, where $r$ is the homogeneous norm and $dy$ is the normalized Haar measure on $G$. The operator $I_{\alpha}$ plays a fundamental role in harmonic analysis, e.g.,  in the theory of Sobolev embeddings, in the theory of sublaplacians on nilpotent groups etc. Weighted estimates for multiple Riesz potentials can be applied, for example, to establish Sobolev and Poincar\'e inequalities on product spaces (see, e.g., \cite{ToPr}).

Let  $G_1$ and $G_2$ be homogeneous groups with homogeneous norms $r_1$ and $r_2$ and homogeneous dimensions $Q_1$ and $Q_2$ respectively. We define the potential operator on $G_1\times G_2$ as follows
$$ I_{\alpha, \beta} f(x,y) = \iint\limits_{G_1\times G_2} f(t,\tau) \big(r_1(xt^{-1})\big)^{\alpha-Q_1} \big(r_2(y\tau^{-1})\big)^{\beta-Q_2}dtd\tau, \;\;\; (x,y)\in G_1\times G_2, \;\; 0<\alpha<Q_1,\; 0<\beta<Q_2. $$

Our aim is to derive two-weight criteria for $I_{\alpha}$ on the cone of radially decreasing functions on $G$. The same problem is also studied for the potential operator with product kernels $I_{\alpha, \beta}$ defined on a product of two homogeneous groups, where  only the right--hand side weight is of product type. As far as we know the derived results for $I_{\alpha, \beta}$ are new even in the case of  Euclidean spaces. The proofs of the main results are based on E. Sawyer (see \cite{Saw})  type  duality theorem which is also true for homogeneous groups (see Propositions  \ref{Saw-Dua} and  \ref{BHP2} below) and  Hardy type two-weight inequalities in homogeneous groups.
Analogous results for multiple potential  operators defined  on ${\Bbb{R}}^n_+$ with respect to the cone of non-negative decreasing functions on ${\Bbb{R}}^n_+$ were studied in \cite{MMS}, \cite{MM}. It should be emphasized that the two-weight problem for multiple Hardy operator for the cone of decreasing functions on ${\Bbb{R}}^n_+$ was investigated by S. Barza, H. P. Heinig and L. -E. Persson \cite{BHP} under the restriction that both weights are of product type.

Historically the one-weight inequality for the classical Hardy operator was characterize by M. A. Arino and B. Muckenhoupt \cite{AM} under the so called  $B_p$ condition. The same problem for multiple Hardy transform was studied by N. Arcozzi,  S. Barza, J. L.  Garcia-Domingo  and J. Soria \cite{ABGS}. This problem in the the two-weight setting was solved by E. Sawyer \cite{Saw}. Some sufficient conditions guaranteeing  the two--weight inequality for the Riesz potential $I_{\alpha}$ on ${\Bbb{R}}^n$ was given by Y. Rakotondratsimba \cite{Rak}. In particular, the author showed that $I_{\alpha}$ is bounded from $L^p_{dec,r}(w, {\Bbb{R}}^n)$ to $L^q(v, {\Bbb{R}}^n)$ if the weighted  Hardy operators $({\mathcal{H}}f)(x)=\frac{1}{|x|^{n-\alpha}} \int\limits_{|y|<|x|} f(y) dy $ and $({\mathcal{H}}'f)(x)= \int\limits_{|y|>|x|} \frac{f(y)}{|y|^{n-\alpha}} dy$ are bounded from $L^p(w, {\Bbb{R}}^n)$ to $L^q(v,{\Bbb{R}}^n)$. In fact the author studied  the problem on the cone of monotone decreasing functions.

Now we give some comments regarding the notation: in the sequel under the symbol $A\approx B$ we mean that there are positive  constants $c_1$ and $c_2$  (depending on appropriate parameters) such that $ c_1A \leq B \leq c_2 A$;  $A \ll B$ means that there is a positive constant $c$ such that $A \leq  c B$; integral over a product set $E_1\times E_2$ from $g$ will be denoted by $\iint\limits_{E_1\times E_2} g(x,y) dx dy$ or $\int_{E_1} \int_{E_2} g(x,y) dx dy$; for a weight functions $w$ and $w_i$  on $G$, by the symbols $W(t)$ and $W_i(t)$ will be denoted the integrals  $\int\limits_{B(e,t)} w(x)dx $ and $\int\limits_{B(e_i,t)} w_i(x)dx $ respectively; for a weight $w$ on $G_1\times G_2$, we denote  $W(t,\tau):= \int\limits_{B(e_1,t)\times B(e_2,\tau)} w(x,y) dx dy$, where $e_1$ and $e_1$ are neutral elements in $G_1$ and $G_2$ respectively. Finally we mention that constants (often different constants in one and the same lines of inequalities) will be denoted by $c$ or $C$. The symbol $p'$ stands for  the conjugate number of $p$:  $p'=p/(p-1)$, where $1<p<\infty$.

\section{Preliminaries}

We begin this section with the statements regarding polar coordinates in $G$ (see e.g., \cite{FS}, P. 14).
\vskip+0.2cm
\begin{propa}\label{polar}  Let $G$ be a homogeneous group and let
$S=\{x\in G:r(x)=1\}$. There is a (unique) Radon measure $\sigma$ on
$S$ such that for all $u\in L^1(G)$,
$$\int \limits_G u(x)dx=\int\limits_0^\infty \int\limits_S
u(\delta_t\overline y)t^{Q-1}d\sigma(\overline y)dt.$$
\end{propa}

Let $a$ be a positive number. The two--weight inequality for the Hardy-type transforms
$$ (H^{a}f)(x)= \int\limits_{B(e, a r(x))} f(y) dy, \;\;\;\; x\in G, $$

$$ (\widetilde{H}^{a}f(x) = \int\limits_{G\setminus B(e, a r(x))} f(y) dy, \;\;\;\; x\in G, $$
reeds as follows (see \cite{EKM}, Ch.1 for more general case, in particular for quasi-metric measure spaces):

\begin{thma}\label{Hardy-G}  Let $1<p\leq q<\infty$ and let $a$ be a positive number. Then

\rm{(i)}

The operator $H^a$ is bounded from $L^p(u_1, G)$ to $L^q(u_2,G)$ if and only if
$$ \sup_{t>0}  \bigg( \int_{G\setminus B(e,t)} u_2(x) dx \bigg)^{1/q} \bigg( \int\limits_{B(e,at)} u_1^{1-p'}(x) dx \bigg)^{1/p'}<\infty. $$

\rm{(ii)}

The operator $\widetilde{H}^a$ is bounded from $L^p(u_1, G)$ to $L^q(u_2,G)$ if and only if
$$ \sup_{t>0}  \bigg( \int_{B(e,t)} u_2(x) dx \bigg)^{1/q} \bigg( \int\limits_{G\setminus B(e,at)} u_1^{1-p'}(x) dx \bigg)^{1/p'}<\infty. $$

\end{thma}
We refer also to \cite{DHK} for the Hardy inequality written for balls with center at the origin.
\vskip+0.2cm

In the sequel we denote $H^{1}$ by $H$.

\vskip+0.2cm

The following statement for Euclidean spaces was derived by S. Barza, M. Johansson and L. -E. Persson \cite{BJP}.
\begin{propa}\label{Duality} Let $w$ be a weight function on $G$ and let $1<p<\infty$. If $f\in {\mathcal{DR}}(G)$ and $g \geq 0$, then
$$ \sup_{f \downarrow r} \frac{\int\limits_{G} f(x)g(x) dx  }{\Big(\int\limits_G f(x)^p w(x) dx \Big)^{1/p}} \approx \|w\|_{L^1(G)}^{-1/p} \|g\|_{L^1(G)}+ \bigg( \int\limits_{G} H^{p'}(r(x)) W^{-p'}(r(x)) w(x) dx \bigg)^{1/p'}, $$
where $H(t)=\int\limits_{B(e,t)}g(x) dx$, $W(t)=\int\limits_{B(e,t)}w(x) dx$.
\end{propa}

The proof of  Proposition \ref{Duality} repeats the arguments (for ${\Bbb{R}}^n$) used in the proof of Theorem 3.1 of \cite{BJP} taking  Proposition \ref{polar} and the following lemma into account.

\begin{lema} let $1<p<\infty$. For a weight function $w$, the inequality
$$ \int\limits_G w(x) \bigg( \int\limits_{G\setminus B(e, r(x))} f(y) dy\bigg)^p dx \leq p \int\limits_G f^p(x)  W^p(r(x)) w^{1-p}(x) dx , \;\;\;\; f\geq 0,$$
holds.
\end{lema}
{\em Proof} of this lemma is based on  Theorem \ref{Hardy-G} (part (ii)) taking $a=1$, $p=q$, $u_2(x)= v(x)$, $u_1= w^{1-p}(x) W^p(r(x))$ there.  Details are omitted. $\;\;\; \Box$

\vskip+0.2cm

\begin{cora}\label{Duality-1} Let the conditions of Proposition \ref{Duality} be satisfied and let $\int\limits_{G}w(x) dx=\infty$. Then
the following relation holds:
$$ \sup_{f \downarrow r} \frac{\int\limits_{G} f(x)g(x) dx  }{\Big(\int\limits_G f^p(x) w(x)dx \Big)^{1/p}} \approx  \bigg( \int\limits_{G} H^{p'}(r(x)) W(r(x)) w(x) dx \bigg)^{1/p}.$$
\end{cora}

Corollary  \ref{Duality-1} implies the following duality result which follows by the standard way (see \cite{Saw}, \cite{BJP} for details).
\begin{propa}\label{Saw-Dua} Let $1< p,q<\infty$ and let $v,w$ be weight functions on $G$ with $\int\limits_{G} w(x) dx =\infty$. Then the integral operator $T$ defined on functions on $G$ is bounded from $L^p_{dec,r}(w, G)$ to $L^q(v, G)$ if and only if
\begin{equation}
\bigg( \int\limits_{G} \bigg( \int\limits_{B(e, r(x))} (T^*g)(y) dy \bigg)^{p'} W^{-p'}(r(x)) w(x) dx \bigg)^{1/p'} \leq C \bigg( \int\limits_{G} g^{q'}(x) v^{1-q'}(x) dx \bigg)^{1/q'}
\end{equation}
holds for every positive measurable $g$ on $G$.
\end{propa}

The next statement yields the criteria for the two--weight boundedness of the  operator $H$ on the cone ${\mathcal{DR}}(G)$. In particular the following statement is true:

\begin{thma}\label{Hardy-Inequality-G} Let $1<p\leq q <\infty$ and let $v$ and $w$ be weights on $G$ such that $\|w\|_{L^1(G)}=\infty$. Then $H$ is bounded from $L^p_{dec,r}(w,G)$ to $L^q_{v}(v,G)$ if and only if

\rm{(i)}
$$ \sup_{t>0} \bigg( \int\limits_{B(e,t)} w(x) dx \bigg)^{-1/p} \bigg( \int\limits_{B(e,t)} v(x) r^{Qq}(x) dx\bigg)^{1/q}<\infty; $$
\rm{(ii)}
$$ \sup_{t>0} \bigg( \int\limits_{B(e,t)} r^{Qp'}(x) W^{-p'}(r(x)) w(x) dx \bigg)^{1/p'} \bigg( \int\limits_{G\setminus B(e,t)} v(x) dx \bigg)^{1/q} <\infty. $$
\end{thma}

{\em Proof} of this statement follows by the standard way applying Proposition  \ref{Saw-Dua} (see e.g. \cite{Saw}, \cite{BJP}). $\Box$.
\vskip+0.2cm

\begin{defn} Let $\rho$ be a locally integrable a.e.
positive function on $G$. We say that $\rho$ satisfies the  doubling
condition at $e$ ( $\rho \in DC(G)$ )
if there is a positive constant $b>1$ such that for all $t>0$ the following inequality holds:

$$  \int\limits_{B(e, 2t)} \rho(x) dx \leq b    \int\limits_{B(e,t)} \rho(x)  dx.$$

Further, we say that $w\in  DC^{\gamma, p}(G)$, where $1<p<\infty$, $0<\gamma<Q/p$, if there is a positive constant $b$ such that for all $t>0$
$$  \int\limits_{G\setminus B(e, t)} r^{\gamma p'}(x) W^{-p'}(r(x))w(x)  dx \leq b    \int\limits_{G\setminus B(e,2t)} r^{\gamma p'}(x) W^{-p'}(r(x))w(x)  dx.$$
\end{defn}

\begin{rem}\label{Rem} It is also to check that under the assumption  $1<p<\infty$, $0<\gamma<Q/p$ the condition $w\in  DC^{\gamma, p}(G)$ is satisfied for $w\equiv \; const$.
\end{rem}

\begin{defn} We say that a locally integrable a.e. positive function $\rho$ on $G_1\times G_2$ satisfies the  doubling
condition with respect to the second variable ( $\rho \in DC^{(s)}(y)$ )
uniformly to the first one if there is a positive constant $c$ such that for all $t>0$ and
almost every $x\in G_1$ the following inequality holds:

$$\int\limits_{B(e_2, 2t)} \rho(x,y) dy \leq c  \int\limits_{B(e_2,t)} \rho(x,y)  dy.$$

Analogously is defined the class of weights $DC^{(s)}(x)$.
\end{defn}

\section{Riesz Potentials on $G$}

The main result of this section reeds as follows:

\begin{thm}\label{Main-Theorem} Let $1<p\leq q<\infty$ and let $v$ and $w$ be weights such that either $w\in DC^{\alpha, p}(G) $ or $v\in DC(G)$; let $\|w\|_{L^1(G)}=\infty$. Then the operator $I_{\alpha}$ is
bounded from $L^p_{dec,r}(w,G)$ to $L^q(v, G)$ if and only if

\rm{(i)}
\begin{equation}\label{F1}
\sup_{t>0} \bigg(\int\limits_{B(e,t)}  w(x) dx \bigg)^{-1/p}  \bigg(\int\limits_{B(e,t)}   r^{\alpha q}(x) v(x) dx \bigg)^{1/q}< \infty;
\end{equation}

\rm{(ii)}
\begin{equation}\label{F2}
\sup_{t>0} \bigg(\int\limits_{B(e,t)}  r^{p'Q}(x)W^{-p'}(r(x)) w(x)
dx \bigg)^{1/p'} \bigg(\int_{G\setminus B(e, t)}   r^{(\alpha-Q)q}(x) v(x)dx \bigg)^{1/q} <\infty;
\end{equation}

\rm{(iii)}
\begin{equation}\label{F3}
\sup_{t>0} \bigg(\int\limits_{B(e,t)}  v(x) dx \bigg)^{1/q} \bigg(\int_{G\setminus B(e, t)} r^{\alpha p'}(x) W^{-p'}(r(x)) w(x) dx \bigg)^{1/p'} <\infty.
\end{equation}

\end{thm}

To prove this result we need to prove some auxiliary statements.

\begin{lem}\label{estimate}
Let $0<\alpha<Q$ and let $c_o$ be the constant from the triangle inequality of $r$. Then there is a positive constant $c$ depending only on $Q$, $\alpha$ and $c_o$ such that for all $s\in B(e, r(x)/2)$,
\begin{equation}\label{est}
I(x,y):= \int\limits_{B(e, r(x))\setminus B(e, 2c_0 r(y))} r(ty^{-1})^{\alpha-Q} dt \leq c r(xy^{-1})^{\alpha}.
\end{equation}
\end{lem}
\begin{proof}
We have
$$ I(x,y)= \int\limits_0^{\infty} |\{ t\in G: r(ty^{-1})^{\alpha-Q}>\lambda\}\cap B(e, r(x))\setminus B(e, 2c_0 r(y))|d \lambda = \int\limits_{0}^{r(xy^{-1})^{\alpha-Q}} (\cdots) + \int\limits_{r(xy^{-1})^{\alpha-Q}}^{\infty}(\cdots) =: I^{(1)}(x,y) + I^{(2)}(x,y). $$
Observe that, by the triangle inequality for $r$, we have $r^Q(x)\leq c^{Q}_0 2^{Q-1}( r^Q(xy^{-1}) + r^Q(y))$. This implies that $r^Q(x)- (2c_0)^Q r^Q(y) \leq c^{Q}_0 2^{Q-1} r^Q(xy^{-1})$. Hence,
$$ I^{(1)}(x,y) \leq r(xy^{-1})^{\alpha-Q} |B(e, r(x))\setminus B(e, 2c_0 r(y))| =r(xy^{-1})^{\alpha-Q} \Big( r^Q(x)- (2c_0)^Q r^Q(y)\Big) \leq c r(xy^{-1})^{\alpha}.$$
Further, it is easy to see that
$$ I^{(2)}(x,y) \leq c r(xy^{-1})^{\alpha}.$$

Finally we have \eqref{est}.
\end{proof}

Let us introduce the following potential operators
$$ (J_{\alpha}f)(x)= \int\limits_{B(e,2c_0 r(x))}  f(y) r^{\alpha-Q}(xy^{-1}) dy, \;\;\; (S_{\alpha}f)(x)= \int\limits_{G\setminus B(e,2c_0 r(x))}  f(y) r^{\alpha-Q}(xy^{-1}) dy,\;\;\;\ x\in G, \; 0<\alpha<Q. $$

It is easy to see that

\begin{equation}\label{representation}
 I_{\alpha}f = J_{\alpha}f + S_{\alpha}f.
\end{equation}

We need also to introduce the following weighted Hardy operator
$$ (H_{\alpha}f)(x)= r(x)^{\alpha-Q} (Hf)(x). $$

\begin{prop}\label{main1}
 The following relation holds for all $f\in {\mathcal{DR}}(G)$
\begin{equation}
J_{\alpha}f \approx H_{\alpha}f.
\end{equation}
\end{prop}
\begin{proof}
We have

$$ (J_{\alpha}f)(x) = \int\limits_{B(e, r(x)/ 2c_0)} f(y) r^{\alpha-Q}(xy^{-1}) dy  + \int\limits_{B(e, 2c_0 r(x))\setminus B \big(e, r(x)/(2c_0)\big)} f(y) r^{\alpha-Q}(xy^{-1}) dy=: (J^{(1)}_{\alpha}f)(x) + (J^{(2)}_{\alpha}f)(x). $$
If $y\in B(e, r(x)/ 2c_0)$, then $r(x) \leq c_0 (r(xy^{-1}) +r(y)) \leq c_0 r(xy^{-1}) + r(x)/2 $. Hence $r(x) \leq 2 c_0 (r(xy^{-1})$. Consequently,
$$(J^{(1)}_{\alpha}f)(x) \leq c (H_{\alpha}f)(x).$$
Applying now the fact that $f\in DR (G)$ we see that
\begin{eqnarray*} (J^{(2)}_{\alpha}f)(x)&\leq& f(r(x)/2c_0) \int\limits_{B(e, r(x)/ 2c_0)\setminus B(e, 2c_0 r(x))} r^{\alpha-Q}(xy^{-1}) dy \leq c f(r(x)/2c_0) r(x)^{\alpha} \leq c (H_{\alpha}f)(x).
\end{eqnarray*}
\end{proof}

\begin{lem}\label{main2}Let $1<p\leq q<\infty$ and let $v$ and $w$ be weights on $G$ such that $\| w \|_{L^1(G)}=\infty$. Then the operator $S_{\alpha}$ is bounded from $L^p_{dec,r}(w,G)$ to $L^q(v,G)$ if

$$ \sup_{t>0}  \bigg( \int\limits_{G\setminus B(e,t)} r^{\alpha p'}(x) W^{-p'}(r(x))w(x) dx \bigg)^{1/p'} \bigg( \int\limits_{B\big(e,t/(2c_0)\big)}  v(x) dx \bigg)^{1/q}<\infty. $$\
Conversely, if $S_{\alpha}$ is bounded from $L^p_{dec,r}(w,G)$ to $L^q(v,G)$, then the condition
$$ \sup_{t>0}  \bigg( \int_{G\setminus B(e,t)} r^{\alpha p'}(x) W^{-p'}(x)w(x) dx \bigg)^{1/q} \bigg( \int\limits_{B\big(e,t/(4c_0)\big)}  v(x) dx \bigg)^{1/p'}<\infty $$
is satisfied. Furthermore, if either $w\in DC^{\alpha, p}$ or $v\in DC(G)$, then  the operator $S_{\alpha}$ is bounded from $L^p_{dec,r}(w,G)$ to $L^q(v,G)$ if and only if
$$ \sup_{t>0}  \bigg( \int_{G\setminus B(e,t)} r^{\alpha p'}(x) W^{-p'}(r(x))w(x) dx \bigg)^{1/q} \bigg( \int\limits_{B(e,t)} v(x) dx \bigg)^{1/p'}<\infty. $$
\end{lem}
\begin{proof}
Applying Proposition \ref{Saw-Dua}, $S_{\alpha}$ is bounded from $L^p_{dec,r}(w,G)$ to $L^q(v,G)$ if and only if
$$ \bigg( \int\limits_G \bigg( \int\limits_{B(e, r(x))}  (S^*_{\alpha}f)(y) dy\bigg)^{p'} W^{-p'}(r(x)) w(x)dx \bigg)^{1/p'} \leq c\bigg( \int\limits_G g^{q'}(x) v^{1-q'}(x) dx \bigg)^{1/q'}, $$
where
$$ (S^*_{\alpha}f)(x)= \int\limits_{B\big(e, r(x)/(2c_0)\big)}  f(y) r^{\alpha-Q}(xy^{-1}) dy.$$

Now we show that
\begin{equation}\label{two-sided}
c_1 r^{\alpha}(x) \int\limits_{B\big(e, r(x)/ (4c_0)\big)} g(s) ds \leq  \int\limits_{B(e, r(x))}  (S^*_{\alpha}g)(y) dy   \leq c_2 r^{\alpha}(x) \int\limits_{B\big(e, r(x)/ (2c_0)\big)}g(s) ds,\;\;\; g\geq 0.
\end{equation}
To prove the right-hand side estimate in \eqref{two-sided} observe that by Tonelli's theorem and  Lemma \ref{estimate} we have that
\begin{eqnarray*}
\int\limits_{B(e, r(x))}  (S^*_{\alpha}g)(y) dy &=& \int\limits_{B\big(e, r(x)/(2c_0)\big)} f(s)  \bigg( \int\limits_{B\big(e, r(x)\big) \setminus B \big(e, 2c_0 r(s)\big)} r^{\alpha-Q}(sy^{-1}) dy \bigg) ds \\
 &\leq &   c_2 r(x)^{\alpha}  \int\limits_{B(e, r(x)/(2c_0))} f(s) ds .
\end{eqnarray*}
On the other hand,

\begin{eqnarray*}
\int\limits_{B(e, r(x))}  (S^*_{\alpha}g)(y) dy &\geq& c r^{\alpha-Q}(x) \bigg( \int\limits_{B(e, r(x))\setminus B(e, r(x)/2)}  \bigg( \int\limits_{B\big(e, r(y)/(2c_0)\big)} f(s) ds \bigg) dy\bigg)  \\
 &\geq& c_1 r^{\alpha}(x) \bigg( \int\limits_{B\big(e,   r(x)/(4c_0)\big) } f(s) ds   \bigg) .
\end{eqnarray*}
Thus, Theorem \ref{Hardy-G} completes the proof.
\end{proof}

{\em Proof} of Theorem \ref{Main-Theorem}.  By \eqref{representation} it is enough to estimate the terms with
$J_{\alpha}f$ and $S_{\alpha}f. $
By applying  Proposition \ref{main1} and Theorem \ref{Hardy-Inequality-G} we have that $J_{\alpha}$ is bounded from $L^p_{dec,r}(w,G)$ to $L^q(v,G)$ if and only if the conditions (ii) and (iii) are satisfied. Now by Lemma \ref{main2} and the equality (which is a consequence of Proposition \ref{polar})
 $$\bigg( \int\limits_{G\setminus B(e,t)} W(r(x)) w(x) dx\bigg)^{1/p'}= \bigg(\int\limits_{B(e,t)}  w(x) dx \bigg)^{-1/p} $$
 we have that $S_{\alpha}$  is bounded from $L^p_{dec,r}(w,G)$ to $L^q(v,G)$ if and only if (i) is satisfied. $\Box$

\section{Multiple Potentials on $G_1 \times G_2$}

Let us now investigate the two--weight problem for the operator $I_{\alpha, \alpha_2}$ on the cone ${\mathcal{DR}}(G_1\times G_2)$. In the sequel without loss of generality we denote  the triangle inequality constants for $G_1$ and $G_2$ by one and the same symbol $c_0$.

The following statement can be derived just in the same way as Theorem 3.1 was obtained  in \cite{BHP}. The proof is omitted because to avoid repeating  those arguments.
\vskip+0.2cm
\begin{propa}\label{BHP1} Let $1<p<\infty$ and let $w(x,y)= w_1(x)w_2(y)$ be a product weight on $G_1\times G_2$. Then the following relation
$$\sup_{0\leq f \downarrow r}\frac{\iint\limits_{G_1\times G_2} f(x,y)g(x,y) dxdy}{ \bigg(\iint\limits_{G_1\times G_2} f^{p}(x,y) w(x,y)\bigg)^{1/p}} \approx \sum_{i=1}^4 I_k,$$
holds for a non-negative measurable function $g$, where
$$ I_1:= \|w\|_{L^1(G_1\times G_2)}^{-1/p} \|g\|_{L^1(G_1 \times G_2)}, $$

$$  I_2:= \| w_2\|^{-1/p}_{L^1(G_1)} \bigg( \int\limits_{G_1} \int\limits_{B(e_1,r_1(x))} \| g(t,\cdot)\|_{L^1(G_2)}dt \bigg)^{p'} W_1^{-p'}(r_1(x))w_1(x) dx\bigg)^{1/p'}, $$

$$  I_2:= \| w_1\|^{-1/p}_{L^1(G_1)} \bigg( \int\limits_{G_2} \int\limits_{B(e_2,r_2(y))} \| g(\cdot, \tau)\|_{L^1(G_1)}d\tau \bigg)^{p'} W_2^{-p'}(r_2(y))w_2(y) dy\bigg)^{1/p'}, $$

$$ I_4:=\bigg( \int\limits_{G_1\times G_2} \bigg( \int\limits_{G_1\times G_2} g(t,\tau) dt d\tau \bigg)^{p'} W^{-p'}(r_1(x),r_2(y))w(x,y) dx dy\bigg)^{1/p'}.$$
\end{propa}

Applying  Proposition \ref{BHP1}  together with the duality arguments we can get the following statement (cf. \cite{BHP}).
\begin{propa}\label{BHP2}
Let $1<p<\infty$ and let  $v$ and $w$ be weights on $G_1\times G_2$ such that $w(x,y)= w_1(x) w_2(y)$, $\|w\|_{L^1(G_1\times G_2)}=\infty$. Then an integral operator $T$ defined for  functions from  ${\mathcal{DR}}(G_1\times G_2)$ is bounded from $L^p_{dec, r}(w, G_1\times G_2)$ to $L^p(v, G_1\times G_2)$ if and only if for all non-negative measurable $g$ on $G_1 \times G_2$,
$$ \bigg( \iint\limits_{G_1\times G_2} \bigg( \iint\limits_{B(e_1, r_1(x))\times B(e_2, r_2(y))} (T^*g)(t,\tau)dt d\tau\bigg)^{p'} W^{-p'}(x,y)  w(x,y) dxdy\bigg)^{1/p'} \leq C \bigg( \iint\limits_{G_1\times G_2} g^{q'}(x,y) v^{1-q'}(x,y) dx dy\bigg)^{1/q'}.$$
\end{propa}

The next statements deals with the double Hardy--type operators defined on $G_1\times G_2$

$$ (H^{a,b}f)(x,y)= \int\limits_{B(e_1, a r_1(x))} \int\limits_{B(e_2, b r_2(x)) } f(t, \tau) dt d\tau, \;\;\;\; (x,y)\in G_1 \times G_2, $$

$$ (\tilde{H}^{a,b}f)(x,y)= \int\limits_{G_1\setminus B(e_1, a r_1(x))} \int\limits_{G_2\setminus B(e_2, b r_2(x))} f(t, \tau) dt d\tau, \;\;\;\; (x,y)\in G_1 \times G_2, $$

$$
(H_1^{a,b}f)(x,y)= \int\limits_{B(e_1, a r_1(x))}  \int\limits_{G_2 \setminus B(e_2, b r_2(y))}   f(t, \tau) dt d\tau, \;\;\;\; (x,y)\in G_1 \times G_2, $$

$$
(H_2^{a,b}f)(x,y)= \int\limits_{G_1\setminus B(e_1, a r_1 (x))}  \int\limits_{B(e_2, b r_2(y))}   f(t, \tau) dt d\tau, \;\;\;\; (x,y)\in G_1 \times G_2. $$

\begin{prop}\label{Hardy-G_1-G_2}  Let $1<p\leq q<\infty$. Suppose that $v$ and $w$ be weights on $G_1\times G_2$ such that either $w(x,y)= w_1(x) w_2(y)$ or $v(x,y)= v_1(x) v_2(y)$. Then

 \rm{(i)} The operator  $H^{a,b}$ is bounded from $L^p(w, G_1\times G_2)$ to $L^q(v, G_1\times G_2)$ if and only if
$$ A:= \sup_{t>0, \tau>0}  \bigg( \int\limits_{G_1\setminus B(e_1,t)}\int\limits_{G_2\setminus B(e_2,\tau)} v(x,y) dxdy \bigg)^{1/q} \bigg( \int\limits_{B(e_1,at)} \int\limits_{B(e_2,b\tau)} w^{1-p'}(x,y) dxdy \bigg)^{1/p'}<\infty. $$

\rm{(ii)} The operator  $\tilde{H}^{a,b}$ is bounded from $L^p(w, G_1\times G_2)$ to $L^q(v, G_1\times G_2)$ if and only if
$$ \sup_{t>0, \tau>0}  \bigg( \int\limits_{B(e_1,t)} \int\limits_{B(e_2,\tau)} v(x,y) dxdy \bigg)^{1/q} \bigg(  \int\limits_{G_1\setminus B(e_1,at)}\int\limits_{G_2\setminus B(e_2, b \tau)} w^{1-p'}(x,y) dxdy \bigg)^{1/p'}<\infty. $$

\rm{(iii)} The operator  $H_1^{a,b}$ is bounded from $L^p(w, G_1\times G_2)$ to $L^q(v, G_1\times G_2)$ if and only if
$$ \sup_{t>0, \tau>0}  \bigg( \int\limits_{G_1\setminus B(e_1,t)} \int\limits_{B(e_2, \tau)} v(x,y) dxdy \bigg)^{1/q} \bigg(  \int\limits_{B(e_1,at)}\int\limits_{G_2\setminus B(e_2, b \tau)} w^{1-p'}(x,y) dxdy \bigg)^{1/p'}<\infty. $$

\rm{(iv)} The operator  $H_2^{a,b}$ is bounded from $L^p(w, G_1\times G_2)$ to $L^q(v, G_1\times G_2)$ if and only if
$$ \sup_{t>0, \tau>0}  \bigg( \int\limits_{B(e_1,t) } \int\limits_{G_2\setminus B(e_2,\tau )} v(x,y) dxdy \bigg)^{1/q} \bigg(  \int\limits_{G_1\setminus B(e_1,at)}\int\limits_{ B(e_2, b \tau)} w^{1-p'}(x,y) dxdy \bigg)^{1/p'}<\infty. $$
\end{prop}

\begin{proof}  Let $w(x,y)= w_1(x) w_2(y)$. Then the proposition  follows  in the same way as the appropriate statements regarding the Hardy operators defined on ${\Bbb{R}}_+^2$ in \cite{MeJFSA}, \cite{KoMe3} (see also Theorem 1.1.6 of \cite{KMP}). If $v$ is  a product weight, i.e.  $v(x,y)= v_1(x) v_2(y)$, then the result follows from the duality arguments.
We give the proof, for example,  for $H^{a,b}$ in the case when $w(x,y)= w_1(x) w_2(y)$.

First suppose that $S:=\int\limits_{G_2}w^{1-p'}_2(y)dy =\infty$. Let $\{x_k\}_{k=-\infty}^{+\infty}$ be a
sequence of positive numbers for which the equality

\begin{equation}\label{1.1.16}
2^k=\int\limits_{B(e_2, b x_k)} w_2^{1-p'}(y) dy
\end{equation}
holds for all $k\in {\mathbb{Z}}$. This equality  follows because of the continuity in $t$ of the integral over the ball $B(e_2, bt)$.  It is clear that $\{x_k\}$ is increasing
and ${\mathbb{R}}_+ = \cup_{k\in {\mathbb{Z}}}[x_k, x_{k+1})$. Moreover, it is easy to verify that
$$ 2^k = \int\limits_{B(e_2, b x_{k+1})\setminus B(e_2, b x_{k})} w_2^{1-p'}(y)dy. $$  Let $f\geq 0$. We have that
\begin{eqnarray*} && \|H^{a,b}f\|^q_{L^q_{v}(G_1\times G_2)}= \iint\limits_{G_1\times G_2}  v(x,y) \big(H^{a,b} f\big)^q (x,y) dxdy \\
 &\leq& \sum_{k \in {\mathbb{Z}}} \int\limits_{G_1} \int\limits_{B(e_2, x_{k+1})\setminus B(e_2,  x_{k})} v(x,y) \bigg( \iint\limits_{ B\big(e_1, a r_1(x)\big)\times B\big(e_2, b r_2(x)\big)}
f(t,\tau)dt d\tau \bigg)^q dxdy
\\ &\leq& \sum_{k \in {\mathbb{Z}}} \int\limits_{G_1}  \bigg(\int\limits_{B(e_2, x_{k+1})\setminus B(e_2,  x_{k})} v(x,y) dy\bigg)\bigg( \int\limits_{B(e_1, ar_1(x))} \bigg(\int\limits_{B(e_2, bx_{k+1})}
f(t,\tau)d\tau\bigg)dt \bigg)^q dx \\ &=& \sum_{k\in {\mathbb{Z}}} \int\limits_{G_1}  V_k(x)\bigg( \int\limits_{B(e_1, ar_1(x))} F_k(t) dt \bigg)^q dx,
\end{eqnarray*}
where
$$V_k(x):=\int\limits_{B(e_2,  x_{k+1})\setminus B(e_2,  x_{k})} v(x,y) dy;\;\; F_k(t):= \int\limits_{B(e_2,  b x_{k+1})} f(t,\tau)d\tau.$$

It is obvious that
$$ A^q \geq \sup_{\substack{a>0 \\ j\in {\mathbb{Z}}}} \bigg(\int\limits_{G_1\setminus B(e_1, t)}v_j (y) dy\bigg) \bigg(\iint\limits_{B(e_1, at)\times B(e_2, b x_j)} w^{1-p'}(x,y)dxdy\bigg)^{q/p'}. $$
Hence, by Theorem A
$$ \|H^{a,b}f\|^q_{L^q_{v}(G_1 \times G_2)} \leq  c A^q \sum_{j\in {\mathbb{Z}}}\bigg[ \int\limits_{G_1} w_1(x) \bigg(\int\limits_{B(e_2, bx_j)} w_{2}^{1-p'}(y)dy\bigg)^{1-p} (F_k(x))^p dx \bigg]^{q/p} $$
$$ \leq cA^q \bigg[ \int\limits_{G_1} w_1(x) \sum_{j\in Z} \bigg( \int\limits_{B(e_2, b x_j)} w_{2}^{1-p'}(y)dy\bigg)^{1-p} \\  \bigg(\sum_{k=-\infty}^{j} \int\limits_{B(e_2, b x_{k+1}) \setminus B(e_2, b x_{k})}
f(x,\tau)d\tau\bigg)^p dx \bigg]^{q/p}.
$$
 On the other hand,
\eqref{1.1.16} yields that
\begin{eqnarray*} && \sum_{k=n}^{+\infty} \bigg( \int\limits_{B(e_2, b x_k)} w_2^{1-p'}(y) dy \bigg)^{1-p} \bigg(\sum_{k=-\infty}^{n} \int\limits_{B(e_2, b x_{k+1})\setminus B(e_2, b x_{k})}w_2^{1-p'}(y) dy \bigg)^{p-1}\\ & =& \sum_{k=n}^{+\infty} \bigg( \int\limits_{B(e_2, b x_k)} w_2^{1-p'}(y) dy \bigg)^{1-p} \bigg( \int\limits_{B(e_2, b x_{n+1})}w_2^{1-p'}(y) dy \bigg)^{p-1}\!\!\!=
\Big(\sum_{k=n}^{+\infty}2^{k(1-p)}\Big)2^{(n+1)(p-1)}\leq c
\end{eqnarray*}
for all $n\in {\mathbb{Z}}$. Hence by the discrete Hardy inequality (see e.g. \cite{Ben}) and H\"{o}lder's
inequality we have
\begin{eqnarray*} \|H^{a,b}f\|^q_{L^q_{v}(G_1 \times G_2)} &\leq& c A^q \bigg[ \int\limits_{G_1} w_1(x) \sum_{j\in {\mathbb{Z}}} \bigg( \int\limits_{B(e_2, b x_{j+1})\setminus B(e_2, b x_{j})}  w_{2}^{1-p'}(y)dy\bigg)^{1-p} \bigg(\int\limits_{B(e_2, b x_{j+1})\setminus B(e_2, bx_{j})} f(x,\tau)d\tau\bigg)^p dx \bigg]^{q/p} \\ &\leq& c A^q \bigg[ \int\limits_{G_1}  w_1(x) \sum_{j\in {\mathbb{Z}}}  \bigg( \int\limits_{B(e_2, b x_{j+1})\setminus B(e_2, b x_{j})}
w_2(\tau)f^p(x,\tau)d\tau\bigg) dx \bigg]^{q/p}=  c A^q \|f\|_{L^p_w(G_1 \times G_2)}^q.
\end{eqnarray*}

If  $S<\infty$, then without loss of generality we can assume that
$S=1$. In this case we  choose the sequence $\{ x_k \}_{k=-\infty}^{0}$ for which (\ref{1.1.16}) holds for all $k\in {\mathbb{Z}}_-$.
Arguing as in the case $S=\infty$ and using  slight modification of the discrete Hardy inequality (see also \cite{KMP}, Chapter 1 for similar arguments), we finally obtain the desired result.

Finally we notice that the  part (i) can be also proved if we first  establish the  boundedness of the operator $({\mathcal{H}}^{a,b}\varphi)(t,\tau) =\int\limits_0^{at}\int\limits_0^{b\tau} \varphi(s,r) dsdr $ in the spirit of Theorem 1.1.6 in \cite{KMP} and then pass to the case of $G_1 \times G_2$ by Proposition \ref{polar}.
\end{proof}

\vskip+0.2cm

The next statement will be useful for us.

\begin{prop}\label{double-Hardy}  Let $1<p\leq q<\infty$. Assume that $v$ and $w$ are weights on $G_1\times G_2$. Suppose that $w(x,y)=  w_1(x)w_2(y)$ and that $W_i(\infty)=\infty$, $i=1,2$. Then  the operator $H^{1,1}$ is bounded from $L^p_{dec,r}(w, G_1\times G_2)$ to $L^q(v, G_1 \times G_2)$ if and only if the following four conditions are satisfied:

${\rm{(i)}}$
$$ \sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)}  w(x,  y) dx dy \bigg)^{-1/p} \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)}  r_1^{Q_1 q}(x) r_2(y) ^{Q_2 q} v(x,y) dx dy\bigg)^{1/q}< \infty;$$

${\rm{(ii)}}$
$$\sup_{a_1, a_2>0} \bigg( \int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)}   r_1^{Q_1 p'} (x) r_2(y)^{Q_2 p'} W^{-p'}(r_1(x), r_2(y)) w(x,y)
dx dy  \bigg)^{1/p'} \bigg(\int\limits_{G_1\setminus B(e_1, a_1)}  \int\limits_{G_2 \setminus B(e_2, a_2)}  v(x,y)dx dy \bigg)^{1/q} <\infty; $$

${\rm{(iii)}}$
$$\sup_{a_1, a_2>0} \bigg( \int\limits_{B(e_1, a_1)} w_1(r_1(x)) dx\bigg)^{-1/p} \bigg(\int\limits_{B(e_2, a_2)} r_2(y)^{Q_2p'} W_2^{-p'}(r_2(y))w_2(y)dy\bigg)^{1/p'} \bigg(\int\limits_{B(e_1, a_1)} \int\limits_{G_2\setminus B(e_2, a_2)} r_1(x)^{Q_1 q} v(x,y)dx dy \bigg)^{1/q} <\infty; $$

${\rm{(iv)}}$
$$\sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)} r_1(x)^{Q_1p'}W_1^{-p'}(r_1(x))w_1(x)dt_1\bigg)^{1/p'}\bigg( \int\limits_{B(e_2, a_2)} w_2(y)d y\bigg)^{-1/p}  \bigg(\int\limits_{G_1\setminus B(e_1, a_1)} \int\limits_{B(e_2, a_2)}   r_2(y)^{Q_2q} v(x,y)dx dy \bigg)^{1/q} <\infty. $$
\end{prop}

\begin{proof} We follow the proof of Theorem 5.3 in \cite{BHP}. First of all observe that by Proposition \ref{BHP2},  if $w$ is a product
weight, i.e., $w(x_1,  x_2)= w_1(x_1) w_2(x_2)$, such that $W_i(\infty)=\infty$, $i=1, 2$, and $v$ is any weight on
$G_1\times G_2$, then $H^{1,1}$ is bounded from $L^p_{dec,r}(w, G_1)$ to $L^q(v, G_2)$ if and only if

\begin{equation*}\bigg( \iint\limits_{G_1 \times G_2} \bigg( \int\limits_{B(e_1, r_1(x))}    \int\limits_{B(e_2, r_2(x))} \bigg[ \int\limits_{G_1 \setminus B(e_1, r_1(t))}  \int\limits_{G_1 \setminus B(e_2, r_2(\tau))} g(s,  \varepsilon) ds d\varepsilon\bigg] dt d\tau\bigg)^{p'} W^{-p'}(r_1(x), r_2(y)) w(x,y)dx dy \bigg)^{1/p'}
\end{equation*}
\begin{equation}\label{dual}\leq
c \bigg( \iint\limits_{G_1\times G_2} g^{q'}(x,y) v^{1-q'}(x,y) dx dy  \bigg)^{1/q'}, \;\; g\geq 0.
\end{equation}
Further, we have that

\begin{eqnarray*} && \iint\limits_{B(e_1, r_1(x))\times B(e_2, r_2(x))}  \bigg( \int\limits_{G_1 \setminus B(e_1, r_1(t))}  \int\limits_{G_2 \setminus B(e_2, r_2(t))} g(s,\varepsilon) ds d\varepsilon\bigg) dt d\tau \\
 &=& \int\limits_{B(e_1, r_1(x))}    \int\limits_{B(e_2, r_2(x))}  r_1^{Q_1}(t) r_2^{Q_2}(\tau) g(t,\tau) dt d\tau  +  r_1^{Q_1}(x) \int\limits_{G_1 \setminus B(e_1, r_1(x))} \int\limits_{B(e_2, r_2(y))}  r_2^{Q_2}(\tau) g(t, \tau) dt d\tau \\ & +& r_2^{Q_2}(y) \int\limits_{B(e_1, r_1(x))} \int\limits_{G_2 \setminus B(e_2, r_2(y))}  r_1^{Q_1}(t)  g(t, \tau) dt d\tau \\&+&  r_1^{Q_1}(x)r_2^{Q_2}(y) \int\limits_{G_1 \setminus B(e_1, r_1(x))}\int\limits_{G_2 \setminus B(e_2, r_2(y))} g(t, \tau) dt d\tau \\ &=:& I^{(1)}(x,y) + I^{(2)}(x,y)+ I^{(3)}(x,y)+ I^{(4)}(x,y).
 \end{eqnarray*}

It is obvious that \eqref{dual}  holds  if and only if
\begin{equation}\label{j-inequality} \bigg( \iint\limits_{G_1\times G_2} (I^{(j)})^{p'}(x,y) W^{-p'}(r_1(x), r_2(y))w(x,y) dxdy\bigg)^{1/p'}
\leq c \bigg( \iint\limits_{G_1\times G_2} g^{q'}(x,y)  v^{1-q'}(x,y)dx dy \bigg)^{1/q'}
\end{equation}
for $j= 1, 2, 3,4. $ By using  Proposition \ref{Hardy-G_1-G_2} (Part (i)) we find that
$$\Bigg( \iint\limits_{G_1 \times G_2} (I^{(1)})^{p'}(x,y) W^{-p'}(r_1(x), r_2(y))w(x,y) dxdy \Bigg)^{1/p'}\leq c  \Bigg( \iint\limits_{G_1 \times G_2}  g^{q'}(x,y) v^{1-q'}(x,y) dx dy \Bigg)^{1/q'} $$
if and only if

 \begin{eqnarray*} && \bigg( \int\limits_{G_1 \setminus B(e_1, t)} \int\limits_{G_2 \setminus B(e_2, \tau)}  W^{-p'}(r_1(x), r_2(y)) w(x,y)dx dy  \bigg)^{1/p'} \bigg( \iint\limits_{B(e_1, t)\times B(e_2,\tau)} \bigg( \frac{v^{1-q'}(x,y)}{r_1^{Q_1 q'}(x) r_2^{Q_2 q'}(y)}\bigg)^{1-q} dxdy \bigg)^{1/q} \\ &=& c_p \bigg( \iint\limits_{B(e_1, t)\times B(e_2, \tau)} w(x,y) dx dy \bigg)^{-1/p} \bigg( \iint\limits_{B(e_1, t)\times B(e_2, \tau)}  v(x,y) r_1^{Q_1 q}(x)r_2^{Q_2 q}(y) dx dy \bigg)^{1/q} \leq C.
 \end{eqnarray*}
In the latter equality we used the equality

$$  \bigg( \int\limits_{G_i \setminus B(e_i, t)}  W_i^{-p'}\big(r_i(x)\big) w_i(x) dx  \bigg)^{1/p'}=\bigg( \int\limits_{B(e_i, t)} w_i(x) dx \bigg)^{-1/p}, \;\;\ i=1,2,$$
which is direct consequence of integration by parts and Proposition \ref{polar}. Taking now Proposition  \ref{Hardy-G_1-G_2} (Part (ii)) into account  we find that \eqref{j-inequality} holds for $j=4$ if and only if condition (ii) is satisfied, while
Proposition \ref{Hardy-G_1-G_2} (Parts (iii) and (iv)) and the following observation:

\begin{eqnarray*}&& \sup_{a_1, a_2>0} \bigg( \int\limits_{G_1 \setminus B(e_1, a_1)} w_1(x) W_1^{-p'}(r_1(x))dx\bigg)^{1/p'}
\bigg(\int\limits_{B(e_2, a_2)} r_2^{p' Q_2}(y) W_2^{-p'}(r_2(y))w_2(y)dy\bigg)^{1/p'} \\ &\times& \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{G_2 \setminus B(e_2, a_2)} r_1^{Q_1q}(x) v(x,y)dx dy \bigg)^{1/q}\\ &=& c_p \sup_{a_1, a_2>0} \bigg( \int\limits_{B(e_1, a_1)} w_1(x) dx \bigg)^{-1/p} \bigg(\int\limits_{B(e_2, a_2)} r_2^{Q_2p'}(y)  W_2^{-p'}(r_2(y)) w_2(y)dy\bigg)^{1/p'}
\\&\times& \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{G_2\setminus B(e_2, a_2)} r_1^{Qq}(x)  v(x,y)dx dy \bigg)^{1/q} <\infty;
\end{eqnarray*}

\begin{eqnarray*} && \sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)}  r_1^{Q_1p'}(x) W_1^{-p'}(r_1(x))w_1(x)dx \bigg)^{1/p'}\bigg( \int\limits_{G_2 \setminus B(e_2, a_2)} w_2(y)W_2^{-p'}(r_2(y)) dy \bigg)^{1/p'}
\\
&\times & \bigg(\int\limits_{G_1 \setminus B(e_1, a_1)} \int\limits_{B(e_2, a_2)}   r_2^{Q_2 q}(y)  v(x,y)dx dy  \bigg)^{1/q}  \\ &=& c_p\sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)} r_1^{Q_1 p'}(x) W_1^{-p'}(r_1(x))w_1(x)dx \bigg)^{1/p'}
\bigg( \int\limits_{B(e_2, a_2)} w_2(t_2) dt_2\bigg)^{-1/p}   \\
&\times & \bigg(\int\limits_{G_1\setminus B(e_1, a_1)}\int\limits_{B(e_2, a_2)}    r_2^{Q_2 q}(y)  v(x,y)dx dy  \bigg)^{1/q} <\infty
\end{eqnarray*}
yield \eqref{j-inequality} for $j=2,3$.
\end{proof}

Let
$$(J_{\alpha_1,\alpha_2}f)(x,y)= \int\limits_{B \big(e_1, 2 c_0 r_1(x)\big)}\int\limits_{B \big(e_2, 2 c_0 r_2(y)\big)} f(t,\tau) r_1(xt^{-1})^{\alpha_1-Q_1} r_2(y\tau^{-1})^{\alpha_2-Q_2} dt d\tau, $$

$$(J_{\alpha_1}S_{\alpha_2}f)(x,y)= \int\limits_{B \big(e_1, 2 c_0 r_1(x)\big)}\int\limits_{G_2 \setminus B\big(e_2, 2 c_0 r_2(y)\big)} f(t,\tau) r_1(xt^{-1})^{\alpha_1-Q_1} r_2(y\tau^{-1})^{\alpha_2-Q_2} dt d\tau, $$

$$(S_{\alpha_1}J_{\alpha_2}f)(x,y)= \int\limits_{G_1 \setminus B \big(e_1, 2 c_0 r_1(x)\big)}\int\limits_{ B\big(e_2, 2 c_0 r_2(y)\big)} f(t,\tau) r_1(xt^{-1})^{\alpha_1-Q_1} r_2(y\tau^{-1})^{\alpha_2-Q_2} dt d\tau, $$

$$(S_{\alpha_1, \alpha_2}f) (x,y) = \int\limits_{G_1 \setminus B \big(e_1, 2 c_0 r_1(x)\big)}\int\limits_{G_2\setminus B \big(e_2, 2 c_0 r_2(y)\big)} f(t,\tau) r_1(xt^{-1})^{\alpha_1-Q_1} r_2(y\tau^{-1})^{\alpha_2-Q_2} dt d\tau, $$
where $c_0$ is the constant from the triangle inequality for the homogeneous norms $r_1$ and $r_2$.

It is obvious that
\begin{equation}\label{rep}
I_{\alpha_1, \alpha_2}f = J_{\alpha_1,\alpha_2}f+ J_{\alpha_1}S_{\alpha_2}f +S_{\alpha_1}J_{\alpha_2}f +S_{\alpha_1, \alpha_2}f.
\end{equation}

Now we formulate the main result of this section.
\begin{thm}\label{Main-theorem}   Let $1<p\leq q<\infty$. Assume that $v$ and $w$ are weights on $G_1\times G_2$ such that $w(x,y)=  w_1(x)w_2(y)$. Suppose that either $w_i\in DC^{\alpha_i,p}$, $i=1,2$, or $v\in DC(x) \cap DC(y)$.  Then  the operator $I_{\alpha_1, \alpha_2}$ is bounded from $L^p_{dec,r}(w, G_1\times G_2)$ to $L^q(v, G_1 \times G_2)$ if and only if the following  conditions are satisfied:

{\rm{(i)}}

$$ A_1:= \sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)} w(x,y) dx dy\bigg)^{-1/p}  \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)} \Big( r_1^{\alpha_1}(x)  r_2^{\alpha_2}(y) \Big)^qv(x,y) dx dy \bigg)^{1/q}< \infty; $$

{\rm{(ii)}}
$$A_2:= \sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)}  r_1^{Q_1 p'}(x)r_2^{Q_2 p'}(y)W^{-p'}(r_1(x), r_2(y)) w(x,y)
dx dy  \bigg)^{1/p'} $$
$$ \times \bigg(\int\limits_{G_1 \setminus B(e_1, a_1)}  \int\limits_{G_2 \setminus B(e_2, a_2)} \Big( r_1^{\alpha_1 -Q_1}(x) r_2^{\alpha_2 -Q_2}(y)\Big)^qv(x,y )dx dy \bigg)^{1/q} <\infty;$$

{\rm{(iii)}}
$$A_3:= \sup_{a_1, a_2>0} \bigg( \int\limits_{B(e_1, a_1)} w_1(x) dx\bigg)^{-1/p} \bigg(\int\limits_{B(e_2, a_2)} r_2^{Q_2 p'}(y) W_2^{-p'}(r_2(y))w_2(y)dy\bigg)^{1/p'} $$
$$\times \bigg(\int\limits_{B(e_1, a_1)}   \int\limits_{G_2\setminus B(e_2, a_2)} r_1^{\alpha_1 q}(x) r_2^{  q(\alpha_2 -Q_2)}(y) v(x,y)dx dy \bigg)^{1/q} <\infty; $$
\rm{(iv)}
$$A_4:= \sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)}   r_1^{ Q_1 p'} (x) W_1^{-p'}(r_1(x))w_1(x)dx \bigg)^{1/p'}\bigg( \int\limits_{B(e_2, a_2)} w_2(y)dy \bigg)^{-1/p}  $$
$$ \bigg(\int\limits_{G_1\setminus B(e_1, a_1)}\int\limits_{B(e_2, a_2)}   r_1^{q(\alpha_1 -Q_1)}(x) r_2^{q\alpha_2} (y) v(x,y)dx dy \bigg)^{1/q} <\infty. $$

\rm{(v)}
$$ A_5:= \sup_{a_1, a_2>0} \bigg(\int\limits_{ G_1 \setminus B(e_1, a_1)}  \int\limits_{G_2\setminus B(e_2, a_2)}  r_1^{\alpha_1 p'}(x) r_2^{\alpha_2 p'}(y) W^{-p'}(r_1(x), r_2(y)) w(x,y) dx dy\bigg)^{1/p'}  $$
$$ \times \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)}  v(x,y) dx dy \bigg)^{1/q}< \infty; $$

\rm{(vi)}
$$ A_6:=\sup_{a_1, a_2>0} \bigg(\int\limits_{ B(e_1, a_1)}  w_1(x) dx\bigg)^{-1/p} \bigg(\int\limits_{G_2\setminus B(e_2, a_2)}  r_2^{\alpha_2 p'}(y) W_2^{-p'}(r_2(y)) w_2(y) dy\bigg)^{1/p'}  $$
$$\times \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)} r_1^{\alpha_1 q}(x) v(x,y) dx dy \bigg)^{1/q}< \infty; $$

\rm{(vii)}
$$ A_7:= \sup_{a_1, a_2>0} \bigg(\int\limits_{ B(e_1, a_1)}  r_1^{Q_1 p'}(x)  W_1^{-p'}(r_1(x)) w_1(x) dx\bigg)^{1/p'} \bigg( \int\limits_{G_2\setminus B(e_2, a_2)}  r_2^{\alpha_2 p'}(y) W_2^{-p'}(r_2(y)) w_2(y)  dy\bigg)^{1/p'}$$
  $$\times \bigg(\int\limits_{G_1\setminus B(e_1, a_1)}  \int\limits_{B(e_2, a_2)} r_1^{(\alpha_1 -Q_1)q}(x) v(x,y) dx dy \bigg)^{1/q}< \infty; $$

\rm{(viii)}
$$ A_8:= \sup_{a_1, a_2>0} \bigg(\int\limits_{G_2\setminus B(e_1, a_1)} r_1^{\alpha_1 p'}(x) W_1^{-p'}(r_1(x)) w_1(x) dx\bigg)^{-1/p} \bigg( \int\limits_{B(e_2, a_2)} w_2(y)  dy\bigg)^{1/p'}   $$
$$\times \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)} r_2^{\alpha_2 q}(x) v(x,y) dx dy \bigg)^{1/q}< \infty; $$

\rm{(ix)}
$$ A_9:= \sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1) }  r_2^{Q_2 p'}(y)  W_2^{-p'}(r_2(y)) w_2(y) dy\bigg)^{1/p'} \bigg(\int\limits_{G_1 \setminus B(e_1, a_1) }  r_1^{\alpha_1 p'}(x) W_1^{-p'}(r_1(x)) w_1(x)  dx\bigg)^{1/p'} $$
 $$\times \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{G_2 \setminus B(e_2, a_2)} r_2^{(\alpha_2 -Q_2)q}(y) v(x,y) dx dy \bigg)^{1/q}< \infty. $$

\end{thm}

\begin{proof}

Let us assume that $v\in DC(x) \cap DC(y)$. The case when $w_i\in DC^{\alpha_i,p}(G_i)$, $i=1,2$  follows analogously. By using representation \eqref{rep} we have to investigate the boundedness of the operators  $J_{\alpha_1,\alpha_2}f$,  $J_{\alpha_1}S_{\alpha_2}f$, $S_{\alpha_1}J_{\alpha_2}f$, $S_{\alpha_1, \alpha_2}f$ separately.

Since $f\in \mathcal{DR}(G_1\times G_2)$ by using the arguments of the proof of Proposition \ref{main1} it can be checked that
$$  (J_{\alpha_1,\alpha_2}f)(x,y)  \approx r_1^{\alpha_1-Q_1}(x)  r_2^{\alpha_2-Q_2}(y)  \iint\limits_{B(e_1, r_1(x))\times B(e_2, r_2(y))} f(t,\tau) dt d\tau $$
(see also \cite {MMS} for similar estimate in the case of the multiple one-sided potentials on ${\Bbb{R}}_+^2$).
Hence, by Proposition \ref{double-Hardy} we have that  $J_{\alpha_1,\alpha_2}$ is bounded from $L^p_{dec,r}(w, G_1\times G_2)$ to $L^q(v, G_1 \times G_2)$ if and only if conditions (i)- (iv) hold.

Observe that the dual to $S_{\alpha_1, \alpha_2}$ is given by
$$(S^{*}_{\alpha_1, \alpha_2} g)(x,y)=  \iint\limits_{B(e_1, r_1(x)/(2c_0))\times B(e_2, r_2(y)/(2c_0))}g(t,\tau) r_1^{\alpha_1-Q_1}(xt^{-1}) r_2^{\alpha_2-Q_2}(y\tau^{-1})dt d\tau. $$
Further, Tonelli's theorem together with  Lemma \ref{estimate} for both variables implies that there are positive constants $c_1$ and $c_2$ such that for all $(x,y)\in G_1 \times G_2$ for the dual  (see also the proof of Lemma  \ref{main2})

\begin{eqnarray*} &&
r_1^{\alpha_1}(x) r_2^{\alpha_2}(y) \iint\limits_{B(e_1, r_1(x)/(4c_0))\times B(e_2, r_2(y)/(4c_0))}g(t,\tau) dt d\tau \leq c_1 \iint\limits_{B(e_1, r_1(x))\times B(e_2, r_2(y))} \big( S^*_{\alpha_1, \alpha_2} g\big) (t, \tau) dt d\tau \\
&\leq& c_2
r_1^{\alpha_1}(x) r_2^{\alpha_2}(y) \iint\limits_{B(e_1, r_1(x)/(2c_0))\times B(e_2, r_2(y)/(2c_0))}g(t,\tau) dt d\tau.
\end{eqnarray*}
Applying Propositions \ref{Hardy-G_1-G_2} and  \ref{double-Hardy} with the condition that $v\in DC(G_1\times G_2)$ we find that the operator  $S_{\alpha_1, \alpha_2}$ is bounded from $L^p_{dec,r}(w, G_1\times G_2)$ to $L^q(v, G_1 \times G_2)$ if and only if condition (v) is satisfied.

Further, observe that due to the fact that $f$ is radially decreasing with respect to the first variable we have

$$ (J_{\alpha_1} S_{\alpha_2} f)(x,y)\approx ({\mathcal{H}}_{\alpha_1}  S_{\alpha_2} f)(x,y),$$
where
$$({\mathcal{H}}_{\alpha_1}  S_{\alpha_2} f)(x,y) = r_1^{\alpha_1-Q_1}(x) \int\limits_{B\big(e_1, 2 c_0 r_1(x)\big)}\int\limits_{ G_2 \setminus B\big(e_2, 2 c_0 r_2(y)\big)} f(t,\tau)  r_2(y\tau^{-1})^{\alpha_2-Q_2} dt d\tau. $$

Dual of ${\mathcal{H}}_{\alpha_1}  S_{\alpha_2}$ is given by
$$ \big( {\mathcal{H}}^{*}_{\alpha_1}  S^{*}_{\alpha_2} g\big)(t,\tau) =   \int\limits_{G_1\setminus B(e_1, r(t))}
\int\limits_{B(e_2, r(\tau)/2c_0)}r_1^{\alpha_1-Q_1}(s) r_2^{\alpha_2-Q_2}(\varepsilon \tau ^{-1}) f(s,\varepsilon) ds d\varepsilon. $$
Further, we have
 \begin{eqnarray*}
 T(x,y)&:=&  \iint\limits_{B(e_1, r_1(x))\times B(e_2, r_2(y))} ({\mathcal{H}}^{*}_{\alpha_1}  S^{*}_{\alpha_2} g)(t,\tau) dt d\tau \\
 &=& \iint\limits_{B(e_1, r_1(x))\times B(e_2, r_2(y))} \bigg( \int\limits_{B(e_1, r_1(x))\setminus B(e_1, r(t))}
\int\limits_{B(e_2, r(\tau)/2c_0)}r_1^{\alpha_1-Q_1}(s) r_2^{\alpha_2-Q_2}(\tau \varepsilon^{-1}) f(s,\varepsilon) ds d\varepsilon\bigg) dt d\tau \\
&+&  \iint\limits_{B(e_1, r_1(x))\times B(e_2, r_2(y))} \bigg( \int\limits_{G_1\setminus B(e_1, r_1(x))}
\int\limits_{B\big(e_2, r(\tau)/(2c_0)\big)}r_1^{\alpha_1-Q_1}(s) r_2^{\alpha_2-Q_2}(\tau \varepsilon^{-1}) f(s,\varepsilon) ds
d\varepsilon\bigg) dt d\tau \\
&=:&  T_1(x,y)+T_2(x,y).
\end{eqnarray*}
 Tonelli's theorem for $G_1$ , the inequality $r_2^{\alpha_2-Q_2}(\tau \varepsilon^{-1})\geq c r_2^{\alpha_2-Q_2}(y)$ for $\tau\in B(e_2, r(y))$, $\varepsilon \in B\big(e_2, r(\tau)/(2c_0)\big)$, and the fact that the integral $\int\limits_{B(e_1, \tau)} f(s, \varepsilon) ds $ is decreasing in  $\tau$ uniformly  to $\varepsilon$ yield that
\begin{eqnarray*} T_1(x,y) &\geq& c r_2^{\alpha_2-Q_2}(y) \int\limits_{B(e_1, r_1(x))} \int\limits_{B(e_2, r_2(y))\setminus B(e_2, r_2(y)/2)} \bigg( \int\limits_{B(e_1, r_1(x))\setminus B(e_1, r(t))}
\int\limits_{B\big(e_2, r_2(y)/(4c_0)\big)}r_1^{\alpha_1-Q_1}(s)  f(s,\varepsilon) ds d\varepsilon\bigg) dt d\tau \\
&= & c  r_2^{\alpha_2}(y) \int\limits_{B(e_1, r_1(x))}  \bigg( \int\limits_{B(e_1, r_1(x))\setminus B(e_1, r(t))}\bigg( r_1^{\alpha_1-Q_1}(s) \bigg( \int\limits_{B\big(e_2, r_2(y)/(4c_0)\big)}   f(s,\varepsilon)  d\varepsilon\bigg)  ds \bigg) dt \\
&=& c  r_2^{\alpha_2}(y) \int\limits_{B(e_1, r_1(x))}  \bigg( \int\limits_{B(e_1, r_1(x))\setminus B(e_1, r(t))} F(s,y)  ds \bigg) dt = c  r_2^{\alpha_2}(y) \int\limits_{B(e_1, r_1(x))}  F(s,y) \bigg( \int\limits_{ B(e_1, r(s))} dt \bigg)ds \\ &=& c  r_2^{\alpha_2}(y) \int\limits_{B(e_1, r_1(x))}    \int\limits_{B\big(e_2, r_2(y)/(4c_0)\big)}   r_1^{\alpha_1}(s) f(t,\tau)  d\varepsilon  ds .
\end{eqnarray*}
Here we used the notation
$$F(s,y): = \int\limits_{B\big(e_2, r_2(y)/(4c_0)\big)}   f(s,\varepsilon)  d\varepsilon.$$

Taking into account that the function $\int\limits_{B(e_2, 2c_0\lambda)} f(s, \varepsilon) d\varepsilon $ is decreasing in $\lambda$ uniformly to $s$, the inequality  $r_2(\tau\varepsilon^{-1})\leq c r_2(y)$ for $\tau\in B(e_2, r(y))$, $\varepsilon \in B\big(e_2, r(\tau)/(2c_0)\big)$, and Tonelli's theorem for $G_1$ we find that

$$  T_2(x,y) \geq c  r_1^{Q_1}(x)  r_2^{\alpha_2}(y) \int\limits_{G_1\setminus B(e_1, r_1(x))}    \int\limits_{B\big(e_2, r_2(y)/(4c_0)\big)}   r_1^{\alpha_1-Q_1}(s) f(t,\tau)  d\varepsilon  ds . $$

To get the upper estimate, observe that  Tonelli's theorem for $G_1\times G_2$ and Lemma \ref{estimate} for $r_2$ yield that
\begin{eqnarray*}
T_1(x,y) &\leq&  \int\limits_{B(e_1, r_1(x))} \int\limits_{B\big(e_2, r_2(y)/(2c_0)\big)}  r_1^{\alpha_1-Q_1}(s)f(s,\varepsilon)  \bigg( \int\limits_{B(e_1, r_1(s))}
\int\limits_{B\big(e_2, r_2(y)\big)\setminus B\big(e_2, 2c_0 r_2(\varepsilon)\big)}r_2^{\alpha_2-Q_2}(\tau\varepsilon^{-1})   dt d\tau \bigg) ds  d\varepsilon \\
&\leq& c r_2^{\alpha_2}(y)  \iint\limits_{B\big(e_1, r_1(x)\big)\times B\big(e_2, r_2(y)/(2c_0)\big)}  r_1^{\alpha_1}(s) f(s,\varepsilon) ds d\varepsilon.
\end{eqnarray*}

Similarly,
$$ T_2(x,y) \leq  c r_1^{Q_1}(x) r_2^{\alpha_2}(y) \iint\limits_{G_1 \setminus B\big(e_1, r_1(x)\big)\times B\big(e_2, r_2(y)/(2c_0)\big)}  r_1^{\alpha_1-Q_1}(s) f(s,\varepsilon)  ds d\varepsilon. $$

Summarazing these estimates we see that there are positive constants $c_1$ and $c_2$ depending only on $\alpha_1$, $\alpha_2$, $Q_1$ and $Q_2$ such that

\begin{eqnarray*}
&& r_2^{\alpha_2}(y)   \iint\limits_{B\big(e_1, r_1(x)\big)\times B\big(e_2, r_2(y)/(4c_0)\big)}  r_1^{\alpha_1}(s) f(s,\varepsilon) ds d\varepsilon \\
&+& r_1^{Q_1}(x) r_2^{\alpha_2}(y)   \iint\limits_{G_1 \setminus B\big(e_1, r_1(x)\big)\times B\big(e_2, r_1(y)/(4c_0)\big)}  r_1^{\alpha_1-Q_1}(s) f(s,\varepsilon)  ds d\varepsilon.
 \\
 &\leq & c_1 T(x,y) \leq r_2^{\alpha_2}(y)   \iint\limits_{B\big(e_1, r_1(x)\big)\times B\big(e_2, r2(y)/(2c_0)\big)}  r_1^{\alpha_1}(s) f(s,\varepsilon)  ds d\varepsilon \\
 &+& r_1^{Q_1}(x) r_2^{\alpha_2}(y)   \iint\limits_{G_1 \setminus B\big(e_1, r_1(x)\big)\times B\big(e_2, r_1(y)/(2c_0)\big)}  r_1^{\alpha_1-Q_1}(s) f(s,\varepsilon)  ds d\varepsilon.
\end{eqnarray*}

Taking Propositions \ref{Hardy-G_1-G_2} and  \ref{BHP2} into account together with the doubling condition for $v$ with respect to the second variable we see that the operator $J_{\alpha_1} S_{\alpha_2}$ is bounded from $L^p_{dec,r}(w,G_1)$ to $L^q(v,G_2)$ if and only if the conditions (vi) and (vii) are satisfied.

By the similar manner (changing the roles of the first and second variables) we can get that $S_{\alpha_1} J_{\alpha_2}$ is bounded from $L^p_{dec,r}(w,G_1)$ to $L^q(v,G_2)$ if and only if the conditions (viii) and (ix) are satisfied.

 \end{proof}

 Theorem \ref{Main-theorem} and Remark \ref{Rem}   imply  criteria for the trace inequality for $I_{\alpha_1, \alpha_2}$. Namely the following statement holds:

\begin{thm}\label{Main-theorem-1}   Let $1<p\leq q<\infty$ and let $0<\alpha_i<Q_i/p$, $i=1,2$. Then $I_{\alpha_1, \alpha_2}$ is bounded from $L^p_{dec,r}(G_1\times G_2)$ to $L^q(v, G_1 \times G_2)$ if and only if the following  condition holds

$$ B:= \sup_{a_1, a_2>0} \bigg(\int\limits_{B(e_1, a_1)}  \int\limits_{B(e_2, a_2)} v(x,y) dx dy\bigg)^{1/q}  a_1^{\alpha_1-Q_1/p} a_2^{\alpha_2-Q_2/p} < \infty. $$

\end{thm}

\begin{proof}

Sufficiency is a consequence of the inequality $\max \{ A_1, \cdots, A_9\}\leq c B$, while necessity follows immediately by taking the test function $f_{a_1, a_2}(x,y) = \chi_{B(e_1, a_1)}(x) \chi_{B(e_2, a_2)}(y)$, $a_1, a_2>0$.
\end{proof}

\vskip+0.2cm

\section*{Acknowledgements}
The first author is grateful to Professor V. Kokilashvili for drawing his attention to the two-weight problem for multiple Riesz potentials.

The first author was partially supported by  the Shota Rustaveli National
Science Foundation Grant (Contract Numbers:  D/13-23 and 31/47).

config: arxiv


TITLE: What is the first eigenvalue of $p$-Laplacian on unit sphere $S^n$?
QUESTION [11 upvotes]: We know that the first eigenvalue of Laplacian on the Riemannian unit sphere $S^n$ is $n$, then what is the explicit expression for the first eigenvalue of $p$-Laplacian on $S^n$?
The $p$-Laplacian eigevalue equation is
$-div(|\nabla u|^{p-2}\nabla u)=\lambda| u|^{p-2}u$,   for any $p>1$.

REPLY [5 votes]: For the case $n=1$, it is well known that for the domain $\Omega = (a,b)$, the first eigenvalue for the $p$-Laplacian is given by
$$\lambda_1 = (p-1)(\frac{\pi_p}{b-a}),$$
where $$\pi_p = 2\int_0^1 \frac{ds}{\sqrt[p]{1-s^p}}.$$
For higher dimensions (n>1) where $p>1, p \neq 2$, the explicit expression is not known even for simple domains such as the sphere and the square. However, it is known that $\lambda_1$ is positive, simple, and the corresponding eigenfunction does not change sign.
Several numerical algorithms exist for computing the first eigenvalue of the $p$-Laplacian. For an iterative method, see http://arxiv.org/pdf/1011.3172.pdf. For an gradient descent algorithm based on the Rayleigh quotient formulation, see http://arxiv.org/pdf/1106.0602.pdf.


TITLE: Is there a name for the set $E(x, n)=\{x^p \mid p\in \mathbb N \land \space 0\leq p \leq n-1\}$ with $x\in G $, $G$ a group?
QUESTION [0 upvotes]: Source : Reversat & Bigonnet , Algèbre pour la licence ( = Undergraduate abstract algebra).

The authors of the above mentionned  book define the order of element $x\in G$ as the cardinal of  $\langle x\rangle$ (i.e. $\langle \{x\}\rangle $)  , that is, the cardinal of the subgroup generated by the set $\{x\} \subseteq G$ .

Note: So $\langle x\rangle$  is  (by definition)  the smallest subgroup of $G$ that contains $\{x\}$.
In symbols: $$ O(x)= Card (\langle x\rangle).$$

Then they proceed to show that this definition of $O(x)$ is equivalent to the ( maybe more common) one:

$$ O(x)= n  \iff ( x^n = e \land n= \min \{p\in \mathbb N^{*} \mid x^p = e\}) $$
with $e$ as the identity element of $G$.

In order to prove this equivalence, they first define the set $E(x, m)$ as follows:

$$ E(x, n)=\{x^p \space \mid p\in \mathbb N \land \space 0\leq p \leq n-1\}.$$
Example: $E(x,4)= \{x^0, x^1, x^2, x^3 \}$ .

What I'd like to know is whether there is a common name for $E(x, n)$, and whether there is a symbol communly used to denote this set.

REPLY [1 votes]: A common name for $E(x,n)$, considered as a group, is $C_n$, the cyclic group of order $n$. So
$$
C_n=\{e,x, x^2,\ldots ,x^{n-1}\},
$$
where $x^n=x^0=e$.
In the context of your question, $C_n\cong \langle x\rangle$, where $x$ has order $n$, e.g., $x^n=e$, but $x^m\neq e$ for $1\le m<n$.
It is also common to write this by the presentation
$$
C_n=\langle x\mid x^n=e\rangle.
$$
The "C" stands for "cyclic". Another example of a group presentation is, for the dihedral group $D_n$,
$$
D_n=\{r,s\mid r^n=s^2=e, srs=r^{-1}\rangle,
$$
for $n\ge 3$. This is not a cyclic group in general.


TITLE: How prove the integral inequality
QUESTION [2 upvotes]: if $f(x)$ is second-order differentiable and $f(0)=f(1)=f′(1)=0$ $f′(1)=1$ Prove $\int^1_0(f′′(x))^2dx)≥4$
I have some trial but failed

REPLY [2 votes]: For symmetry reasons it does not make a difference whether we assume $$f'(0)=0,\quad f'(1)=1\ ,\tag{1}$$ or the other way around. So I shall assume $(1)$.
Le t $f_0$ be the admissible $f$ for which $$J(f):=\int_0^1\bigl(f''(x)\bigr)^2\ dx$$ is minimal, let $u(\cdot)$ be any function fulfilling
$$u(0)=u(1)=u'(0)=u'(1)=0\ ,\tag{2}$$
and consider the function $f_\epsilon:=f_0+\epsilon u$. Then
$$J(f_\epsilon)=\int_0^1\bigl(f_0''^2+2\epsilon f_0'' u''+ u''^2)\ dx\ .$$
Since $f_0$ is the optimal function we should have
$${d\over d\epsilon}J(f_\epsilon)\biggr|_{\epsilon=0}=0\ .$$
This means $$0=\int_0^1 f_0''\>u''\ dx=\ldots=\int_0^1 f_0^{(4)}(x)u(x)\ dx\ ,$$
and this should be true for all $u$ fulfilling $(2)$. Therefore we conjecture that $f_0$ is a polynomial of degree $\leq3$, and the given boundary conditions then enforce
$$f_0(x)=x^3-x^2\ .\tag{3}$$
In order to prove that this is indeed the optimal function we argue as follows: We can write any $f$ fulfilling the given boundary conditions in the form
$$f(x)=f_0(x)+u(x)$$
with an $u$ satisfying $(2)$ and $f_0$ as in $(3)$. We then obtain
$$\eqalign{\int_0^1 f''^2(x)\ dx&=\int_0^1\bigl((6x-2)^2+(6x-2)u''(x)+u''^2(x)\bigr)\ dx\cr &=\mathstrut\ldots\cr &=\int_0^1\bigl((6x-2)^2+u''^2(x)\bigr)\ dx\geq4\ ,\cr}$$
with equality only if $u''(x)\equiv0$, or $u(x)\equiv0$.


TITLE: variance of the number of fixed points for a permutation group
QUESTION [3 upvotes]: It is reasonably well-known that the variance of the number of fixed points for $S_n$ equals $1.$ Now, what about other transitive permutation groups on $\{1, \dotsc, n\}?$ Presumably much is known. I am particularly curious about which groups have the highest variance...

REPLY [13 votes]: Let $\pi(g)$ be the number of fixed points of a permutation $g$ on $\Omega=\{1,2,\dots,n\}$, and $G$ be a transitive permutation group on $\Omega$. Then, by the orbit counting formula (usually attributed to Burnside, but previously known by Cauchy and Frobenius too), the variance $V(\pi)=E(\pi^2)-E(\pi)^2=E(\pi^2)-1=\frac{1}{|G|}\sum_{g\in G}\pi(g)^2-1$ is $r-1$, where $r$ is the number of orbits of $G$ on $\Omega\times\Omega$. Note that $r$ is also the number of orbits of a point stabilizer $G_1$ on $\Omega$. In particular, $V(\pi)\le n-1$, with equality if and only if $|G|=|\Omega|=n$.


TITLE: Dirichlet to Neumann operator for a nonlocal ODE
QUESTION [3 upvotes]: Consider the following nonlocal ODEs on $[1,\infty)$.
#1)
$$\begin{align}
r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\
f(1) &= \alpha \\
\lim_{r\to \infty} f(r) &= 0
\end{align}$$
#2
$$\begin{align}
r^2 f''(r) + 2rf'(r)-l(l+1) f(r) &= -\frac{(f'(1) + f(1))}{r^2}\\
f'(1) &= \beta \\
\lim_{r\to \infty} f(r) &= 0
\end{align}$$
where $l$ is a positive integer, $\alpha, \beta \in \mathbb{R}$.
Both ODEs are uniquely solvable, and so we can define the Dirichlet to Neumann operator $T: \alpha \mapsto \beta$. I am trying to prove some properties of $T$.
Define the following norm $\lVert \cdot \lVert$:
$$\lVert f \lVert^2 := \int_1^{\infty} r^2 f'(r)^2 dr + l(l+1) \int_1^{\infty} f(r)^2 dr$$
It can be shown that $\lVert f \lVert \leq C |f'(1)|$ for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. It then follows that $|f(1)| \leq |\int_1^{\infty} f'(r) dr| \leq C' \sqrt{\int_1^{\infty}r^2 f'^2} \leq C' \lVert f \lVert \leq C'C|f'(1)|$
And so we have for any $f$ solving the above ode,
$$|f(1)| \leq \bar C |f'(1)|$$
for some $\bar C$ that is independent of $f$ and $l$.
However, the other way around is not true. In fact, it holds that
$$|f'(1)| \leq C \sqrt{l(l+1)}|f(1)|$$
for any $f$ solving the above ODE, where $C$ is independent of $f$ and $l$. I don't know how to prove this inequality. A weaker inequality is the following,
$$\lVert f \lVert \leq C \sqrt{l(l+1)} |f(1)|$$
which I also was not able to prove.
Any help is appreciated.

REPLY [3 votes]: You have a linear ODE with explicit coefficients: the solutions can actually be written down explicitly via variation of constants. To summarize the result1 when $\ell > 1$, let $g_{a,b}(r)$ be given by
$$ g_{a,b}(r) = \frac{a}{r^{\ell+1}} + \frac{b}{r^2} $$
you find that
$$ r^2 g_{a,b}'' + 2r g_{a,b}' - \ell(\ell+1) g_{a,b} = -\frac{(\ell+2)(\ell-1) b}{r^2} $$
As
$$ g_{a,b}(1) = a + b , \qquad g'_{a,b}(1) = -(\ell+1) a - 2b $$
for $g_{a,b}$ to solve the equation you indicated requires $(\ell+2)\ell-1)b = g_{a,b}(1) + g'_{a,b}(1)$. So we need (when $\ell > 1$)
$$   - \ell a - b = (\ell+2)(\ell - 1)b \implies a = \frac{1 - \ell - \ell^2}{\ell} b $$
which yields
$$ \alpha = \frac{1-\ell^2}{\ell} b, \qquad \beta = - \frac{b}{\ell}(1 + 2\ell - 2\ell^2 - \ell^3) $$
so the Dirichlet to Neumann map has norm exactly
$$ \frac{\ell^3 - 2\ell^2 + 2\ell + 1}{\ell^2 - 1} = O(\ell) = O(\sqrt{\ell(\ell+1)})$$

1 The cases where $\ell = 0$ and $1$ requires more care.
When $\ell = 0$, the equation you wrote is only solvable when $b = 0$, and the Dirichlet-to-Neumann map has norm 1.
When $\ell = 1$, variation of constants gives
$$ g_{a,b} = \frac{1}{r^2}(a + b\ln(r)) $$
for which
$$ r^2 g_{a,b}'' + 2r g'_{a,b} - 2 g_{a,b} = -\frac{3b}{r^2} $$
which then requires $a = -2 b$, and hence $\alpha = -2b$ and $\beta = 5b$, and the D2N map has norm $5/2$.


TITLE: What are the odds of drawing the same card $3$ times in a row in a $4$ card deck ( $3$ of the same card and $1$ joker )
QUESTION [1 upvotes]: I made the question simple but there are $2$ things that i'd like to know:
In a deck of $4$ randomly shuffled cards with $3$ aces and $1$ joker, what are the odds of drawing $3$ aces in a row, and what are the odds of the joker being the last of those cards (in the same conditions, so $4$ cards randomly shuffled, $3$ of which are aces and $1$ is a joker).  And if they are any different, why ( with the steps or method for calculating them would be better ).
It's a dumb question probably, but i feel like i am missing something or doing something wrong
Thanks
EDIT:
So to clear some doubts, what i mean by drawing is taking a card out of the deck and not placing it back, so every time i draw i find myself with $1$ less card in the deck i am drawing from.
Also, for the first part of the question, i want to know the odds of drawing $3$ aces in a row and what it takes to calculate that.
The second part, is referring to how many odds i have of having the joker as the last card in the deck and why is that different from saying drawing $3$ aces in a row (If there is any difference). Am i being clear? Sorry if i am not. I will clarify further if needed.

REPLY [1 votes]: Drawing(but not replacing) from a shuffled deck of $4$ cards consisting of $3$ aces and a joker, what is the probability that

The first three are aces.
The last one to be drawn is the joker. Are they different? If yes, why?

\begin{align*}
P(\text{the joker is the last card to be drawn})&=P(\text{three aces are drawn one after the other})\\
&=P(\text{first is an ace})\cdot P(\text{second is also an ace})\cdot P(\text{third is also an ace})\\
&=\frac34\cdot \frac 23\cdot \frac 12=\frac14\\\end{align*}
They are not different because drawing three cards in a row with a certain probability leaves the joker as the last card definitely.


TITLE: Sound waves through a glass filled with liquid
QUESTION [1 upvotes]: I was pondering about the following 'experiment', and was curious about the formulae behind it.
Imagine having a glass filled with a liquid. On one side there is a sensitive receiver that records all the vibrations and is attached to the outside. If I would then tap on the opposite side I would in send an impulse to the glass side, which would transmit the waves to the liquid, which would travel to the opposite side and will be transmitted to the glass and finally the receiver. Of course the glass is higher density so some waves will reach the receiver earlier.
I have some questions here:

I am used to dynamical mechanical systems involving eigenmotions and the likes. This is no different I presume? I am sending an impulse and recording the output, so in fact I have a transfer function. What I am wondering is is what the eigenfrequency of a solid/fluid is when being excited. Will this be the speed of sound through the solid/fluid?
What happens at the interface between the glass and the fluid? I can imagine the wave moving with $V_{glass}$ through the glass, however how is this transmitted to the fluid? What will $V_{fluid}$ be? Will it be equal to $V_{glass}$? I can imagine that the fluid will attenuate the higher velocity movement, because high frequency sound is attenuated by a fluid, but that is sound... and is this sound?

Can someone point me into the right direction?

REPLY [2 votes]: I think the question is too complex to give a satisfactory answer without deep study of the problem.  The biggest problem is that there are several ways how vibrations (sound waves) travel from the source to the receiver.  The path of the excitation: the source - glass - fluid - glass - the receiver you've described is only one one possible path.  Some of the excitation shall travell through glass only, e.g. bottom or side of the glass.
So if you want to have exact solution you should find standing sound wave (eigenfunction) solution of the whole system glass + fluid, which is far from being trivial.  Of course, in this case you must also consider dissipation, because you have constant excitation and without the dissipation the vibrations would continue to grow to infinity.
The second possible, more engineering way to get the answer is to disregard sidepaths of excitations and calculate in terms of traveling sound wave.  At one end you have the source, excitations first travel through glass, then difract on the border between glass and liquid, travel throught liquid, again difract on the border between glass and liquid and finally travel through the glass to the receiver, where all of the wave energy is absorbed.
In both cases you need speed of sound in medium, which is not hard to obtain:
$$v^2 = \frac{E}{\rho} = \frac{K}{\rho},$$
where $\rho$ is density of the medium, while $E$ is Young's modulus (solid) and $K$ is bulk modulus (liquid).  In the second case you also need transmitivity, which is of course dependent on the incident angle and you can find the right equations by peaking at Fresnel equations.
I hope I understood you right and that my answer helps.


TITLE: Find all real numbers $c$ satisfying the following condition: For all $n \in \mathbb{N}$, we have $n^c \in \mathbb{N}$.
QUESTION [6 upvotes]: Find all real numbers $c$ satisfying the following condition:
For all $n \in \mathbb{N}$, we have $n^c \in \mathbb{N}$.
My attempt: Clearly all $c \in \mathbb{N}$ works while negative integer $c$ does not work.
Suppose $c=\frac{p}{q} \in \mathbb{Q}$ such that $n^c \in \mathbb{N}$. Then denote $\{ x \}$ as fractional part of $x$ while denote $\lfloor x \rfloor$ as integer part of $x$. Then we have $n^{\frac{p}{q}}=n^{\lfloor \frac{p}{q} \rfloor} n^{\{ \frac{p}{q} \}} \in \mathbb{N} \Rightarrow n^{\{ \frac{p}{q} \}} \in \mathbb{Q}$, which is a contradiction. So $c$ is not rational number.
I have trouble proving $c$ is not irrational number. Can anyone give some hint?

REPLY [1 votes]: If $0<c<1,$ then $1<2^c<2,$ so $c$ doesn't have the property. Suppose now $1<c<2.$ Define $f(x) = (x+1)^c-x^c.$ Because $x^c$ is strictly convex, $f(x)-f(x-1)> 0$ for $x\ge 1.$ That's because it equals
$$\frac{(x+1)^c - x^c}{(x+1)-x}-\frac{x^c - (x-1)^c}{x-(x-1)},$$
i.e., the difference in the slopes of successive chords on the graph of a strictly convex function. On the other hand, the MVT shows $f(x)-f(x-1) = f'(y_x) = c[(y_x+1)^{c-1}-y_x^{c-1}].$ Because $0<c-1<1,$ this latter difference $\to 0$ as $x\to \infty.$ If $c$ had the property in question, then we have $f(n+1)-f(n)$ is always a positive integer and yet $\to 0,$ contradiction.
Well, that's a start ...


TITLE: Inner product of fractional Sobolev space.
QUESTION [3 upvotes]: I have been reading up on fractional sobolev spaces (Hitchhikers guide to fractional sobolev spaces https://arxiv.org/pdf/1104.4345.pdf).
Let me denote $H^s(\Omega)=W^{s,2}(\Omega)$ where $\Omega\subset\mathbb{R}^n$ is suitably regular. The norm for $H^{\frac{1}{2}}(\Omega)$ is defined as,
$\|u\|_{H^{\frac{1}{2}(\Omega)}}^2=\int_\Omega |u|^2\ dx+\int_\Omega\int_\Omega \frac{|u(x)-u(y)|^2}{|x-y|^{n+1}}\ dxdy$
The second term is similar to Holder continuity. I want to ask what is the inner product for $H^{\frac{1}{2}}(\Omega)$ Is it.
$\langle u,v\rangle_{H^{\frac{1}{2}}(\Omega)}=\int_\Omega uv\ dx+\int_\Omega\int_\Omega \frac{|u(x)-u(y)||v(x)-v(y)|}{|x-y|^{n+1}}\ dxdy$
I believe in the above inner product the second term on the right is correct. Please correct me if I am wrong. Can someone please explain why this is the case intuitively? Thanks for any help.

REPLY [3 votes]: Whenever you have a (real) Hilbert space with norm $\|\cdot\|$, and you're not sure what the inner product is, you can just check the polarization identity:
$$ \langle u,v\rangle = \frac14\left(\|u+v\|^2-\|u-v\|^2\right).$$
This will give
$$\langle u,v\rangle_{H^{\frac{1}{2}}(\Omega)}=\int_\Omega uv\ dx+\int_\Omega\int_\Omega \frac{(u(x)-u(y))\cdot (v(x)-v(y))}{|x-y|^{n+1}}\ dxdy.$$
Another thing, an inner product is supposed to be bilinear in $u,v$; your guess isn't.