text:
\section{Introduction}

The classical Hardy--Littlewood--Sobolev inequality on $\R^n$ states that for any $n\geq 1$, $p,r > 1$, and $\lambda \in (0,n)$ satisfying the so-called balance condition
\begin{equation*}\label{eq-HLS-BC}
1 /p + 1 /r +\lambda /n =2,
\end{equation*}
the following convolution-type inequality holds
\begin{equation}\label{eq-HLS}
\Big |\iint_{\R^n \times \R^n} \frac{f(x)g(y)}{|x-y|^\lambda} dx dy\Big|
\lesssim \| f \| _{L^p(\R^n )} \| g\| _{L^r(\R^n)}
\end{equation}
for any $f\in L^p(\R^n)$ and any $g\in L^r(\R^n)$. The case $n=1$ of \eqref{eq-HLS} was first appeared in a work due to Hardy and Littlewood in 1928; see \cite{hl1928}. Then, in 1938 Sobolev generalized the inequality \eqref{eq-HLS} for arbitrary $n$ in \cite{sobolev1938}. The sharp form of \eqref{eq-HLS} as well as the existence of a pair of the optimal functions was eventually obtained by Lieb in \cite{Lieb} via a symmetric decreasing rearrangement technique; see also \cite{Lions} for a different argument based on concentration-compactness principle.

In 1958, Stein and Weiss proved the so-called doubly weighted Hardy--Littlewood--Sobolev inequality, which generalizes the HLS inequality \eqref{eq-HLS} by inserting two weights $|x|^{-\alpha}$ and $|y|^{-\beta}$ into the integrand $f(x)|x-y|^{-\lambda}g(y)$. To be more precise, it was proved in \cite{SW} that for $n \geq 1$, $p,r>1$,
\begin{equation}\label{eq-SW-L}
0<\lambda <n
\end{equation}
and
\begin{equation}\label{eq-SW-AB}
\alpha < n(p-1)/p, \quad \beta < n (r-1)/r
\end{equation}
with
\begin{equation}\label{eq-SW-A+B}
\alpha + \beta \geq 0,
\end{equation}
the following doubly weighted inequality holds
\begin{equation}\label{eq-SW}
\Big |\iint_{\R^n \times \R^n} \frac{f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dx dy\Big|
\lesssim \| f \| _{L^p(\R^n )} \| g\| _{L^r(\R^n)}
\end{equation}
for any $f\in L^p(\R^n)$ and any $g\in L^r(\R^n)$ under the following conditions
\begin{equation}\label{eq-SW-Key}
1/p+1/r \geq 1
\end{equation}
and the balance condition
\begin{equation}\label{eq-SW-BC}
1 / p + 1/ r + (\lambda +\alpha+\beta)/n =2.
\end{equation}
In the literature, the inequality \eqref{eq-SW} is widely known as the Stein--Weiss inequality. Apparently, the SW inequality \eqref{eq-SW} becomes the classical HLS inequality \eqref{eq-HLS} if $\alpha = \beta =0$. In the original work of Stein and Weiss \cite{SW}, the condition \eqref{eq-SW-Key} is replaced by the following condition
\[
p \leq q := \Big(1 - \frac 1 r \Big)^{-1}.
\]
However, it is easy to verify that the above condition $p \leq q$ is actually equivalent to \eqref{eq-SW-Key}. Compared to the situation of \eqref{eq-HLS}, in the case of \eqref{eq-SW} there are three extra conditions for $\alpha$ and $\beta$: \eqref{eq-SW-AB}, \eqref{eq-SW-A+B}, and \eqref{eq-SW-Key}. It is clear that these three conditions are automatically satisfied if $\alpha=\beta=0$.

Our motivation of writing this note comes from the fact that there was no further information on the conditions \eqref{eq-SW-AB}, \eqref{eq-SW-A+B}, and \eqref{eq-SW-Key} in the work of Stein and Weiss. It is worth noting that in the case $n=1$, Hardy and Littewood showed that the three conditions \eqref{eq-SW-AB}, \eqref{eq-SW-A+B}, and \eqref{eq-SW-Key} are \textit{necessary}. From this difference, a natural question is that \textit{whether these conditions are also necessary}.

In the existing literature, these three conditions are often assumed, without justification, in many works such as \cite{HLZ12} for the case of the Heisenberg group, \cite{Dou2016} for the case of the upper half space, \cite{CLLT} for the case with the fractional Poisson kernel, and \cite{KRS} for the case of homogeneous Lie groups.

Our first observation in this note is the following which indicates the necessity of all conditions appearing in the SW inequality \eqref{eq-SW}.

\begin{theorem}\label{thm-MAIN}
All the conditions \eqref{eq-SW-L}, \eqref{eq-SW-AB}, \eqref{eq-SW-A+B}, \eqref{eq-SW-Key}, and \eqref{eq-SW-BC} are necessary for the Stein--Weiss inequality \eqref{eq-SW}.
\end{theorem}

We prove Theorem \ref{thm-MAIN} in section \ref{sec-Proof} by constructing counter-examples whenever one of these conditions is violated. Clearly, our result now concludes that the SW inequality \eqref{eq-SW} in its full generality. A careful examination of the necessary conditions for \eqref{eq-SW} reveals that the quantity $\alpha+\beta$ cannot be too positive. In fact, the size of $\alpha +\beta$ is implicitly encoded in \eqref{eq-SW-BC} because it follows from \eqref{eq-SW-BC} that
\[
\alpha+\beta \leq n -\lambda,
\]
which cannot be obtained directly from \eqref{eq-SW-AB}. Furthermore, it is clear that we have the following
\[
\text{\eqref{eq-SW-A+B}, \; \eqref{eq-SW-Key}, \; and \; \eqref{eq-SW-BC}} \quad \longrightarrow \quad \text{\eqref{eq-SW-L}}.
\]
Therefore, to conclude the necessity of all conditions for the SW inequality \eqref{eq-SW}, we must not use the other condition as a means.

To further illustrate the idea of constructing counter-examples when proving Theorem \ref{thm-MAIN}, let us consider the doubly weighted Hardy--Littlewood--Sobolev inequality on the upper half space $\R_+^n$ which is defined as follows
$$\R_+^n = \{ (x',x_n) : x_n > 0\}.$$

In 2016, Dou proved the so-called the doubly weighted HLS inequality on $\R^{n-1} \times \R_+^n$; see \cite{Dou2016} and \cite{DZ-IMRN}. To be precise, given $n \geq 2$, $p,r>1$,
\begin{equation}\label{eq-SW-L-u}
0<\lambda <n-1
\end{equation}
and
\begin{equation}\label{eq-SW-AB-u}
\alpha < (n-1)(p-1)/p, \quad \beta < n (r-1)/r
\end{equation}
with
\begin{equation}\label{eq-SW-A+B-u}
\alpha + \beta \geq 0,
\end{equation}
the following inequality holds
\begin{equation}\label{eq-SW-u}
\Big |\iint_{\R_+^n \times \R^{n-1}} \frac{f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dx dy\Big|
\lesssim \| f \| _{L^p(\R^{n-1} )} \| g\| _{L^r(\R_+^n)}
\end{equation}
for any $f\in L^p(\R^{n-1})$ and any $g\in L^r(\R_+^n)$ under the following conditions
\begin{equation}\label{eq-SW-Key-u}
 1 / p + 1 / r \geq 1
\end{equation}
and the balance condition
\begin{equation}\label{eq-SW-BC-u}
\frac{n-1}n \frac 1 p + \frac 1 r + \frac{\lambda +\alpha+\beta + 1}n =2.
\end{equation}
Here we identify $\R^{n-1}$ as the boundary of $\R_+^n$ and the `distance' $|x-y|$ is understood as follows
$$|x-y| = \sqrt{|x-y'|^2 + y_n^2}.$$
Compared to the classical SW inequality \eqref{eq-SW} on $\R^n \times \R^n$, the two conditions \eqref{eq-SW-A+B-u} and \eqref{eq-SW-Key-u} remains the same as those of \eqref{eq-SW}, only the three conditions \eqref{eq-SW-L-u}, \eqref{eq-SW-AB-u}, and \eqref{eq-SW-BC-u} change. However, we \textit{no longer} have the following
\[
\text{\eqref{eq-SW-A+B-u},\; \eqref{eq-SW-Key-u}, \; and \; \eqref{eq-SW-BC-u}} \quad \longrightarrow \quad \text{\eqref{eq-SW-L-u}}
\]
as in the previous case. This suggests that one of the three conditions \eqref{eq-SW-L-u}, \eqref{eq-SW-AB-u}, and \eqref{eq-SW-BC-u} should not be necessary. Based on this intriguing observation, we are forced to further investigate this by seek for the necessity of the above conditions. The following result is what we obtain.

\begin{theorem}\label{thm-MAIN-u}
All the conditions \eqref{eq-SW-L-u} replaced by $0<\lambda<n-1/r$, \eqref{eq-SW-AB-u}, \eqref{eq-SW-A+B-u}, \eqref{eq-SW-Key-u}, and \eqref{eq-SW-BC-u} are necessary for the Stein--Weiss inequality \eqref{eq-SW-u}.
\end{theorem}

We prove Theorem \ref{thm-MAIN-u} in section \ref{sec-Proof-u} below. The new threshold $n-1/r$ for $\lambda$ appears by means of the identity \eqref{eq-SW-BC-u-n1} below, which as far as we know has not been observed before. As can be seen from Theorem \ref{thm-MAIN-u}, we cannot say anything if
$$n-1 \leq \lambda \leq n-1/r.$$
In fact, one could expect by seeing the condition \eqref{eq-SW-L-u} used in \cite{Dou2016} that one still needs $\lambda < n-1$.
From this fundamental difference, it is natural to ask whether or not the SW inequality \eqref{eq-SW-u} on $\R_+^n$ is \textit{still valid} for the following new range of the parameter $\lambda$
$$n-1 \leq \lambda < n-1/r.$$
In other words, if we fix $0<\lambda<n$, this is equivalent to saying that the inequality \eqref{eq-SW-u} remains valid so long as
\[
r > \max \Big\{ 1, \frac 1{n-\lambda} \Big\}.
\]
It turns out that this is indeed the case. In the paper \cite{NNN}, served as the second in the set, we address this issue. To be more precise, we prove in \cite{NNN} the inequality \eqref{eq-SW-m} below, namely we are interested in the validity of the following inequality
\begin{equation}\label{eq-SW-m}
\Big |\iint_{\R^n \times \R^{n-k}} \frac{f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dx dy\Big|
\lesssim \| f \| _{L^p(\R^{n-k} )} \| g\| _{L^r(\R^n)}
\end{equation}
for any $f\in L^p(\R^{n-k})$ and any $g\in L^r(\R^n)$. Here we assume $n > k \geq 0$. Clearly, the above inequality \eqref{eq-SW-m} includes \eqref{eq-SW-u} as a special case if we choose $k=1$.
As far as we know, the inequality \eqref{eq-SW-m} is not yet studied. Although the two inequalities \eqref{eq-SW-u} and \eqref{eq-SW-m} look rather similar, the motivation of working on the inequality \eqref{eq-SW-m} is not just to create an artificial one, but it stems from an intriguing connection between isoperimetric inequalities and HLS inequalities with Poisson-type kernel on $\R^{n-1} \times \R_+^n$; see \cite{HWY08} for the intriguing connection between these inequalities. It is shown in \cite{NNN} that the inequality \eqref{eq-SW-m} provides more information on HLS inequalities with Poisson-type kernel on $\R^{n-1} \times \R_+^n$ and this is the key observation to work on \eqref{eq-SW-m}.

Toward the validity of the inequality \eqref{eq-SW-m}, we aim to address the necessity for the inequality \eqref{eq-SW-m}, leading to the content of the next result of the present paper. Inspired by the set of conditions for \eqref{eq-SW-u}, our set of assumptions for \eqref{eq-SW-m} is as follows. First we let $p,r>1$,
\begin{equation}\label{eq-SW-L-m}
0<\lambda <n-k/r ,
\end{equation}
and
\begin{equation}\label{eq-SW-AB-m}
\alpha < (n-k)(p-1)/p, \quad \beta < n (r-1)/r,
\end{equation}
with
\begin{equation}\label{eq-SW-A+B-m}
\alpha + \beta \geq 0.
\end{equation}
We also assume the following conditions
\begin{equation}\label{eq-SW-Key-m}
 1 / p + 1 / r \geq 1
\end{equation}
and
\begin{equation}\label{eq-SW-BC-m}
\frac{n-k}n \frac 1 p + \frac 1 r + \frac{\lambda +\alpha+\beta + k}n =2.
\end{equation}
Now we identify necessary conditions for the inequality in the same fashion of the previous two inequalities \eqref{eq-SW} and \eqref{eq-SW-u}. We shall prove the following necessity result.

\begin{theorem}\label{thm-MAIN-m}
All the conditions \eqref{eq-SW-L-m}, \eqref{eq-SW-AB-m}, \eqref{eq-SW-A+B-m}, \eqref{eq-SW-Key-m}, and \eqref{eq-SW-BC-m} are necessary for the validity of the Stein--Weiss inequality \eqref{eq-SW-m}.
\end{theorem}

We prove Theorem \ref{thm-MAIN-m} in section \ref{sec-Proof-m} below. As in the case of the upper half space, we obviously do \textit{not} have the following
\[
\text{\eqref{eq-SW-A+B-m},\; \eqref{eq-SW-Key-m}, \; and \; \eqref{eq-SW-BC-m}} \quad \longrightarrow \quad \text{\eqref{eq-SW-L-m}}.
\]
In \cite{NNN}, we investigate the validity of \eqref{eq-SW-m} under the above necessary conditions. To be more precise, instead of proving \eqref{eq-SW-m} in its form, we simply consider \eqref{eq-SW-m} with $\alpha=0$ and the weight $|y|^{-\beta}$ replaced by $|y''|^{-\beta}$, namely
\begin{equation}\label{eq-SW-m-simpler}
\Big |\iint_{\R^n \times \R^{n-k}} \frac{f(x)g(y)}{ |x-y|^\lambda |y''|^\beta} dx dy\Big|
\lesssim \| f \| _{L^p(\R^{n-k} )} \| g\| _{L^r(\R^n)}
\end{equation}
with $y=(y',y'') \in \R^{n-k} \times \R^k$. Clearly, a special case of the inequality \eqref{eq-SW-m-simpler} is the following HLS inequality on $\R^{n-k} \times \R^n$ with $n > k \geq 0$
\begin{equation}\label{eq-HLS-m}
\Big |\iint_{\R^n \times \R^{n-k}} \frac{f(x)g(y)}{ |x-y|^\lambda } dx dy\Big|
\lesssim \| f \| _{L^p(\R^{n-k} )} \| g\| _{L^r(\R^n)}.
\end{equation}
Surprisingly, with the new technique introduced in \cite{NNN}, the validity of \eqref{eq-SW-m} can be quickly derived from \eqref{eq-HLS-m}. This provides a very short proof of the SW inequality \eqref{eq-SW-u} on the upper half space $\R_+^n$. For the classical approach of Stein--Weiss applied to the case of the upper half space $\R_+^n$ with the limitation $0<\lambda<n-1$, we refer the reader to the work \cite{Dou2016}. We expect that the classical approach of Stein--Weiss can also be applied to the case of \eqref{eq-SW-m} giving another the proof for $0<\lambda<n-k$, but we are not so sure if this approach can cover the range $n-k \leq \lambda < n-k/r$. We also expect that the approach in \cite{SDY2014}, which is based on interpolation of Lorentz spaces, can also be applied in this case. Hence, we might have another proof of \eqref{eq-SW-m}.

Before closing this section, we have three comments. First, we can easily show that all the conditions for the Stein--Weiss inequality
\[
\Big |\iint_{\mathbb G \times \mathbb G} \frac{f(x)g(y)}{|x|^\alpha |y^{-1}x|^\lambda |y|^\beta} dx dy\Big|
\lesssim \| f \| _{L^p (\mathbb G) } \| g\| _{L^r (\mathbb G) }
\]
on the product of the homogeneous Lie group $\mathbb G$ are also necessary; see \cite{KRS} for the precise statement of the inequality. Since the argument in the proof of Theorem \ref{thm-MAIN} can be used for the context of homogeneous Lie groups directly without any difficulty, we omit the details.

For the second comment, over the last few years, there have been intensive studies on reverse cases of the classical HLS inequality and well as the classical SW inequality. For example, the authors in \cite{CLLT-TAMS} consider the reverse SW inequality on $\R^n \times \R^n$, following the case on $\R^{n-1} \times \R_+^n$ in \cite{CLT-ANS}. These inequalities can formally be written as follows
\begin{equation}\label{eq-SW-r}
\Big |\iint \frac{f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dx dy\Big|
\gtrsim \| f \| _{L^p } \| g\| _{L^r }
\end{equation}
with $\lambda <0$ and $p, r \in (0,1)$. Accordingly, the above inequality also enjoys the same balance condition as in \eqref{eq-SW-BC} for the case $\R^n \times \R^n$ and \eqref{eq-SW-BC-u} for case $\R^{n-1} \times \R_+^n$. However, unlike the inequalities \eqref{eq-SW} and \eqref{eq-SW-u}, the reverse case \eqref{eq-SW-r} considered in the above mentioned papers does not exhibit similar `necessary' conditions except a suitable balance condition and the natural assumption $\lambda <0$. Using the balance condition and due to the fact that $\lambda<0$, there are foreseen conditions
\[
\alpha + \beta > n [ (p-1)/p + (r-1)/r ]
\]
for the case $\R^n \times \R^n$ and
\[
\alpha + \beta > n (p-1)/p + (n-1)(r-1)/r
\]
for the case $\R^{n-1} \times \R_+^n$. Consequently, to verify the validity of \eqref{eq-SW-r} it is natural to assume that at least one of the following two conditions
\[
\alpha >n(p-1)/p, \quad \beta > n(r-1)/r
\]
for the case $\R^n \times \R^n$ and
\[
\alpha > n(p-1)/p , \quad \beta > (n-1)(r-1)/r
\]
for the case $\R^{n-1} \times \R_+^n$ occurs.

However, we can say more about the two sets of conditions above. Following the argument proving the necessity of the conditions \eqref{eq-SW-AB} and \eqref{eq-SW-AB-u} established in Theorems \ref{thm-MAIN} and \ref{thm-MAIN-u} above, it is quite clear that if at least one of the two conditions for $\alpha$ and $\beta$ for each case does not not occur, the left hand side of \eqref{eq-SW-r} diverges, leading to the triviality of \eqref{eq-SW-r}. This sheds light on why the authors in \cite[Theorem 1]{CLLT-TAMS} can assume that the following conditions
\[
0 \geq \alpha > n(p-1)/p, \quad 0 \geq \beta > n(r-1)/r.
\]
However, we suspect that the negativity of both $\alpha$ and $\beta$ seems to be not necessary.

As the final remark before closing this section, we would like to note that during the preparation of this work, it has come to our attention that the necessity of all conditions for the SW inequality \eqref{eq-SW} was already studied in \cite{SDY2014}. Nevertheless and very fortunate, the argument and examples demonstrated in \eqref{eq-SW} are essentially different from ours. We also expect that our finding in this note could be useful and applicable for other cases.

This note is organized as follows:

\renewcommand{\baselinestretch}{0.75}\normalsize
\tableofcontents
\renewcommand{\baselinestretch}{1.0}\normalsize

From now on, by $B_r^\ell$ we mean the open ball in $\R^\ell$ centered at the origin with radius $r$. Since we are also interested in the upper half space $\R_+^n$, for simplicity by $B_r^n$ we also mean $\R_+^n \cap B_r^n$ if no confusion occurs.

\section{Necessity of conditions for (\ref{eq-SW}): proof of Theorem \ref{thm-MAIN}}
\label{sec-Proof}

In this section, we give a proof of Theorem \ref{thm-MAIN}. As mentioned earlier, the result is already known, thanks to \cite{SDY2014}. Hence, we aim to provide new examples to support the finding, especially to be able to consider the case of the upper half space $\R^{n-1} \times \R_+^n$ as well as $\R^{n-k} \times \R^n$.

Compared to the examples constructed in \cite{SDY2014}, our main contribution in this section is a new example constructed in subsection \ref{subsec-A+B} below, which is inspired by a similar construction due to Hardy and Littewood.

\subsection{The necessity of (\ref{eq-SW-L}), (\ref{eq-SW-AB}), and (\ref{eq-SW-BC})}
\label{subsec-L}

\subsubsection{The necessity of $0<\lambda<n$}

We start with the necessity of \eqref{eq-SW-L}, which is quite obvious and well-known. Indeed, first we rule out the case $\lambda \leq 0$. Clearly for any each fixed $x$, near the infinity the quantity $|x-y|^{-\lambda}$ is as large as we want if $\lambda < 0$. Hence \eqref{eq-SW} cannot hold if $\lambda < 0$. The case $\lambda = 0$ cannot occur because
\[
\iint_{\R^n \times \R^n} \frac{dxdy}{|x|^\alpha |y|^\beta} > \iint_{\{ 1 \leq |x| \leq 2\} \times \R^n} \frac{dxdy}{|x|^\alpha |y|^\beta} = +\infty
\]
regardless of $\beta$. It is worth noticing that the case $\lambda = 0$ is often called the limiting case and in this scenario, we expect that there is a log-SW inequality in the same fashion of the log-HLS inequality; see \cite[Corollary 5.3]{DZ-IMRN}. Clearly, by seeing the necessity of the conditions \eqref{eq-SW-A+B} and \eqref{eq-SW-Key}, we must have $\lambda \leq n$. However, we still need to rule out the case $\lambda = n$. To rule out the case $\lambda \geq n$, we simply choose
\[
f = g = \chi_{B_6 \setminus B_1}.
\]
and show that the two terms $|x|^{-\alpha}$ and $|y|^{-\beta}$ are negligible. Then, by Tonelli's theorem, it is easy to see that
\begin{align*}
+\infty
&> \iint_{\R^n \times \R^n} \frac{\chi_{B_6 \setminus B_1} (x) \chi_{B_6 \setminus B_1} (y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dx dy \\
&\gtrsim \int_{B_4 \setminus B_2} \Big( \int_{B_{|y|/2}(y)} \frac{dx}{|x-y|^\lambda} \Big) dy \\
\overset{|y| \geq 2} &\geq \int_{B_4 \setminus B_2} \Big( \int_{B_1 (y)} \frac{dx}{|x-y|^\lambda} \Big) dy
= +\infty.
\end{align*}
This completes the proof of the necessity of $\lambda < n$.

\begin{remark}\label{rmk-SW-L}
In the above argument, if we use the dual version of \eqref{eq-SW-u}, namely
\begin{equation}\label{eq-Duality-f}
\| f \|_{L^p(\R^n)}^q \geq \int_{\R^n} \Big( \int_{\R^n} \frac{f(x)}{ |x|^\alpha |x-y|^\lambda |y|^\beta } dx \Big)^q dy
\end{equation}
with $q = r/(r-1)$, then the above argument still works. Indeed, we easily obtain
\begin{align*}
\int_{\R^n} \Big( \int_{\R^n} \frac{ \chi_{B_6 \setminus B_1} (x)}{ |x|^\alpha |x-y|^\lambda |y|^\beta } dx \Big)^q dy
&\gtrsim \int_{B_4 \setminus B_2} \Big( \int_{B_{|y|/2}(y)} \frac{dx}{|x|^\alpha |x-y|^\lambda} \Big)^q dy \\
&\gtrsim \int_{B_4 \setminus B_2} \Big( \int_{B_1 (y)} \frac{dx}{|x-y|^\lambda} \Big)^q dy
= +\infty
\end{align*}
if $\lambda \geq n$.
\end{remark}

\subsubsection{The necessity of $\beta < n(r-1)/r$}
\label{subsubsec-beta<}

First, we prove the necessity of the condition $\beta < n(r-1)/r$. By way of contradiction, we assume
$$\beta \geq n(r-1)/r=n/q.$$
We shall obtain contradiction from \eqref{eq-Duality-f}. Then we choose $f = \chi_{B_3 \setminus B_2}$ and show that the two terms $|x|^{-\alpha}$ and $|x-y|^{-\lambda}$ in \eqref{eq-Duality-f} are negligible. Indeed, we can estimate
\begin{align*}
\int_{\R^n} \Big( \int_{\R^n} \frac{\chi_{B_3 \setminus B_2 }(x)}{ |x|^\alpha |x-y|^\lambda |y|^\beta } dx \Big)^q dy
&\geq \int_{B_1 }
\Big[ \int_{B_3 \setminus B_2} \frac {dx }{ |x|^\alpha |x-y|^\lambda} \Big]^q
\frac{dy}{|y|^{\beta q}}
= +\infty.
\end{align*}
This violates \eqref{eq-Duality-f}, hence concluding the necessity of $\beta < n(r-1)/r$. Notice that in the above estimate, we also need the uniformly lower bound of $\int_{B_3 \setminus B_2 } |x|^{-\alpha} |x-y|^{-\lambda}dx$ but this is quite obvious because $2 \leq |x| \leq 3$ and $$1 \leq |x|-|y| \leq |x-y| \leq |x| + |y| \leq 4.$$

\subsubsection{The necessity of $\alpha < n(p-1)/p$}
\label{subsubsec-alpha<}

Now we establish the necessity of the condition $\alpha < n(p-1)/p$. In this case, we use the following equivalent form of \eqref{eq-SW}
\begin{equation}\label{eq-Duality-g}
\| g \|_{L^p(\R^n)}^q \geq \int_{\R^n} \Big( \int_{\R^n} \frac{g(y)}{ |x|^\alpha |x-y|^\lambda |y|^\beta } dy \Big)^q dx
\end{equation}
with $q = p/(p-1)$. By contradiction, we assume
$$\alpha \geq n(p-1)/p=n/q.$$
Then we choose $g = \chi_{B_3 \setminus B_2}$ and show that the two terms $|x-y|^{-\lambda}$ and $|y|^{-\beta}$ are negligible. Indeed, we can estimate
\begin{align*}
\int_{\R^n} \Big( \int_{\R^n} \frac{\chi_{B_3 \setminus B_2}(y) }{ |x|^\alpha |x-y|^\lambda |y|^\beta } dy \Big)^q dx
&\geq \int_{B_1 }
\Big[ \int_{B_3 \setminus B_2} \frac {dy }{ |x-y|^\lambda |y|^\beta } \Big]^q
\frac{dx}{|x|^{\alpha q}} = +\infty.
\end{align*}
This violates \eqref{eq-Duality-g}, hence concluding the necessity of $\alpha < n(p-1)/p$.

\subsubsection{The necessity of $1/p + 1/r + (\lambda + \alpha + \beta)/n =2$}
\label{subsubsec-=}

This part is well-known due to the scaling invariance of the inequality. Since its argument is short, we include its proof for completeness. Fix two positive functions $f$ and $g$ such that $\|f\|_{L^p(\R )}<+\infty$ and $\|g\|_{L^r(\R^n)}<+\infty$. Denote
\begin{equation}\label{eq-FeGe}
f_\epsilon (x) = \epsilon^{-\frac np} f \big( \frac x \epsilon \big),
\quad
g_\epsilon (y) = \epsilon^{-\frac nr} g \big( \frac y \epsilon \big).
\end{equation}
Obviously,
\[
\|f_\epsilon\|_{L^p(\R )} = \|f\|_{L^p(\R )}<+\infty, \quad
\|g_\epsilon\|_{L^r(\R^n)} = \|g\|_{L^r(\R^n)}<+\infty.
\]
A simple change of variables gives
\begin{align*}
\iint_{\R^n \times \R^n } \frac{f_\epsilon (x) g_\epsilon (y) dx dy}{|x|^\alpha |x-y|^\lambda |y|^\beta }
&=\iint_{\R^n \times \R^n } \frac{f (\frac x \epsilon ) g (\frac y \epsilon ) dx dy }{\epsilon^{\frac np + \frac nr + \lambda + \alpha + \beta} |\frac x\epsilon|^\alpha |\frac x\epsilon - \frac y \epsilon |^\lambda |\frac{y}\epsilon|^\beta } \\
&=\big( \frac 1 \epsilon \big)^{( \frac 1p + \frac 1r + \frac{\lambda + \alpha + \beta}n - 2 ) n}\iint_{\R^n \times \R^n } \frac{f (x ) g (y ) dx dy}{|x|^\alpha |x -y |^\lambda |y |^\beta } .
\end{align*}
Hence, if the balance condition \eqref{eq-SW-BC} does not hold, we simply send $\epsilon$ to zero or to infinity to get a contradiction.

\begin{remark}
J. Dou kindly informs us that the necessity of \eqref{eq-SW-BC} can also be derived from a suitable Pohozaev-type identity. We leave the detail for the interested readers.
\end{remark}

\subsection{The necessity of (\ref{eq-SW-A+B})}
\label{subsec-A+B}

To verify the necessity of $\alpha + \beta \geq 0$ for the SW inequality on $\R^n$, we construct an example of non-negative functions $f$ and $g$ in such a way that $\|f\|_{L^p(\R^n)} <+\infty$, $\|g\|_{L^r(\R^n)} <+\infty$, but
\[
\iint_{\R^n \times \R^n} \frac{f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dx dy = +\infty.
\]
Our example is inspired by a similar example in 1-dimensional case due to Hardy and Littewood; see \cite[page 580]{hl1928}. Roughly speaking, we construct an example in such a way that the following two properties are satisfied:
\begin{itemize}
 \item $|x-y|^{-\lambda}$ is bounded from below away from zero and
 \item $|x|$ and $|y|$ are compatible (near infinity).
\end{itemize}
For a general point $x \in \R^n$, we often write $x=(x',x_n) \in \R^{n-1} \times \R$. Now fix any small $\epsilon > 0$ and define
\[
f(x)=
\left\{
\begin{aligned}
& (\log_2 x_n)^{-\frac 1p - \epsilon} &&\text{if } \; 2^m \leq x_n \leq 2^m+1, \; m \geq 1,\\
& & & \text{and } \; |x'| \leq 1,\\
&0 &&\text{otherwise},
\end{aligned}
\right.
\]
and
\[
g(y)=
\left\{
\begin{aligned}
& (\log_2 y_n)^{-\frac 1r - \epsilon} &&\text{if } \; 2^m \leq y_n \leq 2^m+1, \; m \geq 1, \\
& & & \text{and } \; |y'| \leq 1,\\
&0 &&\text{otherwise}.
\end{aligned}
\right.
\]
Intuitively, the support of $f$ and $g$ consists of \textit{disjoint cylinders} concentrated along the half axis $x_n$. The condition $|x'| \leq 1$ is used to make the functions $f$ and $g$ essentially depending only on the last coordinate. Observe that
\begin{align*}
\int_{\R^n} f(x)^p dx &=\sum_{m \geq 1} \int_{2^m}^{2^m+1}\Big( \int_{|x'| \leq 1} \frac 1{(\log_2 x_n)^{1+ p\epsilon}} dx' \Big) dx_n\\
&\leq |B_1^{n-1} | \sum_{m \geq 1} \frac{1}{m^{1+ p\epsilon}}
<+\infty.
\end{align*}
Hence $f \in L^p(\R^n)$. Similarly, we also obtain $g \in L^r(\R^n)$. Now we estimate the double integral as follows
\begin{equation}\label{eq-Double-1}
\begin{aligned}
\iint_{\R^n \times \R^n} & \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy \\
&\geq\sum_{m \geq 1} \iint_{B_1^{n-1} \times [2^m, 2^m+1]} \Big( \iint_{B_1^{n-1} \times [2^m, 2^m+1]} \frac {g(y) dy}{ |x-y|^\lambda |y|^\beta} \Big) \frac{f(x) dx}{|x|^\alpha}.
\end{aligned}
\end{equation}
For each $m \geq 1$, let $x$ and $y$ be such that
\begin{equation}\label{eq-Strip}
x,y \in \overline{B_1^{n-1}} \times [2^m, 2^m+1].
\end{equation}
We need to estimate $|x|^{-\alpha}$, $|y|^{-\beta}$, and $|x-y|^{-\lambda}$ from below. Apparently, we easily bound $|x-y|$ from the above as follows
\[
|x-y|^2 =|x'-y'|^2 + |x_n-y_n|^2 \leq 2(|x'|^2+|y'|^2) + 1 \leq 5.
\]
This together with $\lambda > 0$ allows us to estimate
\[
|x-y|^{-\lambda} \geq 3^{-\lambda}
\]
provided \eqref{eq-Strip} holds for each $m \geq 1$. Since $|x'| \leq 1$ and $x_n \geq 2$, it is easy to verify that
\[
x_n \leq \sqrt{|x'|^2+x_n^2} \leq 2x_n.
\]
Similarly, we get
\[
y_n \leq \sqrt{|y'|^2+y_n^2} \leq 2y_n.
\]
Hence, regardless of the sign of $\alpha$ and $\beta$, we always have
\[
|x|^{-\alpha} \gtrsim x_n^{-\alpha} \quad \text{and} \quad
|y|^{-\beta} \gtrsim y_n^{-\beta}.
\]
Putting the above estimates together, we can further estimate \eqref{eq-Double-1} as follows
\begin{equation}\label{eq-Double-2}
\begin{aligned}
\iint_{\R^n \times \R^n} & \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy\\
& \gtrsim \sum_{m \geq 1} \Big( \iint_{B_1^{n-1} \times [2^m, 2^m+1]} \frac{f(x) dx}{|x|^\alpha} \Big) \Big( \iint_{B_1^{n-1} \times [2^m, 2^m+1]} \frac {g(y) dy}{ |y|^\beta} \Big) \\
& = |B_1^{n-1}|^2 \sum_{m \geq 1} \Big( \int_{2^m}^{2^m+1} \frac {dx_n}{x_n^\alpha (\log_2 x_n)^{\frac 1p +\epsilon}} \Big) \Big( \int_{2^m}^{2^m+1} \frac {dy_n}{y_n^\beta (\log_2 y_n)^{\frac 1r +\epsilon}} \Big) \\
& \gtrsim\sum_{m \geq 1} \frac 1{(m+1)^{\frac 1p +\frac 1r +2\epsilon}}
\Big( \int_{2^m}^{2^m+1} \frac {dx_n}{x_n^\alpha } \Big) \Big( \int_{2^m}^{2^m+1} \frac {dy_n}{y_n^\beta } \Big).
\end{aligned}
\end{equation}
Notice that for $2^m \leq x_n \leq 2^m+1$ we always have
\[
\frac 1{x_n^\alpha} \geq
\left\{
\begin{aligned}
& \frac 1{2^\alpha} \frac 1{(2^m)^\alpha} &&\text{if } \; \alpha \geq 0,\\
& \frac 1{(2^m)^\alpha} &&\text{if } \; \alpha < 0.
\end{aligned}
\right.
\]
Similarly, we also obtain $y_n^{-\beta} \gtrsim (2^m)^{-\beta}$ provided $2^m \leq y_n \leq 2^m+1$. Hence
\[
\Big( \int_{2^m}^{2^m+1} \frac {dx_n}{x_n^\alpha } \Big) \Big( \int_{2^m}^{2^m+1} \frac {dy_n}{y_n^\alpha } \Big)
\geq \frac 1{(2^m)^{\alpha + \beta}}.
\]
Thus, recalling \eqref{eq-Double-2} we arrive at
\begin{align*}
\iint_{\R^n \times \R^n} & \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy
\gtrsim \sum_{m \geq 1} \frac 1{(2^m)^{\alpha + \beta}} \frac 1{(m+1)^{\frac 1p +\frac 1r + 2\epsilon}} .
\end{align*}
Obviously, one can easily see that the above sum diverges if $\alpha +\beta < 0$. Hence, the necessity of \eqref{eq-SW-AB} is proved.

\subsection{The necessity of (\ref{eq-SW-Key})}
\label{subsec-P+R}

Now we establish the necessity of the condition \eqref{eq-SW-Key}. Indeed, suppose $1/p+1/r<1$. Then
\[
q := \Big(1- \frac 1r \Big)^{-1} < p.
\]
(Note that this is also equivalent to $\lambda+\alpha+\beta>n$.) In view of \eqref{eq-Duality-f}, it suffices to show that the left hand side of \eqref{eq-Duality-f} becomes infinity for suitable $f \in L^p(\R^n)$. By modifying the counter-example constructed in subsection \ref{subsec-A+B} above, we now choose
\[
f(x)=
\left\{
\begin{aligned}
& |x|^{-\frac np} (\log |x|)^{-\frac 1 q} & & \text{if } |x| \geq 2,\\
& 0 & & \text{otherwise}.
\end{aligned}
\right.
\]
It is not hard to see that
\[
\int_{\R^n} f(x)^p dx = |\mathbb S^{n-1}| \int_2^{+\infty} \frac{d\rho}{\rho (\log \rho)^{p/q}} < +\infty,
\]
thanks to $p>q$. Hence, $f \in L^p(\R^n)$. To obtain a contradiction, we estimate the double integral near the region $\{|x-y| = 0\}$; see \cite[page 580]{hl1928}. By \eqref{eq-SW} we have
\begin{align*}
+\infty > \|f\|_{L^p(\R^n)}^q &> \int_{\R^n } \Big( \int_{\R^n} \frac{ f(x) }{|x|^\alpha |x-y|^\lambda |y|^\beta} dx \Big)^q dy\\
&\geq
\int_{|y| \geq 4} \Big( \int_{|y|/2 \leq |x| \leq 2|y|} \frac{dx}{|x|^{\alpha + \frac np} (\log |x|)^{ \frac 1 q} |x-y|^\lambda } \Big)^q \frac {dy}{|y|^{\beta q}}.
\end{align*}
As $\lambda > 0$ and $|x| \leq 2|y|$, we have $|x-y|^{\lambda} \leq 3^\lambda |y|^\lambda$, which then yields
\[
|x|^{\alpha + \frac np} |x-y|^{\lambda} \leq 2^{\alpha + \frac np} 3^\lambda |y|^{\lambda + \alpha + \frac np}
\]
for any $|x|/2 \leq |y| \leq 2|x|$. (We need the extra condition $|y| \geq |x|/2$ because $\alpha + n/p$ could be negative.) Hence, together with
\[
0< \log |x| \leq \log (2|y|) \leq \log 2 + \log |y| \leq 2\log |y|,
\]
we arrive at
\[
\frac 1{|x|^{\alpha + \frac np} (\log |x|)^{ \frac 1 q} |x-y|^\lambda }
\gtrsim \frac 1{|y|^{ \lambda + \alpha + \frac np} (\log |y|)^{ \frac 1 q} }.
\]
Keep in mind that $q>1$. Thus,
\begin{align*}
+\infty > \|f\|_{L^p(\R^n)}^q
&\gtrsim
\int_{|y| \geq 4} \Big( \int_{|y|/2 \leq |x| \leq 2|y|} dx \Big)^q \frac{ dy}{|y|^{(\lambda+\alpha +\beta+ \frac np ) q } \log |y|}\\
& \gtrsim \int_{|y| \geq 4} \frac{ dy}{|y|^n \log |y|}
= +\infty,
\end{align*}
thanks to
\begin{align*}
\Big(\lambda + \alpha +\beta + \frac np \Big) q &= \Big(\frac{\lambda + \alpha +\beta}n + \frac 1p \Big) n q
= \Big( 1 - \frac 1r + 1 \Big) n q = n + nq.
\end{align*}
In the above estimate, we have used the fact that $\int_{|y|/2 \leq |x| \leq 2|y|} dx \gtrsim |y|^n$, thanks to $|y| \geq 4$. Hence, we obtain a contradiction.

\section{Necessity of conditions for (\ref{eq-SW-u}) on $\R_+^n \times \R^{n-1}$: proof of Theorem \ref{thm-MAIN-u}}
\label{sec-Proof-u}

In this section, we show that our argument performed in section \ref{sec-Proof} above can be applied to the case of the upper half space $\R_+^n$ with some changes. Although there are some similarities between the previous section and this section, for a number of reasons, the analysis in this section is considerably more difficult than that in section \ref{sec-Proof}.

For a general point $x \in \R^{n-1}$, we write $$x=(x',x_{n-1}) \in \R^{n-2} \times \R.$$ Hence for $y \in \R^n$ we can also write $$y = (y', y_{n-1}, y_n) \in \R^{n-2} \times \R \times \R.$$

\subsection{The necessity of $0<\lambda<n-1/r$}
\label{subsec-L-u}

Employing a similar argument as in subsection \ref{subsec-L}, we can easily rule out the case $\lambda \leq 0$. Now we rule out the case $\lambda \geq n-1/r$. Compared to the case of \eqref{eq-SW-L}, this case is non-trivial due to the presence of the factional number $n-1/r$. Our crucial observation can be described as follows. Still denote
\[
q := \Big(1-\frac 1r\Big)^{-1},
\]
the condition
\[
\lambda \geq n - 1/r \quad \text{is \textit{equivalent} to} \quad \lambda - 1/q \geq n-1.
\]
It is important to note that the balance condition \eqref{eq-SW-BC-u} can be rewritten as
\begin{equation}\label{eq-SW-BC-u-n1}
\frac 1p + \frac 1r + \frac{\lambda + \alpha +\beta - \frac 1q}{n-1} = 2.
\end{equation}
Hence, the two conditions $\lambda - 1/q \geq n-1$ and $\alpha + \beta \geq 0$ basically say that we must have
\[
1/p + 1/r \leq 1,
\]
which should not occur by seeing \eqref{eq-SW-Key-u}. However, we still need to rule out the case $1/p+1/r=1$, which corresponds to the case both
\[
\lambda - 1/q = n-1 \quad \text{and} \quad \alpha + \beta = 0
\]
occur simultaneously by seeing \eqref{eq-SW-BC-u-n1}. This is a delicate issue, as we shall soon see.

The difficulty comes from the presence of the fractional number $1/q$. Hence we need some new idea. It is worth noticing that our argument below works for all $\lambda \geq n-1/r$, not just $\lambda = n-1/r$, and it does not depend on other conditions for the inequality, as expected.

To overcome this difficulty and in view of Remark \ref{rmk-SW-L}, we need to use the dual version of \eqref{eq-SW-u}. Clearly, for some non-negative function $f \in L^p(\R^{n-1})$ to be determined later, the inequality \eqref{eq-SW-BC-u} is equivalent to
\begin{align*}
+\infty > \| f \|_{L^p(\R^{n-1})}^q &\gtrsim \int_{\R_+^n} \Big( \int_{\R^{n-1}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda |y|^\beta } dx \Big)^q dy \\
&=\int_{\R^{n-1}} \int_0^{+\infty} \Big( \frac 1{ |y|^\beta } \int_{\R^{n-1}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda } dx \Big)^q dy_n dy',
\end{align*}
with $q=r/(r-1)$. Using Fubini's theorem and the non-decreasing of the integral $ \int_{B_{\rho}^{n-1} (y')} |x|^{-\alpha } f(x)dx$ in $\rho$, we easily get
\begin{align*}
\int_{\R^{n-1}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda } dx
&=\lambda \int_{\R^{n-1}} \Big( \int_{|x-y|}^{+\infty} \frac{d\rho}{\rho^{\lambda +1}} \Big) \frac{f(x) }{ |x|^\alpha } dx\\
&=\lambda \int_{y_n}^{+\infty} \Big( \int_{B_{\rho - y_n}^{n-1} (y')} \frac{f(x) }{ |x|^\alpha } dx \Big) \frac{d\rho}{\rho^{\lambda +1}}\\
&\gtrsim \frac 1{y_n^\lambda } \int_{B_{y_n}^{n-1} (y')} \frac{f(x) }{ |x|^\alpha } dx.
\end{align*}
Hence, we obtain
\begin{align*}
 \int_{\R^{n-1}} \int_0^{+\infty} & \Big( \frac 1{ |y|^\beta } \int_{\R^{n-1}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda } dx \Big)^q dy_n dy'\\
&\geq \int_{B_4^{n-1} \setminus B_2^{n-1}} \int_0^1 \Big( \frac 1{y_n^{ \lambda -1/q} |y|^\beta } \int_{B_{y_n}^{n-1} (y')} \frac{f(x) }{ |x|^\alpha } dx \Big)^q \frac{dy_n}{y_n} dy' .
\end{align*}
For the last line in the above computation, thanks to $|y'| \geq 2$ and $0\leq y_n \leq 1$, we know that
\[
B_1^{n-1} \subset B_{y_n}^{n-1} (y') \subset B_5^{n-1}
\]
and that $2 \leq |y| \leq \sqrt 5$. Hence, if we choose $f = \chi_{B_6^{n-1}}$, then we can bound
\[
\frac 1{|y|^\beta } \int_{B_{y_n}^{n-1} (y')} \frac{f(x) }{ |x|^\alpha } dx
=\frac 1{|y|^\beta } \int_{B_{y_n}^{n-1} (y')} \frac 1{ |x|^\alpha } dx
\gtrsim y_n^{n-1},
\]
which yields
\[
\int_0^1 \Big( \frac 1{y_n^{\lambda - 1/q}} \frac 1{|y|^\beta } \int_{B_{y_n}^{n-1} (y')} \frac{f(x) }{ |x|^\alpha } dx \Big)^q \frac{dy_n}{y_n }
\gtrsim
\int_0^1 \frac{dy_n}{y_n^{(\lambda -1/q + 1- n)q +1} }.
\]
However, the integral on the right hand side of the preceding inequality diverges if $$\lambda - 1/q \geq n-1.$$ Hence, we necessarily have $\lambda - 1/q < n-1$. This completes the proof.

\begin{remark}
Now we have some remarks.

\begin{itemize}

 \item The identity \eqref{eq-SW-BC-u-n1} reveals that we can transform the HLS inequality on $\R_+^n$ to a suitable HLS inequality on $\R^{n-1}$. This point is fully exploited in \cite{NNN}.

 \item Suppose that we only assume $\lambda \geq n-1$, instead of $\lambda \geq n-1/r$. In this scenario, the key estimate \eqref{eq-SW-u-qq} is no longer true because we do not necessarily have $$\alpha + \beta \geq 1/q.$$

 \item As mentioned in the proof, the above argument works for all $$\lambda \geq n-1/r$$ regardless of $\alpha$ and $\beta$, and this could confuse us why this condition does not depend on $\alpha$ and $\beta$. Nevertheless, seeing \eqref{eq-SW-BC-u-n1} and \eqref{eq-SW-Key-u}, by a simple contradiction argument, we are led to the following inequality for the validity of \eqref{eq-SW-u}
$$\lambda + \alpha + \beta \leq n-1/r,$$
which now depends on $\alpha$ and $\beta$. The above condition tells us that the closer to $n-1/r$ the parameter $\lambda$ is, the smaller the sum $\alpha+\beta$ is.
\end{itemize}
\end{remark}

\subsection{The necessity of (\ref{eq-SW-AB-u}) and (\ref{eq-SW-BC-u})}

In this part we prove the necessity of \eqref{eq-SW-AB-u} and \eqref{eq-SW-BC-u}, whose proof makes use of the same idea used in sections \ref{subsubsec-alpha<}, \ref{subsubsec-beta<}, and \ref{subsubsec-=}. However, to be able to consider the case of $\R^{n-k}\times \R^n$ in Section \ref{sec-Proof-m} below, we provide a quick argument for completeness.

\subsubsection{The necessity of $\beta < n(r-1)/r$}
\label{subsubsec-b-u}

Following the idea used in section \ref{subsubsec-beta<}, to see why $\beta < n(r-1)/r$ is necessary, we simply choose $f = \chi_{B_3^{n-1} \setminus B_2^{n-1}}$ and estimate
\begin{align*}
\int_{\R_+^n} \Big( \int_{\R^{n-1}} & \frac{\chi_{B_3^{n-1} \setminus B_2^{n-1} }(x) dx }{ |x|^\alpha |x-y|^\lambda |y|^\beta } \Big)^q dy
\geq \int_{ B_1^n }
\Big[ \int_{B_3^{n-1} \setminus B_2^{n-1}} \frac {dx }{ |x|^\alpha |x-y|^\lambda} \Big]^q
\frac{dy}{|y|^{\beta q}}
= +\infty
\end{align*}
if $\beta \geq n(r-1)/r=n/q$. Here $q=(1-1/r)^{-1}$ and we need the estimate $$|x-y| \leq \sqrt{2|x|^2 + 3|y|^2} \leq \sqrt{19}$$ to bound the integral of $|x|^{-\alpha}|x-y|^{-\lambda}$ from below away from zero.

\subsubsection{The necessity of $\alpha < (n-1)(p-1)/p$}
\label{subsubsec-a-u}

Now to see why $\alpha < (n-1)(p-1)/p$ is necessary, we choose $g = \chi_{ B_3^n \setminus B_2^n }$ and estimate
\begin{align*}
\int_{\R^{n-1}} \Big( \int_{\R_+^n} & \frac{\chi_{ B_3^n \setminus B_2^n }(y) dy }{ |x|^\alpha |x-y|^\lambda |y|^\beta } \Big)^q dx
\geq \int_{B_1^{n-1} }
\Big[ \int_{ B_3^n \setminus B_2^n } \frac {dy }{ |x-y|^\lambda |y|^\beta} \Big]^q
\frac{dx}{|x|^{\alpha q}}
= +\infty,
\end{align*}
thanks to $\alpha \geq (n-1)(p-1)/p=(n-1)/q$. Here $q=(1-1/p)^{-1}$ and, as before, we need $$|x-y| \leq \sqrt{2|x|^2 + 3|y|^2} \leq \sqrt{29}$$ to bound the integral of $|x-y|^{-\lambda} |y|^{-\beta}$ from below away from zero.

\subsubsection{The necessity of $(n-1)/(np) + 1/r + (\lambda + \alpha + \beta + 1)/n =2$}

Still using \eqref{eq-FeGe}, we now obtain
\begin{align*}
\iint_{\R_+^n \times \R^{n-1} } & \frac{f_\epsilon (x) g_\epsilon (y) dx dy}{|x|^\alpha |x-y|^\lambda |y|^\beta } \\
&=\iint_{\R_+^n \times \R^{n-1} } \frac{f (\frac x \epsilon ) g (\frac y \epsilon )dx dy }{\epsilon^{\frac {n-1}p + \frac nr + \lambda + \alpha + \beta} |\frac x\epsilon|^\alpha |\frac x\epsilon - \frac y \epsilon |^\lambda |\frac{y}\epsilon|^\beta } \\
&=\big( \frac 1 \epsilon \big)^{(\frac{n-1}n \frac 1p + \frac 1r + \frac{\lambda + \alpha + \beta + 1}n - 2 ) n}\iint_{\R_+^n \times \R^{n-1} } \frac{f (x ) g (y ) dx dy}{|x|^\alpha |x -y |^\lambda |y |^\beta } .
\end{align*}
From this we obtain the necessity of \eqref{eq-SW-BC-u}.

\subsection{The necessity of (\ref{eq-SW-A+B-u})}
\label{subsec-A+B-u}

As before, to verify the necessity of $\alpha + \beta \geq 0$ for the SW inequality on $\R_+^n$, we construct an example of non-negative functions $f$ and $g$ in such a way that $\|f\|_{L^p(\R^{n-1})} <+\infty$, $\|g\|_{L^r(\R_+^n)} <+\infty$, but
\[
\iint_{\R_+^n \times \R^{n-1}} \frac{f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dx dy = +\infty.
\]
To this purpose, we modify our example constructed in subsection \ref{subsec-A+B}. The idea consists of the following two points
\begin{itemize}
 \item to isolate $|x-y|$, as in the previous case, and
 \item to make sure that $|x|$ and $|y|$ behave similarly, which can be done if we isolate $y_n$.
\end{itemize}
Now fix any small $\epsilon > 0$ and define
\[
f(x)=
\left\{
\begin{aligned}
& (\log_2 x_{n-1})^{-\frac 1p - \epsilon} &&\text{if } \; 2^m \leq x_{n-1} \leq 2^m+1, \; m \geq 1, \\
& & & \text{and } \; |x'| \leq 1,\\
&0 &&\text{otherwise},
\end{aligned}
\right.
\]
and
\[
g(y)=
\left\{
\begin{aligned}
& (\log_2 y_{n-1})^{-\frac 1r - \epsilon} y_n &&\text{if } \; 2^m \leq y_{n-1} \leq 2^m+1, \; m \geq 1,\\
& & & \text{and } \; |y'| \leq 1, \; 1 < y_n < 2,\\
&0 &&\text{otherwise}.
\end{aligned}
\right.
\]
Intuitively, the support of $f$ and $g$ still consists of \textit{disjoint cylinders} concentrated along the half axis $x_{n-1}$. However, to isolate the extra dimension $y_n$, the support of $g$ is chosen as a `slice' of that of $f$. From the above, we obviously have $f \in L^p(\R^{n-1})$. Observe that
\begin{align*}
\int_{\R_+^n} g(y)^r dy &=\sum_{m \geq 1} \int_1^2 \Big( \int_{2^m}^{2^m+1}\Big( \int_{|y''| \leq 1} \frac 1{(\log_2 y_{n-1})^{1+ r\epsilon}} dy' \Big) dy_{n-1} \Big) dy_n\\
&\leq |B_1^{n-2} | \sum_{m \geq 1} \frac{1}{m^{1+ r\epsilon}}
<+\infty,
\end{align*}
which implies $g \in L^r(\R_+^n)$. Now we estimate the double integral as follows
\begin{equation}\label{eq-Double-1-u}
\begin{aligned}
\iint_{\R+^n \times \R^{n-1}} & \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy \\
\geq \sum_{m \geq 1}& \iint_{B_1^{n-2} \times [2^m, 2^m+1]} \Big( \iiint_{B_1^{n-2} \times [2^m, 2^m+1] \times [1,2]} \frac {g(y) dy}{ |x-y|^\lambda |y|^\beta} \Big) \frac{f(x) dx}{|x|^\alpha}.
\end{aligned}
\end{equation}
For each $m \geq 1$, let $x$ and $y$ be such that
\begin{equation}\label{eq-Strip-u}
x,(y',y_{n-1}) \in \overline{B_1^{n-2}} \times [2^m, 2^m+1], \quad 1<y_n<2.
\end{equation}
In view of \eqref{eq-Double-1-u}, we need to estimate $|x|^{-\alpha}$, $|y|^{-\beta}$, and $|x-y|^{-\lambda}$ from below. Apparently, we can easily bound $|x-y|$ from the above as follows
\begin{align*}
|x-y|^2 &=|x''-y''|^2 +|x_{n-1}-y_{n-1}|^2 + y_n^2
\leq 9.
\end{align*}
This together with $\lambda > 0$ allows us to estimate $|x-y|^{-\lambda} \geq 3^{-\lambda}$ provided \eqref{eq-Strip-u} holds for each $m \geq 1$. Since $|x'| \leq 1$ and $x_{n-1} \geq 2$, we also have
\[
x_{n-1} \leq \sqrt{|x'|^2+x_{n-1}^2} \leq 2x_{n-1}.
\]
Similarly, as $|y'| \leq 1$, $y_{n-1} \geq 2$, and $1 \leq y_n \leq 2$ we get
\[
y_{n-1} \leq \sqrt{|y'|^2+y_{n-1}^2 + y_n^2} \leq 2y_{n-1}.
\]
Hence, regardless of the sign of $\alpha$ and $\beta$, we always have
\[
|x|^{-\alpha} \gtrsim x_{n-1}^{-\alpha}, \quad |y|^{-\beta} \gtrsim y_{n-1}^{-\beta}.
\]
Putting the above estimates together, we can further estimate \eqref{eq-Double-1-u} as follows
\begin{equation}\label{eq-Double-2-u}
\begin{aligned}
 \iint_{\R_+^n \times \R^{n-1}}& \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy\\
\gtrsim& \sum_{m \geq 1} \Big( \iint_{B_1^{n-2} \times [2^m, 2^m+1]} \frac{f(x) dx}{|x|^\alpha} \Big) \Big( \iiint_{B_1^{n-2} \times [2^m, 2^m+1] \times [1,2]} \frac {g(y) dy}{ |y|^\beta} \Big) \\
 =& |B_1^{n-2}|^2
 \sum_{m \geq 1} \Big( \int_{2^m}^{2^m+1} \frac {dx_{n-1}}{x_{n-1}^\alpha (\log_2 x_{n-1})^{\frac 1p +\epsilon}} \Big) \\
 & \qquad \qquad \times \Big( \int_1^2 y_n dy_n \Big)\Big( \int_{2^m}^{2^m+1} \frac {dy_{n-1}}{y_{n-1}^\beta (\log_2 y_{n-1})^{\frac 1r +\epsilon}} \Big) \\
 \gtrsim& \sum_{m \geq 1} \frac 1{(m+1)^{\frac 1p +\frac 1r +2\epsilon}}
 \Big( \int_{2^m}^{2^m+1} \frac {dx_{n-1}}{x_{n-1}^\alpha } \Big) \Big( \int_{2^m}^{2^m+1} \frac {dy_{n-1}}{y_{n-1}^\beta } \Big).
\end{aligned}
\end{equation}
Arguing as in section \ref{subsec-A+B}, we know that
\[
\Big( \int_{2^m}^{2^m+1} \frac {dx_{n-1}}{x_{n-1}^\alpha } \Big) \Big( \int_{2^m}^{2^m+1} \frac {dy_{n-1}}{y_{n-1}^\beta } \Big)
\geq \frac 1{(2^m)^{\alpha + \beta}}
\]
provided \eqref{eq-Strip-u} holds. Thus, recalling \eqref{eq-Double-2-u} we arrive at
\begin{align*}
\iint_{\R_+^n \times \R^{n-1}} & \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy
\gtrsim \sum_{m \geq 1} \frac 1{(2^m)^{\alpha + \beta}} \frac 1{(m+1)^{\frac 1p +\frac 1r + 2\epsilon}} .
\end{align*}
From this we necessarily have $\alpha +\beta \geq 0$. This proves the necessity of \eqref{eq-SW-AB-u}.

\subsection{The necessity of (\ref{eq-SW-Key-u})}
\label{subsec-P+R-u}

Now we establish the necessity of the condition $1/p+1/r \geq 1$. The example provided below is essentially the same as that constructed in section \ref{subsec-P+R} with some necessary changes due to the fact that we now work on $\R_+^n \times \R^{n-1}$. To this purpose, by duality, we obtain an equivalent form of \eqref{eq-SW-u} as follows
\begin{equation}\label{eq-SW-u-n2}
\int_{\R^{n-1}} \Big( \int_{\R_+^n} \frac{g (y) }{|x|^\alpha |x-y|^\lambda |y|^{\beta }} dy \Big)^q dx
\lesssim \| g\|_{L^p(\R_+^n)}
\end{equation}
with
\begin{equation}\label{eq-SW-u-qq}
q = \Big(1- \frac 1p \Big)^{-1} \overset{\eqref{eq-SW-BC-u-n1}}{=}
\Big(\frac 1 r + \frac {\lambda + \alpha + \beta - \frac 1q }{n-1} -1 \Big)^{-1}
<r,
\end{equation}
thanks to $\lambda + \alpha + \beta - 1/q >n-1$. Now we choose
\[
g(y)=
\left\{
\begin{aligned}
& |y|^{-\frac nr } (\log |y|)^{- \frac 1 q} & & \text{if } |y| \geq 2,\\
& 0 & & \text{otherwise}.
\end{aligned}
\right.
\]
As in subsection \ref{subsec-P+R}, we clearly have
\[
\int_{\R^n} g(y )^r dy = |\mathbb S^{n-1}| \int_2^{+\infty} \frac{d\rho}{\rho (\log \rho)^{ \frac r q}} < +\infty,
\]
thanks to $r > q$. Hence, $ g \in L^r (\R_+^n)$. Clearly, by \eqref{eq-SW-u-n2} we have
\begin{align*}
+\infty > \| g\|_{L^p(\R_+^n)}^q &> \int_{\R^{n-1}} \Big( \int_{\R_+^n} \frac{g (y) }{|x|^\alpha |x-y|^\lambda |y|^\beta} dy \Big)^q dx\\
&\geq
\int_{|x| \geq 4} \Big( \int_{|x|/2 \leq |y| \leq 2|x|} \frac{ dy }{|y|^{\beta + \frac nr} (\log |y|)^{\frac 1 q } |x-y|^\lambda } \Big)^q \frac{ dx}{|x|^{\alpha q} }.
\end{align*}
As $\lambda > 0$ and $|y| \leq 2|x|$, we can bound $|x-y|$ from the above as follows
\[
|x-y|^2 = |x-y'|^2 + y_n^2 \leq 2|x|^2 + 2|y'|^2 + y_n^2
\leq 6|x|^2,
\]
which, together with $\lambda>0$, implies
\[
|y|^{\beta + \frac nr} |x-y|^{\lambda} \lesssim |x|^{\lambda +\beta + \frac n r}
\]
for any $|x|/2 \leq |y| \leq 2|x|$. Hence, together with
\[
0< \log |y| \leq \log (2|x|) \leq \log 2 + \log |x| \leq 2\log |x|,
\]
we know that
\[
\frac 1{|y|^{\beta + \frac nr} (\log |y|)^{\frac 1 q } |x-y|^\lambda }
\gtrsim \frac 1{|x|^{ \lambda + \beta + \frac nr } (\log |x|)^{ \frac 1 q } }.
\]
Observe that
\begin{equation}\label{eq-SW-u-Exponent}
\begin{aligned}
\Big(\lambda + \alpha +\beta + \frac nr \Big) q
= \Big(\frac{n-1}n \big( 1 - \frac 1p \big) + 1 \Big) n q = n-1 + n q.
\end{aligned}
\end{equation}
Hence
\begin{align*}
+\infty > \| g \|_{L^p(\R_+^n )}^q
&\gtrsim
\int_{|x| \geq 4} \Big( \int_{|x|/2 \leq |y| \leq 2|x|} dy \Big)^q
\frac {dx}{|x|^{( \lambda + \alpha +\beta + \frac nr )q } (\log |x|) } \\
&=
\int_{|x| \geq 4} \frac 1{|x|^{n-1} \log |x| }
= +\infty,
\end{align*}
thanks to $\int_{|x|/2 \leq |y| \leq 2|x|} dy \sim |x|^n$. Hence, we obtain the necessity of \eqref{eq-SW-Key-u} as claimed.

\section{Necessity of conditions for (\ref{eq-SW-m}) on $\R^{n-k} \times \R^n$: proof of Theorem \ref{thm-MAIN-m}}
\label{sec-Proof-m}

In this section, we show that our argument performed in section \ref{sec-Proof-u} above can be further applied to the case of $\R^{n-k} \times \R^n$ with some changes due to the fact that we are no longer in $\R^{n-1} \times (0, +\infty)$. The key idea is to transform the higher-dimensional space $\R^k$ into the one-dimensional case $(0,+\infty)$. We also note that if we let $k=0$, then Theorem \ref{thm-MAIN-m} becomes Theorem \ref{thm-MAIN}. However, to further illustrate the method, we shall use the condition $k \geq 1$ in several places.

For clarity and convenience, for a general point $x \in \R^{n-k}$, we write
\[
x=(x',x_{n-k}) \in \R^{n-k-1} \times \R.
\]
Hence, for $y \in \R^n$ we shall write
\[
y = (y', y_{n-k}, y'') \in \R^{n-k-1} \times \R \times \R^k.
\]

\subsection{The necessity of (\ref{eq-SW-L-m})}
\label{subsubsec-L-m}

Employing a similar argument as in subsection \ref{subsec-L-u}, we can easily rule out the case $\lambda \leq 0$. Now we rule out the case $\lambda \geq n-k/r$. For some non-negative function $f \in L^p(\R^{n-k})$ to be determined later, the inequality \eqref{eq-SW-BC-m} is equivalent to
\begin{align*}
\| f \|_{L^p(\R^{n-k})}^q &\gtrsim \int_{\R^n} \Big( \int_{\R^{n-k}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda |y|^\beta } dx \Big)^q dy \\
&=|\mathbb S^{k-1}| \int_{\R^{n-k}}\Big[ \int_0^{+\infty} \Big( \frac 1{ |y|^\beta } \int_{\R^{n-k}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda } dx \Big)^q \rho^{k-1} d\rho \Big] dy',
\end{align*}
with $q=r/(r-1)$ and $\rho = |y''|$. Now we need some estimate for the integral $\int_{\R^{n-k}}$ which costs us some energy due to the fact that $k \geq 1$. Still by Fubini's theorem and the non-decreasing of
$$ \rho \mapsto \int_{B_{\rho}^{n-k} (y',y_{n-k})} |x|^{-\alpha } f(x)dx,$$
we easily get
\begin{align*}
\int_{\R^{n-k}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda } dx
&=\lambda \int_{\R^{n-k}} \Big( \int_{|x-y|}^{+\infty} \frac{d\rho}{\rho^{\lambda +1}} \Big) \frac{f(x) }{ |x|^\alpha } dx\\
&=\lambda \int_{|y''|}^{+\infty} \Big( \int_{B_{\rho - |y''|}^{n-k} (y',y_{n-k})} \frac{f(x) }{ |x|^\alpha } dx \Big) \frac{d\rho}{\rho^{\lambda +1}}\\
&\gtrsim \int_{2|y''|}^{+\infty} \Big( \int_{B_{\rho - |y''|}^{n-k} (y',y_{n-k})} \frac{f(x) }{ |x|^\alpha } dx \Big) \frac{d\rho}{\rho^{\lambda +1}}\\
& \geq \Big( \int_{B_{|y''|}^{n-k} (y',y_{n-k})} \frac{f(x) }{ |x|^\alpha } dx \Big) \int_{2|y''|}^{+\infty} \frac{d\rho}{\rho^{\lambda +1}}\\
&\gtrsim \frac 1{|y''|^\lambda } \int_{B_{|y''|}^{n-k} (y',y_{n-k})} \frac{f(x) }{ |x|^\alpha } dx.
\end{align*}
Hence, we obtain
\begin{align*}
\| f \|_{L^p(\R^{n-k})}^q
&\gtrsim |\mathbb S^{k-1}| \int_{\R^{n-k}} \Big[\int_0^{+\infty} \Big( \frac 1{ |y|^{\beta } } \int_{\R^{n-k}} \frac{f(x) }{ |x|^\alpha |x-y|^\lambda} dx \Big)^q \rho^{k-1} d\rho \Big] dy' dy_{n-k}\\
&\gtrsim \int_{B_4^{n-k} \setminus B_2^{n-k}} \Big[ \int_0^1 \Big( \frac 1{\rho^{ \lambda - k/q} |y|^\beta } \int_{B_{\rho}^{n-k} (y', y_{n-k})} \frac{f(x) }{ |x|^\alpha } dx \Big)^q \frac{d\rho}{\rho} \Big] dy' dy_{n-k} .
\end{align*}
For the last line in the above computation, thanks to $|(y', y_{n-k})|^2 \geq 4$ and $0\leq |y''| \leq 1$, we know that
\[
B_1^{n-k} \subset B_{\rho}^{n-k} (y', y_{n-k}) \subset B_5^{n-k}
\]
and that $2 \leq |y| \leq \sqrt 5$. Hence, if we choose $f = \chi_{B_6^{n-1}}$, then we can bound
\[
\frac 1{|y|^\beta } \int_{B_{\rho}^{n-k} (y', y_{n-k})} \frac{f(x) }{ |x|^\alpha } dx
 \gtrsim \rho^{n-k},
\]
which yields
\[
\int_0^1 \Big( \frac 1{\rho^{\lambda - k/q}} \frac 1{|y|^\beta } \int_{B_{\rho}^{n-k} (y', y_{n-k})} \frac{f(x) }{ |x|^\alpha } dx \Big)^q \frac{d \rho}{\rho }
\gtrsim
\int_0^1 \frac{d\rho}{\rho^{(\lambda -k/q -(n- k))q +1} }.
\]
However, the integral on the right hand side of the preceding inequality diverges if
\[
\lambda - k/q \geq n-k.
\]
Hence, we necessarily have $\lambda - k/q < n-k$. This completes the proof.

\begin{remark}
The necessity of the condition $\lambda - k/q < n-k$ for \eqref{eq-SW-m} is quite different from that of \eqref{eq-SW-m-simpler} as \eqref{eq-SW-m-simpler} holds for any $\lambda > 0$; see \cite{NNN}.
\end{remark}

\subsection{The necessity of (\ref{eq-SW-BC-m}) and (\ref{eq-SW-AB-m})}

As before, the balance condition \eqref{eq-SW-BC-m} follows from standard computation. In addition, the condition \eqref{eq-SW-AB-m} can also be proved by using the idea in subsections \ref{subsubsec-a-u} and \ref{subsubsec-b-u}. Hence we omit the details.

\subsection{The necessity of (\ref{eq-SW-A+B-m})}
\label{subsec-A+B-m}

To verify the necessity of $\alpha + \beta \geq 0$ for the inequality \eqref{eq-SW-m} on $\R^{n-k} \times \R^n$, we follows the construction in subsection \ref{subsec-A+B-u}. Now fix any small $\epsilon > 0$ and define
\[
f(x)=
\left\{
\begin{aligned}
& (\log_2 x_{n-k})^{-\frac 1p - \epsilon} &&\text{if } \; 2^m \leq x_{n-k} \leq 2^m+1, \; m \geq 1, \\
& & & \text{and } \; |x'| \leq 1,\\
&0 &&\text{otherwise},
\end{aligned}
\right.
\]
and
\[
g(y)=
\left\{
\begin{aligned}
& (\log_2 y_{n-k})^{-\frac 1r - \epsilon} |y''| &&\text{if } \; 2^m \leq y_{n-k} \leq 2^m+1, \; m \geq 1,\\
& & & \text{and } \; |y'| \leq 1, \; 1 < |y''| < 2,\\
&0 &&\text{otherwise}.
\end{aligned}
\right.
\]
From the above construction, it is clear that the quantity $|y''|$ serves a similar role as $y_n$ in the case of $\R^{n-1} \times \R_+^n$. Now we obviously have $f \in L^p(\R^{n-k})$. Recall that $\int_{\R^k} = |\mathbb S^{k-1}| \int_0^{+\infty}$. Then for the function $g$ we have
\begin{align*}
\int_{\R^n} g(y)^r dy &=|\mathbb S^{k-1}| \sum_{m \geq 1} \int_1^2 \Big( \int_{2^m}^{2^m+1}\Big( \int_{|y'| \leq 1} \frac 1{(\log_2 y_{n-k})^{1+ r\epsilon}} dy' \Big) dy_{n-k} \Big) d\rho\\
&\leq |\mathbb S^{k-1}| |B_1^{n-k-1} | \sum_{m \geq 1} \frac{1}{m^{1+ r\epsilon}}
<+\infty,
\end{align*}
which implies $g \in L^r(\R^n)$. Now we estimate the double integral as follows
\begin{equation}\label{eq-Double-1-m}
\begin{aligned}
\iint_{\R^n \times \R^{n-k}} & \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy \\
\geq \sum_{m \geq 1}& \iint_{B_1^{n-k-1} \times [2^m, 2^m+1] } \Big( \iiint_{B_1^{n-k-1} \times [2^m, 2^m+1] \times [1, 2]} \frac {g(y) dy}{ |x-y|^\lambda |y|^\beta} \Big) \frac{f(x) dx}{|x|^\alpha}.
\end{aligned}
\end{equation}
For each $m \geq 1$, let $x$ and $y$ be such that
\begin{equation*}\label{eq-Strip-m}
x, (y',y_{n-k}) \in \overline{B_1^{n-k-1}} \times [2^m, 2^m+1], \quad 1<|y''|<2.
\end{equation*}
As before, we need to estimate $|x|^{-\alpha}$, $|y|^{-\beta}$, and $|x-y|^{-\lambda}$ from below. Apparently, we can easily bound $|x-y|$ from the above as follows
\begin{align*}
|x-y|^2 &=|x'-y'|^2 +|x_{n-k}-y_{n-k}|^2 + |y''|^2
\leq 9,
\end{align*}
giving $|x-y|^{-\lambda} \geq 3^{-\lambda}$. Now as $|x'| \leq 1$ and $x_{n-k} \geq 2$, we also have $x_{n-k} \leq |x \leq 2x_{n-k}$, giving $$|x|^{-\alpha} \gtrsim x_{n-k}^{-\alpha}.$$ Similarly, as $|y'| \leq 1$, $y_{n-k} \geq 2$, and $1 \leq |y''| \leq 2$ we get $y_{n-k} \leq |y|\leq 2y_{n-k}$, giving $$|y|^{-\beta} \gtrsim y_{n-k}^{-\beta}.$$ Putting the above estimates together and similar to \eqref{eq-Double-2-u} we can further estimate \eqref{eq-Double-1-m} as follows
\begin{equation*}\label{eq-Double-2-m}
\begin{aligned}
 \iint_{\R^n \times \R^{n-k}}& \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy\\
\gtrsim& \sum_{m \geq 1} \Big( \int_{2^m}^{2^m+1} \frac {dx_{n-k}}{x_{n-k}^\alpha (\log_2 x_{n-k})^{\frac 1p +\epsilon}} \Big) \\
 & \qquad \times \Big( |\mathbb S^{k-1}| \int_1^2 \rho^k d \rho \Big)\Big( \int_{2^m}^{2^m+1} \frac {dy_{n-k}}{y_{n-k}^\beta (\log_2 y_{n-k})^{\frac 1r +\epsilon}} \Big) \\
 \gtrsim& \sum_{m \geq 1} \frac 1{(m+1)^{\frac 1p +\frac 1r +2\epsilon}}
 \Big( \int_{2^m}^{2^m+1} \frac {dx_{n-k}}{x_{n-k}^\alpha } \Big) \Big( \int_{2^m}^{2^m+1} \frac {dy_{n-k}}{y_{n-k}^\beta } \Big) .
\end{aligned}
\end{equation*}
Thus, we arrive at
\begin{align*}
\iint_{\R^n \times \R^{n-k}} & \frac {f(x)g(y)}{|x|^\alpha |x-y|^\lambda |y|^\beta} dxdy
\gtrsim \sum_{m \geq 1} \frac 1{(2^m)^{\alpha + \beta}} \frac 1{(m+1)^{\frac 1p +\frac 1r + 2\epsilon}} .
\end{align*}
This proves the necessity of \eqref{eq-SW-AB-m}, thanks to \eqref{eq-SW-m}.

\subsection{The necessity of (\ref{eq-SW-Key-m})}
\label{subsec-P+R-m}

Now we establish the necessity of the condition $1/p+1/r \geq 1$ by following the construction in subsection \ref{subsec-P+R-u} with some necessary changes due to the fact that we now work on $\R^{n-k} \times \R^n$. To this purpose, by duality, we obtain an equivalent form of \eqref{eq-SW-m} as follows
\begin{equation}\label{eq-SW-m-n2}
\int_{\R^{n-k}} \Big( \int_{\R^n} \frac{g (y) }{|x|^\alpha |x-y|^\lambda |y|^{\beta }} dy \Big)^q dx
\lesssim \| g\|_{L^p(\R^n)}^q
\end{equation}
with $q = (1- 1/p)^{-1} < r$, thanks to $\lambda + \alpha + \beta - 1/q >n-k$. Now we choose
\[
g(y)=
\left\{
\begin{aligned}
& |y|^{-\frac nr } (\log |y|)^{- \frac 1 q} & & \text{if } |y| \geq 2,\\
& 0 & & \text{otherwise}.
\end{aligned}
\right.
\]
We clearly have $ g \in L^r (\R^n)$, thanks to $r > q$. Clearly, by \eqref{eq-SW-u-n2} we have
\begin{align*}
+\infty > \| g\|_{L^p(\R^n)}^q &> \int_{\R^{n-k}} \Big( \int_{\R^n} \frac{g (y) }{|x|^\alpha |x-y|^\lambda |y|^\beta} dy \Big)^q dx\\
&\geq
\int_{|x| \geq 4} \Big( \int_{|x|/2 \leq |y| \leq 2|x|} \frac{ dy }{|y|^{\beta + \frac nr} (\log |y|)^{\frac 1 q } |x-y|^\lambda } \Big)^q \frac{ dx}{|x|^{\alpha q} }.
\end{align*}
As $\lambda > 0$ and $|y| \leq 2|x|$, we can bound $|x-y|$ from the above as follows
\begin{align*}
|x-y|^2 &= |x'-y'|^2 + |x_{n-k} - x_{n-k}|^2 + |y''|^2 \\
& \leq 2|x|^2 + 2|y'|^2 + 2 y_n^2 + |y''|^2 \\%\leq 2(|x|^2 + |y|^2)
&\leq 6|x|^2,
\end{align*}
which implies
\[
|y|^{\beta + \frac nr} |x-y|^{\lambda} \lesssim |x|^{\lambda +\beta + \frac n r}
\]
for any $|x|/2 \leq |y| \leq 2|x|$. Hence, we know that
\[
\frac 1{|y|^{\beta + \frac nr} (\log |y|)^{\frac 1 q } |x-y|^\lambda }
\gtrsim \frac 1{|x|^{ \lambda + \beta + \frac nr } (\log |x|)^{ \frac 1 q } }.
\]
Hence, together with \eqref{eq-SW-m-n2} we can estimate
\begin{align*}
+\infty > \| g \|_{L^p(\R^n )}^q
&\gtrsim
\int_{|x| \geq 4} \Big( \int_{|x|/2 \leq |y| \leq 2|x|} dy \Big)^q
\frac {dx}{|x|^{( \lambda + \alpha +\beta + \frac nr )q } (\log |x|) } \\
&=
\int_{|x| \geq 4} \frac 1{|x|^{n-k} \log |x| } = +\infty,
\end{align*}
thanks to $ (\lambda + \alpha +\beta + n/r ) q = n-k + n q$. This proves the necessity of \eqref{eq-SW-Key-m} as claimed.

\section*{Acknowledgments}

This project was initiated while the author was staying at the University of Tokyo. Funding received including the Tosio Kato Fellowship awarded in 2018 is greatly acknowledged. The author also benefited from the Vietnam Institute for Advanced Study in Mathematics (VIASM) during his visit in 2021 and from the Vietnam National University, Hanoi (VNU) under project number QG.21.01. Last but not least, we thank Jingbo Dou and Quoc Hung Nguyen for useful discussion during the preparation of the note.


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text:
The Battle of Noirmoutier was a confrontation in the War in the Vendée which took place on 3 January 1794 between the Republicans and the Vendeans for control of the island of Noirmoutier.

Situated in the Bay of Bourgneuf and linked to the mainland by the Gois causeway, the island was occupied on 12 October 1793 by General Charette's Vendean troops. This capture immediately alarmed the Committee of Public Safety, which feared British intervention on behalf of the royalists. However, Charette did not attempt to send a schooner to Great Britain until December to make contact with the London government, and he received no reply for several months.

The republicans landed on Noirmoutier on the morning of 3 January 1794 and took the whole of its southern part after fighting at Pointe de la Fosse and the village of Barbâtre. They then marched in the afternoon towards the town of Noirmoutier, where they received the capitulation of the Vendeans, who laid down their arms against the promise of their lives.

Accepted by General Haxo, the capitulation was not respected by the representatives on a mission, Prieur de la Marne, Turreau and Bourbotte, who had 1,200 to 1,500 prisoners shot in a few days, including Maurice d'Elbée, the former generalissimo of the Catholic and Royal Army.

The recapture of Noirmoutier followed the battle of Savenay a few days later, which saw the destruction of the Vendean forces engaged in the Galerne's campaign. After these two victories, which were considered decisive, the Republican commanders believed that the War in the Vendée was on the point of being over. The Vendeans no longer control any territory and the last insurgent bands are numbered by only a few hundred or a few thousand men, hunted down in the woods and in the countryside. In mid-January, General Turreau launched his "" with the intention of giving the final blow to the Vendean insurrection.

Context

Capture of Noirmoutier by the Vendeans

The island of Noirmoutier changed hands several times during the War in the Vendée. Located in the bay of Bourgneuf not far from the island of Bouin, accessible from land by the Gois causeway, a 5 kilometres long submersible road that can be walked on during low tides, which connects the island to the mainland by the port of Cronière, in the commune of Beauvoir-sur-Mer. Noirmoutier is divided into two parishes: to the north, the town of Noirmoutier-en-l'Île, also called Saint-Philbert de Noirmoutier, with a population of 3,675 in 1791, and to the south the village of Barbâtre, with a population of 2,221 in 1791. The north of the island, from La Guérinière to l'Herbaudière, consists of salt marshes. The south, from La Guérinière to Pointe de la Fosse, is covered with sand, arable land and vines.

Noirmoutier first fell to the insurgents on 19 March 1793, when peasant farmers led by Joseph Guerry de La Fortinière crossed the Gois causeway and occupied the island without resistance.

However, in April, the republicans retook the towns of Challans and Machecoul and reached the outskirts of the island. After summons from General Beysser and the landing of 200 men from the ship Le Superbe, from the squadron of the Rear Admiral Villaret de Joyeuse, the inhabitants of Noirmoutier made their submission on 29 April. The insurgent leaders René Augustin Guerry and Rorthais des Chataigners were arrested and sent to Nantes, but Guerry de la Fortinière, Tinguy and the knight of Régnier managed to escape.

However, in June, the Vendeans recaptured Challans and Machecoul and were once again able to threaten Noirmoutier. After a first unsuccessful attempt on 30 September, General Charette's army, aided by the inhabitants of Barbâtre, took the island on 12 October. The weak Republican garrison offered little resistance and capitulated. Charette formed a royalist administration in Noirmoutier and left some of his troops there before leaving after three days. The Republican prisoners were sent to Bouin where the local chief, François Pajot, had several hundreds of them massacred on 17 and 18 October.

Republican attack against Charette

In October 1793, the military situation in the Vendée was to the advantage of the Republicans. The army of the west concentrated its efforts on the army of Anjou and Haut-Poitou, commanded by the generalissimo Maurice d'Elbée, which was defeated on 17 October at the battle of Cholet. The "Galerne campaign" then began: the bulk of the royalist and republican forces moved north of the Loire and clashed until December in Maine, Brittany and Normandy. Despite the several appeals for help sent by d'Elbée, Charette and the other chiefs of Pays de Retz and Bas-Poitou remained aloof from these events.

Meanwhile, in Paris, the news of the capture of the island of Noirmoutier aroused the concern of the Committee of Public Safety, which feared that it would allow the Vendeans to receive help from the British. On 18 October, the executive council received a decree signed by Barère, Prieur de la Côte d'Or, Collot d'Herbois, Billaud-Varenne, Robespierre and Hérault de Séchelles ordering it to "take all necessary measures to attack the island of Noirmoutier as soon as possible, to chase away the brigands and to ensure possession of the island by the Republic". On October 21, the Committee of public safety enjoins the representatives on a mission, Prieur de la Marne and Jeanbon Saint-André "to take again the island or to swallow it in the sea".

On 2 November 1793, the council of war of the army of the West charged Brigadier General Nicolas Haxo with forming a corps of 5,000 to 6,000 men to retake the island of Noirmoutier. He was ordered to attack and defeat Charette "wherever he might encounter him, pursuing him into Noirmoutier itself". Haxo planned his offensive and left Nantes on 21 and 22 November with two columns commanded by himself and his second-in-command, Adjutant-General Nicolas Louis Jordy. At the same time, another column commanded by General Dutruy and Lieutenant-Colonel Aubertin set off from Les Sables-d'Olonne. In the days that followed, Haxo and Jordy took Port-Saint-Père, Sainte-Pazanne, Bourgneuf-en-Retz, Machecoul and Legé, while Dutruy and Aubertin took La Roche-sur-Yon, Aizenay, Le Poiré-sur-Vie, Palluau and Challans. Defeated by Aubertin at La Garnache on 27 November, Charette tried to take refuge in Noirmoutier, but he found the Gois causeway impracticable because of the high tide and was forced to lock himself in the island of Bouin, where he was soon surrounded.

On 4 December, Charette embarked for the island of Noirmoutier. He entrusted his aide-de-camp, Joseph Hervouët de La Robrie, with the mission of going to England to request for help. La Robrie embarked on a 60 barrels schooner, Le Dauphin, commanded by Louis François Lefebvre. But because of unfavourable winds or the presence of republican ships, he can only set sail during the night of 23 to 24 December.

On 5 December, Charette returned to Bouin. On the 6th, the columns of Jordy and Aubertin launched the assault on the island. In a few hours, the Republicans broke through the Vendean defences and freed several hundred patriot prisoners, but Charette only just escaped annihilation by fleeing through the marshes with some of his men.

He then joined the forces of Jean-Baptiste Joly and Jean Savin, with whom he unsuccessfully attacked the town of Legé on 8 December, but crushed the garrison of the camp of L'Oie on the 11th. On 12 December, at Les Herbiers, Charette was elected general-in-chief of the "Catholic and Royal Army of Bas-Poitou". He then decided to go to Anjou and Haut-Poitou to revive the insurrection there. In a few days, he crossed Le Boupère, Pouzauges, Cerizay and Châtillon, then reached Maulévrier.

Meanwhile, the Vendean forces engaged in the Galerne's campaign approached the Loire with the intention of regaining the Vendée. On 3 and 4 December, they attacked Angers, without success. On the 16th, a thousand Vendeansled by Generalissimo Henri de La Rochejaquelein, the successor of Maurice d'Elbée, used boats to cross the river at Ancenis, but the arrival of Republican gunboats prevented the rest of the army from crossing. These events forced Haxo to suspend the attack on Noirmoutier and to loosen the stranglehold on Charette. He sent reinforcements against La Rochejaquelein and was only able to send 2,400 men in pursuit of Charette under the command of Adjutant-General Dufour.

However, Charette's expedition was unsuccessful because with the return of Henri de La Rochejaquelein, the insurgent regions of Anjou and Haut-Poitou came back under his authority. After meeting La Rochejaquelein at Maulévrier on 22 December, Charette turned back and returned to Les Herbiers.

Prelude

On 29 December 1793, General Louis-Marie Turreau arrived in Nantes to take command of the Army of the West. Six days earlier, the Vendée army north of the Loire had been destroyed at the battle of Savenay. Turreau then gave the order to Haxo to launch the attack against the island of Noirmoutier.

From the end of December, exchanges of fire opposed the Republican ships and the Vendean coastal batteries. On 29 December, three frigates were hit. On 30 December, a cannonade took place on the Bois de la Chaize side. LaNymphe, a 26-gun frigate commanded by Lieutenant Pitot, the corvette Le Fabius and the gunboat L'Île-Dieu engaged in combat in the early afternoon, but it turned out badly for the Republicans. The Vendeans fired red balls, and one of them hit the mizzen mast of the Fabius. La Nymphe, with on board the representatives on a mission Prieur de la Marne, Turreau and Guermeur, received seven balls and ran aground after three hours of cannonade. The yard of grand hunier was cleanly broken, and the crew counted two dead and two to five wounded. As for the hull, it remains submerged. The remains of the wreck were discovered in the Bay of Bourgneuf in 2014.

On 31 December, the movements of the barges between Bouin and Barbâtre alerted the besieged who expected an imminent attack. Throughout the end of December, the Vendeans fortified the various artillery batteries scattered around the island. Hyacinthe Hervouët de La Robrie and Dubois de La Guignardière developed a system of lights to warn the town of Noirmoutier in case of an attack. The stones and the beacons of the Gois causeway were destroyed. A messenger was also sent to Charette to ask for help.

On 31 December, the troops of Adjutant-General Carpantier left Machecoul and moved on to Challans to secure General Haxo's rear during the attack on Noirmoutier. But the same day, Charette stormed Machecoul, where only 200 to 300 men had been left. Informed, Haxo proposed to delay the expedition but Turreau refused and gave the order to continue. On the 2nd January 1794, a thousand men commanded by Carpantier retook Machecoul and put Charette's troop to flight. The latter attempted a counter-attack the following day, but was repulsed and retreated to La Copechagnière.

Opposing forces

Republican Army

The Republicans mobilised 5,000 men for the operation, but only 3,000 to 3,200 took part in the fighting on the island. La Gazette de France of 22 Nivôse Year II (11 January 1794) gave the precise strength of 3,112 men, made up of 322 men from a battalion of volunteers from the Meurthe, 460 men from the 109th infantry regiment, 400 men from the 11th battalion of the Orleans formation, 200 men from a detachment from the Bec d'Ambès, 68 men from a detachment from the Charente, 46 men from a detachment from Ille-et-Vilaine, 60 men from a detachment of the 77th infantry regiment, 420 men from the 57th infantry regiment, 146 men from the Grenadiers réunis, 201 men from a battalion of grenadiers from the Ardèche, 100 men from a battalion from the Marne, 153 riflemen from the Loire-Inférieure, the Deux-Sèvres, Challans, Beaulieu and Apremont, 386 men from the 39th infantry regiment and 150 men from the 3rd battalion of volunteers from Lot-et-Garonne.

Adjutant-General Jordy, at the head of 1,500 men transported by 19 barges, led the first wave of the assault. In his "Précis historique", he indicated that his troop was made up of the 3rd Battalion of Vosges Volunteers, the 10th Battalion of Meurthe Volunteers, the 38th, the 57th and a detachment of the 109th Infantry Regiment, as well as a company of foot artillery, which had no guns with it but had to use those that would be taken at Noirmoutier. General Haxo, with 700 to 900 men, and nine gabares, was at the head of the second assault wave at the port of La Cronière, in Beauvoir-sur-Mer. A third force, made up of seven barges, was to move to the north of the island to create a diversion. A column was also in reserve under the orders of Lieutenant-Colonel Aubertin. In his memoirs, Aubertin states that his column was made up of 455 men from the 11th battalion of the Orléans formation, detachments from the 109th and 110th infantry regiments, as well as some other corps, Mayençais grenadiers and artillerymen from the 1st battalion of Bas-Rhin volunteers.

The operations were supervised by General Louis-Marie Turreau, commander-in-chief of the army of the West, who had arrived in Beauvoir-sur-Mer on 1 January, and by three representatives on a mission: Prieur de la Marne, who was also a member of the Committee of Public Safety, Louis Turreau, the cousin of the general in chief, and Pierre Bourbotte.

Vendean Army
The Vendean garrison left by Charette to defend the island of Noirmoutier was 1,500 men strong according to the royalists Lucas de La Championnière and Le Bouvier-Desmortiers, 1,800 for General Turreau and 2,000 according to the anonymous memoirs of the military administrator. These numbers are variously taken up by historians: Jean-Joël Brégeon and Gérard Guicheteau give 1,500, Yves Gras and Jean Tabeur 1,800, Émile Gabory and Lionel Dumarcet 2,000. This garrison was under the orders of Alexandre Pineau du Pavillon. René de Tinguy was governor of the island, Benjamin Dubois de La Guignardière, former captain of Soullans' national guard, was commander of the main town, Louis Vasselot de Reigner was commander of the castle and Bernard Massip was second in command of the garrison. Hyacinthe Hervouët de La Robrie commanded 500 men in Barbâtre. Other leaders included Louis Savin, Jean Jodet, Pierre Gouin, Gazette de La Limousinière, Joseph Béthuis, Loizeau, Barraud père and fils de La Garnache and the parish captains Barraud de Saint-Hilaire-de-Riez and Barraud du Perrier.

Maurice d'Elbée, former generalissimo of the Catholic and Royal Army, is also present in Noirmoutier after having taken refuge there on the 2nd or 3rd November because of his serious wounds received at the battle of Cholet. He was accompanied by his wife, Marguerite-Charlotte Duhoux d'Hauterive; his brother-in-law, Pierre Duhoux d'Hauterive; and his friend Pierre Prosper Gouffier de Boisy. The place of his asylum is not well known. According to his grand-nephew, Charles-Maurice d'Elbée, and the Marquise de Bonchamps, he was lodged in the Jacobsen Hotel, then in the house of Madame Mourain. According to François Piet, he lived in a house called La Maduère.

Course of events

Sources
The course of events is known from various Republican sources. Adjutant-General Jordy sent a report to the Minister of War, and then wrote a more detailed account in his "Précis historique de mes actions civiles et militaires, par Jordy". The events are also evoked in the memoirs of General Turreau, Adjutant-General Aubertin and in the anonymous memoirs of a "military administrator". Testimonies are also left by two republican soldiers: André Amblard, a volunteer from Mayence, who wrote a logbook, and Auguste Dalicel, a brigadier in the 3rd battalion of volunteers from Lot-et-Garonne, who reported the fighting in a letter to his uncle. They both took part in the landings, under Jordy's orders. The most detailed account is given by Captain François Piet, then aged 19, born in Montmédy, aide-de-camp to General Dutruy. who wrote his "Memoirs left to my son" in 1806 During the battle, he fought in General Haxo's column. Appointed as a war commissioner in Noirmoutier in 1795, he then settled on the island where he worked as a notary during the Empire and the Restoration, before becoming a justice of the peace during the July Monarchy. His memoirs were published by his son, Jules Piet, in 1863.

Disembarkation at Pointe de la Fosse

On 2 January, Jordy had his soldiers embark on the barges at La Barre-de-Monts and L'Époids, between Bouin and Beauvoir-sur-Mer. At midnight, the flotilla set sail. The place chosen for the landing was between the Pointe de la Fosse, at the southern end of the island, and the Gois causeway.

While a small number of the barges sailed towards La Chaise and the port of Noirmoutier to create a diversion, Jordy headed for La Fosse. Despite the darkness, the Vendeans saw the Republican ships and at around 5 or 6 o'clock in the morning the eleven cannons of the artillery battery at La Fosse opened fire. About half a league north of the tip of the point, Jordy placed his barges in line, facing the shore: those in the centre responded to the fire of the Vendeans in order to attract their attention, while those on the wings began to unload the troops. A rocket gave the signal. As the republicans had no boats, the tide was low and the sea shallow, the infantrymen threw themselves directly into the water. In order to avoid fratricidal fire, they were ordered to fight only with the bayonet. Several barges ran aground on the beach during the operation. Jordy was wounded in the thigh and leg before he could leave his ship. He reached the shore, carried by grenadiers who made an improvised stretcher with their rifles. However, his manoeuvre succeeded and the Vendéens retreated so as not to find themselves surrounded. At 7 o'clock in the morning, all his troops had disembarked. The Fosse battery was taken in reverse and the Republican artillerymen seized the cannons. In his letter, Brigadier Dalicel wrote: "We were in firing range until eight o'clock. When we set foot on the ground, our people fell like cardboard capuchins. Once we were five or six hundred on the island, we made carnage of everything. On the side where we made the descent, we filled a salt marsh with the dead; there may well have been 900 of them, more of ours than of theirs.

Fighting at Barbâtre

The republicans then marched on the village of Barbâtre, held by the forces of Hyacinthe de La Robrie with a battery of four 36-pounder guns, and took it after a one-hour battle. All the men who were found in Barbâtre were massacred. According to the memoirs of François Piet: "peaceful people, fathers of families, old men who had stayed in their houses, millers who had not wanted to leave their mills, became the victims of the soldier's fury".

On his side, General Haxo began to cross the Gois causeway at about 9 o'clock in the morning, followed by the column of Aubertin. In spite of the low tide, the soldiers are up to their knees in water and the crossing is carried out slowly. Around noon or one o'clock in the afternoon, Haxo arrived at Barbâtre, where he joined up with Jordy, whose troops were at rest. Shortly before giving up his command, Jordy was seriously wounded by a Biscayan, who pierced his left parietal bone.

Capitulation of the Vendeans
After joining forces with Jordy, Haxo, now at the head of 3,000 men, marched on the town of Noirmoutier, in the north of the island. Near La Guérinière, on the narrowest part of the island, he encountered an entrenchment equipped with artillery. However, the Republicans took advantage of the low tide to bypass it on the right and left, and the Vendeans retreated without firing a single cannon shot. Haxo's troops then advanced to the outskirts of the town without encountering any resistance.

At three o'clock in the afternoon, two or three Vendéen horsemen presented themselves at the bridge of the Corbe, to hand over an offer of capitulation in exchange for the life of the whole garrison. General Haxo welcomed the offer, but the representatives on mission, Prieur de la Marne, Turreau and Bourbotte, expressed their opposition. However, on the insistence of Haxo, the representatives let it go.

Governor Tinguy, a conciliator, seemed to believe in keeping his word. If some leaders, such as Hyacinthe de La Robrie and Benjamin Dubois de La Guignardière, were of the opinion that they should fight to the end, the vast majority of the defenders preferred to surrender. The republicans enter the town without encountering any resistance. The Vendeans gather in the square and throw their rifles in heaps

The generalissimo Maurice d'Elbée is quickly discovered and taken prisoner. Dubois de La Guignardière is wounded and captured. According to Piet, his hiding place is denounced by Foré, a republican gunner to whom he had saved his life during the capture of the island by Charette. Only one leader, Hyacinthe de La Robrie, barely managed to escape and leave the island after having remained hidden for several days in the marshes of l'Épine.

Losses
On the evening of 3 January, the representatives on mission, Prieur de la Marne, Turreau and Bourbotte, stated in a letter to the National Convention that the expedition had only cost the Republic two men killed and ten to twelve wounded, including "the brave Jordy". In his report to the Minister of War, General Turreau also only mentions the loss of "ten men, most of whom are only wounded". and in his memoirs "ten or twelve men and some wounded" Jordy, on the other hand, puts Republican losses at 130 killed and 200 wounded in his report to the Minister of War. Jordy himself was seriously wounded and had to undergo trepanning. The day after the battle, he was promoted to the rank of brigadier general, while Aubertin was raised to the rank of adjutant-general.

On the side of the Vendéens, the number of combatants killed in battle is put at 400 by General Turreau, 500 by the representatives Prieur de la Marne, Turreau and Bourbotte, as well as by the "military administrator" and 600 by the representative Guermeur. The number of prisoners was 600 according to François Piet, 600 to 700 initially and 1,200 in total, according to Adjutant-General Aubertin, 1,000 according to General Turreau, 1,100 according to Dalicel 1,000 to 1,200 according to the representative Guermeur, 1,200 according to representatives Prieur, Turreau and Bourbotte 1,500 according to the "military administrator", 2,000 according to the Revolutionary Committee of the "île de la Montagne" and 3,000 according to the volunteer Amblard. However, Turreau and Bourbotte announced on 8 January that 300 more "brigands" had been taken after battues and searches in the woods and underground passages. For the historian Alain Gérard, the estimates of 1,100 to 1,200 prisoners seem the most likely. In 1975, in his dissertation for his DES in legal history, Les Commissions de Noirmoutier. Janvier 1794 à août 1794, Jean-Louis Pellet recommended a range of 1,200 to 1,500 prisoners shot.

The republicans also took 19 cannons and 11 pierriers according to General Turreau. Prieur, Turreau and Bourbotte reported the capture of around 50 cannons and 700 to 800 rifles, Guermeur 52 cannons and 800 rifles and Jordy 80 pieces of large artillery and 23 medium and small ones.

Execution of prisoners

The prisoners are locked up in the churches of the city. The capitulation is not respected by the representatives in mission who decide to make pass all the garrison by the weapons in spite of a vain attempt of Haxo to oppose it. The evening of the capture of the island, they write to the Committee of Public Safety: "a military commission, which we have just created, will make prompt justice of all these traitors". The executions began on 4 January, and were carried out for the most part in two or three days. Only 70 men from Notre-Dame-de-Monts managed to convince the commission that they had been forcibly enlisted and were spared. The prisoners were taken in groups from the church of Saint-Philbert and taken to the place known as "la Vache", in the district of Banzeau, where they were shot at close range. The street used by the condemned prisoners would be named "Rue des Martyrs" many years later.

François Piet writes in his memoirs:

On January 4, the republican troops start to search the island and according to Turreau and Bourbotte "a flood of priests, women of emigrants" and many leaders are arrested. On 8, the representatives in mission announce to the Committee of Public Safety the creation of a military commission to judge these prisoners. This one sits at the Jacobsen hotel. The instruction is very summary, the judges being satisfied to record the identity of the captives and to obtain from them various information. Tinguy, Pineau du Pavillon, Dubois, Savin, Reigner and the various Vendéen chiefs were shot at a date that is impossible to determine. Eight clergymen were also among the victims, including René-Charles Lusson, vicar of Saint-Georges-de-Montaigu and chaplain to the army of the Centre.

The republicans also discovered in Noirmoutier Lieutenant-Colonel Jean-Conrad Wieland, who commanded the republican garrison when Charette took the island. He was soon accused of treason, especially as a letter he had written to Commandant Pineau du Pavillon indicated that he had maintained good relations with the royalist staff during the occupation of the island by the Vendeans. The representatives on mission then decided to have Wieland executed, without trial according to Piet's memoirs.

Maurice d'Elbée was interrogated by the representatives on mission and by General Turreau at an uncertain date. The minutes of his interrogation were written by François Piet. The former generalissimo was executed between 6 and 9 January. Unable to walk, he was carried in an armchair to the Place d'Armes, accompanied by Pierre Duhoux d'Hauterive and Pierre Prosper Gouffier de Boisy. Jean-Conrad Wieland was added to them despite the denials of his co-accused who tried in vain to clear him. The four men were then tied to posts. and shot

A few more executions took place in the following weeks. On 29 January, d'Elbée's wife, Marguerite-Charlotte Duhoux d'Hauterive, and Victoire Élisabeth Mourain de L'Herbaudière, née Jacobsen, were shot after being sentenced to death by the military commission.

Most of the bodies are buried in the dunes of La Claire and near Le Vieil and La Madeleine. Others, like that of d'Elbée, were thrown into the moat of the castle. From 5 January to 25 April, up to 25 gravediggers, led by the citizen Honoré Aubert, who had become "responsible for the health of the air", were employed to "recover the corpses".

In 1950, on the initiative of Abbé Raimond, a chapel called Notre-Dame de la Pitié, or "Chapelle des Martyrs", was built on the site of the Banzeau shootings.

Destructions at Barbâtre
The same evening of the capture of the island, the representatives Prieur de la Marne, Turreau and Bourbotte decide to rename Noirmoutier to "île de la Montagne". The island of Bouin becomes the "Marat island". On 8 January, they declared the inhabitants of Barbâtre "traitors to the Fatherland" and ordered the destruction of all the houses in the town, except those "suitable for public establishments and the defence of the coast". According to Piet, the order of destruction was only carried out "as regards the church and the houses bordering the plain on the eastern side". In total, 87 houses were destroyed and 200 to 300 inhabitants of Barbâtre emigrated to Noirmoutier-en-l'Île. The population of Barbâtre fell from 2,221 inhabitants in 1791 to 1,421 in 1804.

The prison of Noirmoutier
After the fighting and massacres of January, Noirmoutier experienced a period of calm. According to the memoirs of François Piet: "A few months passed without anything disturbing the peace we enjoyed. [...] A large garrison left us without worrying about new ventures on the part of the Vendéens. [...] Country parties, music, balls, were a diversion from past misfortunes, and a stunner for future ones".

However, the island was then used as a prison, where male and female Vendéen prisoners arrested on the continent were sent. A new military commission was then established in Noirmoutier due to the dissolution of the commissions in Sables-d'Olonne and Fontenay-le-Comte. It was made up of Collinet, president, Simon, Foré and Tyroco, judges, and François Piet, accuser. From 5 May to 14 June 1794, it pronounced at least 28 death sentences and a large number of acquittals. The condemned were shot on the dunes between Le Vieil and La Claire and buried on the spot. However, according to Piet, about fifty prisoners sent by the Revolutionary Committee of Les Sables, as soon as they disembarked at La Chaise, were shot without trial on Tyroco's orders. Following a decree by representatives Bourbotte and Bô, the Collinet commission was replaced by members of the Angers commission - made up of Antoine Félix, president; Laporte, vice-president; Obrumier and Goupil, judges; and Hudoux, accuser - which pronounced 25 death sentences, 18 deportations and 600 acquittals between 17 June and 14 August 1794. The last executions on the island took place on 3 August at La Claire: 22 people, including 13 women, were shot. In total, at least 1,300 people were imprisoned in Noirmoutier during 1794 and between 128 and 400 died there.

Consequences
On the evening of January 3, 1794, just before embarking for Nantes, general Louis-Marie Turreau writes from Noirmoutier to the minister of war Jean-Baptiste Bouchotte: "The war of the Vendée is not completely finished, but it should not give any more concern". On their side, the representatives in mission Prieur de la Marne, Turreau and Bourbotte write the same day to the Committee of Public Safety: "The victory is with us. The resumption of the important port of Noirmoutier, which was the last entrenchment and the last hope of the rebels of the Vendée, gives us the insurance to see soon completely finished this infamous war; it removes to the brigands any communication by sea with the perfidious English; it makes the Republic mistress of a country fertile in subsistence".

The Vendée insurrection then seemed to have expired. In the Bas-Poitou, the column of Adjutant-General Joba crushed the 3,000 men of Charette and Joly at Saint-Fulgent on 10 January, then dislodged them the next day from the forest of Grasla. Wounded, Charette spent the rest of January hiding with several hundred men in the convent of Val de Morière, then in Saligny and in the forest of Grasla. In Pays de Retz, General Haxo routed La Cathelinière's forces on 12 January, who had taken refuge in the Princé forest. Wounded, La Cathelinière is captured in February and executed in March. In Anjou, the few hundred men gathered by Henri de La Rochejaquelein and Jean-Nicolas Stofflet are dispersed on January 1st at Les Cerqueux by the troops of General Grignon.

Repression intensified in the Vendée in early 1794. On 11 December 1793, the representative on a mission, Jean-Baptiste Carrier, had written to the Committee of Public Safety: "As soon as the news of the capture of Noirmoutier reaches me, I will send an imperative order to Generals Dutruy and Haxo to put to death in all the insurgent countries all the individuals of all sexes who are found there, without distinction, and to complete the burning of everything". In Nantes, General Louis-Marie Turreau planned a general offensive aimed at destroying the last royalist bands and setting fire to the entire insurgent territory. In mid-January, he had his "infernal columns" set in motion.

Notes and references

Notes

References

Bibliography

1794 in France
Battles involving France
Battles of the War in the Vendée
Battles in Pays de la Loire
Conflicts in 1794
History of Vendée

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Strevelshoek is a hamlet in the Dutch province of South Holland. It is located about 1 km northwest of the village of Rijsoord, on the south bank of the Waaltje. The hamlet is not mentioned on recent topographical maps.

Strevelshoek was a separate municipality between 1817 and 1846, when it merged with Rijsoord. According to the 19th-century historian A.J. van der Aa, there were five houses in the municipality around 1845, four of which made up the hamlet Strevelshoek. The municipality had about 40 inhabitants. The area is now part of the municipality of Ridderkerk.

References

Ridderkerk
Populated places in South Holland
Former municipalities of South Holland


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October 15, 2015 by ajohnson
Review| Pan: When Bad Things Happen to Good Ideas
"Resoundingly god-awful" and "theme-park ride from hell," are just a couple of phrases popular critics are throwing around regarding Joe Wright's adaptation of Peter Pan. I am quite a fan of Wright's films, and I also know that every director has at least one film that doesn't quite happen the way it should. While I don't share the severity of loathing and repulsion radiating from other critics, I must say the film is quite a shipwreck. Pun entirely intended.
Pan is not entirely without merit. Its ideas for telling the Peter Pan story are incredibly unique. It's not Peter Pan as we know it, it's a prequel. As the narrator tells us in the opening scene, it's a story where friends start out as enemies and enemies start out as friends. These themes and ideas provide a solid foundation, but the house built upon that foundation severely lacks the necessary strength and function. One of the things that grabbed me from the trailer is the notion of Hook being a good guy and wanting to see where/why he goes wrong. Sadly, this is one of many ideas that never fully develops.
In J.M. Barrie's original story, we never learn who Peter really is or where he comes from. It may be that it was intended to be ambiguous, but I appreciate that Pan at least tries to bring some new things to the table. Writer Jason Fuchs crafts a fairy tale of Peter's origin as the product of the love between a great warrior woman and a fairy prince. Peter (played actually very well by adorable newcomer Levi Miller) is more than just a mischievous boy who never grows up. Even in the earliest scenes at the run-down, abusive orphanage we see his potential for leadership. It's in this aspect I find this version of Peter ultimately more interesting than the character in the source material.
Though many critics gripe about the noisy design and special effects, I think they are the more redeemable qualities of the film. I totally dig the Shakespearean/Steam Punk panache we get from Captain Blackbeard, and absolutely entranced by the crystalized, stalactitical realm of the fairy kingdom. Are the costumes, sets, and effects flamboyant and absurd? Absolutely! It's Neverland, that's exactly what anyone with an imagination should expect. It's a place where pirates, Native Americans, fairies, and mermaids all exist together. Neverland, like Wonderland, is anything you can possibly dream it up to be.
Now, where the good ideas go bad…
The first thing that leaps to mind is the horrendously gimmicky musical numbers that Blackbeard and his miners sing in the beginning. First of all, Nirvana's "Smells Like Teen Spirit" was already redone as a musical number much more effectively in 2001's Moulin Rouge. Additionally, these 90s tunes they chose are totally jarring, especially given that no other singing takes place in the film. It seems like Joe Wright is trying to be Baz Luhrmann here, but to what purpose? He's already established himself as a brilliant filmmaker. Copying another person's style so overtly is a rookie mistake.
Speaking of overt, the dialogue is most definitely so. Particularly in the second half, everything spoken is so completely on-the-nose even a five year-old might find their intelligence insulted. In the last moments of the film, Peter and Hook (friends at this point) have a quick exchange where Peter asks Hook "We'll always be friends, right?" and Hook says something like, "Sure, what could possibly go wrong?" Most people watching this movie know the story inside out, there's certainly been enough reboots of it by now. It was completely unnecessary to drive this point like a nail through our brains. Hook and Peter later become enemies. WE GET IT.
I quite like Garrett Hedlund, who came off like a Han Solo of sorts in his role as James Hook, but his scenes with Tiger Lily (Rooney Mara) were nauseatingly confectionary. They "fell in love" so fast it set a record even for fairy tales. Hook even had to explain to the audience he was flirting with her, as if her forced giggles weren't enough of a tip-off. With such a sensational cast, including the charismatic Hugh Jackman as Blackbeard, the performances should have been a delight. I do not believe it was a lack of ability that made the acting so exaggerated, but more of a lack decent material to work with.
Pan cannot decide whether or not it wants to be a dark comedy/fantasy, or a slapstick family movie. One minute Blackbeard is kicking children off planks to certain death, the next we are bombarded with corny hijinks including a duel on a trampoline… seriously. I personally would have preferred the darker route as this would have aided the maturity level of the story. However, the more pressing matter is the lack of consistency in tone and plot.
I am not one who particularly enjoys writing reviews or analyses on a negative slant. There's enough of that already out there. However, when it comes to good ideas that go bad, I feel an obligation to speak up. This movie could have and should have been better. I would not label Pan as "bad" in the typical sense. In fact, the audience at my theater was laughing and enjoying it quite a bit at times. However it is profoundly disappointing to anyone who desired it to be more than yet another serving of cinematic junk food.
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