J Fourier Anal Appl https://doi.org/10.1007/s00041-018-9613-7

The Multilinear Littlewood–Paley Operators with Minimal Regularity Conditions Mingming Cao1 · Kôzô Yabuta2

Received: 31 October 2017 / Revised: 5 March 2018 © Springer Science+Business Media, LLC, part of Springer Nature 2018

Abstract We investigate the quantitative weighted estimates for a large class of the multilinear Littlewood–Paley square operators. Our kernels satisfy the minimal regularity assumption, called L r -Hörmander condition. We respectively establish the pointwise sparse domination for the multilinear square functions and their iterated commutators. Based on them, we obtain the strong type quantitative bounds and endpoint estimates. We recover lots of known weighted inequalities for Littlewood–Paley operators. Significantly, the approach is dyadic, quite elementary and simpler than that presented previously. Keywords Multilinear · Sparse operators · Littlewood–Paley functions · Quantitative weighted estimates Mathematics Subject Classification Primary 42B25 · Secondary 42B20

Communicated by Loukas Grafakos.

B

Mingming Cao [email protected] Kôzô Yabuta [email protected]

1

Laboratory of Mathematics and Complex Systems, Ministry of Education, School of Mathematical Sciences, Beijing Normal University, Beijing 100875, People’s Republic of China

2

Research Center for Mathematical Sciences, Kwansei Gakuin University, Gakuen 2-1, Sanda 669-1337, Japan

J Fourier Anal Appl

1 Introduction Sharp weighted norm estimates have been a subject of many recent research papers in harmonic analysis. One important feature of these estimates is the fact that they imply the rate of growth of the norm of operators. The sharp dependence for the Hardy– Littlewood maximal function was first given by Buckley [3]. The optimal bound for Calderón–Zygmund operators was proven by Hytönen [23]. The author showed that any Calderón–Zygmund operator has a representation in terms of positive dyadic operators. In some sense, Hytönen’s representation theorem is the best since it is an identity. Later, a better class of dyadic operators, called sparse operators, was first introduced by Lerner [31]. But the sparse control of Calderón–Zygmund operators is of Banachnorm type. Soon after, it was improved by pointwise domination in [10,28,33,34]. There is a large literature adopting the ideology of pointwise bound in both linear and multilinear cases, Dini conditions and rough homogeneous kernels setting, for example [13,27,36–39]. Despite all this, the pointwise estimates do not always hold. Instead of them, sparse bilinear forms were first used to derive optimal norm estimates for singular non-integral operators [2]. It provides us a new approach to investigate certain much more ’singular’ operators, such as Bochner–Riesz means, rough singular integrals, Hilbert transforms along curves, oscillatory integrals. Since then many remarkable publications came to enrich the literature on this subject, see [1,9,11,12,29]. The purpose of this current work is to study quantitative weighted inequalities for multilinear Littlewood–Paley square functions. To fix ideas let us present some notion. We define the multilinear Littlewood–Paley g-function and its iterated commutators gb by setting ˆ

2 dt 1/2    , G t ( f )(x) t 0 ˆ ∞  2 dt 1/2  b   gb ( f)(x) = , G t ( f )(x) t 0 g( f)(x) =

∞

where b = b() = (b1 , . . . , b ) (1 ≤  ≤ m) and G t is an m-linear operator from S(Rn ) × · · · × S(Rn ) to the set of measurable functions on Rn , with kernel function K t (x, y) defined off the diagonal x = y1 = · · · = ym in (Rn )m+1 , satisfying G t ( f)(x) =

ˆ (Rn )m

K t (x, y)

m  j=1

f j (y j )d y, x ∈ /

m 

supp f j ,

j=1

and G bt ( f)(x) = [b1 , [b2 , . . . [b , G t ] . . .]2 ]1 ( f)(x) ˆ  m   (bi (x) − bi (yi ))K t (x, y) f j (y j )d y, = (Rn )m i=1

j=1

J Fourier Anal Appl

 for any x ∈ / mj=1 supp f j . We will seek the minimal regularity assumptions on the kernels K t . Moreover, motivated by the above works, we will adopt the strategy of sparse control for g and gb . Essentially, it will reduce the original problem to consider the following type of multilinear operators associated with a sparse family S of cubes ⎛

⎞1/γ m

 γ AS , p0 ,γ ( f)(x) := ⎝  f i  Q, p0 1 Q (x)⎠ , Q∈S i=1

  f  Q, p0 :=

| f (x)| p0 d x

1/ p0

.

Q

where γ > 0 and p0 ≥ 1. We will omit the subscript when p0 = 1 or γ = 1. The A p –A∞ estimates for AS , p0 ,γ in two-weight setting have been obtained in [13,21]. Moreover, sparse bound for commutators of Calderón–Zygmund operators was first given by Lerner et al. [36]. The main contributions of the current paper are the following. First, the minimal regularity conditions for the multilinear kernels are formulated along the lines of [37] and [48], establishing the appropriate class of multilinear square operators. Second, we prove pointwise sparse domination theorems for multilinear square operators and their iterated commutators. Thus, we improve the previous pointwise estimates by sharp maximal functions. Additionally, it possibly enables us to investigate two-weight quantitative estimates for multilinear iterated commutators just as the results in [20]. This will be our further topics. Finally, we obtain quantitative weighted inequalities and endpoint estimates. It is the first time to do such in terms of iterated commutators of multilinear square functions. Definition 1.1 Let 1 ≤ r < ∞. We say that the function {K t } satisfies the m-linear L r -Hörmander condition if there holds that Kr := sup sup

∞

Q x,x ∈ 1 Q k=1 2

⎛ ×⎝

m

|2k Q| r

ˆ

ˆ (2k Q)m \(2k−1 Q)m

0

∞

dt |K t (x, y) − K t (x , y)|2 t

 r 2

⎞ 1 r

d y⎠

< ∞.

When r = 1, the above formula is understood as K1 := sup sup

∞

Q x,x ∈ 1 Q k=1 2

ˆ ×

0

∞

|2k Q|m

esssup y∈(2k Q)m \(2k−1 Q)m

|K t (x, y) − K t (x , y)|2

dt t

1 2

< ∞.

We will write Hr for the class of kernels satisfying the L r -Hörmander condition.

J Fourier Anal Appl

Throughout this article, the notation τm will always denote the set {1, . . . , m}. The symbol |τ | denotes the number of the elements in τ ⊂ τm . For any subset τ of τm , we set τ = τm \ τ to be the complementary set. Some other notations and definitions will be given in the next section. We are ready now to state our main results. Theorem 1.2 Let 1p = p11 + · · · + p1m with 1 ≤ r < pi < ∞, i = 1, . . . , m. Assume that g is bounded from L r ×· · ·×L r to L r/m,∞ and its kernel {K t } satisfies the m-linear 1−( pi /r ) L r -Hörmander condition. Set σi = ωi , i = 1, . . . , m. If b = (b1 , . . . , b ) ∈ B M O  (1 ≤  ≤ m) and ω  ∈ A p/r , then we have max {1, 1p (

1≤i≤m  A p/r g( f) L p (νω ) ≤ cm,n,r, p C [ω]

pi r )}

×

m 

 f i  L pi (ωi ) ,

(1.1)

i=1 max {1, 1p (

gb ( f) L p (νω ) ≤ cm,n,r, p C [ω]  A p/r

1≤i≤m

×

 

b j  B M O

j=1

m 

pi r )}

⎛

×⎝

τ ⊂τ

|τ |

[νω ] A∞



⎞ [σ j ] A∞ ⎠

j∈τ \τ

 f i  L pi (ωi ) ,

(1.2)

i=1

where C = Kr + ||g|| L r ×···×L r →L r/m,∞ . Moreover, the exponent on [ω]  A p is the best possible. Theorem 1.3 Assume that g is bounded from L r × · · · × L r to L r/m,∞ and its kernel  ∈ A1 , then we {K t } satisfies the m-linear L r -Hörmander condition, 1 ≤ r < ∞. If ω have 1/m m ˆ

   | f i (x)|r m  ≤C ωi d x . (1.3) νω x; |g( f )(x)| > λ λr Rn i=1

If b = (b1 , . . . , b ) ∈ νω



B M O

x; |gb ( f)(x)| > λm

(1 ≤  ≤ m) and ω  ∈ A1 , then for any λ > 0



≤ Cb,r,

m ˆ  i=1

Rn

 r,

 1/m | f i (x)| ωi d x , λ

(1.4)

where r, (t) = t r (1 + log+ t)r  , Moreover, the result is sharp in the sense that  can not be replaced by any nonnegative number k with k < . In order to demonstrate the above results, we establish the pointwise sparse domination theorems for g and gb respectively. Theorem 1.4 Let 1 ≤ r < ∞. Assume that g is bounded from L r × · · · × L r to L r/m,∞ and its kernel {K t } satisfies the m-linear L r -Hörmander condition. Then, for any compactly supported functions f i ∈ L r (Rn ), i = 1, . . . , m, there exists a sparse family S such that for a.e. x ∈ Rn ,

J Fourier Anal Appl

|g( f)(x)| ≤ cm,n,r C

m



 f i  Q,r 1 Q (x).

(1.5)

Q∈S i=1

Theorem 1.5 Let 1 ≤ r < ∞ and 1 ≤  ≤ m. Assume that g is bounded from L r × · · · × L r to L r/m,∞ and its kernel {K t } satisfies the m-linear L r -Hörmander condition. Then, for any compactly supported functions f i ∈ L r (Rn ), i = 1, . . . , m, there exist 3n sparse families S j such that for a.e. x ∈ Rn , n

gb ( f)(x) ≤ cm,n,r C

3

j=1 τ ⊂τ



where bS ,τ ( f) =

Q∈S

λbQ,τ 1 Q , and

λbQ,τ (x) = ×

bS j ,τ ( f)(x),

 j∈τ \τ



|bi (x) − bi,Q | f i  Q,r

i∈τ

(b j − b j,Q ) f j  Q,r ×



 f k  Q,r .

k∈τm \τ

We note here that our results cover not only Littlewood–Paley operators but also singular integral operators, as are discussed in Sects. 5 and 6. This article is organized as follows. Necessary definitions and notation will be presented in Sect. 2. Section 3 is devoted to demonstrating quantitative weighted inequalities and endpoint estimates for commutators. The proof of sparse domination theorems will be given in Sect. 4. After that, in Sect. 5, we will prove the sharpness of Theorems 1.2 and 1.3 via some examples. In Sect. 6, after comparing with different known kernels conditions, we recover the previous weighted inequalities for Littlewood–Paley operators. Moreover, we deeply discuss the improvement of the pointwise sparse domination Theorem 1.4. Finally, in Appendix, we present some careful observations about the inverse of L p (log L)a functions.

2 Preliminaries 2.1 Dyadic Grids Denote by (Q) the sidelength of the cube Q. Given a cube Q 0 ⊂ Rn , let D(Q 0 ) denote the set of all dyadic cubes with respect to Q 0 , that is, the cubes obtained by repeated subdivision of Q 0 and each of its descendants into 2n congruent subcubes. Definition 2.1 A collection, D of cubes is said to be a dyadic grid if it satisfies (1) For any Q ∈ D, (Q) = 2k for some k ∈ Z. (2) For any Q, Q ∈ D, Q ∩ Q = {Q, Q , ∅}. (3) The family Dk = {Q ∈ D; (Q) = 2k } forms a partition of Rn for any k ∈ Z.

J Fourier Anal Appl

Definition 2.2 A subset S of a dyadic grid is said to be η-sparse, 0 < η < 1, if for every Q ∈ S, there exists a measurable set E Q ⊂ Q such that |E Q | ≥ η|Q|, and the sets {E Q } Q∈S are pairwise disjoint. The following lemma given by Hytönen et al. [25, Lemma 2.5] enables us to clearly understand the structure of dyadic grids. Lemma 2.3 There are 3n dyadic grids Du , u ∈ {0, 13 , 23 }n such that for any cube Q ⊂ Rn and k ∈ N, one can find a cube R ∈ Du satisfying Q ⊂ R, 2k Q ⊂ R (k) and 3(Q) < (R) ≤ 6(Q), where     Du ≡ 2−k [0, 1)n + m + (−1)k u ; k ∈ Z, m ∈ Zn . 2.2 Multiple Weights Definition 2.4 Let 1 ≤ p1 , . . . , pm < ∞. Given ω  = (ω1 , . . . , ωm ), where each ωi is a nonnegative function on Rn , we say that ω  satisfies the A p condition if  [ω]  A p := sup Q

where νω = (inf Q ωi )−1 .

m

p/ p j . j=1 ω j

Q

νω

 m  i=1



 p/ p i

ωi 1− pi

< ∞,

Q

When pi = 1, (

ffl



Q



ωi 1− pi )1/ pi is understood as

The above multiple weights were introduced by Lerner et al. [35]. If m = 1, the multiple A p weights coincide with the classical Muckenhoupt A p weights. In the linear case, the A1 condition is given by [ω] A1 := sup

x∈Rn

Mω(x) < ∞, ω(x)

where Mω is the Hardy–Littlewood maximal function of ω. Also we define the A∞ constant by ˆ 1 [ω] A∞ := sup M(ω1 Q )d x. Q ω(Q) Q Then we introduce the sharp reverse Hölder’s property of A∞ weights, which was proved in [26]. Lemma 2.5 Let ω ∈ A∞ . Then there holds  ω(x)rω d x Q

where rω = 1 +

1 νn [ω] A∞

1/rω

≤2

ω(x)d x, Q

and νn = 2n+11 . Notice that the conjugate rω  [ω] A∞ .

J Fourier Anal Appl

Lemma 2.6 Let

1 p

=

1 p1

+ ··· +

1 pm

with 1 ≤ p1 , . . . , pm < ∞. Then we have 1− pi

ω  ∈ A p if and only if νω ∈ Amp and ωi 1− pi

In the case pi = 1, the condition ωi

∈ Ampi . 1/m

∈ Ampi is understood as ωi

∈ A1 .

The above characterization of multiple weights was given in [35]. We refer the reader to [6] for the further discussion on the multiple weights. 2.3 Sharp Maximal Functions Given δ > 0, the maximal function Mδ is defined by δ

Mδ f (x) = (M(| f | ))

 1/δ

δ

(x) = sup Qx

| f (y)| dy

1/δ .

Q

In addition, M is the sharp maximal function of Fefferman and Stein, ˆ ˆ 1 1

M f (x) = sup inf | f (y) − c|dy ≈ sup | f (y) − f Q |dy x∈Q c |Q| Q x∈Q |Q| Q and

Mδ f (x) = M (| f |δ )1/δ (x), ffl

where f Q = Q f (x)d x. In this paper, we will use MD,δ to denote the dyadic version of sharp maximal function with respect to D. The following inequalities, due to Fefferman and Stein [14], are very powerful to obtain weighted norm inequalities, especially the weak-type estimates. The applications of them for singular integrals, square functions and their corresponding commutators can be found in [7,35,44,48,49]. Lemma 2.7 Suppose that 0 < p, δ < ∞ and ω ∈ A∞ . (i) There is a constant C depending only on [ω] A∞ such that   Mδ ( f ) L p (ω) < C  Mδ ( f ) L p (ω) , for any function f for which the left-hand side is finite. (ii) If ϕ : (0, ∞) → (0, ∞) is doubling, there is a constant C depending on [ω] A∞ and the doubling constant of φ such that

   

sup ϕ(λ)ω x; Mδ f (x) > λ < C sup ϕ(λ)ω x; Mδ f (x) > λ , λ>0

λ>0

for every function f such that the left-hand side is finite.

J Fourier Anal Appl

2.4 Young Functions and Orlicz Spaces A function φ : [0, ∞) → [0, ∞) is said to be a Young function, if φ is continuous, convex, increasing function such that φ(0) = 0 and φ(t)/t → ∞ as t → ∞. The φ-norm of a function f over a set E with finite measure is defined by   f φ,E = inf λ > 0;

φ E



| f (x)|  dx ≤ 1 . λ

For a given Young function φ, one can define a complementary function ¯ φ(s) = sup{st − φ(t)}, s ≥ 0. t>0

Then the following Hölder’s inequality holds

E

| f (x)g(x)|d x ≤ 2 f φ,E gφ,E ¯ .

Moreover, the more generalized Hölder’s inequality on Orlicz spaces due to O’Neil [42] holds. Lemma 2.8 If A, B and C are Young functions satisfying A−1 (t)C −1 (t) ≤ B −1 (t), for any t > 0, then for all functions f, g and any measurable set E ⊂ Rn , it holds  f gB,E ≤ 2 f A,E gC ,E . Here A−1 (x) = inf{y ∈ R : A(y) > x} for 0 ≤ x ≤ ∞, where inf ∅ = ∞. Some other inequalities are also necessary. Lemma 2.9 Let r > 0 and b ∈ B M O. Then there is a constant C > 0 independent of b such that the following inequalities hold 

C −1

| f | ≤  f  L(log L),Q ≤ C

Q



1 r

|b − b Q |r

sup



1 r +1

;

(2.1)

Q

≤ Cb B M O ;

(2.2)

≤ CbrB M O ;

(2.3)

Q

Q

|b − b Q |r 

1

exp L r ,Q

Q

where

| f |r +1

1 s

=

1 s1

| f 1 . . . f k g| ≤ C f 1 exp L s1 ,Q · · ·  f k exp L sk ,Q g + ··· +

1 sk

with s1 , . . . , sk ≥ 1.

1

L(log L) s ,Q

,

(2.4)

J Fourier Anal Appl

The inequalities (2.9) and (2.9) can be found in [15]. The last two inequalities were proved in [43]. We will also employ several times the Kolmogorov’s inequality  f  L p (Q, d x ) ≤ c p,q  f  L q,∞ (Q, d x ) , |Q|

(2.5)

|Q|

for 0 < p < q < ∞. Additionally, the following another version of Kolmogorov’s inequality is very useful to us. The proof is trivial. See, for example [22]. Lemma 2.10 Let s ≥ 0 and 1 ≤ q < ∞. If the operator T satisfies 

ˆ |{x; |T f (x)| > λ}| 

Rn

s

| f (x)|q λq

 d x,

where s (t) = t (1 + log+ t)s , then there exists a positive constant c p,q such that for any 0 < p < q, set E with finite measure, and appropriate function f with supp( f ) ⊂ E,  1/ p 1/q p |T f (x)| d x ≤ c p,q  f q  L(log L)s ,E . E

m , we define the multilinear Orlicz Given a sequence of Young functions { i }i=1 maximal operator by m   f i  i ,Q . M  ( f)(x) := sup Qx i=1

In particular, if i (t) = t, i = 1, . . . , m, we by M denote M  . m be a sequence of submultiplicative Young function. If ν ∈ Lemma 2.11 Let { i }i=1 ω  A1 , then there holds

νω



x; M  ( f)(x)>λm





m ˆ  i=1



Rn

1 ◦ · · · ◦ m

 1/m | f i (x)| ωi (x)d x . λ (2.6)

The proof of Lemma 2.11 is based on the following. Lemma 2.12 Let m ∈ N and E be any set. Let i be a submultiplicative Young function, i = 1, . . . , m. Then there is a constant c such that whenever 1<

m 

 f i  i ,E

i=1

holds, then m  i=1

 f i  i ,E ≤ c

m  i=1

1 ◦ · · · ◦ m (| f i (x)|) d x. E

J Fourier Anal Appl

Lemma 2.12 is a general version of Lemma 6.2 in [17]. The induction provides us an effective way to show it. The proof follows the scheme of that in [17]. Proof of Lemma 2.11 Introduce the notations   E λ,k = {x ∈ E λ ; |x| ≤ k} and E λ = x ∈ Rn ; M  ( f)(x) > λm . By the monotone convergence theorem, it suffices to bound E λ,k . For any x ∈ E λ,k , there is a cube Q x such that λ < m

m 

 f j  j ,Q .

j=1

Hence, {Q x }x∈E λ,k is a family of cubes covering E λ,k . Using a covering argument, we obtain a finite family of disjoint cubes {Q i } whose dilations cover E λ,k such that |E λ,k | 



|Q i |

λm <

and

m 

i

 f j  j ,Q i .

j=1

Lemma 2.12 implies that 1≤

m m    fj     λ j ,Q i j=1

j=1

 1 ◦ · · · ◦ m Qi

 | f j (x)| d x. λ

Thus we obtain

νω (E λ,k )m 



m νω (Q i )

⎛ ⎝



i

i

⎛ m ˆ

 ⎝ i

≤ ≤

j=1

m ˆ  j=1 i m ˆ  n j=1 R

This completes the proof.

|Q i |

m  j=1

 1 ◦ · · · ◦ m Qi

 1 ◦ · · · ◦ m

Qi

 1 ◦ · · · ◦ m

⎞m inf ω j (x)1/m ⎠

x∈Q i

⎞  1/m m | f j (x)| ⎠ w j (x)d x λ

 | f j (x)| w j (x)d x λ

 | f j (x)| w j (x)d x. λ  

J Fourier Anal Appl

3 The Multilinear Weighted Estimates In this section, we will show Theorems 1.2 and 1.3. We will focus on the estimates for the multilinear iterated commutators, since the weighted inequalities for the multilinear sparse operators are essentially contained in [21,39]. Thus, the inequality (1.1) immediately follows from Theorem 1.4. For the sake of simplicity, we only prove the case  = m. 3.1 Strong Type Bounds In order to show (1.2), we first present a useful lemma. Lemma 3.1 Let s > 1, t > 0, and ω ∈ A∞ . Then there holds 1/s

ω1/s  L s (log L)st ,Q  [ω]tA∞ ω Q ,  f ω L(log L)t ,Q 

[ω]tA∞

(3.1)

inf Mω (| f | )(x) s

1/s

x∈Q

ω Q ,

(3.2)

Proof The inequality (3.1) follows from Lemma 2.5. Indeed, ω1/s sL s (log L)st ,Q  ω L(log L)st ,Q   1  1+ (rω − 1)st

ω(x) d x rω

1/rω

Q

 [ω]stA∞ ω Q . Now we show the inequality (3.2). By Lemma 2.8 and (3.1), we have   f ω L(log L)t ,Q 

1/s | f (x)|s ωd x Q

 [ω]tA∞



1 ω(Q)



ω1/s  L s (log L)s t ,Q 1/s

ˆ | f (x)|s ωd x

ω Q

Q

 [ω]tA∞ inf Mω (| f |s )(x)1/s ω Q . x∈Q

It is worth pointing out that in order to apply Lemma 2.8, we need the inverse of L log L functions, which are contained in Propositions 7.1 and 7.5. This completes the proof.   By Theorem 1.5, the inequality (1.2) immediately follows from the following. 1− p /r

Theorem 3.2 Let 1p = p11 + · · · + p1m with 1 ≤ r < pi < ∞, and ωi = σi i ,  ∈ A p/r , then for any τ ⊂ τm and a sparse family i = 1, . . . , m. If b ∈ B M O m and ω S

J Fourier Anal Appl

 b   

S ,τ L p1 (ω1 )×···×L pm (ωm )→L p (νω )





|τ | [νω ] A∞

max {1, 1p (

1≤i≤m

j∈τ

1/ p j

Proof Changing f j to f j ω j the estimate:

[σ j ] A∞ [ω]  A p/r

pi r )}

m 

bi  B M O .

i=1

−1/ p j

σj

  b  ( fσ 1/r ) p S ,τ L (ν

ω )



, we see that our desired result is equivalent to



|τ | [νω ] A∞

×

m 

j∈τ

max {1, 1p (

1≤i≤m

[σ j ] A∞ [ω]  A p/r

pi r )}

bi  B M O  f i  L pi (σi ) ,

i=1

where ( fσ 1/r ) := ( f 1 σ1 , . . . , f m σm ). Let us begin with showing the case p > 1. Let g ∈ L p (νω ) be a nonnegative function satisfying ||g|| L p (νω ) = 1. Applying (2.9) and (2.9), we have 1/r

ˆ

1/r

bS ,τ ( fσ 1/r )(x)g(x)νω d x  

 1/r  f i σi Q,r |bi (x) − bi,Q |g(x)νω d x  |Q|

Rn

Q i∈τ

Q∈S

  (b j − b j,Q )r 1/r ×

1 exp L r ,Q

j∈τ





Q∈S

×

|Q|gνω  L(log L)|τ | ,Q

  (b j − b j,Q )r 1/r



Q∈S

⎛ ⎝

i∈τ 1



1/r 

f i σi

i∈τ

Q,r

L(log L)r ,Q

j



exp L r ,Q

j∈τ c



 r 1/r  f σj f i σ 1/r

j

× |Q|gνω  L(log L)|τ | ,Q

m 

Q,r

bi − bi,Q exp L ,Q

 r 1/r  f σj

L(log L)r ,Q

j

   f r σ j 1/r j∈τ



L(log L)r ,Q

⎞ ⎠

bi  B M O .

i=1

Now we analyze the inner terms. We have 

1/r r Q,r

f i σi

  ≤ inf Mσi ( f ir )(x)σi  Q ≤ Mσi ( f ir ) · σi Q . x∈Q

J Fourier Anal Appl

The inequality (3.1) implies that 

 r 1/r  f σj

L(log L)r ,Q

j

 [σ j ] A∞

1 |Q|

1/r

ˆ Mσ j (| f j |

rsj

Q

)(x)

1/s j

σjdx

,

(3.3)

where 1 < r s j < p j , j ∈ τ . In addition, it yields by (3.1) again |τ |

|Q|gνω  L(log L)|τ | ,Q  [νω ] A∞ inf Mνω (|g|s )(x)1/s νω (Q), x∈Q

where 1 < s < p . Collecting these estimates, we obtain ˆ Rn

bS ,τ ( fσ 1/r )(x)g(x)νω d x |τ |

 [νω ] A∞ ×





[σ j ] A∞

j∈τ c

m  i=1

Mσ j (| f j |

)

r s j 1/s j

j∈τ c |τ |

 [νω ] A∞



ˆ bi  B M O

[σ j ] A∞

j∈τ c

m  i=1

· σj

1/r Q

Rn

 1/r Mσi ( f ir ) · σi  Q Q∈S i∈τ

1 Q (x) · Mνω (|g|s )(x)1/s νω d x

bi  B M O Mνω (|g|s )1/s  L p (νω )

   1/r r    1/r r r s j 1/s j M × M ( f ) · σ  (| f | ) · σ 1Q  σi i i Q σj j j   p Q∈S

Q

j∈τ

i∈τ

L

r

(νω )

We need the sharp estimate (4.1) in [5] AS ,γ ( fσ ) L p (νω ) 

max [ω]  A p

1 γ

,

p1 pm p ,..., p



m 

 f i  L pi (σi ) ,

i=1

for any ω  ∈ A p . Accordingly, we deduce that ˆ

bS ,τ ( fσ 1/r )(x)g(x)νω d x

Rn



|τ | [νω ] A∞

×

m 

 j∈τ

bi  B M O

i=1



|τ | [νω ] A∞



max r, rp ( 1/s [σ j ] A∞ |g|s  p /s [ω]  Aip/r (νω ) L

   Mσ ( f r )1/rpi i i L

i∈τ

 j∈τ c



max 1, 1p (

[σ j ] A∞ [ω]  A p/r i

r

(σi )

pi r )



pi r )



× r1

   Mσ (| f j |r s j )1/s j 1/rp j j j∈τ

m  i=1

L

bi  B M O  f i  L pi (σi ) .

r

(σ j )

.

J Fourier Anal Appl

Next, we turn our attention to the case 0 < p ≤ 1. Using (2.9), (2.9) and (3.1), we have   b  ( fσ 1/r ) p p S ,τ L (νω )  

ˆ   1/r  p 1/r  p p f i σi Q,r |bi (x) − bi,Q | νω d x × (b j − b j,Q ) f j σ j Q,r ≤ Q∈S





Q∈S

Q i∈τ

j∈τ

  p/r  |Q|νω  L(log L) p|τ | ,Q  f ir σi  Q (bi − bi,Q ) p 

  (b j − b j,Q )r  p/r ×



Q∈S

1

exp L r ,Q

j∈τ

1 p

exp L

i∈τ

,Q

 r  p/r  f σj

L(log L)r ,Q

j

⎛ ⎞      p/r p/r ⎝ f ir σi Q Mσ j (| f j |r s j )1/s j · σ j Q ⎠ νω (Q) j∈τ

i∈τ p|τ |

× [νω ] A∞

 j∈τ

p

[σ j ] A∞

m 

p

bi  B M O .

i=1

Following the same technique used in the proof of Theorem 1.2 [39], we obtain p  bS ,τ ( fσ 1/r ) L p (νω )



p|τ | [νω ] A∞

×

 i∈τ

 j∈τ c



p|τ |

 j∈τ c



m 

p

bi  B M O

i=1

p /r  f ir  i pi L r (σi )

 [νω ] A∞

pi r )

max ( p [σ j ] A∞ [ω]  Aip/r

   Mσ (| f j |rs j )1/s j  p jp/r j j L

j∈τ c

p



max (

[σ j ] A∞ [ω]  Aip/r

pi r )



m  i=1

p

r

(σ j )

bi  B M O

m 

p

 f i  L pi (σi ) .

i=1

 

So far, we have proved our theorem. 3.2 Endpoint Bounds

In order to show the endpoint behavior, we first introduce some notation. Given τ ∈ τm and r ∈ [1, ∞), we define AτS ,L(log L)r ( f)(x) := MτL(log L)r ( f)(x)

  1/r  f i  Q,r ×  f jr  L(log L)r ,Q 1 Q (x),

Q∈S i∈τ

:= sup



 fi  Q ×

Q i∈τ

j∈τ



 f j  L(log L)r ,Q 1 Q (x).

j∈τ

A simple calculation gives that for 0 < δ ≤ 1



MD,δ AτS ,L(log L)r ( f) (x)  MτL(log L)r ( fr )(x)1/r .

(3.4)

J Fourier Anal Appl

Together with Lemma 2.11, this implies νω



x; AτS ,L(log L)r ( f)(x)

>λ

m





m ˆ  Rn

i=1

r(|τ |)



1/m  | f i (x)|r , ωi d x λr (3.5)

for any νω ∈ A1 . Hence, the inequality (1.3) follows from Theorem 1.4. Proof of (1.3). We claim that there holds νω



x; |gb ( f)(x)| > λm



≤ Cb

m ˆ

 τ ⊂τ i=1

Rn

r(−|τ |)



 1/m | f i (x)|r ω d x , i λr (3.6)

j

!" # where r (t) = and = r ◦ · · · ◦ r . Then (1.3) immediately follows from (3.2). To check this, we proceed as follows. Since t (1 + log+ t)r

( j) r

r ◦ r (t) = t (1 + log+ t)r (1 + log+ {t (1 + log+ t)r })r , we get t (1 + log+ t)2r ≤ r ◦ r (t) ≤ t (1 + log+ t)r (1 + log+ {t r +1 })r ≤ (r + 1)r t (1 + log+ t)r (1 + log+ t)r = (r + 1)r t (1 + log+ t)2r . Inductively, we have ( j)

t (1 + log+ t) jr ≤ r (t) ≤ C j,r t (1 + log+ t) jr , ( j)

that is, r

≡ jr . Moreover

t r (1 + log+ t) jr ≤ jr (t r ) = t r (1 + log+ t r ) jr = t r (1 + r log+ t) jr ≤ r jr t r (1 + log+ t) jr . So, we get for τ ⊂ τ r(−|τ |) (t r ) ≤ r  (t r ) ≤ Cr, t r (1 + log+ t)r  = Cr, r, (t). Thus, what is left to do is to show (3.6). We here only present the proof in the case   = m. Note that bS ,τ ( f)(x)  b,τ S ,τ ( f )(x), where

J Fourier Anal Appl

 b,σ S ,τ ( f )(x) :=



|bi (x) − bi,Q |

Q∈S i∈σ

×

  1/r  f i  Q,r × b j  B M O  f jr  L(log L)r ,Q . j∈τ

i∈τ

Accordingly, following the scheme of the proof in [35, Theorem 3.16], we deduce (3.6) from Theorem 3.3, (3.4), Lemmas 2.7 and 2.11.   Theorem 3.3 Let 1 ≤ r < ∞ and 0 < δ < min{, m1 }. Then there holds that m   



) (x)  MD,δ b,τ ( f bi  B M O MτL(log L)r ( fr )(x)1/r + M AτS ,L(log L)r ( f) (x) S ,τ i=1

+



σ τ i∈σ



) (x). bi  B M O M b,σ ( f S ,τ

Proof In order to embody an iterative procedure and avoid cumbersome notations, we only prove the case m = 3 and τ = {1, 2}. We may assume that b j  B M O = 1, j ∈ τ . For a fixed cube Q ∈ D containing x, denote c=



|bi,Q − bi,Q | f i  Q,r ×



1/r

j∈τ

Q∈S i∈τ Q⊃Q

b j  B M O  f jr  L(log L)r ,Q .

Then by inserting bi,Q we obtain  Q

  1/δ  δ  b,τ   ≤  S ,τ ( f )(x) − cδ  d x

Q

 δ 1/δ  b   S ,τ ( f)(x) − c d x

≤ N1 + N2 + N3 + N4 , where 

 2

N1 =

Q

− bi,Q |

Q∈S

⎛ N2 = ⎝

|bi (x)

i=1

Q

−

1/r

 f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

δ

⎞1 δ

dx⎠ ,



|b1 (x) − b1,Q | |b2,Q Q∈S

1/r b2,Q | f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

1

δ

δ

dx

,

J Fourier Anal Appl

⎛ N3 = ⎝



|b2 (x) − b2,Q | |b1,Q

Q

Q∈S 1/r

− b1,Q | f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x) ⎛ N4 = ⎝ Q

δ

,

dx

  2   |bi,Q  Q∈S i=1

1

δ

 1/r bi,Q | f i  Q,r  f 3r  L(log L)r ,Q 1 Q (x) − c

−

1

δ

δ

.

dx

We will treat them separately. Selecting s1 , s2 , s3 ∈ (1, ∞) such that δs3 < , δs1 <  and s11 + s12 + s13 = 1. We have by Hölder’s inequality and (2.2)  N2 ≤

Q

|b1 (x) − b1,Q |δs1

⎛ ×⎝



Q

Q∈S



Q

⎛ + b1  B M O ⎝

1 δs1

1/r |b2,Q − b2,Q | f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

⎛  b1  B M O ⎝



Q∈S

|b2 (x) − b2,Q |



⎜ ×⎝

⎛ Q

⎝

Q∈S

⎞ dx⎠

Q∈S

1 δs1

δs

1/r  f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

 

b,{2}  b1  B M O Mδs1 S ,τ ( f) (x) + b1  B M O ⎛

1

1/r |b2 (x) − b2,Q | f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

 Q

δs

Q

|b2 (x) − b2,Q |δs2 d x ⎞δs3

 f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)⎠ 1/r

⎞ ⎟ dx⎠



1

⎞ dx⎠

δs

1

1 δs1

⎞ dx⎠

1 δs1

1 δs2

1 δs3

2

   b,{2} {3}  b1  B M O Mδs1 S ,τ ( f) (x) + bi  B M O Mδs3 AS ,L(log L)r ( f) (x) i=1 2

   b,{2} {3} bi  B M O M AS ,L(log L)r ( f) (x). ≤ b1  B M O M S ,τ ( f) (x) + i=1

Symmetrically, we have 2

   b,{1} {3} N3  b2  B M O M S ,τ ( f) (x) + bi  B M O M AS ,L(log L)r ( f) (x). i=1

J Fourier Anal Appl

In addition, it easily follows from Hölder’s inequality and (2.9) that N1 

2 

 bi  B M O M AτS ,L(log L)r ( f) (x).

i=1

It remains to analyze the last term. We dominate ⎛ N4 = ⎝

  2 Q

|b

i,Q

Q⊂Q i=1

1/r − bi,Q | f i  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

δ

⎞1 δ

dx⎠

≤ N41 + N42 + N43 + N44 , where ⎛ N41 = ⎝

 2 Q

⎛ N42 = ⎝

|b1 (x) − b1,Q |

Q



|b2 (x) − b2,Q |



Q

Q⊂Q i=1

δ

⎞1 δ

dx⎠ ,

1/r |b1 (x) − b1,Q | f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

Q⊂Q

  2

1/r  f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

1/r |b2 (x) − b2,Q | f 1  Q,r  f 2  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

Q⊂Q



⎛ N44 = ⎝

Q⊂Q



⎛ N43 = ⎝

|bi (x) − bi,Q |

i=1

Q



1/r |bi (x) − bi,Q | f i  Q,r  f 3r  L(log L)r ,Q 1 Q (x)

δ

δ

δ

⎞1 δ

dx⎠ , ⎞1 δ

dx⎠ ,

⎞1 δ

dx⎠ .

We begin with controlling N41 . Taking δi ∈ (0, ∞) and γi ∈ (0, 1) such that 1δ = 1 1 1 1 1 1 1 δ1 + δ2 + δ3 and δ3 = γ1 + γ2 + γ3 . Then Hölder’s inequality and (2.9) imply that  N41 ≤

Q

|b1 (x) − b

1,Q

⎛ ×⎝



2 



Q

bi  B M O

i=1

Q

Q⊂Q

δ1

Q

|b2 (x) − b

2,Q

1/r

2 



1  |

δ2

1

 f 1  Q,r  f 2  Q,r  f 3r  L(log)r ,Q 1 Q (x)

i=1

 ×

Q⊂Q

|

δ1

⎛ ⎝



Q

 f i  Q,r 1 Q (x)

γi

Q⊂Q

1/r  f 3r  L(log)r ,Q 1 Q (x)

γ3

1 dx

γ3

.

δ2

 δ3

⎞1

δ3

dx⎠

⎞1

γi

dx⎠

J Fourier Anal Appl



Observing that (3.2) implies that the dyadic operator f → weak (r, r ) type, and   ˆ

  r 1/r  x;  f  L(log)r ,Q 1 Q (x) > λ   

Rn

Q∈S

 r

Q∈S  f  Q,r 1 Q (x)

| f (x)|r λr

is of

 d x.

Together with Lemma 2.10, it yields that

N41  ≤

2  i=1 2 

1/r

bi  B M O  f i  Q ,r ×  f 3r  L(log L)r ,Q

bi  B M O MτL(log L)r ( f)(x).

i=1

Similarly, combining Hölder’s inequality, (2.9) and Lemma 2.10, we deduce that for any ti ∈ (0, δ −1 ) with 1 = t11 + t12 + t13 ⎛ N42  b1  B M O ⎝ ⎛ ×⎝

 ≤

2  i=1 2 

Q



Q

⎛ ×⎝



⎞

1 δt1

dx⎠

Q⊂Q

|b2 (x) − b2,Q | f 2  Q,r 1 Q (x)

δt2

⎞

1 δt2

dx⎠

Q⊂Q



Q

 f 1  Q,r 1 Q (x)

δt1

Q⊂Q

1/r

 f 3r  L(log L)r ,Q 1 Q (x)

δt3

⎞

1 δt3

dx⎠

1/r

bi  B M O  f i  Q ,r ×  f 3r  L(log L)r ,Q

bi  B M O MτL(log L)r ( f)(x).

i=1 r r,∞  ||b|| B M O , We just used the fact that if m = 1 and τ = {1}, then b,τ S ,τ  L →L which was proved in [36, Lemma 4.2]. Since N43 is symmetric to N42 , it yields that

N43 

2  i=1



bi  B M O MτL(log L)r ( f)(x).

J Fourier Anal Appl

Finally, applying the techniques in the estimates of N42 again, we obtain

N44 ≤

2 

⎛ ⎝

i=1



Q

⎛ ×⎝



2 

⎞

1 δti

⎠

Q⊂Q



Q

|bi (x) − bi,Q | f i  Q,r 1 Q (x)

δti

Q⊂Q

1/r

 f 3r  L(log L)r ,Q 1 Q (x)

δt3

⎞

1 δt3

dx⎠

bi  B M O MτL(log L)r ( f)(x).

i=1

Consequently, collecting the above estimates, we complete the proof.

 

4 Sparse Domination The goal of this section is to demonstrate sparse domination Theorems 1.4 and 1.5. We will combine the some ideas from [36,37]. 4.1 Sparse Bounds for Square Function Now to show the sparse domination theorem, we need a multilinear analogue of grand maximal truncated operator. Given an operator T , define MT ( f)(x) := sup esssup |T ( f)(ξ ) − T ( f · 13Q )(ξ )|. Qx

ξ ∈Q

Given a cube Q 0 , for x ∈ Q 0 , we also define a local version of MT by MT,Q o ( f)(x) :=

sup Qx,Q⊂Q 0

esssup |T ( f · 13Q 0 )(ξ ) − T ( f · 13Q )(ξ )|. ξ ∈Q

Based on the weak-type boundedness of the above maximal operator, Li [37] established the sparse pointwise domination for a class of more general multilinear operators. Theorem A Let 1 ≤ q ≤ r < ∞. Assume that T is bounded from L q × · · · × L q to L q/m,∞ and MT is bounded from L r × · · · × L r to L r/m,∞ . Then, for compactly supported functions f i ∈ L r (Rn ), i = 1, . . . , m, there exists a sparse family S such that for a.e. x ∈ Rn , |T ( f)(x)| ≤ cn,r,q C T

m

  f i  Q,r 1 Q (x), Q∈S i=1

J Fourier Anal Appl

where C T = ||T || L q ×···×L q →L q/m,∞ +||MT || L r ×···×L r →L r/m,∞ . Moreover, there holds that for a.e. x ∈ Q 0 m       | f i (x)| + MT,Q 0 ( f)(x). T ( f · 13Q 0 )(x) ≤ cn ||T || L q ×···×L q →L q/m,∞

(4.1)

i=1

Note that the above operator T may be a singular non-integral operator. It also does not require T to be multi-sublinear. Lemma 4.1 Let 1 ≤ r < ∞. Under the assumption of Theorem 1.4, there holds that ||Mg || L r ×···×L r →L r/m,∞  1.

(4.2)

More specifically, for any x ∈ Rn and δ ∈ (0, 1), Mg ( f)(x) ≤ CM(| f|r )(x) r + Mr δ/m (g( f))(x), 1

(4.3)

where C = Kr + cm,n,r,δ ||g|| L r ×···×L r →L r/m,∞ . Consequently, Theorem 1.4 follows from Theorem A and Lemma 4.1. Proof of Lemma 4.1 It suffices to show (4.1), since (4.1) is implied by (4.1) and the 1 facts M(| f|r )(x) r is bounded from L r × · · · × L r to L r/m,∞ and Mr δ/m is bounded on L r/m,∞ . Let x, ξ, ξ ∈ Q. Since g is sublinear, we deduce that |g( f)(ξ ) − g( f · 13Q )(ξ )| ˆ ∞  2 dt 1/2     ≤ G t ( f )(ξ ) − G t ( f · 13Q )(ξ ) t 0 ˆ ∞  2 dt 1/2 ˆ ∞  2 1/2     dt   ≤ ( f )(ξ ) + ( f · 1 )(ξ ) G t G t   3Q t t 0 0 ⎛ 2 ⎞1/2  ˆ ∞ ˆ m  dt ⎠ +⎝ |K t (ξ, y) − K t (ξ , y)| | f i (yi )|d y n m m t (R ) \(3Q) 0 i=1

:= g( f)(ξ ) + g( f · 13Q )(ξ ) + E. It follows from Minkowski’s inequality and L r -Hörmander condition that E≤

≤

ˆ

∞ ˆ

k+1 m k m k=0 (2 3Q) \(2 3Q)

∞

k=0

⎛

|2k+1 3Q| ⎝ m r

∞

|K t (ξ, y) − K t (ξ , y)|2

0

ˆ (2k+1 3Q)m \(2k 3Q)m

ˆ 0

∞

dt t

1/2  m

| f i (yi )|d y

i=1

dt |K t (ξ, y) − K t (ξ , y)|2 t

 r 2

⎞ 1 r

d y⎠

J Fourier Anal Appl

 ×

1 |2k+1 3Q|m

ˆ

m 

(2k+1 3Q)m i=1

1 r

| f i (yi )| d y r

≤ Kr · M( fr )(x) . 1 r

Taking L r δ/m average over ξ ∈ Q, we have by Kolmogorov’s inequality (2.5) |g( f)(ξ ) − g( f · 13Q )(ξ )| ≤ Kr · M( fr )(x) r + g( f · 13Q ) L r δ/m (Q, dξ ) + 1



|Q|

Q

 rmδ   rδ   m g( f )(ξ ) dξ

≤ Kr · M( fr )(x) + cm,r,δ g( f · 13Q ) L r/m,∞ (Q, dξ ) + Mr δ/m (g( f))(x). 1 r

|Q|

By the assumption that g is bounded from L r × · · · × L r to L r/m,∞ , we obtain that |g( f)(ξ ) − g( f · 13Q )(ξ )| ≤ Kr · M( fr )(x) r + Mr δ/m (g( f))(x) 1 ˆ m   r 1 r + g L r ×···×L r →L r/m,∞ | fi | |Q| 3Q i=1   1 ≤ Kr + cm,n,r,δ M( fr )(x) r + Mr δ/m (g( f))(x). 1

 

This shows (4.1).

4.2 Sparse Bounds for Iterated Commutators The remainder of this section is devoted to the proof of Theorem 1.5. We shall show only the case  = m. The other cases are somewhat complicated but similar. We will divide the proof in two steps, as follows. Note that by Lemma 2.3, there exist 3n dyadic grids D j such that for any cube Q ⊂ n R , one can find a cube R Q ∈ D j for some j, with 3Q ⊂ R Q and (R Q ) ≤ 12(Q). • Step 1 Let Q 0 ⊂ Rn be a fixed dyadic cube.  We shall show that there exists a disjoint family of cubes {Q j } ⊂ D(Q 0 ) with j |Q j | ≤ 21 |Q 0 | such that |gb ( f · 13Q 0 )(x)|1 Q 0 ⎛ ⎞

  1 1  r ⎝ |bi (x) − bi,R Q || f i |r  r × ⎠ 1 Q0 |(b j − b j,R Q 0 ) f j |r 3Q  3Q 0 0 0 τ ⊂τm

+

j

i∈τ

|gb ( f · 13Q j )(x)|1 Q j .

j∈τ

(4.4)

J Fourier Anal Appl

&

Denote E =

τ ⊂τm

(E τ1 ∪ E τ2 ), where

 E τ1

:= x ∈ Q 0 :

m 

| f i (x)|

 j∈τ

i=1

> αn

 i∈τ



E τ2 := x ∈ Q 0 :Mg,Q 0 > αn



|(b j,R Q 0 − b j (x))| 1

r | f i |r 3Q 0

 j∈τ

|(b j,R Q 0 − b j ) f j |r

 3Q 0 ,

 r1

 f i |i∈τ , (b j,R Q 0 − b j ) f j | j∈τ c (x) 1 r

| f i | 3Q 0 r

i∈τ

 j∈τ

|(b j,R Q 0

 − b j ) f j | 3Q 0 . r

 r1

1 Then, by Lemma 4.1, one can choose αn large enough such that |E| ≤ 2n+2 |Q 0 |. 1 , we Moreover, applying Calderón–Zygmund decomposition to 1 E on Q 0 at λ = 2n+1 obtain pairwise disjoint cubes {Q j } ⊂ D(Q 0 ) such that

1

2

|Q j | ≤ |Q j ∩ E| ≤ n+1

' 1 |Q j |, |E \ Q j | = 0. 2 j

Moreover it follows that

 j

|Q j | ≤ 21 |Q 0 | and Q j ∩ E c = ∅. We split

|gb ( f · 13Q 0 )(x)|1 Q 0 ≤ |gb ( f · 13Q 0 )(x)|1 Q 0 \∪ j Q j +

  gb ( f · 13Q )(x)1 Q j j j



  + gb ( f · 13Q 0 )(x) − gb ( f · 13Q j )(x) 1 Q j . j

We will focus on dominating the first and last terms, since the second one is desired. Note that G bt ( f)(x) = ˆ

 τ ⊂τm i∈τ

(Rn )m

=

(bi (x) − bi,R Q 0 )

K t (x, y)

 τ ⊂τm i∈τ

 i∈τ

f i (yi )

 j∈τ

(b j,R Q 0 − b j (y j )) f j (y j )d y

(bi (x) − bi,R Q 0 )G t

 f i |i∈τ , (b j,R Q 0 − b j ) f j | j∈τ (x).

J Fourier Anal Appl

By the inequality (4.1) and definition of E, we deduce that for x ∈ (Q 0 \ E c ) \

& j

Qj

     g 13Q 0 f i |i∈τ , 13Q 0 (b j,R Q 0 − b j ) f j | j∈τ (x) ≤ cn ||g|| L r ×···×L r →L r/m,∞

 j∈τ

|b j (x) − b j,R Q 0 |

m 

| f i (x)|

i=1

+ Mg,Q 0 ( f i |i∈τ , (b j,R Q 0 − b j ) f j | j∈τ )(x)   1 1 r r r | f i |r 3Q |(b − b ) f |  .  j,R j j Q 3Q 0 0 0 j∈τ

i∈τ

This implies that for a.e. x ∈ Q 0 \

& j

Qj

|gb ( f · 13Q 0 )(x)|1 Q 0 \∪ j Q j   

   ≤ |bi (x) − bi,R Q 0 | g 13Q 0 f i |i∈τ , 13Q 0 (b j,R Q 0 − b j ) f j | j∈τ (x) τ ⊂τm i∈τ





τ ⊂τm i∈τ

1

r |bi (x) − bi,R Q 0 || f i |r 3Q 0

 j∈τ

1

r |(b j,R Q 0 − b j ) f j |r 3Q 1 . 0 Q0

For the last term, it follows from the definition of E τ2 and Q j ∩ E c = ∅ that 

  gb ( f · 13Q 0 )(x) − gb ( f · 13Q j )(x) 1 Q j j

≤

 j τ ⊂τm i∈τ

   |bi (x) − bi,R Q 0 |g 13Q 0 f i |i∈τ , 13Q 0 (bk,R Q 0 − bk ) f k |k∈τ (x)

 

 − g 13Q j f i |i∈τ , 13Q j (bk,R Q 0 − bk ) f k |k∈τ (x)1 Q j



 ≤ |bi (x) − bi,R Q 0 | essinf Mg,Q 0 f i |i∈τ , (b j,R Q 0 − b j ) f j | j∈τ (ξ )1 Q j j τ ⊂τm i∈τ



 τ ⊂τm i∈τ

ξ ∈Q j 1

r |bi (x) − bi,R Q 0 || f i |r 3Q 0

 j∈τ

1

r |(b j,R Q 0 − b j ) f j |r 3Q 1 . 0 Q0

Collecting the above estimates, we complete the proof of (4.4). • Step 2 Iterating the above procedure, we obtain a 21 -sparse family F ⊂ D(Q 0 ) such that for a.e. x ∈ Q 0 |gb ( f · 13Q 0 )(x)| ⎛ ⎞

  1  r  r1  r ⎠ ⎝ |bi (x) − bi,R Q | | f i |  |(b j − b j,R Q ) f j |r 3Q 1Q . 3Q × Q∈F τ ⊂τm

i∈τ

j∈τ

(4.5)

J Fourier Anal Appl

We shall transfer the local result to the global setting. The following argument follows exactly the same scheme of proof of [33, Theorem 3.1]. However,we present the details for the sake of completeness.  j. Take a partition of Rn by cubes  Q j such that i supp( f i ) ⊂ 6Q j for each Indeed, take a cube Q 0 such that i supp( f i ) ⊂ Q 0 and cover 3Q 0 \ Q 0 by 3n − 1 congruent cubes Q j . Each of them satisfies Q 0 ⊂ 3Q j . Next, in the same way cover 9Q 0 \ 3Q 0 , and so on. The union of resulting cubes, including Q 0 , will satisfy the desired property. a 21 -sparse family Having such a partition, apply (4.5) to each Q j . We obtain & F j ⊂ D(Q j ) such that (4.5) holds. Therefore, setting F = j F j , we obtain that F is a 21 -sparse family and for a.e. x ∈ Rn |gb ( f)(x)| 



Q∈F τ ⊂τm

⎛ ⎞   1 1 r r r ⎠ ⎝ |bi (x) − bi,R Q || f i |  r × |(b j − b j,R Q ) f j | 3Q 1Q . 3Q j∈τ

i∈τ

(4.6) Since 3Q ⊂ R Q and |R Q | ≤ 12n |Q|, we obtain | f i |3Q ≤ cn | f i | R Q . Further,   setting S j := R Q ∈ D j : Q ∈ F , and using that F is 21 -sparse, we obtain that each 1 family S j is 2·12 n -sparse. It follows from (4.6) that n

|gb ( f)(x)| ≤ cn

3



j=1 Q∈S j τ ⊂τm

⎛ ⎝



1 r

|bi (x) − bi,Q || f i |  Q × r

i∈τ

 j∈τ

⎞ 1 r

|(b j − b j,Q ) f j |  Q ⎠ 1 Q . r

 

This completes the proof.

5 Sharpness of Theorems 1.2 and 1.3 In this section, we will give some examples to show our estimates (1.1), (1.2) and (1.3) are sharp. For an m-linear Calderón–Zygmund kernel K determined in Sects. 5.1 and 5.2, we denote T ( f)(x) = p.v.

ˆ (Rn )m

K (x, y)

m 

f j (y j ) d y.

(5.1)

j=1

Then, T is an m-linear Calderón–Zygmund singular operator and bounded from L 1 (Rn ) × · · · × L 1 (Rn ) to L 1/m,∞ (Rn ) (see Corollary 2 in [16]). If we set ( K t (x, y) =

K (x, y), 0,

1 ≤ t ≤ e, otherwise.

J Fourier Anal Appl

then it follows from Proposition 6.4 that K t ∈ K2  H1 . In other words, K t satisfies the m-linear L 1 -Hörmander condtion. Furthermore, we have g( f)(x) =

ˆ

ˆ   gb ( f )(x) = 

0

∞

dt |G t ( f)(x)|2 t

1/2

= |T ( f)(x)|,

  m    (bi (x) − bi (yi ))K (x, y) f j (y j ) d y.

(Rn )m i=1

(5.2) (5.3)

j=1

5.1 Examples for Theorem 1.2 We first show that the inequality (1.2) is sharp. In the equation (5), let m j=1 (x 1 − (y j )1 )  K (x, y) = , m ( j=1 |x − y j |2 )(mn+1)/2

(5.4)

where (y j )1 denotes the first coordinate of y j . By (5) and Theorem 1.2 (1.2) (r = 1), it holds that max {1,

T ( f) L p (νω ) = g( f) L p (νω )  [ω]  A p

1≤i≤m

pi p

}

×

m 

 f i  L pi (ωi )

(5.5)

i=1

for all f i ∈ L pi (ωi ) and ω  ∈ A p . On the other hand, it was shown in [39, Theorem 1.4] that there exist functions f and weights ω  ∈ A p such that max {1,

T ( f) L p (νω )  [ω]  A p

1≤i≤m

pi p

}

×

m 

 f i  L pi (ωi ) .

(5.6)

i=1

Consequently, the inequalities (5.1) and (5.1) imply that the exponent on [ω]  A p is the best possible. Next, we demonstrate the inequality (1.2) is sharp. We follow the scheme of the proofs in [8,39] to track the precise constants. We here only consider the setting  = 1 and m = 2. We focus on the case max{ p1 , p2 } ≥ p, since the other case can be deal with using a duality argument. Without loss of generality, we suppose that p1 = max{ p1 , p2 }. Let the kernel K be the same as (5.1), and U = {x ∈ Rn ; |x| ≤ 1, 0 < xi < x1 , i = 2, . . . , n}, V = {x ∈ Rn ; |x| ≤ 1, xi ≤ 0, i = 1, . . . , n}.

J Fourier Anal Appl

Then for any x ∈ U and y j ∈ V with |y j | ≤ |x|, we have |x| 2|x| ≥ |x − y j | ≥ x1 − (y j )1 = |x1 − (y j )1 | ≥ |x1 | > √ . n

(5.7)

For 0 <  < 1, let b1 (x) = log |x|, ω1 (x) = |x|(n−)( p1 −1) , ω2 (x) = 1, f 1 (x) = |x|−n 1V (x), f 2 (x) = |x|(−n)/ p2 1V (x). A simple calculation gives that  f 1  L p1 (ω1 )  

− p1

1

,  f 2  L p2 (ω2 )  

− p1

2

, and [ω]  A p  

−

p p1

.

Additionally, applying the integration by parts k times, we obtain ˆ

θ

log

0

1 k β−1 θβ t dt  k+1 , k ≥ 0, and θ, β ∈ (0, 1], t β

which immediately implies that ˆ |y|≤|x|

log

|x| k |y|

ˆ

α

|x| k

|x|

· |y| dy 

log 0

ˆ

= |x|n+α

r 1

log 0

· r n+α−1 dr

1 k

· t n+α−1 dt 

t

|x|n+α . (n + α)k+1

Consequently, together with (5) and (5.1), it yields that for any x ∈ U 

|x|  −n  · K (x, y1 , y2 )|y1 |−n |y2 | p2 dy1 dy2  |y1 | V ×V ˆ ˆ

|x|  −n 1 −n  · |y log | dy × |y2 | p2 dy2 1 1 2n |x| |y1 | |y1 |≤|x| |y2 |≤|x|

ˆ  gb ( f)(x) = 

  −2 |x|

log

(−n)(1+ p1 ) 2

,

which gives that gb ( f) L p (νω )   −2 =

−2

ˆ |x|

(−n) p(1+ p1 − 2

U

ˆ

|x|≤1

|x|

(−n)

1 ) p1

1/ p dx

1/ p dx



−(2+ 1p )

.

J Fourier Anal Appl

Moreover, note that if ω  ∈ A p , then there holds that pi p

 A p and [σi ] Amp ≤ [ω]  A p , i = 1, . . . , m, [νω ] Amp ≤ [ω] i

which was essentially contained in the proof of [35, Theorem 3.6]. Hence, we deduce that

 |τ | [νω ] A∞ [σ j ] A∞ = [σ1 ] A∞ + [νω ] A∞ τ ⊂τ

j∈τ \τ

p1 p

≤ [σ1 ] Amp + [νω ] Amp ≤ 2[ω]  A p   −1 . 1

Collecting the above estimates, we obtain that gb ( f) L p (νω )  

−(2+ 1p )

⎛

×⎝

τ ⊂τ

p1 p

 [ω]  A p

|τ | [νω ] A∞



⎞ [σ j ] A∞ ⎠ ×

j∈τ \τ

2 

 f i  L pi (ωi ) .

i=1

This shows that the inequality (1.2) is sharp. 5.2 Examples for Theorem 1.3 In the equation (5), we take n

j=1 (x j − (y1 ) j ) K (x, y) = n . ( j=1 |x − y j |2 )(mn+1)/2

By Theorem 1.3, we have: For b = (b1 , . . . , b ) ∈ B M O  (1 ≤  ≤ m) and ω  ∈ A1 , then for any λ > 0 νω



x; |gb ( f)(x)| > λm



≤ Cb,

m ˆ  i=1

Rn

 1,

 1/m | f i (x)| ωi d x , λ

where r, (t) = t r (1 + log+ t)r  . We will see that we cannot replace one of 1, by 1,s with 0 ≤ s <  in the case ω j ≡ 1 (1 ≤ j ≤ m). Set b j (x) = log(1 + |x|) ∈ BMO(Rn ), f j (x) = 1[0,1/√n]n (x), 1 ≤ j ≤ m.

J Fourier Anal Appl

√ For x ∈ [10, ∞)n and y j ∈ Rn with |y j | ≤ 1/ n it holds n K (x, y) >

√

j=1 (x j − 1/ n) √ mn+1 mn+1

|x|

n

√  √ (1 − 1/(10 n)) nj=1 (x j − 1/ n) > √ mn+1 mn+1 n |x| √ √ (1 − 1/(10 n)) n) > , |x|mn

and b j (x) − b j (y j ) = log(1 + |x|) − log(1 + |y j |) >

1 log(1 + |x|). 2

So, for x ∈ [10, ∞)n we get gb ( f)(x) > c0

  2 log(1 + |x|) , |x|mn

where c0 is a positive constant. Thus,       2 log(1 + |x|) m  |{x ∈ Rn : gb ( f)(x) > λm }| ≥  x ∈ [10, ∞)n : c0 > λ  mn |x|   1 

m      λ mn =  x ∈ [10, ∞)n : 2 log(1 + |x|) mn > |x| . c 0

For c0 > λm , let t1 = Then

c  0 λm

1 mn

c  1   0 mn mn 1 + log m . λ



c  1     mn 0 mn 2 log(1 + t1 ) mn > 2 log 1 + m , λ

and

λm 

1 mn

c0

c  1   0 mn mn t1 = 1 + log m . λ

Since we easily see that 2 log(1 + t) > 1 + log t for t > e − 1, we have for λ satisfying

m 1 mn λ > e − 1, c0

λm  1   mn 2 log(1 + t1 ) mn > t1 . c0 From this it follows 



2 log(1 + t)

 mn

>

λm  c0

1 mn

t for 0 < t < t1 .

J Fourier Anal Appl

So, we have |{x ∈ Rn : gb ( f)(x) > λm }| > |{x ∈ [10, ∞)n : |x| ≤ t1 }| 1



t1n

1

cm 1   cm n m = 0 1 + log 0 . λ λ

(5.8)

Now, suppose that the inequality (1.4) were to hold replacing one of 1, , say for j = 1, by 1,s with 0 ≤ s <  in the case ω j ≡ 1 (1 ≤ j ≤ m). Then, letting f 1 → f 1 /λm−1 , f 2 → λ f 2 , . . . , f m → λ f m , we get via submultiplicativity of 1,s      x; |gb ( f)(x)| > λm  ˆ   1/m  1/m m ˆ | f 1 (x)| ≤ Cb, d x 1,s | f (x)|)d x 1, i λm Rn Rn i=2

ˆ ≤ Cb, 1,s (1/λm )1/m

Rn

1,s (| f 1 (x)|)d x

1/m  m ˆ i=2

Cb, 1  ms 1 + log m ≤ . λ λ

Rn

1, | f i (x)|)d x

1/m

This leads to a contradiction with (5.8) by letting λ → 0 because of 0 ≤ s < .

6 Final Remarks In this section, we will point out that our main theorems in Sect. 1 recover many known results about the multilinear Littlewood–Paley operators. 6.1 Different Regularity Conditions Definition 6.1 For any t ∈ (0, ∞), let K t (x, y) be a locally integrable function defined on (Rn )m+1 . K t is called a multilinear Littlewood–Paley kernel, written K t ∈ K1 , if for some positive constants A, γ , δ, and B > 1, |K t (x, y)| ≤ |K t (x, y) − K t (x , y)| ≤ whenever |x − x | ≤

1 max |x B 1≤ j≤m

(t +

At δ , mn+δ j=1 |x − y j |)

m

At δ |x − x |γ m , (t + j=1 |x − y j |)mn+δ+γ

− y j |, and

|K t (x, y) − K t (x, y1 , . . . , yi , . . . , ym )| ≤

At δ |yi − yi |γ m , (t + j=1 |x − y j |)mn+δ+γ

J Fourier Anal Appl

whenever |yi − yi | ≤

1 B |x

− yi |, for all 1 ≤ i ≤ m.

The class K1 is the classical multilinear Littlewood–Paley kernel. It was investigated by many authors, such as [7,45,49] and [19]. After that, it was improved by Xue and Yan [48] to the integral type kernel condition below. Definition 6.2 For any t ∈ (0, ∞), let K t (x, y) be a locally integrable function defined on (Rn )m+1 off the diagonal x = y1 = · · · = ym . We say K t satisfies the integral condition of C-Z type, written K t ∈ K2 , if for some positive constants γ , A, and B > 1, the following inequalities hold: ˆ ˆ

∞

|K t (x, y)|2

0 ∞



|K t (x, y) − K t (x , y)|

0

whenever |x − x | ≤ ˆ

∞

1 max {|x B 1≤ j≤m

0 1 B |x

2 dt

1 2

1

t

2

A ≤ m , ( j=1 |x − y j |)mn A|x − x |γ ≤ m , ( j=1 |x − y j |)mn+γ

− y j |}, and

|K t (x, y) − K t (x, y1 , . . . ,

whenever |yi − yi | ≤

dt t

dt yi , . . . , ym )|2

1 2

t

A|yi − yi |γ ≤ m ( j=1 |x − y j |)mn+γ

− yi |, for all 1 ≤ i ≤ m.

Recently, the authors in [46] introduced the following weaker regularity conditions associated with square functions. Essentially, it goes back to Bui and Duong’s condition in [4]. Definition 6.3 Let 1 ≤ r < ∞. Denote S j (Q) = 2 j Q \ 2 j−1 Q if j ≥ 1, and S0 (Q) = Q. We say that K t satisfies regularity assumptions of B-D type, written K t ∈ K3 , if the following conditions holds: (H 2) There exist a positive constant C such that ⎛ ˆ ⎝

ˆ

ˆ S jm (Q)

···

S j1 (Q)

∞

0

dt |K t (x, y)|2 t

 r 2

⎞ 1 r

d y⎠

≤C

2−mn j0 /r , |Q|m/r

for all balls Q with center at x and ( j1 , . . . , jm ) = 0, where j0 = max { jk }. 1≤k≤m

(H 3) There exist δ > n/r so that ⎛ ⎝

ˆ

ˆ

ˆ S jm (Q)

≤C

···

S j1 (Q)

0

|x − x |m(δ−n/r ) , 2mδ j0 |Q|mδ/n

∞

|K t (x, y) − K t (x , y)|2

dt t

 r 2

⎞ 1 r

d y⎠

J Fourier Anal Appl

for all balls Q, x, z ∈ 21 Q and ( j1 , . . . , jm ) = 0, where j0 = max { jk }. 1≤k≤m

Proposition 6.4 For any 1 ≤ r1 < r2 < ∞, there holds that K1  K2  K3  Hr1  Hr2 . Proof Each inclusion relation is straightforward. We will give some examples to complete the proof. (1) We show that there exist {K t(0) } ∈ K2 \ K1 . Set t 1(0,2] (t), |t − 1|1/4 n j=1 (x j − (y1 ) j ) K (0) (x, y) = n , ( j=1 |x − y j |2 )(mn+1)/2 h(t) =

(0)

K t (x, y) = h(t)K (0) (x, y). Then we have 1 2   K t(0) (x, y)2 dt = C|K (0) (x, y)|, t 0 ˆ ∞  2 dt  21  (0)  (0) = C|K (0) (x, y) − K (0) (x , y)|, K t (x, y) − K t (x , y) t 0

ˆ

∞

and ˆ 0

  K t(0) (x, y) − K t(0) (x, y1 , . . . , y , . . . , ym )2 dt i t

∞

1 2

= C|K (0) (x, y) − K (0) (x, y1 , . . . , yi , . . . , ym )|. (0)

So, since K (0) (x, y) is a Calderón–Zygmund kernel, {K t } belongs to K2 , but clearly it does not belong to K1 . (1) (1) (2) We present an example {K t } such that {K t } ∈ K3 \ K2 . We only consider the case m = 1. Let 1 < r ≤ 2, n/r < s < n, and K (1) (x) = G s (x − e1 ), where e1 = (1, 0, . . . , 0) and G s (x) is the kernel of the Bessel potential of order )s (ξ ) = (1 + |ξ |2 )s/2 (see Stein’s book [47, p. 132]). s, i.e. G Next, let (1) K t (x, y) = 1[1,e] (t)K (1) (x − y). Then

ˆ 0

2 dt  21  (1)  (x, y) = |K (1) (x − y)|, K t  t

∞

J Fourier Anal Appl

and ˆ

2 dt  21   (1) (1) (x, y  ) − K (x , y) = |K (1) (x − y) − K (1) (x − y)|.  K t t t

∞

0

We know that ˆ

2R



|K (1) (y)|r dy

1/r

≤ C R −n/r , R > 0, and

R

ˆ

2R



|K (1) (y + h) − K (1) (y)|r dy

R

≤ C R −n/r

|h| s−n/r R

1/r

, |h| < R/2, R > 0,

cf. [50, Proposition 2]. From this we see that {K t(1) } belongs to K3 , but clearly it does not belong to K2 , because |K (1) (e1 )| = +∞. (3) We next demonstrate K3  Hr . Let K  be the kernel function in Remark 3.5 in [37]. Denote (2,) K t (x, y) = K  (x − y)1[1,e] (t), t > 0. Then there hold ˆ 0

2 dt  21  (2,)  = |K  (x − y)|, K t (x, y) t

∞

and ˆ 0

2 dt  21   (2,) (2,) = |K  (x − y) − K  (x − y)|.  K t (x, y) − K t (x , y) t

∞

Hence, K t(2,) satisfies Hr condition uniformly in  ∈ N i.e. sup∈N Kr (K t(2,) ) < ∞. But its constant in (H3) of Definition 6.3 tends to ∞ as  → ∞, as Li [37] commented. This shows the strict inclusion K3  Hr . (4) For 1 < r < ∞ and β > 0, write K (r ) (x) = and

−

e 1 log |x − 4e1 | |x − 4e1 |n/r (r )

1+β r

1(0,1) (|x − 4e1 |),

K t (x) = K (r ) (x − y)1[1,e] (t), t > 0.

J Fourier Anal Appl

Then we deduce that ˆ

2 dt  21  (r )  = |K (r ) (x − y)|,  K t (x, y) t

∞

0

and ˆ 0

2 dt  21  (r )  (r ) = |K (r ) (x − y) − K (r ) (x − y)|. K t (x, y) − K t (x , y) t

∞

(r )

(r )

Hence, {K t } belongs to Hr (cf. [41, Lemma 6.1]). But, {K t } does not belong to Hs for any 1 ≤ s < r , because ˆ

ˆ 2<|y|∞ ≤4

ˆ

=

0





2<|y|∞ ≤4

ˆ =

2 dt  s2   (r ) (r ) K (e , y) − K (−e , y) dy   t 1 1 t t

∞

|K (r ) (e1 − y) − K (r ) (−e1 − y)|s dy

2<|y|∞ ≤4

|K (r ) (e1 − y)|s dy = +∞.

So far, we have completed the proof.

 

6.2 Lusin Area Integral S and g∗λ -Function We define the multilinear Lusin area integral S and Littlewood–Paley g∗λ -function by setting  2 dzdt  21   , G t ( f)(z) n+1 t (x) ¨ 1  nλ  2 dzdt 2 t   ∗  gλ ( f )(x) := . G t ( f)(z) n+1 t + |x − z| t Rn+1 + S( f)(x) :=

¨

We also define their corresponding iterated commutators when G t ( f) is replaced by G bt ( f). Additionally, in order to establish quantitative weighted estimates and sparse domination theorems for S and g∗λ , we may introduce the admissible m-linear L r Hörmander conditions. For example, in terms of g∗λ , the L r -Hörmander condition can be formulated as follows. Definition 6.5 Let 1 ≤ r < ∞. We say that the function {K t } satisfies the m-linear L r∗ -Hörmander condition if there holds that

J Fourier Anal Appl

Kr∗ := sup

∞

sup

Q x,x ∈ 1 Q k=1 2

¨ ×

Rn+1 +

m

ˆ

|2k Q| r

(2k Q)m \(2k−1 Q)m

t nλ dzdt |K t (x − z, y) − K t (x − z, y)|2 n+1 t + |z| t

 r 2

d y

 1 r

< ∞.

When r = 1, the above formula is understood as K1∗ := sup sup

∞

Q x,x ∈ 1 Q k=1 2

¨ ×

Rn+1 +

|2k Q|m

esssup y∈(2k Q)m \(2k−1 Q)m

dzdt t nλ |K t (x − z, y) − K t (x − z, y)|2 n+1 t + |z| t

 r 2

< ∞.

A completely analogous argument to that of the previous sections shows the following. Theorem 6.6 Let λ > 2m. All the estimates in the Sect. 1 hold for g∗λ and S if their kernel {K t } satisfies admissible m-linear L r -Hörmander conditions. Remark 6.7 Combining Theorems 1.2, 1.3, 6.6 and Proposition 6.4, we do improve and recover all the results in [7,45,46,48,49]. And, more remarkable, we do not assume that g, g∗λ or S can be extended to be a bounded operator from L q1 × · · · × L qm to L q for some 1 ≤ q1 , . . . , qm < ∞ with q1 = q11 + · · · + q1m . Although we need the assumption g, g∗λ and S are is bounded from L r ×· · ·× L r to L r/m,∞ , it is much weaker actually. Indeed, if K t ∈ K3  H1 , the endpoint weak type boundedness follows from the strong type L q1 × · · · × L qm → L q bounds. This was shown in Theorem 1.1 [48]. Remark 6.8 We define the iterated commutators Tb of singular integrals T by ˆ  Tb ( f )(x) := G bt ( f)(x)d x. (Rn )m

The admissible multilinear L r -Hörmander condition was first used by Li [37] Kr := sup sup

∞

Q x,x ∈ 1 Q k=1 2

m

|2k Q| r

ˆ (2k Q)m \(2k−1 Q)m



|K t (x, y) − K t (x , y)|r d y

 1 r

< ∞.

(In linear case, the above condition was introduced in [41] to investigate Coifman’s inequality for singular integrals.) Theorems 1.2 and 1.3 still hold when gb is replaced by Tb . Thus, we not only weaken the assumptions in [44] but also obtain the quantitative weighed inequality. 6.3 Discussion About Sparse Domination In view of the sublinear sparse domination for the classical Littlewood–Paley square functions, we hope to improve the inequality (1.5) to the following:

J Fourier Anal Appl

⎛ |g( f)(x)|  ⎝

m



⎞1/2  f i 2Q,r 1 Q (x)⎠

.

(6.1)

Q∈S i=1

An analogue of (6.3) indeed holds in linear case [30,32], and in the m-linear case [5]. It is worth pointing out that our kernels in the current paper satisfy the L r -Hörmander condition. It is strictly weaker than the known kernel conditions, which was proved in Proposition 6.4. Although we have tried many ways, we failed. The biggest obstacle is that we know nothing about the size condition of kernels. This is the main difference from the classical case. It also leads us to deal with the global estimates roughly and integrally, see the proof of Lemma 4.1. Under the L r -Hörmander condition, due to technical reasons, we have to deal with the m-linear square functions applying those techniques used to investigate singular integrals. Despite all this, we consider the inequality (1.4) as a compromise under the L r Hörmander condition. This follows from the following three facts. (•) The improved inequality (6.3) holds for the kernel K t satisfies L r -Hörmander condition and the following conditions:   supp K t ⊂ (x, y) : sup |x − yi | ≤ t , |K t (x, y)| 

1≤i≤m tδ

(t +

m

i=1 |x

− yi |)mn+δ

.

(6.2) (6.3)

We will apply the strategy in the proof of Theorem A and Lemma 4.1, in which we will see what the barrier we encounter. The core is to show the weak type bound for MT , which follows from its pointwise domination. Our object is not Mg but Mg2 . Now we establish the pointwise control for Mg2 . Let x, ξ, ξ ∈ Q. It suffices to estimate the following ˆ    g( f)(ξ )2 − g( f · 13Q )(ξ )2  =  

0

∞

  2 2 dt    . G t ( f )(ξ ) − G t ( f · 13Q )(ξ ) t 

Note that the support condition (6.2) implies that {y : sup1≤i≤m |ξ − yi | ≤ t} ⊂ (3Q)m for any 0 < t ≤ (Q). It yields that ˆ

(Q)

(G t ( f)(ξ )2 − G t ( f · 13Q )(ξ )2 )

0

dt = 0. t

Additionally, there holds ˆ   

  2 2 dt    (G t ( f )(ξ ) − G t ( f · 13Q )(ξ ) )  t (Q) ˆ ∞ ˆ ∞ dt dt ≤2 |G t ( f)(ξ ) − G t ( f · 13Q )(ξ )|2 + 3 |G t ( f · 13Q )(ξ )|2 . t t 0 (Q) ∞

J Fourier Anal Appl

The second term is bounded by maximal function. Indeed, it follows from the size condition (6.3) that ˆ

∞

dt  |G t ( f · 13Q )(ξ )|2 t (Q)

ˆ

∞

(Q)

1 dt t 2mn t



m ˆ 

2 | f i |dy

i=1 3Q

m 2  1 ˆ  | f i |dy ≤ M( f)(x)2 . |3Q| 3Q i=1

Let us turn our attention to the first term. ˆ ∞ dt |G t ( f)(ξ ) − G t ( f · 13Q )(ξ )|2 t 0 ˆ ∞ ˆ ∞ dt dt ≤ |G t ( f)(ξ )|2 + G t ( f · 13Q )(ξ )|2 t t 0 0 2 ˆ ∞ ˆ m  dt + |K t (ξ, y) − K t (ξ , y)| | f i (yi )|d y t (Rn )m \(3Q)m 0 i=1

=: g( f)(ξ )2 + g( f · 13Q )(ξ )2 + E. The remaining calculations are similar to those in the proof of Lemma 4.1. Thus, taking  = 1/2, we deduce that Mg2 ( f)(x)  M( f)(x)2 + M (g( f)2 )(x) = M( f)(x)2 + M(g( f))(x)2 . Consequently, we derive the inequality (6.1) following the same scheme of proof in Theorem A. (•) An analogue of sparse domination (6.3) will be obtained if K t satisfies size condition (6.3) and log-Dini regular condition defined by ˆ

∞

ωlog-Dini ≡ 0

  1 dt < ∞. ω(t) 1 + log t t

Our strategy is to apply the local mean oscillation formula in [34], which developed a new version of Lerner’s formula that allows to obtain a pointwise control of any function satisfying minimum assumptions in terms of its oscillations. Refining the argument in [34, Section 9] and [32, Lemma 3.1], we obtain g( f)(x)2 

∞

k=0

ω(2−k )

m

 Q∈S i=1

| f i |22k Q 1 Q (x).

It is worth mentioning that in order to obtain the above inequality, we need the endpoint weak type estimate g : L 1 (Rn ) × · · · × L 1 (Rn ) → L 1/m (Rn ). Arguing as in the proof of [18, Theorem 1.1], we arrive at

J Fourier Anal Appl

g( f)(x)2 

∞

ω(2−k )(k + 1)

Q∈Sk i=1

k=0

≤

∞

m

 | f i |2Q 1 Q (x) n

ω(2

−k

)(k + 1)

3 m



| f i |2Q 1 Q (x)

j=1 Q∈S j i=1

k=0 3n

 ωlog-Dini

m



| f i |2Q 1 Q (x),

j=1 Q∈S j i=1 3 ∞ 3 where {Sk }∞ k=0 and {S j } j=1 are sparse collections such that {Sk }k=0 ⊂ {S j } j=1 . (•) It is an open problem to obtain the inequality (6.1) for 1-linear square functions satisfying the Dini condition: n

n

ˆ ωDini ≡ 0

∞

ω(t)

dt < ∞. t

Other weighted inequalities and sparse domination for square functions involving Dini condition can be found in [51]. The known technique seems to be to invalid in terms of Dini condition. We conjecture a possibility to solve this problem would be to give a modification of the technique in [33]. That will be our further topic. As already mentioned in the introduction prat, our motivation of this paper is to weaken the regularity of kernels and establish the sharp weighted estimates for the multilinear Littlewood–Paley square functions and their iterated commutators in terms of weights norm. The sharpness of Theorem 1.2 means that the inequality (1.5) does not affect the sharp weighted norm estimates. Therefore, so far, it seems that the sparse domination (1.5) is sharp. Additionally, it shows the dyadic approach we used is simpler than the classical one involving sharp maximal functions. Acknowledgements The authors would like to thank the referee for valuable suggestions which have improved the quality of this paper.

Appendix In this section, we will present some observations about L p (log L)a functions, which will help us to comprehend the inverse of L p (log L)a functions. Those can not be found in any other references. Henceforth, given a Young function A, we mean that A−1 (x) = inf{y ∈ R : A(y) > x} for x ∈ [0, ∞], where inf ∅ = ∞. Proposition 7.1 Let −∞ < a < ∞. Let (x) = x(1 + log+ x)a . Then −1 (x) ≈ x(1 + log+ x)−a on [0, ∞] if a ≥ −1, and on [0, 1] ∪ [e−a−1 , ∞] if a < −1. Proof Let (x) = x(1 + log+ x)−a . (i) The case a = 0. In this case tivially −1 (t) = (t).

J Fourier Anal Appl

(ii) The case a > 0. For 0 ≤ t ≤ 1 we have ((t)) = t, which means that

−1 (t) = (t) for 0 ≤ t ≤ 1.

For t > 1, there holds  a ((t)) = t (1 + log+ t)−a 1 + log+ (t (1 + log+ t)−a )  a < t (1 + log+ t)−a 1 + log+ t = t, which combined with the increasingness of (t) yields that −1 (t) ≥ (t) for 1 < t < ∞. Next, let A ≥ max{2a , (2a)a }. Then a  (A(t)) = At (1 + log+ t)−a 1 + log+ (At (1 + log+ t)−a ) .

(7.1)

In the case 0 < a ≤ 21 , we see that (1 + log+ t)a ≤ (1 + log+ t)1/2 ≤

√

t.

This and (7.1) yield that √ a  (A(t)) > At (1 + log+ t)−a 1 + log+ t)  a A A ≥ a t (1 + log+ t)−a 1 + log+ t) = a t > t. 2 2 Hence, we deduce that −1 (t) ≤ A(t) for 1 < t < ∞. In the case a > 21 , let

1

1

f (t) = A a t 2a − 1 − log t. 1

Then f (1) = A a − 1 > 0, and noting A ≥ (2a)a , we get 1 1 A a 1 −1 1 2a  Aa 1 2a f (t) = t 2a − 1 > 0 for t > 1. − = t 2a t 2at Aa



Hence we have

√ A t > (1 + log+ t)a for t > 1,

(7.2)

J Fourier Anal Appl

and so combinig (7.1) we get √ a  (A(t)) > At (1 + log+ t)−a 1 + log+ t) . So, as in the case 0 < a ≤ 21 , we have (7.2). For t = ∞ we have trivially −1 (∞) = (∞) = ∞. All together we obtain (t) ≤ −1 (t) ≤ A(t) for 0 ≤ t ≤ ∞. (iii) The case a < 0. On [0, 1], (t) = 1, and so the conclusion is clear. On (1, ∞) we have (1 + log+ t)−a ((t)) = t  −a 1 + log+ (t (1 + log+ t)−a ) a  < t (1 + log+ t)−a 1 + log+ t = t. Since (t) is increasing on [e−a−1 , ∞], we see that −1 (t) ≥ (t) on [e−a−1 , ∞).

(7.3)

For A ≥ (−2a)−a (A(t)) = At 

(1 + log+ t)−a 1 + log+ (At (1 + log+ t)−a )

−a .

(7.4)

In (ii), we showed that if A ≥ (−2a)−a √ A t > (1 + log+ t)−a for t > 1. This combined with (7.4) gives (1 + log+ t)−a A (A(t)) = At  ≤  −a t. −a 3 1 + log+ (A2 t 3/2 ) 2 (1 + 2 log A) So, if A ≥ (−3a)−a , we have  3

A

2 (1+2 log A)

−a ≥ 1, and

(A(t)) ≥ t for t > 1. Thus, since (t) is increasing on [e−a−1 , ∞], we get −1 (t) ≤ A(t) for t ≥ e−a−1 ,

(7.5)

J Fourier Anal Appl

if A ≥ max{(−2a)−a , (−3a)−a }. By (7.3) and (7.5), we obtain (t) ≤ −1 (t) ≤ A(t) for t ≥ e−a−1 . This completes the proof of Proposition 7.1.

 

Remark 7.2 Let (t) = t (1 + log+ t)a and a ∈ R. If a ≥ 0, (t) is increasing and convex on [0, ∞]. If −1 ≤ a < 0, (t) is increasing on [0, ∞]. If a < −1, then (t) is increasing on [0, 1] ∪ [e−a−1 , ∞], and is decreasing on [1, e−a−1 ]. Moreover, for a < 0, (t) = 0 on (0, 1), > 0 on (1, e−a ) and < 0 on (e−a , ∞). Proposition 7.3 Let −∞ < a < ∞ and p > 0. Let p (x) = x p (1 + log+ x)a and ˜ p (x) = x p (1 + log+ x p )a . Then there exist A1 , A2 ≥ 1 such that ˜ p (A1 x), 0 ≤ x < ∞, p (x) ≤ and

˜ p (x) ≤ p (A2 x), 0 ≤ x < ∞.

˜ p (x) = x, and so we have only to treat the Proof If x ∈ [0, 1] or a = 0, p (x) = case x > 1 and a = 0. We consider the following four cases: p ≥ 1 and a > 0, p ≥ 1 and a < 0, 0 < p < 1 and a > 0, and 0 < p < 1 and a < 0. ˜ p (x). And Case 1 p ≥ 1 and a > 0. Clearly p (x) ≤ ˜ p (x) = x p (1 + p log x)a ≤ pa x a (1 + log x)a ≤ ( pa/ p x)a (1 + log( pa/ p x))a = p ( pa/ p x). ˜ p (x). And Case 2 p ≥ 1 and a < 0. Clearly p (x) ≥ ( p A)−a x p ( p A)−a x p ≤ −a ( p A + p A log x) ( p A + p log x)−a ( p A)−a x p ≤ ( p A − p log( p A)−a/ p + p log( p A)−a/ p x)−a ( p A)−a x p = . ( p A − p log( p A)−a/ p + log(( p A)−a/ p x) p )−a

p (x) =

There exists A0 > 0 such that p A0 − p log( p A0 )−a/ p ≥ 1. In this case p A0 ≥ 1. Hence, letting A1 = ( p A0 )−a/ p > 1 we have ˜ p (A1 x). p (x) ≤

J Fourier Anal Appl

˜ p (x). And Case 3 0 < p < 1 and a > 0. It is easy to see that p (x) ≥ p (x) = x p (1 + log x)a =

1 a

x p ( p + p log x)a p a  

1 a  1 a/ p p ≤ x p + p log x p p    

1 a/ p  p a 1 a/ p 1 ˜p ≤ (( )a/ p x) p 1 + log x = x . p p p

Case 4 0 < p < 1 and a < 0. There holds ˜ p (x) =

xp xp ≥ = p (x). (1 + p log+ x)−a (1 + log+ x)−a

And for A > 1 ˜ p (x) = ≤ ≤

xp (1 + p log+ x)−a ( 1p )−a x p (1 + log x)−a ( Ap )−a x p (A + log x)−a

= =

( Ap )−a x p (A + A log x)−a (( Ap )−a/ p x) p (A +

a p

log

A p

+ log(( Ap )−a/ p x))−a

.

There exists A4 > 1 such that A4 +

A4 a log ≥ 1. p p

In this case A4 / p ≥ 1. Hence, letting A2 = (A4 / p)−a/ p > 1 we have ˜ p (x) ≤ p (A2 x). Lemma 7.4 Let a1 ≥ 0 and let 1 , 2 be nonnegative, increasing and continuous functions on [a1 , ∞). If for some A1 ≥ 1 1 (t) ≤ 2 (A1 t), a1 ≤ t < ∞,

(7.6)

−1 −1 2 (t) ≤ A1 1 (t) for max{ 1 (a1 ), 2 (a1 )} ≤ t < ∞.

(7.7)

then it holds

Proof From (7.6) we get for t ≥ 1 (a1 ) −1 2 (A1 −1 1 (t)) ≥ 1 ( 1 (t)) = t,

J Fourier Anal Appl

which implies that −1 A1 −1 1 (t) ≥ 2 (t) for t ≥ max{ 1 (a1 ), 2 (a1 )}.

Proposition 7.5 Let −∞ < a < ∞ and p > 1. Let p (x) = x p (1 + log+ x)a . Then 1/ p (1 + log+ x)−a/ p on [0, ∞] if a ≥ −1, and on [0, 1] ∪ [−ae−a−1 , ∞] −1 p (x) ≈ x if a < −1. Proof One can easily check that p (t) is increasing on [0, ∞) if a ≥ − p, and is ˜ p (t) = t p (1 + log+ t p )a . We know that increasing on [0, 1] ∪ [e−a/ p−1 , ∞). Let ˜ p (t) is increasing on [0, ∞) if a ≥ −1, and is increasing on [0, 1] ∪ [e−(a+1)/ p , ∞). ˜ −1 Hence, if a ≥ 1, by Proposition 7.3 and Lemma 7.4 we see that −1 p (t) ≈ p (t) on [0, ∞). −1 1/ p , where (t) = t (1 + log+ t)a . Hence ˜ −1 Here we see easily that p = ( (t)) by Proposition 7.1 we obtain 1/ p −1 (1 + log+ t)−a/ p on [0, ∞]. p (t) ≈ t

In the case a < −1, on [0, 1] it is trivial. Next, we observe that e−a/ p−1 < e−(a+1)/ p − a −1 ˜ p (e−(a+1)/ p ). By the proof of and p (e−a/ p−1 ) = − ap e p < −ae−(a+1) = ˜ p (t) ≤ p (t) ≤ ˜ p (At) (0 ≤ t < ∞), for some Proposition 7.3, we know that −1 (t) on [−ae−(a+1) , ∞). ˜ (t) ≈ A > 1. So, by Lemma 7.4 we see that −1 p p Remark 7.6 In genera, the following does not hold: Let a1 ≥ 0 and let 1 , 2 be nonnegative, increasing and continuous functions on [a1 , ∞). −1 Then, from 1 (t) ≈ 2 (t) it follows −1 1 (t) ≈ 2 (t). In fact, let 1 (t) = 2 t log(1 + t) and 2 (t) = log(1 + t) . Then 1 (t) ≈ 2 (t), but −1 1 (t) = e − 1 and t/2 − 1. And there exists no C > 0 such that −1 (t) ≤ C −1 (t). −1 2 (t) = e 1 2 However, if we suppose the convexty in addition, then we get a good result. Lemma 7.7 Let 1 , 2 be nonnegative, increasing and convex functions on [0, B]. If for some A1 , A2 ≥ 1 2 (t) ≤ 1 (t) ≤ A2 2 (t), 0 ≤ t < ∞, A1 then it holds 1 (t) ≤ 2 (A2 t) and 2 (t) ≤ 1 (A1 t) 0 ≤ t < ∞. Proof Since 2 (0) = 0 and 2 is convex on [0, B], we have 2 (t) 2 (A2 t) ≤ , 0 ≤ t < ∞, t A2 t

J Fourier Anal Appl

and hence A2 2 (t) ≤ 2 (A2 t), 0 ≤ t < ∞. Thus we get 1 (t) ≤ 2 (A2 t) 0 ≤ t < ∞. Similarly we have 2 (t) ≤ 1 (A1 t) 0 ≤ t < ∞.

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