Let b be a locally integrable function on R n and T be an integral operator. For a suitable function f, the commutator generated by b and T is defined by [ b , T ] f = b T ( f ) − T ( b f ) . The investigation of the commutator begins with Coifman-Rochberg-Weiss pioneering study and classical result (see [1] ). The major reason for considering the problem of commutators is that the boundedness of commutator can produce some characterizations of function spaces (see [1] [2] ). Now, with the development of the Calderón-Zygmund singular integral operators, their commutators and multilinear operators have been well studied (see [1] [3] - [7] ). In [8] , Hu and Yang proved a variant sharp function estimate for the multilinear singular integral operators. In [9] [10] [11] [12] , C. Pérez, G. Pradolini and R. Trujillo-Gonzalez obtained a sharp weighted estimate for the singular integral operators and their commutators. The boundedness of the pseudo-differential operators was studied by many authors (see [13] - [21] ). In [15] , the boundedness of the commutators associated to the pseudo-differential operators is obtained. The main purpose of this paper is to study the multilinear pseudo-differential operators as follows.

We say a symbol σ ( x , ξ ) is in the class S ρ , δ m or σ ∈ S ρ , δ m , if for x , ξ ∈ R n ,

| ∂ α ∂ x α ∂ β ∂ ξ β σ ( x , ξ ) | ≤ C α , β ( 1 + | ξ | ) m − ρ | β | + δ | α | .

A pseudo-differential operator with symbol σ ( x , ξ ) ∈ S ρ , δ m is defined by

T ( f ) ( x ) = ∫ R n     e 2 π i x ⋅ ξ σ ( x , ξ ) f ^ ( ξ ) d ξ ,

where f is a Schwartz function and f ^ denotes the Fourier transform of f. We know there exists a kernel K ( x , y ) such that

T ( f ) ( x ) = ∫ R n     K ( x , x − y ) f ( y ) d y ,

where, formally,

K ( x , y ) = ∫ R n     e 2 π i ( x − y ) ⋅ ξ σ ( x , ξ ) d ξ .

In [14] , the boundedness of the pseudo-differential operators with symbol σ ∈ S 1 − θ , δ − β ( β < n θ / 2 , 0 ≤ δ < 1 − θ ) is obtained. In [14] , the boundedness of the pseudo-differential operators with symbol of order 0 and − ∞ is obtained. In [17] , the sharp function estimate of the pseudo-differential operators with symbol σ ∈ S 1 − θ , δ − n θ / 2 ( 0 < θ < 1 , 0 ≤ δ < 1 − θ ) is obtained. In [15] , the boundedness of the pseudo-differential operators and their commutators with symbol σ ∈ S 1 − θ , δ − n θ / 2 ( 0 < θ < 1 , 0 ≤ δ < 1 − θ ) is obtained. Our results are motivated by these papers.

Suppose T is a pseudo-differential operator with symbol σ ( x , ξ ) ∈ S ρ , δ m . Let m j be the positive integers ( j = 1 , ⋯ , l ), m 1 + ⋯ + m l = L and b j be the functions on R n ( j = 1 , ⋯ , l ). Set, for 1 ≤ j ≤ m ,

R m j + 1 ( b j ; x , y ) = b j ( x ) − ∑ | α | ≤ m j 1 α ! D α b j ( y ) ( x − y ) α .

The multilinear operator associated to T is defined by

T b ( f ) ( x ) = ∫ R n ∏ j = 1 l R m j + 1 ( b j ; x , y ) | x − y | L K ( x , x − y ) f ( y ) d y .

Note that when L = 0 , T b is just the multilinear commutator of T and b j (see [11] ). While when L > 0 , T b is non-trivial generalizations of the commutator. It is well known that multilinear operators are of great interest in harmonic analysis and have been widely studied by many authors. Hu and Yang (see [8] ) proved a variant sharp estimate for the multilinear singular integral operators. In [11] , Pérez and Trujillo-Gonzalez prove a sharp estimate for the multilinear commutator when b j ∈ O s c e x p L r j ( R n ) . The main purpose of this paper is to prove a sharp function inequality for the multilinear operators associated to the pseudo-differential operators with symbol σ ∈ S 1 − θ , δ − n θ / 2 ( 0 < θ < 1 , 0 ≤ δ < 1 − θ ) when D α b j ∈ B M O ( R n ) for all α with | α | = m j . As the application, we obtain the L p ( p > 1 ) norm inequality for the multilinear operators.

First, let us introduce some notations. Throughout this paper, Q = Q ( x , d ) will denote a cube of R n with sides parallel to the axes, whose center is x and side length is d. For a locally integrable function b, the sharp function of b is defined by

b # ( x ) = sup Q ∋ x 1 | Q | ∫ Q | b ( y ) − b Q | d y ,

where, and in what follows, b Q = | Q | − 1 ∫ Q     b ( x ) d x . It is well-known that (see [22] [23] [24] )

b # ( x ) ≈ sup Q ∋ x inf c ∈ C 1 | Q | ∫ Q | b ( y ) − c | d y

and

‖ b − b 2 k Q ‖ B M O ≤ C k ‖ b ‖ B M O       for     k ≥ 1.

We say that b belongs to B M O ( R n ) if b # belongs to L ∞ ( R n ) and ‖ b ‖ B M O = ‖ b # ‖ L ∞ . Let M be the Hardy-Littlewood maximal operator defined by

M ( f ) ( x ) = sup x ∈ Q 1 | Q | ∫ Q | f ( y ) | d y ,

we write that M p ( f ) = ( M ( f p ) ) 1 / p for 0 < p < ∞ .

We shall prove the following theorems.

Theorem 1. Suppose T is the pseudo-differential operator with symbol σ ∈ S 1 − θ , δ − n θ / 2 ( 0 < θ < 1 , 0 ≤ δ < 1 − θ ) . Let D α b j ∈ B M O ( R n ) for all α with | α | = m j and j = 1 , ⋯ , l . Then there exists a constant C > 0 such that for every f ∈ C 0 ∞ ( R n ) , 2 < r < ∞ and x ˜ ∈ R n ,

( T b ( f ) ) # ( x ˜ ) ≤ C ∏ j = 1 l ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) M r ( f ) ( x ˜ ) .

Theorem 2. Suppose T is the pseudo-differential operator with symbol σ ∈ S 1 − θ , δ − n θ / 2 ( 0 < θ < 1 , 0 ≤ δ < 1 − θ ) . Let D α b j ∈ B M O ( R n ) for all α with | α | = m j and j = 1 , ⋯ , l .

1) If w ∈ A ∞ and 2 < p < ∞ , then

‖ T b ( f ) ‖ L p ( w ) ≤ C ∏ j = 1 l ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) ‖ M r ( f ) ‖ L p ( w ) .

2) If w ∈ A 1 and 2 < p < ∞ , then

‖ T b ( f ) ‖ L p ( w ) ≤ C ∏ j = 1 l ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) ‖ f ‖ L p ( w ) .

To prove the theorems, we need the following lemmas.

Lemma 1. (see [4] ) Let b be a function on R n and D α b ∈ L q ( R n ) for all α with | α | = L and some q > n . Then

| R L ( b ; x , y ) | ≤ C | x − y | L ∑ | α | = L ( 1 | Q ˜ ( x , y ) | ∫ Q ˜ ( x , y ) | D α b ( z ) | q d z ) 1 / q ,

where Q ˜ is the cube centered at x and having side length 5 n | x − y | .

Lemma 2. (see [13] ) Let T be the pseudo-differential operator with symbol σ ∈ S 1 − θ , δ − n θ / 2 ( 0 < θ < 1 , 0 ≤ δ < 1 − θ ) . Then, for every f ∈ L p ( R n ) , 1 < p < ∞ ,

‖ T ( f ) ‖ L p ≤ C ‖ f ‖ L p .

Lemma 3. (see [13] ) Let σ ∈ S 1 − θ , δ − n θ / 2 ( 0 < θ < 1 , 0 ≤ δ < 1 − θ ) and K be the kernel of the pseudo-differential operator T with symbol σ ∈ S 1 − θ , δ − n θ / 2 . Then, for | x 0 − x | ≤ d < 1 and k ≥ 1 ,

( ∫ ( 2 k d ) 1 − θ ≤ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | K ( x , x − y ) − K ( x 0 , x 0 − y ) | 2 d y ) 1 / 2 ≤ C | x 0 − x | ( 1 − θ ) ( m − n / 2 ) ( 2 k d ) m ( 1 − θ ) ,

provided m is an integer such that n / 2 < m < n / 2 + 1 / ( 1 − θ ) .

Lemma 4. (see [13] ) Let σ ∈ S ρ , δ 0 ( 0 < ρ < 1 ) and

K ( x , w ) = ∫ R n     e 2 π i w ⋅ ξ σ ( x , ξ ) d ξ .

Then, for | w | ≥ 1 / 4 and any integer N ≥ 1 ,

| K ( x , w ) | ≤ C N | w | − 2 N .

Proof of Theorem 1. It suffices to prove for f ∈ C 0 ∞ ( R n ) and some constant C 0 , the following inequality holds:

1 | Q | ∫ Q | T b ( f ) ( x ) − C 0 | d x ≤ C ∏ j = 1 l ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) M r ( f ) ( x ) .

Without loss of generality, we may assume l = 2 . Fix a cube Q = Q ( x 0 , d ) and x ˜ ∈ Q . We consider the following two cases:

Case 1. d ≤ 1 . In this case, let Q ∗ be the cube concentric with Q of side length d 1 − θ . Let Q ˜ = 5 n Q ∗ and b ˜ j ( x ) = b j ( x ) − ∑ | α | = m j 1 α ! ( D α b j ) Q ˜ x α , then R m j ( b j ; x , y ) = R m j ( b ˜ j ; x , y ) and D α b ˜ j = D α b j − ( D α b j ) Q ˜ for | α | = m j . We write, for f = f χ Q ˜ + f χ R n \ Q ˜ = f 1 + f 2 ,

T b ( f ) ( x ) = ∫ R n ∏ j = 1 2 R m j + 1 ( b ˜ j ; x , y ) | x − y | L K ( x , x − y ) f ( y ) d y = ∫ R n ∏ j = 1 2 R m j ( b ˜ j ; x , y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y     − ∑ | α 1 | = m 1 1 α 1 ! ∫ R n R m 2 ( b ˜ 2 ; x , y ) ( x − y ) α 1 D α 1 b ˜ 1 ( y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y     − ∑ | α 2 | = m 2 1 α 2 ! ∫ R n R m 1 ( b ˜ 1 ; x , y ) ( x − y ) α 2 D α 2 b ˜ 2 ( y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y     + ∑ | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! ∫ R n ( x − y ) α 1 + α 2 D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y     + ∫ R n ∏ j = 1 2 R m j + 1 ( b ˜ j ; x , y ) | x − y | L K ( x , x − y ) f 2 ( y ) d y ,

then

1 | Q | ∫ Q | T b ( f ) ( x ) − T b ˜ ( f 2 ) ( x 0 ) | d x ≤ 1 | Q | ∫ Q | ∫ R n ∏ j = 1 2 R m j ( b ˜ j ; x , y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y | d x       + C | Q | ∫ Q | ∑ | α 1 | = m 1 ∫ R n R m 2 ( b ˜ 2 ; x , y ) ( x − y ) α 1 | x − y | L D α 1 b ˜ 1 ( y ) K ( x , x − y ) f 1 ( y ) d y | d x       + C | Q | ∫ Q | ∑ | α 2 | = m 2 ∫ R n R m 1 ( b ˜ 1 ; x , y ) ( x − y ) α 2 | x − y | L D α 2 b ˜ 2 ( y ) K ( x , x − y ) f 1 ( y ) d y | d x

      + C | Q | ∫ Q | ∑ | α 1 | = m 1 , | α 2 | = m 2 ∫ R n ( x − y ) α 1 + α 2 D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y | d x       + 1 | Q | ∫ Q | T b ˜ ( f 2 ) ( x ) − T b ˜ ( f 2 ) ( x 0 ) | d x : = I 1 + I 2 + I 3 + I 4 + I 5 .

Now, let us estimate I 1 , I 2 , I 3 , I 4 and I 5 , respectively. First, for x ∈ Q and y ∈ Q ˜ , by Lemma 1, we get

R L ( b ˜ j ; x , y ) ≤ C | x − y | L ∑ | α j | = L ‖ D α j b j ‖ B M O .

Now, let σ ( x , ξ ) = σ ( x , ξ ) | ξ | n θ / 2 | ξ | − n θ / 2 = q ( x , ξ ) | ξ | − n θ / 2 , we have q ( x , ξ ) ∈ S 1 − θ , δ 0 , set S be the pseudo-differential operator with symbol q ( x , ξ ) , by the Hardy-Littlewood-Soboleve fractional integration theorem and the L 2 -boundedness of S (see [13] ), we obtain, for 1 / p = 1 / 2 − θ / 2 ,

I 1 ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) 1 | Q | ∫ Q | T ( f 1 ) ( x ) | d x ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) ( 1 | Q | ∫ Q | T ( f 1 ) ( x ) | p d x ) 1 / p ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) | Q | − 1 / p ( ∫ R n | S ( f 1 ) ( x ) | 2 d x ) 1 / 2 ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) | Q | − 1 / p ( ∫ R n | f 1 ( x ) | 2 d x ) 1 / 2

≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) | Q ˜ | 1 / 2 | Q | 1 / p ( 1 | Q ˜ | ∫ Q ˜ | f ( x ) | 2 d x ) 1 / 2 ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) ( 1 | Q ˜ | ∫ Q ˜ | f ( x ) | r d x ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) M r ( f ) ( x ˜ ) .

For I 2 , by Lemma 1 and Hölder’s inequality, we get, for 1 / r + 1 / r ′ = 1 / 2 ,

I 2 ≤ C ∑ | α 2 | = m 2 ‖ D α 2 b 2 ‖ B M O ∑ | α 1 | = m 1 ( 1 | Q | ∫ Q | T ( D α 1 b ˜ 1 f 1 ) ( x ) | p d x ) 1 / p ≤ C ∑ | α 2 | = m 2 ‖ D α 2 b 2 ‖ B M O ∑ | α 1 | = m 1 | Q | − 1 / p ( ∫ R n | S ( D α 1 b ˜ 1 f 1 ) ( x ) | 2 d x ) 1 / 2 ≤ C ∑ | α 2 | = m 2 ‖ D α 2 b 2 ‖ B M O ∑ | α 1 | = m 1 | Q | − 1 / p ( ∫ R n | D α 1 b ˜ 1 ( x ) f 1 ( x ) | 2 d x ) 1 / 2 ≤ C ∑ | α 2 | = m 2 ‖ D α 2 b 2 ‖ B M O | Q ˜ | 1 / 2 | Q | 1 / p ∑ | α 1 | = m 1 ( 1 | Q ˜ | ∫ Q ˜ | D α 1 b 1 ( x ) − ( D α b j ) Q ˜ | r ′ d x ) 1 / r ′     × ( 1 | Q ˜ | ∫ Q ˜ | f ( x ) | r d x ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

For I 3 , similar to the proof of I 2 , we get

I 3 ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

Similarly, for I 4 , taking r 1 , r 2 > 1 such that 1 / r + 1 / r 1 + 1 / r 2 = 1 / 2 , we obtain

I 4 ≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2 ( 1 | Q | ∫ Q | T ( D α 1 b ˜ 1 D α 2 b ˜ 2 f 1 ) ( x ) | p d x ) 1 / p ≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2 | Q | − 1 / p ( ∫ R n | S ( D α 1 b ˜ 1 D α 2 b ˜ 2 f 1 ) ( x ) | 2 d x ) 1 / 2 ≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2 | Q | − 1 / p ( ∫ R n | D α 1 b ˜ 1 ( x ) D α 2 b ˜ 2 ( x ) f 1 ( x ) | 2 d x ) 1 / 2 ≤ C | Q ˜ | 1 / 2 | Q | 1 / p ∑ | α 1 | = m 1 , | α 2 | = m 2 ∏ j = 1 2 ( 1 | Q ˜ | ∫ Q ˜ | D α j b ˜ j ( x ) | r j d x ) 1 / r j ( 1 | Q ˜ | ∫ Q ˜ | f ( x ) | r d x ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

For I 5 , we write

T b ˜ ( f 2 ) ( x ) − T b ˜ ( f 2 ) ( x 0 ) = ∫ R n ( K ( x , x − y ) | x − y | L − K ( x 0 , x 0 − y ) | x 0 − y | L ) ∏ j = 1 2     R m j ( b ˜ j ; x , y ) f 2 ( y ) d y     + ∫ R n ( R m 1 ( b ˜ 1 ; x , y ) − R m 1 ( b ˜ 1 ; x 0 , y ) ) R m 2 ( b ˜ 2 ; x , y ) | x 0 − y | L K ( x 0 , x 0 − y ) f 2 ( y ) d y     + ∫ R n ( R m 2 ( b ˜ 2 ; x , y ) − R m 2 ( b ˜ 2 ; x 0 , y ) ) R m 1 ( b ˜ 1 ; x 0 , y ) | x 0 − y | L K ( x 0 , x 0 − y ) f 2 ( y ) d y

    − ∑ | α 1 | = m 1 1 α 1 ! ∫ R n [ R m 2 ( b ˜ 2 ; x , y ) ( x − y ) α 1 | x − y | L K ( x , x − y )     − R m 2 ( b ˜ 2 ; x 0 , y ) ( x 0 − y ) α 1 | x 0 − y | L K ( x 0 , x 0 − y ) ] D α 1 b ˜ 1 ( y ) f 2 ( y ) d y     − ∑ | α 2 | = m 2 1 α 2 ! ∫ R n [ R m 1 ( b ˜ 1 ; x , y ) ( x − y ) α 2 | x − y | L K ( x , x − y )     − R m 1 ( b ˜ 1 ; x 0 , y ) ( x 0 − y ) α 2 | x 0 − y | L K ( x 0 , x 0 − y ) ] D α 2 b ˜ 2 ( y ) f 2 ( y ) d y

    + ∑ | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! ∫ R n [ ( x − y ) α 1 + α 2 | x − y | L K ( x , x − y )     − ( x 0 − y ) α 1 + α 2 | x 0 − y | L K ( x 0 , x 0 − y ) ] D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) f 2 ( y ) d y = I 5 ( 1 ) + I 5 ( 2 ) + I 5 ( 3 ) + I 5 ( 4 ) + I 5 ( 5 ) + I 5 ( 6 ) .

By Lemma 1 and the following inequality (see [24] )

| b Q 1 − b Q 2 | ≤ C log ( | Q 2 | / | Q 1 | ) ‖ b ‖ B M O       for     Q 1 ⊂ Q 2 ,

we know that, for x ∈ Q and y ∈ Q ( x 0 , ( 2 k + 1 d ) 1 − θ ) \ Q ( x 0 , ( 2 k d ) 1 − θ ) ,

| R L ( b ˜ ; x , y ) | ≤ C | x − y | L ∑ | α | = L ( ‖ D α b ‖ B M O + | ( D α b ) Q ˜ ( x , y ) − ( D α b ) Q ˜ | ) ≤ C k | x − y | L ∑ | α | = L ‖ D α b ‖ B M O .

Note that | x − y | ~ | x 0 − y | for x ∈ Q ˜ and y ∈ R n \ Q ˜ , we obtain

| I 5 ( 1 ) | ≤ ∑ k = 0 ∞     k 2 ∫ ( 2 k d ) 1 − θ ≤ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | K ( x , x − y ) − K ( x 0 , x 0 − y ) |     × 1 | x − y | m ∏ j = 1 2 | R m j ( b ˜ j ; x , y ) | | f ( y ) | d y     + ∑ k = 0 ∞     k 2 ∫ ( 2 k d ) 1 − θ ≤ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | 1 | x − y | L − 1 | x 0 − y | L |     × | K ( x 0 , x 0 − y ) | ∏ j = 1 2 | R m j ( b ˜ j ; x , y ) | | f ( y ) | d y

≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 0 ∞     k 2 ( ∫ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | f ( y ) | 2 d y ) 1 / 2       × ( ∫ ( 2 k d ) 1 − θ ≤ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | K ( x , x − y ) − K ( x 0 , x 0 − y ) | 2 d y ) 1 / 2       + C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 0 ∞     k 2 ( ∫ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | f ( y ) | 2 d y ) 1 / 2       × ( ∫ ( 2 k d ) 1 − θ ≤ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | x 0 − x | 2 | x 0 − y | 2 | K ( x 0 , x 0 − y ) | 2 d y ) 1 / 2 ,

for the second term above, similar to the proof of Lemma 2.1 in [13] , we have

( ∫ ( 2 k d ) 1 − θ ≤ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | x 0 − x | 2 | x 0 − y | 2 | K ( x 0 , x 0 − y ) | 2 d y ) 1 / 2 ≤ C | x 0 − x | ( 1 − θ ) ( m − n / 2 ) ( 2 k d ) m ( 1 − θ ) ,

thus, by Lemma 3 and recall that n / 2 < m ,

| I 5 ( 1 ) | ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 0 ∞     k 2 d ( 1 − θ ) ( m − n / 2 ) ( 2 k d ) m ( 1 − θ ) ( ∫ | y − x 0 | < ( 2 k + 1 d ) 1 − θ | f ( y ) | 2 d y ) 1 / 2 ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 1 ∞     k 2 2 k ( 1 − θ ) ( n / 2 − m ) ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | f ( y ) | r d y ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 1 ∞     k 2 2 k ( 1 − θ ) ( n / 2 − m ) M r ( f ) ( x ˜ ) ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

For I 5 ( 2 ) , by the formula (see [4] ):

R L ( b ˜ ; x , y ) − R L ( b ˜ ; x 0 , y ) = ∑ | β | < L 1 β ! R L − | β | ( D β b ˜ ; x , x 0 ) ( x − y ) β

and Lemma 1, we have

| R L ( b ˜ ; x , y ) − R L ( b ˜ ; x 0 , y ) | ≤ C ∑ | β | < L     ∑ | α | = L | x − x 0 | L − | β | | x − y | | β | ‖ D α b ‖ B M O ,

thus

| I 5 ( 2 ) | ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 0 ∞ ∫ ( 2 k d ) 1 − θ ≤ | y − x 0 | < ( 2 k + 1 d ) 1 − θ k | x − x 0 | | x 0 − y | | K ( x 0 , x 0 − y ) | | f ( y ) | d y ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 1 ∞     k 2 k ( 1 − θ ) ( n / 2 − m ) ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | f ( y ) | r d y ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

Similarly,

| I 5 ( 3 ) | ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

For I 5 ( 4 ) , recall that | b 2 k Q − b 2 Q | ≤ C k ‖ b ‖ B M O , similar to the proofs of I 5 ( 1 ) and I 5 ( 2 ) , we get, for 1 / r + 1 / r ′ = 1 / 2

| I 5 ( 4 ) | ≤ C ∑ | α 1 | = m 1 ∫ R n | ( x − y ) α 1 | x − y | L − ( x 0 − y ) α 1 | x 0 − y | L | | R m 2 ( b ˜ 2 ; x , y ) | | K ( x , x − y ) | | D α 1 b ˜ 1 ( y ) | | f 2 ( y ) | d y     + C ∑ | α 1 | = m 1 ∫ R n | R m 2 ( b ˜ 2 ; x , y ) − R m 2 ( b ˜ 2 ; x 0 , y ) | | ( x 0 − y ) α 1 | | x 0 − y | L | K ( x , x − y ) | | D α 1 b ˜ 1 ( y ) | | f 2 ( y ) | d y     + C ∑ | α 1 | = m 1 ∫ R n | K ( x , x − y ) − K ( x 0 , x 0 − y ) | | ( x 0 − y ) α 1 | | x 0 − y | L | R m 2 ( b ˜ 2 ; x 0 , y ) | | D α 1 b ˜ 1 ( y ) | | f 2 ( y ) | d y ≤ C ∑ | α | = m 2 ‖ D α b 2 ‖ B M O ∑ | α 1 | = m 1 ∑ k = 1 ∞     k 2 k ( 1 − θ ) ( n / 2 − m )     × ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | f ( y ) D α 1 b ˜ 1 ( y ) | 2 d y ) 1 / 2

≤ C ∑ | α | = m 2 ‖ D α b 2 ‖ B M O ∑ k = 1 ∞     k 2 k ( 1 − θ ) ( n / 2 − m ) ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | f ( y ) | r d y ) 1 / r     × ∑ | α 1 | = m 1 ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | D α 1 b 1 ( y ) − ( D α 1 b 1 ) Q ˜ | r ′ d y ) 1 / r ′ ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 1 ∞     k 2 2 k ( 1 − θ ) ( n / 2 − m ) M r ( f ) ( x ˜ ) ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

Similarly,

| I 5 ( 5 ) | ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

For I 5 ( 6 ) , similar to the proof of I 5 ( 1 ) , we get, for 1 / r + 1 / r 1 + 1 / r 2 = 1 / 2 ,

| I 5 ( 6 ) | ≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2     ∫ R n | ( x − y ) α 1 + α 2 | x − y | L − ( x 0 − y ) α 1 + α 2 | x 0 − y | L |   × | K ( x , x − y ) | | D α 1 b ˜ 1 ( y ) | | D α 2 b ˜ 2 ( y ) | | f 2 ( y ) | d y   + C ∑ | α 1 | = m 1 , | α 2 | = m 2     ∫ R n | K ( x , x − y ) − K ( x 0 , x 0 − y ) | | ( x 0 − y ) α 1 + α 2 | | x 0 − y | L   × | D α 1 b ˜ 1 ( y ) | | D α 2 b ˜ 2 ( y ) | | f 2 ( y ) | d y

≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2     ∑ k = 1 ∞     2 k ( 1 − θ ) ( n / 2 − m )     × ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | f ( y ) D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) | 2 d y ) 1 / 2

≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2     ∑ k = 1 ∞     2 k ( 1 − θ ) ( n / 2 − m ) ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | f ( y ) | r d y ) 1 / r     × ∏ j = 1 2 ( 1 | Q ( x 0 , ( 2 k d ) 1 − θ ) | ∫ Q ( x 0 , ( 2 k d ) 1 − θ ) | D α j b j ( y ) − ( D α j b j ) Q ˜ | r j d y ) 1 / r j ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

Thus

| I 5 | ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

Case 2. d > 1 . In this case, let Q ˜ = 5 n Q and b ˜ j ( x ) = b j ( x ) − ∑ | α | = m j 1 α ! ( D α b j ) Q ˜ x α , then R m j ( b j ; x , y ) = R m j ( b ˜ j ; x , y ) and D α b ˜ j = D α b j − ( D α b j ) Q ˜ for | α | = m j . Write, for f = f χ Q ˜ + f χ R n \ Q ˜ = f 1 + f 2 ,

1 | Q | ∫ Q | T b ( f ) ( x ) | d x ≤ 1 | Q | ∫ Q | ∫ R n ∏ j = 1 2 R m j ( b ˜ j ; x , y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y | d x     + C | Q | ∫ Q | ∑ | α 1 | = m 1 ∫ R n R m 2 ( b ˜ 2 ; x , y ) ( x − y ) α 1 | x − y | L D α 1 b ˜ 1 ( y ) K ( x , x − y ) f 1 ( y ) d y | d x     + C | Q | ∫ Q | ∑ | α 2 | = m 2 ∫ R n R m 1 ( b ˜ 1 ; x , y ) ( x − y ) α 2 | x − y | L D α 2 b ˜ 2 ( y ) K ( x , x − y ) f 1 ( y ) d y | d x

    + C | Q | ∫ Q | ∑ | α 1 | = m 1 , | α 2 | = m 2 ∫ R n ( x − y ) α 1 + α 2 D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) | x − y | L K ( x , x − y ) f 1 ( y ) d y | d x     + 1 | Q | ∫ Q | T b ˜ ( f 2 ) ( x ) | d x : = J 1 + J 2 + J 3 + J 4 + J 5 .

Similar to the proof of I 1 , I 2 , I 3 and I 4 , we get, by the L p ( 1 < p < ∞ ) -boundedness of T (see Lemma 2),

J 1 ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) ( 1 | Q | ∫ R n | T ( f 1 ) ( x ) | r d x ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) | Q | − 1 / r ( ∫ R n | f 1 ( x ) | r d x ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) ( 1 | Q ˜ | ∫ Q ˜ | f ( x ) | r d x ) 1 / r

≤ C ∏ j = 1 2 ( ∑ | α j | = m j ‖ D α j b j ‖ B M O ) M r ( f ) ( x ˜ ) ;

J 2 ≤ C ∑ | α 2 | = m 2 ‖ D α 2 b 2 ‖ B M O ∑ | α 1 | = m 1 ( 1 | Q | ∫ R n | T ( D α 1 b ˜ 1 f 1 ) ( x ) | 2 d x ) 1 / 2 ≤ C ∑ | α 2 | = m 2 ‖ D α 2 b 2 ‖ B M O ∑ | α 1 | = m 1 | Q | − 1 / 2 ( ∫ R n | D α 1 b ˜ 1 ( x ) f 1 ( x ) | 2 d x ) 1 / 2 ≤ C ∑ | α 2 | = m 2 ‖ D α 2 b 2 ‖ B M O ∑ | α 1 | = m 1 ( 1 | Q ˜ | ∫ Q ˜ | D α 1 b 1 ( x ) − ( D α b 1 ) Q ˜ | r ′ d x ) 1 / r ′     × ( 1 | Q ˜ | ∫ Q ˜ | f ( x ) | r d x ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) ;

J 3 ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) ;

J 4 ≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2 | Q | − 1 / 2 ( ∫ R n | T ( D α 1 b ˜ 1 D α 2 b ˜ 2 f 1 ) ( x ) | 2 d x ) 1 / 2 ≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2 | Q | − 1 / 2 ( ∫ R n | D α 1 b ˜ 1 ( x ) D α 2 b ˜ 2 ( x ) f 1 ( x ) | 2 d x ) 1 / 2 ≤ C ∑ | α 1 | = m 1 , | α 2 | = m 2     ∏ j = 1 2 ( 1 | Q ˜ | ∫ Q ˜ | D α j b ˜ j ( x ) | r j d x ) 1 / r j ( 1 | Q ˜ | ∫ Q ˜ | f ( x ) | r d x ) 1 / r ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

For J 5 , we write

T b ˜ ( f 2 ) ( x ) = ∫ R n ∏ j = 1 2 R m j ( b ˜ j ; x , y ) | x − y | L K ( x , x − y ) f 2 ( y ) d y   − ∑ | α 1 | = m 1 1 α 1 ! ∫ R n R m 2 ( b ˜ 2 ; x , y ) ( x − y ) α 1 | x − y | L K ( x , x − y ) D α 1 b ˜ 1 ( y ) f 2 ( y ) d y   − ∑ | α 2 | = m 2 1 α 2 ! ∫ R n R m 1 ( b ˜ 1 ; x , y ) ( x − y ) α 2 | x − y | L K ( x , x − y ) D α 2 b ˜ 2 ( y ) f 2 ( y ) d y   + ∑ | α 1 | = m 1 , | α 2 | = m 2 1 α 1 ! α 2 ! ∫ R n ( x − y ) α 1 + α 2 | x − y | L K ( x , x − y ) D α 1 b ˜ 1 ( y ) D α 2 b ˜ 2 ( y ) f 2 ( y ) d y ,

similar to the proof of I 5 and by using lemma 4, we get

| T b ˜ ( f 2 ) ( x ) | ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 0 ∞     k 2 ∫ 2 k + 1 Q ˜ \ 2 k Q ˜ | x − y | − 2 n | f ( y ) | d y     + C ∑ | α | = m 2 ‖ D α b 2 ‖ B M O ∑ | α 1 | = m 1     ∑ k = 0 ∞     k ∫ 2 k + 1 Q ˜ \ 2 k Q ˜ | x − y | − 2 n | D α 1 b ˜ 1 ( y ) | | f ( y ) | d y

    + C ∑ | α | = m 1 ‖ D α b 1 ‖ B M O ∑ | α 2 | = m 2     ∑ k = 0 ∞     k ∫ 2 k + 1 Q ˜ \ 2 k Q ˜ | x − y | − 2 n | D α 2 b ˜ 2 ( y ) | | f ( y ) | d y     + C ∑ | α 1 | = m 1 , | α 2 | = m 2     ∑ k = 0 ∞     ∫ 2 k + 1 Q ˜ \ 2 k Q ˜ | x − y | − 2 n | D α 1 b ˜ 1 ( y ) | | D α 2 b ˜ 2 ( y ) | | f ( y ) | d y

≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) d − n ∑ k = 1 ∞     k 2 2 − k n ( 1 | 2 k Q ˜ | ∫ 2 k Q ˜ | f ( y ) | r d y ) 1 / r     + C ∑ | α | = m 2 ‖ D α b 2 ‖ B M O d − n ∑ k = 1 ∞     k 2 − k n ( 1 | 2 k Q ˜ | ∫ 2 k Q ˜ | f ( y ) | r d y ) 1 / r     × ∑ | α 1 | = m 1 ( 1 | 2 k Q ˜ | ∫ 2 k Q ˜ | D α 1 b 1 ( y ) − ( D α 1 b 1 ) Q ˜ | r ′ d y ) 1 / r ′     + C ∑ | α | = m 1 ‖ D α b 1 ‖ B M O d − n ∑ k = 1 ∞     k 2 − k n ( 1 | 2 k Q ˜ | ∫ 2 k Q ˜ | f ( y ) | r d y ) 1 / r     × ∑ | α 2 | = m 2 ( 1 | 2 k Q ˜ | ∫ 2 k Q ˜ | D α 2 b 2 ( y ) − ( D α 1 b 2 ) Q ˜ | r ′ d y ) 1 / r ′

    + C ∑ | α 1 | = m 1 , | α 2 | = m 2 d − n ∑ k = 1 ∞     2 − k n ( 1 | 2 k Q ˜ | ∫ 2 k Q ˜ | f ( y ) | r d y ) 1 / r     × ∏ j = 1 2 ( 1 | 2 k Q ˜ | ∫ 2 k Q ˜ | D α j b j ( y ) − ( D α j b j ) Q ˜ | r j d y ) 1 / r j ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ∑ k = 1 ∞     k 2 2 − k n M r ( f ) ( x ˜ ) ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) ,

thus

| J 5 | ≤ C ∏ j = 1 2 ( ∑ | α | = m j ‖ D α b j ‖ B M O ) M r ( f ) ( x ˜ ) .

This completes the proof of Theorem 1.

Proof of Theorem 2. (a) follows from Theorem 1. For (b), Choose 1 < r < p in Theorem 1, we get

‖ T b ( f ) ‖ L p ≤ ‖ M ( T b ( f ) ) ‖ L p ≤ C ‖ ( T b ( f ) ) # ‖ L p ≤ C ∏ j = 1 l ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ‖ M r ( f ) ‖ L p ≤ C ∏ j = 1 l ( ∑ | α | = m j ‖ D α b j ‖ B M O ) ‖ f ‖ L p .

This finishes the proof.

The authors would like to express their gratitude to the referee for his/her valuable comments and suggestions.

Project supported by Scientific Research Fund of Hunan Provincial Education Departments (19C1037).

The authors declare no conflicts of interest regarding the publication of this paper.

Zeng, J.S., Chen, R.P., Liu, J.G., Wang, X.J. and Liu, C.S. (2023) Boundedness for Multilinear Operators of Pseudo-Differential Operators. Advances in Pure Mathematics, 13, 449-462. https://doi.org/10.4236/apm.2023.137029