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HARDY SPACE ESTIMATES FOR BILINEAR SQUARE FUNCTIONS ANDCALDERON-ZYGMUND OPERATORS

JAROD HART AND GUOZHEN LU

ABSTRACT. In this work we prove Hardy space estimates for bilinear Littlewood-Paley-Stein squarefunction and Calderon-Zygmund operators. Sufficient Carleson measure type conditions are givenfor square functions to be bounded from H p1 ×H p2 into Lp for indices smaller than 1, and suffi-cient BMO type conditions are given for a bilinear Calderon-Zygmund operator to be bounded fromH p1 ×H p2 into H p for indices smaller than 1. Subtle difficulties arise in the bilinear nature of theseproblems that are related to frequency properties of products of functions. Moreover, three typesof bilinear paraproducts are defined and shown to be bounded from H p1 ×H p2 into H p for indicessmaller than 1. The first is a bilinear Bony type paraproduct that was defined in [33]. The sec-ond is a paraproduct that resembles the product of two Hardy space functions. The third class ofparaproducts are operators given by sums of molecules, which were introduced in [2].

1. INTRODUCTION

There is a rich theory of Hardy spaces in harmonic analysis. Some of the early groundbreakingwork in the area came from Stein, Weiss, Coifman, and C. Fefferman, among many others, see forexample [49, 48, 18, 10]. In more recent years, Hardy space theory has been studied in product(aka multiparameter) settings and to a lesser extent in multlinear settings. The challenges in theseareas are formidable. There are many instances where results from the classical theory, which onemay initially expect to hold in the product and multilinear setting, fail. This phenomenon has beenobserved many times in the multiparameter setting, for example in the absence of a weak (1,1)estimate for the strong maximal operator (see for example [37]) and the difference between variousdefinitions of Hardy and BMO spaces in the product setting (see for example [6, 7, 39, 19, 40, 31]).This type of difficulty presents in multilinear analysis as well; of interest in this work are the failureof some boundedness properties for mutlilinear operators on products of Hardy spaces.

The study of multilinear Calderon-Zygmund operators was initiated by Coifman and Meyer[12, 13, 14]. A fruitful theory has grown around these operators, see for example [9, 26, 27, 23,2, 43, 29, 36, 34, 41]. A multilinear and multi-parameter version of the Coifman-Meyer Fouriermultiplier theorem was established in [44, 45] using time-frequency analysis (see also [8] usingthe Littlewood-Paley analysis), and a pseudo-differential analogue was carried out in [15]. Therehas also been some work done for the operators in the context of distributions spaces (Triebel-Lizorkin and Besov spaces), see for example [28, 1, 43, 2]. For appropriate indices, some of thesedistribution spaces coincide with Hardy spaces, which are the focus of this work. One of the mainresults we prove in this article is a bilinear T 1 type theorem that extends the bilinear T 1 theorem

Date: June 9, 2015.2010 Mathematics Subject Classification. 42B20, 42B25, 42B30.Key words and phrases. Square Function, Littlewood-Paley-Stein, Bilinear, Calderon-Zygmund Operators.Hart was partially supported by an AMS-Simons Travel Grant. Lu was supported by NSF grant #DMS1301595.

1

2 JAROD HART AND GUOZHEN LU

from [9, 26, 27, 33] to a Hardy space setting. This work also provides bilinear versions of thelinear results in [18, 50, 22, 20, 35]; in particular there is a close parallel with the Hardy spaceresults in [35]. There has also been a considerable development of bilinear Littlewood-Paley-Steintheory in recent years, see for example [42, 43, 32, 29, 25, 5, 24]. All of these articles deal withLittlewood-Paley-Stein operator mapping properties from Lp1 ×Lp2 into Lp for p1, p2 > 1. Herewe extend this to H p1×H p2 to Lp boundedness properties for indices 0 < p1, p2 ≤ 1.

To highlight a challenge that arises in bilinear Hardy space theory that is not present in thelinear theory, consider the following problem, to which we give a solution in this work. Given abilinear operator T acting on a product of Hardy spaces H p1×H p2 , what conditions are sufficientfor T to map H p1 ×H p2 into H p? At the root of this problem is a simple, but menacing, fact.Given a nonzero real-valued function f ∈ H1 (with sufficient decay so that f ∈ L1/2), it followsthat f · f = f 2 /∈ H1/2. Note that it is that fact that f 2 ≥ 0 has non-zero integral that bars it frommembership in H1/2 (any integrable function in H1/2 must have mean zero). This is a failure of abilinear analogue of classical theory in the following sense. The operator P( f1, f2)(x) = f1(x) f2(x)is analogous to the identity operator I f (x) = f (x) in the linear theory. Clearly the identity operatorI is bounded on any reasonable function space, including H p for all 0 < p ≤ 1, and the bilinearoperator P enjoys many similar properties to I. For example P is bounded from Lp1×Lp2 into Lp

for all 0 < p1, p2, p < ∞ satisfying 1p1+ 1

p2= 1

p (this is just Holder’s inequality), and by a result in[23], P is also bounded from H p1 ×H p2 into Lp for any 0 < p1, p2, p ≤ 1 such that 1

p1+ 1

p2= 1

p .Although, the product operator P fails to satisfy many Hardy space bounds that one might initiallyexpect, and this presents a much more difficult problem. Namely, P is not bounded from H p1×H p2

into H p, whenever 0 < p1, p2, p ≤ 1 and 1p1+ 1

p2= 1

p . The product structure of bilinear operatorsseverely complicates oscillatory properties of functions. For this reason, addressing Hardy spaceH p1×H p2 to H p bounds for bilinear operators is connected to the deep mathematical problem ofunderstanding the oscillatory behavior of products of functions. One result in this work (Theorem2.5) is a paraproduct operator Π( f1, f2) that resembles the product P( f1, f2) = f1 · f2 in somesenses, but satisfies Π( f1, f2) ∈ H1/2 for all f1, f2 ∈ H1, along with other properties.

The example in the last paragraph gives some initial insight into what we can expect as far asresults for the bilinear singular integral operators we work with in this article. Consider againthe pointwise product operator P( f1, f2)(x) = f1(x) f2(x). This is arguably the simplest bilinearoperator that one can consider, but it is not bounded from H p1×H p2 into H p for 0 < p≤ 1. So theconditions we impose on our bilinear operators below must not be verified by P. The appropriatebilinear operators to have the Hardy space boundedness properties must “improve” the functionsf1 and f2 in the following sense. Formally, f1 ∈ H p1 and f2 ∈ H p2 must have some regularityand vanishing moment properties up to order (comparable to) 1/p1 and 1/p2 respectively. Sincewe wish to obtain T ( f1, f2) ∈ H p, T ( f1, f2) must satisfy better regularity and vanishing momentproperties than either f1 or f2 (roughly speaking, up to order 1/p = 1/p1 +1/p2). This precludesthe typical paradigm for interpreting the regularity properties for inputs and outputs of a bilinearCalderon-Zygmund operators as set by the product operator P. That is, one typically considersT ( f1, f2) to satisfy the regularity properties that f1(x) · f2(x) would satisfy; which means that ageneral bilinear Calderon-Zygmund operator cannot be smoothing. We look at a smaller class ofbilinear Calderon-Zygmund operators that are smoothing in some sense; hence from this point ofview we can again rule out the product operator P as a representative example. An example of a

HARDY SPACE ESTIMATES FOR BILINEAR SQUARE FUNCTIONS AND CALDERON-ZYGMUND OPERATORS 3

smoothing bilinear Calderon-Zygmund operator was given in [2] by a certain paraproduct, in whichthe authors prove boundedness properties on homogeneous Sobolev spaces, from Lp1

s × Lp2t to Lp

s+t .We consider these smoothing operators since the output has regularity s+ t, whereas the inputs,f1 ∈ Lp1

s and f2 ∈ Lp2t , have lesser regularity, s and t respectively. A general Calderon-Zygmund

operator that is bounded L2×L2 to L1 does not satisfy this type of smoothing boundedness property.Later we apply our main Calderon-Zygmund operator result (Theorem 2.2) to the paraproductsfrom [2] to prove that they are also bounded from H p1×H p2 into H p under appropriate conditions,see Theorem 2.6.

There has been work done on bilinear operators that are related to the Hardy space mappingproblem we are considering that also hint at what conditions may be needed for bilinear operatorsto be bounded in H p for 0 < p ≤ 1. In [23], Grafakos and Kalton give conditions for a bilinearCalderon-Zygmund operator to be bounded from H p1 ×H p2 into Lp (we restrict to the bilinearsetting for this discussion). In fact, they only require that T is bounded from L2×L2 into L1 andadditional kernel regularity for this conclusion. There is no further cancellation required for Tbeyond what is necessary for Calderon-Zygmund operators to be bounded on Lebesgue spaceswith indices larger than 1, see for example [9, 26, 27, 33]. Note that the product operator still fallsinto the class of operators to which the results in [23] apply; as previously mentioned P is boundedfrom H p1×H p2 into Lp for indices smaller than 1. This highlights the drastic difference betweenthe work in [23] and this article. Another work that is closely related to this article is [36]. In thatwork, the authors prove some H p1×H p2 to H p estimates when p≤ 1. Their result, which is for pclose to 1, relies heavily on atomic decompositions of Hardy spaces and the boundedness resultsfrom [23]. We give sufficient conditions for a bilinear operator T to be decomposed in terms ofLittelwood-Paley-Stein theory, and prove estimates for Hardy spaces with indices p ranging allthe way down to zero without using atomic decompositions. This is inspired by the works in themulti-parameter Hardy space theory in [30, 31] where a discrete Littlewood-Paley theory is carriedout to prove boundedness of singular integral operators on multi-parameter Hardy spaces.

2. MAIN RESULTS

We take a moment to describe our general approach to the results in this article. Our primarygoal is to prove H p1×H p2 to H p estimates for bilinear Calderon-Zygmund operators. The way weapproach this is to decompose T into smooth truncation operators Θk = QkT for k ∈ Z, and reducethe H p estimates for T to estimates in for Θk using the Littlewood-Paley characterization of H p

from [18]. That is, we choose Qk to be Littlewood-Paley-Stein type operators, sufficient for thefollowing semi-norm equivalence,

||T ( f1, f2)||H p ≈

∣∣∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|QkT ( f1, f2)|2

) 12∣∣∣∣∣∣∣∣∣∣∣∣Lp

=

∣∣∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|Θk( f1, f2)|2

) 12∣∣∣∣∣∣∣∣∣∣∣∣Lp

= ||SΘ( f1, f2)||Lp,

where SΘ is the square function associated to the collection Θk = QkT . These operators Θk fork ∈ Z define what we call a collection of bilinear Littlewood-Paley-Stein operators. In this waywe reduce our H p1 ×H p2 to H p estimates for T to H p1 ×H p2 to Lp estimates for SΘ. The latterestimates for SΘ are the square function type estimates that we will prove, thereby yielding theCalderon-Zygmund operator estimates as well. We also obtain paraproduct boundedness propertiesby applying our Calderon-Zygmund operator estimates.


2.1. Bilinear Littlewood-Paley-Stein Theory. Given kernel functions θk : R3n → C for k ∈ Z,define

Θk( f1, f2)(x) =∫R2n

θk(x,y1,y2) f1(y1) f2(y2)dy1 dy2

for appropriate functions f1, f2 : Rn→ C. Define the square function associated to {Θk}

SΘ( f1, f2)(x) =

(∑k∈Z|Θk( f1, f2)(x)|2

) 12

.

We say that a collection of operators Θk for k ∈Z is a collection of bilinear Littlewood-Paley-Steinoperators with decay and smoothness (N,L), written {Θk} ∈ BLPSO(N,L), for an integer L ≥ 0and N > 0 if there exists a constant C such that

|Dα1 Dβ

2θk(x,y1,y2)| ≤C2(|α|+|β|)kΦNk (x− y1,x− y2) for all |α|, |β| ≤ L.(2.1)

Here we use the notation ΦNk (x,y) = 22kn(1 + 2k|x|+ 2k|y|)−N for N > 0, x,y ∈ Rn, and k ∈

Z. We also use the notation Dα0 F(x,y1,y2) = ∂α

x F(x,y1,y2), Dα1 F(x,y1,y2) = ∂α

y1F(x,y1,y2), and

Dα2 F(x,y1,y2) = ∂α

y2F(x,y1,y2) for F : R3n→ C and α ∈ Nn

0.Given {Θk} ∈ BLPSO(N,L) and α,β ∈ Nn

0 with |α|+ |β| < N − 2n, define the (α,β) ordermoment function associated to {Θk} by

[[Θk]]α,β (x) = 2k(|α|+|β|)∫R2n

θk(x,y1,y2)(x− y1)α(x− y1)

βdy1 dy2

for k ∈ Z and x ∈ Rn. It is worth noting that [[Θk]]0(x) = Θk(1,1)(x), which is an object that isclosely related to boundedness properties of SΘ, see for example [42, 43, 32, 25, 29, 24]. Our mainsquare function boundedness result is the following theorem.

Theorem 2.1. Let L≥ 1 be an integer and N = 2n+L(2n+L+5)/2. If {Θk} ∈ BLPSO(N,L) and

dµ(x, t) = ∑k∈Z

∑|α|,|β|≤L−1

|[[Θk]]α,β(x)|2δt=2−k dx(2.2)

is a Carleson measure, then SΘ can be extended to a bounded operator from H p1×H p2 into Lp forall n

2n+L < p≤ 1 and nn+L < p1, p2 ≤ 1 satisfying 1

p1+ 1

p2= 1

p .

Here δt=2−k is the delta point mass measure on (0,∞) concentrated at 2−k. That is, for a setE ⊂ (0,∞), the measure δt=2−k is defined by δt=2−k(E)= 1 if 2−k ∈E and δt=2−k(E)= 0 if 2−k /∈E.Also, a non-negative measure dµ(x, t) on Rn+1

+ = Rn× (0,∞) is a Carleson measure if there is aC > 0 such that dµ(Q× (0, `(Q)))≤C|Q| for all cubes Q⊂Rn, where `(Q) is the side length of Qand |Q| is the Lebesgue measure of Q.

2.2. Bilinear Calderon-Zygmund Theory. Let S = S (Rn) be the Schwartz class of smooth,rapidly decreasing functions endowed with the standard Schwartz semi-norm topology, and letS ′ =S ′(Rn) be its continuous dual, the class of tempered distributions. We say that a continuousbilinear operator T from S ×S into S ′ is a bilinear Calderon-Zygmund operator with smooth-ness M, written T ∈ BCZO(M), if T has function kernel K : R3n\{(x,x,x) : x ∈ Rn} → C such


that

〈T ( f1, f2),g〉=∫R3n

K(x,y1,y2) f1(y1) f2(y2)g(x)dy1 dy2 dx

whenever f1, f2,g ∈ C∞0 have disjoint support, supp( f1)∩ supp( f2)∩ supp(g) = /0, and there is a

constant C > 0 such that the kernel function K satisfies

|Dα0 Dβ

1Dµ2K(x,y1,y2)| ≤

C(|x− y1|+ |x− y2|)2n+|α|+|β|+|µ| for all |α|, |β|, |µ| ≤M

for x,y1,y2 ∈ Rn such that |x− y1|+ |x− y2|> 0. Define the transposes of T by the dual relations

〈T ( f1, f2),g〉=⟨T ∗1(g, f2), f1

⟩=⟨T ∗2( f1,g), f2

⟩for f1, f2,g ∈S , which also satisfy T ∗ j ∈ BCZO(M) for j = 1,2 whenever T ∈ BCZO(M). Wealso define moment distributions for an operator T ∈ BCZO(M), but we require some notation first.For an integer M ≥ 0, define the collections of functions

OM = OM(Rn) =

{f ∈C∞(Rn) : sup

x∈Rn| f (x)| · (1+ |x|)−M < ∞

}and

DM = DM(Rn) =

{f ∈C∞

0 (Rn) :∫Rn

f (x)xαdx = 0 for all |α| ≤M}.

Define the topology of DM by the sequential characterization, for fk, f ∈DM for k ∈ N, fk→ f inDM if there exists a compact set K such that supp( fk),supp( f )⊂ K for all k ∈ N and

limk→∞||Dα fk−Dα f ||L∞ = 0

for all α ∈ Nn0. Then D ′M is defined to be all linear functionals W : DM→ C such that

fk→ f in DM implies 〈W, fk〉 → 〈W, f 〉 .Let η ∈C∞

0 (Rn) be supported in B(0,2) and η(x) = 1 for x ∈ B(0,1). Define for R > 0, ηR(x) =η(x/R). We reserve the notation ηR for functions constructed in this way. In [1], Benyi definedT ( f1, f2) for f1 ∈ OM1 and f2 ∈ OM2 where T is a bilinear singular integral operator. We give anequivalent definition. Let T be a BCZO(M + 1) and f1 ∈ OM1 and f2 ∈ OM2 for some integersM1,M2 ≥ 0 such that M1 +M2 ≤M. For ψ ∈DM, define

〈T ( f1, f2),ψ〉= limR→∞〈T (ηR f1,ηR f2),ψ〉 .

Also, for f1 ∈ OM, f2 ∈C∞0 , and ψ ∈DM, define

〈T ( f1, f2),ψ〉= limR→∞〈T (ηR f1, f2),ψ〉 .

These limits exist based on the kernel representation and kernel properties for T ∈ BCZO(M+1),see [1] for proof of this fact. Now we define the moment distribution [[T ]]α,β ∈ D ′|α|+|β| for T ∈BCZO(M+1) and α,β ∈ Nn

0 with |α|+ |β| ≤M by⟨[[T ]]α,β,ψ

⟩= lim

R→∞

∫R3n

K (x,y1,y2)(x− y1)α

ηR(y1)(x− y2)βηR(y2)ψ(x)dy1 dy2 dx

for ψ ∈ DM. Here K ∈ S ′(R3n) is the distribution kernel of T , and the integral above is in-terpreted as a dual pairing between K ∈ S ′(R3n) and elements of S (R3n). This distributionalmoment associated to T generalizes the notion of T (1,1) as used in [9, 26, 27, 33] in the sense that


〈[[T ]]0,0,ψ〉= 〈T (1,1),ψ〉 for all ψ ∈D0. We will also use a generalized notion of BMO to extendthe cancellation conditions T (1,1),T ∗1(1,1),T ∗2(1,1)∈BMO, which were used in the bilinear T 1theorems from [9, 26, 27, 33]. Let M ≥ 0 be an integer and F ∈D ′M/P , distributions D ′M modulopolynomials. We say that F ∈ BMOM if the non-negative measure on Rn+1

+ ,

∑k∈Z

22Mk|QkF(x)|2dxδt=2−k

is a Carleson measure. This definition agrees with the classical definition of BMO by the charac-terization of BMO in terms of Carleson measures in [4, 38]. It should be noted that we definedthis polynomial growth BMOM space in [35], and a similar polynomial growth BMOM was definedby Youssfi [51]. We use this polynomial growth BMOM to quantify cancellation conditions foroperators T ∈ BCZO(M) in the following result.

Theorem 2.2. Assume that T ∈ BCZO(M), where M = L(2n+L+ 5)/2 for some integer L ≥ 1,and that T is bounded from L2×L2 into L1. If

T ∗1(xα,ψ) = T ∗2(ψ,xα) = 0 in D ′|α| for all |α| ≤ 2L+n and ψ ∈D2L+n,(2.3)

[[T ]]α,β ∈ BMO|α|+|β| for all |α|, |β| ≤ L−1,(2.4)

then T can be extended to bounded operator from H p1 ×H p2 into H p for all n2n+L < p ≤ 1 and

nn+L < p1, p2 ≤ 1 satisfying 1

p1+ 1

p2= 1

p .

We prove Theorem 2.2 by decomposing an operator T ∈BCZO(M) into a collection of operators{Θk} ∈ BLPSO(2n+M,L) in Theorem 2.1, where L depends on the regularity parameter M. Thisdecomposition of T into a collection of bilinear Littlewood-Paley-Stein operators is stated preciselyin the next theorem.

Theorem 2.3. Let L≥ 1 be an integer, N > 2n, and T ∈ BCZO(M) for some integer M ≥max(N−2n,2L+1). Also assume that T is bounded from L2×L2 into L1. If T satisfies (2.3), then {Θk}={QkT} ∈ BLPSO(N,L) for any ψ ∈DM, where Qk f = ψk ∗ f and ψk(x) = 2knψ(2kx).

2.3. Applications to Paraproducts. In this article, we use Theorem 2.2 to prove that three typesof paraproducts are bounded on Hardy spaces. The first is a bilinear Bony type paraproduct, whichis a bilinear version of Bony’s paraproduct in [3] and was originally introduced in [33]. It isconstructed as follows.

Let ψ ∈DM and ϕ ∈C∞0 , and define ψk(x) = 2knψ(2kx), ϕk(x) = 2knϕ(2kx), Qk f = ψk ∗ f , and

Pk f = ϕk ∗ f . For b ∈ BMO, define the bilinear Bony paraproduct

Πb( f1, f2)(x) = ∑j∈Z

Q j(Q jb ·Pj f1 ·Pj f2

)(x).(2.5)

If ψ and ϕ are chosen appropriately, then this paraproduct satisfies Πb(1,1) = b, see [33]. We arenot concerned with this particular property here, so we allow for a more general selection of ψ andϕ. We will apply Theorem 2.2 to Πb to prove the following theorem.

Theorem 2.4. Let L be a non-negative integer, b ∈ BMO, ψ ∈ D3L+n, and ϕ ∈ C∞0 , then Πb, as

defined in (2.5), is bounded from H p1 ×H p2 into H p for all n2n+L < p ≤ 1 and n

n+L < p1, p2 ≤ 1satisfying 1

p = 1p1+ 1

p2.


The motivation for our second paraproduct operator is to find a replacement for the productf1(x) f2(x) that is in H p for 0 < p ≤ 1 for appropriate f1 and f2. Once again, we construct thisoperator a bit more generally. We will use the following definition for the Fourier transform; forf ∈ L1, define

F [ f ](ξ) = f (ξ) =∫Rn

f (x)e−ixξdx.

Let ψ ∈S such that its Fourier transform ψ is supported away from the origin, and let ϕ ∈S . Letψk, ϕk, Qk, and Pk be as above. Define

Π( f1, f2)(x) = ∑k∈Z

Qk (Pk f1 ·Pk f2)(x).(2.6)

We prove the following Hardy space bounds for Π.

Theorem 2.5. Let Π( f1, f2) be as in (2.6). Then Π is bounded from H p1 ×H p2 into H p for all0 < p1, p2, p≤ 1 satisfying 1

p = 1p1+ 1

p2.

In order to construct the paraproduct Π( f1, f2) to resemble the product f1 · f2, we choose ψ andϕ in the following way. Let ϕ ∈S such that ϕ(ξ) = 1 for |ξ| ≤ 1 and supp(ϕ) ⊂ B(0,2). Defineψ(x) = 2−2nϕ(2−2x)−2−3nϕ(2−3x). With this choice of ψ and ϕ, it follows that Qk =PkQk. Underthese conditions, it also follows that

Π(1, f2)(x) = ∑k∈Z

Qk (Pk(1) ·Pk f2)(x) = ∑k∈Z

QkPk f2(x) = ∑k∈Z

Qk f2(x) = f2(x) and

Π( f1,1)(x) = ∑k∈Z

Qk (Pk f1 ·Pk(1))(x) = ∑k∈Z

QkPk f1(x) = ∑k∈Z

Qk f1(x) = f1(x)

in H p for all f1, f2 ∈ H p ∩ L2. This gives precise meaning to how Π( f1, f2)(x) “resembles” theproduct f1(x) · f2(x).

We should remark here that the constructions in Theorems 2.4 and 2.5 use slightly differenttechniques, but are interchangeable in some senses. In Theorem 2.4, we choose the convolutionkernels ψk ∈D3L+n, and conclude bounds for Hardy spaces with indices bounded below by n

2n+Land n

n+L , where L can be taken arbitrarily large. One can construct the bilinear Bony paraproductΠb with ψ ∈S such that ψ is supported away from the origin, and the conclusion is strengthenedto all 0 < p1, p2, p ≤ 1 similar to Theorem 2.5. Similarly, the paraproduct Π can be constructedwith functions ψk ∈ DM, and obtain the same conclusion as in Theorem 2.4 with the appropriatelower bounds for 0 < p1, p2, p≤ 1.

Finally, we consider a class of paraproducts defined in [2]. For a dyadic cube Q, a functionφQ : Rn → C is an (M,N)-smooth molecule associated to Q if there exists a constant C = CM,Nindependent of Q such that

|DαφQ(x)| ≤C

`(Q)−n/2`(Q)−|α|

(1+ `(Q)−1|x− xQ|)N

for all |α| ≤M, where xQ denotes the lower-left corner of Q. A family of (M,N)-smooth moleculesψQ indexed by dyadic cubes Q is an (M,N,L)-smooth family of molecules with cancellation if ψQis an (M,N)-smooth molecule associated to Q and∫

RnψQ(x)xαdx = 0(2.7)


for all |α| ≤ L. Let φ1Q,φ

2Q,φ

3Q be three families of molecules indexed by dyadic cubes Q, and define

the paraproduct T , as in [2], by

T ( f1, f2)(x) = ∑Q

⟨f1,φ

1Q⟩⟨

f2,φ2Q⟩

φ3Q(x).

The sum here indexed by Q is over all dyadic cubes. In [2], the authors prove the boundedness ofthese operators on many different function spaces, for example from Lp1×Lp2→ Lp, H p1×H p2→Lp, L∞×L∞→ BMO, as well as a number of estimates for weak Lp spaces and weighted spaces.Here we extend their results to boundedness properties from products of Hardy spaces into Hardyspaces.

Theorem 2.6. Let L ≥ 0 be an integer, M = L(2n+L+ 5)/2, and N > 10n+ 15M + 5. Assumethat φ1

Q,φ2Q,φ

3Q are three families of (M,N)-smooth molecules. Furthermore assume that φ3

Q is acollection of (M,N,2L+n)-smooth molecules with cancellation, and either φ1

Q or φ2Q is a family of

(M,N,L−1)-smooth molecules with cancellation. Then T can be extended to a bounded operatorfrom H p1×H p2 into H p for all n

2n+L < p≤ 1 and nn+L such that 1

p1+ 1

p2= 1

p .

The selection of M and N in this result are not optimal. Here we use the selection of parametersthat are sufficient to show that T ∈ BCZO(M) based on the work in [2], in which the authors notethe parameters are not chosen to be optimal.

This article is organized in the following way. In Section 3, we establish some notation, and statesome preliminary results. Section 4 is dedicated to proving the bilinear square function estimatesin Theorem 2.1. Section 5 is used to prove our bilinear Calderon-Zygmund operator estimates inTheorems 2.2 and 2.3. Finally in Section 6, we apply Theorem 2.2 to the paraproduct operators inTheorems 2.4, 2.5, and 2.6.

3. PRELIMINARIES

We use the notation A. B to mean that A≤CB for some constant C. The constant C is allowedto depend on the ambient dimension, smoothness and decay parameters of our operators, indicesof function spaces etc.; in context, the dependence of the constants is clear.

We will use the following Frazier and Jawerth type discrete Calderon reproducing formula [21](see also [30] for a multiparameter formulation of this reproducing formula): there exist φ j, φ j ∈Sfor j ∈ Z with infinite vanishing moment such that

f (x) = ∑j∈Z

∑`(Q)=2−( j+N0)

|Q|φ j(x− cQ)φ j ∗ f (cQ) in L2(3.1)

for f ∈ L2. The summation in Q here is over all dyadic cubes with side length `(Q) = 2−( j+N0) forsome N0 > n, and cQ denotes the center of cube Q.

We will also use a more traditional formulation of Calderon’s reproducing formula: fix ϕ ∈C∞

0 (B(0,1)) such that

∑k∈Z

Qk f = f in L2(3.2)

for f ∈ L2, where ψ(x) = 2nϕ(2x)−ϕ(x), ψk(x) = 2knψ(2kx), and Qk f = ψk ∗ f . Furthermore, wecan assume that ψ has an arbitrarily large, but fixed, number of vanishing moments.


There are many equivalent definitions of the real Hardy spaces H p = H p(Rn) for 0 < p < ∞.We use the following one. Define the non-tangential maximal function

N ϕ f (x) = supt>0

sup|x−y|≤t

∣∣∣∣∫Rnt−n

ϕ(t−1(y−u))∗ f (u)du∣∣∣∣ ,

where ϕ ∈S with non-zero integral. It was proved by Fefferman and Stein in [18] that one candefine || f ||H p = ||N ϕ f ||Lp to obtain the classical real Hardy spaces H p for 0 < p < ∞. It was alsoproved in [18] that for any ϕ ∈S and f ∈ H p for 0 < p < ∞,∣∣∣∣∣∣∣∣sup

k∈Z|ϕk ∗ f |

∣∣∣∣∣∣∣∣Lp. || f ||H p.

We will use a number of equivalent quasi-norms for H p. Let ψ∈DM for some integer M > n(1/p−1), and let ψk and Qk be as above. For f ∈S ′/P (tempered distributions modulo polynomials),f ∈ H p if and only if ∣∣∣∣∣∣

∣∣∣∣∣∣(

∑k∈Z|Qk f |2

) 12∣∣∣∣∣∣∣∣∣∣∣∣Lp

< ∞,

and this quantity is comparable to || f ||H p . The space H p can also be characterized by the operatorsφ j and φ j from the discrete Littlewood-Paley-Stein decomposition in (3.1). This characterizationis given by the following, which can be found in [30]. Given 0 < p < ∞∣∣∣∣∣∣∣

∣∣∣∣∣∣∣∑

j∈Z∑

`(Q)=2−( j+N0)

|φ j ∗ f (cQ)|2χQ

12∣∣∣∣∣∣∣∣∣∣∣∣∣∣Lp

≈ || f ||H p,

where χE(x) = 1 for x ∈ E and χE(x) = 0 for x /∈ E for a subset E ⊂Rn. For a continuous functionf ∈ L1

loc(Rn) and 0 < r < ∞, define

M rj f (x) =

M

∑`(Q)=2−( j+N0)

f (cQ)χQ

r (x)

1r

,(3.3)

where M is the Hardy-Littlewood maximal operator. The following estimate was also proved byHan and Lu in [30].

Proposition 3.1. For 0 < r < p≤ 1 and f ∈ H p∣∣∣∣∣∣∣∣∣∣∣∣(

∑j∈Z

(M r

j (φ j ∗ f ))2

) 12∣∣∣∣∣∣∣∣∣∣∣∣Lp

. || f ||H p,

where M rj is defined as in (3.3).

The next result is a reformulation of an estimate proved by Han and Lu in [30]; this version ofthe result was proved in [35].


Proposition 3.2. Let f : Rn → C a non-negative continuous function, ν > 0, and nn+ν

< r ≤ 1.Then

∑`(Q)=2−( j+N0)

|Q|Φn+ν

min( j,k)(x− cQ) f (cQ). 2max(0, j−k)(N−n)M rj f (x)

for all x ∈ Rn, where M rj is defined in (3.3) and the summation indexed by `(Q) = 2−( j+N0) is the

sum over all dyadic cubes with side length 2−( j+N0) and cQ denotes the center of cube Q.

The next result is proved using some well-known techniques for Carleson measure. The proofcan be found in [35].

Proposition 3.3. Suppose


µk(x)δt=2−k dx(3.4)

is a Carleson measure, where µk is a non negative, locally integrable function for all k ∈ Z. Alsolet ϕ ∈S , and define Pk f = ϕk ∗ f , where ϕk(x) = 2knϕ(2kx) for k ∈ Z. Then∣∣∣∣∣∣

∣∣∣∣∣∣(

∑k∈Z|Pk f |pµk

) 1p∣∣∣∣∣∣∣∣∣∣∣∣Lp

. || f ||H p for all 0 < p < ∞

and ∣∣∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|Pk f |2µk

) 12∣∣∣∣∣∣∣∣∣∣∣∣Lp

. || f ||H p for all 0 < p≤ 2.

We will also need Hardy space estimates for linear Littlewood-Paley-Stein square function op-erators that was proved in [35]. First we set some notation for linear Littlewood-Paley-Stein oper-ators. Given kernel functions λk : R2n→ C for k ∈ Z, define

Λk f (x) =∫Rn

λk(x,y) f (y)dy

for appropriate functions f : Rn→ C. Define the square function associated to {Λk}

SΛ f (x) =

(∑k∈Z|Λk f (x)|2

) 12

.

We say that a collection of operators Λk for k ∈ Z is a collection of Littlewood-Paley-Stein op-erators with decay and smoothness (N,L), written {Λk} ∈ LPSO(N,L), for an integer L ≥ 0 andN > 0 if there exists a constant C such that

|Dα1 λk(x,y)| ≤C2|α|kΦ

Nk (x− y) for all |α| ≤ L.(3.5)

Here we use the notation ΦNk (x) = 2kn(1 + 2k|x|)−N for N > 0, x ∈ Rn, and k ∈ Z. We also

write Dα0 F(x,y) = ∂α

x F(x,y) and Dα1 F(x,y) = ∂yF(x,y) for F : R2n → C and α ∈ Nn

0. This is aslightly different definition for LPSO(N,L) than what was used in [35]. In that article, we definedLPSO(N,L+ δ) for integers L ≥ 0 and 0 < δ ≤ 1 in terms of δ-Holder conditions on λk(x,y) inthe y variable. Here we require {Λk} ∈ LPSO(N,L) to have kernel that is L times differentiable iny instead of L− 1 times differentiable with Lipschitz L− 1 order derivatives as in [35]. It is not


hard to see that the definition we use for LPSO(N,L) here is contained in the class defined in [35];hence all results from [35] for LPSO(N,L) are still applicable in the current work.

Given {Λk} ∈ LPSO(N,L) and α ∈ Nn0 with |α|< N−n, define

[[Λk]]α (x) = 2k|α|∫Rn

λk(x,y)(x− y)αdy

for k ∈ Z and x ∈ Rn. The next theorem was also proved in [35].

Theorem 3.4. Let {Λk} ∈ LPSO(n+2L,L) for some integer L≥ 1. If

dµα(x, t) = ∑k∈Z|[[Λk]]α(x)|2δt=2−k dx(3.6)

is a Carleson measure for all α ∈ Nn0 with |α| ≤ L−1, then SΛ is bounded from H p into Lp for all

nn+L < p≤ 1.

Note that we use Λk to denote linear operators and Θk to denote bilinear operators. We will keepthis convention throughout the paper.

4. HARDY SPACE ESTIMATES FOR BILINEAR SQUARE FUNCTIONS

To start this section, we prove a reduced version of Theorem 2.1, where we strengthen thecancellation assumptions on Θk from the ones in (2.2) to the conditions in the next lemma. Oncewe establish Lemma 4.1, we extend to the general situation in Theorem 2.1 by using a paraproducttype decomposition.

Lemma 4.1. Let {Θk} ∈ BLPSO(N,L), where N = 2n+3L for some integer L≥ 1. If∫Rn

θk(x,y1,y2)yα1 dy1 =

∫Rn

θk(x,y1,y2)yα2 dy2 = 0

for all k ∈ Z and |α| ≤ L−1, then SΘ can be extended to a bounded operator from H p1×H p2 intoLp for all n

2n+L < p≤ 1 and nn+L < p1, p2 ≤ 1 such that 1

p = 1p1+ 1

p2.

Proof. Let p, p1, p2 be as above. Choose ν such that np − 2n < ν < L, which is possible since

n2n+L < p. Since ν > n

p − 2n and 1p1+ 1

p2= 1

p , there exists s ∈ (0,1) such that sν > np1− n and

(1− s)ν > np2−n. Finally fix r1,r2 > 0 such that n

n+sν< r1 < p1 and n

n+(1−s)ν < r2 < p2. Note thatthis also implies that n+ sν≤ n/p1 < n+L and n+(1− s)ν≤ n/p2 < n+L, which will be usedlater to conclude that Φ

n+Lk (x)≤Φ

n+sν

k (x) and Φn+Lk (x)≤Φ

n+(1−s)νk (x).

By density, it is sufficient to prove the appropriate estimate for fi ∈ H pi ∩L2 for i = 1,2. Wedecompose Θk( f1, f2) using (3.1) for each f1 and f2,

Θk( f1, f2)(x) = ∑j1, j2∈Z

∑Q1,Q2

|Q1| |Q2|φ j1 ∗ f1(cQ1)φ j2 ∗ f2(cQ2)Θk(φcQ1j1 ,φ

cQ2j2 )(x)

= ∑j1, j2∈Z

∑Q1,Q2

|Q1| |Q2|φ j1 ∗ f1(cQ1)φ j2 ∗ f2(cQ2)∫R2n

θk(x,y1,y2)φcQ1j1 (y1)φ

cQ2j2 (y2)dy1 dy2.

The summation in Q1 and Q2 are over all dyadic cubes with side lengths `(Q1) = 2−( j1+N0) and`(Q2) = 2−( j2+N0) respectively. Now we establish an almost orthogonally estimate for the integralterm in the previous equation. Using the vanishing moment properties of θk and the regularity of


φ j1 , we obtain the following.∣∣∣∣∫R2nθk(x,y1,y2)φ

cQ1j1 (y1)φ

cQ2j2 (y2)dy1 dy2

∣∣∣∣=

∣∣∣∣∣∫R2n

θk(x,y1,y2)

(φ

cQ1j1 (y1)− ∑

|α|≤L−1

DαφcQ1j1 (x)

α!(y1− x)α

)φ

cQ2j2 (y2)dy1 dy2

∣∣∣∣∣.

∫R2n

ΦNk (x− y1,x− y2)(2 j1|x− y1|)L

×(

Φn+Lj1 (y1− cQ1)+Φ

n+Lj1 (x− cQ1)

)Φ

n+Lj2 (y2− cQ2)dy1 dy2

. 2L( j1−k)∫R2n

Φ2n+2Lk (x− y1,x− y2)

×(

Φn+Lj1 (y1− cQ1)+Φ

n+Lj1 (x− cQ1)

)Φ

n+Lj2 (y2− cQ2)dy1 dy2

. 2L( j1−k)Φ

n+sν

min( j1,k)(x− cQ1)Φ

n+(1−s)νmin( j2,k)

(x− y2).

A similar estimate holds for 2L( j2−k) in place of 2L( j1−k). Using the vanishing moment propertiesof φ j1 , we have the following estimate,∣∣∣∣∫R2n

θk(x,y1,y2)φcQ1j1 (y1)φ

cQ2j2 (y2)dy1 dy2

∣∣∣∣=

∣∣∣∣∣∫R2n

(θk(x,y1,y2)− ∑

|α|≤L−1

Dα1 θk(x,cQ1,y2)

α!(y1− cQ1)

α

)φ

cQ1j1 (y1)φ

cQ2j2 (y2)dy1 dy2

∣∣∣∣∣.

∫R2n

ΦNk (x− y1,x− y2)(2k|y1− cQ1|)

LΦ

N+Lj1 (y1− cQ1)Φ

Nj2(y2− cQ2)dy1 dy2

×∫R2n

ΦNk (x− cQ1,x− y2)(2k|y1− cQ1|)

LΦ

N+Lj1 (y1− cQ1)Φ


. 2L(k− j1)∫R2n

ΦNk (x− y1,x− y2)Φ

Nj1(y1− cQ1)Φ


. 2L(k− j1)Φn+sν



(x− cQ2).

Once again, this estimate holds with 2L(k− j2) in place of 2L(k− j1). Therefore∣∣∣∣∫R2nθk(x,y1,y2)φ

cQ1j1 (y1)φ

cQ2j2 (y2)dy1 dy2

∣∣∣∣. 2−Lmax(| j1−k|,| j2−k|)

Φn+sν



(x− cQ2).


Applying Proposition 3.2 yields

|Θk( f1, f2)(x)|. ∑j1, j2∈Z

∑Q1,Q2

|Q1| |Q2|φ j1 ∗ f1(cQ1)φ j2 ∗ f2(cQ2)

×2−Lmax(| j1−k|,| j2−k|)Φ

n+sν



(x− cQ2)

. ∑j1, j2∈Z

2−Lmax(| j1−k|,| j2−k|)2sνmax(0,k− j1)2(1−s)νmax(0,k− j2)M r1j1 (φ j1 ∗ f1)(x)M r2

j2 (φ j2 ∗ f2)(x)

. ∑j1, j2∈Z

2−ε max(| j1−k|,| j2−k|)M r1j1 (φ j1 ∗ f1)(x)M r2

j2 (φ j2 ∗ f2)(x),

where ε = L−ν > 0. Applying Proposition 3.1 for both M r1j1 (φ j1 ∗ f1) and M r2

j2 (φ j2 ∗ f2) yields theappropriate estimate below,

||SΘ( f1, f2)||Lp .

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∑

k∈Z

[∑

j1, j2∈Z2−

ε

2 | j1−k|2−ε

2 | j2−k|M r1j1 (φ j1 ∗ f1)M r2

j2 (φ j2 ∗ f2)

]2 1

2∣∣∣∣∣∣∣∣∣∣∣∣∣∣Lp

.

∣∣∣∣∣∣∣∣∣∣∣∣(

∑j1, j2,k∈Z

2−ε

2 | j1−k|2−ε

2 | j2−k|[M r1

j1 (φ j1 ∗ f1)M r2j2 (φ j2 ∗ f2)

]2) 1

2∣∣∣∣∣∣∣∣∣∣∣∣Lp

.

∣∣∣∣∣∣∣∣∣∣∣∣(

∑j1∈Z

[M r1

j1 (φ j1 ∗ f1)]2) 1

2∣∣∣∣∣∣∣∣∣∣∣∣Lp1

∣∣∣∣∣∣∣∣∣∣∣∣(

∑j2∈Z

[M r2

j2 (φ j2 ∗ f2)]2) 1

2∣∣∣∣∣∣∣∣∣∣∣∣Lp2

. || f1||H p1 || f2||H p2 .

This completes the proof of Lemma 4.1. �

Next we construct paraproducts to decompose Θk. These are the same operators that wereconstructed in [35], although we modify the decomposition to fit the bilinear framework. Fix anapproximation to identity operator Pk f = ϕk ∗ f , where ϕk(x) = 2knϕ(2kx) and ϕ∈S with integral1. Define for α,β ∈ Nn

0

Mα,β =

(−1)|α|

β!(β−α)!

∫Rn

ϕ(y)yβ−αdy α≤ β

0 α 6≤ β

.

Here we say α≤ β for α = (α1, ...,αn),β = (β1, ...,βn) ∈Nn0 if αi ≤ βi for all i = 1, ...,n. It is clear

that |Mα,β|< ∞ for all α,β ∈ Nn0 since ϕ ∈S . It is not hard to verify that when |α|= |β|,

Mα,β =

{(−1)|β|β! α = β

0 α 6= β and |α|= |β| .(4.1)

Consider the operators PkDα f defined for f ∈S ′, where Dα is the distributional derivative on S ′.Hence PkDα f (x) is well defined for f ∈S ′ since PkDα f (x) =

⟨ϕx

k,Dα f⟩= (−1)|α|

⟨Dα(ϕx

k), f⟩

and Dα(ϕxk) ∈S . In fact, this gives a kernel representation for PkDα; estimates for this kernel are


addressed in the proof of Proposition 4.3. Also,

[[PkDα]]β(x) = 2|β|k∫Rn

ϕk(x− y)∂αy ((x− y)β)dy = 2k|α|Mα,β.

For {Θk} ∈ BLPSO(N,L) and k ∈ Z, define

Θ(0)k ( f1, f2)(x) = Θk( f1, f2)(x)− [[Θk(·, f2)]]0(x) ·Pk f1(x)

− [[Θk( f1, ·)]]0(x) ·Pk f2(x)+ [[Θk]]0,0(x) ·Pk f1(x)Pk f2(x), and(4.2)

Θ(m)k ( f1, f2)(x) = Θ

(m−1)k ( f1, f2)(x)− ∑

|α|=m(−1)|α|

[[Θ(m−1)k (·, f2)]]α(x)

α!·2−k|α|PkDα f1(x)

− ∑|β|=m

(−1)|β|[[Θ

(m−1)k ( f1, ·)]]β(x)

β!·2−k|β|PkDβ f2(x)(4.3)

+ ∑|α|=|β|=m

(−1)|α|+|β|[[Θ

(m−1)k ]]α,β(x)

α!β!·2−k(|α|+|β|)PkDα f1(x)PkDβ f2(x)

for 1≤ m≤ L.

Lemma 4.2. Let {Θk} ∈ BLPSO(N,L) for some integer L≥ 1 and N > 2n. Also let 0≤M < N−nbe an integer. Then ∫

Rnθk(x,y1,y2)yα

1 dy1 = 0(4.4)

for all x,y2 ∈Rn and |α| ≤M if and only if [[Θk(·, f )]]α(x) = 0 for all x∈Rn, |α| ≤M, and f ∈C∞0 .

Likewise for integrals in y2 and [[Θk( f , ·)]]α.

Here we define [[Θk(·, f )]]α by applying the definition of [[ · ]]α for linear operator to the linearoperator g 7→Θk(g, f ) with f fixed. A similar notation is used for [[Θk( f , ·)]]α.

Proof. Note that the condition 0≤M < N−n implies that [[Θk(·, f )]]α is well defined for |α| ≤M.Assume that (4.4) holds. Then for any |α| ≤M

[[Θk(·, f )]]α = 2|α|k∫R2n

θk(x,y1,y2)(x− y1)α f (y2)dy1 dy2

= 2|α|k ∑µ+ν=α

(−1)|ν|cµ,ν

∫Rn

(∫Rn

θk(x,y1,y2)yν1dy1

)xµ f (y2)dy2 = 0.

Here cµ,ν are binomial coefficients. Now assume that [[Θk(·, f )]]α(x) = 0 for all x ∈ Rn, |α| ≤M,and f ∈C∞

0 . Let ϕ∈C∞0 be radial with integral 1 and define ϕk(x)= 2knϕ(2kx). The radial condition


here is not strictly necessary, but it simplifies notation. Then for |α| ≤M, we have∫Rn

θk(x,y1,y2)yα1 dy1 = ∑

µ+ν=α

∫Rn

θk(x,y1,y2)(−1)|α|+|µ|xµ(x− y1)νdy1

= ∑µ+ν=α

(−1)|α|+|µ|xµ limN→∞

∫R2n

θk(x,y1,u)(x− y1)νϕN(u− y2)dy1 du

= ∑µ+ν=α

(−1)|α|+|µ|xµ limN→∞

[[Θk(·,ϕy1N )]]ν(x) = 0.

We use that θk(x,y1,y2) is a bounded L1(Rn) continuous function in x for y1 6= y2 to use theapproximation to identity property ϕN ∗ θk(x, ·,y2)(y2) → θk(x,y1,y2) as N → ∞ pointwise fory1 6= y2 and in Lp(Rn) for 1 < p < ∞. This completes the proof. �

Proposition 4.3. Let {Θk} ∈ BLPSO(N,L) for some integer L≥ 1 and N ≥ 2n+L(2n+L−1)/2,and assume that


∑|α|,|β|≤L−1

|[[Θk]]α,β(x)|2δt=2−k dx(4.5)

is a Carleson measure. Also let Θ(m)k be as in as in (4.2) and (4.3) for 0≤m≤ L−1. Then Θ

(m)k ∈

BLPSO(Nm,L) where Nm = N−L(2n+L−1)/2, and they satisfy the following for 0≤m≤ L−1:

(1) For all α ∈ Nn0 with |α| ≤ m≤ L−1, we have∫Rn

θ(m)k (x,y1,y2)yα

1 dy1 =∫Rn

θ(m)k (x,y1,y2)yα

2 dy2 = 0.

(2) dµ(x, t) is a Carleson measure, where dµ is defined by


L−1

∑m=0

∑|α|,|β|≤L−1

|[[Θ(m)k ]]α,β(x)|2δt=2−k dx.

Proof. We will show that {Θ(m)k } ∈ BLPSO(Nm,L) by induction. First we check that {Θ(0)

k } ∈BLPSO(N0,L) = BLPSO(N − n,L). Using the definition in (4.2), it is sufficient to show thatΘk( f1, f2), [[Θk(·, f2)]]0 ·Pk f1, [[Θk( f1, ·)]]0 ·Pk f2, and [[Θk]]0,0 ·Pk f1 ·Pk f2 each define operatorsof type BLPSO(N0,L). The first and last terms trivially satisfy these properties; note that {Θk} ∈BLPSO(N,L)⊂BLPSO(N0,L) by assumption and [[Θk]]0,0 ·Pk f1 ·Pk f2 is in any BPLSO(N,L) classsince [[Θk]]0,0. 1 and the convolution kernel of Pk is in Schwartz class S . We consider the secondterm [[Θk(·, f2)]]0 ·Pk f1, whose kernel is

ϕk(x− y1)∫Rn

θk(x,u1,y2)du.


It is not hard to verify that this kernel satisfies the appropriate estimates to be a member ofBLPSO(N0,L). The kernel condition (2.1) is verified by the following. For |α|, |β| ≤ L∣∣∣∣Dα

1 Dβ

2

(ϕk(x− y1)

∫Rn

θk(x,u1,y2)du1

)∣∣∣∣= ∣∣∣∣∂αy1

ϕk(x− y1)∫Rn

∂βy2

θk(x,u1,y2)du1

∣∣∣∣. 2|α|kΦ

N−nk (x− y1)

∫Rn

2|β|kΦNk (x−u1,x− y2)du1

. 2(|α|+|β|)kΦN−nk (x− y1)Φ

N−nk (x− y2)

≤ 2(|α|+|β|)kΦN0k (x− y1,x− y2).

Here we use that

ΦRk (x)Φ

Rk (y) =

2kn

(1+2k|x|)R2kn

(1+2k|y|)R ≤22kn

(1+2k|x|+2k|y|)R = ΦRk (x,y).

By symmetry [[Θk( f1, ·)]]0 ·Pk f2 defines an operator of type BLPSO(N0,L) as well. Now we pro-ceed by induction. Assume that {Θ(m−1)

k } ∈ BLPSO(Nm−1,L). Similar to the reduction for the m=

0 case, this can be easily reduced to showing that [[Θ(m−1)k (·, f2)]]α ·2−k|α|PkDα f1 for |α|= m and

the symmetric term define a collections of type BLPSO(Nm,L). The kernel of [[Θ(m−1)k (·, f2)]]α ·

2−k|α|PkDα f1 is

(−1)|α|(Dαϕ)k(x− y1)

∫Rn

θ(m−1)k (x,u,y2)2k|α|(x−u)αdu.

The kernel condition (2.1) for [[Θ(m−1)k (·, f2)]]α · 2−k|α|PkDα f1 is verified as follows using the in-

ductive hypothesis. For |µ|, |ν| ≤ L and |α|= m∣∣∣∣Dµ1Dν

2

((−1)|α|(Dα

ϕ)k(x− y1)∫Rn

θ(m−1)k (x,u,y2)2k|α|(u− y1)

αdu)∣∣∣∣

. 2|µ|k|(Dα+µϕ)k(x− y1)|

∫Rn|Dν

2θ(m−1)k (x,u,y2)2k|α|(u− y1)

α|du

. 2|µ|kΦNmk (x− y1)

∫Rn

2|ν|kΦNm−1k (x−u,x− y2)2k|α|(u− y1)

α|du

≤ 2(|µ|+|ν|)kΦNmk (x− y1)

∫Rn

ΦNm−1−|α|k (x−u,x− y2)du

≤ 2(|µ|+|ν|)kΦNmk (x− y1)Φ

Nm−1−n−mk (x− y2)

≤ 2(|µ|+|ν|)kΦNmk (x− y1,x− y2).

This gives the appropriate formula Nm, defined recursively by N0 = N−n and Nm = Nm−1−n−mfor m≥ 1. This yields the formula Nm = N−(m+1)(2n+m)/2. Since Nm ≤ Nm−1 for all 1≤m≤L− 1, this gives the requirement NL−1 = N−L(2n+L− 1)/2 as in the statement of Proposition4.3.

Since {Θk} ∈ BLPSO(N,L) for some N ≥ 2n + 2L, we know that [[Θk]]α,β(x) exists for all

|α|, |β| ≤ L− 1. We prove (1) by induction and using the reduction in Lemma 4.2 for each Θ(m)k :


the m = 0 case for (1) is not hard to verify; for all f1 ∈C∞0

[[Θ(0)k ( f1, ·)]]0(x) = [[Θk( f1, ·)]]0(x)− [[Θk( f1, ·)]]0(x)

− [[Θk]]0,0(x)Pk f1(x)+ [[Θk]]0,0(x)Pk f1(x) = 0,

and likewise [[Θ(0)k (·, f2)]]0 = 0 for all f2 ∈ C∞

0 . Now assume that (1) holds for m− 1. Then for|µ| ≤ m−1 and f1 ∈C∞

0

[[Θ(m)k ( f1, ·)]]µ = [[Θ

(m−1)k ( f1, ·)]]µ− ∑

|α|=m(−1)|α|

[[Θ(m−1)k ]]α,µ

α!·2−k|α|PkDα f1

− ∑|β|=m

(−1)|β|[[Θ

(m−1)k ( f1, ·)]]β

β!·2−k|β|[[PkDβ]]µ

+ ∑|α|=|β|=m

(−1)|α|+|β|[[Θ

(m−1)k ]]α,βα!β!

·2−k(|α|+|β|)PkDα f1[[PkDβ]]µ = 0

by the inductive hypothesis and the fact that [[PkDα]]µ = [[PkDβ]]µ = 0 for |µ| ≤m−1 < m = |α|=|β|. For |µ|= m,

[[Θ(m)k ( f1, ·)]]µ = [[Θ

(m−1)k ( f1, ·)]]µ− ∑

|α|=m(−1)|α|

[[Θ(m−1)k ]]α,µ


− ∑|β|=m

(−1)|β|[[Θ

(m−1)k ( f1, ·)]]β

β!·2−k|β|[[PkDβ f2]]µ

+ ∑|α|=|β|=m

(−1)|α|+|β|[[Θ

(m−1)k ]]α,βα!β!

·2−k(|α|+|β|)PkDα f1[[PkDβ f2]]µ

= [[Θ(m−1)k ( f1, ·)]]µ− ∑

|α|=m(−1)|α|

[[Θ(m−1)k ]]α,µ


− [[Θ(m−1)k ( f1, ·)]]µ + ∑

|α|=m(−1)|α|

[[Θ(m−1)k ]]α,µ

α!·2−k|α|PkDα f1 = 0,

where the summations in β collapse using (4.1) and that Mβ,µ = (−1)|β| when µ = β and Mβ,µ = 0when |β| = |µ| but β 6= µ. By symmetry the same holds for [[Θk(·, f2)]]µ and hence by inductionand Lemma 4.2 this verifies (1) for all m ≤ L− 1. Given the Carleson measure assumption fordµ(x, t) in (4.5), one can easily prove (2) if the following statement holds: for all 0≤ m≤ L−1

∑|α|,|β|≤L−1

|[[Θ(m)k ]]α,β(x)| ≤C2(m+1)

0 ∑|α|,|β|≤L−1

|[[Θk]]α,β(x)|,(4.6)

where C0 = ∑|α|,|β|≤L−1

(1+ |Mα,β|).


We verify (4.6) by induction. For m = 0, let |α|, |β| ≤ L−1, and it follows that

[[Θ(0)k ]]α,β = [[Θk]]α,β− [[Θk]]0,β[[Pk]]α− [[Θk]]α,0[[Pk]]β +[[Θk]]0,0[[Pk]]α[[Pk]]β

Then

∑|α|,|β|≤L−1

|[[Θ(0)k ]]α,β| ≤ ∑

|α|,|β|≤L−1|[[Θk]]α,β|+ |[[Θk]]0,βM0,α|

+ ∑|α|,|β|≤L−1

|[[Θk]]α,0M0,β|+ |[[Θk]]0,0M0,αM0,β|

≤

(∑

|α|,|β|≤L−1(1+ |M0,α|+ |M0,β|+ |M0,αM0,β|)

)(∑

|µ|,|ν|≤L−1|[[Θk]]α,β|

)

≤

(∑

|α|,|β|≤L−1(1+ |M0,α|)(1+ |M0,β|)

)(∑

|µ|,|ν|≤L−1|[[Θk]]α,β|

)≤C2

0 ∑|µ|,|ν|≤L−1

|[[Θk]]α,β|.

Now assume that (4.6) holds for m−1, and consider

∑|α|,|β|≤L−1

|[[Θ(m)k ]]α,β| ≤ ∑

|α|,|β|≤L−1|[[Θ(m−1)

k ]]α,β|+ ∑|α|,|β|≤L−1

∑|µ|=m

|[[Θ(m−1)k ]]0,βMµ,α|

+ ∑|α|,|β|≤L−1

∑|ν|=m

|[[Θ(m−1)k ]]α,0Mν,β|+ ∑

|α|,|β|≤L−1∑

|µ|=|ν|=m|[[Θ(m−1)

k ]]0,0Mµ,αMν,β|

≤

(∑

|α|,|β|≤L−1

[1+ ∑

|µ|=m|Mµ,α|+ ∑

|ν|=m|Mν,β|+ ∑

|µ|=|ν|=m|Mµ,αMν,β|

])

×

(∑

|α|,|β|≤L−1|[[Θ(m−1)

k ]]α,β|

)

≤

(∑

|α|,|β|≤L−1∑

|µ|,|ν|≤m(1+ |Mµ,α|)(1+ |Mν,β|)

)(∑

|α|,|β|≤L−1|[[Θ(m−1)

k ]]α,β|

)≤C2

0 ∑|α|,|β|≤L−1

|[[Θ(m−1)k ]]α,β| ≤C2(m+1)

0 ∑|α|,|β|≤L−1

|[[Θk]]α,β|

We use the inductive hypothesis in the last inequality here to bound the [[Θ(m−1)]]α,β. Then byinduction, the estimate in (4.6) holds for all 0≤ m≤ L−1, and this completes the proof. �

Define the linear operators

Λ1,0,αk f (x) = [[Θk( f , ·)]]α(x), Λ

2,0,αk f (x) = [[Θk(·, f )]]α(x),(4.7)

Λ1,m,αk f (x) = [[Θ

(m−1)k ( f , ·)]]α(x), and Λ

2,m,αk f (x) = [[Θ

(m−1)k (·, f )]]α(x)(4.8)

for 1≤ m≤ L−1.


Lemma 4.4. Let L ≥ 1 be an integer, and assume that {Θk} ∈ BLPSO(N,L), where N = 2n+L(2n+L+ 3)/2. Define Θ

(m)k and Λ

i,m,αk for i = 1,2, 0 ≤ m ≤ L− 1, and |α| ≤ L− 1 as in (4.2),

(4.3), (4.7), and (4.8). Then {Λi,m,αk } ∈ LPSO(n+ 2L,L) for each i = 1,2, 0 ≤ m ≤ L− 1, and

|α| ≤ L−1. Furthermore, if Θk satisfies the Carleson condition in (4.5), then

dν(x, t) = ∑k∈Z

2

∑i=1

L−1

∑m=0

∑|α|,|β|≤L−1

|[[Λi,m,αk ]]β(x)|2δt=2−k dx(4.9)

is a Carleson measure and SΛi,m,α is bounded from H p into Lp for all nn+L < p ≤ 1, i = 1,2, m =

0,1, ...,L−1, and |α| ≤ L, where SΛi,m,α is the square function operator associated to {Λi,m,α}

SΛi,m,α f (x) =

(∑k∈Z|Λi,m,α

k f (x)|2) 1

2

.

Proof. We first look at Λ1,0,αk f = [[Θk( f , ·)]]α for i = 1,2 and |α| ≤ L− 1. We wish to show that

{Λ1,0,αk } ∈ LPSO(n+2L,L). The kernel of Λ

1,0,αk is bounded in the following way. For |β| ≤ L

|Dβ

1λ1,0,αk (x,y)| ≤

∫Rn|Dβ

1θk(x,y,y2)2|α|k(x− y2)α|dy2

. 2|β|k∫Rn

ΦNk (x− y,x− y2)(2k|x− y2|)|α|dy2

. 2|β|k∫Rn

ΦN−|α|k (x− y,x− y2)dy2

. 2|β|kΦN−n−(L−1)k (x− y)≤ 2|β|kΦ

n+2Lk (x− y).

Here we use that our hypotheses on N imply that N ≥ 2n+ 3L− 1. Then {Λ1,0,αk } ∈ LPSO(n+

2L,L) for all |α| ≤ L− 1, and by symmetry it follows that {Λ2,0,αk } ∈ LPSO(n+ 2L,L) for |α| ≤

L−1 as well. If L = 1, this completes the estimates.Now we continue to estimate these operators for 1≤m≤ L−1, where L≥ 2. Fix 0≤m≤ L−1

and |α| ≤ L−1. The kernel λ1,m,αk f (x) is given by

λ1,m,αk (x,y) =

∫Rn

θ(m−1)k (x,y,y2)(x− y2)

αdy2.

By Proposition 4.3, we know that {Θ(m−1)k } ∈ BLPSO(Nm−1,L) where Nm−1 = N−m(2n+m−

1)/2, and we now show that {Λ1,m,αk } ∈ LPSO(n+ 2L,L). We check the kernel conditions for

λ1,m,αk hold. For 1≤ m≤ L−1, |α| ≤ L−1, and |β| ≤ L

|Dβ

1λ1,m,αk (x,y)| ≤

∫Rn|Dβ

1θ(m−1)k (x,y,y2)2|α|k(x− y2)

α|dy2

≤ 2|β|k∫Rn

ΦNm−1k (x− y,x− y2)2|α|k|(x− y2)

α|dy2

≤ 2|β|k∫Rn

ΦNm−1−|α|k (x− y,x− y2)dy2

≤ 2|β|kΦNm−1−n−(L−1)k (x− y)≤ 2|β|kΦ

n+2Lk (x− y).


The last inequality is given by our assumption that N ≥ 2n+L(2n+L+ 3)/2 since for 0 ≤ m ≤L−1, we have

Nm−1−n− (L−1)≥ NL−2−n− (L−1) = NL−1 = N−L(2n+L−1)/2

≥ 2n+L(2n+L+3)/2−L(2n+L−1)/2 = 2n+2L≥ n+2L.

Also, by (2) in Proposition 4.3 it follows that dν(x, t) defined in (4.9) is a Carleson measure; thisis because

[[Λ1,0,αk ]]β = [[Θk]]β,α, [[Λ2,0,α

k ]]β = [[Θk]]α,β,

[[Λ1,m,αk ]]β = [[Θ

(m−1)k ]]β,α, and [[Λ2,m,α

k ]]β = [[Θ(m−1)k ]]α,β for 0 < m≤ L−1.

By Theorem 2.1 from [35], it follows that SΛi,m,α is bounded from H p into Lp for all nn+L < p ≤

1. �

Now we use Lemma 4.1 and the paraproduct operators Θ(m)k along with Propositions 3.3 and 4.3

to prove Theorem 2.1.

Proof of Theorem 2.1. Let n2n+L < p ≤ 1 and n

n+L < p1, p2 ≤ 1 such that 1p = 1

p1+ 1

p2. Using the

definitions in (4.2), (4.3), (4.7), and (4.8), we can rewrite Θ(m)k in terms of Λi,m,α in the following

way;

Θ(0)k ( f1, f2)(x) = Θk( f1, f2)(x)−Λ

2,0,0k f2(x) ·Pk f1(x)−Λ

1,0,0k f1(x) ·Pk f2(x)

+ [[Θk]]0,0(x) ·Pk f1(x)Pk f2(x)

= Θ(m−1)k ( f1, f2)(x)− ∑

|α|=m

(−1)|α|

α!Λ

2,m,αk f2(x) ·2−k|α|PkDα f1(x)

− ∑|β|=m

(−1)|β|

β!Λ

1,m,βk f1(x) ·2−k|β|PkDβ f2(x)

+ ∑|α|=|β|=m

(−1)|α|+|β|

α!β![[Θ

(m−1)k ]]α,β(x) ·2−k(|α|+|β|)PkDα f1(x)PkDβ f2(x).

Then Θk is bounded in the following way;

|Θk( f1, f2)| ≤ |Θ(0)k ( f1, f2)|+ |Λ1,0,0

k f1Pk f2|+ |Λ2,0,0k f2Pk f1|+ |[[Θk]]0,0Pk f1Pk f2|

≤ |Θ(1)k ( f1, f2)|+ |Λ1,0,0

k f1Pk f2|+ |Pk f1Λ2,0,0k f2|+ |[[Θk]]0,0Pk f1Pk f2|

+ ∑|α|=1|2−k|α|PkDα f1Λ

2,1,αk f2|+ ∑

|β|=1|Λ1,1,β

k f12−k|β|PkDβ f2|

+ ∑|α|=|β|=1

|[[Θ(0)k ]]α,β2−k|α|PkDα f12−k|β|PkDβ f2|


≤ |Θ(L−1)k ( f1, f2)|+

L−1

∑m=0

∑|α|=m

|2−k|α|PkDα f1Λ2,m,αk f2|(4.10)

+L−1

∑m=0

∑|β|=m

|Λ1,m,βk f12−k|β|PkDβ f2|

+L−1

∑m=0

∑|α|=|β|=m

|[[Θ(m)k ]]α,β2−k|α|PkDα f12−k|β|PkDβ f2|+ |[[Θk]]0,0Pk f1Pk f2|.

We verify that the square functions associated to each of the terms on the right hand side of (4.10)is bounded by a constant times || f1||H p1 || f2||H p2 .

The Θ(L−1)k ( f1, f2) term: First we note that since {Θk} ∈ BLPSO(N,L) for N = 2n+L(2n+L+

5)/2, by Proposition 4.3 it follows that {Θ(L−1)k } ∈ BLPSO(NL−1,L) where

NL−1 = N−L(2n+L−1)/2 = 2n+L(2n+L+5)/2−L(2n+L−1)/2 = 2n+3L.

This inequality for NL−1 is the binding restriction for the requirement N = 2n+L(2n+L+5)/2.The other terms require N to be large, but none as large as the requirement for {Θ(L−1)

k } to be amember of BLPSO(2n+3L,L). Also by property (1) in Proposition 4.3, we have that∫

Rnθ(L−1)k (x,y1,y2)yα

1 dy1 =∫Rn

θ(L−1)k (x,y1,y2)yα

2 dy2 = 0

for all |α| ≤ L−1. By Lemma 4.1, it follows that

||SΘ(L−1)( f1, f2)||Lp =

∣∣∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|Θ(L−1)

k ( f1, f2)|2) 1

2∣∣∣∣∣∣∣∣∣∣∣∣Lp

. || f1||H p1 || f2||H p2

since n2n+L < p≤ 1 and n

n+L < p1, p2 ≤ 1.

The 2−k|α|PkDα f1Λ2,m,αk f2 terms: By Lemma 4.4, we know that SΛi,m,α is bounded from Hq into Lq

for any nn+L < q ≤ 1 and each 0 ≤ m ≤ L− 1, i = 1,2, and |α| ≤ L− 1. It is also not hard to see

that 2−k|α|PkDα f (x) = (−1)|α|(Dαϕ)k ∗ f (x), where Dαϕ ∈S since ϕ ∈S . Therefore∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|2−k|α|PkDα f1Λ

2,m,αk f2|2

)∣∣∣∣∣∣∣∣∣∣Lp

≤∣∣∣∣∣∣∣∣sup

k∈Z|(Dα

ϕ)k ∗ f1|∣∣∣∣∣∣∣∣

Lp1

∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|Λ2,m,α

k f1|2)∣∣∣∣∣∣∣∣∣∣Lp1

. || f1||H p1 || f2||H p2 .

Note that here we use that nn+L < p1 ≤ 1. Then the Λ

1,m,βk f12−k|β|PkDβ f2 terms are bounded ap-

propriately by a similar argument and the assumption nn+L < p2 ≤ 1.


The [[Θ(m)k ]]α,β2−k|α|PkDα f12−k|β|PkDβ f2 terms: For these terms we use the Carleson measure prop-

erty (2) from Proposition 4.3. Also using Proposition 3.3, it follows that∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|[[Θk]]α,β2−k|α|PkDα f12−k|β|PkDβ f2|2

)∣∣∣∣∣∣∣∣∣∣Lp

≤

∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|[[Θ(m)

k ]]α,β(Dα

ϕ)k ∗ f1|2)∣∣∣∣∣∣∣∣∣∣Lp1

∣∣∣∣∣∣∣∣supk∈Z|(Dβ

ϕ)k ∗ f2|∣∣∣∣∣∣∣∣

Lp2

. || f1||H p1 || f2||H p2

for 0≤m≤ L−1. In fact, this estimate holds for p, p1, p2 that range all the way down to zero, butthe bounds for the other terms require the assumed lower bounds for p, p1, p2.

The [[Θk]]0,0Pk f1Pk f2 term: This term is bounded in the same way as the ones from the previouscase. By Proposition 3.3 and the assumed Carleson measure properties for [[Θk]]0,0 it follows that∣∣∣∣∣

∣∣∣∣∣(

∑k∈Z|[[Θk]]0,0Pk f1Pk f2|2

)∣∣∣∣∣∣∣∣∣∣Lp

≤

∣∣∣∣∣∣∣∣∣∣(

∑k∈Z|[[Θk]]0,0Pk f1|2

)∣∣∣∣∣∣∣∣∣∣Lp1

∣∣∣∣∣∣∣∣supk∈Z|Pk f2|

∣∣∣∣∣∣∣∣Lp2

. || f1||H p1 || f2||H p2 .

Therefore the H p norm of the square function associated to the right hand side of (4.10) is boundedby a constant times || f1||H p1 || f2||H p2 as desired, and hence SΘ is bounded from H p1 ×H p2 intoLp. �

5. HARDY SPACE BOUNDS FOR SINGULAR INTEGRAL OPERATORS

As described above, we will apply Theorem 2.1 to prove Theorem 2.2. In order to do so, weprove the decomposition result in Theorem 2.3. First we prove a quick lemma that show that if(2.3) holds for ψ ∈D2L+n as stated, then we can replace ψ ∈D2L+n by ϕ ∈C∞

0 in (2.3).

Lemma 5.1. Assume that T ∈ BCZO(M) for an integer M = L(2n+ L+ 5)/2 for some integerL≥ 1, and that T is bounded from L2×L2 into L1. If T ∗1(xα,ψ) = 0 in D ′|α| for all ψ ∈D|α|, thenT ∗1(xα,ϕ) = 0 for all ϕ ∈C∞

0 . Likewise for T ∗2(ψ,xα) and T ∗2(ϕ,xα).

Proof. Let ϕ ∈C∞0 and ψ ∈D2L+n such that

f = ∑k∈Z

ψk ∗ f

in L2 for f ∈ L2. Then for |α| ≤ 2L+n and φ ∈D|α|, we have⟨T ∗1(xα,ϕ),φ

⟩= lim

R→∞

∫Rn

T (φ,ϕ)(x)xαηR(x)dx

= limR→∞

∑k∈Z

∫Rn

T (φ,ψk ∗ϕ)(x)xαηR(x)dx

= ∑k∈Z

limR→∞

∫Rn

T (φ,ψk ∗ϕ)(x)xαηR(x)dx = 0.

This holds if the series above is absolutely convergent. Using the L2×L2 to L1 bound and kernelrepresentation for T , it is not hard to show that there exists a constant Cϕ,T < ∞ (independent of


k ∈ Z and R for R sufficiently large) such that∣∣∣∣∫RnT (φ,ψk ∗ϕ)(x)xα

ηR(x)dx∣∣∣∣≤Cφ,T ||ψk ∗ϕ||L2.

Since ψ ∈ D2L+n ⊂ Dn, it follows that ||ψk ∗ ϕ||L2 . min(2kn/2,2−kn/2). This estimate is easywhen k ≤ 0 since ||ψk ∗ϕ||L2 ≤ ||ψk||L2||ϕ||L1 . 2kn/2. When k > 0, we use ψ ∈ Dn to concludethat |ψk ∗ϕ(x)|. 2−nk (Φn+1

k (x)+Φn+10 (x)

). Hence for k > 0, we have

||ψk ∗ϕ||L2 . 2−nk (||Φn+1k ||L2 + ||Φn+1

0 ||L2). 2−nk

(2nk/2 +1

). 2−nk/2.

Then the above sum is absolutely convergent, and the proof of the lemma is complete. �

Proof of Theorem 2.3. Assume that T and ψ are as in the statement of Theorem 2.3. Withoutloss of generality we assume that supp(ψ) ⊂ B(0,1). To compute the kernel for Θk f = QkT , weconsider the following decomposition: Let ϕ ∈C∞

0 (B(0,1)) with integral 1 such that∫Rn

ϕ(x)xαdx = 0

for 0 < |α| ≤M. Define ψk(x) = ϕk+1(x)−ϕk(x), and it follows that ψ ∈DM. Then

Θk( f1, f2)(x) = limR→∞

R−1

∑`1=k

⟨T (Q`1 f1,PR f2),ψ

xk

⟩+ 〈T (Pk f1,PR f2),ψ

xk〉

= limR→∞

R−1

∑`1=k

R−1

∑`2=k

⟨T (Q`1 f1, Q`2 f2),ψ

xk

⟩+

R−1

∑`1=k

⟨T (Q`1 f1,Pk f2),ψ

xk

⟩+

R−1

∑`2=k

⟨T (Pk f1, Q`2 f2),ψ

xk

⟩+ 〈T (Pk f1,Pk f2),ψ

xk〉

= limR→∞

∫R2n

(R−1

∑`1=k

R−1

∑`2=k

I`1,`2,k(x,y1,y2)+R−1

∑`1=k

II`1,k(x,y1,y2)(5.1)

+R−1

∑`2=k

III`2,k(x,y1,y2)+ IVk(x,y1,y2)

)f1(y1) f2(y2)dy1 dy2

where

I`1,`2,k(x,y1,y2) =⟨

T (ψy1`1, ψ

y2`2),ψx

k

⟩, II`2,k(x,y1,y2) =

⟨T (ϕy1

k , ψy2`2),ψx

k

⟩,

III`1,k(x,y1,y2) =⟨

T (ψy1`1,ϕ

y2k ),ψx

k

⟩, and IVk(x,y1,y2) =

⟨T (ϕy1

k ,ϕy2k ),ψx

k⟩.


To show that Θk satisfies (2.1), we first assume that |x− y1|+ |x− y2| > 22−k. Without loss ofgenerality, we assume that `1 ≥ `2. For |α|, |β| ≤ L we have∣∣∣Dα

1 Dβ

2I`1,`2,k(x,y1,y2)∣∣∣= 2|α|`1+|β|`2

∣∣∣⟨T ((Dα1 ψ)

y1`1,(Dβ

2ψ)y2`2),ψx

k

⟩∣∣∣= 2|α|`1+|β|`2

∣∣∣∣∣∫R3n

(K(u,v1,v2)− ∑

|µ|≤M−1

Dµ1K(u,y1,v2)

µ!(v1− y1)

µ

)

× (Dαψ)

y1`1(v1)(Dβ

ψ)y2`2(v2)ψ

xk(u)dudv1 dv2

∣∣∣∣. 2|α|`1+|β|`2

∫R3n

|v1− y1|M

(|u− v1|+ |u− v2|)2n+M |(Dα

ψ)y1`1(v1)(Dβ

ψ)y2`2(v2)ψ

xk(u)|dudv1 dv2

. 2−`1(M−|α|−|β|) 2(2n+M)k

(1+2k|x− y1|+2k|x− y2|)2n+M

. 2(k−max(`1,`2))(M−|α|−|β|)2(|α|+|β|)kΦ2n+Mk (x− y1,x− y2).

Then∞

∑`1=k

∞

∑`2=k

∣∣∣Dα1 Dβ

2I`1,`2,k(x,y1,y2)∣∣∣

. 2(|α|+|β|)kΦ2n+Mk (x− y1,x− y2)

∞

∑`1=k

∞

∑`2=k

2(k−`1)M−|α|−|β|

2 2(k−`1)M−|α|−|β|

2

. 2(|α|+|β|)kΦNk (x− y1,x− y2).

The second term satisfies a similar estimate∣∣∣Dα1 Dβ

2II`1,k(x,y1,y2)∣∣∣= 2|α|`12|β|k

∣∣∣∣∫R3nK(u,v1,v2)(Dα

ψ)y1`1(v1)(Dβ

ϕ)y2k (v2)ψ

xk(u)dudv1 dv2

∣∣∣∣= 2|α|`12|β|k

∣∣∣∣∣∫R3n

(K(u,v1,v2)− ∑

|µ|≤M−1

Dµ1K(u,y1,v2)

µ!(v1− y1)

µ

)

× (Dαψ)

y1`1(v1)(Dβ

ϕ)y2k (v2)ψ

xk(u)dudv1 dv2

∣∣∣∣. 2|α|`12|β|k

∫R3n

|v1− y1|M

(|u− v1|+ |u− v2|)2n+M |(Dα

ψ)y1`1(v1)(Dβ

ϕ)y2k (v2)ψ

xk(u)|dudv1 dv2

. 2(|α|+|β|)k2(k−`1)(M−|α|)Φ2n+Mk (x− y1,x− y2).

Again summing over `1 ≥ k, we obtain∞

∑`1=k

∣∣∣Dα1 Dβ

2II`1,k(x,y1,y2)∣∣∣. 2(|α|+|β|)kΦ

2n+Mk (x− y1,x− y2)

∞

∑`1=k

2(k−`1)(M−|α|)

. 2(|α|+|β|)kΦNk (x− y1,x− y2).


The third term III`2,k(x,y1,y2) is also bounded by a constant times 2(|α|+|β|)kΦNk (x− y1,x− y2) by

symmetry. We estimate the last term.

|Dα1 Dβ

2IVk(x,y1,y2)|= 2(|α|+|β|)k∣∣∣⟨T ((Dα

ϕ)y1k ,(Dβ

ϕ)y2k ),ψx

k

⟩∣∣∣= 2(|α|+|β|)k

∣∣∣∣∣∫R3n

(K(u,v1,v2)− ∑

|µ|≤M−1

Dµ0K(x,v1,v2)

µ!(u− x)µ

)

× (Dαϕ)

y1k (v1)(Dβ

ϕ)y2k (v2)ψ

xk(u)dudv1 dv2

∣∣∣∣. 2(|α|+|β|)k

∫R3n

|u− x|M

(|x− v1|+ |x− v2|)2n+M |(Dα

ϕ)y1k (v1)(Dβ

ϕ)y2k (v2)ψ

xk(u)|dudv1 dv2

. 2(|α|+|β|)k2−Mk

(2−k + |x− y1|+ |x− y2|)2n+M = 2(|α|+|β|)kΦNk (x− y1,x− y2).

So we that that θk(x,y1,y2) as given in (5.1) satisfies (2.1) for |x− y1|+ |x− y2| ≥ 22−k. Now weassume that |x− y1|+ |x− y2| ≤ 22−k. In this situation 22kn .ΦN

k (x− y1,x− y2), so it is sufficientto prove (2.1) with 22kn in place of ΦN

k (x− y1,x− y2). We again use the decomposition in (5.1).To bound I`1,`2,k, we assume without loss of generality that `1 ≥ `2. Since we have assumed that⟨

T (ψy1`1, ψ

y2`2),xµ

⟩= 0 for |µ| ≤ 2L+n, we can write∣∣∣Dα

1 Dβ

2

⟨T (ψy1

`1, ψ

y2`2),ψx

k

⟩∣∣∣≤ |A`1,`2,k(x,y1,y2)|+ |B`1,`2,k(x,y1,y2)|, where

A`1,`2,k(x,y1,y2) = 2`1|α|+`2|β|∫|u−y1|≤21−`1

T ((Dαψ)

y1`1,(Dβ

ψ)y2`2)(u)

×

(ψ

xk(u)− ∑

|α|≤2L+n

Dαψxk(y1)

α!(u− y1)

α

)du,

B`1,`2,k(x,y1,y2) = 2`1|α|+`2|β|∫|u−y1|>21−`1

T ((Dαψ)

y1`1,(Dβ

ψ)y2`2)(u)

×

(ψ

xk(u)− ∑

|α|≤2L+n

Dαψxk(y1)

α!(u− y1)

α

)du.

The A`1,`2,k term is bounded as follows,

|A`1,`2,k(x,y1,y2)| ≤ 2`1|α|+`2|β|||T ((Dαψ)

y1`1,(Dβ

ψ)y2`2) ·χB(y1,21−`1)||L1

×

∣∣∣∣∣∣∣∣∣∣(

ψxk− ∑|α|≤2L+n

Dαψxk(y1)

α!(·− y1)

α

)·χB(y1,21−`1)

∣∣∣∣∣∣∣∣∣∣L∞

. 2`1|α|+`2|β|2−`1n/2||T ((Dαψ)

y1`1,(Dβ

ψ)y2`2)||L22(n+2L+1)(k−`1)2kn

. 2`1|α|+`2|β|2−`1n/2||(Dαψ)

y1`1||L2 ||(Dβ

ψ)y2`2||L∞2(n+2L+1)(k−`1)2kn

. 2`1|α|+`2|β|2(n+2L+1)(k−`1)2kn2`2n . 2k(|α|+|β|)2k−max(`1,`2).


Let 0 < δ < 1. The B`1,`2,k term is bounded using the kernel representation of T , which is viablesince |u− y1|> 21−`1 implies u /∈ supp(ψy1

`1). Since ψ`1 ∈D2L, we have

|B`1,`2,k(x,y1,y2)|= 2`1|α|+`2|β|∣∣∣∣∫|u−y1|>21−`1

∫R2n

K(u,v1,v2)(Dαψ)

y1`1(v1)(Dβ

ψ)y2`2(v2)dv1 v2

×

(ψ

xk(u)− ∑

|α|≤2L+n

Dαψxk(y1)

α!(u− y1)

α

)du

= 2`1|α|+`2|β|

∣∣∣∣∣∫|u−y1|>21−`1

∫R2n

(K(u,v1,v2)− ∑

|µ|≤2L

K(u,y1,v2)

µ!(v1− y1)

µ

)× (Dα

ψ)y1`1(v1)(Dβ

ψ)y2`2(v2)dv1 v2

×

(ψ

xk(u)− ∑

|α|≤2L+n

Dαψxk(y1)

α!(u− y1)

α

)du

. 2`1|α|+`2|β|∞

∑m=1

∫2m−`1<|u−y1|≤2m+1−`1

∫R2n

|v1− y1|2L+1

(|u− y1|+ |u− v2|)2n+2L+1

×|(Dαψ)

y1`1(v1)(Dβ

ψ)y2`2(v2)|

∣∣∣∣∣ψxk(u)− ∑

|α|≤2L+n

Dαψxk(y1)

α!(u− y1)

α

∣∣∣∣∣dudv1 dv2

. 2`1|α|+`2|β|∞

∑m=1

2n(m−`1)2−(2L+1)`1

2(2n+2L+1)(m−`1)2kn2(n+2L+δ)(k+m−`1)

. 2`1|α|+`2|β|22L(k−`1)2δ(k−`1)22kn∞

∑m=1

2−(1−δ)m

. 2k(|α|+|β|)2δ(k−`1)22kn

Recall that by symmetry, we assume that `1 ≥ `2, so 2δ(k−`1) = 2δ(k−max(`1,`2)). It follows thatR−1

∑`1=k+1

R−1

∑`1=k+1

∣∣∣Dα1 Dβ

2I`1,`2,k(x,y1,y2)∣∣∣. 2k(|α|+|β|)22kn

∑`1,`2>k

2δ(k−max(`1,`2)) . 2k(|α|+|β|)22kn

for all |α|, |β| ≤ L. The estimate for II`1,`2,k is obtained by essentially the same argument. That is,

we estimate II`2,k in the following way using the hypothesis⟨

T (ψy1`1, ψ

y2`2),xµ

⟩= 0 for |µ| ≤ 2L+n,∣∣∣Dα

1 Dβ

2

⟨T (ϕy1

k , ψy2`2),ψx

k

⟩∣∣∣≤ |A`2,k(x,y1,y2)|+ |B`2,k(x,y1,y2)|,

where

A`2,k(x,y1,y2) = 2k|α|+`2|β|∫|u−y1|≤21−`2

T ((Dαϕ)

y1k ,(Dβ

ψ)y2`2)(u)

×

(ψ

xk(u)− ∑

|α|≤2L+n

Dαψxk(y1)

α!(u− y1)

α

)du


and

B`2,k(x,y1,y2) = 2k|α|+`2|β|∫|u−y1|>21−`2

T ((Dαϕ)

y1k ,(Dβ

ψ)y2`2)(u)

×

(ψ

xk(u)− ∑

|α|≤2L+n

Dαψxk(y1)

α!(u− y1)

α

)du.

By the argument above, we obtain the estimateR−1

∑`2=k+1

∣∣∣Dα1 Dβ

2II`2,k(x,y1,y2)∣∣∣. 2k(|α|+|β|)22kn

∑`2>k

2δ(k−`2) . 2k(|α|+|β|)22kn.

A similar argument can be applied to III`1,k as well. The estimate for IVk trivially follows from theL2×L2 to L1 boundedness of T ,

|Dα1 Dβ

2⟨T (ϕy1

k ,ϕy2k ),ψx

k⟩|= 2k(|α|+|β|)

∣∣∣⟨T ((Dα1 ϕ)

y1k ,(Dβ

2ϕ)y2k ),ψx

k

⟩∣∣∣. 2k(|α|+|β|)22nk.

Therefore all of the sums on the right hand side of (5.1) converge absolutely, and hence

θk(x,y1,y2) =∞

∑`1=k

∞

∑`2=k

I`1,`2,k(x,y1,y2)+∞

∑`2=k

II`2,k(x,y1,y2)+∞

∑`1=k

III`1,k(x,y1,y2)+ IVk(x,y1,y2)

satisfies (2.1). Hence {Θk}= {QkT} ∈ BLPSO(N,L). �

The proof of Theorem 2.2 follows immediately from Theorems 2.1 and 2.3.

6. APPLICATIONS TO PARAPRODUCT

6.1. Bilinear Bony Paraproducts. Let b ∈ BMO, and recall the bilinear Bony paraproduct

Πb( f1, f2)(x) = ∑j∈Z

Q j(Q jb ·Pj f1 ·Pj f2

)(x)

Now we prove Theorem 2.4

Proof. It is easily shown that Πb ∈ BCZO(M) for all M ∈ N. So it is sufficient to show conditions(2.3) and (2.4) from Theorem 2.2. We first show (2.3). For f1 ∈D2L+n, f2 ∈D|α|, and |α| ≤ 2L+n

〈Πb( f1, f2),xα〉= limR→∞

∑j∈Z

∫Rn

Q jb(u)Pj f1(u)Pj f2(u)Q j(ηRxα)(u)du

= ∑j∈Z

∫Rn

Q jb(u)Pj f1(u)Pj f2(u)Q j(xα)(u)du = 0,


and likewise 〈Πb( f2, f1),xα〉 = 0. The BMO|α|+|β| conditions in (2.4) are more difficult to verify,but hold nonetheless. For |α|, |β| ≤ L−1, we first compute

2k(|α|+|β|)Qk[[Πb]]α,β(x) = 2k(|α|+|β|) ⟨[[Πb]]α,β,ψxk⟩

= limR→∞

2k(|α|+|β|)∑j∈Z

∫R4n

ψ j(u− v)Q jb(v)ϕ j(v− y1)ϕ j(v− y2)ηR(y1)ηR(y2)

× (u− y1)α(u− y2)

βψ

xk(u)dudvdy1 dy2

= limR→∞

∑µ≤α

∑ν≤β

cα,µcβ,ν2k(|α|+|β|)∑j∈Z

∫R4n

ψ j(u− v)Q jb(v)ϕ j(v− y1)ϕ j(v− y2)ηR(y1)ηR(y2)

× (u− v)µ+ν(v− y1)α−µ(v− y2)

β−νψ

xk(u)dudvdy1 dy2

= ∑µ≤α

∑ν≤β

cα,µcβ,νCα−µCβ−ν2(k− j)(|α|+|β|)∑j∈Z

∫R2n

ψ(µ+ν)j (u− v)Q jb(v)ψx

k(u)dudv

= ∑µ≤α

∑ν≤β

cα,µcβ,νCα−µCβ−ν2(k− j)(|α|+|β|)∑j∈Z

QkQ(µ+ν)j Q jb(x)

= ∑µ≤α

∑ν≤β

cα,µcβ,νCα−µCβ−νWk ∗ (TV (µ+ν)b)(x),

where cα,µ are binomial coefficients, Cµ =∫Rn ϕ(x)xµdx, and V , W , and TV are defined via V (µ+ν)(ξ)=

|ξ||α|+|β|ψ(µ+ν)(ξ)ψ(ξ), W (ξ)= |ξ|−(|α|+|β|)ψ(ξ), V (µ+ν)j (x)= 2 jnV (µ+ν)(2 jx), Wj(x)= 2 jnW (2 jx),

and

TV (µ+ν) f (x) = ∑j∈Z

V (µ+ν)j ∗ f (x).

This is verified by the following computation.

2(k− j)(|α|+|β|)F[QkQ(µ+ν)

j Q j f](ξ) = 2(k− j)(|α|+|β|)

ψ(2−kξ)ψ(µ+ν)(2− j

ξ)ψ(2− jξ) f (ξ)

= (2−k|ξ|)−(|α|+|β|)(2− j|ξ|)|α|+|β|ψ(2−kξ)ψ(µ+ν)(2− j

ξ)ψ(2− jξ) f (ξ)

= W (2−kξ)V (µ+ν)(2− j

ξ) f (ξ)

= F[Wk ∗V (µ+ν)

j ∗ f](ξ).

It was shown in [35] that TV (µ+ν) is bounded on BMO. Then TV (µ+ν)b ∈ BMO with ||TV (µ+ν)b||BMO .||b||BMO, and

∑k∈Z

22k(|α|+|β|)|Qk[[Πb]]α,β(x)|2δt=2−k dx

= ∑k∈Z

∣∣∣∣∣∑µ≤α

∑ν≤β

cα,µcβ,νCα−µCβ−νWk ∗ (TV (µ+ν)b)(x)

∣∣∣∣∣2

δt=2−k dx

is a Carleson measure since TV (µ+ν)b ∈ BMO and Wk(x) = 2knW (2kx) has integral zero. �


6.2. Product Function Paraproducts. Now we prove Theorem 2.5. First recall the definition ofΠ( f1, f2).

Π( f1, f2) = ∑k∈Z

Πk( f1, f2), where Πk( f1, f2) = Qk (Pk f1 ·Pk f2) .

Proof of Theorem 2.5. It is easy to verify that Π ∈ BCZO(M) for all M ≥ 1. Then to show that Π

is bounded as described in Theorem 2.5, it is sufficient to show (2.3) and (2.4) from Theorem 2.2.The kernel of Πk is given by

πk(x,y1,y2) =∫Rn

ψk(x−u)ϕk(u− y1)ϕk(u− y2)du.

We first prove (2.3); for α ∈ Nn0, f1 ∈D|α|, and f2 ∈D2L+n, we have

〈Π( f1, f2),xα〉= limR→∞

∑k∈Z

∫R4n

ψk(x−u)ϕk(u− y1)ϕk(u− y2) f1(y1) f2(y2)ηR(x)xαdudxdy1 dy2

= limR→∞

∑k∈Z

∫R2n

ϕk(u− y1)ϕk(u− y2) f1(y1) f2(y2)

(∫Rn

ψk(x−u)ηR(x)xαdx)

dudy1 dy2 = 0

To prove (2.4), let α,β ∈ Nn0 and f ∈D|α|+|β|. It follows that⟨

[[Π]]α,β, f⟩= lim

R→∞∑k∈Z

∫R4n

ψk(x−u)ϕk(u− y1)ϕk(u− y2)ηR(y1)ηR(y2)

× (x− y1)α(x− y2)

β f (x)dudxdy1 dy2

= limR→∞

∑k∈Z

∑µ1+ν1=α

∑µ2+ν2=β

cµ1,ν1cµ2,ν2

∫R4n

ψk(x−u)ϕk(u− y1)ϕk(u− y2)ηR(y1)ηR(y2)

× (x−u)µ1(u− y1)ν1(x−u)µ2(u− y2)

ν2 f (x)dudxdy1 dy2

= limR→∞

∑k∈Z

∑µ1+ν1=α

∑µ2+ν2=β

cµ1,ν1cµ2,ν2

∫R2n

ψk(x−u)(x−u)µ1+µ2 f (x)

×(∫

Rnϕk(u− y1)(u− y1)

ν1ηR(y1)dy1

)×(∫

Rnϕk(u− y2)(u− y2)

ν2ηR(y2)dy2

)dudx

= ∑k∈Z

∑µ1+ν1=α

∑µ2+ν2=β

cµ1,ν1cµ2,ν2Cν1Cν2

∫Rn

(∫Rn

ψk(u)uµ1+µ2du)

f (x)dx = 0.

Here we define Cν for ν ∈ Nn0

Cν =∫Rn

ϕ(x)xνdx,

and cν,µ are binomial coefficients. Therefore Π is bounded from H p1×H p2 into H p for all n2n+L <

p≤ 1 and nn+L < p1, p2 ≤ 1, as in the statement of Theorem 2.5. �


6.3. Molecular Parparoducts. Finally, we address the paraproducts in Theorem 2.6. Recall theparaproduct operators

T ( f1, f2)(x) = ∑Q|Q|−1/2 ⟨ f1,φ

1Q⟩⟨

f2,φ2Q⟩

φ3Q(x),

where φ1Q,φ

2Q,φ

3Q are (M,N)-smooth molecules indexed by dyadic cubes.

Proof of Theorem 2.6. It is easy to see that T ∈ BCZO(M) and is bounded from L2×L2 into L1,which was proved in [2]. So we must verify conditions (2.3) and (2.4) of Theorem 2.2. As wasshown in [2], the kernel of T is

K(x,y1,y2) = ∑Q|Q|−1/2

φ1Q(y1)φ

2Q(y2)φ

3Q(x).

For f1 ∈D|α| and f2 ∈D2L+n with |α| ≤ L, we have

〈T ( f1, f2),xα〉= limR→∞

∑Q|Q|−1/2 ⟨ f1,φ

1Q⟩⟨

f2,φ2Q⟩⟨

φ3Q,x

αηR⟩

= ∑Q|Q|−1/2 ⟨ f1,φ

1Q⟩⟨

f2,φ2Q⟩∫

Rnφ

3Q(x)x

αdx = 0.

It is not hard to show that this series is absolutely summable independent of R when f1 ∈ D|α|and f2 ∈C∞

0 ; hence we can bring the limit in R inside the sum. A similar computation shows that〈T ( f1, f2),xα〉= 0 for f1 ∈D2L+n and f2 ∈D|α|. To show (2.4), we first verify that we can use analternate formula to compute [[T ]]α,β. For α,β ∈Nn

0 with |α|+ |β| ≤ L−1 and ψ ∈D|α|+|β|, fix R0large enough so that supp(ψ)⊂ B(0,R0/4). Then for S,R > R0, we have∫

R3nK (x,y1,y2)(x− y1)

α(x− y2)βηR(y1)ηS(y2)ψ(x)dy1 dy2 dx

=∫R3n

K(x,y1,y2)(x− y1)α(x− y2)

β(ηR(y1)ηS(y2)−ηR0(y1)ηR0(y1))ψ(x)dy1 dy2 dx

+∫R3n

K (x,y1,y2)(x− y1)α(x− y2)

βηR0(y1)ηR0(y1)ψ(x)dy1 dy2 dx,

where the integrand of the first integral is bounded by an L1(R3n) function independent of both Rand S. Then the limit as S,R→ ∞ exists (as a two dimensional limit), and

limR→∞

limS→∞

∫R3n

K (x,y1,y2)(x− y1)α(x− y2)

βηR(y1)ηS(y2)ψ(x)dy1 dy2 dx =

⟨[[T ]]α,β,ψ

⟩.

Therefore⟨[[T ]]α,β,ψ

⟩= lim

R→∞limS→∞

∑Q|Q|−1/2

∫R3n

φ1Q(y1)(x− y1)

αηS(y1)φ

2Q(y2)(x− y2)

βηR(y2)φ

3Q(x)ψ(x)dy1 dy2 dx

= limR→∞

∑Q|Q|−1/2

∫R2n

(∫Rn

φ1Q(y1)(x− y1)

αdy1

)φ

2Q(y2)(x− y2)

βηR(y2)φ

3Q(x)ψ(x)dy2 dx = 0.

Then we apply Theorem 2.2 to complete the proof. �


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DEPARTMENT OF MATHEMATICS, WAYNE STATE UNIVERSITY, DETROIT, MI 48202E-mail address: [email protected]

DEPARTMENT OF MATHEMATICS, WAYNE STATE UNIVERSITY, DETROIT, MI 48202E-mail address: [email protected]

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