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BOCHNER–RIESZ MEANS ON EUCLIDEAN SPACES

8745hc_9789814458764_tp.indd 1

21/6/13 9:23 AM

This page intentionally left blank

BOCHNER–RIESZ MEANS ON EUCLIDEAN SPACES Shanzhen Lu Beijing Normal University, China

Dunyan Yan University of Chinese Academy of Sciences, China

World Scientific NEW JERSEY

8745hc_9789814458764_tp.indd 2

•

LONDON

•

SINGAPORE

•

BEIJING

•

SHANGHAI

•

HONG KONG

•

TA I P E I

•

CHENNAI

21/6/13 9:23 AM

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

BOCHNER–RIESZ MEANS ON EUCLIDEAN SPACES Copyright © 2013 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

ISBN 978-981-4458-76-4

In-house Editor: Angeline Fong

Printed in Singapore

Contents Preface

vii

1 An 1.1 1.2 1.3 1.4

1 3 10 15 34

introduction to multiple Fourier series Basic properties of multiple Fourier series . . Poisson summation formula . . . . . . . . . . Convergence and the opposite results . . . . . Linear summation . . . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

2 Bochner-Riesz means of multiple Fourier integral 2.1 Localization principle and classic results on ﬁxed-point convergence . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Lp -convergence . . . . . . . . . . . . . . . . . . . . . . . 2.3 Some basic facts on multipliers . . . . . . . . . . . . . . 2.4 The disc conjecture and Feﬀerman theorem . . . . . . . 2.5 The Lp -boundedness of Bochner-Riesz operator Tα with α>0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem 2.7 Kakeya maximal function . . . . . . . . . . . . . . . . . 2.8 The restriction theorem of the Fourier transform . . . . 2.9 The case of radial functions . . . . . . . . . . . . . . . . 2.10 Almost everywhere convergence . . . . . . . . . . . . . 2.11 Commutator of Bochner-Riesz operator . . . . . . . . .

. . . .

. . . .

41 . . . .

. . . .

. . . .

. . . . . . .

. . . . . . .

. 59 . 61 . 78 . 89 . 97 . 105 . 129

. . . . . .

141 141 146 167 177 194 208

3 Bochner-Riesz means of multiple Fourier series 3.1 The case of being over the critical index . . . . . . . . . . 3.2 The case of the critical index (general discussion) . . . . . 3.3 The convergence at ﬁxed point . . . . . . . . . . . . . . . 3.4 Lp approximation . . . . . . . . . . . . . . . . . . . . . . 3.5 Almost everywhere convergence (the critical index) . . . . 3.6 Spaces related to the a.e. convergence of the Fourier series v

. . . .

. . . . . .

41 45 48 51

vi

Contents 3.7 3.8 3.9 3.10

The uniform convergence and approximation . . . . . (C, 1) means . . . . . . . . . . . . . . . . . . . . . . . . The saturation problem of the uniform approximation Strong summation . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

4 The conjugate Fourier integral and series 4.1 The conjugate integral and the estimate of the kernel . . 4.2 Convergence of Bochner-Riesz means for conjugate Fourier integral . . . . . . . . . . . . . . . . . . . . . . . . 4.3 The conjugate Fourier series . . . . . . . . . . . . . . . . . 4.4 Kernel of Bochner-Riesz means of conjugate Fourier series 4.5 The maximal operator of the conjugate partial sum . . . . 4.6 The relations between the conjugate series and integral . . 4.7 Convergence of Bochner-Riesz means of conjugate Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 (C, 1) means in the conjugate case . . . . . . . . . . . . . 4.9 The strong summation of the conjugate Fourier series . . 4.10 Approximation of continuous functions . . . . . . . . . . .

. . . .

. . . .

244 251 259 280

293 . . 293 . . . . .

. . . . .

303 309 316 319 324

. . . .

. . . .

332 334 337 347

Bibliography

367

Index

375

Preface This book mainly concerns with the Bochner-Riesz means of multiple Fourier integral and series, which shall be simply termed the Bochner-Riesz means for short. Bochner-Riesz means is an important branch of multiple Fourier analysis initiated in the 1930s by Bochner. At that time, Bochner and his fellows restricted their studies to the cases when its degree is greater than the critical index. In the late of 1940s, Minde Cheng, Bochner’s student, carried out a systematic research about the Bochner-Riesz means. In the late 50s and early 60s, Stein contributed signiﬁcantly to the research where the degree is at and below the critical index. In the same period, Minde Cheng and his students also started the research on approximation of functions by the Bochner-Riesz means. Since the 1970s, inspired by Stein’s work, many researchers in Europe, America, Soviet Union and China have joined the study. A number of important or valuable results are achieved in this ﬁeld. Although the development in this ﬁeld took nearly half of a century with many achievements, there still remain many fundamental problems waiting to be resolved. To introduce these results and some remaining problems to Chinese scholars, Shanzhen Lu and Kunyang Wang published a monograph “Bochner-Riesz Means” in Chinese in 1988. We note that there have been some important progresses in this ﬁeld since then. To supplement the previous book with these important results, we wish to publish a new monograph in English with the present title. The book is aimed at giving a systematical introduction to fundamental theories of the Bochner-Riesz means and important achievements attained through the latest 50 years. Numerous results illustrate that the BochnerRiesz means oﬀer the most natural way for the extension of Fourier series from the case of single variable to that of several variables. This book consists of four chapters. The ﬁrst chapter is a brief introduction to the theory of multiple Fourier series. Chapter 2 and 3 contain main topics of this book which are closely related to each other. Chapter 2 introduces the Bochnervii

viii

Preface

Riesz means of multiple Fourier integral, including Feﬀerman theorem which negates the Disc multiplier’s conjecture and the famous Carleson-Sj¨ olin theorem. For almost everywhere convergence of the Bochner-Riesz means below the critical index, we introduce Carbery-Rubio de Francia-Vega’s work. Some recent results on commutators of the Bochner-Riesz means are also included in the last section of this chapter. Chapter 3 concerns with the Bochner-Riesz means of multiple Fourier series, including the theory and application of a class of function space (block space), developed by Taibleson, Weiss and other mathematicians, which is closely related to almost everywhere convergence of the Bochner-Riesz means. In addition, a class of function space named spaces generated by smooth blocks is also introduced in this chapter. Such spaces are closely related to the rate of convergence of the Bochner-Riesz means. Chapter 4 discusses the Bochner-Riesz means of conjugate Fourier integral and conjugate Fourier series, where the concepts of conjugate integral and conjugate series are based on the theory of singular integral by Calder´on-Zygmund. We would like to thank Kunyang Wang whose research work contributes nicely to this book. We have to give our thanks to Bolin Ma who provides some useful materials related to new progress in this ﬁeld since 1988. In addition, Shanzhen Lu wishes to express his gratitude to Guido Weiss for their collaboration in the early 1980s. This book is dedicated to Guido Weiss on the occasion of his 85th birthday. The book is also in memory of Minde Cheng, Yongsheng Sun and Mitchell H. Taibleson.

Shanzhen Lu (Beijing Normal University) Dunyan Yan (University of Chinese Academy of Sciences) December, 2012

Chapter 1

An introduction to multiple Fourier series

Let Q := [−π, π)n . Sometimes [−π, π)n is denoted by Qn for emphasizing n-dimension. We denote by Lp (Q) the set of all complex-valued measurable functions whose modulus to the p-th power is integrable on Q and 2π-periodic in each of its variables for 1 ≤ p < ∞. The space Lp (Q) is a Banach space with the norm

1 |f (x)| dx

f p =

p

p

.

Q

We denote a set of all continuous functions on Q having period 2π in each of its variables by C(Q), which is a Banach space with the uniform norm f C = max{|f (x)| : x ∈ Q}. Let f ∈ L(Q) := L1 (Q). The Fourier coeﬃcients of f are Cm

1 := Cm (f ) = (2π)n

f (x)e−ix·m dx, m ∈ Zn .

(1.0.1)

Q

We denote the Fourier series of f by σ(f ) as follow: σ(f )(x) ∼

m∈Zn

1

Cm (f )eim·x .

(1.0.2)

2

C1. An introduction to multiple Fourier series

According to diﬀerent methods of summation, partial sum of Fourier series has various forms. The rectangular and spherical partial sums are the two most common forms. The rectangular partial sum is deﬁned by Cl (f )eil·x , m ∈ Zn+ , (1.0.3) Sm (f ; x) = 0≤|lj |≤mj ,j=1,...,n

which can be expressed in the integral form deﬁned as 1 Sm (f ; x) = n f (x − t)Dm1 (t1 ) · · · Dmn (tn )dt, m ∈ Zn+ , π Q where

(1.0.4)

sin v + 12 u Dv (u) = 2 sin 12 u

is the Dirichlet kernel. The spherical partial sum is deﬁned by 0 SR (f ; x) = Cm (f )eim·x ,

R > 0,

|m|
where |m| = (m21 + · · · + m2n ) 2 . Obviously, there exist various ways of only taking the limit when we discuss the convergence or summation of the multiple Fourier series. We usually discuss several kinds of limits. The rectangular limit is deﬁned by lim

m1 ,..., mn →∞

Sm (f ; x).

The restricted rectangular limit is deﬁned as the rectangular limit, but we need to restrict mj 1 ,λ ∈ mk λ as m1 , . . . , mn → ∞ for some ﬁxed constant λ ≥ 1. The spherical limit which we will mainly discuss is deﬁned as lim SR (f ; x).

R→∞

For the rectangular sum, we introduce the arithmetic means (also called by the Fej´er means). The rectangular Fej´er means of f is the trigonometric polynomial σm (f ; x) =

m1 mn 1 ··· Sj1 ,...,jn (f ; x). (m1 + 1) · · · (mn + 1) j1 =0

jn =0

(1.0.5)

1.1 Basic properties of multiple Fourier series

3

Substituting the equation (1.0.4) into the equation (1.0.5), we can obtain the integral form 1 f (x − t)Km1 (t1 ) · · · Kmn (tn )dt, (1.0.6) σm (f ; x) = n π Q where

1 1 Kj (u) = Dl (u) = j+1 2(j + 1) j

l=0

sin j+1 2 u sin u2

2

is the Fej´er kernel. For the spherical summation, we will mainly discuss the Bochner-Riesz means, that is, the following spherical trigonometric polynomial of R-th order α |m|2 α 1− 2 Cm (f )eim·x , R > 0, (1.0.7) SR (f ; x) = R |m|
where the power α is any complex number whose real part is larger than −1. α , we always restrict R > 0. For the We remark that, in the deﬁnition of SR α α sake of convenience, the notation S0 denotes lim SR , that is, S0α (f ; x) = R→0+

C0 (f ). In many aspects, it is more suitable for us to regard the spherical sum theory of multiple trigonometric series as the extension of the theory with one variable. For example, the uniqueness theory of the spherical summation of multiple trigonometric series, such as the foundational results, obtained by Cheng [Che1] illustrated these facts. As far as the convergence and approximation theory of the Fourier series is concerned, a series of facts we will introduce in the following also reﬂect this point. It is one of the reasons why we are more interested in the theory of spherical summation.

1.1

Basic properties of multiple Fourier series

In order to discuss the Fourier series, we begin with proving a basic property of Lp (Q), that is, the density of trigonometric polynomials. For the sake of convenience, we denote C(Q) by L∞ (Q).

4

C1. An introduction to multiple Fourier series

Theorem 1.1.1 The set of the trigonometric polynomials of n variables are dense in Lp (Q) for all 1 ≤ p ≤ +∞. Proof. By the fundamental facts in real analysis, we all know that C(Q) is a dense subset of Lp (Q). Hence, we ﬁrst prove that the set of trigonometric polynomials is dense in C(Q) by using the Stone theorem, and then ﬁnish the proof by noticing that the distance in Lp is not larger than the distance in C(Q). Here we complete the proof of the theorem by proving lim

m1 ,...,mn →∞

f − σm (f )p = 0.

(1.1.1)

For the sake of simplicity, let n = 2. By the equation (1.0.6), we have that f − σm1 ,m2 (f )p 1 = 2 f (x1 + t1 , x2 + t2 ) − f (x1 , x2 ) Km1 (t1 )Km2 (t2 )dt1 dt2 π Q p 1 f (· + t1 , · + t2 ) − f (·, ·) Km1 (t1 )Km2 (t2 )dt1 dt2 . ≤ 2 p π Q We denote the Lp modulus of continuity of f by ω(f ; μ, ν)p , that is, ω(f ; μ, ν)p =

sup |h|≤μ,|τ |≤ν

f (· + h, · + τ ) − f (·, ·)p .

Hence, it implies that f − σm1 ,m2 (f )p ≤

4 π2

π

π

ω(f ; t1 , t2 )p Km1 (t1 )Km2 (t2 )dt1 dt2 .

0

0

μ=

log(m1 + 1) m1 + 1

Set

and

log(m2 + 1) . m2 + 1 Let m1 , m2 > 0. We conclude that μ ν π ν μ π π π 4 + + + f − σm1 ,m2 (f )p ≤ 2 π 0 0 μ 0 0 ν μ ν ν=

× ω(f ; t1 , t2 )p Km1 (t1 )Km2 (t2 )dt1 dt2 4 = 2 (I1 + I2 + I3 + I4 ). π

1.1 Basic properties of multiple Fourier series I1 ≤ ω(f ; μ, ν)p

π

0

0

π

Km1 (t1 )Km2 (t2 )dt1 dt2 ≤

5 π2 ω(f ; μ, ν)p . 4

Using the property of modulus of continuity, we have t1 ω(f ; t1 , t2 )p ≤ ω(f ; μ, t2 )p +1 . μ Thus, it implies that π dt1 I2 =

ν

ω(f ; t1 , t2 )p Km1 (t1 )Km2 (t2 )dt2 π π t1 ≤ ω(f ; μ, ν)p Km2 (t2 )dt2 Km1 (t1 )dt1 1+ μ μ 0 π t1 dt1 ≤ Cω(f ; μ, ν)p 2 +1 μ μ (m1 + 1)t1 μ

0

≤ Cω(f ; μ, ν)p . Similarly, we have I3 ≤ Cω(f ; μ, ν)p . It follows that π dt1 I4 =

π

dt2 ω(f ; t1 , t2 )p Km1 (t1 )Km2 (t2 ) π π t1 t2 Km1 (t1 )dt1 Km2 (t2 )dt2 1+ 1+ ≤ ω(f ; μ, ν)p μ ν μ ν μ

ν

≤ Cω(f ; μ, ν)p . Consequently, we have that

log(m1 + 1) log(m2 + 1) , f − σm1 ,m2 (f )p ≤ Cω f ; m1 + 1 m2 + 1

,

(1.1.2)

p

for m1 , m2 > 0. This completes the proof of Theorem 1.1.1.

Remark 1.1.1 From the proof of the above theorem, we actually transformed the multiple problem into the repeated estimate of unitary variable. That is to say, the proof is trivial in essence. In many cases, multiple rectangular problem can be simpliﬁed into repeated unitary variable problem.

6

C1. An introduction to multiple Fourier series

Consequently, many results in the unitary case can naturally be extended to the multiple rectangular case. In this case, we can not tell any essential diﬀerence between the unitary and the multiple. We have very little interest in this situation. Theorem 1.1.2 Let f ∈ L(Q). If Cm (f ) = 0, f or any m ∈ Zn , then we have f (x) = 0, for a.e. x ∈ Q. Proof. From the equations 1 Cm (f ) = (2π)n

f (x)e−im·x dx = 0, Q

with every m ∈ Zn , we conclude that f (x)P (x)dx = 0,

(1.1.3)

Q

for any trigonometric polynomial P (x). For any ﬁxed g ∈ C(Q) and any positive number ε, by Theorem 1.1.1, there exists a trigonometric polynomial Pε such that (1.1.4) g − Pε C < ε. It follows from the equations (1.1.3) and (1.1.4) that f (x)g(x)dx = f (x)(g(x) − Pε (x))dx ≤ ε |f (x)|dx. Q

Then this implies that

Q

Q

f (x)g(x)dx = 0 Q

for any g ∈ C(Q) which implies f = 0. This completes the proof of the theorem.

Corollary 1.1.1 Let f ∈ L(Q) and

|Cm (f )| < +∞. Then there exists

m

g ∈ C(Q) such that f (x) = g(x) for almost everywhere x ∈ Q. For the space L2 (Q), the following basic theorem is the same as the one for the case n = 1.

1.1 Basic properties of multiple Fourier series

7

Theorem 1.1.3 Let f ∈ L2 (Q). Then we have

|Cm (f )|2 =

m∈Zn

1 f 22 . (2π)n

(1.1.5)

Proof. Since the proof is the same as the proof of one dimensional case, we brieﬂy write down the proof just for the review purpose here. n 2 |Cm (f )| = f (x)SR (f, x)dx (2π) Q

|m|
≤ f 2 SR (f )2 ⎛ = f 2 ⎝(2π)n

⎞1 2

2⎠

|Cm (f )|

.

|m|
This is equivalent to

|Cm (f )|2 ≤

|m|
1 f 22 (2π)n

for any R > 0. Thus SR (f ) is a Cauchy sequence of L2 (Q). Consequently, there exists g ∈ L2 (Q) such that SR (f ) → g in the sense of L2 norm. So we have σ(g) = σ(f ). It follows from Theorem 1.1.2 that g = f . On the other hand, since SR (f ) converges to f in L2 as R goes to ∞, we have SR (f )2 → f 2 . It follows that

n

(2π)

1 2

|Cm (f )|

2

= f 2 .

m∈Zn

This is our desired result.

Corollary 1.1.2 Suppose f ∈ C (k) (Q) with k > n2 . Then we have

|Cm (f )| < +∞.

8

C1. An introduction to multiple Fourier series

Proof. Let α = (α1 , . . . , αn ) ∈ Zn+ with |α|1 = α1 + · · · + αn ≤ k. Then we have ∂ α1 +···+αn f (x) Dα (f )(x) = ∈ C(Q). ∂xα1 1 · · · ∂xαnn By the integration by parts, we obtain that

Dα (f )(x)e−im·x dx = (im)α Q

f (x)e−im·x dx Q

α

= (im) (2π)n Cm (f ) for m ∈ Zn . According to Theorem 1.1.3, we have

|α|1 =k

|Cm (f )|2 |(im)α |2

< +∞.

m∈Zn

That is,

⎛ |Cm (f )|2 ⎝

m∈Zn

⎞

|m1 |2α1 · · · |mn |2αn ⎠ < +∞.

(1.1.6)

|α|1 =k

Obviously, there exists a constant C = Ck,n > 0 such that k |m|2k = |m1 |2 + · · · + |mn |2 ≤ C |m1 |2α1 · · · |mn |2αn .

(1.1.7)

|α|=k

Set

⎛ bm

=⎝

⎞1 2

2α1

|m1 |

2αn ⎠

· · · |mn |

.

|α|1 =k

Then, the inequalities (1.1.6) and (1.1.7) can be written as

|Cm (f )|2 b2m < +∞,

(1.1.8)

|m|k ≤ Cbm .

(1.1.9)

m∈Zn

and

1.1 Basic properties of multiple Fourier series

9

It implies from (1.1.9) that

|Cm (f )| ≤ C

m=0

|Cm (f )|

m=0

⎛

≤C⎝

bm |m|k ⎞1 ⎛ 2

|Cm (f )|2 b2m ⎠ ⎝

m=0

Since k > n2 , we have

m=0

m=0

⎞1 2

(1.1.10)

1 ⎠ . |m|2k

1 < +∞. |m|2k

Thus from (1.1.8) and (1.1.10), we can deduce that |Cm (f )| < +∞.

It should be pointed out that Corollary 1.1.2 gives a suﬃcient condition for the Fourier series to be absolutely convergent. Of course, this result is rather sketchy. Corollary 1.1.3 (Riemann-Lebesgue Theorem) If f ∈ L(Q), then we have lim |Cm (f )| = 0. |m|→∞

Proof. For any ε > 0, there exists g ∈ C(Q) such that the function h = f − g satisﬁes h1 < ε. Since there hold Cm (f ) = Cm (g) + Cm (h) and |Cm (h)| ≤ h1 , we have that lim |Cm (f )| ≤ lim |Cm (g)| + ε.

|m|→∞

|m|→∞

Theorem 1.1.3 impies that lim |Cm (g)| = 0.

|m|→∞

10

C1. An introduction to multiple Fourier series

Hence, we obtain lim |Cm (f )| ≤ ε.

|m|→∞

This completes the proof.

1.2

Poisson summation formula

Poisson summation formula is an important tool for investigating multiple Fourier series. Let f ∈ L(Rn ). The formal series f (x + 2πm), x ∈ Rn (1.2.1) m∈Zn

is called the periodization of the function f . If the series (1.2.1) converges to a function in some sense (some metric sense or some limit sense), then this sum function is undoubtedly 2π-periodic. Theorem 1.2.1 Let f ∈ L(Rn ). The series (1.2.1) converges absolutely almost everywhere and the sum function belongs to L(Q), with the Fourier series being f(m)eim·x . (1.2.2) m∈Zn

That is to say, f(m), m ∈ Zn , are the Fourier coeﬃcients of f (x + 2πm). m∈Zn

Proof. By Levi’s theorem, we have |f (x + 2πm)|dx |f (x + 2πm)|dx = Q Q = |f (x)|dx m =

Rn

Q+2πm

|f (x)|dx < +∞,

1.2 Poisson summation formula and 1 (2π)n = = =

11

Q

1 (2π)n 1 (2π)n 1 (2π)n

1 (2π)n = f(m).

=

f (x + 2πm ) e−im·x dx

m ∈Zn

m

m

m

Rn

f (x + 2πm )e−im·x dx Q

Q+2πm

Q+2πm

f (x)e−im·(x−2πm ) dx f (x)e−im·x dx

f (x)e−im·x dx

This completes the proof Theorem 1.2.1.

Theorem 1.2.2 Let f ∈ C(Rn ). If there exist δ > 0 and A > 0 such that |f (x)| ≤ A(1 + |x|)−n−δ , then we have

|f(y)| ≤ A(1 + |y|)−n−δ ,

f (x + 2πm) =

m∈Zn

f(m)eim·x

(1.2.3)

(1.2.4)

m∈Zn

for all x ∈ Rn . The equality 1.2.4 is the so called Poisson summation formula. Proof. By the assumption (1.2.3), we have that f ∈ L(Rn ) and |f(m)| < +∞. m∈Zn

According to Theorem 1.2.1, we get f(m)eim·x , f (x + 2πm) = m∈Zn

m∈Zn

for a.e. x ∈ Rn . Since the functions on the both sides of (1.2.3) are continuous, the above equation holds everywhere, that is, (1.2.4) holds.

12

C1. An introduction to multiple Fourier series We need the following formulas which need weaker assumptions.

Theorem 1.2.3 Suppose that f ∈ L(Rn ) is continuous except at x = 0. Let f (x)eix·y dx. (1.2.5) g(y) = Rn

If the following three conditions (i)

(ii)

|f (m)| < +∞;

|g(y)| = o (iii) the series

1 |y|n−1

∞

(1.2.6)

, |y| → ∞;

g(x + 2πm)

(1.2.7)

(1.2.8)

j=0 m∈Sj

is locally uniformly convergent on Rn , where Sj := {x ∈ Zn : j ≤ |x| < j + 1}, are satisﬁed, then ∞

g(x + 2πm) =

f (m)eim·x + C0 ,

(1.2.9)

j=0 m∈Sj

holds for any x ∈ Rn , where −n

C0 = (2π)

lim

ρ→∞ |y|<ρ

g(y)dy.

Proof. By the deﬁnition (1.2.5), we have g ∈ C(Rn ). According to the condition (iii), we obtain that the series (1.2.8) is a continuous 2π-periodic function, which we denote by G(x). We now compute the Fourier coeﬃcients of G Cm (G) = (2π)−n

G(x)e−im·x dx.

Q

1.2 Poisson summation formula

13

It easily follows from the condition (iii) that Cm (G) =

∞ j=0 μ∈Sj

Let ρ > 2nπ + 1 and N =

ρ 2π

1 (2π)n

g(x)e−im·x dx. Q+2πμ

. We obviously have

N −im·x −im·x g(x)e dx − g(x)e dx |x|<ρ j=0 μ∈Sj Q+2πμ ≤ |g(x)|dx.

(1.2.10)

ρ−2nπ<|x|<ρ+2nπ

The condition (ii) shows that the quantity on the right side of (1.2.10) is o(1) as ρ → ∞. Hence, we have 1 g(x)e−im·x dx Cm (G) = lim ρ→∞ (2π)n |x|<ρ (1.2.11) 1 im·x = lim g(−x) e dx. ρ→∞ |x|<ρ (2π)n Specially, C0 (G) = C0 . By (1.2.5), we get that 1 g(−x) = f(x). (2π)n Since f is continuous except at x = 0, we have for each m = 0, f(x)eim·x e−ε|x| dx = f (m). (1.2.12) lim ε→0+

Rn

Let ϕ(t) =

|x|=t

and

f (x)eim·x dσ(x)

Φρ (t) =

ϕ(τ )dτ, ρ

for 0 < ρ ≤ t < ∞.

t

14

C1. An introduction to multiple Fourier series According to (1.2.11) and (1.2.12), we get respectively ρ ϕ(t)dt = Cm (G) lim ρ→∞ 0

and

∞

lim

ε→0 0

ϕ(t)e−εt dt = f (m),

(1.2.13)

(1.2.14)

for m = 0. By (1.2.13), we obtain that for any η > 0, there exists M > 0 such that |Φρ (t)| < η, if M ≤ ρ ≤ t < ∞. Since there holds ∞ ∞ ∞ −εt −εt ϕ(t)e dt = Φρ (t)e + ε Φρ (t)e−εt dt ρ ρ ρ ∞ = Φρ (t)εe−εt dt, ρ

we have that

∞

−εt

ϕ(t)e ρ

dt ≤ η

∞

εe−εt dt = ηe−ερ ,

ρ

which implies lim ε→0

∞

−εt

ϕ(t)e

ρ

dt < η, M ≤ ρ ≤ t < ∞.

By (1.2.13), we can choose ρ > M big enough, such that ρ Cm (G) − ϕ(t)dt < η. 0

By making use of (1.2.14), we have that for m = 0, ∞ ρ ϕ(t)dt + lim ϕ(t)e−εt dt. f (m) = 0

ε→0 ρ

Hence, we have |Cm (G) − f (m)| < 2η,

m = 0,

which leads to Cm (G) = f (m), m = 0.

1.3 Convergence and the opposite results

15

Now, by the condition (i), we know that the function on the right side of (1.2.9) is a continuous periodic function, whose Fourier coeﬃcient is Cm (G). So this function is G(x), that is to say, the equation (1.2.9) holds. This completes the proof.

Remark 1.2.1 From the proof, we can conclude that the continuity of f is not intrinsic, as long as f is continuous at non-zero grid points. Actually the conditions can be much weaker, as long as those points are Lebesgue points. Remark 1.2.2 We can make an explanation about Theorem 1.2.3 in aspect of form: the function g deﬁned by (1.2.5) can be regarded as the Fourier inverse transform of f , then gˆ = f . Now we can trace back to the equation (1.2.4) in Theorem 1.2.2, just executing the summation in spherical method for the left side. The three theorems in this section are all called the Poisson summation formulas.

1.3

Convergence and the opposite results

For the convergence of Fourier series, the most wonderful result in the unitary case has been established by Carleson-Hunt. They have conﬁrmed the conjecture posed by Lusin, that is, f ∈ Lp ([−π, π)) ⇒ Sm (f, x) → f (x) holds for a.e. x ∈ [−π, π) and 1 < p < ∞, as m → ∞. The case of multi-dimension is much more complicated. The convergence of the multi-dimensional rectangular partial sum is not good. Feﬀerman [Fe2] proved that there exists a continuous periodic binary function f (x, y), whose rectangular Fourier partial sum Sm,n (f ; x, y) diverges everywhere as m → ∞ and n → ∞. However, it is much better when we take the square limit. At this time, the multi-dimensional analogy of the conjecture still holds. In the 1970s, many researchers have proven that f ∈ Lp (Q) ⇒ Sk,...,k (f ; x) → f (x) holds for a.e. x ∈ Q and 1 < p < ∞, as k → ∞.

16

C1. An introduction to multiple Fourier series Later, Sj¨olin [Sj1] moved forward with the above results and proved that f ∈ L(log+ L)n log+ log + L(Q) ⇒ Sk,...,k (f ; x) → f (x)

holds for a.e. x ∈ Q, as k → ∞. Here, we prove the following theorem about the convergence of the spherical sum.

2n Theorem 1.3.1 If n > 1, then for every p ∈ 1, n+1 , there exists f ∈ p L (Q) such that 0 (f ; x) = +∞, (1.3.1) lim sup SR R→∞

for a.e. x ∈ Q. Definition 1.3.1 We call a set A in R a linearly independent set if for any ﬁnite elements of A, a1 , . . . , am , the linear combination with integer coeﬃcients p1 a1 + · · · + pm am equals to zero if and only if the coeﬃcients p1 = p2 = · · · = pm = 0. Definition 1.3.2 Let x ∈ Rn . If {|x + 2πm| : m ∈ Zn } is a linearly independent set, then we say x ∈ S and denote by S c = Rn \S. Lemma 1.3.1 Let n > 1. Then |S c | = 0. Proof. Arbitrarily arrange the elements of Zn as {mj }∞ j=1 . Let A = {A = (a1 , . . . , ak ) : aj ∈ Zn , j = 1, . . . , k, k ∈ N}. Choose any A ∈ A such that A = 0 := (0, . . . , 0) and let ΦA (y) =

k

aj |y + 2πmj |.

j=1

It is obvious that ΦA (y) is an analytic function deﬁned on simply connected domain Rn \2πZn and continuous on Rn . For A = 0, then ΦA is not zero function (notice that ΦA (y) is unbounded near some points). Denote

1.3 Convergence and the opposite results

17

the set of all the zero points of ΦA by ZA . By the theory of analytic function, we have |ZA | = 0. Then we also have Z A = 0, A∈A

together with the fact Sc =

ZA ,

A∈A

this leads to

|S c

| = 0.

Lemma 1.3.2 Suppose that K(t) is a real bounded measurable function deﬁned on (1, ∞) and for every λ ≥ 0, there always exists 1 T lim K(t)eiλt dt = b(λ). (1.3.2) T →+∞ T 1 If the set {λ ≥ 0 : b(λ) = 0} is countable, denoted by {λj }∞ j=1 , and linearly independent, then we have ∞

|b(λj )| < ∞.

(1.3.3)

j=1

Proof. Let b(λj ) = |b(λj )|eiμj and A(t) =

N

(1 + cos(λj t − μj )).

j=1

It is obvious that A(t) ≥ 0. By the linear independence of {λj }∞ j=1 , we get 1 lim T →∞ T Since we have T A(t)K(t)dt = 1

1

T

T

A(t)dt = 1. 1

K(t) 1 + cos(λ1 t − μ1 ) + · · · + cos(λN t − μN )

+ cos(λ1 t − μ1 ) · cos(λ2 t − μ2 ) + · · · + cos(λ1 t − μ1 ) · · · cos(λN t − μN ) dt,

18

C1. An introduction to multiple Fourier series

by deﬁnition of {λj }∞ j=1 , it follows that 1 T →∞ T

lim

1

T

1 T →∞ T

K(t)A(t)dt = lim

=

T

K(t) 1

N

cos(λj t − μj )dt

j=1

N

Reb(λj ) · cos μj + Imb(λj ) · sin μj

j=1

=

N

|b(λj )|.

j=1

Consequently, we have N j=1

1 |b(λj )| ≤ lim T →∞ T

T 1

|K(t)|A(t)dt

≤ sup{|K(t)| : t > 0} < +∞. Thus (1.3.3) holds and we complete the proof of Lemma 1.3.2.

We deﬁne α (x) DR

=

|m|
|m|2 1− 2 R

α eim·x ,

(1.3.4)

for Reα > −1. Lemma 1.3.3 If Reα >

n−1 2 ,

then we have

n 2

α DR (x) = 2α Γ(α + 1) 2π R

n −α 2

J n +α (Rrm ) 2 n

m∈Zn

+α

2 rm

where rm = |x + 2πm|. Proof. In Section 1.2, we introduce α 1 − |x|2 , α Φ (x) = 0,

|x| < 1, |x| ≥ 1,

,

(1.3.5)

1.3 Convergence and the opposite results

19

and obtain that α 1) J n2 +α (|x|) α (x) = 2 Γ(α + B α (x) := Φ . n n (2π) 2 |x| 2 +α n−1 2 ,

Due to the fact that Reα >

we have B α ∈ L(Rn ) which leads to

α (x). Φα (x) = F ((2π)n B α (−y)) (x) = (2π)n B According to Theorem 1.2.2, we have that F ((2π)n B α (Ry)Rn ) (m)eim·x = (2π)n B α (R(x + 2πm))Rn , m∈Zn

m∈Zn

that is, m∈Zn

Φα

m R

n

eim·x = 2α Γ(α + 1)(2π) 2 Rn

J n +α (R|x + 2πm|) 2 n

m∈Zn

(R|x + 2πm|) 2 +α

,

which is just the equation (1.3.5). This completes the proof.

Lemma 1.3.4 Let x ∈ Rn with x = 2πm, f or any m ∈ Zn . Then, for λ ≥ 0, we have ⎧ ⎨ C , if λ = r , f or m ∈ Zn , 1 T n−1 m iλR 2 DR (x)e dR = λn lim (1.3.6) ⎩ T →∞ T 1 0, if λ = rm , f or m ∈ Zn ,

where n−1

C=2 Proof. Let α ∈

n−1 2

π

n−1 2

Γ

n

= Cα,n R

n−1 −α 2

ei

πn 2

.

J n +α (Rrm ) 2

cos(Rrm + θn ) n+1 +α 2

n+1

rm2

n

+α

2 rm

rm cos(Rrm + θn ) n−1 −α 2

, n+1 2 . By Lemma 1.3.3, we get

α (x) = 2α Γ(α + 1)(2π)n/2 R 2 −α DR

= Cα,n R

n+1 2

+α

+R

n−3 −α 2

O

1 +O , R

1 n+3

rm2

+α

(1.3.7)

20

C1. An introduction to multiple Fourier series

as R → ∞, where Cα,n = 2

n+1 +α 2

π

n−1 2

π π n Γ(α + 1), θ = − ( + α) − , 2 2 4 n−1 2 .

and ‘O’ is uniformly valid for α > 1 T

T 1

α DR (x)eiλR dR

= Cα,n

+O

Thus, we have 1 T

1 log T T

T

1

cos(Rrm + θ) Rα−

n−1 2

n+1

rm2

+α

iλR

e

.

dR (1.3.8)

Let ϕαT (m, λ)

1 = T

T

cos(Rrm + θ) Rα− 1

1

1 = 2T

T 1

R

n−1 2

eiλR dR

ei[(λ+rm )R+θ] + ei[(λ−rm )R−θ] dR.

α− n−1 2

By the integration by parts, we have that 1 log T = O (1.3.9) |λ − rm | T n+1 holds uniformly for α ∈ n−1 and λ = rm , as T → ∞. 2 , 2 + Therefore, if λ ∈ R \{rm : m ∈ Zn }, then we have ⎛ ⎞ 1 1 T α log T iλR . DR (x)e dR = ⎝Cα,n + 1⎠ O n+1 +α T 1 T 2 m∈Zn |λ − rm |rm |ϕαT (m, λ)|

By taking α →

n−1 2 ,

1 T

we get

T

n−1 2

DR

1

iλR

(x)e

Let λ = rm . Then we have for α ∈ ϕαT (m, rm )

1 = 2T

T

R

n−1 −α 2

1

1 = e−iθ 2

T 1 n+1 2 −α

n−1

−iθ

e

dR = O

2

T

.

(1.3.10)

, , n+1 2

1 dR + 2T

n+1 −α 2

log T T

−1

T

R 1

+O

n−1 −α 2

log T T

ei(2rm R+θ) dR 1 rm

(1.3.11)

1.3 Convergence and the opposite results

21

holds uniformly. Substituting (1.3.9) and (1.3.11) into (1.3.8), we obtain 1 T

T 1

n+1

T 2 −α − 1 Cα,n 1 n+1 n+1 +α T 2 −α rm2 ⎛ ⎞ log T ⎝ 1 Cα,n +O + 1⎠ . n+1 +α T 2 n l∈Z ,l=m |rl − rm |rl

1 α DR (x)eirm R dR = e−iθ 2

By taking α → 1 T

1

T

n−1 2 , n−1 2

we have irm R

DR (x)e

C T −1 +O dR = n rm T

log T T

.

(1.3.12)

Combining (1.3.10) with (1.3.12), we obtain (1.3.6). This completes the proof of Lemma 1.3.4.

Lemma 1.3.5 Let x0 ∈ S. Then we have n−1 0 2 lim sup DR (x ) = +∞.

(1.3.13)

R→∞

n−1

Proof.

Suppose that (1.3.13) is not valid, that is, DR2 (x0 ) is bounded n−1

about R on (0, ∞). In fact, DR2 (x0 ) is a real measurable function about R. By Lemma 1.3.4, we get that 1 T n−1 DR2 (x0 )eiλR dR lim T →∞ T 1 exists everywhere about λ ≥ 0 and the set of non-zero points is {rm = |x0 + 2πm| : m ∈ Zn }. For x0 ∈ S, the set {rm : m ∈ Zn } is linearly independent. According to Lemma 1.3.2, we have |bm | < ∞, m∈Zn

where bm

1 = lim T →∞ T

T 1

n−1

DR2 (x0 )eirm R dR =

C , n rm

22

C1. An introduction to multiple Fourier series

for C = 0. However, it is evident that 1 1 = = +∞. n 0 + 2πm|n r |x n m n

m∈Z

m∈Z

This contradiction shows that (1.3.13) holds.

Lemma 1.3.6 (M. Riesz convexity theorem) Let x0 ∈ Rn and t > 0. We deﬁne 0 α Aα (t) = tα D√ (x0 ) = (t − |m|2 )α eim·x . t √ |m|< t

Suppose that V (t), W (t) are both positive increasing functions on (0, +∞) and let Uθ (t) = (V (t))1−θ (W (t))θ , for 0 ≤ θ ≤ 1. If

|A0 (t)| ≤ V (t), |Aα (t)| ≤ W (t),

for α > β > 0, then we have |Aβ (t)| ≤ CU β (t), α

where C = Cα,β is a constant. Proof. Because of |A0 (t)| ≤ V (t), V (t) is a non-negative increasing function and t β (t − s)β−1 A0 (s)ds. (1.3.14) A (t) = β 0

Then we get that |Aβ (t)| ≤ tβ V (t). Therefore, if

t ≤ β

or that is,

t≤

W (t) V (t)

W (t) V (t)

β

α

,

1

α

,

1.3 Convergence and the opposite results

23

then the theorem turns to be trivial. Hence, we can assume that there exists ξ(t) > 0 such that 1 W (t) α . t−ξ = V (t) On one hand, for the case 0 < α ≤ 1, we have t β (t − s)β−1 A0 (s)ds A (t) = β 0 ξ t =β + (t − s)β−1 A0 (s)ds 0

ξ

= J1 + J2 . For J2 , we have the estimate t β−1 0 (t − s) A (s)ds ≤ (t − ξ)β V (t) = U β (t). |J2 | = β α

ξ

Since we have

ξ

(t − s)β−α (t − s)α−1 A0 (s)ds 0 ξ β−α = β(t − ξ) (t − s)α−1 A0 (s)ds, 0 ≤ u ≤ ξ.

J1 = β

u

β = (t − ξ)β−α (Aα (ξ) − Aα (u)), α we have the estimate of J1 as β β |J1 | ≤ 2 (t − ξ)β−α W (t) = 2 U β (t). α α α Consequently, we get the conclusion |Aβ (t)| ≤ 3U β (t). α

On the other hand, for the case α > 1 and h = [α], let ζ=

t−ξ , α − h = δ ∈ [0, 1). h+1

We ﬁrst assume that β ∈ Z+ and use induction to prove the statement. Clearly the conclusion holds for β = 0.

24

C1. An introduction to multiple Fourier series We assume that β ∈ N and

Aβ−1 (t) = O U β−1 (t) . α

Then, for ξ ≤ t < t, it follows that

A (t) − A (t ) = β β

β

t

Aβ−1 (s)ds

t

= O (t − t ) · U β−1 (t) α 1 β−1 β−1 W (t) α · W (t) α V (t)t− α =O V (t)

= O U β (t) . α

(1.3.15)

In the following, we deﬁne the ﬁnite diﬀerence. For m ∈ Z+ , m ≤ h, h m −ζ A (t)

m ν h = (−1)ν Cm A (t − νζ). ν=0

For δ ∈ (0, 1), we deﬁne δ−ζ f (t) = δ

t

(t − s)δ−1 f (s)ds.

t−ζ

It is obvious that m δ δ−ζ · m −ς = −ζ · −ς . m+δ . Since we have We denote the operation by −ζ

Ah−1 (t) = =

√ |m|< t

m∈Zn

(t − |m|2 )h−1 eim·x

0

0

χ[0,√t) (|m|)(t − |m|2 )h−1 eim·x ,

1.3 Convergence and the opposite results

25

we conclude that

t

Ah−1 (s)ds t−ζ

⎛

=⎝

=

√ |m|< t−ζ

√ |m|< t−ζ

+ =

+ t

⎠

√ t−ζ≤|m|

t

0

χ[0,√s) (|m|)(s − |m|2 )h−1 eim·x ds

t−ζ

(s − |m|2 )h−1 dseim·x

0

t−ζ

√

⎞

t

√ |m|2 t−ζ≤|m|< t

(s − |m|2 )h−1 ds eim·x

0

1 h 1 (A (t) − Ah (t − ζ)) = −ζ Ah (t). h h

Then we obtain by induction that h m −ζ A (t)

Γ(h + 1) = Γ(h − m + 1)

t

dt1 t−ζ

t1 t1 −ζ

dt2 · · ·

tm−1

tm−1 −ζ

Ah−m (tm )dtm

Γ(h + 1) m h−m ζ A = (t) Γ(h − m + 1) tm−1 t dt1 · · · Ah−m (tm ) − Ah−m (t) dtm , + tm−1 −ζ

t−ζ

and m+δ h −ζ A (t) =

Γ(h + 1) Γ(h + 1) ζ m δ−ζ Ah−m (t) + δ Γ(h − m + 1) Γ(h − m + 1) −ζ

tm−1 t h−m h−m × dt1 · · · (tm ) − A (t) dtm . A t−ζ

tm−1 −ζ

In the above equation, the symbol Ah−m (t) denotes the value of Ah−m at t, which does not participate in the operation of ﬁnite diﬀerence. We hence get δ−ζ (Ah−m (t)) = Ah−m (t)ζ δ . Substituting this into the above equation and transposing the terms, we have

26

C1. An introduction to multiple Fourier series

ζ m+δ Ah−m (t) =

Γ(h − m + 1) m+δ h −ζ A (t) Γ(h + 1)

tm−1 t δ h−m h−m dt1 · · · (tm ) − A (t) dtm . A − −ζ tm−1 −ζ

t−ζ

(1.3.16) Substitute m = h − β into (1.3.16) and note β ∈ Z+ . If δ = 0, then h = [α] = α. By the condition |Aα (f )| ≤ W (t), we have that m+δ h A (t) = O(W (t)). −ζ If δ ∈ (0, 1), then we ﬁrst consider δ−ζ Ah (t). Let 0 ≤ s < u < t and θ(t, u; v) =

(t − u)δ , v ∈ (−∞, u). Γ(δ)Γ(1 − δ)(t − v)(u − v)δ

Then we get

u

θ(t, u; v)(v − s)

δ−1

s

u (v − s)δ−1 (t − u)δ dv = dv Γ(δ)Γ(1 − δ) s (t − v)(u − v)δ 1 vδ−1 (t − u)δ

dv. = t−s Γ(δ)Γ(1 − δ)(u − s) 0 − v (1 − v)δ u−s

(1.3.17) Set a =

u−s t−s

∈ (0, 1). By the expansion ∞

1 = (av)k , 1 − av k=0

we obtain that

1 0

t−s u−s

∞ 1 vδ−1 dv = a vk+δ−1 (1 − v)−δ dvak δ − v (1 − v) k=0 0

=a

∞ Γ(k + δ)Γ(1 − δ) k=0

Γ(k + 1)

ak

= aΓ(δ)Γ(δ)(1 − a)−δ = Γ(δ)Γ(1 − δ)

(u − s)(t − s)δ−1 . (t − u)δ

1.3 Convergence and the opposite results

27

Substituting this into (1.3.17), we have (t − s)

δ−1

u

=

θ(t, u; v)(v − s)δ−1 dv.

(1.3.18)

s

For 0 < u < t, by making use of (1.3.18), we get

u 0

A (s)ds = θ(t, u; v)(v − s) dvA (s) ds 0 s v (1.3.19) u = θ(t, u; v) (v − s)δ−1 Ah (s) dv.

(t − s)

δ−1

u u

h

δ−1

0

h

0

It is easy to deduce the following formula from the equation (1.3.14), h+δ

A

Γ(h + δ + 1) (t) = Γ(h + 1)Γ(δ)

t 0

(t − s)δ−1 Ah (s)ds,

(1.3.20)

where h, δ and t are arbitrary positive numbers, which yields v Γ(h + 1)Γ(δ) α A (v), α = h + δ. (v − s)δ−1 Ah (s)ds = Γ(h + δ + 1) 0

(1.3.21)

Substituting this into (1.3.19), then we get

u

(t − s)

δ−1

0

Γ(h + 1)Γ(δ) A (s)ds = Γ(h + δ + 1)

u

h

θ(t, u; v)Aα (v)dv.

(1.3.22)

0

Therefore, we get the conclusion that δ−ζ Ah (t)

t

(t − s)δ−1 Ah (s)ds

=δ t−ζ t

=δ 0

(t − s)

δ−1

A (s)ds − δ

t−ζ

h

0

(t − s)δ−1 Ah (s)ds.

By (1.3.20) and (1.3.21), we derive from the above equation that δ−ζ Ah (t)

Γ(h + 1)Γ(δ + 1) = Γ(h + δ + 1)

A (t) −

t−ζ

α

θ(t, t − ζ; v)A (v)dv . α

0

(1.3.23) Due to the condition |Aα (v)| ≤ W (v) ≤ W (t), 0 ≤ v ≤ t,

28

C1. An introduction to multiple Fourier series

we have

u

u θ(t, u; v)A (v)dv ≤ W (t) θ(t, u; v)dv 0 u θ(t, u; v)dv ≤ W (t) α

0

(1.3.24)

−∞

= W (t), where we use the fact θ(t, u; v) > 0 and

u −∞

(t − u)δ dv = Γ(δ)Γ(1 − δ)(t − v)(u − v)δ

∞ 0

1 ds = 1. Γ(δ)Γ(1 − δ)(1 + s)sδ

Combining (1.3.23) and (1.3.24), we have δ−ζ Ah (t) = O(W (t)). Consequently, taking account of m = h − β, α = h + δ, we have

α−β h δ h A (t) = h−β A (t) = O(W (t)). −ζ −ζ −ζ

By the above discussion, the ﬁrst term on the right side of (1.3.16) is O(W (t)). In the integral of the second term on the right side of (1.3.16), it is obvious that t ≥ tm ≥ t − mζ. Since ζ= and mτ =

t−ξ h+1

h−β (t − ξ) < t − ξ, h+1

we have ξ < tm < t. Therefore, by (1.3.15), we have

h−m (tm ) − Ah−m (t) = O U β (t) A α

1.3 Convergence and the opposite results

29

and

t t−τ

dt1 · · ·

tm−1 tm−1 −τ

Aβ (tm ) − Aβ (t) dtm = O U β (t) τ m , α

where β = h − m.

The second term on the right side of (1.3.16) is O U β (t) τ α−β , then α we have

τ α−β Aβ (t) = O(W (t)) + τ α−β O U β (t) . α

By noticing τ=

1 (t − ξ) h+1

and

t−ξ =

we get

W (t) V (t)

1

α

,

Aβ (t) = O U β (t) . α

This is included in case (ii) when β ∈ Z+ . Now assume that β is not an integer and let β = [β] + γ, 0 < γ < 1. If [β] < [α], then by the proven result, we have that

A[β] (t) = O U [β] (t) α

and

A[β]+1 (t) = O U [β]+1 (t) . α

Paying attention to (1.3.20), we apply the interpolation result between [β] A (t) and A[β]+1(t) as in case (i), and then have

γ U 1−γ U (t) . Aβ (t) = O (t) = O U β [β] [β]+1 α

α

α

If [β] = [α], then we directly interpolate between A[β] and Aα and get

γ γ Aβ (t) = O (U [β] (t))1− δ (W (t)) δ , α

where δ = α − [α] ≥ γ > 0.

30

C1. An introduction to multiple Fourier series Consequently, we conclude that [α] γ γ 1− [α] (1− γδ ) α W (t) α (1− δ )+ δ A (t) = O V (t)

β β = O V (t)1− α W (t) α

= O U β (t) . β

α

This completes the proof of Lemma 1.3.6.

Remark 1.3.1 Lemma 1.3.6 is actually a part of Theorem 1.71 by Chandrasekharan and Minakshisundaram [CM1]. Lemma 1.3.7 Let x0 ∈ S. Then we have 1

sup R>1

R

n−1 2

0 0 DR (x ) = +∞.

(1.3.25)

Proof. Suppose that (1.3.25) is false, that is, there exists M > 0 such that 0 0 DR (x ) ≤ M R n−1 2 , R ≥ 1. We use the notations in Lemma 1.3.6, that is, 0 0 A (t) = D√ (x0 ) ≤ M t n−1 4 (t ≥ 1). t On the other hand, by (1.3.7), we get that n+1 n+1 n+1 n 2 2 √2 0 (t) = t D t (x ) ≤ Ct 2 . A Then according to Lemma 1.3.6 and interpolating at β = that n−1 n−1 n−1 n n−1 n−1 2 (t) ≤ Ct 4 (1− n+1 ) t 2 · n+1 = Ct 2 , A

n−1 2 ,

we obtain

for t > 1, which contradicts with the conclusion in Lemma 1.3.5. And thus (1.3.25) holds.

1.3 Convergence and the opposite results Lemma 1.3.8 Let x0 ∈ S. Then there holds 1 im·x 0 sup n−1 e = +∞. 2 R>0 |m| 0<|m|
31

(1.3.26)

Proof. We suppose that the item on the left side of (1.3.26) is a ﬁnite number M . Let eim·x0 σt = n−1 , 2 |m| 0<|m| 0, which is left-continuous step function on (0, +∞). It obviously follows that R n−1 im·x0 e = t 2 dσt 1≤|m|
1

n − 1 R n−3 =R σR − σ1 − t 2 σt dt 2 1 n−1 n − 1 R n−3 2 =R σR − t 2 σt dt. 2 1 n−1 2

Then we have n−1 n−3 n−1 n−1 R im·x0 e M t 2 dt ≤ 2M R 2 . ≤ MR 2 + 2 1 1≤|m|
n−1 2

0 |DR (x0 )| ≤ 2M,

for R ≥ 1, which contradicts with Lemma 1.3.7 and thus (1.3.26) is valid. Corollary 1.3.1 The equality (1.3.26) is valid almost everywhere on Rn . Lemma 1.3.9 If 0 < α < n, then the series m=0

1 im·x e |m|α

(1.3.27)

is the Fourier series of the function F = Fα , where Fα has the same singular 1 and Fα ∈ C(Q\{0}). properties at x = 0 as |x|n−α

32

C1. An introduction to multiple Fourier series

Proof. Let η ∈ C ∞ (Rn ) satisfying 1 , if |x| ≥ 1 , η(x) = 0 , if |x| < 12 . Deﬁne G(x) = If x = 0, we have G(x) =

η(x) . |x|α

1 1 + (η(x) − 1) α . α |x| |x|

By a theorem on the the Fourier transforms of radial functions (see [SW1]), we get that as a generalized slowly increasing function, the Fourier transform of |x|1α is 1 1 (y) = γα n−α , F (1.3.28) α |x| |y| where γα =

1 Γ( n−α 2 ) . α n/2 Γ(α/2) 2 π

On the other hand, (η(x) − 1)

1 |x|α

is an integrable function supported on {x : |x| < 1}, whose normal Fourier transform denoted by b1 , and b1 ∈ C0∞ (Rn ), where C0∞ (Rn ) refers to a class of inﬁnitely diﬀerentiable functions whose derivatives of any order all vanish at inﬁnity. We deﬁne γα n f (y) = (2π) + b1 (y) . (1.3.29) |y|n−α For any β = (β1 , . . . , βn ) ∈ Zn+ and γ = (γ1 , . . . , γn ) ∈ Zn+ , the Fourier transform is as follows: " " ! ! F (−ix)γ (Dβ f )(x) (y) = D γ F (D β f ) (y) " ! = D γ (iy)β F (f ) (y) " ! = D γ (iy)β G(y) . Here, we utilize a theorem on the operation law of the Fourier transform (see [SW1]).

1.3 Convergence and the opposite results

33

Since G(y)(iy)β vanishes when |y| < 1/2 and equals to (iy)β when |y| > 1, for any β ∈ Zn+, as long as γ ∈ Zn+ big enough, we can make sure that

D γ (iy)β G(y) ∈ L(Rn ). Then we have

# "$ !

(2π)n F Dγ (iy)β G(y) (−x) = (−ix)γ Dβ f (x).

The left side of the above equation is the normal the Fourier transform of the function in L(Rn ) and thus it is bounded. Taking β = 0, we get that (1.3.30) |f (x)| = O |x|−N , as |x| → ∞, for any N > 0. So we obtain that f ∈ L(Rn ) and hence it follows from f = G in normal sense that f(0) = G(0) = 0 and

f(m) =

From (1.3.30),

1 |m| ≥ 1. |m|α

f (x + 2πm)

m=0

is uniformly convergent on Q. Set b(x) = (2π)n b1 (x) +

f (x + 2πm),

m=0

which leads to b ∈ C(Rn ). Moreover, by Theorem 1.2.1, the Fourier series of the function in L(Q) f (x + 2πm) Fα (x) := is σ(Fα )(x) =

f(m)eim·x =

m=0

1 im·x e . |m|α

Obviously, if x = 0, we have F (x) = Fα (x) =

(2π)n γα + b(x). |x|n−α

(1.3.31)

34

C1. An introduction to multiple Fourier series

This completes the proof.

The proof of Theorem 1.3.1. Suppose that n > 1 and denote the value n−1 of (2π)n γα at α = n−1 2 by γ. Taking α = 2 , we have γ

F (x) = where b ∈ have

C(Rn

|x|

n+1 2

+ b(x),

2n )). Then for p ∈ 1, n+1 , we have F ∈ Lp (Q). Since we

0 SR (F ; x) =

0<|m|
1 |m|

n−1 2

eim·x ,

by the above equation and Corollary 1.3.1. This ﬁnishes the proof of Theorem 1.3.1. At present, it is still an unresolved problem whether the Fourier series of Lp (Q), p ≥ 2, n > 1 is convergent almost everywhere.

1.4

Linear summation

Similar to the unitary case, in order to investigate functions by the method of Fourier series which has bad convergence property, we introduce various kinds of linear means. We only discuss some special linear means here. Suppose that Φ ∈ C(Rn ) satisﬁes (i) |Φ(x)| ≤ A(1 + |x|)−n−δ , δ > 0; (ii) Φ(0) = 1. For f ∈ L(Q) and ε > 0, we deﬁne the series by σεΦ (f )(x) =

Φ(εm)Cm (f )eim·x ,

(1.4.1)

m∈Zn

which is an absolutely convergent series by the condition (i), which is called the Φ means of σ(f ). We consider the relation between σεΦ (f )(x) and f (x) as ε → 0+ .

1.4 Linear summation

35

Theorem 1.4.1 Suppose f ∈ Lp (Q) (1 ≤ p ≤ ∞). If Φ also satisﬁes the following condition: (iii) |Φ(y)| ≤ A(1 + |y|)−n−δ , δ > 0, then (1.4.1) converges to f in the norm of Lp (Q) and converges to f at the Lebesgue points of f . Proof. By the condition (iii), we have that the function ϕ(y) := (2π)n Φ(−y) (∈ C(Rn ))

(1.4.2)

belongs to L(Rn ). It follows that ϕ = Φ. By letting ϕε (y) = ε−n ϕ(ε−1 y), we get ϕ ε (x) = Φ(εx). Since ϕε satisﬁes the conditions of Theorem 1.2.2, we have ϕ0ε (x + 2πm) = Φ(εm)eim·x ∈ C(Q). (1.4.3) Kε (x) := Taking convolution on Q, we get that 1 f (y)Kε (x − y)dy f ∗ Kε (x) := (2π)n Q 1 −im·y im·x f (y)e dyΦ(εm)e = (2π)n Q m∈Zn = Cm (f )Φ(εm)eim·x =

m∈Zn σεΦ (f )(x).

(1.4.4)

By the generalized Minkowski’s inequality, we have Φ σε (f ) ≤ 1 f p Kε 1 . p (2π)n It easily follows that σεΦ is a bounded linear operator from Lp (Q) to Lp (Q) and its operator norm satisﬁes that Φ σε ≤ 1 Kε 1 (2π)n 1 ≤ |ϕε (x + 2πm)|dx (2π)n m∈Zn Q 1 |ϕε (x)|dx = (2π)n Rn 1 |ϕ(x)|dx. (1.4.5) = (2π)n Rn

36

C1. An introduction to multiple Fourier series

Furthermore, for any triangle polynomial T (x), by the condition (ii), we have lim Φ(εm)Cm (T )eim·x lim σεΦ (T )(x) = ε→0+

m∈Zn

ε→0+

= T (x) holds uniformly for x ∈ Q. Therefore, σεΦ converges to the identity operator in the Lp (Q) norm. Then we obtain σεΦ (f ) → f, in the sense of Lp norm, as ε → 0+ , for any f ∈ Lp (Q). Now we turn to proving the convergence of σεΦ at the Lebesgue points. We might as well take the origin point into consideration. By (1.4.3) and (1.4.4), we obtain σεΦ (f )(0)

1 = f (y)Kε (−y)dy (2π)n Q 1 f (y)ϕε (−y)dy = (2π)n Q 1 + f (y) ϕε (−y + 2πm)dy. (2π)n Q m=0

For m = 0, y ∈ Q, we have

−n −y + 2πm |ϕε (−y + 2πm)| = ε ϕ ε −y + 2πm −n−δ −n ≤ Aε 1+ ε = Aεδ (ε + | − y + 2πm|)−n−δ ≤A

which yields

εδ , |m|n+δ

1 ϕε (−y + 2πm) ≤ Aεδ ≤ Cεδ . n+δ |m| m=0 m=0

It follows from the above estimate that 1 f (y)ϕε (−y)dy + O(εδ ). σεΦ (f )(0) = |Q| Q

1.4 Linear summation

37

By the condition (ii), that is,

1 (2π)n we get 1 |Q|

Rn

ϕ(y)dy = 1,

Rn

ϕε (−y)dy = 1.

Hence, we have σεΦ (f )(0) − f (0) =

1 |Q|

Rn

(f (y)xQ (y) − f (0))ϕε (−y)dy + O(εδ ).

For the kernel ϕε , we have the following estimate

|ϕε (−y)| ≤ Cε−n 1 +

|y| ε

−n−δ

⎧ −n ⎪ ⎨ Cε , ≤ εδ ⎪ ⎩ C n+δ , |y|

if |y| < ε, if |y| ≥ ε.

Since x = 0 is the Lebesgue point of f , for any η > 0, there exists 0 < σ < 1 such that, for r ∈ (0, σ), we have 1 rn

|y|
|f (y) − f (0)|dy < η.

Letting ε > σ, we have (f (y)xQ (y) − f (0)) ϕε (−y)dy

(f (y)xQ (y) − f (0)) ϕε (−y)dy + + = Rn

|y|<ε

ε≤|y|<σ

|y|≥σ

:= I1 + I2 + I3 . We have the estimates of I1 , I2 and I3 as |I1 | ≤

|y|<ε

|f (y) − f (0)|

C dy < Cη, εn

38

C1. An introduction to multiple Fourier series

Cεδ |f (y) − f (0)| n+δ dy |y| ε≤|y|<σ

δ 1 δ |f (y) − f (0)|dσ(y) dr = Cε r n+δ |y|=τ ε

r r=σ 1 δ |f (y) − f (0)|dσ(y)dτ = Cε rn+δ 0 |y|=τ r=ε

δ dr + (n + δ) |f (y) − f (0)|dy n+δ+1 r ε |y|
|I2 | ≤

and

Cεδ Cεδ |I3 | ≤ |f (y)| n+δ dy + |f (0)| n+δ dy |y| |y| |y|≥σ, y∈Q |y|>σ 1 dy ≤ Cf 1 n+δ εδ + |f (0)|Cεδ n+δ σ |y|>σ |y| = O(εδ ).

Due to the above discussion, we have lim sup σεΦ (f )(0) − f (0) ≤ Cη, ε→0+

for any η > 0. This implies lim σεΦ (f )(0) = f (0).

ε→0+

This completes the proof. Definition 1.4.1 For f ∈ L(Q), Aε (f )(x) =

e−ε|m| Cm (f )eim·x

(1.4.6)

m∈Zn

is called the Abel-Poisson means of f . We denote by e−ε = r ∈ (0, 1), then (1.4.6) can be rewritten as P (f )(r; x) = r |m| Cm (f )eim·x . (1.4.7) m∈Zn

1.4 Linear summation

39

Definition 1.4.2 For f ∈ L(Q), 2 e−|εm| Cm (f )eim·x Wε (f ; x) =

(1.4.8)

m∈Zn

is called the Gauss-Weierstrass means of f . Definition 1.4.3 Suppose that α is a complex number whose real part is bigger than −1 and f ∈ L(Q). The triangle polynomial α |m|2 α 1− 2 SR (f ; x) = Cm (f )eim·x (1.4.9) R |m|
is called the Bochner-Riesz means of f and α is called its degree (or index). Substituting the integral expression of Fourier coeﬃcients Cm (f ) into α (f ; x) as (1.4.9), we get the integral form of SR 1 α α SR (f ; x) = f (y)DR (x − y)dy, (1.4.10) (2π)n Q α (x) is deﬁned as in (1.3.4). where the kernel DR We all know that the kernel of Abel-Poisson means Φ(x) = e−|x| is the Fourier transform of the function

− n+1 2 ϕ(y) = Cn 1 + |y|2 , satisfying the conditions in Theorem 1.4.1. Therefore, Aε (f )(x) is convergent in Lp norm and also convergent at Lebesgue points. The Gauss2 Weierstrass kernel e−|x| is the Fourier transform of the function n

1

2

ϕ(y) = π 2 e− 4 |y| , also satisfying the conditions in Theorem 1.4.1. So Wε (f ) is also convergent in Lp norm and convergent at Lebesgue points at the same time. As far as Bochner-Riesz means is concerned, the Fourier transform of the kernel |x| < 1, (1 − |x|2 )α , α Φ (x) = 0, |x| ≥ 1, is B α (x) =

2α Γ(α + 1) J n2 +α (x) . n n (2π) 2 |x| 2 +α

40

C1. An introduction to multiple Fourier series

α If α > α0 = n−1 2 , or more generally, Reα > α0 , then we have B ∈ n α n α L(R ). Consequently, Φ is the Fourier transform of (2π) B and satisﬁes the conditions in Theorem 1.4.1. However, for the case of Reα ≤ α0 , Φα is no longer the Fourier transform of the integrable functions. The index (or order) α0 = n−1 2 is called the critical index both in Bochner-Riesz summation of the Fourier series and the Fourier integral. The index whose real part is bigger than α0 is called over the critical index. For the indivisible intrinsic relations between the Fourier integral and the Fourier series, we will discuss the Fourier integral ﬁrstly in the next chapter and focus on the discussion below the critical index. Chapter 3 will get back to the Bochner-Riesz means of the series and focus on the discussion for the critical index situations.

Chapter 2

Bochner-Riesz means of multiple Fourier integral

2.1

Localization principle and classic results on ﬁxed-point convergence

Suppose that f ∈ L(Rn ) and fˆ is the the Fourier transform of f . Then the Bochner-Riesz means of the Fourier integral of f is deﬁned by α α f (x + y)BR (y)dy, (2.1.1) BR (f ; x) = Rn

α for Reα > −1, where the kernel BR (y) is α (y) = BR

2α Γ(α + 1) J n2 +α (R|y|) n R , n n (2π) 2 (R|y|) 2 +α

(2.1.2)

and the index α0 = n−1 2 is called the critical index. In this section, we mainly consider the case of n−1 n−3 <α≤ . 2 2 Suppose that f is locally integrable on Rn , that is, f is integrable on any bounded set of Rn , and we denote it by f ∈ Lloc (Rn ). For each x ∈ Rn , we 41

42

C2. Bochner-Riesz means of multiple Fourier integral

deﬁne

1 fx (t) = ωn

1 f (x + tξ)dσ(ξ) = n−1 ω t n−1 n S

f (x + ξ)dσ(ξ) (t > 0). |ξ|=t

Then equation (2.1.1) can be represented by α (f ; x) BR

2α+1 Γ(α + 1) n R = n/2 2 Γ(n/2)

∞ 0

fx (t)

J n2 +α (Rt) n

(Rt) 2 +α

tn−1 dt.

(2.1.3)

α0 Bochner [Bo1] established the localization principle of BR (f ; x) convergn n ing at a point, where f ∈ L(R ). That is, if f ∈ L(R ), and there exists a real number δ > 0 such that

f (x) = 0,

for |x − x0 | < δ,

then we have α0 lim BR (f ; x0 ) = 0.

R→∞

The above conclusion is included in the following convergence theorem. Theorem 2.1.1 (Bochner) Let f ∈ L(Rn ) and α = α0 − β, with 0 ≤ β < 1. If there exists a number s such that t τ n−1 |fx (τ ) − s|dτ = 0, (2.1.4) lim t−n t→0+

0

then α (f ; x) = s lim BR

(2.1.5)

R→∞

holds if and only if lim Rβ

R→∞

∞

R−1

(fx (t) − s)

cos tR −

(n−β)π 2 1−β t

dt = 0.

(2.1.6)

Theorem 2.1.2 Suppose that (2.1.4) holds. Then α0 (f ; x) = s lim BR

R→∞

holds if and only if for some positive number η, we have n cos tR − nπ 2 dt = 0. lim (fx (t) − s) R→∞ R−1 t

(2.1.7)

(2.1.8)

2.1 Localization principle and classic results on ﬁxed-point convergence 43 Proof. Set

n −1 . Cn,α = 2α+1 Γ(α + 1) 2n/2 Γ( ) 2 It follows from the equality (2.1.3) that J 1 (t) ∞ n− 2 −β t α fx ( ) − s BR (f ; x) − s = Cn,α dt 1 R 0 t 2 −β 1 1 t fx ( ) − s tβ− 2 Jn− 1 −β (t)dt = Cn,α 2 R 0 ∞ 1 t fx ( ) − s tβ− 2 Jn− 1 −β (t)dt. (2.1.9) +Cn,α 2 R 1

1 By Jn− 1 −β (t) = O tn− 2 −β as t → 0, we immediately have 0

1

2

1 n−1 t t β− 12 −s t fx Jn− 1 −β (t)dt = O dt . fx R − s t 2 R 0

By (2.1.4), we easily have 1 1 t fx − s tβ− 2 Jn− 1 −β (t)dt = o(1), 2 R 0 as R → ∞. From the equation Jν (t) =

&

2 νπ π cos t − − + O t−3/2 , πt 2 4

as t → +∞, we have ∞

1 t − s tβ− 2 Jn− 1 −β (t)dt fx 2 R 1

& cos t − n−β π ∞ 2 t 2 fx = dt −s 1−β R π t 1 ∞ dt t +O fx R − s t2−β , 1

where the second term can be estimated as t ∞ τ n−1 ∞ fx t − s dt = t−(2−β+n−1) − s f dτ 1 τ x 2−β R t R 1 0 ∞ t τ n−1 −(n+2−β) +O t dτ dt . − s τ fx R 1 0

44

C2. Bochner-Riesz means of multiple Fourier integral It follows from f ∈ L(Rn ) that ∞ |fx (t)|tn−1 dt = O(1). 0

Together with (2.1.4) and β < 1, this leads to ∞ ∞ tn fx ( t ) − s dt = o(1) + O dt = o(1), R t2−β tn+2−β 1 1 as R → ∞. Consequently, we get that & α (f ; x) − s = Cn,α BR

2 π

∞

1

cos t − n−β π 2 t dt + o(1), −s fx 1−β R t (2.1.10)

as R → ∞. This implies Theorem 2.1.1 holds. Substituting β = 0 into (2.1.10), and letting η > 0, R > η1 , we conclude that & ηR cos t − n2 π t 2 α dt fx ( ) − s BR (f ; x) − s = Cn,α π 1 R t & ∞ cos t − n2 π t 2 dt + o(1), + Cn,α fx ( ) − s π ηR R t as R → ∞. Since fx (t)t−1 ∈ L(η, ∞), by the Riemann-Lebesgue theorem, we obtain ∞ t cos t − n2 π dt = o(1) fx R t ηR and

∞ ηR

cos t − n2 π dt = o(1), t

as R → ∞. Therefore when R → ∞, we immediately have & η cos Rt − n2 π 2 α BR (f ; x) − s = Cn,α dt + o(1). (fx (t) − s) π R−1 t This completes the proof of Theorem 2.1.2.

2.2 Lp -convergence

45

∞

denotes

Remark 2.1.1 In the proof, the notation 1

lim

A→+∞ 1

A

.

Remark 2.1.2 The condition (2.1.4) is essentially the one for the Lebesgue points. Remark 2.1.3 The localization principle can be deduced from Theorem 2.1.2. α If α < α0 , the localization property of BR (f ; x) at some points, where n f ∈ L(R ) is not valid. For the cases when the order is lower than the critα (f ; x) of f ∈ L1 (Rn ) still has the ical order, Pan [P1] has proven that BR m localization property of convergence point. More accurately, if f ∈ L1m (Rn ) with n ≥ 2m − 1, and n − (2m + 1) , α> 2 α (f ; x) has the localization property of any point. Here, we denote then BR the Sobolev space L1m (Rn ) by ⎧ ⎫ ⎨ ⎬ ∂αf <∞ . L1m (Rn ) = f ∈ L(Rn ) : ∂xα ⎩ ⎭ |α|≤m

2.2

1

Lp-convergence Let f ∈ Lp (Rn ). In this section, we consider whether or not α (f ) − f p = 0 lim BR

R→∞

holds for any f ∈ Lp (Rn ). Next we ﬁrst point out that the Lp convergence of the Bochner-Riesz means of the Fourier integral of f is equivalent to the Lp -boundedness of α. the operator BR Theorem 2.2.1 Let f ∈ Lp (Rn ). The equality α lim BR (f ) − f p = 0

R→∞

holds if and only if there exists a constant Cp such that the inequality α (f )p ≤ Cp f p BR

holds, where Cp is independent of R.

46

C2. Bochner-Riesz means of multiple Fourier integral

Proof. Assume α lim BR (f ) − f p = 0

R→∞

α (f ) has bound with for every f ∈ Lp (Rn ). Thus, we obtain that BR p respect to R. It follows from the uniform boundedness theorem in functional analysis that α (f )p ≤ Cp f p . BR

Conversely, we suppose that there exists a constant Cp such that the inequality α BR (f )p ≤ Cp f p holds for any f ∈ Lp (Rn ). Notice that the space D(Rn ), the class of inﬁnitely diﬀerential functions which have compact support, is dense in Lp (Rn ). It suﬃces to prove that α (f ) − f p = 0 lim BR

R→∞

holds for every f ∈ D(Rn ). Since f ∈ D(Rn ) ⊂ S (Rn ), we have fˆ ∈ S (Rn ), where S (Rn ) denotes the Schwartz class. Combining Lebesgue’s control convergence theorem with the inversion formula of the Fourier transform of f , we have that

y α dy = f (x), fˆ(y)eiy·x Φα lim BR (f ; x) = lim R→∞ R→∞ Rn R for a.e. x ∈ Rn . Therefore, in order to prove α lim |BR (f ; x) − f (x)|p dx = 0, R→∞ Rn

α (f ; x) − f (x)|p can be controlled by an integrable we need to prove that |BR function. This can be easily done, (see Pan [P1]).

After summing up the Lp convergence of Bochner-Riesz means of Fourier integral of f to the Lp boundedness of Bochner-Riesz means, we can deduce the question to the Lp -boundedness of another simpler operator. The operator can be deﬁned just by the Fourier transform. Let f ∈ S (Rn ), we deﬁne the Bochner-Riesz operator Tα α f (x) = Φα (x)fˆ(x), T

2.2 Lp -convergence

47

where Φα (x) is deﬁned in Section 1.2. The following theorem shows why the Lp boundedness of Bochner-Riesz spherical means can be transformed into the Lp boundedness of the BochnerRiesz spherical operator. Theorem 2.2.2 The operator Tα has a bounded extension on Lp (Rn ) if and only if there exists a constant Cp such that the inequality α BR (f )p ≤ Cp f p

holds for every f ∈ Lp (Rn ). Proof. Suppose that Tα has a bounded extension on Lp (Rn ). For f ∈ α (x). Therefore, S (Rn ), it is easy to see that Tα f = B α ∗f , where B α (x) = Φ we have B1α (f )p ≤ Cp f p , f ∈ Lp (Rn ). By the change of variables x = Rx, u =

y , R

u and letting g(u) = f ( R ), we get that

α (f ; x) BR

= |u|<1

fˆ(uR)(1 − |u|2 )α eix·u Rn du

=

u (1 − |u|2 )α eix·u du f R |u|<1

= B1α (g; x). Hence, we have α (f ; x)pp BR

1 = n |B α (g; x)|p dx R Rn 1 Cp ≤ n |g(x)|p dx R Rn = Cp f pp .

The proof for the other part of the theorem is trivial. This completes the proof of Theorem 2.2.2.

48

2.3

C2. Bochner-Riesz means of multiple Fourier integral

Some basic facts on multipliers

Bochner-Riesz operator is a kind of multipliers. In order to investigate the Bochner-Riesz operator, we begin with establishing some fundamental properties. Definition 2.3.1 Let m ∈ L∞ (Rn ). The linear operator Tm deﬁned by * ˆ f ∈ L2 (Rn ) Lp (Rn ) T m f (x) = m(x)f (x), is called a multiplier. Moreover, if Tm satisﬁes Tm f p ≤ Cp f p ,

f ∈ L2 (Rn )

*

Lp (Rn ),

then we call Tm the Lp -bounded multiplier with the multiplier function m(x), and the minimal constant Cp satisfying the above inequality is called the norm of Tm , denoted by Tm p . Theorem 2.3.1 Let p

1 p

+ p1 = 1. Then Tm is a Lp -bounded multiplier if and

only if Tm is a L -bounded multiplier and Tm p = Tm p . Proof. Assume that Tm is a Lp -bounded multiplier. By the property of the Fourier transform, we have that < Tm f, g > = < mfˆ, gˆ > = < f, Tm g >, which implies Tm f p = sup | < Tm f, g > | ≤ Tm p f p .

g p ≤1

Therefore, Tm is also a Lp -bounded multiplier and Tm p ≤ Tm p . Replacing p by p , similar discussions lead to Tm p ≤ Tm p . This completes the proof.

2.3 Some basic facts on multipliers

49

Theorem 2.3.2 If Tm is a L2 -bounded multiplier, then Tm 2 = m∞ . Proof. that

Firstly, by the fact m ∈ L∞ and the Plancherel theorem, we have Tm f 2 = mfˆ2 ≤ m∞ f 2 ,

which yields Tm 2 ≤ m∞ . Next, to show the inverse inequality Tm 2 ≥ m∞ , we denote by E = {x : |m(x)| > Tm 2 }. Suppose that |E| = 0 and let fˆ = χE . By the inequality 1 1 Tm f 2 |E| 2 < mfˆ2 = Tm f 2 ≤ Tm 2 |E| 2 ,

we can deduce a contradiction. Hence, we have |E| = 0.

Theorem 2.3.3 Let 1 < p < ∞. If Tm is a Lp -bounded multiplier, then we have Tm 2 ≤ Tm p . Proof. Without loss of generality, we assume that 1 < p < 2. By the assumption and Theorem 2.3.1, we obtain Tm f p ≤ Tm p f p := k0 f p and Tm f p ≤ Tm p f p = Tm p f p := k1 f p respectively. According to the Riesz-Thorin convexity theorem, we can get that there exists 0 < t < 1 such that Tm f 2 ≤ k01−t k1t f 2 = Tm p f 2 , which leads to Tm 2 ≤ Tm p .

50

C2. Bochner-Riesz means of multiple Fourier integral

Theorem 2.3.4 Let 1 < p < ∞. If Tm is a Lp (Rn+1 )-bounded multiplier, then Tm is also a Lp (Rn )-bounded multiplier. Remark 2.3.1 To express the meaning of Theorem 2.3.4 more accurately, we rewrite Tm f (x) as Tm(x) f (x), where x = (ξ, η) ∈ Rn+1 , ξ ∈ R and η ∈ Rn . Then Theorem 2.3.4 shows that if the multiplier Tm satisﬁes Tm f Lp (Rn+1 ) ≤ Tm p f Lp (Rn+1 ) , then the following inequality Tm(ξ,.) gLp (Rn ) ≤ Cp gLp (Rn ) holds for a.e. ξ ∈ R and Cp ≤ Tm p . Proof. Let

1 p

+

1 q

= 1. By the property of Fourier transform, we get

Rn+1

Tm f (x)α(x)dx =

Rn+1

m(x)fˆ(x) α(x)dx.

Then we have ˆ ≤ Tm f Lp (Rn+1 ) αLq (Rn+1 ) . m(x) f (x) α (x)dx n+1 R

Let f (x) = h(ξ)g(η) and α(x) = γ(ξ)β(η), where h, γ ∈ S (R) and g, β ∈ S (Rn ). The above inequality can be rewritten as m(ξ, η)ˆ g (η) β(η)dη h(ξ) γ (ξ)dξ n R

R

≤ Tm p hLp (R) gLp (Rn ) γLq (R) βLq (Rn ) .

Set M (ξ) =

Rn

m(ξ, η)ˆ g (η)β(η)dη

and deﬁne the operator TM by T M h(ξ) = M (ξ)h(ξ).

2.4 The disc conjecture and Feﬀerman theorem

51

Then we rewrite the above inequality as TM h(ξ)γ(ξ)dξ ≤ Tm p hLp (R) gLp (Rn ) βLq (Rn ) γLq (R) , R

which implies that TM hLp (R) ≤ Tm p gLp (Rn ) βLq (Rn ) hLp (R) . Hence, we have TM p ≤ Tm p gLp (Rn ) βLq (Rn ) . By Theorem 2.3.2, Theorem 2.3.3 and the above inequality, we obtain g (η)β(η)dη ≤ Tm p gLp (Rn ) βLq (Rn ) , n m(ξ, η)ˆ R

which can be rewritten as n Tm(ξ,η) g(η)β(η)dη ≤ Tm p gLp (Rn ) βLq (Rn ) .

(2.3.1)

R

It obviously follows from the inequality 2.3.1 that the conclusion of this theorem is true.

2.4

The disc conjecture and Feﬀerman theorem

In Section 2.2, we have deﬁned the Bochner-Riesz operator Tα by α ˆ T+ α f (x) = Φ (x)f (x),

for f ∈ S (Rn ). We notice that when α = 0, Φ0 (x) is just the characteristic function of the unit ball. The conjecture that the corresponding operator T0 is bounded 2n 2n < p < n−1 is just the disc conjecture which is referred on Lp (Rn ) with n+1 for a long time. The conjecture is based on the fact that the conclusion is correct when n = 1. However, for the case of high dimension (n ≥ 2), Feﬀerman [Fe3] published an amazing result: T0 is only bounded on L2 (Rn ). Therefore, except for p = 2, the disc conjecture is false. In the following, we ﬁrst introduce the result of Feﬀerman.

52

C2. Bochner-Riesz means of multiple Fourier integral

Theorem 2.4.1 Let n ≥ 2. The operator T0 is bounded on Lp (Rn ) if and only if p = 2. For p = 2, by the deﬁnition of the multiplier T0 , it is obvious that T0 is bounded on L2 (Rn ). So the most essential part of the proof of the theorem lies in giving the conclusion that T0 is unbounded in Lp (Rn ), p = 2. Before proving, we need some lemmas. Lemma 2.4.1 Suppose that p > 2, vj , 1 ≤ j ≤ k, is the unit vector in R2 and , Lj = x ∈ R2 : x · vj ≥ 0 is a half-plane. Deﬁne the operator Tj by ˆ + T j f (x) = χLj (x)f (x),

f ∈ S,

where χE (x) is the characteristic function of the set E. If T0 is bounded on Lp (R2 ), then for any fj ∈ Lp (R2 ) (1 ≤ j ≤ k), we have ⎛ ⎛ ⎞1 ⎞1 k k 2 2 ⎝ ⎝ 2⎠ 2⎠ |Tj fj | |fj | ≤ Cp . j=1 j=1 p

p

Proof. Let Djr be a circle, centered at rvj , with a radius of r. Obviously, Djr → Lj as r → ∞. We deﬁne the multiplier Tjr by r ˆ r T+ j f (x) = χDj (x)f (x),

f ∈ S.

For f ∈ S , by the equation Tj f (x) − Tjr f (x) = =

2

R R2

r f (y) eix·y dy + + f (y) − T T j j χLj (y) − χDjr (y) fˆ(y)eix·y dy

and Lebesgue’s dominated convergence theorem, we obtain that Tj f (x) = lim Tjr f (x). r→∞

2.4 The disc conjecture and Feﬀerman theorem

53

Combining the above equation with Fatou lemma, we obtain 1 1 2 2 r 2 2 |T |T f | ≤ lim inf f | j j j j . r→∞

p

p

Therefore, in order to prove Lemma 2.4.1, we merely need to prove that 1 r 2 12 2 2 |fj | |Tj fj | ≤ Cp , p

p

where Cp is independent of r. Similar to the proof of Theorem 2.2.2, it suﬃces to show that 1 1 1 2 2 2 2 f | ≤ C | |f |T p j j j . p

p

Let D = {x : |x| < 1}. Then Tj1 fj (x) can be represented as 1 χD1 (y)fj (y)eix·y dy Tj fj (x) = j 2 R = χD (y − vj )fj (y)eix·y dy R2 χD (z)fj (vj + z)eix·z dz = eix·vj R2

ix·vj

=e

T0 (e−ivj ·z fj (z))(x).

Hence, we have 1 2 12 2 1 2 iv ·y j T0 (e = . fj (y)) |Tj fj | p

p

If we can prove 1 1 2 2 2 2 g | ≤ C | |g |T 0 j p j , p

p

then letting gj (y) = eivj ·y fj (y), we get that 1 1 iv ·y T0 (eivj ·y fj (y))2 2 ≤ Cp e j fj (y)2 2 p p 1 2 |fj (y)|2 = Cp . p

54

C2. Bochner-Riesz means of multiple Fourier integral

We complete the proof of this lemma. Now we prove the inequality 1 1 2 2 2 2 ≤ Cp . |gj | |T0 gj | p

Denote by

p

# $ Sk−1 = x ∈ Rk : |x| = 1

and let ω = (ω1 , ω2 , . . . , ωk ) ∈ Sk−1. Since T0 is bounded on Lp (R2 ), we have p ⎛ ⎞p k k dx ⎠ ⎝ ω T g (x) dx = ω g (x) T j 0 j 0 j j R2 j=1 R2 j=1 p k p ≤ Cp ωj gj (x) dx. R2 j=1 Set

A(x) = T0 g1 (x), T0 g2 (x), . . . , T0 gk (x)

and

B(x) = g1 (x), g2 (x), . . . , gk (x) .

The above inequality can be rewritten as p p | < ω, A(x) > | dx ≤ Cp R2

R2

| < ω, B(x) > |p dx,

which can also be rewritten as . 0p 1 p A(x) 2 2 |T ω, g (x)| dx 0 j 1 / 2 2 R ( |T0 gj (x)| ) 2 . 0p 1 p B(x) 2 2 ω, (x)| |g ≤ Cpp dx. j 1 / R2 ( |gj (x)|2 ) 2 We notice that A(x) k−1 1 ∈ S / ( |T0 gj (x)|2 ) 2

2.4 The disc conjecture and Feﬀerman theorem

55

and B(x) k−1 . 1 ∈ S / 2 2 ( |gj (x)| ) Taking the integral about ω on Sk−1 to both sides of the above inequality, we can get that R

1 p 2 2 p |T g (x)| dx ≤ C 0 j p 2

R

1 p 2 2 dx. |gj (x)| 2

This ﬁnishes the proof Lemma 2.4.1.

Lemma 2.4.2 Suppose that the longer edge of the rectangle Rj is parallel 1j two rectangles which have communal edge to the vector vj . We denote by R with the shorter edge of Rj and the same size as Rj . Let fj = HRj , then we have |Tj fj (x)| ≥ C > 0,

1j , when x ∈ R

where C is some positive constant.

Figure 2.1.

Proof. Without loss of generality, suppose that vj is horizontal and then construct the coordinate system as Figure 2.1.

56

C2. Bochner-Riesz means of multiple Fourier integral Let Rj = [0, a] × [0, b], and

fj (y)eix·y dy

Tj fj (x) =

Lj

=

=

Lj ∞

Rj

e

−iy·z

dz eix·y dy

1 ix1 y1 − ei(x1 −a)y1 dy1 e iy1 0 +∞ 1 ix2 y2 × − ei(x2 −b)y2 dy2 . e −∞ iy2

It is easy to compute that

∞ 0

∞ 1 ix1 y1 cos(x1 y1 ) − cos(x1 − a)y1 i(x1 −a)y1 −e dy1 e dy1 ≥ iy1 y1 0 x1 = log x1 − a ≥ log 2,

1j , and for x = (x1 , x2 ) ∈ R

+∞

−∞

+∞ 1 ix2 y2 sin(x2 y2 ) − sin[(x2 − b)y2 ] i(x2 −b)y2 −e dy2 e dy2 = iy2 y2 −∞ = 2π.

Combining with all the above equations, we get the conclusion of the lemma.

Lemma 2.4.3 For any given η > 0, there exists a measurable set E ⊂ R2 and ﬁnite rectangles Rj (1 ≤ j ≤ k) which do not intersect with each other, satisfying the following two conditions: 2 1j (C being a positive constant); 1j ≥ C R (I) E R (II) |E| ≤ η

k j=1

|Rj |,

1j are deﬁned as in Lemma 2.4.2. where R

2.4 The disc conjecture and Feﬀerman theorem

57

Proof. In order to construct E, we ﬁrst introduce a notation. Give a triangle ABC with the bottom edge AB being horizontal and the height being h. Then we extend AC to A and BC to B , such that the distances from the point A, B to the edge AB are both equal to h (h > h). Joining the midpoint D of AB with A , and D with B , we get two triangles ADA and BDB . We call the two triangles as the ones which grow out of the original triangle ABC from the height of h to h . Now we proceed to 0 construct the measurable set E. Give an √ equilateral triangle , with the bottom edge [0, 1] and the height h0 = 3/2. Let h0 < h1 < · · · < hm . The triangle obtain 0 grows from the height h0 to h1 and we get two triangles and , while and grow from the height h1 to h2 and we get four triangles in all. Thus we make an induction by this way until the m-th time, 2m triangles in all, denoted by j , 1 ≤ j ≤ 2m . Set m

E=

2

j ,

j=1

and let

√ √ 3 3 1 1 1 1+ , . . . , hm = 1 + + ··· + . h1 = 2 2 2 2 m+1

It is easy to check that |E| ≤ 17. We next construct ﬁnite disjoint rectangles Rj (1 ≤ j ≤ k). We divide [0, 1] equally into 2m intervals and let I be anyone of them. It is obvious that I must be the bottom edge of some j . We denote j by (I) and the vertex of j as P (I). The region I formed by the extension lines of two edges of (I) is called the below extension region of (I). Make a rectangle R(I) in this extension region as in Figure 2.2, such that the longer edge of R(I) is log m.

Obviously, the shorter edge of R(I) is ∼ |I|. Then, we have |R(I)| ∼ 2−m log m. is the union of two rectangles which share the same As in Lemma 2.4.2, R(I) edge with the shorter edge of R(I) and has the same size as R(I). Though cannot include (I), by the construction we can see that R(I) 1 R(I) ∩ (I) ≥ R(I) . 10

58

C2. Bochner-Riesz means of multiple Fourier integral

Figure 2.2.

Denote the 2m equally divided intervals of [0, 1] by I1 , I2 , . . . , I2m . For every Ij with 1 ≤ j ≤ 2m , we get R(Ij ) correspondingly. From the last equation above, we have that the ﬁnite rectangles {R(Ij ) : 1 ≤ j ≤ 2m } satisfy conclusion I in the lemma. Next, if we choose m big enough such that log m > 17/η, then there holds m

|E| ≤ 17 < η log m ∼ η

2

|R(Ij )|.

j=1

Thus conclusion II of the lemma is also met. Finally, for diﬀerent Ii and Ij , we have P (Ii ) = P (Ij ). Then the below extension regions of (Ii ) and (Ij ) are disjoint and so are R(Ii ) and R(Ij ). This completes the proof. Next we will give the proof of Theorem 2.4.1. Proof. By Theorem 2.3.4, it suﬃces to show that T0 is unbounded on Lp (R2 ), p = 2. According to Theorem 2.3.1, we merely need to prove that T0 is unbounded on Lp (R2 ) with p > 2. If T0 is bounded on Lp (R2 ). By Lemma 2.4.1, we get that 1 1 2 2 2 2 |fj | |Tj fj | ≤ Cp . p

p

2.5 The Lp -boundedness of Bochner-Riesz operator Tα with α > 0

59

Notice that p > 2. By H¨older’s inequality, we have ⎛ ⎞ p−2 ⎝ |Tj fj (x)|2 ⎠ dx ≤ Cp |E| p

E

j

2 2 12 ( |fj | ) , j p

where E is the measurable set in Lemma 2.4.3. Let fj = χRj , where Rj is the rectangle in Lemma 2.4.3. Notice that diﬀerent Rj disjoint. Then we get from the above inequality and Lemma 2.4.3 that ⎛ ⎞ 2 p−2 Tj χR (x) ⎠ dx ≤ Cp η p ⎝ |Rj |. j E

j

On the other hand, by Lemmas 2.4.2 and 2.4.3, we obtain ⎛ ⎞ 2 Tj χR (x) ⎠ dx ≥ Tj χR (x)2 dx ≥ C ⎝ |Rj |. j j E

j

j

E

j R

j

Obviously, the last two inequality contradicts with each other and this completes the proof.

2.5

The Lp -boundedness of Bochner-Riesz operator Tα with α > 0

In the last section, we obtained the conclusions about Lp -boundedness of the Bochner-Riesz operator T0 . We now turn to investigating the Lp boundedness of the operators Tα (α > 0). In this ﬁeld, Herz [He1] gave a necessary condition for Tα (α > 0) being bounded on Lp (Rn ). p n Theorem 2.5.1 Let 0 < α < n−1 2 and p > 1. If Tα is bounded on L (R ), then we have 2n 2n
Proof. Give any f0 ∈ D, satisfying 1, f0 (x) = 0,

|x| ≤ 1; |x| ≥ 2.

60

C2. Bochner-Riesz means of multiple Fourier integral

Then f0 ∈ S and by the conditions of the theorem, we get that Tα f0 ∈ Lp . Let

Φ(x) = Tα f0 (x).

Taking the Fourier transform on both sides of the equation, we have Φ(x) = Φα (x), which leads to Φ(x) = B α (x). By the expression of B α (x), we get that Tα f0 ∈ Lp if and only if

1 n +α+ 2 2

p > n.

2n Therefore, if Tα is bounded on Lp (Rn ), then we must have p > n+1+2α . We 2n 2n notice that n+1+2α and n−1−2α are a pair of conjugate indexes. Hence by 2n . Theorem 2.3.1, we have that p < n−1−2α This ﬁnishes the proof Theorem 2.5.1.

By the conclusion of Theorem 2.5.1, one naturally raises the following question: if α > 0 and 2n 2n
2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

2.6

61

Oscillatory integral and proof of Carleson-Sj¨ olin theorem

2.6.1 Oscillatory integrals. By the deﬁnition of the Bochner-Riesz operator, we rewrite Tα f (x) as B α (x − y)f (y)dy, Tα f (x) = Rn

where B α (t) =

2α Γ(α + 1) J n2 +α (|t|) · . n (2π)n/2 |t| 2 +α

According to the asymptotic property of Bessel function, we have 1

2 2 πm π Jm (t) = − + r(t), t → ∞, cos t − πt 2 4 where r(t) = O(t−3/2 ) (when t → ∞). Then Tα f (x) can be represented as 1 π(n + 1 + 2α) cos |x − y| − Tα f (x) = C n 1 f (y)dy 4 |x−y|≥1 |x − y| 2 +α+ 2 1 r(|x − y|) f (y)dy +C n |x − y| 2 +α |x−y|≥1 J n2 +α (|x − y|) 2α Γ(α + 1) + f (y)dy n +α (2π)n/2 |x−y|<1 |x − y| 2 := P f (x) + Qf (x) + Rf (x), where the constant C only depends on n and α. By the asymptotic estimate Jm (t) ∼ Ctm , as t → 0, we have |Rf (x)| ≤ CM f (x), where M f is the Hardy-Littlewood maximal function of f . Next, noticing r(t) = O(t−3/2 ), as t → ∞, it follows that |f (y)| |Qf (x)| ≤ C dy. 3 n |x−y|≥1 |x − y| 2 + 2 +α Let p > that

2n n+1+2α .

Then by Minkowski’s inequality of integral, we can get Qf p ≤ Cp f p .

62

C2. Bochner-Riesz means of multiple Fourier integral

2n , the boundedness of Tα Therefore, under the assumption that p > n+1+2α p n on L (R ) is equivalent to the boundedness of the operator P on Lp (Rn ). Due to the expression of P f (x), we might as well take it for granted that n 1 ei|x−y| |x − y|−( 2 +α+ 2 ) f (y)dy. P f (x) = |x−y|≥1

But if p >

2n n+1+2α ,

it is easy to see that there holds 1 i|x−y| −( n +α+ ) 2 2 e |x − y| f (y)dy ≤ CM f (x). |x−y|<1

Hence we get that

P f (x) =

n

Rn

1

ei|x−y| |x − y|−( 2 +α+ 2 ) f (y)dy.

Obviously, P is a class of convolution operator with oscillatory kernel s. For this reason, we will discuss the operators, called the oscillatory integrals. Suppose that x ∈ Rn , ξ ∈ Rn , ∇i g(x, ξ) = (

∂g ∂g ∂g , ,..., ), ∂ξ1 ∂ξ2 ∂ξn

and det

∂ 2 g(x, ξ) ∂xi ∂ξj

is the determinant with the elements of 2 ∂ g(x, ξ) : 1 ≤ i ≤ n, 1 ≤ j ≤ n . ∂xi ∂ξj In this section, we ﬁrst study the oscillatory integral deﬁned by Pλ f (ξ) = eiλϕ(x,ξ) ψ(x, ξ)f (x)dx, ξ ∈ Rn , Rn

where ψ ∈ D(Rn × Rn ), ϕ is a real valued smooth function, and 2 ∂ ϕ(x, ξ) = 0. det ∂xi ∂ξj Theorem 2.6.1 For any λ > 0, Pλ is of type (2.2) and satisﬁes n

Pλ f 2 ≤ Cλ− 2 f 2 , where the constant C is independent of λ and f .

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

63

Proof. Firstly, we rewrite Pλ f 22 as 2 |Pλ f (ξ)| dξ = eiλ(ϕ(x,ξ)−ϕ(y,ξ)) ψ(x, ξ)ψ(y, ξ)f (x)f (y)dxdydξ n n n n R R R R Kλ (x, y)f (x)f (y)dxdy, = Rn

Rn

where Kλ (x, y) =

Rn

eiλ(ϕ(x,ξ)−ϕ(y,ξ)) ψ(x, ξ)ψ(y, ξ)dξ.

Obviously, we have |Kλ (x, y)| ≤ C, where C is independent of x, y and λ. Therefore, if we can prove |Kλ (x, y)| ≤

C , (λ|x − y|)(n+1)

(2.6.1)

then we can easily obtain that C |Kλ (x, y)|dx ≤ n sup λ y Rn

and sup x

Rn

|Kλ (x, y)|dy ≤

C . λn

Consequently, by the Schwarz inequality, we get that

2

Rn

|Pλ f (ξ)| dx ≤

Rn

Rn

× −n

≤Cλ

1

2

|Kλ (x, y)||f (x)| dxdy

Rn Rn f 22 .

2

2

|Kλ (x, y)||f (y)| dxdy

1 2

Thus we complete the proof of the theorem. Next it suﬃces to prove the inequality (2.6.1). Choose a vector a = (a1 , a2 , . . . , an ) satisfying |a| ≤ 1. It easily implies that

a, ∇ξ eiλ(ϕ(x,ξ)−ϕ(y,ξ)) = iλ a, ∇ξ (ϕ(x, ξ) − ϕ(y, ξ)) eiλ(ϕ(x,ξ)−ϕ(y,ξ)) . Let

M := a, ∇ξ ϕ(x, ξ) − ϕ(y, ξ) ,

64

C2. Bochner-Riesz means of multiple Fourier integral

then we have Kλ (x, y) =

Rn

1 (a, ∇ξ )eiλ(ϕ(x,ξ)−ϕ(y,ξ)) ψ(x, ξ)ψ(y, ξ)dξ. iλM

According to the ﬁnite increment formula for functions of multiple variables, we obtain that

- , M = a, ∇ξ ϕ(x1 , . . . , xn ; ξ1 , . . . , ξn ) − ϕ(y1 , . . . , yn ; ξ1 , . . . , ξn ) ∂ϕ ∂ϕ = a, ∇ξ (x1 − y1 ) + · · · + (xn − yn ) ∂x1 ∂xn x=x

· (x − y) , = a, ∇ξ ∇x ϕ x=x

where the component of x = (x1 , . . . , xn ), say, xi is between xi and yi. Together with the condition, this implies that the matrix ⎛ ∂2ϕ ⎞ ∂2ϕ ∂x1 ∂ξ1 · · · ∂x1 ∂ξn ⎜ ⎟ ∇ ξ ∇x ϕ = ⎝ · · · · · · · · · ⎠ 2 2 ∂ ϕ ∂ ϕ ∂xn ∂ξ1 · · · ∂xn ∂ξn is a non-degenerate matrix. Due to the smoothness of ϕ, there exists C > 0, such that |det(∇ξ ∇x ϕ)| ≥ C. Let b = ∇ξ ∇x ϕ · (x − y), then we have which implies

x − y = (∇ξ ∇x ϕ)−1 b, |x − y| ≤ (∇ξ ∇x ϕ)−1 |b|,

where the norm of matrix A is deﬁned as |Ax| . |x|=0 |x|

A = sup For A = (aij ), it is obvious that

A ≤ C max

1≤j≤n

n i=1

|aij |.

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

65

Now, it follows from the components of the matrix (∇ξ ∇x ϕ)−1 and |det(∇ξ ∇x ϕ)| ≥ C > 0, that

(∇ξ ∇x ϕ)−1 ≤ C.

Hence, we have |b| ≥ C|x − y|. Choose a with |a| = 1 such that M = ∇ξ ∇x ϕ

x=x

(x − y) ,

and thus we have M ≥ C|x − y|. By making use of partial integration to the last expression in Kλ (x, y) and the fact |∇ξ M | C C 1 ≤ ≤ , , 2 M |x − y| M |x − y| and ψ ∈ D(Rn × Rn ), we get that |Kλ (x, y)| ≤

C . λ|x − y|

By repeating the above arguments for (n + 1) times, we get the inequality (2.6.1). Since the operator Pλ is of type (1, ∞), combining Theorem 2.6.1 with the Riesz-Thorin convex theorem, we can get the following theorem. Theorem 2.6.2 Let 1 ≤ p ≤ 2 and 1p + 1q = 1. Then Pλ is of type (p, q) and satisﬁes Pλ f q ≤ Cλ−n/q f p . In order to prove Carleson-Sj¨ olin theorem, we should turn to investigatp 2 ing the L (R )-boundedness of the oscillatory integral Pλ f (ξ) = eiλϕ(x,ξ) ψ(x, ξ)f (x)dx R2

for ξ ∈ R2 .

66

C2. Bochner-Riesz means of multiple Fourier integral

To this end, we begin with the oscillatory integral which is closely related to the above. Let x = (x1 , x2 ) ∈ R2 and t ∈ R, we consider the oscillatory integral (1)

Pλ f (x) = where Ψ ∈

D(R2

eiλΦ(x,t) Ψ(x, t)f (t)dt, R

× R), and Φ is a real and smooth function satisfying ⎞ ⎛ 2 3 ∂ Φ

⎜ ∂x1 ∂t det ⎝

∂ Φ ∂x1 ∂t2

∂3Φ ∂2Φ ∂x2 ∂t ∂x2 ∂t2

Theorem 2.6.3 Let

3 1 q+p

⎟ ⎠ = 0.

(1)

= 1 and q > 4. Then Pλ

(1) Pλ f

Lq (R2 )

satisﬁes the inequality

≤ Cq λ−2/q f Lp (R) .

Proof. We ﬁrst write that

2 (1) Pλ f (x) = eiλ{Φ(x,s)+Φ(x,t)} Ψ(x, s)Ψ(x, t)f (s)f (t)dsdt, R

R

Let y1 = s + t, y2 = st and set y = (y1 , y2 ). We put ϕ(x, y) = Φ(x, s) + Φ(x, t), ψ(x, y) = Ψ(x, s)Ψ(x, t), and let F (y) = Jf (s)f (t), where J is the Jacobi determinant of the transformation, that is, J= Then we have

2 (1) Pλ f (x) =

R2

D(t, s) . D(y1 , y2 )

eiλϕ(x,y) F (y)ψ(x, y)dy := Pλ F (x).

It is easy to see that ψ ∈ D(R2 × R2 ). If we can prove that det(∇x ∇y ϕ) = 0,

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

67

then from Theorem 2.6.2, we can get R2

where

1 r

+

1 r

2r 1/r (1) = Pλ F r ≤ Cλ−2/r F r , Pλ f (x) dx

= 1 and 1 < r < 2. From the equation,

F r =

1/r

R2

|F (y)|r dy

= R

R

|f (s)f (t)J|r

D(y1 , y2 ) dsdt D(t, s)

1/r

D(y1 , y2 ) D(t, s) = |s − t|.

and

We have that F r =

R

=

R

R

1−r

|f (s)| |f (t)| |s − t| r

|f (t)|

r

r

dsdt

−1+(2−r)

|f (s)| |s − t| r

R

1/r

1/r

ds dt

≤ C (f u I2−r (f r )u )1/r , r

where u1 + u1 = 1 and Iα (g) is the Riesz potential operator. Then by the boundedness of the Riesz potential operator (cf. Stein [St4]), we obtain that Iα (g)u ≤ Cgu with 0 < α < 1. This leads to F r ≤ Cf r 2/r u . Consequently, we have R2

(1) |Pλ f (x)|2r dx

1/2r

−1/r

≤ Cλ

1/ru |f (t)| dt ru

R

By letting p = ur, q = 2r , and u = we get the conclusion of the proof of Theorem. Hence, it suﬃces to check det(∇x ∇y ϕ) = 0.

2 , 3−r

.

68

C2. Bochner-Riesz means of multiple Fourier integral

Let z = (s, t). Then we obtain that det(∇y ∇x ϕ) = det(∇z ∇x ϕ)|t − s|−1 ⎛ ∂ 2 Φ(x,s)

and

⎜ det(∇z ∇x ϕ) = det ⎝

∂x1 ∂s

∂ 2 Φ(x,t) ∂x1 ∂t

∂ 2 Φ(x,s)

∂ 2 Φ(x,t)

∂x2 ∂s

∂x2 ∂t

⎛ ∂ 2 Φ(x,s) ⎜ = det ⎝

⎞ ⎟ ⎠

∂ 2 Φ(x,t) ∂x1 ∂t

−

∂ 2 Φ(x,s) ∂x1 ∂s

∂ 2 Φ(x,s) ∂ 2 Φ(x,t) ∂x2 ∂s ∂x2 ∂t

−

∂ 2 Φ(x,s) ∂x2 ∂s

∂x1 ∂s

⎛ ⎜ ≈ (t − s)det ⎝

∂3Φ ∂2Φ ∂x1 ∂s ∂x1 ∂s2 ∂2Φ ∂3Φ ∂x2 ∂s ∂x2 ∂s2

⎞ ⎟ ⎠

⎞ ⎟ ⎠

,

s=s

where s is between s and t. Combining the condition of the theorem with the above two equations, we get that det(∇x ∇y ϕ) = 0. This completes the proof Theorem 2.6.3. From the proof of Theorem 2.6.1, we can see that if suppf ⊂ S1 , suppψ ⊂ S1 and ξ ∈ S2 , where S1 and S2 are both bounded closed sets, then it follows from the condition that 2 ∂ ϕ(x, ξ) = 0 det ∂xi ∂ξj on S1 × S2 that

n

Pλ f L2 (S2 ) ≤ Cλ− 2 f L2 (S1 ) .

We denote by S1 a bounded closed set in R, and by S2 a bounded closed set in R2 . Let (1) Pλ f (x) = eiλΦ(x,t) Ψ(x, t)f (t)dt, x ∈ S2 , S1

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

69

where Ψ ∈ D(R2 × R), suppΨ ⊂ S2 × S1 , Φ is real and smooth, which satisﬁes ⎞ ⎛ 2 3 ∂ Φ

⎜ ∂x1 ∂t det ⎝ ∂2Φ

∂ Φ ∂x1 ∂t2 ∂3Φ

⎟ ⎠ = 0.

∂x2 ∂t ∂x2 ∂t2

on S2 × S1 . Then we have the following corollary. Corollary 2.6.1 Let

3 q

+

1 p

= 1 and q > 4. Then Pλ

(1) P f λ

Lq (S2 )

(1)

satisﬁes

≤ Cq λ−2/q f Lp (S1 ) .

Set S1 = I = [0, 1] and 3 5 3 5 3 1 3 5 S2 = − , × , − , × − ,− . 2 2 2 2 2 2 2 2 Deﬁne (2) Pλ,s f (x)

=

2 +(x

eiλ{(x1 −t)

1 2 2 −s) } 2

Ψ(x, t)f (t)dt, x ∈ S2 ,

I

where Ψ ∈ D(R2 × R), and suppΨ ⊂ S2 × I. We have the following theorem. Theorem 2.6.4 For any s ∈ [0, 1], the following inequality 1 (2) ≤ Cε λ− 2 +ε f L4 (I) Pλ,s f 4 L (S2 )

holds for any ε > 0. Here Cε depends only on ε. Proof. Let

-1 , Φ(x, t) = (x1 − t)2 + (x2 − s)2 2 .

Then we have

(2)

Pλ,s f (x) = Pλ Since

⎛ ⎜ det ⎝

∂2Φ ∂3Φ ∂x1 ∂t ∂x1 ∂t2 ∂3Φ ∂2Φ ∂x2 ∂t ∂x2 ∂t2

(1)

f (x).

⎞ ⎟ (x2 − s)3 , ⎠= Φ6

70

C2. Bochner-Riesz means of multiple Fourier integral

0 ≤ s ≤ 1 and (x1 , x2 ) ∈ S2 , the above determinant is not equal to 0 on S2 × I. By Corollary 2.6.1, we directly obtain that (2)

Pλ,s f Lq (S2 ) ≤ Cq λ−2/q f Lp (I) , which follows the conclusion of the theorem. This completes the proof. If we denote by (1) S2

and (2) S2

3 5 3 5 1 3 3 5 × , − , × − , , = − , 2 2 2 2 2 2 2 2

3 5 3 5 3 5 3 1 , × − , , = − ,− × − , 2 2 2 2 2 2 2 2

and let (1)

S1 = I 2 , S2 = S2

(2)

S2 ,

then it is easy to check that S2 consists of the union of 12 squares each in size of I 2 , and for x ∈ S1 and y ∈ S2 , we have 1 |x − y| ≥ . 2 For convenience, we denote by S2 = F (I 2 ) and set (3) Pλ f (x)

=

I2

eiλ|x−y| Ψ(x, y)f (y)dy,

for x ∈ F (I 2 ), where Ψ ∈ D(R2 × R2 ) and suppΨ ⊂ F (I 2 ) × I 2 . We have the following theorem. Theorem 2.6.5 The following inequality (3) Pλ f

L4 (F (I 2 ))

holds for any ε > 0.

1

≤ Cε λ− 2 +ε f L4 (I 2 )

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

71

Proof. Let x = (x1 , x2 ) ∈ F (I 2 ) and y = (t, τ ) ∈ I 2 . Then by the Minkowski inequality of integral, we get that (3) Pλ f 4 L (F (I 2 )) (3) (3) ≤ Pλ f 4 (1) + Pλ f 4 (2) ≤ I

L (S2 )

S2

I

+ I

L (S2 )

4 14 1 2 +(x −τ )2 ] 2 iλ[(x −t) 1 2 e Ψ(x, t, τ )f (t, τ )dt dx dτ (1) 4 14 1 2 +(x −τ )2 ] 2 iλ[(x −t) 1 2 e Ψ(x, t, τ )f (t, τ )dτ dx dt. (2)

S2

I

Applying Theorem 2.6.4 to the two terms on the right side of the above equation, we get that (3) Pλ f 4 L (F (I 2 )) 1 1 4 4 1 ≤ Cε λ− 2 +ε |f (t, τ )|4 dt dτ + |f (t, τ )|4 dt dt . I

I

I

I

Hence from H¨older’s inequality we get the conclusion of the theorem. For the operator (4) Pλ f (x)

=

I2

eiλ|x−y|

1 3

|x − y| 2 +α

f (y)dy,

for x ∈ F (I 2 ), we can use the operator (3) P λ f (x)

=

I2

eiλ|x−y|

1 3

|x − y| 2 +α

ψ0 (x, y)f (y)dy

to approximate it arbitrarily (with diﬀerent ψ0 chosen), where suppψ0 ⊂ F (I 2 ) × I 2 . For x ∈ D1 ⊂ F (I 2 ) and y ∈ D2 ⊂ I 2 , we let ψ0 (x, y) = 1,

72

C2. Bochner-Riesz means of multiple Fourier integral

and D1 and D2 can arbitrarily approximate to F (I 2 ) and I 2 , respectively. Noticing the smoothness of 1 3

|x − y| 2 +α (3)

(4)

on F (I 2 ) × I 2 , we see that if Pλ is replaced by Pλ , Theorem 2.6.5 still holds. Now let 1 (4) Sλ f (x) = λ 2 −α Pλ f (x), for x ∈ F (I 2 ). By Theorem 2.6.5, we have that Sλ f L4 (F (I 2 )) ≤ Cε λ−α+ε f L4 (I 2 ) . Hence, by taking ε small enough, we obtain the following corollary. Corollary 2.6.2 There exists a positive constant δ > 0 such that the inequality Sλ f L4 (F (I 2 )) ≤ Cλ−δ f L4 (I 2 ) holds. Remark 2.6.1 It is easy to see that F (I 2 ) = [−1.5, 2.5]2 \ [−0.5, 1.5]2 . If F (I 2 ) is replaced by the set F (I 2 ) = [−2.5, 3.5]2 \ [−1.5, 2.5]2 , we can still obtain the same result as Corollary 2.6.2. 2.6.2 Proof of Carleson-Sj¨ olin theorem. 4 , the In Section 2.6.1, we mentioned that, under the condition p > 3+2α p boundedness of Tα is equivalent to the L -boundedness of the operator 1 ei|x−y| f (y)dy, P f (x) = 3 |x−y|≥1 |x − y| 2 +α

for x ∈ R2 .

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

73

It is not diﬃcult to prove that the integral 1 ei|x−y| f (y)dy 3 |x−y|<1 |x − y| 2 +α can be controlled by some constant times of M f (x). Thus, we might as well take it for granted that 1 ei|x−y| f (y)dy, x ∈ R2 . Tα f (x) = |x − y|3/2+α R2 In order to prove the boundedness of Tα on Lp (R2 ), we ﬁnd the key is the case that p = 4. By the boundedness of Tα on L4 (R2 ) and the theory of interpolation and duality, we can easily obtain that Tα is bounded on Lp (R2 ) where 4 4
4 1 i|x−y| e f (y)dy dx ≤ C |f (x)|4 dx, 3 +α 2 2 (λI)2 (λI)2 (λI) |x − y|

where λI = [0, λ]. By some change of variables, the inequality above turns out to be 4 1 12 −α iλ|x−y| e F (y)dy dx ≤ C |F (x)|4 dx. λ 3 +α 2 2 I2 I2 I |x − y| Then, the proof of Carleson-Sj¨olin theorem can be reduced to the proof of the following basic facts. Proposition 2.6.1 There exists a constant C > 0 such that the inequality Sλ f L4 (I 2 ) ≤ Cf L4 (I 2 ) holds. Proof. Before the proof of the proposition, we recall the meaning of the set F (I 2 ) in section 6.1. Let F1 and F2 be two squares whose centers are the same as I 2 and edges both parallel to the axes, with lengths 2 and 4. Then we have F (I 2 ) = F2 \ F1 . In the following, for any square ω with edges paralleling to the axes, we similarly deﬁne F (ω). We denote by F1 (ω)

74

C2. Bochner-Riesz means of multiple Fourier integral

and F2 (ω) two squares, whose centers are the same as ω and edges are both parallel to the axes, with lengths 2 and 4 times of the one of ω. Then, we deﬁne F (ω) = F2 (ω) \ F1 (ω). Obviously, for x ∈ ω and y ∈ F (ω), we have |x − y| ≥ 12 × (the length of ω). In addition, any square ω whose edges are parallel to the axes, we denote by 12 ω the square which has the same center as ω and edges paralleling to the axes, with half length of ω. Similarly, we deﬁne F (ω) = F2 (ω) \ F1 (ω), where F1 (ω) and F2 (ω) are two squares, whose centers are the same as ω and edges are both parallel to the axes, with lengths 4 and 6 times of the one of ω. Let Ωk be all of the binary squares in the square of [−2, 2] × [−2, 2] with length 2−k and Ω∗k be a collection of all the squares which consist of all four squares belonging to Ωk . It is obvious that the edge length of the square in Ω∗k is 2−k+1 . For any x ∈ I 2 , but not in the boundary of any binary squares, noticing that the measure of the set consisted by all those boundaries is zero in two dimensional sense, obviously there exists an unique square in Ω∗k , such that the half small square which is concentrically contracted from the original square contains x. We denote the unique square by ωk∗ (x). Then, x ∈ 12 ωk∗ (x). Since 1 I2 = [−2, 2] × [−2, 2] , 2 we denote by ∗ ω−1 (x) = [−2, 2] × [−2, 2]. Set A(x, D) = λ

1 −α 2

eiλ|x−y| D

and

f (y) dy, x ∈ I 2 , |x − y|3/2+α

∗ Ak (x) = A x, [ωk−1 (x) \ ωk∗ (x)] ∩ I 2 , k ≥ 0.

Then we have Sλ f (x) =

kN

Ak (x) + A(x, ωk∗N (x)),

k=0

where kN satisﬁes

2−(kN +1) < λ−1 ≤ 2−kN .

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

75

We notice that if k is large enough, then we have ∗ (x) \ ωk∗ (x) ⊂ I 2 . ωk−1

Thus we might as well take it for granted that ∗ (x) \ ωk∗ (x)). Ak (x) = A(x, ωk−1 ∗ (x) \ ω ∗ (x) is the union of By the construction of ωk∗ (x), we have that ωk−1 k 12 squares in the same size, with length 2−k . The arrangement of the 12 squares at least falls into one of the following three cases. Case 1. (see Figure 2.3)

Figure 2.3: Case 1.

The 12 squares share either the same edges or points with ωk∗ (x), which are denoted by Di , 1 ≤ i ≤ 12. Obviously, for any i, 1 ≤ i ≤ 12, we have 1 F (Di ) ⊃ ωk∗ (x). 2 Thus, we have x ∈ F (Di ). Then we have Ak (x) =

12

A(x, Di )χF (Di ) (x).

i=1

It follows that |Ak (x)|4 ≤ C

12 i=1

|A(x, Di )|4 χF (Di ) (x)

76 and

C2. Bochner-Riesz means of multiple Fourier integral I2

|Ak (x)|4 dx ≤ C

12 i=1

F (Di )

|A(x, Di )|4 dx.

Applying Corollary 2.6.2 to each integral on the right side of the above inequality and making change of variables, we conclude that

12 −k −4δ |Ak (x)| dx ≤ C (λ2 ) 4

I2

i=1

≤ Cλ−4δ 24kδ

|f (x)|4 dx Di

|f (x)|4 dx.

I2

Case 2. Among all the 12 squares, there are 8 squares sharing either the same edges or points with ωk∗ (x) (see Figure 2.4: Case 2 and Case 3), which are denoted by Di , 1 ≤ i ≤ 8. The other 4 squares have no common points with ωk∗ (x), denoted by Dj , 9 ≤ j ≤ 12. It is evident that for any Dj , 9 ≤ j ≤ 12, we have 1 F (Dj ) ⊃ ωk∗ (x). 2 Then we have Ak (x) =

8 i=1

A(x, Di )χF (Di ) (x) +

12

A(x, Dj )χF (Dj ) (x).

j=9

Figures 2.4: Case 2 and Case 3.

Noticing the remark at the end of Section 2.6.1 and by the similar discussion in Case 1, we have that |Ak (x)|4 dx ≤ Cλ−4δ 24kδ |f (x)|4 dx. I2

I2

2.6 Oscillatory integral and proof of Carleson-Sj¨olin theorem

77

Case 3. Among all the 12 squares, there are only 5 ones sharing either the same edges or points with ωk∗ (x), (see Figure 2.4: Case 2 and Case 3). In this case, the proof is the same as that in Case 2. Consequently, for any case, we always have that Ak L4 (I 2 ) ≤ Cλ−δ 2kδ f L4 (I 2 ) . Then we have

Sλ f L4 (I 2 ) ≤ C

λ−δ

≤C

kN

2kδ f L4 (I 2 ) +

k=0

f L4 (I 2 ) +

I2

I2

A(x, ω ∗ (x))4 dx kN

A(x, ω ∗ (x))4 dx kN

1

1 4

.

Therefore, it suﬃces to prove that 1 4 A(x, ω ∗ (x))4 dx ≤ Cf L4 (I 2 ) . kN I2

In fact, we have A(x, ω ∗ (x)) ≤ λ 12 −α kN

ωk∗

N

1

≤ λ 2 −α

∞ i=0

We notice that for

|f (y)| dy |x − y|3/2+α |f (y)| dy. |x − y|3/2+α

ωk∗ +i \ωk∗ +i+1 N N

y ∈ ωk∗N +i (x) \ ωk∗N +i+1 (x)

and x ∈ 12 ωk∗N +i+1 (x), there holds |x − y| ≥ 2−(kN +i+2) , which yields ∞ A(x, ω ∗ (x)) ≤ λ 12 −α 2(3/2+α)(kN +i+2) kN

≤C

ωk∗

N +i

i=0 ∞

1

2−( 2 −α)i M f (x)

i=0

≤ CM f (x). This ﬁnishes the proof of Theorem 2.5.2.

(x)

|f (y)|dy

4

78

C2. Bochner-Riesz means of multiple Fourier integral

Remark 2.6.2 Let us recall the proof of Carleson-Sj¨ olin theorem in this section. Firstly, the conclusions of Theorems 2.6.1 to 2.6.3 cannot be applied directly to the proof of Carleson-Sj¨ olin theorem. The reason for that lies in the fact that when ϕ(x, y) = |x − y| is restricted, the condition det

∂ 2 ϕ(x, y) ∂xi ∂yi

= 0

does not hold. But when n = 2, a special method could be utilized to conquer the diﬃculty. Therefore, the method of proving in this section is not suitable for the case n > 2. To this end, ones are trying to ﬁnd out new methods of Carleson-Sj¨ olin theorem, such that the new one is feasible in solving the following speculation: Tα , α > 0, is bounded on Lp (Rn ), where n > 2 and and 2n 2n 2, yet they oﬀer a powerful tool in solving other problems. For this reason, we will introduce the methods of Cordoba and Tomas-Stein.

2.7

Kakeya maximal function

To investigate the Lp -boundedness of the operator Tα , we can use the method for proving the Lp -boundedness of partial sum operator in the case of one dimension. It is well known that in the case of one dimension, the Hilbert transform plays an important role in solving questions and the estimate of Hilbert transform can be reduced to Hardy-Littlewood maximal function. However it is diﬀerent when it comes to the cases of high dimension. For instance, when Cordoba considered the case of two-dimension, what he met with was not Hardy-Littlewood maximal function but the maximal function about the

2.7 Kakeya maximal function

79

rectangle with some eccentricity, which is called Kakeya maximal function. The Kakeya maximal function is deﬁned as follows. If given δ > 0 and N > 0, we denote by R = {rectangle R : |R| = N δ × δ}, that is, the length of the longer edge of R is N δ and the shorter is δ. Let the direction of the long edge be the direction of the rectangle. Then we deﬁne 1 |f (y)|dy, MN f (x) = sup R x |R| R where sup is taken over all the R ∈ R including x, and N is the eccentricity. Cordoba [Co1] has proven the following basic estimate on MN .

Theorem 2.7.1 There exists a constant C independent of δ and N , such that 1 MN f 2 ≤ C(log 3N ) 2 f 2 for all f ∈ L2 (R2 ). Proof. Since the proof of Theorem 2.7.1 is rather long, it will be convenient to divide it into seven steps. Step I. Restrict the direction of the rectangle. Divide the range of the direction of rectangle [0, 2π) into eight parts as 8 7π π !π π , ··· , 2π = [0, 2π) = 0, Ii . 4 4 2 4 !

i=1

i f (x), with the requireSimilar to the deﬁnition of MN f (x), we deﬁne MN ment that the direction of rectangle R must be inside Ii . Then we have

MN f (x) ≤

8

i MN f (x).

i=1 i Since the method of dealing with all these MN f (x) is the same, we merely need to prove that 1 1 f 2 ≤ C(log 3N ) 2 f 2 MN 1 f by M f . for any f ∈ L2 (R2 ). For convenience, we still denote MN N

Step II. Restrict the support of f .

80

C2. Bochner-Riesz means of multiple Fourier integral

Dividing the plane into squares with the length of the edge of N δ along the axas as Qα , R2 = 2 where the length of the edge of Qα is N δ and Qα Qβ = ∅ for α = β. Set fα (x) = f (x)χQα (x). It is obvious that f (x) =

fα (x),

for a.e. x ∈ R2 . Denoted by Q∗ = 3Q, the length of the edge of Q∗ is 3 times of Q and the two have the same center. 2 When x ∈ Q∗α , and Q∗α Q∗β = ∅, we have MN fβ (x) = 0. Consequently, there exist at most 52 terms in the sum Therefore, it follows that 2 MN fα (x) ≤ 54 |MN fα (x)|2 . |MN f (x)|2 ≤ α α

/

α MN fα (x).

(2.7.1)

For each fα , if the inequality 1

MN fα 2 ≤ C(log 3N ) 2 fα 2 holds, then the conclusion of the theorem immediately follows from (2.7.1). Consequently, the proof of the theorem can be reduced to prove that if suppf ⊂ Q and the edge length of Q is N δ, then 1

MN f 2 ≤ C(log 3N ) 2 f 2

(2.7.2)

holds. It is obvious that, for such f , we must have MN f (x) = 0 for x ∈ (Q∗ )c . Step III. The linearization of the maximal operator.

(2.7.3)

2.7 Kakeya maximal function

81

Suppose that suppf ⊂ the square Q, the edge length of Q is N δ, and Q∗ = 3Q. Again by the straight line along the axas, we divide Q∗ into 9N 2 squares Qi,p (1 ≤ i ≤ 3N, 1 ≤ p ≤ 3N ) of the edge length δ. For any 1 x ∈ Qi,p , by the deﬁnition of MN f , there exists a rectangle Ri,p (x) such that 2 |f (y)|dy MN f (x) ≤ |Ri,p (x)| Ri,p (x) for πx ∈ Qi,p , where the direction of the rectangle Ri,p (x) belongs to I1 = 0, 4 , and |Ri,p (x)| = N δ × δ. By the deﬁnition of the supremum, there exists a , Ri,p ∈ Ri,p (x) : x ∈ Qi,p ,

such that

sup x∈Qi,p

Ri,p (x)

|f (y)|dy ≤ 2

|f (y)|dy, Ri,p

where the direction of Ri,p belongs to I1 , and |Ri,p | = N δ × δ. It follows from the above discussion that 4 MN f (x) ≤ |f (y)|dy, |Ri,p | Ri,p for x ∈ Qi,p . Together with (2.7.3), the above inequality leads to

1 |f (y)|dy χQi,p (x). MN f (x) ≤ 4 |Ri,p | Ri,p i,p

2 It is obvious that there holds, Ri,p Qi,p = ∅. For ﬁxed f , if we deﬁne the linear operator g → Tf (g) as

1 Tf (g)(x) = g(y)dy χQi,p (x), |Ri,p | Ri,p i,p

then, in order to prove (2.7.2), it suﬃces to show that 1

Tf (g)2 ≤ C(log 3N ) 2 g2 , f or any g ∈ L2 (Q∗ ).

(2.7.4)

Step IV. Transfer into the estimate for the conjugate operator Tf∗ . Let h, g ∈ L2 (Q∗ ). If we denote the conjugate operator of Tf by Tf∗ , then by the equation g(y)Tf∗ (h)(y)dy = Tf (g)(x)h(x)dx. Q∗

Q∗

82

C2. Bochner-Riesz means of multiple Fourier integral

We easily have Tf∗ (h)(y)

=

i,p

1 |Ri,p |

h(y)dy Qi,p

χRi,p (x).

By the fact that Tf∗ = Tf , to get (2.7.4), it suﬃces to show 1

Tf∗ (h)2 ≤ C(log 3N ) 2 h2 , f or any h ∈ L2 (Q∗ ).

(2.7.5)

Step V. The proof of the equation (2.7.5). Q∗

Let h ∈ L2 (Q∗ ). Now with straight lines parallel to the y-axis, we divide into 3N rectangles, that is, Q∗ =

3N

Ei , |Ei | = δ × 3N δ.

i=1

Let hi (x) = h(x)χEi (x), then we have

3N

h(x) =

hi (x).

i=1

It is easy to prove that if the inequality 1

1

Tf∗ (hi )2 ≤ CN − 2 (log 3N ) 2 hi 2 , (1 ≤ i ≤ 3N )

(2.7.6)

holds, then, (2.7.5) is also holds. In fact, this can be obtained from the following inequality Tf∗ (h)2

≤

3N

Tf∗ (hi )2

i=1 1

1

≤ CN − 2 (log 3N ) 2

3N

hi 2

i=1

≤ CN

− 21

(log 3N )

1 2

i=1 1 2

12 3N 12 |h(x)|2 dx 1

3N

≤ C(log 3N ) h2 .

Ei

i=1

2.7 Kakeya maximal function

83

Step VI. The proof of the equation (2.7.6). We divide Ei into 3N squares with the edge length δ:

Ei =

3N

Qi,p .

p=1

Let hi,p (x) = hi (x)χQi,p . We easily have hi (x) =

3N

hi,p (x).

p=1

Since there holds Tf∗ (hi )(x) =

p

1 |Ri,p |

we have |Tf∗ (hi )(x)| ≤

hi,p (y)dy Qi,p

χRi,p (x),

1 hi,p 2 χRi,p (x), Nδ p

which yields Tf∗ (hi )22

1 ≤ (δN )2

2 hi,p 2 χRi,p (x)

dx

p

* 1 R h h = R . i,p 2 i,q 2 i,p i,q 2 (δN ) p,q It is not diﬃcult to prove that if the inequality * Ri,p Ri,q ≤ CN δ 2 /(|p − q| + 1)

(2.7.7)

holds, then we can get (2.7.6). This can be obtained from the following

84

C2. Bochner-Riesz means of multiple Fourier integral

equation Tf∗ (hi )22 ≤

C N

1≤p, q≤3N

hi,p 2 hi,q 2 |p − q| + 1

⎛ ⎞ 3N 3N −1 C ⎝ 1 hi,p 22 + hi,p 2 hi,q 2 ⎠ = N 1+ν p=1 ν=1 |p−q|=ν ⎧ ⎛ ⎞⎫ 3N −1 3ν ⎨ ⎬ 1 ⎝ C 2 2⎠ hi 2 + ≤ hi,p 2 ⎭ N⎩ 1+ν ν=1

≤

p=1

C (log 3N )hi 22 . N

Step VII. The proof of (2.7.7). 2 Here, Ri,p Qi,p = ∅, the direction of Ri,p is inside I1 , and |Ri,p | = δ × N δ. We might as well prove (2.7.7) for q = 1. Suppose that the direction of Ri,1 2 and Ri,p forms an angle of θ, and Ri,1 Ri,p is a parallelogram ABCD (see Figure 2.5).

Figure 2.5.

Then we have * δ2 . Ri,1 Ri,p = δ × |AB| = sin θ

2.7 Kakeya maximal function

85

We consider the case: θ > α, where α is just the angle leaning toward the edge EF in the triangle EF G in Figure 2.6. Here, F G and HF represent Ri,p and Ri,1 , respectively.

Figure 2.6.

Hence, we have |Ri,1 ∩ Ri,p | ≤

δ2 N δ2 . = sin α (p − 1) sin π4

Thus (2.7.7) is valid (when q = 1), then we could duplicate the proof for q in general case. This ﬁnishes the proof of Theorem 2.7.1. Cordoba (see [Co1] and [Co2]) gave another proof of Carleson-Sj¨ olin theorem by the method of geometric proof and Kakeya maximal function. We only point out why Cordoba’s proof is related with the Kakeya maximal function brieﬂy. It is well known that the key point of proving CarlesonSj¨olin theorem lies in the estimate for the case p = 4, while for other p, we can prove by the methods of interpolation and duality. The estimate for the case when p = 4 is carried out in the following way. Firstly, by segmentation along the radial diameters, we divide the unit disc {|x| < 1} into the union 5 of a series of loops k Dk and give some smooth decomposition of (1−|x|2 )α+ as mk (x), (1 − |x|2 )α+ = k

where mk is smooth and supp mk ⊂ Dk . Deﬁne k ˆ T+ α f = mk f .

86

C2. Bochner-Riesz means of multiple Fourier integral

Then the estimate for Tα can be transfered into the one for Tαk . These are all multiplier operators in corresponding with the smooth multiplier function whose support is in a loop. By dilations, we merely need to estimate one of the Tαk s. The multiplier operator chosen by Cordoba is deﬁned as follows. Suppose that φ6 is a smooth function in R, and supp φ6 ⊂ [−1, 1], |ξ| − 1 φ(ξ) = φ6 δ Tαk

for ξ ∈ R2 . It is easy to see that , supp φ ⊂ ξ ∈ R2 : 1 − δ ≤ |ξ| ≤ 1 + δ . Deﬁne

ˆ + S δ f (ξ) = φ(ξ)f (ξ)

for f ∈ D. Then the estimate for Tα f 4 is transfered into the one for Sδ f 4 . Secondly, let ξ = |ξ|eiθ and through the partition of the argument θ, we give a smooth decomposition of φ(ξ) as mj (ξ). φ(ξ) = j

We also deﬁne the multiplier Tj by 1

Tj f (ξ) = mj (ξ)fˆ(ξ),

(0 ≤ j ≤ [δ − 2 ] − 1),

and then we transfer the estimate for Sδ f 4 into the one for We notice that suppmj is a small part of the loop

/ j

Tj f 4 .

{ξ : 1 − δ ≤ |ξ| ≤ 1 + δ} 1

and the central angle which is leaning toward the small part is 2πδ 2 . Now, we write Tj as a convolution operator, that is, Tj f = m j ∗ f. Since m j (x1 , x2 ) ≈

1+δ

1−δ

1

2π(j+1)δ 2 1 2πjδ 2

mj (ξ1 , ξ2 )e−i(ξ1 x1 +ξ2 x2 ) dξ1 dξ2 ,

2.7 Kakeya maximal function

87

by partial integration, we obtain a control function of m j 1

|m j (x1 , x2 )| ≤ Cp,q δ3/2 |δx1 |−p |δ 2 x2 |−q

(p ≥ 0, q ≥ 0).

From the above inequality, we verify that

|m j (x1 , x2 )| ≤ C

∞ ν=0

where

2−ν

1 χR (x1 , x2 ), |Rν,j | ν,j

# $ 1 Rν,j = (x1 , x2 ) : |x1 | ≤ 2ν δ −1 , |x2 | ≤ 2ν δ− 2 . 1

Obviously, Rν,j is a rectangle with the eccentricity being δ− 2 . Then, by the method for the case of L2 and the inequality ⎛ 4 ⎞1 4 2 2⎠ ≤C ⎝ T f |T f | , j j j j 4

4

we have that 4 ⎛ ⎞2 ⎝ Tj f |(m j ∗ f )(x)|2 ⎠ dx ≤C 2 R j j 4 ⎧ 2 ⎫2 ⎨ ⎬ χRν,j ∗ |f |(x) 2−ν dx ≤C ⎭ |Rν,j | R2 ⎩ j ν ⎧ ⎫1 4 2 2 ⎨ ⎬ χRν,j ∗ |f | 2−ν ≤C . ⎩ ⎭ |R | ν,j ν j 4

Consequently, we face the estimate of the norm on the right side of the above

88

C2. Bochner-Riesz means of multiple Fourier integral

inequality. It is easy to see that ⎧ ⎫1 2 2 2⎬ 2 ⎨ χ χRν,j Rν,j ∗ |f | ∗ |f | = ⎭ ⎩ |Rν,j | |Rν,j | j j 2 4 2 χRν,j = sup ∗ |f |(x) ω(x)dx |Rν,j |

ω 2 ≤1 R2 j χRν,j 2 |f (y)| ≤ sup ∗ ω (y)dy |Rν,j |

ω 2 ≤1 j R2 |f (y)|2 Mδ−1/2 (ω)(y)dy, ≤ sup

ω 2 ≤1 j

where M − 12

1

δ− 2

R2

(ω)(y) is the Kakeya maximal function with eccentricity being

δ . In this way, in Cordoba’s proof, the solution to the problem is closely related with the Kakeya maximal function. It is worthy of being pointed out that Cordoba and Lopez-Melero both used the same methods in [CoL1] to extend the Carleson-Sj¨ olin theorem to the case of vector-valued functions, and get the following theorem. Theorem 2.7.2 Let 0 < α < 12 . If 4 4
Lp (R2 )

⎛ ⎞1 2 ⎝ 2⎠ ≤ Cp,α |fj | j

,

Lp (R2 )

for all f ∈ S . Remark 2.7.1 Adding into Theorem 2.7.2 the condition that {Rj }∞ 1 is a lacunary sequence, Igari [Ig1] independently acquired the results as well. When α = 0, n ≥ 2, the following result similarily to Theorem 2.7.2 is obtained by Carbey, Rubio de Francia, Vega [CRV1] and independently by Ma, Liu, Lu [MLL1].

2.8 The restriction theorem of the Fourier transform

89

2n . If {Rj }∞ Theorem 2.7.3 Let 2 ≤ p < n−1 1 is a lacunary sequence, then ⎛ ⎛ ⎞1 ⎞1 2 2 ⎝ ⎝ 0 2⎠ 2⎠ |BR f | ≤ C |f | p,α j j j p n j p n j L (R )

L (R )

for f ∈ S (Rn ).

2.8

The restriction theorem of the Fourier transform

For any n with n ≥ 3 and 2n 2n
for f ∈ S . We say that the (Lp , L2 ) restriction theorem of the Fourier transform holds. The result due to Stein illustrates that: under the condition that the , L2 ) restriction theorem of the Fourier transform holds, the Feﬀerman’s conjecture above is valid. By the method of duality, we merely need to 2n < p < 2. consider the case when n+1+2α

(Lp

Theorem 2.8.1 Suppose that 2n < p < 2, n + 1 + 2α if the (Lp , L2 ) restriction theorem of the Fourier transform holds, then α (f )p ≤ Cp,α f p . BR

90

C2. Bochner-Riesz means of multiple Fourier integral

Proof. It is easy to prove that there exists a radial function ψ0 , ψ ∈ D(Rn ), where suppψ is around |x| = 1, satisfying ψ0 (x) +

∞

ψ(2−j x) = 1.

j=1

Let

ψj (x) = ψ(2−j x) (j = 1, 2, . . .)

and α α (x) = BR (x) · ψj (x) (j = 0, 1, . . .). Bj,R

Then we have α BR (f )(x) =

α (Bj,R ∗ f )(x) := (Tj f )(x). j

j

Obviously, it suﬃces to prove for some ε > 0, there holds |Tj f (x)|p dx ≤ C p 2−jpε |f (x)|p dx, Rn

Rn

(2.8.1)

where C is independent of f, ε. Now we separate Rn into the union of countable n-dimensional cubes, with the edges parallel to the axas and the edge length being 2j . Denote any 6 which one of the cubes by Q and the constant times extension of Q by Q, α share the same center with Q. According to the fact that Bj,R is around |x| = 2j , then in order to prove (2.8.1), we merely need to prove that |Tj f (x)|p dx ≤ C p 2−jpε |f (x)|2 dx. (2.8.2) Q

Q

By the translation invariance, we might as well take it for granted that Q is centered at O . It is not diﬃcult to see that we might as well deem Q as a ball, centered at O , with a diameter being 2j . To this end, we suppose that , suppf ⊂ Q = x : |x| < 2j . Let

2 p

and r be a pair of conjugate indexes, that is, 1 2 p

By H¨older’s inequality, we have

+

1 = 1. r

2.8 The restriction theorem of the Fourier transform

|Tj f (x)| dx ≤

2

p

Q

Q

≤ C2

nj r

91

p/2

|Tj f (x)| dx

2

Rn

6 1/r · |Q| p/2

|Tj f (x)| dx

.

(2.8.3)

It follows from the Plancherel theorem that α (x)|2 · |fˆ(x)|2 dx |Tj f (x)|2 dx = |B j,R n n R R α (x)|2 · |fˆ(x)|2 dx = |B j,R |x|≤α α (x)|2 |fˆ(x)|2 dx, |B + j,R |x|>α

where α is to be determined. Next, we make estimates of the two terms on the right side of the above inequality. Firstly, we have α 2 ˆ 2 α 2 |Bj,R (x)| |f (x)| dx ≤ sup |Bj,R (x)| |fˆ(x)|2 dx. |x|≤α

|x|≤α

According to the fact

α j,α R ∗ ψj )(x) B (x) = (B

and ψ(0) = we obtain α j,α B (x)

Since

|x|≤α

=

Rn

Rn

ψ(u)du = 0,

$ # R (y) − 1 ψ 2j (x − y) 2nj dy. B α R (y) = 1 lim B

y→0

and ψ ∈ D(Rn ) ⊂ S (Rn ), there exists δ > 0 small enough such that α (x)| ≤ C α (y)ψ 2j (x − y) 2nj dy |B B j,R R |y|≥δ j ≤C ψ(2 (x − y)) 2nj dy. |y|≥δ

92

C2. Bochner-Riesz means of multiple Fourier integral

By letting α ≤ δ2 , we obtain α j,R (x)| ≤ C sup |B

1

|u|≥2j−1 δ

|x|≤α

|ψ(u)|du ≤ C2−j(α+ 2 ) .

The reason being on hold for the last inequality on the right side of the above inequalities is ψ ∈ S . Choosing p > 2 such that 1/p + 1/p = 1, by Hausdorﬀ-Young’s inequality, we have 1

|x|≤α

which implies

|fˆ(x)| dx

|x|≤α

2

2

≤

1/p p

|x|≤α

|fˆ(x)| dx

≤ Cf p ,

2 1 α Bj,R (x)|2 |fˆ(x) dx ≤ C2−2j(α+ 2 ) f 2p .

α (x) is a radial function, then we get Secondly, for B j,R

∞ α α j,R j,R |B (x)|2 |fˆ(x)|2 dx = ρn−1 |B (ρ)|2 |x|≥α

|ξ|=1

α

|fˆ(ρξ)|2 dξ

dρ.

Thus, by the (Lp , L2 ) restriction theorem of the Fourier transform, we have α j,R |B (x)|2 |fˆ(x)|2 dx |x|≥α

p 2/p f y dy ρ dρ ≤C ρ α Rn ∞ 2n α (ρ)|2 dρ f 2 ρ−n−1+ p |B =C j,R p α∞ α j,R ≤C |B (ρ)|2 ρn−1 dρ f 2p α α 2 ≤C |Bj,R (x)| dx f 2p n

R

∞

≤C

≤C

−n−1

|x|≥2j

α j,R |B (ρ)|2

α |BR (x)|2 dx f 2p

dx |x|≥2j 1

|x|n+2α+1

≤ C2−2j(α+ 2 ) f 2p .

f 2p

2.8 The restriction theorem of the Fourier transform

It follows

Rn

93

1

|Tj f (x)|2 dx ≤ C2−2j(α+ 2 ) f 2p .

Together with (2.8.3), the above inequality leads to 1 |Tj f (x)|p dx ≤ C p 2nj/r · 2−pj(α+ 2 ) f pp Q

= C p 2−n( Since

n+1+2α −1)pj 2n

f pp .

2n < p, n + 1 + 2α

we take

n + 1 + 2α 1 − 2n p

ε=n

,

and get (2.8.2), which completes the proof of the theorem.

Remark 2.8.1 The converse of Theorem 2.8.1 is also valid. This is proven by Tao [Ta1] in 1999. That means the Bochner-Riesz conjecture is equivalent to the restriction theorem of the Fourier transform. By Theorem 2.8.1, we are interested in the following problem. For what kind of p, (Lp , L2 )-restriction of the Fourier transforms holds. The following result is due to Tomas [To1]. Theorem 2.8.2 Let 1 ≤ p <

|ξ|=1

2n+2 n+3 .

Then we have

1 2

|fˆ(ξ)|2 dσ(ξ)

≤ Ap f Lp (Rn ) ,

Proof. Actually, we have 2 ˆ |f (ξ)| dσ(ξ) =

Rn

|ξ|=1

Set dξ(x) =

(f ∗ f )(x)

e−ix·ξ dσ(ξ)dx. |ξ|=1

e−ix·ξ dσ(ξ). |ξ|=1

f ∈ S.

94

C2. Bochner-Riesz means of multiple Fourier integral

Then we have |ξ|=1

f (x) · ( dξ ∗ f )(x)dx 1 1 + =1 . ≤ f p dξ ∗ f p p p

|fˆ(ξ)|2 dσ(ξ) =

Rn

It suﬃces to prove that dξ ∗ f ≤ Cp f p .

(2.8.4)

p

Now suppose that the radial function K(x) ∈ S , satisfying K(x) = 1, when |x| ≤ 100, # x x $ Tk (x) = K( k ) − K( k−1 ) dξ(x). 2 2 In order to prove (2.8.4), it suﬃces to show that there exists a positive number ε = ε(ρ), such that and set

Tk ∗ f p ≤ Cp 2−εk f p .

(2.8.5)

We notice that there holds (n−2) e−ix·ξ dσ(ξ) = Cn |x|− 2 J n−2 (|x|), dξ(x) = 2

|ξ|=1

and get that n Tk (y)f (x − y)dy R y

y − n−1 − K |y| 2 |f (x − y)|dy ≤ Cn K k k−1 2 2 k−1 |y|≥100×2 ≤ Cn 2−

k(n−1) 2

f 1 .

Hence, we obtain Tk ∗ f ∞ ≤ Cn 2− In addition, since

k(n−1) 2

f 1 .

(2.8.6)

Tk ∗ f 2 ≤ Tk ∞ f 2 ,

then we have Tk ∗ f 2 ≤ C2k f 2 .

(2.8.7)

2.8 The restriction theorem of the Fourier transform

95

By (2.8.6), (2.8.7) and Riesz-Thorin’s convexity theorem, we get the conclusion that

Tk ∗ f p ≤ C 2−

k(n−1) 2

1−t

(n+1)(1−t) − −1 k 2

2k

t

f p

= C2 where

1 t =1−t+ . p 2

We notice that if 1≤p< then

Let ε =

f p ,

2n + 2 , n+3

2n + 2 (n + 1)(1 − t) −1= − n − 2 > 0. 2 p 2n+2 p

− n − 2. Then (2.8.5) holds and this ﬁnishes the proof.

Remark 2.8.2 The condition in Theorem 2.8.2 that 1 ≤ p < improved by Stein to 1 ≤ p ≤ 2n+2 n+3 .

2n+2 n+3

has been

Combining Theorem 2.8.2 with Theorem 2.8.1, we can get the next corollary. Corollary 2.8.1 Let 2n 2n

n−1 , 2n + 2

then we have α (f )p ≤ Cp,α f p . BR

Proof. when

By the method of duality, we merely need to consider the case 2n 

n−1 , 2n + 2

96

C2. Bochner-Riesz means of multiple Fourier integral

we have 2n + 2 2n < . n + 1 + 2α n+3 Then, by Theorems 2.8.1 and 2.8.2, we know that when 2n 2n + 2
2n + 2 ≤ p < 2, n+3 the conclusion of the boundedness of the operator α (f ) f → BR

on Lp (Rn ) is obtained, and this ﬁnishes the proof.

Remark 2.8.3 We point out that after some proper revisions about the proof of Theorem 2.8.1 in this section, Feﬀerman [Fe4] gave a new proof of Carleson-Sj¨ olin theorem. Limited by the space, here we omit the proof in details. Many researchers take the restriction theorem of the Fourier transform as the starting point of the study on the Bochner-Riesz conjecture (see [TVV1], [Le1],[BG]). Other researchers regard the Kakeya maximal function as the starting point to study the Bochner-Riesz conjecture (see [Bou1], [Wo1]). Here the remarkable result that we state is due to Bourgain and Guth [BG]. If n ≥ 3, 0 < α < n−1 2 and ⎧ 4n+3 2 4n−3 , n = 0 (mod 3), ⎪ ⎪ ⎪ ⎪ ⎨ max(p, p ) > 2n+1 n−1 , n = 1 (mod 3), ⎪ ⎪ ⎪ ⎪ ⎩ 4n+4 2n−1 , n = 2 (mod 3), then the Bochner-Riesz conjecture is true.

2.9 The case of radial functions

2.9

97

The case of radial functions

In this section, we will use the notation Lp (Rn , r) to denote the collection of all the functions f which satisfy f ∈ Lp (Rn ) and f (x) = ϕ(|x|), and we call it the radial functions class of Lp (Rn ). In Section 2.4, we have proven that Bochner-Riesz operator T0 is unbounded on Lp (Rn ), where 2n 2n 0, then it is natural to guess that when n > 2, Tα should be bounded on Lp (Rn ) for 2n 2n
0 f p ≤ Cp,n f p BR

holds. Theorem 2.9.2 Suppose that f ∈ Lp (Rn , r), 0 < α < 2n 2n
n−1 2 ,

and

98

C2. Bochner-Riesz means of multiple Fourier integral

We merely give the proof of Theorem 2.9.2. From the proof of Theorem 2.2.2 in Section 2.2, it suﬃces to prove that B1α (f )p ≤ Cp f p , for f ∈ Lp (Rn , r), where p, α satisfy the assumptions of Theorem 2.9.2. Similarly, since the class of radial simple functions having compact support is dense in Lp (Rn , r), we merely need to prove the above inequality for any simple function f ∈ Lp (Rn , r). Assume that f ∈ Lp (Rn , r) is a simple function. By the equation (2.1.1) in Section 2.1, we have B1α (f ; x) = Cα

Rn

f (t)

α(z)

We consider a set of operators {B1

J n2 +α (|x − t|) n

|x − t| 2 +α

dt.

}, where

α(z) = α0 (1 − z) + ε, ε > 0, α0 =

n−1 2

and 0 < Re(z) < 1. According to the following three properties of the Bessel functions, (i)

t ζ 1 ζ− 1 1 2 2 Jζ (t) = 1 1 − u2 cos utdu, Re(ζ) > − , 1 2 Γ 2 Γ ζ+2 0

(ii)

|Jξ+iη (t)| ≤ Aξ eπ|η| t−1/2 ,

(iii)

|Jξ+iη (t)| ≤ Aξ e

t ≥ 0, ξ ≥ 0, t > 0, ξ ≥ 0.

π|η| ζ

t ,

α(z)

It is easy to check that the operator family {B1 } is an admissible analytic family (see Stein and Weiss [SW1]). In order to use the interpolation theorem of the analytic family of operators, we take p0 = q0 = 1, p1 = q1 >

2n , n+1

and point out that the inequality α(iy)

|B1 holds, where

f |1 ≤ Aξ eπη f 1

η ξ + iη = α(iy) + . 2

(2.9.1)

2.9 The case of radial functions In fact, we have

99

α(iy) α(iy) (x), B1 f (x) = f ∗ B1

where α(iy)

B1

n

(x) = C|x|−α(iy)− 2 Jα(iy)+ n2 (|x|).

Obviously, there holds α(iy)

B1 Then we have

α(iy) α(iy) B1 f ≤ B1 f 1 ≤ Aξ eπ|η| f 1 , 1

1

which leads to (2.9.1). Next, we need to prove α(1+iy) f B 1

p1

for

1 ≤ Aξ eπ|η| .

≤ Ap |η|ε−1 + 1 f p1 ,

(2.9.2)

2n n+1

< p1 ≤ 2, where η = −α0 y. Since α(1 + iy) = ε + iη, then (2.9.2) is transfered into B1ε+iη f p1 ≤ Ap1 |η|ε−1 + 1 f p1 .

(2.9.3)

For any f ∈ Lp1 (Rn , r), we ﬁrst give an expression of (B1ε+iη f )(x) as

B1ε+iη f (x) ∞ 1 (2.9.4) ε+iη n−2 n 1 − r2 f (t)t 2 J n−2 (r|x|)J n−2 (rt)rdrdt. = C|x|− 2 0

2

0

2

In fact, we have that if f (x) is a radial function, then so is fˆ(x). Denote by fˆ(x) = F (|x|) = F (r), and through the computation in polar coordinates, we can get that π ∞ f (t) e−irt cos θ sinn−2 θdθ tn−1 dt, F (r) = (n − 1)ωn−1 0

0

where ωn−1 is the surface area of the unit sphere in Rn . Since there holds

ε+iη 1 B1ε+iη f (x) = fˆ(y)eix·y dy, 1 − |y|2 n (2π) |y|<1

100

C2. Bochner-Riesz means of multiple Fourier integral

if we also put the computation for the integral on the right side of the above equation under the polar coordinates, we have that

ε+iη n−1 (n − 1)ωn−1 1 ε+iη 1 − r2 r F (r) B1 f (x) = n (2π) 0 π ei|x|r cos ϕ sinn−2 ϕdϕ dr × (n

=

0 2 − 1)2 ωn−1 (2π)n

×

∞

1

0

1 − r2 π

f (t)

×

−irt cos θ

e

0

ε+iη

rn−1 sin

n−2

θdθ tn−1 dt

0 π

0

ei|x|r cos ϕ sinn−2 ϕdϕ dr.

Thus, by the deformation of the property (i) of the Bessel function, we have z ζ π 2 e±iz cos ϕ sin2ζ ϕdϕ, Jζ (z) = Γ ζ + 12 Γ 12 0 and

1 Re(ζ + ) > 0, 2

we immediately have (2.9.4). By the relationship of two Bessel functions (see Welland [We1]) d 1 d rJν (|x|r)Jν (tr) = rtJν (|x|r) Jν (y) 2 2 |x| − t dr dy y=tr d −r|x|Jν (tr) Jν (y) dy y=|x|r

and

νJν (y) d Jν (y) = − Jν+1 (y), dy y

we conclude that 1 (1 − r)ε+iη J n−2 (|x|r)J n−2 (tr)rdr = 2

0

where

I1 =

0

1

1 − r2

2

|x|2

1 (I2 − I1 ), − t2

ε+iη d {rtJν (|x|r)Jν+1 (tr)}dr dr

2.9 The case of radial functions and

I2 =

1

0

1 − r2

101

ε+iη d {r|x|Jν (tr)Jν+1 (|x|r)}dr. dr

By the integration by part, we have I2 = 2(ε + iη)

1 0

ε−1+iη r 2 |x| 1 − r 2 Jν (tr)Jν+1 (|x|r)dr.

Due to the property (ii) of Bessel function, we easily have that & |I2 | ≤ Cε−1 (|η| + ε)

|x| , t

where C is a constant. Similarly, we have 7 |I1 | ≤ Cε

−1

(|η| + ε)

t . |x|

Thus, it follows from (2.9.4) and the estimate above that 7

& |f (t)|tn/2 |x| t + dt 2 2 |x| − t t |x| 0 7

& ∞ |f (t)|tn/2 |x| t + dt + 2 2 t |x| |x| t − |x| n−1 |x| n−3 |f (t)|t 2 −1 ≤ Cε (|η| + ε) |x|− 2 dt |x|2 − t2 0 n−1 n−1 ∞ |x| |f (t)|t 2 |f (t)|t 2 − n−3 − n−1 2 2 dt + |x| dt + |x| 2 2 |x|2 − t2 |x| t − |x| 0 n+1 ∞ |f (t)|t 2 − n−1 dt . + |x| 2 2 2 |x| t − |x|

n−2 ε+iη (B1 f )(x) ≤ Cε−1 (|η| + ε)|x|− 2

|x|

For any integral on the right side of the inequality above, if we take it for granted that f equals to zero in the complement of the region of the integration, then in order to prove (2.9.3), we merely need to prove that for 2n < p1 ≤ 2, n+1

102

C2. Bochner-Riesz means of multiple Fourier integral

there holds Ti f p1 ≤ Ap1 f p1 ,

(2.9.5)

for i = 1, 2 and f ∈ Lp1 (Rn , r), where − n−3 2

T1 f (x) = |x|

|f (t)|t 2 dt, |x|2 − t2

∞

|f (t)|t 2 dt. |x|2 − t2

0

and − n−1 2

T2 f (x) = |x|

0

n−1

∞

n−1

Since there holds

p1 ∞ |f (t)|t n−1 2 = |x| dt dx 2 − t2 |x| n 0 R p1 ∞ ∞ |f (t)|t n−1 (n−3)p1 2 n−1− 2 =C r dt dr, 2 − t2 r 0 0

T1 f pp11

− n−3 p1 2

by the change of variables r 2 = σ, t2 = τ , we get that p1 1 ∞ ∞ n−2 − n−2 τ 2p |f (τ 2 )| τ n−1 4 2p p1 dτ dσ. T1 f p1 = C σ − τ σ 0 0 By another change of variables r 2 = σ, we have ∞ ∞# $p 1 n−2 1 |f (r)|p1 r n−1 dr = C dτ. τ 2p |f (τ 2 )| f pp11 = C 0

0

Let φ(τ ) = τ

and

∞

ψ(σ) = 0

n−2 2p

1 f (τ 2 )

− n−2 φ(τ ) τ n−3 4 2p dτ. σ−τ σ

Then the inequality T1 f p1 ≤ Ap1 f p1 can be reduced to prove ψp1 ≤ Ap1 φp1 . Now we separate ψ into ψ = ψ1 + ψ2 , where ∞ φ(τ ) dτ ψ1 (σ) = σ−τ 0

2.9 The case of radial functions and

ψ2 (τ ) =

103 τ n−3 − n−2

∞

φ(τ )

σ

4

2p

−1

dτ. σ−τ By the Lp1 -boundedness of the Hilbert transform with p1 > 1, it suﬃces to estimate ψ2 . Let n−3 n−2 τ 4 − 2p1 −1 −1 , K(σ, τ ) = (σ − τ ) σ we have ∞ φ(τ )K(στ )dτ. ψ2 (σ) = 0

0

It is obvious that in order to prove ψ2 p1 ≤ Ap1 φp1 , we merely need to prove that, for any h ∈ Lp1 and hp1 ≤ 1, ∞ ψ2 (σ)h(σ)dσ ≤ Ap1 φp1 0

holds. In fact, it follows that ∞ ∞ ∞ ψ2 (σ)h(σ)dσ = K(σ, τ )φ(τ )h(σ)dτ dσ 0 0 0 ∞ ∞

τ 1 1 p1 p p1 1 |φ(τ )| ≤ K (σ, τ ) σ 0 0

σ 1 1 p1 p p 1 K 1 (σ, τ ) dτ dσ × |h(σ)| τ ∞ ∞ 1/p1

τ 1 p1 p1 |φ(τ )| |K(σ, τ )| dτ dσ ≤ σ 0 0 ∞ ∞ 1/p1

σ 1 p1 p1 × |h(σ)| |K(σ, τ )| dτ dσ . τ 0 0 By a change of variables α = στ , we obtain that 1 ∞ τ n−3 ∞ − n−2

τ 1 4 2p1 τ p − 1 p1 σ 1 dσ |K(σ, τ )| dσ = σ σ σ−τ 0 0 ∞ n−3 − n−2 α 4 2p1 − 1 − p11 = dα. α α−1 0

104

C2. Bochner-Riesz means of multiple Fourier integral

It is easy to check that the integral on the right side of the above equation converges when 2n < p1 ≤ 2. n+1 Similarly, when 2n < p1 ≤ 2, n+1 we have that

∞ 0

Hence we get

|K(σ, τ )|

∞

0

σ τ

1 p1

dτ < c < ∞.

ψ2 (σ)h(σ)dσ ≤ Ap1 φp1 ,

provided that hp1 ≤ 1. This implies T1 f p1 ≤ Ap1 f p1 , for

2n < p1 ≤ 2. n+1

Similarly, we can get the similar inequality for T2 . Consequently, (2.9.3) holds and then (2.9.2) is valid. Now we choose p0 = q0 = 1, p1 = q1 =

2n +ε n+1

and

n−1 (1 − z) + ε. 2 By the interpolation theorem on analytic families of operators, (2.9.1) and (2.9.2), we get that α(t) B1 f ≤ Ap f p , α(z) =

p

where

t 1 1−t + , 0 < t < 1. = p p0 p1

Thus we have 1 1 1 1 1 α(t) − ε = (1 − t) 1 − + 1− + = n−1 p p1 p1 p1 p1 2

2.10 Almost everywhere convergence

105

and

n−1 + ε. 2 Since ε is arbitrary, it is not hard to get the conclusion of Theorem 2.9.2. ε ≤ α(t) ≤

2.10

Almost everywhere convergence

In this section, we will study the almost everywhere convergence of the Bochner-Riesz means of Fourier integral of f . That is to say whether or not α lim BR (f )(x)

R→∞

exists and is ﬁnite for almost every x ∈ Rn and for all f ∈ Lp (Rn ), 1 ≤ p < ∞. Since the operator α (f ) f → BR is linear, thus we discuss the property of the almost everywhere convergence of any family of linear operators f → TR f . Suppose that {TR : R > 0} is a family of linear operators and each TR maps Lp (Rn ), 1 ≤ p ≤ ∞, to a measurable function space M (Rn ). Besides, we also assume that, for any g ∈ D(Rn ), lim TR g(x)

R→∞

exits and is ﬁnite for a.e. x ∈ Rn . To guarantee the almost everywhere convergence of TR f (x) when R → ∞, it suﬃces to prove that x ∈ Rn : lim sup |TR1 f (x) − TR2 f (x)| > λ = 0 R1 ,R2 →∞ is valid for every λ > 0. However, for any f ∈ Lp (Rn ), we choose g ∈ D(Rn ) and set h = f − g. Obviously, for any λ > 0, we have n x ∈ R : lim sup |TR1 f (x) − TR2 f (x)| > λ R1 ,R2 →∞ ≤ x ∈ Rn : lim sup |TR1 h(x) − TR2 h(x)| > λ . R1 ,R2 →∞

106

C2. Bochner-Riesz means of multiple Fourier integral

Thus, in order to prove the a.e. convergence of TR f (x), it suﬃces to show that n : lim sup |T h(x) − T h(x)| > λ x ∈ R → 0, R1 R2 R1 ,R2 →∞ as hp → 0. We denote by T∗ f (x) = sup |TR f (x)|, R>0

and notice that λ n n . x ∈ R : lim sup |TR1 h(x) − TR2 h(x)| > λ ≤ x ∈ R : T∗ h(x) > 2 R1 ,R2 →∞ Thus, to guarantee the a.e. convergence of TR f (x), we merely need to prove that |{x ∈ Rn : T∗ f (x) > λ}| → 0, as f p → 0 for any ﬁxed λ > 0. It is necessary to point out that the validity of the last relationship above is closely related to the type of the operator T∗ . In the following, we begin with giving the deﬁnition of the type of the operator T . Definition 2.10.1 Let 1 ≤ p, q ≤ ∞. The operator T maps Lp (Rn ) to Lq (Rn ). (i) If the operator T satisﬁes the inequality T f q ≤ Cp,q f p , then we call T to be of type (p, q) or strong type (p, q). (ii) If the operator T satisﬁes the inequality f p q n |{x ∈ R : T f (x) > λ}| ≤ Cp,q , λ then we call T an operator of weak type (p, q), where Cp,q is independent of f. By Deﬁnition 2.10.1, if T∗ is of weak type (p, q), then we have that {x ∈ Rn : T∗ f (x) > λ} → 0, as f p → 0. Thus, we obtain the following important facts.

2.10 Almost everywhere convergence

107

Theorem 2.10.1 Suppose that {TR : R > 0} is a family of linear operators and each TR maps Lp (Rn ) (1 ≤ p ≤ ∞) to the measurable function space M (Rn ). Moreover, we also assume that for any g ∈ D(Rn ), lim TR g(x) R→∞

exits and is ﬁnite for a.e x. If the maximal operator T∗ f (x) = sup |TR f (x)| R>0

is of weak type (p, q), then for every f ∈ Lp (Rn ), lim TR f (x) exists and is R→∞

ﬁnite almost everywhere. Here q ≥ 1. Corollary 2.10.1 Let α >

n−1 2 .

If f ∈ Lp (Rn ) (1 ≤ p < ∞), then

α lim BR f (x) = f (x)

R→∞

holds for a.e. x ∈ Rn . Proof. For α >

n−1 2 ,

let α B∗α (f ; x) = sup |BR (f ; x)|, R>0

we ﬁrst prove B∗α (f ; x) ≤ CM f (x), where M f is the Hardy-Littlewood maximal function of f and C is independent of f . In fact, the equation (2.1.3) int Section 2.1 of this chapter reveals that ∞ J n +α (Rt) α n BR f (x) = Cn,α R fx (t) 2 n +α tn−1 dt. (Rt) 2 0 Separating the integration region on the the right side of the above integral as 1 1 ,∞ , (0, ∞) = 0, R R and on 0, R1 and R1 , ∞ , using Jm (f ) ∼ Cr m, as r → 0, and Jm (r) = 1 O(r − 2 ), when r → ∞, we have that 1 α f (x)| ≤ C |BR

R

Rn 0

|fx (t)|tn−1 dt + R

n−1 −α 2

∞

1 R

|fx (t)|t

n−1 −α−1 2

dt .

108

C2. Bochner-Riesz means of multiple Fourier integral

It easily follows that R

1 R

n 0

|fx (t)|tn−1 dt ≤

By partial integration, we get that ∞ n−1 n−1 R 2 −α |fx (t)|t 2 −α−1 dt ≤ 1 R

M f (x) . n

C M f (x). α − n−1 2

Hence, we get the conclusion that B∗α (f ; x) ≤ CM f (x). According to the well-known property of the Hardy-Littlewood maximal function, B∗α is of type (p, p) for 1 < p < ∞ and B∗α is of weak type (1, 1). Besides, Section 2.2 in this chapter has pointed out that: for any f ∈ D(Rn ), α (f ; x) = f (x) lim BR

R→∞

holds for a.e x. Hence, from Theorem 2.10.1, we can directly deduce that for any f ∈ Lp (Rn ) with 1 ≤ p < ∞, α (f ; x) lim BR

R→∞

exits and is ﬁnite almost everywhere. To prove α (f ; x) = f (x) a.e. lim BR

R→∞

holds, it suﬃces to show that there exists a subsequence {Rk }, such that lim Bkα (f ; x) = f (x)

k→∞

for a.e. x ∈ Rn . α n α In fact, when α > n−1 2 , the kernel BR (x) ∈ L(R ) and BR is independent of R. By the property of convolution, we have α (f )p ≤ Cp f p . BR

Due to Theorem 2.2.1, we have α lim BR (f ) − f p = 0.

R→∞

2.10 Almost everywhere convergence

109

Thus there exists a subsequence {Rk } such that α (f ; x) = f (x), lim BR k

k→∞

a.e.

and this completes the proof of the corollary. For the case when α ≤ follow.

n−1 2 ,

there are some basic results due to Stein as

Theorem 2.10.2 Let 1 < p < ∞, n ≥ 2 and 1 1 n−1 . (n − 1) − < α ≤ 2 p 2 Then we have B∗α (f )p ≤ Cp,n,αf p and α (f ; x) = f (x) lim BR

R→∞

for a.e. x ∈ Rn . Obviously, the second assertion is directly obtained from Theorem 2.10.1 and the ﬁrst assertion. As far as the ﬁrst assertion is concerned, its proof is the same as the case of the series in the following and thus we omit the proof here. One can see Theorems 3.4.1 and 3.4.3 in Section 3.4 of Chapter 3. In Theorem 2.10.2, when n = 2, the ﬁrst assertion shows that if 2 2

2 2 , 1 + 2α 1 − 2α

⊂

4 4 , 3 + 2α 1 − 2α

(2.10.1) ,

and applying Carleson-Sj¨ olin theorem, that is Theorem 2.5.2, in Section 2.5, one naturally anticipates the following results. When 4 4
110

C2. Bochner-Riesz means of multiple Fourier integral

the inequality (2.10.1) should hold. Though the conjecture is still not proven completely, yet Carbery [Ca1] has promoted the conclusion of Theorem 2.10.2 in an essential step for the case n = 2. If we replace the normal maximal operator α f )(x)| sup |(BR R>0

by the following maximal operator with the lacunary form α f )(x)|, sup |(BR j j

then the conjecture has been veriﬁed by Igari [Ig1] and independently by Cordoba and Lopez-Melero [CoL1]. Theorem 2.10.3 Suppose that there exists a positive number q > 1 such that the sequence {Rj }∞ j=1 satisﬁes Rj+1 ≥q Rj for any j ∈ N. If

we have

4 4
Particularly, α (f )(x) = f (x) lim BR j

j→∞

holds for a.e. x ∈ Rn . Proof. The proof of Theorem 2.10.3 mainly depends on Theorem 2.7.2 about the vector valued function and the Littlewood-Paley decomposition theorem (see Stein [St4]). Firstly, it is easy to prove that there exists φ ∈ D(R) such that suppφ ⊂ (1, 3) and

∞ n=−∞

φ 2−n ρ = 1

2.10 Almost everywhere convergence with ρ > 0. Set

111

φj (x) = φ

and

ψj (x) =

2|x| Rj

1 − φj (x),

|x| ≤ Rj ,

0,

otherwise.

Clearly we obtain that

α (x) BR j

α |x|2 = 1− 2 Rj +

α

α |x|2 |x|2 j (x). = 1− 2 ∗ φj (x) + 1 − 2 ∗ψ Rj Rj +

Put

η

|x| Rj

=

+

|x|2 1− 2 Rj

α ∗ ψj (x), +

then we have η ∈ D(R) and

α |x|2 j (x) = η |x| = ( ∗ψ η ) 1 (x). 1− 2 Rj Rj Rj +

Thus, we get

α |x|2 j (x) ∗ f (x) = ( 1− 2 ∗ψ η ) 1 ∗ f (x). Rj Rj +

By the property of the convolution operator ([SW1]), it follows that

2 α |x| ∗ ψj (x) ∗ f (x) ≤ CM f (x). sup 1 − 2 Rj j + Hence we conclude that

112

C2. Bochner-Riesz means of multiple Fourier integral

α 2 |x| α 1 − f (x)| ≤ CM f (x) + sup · φ ∗ f (x) sup |BR j j Rj2 j j + ⎞1 ⎛ 2 2 α ≤ CM f (x) + ⎝ (φj ∗ f )(x) ⎠ . BR j j

From the inequality above and Theorem 2.7.2 in Section 2.7, we can obtain that when 4 4
⎛ ⎞ 1 2 α ⎠ (f )| ≤ C ⎝f p + |φj ∗ f |2 sup |BR . j i

p

p

Consequently, by the theorem with the Littlewood-Paley decomposition (see Stein [St4]), we can deduce the conclusion of the theorem. This completes the proof. From the conditions of Theorem 2.10.3, we also know that when p = 2, we have B∗α (f )2 ≤ Cn,α f 2 (α > 0), where we require the order α > 0. Naturally, one hopes the inequality above can still hold for the case when α = 0. However, the problem is still an open problem until now. Similarly, if we replace the normal maximal operator 0 f (x)| sup |BR

R>0

0 f (x)| in the lacunary form, then, the with the maximal operator supj |BR j above guess is correct, Lu [Lu8] gives a proof as follows.

Theorem 2.10.4 Let f ∈ L2 (Rn ). If there exists a positive number q > 1, Rj+1 such that the positive number sequence {Rj }∞ 1 satisﬁes Rj ≥ q, ∀ j ∈ N , then we have 0 (f )|2 ≤ Cf 2 . sup |BR j j

In particular, we have 0 (f )(x) = f (x), lim BR j

j→∞

a.e. x.

2.10 Almost everywhere convergence

113

The proof of the theorem is similar to the one in the case of the series, which can be checked out at Theorem 3.5.6 in Chapter 3. Remark 2.10.1 In 1988, Carbery, Rubio de Francia and Vega carried The2n . If orem 2.10.4’ to the ﬁnal form in [CRV1]. Let f ∈ Lp (Rn ), 2 ≤ p < n−1 there exists a positive number q > 1, such that the positive number sequence Rj+1 {Rj }∞ 1 satisﬁes Rj ≥ q for all j ∈ N, then we have 0 lim BR (f )(x) = f (x) j

j→∞

for a.e. x ∈ Rn . In addition, it should be pointed out that the range of p above is best (see [CSo1]). As pointed out as previous chapters, the results in Theorem 2.10.3 are not best. When n > 1 and 0 < α < n−1 2 , the convergence of the Bochnern Riesz means almost everywhere in R has a conjecture as follows: If n > p n 1, < α < n−1 2 , f ∈ L (R ), and 2n 2n
R→∞

for a.e. x ∈ Rn . The Bochner-Riesz conjecture has not been solved yet as of now. But when p ≥ 2, Carbery-Rubio de Francia-Vega [CRV1] gave an aﬃrmative answer to the conjecture above. Let us now state this result as follows. Theorem 2.10.5 Let 0 < λ ≤ 12 (n − 1), n ≥ 2. If f ∈ Lp (Rn ) and 2 ≤ p < pλ , then λ f (x) = f (x) (2.10.2) lim BR R→∞

holds for a.e. x ∈

Rn ,

where pλ =

2n n−1−2λ .

λ Denote B∗λ as the corresponding maximal operator of BR , that is, λ f (x) . B∗λ f (x) = sup BR R>0

Consequently, the proof of Theorem 2.10.2 can be attributed to doing research on the maximal operator of B∗λ . But now, we will study it through

114

C2. Bochner-Riesz means of multiple Fourier integral

the decomposition of space, not discussing the boundedness of the maximal operator of type (p, p) as usual. Thus we convert the consideration of the boundedness of the operator into the problem of the boundedness of the square integral space with a power weighted function. For the sake of conciseness, we denote L2 (|x|−α ) = L2 (Rn , |x|−α dx), and Lp = Lp (Rn ) for 1 < p < ∞. Thus when 2 ≤ p < pλ for any f ∈ Lp (Rn ), there exists 2 , 0 ≤ α < 1 + 2λ = n 1 − pλ such that Lp ⊂ L2 + L2 (|x|−α ). In fact, we set χ{x:|x|<1} to be the characteristic function of the unit ball, then (2.10.3) f = f χ{x:|x|<1} + f (1 − χ{x:|x|<1}). By H¨older’s inequality, it is easy to check that f χ{x:|x|<1} ∈ L2 (Rn ) and f (1 − χ|x|<1 ) ∈ L2 (|x|−α ). Thus Theorem 2.10.5 can be derived from the boundedness of the maximal operator with a weighted function. Theorem 2.10.6 Let λ > 0 and 0 ≤ α < 1 + 2λ ≤ n. For any Schwartz function f , we have 2 λ −α |f (x)|2 |x|−α dx. (2.10.4) B∗ f (x) |x| dx ≤ Cα,λ Rn

Rn

It is easy to see that Theorem 2.10.6 includes the consequences of Theorem 2.10.5. Actually, because of the decomposition formula (2.10.3), for any f ∈ Lp , we have f = f 1 + f2 , where f1 ∈ L2 (Rn ) and f2 ∈ L2 (|x|−α ). If Theorem 2.10.6 has been proven, then it follows from Theorem 2.10.6 that λ lim BR f1 (x) = f1 (x) R→∞

holds for almost everywhere x ∈ Rn .

2.10 Almost everywhere convergence

115

Since the measure |x|−α dx is absolutely continuous with respect to the Lebesgue measure, we have λ lim BR f2 (x) = f2 (x)

R→∞

for almost any x ∈ Rn . Consequently, next our main task is to show Theorem 2.10.6. For a given small positive real number δ > 0, we denote D as the diﬀerential operator. Set mδ ∈ C ∞ satisfying supp mδ ⊂ [1 − δ, 1], and 0 ≤ mδ (t) ≤ 1,

k δ D m (t) ≤ Cδ−k

(2.10.5)

for every positive integer k. We deﬁne the operator Stδ as δ δ ˆ S+ t f (ξ) = m (t|ξ|)f (ξ).

Next we give a fundamental estimate of the operator Stδ . Lemma 2.10.1 For any δ > 0 and 0 ≤ α < n, we have that Rn

2 δ St f (x) ≤ Cα Aα (δ)

2 1

Rn

|f (x)|2

dx , |x|α

(2.10.6)

where Cα is a constant which is independent of δ, and ⎧ 2−α , 1 < α < n, ⎨δ Aα (δ) = δ| log δ|, α = 1, ⎩ δ, 0 ≤ α < 1. Assume the inequality (2.10.6) holds. Then we will prove Theorem 2.10.6 is valid. We denote the corresponding maximal operator of Stδ by S∗δ , that is, S∗δ f (x) = sup Stδ f (x) . t>0

Similar to Section 2.7 in this chapter, we have the following decomposition ∞ −k 2 λ 2−kλ m2 (|ξ|), 1 − |ξ| + = k=0

116

C2. Bochner-Riesz means of multiple Fourier integral

thus we have B∗λ f (x)

≤

∞

2−kλ S∗2−k f (x).

k=0

Clearly we have that the terms about k = 0, 1 can be controlled by the Hardy-Littlewood maximal function, and when −n < α < n, the HardyLittlewood maximal function operator is bounded on L2 (|x|α ). Thus we mainly pay attention to the terms that k ≥ 1. To research the above maximal operator, we introduce the square operator ∞ 1 2 δ δ 2 dt |St f (x)| G f (x) = t 0 6δ , whose corresponding multiplier is and the operator G m 6 δ (t) = δt

d δ m (t) dt

6 δ (t) satisﬁes (2.10.5) similar to instead of mδ (t). It is easy to check that m δ m (t). By the diﬀerential integral fundamental theorem, we have ∞ δ d δ 2 δ 6δ f (x). 2 St f (x) St f (x) dt ≤ 2δ −1 Gδ f (x)G (S∗ f (x)) ≤ dt 0 Next we will consider the estimate about Gδ . Let a function ψ satisfy ψ(0) = 0, and when

1 2

≤ |ξ| ≤ 2, we have = 1. ψ(ξ)

Then, we denote

ψk (x) = 2−kn ψ 2−k x .

Then when 2k−1 ≤ t ≤ 2k , the support of mδ (t|ξ|) is the subset of the set $ # ξ : 2−k (1 − δ) ≤ |ξ| ≤ 2−k+1 , and when ξ is in the set, we have

k (ξ) = ψ 2k ξ = 1. ψ

2.10 Almost everywhere convergence

117

Therefore, we have δ δ ˆ S+ t f (ξ) = m (t|ξ|)f (ξ) k (ξ)fˆ(ξ) = mδ (t|ξ|)ψ

= Stδ (ψ k ∗ f )(ξ). Because of Lemma 2.10.1, we obtain

Rn

2 dt dx dx δ S f (x) ≤ C A (δ) |ψk ∗ f (x)|2 α . t α α α t |x| |x| k−1 n 2 R 2k

About the summation of k ∈ Z, using the fact that when −n < α < n, the operator 1

2 f→ |ψk ∗ f |2 is bounded on L2 (|x|−α ) (see [GR1]), then we get 2 dx dx δ ≤ Cα Aα (δ) |f (x)|2 α . G f (x) α |x| |x| Rn Rn Consequently we get 2 dx dx δ −1 ≤ Cα Aα (δ)δ |f (x)|2 α . S∗ f (x) α |x| |x| Rn Rn

(2.10.7)

Set δ = 2−k , then we get the boundedness of B∗λ on L2 (|x|−α ). Here we just choose 0 ≤ α ≤ 1 when λ > 0; or 1 < α < n when λ > 12 (α − 1), thus we have the conclusion of Theorem 2.10.6. Now we turn to the proof of the fundamental estimate. Here we consider the duality operator of Stδ . In Lemma 2.10.1, we need to show the boundedness of the operator δ 2 −α 2 n dt dx . St : L (|x| ) → L [1, 2] × R , t |x|−α It is easy to note that the duality spaces of L2 (|x|−α ) and 2 n dt dx L [1, 2] × R , t |x|−α are L2 (|x|α ) and 2

L

dt dx [1, 2] × R , t |x|α n

.

118

C2. Bochner-Riesz means of multiple Fourier integral

For gt ∈ L

2

dt dx [1, 2] × R , t |x|α

n

,

and f ∈ L2 (|x|−α ), denote 6 δ (x) = K δ (−x). K t t We conclude that (f, (Stδ )∗ gt ) = (Stδ f, gt ) 2 dt Stδ f (x)gt (x) dx = t Rn 1 2 dt Ktδ (x − y)f (y)dygt (x) dx = t Rn 1 Rn 2 dt = Ktδ (x − y)gt (x) dxf (y)dy t n n 1 R R 2 6 δ (y − x)gt (x) dt dxf (y)dy, K = t t Rn Rn 1 then we obtain (Stδ )∗ gt (x)

2

= 1

Stδ gt (x)

dt . t

Because of the property of the duality, the proof of Lemma 2.10.1 is equivalent to proving the following inequality,

n

R

2 1

2

dt Stδ gt (x) t

|x| dx ≤ Cα Aα (δ) α

Rn

1

2

|gt (x)|2

dt α |x| dx. (2.10.8) t

In order to prove (2.10.8), we now introduce the Sobolev spaces with the fractional order. When 0 < β < 1, we deﬁne the diﬀerential operator with the fractional order as follows D f (x) = β

Rn

|f (x + y) − f (x)|2 dy |y|2β |y|n

12 .

When β ≥ 1, we deﬁne D β by the diﬀerence with a higher order which is not smaller than [β] + 1. We also deﬁne the Sobolev spaces with the homogeneous order L2β (Rn ) as follow # $ L2β (Rn ) = f : Dβ fˆ < ∞ . 2

2.10 Almost everywhere convergence

119

Due to the Plancherel theorem, we have that |f (x + y) − f (x)|2 dy β 2 dx D f = |y|2β |y|n 2 Rn Rn 1 dy = |f (x + y) − f (x)|2 dx 2β n |y| |y| Rn Rn −i2πy·ξ 2 |e − 1| ˆ 2 = dy f (ξ) dξ. |y|2β+n Rn Rn

(2.10.9)

Now we consider the integral in (2.10.9) I(ξ) =

Rn

2 −i2πy·ξ e − 1 dy. |y|2β+n

Set ρ is a rotational transform with the original point as the circle, and ρt as its transpose. Since a rotational transform is orthonormal transform with the determinant 1, then we have det(ρ) = det ρt = 1. Thus we have

|e−i2πy·ρξ − 1|2 dy |y|2β+n n R t |e−i2πρ y·ξ − 1|2 = dy |y|2β+n Rn −i2πy·ξ |e − 1|2 dy = |y|2β+n Rn = I(ξ).

I(ρξ) =

This means that I(ξ) is rotation-invariant. Thus I(ξ) = I0 (|ξ|). Next set ξ = |ξ|η, where η is a unit vector. We have

|e−i2πy·|ξ|η − 1|2 dy |y|2β+n Rn −i2π|ξ|y·η |e − 1|2 dy = 2β+n Rn |y| |e−i2πy·η − 1|2 = |ξ|2β dy |y|2β+n Rn = |ξ|2β I(η).

I(ξ) =

Then we will prove that I(η) is a constant.

120

C2. Bochner-Riesz means of multiple Fourier integral

Since e−i2πy·η ≤ 2 and e−i2πy·η − 1 ≤ C|y|, with the polar coordinate expression, we have −i2πy·η −i2πy·η 2 2 e e − 1 − 1 dy + dy |I(η)| ≤ |y|≥1 |y|≤1 |y|2β+n |y|2β+n 1 ∞ 2 r 1 ≤C r n−1 dr + C rn−1 dr 2β+n 2β+n r r 0 1 < +∞. Now let us turn to (2.10.9). We obtain β 2 |fˆ(ξ)|2 |ξ|2β dξ. D f = C 2

Rn

(2.10.10)

Due to the Plancherel theorem and (2.10.10), (2.10.8) is equivalent to 2 2 1α 2 δ 1 dt dt D 2 m (t| · |)ˆ gt (·) ≤ Cα Aα (δ) |D 2 α gˆt (ξ)|2 dξ. (2.10.11) t 2 t 1 Rn 1 If we calculate the diﬀerence with the deﬁnition of Dβ and use the Cauchy-Schwartz’s inequality with the variable t, the integral of the nonzero part of the function with the measure dtt being at least Cδ. Thus the left side of (2.10.11) is controlled by 2 2 dt 12 α δ Cδ gt (ξ)) dξ . D (m (t|ξ|)ˆ t n 1 R Consequently, if we can prove that mδ (t·) is a multiplier in the pointwise in the homogeneous Sobolev spaces, and the constant is uniformly controlled 1

1

by Cα δ − 2 Aα2 (δ), provided that 1 ≤ t ≤ 2. Thus Lemma 2.10.1 will be proven. Notice the homogeneity property, we merely need to prove the inequality when t = 1. Proposition 2.10.1 If 0 ≤ β < n2 , we have δ 2 m f 2 ≤ Cβ A2β (δ)δ−1 f 2L2 . Lβ

β

Equivalently, if 0 ≤ α < n, we have 2 dx dx δ −1 ≤ C A (δ)δ |f (x)|2 α . S f (x) α α α |x| |x| Rn Rn

2.10 Almost everywhere convergence

121

To prove Proposition 2.10.1, we need the following estimate. Proposition 2.10.2 If 0 < ε < 12 , we have |f (x)|2 dx ≤ Cβ Bβ (ε)f 2L2 , β

||x|−1|≤ε

⎧ 2β ε , 0 ≤ β < 12 , ⎪ ⎪ ⎪ ⎪ ⎨ ε| log ε|, β = 12 , Bβ (ε) = ⎪ ⎪ ⎪ ⎪ ⎩ 1 1 ε, 2 < β < 2 n.

where

Proof. Taking the Fourier transform, Proposition 2.10.2 is equivalent to ˆ 2 dx ≤ C B (ε) |f (x)|2 |x|2β dx. f (x) β β Rn

||x|−1|≤ε

By the property of the duality, for the function g whose support set is a subset of {x : ||x| − 1| ≤ ε}, then the above inequality is also equivalent to dx |ˆ g (x)|2 2β ≤ Cβ Bβ (ε) |g(x)|2 dx. |x| n R ||x|−1|≤ε Using the formula of the inner product, the convolution formula and H¨ older’s inequality, when 0 < β < 12 n, it follows 1 1 (y) = C n−2β . | · |2β |y| Thus we have dx 2 dx |ˆ g (x)| = (g ∗ g6)(x) 2β 2β |x| |x| Rn R n 1 = g∗6 g (x) n−2β dx |x| Rn 1 g(x)g(y) dxdy = n−2β |x − y| ||x|−1|≤ε ||y|−1|≤ε 1 ≤C |g(x)|dx dy g(y) sup n−2β x |x ||y|−1|≤ε − y| ||x|−1|≤ε 2β−n |g(x)|dxg2 ≤C sup |x − ·| ≤

||x|−1|≤ε Cg22 sup |x x

x

2β−n

− ·|

L1 (||y|−1|≤ε) .

L1 (||y|−1|≤ε)

122

C2. Bochner-Riesz means of multiple Fourier integral

By the change of the variable v = (v1 , v ), v1 ∈ R, v ∈ Rn−1, we have the following estimate dv ≤ Cβ Bβ (ε). n−2β |v1 |≤ε,|v |≤1 |v| It is easy to have that |v1 |≤ε,|v |≤ε

and |v1 |≤ε,ε≤|v |≤1

dv ≈ |v|n−2β

dv ≈ |v|n−2β

0

ε

rn−1 dr ≈ ε2β ≤ Bβ (ε), r n−2β

ε

0

dv1

|v |≥ε

dv ≈ε |v |n−2β

1 ε

r n−2 dr ≈ Bβ (ε). rn−2β

Next we will give some estimates about the kernel of the operator S δ . Denote K(x) = K δ (x) as its kernel, such that K(ξ) = mδ (|ξ|). Suppose ϕ ∈ C ∞ (R), whose support set is [−1, 2], and satisﬁes ϕ(t) = 1 when 0 ≤ t ≤ 1. Set ⎧ j = 0, ⎨ ϕ(|x|), hj (x) = ⎩ ϕ(2j |x|) − ϕ(2j−1 |x|), j ≥ 1. Thus we get the binary unit decomposition {hj (x)}. Denote Kj (x) = K(x)hj (δx). We have the decomposition K(x) =

∞

Kj (x).

j=0

j are radial functions, when |ξ| = r, we can denote Since Kj and K j (ξ). j (r) = K K The following lemma will give an elaborate estimate about the kernel.

2.10 Almost everywhere convergence

123

Proposition 2.10.3 For any positive integer m, if 0 ≤ α < n, we have ∞ j (r)|rα−1 dr ≤ C2−mj δ. |K 0

Proof. that

j = mδ ∗ h Denote h(x) = ϕ(2|x|) − ϕ(|x|). Since K j (δ·), we have j (ξ) = 2nj δ−n mδ ∗ h 2j δ−1 · (ξ) K = 2nj δ−n mδ (ξ − η) h(2j δ −1 η)dη Rn mδ (ξ − 2−j ξη) h(η)dη. = Rn

Due to the fact that h equals 0 in the neighborhood of the origin, for any multi index a = (a1 , a2 , . . . , an ), we have h(η)dη = 0. ηa Rn

Expanding mδ at ξ in Taylor’s series, we have Kj (ξ) = Rm (ξ, η) h(η)dη, Rn

where the remaining term Rm satisﬁes a δ |Rm (ξ, η)| ≤ D m |2−j δη|m ≤ C2−mj |η|m . |a|=m

∞

Here h is a Schwartz function, the integral of the product of h and |η|m is n bounded, then for any positive integer m, ξ ∈ R , we have (2.10.12) Kj (ξ) ≤ C2−mj . j (ξ), if |ξ| < 1 , then On the other hand, in the integral of the deﬁnition of K 2 we have |η| > 2j−2 δ −1 . Using the fact that h is a Schwartz function again, for any positive integer m and |ξ| < 12 , we have K (ξ) j ≤ C2−mj δ m .

124

C2. Bochner-Riesz means of multiple Fourier integral

Consequently, we obtain 1 2 α−1 −mj m |Kj (r)|r dr ≤ C2 δ 0

0

1 2

r α−1 dr ≤ C2−mj δ.

For the other part, set S = {ξ ∈ Rn : 1 − 2δ < |ξ| < 1 + 2δ}, we have ∞ Kj (r) rα−1 dr ≤ C Kj (ξ) dξ 1 n 2

R =C + Kj (ξ) dξ n S R \S −mj δ+C mδ (ξ − 2−j δη)dξdη h(η) ≤ C2 j n |η|≥2 R \S δ −mj ≤ C2 δ + C m h(η) dη 1

≤ C2

−mj

|η|≥2j

δ.

Proposition 2.10.4 If β > 0, we have ⎧ −β δ , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ −(β+ 1 ) 1 2 β δ |D m (|ξ|)| ≤ Cβ , δ 2 |ξ| − 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ 12 −(β+ 12 n) δ |ξ| ,

||ξ| − 1| ≤ 2δ, 0 ≤ |ξ| ≤ 2, |ξ| ≥ 2.

Proof. We just prove the case of 0 < β < 1, the case β ≥ 1 one can get the similar estimate about the diﬀerence with the higher order. Firstly, considering the case when ||ξ| − 1| ≤ 2δ, we divide the integral about D β in its deﬁnition into two parts: |η| ≤ δ and |η| > δ. We have |η| δ . m (|ξ + η|) − mδ (|ξ|) ≤ |η|∇mδ ∞ ≤ δ Thus we conclude that δ m (|ξ + η|) − mδ (|ξ|)2 dη |η|2 −2 ≤ δ dη 2β+n |η|2β |η|n |η|≤δ |η|≤δ |η| δ = Cδ −2 r−2β+1 dr 0

= Cδ−2β .

2.10 Almost everywhere convergence

125

When |η| ≥ δ, we have δ m (|ξ + η|) − mδ (|ξ|)2 dη dη ≤C 2β n n+2β |η| |η| |η|≥δ |η|≥δ |η| ∞ dr =C 1+2β r δ = Cδ −2β . Secondly, we consider the case of |ξ| ≥ 2. Noting that the support set of mδ is [1 − δ, 1], we conclude that 2 δ 2

m (|ξ + η|) − mδ (|ξ|) dη β δ D m (|ξ|) = |η|2β |η|n Rn |mδ (|ξ + η|)|2 = dη |η|n+2β Rn |mδ (|η|)|2 dη = n+2β Rn |η − ξ| |mδ (|η|)|2 = dη n+2β 1−δ≤|η|≤1 |η − ξ| 1 1 ≤ C n+2β r n−1dr |ξ| 1−δ 1 1 dr ≤ C n+2β |ξ| 1−δ δ ≤ C n+2β . |ξ| Finally we consider the case when |ξ| ≤ 2 and ||ξ| − 1| ≥ 2δ. Think of the support set, we should estimate the integral with the variable ξ dη . n+2β {η:||ξ−η|−1|≤δ} |η| If |ξ| ≈ 2−j , we consider the integral on the rings |η| ≈ 2−j+r (r ≥ 0) respectively. The integral area has a measure O(δ2−(j−r)(n−1) ), the integral function has the size about O(2(j−r)(n+2β) ), then the integral on this ring is controlled by O(δ2(j−r)(2β+1) ). Then take the summation about r, we get 1

1

1

1

|D β mδ (|ξ|)| ≤ Cδ 2 2j(β+ 2 ) ≈ Cδ 2 ||ξ| − 1|−(β+ 2 ) .

126

C2. Bochner-Riesz means of multiple Fourier integral

Let us come to the proof of Proposition 2.10.1. We consider the two cases 1 < α < n and 0 < α < 2, respectively. Firstly, we think about the ﬁrst case. Assume j ≥ 0. By Proposition 2.10.2, we have 2 2 |Kj ∗ f (x)| dx = Kj (x) |fˆ(ξ)|2 dξ n n R R∞ 2 = r n−1 K (r) |fˆ(rω)|2 dωdr j n−1 0 S ∞ 2 2 α α−n n−1 |f (x)| |x| dx r r ≤C Kj (r) dr. Rn

0

j ∞ ≤ 1, and Proposition 2.10.3 for any positive integer m, we Due to K obtain 2 −mj |Kj ∗ f (x)| dx ≤ C2 δ |f (x)|2 |x|α dx. Rn

Rn

Using the property of the duality, we have 2 dx −mj |Kj ∗ f (x)| ≤ C2 δ |f (x)|2 dx. |x|α Rn Rn Applying the above inequality to f , who has the support {x : |x| ≤ C2j δ −1 }, we have dx 2 dx (α−m)j 1−α |Kj ∗ f (x)| ≤ C2 δ |f (x)|2 α . α |x| |x| Rn Rn Choose m ≥ α + 1, then we have dx dx |Kj ∗ f (x)|2 α ≤ C2−j δ 1−α |f (x)|2 α . |x| |x| n n R R

(2.10.13)

We will prove (2.10.13) is true for any f . Notice that Kj (x) = K(x)hj (x), and the support of hj is in the ball with the origin as its center and 2j+2 δ −1 as its radius, then Kj has the similar ball. We decompose f as f=

fi

i∈Z

where fi = f χQi , and the center of Q0 is the origin, Qi has the side length 10 · 2j δ−1 , which does not intersect with each other and {Qi } covers Rn .

2.10 Almost everywhere convergence

127

Thus the support of Kj ∗ fQi shares the same center with Qi , but its radius is C2j δ−1 , where the constant C is independent of j, i, δ. Then for any i, Kj ∗ fQi has compact support, and they satisfy Kj ∗ fi(x) · Kj ∗ fi (x) = 0, provided that |i − i | > C. Consequently we have

2 dx Kj ∗ fi(x) |x|α Rn i dx = Kj ∗ fi (x) Kj ∗ fi (x) α Rn i,i |x| dx Kj ∗ fi (x) Kj ∗ fi (x) α = Rn i |x| |≤C |i−i |Kj ∗ fi (x)|2 + |Kj ∗ fi (x)|2 dx ≤ 2 |x|α Rn i |i−i |≤C dx ≤C |Kj ∗ fi (x)|2 α . |x| Rn

dx |Kj ∗ f (x)|2 α = |x| Rn

i

It suﬃces to prove that (2.10.13) is valid for any i ∈ N. The case of i = 0 has been proven. When i ≥ 1, the term |x|−α can be roughly regarded as a constant in the integral. So we merely need to prove 2 −j 1−α |Kj ∗ f (x)| dx ≤ C2 δ |f (x)|2 dx. Rn

Noting that α > 1, δ ≤

Rn

1 2

and (2.10.12), we have 1 1 Kj ≤ C2− 2 j δ 2 (1−α) . ∞

Applying the Plancherel theorem, we get the proof of Proposition 2.10.1 for the case of 1 < α < n. For the case 0 < α < 2, we can use the similar method as before. But we choose another technique to consider it. When 0 < β < 1, by the deﬁnition of D β , it implies from Leibniz’s law that β β β D D (gh)(x) ≤ g h(x) + |h(x)| D g(x) . ∞

128

C2. Bochner-Riesz means of multiple Fourier integral

Then we have δ f m

L2β

≤ C mδ D β f + f D β mδ . ∞

2

2

Next we merely need to show 2 |f (x)|2 D β mδ (x) dx ≤ CA2β (δ)δ−1 f 2L2 . β

Rn

(2.10.14)

Due to Proposition 2.10.4, the left side of (2.10.14) can be controlled by the following estimate Cδ−2β

2

||x|−1|≤2δ

|f (x)| dx + δ

log δ k=1

2k(β+ 12 )

2

||x|−1|≤2−k

|f (x)|2 dx

dx |f (x)| +δ |x|n+2β

|x|≥2 log

δ −2β 2k(β+ 12 ) −k ≤C δ Bβ (δ) + δ 2 Bβ 2 f 2L2 + Cδ

2

∞ k=1

≤ CA2β (δ)δ

−1

β

k=1

|x|∼2k

f 2L2 β

|f (x)|2 dx + Cδ

≤ CA2β (δ)δ−1 f 2L2 .

∞ k=1

2−k(n+2β) 22kβ f 2L2

β

β

Remark 2.10.2 Theorem 2.10.5 only conﬁrms the conjecture on a.e. convergence of Bochner-Riesz means of Fourier integrals below the critical index for p ≥ 2. The conclusion of Theorem 2.10.5 is also obtained by Christ n−1 . For the case p < 2, the [Chr1] under the restrict condition α > 2(n+1) conjecture is still open. However, we have to mention two remarkable results in this direction. In 2002, Tao [Ta2] proved that if n = 2, 1 max{ 4p − 38 , 6p − 23 }, then the conclusion of the conjecture holds. 2n ≤ p < n−1−2α , then the In 2004, Lee [Le1] proved that if n ≥ 3 and 2n+4 n conclusion of the conjecture also holds. Remark 2.10.3 Let Ω ⊂ Rn , and |Ω| < ∞. We also assume that {TR : R > 0} is a family of linear operators, with each of TR mapping Lp (Ω) with 1 ≤ p ≤ ∞ to the measurable function space M (Ω). With the method

2.11 Commutator of Bochner-Riesz operator

129

completely the same as the one in Theorem 2.10.1, we know that, if the maximal operator T ∗ f (x) = sup |TR f (x)| R>0

is of weak type (p, p), then for any f ∈ Lp (Ω), lim TR f (x)

R→∞

exists almost every x ∈ Rn . Stein [St3] pointed out that, for the case 1 ≤ p ≤ 2, the converse proposition also holds. That is to say, if for any f ∈ Lp (Ω) with 1 ≤ p ≤ 2, lim TR f (x) R→∞

exists almost every x ∈ Rn . Thus T∗ is of weak type (p, p). The fact can be applied to the case of the series.

2.11

Commutator of Bochner-Riesz operator

After the work on commutators of singular integrals, which is due to Coifman, Rochberg and Weiss [CRW1], was published, it has largely promoted the research of this ﬁeld. However, without united method for Lp boundedness of commutators of other operators, its commutator for each operator is studied independently. In 1993, Alvarez, Bagby, Kurtz and Perez [ABKP1] got a judge rule for the boundedness of the commutators generated by general linear operator T : Lpω → Lpω ⇒ [b, T ] : Lp → Lp . for ω ∈ Aq and 1 < p, q < ∞. Thus, it is easy to know that when the weighted norm inequality of a linear operator has been established, the commutator generated by this one n−1 and a BMO function is then bounded on Lp (Rn ). For example, let B 2 be the Bochner-Riesz operator at the critical index. In 1992, Shi and Sun n−1 [ShS1] established the Lpω boundedness of B 2 , where ω ∈ Ap . From the n−1 judge rule above, the commutator generated by a BMO function and B 2 is bounded on Lp (Rn ). On the other hand, it should be noticed that there are lots of important operators which do not satisfy the condition of the above judge rule. For

130

C2. Bochner-Riesz means of multiple Fourier integral

example, the Bochner-Riesz operator below the critical index is one of those operators. Let us consider the case b ∈ BMO(Rn ). It is natural to expect that the commutator [b, Brα ] enjoy the same Lp convergence as Brα as r → ∞, or [b, B α ] enjoy the same Lp -boundedness of B α . In this case, the following problem on the Lp boundedness of their commutators is raised naturally. Question 1 Let b ∈ BMO(Rn ). The commutator [b, B α ] is bounded on Lp (Rn ) if and only if 2n 2n 1. If the commutator [b, B ] is bounded on L (R ), then we have 2n 2n

2n . n + 1 + 2α

Fix ψ ∈ Cc∞ (Rn ), 0 ≤ ψ(x) ≤ 1 such that ⎧ ⎨ 1, |x| ≤ 1 , ψ(x) = ⎩ 0, |x| ≥ 2 . It follows immediately that ψ(x) +

∞ −j ψ 2 x − ψ 2−j+1 x = 1, j=1

(2.11.1)

2.11 Commutator of Bochner-Riesz operator

131

for x ∈ Rn . Write ϕ0 (x) = ψ(x), ϕj (x) = ψ 2−j x − ψ 2−j+1 x , for j ∈ N, and B α (x) = Cα

J n2 +α (|x|) n

|x| 2 +α

,

where Jγ (t) denotes the Bessel function of order γ. Thus we have α

B f (x) =

∞ j=0

(B ϕj ) ∗ f (x) = α

∞

Tj f (x).

j=0

Noting suppϕ0 ⊂ {x : |x| ≤ 2} and

, suppϕj ∈ x : 2j−1 ≤ |x| ≤ 2j+1

for j ≥ 1, we have , supp[b, Tj ]f ⊂ x : 2j−1 − 1 ≤ |x| ≤ 2j+1 + 1 and supp[b, T0 ]f ⊂ {x : |x| ≤ 3}. Since [b, B α ]f Lp (Rn ) ≤ C < +∞, we conclude that p ∞ [b, Tj ]f (x) dx = Rn j=0 p ∞ = [b, Tj ]f (x) dx |x|<4 j=0 p ∞ ∞ [b, Tj ]f (x) dx + i i+1 i=2 2 ≤|x|<2 j=0

[b, B α ]f pLp (Rn )

132

C2. Bochner-Riesz means of multiple Fourier integral p 3 = [b, Tj ]f (x) dx |x|<4 j=0 p ∞ i+2 [b, Tj ]f (x) dx. + i i+1 i=2 2 ≤|x|<2 j=i−1

It follows that p i+2 [b, Tj ]f (x) dx ≤ C < +∞. 2i ≤|x|<2i+1 j=i−1

i≥6

Set

i+2

φi (x) =

(2.11.2)

ϕj (x).

j=i−1

We have

, suppφi ⊂ x : 2i−2 ≤ |x| ≤ 2i+3 .

Noting that

Jγ (t) =

as t → +∞, and

2 tπ

1 2

πγ π cos t − − + r(t), 2 4 3

r(t) = O(t− 2 ),

(2.11.3)

(2.11.4)

as t → +∞, respectively, we conclude that i+2

1 2 2 π(n + 1 + 2α) [b, Tj ]f (x) = Cα cos |x − y| − π 4 |x−y|≥1

j=i−1

×

φi (x − y)

|x − y| + Cα

n+1 +α 2

[b(x) − b(y)]f (y)dy

r(|x − y|)φi (x − y)

[b(x) − b(y)]f (y)dy n |x − y| 2 +α J n2 +α (|x − y|)φi (x − y) + Cα [b(x) − b(y)]f (y)dy n |x − y| 2 +α |x−y|≤1 =: Pi f (x) + Qi f (x) + Ri f (x). |x−y|≥1

For each i ≥ 6, we have Ri f (x) = 0.

2.11 Commutator of Bochner-Riesz operator Set mi (b) = C2−(i+1)n

133

|x|≤2i+1

b(x)dx

and

b(x)dx.

m(b) = C |x|≤1

Noting that supp f ⊂ {x : |x| ≤ 1}, it follows that n+3 |x − y|−( 2 +α) |b(x) − b(y)||φi (x − y)f (y)|dy |Qi f (x)| ≤ Cα |x−y|≥1

≤ 2Cα |b(x) − mi (b)| n+3 |x − y|−( 2 +α) |mi (b) − b(y)|f (y)dy × 2i−2 ≤|x−y|≤2i+3 −(i−2)( n+3 +α ) 2 ≤ 2Cα 2 |b(x) − mi (b)| |mi (b) − b(y)|dy |y|≤1

≤ Cα 2−i(

n+3 +α 2

) |b(x) − m (b)||m (b) − m(b)| i i n+3 |m(b) − b(y)|dy +Cα 2−i( 2 +α) |b(x) − mi (b)| |y|≤1

−i( n+3 +α) n 2

≤ Cα 2

≤ Cα,n 2−i(

2 ibBMO |b(x) − mi (b)| ) 2 2i b |b(x) − m (b)|.

n+3 +α 2

BMO

i

Therefore, we obtain |Qi f (x) + Ri f (x)|p dx p −ip( n+2 +α) p 2 bBMO ≤ Cα,n 2 2i ≤|x|<2i+1

|x|<2i+1

≤ C1 2−i[(

n+2 +α 2

)p−n] .

|b(x) − mi (b)|p dx (2.11.5)

Let

π π(n + 1 + 2α) π π(n + 1 + 2α) Ik = 2kπ − + + 1, 2kπ + + −1 3 4 3 4 and

π π(n + 1 + 2α) π π(n + 1 + 2α) − 1, 2(k + 1)π − + +1 . Ak = 2kπ + + 3 4 3 4

134

C2. Bochner-Riesz means of multiple Fourier integral

Since |Ik | + |Ak | = 2π <

2i − 2 , 3

for i ≥ 6, we have , Mi = k ∈ N : Ik ⊂ 2i + 1, 2i+1 − 1 = ∅, for i ≥ 6. From the fact |Ik | = C|Ak | with the constant C being independent of k, it follows that dx = C rn−1 dr |x|∈Ak

Ak

n−1 π π(n + 1 + 2α) |Ak | + +1 3 4 n−1 π π(n + 1 + 2α) +1 ≤ C2n−1 2kπ − + |Ik | 3 4 r n−1dr ≤C Ik =C dx. (2.11.6) ≤ C 2(k + 1)π −

|x|∈Ik

Similar to the estimate above, we have dx ≤ dx ≤ C |x|∈Ik−1

|x|∈Ik

dx,

|x|∈Ik−1

where C is independent of k. For each |x| ∈ Ik for k ∈ Mi , |x − y| ≤ 1 easily implies 2i ≤ |x| − 1 ≤ |y| ≤ |x| + 1 ≤ 2i+1 , 2kπ −

π π(n + 1 + 2α) π ≤ |y| − ≤ 2kπ + , 3 4 3

and log |x| − log |x − y| = log

|x| ≥ 1. |x − y|

Furthermore, it follows from the deﬁnition φi that φi (y) = 1 for 2i ≤ |y| ≤ 2i+1 .

(2.11.7)

2.11 Commutator of Bochner-Riesz operator

135

Hence, for |x| ∈ Ik , and noting that suppf ⊂ {y : |y| ≤ 1}, we conclude that 1 2 2 π(n + 1 + 2α) Pi f (x) = Cα cos |x − y| − π 4 |x−y|≥1 φi (x − y)

[b(x) − b(y)]f (y)dy n+1 |x − y| 2 +α 1 2 2 π(n + 1 + 2α) = Cα cos |x − y| − π 4 |y|≤1 ×

×

φi (x − y)

[b(x) − b(y)]f (y)dy n+1 |x − y| 2 +α 1 π(n + 1 + 2α) 2 2 Cα cos |y| − = π 4 |x−y|≤1 ×

φi (y) n+1 +α 2

|y| 1

[log |x| − log |x − y|]f (x − y)dy

n+1 1 2 Cα 2−i( 2 +α) (log |x| − log |x − y|)f (x − y)dy 2π |x−y|≤1 1 2 2 −i( n+1 +α) −2 2 Cα 2 f (x − y)dy ≥2 π |x−y|≤1

≥

≥ C2 2−i(

n+1 +α 2

).

Therefore, using (2.11.6) and (2.11.7), we have that 2i ≤|x|<2i+1

|Pi f (x)|p dx ≥

C2 2−pi(

n+1 +α 2

)

dx |x|∈Ik

k∈Mi

≥ C2 2−pi(

n+1 +α 2

≥ C2 2−pi(

n+1 +α 2

)

k∈Mi

≥ C2 2−i((

)

2i ≤|x|≤2i+1

n+1 +α)p−n) 2

.

dx

|x|∈Ik ∪Ak

dx

136

C2. Bochner-Riesz means of multiple Fourier integral

Using (2.11.5) and (2.11.8), for i ≥ 6, we have p ⎞1/p ⎛ i+2 dx⎠ ⎝ [b, T ]f (x) j 2i ≤|x|<2i+1 j=i−1

1/p =

≥

2i ≤|x|<2i+1

|Pi f (x) + Qi f (x) + Ri f (x)|p dx 1/p |Pi f (x)| dx

2i ≤|x|<2i+1

−

p

1/p |Qi f (x) + Ri f (x)| dx p

2i ≤|x|<2i+1

1/p 1/p n+1 n+2 ≥ C2 2−i(( 2 +α)p−n) − C1 2−i(( 2 +α)p−n)

n+1 1/p 1/p = 2−i(( 2 +α)p−n)/p C2 − C1 2−i/2 , where C1 , C2 are two constants independent of f and i. Consequently, if (2.11.1) holds, we must have n+1 + α p − n > 0, 2 which implies (2.11.1). For the suﬃciency in Question 1, Hu and Lu [HL1] gave an aﬃrmative result when n = 2. Theorem 2.11.2 Let 0 < α < 1/2 and b ∈ BMO(R2 ). If 4/(3 + 2α) < p < 4/(1 − 2α), then the commutator [b, B α ] is bounded on Lp (R2 ) with the norm being smaller than C(p)bBMO . Proof. Write

α

[b, B ]f (x) =

R2

b(x) − b(y) B α (x − y)f (y)dy,

where B α (x) = Cα

J1+α (|x|) , |x|1+α

and Jβ (t) denotes the Bessel function of order β. Noting that πβ π − 21 Jβ (t) = Ct cos t − − + r(t), 2 4

2.11 Commutator of Bochner-Riesz operator

137

3 r(t) = O t− 2 ,

and

as t → ∞, we hence have that [b, B α ]f (x) f (y) π(3 + 2α) b(x) − b(y) =C cos |x − y| − dy 3 4 |x−y|≥1 |x − y| 2 +α f (y) r(|x − y|) b(x) − b(y) dy +C |x − y|1+α |x−y|≥1 J1+α (|x − y|) (b(x) − b(y))f (y)dy +C 1+α |x−y|≤1 |x − y| = P f (x) + Qf (x) + Rf (x).

(2.11.8)

Since |Jβ (t)| ≤ Cβ |t|β , as t → 0, it follows that J1+α (|x − y|) f (y)dy ≤ CM f (x), |x−y|≤1 |x − y|1+α where M f denotes the Hardy-Littlewood maximal function of f . Thus by the weighted estimate for M (see[GR1]), we have Rf p ≤ CbBMO f p , for 1 0, we have ∞ f (y) |f (y)| r(|x − y|) dy ≤ C dy 1+α |x−y|≥1 |x − y| k−1 ≤|x−y|<2k |x − y|5/2+α k=1 2 ≤ CM f (x). This implies that Qf p ≤ CbBMO f p , for 1 < p < ∞. Obviously, the Lp norm of the operator P deﬁned by (2.11.8) can be controlled by that of the operator P6 deﬁned by f (y) 6 P f (x) = ei|x−y| (b(x) − b(y)) dy. 3 |x−y|≥1 |x − y| 2 +α

138

C2. Bochner-Riesz means of multiple Fourier integral

Furthermore, since α < 12 and f (y) i|x−y| e dy ≤ CM f (x), 3 +α |x−y|<1 |x − y| 2 We have already known that f (y) ei|x−y| (b(x) − b(y)) dy ≤ CbBMO f p 3 +α |x−y|<1 |x − y| 2 p

for 1 < p < ∞. Actually we may thus view the operator P as f (y) ei|x−y| (b(x) − b(y)) dy. P f (x) = 3 R2 |x − y| 2 +α

(2.11.9)

By Stein’s interpolation theorem (see [SW1]), to prove Theorem 2.11.2, it is enough to show that for any 0 < α < 12 , the operator P deﬁned by (2.11.9) is bounded on L4 (R2 ). Denote I = [0, 1], I 2 = I × I, and F (I 2 ) = [−1.5, 2.5]2 \[−0.5, 1.5]2 . For ﬁxed λ > 0, deﬁne f (y) λ P f (x) = eiλ|x−y| dy 3 I2 |x − y| 2 +α and the corresponding commutator λ eiλ|x−y| (b(x) − b(y)) Pb f (x) = I2

Set

f (y) 3

|x − y| 2 +α

1

S λ f (x) = λ 2 −α P λ f (x)

and

1

Sbλ f (x) = λ 2 −α Pbλ f (x).

Note that if b ∈ BMO(Rn ), then we can easily have b(t(·)) ∈ BMO(Rn ) and b(t(·))BMO = bBMO for any t > 0.

dy.

2.11 Commutator of Bochner-Riesz operator

139

By the same argument as in Section 2.6, we can conclude that the proof of Theorem 2.11.2 can be reduced to the following inequality λ ≤ Cλ−δ bBMO f L4 (I 2 ) (2.11.10) Sb f 4 2 L (F (I ))

for some positive constant δ > 0. Now we prove (2.11.10). Let s be a small positive constant which will be determined later. Set 0 < r < 12 and σ > 0 such that 1 1 r = − . 4+σ 4 2 Observe that if x ∈ F (I 2 ), then we have λ |f (y)|dy P f (x) ≤ C 2 I 1 |f (y)χI 2 (y)|dy ≤ Cr 2−r R2 |x − y| = Cr Ir (f χI 2 )(x), where χI 2 is the characteristic function of I 2 , and Ir is the usual fractional integral operator of order r. By the Hardy-Littlewood-Sobolev theorem, it follows that 1 1 λ −α λ 2 ≤ Cλ f S f 4+σ 4+σ 2 ≤ Cλ 2 −α f L4 (I 2 ) . (2.11.11) P 2 L

(F (I ))

L

(R )

Similarly, if σ is small enough, we have 1 λ −α 2 ≤ Cλ f L4−σ (I 2 ) . S f 4 2

(2.11.12)

By the key estimate used in [CS1], we have λ ≤ Cλ−ε f L4 (I 2 ) , S f 4 2

(2.11.13)

L (F (I ))

L (F (I ))

where ε > 0. An interpolation between the inequalities (2.11.11) and (2.11.13) yields λ f ≤ Cλ−ε+(1/2−α+ε)s f L4 (I 2 ) (2.11.14) S 4+sσ 2 L

with 0 < s < 1.

(F (I ))

140

C2. Bochner-Riesz means of multiple Fourier integral

On the other hand, interpolation between the inequalities (2.11.12) and (2.11.13) gives λ ≤ Cλ−ε+(1/2−α+ε)s f L4−sσ (I 2 ) . (2.11.15) S f 4 2 L (F (I ))

We can also get by the inequalities (2.11.14) and (2.11.15) that λ ≤ Cλ−ε+(1/2−α+ε)s f L4−s2 σ (I 2 ) . S f 4+s2 σ 2 L

(F (I ))

(2.11.16)

Let φ(x) ∈ C0∞ (R2 ) such that φ(x) = 1 if |x| ≤ 50 and suppφ ⊂ {x : |x| ≤ 100}. Denote

6b(y) = [b(y) − m10I 2 (b)]φ(y),

where m10I 2 (b) denotes the mean value of b on 10I 2 . Obviously, if x ∈ F (I 2 ), then

Sbλ f (x) = 6b(x)S λ f (x) + S λ 6bf (x) = I(x) + II(x). For the ﬁrst term, we have IL4 (F (I 2 )) ≤ 6b

Lq (R2 )

λ S f

L4+sσ (F (I 2 ))

≤ C(σ, s)bBMO λ−ε+(1/2−α+ε)s f L4 (I 2 ) , where 1/q = 1/4 − 1/(4 + sσ), and the second inequality follows from the inequality (2.11.14) and the fact 6 b q 2 ≤ C(s, σ)bBMO . L (R )

The estimate for the second term follows from the inequality (2.11.15) by IIL4 (F (I 2 )) ≤ Cλ−ε+(1/2−α+ε)s 6bf L4−sσ (I 2 ) ≤ Cλ−ε+(1/2−α+ε)s bBMO f L4 (I 2 ) . Choose s so small that δ = ε − (1/2 − α + ε)s > 0. Combining the estimates above we get λ ≤ Cλ−δ bBMO f L4 (I 2 ) . Sb f 4 2 L (F (I ))

This concludes the proof of the lemma.

Chapter 3

Bochner-Riesz means of multiple Fourier series 3.1

The case of being over the critical index

3.1.1 Bochner formula For a locally integrable function f , we deﬁne 1 fx (t) = f (x − tξ)dσ(ξ) ωn−1 Sn−1 for t ≥ 0, where

n

ωn−1 =

2π 2 . Γ n2

Theorem 3.1.1 (Bochner) If f ∈ L(Q) and Reα > n

α (f, x) SR

(3.1.1)

2α+1− 2 Γ(α + 1) n −α = R2 Γ( n2 )

Proof. Denote α

Φ (x) =

0

⎧ 2 α ⎪ ⎨ (1 − |x| ) ⎪ ⎩

B α (x) =

∞

fx (t)

n−1 2 ,

J n2 +α (Rt) n

tα− 2 +1

then we have dt.

(3.1.2)

|x| < 1, (3.1.3) |x| ≥ 1,

0

2α Γ(α + 1) J n2 +α (|x|) , n n (2π) 2 |x| 2 +α 141

142

C3. Bochner-Riesz means of multiple Fourier series

and n

n

ϕα (x) = (2π)n B α (x) = 2α+ 2 π 2 Γ(α + 1)

J n2 +α (|x|) n

|x| 2 +α

.

At the end of Section 1.4, we have mentioned that, if Reα > we have α = Φ, ϕ

(3.1.4) n−1 2 ,

then

which satisﬁes the conditions of Theorem 1.4.1 in Chapter 1. Hence, by (1.4.3) and (1.4.4) in Chapter 1, we obtain that

m α Cm (f )eim·x SR (f ; x) = Φα R m∈Zn

=f∗ ϕα1 (y + 2πm) (x) R 1 = f (x − y) ϕα1 (y + 2πm)dy (2π)n Q R m∈Zn 1 f (x − y)ϕα1 (y)dy = n (2π) R m∈Zn Q+2πm 1 = f (x − y)ϕα1 (y)dy (2π)n Rn R ∞ 1 α f (x − tξ)ϕ 1 (tξ)dσ(ξ) tn−1 dt. = (2π)n 0 R n−1 S Substitute (3.1.4) into ϕα1 (u) = Rn ϕα (Ru), and then it follows R

α SR (f ; x)

∞ J n (Rt) n 1 α+ n n n−1 2 +α 2 2 = 2 π Γ(α + 1)ω f (t)R t dt n n x (2π)n (Rt) 2 +α 0 J n2 +α (Rt) n Γ(α + 1) ∞ α+1− n 2 =2 f (t) dtR 2 −α . n x n α− +1 Γ( 2 ) t 2 0

3.1.2 The localization theorem Now we formulate the localization theorem. Theorem 3.1.2 Suppose that f ∈ L(Q), and vanishes at the ball B(0, ε) = {x ∈ Rn : |x| < ε} for some ε > 0. If Reα > n−1 2 , then we have

n−1 α (f ; 0) = O R 2 −Reα (3.1.5) SR

3.1 The case of being over the critical index

143

as R → ∞. Proof. By (3.1.2), we have α (f ; 0) SR

= CR

n −α 2

∞

f0 (t)

J n2 +α (Rt) n

tα− 2 +1

ε

dt.

Using the asymptotic formula C n J 2 +α (Rt) ≤ √ , Rt for t >

1 R,

when R > 1ε , we can obtain α (f ; 0)| |SR

≤ CR

n −Reα 2

∞ ε

Denote β = Reα −

√

|f0 (t)| R tReα−

n−1 +1 2

dt.

n−1 >0 2

and |f0 (t)|tn−1 = g(t). We have α |SR (f ; 0)| ≤ CR−β

∞ ε

g(t) dt. tβ+n

(3.1.6)

Since it follows that ∞ A g(t) g(t) dt = lim dt β+n A→+∞ ε tβ+n t ε t A 8t A g(τ )dτ 1 ε g(τ )dτ + (β + n) dt = lim A→+∞ tβ+n ε tβ+n+1 ε ε ∞ O(tn ) = (β + n) dt tβ+n+1 ε = O(1), we have α (f ; 0) = O(R−β ) SR

as R → ∞, which completes the proof of Theorem 3.1.2.

144

C3. Bochner-Riesz means of multiple Fourier series

Remark 3.1.1 Here, the symbol ‘ O’ depends on β. As β → 0+ , ‘ O’ increases as β1 . 3.1.3 The maximal operator S∗α Definition 3.1.1 Let f ∈ L(Q). Deﬁne α (f ; x)|. S∗α (f )(x) = sup |SR

(3.1.7)

R>0

We denote by M f the Hardy-Littlewood maximal operator of a locally integrable function f as 1 |f (x − y)|dy. (3.1.8) M f (x) = sup n r>0 r |y|
1 α M f (x) (3.1.9) S∗ (f )(x) ≤ C 1 + Reα − n−1 2

Theorem 3.1.3 If Reα >

n−1 2

holds, where the positive number C ≤ |2α Γ(α + 1)|e2π|Imα| Cn . n−1 = g(t), then Proof. Denote again Reα − n−1 2 = β > 0, |fx (t)|t r g(t)dt ≤ Cn |f (x − y)|dy ≤ Cn r n M f (x). 0

|y|
By the Bochner formula, we have

α SR (f ; x) = Cn 2α Γ(α + 1)R

n −α 2

R−1 0

+

∞

R−1

fx (t)tn−1

J n2 +α (Rt) n

tα+ 2

dt

= I1 + I2 . Next we will estimate I1 and I2 , respectively. For I1 , it follows that |I1 | ≤ Cn |2 Γ(α + 1)|R α

R−1

n 0

g(t)dt ≤ Cn |2α Γ(α + 1)|M f (x).

(3.1.10)

3.1 The case of being over the critical index

145

On the other hand, we have J n +α (Rt) 3π 1 2 Cn e 2 |Imα| , ≤ √ n n+1 +α +Reα t2 Rt 2 for t > R−1 . 3π About the appearance of the constant e 2 |Imα| , which is concerned about the imaginary part of the order in the asymptotic equation of the multiordered Bessel function, we omit the details here. One can refer to the related contents in Watson [Wat1]. Hence |I2 | ≤ Cn e2π|Imα| R−β ≤ Cn e

2π|Imα|

∞ R−1

M f (x)

dt tn+β dt(n + β) t1+β Rβ

g(t) ∞

R−1

1 ≤ Cn (1 + )e2π|Imα| M f (x). β

(3.1.11)

Combining (3.1.10) with (3.1.11), we immediately obtain (3.1.9). The proof of Theorem 3.1.3 has been completed.

α Corollary 3.1.1 If Reα > n−1 2 , then the operator S∗ is of type (p, p) for 1 λ}| ≤

Cn f 1 , λ

(3.1.13)

for any λ > 0. We point out that (3.1.12) and (3.1.13) are direct consequences of Theorem 3.1.3 and the estimate of the Hardy-Littlewood maximal operator. One can refer to Section 2.3 in Stein and Weiss [SW1]. Although they discussed functions deﬁned on Rn , the conclusions are completely the same as that of Q.

146

3.2

C3. Bochner-Riesz means of multiple Fourier series

The case of the critical index (general discussion)

3.2.1 Localization problems

For the Bochner-Riesz means at the critical index, the localization principle does not hold on L(Qn ) with n > 1 any more. It was proven by Bochner [Bo1].

Theorem 3.2.1 (Bochner) If n > 1, then there exists f ∈ L(Qn ) which vanishes on B(0, δ) for some δ > 0 such that n−1

lim sup SR 2 (f ; 0) = +∞. R→+∞

n−1

Proof. By Lemma 1.3.5 in Chapter 1, for almost every y, DR2 (y) is unbounded about R. Deﬁne , Lδ = f ∈ L(Qn ) : f vanishes at B(0, δ) . Let δ < 1. Obviously Lδ is a closed subspace of L(Qn ). Deﬁne a linear functional FR (R > 0) on Lδ , n−1

FR (f ) = SR 2 (f ; 0) for f ∈ Lδ . It follows that FR =

sup

f ∈Lδ , f 1 =1

|FR (f )|

n−1 1 f (y)DR2 (−y)dy = sup (2π)n Qn f ∈Lδ , f 1 =1 n−1 1 2 . D = sup (−y) (2π)n y∈Qn \B(0,δ) R Since {FR : R > 0} is unbounded, it implies from the uniform boundedness theorem that there exists a f ∈ Lδ such that {FR (f ) : R > 0} is unbounded. This ﬁnishes the proof of Theorem 3.2.1.

3.2 The case of the critical index (general discussion)

147

3.2.2 An example of being divergent almost everywhere

Theorem 3.2.2 (Stein) There exists f ∈ L(Qn ) (n > 1), such that n−1 2 lim sup SR (f ; x) = +∞ R→+∞

for a.e. x ∈ Qn , and f can be constructed so that it is supported in an arbitrary small given neighborhood of the origin. It should be pointed out that the support discussed here is restricted in Qn . Proof. Denote B(0, 1) = {x ∈ Rn : |x| ≤ 1}. Choose a function ψ, satisfying 1 ∞ n suppψ ⊂ B, ψ ∈ C (R ), ψ(x)dx = 1. (3.2.1) (2π)n Rn By (3.2.1), it follows from the properties of the the Fourier transform of ψ that ˆ ψ(0) = 1; (ii) ψ(y) ≤ |y|−2k k ψ (i)

L(Rn )

for all k ∈ N and y = 0,

where is the Laplace operator and k = (k−1 ). Choose ε ∈ (0, 1). Deﬁne ψε (x) = ε−n ψ(ε−1 x). The periodization of ψε is ϕε (x) :=

ψε (x + 2πm).

m∈Zn

Since suppψε = εB ⊂ Qn , then ϕε is exactly the 2π-periodic extension of ψε when restricted on Qn . It follows that ϕε C = ψε C = ε−n ψC . By the Poisson summation formula (see Section 1.2), im·x ε (m)eim·x = σ(ϕε )(x) ∼ ψ ψ(εm)e .

(3.2.2)

148

C3. Bochner-Riesz means of multiple Fourier series

Since we have

⎛

|ψ(εm)| =⎝

+

|m|<ε−1

⎞

⎠ |ψ(εm)|

|m|≥ε−1

≤ ψL(Rn )

1 + n ψL(Rn )

|m|<ε−1

|m|≥ε−1

1 |εm|2n

≤ Cε−n , Hence, it follows that

n−1 S 2 (ϕε ) ≤ Cε−n . R

(3.2.3)

C

And it is obvious that n−1

n−1

lim SR 2 (ϕε ; x) = DR2 (x)

ε→0+

(3.2.4)

is valid for (x, R) on Q × (0, R0 ) uniformly for any R0 > 0. By Lemma 1.3.5 in Chapter 1, as long as x0 ∈ S , we have n−1 0 2 lim sup DR (x ) = +∞. R→∞

The condition of x0 ∈ S actually implies that {|x0 + 2πm| : m ∈ Zn } is a linearly independent set (see Deﬁnitions 1.3.1 and 1.3.2 in Chapter 1), while S is a full measurable set, when n > 1. For k, l ∈ N, deﬁne * n−1 2 Ek,l = x ∈ S Q : sup DR (x) > l . 0
5 2 Q. Thus Then it is evident that Ek,l ⊂ Ek+1,l and ∞ k=1 Ek,l ⊂ S ∞ * Ek,l = S Q = (2π)n , lim |Ek,l | = k→∞ k=1

for all l ∈ N. According to the above argument, for any A > 0, m ∈ N, choose l ≥ A+1 and k = k(A, m, l) large enough such that 1 |Ek,l | > 1 − |Q|. m+1

3.2 The case of the critical index (general discussion)

149

Then, if x ∈ Ek,l , we have n−1 2 sup DR (x) > A + 1.

0
Thus, according to (3.2.4), there exists ε0 = ε0 (k) > 0, such that if x ∈ Ek,l and ε ∈ (0, ε0 ], n−1 2 sup SR (ϕε ; x) > A. (3.2.5) 0
For any ε, δ : 0 < ε < δ < 1, we have 1 |ψ(εm) − ψ(δm)| ≤ |ψ(x)||e−imεx − e−imδx |dx (2π)n Rn 1 |ε − δ||m| |ψ(x)||x|dx ≤ |Q| n R n−1 2 ≤ C|m|(δ − ε) SR (ϕε − ϕδ ) C ≤ |ψ(εm) − ψ(δm)| ≤ CδRn+1 . |m|
Consequently, for arbitrary R0 > 0, there exists δ = δ(R0 ) ∈ (0, 1) such that if ε , ε ∈ (0, δ], then n−1 2 (3.2.6) sup SR (ϕε − ϕδ ) < 1. 0
Finally,

C

ψε (x + 2πm) ϕε L(Q) =

L(Q)

= ψε L(Rn )

= ψL(Rn ) < +∞; thus for any sequence {εj }∞ j=0 (εj > 0), ∞ 1 ϕε (x) 2j j

(3.2.7)

j=0

absolutely converges in L(Q)-norm. Now we take any ε0 ∈ (0, 1), R0 > 0 and let δ0 = ε0 . By (3.2.6), there exists δ1 < ε0 = δ0 such that ∀ ε ∈ (0, δ1 ], n−1 2 sup SR (ϕε − ϕδ1 ) < 1. 0
C

150

C3. Bochner-Riesz means of multiple Fourier series According to (3.2.3), we deﬁne −1 n−1 2 A1 = sup 2 SR (ϕδ1 ) , A1 < +∞. R>0

C

2

By (3.2.5), there exists a set E1 ⊂ S Q, R1 > R0 , ε1 ∈ (0, δ1 ) such that ⎧ 1 ⎪ ⎪ ⎨ |E1 | > (1 − 2 )|Q|, n−1 ⎪ −1 2 ⎪ ∀ x ∈ E1 , sup 2 ⎩ SR (ϕε1 ; x) > A1 + 2. 0
Repeat the above steps. By (3.2.6), there exists δ2 < ε1 < δ1 such that for each ε ∈ (0, δ2 ], n−1 2 sup SR (ϕε − ϕδ2 ) < 1. 0
C

By (3.2.3), set ⎛ ⎞ 1 n−1 −j −2 2 ⎝ 2 (ϕεj − ϕδj ) − 2 ϕδ2 ⎠ A2 = sup SR < +∞. R>0 j=0 C

2

By (3.2.5), there exists a set E2 ⊂ S Q, R2 > R1 and ε2 ∈ (0, δ2 ) such that ⎧ 1 ⎪ ⎪ ⎪ ⎨ |E2 | > 1 − 2 + 1 |Q|, −2 n−1 ⎪ ⎪ 2 ⎪ ⎩ ∀ x ∈ E2 , sup 2 SR (ϕε2 ; x) > A2 + 3. 0
Repeating these steps many times, we have ε0 ≥ ε1 > ε2 > · · · > εj > 0, δ0 ≥ δ1 > δ2 > · · · > δj > 0, 0 < R0 < R1 < R2 < · · · < Rj < · · · , for j ∈ N, and

⎛ ⎞ k−1 n−1 −j −k 2 ⎝ ⎠ 2 (ϕεj − ϕδj ) − 2 ϕδk Ak = sup SR < +∞, R>0 j=0 C

Ek ⊂ S

*

Q,

3.2 The case of the critical index (general discussion)

151

for k = 1, 2, 3, . . ., such that |Ek | ≥ and sup

0
1 1− k+1

|Q|

n−1 S 2 (ϕε − ϕδ ) < 1 k k R C

for k ∈ N, and for any x ∈ Ek for k ∈ N, −k n−1 2 sup 2 SR (ϕεk ; x) > Ak + k + 1. 0
By (3.2.7), deﬁne

∞ 1 f= (ϕεj − ϕδj ) 2j j=1

in the sense of L(Q) norm. Restricted to Q, we can see suppf ⊂ ε0 B, and for any k ∈ N, and n−1 2

SR

n−1 2

(f ) = SR

⎛ ⎞ k−1 ⎝ 2−j (ϕεj − ϕδj ) − 2−k ϕδk ⎠ j=0 n−1

+ 2−k SR 2 (ϕεk ) +

∞

m−1

2−j SR 2 (ϕεj − ϕδj ).

j=k+1

When x ∈ Ek , we have n−1 2 sup SR (f ; x) ≥ 0
−

−k n−1 2 S 2 (ϕε ; x) − Ak k R

sup

R∈(0,Rk ) ∞ −j

2

j=k+1

sup

0
> k. It immediately follows that n−1 sup SR 2 (f ; x) = +∞,

R>0

n−1 S 2 (ϕε − ϕδ ) j j R

C

152

C3. Bochner-Riesz means of multiple Fourier series

for x∈

∞ ∞ *

Ek .

l=1 k=l

However, we have

∞ ∞ ∞ * Ek = lim Ek = |Q|. l→∞ l=1 k=l

k=l

This completes the proof of Theorem 3.2.2.

Theorem 3.2.2 can be viewed as the multi-dimensional analogy of the classic Kolmogorov Theorem in the case of one variable where we draw an n−1

analogy between SR 2 and partial Fourier sum of one variable. Many facts show that such an analogy is quite appropriate. We have seen that the conclusion of Theorem 3.2.2 is stronger than that of Theorem 3.2.1, and the former gave such a function which is integrable on n−1

Q and vanishes at some neighborhood of a point, where SR 2 is unbounded almost everywhere. 3.2.3 The relation between the series and the integral Let f ∈ L(Qn ). Deﬁne f6 = f χQ , then f6 deﬁned on Rn is supported in α 6 (f ; x) as the Bochner-Riesz means of the Fourier Qn , f6 ∈ L(Rn ). Denote BR 6 integral of f . Stein [St2] established the following important theorem, which n−1

n−1

joined SR 2 (f ) with the convergence property of BR 2 (f ; x) which gave a n−1

new way in the research of SR 2 (f ). Compared with the case of the series, lots of problems are much easily handled with after they are transfered into the Fourier integral. Theorem 3.2.3 (Stein) Suppose that f ∈ L log+ L(Qn ) with n > 1, and G is a closed set in Qn . Then we have n−1 n−1 2 2 6 lim max SR (f ; x) − BR (f ; x) = 0. R→∞ x∈G

To prove Theorem 3.2.3, some lemmas are needed. Firstly, set D = {σ + iτ : 0 ≤ σ ≤ 1, − ∞ < τ < +∞}.

3.2 The case of the critical index (general discussion)

153

For z ∈ D and R > 0, deﬁne n−1 +z 2

DR

(u) =

|m|
HR and

n−1

R2

(u) =

+z

|m|2 1− 2 R

|ξ|
(u) = DR2

|ξ|2 1− 2 R

+z

n−1 +z 2

n−1 +z 2

n−1

(u) − HR 2

eimu ,

eiξu dξ,

+z

(u),

for u ∈ Qn . Obviously, we have that n−1 +σ 2 2 n−1 +z |m| ≤ D 2 ≤ CRn , 1− 2 R R

(3.2.8)

|m|
n−1 +z 2 H ≤ CRn . R

and

(3.2.9)

In the following, we set R > 1 and q > 2. We deﬁne the linear operator Tz : L(Qn ) → C(Qn ) as Tz (f )(x) = R

1 (z−1+ 2q ) 2

n−1

Q

f (x − y)R2

+ 12 (z−1+ 2q )

(y)dy.

For any f, g ∈ L(Q), the function Tz (f )(x)g(x)dx F (z) = Q

is analytic on D, and 1

1

|Tz (f )(x)| ≤ R q

|f (x − y)|CRn dy ≤ CRn+ 2 f L(Q) , Q 1

|F (z)| ≤ CRn+ 2 f L(Q) gL(Q) (independent of z). We can see that {Tz }z∈D is an allowed family (see [SW1]).

154

C3. Bochner-Riesz means of multiple Fourier series As the operator from Lr (Q) with 1 ≤ r < ∞ to L∞ (Q), the norm of Tz

is Tz r,∞ = sup Tz (f )∞

f r =1

n−1 1 2 + 21 (z−1+ 2q ) 2 = sup sup f (x − y) · R (y)dy R 2 (z−1+ q )

f r =1 x∈Q Q n−1 1 1 2 + 2 (z−1+ 2q ) (σ−1+ ) 2 , q = R2 R r

where σ = Rez ∈ [0, 1], r = Tiτ 2,∞ and

and r = +∞, if r = 1. Then we have n−1 1 1 + 2 (−1+ 2q )+i τ2 (−1+ 2q ) 2 , 2 =R (3.2.10) R r r−1 ,

2

n−1 1 τ 1 + q +i 2 2 . T1+iτ 1,∞ = R q R

(3.2.11)

∞

With proper estimates of the two quantities of (3.2.10) and (3.2.11) as well as the complex interpolation theorem of Stein (see Stein and Weiss [SW1]), we can interpolate at t = 1 − 2q to get an appropriate estimate of the quantity n−1 2 (3.2.12) Tt q ,∞ ≤ R . q

Our purpose is to get the following inequality n−1 2 ≤ M q, R

(3.2.13)

q

for q ≥ 1 and R > 1. The following assertion holds.

Lemma 3.2.1 If (3.2.13) holds, then Theorem 3.2.3 is valid. Proof. Suppose that (3.2.13) holds. Choose a = e

n−1 a R 2 (u)

=

k ∞ ak n−1 2 . (u) R k! k=0

By Levi’s theorem, we have

1 2eM

for M > 0.

3.2 The case of the critical index (general discussion) e

n−1 a R 2 (u)

Q

155

k n−1 ∞ ak 2 R (u) du du = k! Q k=0

≤ |Q| + = |Q| +

∞ ak k=1 ∞ k=1

k! 1 k!

(M k)k

k 2e

k .

By the Stirling formula, we conclude that the right side of the above inequality is less than 1 + |Q|. On the other hand, if we write Φ(μ) = eμ − μ − 1 and Ψ(ν) = (ν + 1) log(ν + 1) for μ, ν ≥ 0, then Φ, Ψ is a pair of conjugate functions in the sense of Young (see Zygmund [Zy1]). Then μν ≤ Φ(μ) + Ψ(ν) holds for every μ, ν ≥ 0. It follows that n−1 n−1 2 2 a R (y)f (x − y) ≤ Φ aR (y) + Ψ(|f (x − y)|) n−1 a R 2 (y)

≤e

+ (|f (x − y)| + 1) log(|f (x − y)| + 1).

Hence, we have n−1 + 2 (y)f (x − y) dy ≤ C |f | log |f |dx + 1 . R Q

Q

(3.2.14)

156

C3. Bochner-Riesz means of multiple Fourier series

By the deﬁnition, we conclude that n−1

n−1

n−1 1 f (y)DR2 (x − y)dy n (2π) Q n−1 1 2 + f (y)H R (x − y)dy (2π)n Q n−1 1 f (x − y)R2 (y)dy = n (2π) Q n−1 1 2 f (x − y)H + R (y)dy (2π)n Q n−1 1 2 − (f χ )(x − y)H (y)dy. Q R (2π)n Rn

SR 2 (f ; x) − BR 2 (f6; x) =

Let δ = dis(G, Qc ) > 0. When x ∈ G, if |y| < δ, then x − y ∈ Q. Thus we have n−1 n−1 n−1 1 2 2 6 f (x − y)R2 (y)dy SR (f ; x) − BR (f ; x) = n (2π) Q n−1 1 2 + f (x − y)H R (y)dy (2π)n y∈Q,|y|>δ n−1 1 (f χQ )(x − y)HR 2 (y)dy − n (2π) |y|>δ = I1 + I2 + I3 .

(3.2.15)

From Chapter 2, we have n

α (y) = (2π) 2 2α Γ(α + 1) HR

J n2 +α (R|y|) (R|y|)

n +α 2

· Rn

for Reα > −1. Put α = n−1 2 . By the asymptotic equation of Bessel function, we have that n−1 H 2 (y) ≤ Cn |y|−n ≤ Cn , R δn for |y| > δ, R ≥ 1. It follows that |I2 | ≤ C and |I3 | ≤ C

1 f L(Q) , δn

1 f χQ L(Rn ) = Cδ−n f L(Q) . δn

3.2 The case of the critical index (general discussion)

157

Besides, the estimate of I1 is contained in (3.2.14). Thus we have n−1 n−1 + 2 2 |f | log |f |dx + 1 . (3.2.16) max SR (f ; x) − BR (f6; x) ≤ Cδ x∈G

Q

Combining the above results and the fact that trigonometric polynomials are dense in L(Q), by a standard argument, the proof of Theorem 3.2.3 is concluded.

It suﬃces to prove (3.2.13). We shall ﬁrst get the estimate of (3.2.11). Lemma 3.2.2 Let σ ∈ (0, 1) and τ ∈ (−∞, ∞). We have n−1 +σ+iτ σ ≤ C eπ|τ | , R R2 σ ∞

(3.2.17)

for R > 1. Proof. By the expression n−1 +z 2

HR

n 2

(u) = (2π) 2

n−1 +z 2

Γ

n+1 +z 2

J

n− 12 +z (R|u|)

(R|u|)

n− 12 +z

Rn ,

(3.2.18)

for Rez > − n+1 2 , and the formula (see Watson [Wat1]) & 2 1 1 π Jν (t) = cos t − νπ − + O eπ|Imν| 3 , πt 2 4 t2 t ≥ 1, where the real part of ν takes values within a bounded set, we have 1

and

|Jν (t)| ≤ Ceπ|Imν| t− 2

(3.2.19)

n−1 +z 2 H (u) ≤ Ceπ|τ | /(Rσ |u|n+σ ), R

(3.2.20)

for σ > 0, if Rez ∈ (0, 1]. By Theorem 1.2.2, we have n−1

DR2

+z

(u) =

n−1

+z

n−1

+z

HR 2

(m)eim·x

m∈Zn

=

m∈Zn

HR 2

(u + 2πm),

158

C3. Bochner-Riesz means of multiple Fourier series

for Rez = σ > 0. It follows that n−1

R 2

+z

(u) =

n−1

HR 2

+z

(u + 2πm),

(3.2.21)

m=0

for Rez = σ > 0. Substituting the above equation into (3.2.20), when u ∈ Qn , we can obtain that n−1 +z 1 ≤ Ceπ|τ | R−σ · 2 (u) R |u + 2πm|n+σ m=0

C π|τ | −σ e R , σ

≤

for 0 < σ ≤ 1, z = σ + iτ . This completes the proof of Lemma 3.2.2. Let σ =

1 q

with q > 2. By Lemma 3.2.2 and (3.2.11), we have that T1+iτ 1,∞ ≤ Ceπ|τ | q := M1 (τ ).

(3.2.22)

In the following, we shall estimate (3.2.10). Choose a function ψ(t) ∈ C 1 [0, 1] such that 1 ψ(t)dt = 1 0

and

1

tk ψ(t)dt = 0, 0

for k = 1, 2, . . . , n − 1. Deﬁne 9 D R 9 H R

n−1 +z 2

1

(x) =

1+ 0

n−1 +z 2

and 9 R

(x) =

n−1 +z 2

0

1

t R

t 1+ R

9 (x) = D R

n−1+2z

n−1

2 ψ(t)DR+t

n−1+2z

n−1 +z 2

+z

n−1

2 ψ(t)HR+t

9 (x) − H R

n−1 +z 2

(x)dt,

+z

(x)dt

(x).

3.2 The case of the critical index (general discussion) Lemma 3.2.3 If −1 < σ ≤ 0, then n−1 +σ+iτ 2 9 R

∞

≤

159

C π|τ | −σ e R 1+σ

(3.2.23)

holds. Proof. Let us deﬁne a function of z 1 n−1 t n−1+2z +z 2 1+ F (z) = ψ(t)HR+t (x + 2πm)dt. R 0

(3.2.24)

m=0

When Rez > 0, by (3.2.20) and (3.2.21), we have 9 F (z) = R

n−1 +z 2

(x).

(3.2.25)

n−1

n−1

+z +z 2 9 It is easy to notice that both (x) and R2 (x) are analytic R n+1 in the region Rez > − 2 . We shall now prove that F (z) is analytic, when Rez = σ > −1. Denote n n−1 n+1 +z 2 2 +z . C(z) = (2π) 2 Γ 2

By (3.2.18), we have 9 H R

n−1 +z 2

1

(y) = C(z)

0

t 1+ R

n−1+2z ψ(t)

Jn− 1 +z (R + t)|y| 2

1

1

(R + t)− 2 +z |y|n− 2 +z

dt,

which implies that it is analytic when Rez > − n+1 2 . Using the formula Jν (t) d Jν−1 (t) = ν−1 − ν−1 dt t t and integration by parts, we have that 9 H R

n−1 +z 2

t=1 −(R + t)2n−2+2z Jn− 32 +z ((R + t)|y|) (y) = n−1+2z R |y|2 [(R + t)|y|]n−3/2+z t=0 1 1 (2n − 2 + 2z)(R + t)2n−3+2z ψ(t) + 2 |y| 0 J 3 ((R + t)|y|) n− 2 +z + (R + t)2n−2+2z ψ (t) dt , 3 |(R + t)y|n− 2 +z C(z)

160

C3. Bochner-Riesz means of multiple Fourier series

and n−1 +z 2 H 9 (y) ≤ R

|C(z)| Ceπ|τ | Rn−1+σ · Rn−1+2σ |y|2 |y|n−1+σ 1 1 Ceπ|τ | 2n−3+2σ 2n−2+2σ R + 2 |ψ(t)| + R |ψ (t)| dt |y| 0 |Ry|n−1+σ ≤ Cσ eπ|τ | R−σ /|y|n+1+σ .

(3.2.26)

n−1 / +z 2 9 (x + 2πm) is uniformly Therefore, for x ∈ Q and R ≥ 1, m=0 H R convergent in every compact subset of the half plane Rez = σ > −1. Thus F (z) is analytic in the half plane. Consequently, as analytic functions deﬁned on Rez > −1, both sides of (3.2.25) coincide in Rez > −1. Invoking (3.2.26), we can get n−1 n−1 +z +z 2 2 9 H 9 (x) ≤ (x + 2πm) |F (z)| = R R

≤ Ceπ|τ | R−σ

m=0

≤C

m=0

1 |x + 2πm|n+1+σ

1 π|τ | −σ e R , 1+σ

for −1 < σ ≤ 0, x ∈ Q and R ≥ 1. This completes the proof of Lemma 3.2.3. Lemma 3.2.4 If −1 < σ ≤ 0, then σ+iτ + n−1 σ+iτ + n−1 2 2 9 (x) − HR (x) ≤ Cn e2π|τ | R−σ . sup HR

(3.2.27)

x∈Q

Proof. We conclude that 9 H R

n−1 +z 2

n−1 +z 2

(x) − HR

1

(x) = 0

:

t 1+ R

n−1+2z

n−1

2 HR+t

+z

(x)

−HR (x) ψ(t)dt 1 n−1 1 +z 2 (R + u)n−1+2z HR+t (x) = n−1+2z R 0 n−1 +z n−1+2z 2 HR (x) ψ(u)du. −R n−1 +z 2

3.2 The case of the critical index (general discussion)

161

n−1

2 (x). Then, we have Write ϕ(u) = (R + u)n−1+2z HR+u

ϕ(u) = C(z)(R + u)2n−1+2z Denote ν = n −

1 2

Jn− 1 +z ((R + u)|x|) 2

1

[ (R + u)|x|]n− 2 +z

.

+ z, y = (R + u)|x|, 0 < u < 1. We have

Jν+1 (y) Jν+1 (y) d Jν (y) =− |x| = − ν+1 y|x| ν ν du y y y and d2 Jν (y) Jν+2 (y) Jν+1 (y) 2 = y|x|2 − |x| 2 ν ν+1 du y y y ν+1 Jν+2 (y) Jν+1 (y) |x|2 . = − yν y ν+1 By an induction argument, it is easy to show that dk duk

Jν (y) yν

Jν+k (y) (k) Jν+k−1 (y) + a1 ν y y ν+1 (k) Jν+1 (y) + · · · + ak−1 ν+k−1 |x|k , y (k)

= a0

(3.2.28)

for k ∈ N, and k Ck eπ|τ | k d Ck eπ|τ | |x|k (y) J ν ≤ |x| ≤ . duk yν |y|n+σ (R|x|)n+σ

(3.2.29)

By Taylor’s formula, we obtain that ϕ(u) = ϕ(0)+ϕ (0)u+· · ·+

1 1 ϕ(n−1) (0)un−1 + ϕ(n) (θu )un , (3.2.30) (n − 1)! n!

for some 0 < θ < u. In addition, we use the orthogonal condition of ψ(u) to obtain 9 H R

n−1 +z 2

n−1 +z 2

(x) − HR

1

1

[ϕ(u) − ϕ(0)]ψ(u)du Rn−1+2z 0 1 1 1 (n) ϕ (θu )un ψ(u)du. = n−1+2z R n! 0

(x) =

162

C3. Bochner-Riesz means of multiple Fourier series

Hence we obtain n−1 n−1 +z +z 2 2 ≤ H 9 (x) − H (x) R R

1 Rn−1+2σ

ϕ(n) C[0,1] .

(3.2.31)

By Leibniz’ formula, we obtain that ϕ(n) (θ) = C(z)

n k=0

Cnk

dk Jν (y) dn−k · n−k (R + u)2n−1+2z duk y ν du

. u=θ

Since n−k d 2n−1+2z n n−1+2σ+k (R + u) , u ∈ [0, 1], dun−k ≤ Cn (1 + |τ | )R

(3.2.32)

if we substitute this into the (3.2.29), then we get that |x| > R−1 , |ϕ(n) (θ)| ≤ Cn (1 + |τ |n )eπ|τ |

n |x|k Rn−1+2σ+k

(R|x|)n+σ

k=0

≤ Cn e2π|τ | Rn−1+σ .

(3.2.33)

If |x| ≤ R−1 , it follows immediately from (3.2.28) that k d Jν (y) Cn π|τ | π|τ | k duk y ν ≤ Cn e |x| ≤ Rk e .

(3.2.34)

Invoking this estimate and (3.2.32), we obtain that |ϕ(n) (θ)| ≤ Cn eπ|τ | ≤ Cn e

n

Cnk ·

k=0 π|τ | n−1+2σ

R

Rn−1+2σ+k Rk

(3.2.35)

,

for |x| ≤ R−1 . Observe that σ ∈ (−1, 0], we have that (3.2.33) holds for every x ∈ Qn . By (3.2.31) and (3.2.33), the proof of Lemma 3.2.4 is complete.

Lemma 3.2.5 If − n−1 2 ≤ σ ≤ 0, then we have n−1 n−1 +σ+iτ +σ+iτ 2 9 ≤ Cn eπ|τ | R−σ , D 2 − DR R 2

where R ≥ 1.

(3.2.36)

3.2 The case of the critical index (general discussion)

163

Proof. We conclude that n−1

n−1

+σ+iτ +σ+iτ 2 9 (x) − D (x) DR2 R 1 # n−1 u n−1+2z |m|2 1+ = (1 − ) 2 +z eim·x 2 R (R + u) 0

−

|m|
=

1

1#

(R + u)2 − |m|2

n−1 +z 2

Rn−1+2z 0 |m|
=

|m|
n−1 +z $ |m|2 2 1− 2 eim·x ψ(u)du R

1 Rn−1+2z

1

eim·x

# n−1 [(R + u)2 − |m|2 ] 2 +z

0 |m|
n−1

−(R2 − |m|2 ) 2 1 1 + n−1+2z R 0

+z

$ eim·x ψ(u)du

(R + u)2 − |m|2

n−1 +z 2

R≤|m|
×e

im·x

χ(|m|−R,1] (u)ψ(u)du.

Deﬁne

λm

⎧ 1 n−1 +z 2 n−1 +z ⎪ ⎪ (R + u)2 − |m|2 2 − R − |m|2 2 ⎪ ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ × ψ(u)du, |m| < R, ⎪ ⎨ 1 n−1 +z = ⎪ (R + u)2 − |m|2 2 χ(|m|−R,1] (u)ψ(u)du, ⎪ ⎪ 0 ⎪ ⎪ ⎪ ⎪ ⎪ R ≤ |m| < R + 1, ⎪ ⎪ ⎩ 0, |m| ≥ R + 1.

Then 9 D R and

n−1 +z 2

n−1

(x) − DR 2

n−1 n−1 +z +z 2 2 D 9 = − DR R 2

+z

(x) =

1 Rn−1+2z

λm eim·x ,

m∈Zn

1 1 2 2 ; |λ | . m Rn−1+2σ |Q| 1

(3.2.37)

164

C3. Bochner-Riesz means of multiple Fourier series

When R − 1 ≤ |m| < R + 1, for σ ≥ − n−1 2 , it is obvious that |λm | ≤ C[(R + 1)2 − (R − 1)2 ]

n−1 +σ 2

≤ CR

n−1 +σ 2

.

Hence

|λm |2 =

R−1≤|m|<+∞

|λm |2 ≤ Cn R2(n−1+σ) .

(3.2.38)

R−1≤|m|
When |m| < R − 1, with the Taylor formula and the orthogonal property of ψ(u) and {u, . . . , un−1 }, we can get that λm =

1 0

1 dn 2 2 n−1 +z 2 [ (R + t) − |m| ] un ψ(u)du (0 < θu < u ≤ 1). n! dtn t=θu

However k d 2 2 n−1 +z k k 2 2 n−1 +σ−k 2 [ (R + t) − |m| ] , dtk ≤Cn (1 + |z|) R (R − |m| ) 2 k = 1, 2, . . . , n, 0 < t < 1. And it follows that

|λm |2 ≤ Cn eπ|τ | R2n

|m|
|m|
π|τ |

≤ Cn e

R

2n

1 Rn+1−2σ ∞

≤ Cn eπ|τ | Rn−1+2σ π|τ |

≤ Cn e

R

(R2

2(n−1+σ)

1

1 − |m|2 )n+1−2σ

R−1

tn−1 dt (R − t)n+1−2σ 0 Rn−1 dt tn+1−2σ

.

(3.2.39)

Combining (3.2.38), (3.2.39) and then substituting the above two into (3.2.37), we can obtain (3.2.36). This ﬁnishes the proof of Lemma 3.2.5.

Remark 3.2.1 Considering the above case, if n > 1, then fore, Lemma 3.2.5 is still valid in the case − 12 ≤ σ ≤ 0.

n−1 2

≥ 12 . There-

3.2 The case of the critical index (general discussion) Lemma 3.2.6 If − 12 ≤ σ ≤ 0, then we have n−1 +σ+iτ 2 ≤ Cn eπ|τ | R−σ , R

165

(3.2.40)

2

for R ≥ 1 and n > 1. Proof. By the deﬁnition, we have n−1 n−1 n−1 n−1 n−1 n−1 +z +z +z +z +z +z 2 2 2 2 2 2 9 9 9 + HR + = DR − DR − HR . R R Therefore, Lemma 3.2.6 is a direct consequence of Lemmas 3.2.3, 3.2.4 and 3.2.5.

Lemma 3.2.7 If n > 1, then (3.2.13) is valid. Proof. Take q > 2, R > 1 and write (3.2.22) as T1+iτ 1,∞ ≤ M1 (τ ) = Cn eπ|τ | q.

Substituting σ = 12 −1 + 2q ∈ − 12 , 0 into (3.2.40), and invoking (3.2.10), we can get that n−1 +σ+i τ π σ |τ | 2 2 (3.2.41) Tiτ 2,∞ = R R ≤ Cn e 2 := M0 (τ ). 2

By the interpolation theorem of Stein, we interpolate at t = 1 − 2q and then get n−1 2 T1− 2 q,∞ = R q q ∞ 1 log M1 (τ ) log M0 (τ ) ≤ exp sin πt + dτ 2 chπτ + cos πt −∞ chπτ − cos πt ≤ Cn q. This completes the proof of Lemma 3.2.7, and then that of Theorem 3.2.3. From Theorem 3.2.3, we can directly acquire a corollary about the localization problem.

166

C3. Bochner-Riesz means of multiple Fourier series

Corollary 3.2.1 Let f ∈ L log+ L(Qn ), n > 1. If f vanishes at the ball B(x, δ) for some δ > 0, then n−1

lim SR 2 (f ; x) = 0

R→∞

holds.

Proof. Because the convergence of the Bochner-Riesz means at the critical index of Fourier integral is a local property, in other words, the localization principle is valid (see Chapter 2), then by Theorem 3.2.3 we can transfer the problem into the integral and then obtain the conclusion immediately. This ﬁnishes the proof.

3.2.4 The order of Lebesgue constant In the above, we have mentioned that it is reasonable to draw an analogy n−1

between SR 2

and the Fourier partial sum with n = 1; and between the

kernel of SR

deﬁned as

n−1 2

n−1 2

DR

(x) =

|m|
|m|2 1− 2 R

n−1 2

eim·x

n−1

1 2 and the Dirichlet kernel with unitary variable. Then we call (2π) L(Qn ) n DR as Lebesgue constant. It is meaningful to ﬁgure out the order which shows how it increases as R → ∞. By Lemma 3.2.7, it follows that

n−1 2 ≤ M, R 1

so we have that

n−1 n−1 D 2 ≤ H 2 + M. R R 1

1

3.3 The convergence at ﬁxed point

167

A simple computation gives Jn− 1 (R|y|) n−1 2 H 2 = Cn Rn dy 1 R n− 2 Q (R|y|) 1 √nπ | cos(Rt − nπ 2 )| dt ≤ Cn t 1/R

+O Rn dy + O

√ nπ

1/R

|y|<1/R

dt Rt2

(3.2.42)

= O(log R) for R > 2. Therefore, we have

n−1 2 D R = O(log R),

(3.2.43)

1

as R → ∞. On the other hand, since n−1 n−1 n−1 n−1 D 2 ≥ H 2 − 2 ≥ H 2 − M, R R R R 1

1

1

it is easy to see that (3.2.42) is accurate, so is (3.2.43) in the sense of order. Remark 3.2.2 It is well known that the estimate of Lebesgue constant is very important in the study of convergence and approximation of linear means of Fourier series. Further reference on the estimates of Lebesgue constant can see Trigub [Tr1].

3.3

The convergence at ﬁxed point

Theorem 3.3.1 Let f ∈ L log+ L(Qn ), n ≥ 2, satisﬁes t un−1 (fx (u) − f (x)) du = o(tn ) (t → 0),

(3.3.1)

0

and

t

η

|fx (u + t) − fx u| du = o(1) (t → 0, η > 0). u

Then we have

(3.3.2)

n−1

lim SR 2 (f ; x) = f (x).

R→∞

(3.3.3)

168

C3. Bochner-Riesz means of multiple Fourier series

Proof. We might as well take it for granted that x = 0 and f (0) = 0. Let g(y) = f (y)χQ (y), y ∈ Rn . By Theorem 3.2.3 n−1 n−1 2 2 lim SR (f ; 0) − BR (g; 0) = 0, R→∞

it suﬃces to show that

n−1

lim BR 2 (g; 0) = 0.

(3.3.4)

R→∞

It is well known that n−1 2

BR

Jn− 1 (R|y|) Γ( n+1 2 ) 2 (g; 0) = √ g(y) Rn dy n− 12 2π n/2 Q (R|y|) √ n+1 ∞ Jn− 1 (Rt) 2Γ( 2 ) 2 = g0 (t)tn−1 Rn dt n− 12 Γ(n/2) 0 (Rt)

π ∞ R + . = cn π R

0

Denote

t

ϕ(t) = 0

g0 (τ )τ n−1 dτ.

By the condition of (3.3.1), we have ϕ(t) = o(tn ),

(3.3.5)

as t → 0. Denote Vν (t) = Jνtν(t) , then by integration by parts we can get that π R g0 (t)tn−1 Vn− 1 (Rt)Rn dt 2

0

π = ϕ(t)Vn− 1 (Rt)R −

π R

n R

0

2

π/R

ϕ(t)

= o(1) +

Jn− 1 (Rt)

0

= o(1) +

0

2

(Rt) π/R

0

n+1 ϕ(t)Vn− dt 1 (Rt)R

n− 12

2

Rn+1 dt

o(tn )O(Rt)Rn+1 dt = o(1) (R → ∞).

It follows that n−1 2

BR

(g; 0) = o(1) + Cn

∞ π R

g0 (t)tn−1

Jn− 1 (Rt) 2

n− 12

(Rt)

Rn dt.

(3.3.6)

3.3 The convergence at ﬁxed point Let Kn (u) =

and

169

&

2 ! nπ cos u − 1 2 πu 2 un− 2

nπ n(n − 1) sin u − − 2u 2 1

Jn− 1 (u) −

&

∞

2 cos(Rt − nπ 2 ) n R dt, n π π (Rt) R

& ∞ 2 n(n − 1) sin(Rt − nπ 2 ) n g0 (t)tn−1 − R dt, I2 = n+1 π π 2 (Rt) R ∞ I3 = g0 (t)tn−1 · Kn (Rt) · Rn dt. I1 =

g0 (t)t

n−1

·

π R

It follows from (3.3.6) that n−1

BR 2 (g; 0) = Cn (I1 + I2 + I3 ) + o(1) (R → ∞). & ∞ cos(Rt − nπ 2 2 ) dt g0 (t) I1 = π π t & R & cos(Rt − nπ cos(Rt − 2 η 2 ∞ 2 ) = g0 (t) g0 (t) dt + π π t π η t

(3.3.7)

nπ 2 )

dt,

R

where the second part is inﬁnitely small, since g0 (t)t−1 ∈ L((η, ∞)), by the Riemann-Lebesgue lemma. It follows that & cos(Rt − nπ 2 η 2 ) dt + o(1) (R → ∞). I1 = g0 (t) · π π t R

Obviously,

η π R

=

g0 (t) · 1 2

η

cos(Rt − t

π

+

1 2

−

1 2

R

π η− R

0

dt

π ) g0 (t) g0 (t + R π − cos(Rt − n)dt π t t+ R 2

η

π R

nπ 2 )

π g0 (t + R ) nπ cos(Rt − )dt π t+ R 2

π g0 (t + R ) π cos(Rt − n)dt. π t+ R 2

170

C3. Bochner-Riesz means of multiple Fourier series

By integration by parts and (3.3.5), we have π π

R g0 (t + π R) n dt cos Rt − − π t+ R 2 0 cos(Rt − π2 n) 2π R = ϕ(t) π tn R 2π R R sin(Rt − π2 n) n cos(Rt − π2 n) dt + ϕ(t) + π tn tn+1 R

= o(1), as R → ∞. By the absolute continuity of integration, it follows that η g (t + π ) nπ 1 0 R cos(Rt − |g(y)|dy = o(1). )dt ≤ π η− π t + R ηn η<|y|<η+ π 2 R

R

Besides, we have η π R

=

π ) g0 (t) g0 (t + R π − cos(Rt − n)dt π t t+ R 2

π g0 (t) − g0 (t + R ) π cos(Rt − n)dt π t 2 R η 1 1 π π + − cos(Rt − n)dt. g t+ π π R t t+ R 2 η

R

Making integration by parts to the second term on the right side of the above equality and (3.3.5), we can obtain that η+ π R 1 π π g(t) · cos(Rt − n)dt − π 2π (t − R )t R 2 R π cos(Rt − π2 n) η+ Rπ = −ϕ(t) π R tn (t − R ) 2π R η+ π R cos(Rt − π2 n) π + ϕ(t) dt π 2π tn (t − R ) R R

= o(1). From (3.3.2), it follows that η π g0 (t) − g0 (t + R ) π cos(Rt − n)dt = o(1) (R → ∞). π t 2 R

3.3 The convergence at ﬁxed point

171

In another words, we get I1 = o(1) as R → ∞. Similarly, we can have I2 = o(1). In order to estimate I3 , we shall now consider

&

Jn− 1 (u) 2 nπ n(n − 1) nπ 1 2 + Kn (u) = sin u − + cos u − 1 π 2 2u 2 un un− 2 &

cos u − nπ 2 (n + 1)n(n − 1) nπ 2 n + − sin u − π un+1 2un+2 2

& nπ nπ Jn+ 1 (u) sin u − cos u − 2 n(n + 1) 2 2 2 =− + + n n+1 n− 12 π u 2 u u 1 . (3.3.8) +O un+2 From the asymptotic formula (see Watson [Wat1]), it follows that & &

π 2 cos(u − π2 ν − π4 ) 2 (ν + 12 )(ν − 12 ) π √ sin u − − ν − Jν (u) = π u π 2 4 2u3/2 1 +O , (3.3.9) 5/2 u for u ≥ 1. We thus have that Kn (u)

=O

1

un+2

.

(3.3.10)

Now we proceed to compute I3 . In the method of integration by parts, with (3.3.1) and (3.3.10), we get that ∞ ∞ n ϕ(t)Kn (Rt)Rn+1 dt I3 = ϕ(t)Kn (Rt)R − = o(1) −

∞ π R

π R

n

o(t )O

π R

1 (Rt)n+2

Rn+1 dt

= o(1). Thus (3.3.4) holds and this ﬁnishes the proof. The above Theorem 3.3.1 was published in the essay by Lu [Lu1]. At the same time, Lu also proved it.

172

C3. Bochner-Riesz means of multiple Fourier series

Theorem 3.3.2 Let f ∈ L log+ L(Qn ), n ≥ 2. If f satisﬁes the following conditions at x0 ,

t

(i) 0

un−1 (fx0 (u) − f (x0 ))du = o(tn ), as t → 0,

(ii) The function Qx0 (t) := t(fx0 (t) − f (x0 )) has bounded variation on [0, η] with η > 0, and t < (Qx0 ) = O(t), as t → 0+ , 0

then (3.3.3) holds at x0 . Proof. In fact, condition (i) is (3.3.1). Make the same assumptions as that of 3.3.1. We can see that it only suﬃces to show that η cos(Rt − πn 2 ) g0 (t) lim dt = 0. (3.3.11) R→∞ π t R

By (3.3.5), for every ε > 0, there exists δ > 0, such that if 0 < t ≤ δ, |ϕ(t)t−n | < ε.

(3.3.12)

Of course we can take it for granted that δ < η. Now let R > we denote τ = √πεR .

τ π R

cos(Rt − g0 (t) t

πn 2 )dt

τ cos Rt − πn 2 = ϕ(t) π tn R τ R sin(Rt − + ϕ(t) π tn R

nπ 2 )

n cos(Rt − + tn+1

√π εδ

and

nπ 2 )

dt.

Since 0 < τ < δ, we substitute (3.3.12) into the above inequality and get τ τ cos(Rt − πn ε 2 ) εR + n dt ≤ ε + dt g (t) 0 π π t t R

R n (3.3.13) ετ R ≤ ε+ 1+ π √ ≤ Cn ε.

3.3 The convergence at ﬁxed point

173

Besides, with the formula θ0 (t) 1 g0 (t) = 2 dθ0 (t) − 2 3 dt, d t t t and (θ0 (t) = tg0 (t)), we denote the total variation as t < (θ0 ) = h(t). 0

Then, we have η cos(Rt − g0 (t) t τ

Therefore, η cos(Rt − g0 (t) t τ

nπ 2 )

dt =

1 πn g0 (t) η sin Rt − · R 2 t τ η sin(Rt − πn 2 ) − dθ0 (t) Rt2 τ η sin(Rt − πn 2 ) +2 θ0 (t)dt. 3 Rt τ

η η 1 h(t) 1 g0 (η) 1 h(τ ) + dh(t) + 2 dt dt ≤ + 2 2 3 η R R τ τ Rt τ Rt η dt 1 1 =O +O +O 2 R Rτ τ Rt √ ≤ M ε, (3.3.14) where M is independent of ε, R with R > √πεδ . Combining (3.3.13) with (3.3.14), we can obtain (3.3.11), which ﬁnishes the proof. nπ 2 )

In the following, we will prove a suﬃcient condition which guarantees (3.3.2) to be valid which was proven by Lu [Lu5]. Let us ﬁrst introduce some concepts. Definition 3.3.1 Deﬁne a ﬁnitely valued function f (t) on a ﬁnite interval [a, b]. If for any open interval of [a, b], that is, In = [an , bn ], n = 1, 2, . . . , which does not intersect with each other, the following holds: HV(f ; [a, b]) := sup

∞ |f (bn ) − f (an )|

{In } n=1

n

< ∞,

174

C3. Bochner-Riesz means of multiple Fourier series

then we say f is of harmonic bounded variation on [a, b], denoted by f ∈ HBV[a,b] . From the deﬁnition, we can see that if f ∈ HBV[a,b] , then f (x0 +0) exists and is ﬁnite at every point x0 on (a, b]. In fact, if lim inf f (x) = α < lim sup f (x) = β, x→x+ 0

x→x+ 0

then we choose the point sequence b1 > a1 > b2 > a2 > · · · , an → x+ 0, such that f (bn ) > 34 β + 14 α, f (an ) < 14 β + 34 α. Therefore, ∞ f (bn ) − f (an ) n=1

n

≥

∞ 11 (β − α) = +∞, n2

n=1

which contradicts the fact f ∈ HBV[a,b] . Similarly, f (x0 − 0) exists and is ﬁnite (x0 ∈ [a, b)). Besides, the following relation is quite evident, BV[a,b] ⊂ HBV[a,b] ⊂ B[a,b] , where BV represents for the class of usual bounded variation functions, and B means the class of bounded functions. Lemma 3.3.1 Let f ∈ L(Qn ) with n ≥ 2. If fx (t) is of bounded variation at x on [0, η] (η > 0), then (3.3.2) holds. Proof. Firstly, we will prove that, for any two positive numbers η1 , η2 with η1 < η2 , η1 η2 |fx (t + h) − fx (t)| |fx (t + h) − fx (t)| dt = lim sup dt. lim sup t t h→0 h→0 h h (3.3.15) In fact, η1 η2 |fx (t + h) − fx (t)| |fx (t + h) − fx (t)| dt ≤ dt t t h hη1 |fx (t + h) − fx (t)| dt ≤ (3.3.16) t h 1 η2 |fx (t + h) − fx (t)|dt, + η1 η1

3.3 The convergence at ﬁxed point

175

for h < η1 . Since fx (t) ∈ L(η1 , η1 + η2 ), by the property of the norm of the integration, the second term on the right side of the above inequality is a ﬁnitely small quantity (h → 0+ ). Hence we can get (3.3.15). If (3.3.2) is not valid, then according to (3.3.15), there exists a positive number ε, such that for every δ > 0, lim sup h→0+

h

δ

|fx (t + h) − fx (t)| dt > ε. t

(3.3.17)

Therefore, we could choose such a sequence by induction: η ≥ δ1 > h1 > δ2 > h2 > · · · such that, for n ∈ N, δn = λn hn ,

(3.3.18)

hn ≥ 2δn+1 ,

(3.3.19)

for λn ∈ N, λn ≥ 2,

δn hn

and

λ n −1 j=1

|fx (t + hn ) − fx (t)| dt > ε, t

1 1 |fx (t + h) − fx (t)| < ε, sup j 0
(3.3.20)

(3.3.21)

for n = 2, 3, . . . . By (3.3.16), (3.3.17) and the existence of fx (0 + 0), there exists such a sequence. Thus we have

|fx (t + hn ) − fx (t)| dt t hn λ n −1 hn |fx (t + (j + 1)hn ) − fx (t + jhn )| dt = t + jhn j=1 0 ⎛ ⎞ hn λ n −1 1 1 ⎝ |fx (t + (j + 1)hn ) − fx (t + jhn )|⎠ dt. ≤ hn 0 j δn

ε<

j=1

Since fx (t) is a bounded function, there exists θn ∈ (0, hn ), such that

176

C3. Bochner-Riesz means of multiple Fourier series

ε≤

λ n −1 j=1

1 |fx (θn + (j + 1)hn ) − fx (θn + jhn )| j

λ n −1

=

j=λn−1 −1

1 |fx (θn + (j + 1)hn ) − fx (θn + jhn )| j

λn−1 −1

1 |fx (θn + (j + 1)hn ) − fx (θn + jhn )|. j

+

j=1

It follows that λ n −1 1 1 ε< |fx (θn + (j + 1)hn ) − fx (θn + jhn )|, 2 j j=λn−1

for n = 2, 3, . . . . And thus ∞ λ n −1 1 |fx (θn + (j + 1)hn ) − fx (θn + jhn )| = +∞, j n=2 j=λn−1

which contradicts the fact fx (t) ∈ HBV[0,η] , and this ﬁnishes the proof. From Theorem 3.3.1 and Lemma 3.3.1, we can deduce the main result of Lu [Lu5] immediately. Theorem 3.3.3 Let f ∈ L log+ L(Qn ), for n ≥ 2. If fx (t) is of harmonic bounded variation on some interval [0, η] with η > 0, and lim fx (t) = f (x),

t→0+

(3.3.22)

then (3.3.3) is valid. Obviously, if we remove the assumption (3.3.22), we can obtain n−1

lim SR 2 (f ; x) = fx (0 + 0),

R→∞

which is just the original form in Lu’s theorem [Lu5]. At the end of this section, we will point out that the proof methods of several suﬃcient conditions to guarantee convergence at ﬁxed points are

3.4 Lp approximation

177

similar. They all transfer the problems into the Fourier integral under the condition that L log L is integrable, where Theorem 3.2.3 weighs a lot. At the same time, we notice that these conclusions are undoubtedly applicable for the Bochner-Riesz means of Fourier integral at the critical index, and in the case of integral, as the condition of convergence, we do not need the local L log L-integrability any more. As far as the convergence problem of the Bochner-Riesz means of the multiple Fourier series is concerned, Wang [Wan1] and Chang [Chan1] also dedicated in the=study while Shi [Sh1] investigated the Fourier series of functions of the BMV class.

3.4

Lp approximation

When f ∈ Lp (Qn ), n > 1, we shall consider the convergence (Lp convergence and a.e.-convergence ) of the Bochner-Riesz means of the order n−1 2 α > αp := 2 p − 1. We shall see that for functions of Lp class, when of the Bochner-Riesz means all lose critical 1 < p < ∞, the index n−1 2 p meanings in problems of L -convergence and a.e.-convergence. We think that for Lp - functions and when 1 < p ≤ 2, under the conclusion in Theorem 3.10.3, the order of the Bochner-Riesz means might be regarded

critical n−1 2 as αp = 2 p − 1 . At this time, it makes sense to consider the properties of the αp order of the Bochner-Riesz means, however, it is not an easy task. Remark 3.4.1 The study on approximation of Bochner-Riesz means is usually based on their convergence. To study approximation of Bochner-Riesz means in Lp with 1 < p < ∞, we have to consider Theorem 3.10.3 as our starting point. In this sense, we temporarily regard the critical order of the Bochner-Riesz means in Lp approximation with 1 < p < ∞ as 2 αp = n−1 2 p − 1 . In fact, according to the conjecture in Section 2.2, the critical order of the Bochner-Riesz means in Lp convergence or approxima tion should be αp = n 1p − 12 − 12 (see Sogge [S1], Stein [St5] or Davis-Chang [DC1]). 3.4.1 The estimate of the maximal operator Let f ∈ L(Qn ) with n > 1. Let α ∈ C and Reα > −1. S∗α (f ) is the maximal operator given by Deﬁnition 3.1.1. The estimate of S∗α above the critical index n−1 2 has been ﬁgured out by Theorem 3.1.3. Now we assume

178

C3. Bochner-Riesz means of multiple Fourier series

f ∈ Lp , 1 0

1 R

R 0

−

dR α SR (f ; x)|2

1 2

R

|Suα (f ; x)|2 du

1 2

.

Lemma 3.4.1 If − 12 < Reα < n and τ = Imα, then we have ⎛ ⎞ 1 ⎠ f 2 , M α (f )2 ≤ An e2π|τ | ⎝1 + > 1 Reα + 2 ⎛

and

(3.4.1)

⎞

1 ⎠ f 2 . Gα (f )2 ≤ An ⎝1 + > 1 Reα + 2

(3.4.2)

Proof. We conclude that ∞ dR α 2 α+1 α |SR (f ; x) − SR (f ; x)|2 dx G (f )2 = R 0 Q α 2 ∞ α+1 |m|2 |m|2 dR − 1− 2 = 1− 2 |Cm (f )|2 |Q| R R R 0 |m|
2Reα |m|2 |m|4 1 1− 2 |Q| · 4 |Cm (f )|2 dR R R R 0 |m|
1 A 1+ f 22 . Reα + 12

=

= = = ≤

∞

3.4 Lp approximation

179

Consequently, we can get (3.4.2). In order to prove (3.4.1), we choose n+1 n+1 +1> , k= 2 2 and we have Reα + k > Since

1 R

R 0

1 2

|Suα (f )|2 du

≤

k 1 R

j=1

+ ≤

k

1 R

R 0

n . 2

|Suα+j−1 (f ) −

R 0

12

Suα+j (f )|2 du

|Suα+k (f )|2 du

1 2

Gα+j−1 (f ) + S∗α+k (f ),

j=1

it follows from (3.4.2) and (3.1.12) that (3.4.1) holds, which completes the proof.

Lemma 3.4.2 Let β ∈ C, Reβ > 0, δ > −1, then R 2 1 β+δ (R2 − r 2 )β−1 r 2δ+1 Srδ (f ; x)dr. (3.4.3) · SR (f ; x) = B(β, δ + 1) R2(β+δ) 0 Proof. Set s = xt ,

1

B(β, δ + 1) =

0 t

=

0 t

= 0

(1 − s)β−1 sδ ds

(1 − 1 tβ+δ

x β−1 xδ dx ) t tδ+1 (t − x)β−1 xδ dx

for every t > 0. For any r > 0, let x = u2 − r 2 (u ≥ r). We have B(β, δ + 1) =

1 tβ+δ

r

√

r2 +t

(t + r 2 − u2 )β−1 (u2 − r 2 )δ 2udu.

180

C3. Bochner-Riesz means of multiple Fourier series √

Substituting R =

r 2 + t into the above equation, we have that

B(β, δ + 1) =

t

β+δ

2

2 β+δ

= (R − r )

2 tβ+δ

R

2

2 β−1

(R − u )

r

2 = B(β, δ + 1)

R

2

r2 1− 2 u

δ

u2δ+1 du,

2 β−1 2δ+1

(R − u )

u

r

1−

r2 u2

δ du

and β+δ δ R 2 1 r2 r2 2 2 β−1 2δ+1 = (R − u ) u du. 1 − 1− 2 R B(β, δ + 1) R2(β+δ) r u2 It follows that β+δ (f ; x) SR

=

# |m|
2 1 B(β, δ + 1) R2(β+δ)

δ $ |m|2 (R2 − u2 )β−1 u2δ+1 1 − 2 du Cm (f )eim·x u |m| R 2 = (R2 − u2 )β−1 u2δ+1 · B(β, δ + 1)R2(β+δ) 0 ⎞ ⎛ 2 δ |m| ⎝ χ(|m|,R) (u)Cm (f )eim·x ⎠ du 1− 2 u |m|
×

R

This completes the proof.

Lemma 3.4.3 Let Reβ > 0, δ > −1. Then we have S∗β+δ (f )(x) ≤ for σ = Reβ.

B(σ, δ + 1) δ S (f )(x) |B(β, δ + 1)| ∗

(3.4.4)

3.4 Lp approximation

181

Proof. By (3.4.3) and the following estimate

1 |R2(β+δ) |

R 0

1

|R2 − u2 |β−1 u2δ+1 du R

(R2 − u2 )σ−1 u2δ+1 du R2(σ+δ) 0 1 1 = 2(σ+δ) · R2(σ+δ) · (1 − t2 )σ−1 t2δ+1 dt R 0 1 = B(σ, δ + 1), 2 =

we can get (3.4.4), which ﬁnishes the proof.

Lemma 3.4.4 Let Reβ >

1 2

and δ > − 34 . Then

; S∗β+δ (f )(x)

≤

2B(2σ − 1, 2δ + 3/2) δ M (f )(x), |B(β, δ + 1)|

(3.4.5)

where σ = Reβ. Proof. By (3.4.3) and H¨ older’s inequality, we can get that β+δ SR (f )(x) ≤

R 1 2 (R2 − u2 )2(σ−1) u4δ+2 du |B(β, δ + 1)| R2(σ+δ) 0 1 R 2 δ 2 × |Su (f ; x)| du 0

1 2 · ≤ |B(β, δ + 1)| R2(σ+δ)− 12 × M δ (f )(x).

1 2

B(2σ − 1, 2δ + 3/2) 4(σ+δ)−1 R 2

1 2

And thus we can obtain (3.4.5), which completes the proof.

Corollary 3.4.1 Let f ∈ L2 (Qn ) with n > 1 and 0 < Reα < n + 1. Then we ahve 1 (3.4.6) S∗α (f )2 ≤ An eAn |τ | f 2 , σ for σ = Reα and τ = Imα.

182

C3. Bochner-Riesz means of multiple Fourier series + iτ, δ = σ−1 2 . By (3.4.5), we have > 2B(σ, σ + 12 ) σ−1 α M 2 (f )(x). S∗ (f )(x) ≤ σ+1 σ+1 B + iτ,

Proof. Choose β =

σ+1 2

2

2

Now, we quote a formula on Gamma functions (see Bateman [Ba1]): −iγy

Γ(x + iy) = Γ(x)e

−1

x(x + iy)

∞ k=1

y

ei k 1+

iy k+x

,

(3.4.7)

where x > 0, y ∈ R, γ is the Euler number. We have that |Γ(σ + 1 + iτ )| 1 σ+1 σ+1 σ+1 σ+1 = Γ B + iτ, + iτ Γ 2 2 2 2 ∞ Γ(σ + 1)(σ + 1) (σ + 1 + iτ )−1 1 + = Γ σ+1 2 · σ+1 σ+1 + iτ −1 k=1 1 + 2 2 2 ⎛⎧ ⎫1 ⎞ τ2 ⎪ ⎪2 ∞ 1+ σ+1 2 ⎬ ⎟ ⎜⎨ k+ ( 2 ) ⎟ = O⎜ ⎝⎪ τ2 ⎪ ⎠ 1+ ⎩ 2 ⎭ k=1

iτ k+ σ+1 2 iτ k+σ+1

(k+σ+1)

⎛ 12 ⎞ ∞ 2 k + σ+1 2 + τ 2 (k + σ + 1) 2 ⎠ = O⎝ σ+1 2 (k + σ + 1)2 + τ 2 k=1 k + 2 ⎞ ⎛? @∞ 2 @ 4nτ ⎠ 1+ 3 = O ⎝A k + kτ 2 k=1 ⎛? ⎞ @∞ @ 2n|τ | ⎠ = O ⎝A 1+ . k2 k=1

Using the following fact 2nτ 1+ 2 = k we have that

2nτ 1+ 2 k

k2 2nτ

2nτ k2

2nτ

≤ e k2 ,

1 σ+1 ≤ An eAn |τ | σ+1 B + iτ, 2 2

(3.4.8)

3.4 Lp approximation

183

for 0 < σ < n + 1. Hence σ−1 1 S∗α (f )(x) ≤ An √ eAn |τ | M 2 (f )(x). σ

(3.4.9)

Then, by (3.4.1) we can get (3.4.6). This completes the proof.

Theorem 3.4.1 (Stein) If 1 1, n−1 2 ≥ α > αp := then we have S∗α (f )p ≤ An,α,p f p .

n−1 2

2 p

−1 ,

The proof of the above theorem needs the interpolation theorem of the analytic family of operators (see Stein and Weiss [SW1]). The steps of the proof are as follows: Firstly, by Corollary 3.4.1, S∗α of positive order (Reα > 0) is of type (2, 2); secondly, by Corollary 3.1.1, S∗α of the order n−1 higher than n−1 2 (Reα > 2 ) is of type (q, q) (1 < q ≤ ∞). Therefore, for any given p ∈ (1, 2) and 0 < α ≤ n−1 2 , we assume that μ0 is taken close to zero, 0 < μ0 < α, while we choose μ1 > n−1 2 . Make an analytic family δ(z)

S∗ , δ(z) = μ0 (1 − z) + μ1 z (which is not a family of linear operators, for further treatments needed), then, by Reδ(iτ ) = μ0 > 0 and Reδ(1 + iτ ) = δ(iτ ) μ1 > n−1 in type (2, 2) as well as that of 2 , we can get an estimate of S∗ S δ(Hiδ) in type (p1 , p1 ), where p1 is some chosen positive number, 1 < p1 < p. Then we interpolate at t, which is decided by 1−t t 1 + = , 2 p1 p that is, t=

1 1 p − 2 1 1 p1 − 2

∈ (0, 1).

(3.4.10)

μ0 , μ1 must satisfy the condition that δ(t) = μ0 (1 − t) + μ1 t = α.

(3.4.11)

Hence we can induce that S∗α is of type (p, p). Actually, we can select p1 ∈ (1, p) ﬁrstly, and then give the option of t to (3.4.10). Substituting the value of t into (3.4.11), then we can choose proper μ0 , μ1 by (3.4.11).

184

C3. Bochner-Riesz means of multiple Fourier series

Here we have to point out that: if α is not given ahead, then by (3.4.11), since μ0 > 0, μ1 > n−1 2 , we have n−1 t 2 1 1 n−1 p − 2 = 2 p1 − 12 1 n−1 2 ≥ −1 2 p := αp .

α >

That is to say that the index α, which is applicable for the method of interpolation, must be greater than αp . Proof of Theorem 3.4.1. When p = 2, the conclusion of Theorem 3.4.1 is contained in Corollary 3.4.1. If we assume 1 2 −1 2

and μ1 <

n−1 1 . 2 2−p

It follows that μ0 =

1 p1 1 p1

− −

1 2 1 p

1 α + αp α− 2 2 p1 − 1

3α + αp p(2 − p) (α − αp ) 2(p − 1) 4(α + αp ) n−1 3 < · , 2 4 =

and μ0 > 0.

3.4 Lp approximation

185

Let δ(z) = μ0 (1 − z) + μ1 z, 0 ≤ Rez = σ ≤ 1, Imz = τ . By Corollary 3.4.1, we have that |Imδ(iτ )| = (μ1 − μ0 )|τ | < and

n−1 1 |τ |, 2 2−p

δ(iτ ) A 1 |τ | 1 f 2 . S∗ (f ) ≤ An e n 2−p μ0 2

(3.4.12)

Similarly, from Corollary 3.1.1, we have |Imδ(1 + iτ )| = (μ1 − μ0 )|τ | <

n−1 1 × · |τ |, 2 2−p

and therefore δ(1+iτ ) (f ) S∗

1 An 2−p |τ |

p1

≤ An e

p1 1 n−1 + 1 p − 1 f p1 . μ1 − 2 1

(3.4.13)

Denote by F the collection of all non-negative bounded measurable functions deﬁned on Q. Choose any R = R(x) ∈ F . The mapping δ(z) Tz = TzR : f (x) → SR(x) (f ; x), δ(z)

for f ∈ L(Q)), is linear. The image of the mapping SR(x) (f ; x) is a bounded measurable function deﬁned on Q, and the family {Tz : 0 ≤ Rez ≤ 1} is admissible. Obviously, by (3.4.12) and (3.4.13), we can deduce that δ(z)

|Tz (f )(x)| ≤ SR (f ; x).

Tiτ (f )2 ≤ M0 (τ )f 2 ,

(3.4.14)

T1+iτ (f )p1 ≤ M1 (τ )f p1 ,

(3.4.15)

M0 (τ ) = An M1 (τ ) = An

(z = σ + iτ ),

1 1 p−1 · eAn 2−p |τ | , 2 − p α − αp

1 α + αp A 1 |τ | · e n 2−p . +1 μ 1 − α1 (α − αp )(p − 1)

(3.4.16)

(3.4.17)

186

C3. Bochner-Riesz means of multiple Fourier series By Stein’s interpolation theorem, for t =

1 −1 p 2 1 − 12 p1

, we get that

Tt (f )p ≤ Mt f p , Mt = exp

1 sin tπ 2

∞

−∞

(3.4.18)

log M1 (τ ) log M0 (τ ) + chπτ − cos πt chπτ + cos πτ

dτ

. (3.4.19)

And (3.4.18) gives that α (f ; x)p ≤ An,α,p f p , SR(x)

∀ R ∈ F.

(3.4.20)

Choose R0 > 1 arbitrarily. Line all the rational numbers in [0, R0 ) as a sequence of {Rj }∞ j=1 . Obviously, α α sup |SR (f ; x)| = sup{|SR (f ; x)|}. j

0
Let

j∈N

Ej =

x ∈ Qn :

α α sup |SR (f ; x)| ≤ |SR (f ; x)| + j

0
Deﬁne F1 = E1 , Fj+1 = Ej+1 \

j

1 R0

, j ∈ N.

Ek , j ∈ N,

k=1

and R0 (x) = Rj , when x ∈ Fj , j ∈ N. Then R0 (x) ∈ F and α α (f ; x)| ≤ |SR (f ; x)| + sup |SR 0 (x)

0
1 . R0

By the Fatou lemma, we can deduce that α S∗α (f )p ≤ lim inf SR (f ; x)p . 0 (x) R0 →+∞

Thus we have transfered from (3.4.20) to S∗α (f )p ≤ An,α,p f p for all f ∈ Lp (Qn ), which completes the proof of Theorem 3.4.1.

(3.4.21)

3.4 Lp approximation

187

If we restrict p ∈ (1, 3/2], n > 1, α = n−1 2 , and in the proof of Theorem 3.4.1, we choose p+1 n−1 , μ0 = p1 = 2 4 and n − 1 1 + 2p − p2 μ1 = , 2 2 + p − p2 then by the method of interpolation, we can obtain that: Theorem 3.4.2 If 1 1, then we have n−1 1 S∗ 2 (f ) ≤ An f p . (p − 1)2 p

(3.4.22)

Proof. From the proof of Theorem 3.4.1, we can see that: the constant in (3.4.21) actually can be chosen as Mt in (3.4.19). When 1 < p ≤ 3/2, and α = α1 = n−1 2 , with the above interpolation method, we can compute the corresponding M0 (τ ) and M1 (τ ) as M0 (τ ) = An eAn |τ | ,

M1 (τ ) = An

1 eAn |τ | . (p − 1)2

With the help of 1 sin πt 2 and 1 sin πt 2

∞

−∞

∞ −∞

1 1 + chπτ − cos πt chπτ + cos πt

|τ | |τ | + chπτ − cos πt chπτ + cos πt

we can get that log

Mt = e

An +An (p−1)2

≤ An

dτ = 1

dτ ≤ A < +∞,

1 . (p − 1)2

It follows (3.4.22), which completes the proof. Similar to the proof of Theorem 3.4.1, if we interpolate between 2 and +∞, then we can get the conclusion of p > 2. At this time, the critical n−1 2 p index for L is αp = 2 1 − p . Detailed steps are much simpler. And t is decided by the interpolation formula t 1 1−t + = , 2 ∞ p

188

C3. Bochner-Riesz means of multiple Fourier series

that is, t = 1 − 2p . And then we choose μ0 > 0, μ1 > α1 =

n−1 2 ,

satisfying

μ0 (1 − t) + μ1 t = α. 2 By the above equation, as long as α > αp = n−1 2 (1 − p ), we can make it, and that is where the meaning of the critical property of αp lies. We can take

μ1 =

α + αp α − αp α + αp = p, μ0 = p, 2t 2(p − 2) 4

where

n−1 . 2 Making an estimate in type (2, 2) with (3.4.6), we can get 1 μ0 +iτ (μ1 −μ0 ) (f ) ≤ An eAn (μ1 −μ0 )|τ | f 2 . S∗ μ0 2 α p < α ≤ α1 =

(3.4.23)

Make an estimate in type (∞, ∞) with Corollary (3.1.1) (see (3.1.12)) 1 μ1 +iτ (μ1 −μ0 ) An (μ1 −μ0 )|τ | (f ) ≤ An e + 1 f ∞ . (3.4.24) S∗ μ 1 − α1 ∞ Therefore, we can easily obtain the following theorem. Theorem 3.4.3 Let p > 2, n > 1, inequality (3.4.21) still holds.

n−1 2

≥ α > αp :=

n−1 2

1−

2 p

. The

In detail, we substitute with α = α1 = n−1 2 , and we can see from (3.4.23), p−1 n−1 (3.4.24) the corresponding quantities: μ1 = n−1 2 p−2 , μ0 = 4 , μ1 − μ0 = p n−1 4 p−2 , and p An p−2 |τ |

,

p An p−2 |τ |

p.

M0 (τ ) = An e M1 (τ ) = An e

And thus we get the following statement. Theorem 3.4.4 Let p > 3, n > 1, then n−1 S∗ 2 (f ) ≤ An pf p .

(3.4.25)

p

Combining (3.4.22) with (3.4.25), we have 3 n−1 S∗ 2 (f ) ≤ An p f p (p − 1)2 p for 1 < p < ∞.

(3.4.26)

3.4 Lp approximation

189

3.4.2 The Lp approximation With the estimate of the maximal operator S∗α in type (p, p), 1 < p < ∞, α (f ) is a natural corollary. Now let us compute it the Lp -convergence of SR more accurately and then we can acquire the approximation order characterized by the modulus of continuity. Let f ∈ Lp (Qn ). Deﬁne ω(f ; t)p = sup f (· + h) − f (·)p (t ≥ 0) |h|≤t

as the modulus of continuity of f in Lp . Denote by ER (f )p the best approximation of trigonometric polynomial f with order below R ( spherical order ) in the norm of Lp . If gR (x) is the best Lp approximation trigonometric polynomial with order R, then by the well-known theorem about the relation between trigonometric polynomial and the norm of its derivative, we can get 1 2 , (3.4.27) gR p ≤ Cn,p R ω2 f ; R p where ω2 represents for 2-ordered the modulus of continuity, and its deﬁnition is as follows ω2 (f ; t)p = sup f (· + h) + f (· − h) − 2f (·)p |h|
and is the well-known Laplace operator. Having got all the above preparations, it is time for us to prove the following theorem on the Lp approximation: 2 − 1 Theorem 3.4.5 If n > 1, 1 αp := n−1 , then we 2 p have 1 α . (3.4.28) SR (f ) − f p ≤ Cn,α,p · ω2 f ; R p Proof. Suppose that R > 0, g = gR is the best R-ordered Lp approximation trigonometric polynomial. Then, α α α SR (f ) − f p ≤ SR (f ) − SR (g)p +

n

α+j S α+j−1 (g) − SR (g)p

j=1

+

α+n SR (g)

− gp + g − f p . (3.4.29)

190

C3. Bochner-Riesz means of multiple Fourier series

By Theorem 3.4.1 (and (3.4.3)), we can get that α α (f ) − SR (g)p ≤ S∗α (f − g)p ≤ Cn,α,p f − gp , SR

and with the Jackson inequality f − gp = ER (f )p ≤ Cn,p

1 ω2 f ; R p

(3.4.30)

we have α SR (f )

−

α SR (g)p

≤ Cn,α,p f − gp ≤ Cn,α,p ω2

1 f; R

.

(3.4.31)

p

Besides α+j−1 (g) SR

−

α+j SR (g)

=

|m|
=

|m|2 1− 2 R

−1 α+j−1 S (g) R2 R

α+j−1

|m|2 Cm (g)eim·x R2

∀ j.

With the estimate in the strong type and (3.4.27), it follows that 1 α+j−1 α+j (g) − SR (g) ≤ Cn,α,p 2 gp SR R p 1 ≤ Cn,α,p ω2 f ; , R p for j = 1, 2, . . . , n. Finally, it is obvious that α + n > that when β > n−1 2 ,

n−1 2 .

(3.4.32)

In the next lemma, we will prove

1 β (f ) − f p ≤ Cn,β,p · ω2 f ; . SR R p

(3.4.33)

Then, substituting β = α + n into (3.4.33), and combining (3.4.32) with (3.4.31), we put them together into (3.4.29), and thus we get (3.4.28). This ﬁnishes the proof of the theorem.

Lemma 3.4.5 If β >

n−1 2 ,

1 < p < ∞, then (3.4.33) is valid.

3.4 Lp approximation

191

Proof. By Bochner’s formula (1.2), α (f ; x) − SR

f (x) = Cn,β · R

n −β 2

∞ 0

Denote ϕx (t) = fx (t) − f (x) =

1

[fx (t) − f (x)]

J n2 +β Rt n

tβ+1− 2

dt.

(3.4.34)

ωn−1

Sn−1

[f (x − tξ) − f (x)]dσ(ξ).

Divide the sphere Sn−1 into two parts: S + = {ξ : |ξ| = 1, ξ1 > 0} and

S − = {ξ : |ξ| = 1, ξ1 < 0},

then ϕx (t) =

1 ωn−1

S+

[f (x + tξ) + f (x − tξ) − 2f (x)]dσ(ξ).

Rewrite (3.4.34) as α (f ; x) − f (x) SR ∞ J n +β (t) t tn−1 2 n +β dt = Cn,β · ϕx R t2 0 ∞ 1 n n (t) J (t) J +β +β t t tn−1 · 2 n +β dt + tn−1 2 n +β dt = Cn,β ϕx ϕx R R t2 t2 0 1

= I1 + I2 . (3.4.35)

I1 p ≤ ≤ ≤ ≤

p 1 1/p t n−1 ϕx dt · dx t C R 0 Q p 1 1/p t n−1 C dx dt ϕ t x R 0 Q p 1 t t dx C f x + ξ + f x − ξ − 2f (x) R R 0 S+ Q 1 . Cn,β,p · ω2 f ; R p

1 p

dσ(ξ)dt

(3.4.36)

192

C3. Bochner-Riesz means of multiple Fourier series

In order to estimate the Lp norm of I2 , with the asymptotic formula (3.3.8) of Bessel functions, we separate I2 into three terms. In the following, we omit the constants and write down that ∞ ∞ t cos(t − θ) t sin(t − θ) ϕx ϕx I2 = 3 dt + n 5 dt β− n + R R 2 2 1 tβ− 2 + 2 t 1 ∞ 1 t (3.4.37) + ·O dt ϕx n 7 R 1 tβ− 2 + 2 = J1 + J2 + J3 . Let δ = β −

n−1 2

Then we have J1 =

∞

1

=

> 0, and we denote ϕx

t R

1

3 β− n 2 +2

t

= ϕx

t R

1 t1+δ

h(t) cos(t − θ)dt

1 1 2 3π h(t) cos(t − θ)dt − π h(t + π) 1 1−π 8 1 1 + π h(t + π) + h(t + π) · cos(t − θ)dt, 4 2

1 8

= h(t).

∞

(3.4.38)

where ‘π ’ is a diﬀerential operation for the step of π: π h(t) = h(t) − h(t + π), 2π = π (π ), 3π = π (2π ). Then just the same as the estimate of I1 p , we can obtain that p 1/p 1 1 1 2 π h(t + π) + π h(t + π) + h(t + π) cos(t − θ)dt dx 4 2 Q 1−π 8 1 ≤ Cn,β,p · ω2 f ; . (3.4.39) R p

1

We substitute u(t) = ϕx

t R

and v(t) =

1 t1+δ

into the formula

3π (u(t)v(t)) = 3π v(t)v(t) + 32π u(t + π)π v(t) + 3π u(t + 2π)2π v(t) + u(t + 3π)3π v(t), and with the notice of the fact that when t ≥ 1,

3.4 Lp approximation

193

3π u(t)Lp (Q) = π u(t)Lp (Q) = = π v(t) = 3π v(t) =

1 O 2π u(t)Lp (Q) = O ω2 f ; , R p

t O u(t)Lp (Q) = O ω2 f ; R p

1 , O t 2 ω2 f ; R p 1 1 2 O 2+δ , π v(t) = O 3+δ , t t 1 O 4+δ , t

we can acquire that

∞

1

3π h(t) cos(t

∞

=O

− θ)dt p L (Q) 1

ω2 f ; R

p

t1+δ

1 . = O ω2 f ; R p

∞

dt +

1

ω2 f ; Rt p t3+δ

1

Consequently, we have

J1 p = O ω2

1 f; R

. p

For J2 , we have the same estimate. Finally, we conclude that ∞ 1 t ϕx O 3+δ dt J3 p = R t 1 p

∞ ϕx t 1 dt =O R p t3+δ 1

1 = O ω2 f ; . R p

dt

194

C3. Bochner-Riesz means of multiple Fourier series

Combining all of the above results together, Lemma 3.4.5 can be proved. We notice that α1 = α∞ = n−1 2 , and it is obvious that Lemma 3.4.5 is valid for p = 1 and p = ∞. It is to be noted that we have taken L∞ (Q) to represent for C(Q), and thus Theorem 3.4.5 is also valid for p = 1 and p = ∞. By the way, we would like to point out that the above approximation results in the case of p = ∞, that is, in the situation of C(Q), were ﬁrstly given by Cheng and Chen [CheC1](in the form of unitary ordered continuous norm). For the case of Lp with 1 ≤ p < ∞, Wang [Wa6] considered the uniform approximation problems with the order of α > n−3 2 . We have seen that Theorem 3.4.5 only ﬁgured out the order of Lp approximation above the critical order αp . As far as the case of being below or equaling to the order of αp , there is no result in that aspect.

3.5

Almost everywhere convergence (the critical index)

First of all, we will consider the problem of almost everywhere convern gence in the case at the critical index (α1 = n−1 2 ) in the context of L(Q ). Theorem 3.5.1 (Stein) If n > 1 and f ∈ L(log+ L)2 (Qn ), then we have n−1 S∗ 2 (f ) ≤ An |f |(log + |f |)2 dx + 1 . 1

Q

Proof. According to Theorem 3.4.2, we have n−1 S∗ 2 (g) ≤ An gp , (p − 1)2 p for 1 < p < 2. Let and

for k = 1, 2, 3, . . . .

, E0 = x ∈ Qn : |f (x)| < 1 , # $ Ek = x ∈ Qn : 2k−1 ≤ |f (x)| < 2k ,

(3.5.1)

3.5 Almost everywhere convergence (the critical index)

195

Set fk = χEk f , and make a periodical extension, then f (x) =

∞

fk (x).

(3.5.2)

k=0

We ﬁrst assume that f is a simple function, then the summation of (3.5.2) is actually ﬁnite. Therefore, ∞ n−1 n−1 2 S∗ 2 (f ) ≤ S (fk ), ∗ 1

k=0

1

k=0

∞ n−1 n−1 S∗ 2 (f ) ≤ S∗ 2 (fk ) .

(3.5.3)

1

We take g = fk , p = 1 + k1 , k = 1, 2, . . . , and then substitute them into (3.5.1) n−1 n−1 2 2 S∗ (fk ) ≤ An S∗ (fk ) 1

p

2

≤ An k fk p k

≤ An k2 2k |Ek | k+1 . By Young’s inequality, 1 1 1 1 1 + = 1, p = 1 + , q = k + 1 a · b ≤ ap + bq p q p q k k+1 k k 1 1 1 1 k k+1 | + , 4|E 4|Ek | k+1 · ≤ k 1 k+1 4 k+14 1+ k we can deduce that 2 n−1 1 2 S∗ (fk ) ≤ An · k 2 · 2k |Ek | + k , k + 1 2k 1 for k = 1, 2, . . . . At the same time, we have n−1 n−1 S∗ 2 (f0 ) ≤ An S∗ 2 (f0 ) ≤ An f0 2 ≤ An . 1

2

196

C3. Bochner-Riesz means of multiple Fourier series

Substituting all the above into (3.5.3), we obtain ∞ n−1 S∗ 2 (f ) ≤ An k2 2k |Ek | + 1 1 k=1 ∞ ≤ An (log |fk (x)|)2 |fk (x)|dx + 1 k=1

Q

|f (x)|(log+ |f (x)|)2 dx + 1 ,

= An

(3.5.4)

Q

which is valid for every simple functions. Starting from here, it is not diﬃcult to move onto the general situation. And that is the end of the proof.

Theorem 3.5.2 (Stein) If f ∈ L(log + L)2 (Qn ) and n > 1, then we have n−1

lim SR 2 (f ; x) = f (x),

(3.5.5)

R→∞

for a.e. x ∈ Qn . Proof. The proof method is quite common: with the help of Theorem 3.5.1, we can choose good functions as the approximation of f in L(Qn ), and then pass to the limit( ones may give a proof as an exercise). n−1

Theorem 3.2.2 in Section 2 asserts that: SR 2 (f ) may be divergent almost everywhere in L(Qn ). Yet, Theorem 3.5.2 ensures that if f ∈ n−1

L(log + L)2 (Qn ), then SR 2 (f ) must be convergent almost everywhere. It is well known that in the case of single variable, suﬃcient conditions that guarantee a.e.-convergence of the Fourier series are listed as follows: (i) Marcinkiewicz’s result: if

h 0

1 |f (x + t) − f (x)|dt = O h/ log h

(h → 0)

(3.5.6)

holds for every points in [0, 2π], then Sk (f ; x) → f (x) for a.e. x ∈ [0, 2π]; (ii) Salem’s result: if 0

h

1 {f (x + t) − f (x − t)}dt = o h/ log h

(h → 0)

(3.5.7)

3.5 Almost everywhere convergence (the critical index)

197

uniformly holds for x ∈ [0, 2π], then Sk (f ; x) → f (x) for a.e. x ∈ [0, 2π]. In 1965, Chang [Cha2] made an extension of Marcinkiewicz-type criterion to the case of multi-variables. He proved Theorem 3.5.3 Let f ∈ L log+ L(Qn ), n > 1. If f satisﬁes

1 −1 1 |f (x + y) − f (x)|dy = O log (h → 0), hn |y|
(3.5.8)

n−1

at every point of the positive measure set E ⊂ Qn , then SR 2 (f ; x) is a.e. convergent to f (x) on E. Limited by space, we omit the proof of this theorem here. In the following, we will formulate the Salem-type condition given by Lu [Lu2]. Lu deﬁned (f (y) − f (x))dy, (3.5.9) F (x, r) = B(x,r)

as the spherical integral of f (x) in [Lu2]. Theorem 3.5.4 Suppose that f ∈ L log+ L(Qn ) with n > 1 and x ∈ Qn is the Lebesgue point of f . At the point of x, if the condition $ h 1 # F (x, r + 2h) + F (x, r) − 2F (x, r + h) = o (3.5.10) rn−1 log 1/h is uniformly valid for r ∈ [h, r0 ], as h → 0+ , then we have n−1

lim SR 2 (f ; x) = f (x).

R→∞

It is easy to see that if we let F (x) =

x

f (t)dt 0

then (3.5.7) can be rewritten as F (x + h) + F (x − h) − 2F (x) = o as h → 0+ .

h log 1/h

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C3. Bochner-Riesz means of multiple Fourier series

In this sense, we can say that (3.5.10) is the multi-dimensional analogy of (3.5.7). Proof of Theorem 3.5.4. Since f ∈ L log+ L(Qn ), then with Stein’s theorem, that is, Theorem 3.2.3, we can transfer the problem of series into that of the integral. Suppose that the point x considered is the origin which is the Lebesgue point of f . Let g = f χQn . Then g ∈ L(Rn ), and x = 0 is the Lebesgue point of g. It is enough to prove n−1

BR 2 (g; 0) → g(0). Of course, we could premise f (0) = 0, that is, g(0) = 0. We denote the spherical integral of g at the origin as g(y)dy. G(r) = B(0,r)

Let ϕ(t) =

1 ωn−1

g(ξ)dσ(ξ) |ξ|=t

for t > 0. Then

r

ϕ(t)dt.

G(r) = ωn−1

0

By (3.5.10), with the notice of G(r + 2h) + G(r) − 2G(r + h) = ωn−1

0

h,

ϕ(r + h + t) − ϕ(r + h − t) dt,

we can get that h 1 {ϕ(r + h + t) − ϕ(r + h − t)}dt = o(h/ log 1/h) r n−1 0

(3.5.11)

uniformly holds for r ∈ [h, r0 ], as h → 0+ . Since x = 0 is the Lebesgue point of g, and by the suﬃcient condition n−1

(see Chapter 2) about the convergence of BR 2 (g) at Lebesgue points, it only suﬃces to prove that r0 cos(tR − θ) lim ϕ(t) dt = 0 (3.5.12) R→∞ π+θ tn R

3.5 Almost everywhere convergence (the critical index)

199

holds for θ = nπ 2 . 2π+|θ| Let R > r0 and m = [ r0 R−θ π ]. Then, we have

m∈

r0 R − θ r0 R − θ − 1, π π

and m > 2. Divide the integral of (3.5.12) into two parts as follows,

mπ+θ r0 r0 R cos(tR − θ) cos(tR − θ) ϕ(t) dt = + dt. ϕ(t) n π+θ mπ+θ π+θ t tn R

R

R

Therefore, it follows from the absolute continuity of the integration that r0 r0 cos(tR − θ) 2n ϕ(t) dt ≤ |ϕ(t)|dt = o(1). mπ+θ tn r0n r0 − π R

R

Hence, we merely need to prove IR :=

mπ+θ R π+θ R

ϕ(t)

cos(tR − θ) dt = o(1). tn

We have that IR =

m−1 (j+1)π

ϕ

jπ

j=1

=

t+θ R

m−1

j+1

(−1)

R

n−1

cos t dtRn−1 (t + θ)n

(j + 12 )π + θ + t sin t n dt. ϕ 1 R j + 2 )π + θ + t

π 2

− π2

j=1

Denote hj = j + 12 π + θ. It follows that IR =

m−1

j+1

(−1)

R

π 2

n+1 0

j=1

−n

(hj + t)

hj + t hj − t −n ϕ − (hj − t) ϕ R R

× sin tdt. Set

hj + t [(hj + t) − (hj − t) ]ϕ sin tdt, αj = R R 0 π 2 hj − t hj − t −n n−1 −n [(hj − t) − hj ] ϕ −ϕ sin tdt βj = R R R 0

n−1

π 2

−n

−n

200

C3. Bochner-Riesz means of multiple Fourier series

and

γj =

R hj

n

1 R

hj − t hj + t −ϕ sin tdt. ϕ R R

π 2

0

Then we have

m−1

IR =

(−1)j+1 (αj + βj + γj ).

j=1

Since Rn−1 |αj | ≤ Cn n+1 j we have m−1

π 2

0

m−2 j=1

(j+1)π+θ n R ϕ hj + t dt ≤ Cn R |ϕ(t)|dt, n+1 jπ+θ R j R

⎛

|αj | ≤ Cn ⎝

j=1

n

R j n+2

(j+1)π+θ R

0

|ϕ(t)|dt +

Rn (m − 1)n+1

mπ+θ R

0

⎞ |ϕ(t)|dt⎠ .

Noting that x = 0 is the Lebesgue point of g, and g(0) = 0, we have that r |ϕ(t)|dt = o(r n ), 0

as r → 0. It clearly follows that

m−1

|αj | = o(1)

j=1

as R → ∞. Similarly, we also get

m−1

|βj | = o(1),

j=1

as R → ∞. Therefore, we immediately have IR =

m−1

(−1)j+1 γj + o(1),

(3.5.13)

j=1

as R → ∞. Denote 1 Fj (s) = R

s

ϕ 0

hj + t R

−ϕ

hj − t R

dt

3.5 Almost everywhere convergence (the critical index)

201

for s ∈ 0, π2 . By the condition of (3.5.11), we can get that if s ∈ 0, π2 , then n−1 s/R hj − s hj − s R s s ϕ + +t −ϕ + −t dt hj − s R R R R 0 n−1 R Fj (s) = hj − s 1 s =o R log R/s holds uniformly about j = 1, 2, . . . , m − 1, as R → ∞. From the above it follows that for s ∈ 0, π2 , Fj (s) = o

j n−1 Rn log R

is valid uniformly about j ∈ {1, 2, . . . , m − 1}, as R → +∞. Therefore, by the method of integration by parts, we can deduce that π π n 2 2 R 1 Fj (t) cos tdt = o Fj (t) sin t − γj = hj j log R 0

0

uniformly about j ∈ {1, 2, . . . , m − 1}, as R → ∞. And thus we have m−1 j=1

j+1

(−1)

γj = o

1 j log R

m−1 j=1

1 = o(1), j

(3.5.14)

as R → ∞. We substitute the formula (3.5.14) into (3.5.13), and then obtain IR = o(1). This ﬁnishes the proof of Theorem 3.5.4. Of course, for a function belonging to L log+ L(Qn ) with n > 1, if the condition of (3.5.10) is valid at every Lebesgue point, then the Bochner-Riesz means at the critical index converges almost everywhere. Fan [Fa1] extended Theorem 3.5.4 to the case of classical groups. Now let us turn our attention to discuss the problems of always almost everywhere convergence at the critical index α2 = 0 on L2 (Qn ). Only a few results can be obtained in this aspect. Mathematicians have been in0 always almost vestigating for a long time about the problem whether SR

202

C3. Bochner-Riesz means of multiple Fourier series

everywhere converges on L2 (Qn ), n > 1, yet no any results were acquired. In recent years, the majority of researchers are prone to negative conclusions. In the following, we will introduce a suﬃcient condition by Golubov [Go1]. Firstly, let us introduce some notations: Let f ∈ L2 (Qn ), n > 1, and we denote ⎛ ⎞1 2 2 Aν = ⎝ |Cm (f )| ⎠ , |m|2 =ν

for ν = 0, 1, 2, . . .. If {m ∈ Zn : |m|2 = ν} = ∅, then we take it for granted that Aν = 0. Suppose that α is a non-negative and non-increasing function deﬁned on (0, 1], and we denote 1 α(t)dt, ω(ν) = √1 ν

for ν ∈ N. Theorem 3.5.5 Let f ∈ L(Qn ) with n > 1. If

δ

t α(t)dt = O δ 4 4

1

α(t)dt ,

(3.5.15)

α(t)fx (t) − f (x)22 dt < +∞,

(3.5.16)

0

δ

as δ → 0+ , then the two conditions 0

1

and

∞

A2ν ω(ν) < ∞

(3.5.17)

ν=1

are equivalent. Proof. Deﬁne fx (t) = n/2

1 ωn−1

Sn−1

f (x + tξ)dσ(ξ),

2π where ωn−1 = Γ(n/2) , and for ﬁxed t, as a function about x. Obviously we 2 n have fx (t) ∈ L (Q ).

3.5 Almost everywhere convergence (the critical index)

203

The Fourier coeﬃcients of fx (t) are 1 1 1 −im·x fx (t)e dx = f (x + tξ)e−im·x dx dσ(ξ) (2π)n Q ωn−1 Sn−1 (2π)n Q 1 Cm (f )eit·mξ dσ(ξ). = ωn−1 Sn−1 n 2

Let k =

− 1. By the formula

n

eitm·ξ dσ(ξ) = (2π) 2

Sn−1

J n2 −1 (t|m|) n

(t|m|) 2 −1

,

we can obtain that 1 Jk (t|m|) fx (t)e−im·x dx = 2k Γ(k + 1)Cm (f ) . n (2π) Q (t|m|)k

(3.5.18)

Hence,

σ(fy (t))(x) =

2k Γ(k + 1)Cm (f )

m

Jk (t|m|) im·x e . (t|m|)k

(3.5.19)

By the Parseval equation and the fact that Jk (u) k = 1, 2 Γ(k + 1) k u u=0 we conclude that fx (t) −

f (x)22

n

= (2π)

2 Jk (t|m|) |Cm (f )| 2 Γ(k + 1) − 1 dt, k (t|m|)

2 k

m=0

and 0

1

α(t)fx (t) − f (x)22 dt

= (2π)

n

∞ ν=1 |m|2 =ν

2

|Cm (f )|

0

1

2 k Jk (t|m|) dt α(t) 2 Γ(k + 1) − 1 (t|m|)k

2 √ 1 ∞ k J (t ν) k n 2 Aν α(t) 2 Γ(k + 1) √ k − 1 dt. = (2π) (t ν) ν=1

0

(3.5.20)

204

C3. Bochner-Riesz means of multiple Fourier series Assume that (3.5.16) holds. Since & Jk (t) =

2 π π 1 cos kt − k − +O 3 , πt 2 4 t2

as t → +∞, we know that there exists M ≥ 1, such that 2 √ k 1 ν) 2 Γ(k + 1) Jk (t √ k − 1 > , 2 (t ν) √ if t ν ≥ M . Then we have

A2ν

ν≥M 2

1 M √ ν

α(t)dt < +∞.

It follows that ω(ν) =

1 √1 ν

= ≤ ≤ ≤

α(t)dt M √ ν

√1 ν

+

1 M √ ν

1 M √ ν

α(t)dt

α(t)dt +

1 M √ ν

1 M √ ν

α(t)dt + ν

M √ ν √1 ν

2

√ ( νt)4 α(t)dt

M √ ν

0

α(t)dt + CM 4

t4 α(t)dt 1

M √ ν

α(t)dt.

Consequently, (3.5.17) holds. On the other hand, assume that (3.5.17) holds. By 2k Γ(k + 1)Jk (t) = 1 + O(t2 ), tk

3.5 Almost everywhere convergence (the critical index)

205

as t → 0+ , we can get that k 2 √ 2 Γ(k + 1)Jk (t ν) dt √ α(t) − 1 (t ν)k 0

1 2 √ √ 1 2k Γ(k + 1)Jk (t ν) ν √ k = + − 1 dt α(t) (t ν) √1 0 ν 1 √1 ν α(t)t4 ν 2 dt + C α(t)dt ≤ C

1

√1 ν

0

= O ω(ν) .

It hence follows from the above estimates and (3.5.20) that (3.5.16) holds. This completes the proof of Theorem 3.5.5.

We can deduce two suﬃcient conditions of almost everywhere convergence from Theorem 3.5.5.

Corollary 3.5.1 Let f ∈ L2 (Qn ) with n > 1. If

1

0

ω(f ; t)2

2 log 1 t

t

dt < ∞,

(3.5.21)

then we have 0 (f ; x) = f (x), lim SR

R→∞

(3.5.22)

for a.e. x ∈ Qn . Proof. By fx (t) −

f (x)22

2 1 f (x + tξ) − f (x) dσ(ξ) dx = Qn ωn−1 Sn−1 2 ≤ C ω(f ; t)2 ,

and the inequality (3.5.21), we can easily obtain

1 0

1 1 fx (t) − f (x)22 log dt < ∞. t t

(3.5.23)

206

C3. Bochner-Riesz means of multiple Fourier series

Denote

ω(ν) =

1 √1 ν

1 1 1 log dt = log2 ν, t t 8

for ν ∈ N. We obviously have that 1 δ 1 1 1 3 4 t log dt = O δ log dt . t t 0 δ t Let α(t) = 1t log 1t , for t ∈ (0, 1]. Clearly the function α satisﬁes the condition (3.5.15). And thus from Theorem 3.5.5, we can deduce that ∞

A2ν log2 ν < ∞.

(3.5.24)

ν=1

Let ϕν (x) =

1 Cm (f )eim·x . Aν 2 |m| =ν

By (3.5.24) and the Menshoﬀ-Rademacher theorem in theory of orthogonal /∞ series (see [KS1]), it implies that ν=0 Aν ϕν (x) is convergent for a.e. x ∈ Qn . That is to say, (3.5.22) is valid. This ﬁnishes the proof of Corollary 3.5.1.

Corollary 3.5.2 Let f ∈ L2 (Qn ) (n > 1). If there exists ε > 0, such that 1 (3.5.25) ω(f ; δ)2 = O (log 1/δ)1+ε as δ → 0+ , then (3.5.22) holds. Obviously, (3.5.25) implies (3.5.21). Thus Corollary 3.5.2 is a special case of Corollary 3.5.1. Besides, Lu [Lu8] also obtained a result on the a.e.-convergence of lacunary sequence of the spherical Fourier partial sum for L2 . Theorem 3.5.6 Let f ∈ L2 (Qn ) with n > 1. If R1 > 0, and for all j ∈ N, then 0 (f ; x) = f (x) lim SR j j→∞

holds for a.e. x ∈ Qn .

Rj+1 Rj

≥ q > 1, (3.5.26)

3.5 Almost everywhere convergence (the critical index) Proof. Let

207

0 S∗ (f )(x) = sup SR (f ; x) . j j∈N

Next it suﬃces to prove that S∗ is of type (2, 2). We have that :

n B 2 |m| 0 n SR (f ; x) = SR (f ; x) + 1− 1− 2 Cm (f )eim·x , j j Rj |m|
and 2 ⎫ ⎧ :

n B ⎬ ∞ ⎨ 2 |m| 2 2 n im·x . [S∗ (f )(x)] ≤ 2 S∗ (f )(x) + 1− 1− 2 Cm (f )e ⎭ ⎩ R j j=1 |m|
It follows that

n 2 ∞ 2 |m| S∗ (f )22 ≤ 2S∗n (f )22 + 2 (2π)n 1 − 1 − 2 |Cm (f )|2 R j j=1 |m|
j:Rj >|m|

By the property of {Rj }, we have j:Rj >|m|

1 =O Rj4

1 |m|4

for m = 0. Therefore, we have S∗ (f )2 ≤ Cf 2 . This ﬁnishes the proof of Theorem 3.5.6.

Among the results about the a.e. convergence, Stein’s theorem, i.e., Theorem 3.5.2, is the most beautiful one, which was published in 1958. Since then, many scholars were searching for the possibility of ﬁnding the subclass

208

C3. Bochner-Riesz means of multiple Fourier series

of L(Qn ), such that it is no smaller than L(log+ L)2 which guarantees the a.e. convergence of the Bochner-Riesz means at the critical index. It did not get any great development until the proposal of concepts of entropy put forward by Feﬀerman [Fef1] in 1978 and that of Block spaces by Taibleson and Weiss [TW1] later on. Entropy space and the space generated by blocks both includes the Dini function class. To some extent, we can claim that, to the knowledge of all that we know at present, the block space is the biggest function space where the Bochner-Riesz means of the Fourier series at the critical index is a.e. convergent. In the next section, we will give a brief introduction about all those concepts mentioned above and some main results.

3.6

Spaces related to the a.e. convergence of the Fourier series

3.6.1 The concept of the entropy class Let S is a measurable subset of Qn . Deﬁne the entropy of S as |Qn |e |Ik | log E(S) = inf : Ik ⊃ S , |Ik | k

where Ik represents for the cube in Qn , and {Ik } is a sequence of the cubes. By the deﬁnition above, the entropy of a set is not only related with the measure of the set, but also with its geometric shape. Feﬀerman [Fef1] put forward with such a kind of sequence of sets on one-dimensional Euclidean space, whose measure is zero, and yet entropy tends to 1. Deﬁne the entropy of the integrable function f on Qn as follows: ∞ E({x ∈ Qn : |f (x)| < λ})dλ. J(f ) = 0

We have to point out that the entropies of two a.e. equal functions are not necessarily the same. We deﬁne a function class denoted by J as follows. f ∈ J ⇐⇒ ∃ g, g = f, a.e., and J(g) < +∞. If n = 1, it is easy to prove J(f + g) ≤ J(f ) + J(g).

3.6 Spaces related to the a.e. convergence of the Fourier series

209

However if n > 1, E(Q1

*

Q2 ) + E(Q1

Q2 ) ≤ E(Q1 ) + E(Q2 )

is not necessarily valid, and then the Minkowski inequality no longer holds. But if we restrict the ‘inf’ in the deﬁnition of the entropy of a set is only taken over all covers of binary cubes Ik , then the entropy of a function is a norm and this deﬁnition is equivalent to the original one. In order to illustrate the size of J, let us formulate the concept of the Dini class denoted by D. Let f ∈ L(Qn ). Deﬁne |f (x) − f (y)| dxdy. f D = f L(Qn ) + |x − y|n x,y∈Qn f ∈ D ⇐⇒ f D < +∞. Feﬀerman [Fef1] proved the following result for the case of n = 1: Lemma 3.6.1 J(f ) ≤ Cf D , and thus we have D ⊂ J. Since the proof is tedious and troublesome, we do not quote the proof here. Lemma 3.6.2 J is a subset of L log L, that is, J ⊂ L log L. Proof. Let f ∈ J. Without loss of generality, we assume that f is nonnegative. Let Ek = {x ∈ Qn : f (x) > 2k } for k ∈ Z. We conclude that

∞

J(f ) = 0

=

k∈Z

≥

E({x ∈ Qn : f (x) > α})dα

2k−1

E({x ∈ Qn : f (x) > α})dα

2k−1 E(Ek )

k∈Z

≥

2k

1 k |Qn |e . 2 |Ek | log 2 |Ek | k∈Z

210

C3. Bochner-Riesz means of multiple Fourier series

Here we use the property of the entropy of a set E(S) ≥ |S| log

|Qn |e . |S|

We write # $ Fk = Ek − Ek+1 = x ∈ Qn : 2k < f (x) ≤ 2k+1 . Then we can get

∞

J(f ) ≥

1 k+1 |Qn |e |Fk |. 2 log 4 |Ek | k=0

By |Ek | ≤ 2−k f 1 , we have ∞ 1 k+1 |Qn |e k+1 log 2 |Fk | J(f ) ≥ 2 + log 4 2f 1 k=0 |Qn |e 1 + f (x) log f (x)dx + log f (x)dx . ≥ 4 2f 1 E0 Qn This implies f ∈ L log+ L. Consequently, we have the inclusion relationship, D ⊂ J ⊂ L log+ L.

Feﬀerman [Fef1] gave a conjecture as follows. “It seems that the entropy has close relation with the pointwise convergence of the Fourier series.” Taibleson and Weiss [TW1] and Lu-Taibleson-Weiss [LTW1] veriﬁed the conjecture for the casees of unitary and multiple variables, respectively. Theorem 3.6.1 If f ∈ J, then n−1

SR 2 (f )(x) = f (x) holds for a.e. x ∈ Qn with n ∈ N. Since Theorem 3.6.1 is included in Theorem 3.6.5 and Theorem 3.6.6 in the following, we do not give the proof here.

3.6 Spaces related to the a.e. convergence of the Fourier series

211

3.6.2 The concept of the block The block becomes more and more important, since it can solve the n−1

problem of the a.e. convergence of SR 2 well and the block space is bigger than the entropy space. Definition 3.6.1 Let 1 < q ≤ ∞. We say that measurable function b is a q-block, if there exists a block Q, where Q is a cube whose edge is parallel to the axis and Q ⊂ Qn , such that (i)

supp b ⊂ Q,

(ii)

−1 bLq ≤ |Q| q .

1

Definition 3.6.2 If f ∈ L(Qn ) can be expressed as f (x) =

∞

ck bk (x),

(3.6.1)

k=1

where bk is a q-block, and the group of coeﬃcients {ck } := c satisﬁes / ∞ j |cj | N (c) := |ck | 1 + log < +∞, (3.6.2) |ck | k=1

then we say f ∈ Bq . We deﬁne the number as follows, # $ such that (3.6.1) holds . Nq (f ) = inf N (c) : c = {ck }∞ k=1 By (3.6.2) and the fact that bk is a q-block, we say that the series of the right side of (3.6.1) must be absolutely convergent in the norm of L1 . In addition, by the condition of (ii) about the size of q-block, we have f 1 ≤

∞ k=1

|ck |bq |Q|

1− 1q

≤

∞

|ck | ≤ N (c).

(3.6.3)

k=1

Besides, it is obvious that Bq is a linear space. Proposition 3.6.1 If f ∈ Bq , and g ∈ Bq , then we have Nq (f + g) ≤ (1 + log 2) Nq (f ) + Nq (g) .

(3.6.4)

212

C3. Bochner-Riesz means of multiple Fourier series

Proof. Let ϕ(x) = 1 − x log x − (1 − x) log(1 − x), for 0 < x < 1, and ϕ(0) = ϕ(1) = 1. We easily have ϕ (x) = log

1−x . x

Obviously ϕ has the maximum at x = 12 and 1 = 1 + log 2. ϕ 2 For every ε > 0, choose f= and g=

m k bk

nj cj

such that N (m) < Nq (f ) + ε and Let A =

/

N (n) < Nq (g) + ε. / A |nj | and x = A+B . We thus have that |mk |, B =

A+B A+B + |nk | log + (A + B) |mk | |nk | A 1 = (A + B) + |mk | log + log x |mk | B 1 + log + |nk | log 1−x |nk | A B = (A + B)ϕ(x) + |mk | log + |nk | log |mk | |nk | A B ≤ (1 + log 2)(A + B) + |mk | log + |nk | log |mk | |nk | ≤ (1 + log 2) N (m) + N (n) ≤ (1 + log 2) Nq (f ) + Nq (g) + 2ε .

N q(f + g) ≤

|mk | log

Consequently, we immediately get (3.6.4). This completes the proof of Proposition 3.6.1.

3.6 Spaces related to the a.e. convergence of the Fourier series

213

Besides, it is obvious that Nq has absolute homogeneity. That is to say, Nq (af ) = |a|Nq (f )

(3.6.5)

for a ∈ R and f ∈Bq . According to (3.6.4), (3.6.5) and the obvious property, Nq (f ) = 0 ⇐⇒ f = 0, we call Nq as the quasi-norm, inherited from Taibleson and Weiss [TW1]. Its diﬀerence from the norm is that the inequality (3.6.4) holds instead of Minkowski’s inequality. Proposition 3.6.2 Bq forms a complete space if it is normed with the quasi-norm Nq . Proof. Assume that {fk }∞ k=0 is a Cauchy sequence in Bq in the norm of quasi-norm Nq . Then we can take a subsequence {gk }∞ k=1 from the sequence, such that (3.6.6) Nq (gk+1 − gk ) < 2−k , for k = 1, 2, . . .. It follows from the inequality (3.6.3) that {fk } is also a Cauchy sequence in L(Qn ). Then by the completeness of L1 , there exists f ∈ L(Qn ), such that ∞ (gk+1 − gk ) (3.6.7) f = g1 + k=1

L1

in the sense of norm. Take some proper decomposition of gk+1 − gk in block gk+1 − gk =

(k) (k)

c j bj ,

such that N (c(k) ) < 2−k . Assume that g1 has the following decomposition cj bj . g1 = Then we have f=

cj bj +

∞ ∞ k=1 j=1

(k) (k)

cj bj .

(3.6.8)

214

C3. Bochner-Riesz means of multiple Fourier series

Therefore, we have f ∈ Bq . Both (3.6.7) and (3.6.8) implies that ⎫ ⎧ ∞ ⎬ ⎨ (gj+1 − gj ) −→ 0, Nq (f − gk ) = Nq ⎭ ⎩ j=k+1

as k → ∞, which leads to gk → f and fk → f in Bq , as k → ∞. This ﬁnishes the proof of Proposition 3.6.2.

Proposition 3.6.3 Let f ∈ Bq . If g is a measurable function deﬁned on Qn , and |g| ≤ |f |, then we have g ∈ Bq and Nq (g) ≤ Nq (f ). Proof. Since |g| ≤ |f |, there exists a measurable function r(x), |r| ≤ 1, such that g = rf . Suppose that f has a decomposition as f= m j bj . By the fact that rbj is also a block, then g= mj (rbj ) is a decomposition of g. And therefore g ∈ Bq and Nq (g) ≤ Nq (f ). From the above propositions, we can easily deduce that f ∈ Bq implies |f | ∈ Bq and f, g ∈ Bq implies sup{f, g} ∈ Bq and inf{f, g} ∈ Bq . In Deﬁnition 3.6.2, under the condition of (3.6.2), the series (3.6.1) is not only convergent in the norm of L1 , but also a.e. convergent, which is contained in the following iteration principle. The principle of iteration is constructed by Stein and Weiss [SWe1], which is the foundation of the application of the whole block theory. Lemma 3.6.3 If {fk } is a sequence of measurable functions, such that |{x : |fk (x)| > λ}| ≤ λ−1

(3.6.9)

for every λ > 0, and m = {mk }∞ k=1 is a sequence of numbers, then ∞ mk fk (x) > λ ≤ 4N (m)λ−1 (3.6.10) x: k=1

holds for every λ > 0.

3.6 Spaces related to the a.e. convergence of the Fourier series

215

Proof. Without loss of generality, we / might as well think that mk > 0 and fk (x) ≥ 0, for all k. Denote M = mk . Let λ > 0 and deﬁne ⎧ λ ⎪ ⎨ fk (x) if fk (x) < , 2M lk (x) = λ ⎪ ⎩0 , if fk (x) ≥ 2M ⎧ λ ⎪ ⎪ if fk (x) > , ⎨ fk (x) 2mk uk (x) = λ ⎪ ⎪ ⎩0 , if fk (x) ≤ 2mk and vk (x) = fk (x) − lk (x) − uk (x) ⎧ λ λ ⎪ ⎪ ≤ fk (x) ≤ if , ⎨ fk (x) 2M 2mk = λ λ ⎪ ⎪ ⎩0 / , if fk (x) ∈ . 2M 2mk In addition, we also denote l(x) = u(x) = and v(x) =

mk lk (x), mk uk (x), mk vk (x).

We have that {x : u(x) > 0} =

{x : uk (x) > 0} =

k

and |{x : u(x) > 0}| ≤

k

x : fk (x) > λ 2mk k

Since

we clearly have

x : fk (x) >

λ x : l(x) > 2

x : l(x) > λ = 0. 2

,

2mk 2M ≤ = . λ λ

λ ⊂ x : lk (x) > 2M k

λ 2mk

k

= ∅,

216

C3. Bochner-Riesz means of multiple Fourier series

Let μk (y) = |{x : fk (x) > y}| for y > 0. We conclude that v(x)dx = mk vk (x)dx =

= ≤

mk

λ 2mk λ 2M

mk mk

−ydμk (y)

λ 2mk λ 2M

λ 2mk μk (y)dy − yμk (y)

M log +1 mk

λ 2M

= N (m). Therefore it follows that x : v(x) > λ 2

2 2 ≤ v(x)dx ≤ N (m). λ λ λ {x:v(x)> } 2

Since # $ x: mk fk (x) > λ = {x : l(x) + u(x) + v(x) > λ} λ ⊂ {x : u(x) > 0} x : v(x) > 2

λ x : l(x) > 2

we have # $ λ mk fk (x) > λ ≤ |{x : u(x) > 0}| + x : v(x) > x: 2 2M 2 ≤ + N (m) λ λ ≤ 4N (m)λ−1 . This completes the proof.

3.6.3 The structure of the block space In this section, we will study the relation between Bq and Lq , the inclusion relations between Bq corresponding to diﬀerent q ∈ (1, ∞) and the properties of Bq as a complete linear distance space in the norm of Nq . The conjugate space of Bq is a collection of continuous linear functional deﬁned on Bq , denoted by Bq . The conclusion is as follows.

,

3.6 Spaces related to the a.e. convergence of the Fourier series

217

Theorem 3.6.2 If 1 ≤ p < q ≤ ∞, then we have (i) Lq ⊂ Bq , / Bq ; that is, we have Bp Bq and (ii) there exists f ∈ Bp , but f ∈ B1 = L 1 .

Theorem 3.6.3 Let 1 < q < ∞. Then Bq = L∞ (Qn ). ∗ By Theorem 3.6.3 and L1 = L∞ , we can see that Bq is indeed very close to L1 . Lemma 3.6.4 Suppose that there exists two sequences of numbers: m = {mk } and l = {lk }, satisfying |mk | ≤ |lk |, for all k. Then N (m) ≤ N (l) holds. Proof. We might as well take it for granted that 0 < mk ≤ lk . In fact, it suﬃces to show that, for every ε ≥ 0, for the sequence of numbers m(ε) = {m1 + ε, m2 , m3 , . . .}, we have N (m(ε)) ≥ N (m). Denote ϕ(ε) = N (m(ε)). We merely need to show that ϕ(ε) is monotonically increasing. Denote ∞ mj . A(ε) = ε + j=1

We have

A(ε) ϕ(ε) = (m1 + ε) 1 + log m1 + ε

+

∞

mk

k=2

A(ε) 1 + log mk

.

Therefore, we conclude that A(ε) ϕ (ε) = 1 + log + (m1 + ε) m1 + ε

1 1 − A(ε) m1 + ε

A(ε) = 1 + log > 0, m1 + ε for any ε ≥ 0. Consequently, the proof of the lemma is ﬁnished.

+

∞ k=2

m1

1 A(ε)

218

C3. Bochner-Riesz means of multiple Fourier series

Lemma 3.6.5 Suppose that ν ∈ N, 1 < q ≤ ∞ and 1 N = ν(1 + log ν)q , δ = ν−1 1 − N1 , and the intervals 1 λ, Ii = (i − 1)δ, (i − 1)δ + N

1 q

+

1 q

for i = 1, 2, . . . , ν and 0 < λ < 1. If the function bi (x) = / deﬁned on Q1 , and f = νi=1 bi , then we have

= 1. Let

1 χ (x) |Ii | Ii

is

Nq (f ) = ν(1 + log ν). Proof. Every bi is a q-block, and 1 −1 bi q = |Ii |−q+1 q = |Ii | q , so we have Nq (f ) ≤ ν(1 + log ν). Now / we assume that there exists another decomposition in blocks as f = mk ck , where ck is a q-block. It suﬃces to show that ν(1 + log ν) ≤ N (m). − q1

Let Jk is a interval of Q1 such that suppck ⊂ Jk and ck q ≤ |Jk | − 1q

If ck q < |Jk | such that

ck q

, of course ck q > 0, then we can expand ck into new − q1

= |Jk |

.

ck

, at the same time, we shrink mk into mk . Thus in 1

the new decomposition of f , it always holds ck q = |Jk |− q . Yet by Lemma 3.6.4, N (m ) ≤ N (m), then our proof can switch to prove ν(1 + log ν) ≤ −

1

N (m ). For this reason, we can assume ck q = |Jk | q , for all k ∈ N. If every Jk can be contained in some Ii , then the proof is end at once. In this case, we can write bi =

k

mik bik , f =

ν i=1

mik bik ,

k

where each bik is a q-block whose support is contained in Jki ⊂ Ii . It follows bi (x)dx = 1 ≤ |mik |, Q1

k

1i }, such that for i = 1, 2, . . . , ν. Thus, we can choose {m k 1i i 0 ≤ m k ≤ mk

(3.6.11)

3.6 Spaces related to the a.e. convergence of the Fourier series and

219

|mik | = 1,

k

for i = 1, 2, . . . , ν. Using Lemma 3.6.4, we have ⎛ ⎞

# $ , - ν 1i ⎝ 1i = N m mk 1 + log ⎠ ≤ N mik = N (m). k 1i m i,k k 1i Since m k ≤ 1 and ν log ≥ log ν, 1i m k

we have N

# $ 1i 1i ≥ m mk (1 + log ν) = ν(1 + log ν), k i

k

which leads to N (m) ≥ ν(1 + log ν). / Hence, it remains to rewrite the decomposition f = k mk ck as f = / k mk ck , such that the interval Jk corresponding to the support of ck is contained in some Ii while N (m ) ≤ N (m) still holds. The rewriting step is carried out in the method of separating mk ck into m1k c1k + · · · + mνk cνk one by −

1

one, where the support of cik is contained in Jki ⊂ Ii and cik q = |Jki | q still holds. And it only suﬃces to show that such a decomposition is applicable for m1 c1 . 2 Ii | > 0 for i = Suppose that the interval J1 related to c1 satisﬁes |J 1 2 μ, . . . , μ + l, 1 ≤ μ < μ + 1 ≤ μ + l ≤ ν, and |J1 Ii | = 0 for i < μ or i > μ + l. 2 Denote Lj = J1 Ij , j = μ, . . . , μ + l. Set 1

μj = |Lj | q b1 χLj q j j and b1 χLj = bj1 for j = μ, . . . , μ + l, and aj = μ−1 j b1 . Since b1 q > 0, we have μj > 0. Obviously aj is a q-block. By the deﬁnition of Ii , we have μ+l 1 |J1 | = λ |Li | + l δ − N i=μ

=

μ+l i=μ

|Li | + l

" λ ν ! (1 + log ν)q − 1 . ν−1 N

220

C3. Bochner-Riesz means of multiple Fourier series

Since θ(t) =

t (1 + log t)q − 1 t−1

is monotonically increasing for t > 1 and 1 ≤ q < ∞, we have l

" " l+1! ν ! (1 + log ν)q − 1 ≥ l (1 + log(l + 1))q − 1 , ν−1 l

and |J1 | ≥

μ+l i=μ

! " λ |Li | + (l + 1) (1 + log(l + 1))q − 1 N

λ λ + |Lμ | + |Lμ+l | − 2 . N N It follows from the H¨older’s inequality that = (l + 1)[1 + log(l + 1)]q

μ+l j=μ

μj =

μ+l

1

|Lj | q bi1 q

j=μ

⎞1 ⎛ ⎞ 1 ⎛ q q μ+l μ+l j q ≤ ⎝ |Lj |⎠ ⎝ b1 q ⎠ j=μ

j=μ

⎛ ⎞ μ+l = ⎝ |Lj |⎠

1 q

b1 q

j=μ

⎛ ⎞ 1 q μ+l −1 ⎝ ⎠ = |Lj | |J1 | q j=μ

⎛ μ+l ⎝ ≤

|Lj | (l + 1)(1 + log(l + 1))q Nλ + |Lμ | + |Lμ+l | − j=μ

=

1−α [1 + log(l + 1)]q − α

where α= Since

1 q

,

N 2λ − |Lμ ||Lμ+l | ≥ 0. N (l + 1)λ

1 1−α , ≤ q [1 + log(l + 1)] − α [1 + log(l + 1)]q

⎞ 1 q

2λ N

⎠

3.6 Spaces related to the a.e. convergence of the Fourier series we have

μ+l

μj ≤

j=μ

1 . 1 + log(l + 1)

We know f=

∞

mk ck +

⎛ m 1 · μ j a j = m1 ⎝

j=μ

k=2

where mk = Denote

μ+l

221

∞

(3.6.12)

mk ck +

μ+l

⎞ μj aj ⎠ ,

j=μ

k=2

mk m1 .

A=

∞

|mk |

k=2

and B=

μ+l

μj .

j=μ

It suﬃces to show that ∞

μ+l

A+B + μj 1 + log μj j=μ k=2 ∞ 1+A ≤ 1 + log(1 + A) + |mk | 1 + log |mk | |mk |

A+B 1 + log |mk |

k=2

= 1 + log(1 + A) + A[1 + log(1 + A)] −

∞

|mk | log |mk |.

k=2

That is, (1 + A)[1 + log(1 + A)] ≥ (A + B)[log(A + B) + 1] −

μ+l

μj log μj . (3.6.13)

j=μ

Since 0

A

2 + log(1 + x) dx = (1 + A)(1 + log(1 + A)) − 1 A ≥ 2 + log(B + x) dx 0 = (B + A) log(B + A) + 1 − B(log B + 1),

222

C3. Bochner-Riesz means of multiple Fourier series

where we use B ≤

1 1+log(l+1)

< 1 in (3.6.12). We merely need to prove

1 − B(log B + 1) ≥ −

μ+l

μj log μj .

(3.6.14)

j=μ

/ μ Denote ϕj = Bj . Since μ+l j=μ ϕj = 1, by the convexity of the exponential function, we can get ⎛ ⎞ μ+l μ+l 1⎠ 1 exp ⎝ = l + 1. ≤ ϕj log ϕj exp log ϕj ϕj j=μ

j=μ

Therefore, we have μ+l

μj log

j=μ

and −

μ+l

B ≤ B log(l + 1) μj

μj log μj ≤ B log(l + 1) − B log B.

j=μ

We substitute the above results into (3.6.12) to obtain B≤

1 . 1 + log(l + 1)

This is equivalent to B log(l + 1) ≤ 1 − B. We can immediately obtain (3.6.14). This completes the proof of Lemma 3.6.5. The function in Lemma 3.6.5 is f=

ν 1 χI , |Ii | i i=1

where

λ = λν −1 (1 + log ν)−q . N 5 Denote S = νi=1 Ii . We obviously have S ⊂ [0, λ]. Set |Ii | =

χS = λν −1 (1 + log ν)−q f.

3.6 Spaces related to the a.e. convergence of the Fourier series It follows and

|S| = λ(1 + log ν)−q ,

223

(3.6.15)

Nq (χS ) = λ(1 + log ν)1−q .

(3.6.16)

According to the property of translation invariance of the measure and the block, we can have from (3.6.15), (3.6.16) that Lemma 3.6.5 implies Lemma 3.6.6. Lemma 3.6.6 In any interval I with the length of λ for 0 < λ < 1 in Q1 , there exists a subset S such that |S| = |I|(1 + log ν)−q and

(3.6.17)

Nq (χS ) = |I|(1 + log ν)1−q ,

(3.6.18)

where ν is a previously given arbitrary natural number with ν > 1, and q satisﬁes 1 ≤ q < ∞ with 1q + q1 = 1. Proof of Theorem 3.6.2. Firstly, the statement (i) is evident, because each function in Lq (Qn ) is the multiple of a q-block. For the statement (ii), we can only give a proof in the one-dimensional case with the help of Lemma 3.6.6. It is rather more troublesome for the case n > 1. We consider the case of q = ∞ at ﬁrst. In this case, q = 1. Choose a series of intervals Ik in [0, 1] which is not intersected with each other, such that |Ik | = 2−k , for k ∈ N. By Lemma 3.6.6, we can select the set Sk in Ik , such that |Sk | = |Ik |(1 + log nk )−1 , 22k

n k ≥ ee and

N∞ (χSk ) = |Ik | = 2−k .

Let f=

∞

22k χSk .

k=1

Proposition 3.6.3 implies that N∞ (f ) ≥ 22k N∞ (χSk ) = 2k → ∞.

224

C3. Bochner-Riesz means of multiple Fourier series

This means f ∈ / B∞ . However, we have

f (x)

Q1

e

dx =

∞ k=1

2k

e2

=

Sk

∞

2k

e2 2−k (1 + log nk )−1 ≤

k=1

∞

2−k = 1.

k=1

This implies f ∈ Lp (Q1 ) for 1 ≤ p < ∞. Now we suppose that 1 ≤ p < q < ∞. Choose β> and

ν∈

and take

q−1 q−p

pβ − 1 β − 1 , q q −1

! γk " nk = e2 + 1.

Let the intervals Ik ⊂ [0, 1] with |Ik | = 2−k , and those Ik do not intersect with each other. By Lemma 3.6.6, we can take Sk ⊂ Ik , such that |Sk | = |Ik |(1 + log nk )−q and

Let f =

/

Nq (χSk ) = |Ik |(1 + log nk )1−q . 2βk χSk . We have

f pp =

2(pβ−1)k (1 + log nk )−q ≤

2(pβ−1)k−γq k < +∞,

and Nq (f ) ≥ 2βk Nq (χSk ) ≥ 2βk 2−k (2 log nk )1−q

≥ 2(β−1)k 2(γk+2)(1−q )

= 41−q 2[(β−1)−(q −1)γ]k → +∞, as k → ∞. This shows that f ∈ Lp , but f ∈ / Bq . In addition, since any q-block must be p-block, then we have Bp ⊃ Bq .

3.6 Spaces related to the a.e. convergence of the Fourier series

225

The statement (ii) in Theorem 3.6.2 is proven for n = 1. One can give the proof in the case of n > 1 for practise. Before proving Theorem 3.6.3, we would like to clarify three facts as follows. Fact I. It is easy to check directly that the continuity of the linear functional l on Bq about the topology generated by the quasi-norm Nq is equivalent to its boundedness. That is to say, the continuity of l is equivalent to the fact that there exists K ≥ 0, such that |l(f )| ≤ KNq (f ), for f ∈ Bq . We denote the minimum of K as l. With this norm, Bq becomes a Banach space. Fact II. We have mentioned that f 1 ≤ Nq (f ), for f ∈ Bq . Fact III. Let f ∈ Lq . We have

1 −1 f b := f q |Qn | q is a q-block, and thus

1

Nq (f ) ≤ |Qn | q f q . Proof of Theorem 3.6.3. Let l ∈ Bq . Restricted on Lq , if {fk } ⊂ Lq and fk → 0 in Lq . Fact III implies that fk → 0 in B q . Thus we have l(fk ) −→ 0. This shows that l restricted on Lq is a bounded linear functional on Lq . Since 1 < q < ∞, and (Lq )∗ = Lq , there exists an unique g ∈ Lq , such that l(f ) = gf, (3.6.19) for all f ∈ Lq . Let N > l and E = {x ∈ Qn : |g(x)| > N }. If |E| > 0, since almost every point of E is its Lebesgue point, then there exists a cube Q ⊂ Qn , such that l |Q ∩ E| < ≤ 1. N |Q|

226

C3. Bochner-Riesz means of multiple Fourier series

Let b = |Q|−1 sgngχQ E . b is a q-block. Therefore, we have l(b) =

1 |Q|

Q

|g| ≥ N E

2 |Q E| > l, |Q|

which contradicts with the following inequality |l(b)| ≤ lNq (b) ≤ l. This implies |E| = 0. That is to say, g ∈ L∞ and g∞ ≤ l.

(3.6.20)

Since g ∈ L∞ , we can deﬁne h(f ) =

gf

for f ∈ Bq . Obviously, we have h ∈ Bq . However, by (3.6.19), we know that h = l on Lq . Since Lq is dense in Bq , we have that h, as an element of Bq , equals to l. That is to say, (3.6.19) is actually valid on Bq . It follows from (3.6.19) that |l(f )| ≤ g∞ f 1 , for f ∈ Bq . Fact II implies f 1 ≤ Nq (f ). It follows |l(f )| ≤ g∞ Nq (f ), for f ∈ Bq , which yields that l ≤ g∞ .

(3.6.21)

Combining (3.6.20) with (3.6.21), we immediately have l = g∞ . We can have that Bq can be embedded into L∞ with the same norm. This completes the proof of Theorem 3.6.3. Remark 3.6.1 Theorem 3.6.9 is due to Meyer, Taibleson and Weiss [MTW1] or [LTW2].

3.6 Spaces related to the a.e. convergence of the Fourier series

227

3.6.4 B∞ ⊃J Theorem 3.6.4 If J(f ) < ∞, then we have N∞ (f ) < ∞. Proof. Let

# $ Sk = x ∈ Qn : 2k−1 < |f (x)| ≤ 2k ,

We have

J(f ) = =

∞

0

E({x ∈ Qn : |f (x)| > λ})dλ

∞ k=−∞

Since

2k

2k−1

for k ∈ Z.

2k 2k−1

E({x ∈ Qn : |f (x)| > λ})dλ.

(3.6.22)

E({|f | > λ})dλ ≥ (2k − 2k−1 )E({|f | > 2k }) ≥ 2k−1 E(Sk+1 ) 1 = 2k+1 E(Sk+1 ), 4

we have

∞ 1 k 2 E(Sk ). J(f ) ≥ 4

(3.6.23)

k=−∞

For any ﬁxed k > 0, we choose a sequence of cubes {Ik,l }∞ l=1 such that each cube 5 has the positive measure and does not intersect with each other, Sk ⊂ l Ik,l and E(Sk ) ≤

l

Let

|Ik,l | log

|Qn |e 1 ≤ E(Sk ) + k . |Ik,l | 4

(3.6.24)

−1 bk,l (x) = f (x)χSk Ik,l (x) 2k |Ik,l | ,

for k ∈ N and l ∈ N. We thus have bk,l is a ∞-block whose support is contained in Ik,l . Denote mk,l = 2k |Ik,l | and ⎧ ⎨ 1 f (x), if |f (x)| ≤ 1, b0 (x) = |Qn | ⎩ 0, if |f (x)| > 1.

228

C3. Bochner-Riesz means of multiple Fourier series

We have that b0 is a ∞-block whose support is contained in Qn . We can get a decomposition of blocks as f (x) =

∞ ∞

mk,l bk,l (x) + |Qn |b0 (x).

(3.6.25)

k=1 l=1

By (3.6.24), we have that A := ≤

∞ ∞

mk,l k=1 l=1 ∞ ∞ k k=1

≤

|Ik,l | log

2

∞

l=1

|Qn |e |Ik,l |

2k E(Sk ) + 1.

k=1

By (3.6.23), we have A ≤ 4J(f ) + 1.

(3.6.26)

It follows that ∞ ∞ A + |Qn | A + |Qn | n N∞ (f ) ≤ mk,l log + 1 + |Q | 1 + log mk,l |Qn | k=1 l=1 ∞ ∞ A + |Qn | A + |Qn | + = A + |Qn | 1 + log m log . k,l |Qn | mk,l k=1 l=1 (3.6.27) Notice |Qn |e A + |Qn | A + |Qn | = log + log − k log 2 log mk,l |Ik,l | |Qn |e A + |Qn | |Qn |e + log , ≤ log |Ik,l | |Qn |e for k ∈ N. We can have that ∞ ∞ ∞ ∞ A + |Qn | |Qn |e A + |Qn | k mk,l log ≤ 2 |Ik,l | log + log A mk,l |Ik,l | |Qn |e k=1 l=1

k=1

l=1

A + |Qn | + 4J(f ) + 1. ≤ A log |Qn |e

(3.6.28)

Substituting (3.6.28) into (3.6.27) yields that N∞ (f ) ≤ (A + |Qn |) log

A+|Qn | |Qn |

+ (1 + |Qn |) + 4J(f ) n ) ≤ [1 + |Qn | + 4J(f )] log 1+|Q|Q|+4J(f + 1 . n|

(3.6.29)

3.6 Spaces related to the a.e. convergence of the Fourier series

229

This ﬁnishes the proof of Theorem 3.6.4.

3.6.5 The convergence of the Fourier series on Bq For the case of one-dimension, Taibleson and Weiss [TW1] proved the following theorem. Theorem 3.6.5 Let 1 < q ≤ ∞. If f ∈ Bq (Q1 ), then Sk (f, x) −→ f (x) holds for a.e. x ∈ Q1 . Proof. For the case of one-dimension, we merely need to consider in the case of R ∈ Z+ . At this time, we have 1 π f (x − t)DR (t)dt, SR (f, x) = π −π for R ∈ Z+ , where

sin R + 12 t DR (t) = 2 sin( 2t )

is the Dirichlet kernel. We ﬁrst prove that there exists a constant C = Cq > 0, such that, for each q-block b, , - x ∈ Q1 : S∗ (b)(x) > λ ≤ C , (3.6.30) λ for λ > 0, where S∗ is the maximal operator of the partial sum deﬁned as S∗ (f )(x) = sup |SR (f ; x)|. R≥0

−

1

Suppose that suppb ⊂ I such that bq ≤ |I| q , where I is an interval in Q1 . It suﬃces to consider under the situation 1 < q < ∞. It is well-known that S∗ is of type (q, q), and thus, we have q , - 1 1 x ∈ Q1 : S∗ (b)(x) > λ ≤ Cq bq ≤ Cq . q−1 λ (|I|λ) λ Hence, when λ ≥ |I|−1 , we have |{S∗ (b) > λ}| ≤

Cq . λ

230

C3. Bochner-Riesz means of multiple Fourier series

, We substitute x ∈ Q1 : S∗ (b)(x) > λ by {S∗ (b) > λ} for simpleness, and we will adopt the brief form in the following. C , we Let λ < |I|−1 and deﬁne ρ(x, I) = inf t∈I |x − t|. Since |DR (t)| ≤ |t| have that 1 C C |SR (b, x)| = b(t)DR (x − t)dt ≤ b1 = . π I ρ(x, I) ρ(x, I) And it follows that S∗ (b)(x) ≤

C . ρ(x, I)

Consequently, we conclude that C |{S∗ (b) > λ}| ≤ >λ ρ(x, I) Taking |I| <

1 λ

into consideration, we ρ(x, I) < C λ

= ρ(x, I) < C λ

.

can get that C < . λ

This means that (3.6.30) is valid. Since f ∈ Bq , and by the decomposition in blocks as well as (3.6.30), we can use the iteration principle of Lemma 3.6.3 to deduce that |{x : |x| < π, S∗ (f ) > λ}| ≤

CNq (f ) , λ

for all λ > 0. Lu-Taibleson-Weiss [LTW1] proved the conclusions in the case of multidimension. Theorem 3.6.6 Let n > 1 and 1 < q ≤ ∞. If f ∈ Bq

2

L log+ L(Qn ), then

n−1

SR 2 (f ; x) converge to f (x) for a.e. x ∈ Qn . By Theorem 3.2.3 and the condition that f ∈ L log+ L(Qn ), it n−1 suﬃces to prove that BR 2 (f6; x) → f (x). It is well-known n−1 n−1 1 f (y)HR 2 (x − y)dy, BR 2 f6; x = n (2π) Qn Proof.

3.6 Spaces related to the a.e. convergence of the Fourier series where n−1 2

HR

n 2

(u) = (2π) 2

n−1 2

Γ

n+1 2

J

n− 12 (R|u|) n− 12

231

Rn .

(R|u|)

Similar to the method of the proof in Theorem 3.6.5, we merely need to construct the weak-type inequality |{B∗ (f )(x) > λ}| <

C , λ

(3.6.31)

for λ > 0, where the maximal operator B∗ is deﬁned by n−1 2 B∗ (f )(x) = sup BR (f ; x) . R>0

Since f ∈ Bq , so does f6. By the decomposition in blocks and the iteration principle, it suﬃces to establish (3.6.31) for the q-block b. Here, we directly quote the conclusion that the maximal operator B∗ is of type (q, q) with q ∈ (1, ∞). Thus we merely need notice that the kernel n−1

HR 2 (u) satisﬁes the following inequality n−1 H 2 (u) ≤ C . |u|n R Using the method of Theorem 3.6.5, we can complete the proof. This ﬁnishes the proof of Theorem 3.6.6. In summary, we can have that the function classes (or spaces) mentioned above have the inclusion relationships as follows. D ⊂ J ⊂ L log+ L and J ⊂ B∞

*

L log+ L ⊂ Bq

*

L log+ L ⊂ L,

where 1 < q < ∞. Theorem 3.6.6 established the almost everywhere convergence property of the Bochner-Riesz means of multiple Fourier series at the critical index on the function class of (∪q>1 Bq ) ∩ L log+ L. It is a beautiful result, which includes Theorem 3.6.1. By the way, we have to mention that Theorem 3.6.1 was independently proven by Sato [Sa1]. In Sato’s paper, he also constructed 5 such a function f ∈ L(Q1 ), J(f ) < ∞, but f ∈ / (L log + L log+ log+ L D). The example illustrates that D is a proper subset of J, and on the other

232

C3. Bochner-Riesz means of multiple Fourier series

hand, Sj¨ olin’s theorem about the a.e.-convergence in [Sj1] cannot contain Theorem 3.6.1. Finally, we want to point out that the block space is not only applicable for the research of the a.e.-convergence problem about the triangle Fourier series, but for the Walsh-Fourier series as well. Wang [Wa4] have a notation about this. In the following, we introduce certain building blocks called smooth blocks, and deﬁne spaces generated by smooth blocks. A smooth block is obtained by imposing certain smoothness on a block. The reason for studying spaces generated by smooth blocks is to investigate the relation between the smoothness imposed on the blocks and the rate of convergence of Bochner-Riesz means of multiple Fourier integral at the critical index. Let us now turn to the deﬁnition of smooth blocks. A (q, λ)-block, 1 < q ≤ ∞, is a function b that is supported on a cube Q satisfying 1

bLq ≤ |Q| q λ

−1

,

(Rn ) be the function where Lqλ denotes the Bessel potential space. Let Bqλ/ space that consists of all functions f of the form, f = k mk bk , where each bk is a (q, λ)-block, and N ({mk }) < ∞. Note that Bq0 (Rn ) = Bq (Rn ). A natural conjecture on the relation between the smoothness and the rate of convergence is formulated as follows. For 0 ≤ λ < 2, f ∈ Bqλ (Rn ) implies n−1 1 2 BR f (x) − f (x) = o Rλ for a.e. x ∈ Rn , as R → ∞. Lu and Wang in [LW1] only obtain an aﬃrmative answer for λ = 1. Their results are stated as follows. Theorem 3.6.7 If f ∈ Bq1 (Rn ) with 1 < q ≤ ∞, then n−1 1 2 BR f (x) − f (x) = o R holds for a.e. x ∈ Rn , as R → ∞. The proof of Theorem 3.6.7 is based on Lp -estimates of a maximal operator. Let Mλα f be the maximal function deﬁned by α f )(x) − f (x)} . (Mλα f )(x) = sup Rλ {(BR R>0

3.6 Spaces related to the a.e. convergence of the Fourier series

233

Theorem 3.6.8 Let 0 ≤ λ ≤ 2, 1 2 p If f ∈ Lpλ (Rn ), then we have Mλα f p ≤ Cf Lp . λ

To prove Theorem 3.6.8, we need some lemmas. Lemma 3.6.7 Let 1 then we have 2 M2α f p ≤ Cn,σ,p eπ|τ | f Lp2 .

n−2 2 .

If f ∈ Lp2 (Rn ),

Proof. We write α f )(x) − f (x) R2 (BR J(n/2)+α (R|y|) f (x + y) + f (x − y) − 2f (x) Rn+2 = Cn,α dy, (R|y|)(n/2)+α Rn where

n |Cn,α | = 2α−(n−2) π − 2 Γ(α + 1) ≤ Cn,σ .

Let g(x, t) =

Sn−1

f (x + ty) + f (x − ty) − 2f (x) ds(y),

where ds is surface measure on Sn−1. Thus we have ∞ α J(n/2)+α (t) t 2 2 R (BR f )(x) − f (x) = Cn,α tn−1 (n/2)+α dt. (3.6.32) R g x, R t 0 Denote

A(t) =

∞

r n−1

J(n/2)+α (t) t(n/2)+α

t

and

∞

A(r)dr.

B(t) = t

Clearly we have that

t g x, R

t=0

=0

dr

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C3. Bochner-Riesz means of multiple Fourier series

and

d dt

t = 0. g x, R t=0

Thus, it follows from (3.6.32) and integration by parts that ∞ 2 α t 2 2 d B(t)dt. R (BR f )(x) − f (x) = Cn,α R 2 g x, dt R 0

Let gij (x, t) =

Sn−1

where Dij f (x) =

|Dij f (x + ty)|ds(y),

∂ ∂xj

We hence have (M2α f )(x)

≤ Cn,σ

n

∂ ∂xi

∞

sup

gij

i,j=1 R>0 0

x,

f (x).

t R

|B(t)|dt,

1 1 2 2 cos(t − θ) + O σ+2−(n−1)/2 , B(t) = − π tα+1−(n−1)/2 t for t ≥ 1, and 2 |B(t)| ≤ Cn,σ e|τ | ,

(3.6.33)

(3.6.34)

(3.6.35)

for 0 < t < 1, where π(n + 2α + 1) 4 1 1 O ≤ Cn,σ eπ|τ | . σ+2−(n−1)/2 σ+2−(n−1)/2 t t θ=

and

It follows from (3.6.35) that 1 1 t t 2 gij x, gij x, |B(t)|dt ≤ Cn,σ e|τ | dt R R 0 0 |τ |2 |y|1−n |Dij f (x + y)|dy ≤ Cn,σ e R |y|<1/R

|τ |2

≤ Cn,σ e

M (Dij f )(x),

where M f is the Hardy-Littlewood maximal function of f . Thus, we have 1 t 2 sup |B(t)|dt gij x, ≤ Cn,σ,p e|τ | f Lp2 . (3.6.36) R R>0

0

p

3.6 Spaces related to the a.e. convergence of the Fourier series

235

Meanwhile, by(3.6.34), we have ∞ ∞ n−1 t t π|τ | gij x, gij x, |B(t)|dt ≤ Cn,σ e t 2 −σ−1 dt R R 1 1 ∞ n−1 n−1 gij (x, t)t 2 −σ−1 dt. = Cn,σ eπ|τ | R 2−σ 1 R

Using integration by parts and the following inequality t τ n−1 gij (x, τ )dτ ≤ Cn tn M (Dij f )(x), 0

we have

1

∞

t gij x, |B(t)|dt ≤ Cn,σ eπ|τ | M (Dij f )(x). R

Thus we obtain ∞ t π|τ | sup gij x, |B(t)|dt ≤ Cn,σ,p e f Lp2 . R R>0 1 p

(3.6.37)

Consequently, the conclusion of Lemma 3.6.7 is immediately derived from (3.6.33), (3.6.36) and (3.6.37).

Lemma 3.6.8 Let 0 ≤ λ ≤ 2, 1 (n − 1)/2. If f ∈ Lpλ (Rn ), then 2

Mλα f p ≤ Cn,σ,λ,p eπ|τ | f Lp . λ

Proof. Let {rk } be a sequence consisting of all positive rational numbers and Λk = {r1 , ..., rk }. Deﬁne

Fλα,k f (x) = sup rjλ (Brαj f )(x) − f (x) . rj ∈Λk

Thus we have

Fλα,k f (x) ≤ Fλα,k+1 f (x)

and (Mλα f )(x) = lim

k→∞

Fix f ∈

Lpλ (Rn )

Fλα,k f (x).

and k. Let Sj with 1 ≤ j ≤ k be a set such that, for x ∈ Sj ,

Fλα,k f (x) = rjλ (Brαj f )(x) − f (x) ,

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C3. Bochner-Riesz means of multiple Fourier series

Fλα,k f (x) > riλ |(Brαi f )(x) − f (x)|,

and

for i < j. It is easy to note that the sets Sj do not intersect each other. Let Ω = {z ∈ C : 0 ≤ Rez ≤ 1}. Deﬁne

ψj (x) = sign (Brαj f )(x) − f (x) ,

and (Tz g)(x) =

rj ∈Λk

rj2z χSj (x) Brαj (J2z g)(x) − (J2z g)(x) ψj (x).

Then {Tz } is a family of linear operators. It is easy to verify that {Tz } is an admissible family of operators (see [SW1]). Now we can write f = Gλ ∗ g, where ˆ λ (x) = 1 + 4π 2 |x|2 −λ/2 . G Since f ∈ Lpλ , a multiplier theorem (see [St4]) implies that Jiη gp ≤ Cn,p P (η)gp , where P is a polynomial of degree k > n/2. Note α > (n − 2)/2. We have Tiη gp ≤ F0α,k (J2iη g) ≤

p

M0α (J2iη g)p

≤ Cn,σ,p e2π|τ | J2iη gp ≤ Cn,σ,p e2π|τ | |P (2η)|gp . On the other hand, it follows from Lemma 3.6.7 that T1+iη gp ≤ F2α,k (J2+2iη g) p

≤ M2α (J2+2iη g)p 2

≤ Cn,σ,p e|τ | J2+2iη gp 2

≤ Cn,σ,p e|τ | J2iη gp 2

≤ Cn,σ,p e|τ | |P (2η)|gp .

3.6 Spaces related to the a.e. convergence of the Fourier series

237

Using the Stein’s interpolation theorem of analytic operators, we have Fλα,k f p = Tλ/2 gp 2

≤ Cn,σ,p e|τ | gp 2

≤ Cn,σ,λ,p e|τ | f Lp . λ

Finally, by Lebesgue’s monotonic convergence theorem, we obtain the conclusion of Lemma 3.6.8. Let

α+1 α (Nλα f ) (x) = sup Rλ (BR f )(x) − (BR f )(x) . R>0

Lemma 3.6.9 Let α = σ + iτ, σ > 0, and 0 ≤ λ ≤ 2. If f ∈ L2λ (Rn ), then we have Nλα f 2 ⎧ √ √ √ ⎪ ⎨ Cn exp{Cn (σ + |τ |)}1/ σ(1 + 1/ 2 − λ + 1/ σ)f L2λ , 0 ≤ λ < 2, ≤ ⎪ ⎩ Cn exp{Cn (σ + |τ |)}1/σf 2 , λ = 2. L2 Proof. Let β ∈ C, Re(β) > 12 , δ > − 12 , and 0 ≤ λ < 2, Using the formula at the page 278 in [SW1], taking the the Fourier transform, and doing some algebra, we get the identity:

β+δ+1 β+δ BR f (x) − BR f (x) R ! 2 2 1 2 β−1 2δ+3 δ+1 − u u f (x) = B R u B(β, δ + 1) R2(β+δ+1) 0

" − Buδ f (x) du, where B(β, δ) =

Γ(β)Γ(δ) . Γ(β + δ)

238

C3. Bochner-Riesz means of multiple Fourier series

Thus we conclude that

β+δ+1 β+δ f (x) − BR f (x) R λ BR ≤

1 2 2(Reβ+δ+1)−λ |B (β, δ + 1) | R R 12 2 2 2Reβ−2 4δ+7−2λ × u du R −u ×

0

R

u 0

2 ≤ |B(β, δ + 1)| =

2 12

δ+1 δ Bu f (x) − Bu f (x) du

2λ−1

1

0

1−t

2 2Reβ−2 4δ+7−2λ t

1 2

dt

Gδλ f (x)

1 {2B(2Reβ − 1, 2δ − λ + 4)} 2 δ Gλ f (x), B(β, δ + 1)

where

δ Gλ f (x) =

∞

u

0

Let δ =

σ−1 2

2 12 δ+1 δ . Bu f (x) − (Bu f )(x) du

2λ−1

> − 12 , β = α − δ =

σ+1 2

+ iτ . We conclude that 1

Nλα f 2

[2B(σ, σ − λ + 3)] 2 ≤ σ+1 |B( σ+1 2 + iτ, 2 )|

It follows from the Plancherel theorem that σ−1 2 G 2 f λ 2 ∞

σ−1 G 2 f . λ 2

2 σ−1 σ+1 2 2 = u Bu f (x) − Bu f (x) dudx Rn 0 σ+1 σ−1 2 ∞ 2 2 2 2 |y| |y| ˆ 2 u2λ−1 − 1− 2 = 1− 2 |f (y)| dydu u u 0 |y|
∞ 2 σ−1 |y| |y|4 |fˆ(y)|2 u2λ−5 1 − 2 du dy = u |y| Rn

2λ−1

3.6 Spaces related to the a.e. convergence of the Fourier series

239

2 1 σ−1 1−λ |y| f (y) dy (1 − v) v dv n 0 R 1 1 ≤C 1+ + f 2L2 . λ σ 2−λ 1 = 2

Note that

2λ ˆ

σ+1 σ+1 + iτ, B 2 2

and

≤ Cn exp{Cn (σ + |τ |)}

−1

1 Cn 2B(σ, σ − λ + 3) 2 ≤ 1 . σ2

Hence we have Nλα f 2

1 ≤ Cn exp{Cn (σ + |τ |)} √ σ

1 1 1+ √ + √ σ 2−λ

f L2 , λ

for 0 ≤ λ < 2. When λ = 2, we have α+1 α α R2 (BR f )(x) − (BR f )(x) = −BR (Δf )(x). Thus, by the method similar to the proof of Lemma 5.10 in [SW1], we have α α N2 f 2 = sup BR (Δf ) R>0

2

≤ Cn eCn (σ+|τ |) (1/σ)Δf 2 ≤ Cn eCn (σ+|τ |) (1/σ)f L2 . 2

Next let us turn to the proof of Theorem 3.6.8. Set f = Gλ ∗ g ∈ Lp (Rn ). For σ > 0, we choose a k ∈ N, such that σ + k > (n − 1)/2. Thus, by Lemma 3.6.7, 3.6.8, and the inequality (Mλα f )(x) ≤

k−1 (Nλα+1 f )(x) + (Mλα+k f )(x), j=0

it follows that 2

2

Mλα f 2 ≤ Cn,σ,λ e|τ | f L2 = Cn,σ,λ e|τ | g2 . λ

(3.6.38)

240

C3. Bochner-Riesz means of multiple Fourier series

Now, let p1 > 1 and σ > (n − 1)/2. It follows from Lemma 3.6.7 that 2

2

Mλα f p1 ≤ Cn,σ,λ,p1 e|τ | f Lp1 = Cn,σ,λ,p1 e|τ | gp1 . λ

(3.6.39)

Consider p1 ≤ p ≤ 2, and write 1/p = (1 − t)/2 + t/p1 . Let μ0 > 0, μ1 > (n − 1)/2, and δ(z) = μ0 (1 − z) + μ1 z for 0 ≤ Rez ≤ 1. If μ0 → 0, μ1 → (n − 1)/2, and p1 → 1, then we have δ(t) → Therefore, if σ>

n−1

. 2 2p − 1

n−1

, 2 2p − 1

then there exist μ0 , μ1 and p1 satisfying the above condition such that δ(t) = σ, where 1 1 p − 2 t= 1 1. p − 2 1

Let such μ0 , μ1 , and p1 be ﬁxed later. Let {Gj } be a sequence consisting of all positive rational numbers. Denote Ak = {R1 , ..., Rk }, and $

# α f )(x) − f (x)| . Fλα,k f (x) = sup Rλ |(BR R∈Ak

We have

Fλα,k f (x) ≤ Fλα,k+1 f (x),

and (Mλα f ) (x) = lim

k→∞

Fλα,k f (x).

For 1 ≤ j ≤ k, let Ej =

$ # α x ∈ Rn : sup Rλ |(Bλα f )(x) − f (x)| = Rjλ |(BR f )(x) − f (x)| , j R∈Ak

and F1 = E1 , Fj = Ej \ (Tz h)(x) =

k j=1

5j−1 i=1

Ei for j = 2, ..., k. Deﬁne

δ(z) Rjλ χFj (x) BRj (Gλ ∗ h)(x) − (Gλ ∗ h)(x) ψj (x),

3.6 Spaces related to the a.e. convergence of the Fourier series

241

σ ψj (x) = sign BR (G ∗ g)(x) − G ∗ g(x) . λ λ j

where

It is easy to verify that {Tz } is an admissible family of linear operator (see [SW1].) Using (3.6.38) and (3.6.39) implies that 2 2 δ(iτ ),k δ(iτ ) f ≤ Mλ f ≤ Cn,μ0 ,λ e(μ1 −μ0 ) τ g2 , Tiτ g2 ≤ Fλ 2

and

2

δ(1+iτ ),k f T1+iτ gp1 ≤ Fλ

p1

2τ 2

≤ Cn,μ1 ,λ,p1 e(μ1 −μ0 )

gp1 .

Thus, by the Stein’s interpolation theorem of analytic operators, we have that δ(t),k σ,k f = Tt gp ≤ Cgp = Cf Lp . Fλ f = Fλ λ p

p

It follows from the monotonic convergence theorem that Mλσ f p ≤ Cf Lp , λ

for 1 < p ≤ 2. We can obtain the similar estimate for Mλα , α = σ + iτ , as in [SW1]. Finally, it should be pointed out that the proof in the case of 2 < p < ∞ is similar to the above. To prove Theorem 3.6.7, we ﬁrst need to establish a weak type estimate (n−1)/2 of the maximal operator M1 on any block. Lemma 3.6.10 If b is a (q, 1)-block, then we have # $ (n−1)/2 b (x) > λ ≤ Cλ−1 , x : M1 where C is independent of λ and b. Proof. We write # $ (n−1)/2 R BR b (x) − b(x) Jn− 1 (R|y|) n+1 2 [b(x + y) − b(x)] = CR 1 dy n R (R|y|)n− 2 ∞ n+1 = CR [b(x + tξ) − b(x)]dσ(ξ) 0

∞

= CR 0

Sn−1

" ! τ b x + ξ − b(x) dσ(ξ) R Sn−1

Jn− 1 (Rt) 2

n− 12

(Rt) Jn− 1 (τ ) 2

n− 12

(τ )

tn−1 dt

τ n−1 dτ.

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C3. Bochner-Riesz means of multiple Fourier series

/ L1 (Rn ), these integrals should be interpreted as Since t−1/2 Jn− 1 (t) ∈ 2 T and lim , respectively. lim

T →+∞ |y|≤T

T →+∞ 0

Denote g(x, τ ) =

Sn−1

[b(x + τ ξ) − b(x)]dσ(ξ).

We have # $ (n−1)/2 R BR b (x) − b(x) = CR

∞

g(x, τ /R)

0

Jn− 1 (τ ) 2

n− 12

τ n−1 dτ.

(τ )

Using integration by parts, we obtain $ # (n−1)/2 b (x) − b(x) = CR R BR where

∞

A(τ ) = τ

xn−1

∞ 0

d g(x, τ /R)A(τ )dτ, dτ

Jn− 1 (x) 2

n− 12

dx.

(x)

By the properties of Bessel function, it follows that |A(τ )| ≤ Cτ for τ ≥ 1, and |A(τ )| ≤ C for 0 < τ < 1. Thus, we have # $ (n−1)/2 b (x) − b(x) R BR ∞

τ ≤C ∇x b x + ξ dσ(ξ) |A(τ )|dτ R n−1 0 ∞ S =C |∇x b (x + tξ)| dσ(ξ) |A(Rt)|Rdt n−1 0 ∞ S |∇x b (x + tξ)| dσ(ξ) t−1 dt ≤C n−1 S 0 dy =C |∇x b(x + y)| n n |y| R |∇b(u)| =C du. n Rn |u − x| It follows that

(n−1)/2 M1 b (x) ≤ C

Q

where supp b ⊂ Q.

|∇b(u)| du, |u − x|n

(3.6.40)

3.6 Spaces related to the a.e. convergence of the Fourier series

243

˜ = 2Q. Then it follows from (3.6.40) that Let Q

(n−1)/2 b (x) ≤ M1

C , |x − xQ |n

˜ where xQ is the center of Q. Thus, we have provided that x ∈ / Q, #

$ ˜ : M (n−1)/2 b (x) > λ, λ ≤ 1/|Q| ≤ Cλ−1 . /Q x∈ 1

(3.6.41)

It follows from Theorem 3.6.8 that (n−1)/2

M1

f q ≤ Cf Lq1 ,

(3.6.42)

for 1 < q < ∞. Thus, we have # $ (n−1)/2 n b)(x) > λ, λ > 1/|Q| ≤ C x ∈ R : (M1

bLq1 λ

q ≤ Cλ−1 .

(3.6.43) Combining (3.6.41) with (3.6.43) yields the conclusion of Lemma 3.6.10. Proof of Theorem 3.6.7.

f (x) =

mk bk (x) =

k

N

Let f ∈ Bq1 (Rn ). We have mk bk (x) +

k=1

∞

mk bk (x) := g(x) + h(x),

k=N +1

where each bk is a (q, 1)-block and N ({mk }) < ∞. To complete the proof of Theorem 3.6.7, we must prove # $ (n−1)/2 x : lim sup R (B f )(x) − f (x) > λ = 0. R R→∞

Since (3.6.42) implies # $ 1 (n−1)/2 BR g (x) − g(x) = o lim R→∞ R for a.e. x ∈ Rn and g ∈ Lq1 (Rn ), we have # $ x : lim sup R B (n−1)/2 g (x) − g(x) > λ/2 R R→∞

= 0.

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C3. Bochner-Riesz means of multiple Fourier series

Thus, we obtain that # $ x : lim sup R B (n−1)/2 f (x) − f (x) > λ R R→∞ λ (n−1)/2 ≤ x : MR h (x) > 2 /∞ ∞ −1 s=1 |ms | . ≤ Cλ |mk | 1 + log |mk |

k=N +1

This completes the proof of Theorem 3.6.7.

3.7

The uniform convergence and approximation

Concerning the problems of the convergence and approximation in the scale of uniform, here, using C(Rn ) instead of L∞ (Rn ), we have mentioned in Section 3.4. However, the order is α > α∞ = n−1 2 . In this section, we mainly discuss the situation of the critical index. Parallel to Theorem 3.5.4 about the pointwise convergence, Lu [Lu2] obtained the following result. Theorem 3.7.1 Let f ∈ C(Qn ) with n ≥ 2. The following two assertions hold. (a) If the condition (3.5.10) uniformly holds for x ∈ Qn and r ∈ [h, r0 ], then we have n−1 2 S R (f ) − f → 0 C

as R → ∞. (b) If

ω 6 (f ; δ) = o

1 log 1δ

as δ → 0+ , then the condition of (a) is satisﬁed, where , ω 6 (f ; δ) = sup |fx (t + h) − fx (t)| : x ∈ Qn , 0 ≤ h ≤ δ, t > 0 is the modulus of continuity introduced by Golubov [Go1].

3.7 The uniform convergence and approximation

245

Proof. For any ﬁxed point x0 ∈ Qn , we denote 1 Qx0 = x : x − x0 ∈ Qn . 2 Obviously, the following propositions are equivalent: n−1 2 (I) SR (f ) − f → 0, as R → ∞. C n−1 2 → 0, as R → ∞, for x0 ∈ Qn . (II) SR (f ) − f C(Q0 )

(III) For x0 ∈

Qn ,

deﬁne

g(x) = g(x0 ) (x) =

⎧ ⎪ ⎨ f (x), ⎪ ⎩

0,

if x − x0 ∈ Qn , if x − x0 ∈ / Qn .

n−1 2 sup BR (g; x) − g(x) → 0, as R → ∞.

x∈Q0

(IV) For x0 ∈ Qn , we have √ 2 nπ cos tR − [gx (t) − g(x)] lim sup R→∞ s∈Q0 π t R

nπ 2

dt = 0.

Here gC(Q0 ) = sup{|g(x)| : x ∈ Q0 }. We have to point out that it is quite obvious that (I) ⇐⇒ (II) and (II) ⇐⇒ (III) follows from Stein’s theorem (see Theorem 3.2.3). (III) ⇐⇒ n−1

(IV) follows from the following facts, the integral expression of BR 2 (g), the uniform continuity of g on Q0 , the asymptotic formula of Bessel functions √ and the conclusion that when t > 2 nπ, gx (t) = 0, for x ∈ Q0 . Hence, to prove the conclusion of (I), it suﬃces to prove (IV). We may take it for granted that x0 = 0, and in the condition (3.5.10), r0 ∈ 0, 12 . Firstly, we will show that √ 2 nπ cos tR − nπ 2 dt = 0. (3.7.1) lim sup [gx (t) − g(x)] R→∞ x∈Q − 1 Qn r0 t 0 2

Now let x0 = 0 and g(x) = f (x)χQn (x).

246

C3. Bochner-Riesz means of multiple Fourier series

We have |g(x)| ≤ maxn |f (x)| = M. x∈Q

Therefore, we have √ nπ cos tR − g(x) r0 t

nπ 2

R2√nπ nπ cos u − 2 dt = g(x) du u Rr0 √ R2 nπ cos u − nπ 2 du → 0. ≤ M Rr0 u

(3.7.2)

For any ε > 0, there exists ϕ ∈ C ∞ (Rn ) with |ϕ| ≤ 1, such that ϕ(x) = 1, if x ∈ Qn , and ϕ(x) = 0, if x ∈ / (1 + ε) · Qn . Let h = f ϕ. Then we have √ 2 nπ cos tR − nπ 2 [gx (t) − hx (t)] dt r0 t √ 2 nπ dt |g(x + tξ) − h(x + tξ)|dσ(ξ) ≤ t n−1 r0 S 1 ≤ n |g(y) − h(y)|dy r0 r0 <|y−x|<2√nπ M M ≤ n |(1 + ε)Qn − Qn | = n ((1 + ε)n − 1) |Qn | r0 r0 < Cε, (3.7.3) where we may assume that 0 < ε < 1. Since h is uniformly continuous, if R is big enough, we have √ nπ n cos tR − 2 dt < ε, hx (t) r0 t for x ∈ Rn . By (3.7.3) and (3.7.4), we can easily get (3.7.2). Now it remains to prove the following equality. r0 nπ cos tR − 2 dt = 0. lim sup [gx (t) − g(x)] R→∞ x∈ 1 Qn π t

(3.7.4)

(3.7.5)

R

2

Denote the spherical integral of g centered at x with radius r by r G(x, r) = [g(y) − g(x)]dy = ωn−1 [gx (t) − g(x)]tn−1 dt. B(x,r)

0

3.7 The uniform convergence and approximation

247

Set ϕx (t) = [gx (t) − g(x)]t

n−1

=

1 ωn−1

|ξ|=t

Then G(x, r) = ωn−1

[g(x + ξ) − g(x)]dσ(ξ).

r

0

ϕx (t)dt.

Since g and f coincide in Qn , for x ∈ 12 Qn and r ∈ [h, r0 ], 0 < h < r0 ≤ 12 , the following equality

h h 1 ϕx (r + h + t) − ϕx (r + h − t) dt = o (3.7.6) r n−1 0 log h1 holds uniformly, as h → 0+ . The remaining arguments are the same as that of Theorem 3.5.4. However, the estimate we can get at this time is uniformly valid for x ∈ 12 Qn . Thus we have proven (3.7.5), from which we can induce the conclusion of (I) directly. In order to prove (II), it suﬃces to show that when

ω 6 (f ; δ) = o

1 log 1δ

as δ → 0+ ,

h#

n−1 f (r + h + t) − f (x) (r + h + t) x r n−1 0 $ −(r + h − t)n−1 fx (r + h − t) − f (x) dt

h =o log h1 1

(3.7.7)

holds uniformly for x ∈ Qn and r ∈ [h, r0 ], as h → 0+ . The integral on the left side of (3.7.7) can be divided into the sum of the following three terms: J1 =

0

h

[fx (r + h + t) − fx (r + h − t)]dt,

248

C3. Bochner-Riesz means of multiple Fourier series J2 =

and

h 0

J3 = −

:

0

h

h+t 1+ r

:

B

n−1

h+t 1− r

− 1 [fx (r + h + t) − f (x)]dt, n−1

B − 1 [fx (r + h − t) − f (x)]dt.

Obviously, we can have that

h h , J1 = O ω 6 (f, 2t) dt = o log h1 0 J2 =

h

O

t

O(6 ω (f ; r + h + t))dt r h h r O 1+ ·ω 6 f ; h dt = r h 0

h =o log h1 0

and J3 = o

h log h1

hold uniformly for x ∈ Qn and r ∈ [h, r0 ]. This completes the proof of Theorem 3.7.1. The conclusion of (II) shows that Theorem 3.7.1 is stronger than the corresponding result obtained by Golubov. Of course, the condition (3.5.10) in Theorem 3.7.1 (and Theorem 3.5.4) can be weakened into unilateral condition. The unilateral condition means that: if |A(t)| = o(α(t)) as t → t0 , then A(t) ≤ ε(t)α(t) or A(t) ≥ −ε(t)α(t), where ε(t) ≥ 0 and ε(t) = o(1) as t → t0 . Jiang has obtained such results and extended these conclusions to the situation of the conjugate series (see Jiang [J1]). Making use of the Lebesgue constant, one can deduce the following results immediately (see Wang [Wa6]). Theorem 3.7.2 Let f ∈ C(Qn ) with n > 1. Then we have n−1 2 = O(log R)ω2 f ; 1 , S (f ) − f R R C where ω2 is the 2-ordered continuous norm of f .

(3.7.8)

3.7 The uniform convergence and approximation

249

Proof. Denote g by the best R-ordered triangle polynomial approximation of f . We have that n−1

n−1

n−1

SR 2 (f ) − f = SR 2 (f − g) − (f − g) + SR 2 (g) − g. It follows that n−1 n−1 S 2 (f ) − f ≤ LR f − gC + S 2 (g) − g , R R C

C

n−1

where LR is the Lebesgue constant of SR 2 , and LR = O(log R). At the same time, just as we do in Section 3.4, 1 , f − gC = ER (f )c ≤ Cω2 f ; R we conclude that n−1 S 2 (g) − g R

≤ C

= ≤ ≤

n+1 n−1 n+1 S 2 (g) − S 2 (g) + S 2 (g) − g R R R C C n−1 1 1 S 2 (g) + Cω2 g; R2 R R C 1 1 LR gC + Cω2 f ; R2 R 1 , C log R · ω2 f ; R

for R > 2. In the above estimate, we have utilized the relation between the modulus of continuity of the derivative of the best triangle polynomial approximation and that of the function which is to be approximated. This relation can be easily extended from the corresponding result in the unitary case. Combining all the above results, we can have (3.7.8) which concludes the proof.

Remark 3.7.1 From the proof, we can see that if n is an odd number with n ≥ 3, there is a more accurate estimate n−1 f − S 2 (f ) = O(log R)ER (f ) + O ω2 f ; 1 . (3.7.9) R R C

250

C3. Bochner-Riesz means of multiple Fourier series

For functions with higher derivatives, the approximation order can be increased. Definition 3.7.1 Let f ∈ C(Qn ). If f is absolutely continuous with respect ∂f are essentially bounded on Qn , for to all xj and its partial derivative ∂x j j = 1, 2, . . . , n, then we call f ∈ W 1 L∞ , where L∞ does not refer to C, but ∂f is essentially bounded). If ∂x ∈ W 1 L∞ , j = 1, 2, . . . , n, then it implies that j f ∈ W 2 L∞ . Besides, Lipα refers to the function class of ω(f ; δ) = O(δ α ) as δ → 0+ where δ ∈ (0, 1]. Here it should be pointed out that we limit to consider the periodical functions only. It is obvious that W 1 L∞ = Lip1. 2f Obviously, if f ∈ W 2 L∞ , then ∂x∂i ∂x is essentially bounded, for i, j = j 1, 2, . . . , n. We also denote the subclass of the functions in C(Qn ) whose j-th partial derivatives are continuous in C j (Qn ). Theorem 3.7.3 If f ∈ W 2 L∞ , then we have n−1 log R 2 S R (f ) − f = O R2 C and

n−1 2

SR

(f ; x) − f (x) = O

1 R2

(3.7.10)

(3.7.11)

for a.e. x ∈ Qn . Proof. Since f ∈ W 2 L∞ , we have M = M f := f ∞

and ω2

2 2 ∂ f ∂ f = ∂x2 + · · · + ∂x2 < ∞, n ∞ 1

1 f; R

=O

1 R2

.

Then, n−1

n−1

n+1

n+1

SR 2 (f ; x) − f (x) = SR 2 (f ; x) − SR 2 (f ; x) + SR 2 (f ; x) − f (x). For

n−1

n+1

SR 2 (f ; x) − SR 2 (f ; x) =

−1 n−1 S 2 (f ; x), R2 R

3.8 (C, 1) means

251

we have n+1 n−1 1 log R 2 2 S , R (f ) − SR (f ) = R2 O(log R) · f ∞ = O R2 C and

n+1 n−1 n−1 S 2 (f ; x) − S 2 (f ; x) ≤ 1 S∗ 2 (f ; x), R R R2

(3.7.12)

(3.7.13)

n−1

where S∗ 2 is the maximal operator, and since it is type (p, p) with 1 < p < n−1

∞, we can get that S∗ 2 (f ; x) is ﬁnite almost everywhere. Besides of the above, by the result in Section 3.4, n+1 1 S 2 (f ) − f = O ω2 f ; 1 =O . R R R2 C

(3.7.14)

Therefore, combining (3.7.12) with (3.7.14), we can get (3.7.10), (3.7.13) and (3.7.14) together giving (3.7.11). This completes the proof of Theorem 3.7.3.

3.8

(C, 1) means n−1

Since the role of SR 2 is the similar as that of Fourier partial sum in unitary variable in the spaces of L1 (Qn ) and C(Qn ), it is natural to ask whether n−1 1 R n−1 2 Sr 2 dr σR := R 0 is considerably equivalent to the Fej´er means or not. And also, we can consider 1 2R n−1 Sr 2 dr. VR = R R n−1

As an analogy of the Vall´ee Poussin means, we call σR2 as (C, 1) means. Jiang discussed the approximation problem of continuous functions by the (C, 1) means (see Jiang [J1] or [J2]).

252

C3. Bochner-Riesz means of multiple Fourier series

Lemma 3.8.1 Let Reα > α σR (f ; x)

1 := R =

n−3 2

and f ∈ L(Qn ). Then we have

R

Srα (f ; x)dr

0 α+1− n 2

2

Γ

Γ(α + 1) n

∞

0

2

1 fx (t) R

R

J n2 +α (tr) (tr)

0

n +α 2

r n drtn−1dt. (3.8.1)

Proof. Denote

n

2α+1− 2 Γ(α + 1) . Cn (α) = Γ n2 Clearly Cn is analytic function with respect to α in the domain Reα > −1. By the Bochner formula (see (3.1.2)), if Reα > n−1 2 , we have Srα (f ; x) = Cn (α)r n

∞ 0

fx (t)

J n2 +α (rt) (rt)

n +α 2

tn−1 dt,

for r > 0. Thus we immediately deduce that (3.8.1) holds, provided that Reα > n−1 2 . Denote 1 R J n2 +α (tr) n n−1 α r drt . KR (t) = n R 0 (tr) 2 +α This is equivalent to α KR (t)

By the formula

1 = 2 t R

tR 0

n

J n2 +α (s)s 2 −α ds.

(3.8.2)

dν t Jν (t) = tν Jν−1 (t), dt

and integration by parts, we can have α (t) KR

tR n n 1 = 2 s−1−α+ 2 J n2 +α+1 (s)ds (tR) 2 −α J n2 +α+1 (tR) + (1 + 2α) t R 0 n n 1 = 2 (tR) 2 −α J n2 +α+1 (tR) + (1 + 2α)(tR) 2 −1−α J n2 +α+2 (tR) t R tR n s−2−α+ 2 J n2 +α+2 (s)ds . (3.8.3) + (1 + 2α)(3 + 2α) 0

3.8 (C, 1) means Denote α = α (t)| |KR

253

n−3 2

+ δ + iτ, for δ > 0 and τ ∈ R. When t ≥ R−1 , we have

tR 1 1 1 ≤ Me + + (1 + s)−1−δ ds t (tR)δ (tR)1+δ tR 0 log(1 + tR) 1 2π|τ | 1 + ≤ Me . (3.8.4) δ t (tR) tR 2π|τ | 1

When 0 < t < R−1 , we have α (t)| ≤ M e2π|τ | tn−1 Rn . |KR

(3.8.5)

According to (3.8.4) and (3.8.5), if we ﬁx x and R, then the following integral ∞ α F (α) := Cn (α) fx (t)KR (t)dt 0

is uniformly convergent about α on the compact subset of Reα > n−3 2 . It is α obvious that KR (t) is an analytic function with respect to α. Consequently, F (α) is analytic on Reα > n−3 2 . α Meanwhile, σR (f ; x) is obviously analytic about α. α Since Reα > n−1 2 , F (α) = σR (f ; x), as the analytic function on Reα > n−3 2 , the equation is valid on this whole scale. This ﬁnishes the proof of Lemma 3.8.1.

Theorem 3.8.1 Let f ∈ C(Qn ) and Reα > α (f ; x) − f (x) σR = Cn (α)λn (α)

∞

−2

t 1

n−3 2 .

Then

1 t − f (x) lt + O ω 6 f; fx R R

(3.8.6)

holds uniformly about x, where the deﬁnition of ω 6 can be checked in II of Theorem 3.7.1, and n 2 2 −α−1 Γ n+1 2 λn (α) = (1 + 2α), Γ(α + 3/2) hence,

Γ(α + 1) Γ n+1 2 (1 + 2α). Cn (α)λn (α) = n Γ 2 Γ(α + 3/2)

(3.8.7)

254

C3. Bochner-Riesz means of multiple Fourier series

Proof. By (3.8.1), we have α (f ; x) σR

− f (x) = Cn (α)

∞

α [fx (t) − f (x)]KR (t)dt ∞ 1 t − f (x) Ktα (1)dt. = Cn (α) fx R t 0 0

(3.8.8)

It follows from (3.8.3) that n

n

tKtα (1) = t 2 −α J n2 +α+1 (t) + (1 + 2α)t 2 −α−1 J n2 +α+2 (t) t n + (1 + 2α)(2 + 2α) s 2 −α−2 J n2 +α+2 (s)ds.

(3.8.9)

0

Thus we have 1 1 1 1 α t ω 6 f; − f (x) Kt (1)dt ≤ An,α tn−1 dt fx R t R 0 0 1 ≤ An,α ω . 6 f; R When t ≥ 1, we still write α = n−3 2 + δ + iτ, δ > 0. We have & n 1 2 cos(t − θ) −α +O δ t 2 J n2 +α+1 (t) = δ−1+iτ π t t

and t

n −α−1 2

J n2 +α+2 (t) = O

where θ = π2 n2 + α + 1 + π4 . Consequently, we have t n s 2 −α−2 J n2 +α+2 (s)ds 0 ∞ n −α−2 n 2 = s J 2 +α+2 (s)ds − 0

∞ t

1 tδ

,

n

s 2 −α−2 J n2 +α+2 (s)ds.

Substituting the two equalities n ∞ 2 2 −α−2 Γ n+1 n −α−2 2 s2 J n2 +α+2 (s)ds = Γ α + 52 0 and

∞

s t

n −α−2 2

J

n +α+2 2

1 (s)ds = O δ t

(3.8.10)

3.8 (C, 1) means

255

into (3.8.9) yields that & tKtα (1)

=

n 1 2 cos(t − θ) 2 2 −α−2 Γ n+1 2 + +O δ . 5 π tδ−1+iτ t Γ α+ 2

Therefore, we conclude that ∞ 1 t − f (x) Ktα (1)dt fx Cn (α) R t 1 Γ n+1 n −α−2 2 = Cn (α)(1 + 2α)(3 + 2α)2 2 Γ (α + 5/2) & ∞ 2 ∞ t t −2 × − f (x) t dt + Cn (α) − f (x) fx fx R π 1 R 1 cos(t − θ) 1 . × δ+1+iτ dt + O ω 6 f; t R It is obvious that ∞ cos(t − θ) t 1 − f (x) . fx dt = O ω 6 f; R tδ+1+iτ R 1 We substitute the above into (3.8.8), and immediately obtain (3.8.6). This completes the proof of Theorem 3.8.1.

Remark 3.8.1 Theorem 3.8.1 is also valid for n = 1. When n = 1 and α = 0, we can get the well-known Fej´er approximation estimate. It seems 1 interesting λ1 − 12 = 0, when n = 1 and α = n−2 2 = − 2 . Then we can get the uniform estimate from (3.8.6) 1 −1 . σR 2 (f ; x) − f (x) = O ω f ; R Generally, the result is better than the one in the case α > 0. From Theorem 3.8.1, we can directly obtain the following corollary. Corollary 3.8.1 Let f ∈ C(Qn ) and Reα > n−3 2 . Then we have Γ(α + 1) Γ n+1 α 2 σR (f ) − f C ≤ (1 + 2α) Γ(α + 32 ) Γ( n2 ) 1 1 1 R ω 6 f; dr + O ω 6 f; . × R 0 r R

(3.8.11)

256

C3. Bochner-Riesz means of multiple Fourier series Suppose that ω(δ) is a modulus of continuity. Deﬁne a function class , 6 (f ; δ) ≤ ω(δ) . H ω = f ∈ C(Qn ) : ω

As far as the approximation on H ω is concerned, there is the following theorem (see [J1]). Theorem 3.8.2 There exists constants C1 > C2 > 0, only related to the variable number n, such that n−1 C2 R 1 C1 R 1 2 dr ≤ sup σR (f ) − f ≤ dr (3.8.12) ω ω ω R 0 r R r f ∈H 0 C for R > 0. Proof. It only suﬃces to prove the left half of the above inequality. Let us take it for granted that ω is upper convex. Take ω(|x|) 0 ≤ |x| ≤ π, f0 (x) = ω(π) x ∈ Qn \ B(0, π). Obviously, f0 ∈ H ω . According to (3.8.6), we have

2 R n+1 n−1 Γ 1 1 2 2 − f (x) dr fx σR (f ; x) − f (x) = 2 R 0 r Γ n2 1 +O ω 6 f; . R Thus we have R n−1 1 1 σ 2 (f ; x) − f (x) ≥ 2 1 − f (x) dr − A6 ω f; , fx R R r R 0 (3.8.13) where A > 0. Substituting into the inequality (3.8.13) with f0 and x = 0, it follows that R n−1 1 1 σ 2 (f0 ; 0) − f0 (0) ≥ 2 1 ω dr − Aω . (3.8.14) R R 0 r R If

1 1 1 R 1 ω < dr ω R AR 0 r

3.8 (C, 1) means

257

holds for ﬁxed R > 0, then the above inequality shows that R n−1 1 σ 2 (f0 ; 0) − f0 (0) > 1 dr. ω R R r 0 1 1 1 1 R ω ω ≥ dr R AR 0 r

If

holds for ﬁxed R ≥ 2, then, setting R 1 1 R 1 drei[ 2 ]x1 , ω fR (x) = 3A R 0 r it follows that

R , 0≤r≤ 2 n−1 n−1 Sr 2 (fR ; x) = R

2 R 2 ⎪ ] [ 1 R 1 1 ⎪ i[ R x1 2 ] ⎪ 2 ⎪ ω e , r> dr 1 − 2 . ⎩ 3A R r r 2 0 ⎧ ⎪ ⎪ 0, ⎪ ⎪ ⎨

Therefore, we conclude that n−1

σR2 (fR ; 0) − fR (0) ⎡ ⎤ R 2 n−1 R R

2 1 1 1 = 1 − 22 dr · ⎣ ω dr − 1⎦ 3AR 0 r R [R] r 2 R 1 R 1 1 dr R− −1 ω ≤ 3AR 0 r R 2 1 1 1 R ω <− dr. 12A R 0 r When 0 < δ ≤

1

[ R2 ]

, we have

1 1 R 1 R ω dr ·δ ω(fR ; δ) ≤ 3A R 0 r 2 1 R 1 δ ≤ ω 3 R 2 1 R 1 +1 δ ≤ ω(δ) 3 δR 2 ≤ ω(δ).

258

C3. Bochner-Riesz means of multiple Fourier series

When δ >

1 , [R ] 2

we obtain

2 1 ω(fR ; δ) ≤ 3 A

0

R

2 1 1 dr ≤ ω ≤ ω(δ), ω r 3 R

which implies fR ∈ The above results imply that when R ≥ 2, n−1 1 1 R 1 2 ω sup σR (f ) − f ≥ dr, ω 12A R r f ∈H 0 C Hω.

where the constant A only depends on n, and we take it for granted that A ≥ 1. When 0 < R < 2, if 1 1 1 R 1 ω ≥ dr, ω R A R 0 r then we take f (x) =

ω(1) i2x1 . 4 e

Thus we have

⎧ ω(1) ⎪ ⎨ · 2δ < ω(δ) 0 < δ ≤ 1, 4 ω(f ; δ) ≤ ⎪ ⎩ ω(1) · 2 < ω(δ) δ > 1. 4 Therefore, f ∈ H ω . Meanwhile, when 0 < R < 2, we have that n−1 2 σ R (f ; 0) − f (0) = f (0) 1 ω(1) 4 √ 1 ≥ √ ω(2 nπ) 16 nπ 1 1 R 1 ≥ √ dr. ω r 16 nπ R 0 =

This completes the proof of Theorem 3.8.2. α From Lemma 3.8.1 and the estimate of the kernel KR (t), we can directly get the estimate of the maximal operator α σ∗α (f )(x) := sup |σR (f ; x)| R>0

for Reα >

n−3 2 .

3.9 The saturation problem of the uniform approximation Theorem 3.8.3 If Reα >

n−3 2

259

and f ∈ L(Qn ), then

σ∗α (f )(x) ≤ Cn,α M f (x) holds, where M is the Hardy-Littlewood maximal operator.

3.9

The saturation problem of the uniform approximation

Let f ∈ C(Qn ). Using the non-zero-ordered Bochner-Riesz means to approximate to f , we have that the order of the best approximation of f is R−2 as R → ∞, except for the trivial case when f is a constant. That is to say, the saturation of the uniform approximation is R−2 . It is well-known that if α (f ) − f C = o R−2 SR holds for α = 0, then we have 1 2 lim R [S α (f ; x) − f (x)] e−im·x dx = 0, R→∞ (2π)n Qn R for every m ∈ Zn . That is, α |m|2 lim R2 1 − 2 − 1 αm (f ) = 0 R→∞ R holds. Since α = 0, from the following equality α |m|2 2 1− 2 − 1 = −|m|2 α, lim R R→∞ R we can easily conclude that αm (f ) = 0, if m = 0. consequently, f is a constant. The ﬁrst question which we will discuss is how to identify the saturation α }. In [J1], Jiang gave the characterization of the saturation class class of {SR n−1 if α > 2 . The second question is to investigate the saturation problem of the opα } given by the previous section. In the case of α = n−1 and the erator {σR 2

260

C3. Bochner-Riesz means of multiple Fourier series

dimension n > 1, Jiang [J1] also obtained the characterization of the saturation class. He carried out general discussion about the saturation problem, referring to Lp , 1 ≤ p < ∞ and C. Here, we only discuss in the context of C(Qn ). The same steps can be applied to the situation of Lp . Definition 3.9.1 Suppose that {TR } with R > 0 and R → ∞ is a family of bounded linear operators mapping from C(Qn ) to itself, and ϕ(·) is a positive function monotonically decreasing to zero. We say {TR } is saturated on C(Qn ), with the saturation of ϕ(R), if the following two conditions hold. (1) f − TR (f )C = o(ϕ(R)) ⇐⇒ f = constant, (2) there exists f0 ∈ C(Qn ), f0 is not a constant, and f0 − TR (f0 )C = O ϕ(R) . We denote the collection of all f0 which satisﬁes the condition (2) by F (T, C), and call it the saturation class of {TR }. n Now suppose that, ) and the ﬁxed λR (m) with m ∈ Zn and / for f ∈ C(Qim·x is uniformly convergent with respect R > 0, the series λR (m)Cm (f )e to x, where Cm (f ) is the coeﬃcient of f . Thus it is the Fourier series of the function g in C(Qn ), We denote g by TR (f ). In this way, we have deﬁned a linear operator TR , the multiplier operator, which is identiﬁed by {λR (m)} deﬁned by λR (m)Cm (f )eim·x (3.9.1) TR (f )(x) =

for R > 0. Definition 3.9.2 Suppose that {ψ(m)}m∈Zn is a sequence of numbers. # V C, {ψ(m)} = f ∈ C(Qn ) : f is not a constant, and there exists an

$ essentially bounded function g, such that ψ(m)Cm (f ) = Cm (g), ∀ m ∈ Zn .

Theorem 3.9.1 Suppose that the linear operator TR deﬁned by (3.9.1) is bounded, λR (0) = 1, and there exists a positive function ϕ(R) which is monotonically decreasing to zero, such that 1 − λR (m) = ψ(m) = 0, R→∞ ϕ(R) lim

(3.9.2)

3.9 The saturation problem of the uniform approximation

261

for m ∈ Zk with m = 0. Then the following two conclusions I and II are satisﬁed. (I) {TR } is saturated on C(Qn ), with the saturation of ϕ(R). (II) Let f ∈ C(Qn ). We have the implication relationship for the following three propositions. (a) f is not a constant and there uniformly holds about R im·x λR (m)ψ(m)Cm (f )e = O(1), ∞

(b) f ∈ V (C, {ψ(m)}), (c) f ∈ F (T, C). Then, we have (a) =⇒ (b). If we also know that the norm of the operator satisﬁes TR ≤ M < +∞, (3.9.3) for R > 0, then we have (c) =⇒ (a).

Proof. Choose f ∈ C(Qn ) such that f − TR (f )C = o ϕ(R) . Then for m ∈ Zn with m = 0, we have [f (x) − TR (f )(x)]e−imx dx Cm (f )(1 − λR (m)) = (2π)−n Qn

= o(ϕ(R)), as R → ∞. By the condition (3.9.2), we have Cm (f ) = 0 for m = 0, and thus f = constant. On the other hand, we choose f0 (x) = eix1 and denote e1 = (1, 0, . . . , 0). Then it follows that f0 − TR (f0 )C = |1 − λR (e1 )| = O(ϕ(R)), as R → ∞, which implies that {TR } is saturated, with the saturation of ϕ(R). Next we show the conclusions II. Assume (a) is valid. Denote λR (m)ψ(m)Cm (f )eim·x . FR (x) = Since L∞ is weak ∗ sequentially compact, as the conjugate space of L1 , in the bounded family {FR } of L∞ , there exists a subsequence {FRj }∞ j=1 , with Rj → +∞ as j → ∞, such that FRj → g ∈ L∞ (Qn )

262

C3. Bochner-Riesz means of multiple Fourier series

in the sense of weak ∗ topology. Thus we have −n −im·x −n FRj (x)e dx = (2π) lim (2π) j→∞

Qn

Qn

g(x)e−im·x dx = Cm (g),

for all m ∈ Zn . That is, lim λRj (m)ψ(m)Cm (f ) = Cm (g).

j→∞

It follows from (3.9.2) that λR (m) → 1 with m = 0 as R → ∞ and λR (0) = 1. Hence, we have ψ(m)Cm (f ) = Cm (g), for all m ∈ Zn , which shows (b) is valid. Assume (3.9.3) holds. Let f ∈ F (T, C), that is to say, (c) is valid. Since f − TR (f )C = O ϕ(R) , we have that TR (f − TR (f ))C = O(ϕ(R)). uniformly holds for R > R > 0. That is to say, 1 − λR (m) im·x λR (m) Cm (f )e ≤ M < +∞. ϕ(R) C For the ﬁxed R , the function hR (x) :=

λR (m)

1 − λR (m) Cm (f )eim·x ϕ(R)

is also a bounded family in the norm of C(Qn ), for R > R . By the property of being ∗ weak sequentially compact, there exists a monotone increasing sequence Rj → ∞, such that hRj → h∗R ∈ L∞ (Qn ) in the sense of ∗ weak topology, keeping h∗R ∞ ≤ M for R > 0. Since 1 − λRj (m) = ψ(m), lim j→∞ ϕ(Rj ) we have

λR (m)ψ(m)Cm (f ) = Cm (h∗R ).

3.9 The saturation problem of the uniform approximation That is to say,

263

λR (m)ψ(m)Cm (f )eim·x

is the Fourier series of h∗R in L∞ , while h∗R ∞ ≤ M , for R > 0, which is exactly the conclusion (a). This completes the proof of Theorem 3.9.1. Now we investigate two concrete operators. Firstly, we investigate the Abel-Poisson means. Let f ∈ L(Qn ). Deﬁne the Abel-Poisson means of f as Pε (f ; x) =

e−ε|m| Cm (f )eim·x ,

(3.9.4)

m∈Zn

for ε > 0. Obviously we have that 1 Pε (f ; x) = (2π)n where Pε (y) =

Qn

f (x − y)Pε (y)dy,

(3.9.5)

e−ε|m| eim·y .

m∈Zn

We all know that the the Fourier transform of the function ψ(x) = e−|x| , x ∈ Rn , is Γ n+1 n 2 (3.9.6) ψ(y) = n+1 := P (y) ∈ L(R ). 2 2 [π(1 + |y| )] Thus we conclude that P (y)eix·y dy 1 [(2π)n P (y)]e−ix·y dy = (2π)n Rn = F (2π)n P (x).

ψ(x) =

It follows that

Rn

ψ(εx) = F (2π)n ε−n P (ε−1 y) (x).

264

C3. Bochner-Riesz means of multiple Fourier series

Since the function (2π)n ε−n P (ε−1 y) satisﬁes the condition of the Poisson summation formula, the kernel Pε here has the following expression, y + 2πm n −n (2π) ε P Pε (y) = ε m∈Zn − n+1 n−1 n+1 2 . (3.9.7) = 2n π 2 Γ ε ε2 + |y + 2πm|2 2 We notice that the kernel is positive and y + 2πm n −n Pε (y) = (2π) ε P dy ε Qn Qn m

y P dy = (2π)n ε−n ε n R n P (y)dy = (2π) n

Rn

= (2π) . Therefore, the operator Pε is uniformly bounded on L∞ . For f ∈ L∞ (Qn ), we have 1 Pε (y)dy = f ∞ . (3.9.8) |Pε (f ; x)| ≤ f ∞ (2π)n Qn From (3.9.5), (3.9.6) and (3.9.7), we can obtain that ∞ 2Γ n+1 tn−1 2 fx (tε)dt Pε (f ; x) = √ n nΓ 2 0 (1 + t2 ) n+1 2 and

∞ 2Γ n+1 tn−1 2 Pε (f ; x) − f (x) = √ n n+1 fx (tε) − f (x) dt. nΓ 2 0 (1 + t2 ) 2

for f ∈ L(Qn ). Consequently, we can immediately have that ∞ 2 tn−1 Pε (f ) − f C ≤ n 1 6 (f ; tε)dt. n+1 ω B 2 , 2 0 (1 + t2 ) 2 It follows that

for ε < e−1 .

1 Pε (f ) − f C ≤ An ω f ; ε log ε

(3.9.9)

(3.9.10)

,

(3.9.11)

3.9 The saturation problem of the uniform approximation

265

Theorem 3.9.2 The Abel-Poisson means {Pε } is saturated in C(Qn ), with a saturation of ε, ε → 0+ , and the following three properties are equivalent. / (a) e−ε|m| |m|Cm (f )eim·x = O(1). C

(b) there exists g ∈ L∞ (Qn ), such that |m|Cm (f ) = Cm (g), for m ∈ Zn . (c) f ∈ F (P, C) or f is a constant. Proof. Since lim

ε→0+

1 − e−ε|m| = |m|, ε

for m ∈ Zn , by Theorem 3.9.1, we can have that the saturation of {Pε } is ε. In addition, by (3.9.8) we can see that (3.9.3) is also satisﬁed. And thus by Theorem 3.9.1, it follows that (c) =⇒ (a) and (a) =⇒ (b). Assume that (b) is valid. By (3.9.8) and |Pε (g; x)| ≤ g∞ , it follows that ε ε 1 1 im·x Pη (g; x)dη e dx = Pη (g; x)eim·x dx dη n (2π)n Qn (2π) n 0 0 Q ε e−η|m| Cm (g)dη, (3.9.12) = 0

for m ∈ Zn . On the other hand, we have

1 −imx −ε|m| [f (x) − P (f ; x)]e dx = 1 − e Cm (f ) ε (2π)n Qn ε e−η|m| dη, = Cm (f )|m|

(3.9.13)

0

for m ∈ Zn . By the condition that |m|Cm (f ) = Cm (g), (3.9.12) and (3.9.13), we can deduce that ε f (x) − Pε (f ; x) =

for ε > 0.

0

Pη (g; x)dη,

266

C3. Bochner-Riesz means of multiple Fourier series It follows that f − Pε (f )C ≤ g∞ ε,

which follows (c) is valid. This ﬁnishes the proof of Theorem 3.9.2. Another example is the Gauss-Weierstrass means. Deﬁne − |m|2 WR (f ; x) = e R2 Cm (f )eim·x ,

(3.9.14)

for f ∈ L(Qn ). By the method similar as the Abel-Poisson means, we can obtain the expression as follows, ∞ 1 2 t 1 − t n−1 dt WR (f ; x) = n−1 n e 4 t fx R 2 Γ 2 0 √ ∞ 1 2 t −t n −1 2 e t fx dt. = Γ( n2 ) 0 R

(3.9.15)

Theorem 3.9.3 The Gauss-Weierstrass means {WR } is saturated in C(Qn ), with the saturation of R−2 , and the following three are equivalent. / − |M |22 2 im·x = O(1), and f is not a constant, |R| |m| C (f )e (a) m e C , 2 - (b) f ∈ V C, |m| , (c) f ∈ F (W, C). For the application later on, we can deﬁne the operator Vεl for ε > 0 and l = 1, 2 as follows Vεl (f ; x) =

n+1 e−(ε|m|)l 1 − (ε|m|)l Cm (f )eim·x .

Lemma 3.9.1 Vεl is an uniformly bounded linear operator mapping from C(Qn ) to itself. That is to say, for any ε > 0, the operator norm satisﬁes Vεl ≤ M < +∞.

(3.9.16)

3.9 The saturation problem of the uniform approximation

267

Proof. By the Poisson summation formula, it suﬃces to show that

l F e−|x| (1 − |x|l )n+1 (y) ∈ L(Rn ). However, by the results about the the Fourier transform of the radial function, we merely need to prove the unitary-variable function ∞

n+1 n n l e−t 1 − tl t 2 J n2 −1 (ts)dt ∈ L(R+ ) . s2 0

Denote ϕl (s) =

∞

0

l

n

e−t (1 − tl )n+1 t 2 J n2 −1 (ts)dt,

for s ≥ 0. It suﬃces to show that there exists a positive number η, such that 1 (3.9.17) ϕ(s) = O − n/2+1+η , s as s → ∞. We ﬁrst consider the case l = 1. Denote ∞ n Ij (s) = e−t t 2 +j J n2 −1 (ts)dt, 0

for j = 0, 1, . . . , n. Directly using a formula in [Ba2], we can have that

n − n+1 n 2 = O s−( 2 +2) , I0 (s) = Cn s 2 −1 1 + s2 as s → +∞. We can get that Γ(n + j) −ν Pν+1+j Ij (s) = 1 n +1+j ( ) 2 (1 + s ) 2 2

1

; 1 + y2

,

−ν is the Legendre function. for j = 1, 2, . . . , n, where ν = n2 − 1 and Pν+1+j Thus we have

n Ij (s) = O s−( 2 +1+j) ,

for j = 1, 2, . . . , n. Since ϕ1 (s) =

n+1

j C1+n (−1)j Ij (s),

j=0

we have

n ϕ1 (s) = O s− 2 −2

268

C3. Bochner-Riesz means of multiple Fourier series

as s → +∞. Now we consider the case of l = 2. Denote n+1 2 h0 (t) = e−t 1 − t2 and hj+1 (t) = for j = 0, 1, 2, . . .. Obviously, we have and

hj (t) t

,

hj (t) ∈ C ∞ [0, ∞) 2

|hj (t)| ≤ Mn,j e−t

1 + t2n ,

for j = 0, 1, 2, . . .. Using the formula d ν+1 [t Jν+1 (ts)] = tν+1 Jν (ts)s, dt we can obtain that ϕ2 (s) = (−s)−j

∞ 0

n

hj (t)t 2 +j J n2 +j−1(ts)dt,

for j = 0, 1, 2, . . .. Consequently, it follows that ϕ2 (s) = O(s−j ) holds for every natural number j. Of course, we have

n ϕ2 (s) = O s− 2 −2 as → +∞, which gets (3.9.17) and thus completes the proof of Lemma 3.9.1.

Lemma 3.9.2 Deﬁne a linear operator Hεl with ε > 0 and l = 1, 2 on L∞ (Qn ) as "n+1 ! 1 l −(ε|m|)l e Cm (f )eim·x 1 − 1 − (ε|m|)l Hε (f ; x) = (ε|m|)l m=0

+ (n + 1)C0 (f ).

(3.9.18)

3.9 The saturation problem of the uniform approximation

269

Then, the norm of Hεl is uniformly bounded from L∞ (Qn ) to C(Qn ), that is, l (3.9.19) Hε (f ) ≤ M f ∞ , C

for ε > 0, where the positive number M is independent of f , l and ε. Proof. By the Poisson summation formula, it suﬃces to show that "n+1 ! l 1 F e−|x| 1 − 1 − |x|l (·) ∈ L(Rn ). |x|l Hence, we merely need to prove that ∞ l n+1 n n l 1 − (1 − t ) ψl (s) := e−t t 2 J n2 −1 (ts)dts 2 ∈ L(R+ ). l t 0 It follows from Lemma 3.9.1 that n

ψ1 (s) = s 2

n

j+1 Cn+1 (−1)j Ij (s).

j=0

Therefore, we have

ψ1 (s) = O s−2 ,

as s → +∞. This follows ψ1 ∈ L(R+ ). Similar to the estimate of ϕ2 in Lemma 3.9.1, we can get that ψ2 (s) = O(s−j ) for all natural number j as s → +∞. And therefore, ψ2 ∈ L(R+ ). This completes the proof of Lemma 3.9.2. Theorem 3.9.4 A family of operators {Vεl } with l = 1, 2 and ε → 0 is saturated on C, with a saturation of εl . The saturation class is $

# F (V l , C) = V C, |m|l . Proof. It is obvious that

n+1 l 1 − e−(ε|m|) 1 − (ε|m|)l lim = (n + 2)|m|l . ε→0 εl

Therefore, according to Theorem 3.9.1, we have the saturation of {Vεl } is εl . By Lemma 3.9.1, we know that the condition (3.9.3) is valid (see (3.9.16)). Then by Theorem 3.9.1, we have

# $ F (V l , C) ⊂ V C, |m|l .

270

C3. Bochner-Riesz means of multiple Fourier series

Now suppose that f ∈ V (C, {|m|}) in the case of l = 1. By Theorem 3.9.2, we have f − Pε (f )C = O(ε) as ε → 0. We conclude that

, e−ε|m| 1 − (1 − ε|m|)n+1 Cm (f )eim·x , - 1 =ε e−ε|m| 1 − (1 − ε|m|)n+1 Cm (g)eim·x , ε|m|

Pε (f ; x) − Vεl (f ; x) =

m=0

where g ∈ L∞ (Qn ). By the deﬁnition of Hεl , we can write Pε (f ; x) − Vε1 (f ; x) = ε Hε1 (g; x) − (n + 1)C0 (g) . By Lemma 3.9.2, we can get Pε (f ) − V 1 (f ) ≤ εM g∞ , ε C

(3.9.20)

which follows f − Vε1 (f ) ≤ f − Pε (f )C + Pε (f ) − Vε1 (f ) = O(ε). C Thus we have that f ∈ F (V 1 , C), And this implies F (V 1 , C) = V (C, {|m|}). In the case of l = 2, we need the help of Theorem 3.9.3, and the other steps are the same as that of l = 1. This completes the proof of Theorem 3.9.4. 1l : Lemma 3.9.3 Let 0 < ε ≤ 1 and l = 1, 2. Deﬁne an operator V ε

n+1 1 l 1l (f ; x) = e−|εm| 1 − |εm|l Cm (f )eim·x . V ε |εm|l |m|<1/ε

1l is an uniformly bounded linear operator mapping from L∞ (Qn ) to Then V ε n C(Q ). That is, 1l (3.9.21) Vε (f ) ≤ M f ∞ , C

for every ε ∈ (0, 1].

3.9 The saturation problem of the uniform approximation

271

Proof. It suﬃces to show that ∞

n+1 1 n n l e−t 1 − tl t 2 J n2 −1 (ts)dt ∈ L(R+ ). ϕ6l (s) := s 2 l t 1 Denote l

h0 (t) = e−t

n+1 1 1 − tl tl

and

hj (t)

hj+1 (t) =

t

,

for j = 0, 1 . . . , n. Obviously we have hj ∈ C ∞ [1, ∞), hj (1) = 0 and l

|hj (t)| ≤ Mn,j e−t (1 + t2n ), for j = 0, 1, . . . , n + 1. After executing the partial integration for n + 1 times, we can have that ∞ n n −(n+1) 6 l (−1)n+1 hn+1 (t)t 2 +n+1 J n2 +n (ts)dt. ϕ (s) = s 2 1

This implies that ϕ6l (s) ∈ L(R+ ). This is the end of the proof. n−1

Now we begin to consider the (C, 1) means σR 2 . According to the deﬁnition, n−1 2

σR

1 (f ; x) = R =

R 0

|m|
|m|
|m|2 1− 2 r

im·x

Cm (f )e

1 R

n−1 2

R

0

1 R |m|2 1− 2 = R |m| r

Cm (f )eim·x dr

|m|2 1− 2 r

n−1 2

n−1 2 χ(|m|, R)(r)dr

drCm (f )eim·x ,

|m|
we have

⎧ n−1 ⎪ 1 R |m|2 2 ⎪ ⎪ ⎨ 1− 2 dr, |m| < R, r λR (m) = R |m| ⎪ ⎪ ⎪ ⎩ 0, |m| ≥ R.

(3.9.22)

272

C3. Bochner-Riesz means of multiple Fourier series If n = 1, then we have λR (m) = 1 −

for |m| < R, and

|m| , R

lim R 1 − λR (m) = |m|.

R→∞

If n > 1, then we can compute it directly by the method of integration by parts. When |m| < R, we have λR (m) = =

|m|2 1− 2 R |m|2 1− 2 R

n−1 2 n−1 2

n−3 |m| 1 (1 − t2 ) 2 dt R |m| R 1 n−3 1 n − 1 |m| − (1 − t) 2 t− 2 dt. 2 |m| 2 R 2 − (n − 1)

R

Thus it follows that 1 n−3 1 n−1 |m| (1 − t) 2 · t− 2 dt lim R(1 − λR (m)) = R→∞ 2 n+1 0 1 |m|Γ 2 Γ 2 , = Γ n2 for every n ∈ N.

n−1 From the above results and Theorem 3.9.1, we can think that σR 2

is saturated in C(Qn ) with a saturation of R−1 as R → ∞. We denote its saturation class as F (σ, C). Lemma 3.9.4 Let ξ(t) = t

−1

1−t

1

2

(1 − r )

n−1 2

r

−2

t

for 0 < t ≤ 1,

η(t) = e−t (1 − t)n+1

for 0 ≤ t ≤ 1, and ϕ(t) =

ξ(t)η(t), 0,

0 ≤ t < 1, t ≥ 1.

dr

3.9 The saturation problem of the uniform approximation

273

1l as, Deﬁne an operator H ε

1l (f ; x) = H ε

ϕ(ε|m|)Cm (f )eim·x ,

1l is an uniformly bounded linear operator mapping from for ε > 0. Then, H ε ∞ n n L (Q ) to C(Q ). That is, 1l (3.9.23) Hε (f ) ≤ M f ∞ , C

for ε > 0. Proof. It suﬃces to show that ∞ n n ϕ(t)t 2 J n2 −1 (ts)dt ∈ L(R+ ). h(s) := s 2 0

Consider h(s) = s

n 2

1

0

Since

ξ (t) = t

−2

n

ξ(t) η(t)t 2 J n2 −1 (ts)dt.

1−t

2

n−1 2

∞ −1 = aj t2j , j=0

∞

for 0 ≤ t < 1, we can see that ξ ∈ C [0, 1). On the other hand, since ϕ

(k)

(t) =

k

Ckj ξ (j) (t)η (k−j) (t),

(3.9.24)

j=0

for k ∈ Z+ , and η ∈ C ∞ [0, 1), we have ϕ ∈ C ∞ [0, 1). It is obvious that when k = 0, 1, . . . , n, we have ϕ(k) (1) = lim ϕ(k) (t) = 0 t→1−

and

ϕ(n+1) (1) = lim ϕ(n+1) (t). t→1−

Hence, we have ϕ ∈ C n+1 [0, 1] and ϕ(k) (1) = 0 for k = 0, 1, . . . , n. Deﬁne h0 (t) = ϕ(t) and hj+1 (t) = hj (t)t−1 for j = 0, 1, . . . , n. Obviously, when j = 1, 2, . . . , n, we have hj (t) =

ϕ(j) (t) ϕ(j−1) (t) ϕ (t) + bj,1 + · · · + bj,j−1 2j−1 . j j+1 t t t

274

C3. Bochner-Riesz means of multiple Fourier series

Hence, it follows that

hj (t) = O t−2j ,

as t → 0+ , and hj (1) = 0, for j = 0, 1, . . . , n. Since n t 2 +j J n2 +j (ts) = O tn+2j , as s > 0, we can have hj (t) t

n +j 2

1 J n2 +j (ts) = 0, 0

for j = 0, 1, . . . , n. Thus by the method of integration by parts for n + 1 times, we can deduce that n (−1)n+1 1 h(s) = hn+1 (t) t 2 +n+1 J n2 +n (ts)dt n +1 s2 0 1 =O , n s 2 +1 as s → +∞. Therefore, we have h ∈ L(R+ ), and ﬁnish the proof of the lemma. Lemma 3.9.5 Deﬁne an operator 1R as n−1

1R (f ; x) = V 11 (f ; x) − σR2 R

If f ∈ V (C, {|m|}), then we have 1 R (f ) = O C

V 11 (f ); x . R

(3.9.25)

1 . R

Proof. By the deﬁnition, we have |m| |m| n+1 1 − R e Cm (f )eim·x R (f ; x) = 1− R |m| |m| n+1 − R 1− − λR (m)e Cm (f )eim·x , R |m|
where λR (m) is given by (3.9.22). It follows from direct computation that |m| 1 1 − r2 r −2 dr, λR (m) = |m| R R

3.9 The saturation problem of the uniform approximation

275

for 0 < |m| < R, and λR (0) = 1. Since f ∈ V (C, {|m|}), then there exists g ∈ L∞ (Qn ), such that |m|Cm (f ) = Cm (g), for m ∈ Zn . We conclude that 1R (f ; x)

|m| n+1 R 1 − |m| Cm (g)eim·x = e R 1− R R |m| |m|>R − |m| 1 |m| n+1 R R + Cm (g)eim·x 1 − λR (m) e 1− R R |m| 0<|m|
for R ≥ 1. By Lemma 3.9.3 and 3.9.4, we can obtain that 1 1R (f )C = O . R This completes the proof of Lemma 3.9.5. n−1 Theorem 3.9.5 The saturation of the family of operators σR2

is R−1 ,

and the saturation class is F (σ, C) = V (C, {|m|}). Proof. We have calculated the saturation in the previous. Concerned about the saturation class, by Theorem 3.9.1, we can notice that F (σ, C) ⊂ V C, {|m|}m∈Zn . Next we will prove the reverse inclusion relationship. Assume f ∈ V (C, {|m|}). We conclude that n−1 n−1 f − σR 2 (f ) = f − V 11 (f ) + V 11 (f ) − σR 2 V 11 (f ) R R R n−1 n−1 + σR2 V 11 (f ) − σR2 (f ) R n−1 1 = f − V 1 (f ) + 1R (f ) + σR2 V 11 (f ) − f . R

R

276

C3. Bochner-Riesz means of multiple Fourier series

It follows from Theorem 3.9.4 that 1 1 . f − V 1 (f ) = O R R C n−1

By the uniform boundedness of σR 2 , we can have

n−1 1 1 1 σ 2 V 1 (f ) − f = O =O . V 1 (f ) − f R R R R C C By Lemma 3.9.5, we immediately have 1 R (f ) = O C

1 . R

In another words, f ∈ F (σ, C), which ﬁnishes the proof of the theorem.

Now let us consider the saturation class of the Bochner-Riesz means with α > n−1 2 , that is the case over the critical index.

α {SR }

Lemma 3.9.6 Let α >

n−1 2 ,

α ξ(t) = t−2 1 − 1 − t2

for 0 < t ≤ 1, and

ξ(0) = ξ(0+ ) = a.

Set

2

η(t) = e−t

for 0 ≤ t ≤ 1, and

1 − t2

n+1

⎧ ⎨ ξ(t)η(t), 0 ≤ t < 1, ϕ(t) =

⎩

12 as Deﬁne the operator H ε 12 (f ; x) = H ε

t ≥ 1.

0,

ϕ(ε|m|)Cm (f )eim·x ,

12 is a uniformly bounded linear operator mapping from for ε > 0. Then H ε L∞ (Qn ) to C(Qn ). That is, 12 (3.9.26) Hε (f ) ≤ M f ∞ , C

for ε > 0.

3.9 The saturation problem of the uniform approximation

277

The proof of lemma 3.9.6 is the same as that of Lemma 3.9.4 and thus we omit the proof here. Lemma 3.9.7 Deﬁne an operator 2R as

α V 12 (f ); x . 2R (f ; x) = V 12 (f ; x) − SR R

R

- , If f ∈ V C, |m|2 , then we have 2 (f ) = O R C

1 R2

.

The steps of the proof are the same as that of Lemma 3.9.5 with the help of Lemmas 3.9.3 and 3.9.6, so we skip without the proof here. Theorem 3.9.6 Let α > n−1 2 . The saturation of the family of the operators α } is R−2 , and the saturation class is V (C, {|m|2 }). {SR We have to point out that when α > n−1 2 , Theorem 3.9.6 gave a characα }. According to Theorem 3.9.1, this terization of the saturation class of {SR conclusion can be written as, f ∈ F (S α , C) if and only if f is not a constant and α 2 |m| 2 im·x 1 − |m| C (f )e (3.9.27) m = O(1), R2 |m|
C

for R > 0. This means the condition (3.9.27) can be considered as a characterization of the saturation class of the Riesz means with a positive order α > 0. The above discussion is also suitable for the generalized Riesz means l,α SR (f ; x) for l ∈ N (see [J1]). The deﬁnition of the generalized Riesz means is as follows, α |m|l l,α 1− l Cm (f )eim·x . SR (f ; x) = R |m|
Cheng and Chen [ChC1] discussed about the uniform approximation probl,α with l ∈ N. Wang [Wa5] went further to carry out more general lem of SR discussions about the operator. Finally, we will characterize the constructive property of the functions in F (S α , C) with α > n−1 2 .

278

C3. Bochner-Riesz means of multiple Fourier series

Theorem 3.9.7 Let f ∈ C(Qn ), f is not a constant and α > n−1 2 . Then f ∈ F (S α , C) if and only if either of the following two conditions holds. α (f ) = O(1), for R ≥ 0; (a) SR C d d (b) fx (t) = sup fx (t) = O(t), for t > 0. dt x∈Qn dt C Proof. In fact, the condition (a) is exactly (3.9.27). Obviously it is the necessary and suﬃcient condition for f ∈ F (S α , C) with α > n−1 2 . Now let f ∈ F (S α , C) ﬁrstly. It follows from the property of uniform convergence that f (x) = S1α (f ; x) +

∞ S2αj+1 (f ; x) − S2αj (f ; x)

(3.9.28)

j=0

uniformly holds. By the well-known Bernstein inequality, we have ∂ α α j+1 α S2j+1 (f ) − S2αj (f ) C ∂xk S2j+1 (f ; x) − S2j (f ; x) ≤ 2 C ≤ 2j+1 S2αj+1 (f ) − f C + f − S2αj (f ) C = M 2−j , for j ∈ Z+ and k = 1, 2, . . . , n. Then the series of the right side of (3.9.28) are uniformly convergent. Thus we have ∞ " ∂f (x) ∂ α ∂ ! α = S1 (f ; x) + (3.9.29) S2j+1 (f ; x) − S2αj (f ; x) ∂xk ∂xk ∂xk j=0

uniformly with respect to x. This implies f ∈ C 1(Qn ). Thus we have Γ(n/2) d fx (t) = f (x + tξ)dσ(ξ) 2π n/2 dt |ξ|=1 n ∂f (u) Γ(n/2) ξj dσ(ξ). = ∂uj 2π n/2 |ξ|=1 j=1

Set

IR (t) =

n ∂f (u) |ξ|=1 j=1

∂uj

u=x+tξ

α ∂SR (f ; u) − ∂uj

u=x+tξ

3.9 The saturation problem of the uniform approximation and

JR (t) =

n α ∂SR (f ; u) ∂uj

|ξ|=1 j=1

279

ξj dσ(ξ).

u=x+tξ

Making use of the Green formula, we have α JR (t) = SR (f ; x + tξ)tdξ. |ξ|<1

It follows from (a) that |JR (t)| ≤ M t, for t ≥ 0. By (3.9.29), we can obtain that

fx (t) ≤ lim sup Γ(n/2) I2j (t) + J2j (t) 2π n/2 j→∞ ≤ Cn lim sup |J2j (t)| j→∞

≤ M t, for any x. this implies that (b) holds. On the contrary, we assume that (b) holds. It follows from the Bochner formula that ∞ n t α − f (x) J n2 +α (t)t 2 −α−1 dt, SR (f ; x) − f (x) = C fx R 0 and the mean value formula 2 t t t θ − f (x) = fx · =O , fx R R R R2 and setting α =

n 2

+ 4, we have that

α |SR (f ; x)

− f (x)| ≤ M

∞ 0

|Jn+4 (t)|t−3 dt

1 R2

1 ≤ M 2. R n

Hence, we have f ∈ F (S 2 +4 , C). By Theorem 3.9.6, for every α > is F (S α , C). This completes the proof of Theorem 3.9.7.

n−1 2 ,

so

280

C3. Bochner-Riesz means of multiple Fourier series

3.10

Strong summation

The strong summation problem is to ﬁnd certain conditions for the limit q 1 R n−1 2 Sr (f ; x) − f (x) dr = 0 (3.10.1) lim R→∞ R 0 to hold, where q > 0. When n > 1, Bochner and Chandrasekharan [BoC1] initiated to investigate the problem. They obtained that the validity of (3.10.1) is a local property of the functions, that is to say, if f vanishes at the neighborhood of x0 , then (3.10.1) holds for q = 2 at the point of x0 . Besides of the above, they also characterized a suﬃcient condition for (3.10.1) to hold in the case q = 2, that is to say, if f ∈ L2 (B(x0 , η)) for some η > 0 and t |fx0 (τ ) − f (x0 )|2 dτ = o(t), (3.10.2) 0

as t → 0, then the equality (3.10.1) holds for q = 2 at x0 . In 1958, using the method of interpolation of operators, Stein proved that when 1 n−1 2 −1 − 1− δ> 2 p p and 1 0 0 is of type (p, p). And thus we can deduce that 2 1 R δ lim Su (f ; x) − f (x) du = 0 R→∞ R 0 holds almost everywhere x ∈ Qn for f ∈ Lp (Qn ). The case of L1 is especially important. However, the related conclusion has not been proven yet. In 1985, Lu [Lu7] achieved breakthrough result about the strong summation problem. Theorem 3.10.1 Let f ∈ L(Qn ) with n ≥ 2. If there exists δ > 0, such that f vanishes in B(x0 , δ), then we have q 1 R n−1 2 S lim (f ; x ) r 0 dr = 0, R→∞ R 0 for every q > 0.

3.10 Strong summation

281

Proof. Choose β ∈ 0, 12 . We have q q 1 R n−1 1 R n−1 +β 2 2 dr, (f ; x ) dx = lim (f ; x ) S S r r 0 0 R 1 β→0+ R 1 for R > 1. Consequently, it suﬃces to show that q R n−1 +β 2 Sr sup (f ; x0 ) dx = o(R), β∈(0, 12 ) 1

(3.10.3)

for R → ∞. By the Bochner formula n+1 ∞ 1 J 1 (t) n−1 n− 2 +β t +β 2 +β Γ 2 +β 2 (f ; x0 ) = f x0 dt. SR 1 n Γ( 2 ) R Rδ t 2 +β We might as well take it for granted that x0 = 0. Denote 1 n+1 2 +β Γ 2 +β . Cβ = Γ( n2 ) Obviously, 0 < Cβ ≤ n2 , for β ∈ 0, 12 . By the asymptotic formula &

2 π π 1 cos t − ν − +O 3 , Jν (t) = πt 2 4 t2

where the remainder O 13 satisﬁes t2

O 13 ≤ Mn · 13 t2 t2 and ν = n − 12 + β with β ∈ 0, 12 , we can get that & ∞ n−1 t cos t − π2 (n + β) 2 +β 2 Cβ (f ; 0) = f0 dt + Irβ , Sr 1+β π r t rδ where

β ≤ M I r n

f0 t 1 dt ≤ Mn 1 . r t2 r

∞ rδ

Thus, it suﬃces to prove q R ∞ π cos t − (n + β) t 2 sup f0 dt dr = o(R). 1+β r t 1 rδ β∈(0, 2 ] 1

(3.10.4)

282

C3. Bochner-Riesz means of multiple Fourier series

π cos t − (n + β) = aβ cos t − bβ sin t, 2 with |aβ | + |bβ | ≤ 2, we merely need to prove the following two equations, q R ∞ cos rt f0 (t) 1+β β dt dr = o(R) (3.10.5) sup t r δ β∈(0, 12 ) 1 Since

and

sup β∈(0, 12 ) 1

R ∞ δ

sin rt q f0 (t) 1+β β dt dr = o(R). t r

(3.10.6)

Next we only prove (3.10.5), for the proof of either equation above is quite the same. In addition, when 0 < q < q , it follows from the H¨ older’s inequality that R qq R 1− q q q |h(t)| dt ≤ |h(t)| dt (R − 1) q . 1

1

To prove (3.10.5) is valid for every q > 0, we merely need to prove its validity for every index in the form of q = 2N for N = 1, 2, · · · . To this end, we assume q = 2N . Without loss of generality, we may assume x0 = 0. Since the integral on the left side of (3.10.5) is R ∞ cos rt q f0 (t) 1+β β dt dr t r 1 δ R ∞ ∞ f0 (t1 ) · · · f0 (tq ) 1 cos rt dr dt1 · · · dtq , ··· · · · cos rt = 1 q qβ (t1 · · · tq )1+β δ δ 1 r the second mean value theorem implies that ξ R 1 (cos rt · · · cos rt )dr = (cos rt1 · · · cos rtq )dr 1 q qβ 1 r 1 := a(ξ; t1 , . . . , tq ), with ξ ∈ (1, R). We line up all the q-dimensional vectors whose coordinate is either 1 or q −1 to denote by {ek }2k=1 , where ek = (ek,1 , . . . , ek,q ), and ek,j = 1 or − 1 for j = 1, 2, . . . , q. Denote by t = (t1 , . . . , tq ) and (ek · t) = ek1 t1 + · · · + ekq tq . Then, we have 2q −q cos r(ek · t). cos rt1 · · · cos rtq = 2 k=1

3.10 Strong summation

283

Thus it follows that q

−q

a(ξ; t1 , . . . , tq ) = 2

2

1 sin ξ(ek · t) − sin(ek · t) . (ek · t)

k=1

We have that R 2q

1 MR (|ek · t|) −q cos rt1 · · · cos rtq dr ≤ 2 sup , qβ r |ek · t| 1 β∈(0, 2 ) 1 k=1

(3.10.7)

where MR (|ek · t|) ≤ min(2, 2R|ek · t|). Hence, in order to prove (3.10.5), it suﬃces to show that ∞ ∞ |f0 (t1 ) · · · f0 (tq )| MR (|ek · t|) ··· dt1 · · · dtq = o(R) t 1 · · · tq |ek · t| δ δ

(3.10.8)

(3.10.9)

holds for every k ∈ (1, . . . , 2q ). Denote the left side of the above equation by IR . It follows ∞

IR =

(n1 +1)δ

(nq +1)δ

···

n1 ,...,nq =1 n1 δ

nq δ

|f0 (t1 ) · · · f0 (tq )| t1 · · · tq

MR (|ek · t|) dt1 · · · dtq |ek · t| (n1 +1)δ (nq +1)δ ∞ 1 δ−q ··· |f0 (t1 ) · · · f0 (tq )| n 1 · · · nq n1 δ nq δ

× ≤

n1 ,...,nq =1

×

MR (|ek · t|) dt1 · · · dtq . |ek · t|

Set τn1 ···nq (R) =

(n1 +1)δ n1 δ

···

(nq +1)δ

|f0 (t1 ) · · · f0 (tq )| ·

nq δ

We might as well take it for granted that a

q−a

G HI J G HI J ek = (1, . . . , 1, −1, . . . , −1). Let us ﬁrst prove that

MR (|ek · t|) dt1 · · · dtq . |ek · t| (3.10.10)

284

C3. Bochner-Riesz means of multiple Fourier series I

We uniformly have τn1 ···nq (R) = o(R),

as R → ∞, for all (n1 · · · nq ) ∈ Nq ; II When |n1 + · · · + nα − (nα+1 + · · · + nq )| > 2qδ, −1 τn1 ···nq (R) ≤ C n1 + · · · + nα − (nα+1 + · · · + nq ) holds. Making changes of variables as follows t1 = s1 − s2 − · · · − sα + sα+1 + · · · + sq , tj = sj ,

j = 2, . . . , q,

we can get that τn1 ···nq (R) =

(n2 +1)δ

···

n2 δ

(nq +1)δ

|f0 (s2 ) · · · f0 (sq )|

nq δ

× hn1 (s2 , . . . , sq ; R)ds2 · · · dsq , where hn1 (s2 , . . . , sq ; R) =

|f0 (s1 − · · · − sα + · · · + sq )| En1

MR (|s1 |) ds1 . |s1 |

Set # $ En1 = s1 : n1 δ < s1 − s2 − · · · − sα + sα+1 + · · · + sq < (n1 + 1)δ . If we denote u = s1 + s2 + · · · + sα − (sα+1 + · · · + sq ), then we have

, En1 = s1 : n1 δ < s1 − u < (n1 + 1)δ .

Thus, we have hn1 (s2 , . . . , sq ; R) =

(n1 +1)δ+u n1 δ+u

|f0 (s1 − u)|

MR (|s1 |) ds1 . |s1 |

3.10 Strong summation

285

Since f ∈ L(Qn ), associate to ε > 0, there exists ξ > 0 such that for any set E ⊂ Rn , if |E| < ξ, then |f (x)|dx < ε E

holds. Geometrically, there exists η0 > 0, such that for every m ∈ Zn and every γ ≥ 0, the measure of the set Em,γ = {x : x ∈ Qn + 2πm, γ < |x| < γ + η0 } is always smaller than ξ. It follows that |f (x)|dx < ε.

(3.10.11)

Em,γ

Now we set η0 < 12 δ. We will prove that, for any γ ≥ 0 and any η > 0, if η ≤ η0 , then we have γ+η |f0 (t)|dt ≤ Cε, (3.10.12) γ

where C is independedent of ε, γ and η. Since f vanishes in B(0, δ), when γ < 12 δ, then the left side of (3.10.12) became 0. Assume γ ≥ 12 δ, meanwhile, 2γ > η0 + γ. Thus we conclude that γ+η γ+η0 1 |f0 (t)|dt ≤ n−1 |f0 (t)|tn−1 dt γ γ γ 2n−1 ≤ |f (x)|dx (γ + η0 )n−1 γ<|x|<γ+η0 2n−1 |f (x)|dx. = (γ + η)n−1 m Em,γ Since the order of the number of the integer points falling in the spherical shell γ < |x| < γ +η0 and taking the form of 2πm with m ∈ Zn is the same as that of (γ + η0 )n−1 , we say that the number of m which guarantees Em,δ = ∅ shares the same order with (γ + η0 )n−1 . Hence, by (3.10.11), we can obtain (3.10.12). In this way, combining (3.10.12) with (3.10.8) yields that MR (|s1 |) ds1 < CRε. |f0 (s1 − u)| (3.10.13) |s1 | |s1 |< 1 η0 2

286

C3. Bochner-Riesz means of multiple Fourier series

Meanwhile, we also have u+n1 δ<s1 <(n1 +1)δ+u,|s1 |> 21 η0

|f0 (s1 − u)|

4 (n1 +1)δ ≤ |f0 (t)|dt η0 n1 δ C ≤ f 1 . η0

MR (|s1 |) ds1 |s1 |

(3.10.14)

Combining the two inequalities (3.10.13) with (3.10.14), we have that hn1 (s2 , . . . , sq ; R) = o(R) uniformly holds for all the parameters. Consequently, the conclusion I holds. n1 + · · · + nα − (nα+1 + · · · + nq ) > 2qδ,

When for

(t1 , . . . , tq ) ∈

q

[nj δ, (nj + 1)δ] ,

j=1

we must have t1 + · · · + tα − (tα+1 + · · · + tq ) > 1 n1 + · · · + nα − (nα+1 + · · · + nq ). 2 Therefore, the equality (3.10.10) directly implies the validity of the conclusion II. The remainder is to estimate 1 δ−q τn ···n (R). (3.10.15) IR ≤ n 1 · · · nq 1 q We decompose/ the / summation on the right side of the inequality 3.10.15 into two parts 1 , 2 as 1

and

2

=

δ−q

1 τn ···n (R) n 1 · · · nq 1 q

δ −q

1 τn ···n (R). n 1 · · · nq 1 q

|n1 +···+nα −(nα+1 +···+nq )|≤2qδ

=

|n1 +···+nα −(nα+1 +···+nq )|>2qδ

3.10 Strong summation

287

Applying the estimate τn1 ···nq (R) = o(R) in the conclusion I into can easily get that 1

= o(R) ·

μα−2 −1

μ−1 ∞

···

μ=α μ1 =α−1

1,

we

μα−1 =1 ν≥q−α,|ν−μ|≤2qδ

νq−α−2 −1

ν=1

/

1 1 1 ··· · ··· μ − μ1 μ 1 − μ2 μα−2 − μα−1 ν1 =q−α−1 νq−α−1 =1 1 1 1 1 ··· . ν − ν1 ν1 − ν2 νq−α−2 − νq−α−1 νq−α−1

1 μα−1

When α > 1 and μ ≥ α, we have μα−2 =1

μ−1

···

μ1 =α−1

μα−1

1 logα−1 (1 + μ) 1 1 . ··· ≤C μ − μ1 μ 1 − μ2 μα−2 − μα−1 μ =1 (3.10.16)

Thus we have that 1

= o(R)

= o(R)

∞ logα−1 (1 + μ) μ μ=α ∞

log

ν≥q−α,|ν−μ|≤2qδ

logq−α−1 (1 + ν) ν

q−2

μ=1

(1 + μ) μ2

= o(R).

(3.10.17)

On the other hand, using the conclusion II implies that 2

≤C

|n1 +···+nα −(nα+1 +···+nq )|>2qδ

× =C

1 |n1 + · · · + nα − (nα+1 + · · · + nq )|

μ−1 ∞ μ=α μ1 =α−1

1 n1 · · · n q

μα−2 =1

···

ν−1

μα−1 =1 ν≥q−α,|ν−μ|>2qδ ν1 =q−α−1

νq−α−2 −1

···

νq−α−1 =1

1 1 1 ··· μ − μ1 μα−2 − μα−1 μα−1 1 1 1 1 . ··· × ν − ν1 νq−α−2 − νq−α−1 νq−α−1 |μ − ν|

288

C3. Bochner-Riesz means of multiple Fourier series

Applying (3.10.16) into the above inequality, we can get that 2

∞ logα−1 (1 + μ) μ μ=α

1 logq−α−1 (1 + ν) |ν − μ| ν ν≥q−α,|ν−μ|≥2qδ ⎛ ⎞ ∞ logα−1 (1 + μ) ⎝ ⎠ 1 + ≤C μ |ν − μ| μ=α ≤C

ν≥μ+2qδ

×

log

q−α−1

(1 + ν)

ν

q−α≤ν<μ−2qδ

.

For μ ≥ 1, since we have

1 logq−α−1 (1 + ν) |ν − μ| ν ν>μ+2qδ ⎛ ⎞ 2μ ∞ q−α−1 (1 + ν) ⎠ 1 log + ≤⎝ ν−μ ν ν=μ+1

≤C

ν=2μ+1

1 logq−α (1 + μ) μ

q−α≤ν<μ−2qδ

μ−1

≤ logq−α−1 (1 + μ)

ν=1

1 logq−α−1 (1 + ν) |ν − μ| ν

1 ν(μ − ν)

1 ≤ C logq−α (1 + μ), μ it follows that

2

≤C

∞ logq−1 (1 + μ) < +∞. 2 μ μ=α

(3.10.18)

Thus Both (3.10.17) and (3.10.18) yield that (3.10.9) holds. This completes the proof of Theorem 3.10.1.

Theorem 3.10.2 Let f ∈ L(Qn ) with n ≥ 2. If there exists δ > 0 and p > 1, such that, for f ∈ Lp (B(x0 , δ)) with some δ > 0, 0

t

τ n−1 {fx0 (τ ) − f (x0 )}dτ = o(tn ),

3.10 Strong summation as t → 0+ , and

289

t 0

then 1 lim R→∞ R

|fx0 (τ ) − f (x0 )|p dτ = O(t),

q n−1 2 Sr (f ; x0 ) − f (x0 ) dr = 0

R

0

holds for every q > 0. By Theorem 3.2.3, the proof of Theorem 3.10.2 can be ascribed to that of the theorem about the Fourier integral. Theorem 3.10.3 Let g ∈ L(Rn ) with n ≥ 2 and g ∈ Lp (B(x0 , δ)) for some δ > 0 and p > 1. If t , τ n−1 gx0 (τ ) − g(x0 ) dτ = o(tn ), Φ1 (t) := 0

as t → 0+ , and Φ2 (t) := then 1 lim R→∞ R

t 0

R 0

|gx0 (τ ) − g(x0 )|p dτ = O(t),

q n−1 2 Br (g; x0 ) − g(x0 ) dr = 0 n−1

holds for every q > 0, where Br 2 an order of n−1 2 .

is the Bochner-Riesz means of g with

Proof. We might as well take it for granted that 1 < p < 2, 1p + 1q = 1 and x0 = 0, g(x0 ) = 0. According to the discussion in Chapter 2 about the Fourier integral, we have that ∞ n−1 cos rt − nπ 2 2 lim Br (g; 0) − C dt = 0. g0 (t) r→∞ 1 t r

Thus it suﬃces to show that cos rt − 1 R ∞ lim g0 (t) R→∞ R 1 1 t r

nπ 2

q dt dr = 0.

290

C3. Bochner-Riesz means of multiple Fourier series

nπ nπ cos rt − = Re e−i 2 eirt , 2 we merely need to prove q eirt 1 R ∞ dt dr = 0. g0 (t) lim R→∞ R 1 1 t

Since

(3.10.19)

r

For every ε ∈ (0, 1), we have

1 εr 1 r

1 1 εr 1 irt εr 1 irt ir irt g0 (t) irt ϕ(t) 2 e − e e dt = ϕ(t) e + dt, 1 t t t t 1 r

r

where ϕ(t) =

0

t

g0 (τ )dτ = Φ1 (t)

1 tn−1

+ (n − 1)

t 0

Φ1 (τ ) dτ = o(t), τn

as t → 0. Thus it follows that

1 εr 1 r

1 1 εr εr g0 (t) irt tdt + r dt e dt = o(1) + o(1) 1 1 t r r 1 1 = o(1) 1 + log + ε ε

(3.10.20)

= o(1), as r → +∞. Therefore, it follows from (3.10.20) that q 1 irt 1 R ∞ α := lim sup g0 (t) e dt dr t R→∞ R 1 1r q 1 1 irt 1 R εr q g0 (t) e dt dr ≤ 2 lim sup t R→∞ R 1 1r q 1 irt 1 R ∞ q + 2 lim sup g0 (t) e dt dt 1 t R→∞ R 1 εr q 1 R ∞ 1 irt q = 2 lim sup g0 (t) e dt dr. t R→∞ R 1 1 εr

(3.10.21)

3.10 Strong summation

291

Denote

t irτ

Φ(t, r) =

g0 (τ )e

0

dτ = ϕ(t)e

irτ

− ir

t 0

ϕ(τ )eirt dτ.

Since ϕ(t) = o(t), it follows that Φ(t, r) = o(t) + ir · o t2 , as t → 0. Meanwhile, we have |Φ(t, r)| ≤

t 0

|g0 (τ )|dτ ≤ M,

for any r > 0 and t > 0. Thus we have ∞ ∞ 1 1 g0 (t)eirt dt = o(1) + Φ(t, r) 2 dt. 1 1 t t εr

εr

Therefore, we conclude that ∞ Φ(t, r) q dt dr 2 1 t 1 εr q 1 R δ Φ(t, r) ≤ Cq lim sup dt dr 1 1 t2 R→∞ R εr εR R ∞ 1 Φ(t, r) q dt + Cq lim sup dr. t2 R→∞ R 1 δ

1 α ≤ 2 lim sup R R→∞

R

q

The integral of the second term on the right side of the above inequality can be rewritten as 1 R h(r)dr, R 1 where

h(r) =

δ

∞

Φ(t, r) q dt . t2

Notice that Φ(t, r) → 0, as r → ∞, and |Φ(t, r)| ≤ M . It follows from the control convergence theorem that h(r) → 0, as r → ∞. Thus we have 1 R h(r)dr → 0, R 1

292

C3. Bochner-Riesz means of multiple Fourier series

as R → ∞. Hence, we have that 1 α ≤ Cq lim sup R→∞ R

q Φ(t, r) dt dr 1 t2 εR 1 ⎫q

q ⎬ R 1 q |Φ(t, r)| dr dt . 1 ⎭ t2

R δ 1 εδ

⎧ 1 ⎨ ≤ Cq lim sup R→∞ R ⎩

δ 1 εR

εδ

Applying Hausdorﬀ-Young’s inequality to the function g0 χ[0,t) with 0 < t < δ in Lp ((0, δ)), we can get that

1 |Φ(t, r)| dr q

R

for t ∈ (0, δ). Consequently, we have

1 α ≤ Cq lim sup R→∞ R

q

≤

δ 1 εR

t

1 −2 p

t

|g0 (τ )| dτ p

0

1p

1 = O tp ,

q dt

≤ Cq lim sup R→∞

1 εR = Cq ε. R

Since ε is arbitrary, this leads to α = 0, and (3.10.19) is proven. This completes the proof of Theorem 3.10.3.

Chapter 4

The conjugate Fourier integral and series

4.1

The conjugate integral and the estimate of the kernel

The conjugate integral considered here is based on the singular integral theory by Calder´ on and Zygmund [CZ1]. We use normal notations. Let n denote the number of dimensions and (n) (n) P (·) ∈ Ak for k ≥ 1. Here Ak denotes the collection of all the homogeneous harmonic polynomials of degree k with n variables. We consider the kernel K(x) = P (x)|x|−n−k , x = 0, whose principal value Fourier transform can be obtained as K k Γ P (y) 2 K(y) = (−i)k n , n+k · n |y|k π22 Γ 2 for y = 0, and K(0) = 0. For any f ∈ L(Rn ), we deﬁne, x ∈ Rn , 1 6 fε (x) = f (x − y)K(y)dy, (2π)n |y|>ε for ε > 0, and

f6(x) = lim f6ε (x). ε→0+

(4.1.1)

(4.1.2)

The function f6 is the extension of the Hilbert transform of univariate function, which is called the multiple Hilbert transform by Calder´on and 293

294

C4. The conjugate Fourier integral and series

Zygmund. Here, we prefer to call f6 the Calder´ on-Zygmund transform of f . It is well known that the Calder´ on-Zygmund transform is of type (p, p) for 1 < p < ∞, as well as weak type (1, 1) (see Stein [St4] or [St5]). The conjugate Fourier integral of f is the following integral Rn

ix·y f(y)K(y)e dy.

(4.1.3)

From (4.1.1) and (4.1.2), it is easy to see that f6 is the convolution of f and should be (f ∗ K) = f6 and K in the sense of principal value. Formally, fˆK then (4.1.3) is the inversion of the Fourier transform of f6. So we should turn back to f6. However, these discussions are formal. In fact, we can only discuss the relation between some kind of linear means of (4.1.3) and f6. We deﬁne the Bochner-Riesz means of the conjugate Fourier integral (4.1.3) as α σ1 R (f ; x) =

α |y|2 fˆ(y)K(y) 1− 2 eix·y dy, R |y|
(4.1.4)

where the index α can be taken as some complex whose real part is bigger than −1 and R > 0. (n)

Lemma 4.1.1 Let P ∈ Ak we have

P |y|<1

y |y|

with k ≥ 1, n ≥ 2 and K(x) =

P (x) . |x|n+k

(1 − |y|2 )α0 +β eix·y dy = ik P (x)E(n, β, k, |x|),

Then

(4.1.5)

where n ∞

u 2j (−1)j Γ n+k π2 2 + j Γ(α0 + β + 1) n n+k , E(n, β, k, u) = k 2 2 j!Γ + k + j Γ + α + β + 1 + j 0 2 2 j=0 (4.1.6) n+1 α0 = n−1 and β is a complex number whose real part is bigger than − 2 2 . This lemma was proven by Chang [Cha1]. For the sake of integrality, we give the proof in the following.

4.1 The conjugate integral and the estimate of the kernel

295

Proof. We have that y (1 − |y|2 )α0 +β eix·y dy P |y| |y|<1 B 1 : ξ (1 − t2 )α0 +β eix·ξ dσ(ξ) dt P = |ξ| 0 |ξ|=t 1 2 α0 +β n−1 (1 − t ) t P (ξ)eitx·ξ dσ(ξ)dt. = 0

|ξ|=1

By the property of transforms of radial functions (see [SW1]), we can get that y (1 − |y|2 )α0 +β eix·y dy P |y| |y|<1 1 n J n +k−1 (t|x|) x 2 α0 +β n−1 k 2 2 (1 − t ) t (i) (2π) P dt. (4.1.7) = n −1 |x| 2 (t|x|) 0 By the series expression of the Bessel functions and changing the order of integration and summation, we have that ∞ n (−1)j x k i (2π) 2 P |x| j! Γ j + n2 + k j=0 2j+ n +k−1 1 2 1 2 α0 +β n−1 t|x| 1−t t dt × n 2 (t|x|) 2 −1 0 ∞ n (−1)j = ik (2π) 2 P (x) j! Γ j + n2 + k j=0 2j 1 1 2 α0 +β n+k−1 t|x| t dt n +k−1 × 1−t 2 2 2 0 2j ∞ j n (−1) |x| = ik π 2 2−k n 2 j! Γ j + 2 + k − 1 j=0 1 α0 +β n+k−1+2j × t 2dt. 1 − t2 0

Noticing the fact that 1 1 n+k (1 − t2 )α0 +β tn+k−1+2j 2dt = (1 − t)α0 +β t 2 +j−1dt 0 0 n+k +j , = B α0 + β + 1, 2

296

C4. The conjugate Fourier integral and series

we can get the lemma proven. In addition, for u ≥ 1, Chang [Cha1] also gave an asymptotic estimate of the function E(n, β, k, u). That is, E(n, β, k, u) =

R2 (β, u) c(n, k) R1 (β, u) + n+k+1 + n+k+1+β un+k u u cos u − π2 (n + k + β) +A(n, β) , un+k+β

(4.1.8)

where both R1 and R2 are bounded about u. The estimate is enough for the case when |Reβ| ≤ 1. However, for further need, it is necessary to give a more accurate estimate. Lemma 4.1.2 Let β = σ + iτ and − n+1 2 + h ≤ σ ≤ G with 0 < h < G < ∞. For u ≥ 1, we have c(n, k) G(n, β, k, u) H(n, β, k, u) + + n+k+2 un+k un+k+2+β

u n+1 + β B(n, β) √ + n+k+β uJn+k+β− 1 (u) + k 2√ Jn+k+β+ 1 (u) , 2 2 u u (4.1.9) where G and H are both inﬁnitely diﬀerentiable with respect to u, Γ n+k n n 2 , c(n, k) = π 2 2 Γ(k/2) n n+1 n− 12 +β 2 Γ B(n, β) = π 2 +β , 2 E(n, β, k, u) =

and |τ | H(n, β, k, u) ≤ An,k,h,G e 3π 2 .

G(n, β, k, u) ≤ An,k,h,Ge 32 π|τ | , Proof. For ξ > −(n + k), let ϕξ (x) =

∞

0

n

e−xt t 2 +ξ J n2 +k−1 (t)dt,

for 0 < x ≤ 1, and fξ (u) =

0

∞

n

e−t t 2 +ξ J n2 +k−1 (ut)dt.

4.1 The conjugate integral and the estimate of the kernel Clearly we have

297

1 . fξ (u) = n +1+ξ ϕξ u u2 1

From a formula in Bateman’s book (see [Ba2], page 29), we have that fξ (u) =

1 1 + u2

n +ξ+1 2

Γ(n + k +

−( n +k−1) ξ)P n +ξ2 2

√

1 1 + u2

,

where Pμν is the Legendre function (see [Ba1]). It follows from the above equation that n −( n +k−1) (0) = ϕξ (0+ ). lim u 2 +ξ+1fξ (u) = Γ(n + k + ξ)P n +ξ2

u→∞

2

(4.1.10)

For x > 0, it is obvious that ∞ n ϕξ (x) = − e−xt t 2 +ξ+1 J n2 +k−1 (t)dt = −ϕξ+1 (x), 0

which leads to

ϕξ (0+ ) = −ϕξ+1 (0+ ).

(4.1.11)

Hence, by the L’Hospital rule, we have n

n

n

fξ (u) = ϕξ (0+ )u−( 2 +ξ+1) + ϕξ (0+ )u−( 2 +ξ+2) gξ (u)u−( 2 +ξ+3) ,

(4.1.12)

where |gξ (u)| ≤ An,k,ξ . ∞ such that Choose ψ ∈ C[0,∞) ⎧ 1 ⎪ ⎨ 1, 0 ≤ t ≤ , 3 ψ(t) = 2 ⎪ ⎩ 0, t≥ . 3 We construct a polynomial Q(t) =

2n+3k+6

aj t j

j=0

satisfying

$ dm # −t 2 n−1 +β ψ(t) = 0, e Q(t) − (1 − t ) 2 dtm t=0

for m = 0, 1, . . . , q.

(4.1.13)

298

C4. The conjugate Fourier integral and series Obviously we have a0 = a1 = 1. Let ∞ n−1 +β n I1 (u) = ψ(t) 1 − t2 2 t 2 J n2 +k−1 (ut)dt, 0

I1∗ (u)

∞

= 0

n

e−t Q(t)t 2 J n2 +k−1 (ut)dt,

and set 1 (u) = I1∗ (u) − I1 (u) ∞ n−1 +β n e−t Q(t) − 1 − t2 2 = ψ(t) t−k t 2 +k J n2 +k−1 (ut)dt. 0

(4.1.14) We denote by n−1 2 2 +β ψ(t) t−k . ε(t) = e Q(t) − 1 − t

−t

By the condition (4.1.13), it is clear that t = 0 is the zero point of ε(t) of degree q − k = 2(n + k + 3), which implies ε(m) (0) = 0, for m = 0, 1, . . . , 2(n + k + 3). For ε(t) ∈ C n+k+3 [0, ∞), we deﬁne an operator A by A(ε)(t) = and A

m+1

ε (t) t

m A (ε) (t) (ε)(t) = , t

for 0 ≤ m < n + k + 3. Obviously, for the index λ = n + k + 3, we have Aλ (ε)(t) = ε(λ) (t)t−λ + a1 ε(λ−1) (t)t−λ−1 + · · · + a−2λ+1 ε (t)t−2λ+1 . Therefore, Aλ (ε)(t) is bounded in the neighborhood of the origin and for t ≥ 1, Aλ (ε)(t) is not bigger than the product of a polynomial about t and 2 e−t . Moreover, for m ∈ [0, λ) Z, Am (ε)(t) also satisﬁes = 0. Am (ε)(t) t=0

4.1 The conjugate integral and the estimate of the kernel

299

Hence, by the formula d ν t Jν (ut) = utν Jν−1 (ut), dt and λ times the integration by parts, we get, for u > 0, ∞ n λ 1 Aλ (ε)(t)t 2 +k+λ J n2 +k+λ−1 (ut)dt. 1 (u) = (−1) λ u 0 According to the deﬁnition of ε(t), we know that there exists a positive number M = M (n, k, h, G) such that ∞ n λ A (ε)(t) t 2 +k+λ J n2 +k+λ−1 (ut) dt ≤ M (1 + |τ |)M . 0

Consequently, we have |1 (u)| ≤ M (1 + |τ |)M u−(n+k+3) ,

(4.1.15)

for u ≥ 1. We note I1∗ (u)

= f0 (u) + f1 (u) +

q

aj fj (u).

j=2

By (4.1.12), we have n

n

n

n

n

n

f0 (u) = ϕ0 (0+ )u−( 2 +1) + ϕ0 (0+ )u−( 2 +2) + g0 (u)u−( 2 +3) , f1 (u) = ϕ1 (0+ )u−( 2 +2) + ϕ1 (0+ )u−( 2 +3) + g1 (u)u−( 2 +4) , and

q

n

aj fj (u) = G(u) u−( 2 +3)

j=2

with |G(u)| ≤ An,k,β ≤ An,k,h,Ge

3π |τ | 2

.

Thus we have n n n I1∗ (u) = ϕ0 (0+ )u−( 2 +1) + ϕ0 (0+ ) + ϕ1 (0+ ) u−( 2 +2) + G1 (u)u−( 2 +3) , and |G1 (u)| ≤ An,k,h,G e

3π |τ | 2

.

(4.1.16)

300

C4. The conjugate Fourier integral and series

Combining (4.1.11) with the fact ϕ0 (0+ ) + ϕ1 (0+ ) = 0, we get n

n

I1∗ (u) = ϕ0 (0+ )u−( 2 +1) + G1 (u)u−( 2 +3) .

(4.1.17)

Now we again construct a polynomial Q(t) =

q

bj tj

j=0

such that the function η(t) := Q 1 − t2 − t−k 1 − ψ(t) satisﬁes η (m) (1) = 0, for m = 0, 1, . . . , q, where q = 2n + 3k + 6. It is easy to calculate that b0 = 1 and b1 = 12 k. Set 1

I2 (u) = I2∗ (u) =

0 1 0

(1 − ψ(t))(1 − t2 )

n−1 +β 2

n

Q(1 − t2 )t 2 +k (1 − t2 )

n

t 2 J n2 +k−1 (ut)dt,

n−1 +β 2

J n2 +k−1(ut)dt

and 2 (u) = I2∗ (u) − I2 (u) 1 n−1 n η(t)(1 − t2 ) 2 +β t 2 +k J n2 +k−1 (ut)dt. = 0

According to the property of η(t), we have that |2 (u)| ≤ M (1 + |τ |)M u−(n+k+3) , for u ≥ 1. We clearly have that I2∗ (u)

=

q j=0

bj

0

1

(1 − t2 )

n−1 +β+j 2

n

t 2 +k J n2 +k−1 (ut)dt.

By a formula in Bateman’s book (see [Ba2], page 26), we have 1 n−1 n (1 − t2 ) 2 +β+j t 2 +k J n2 +k−1 (ut)dt 0 n−1 n−1 n+1 +β+j 2 Γ + β + j Jn− 1 +k+β+j (u)u−( 2 +β+j ) . =2 2 2

(4.1.18)

4.1 The conjugate integral and the estimate of the kernel

301

By the asymptotic formula of the Bessel functions, we have n+1 n−1 n+1 ∗ +β 2 + β u−( 2 +β ) Γ I2 (u) = 2 2 n+1 + β H(u) Jn− 1 +k+β (u) + k 2 Jn+ 1 +k+β (u) + n +3+β , 2 2 u u2 and |H(u)| ≤ An,k,h,G e

3π |τ | 2

.

Therefore, we have n n−1 n+1 +β 2 Γ I2 (u) = 2 + β u−( 2 +1+β ) 2 B : n+1 √ + β H(u) u Jn− 1 +k+β (u) + k 2√ Jn+ 1 +k+β (u) + n +3+β − 2 (u). 2 2 u u2 (4.1.19) From an equation in Bateman’s book (see [Ba1], page 145), it follows that −( n +k−1) 2

Pn 2

−( n +k−1) 2

(0) = 2

Substituting

1 2

π cos

k−1 π 2

Γ 1 − k2 . Γ n+k+1 2

k π k−1 π = sin π = 2 2 Γ 1 − k2 Γ k2

cos

into the above equation, we have −( n +k−1) 2

Pn 2

−( n +k−1) √ 2

(0) = 2

−1 k n+k+1 π Γ . Γ 2 2

Again substituting this into (4.1.9), we obtain √

+

ϕ0 (0 ) =

2

π

n +k−1 2

Γ(n + k) n+k+1 k . Γ 2 Γ 2

Applying the formula Γ(n + k) = 2

n+k−1 − 21

π

Γ

n+k 2

n+k+1 Γ 2

(4.1.20)

302

C4. The conjugate Fourier integral and series

in Bateman’s book (see [Ba1], page 5), we can obtain that n 2 2 Γ n+k + 2 . ϕ0 (0 ) = Γ k2

(4.1.21)

From (4.1.5) and (4.1.7), we can see that n (2π) 2 E(n, β, k, u) = n +k−1 I1 (u) + I2 (u) . u2

By combining (4.1.15) with (4.1.19) together and replacing them into (4.1.21), we get Lemma 4.1.2. Substituting the expression of fˆ into (4.1.4) and changing the order of the integration, we have that 2 α 1 |y| α f (u) ei(x−u)·y dydu K(y) 1− 2 σ1 R (f ; x) = (2π)n Rn R |y|
Γ 22n π

k

3π 2

2

Γ

n+k ,

(4.1.22)

2

and ψx (t) =

Sn−1

for t > 0. Thus, it follows that α0 +β σ1 (f ; x) = b(n, k)Rn+k R

f (x − tξ)P (ξ)dσ(ξ),

∞ 0

ψx (t)E(n, β, k, Rt) tn+k−1 dt.

(4.1.23)

(4.1.24)

α0 +β . We call the function E(n, β, k, Rt) the kernel of the operator σ1 R

4.2 Convergence of Bochner-Riesz means for conjugate Fourier integral 303

4.2

Convergence of Bochner-Riesz means for conjugate Fourier integral

Let us ﬁrst prove a proposition parallel to the case of non-conjugate, which was proven by Lu [Lu3]. Theorem 4.2.1 Let f ∈ L(Rn ) with n ≥ 2, and α = 1. If P ∈

(n) Ak

n−1 2

− β with 0 ≤ β <

with k ≥ 1 and ψx (t) deﬁned as in (4.1.23) satisﬁes that t |ψx0 (τ )|τ n−1 dτ = o(tn ), (4.2.1) 0

as t →

0+ ,

then, the equation # $ α (f ; x ) − f1 lim σ1 =0 1 (x0 ) 0 R R

R→∞

holds if and only if +∞ cos Rt − π2 (n + k − β) β ψx0 (t) dt = 0. lim R 1 R→∞ t1−β

(4.2.2)

(4.2.3)

R

Proof. It follows from (4.1.1) and (4.1.23) that ∞ 1 1 1 ψx (t)dt. f 1 (x) = n R 1 (2π) t

(4.2.4)

R

By (4.1.24), we have σ1 R

α0 −β

∞

ψx0 (t)E(n, −β, k, Rt) tn+k−1 dt

0 1 ∞ R + ψx0 (t)E(n, −β, k, Rt) tn+k−1 dt = b(n, k)Rn+k

(f ; x0 ) = b(n, k)R

n+k

0

1 R

= I1 + I2 .

(4.2.5)

Since |E(n, −β, k, Rt)| ≤ Mn,k,β < +∞, for t ≥ 0, by condition (4.2.1), we have 1 R n+k ψx0 (t)E(n, −β, k, Rt)tn+k−1 dt I1 = b(n, k)R 0

= o(1), as R → +∞.

304

C4. The conjugate Fourier integral and series

To estimate I2 , we denote by 1 − β = δ > 0. By (4.1.9) and the asymptotic formula of the Bessel functions &

2 π π 1 cos u − ν − +O Jν (u) = , 3 πu 2 4 u2 for u ≥ 1, we have E(n, −β, k, Rt) =

c(n, k) B(n, −β) π (n + R − β) + cos Rt − (Rt)n+k (Rt)n+k−β 2 1 +O . (Rt)n+k+δ

Thus, we can obtain the estimate ∞ −1 β ψx0 (t)t dt + b(n, k)B(n, −β)R I2 = b(n, k)c(n, k) 1 R

∞ 1 R

ψx0 (t)

∞ cos Rt − π2 (n + k − β) 1 1 dt + O ψx0 (t) 1+δ dt . t1−β Rδ 1 t R

(4.2.6) By the condition (4.2.1), we have

∞ 1 1 O ψx0 (t) 1+δ dt = o(1) Rδ 1 t R

as R → ∞. Since b(n, k) =

Γ(k/2) 3n 22n π 2 Γ n+k 2

and Γ c(n, k) = π 2n n 2

we have b(n, k)c(n, k) =

n+k 2

Γ( k2 )

,

1 . (2π)n

Therefore, it follows from (4.2.4)–(4.2.6) that α0 −β (f ; x0 ) − f11 (x0 ) = b(n, k)B(n, −β)Rβ σ1 R R

as R → +∞.

∞ 1 R

ψx0 (t)

cos Rt − π2 (n + k − β) dt + o(1), t1−β

(4.2.7)

4.2 Convergence of Bochner-Riesz means for conjugate Fourier integral 305 Together with the fact that b(n, k)B(n, −β) = 0, this leads to the conclusion of Theorem 4.2.1. Theorem 4.2.1 above improves the corresponding results by Lippman [Li1] and Golubov [Go2]. Remark 4.2.1 The condition (4.2.1) of Theorem 4.2.1 can be replaced by the following weaker one t ψx0 (τ )τ n−1 dτ = o(tn ), 0

as →

0+ .

Theorem 4.2.2 Let f ∈ L(Rn ) with n ≥ 2 satisﬁes the condition (4.2.1) and ∞ |ψx0 (τ + η) − ψx0 (τ )| dτ = o(1), (4.2.8) τ η as η → 0. If β = 0, then (4.2.2) holds. Proof. Let θ = − π2 (n + k). By Theorem 4.2.1, we merely need to verify that if β = 0, then (4.2.3) holds. That is, ∞ cos(Rt + θ) ψx0 (t) dt = o(1), (4.2.9) I(R) := 1 t R

as R → ∞. By the periodicity of cosine functions and the condition (4.2.1), we have that 1+π R cos(Rt + θ) 1 ψx0 (t) I(R) = dt 2 1 t R

1 π −1 1 ∞ cos(Rt + θ)dt ψx0 t + + π + ψx0 (t) 2 1 R t+ R t R

π 1 1 ∞ 1 = o(1) + − ψx0 t + π 2 1 R t t+ R R π ψx t + R − ψx0 (t) cos(Rt + θ)dt − 0 t ∞ π ψx0 t + R π cos(Rt + θ)dt = o(1) + π 2R 1 t t + R R π − ψx0 (t) 1 ∞ ψx0 t + R cos(Rt + θ)dt. (4.2.10) − 2 1 t R

306

C4. The conjugate Fourier integral and series

It follows from the condition (4.2.1) that π 2π ∞ |ψx0 (t)| 1 ∞ ψx0 t + R dt ≤ dt π R 1 R 1 t2 t t+ R R R t τ n−1 |ψx0 (τ )|dτ 2π ∞ ≤ (n + 1) 0 dt R 1 tn+2 R

= o(1). It follows from the condition that ∞ π − ψx0 (t) ψx0 t + R cos(Rt + θ)dt = o(1). 1 t R

Then (4.2.9) is proven and thus we have proven this theorem.

Corollary 4.2.1 Let f ∈ L(Rn ) with n ≥ 2. If there exists η > 0 such that η −1 t |f (x0 + tξ) − f (x0 )|dσ(ξ)dt < ∞, (4.2.11) 0

Sn−1

then

σ1 R

lim

R→∞

n−1 2

(f ; x0 ) − f61 (x0 ) R

=0

(4.2.12)

holds. Proof. It is obvious that (4.2.1) can be deduced from (4.2.11). For any ε > 0, by (4.2.11), there exists δ1 > 0 such that 2δ1 −1 t |f (x0 + tξ) − f (x0 )|dσ(ξ)dt < ε. (4.2.13) 0

Sn−1

Since f ∈ L(Rn ), ψx0 (t), as a function of t, is in L((δ1 , +∞)). Hence, by the continuous property of integral, we have that there exists δ > 0 with δ ≤ δ1 , such that when 0 < η < δ, we have ∞ −1 δ1 |ψx0 (t + η) − ψx0 (t)|dt < ε. δ1

Thus it follows that

∞ δ1

|ψx0 (t + η) − ψx0 (t)| dt < ε. t

(4.2.14)

4.2 Convergence of Bochner-Riesz means for conjugate Fourier integral 307 On the other hand, when 0 < η < δ, by the condition (4.2.13), we conclude that δ1 |ψx0 (t + η) − ψx0 (t)| dt t η δ1 δ1 |ψx0 (t + η)| |ψx0 (t)| dt + dt ≤ t t η η 2δ1 δ1 |ψx0 (t)| |ψx0 (t)| ≤2 dt + dt t t 2η η 2δ1 |ψx0 (t)| dt ≤3 t 0 2δ1 ≤ 3 max |P (ξ)| t−1 |f (χx0 + tξ) − f (x0 )|dσ(ξ)dt ξ∈Sn−1

Sn−1

0

≤ M ε,

(4.2.15)

where M is a constant. Combining (4.2.14) with (4.2.15) yields ∞ |ψx0 (t + η) − ψx0 (t)| dt ≤ (M + 1)ε. t η This implies the validity of (4.2.8). This completes the proof of Corollary 4.2.1.

Corollary 4.2.1 gives the Dini type criteria for the convergence of the conjugate Bochner-Riesz means, which was ﬁrst established by Lippman [Li1]. Other literatures in the research of the Bochner-Riesz means of the conjugate Fourier integral include the works by Wu [Wu1]. n−1 For the conjugate means of the critical index σ1 R 2 (f ; x), we can regard it as the multiple extension of the one variable means with an order of zero, that is, fˆ(y) k(y)eixy dy. |y|
Thus we call 1 R

R 0

σ 1u

n−1 2

(f ; x)du

the (C, 1) means of the conjugate Fourier integral. Lu established the following theorem in [Lu4].

308

C4. The conjugate Fourier integral and series

Theorem 4.2.3 Let f ∈ L(Rn ) with n ≥ 2. If the condition (4.2.1) holds at x0 , then we have R n−1 1 σ 1u 2 (f ; x0 )du − f61 (x0 ) = 0. (4.2.16) lim R R→∞ R 0 Proof. By setting θ = − π2 (n + k) and (4.2.7), we have that 1 R

R

n−1 2

(f ; x0 )du − f61 (x0 ) R R ∞ 1 = b(n, k)B(n, 0) ψx0 (t) 1 R 0 u 1 R 6 cos(ut + θ) dtdu + f 1 (x0 )du − f61 (x0 ) + o(1). × R t R 0 u 0

σ 1u

By (4.2.4), we conclude that 1 R 6 f 1 (x0 )du − f61 (x0 ) R R 0 u R

∞ ∞ 1 1 ψx0 (t) ψx0 (t) 1 = dt du − dt n n 1 1 R 0 (2π) t (2π) t u R ∞ 1 ∞ ψx0 (t) 1 ψx0 (t) 1 dt − dt R− = 1 (2π)n R 1 t t t R R 1 1 ∞ ψx0 (t) =− dt. (2π)n R 1 t2

(4.2.17)

(4.2.18)

R

Let

t

F (t) = 0

|ψx0 (τ )|τ n−1 dτ.

Integration by parts implies that ∞ ∞ ∞ |ψx0 (t)| F (t) −(n+1) dt = F (t)t + (n + 1) dt. 2 1 1 t tn+2 1 R

R

R

By the condition (4.2.1), that is, F (t) = o(tn ) as t → 0+ , we have that 1 ∞ |ψx0 (t)| dt = o(1), (4.2.19) R 1 t2 R

as R → ∞.

4.3 The conjugate Fourier series

309

In addition, it follows that ∞ cos(ut + θ) 1 R ψx0 (t) dt du 1 R 0 t u R 1 ∞ ψx0 (t) dt = cos(ut + θ)du 1 R 1 t R t 1 ∞ sin(Rt + θ) − sin(1 + θ) ψx0 (t) dt. = R 1 t t

(4.2.20)

R

Combining (4.2.17)–(4.2.20) yields that Theorem 4.2.3 holds. It is easy to see that if 0 ≤ β < 1, then, under the condition of Theorem 4.2.3, we have R n−1 1 lim σ 1u 2 −β (f ; x0 )du − f61 (x0 ) = 0. (4.2.21) R R→∞ R 0 Since the condition (4.2.1) is obviously valid at the Lebesgue points, we can deduce from the weak (1, 1) type of the Calder´on-Zygmund transform that the limit in (4.2.1) exists almost everywhere ﬁnite. Therefore, as a corollary, we can get the following theorem. Theorem 4.2.4 If f ∈ L(Rn ), then we have 1 R n−1 σ 1u 2 (f ; x)du = f6(x), lim R→∞ R 0

(4.2.22)

for a.e. x ∈ Rn . n−1

n−1

Of course, if we replace σ 1u 2 by σ 1u 2 −β with 0 ≤ β < 1, then the conclusion still holds. At last, we have to point out that, from Theorem 4.2.1, it is easy to see α0 that for the critical index α0 , the convergence of σ1 R (f ) only depends on the local property of the functions, while things are diﬀerent for the order below the critical index.

4.3

The conjugate Fourier series

Suppose that f ∈ L(Qn ) and its Fourier series is σ(f )(x) = Cm (f )eim·x .

(4.3.1)

310

C4. The conjugate Fourier integral and series

For n = 1, it is well known that σ 6(f )(x) =

(−isgn m)Cm (f )eim·x ,

(4.3.2)

m=0

is called the conjugate series of σ(f ). By iterated superposition, this notion can be extended from the case of one variable to the case of multiple variables. For a function of n variables, there are 2n − 1 conjugate series and the status of these conjugate series are quite asymmetric, or we say their extent of the conjugation are diﬀerent. Thus, the above factors complicate the research of the conjugate series. Especially, when we investigate the series by the spherical limit, the notion of conjugation is not quite adequate for the case of multiple variables. Hence, we will study the conjugate in the sense of Calder´ on-Zygmund. Here, we only consider the conjugate series in correspondence with the spherical harmonic kernel which is parallel to the concept of the conjugate integral in the previous two sections. is the Fourier transform of the Suppose that K(x) = P (x)|x|−n−k and K principal value of K. We deﬁne by σ 6(f )(x) =

im·x K(m)C m (f )e

(4.3.3)

and call it the conjugate series of σ(f ) about the kernel K, or the conjugate Fourier series of f for brief. If n = 1, then P (x) is a homogenous function of degree 1. That is, P (x) = cx, where c is a nonzero constant. Then by the in Section 4.1, we have expression of K c K(y) = (−i) sgn(y). 2

(4.3.4)

If we choose c = 2, then (4.3.3) coincides with (4.3.2). Since we are familiar with (4.3.2), we only discuss the case when n ≥ 2 in the following. Let m ∈ Zn . We deﬁne 1 Im = K(x + 2πm)dx. (4.3.5) (2π)n Qn If m = 0, the above integral should be put in the sense of principal value. That is to say, 1 1 I0 = lim K(x)dx = K(x)dx, (4.3.6) (2π)n ε→0+ Qn \B(0,ε) (2π)n Qn \B(0,1)

4.3 The conjugate Fourier series

311

where we use

Sn−1

P (ξ)dσ(ξ) = 0.

Let x, y ∈ Q := Qn and m ∈ Zn , m = 0. We set ξm =

x + 2πm y + 2πm , ηm = . x + 2πm y + 2πm

Then, we have K(x + 2πm) − Im =

1 |Q|

Q

P (ξm ) P (ηm ) − |ξm |n |ηm |n

dy.

Thus it follows that |K(x + 2πm) − Im | 1 1 1 |P (ξm ) − P (ηm )| 1 dy + − |P (ηm )|dy. ≤ |Q| Q |ξm |n |Q| Q |ξm |n |ηm |n Since we have

|P (ξm ) − P (ηm )| ≤ |∇P (θ)||ξm − ηm | = O and

1 |m|

1 1 1 − = O , |ξ |n |η |n |m|n+1 m m

it follows that |K(x + 2πm) − Im | = O

1 |m|n+1

,

(4.3.7)

for m = 0. We deﬁne the periodization of K by K(x + 2πm) − Im , K ∗ (x) = m∈Zn

for x ∈ Q and x = 0. By (4.3.7), if we get rid of the term of m = 0 in the series on the right side of the above equation, then the series is absolutely and uniformly convergent. Therefore, [K(x + 2πm) − Im ] K ∗ (x) = K(x) − I0 + m=0

is continuous everywhere on Q except the origin.

312

C4. The conjugate Fourier integral and series We deﬁne fε∗ by ⎧ √ ⎨ 1 f (x − y)K ∗ (y)dy, if 0 < ε < nπ ∗ n fε (x) = (2π) y∈Q,|y|>ε √ ⎩ 0, if ε ≥ nπ.

We call fε∗ the truncated conjugate function of f relative to the kernel K. The limit of fε∗ is deﬁned by f ∗ (x) = lim fε∗ (x), ε→0+

√ which is called the conjugate function of f . Let ε nπ > 0. Thus we have 1 ∗ fε (x) = f (x − y)(K(y) − I0 )dy (2π)n y∈Q,|y|>ε 1 [K(y + 2πm) − Im ]f (x − y)dy + (2π)n y∈Q m=0 −1 f (x − y)[K(y + 2πm) − Im ]dy + (2π)n |y|<ε m=0 1 f (x − y)(K(y) − I )dy + + . = 0 1 2 (2π)n y∈Q,|y|>ε / For the item 1 in the above equation, let , Sj = x ∈ Zn : j ≤ |x| < j + 1 , for j = 0, 1, 2, . . . , and by the spherical summation, we have ∞ 1 = f (x − y)K(y)dy − Im C0 (f ) . 1 (2π)n Q+2πm j=1 m∈Sj

Let N ∈ N and N > 10n. We set EN =

N

(Q + 2πm)

j=1 m∈Sj

and

# √ $ / Q, |y| ≤ 2π(N − n) . DN = y : y ∈

For any y ∈ DN , there exists m ∈ Zn such that y ∈ Q + 2πm. However, √ since y ∈ / Q, m = 0, by |y − 2πm| ≤ nπ, we have that √ 2π|m| ≤ |y| + nπ < 2N π, 0 < |m| < N.

4.3 The conjugate Fourier series

313

Therefore, we have y ∈ EN , which implies DN ⊂ EN . √ On the other hand, for any y ∈ EN , we have |y| ≤ 2πN + nπ. We √ √ denote by rN = 2π(N − n) and ρN = 2π(N + n). Then we have EN ⊂ B(0, ρN ), DN = B(0, rN ) \ Q and {EN \ Dn } ⊂ {B(0, ρN ) \ B(0, rN )}. From the above discussion, we obtain that N f (x − y)K(y)dy − f (x − y)K(y)dy DN j=1 m∈Sj Q+2πm |f (x − y)K(y)|dy. (4.3.8) ≤ rN <|y|<ρN

Since for |y| ∈ (rN , ρN ), we have |K(y)| = O

1 Nn

and

|f (x − y)|dy = O(N n−1 )f L(Q) . rN <|y|<ρN

Then the item on the right side of (4.3.8) is O Meanwhile, we have N j=1 m∈Sj

where

1 N

.

N 1 K(y)dy |Q| j=1 m∈Sj Q+2πm 1 = K(y)dy |Q| EN 1 1 K(y)dy + K(y)dy, = |Q| DN |Q| EN \DN

Im =

C 1 . K(y)dy ≤ dy = O n EN \DN N rN <|y|<ρN |y|

(4.3.9)

314

C4. The conjugate Fourier integral and series By the harmonicity, we have K(y)dy DN K(y)dy + √ K(y)dy = √ B(o, nπ)\Q nπ<|y|
(4.3.10)

B(o, nπ)\Q

It follows from (4.3.8),(4.3.9) and (4.3.10) that = lim f (x − y)K(y)dy − C0 (f ) 1

N →∞

=

DN ∞

√

K(y)dy

B(o, nπ)\Q

1 1 f (x − tξ)P (ξ)dσ(ξ) dt (2π)n √nπ t Sn−1 1 [f (x − y) − C0 (f )]K(y)dy. + (2π)n B(o,√nπ)\Q

1 |Q|

(4.3.11)

According to (4.3.7), we have ≤c 2

m=0

1 |m|n+1

|y|<ε

|f (x − y)|dy.

Combining the fact that f ∈ L(Q) with the absolute continuity of integration, we have = o(1), (4.3.12) 2

0+

uniformly holds about x, as ε → . Consequently, we have ∞ 1 1 ∗ f (x − tξ)P (ξ)dσ(ξ)dt fε (x) = n (2π) ε t Sn−1 1 C0 (f )K(y)dy + o(1). − (2π)n B(o,√nπ)\B(o,ε) That is, fε∗ (x)

1 = (2π)n

∞ ε

1 t

Sn−1

f (x − tξ)P (ξ)dσ(ξ)dt + o(1),

(4.3.13)

4.3 The conjugate Fourier series

315

holds uniformly about x, as ε → 0+ . Thus, we obtain the expression of the conjugate function ρ 1 1 f ∗ (x) = lim f (x − tξ)P (ξ)dσ(ξ)dt n (2π) ε→0+ , ρ→+∞ ε t Sn−1 1 lim f (x − y)K(y)dy. (4.3.14) = (2π)n ε→0+ , ρ→+∞ ε<|y|<ρ We can notice that fε∗(x) is the convolution of an integrable function and a continuous function on Qn and so is continuous. Hence, the upper limit of the integral on the right side of (4.3.13) tends to ∞, which is a limit procedure of everywhere converging. And thus (4.3.14), which is worth being discussed, is the limit procedure when ε → 0+ . The existence of the conjugate functions and related problems were solved by Calder´on and Zygmund [CZ1], [CZ2]. Let us now formulate the following theorem. Theorem 4.3.1 The truncated conjugate function is deﬁned as above. We also deﬁne the conjugate maximal operator ϕ by ϕ(f )(x) = sup |fε∗ (x)|, ε>0

for f ∈ L(Qn ). Then, we have ϕ(f )p ≤ c

p2 f p , p−1

for 1 < p < ∞, and ϕ is weak-type (1, 1) and satisﬁes |f | log+ |f |dx + 1 . ϕ(f )1 ≤ c Q

According to (4.3.14), for convenience, we sometimes replace fε∗ by ∞ 1 1 f ε (x) = f (x − tξ)P (ξ)dσ(ξ)dt. (4.3.15) n (2π) ε t Sn−1 In the following, we give another basic conclusion. Theorem 4.3.2 For f ∗ ∈ L(Q), the series (4.3.2) is the Fourier series of f ∗ , that is, Cm (f ∗ ) = K(m)C m (f ), for m ∈ Zn .

316

C4. The conjugate Fourier integral and series

This theorem can be proven by the method for the case of one variable (see [Zy1]) and we omit the proof here. We will consider the convergence and approximation of the BochnerRiesz means of the series (4.3.2) α |m|2 α im·x 6 , (4.3.16) K(m)C 1− 2 SR (f ; x) = m (f )e R |m|
where α is a complex number with real part being bigger than −1.

4.4

Kernel of Bochner-Riesz means of conjugate Fourier series α By (4.3.16), we deﬁne S6R (f ; x) as 1 α 6 α (y)dy, 6 f (x − y)D SR (f ; x) = R |Q| Q

6 α (·) has the following expression where the kernel D R α |m|2 α 6 K(m) 1 − 2 DR (y) = eim·x . R

(4.4.1)

(4.4.2)

|m|
In Section 4.1, we calculate that (see Lemma 4.1.1 and 4.1.2) y (1 − |y|2 )α0 +β eix·y dy = i−k P (x) · E(n, β, k, |x|), P |y| |y|<1 where n ≥ 2, k ≥ 1, α0 =

n−1 2 ,

(4.4.3)

Re β > − n+1 2 and

C(n, k) G(n, β, k, u) H(n, β, k, u) + + n+k+2+β n+k+2 un+k u

u n+1 √ + β B(n, β) Jn+k+β+ 1 (u) . + n+k+β uJn+k+β− 1 (u) + k 2√ 2 2 u u (4.4.4) Let us now turn to verifying that for Re β > 0, the function

E(n, β, k, u) =

f (y) = P

y |y|

|y|2 1− 2 R

α0 +β χ(0,R) (|y|)

satisﬁes the condition of Theorem 1.2.3 in Chapter 1 for y = 0.

4.4 Kernel of Bochner-Riesz means of conjugate Fourier series

317

Firstly, since Re β > 0, f is continuous except at the origin and (i)

|f (m)| =

m=0

|f (m)| < +∞.

0<|m|
By (4.4.3), we have that f (x)eix·y dx = ik Rn+k P (y)E(n, β, k, R|y|). g(y) :=

(4.4.5)

According to the harmonicity of P , we get that ρ g(y)dy = ik Rn+k E(n, β, k, Rt) P (ξ)dσ(ξ)dt = 0,

(4.4.6)

Rn

|y|<ρ

0

|ξ|=t

ρ > 0, which yields 1 lim (2π)n ρ→∞

|y|<ρ

g(y)dy = 0.

(4.4.7)

So due to the asymptotic formula (4.4.4), we obtain (ii) |g(y)| = O(|y|−n ), as |y| → ∞. Finally, deﬁne Am

1 = |Q|

g(x + 2πm)dx, Q

for m ∈ Zn . For N ∈ N with N > 10n, let ρ = 2πN . Combining (4.4.6) with (ii), we have that N N 1 Am = Am − g(x)dx |Q| |x|<ρ j=0 m∈Sj j=0 m∈Sj 1 ≤ |g(x)|dx |Q| ρ−2πn<|x|<ρ+2πn 1 N n−1 =O Nn 1 , =O N which implies

∞ j=0 m∈Sj

Am = 0.

(4.4.8)

318

C4. The conjugate Fourier integral and series

Since we have N

N N g(x + 2πm) − Am + g(x + 2πm) = Am ,

j=0 m∈Sj

j=0 m∈Sj

j=0 m∈Sj

to prove the locally uniform convergence of the left side of the above equation about x as N → ∞, it suﬃces to show that N

g(x + 2πm) − Am

(4.4.9)

j=0 m∈Sj

is locally and uniformly convergent. Let |x| ≤ M . When |m| ≥ 2(M + 1), we have 1 |g(x) − g(y)|dy. |g(x + 2πm) − Am | ≤ |Q| Q+2πm By (4.4.4), we get that g(y) = ik C(n, k)

P (y) −n−δ + O |y| , |y|n+k

for |y| > 1, where δ = min(1, Re β) > 0. Thus, for x, y ∈ Q + 2πm, we have 1 1 =O . |g(x) − g(y)| = O |y|n+δ |m|n+δ Therefore, (4.4.9) is uniformly convergent for|x| ≤ M . Since the conditions for the Poisson summation formula are all satisﬁed, we have, for Re β > 0, that ∞

ik Rn+k P (x + 2πm)E(n, β, k, R|x + 2πm|)

j=0 m∈Sj

=

P

|m|
m |m|

|m|2 1− 2 R

α0 +β eim·x .

(4.4.10)

Multiplying both sides of the above equation by the following constant (−i)k

Γ( k2 ) 1 , n+k = (−i)k C(n, k) π 2n Γ 2 n 2

we can obtain the following result.

4.5 The maximal operator of the conjugate partial sum

319

Theorem 4.4.1 For Re β > 0, we have 6 α0 +β (y) = D R

∞ Γ( k2 )Rn+k P (y + 2πm)E(n, β, k, R|y + 2πm|). n+k n π 2 2n Γ 2 j=0 m∈S j

(4.4.11) Here, the dimension n is bigger than 1. The above expression was established by Chang [Cha1]. By the uniform convergence of the series on the right side of (4.4.11), we directly obtain the following result from Theorem 4.4.1. Proposition 4.4.1 For Re β > 0, we have →∞ 1 t 1 α0 +β 6 tn+k−1 E(n, β, k, t)dt, SR (f ; x) = ψx |Q| C(n, k) 0 R 8ρ 8∞ where we use the notation 0 = limρ→∞ 0 and ψx (s) = f (x − sξ)P (ξ)dσ(ξ).

(4.4.12)

|ξ|=1

Remark 4.4.1 When Re β ≤ 0, the above equation does not hold since we cannot prove the locally uniform convergence of the series (4.4.9). From this, we can see that the index α0 = n−1 also has the critical properties 2 for the conjugate Bochner-Riesz means. Later on, a series of facts will be supplemented to illustrate it.

4.5

The maximal operator of the conjugate partial sum

Our purpose is to give an estimate of the type of the maximal operator at the critical index n−1 n−1 2 2 6 6 S∗ (f )(x) := sup SR (f ; x) . (4.5.1) R>0

Lemma 4.5.1 Let β = σ + iτ, σ > 0 and τ ∈ R. If we deﬁne n−1 +β (f ; x) − f 1 (x) , M β (f )(x) = sup S6R 2 R R>0

320

C4. The conjugate Fourier integral and series

then σ 2π|τ | 2 (1

M (f )(x) ≤ An (P )e β

+ σ)Γ n+1 2 +σ M (f )(x) σ

(4.5.2)

holds. Here, we replace the truncated conjugate function introduced at the beginning of Section 4.3 by f ε deﬁned as in (4.3.15). We denote by M the Hardy-Littlewood maximal operator and by An (P ) a constant only related to n and the function P . Proof. By (4.3.15), we have that 1 f 1 (x) = R |Q|

1

∞

1 ψx t

t dt. R

(4.5.3)

According to (4.4.12), we get that 1 1 t 1 α0 +β 6 tn+k−1 E(n, β, k, t)dt SR (f ; x) = ψx |Q| C(n, k) 0 R ∞ 1 t 1 tn+k−1 E(n, β, k, t)dt. ψx + |Q| C(n, k) 1 R We denote the ﬁrst term on the right side of the above equation by AR (x). Obviously, it follows that |AR (x)| ≤ An (P )M (f )(x).

(4.5.4)

Thus, we have that ∞ t 1 α0 +β 6 ψx SR (f ; x) − f 1 (x) = AR (x) + t−1 R |Q| 1 R n+k t E(n, β, k, t) − 1 dt. C(n, k)

(4.5.5)

Due to (4.4.4), we obtain that n+k t 1 1 2π|τ | + . C(n, k) E(n, β, k, t) − 1 ≤ An,k e t2 tσ

(4.5.6)

4.5 The maximal operator of the conjugate partial sum

321

It is easy to calculate that

∞ 2j+1 dt t ψx ψx t dt = 1+σ R t R t1+σ 2j

∞ 1

j=0

∞ 1 ≤ An (P ) M f (x) 2σj

(4.5.7)

j=0

1+σ M f (x). ≤ An (P ) σ It follows from (4.5.4), (4.5.5), (4.5.6) and (4.5.7) that (4.5.2) naturally holds. This completes the proof of Lemma 4.5.1.

n−1 Lemma 4.5.2 Let β = σ + iτ, σ and τ ∈ R. If n+1 2 > σ > − 2 , then we have n−1 +β e2π|τ | S6∗ 2 (f ) (4.5.8) ≤ An (P ) σ + n−1 f 2 . 2

2

Proof. According to Theorem 4.3.1, when f ∈ L2 (Qn ), we have f ∗ ∈ 2 n L (Q ) and f ∗ 2 ≤ An (P )f 2 . By Theorem 4.3.2, if f ∗ ∈ L2 (Qn ), we have n−1

S6R 2

+β

n−1

(f ) = SR 2

+β

(f ∗ ).

Thus we have that n−1 +β n−1 +β ∗ 2 S6∗ 2 (f ) = S∗ (f ) 2

2

1 ≤ An e f ∗ 2 n−1 σ+ 2 1 ≤ An (P ) e2π|τ | f 2 . σ + n−1 2 2π|τ |

This completes the proof of Lemma 4.5.2. Since An (P ) does not play any role in our discussion, we can rewrite it as A for short.

322

C4. The conjugate Fourier integral and series

According to Lemma 4.5.1 and 4.5.2, and by Stein’s interpolation theorem, we can imitate the ways of dealing with the Bochner-Riesz operator in Section 3.5. Theorem 4.5.1 If 1 < p < ∞, then we have 3 n−1 S6∗ 2 (f ) ≤ A p f p . (p − 1)2 p

(4.5.9)

In the following, we will start from the estimate of the type of the maximal operator on the partial sum and then deduce another applicable type of operators. Definition 4.5.1 Suppose that P is a homogeneous harmonic polynomial of degree k, where k is a nonnegative integer. For f ∈ L(Q), let P (f )(x; t) = f (x − tξ)P (ξ)dσ(ξ), (4.5.10) Sn−1

for t ≥ 0. We deﬁne an operator T P by ∞ t cos t − π2 (n + k + β) P (f ) x; dt . T P (f )(x) = sup lim sup R t1+β R>1 β→0+ 1 (4.5.11) Lemma 4.5.3 Let 1 0. By choosing β ∈ (0, 1) and (4.5.5), we have that ∞ 1 t α +β 0 ψx S6R (f ; x) = f 1 (x) + AR (x) + t−1 R |Q| 1 R n+k t × E(n, β, k, t) − 1 dt. c(n, k) It follows from (4.4.4) that & 1 B(n, β) 2 + t2 C(n, k) π cos t − π2 (n + k + β) 1 + O 1+β . × tβ t

tn+k E(n, β, k, t) − 1 = O C(n, k)

4.5 The maximal operator of the conjugate partial sum Therefore, we get that & t cos t − π2 (n + k + β) 1 B(n, β) 2 ∞ ψx dt |Q| C(n, k) π 1 R t1+β ∞ 1 t α0 +β 6 ψx = SR (f ; x) − f 1 (x) − AR (x) + O 2 dt, R R t 1

where n 2

n− 12 +β

B(n, β) = π 2

Γ

(4.5.13)

n+1 +β . 2

Combining (4.5.4) with (4.5.7), we obtain that ∞ t 1 ψx dt ≤ AM f (x). |AR (x)| + R t2 1 Set

323

(4.5.14)

N (f )(x) = sup f 1 (x) . R>0

R

Then, it follows from (4.5.13) that ∞ t cos t − π2 (n + k + β) P (f ) x; dt lim R t1+β β→0+ 1 $ # ≤ A S6∗α0 (f )(x) + N (f )(x) + M (f )(x) . Obviously taking sup for R on the left side of the above inequality equals to TP(f )(x). By making use of the estimate about the type of S6∗α0 in Theorem 4.5.1 and that of the operator ϕ in Theorem 4.3.1, we have the equivalence of ϕ and N (cf. (4.3.13)). In addition, we use the estimate about the type of the operator M to get (4.5.12). n−1 n−1 +β +β with SR 2 in the above arguFor the case k = 0, we replace S6R 2 ment and utilize the estimate about the type of S∗α0 (in the same form as Theorem 4.5.1) to carry out the similar discussions. We note that for the case when k = 0, P (·) is a constant and then we have P (f )(x; t) = Cfx (t). This is the end of the proof of Lemma 4.5.3.

Lemma 4.5.4 If the operator T deﬁned on L(Q) satisﬁes T (f )p ≤ A0 pf p ,

324

C4. The conjugate Fourier integral and series

for 2 ≤ p < ∞, then there exists a constant A only depending on A0 such that − λ (4.5.15) |{x ∈ Q : T (f )(x) > λ}| ≤ Ae Af ∞ , for any λ > 0. Proof. Let f ∈ L∞ (Q) and set , Eλ = x ∈ Q : T (f )(x) > λ . Then, we have

T (f )(x) p A0 pf p p , |Eλ | ≤ dx ≤ λ λ Eλ

for p ≥ 2. If λ > 2A0 f ∞ e, by choosing p = |Eλ | ≤ e

λ A0 f ∞ e , λ 0 f ∞

−A

then we immediately have

e.

Hence, we merely need to take A = |Q|e2 + A0 e and then (4.5.15) holds. This completes the proof of Lemma 4.5.4.

4.6

The relations between the conjugate series and integral

For the non-conjugate case, by the Stein’s theorem (see Section 3.2) about the relation between the series and the integral, we can transfer the problem of series into that of integral in many cases. This facilitates out research of the series. For the conjugate case, we have mentioned the conjugate integral in the previous two sections of this chapter. It is easy to see that the research of the conjugate integral is more convenient than that of the conjugate series. As far as the case at the critical index is concerned, the Bochner-Riesz means of the series has no integral expression. While for the integration, the formula (4.1.24) is valid in the large domain of Re β > − n+1 2 . Therefore, it is necessary to extend Stein’s theorem to the conjugate case. Chang [Cha1] just did such a work in this ﬁeld. (n) Let P (x) ∈ Ak , k ≥ 1 and K(x) = P (x)|x|−n−k . In Section 4.4, we 6 α (y) (see (4.4.2)). Now, we denote by have discussed the kernel D R α +s P (m) |m|2 0 S eim·x , (4.6.1) DR (y) = 1− 2 |m|k R 0<|m|
4.6 The relations between the conjugate series and integral

325

n+1 where α0 = n−1 2 and s is a complex with Res > − 2 . We also denote the kernel of the conjugate integral by α +s |x|2 0 P (x) s 1− 2 HR (y) = eix·y dx. (4.6.2) k R |x|
We easily have Γ( k2 ) s (y) DR π 2n · Γ n+k 2 α +s |m|2 0 = k(m) 1 − 2 eim·y , R

6 α0 +s (y) = (−i)k D

n 2

|m|
and 6 α0 +s (y) := H R

α +s |x|2 0 eix·y dx K(x) 1 − 2 R |x|

= (−i)k Set = (−i)k Cn,k

Γ( k2 ) s n+k HR (y). n n 2 π 2 Γ 2

Γ(k/2) 1 , n+k = (−i)k C(n, k) π 2n Γ 2 n 2

where C(n, k) was introduced in Section 4.1 (see Lemma 4.1.2). For f ∈ L(Qn ), we have 1 α0 +s 6 α0 +s (y)dy, S6R (f ; x) = f (x − y)D R |Q| Q while for g ∈ L(Rn ), we also have α0 +s σ 6R (g; x) =

1 |Q|

Rn

6 α0 +s (y)dy. g(x − y)H R

Set s s (x) − HR (x). sR (x) = DR

(4.6.3)

Similar to the non-conjugate case, we aim at proving the following theorem. Theorem 4.6.1 There exists a constant A such that sup 0R p ≤ Ap, for 1 ≤ p < +∞.

(4.6.4)

326

C4. The conjugate Fourier integral and series From (4.6.4), we can deduce that there exists a constant α > 0 such that sup eα|R (x)| dx ≤ A. (4.6.5) R>0 Q

Furthermore, the inequality (4.6.5) yields the next desired result. Theorem 4.6.2 Suppose that f ∈ L log + L(Qn ) with n > 1, g = f χQ and G is a closed subset of Q. Then we have α0 α0 (f ; x) − σ 6R (g; x) − C(x) = 0, (4.6.6) lim sup S6R R→∞ x∈G

where C(x) =

1 (2π)n

Qn

f (y) K ∗ (x − y) − K(x − y) dy.

(4.6.7)

In the following, we will brieﬂy formulate the steps of the proof, without giving any detailed proof like the non-conjugate case. Step I. Let σ = Res ∈ (0, 1]. By (4.4.11), we have s DR (x)

k

=i

∞

P (x + 2πm)E(n, s, k, R|x + 2πm|)Rn+k .

(4.6.8)

j=0 m∈Sj

Combining (4.6.2) with (4.1.5), we have that s (x) = ik P (x)E(n, s, k, R|x|)Rn+k . HR

(4.6.9)

Subtracting (4.6.9) from (4.6.8), we get that sR (x)

k

=i

∞

P (x + 2πm)E(n, s, k, R|x + 2πm|)Rn+k .

(4.6.10)

j=1 m∈Sj

Let

∞ P (x + 2πm) , F (x) = i C(n, k) |x + 2πm|n+k k

(4.6.11)

j=1 m∈Sj

for x ∈ Q. −n−k = K(x) and the In (4.4.8) and (4.4.9), / we replace g(x) by P (x)|x| /∞ summation by j=1 m∈Sj . Similar arguments show that the series in (4.6.11) is locally and uniformly convergent and thus F ∈ C(Q).

4.6 The relations between the conjugate series and integral

327

Lemma 4.6.1 Let s = σ + iτ , 0 < σ ≤ 1 and τ ∈ R. Then we have sup |sR (x) − F (x)| ≤

x∈Q

Ae2π|τ | . σRσ

(4.6.12)

Proof. By (4.4.4) we have that |sR (x) − F (x)| ≤ Ae2π|τ |

|m|=0

1 Rσ |m|n+σ

,

for x ∈ Q. This completes the proof of Lemma 4.6.1. Step II. Let − 12 ≤ Res = σ ≤ 0. We choose ψ ∈ C 1 ([0, 1]) satisfying 1 ψ(u)uj du = 0, 0

for j = 0, 1, . . . , n − 1. We deﬁne some modiﬁed quantities as follows 1 u n−1+2s 6 s (x) = ψ(u)sR+u (x)du, 1 + R R 0 1 u n−1+2s s s 6R (x) = ψ(u)HR+u (x)du, 1+ D R 0 and s 6R (x) = H

1

1+ 0

u n−1+2s s ψ(u)HR+u (x)du. R

Lemma 4.6.2 Let S = σ + iτ , − 12 ≤ σ ≤ 0 and τ ∈ R. Then we have 6s sup (x) (4.6.13) ≤ Ae2π|τ | R−σ . R x∈Q

Proof. By making use of the uniform convergence of (4.6.10) for σ > 0, we have 1

u n−1+2s s 6 R (x) = sR+u (x) 1 + ψ(u)du R 0 1 ∞ k P (x + 2πm) E(n, s, k, (R + u)|x + 2πm|) =i j=1 m∈Sj

×(R + u)n+k ψ(u)du.

0

(4.6.14)

328

C4. The conjugate Fourier integral and series

Then we will verify the fact that the series on the right side of the above equation is internally closed uniformly convergent about s when σ ∈ (−1, 1). And thus the series are analytic with respect to s. Since the analyticity of 6 s (x) about s is evident, the equation (4.6.13) ia valid when σ > −1. R The key point lies in the fact that when the series which execute summation about m with m = 0 is added in some terms and makes the integral about some parameter, u ∈ [0, 1], for one time, then its order characterizing the increase about |m| will decrease by 1. This is equivalent to multiplication / u|m| by the factor |m|−1 . So does the case of the series m=0 cos|m| n . Thus we immediately obtain (4.6.13) from (4.6.14), because it has a ﬁne property of convergence on the right side. This completes the proof of Lemma 4.6.2. Lemma 4.6.3 Let s = σ + iτ , − 12 ≤ σ ≤ 0 and τ ∈ R. Then we have 6s s (x) − H (x) (4.6.15) sup H ≤ AR−σ e2π|τ | . R R x∈Q

Proof. By the deﬁnition, we have that 1# $ s s −(n−1+2s) s s 6 HR (x) − HR (x) = R (x) − Rn−1+2s HR (x) (R + u)n−1+2s HR+u 0

× ψ(u)du. By the Taylor formula, we have that s s (x) − Rn−1+2s HR (x) = (R + u)n−1+2s HR+u

n−1

aj uj + E(n),

j=1

where the remainder E(n) satisﬁes n d 1 n−1+2s s sup n {(R + u) |E(n)| ≤ HR+u (x)} . n! 0≤u≤1 du

(4.6.16)

It follows from the orthogonal condition of ψ that 1 s s −(n−1+2s) 6 E(n)ψ(u)du, HR (x) − HR (x) = R 0

which implies 6s s (x) ≤ HR (x) − HR

A Rn−1+2σ

0

1

|E(n)|du.

(4.6.17)

4.6 The relations between the conjugate series and integral

329

Set s f (u) = (R+u)n−1+2s HR+u (x) = ik P (x)(R+u)2n+k−1+2s E(n, s, k, (R+u)|x|).

Applying the Leibniz formula, we have f

(n)

k

(u) = i P (x)

n

Aj (R + u)2n+k−1+2s−j |x|n−j E (n−j) n, s, k, (R + u)|x| ,

j=0

where Aj = Cnj 2n + k − 1 + 2s · · · 2n + k − 1 + 2s − j + 1 . Thus we obtain n (n) Aeπ|τ |/2 (R + u)2n+k−1+2σ−j |x|n+k−j f (u) ≤ j=0

× E (n−j) (n, s, k, (R + u)|x|) , for |σ| < 1. Since n+1 2j ∞ (−1)j Γ n+k + j Γ + β π n/2 t 2 2 n n+k n+1 E(n, β, k, t) = k , 2 2 j! Γ 2 + k + j Γ 2 + 2 + β + j j=0 E is analytic about t. If we take a derivative with respect to t, then we have 1 E n, β, k, t = − tE n + 2, β − 1, k, t . 2π By induction, we get that, for ν = 1, 2, . . ., E (2ν−1) (n, β, k, t) =

ν−1

aν,j t2j+1 E(n + 2(ν + j), β − (ν + j), k, t),

j=0

and E

(2ν)

(n, β, k, t) =

ν

bν,j t2j E(n + 2(ν + j), β − (ν + j), k, t),

j=0

where |aν,j | + |bν,j | ≤ Aν .

(4.6.18)

330

C4. The conjugate Fourier integral and series

For σ ∈ (−1, 0], it follows from (4.6.18) and the asymptotic formula (4.4.4) that 1 (ν) E (n, s, k, t) ≤ Aν n+k+σ , t for t > 0 and ν = 0, 1, 2, . . . . Since x ∈ Q, we have (n) f (u) ≤ Ae2π|τ | Rn−1+σ , for u ∈ [0, 1]. By (4.6.16), we have |E(n)| ≤ Ae2π|τ | Rn−1+σ . By substituting this into (4.6.17), we get (4.6.15). This ﬁnishes the proof of Lemma 4.6.3. Lemma 4.6.4 Let s = σ + iτ , − 12 ≤ σ ≤ 0 and τ ∈ R. Then we have 6s s (4.6.19) ≤ Ae2π|τ | R−σ , DR − DR 2

for any R > 0. Here we denote by · 2 the L2 norm on Qn . Proof. The proof of this lemma can be ascribed to the fact in the nonconjugate case, which has already been obtained. Let α +s |m|2 0 λm = 1 − 2 χ[0,R) (|m|) R and 6m = λ

1

0

1−

|m|2 (R + u)2

α0 +s

u n−1+2s χ[0,R+u) (|m|)ψ(u) 1 + du. R

For the non-conjugate case, we have

2 6m − λm λ

1 2

≤ Ae2π|τ | R−σ .

m

s (x) and D 6m , 6 s are P m λm and P m λ The Fourier coeﬃcients of DR R |m| |m| m = 0, respectively, and their coeﬃcients of order zero are both zero. Thus

4.6 The relations between the conjugate series and integral

we immediately get (4.6.19) by the boundedness of P

m |m|

331 with m = 0.

Combining Lemma 4.6.2, 4.6.3 and 4.6.4, we obtain the L2 estimate sR (x) − F (x) 2 ≤ Ae2π|τ | R−σ ,

(4.6.20)

for − 12 ≤ σ ≤ 0. Step III. Similar to the non-conjugate case, by the L∞ estimate (4.6.12), L2 estimate (4.6.20) and the method of the complex interpolation, we have sup 0R (x) − F (x)p ≤ Ap,

R>0

for 2 ≤ p < ∞. From the above discussion, we have proven Theorem 4.6.1.

n−1

6 2 . Finally, we deduce the Lebesgue constant of the conjugate kernel D R That is, n−1 n−1 6 2 := 6 R := L D 6 2 (4.6.21) L R R (y) dy. Q

Theorem 4.6.3 There exists a constant α , only depending on the dimension n and the spherical harmonic function P , such that LR = α log R + O(1),

(4.6.22)

as R → ∞. Proof. Since 6 α0 (y) = H 6 α0 (y) + (−i)k D R R

1 0 (y), C(n, k) R

6 α0 LR = HR (y) dy + r,

we have

Q

where |r| ≤ A

0 (y) dy = O(1). R

Q

However, it follows that 1 6 α0 (y)|dy = A |H |E(n, o, k, Rt)|Rn+k tn+k−1 dt + O(1) R Q

=A

0

#

$ 1 π t α + β cos Rt − 2 (n + k) dt + O(1)

1 1 R

= α log R + O(1).

(4.6.23)

332

C4. The conjugate Fourier integral and series

By Lemma 4.1.2, we have

n 2n π 2 Γ n+k 2 α = C(n, k) = Γ k2 &

n 2n π 2 Γ n+1 2 2 √ . B(n, 0) = β= π π By the deﬁnition of β-function, we conclude that if z ≥ y > 0 and x > 0, then B(z, x) ≤ B(y, x). On the other hand, by the result (see [Ba1]),

and

B(x, y)B(x + y, z) = B(z, x)B(x + z, y), we have B(x + y, z) =

B(z, x) B(x + z, y) ≤ B(x + z, y), B(y, x)

for z ≥ y > 0. Consequently, it follows that n+k 1 n+1 k , ≤B , . B 2 2 2 2 That is, Γ

n+k

2 Γ k2

Γ n+1 2 ≥ Γ 12

for k ≥ 1. This implies α ≥ β. Here the equal sign holds only if k = 1. This completes the proof of Theorem 4.6.3.

4.7

Convergence of Bochner-Riesz means of conjugate Fourier series

As an application of the theorems in Section 4.6, we will give, in this section, the results about the a.e.-convergence of the conjugate series due to Lu [Lu6], which is parallel to the results in Section 3.6. 2 Theorem 4.7.1 Let 1 < q ≤ ∞. If f ∈ Bq L log+ L(Qn ) with n > 1, (n) P ∈ Ak , k ≥ 1, then by taking the conjugation about P , we have n−1

lim S6R 2 (f ; x) = f ∗ (x)

R→∞

for a.e. x ∈ Qn .

(4.7.1)

4.7 Convergence of Bochner-Riesz means of conjugate Fourier series Proof. that

333

Let g = f χQ . Since f ∈ L log+ L(Q), by Theorem 4.6.2, we have n−1 n−1 lim S6R 2 (f ; x) − σ 6R2 (g; x) − C(x) = 0, R→∞

for any x ∈ Q◦ , where 1 C(x) = n |Q |

Qn

K ∗ (x − y) − K(x − y) f (y)dy.

Since K ∗ (u) − K(u) is internally and closed uniformly continuous on the set 2Qn , the function C is continuous on (Qn )◦ . In addition, we have C(x) = f ∗ (x) − g6(x), for almost everywhere x ∈ Qn . It suﬃces to show that n−1

g (x) σ 6R2 (g; x) −→ 6 for a.e. x ∈ Qn . And therefore we merely need to prove that n−1 2 6R (g; x) σ 6∗ (g)(x) := sup σ R>0

satisﬁes the weak-type estimate 1 {x : σ 6∗ (g)(x) > λ} < ANq (g). λ

(4.7.2)

By the superposition principle, it suﬃces to prove this for each q-block. Suppose that b is a q-block for 1 < q < ∞. As what we have mentioned before in Theorem 4.3.1, it has been proven in the theory of the Calder´onZygmund singular integral that the Calder´on-Zygmund transform or Hilbert transform is of strong-type (q, q) with 1 < q < ∞. And thus 6b ∈ Lq (Rn ). Here, we introduce the following facts without proof that, for b ∈ Lq with 1 < q < ∞, we have

α0 6 α0 σR (b). b =σ 6R Thus we have

6 σ∗ (b)q = σ∗ 6b ≤ Aq 6b ≤ Aq bq . q

By noticing the kernel 6 α0 (y) = O H R

q

1 |y|n

(4.7.3)

334

C4. The conjugate Fourier integral and series

together with (4.7.3), we can deduce the validity of the equation (4.7.2) for the q-block b, just as we have done in the non-conjugate case. And therefore, we have ﬁnished the proof of the theorem. We have to point out that, similar to the non-conjugate case, the conn−1 vergence of the S6R 2 (f ; x) also does not satisfy the localization principle, which was proven by Lippman [Li1]. For the similarity between its proof and that of the non-conjugate case, we omit the detailed proof here.

4.8

(C, 1) means in the conjugate case n−1

The arithmetic mean of the conjugate Bochner-Riesz means S6R 2 (f ; x) of the critical order 1 R 6 n−1 (4.8.1) Su 2 (f ; x)du R 0 is called the conjugate (C, 1) mean. It should have some good convergence properties. α0 +β Let β ∈ (0, 1). We ﬁrst consider SR and then turn to (4.8.1) by taking limit β → 0+ . (n)

Theorem 4.8.1 Let f ∈ L(Qn ) with n > 1, and P ∈ Ak f (x − ty)P (y)dσ(y). ψx (t) =

with k ≥ 1. Set

Sn−1

If

t

0

|ψx (τ )|dτ = o(t),

holds, as t → 0+ , then we have

lim S6α0 +β (f ; x) − f 1 (x) = 0, R→∞

R

R

(4.8.2)

(4.8.3)

(4.8.4)

for β > 0. Proof. By (4.3.15) and (4.4.12) (also in (4.5.5)), we conclude that α0 +β (f ; x) − f 1 (x) S6R R 1 1 t 1 tn+k−1 E(n, β, k, t)dt = ψx (4.8.5) |Q| C(n, k) 0 R ∞ 1 1 t C(n, k) ψx + tn+k−1 E(n, β, k, t) − n+k dt. |Q| C(n, k) 1 R t

4.8 (C, 1) means in the conjugate case

335

We denote the ﬁrst term on the right side of the equation (4.8.5) by AR (x). From the expression of E in (4.1.6), we have that when β > 0, |E(n, β, k, t)| ≤ A uniformly holds about β and t ∈ [0, 1]. And therefore, under the condition of (4.8.3), there uniformly holds about β AR (x) = o(1),

(4.8.6)

as R → ∞. By Lemma 4.1.2, we have that when β ∈ (0, 1), with respect to β, cos t − π2 (n + k + β) 1 C(n, k) E(n, β, k, t) − n+k = O n+k+1 + A(n, β) t t tn+k+β uniformly holds, where A(n, β) = π

n−1 2

n+β

2

Γ

n+1 +β . 2

Thus, it follows from the condition (4.8.3) that ∞ t C(n, k) n+k−1 t E(n, β, k, t) − n+k dt ψx R t 1 ∞ π t cos t − 2 (n + k + β) ψx dt + o(1), = A(n, β) R t1+β 1 as R → ∞. Let 1 2β Γ 1 1 √ Aβ = A(n, β) = |Q| C(n, k) (2π)n π

n+1 2

Γ

(4.8.7)

+ β Γ k2 n+k . 2

By substituting (4.8.6) and (4.8.7) into (4.8.5), we obtain that α0 +β (f ; x) − f 1 (x) S6R R ∞ t cos t − π2 (n + k + β) ψx dt + o(1) = Aβ R t1+β 1 cos Rt − π2 (n + k + β) Aβ ∞ = β ψx (t) dt + o(1), 1 R t1+β

(4.8.8)

R

holds uniformly about β ∈ (0, 1), as R → ∞. Since the ﬁrst term on the right side of (4.8.8) is obviously o(1), we have actually proved Theorem 4.8.1.

336

C4. The conjugate Fourier integral and series

Remark 4.8.1 From the proof of Theorem 4.8.1, we can weaken the condition (4.8.3) into the following

t 0

ψx (τ )dτ = o(t),

as t → 0+ . If f vanishes in B(x; δ) with δ > 0, then (4.8.3) is valid. And thus Theorem 4.8.1 contains the localization principle over the critical index. Our main results are formulated as follows. Theorem 4.8.2 Let f ∈ L(Qn ) with n > 1. If (4.8.3) holds, then we have 1 lim R→∞ R

R

n−1 ∗ 2 6 Su (f ; x) − f 1 (x) du = 0. u

0

(4.8.9)

Proof. Let β ∈ (0, 1). By (4.8.8), when R → ∞, we have that

α0 +β ∗ 6 (f ; x) − f 1 (x) du Su u 0 ∞ ψx (t) 1 R cos(ut − θ) du dt + o(1), = Aβ 1 t1+β R 1 uβ

1 R

R

R

(4.8.10)

t

where θ = π2 (n + k + β) and o(1) is uniformly valid for β ∈ (0, 1). Set R 1 MR (β, t, τ ) = t cos(ut − θ)du, (4.8.11) β (ut) τ for β > 0, τ ∈ (0, ∞) with tτ ≥ 1. By the mean value theorem of integral, we have t MR (β, t, τ ) = (tτ )β

τ

ξ

1 cos(ut − θ)du = (τ t)β

τξ

cos(u − θ)du.

tτ

Since |MR (β, t, τ )| ≤ 2

(4.8.12)

4.9 The strong summation of the conjugate Fourier series holds for β > 0, τ ∈ (0, ∞) with tτ ≥ 1, we conclude that ∞ ψx (t) 1 R cos(ut − θ) sup Aβ dudt 1+β R 1 β 1 t u β∈(0,1) R t ∞ |ψx (t)| 1 sup MR β, t, t−1 dt ≤A 2 1 t R β∈(0,1) R ∞ 1 |ψx (t)| dt ≤A R 1 t2

337

(4.8.13)

R

= o(1), as R → ∞. Substituting (4.8.13) into (4.8.10) and taking β → 0+ , we get (4.8.9). This completes the proof. Since the condition (4.8.3) is valid for a.e. x ∈ Rn , then so is (4.8.9). In addition, since fε∗ (x) → f ∗ (x), for a.e. x ∈ Rn , then Theorem 4.8.2 shows that 1 R 6 n−1 lim Su 2 (f ; x)du = f ∗ (x) R→∞ R 0 holds for a.e. x ∈ Rn . The above results were obtained by Wang [Wa2].

4.9

The strong summation of the conjugate Fourier series

Because (4.8.9) is valid for a.e. x ∈ Rn , it is virtually natural to ask whether there is any possibility to strengthen it to the strong limit. That is the strong summation problem, which is far from being completely solved, like the non-conjugate case. Here we will give the suﬃcient condition for the strong summation at a ﬁxed point. This parallels to the non-conjugate case. We shall prove this result in three steps. We ﬁrst describe the localized results and then transform the problem to the conjugate integral by the condition of the locally p integral with p > 1 as well as the results in Section 4.6. Finally, we prove the conclusion about the strong summation for the conjugate integral. Step I. The localization.

338

C4. The conjugate Fourier integral and series

Theorem 4.9.1 Let f ∈ L(Qn ) with n > 1. If f vanishes in B(x, δ) with δ > 0 and q > 0, then we have q 1 R 6 n−1 2 (4.9.1) Su (f ; x) − f 1 (x) du = 0. lim u R→∞ R 0 Proof. Let β ∈ (0, 1). By (4.8.8), we merely need to prove that 1 lim sup sup R→∞ β∈(0,1) R where

β

I (u) =

∞ 1 u

R 1 δ

|I β (u)|q du = 0,

cos ut − π2 (n + k + β) ψx (t) dt. uβ t1+β

(4.9.2)

(4.9.3)

For u > δ−1 , this implies u−1 < δ. At the same time, ψx (t) equals to zero when t < δ. Thus I β (u) in (4.9.3) can be rewritten as ∞ cos ut − π2 (n + k + β) β I (u) = ψx (t) dt, (4.9.4) uβ t1+β δ for δ−1 ≤ u ≤ R. In addition, by H¨older’s inequality, it suﬃces to prove q = 2λ with λ ∈ N. Since cos(ut − θ) = cos ut cos θ + sin ut sin θ, we merely need to show that 1 lim sup sup R→∞ β∈(0,1) R

β q J (u) du = 0,

R I δ

where q = 2λ and J β is taken as the following two functions ∞ cos ut ψx (t) β 1+β dt u t δ and

∞

ψx (t) δ

sin ut dt. uβ t1+β

(4.9.5)

(4.9.6)

(4.9.7)

Since the method of the proof for the two diﬀerent cases are the same, we choose J β as in (4.9.6). Since q = 2λ and J β (u) is a real function, we have

q β q β (u) = J (u) . J

4.9 The strong summation of the conjugate Fourier series We have 1 R β q J (u) du R 1 δ

=

1 R

⎞ q ψx (t1 ) · · · ψx (tq ) ⎝ 1 cos utj ⎠ du dt1 · · · dtq . 1+β (t1 · · · tq ) uqβ 1/δ

(δ, ∞)q

339

⎛

R

j=1

Line up all the q-dimensional vectors whose components are taken as either 1 or −1. We denote these vectors by {el }2l=1 , where el = (el,1 , . . . , el,q ), el,j = 1 or − 1, j = 1, 2, . . . , q. By the inner product el · t = el,1 t1 + · · · + el,q tq , we have q j=1

2q 1 cos utj = q · cos(el · t)u, t l=1

where t = (t1 , . . . , tq ) ∈ Rq . According to the mean value formula of integral, we have that, for (el · t) = 0, R 1 cos u(el · t)du 1 uqβ δ ξ cos(el · t)udu = δqβ 1 δ

= δqβ (el · t)−1 sin ξ(el · t) − sin δ−1 (el · t) , for ξ ∈ (δ −1 , R). Thus we have R 1 A cos u(el · t)du ≤ , 1 uqβ |el · t| + R−1 δ

for β ∈ (0, 1), t ∈ Rq . Set h(tj ) = |ψx (tj )|,

340

C4. The conjugate Fourier integral and series

for j = 1, 2, . . . , q. Then it suﬃces to show that ⎛ ⎞ q −1 1 h(tj ) ⎠ ⎝ |el · t| + R−1 dt = 0, lim R→∞ R (δ, ∞)q tj

(4.9.8)

j=1

for l = 1, . . . , 2q . Let el · t = t1 + · · · + tα − (tα+1 + · · · + tq ). We divide the integral into (δ, ∞)q

=

∞

,

m1 ···mq =1 [mδ, (m+1)δ]

where [mδ, (m + 1)δ] =

q

[mj δ, (mj + 1)δ] := m .

j=1

Set τm (R) =

m

h(t1 ) · · · h(tq )dt1 · · · dtq . |t1 + · · · + tα − (tα+1 + · · · + tq )| + R−1

We can ascribe the problem into proving m∈Nq

1 τm (R) = o(R). m1 · · · mq

(4.9.9)

Clearly we have that

a+δ

h(s)ds ≤ Aδ f L(Q) ,

(4.9.10)

a

for any a ≥ δ, and for any ε > 0, there exists η > 0 such that

a+η

h(s)ds < Aε,

(4.9.11)

a

for a ≥ δ. Since f vanishes in B(x; δ), the above restriction a ≥ δ can be canceled. These two things are based on the periodicity and the local integrability of f. We ﬁrst prove the following results.

4.9 The strong summation of the conjugate Fourier series

341

(a) τm (R) = o(R) uniformly holds for m ∈ Nq . (b) If |m1 + · · · + mα − (mα+1 + · · · + mq )| > 2qδ, we have t1 + · · · + tα − (tα+1 + · · · + tq ) ≥ δ m1 + · · · + mα − (mα+1 + · · · + mq ), 2 for t ∈ m , which implies τm (R) ≤

A . |m1 + · · · + mα − (mα+1 + · · · + mq )|

Since (b) is obvious, we will prove (a) in the following. Let s = −(t2 + · · · + tα ) + tα+1 + · · · + tq . Then we have [m1 δ,(m1 +1)δ]

=

h(t1 ) dt1 |t1 − s| + R−1

{t1 :|t1 −s|< η2 }∩[m1 δ, (m1 +1)δ]

+

h(t1 ) dt1 |t1 − s| + R−1

{t1 :|t1 −s|≥ η2 }∩[m1 δ, (m1 +1)δ]

h(t1 ) dt1 |t1 − s| + R−1

:= I1 + I2 . It follows from (4.9.11) and (4.9.10) that I1 < ARε and

1 I2 < A f L(Q) . η

Thus it follows that (m2 +1)δ (mq +1)δ 1 h(t2 )dt2 · · · h(tq )dtq τm (R) ≤ A Rε + η m2 δ mq δ 1 ≤ A Rε + η holds uniformly for m ∈ Nq . This implies (a).

342

C4. The conjugate Fourier integral and series Now we rewrite the sum on the left side of the equation (4.9.9) as m∈Nq

=

+

|m1 +···+mα −(mα+1 +···+mq )|<2qδ

other

:= σ1 + σ2 . m

Applying the estimate of (b) in σ1 , we get σ1 = O(1). And by the estimate of (a) in σ2 we have σ2 = o(R). One can check the detailed steps in the computation of (3.10.17) and (3.10.18) in Chapter 3). Hence (4.9.9) holds. This ﬁnishes the proof of Theorem 4.9.1.

Step II. The strong summation of the conjugate Fourier integral. Theorem 4.9.2 Let g ∈ L(Rn ) with n > 1. We have, at the point x, t

0

|y|=1

g(x − τ y)P (y)dσ(y) dτ = o(t),

(4.9.12)

as t → 0. In addition, if g is r integrable on x + B(0; δ) with δ > 0 and r > 1, and r t g(x − τ y)P (y)dσ(y) (4.9.13) dτ = O(t), 0 |y|=1 then we have 1 lim R→∞ R

R 0

q α0 6u (g; x) − 6 g 1 (x) du = 0, σ u

(4.9.14)

for any q > 0. Proof. We might as well take it for granted that x = 0 and r < 2. It r suﬃces to prove for the conjugate number q = r−1 with the index of r. Deﬁne g(−ty)P (y)dσ(y) ψ(t) = |y|=1

and

t

ψ(τ )dτ.

h(t) = 0

Then the conditions (4.9.12) and (4.9.13) are represented as h(t) = o(t)

4.9 The strong summation of the conjugate Fourier series

and

t 0

343

|ψ(τ )|r dτ = O(t).

According to (4.1.24), we have α0 σ 6R (g; 0) =

1 Rn+k (2π)n

∞ 0

By the deﬁnition 1 g6 1 (0) = R (2π)n we also obtain α0 (g; 0) σ 6R

1 − g6 1 (0) = R (2π)n

1 + (2π)n := I1 + I2 .

∞ 1 R

ψ(t)t

1 R

0

n+k−1

E(n, 0, k, Rt) dt. C(n, k)

ψ(t)t−1 dt,

∞ 1 R

ψ(t)tn+k−1

R

ψ(t)tn+k−1

n+k E(n, 0, k, Rt)

C(n, k)

−

1

tn+k

dt

1 Rn+k E(n, 0, k, Rt)dt C(n, k) (4.9.15)

By integration by parts, we have that R−1 1 n+k−1 n+k I2 = h(t)t R E(n, 0, k, Rt) (2π)n C(n, k) 0 1/R ! " d n+k−1 n+k t − h(t) R E(n, 0, k, Rt) dt . dt 0 By substituting h(t) = o(t) into the above equation and (4.6.18), we have that 1/R d I2 = o(1) + A h(t) (E(n, 0, k, Rt))tn+k−1 Rn+k dt dt 0 1/R h(t)tn+k E(n + 2, −1, k, Rt)Rn+k+2 dt = o(1) + A 0

= o(1).

(4.9.16)

Again, integration by parts for I1 , we have ∞ 1 1 n+k−1 n+k E(n, 0, k, Rt) − I1 = R h(t)t (2π)n C(n, k) (Rt)n+k 1 R ∞ d 1 E(n, 0, k, Rt) tn+k−1 − dt . + h(t)Rn+k 1 dt C(n, k) (Rt)n+k R

344

C4. The conjugate Fourier integral and series

By the facts |h(t)| ≤ M < +∞, as t → 0, and

h(t) = o(t)

E(n, 0, k, Rt) = O

1 (Rt)n+k

,

for t ≥ R1 , we have that the ﬁrst term on the right side is o(1). According to the formula (4.6.18) and the expression (4.4.4) of the function E, we calculate that d n+k−1 E(n, 0, k, Rt) 1 t − Rn+k dt C(n, k) (Rt)n+k 1 sin Rt cos Rt R sin Rt R cos Rt + A4 +O . = A1 2 + A2 2 + A3 t t t t Rt3 From the above equation and the conditions h(t) = o(t), as t → 0, and |h(t)| ≤ M , we get that

∞ ψ(t) ψ(t) cos Rtdt + B2 sin Rtdt t t −1 −1 R R ∞ ∞ ψ(t) ψ(t) cos Rtdt + B4 sin Rtdt + o(1). + B3 2 2 R−1 Rt R−1 Rt ∞

I 1 = B1

We denote the four integrals on the right side of the above equation by J1 (R), J2 (R), J3 (R), J4 (R), respectively. It follows from (4.9.15) and (4.9.16) that 4 α0 σ 6R (g; 0) − g6 1 (0) = Bν Jν (R) + o(1). R

ν=1

Thus the problem can be ascribed into proving 1 R→∞ R

R

lim

0

|Jν (u)|q du = 0,

(4.9.17)

for ν = 1, 2, 3, 4. In the following, we only write down the proof for the case ν = 1 and the proofs for ν = 2, 3, 4 are similar. Set 1 R α = lim sup |J1 (u)|q du. R→∞ R 0

4.9 The strong summation of the conjugate Fourier series

345

For any ε ∈ (0, 1), we have that

1 εu 1 u

1 1 εu ψ(t) cos ut εu cos ut u sin ut cos utdt = h(t) dt + h(t) + 1 t t 1 t2 t u

u

1 1 = o(1) + o(1) log + o(1) ε ε = o(1), as R → ∞. Hence, we have q 1 εu ψ(t) cos ut dt du 1 t 0 u q 1 R ∞ ψ(t) q +2 lim sup cos utdt du 1 R t R→∞ 0 εu q 1 R ∞ ψ(t) q cos utdt du. = 2 lim sup 1 R t R→∞ 0

1 α ≤ 2q lim sup R→∞ R

R

(4.9.18)

εu

Let

t

Φ(u, t) = 0

ψ(τ ) cos uτ dτ.

Obviously we have that that 0

= o(t) + uo(t2 ), as t → 0, and

|Φ(u, t)| ≤

t 0

t

h(τ ) sin uτ dτ

Φ(u, t) = h(t) cos ut + u

(4.9.19)

|ψ(τ )|dτ ≤ M < +∞,

for any u, t. Therefore, we have

∞ 1 εu

∞ ψ(t) 1 ∞ cos utdt = Φ(u, t) + Φ(u, t)t−2 dt 1 t t 1 εu εu ∞ Φ(u, t)t−2 dt, = o(1) + 1 εu

(4.9.20)

346

C4. The conjugate Fourier integral and series

as u → ∞. This implies 1 α ≤ Aq lim sup · R→∞ R 1 ≤ Aq lim sup R R→∞

R 0

1 εδ

1 +Aq lim sup R R→∞

1 εu

∞ 1 εu

0

1 +Aq lim sup R→∞ R

∞

R

1 εδ

|Φ(u, t)| dt t2 |Φ(u, t)| dt t2

δ 1 εu

R ∞

1 εδ

δ

q du q

|Φ(u, t)| dt t2

du q

|Φ(u, t)| dt t2

du q du.

Since |Φ(u, t)| ≤ M , the ﬁrst term on the right side is obviously zero. For ﬁxed t, by the Riemann-Lebesgue theorem, we get that limu→∞ Φ(u, t) = 0, which leads to the third term is also zero. By the controlled convergence theorem, we have that q ∞ |Φ(u, t)| dt = 0. lim u→∞ t2 δ Then, according to the generalized Minkowski’s inequality, we get that q R δ 1 |Φ(u, t)| dt du α ≤ Aq lim sup 1 1 t2 R→∞ R εδ εu ⎧ ⎫ R 1q ⎬q δ ⎨ 1 1 ≤ Aq lim sup |Φ(u, t)|q du dt . 2 ⎭ R→∞ R ⎩ 1 t 0 εR

It follows from the Hausdorﬀ-Young’s inequality that 1q 1r t R 1 q r |Φ(u, t)| du ≤ Aq |ψ(τ )| dτ ≤ Aq t r , 0

which implies 1 α ≤ Aq lim sup R R→∞

(4.9.12)

0

δ 1 εR

1 2− 1r

t

q ≤ Aq

1 q 1 (εR)(1− r ) = Aq ε. R

Together with the arbitrary of ε, this yields α = 0. This completes the proof. Step III. Combining Theorem 4.9.1, 4.9.2 with 4.6.2, we have the next result.

4.10 Approximation of continuous functions

347

Theorem 4.9.3 Let f ∈ L(Qn ), n > 1, and f is r (r > 1) integrable on B(x, δ) with δ > 0. If the conditions (4.9.12) and (4.9.13) hold for f , then for any q > 0, we have q 1 R 6 n−1 2 Su (f ; x) − f 1 (x) du = 0. lim u R→∞ R 0 The above results about the strong summation were done by Wang [Wa7].

4.10

Approximation of continuous functions

For the case of one dimension, the approximation of continuous functions and their conjugate functions through their partial sum in Fourier series deﬁned on set of total measure, was solved by Oskolkov [O1]. Now we turn to discuss the case of higher dimension. We have such a result that let f ∈ Lipα n−1

(0 < α < 1), if we consider uniform approximation by SR 2 (f ), then we only R get the order of approximation as log Rα . If we restrict functions on set of total measure, then the order of the a.e. approximation can be O (log log R/Rα ), and we can get the same results about the conjugate functions. Recall the deﬁnitions of the classes of functions W 1 L∞ = Lip1 and W 2 L∞ in Section 3.7. There we have proven that if f ∈ W 2 L∞ , there holds n−1 1 2 SR (f ; x) − f (x) = O (4.10.1) R2 a.e. x ∈ Rn . Now we take further research on the problem of the approximation of some functions which include the conjugate functions. Lemma 4.10.1 If f ∈ W 1 L∞ , then we have ⎫ ⎧ n ⎬ ⎨ n+1 α 2 ) 1 S 0 (f ; x) − f (x) ≤ 2Γ(n+1 T P (f )(x) + θ (f )(x) , (4.10.2) j j R R ⎭ R⎩ π 2 j=1

where Pj (x) = xj , T Pj is the operator deﬁned as in Deﬁnition 4.5.1 and ∂f the formulas (4.5.10) and (4.5.11), and fj = ∂x for (j = 1, 2, . . . , n). The j remainder term θR (f )(x) satisﬁes (I) 0 ≤ θR (f )(x) ≤ A · max{fj ∞ : j = 1, 2, . . . , n}, (II) lim θR (f )(x) = 0, for a.e. x ∈ Qn . R→∞

348

C4. The conjugate Fourier integral and series

Proof. Let β ∈ (0, 1). Due to Bochner’s formula, we get n−1 +β 2

SR

∞

(f ; x) − f (x) = Cn,β

0

where

J 1 (t) t n− 2 +β − f (x) fx dt, 1 R t 2 +β

1

Cn,β Set

2 2 +β Γ( n+1 2 + β) = . Γ(n/2)

∞

Fβ (t) =

1

s− 2 −β Jn− 1 +β (s)ds. 2

t

Then, for t ≥ 1, we conclude that & ∞# cos(s − π2 (n + β)) 2 Fβ (t) = π 2 s1+β $ π sin(s − 2 (n + β)) 1 ds −An,β + O s2+β s3 & 1 2 cos(t − π2 (n + 1 + β)) , = + O π t2 t1+β

(4.10.3)

as t → ∞, where “O” holds uniformly about β ∈ (0, 1). For 0 ≤ t < 1, we have 1 1 Fβ (t) = s− 2 −β Jn− 1 +β (s)ds + Fβ (1) = O(1), (4.10.4) 2

t

where “O” holds uniformly about β ∈ (0, 1). Then it follows from integration by parts that 0

∞

J 1 (t) ∞ t t d n− 2 +β dt fx ( ) − f (x) dt = − Fβ (t) fx 1 +β R dt R 0 t2 1 ∞ t d dt Fβ (t) fx =− + dt R 0 0 = I1 + I 2 .

Applying (4.10.4), we have |I1 | ≤ A

d fx t dt. dt R

1 0

4.10 Approximation of continuous functions

349

It is easy to see that d fx dt

n t t 1 1 fj x − ξ (−ξj )dσ(ξ) = R R ωn−1 Sn−1 R j=1 n t 1 1 Pj (fj ) x; =− . (4.10.5) ωn−1 R R i=1

Obviously, if fj (x) is ﬁnite, then we have t = Pj fj x; t fj x − ξ − fj (x) ξj dσ(ξ) R R Sn−1 t fj x − ξ − fj (x) dσ(ξ). ≤ R Sn−1 If x belongs to the intersection of the sets of the Lebesgue points of ∂f , for j = 1, 2, . . . , n, we denote by fj = ∂x j s

Φj (x; s) =

0

|ξ|=t

|fj (x − ξ) − fj (x)|dσ(ξ)dt.

Then, we have Φj (x; s) = o(sn ), as s → 0+ . Consequently, we have 0

1 R 1 t |fj (x − ξ) − fj (x)|dσ(ξ)dt dt ≤ R Pj (fj ) x; R tn−1 |ξ|=t 0 1 1 R R Φj (x; t) 1 dt = R n−1 Φj (x; t) + (n − 1) t tn 0 0

1

= o(1), as R → ∞. Thus we have I1 = o( R1 ). Combining (4.10.3) with (4.10.5), we have & ωn−1 I2 = −

2 1 πR n

i=1

∞ 1

t Pj (fj ) x; R

cos t − π2 (k + 1 + β) dt + I2 , 1+β t

350

C4. The conjugate Fourier integral and series

where

d fx t 1 dt dt R t2 1 n 1 ∞ −1 −n−1 ≤A R t |fj (x − ξ) − fj (x)|dσ(ξ)dt 1 R |ξ|=t j=1 R n 1 ∞ −1 −n−2 R t Φj (x; t)dt ≤A 1 R j=1 R 1 . =o R

I2 ≤ A

∞

&

Set θR (f ; x) =

π R lim |I1 + I2 |. + 2 β→0

We have ⎞ ⎛ & n n−1 1 2 S 2 (f ; x) − f (x) ≤ ⎝ T Pj (fj )(x) + θR (f )(x)⎠ . R π Cn,0 R j=1

Here when x belongs to the intersection of the sets of the Lebesgue points of fj , then θR (f ) equals to o(1), as R → ∞. Thus the lemma is proven. Suppose P (x) is the homogeneous harmonic polynomial with order k, then for every j ∈ {1, . . . , n}, xj P (x) ∈ Pk+1 . Due to the decomposition theorem on harmonic polynomials (see [SW1]), xj P (x) can be uniquely decomposed into the following form xj P (x) = Pj,k+1(x) + Pj,k−1(x)|x|2 + · · · + Pj,k−1−2νk (x)|x|2νk ,

(4.10.6)

where νk = [ k+1 2 ] and Pj,k+1−z l ∈ Ak+1−2l , for l = 0, 1, 2, . . . , νk . Lemma 4.10.2 If f ∈ W 1 L∞ , then we have ⎞ ⎛ νk n n−1 1 S6 2 (f ; x) − f ∗ (x) ≤ A ⎝ T Pj,k+1−2l (fj )(x) + θ1 R (f )(x)⎠ , R R j=1 l=0

where T Pj,k+1−2l is the operator deﬁned (4.5.11) in Deﬁnition 4.5.1 to Pj,k+1−2l which came from (4.10.6). Here P (x) is the homogeneous harmonic polynomial about conjugate function with order k, k ≥ 1, fj is the 6 )(x) satisﬁes j-th partial derivative on f , and the remainder term θ(f , 0 ≤ θ1 R (f )(x) ≤ A max fj ∞ : j = 1, 2, . . . , n .

4.10 Approximation of continuous functions

351

Proof. Let β ∈ (0, 1). Combining (4.3.14) with (4.4.12), we get ∞ 1 t α0 +β ∗ 6 tn+h−1 P (f ) x; SR (f ; x) − f (x) = |Q| 0 R E(n, β, k, t) 1 × − n+k dt. C(n, k) t Set

∞

Gβ (t) =

n+k−1

s t

1 E(n, β, k, s) − n+k C(n, k) s

(4.10.7)

ds,

for t > 0. Due to the asymptotic formula (4.4.4), we have

cos s − π2 (n + k + β) E(n, β, k, s) −n−k −n−k−2 + A1 (β) −s =O s C(n, k) sn+k+β π sin(s − 2 (n + k + β)) +A2 (β) . sn+k+1+β By making use of the integration by parts, we have that ∞ cos(s − θ) sin(t − θ) cos(t − θ) 1 ds = − + (1 + β) +O 2 , 1+β 1+β 2+β s t t t t and

∞ t

sin(s − θ) ds = O s2+β

1 t2

hold uniformly about β ∈ [0, 1]. Consequently, we have sin(t − θ) +O Gβ (t) = −A1 (β) t1+β and

1 t2

π (n + k + β), 2

θ=

(4.10.8)

for t ≥ 1. While, for 0 < t < 1, we get

1 n+k−1

Gβ (t) = Gβ (1) +

s t

= O(1) + log t

1 E(n, β, k, s) − n+k C(n, k) s

ds (4.10.9)

352

C4. The conjugate Fourier integral and series

uniformly holds about β ∈ [0, 1]. Applying the integration by parts to (4.10.7), we have that 1 ∞ 1 t d α +β ∗ 0 + P (f ) x; Gβ (t)dt S6R (f ; x) − f (x) = |Q| dt R 0 1 = I 1 + I2 , where d dt

n t 1 t P (f ) x; =− fj x − ξ ξj P (ξ)dσ(ξ). R R R Sn−1 j=1

By (4.10.6), we obtain that ξj P (ξ) =

νk

Pj,k+1−2l (ξ)

l=0

holds on Sn−1. Then it follows that n νk d t 1 t fj x − ξ Pj,k+1−2l (ξ)dσ(ξ) P (f ) x; =− dt R R R n−1 j=1 l=0 S n νk 1 t =− Pj,k+1−2l (fj ) x; . (4.10.10) R R j=1 l=0

It is easy to see that $ # d t ≤ 1 A max fj ∞ : j = 1, 2, . . . , n . P (f ) x; R dt R

(4.10.11)

Concisely, we write M = max{fj ∞ : j = 1, 2, . . . , n}. Thus combining with (4.10.9), we get 1 A A (4.10.12) |I1 | ≤ M (1 + | log t|)dt = M. R R 0 Due to (4.10.8) and (4.10.9), we obtain that n νk ∞ A1 (β) t cos t − π2 (n + k + 1 + β) I2 = Pj,k+1−2l (fj ) x; dt R R t1+β j=1 l=0 1 ∞ M 1 O + dt (4.10.13) R t2 1

4.10 Approximation of continuous functions uniformly holds about β. By Deﬁnition 5.1, we have ⎞ ⎛ νk n A T Pj,k+1−2l (fj )(x) + M ⎠ . lim sup |I2 | ≤ ⎝ R + β→0

353

(4.10.14)

j=1 l=0

Thus combining (4.10.13) with (4.10.14), we get the proof Lemma 4.10.2.

Lemma 4.10.3 (A) If f ∈ W 1 L∞ , then we have 1 ∗ . ω(f ; δ) = O δ log δ (B) If f ∈ W 2 L∞ , then we have 1 2 . ω2 (f ; δ) = O δ log δ ∗

Here ω and ω2 denote the modulus of continuity with the ﬁrst and second order, respectively. Proof. According to the deﬁnition, we have 1 ∗ f (x) = f (x − y)K ∗ (y)dy |Q| Q 1 f (x − y)K(y)dy = |Q| Q ⎞ ⎛ 1 + K(y + 2πm) − Im − I0 ⎠ dy f (x − y) ⎝ |Q| Q m=0

= g1 (x) + g2 (x). It is easy to see that [K(y + 2πm) − Im ] − I0 ≤ A, m=0 for y ∈ Q. Then we have ω(g2 ; δ) ≤ Aω(f ; δ)

354

C4. The conjugate Fourier integral and series

and ω2 (g2 ; δ) ≤ Aω2 (f ; δ). Set h ∈ Q with 0 < |h| ≤ 1/2. Then we have |h| 1 1 1 P (f )(x; t) dt + f (x − y)K(y)dy |Q| 0+ t |Q| |y|>|h|,y∈Q = g1,1 (x) + g1,2 (x).

g1 (x) =

Applying the integration by parts, we have ⎛ ⎞ n |h| |h| 1 ⎝ P (f )(x; t) log t − log t Pj (fj )(x; t)dt⎠ , g1,1 = |Q| 0 0 j=1

where Pj (x) is the polynomial −xj P (x). Then it follows that ⎛ ⎞ n |h| 1 ⎝ g1,1 (x) = P (f )(x; |h|) log |h| − log tPj (fj )(x; t)dt⎠ . |Q| 0 j=1

For any t > 0, we have |P (f )(x + h; t) − P (f )(x; t)| ≤ Aω(f ; |h|) and |P (f )(x + h; t) + P (f )(x − h; t) − 2P (f )(x; t)| ≤ Aω2 (f ; |h|). Thus, we have that

1 +A M ω(g1,1 ; |h|) ≤ Aω(f ; |h|) log |h| 1 , = O |h| log |h|

|h| 0

1 log dt t

for f ∈ W 1 L∞ , and 1 + AM ω2 (g1,1 ; |h|) ≤ Aω2 (f ; |h|) log |h| 1 , = O |h|2 log |h| for f ∈ W 2 L∞ .

0

|h|

1 log dt · |h| t

4.10 Approximation of continuous functions

355

For the term g1,2 , it is easy to get that ω(f ; |h|)|K(y)|dy ω(g1,2 ; |h|) ≤ |y|>|h| 1 , = O |h| log |h| and

ω2 (g1,2 ; |h|) ≤

ω2 (f ; |h|)|K(y)|dy 1 2 . = O |h| log |h| |y|>|h|

Combining these results, we get the proof of Lemma 4.10.3.

Remark 4.10.1 From the above proof, we see that for the general kernel K(x) = Ω(x/|x|) |x|n , which satisﬁes the conditions Sn−1

and

0

1

Ω(ξ)dσ(ξ) = 0

ω(Ω; δ) dδ < ∞, δ

Lemma 4.10.3 is still true.

Theorem 4.10.1 If f ∈ W 2 L∞ (Qn ), then we have n−1 1 2 S R (f ; x) − f (x) = O R2

(4.10.15)

for a.e. x ∈ Qn , n−1 S 2 (f ) − f = O log R R R2

(4.10.16)

c

and

n−1 S6 2 (f ) − f ∗ = O log R . R R2 c

(4.10.17)

356

C4. The conjugate Fourier integral and series

Proof. (4.10.15) and (4.10.16) have been proven in Theorem 3.7.3. For (4.10.17), it should point out that the Lebesgue constant of the conjugate function is also O(log R), then (4.10.17) can be obtained in the same way as (4.10.16). Theorem 4.10.2 If f ∈ W 1 L∞ (Qn ), then both equalities n−1 S 2 (f ; x) − f (x) = o 1 R R n−1 S6 2 (f ; x) − f ∗ (x) = O 1 R R

and

(4.10.18)

(4.10.19)

hold for a.e. x ∈ Qn . Proof. For any ε > 0, we denote the Steklov function of f by 1 f (x + y)dy. fε (x) = (2ε)n (−ε,ε)n It easily follows that 1 ∂fε (x) ∂ = f (x + y)dy ∂xj (2ε)n (−ε,ε)n ∂xj 1 f (x + y¯ + εej ) − f (x + y¯ − εej ) d¯ y, = n (2ε) (−ε,ε)n−1 where y¯ = (y1 , . . . , yj−1 , 0, yj+1 , . . . , yn ), ej = (0, . . . , 0, 1, 0, . . . , 0), and d¯ y = dy1 · · · dyj−1 dyj+1 · · · dyn . Obviously, we have ∂fε ∈ W 1 L∞ ∂xj for j = 1, 2, . . . , n, and 2 ∂ fε ≤ 1 max ∂f : j = 1, 2, . . . , n . ∂xj ∂x ε ∂xj ∞ l ∞

4.10 Approximation of continuous functions

357

It evidently follows from Theorem 4.10.1 that n−1 log R 2 . R SR (fε ) − fε = O R c

(4.10.20)

We set g = f − fε and deﬁne the operator γ by n−1 2 γ(f )(x) = lim sup R SR (f ; x) − f (x) . R→∞

By the inequality n−1 n−1 n−1 2 2 2 R SR (f ; x) − f (x) ≤ R SR (g; x) − g(x) + R SR (fε ; x) − fε (x) and (4.10.20), we have γ(f )(x) ≤ γ(g)(x). By the fact g ∈ W 1 L∞ and conclusion (II) in Lemma 8.1, we obtain γ(g)(x) ≤ A

n

T Pj (gj )(x)

j=1

for a.e. x ∈ Qn . Using the estimate about the type of the operator T P in Section 4.5, we have T Pj (gj )2 ≤ Agj 2 . Since ∂f (x) ∂fε (x) 1 − gj (x) = ∂xj ∂xj (2ε)n

(−ε,ε)n

∂f (x) ∂f (x + y) − ∂xj ∂xj

we have 7 1 |fj (x) − fj (x + y)|2 dx · dy gj 2 ≤ (2ε)n (−ε,ε)n n Q √ ≤ ω fj ; nε L2 (Qn ) . Finally, we get γ(f )2 ≤ A

n √ ω fj ; nε L2 (Qn ) . j=1

dy,

358

C4. The conjugate Fourier integral and series

Taking ε → 0+ , we get γ(f )(x) = 0 for a.e. x ∈ Qn . Thus, we get (4.10.18) proven. As (4.10.19) is the direct result from Lemma 4.10.2. Finally, we get the Theorem proven.

The following Theorem is our main goal. Theorem 4.10.3 Suppose that f ∈ C(Qn ) and f ∈ / W 1 L∞ . Let ω(δ) = ω(f ; δ) and ω ¯ (δ) =

δ ω(1), 0 < δ ≤ 1. ω(δ)

We deﬁne δ0 = 1 and δm+1 = min δ : max

ω(δ) ω ¯ (δ) , ω(δm ) ω ¯ (δm )

=

1 6

and the function Ω(δ) by Ω(δ) = 6−m , δ ∈ (δm+1 , δm ], for m = 0, 1, 2, . . .. We also denote n−1 2 n−1 ∗ 2 6 ρR (f ; x) = max SR (f ; x) − f (x) , SR (f ; x) − f (x) . Then we have 9 1 log log , for R ≥ 1, where C(x) is (I) ρR (f ; x) ≤ C(x)ω R Ω(1/R) a.e. ﬁnite, and , - λ x ∈ Qn : C(x) > λ ≤ Ae− A for λ > 0. 9 1 log log (II) lim sup ρR (f ; x) ω R Ω(1/R) R→∞

−1

≤ A for a.e. x ∈ Qn .

Proof. Firstly, it follows from the fact f ∈ / W 1 L∞ that lim ω(δ)δ−1 = +∞.

δ→0+

(4.10.21)

4.10 Approximation of continuous functions

359

Thus the above deﬁnitions make sense. Set Qf = x ∈ Q : lim ρR (f ; x) = 0 . R→∞

It is easy to see that |Qf | = |Q|. For any λ > 0, we deﬁne 9 1 log log GR (λ) = x ∈ Qf : ρR (f ; x) > λω R Ω(1/R) for R ≥ 1, and Δm (λ) =

,

GR (λ),

−1 −1 R∈[δm ,δm+1 )

for m = 0, 1, 2, . . .. Then, for any m ∈ Z+ , one of the following two equations 1 ω(δm+1 ) = ω(δm ) 6 and

1 δm δm+1 = ω(δm+1 ) 6 ω(δm )

must holds at least. We divide this problem into two cases. Case 1. Let ω(δm+1 ) = 16 ω(δm ). We choose g as the optimal approximate −1 trigonometric polynomial for f with the order δm , and set f − g = h. Let −1 −1 R ∈ [δm , δm+1 ). Then we have 1 hc = Eδm . −1 (f ) ≤ Aω(δm ) ≤ Aω R It obviously follows that ρR (f ; x) ≤ ρR (g; x) + ρR (h; x) and

h ρR (h; x) ≤ hc ρR ;x hc n−1 n−1 1 h h 2 2 6 ; x + S∗ ;x . ≤ Aω S∗ R hc hc

360

C4. The conjugate Fourier integral and series

In addition, we have that n+1 n−1 n+1 n−1 S 2 (g; x) − g(x) ≤ S 2 (g; x) − S 2 (g; x) + S 2 (g; x) − g(x) R R R R 1 1 n−1 ≤ 2 SR 2 (−Δg; x) + Aω g; R R n−1 Δg 1 1 2 . ≤ 2 Δgc S∗ ; x + Aω R Δgc R 1 −2 ω2 (g; δm ) ≤ AR2 ω2 f ; Δgc ≤ Aδm , R

Since

we have n−1 n−1 Δg 2 S 2 (g; x) − g(x) ≤ Aω 1 S∗ ;x + 1 . R R Δgc

(4.10.22)

For the conjugate case, similarly we get that n−1 n−1 Δg S6 2 (g; x) − g ∗ (x) ≤ Aω 1 S6∗ 2 ;x R R Δgc n−1 ∗ 2 6 + SR (g; x) − g (x) . We conclude that n−1 2

S6R

∞ 1 t 1 n+k−1 E(n, 1, k, t) (g; x) − g (x) = P (g) x; t − n+k dt |Q| 0 R C(n, k) t 1 t 1 1 n+k−1 E(n, 1, k, t) P (g) x; t − n+k dt = |Q| 0 R C(n, k) t ∞ 1 t 1 n+k−1 E(n, 1, k, t) t − n+k dt P (g) x; = |Q| 1 R C(n, k) t ∗

:= I1 + I2 .

(4.10.23)

By choosing M = max{gj : j = 1, 2, . . . , n}, we have that

t |P (g)| x; R

t = g x − ξ − g(x) P (ξ)dσ(ξ) |ξ|=1 R ≤ AM

t . R

4.10 Approximation of continuous functions Applying the condition ω(δm ) = 6ω(δm+1 ), we obtain 1 −1 1 1 . |I1 | ≤ A M ≤ A δm ω(g; δm ) ≤ Aω R R R

361

(4.10.24)

According to the expansion formula

1 π E(n, 1, k, t) 1 − n+k = O n+k+2 +At−(n+k+1) cos t − (n + k + 1) , C(n, k) t t 2 we have

t π P g; I2 = A t−2 cos t − (n + k + 1) dt R 2 1 ∞ t 1 P g; O 3 dt + R t 1 = J + τ.

∞

It easily follows that ∞ t 1 1 P g; O 3 dt = O ω(g; ) , τ= R t R 1 and

1 1 |τ | ≤ Aω g; ≤ Aω(g; δm ) ≤ 6Aω(δm+1 ) ≤ Aω . R R

(4.10.25)

Here we denote by θ = π2 (n + k + 1). By the property of cosine function, we have t+π cos(t − θ) t 1 ∞ − P (g) x; P (g) x; J =A dt 2 1 R R t2 1 t+π 1 1 ∞ cos(t − θ)dt P (g) x; − +A 2 1 R t2 (t + π)2 1 1 t + π cos(t − θ) −A P (g) x; dt, 2 1−π R (t + π)2 which implies

1 1 ≤ Aω . |J| ≤ Aω g; R R

(4.10.26)

Combining (4.10.23), (4.10.24), (4.10.25) with (4.10.26), we get that n+1 S6 2 (g) − g∗ ≤ Aω 1 . R R c

362

C4. The conjugate Fourier integral and series

Consequently, we have n+1 n+1 1 Δg ∗ 2 2 6 S6 ;x + 1 . R (g; x) − g (x) ≤ Aω( R ) S∗ Δgc

(4.10.27)

¯ Due to (4.10.22) Concisely, for any h ∈ C(Q), we denote by h/hc = h. and (4.10.27), we have n−1 n−1 1 S∗ 2 (Δg; x) + S6∗ 2 (Δg; x) + 1 . ρR (g; x) ≤ Aω R Thus, if we write

n−1

n−1

S∗ = S∗ 2 + S6∗ 2 for short, then we have ρR (f ; x) ≤ Aω

1 , ¯ S∗ (h; x) + S∗ (Δg; x) + 1 . R n−1

(4.10.28) n−1

Because of the estimate of S6∗ 2 in Section 4.5 and the estimate of S∗ 2 in Section 3.4, we get the similar estimate of S∗ . Using the Theorem 4.5.1, for any f ∈ L∞ (Qn ), we have that , λ - x ∈ Q : S∗ (f )(x) > λ log(m + 2) ≤ A(m + 2)− Af ∞ .

(4.10.29)

−1 , δ −1 ), there holds Ω( 1 ) = 6−m . Then we have Notice that when R ∈ [δm m+1 R 1 GR (λ) ⊂ x ∈ Qf : ρR (f ; x) > λω log(2 + m) := Em (λ), R

which yields Δm (λ) ⊂ Em (λ). Then due to (8.29), we obtain λ

|Δm (λ)| ≤ A(m + 2)− A . Case 2.

If

(4.10.30)

1 δm δm+1 = , ω(δm+1 ) 6 ω(δm )

−1 −1 by the property of the modulus of continuity, when R ∈ [δm , δm+1 ), we have ω(δm ) 1 ω(δm+1 ) =6 ≤ 12Rω . (4.10.31) δm+1 δm R

4.10 Approximation of continuous functions

363

Now we let g(x) be the best approximate trigonometric polynomial for f (x) −1 with the order δm+1 , and set h = f − g. Similar to Case 1, when x ∈ Qf , we have n−1 n−1 n−1 ¯ x) + S 2 (g; x) − g(x) . S 2 (f ; x) − f (x) ≤ Aω 1 S∗ 2 (h; R R R Applying Lemma 10.1, we get ⎫ ⎧ n ⎬ ⎨ n−1 S 2 (g; x) − g(x) ≤ A T P (g )(x) + θ (g)(x) , j j R R⎩ R ⎭ j=1

where θR satisﬁes |θR (g)(x)| ≤ M ≤

−1 Aδm+1 ω(δm+1 )

1 ≤ ARω . R

On the other hand, we have T Pj (gj )(x) = gj c T Pj (g j )(x) 1 T Pj (g j )(x). ≤ ARω R Hence we can get ⎧ ⎫ ⎨ n−1 n ⎬ n−1 2 ¯ S 2 (f ; x) − f (x) ≤ Aω 1 S ( h; x) + T P (g )(x) + 1 . ∗ j j R ⎭ R ⎩ j=1

(4.10.32) Similarly, when x ∈ Qf , we have n−1 n−1 n−1 1 ∗ ∗ 2 2 2 ¯ 6 6 S6 R (f ; x) − f (x) ≤ Aω R S∗ (h; x) + SR (g; x) − g (x) . Due to Lemma 10.2, we have ⎫ ⎧ νk n ⎬ ⎨ n−1 A 1 S6 2 (g; x) − g∗ (x) ≤ T P (g )(x) + θ (g(x)) l j R j,k+1−2 R ⎭ R⎩ j=1 l=0 ⎫ ⎧ n νk ⎬ 1 ⎨ T Pj,k+1−2l (g¯j )(x) + 1 , ≤A m ⎭ R ⎩ j=1 l=0

364

C4. The conjugate Fourier integral and series

where M = max{gj ∞ : j = 1, 2, . . . , n}. It easily implies that 1 −1 ω(δm+1 ) ≤ ARω M ≤ Aδm+1 . R Consequently, we have n−1 S6 2 (f ; x) − f ∗ (x) R ⎛ ⎞ νk n n−1 1 ⎝6 2 ¯ S∗ (h; x) + T Pj,k+1−2l ((gj )(x)) + 1⎠ . ≤ Aω R

(4.10.33)

j=1 l=0

Combining (4.10.32) with (4.10.33) and applying the estimate on the type of the operator T P , similar to Case 1 , we have that the inequality (4.10.33) in Case 2 that still holds. Thus we have ∞ ∞ λ Δ (λ) ≤ A (m + 2)− A . m m=0

m=0

If λ ≥ 3A, by the estimate −1 ∞ λ λ λ λ −A x dx = 2− A +1 < e− A log 2 , −1 A 2 we have

∞

λ

λ

(m + 2)− A ≤ e− A ,

m=0

which implies

∞ λ Δm (λ) ≤ Ae− A m=0

for any λ > 0, where A = An (P ). We deﬁne −1 9 1 log log . C(x) = sup ρR (f ; x) ω R Ω(1/R) R≥1 Then for any λ > 0, it is easy to have that {x ∈ Qf : C(x) > λ} ⊂

∞ m=0

Δm (λ).

(4.10.34)

4.10 Approximation of continuous functions

365

Hence, we have λ

|{x ∈ Qf : C(x) > λ}| = |{x ∈ Q : C(x) > λ}| ≤ Ae− A . Thus we have obtained the conclusion (I). Denote the constant A = An (P ) in (4.10.30) by A0 and let λ0 = 2A0 . We have 1 9 x ∈ Qf : lim sup ρR (f ; x) ω log log > λ0 ⊂ lim sup Δm (λ0 ). R Ω(1/R) m→∞ R→∞ From the estimate ∞ lim sup Δm (λ0 ) = lim Δm (λ0 ) m→∞ m→∞ j=m ≤ lim

m→∞

∞

A0 (j + 2)−2

j=m

= 0, it follows that

1 9 log log lim sup ρR (f ; x) ω R Ω(1/R) R→∞

−1

≤ 2A0

for a.e. x ∈ Qn . Thus we get the conclusion (II).

As a direct consequence of Theorem 4.10.3, we have the next theorem. Theorem 4.10.4 If f ∈ Lipα, 0 < α < 1, that is, ω(f ; t) = O(tα ), then we have n−1 n−1 S 2 (f ; x) − f (x) + S6 2 (f ; x) − f ∗ (x) ≤ C(x)R−α log log R, R R for R ≥ 9, where C(x) is ﬁnite for a.e. x ∈ Qn and satisﬁes λ

|{x ∈ Q : C(x) > λ}| < Ae− A for any λ > 0, and lim sup ρR (f ; x)Rα (log log R)−1 ≤ A R→∞

for a.e. x ∈

Qn .

366

C4. The conjugate Fourier integral and series

By similar discussions, we easily get the following theorem which is described by the modulus of continuity of second order. Here we omit the proof. Theorem 4.10.5 Suppose that f ∈ C(Q) satisﬁes lim

δ→0+

ω2 (f ; δ) = +∞. δ2

If we replace w in Theorem 4.10.3 by ω2 , then the conclusion is still true.

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Index Bq , 211, 213, 214, 216, 217, 224–226, 229– 232, 332, 371 BMO, 129, 130, 136–140 L log L, 177 L1 , 211, 261, 280 L∞ , 226, 250, 261, 263, 264, 331 Lq , 225, 226

conjugate function, 312, 315, 320, 347, 350, 356, 373 critical index, vii, viii, 40, 128, 130, 141, 146, 166, 177, 187, 194, 201, 208, 231, 244, 307, 309, 319, 324, 336, 370 Dirichlet kernel, 2, 229 disc conjecture, 51 duality, 85

a.e. approximation, 347 Abel-Poisson means, 38, 39, 263, 265, 266 almost everywhere convergence, viii, 105, 194, 368, 370–373

entropy, 208–211 equivalence, 323 Euler number, 182 exponential sums, 368, 372

Banach space, 1, 225 Bessel function, 61, 98, 100, 101, 131, 136, 145, 192, 242, 245, 295, 301, 304, 373 Bessel potential space, 232 block, viii, 208, 211, 213, 214, 216, 218, 219, 223–232, 241, 243, 333, 334, 371 Bochner-Riesz, 48, 59, 61, 334, 368–372, 374 Bochner-Riesz conjecture, 96, 113, 372 Bochner-Riesz means, vii, viii, 3, 39–41, 45–47, 105, 113, 128, 141, 146, 152, 166, 177, 201, 208, 231, 259, 276, 289, 294, 303, 307, 316, 319, 324, 332, 368, 370, 371, 374

Feﬀerman, 15, 51, 60, 89, 96, 208–210, 369 Feﬀerman theorem, viii, 51 Fej´er mean, 2 Fourier series, 373 Fourier transform, 237, 263, 267, 294, 310, 370, 373 fractional integral operator, 139 H¨ older’s inequality, 114, 220, 282, 338 Hardy-Littlewood maximal function, 61, 107, 116, 137, 234 Hardy-Littlewood maximal operator, 144, 145, 259 Hardy-Littlewood-Sobolev theorem, 139 Hausdorﬀ-Young’s inequality, 92, 292, 346 Hilbert transform, 78, 103, 293

Calder´ on-Zygmund, viii, 294, 309, 310, 333 Carleson-Sj¨ olin theorem, viii, 61, 65, 72, 73, 78, 85, 96, 97, 109 Cauchy sequence, 7, 213 commutators, viii, 129, 130, 367, 371 conjugate Fourier integral, 293, 294, 303, 307, 342, 370 conjugate Fourier series, 309, 310, 316, 332, 337, 370

interpolation, 73, 98 interpolation of operators, 280 Kakeya maximal function, 78, 79, 85, 88, 96, 368

375

376 kernel, 3, 37, 39, 41, 62, 108, 122, 166, 231, 258, 264, 293, 302, 310, 312, 316, 324, 325, 331, 333, 355 Kolmogorov, 152 Lebesgue, 9, 15, 35–37, 39, 44–46, 52, 115, 166, 167, 169, 197, 198, 200, 201, 225, 237, 248, 249, 309, 331, 346, 349, 350, 356, 371, 373 Lebesgue constant, 166, 167, 248, 249, 331 Lebesgue point, 15, 35–37, 39, 198, 200, 201, 225, 350 Leibniz formula, 329 linear operator, 48, 81, 105, 107, 128, 129, 183, 236, 241, 260, 266, 268, 367 localization, 42, 45, 146, 165, 166, 334, 336, 337, 367, 368, 371 maximal function, 78, 79, 85, 88, 96, 232, 368, 371, 374 Minkowski’s inequality, 35, 61, 346 multiple Fourier integral, vii, 41, 371 multiple Fourier series, vii, viii, 2, 3, 10, 141, 231, 367–370, 372, 373 multiplier, viii, 48–50, 52, 86, 116, 120, 236, 260, 368, 369 orthogonality, 372 oscillatory integral, 60–62, 66, 367, 368, 372 Plancherel theorem, 49, 91, 119, 120, 127, 238 Poisson summation formula, 10, 11, 147, 264, 267, 269, 318

Index quasi-norm, 213, 225 radial function, 32, 90, 92, 94, 97, 122, 267, 295, 374 rectangular partial sum, 2, 15 restriction conjecture, 372 restriction theorem, 78, 89, 92, 93, 96 Riesz potential, 67 saturation, 259–261, 265, 266, 269, 275–277, 370 Schwartz function, 114, 123 singular integrals, 129 Sobolev, 45, 118, 120, 368 spherical partial sum, 2 spherical Riesz means, 369–371, 373 Stein, vii, 67, 78, 89, 95, 98, 109, 110, 129, 138, 145, 147, 152, 154, 177, 183, 186, 194, 196, 198, 214, 237, 241, 245, 280, 294, 324, 369, 372 strong summability, 371, 373 strong summation, 280, 337, 342, 347 strong type, 106, 190, 333

272,

112, 165, 207, 322,

the the Fourier transform, 32 Trigonometric series, 374 weak type, 106–108, 129, 241, 294, 315, 333 Weiss, viii, 98, 129, 145, 154, 183, 208, 210, 213, 214, 226, 229, 230, 368, 371, 372 Zygmund, viii, 155, 293, 294, 309, 315, 367, 372