Weighted vector-valued bounds for the singular integral operators with ...

arXiv:1607.05586v2 [math.CA] 20 Nov 2016

WEIGHTED VECTOR-VALUED BOUNDS FOR THE SINGULAR INTEGRAL OPERATORS WITH NONSMOOTH KERNELS GUOEN HU Abstract. Let T be a singular integral operator with non-smooth kernel which was introduced by Duong and McIntosh. In this paper, we prove that this operator and its corresponding grand maximal operator satisfy certain weak type endpoint vector-valued estimate of L log L type. As an application we established a refined weighted vector-valued bound for this operator.

1. Introduction We will work on Rn , n ≥ 1. Let Ap (Rn ) (p ∈ (1, ∞)) be the weight functions class of Muckenhoupt, that is, w ∈ Ap (Rn ) if w is nonnegative and locally integrable, and  1 Z  1 Z p−1 1 [w]Ap := sup < ∞, w(x)dx w− p−1 (x)dx |Q| Q |Q| Q Q where the supremum is taken over all cubes in Rn , [w]Ap is called the Ap constant of w, see [6] for properties of Ap (Rn ). In the last several years, there has been significant progress in the study of sharp weighted bounds with Ap weights for the classical operators in Harmonic Analysis. The study was begin by Buckley [1], who proved that if p ∈ (1, ∞) and w ∈ Ap (Rn ), then the Hasrdy-Littlewood maximal operator M satisfies 1

(1.1)

kf kLp(Rn , w) , kM f kLp(Rn , w) .n, p [w]Ap−1 p

Moreover, the estimate (1.1) is sharp since the exponent 1/(p − 1) can not be replaced by a smaller one. Hyt¨onen and P´erez [9] improved the estimate (1.1), and showed that
1 1 kM f kLp(Rn , w) .n, p [w]Ap [w− p−1 ]A∞ p kf kLp(Rn , w) . (1.2) where and in the following, for a weight u, [u]A∞ is defined by Z 1 M (uχQ )(x)dx. [u]A∞ = sup Q⊂Rn u(Q) Q 1

1

It is well known that for w ∈ Ap (Rn ), [w− p−1 ]A∞ . [w]Ap−1 . Thus, (1.2) is more p subtle than (1.1). The sharp dependence of the weighted estimates of singular integral operators in terms of the Ap (Rn ) constant was first considered by Petermichl [14, 15], who 2010 Mathematics Subject Classification. 42B20. Key words and phrases. weighted vector-valued inequality, singular integral operator, nonsmooth kernel, sparse operator, sharp maximal operator, The research was supported by the NNSF of China under grant #11371370. 1

2

GUOEN HU

solved this question for Hilbert transform and Riesz transform. Hyt¨onen [7] proved that for a Calder´ on-Zygmund operator T and w ∈ A2 (Rn ), (1.3)

kT f kL2(Rn , w) .n [w]A2 kf kL2(Rn , w) .

This solved the so-called A2 conjecture. Combining the estimate (1.3) and the extrapolation theorem in [3], we know that for a Calder´ on-Zygmund operator T , p ∈ (1, ∞) and w ∈ Ap (Rn ), (1.4)

max{1,

kT f kLp(Rn , w) .n, p [w]Ap

1 p−1 }

kf kLp(Rn , w) .

In [12], Lerner gave a much simple proof of (1.4) by controlling the Calder´ onZygmund operator using sparse operators. Lerner [12] proved that Theorem 1.1. Let T be a sublinear operator and MT be the corresponding grand maximal operator defined by MT f (x) = sup ess sup |T (f χRn \3Q )(ξ)|. Q∋x

ξ∈Q

Suppose that both T and MT are bounded from L1 (Rn ) to L1, ∞ (Rn ). Then for p ∈ (1, ∞) and w ∈ Ap (Rn ), T satisfies (1.4). Let p, r ∈ (0, ∞] and w be a weight. As usual, for a sequence of numbers
P r 1/r . The space Lp (lr ; Rn , w) is defined {ak }∞ k=1 , we denote k{ak }klr = k |ak | as  Lp (lr ; Rn , w) = {fk }∞ k=1 : k{fk }kLp (lr ; Rn , w) < ∞ where Z  k{fk }kLp(lr ; Rn , w) =

Rn

k{fk (x)}kplr w(x) dx

1/p

.

When w ≡ 1, we denote k{fk }kLp(lr ; Rn , w) by k{fk }kLp (lr ; Rn ) for simplicity. Hu [10] extended Lerner’s result to the vector-valued case, proved that Theorem 1.2. Let T be a sublinear operator and MT be the corresponding grand maximal operator. Suppose that for some q ∈ (1, ∞),  y ∈ Rn : k{T fk (y)}klq + k{MT fk (y)}klq > λ Z  k{fk (y)}klq k{fk (y)}klq  . dy. log 1 + λ λ Rn Then for all p ∈ (1, ∞) and w ∈ Ap (Rn ), 1 1 1


− 1 β 1 − p−1 p p p′

{T fk } p q n p−1 ] [w ] + [w] . [w] n, p Ap A∞ A∞ k{fk }kLp (lq ; Rn , w) . A∞ [w L (l ;R ,w)

Let T be a L2 (Rn ) bounded linear operator with kernel K in the sense that for all f ∈ L2 (Rn ) with compact support and a. e. x ∈ Rn \supp f , Z (1.5) T f (x) = K(x, y)f (y)dy. Rn

where K is a measurable function on Rn × Rn \{(x, y) : x = y}. To obtain a weak (1, 1) estimate for certain Riesz transforms, and Lp boundedness with p ∈ (1, ∞) of holomorphic functional calculi of linear elliptic operators on irregular domains, Duong and McIntosh [4] introduced singular integral operators with nonsmooth kernels on spaces of homogeneous type via the following generalized approximation to the identity.

SINGULAR INTEGRAL OPERATORS

3

Definition 1.3. A family of operators {At }t>0 is said to be an approximation to the identity, if for every t > 0, At can be represented by the kernel at in the following sense: for every function u ∈ Lp (Rn ) with p ∈ [1, ∞] and almost everywhere x ∈ Rn , Z At u(x) =

at (x, y)u(y)dy,

Rn

and the kernel at satisfies that for all x, y ∈ Rn and t > 0,  |x − y|  (1.6) , |at (x, y)| ≤ ht (x, y) = t−n/s h t1/s where s > 0 is a constant and h is a positive, bounded and decreasing function such that for some constant η > 0, (1.7)

lim rn+η h(r) = 0.

r→∞

Assumption 1.4. There exists an approximation to the identity {At }t>0 such that the composite operator T At has an associated kernel Kt in the sense of (1.6), and there exists a positive constant c1 such that for all y ∈ Rn and t > 0, Z K(x, y) − Kt (x, y)|dx . 1. 1 |x−y|≥c1 t s

An L2 (Rn ) bounded linear operator with kernel K satisfying Assumption 1.4 is called a singular integral operator with nonsmooth kernel, since K does not enjoy smoothness in space variables. Duong and McIntosh [4] proved that if T is an L2 (Rn ) bounded linear operator with kernel K, and satisfies Assumption 1.4, then T is bounded from L1 (Rn ) to L1, ∞ (Rn ). To consider the weighted estimates with Ap (Rn ) boundedness of singular integral operators with non-smooth kernel, Martell [13] introduced the following assumptions. Assumption 1.5. There exists an approximation to the identity {Dt }t>0 such that the composite operator Dt T has an associated kernel K t in the sense of (1.6), and there exist positive constants c2 and α ∈ (0, 1], such that for all t > 0 and 1 x, y ∈ Rn with |x − y| ≥ c2 t s , |K(x, y) − K t (x, y)| .

tα/s . |x − y|n+α

Martell [13] proved that if T is an L2 (Rn ) bounded linear operator, satisfies Assumption 1.4 and Assumption 1.5, then for any p ∈ (1, ∞) and w ∈ Ap (Rn ), T is bounded on Lp (Rn , w). The first purpose of this paper is to establish the endpoint vector-valued estimates for the corresponding grand maximal operator of singular integral operators with nonsmooth kernels. Our main result can be stated as follows. Theorem 1.6. Let T be an L2 (Rn ) bounded linear operator with kernel K as in (1.5). Suppose that T satisfies Assumption 1.4 and Assumption 1.5. Then for each λ > 0, {x ∈ Rn : k{T fk (x)}klq + k{MT fk (x)}klq > λ} (1.8) Z k{fk (x)}klq
k{fk (x)}klq dx. log 1 + . λ λ n R

4

GUOEN HU

If we further assume that the kernels {K t }t>0 in Assumption 1.5 also satisfy that 1 for all t > 0 and x, y ∈ Rn with |x − y| ≤ c2 t s , n

|K t (x, y)| . t− s ,

(1.9)

then (1.8) is also true for T ∗ , here and in the following, T ∗ is the maximal singular integral operator defined by T ∗ f (x) = sup |Tǫ f (x)|, ǫ>0

with Tǫ f (x) =

Z

K(x, y)f (y)dy.

|x−y|>ǫ

As a consequence of Theorem 1.6 and Theorem 1.1, we obtain the following weighted vector-valued bounds for T and T ∗ . Corollary 1.7. Let T be an L2 (Rn ) bounded linear operator with kernel K in the sense of (1.9). Suppose that T satisfies Assumption 1.4 and Assumption 1.5. Then for p ∈ (1, ∞) and w ∈ Ap (Rn ), 1 1 1

′
(1.10) {T fk } Lp (lq ; Rn ,w) .n, p [w]Ap p [σ]Ap ∞ + [w]Ap ∞ [σ]A∞ k{fk }kLp (lq ; Rn , w) , 1

with σ = w− p−1 . Moreover, if the kernels {K t }t>0 in Assumption 1.5 satisfy (1.9), then the weighted estimate (1.10) also holds for T ∗ .

Remark 1.8. We do not know if the weighted bound in (1.10) is sharp. In what follows, C always denotes a positive constant that is independent of the main parameters involved but whose value may differ from line to line. We use the symbol A . B to denote that there exists a positive constant C such that A ≤ CB. Constant with subscript such as C1 , does not change in different occurrences. For any set E ⊂ Rn , χE denotes its characteristic function. For a cube Q ⊂ Rn and λ ∈ (0, ∞), we use ℓ(Q) (diamQ) to denote the side length (diamter) of Q, and λQ to denote the cube with the same center as Q and whose side length is λ times that of Q. For x ∈ Rn and r > 0, B(x, r) denotes the ball centered at x and having radius r. 2. Proof of Theorem 1.6 We begin with some preliminary lemmas. Lemma 2.1. Let q, p0 ∈ (1, ∞), ̺ ∈ [0, ∞) and S be a sublinear operator. Suppose that k{Sfk }kLp0 (lq ; Rn ) . k{fk }kLp0 (lq ; Rn ) , and for all λ > 0, Z {x ∈ Rn : k{Sfk (x)}klq > λ} . n

Rn

 k{fk }klq  k{fk }klq log̺ 1 + dx. λ λ

Then for cubes Q2 ⊂ Q1 ⊂ R , Z



1

{S(fk χQ2 )(x)} q dx . k{fk }klq , l L(log L)̺+1 , Q2 |Q1 | Q1

SINGULAR INTEGRAL OPERATORS

5

here and in the following, for β ∈ [0, ∞), Z o  n |f (y)|  |f (y)| 1 dy ≤ 1 . logβ 1 + kf kL(log L)β , Q = inf λ > 0 : |Q| Q λ λ Proof. Lemma 2.1 is a generalization of Lemma 3.1 in [11].

Their proofs are very similar. By homogeneity, we may assume that k{fk }klq L(log L)̺+1 , Q2 = 1, which implies that Z
k{fk (x)}klq log̺+1 1 + k{fk (x)}klq dx ≤ |Q2 |. Q2

p0 −1  For each fixed λ > 0, set Ωλ = x ∈ Rn : k{fk (x)}klq > λ 2p0 . Decompose fk as

fk (x) = fk (x)χΩλ (x) + fk (x)χRn \Ωλ (x) = fk1 (x) + fk2 (x). p0 −1

It is obvious that k{fk2 }kL∞ (lq ; Rn ) ≤ λ 2p0 . A trivial computation leads to that Z ∞ {x ∈ Rn : k{S(f 2χQ2 )(x)}klq > λ/2} dλ k 1 Z ∞Z

2

{fk (x)} pq0 dxλ−p0 dλ . l 1 Q2 Z ∞ Z (p −1)2

2

−p0 + 02p

{f (x)} q dx 0 dλ . |Q2 |. λ . k l 1

Q2

On the other hand, Z ∞ {x ∈ Rn : k{S(fk1 χQ2 )(x)}klq > λ/2} dλ 1 Z ∞Z

1




{f (x)} q log̺ 1 + {f 1 (x)} q dxλ−1 dλ . k k l l 1

.

Q2

Z

Q2

.

Z

Q2

1




{fk (x)} q log̺ 1 + {fk1 (x)} q l l

Z

2p0 p −1

k{fk (x)}klq0 1

1 dλdx λ

1




{fk (x)} q log̺+1 1 + {fk1 (x)} q dx. l l

Combining the estimates above then yields Z ∞ {x ∈ Q1 : k{S(fk χQ2 )(x)}klq > λ} dλ 0

.

Z

1

0

+ +

Z

{x ∈ Q1 : k{S(fk χQ2 )(x)}klq > λ} dλ ∞

Z1 ∞ 1

. |Q1 |.

{x ∈ Rn : k{S(fk1 χQ2 )(x)}klq > λ/2} dλ

{x ∈ Rn : k{S(fk2 χQ2 )(x)}klq > λ/2} dλ

This completes the proof of Lemma 2.1.



6

GUOEN HU

Recall that the standard dyadic grid in Rn consists of all cubes of the form 2−k ([0, 1)n + j), k ∈ Z, j ∈ Zn . Denote the standard grid by D. For a fixed cube Q, denote by D(Q) the set of dyadic cubes with respect to Q, that is, the cubes from D(Q) are formed by repeating subdivision of Q and each of descendants into 2n congruent subcubes. As usual, by a general dyadic grid D, we mean a collection of cube with the following properties: (i) for any cube Q ∈ D, it side length ℓ(Q) is of the form 2k for some k ∈ Z; (ii) for any cubes Q1 , Q2 ∈ D, Q1 ∩ Q2 ∈ {Q1 , Q2 , ∅}; (iii) for each k ∈ Z, the cubes of side length 2k form a partition of Rn . By the one-third trick, (see [8, Lemma 2.5]), there exist dyadic grids D1 , . . . , D3n , such that for each cube Q ⊂ Rn , there exists a cube I ∈ Dj for some j, Q ⊂ I and ℓ(Q) ≈ ℓ(I). Let {Dt }t>0 be an approximation to the identity. Associated with {Dt }t>0 , ♯ define the sharp maximal operator MD by Z 1 ♯ |f (y) − DtB f (y)|dy, f ∈ Lp (Rn ), p ∈ [1, ∞) MD f (x) = sup B∋x |B| B s with tB = rB and s the constant appeared in (1.6), the supremum is taken over all balls in Rn . This operator was introduced by Martell [13] and plays an important role in the weighted estimates for singular integral operators with non-smooth kernels. Let q ∈ (1, ∞), {fk } ⊂ Lp0 (Rn ) for some p0 ∈ [1, ∞], set Z

1 ♯

{|fk (y) − DtB fk (y)|} q dy. MD ({fk })(x) = sup l B∋x |B| B

Lemma 2.2. Let λ > 0, {fk } ⊂ L1 (Rn ) with compact support, B ⊂ Rn be a cube such that there exists x0 ∈ B with M (k{fk }klq )(x0 ) < λ. Then, for every ζ ∈ (0, 1), we can find γ > 0 (independent of λ, B, f , x0 ), such that (2.1)

♯ ({fk })(x) ≤ γλ}| ≤ ζ|B|, |{x ∈ B : M (k{fk }klq )(x) > Aλ, MD

where A > 1 is a fixed constant which only depends on the approximation of the identity {Dt }t>0 . Proof. Let A ∈ (1, ∞) be a constant which will be chosen later. For λ > 0, set
♯ Eλ = {x ∈ B : M k{fk }klq (x) > Aλ, MD ({fk })(x) ≤ γλ}.

We assume that there exists xE ∈ Eλ , for otherwise there is nothing to prove. As in the proof of Proposition 4.1 in [13] (see also the proof of Lemma 2.6 of [11]), we e = 2−2n A, can verify that for each x ∈ Eλ and A e M (k{fk }klq χ4B )(x) > Aλ.

s Now let t = r16B . For y ∈ 4B, write

|Dt (fk χ16B )(y)| ≤

Z

|ht (y, z)fk (z)|dz.

16B

By Minkowski’s inequality, we deduce that Z

{|Dt (fk χ16B )(y)|} q ≤ |ht (y, z)|k{fk (z)}klq dz . M (k{fk }klq )(x0 ), l 16B

SINGULAR INTEGRAL OPERATORS

7

since h is bounded on [0, ∞). Also, we have that for y ∈ 4B, ∞ Z X

{|Dt (fk χRn \16B )(y)|} q ≤ |ht (y, z)|k{fk (z)}klq dz l .

l=4 ∞ X l=4

2l+1 B\2l B

1 |B|

Z

h(2l+4 )k{fk (z)}klq dz

2l+1 B\2l B

. M (k{fk }klq )(x0 ). This, in turn implies that for all y ∈ Rn ,
M k{(Dt fk )χ4B }klq (x) . M (k{fk }klq )(x0 ) ≤ C1 λ,

with C1 > 0 a constant. Therefore, for each x ∈ Eλ ,


M k{fk χ4B }klq (x) ≤ M k{(fk − Dt fk )χ4B }klq (x) + M k{(Dt fk )χ4B }klq (x)
≤ M k{(fk − Dt fk )χ4B }klq (x) + C1 λ.

e = C1 + 1. It then follows that We choose A > 1 such that A
Eλ ⊂ {x ∈ B : M k{(fk − Dt fk )χ4B }klq (x) > λ}.

This, via the weak type (1, 1) estimate of M , tells us that Z ♯ |Eλ | ≤ C2 λ−1 kfk (y)−Dt fk (y)}klq dy ≤ C2 16n λ−1 |B|MD ({fk })(xE ) ≤ C2 16n γ|B|. 4B

For each ζ ∈ (0, 1), let γ = ζ(2C2 16n )−1 . The inequality (2.1) holds for this γ.  As in the proof of the Fefferman-Stein inequality (see [6, pp 150-151], or the proof of Theorem 2.2 in [11]), we can deduce from Lemma 2.2 that Corollary 2.3. Let Φ be an increasing function on [0, ∞) satisfying that Φ(2t) ≤ CΦ(t), t ∈ [0, ∞). {Dt }t>0 be an approximation to the identity as in Definition 1.4. Let {fk } be a sequence of functions such that for any R > 0, sup Φ(λ)|{x ∈ Rn : M (k{fk }klq )(x) > λ}| < ∞.

0 λ}| . sup Φ(λ)|{x ∈ Rn : MD λ>0

λ>0

Lemma 2.4. Let T be an L2 (Rn ) bounded linear operator with kernel K as in (1.5). Suppose that T satisfies Assumption 1.4. Then for any q ∈ (1, ∞), T is bounded from L1 (lq ; Rn ) to L1, ∞ (lq ; Rn ) Proof. We only consider the case c1 = 2. The other cases can be treated in the same way. For λ > 0, by applying the Calder´ on-Zygmund decomposition to k{fk }klq at level λ, we obtain a sequence of cubes {Ql } with disjoint interiors, such that Z

1

{fk (x)} q dx . λ, λ< l |Ql | Ql and k{fk (x)}klq . λ for a. e. x ∈ Rn \ ∪l Ql . For each fixed k, set fk1 (x) = fk (x)χRn \(∪l Ql ) (x),

8

GUOEN HU

X

fk2 (x) =

X

AtQl bk, l (x), fk3 (x) =

l

l


bk, l (x) − AtQl bk, l (x) χQl (x),

 with bk, l (y) = fk (y)χQl (y), tQl = {ℓ(Ql )} . By the fact that fk1 L∞ (lq ; Rn ) . λ, we deduce that

1 q

{fk } q q n . λq−1 {fk } 1 q n . L (l ; R ) L (l ; R ) s

Recalling that T is bounded on Lq (Rn ), we have that n o



q

x ∈ Rn : T fk1 (x) lq > λ/3 . λ−q {fk1 } Lq (lq ; Rn ) . λ−1 {fk } L1 (lq ; Rn ) . (2.2)

On the other hand, we get from (1.5) and (1.6) that Z Z Z vk (y)AtQ bk, l (y) dy ≤ htQl (z, y)|vk (z)|dzdy |b (z)| k, l l Rn Ql Rn Z |bk, l (z)|dz inf M vk (y). . y∈Ql

Ql

A straightforward computation involving Minkowski’s inequality gives us that X 1/q Z  X 1/q kbk, l kqL1 (Rn ) ≤ dy . λ|Ql |. |fk (y)|q Ql

k

k

Therefore, by Minkowski’s inequality and the vector-valued inequality of the HardyLittlewood maximal operator M (see [5]),

 X X q 1/q



AtQl bk, l

q n k

L (R )

l

≤

XXZ

sup

k{vk }kLq′ (lq′ ; Rn ) ≤1 k

sup

. k{vk }k

≤1 p′ r′ L m (l m ; Rn )

sup

.

k{vk }kLq′ (lq′ ; Rn ) ≤1

sup

.

k{vk }kLq′ (lq′ ; Rn ) ≤1

.λ

k

l

l

XZ

sup q−1 q

l

XXZ

vk (y)AtQ bk, l (y) dy l

|bk, l (z)|dz inf M vk (y)

l

k

Ql

Z

∪j Qj

1

{fk } q 1 q n . L (l ; R )

y∈Ql

Ql

XnXZ

k{vk }kLq′ (lq′ ; Rn ) ≤1

.λ

Rn

|bk, l (z)|dz

Ql

q o 1q

 inf M vk (y) lq′

y∈Qlm





bk, l (z) q dz inf M vk (y) q′ l l y∈Ql



M vk (y) q′ dy l

This, along with the fact that T is bounded from Lq (Rn ), leads to that 



x ∈ Rn : T f 2 (x) q > λ/3 . λ−1 {fk } 1 q n . (2.3) k l L (l ; R )

We turn our attention to T fk2 . Let Ω = ∪l 4nQl . It is obvious that |Ω| . λ k{fk }kL1 (lq ; Rn ) . For each x ∈ Rn \Ω, write Z 3 X T fk (x) ≤ K(x; y) − KAt (x; y) |bk,l (y)|dy Q −1

l

Rn

l

SINGULAR INTEGRAL OPERATORS

9

Applying Minkowski’s inequality twice, we obtain XZ

K(x; y) − KAt (x; y) k{bk,l (y)}klq dy

{T f 3(x)} q ≤ k l Q Therefore, (2.4)

l

Rn

l



 x ∈ Rn \Ω : T fk3 (x) q > λ/3 l XZ Z K(x; y) − KAt (x; y) dxk{bk,l (y)}klq dy . λ−1 Q Rn

l

l

Rn \4nQl

. λ−1 k{fk }kL1 (lq ; Rn ) .

Combining the inequalities (2.2)-(2.4) leads to our conclusion.



Lemma 2.5. Let T be the singular integral operator in Theorem 1.2, then for each n p0 N ∈ Rn and functions {fk }N k=1 ⊂ L (R ) for some p0 ∈ [1, ∞), ♯ MD ({T fk })(x) . ML log L (k{fk }klq )(x) + k{M fk (x)}klq . s Proof. Let x ∈ Rn , B be a ball containing x and tB = rB . Write Z

1

{|T fk (y) − DtB T fk (y)|} q dy ≤ E1 + E2 + E3 , l |B| B

with

Z 1 k{T (fk χ4B )(y)}klq dy, |B| B Z 1 E2 = k{DtB T (f χ4B )(y)}klq dy, |B| B E1 =

and

Z

1

{|T (fk χRn \4B )(y) − DtB T (fk χRn \4B )(y)|} q dy. E3 = l |B| B q n q q Recall that T is bounded on L (R ) (and so is bounded on L (l ; Rn )). Thus by Lemma 2.1 and Lemma 2.4,

E1 . k{fk }klq . ML log L (k{fk }klq )(x). L log L, 4B

On the other hand, it follows from Minkowski’s inequality that Z |htB (y, z)|k{T (fk χ4B )(z)}klq dz k{DtB T (fk χ4B )(y)}klq . Rn

Let F0 =

Z

|htB (y, z)|k{T (fk χ4B )(z)}klq dz

16B

and for j ∈ N, Fj =

Z

|htB (y, z)|k{T (fk χ4B )(z)}klq dz.

2j+5 B\2j+4 B

By the estimate (1.7) and Lemma 2.1, we know that

F0 ≤ k{fk }klq L log L, 4B ,

and

1 Fj ≤ h(2j ) |B|

Z

2j+5 B

k{T (fk χ4B )(z)}klq dz . 2−δj k{fk }klq L log L, 4B .

10

GUOEN HU

This, in turn gives us that

E2 . k{fk }klq L log L, 4B .

Finally, Assumption 1.5 tells us that for each k and y ∈ B, T (fk χRn \4B )(y) − DtB T (fk χRn \4B (y) . M fk (x), which implies that

E3 . k{M fk (x)}klq . Combining the estimates for E1 , E2 and E3 then leads to our desired conclusion.  Let D be a dyadic grid. Associated with D, define the sharp maximal function as Z 1 ♯ |f (y) − c|dy. MD f (x) = sup inf Q∋x c∈C |Q| Q

♯ MD

Q∈D



1/δ ♯ ♯ δ For δ ∈ (0, 1), let MD, . Repeating the argument in [16, δ f (x) = MD (|f | )(x) p. 153], we can verify that if Φ is a increasing function on [0, ∞) which satisfies that Φ(2t) ≤ CΦ(t), t ∈ [0, ∞), then (2.5)

♯ sup Φ(λ)|{x ∈ Rn : |h(x)| > λ}| . sup Φ(λ)|{x ∈ Rn : MD,δ h(x) > λ}|, λ>0

λ>0

n

provided that supλ>0 Φ(λ)|{x ∈ R : MD, δ h(x) > λ}| < ∞. Lemma 2.6. Under the assumption of Theorem 1.2, for each λ > 0, Z  k{fk }klq  k{fk }klq {x ∈ Rn : k{M T fk (x)}klq > λ} . dx. log 1 + λ λ Rn

Proof. By the well known one-third trick (see [8, Lemma 2.5]), we only need to prove that, for each dyadic grid D, the inequality 

 x ∈ Rn : MD (T fk )(x) q > 1 (2.6) l Z
k{fk (x)}klq log 1 + k{fk (x)}klq dx. . Rn

holds when {fk } is finite. As in the proof of Lemma 8.1 in [2], we can very that for each cube Q ∈ D, δ ∈ (0, 1), δ  δ1  1 Z  1 Z  δ1 . inf k{MD (fk χQ )}kδlq k{MD fk (y)}klq − c dy c∈C |Q| Q |Q| Q . hk{fk χQ }klq iQ ,

where in the last inequality, we invoked the fact that MD is bounded from L1 (lq ; Rn ) to L1, ∞ (lq ; Rn ). This, in turn, implies that

♯ MD, (2.7) δ k{MD fk }klq (x) . MD k{fk }klq (x).

SINGULAR INTEGRAL OPERATORS

11

Let Φ(t) = t log−1 (1 + t−1 ). It follows from (2.5), (2.7), (2.1) and Lemma 2.5 that {x ∈ Rn : k{MD T fk (x)}klq > 1}
k{MD T fk }klq (x) > t} . sup Φ(t) {x ∈ Rn : M ♯ D, δ

t>0


. sup Φ(t) {x ∈ Rn : M k{T fk }klq (x) > λ} t>0

♯ . sup Φ(t)|{x ∈ Rn : MD ({T fk })(x) > t}| t>0 . sup Φ(t) {x ∈ Rn : ML log L (k{fk }klq )(x) + k{M fk (x)}klq > t} t>0 Z
k{fk (x)}klq log(1 + k{fk (x)}klq dx, . Rn

where in the last inequality, we have invoked the fact that M is bounded from L1 (lq ; Rn ) to L1, ∞ (lq ; Rn ) (see [5]). This establish (2.6) and completes the proof of Lemma 2.6.  Proof of Theorem 1.6. Let q ∈ (1, ∞). We know by Lemma 2.4 that T is bounded from L1 (lq ; Rn ) to L1, ∞ (lq ; Rn ). On the other hand, it was proved in [4] (see also [13]) that under the assumption of Theorem 1.6, T ∗ f (x) ≤ M T f (x) + M f (x). Thus by Lemma 2.6, for each λ > 0, {x ∈ Rn : k{T ∗ fk (x)}klq > λ} .

Z

Rn

k{fk (x)}klq
k{fk (x)}klq dx. log 1 + λ λ

Therefore, it suffices to consider MT and MT ∗ . On the other hand, it was proved in that maximal operator ML log L satisfies that 

 x ∈ Rn : ML log L fk (x) q > λ l Z  k{fk (x)}klq k{fk (x)}klq  . dx. log 1 + λ λ Rn Thus, by Lemma 2.6, our proof is now reduced to proving that the inequalities (2.8)

MT f (x) . M T f (x) + ML log L f (x).

and (2.9)

MT ∗ f (x) . M T f (x) + ML log L f (x).

hold. Without loss of generality, we assume that c2 > 1.
s 1 Let Q ⊂ Rn be a cube and x, ξ ∈ Q. Set tQ = c2 √ ℓ(Q) and write n T (f χRn \3Q )(ξ)

= DtQ T f (ξ) − DtQ T (f χ3Q )(ξ)   + T (f χRn \3Q )(ξ) − DtQ T (f χRn\3Q )(ξ) .

12

GUOEN HU

A trivial computation involving (1.8) leads to that |DtQ T f (ξ)| .

|Q|−1

∞ Z X

s