奇異積分的二次平均以及最加加權上界

୯ҥᆵ᡼εᏢ౛ᏢଣኧᏢ܌

ᅺγፕЎ

Department of Mathematics College of Science

National Taiwan University Master Thesis

⤮䕗䨴⇭䙫ṳ㬈⹚✮Ọ⎱㛧⊇⊇㫱ᷱ䔳

Representing singular kernel as dyadic average and sharp A2 bound

ࢫඵ௘

Chih-Chieh Hung

ࡰᏤ௲௤Ǻ؇ߪᝄ ௲௤

Advisor: Professor Chun-Yen Shen

ύ๮҇୯ 109 ԃ 6 Д

June, 2020

謝辭

兩年的碩士生涯一轉眼即將結束，回顧這兩年自己離成為一個數學家的目標又 往前了好多步。

首先並且也是最想感謝的是我的指導教授沈俊嚴老師，老師在各方面真的給予我很 大的幫助，回顧過往兩年，老師投入很多時間在栽培我們這些學生身上，除了平常 每周的討論會，老師也時常帶我們參加各種學術會議。在研究方面，老師除了在調 和分析方面給予詳細的指導，在加性組合與幾何測度論方面也有許多介紹，但我個 人時間有限，很可惜沒在後兩個方面深入研究。而在其他方面，老師也提供他作為 過來人的心得，這些心得讓我能在擔憂未來的時候靜下心來，繼續前進，真的很謝 謝老師的指導!

接著想感謝中央大學的俞韋亘老師，俞老師總是在我對於未來很擔心迷惘的時 候給予我不少的意見與方向，我也從俞老師那邊學了一些關於組合的數學，十分謝 謝老師!

最後我想感謝我的家人與朋友的陪伴，不論是在我情緒很低潮的時候，又或是 在我因研究進展感到開心的時候。

洪智捷一零九年七月十一日於台灣大學

中文摘要

在調和分析中一個重要的核心問題是研究奇異積分算子的最佳加權上界問題，而此 問題相當於研究奇異積分算子在 L2 加權的有界性。

在 2000 年，S. Petermichl 使用哈爾小波平均來表示希爾伯特轉換的核，此方法後 來被發現是研究此問題的重大突破，爾後里斯轉換 (Riesz transform) 的核，甚至 一般奇異積分算子的核也被找出類似的表示方法。在此基礎之上，S.Petermichl 於 2007 解決希爾伯特轉換的最佳加權上界問題，T. Hytonen 則於 2012 解決一般奇 異積分最佳加權上界問題。

本篇論文會先介紹如何使用哈爾小波平均來表示希爾伯特轉換的核 (2000, S.

Petermichl)，此方法雖然簡單卻隱含對希爾伯特轉換深刻的觀察。接著我們會介紹 如何使用哈爾小波平均來表示里斯轉換的核 (2002, S. Petermichl, S. Treil and A.

Volberg)，這不單單只是推廣希爾伯特轉換的結果到高維度，而是將前方法作一個 統整與重新表示，找出一個推廣到高維度的方式，而這證明過程中，出現一個特殊 積分不等於零的假設，雖然最後作者提出另一條路徑解決，但原本特殊積分不等於 0 的問題在維度大於 2 還是未知的，本篇論文中我們解決積分非零的問題在維度等 於 3 的時候。最後我們介紹如何表示一般的奇異積分算子，並解決最佳加權上界的 問題 (2012, T. Hytonen)。

關鍵字: 卡德隆-吉格曼算子, 希爾伯特轉換, 里斯轉換, 二次平均, 最佳加權上界, 哈 爾偏移算子.

Abstract

A central research problem in the area of Harmonic analysis is to prove the sharp weighted bound for singular integrals. In 2000 S.Petermichl used dyadic averages of Haar shifts to represent the kernel of Hilbert transform which in turn enabled her to obtain the sharp A2 bound for Hilbert transform.

Shortly after, the kernels of Riesz transforms were also obtained via averages of Haar shifts and finally the full generality was made by T. Hytonen who solved the longstanding A2 conjecture for singular integrals. In this dissertation, we first introduce how to use the averages of Haar shifts to represent the kernel of Hilbert transform (2000, S.Petermichl). Second, we will introduce how to represent the kernels of Riesz transforms via dyadic averages of Haar shifts (2002,S. Petermichl, S. Treil and A. Volberg). This result not only extends Petermichl’s ideas to higher dimensions, but also explicitly constructs the Haar shifts for Riesz transforms. However in order to make the result nondegenerate an integral that arises in the process of averaging Haar shifts must be nonzero.

S. Petermichl, S. Treil and A. Volberg provided a proof to show the integral is nonzero in dimension two but for other dimensions the problem remains unknown. A new part of this dissertation is to prove the integral is nonzero in dimension three. Finally we also discuss the breakthrough work of T. Hytonen in 2012 that solves the A2 conjecture for singular integrals.

key words: Calderón-Zygmund operator , Hilbert transform, Riesz transform, dyadic average, sharp A₂ bound, Harr shift operator.

Contents

1 Introduction 5

2 Hilbert transform 6

3 Riesz transform 10

4 An integral arising from dyadic average of Riesz transforms 26

5 General Calderon-Zygmund operators and sharp A2 bound 58

6 References 105

1 Introduction

One of the important milestones that appears in the area of Harmonic analysis in the past decades is the appearance of dyadic Haar shifts. It not only con- nects the continuous singular integrals with dyadic operators but also enables people to resolve the longstanding A₂ conjecture concerning with the sharp weighted bound for Calderón-Zygmund singular integrals. More precisely, the breakthrough work of Petermichl [12] showed that the kernel of the Hilbert transform is actually an average of some certain dyadic operators:

c₀

t− x = lim

L→∞

1 2logL

Z L

1 L

Rlim→∞

1 2R

Z R

−R

X

I∈D^α,r

hI(t)(hI−(x)−h^I+(x))dαdr. (1.1) Therefore the sharp weighted bound for Hilbert transform can be reduced to proving a uniform sharp weighted bound for above dyadic operators which are called dyadic Haar shifts. Such representation of dyadic average for Hilbert transform kernel later was generalized by Stefanie Petermichl, Sergei Treil, Alexander Volberg, [13] to a slightly wider class of kernels but still restricted on one dimensional singular integrals. Finally the full generality was made by Hytönen in [6] who showed that any Calderón-Zygmund operator is a simple variant of dyadic averages, and part of the work was built on a previous result obtained by Hytönen, Perez, Treil and Volberg [5].

Moreover as shown in the work of S. Petermichi that an explicit dyadic Haar shift can actually be given for Hilbert transform. As a result it may be also expected that some explicit dyadic Haar shifts can also be given for Riesz transforms.

Indeed, it was shown in the work of Petermichl, Treil, Volberg [13] that each component of the kernel of Riesz transforms can be explicitly represented by an average of dyadic shift.

In this dissertation, we will go through the history of dyadic averages of Haar shifts for singular integrals and give a proof to a question posed in [13]. More precisely, in section 2 we will demonstrate how to use Haar shifts to represent the kernel of Hilbert transform. In section 3, we also illustrate another way to represent the kernel of Hilbert transform and extend the method to Riesz transforms that are vector singular integrals in higher dimensions. In section 4, we prove a new result that shows an integral arising from averages of Haar shifts for Riesz transforms is nonzero in dimension three. In last section, we discuss the work of T. Hytonen who showed that any Cardelon-Zygmund operator is a simple variant of averages of haar shifts and gave the solution of the longstanding A₂ conjecture.

2 Hilbert transform

Hilbert transform as dyadic operator This part mainly comes from [12].

It connects the discrete Haar shift with continuous singular kernel, _x¹. We first introduce a variety of dyadic grids in R. The basic dyadic grid, starting at 0 with intervals of length 1· 2ⁿ, will be denoted by D0¹ i.e.

D¹0 := {2^k([0, 1) + m) : k ∈ Z, m ∈ Z}.

hJ is the Haar function for J ∈ D0¹, namely hJ := 1

p|J|(χJ−− χ^{J +}),

where J− is the left half of J and J + is the right half of J.

We obtain a variation of D¹0 by first shifting the starting point 0 to α ∈ R and secondly choosing intervals of length r · 2ⁿ for a positive r. The resulting grid is called D^1,α, and the corresponding Haar functions h_J are chosen so that they are still normalized in L².

Since Haar functions forms a basis in L²(_R), for f ∈ L²(_R) we have f (x) = ^X

I∈D^α,r

hf, h^Iih^I(x), ∀α ∈ R, r > 0.

We define for such α, r a dyadic shift operator S^α,r by

(S^α,rf )(x) = Σ_I∈D^α,rhf, hIi hI₋(x) − hI+(x)
. It’s L² operator norm is √

2 and its representing kernel is K^α,r(t, x) = ^X

I∈D^α,r

h_I(t) h_I−(x)− hI+(x)
. (2.1) Lemma 1. The convergence of sum (see above) is uniform for |x − t| ≥ δ ^for every δ > 0. For x 6= t ^let

K(t, x) = lim

L→∞

1 2logL

Z L 1/L

Rlim→∞

1 2R

Z R

−R

K^α,r(t, x)dαdr r .

The limits exist pointwise and the convergence is bounded |x − t| ≥ δ ^{for every} δ > 0 and K(t, x) = _t−x^c0 for some c₀ > 0.

Proof. It is easy to see that P

I∈D^α,r|h^I(t) (h_I−(x)− h^I+(x))| ≤ 2√

2/|t − x|, ∀α ∈ R and ∀r > 0. 0. In particular, the sum converges absolutely and

uniformly |x − t| ≥ δ ^{for every} δ > 0. > 0. The existence of the limits is due to fact that summands repeat for different dyadic grids. The main point is to show |K(t, x) = c⁰/(t− x)| ^with c₀ 6= 0. 0. It is enough to prove the following properties of K(t, x):

1 translation invariance, i.e., K(t, x) = K(t + c, x + c),∀c ∈ R, so K(t, x) = K(t− x)^;

2 antisymmetry, i.e, K(t, x) = −K(−t, −x), ^so K(x− t) = −K(t − x)^; 3 dilation invariance, i.e., K(t, x) = λK(λt, λx),∀λ > 0^;

4 K(1) = c₀ > 0.

In order to check the first three properties we observe the following simple rela- tionships between the Haar functions of different dyadic grids for translations, reflections and dilations:

For any interval I ∈ D^α,r there exists an interval of the same length in D^α−c,r so that h^α,r_I (t + c) = h^α_I^−c,r(t). In a similar sense h^α,r_I (−t) = −h^−α,rI (t) when changing grids from D^{α,r to} D^{−α,r and} h^α,r_I (λt) = λ^−1/2h^α/λ,r/λ_I (t) when chang- ing from D^{α,r to} D^α/λ,rλ.

Using these facts, the proof of the first three properties are simple computations, mainly involving changes of integration variables. Note that these properties show that K(t, x) = _t−x^c0 , we turn to the essential part to show that c₎ 6= 0^. The product h_I(t) (h_I−(x)− hI+(x)) 6= 0 if and only if the point (t, x) lies in this square I×I. Its value is±√

2/|I|, where the correct sign is indicated inside

the smaller rectangles. Let us first compute K_n^r(t, x) := lim

R→∞

1 2R

Z R

−R

X

I∈D^α,r

|I|=r2ⁿ

h_I(t) (h_I−(x)− h^I+(x)) dα, (2.2)

for fixed r > 0 and n∈ Z and assuming t > x. Due to the averaging process in α, this is only going to depend on t− x^{. If:}

t− x = 0 ^{, then} K_n^r(t, x) = 0 and similarly;

t− x = |I|/4 ^{, then} K_n^r(t, x) = 3/4·√ 2/|I|^; t− x = |I|/2 ^{, then} K_n^r(t, x) = 0;

t− x = 3|I|/4 ^{, then} K_n^r(t, x) = −1/4 ·√ 2/|I|^; t− x ≥ |I| ^{, then} K_n^r(t, x) = 0.

Now we compute

K_n^r(t, x) := lim

R→∞

1 2R

Z R

−RK^α,r(t, x)dα

= ^X

n∈Z Rlim→∞

1 2R

Z R

−R

X

I∈D^α,r

|I|=r2ⁿ

h_I(t) (h_I−(x)− h^I+(x)) dα.

So we compute K^r(t, x) using K_n^r(t, x) for different values of n and summing over n ∈ Z. It suffices to compute K^r(t, x) for values t − x = 3/4 · r2ⁿ andt− x = ·r2^n:

K^r(3

4r2ⁿ) = −1 4

√2 r2ⁿ + 3

16

√2 r2ⁿ + 9

64

√2 r2ⁿ

 1 + 1

4 + 1

16 + . . .



=

√2

8r2ⁿ, (2.3) K^r(r2ⁿ) = 3

16

√2 r2ⁿ

 1 + 1

4 + 1

16 + . . .



=

√2

4r2ⁿ. (2.4)

The above equations imply that 3√

2

32(t− x) ≤ K^r(t− x) ≤

√2

4(t− x) ∀r > 0. ^(2.5)

From above, it is clear that c₀ > 0. _□

3 Riesz transform

The “simplest”operator whose average is the Hilbert transform This part comes from [13]. It uses average technique to generalize the method in section 1 to the n-dimensional Riesz kernels. Let L denote a dyadic lattice in R. By L(k) we understand the dyadic grid of intervals from L having length 2^−k, k ∈ Z. For the convenience we would like to use the notations D =: L(0)^. We consider first such a dyadic lattice that the grid D has the point 0 as one of the end-points of its intervals. To emphasize that we write D0. Later we will have D^{t —the point} t plays the role of 0.

Let us consider the following linear operation f → ϕ(x) := ^X

I∈D0

hf, hIiχI(x).

Here h_I denotes the Haar function of the interval I, that is

h_I(x) =









|I|−1^1/2 , for x ∈ I−

|I|1^1/2 , forx ∈ I⁺,

and I₋, I₊ are left and right halves of the interval I correspondingly. Symbol χI as usual stands for the characteristic function of the interval I.

This linear operation will be our main building block, so it deserves a name P.

Actually, we will call it P0, thus ϕ₀(x) := _P0f := ^P_I∈D

0hf, h^Iiχ^I(x). Index 0 indicates the end-point of one of the intervals from D⁰. So similarly we consider

ϕt(x) := _Ptf

defined exactly as before, but with respect to the grid Dt of unit intervals such that the end-point of one of them is in t ∈ R.

Notice that the family of grids D^t, t ∈ R, can be naturally provided with the structure of probability space. This space is (_R/_Z, dt) = ((−1, 0], dt)^{. As} usual we can use the letter ω for a point from (−1, 0]^{, and} dP (ω) denotes the probability —in this case just Lebesgue measure on the interval (−1, 0]^{. We} want to fix x ∈ R and to write a nice formula for

E(ϕ_ω(x)dP (ω)) .

So we want to average operatorsP^ω. It can be noticed immediately thatEP^ω is a convolution operator. In fact, let us denote by La the shift operator: L_a(f )(x) = f (x + a). Then obviously

Pt−aL_a = L_aPt.

Applying averaging (and the fact that our dP (ω) is invariant with respect to the natural shift on R/Z induced by the shift on R) we immediately get

EPωL_a = L_aEPω. (3.1)

So the average operator EPω is a convolution operator, we will write this as follows

E(ϕ_ω(x)dP (ω)) =_EPω(x) = F₀ ∗ f(x). ^(3.2)

It is easy to compute F0. By the definition of ϕt(x) one can write ϕ_t(x) =

Z

f (s)h^t−12(s)ds, x− 1

2 < t− 1

2 < x + 1

2, (3.3)

where

h^t(s) =









−1 ^{, for} s∈ (t − ¹₂, t) +1 , fors ∈ (t, t + ¹₂).

But h^t(s) = k₀(t− x), ^where

k₀(s) =









+1 , for s ∈ (−¹₂, 0)

−1 ^{, for}s ∈ (0,¹₂).

So (3.3) can be rewritten as follows ϕ_t+¹

2(x) = Z

f (s)k0(t− s)ds, x − 1

2 < t < x + 1

2. (3.4)

Thus comparing this with (3.2) ) (and using again the shift invariance ofdP (ω)) we get

F₀ ∗ f(x) = E(ϕ_ω(x)dP (ω))

= _E

 ϕ_ω+¹

2(x)dP (ω)



=

Z _x+¹

2

x−¹₂

Z

f (x)k₀(t− x)ds

 dt.

From which we get the formula for F₀: F₀(x) =

Z x+¹₂ x−¹

k₀(t)dt = k₀ ∗ χ⁰(x), (3.5)

where χ0 is the characteristic function of the unit interval (−1/2, 1/2)^.

Let us start over the beginning of this section with one slight difference — we rescale all our operators, and now P^ρt, ϕ^ρ_t, F₀^ρ, k^ρ₀ are precisely as above, but when the unit length intervals are replaced by intervals of length ρ > 0. We just change the scale —nothing else. In particular,

ϕ^ρ₀(x) := _P^ρ₀f := ^X

I∈D^ρ0

hf, h^Iiχ^I(x)/√ ρ

where D0^ρ is the grid of intervals of length ρ such that 0 is the end-point of two intervals from this grid. We want to remind that h_I here is always normalized in L².

Again we have a natural probability space of all grids of intervals of size ρ:

R/ρ_Z;_ρ¹dt|(−ρ, 0]^.

ϕ^ρ_t(x) := _P^ρ_tf := ^X

I∈D^ρt

hf, h^Iiχ^I(x)/√ρ.

Averaging over all grids of intervals of size ρ makes P^ρt a convolution operator

—there is no difference with our reasoning above. It is easy to see that this is the convolution operator with the kernel

F_t^ρ(x) := 1 ρ

Z x+^rho₂ x−^ρ₂

1 ρk₀

t ρ



dt = ρF₀

x ρ



. (3.6)

The first ¹_ρis because of the form our probability has. The second ¹_ρbecause we should average a function normalized in L¹.

Let us now consider all convolution operators with kernels F₀^ρ. Let us fix r ∈

[1, 2) and let us take a look at the convolution operator with kernel F_r =

X∞ n=−∞

F₀^2nr. (3.7)

The grids Dt^2nr (t is fixed) can be united into a “dyadic”lattice L^rt . Here t means the reference point —one of the end-point of intervals from our lattice, and r means the length of one of the intervals of the lattice—let us call r the calibre of the lattice. Obviously the convolution operator with the kernel Fr is the averaging over all “dyadic”lattices (not grids!) L^rt of fixed calibre r of the operators given by

PL^rtf = ^X

I∈L^rt

hf, hIiχI(x)/

q|I|

F_r ∗ f =EPL^rtf.

This is just because the kernelFr is the sum of kernels, each of which appeared as averaging of the grid opearators assigned to grids of size 2ⁿr,n = 0,±1, ±2, . . . , where we summed up over the grids, and the lattice of calibre r is the union of such grids.

Now let us finally average over r ∈ [1, 2)^: F (x) :=

Z 2 1

F_r(x)dr r . Now we have from one side

F ∗ f = (^AveragePL) f, (3.8) where averaging is performed over all lattices L^rt.

where averaging is performed over all lattices Φ. F (x) =

Z 2 1

F_r(x)dr

r ^(3.9)

= Z 2

1

X∞ n=−∞

F₀^2nrdr r =

Z _∞

0

F₀^ρdρ ρ =

Z _∞

0

F0

x ρ

 dρ

ρ². (3.10) We used (3.6) here. Finally we have

F (x) = −1 x

Z _∞

0

F₀(t)dt = 1 4

1

x. (3.11)

Theorem 3.1. Averaging of operators PL^rt over both parameters t and r is equal to one quarter of kernel of the Hilbert transform.

We have a good thing:

The Hilbert transform is the averaging over the family of lattices of very simple operators What is the dyadic shift? The function that generated everything in the first section was function F0 —the kernel of the convolution operator which is the averaging of grid operators Pt. It is easy to see that F₀(x± 1) ^are also kernels of the convolution operators which are the averagings of some grid operators. Given f, let us consider ϕ_t(x) as above and also

ϕt(x + 1) = ^X

I∈Dt

hf, h^I−1iχ^I(x) = ^X

I∈Dt

hf, h^Iiχ^I+1(x) =: _P⁺_t (f )

ϕ_t(x− 1) = ^X

I∈Dt

hf, h^I+1iχ^I(x) = ^X

I∈Dt

hf, h^Iiχ^I−1(x) =:_P⁻_t (f )

So we test f on h_I and put the result on I ± 1. What if we average these operators? Repeating (3.2) we get

Z 1 0

P^±t dt



f = F₀(x∓ 1) ∗ f. ^(3.12)

Consider

S(x) := F₀(x)− 1

2[F₀(x + 1) + F₀(x− 1)] . ^(3.13) Supposedly S is a kernel of a convolution operator corresponding to averaging over grids of a certain grid operator (we will show which one). If we build S_ρ as before for all calibres, we can consider again S_r := ^P∞_n=−∞S^2nr. Operators S_r are averagings over all lattices of calibre r of the operators which are sums of our hypothetical grid operators. Averaging over r ∈ [1, 2) with respect to the measure dr/r, we will get the operator with kernel

Z ₂

1

S_r(x)dr r =

Z _∞

0

S(x ρ)dρ

ρ² = 1 x

Z _∞

0

S(t)dt = 1 4

1

x. (3.14) So we are left to invent a simple “grid”operator, whose average will give us S(x).

Theorem 3.2. Let D⁽²⁾t be a grid of intervals of length 2 such that t is the end-point. Consider operators

f → ^X

J∈D⁽²⁾t

hf, hJ−iχJ +

f → ^X

J∈D⁽²⁾t

hf, h^{J +}iχ^J−

f → ^X

J∈D⁽²⁾t

hf, hJ−iχJ−

f → ^X

J∈D⁽²⁾t

hf, h^{J +}iχ^{J +}

The averaging over t of the first operator gives a convolution with kernel ¹₂F₀(x− 1), the averaging over t of the second operator gives a convolution with kernel

1

2F₀(x + 1), and the averaging over t of the third and the fourth operator gives a convolution with kernel ¹₂F0(x) each.

Proof. Let us call the first operator Ht, and let us average EHt it over its probability space , R/_Z;¹₂dt|(−2, 0]
. Instead of considering the grid of intervals of length 2 let us consider the grid of intervals of length 1 —we call it Dt¹ . Consider operators

A_t :→ ^X

I is odd,I∈Dt¹

hf, hIiχI+1

Bt :→ ^X

I is even,I∈Dt¹

hf, h^Iiχ^I+1.

Clearly A_t+1 = B_t. Also it is clear that A_t + B_t = _P⁺_t , where the last operator is our grid operator from the beginning of this Section.

EH_t = 1 2

Z 1 0

(A_t + A_t+1) = 1 2

Z 1 0

(A_t + B_t) = 1 2

Z 1 0

P⁺t dt.

From (3.12) we get that

EH_t = 1

2F₀(x− 1) ∗ .

Similarly, if we call the second operator G_t we get from (3.12) EGt = 1

2F0(x + 1)∗ .

Using (3.2) and (3.5) we show that averagings of the third and the fourth op- erators give us convolution operator with kernel ¹₂F₀. The theorem is proved.

□

Theorem 3.3. Let us consider the following grid operator

f → ^X

J∈Dt⁽²⁾

hf, h^Ji,

where

t ∈



R/_Z; 1

2dt|(−2, 0]

 .

Then its averaging is the convolution operator with kernel ^√1

2S(x). Proof. We weite hJ as ^√1

2 (−χ^J−+ χJ +). Then it is an obvious algebraic remark that

√2 our operator= third operator of Theorem 3.2

+third operator of Theorem 3.2

−third operator of Theorem 3.2

−third operator of Theorem 3.2

Averaging this and using Theorem 3.2 finishes the proof. □

As in the previous section, given the latticeL = L^rt, we can consider the lattice operator

K^Lf := ^X

J∈L

hf, h^{J +} − h^J−ih^J amalgamated from the grid operators of Theorem 3.2.

This operator is called the dyadic shift. It has been proved that averaging of dyadic shifts over all lattices gives us operator which is proportional to the Hilbert transform (we certainly mean that coefficient of proportionality is not zero).

Let us reproduce this result. Fixing r and averaging over lattices with fixed calibre r (we leave for the reader to invent the natural probability space of all lattices with fixed calibre r) we get the convolution operator with the kernel

√1 2

X∞ n=−∞

1 2ⁿrS

 x 2ⁿr



=: 1

√2S_r(x).

Averaging convolution operators with kernels ^√1

2Sr over [1, 2); ^dr_r
, we get the operator with the kernel ¹₄^√1

2 1

x. So we get averaging of the shift operators over all lattices of all calibres = ¹

4√

2 kernel of the Hilbert transform.

Planar case We can and will reason by analogy. We have latticesL^ρt of squares, where t now is in Ω^ρ := _R²/ρZ with normalized Lebesgue measure (Lebesgue measure on the torus Ω^ρ divided by ρ²). We have the main grid operator

Ptf := ^X

Q∈Dt

hf, h^Qiχ^Q

where D^t is a grid of unit squares such that t ∈ R² is a vertex for 4 of them, where

h_Q(x) =

















|Q|−1^1/2 , for x ∈ Q^l

1

|Q|^1/2 , forx ∈ Q^r, 0, otherwise

Here Ql, Qr are left and right halves of Q, functionhQ is normalized in L². We consider the same type of grid operators for grids Dt^ρ of squares of side ρ —the only change is that we divide χ_Q by ρ to make it normalized in L².

Let us denote by k0 the function −h^Q0, where Q0 is the unit square centered at 0. Also χ₀ denotes the characteristic function of this square. Consider

Φ₀ := χ₀ ∗ k⁰,

Φ^ρ₀(x) := 1 ρ²

1 ρ²χ0

· ρ



∗ k⁰

· ρ



= 1 ρ²Φ0

x ρ

 .

Exactly as before (in one dimensional case) function Φ₀ is the kernel of the convolution operator, which appears as averaging of Pt over Ω¹. Function Φ^ρ₀ is the kernel of the convolution operator, which appears as averaging of P^ρt over Ω^ρ.

Again, we can consider kernel

k(x) :=

Z _∞

0

Φ^ρ₀(x)dρ ρ =

ω

 x

|x|



|x|² .

And it is very easy to see that ω is an odd non-zero function on the unit circle. Literally as before we can see that k is the convolution operator which is the average with respect to measure ^dr_r |[1, 2) of the convolution operators with kernels

k_r(x) :=

X∞ n=−∞

Φ^r₀^·2n(x).

In its turn, k_r is the average of the lattice operators which are sums of corre- sponding grid operators, here are those lattice operators:

PL^r := ^X

Q∈L^r

hf, h^Qiχ^Q/

q|Q|.

Here r is fixed and denotes the calibre of the lattice. The averaging over the lattices of this fixed calibre gives us the convolution operator with kernel k_r. So the averaging over the calibres (= ^R₁². . .^dr_r ) gives us the averaging over all lattices, over all calibres. As a result we get the convolution operator with kernel k = ^ω(_|x|^x )

|x|² .

Again we would like to repeat all this but with slightly different lattice opera- tors —just because there are nicer ones and because PL^r are not L² bounded.

Another problem we face now is that k is not necessarily a kernel of a Riesz transform. So we will need to work a bit more than in the one-dimensional case to obtain the Riesz transform kernel.

For a square Q consider its partition to 4 equal squares and let us call them Q^nw, Q^ne, Q^sw, Q^se according to northwest, northeast,. . .. Let us consider the following grid operator

f → ^X

Q∈D⁽²⁾t

hf, h^Qne + h_Q^se − h^Qnw − h^Qswih^Q,

t∈ Ω⁽²⁾ :=



R²/2_Z²;1

4 Lebesgue measure

 . Consider also the function (x = (x₁, x₂))

S(x₁, x₂) = Φ₀(x₁, x₂)− 1

2Φ₀(x₁ + 1, x₂)− 1

2Φ₀(x₁ − 1, x2) + 1

2Φ₀(x₁, x₂ + 1)− 1

4Φ₀(x₁ + 1, x₂ + 1)− 1

4Φ₀(x₁ − 1, x2) (3.15) + 1

2Φ₀(x₁, x₂ − 1) − 1

4Φ₀(x₁ + 1, x₂ − 1) − 1

4Φ₀(x₁ − 1, x2 − 1).

Theorem 3.4. The averaging of the grid operator above over Ω⁽²⁾ gives the convolution operator with kernel ¹₂S(x).

The proof is literally the same as the proof of Theorem 3.3.

Let us start with one observation about (3.15). Function Φ₀ is the convolution χ₀ ∗ k0. But both functions χ₀ and k₀ are products of functions of one variable Φ₀(x₁, x₂) = f₀(x₂) · F0(x₁). Moreover, function f₀ is nonnegative. Actually f0(x2) is a convolution square of the characteristic function of the unit interval centered at 0. Formula (3.15) now looks like

S(x₁, x₂) =



f₀(x₂) + 1

2f₀(x₂ + 1) + 1

2f₀(x₂ − 1)



×



F₀(x₁)− 1

2F₀(x₁ + 1)− 1

2F₀(x₁ − 1)

 .

For the future purposes we can say what happens in n > 2 case easily. We get S_n(x) = S_n(x₁, x₂, . . . , x_n) and

S_n(x) =



f₀(x₂) + 1

2f₀(x₂ + 1) + 1

2f₀(x₂ − 1)



(3.16)

× Yn i=2



F₀(x₁)− 1

2F₀(x₁ + 1)− 1

2F₀(x₁ − 1)

 . As in the previous section this S generates kernel s by formula

s(x) = Z _∞

0

1 ρⁿS

x ρ

 dρ ρ =

ξ(_|x|^x )

|x|ⁿ .

And it is very easy to see thatξ_n is an odd non-zero function on the unit sphere.

We will show it below. Literally as before we can see that s is the convolution operator which is the average with respect to measure ^dr_r |[1, 2)of the convolution operators with kernels

s_r(x) :=

X∞ n=−∞

S₀^r·2n(x).

In its turn, sr is the average of the lattice operators which are sums of corre- sponding grid operators, here are those lattice operators:

S_L^r := ^X

Q∈L^r

hf, h^Qne + h_Q^se − h^Qnw − h^Qswih^Q. (3.17) Here r is fixed and denotes the calibre of the lattice. The averaging over the lattices of this fixed calibre gives us the convolution operator with kernel s_r. So the averaging over the calibres (= ^R₁². . .^dr_r ) gives us the averaging over all lattices, over all calibres. As a result we get the convolution operator with kernel s = ^ξn(_|x|^x)

|x|ⁿ .

Let Sⁿ⁻¹ denote as always the boundary sphere of the n-dimensional unit ball.

Denote by S₊ⁿ⁻¹ the right half sphere —the half that lies in {x ∈ R : x₁ > 0}^. Let e₁ be a unit vector in the direction of coordinate axis x₁. Let σ denote Lebesgue measure of Sⁿ⁻¹. It would be important to prove

Z

S₊ⁿ⁻¹

ξ_n(ω)hω, e¹idσ(ω) < 0. ^(3.18) For n = 2 we can just prove that ξ₂(ω) < 0 for any ω ∈ S+¹. Then (3.18) follows immediately. To do this we use formula (3.16) and notice that f0(x) + ¹₂f0(x + 1) + ¹₂f0(x − 1) = (1 − ¹₂x)+. Then the fact that ξ2(ω) < 0 follows from the following lemma.

Lemma 2. For any k ∈ [0, ∞) ^{we have} Z 2

0

 1− 1

2kx



+



F₀(x) − 1

2F₀(x− 1)



xdx < 0.

Proof. Ifk ≥ 2then the first factor vanishes everywhere where the second factor is positive. So we are done for such k. For 0 ≤ k ≤ 1^{, we have} (1− ¹₂kx)₊ =

(1−¹₂kx) on [0,2], and we can make an easy calculation of the integral. For the range 1 < k < 2 the calculation becomes unpleasant, but still straightforward, we skip it just to avoid direct and simple calculations. □

For n = 2, ω can be identified with a point of [−π, π).Under this identification the kernel ξ2 becomes an even function skew symmetric on [0, π] with respect to the pointπ/2. Rotation of the kernelξ₂(ω) means just the new kernel ξ₂(ω−ϕ)^. Then

(ξ2 ∗^cos) (ϕ) = cosϕ·

Z π

−πξ2(s)cossds



(3.19)

= cosϕ·

Z

S₁

ξ₂(ω)hω, e¹idσ(ω)



= c₂cosϕ, and constant A2 :=

Rπ

−π| cos s|ds

|c2| . Notice that rotation of kernel ξ2 corresponds to rotation of dyadic lattices on the plane. We have just proved the following theorem.

Theorem 3.5. The Riesz transform _|x|^x13∗is the operator integralc⁻¹₂ ^R cosψ^ξ2(^Uψ x

|x|)

|x|³ ∗ dψ. In particular, this means that operator with the kernel A⁻¹_{2 |x|}^x13 lies in the closed convex hull (in the weak operator topology) of the planar dyadic shifts.

Thus, uniform boundedness of dyadic shift operators in any Banach space im- plies the boundedness of the Riesz transform in the same space.

For the case n > 2 we again start with (3.18). Let us average ξ_n with respect to all rotations that leave e₁ fixed. We get a new function η_n(ω) = f (hω, e1i)^. Obviously,

Z

Sⁿ⁻¹

f (hω, e¹i)hω, e¹idσ(ω) < 0. ^(3.20)

Let SO is the group of orthogonal rotations of Sⁿ⁻¹.

Let us calculate c_n = ^R_SOf (hUe¹, e₁i)hUe¹, e₁idU. Obviously, c_n =

Z

Sⁿ⁻¹

f (hω, e1i)hω, e1idσ(ω) 6= 0,

because of (3.20). Now let us consider the rotated functions f (hUe¹, e1i)^{. Con-} sider

g(ω) = Z

SO

f (hUω, e¹i)hUe¹, e₁idU.

Then it is clear that g(Rω) = g(ω) for every R ∈ SO ^{that fixes} e₁. In fact, g(Rω) =

Z

SO

f (hURω, e¹i)hUe¹, e1idU

= Z

SO

f (hV ω, e¹i)hV R^∗e1, e1idV

= Z

SO

f (hV ω, e¹i)hV e¹, e₁idV = g(ω).

On the other hand, it easy to see that g(ω) =

Z

Sⁿ⁻¹

f (hω, ξi)hξ, e¹idσ(ω). ^(3.21) Such a function (as we saw) depends only on hω, e1i. But moreover, it can be written asR

Sⁿ⁻¹ f (he¹, ξi)hξ, ωidσ(ω). This is a restriction of a linear polynomial onto the sphere. This linear polynomial depends on hω, e¹i only, and, thus, is c· hω, e¹i. The constant c is just our c_n. One can see that by plugging ω = e₁ into our formula (3.21) for g(ω).

Consider A_n :=

R

SO|hUe1,e1i|dU

||cn . Notice that rotation of kernel ξ_n corresponds to rotation of dyadic lattices on the plane. We have just proved the following theorem.

Theorem 3.6. The Riesz transform _|x|^xn+1¹ ∗ is the operator integral

c⁻¹_n Z

SO

hUe1, e₁iη_n

 U_|x|^x



|x|ⁿ⁺¹ ∗ dU.

In particular, this means that operator with the kernelA⁻¹_{n |x|}^xn+1¹ lies in the closed convex hull (in the weak operator topology) of the planar dyadic shifts. Thus, uniform boundedness of dyadic shift operators in any Banach space implies the boundedness of the Riesz transform in the same space.

4 An integral arising from dyadic average of Riesz transforms

Introduction The question that was risen in their work [13] is whether the following integral is zero or not (the detail definitions of some notations in this integral are given in next section):

Z

S₊ⁿ⁻¹

< ω, e₁ > ξ_n(ω)dσ(ω). (4.1) They were able to show the integral is nonzero (in fact it is negative) when dimension n = 2 but for dimension n ≥ 3 the problem remains unsolved.

Therefore the purpose of this section is to show the above integral when di- mension n = 3 is also negative. This was done via a careful and an efficient decomposition for the integral. For some terms in our decomposition we are able to show explicitly that their values are negative. For some other terms we are able to prove an upper bound. Combining all the estimates shows the integral is indeed negative. Now let us mention the difficulties of the integral

for dimension n = 3. First, the integrand functions in the integral are piece- wise defined on some compact intervals, and the range of the integration is only half-sphere. Secondly, after we carefully analyse the integrand functions, one of the main difficulties then arises due to the mutual overlaps of their supports.

More precisely, after using the sphere coordinates in the integral, the supports of the functions will create several difficulties since the behaviors of the points in these supports will now depend on the values of some complicated trigonom- etry functions. For these difficulties, it requires us to very carefully distinguish the range of the integrals in our decomposition. Finally for several integrals in our decomposition, we are able to show that their exact definite integrals can be computed. For the other integrals, we are not able to find their exact definite integrals, but we are able to find their upper bounds whose values can be explicitly estimated.

Preliminary In this section we first introduce some notations that will be used

frequently in this paper. Let F0, F, f0, and f be defined as followings.

F₀(x) =

















1

2 − |x + ¹₂| −1 ≤ x < 0

−(¹₂ − |x − ¹₂|) 0 ≤ x < 1

0 otherwise

;

F (x) = F₀(x)− 1

2F₀(x + 1)− 1

2F₀(x− 1);

f0(x) =









1− |x| ^if |x| ≤ 1

0 otherwise

;

f (x) = f0(x) + 1

2f0(x + 1) + 1

2f0(x− 1).

Note that F is an odd function so that we may only describe F on x ≥ 0^{, i.e.}

F (x) =















−3 4 + 3

2|x − 1

2| if |x − 1 2| ≤ 1

2 1

4 − 1

2|x − 3

2| if |x − 3 2| ≤ 1

2

0 x ≥ 2

and f (x) = 1− ¹₂|x| ^if |x| ≤ 2 ^and 0 otherwise. See below for their graphs.

(a) F0 (b) F

Figure 1: F , and F₀

(a) f0 (b) f

Figure 2: f , and f₀