Endpoint estimates for multilinear commutator of Marcinkiewicz operator

159 Kragujevac J. Math. 31 (2008) 159–176.

ENDPOINT ESTIMATES FOR MULTILINEAR COMMUTATOR OF MARCINKIEWICZ OPERATOR Yu Wenxin and Liu Lanzhe Department of Mathematics, Changsha University of Science and Technology, Changsha, 410076, P.R.of China (e-mail: [email protected])

(Received March 15, 2007)

Abstract. In this paper, we prove the endpoint estimates for the multilinear commutator of Marcinkiewicz operator.

1. INTRODUCTION and PRELIMINARIES As the development of singular integral operators, their commutators have been well studied. Let b ∈ BM O(Rn ) and T be the Calderón-Zygmund operator, the commutator [b, T ] generated by b and T is defined by [b, T ](f )(x) = b(x)T (f )(x) − T (bf )(x). A classical result of Coifman, Rochberb and Weiss (see [3]) proved that the commutator [b, T ] is bounded on Lp (Rn ), (1 < p < ∞). In [2, 5], the boundedness properties of the commutators for the extreme values of p are obtained. And note that [b, T ]

160 is not bounded for the end point boundedness (that is p = 1 and p = ∞). In this paper, we will introduce the multilinear commutator of Marcinkiewicz operator and prove the boundedness properties of the operator for the extreme cases. First let us introduce some notations (see [1, 4, 7, 8]). In this paper, Q = Q(x, r) will denote a cube of Rn with sides parallel to the axes and center at x and edge is r. For a cube Q and a locally integrable function f , let fQ = |Q|−1 f (Q) =

R

Q

R Q

f (x)dx and

f (x)dx, the sharp function of f is defined by 1 Z |f (y) − fQ |dy. f (x) = sup x∈Q |Q| Q #

f is said to belong to BM O(Rn ) if f # ∈ L∞ (Rn ) and define ||f ||BM O = ||f # ||L∞ . We nave |f2Q − fQ | ≤ C||f ||BM O and ||f − f2k Q ||BM O ≤ Ck||f ||BM O for k ≥ 1 (see [4, 8]). We also define the central BM O space by CM O(Rn ), which is the space of those functions f ∈ Lloc (Rn ) such that Z

||f ||CM O = sup |Q(0, r)| r>1

−1 Q

|f (y) − fQ |dy < ∞.

It is well-known that Z

||f ||CM O ≈ sup inf |Q(0, r)| r>1 c∈C

−1 Q

|f (x) − c|dx.

Let M be the Hardy-Littlewood maximal operator, that is 1 Z M (f )(x) = sup |f (y)|dy. x∈Q |Q| Q

Definition 1. A function a is called a H 1 (Rn )−atom, if there exists a cube Q, such that 1◦ ) supp a ⊂ Q = Q(x0 , r), 2◦ ) ||a||L∞ ≤ |Q|−1 , 3◦ )

R

Rn

a(x)dx = 0.

It is well known that the Hardy space H 1 (Rn ) has the atomic decomposition characterization (see [4], [8]).

161 The Ap weight is defined by (see [4])  

Ã

1 Z Ap = w : sup w(x)dx  |Q| Q Q

!Ã

1 Z w(x)−1/(p−1) dx |Q| Q

 

!p−1

< ∞ , 1 < p < ∞, 

and A1 = {w : M (w)(x) ≤ Cw(x), a.e.}. Definition 2. Let 0 < δ < n and 1 < p < n/δ. We shall call Bpδ (Rn ) the space of those functions f on Rn such that ||f ||Bpδ = sup r−n(1/p−δ/n) ||f χQ(0,r) ||Lp < ∞. r>1

Definition 3. Let 0 < δ < n, 0 < γ ≤ 1 and Ω be homogeneous of degree zero on Rn such that

R S n−1

Ω(x0 )dσ(x0 ) = 0. Assume that Ω ∈ Lipγ (S n−1 ), that is there exists

a constant M > 0 such that for any x, y ∈ S n−1 , |Ω(x) − Ω(y)| ≤ M |x − y|γ . Let m be the positive integer and bj be the locally integrable functions on Rn (j = 1, · · ·, m). The Marcinkiewicz multilinear commutator is defined by ~ µbδ (f )(x)

where ~

=

∞

dt ~ |Ftb (f )(x)|2 3

!1/2

,

t

0



Z

Ftb (f )(x) =

ÃZ

|x−y|≤t



m Ω(x − y)  Y (bj (x) − bj (y)) f (y)dy. |x − y|n−1−δ j=1

Set

Z

Ft (f )(x) = we also define that

|x−y|≤t

ÃZ

µδ (f )(x) =

Ω(x − y) f (y)dy, |x − y|n−1−δ

0

∞

dt |Ft (f )(x)|2 3 t

!1/2

,

which is the Marcinkiewicz operator (see [6, 10]). n

Let H be the space H = h : ||h|| = (

R∞ 0

1/2

|h(t)|2 dt/t3 )

o

< ∞ . Then, it is clear

that ~

~

µδ (f )(x) = ||Ft (f )(x)|| and µbδ (f )(x) = ||Ftb (f )(x)||.

162 ~

Note that when b1 = . . . = bm , µbδ is just the m order commutator, and, when ~

m = 1 and δ = 0, µbδ is just the commutator generated by Marcinkiewicz operator and b (see [9, 10]). It is well known that commutators are of great interest in harmonic analysis and have been widely studied by many authors (see [1, 2, 3, 5, 6, 7]). Our main purpose is to establish the boundedness properties of the operator for the extreme cases. Given a positive integer m and 1 ≤ j ≤ m, we denote by Cjm the family of all finite subsets σ = {σ(1), . . . , σ(j)} of {1, . . . , m} of j different elements. For σ ∈ Cjm , set σ c = {1, . . . , m} \ σ. For ~b = (b1 , . . . , bm ) and σ = {σ(1), . . . , σ(j)} ∈ C m , set j

~bσ = (bσ(1) , . . . , bσ(j) ), bσ = bσ(1) . . . bσ(j) and ||~bσ ||BM O = ||bσ(1) ||BM O . . . ||bσ(j) ||BM O . 2. THEOREMS AND PROOFS We begin with some preliminaries lemmas. Lemma 1. Let 1 < r < ∞, bj ∈ BM O(Rn ) for j = 1, . . . , k and k ∈ N . Then, we have

and

k k Y 1 Z Y |bj (y) − (bj )Q |dy ≤ C ||bj ||BM O |Q| Q j=1 j=1



1/r

k 1 Z Y  |bj (y) − (bj )Q |r dy  |Q| Q j=1

≤C

k Y

||bj ||BM O .

j=1

Proof. In fact, we just need to choose pj > 1 and qj > 1 , where 1 ≤ j ≤ k, such that 1/p1 + . . . + 1/pk = 1 and r/q1 + . . . + r/qk = 1. After that, using the Hölder’s inequality with exponent 1/p1 + . . . + 1/pk = 1 and r/q1 + . . . + r/qk = 1 respectively, we may get the results by [4] and [8].

2

Lemma 2. [10] Let w ∈ A1 , 0 < δ < n, 1 < p < n/δ and 1/q = 1/p − δ/n. Then µδ is bounded from Lp (w) to Lq (w). Lemma 3. Let w ∈ Ap , 1 < p < ∞, then wχQ0 ∈ Ap for any cube Q0 , where χQ0 denotes the characteristic function of the cube Q0 .

163 Proof. By definition, we have Ã

1 Z sup w(x)dx |Q| Q Q

!Ã

1 Z w(x)−1/(p−1) dx |Q| Q

!p−1

< ∞,

thus Ã

1 Z sup w(x)χQ0 (x)dx |Q| Q Q Ã

1 Z ≤ sup w(x)dx |Q| Q Q

!Ã

!Ã

1 Z (w(x)χQ0 (x))−1/(p−1) dx |Q| Q

1 Z w(x)−1/(p−1) dx |Q| Q

!p−1

!p−1

< ∞,

that is wχQ0 ∈ Ap .

2

Theorem 1. Let 0 < δ < n and ~b = (b1 , . . . , bm ) with bj ∈ BM O(Rn ) for ~

1 ≤ j ≤ m. Then µbδ is bounded from Ln/δ to BM O(Rn ). Proof. It is only to prove that there exist a constant CQ such that 1 Z ~ |µb (f )(x) − CQ |dx ≤ C||f ||Ln/δ . |Q| Q δ Fix a cube Q, Q = Q(x0 , r), we decompose f into f = f1 + f2 with f1 = f χ2Q , f2 = f χ(Rn \2Q) . When m = 1, set (b1 )Q = |Q|−1

R

Q b1 (y)dy,

we have

Ftb1 (f )(x) = (b1 (x) − (b1 )Q )Ft (f )(x) − Ft ((b1 − (b1 )Q )f1 )(x) − Ft ((b1 − (b1 )Q )f2 )(x), so |µbδ1 (f )(x) − µδ (((b1 )Q − b1 )f2 )(x0 )| ¯ ¯

¯ ¯

= ¯||Ftb1 (f )(x)|| − ||Ft (((b1 )Q − b1 )f2 )(x0 )||¯ ≤ ||Ftb1 (f )(x) − Ft (((b1 )Q − b1 )f2 )(x0 )|| ≤ ||(b1 (x) − (b1 )Q )Ft (f )(x)|| + ||Ft ((b1 − (b1 )Q )f1 )(x)|| +||Ft ((b1 − (b1 )Q )f2 )(x) − Ft ((b1 − (b1 )Q )f2 )(x0 )|| = A(x) + B(x) + C(x). For A(x), set 1 < p < n/δ, 1/q = 1/p − δ/n and 1/q + 1/q 0 = 1, by the Hölder’s inequality and Lemma 2.3, we get

164

!1/q0 Ã

Ã

1 Z |A(x)|dx ≤ Q Q

1 Z 1 Z 0 |b1 (x) − (b1 )Q |q dx |µδ (f )(x)|q χQ (x)dx |Q| Q |Q| Rn µZ ¶1/p 1 p |f (x)| χQ (x)dx ≤ C||b1 ||BM O q |Q| Rn 1 ≤ C||b1 ||BM O q ||f ||Ln/δ |Q|(1−(δp/n))/p |Q| ≤ C||b1 ||BM O ||f ||Ln/δ .

!1/q

For B(x), taking 1 < r < n/δ and 1/s = 1/r − δ/n, by the Hölder’s inequality, we have 1 Z |B(x)|dx ≤ |Q| Q

Ã

1 Z (µδ ((b1 (x) − (b1 )Q )f1 )(x))s dx |Q| Rn

!1/s

≤ C|Q|−1/s ||(b1 (x) − (b1 )Q )f χ2Q ||Lr Ã

1 Z |b1 (x) − (b1 )Q |s dx ≤ C |2Q| 2Q ≤ C||b1 ||BM O ||f ||Ln/δ .

!1/s

||f ||Ln/δ

For C(x), note that r ≤ |x − y| and |x0 − y| ≈ |x − y| for y ∈ (2Q)c , we have ¯Z ¯ Ω(x − y)f2 (y) ¯ (b1 (y) − (b1 )Q )dy ¯ ¯ 0 |x−y|≤t |x − y|n−1−δ 1/2 ¯2 Z ¯ dt Ω(x0 − y)f2 (y) ¯ − (b1 (y) − (b1 )Q )dy ¯¯ 3  t |x0 −y|≤t |x0 − y|n−1−δ

ÃZ

C(x) =



≤ 

∞

Z ∞ "Z 0



+

|x−y|≤t,|x0 −y|>t

Z ∞ "Z 0

ÃZ

+

|Ω(x − y)||f2 (y)| |b1 (y) − (b1 )Q |dy |x − y|n−1−δ

0

∞

|x−y|>t,|x0 −y|≤t

× |b1 (y) − (b1 )Q ||f2 (y)|dy]2 ≡ I1 + I2 + I3 , thus

dt t3

1/2

dt  t3

|Ω(x0 − y)||f2 (y)| |b1 (y) − (b1 )Q | dy |x0 − y|n−1−δ

¯ ¯ ¯ |Ω(x − y)| |Ω(x0 − y)| ¯¯ ¯ − ¯ ¯ |x0 − y|n−1−δ ¯ |x−y|≤t,|x0 −y|≤t ¯ |x − y|n−1−δ !1/2

"Z

#2

#2

1/2

dt  t3

165

ÃZ !1/2 |f (y)| dt ≤ C |(b1 (y) − (b1 )Q )| dy |x − y|n−1−δ |x−y|≤t 1, set ~bQ = ((b1 )Q , . . . , (bm )Q ) ∈ Rn , where (bj )Q = |Q|−1 Q bj (y)dy, 1 ≤ j ≤ m, we have ~

Ftb (f )(x) = (b1 (x) − (b1 )Q ) · · · (bm (x) − (bm )Q )Ft (f )(x) +(−1)m Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f )(x) +

m−1 X

X

j=1

σ∈Cjm

(−1)m−j (~b(x) − ~bQ )σ

Z Rn

(~b(y) − ~bQ )σc ψt (x − y)f (y)dy

= (b1 (x) − (b1 )Q ) · · · (bm (x) − (bm )Q )Ft (f )(x) +(−1)m Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f1 )(x) +(−1)m Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f2 )(x) +

m−1 X

X

j=1

σ∈Cjm

(−1)m−j (~b(x) − ~bQ )σ Ft ((~b − ~bQ )σc f )(x),

thus ~

|µbδ (f )(x) − µδ (((b1 )Q − b1 ) · · · ((bm )Q − bm )f2 )(x0 )| ≤ ||(b1 (x) − (b1 )Q ) · · · (bm (x) − (bm )Q )Ft (f )(x)|| +

m−1 X

X

||(~b(x) − ~bQ )σ Ft ((~b − ~bQ )σc f )(x)||

j=1 σ∈Cjm

+||Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f1 )(x)|| +||Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f2 )(x) − Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f2 )(x0 )|| = S1 (x) + S2 (x) + S3 (x) + S4 (x). For S1 (x), taking 1 < p < n/δ, and 1/q = 1/p − δ/n, by the Hölder’s inequality and Lemma 1,2,3, we have 1 Z S1 (x)dx |Q| Q



1/q0 Ã

m 1 Z Y 0 ≤  | (bj (x) − (bj )Q )|q dx |Q| Q j=1

≤ C||~b||BM O |Q|−1/q

µZ

≤ C||~b||BM O ||f ||Ln/δ .

Q

|f (x)|p dx

¶1/p

1 Z |µδ (f )(x)|q dx |Q| Q

|Q|(1−(δp/n))/p

!1/q

167 For S2 (x), taking 1 < p < n/δ and 1/q = 1/p − δ/n, then 1 Z S2 (x)dx |Q| Q ≤ C

m−1 X

Ã

X

1 Z ~ 0 |(b(x) − ~bQ )σ |q dx |Q| Q

j=1 σ∈Cjm

≤ C

m−1 X

||~bσ ||BM O |Q|1/q

j=1

≤ C

m−1 X

Rn

Ã

X

||~bσ ||BM O

j=1 σ∈Cjm

≤ C

µZ

m−1 X

X

j=1

σ∈Cjm

!1/q0 Ã

1 Z |µδ ((~b − ~bQ )σc )f )(x)|q dx |Q| Q

~ − ~bQ )σc )f (x)|p χQ (x)dx |(b(x)

1 Z ~ |(b(x) − ~bQ )σc |q dx |Q| Q

!1/q

¶1/p

!1/q

||f ||Ln/δ

||~bσ ||BM O ||~bσc ||BM O ||f ||Ln/δ

≤ C||~b||BM O ||f ||Ln/δ . For S3 (x), taking 1 < p < n/δ and 1/q = 1/p − δ/n, we get 1 Z S3 (x)dx |Q| Q

Ã

≤

1 Z |µδ ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f1 )(x)|q dx |Q| Q

!1/q

≤ C|Q|−1/q ||((b1 (x) − (b1 )Q ) · · · (bm (x) − (bm )Q )f1 (x)||Lp Ã

1 Z ≤ C |(b1 − (b1 )Q ) · · · (bm − (bm )Q )|q dx |2Q| 2Q ≤ C||~b||BM O ||f ||Ln/δ .

!1/q

||f ||Ln/δ

For S4 (x), we have S4 (x) 

≤

1/2 ¯ ¯ 2 ¯m ¯ Y ¯ ¯ |Ω(x − y)||f2 (y)| ¯ dt  ¯   ¯ (bj (y) − (bj )2Q )¯ dy n−1−δ |x − y| t3 |x−y|≤t,|x0 −y|>t ¯j=1 ¯

 Z Z  ∞  0



1/2 ¯ ¯ 2 ¯m ¯ Y ¯ |Ω(x0 − y)||f2 (y)| ¯¯ dt  ¯   ¯ (bj (y) − (bj )2Q )¯ dy n−1−δ |x0 − y| t3 |x−y|>t,|x0 −y|≤t 0 ¯j=1 ¯ ¯ ¯ ÃZ " ¯ |Ω(x − y)| ∞ Z |Ω(x0 − y)| ¯¯ ¯ − + ¯ ¯ |x0 − y|n−1−δ ¯ 0 |x−y|≤t,|x0 −y|≤t ¯ |x − y|n−1−δ  Z Z  ∞ +

168 1/2 ¯ ¯ 2 ¯Y ¯ ¯m ¯ dt  × ¯¯ (bj (y) − (bj )2Q )¯¯ |f2 (y)|dy  3  t ¯j=1 ¯

≡ V1 + V2 + V3 , thus ¯ ¯ ÃZ !1/2 ¯Y ¯ ¯m ¯ |f (y)| dt ¯ ¯ dy C ¯ (bj (y) − (bj )2Q )¯ (2Q)c ¯j=1 |x−y|≤t 1. Fix a cube Q = Q(0, d) with d > 1. Set f1 = f χ2Q , f2 = f χRn \2Q and ~bQ = ((b1 )Q , . . . , (bm )Q ), where (bj )Q = |Q|−1

R

Q

|bj (y)|dy, 1 ≤ j ≤ m, we have

~

|µbδ (f )(x) − µδ (((b1 )Q − b1 ) · · · ((bm )Q − bm )f2 )(x0 )| ≤ ||(b1 (x) − (b1 )Q ) · · · (bm (x) − (bm )Q )Ft (f )(x)|| +

m−1 X

X

j=1

σ∈Cjm

||(~b(x) − ~bQ )σ Ft ((~b − ~bQ )σc f )(x)||

+||Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f1 )(x)|| +||Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f2 )(x) − Ft ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f2 )(x0 )|| = H1 (x) + H2 (x) + H3 (x) + H4 (x). Taking 1 < p < n/δ, 1/s = 1/r − δ/n, by the Hölder’s inequality and Lemma 1,2,3, we have 1 Z H1 (x)dx |Q| Q



1/q0 Ã

m 1 Z Y 0 ≤  | (bj (x) − (bj )Q )|q dx |Q| Q j=1

≤ C||~b||BM O |Q|−1/q

µZ p

Q

1 Z |µδ (f )(x)|q dx |Q| Q

!1/q

¶1/p

|f (x)| dx

≤ C||~b||BM O d−n(1/p−δ/n) ||f χQ ||Lp ≤ C||~b||BM O ||f ||Bpδ . For H2 (x), taking 1 < p < n/δ, 1/s = 1/r − δ/n, and 1/s0 + 1/s = 1, then 1 Z H2 (x)dx |Q| Q ≤C

m−1 X

X

j=1 σ∈Cjm

Ã

1 Z ~ 0 |(b(x) − ~bQ )σ |s dx |Q| Q

!1/s0 Ã

1 Z |µδ ((~b − ~bQ )σc )f )(x)|s dx |Q| Q

!1/s

170

≤C

m−1 X

||~bσ ||BM O |Q|−1/s

j=1

≤C

m−1 X

X

Ã

||~bσ ||BM O

j=1 σ∈Cjm

≤C

m−1 X

X

µZ Rn

~ − ~bQ )σc )f (x)|r χQ (x)dx |(b(x)

1 Z ~ |(b(x) − ~bQ )σc |pr/(p−r) dx |Q| Q

¶1/r

!(p−r)/pr

|Q|(δ/n−1/p) ||f χQ ||Lp

||~bσ ||BM O ||~bσc ||BM O d−n(1/p−δ/n) ||f χQ ||Lp

j=1 σ∈Cjm

≤ C||~b||BM O ||f ||Bpδ . For H3 (x), taking 1 < p < n/δ , 1/s = 1/r − δ/n and 1/s0 + 1/s = 1, we get 1 Z H3 (x)dx |Q| Q

Ã

≤

1 Z |µδ ((b1 − (b1 )Q ) · · · (bm − (bm )Q )f1 )(x)|s dx |Q| Q

!1/s

≤ C|Q|−1/s ||((b1 (x) − (b1 )Q ) · · · (bm (x) − (bm )Q )f χ2Q ||Lr Ã

1 Z ≤C |(b1 − (b1 )Q ) · · · (bm − (bm )Q )|pr/(p−r) dx |2Q| 2Q ≤ C||~b||BM O ||f ||Bpδ .

!(p−r)/pr

d−n(1/p−δ/n) ||f χ2Q ||Lp

For H4 (x), we have   Z ∞ Z   

H4 (x) ≤

0

1/2 ¯ ¯ 2 ¯m ¯ ¯ |Ω(x − y)||f2 (y)| ¯¯ Y dt  ¯   ¯ (bj (y) − (bj )2Q )¯ dy n−1−δ |x − y| t3 |x−y|≤t,|x0 −y|>t ¯j=1 ¯

  Z ∞ Z  + 

1/2 ¯ ¯ 2 ¯Y ¯ m ¯ |Ω(x0 − y)||f2 (y)| ¯¯ dt  ¯   ¯ (bj (y) − (bj )2Q )¯ dy n−1−δ |x0 − y| t3 0 |x−y|>t,|x0 −y|≤t ¯j=1 ¯ ¯ ¯ ÃZ " ¯ |Ω(x − y)| ∞ Z |Ω(x0 − y)| ¯¯ ¯ − ¯ + ¯ |x0 − y|n−1−δ ¯ 0 |x−y|≤t,|x0 −y|≤t ¯ |x − y|n−1−δ 1/2 ¯ ¯ 2 ¯m ¯ Y ¯ ¯ dt × ¯¯ (bj (y) − (bj )2Q )¯¯ |f2 (y)|dy  3   t ¯j=1 ¯

≡ M1 + M2 + M3 , thus M1 ≤

¯ ¯ ¯m ¯ ¯Y ¯ |Q|1/2n |f (y)| ¯ ¯ dy C (b (y) − (b ) ) j j 2Q ¯ ¯ k+1 Q\2k Q ¯ 2 ¯ |x0 − y|n+1/2−δ j=1 k=1 ∞ Z X

171 ¯

¯

¯Y ¯ Z ¯m ¯ 1 −k/2 ¯ ≤ C 2 (bj (y) − (bj )2Q )¯¯ |f (y)|dy ¯ k+1 1−δ/n |2 Q| 2k+1 Q ¯j=1 ¯ k=1 ∞ X



¯ ¯

¯p/(p−1) ¯

Z m ¯ ¯Y 1  ¯ (bj (y) − (bj )2Q )¯ ≤ C 2−k/2 k+1 1−δ/n  ¯ ¯ |2 Q| 2k+1 Q ¯j=1 ¯ k=1 ∞ X

µZ

× ≤

2k+1 Q

|f (y)|p dy

(p−1)/p 

dy 

¶1/p



¯ ¯p/(p−1) (p−1)/p ¯Y ¯ m ¯ ¯ 1  −k/2  ¯ dy  C 2 (bj (y) − (bj )2Q )¯¯  k+1 ¯ k+1 |2 Q| 2 Q ¯j=1 ¯ k=1 ∞ X

Z

×|2k+1 Q|−(1/p−δ/n) ||f χ2k+1 Q ||Lp ≤ C||~b||BM O ||f ||Bpδ ; similarly, we have M2 ≤ C||~b||BM O ||f ||Bpδ . We now estimate V3 . We gain ¯ ¯Ã ! ¯m ¯ ¯Y ¯ |Q|γ/n |Q|1/n ¯ ¯ M3 ≤ C + |f (y)|dy ¯ (bj (y) − (bj )2Q )¯ k+1 Q\2k Q ¯ |x0 − y|n+γ−δ ¯ |x0 − y|n+1−δ j=1 k=1 2 ¯ ¯ ¯m ¯ Z ∞ X Y ¯ ¯ 1 −k −kγ ¯ ≤C (2 + 2 ) k+1 1−δ/n (bj (y) − (bj )2Q )¯¯ |f (y)|dy ¯ |2 Q| 2k+1 Q ¯j=1 ¯ k=1  ¯ ¯p/(p−1) (p−1)/p ¯m ¯ Z ∞ X Y ¯ ¯ 1   ¯ (bj (y) − (bj )2Q )¯ ≤C (2−k + 2−kγ )  k+1 dy  ¯ ¯ k |2 Q| 2 Q ¯j=1 ¯ k=1 ∞ Z X

×|2k+1 Q|−(1/p−δ/n) ||f χ2k+1 Q ||Lp ≤ C||~b||BM O ||f ||Bpδ . This completes the total proof of Theorem 2.

2

Theorem 3. Let 0 < δ < n and ~b = (b1 , . . . , bm ) with bj ∈ BM O(Rn ) for 1 ≤ j ≤ m. If for any H 1 (Rn )−atom a supported on certain cube Q and u ∈ Q, there is m X Z X j=1 σ∈Cjm ~

(2Q)c

¯¯Z ¯¯!n/(n−δ) ¯¯ ¯¯ Ω(x − u) ¯¯ ¯¯ |(b(x) − bQ )σc | ¯¯ (~b(y) − ~bQ )σ a(y)dy ¯¯ dx ≤ C, n−1−δ ¯¯ Q ¯¯ |x − u|

Ã

then µbδ is bounded from H 1 (Rn ) to Ln/(n−δ) (Rn ).

172 Proof. Let a be an atom supported in some cube Q. We write Z Rn

~ |µbδ (a)(x)|n/(n−δ) dx

Z

=

2Q

~ |µbδ (a)(x)|n/(n−δ) dx

Z

+

~

(2Q)c

|µbδ (a)(x)|n/(n−δ) dx = I + II.

For I, taking 1 < p < n/δ and 1/q = 1/p − δ/n, we have ~

n/(n−δ)

I ≤ ||µbδ (a)||Lq

n/(n−δ)

|2Q|1−n/((n−δ)q) ≤ C||a||Lp

|Q|1−n/((n−δ)q) ≤ C.

~

For II, we first calculate Ftb (a)(x), we have ~ |Ftb (a)(x)|

≤

¯ ¯ ¯Y ¯ Z ¯m ¯ Ω(x − y) ¯ (bj (x) − (bj )Q ) a(y)dy ¯¯ ¯ n−1−δ |x−y|≤t |x − y| ¯j=1 ¯ Z m X X

|(~b(x) − ~bQ )σc

+

j=1 σ∈Cjm

|x−y|≤t

(

Ω(x − u) Ω(x − y) − ) n−1−δ |x − y| |x − u|n−1−δ

×(~b(y) − ~bQ )σ a(y)dy|

¯ ¯ Z m X ¯ ¯ X Ω(x − u) ¯~ ~b(y) − ~bQ )σ a(y)dy ¯¯ + ( ¯(b(x) − ~bQ )σc ¯ ¯ |x−y|≤t |x − u|n−1−δ j=1 σ∈C m j

= ν1 + ν2 + ν3 , ~ µbδ (a)(x)

=

~ ||Ftb (a)(x)||

≤

ÃZ ∞ 0

dt |ν1 |2 3 t

!1/2 ÃZ ∞

+

0

dt |ν2 |2 3 t

!1/2 ÃZ ∞

+

0

dt |ν3 |2 3 t

!1/2

= A(x) + B(x) + C(x). n+1 For A(x), denoting Γ(x) = {(z, t) ∈ R+ : |x−z| < t} and the characteristic function

of Γ(x) by χΓ(x) , we have A(x) =

≤

1/2  ¯2 ! Z ∞ ¯¯Z Ã m ¯ dt Y χ Ω(x − u) χ Ω(x − y) Γ(u) Γ(y) ¯ ¯   − a(y)dy ¯ |bj (x) − (bj )Q | ¯ ¯ t3 |x − y|n−1−δ |x − u|n−1−δ 0 ¯ Rn j=1 ¯1/2  ! !2 Ã ¯ Z ∞ ¯¯Z ¯ Ω(x − u) dt Ω(x − y) ¯ ¯  ¯ a(y) − a(y)dy χΓ(y) n−1−δ n−1−δ 3 |x − y| |x − u| t ¯¯ 0 ¯ Rn

×

m Y

|bj (x) − (bj )Q |

j=1

¯1/2 !2 ¯ Z ∞ ¯¯Z m Y ¯ (χ − χ )Ω(x − u) dt Γ(y) Γ(u) ¯ ¯  a(y)dy + ¯ |bj (x) − (bj )Q | |x − u|n−1−δ t3 ¯¯ j=1 0 ¯ Rn 

= A1 (x) + A2 (x),

173 by the following inequality: ¯ ¯ Ã ! ¯ Ω(x − y) Ω(x − u) ¯¯ |y − u| |y − u|γ ¯ − ¯≤C + , ¯ ¯ |x − y|n−1−δ |x − u|n−1−δ ¯ |x − u|n−δ |y − u|n−1−δ+γ

we have A1 (x) 

≤ C

Z ∞ "Z

≤C

Rn

0

Ã

Z

Ã

Rn

|y − u| |y − u|γ + |x − u|n−δ |x − u|n−1−δ+γ

|y − u|γ |y − u| + |x − u|n−δ |x − u|n−1−δ+γ

Ã

|Q|1/n+1 |Q|γ/n+1 + ≤C |x − u|n+1−δ |x − u|n−δ+γ

!

!

#2

|a(y)|dy ÃZ

|a(y)|

!

m Y

||a||L∞

∞

0

dt t3

1/2

dt  t3

!1/2

dy

m Y

|bj (x) − (bj )Q |

j=1 m Y

|bj (x) − (bj )Q |

j=1

|bj (x) − (bj )Q |,

j=1

thus ÃZ

!(n−δ)/n n/(n−δ)

(2Q)c

(A1 (x))

≤ C||a||L∞

×

m Y

dx ÃÃ

"∞ Z X k+1 Q\2k Q k=1 2

|Q|γ/n+1 |Q|1/n+1 + |x − u|n+1−δ |x − u|n+γ−δ

!

(n−δ)/n

n/(n−δ)



|bj (x) − (bj )Q |

dx

j=1

≤ C|Q|1+1/n ||a||L∞

 Z ∞ X  

 

2k+1 Q

k=1



+C|Q|1+γ/n ||a||L∞

≤C

∞ X



(2−k + 2−kγ ) 

k=1

≤ C||~b||BM O . For A2 (x), we have



2k+1 Q



1 |2k+1 Q|

|2 Q|

δ/n

|2k Q|(n+1)/n 

Z ∞ X   k=1

k

k

|2 Q|

m Y

  2k+1 Q

m Y j=1

dx 

|bj (x) − (bj )Q |

j=1

δ/n

|2k Q|(n+γ)/n

Z

(n−δ)/n

n/(n−δ)

m Y

(n−δ)/n

n/(n−δ)



|bj (x) − (bj )Q |

dx

j=1

n/(n−δ)

|bj (x) − (bj )Q |

(n−δ)/n 

dx

174 A2 (x) 

≤ C

Z ∞ ÃZ Rn

0

|χΓ(y) − χΓ(u) | |a(y)|dy |x − u|n−1−δ

!2

1/2

dt  t3

m Y

|bj (x) − (bj )Q |

j=1

¯Z ¯1/2 ¯ 1 dt Z 1 dt ¯¯ ¯ ¯ ≤ C − ¯ |a(y)|dy Rn ¯ |x−z|≤t |x − y|2n−2−2δ t3 |u−z|≤t |x − u|2n−2−2δ t3 ¯ m Y Z

×

|bj (x) − (bj )Q |

j=1

¯ ¯ !1/2 ¯ ¯ dt 1 1 ¯ ¯ ¯ ¯ 3 − |a(y)|dy 2n−2−2δ 2n−2−2δ ¯ ¯t |x| |x|≤t,|x+y−u|≤t |x + y − u|

ÃZ

Z

≤ C ×

Rn m Y

|bj (x) − (bj )Q |

j=1

ÃZ

Z

≤ C

Rn

|x|≤t,|x+y−u|≤t

Z

≤ C

dt |y − u| 2n−1−2δ |x + y − u| t3

!1/2

|a(y)|dy

m Y

|bj (x) − (bj )Q |

j=1

m Y |y − u|1/2 |a(y)|dy |bj (x) − (bj )Q | |x + y − u|n+1/2−δ j=1

Rn

m Y

≤ C|Q|1/2n |x − u|−(n+1/2−δ)

|bj (x) − (bj )Q |,

j=1

thus ÃZ

!(n−δ)/n n/(n−δ)

(2Q)c

(A2 (x)) 

≤ C

∞ X k=1



Z

1



2−k/2 

dx

|2k+1 Q|

m Y

 2k+1 Q

n/(n−δ)

|bj (x) − (bj )Q |

(n−δ)/n 

dx

j=1

≤ C||~b||BM O . For B(x), we have m X X

 Z ∞ ÃZ ~ ~ (b(x) − bQ )σc  0

j=1 σ∈Cjm

=

m X X j=1

|x−y|≤t

Ω(x − y) a(y)(~b(y) − ~bQ )σ dy |x − y|n−1−δ

!2

(~b(x) − ~bQ )σc

σ∈Cjm



×

Z ∞ ÃZ 0

1/2

dt  t3

Rn

Ã

χΓ(y)

!

|Ω(x − y)| |Ω(x − u)| − a(y)(~b(y) − ~bQ )σ dy n−1−δ |x − y| |x − u|n−1−δ

!2

1/2

dt  t3

175 m X X

+

j=1

(~b(x) − ~bQ )σc

σ∈Cjm



×

Z ∞ ÃZ Rn

0

(χΓ(y) − χΓ(u) )Ω(x − u) a(y)(~b(y) − ~bQ )σ dy |x − u|n−1−δ

!2

1/2

dt  t3

,

similarly, we get |B(x)| m X X

≤C

Ã

|(~b(x) − ~bQ )σc |

j=1 σ∈Cjm

!

|Q|1/n |Q|γ/n |Q|1/2n + + ||~bσ ||BM O , |x − u|n+1−δ |x − u|n−δ+γ |x − u|n+1/2−δ

thus ÃZ

!(n−δ)/n

(B(x))n/(n−δ) dx

(2Q)c m X X

≤ C

j=1

Ã

×

σ∈Cjm

∞ X

(2−k + 2−kγ + 2−k/2 )

k=1

1

|2k+1 Q| ≤ C||~b||BM O .

Z 2k+1 Q

!(n−δ)/n ¯ ¯n/(n−δ) ¯~ ¯ ¯(b(x) − ~bQ )σc ¯ dx ||~bσ ||BM O

So, if m X Z X j=1 σ∈Cjm

(2Q)c

¯¯!n/(n−δ) ¯¯Z ¯¯ ¯¯ Ω(x − u) ¯¯ ¯¯ |(b(x) − bQ )σc | ¯¯ (~b(y) − ~bQ )σ a(y)dy ¯¯ dx ≤ C, n−1−δ ¯¯ ¯¯ Q |x − u|

Ã

then

Z

~

Rn

|µbδ (a)(x)|n/(n−δ) dx ≤ C.

This completes the proof of the Theorem 3.

2

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