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 $m(\xi ,\eta )$ defined on ${\mathbb{R}}^n\times {\mathbb{R}}^n$ is a bilinear multiplier on ${\mathbb{R}}^n$ of type $(p_1,{\omega }_1;p_2,{\omega }_2;p_3,{\omega }_3)$ (shortly $({\omega }_1,{\omega }_2,{\omega }_3)$) if

$$B_m(f,g)(x)={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta$$

is a bounded bilinear operator from $L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)\times L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$ to $L_{{\omega }_3}^{p_3}({\mathbb{R}}^n)$. We denote by $BM(p_1,{\omega }_1;p_2,{\omega }_2;p_3,{\omega }_3)$ (shortly $BM({\omega }_1,{\omega }_2,{\omega }_3)$) the vector space of bilinear multipliers of type $({\omega }_1,{\omega }_2,{\omega }_3)$.

In this paper first we discuss some properties of the space $BM({\omega }_1,{\omega }_2,{\omega }_3)$. Furthermore, we give some examples of bilinear multipliers.

At the end of this paper, by using variable exponent Lebesgue spaces $L^{p_1(x)}({\mathbb{R}}^n)$, $L^{p_2(x)}({\mathbb{R}}^n)$ and $L^{p_3(x)}({\mathbb{R}}^n)$, we define the space of bilinear multipliers from $L^{p_1(x)}({\mathbb{R}}^n)\times L^{p_2(x)}({\mathbb{R}}^n)$ to $L^{p_3(x)}({\mathbb{R}}^n)$ and discuss some properties of this space.

MSC:42A45, 42B15, 42B35.

1 Introduction

Throughout this paper $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$, $C_c({\mathbb{R}}^n)$ and $S({\mathbb{R}}^n)$ 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 $1\le p\le \mathrm{\infty }$, $L^p({\mathbb{R}}^n)$ denotes the usual Lebesgue space. A continuous function ω satisfying $1\le \omega (x)$ and $\omega (x+y)\le \omega (x)\omega (y)$ for $x,y\in {\mathbb{R}}^n$ will be called a weight function on ${\mathbb{R}}^n$. If ${\omega }_1(x)\le {\omega }_2(x)$ 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 ${\upsilon }_s(x)={(1+|x|)}^s$ 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

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

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

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

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

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

or

$${\parallel f\parallel }_{\mathrm{\infty },\omega }={\parallel f\omega \parallel }_{\mathrm{\infty }}=\underset{x\in {\mathbb{R}}^n}{ess\hspace{0.17em}sup}|f(x)\omega (x)|,p=\mathrm{\infty },\text{[1, 2]}.$$

The dual of the space $L_{\omega }^p({\mathbb{R}}^n)$ is the space $L_{{\omega }^{-1}}^q({\mathbb{R}}^n)$, where $\frac{1}{p}+\frac{1}{q}=1$ and ${\omega }^{-1}(x)=\frac{1}{\omega (x)}$. For $f\in L^1({\mathbb{R}}^n)$, the Fourier transform of f is denoted by $\hat{f}$. We know that $\hat{f}$ is a continuous function on ${\mathbb{R}}^n$, which vanishes at infinity, and it has the inequality ${\parallel \hat{f}\parallel }_{\mathrm{\infty }}\le {\parallel f\parallel }_1$ [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 $T_{x}f(y)=f(y-x)$, $M_{x}f(y)=e^{2\pi i〈x,y〉}f(y)$ and $D_t^{p}f(y)=t^{-\frac{n}{p}}f(\frac{y}{t})$ respectively for $x,y\in {\mathbb{R}}^n$, $0<p,t<\mathrm{\infty }$. With this notation out of the way one has, for $1\le p\le \mathrm{\infty }$ and $\frac{1}{p}+\frac{1}{p^{\mathrm{\prime }}}=1$,

$$(T_{x}f)ˆ(\xi )=M_{-x}\hat{f}(\xi ),(M_{x}f)ˆ(\xi )=T_x\hat{f}(\xi ),\left(D_t^{p}f\right)ˆ(\xi )=D_{t^{-1}}^{p^{\mathrm{\prime }}}\hat{f}(\xi ).$$

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

$${\parallel \mu \parallel }_{\omega }={\int }_{{\mathbb{R}}^n}\omega d|\mu |<\mathrm{\infty }.$$

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

$$p_{\ast }=\underset{x\in {\mathbb{R}}^n}{ess\hspace{0.17em}inf}p(x),p^{\ast }=\underset{x\in {\mathbb{R}}^n}{ess\hspace{0.17em}\; sup}p(x).$$

We shall also use the notation

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

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

$${\varrho }_p(\lambda f)={\int }_{{\mathbb{R}}^n\mathrm{\setminus }{\mathrm{\Omega }}_{\mathrm{\infty }}}|\lambda f(x){|}^{p(x)}dx+\underset{x\in {\mathrm{\Omega }}_{\mathrm{\infty }}}{ess\hspace{0.17em}\; sup}(\lambda f(x))<\mathrm{\infty }$$

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

$${\varrho }_p(\lambda f)={\int }_{{\mathbb{R}}^n\mathrm{\setminus }{\mathrm{\Omega }}_{\mathrm{\infty }}}|\lambda f(x){|}^{p(x)}dx,\text{[6, 7]}.$$

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

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

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

2 The bilinear multipliers space $BM(p_1,{\omega }_1;p_2,{\omega }_2;p_3,{\omega }_3)$

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

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

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

(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 $S({\mathbb{R}}^n)\subset L_m^p({\mathbb{R}}^n)$. Hence, by (2.1), we obtain $S({\mathbb{R}}^n)\subset L_m^p({\mathbb{R}}^n)\subset L_{\omega }^p({\mathbb{R}}^n)$.

Now, we show that $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ is dense $L_m^p({\mathbb{R}}^n)$. Let $f\in L_w^p({\mathbb{R}}^n)$ be given. Then $fm\in L^p({\mathbb{R}}^n)$. Since $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ is dense $L^p({\mathbb{R}}^n)$ by [6], for given $\epsilon >0$, there exists $g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ 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 $g{m}^{-1}\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. Thus, we have $\overline{C_c^{\mathrm{\infty }}({\mathbb{R}}^n)}=L_m^p({\mathbb{R}}^n)$. By using the inclusion $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)\subset S({\mathbb{R}}^n)\subset L_m^p({\mathbb{R}}^n)$, we obtain $\overline{S({\mathbb{R}}^n)}=L_m^p({\mathbb{R}}^n)$.

Now, take any $f\in L_{\omega }^p({\mathbb{R}}^n)$. Since $\overline{C_c({\mathbb{R}}^n)}=\overline{L_m^p({\mathbb{R}}^n)}=L_{\omega }^p({\mathbb{R}}^n)$, there exists $g\in L_m^p({\mathbb{R}}^n)$ such that

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

(2.3)

Furthermore, since $S({\mathbb{R}}^n)$ is dense $L_m^p({\mathbb{R}}^n)$, there exists $h\in S({\mathbb{R}}^n)$ 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 $\overline{S({\mathbb{R}}^n)}=L_{\omega }^p({\mathbb{R}}^n)$. □

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 $m(\xi ,\eta )$ is a bounded function on ${\mathbb{R}}^n\times {\mathbb{R}}^n$. Define

$$B_m(f,g)(x)={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta$$

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

m is said to be a bilinear multiplier on ${\mathbb{R}}^n$ of type $(p_1,{\omega }_1;p_2,{\omega }_2;p_3,{\omega }_3)$ (shortly $({\omega }_1,{\omega }_2,{\omega }_3)$) if there exists $C>0$ such that

$${\parallel B_m(f,g)\parallel }_{p_3,{\omega }_3}\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}$$

for all $f,g\in S({\mathbb{R}}^n)$. That means $B_m$ extends to a bounded bilinear operator from $L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)\times L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$ to $L_{{\omega }_3}^{p_3}({\mathbb{R}}^n)$.

We denote by $BM(p_1,{\omega }_1;p_2,{\omega }_2;p_3,{\omega }_3)$ (shortly $BM({\omega }_1,{\omega }_2,{\omega }_3)$) the space of all bilinear multipliers of type $({\omega }_1,{\omega }_2,{\omega }_3)$ and ${\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}=\parallel B_m\parallel$.

Theorem 2.1 Let $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p_3}$ and ${\omega }_3\le {\omega }_1$. If $K\in L_{{\omega }_3}^1({\mathbb{R}}^n)$, then $m(\xi ,\eta )=\hat{K}(\xi -\eta )$ defines a bilinear multiplier and ${\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\le {\parallel K\parallel }_{1,{\omega }_3}$.

Proof For $f,g\in S({\mathbb{R}}^n)$, we have $f(x-y)={\int }_{{\mathbb{R}}^n}\hat{f}(\xi )e^{2\pi i〈x-y,\xi 〉}d\xi$ and $g(x+y)={\int }_{{\mathbb{R}}^n}\hat{g}(\eta )e^{2\pi i〈x+y,\eta 〉}d\eta$. Thus, by the Fubini theorem, we write

$$\begin{array}{rl}{B}_m(f,g)(x)& ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{K}(\xi -\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )({\int }_{{\mathbb{R}}^n}K(y)e^{-2\pi i〈\xi -\eta ,y〉}dy)e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )K(y)e^{-2\pi i〈\xi -\eta ,y〉}{e}^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta dy.\end{array}$$

(2.5)

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

$$\begin{array}{rl}{B}_m(f,g)(x)& ={\int }_{{\mathbb{R}}^n}({\int }_{{\mathbb{R}}^n}\hat{f}(\xi )e^{2\pi i〈x-y,\xi 〉})({\int }_{{\mathbb{R}}^n}\hat{g}(\eta )e^{2\pi i〈x+y,\eta 〉}d\eta )K(y)dy\\ ={\int }_{{\mathbb{R}}^n}f(x-y)g(x+y)K(y)dy.\end{array}$$

(2.6)

Since ${\omega }_3\le {\omega }_1$, then

$${\parallel f(x-y){\omega }_3\parallel }_{p_1}\le {\omega }_3(y){\parallel f\parallel }_{p_1,{\omega }_1}$$

(2.7)

and hence $f(x-y){\omega }_3\in L^{p_1}({\mathbb{R}}^n)$. Therefore from (2.6) and the Minkowski inequality, we write

$$\begin{array}{rl}{\parallel B_m(f,g)\parallel }_{p_3,{\omega }_3}& \le {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\parallel f(x-y)g(x+y)\parallel }_{p_3,{\omega }_3}|K(y)|dy\\ ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\parallel f(x-y)g(x+y){\omega }_3\parallel }_{p_3}|K(y)|dy.\end{array}$$

(2.8)

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

$$\begin{array}{rcl}{\parallel B_m(f,g)\parallel }_{p_3,{\omega }_3}& \le & {\int }_{{\mathbb{R}}^n}{\parallel f(x-y){\omega }_3\parallel }_{p_1}{\parallel g(x+y)\parallel }_{p_2}{\omega }_3(y)|K(y)|dy\\ \le & {\int }_{{\mathbb{R}}^n}{\parallel f\parallel }_{p_1}{\parallel g\parallel }_{p_2}{\omega }_3(y)|K(y)|dy\\ \le & {\int }_{{\mathbb{R}}^n}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\omega }_3(y)|K(y)|dy\\ =& {\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel K\parallel }_{1,{\omega }_3}.\end{array}$$

(2.9)

If we set $C={\parallel K\parallel }_{1,{\omega }_3}$, we obtain

$${\parallel B_m(f,g)\parallel }_{p_3,{\omega }_3}\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}.$$

Then $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$. Consequently, using (2.9), we have

$$\begin{array}{rcl}{\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}& =& sup\{\frac{{\parallel B_m(f,g)\parallel }_{p_3,{\omega }_3}}{{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}}:{\parallel f\parallel }_{p_1,{\omega }_1}\le 1,{\parallel g\parallel }_{p_2,{\omega }_2}\le 1\}\\ \le & {\parallel K\parallel }_{1,{\omega }_3}.\end{array}$$

 □

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 $\tilde{M}({\omega }_1,{\omega }_2,{\omega }_3)$ the space of measurable functions $M:{\mathbb{R}}^n\to \mathbb{C}$ such that $m(\xi ,\eta )=M(\xi -\eta )\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, that is to say,

$$B_M(f,g)(x)={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )M(\xi -\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta$$

extends to a bounded bilinear map from $L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)\times L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$ to $L_{{\omega }_3}^{p_3}({\mathbb{R}}^n)$. We denote ${\parallel M\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}=\parallel B_M\parallel$.

Theorem 2.2 Let $p_3\ge 1$ and ${\omega }_3(-x)={\omega }_3(x)$. Then $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ if and only if there exists $C>0$ such that

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}$$

for all $f,g,h\in S({\mathbb{R}}^n)$, where $\frac{1}{p_3}+\frac{1}{p_3^{\mathrm{\prime }}}=1$.

Proof Let $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$. We take any $f,g,h\in S({\mathbb{R}}^n)$. From the Fubini theorem, we write

(2.10)

where ${\tilde{B}}_m(f,g)(y)=B_m(f,g)(-y)$. On the other hand, since $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, then $B_m(f,g)\in L_{{\omega }_3}^{p_3}({\mathbb{R}}^n)$. Thus we obtain ${\tilde{B}}_m(f,g)\in L_{{\omega }_3}^{p_3}({\mathbb{R}}^n)$. Also, $h\in S({\mathbb{R}}^n)\subset L^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)\subset L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)$. Hence, using the Hölder inequality and the inequality (2.10), we write

$$\begin{array}{r}|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\\ \le {\int }_{{\mathbb{R}}^n}|h(y){\omega }_3^{-1}(y)||{\tilde{B}}_m(f,g)(y){\omega }_3(y)|dy\\ \le {\parallel {\tilde{B}}_m(f,g)\parallel }_{p_3,{\omega }_3}{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}.\end{array}$$

(2.11)

Moreover, since $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, there exists $C>0$ such that

$${\parallel B_m(f,g)\parallel }_{p_3,{\omega }_3}\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}.$$

(2.12)

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

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}.$$

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

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}$$

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

$$|{\int }_{{\mathbb{R}}^n}h(y){\tilde{B}}_m(f,g)(y)dy|\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}.$$

(2.13)

Define a function l from $S({\mathbb{R}}^n)\subset L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)$ to ℂ such that

$$\ell (h)={\int }_{{\mathbb{R}}^n}h(y){\tilde{B}}_m(f,g)(y)dy.$$

It is clear that the function ℓ is linear and bounded by (2.13). By using $\overline{C_c({\mathbb{R}}^n)}=L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)$ in [10], it is easy to show that $\overline{C_c^{\mathrm{\infty }}({\mathbb{R}}^n)}=L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)$. So, by the inclusion $C_c^{\mathrm{\infty }}({\mathbb{R}}^n)\subset S({\mathbb{R}}^n)\subset L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)$, we have $\overline{S({\mathbb{R}}^n)}=L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)$. Thus ℓ extends to a bounded function from $L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n)$ to ℂ. Then $\ell \in {(L_{{\omega }_3^{-1}}^{p_3^{\mathrm{\prime }}}({\mathbb{R}}^n))}^{\ast }=L_{{\omega }_3}^{p_3}({\mathbb{R}}^n)$ and by (2.13), we have

$$\begin{array}{rcl}{\parallel B_m(f,g)\parallel }_{p_3,{\omega }_3}& =& {\parallel {\tilde{B}}_m(f,g)\parallel }_{p_3,{\omega }_3}=\parallel \ell \parallel =\underset{{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}\le 1}{sup}\frac{|l(h)|}{{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}}\\ \le & \underset{{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}\le 1}{sup}\frac{C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}}{{\parallel h\parallel }_{p_3^{\mathrm{\prime }},{\omega }_3^{-1}}}\le C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}.\end{array}$$

Hence, we obtain $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$. □

Theorem 2.3 Let $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p_3}$, $p_3\ge 1$ and ${\upsilon }_s(x)={(1+|x|)}^s$, $s\ge 0$ be a weight function of polynomial type such that ${\upsilon }_s\le {\omega }_1$. If $\mu \in M({\upsilon }_s)$ and $m(\xi ,\eta )=\hat{\mu }(\alpha \xi +\beta \eta )$ for $\alpha ,\beta \in \mathbb{R}$, then $m\in BM({\omega }_1,{\omega }_2,{\upsilon }_s)$. Moreover,

$$\begin{array}{c}{\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\upsilon }_s)}\le {\parallel \mu \parallel }_{{\upsilon }_s}\mathit{\text{if}}|\alpha |\le 1,\hfill \\ {\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\upsilon }_s)}\le |\alpha {|}^s{\parallel \mu \parallel }_{{\upsilon }_s}\mathit{\text{if}}|\alpha |>1.\hfill \end{array}$$

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

$$\begin{array}{rcl}{B}_m(f,g)(x)& =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{\mu }(\alpha \xi +\beta \eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\{{\int }_{{\mathbb{R}}^n}{e}^{-2\pi i〈\alpha \xi +\beta \eta ,t〉}d\mu (t)\}{e}^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}\{{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )e^{2\pi i〈x-\alpha t,\xi 〉}d\xi \}\{{\int }_{{\mathbb{R}}^n}\hat{g}(\eta )e^{2\pi i〈x-\beta t,\eta 〉}d\eta \}d\mu (t)\\ =& {\int }_{{\mathbb{R}}^n}f(x-\alpha t)g(x-\beta t)d\mu (t).\end{array}$$

(2.14)

On the other hand, by the assumption ${\upsilon }_s\le {\omega }_1$, it is easy to see that $f(x-\alpha t){\upsilon }_s\in L^{p_1}({\mathbb{R}}^n)$ and

$${\parallel f(x-\alpha t){\upsilon }_s\parallel }_{p_1}\le {\upsilon }_s(\alpha t){\parallel f\parallel }_{p_1,{\omega }_1}.$$

(2.15)

Also, $g(x-\beta t)\in L^{p_2}({\mathbb{R}}^n)$. Then, by (2.14), (2.15) and the generalized Hölder inequality, we have

$$\begin{array}{rcl}{\parallel B_m(f,g)\parallel }_{p_3,{\upsilon }_s}& \le & {\int }_{{\mathbb{R}}^n}{\parallel f(x-\alpha t)g(x-\beta t)\parallel }_{p_3,{\upsilon }_s}d|\mu |(t)\\ \le & {\int }_{{\mathbb{R}}^n}{\parallel f(x-\alpha t){\upsilon }_s\parallel }_{p_1}{\parallel g(x-\beta t)\parallel }_{p_2}d|\mu |(t)\\ \le & {\int }_{{\mathbb{R}}^n}{\upsilon }_s(\alpha t){\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2}d|\mu |(t)\\ \le & {\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\int }_{{\mathbb{R}}^n}{\upsilon }_s(\alpha t)d|\mu |(t).\end{array}$$

(2.16)

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

$$\begin{array}{rcl}{\int }_{{\mathbb{R}}^n}{\upsilon }_s(\alpha t)d|\mu |(t)& =& {\int }_{{\mathbb{R}}^n}{(1+|\alpha t|)}^{s}d|\mu |(t)\\ \le & {\int }_{{\mathbb{R}}^n}{(1+|t|)}^{s}d|\mu |(t)={\parallel \mu \parallel }_{{\upsilon }_s}.\end{array}$$

Hence by (2.16)

$${\parallel B_m(f,g)\parallel }_{p_3,{\upsilon }_s}\le {\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel \mu \parallel }_{{\upsilon }_s}.$$

(2.17)

Thus $m\in BM({\omega }_1,{\omega }_2,{\upsilon }_s)$ and by (2.17), we have

$${\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\upsilon }_s)}=sup\{\frac{{\parallel B_m(f,g)\parallel }_{p_3,{\upsilon }_s}}{{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}}:{\parallel f\parallel }_{p_1,{\omega }_1}\le 1,{\parallel g\parallel }_{p_2,{\omega }_2}\le 1\}\le {\parallel \mu \parallel }_{{\upsilon }_s}.$$

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

$$\begin{array}{rcl}{\int }_{{\mathbb{R}}^n}{\upsilon }_s(\alpha t)d|\mu |(t)& <& {\int }_{{\mathbb{R}}^n}{(|\alpha |+|\alpha ||t|)}^{s}d|\mu |(t)\\ =& |\alpha {|}^s{\int }_{{\mathbb{R}}^n}{\upsilon }_s(t)d|\mu |(t)=|\alpha {|}^s{\parallel \mu \parallel }_{{\upsilon }_s}.\end{array}$$

Again, by (2.16) we have

$${\parallel B_m(f,g)\parallel }_{p_3,{\upsilon }_s}\le |\alpha {|}^s{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel \mu \parallel }_{{\upsilon }_s}.$$

(2.18)

Hence, we obtain $m\in BM({\omega }_1,{\omega }_2,{\upsilon }_s)$ and by (2.18)

$${\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\upsilon }_s)}=sup\{\frac{{\parallel B_m(f,g)\parallel }_{p_3,{\upsilon }_s}}{{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}}:{\parallel f\parallel }_{p_1,{\omega }_1}\le 1,{\parallel g\parallel }_{p_2,{\omega }_2}\le 1\}\le |\alpha {|}^s{\parallel \mu \parallel }_{{\upsilon }_s}.$$

 □

Theorem 2.4 Let $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$.

• (a) $T_{({\xi }_0,{\eta }_0)}m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ for each $({\xi }_0,{\eta }_0)\in {\mathbb{R}}^{2n}$ and

  $${\parallel T_{({\xi }_0,{\eta }_0)}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}={\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.$$

• (b) $M_{({\xi }_0,{\eta }_0)}m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ for each $({\xi }_0,{\eta }_0)\in {\mathbb{R}}^{2n}$ and

  $${\parallel M_{({\xi }_0,{\eta }_0)}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\le {\omega }_1(-{\xi }_0){\omega }_2(-{\eta }_0){\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.$$

Proof (a) Let us take any $f\in L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$ and $g\in L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$. If we say that $\xi -{\xi }_0=u$ and $\eta -{\eta }_0=v$, then

$$\begin{array}{rl}{B}_{T_{({\xi }_0,{\eta }_0)}m}(f,g)(x)& ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )T_{({\xi }_0,{\eta }_0)}m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(u+{\xi }_0)\hat{g}(v+{\eta }_0)m(u,v)e^{2\pi i〈u+{\xi }_0,x〉}{e}^{2\pi i〈v+{\eta }_0,x〉}dudv\\ ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{T}_{-{\xi }_0}\hat{f}(u)T_{-{\eta }_0}\hat{g}(v)m(u,v)e^{2\pi i〈{\xi }_0+{\eta }_0,x〉}{e}^{2\pi i〈u+v,x〉}dudv.\end{array}$$

(2.19)

By (2.19), we have

$$\begin{array}{rcl}{B}_{T_{({\xi }_0,{\eta }_0)}m}(f,g)(x)& =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{T}_{-{\xi }_0}\hat{f}(u)T_{-{\eta }_0}\hat{g}(v)e^{2\pi i〈{\xi }_0+{\eta }_0,x〉}{e}^{2\pi i〈u+v,x〉}m(u,v)dudv\\ =& e^{2\pi i〈{\xi }_0+{\eta }_0,x〉}{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}(M_{-{\xi }_0}f)ˆ(u)(M_{-{\eta }_0}g)ˆ(v)m(u,v)e^{2\pi i〈u+v,x〉}dudv\\ =& e^{2\pi i〈{\xi }_0+{\eta }_0,x〉}{B}_m(M_{-{\xi }_0}f,M_{-{\eta }_0}g)(x).\end{array}$$

(2.20)

Since $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, ${\parallel M_{-{\xi }_0}f\parallel }_{p_1,{\omega }_1}={\parallel f\parallel }_{p_1,{\omega }_1}$ and ${\parallel M_{-{\eta }_0}g\parallel }_{p_2,{\omega }_2}={\parallel g\parallel }_{p_2,{\omega }_2}$ are satisfied for all $f\in L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$ and $g\in L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$. Hence, by (2.20), we have

$$\begin{array}{rcl}{\parallel B_{T_{({\xi }_0,{\eta }_0)}m}(f,g)\parallel }_{p_3,{\omega }_3}& =& {\parallel e^{2\pi i〈{\xi }_0+{\eta }_0,x〉}{B}_m(M_{-{\xi }_0}f,M_{-{\eta }_0}g)\parallel }_{p_3,{\omega }_3}\\ \le & C{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}\end{array}$$

for some $C>0$. Thus $T_{({\xi }_0,{\eta }_0)m}\in BM({\omega }_1,{\omega }_2,{\omega }_3)$. Also, we obtain

$$\begin{array}{c}{\parallel T_{({\xi }_0,{\eta }_0)}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\hfill \\ =\parallel B_{T_{({\xi }_0,{\eta }_0)}m}\parallel \hfill \\ =sup\{\frac{{\parallel B_{T_{({\xi }_0,{\eta }_0)}m}(f,g)\parallel }_{p_3,{\omega }_3}}{{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}}:{\parallel f\parallel }_{p_1,{\omega }_1}\le 1,{\parallel g\parallel }_{p_2,{\omega }_2}\le 1\}\hfill \\ =sup\{\frac{{\parallel B_m(M_{-{\xi }_0}f,M_{-{\eta }_0}g)\parallel }_{p_3,{\omega }_3}}{{\parallel M_{-{\xi }_0}f\parallel }_{p_1,{\omega }_1}{\parallel M_{-{\eta }_0}g\parallel }_{p_2,{\omega }_2}}:{\parallel M_{-{\xi }_0}f\parallel }_{p_1,{\omega }_1}\le 1,{\parallel M_{-{\eta }_0}g\parallel }_{p_2,{\omega }_2}\le 1\}\hfill \\ =\parallel B_m\parallel ={\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.\hfill \end{array}$$

• (b) Let us rewrite the value $B_m(f,g)$ as follows:

  $$\begin{array}{rcl}{B}_{M_{({\xi }_0,{\eta }_0)}m}(f,g)(x)& =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )M_{({\xi }_0,{\eta }_0)}m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )e^{2\pi i〈({\xi }_0,{\eta }_0),(\xi ,\eta )〉}m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{M}_{{\xi }_0}\hat{f}(\xi )M_{{\eta }_0}\hat{g}(\eta )m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}(T_{-{\xi }_0}f)ˆ(\xi )(T_{-{\eta }_0}g)ˆ(\eta )m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ =& B_m(T_{-{\xi }_0}f,T_{-{\eta }_0}g)(x).\end{array}$$

  (2.21)

Also, the inequalities ${\parallel T_{-{\xi }_0}f\parallel }_{p_1,{\omega }_1}\le {\omega }_1(-{\xi }_0){\parallel f\parallel }_{p_1,{\omega }_1}$ and ${\parallel T_{-{\eta }_0}g\parallel }_{p_2,{\omega }_2}\le {\omega }_2(-{\eta }_0){\parallel g\parallel }_{p_2,{\omega }_2}$ are satisfied for all $f\in L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$, $g\in L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$. Hence, since $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, by (2.21) we have

$$\begin{array}{rcl}{\parallel B_{M_{({\xi }_0,{\eta }_0)}m}(f,g)\parallel }_{p_3,{\omega }_3}& =& {\parallel B_m(T_{-{\xi }_0}f,T_{-{\eta }_0}g)\parallel }_{p_3,{\omega }_3}\le \parallel B_m\parallel {\parallel T_{-{\xi }_0}f\parallel }_{p_1,{\omega }_1}{\parallel T_{-{\eta }_0}g\parallel }_{p_2,{\omega }_2}\\ \le & {\omega }_1(-{\xi }_0){\omega }_2(-{\eta }_0)\parallel B_m\parallel {\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}.\end{array}$$

(2.22)

Then $M_{({\xi }_0,{\eta }_0)}m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, and by (2.22) we obtain

$$\begin{array}{rcl}{\parallel M_{({\xi }_0,{\eta }_0)}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}& =& sup\{\frac{{\parallel B_{M_{({\xi }_0,{\eta }_0)}m}(f,g)\parallel }_{p_3,{\omega }_3}}{{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}}:{\parallel f\parallel }_{p_1,{\omega }_1}\le 1,{\parallel g\parallel }_{p_2,{\omega }_2}\le 1\}\\ \le & {\omega }_1(-{\xi }_0){\omega }_2(-{\eta }_0){\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.\end{array}$$

 □

Lemma 2.2 If ${\upsilon }_s$ is a polynomial-type weight function and $f\in L_{{\upsilon }_s}^p({\mathbb{R}}^n)$, then $D_t^{p}f\in L_{{\upsilon }_s}^p({\mathbb{R}}^n)$. Moreover,

$$\begin{array}{c}{\parallel D_t^{p}f\parallel }_{p,{\upsilon }_s}\le {\parallel f\parallel }_{p,{\upsilon }_s}\mathit{\text{if}}t\le 1,\hfill \\ {\parallel D_t^{p}f\parallel }_{p,{\upsilon }_s}<t^s{\parallel f\parallel }_{p,{\upsilon }_s}\mathit{\text{if}}t>1.\hfill \end{array}$$

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

$$\begin{array}{rcl}{\parallel D_t^{p}f\parallel }_{p,{\upsilon }_s}& =& {\left\{{\int }_{{\mathbb{R}}^n}\right|D_t^{p}f(x){|}^p{\upsilon }_s{(x)}^{p}dx\}}^{\frac{1}{p}}\\ =& {\left\{{\int }_{{\mathbb{R}}^n}\right|t^{-\frac{n}{p}}f\left(\frac{x}{t}\right){|}^p{(1+|x|)}^{sp}dx\}}^{\frac{1}{p}}={\left\{{\int }_{{\mathbb{R}}^n}\right|f(u){|}^p{(1+|ut|)}^{sp}du\}}^{\frac{1}{p}}\\ \le & {\left\{{\int }_{{\mathbb{R}}^n}\right|f(u){|}^p{(1+|u|)}^{sp}du\}}^{\frac{1}{p}}\\ =& {\parallel f\parallel }_{p,{\upsilon }_s}<\mathrm{\infty }.\end{array}$$

(2.23)

Thus we have $D_t^{p}f\in L_{{\upsilon }_s}^p({\mathbb{R}}^n)$ 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)

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

Hence $D_t^{p}f\in L_{{\upsilon }_s}^p({\mathbb{R}}^n)$, 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 $m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$. If $\frac{2}{q}=\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}$ and $0<t<\mathrm{\infty }$, then $D_t^{q}m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$. Moreover, then

$$\begin{array}{c}{\parallel D_t^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}\le {\left(\frac{1}{t}\right)}^{s_3}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}\mathit{\text{if}}t\le 1,\hfill \\ {\parallel D_t^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}<t^{s_1+s_2}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}\mathit{\text{if}}t>1.\hfill \end{array}$$

Proof Let $f\in L_{{\upsilon }_{s_1}}^{p_1}({\mathbb{R}}^n)$ and $g\in L_{{\upsilon }_{s_2}}^{p_2}({\mathbb{R}}^n)$ be given. We know by Lemma 2.2 that $D_t^{p_1}f\in L_{{\upsilon }_{s_1}}^{p_1}({\mathbb{R}}^n)$ and $D_t^{p_2}g\in L_{{\upsilon }_{s_2}}^{p_2}({\mathbb{R}}^n)$. If we get $\frac{\xi }{t}=u$ and $\frac{\eta }{t}=v$, we obtain

$$\begin{array}{rcl}{B}_{D_t^{q}m}(f,g)(x)& =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )D_t^{q}m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(tu)\hat{g}(tv)t^{-\frac{2n}{q}}m(u,v)e^{2\pi i〈u+v,tx〉}{t}^{2n}dudv.\end{array}$$

Hence, from the equality $\frac{2}{q}=\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}$, we have

$$\begin{array}{rcl}{B}_{D_t^{q}m}(f,g)(x)& =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(tu)\hat{g}(tv)t^{-n(\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3})}m(u,v)e^{2\pi i〈u+v,tx〉}{t}^{2n}dudv\\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{t}^{-n(1-\frac{1}{p_1^{\mathrm{\prime }}})}\hat{f}(tu)t^{-n(1-\frac{1}{p_1^{\mathrm{\prime }}})}\hat{g}(tv)t^{\frac{n}{p_3}}m(u,v)e^{2\pi i〈u+v,tx〉}{t}^{2n}dudv\\ =& t^{\frac{n}{p_3}}{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{D}_{t^{-1}}^{p_1^{\mathrm{\prime }}}\hat{f}(u)D_{t^{-1}}^{p_2^{\mathrm{\prime }}}\hat{g}(v)m(u,v)e^{2\pi i〈u+v,tx〉}dudv\\ =& t^{\frac{n}{p_3}}{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\left(D_t^{p_1}f\right)ˆ(u)\left(D_t^{p_2}g\right)ˆ(v)m(u,v)e^{2\pi i〈u+v,tx〉}dudv\\ =& D_{t^{-1}}^{p_3}{B}_m(D_t^{p_1}f,D_t^{p_2}g)(x).\end{array}$$

(2.24)

Assume that $t\le 1$. Since $m\in B_m({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$, by Lemma 2.2 and using equality (2.24), we obtain

$$\begin{array}{rcl}{\parallel B_{D_t^{q}m}(f,g)\parallel }_{p_3,{\upsilon }_{s_3}}& =& {\parallel D_{t^{-1}}^{p_3}{B}_m(D_t^{p_1}f,D_t^{p_2}g)(x)\parallel }_{p_3,{\upsilon }_{s_3}}\\ \le & {\left(\frac{1}{t}\right)}^{s_3}{\parallel B_m(D_t^{p_1}f,D_t^{p_2}g)(x)\parallel }_{p_3,{\upsilon }_{s_3}}\\ \le & {\left(\frac{1}{t}\right)}^{s_3}\parallel B_m\parallel {\parallel D_t^{p_1}f\parallel }_{p,{\upsilon }_{s_1}}{\parallel D_t^{p_2}g\parallel }_{p,{\upsilon }_{s_2}}\\ \le & {\left(\frac{1}{t}\right)}^{s_3}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}.\end{array}$$

(2.25)

Then $D_t^{q}m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$, and by (2.25)

$${\parallel D_t^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}\le {\left(\frac{1}{t}\right)}^{s_3}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}.$$

Now let $t>1$. Again, since $m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$, by Lemma 2.2 and using equality (2.24), we obtain

$$\begin{array}{rcl}{\parallel B_{D_t^{q}m}(f,g)\parallel }_{p_3,{\upsilon }_{s_3}}& <& {\parallel B_m(D_t^{p_1}f,D_t^{p_2}g)\parallel }_{p_3,{\upsilon }_{s_3}}\\ \le & \parallel B_m\parallel {\parallel D_t^{p_1}f\parallel }_{p,{\upsilon }_{s_1}}{\parallel D_t^{p_2}g\parallel }_{p,{\upsilon }_{s_2}}\\ <& t^{s_1+s_2}\parallel B_m\parallel {\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}\\ =& t^{s_1+s_2}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}.\end{array}$$

(2.26)

Thus $D_t^{q}m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$ and by (2.26)

$${\parallel D_t^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}<t^{s_1+s_2}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}.$$

 □

Theorem 2.6 Let ${\upsilon }_{s_1}$, ${\upsilon }_{s_2}$, ${\upsilon }_{s_3}$ be weight functions of polynomial type and let $m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$ such that $m(t\xi ,t\eta )=m(\xi ,\eta )$ for any $t>0$, where $\frac{2}{q}=\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}$. Then

$$\begin{array}{c}\frac{2}{q}<\frac{s_3}{n}\mathit{\text{if}}t<1,\hfill \\ \frac{2}{q}>-\frac{s_1+s_2}{n}\mathit{\text{if}}t>1.\hfill \end{array}$$

Proof Take any $f\in L_{{\upsilon }_{s_1}}^{p_1}({\mathbb{R}}^n)$, $g\in L_{{\upsilon }_{s_2}}^{p_2}({\mathbb{R}}^n)$. It is known by Theorem 2.5 that

$$B_{D_t^{q}m}(f,g)(x)=D_{t^{-1}}^{p_3}{B}_m(D_t^{p_1}f,D_t^{p_2}g)(x),x\in {\mathbb{R}}^n.$$

(2.27)

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

(2.28)

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

$$B_m(f,g)(x)=t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})}{B}_{D_t^{q}m}(f,g)(x).$$

Since $D_t^{q}m=m$ for $t=1$, we let $t\ne 1$. Assume first that $t<1$. Also, since $m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$, by Theorem 2.5 we have $D_t^{q}m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$ and ${\parallel D_t^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}<{(\frac{1}{t})}^{s_3}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}$. Then by (2.28)

$$\begin{array}{rcl}{\parallel B_m(f,g)(x)\parallel }_{p_3,{\upsilon }_{s_3}}& =& t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})}{\parallel B_{D_t^q}(f,g)(x)\parallel }_{p_3,{\upsilon }_{s_3}}\\ \le & t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})}{\parallel D_t^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}\\ <& t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})-s_3}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}\\ =& t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})-s_3}\parallel B_m\parallel {\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}.\end{array}$$

Thus,

$$\parallel B_m\parallel <t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})-s_3}\parallel B_m\parallel =t^{\frac{2n}{q}-s_3}\parallel B_m\parallel .$$

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 $D_t^{q}m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$ and ${\parallel D_t^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}<t^{s_1+s_2}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}$. Similarly,

$${\parallel B_m(f,g)(x)\parallel }_{p_3,{\upsilon }_{s_3}}<t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})+s_1+s_2}\parallel B_m\parallel {\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}.$$

Thus, we have

$$\parallel B_m\parallel <t^{-n(\frac{1}{p_3}-\frac{1}{p_1}-\frac{1}{p_2})+s_1+s_2}\parallel B_m\parallel =t^{\frac{2n}{q}+s_1+s_2}\parallel B_m\parallel .$$

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 $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ and $p_3\ge 1$.

• (a) If $\mathrm{\Phi }\in L^1({\mathbb{R}}^n)$, then $\mathrm{\Phi }\ast m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ and

  $${\parallel \mathrm{\Phi }\ast m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\le {\parallel \mathrm{\Phi }\parallel }_1{\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.$$

• (b) If $\mathrm{\Phi }\in L_{\omega }^1({\mathbb{R}}^n)$ such that $\omega (u,\upsilon )={\omega }_1(u){\omega }_2(\upsilon )$, then $\hat{\mathrm{\Phi }}m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ and

  $${\parallel \hat{\mathrm{\Phi }}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\le {\parallel \mathrm{\Phi }\parallel }_{1,\omega }{\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.$$

Proof (a) Let $f\in L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$ and $g\in L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$. Since $L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)\subset L^{p_1}({\mathbb{R}}^n)$ and $L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)\subset L^{p_2}({\mathbb{R}}^n)$, then by Proposition 2.5 in [11]

$$B_{\mathrm{\Phi }\ast m}(f,g)(x)={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(u,v)B_{T_{(u,v)}m}(f,g)(x)dudv.$$

Also, since $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, we have $T_{(u,v)}m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ by Theorem 2.4. So, we write

$$\begin{array}{rl}{\parallel B_{\mathrm{\Phi }\ast m}(f,g)\parallel }_{p_3,{\omega }_3}& \le {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\parallel \mathrm{\Phi }(u,v)B_{T_{(u,v)}m}(f,g)\parallel }_{p_3,{\omega }_3}dudv\\ \le {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|\mathrm{\Phi }(u,v)|{\parallel T_{(u,v)}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}dudv\\ ={\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel \mathrm{\Phi }\parallel }_1{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}<\mathrm{\infty }.\end{array}$$

(2.29)

Hence $\mathrm{\Phi }\ast m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$. Finally, by (2.29), we obtain

$${\parallel \mathrm{\Phi }\ast m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\le {\parallel \mathrm{\Phi }\parallel }_1{\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.$$

• (b) Let $\mathrm{\Phi }\in L_{\omega }^1({\mathbb{R}}^n)$. Take any $f\in L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$ and $g\in L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$. It is known by Proposition 2.5 in [11] that the equality

  $$B_{\hat{\mathrm{\Phi }}m}(f,g)(x)={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(u,v)B_{M_{(-u,-v)}m}(f,g)(x)dudv.$$

Since $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$, by Theorem 2.4 we have $M_{(-u,-v)}m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ and

$${\parallel M_{(-u,-v)}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\le {\omega }_1(u){\omega }_2(\upsilon ){\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.$$

Then we write

$$\begin{array}{rcl}{\parallel B_{\hat{\mathrm{\Phi }}m}(f,g)\parallel }_{p_3,{\omega }_3}& \le & {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}{\parallel \mathrm{\Phi }(u,v)B_{M_{(-u,-v)}m}(f,g)\parallel }_{p_3,{\omega }_3}dudv\\ \le & {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|\mathrm{\Phi }(u,v)|{\parallel M_{(-u,-v)}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}dudv\\ \le & {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|\mathrm{\Phi }(u,v)|{\omega }_1(u){\omega }_2(\upsilon ){\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}dudv\\ =& {\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}{\parallel \mathrm{\Phi }\parallel }_{1,\omega }.\end{array}$$

(2.30)

Thus from (2.30), we obtain $\hat{\mathrm{\Phi }}m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$ and

$${\parallel \hat{\mathrm{\Phi }}m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}\le {\parallel \mathrm{\Phi }\parallel }_{1,\omega }{\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}.$$

 □

Theorem 2.8 Let ${\upsilon }_{s_1}$, ${\upsilon }_{s_2}$, ${\upsilon }_{s_3}$ be weight functions of polynomial type and let $m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$. If $\mathrm{\Psi }\in L^1({\mathbb{R}}^{+},t^{-\frac{2n}{q}}dt)$ such that $\frac{2}{q}=\frac{1}{p_1}+\frac{1}{p_2}-\frac{1}{p_3}$, then $m_{\mathrm{\Psi }}(\xi ,\eta )={\int }_0^{\mathrm{\infty }}m(t\xi ,t\eta )\mathrm{\Psi }(t)dt\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$. Moreover,

$${\parallel m_{\mathrm{\Psi }}\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}<{\parallel \mathrm{\Psi }\parallel }_{L^1({\mathbb{R}}^{+},t^{-\frac{2n}{q}}dt)}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}.$$

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

$$\begin{array}{rcl}{B}_{m_{\mathrm{\Psi }}}(f,g)(x)& =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )m_{\mathrm{\Psi }}(\xi ,\eta )e^{2\pi i〈u+v,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\{{\int }_0^{\mathrm{\infty }}m(t\xi ,t\eta )\mathrm{\Psi }(t)dt\}{e}^{2\pi i〈u+v,x〉}d\xi d\eta \\ =& {\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\{{\int }_0^{\mathrm{\infty }}{D}_{t^{-1}}^{q}m(\xi ,\eta )\mathrm{\Psi }(t)t^{-\frac{2n}{q}}dt\}{e}^{2\pi i〈u+v,x〉}d\xi d\eta \\ =& {\int }_0^{\mathrm{\infty }}{B}_{D_{t^{-1}}^{q}m}(f,g)\mathrm{\Psi }(t)t^{-\frac{2n}{q}}dt.\end{array}$$

Since $m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$, $D_{t^{-1}}^{q}m\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$ by Theorem 2.5, thus we observe that

$$\begin{array}{rcl}{\parallel B_{m_{\mathrm{\Psi }}}(f,g)(x)\parallel }_{p_3,{\upsilon }_{s_3}}& \le & {\int }_0^{\mathrm{\infty }}{\parallel B_{D_{t^{-1}}^{q}m}(f,g)\parallel }_{p_3,{\upsilon }_{s_3}}|\mathrm{\Psi }(t)|t^{-\frac{2n}{q}}dt\\ \le & {\int }_0^{\mathrm{\infty }}\parallel B_{D_{t^{-1}}^{q}m}\parallel {\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}|\mathrm{\Psi }(t)|t^{-\frac{2n}{q}}dt\\ =& {\int }_0^{\mathrm{\infty }}{\parallel D_{t^{-1}}^{q}m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}|\mathrm{\Psi }(t)|t^{-\frac{2n}{q}}dt\\ <& {\int }_0^1{t}^{s_3}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}|\mathrm{\Psi }(t)|t^{-\frac{2n}{q}}dt\\ +{\int }_1^{\mathrm{\infty }}{t}^{-s_1-s_2}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}|\mathrm{\Psi }(t)|t^{-\frac{2n}{q}}dt\\ =& {\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}\\ \times \left\{{\int }_0^1{t}^{s_3}\right|\mathrm{\Psi }(t)|t^{-\frac{2n}{q}}dt+{\int }_1^{\mathrm{\infty }}{t}^{-s_1-s_2}|\mathrm{\Psi }(t)\left|t^{-\frac{2n}{q}}dt\right\}.\end{array}$$

(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)

$${\parallel B_{m_{\mathrm{\Psi }}}(f,g)(x)\parallel }_{p_3,{\upsilon }_{s_3}}<{\parallel \mathrm{\Psi }\parallel }_{L^1({\mathbb{R}}^{+},t^{-\frac{2n}{q}}dt)}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}{\parallel f\parallel }_{p_1,{\upsilon }_{s_1}}{\parallel g\parallel }_{p_2,{\upsilon }_{s_2}}.$$

Hence, $m_{\mathrm{\Psi }}\in BM({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})$ and

$${\parallel m_{\mathrm{\Psi }}\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}<{\parallel \mathrm{\Psi }\parallel }_{L^1({\mathbb{R}}^{+},t^{-\frac{2n}{q}}dt)}{\parallel m\parallel }_{({\upsilon }_{s_1},{\upsilon }_{s_2},{\upsilon }_{s_3})}.$$

 □

Theorem 2.9 Let $p_3\ge 1$ and $m\in BM({\omega }_1,{\omega }_2,{\omega }_3)$. If $Q_1$, $Q_2$ are bounded measurable sets in ${\mathbb{R}}^n$, then

$$h(\xi ,\eta )=\frac{1}{\mu (Q_1\times Q_2)}\int {\int }_{Q_1\times Q_2}m(\xi +u,\eta +v)dudv\in BM({\omega }_1,{\omega }_2,{\omega }_3).$$

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

$$\begin{array}{c}{B}_h(f,g)(x)\hfill \\ ={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )h(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \hfill \\ =\frac{1}{\mu (Q_1\times Q_2)}\int {\int }_{Q_1\times Q_2}\{{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )m(\xi +u,\eta +v)e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta \}dudv\hfill \\ =\frac{1}{\mu (Q_1\times Q_2)}\int {\int }_{Q_1\times Q_2}{B}_{T_{(-u,-v)m}}(f,g)(x)dudv.\hfill \end{array}$$

By using Theorem 2.4, we have

$$\begin{array}{rcl}{\parallel B_h(f,g)\parallel }_{p_3,{\omega }_3}& \le & \frac{1}{\mu (Q_1\times Q_2)}\int {\int }_{Q_1\times Q_2}{\parallel B_{T_{(-u,-v)m}}(f,g)\parallel }_{p_3,{\omega }_3}dudv\\ \le & \frac{1}{\mu (Q_1\times Q_2)}\int {\int }_{Q_1\times Q_2}{\parallel T_{(-u,-v)m}\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}dudv\\ =& \frac{1}{\mu (Q_1\times Q_2)}{\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}\mu (Q_1\times Q_2)\\ =& {\parallel m\parallel }_{({\omega }_1,{\omega }_2,{\omega }_3)}{\parallel f\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{p_2,{\omega }_2}.\end{array}$$

Hence, we obtain $h(\xi ,\eta )\in BM({\omega }_1,{\omega }_2,{\omega }_3)$. □

Theorem 2.10 Let $\omega (u,v)={\omega }_1(u){\omega }_2(v)$, ${\omega }_3\le {\omega }_1$, ${\omega }_3(-u)={\omega }_3(u)$ and $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p_3}\le 1$. Assume that $\mathrm{\Phi }\in L_{\omega }^1({\mathbb{R}}^{2n})$, ${\mathrm{\Psi }}_1\in L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$ and ${\mathrm{\Psi }}_2\in L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$. If $m(\xi ,\eta )={\hat{\mathrm{\Psi }}}_1(\xi )\hat{\mathrm{\Phi }}(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )$, then $m\in BM(1,{\omega }_1;1,{\omega }_2;p_3,{\omega }_3)$.

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

$$\begin{array}{r}|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\\ =|{\int }_{{\mathbb{R}}^n}h(y)\{{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta ){\hat{\mathrm{\Psi }}}_1(\xi )\hat{\mathrm{\Phi }}(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )e^{-2\pi i〈\xi +\eta ,x〉}d\xi d\eta \}dy|\\ \le {\int }_{{\mathbb{R}}^n}|h(y){\omega }_3^{-1}(y)B_{\hat{\mathrm{\Phi }}}(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)(-y){\omega }_3(y)|dy.\end{array}$$

(2.32)

Since the spaces $L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$ and $L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$ are Banach convolution module over the spaces $L_{{\omega }_1}^1({\mathbb{R}}^n)$, $L_{{\omega }_2}^1({\mathbb{R}}^n)$ respectively, we write $f\ast {\mathrm{\Psi }}_1\in L_{{\omega }_1}^{p_1}({\mathbb{R}}^n)$ and $g\ast {\mathrm{\Psi }}_2\in L_{{\omega }_2}^{p_2}({\mathbb{R}}^n)$. Also, by Theorem 2.7, $\hat{\mathrm{\Phi }}\in BM(p_1,{\omega }_1;p_2,{\omega }_2;p_3,{\omega }_3)$. Therefore we obtain $B_{\hat{\mathrm{\Phi }}}(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)\in L_{{\omega }_3}^{p_3}({\mathbb{R}}^n)$. By using the Hölder inequality and the inequality (2.32), we find

$$\begin{array}{c}|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\hfill \\ \le {\parallel h\parallel }_{p_3^{-1},{\omega }_3^{-1}}{\parallel B_{\hat{\mathrm{\Phi }}}(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)\parallel }_{p_3,{\omega }_3}\hfill \\ \le {\parallel h\parallel }_{p_3^{-1},{\omega }_3^{-1}}\parallel B_{\hat{\mathrm{\Phi }}}\parallel {\parallel f\parallel }_{1,{\omega }_1}{\parallel {\mathrm{\Psi }}_1\parallel }_{p_1,{\omega }_1}{\parallel g\parallel }_{1,{\omega }_2}{\parallel {\mathrm{\Psi }}_2\parallel }_{p_2,{\omega }_2}.\hfill \end{array}$$

If we say $C=\parallel B_{\hat{\mathrm{\Phi }}}\parallel {\parallel {\mathrm{\Psi }}_1\parallel }_{p_1,{\omega }_1}{\parallel {\mathrm{\Psi }}_2\parallel }_{p_2,{\omega }_2}$, then we obtain

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\le C{\parallel f\parallel }_{1,\omega }{\parallel g\parallel }_{1,{\omega }_2}{\parallel h\parallel }_{p_3^{-1},{\omega }_3^{-1}},$$

which means $m\in BM(1,{\omega }_1;1,{\omega }_2;p_3,{\omega }_3)$. □

The following theorem can be proved easily by using the technique of the proof in Theorem 2.10.

Theorem 2.11 Let $\omega (u,v)={\omega }_1(u){\omega }_2(v)$, ${\omega }_3\le {\omega }_1$, ${\omega }_3(-u)={\omega }_3(u)$ and $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p_3}\le 1$. If $m(\xi ,\eta )={\hat{\mathrm{\Psi }}}_1(\xi )\hat{\mathrm{\Phi }}(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )$ such that $\mathrm{\Phi }\in L_{\omega }^1({\mathbb{R}}^{2n})$, ${\mathrm{\Psi }}_1\in L_{{\omega }_1}^1({\mathbb{R}}^n)$ and ${\mathrm{\Psi }}_2\in L_{{\omega }_2}^1({\mathbb{R}}^n)$, then $m\in BM(p_1,{\omega }_1;p_2,{\omega }_2;p_3,{\omega }_3)$.

3 The bilinear multipliers space $BM(p_1(x),p_2(x),p_3(x))$

Definition 3.1 Let $p_1(x),p_2(x),p_3(x)\in P({\mathbb{R}}^n)$ and let $p_1^{\ast }<\mathrm{\infty }$, $p_2^{\ast }<\mathrm{\infty }$, $p_3^{\ast }<\mathrm{\infty }$. Assume that $m(\xi ,\eta )$ is a bounded function on ${\mathbb{R}}^n\times {\mathbb{R}}^n$. Define

$$B_m(f,g)(x)={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )m(\xi ,\eta )e^{2\pi i〈\xi +\eta ,x〉}d\xi d\eta$$

for all $f,g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$.

m is said to be a bilinear multiplier on ${\mathbb{R}}^n$ of type $(p_1(x),p_2(x),p_3(x))$ if there exists $C>0$ such that

$${\parallel B_m(f,g)\parallel }_{p_3(x)}\le C{\parallel f\parallel }_{p_1(x)}{\parallel g\parallel }_{p_2(x)}$$

for all $f,g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$, i.e., $B_m$ extends to a bounded bilinear operator from $L^{p_1(x)}({\mathbb{R}}^n)\times L^{p_2(x)}({\mathbb{R}}^n)$ to $L^{p_3(x)}({\mathbb{R}}^n)$. We denote by $BM(p_1(x),p_2(x),p_3(x))$ the space of bilinear multipliers of type $(p_1(x),p_2(x),p_3(x))$ and ${\parallel m\parallel }_{(p_1(x),p_2(x),p_3(x))}=\parallel B_m\parallel$.

The following theorem can be proved easily by using the technique of the proof in Theorem 2.2.

Theorem 3.1 Let $p_3(-x)=p_3(x)$ and $\frac{1}{p_3(x)}+\frac{1}{q(x)}=1$ for all $x\in {\mathbb{R}}^n$. Then $m\in BM(p_1(x),p_2(x),p_3(x))$ if and only if there exists $C>0$ such that

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )m(\xi ,\eta )d\xi d\eta |\le C{\parallel f\parallel }_{p_1(x)}{\parallel g\parallel }_{p_2(x)}{\parallel h\parallel }_{q(x)}$$

for all $f,g,h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$.

Theorem 3.2 Let $\frac{1}{p(x)}+\frac{1}{q(x)}=\frac{1}{r}$. If $\mathrm{\Phi }\in L^1({\mathbb{R}}^n)$, then $m(\xi ,\eta )=\hat{\mathrm{\Phi }}(\xi +\eta )\in BM(p(x),q(x),r)$.

Proof Take any $f,g,h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. Then

$$\begin{array}{r}|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )\hat{\mathrm{\Phi }}(\xi +\eta )d\xi d\eta |\\ =|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )(h\ast \mathrm{\Phi })ˆ(\xi +\eta )d\xi d\eta |\\ =|{\int }_{{\mathbb{R}}^n}(h\ast \mathrm{\Phi })(x)\{{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )e^{-2\pi i〈\xi +\eta ,x〉}d\xi d\eta \}dx|\\ \le {\int }_{{\mathbb{R}}^n}|(h\ast \mathrm{\Phi })(x)||{\tilde{B}}_1(f,g)(x)|dx.\end{array}$$

(3.1)

Since the space $L^{r^{\mathrm{\prime }}}({\mathbb{R}}^n)$ is the Banach convolution module over $L^1({\mathbb{R}}^n)$ such that $\frac{1}{r}+\frac{1}{r^{\mathrm{\prime }}}=1$, we write $h\ast \mathrm{\Phi }\in L^{r^{\mathrm{\prime }}}({\mathbb{R}}^n)$. Also, we have $1\in BM(p(x),q(x),r)$. Then by (3.1), we find $C_1>0$ such that

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )\hat{\mathrm{\Phi }}(\xi +\eta )d\xi d\eta |\le C_1{\parallel h\parallel }_{r^{\mathrm{\prime }}}{\parallel \mathrm{\Phi }\parallel }_1{\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}.$$

If we set $C=C_1{\parallel \mathrm{\Phi }\parallel }_1$, we obtain

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta )\hat{\mathrm{\Phi }}(\xi +\eta )d\xi d\eta |\le C{\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}{\parallel h\parallel }_{r^{\mathrm{\prime }}}$$

and $m(\xi ,\eta )=\hat{\mathrm{\Phi }}(\xi +\eta )\in BM(p(x),q(x),r)$. □

Theorem 3.3 If $m\in BM(p_1(x),p_2(x),p_3(x))$, then $T_{({\xi }_0,{\eta }_0)}m\in BM(p_1(x),p_2(x),p_3(x))$ and

$${\parallel T_{({\xi }_0,{\eta }_0)}m\parallel }_{(p_1(x),p_2(x),p_3(x))}={\parallel m\parallel }_{(p_1(x),p_2(x),p_3(x))}$$

for all $({\xi }_0,{\eta }_0)\in {\mathbb{R}}^{2n}$.

Proof Let us take any $f,g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. By the proof of (a) Theorem 2.4, we know that

$$B_{T_{({\xi }_0,{\eta }_0)}m}(f,g)(x)=e^{2\pi i〈{\xi }_0+{\eta }_0,x〉}{B}_m(M_{-{\xi }_0}f,M_{-{\eta }_0}g)(x),x\in {\mathbb{R}}^n.$$

(3.2)

By Lemma 5 in [8], we know ${\parallel M_{-{\xi }_0}f\parallel }_{p_1(x)}={\parallel f\parallel }_{p_1(x)}$and ${\parallel M_{-{\eta }_0}g\parallel }_{p_2(x)}={\parallel g\parallel }_{p_2(x)}$. Since $m\in BM(p_1(x),p_2(x),p_3(x))$, by (3.2), there exists $C>0$ such that

$${\parallel B_{T_{({\xi }_0,{\eta }_0)}m}(f,g)\parallel }_{p_3(x)}={\parallel B_m(M_{-{\xi }_0}f,M_{-{\eta }_0}g)\parallel }_{p_3(x)}\le C{\parallel f\parallel }_{p_1(x)}{\parallel g\parallel }_{p_2(x)}.$$

Thus $T_{({\xi }_0,{\eta }_0)}m\in BM(p_1(x),p_2(x),p_3(x))$. Moreover, by using the same technique as in the proof of Theorem 2.4, we obtain

$${\parallel T_{({\xi }_0,{\eta }_0)}m\parallel }_{(p_1(x),p_2(x),p_3(x))}={\parallel m\parallel }_{(p_1(x),p_2(x),p_3(x))}.$$

 □

Theorem 3.4 Let $\frac{1}{p(x)}+\frac{1}{q(x)}=\frac{1}{r(x)}$. If $m\in BM(p(x),q(x),r(x))$, then $\mathrm{\Phi }\ast m\in BM(p(x),q(x),r(x))$ and there exists $C>0$ such that

$${\parallel \mathrm{\Phi }\ast m\parallel }_{(p(x),q(x),r(x))}\le C{\parallel \mathrm{\Phi }\parallel }_1{\parallel m\parallel }_{(p(x),q(x),r(x))}$$

for all $\mathrm{\Phi }\in L^1({\mathbb{R}}^{2n})$.

Proof Take any $f,g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. By Proposition 2.5 in [11], we know that

$$B_{\mathrm{\Phi }\ast m}(f,g)(x)={\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(u,v)B_{T_{({\xi }_0,{\eta }_0)}m}(f,g)(x)dudv.$$

(3.3)

Since $m\in BM(p(x),q(x),r(x))$, then $T_{({\xi }_0,{\eta }_0)}m\in BM(p(x),q(x),r(x))$ and

$${\parallel T_{({\xi }_0,{\eta }_0)}m\parallel }_{(p(x),q(x),r(x))}={\parallel m\parallel }_{(p(x),q(x),r(x))}$$

by Theorem 3.3. Using (3.3) and the Minkowski inequality for a variable exponent Lebesgue space [12], we find $C>0$ such that

$$\begin{array}{rcl}{\parallel B_{\mathrm{\Phi }\ast m}(f,g)\parallel }_{r(x)}& \le & C{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|\mathrm{\Phi }(u,v)|{\parallel B_{T_{({\xi }_0,{\eta }_0)}m}(f,g)\parallel }_{r(x)}dudv\\ \le & C{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|\mathrm{\Phi }(u,v)|\parallel B_{T_{({\xi }_0,{\eta }_0)}m}\parallel {\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}dudv\\ =& C{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}|\mathrm{\Phi }(u,v)|{\parallel m\parallel }_{(p(x),q(x),r(x))}{\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}dudv\\ =& C{\parallel m\parallel }_{(p(x),q(x),r(x))}{\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}{\parallel \mathrm{\Phi }\parallel }_1.\end{array}$$

(3.4)

Hence $\mathrm{\Phi }\ast m\in BM(p(x),q(x),r(x))$ and by (3.4), we have

$${\parallel \mathrm{\Phi }\ast m\parallel }_{(p(x),q(x),r(x))}\le C{\parallel \mathrm{\Phi }\parallel }_1{\parallel m\parallel }_{(p(x),q(x),r(x))}.$$

 □

Theorem 3.5 Let $r(-x)=r(x)$.

• (a) If ${\mathrm{\Psi }}_1\in L^p({\mathbb{R}}^n)$, ${\mathrm{\Psi }}_2\in L^q({\mathbb{R}}^n)$ and $m\in BM(p,q,r(x))$, then ${\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )\in BM(1,1,r(x))$.

• (b) If ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2\in L^1({\mathbb{R}}^n)$ and $m\in BM(p,q,r(x))$, then ${\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )\in BM(p,q,r(x))$.

• (c) If ${\mathrm{\Psi }}_1\in L^p({\mathbb{R}}^n)$ and $m\in BM(p,q,r(x))$, then ${\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta )\in BM(1,q(x),r(x))$.

• (d) If ${\mathrm{\Psi }}_1\in L^1({\mathbb{R}}^n)$ and $m\in BM(p,q,r(x))$, then ${\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta )\in BM(p,q(x),r(x))$.

Proof (a) Let $f,g,h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ be given. Then

(3.5)

Since the spaces $L^p({\mathbb{R}}^n)$ and $L^q({\mathbb{R}}^n)$ are Banach convolution module over $L^1({\mathbb{R}}^n)$, we have $f\ast {\mathrm{\Psi }}_1\in L^p({\mathbb{R}}^n)$ and $g\ast {\mathrm{\Psi }}_2\in L^q({\mathbb{R}}^n)$. Also, since $m\in BM(p,q,r(x))$, we write $B_m(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)(y)\in L^{r(x)}({\mathbb{R}}^n)$. Then, by the equality

$${\parallel {\tilde{B}}_m(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)(y)\parallel }_{r(x)}={\parallel B_m(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)(y)\parallel }_{r(x)},$$

the Hölder inequality and the inequality (3.5), we have

$$\begin{array}{c}|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta ){\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )d\xi d\eta |\hfill \\ \le {\parallel h\parallel }_{r^{\mathrm{\prime }}(x)}{\parallel B_m(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)(y)\parallel }_{r(x)}\hfill \\ \le {\parallel h\parallel }_{r^{\mathrm{\prime }}(x)}\parallel B_m\parallel {\parallel f\parallel }_1{\parallel {\mathrm{\Psi }}_1\parallel }_p{\parallel g\parallel }_1{\parallel {\mathrm{\Psi }}_2\parallel }_q.\hfill \end{array}$$

If we say $C=\parallel B_m\parallel {\parallel {\mathrm{\Psi }}_1\parallel }_p{\parallel {\mathrm{\Psi }}_2\parallel }_q$, we obtain

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta ){\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )d\xi d\eta |\le C{\parallel f\parallel }_1{\parallel g\parallel }_1{\parallel h\parallel }_{r^{\mathrm{\prime }}(x)}.$$

Hence, ${\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )\in BM(1,1,r(x))$.

• (b) Take any $f,g,h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. By (a), we know that

  $$\begin{array}{c}|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta ){\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )d\xi d\eta |\hfill \\ \le {\int }_{{\mathbb{R}}^n}|h(y)||{\tilde{B}}_m(f\ast {\mathrm{\Psi }}_1,g\ast {\mathrm{\Psi }}_2)(y)|dx.\hfill \end{array}$$

Similarly, if we say $C=\parallel B_m\parallel {\parallel {\mathrm{\Psi }}_1\parallel }_1{\parallel {\mathrm{\Psi }}_2\parallel }_1$, we obtain

$$|{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\hat{f}(\xi )\hat{g}(\eta )\hat{h}(\xi +\eta ){\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )d\xi d\eta |\le C{\parallel f\parallel }_p{\parallel g\parallel }_q{\parallel h\parallel }_{r^{\mathrm{\prime }}(x)},$$

which means ${\hat{\mathrm{\Psi }}}_1(\xi )m(\xi ,\eta ){\hat{\mathrm{\Psi }}}_2(\eta )\in BM(p,q,r(x))$.

In this theorem, (c) and (d) can be proved easily by using the technique of the proof in (a) and (b), respectively. □

Theorem 3.6 Let $m\in BM(p_1(x),p_2(x),p_3(x))$. If $Q_1,Q_2\subset {\mathbb{R}}^n$ are bounded sets, then

$$h(\xi ,\eta )=\frac{1}{\mu (Q_1\times Q_2)}\int {\int }_{Q_1\times Q_2}m(\xi +u,\eta +v)dudv\in BM(p_1(x),p_2(x),p_3(x)).$$

Proof Let $f,g,h\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ be given. By using the Fubini theorem, we can easily prove the following equality:

$$B_h(f,g)(x)=\frac{1}{\mu (Q_1\times Q_2)}\int {\int }_{Q_1\times Q_2}{B}_{T_{(-u,-v)m}}(f,g)(x)dudv.$$

Also, by the Minkowski inequality and Theorem 3.3, we find $C_0>0$ such that

$$\begin{array}{c}{\parallel B_h(f,g)\parallel }_{p_3(x)}\hfill \\ \le \frac{1}{\mu (Q_1\times Q_2)}{C}_0\int {\int }_{Q_1\times Q_2}{\parallel B_{T_{(-u,-v)m}}(f,g)\parallel }_{p_3(x)}dudv\hfill \\ \le \frac{1}{\mu (Q_1\times Q_2)}{C}_0\int {\int }_{Q_1\times Q_2}{\parallel T_{(-u,-v)m}\parallel }_{(p_1(x),p_2(x),p_3(x))}{\parallel f\parallel }_{p_1(x)}{\parallel g\parallel }_{p_2(x)}dudv\hfill \\ =\frac{1}{\mu (Q_1\times Q_2)}{C}_0{\parallel m\parallel }_{(p_1(x),p_2(x),p_3(x))}{\parallel f\parallel }_{p_1(x)}{\parallel g\parallel }_{p_2(x)}\mu (Q_1\times Q_2)\hfill \\ =C_0{\parallel m\parallel }_{(p_1(x),p_2(x),p_3(x))}{\parallel f\parallel }_{p_1(x)}{\parallel g\parallel }_{p_2(x)}.\hfill \end{array}$$

If we say $C=C_0{\parallel m\parallel }_{(p_1(x),p_2(x),p_3(x))}$, we obtain

$${\parallel B_h\parallel }_{(p_1(x),p_2(x),p_3(x))}\le C{\parallel f\parallel }_{p_1(x)}{\parallel g\parallel }_{p_2(x)}.$$

Hence $h(\xi ,\eta )\in BM(p_1(x),p_2(x),p_3(x))$. □

Theorem 3.7

• (a) If $L^{s(x)}({\mathbb{R}}^n)\subset L^{r(x)}({\mathbb{R}}^n)$ and $m\in BM(p(x),q(x),s(x))$, then $m\in BM(p(x),q(x),r(x))$.

• (b) If $L^{s(x)}({\mathbb{R}}^n)\subset L^{r(x)}({\mathbb{R}}^n)$, $L^{p(x)}({\mathbb{R}}^n)\subset L^{k(x)}({\mathbb{R}}^n)$, $L^{q(x)}({\mathbb{R}}^n)\subset L^{t(x)}({\mathbb{R}}^n)$ and $m\in BM(k(x),t(x),s(x))$, then $m\in BM(p(x),q(x),r(x))$.

Proof (a) Take any $f,g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. Since $m\in BM(p(x),q(x),s(x))$, there exists $C_1>0$ such that

$${\parallel B_m(f,g)\parallel }_{s(x)}\le C_1{\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}.$$

(3.6)

Also, since $L^{s(x)}({\mathbb{R}}^n)\subset L^{r(x)}({\mathbb{R}}^n)$, there exists $C_2>0$ such that

$${\parallel B_m(f,g)\parallel }_{r(x)}\le C_2{\parallel B_m(f,g)\parallel }_{s(x)}.$$

(3.7)

If we set $C=C_1{C}_2$ and combine the inequalities (3.6) and (3.7), we have

$${\parallel B_m(f,g)\parallel }_{r(x)}\le C{\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}.$$

Therefore $m\in BM(p(x),q(x),r(x))$.

• (b) Let us take any $f,g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$. Since $m\in BM(k(x),t(x),s(x))$, there exists $C_3>0$ such that

  $${\parallel B_m(f,g)\parallel }_{s(x)}\le C_3{\parallel f\parallel }_{k(x)}{\parallel g\parallel }_{t(x)}.$$

  (3.8)

By using the inclusions $L^{s(x)}({\mathbb{R}}^n)\subset L^{r(x)}({\mathbb{R}}^n)$, $L^{p(x)}({\mathbb{R}}^n)\subset L^{k(x)}({\mathbb{R}}^n)$ and $L^{q(x)}({\mathbb{R}}^n)\subset L^{t(x)}({\mathbb{R}}^n)$, we find $C_4,C_5,C_6>0$ such that

(3.9)

(3.10)

and

$${\parallel g\parallel }_{t(x)}\le C_6{\parallel g\parallel }_{q(x)}.$$

(3.11)

If we set $C=C_3{C}_4{C}_5{C}_6$ and combine the inequalities (3.8), (3.9), (3.10) and (3.11), we have

$${\parallel B_m(f,g)\parallel }_{r(x)}\le C{\parallel f\parallel }_{p(x)}{\parallel g\parallel }_{q(x)}.$$

Then $m\in BM(p(x),q(x),r(x))$. □