Open Access

Bilinear multipliers of weighted Lebesgue spaces and variable exponent Lebesgue spaces

Journal of Inequalities and Applications20132013:259

https://doi.org/10.1186/1029-242X-2013-259

Received: 23 November 2012

Accepted: 4 May 2013

Published: 23 May 2013

Abstract

Let 1 p 1 , p 2 < , 0 < p 3 and ω 1 , ω 2 , ω 3 be weight functions on R n . Assume that ω 1 , ω 2 are slowly increasing functions.

We say that a bounded function m ( ξ , η ) defined on R n × R n is a bilinear multiplier on R n of type ( p 1 , ω 1 ; p 2 , ω 2 ; p 3 , ω 3 ) (shortly ( ω 1 , ω 2 , ω 3 ) ) if

B m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η

is a bounded bilinear operator from L ω 1 p 1 ( R n ) × L ω 2 p 2 ( R n ) to L ω 3 p 3 ( R n ) . We denote by BM ( p 1 , ω 1 ; p 2 , ω 2 ; p 3 , ω 3 ) (shortly BM ( ω 1 , ω 2 , ω 3 ) ) the vector space of bilinear multipliers of type ( ω 1 , ω 2 , ω 3 ) .

In this paper first we discuss some properties of the space BM ( ω 1 , ω 2 , ω 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 ) ( R n ) , L p 2 ( x ) ( R n ) and L p 3 ( x ) ( R n ) , we define the space of bilinear multipliers from L p 1 ( x ) ( R n ) × L p 2 ( x ) ( R n ) to L p 3 ( x ) ( R n ) and discuss some properties of this space.

MSC:42A45, 42B15, 42B35.

Keywords

bilinear multipliersweighted Lebesgue spacevariable exponent Lebesgue space

1 Introduction

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

for all x 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 p , we set
L ω p ( R n ) = { f : f ω L p ( R n ) } .
It is known that L ω p ( R n ) is a Banach space under the norm
f p , ω = f ω p = { R n | f ( x ) ω ( x ) | p d x } 1 p , 1 p <
or
f , ω = f ω = ess sup x R n | f ( x ) ω ( x ) | , p = , [1, 2] .
The dual of the space L ω p ( R n ) is the space L ω 1 q ( R n ) , where 1 p + 1 q = 1 and ω 1 ( x ) = 1 ω ( x ) . For f L 1 ( R n ) , the Fourier transform of f is denoted by f ˆ . We know that f ˆ is a continuous function on R n , which vanishes at infinity, and it has the inequality f ˆ f 1 [3, 4]. Let f be a measurable function on 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 π i x , y f ( y ) and D t p f ( y ) = t n p f ( y t ) respectively for x , y R n , 0 < p , t < . With this notation out of the way one has, for 1 p and 1 p + 1 p = 1 ,
( T x f ) ˆ ( ξ ) = M x f ˆ ( ξ ) , ( M x f ) ˆ ( ξ ) = T x f ˆ ( ξ ) , ( D t p f ) ˆ ( ξ ) = D t 1 p f ˆ ( ξ ) .
We denote by M ( R n ) the space of bounded regular Borel measures, by M ( ω ) the space of μ in M ( R n ) such that
μ ω = R n ω d | μ | < .
If μ M ( R n ) , the Fourier-Stieltjes transform of μ is denoted by μ ˆ [5]. In this paper, P ( R n ) denotes the family of all measurable functions p : R n [ 1 , ) . We put
p = ess inf x R n p ( x ) , p = ess  sup x R n p ( x ) .
We shall also use the notation
Ω = { x R n : p ( x ) = } .
The generalized Lebesgue space (or the variable exponent Lebesgue space) L p ( x ) ( R n ) is defined to be a space of (equivalence classes) measurable functions f such that
ϱ p ( λ f ) = R n Ω | λ f ( x ) | p ( x ) d x + ess  sup x Ω ( λ f ( x ) ) <
for some λ = λ ( f ) > 0 . If p < , then
ϱ p ( λ f ) = R n Ω | λ f ( x ) | p ( x ) d x , [6, 7] .
It is known by Theorem 2.5 in [6] that L p ( x ) ( R n ) is a Banach space with the Luxemburg norm
f p ( x ) = inf { λ > 0 : ϱ p ( f λ ) 1 } .

If p < , then C c ( R n ) is dense in L p ( x ) ( R n ) . Also, if p ( x ) = p is a constant function, then the above norm p ( x ) coincides with the usual norm p . The vector space of locally integrable functions on R n is denoted by L loc 1 ( R n ) . The space L p ( x ) ( R n ) is a solid space, that is, if f L p ( x ) ( R n ) is given and g L loc 1 ( R n ) satisfies | g ( x ) | | f ( x ) | a.e., then g L p ( x ) ( R n ) and g p ( x ) f p ( x ) by [8]. In this paper we assume that p < .

2 The bilinear multipliers space BM ( p 1 , ω 1 ; p 2 , ω 2 ; p 3 , ω 3 )

Lemma 2.1 Let 1 p < and let ω be a slowly increasing weight function. Then S ( R n ) is dense in L ω p ( R n ) .

Proof Let f S ( R n ) be given. Since ω is a slowly increasing weight function, there exist C > 0 and N N such that
| ω ( x ) | C ( 1 + x 2 ) N = m ( x )
(2.1)

for all x R n . Also, since m is a polynomial, then by Proposition 19.2.2 in [9], we have S ( R n ) L m p ( R n ) . Hence, by (2.1), we obtain S ( R n ) L m p ( R n ) L ω p ( R n ) .

Now, we show that C c ( R n ) is dense L m p ( R n ) . Let f L w p ( R n ) be given. Then f m L p ( R n ) . Since C c ( R n ) is dense L p ( R n ) by [6], for given ε > 0 , there exists g C c ( R n ) such that
f m g p < ε .
(2.2)
Therefore, by using the inequality (2.2), we write
f m g p = f g m 1 p , m < ε .

Also, since m 0 and m is a polynomial, we have g m 1 C c ( R n ) . Thus, we have C c ( R n ) ¯ = L m p ( R n ) . By using the inclusion C c ( R n ) S ( R n ) L m p ( R n ) , we obtain S ( R n ) ¯ = L m p ( R n ) .

Now, take any f L ω p ( R n ) . Since C c ( R n ) ¯ = L m p ( R n ) ¯ = L ω p ( R n ) , there exists g L m p ( R n ) such that
f g p , ω < ε 2 .
(2.3)
Furthermore, since S ( R n ) is dense L m p ( R n ) , there exists h S ( R n ) such that
g h p , m < ε 2 .
(2.4)
Combining the inequalities (2.3) and (2.4), we have
f g p , ω f g p , ω + h g p , ω f g p , ω + h g p , m < ε ,

which means S ( R n ) ¯ = L ω p ( R n ) . □

Definition 2.1 Let 1 p 1 , p 2 < , 0 < p 3 and ω 1 , ω 2 , ω 3 be weight functions on R n . Assume that ω 1 , ω 2 are slowly increasing functions and m ( ξ , η ) is a bounded function on R n × R n . Define
B m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η

for all f , g S ( R n ) .

m is said to be a bilinear multiplier on R n of type ( p 1 , ω 1 ; p 2 , ω 2 ; p 3 , ω 3 ) (shortly ( ω 1 , ω 2 , ω 3 ) ) if there exists C > 0 such that
B m ( f , g ) p 3 , ω 3 C f p 1 , ω 1 g p 2 , ω 2

for all f , g S ( R n ) . That means B m extends to a bounded bilinear operator from L ω 1 p 1 ( R n ) × L ω 2 p 2 ( R n ) to L ω 3 p 3 ( R n ) .

We denote by BM ( p 1 , ω 1 ; p 2 , ω 2 ; p 3 , ω 3 ) (shortly BM ( ω 1 , ω 2 , ω 3 ) ) the space of all bilinear multipliers of type ( ω 1 , ω 2 , ω 3 ) and m ( ω 1 , ω 2 , ω 3 ) = B m .

Theorem 2.1 Let 1 p 1 + 1 p 2 = 1 p 3 and ω 3 ω 1 . If K L ω 3 1 ( R n ) , then m ( ξ , η ) = K ˆ ( ξ η ) defines a bilinear multiplier and m ( ω 1 , ω 2 , ω 3 ) K 1 , ω 3 .

Proof For f , g S ( R n ) , we have f ( x y ) = R n f ˆ ( ξ ) e 2 π i x y , ξ d ξ and g ( x + y ) = R n g ˆ ( η ) e 2 π i x + y , η d η . Thus, by the Fubini theorem, we write
B m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) K ˆ ( ξ η ) e 2 π i ξ + η , x d ξ d η = R n R n f ˆ ( ξ ) g ˆ ( η ) ( R n K ( y ) e 2 π i ξ η , y d y ) e 2 π i ξ + η , x d ξ d η = R n R n R n f ˆ ( ξ ) g ˆ ( η ) K ( y ) e 2 π i ξ η , y e 2 π i ξ + η , x d ξ d η d y .
(2.5)
Since f , g S ( R n ) , we have f ˆ , g ˆ S ( R n ) L 1 ( R n ) . Hence, by (2.5), we obtain
B m ( f , g ) ( x ) = R n ( R n f ˆ ( ξ ) e 2 π i x y , ξ ) ( R n g ˆ ( η ) e 2 π i x + y , η d η ) K ( y ) d y = R n f ( x y ) g ( x + y ) K ( y ) d y .
(2.6)
Since ω 3 ω 1 , then
f ( x y ) ω 3 p 1 ω 3 ( y ) f p 1 , ω 1
(2.7)
and hence f ( x y ) ω 3 L p 1 ( R n ) . Therefore from (2.6) and the Minkowski inequality, we write
B m ( f , g ) p 3 , ω 3 R n R n f ( x y ) g ( x + y ) p 3 , ω 3 | K ( y ) | d y = R n R n f ( x y ) g ( x + y ) ω 3 p 3 | K ( y ) | d y .
(2.8)
Hence, using the generalized Hölder inequality and combining (2.7), (2.8), we have
B m ( f , g ) p 3 , ω 3 R n f ( x y ) ω 3 p 1 g ( x + y ) p 2 ω 3 ( y ) | K ( y ) | d y R n f p 1 g p 2 ω 3 ( y ) | K ( y ) | d y R n f p 1 , ω 1 g p 2 , ω 2 ω 3 ( y ) | K ( y ) | d y = f p 1 , ω 1 g p 2 , ω 2 K 1 , ω 3 .
(2.9)
If we set C = K 1 , ω 3 , we obtain
B m ( f , g ) p 3 , ω 3 C f p 1 , ω 1 g p 2 , ω 2 .
Then m BM ( ω 1 , ω 2 , ω 3 ) . Consequently, using (2.9), we have
m ( ω 1 , ω 2 , ω 3 ) = sup { B m ( f , g ) p 3 , ω 3 f p 1 , ω 1 g p 2 , ω 2 : f p 1 , ω 1 1 , g p 2 , ω 2 1 } K 1 , ω 3 .

 □

Definition 2.2 Let 1 p 1 , p 2 < , 0 < p 3 and ω 1 , ω 2 , ω 3 be weight functions on R n . Suppose that ω 1 , ω 2 are slowly increasing functions. We denote by M ˜ ( ω 1 , ω 2 , ω 3 ) the space of measurable functions M : R n C such that m ( ξ , η ) = M ( ξ η ) BM ( ω 1 , ω 2 , ω 3 ) , that is to say,
B M ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) M ( ξ η ) e 2 π i ξ + η , x d ξ d η

extends to a bounded bilinear map from L ω 1 p 1 ( R n ) × L ω 2 p 2 ( R n ) to L ω 3 p 3 ( R n ) . We denote M ( ω 1 , ω 2 , ω 3 ) = B M .

Theorem 2.2 Let p 3 1 and ω 3 ( x ) = ω 3 ( x ) . Then m BM ( ω 1 , ω 2 , ω 3 ) if and only if there exists C > 0 such that
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | C f p 1 , ω 1 g p 2 , ω 2 h p 3 , ω 3 1

for all f , g , h S ( R n ) , where 1 p 3 + 1 p 3 = 1 .

Proof Let m BM ( ω 1 , ω 2 , ω 3 ) . We take any f , g , h S ( R n ) . From the Fubini theorem, we write
(2.10)
where B ˜ m ( f , g ) ( y ) = B m ( f , g ) ( y ) . On the other hand, since m BM ( ω 1 , ω 2 , ω 3 ) , then B m ( f , g ) L ω 3 p 3 ( R n ) . Thus we obtain B ˜ m ( f , g ) L ω 3 p 3 ( R n ) . Also, h S ( R n ) L p 3 ( R n ) L ω 3 1 p 3 ( R n ) . Hence, using the Hölder inequality and the inequality (2.10), we write
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | R n | h ( y ) ω 3 1 ( y ) | | B ˜ m ( f , g ) ( y ) ω 3 ( y ) | d y B ˜ m ( f , g ) p 3 , ω 3 h p 3 , ω 3 1 .
(2.11)
Moreover, since m BM ( ω 1 , ω 2 , ω 3 ) , there exists C > 0 such that
B m ( f , g ) p 3 , ω 3 C f p 1 , ω 1 g p 2 , ω 2 .
(2.12)
If we combine (2.11) and (2.12), we write
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | C f p 1 , ω 1 g p 2 , ω 2 h p 3 , ω 3 1 .
For the proof of converse, assume that there exists a constant C > 0 such that
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | C f p 1 , ω 1 g p 2 , ω 2 h p 3 , ω 3 1
for all f , g , h S ( R n ) . From the assumption and (2.10), we write
| R n h ( y ) B ˜ m ( f , g ) ( y ) d y | C f p 1 , ω 1 g p 2 , ω 2 h p 3 , ω 3 1 .
(2.13)
Define a function l from S ( R n ) L ω 3 1 p 3 ( R n ) to such that
( h ) = R n h ( y ) B ˜ m ( f , g ) ( y ) d y .
It is clear that the function is linear and bounded by (2.13). By using C c ( R n ) ¯ = L ω 3 1 p 3 ( R n ) in [10], it is easy to show that C c ( R n ) ¯ = L ω 3 1 p 3 ( R n ) . So, by the inclusion C c ( R n ) S ( R n ) L ω 3 1 p 3 ( R n ) , we have S ( R n ) ¯ = L ω 3 1 p 3 ( R n ) . Thus extends to a bounded function from L ω 3 1 p 3 ( R n ) to . Then ( L ω 3 1 p 3 ( R n ) ) = L ω 3 p 3 ( R n ) and by (2.13), we have
B m ( f , g ) p 3 , ω 3 = B ˜ m ( f , g ) p 3 , ω 3 = = sup h p 3 , ω 3 1 1 | l ( h ) | h p 3 , ω 3 1 sup h p 3 , ω 3 1 1 C f p 1 , ω 1 g p 2 , ω 2 h p 3 , ω 3 1 h p 3 , ω 3 1 C f p 1 , ω 1 g p 2 , ω 2 .

Hence, we obtain m BM ( ω 1 , ω 2 , ω 3 ) . □

Theorem 2.3 Let 1 p 1 + 1 p 2 = 1 p 3 , p 3 1 and υ s ( x ) = ( 1 + | x | ) s , s 0 be a weight function of polynomial type such that υ s ω 1 . If μ M ( υ s ) and m ( ξ , η ) = μ ˆ ( α ξ + β η ) for α , β R , then m BM ( ω 1 , ω 2 , υ s ) . Moreover,
m ( ω 1 , ω 2 , υ s ) μ υ s if | α | 1 , m ( ω 1 , ω 2 , υ s ) | α | s μ υ s if | α | > 1 .
Proof Let f , g , S ( R n ) . Then
B m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) μ ˆ ( α ξ + β η ) e 2 π i ξ + η , x d ξ d η = R n R n f ˆ ( ξ ) g ˆ ( η ) { R n e 2 π i α ξ + β η , t d μ ( t ) } e 2 π i ξ + η , x d ξ d η = R n { R n f ˆ ( ξ ) e 2 π i x α t , ξ d ξ } { R n g ˆ ( η ) e 2 π i x β t , η d η } d μ ( t ) = R n f ( x α t ) g ( x β t ) d μ ( t ) .
(2.14)
On the other hand, by the assumption υ s ω 1 , it is easy to see that f ( x α t ) υ s L p 1 ( R n ) and
f ( x α t ) υ s p 1 υ s ( α t ) f p 1 , ω 1 .
(2.15)
Also, g ( x β t ) L p 2 ( R n ) . Then, by (2.14), (2.15) and the generalized Hölder inequality, we have
B m ( f , g ) p 3 , υ s R n f ( x α t ) g ( x β t ) p 3 , υ s d | μ | ( t ) R n f ( x α t ) υ s p 1 g ( x β t ) p 2 d | μ | ( t ) R n υ s ( α t ) f p 1 , ω 1 g p 2 d | μ | ( t ) f p 1 , ω 1 g p 2 , ω 2 R n υ s ( α t ) d | μ | ( t ) .
(2.16)
Now, suppose that | α | 1 . Then we write
R n υ s ( α t ) d | μ | ( t ) = R n ( 1 + | α t | ) s d | μ | ( t ) R n ( 1 + | t | ) s d | μ | ( t ) = μ υ s .
Hence by (2.16)
B m ( f , g ) p 3 , υ s f p 1 , ω 1 g p 2 , ω 2 μ υ s .
(2.17)
Thus m BM ( ω 1 , ω 2 , υ s ) and by (2.17), we have
m ( ω 1 , ω 2 , υ s ) = sup { B m ( f , g ) p 3 , υ s f p 1 , ω 1 g p 2 , ω 2 : f p 1 , ω 1 1 , g p 2 , ω 2 1 } μ υ s .
Similarly, if | α | > 1 , then we write
R n υ s ( α t ) d | μ | ( t ) < R n ( | α | + | α | | t | ) s d | μ | ( t ) = | α | s R n υ s ( t ) d | μ | ( t ) = | α | s μ υ s .
Again, by (2.16) we have
B m ( f , g ) p 3 , υ s | α | s f p 1 , ω 1 g p 2 , ω 2 μ υ s .
(2.18)
Hence, we obtain m BM ( ω 1 , ω 2 , υ s ) and by (2.18)
m ( ω 1 , ω 2 , υ s ) = sup { B m ( f , g ) p 3 , υ s f p 1 , ω 1 g p 2 , ω 2 : f p 1 , ω 1 1 , g p 2 , ω 2 1 } | α | s μ υ s .

 □

Theorem 2.4 Let m BM ( ω 1 , ω 2 , ω 3 ) .
  1. (a)
    T ( ξ 0 , η 0 ) m BM ( ω 1 , ω 2 , ω 3 ) for each ( ξ 0 , η 0 ) R 2 n and
    T ( ξ 0 , η 0 ) m ( ω 1 , ω 2 , ω 3 ) = m ( ω 1 , ω 2 , ω 3 ) .
     
  2. (b)
    M ( ξ 0 , η 0 ) m BM ( ω 1 , ω 2 , ω 3 ) for each ( ξ 0 , η 0 ) R 2 n and
    M ( ξ 0 , η 0 ) m ( ω 1 , ω 2 , ω 3 ) ω 1 ( ξ 0 ) ω 2 ( η 0 ) m ( ω 1 , ω 2 , ω 3 ) .
     
Proof (a) Let us take any f L ω 1 p 1 ( R n ) and g L ω 2 p 2 ( R n ) . If we say that ξ ξ 0 = u and η η 0 = v , then
B T ( ξ 0 , η 0 ) m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) T ( ξ 0 , η 0 ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η = R n R n f ˆ ( u + ξ 0 ) g ˆ ( v + η 0 ) m ( u , v ) e 2 π i u + ξ 0 , x e 2 π i v + η 0 , x d u d v = R n R n T ξ 0 f ˆ ( u ) T η 0 g ˆ ( v ) m ( u , v ) e 2 π i ξ 0 + η 0 , x e 2 π i u + v , x d u d v .
(2.19)
By (2.19), we have
B T ( ξ 0 , η 0 ) m ( f , g ) ( x ) = R n R n T ξ 0 f ˆ ( u ) T η 0 g ˆ ( v ) e 2 π i ξ 0 + η 0 , x e 2 π i u + v , x m ( u , v ) d u d v = e 2 π i ξ 0 + η 0 , x R n R n ( M ξ 0 f ) ˆ ( u ) ( M η 0 g ) ˆ ( v ) m ( u , v ) e 2 π i u + v , x d u d v = e 2 π i ξ 0 + η 0 , x B m ( M ξ 0 f , M η 0 g ) ( x ) .
(2.20)
Since m BM ( ω 1 , ω 2 , ω 3 ) , M ξ 0 f p 1 , ω 1 = f p 1 , ω 1 and M η 0 g p 2 , ω 2 = g p 2 , ω 2 are satisfied for all f L ω 1 p 1 ( R n ) and g L ω 2 p 2 ( R n ) . Hence, by (2.20), we have
B T ( ξ 0 , η 0 ) m ( f , g ) p 3 , ω 3 = e 2 π i ξ 0 + η 0 , x B m ( M ξ 0 f , M η 0 g ) p 3 , ω 3 C f p 1 , ω 1 g p 2 , ω 2
for some C > 0 . Thus T ( ξ 0 , η 0 ) m BM ( ω 1 , ω 2 , ω 3 ) . Also, we obtain
T ( ξ 0 , η 0 ) m ( ω 1 , ω 2 , ω 3 ) = B T ( ξ 0 , η 0 ) m = sup { B T ( ξ 0 , η 0 ) m ( f , g ) p 3 , ω 3 f p 1 , ω 1 g p 2 , ω 2 : f p 1 , ω 1 1 , g p 2 , ω 2 1 } = sup { B m ( M ξ 0 f , M η 0 g ) p 3 , ω 3 M ξ 0 f p 1 , ω 1 M η 0 g p 2 , ω 2 : M ξ 0 f p 1 , ω 1 1 , M η 0 g p 2 , ω 2 1 } = B m = m ( ω 1 , ω 2 , ω 3 ) .
  1. (b)
    Let us rewrite the value B m ( f , g ) as follows:
    B M ( ξ 0 , η 0 ) m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) M ( ξ 0 , η 0 ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η = R n R n f ˆ ( ξ ) g ˆ ( η ) e 2 π i ( ξ 0 , η 0 ) , ( ξ , η ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η = R n R n M ξ 0 f ˆ ( ξ ) M η 0 g ˆ ( η ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η = R n R n ( T ξ 0 f ) ˆ ( ξ ) ( T η 0 g ) ˆ ( η ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η = B m ( T ξ 0 f , T η 0 g ) ( x ) .
    (2.21)
     
Also, the inequalities T ξ 0 f p 1 , ω 1 ω 1 ( ξ 0 ) f p 1 , ω 1 and T η 0 g p 2 , ω 2 ω 2 ( η 0 ) g p 2 , ω 2 are satisfied for all f L ω 1 p 1 ( R n ) , g L ω 2 p 2 ( R n ) . Hence, since m BM ( ω 1 , ω 2 , ω 3 ) , by (2.21) we have
B M ( ξ 0 , η 0 ) m ( f , g ) p 3 , ω 3 = B m ( T ξ 0 f , T η 0 g ) p 3 , ω 3 B m T ξ 0 f p 1 , ω 1 T η 0 g p 2 , ω 2 ω 1 ( ξ 0 ) ω 2 ( η 0 ) B m f p 1 , ω 1 g p 2 , ω 2 .
(2.22)
Then M ( ξ 0 , η 0 ) m BM ( ω 1 , ω 2 , ω 3 ) , and by (2.22) we obtain
M ( ξ 0 , η 0 ) m ( ω 1 , ω 2 , ω 3 ) = sup { B M ( ξ 0 , η 0 ) m ( f , g ) p 3 , ω 3 f p 1 , ω 1 g p 2 , ω 2 : f p 1 , ω 1 1 , g p 2 , ω 2 1 } ω 1 ( ξ 0 ) ω 2 ( η 0 ) m ( ω 1 , ω 2 , ω 3 ) .

 □

Lemma 2.2 If υ s is a polynomial-type weight function and f L υ s p ( R n ) , then D t p f L υ s p ( R n ) . Moreover,
D t p f p , υ s f p , υ s if t 1 , D t p f p , υ s < t s f p , υ s if t > 1 .
Proof Let υ s be a polynomial-type weight function and f L υ s p ( R n ) . Assume that t 1 . If we get x t = u ,
D t p f p , υ s = { R n | D t p f ( x ) | p υ s ( x ) p d x } 1 p = { R n | t n p f ( x t ) | p ( 1 + | x | ) s p d x } 1 p = { R n | f ( u ) | p ( 1 + | u t | ) s p d u } 1 p { R n | f ( u ) | p ( 1 + | u | ) s p d u } 1 p = f p , υ s < .
(2.23)

Thus we have D t p f L υ s p ( R n ) and D t p f p , υ s f p , υ s .

Now, assume that t > 1 . Similarly by (2.23)
D t p f p , υ s = { R n | f ( u ) | p ( 1 + | u t | ) s p d u } 1 p < { R n | f ( u ) | p ( t + | u t | ) s p d u } 1 p = t s { R n | f ( u ) | p ( 1 + | u | ) s p d u } 1 p = t s f p , υ s < .

Hence D t p f L υ s p ( R n ) , and we also have D t p f p , υ s < t s f p , υ s . □

Theorem 2.5 Let υ s 1 , υ s 2 , υ s 3 be weight functions of polynomial type and let m BM ( υ s 1 , υ s 2 , υ s 3 ) . If 2 q = 1 p 1 + 1 p 2 1 p 3 and 0 < t < , then D t q m BM ( υ s 1 , υ s 2 , υ s 3 ) . Moreover, then
D t q m ( υ s 1 , υ s 2 , υ s 3 ) ( 1 t ) s 3 m ( υ s 1 , υ s 2 , υ s 3 ) if t 1 , D t q m ( υ s 1 , υ s 2 , υ s 3 ) < t s 1 + s 2 m ( υ s 1 , υ s 2 , υ s 3 ) if t > 1 .
Proof Let f L υ s 1 p 1 ( R n ) and g L υ s 2 p 2 ( R n ) be given. We know by Lemma 2.2 that D t p 1 f L υ s 1 p 1 ( R n ) and D t p 2 g L υ s 2 p 2 ( R n ) . If we get ξ t = u and η t = v , we obtain
B D t q m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) D t q m ( ξ , η ) e 2 π i ξ + η , x d ξ d η = R n R n f ˆ ( t u ) g ˆ ( t v ) t 2 n q m ( u , v ) e 2 π i u + v , t x t 2 n d u d v .
Hence, from the equality 2 q = 1 p 1 + 1 p 2 1 p 3 , we have
B D t q m ( f , g ) ( x ) = R n R n f ˆ ( t u ) g ˆ ( t v ) t n ( 1 p 1 + 1 p 2 1 p 3 ) m ( u , v ) e 2 π i u + v , t x t 2 n d u d v = R n R n t n ( 1 1 p 1 ) f ˆ ( t u ) t n ( 1 1 p 1 ) g ˆ ( t v ) t n p 3 m ( u , v ) e 2 π i u + v , t x t 2 n d u d v = t n p 3 R n R n D t 1 p 1 f ˆ ( u ) D t 1 p 2 g ˆ ( v ) m ( u , v ) e 2 π i u + v , t x d u d v = t n p 3 R n R n ( D t p 1 f ) ˆ ( u ) ( D t p 2 g ) ˆ ( v ) m ( u , v ) e 2 π i u + v , t x d u d v = D t 1 p 3 B m ( D t p 1 f , D t p 2 g ) ( x ) .
(2.24)
Assume that t 1 . Since m B m ( υ s 1 , υ s 2 , υ s 3 ) , by Lemma 2.2 and using equality (2.24), we obtain
B D t q m ( f , g ) p 3 , υ s 3 = D t 1 p 3 B m ( D t p 1 f , D t p 2 g ) ( x ) p 3 , υ s 3 ( 1 t ) s 3 B m ( D t p 1 f , D t p 2 g ) ( x ) p 3 , υ s 3 ( 1 t ) s 3 B m D t p 1 f p , υ s 1 D t p 2 g p , υ s 2 ( 1 t ) s 3 m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 .
(2.25)
Then D t q m BM ( υ s 1 , υ s 2 , υ s 3 ) , and by (2.25)
D t q m ( υ s 1 , υ s 2 , υ s 3 ) ( 1 t ) s 3 m ( υ s 1 , υ s 2 , υ s 3 ) .
Now let t > 1 . Again, since m BM ( υ s 1 , υ s 2 , υ s 3 ) , by Lemma 2.2 and using equality (2.24), we obtain
B D t q m ( f , g ) p 3 , υ s 3 < B m ( D t p 1 f , D t p 2 g ) p 3 , υ s 3 B m D t p 1 f p , υ s 1 D t p 2 g p , υ s 2 < t s 1 + s 2 B m f p 1 , υ s 1 g p 2 , υ s 2 = t s 1 + s 2 m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 .
(2.26)
Thus D t q m BM ( υ s 1 , υ s 2 , υ s 3 ) and by (2.26)
D t q m ( υ s 1 , υ s 2 , υ s 3 ) < t s 1 + s 2 m ( υ s 1 , υ s 2 , υ s 3 ) .

 □

Theorem 2.6 Let υ s 1 , υ s 2 , υ s 3 be weight functions of polynomial type and let m BM ( υ s 1 , υ s 2 , υ s 3 ) such that m ( t ξ , t η ) = m ( ξ , η ) for any t > 0 , where 2 q = 1 p 1 + 1 p 2 1 p 3 . Then
2 q < s 3 n if t < 1 , 2 q > s 1 + s 2 n if t > 1 .
Proof Take any f L υ s 1 p 1 ( R n ) , g L υ s 2 p 2 ( 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 R n .
(2.27)
On the other hand, using m ( t ξ , t η ) = m ( ξ , η ) and changing the variables t u = ξ , t v = η , we note that
(2.28)
Hence by (2.27) and (2.28), we have
B m ( f , g ) ( x ) = t n ( 1 p 3 1 p 1 1 p 2 ) B D t q m ( f , g ) ( x ) .
Since D t q m = m for t = 1 , we let t 1 . Assume first that t < 1 . Also, since m BM ( υ s 1 , υ s 2 , υ s 3 ) , by Theorem 2.5 we have D t q m BM ( υ s 1 , υ s 2 , υ s 3 ) and D t q m ( υ s 1 , υ s 2 , υ s 3 ) < ( 1 t ) s 3 m ( υ s 1 , υ s 2 , υ s 3 ) . Then by (2.28)
B m ( f , g ) ( x ) p 3 , υ s 3 = t n ( 1 p 3 1 p 1 1 p 2 ) B D t q ( f , g ) ( x ) p 3 , υ s 3 t n ( 1 p 3 1 p 1 1 p 2 ) D t q m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 < t n ( 1 p 3 1 p 1 1 p 2 ) s 3 m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 = t n ( 1 p 3 1 p 1 1 p 2 ) s 3 B m f p 1 , υ s 1 g p 2 , υ s 2 .
Thus,
B m < t n ( 1 p 3 1 p 1 1 p 2 ) s 3 B m = t 2 n q s 3 B m .

Hence 1 < t 2 n q s 3 . Since t < 1 , we have 2 n q s 3 < 0 . Thus, we write 2 q < s 3 n .

Assume now that t > 1 . Again, by Theorem 2.5, we have D t q m BM ( υ s 1 , υ s 2 , υ s 3 ) and D t q m ( υ s 1 , υ s 2 , υ s 3 ) < t s 1 + s 2 m ( υ s 1 , υ s 2 , υ s 3 ) . Similarly,
B m ( f , g ) ( x ) p 3 , υ s 3 < t n ( 1 p 3 1 p 1 1 p 2 ) + s 1 + s 2 B m f p 1 , υ s 1 g p 2 , υ s 2 .
Thus, we have
B m < t n ( 1 p 3 1 p 1 1 p 2 ) + s 1 + s 2 B m = t 2 n q + s 1 + s 2 B m .

Hence 1 < t 2 n q + s 1 + s 2 Since t > 1 , we have 2 n q + s 1 + s 2 > 0 . Thus, we write 2 q > s 1 + s 2 n . □

Theorem 2.7 Let m BM ( ω 1 , ω 2 , ω 3 ) and p 3 1 .
  1. (a)
    If Φ L 1 ( R n ) , then Φ m BM ( ω 1 , ω 2 , ω 3 ) and
    Φ m ( ω 1 , ω 2 , ω 3 ) Φ 1 m ( ω 1 , ω 2 , ω 3 ) .
     
  2. (b)
    If Φ L ω 1 ( R n ) such that ω ( u , υ ) = ω 1 ( u ) ω 2 ( υ ) , then Φ ˆ m BM ( ω 1 , ω 2 , ω 3 ) and
    Φ ˆ m ( ω 1 , ω 2 , ω 3 ) Φ 1 , ω m ( ω 1 , ω 2 , ω 3 ) .
     
Proof (a) Let f L ω 1 p 1 ( R n ) and g L ω 2 p 2 ( R n ) . Since L ω 1 p 1 ( R n ) L p 1 ( R n ) and L ω 2 p 2 ( R n ) L p 2 ( R n ) , then by Proposition 2.5 in [11]
B Φ m ( f , g ) ( x ) = R n R n Φ ( u , v ) B T ( u , v ) m ( f , g ) ( x ) d u d v .
Also, since m BM ( ω 1 , ω 2 , ω 3 ) , we have T ( u , v ) m BM ( ω 1 , ω 2 , ω 3 ) by Theorem 2.4. So, we write
B Φ m ( f , g ) p 3 , ω 3 R n R n Φ ( u , v ) B T ( u , v ) m ( f , g ) p 3 , ω 3 d u d v R n R n | Φ ( u , v ) | T ( u , v ) m ( ω 1 , ω 2 , ω 3 ) f p 1 , ω 1 g p 2 , ω 2 d u d v = m ( ω 1 , ω 2 , ω 3 ) Φ 1 f p 1 , ω 1 g p 2 , ω 2 < .
(2.29)
Hence Φ m BM ( ω 1 , ω 2 , ω 3 ) . Finally, by (2.29), we obtain
Φ m ( ω 1 , ω 2 , ω 3 ) Φ 1 m ( ω 1 , ω 2 , ω 3 ) .
  1. (b)
    Let Φ L ω 1 ( R n ) . Take any f L ω 1 p 1 ( R n ) and g L ω 2 p 2 ( R n ) . It is known by Proposition 2.5 in [11] that the equality
    B Φ ˆ m ( f , g ) ( x ) = R n R n Φ ( u , v ) B M ( u , v ) m ( f , g ) ( x ) d u d v .
     
Since m BM ( ω 1 , ω 2 , ω 3 ) , by Theorem 2.4 we have M ( u , v ) m BM ( ω 1 , ω 2 , ω 3 ) and
M ( u , v ) m ( ω 1 , ω 2 , ω 3 ) ω 1 ( u ) ω 2 ( υ ) m ( ω 1 , ω 2 , ω 3 ) .
Then we write
B Φ ˆ m ( f , g ) p 3 , ω 3 R n R n Φ ( u , v ) B M ( u , v ) m ( f , g ) p 3 , ω 3 d u d v R n R n | Φ ( u , v ) | M ( u , v ) m ( ω 1 , ω 2 , ω 3 ) f p 1 , ω 1 g p 2 , ω 2 d u d v R n R n | Φ ( u , v ) | ω 1 ( u ) ω 2 ( υ ) m ( ω 1 , ω 2 , ω 3 ) f p 1 , ω 1 g p 2 , ω 2 d u d v = m ( ω 1 , ω 2 , ω 3 ) f p 1 , ω 1 g p 2 , ω 2 Φ 1 , ω .
(2.30)
Thus from (2.30), we obtain Φ ˆ m BM ( ω 1 , ω 2 , ω 3 ) and
Φ ˆ m ( ω 1 , ω 2 , ω 3 ) Φ 1 , ω m ( ω 1 , ω 2 , ω 3 ) .

 □

Theorem 2.8 Let υ s 1 , υ s 2 , υ s 3 be weight functions of polynomial type and let m BM ( υ s 1 , υ s 2 , υ s 3 ) . If Ψ L 1 ( R + , t 2 n q d t ) such that 2 q = 1 p 1 + 1 p 2 1 p 3 , then m Ψ ( ξ , η ) = 0 m ( t ξ , t η ) Ψ ( t ) d t BM ( υ s 1 , υ s 2 , υ s 3 ) . Moreover,
m Ψ ( υ s 1 , υ s 2 , υ s 3 ) < Ψ L 1 ( R + , t 2 n q d t ) m ( υ s 1 , υ s 2 , υ s 3 ) .
Proof Let us take f , g S ( R n ) . Then
B m Ψ ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) m Ψ ( ξ , η ) e 2 π i u + v , x d ξ d η = R n R n f ˆ ( ξ ) g ˆ ( η ) { 0 m ( t ξ , t η ) Ψ ( t ) d t } e 2 π i u + v , x d ξ d η = R n R n f ˆ ( ξ ) g ˆ ( η ) { 0 D t 1 q m ( ξ , η ) Ψ ( t ) t 2 n q d t } e 2 π i u + v , x d ξ d η = 0 B D t 1 q m ( f , g ) Ψ ( t ) t 2 n q d t .
Since m BM ( υ s 1 , υ s 2 , υ s 3 ) , D t 1 q m BM ( υ s 1 , υ s 2 , υ s 3 ) by Theorem 2.5, thus we observe that
B m Ψ ( f , g ) ( x ) p 3 , υ s 3 0 B D t 1 q m ( f , g ) p 3 , υ s 3 | Ψ ( t ) | t 2 n q d t 0 B D t 1 q m f p 1 , υ s 1 g p 2 , υ s 2 | Ψ ( t ) | t 2 n q d t = 0 D t 1 q m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 | Ψ ( t ) | t 2 n q d t < 0 1 t s 3 m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 | Ψ ( t ) | t 2 n q d t + 1 t s 1 s 2 m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 | Ψ ( t ) | t 2 n q d t = m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 × { 0 1 t s 3 | Ψ ( t ) | t 2 n q d t + 1 t s 1 s 2 | Ψ ( t ) | t 2 n q d t } .
(2.31)
Also, since t s 3 1 for s 3 0 , t 1 and t s 1 s 2 < 1 for s 1 s 2 0 , t > 1 , by (2.31)
B m Ψ ( f , g ) ( x ) p 3 , υ s 3 < Ψ L 1 ( R + , t 2 n q d t ) m ( υ s 1 , υ s 2 , υ s 3 ) f p 1 , υ s 1 g p 2 , υ s 2 .
Hence, m Ψ BM ( υ s 1 , υ s 2 , υ s 3 ) and
m Ψ ( υ s 1 , υ s 2 , υ s 3 ) < Ψ L 1 ( R + , t 2 n q d t ) m ( υ s 1 , υ s 2 , υ s 3 ) .

 □

Theorem 2.9 Let p 3 1 and m BM ( ω 1 , ω 2 , ω 3 ) . If Q 1 , Q 2 are bounded measurable sets in R n , then
h ( ξ , η ) = 1 μ ( Q 1 × Q 2 ) Q 1 × Q 2 m ( ξ + u , η + v ) d u d v BM ( ω 1 , ω 2 , ω 3 ) .
Proof Take any f , g S ( R n ) . Then we write
B h ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) h ( ξ , η ) e 2 π i ξ + η , x d ξ d η = 1 μ ( Q 1 × Q 2 ) Q 1 × Q 2 { R n R n f ˆ ( ξ ) g ˆ ( η ) m ( ξ + u , η + v ) e 2 π i ξ + η , x d ξ d η } d u d v = 1 μ ( Q 1 × Q 2 ) Q 1 × Q 2 B T ( u , v ) m ( f , g ) ( x ) d u d v .
By using Theorem 2.4, we have
B h ( f , g ) p 3 , ω 3 1 μ ( Q 1 × Q 2 ) Q 1 × Q 2 B T ( u , v ) m ( f , g ) p 3 , ω 3 d u d v 1 μ ( Q 1 × Q 2 ) Q 1 × Q 2 T ( u , v ) m ( ω 1 , ω 2 , ω 3 ) f p 1 , ω 1 g p 2 , ω 2 d u d v = 1 μ ( Q 1 × Q 2 ) m ( ω 1 , ω 2 , ω 3 ) f p 1 , ω 1 g p 2 , ω 2 μ ( Q 1 × Q 2 ) = m ( ω 1 , ω 2 , ω 3 ) f p 1 , ω 1 g p 2 , ω 2 .

Hence, we obtain h ( ξ , η ) BM ( ω 1 , ω 2 , ω 3 ) . □

Theorem 2.10 Let ω ( u , v ) = ω 1 ( u ) ω 2 ( v ) , ω 3 ω 1 , ω 3 ( u ) = ω 3 ( u ) and 1 p 1 + 1 p 2 = 1 p 3 1 . Assume that Φ L ω 1 ( R 2 n ) , Ψ 1 L ω 1 p 1 ( R n ) and Ψ 2 L ω 2 p 2 ( R n ) . If m ( ξ , η ) = Ψ ˆ 1 ( ξ ) Φ ˆ ( ξ , η ) Ψ ˆ 2 ( η ) , then m BM ( 1 , ω 1 ; 1 , ω 2 ; p 3 , ω 3 ) .

Proof For the proof we will use Theorem 2.2. Take any f , g , h S ( R n ) . Then
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | = | R n h ( y ) { R n R n f ˆ ( ξ ) g ˆ ( η ) Ψ ˆ 1 ( ξ ) Φ ˆ ( ξ , η ) Ψ ˆ 2 ( η ) e 2 π i ξ + η , x d ξ d η } d y | R n | h ( y ) ω 3 1 ( y ) B Φ ˆ ( f Ψ 1 , g Ψ 2 ) ( y ) ω 3 ( y ) | d y .
(2.32)
Since the spaces L ω 1 p 1 ( R n ) and L ω 2 p 2 ( R n ) are Banach convolution module over the spaces L ω 1 1 ( R n ) , L ω 2 1 ( R n ) respectively, we write f Ψ 1 L ω 1 p 1 ( R n ) and g Ψ 2 L ω 2 p 2 ( R n ) . Also, by Theorem 2.7, Φ ˆ BM ( p 1 , ω 1 ; p 2 , ω 2 ; p 3 , ω 3 ) . Therefore we obtain B Φ ˆ ( f Ψ 1 , g Ψ 2 ) L ω 3 p 3 ( R n ) . By using the Hölder inequality and the inequality (2.32), we find
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | h p 3 1 , ω 3 1 B Φ ˆ ( f Ψ 1 , g Ψ 2 ) p 3 , ω 3 h p 3 1 , ω 3 1 B Φ ˆ f 1 , ω 1 Ψ 1 p 1 , ω 1 g 1 , ω 2 Ψ 2 p 2 , ω 2 .
If we say C = B Φ ˆ Ψ 1 p 1 , ω 1 Ψ 2 p 2 , ω 2 , then we obtain
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | C f 1 , ω g 1 , ω 2 h p 3 1 , ω 3 1 ,

which means m BM ( 1 , ω 1 ; 1 , ω 2 ; p 3 , ω 3 ) . □

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

Theorem 2.11 Let ω ( u , v ) = ω 1 ( u ) ω 2 ( v ) , ω 3 ω 1 , ω 3 ( u ) = ω 3 ( u ) and 1 p 1 + 1 p 2 = 1 p 3 1 . If m ( ξ , η ) = Ψ ˆ 1 ( ξ ) Φ ˆ ( ξ , η ) Ψ ˆ 2 ( η ) such that Φ L ω 1 ( R 2 n ) , Ψ 1 L ω 1 1 ( R n ) and Ψ 2 L ω 2 1 ( R n ) , then m BM ( p 1 , ω 1 ; p 2 , ω 2 ; p 3 , ω 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 ) P ( R n ) and let p 1 < , p 2 < , p 3 < . Assume that m ( ξ , η ) is a bounded function on R n × R n . Define
B m ( f , g ) ( x ) = R n R n f ˆ ( ξ ) g ˆ ( η ) m ( ξ , η ) e 2 π i ξ + η , x d ξ d η

for all f , g C c ( R n ) .

m is said to be a bilinear multiplier on R n of type ( p 1 ( x ) , p 2 ( x ) , p 3 ( x ) ) if there exists C > 0 such that
B m ( f , g ) p 3 ( x ) C f p 1 ( x ) g p 2 ( x )

for all f , g C c ( R n ) , i.e., B m extends to a bounded bilinear operator from L p 1 ( x ) ( R n ) × L p 2 ( x ) ( R n ) to L p 3 ( x ) ( 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 m ( p 1 ( x ) , p 2 ( x ) , p 3 ( x ) ) = B m .

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 1 p 3 ( x ) + 1 q ( x ) = 1 for all x R n . Then m BM ( p 1 ( x ) , p 2 ( x ) , p 3 ( x ) ) if and only if there exists C > 0 such that
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) m ( ξ , η ) d ξ d η | C f p 1 ( x ) g p 2 ( x ) h q ( x )

for all f , g , h C c ( R n ) .

Theorem 3.2 Let 1 p ( x ) + 1 q ( x ) = 1 r . If Φ L 1 ( R n ) , then m ( ξ , η ) = Φ ˆ ( ξ + η ) BM ( p ( x ) , q ( x ) , r ) .

Proof Take any f , g , h C c ( R n ) . Then
| R n R n f ˆ ( ξ ) g ˆ ( η ) h ˆ ( ξ + η ) Φ ˆ ( ξ + η ) d ξ d η | = | R n R n f ˆ ( ξ ) g ˆ ( η ) ( h Φ ) ˆ ( ξ + η ) d ξ d η | = | R n ( h Φ ) ( x ) { R n R n f ˆ ( ξ ) g ˆ ( η ) e 2 π i ξ + η , x d ξ d η } d x | R n | ( h Φ ) ( x ) | | B ˜ 1 ( f , g ) ( x ) | d x .
(3.1)
Since the space L r ( R