text:
\section{Introduction}
In this paper,
we use the following two symbol classes of $S_{0,0}$-type.
One is the bilinear H\"ormander class $BS^{m}_{0,0}$, $m \in \R$,
consisting of all $\sigma(x,\xi_1,\xi_2) \in C^{\infty}((\R^n)^3)$ such that
\[
|\partial^{\alpha}_{x}\partial^{\beta_1}_{\xi_1}\partial^{\beta_2}_{\xi_2}
\sigma(x,\xi_1,\xi_2)|
\le C_{\alpha,\beta_1,\beta_2}(1+|\xi_1|+|\xi_2|)^m
\]
for all multi-indices $\alpha,\beta_1,\beta_2 \in \N_0^n=\{0,1,2,\dots \}^n$.
The other is $BS^{(m_1,m_2)}_{0,0}$, $m_1,m_2 \in \R$,
consisting of all $\sigma(x,\xi_1,\xi_2) \in C^{\infty}((\R^n)^3)$ such that
\[
|\partial^{\alpha}_{x}\partial^{\beta_1}_{\xi_1}\partial^{\beta_2}_{\xi_2}
\sigma(x,\xi_1,\xi_2)|
\le C_{\alpha,\beta_1,\beta_2}(1+|\xi_1|)^{m_1}(1+|\xi_2|)^{m_2} .
\]
For a symbol $\sigma$,
the bilinear pseudo-differential operators $T_{\sigma}$ is defined by
\[
T_{\sigma}(f_1,f_2)(x)
=\frac{1}{(2\pi)^{2n}}
\int_{(\R^n)^2}e^{ix\cdot(\xi_1+\xi_2)}
\sigma(x,\xi_1,\xi_2)
\widehat{f_1}(\xi_1)\widehat{f_2}(\xi_2)\, d\xi_1d\xi_2
\]
for $f_1,f_2 \in \Sh(\R^n)$.

In the linear case,
the celebrated Calder\'on-Vaillancourt theorem \cite{CV} states that
if a symbol $\sigma(x,\xi) \in C^{\infty}((\R^n)^2)$ satisfies
\begin{equation}\label{linear}
|\partial^{\alpha}_{x}\partial^{\beta}_{\xi}\sigma(x,\xi)|
\le C_{\alpha,\beta},
\end{equation}
then the corresponding pseudo-differential operator
is bounded on $L^2$.
As a bilinear counterpart of this theorem,
we naturally expect that
the condition $\sigma \in BS^0_{0,0}$
ensures the $L^2 \times L^2 \to L^1$ boundedness
of $T_{\sigma}$.
However, B\'enyi-Torres \cite{BT} pointed out
that this boundedness does not hold in general.
Michalowski-Rule-Staubach \cite{MRS}
treated the subcritical case $m<-n/2$,
and then Miyachi-Tomita \cite{MT-IUMJ} showed that
all bilinear pseudo-differential operators
with symbols in $BS^m_{0,0}$ are bounded from
$L^2 \times L^2$ to $L^1$ if and only if $m \le -n/2$.
It should be mentioned that
the stronger $L^2 \times L^2 \to h^1$ boundedness
was proved in \cite{MT-IUMJ}, where $h^1$ is the local Hardy space.
See also Miyachi-Tomita \cite{MT}
for the classes of $S_{\rho,\rho}$-type,
$0 \le \rho<1$.

Kato-Miyachi-Tomita recently proved that
if $m_1,m_2<0$,
$m_1+m_2=-n/2$ and $\sigma \in BS^{(m_1,m_2)}_{0,0}$,
then $T_{\sigma}$ is bounded from $L^2 \times L^2$ to
the amalgam space $(L^2,\ell^1)$
(\cite[Theorem 1.3, Example 1.4]{KMT}).
Here, we remark that
\begin{equation}\label{two-classes}
BS^{-n/2}_{0,0} \subset BS^{(m_1,m_2)}_{0,0},
\quad m_1,m_2 \le 0, \ m_1+m_2=-n/2.
\end{equation}
Hence, for $1 \le p \le 2$,
the $L^2 \times L^2 \to L^p$ boundedness
of $T_{\sigma}$ with $\sigma \in BS^{-n/2}_{0,0}$ follows
from the embedding $(L^2,\ell^1) \hookrightarrow L^p$
(see Section \ref{section2}).
In the linear case,
since the constant function satisfies \eqref{linear},
the $L^2 \to L^p$ boundedness of pseudo-differential operators
with symbols satisfying \eqref{linear} holds
if and only if $p=2$.
On the other hand,
in the bilinear case,
we have a choice to choose target spaces.
Thus, the purpose of this paper
is to determine the ranges of bilinear pseudo-differential operators
with symbols in the critical class $BS^{-n/2}_{0,0}$ on $L^2 \times L^2$
in the framework of Besov spaces $B_{p,q}^s$.

The main result of this paper is the following.

\begin{thm} \label{main1}
Let $m_1, m_2 < 0$ and $m_1+m_2=-n/2$.
If $1 \le p \le 2$, then all bilinear pseudo-differential operators
with symbols in $BS^{(m_1,m_2)}_{0,0}$ are bounded from $L^2(\R^n) \times L^2(\R^n)$
to $B_{p,1}^0(\R^n)$.
\end{thm}

Since $B_{1, 1}^0 \hookrightarrow L^1$
(in fact, $B_{1,1}^0 \hookrightarrow h^1 \hookrightarrow L^1$),
Theorem \ref{main1} is an improvement of the $L^2 \times L^2 \to L^1$ boundedness.

By \eqref{two-classes},
the following means the optimality of Theorem \ref{main1}.

\begin{thm} \label{main2}
Let $0 < p, q \leq \infty$.
Then all bilinear pseudo-differential operators with symbols in $BS^{-n/2}_{0,0}$
are bounded from $L^2(\R^n) \times L^2(\R^n)$ to $B^0_{p, q}(\R^n)$
if and only if $1 \leq p \leq 2$ and $1 \leq q \leq \infty$.
\end{thm}

By the ``only if" part of Theorem \ref{main2}
and the embedding $B_{p,q_1}^s \hookrightarrow B_{p,q_2}^0$
for $0<p,q_1,q_2 \le \infty$ and $s>0$,
if the $L^2 \times L^2 \to B_{p,q}^s$ boundedness
of all $T_{\sigma}$ with $\sigma \in BS^{-n/2}_{0,0}$ holds,
then $s$ must satisfy $s \le 0$.
This is the reason why
we consider the case $s=0$ in Theorems \ref{main1} and \ref{main2}.

The following says that
Theorem \ref{main1} cannot be compared with the result of \cite{KMT}.

\begin{prop}\label{main3}
There is no embedding relation between
the amalgam space $(L^2, \ell^1)(\R^n)$
and the Besov spaces $B_{p,1}^0(\R^n)$, $1 \le p \le 2$.
\end{prop}

Related results to bilinear Fourier multiplier operators of $S_{0,0}$-type
on $L^2 \times L^2$ can be found in
Grafakos-He-Slav\'ikov\'a \cite{GHS}
and Slav\'ikov\'a \cite{Slavikova}.

The contents of this paper are as follows.
In Section \ref{section2},
we give preliminary facts.
In Section \ref{section3},
we give basic estimates used in the proof of Theorem \ref{main1}.
In Sections \ref{section4}, \ref{section5} and \ref{section6},
we prove Theorems \ref{main1}, \ref{main2}
and Proposition \ref{main3}, respectively.

\section{Preliminaries}\label{section2}
For two nonnegative quantities $A$ and $B$,
the notation $A \lesssim B$ means that
$A \le CB$ for some unspecified constant $C>0$,
and $A \approx B$ means that
$A \lesssim B$ and $B \lesssim A$.
For $1 \le p \le \infty$,
$p'$ is the conjugate exponent of $p$,
that is, $1/p+1/p'=1$.
The usual inner product of $f, g \in L^2(\R^n)$ is denoted by
$\la f, g \ra$ .

Let $\Sh(\R^n)$ and $\Sh'(\R^n)$ be the Schwartz space of
rapidly decreasing smooth functions on $\R^n$ and its dual,
the space of tempered distributions, respectively.
We define the Fourier transform $\F f$
and the inverse Fourier transform $\F^{-1}f$
of $f \in \Sh(\R^n)$ by
\[
\F f(\xi)
=\widehat{f}(\xi)
=\int_{\R^n}e^{-ix\cdot\xi} f(x)\, dx
\quad \text{and} \quad
\F^{-1}f(x)
=\frac{1}{(2\pi)^n}
\int_{\R^n}e^{ix\cdot \xi} f(\xi)\, d\xi.
\]
For $m \in L^{\infty}(\R^n)$,
the Fourier multiplier operator $m(D)$ is defined by
$m(D)f=\F^{-1}[m\widehat{f}]$ for $f \in \Sh(\R^n)$.

Let $\{\psi_\ell \}_{\ell \geq 0} $ be a sequence of Schwartz functions on $\R^n$ satisfying
\begin{align} \label{Besov-parti}
\begin{split}
&\supp \psi_0 \subset \{\xi \in \R^n \,:\, |\xi| \leq 2 \},\\
&\supp \psi_\ell \subset \{ \xi \in \R^n \,:\, 2^{\ell -1 }\leq |\xi| \leq 2^{\ell + 1}\},
\quad \ell\geq 1,
\\
&|\partial^{\alpha}\psi_\ell (\xi)| \leq C_\alpha 2^{-\ell|\alpha|}, \quad \ell \geq 0,
\ \alpha \in \N^n_0, \ \xi \in \R^n,
\\
&\sum_{\ell=0}^\infty \psi_\ell(\xi) = 1, \quad \xi \in \R^n.
\end{split}
\end{align}
For $0 < p, q \leq \infty$ and $s \in \R$,
the Besov space $B^s_{p, q}(\R^n)$ consists of all $f \in \Sh'(\R^n)$
such that
\[
\|f\|_{B^s_{p, q}}=
\bigg(  \sum_{\ell=0}^\infty 2^{\ell sq} \|\psi_\ell(D)f\|^q_{L^p} \bigg)^{1/q}<\infty
\]
with usual modification when $q=\infty$.
It is well known that the definition of Besov spaces
$B_{p,q}^s$ is independent of the choice
of $\{\psi_\ell\}_{\ell \geq 0}$ satisfying \eqref{Besov-parti}.
For $0 < p, q_1,q_2 \le \infty$ and $s_1,s_2 \in \R$,
the embedding $B_{p,q_1}^{s_1} \hookrightarrow B_{p,q_2}^{s_2}$ holds
if $s_1=s_2$ and $q_1 \le q_2$, or if $s_1>s_2$.
The dual space of $B^s_{p, q}$
coincides with $B^{-s}_{p^\prime, q^\prime}$,
where $1 \leq p, q < \infty$ and $s \in \R$.
See \cite{Triebel} for more details on Besov spaces.

For $1 \leq p,q \leq \infty$,
the amalgam space
$(L^p,\ell^q)(\R^n)$ consists of all measurable functions $f$ on
$\R^n$ such that
\begin{equation*}
\| f \|_{ (L^p,\ell^q)}
=\bigg\{ \sum_{\nu \in \Z^n}
\bigg( \int_{\nu+[-1/2,1/2]^n} \big| f(x) \big|^p \, dx
\bigg)^{q/p} \bigg\}^{1/q}
< \infty
\end{equation*}
with usual modification when $p$ or $q$ is infinity.
Obviously, $(L^p,\ell^p) = L^p$,
and $(L^{p_1}, \ell^{q_1})
\hookrightarrow (L^{p_2},\ell^{q_2})$
for $p_1 \geq p_2$ and $q_1 \leq q_2$.
In particular,
$(L^2,\ell^1) \hookrightarrow L^r$
for $1 \leq r \leq 2$,
and
the stronger embedding
$(L^2,\ell^1) \hookrightarrow h^1$ holds
in the case $r=1$,
where $h^1$ is the local Hardy space
(see \cite[Section 2.3]{KMT-2}).

We end this section by quoting the following, which is called Schur's lemma
(see, e.g., \cite[Appendix A]{Grafakos-Modern}).

\begin{lem}\label{Schur's lemma}
Let $\{A_{k_1, k_2}\}_{k_1, k_2 \ge 0}$ be a sequence of nonnegative numbers satisfying
\begin{align*}
\sup_{k_1 \ge 0} \sum_{k_2 \ge 0} A_{k_1, k_2} < \infty
\quad \text{and} \quad
\sup_{k_2 \ge 0} \sum_{k_1 \ge 0} A_{k_1, k_2} < \infty.
\end{align*}
Then
\begin{align*}
\sum_{k_1, k_2 \ge 0} A_{k_1, k_2} b_{k_1} c_{k_2}
\lesssim \Big(\sum_{k_1 \ge 0} b_{k_1}^2 \Big)^{1/2}\Big(\sum_{k_2 \ge 0} c_{k_2}^2 \Big)^{1/2}
\end{align*}
for all nonnegative sequences $\{b_{k_1}\}_{k_1 \ge 0}$ and $\{c_{k_2}\}_{k_2 \ge 0}$.
\end{lem}

\section{Basic estimates}\label{section3}
For $R > 0$ and $L > n$, we set
\begin{align*}
  S_R(f)(x) = R^n\int_{\R^n} \frac{|f(y)|}{(1+R|x-y|)^L}\, dy,
\end{align*}
and simply write $S_1(f)(x) = S(f)(x)$.

 The following lemma can be found in the proof of \cite[Lemma 4.2]{Kato}, but we will give the proof for the reader's convenience.

\begin{lem} \label{lem-dyadic-est}
 Let $\varphi \in \Sh(\R^n)$.  Then, we have
 \begin{align*}
  \bigg(\sum_{\nu \in \Z^n} |\varphi(R^{-1}(D -\nu))f(x)|^2 \bigg)^{1/2}
  \lesssim R^{n/2}S_{R}(|f|^2)(x)^{1/2}
 \end{align*}
 for all $R \ge 1$, where $\varphi(R^{-1}(D -\nu))f = \F^{-1}[\varphi(R^{-1}(\cdot -\nu)) \widehat{f}]$.
\end{lem}

\begin{proof}
Since $\R^n = \cup_{\mu \in \Z^n} (2\pi\mu + [-\pi, \pi]^n)$,
we have
 \begin{align*}
 &\varphi(R^{-1}(D-\nu))f(x)
 = R^n \int_{\R^n} e^{i\nu\cdot(x-y)}
   \Phi(R(x-y)) f(y) \,dy \\
  &= R^n e^{i\nu\cdot x}\sum_{\mu \in \Z^n} \int_{2\pi\mu + [-\pi, \pi]^n} e^{-i\nu\cdot y}
   \Phi(R(x-y)) f(y) \,dy\\
  &= R^n e^{i\nu\cdot x}  \int_{[-\pi, \pi]^n} e^{-i\nu\cdot y}
   \Big(
   \sum_{\mu \in \Z^n} \Phi(R(x-y-2\pi\mu)) f(y+2\pi\mu)
   \Big)\,
    dy,
 \end{align*}
where $\Phi = \F^{-1}\varphi$.
Here, $\sum_{\mu \in \Z^n} \Phi(R(x-y-2\pi\mu)) f(y+2\pi\mu)$
is a $(2\pi \Z)^n$-periodic function of the $y$-variable. Hence, it follows
from Parseval's identity that
 \begin{align*}
  &\Big( \sum_{\nu \in \Z^n} |\varphi(R^{-1}(D-\nu))f(x)|^2 \Big)^{1/2} \\
  &=(2\pi)^{n/2} R^n
  \Big(
  \int_{[-\pi, \pi]^n} \Big| \sum_{\mu \in \Z^n}\Phi(R(x-y-2\pi\mu)) f(y+2\pi\mu) \Big|^2 dy
  \Big)^{1/2}.
 \end{align*}
 Since
 \begin{align*}
  \sup_{z \in \R^n}\sum_{\mu \in \Z^n} |\Phi(R(z-2\pi\mu))|
   &\lesssim
   \sup_{z \in \R^n}\sum_{\mu \in \Z^n} (1+R|z-2\pi\mu|)^{-(n+1)}\\
   &\leq
   \sup_{z \in \R^n}\sum_{\mu \in \Z^n} (1+|z-2\pi\mu|)^{-(n+1)}
   \lesssim 1,
 \end{align*}
 where we used the assumption $R \ge 1$ in the second inequality, we have by Schwarz's inequality
 \begin{align*}
  &\Big(
  \int_{[-\pi, \pi]^n} \Big| \sum_{\mu \in \Z^n}\Phi(R(x-y-2\pi\mu)) f(y+2\pi\mu) \Big|^2 dy
  \Big)^{1/2}\\
  &\lesssim
  \Big(
  \int_{[-\pi, \pi]^n}  \sum_{\mu \in \Z^n} |\Phi(R(x-y-2\pi\mu))| |f(y+2\pi\mu)|^2 dy
  \Big)^{1/2}\\
  &=\Big(
      \int_{\R^n} |\Phi(R(x-y))| |f(y)|^2 dy
      \Big)^{1/2}.
 \end{align*}
 Since $\Phi$ is a rapidly decreasing function, this gives the desired estimate.
\end{proof}

In the rest of this section, we assume that $m_1, m_2 \in \R$ and
$\sigma \in BS^{(m_1, m_2)}_{0, 0}$, and  shall consider the decomposition of $\sigma$.
Let $\{\psi_j\}_{j \geq 0}$ be as in (\ref{Besov-parti}), and let $\varphi \in \Sh(\R^n)$
be such that
\begin{align*}
 \supp \varphi \subset [-1, 1]^n
 \quad \text{and} \quad
 \sum_{\nu \in \Z^n} \varphi(\xi- \nu) = 1, \quad \xi \in \R^n.
\end{align*}
Using these functions, we decompose $\sigma$ as
\begin{align} \label{decomp-symb}
\sigma(x, \xi_1, \xi_2)
= \sum_{j \in \N_0} \sum_{\substack{\K \in (\N_0)^2 \\ \K= (k_1, k_2)}} \sigma_{j, \K} (x, \xi_1, \xi_2)
= \sum_{j \in \N_0} \sum_{\substack{\K \in (\N_0)^2 \\ \K= (k_1, k_2)}}  \sum_{\substack{\Nu \in (\Z^n)^2 \\ \Nu= (\nu_1, \nu_2)}} \sigma_{j, \K, \Nu} (x, \xi_1, \xi_2)
\end{align}
with
\begin{align*}
\sigma_{j, \K} (x, \xi_1, \xi_2)
&=  [\psi_j(D_x)\sigma](x, \xi_1, \xi_2)\psi_{k_1}(\xi_1)\psi_{k_2}(\xi_2)\\
&= \Big(
    \int_{\R^n} \F^{-1}\psi_j(y) \sigma(x-y, \xi_1, \xi_2)\, dy
    \Big) \psi_{k_1}(\xi_1)\psi_{k_2}(\xi_2)
\end{align*}
and
\begin{align*}
\sigma_{j, \K, \Nu} (x, \xi_1, \xi_2) =
\sigma_{j, \K} (x, \xi_1, \xi_2)  \varphi(\xi_1-\nu_1) \varphi(\xi_2-\nu_2).
\end{align*}

\begin{lem} \label{lem-symb-est}
$(1)$
For each $\beta_1, \beta_2 \in (\N_0)^n$ and $N \in \N_0$, we have
\begin{align*}
|\pa^{\beta_1}_{\xi_1} \pa^{\beta_2}_{\xi_2} \sigma_{j, \K, \Nu}(x, \xi_1, \xi_2)| \lesssim 2^{k_1m_1 +k_2m_2 -jN}
\end{align*}
for  all $j \in \N_0, \,  \K = (k_1, k_2) \in (\N_0)^2$ and $\Nu \in (\Z^n)^2$.\\
$(2)$ For each $N \in \N_0$, we have
\begin{align*}
|T_{\sigma_{j, \K,\Nu}}(f_1, f_2)(x) | \lesssim 2^{k_1m_1 +k_2m_2 -jN}  S(f_1)(x) S(f_2)(x)
\end{align*}
for all $j \in \N_0, \, \K = (k_1, k_2) \in (\N_0)^2$ and $\Nu \in (\Z^n)^2$.
\end{lem}

\begin{proof}
We first prove the assertion $(1)$.
Let $j \geq 1$. By the moment condition of $\F^{-1}\psi_j$
and Taylor's formula, we have
\begin{align*}
 &\sigma_{j, \K, \Nu}(x, \xi_1, \xi_2) \\
 &= \int_{\R^n} \F^{-1}\psi_j(y)
   \Big(
   \sigma_{\K, \Nu}(x-y, \xi_1, \xi_2)
   - \sum_{|\alpha| < N} \frac{(-y)^{\alpha}}{\alpha!}
   [\partial_x^\alpha
   \sigma_{\K, \Nu}](x, \xi_1, \xi_2)
   \Big) \,dy\\
 &= \int_{\R^n} \F^{-1}\psi_j(y)
   \Big(
   N \sum_{|\alpha| = N}  \frac{(-y)^{\alpha}}{\alpha!}
   \int_0^1 (1-t)^{N-1} [\partial_x^\alpha
   \sigma_{\K, \Nu}](x-ty, \xi_1, \xi_2)\, dt
   \Big) \,dy ,
\end{align*}
where
\[
\sigma_{\K, \Nu}(x, \xi_1, \xi_2)
=\sigma(x,\xi_1,\xi_2)
\psi_{k_1}(\xi_1)\psi_{k_2}(\xi_2)
\varphi(\xi_1-\nu_1) \varphi(\xi_2-\nu_2) .
\]
Since $1 + |\xi_i| \approx 2^{k_i}$ for $\xi_i \in \supp \psi_{k_i}$,
we see that
\begin{align*}
 |\partial_x^\alpha \partial_{\xi_1}^{\beta_1} \partial_{\xi_2}^{\beta_2}
   \sigma_{\K, \Nu}(x, \xi_1, \xi_2)| \lesssim 2^{k_1m_1+k_2m_2}.
\end{align*}
Hence,
\begin{align*}
 &|\partial_{\xi_1}^{\beta_1} \partial_{\xi_2}^{\beta_2}
 \sigma_{j, \K, \Nu}(x, \xi_1, \xi_2)|\\
  &\lesssim \int_{\R^n} |\F^{-1}\psi_j(y)|
   \Big(
   \sum_{|\alpha| = N}  |y^{\alpha}|
   \int_0^1 |[\partial_x^\alpha \partial_{\xi_1}^{\beta_1} \partial_{\xi_2}^{\beta_2}
   \sigma_{\K, \Nu}](x-ty, \xi_1, \xi_2)|\,dt
   \Big) \,dy\\
 &\lesssim 2^{k_1m_1+k_2m_2} \int_{\R^n} |\F^{-1}\psi_j(y)| |y|^N \, dy
 \lesssim 2^{k_1m_1+k_2m_2-jN},
\end{align*}
where we used the inequality $|\F^{-1}\psi_j(y)| \lesssim 2^{jn}(1+2^j|y|)^{-(N+n+1)}$. If we do not use the moment condition in the above argument, we have the same estimate with $j=0$. The proof of the assertion $(1)$ is complete.

Next, we will show the assertion (2). We write
\begin{align*}
 T_{\sigma_{j, \K, \Nu}}(f_1, f_2)(x)
 = \int_{(\R^n)^2}  K_{j, \K, \Nu}(x, x-y_1, x-y_2) f_1(y_1)f_2(y_2)\, dy_1 dy_2,
\end{align*}
where $K_{j, \K, \Nu}$ is defined by
\begin{align*}
  K_{j, \K, \Nu}(x, y_1, y_2) =
  \frac{1}{(2\pi)^{2n}}\int_{(\R^n)^2} e^{i(y_1 \cdot \xi_1+y_2 \cdot \xi_2)} \sigma_{j, \K, \Nu}(x, \xi_1, \xi_2) \, d\xi_1 d\xi_2.
\end{align*}
It follows from integration by parts,
the assertion (1) and the support condition
$\supp \sigma_{j, \K, \Nu}(x,\cdot,\cdot) \subset (\nu_1, \nu_2) + [-1, 1]^{2n}$
for each $x \in \R^n$
that
\begin{align*}
 |K_{j, \K, \Nu}(x, y_1, y_2)| \lesssim 2^{k_1m_1+k_2m_2 -jN}(1+|y_1|)^{-L} (1+|y_2|)^{-L},
\end{align*}
which gives the desired result.
\end{proof}

The following lemma plays a crucial role in the proof of Theorem \ref{main1},
and its essence goes back to \cite[Lemma 3.6]{MT}.

\begin{lem} \label{lem-L2L2Lr-est}
Let $2 \leq r \leq \infty$, and let $\Lambda$ be a finite subset of $\Z^n$.
For each $N \ge 0$, we have
 \begin{multline*}
  \Big(
  \sum_{\nu_1 \in \Lambda} \sum_{\nu_2 \in \Z^n} +
  \sum_{\nu_1 \in \Z^n} \sum_{\nu_2 \in \Lambda} +
  \sum_{\mu \in \Lambda} \sum_{\nu_1+\nu_2 = \mu}
  \Big)
  |\la T_{\sigma_{j, \K, \Nu}}(f_1, f_2), g \ra|
  \\
 \lesssim 2^{k_1 m_1+k_2 m_2 -jN}
 |\Lambda|^{1/2} \|f_1\|_{L^2}\|f_2\|_{L^2} \|g\|_{L^r}
 \end{multline*}
for all $j \in \N_0$ and $\K=(k_1,k_2) \in (\N_0)^2$,
where $|\Lambda|$ is the number of elements of $\Lambda$.
\end{lem}

\begin{proof}
 Let $\widetilde{\varphi}$ be a Schwartz function such that
  $\supp \widetilde{\varphi} \subset [-2, 2]^n$ and $\widetilde{\varphi}(\xi) = 1$ on $ [-1, 1]^n$. For $\nu_i \in \Z^n$, $i=1,2$, $\mu \in \Z^n$ and $j \geq 0$, we set
 \begin{align*}
  f_{i, \nu_i} = \widetilde{\varphi}(D-\nu_i)f_i,
  \quad
  g_{j, \mu} = \widetilde{\varphi}(2^{-(j+2)}(D-\mu))g.
 \end{align*}
 Then, it follows from the identity $\varphi \widetilde{\varphi} = \varphi$ that
 \begin{equation}\label{AAA}
T_{\sigma_{j, \K, \Nu}}(f_1, f_2) = T_{\sigma_{j, \K, \Nu}}( f_{1, \nu_1}, f_{2, \nu_2}) .
 \end{equation}

The Fourier transform of $T_{\sigma_{j, \K, \Nu}}(f_1, f_2)$ can be written as
\begin{align*}
 &\F[T_{\sigma_{j, \K, \Nu}}(f_1, f_2)](\zeta)\\
 &= \frac{1}{(2\pi)^{2n}}\int_{(\R^n)^2}
   \psi_{j}(\zeta- (\xi_1+\xi_2)) [\F_x\sigma](\zeta-(\xi_1+\xi_2), \xi_1, \xi_2)\\
 &\qquad\qquad\qquad \times  \psi_{k_1}(\xi_1)\psi_{k_2}(\xi_2)
   \varphi(\xi_1-\nu_1)\varphi(\xi_2-\nu_2)
   \widehat{f_1}(\xi_1) \widehat{f_2}(\xi_2)\,d\xi_1d\xi_2,
\end{align*}
where $\F_x\sigma$ denotes the partial Fourier transform
of $\sigma(x,\xi_1,\xi_2)$ with respect to  the $x$-variable.
Now, if $\xi_i \in \supp \varphi(\cdot - \nu_i)$ for
$i=1,2 $ and $\zeta - (\xi_1+\xi_2) \in \supp \psi_j $, then
$\zeta \in \nu_1 +\nu_2 + \left[-2^{j+2}, 2^{j+2} \right]^n$. This implies that
\begin{align} \label{supp-uniform}
 \supp \F[T_{\sigma_{j, \K, \Nu}}(f_1, f_2)]
 \subset \nu_1+ \nu_2 +\left[-2^{j+2}, 2^{j+2} \right]^n.
\end{align}
Hence, it follows from \eqref{AAA} that
\begin{equation}\label{trilin-rep}
\begin{split}
\la T_{\sigma_{j, \K, \Nu}}( f_1, f_2), g \ra
&=\la T_{\sigma_{j, \K, \Nu}}( f_1, f_2), g_{j,\nu_1+\nu_2} \ra
\\
&= \la T_{\sigma_{j, \K, \Nu}}( f_{1, \nu_1}, f_{2, \nu_2}), g_{j, \nu_1+\nu_2} \ra.
\end{split}
\end{equation}

First, we consider the sum $\sum_{\nu_1 \in \Lambda} \sum_{\nu_2 \in \Z^n}$.
By \eqref{trilin-rep},
Lemma \ref{lem-symb-est} (2) with $N$ replaced by $N+n/2+n/r$
and Schwarz's inequality, we have
\begin{align*}
&\sum_{\nu_1 \in \Lambda} \sum_{\nu_2 \in \Z^n} |\la T_{\sigma_{j, \K, \Nu}}( f_1, f_2), g\ra|
= \sum_{\nu_1 \in \Lambda} \sum_{\nu_2 \in \Z^n} |\la T_{\sigma_{j, \K, \Nu}}( f_{1, \nu_1}, f_{2, \nu_2}), g_{j, \nu_1+\nu_2}\ra| \\
 &\lesssim 2^{k_1m_1 + k_2m_2-j(N+n/2+n/r)}
 \sum_{\nu_1 \in \Lambda} \sum_{\nu_2 \in \Z^n}
 \int_{\R^n} S(f_{1, \nu_1})(x) S(f_{2, \nu_2})(x) |g_{j, \nu_1+\nu_2}(x)| \, dx  \\
 &\leq 2^{k_1m_1 + k_2m_2-j(N+n/2+n/r)} \\
 &\quad \times
 \sum_{\nu_1 \in \Lambda}
 \int_{\R^n} S(f_{1, \nu_1})(x)
 \Big( \sum_{\nu_2 \in \Z^n} S(f_{2, \nu_2})(x)^2 \Big)^{1/2}
 \Big( \sum_{\nu_2 \in \Z^n} |g_{j, \nu_1+\nu_2}(x)|^2 \Big)^{1/2} \, dx \\
 &\leq 2^{k_1m_1 + k_2m_2-j(N+n/2+n/r)} |\Lambda|^{1/2}\\
 &\quad\times
 \int_{\R^n}
\Big( \sum_{\nu_1 \in \Z^n} S(f_{1, \nu_1})(x)^2 \Big)^{1/2}
\Big(  \sum_{\nu_2 \in \Z^n} S(f_{2, \nu_2})(x)^2 \Big)^{1/2}
\Big( \sum_{\mu \in \Z^n} |g_{j, \mu}(x)|^2\Big)^{1/2} \, dx.
\end{align*}
It follows from Lemma \ref{lem-dyadic-est} that
\begin{align*}
 \Big( \sum_{\nu_1 \in \Z^n} S(f_{1, \nu_1})(x)^2 \Big)^{1/2}
 &\lesssim \Big( \sum_{\nu_1 \in \Z^n} S(|f_{1, \nu_1}|^2)(x) \Big)^{1/2}
  =\Big( S\Big(\sum_{\nu_1 \in \Z^n} |f_{1, \nu_1}|^2 \Big)(x) \Big)^{1/2} \\
 &\lesssim S(S(|f_1|^2))(x)^{1/2}
 \approx S(|f_1|^2)(x)^{1/2}
\end{align*}
and
\begin{align*}
\Big( \sum_{\mu \in \Z^n} |g_{j, \mu}(x)|^2\Big)^{1/2}
&\lesssim 2^{(j+2)n/2}S_{2^{j+2}}(|g|^2)(x)^{1/2}
\\
&\lesssim 2^{jn} \Big( \int_{\R^n} \frac{1}{(1+2^{j}|x-y|)^{Lq}}\, dy \Big)^{1/q}
\Big( \int_{\R^n} |g(y)|^r \,dy \Big)^{1/r}
\\
&\approx 2^{jn(1/2+1/r)}\|g\|_{L^r},
\end{align*}
where we used H\"older's inequality with $1/q+1/r=1/2$
in the second inequality.
Therefore, by Schwarz's inequality and Young's inequality, we obtain
\begin{align*}
&\sum_{\nu_1 \in \Lambda} \sum_{\nu_2 \in \Z^n} |\la T_{\sigma_{j, \K, \Nu}}( f_1, f_2), g \ra|\\
&\lesssim 2^{k_1m_1 + k_2m_2-jN} |\Lambda|^{1/2}
 \Big(\int_{\R^n}S(|f_1|^2)(x)^{1/2} S(|f_2|^2)(x)^{1/2} \, dx \Big)
 \|g\|_{L^r}\\
&\leq 2^{k_1m_1 + k_2m_2-jN} |\Lambda|^{1/2}
\|S(|f_1|^2)^{1/2}\|_{L^2} \|S(|f_2|^2)^{1/2}\|_{L^2}  \|g\|_{L^r}\\
&\approx 2^{k_1m_1 + k_2m_2-jN} |\Lambda|^{1/2}
\|f_1\|_{L^2} \|f_2\|_{L^2} \|g\|_{L^r}.
\end{align*}
In the same way, we can estimate the sum $\sum_{\nu_1 \in \Z^n} \sum_{\nu_2 \in \Lambda}$.

Next, we consider the sum $\sum_{\mu \in \Lambda} \sum_{\nu_1+\nu_2 = \mu}$.
By (\ref{trilin-rep}), Lemma \ref{lem-symb-est} (2) and Schwarz's inequality,
\begin{align*}
 &\sum_{\mu \in \Lambda} \sum_{\nu_1+\nu_2 = \mu} |\la T_{\sigma_{j, \K, \Nu}}( f_1, f_2), g \ra|
 =\sum_{\mu \in \Lambda} \sum_{\nu_1+\nu_2 = \mu} |\la T_{\sigma_{j, \K, \Nu}}( f_{1, \nu_1}, f_{2, \nu_2}), g_{j, \nu_1+\nu_2}\ra|\\
 &\lesssim 2^{k_1m_1 + k_2m_2-j(N+n/2+n/r)}
 \sum_{\mu \in \Lambda} \sum_{\nu_1 \in \Z^n}
 \int_{\R^n} S(f_{1, \nu_1})(x) S(f_{2, \mu-\nu_1})(x) |g_{j, \mu}(x)| \, dx\\
 &\leq 2^{k_1m_1 + k_2m_2-j(N+n/2+n/r)} \\
 &\quad \times
 \sum_{\mu \in \Lambda} \int_{\R^n}
 \Big( \sum_{\nu_1 \in \Z^n} S(f_{1, \nu_1})(x)^2 \Big)^{1/2}
 \Big( \sum_{\nu_1 \in \Z^n} S(f_{2, \mu-\nu_1})(x)^2 \Big)^{1/2}
 |g_{j, \mu}(x)| \, dx \\
 &\leq 2^{k_1m_1 + k_2m_2-j(N+n/2+n/r)} |\Lambda|^{1/2}\\
 &\times
 \int_{\R^n}
\Big( \sum_{\nu_1 \in \Z^n} S(f_{1, \nu_1})(x)^2 \Big)^{1/2}
\Big(  \sum_{\nu_2 \in \Z^n} S(f_{2, \nu_2})(x)^2 \Big)^{1/2}
\Big( \sum_{\mu \in \Z^n} |g_{j, \mu}(x)|^2\Big)^{1/2} \, dx.
\end{align*}
The rest of the proof is the same as before. The proof is complete.
\end{proof}

\section{Proof of Theorem \ref{main1}}\label{section4}
Let $m_1$, $m_2$, $\sigma$ and $p$ be the same
as in Theorem \ref{main1}.
To obtain Theorem \ref{main1}, by duality,
it is sufficient to prove that
\begin{align*}
 |\la T_\sigma(f_1, f_2), g \ra| \lesssim \|f_1\|_{L^2} \|f_2\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}}.
\end{align*}
From (\ref{decomp-symb}), we can write
\begin{align*}
  &\la T_\sigma(f_1, f_2), g \ra
  = \sum_{j\ge 0} \sum_{\K \in (\N_0)^2} \sum_{\Nu \in (\Z^n)^2}
 \la T_{\sigma_{j, \K, \Nu}}(f_1, f_2), g \ra \\
  &=  \sum_{j\ge 0}
  \sum_{k_1 \geq k_2}
  \sum_{\Nu \in (\Z^n)^2}
  \la T_{\sigma_{j, \K, \Nu}}(f_1, f_2), g \ra
  +
  \sum_{j\ge 0}
  \sum_{k_1 < k_2}
  \sum_{\Nu \in (\Z^n)^2}
  \la T_{\sigma_{j, \K, \Nu}}(f_1, f_2), g \ra.
\end{align*}
By symmetry, we only consider the former sum in the last line, because the argument below works for the latter one.

Let $\widetilde{\psi}_k  \in \Sh(\R^n)$, $k \geq 0$, be such that
\begin{align*}
&\supp \widetilde{\psi}_0 \subset \{ |\xi| \leq 4 \}, \quad
\supp \widetilde{\psi}_k \subset \{2^{k-2} \leq |\xi| \leq 2^{k+2}\},\quad k \geq 1,\\
&\widetilde{\psi}_k = 1
\quad \text{on} \quad
\supp \psi_k, \quad k \geq 0.
\end{align*}
Since
$\psi_{k_i}\widetilde{\psi}_{k_i}=\psi_{k_i}$,
it holds that
\begin{align*}
  \la T_{\sigma_{j, \K, \Nu}}(f_1, f_2), g \ra
  =  \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g \ra
\end{align*}
with
$f_{i, k_i} = \widetilde{\psi}_{k_i}(D)f_i$, $i=1,2$.
We  also use the decomposition
\begin{align*}
 g = \sum_{\ell \geq 0} \psi_\ell(D)g = \sum_{\ell \geq 0} g_\ell,
\end{align*}
where $\{\psi_\ell\}_{\ell \geq 0}$ is the same as in (\ref{Besov-parti}).
Then, we can write
\begin{align*}
\sum_{j\ge 0}
  \sum_{k_1 \geq k_2}
  \sum_{\Nu \in (\Z^n)^2}
  \la T_{\sigma_{j, \K, \Nu}}(f_1, f_2), g \ra
= \sum_{j\ge 0}
  \sum_{k_1 \geq k_2}
  \sum_{\ell \geq 0}
  \sum_{\Nu \in (\Z^n)^2}
  \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra.
 \end{align*}
Furthermore, we divide the sum as follows.
\begin{align*}
  &\sum_{j \ge 0}
  \sum_{k_1 \geq k_2}
  \sum_{\ell \geq 0}
  \sum_{\Nu \in (\Z^n)^2}
  \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra\\
 &= \Big( \sum_{\substack{j \geq k_1-3\\ k_1 \geq k_2}}
     + \sum_{\substack{j < k_1-3\\ k_1 \geq k_2}} \Big)
  \sum_{\ell \geq 0}
  \sum_{\Nu \in (\Z^n)^2}
 \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra
 = A_1 + A_2.
\end{align*}

\bigskip
\noindent
{\it Estimate for $A_1$}.
Since $\supp \widehat{g_\ell} \subset \{ 2^{\ell-1} \le |\zeta| \le 2^{\ell+1}\}$, $\ell \ge 1$, and
\begin{align*}
\supp \F[ T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2})]
 \subset \{ |\zeta| \leq 2^{j+6}\}, \quad  j\geq k_1-3, \ k_1 \geq k_2
\end{align*}
 (see the argument around (\ref{supp-uniform})),
 it follows that $\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra = 0$ if $\ell \geq j+7$.
In addition, we see that if $\supp \varphi(\cdot - \nu_2) \cap \supp
\psi_{k_2} = \emptyset$, then $\sigma_{j, \K, \Nu} = 0$, and consequently $\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra = 0$.
From these observations, $A_1$ can be written as
\begin{align*}
A_1
= \sum_{\substack{j \geq k_1-3 \\ \ell \leq j+6, \, k_1 \geq k_2}}
  \sum_{\nu_1 \in \Z^n} \sum_{\nu_2 \in \Lambda_{k_2}}
 \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra,
\end{align*}
where
\begin{align} \label{k2set}
 \Lambda_{k_2} = \{\nu_2 \in \Z^n : \supp \varphi(\cdot - \nu_2) \cap \supp \psi_{k_2} \neq \emptyset \}.
\end{align}
Note that the number of elements of $\Lambda_{k_2}$ satisfies $|\Lambda_{k_2}| \lesssim 2^{k_2n}$.
Using Lemma \ref{lem-L2L2Lr-est} with $r = p^\prime$ and $N \ge 1$, we obtain
\begin{align} \label{A1-est}
\begin{split}
 |A_1|
 &\leq
 \sum_{\substack{j \geq k_1-3 \\ \ell \leq j+6, \, k_1 \geq k_2}}
  \sum_{\nu_1 \in \Z^n} \sum_{\nu_2 \in \Lambda_{k_2}}
   |\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra|\\
 &\lesssim
 \sum_{\substack{j \geq k_1-3 \\ \ell \leq j+6, \, k_1 \geq k_2}}
  2^{k_1 m_1+k_2 m_2 -jN} |\Lambda_{k_2}|^{1/2} \|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2} \|g_\ell\|_{L^{p^\prime}}\\
 &\lesssim
 \sum_{\substack{j \geq k_1-3 \\ \ell \leq j+6, \, k_1 \geq k_2}}
  2^{k_1 m_1+k_2 m_2+k_2n/2 -jN}\|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2} \|g_\ell\|_{L^{p^\prime}}.
\end{split}
\end{align}
By our assumptions $m_1, m_2 < 0$ and $m_1+m_2 = -n/2$,
it follows that $k_1 m_1+k_2 m_2+k_2n/2 \leq k_2 m_1+k_2 m_2+k_2n/2 = 0$ for $k_1 \geq k_2$.
Hence, since $\sum_{k_i \ge 0} |\widetilde{\psi}_{k_i}(\xi_i)|^2 \lesssim 1$, the last quantity in (\ref{A1-est}) is estimated by
\begin{align*}
 &\sum_{\substack{j \geq k_1-3 \\ \ell \leq j+6, \, k_1 \geq k_2}}
  2^{-jN}\|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2} \|g_\ell\|_{L^{p^\prime}}\\
 &\leq
 \sum_{\substack{j \geq k_1-3 \\  k_1 \geq k_2}}
 2^{-jN} (j+7) \|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2}
 \big(\sup_{\ell \ge 0}\|g_\ell\|_{L^{p^\prime}} \big)\\
 &\leq
 \sum_{j \geq 0}
 2^{-jN} (j+7) (j+4)
  \Big( \sum_{k_1 \geq 0} \|f_{1, k_1}\|^2_{L^2} \Big)^{1/2}
  \Big( \sum_{k_2 \geq 0} \|f_{2, k_2}\|^2_{L^2} \Big)^{1/2}
  \|g\|_{B^0_{p^\prime, \infty}} \\
 &\lesssim \|f_1\|_{L^2} \|f_2\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}},
\end{align*}
which gives the desired result.

\bigskip
\noindent
{\it Estimate for $A_2$}.
 We divide $A_2$ as follows.
\begin{align*}
A_2
 &= \Big( \sum_{k_1-3 \leq k_2 \leq k_1} + \sum_{k_2 < k_1-3}  \Big)
  \sum_{j < k_1-3}
  \sum_{\ell \geq 0}
  \sum_{\Nu \in (\Z^n)^2}
 \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra \\
 &= A_{2,1} +A_{2,2}.
\end{align*}

We first consider the estimate for  $A_{2,1}$. Since
\begin{align*}
 \supp \F[ T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2})]
 \subset \{ |\zeta| \leq 2^{k_1+3}\}, \quad j < k_1-3, \ k_2 \le k_1,
\end{align*}
it follows that $\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra = 0$
if $\ell \geq k_1+4$.
Furthermore, by (\ref{supp-uniform}) and
the fact $\supp \widehat{g_\ell} \subset \supp \psi_\ell$,
we see that if
$\left(\nu_1+\nu_2 + \left[-2^{j+2}, 2^{j+2} \right]^n \right) \cap \supp \psi_\ell = \emptyset$,
then $\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra = 0$.
Combining these observations,  we see that $A_{2,1}$ can be written as
\begin{align*}
 A_{2,1}
   =\sum_{\substack{j < k_1-3, \, \ell \le k_1+3 \\ k_1-3 \leq k_2 \leq k_1}}
    \sum_{\mu \in \Lambda_{j, \ell}}
    \sum_{\nu_1 + \nu_2 = \mu}
 \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra,
\end{align*}
where
\begin{align*}
\Lambda_{j, \ell} = \{ \mu \in \Z^n : (\mu + \left[-2^{j+2}, 2^{j+2}\right]^n ) \cap \supp \psi_\ell \neq \emptyset \}.
\end{align*}
The number of elements of $\Lambda_{j, \ell}$ can be estimated by $ |\Lambda_{j, \ell}| \lesssim 2^{(j+\ell)n}$.
Hence, it follows from Lemma \ref{lem-L2L2Lr-est} with $r = p^\prime$ and $N > n/2$ that
\begin{align*}
 |A_{2,1}|
 &\leq \sum_{\substack{j < k_1-3, \, \ell \le k_1+3 \\ k_1-3 \leq k_2 \leq k_1}}
    \sum_{\mu \in \Lambda_{j, \ell}}
    \sum_{\nu_1 + \nu_2 = \mu}
 |\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra|\\
 &\lesssim \sum_{\substack{j < k_1-3, \, \ell \le k_1+3 \\ k_1-3 \leq k_2 \leq k_1}}
    2^{k_1m_1+k_2m_2-jN}  |\Lambda_{j, \ell}|^{1/2}
    \|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2} \|g_\ell\|_{L^{p^\prime}}\\
 &\lesssim \sum_{\substack{j < k_1-3, \, \ell \le k_1+3 \\ k_1-3 \leq k_2 \leq k_1}}
    2^{k_1m_1+k_2m_2-jN} 2^{(j+\ell)n/2}
    \|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}}\\
 &\approx \sum_{ k_1-3 \leq k_2 \leq k_1}
    2^{k_1m_1+k_2m_2} 2^{k_1n/2}
    \|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}}\\
 &= \sum_{ k_1-3 \leq k_2 \leq k_1}
    2^{-k_1m_2+k_2m_2}
    \|f_{1, k_1}\|_{L^2}\|f_{2, k_2}\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}}.
\end{align*}
Since we can write $k_2 = k_1 + \widetilde{k}$ with $\widetilde{k} = -3, -2, -1, 0$ if $k_1-3 \leq k_2 \leq k_1$, the last sum can be written as
\begin{align*}
 &\sum_{ k_1 \ge 0} \sum_{ -3 \le \widetilde{k} \le 0}
    2^{-k_1m_2+(k_1+\widetilde{k})m_2}
    \|f_{1, k_1}\|_{L^2}\|f_{2, k_1+\widetilde{k}}\|_{L^2}\\
 &= \sum_{ k_1 \ge 0} \sum_{ -3 \le \widetilde{k} \le 0}
  2^{\widetilde{k}m_2} \|f_{1, k_1}\|_{L^2}\|f_{2, k_1+\widetilde{k}}\|_{L^2}.
\end{align*}
Hence, by Schwarz's inequality, the right hand side of the above is estimated by
\begin{align*}
\sum_{ -3 \le \widetilde{k} \le 0} 2^{\widetilde{k}m_2}
\Big( \sum_{k_1\ge 0} \|f_{1, k_1}\|^2_{L^2} \Big)^{1/2}
\Big( \sum_{k_1\ge 0} \|f_{2, k_1+\widetilde{k}}\|^2_{L^2} \Big)^{1/2}
\lesssim \|f_1\|_{L^2}\|f_2\|_{L^2},
\end{align*}
which implies the desired estimate.

Next, we consider the estimate for $A_{2, 2}$.
Since
\begin{align*}
  \supp \F[ T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2})]
 \subset \{2^{k_1-2} \le |\zeta| \le 2^{k_1+2}\},
\quad  j < k_1-3, \ k_2 < k_1-3,
\end{align*}it follows that
$\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra = 0$
if $\ell \le k_1-3$ or $k_1+3 \leq \ell$.
As before, if $\supp \varphi(\cdot - \nu_2) \cap \supp \psi_{k_2} = \emptyset$, then $\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra = 0$.
Therefore,  we obtain
\begin{align*}
 A_{2, 2}
 = \sum_{\substack{j < k_1-3 \\ k_2 < k_1-3, \, |\ell-k_1| \le 2}}
    \sum_{\nu_1 \in \Z^n} \sum_{\nu_2 \in  \Lambda_{k_2}}
 \la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra,
\end{align*}
where $\Lambda_{k_2}$ is the same as in (\ref{k2set}).
By applying Lemma \ref{lem-L2L2Lr-est}, it follows that
\begin{align} \label{A22-est}
\begin{split}
|A_{2, 2}|
&\leq \sum_{\substack{j < k_1-3 \\ k_2 < k_1-3, \, |\ell-k_1| \le 2}}
    \sum_{\nu_1 \in \Z^n} \sum_{\nu_2 \in  \Lambda_{k_2}}
 |\la T_{\sigma_{j, \K, \Nu}}(f_{1, k_1}, f_{2, k_2}), g_\ell \ra|\\
&\lesssim \sum_{\substack{j < k_1-3 \\ k_2 < k_1-3, \, |\ell-k_1| \le 2}}
 2^{k_1m_1+k_2m_2-jN} 2^{k_2n/2} \|f_{1, k_1}\|_{L^2} \|f_{2, k_2}\|_{L^2} \|g_{\ell}\|_{L^{p^\prime}}\\
&\lesssim \sum_{k_2 < k_1-3}
 2^{m_1(k_1-k_2)} \|f_{1, k_1}\|_{L^2} \|f_{2, k_2}\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}}\\
&\le \sum_{k_1, k_2 \ge 0}
 2^{m_1|k_1-k_2|} \|f_{1, k_1}\|_{L^2} \|f_{2, k_2}\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}}.
 \end{split}
\end{align}
Here, it follows from our assumption $m_1 < 0$ that
\begin{align*}
 \sup_{k_1 \ge 0} \sum_{k_2 \ge 0} 2^{m_1|k_1-k_2|} < \infty
 \quad
\text{and}
\quad
 \sup_{k_2 \ge 0} \sum_{k_1 \ge 0} 2^{m_1|k_1-k_2|} < \infty.
\end{align*}
Hence, by Lemma \ref{Schur's lemma},
the last quantity in (\ref{A22-est}) can be estimated by
\begin{align*}
 \Big( \sum_{k_1 \ge 0} \|f_{1, k_1}\|^2_{L^2} \Big)^{1/2}
  \Big( \sum_{k_2 \ge 0} \|f_{2, k_2}\|^2_{L^2} \Big)^{1/2}
  \|g\|_{B^0_{p^\prime, \infty}}
\lesssim \|f_1\|_{L^2} \|f_2\|_{L^2} \|g\|_{B^0_{p^\prime, \infty}}.
\end{align*}
This completes the proof of Theorem \ref{main1}.

\section{Proof of Theorem \ref{main2}}\label{section5}

In this section, we shall prove Theorem \ref{main2}. The ``if'' part follows from Theorem \ref{main1}, (\ref{two-classes}) and the embedding
$B^0_{p, 1} \hookrightarrow B^0_{p, q}$,
$1\le q \le \infty$. Thus, let us consider the ``only if'' part.

We assume that the $L^2\times L^2 \to B^0_{p, q}$ boundedness of all $T_\sigma$ with $\sigma \in BS^{-n/2}_{0, 0}$ holds throughout the rest of this section.
We  also take a function $\psi_0 \in \Sh(\R^n)$ satisfying $\psi_0 = 1$ on $\{|\xi| \le 2^{1/4}\}$ and $\supp \psi_0 \subset \{|\xi| \le 2^{3/4}\}$, and set $\psi_\ell (\xi) = \psi_0(2^{-\ell}\xi) - \psi_0(2^{-\ell+1}\xi)$, $\ell \ge 1$. Then $\{\psi_\ell\}_{\ell \ge 0}$ satisfies (\ref{Besov-parti}) with the support conditions replaced by
\begin{align}\label{Besov-parti-2-1}
\begin{split}
 \supp \psi_0 \subset \{ |\xi| \leq 2^{3/4}\}, \quad
\supp \psi_\ell \subset \{2^{\ell -3/4} \leq |\xi| \leq 2^{\ell + 3/4}\}, \quad \ell \ge 1.
\end{split}
\end{align}
Combining this with the condition $\sum_{\ell \ge 0}\psi_{\ell}=1$,
we see that
\begin{equation}\label{Besov-parti-2-2}
\psi_0 = 1
\ \ \text{on} \ \
\{|\xi| \leq 2^{1/4}\},
\quad
\psi_\ell  = 1
\ \ \text{on} \ \
\{2^{\ell -1/4} \leq |\xi| \leq 2^{\ell + 1/4}\},
\ \ \ell \geq 1,
\end{equation}
and use this $\{\psi_\ell\}_{\ell \ge 0}$
as the partition of unity in the definition of Besov spaces.

\bigskip
\noindent
{\it The necessity of the condition $1 \le p \le 2$}.
First, we shall prove that $p \ge 1$.
Let $\sigma \in \Sh((\R^n)^2)$ satisfy $\sigma(\xi_1, \xi_2) = 1$ on $|(\xi_1, \xi_2)| \leq 1$,
and obviously $\sigma \in BS^{-n/2}_{0, 0}$.
For $\varphi \in \Sh(\R^n)$ satisfying
$\supp \varphi \subset \{ |\xi| \leq 1\}$,
we set
\[
\widehat{f_{i, j}}(\xi_i) = 2^{jn/2}\varphi(2^j\xi_i),
\quad i=1,2, \ j \ge 1,
\]
and note that
 $\|f_{i, j}\|_{L^2} \approx 1$.
 Since $|(\xi_1, \xi_2)| \leq 2^{-j+1/2} \leq 1$ for $|\xi_1|, |\xi_2| \leq 2^{-j}$ and $j \ge 1$, it follows that $\sigma(\xi_1, \xi_2) \varphi(2^j\xi_1) \varphi(2^j\xi_2) = \varphi(2^j\xi_1) \varphi(2^j\xi_2)$, and consequently
\[
T_{\sigma}(f_{1, j}, f_{2, j})(x)
= 2^{-jn}([\F^{-1}\varphi](2^{-j}x))^2.
\]
By a change of variables, $\|T_{\sigma}(f_{1, j}, f_{2, j})\|_{L^p} \approx 2^{jn(1/p-1)}$.
Furthermore, we have
\begin{align*}
\begin{split}
 \supp \F[T_\sigma(f_{1, j}, f_{2, j})]
 = \supp [\varphi(2^j \cdot) * \varphi(2^j \cdot)]
 \subset \{ |\zeta| \leq 1\}.
\end{split}
\end{align*}
Hence, it follows from (\ref{Besov-parti-2-1}) and (\ref{Besov-parti-2-2}) that
$\psi_\ell(D)T_\sigma(f_{1, j}, f_{2, j})$ equals $T_\sigma(f_{1, j}, f_{2, j})$ if $\ell=0$, and $0$ otherwise.
From this, we obtain
\begin{align*}
 \|T_\sigma(f_{1, j}, f_{2, j})\|_{B^0_{p, q}}
 = \|T_\sigma(f_{1, j}, f_{2, j})\|_{L^p} \approx 2^{jn(1/p-1)}.
\end{align*}
Therefore, by our assumption, the boundedness of $T_\sigma$ gives
\begin{align*}
2^{jn(1/p-1)} \approx  \|T_\sigma(f_{1, j}, f_{2, j})\|_{B^0_{p, q}} \lesssim  \|f_{1, j}\|_{L^2} \|f_{2, j}\|_{L^2} \approx 1.
\end{align*}
The arbitrariness of $j \ge 1$ implies that $1/p-1 \le 0$, namely $p\ge 1$.

Next we shall prove that $p \leq 2$. Our argument is based on the proof of \cite[Proposition 5.1]{KMT}.
Let $\psi, \varphi, \widetilde{\psi}, \widetilde{\varphi} \in \Sh(\R^n)$ be such that
\begin{align*}
&\supp \psi \subset \{ 2^{-1/8} \leq |\xi_1| \leq 2^{1/8} \}, \quad
\supp \varphi \subset \{|\xi_2| \leq 2^{-j_0}\},\\
&\supp \widetilde{\psi} \subset \{ 2^{-1/4} \leq |\xi_1| \leq 2^{1/4} \},\quad \widetilde{\psi} = 1 \quad \text{on} \quad \supp \psi, \\
&\supp \widetilde{\varphi} \subset \{ |\xi_2| \leq 2^{-j_0+1}\}, \quad
\widetilde{\varphi} = 1 \quad \text{on} \quad \supp \varphi,
\end{align*}
where $j_0$ is a positive integer satisfying
\begin{equation}\label{proof_p2}
2^{-1/4} \le 2^{-1/8} -2^{-j_0}
\quad \text{and} \quad
2^{1/8} + 2^{-j_0} \le 2^{1/4}.
\end{equation}
We set
\begin{align*}
&\sigma(\xi_1, \xi_2) =
\sum_{k \ge 1} 2^{-kn/2} \widetilde{\psi}(2^{-k}\xi_1) \widetilde{\varphi}(2^{-k}\xi_2),\\
 &\widehat{f_{1, j}}(\xi_1) = 2^{-jn/2}\psi(2^{-j}\xi_1), \quad  \widehat{f_{2, j}}(\xi_2) = 2^{-jn/2} \varphi(2^{-j}\xi_2),
 \quad j\ge1.
\end{align*}
It is not  difficult to show that $\sigma \in BS^{-n/2}_{0, 0}$
(since $1+|\xi_1|+|\xi_2| \approx 2^k$
for $\xi_1 \in \mathrm{supp}\, \widetilde{\psi}(2^{-k}\cdot)$
and $\xi_2 \in \mathrm{supp}\, \widetilde{\varphi}(2^{-k}\cdot)$) and
$\|f_{1, j}\|_{L^2}, \, \|f_{2, j}\|_{L^2} \approx 1$.
Now, since  $\widetilde{\psi}(2^{-k}\xi_1) \widetilde{\varphi}(2^{-k}\xi_2) \psi(2^{-j}\xi_1) \varphi(2^{-j}\xi_2)$ equals $\psi(2^{-j}\xi_1) \varphi(2^{-j}\xi_2)$ if $k=j$,
and $0$ otherwise, it follows that
\[
T_{\sigma}(f_{1, j}, f_{2, j})(x)
=2^{jn/2} [\F^{-1}\psi](2^j x) [\F^{-1}\varphi](2^j x).
\]
From this, we have $ \| T_\sigma(f_{1, j}, f_{2, j})\|_{L^p}  \approx 2^{jn(1/2 -1/p)}$.
Moreover, by the support conditions
of $\psi, \varphi$ and \eqref{proof_p2},
we see that
$2^{j-1/4} \le |\xi_1+\xi_2| \le 2^{j+1/4}$
for $\xi_1 \in \supp \psi(2^{-j}\cdot)$
and $\xi_2 \in \supp \varphi(2^{-j}\cdot)$,
and consequently
\begin{align*}
\supp \F[T_\sigma(f_{1, j}, f_{2, j})]
= \supp[\psi(2^{-j}\cdot)*\varphi(2^{-j}\cdot)]
\subset \{2^{j-1/4} \leq |\zeta| \leq 2^{j+1/4}\}.
\end{align*}
Hence,  it follows from (\ref{Besov-parti-2-1}) and (\ref{Besov-parti-2-2}) that
$\psi_\ell(D)T_\sigma(f_{1, j}, f_{2, j})$ equals
$T_\sigma(f_{1, j}, f_{2, j})$ if $\ell=j$, and $0$ otherwise.
This yields that
\begin{align*}
 \|T_\sigma(f_{1, j}, f_{2, j})\|_{B^0_{p, q}}
 = \|T_\sigma(f_{1, j}, f_{2, j})\|_{L^p} \approx  2^{jn(1/2 -1/p)}.
\end{align*}
Therefore, by our assumption, the boundedness of $T_\sigma$ gives
\begin{align*}
 2^{jn(1/2 -1/p)} \approx \|T_\sigma(f_{1, j}, f_{2, j})\|_{B^0_{p, q}} \lesssim  \|f_{1, j}\|_{L^2} \|f_{2, j}\|_{L^2} \approx 1.
\end{align*}
The arbitrariness of $j \ge 1$ implies that $1/2-1/p \le 0$, namely $p\le 2$.

\bigskip
\noindent
{\it The necessity of the condition $1 \le q \le \infty$}.
Finally, we shall prove that $1\leq q \leq \infty$.
The basic idea of this part goes back to
the proofs of \cite[Lemma 6.3]{MT-IUMJ} and \cite[Theorem 1.4]{Park}.
Since there is nothing to prove if $q= \infty$, we may assume that $0<q<\infty$.
By the closed graph theorem,
our assumption implies that
there exists a positive integer $N$ such that
\begin{equation}\label{inequality_operator_norm}
\|T_{\sigma}\|_{L^2 \times L^2 \to B_{p,q}^0}
\lesssim \max_{|\alpha|, |\beta_1|, |\beta_2| \le N}
\|
(1+|\xi_1|+|\xi_2|)^{n/2}
\partial_x^{\alpha}\partial_{\xi_1}^{\beta_1}\partial_{\xi_2}^{\beta_2}
\sigma(x,\xi_1,\xi_2)\|_{L^{\infty}_{x,\xi_1,\xi_2}}
\end{equation}
for all $\sigma \in BS^{-n/2}_{0,0}$,
where $\|T_{\sigma}\|_{L^2 \times L^2 \to B_{p,q}^0}$
denotes the operator norm of $T_{\sigma}$
(see \cite[Lemma 2.6]{BBMNT}).

We use functions
$\varphi, \widetilde{\varphi} \in \Sh(\R^n)$ such that
\begin{align*}
&\supp \varphi \subset \left[-1/4, 1/4\right]^n, \quad |\F^{-1}\varphi| \geq 1
\quad \text{on} \quad [-1, 1]^n,
\\
&\supp \widetilde{\varphi} \subset \left[-1/2, 1/2 \right]^n,
\quad \widetilde{\varphi} = 1 \quad \text{on} \quad \supp \varphi.
\end{align*}
Let $\{r_{\nu}(\omega)\}_{\nu \in \Z^n}$,
$ \omega \in [0,1]^n$,
be a sequence of Rademacher functions
(see \cite[Appendix C]{Grafakos-Classical})
enumerated in such a way that
their index set is $\Z^n$.
For $\omega \in [0,1]^n$ and $\epsilon>0$,
we set
\begin{align*}
&\sigma_{\omega}(\xi_1,\xi_2)
=\sum_{k \ge k_0}\sum_{\nu \in \Lambda_k}\sum_{\nu_1+\nu_2=\nu}
r_{\nu}(\omega)(1+|\nu_1|+|\nu_2|)^{-n/2}
\widetilde{\varphi}(\xi_1-\nu_1)\widetilde{\varphi}(\xi_2-\nu_2),
\\
&\widehat{f_i}(\xi_i)
=\sum_{\mu_i \in \Z^n}\langle \mu_i \rangle^{-n/2}
(\log \langle \mu_i \rangle)^{-(1+\epsilon)/2}\varphi(\xi_i-\mu_i),
\quad i=1,2,
\end{align*}
where $\langle \mu_i \rangle=e+|\mu_i|$,
$k_0$ is a positive integer satisfying
\begin{equation}\label{proof_k0}
2^{k-1/4} \le 2^{k-1/8}-\sqrt{n}/2,
\quad 2^{k+1/8}+\sqrt{n}/2 \le 2^{k+1/4},
\quad k \ge k_0 ,
\end{equation}
and
\[
\Lambda_k=\{\nu \in \Z^n \,:\, 2^{k-1/8} \le |\nu| \le 2^{k+1/8}\}.
\]
The number of elements of $\Lambda_k$ can be estimated by $|\Lambda_k| \approx 2^{kn}$.
It is easy to check that
\begin{equation}\label{inequality_symbol}
\max_{|\alpha|, |\beta_1|, |\beta_2| \le N}
\|
(1+|\xi_1|+|\xi_2|)^{n/2}
\partial_x^{\alpha}\partial_{\xi_1}^{\beta_1}\partial_{\xi_2}^{\beta_2}
\sigma_{\omega}(x,\xi_1,\xi_2)\|_{L^{\infty}_{x,\xi_1,\xi_2}}
\lesssim 1,
\end{equation}
but it should be emphasized that
the implicit constant in this inequality is independent of $\omega \in [0,1]^n$.
Using the facts that
$\{\langle \mu_i \rangle^{-n/2}
(\log \langle \mu_i \rangle)^{-(1+\epsilon)/2}\}$
belongs to $\ell^2(\Z^n)$
and the supports of $\varphi(\cdot-\mu_i)$, $\mu_i \in \Z^n$,
are mutually disjoint,
we also see that
$\|f_i\|_{L^2} \lesssim 1$, $i=1,2$.
Thus,
by \eqref{inequality_operator_norm} and \eqref{inequality_symbol},
\begin{align} \label{norm-estimate}
\|T_{\sigma_\omega}(f_1, f_2)\|_{B^0_{p, q}}
\lesssim  \|f_1\|_{L^2}\|f_2\|_{L^2}
\lesssim 1,
\quad \omega \in [0,1]^n.
\end{align}

Now, since  $ \varphi(\xi_i- \nu_i)\widetilde{\varphi}(\xi_i - \mu_i)$ is equal to  $\varphi(\xi_i- \nu_i)$ if $\nu_i=\mu_i$, and $0$ otherwise, $i= 1, 2$,
it follows that
\begin{align*}
T_{\sigma_\omega}(f_1, f_2)(x)
&= \sum_{k \ge k_0} \sum_{\nu \in \Lambda_k} \sum_{\nu_1+\nu_2 =\nu}
r_{\nu}(\omega)  (1+|\nu_1|+|\nu_2|)^{-n/2}
e^{i(\nu_1+\nu_2) \cdot x} (\F^{-1}\varphi(x))^2
\\
&\qquad \times
\la \nu_1 \ra^{-n/2} ( \log \la \nu_1\ra )^{-(1+\epsilon)/2}
\la \nu_2 \ra^{-n/2} ( \log \la \nu_2\ra )^{-(1+\epsilon)/2}
\\
&=\sum_{k \ge k_0}\sum_{\nu \in \Lambda_k}
r_{\nu}(\omega) d_{\nu} e^{i \nu \cdot x} \Phi(x)^2
\end{align*}
with $\Phi=\F^{-1}\varphi$ and
\begin{align*}
d_{\nu}
&=\sum_{\mu \in \Z^n}
 (1+|\nu-\mu|+|\mu|)^{-n/2}
\\
&\qquad \qquad \times
\la \nu-\mu \ra^{-n/2} ( \log \la \nu-\mu \ra )^{-(1+\epsilon)/2}
\la \mu \ra^{-n/2} ( \log \la \mu \ra )^{-(1+\epsilon)/2} .
\end{align*}
We note that
$\mathrm{supp}\, \varphi*\varphi \subset [-1/2,1/2]^n$ and
\[
\F[T_{\sigma_\omega}(f_1, f_2)](\zeta)
=(2\pi)^{-n}
\sum_{k \ge k_0}\sum_{\nu \in \Lambda_k}
r_{\nu}(\omega) d_{\nu} [\varphi*\varphi](\zeta-\nu) .
\]
By \eqref{proof_k0},
if $\zeta - \nu \in [-1/2, 1/2]^n$, $\nu \in \Lambda_k$ and $k \ge k_0$,
then $2^{k-1/4} \leq |\zeta| \leq 2^{k+1/4}$.
Hence, for $k \ge k_0$ and $\nu \in \Lambda_k$,
it follows from (\ref{Besov-parti-2-1})
and (\ref{Besov-parti-2-2}) that
$\psi_\ell(\zeta) [\varphi* \varphi] (\zeta-\nu)$
is equal to $[\varphi* \varphi] (\zeta-\nu)$
if $k=\ell$, and $0$ otherwise.
This yields that
\[
\psi_\ell(D)T_{\sigma_\omega}(f_1, f_2)(x)
= \sum_{\nu \in \Lambda_\ell}
r_{\nu}(\omega) d_{\nu} e^{i \nu \cdot x} \Phi(x)^2,
\quad \ell \ge k_0.
\]

Raising \eqref{norm-estimate} to the $q$-th power
and integrating over $\omega \in [0,1]^n$,
we have
\begin{align*
\begin{split}
1
&\gtrsim
\int_{[0,1]^n} \|T_{\sigma_\omega}(f_1, f_2)\|_{B^0_{p, q}}^q\, d\omega
=\sum_{\ell \ge 0} \int_{[0, 1]^n}  \|\psi_\ell(D)T_{\sigma_\omega}(f_1, f_2)
\|_{L^p}^q \,d\omega
\\
&\ge
\sum_{\ell \ge k_0}
\left( \int_{\R^n}
\bigg( \int_{[0, 1]^n}
\Big| \sum_{\nu \in \Lambda_\ell}  r_{\nu}(\omega)
d_\nu e^{i\nu \cdot x} \Phi(x)^2 \Big|^{\min\{p, q\}}
d\omega \bigg)^{p/\min\{p, q\}}
 dx \right)^{q/p}
\\
&\ge
\sum_{\ell \ge k_0}
\left( \int_{[-1,1]^n}
\bigg( \int_{[0, 1]^n}
\Big| \sum_{\nu \in \Lambda_\ell}  r_{\nu}(\omega)
d_\nu e^{i\nu \cdot x} \Big|^{\min\{p, q\}}
d\omega \bigg)^{p/\min\{p, q\}}
dx \right)^{q/p} ,
\end{split}
\end{align*}
where
we used H\"older's inequality when $p \leq q$,
or Minkowski's inequality when $p\geq q$
in the second inequality,
and the condition $|\Phi|=|\F^{-1}\varphi| \ge 1$
on $[-1,1]^n$ in the third inequality.
Furthermore,
by Khintchine's inequality
(see, e.g., \cite[Appendix C]{Grafakos-Classical}),
the last quantity is equivalent to
\begin{align*}
\sum_{\ell \ge k_0}
\left( \int_{[-1, 1]^n}
\Big( \sum_{\nu \in \Lambda_{\ell}}
|d_\nu e^{i\nu \cdot x}|^2 \Big)^{p/2} \,dx \right)^{q/p}
=2^{nq/p}\sum_{\ell \ge k_0} \Big( \sum_{\nu \in \Lambda_\ell}   |d_\nu|^2 \Big)^{q/2} .
\end{align*}
Since $|\nu-\mu| \approx |\nu|$
for $|\mu| \le |\nu|/2$, we have
\begin{align*}
|d_\nu|
&\gtrsim \sum_{ |\mu| \le |\nu|/2 }  \la \nu \ra^{-3n/2}
(\log\la \nu \ra)^{-(1+\epsilon)}
\approx \la \nu \ra^{-n/2}  (\log\la \nu \ra)^{-(1+\epsilon)}
\approx 2^{-\ell n/2}\ell^{-(1+\epsilon)}
\end{align*}
for $\nu \in \Lambda_\ell$,
and consequently
\begin{align*}
 \Big( \sum_{\nu \in \Lambda_{\ell}}   |d_\nu |^2 \Big)^{q/2}
 \gtrsim \Big( \sum_{\nu \in \Lambda_{\ell}} 2^{-\ell n} \ell^{-2(1+\epsilon)} \Big)^{q/2}
 \approx \ell^{-q(1+\epsilon)}.
\end{align*}
Combining the above estimates, we obtain
\[
\sum_{\ell \ge k_0} \ell^{-q(1+\epsilon)} \lesssim 1,
\]
which is possible only when $q(1+ \epsilon) > 1$.
Therefore, the arbitrariness of $\epsilon > 0$ implies that $q \ge 1$.
The proof of the ``only if'' part of Theorem \ref{main2} is complete.

\section{The amalgam space $(L^2,\ell^1)$ and Wiener amalgam space $W_{1,2}$}
\label{section6}
In this section,
we shall prove Proposition \ref{main3}.
Let $1 \le p \le 2$.
It is easy to see that the embedding $B_{p,1}^0 \hookrightarrow (L^2,\ell^1)$
does not hold.
In fact, it follows from the embedding
$(L^2,\ell^1) \hookrightarrow L^q$, $1 \le q \le 2$,
(see Section \ref{section2}) that
if $B_{p,1}^0 \hookrightarrow (L^2,\ell^1)$,
then $B_{p,1}^0 \hookrightarrow L^q$ for all $1 \le q \le 2$.
However, since  $B_{p,1}^0 \hookrightarrow L^q$
if and only if $p=q$ (see, e.g., \cite[Corollary 2.2.4/2]{RS}),
this is  a contradiction.
Though we can directly prove that
the embedding $(L^2,\ell^1) \hookrightarrow B_{p,1}^0$
does not hold,
we shall show it by using the characterization of $(L^2,\ell^1)$,
Proposition \ref{amalgam-Wiener} below.
To do this, we recall the definitions
of modulation spaces and Wiener amalgam spaces
based on Feichtinger \cite{Feichtinger} and Gr\"ochenig \cite{Grochenig}.

Fix a function $\phi \in \Sh(\R^n)\setminus \{ 0 \}$
(called the window function).
Then the short-time Fourier transform $V_{\phi}f$ of
$f \in \Sh'(\R^n)$ with respect to $\phi$
is defined by
\[
V_{\phi}f(x,\xi)
=\int_{\R^n}f(t) \overline{\phi(t-x)} e^{-it\cdot \xi}\, dt  ,
\quad  x, \xi \in \R^n,
\]
where the integral is understood in the distributional sense.
For $1\le p,q \le \infty$,
the modulation space $M_{p,q}(\R^n)$
consists of all $f \in \Sh'(\R^n)$
such that
\[
\|f\|_{M_{p,q}}
=\left\{ \int_{\R^n} \left(
\int_{\R^n} |V_{\phi}f(x,\xi)|^{p}\, dx
\right)^{q/p} d\xi \right\}^{1/q}
< \infty
\]
with usual modification when $p=\infty$ or $q=\infty$.
On the other hand,
the Wiener amalgam space $W_{p,q}(\R^n)$ consists of
all $f \in \Sh'(\R^n)$ such that
\[
\|f\|_{W_{p,q}}
=\left\{ \int_{\R^n} \left(
\int_{\R^n} |V_{\phi}f(x,\xi)|^{q}\, d\xi
\right)^{p/q} dx \right\}^{1/p}
<\infty .
\]
These definitions are independent
of the choice of the window function
$\phi \in \Sh(\R^n)\setminus \{ 0 \}$, that is, different window functions yield equivalent norms.
Since $|V_{\phi}f(x,\xi)|=(2\pi)^{n}|V_{\F^{-1}\phi}[\F^{-1}f](-\xi,x)|$,
we see that
\begin{equation}\label{MW-Fourier}
\|f\|_{W_{p,q}} \approx \|\F^{-1}f\|_{M_{q,p}}.
\end{equation}
It is also known that
if $\varphi \in \Sh(\R^n)$
has a compact support and
if there exists a constant $C>0$ such that
$\left|\sum_{\nu \in \Z^n}\varphi(\xi-\nu)\right| \ge C$
for all $\xi \in \R^n$,
then
\begin{equation}\label{MW-equi}
\begin{split}
&\|f\|_{M_{p,q}} \approx
\bigg( \sum_{\nu \in \Z^n}\|\varphi(D-\nu)f\|_{L^p}^q\bigg)^{1/q},
\\
&\|f\|_{W_{p,q}} \approx
\bigg\|\bigg(\sum_{\nu \in \Z^n}|\varphi(D-\nu)f|^q\bigg)^{1/q}\bigg\|_{L^p}
\end{split}
\end{equation}
(see, e.g., \cite[Theorem 3]{Triebel_paper}).

Although the following may be well known to many people,
we shall give a proof for the reader's convenience.

\begin{prop}\label{amalgam-Wiener}
The amalgam space $(L^2,\ell^1)(\R^n)$
coincides with the Wiener amalgam space $W_{1,2}(\R^n)$.
\end{prop}

\begin{proof}
Let $\varphi$ be a nonnegative function in $\Sh(\R^n)$
such that $\mathrm{supp}\, \varphi$ is compact and
$\varphi \ge 1$ on $[-1/2,1/2]^n$.
Then, we have the norm equivalence
\[
\|f\|_{(L^2,\ell^1)}
\approx
\sum_{\nu \in \Z^n}
\|\varphi(\cdot-\nu)f \|_{L^2}
\]
(see, e.g., \cite[Lemma 2.1]{KMT-2}).
Since $\sum_{\nu \in \Z^n}\varphi(\xi-\nu) \ge 1$
for all $\xi \in \R^n$,
by Plancherel's theorem,
\eqref{MW-equi} and \eqref{MW-Fourier},
\begin{align*}
\sum_{\nu \in \Z^n}
\|\varphi(\cdot-\nu)f \|_{L^2}
&\approx \sum_{\nu \in \Z^n}
\|\F^{-1}[\varphi(\cdot-\nu)f ]\|_{L^2}
\\
&= \sum_{\nu \in \Z^n}
\|\varphi(D-\nu)[\F^{-1}f]\|_{L^2}
\approx \|\F^{-1}f\|_{M_{2,1}}
\approx \|f\|_{W_{1,2}}.
\end{align*}
The proof is complete.
\end{proof}

In the rest of this section,
we shall prove that
the embedding $(L^2,\ell^1) \hookrightarrow B_{p,1}^0$
does not hold, where $1 \le p \le 2$.
Let $\phi, \varphi$ be nonnegative functions in $\Sh(\R^n)$ such that
$\mathrm{supp}\, \phi \subset [-1/4,1/4]^n$,
$\mathrm{supp}\, \varphi \subset [-3/4,3/4]^n$
and $\varphi=1$ on $[-1/2,1/2]^n$.
We take a positive integer $N_0$ satisfying
\begin{equation}\label{prop1.3-A}
2^{k-1/4} \le 2^{k}-\sqrt{n}/4,\quad 2^{k}+\sqrt{n}/4 \le 2^{k+1/4},
\quad k \ge N_0,
\end{equation}
and set
\[
f_N(x)=\sum_{k=N_0}^{N}e^{i2^k e_1\cdot x}\F^{-1}\phi(x),
\quad N \ge N_0,
\]
where $e_1=(1,0,\dots,0) \in \R^n$.

We first consider the Besov space norm of $f_N$.
The support condition of $\phi$ and \eqref{prop1.3-A} give
\[
\mathrm{supp}\, \phi(\cdot-2^k e_1)
\subset \{|\xi-2^{k}e_1| \le \sqrt{n}/4\}
\subset \{2^{k-1/4} \le |\xi| \le 2^{k+1/4}\},
\quad k \ge N_0.
\]
Then,
since $\widehat{f_N}(\xi)=\sum_{k=N_0}^{N}\phi(\xi-2^{k}e_1)$,
we see that
$\psi_{\ell}(D)f_N(x)$ is equal to
$e^{i2^\ell e_1\cdot x}\F^{-1}\phi(x)$ if $N_0 \le \ell \le N$,
and $0$ otherwise,
where $\{\psi_{\ell}\}$ is the same as in the beginning of Section \ref{section5},
and we used (\ref{Besov-parti-2-1}) and (\ref{Besov-parti-2-2}).
Hence,
\begin{equation}\label{estimate-Besov}
\|f_N\|_{B_{p,1}^0}
=\sum_{\ell=N_0}^{N}
\|e^{i2^\ell e_1\cdot x}\F^{-1}\phi\|_{L^p}
\approx N
\end{equation}
for sufficiently large $N$.

We next consider the amalgam space norm of $f_N$.
By Proposition \ref{amalgam-Wiener},
it is sufficient to estimate $\|f_N\|_{W_{1,2}}$.
Moreover, since $\mathrm{supp}\, \varphi$ is compact
and $\sum_{\nu \in \Z^n}\varphi(\xi-\nu) \ge 1$,
we have by \eqref{MW-equi}
\[
\|f_N\|_{(L^2,\ell^1)}
\approx \bigg\|\bigg(\sum_{\nu \in \Z^n}
|\varphi(D-\nu)f_N|^2\bigg)^{1/2}\bigg\|_{L^1}.
\]
Noting that
$\mathrm{supp}\, \phi(\cdot-2^k e_1) \subset 2^k e_1+[-1/4,1/4]^n$,
$\mathrm{supp}\, \varphi(\cdot-\nu) \subset \nu+[-3/4,3/4]^n$
and $\varphi(\cdot-\nu)=1$ on $\nu+[-1/4,1/4]^n$,
we see that $\varphi(D-\nu)f_N(x)$ is equal to
$e^{i2^k e_1\cdot x}\F^{-1}\phi(x)$ if $\nu=2^{k}e_1$ and $N_0 \le k \le N$,
and $0$ otherwise.
Therefore,
\begin{equation}\label{estimate-amalgam}
\|f_N\|_{(L^2,\ell^1)}
\approx \bigg\|\bigg(\sum_{k=N_0}^{N}
|e^{i2^k e_1\cdot x}\F^{-1}\phi|^2\bigg)^{1/2}\bigg\|_{L^1}
\approx N^{1/2}
\end{equation}
for sufficiently large $N$.

It follows from \eqref{estimate-Besov} and \eqref{estimate-amalgam} that
if $(L^2,\ell^1) \hookrightarrow B_{p,1}^0$,
then
\[
N \approx \|f_N\|_{B_{p,1}^0} \lesssim \|f_N\|_{(L^2,\ell^1)} \approx N^{1/2}
\]
for all sufficiently large $N$. However, this is a contradiction.
The proof of Proposition \ref{main3} is complete.

\section*{Acknowledgement}
The authors thank the referee for his/her careful reading of the manuscript.
The research of the third author was partially supported
by JSPS KAKENHI Grant Number JP20K03700.

redpajama_set_name: RedPajamaArXiv


text:
Station Name: BEULAH HALT
Date opened: 249.1928
Location: West side of Beulah Hill
Company on opening: Great Western Railway
Date closed completely: 25.10.1937
Company on closing: Great Western Railway
County: Pembrokeshire
OS Grid Ref: SM981284
Date of visit: 12 February 2017
Notes: As road transport became more popular throughout West Wales a number of halts were opened in an effort to boost passenger traffic. Four halts were opened on the North Pembrokeshire branch.
Beulah Halt was built on an embankment and was named after the adjacent Beulah Chapel although it served the village of Little Newcastle. It was provided with a 75ft platform with sleeper edging and a cinder infill. It had a 12ft x 6ft timber shelter and was lit by oil lamps.
The halt was used by an average of 85 - 90 passengers per week but was short-lived. Having opened on 24 September 1928, Beulah and Castlebythe and the two other new halts opened soon after brought some increase in passenger revenue but the North Pembrokeshire line was still running at a loss and closed to passengers nine years later on 25 October 1937.
Ticket from Michael Stewart. Route map drawn by Alan Young
Click here for a brief history of the North Pembrokeshire branch
Morris, John The Railways of Pembrokeshire 1981 HG Walters Ltd
Robertson, Kevin Great Western Railway Halts Volume 1 1990 Irwell Press
See stations on the North Pembrokeshire Branch: Fishguard & Goodwick, Jordanston Halt, Letterston, Martel Bridge Halt, Castlebythe Halt, Puncheston, New Inn Bridge Halt, Rosebush, Maenclochog, Llanycefn & Beag Fair
Looking west at the site of Beulah Halt in July 1974. No photos of Beulah Halt are known to exist.
As the halt was open for only nine years, it was not shown on any large scale Ordnance Survey maps as no such maps were published during its existence. Its approximate position is indicated on this 1907 1:2,500 map.
The bridge abutments on Beulah Hill looking north in September 2016. The halt was to the left.
Photo by Roy Lambeth
Beulah Chapel in September 2016. Although it served the small village of Little Newcastle the halt was named after the chapel which is adjacent to the railway line on Beulah Hill.
Looking north-west at the site of Beulah Halt in February 2017.The course of the old railway is to the right hand side of the blue trailer.
Photo by Richard Martin
Looking up the track from the north side of the bridge to the west where the halt would
have been located.
The west abutment of the Beulah Hill bridge in February 2017. The halt was to the left.
The east abutment of the Beulah Hill bridge and embankment in February 2017.
The east abutment of the Beulah Hill bridge in February 2017.
[Source: Alan Young]

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text:
You Can Get COVID and the Flu at the Same Time
In late February 2020, only a couple of weeks after the pandemic coronavirus disease had even received a formal name, a man with a terrible cough and fever showed up at one of the ProHealth Urgent Care centers in Queens, New York. At that point, no COVID-19 cases had yet been confirmed in New York City, but numbers were on the rise in spots across the country, and the man had recently been at a conference.
An ER-trained physician donned protective goggles, a face mask, and gloves, and went in to swab the man's nose with a giant Q-tip. With any luck at all, this would show he'd gotten sick from some other, more mundane pathogen—and that the timing of his symptoms was just a grim coincidence.
Within a few minutes, the staffers got just the answer they were hoping for: The swab was positive for influenza. But Daniel Griffin, a researcher at Columbia University and the chief of the infectious-disease division at ProHealth, wasn't ready to let down his guard. "Wait a second," he says he told the health-care team, as colleagues prepared to shed their protective gear. "How do we know this gentleman isn't also infected with the pandemic virus?" He suggested that they all assume the worst until they'd had a chance to run some further tests.
In early March, those results finally came in: Yes, the man was positive for the pandemic virus. His wife was, too, and so were their two kids. The whole family had COVID-19 and the flu.
By late summer 2020, many experts were warning of a potential "twindemic" of these two diseases. But their fears were as much about the possibility that COVID-19 and influenza would combine to overwhelm health systems as the possibility that they would combine to overwhelm an individual's immune system. In 2021, we still don't know much about how—or how often—the flu virus and SARS-CoV-2 act in tandem within the same body. A very early study from China in January 2020 found zero cases of coinfection of these two pathogens among 99 COVID-19 patients, but a follow-up, conducted a month later at a COVID-19 hospital, concluded that about one in eight had both diseases at the same time. Whatever the historic prevalence of coinfection, the twindemic never happened last winter. Perhaps due to mask wearing and social distancing, flu numbers in the U.S. were much, much lower than normal during the 2020–21 season.
But with pandemic restrictions relaxing, and fewer people getting flu shots, the same warnings have returned. "I actually think it's more of a risk this year," Griffin told me. Cases like the ones he saw almost two years ago could become far more common. In fact, new research suggests that getting coinfections—not just of COVID-19 and flu, but of many one-two punches of pathogens—may be far more common than we thought. Doctors' understanding of what these coinfections mean for care and treatment remains preliminary, but they could well have important consequences.
Some colds feel worse than others. On occasion they can be a terrible ordeal, knocking us out for days, leaving us miserable; other times they are barely a bother at all. This isn't a great mystery or anything: We know that some cold viruses are nastier than others, and that we can be exposed to larger or smaller amounts of the same virus; we also know that our immune systems can be stronger or weaker at any given time. But what if another, unacknowledged factor is at play? What if the nastiest case of sniffles you ever had was, in fact, two viruses infecting your body at the same time—a double-whammy cold?
Within the past decade, new molecular diagnostics have made such coinfections much easier to identify—and they've turned up some disquieting statistics. Recent screening studies have found that 14 to 70 percent of those hospitalized with flu-like illness test positive for more than one viral pathogen.
Read: We accidentally solved the flu. Now what?
For COVID-19, the acquisition of multiple infections appears to be associated with bad outcomes. People with severe COVID-19 who ended up in the ICU for many days and sometimes weeks were prone to developing additional illnesses—what's known as "superinfection"—while in the hospital. It's hard to shake the images of COVID-19 patients disfigured by mucormycosis, also known as black fungus; but those who end up in the ICU are also susceptible to ventilator-associated pneumonias and sepsis. People sick with flu are similarly prone to bacterial superinfections, which are thought to have caused many of the deaths during the 1918 pandemic, according to Brianne Barker, a professor at Drew University, in New Jersey, who teaches virology and immunology.
Some researchers have estimated that, all told, as many as half of all COVID-19 deaths can be attributed to mixed infections, although others put the number considerably lower. According to an analysis from last spring of more than 100 studies, people who tested positive for both SARS-CoV-2 and a second pathogen had triple the odds of dying compared with those who had only COVID-19. That added risk wasn't just associated with superinfections that came on during a patient's hospital stay; according to this analysis, it applied just as well to those who had coinfections from the time they were first diagnosed with COVID-19.
Those data come from studies of acute coinfections such as COVID-19 and flu—diseases that the body typically clears in a matter of days or weeks. Our bodies also hold a soup of chronic coinfections that never really go away. These include a host of herpes viruses such as cytomegalovirus, a lifelong presence in 50 to 80 percent of U.S. adults; and the varicella-zoster virus, which causes chicken pox and shingles. Another one that sticks with us forever is the Epstein-Barr virus, which can cause mononucleosis ("mono") when a person first is exposed and infects an estimated 90 percent of people worldwide.
These lifelong infections, kept in check by our immune systems, are usually irrelevant to later encounters with a different pathogen. But in certain cases they might be reawakened by a new disease that temporarily weakens our immune defenses. For example, some small studies during the pandemic offered hints that severe COVID-19 is associated with a quickening of latent cytomegalovirus or herpes simplex virus.
On the flip side, chronic infection with lifelong viruses can make people more vulnerable to acute infection. The classic example of this is HIV, which goes after the immune system itself and depletes the body's T cells. Individuals with the virus are far more likely to get tuberculosis, and as a result tuberculosis is a leading cause of death in people living with HIV.
Coinfections aren't always bad news for the patient. In some cases, pathogens are pitted against one another, competing for the same host cells. Take human papillomavirus, which causes cervical cancer and genital warts. A 2018 laboratory study found that a single human skin or cervical cell could be infected simultaneously with two different strains of HPV—and that when this happened, one of the strains ended up less able to generate copies of itself that would go on to infect further cells.
Researchers believe that in some situations a viral infection might also protect you by activating the body's innate immune response, priming it to thwart a second infection that arrives later. In a third scenario, an initial infection might also spark the activation of T cells that happen to recognize a second, similar invading pathogen—an effect known as a "bystander response." That's what happens to people who are infected with both dengue virus and Zika virus, although whether this really offers protection (or whether it might even cause some harm) is unclear.
Christopher Bucklow
Doctors have tried, on rare occasions, to induce coinfections as a treatment. A century ago, before antibiotic drugs were invented, the Austrian neuropsychiatrist Julius Wagner-Jauregg advocated for the use of malaria parasites as a cure for psychosis caused by syphilis. The treatment caused fever in patients, which Wagner-Jauregg found would ameliorate the psychosis. (Quinine would be administered to treat the malaria as soon as the disease had done its job.) Wagner-Jauregg won a Nobel Prize for this discovery, but his work would be overshadowed by his endorsement of eugenics and eventual support for the Nazis. Later scientists took up the idea of using other pathogens—in particular the common parasitic worm Ascaris lumbricoides—against malaria itself.
It's possible that even the dreaded collision of coronavirus and influenza might actually have some benefits for individuals. For a study published in 2018, a group of scientists in Moscow, Idaho, squirted a common-cold-type virus, either a rhinovirus or a mouse coronavirus, up the noses of white mice. When they followed these infections, two days later, with a squirt of a mouse-adapted flu virus, the mice fared better—they lived longer and showed fewer symptoms, such as ruffled fur and labored breathing—than other mice that were never given colds. The researchers suggested that the rhinovirus or coronavirus had set off an "early but controlled" inflammatory response in the animals' lungs, which then helped them defeat the flu virus. For these mice, at least, a lab-induced one-two was protective.
New testing techniques have revealed a combinatorial quagmire of infections in individual patients, even in routine practice. When doctors do a nose or throat swab on a sick child, the results sometimes indicate the presence of multiple pathogens, says Aubrey Cunnington, who is the head of the pediatric-infectious-diseases section at Imperial College London. The kids could turn out to have a mix of rhinovirus, parainfluenza, and respiratory syncytial virus. "We often see two or even three different viruses come up positive," he told me. "Coexisting infections with different organisms, particularly viruses, are the rule rather than the exception."
But what, exactly, do these coinfections mean? The implications of smorgasbord results remain unclear. "We have rather limited understanding of how they interact with one another, with bacteria, and other pathogens, to result in the illness affecting each patient," Cunnington said, adding that children with more than one pathogen don't necessarily seem sicker than others. In other words, having a double-whammy cold may not be especially bad. The same could be true for certain pandemic coinfections: Some research suggests that many COVID-19 patients have bacteria, fungi, or other viruses in their systems at the time of diagnosis, but with perhaps little reason for concern. One study at a local Chinese hospital, for example, found that about 94 percent of COVID-19 patients tested positive for either acute or chronic coinfections, but that rate was even higher (96 percent) for those with mild or asymptomatic illness. Our immune systems can produce different antibodies simultaneously, so they can typically multitask against different pathogens.
It's sometimes hard to tell which of several microbes in a patient is the one causing illness. The new molecular tests are so sensitive, they may pick up on harmless genetic fragments that linger in the body after a pathogen has been cleared—identifying a "coinfection" that is no longer even active. Griffin suggested that multiple-pathogen screens should still be used for patients hospitalized with COVID-19, or with any other grave disease, so that treatment can be tailored to the underlying need. If someone has the flu in addition to COVID-19, for example, a doctor might offer Tamiflu, or antibiotics might be given for a hidden bacterial infection. But Griffin warned that when molecular testing implicates multiple culprits, those results should be validated with further lab tests, such as trying to grow the pathogens in the lab. "You still want to do the cultures," Griffin told me.
Looking for coinfections creates more work for doctors, and interpreting the results when they find multiple pathogens lurking inside someone's organs makes for an even bigger clinical headache. But more and more studies are finding that coinfections are commonplace, so it's important not to ignore them—particularly for patients who are very ill. Not everything in biology abides by Ockham's razor, a principle attributed to the Franciscan friar and philosopher William of Ockham, which argues that, when given several possible solutions, the one with the least complexity is the most likely. "Ockham was not a physician," Griffin said. "You can have more than one thing going on."


Shannon Hannah Mayer became Owner/Director of Off Broadway Children's Theatre in June of 2000. A native of Dothan, Alabama, she began her training in dance at the early age of 8 years old. Shannon received a Bachelor of Arts in Theatre from Brenau Women's College in Gainesville, Georgia. She went on to pursue a career in musical theatre and dance which has given her the opportunity to perform throughout the southeastern United States. Shannon has appeared as a performer for Discovery Cruise Lines, as a performer and Dance Captain for the Savoyards Musical Theatre Company and is the resident choreographer for many north Georgia theatre companies as well as remaining an active performer. She is dedicated to the education of children and the commitment to helping the community.
Christian Mayer is Artistic Director and Owner of the Off Broadway Children's Theatre in Alpharetta, Ga. He has produced and directed over 50 children's productions and has mentored, coached and groomed hundreds of kids. He began his acting career in 1981, and has performed in such notable shows as "House of Blue Leaves", "Two for the Seesaw", "Bent", "Hatful of Rain", "Homefront", "The Mouestrap" "The Elephant Man","Loot" and was lucky enough to have proposed to his wife onstage in "Annie". Christian believes that theatre is a journey and that theatre is about the work of the whole, not the individual.
Erin Greenway studied acting, dance, choreography, and photography in the States and abroad. While in London, she had an apprenticeship with the English National Ballet. While with the Ballet she participated in the Community Education program exposing elementary schools, elderly groups, and children's hospitals to the world of ballet. After moving back to the states, she helped to set up Greensboro, North Carolina's first Arts Advocacy Council, and taught dance to elementary school children with learning disabilities. Companies that Erin has worked with include Spitzer Dance Company, Offshoot Productions, Lee Harper and Dancers, Stage Door Players, Ansley Park Playhouse, Jib Jam Productions, Odd Man Out Theatre, Rosewater Theatre, Performing Arts North Theatre Decatur, Academy Theatre, and extensively with Savage Tree Arts Project. Erin is also the Chair of Fine Arts department at the Brandon Hall School and is an company member of Performing Arts North Repertory Company.
Amy has been teaching and playing in the Atlanta area for many years. She attended Georgia State University, majoring in music education and is proficient on trumpet and piano, as well as voice. Being the music director for Elm Street Cultural Arts Village in Woodstock since 2009, she has worked on over 30 shows and is in high demand as an accompanist in Cherokee County schools and the community.


Birthday special: Taapsee Pannu turns 30 today!
Want to See Tom Hiddleston as Hamlet in London? Good Luck!
Salman Khan and Shah Rukh Khan to clash on the small screen?
Alia Bhatt's 'Raazi' gets a release date!
Did SRK just compare Anushka Sharma and Kajol?


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Its that time of year again that every dog owner loves, the world largest dog show, Crufts. Hooman will always attend the show year each and come home with a truck load of goodies for me, but this year the team at the Kennel Club have very kindly invited me too. How cool is that?
Its my job to give you humans the low down on the event as well as take plenty of pictures whilst I am there, and make plenty of trouble in the press office.
22,000 dogs competing for the world class title of Crufts Best In Show. Spilt into seven groups, Working, Toy, Utility, Hound, Gundogs, Pastoral and most importantly Terriers.
Good Citizen Dog Scheme training ring, the UK's largest dog training programme.
Young Kennel Club, International Junior Handling and Young Kennel Club Groomer of the Year.
Discover Dogs - located in hall 3. With over 200 breeds its the perfect opportunity to speak to experienced dog owners/breeders to find out what breed is best suited to your circumstances.
And of course plenty of shopping. With over 400 trade stands with special show offers and exciting new products for you and your best friend! For a list of all the exhibitors click here.
*Concessions: Children aged 12-15, Students with valid NUS card, Seniors aged 60+. Under 12s go FREE!
Click here to go to Cruft's ticket page.
This cute little video explains the history of Crufts.
The kennel club is celebrating 125 years this year and wants to make sure it the biggest and bestest dog show around. To kick off the celebrating they have very kindly sent us a set of tickets for a giveaway. They are valid for one day of your choice. You will need to purchase a extra tickets for the best in show event as those are sold separately. The giveaway closes on Monday 30 February at midnight. Good luck.
I hope to meet you there so that Lottie can give you lots of cuddles.


As part of this week's party promotions at House of Fraser (more below), I was invited in to try on some of the range of party dresses and jumpsuits in store at the moment and to select my favourite for the upcoming Christmas season.
I was in my element - you know my love of Issa dresses in particular and it was amazing to see this latest drop of frocks from the brand, with longer and more formal options now in store ahead of party season. But I was also impressed this visit with Label Lab and Linea.
Next up, another Issa one and this one is a bit more flattering on the mum tum and over the bust. The purple shade is stunning too for this time of year and I actually think it would work really well in summertime too with a denim jacket.
All the blingarama now with this Biba dress! It's not a great shape for me as I suit a bit more nipping in at the waist but what a Christmas dress this is. It's got those fab sequins that you can push up and down...which is potentially a torture as your kids will want to play with the dress constantly!!
Biba Fully Sequin Shift Dress - I'm wearing the 12 as it's tricky to get on over your head!
Finishing with a slightly less blingy number with Christmas day in mind. This dress would work so hard for you because as well as dressing it up with heels, you can dress it down with an oversized knit and ankle boots. The print is stunning and very seasonal and it covers the arms! Lots of ticks in the box and for me it was the one that stole my heart.
While I was in the store, I also checked out the Charlotte Tilbury counter mainly because I'd heard so much about the Limited Edition Instant Eye palette and their version of an advent calendar - the Naughty & Nice Magic Box which contains 12 products, including 4 full size. A must for any CT fans.
But to my mind, the palette is a more attractive option.
It's got 4 eye looks in one palette though I would mix and match them as the colours are workable together in any combination. As the counter was quiet yesterday, I was offered the option to have my eyes done with the palette and Joanna my MUA used the 'date eye' colours beautifully. It stayed put all day and still looked amazing when I was going to bed and reluctantly taking off.
The other gift I was tempted by was this starter Charlotte Tilbury skincare kit. I asked Joanna what her hero product from the whole brand range is and she chose the Magic Cream. Which retails at an eye watering £70! But the starter kit is a great way to try out this cream, and Charlotte's best selling mask and primer (all in travel sizes). Retails at £39.
But now, more about the Party People activity in House of Fraser stores this week. There already is up to 20% off party wear across many of the brands as part of this promotion, making this a great week to pick up your party outfit for Christmas.
In the Belfast store specifically, there's a fabulous evening planned this Thursday with discounts and treats as well as stylist advice in menswear and womenswear. There will also be gifting stations on all floors so the organised among us can start on their Christmas shopping (not me). The event runs between 6pm and 10pm.
Across all the stores, there's some Christmas charity activity going on in the form of Wishing Trees with all proceeds going to Action for Children. You can buy a decoration, write your wish and hang it in the tree in each store in return for a donation. What a lovely idea.
Disclaimer - I received vouchers to put toward buying the Linea dress above. I was not paid to write this post. In some cases, the links are affiliate meaning House of Fraser pay me a small commission should you purchase (and keep) the item linked to. Thanks for supporting the brands that support this blog.


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Newmarket, Ont. – February 13, 2018 – Southlake Regional Health Centre's Regional Cardiac Care Program has completed its cardiac catheterization (cath) labs redevelopment, reopening three renovated labs with state-of-the-art imaging systems for diagnostic angiograms, angioplasty and structural heart procedures.
This redevelopment has also allowed the three electrophysiology suites – two of which had been temporarily used for cath lab activity – to return to their intended functionality. The Heart Rhythm Interventional Suites will now be dedicated to the delivery of heart rhythm diagnostic studies, ablations and implantable cardiac devices including pacemakers and defibrillators. Now, with six fully functioning labs, there is anticipated growth for both the interventional cardiology and structural heart programs, ensuring greater capacity to deliver vital cardiac services, improving wait times for patients, and enabling the cardiac team to offer more advanced, leading edge procedures – all without overburdening staff or equipment.
Having the third-highest volume in the province, this updated equipment and increased footprint will enable Southlake's Regional Cardiac Care Program to be able to meet the needs of the population served in the Central LHIN, as well as the complex needs of those patients in the North Simcoe Muskoka LHIN. Southlake's partnership with Royal Victoria Hospital will continue to provide care for those more complex cath and PCI procedures requiring surgical back up.
SouthlakeRegional Health Centre is a full-service hospital with a regional, clinically advanced focus. As a regionally designated site, Southlake is responsible for developing and providing advanced levels of care to the more than one million people who reside in York Region, Simcoe County, and as far north as Muskoka. Advanced, specialty services include: arthritis care, cancer care, cardiac care, cataract surgery, child and adolescent eating disorders, mental health services for children, paediatrics and perinatal care, and thoracic surgery. Southlake is proud to have a team of over 3,000 employees, 580 physicians, and more than 900 volunteers. Combined with our commitment to provide the best possible care to patients, our goal is to make Southlake synonymous with healthcare excellence.


Scan any page, and instantly, magical 2D animations begin to play.
'AR TOTO' stories have been carefully crafted to be fun for both parents and children. Each fairy tale-like story details a fun and exciting adventure that also serves to cultivate a child's logical and critical thinking abilities. Each page of the 'AR TOTO' book series springs to life with 2D animation when viewed with a mo-bile device. The adorable characters dance, sing and respond to your child's every touch.
단순히 스마트폰 또는 태블릿을 이용하여 책에 있는 그림을 스캔 함으로써 해당 단어의 3D 애니메이션이 나타나는것을 확인할 수 있습니다. 책 뿐만 아니라 각 스티커들도 장소에 상관없이 테그가 가능합니다.
멀티플 모드를 통해 여러 테그를 한번에 테그 할수 도 있으며, 각 3D 이미지 클릭하여 단어의 스펠링, 파닉스를 학습할 수 있습니다.


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Home News Chelsea earn the right to be beaten by Man City in EFL...
Chelsea earn the right to be beaten by Man City in EFL Cup Final
Admit it, you saw it coming. Didn't you? Oh, come on. You did.
Today's funny football news could start nowhere other than pointing the finger-of-fun at good old Tottenham Hotspur.
Spurs went into their second leg clash with Premier League and city rivals Chelsea 1-0 up on aggregate knowing a draw would mean they'd be the team given the honour of losing to Man City in the EFL Cup Final next month.
Come on, even the most loyal of Tottenham fans saw it coming, surely? Especially when it went to penalties.
If Cesar Azpilicueta scores, you know it's your day and it was Chelsea's day alright.
Eden Hazard and Chelsea played as if Maurizio Sarri's public baiting had an effect. Who'd have thought that, eh? Or maybe they were keen to impress the onlooking and allegedly unfit Gonzalo Higuain. I tell you what, if Higuain is unfit I am about to be carried out in a coffin.
Thierry Henry's stay at AS Monaco didn't last long, did it? He's done such a bad job in France that Monaco have decided to bring back Leonardo Jardim and pretend the whole Henry era never happened. I bet Cesc Fabregas is over the moon having popped down to the tax haven to help an old mate out.
Did he last less time than Gary Neville? No matter, they can work that out when they are reunited in the Sky studios next month.
For us to get a comprehensive survey completed on whether Sky pundits make rubbish managers, we just need Jamie Carragher to take over a club in Germany and see how that goes.
The FA Cup fourth round gets going tonight with a mouthwatering prospect between Arsenal and Manchester United.
Unai Emery believes United were right to sack Jose Mourinho, meaning he has firmly nailed his colours to the same mast as the rest of the world. Emery also feels there is reason to be positive around the Emirates, and I can only presume that comment isn't linked to Hector Bellerin being out for the rest of the season and Petr Cech hanging up his gloves at the end of the season.
Ole Gunnar Solskjaer continues to be everything Mourinho was not and has put Paul Pogba's name forward to be a future long-term captain of United.
Rafa Benitez has been crying out for signings at Newcastle and he has finally got his wish and what a signing it is. In a move that could only remind Geordie fans of the moment they brought Alan Shearer home, Newcastle have signed Lukaku!
Ah, hang on. Wrong one. It's Romelu's brother, Jordan. He's a left back. He's big, strong, quick and got a very average first touch. Must run in the family.
West Ham are still scouring the back pages to decide which of the transfer rumours they should act on. The Hammers are being linked to many Marko Arnautovic replacements, even though the Austrian striker hasn't actually left yet. Today's rumour is Torino's Andrea Belotti although they'd also quite like Celta Vigo's Maxi Gomez who they were linked to in the summer.
Man City didn't get snubbed by Barcelona new boy Frenkie de Jong, alright? They just refused to get involved in a bidding war, OK?
That clears that up, then.
Previous articlePremier League CVs: Charlie Adam, fast food specialist
Next articleBlack & White: FA Cup 4th Round Preview
REVEALED! All the daft things the Premier League managers said before...
How can City beat all the English teams in the Premier...
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Pear Tree Roundup August 27, 2019
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News October 8, 2018
Same old Jose: Never happy, although he shouldn't be


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Q: img wont fit on mobile device i have a problem with my website. i want the background image to rescale corectly on mobile devices so i can see the full image on my mobile phone not just a part of it.
HTML :

Tractari auto intern & international

CSS
.demo-1 .sl-slide .sl-slide-inner,
.demo-2 .sl-slide .sl-slide-inner {
background-repeat: no-repeat;
background-size: contain;
background-position:center;
}

I want to see this on my phone, the full image:
https://gyazo.com/4b3726e24444038f2e6ff824a9f3ec45
not this
https://gyazo.com/f263a347517120a0ad368b7bbd2840ae
thanks in advance

A: So you have a couple of options, but you need to be aware of some constraints first:
The image will either be the width of the screen (and probably significantly shorter than you'd like) or the height of your container (and much wider than the screen).
In order to get the first result (100% wide, any height):
.demo-1 .sl-slide .sl-slide-inner,
.demo-2 .sl-slide .sl-slide-inner {
    background-repeat: no-repeat;
    background-size: 100% auto;
    background-position:center;
}

And in order to get the second one (which is what you're already getting):
    .demo-1 .sl-slide .sl-slide-inner,
    .demo-2 .sl-slide .sl-slide-inner {
        background-repeat: no-repeat;
        background-size: contain OR [auto 100%];
        background-position:center;
    }

But, short of manipulating/changing the image out based on screen size with javascript, there isn't much more control than that.

redpajama_set_name: RedPajamaStackExchange


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Lasers are used to collect hundreds of thousands of points of data a second from the canopy, which are processed to build a three-dimensional picture of the tree revealing its structure and its volume, which allows estimates of mass.
For example, one sycamore tree in Wytham Woods near Oxford was found to have nearly 11km (6.8 miles) of branches, double that of much taller tropical trees measured as part of the study, the researchers said.
It is hoped the information will give a more accurate picture of the amount of carbon absorbed by forests, as part of efforts to reduce greenhouse gases, as well as help predict how trees might respond to climate change.
Dr Mat Disney, from University College London, who pioneered the approach, said that previously tree mass could only be measured by chopping it down and weighing it, which kills the tree, or estimating it manually from the ground.
The new approach, which the scientists have used to analyze trees in places ranging from Gabon and Borneo to the UK and northern California, accurately captures the fine scale 3D structure of the tree.
This can tell researchers about the tree's "life history", including its survival strategy, how it grows and competes with other trees, and because the tree does not have to be cut down, its development can be monitored over time.
Estimating the mass of trees from their volume could help assess more accurately how much carbon is being stored in forests, and consequently the value that should be paid to countries not to cut them down.
Forests are earmarked to provide around a quarter of greenhouse gas emissions reductions under international efforts to cut climate change, but estimates of how much carbon is stored in tropical forests could be as much as 45 billion tonnes out, the researchers said.
There are also concerns about how forests could be affected by the changing climate, and the analysis could help predict the impacts of rising temperatures on trees.
"Going across the tropics, across different climate gradients, temperatures and rainfall, we are able to compare in very fine detail the structural impacts, why are trees slightly taller here, or have different crowns, even though they are similar kinds of forest.
"Understanding what these variations are is crucial to predicting how these trees are likely to respond to things like reduced rainfall and higher temperatures," Dr Disney said.
The research by scientists at University College London and the universities of Oxford, Sonoma State, Ghent and Wageningen, is published in the Royal Society Interface Focus journal.
If you looking to 3D laser scan trees for design planning, preservation, forest management or measuring volume or mass, TruePoint Laser Scanning can help. Accurate 3D measurements are captured in a fast and environmentally friendly manner. Laser scanners are non-contact, they scan from tripods on the ground up to 300 yards, keeping the environment untouched.
TruePoint's team of in-house engineers work quickly, accurately and knowledgably to 3d laser scan and post-process data. We can create custom deliverables, including 3D mesh, 2D drawings, intensity map point cloud files and more. To speak to someone, please contact us today at info@truepointscanning.com or 419-843-7226 (SCAN).


Threaded bushings for grid systems Material: Carbon steel. Version: Tempered to 1100-1300 N/mm². Sample order: K0863.01508305003 Note: Assembly instructions for changing threaded bushing. Inserting the locating and threaded bushing 1. Remove grease from the locating and threaded bushing. 2. Apply adhesive (Loctite 638) in the hole. 3. Apply adhesive (Loctite 638) on the threaded bushing and screw in. 4. Apply adhesive (Loctite 638) to the locating bushing and insert it. If the locating bushing cannot be inserted by hand, please use an assembly tool as shown application example:. 5. Remove any adhesive pressed out by insertion of the locating and threaded bushing before it hardens.


text:
They Are Allowing COVID Positive Staff To Work, While Firing At The Same…..
On Jan. 8, the California Department of Public Health issued guidance for hospital workers that test positive, allowing vaccinated, asymptomatic workers, to return without isolation or testing.
California is now suffering its self-imposed worker shortage after firing thousands of unvaccinated workers. Now, to fill in all the positions from the unvaccinated workers that they fired, they come to an idea to have those COVID positive workers go back to work. Can anyone tell me this idea isn't crazy?
Watch it here: KGET News/Youtube
According to the News Yahoo report:
State officials are attempting to address California's staffing shortage through a sweeping policy change that allows asymptomatic healthcare workers who have tested positive for the coronavirus to return to work immediately. The policy, set to remain in place through Feb. 1, is designed to keep many healthcare workers on the job at a time when hospitals are expecting more patients.
Some experts say California's stance is an unorthodox yet necessary solution to a difficult problem. Yet many healthcare workers and community members say the policy is not only ill-advised, it's potentially dangerous.
According to the guidelines, hospitals should exhaust all other options before resorting to the new policy, and workers who have tested positive for the virus should "preferably be assigned to work with COVID-19 positive patients." The workers must always wear N95 masks.
The announcement was met with outrage by many in the healthcare industry.
The decision is "irresponsible and a huge mistake that will jeopardize everyone's health," said Rosanna Mendez, executive director of SEIU 121RN, a union representing workers in Southern California. "This plan is unscientific and dangerous, and, given what we know about the transmissibility of the new variant, we believe it will put healthcare workers and patients at unnecessary risk."
The decision came shortly before the Supreme Court blocked an OSHA mandate that would have required all employees of private companies to be vaccinated. The Supreme Court upheld the part of the mandate that requires healthcare workers to be vaccinated, however.
Sources: 100 Percent Fed Up, The Gateway Pundit, New Yahoo, KGET.COM
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