A BMO estimate for the multilinear singular integral operator

22:3, 2006, 271-281

Analysis in Theory and Applications

A BMO ESTIMATE FOR THE MULTILINEAR SINGULAR INTEGRAL OPERATOR Qihui Zhang (University of Information Engineering, China) Received Dec. 14, 2005;

Revised June 15, 2006

Abstract The behavior on the space L∞ (Rn ) for the multilinear singular integral operator defined by Z ´ Ω(x − y) ` A(x) − A(y) − ∇A(y)(x − y) f (y) dy TA f (x) = n+1 |x − y| Rn is considered, where Ω is homogeneous of degree zero, integrable on the unit sphere and has vanishing moment of order one, A has derivatives of order one in BMO(Rn ). It is proved that if Ω satisfies some minimum size condition and an L1 -Dini type regularity condition, then for f ∈ L∞ (Rn ), TA f is either infinite almost everywhere or finite almost everywhere, and in the latter case, TA f ∈ BMO(Rn ).

Key words

multilinear singular integral operator, L1 -Dini type regularity condition

AMS(2000) subject classification

42B20

1

Introduction

We will work on Rn , n ≥ 2. Let Ω be homogeneous of degree zero, integrable on the unit sphere S n−1 and satisfy the vanishing condition  Ω(θ)θ dθ = 0.

(1)

S n−1

Let A be a function whose derivatives of order one belong to the space BMO(Rn ). For x, y ∈ Rn , set R(A; x, y) = A(x) − A(y) − ∇A(y)(x − y).

· 272 ·

Analysis in Theory and Applications

22:3, 2006

Define the multilinear singular integral operator TA by  Ω(x − y) R(A; x, y)f (y) dy. TA f (x) = p. v. |x − y|n+1 n R

(2)

A well known result of Cohen[1] states that if Ω ∈ Lip1 (S n−1 ), then TA is a bounded operator on Lp (Rn ) with bound C∇ABMO(Rn ) for 1 < p < ∞, and TA f exists almost everywhere for   q n−1 f∈ Lp (Rn ). Hofmann[2] improved the result of Cohen, and proved that Ω ∈ L (S ) 1 3, ξ∈S n−1

S n−1

then TA is bounded on L2 (Rn ). There are many papers concerning the operator TA , among other things, we refer the references [4,5]. The purpose of this paper is to consider the existence and the behavior on L∞ (Rn ) for the multilinear singular integral operator TA . It is easy to see that if f ∈ L∞ (Rn ), the integral  Ω(x − y) R(A; x, y)f (y) dy n+1 |x−y|> |x − y| may be divergent and hence the above definition no longer makes sense. We now define TA f by  Ω(x − y) R(A; x, y)f (y) dy, (3) TA f (x) = lim n+1 →0  0 and suitable functions f and h,    −1 |f (x)h(x)| dx ≤ C |f (x)| log(2 + a|f (x)|) dx + Ca exp(|h(x)|) dx. E

E

(5)

E

2

Proof of Theorem 1

We begin with some preliminary lemmas. Lemma 1[3] . Let A be a function on Rn with derivatives of order one in Lp (Rn ) for some p > n. Then

 1  1/p |A(x) − A(y)| ≤ C|x − y| |∇A(z)|p dz , y |Ix | Ixy

√ where Ixy is the cube centered at x with sides parallel to the axes and side length is 2 n|x − y|. Lemma 2.

Let Ω be homogeneous of degree zero, integrable on the unit sphere. If there

exists a constant α > 0 such that |y| < αR, then   |Ω(x − y) − Ω(x)| dδ dx ≤ C ω(δ) . n |x| δ R 0,

θ>1

(6)

N in (6)) r

 ∞   N |Ω(x − z) − Ω(x0 − z)| dz k + log r |x − z|n k k−1 B(x, 2 N )\B(x, 2 N) k=1

1−q  ∞   q k + log Nr |Ω(x − z)|∇A(z) − mB(x, 2k N ) (∇A) dz +C k |B(x, 2 N )| B(x, 2k N ) k=1

 1−q ∞   q k + log Nr +C |Ω(x0 − z)|∇A(z) − mB(x, 2k N ) (∇A) dz |B(x, 2k N )| B(x, 2k N )

≤

C

=

E1N + E2N + E3N .

k=1

The term E1N is easy to deal with. In fact, it follows from Lemma 2 that E1N

≤

C

≤

C

∞   k k−1 N ) k=1 Cr/(2 N )