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BOCHNER–RIESZ MEANS ON EUCLIDEAN SPACES

8745hc_9789814458764_tp.indd 1

21/6/13 9:23 AM

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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-Sjölin 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 Calderón-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| 0, (1.0.7) SR (f ; x) = R |m| 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


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


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| 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


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| 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| 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


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| −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)


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


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)


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


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

,


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 ,


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


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


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


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


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 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 Hölder’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


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


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].


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


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)


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


, 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|
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}|
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


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


233

Theorem 3.6.8 Let 0 ≤ λ ≤ 2, 1 < p < ∞, α = σ + iτ , and n − 1 2 . − 1 σ> 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 < p < ∞, α = σ + iτ, and σ > 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

234


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|0

0

p


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 < p < ∞, α = σ + iτ , and σ > (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) ,

236


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 .


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


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| 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


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),


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.

242


/ 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)


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.

244


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


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 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


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 Fejér 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 Vallée 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


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


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 Fejér 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

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


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


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


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)


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)


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


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)


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


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| 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|R − |m| 1 |m| n+1 R R + Cm (g)eim·x 1 − λR (m) e 1− R R |m| 0

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.


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| 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| n−1 2 .

278


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| is F (S α , C). This completes the proof of Theorem 3.9.7.

n−1 2 ,

so

280


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 < p ≤ 2, the maximal operator R 2 12 1 δ Λδ (f ; x) := sup Su (f ; x) du R R>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


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


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 |


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 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


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


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≤νμ+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 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


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)


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


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 Calderón 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| 0. (n)

Lemma 4.1.1 Let P ∈ Ak we have

P |y| 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


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


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| 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|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 / 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 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 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


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


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


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 CalderónZygmund singular integral that the Calderón-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


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


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


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 Hölder’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


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 )| 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)


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


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


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


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


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


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


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 .


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


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


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


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


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 ∞


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


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)


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


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


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


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


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)


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


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 Fejér 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

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