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New York Journal of Mathematics

New York J. Math. 14(2008)145–192.

Uniform estimates for some paraproducts

Xiaochun Li

Abstract. We establishLp×LqtoLrestimates for some general para- products that arise in the study of the bilinear Hilbert transform along curves.

Contents

1. Introduction 145

2. Main results 147

3. A telescoping argument 149

4. Time frequency analysis 153

4.1. Definitions 154

4.2. Reduction 158

4.3. Principal lemmas 159

4.4. The size estimate for a tree 165

4.5. Proof of Lemma 4.11 174

5. Proof of Theorem 2.2 175

5.1. Proof of Lemma 5.1 176

5.2. A truncated trilinear form 181

5.3. Preliminary lemmata 183

5.4. Proof of Lemma 5.10 190

References 191

1. Introduction

It is an important theme of current research in analysis to decompose more complicated operators, such as the Cauchy integral on Lipschitz curves [1], as a sum of simpler operators. This theme has taken special prominence in

Received January 17, 2008.

Mathematics Subject Classification. Primary 42B20, 42B25. Secondary 46B70, 47B38.

Key words and phrases. paraproduct, uniform estimate.

Research was partially supported by NSF grant DMS-0456976.

ISSN 1076-9803/08

145

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multilinear Harmonic Analysis, beginning with the work of Lacey and Thiele [12], which expressed the bilinear Hilbert transforms as a sum of modulated paraproducts. This theme has found much broader application as well.

The bilinear Hilbert transforms have a bilinear symbol given by restriction to a half-plane, with slope that depends upon the transform in question.

In considering more complicated symbols, one is led to paraproducts which have a complicated underlying description. One then seeks certain estimates of these paraproducts that areuniformin the parametrizations. This line of investigation was started in [23], the results of which give a new, multilinear proof of the boundedness of the Calderon commutator, fulfilling a program of study of Calderon [1]. It was further extended in work of the author and Grafakos [8,9,14], in the study of the disc as a bilinear multiplier. Muscalu, Tao and Thiele [16,15,17] gave alternate proofs (and more general proofs) of these results in the multilinear operator setting.

In this paper, we continue this line of study, considering certain uniform estimates that are motivated by an analysis of a bilinear Hilbert transform along polynomial curves. Namely, consider the operators

(1.1) (f, g)−→p.v.

−∞f(x−y)g(x−p(y)) dy y ,

for some polynomial p(y). The study of these operators leads to subtle questions in multilinear analysis, stationary phase methods, and paraprod- ucts. An initial investigation into operators of this type is given in [6], where the polynomial is taken to be a square, and the singular kernel is mollified to ei|y|−β/|y| for someβ > 0. Without this modification, a signif- icant difficulty might be encountered. There is a natural analogue of the bilinear Hilbert transform along parabolas in the ergodic theory setting, that is, the nonconventional ergodic average N1 N−1

n=0 f(Tnx)g(Tn2x). In [7], Furstenberg proved that the characteristic factor of the trilinear ergodic averages N1 N−1

n=0 f(Tan)g(Tbn)h(Tcn) for alla, b, c∈Z is characteristic for the previous nonconventional ergodic average. We are indebted to M. Lacey for bringing these Furstenberg theorems to our attention. Thus a possible method for the bilinear Hilbert transform along a parabola is to understand the trilinear Hilbert transform first. Unfortunately, it turns out the trilinear Hilbert transform is very difficult to handle. It is very interesting to find a proof for the bilinear Hilbert transform along curves without using any information of the trilinear Hilbert transform. It might be possible to obtain such a way by combining time-frequency analysis and the known results for the trilinear oscillatory integrals. This investigation will appear in another paper.

The paraproducts that arise have a richer parametrization than what has been considered before. The question of uniform estimates is the main focus of this article. In the next section, a class of paraproducts are introduced.

They are parametrized by:

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the width of the frequency window associated to the paraproducts, denoted by L1 andL2 below,

the overlap of the frequency window associated to the paraproducts, denoted by M1 and M2 below,

a modulation of the frequency window, denoted by the (lower case) parameters n1, n2,2m below.

Prior results have concentrated on the uniformity of estimates with respect toM1, M2fromLp×LqtoLrforr≥1 andL1 =L2[16]. The principal point of this article is to get the estimates for 1/2 < r <1 and arbitrary L1, L2. Another new point of this article is the (weak) uniformity that we establish inL1, L2 and the modulation parameters 2m (see Theorem2.2below). This novelty is forced upon us by the stationary phase methods that one must use in the analysis of (1.1). One of anticipated applications of our theorems is the bilinear multiplier problems associated to the symbol defined by a characteristic function of a suitable domain with a smooth boundary.

Acknowledgements. The author would like to thank his wife, Helen, and his son, Justin, for being together through the hard times in the past two years. And he is also very thankful to Michael Lacey for his constant support and encouragement.

2. Main results

Letj∈Z,L1, L2 be positive integers andM1, M2 be integers.

ω1,j = [2L1j+M1/2,2·2L1j+M1] and

ω2,j= [2L2j+M2,2L2j+M2].

Let Φ1 be a Schwartz function whose Fourier transform is a standard bump function supported on [1/2,2], and Φ2 be a Schwartz function such that Φ2 is a standard bump function supported on [1,1] and Φ2(0) = 1. For ∈ {1,2} and n1, n2 Z, define Φ,j,n by

Φ,j,n(ξ) =

e2πin(·)Φ(·) ξ 2Lj+M

.

It is clear that Φ,j,n is supported on ω,j. For locally integrable functions f’s, we definef,j’s by

f,j,n(x) =fΦ,j,n(x).

We define a paraproduct to be

(2.1) ΠL1,L2,M1,M2,n1,n2(f1, f2)(x) =

j∈Z

2

=1

f,j,n(x).

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Another paraproduct we should introduce is the following. For ∈ {1,2}, let ω,j denote the set : 2Lj+M/2 ≤ |ξ| ≤ 2·2Lj+M}. Let m be a nonnegative integer and define Φ,j,m by

Φ,j,m(ξ) =

e2πi2m(·)Φ1(·) ξ 2Lj+M

. Letf,j,m be the function defined by

f,j,m(x) =fΦ,j,m(x).

We define a paraproduct to be

(2.2) ΠL1,L2,M1,M2,m(f1, f2)(x) =

j∈Z

2

=1

f,j,m(x).

One reason we study these paraproducts is that one will encounter such paraproducts in the study of the bilinear Hilbert transforms along polyno- mial curves. We have the following uniform estimates for these paraprod- ucts.

Theorem 2.1. For anyp1>1,p2>1with 1/p1+ 1/p2= 1/r, there exists a constant C independent of M1, M2, n1, n2 such that

(2.3) ΠL1,L2,M1,M2,n1,n2(f1, f2)

r ≤C

1 +|n1|10

1 +|n2|10

f1p1f2p2, for all f1∈Lp1 and f2∈Lp2.

Theorem 2.2. Let ΠL1,L2,M1,M2,m(f1, f2) be as in (2.2). Suppose that for allj,

(2.4) 2L2j+M2 2L1j+M1+m.

For anyε >0,p1>1,p2 >1with1/p1+1/p2 = 1/r, there exists a constant C independent of m, M1, M2, L1, L2 such that

(2.5) ΠL1,L2,M1,M2,m(f1, f2)

r≤C2εmf1p1f2p2, for all f1∈Lp1 and f2∈Lp2.

The case when L1 = L2 and r > 1 was proved in [16]. The constant C in Theorem 2.1 may depend on L1, L2. It is easy to see by the following argument that C is O(max{2L1,2L2}). It is possible to get a much bet- ter upper bound such as O

log(1 + max{L2/L1, L1/L2})

by tracking the constants carefully in the proof we will provide. But we do not pursue the sharp constant in this article. The independence ofM1, M2 is the most im- portant issue. In Sections 3,4, we give a proof for Theorem2.1. The proof of Theorem 2.2 will be given in Section 5. By using Theorem 2.1, we get the Lr bound for ΠL1,L2,M1,M2,m with a operator norm O(210m). Unfortu- nately sometimes this is not enough for our application. The desired norm isO(2εm) for a very small positive numberε. It might be possible to remove the condition (2.4) or get the uniform estimate for ΠL1,L2,M1,M2,min which

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the operator norm is independent ofm. The uniform estimate fromL2×L2 toL1is trivial and (2.4) is redundant for this case. In Section5, we see that the uniform estimates for ΠL1,L2,M1,M2,m can be achieved forp1, p2 >2 and 1 < r < 2 (see Proposition 5.7) and (2.4) is superfluous for Theorem 2.2 when p1, p2>2 and 1< r <2 (see Corollary5.3).

3. A telescoping argument

We now start to prove Theorem 2.1. To prove Theorem 2.1, we first introduce a definition of admissible trilinear form. And we should show that by a telescoping argument used in [8, 23], we can reduce the problem to estimates for an admissible trilinear form. And thusLr estimates forr >1 can be obtained by Littlewood–Paley theorem. The r < 1 case is more complicated. We have to use the time frequency analysis to deal with this case in Section 4.

Definition 3.1. An admissible trilinear form is a trilinear form (3.1) ΛL1,L2,M1,M2,n1,n2(f1, f2, f3) =

j∈Z

3

=1

f,j,n(x)dx,

where n3 = 0, f,j,n = fΦ,j,n and Φ,j,n is a function whose Fourier transform is supported onω,j such that:

(1) Eachω,j is an interval inRsuch that the distance from the origin to the interval is not more than 3,j|. And,j}j forms a sequence of lacunary intervals, that is, ,j|/|ω,j+1| ≤ 1/2 for all j Z. More- over, 3,j| ≥Cmax{|ω1,j|,|ω2,j|} for some constant C independent of M1, M2, n1, n2.

(2) There are at least two indices ∈ {1,2,3} such thatΦ,j,n satisfies

(3.2) Φ,j,n(0) = 0

(3.3)

Dα

Φ,j,n

,j|ξ≤ CN(1 +|n|)α (1 +|ξ|)N ,

for allξ Rand all nonnegative integersα, N. If an index in{1,2,3}

satisfies (3.2) and (3.3), we call the index a good index in the trilinear form ΛL1,L2,M1,M2,n1,n2. For the index which is not a good index, we call it a bad index in the trilinear form ΛL1,L2,M1,M2,n1,n2.

(3) If ∈ {2,3} is a bad index, then Φ,j,n satisfies (3.3). Moreover, among the other two good indices=, at least one of them satisfies

,j| ≤ Cmin{|ω1,j|,|ω2,j|,|ω3,j|} for some constant C independent of f1,f2,f3,M1,M2,n1,n2.

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(4) If 1 is a bad index, thenΦ1,j,n1 satisfies (3.4) Φ1,j,n1(x) =

m(j) k=0

Φ1,j+k,n1(x), where m(j) is some nonnegative integer.

Lemma 3.2. Let f3 be a locally integrable function. Then

ΠL1,L2,M1,M2,n1,n2(f1, f2)(x)f3(x)dx

is a sum of finitely many admissible trilinear forms such that the number of admissible trilinear forms in the sum is no more than a constant C inde- pendent of M1, M2, n1, n2.

Proof. For∈ {1,2}, writeω,j as [a,j, b,j]. Ifb2,j< b1,j/16, then|ω2,j|<

1,j|/6 and the distance fromω1,j2,jto the origin is not less than1,j|/4.

In this case, simply letω3,j be a small neighborhood of1,j2,j) and the Fourier transform ofΦ3,jis a suitable bump function adapted toω3,j, then we have the desired lemma. Thus we now only consider the caseb2,j≥b1,j/16.

Let ω3,j be [18b2,j,18b2,j]. And Φ3,j be a Schwartz function such that its Fourier transform is a bump function adapted toω3,j andΦ3,j(ξ) = 1 for all ξ∈[17b2,j,17b2,j]. Then

Π(f1, f2)(x)f3(x)dx=

j∈Z

3

=1

f,j,n(x)dx,

where f3,j,n3(x) =f3Φ3,j(x) and n3 = 0. Let Φ2 be a Schwartz function such that Φ2 is a bump function on [1,1] and Φ2(ξ) = 1 for all ξ [3/4,3/4]. And define Φ2,j by Φ2,j(ξ) = Φ2(ξ/b2,j). Let f2,j = f Φ2,j. We also denote f3,j,n3 by f3,j. We can replacef2,j,n2 by f2,j because

j∈Z

f,j,n1(x)

f2,j,n2−f2,j

(x)f3,j(x)dx

is an admissible trilinear form. Hence the only thing we need to show is that Λ(f1, f2, f3) =

j∈Z

f,j,n1(x)f2,j(x)f3,j(x)dx

is admissible. For any real number x, let [x] denote the largest integer not exceeding x. Let m(j) be the integer defined by

m(j) =

(L2j+M2)(L1j+M1) + 6 L2

.

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By b2,j b1,j/16, we see that m(j) 0. By a telescoping argument, Λ(f1, f2, f3) equals to

j∈Z

f1,j,n1(x)

m(j) k=0

f2,j−k(x)f3,j−k(x)−f2,j−k−1(x)f3,j−k−1(x)

dx,

since

f1,j,n1(x)f2,j−m(j)−1(x)f3,j−m(j)−1(x)dx= 0 due to the following sim- ple fact on the support of Fourier transform of each function in the integrand, i.e.,

suppf1,j,n1+ suppf2,j−m(j)−1

suppf3,j−m(j)−1

=∅. By a change of variables j→j+k, we have that Λ(f1, f2, f3) is equal to

j∈Z m(j)

k=0

f1,j+k,n1(x)

f2,j(x)f3,j(x)−f2,j−1(x)f3,j−1(x)

dx,

wherem(j) is the integer defined by m(j) =

(L2j+M2)(L1j+M1) + 6 L1

.

We write this integral as a sum of three parts Λ1,Λ2,Λ3, where Λ1 =

j∈Z

m(j) k=0

f1,j+k,n1(x)

f2,j(x)

f3,j(x)−f3,j−1(x) dx,

Λ2 =

j∈Z

m(j) k=0

f1,j+k,n1(x)

·

f2,j(x)−f2,j−1(x)

f3,j−1(x)−f3,j−8(x) dx, Λ3 =

j∈Z

m(j) k=0

f1,j+k,n1(x)

f2,j(x)−f2,j−1(x)

f3,j−8(x)dx.

It is clear that Λ2 is an admissible trilinear form. Write Λ1 as Λ11+ Λ12, where

Λ11=

j∈Z

m(j) k=0

f1,j+k,n1(x)

f2,j(x)−f2,j−1(x)

f3,j(x)−f3,j−1(x) dx,

Λ12=

j∈Z

m(j) k=0

f1,j+k,n1(x)

f2,j−1(x)

f3,j(x)−f3,j−1(x) dx.

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Clearly, Λ11 is an admissible trilinear form. Notice that supp

m(j)−10−[L2/L1] k=0

f1,j+k,n1

[0,2−22L2j+M2] = [0,2−22−L2b2,j], and

supp f3,j−f3,j−1

[−18b2,j,18b2,j]\[−16·2−L2b2,j,16·2−L2b2,j].

Thus Λ12 is equal to

j∈Z

m(j) k=m(j)−10−[L2/L1]

f1,j+k,n1(x)

·f2,j−1,n2(x)

f3,j,n3(x)−f3,j−1,n3(x) dx, which is obviously a finite sum of admissible trilinear forms. As for Λ3, observe that

supp

m(j)−100−[L2/L1] k=0

f1,j+k,n1

[0,2−802L2j+M2] = [0,2−802−L2b2,j], and

supp f2,j−f2,j−1

[−b2,j, b2,j]\[2−L2−1b2,j,2−L2−1b2,j].

Thus Λ3 is equal to

j∈Z

m(j) k=m(j)−100−[L2/L1]

f1,j+k,n1(x)

f2,j−f2,j−1(x)

f3,j−8(x)dx, which is a finite sum of admissible trilinear forms.

Lemma 3.3. Let ΛL1,L2,M1,M2,n1,n2 be an admissible trilinear form. Then for any real numbers p1, p2, p3 >1 with1/p1+ 1/p2+ 1/p3 = 1, there exists C independent of M1, M2, n1, n2 such that

(3.5)

ΛL1,L2,M1,M2,n1,n2(f1, f2, f3)≤C(1 +|n1|)10(1 +|n2|)10f1p1f2p2f3p3, for all f1∈Lp1, f2 ∈Lp2 and f3 ∈Lp3.

Proof. If there is no bad index in the trilinear form, take0to be any integer in{1,2,3}. Otherwise, let0be a bad index. Applying the Cauchy–Schwarz inequality, ΛL1,L2,M1,M2,n1,n2 is dominated by

sup

j∈Z f0,j,n

0 =0

j

f,j,n21/2 dx.

Using H¨older’s inequality, we dominate the trilinear form by sup

j∈Z f0,j,n

0 p1

=0

j

f,j,n21/2 p

.

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The Littlewood–Paley theorem yields that for =0

j

f,j,n21/2 p

≤C(1 +|n|)10fp. If0 ∈ {2,3}, then by (3.3), we have

supj∈Z f0,j,n

0(1 +|n0|)10M(f0),

which clearly yields the lemma. We now only need to consider the case 0= 1. It suffices to prove that

(3.6)

sup

j

m(j) k=0

f1Φ1,j+k,n1 p1

≤C(1 +|n1|)10f1p1.

Notice that ω1,j’s are essentailly disjoint intervals and Fourier transform of m(j)

k=0 Φ1,j+k,n1 is supported on a bounded interval depending on j. The left-hand side of (3.6) is less than

C

M

j

f1Φ1,j,n1 p1

. It is easy to verify that

jf1Φ1,j,n1 is a bounded operator onL2 associ- ated to a standard Calder´on–Zygmund kernel by paying at most a cost of (1 +|n1|)10 in the corresponding estimates. Thus by a standard Calder´on–

Zygmund argument, we have for any real numberp >1, there is a constant C independent of M1, M2, n1, n2 such that

j

f Φ1,j,n1

p≤C(1 +|n1|10)fp

holds for allf ∈Lp, which yields (3.6). Therefore we complete the proof of

the lemma.

Combining Lemma3.2 and Lemma 3.3, we obtain (2.3) forp1, p2, r >1.

To finish the proof of Theorem2.1, we need to provide a proof ofLrestimate with 1/2< r≤1 for (2.3), which will be given in Section 4.

4. Time frequency analysis

In this section we prove (2.3) with 1/2 < r 1 for the paraproducts by time frequency analysis, which was used for establishing Lp (uniform) estimates for the bilinear Hilbert transforms in [9,12,13,14,15,16,17,23].

LetF be a measurable set in R. X(F) denotes the set of all measurable functions supported on F such that the L norms of the functions are no more than 1. A function in X(F) can be considered essentially as the characteristic function 1F.

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To obtain Theorem2.1, by Lemma3.3, an interpolation argument in [15], and the scaling invariance, it is sufficient to prove that for any p1, p2 > 1 such that 1/p1+ 1/p2 1 and any measurable set F3 R with |F3| = 1, there exists a subset F3 ⊂F3 such that |F3| ≥1/2 and

(4.1)

ΠL1,L2,M1,M2,n1,n2(f1, f2)(x)f3(x)dx

≤C(1 +|n1|)10(1 +|n2|)10|F1|1/p1|F2|1/p2 holds for all f1 X(F1), f2 X(F2), f3 X(F3), where C is a constant independent off1, f2, f3,M1, M2, n1, n2.

If 2L2j+M2 < 2L1j+M1/8, let ω3,j = [−19·2L1j+M1/8,−2L1j+M1/8] and Φ3,j be a Schwartz function whose Fourier transform is a bump function adapted toω3,jsuch thatΦ3,j(ξ) = 1 for allξ∈[9·2L1j+M1/4,−2L1j+M1/4].

If 2L2j+M2 2L1j+M1/8, letω3,j= [18·2L2j+M2,18·2L2j+M2] and Φ3,j be a Schwartz function whose Fourier transform is a bump function adapted to ω3,j such that Φ3,j(ξ) = 1 for all ξ [−17·2L2j+M2,17·2L2j+M2]. Let n3 = 0, Φ3,j,n3 = Φ3,j, f3,j,n3(x) = f3 Φ3,j,n3(x). Define a trilinear form ΛL1,L2,M1,M2,n1,n2 by

(4.2) ΛL1,L2,M1,M2,n1,n2(f1, f2, f3) =

j∈Z

3

=1

f,j,n(x)dx.

Clearly ΛL1,L2,M1,M2,n1,n2 =

ΠL1,L2,M1,M2,n1,n2(f1, f2)(x)f3(x)dx. Thus to prove (4.1), it suffices to prove the following lemma.

Lemma 4.1. Letp1, p2 >1such that 1/p1+1/p2 1andΛL1,L2,M1,M2,n1,n2 be the trilinear form defined by (4.2). Let F1, F2, F3 be measurable sets in R with |F3|= 1. Then there exists a subset F3 F3 such that |F3| >1/2 and there exists a constantC independent of F1, F2, F3,f1,f2, f3, M1,M2, n1, n2 such that

(4.3) ΛL1,L2,M1,M2,n1,n2(f1, f2, f3)

≤C(1 +|n1|)10(1 +|n2|)10|F1|1/p1|F2|1/p2 holds for all f1 ∈X(F1), f2 ∈X(F2), f3 ∈X(F3).

Lemma4.1and Lemma3.3implies the estimates (2.3) by an interpolation argument in [15]. Therefore we obtain Theorem2.1once we finish a proof of Lemma4.1. The following subsections are devoted to proof of Lemma 4.1.

4.1. Definitions. To prove Lemma4.1, we introduce some definitions first.

Let ψ be a nonnegative Schwartz function such that ψ is supported in [1/100,1/100] and satisfiesψ(0) = 1. Let ψk(x) = 2kψ(2kx) for anyk∈Z. Forj∈Zand∈ {1,2,3}, definekjto be an integer such that,j| ∼2kj.

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Denote min∈{1,2,3}kj by kj. And define

Ikj,n = [2−kjn,2−kj(n+ 1)].

Define

1j,n(x) =1Ikj,n∗ψkj(x).

It is easy to see that

ΛL1,L2,M1,M2,n1,n2(f1, f2, f3) =

j∈Zn∈Z

1j,n(x) 3

=1

f,j,n(x)dx.

For an integer γ with 0 γ < 2100, let Z(γ) be the set of all integers congruent to γ modulo 2100. For SZ(γ)×Zwe define

(4.4) ΛS(f1, f2, f3) =

R(j,n)∈S

1j,n(x) 3

=1

f,j,n(x)dx.

ΛS depends on L1, L2, M1, M1, n1, n2. We suppress this dependence for no- tational convenience. Note that there are finite congruence classes modulo 2100. We will therefore concentrate on proving Lemma 4.1 for the trilinear form ΛS.

In time-frequency space, each functionf,j,n for∈ {1,2,3} corresponds to a box Ikj,n×ω,j. The most difficult situation is when only one of boxes is the Heisenberg box, i.e., |Ik,j,n||ω,j| ∼ 1. In this situation, we can use the John–Nirenberg type argument to get the equivalence of Lp estimates of Littlewood–Paley type square functions for only one of functions. For other two functions, there is no such an equivalence and an extra cost for it has to been paid if one estimates the BMO norm. It turns out that the Lp equivalence for at least one of three functions is the most crucial key to solve the problem. Our proof will heavily rely on this equivalence for one of functions.

Letpbe a positive number close to 1. To obtain the Lemma4.1, it suffices to prove (4.3) for p1 p, p2 p and 1/p1+ 1/p2 1. For simplicity, we only deal with the casen1 =n2 =n3= 0. The general case can be handled in the same way by paying at most a cost of (1 +|n1|)10(1 +|n2|)10 in the constants.

We now start to prove that for n1 = n2 = 0, any 1 < p < 2 and any measurable set F3 with |F3| = 1 in R, there exists a subset F3 of F3 with

|F3| ≥1/2 such that

(4.5) ΛS(f1, f2, f3)≤C|F1|1/p1|F2|1/p2

holds for all p1 p, p2 p with 1/p1 + 1/p2 1, f1 X(F1), f2 X(F2), f3 X(F3), where the constant C is independent of S, F1,F2, F3, f1,f2, f3,M1,M2. Let us introduce some definitions first.

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Definition 4.2. Letp >1. Define the exceptional set Ω by

(4.6) Ω =

3

=1

x∈R:Mp M1F

(x)> C0|F|1/p

whereM f is the Hardy–Littlewood maximal function of f and Mpf equals to

M(|f|p)1/p .

By this definition, for the measurable set F3 with |F3| = 1, we take F3 =F3\Ω. IfC0 is chosen sufficiently large we see that|F3| ≥ |F3|/2.

Definition 4.3. Given S Z(γ) × Z and s = (j, n) S. Let ks = min∈{1,2,3}{kj}. The dyadic interval [2−ksn,2−ks(n+ 1)] is called the time interval of s. We denote it byIs.

Definition 4.4. LetS be a subset ofZ(γ)×Z. We say that S is a convex set inZ(γ)×Zif for anys∈Z(γ)×ZwithIs1 ⊆Is⊆Is2 for somes1, s2 S, we have s∈S.

Definition 4.5. LetTS. If there is t∈Tsuch thatIs⊂Itholds for all s∈ T, then T is called a tree with top t. T is called a maximal tree with toptinS if there does not exist a larger tree inSwith the same top strictly containing T.

Definition 4.6. LetTbe a tree in S. Define scl(T) the set of scale indices of Tby

scl(T) ={j∈Z:∃n∈Z,s.t.(j, n)T}. Forj scl(T), thej-th shadow of Tis defined by

Shj(T) = Is:s= (j, n)T . Define an approximation of1Shj(T) by

1Sh

j(T)(x) =1Shj(T)∗ψkj(x).

Definition 4.7. Let (j, n) =s∈S and ∈ {1,2,3}. And let 1∗∗j,n(x) =

Ikj,n

2kj

(1 + 22kj|x−y|2)200dy.

Define a seminormfj,n by (4.7) f

j,n =f

s = 1

|Is|1/p1∗∗j,nf,j,n

p+ 1

|Is|1/p2−kj1∗∗j,nDf,j,n

p

whereDf,j,n is the derivative off,j,n. Define ζ(j, M, K) by

(4.8) ζ(j, M, K) =

L1j+M1−M26 L2

+

L1 L2

M+K,

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whereL= 2100,K is an integer between−10Land 10LandM is an integer between 0 and 6L. For∈ {2,3}, we define aζ seminormf

j,n,ζ by f

j,n,ζ = (4.9)

fj,n+ sup

M,K

1

|Is|1/p

1∗∗j,nf(j,M,K),0

p+|Is|1∗∗j,nDf,ζ(j,M,K),0

p

. For= 1, let theζ seminormf1

j,n,ζ=f1

j,n.

Definition 4.8. Let T S be a tree and t = (jT, nT) T be the top of T. Denote by IT the time interval of the top of treeT.

(a) In the case 2,j| ≤ |ω1,j|/6 for all j scl(T), define Δ(T) for {1,3} by

(4.10) Δ(T)(x) =

(j,n)∈T

1∗∗j,nf,j,n(x)21/2 .

For = 2, define

(4.11) Δ2(T)(x) =1∗∗jT,nTf2,jT,n(x).

And in this case, for ∈ {1,2,3}, define the-size ofT by

(4.12) size(T) = 1

|IT|1/pΔ(T)

p+f

jT,nT.

(b) In the case2,j|>|ω1,j|/6 for allj∈scl(T), for = 2,3, letf,j,T= f,j,0 ifj scl(T) andf,j,T0 if j /∈scl(T). Define:

Δ(T) =

(j,n)∈T

1∗∗j,n

f,j,T−f,j−L,T

(x)21/2 (4.13)

+

(j,n)∈T

1∗∗j,n

f,j,n−f,j,0

(x)21/2 .

And define Δ1(T) by (4.14) Δ1(T)(x) =

(j,n)∈T

1∗∗j,nf1,j,n1(x)21/2 .

In this case, for ∈ {1,2,3}, define the-size ofT by (4.15) size(T) = 1

|IT|1/pΔ(T)

p+f

jT,nT.

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