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Bilinear multipliers of weighted Lebesgue spaces and variable exponent Lebesgue spaces

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 dξdη

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.

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 1p, L p ( R n ) denotes the usual Lebesgue space. A continuous function ω satisfying 1ω(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 s0. Let f be a measurable function on R n . If there exist C>0 and NN 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 1p, 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 ,1p<

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(yx), 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 1p 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 ωd|μ|<.

If μM( 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 (λf)= R n Ω |λf(x) | p ( x ) dx+ ess  sup x Ω ( λ f ( x ) ) <

for some λ=λ(f)>0. If p <, then

ϱ p (λf)= R n Ω |λf(x) | p ( x ) dx,[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 1p< and let ω be a slowly increasing weight function. Then S( R n ) is dense in L ω p ( R n ).

Proof Let fS( R n ) be given. Since ω is a slowly increasing weight function, there exist C>0 and NN 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 fm 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 m0 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 hS( 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 dξdη

for all f,gS( 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,gS( 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 (ξη) defines a bilinear multiplier and m ( ω 1 , ω 2 , ω 3 ) K 1 , ω 3 .

Proof For f,gS( R n ), we have f(xy)= R n f ˆ (ξ) e 2 π i x y , ξ dξ and g(x+y)= R n g ˆ (η) e 2 π i x + y , η dη. 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,gS( 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(xy) 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 mBM( 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(ξη)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(ξη) e 2 π i ξ + η , x dξdη

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 mBM( 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)dξdη|C f p 1 , ω 1 g p 2 , ω 2 h p 3 , ω 3 1

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

Proof Let mBM( ω 1 , ω 2 , ω 3 ). We take any f,g,hS( 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 mBM( 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, hS( 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 mBM( 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)dξdη|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)dξdη|C f p 1 , ω 1 g p 2 , ω 2 h p 3 , ω 3 1

for all f,g,hS( R n ). From the assumption and (2.10), we write

| R n h(y) B ˜ m (f,g)(y)dy|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)dy.

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 mBM( 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 , s0 be a weight function of polynomial type such that υ s ω 1 . If μM( υ s ) and m(ξ,η)= μ ˆ (αξ+βη) for α,βR, then mBM( 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(xαt) υ s L p 1 ( R n ) and

f ( x α t ) υ s p 1 υ s (αt) f p 1 , ω 1 .
(2.15)

Also, g(xβt) 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 mBM( 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 mBM( 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 mBM( ω 1 , ω 2 , ω 3 ).

  1. (a)

    T ( ξ 0 , η 0 ) mBM( 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 ) mBM( 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 mBM( 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 mBM( 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 ) mBM( 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 t1. 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 mBM( 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 mBM( 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 t1. 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 mBM( 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 mBM( 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 mBM( 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 mBM( υ s 1 , υ s 2 , υ s 3 ) such that m(tξ,tη)=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(tξ,tη)=m(ξ,η) and changing the variables tu=ξ, tv=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 t1. Assume first that t<1. Also, since mBM( υ s 1 , υ s 2 , υ s 3 ), by Theorem 2.5 we have D t q mBM( 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 mBM( 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 mBM( ω 1 , ω 2 , ω 3 ) and p 3 1.

  1. (a)

    If Φ L 1 ( R n ), then ΦmBM( 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 mBM( 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)dudv.

Also, since mBM( ω 1 , ω 2 , ω 3 ), we have T ( u , v ) mBM( 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 ΦmBM( 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)dudv.

Since mBM( ω 1 , ω 2 , ω 3 ), by Theorem 2.4 we have M ( u , v ) mBM( 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 mBM( 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 mBM( υ s 1 , υ s 2 , υ s 3 ). If Ψ L 1 ( R + , t 2 n q dt) such that 2 q = 1 p 1 + 1 p 2 1 p 3 , then m Ψ (ξ,η)= 0 m(tξ,tη)Ψ(t)dtBM( 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,gS( 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 mBM( υ s 1 , υ s 2 , υ s 3 ), D t 1 q mBM( 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, t1 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 mBM( 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)dudvBM( ω 1 , ω 2 , ω 3 ).

Proof Take any f,gS( 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 mBM(1, ω 1 ;1, ω 2 ; p 3 , ω 3 ).

Proof For the proof we will use Theorem 2.2. Take any f,g,hS( 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