# Bilinear multipliers of weighted Lebesgue spaces and variable exponent Lebesgue spaces

## Abstract

Let $1\le {p}_{1},{p}_{2}<\mathrm{\infty }$, $0<{p}_{3}\le \mathrm{\infty }$ and ${\omega }_{1}$, ${\omega }_{2}$, ${\omega }_{3}$ be weight functions on ${\mathbb{R}}^{n}$. Assume that ${\omega }_{1}$, ${\omega }_{2}$ are slowly increasing functions.

We say that a bounded function $m\left(\xi ,\eta \right)$ defined on ${\mathbb{R}}^{n}×{\mathbb{R}}^{n}$ is a bilinear multiplier on ${\mathbb{R}}^{n}$ of type $\left({p}_{1},{\omega }_{1};{p}_{2},{\omega }_{2};{p}_{3},{\omega }_{3}\right)$ (shortly $\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$) if

${B}_{m}\left(f,g\right)\left(x\right)={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta$

is a bounded bilinear operator from ${L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)×{L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$ to ${L}_{{\omega }_{3}}^{{p}_{3}}\left({\mathbb{R}}^{n}\right)$. We denote by $BM\left({p}_{1},{\omega }_{1};{p}_{2},{\omega }_{2};{p}_{3},{\omega }_{3}\right)$ (shortly $BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$) the vector space of bilinear multipliers of type $\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$.

In this paper first we discuss some properties of the space $BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. Furthermore, we give some examples of bilinear multipliers.

At the end of this paper, by using variable exponent Lebesgue spaces ${L}^{{p}_{1}\left(x\right)}\left({\mathbb{R}}^{n}\right)$, ${L}^{{p}_{2}\left(x\right)}\left({\mathbb{R}}^{n}\right)$ and ${L}^{{p}_{3}\left(x\right)}\left({\mathbb{R}}^{n}\right)$, we define the space of bilinear multipliers from ${L}^{{p}_{1}\left(x\right)}\left({\mathbb{R}}^{n}\right)×{L}^{{p}_{2}\left(x\right)}\left({\mathbb{R}}^{n}\right)$ to ${L}^{{p}_{3}\left(x\right)}\left({\mathbb{R}}^{n}\right)$ and discuss some properties of this space.

MSC:42A45, 42B15, 42B35.

## 1 Introduction

Throughout this paper ${C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)$, ${C}_{c}\left({\mathbb{R}}^{n}\right)$ and $S\left({\mathbb{R}}^{n}\right)$ denote the space of infinitely differentiable complex-valued functions with compact support on ${\mathbb{R}}^{n}$, the space of all continuous, complex-valued functions with compact support on ${\mathbb{R}}^{n}$ and the space of infinitely differentiable complex-valued functions on ${\mathbb{R}}^{n}$ rapidly decreasing at infinity, respectively. For $1\le p\le \mathrm{\infty }$, ${L}^{p}\left({\mathbb{R}}^{n}\right)$ denotes the usual Lebesgue space. A continuous function ω satisfying $1\le \omega \left(x\right)$ and $\omega \left(x+y\right)\le \omega \left(x\right)\omega \left(y\right)$ for $x,y\in {\mathbb{R}}^{n}$ will be called a weight function on ${\mathbb{R}}^{n}$. If ${\omega }_{1}\left(x\right)\le {\omega }_{2}\left(x\right)$ for all $x\in {\mathbb{R}}^{n}$, we say that ${\omega }_{1}\le {\omega }_{2}$. We say that a weight function ${\upsilon }_{s}$ is of polynomial type if ${\upsilon }_{s}\left(x\right)={\left(1+|x|\right)}^{s}$ for $s\ge 0$. Let f be a measurable function on ${\mathbb{R}}^{n}$. If there exist $C>0$ and $N\in \mathbb{N}$ such that

$|f\left(x\right)|\le C{\left(1+{x}^{2}\right)}^{N}$

for all $x\in {\mathbb{R}}^{n}$, then f is said to be a slowly increasing function. It is easy to see that polynomial-type weight functions are slowly increasing.

For $1\le p\le \mathrm{\infty }$, we set

${L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)=\left\{f:f\omega \in {L}^{p}\left({\mathbb{R}}^{n}\right)\right\}.$

It is known that ${L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)$ is a Banach space under the norm

${\parallel f\parallel }_{p,\omega }={\parallel f\omega \parallel }_{p}={\left\{{\int }_{{\mathbb{R}}^{n}}|f\left(x\right)\omega \left(x\right){|}^{p}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}},\phantom{\rule{1em}{0ex}}1\le p<\mathrm{\infty }$

or

${\parallel f\parallel }_{\mathrm{\infty },\omega }={\parallel f\omega \parallel }_{\mathrm{\infty }}=\underset{x\in {\mathbb{R}}^{n}}{ess sup}|f\left(x\right)\omega \left(x\right)|,\phantom{\rule{1em}{0ex}}p=\mathrm{\infty },\text{[1, 2]}.$

The dual of the space ${L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)$ is the space ${L}_{{\omega }^{-1}}^{q}\left({\mathbb{R}}^{n}\right)$, where $\frac{1}{p}+\frac{1}{q}=1$ and ${\omega }^{-1}\left(x\right)=\frac{1}{\omega \left(x\right)}$. For $f\in {L}^{1}\left({\mathbb{R}}^{n}\right)$, the Fourier transform of f is denoted by $\stackrel{ˆ}{f}$. We know that $\stackrel{ˆ}{f}$ is a continuous function on ${\mathbb{R}}^{n}$, which vanishes at infinity, and it has the inequality ${\parallel \stackrel{ˆ}{f}\parallel }_{\mathrm{\infty }}\le {\parallel f\parallel }_{1}$ [3, 4]. Let f be a measurable function on ${\mathbb{R}}^{n}$. The translation, character and dilation operators ${T}_{x}$, ${M}_{x}$ and ${D}_{s}$ are defined by ${T}_{x}f\left(y\right)=f\left(y-x\right)$, ${M}_{x}f\left(y\right)={e}^{2\pi i〈x,y〉}f\left(y\right)$ and ${D}_{t}^{p}f\left(y\right)={t}^{-\frac{n}{p}}f\left(\frac{y}{t}\right)$ respectively for $x,y\in {\mathbb{R}}^{n}$, $0. With this notation out of the way one has, for $1\le p\le \mathrm{\infty }$ and $\frac{1}{p}+\frac{1}{{p}^{\mathrm{\prime }}}=1$,

$\left({T}_{x}f\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(\xi \right)={M}_{-x}\stackrel{ˆ}{f}\left(\xi \right),\phantom{\rule{2em}{0ex}}\left({M}_{x}f\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(\xi \right)={T}_{x}\stackrel{ˆ}{f}\left(\xi \right),\phantom{\rule{2em}{0ex}}\left({D}_{t}^{p}f\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(\xi \right)={D}_{{t}^{-1}}^{{p}^{\mathrm{\prime }}}\stackrel{ˆ}{f}\left(\xi \right).$

We denote by $M\left({\mathbb{R}}^{n}\right)$ the space of bounded regular Borel measures, by $M\left(\omega \right)$ the space of μ in $M\left({\mathbb{R}}^{n}\right)$ such that

${\parallel \mu \parallel }_{\omega }={\int }_{{\mathbb{R}}^{n}}\omega \phantom{\rule{0.2em}{0ex}}d|\mu |<\mathrm{\infty }.$

If $\mu \in M\left({\mathbb{R}}^{n}\right)$, the Fourier-Stieltjes transform of μ is denoted by $\stackrel{ˆ}{\mu }$ [5]. In this paper, $P\left({\mathbb{R}}^{n}\right)$ denotes the family of all measurable functions $p:{\mathbb{R}}^{n}\to \left[1,\mathrm{\infty }\right)$. We put

${p}_{\ast }=\underset{x\in {\mathbb{R}}^{n}}{ess inf}p\left(x\right),\phantom{\rule{2em}{0ex}}{p}^{\ast }=\underset{x\in {\mathbb{R}}^{n}}{ess sup}p\left(x\right).$

We shall also use the notation

${\mathrm{\Omega }}_{\mathrm{\infty }}=\left\{x\in {\mathbb{R}}^{n}:p\left(x\right)=\mathrm{\infty }\right\}.$

The generalized Lebesgue space (or the variable exponent Lebesgue space) ${L}^{p\left(x\right)}\left({\mathbb{R}}^{n}\right)$ is defined to be a space of (equivalence classes) measurable functions f such that

${\varrho }_{p}\left(\lambda f\right)={\int }_{{\mathbb{R}}^{n}\mathrm{\setminus }{\mathrm{\Omega }}_{\mathrm{\infty }}}|\lambda f\left(x\right){|}^{p\left(x\right)}\phantom{\rule{0.2em}{0ex}}dx+\underset{x\in {\mathrm{\Omega }}_{\mathrm{\infty }}}{ess sup}\left(\lambda f\left(x\right)\right)<\mathrm{\infty }$

for some $\lambda =\lambda \left(f\right)>0$. If ${p}^{\ast }<\mathrm{\infty }$, then

${\varrho }_{p}\left(\lambda f\right)={\int }_{{\mathbb{R}}^{n}\mathrm{\setminus }{\mathrm{\Omega }}_{\mathrm{\infty }}}|\lambda f\left(x\right){|}^{p\left(x\right)}\phantom{\rule{0.2em}{0ex}}dx,\phantom{\rule{1em}{0ex}}\text{[6, 7]}.$

It is known by Theorem 2.5 in [6] that ${L}^{p\left(x\right)}\left({\mathbb{R}}^{n}\right)$ is a Banach space with the Luxemburg norm

${\parallel f\parallel }_{p\left(x\right)}=inf\left\{\lambda >0:{\varrho }_{p}\left(\frac{f}{\lambda }\right)\le 1\right\}.$

If ${p}^{\ast }<\mathrm{\infty }$, then ${C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)$ is dense in ${L}^{p\left(x\right)}\left({\mathbb{R}}^{n}\right)$. Also, if $p\left(x\right)=p$ is a constant function, then the above norm ${\parallel \cdot \parallel }_{p\left(x\right)}$ coincides with the usual norm ${\parallel \cdot \parallel }_{p}$. The vector space of locally integrable functions on ${\mathbb{R}}^{n}$ is denoted by ${L}_{\mathrm{loc}}^{1}\left({\mathbb{R}}^{n}\right)$. The space ${L}^{p\left(x\right)}\left({\mathbb{R}}^{n}\right)$ is a solid space, that is, if $f\in {L}^{p\left(x\right)}\left({\mathbb{R}}^{n}\right)$ is given and $g\in {L}_{\mathrm{loc}}^{1}\left({\mathbb{R}}^{n}\right)$ satisfies $|g\left(x\right)|\le |f\left(x\right)|$ a.e., then $g\in {L}^{p\left(x\right)}\left({\mathbb{R}}^{n}\right)$ and ${\parallel g\parallel }_{p\left(x\right)}\le {\parallel f\parallel }_{p\left(x\right)}$ by [8]. In this paper we assume that ${p}^{\ast }<\mathrm{\infty }$.

## 2 The bilinear multipliers space $BM\left({p}_{1},{\omega }_{1};{p}_{2},{\omega }_{2};{p}_{3},{\omega }_{3}\right)$

Lemma 2.1 Let $1\le p<\mathrm{\infty }$ and let ω be a slowly increasing weight function. Then $S\left({\mathbb{R}}^{n}\right)$ is dense in ${L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)$.

Proof Let $f\in S\left({\mathbb{R}}^{n}\right)$ be given. Since ω is a slowly increasing weight function, there exist $C>0$ and $N\in \mathbb{N}$ such that

$|\omega \left(x\right)|\le C{\left(1+{x}^{2}\right)}^{N}=m\left(x\right)$
(2.1)

for all $x\in {\mathbb{R}}^{n}$. Also, since m is a polynomial, then by Proposition 19.2.2 in [9], we have $S\left({\mathbb{R}}^{n}\right)\subset {L}_{m}^{p}\left({\mathbb{R}}^{n}\right)$. Hence, by (2.1), we obtain $S\left({\mathbb{R}}^{n}\right)\subset {L}_{m}^{p}\left({\mathbb{R}}^{n}\right)\subset {L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)$.

Now, we show that ${C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)$ is dense ${L}_{m}^{p}\left({\mathbb{R}}^{n}\right)$. Let $f\in {L}_{w}^{p}\left({\mathbb{R}}^{n}\right)$ be given. Then $fm\in {L}^{p}\left({\mathbb{R}}^{n}\right)$. Since ${C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)$ is dense ${L}^{p}\left({\mathbb{R}}^{n}\right)$ by [6], for given $\epsilon >0$, there exists $g\in {C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)$ such that

${\parallel fm-g\parallel }_{p}<\epsilon .$
(2.2)

Therefore, by using the inequality (2.2), we write

${\parallel fm-g\parallel }_{p}={\parallel f-g{m}^{-1}\parallel }_{p,m}<\epsilon .$

Also, since $m\ne 0$ and m is a polynomial, we have $g{m}^{-1}\in {C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)$. Thus, we have $\overline{{C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)}={L}_{m}^{p}\left({\mathbb{R}}^{n}\right)$. By using the inclusion ${C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)\subset S\left({\mathbb{R}}^{n}\right)\subset {L}_{m}^{p}\left({\mathbb{R}}^{n}\right)$, we obtain $\overline{S\left({\mathbb{R}}^{n}\right)}={L}_{m}^{p}\left({\mathbb{R}}^{n}\right)$.

Now, take any $f\in {L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)$. Since $\overline{{C}_{c}\left({\mathbb{R}}^{n}\right)}=\overline{{L}_{m}^{p}\left({\mathbb{R}}^{n}\right)}={L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)$, there exists $g\in {L}_{m}^{p}\left({\mathbb{R}}^{n}\right)$ such that

${\parallel f-g\parallel }_{p,\omega }<\frac{\epsilon }{2}.$
(2.3)

Furthermore, since $S\left({\mathbb{R}}^{n}\right)$ is dense ${L}_{m}^{p}\left({\mathbb{R}}^{n}\right)$, there exists $h\in S\left({\mathbb{R}}^{n}\right)$ such that

${\parallel g-h\parallel }_{p,m}<\frac{\epsilon }{2}.$
(2.4)

Combining the inequalities (2.3) and (2.4), we have

${\parallel f-g\parallel }_{p,\omega }\le {\parallel f-g\parallel }_{p,\omega }+{\parallel h-g\parallel }_{p,\omega }\le {\parallel f-g\parallel }_{p,\omega }+{\parallel h-g\parallel }_{p,m}<\epsilon ,$

which means $\overline{S\left({\mathbb{R}}^{n}\right)}={L}_{\omega }^{p}\left({\mathbb{R}}^{n}\right)$. □

Definition 2.1 Let $1\le {p}_{1},{p}_{2}<\mathrm{\infty }$, $0<{p}_{3}\le \mathrm{\infty }$ and ${\omega }_{1}$, ${\omega }_{2}$, ${\omega }_{3}$ be weight functions on ${\mathbb{R}}^{n}$. Assume that ${\omega }_{1}$, ${\omega }_{2}$ are slowly increasing functions and $m\left(\xi ,\eta \right)$ is a bounded function on ${\mathbb{R}}^{n}×{\mathbb{R}}^{n}$. Define

${B}_{m}\left(f,g\right)\left(x\right)={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta$

for all $f,g\in S\left({\mathbb{R}}^{n}\right)$.

m is said to be a bilinear multiplier on ${\mathbb{R}}^{n}$ of type $\left({p}_{1},{\omega }_{1};{p}_{2},{\omega }_{2};{p}_{3},{\omega }_{3}\right)$ (shortly $\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$) if there exists $C>0$ such that

${\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}$

for all $f,g\in S\left({\mathbb{R}}^{n}\right)$. That means ${B}_{m}$ extends to a bounded bilinear operator from ${L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)×{L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$ to ${L}_{{\omega }_{3}}^{{p}_{3}}\left({\mathbb{R}}^{n}\right)$.

We denote by $BM\left({p}_{1},{\omega }_{1};{p}_{2},{\omega }_{2};{p}_{3},{\omega }_{3}\right)$ (shortly $BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$) the space of all bilinear multipliers of type $\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ and ${\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}=\parallel {B}_{m}\parallel$.

Theorem 2.1 Let $\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}=\frac{1}{{p}_{3}}$ and ${\omega }_{3}\le {\omega }_{1}$. If $K\in {L}_{{\omega }_{3}}^{1}\left({\mathbb{R}}^{n}\right)$, then $m\left(\xi ,\eta \right)=\stackrel{ˆ}{K}\left(\xi -\eta \right)$ defines a bilinear multiplier and ${\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\le {\parallel K\parallel }_{1,{\omega }_{3}}$.

Proof For $f,g\in S\left({\mathbb{R}}^{n}\right)$, we have $f\left(x-y\right)={\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right){e}^{2\pi i〈x-y,\xi 〉}\phantom{\rule{0.2em}{0ex}}d\xi$ and $g\left(x+y\right)={\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{g}\left(\eta \right){e}^{2\pi i〈x+y,\eta 〉}\phantom{\rule{0.2em}{0ex}}d\eta$. Thus, by the Fubini theorem, we write

$\begin{array}{rl}{B}_{m}\left(f,g\right)\left(x\right)& ={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{K}\left(\xi -\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ ={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\left({\int }_{{\mathbb{R}}^{n}}K\left(y\right){e}^{-2\pi i〈\xi -\eta ,y〉}\phantom{\rule{0.2em}{0ex}}dy\right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ ={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)K\left(y\right){e}^{-2\pi i〈\xi -\eta ,y〉}{e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \phantom{\rule{0.2em}{0ex}}dy.\end{array}$
(2.5)

Since $f,g\in S\left({\mathbb{R}}^{n}\right)$, we have $\stackrel{ˆ}{f},\stackrel{ˆ}{g}\in S\left({\mathbb{R}}^{n}\right)\subset {L}^{1}\left({\mathbb{R}}^{n}\right)$. Hence, by (2.5), we obtain

$\begin{array}{rl}{B}_{m}\left(f,g\right)\left(x\right)& ={\int }_{{\mathbb{R}}^{n}}\left({\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right){e}^{2\pi i〈x-y,\xi 〉}\right)\left({\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{g}\left(\eta \right){e}^{2\pi i〈x+y,\eta 〉}\phantom{\rule{0.2em}{0ex}}d\eta \right)K\left(y\right)\phantom{\rule{0.2em}{0ex}}dy\\ ={\int }_{{\mathbb{R}}^{n}}f\left(x-y\right)g\left(x+y\right)K\left(y\right)\phantom{\rule{0.2em}{0ex}}dy.\end{array}$
(2.6)

Since ${\omega }_{3}\le {\omega }_{1}$, then

${\parallel f\left(x-y\right){\omega }_{3}\parallel }_{{p}_{1}}\le {\omega }_{3}\left(y\right){\parallel f\parallel }_{{p}_{1},{\omega }_{1}}$
(2.7)

and hence $f\left(x-y\right){\omega }_{3}\in {L}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$. Therefore from (2.6) and the Minkowski inequality, we write

$\begin{array}{rl}{\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& \le {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{\parallel f\left(x-y\right)g\left(x+y\right)\parallel }_{{p}_{3},{\omega }_{3}}|K\left(y\right)|\phantom{\rule{0.2em}{0ex}}dy\\ ={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{\parallel f\left(x-y\right)g\left(x+y\right){\omega }_{3}\parallel }_{{p}_{3}}|K\left(y\right)|\phantom{\rule{0.2em}{0ex}}dy.\end{array}$
(2.8)

Hence, using the generalized Hölder inequality and combining (2.7), (2.8), we have

$\begin{array}{rcl}{\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& \le & {\int }_{{\mathbb{R}}^{n}}{\parallel f\left(x-y\right){\omega }_{3}\parallel }_{{p}_{1}}{\parallel g\left(x+y\right)\parallel }_{{p}_{2}}{\omega }_{3}\left(y\right)|K\left(y\right)|\phantom{\rule{0.2em}{0ex}}dy\\ \le & {\int }_{{\mathbb{R}}^{n}}{\parallel f\parallel }_{{p}_{1}}{\parallel g\parallel }_{{p}_{2}}{\omega }_{3}\left(y\right)|K\left(y\right)|\phantom{\rule{0.2em}{0ex}}dy\\ \le & {\int }_{{\mathbb{R}}^{n}}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\omega }_{3}\left(y\right)|K\left(y\right)|\phantom{\rule{0.2em}{0ex}}dy\\ =& {\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel K\parallel }_{1,{\omega }_{3}}.\end{array}$
(2.9)

If we set $C={\parallel K\parallel }_{1,{\omega }_{3}}$, we obtain

${\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}.$

Then $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. Consequently, using (2.9), we have

$\begin{array}{rcl}{\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}& =& sup\left\{\frac{{\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}}{{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}}:{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}\le 1,{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\le 1\right\}\\ \le & {\parallel K\parallel }_{1,{\omega }_{3}}.\end{array}$

□

Definition 2.2 Let $1\le {p}_{1},{p}_{2}<\mathrm{\infty }$, $0<{p}_{3}\le \mathrm{\infty }$ and ${\omega }_{1}$, ${\omega }_{2}$, ${\omega }_{3}$ be weight functions on ${\mathbb{R}}^{n}$. Suppose that ${\omega }_{1}$, ${\omega }_{2}$ are slowly increasing functions. We denote by $\stackrel{˜}{M}\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ the space of measurable functions $M:{\mathbb{R}}^{n}\to \mathbb{C}$ such that $m\left(\xi ,\eta \right)=M\left(\xi -\eta \right)\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, that is to say,

${B}_{M}\left(f,g\right)\left(x\right)={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)M\left(\xi -\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta$

extends to a bounded bilinear map from ${L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)×{L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$ to ${L}_{{\omega }_{3}}^{{p}_{3}}\left({\mathbb{R}}^{n}\right)$. We denote ${\parallel M\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}=\parallel {B}_{M}\parallel$.

Theorem 2.2 Let ${p}_{3}\ge 1$ and ${\omega }_{3}\left(-x\right)={\omega }_{3}\left(x\right)$. Then $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ if and only if there exists $C>0$ such that

$|{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{h}\left(\xi +\eta \right)m\left(\xi ,\eta \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta |\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}$

for all $f,g,h\in S\left({\mathbb{R}}^{n}\right)$, where $\frac{1}{{p}_{3}}+\frac{1}{{p}_{3}^{\mathrm{\prime }}}=1$.

Proof Let $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. We take any $f,g,h\in S\left({\mathbb{R}}^{n}\right)$. From the Fubini theorem, we write

(2.10)

where ${\stackrel{˜}{B}}_{m}\left(f,g\right)\left(y\right)={B}_{m}\left(f,g\right)\left(-y\right)$. On the other hand, since $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, then ${B}_{m}\left(f,g\right)\in {L}_{{\omega }_{3}}^{{p}_{3}}\left({\mathbb{R}}^{n}\right)$. Thus we obtain ${\stackrel{˜}{B}}_{m}\left(f,g\right)\in {L}_{{\omega }_{3}}^{{p}_{3}}\left({\mathbb{R}}^{n}\right)$. Also, $h\in S\left({\mathbb{R}}^{n}\right)\subset {L}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)\subset {L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)$. Hence, using the Hölder inequality and the inequality (2.10), we write

$\begin{array}{r}|{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{h}\left(\xi +\eta \right)m\left(\xi ,\eta \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta |\\ \phantom{\rule{1em}{0ex}}\le {\int }_{{\mathbb{R}}^{n}}|h\left(y\right){\omega }_{3}^{-1}\left(y\right)||{\stackrel{˜}{B}}_{m}\left(f,g\right)\left(y\right){\omega }_{3}\left(y\right)|\phantom{\rule{0.2em}{0ex}}dy\\ \phantom{\rule{1em}{0ex}}\le {\parallel {\stackrel{˜}{B}}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}.\end{array}$
(2.11)

Moreover, since $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, there exists $C>0$ such that

${\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}.$
(2.12)

If we combine (2.11) and (2.12), we write

$|{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{h}\left(\xi +\eta \right)m\left(\xi ,\eta \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta |\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}.$

For the proof of converse, assume that there exists a constant $C>0$ such that

$|{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{h}\left(\xi +\eta \right)m\left(\xi ,\eta \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta |\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}$

for all $f,g,h\in S\left({\mathbb{R}}^{n}\right)$. From the assumption and (2.10), we write

$|{\int }_{{\mathbb{R}}^{n}}h\left(y\right){\stackrel{˜}{B}}_{m}\left(f,g\right)\left(y\right)\phantom{\rule{0.2em}{0ex}}dy|\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}.$
(2.13)

Define a function l from $S\left({\mathbb{R}}^{n}\right)\subset {L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)$ to such that

$\ell \left(h\right)={\int }_{{\mathbb{R}}^{n}}h\left(y\right){\stackrel{˜}{B}}_{m}\left(f,g\right)\left(y\right)\phantom{\rule{0.2em}{0ex}}dy.$

It is clear that the function is linear and bounded by (2.13). By using $\overline{{C}_{c}\left({\mathbb{R}}^{n}\right)}={L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)$ in [10], it is easy to show that $\overline{{C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)}={L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)$. So, by the inclusion ${C}_{c}^{\mathrm{\infty }}\left({\mathbb{R}}^{n}\right)\subset S\left({\mathbb{R}}^{n}\right)\subset {L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)$, we have $\overline{S\left({\mathbb{R}}^{n}\right)}={L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)$. Thus extends to a bounded function from ${L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)$ to . Then $\ell \in {\left({L}_{{\omega }_{3}^{-1}}^{{p}_{3}^{\mathrm{\prime }}}\left({\mathbb{R}}^{n}\right)\right)}^{\ast }={L}_{{\omega }_{3}}^{{p}_{3}}\left({\mathbb{R}}^{n}\right)$ and by (2.13), we have

$\begin{array}{rcl}{\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& =& {\parallel {\stackrel{˜}{B}}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}=\parallel \ell \parallel =\underset{{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}\le 1}{sup}\frac{|l\left(h\right)|}{{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}}\\ \le & \underset{{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}\le 1}{sup}\frac{C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}}{{\parallel h\parallel }_{{p}_{3}^{\mathrm{\prime }},{\omega }_{3}^{-1}}}\le C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}.\end{array}$

Hence, we obtain $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. □

Theorem 2.3 Let $\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}=\frac{1}{{p}_{3}}$, ${p}_{3}\ge 1$ and ${\upsilon }_{s}\left(x\right)={\left(1+|x|\right)}^{s}$, $s\ge 0$ be a weight function of polynomial type such that ${\upsilon }_{s}\le {\omega }_{1}$. If $\mu \in M\left({\upsilon }_{s}\right)$ and $m\left(\xi ,\eta \right)=\stackrel{ˆ}{\mu }\left(\alpha \xi +\beta \eta \right)$ for $\alpha ,\beta \in \mathbb{R}$, then $m\in BM\left({\omega }_{1},{\omega }_{2},{\upsilon }_{s}\right)$. Moreover,

$\begin{array}{c}{\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\upsilon }_{s}\right)}\le {\parallel \mu \parallel }_{{\upsilon }_{s}}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}|\alpha |\le 1,\hfill \\ {\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\upsilon }_{s}\right)}\le |\alpha {|}^{s}{\parallel \mu \parallel }_{{\upsilon }_{s}}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}|\alpha |>1.\hfill \end{array}$

Proof Let $f,g,\in S\left({\mathbb{R}}^{n}\right)$. Then

$\begin{array}{rcl}{B}_{m}\left(f,g\right)\left(x\right)& =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{\mu }\left(\alpha \xi +\beta \eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\left\{{\int }_{{\mathbb{R}}^{n}}{e}^{-2\pi i〈\alpha \xi +\beta \eta ,t〉}\phantom{\rule{0.2em}{0ex}}d\mu \left(t\right)\right\}{e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}\left\{{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right){e}^{2\pi i〈x-\alpha t,\xi 〉}\phantom{\rule{0.2em}{0ex}}d\xi \right\}\left\{{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{g}\left(\eta \right){e}^{2\pi i〈x-\beta t,\eta 〉}\phantom{\rule{0.2em}{0ex}}d\eta \right\}\phantom{\rule{0.2em}{0ex}}d\mu \left(t\right)\\ =& {\int }_{{\mathbb{R}}^{n}}f\left(x-\alpha t\right)g\left(x-\beta t\right)\phantom{\rule{0.2em}{0ex}}d\mu \left(t\right).\end{array}$
(2.14)

On the other hand, by the assumption ${\upsilon }_{s}\le {\omega }_{1}$, it is easy to see that $f\left(x-\alpha t\right){\upsilon }_{s}\in {L}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and

${\parallel f\left(x-\alpha t\right){\upsilon }_{s}\parallel }_{{p}_{1}}\le {\upsilon }_{s}\left(\alpha t\right){\parallel f\parallel }_{{p}_{1},{\omega }_{1}}.$
(2.15)

Also, $g\left(x-\beta t\right)\in {L}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. Then, by (2.14), (2.15) and the generalized Hölder inequality, we have

$\begin{array}{rcl}{\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{s}}& \le & {\int }_{{\mathbb{R}}^{n}}{\parallel f\left(x-\alpha t\right)g\left(x-\beta t\right)\parallel }_{{p}_{3},{\upsilon }_{s}}\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)\\ \le & {\int }_{{\mathbb{R}}^{n}}{\parallel f\left(x-\alpha t\right){\upsilon }_{s}\parallel }_{{p}_{1}}{\parallel g\left(x-\beta t\right)\parallel }_{{p}_{2}}\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)\\ \le & {\int }_{{\mathbb{R}}^{n}}{\upsilon }_{s}\left(\alpha t\right){\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2}}\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)\\ \le & {\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\int }_{{\mathbb{R}}^{n}}{\upsilon }_{s}\left(\alpha t\right)\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right).\end{array}$
(2.16)

Now, suppose that $|\alpha |\le 1$. Then we write

$\begin{array}{rcl}{\int }_{{\mathbb{R}}^{n}}{\upsilon }_{s}\left(\alpha t\right)\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)& =& {\int }_{{\mathbb{R}}^{n}}{\left(1+|\alpha t|\right)}^{s}\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)\\ \le & {\int }_{{\mathbb{R}}^{n}}{\left(1+|t|\right)}^{s}\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)={\parallel \mu \parallel }_{{\upsilon }_{s}}.\end{array}$

Hence by (2.16)

${\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{s}}\le {\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel \mu \parallel }_{{\upsilon }_{s}}.$
(2.17)

Thus $m\in BM\left({\omega }_{1},{\omega }_{2},{\upsilon }_{s}\right)$ and by (2.17), we have

${\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\upsilon }_{s}\right)}=sup\left\{\frac{{\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{s}}}{{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}}:{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}\le 1,{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\le 1\right\}\le {\parallel \mu \parallel }_{{\upsilon }_{s}}.$

Similarly, if $|\alpha |>1$, then we write

$\begin{array}{rcl}{\int }_{{\mathbb{R}}^{n}}{\upsilon }_{s}\left(\alpha t\right)\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)& <& {\int }_{{\mathbb{R}}^{n}}{\left(|\alpha |+|\alpha ||t|\right)}^{s}\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)\\ =& |\alpha {|}^{s}{\int }_{{\mathbb{R}}^{n}}{\upsilon }_{s}\left(t\right)\phantom{\rule{0.2em}{0ex}}d|\mu |\left(t\right)=|\alpha {|}^{s}{\parallel \mu \parallel }_{{\upsilon }_{s}}.\end{array}$

Again, by (2.16) we have

${\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{s}}\le |\alpha {|}^{s}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel \mu \parallel }_{{\upsilon }_{s}}.$
(2.18)

Hence, we obtain $m\in BM\left({\omega }_{1},{\omega }_{2},{\upsilon }_{s}\right)$ and by (2.18)

${\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\upsilon }_{s}\right)}=sup\left\{\frac{{\parallel {B}_{m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{s}}}{{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}}:{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}\le 1,{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\le 1\right\}\le |\alpha {|}^{s}{\parallel \mu \parallel }_{{\upsilon }_{s}}.$

□

Theorem 2.4 Let $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$.

1. (a)

${T}_{\left({\xi }_{0},{\eta }_{0}\right)}m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ for each $\left({\xi }_{0},{\eta }_{0}\right)\in {\mathbb{R}}^{2n}$ and

${\parallel {T}_{\left({\xi }_{0},{\eta }_{0}\right)}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}={\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.$
2. (b)

${M}_{\left({\xi }_{0},{\eta }_{0}\right)}m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ for each $\left({\xi }_{0},{\eta }_{0}\right)\in {\mathbb{R}}^{2n}$ and

${\parallel {M}_{\left({\xi }_{0},{\eta }_{0}\right)}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\le {\omega }_{1}\left(-{\xi }_{0}\right){\omega }_{2}\left(-{\eta }_{0}\right){\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.$

Proof (a) Let us take any $f\in {L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and $g\in {L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. If we say that $\xi -{\xi }_{0}=u$ and $\eta -{\eta }_{0}=v$, then

$\begin{array}{rl}{B}_{{T}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\left(f,g\right)\left(x\right)& ={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right){T}_{\left({\xi }_{0},{\eta }_{0}\right)}m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ ={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(u+{\xi }_{0}\right)\stackrel{ˆ}{g}\left(v+{\eta }_{0}\right)m\left(u,v\right){e}^{2\pi i〈u+{\xi }_{0},x〉}{e}^{2\pi i〈v+{\eta }_{0},x〉}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ ={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{T}_{-{\xi }_{0}}\stackrel{ˆ}{f}\left(u\right){T}_{-{\eta }_{0}}\stackrel{ˆ}{g}\left(v\right)m\left(u,v\right){e}^{2\pi i〈{\xi }_{0}+{\eta }_{0},x〉}{e}^{2\pi i〈u+v,x〉}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv.\end{array}$
(2.19)

By (2.19), we have

$\begin{array}{rcl}{B}_{{T}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\left(f,g\right)\left(x\right)& =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{T}_{-{\xi }_{0}}\stackrel{ˆ}{f}\left(u\right){T}_{-{\eta }_{0}}\stackrel{ˆ}{g}\left(v\right){e}^{2\pi i〈{\xi }_{0}+{\eta }_{0},x〉}{e}^{2\pi i〈u+v,x〉}m\left(u,v\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& {e}^{2\pi i〈{\xi }_{0}+{\eta }_{0},x〉}{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\left({M}_{-{\xi }_{0}}f\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(u\right)\left({M}_{-{\eta }_{0}}g\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(v\right)m\left(u,v\right){e}^{2\pi i〈u+v,x〉}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& {e}^{2\pi i〈{\xi }_{0}+{\eta }_{0},x〉}{B}_{m}\left({M}_{-{\xi }_{0}}f,{M}_{-{\eta }_{0}}g\right)\left(x\right).\end{array}$
(2.20)

Since $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, ${\parallel {M}_{-{\xi }_{0}}f\parallel }_{{p}_{1},{\omega }_{1}}={\parallel f\parallel }_{{p}_{1},{\omega }_{1}}$ and ${\parallel {M}_{-{\eta }_{0}}g\parallel }_{{p}_{2},{\omega }_{2}}={\parallel g\parallel }_{{p}_{2},{\omega }_{2}}$ are satisfied for all $f\in {L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and $g\in {L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. Hence, by (2.20), we have

$\begin{array}{rcl}{\parallel {B}_{{T}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& =& {\parallel {e}^{2\pi i〈{\xi }_{0}+{\eta }_{0},x〉}{B}_{m}\left({M}_{-{\xi }_{0}}f,{M}_{-{\eta }_{0}}g\right)\parallel }_{{p}_{3},{\omega }_{3}}\\ \le & C{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\end{array}$

for some $C>0$. Thus ${T}_{\left({\xi }_{0},{\eta }_{0}\right)m}\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. Also, we obtain

$\begin{array}{c}{\parallel {T}_{\left({\xi }_{0},{\eta }_{0}\right)}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {B}_{{T}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=sup\left\{\frac{{\parallel {B}_{{T}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}}{{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}}:{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}\le 1,{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\le 1\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=sup\left\{\frac{{\parallel {B}_{m}\left({M}_{-{\xi }_{0}}f,{M}_{-{\eta }_{0}}g\right)\parallel }_{{p}_{3},{\omega }_{3}}}{{\parallel {M}_{-{\xi }_{0}}f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel {M}_{-{\eta }_{0}}g\parallel }_{{p}_{2},{\omega }_{2}}}:{\parallel {M}_{-{\xi }_{0}}f\parallel }_{{p}_{1},{\omega }_{1}}\le 1,{\parallel {M}_{-{\eta }_{0}}g\parallel }_{{p}_{2},{\omega }_{2}}\le 1\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {B}_{m}\parallel ={\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.\hfill \end{array}$
1. (b)

Let us rewrite the value ${B}_{m}\left(f,g\right)$ as follows:

$\begin{array}{rcl}{B}_{{M}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\left(f,g\right)\left(x\right)& =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right){M}_{\left({\xi }_{0},{\eta }_{0}\right)}m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right){e}^{2\pi i〈\left({\xi }_{0},{\eta }_{0}\right),\left(\xi ,\eta \right)〉}m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{M}_{{\xi }_{0}}\stackrel{ˆ}{f}\left(\xi \right){M}_{{\eta }_{0}}\stackrel{ˆ}{g}\left(\eta \right)m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\left({T}_{-{\xi }_{0}}f\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(\xi \right)\left({T}_{-{\eta }_{0}}g\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(\eta \right)m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {B}_{m}\left({T}_{-{\xi }_{0}}f,{T}_{-{\eta }_{0}}g\right)\left(x\right).\end{array}$
(2.21)

Also, the inequalities ${\parallel {T}_{-{\xi }_{0}}f\parallel }_{{p}_{1},{\omega }_{1}}\le {\omega }_{1}\left(-{\xi }_{0}\right){\parallel f\parallel }_{{p}_{1},{\omega }_{1}}$ and ${\parallel {T}_{-{\eta }_{0}}g\parallel }_{{p}_{2},{\omega }_{2}}\le {\omega }_{2}\left(-{\eta }_{0}\right){\parallel g\parallel }_{{p}_{2},{\omega }_{2}}$ are satisfied for all $f\in {L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$, $g\in {L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. Hence, since $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, by (2.21) we have

$\begin{array}{rcl}{\parallel {B}_{{M}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& =& {\parallel {B}_{m}\left({T}_{-{\xi }_{0}}f,{T}_{-{\eta }_{0}}g\right)\parallel }_{{p}_{3},{\omega }_{3}}\le \parallel {B}_{m}\parallel {\parallel {T}_{-{\xi }_{0}}f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel {T}_{-{\eta }_{0}}g\parallel }_{{p}_{2},{\omega }_{2}}\\ \le & {\omega }_{1}\left(-{\xi }_{0}\right){\omega }_{2}\left(-{\eta }_{0}\right)\parallel {B}_{m}\parallel {\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}.\end{array}$
(2.22)

Then ${M}_{\left({\xi }_{0},{\eta }_{0}\right)}m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, and by (2.22) we obtain

$\begin{array}{rcl}{\parallel {M}_{\left({\xi }_{0},{\eta }_{0}\right)}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}& =& sup\left\{\frac{{\parallel {B}_{{M}_{\left({\xi }_{0},{\eta }_{0}\right)}m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}}{{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}}:{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}\le 1,{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\le 1\right\}\\ \le & {\omega }_{1}\left(-{\xi }_{0}\right){\omega }_{2}\left(-{\eta }_{0}\right){\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.\end{array}$

□

Lemma 2.2 If ${\upsilon }_{s}$ is a polynomial-type weight function and $f\in {L}_{{\upsilon }_{s}}^{p}\left({\mathbb{R}}^{n}\right)$, then ${D}_{t}^{p}f\in {L}_{{\upsilon }_{s}}^{p}\left({\mathbb{R}}^{n}\right)$. Moreover,

$\begin{array}{c}{\parallel {D}_{t}^{p}f\parallel }_{p,{\upsilon }_{s}}\le {\parallel f\parallel }_{p,{\upsilon }_{s}}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}t\le 1,\hfill \\ {\parallel {D}_{t}^{p}f\parallel }_{p,{\upsilon }_{s}}<{t}^{s}{\parallel f\parallel }_{p,{\upsilon }_{s}}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}t>1.\hfill \end{array}$

Proof Let ${\upsilon }_{s}$ be a polynomial-type weight function and $f\in {L}_{{\upsilon }_{s}}^{p}\left({\mathbb{R}}^{n}\right)$. Assume that $t\le 1$. If we get $\frac{x}{t}=u$,

$\begin{array}{rcl}{\parallel {D}_{t}^{p}f\parallel }_{p,{\upsilon }_{s}}& =& {\left\{{\int }_{{\mathbb{R}}^{n}}|{D}_{t}^{p}f\left(x\right){|}^{p}{\upsilon }_{s}{\left(x\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}\\ =& {\left\{{\int }_{{\mathbb{R}}^{n}}|{t}^{-\frac{n}{p}}f\left(\frac{x}{t}\right){|}^{p}{\left(1+|x|\right)}^{sp}\phantom{\rule{0.2em}{0ex}}dx\right\}}^{\frac{1}{p}}={\left\{{\int }_{{\mathbb{R}}^{n}}|f\left(u\right){|}^{p}{\left(1+|ut|\right)}^{sp}\phantom{\rule{0.2em}{0ex}}du\right\}}^{\frac{1}{p}}\\ \le & {\left\{{\int }_{{\mathbb{R}}^{n}}|f\left(u\right){|}^{p}{\left(1+|u|\right)}^{sp}\phantom{\rule{0.2em}{0ex}}du\right\}}^{\frac{1}{p}}\\ =& {\parallel f\parallel }_{p,{\upsilon }_{s}}<\mathrm{\infty }.\end{array}$
(2.23)

Thus we have ${D}_{t}^{p}f\in {L}_{{\upsilon }_{s}}^{p}\left({\mathbb{R}}^{n}\right)$ and ${\parallel {D}_{t}^{p}f\parallel }_{p,{\upsilon }_{s}}\le {\parallel f\parallel }_{p,{\upsilon }_{s}}$.

Now, assume that $t>1$. Similarly by (2.23)

$\begin{array}{rcl}{\parallel {D}_{t}^{p}f\parallel }_{p,{\upsilon }_{s}}& =& {\left\{{\int }_{{\mathbb{R}}^{n}}|f\left(u\right){|}^{p}{\left(1+|ut|\right)}^{sp}\phantom{\rule{0.2em}{0ex}}du\right\}}^{\frac{1}{p}}\\ <& {\left\{{\int }_{{\mathbb{R}}^{n}}|f\left(u\right){|}^{p}{\left(t+|ut|\right)}^{sp}\phantom{\rule{0.2em}{0ex}}du\right\}}^{\frac{1}{p}}={t}^{s}{\left\{{\int }_{{\mathbb{R}}^{n}}|f\left(u\right){|}^{p}{\left(1+|u|\right)}^{sp}\phantom{\rule{0.2em}{0ex}}du\right\}}^{\frac{1}{p}}\\ =& {t}^{s}{\parallel f\parallel }_{p,{\upsilon }_{s}}<\mathrm{\infty }.\end{array}$

Hence ${D}_{t}^{p}f\in {L}_{{\upsilon }_{s}}^{p}\left({\mathbb{R}}^{n}\right)$, and we also have ${\parallel {D}_{t}^{p}f\parallel }_{p,{\upsilon }_{s}}<{t}^{s}{\parallel f\parallel }_{p,{\upsilon }_{s}}$. □

Theorem 2.5 Let ${\upsilon }_{{s}_{1}}$, ${\upsilon }_{{s}_{2}}$, ${\upsilon }_{{s}_{3}}$ be weight functions of polynomial type and let $m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$. If $\frac{2}{q}=\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}-\frac{1}{{p}_{3}}$ and $0, then ${D}_{t}^{q}m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$. Moreover, then

$\begin{array}{c}{\parallel {D}_{t}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}\le {\left(\frac{1}{t}\right)}^{{s}_{3}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}t\le 1,\hfill \\ {\parallel {D}_{t}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}<{t}^{{s}_{1}+{s}_{2}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}t>1.\hfill \end{array}$

Proof Let $f\in {L}_{{\upsilon }_{{s}_{1}}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and $g\in {L}_{{\upsilon }_{{s}_{2}}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$ be given. We know by Lemma 2.2 that ${D}_{t}^{{p}_{1}}f\in {L}_{{\upsilon }_{{s}_{1}}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and ${D}_{t}^{{p}_{2}}g\in {L}_{{\upsilon }_{{s}_{2}}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. If we get $\frac{\xi }{t}=u$ and $\frac{\eta }{t}=v$, we obtain

$\begin{array}{rcl}{B}_{{D}_{t}^{q}m}\left(f,g\right)\left(x\right)& =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right){D}_{t}^{q}m\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(tu\right)\stackrel{ˆ}{g}\left(tv\right){t}^{-\frac{2n}{q}}m\left(u,v\right){e}^{2\pi i〈u+v,tx〉}{t}^{2n}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv.\end{array}$

Hence, from the equality $\frac{2}{q}=\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}-\frac{1}{{p}_{3}}$, we have

$\begin{array}{rcl}{B}_{{D}_{t}^{q}m}\left(f,g\right)\left(x\right)& =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(tu\right)\stackrel{ˆ}{g}\left(tv\right){t}^{-n\left(\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}-\frac{1}{{p}_{3}}\right)}m\left(u,v\right){e}^{2\pi i〈u+v,tx〉}{t}^{2n}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{t}^{-n\left(1-\frac{1}{{p}_{1}^{\mathrm{\prime }}}\right)}\stackrel{ˆ}{f}\left(tu\right){t}^{-n\left(1-\frac{1}{{p}_{1}^{\mathrm{\prime }}}\right)}\stackrel{ˆ}{g}\left(tv\right){t}^{\frac{n}{{p}_{3}}}m\left(u,v\right){e}^{2\pi i〈u+v,tx〉}{t}^{2n}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& {t}^{\frac{n}{{p}_{3}}}{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{D}_{{t}^{-1}}^{{p}_{1}^{\mathrm{\prime }}}\stackrel{ˆ}{f}\left(u\right){D}_{{t}^{-1}}^{{p}_{2}^{\mathrm{\prime }}}\stackrel{ˆ}{g}\left(v\right)m\left(u,v\right){e}^{2\pi i〈u+v,tx〉}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& {t}^{\frac{n}{{p}_{3}}}{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\left({D}_{t}^{{p}_{1}}f\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(u\right)\left({D}_{t}^{{p}_{2}}g\right)\phantom{\rule{0.25em}{0ex}}ˆ\phantom{\rule{0.25em}{0ex}}\left(v\right)m\left(u,v\right){e}^{2\pi i〈u+v,tx〉}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& {D}_{{t}^{-1}}^{{p}_{3}}{B}_{m}\left({D}_{t}^{{p}_{1}}f,{D}_{t}^{{p}_{2}}g\right)\left(x\right).\end{array}$
(2.24)

Assume that $t\le 1$. Since $m\in {B}_{m}\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$, by Lemma 2.2 and using equality (2.24), we obtain

$\begin{array}{rcl}{\parallel {B}_{{D}_{t}^{q}m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}& =& {\parallel {D}_{{t}^{-1}}^{{p}_{3}}{B}_{m}\left({D}_{t}^{{p}_{1}}f,{D}_{t}^{{p}_{2}}g\right)\left(x\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}\\ \le & {\left(\frac{1}{t}\right)}^{{s}_{3}}{\parallel {B}_{m}\left({D}_{t}^{{p}_{1}}f,{D}_{t}^{{p}_{2}}g\right)\left(x\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}\\ \le & {\left(\frac{1}{t}\right)}^{{s}_{3}}\parallel {B}_{m}\parallel {\parallel {D}_{t}^{{p}_{1}}f\parallel }_{p,{\upsilon }_{{s}_{1}}}{\parallel {D}_{t}^{{p}_{2}}g\parallel }_{p,{\upsilon }_{{s}_{2}}}\\ \le & {\left(\frac{1}{t}\right)}^{{s}_{3}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}.\end{array}$
(2.25)

Then ${D}_{t}^{q}m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$, and by (2.25)

${\parallel {D}_{t}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}\le {\left(\frac{1}{t}\right)}^{{s}_{3}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}.$

Now let $t>1$. Again, since $m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$, by Lemma 2.2 and using equality (2.24), we obtain

$\begin{array}{rcl}{\parallel {B}_{{D}_{t}^{q}m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}& <& {\parallel {B}_{m}\left({D}_{t}^{{p}_{1}}f,{D}_{t}^{{p}_{2}}g\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}\\ \le & \parallel {B}_{m}\parallel {\parallel {D}_{t}^{{p}_{1}}f\parallel }_{p,{\upsilon }_{{s}_{1}}}{\parallel {D}_{t}^{{p}_{2}}g\parallel }_{p,{\upsilon }_{{s}_{2}}}\\ <& {t}^{{s}_{1}+{s}_{2}}\parallel {B}_{m}\parallel {\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}\\ =& {t}^{{s}_{1}+{s}_{2}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}.\end{array}$
(2.26)

Thus ${D}_{t}^{q}m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$ and by (2.26)

${\parallel {D}_{t}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}<{t}^{{s}_{1}+{s}_{2}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}.$

□

Theorem 2.6 Let ${\upsilon }_{{s}_{1}}$, ${\upsilon }_{{s}_{2}}$, ${\upsilon }_{{s}_{3}}$ be weight functions of polynomial type and let $m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$ such that $m\left(t\xi ,t\eta \right)=m\left(\xi ,\eta \right)$ for any $t>0$, where $\frac{2}{q}=\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}-\frac{1}{{p}_{3}}$. Then

$\begin{array}{c}\frac{2}{q}<\frac{{s}_{3}}{n}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}t<1,\hfill \\ \frac{2}{q}>-\frac{{s}_{1}+{s}_{2}}{n}\phantom{\rule{1em}{0ex}}\mathit{\text{if}}t>1.\hfill \end{array}$

Proof Take any $f\in {L}_{{\upsilon }_{{s}_{1}}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$, $g\in {L}_{{\upsilon }_{{s}_{2}}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. It is known by Theorem 2.5 that

${B}_{{D}_{t}^{q}m}\left(f,g\right)\left(x\right)={D}_{{t}^{-1}}^{{p}_{3}}{B}_{m}\left({D}_{t}^{{p}_{1}}f,{D}_{t}^{{p}_{2}}g\right)\left(x\right),\phantom{\rule{1em}{0ex}}x\in {\mathbb{R}}^{n}.$
(2.27)

On the other hand, using $m\left(t\xi ,t\eta \right)=m\left(\xi ,\eta \right)$ and changing the variables $tu=\xi$, $tv=\eta$, we note that

(2.28)

Hence by (2.27) and (2.28), we have

${B}_{m}\left(f,g\right)\left(x\right)={t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)}{B}_{{D}_{t}^{q}m}\left(f,g\right)\left(x\right).$

Since ${D}_{t}^{q}m=m$ for $t=1$, we let $t\ne 1$. Assume first that $t<1$. Also, since $m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$, by Theorem 2.5 we have ${D}_{t}^{q}m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$ and ${\parallel {D}_{t}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}<{\left(\frac{1}{t}\right)}^{{s}_{3}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}$. Then by (2.28)

$\begin{array}{rcl}{\parallel {B}_{m}\left(f,g\right)\left(x\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}& =& {t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)}{\parallel {B}_{{D}_{t}^{q}}\left(f,g\right)\left(x\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}\\ \le & {t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)}{\parallel {D}_{t}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}\\ <& {t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)-{s}_{3}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}\\ =& {t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)-{s}_{3}}\parallel {B}_{m}\parallel {\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}.\end{array}$

Thus,

$\parallel {B}_{m}\parallel <{t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)-{s}_{3}}\parallel {B}_{m}\parallel ={t}^{\frac{2n}{q}-{s}_{3}}\parallel {B}_{m}\parallel .$

Hence $1<{t}^{\frac{2n}{q}-{s}_{3}}$. Since $t<1$, we have $\frac{2n}{q}-{s}_{3}<0$. Thus, we write $\frac{2}{q}<\frac{{s}_{3}}{n}$.

Assume now that $t>1$. Again, by Theorem 2.5, we have ${D}_{t}^{q}m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$ and ${\parallel {D}_{t}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}<{t}^{{s}_{1}+{s}_{2}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}$. Similarly,

${\parallel {B}_{m}\left(f,g\right)\left(x\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}<{t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)+{s}_{1}+{s}_{2}}\parallel {B}_{m}\parallel {\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}.$

Thus, we have

$\parallel {B}_{m}\parallel <{t}^{-n\left(\frac{1}{{p}_{3}}-\frac{1}{{p}_{1}}-\frac{1}{{p}_{2}}\right)+{s}_{1}+{s}_{2}}\parallel {B}_{m}\parallel ={t}^{\frac{2n}{q}+{s}_{1}+{s}_{2}}\parallel {B}_{m}\parallel .$

Hence $1<{t}^{\frac{2n}{q}+{s}_{1}+{s}_{2}}$ Since $t>1$, we have $\frac{2n}{q}+{s}_{1}+{s}_{2}>0$. Thus, we write $\frac{2}{q}>-\frac{{s}_{1}+{s}_{2}}{n}$. □

Theorem 2.7 Let $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ and ${p}_{3}\ge 1$.

1. (a)

If $\mathrm{\Phi }\in {L}^{1}\left({\mathbb{R}}^{n}\right)$, then $\mathrm{\Phi }\ast m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ and

${\parallel \mathrm{\Phi }\ast m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\le {\parallel \mathrm{\Phi }\parallel }_{1}{\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.$
2. (b)

If $\mathrm{\Phi }\in {L}_{\omega }^{1}\left({\mathbb{R}}^{n}\right)$ such that $\omega \left(u,\upsilon \right)={\omega }_{1}\left(u\right){\omega }_{2}\left(\upsilon \right)$, then $\stackrel{ˆ}{\mathrm{\Phi }}m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ and

${\parallel \stackrel{ˆ}{\mathrm{\Phi }}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\le {\parallel \mathrm{\Phi }\parallel }_{1,\omega }{\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.$

Proof (a) Let $f\in {L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and $g\in {L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. Since ${L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)\subset {L}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and ${L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)\subset {L}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$, then by Proposition 2.5 in [11]

${B}_{\mathrm{\Phi }\ast m}\left(f,g\right)\left(x\right)={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\mathrm{\Phi }\left(u,v\right){B}_{{T}_{\left(u,v\right)}m}\left(f,g\right)\left(x\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv.$

Also, since $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, we have ${T}_{\left(u,v\right)}m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ by Theorem 2.4. So, we write

$\begin{array}{rl}{\parallel {B}_{\mathrm{\Phi }\ast m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& \le {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{\parallel \mathrm{\Phi }\left(u,v\right){B}_{{T}_{\left(u,v\right)}m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ \le {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}|\mathrm{\Phi }\left(u,v\right)|{\parallel {T}_{\left(u,v\right)}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ ={\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel \mathrm{\Phi }\parallel }_{1}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}<\mathrm{\infty }.\end{array}$
(2.29)

Hence $\mathrm{\Phi }\ast m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. Finally, by (2.29), we obtain

${\parallel \mathrm{\Phi }\ast m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\le {\parallel \mathrm{\Phi }\parallel }_{1}{\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.$
1. (b)

Let $\mathrm{\Phi }\in {L}_{\omega }^{1}\left({\mathbb{R}}^{n}\right)$. Take any $f\in {L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and $g\in {L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. It is known by Proposition 2.5 in [11] that the equality

${B}_{\stackrel{ˆ}{\mathrm{\Phi }}m}\left(f,g\right)\left(x\right)={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\mathrm{\Phi }\left(u,v\right){B}_{{M}_{\left(-u,-v\right)}m}\left(f,g\right)\left(x\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv.$

Since $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$, by Theorem 2.4 we have ${M}_{\left(-u,-v\right)}m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ and

${\parallel {M}_{\left(-u,-v\right)}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\le {\omega }_{1}\left(u\right){\omega }_{2}\left(\upsilon \right){\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.$

Then we write

$\begin{array}{rcl}{\parallel {B}_{\stackrel{ˆ}{\mathrm{\Phi }}m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& \le & {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}{\parallel \mathrm{\Phi }\left(u,v\right){B}_{{M}_{\left(-u,-v\right)}m}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ \le & {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}|\mathrm{\Phi }\left(u,v\right)|{\parallel {M}_{\left(-u,-v\right)}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ \le & {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}|\mathrm{\Phi }\left(u,v\right)|{\omega }_{1}\left(u\right){\omega }_{2}\left(\upsilon \right){\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& {\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}{\parallel \mathrm{\Phi }\parallel }_{1,\omega }.\end{array}$
(2.30)

Thus from (2.30), we obtain $\stackrel{ˆ}{\mathrm{\Phi }}m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$ and

${\parallel \stackrel{ˆ}{\mathrm{\Phi }}m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}\le {\parallel \mathrm{\Phi }\parallel }_{1,\omega }{\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}.$

□

Theorem 2.8 Let ${\upsilon }_{{s}_{1}}$, ${\upsilon }_{{s}_{2}}$, ${\upsilon }_{{s}_{3}}$ be weight functions of polynomial type and let $m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$. If $\mathrm{\Psi }\in {L}^{1}\left({\mathbb{R}}^{+},{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\right)$ such that $\frac{2}{q}=\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}-\frac{1}{{p}_{3}}$, then ${m}_{\mathrm{\Psi }}\left(\xi ,\eta \right)={\int }_{0}^{\mathrm{\infty }}m\left(t\xi ,t\eta \right)\mathrm{\Psi }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$. Moreover,

${\parallel {m}_{\mathrm{\Psi }}\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}<{\parallel \mathrm{\Psi }\parallel }_{{L}^{1}\left({\mathbb{R}}^{+},{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\right)}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}.$

Proof Let us take $f,g\in S\left({\mathbb{R}}^{n}\right)$. Then

$\begin{array}{rcl}{B}_{{m}_{\mathrm{\Psi }}}\left(f,g\right)\left(x\right)& =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right){m}_{\mathrm{\Psi }}\left(\xi ,\eta \right){e}^{2\pi i〈u+v,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\left\{{\int }_{0}^{\mathrm{\infty }}m\left(t\xi ,t\eta \right)\mathrm{\Psi }\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right\}{e}^{2\pi i〈u+v,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\left\{{\int }_{0}^{\mathrm{\infty }}{D}_{{t}^{-1}}^{q}m\left(\xi ,\eta \right)\mathrm{\Psi }\left(t\right){t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\right\}{e}^{2\pi i〈u+v,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \\ =& {\int }_{0}^{\mathrm{\infty }}{B}_{{D}_{{t}^{-1}}^{q}m}\left(f,g\right)\mathrm{\Psi }\left(t\right){t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt.\end{array}$

Since $m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$, ${D}_{{t}^{-1}}^{q}m\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$ by Theorem 2.5, thus we observe that

$\begin{array}{rcl}{\parallel {B}_{{m}_{\mathrm{\Psi }}}\left(f,g\right)\left(x\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}& \le & {\int }_{0}^{\mathrm{\infty }}{\parallel {B}_{{D}_{{t}^{-1}}^{q}m}\left(f,g\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}|\mathrm{\Psi }\left(t\right)|{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\\ \le & {\int }_{0}^{\mathrm{\infty }}\parallel {B}_{{D}_{{t}^{-1}}^{q}m}\parallel {\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}|\mathrm{\Psi }\left(t\right)|{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\\ =& {\int }_{0}^{\mathrm{\infty }}{\parallel {D}_{{t}^{-1}}^{q}m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}|\mathrm{\Psi }\left(t\right)|{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\\ <& {\int }_{0}^{1}{t}^{{s}_{3}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}|\mathrm{\Psi }\left(t\right)|{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\\ +{\int }_{1}^{\mathrm{\infty }}{t}^{-{s}_{1}-{s}_{2}}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}|\mathrm{\Psi }\left(t\right)|{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\\ =& {\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}\\ ×\left\{{\int }_{0}^{1}{t}^{{s}_{3}}|\mathrm{\Psi }\left(t\right)|{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt+{\int }_{1}^{\mathrm{\infty }}{t}^{-{s}_{1}-{s}_{2}}|\mathrm{\Psi }\left(t\right)|{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\right\}.\end{array}$
(2.31)

Also, since ${t}^{{s}_{3}}\le 1$ for ${s}_{3}\ge 0$, $t\le 1$ and ${t}^{-{s}_{1}-{s}_{2}}<1$ for $-{s}_{1}-{s}_{2}\le 0$, $t>1$, by (2.31)

${\parallel {B}_{{m}_{\mathrm{\Psi }}}\left(f,g\right)\left(x\right)\parallel }_{{p}_{3},{\upsilon }_{{s}_{3}}}<{\parallel \mathrm{\Psi }\parallel }_{{L}^{1}\left({\mathbb{R}}^{+},{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\right)}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}{\parallel f\parallel }_{{p}_{1},{\upsilon }_{{s}_{1}}}{\parallel g\parallel }_{{p}_{2},{\upsilon }_{{s}_{2}}}.$

Hence, ${m}_{\mathrm{\Psi }}\in BM\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)$ and

${\parallel {m}_{\mathrm{\Psi }}\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}<{\parallel \mathrm{\Psi }\parallel }_{{L}^{1}\left({\mathbb{R}}^{+},{t}^{-\frac{2n}{q}}\phantom{\rule{0.2em}{0ex}}dt\right)}{\parallel m\parallel }_{\left({\upsilon }_{{s}_{1}},{\upsilon }_{{s}_{2}},{\upsilon }_{{s}_{3}}\right)}.$

□

Theorem 2.9 Let ${p}_{3}\ge 1$ and $m\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. If ${Q}_{1}$, ${Q}_{2}$ are bounded measurable sets in ${\mathbb{R}}^{n}$, then

$h\left(\xi ,\eta \right)=\frac{1}{\mu \left({Q}_{1}×{Q}_{2}\right)}\int {\int }_{{Q}_{1}×{Q}_{2}}m\left(\xi +u,\eta +v\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right).$

Proof Take any $f,g\in S\left({\mathbb{R}}^{n}\right)$. Then we write

$\begin{array}{c}{B}_{h}\left(f,g\right)\left(x\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)h\left(\xi ,\eta \right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mu \left({Q}_{1}×{Q}_{2}\right)}\int {\int }_{{Q}_{1}×{Q}_{2}}\left\{{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)m\left(\xi +u,\eta +v\right){e}^{2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \right\}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\mu \left({Q}_{1}×{Q}_{2}\right)}\int {\int }_{{Q}_{1}×{Q}_{2}}{B}_{{T}_{\left(-u,-v\right)m}}\left(f,g\right)\left(x\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv.\hfill \end{array}$

By using Theorem 2.4, we have

$\begin{array}{rcl}{\parallel {B}_{h}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}& \le & \frac{1}{\mu \left({Q}_{1}×{Q}_{2}\right)}\int {\int }_{{Q}_{1}×{Q}_{2}}{\parallel {B}_{{T}_{\left(-u,-v\right)m}}\left(f,g\right)\parallel }_{{p}_{3},{\omega }_{3}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ \le & \frac{1}{\mu \left({Q}_{1}×{Q}_{2}\right)}\int {\int }_{{Q}_{1}×{Q}_{2}}{\parallel {T}_{\left(-u,-v\right)m}\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& \frac{1}{\mu \left({Q}_{1}×{Q}_{2}\right)}{\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}\mu \left({Q}_{1}×{Q}_{2}\right)\\ =& {\parallel m\parallel }_{\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)}{\parallel f\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{{p}_{2},{\omega }_{2}}.\end{array}$

Hence, we obtain $h\left(\xi ,\eta \right)\in BM\left({\omega }_{1},{\omega }_{2},{\omega }_{3}\right)$. □

Theorem 2.10 Let $\omega \left(u,v\right)={\omega }_{1}\left(u\right){\omega }_{2}\left(v\right)$, ${\omega }_{3}\le {\omega }_{1}$, ${\omega }_{3}\left(-u\right)={\omega }_{3}\left(u\right)$ and $\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}=\frac{1}{{p}_{3}}\le 1$. Assume that $\mathrm{\Phi }\in {L}_{\omega }^{1}\left({\mathbb{R}}^{2n}\right)$, ${\mathrm{\Psi }}_{1}\in {L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and ${\mathrm{\Psi }}_{2}\in {L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. If $m\left(\xi ,\eta \right)={\stackrel{ˆ}{\mathrm{\Psi }}}_{1}\left(\xi \right)\stackrel{ˆ}{\mathrm{\Phi }}\left(\xi ,\eta \right){\stackrel{ˆ}{\mathrm{\Psi }}}_{2}\left(\eta \right)$, then $m\in BM\left(1,{\omega }_{1};1,{\omega }_{2};{p}_{3},{\omega }_{3}\right)$.

Proof For the proof we will use Theorem 2.2. Take any $f,g,h\in S\left({\mathbb{R}}^{n}\right)$. Then

$\begin{array}{r}|{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{h}\left(\xi +\eta \right)m\left(\xi ,\eta \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta |\\ \phantom{\rule{1em}{0ex}}=|{\int }_{{\mathbb{R}}^{n}}h\left(y\right)\left\{{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right){\stackrel{ˆ}{\mathrm{\Psi }}}_{1}\left(\xi \right)\stackrel{ˆ}{\mathrm{\Phi }}\left(\xi ,\eta \right){\stackrel{ˆ}{\mathrm{\Psi }}}_{2}\left(\eta \right){e}^{-2\pi i〈\xi +\eta ,x〉}\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta \right\}\phantom{\rule{0.2em}{0ex}}dy|\\ \phantom{\rule{1em}{0ex}}\le {\int }_{{\mathbb{R}}^{n}}|h\left(y\right){\omega }_{3}^{-1}\left(y\right){B}_{\stackrel{ˆ}{\mathrm{\Phi }}}\left(f\ast {\mathrm{\Psi }}_{1},g\ast {\mathrm{\Psi }}_{2}\right)\left(-y\right){\omega }_{3}\left(y\right)|\phantom{\rule{0.2em}{0ex}}dy.\end{array}$
(2.32)

Since the spaces ${L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and ${L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$ are Banach convolution module over the spaces ${L}_{{\omega }_{1}}^{1}\left({\mathbb{R}}^{n}\right)$, ${L}_{{\omega }_{2}}^{1}\left({\mathbb{R}}^{n}\right)$ respectively, we write $f\ast {\mathrm{\Psi }}_{1}\in {L}_{{\omega }_{1}}^{{p}_{1}}\left({\mathbb{R}}^{n}\right)$ and $g\ast {\mathrm{\Psi }}_{2}\in {L}_{{\omega }_{2}}^{{p}_{2}}\left({\mathbb{R}}^{n}\right)$. Also, by Theorem 2.7, $\stackrel{ˆ}{\mathrm{\Phi }}\in BM\left({p}_{1},{\omega }_{1};{p}_{2},{\omega }_{2};{p}_{3},{\omega }_{3}\right)$. Therefore we obtain ${B}_{\stackrel{ˆ}{\mathrm{\Phi }}}\left(f\ast {\mathrm{\Psi }}_{1},g\ast {\mathrm{\Psi }}_{2}\right)\in {L}_{{\omega }_{3}}^{{p}_{3}}\left({\mathbb{R}}^{n}\right)$. By using the Hölder inequality and the inequality (2.32), we find

$\begin{array}{c}|{\int }_{{\mathbb{R}}^{n}}{\int }_{{\mathbb{R}}^{n}}\stackrel{ˆ}{f}\left(\xi \right)\stackrel{ˆ}{g}\left(\eta \right)\stackrel{ˆ}{h}\left(\xi +\eta \right)m\left(\xi ,\eta \right)\phantom{\rule{0.2em}{0ex}}d\xi \phantom{\rule{0.2em}{0ex}}d\eta |\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel h\parallel }_{{p}_{3}^{-1},{\omega }_{3}^{-1}}{\parallel {B}_{\stackrel{ˆ}{\mathrm{\Phi }}}\left(f\ast {\mathrm{\Psi }}_{1},g\ast {\mathrm{\Psi }}_{2}\right)\parallel }_{{p}_{3},{\omega }_{3}}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel h\parallel }_{{p}_{3}^{-1},{\omega }_{3}^{-1}}\parallel {B}_{\stackrel{ˆ}{\mathrm{\Phi }}}\parallel {\parallel f\parallel }_{1,{\omega }_{1}}{\parallel {\mathrm{\Psi }}_{1}\parallel }_{{p}_{1},{\omega }_{1}}{\parallel g\parallel }_{1,{\omega }_{2}}{\parallel {\mathrm{\Psi }}_{2}\parallel }_{}\hfill \end{array}$