A Real Variable Method for the Cauchy Transform, and Analytic Capacity

Lecture Notes in Mathematics Edited by A. Dold and B. Eckmann

1307 Takafumi Murai

A Real Variable Method for the Cauchy Transform, and Analytic Capacity

Springer-Verlag Berlin Heidelberg NewYork London Paris Tokyo

Author Takafumi Mural Department of Mathematics, College of General Education Nagoya University Nagoya, 464, Japan

Mathematics Subject Classification (t980): Primary 3 0 C 8 5 ; secondary 4 2 A 5 0 ISBN 3-540-19091-0 Springer-Verlag Berlin Heidelberg New York ISBN 0-387-19091-0 Springer-Verlag N e w York Berlin Heidelberg

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. Duplication of this publication or parts thereof is only permitted under the provisions of the German Copyright Law of September 9, 1965, in its version of June 24, 1985, and a copyright fee must always be paid. Violations fall under the prosecution act of the German Copyright Law. © Springer-Verlag Berlin Heidelberg 1988 Printed in Germany Printing and binding: Druckhaus Beltz, Hemsbach/Bergstr. 2146/3140-543210

PREFACE

The purpose of this lecture note is to study the Cauchy transform on curves and analytic capacity.

For a compact set

F

in the complex plane

denotes the Banach space of bounded analytic supremum norm

il-i!H~.

The analytic capacity of

IIflIH~_<_~,

~,(r) = sup{If'(~>l; where

f'(-) : lim

functions

z(f(z)-f(m)).

F

in

~,

~U{o~}-F

H~(F c) (= F c)

with

is defined by

f~H~,

We also define

ilcultH~ =< 1,

y+(F) = s u p t ( ] / 2 r r ) / d>;

Cu~H'(rC),

u >= 0},

where C~(z) = ( 1 / 2 ~ i ) 7 l / ( g - z )

d>(¢)

We a r e c o n c e r n e d w i t h e s t i m a t i n g y ( - ) finite

and y + ( . ) .

finite union of mutually

disjoint smooth arcs. Let

(the generalized

space of functions on

F

length). Let

p(r) = i n f ~ ( E ) / i E I,

transform on

F

~)).

compact s e t s h a v i n g

Hence we assume that i'}

F.

P

is a

denote the l-dimension

LP(F)

(lip! ~)

with respect to the length element

denote the weak L 1 space of functions on

where the infimums

support of

To do t h i s ,

l-dimension Hausdorff measure are critical.

Hausdorff measure

LI(F)

(z¢(the

denote the L p

Idzl,

and let

Put

% ( r ) = i n f ~r+(E)/I~i,

are taken over all compact sets

E

in

F.

The Cauchy(-Hilbert)

is defined by

Hrf(z)

= (i/~)

p . v . IF f ( ¢ ) / ( ~ - z )

id~]

(z~r).

Then we see that p+(F) __< p(F) =< Const p+(F) I/3,

Const p+(F) __< I/]IHFIILI(F),LI(F ) <= Const p+(F), w

where

NHFILI(F),LI(F)

is the norm of

(Theorem D). HenceWthe study of

y(F)

HF

as an operator from

LI(F)

to

is closely related to the study of

Here is a history of the study of the Cauchy transform on Lipschitz According to Professor

Igari,

the L 2 boundedness

LI(F) H F-

graphs.

of the Cauchy transform on

Lipschitz graphs was first conjectured by Professor Zygmund in his lecture at Orsay in 1960's. Let line. Let

C[a]

F = {(x,A(x));

I/{(x-y)+i(A(x)-A(y))}. C[a]

is bounded

(from

a(x) = A'(x),

where

•

is the real

Then the above conjecture means the following assertion: L2(~)

to itself) if

formally expanded in the following the Hilbert

xE~},

denote the singular integral operator defined by a kernel

transform and

Tn[a]

form:

aEL~(~).

The operator

(-~)H + En=0(-i)n Tn[a],

C[a]

where

is H

is

is the singular integral operator defined by a

IV

kernel

(A(x)-A(x))n/(x-y) n+l.

bounded if

aeL~(~)

related to the BMO(R) of functions on Tn[a],

•

Calder6n

theory, where

C[a]

[3] showed that

Tl[a]

is

is the Banach space, modulo constants,

[44] that

Coifman-Meyer

is bounded if

Coifman-McIntosh-Meyer

(Theorem B). David

is already known

BMO(~)

of bounded mean oscillation.

[4] showed that

small, and consequently the affirmative

In 1965, Calderdn

(Theorem A). This theorem is very important and closely

IIC[a]IIL2(R),L2(E)

HF

for continuous

and that the square root is best possible

[18]. Jones-Semmes (See Appendix II.)

with a review of the study of

C[a].

curves

~ Const(l 4 - ~ B M O ( ~ ) )

of Theorem B by complex variable methods. HF

is sufficiently

[7] solved the above conjecture

[17] studied

As a first step of the study of

[8], [9] studied

llallL~(~)

for discontinuous

In CHAP. I,

in

F.

It

(Theorem C)

gives a simple proof

curves

F,

we begin

g proofs of Theorem A will be

given. Once this theorem is known, we can easily deduce Theorem B (cf. CHAP. II), and hence Theorem A is very important if

f, g E L2(~)

(such that (fg)(x)

in the study of

have analytic extensions

f(z), g(z)

limy÷~ f(iy) = limy+~ g(iy) = 0),

functions

is essential

As is easily seen,

to the upper half plane

then the Poisson extension of

to the upper half plane is identical with

of analytic

C[a].

f(z)g(z).

This simple property

in a proof of Theorem A by complex variable

methods. We shall give, in CHAP. I, various interpretations

of this property

the point of view of real analysis

[13]). These proofs

(cf. Coifman-Meyer-Stein

are, of course, mutually very close, but each proof has proper applications interesting

from

and is

in itself.

In CHAP. II, we shall give the proofs of Theorems B and C by perturbation. method

is an improvement

of Calder6n's

perturbation

Our

[4] and David's perturbation

[17]. Put o(C[a]) = sup(i/lli)/llC[a](xlf)(x)l where

XI

intervals

is the characteristic I

function of

and all real-valued

is comparable

to

functions

IIC[a]IIL~(~),BMO(~)

I,

graphs.

necessary.

llfllL~(~) ~ i.

This quantity

for our perturbation. of a primitive

A(x)

CI/iII)/IIC[a](xIf)(x) I dx

(See the figure in § 2.2.) Repeating

is consequently

and the supremum is taken over all with

decomposition

we obtain an a-priori estimate of

and estimating

I f

and convenient

Considering a suitable Calderdn-Zygmund on

dx,

this argument

infinitely many times

infinitely many error terms, we see that the boundedness reduced to the boundedness of

H.

of

C[a]

For the proof, Theorem A is

We shall also give a proof of Theorem A by perturbation

which we use are only the Calder6n-Zygmund

of a(x)

by moderate

decomposition

[45]. Tools

and the covering lemma.

For the proof of Theorem C, we put

~(C[a])

= sup(i/IIl)fiIC[a](xif)(x)i 2 f(x) dx,

where the supremum is taken over all intervals

I

and all real-valued

functions

f

with

0 ~ f ~ i.

Then

o(C[a]) 2 ~ Const $(C[a]).

II C[a](xlf)(x)f(x) dx = 0, a(x),

Since

this quantity behaves like a linear functional of

and this gives an a-priori estimate better than

o(C[a]).

Our method is not

short but very simple, and this is applicable to various kernels. In CHAP. III, we shall study compare

y(-)

HF

for discontinuous graphs

planar Cantor sets are useful to construct various

examples (cfo Denjoy [23], Vitushkin [52]). Let (n$1)

and shall

with integralgeometric quantities. We first give the proof of

Theorem D. As is well-known,

be the union of

Qn-1

F

4n

closed squares with sides of length

with each component of

the component. Put

Q0 = [0,i]~ [0,I]

Qn-i

Q= = ~ n= 0 Qn"

Then

y(Q=) = 0

and

and

(Sl,..,s n 6 ~ )

IQ~I > 0

I'I

try to give grounds to this example. We may consider that Tsl,..,Sn

4 -n

Qn

obtained from

replaced by four squares in the four corners of

This shows that two classes of null sets of y(.)

figure in § 3.3.) Let

and let

(Garnett [28]).

are different. We shall Qn

is a graph. (See the

be the singular integral operator

defined by a kernel i/{(x-y)+i(Asl ,..,an(x) - Asl ,..,an(y))}, • Asl,..,sn~X) = sk

where

(x£ [0,i)).

Sl~-',S n

is comparable to ~ 0 0 an n-tuple (Sl,..,s n) Qm

and

Asl,..,Sn

(x)

=

0

Then we see that max{o(T

to

_ _ ((k-l)/n =< x < k/n, igkin)

); s I .... sn 6 ~ } (Theorem G), and, if we neglect constant multiples,

obtained from a graph

(m = (the integral part of

(log n)/4))

{(x,A 0 0(x)); x E [0,i)} similar ~l,..~Sn is a solution of this extremal

problem. Hence planar Cantor sets are worst curves in a sense. We shall also generalize

Qn"

finite union degree

n,

component

and

A segument

[0,i)

is called a (thick) crank of degree

Fn of

is obtained from a crank Fn_ 1

is less than or equal to

these segments to

~

Fn_ 1

of degree

n-i

replaced by a finite number of segments

parallel to the x-axis such that J

0

and a

of segments parallel to the x-axis is called a (thick) crank of

n

if J

(p=p(J)) Jk

F

2-PIJI

iJk! = 2-PlJ[, (i < k j 2p)

with each

Jl,..,J2p

the distance between and the projections of

are mutually disjoint and contained in the projection of

We shall show that, for any crank

F

of degree

n,

IIHFIIL2(F),L2(F ) ~ Const /~n

and that this estimate is best possible (Theorem E). To prove this, we define singular integral operators

{Tk}~= 0

such that

J.

n+l

T O = (-z)H,

liHrIIL2(F),L2(F)

T n are mutually almost TkIIL2(~),L2(E ) = and { k}k=0 orthogonal. Hence we see that the meaning of ~nn is the central limit theorem.

llz~=0

We define integralgeometric quantities D(z,r) NE(r,0 ) where We put

be the open disk of center (r>0,101%~) i(r,0)

z

Cr (.)

and radius

(0<~
as follows. Let

For a compact set

denotes the (cardinal) number of elements of

is the straight line defined by the equation

E,

ENL(r,0),

x cos 0 + y sin O = r.

VI

Cr (E) = lima+ 0 Cr~S)(E), Cr(C)(E)

= inf f :{f~ Ns{o~=iD(Zk,rk)}(r,8)~

n O{Uk=iD(Zk,rk)}

where

y(E) ~ Const CrI(E),

(cf. Marshall 0<~
U ~ = i D ( Z k , r k) and the infimum is N {D(Zk,rk)}k= 1 of E with radii less than c.

it is interesting

to compare

y(-)

there exists a compact set

sequence of independent

E

such that

y(E ) = i

Yn(X) = Yn_l(X) + Syn_l(x)(X)

This quantity

of degree

n

is comparable

(n~l).

such that to

i/~nn (the central limit theorem) construct

Cr (')

and

Cr ( E )

{Xn}:= 1

the required set

and let

is comparable

space

S O = O,

{Yn}n=O

Then we see that, for Cr (F n)

i/n I-~. and

(n~l), process

= 0

be a

random variables on the standard probability

([O,l),8,Prob) such that Prob(X n = ±i) = 1/2 N S n = Ek= 1 X k (n~l). We define a Galton-Watson

Fn

with

[37]). As an application of Theorem E, we shall show that, for

(Theorem F). For the proof, we use a branching process. Let

a crank

(g>0)

is the boundary of

taken over all finite coverings Since

dr} d~

by

Y0(X) = i,

n~l,

there exists

to Xk= 0 k ~ Prob(Yn=k).

Using the difference of order between

i/n 1-e (the Galton-Watson

process), we

E .

I express my hearty thanks to Professors M.Ohtsuka,

R.R.Coifman,

P.W.Jones

who gave me the chance to lecture during the academic year 1986-1987,

and I am

grateful

to Professors

G.David,

C.Bishop for their variable

express my appreciation

S.Kakutani,

T.Tamagawa,

J.Garnett,

S.Semmes,

comments and suggestions.

to Professor W.H.J.Fuchs

thank to Mrs. Mel D. for typing the manuscr:ipt.

T.Steger,

I especially

for his encouragement. This note is dedicated

I also to the

memory of my mother who died while I was staying at Yale University. New Haven,

July,

1987

CONTENTS

CHAPTER

CHAPTER

CHAPTER

I.

The C a l d e r 6 n

i.i.

Calder6n's

1.2.

Proof of

1.3.

Area

of its boundedness)

........... 1

..............................................

i

..................................................

1

...................................................

2

(1.3)

integral

1.4.

Good ~ i n e q u a l i t i e s

1.5.

BMO

1.6.

The C o i f m a n - M e y e r

.............................................

expression

6

....................................

1.7.

A tent space

...................................................

1.8.

The M c I n t o s h

expression

1.9.

Almost

i.i0.

Interpolation

1 .ii.

Successive

II.

A real v a r i a b l e

2.1.

Coifman-McIntosh-Meyer's

2.2.

Two basic

2.3.

o-function

2.4.

A-priori

2.5.

Proof of T h e o r e m

13 15

..................................................

21

compositions

of kernels

method

.............................

for the Cauchy

transform

on graphs

...... 31 31

...........................................

32

.....................................................

35

estimates

.............................................

A by p e r t u r b a t i o n

2.8.

Proof of

(2.38)

................................................

2.9.

Proof of

(2.39)

................................................

2.10.

Application

III.

Analytic

capacities

3.1.

Relation

between

3.2.

Vitushkin's

3.3.

The Cauchy

3.4.

Proof of

the latter half of T h e o r e m

3.5.

Analytic

capacities

3.6.

Analytic

capacity

of norms of

of

APPENDIX

II.

REFERENCES

E[ °]

(2.38)

of cranks

example,

Proof of T h e o r e m

H

. . . . . . . . . . . . . . . . . . . . . . . . . 53

71 71

example,

Calder6n's

problem

.................................

of fat cranks

79 83

E . . . . . . . . . . . . . . . . . . . . . . . . . . 91

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 quantities

............ 105

F ............................................ ...........................................

B by P . W . J o n e s - S . S e m m e s

112 117

...................... 126

...................................................................

INDEX

61 68

.................................

.......................................... on cranks

55

..................................

and i n t e g r a l g e o m e t r i c

problem

Proof of T h e o r e m

and

Garnett's

problems

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 ,C[']

..........................................

¥(.)

transform

and

39

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Proof of T h e o r e m B by p e r t u r b a t i o n

An e x t r e m a l

24

...............................

principles

theorem

Estimates

I.

ii

...........................................

orthogonality

2.7.

3.7.

9

........................................

2.6.

APPENDIX

4

.............................................................

and extremal

SUBJECT

commutator.( 8 proofs

theorem

................................................................

129 132

CHAPTER I.

THE CALDERON COMMUTATOR

(8 PROOFS OF ITS BOUNDEDNESS)

§i.i.

Calder~n's Theorem (Calder~n [3]) Let

Lp

(i ~ p ~ ®)

denote the

Lp

to the 1-dimension Lebesgue measure I'I-

space on the real line Its norm is denoted by

denote the Banach space, modulo constants, of functions IIflIBMO = sup(I/II I) 7ilf(x)-(f)ildx (finite) intervals

I

and

(f)I

f

on

@

with respect

ll'llp° Let ~

BMO

such that

is finite, where the supremum is taken over all

is the mean of

a ~ L~ , we

f

over

I.

For

We write simply by

T[a]

the operator from

define a kernel (i.i) where

A

T[a](x,y) = {A(x) - A(y)} /(x-y) 2,

is a primitive of

a.

L2

itself defined by the above kernel, i.e., (1.2)

T[a]f(x)

=

lim g+O

7Jx_y I,~

> e T[a](x,y)f(y)dy.

Calder~n showed Theorem A ([3]).

For any

f ~ L 2,

T[a]f(x)

(1.3)

IIall" --< Const llT[a]ll2,2

(1.4)

llT[a]l12,2 ~ Const llalI~ ,

exists a.e.

and

where

llT[a]l12,2

is the norm of

In §1.2, we show (1.3). §1.2.

T[a]

(as an operator from

Ps (x) =

E c ~, 171+

XE

to itself).

In §1.3-i.ii, we show various proofs of (1.4).

Proof of (1.3) (Coifman-Rochberg-Weiss For a set

L2

[15])

denotes the characteristic function of

{fl s(A(s)-A(t))dt}dsl

E.

We put

(x 6 ~,

s > 0),

lim ~ ~ 0

@ / 3 = Const lal

s where

I+ s = (x, x+s) , I_s = (x-s, x).

We have, for almost all

x,

Then

a.e.

to

P (x) = Ill+s ~l_s T[a](s,t) {(s-x) 2 + 2(s-x)(x-t) + (x-t) 2} dt} ds I <= fl+ s [(s-x)2} T[a] XI_ (s)l + 21s-x I IT[a] {(x-')Xl s}(s)l + IT[a]{(x--) 2 X I } (s)l] ds -8 C°nst[s5/211T[a]Xl II + g3/2 llT[a]{(x-')Xl }If -s 2 -e 2

< +

s I/2 l{T[a]{(x-.)2~

}If2] -g

-<

Const llT[a]II2,2 {s5/211XI

+ sl/2N(X-')~ I

-g

I] + s3/211(x-')X 112 _g 2 I_ s

N2 } -<- ConstIIT[a]tI2,2 s

3

,

and hence

lal = Const

lira s_~0

p /3 s

ConstHT[a]ll2, 2

a.e.

Thus we obtain (1.3). §1.3.

Area integral ([3])

• In this section we show the proof of (1.4) by Calderon.

Let

C~O

denote the

totality of infinitely differentiable functions with compact support, the inner product and in

CO

and

Y

= X

(g > 0).

(_~,~)c

(',')

Given real-valued functions

denote a,f,g

~ > 0, we estimate

(TE[a]g,f) = f2Te[a]g(x)f(x)dx, where

Te[a]

is an operator defined by a kernel

assume that e = X[0,~ ).

A(x) = fx a(s)ds.

Then

Ys(x-y) T[a](x,y).

We may

A(x) = f ~ e(x-s)a(s)ds, where

We have

r (x-y) {e(x-s)-e(y-s)}g(y)f(x)dydx]ds.

(T~[a]g,f) = f_~a(s)[ f-~f-i

(x-y) 2

Set f+(z) =

1 2~i

f_~

f(x) x-z

dx

Im z l > 0 )

-

<0

We denote also by

f+(x) (x ~ ~)

We define analogously llf_+II2 < IIfll2

and

the non-tangential limit of

g+(z), g±(x).

llg±N2 _-< Ilgil2. Let

Then

f = f+ - f_,

f±(z), respectively.

g = g+ - g_,

K0(x,y,s) = Y g (x-y){e(x-s)-e(y-s)} /(x-y) 2, + K~ (x,y,s) = {e(x-s)-e(y-s)} /(x-y ± is) 2, K2(x,y,s) = s/{(x-s) 2 + (y-s) 2 + s2} 3/2 Then

+ IK0(x,y,s) - K~(x,y,s) l ~ Const K2(x,y,s),

We have

[(T~[alg,f)I = I f_~ a(s)[ /_i Ko(x,y's){g+(Y) - g-(Y)} f(x)dydx] ds I +

=< If_; a(s)[ 7_= Kl(X,y,s)g+(Y)f(x)dydx] dsl

+ If_; +

a(s)[

f_;

K~(x,y,s)g_(y)f(x)dydx] dsl

Const f_~ la(s) l [f_~ K2(x,y,s) {ig+(Y) I + Ig_(y)I} If(x) l dydx]ds

(=

If_; a(s) kl(S)ds I + If_~ a(S)kl(S)dsl + k~(s), k2(s).

We now estimate =

_

+ Const 7_; Ia(s) Ik2(s)ds,say)"

We have _

Kl(X,y,s)g+(y)dy} dx g+(Y)

=

f ; f(x) -

{e(x-s) f ; -

=-i

f_; f(x) [ f 0

=-i

7~

=

g+(Y) (x_y_is)2

dy

- f= s

2 dy } dx (x-y-is)

g+(s+it)/{(x-is)-(s+it)} 2 dt] dx

g+(s+it) [ f_~ f(x)/{(x-is)-(s+it)}2 dx] dt

oo 2~ f0 f+(s+i(t+s))g+(s+it) dt.

Let F(z) =

-i f;

f~(z+i(t+s)) g+(z+it) dt

(z E U),

where

U = {(x,y); x E ~, Y > 0} . Then F is analytic in U and the non(i/2[i) k~(s). Here is a main lemma necessary for tangential limit F(s) equals the proof of(l.4).

Let

Py(X)

be the Poisson± kernel, i.e., Py(X) = y/{~(x2+y2)}.

For a differentiable function v(x,y) in IVv(x,y)] = {)8v/Ox]2 + l~v/Oy12 }i/2. Lemma i.i ([3]).

For

A(v)(x) = {ff

v E L I,

U, we write

we define

IVv(~,D) l2 d~ d~} I/2 A(x)

(x E ~),

where

v~)

= P

* v~)

and

A(x) = { ~ ) ;

I~ - xl < n}

•

Then

llvllI ~ ConstilA(v)ll1 Once this l e n a

is known, (1.4) is deduced as follows.

F'(z) = f$(z + is)g+(z), Const A(f+)(s) M

M g+(s),

we have where

Since

A(F)(s) & A(f+)(s)m(g+)(s) m(g+)(s) = s u p { I g + ~ ) l ; ( ~ )

6 A(x)}

We have

IIMg+ll2~ ConstlIg+II2 ,

Green's formula shows that

and

(See Lemma 2,3.)

is the non-centered maximal operator (Journ~ [35, p.6]).

IIA(f+)II2 = Constllf~l 2.

Thus we have, by Lemma 1.1,

I~

a(s)k~(s)dsl ~ 2~

Ilall~

IIFJI 1

ConstrlaH, IIA(F)IiI

Const Hal< tlA(f+)Lt 2 !m(g+)lt 2

Const Ila!I~ I!f+II2 !Ig+!I2

Const IlaIl~ llfIl2 Ilgll2 " In the same manner, we have

PFaII.]lfll 2 ilgll 2

I.~i a(s)kl(s)ds I =< Const

We have < k2(s) = f l

s (x_s)2+ ~2

Const

M f(s)

If(x) I [ / Z

V(x-s)

+ s2 (x-s)2+(y-s)2+ s 2 {Ig+ (y) I+Ig-(y)I}dy]dX

{Mg+(s) + Mg (s)} ,

and hence S_~ la(s) Ik2(s) ds <- Const llali~ llfll2 IIglI2 • Consequently

l(Te[a]g,f) i ~

arbitrary, we have (1.4) for

Const IIall~ Ilfll2 Ilgll2 . a 6 CO •

Since

f,g ( cO ,

g > 0

are

In the general case, we can deduce

(1.4) from the boundedness of maximal operators

T • [b]

(b ( Co )

and Fatou's lemma.

(See Lemma 2.5.) §1.4.

Good

k

inequalities ([2], [26], [48])

In this section we give the proof of Lemma i.i by the so-called "good inequalities".

We put

m(v)(x) = sup{Iv(x,y) l; y > 0}.

k

Fixing a sufficiently large

T, We prove

(1.5)

ix; ~(v)(x) > ~x , A(v)(x) ~ ~/~ I (Const/T 2) ix; m(x) > k I

Let

W(k) = {x; re(x) > k},

with a sequence

M k = {Ik}

6(k) = IW(k) l .

Then we can write

W(k) = U]= I Ik

of mutually disjoint open intervals.

sufficient to show that, for each (1.6)

(X > 0).

IE I -<-(Const/2) lli,

I ( Mk ,

It is

where

A(v)(~) ~

X/~

for any

where

for some

Since

a

is the left endpoint of

such that

I. ~ .

We choose Since

Iv(X,Yx) I = sup{Iv(x,y) l; y > yx }

for any

intervals

~ • ~/2.

~{J(x )}

x~I <

~/i0,

(1.8)

f

=

~.

Let

I~- xl < Yx/10}

(See §2.2.)

Green's formula

~(v)(x)

=

Then,

Let

~(x ) = {(~,~);

shows that

- ~ 0-~L~-L~ } ds =

Const ff

OR O/On

(x ~ E).

a finite number of mutually disjoint 5 Z IJ(x )I

QO = {(~'~); ~ E I, 0 < ~ < 2111},

~ > Yx } "

{ ~n ]v12

0 < Yx < 21II

Iv(~,yx) I ~ Iv(x,Yx) I - Const A(v)(x)

There exist

such that IEI ~

R = QO N U A(x ), where

large enough so that the last

x E E, there exists

J(x) = {(~,yx);

(~,yx) ~ J(x), we have

- Const X/T

•

m(v)(a) ~ k, we have

Hence, for any

J(x) = (x - (Yx/5), x + (Yx/5)),

where

A(v)(~) ~ X/~ , we have,

IV(a,y) - v(x,y)] ~ Const A(v)(~) =< Const k/~ ,

Iv(x,y) l ~ 2X (x E I, y ~ 21If).

-

E = ~.

~ ~I; otherwise

quantity in (1.7) is less than

I~

To do this we may assume that

x ~ I, y ~ 2111 ,

(1.7)

~X

X/~}.

E = {x ~ I; m(v)(x) > ~h , A(v)(x)

~ ]VVl 2 d~

dr],

R

is the inner normal derivative and {ff ,

IVvl 2 d~ d~} i / 2

ds

where

is the length element. ~*(x) = {(~,~);

Let

l~-xl < q/lO}.

(x) N R Then a geometric observation shows that is a point which is nearest to

x in

AR(V)(X) ~ A(v)(x ) ~

{x }.

k/~ ,

where

xv

Hence the right-hand side of (1.8) is

dominated by: Const fl ~ (v)(x)2dx~C°nst(X/~)2]I] We divide

oR

~ Const k 2 ]I I.

into the following three parts:

DR 0 =

8R N U J(x ),

DR I = {(~,D);

~ ~ I, ~ = 2111},

0R 2 = oR - (oR 0 U oRI).

~IV v(~,~) I ~

Const

By the definition of

any

X/~

on

(~,D) ~ oR, Iv(~,~)I ~

IfDR ~

oR. ~X

+ Const

~-~-[~ds] 8n =< Const

fOR

X/~ ~

mlVvllvl

Note that

Yx (x ~ E), we have, for

Const ~X •

Thus

ds

Const (k/~) ~k /oR ds ~ Const ~2 ill. Since

Iv(~,O) l G Const k

on

OR I,

These estimates yield t h a t 4 R 0 U o R 2 O~/On ~ 0

on

OR 2, 0H/0n = i

on

we have If0R I On

0R 0

8~ 8n

Ivl 2 ds i ~ Const ~2 iii.

Ivl 2 ds G Const k 2 I I [ . and

Iv(~,~)l

~ ~k12

on

Since 8R 0, we have

2 k21El =< Const~8R 0 ~nn 8~ Ivl 2ds =< Const/oRoU OR 2 8~ ~nn Ivl 2ds =
Consequently

(1.5) holds.

By (1.5), we have, with a constant (1.9) where

~(~I<) +

5(Tk) ~

(Colt 2) 6(k),

6(k) = Ix; A(v)(x) > h i .

quantity in (1.9) by

dk

from

CO,

= 2 CO

We now choose

[Im(v)II1 ~ Const llA(v)lll, which gives

and integrate each

Then we obtain

0 to infinity.

Const IiA(v)llI.

llvll1

This completes the

proof of Lemma i.i. §1.5.

BMO (Fefferman-Stein

[27])

Theorem A is closely related to the theory of BMO [27]. the proof of Theorem A by Fefferman-Stein. d~(x,y)

in

U

is a Carleson measure with constant

/7

In this section, we show

We say that a non-negative measure B

if

d~(x,y) ~ BIIl I ×(0,111)

for any interval

I c R.

The following two facts are elementary. is a Carleson measure

Lemma 1.2 ([27]). with constant Proof.

Let a ( BMO. Then yiV a(x,y) l2~ dx dy 2 Const IIalIBM0 , where a(x,y) = P * a(x). Y

Given an interval

I,

we put

a(1)(x) = (a(x) - (a)i) X ,(x), a(2)(x) = (a(x) - (a)l) X , (x), I I c where

(a)I = (i/IIl)

/I a(y)dy

interval of the same midpoint as

and I

I

is the double of

and of length

21I I .

I, i.e., the (open)

Then

a(x,y) = Py * a(1)(x) + P Y * a(2)(x) + (a) I ( = a(1)(x,y) + a(2)(x,y) + (a)i , John-Nirenberg's (See Lemma 2.5.)

inequality

[32] shows that

Hence we have, with

7f^ Y IVa(1)(x,Y)I 2 dx dy I = Const lla(1)II~ ~ Note that

say).

Ila(1)ll2 ~ Const MalIBMO~--~T •

~ = I × (0,1If),

~ f7

y IVa(1)(x,y)l 2 dx dy U

Const IIalI~M0 IIl.

l(a)i. - (a)iI ~ Const jilaliBMO (j ~ I), where Ij is the interval of J the same midpoint as I and of length 2JlIl. We have, for (x,y) E

[Va(2)(x,y)l < Const fl *c _-< Const

=<

Const

=< Const(

-2

1 (x_s) 2

a(2) (s)i ds

Z j=l

I ljl

Z j=l

t I j I -2 l l j + l I {ll aII BMO + I (a) I. - (a) I t} 3

f

la(y)-(a)iI ds lj+l-I j

E j 2-j) II~IBMO/I I] , j=l

and hence ff^ y ''IVa(2)(x,y)l2 dx dy ~ Const(IIallBMo/lll) 2' I 2 Const Ha]]BMO 111 .

/f^ y dx dy I

Thus

ff^

y

''IVa(x,y)I2 dx dy ~ Const

{ff^

I +

.2

dr^ y I

Let

d~(x,y)

be a Carleson measure with constant

=

Let

B.

f 6 L 2,

flu If(x'y) 12 d~(x,y) < Const BIIfH2

Proof.

Q.E.D.

IVa(2)(x,y)l 2 dx dy} -<_ Const ! a~IBMO III.

Lemma 1.3 ([35, p. 85]). Then, for any

y ''lva(1)(x,y)I2 dx dy

I

(f(x,y) = P

2

W(K) = {(x,y) ~ U; If(x,y) l > k}

y

* f(x)).

, 8(k) = ffw(k) d~(x,y) ( k > 0).

Then the left-hand side of our lemma is dominated by Const f~ k ~

kS(k)dk •

sup{If(~,~)l;

contained in I

of

If

(x,y) ~ W(~), then

Ix-El < ~} ~ c M f(x)

C.

Hence

W(k)

is

W0(k) = U I x (0,1If) , where the union is taken over all components

{x; M f(x) > CX} • 6(k) ~

for some constant

ffw0(k )

Thus

d~(x,y) & BIx; Mf(x) > CX 1 ,

which gives f~ X 8(k)dX

~

B f~

XIx; M f(x) > Ckldk

Const B llMflI~ ~ const B IlfIl~ . We now prove Theorem A. Hf(x) =

~

~ lim ~ 0

Q.E.D.

The Hilbert transform

fls-xl > ~

f(s) s-x

ds .

H

is defined by

For

a, f ~ C~, (i.i0)

where

we have

T[a]f(x) = -~ H(af)(x) + ~ [A,H]f'(x),

[A,H]f' = A(Hf') - H(Af').

Since

llH(af)II2 ~ llall~IlflI2, it is sufficient

II[A,H]f'II2 =< Const llalI~IifIl2 •

to show that (i.ii)

we will prove a better inequality.

II[A,H]f'II2 ~ Const IIalIBM0 llfll2 .

Without loss of generality we may assume that have, for any real-valued function ([A,H]f',g) = ~_~

a,

f

are real-valued.

We

g E C~,

[A,H]f'(x)g(x)dx = (A,Hf''g + f'Hg)

= 4 im(A,f~ g+) = 4 Im(A,F') = -4 Im(a,F), where (1.12) Let in

F(x) = fx_~ f~(s)g+(s)ds = -i f~ f~(x+is)g+(x+is)ds.

a(x,y) = Py * a(x,y), F(x,y) = Py * F(x). U,

we have

~0F (x,y) = f$(x+iy)g+(x+iy).

Since

f$(z), g+(z)

are analytic

Thus Lemmas 1.2, 1.3 and

Parseval's formula yield that l(a,F) l

= Const l//U y

Oa ~ (x,y)

= Const Iff

8a ~-x (x,y)f~(x+iy)g+(x+iy)

Y

OF 8~x (x,y) dx dy l dx dy I

U Const {flU Ylf+(x+iy) 12 dx dy} I/2 {f7 U y IVa(x,y) 121g+(x+iy)12dx dY} I/2 Const IIf+II2 IIalIBMO fig+If2 ~ Const IIalIBMO IIfll2 llgll2 • This completes the proof of Theorem A. Fefferman-Stein essentially

Lemma 1.4 ([27]). Proof.

[27] showed also the following inequality, which is

same as (l.ll). Let

a E BMO.

Then

iI[a,H]ll2,2 ~ Const IIaIIBM0 •

Without loss of generality we may assume that

for any real-valued functions

a

is real-valued.

f,g E C O ,

([a,H]f,g) = (a, Hf'g + fHg) = -4 Im (a,f+g+). Let

G(x) = f+(x)g+(x).

G(x,y) = P

Y

* G(x),

Then Parseval's formula shows that, with

a(x,y) = P

Y

* a(x),

We have,

l(a,f+g+)l

8a y ~- (x,y)

Iff

= l(a,G)I = Const

8G 8~x (x,y) dx dy I

U

{ff Y iVal2 I GI dx dy} 1/2 {77 U Y IVGI 2 I GI-1 dx dy }1/2.

Const Since

U

log IG(x,y) 1

is subharmonic in

U,

2 A log IGI

=

PJ-~4"[

(AIGI _

,~,

)

O,

and hence

~2

= ~,G,

+ ~

=< 2 A]G[.

This shows that flu y IVGI2 IGI-I dx dy ~ 2 flu y Since

IG(x,y) iI/2

is subharmonic in

&IGI dx dy = Const fIG11I.

U,

we have

IG(x,y)l ~ Py * (IGil/2)(x)

Hence Lemmas 1.2 and 1.3 yield that

flu y IVa(x'y) I21G(x'Y)I dx dy

flu

y IVa(x'y) 12 PY * (IGIl/2)(x)2dx dy

2 Const llallBM0 IIGIII. Consequently, we have Q .E.D.

l([a,H]f,g) I _-
§1.6,

The Coifman-Meyer expression (Coifman-Meyer [8]) It is important to understand Theorem A from the point of view of real

analysis.

Coifman-Rochberg-Weiss

functions.

Lemma 1.5 ([8]). =

where

a

-

[A,H]f'(x)

Const f_~ [a_s,H]fs(X)/(l+s2)ds

= k S

* a, f S

= k S

* f, ks(X) = S

We have, for

(a E BMO, f 6 C~),

~s/IXl l+is

and

.

E s = F((l+is)/2)/ {F(-is/2) I S } Proof.

[15] showed Lemma 1.4 without using analytic

Coifman-Meyer gave the following expression.

.

a, f 6 CO,

[A,H]f'(x) = Const i 7 ~ 7_~ ei(~+q)x {sign~]- sign(~-~q)} ^

where

^

a, f

are the Fourier transform of

a, f, respectively.

~(~) i~(~)d~ an

Note that

10

{sign ~] - sign(~+q)} (~]/~) = - {sign ~] - sign(<+~q)} X(0,1)(l~]/~I). I~]/~ I = Constf_~ %_s(~) = k_s(~)a(~) =

I~]/~I is/(l+s2)ds

I~I -is a<~)

[A,H]f'(x) : - Const f ~ a_s(~) ~ s ~ )

d~

(In the case of

and

[i/~

( ID/ ~I =< i),

fs(~) = I~qlis fe),

~

e i~+~)x {sign~

d~]]/(l+s2) ds = -Const a 6 BMO,

Since

we have

- sign~+~])}

~ ~ [a s,H]fs(X)/(l+s2)ds.

f 6 CO, it is necessary to show the convergence of

the quantity in the right-hand side of Lemma 1.5.

This will be shown later in the

proof of Theorem A.)

Q.E.D.

Here is another lemma necessary for the proof of Theorem A. Lemma 1.6 ([8]). Proof.

llasllBMO& Const(l + Isl 3/4) II~IBMO.

Without loss of generality we may assume that

a (I) = (a-(a) I) X i,, a (2) = (a - (a) I) XI, c . a

= a (I) + a (2), where S

S

a (j) = k

S

shows that I

S

* a (j)

s > 0.

(See Lemma 1.2.)

(j = 1,2).

We put Then

John-Nirenberg's inequality

S

IIa(1)II2~ ConstlIallBMO~I/~ , and hence la(1)(x)Idx & s

lla~l)H2~ V ~

=

lla
~ Const [IalIBMO llI.

I Note that with

l~sl

x 0 = (the midpoint of f

In the same manner as in Lemma 1.2, we have,

~ Const(l + ~ ) . I),

la(2)(x) - a(2)(x0 ) I dx I I~sl

f

I fZ I

1 {Ix yl l+is -

Const {IZsl (i + sl/4)} f

=

I

i l+is } Ix0-Y I iX_xoll/4 {fl, c

a(2)(y)dyl dx i iXo_yi5/2

la(y)-(a)lldY} dx

Const (I + s 3/4) !IaIIBMO Ill.

Thus we obtain (la s - (as)ll) I ~ 2(la s - a~2)(Xo)l)l ~ Const (i + s 3/4) IIaIIBM O, which gives the required inequality.

Q.E.D.

Theorem A is deduced from Lemmas 1.4-1.6 as follows. that it is sufficient to show that

Inequality (i.i0) shows

II[A,H]f'II2 ~ Const llall~IIfll 2 • Lemmas 1.4-1.6

11

yield that II[A,H]f'II 2 ~ Const 7 ~ Const f £

II[as,H]f32/(l + s2) ds

I][as,H]ll2, 2 ]Ifsl]2/(I + s2) ds

Const IIfll2 fflIasllBMo/(l + s 2) ds Const IIaHBMO llfll2 ~ Const !lall~!Ifll2 . §1.7.

A tent space (Coifman-Meyer-Stein

[13], [14])

The essential part in the sections 1.3 and 1.5 is the proof of the inequality: IIFII1 ~ Const llfll2 l!gll2 (f, g E L2), where

F(x) = fx~

( = - i /0 f$(x + is)g+(x + is)ds).

Rs

by

Rsh = ~s * h, where

F(x) = Const /~_

Rshs(X) ds/s,

where

denote the operator defined

Then we have

hs(Y) = s f~(y + is)g+(y + is).

From this

introduces tent spaces and generalizes

As seen in the proof of the

tent spaces are very useful. Meyer-Stein's

(s E ~)

~s(X) = s2x/(x 2 + s2) 2.

point of view, Coifman-Meyer-Stein inequality.

Let

f$(s)g+(s) ds

•

Tb

I

theorem (Davld-Journe-Semmes

the above [20]),

The following theorem is a special case of Coifman-

theorem; we rewrite their theorem so that only the proof of

Theorem A can be given. Theorem 1.7 ([13]). with norm

Let

T

be the Banach space of functions

llhllT = IIS(h)lll, where

A(x) = {(y,s);

ly-xl < s}.

hs(Y) = h(y,s). Theorem A

Then

h(y,s)

in

S(h)(x) = {7f~(x) lh(y,s)l 2 ayas/s" ~ i 241/2

For h E T, we put

R(h)(x) = fO Rshs(X)ds/s'

U and where

IIR(h)II1 ~ Const llhllT.

immediately follows from this inequality,

since

risf~(y + is)g+(y + is)IfT ~ IIA(f+)m(g+)I[ I ~ Const JifiI2 prgll2 . Here are two lemmas necessary for the proof. = {(y,s); 0 < s < dis(y,lC)} For an open set components

I

(1.13)

where

supp(')

of

~ c ~,

we write

~ .

We say that

supp(p)

c~,

For an interval

(dis(',')

is the distance).

~ = D ~, where the union is taken over all p E T

ff^ Ip(y,s) I2 I

is a T-atom if, for some interval dy ds

~

I/iii,

S

is the support.

Lemma 1.8 ([13]) •

For any T-atom

I, we write

p,

IIR(p)llI =< Const.

I,

12

Proof.

Let

for any

p

be a T-atom and let

b E L~

b 1 = b Xi,

with

and

I

Then,

llblI~~ i, I (R(p),b)I ~ I (R(p),bl)I + I(R(p),b2)I

b 2 = b XI, c

(I*

is the double of

IRsb2(Y) l ~ Const s/IraI ((y,s) E I), I(R(p),b2)I =

be an interval satisfying (1.13).

Iff^ I

I).

where

Since

we have

p(y,s) Rsb2(Y)

~ s

I

=< (Const/III)

7f^ Ip(y,s)l dy ds I _< (Const/ill) {ff^ ip(y,s)l 2 dy ds }1/2 s I

{ff^ I

s dy ds} I/2 ~ Const.

We have I(R(p),bl)I ~

{ff^

77^ IP(Y,S) IIRsbI(Y)I I

ip(y,s)12

~

I (l~/iil) {ff U Consequently we have Hb]l® ~ i,

/f^ ^ l-g where Proof.

{ff^

iRsbl(y) 12

d y ds }1/2 s

I

iRsbl(Y) l2 dy ds}i/2 s -

I(R(p),b)[ ~ Const.

we obtain

Lemma 1.9 ([13]).

}1/2

s

dYadS

Const.

= Const Ilbli]2/~

Since

b

is arbitrary as long as

HR(p)]]1 ~ Const.

Let

ih(y,s) i2

h E T

and let

dYadS

=< /I-E

Q.E.D. E

be a subset of an interval

I.

Then

S(h)(x) 2 dx,

~ = {x E I; MXE(X) > 1/2} . A geometric observation shows that, for any

Y O ~c ~ ~ , where

Y = [y-s,y+s](C ~).

IY N EI/IY 1 --< MXE(X 0) <-- 1/2,

Let

and hence

fI-E S(h)(x)2dx = fI-E

(y,s) E ~ - ~,

x 0 E Y N ~c.

IY N ECI >-_s.

y c Y

and

Then

This shows that

{ ffa(x) Ih(y's) I2 dy 2ds

} dx

s

>

]h(y,s)12 dy ds

ff^ ^

We now prove Theorem 1.7. ~k = {x; MXEk(X ) > 1/2}

Q.E.D.

S

I--~

Given

h E T, we put

(k = O, ±i, ...).

the totality of components of p(k)(y,s)j = (2-k-I/iI(k) i)j

~k

For each

E k = {x; S(h)(x) > 2k} , i~lj T(k) ~7j=l be k, let

and let

h(y,s) xj(k)(y,s)

(j >-_1),

13

where

X! k) is the characteristic function of ~!k) _ ~!k) J J J Then we obtain the following T-atomic decomposition of h: h(y,s) = p(k)j

Each

Z k =-~

Z j =l

2k+l I£4~(k)jI~ P~k)(y,s)-

is a T-atom, since

ff~!k) I P~k)(y,s)I2

supp(p~ k)) c i(k)j

dYsdS

2-2k-2 II!k) I-23

fl (k) E j

S(h)(x)2 dx =< i/ll~k) l

- k+l

Hence Lemma 1.8 shows that

IIR(h)iII --< k =-~E j=IE 2k+l Ijl(k)

Const <

alia

= 2-2k-2 ll~k) l-2 f/^(k) ^(k) lh(y's) 12 ~y ds I, ~Q. s J J

J

by Lemma 1.9.

~(k) = i!k) N ( j J ~k+l )"

Z k =-~

E j~l

Const llS(h)IIl =

IIR(P~k) )III

2k II~ k)

=

Const

S

2klEk 1

k = -~

Const llhl]T .

This completes the proof of Theorem 1.7. As stated above, Theorem A is deduced from this theorem. ~1.8.

The Mclntosh expression (Coifman-Mclntosh-Meyer

[7])

The proof of Theorem A in this section is a version of the method given in [7] for the proof of Theorem B. of

(See Chapter II .) Here is an interesting expression

T[a].

+~ I ds (a E L ~) , Len=na i.i0 ([7]). T[a] = f_Z I isDMa I + is D s where I is the identity operator, D = -i(8/8x) and M a is the multiplier: f -~ af. Proof.

Let

a(x) = e igx, f(x) = e i~x

r[a]f(x) and

=

(-~i) { ~+~

(~, ~ E ~).

sign(~ + ~) - ~

Then we have sign ~ }

14

J-~

I { I + is D

= 7-Z

{ I + is D

I

1 1 + is(a+

J--

=

i 7

f~

=

7

=

(-~i)

1

{

I I + is D

Ma

i+ 1 1 + is ~

i 1 + is(e + ~)

~

(x)

1

(af) }(x)

~)

f}

{ (1 + i s ( ¢

ds S

is ~ ds s

1 1 + is ~

~ + ~

f-~

as 8

--

}

ds 2s

~ + ~))2

}

ds

(1 + i s ~ ) 2

s

{~ + B sign(~ + B) - ~ sign $}.

Hence

T[a]f

= f -~

I { I + is D

{e iax }~ E

Since

Ma

is complete

I I + is D

the proof

of the integral

ds -s-

in the space of functions

f_Zlf(x) 12/(I + x 2) dx < ~ , the required the convergence

} f

equality

in the right-hand

f

holds. side.

with norm

(It is necessary

Q.E.D.

of Theorem A.)

Let

Ps =

to show

This will be given in

I /(~ + s2D2),

%

= sD/(l + s2D 2)

(s > 0).

In the same manner as

in Lemma 1.2, we have Lemma

i.ii

constant

([7]). Const

Lemma (1.14)

Let a ~ BMO. 2 IIalIBMO •

i.i0 shows

Then

tQsa(~) I

2

dxds

s

is a Carleson measure with

that

T[a] = f - Z

{PsMaPs - i % M a P s - i PsMaQs - QsMa~s

~

ds

= -2i fO ~sMaPs

-s-

- 2i fO

PsMa %

dSs

(= -2i L I - 2i L 2, say).

we see that

%= Hence

Q3s

the integration

S~s by parts

shows

that

} dSs

P s = -2 3- . ~s

L1 = f O

[8Q "

= 870

+ S

{ -% + 2 Ps%}]M a P "

~

Ma Ps dSs - 7 0

Since

IIPsll2,2 -< - Const,

s

8 Ma(SS~s

2 Ps%}

Ps)

ds

~ -

ds -% + 2%Ps }Ma Qs2 s-

say). IIQslI2,2 =< Const

and

70"

jIQsfJl

ds

= Const IIfll2

s

2

]IL12112,2 =< Const llaIl~ • we have, for

Shwartz's inequality shows that ] (g, Lllf) l =

+

-270 {

" = 81"O Q3 s M a Ps dSs +

( = 8 LII + 2 LI2,

{-%

ds

s

If; (~g,

f,g E L 2,

%MaPsf) diss

; 2 2 ds }i/2 ~ 22 ds 1/2 {7 I]~g]12 s{~0 ]]~MaPsfN -}s 2 ds }1/2 Const I]gl]2 {7; I]%MaPsfN 2 ~•

= We see that

{%MaPs}f = (%a)(Psf) + Ps {(Psa)(% f)} (To see this,use

a(x) = e iax, f(x) = e i~x

- %{(%a)(%f)}

(~, ~ E R).)

II%MaPsfI122 ~ Const {Ilall2 ll%fII~ +

.

Hence we have

II(%a)(~sf)II~ }.

Lemma I.ii shows that 7; II%MaPsflI22 dSs =< Const Iiall 2 7; H%fH22 + Const

flU ]Qsa(X) Psf(x) 12 dXsdS

_-< Const

Ilall2 llfIl~ + Const

/7

dSs

]Qsa(X) Ps * f(x)12 U

2

_-< Const {llall which gives

2

+ IIaIIBMO} I]f]l -<_ Const IIall 2 llfll ,

IILIIII2,2 ~ Const llall, . Thus

the dual operator of (1.4).

LI,

we have

IILII]2,2 ~ Const Ilall~ . Since L 2 is IIL2 I12,2 =If LIII2,2 • Consequently, (1.14) gives

•

§1.9.

s

Almost orthogonality (Davld-Journe [19]) David-Journ~ [19] showed the so-called

showed the so-called

Tb

immediately Theorem A. only on For

dx ds s

theorem.

theorem (cf. McIntosh-Meyer Given

5 ; the value of

6 > 0, we use C6

B > 0

C6

David-Journe-Semmes

[40]).

[20]

These theorems give

for various constants depending

differs in general from one occurrence to another.

0 < 5 ~ i, we say that a kernel

kernel if there exists

T1

K(x,y) (x # y; x,y E ~)

such that

is a

8-standard

16

tK(x,y)l

N B/Ix-yl , -<-

IK(x,Y) - K ( x , Y ' ) t

=< B I y - Y ' I 5 / I x - Y i 1+8

We denote by

m 8 (K)

inequalities,

(Ix-x'l

BIx-x'l 6 / t x - y l 1+6

IK(x,y) -K(x',y)l

=< I x - y l / 2 ) ,

(IY-Y'I =<- / x - y I / 2 ) -

the minimum of constants

B

satisfying the above three

For the sake of simplicity, we assume that

(I.15)

K(x,y)

(1.16)

Kl(x) = +

is anti-symmetric, i.e., lim s ~0

lim s~ 0

We write simply

K

fl

fg

K(x,y) = -K(y,x), K(x,y) dy

< Ix-yl < 1 Ix-yl < i/e

K(x,y)dy

exists

a.e.

<

the operator defined by the kernel

K(x,y).

The following

theorem is a special case of the TI theorem; we rewrite the TI theorem so that only the proof of Theorem A can be given. Theorem 1.12 ([19]). (1.16).

Then

Let

K(x,y)

be a

6-standard kernel satisfying (1.15) and

IlK]f2,2 ~ C 8 {IIKIlIBM0 + ~8(K)}

Integration by parts shows that

~l(T[a])

IIHaIIBMO ~ Const IIail~ and

.

T[a]l = ~ H a.

~ Const !fall .

We see that (See eemma 2.5.)

Hence this

theorem immediately yields Theorem A. Lemma 1.13 ([19]). e(x,y)

such that

Proof.

Let ef =

L 2 f;

Then its kernel

For

%{(~b)(Psf) } L(x,y) {//

U

flu Us(X-t)(Vs*b)(t)Vs(t-y)

QQb

d ss

SLIt2, 2 ~ C o n s t

dt ds s

dt ds T } '

Then

v s(x) = sign(x/S)Us(X) L(x,y)

(x ( ~, s > 0)

and

is anti-symmetric and

= b.

s

IVs*b(t)'t 2 dtds/s

we h a v e

ds Ps{(~b)(Qsf)} °~-

m

U = {(t,s); t ( ~, s > 0}.

since

d~Ss + 2 f;

Vs(X-t)(Vs*b)(t)Us(t-y)

Us(X) = (i/s)e -]Xi/s ,

L1. = 2 f~ U

el(L) N Const I]BIIBMO.

is given by

L

where

and

be an operator defined by

L(x,y) = Censt +

b ( BMO, there exists an anti-symmetric l-standard kernel

El = b, []elI2,2 ~ Const IIBIIBM0

is a Carteson

measure with

constant

Const

2

ITbtlBM0 ,

HblIBMo .

I t r e m a i n s to p r o v e

el(L) ~ Const IIbIIBM0 •

I n t h e same manner a s i n

17

Lemma 1.2, we have [L(x,y)l ~

IIVs*bll = ~ Const IIblIBMO. Const HDIIBMO

flu

Us(X-t)Us(t-Y)

d

Const IIBIIBMO S_~

Since

t

(Ix-tl + it_yl) 2

IVs(X)l ~ Us(X), dt ds s

So

e-I/s ds

i

s3

=< Const IIbllBMO/I x-yl Since

lU's(X)I -<- Us(X)/S, 18 ~

IV's<X>I < Us(X)/S

(x # 0), we have

L(x,y) I ~ Const IIBIIBM0 fS U Us(X-t)Us(Y-t)

= Const llbIIBM 0

dt (ix_tl + it_yl)3

f~

~

f0

dt ds 2 s i ~es

-i/s

ds

Const IIbllBMo/IX-yl2 Thus

Q.E.D.

c01(L) =< ConstllbIIBMO . Here is the main tool for the proof of Theorem 1.12.

operator L from any f, g ~ L 2.

L2

Lemma 1.14 (Cotlar's Lemma [16]). itself and let that

{Lk}~=_N

@(t)

be a function from

be anti-self adjoint operators from

llejLkll2,2 ~ o(lJ - kl)

llZ ~=_#kI12,2 =< Const

Let

(j,k = O, il ..... iN).

for

[0,~) L2

to

to itself such

Then

E k= 2N0 ~o(k) •

Proof. The following proof is due to Fefferman [25]. Let for any M g i, IIe2MIII/2M 2,2 = IIell2,2. We have L 2M =

We say that an (ef,g) = -(f,Lg)

to itself is anti-self adjoint if

N

L = E k = _ ~ k.

Then,

% .. -N~kl,...,k2M ~ N LklLk2 " Lk2M

Since HLklLk2 • "" Lk2 M 112,2 ~ PILklLk2ll2,2 " • "]ILk2M_ILk2M II2,2 p(Ikl-k21) .,. @(Ik2M_I - k2M I) and IILkl...Lk2MII2,2 =< IILklIl2,2 IILk2Lk3112,2 -.- IILk2M_2 Lk2M_III2,2 IILk2MII2,2 ~

P(Ik2-k3 I) ... @(IkmM_2-k2M_l I) "P ~

,

18 we have 2

IFLkl..-Lk2 M H 2,2 ~ p(O) p(l kl-k21 )P(I k2-k31 ) ...

p(l k2M_l-k2Ml )"

Thus I}LN2,2 = IIL2~II/2M2,2

{Vp (0) -N~

z ~P(I kl-k21 )P(I k2-k31 ) -.-P(I k2M_l-k2Ml ) }II2M kl,...,k2M =< N

{gp(o)

z

~P(I kl-k21 ).. ,P(I k2M_2-k2M_iI )

-N ~ kl,...,k2M_l =< N 2N

×(2 z Vp(j)) j=0 • .. ~

2N % OV~)

{ ~p(0)(2N+l)(2

}i/2M

2M} 1 / 2M

j=O <

p(0) I/4M (2N+I) I/2M

2N Z

2

I/p(j) .

j=0 Letting

M

tend to infinity, we obtain the required inequality.

We now give the proof of Theorem 1.12. KI ~ BMO,

We may assume that

valued.

Since

e(x,y)

so that

KI = LI, llelI2,2 ~ Const IIKIIIBM0

~8(L)

~ ~I(L)

~ Const IIKII!BMO.

Q.E.D. K(x,y)

is real-

we can define, by Lemma 1.13, an anti-symmetric kernel

is anti-symmetric and satisfies

and

Consider the kernel

K(x,y) - L(x,y).

Then this

(K-L)1 = 0,

~08(K - L) _-< Const {IIKIIIBMO + ¢08(K) } . Hence from the beginning, we assume this, we may assume that 0 =< h(x) < i,

KI = 0,

~5(K) = i.

h(x) = h(-x),

Choosing

and show

IIKII2,2 ~

h ( C~

so that

supp(h) c [-i,i],

C8~8(K).

To do

[]h[]1 = i,

we put Kk(X,Y) = I_: i : K(x-s,y-t){hk(S)hk(t)- hk+l(S)hk+l(t)} ds dt where

hk(X) = 2-~(2-kx).

K = lim k ~ ~ Zk=_NN K k. (1.17)

IKk(X,y) I &

Then

Kk

is anti-self adjoint,

We show that C 8 2-k/{l + Ix-yl2-k} I+6,

Kkl = 0

(k = O,tl .... ), and

19 8 I ~-x

(i.18) If

Ix-yl2-k} 1+8

Kk(X'Y) l < C8 2-2k/{i +

Ix-yl ~ 4 ° 2k,

then we have

IKk(X,Y) l = I 7_~ /_~{K(x-s,y-t)-K(x,Y)}{hk(S)hk(t)-hk+l(S)hk+l(t)}

ds dt I

Isl6 + Itl~_ /_: / ~

c8

ix_ylZ+5

ds dt

{hk(S)hk(t) + hk+l(S)hk+l(t)}

C8 28k/[x_yll+ 8 ~ C8 2-k/ {i + Ix-y[2-k} I+8 Let

I, J

we have

be two intervals in an interval /IN J fIN J K(s,t) ds dt = 0,

(1.19)

I /I

{/J K(s't)ds}dtl =

+ /INJ

L.

Since

K(x,y)

is anti-symmetric,

and hence I/l-(In J) {$J K(s,t)ds} dt

{/J-(INJ) K(s,t)ds}dt

I= <

/ LI

2

Integration by parts and (1.19) show that, if

ds S +tdt =< Const ILl.

Ix-yl < 4 " 2k,

then

IKk(X,Y) I = I /_~ /_~ {/~

/_~

/~

v

/0

K(x-s,y-t)ds dt} {h~(u)h~(v)-h~+l(U>h~+l(V)}

(Const 2k)

{lh~(u)h~(v)I + lh~+l(U)h~+l(V)l}du

du dv I

dv

Const 2-k < C 8 2-k/{l + Ix-yl2-k} I+8 =

Thus (1.17) holds.

If

Ix-yl ~ 4 " 2k,

then

18~ Kk(X,Y) l If _~~ 7 -~"{K(x-s,y-t)-K(x,y)} {h{(S)hk(t) /_~ /_~ C8-Jsl 6 + It15

C 8 2 (8-1)k ~ C 8 2-2k/ {I + Ix-yl2-k} I+8 Ix-y[ < 4 • 2k,

then

' t )} hk+l(S)hk+l(

ds dt I

{lh~(s) lhk(t ) + lh~+l(S) lhk+l(t)} ds dt

Ix_yj1+5

If

-

20

[fi ~(x,Y) [ ~

=

V

[f_~ f_~ {fo f0 K(x-s,y-t)dsdt}{h~(u)h~(v)

- h~+l(U)h~+l(V )} du dv [

Const 2-2k ~ C8 2-2k/{I + ]x-yl2-k} I+6 . Thus (1.18) holds. We now show that, for

(1.20)

k ~ g,

[(KkKe)(x,y)[ -<- C8 Pk,e(x-Y),

where pk,g(s) = 2-(k+~)/2(i + [s[2-g) -(I/2)-8 +2 (~8/2) {2-2k]sll-(8/2)

Let

I be the interval of midpoint

[fic C6

By

+ 2-kls] -8/2} (i + [sl 2-k) -I-8 .

y

Kgl(y) = 0,

Ix-yl/2.

2-k (i + Jx-sI2-k) I+6

Kk(X,s)K (s,y)ds[ ~ C 6 fic 2-g/2 (i + Ix-yl2-g) (I/2)+~

and of length

® f -~

By (1.17), we have

2-g (i + Is-yl2-g) I+8

2-k

(1 + Ix-sl 2-k)i+8 Ix-sl 1/2

ds =< CsPk'~

we have

IfI Kk(X,s)Kg(s,y)ds I ~

IfI {Kk(X,s)-Kk(X,y)}

Kg(s,y)ds I

+

IKk(X,y) f I K~(s,y)dsl

=

+

IKk(X,Y) / cK~(s,y)dsl I

( = Ll(X,y ) + L2(x,y),

Jfl {Kk(X'S) - Kk(X'Y)} Kg(s,y)dsJ say).

By (1.17) and (1.18), we have Ll(X,y) ~ C 8 f I

2-2k I s-Yl (i + Ix-yle-k) I+8

2-~ (i + Is-yI2-g) I+8

2-2k+(gS/2) --< C 8 and

(i + Ix-yl2-k) I+6

fl

ds

Is-y1-6/2 ds _<-C 8

pk,g(x-y)

ds

(x-y).

21

2-k L2(x,y ) -<_ C 6

2 f

(i + Ix-yl2-k) I+ 5

Ic

(i + Is-yl

2_Z)146

ds

2-k+ (g6 / 2 ) C8

Thus (1.20) holds.

Since

!IKkKgll2, 2 ~ C 6 2-(k-Z)/2 Letting

N

flC

(i + Ix-yl 2-k) 146

Is-yl -I-(5/2) ds =< C6 Pk,~(x-y)-

llpk,~II1 ~ C 8 2 -(k-g)/2,

we have

Hence Lemma 1.14 gives

IIZ~=_N Kkll2, 2 ~ C6

tend to infinity, we obtain

IIKII2,2 ~ C 5 .

(N ~ I).

This completes the proof

of Theorem 1.12.

§i. I0.

Interpolation

(Lemarie [36])

In this section, we give a proof of Theorem 1.12 (in the case of by interpolation, which was given by Lemarie. let

E

CO

with respect to the norm

For

denote the Banach space of distributions

Fourier transform of [E ,E_~]

f

a 6~

=

obtained from the completion of

(in the sense of distributions).

{~

K1 = 0)

I~I < i,

lllflll~ = {f_~ I~I ~ I~(~)I 2 d<} I/2,

denote the Banach space of distributions

llfIllnt,~

with

f

in

For E

where

~

0 < a <

I,

~

E_~

is the let

with norm

B(s,f)2 d s2 }1/2 < = , $

where B(s,f) = inf{(IIlgill2+s2ilihllI2 )i/2;_

f = g + h, g ~ E , h E E_~ }-

Lemarie showed the following two facts. Lemma 1.15 ([36]). and

El = 0.

IIIKI~I~,~

Let

K(x,y)

Then, for any

is the norm of

K

constant depending only on

6.

Given

as an operator from ~

We may assume that

INfHI~

=

x, y E [,

21x-y I.

By

c a f_: f ~ we write by

KI = O,

6-standard kernel satisfying

< min{l,26},

and

we have

C

E

C ,6

to itself and

(1.15),

(1.16)

~6(K) , where C

6

is a

8 for various constants depending only on

~8(K) = i. If ( x ) - f ( y ) ~ ix_yil+~ I

IllKIll~,~

6 .

Throughout the proof, we use

Proof. and

be a

0 < a

Note that dx dy

the interval of midpoint

(f E E a) .

(x+y)/2

and of length

22

IKf(x)-

Kf(y)l

=

I f~{K(x,s)

I flK(x,s)(f(s)-f(x)) ifI +

ds -

{K(x,s)

- K(y,s)}

f(s)dsl

f : K(y,s)(f(s)-f(y))dsl

K(x,s)(f(s)-f(x))ds

f

- K(y,s)}

f I K(y,s)(f(s)

- f(y))ds

(f(s) - f(x))ds - (f(x) - f(y))

f

ic

K(y,s)dsl ic

fI If(s) - f(x) I/Is-xl ds +

fI

If(s) - f(Y) I/Is-Yl ds

+ c~ ~ If(s)-f(x)lIx-yl~lls-~l1+s ds ( = Ll(X,y)

+ L2(x,y)

+ C 5 L3(x,y)

+ If(x)-f(y)i

+ L4(x,y) ,

Fie

K(y,s)dsl

say).

Hence 2 IllKfilla

C

=

f : f-:

a

iKf(x)_Kf(y ) [,2 ix_y]l+~ dx dy

4

< =

Choosing

C

Lk(X ,y)

Z k=l

a,5

0 < ~

2

4

f ~ f : -~

< 1/2

-

so that

dx dy

( = C

Ix-Y iI+~ a + 2~

>

i,

say).

we have

fls-xl< 21x-yl If(s)-f(x)12ls_×12(l-~ ), ds

Ll(X,y) 2 =< Ca

6 kE=l L~,

Ix_yll-2~

and hence L 1' =< Ca

f

_®~

f

If(s)-f(x)l 2

-~"

{ J'21x-Yl > I~-xl l~-yl l+a

Is_~12(1-~5

dy } ds dx

= c N1flll~ . In the same manner D 2(1 + 6) Y > 1

L~ ~ C

and

IIIflll~

•

Choosing

(a - 26) + 2(1 + 6) Y < I,

0 < T < i we have

[f(s)-f(x)l 2

L3(x,y)2 £ Cg, 6

fts-xl > t~-yl12

so that

Ix-yl25-2(i+~)~

+ i

is_xi2(l+6)(l-y)

and hence , e3

< Ca, 6

~ If(s)-f(x)I 2 f --® f - ~ is_x 12(1+6) ( l - Y )

= %,5 IIIflll~

•

{flx_yl/2<

Is_xllX-y125-2(l+6)Y-~dY} dsdx

23

Since

lSlC K(y,s)dsl

~

i

$1y-~l > Ix-yll2 K(y's)dsl

tllfllla ( sup

L 4' =< C o n s t

i fly-st

y E ~t,~ >0

Given

Y0 ( R, s > 0,

we have, with

l~c K(Y0's)dsl

=

=

i ~

J = (Y0 - s, Y0 + g)

lSjcK(Y0>s)ds

!_, ijI Sj $j

+

(Const/IJi)

+

(Cs/iJi) Sj { Sj, c

Sj {S , J -J

~

ds

ly-s

Lemma 1.16 ([36]).

For any

i1+6

(I/C)NflI 2 ~ llfilint,¢ ~

0 < a < i,

c a llflI2

f 6 L2~ s > O,

2 gs = -~-

Then

f = gs + hs

IIIKflll~ ~ ca, ~ Illflll~

f dt,

h

= s

More precisely,

2

s -I/a SO

"~

D r + t2D 2

f

dt.

ds

{Iilgslii2a+ S 2 liihsNi2a } --~

--<S 0

_

I ~I i + t2E

at)2

1

s

I~f

dt) 2 S

2

} l~(~)I 2 d E ] ds

IEI a

i'li

a

-lla

S

S°_li= s

= c So E s_Z

i + t2E 2 dt)

1

}

dr)

s

I7(<)I 2

d E ] ds

I~1 a

1 + t2~ ~

ds

ca

, which Q.E.D.

[E ,E_¢] = L 2.

r + t2D 2

s-i/a _ i l l + ( S0 i + t2~ 2

:

{ S j , _ j + S J *c } l

and

4 E I_;

=

i

fJI lfj

dy & C 8 .

as}

S

+ ( SO

K(y,s)ds dyl

we put

IIfI!2Int,a = S; B(s,f) 2 ~

a

= (Y0-2g,Y0 + 2g),

(f (L2).

D S®-I/~ s

J

}dy

2 . Consequently we have Hence L 1 =< Ca, 8 lllfill~ gives the required inequality.

Given

and

Sj {Sjc (K(Y0,S) - K(y,s))ds} dy

<

we have

K(y,s)ds i + 1} .

>

ly0-yl ~

Proof.

+ Const,

S_l 11(O12 {7o (I'_~ -7 t

s

) 2

i +

= C a d'_] If(E) l2 d~ = C a IIfll2 .

t2E2

t -c~ --I~I1-L dt +dO(f0 ds) l+teiGi2

dt} dK

24

Let

f ~ [E ,E_~].

f = gs + hs 2 IIfll2

For each

and =

70

Const

J"

Const

gs ~ Ea' hs E E -
s > 0, we choose

lllgslll2 + s 2 lllhslll2 < 2B(s,f)2 -(l =

Const

fO

Const

70

II

II

{

(i +(t~)2 (t~)2) 2

[ 7 _ ~~ {t a

fll

dt ~-

tl

g-a t

tD I + (tD) 2

[ j'-~

oo

tD 2 I + (tD)

II

+

tD r + (tD) 2

fo

l~lalg _a(~)l2

( IIIg _JII 2

+ t-2c~

I~I-a IE _a(~)l2} d ~ ]

+ t-a

fO

dtt

t

]llh -~1112~- ) dt 1-~

t

Cc~

} ~-

t

{lgt_~(~)12 + lht- a(~)12 } d~ ] dtt

t

ca

_all

h

t

t

([ilgsii[2 + s 2 iiihsiii2a )

-ds2 s

=<

Ca

f0~

B(s 'f)2 --2 as = C-IIfi[2 ~ Int,a " s Q.E.D.

Theorem 1.12 is deduced as follows. ~05(K) = i. for each

We may assume that

We use Lemmas 1.15 and 1.16 with s > 0,

we can choose

gs E E a,

a = 6/2.

hsE

IligsIII2 + IIIhsIiI2-a -<- 2 B(s,f) 2, by Lemma 1.15. [[Kfi[22 -< C6

[[Kfll~nt,a = C6 /0

B(s'Kf)2

E_a

Let

K1 = 0 f ~ L 2.

so that

and Then,

f = gs + hs'

Thus Lemmas 1.15 and 1.16 show that d s2 s

=< C6

J'o

-<- C6

fO

=

§i.ii

(INKgs [[12 + s2 [IIKhsIii2-a) ~ds s 2 + s 2 [IlhslIi2a)--2as < C fo ([[[gsIiia = 6 s

C 6 IIfII2nt,c~ <

B(s'f)2

__ds 2 s

c6 IlfIl~.

Successive compositions of kernels Meyer [41] also gave a proof of Theorem 1,12 from the point of view of

composition of kernels.

Using his method, we show the following lemma which also

yields Theorem 1.12. Lemma 1.17. K1 = 0

Let

K(x,y)

be a

6-standard kernel satisfying (1.15), (1.16),

and

(1.21)

sup x,y E

IK(x,y) I (i + Ix-yI) I+5

< ~.

25

We define kernels

{K(n)(x,y)}~=l

K(1)(x,y) = K(x,y),

by

K(n)(x,y) =

f_~ K(x,s)K(n-l)(s,y)ds

(n a 2),

and define ~(K (n)) = Then, for any

sup { ~

I~I

K (n) Xl(X)dx I ; I

0 < s < 6,

~(K (n)) + ~8_s(K (n)) ~ C n 8~g where

Cs,g

(n a I).

interval}

~6(K) n

is a constant depending only on

8, s.

Postponing the proof later, we now deduce Theorem 1.12 (in the case of KI = 0)

from this lemma.

(This lemma plays the role of Cotlar's lemma.)

Without loss of generality, we may assume that and

K1 = 0.

Using

satisfies (1.21). (1.22)

Kk(X,y)

K(x,y)

is real-valued,

in §1.9, if necessary, we may assume that

~06(K) = I

K(x,y)

We put

~(K) = sup { ~

i

o(l,K,f); f 6 Lreal,l

I

interval},

where, in general, (1.23)

Lreal,~

= {f 6 L~; llflI~--< ~ , f

(1.24)

o(l,K(n),f) =

real-valued }

(6 > 0)

and fl IK(n)(xI f)(x)l dx

(n ~ i).

Then IIKH2,2 <_ Const {o(K) + C8~08(K)} (See Lemma 2.5 in Chapter II.)

For

-<_Const {o(K) + C 5} •

n >-_ I, f ~ Lreal,l

and an interval

we have 1 ~(I'K(2 n-i )'f) =< { 41 ~f~ =

{~

7i

/ -= =IK(2n-l)(xif)(x)l 2 dx}i/2

_2 n 12 f(x)~( )(Xlf)(x)dx} / --< { ~

and hence i Tf ~

~(l,K,f) _<- { ~

i

n -n o(I,K (2) f)}2 .

o(l,K(2n),f)} I/2

I,

26

For a while, we assume that XI f = Zk=IN ~k Xlk

Xlf

(l~kl ~ I)

is a step function, saying

Ik

interval,

Ik D I g

= ~

(k # g)).

Then Shwartz's

inequality and Lemma 1.17 show that {

~-

1

N

~T

<

=

171f(x)

~(l,K(2n),f)} 2 ~ ~

K(2~+I)

% k=l

I:I f(x)

Xlk(X)dxl

N :on+l, :^n+l. kEllflk Kk~ :X I k (x) dx I + ~5(K ~Z )) ~ =

i

~f~

~(K(2n+l))

(2n+l) + Const N

infinity, we have ~k' Ik

~6(K

) ~

N d k%l:l_ik{flk ~ =

(i/II I) ~(l,K)f) ~ C 6.

(i ~ k ~ N),

(i/III) ~(l,K,f) ~ C 6 and all intervals

}dx

2n+l C6 ,

Const N

2n+l 2-n (i/If I) ~(l,K)f) ~ {Const N C 5 } . Letting

which shows that N ~ i,

K <2n+l) (Xlf)(x)dxl

n

tend to

Since this inequality holds for any

we can remove the above assumption, i.e.,

for the given

I, we have

f.

~(K) ~ C6,

We now give the proof of Lemma 1.17.

Taking the supremum over all

f ( e~eal,l

which implies Theorem 1.12. Assuming that

~6(K) = i, we

inductively show that ~(K(n)) wheT e K(x ,y)

CO

+ ~8-e

(K(n))

n ~ CO

(n g I)

is a constant depending only on

is anti-symmetric and

~6(K) = I,

~(K (I)) + ~5_8(K (I)) ~ 0 +

'

6, ~,

show that

C

:I K(n) XI(x)dx 7I

{:I :I } +

n-l.

For the sake of simplicity,

for various constants depending only on

~(K (n)) ~ CC~ -I = 7~ :ic

Since

~5(K (I)) = I.

Suppose that the required inequality holds for we use, from now,

and is determined later.

we have

For an interval

I,

we have

{:I :I K(x's)K(n-I)(s)Y)

dx dy} ds

{II fI } = LI + L2 ' dx

and ILl1 ~ { I I IKXi(x) I2dx} I/2

{I I IK (n-l) ki(x) l2 dx} I/2 .

5, 8.

First we

27

K(n-l)l = 0, we have

Since

fl IK(n-I) El (x) 12 dx

2 71 IK (n-l) X

< =

= fl IK(n-l) XlC(X) I2 dx

(x) l2 dx +

2 71 IK (n-l) X , (x) l2 dx I -I

I*c

_2n-2 2 /I IK(n-l) Xl*c(X) 12dx + 2 C 0 71 (f * I -I

dx~_yT )2 dx

2 fl IK(n-l) X i,c(X) I2 dx + C C_2n-2, 0 Ill = 2 LI0 + C C O2n-2 ii I,

< =

is the double of

I

where

I.

Since

171 K (n-l) Xl,c(X)dx I ~ IfI K (n-l) Xlc(x ) dx I + 171 K (n-l) X , (x)dx I I-I IfI K(n-l) Xl(X)dxl

n-l

+ CO

71 ( 7 , I -I n-1 n-i ~(K (n-l)) III + C C O III < C C O III,

) dx

we have 2n-2 El0 ~ 2 71 IK (n-l) X ,c(X) - (K (n-l) X ,c)112 dx + C C O III I I < =

2 ii12 71 { fl ( fl*C IK(n-l)(x'y)-K(n-l)(s'y)Idy)ds}2 (cc~n-2/lli2) fl {fl ( /l*C

-ix_yll+8_e Ix~slS-¢

_2n-2 dx + C C 0 II

) ds} 2 dx + C C O2n-2 ii 1

CC~ n-2 lIf. Thus

2n-2 fl IK(n-l) Xl (x) 12 dx -<_C C O III.

fl IKXI (x) 12 dx ~ CIII,

and hence

IfI KXi(x)dx I <= C C0-1 III. Since Next we show that estimate of

x~

In the same manner,

Iell =< CCo -I iii. Consequently we have I

is arbitrary, we obtain

~o6_e(K(n)) -<_ C C° -I

In the same manner as in the

ILII, we have sup ~,e> 0

sup ~ E ~,c > 0

Iflx_yI >

g K(n-l)(x,y)dyl

Iflx-y I > s K(x'y) dyl < C.

~(K (n)) < CC0 -I

~ C CO-I ,

28

For

x, y ( ~,

we have = f ~ K(x,s)K(n-l)(s,y)ds

K(n)(x,y)

:

fI i

+ f l 2'

i +

+ fI~

LI

=

i +

L2

l

L3 '

! of midpoint where I 1 is the interval interval of midpoint y and of length

x

I

Ix-yl/2

!

and

t

Ix-yl 12,

and of length

13

=

1

12

(1~ U I~) c

is the We

have

IL~I

ds ---~-~[- ~ c c n-i ___ Co l Si ~ --pT--~--~S o

IL~I

=<

flx-yI

-

_-< c

Ifi, ~

{K(x,s)-K(x,y)} K(n-1)(s,y>dsl

c n-i o

fl ~ .......I s-yj ~ l x-yl 1+5

Isq

K(x,s) {K (n-l) (s,y)-K (n-l) (x,y)

IK(x,Y)IIII~

K ( n - l ) (s, y)

+

i

c c n-1 o ~ c c

i

-T~_yT ds + ] ~ i ~ -

dsl

o-l/l~_yl

and

ILil <--

Thus

IK (n-l) (x,Y) I IfiiK(x,s)dsl

n-1

n-i

Is-xlS-s

1

$ii ~

=< c c o

}dsl+

IK(n)(x,y>l_<- c

ds +

n-I

C C0

c -_< c C 0

ix_yll+~_ ~ I%-1l l x - y l

x, x T, y (

For

c0

/[x-yl.

IX-X'I ~ t x - y I / 2 ,

with

we have K (n)(x,y) - K (n)(x',y) = f_®{K(x,s)-K(x ,s)} K (n-l)(s,y) ds 6 Z k=l

=

I 1I !

where 13

=

(X

Ix-x'i

--

2

31xTyl (x

-

6 Z k=l

=

fl~

Ix-xii

, x+

3 x ~4

, x +

4

II

Lk ,

2

) -

I'~ = (y - Ix-Y~4 ' y + Ix?--~i)4 - I~ -

Ie~l < C C0 1 _n-i

_-<

c ~o

,,

12),

and

I~ = (I~ i

ix_sll+8

, 8-6 /

Ix-x I

I2

(I~' U

~

fl~

i, = (x'

)'

I~-yl

-T-~

1+6-6

,

I~

=

Ix-x'L

(y

2

-

U...U n-i

ds < c c O

'

Ixi0 -x'l I~) e.

x,+ i~i[) 2

, y + ~ i0 ' l We have

,

x-x 18/Ix-yl I+6

"

29

IL~I

= n-i < c co

n-i

C CO

,,l~-x'l 8 f~ ix_ytl_~

]-~cFfds

_ l~-x'l 5

Ix_yll+6 log(C

)

C C~-I Ix-x'18-g/Ix-yl l+8-s,

IL~I ~ Ifl~ {K(x,s)-K(x',s)-K(x,y) + K(x',y)} K(n-i)(s,y) ds j + IK(x,y) - K(x',y) l Ifl~ K(n-l)(s,y)dsl

+

C Cn-~ 0

s~

{ _Isnyl ~

l~-yl ~ Ix'_yl I+8 }

ix_yll+8 +

l

ds

n-i ix_x,18/ix_yll+8

+

Ix-y -_< C C0n-I ix_x,18-S/ix_yjl+8-s EL = fl~

{K(x,s)-K(x',s)} {K(n-l)(s,y) - K(n-l)(x',y)} ds

+ K (n-l)(x',y) ~I~ {K(x,s) - K(x',s)} as = ELI + e~0 , L~ = ]I~ {K(x,s)-K(x',s)} {K(n-l)(s,y) - K(n-l)(x',y)} ds + K(n-l)(x''Y) fl~ {K(x,s)-K(x',s)} ds = e21 + L20 and L~ = fI~ {K(x,s)-K(x',S)} {K (n-l)(s,y) - K (n-l)(x,y)} ds + {K(n-I)(x'Y) - K(n-l)(x''Y)}

fI~' {K(x,s) - K(x',s)} ds

+ K(n-l)(x''Y) 7I~' {K(x,s) - K(x',s)} ds = LII + n12 + L"10. Since

IL~zI S C C~-I $Z~

"Ix-x'j8

Js-x'jS-c " ds

ix,_sl l+~

ix_yl I+6-~

c c o~-I l~_~,ls-~/ix_y11+8-~ ,,

Ie211 ~ C Cg-I fl~ C

~

i

Is-x'l~-~

--ix_yll+8_g

n-i 18-g/ l+8-g C0 Ix-x' Ix-yl ,

as

30

,,

n-i

I s-xJ 6-s

i

]~111 _-< c cO

ds

ix_yll+~_d

Ii~, ] - ~

Ix-x' 16-81I x-yl i+s-5 s C Cn-i 0 and

IL~21 ~ c~-I

Ix-x'16-8 Ix-yPI+5-8

{lli~ K(x,s)ds I +I ,, ds Ii I~'-sl

C C0n-I Ix-x' 16-e/Ix-ylI+6-8 we have, with

I0" = I "1

O

12"

U I~

, .

.

.

X - X t 18-8

IK(n)(x,Y) - K(n)(x',y)I $ ILl0 + L20 + L30 I + C C~ -I n-I = IK(n-l)(x',y)IIYI~ {K(x,s)-K(x',s)}dsl + C C O

-< C Co -I

i ~_y~

I 71~ c

1

{K(x,s)-K(x',s)}dsl .....Ix-x'18

ds + C

+ n-i

x_yll+6_s

_!x-x'[ 8-g

n-i C CO

18-g Jx-x[14_6_8 Ix-Y

Ix-x'l6-8

n-i ix_x,18-S/ix_yll+8-¢ C CO Since

K(n)(x,y)

is either anti-symmetric or symmetric, we have also

IK(n)(x,y) - K(n)(x,y')l for

x, y, y' ( IR with

IY-Y'I ~ Ix-Yl/2.

~(K(n)) + ~8_¢(K(n) ) ~ C Cn-i O CO

ly_y,18-g/ix_yil+8-g ~ C Cn-i O Thus ~8_g(K (n)) ~ C Cn-i O • Consequently,

This shows that

~(K (n)) + ~8 _ s (K(n) ) a CnO

if

is large enough. This completes the proof of Lemma 1.17. In Chapter I, we showed 8 proofs of the boundedness of the CalderSn

commutator

T['].

Since the Calder~n commutator is closely related to analycity

of functions, it seems necessary to give more proofs and to have a unified understanding.

CHAPTER II. A REAL VARIABLE METHOD FOR THE CAUCHY TRANSFORM ON GRAPHS

§2.1.

Coifman-Mclntosh-Meyer's

Theorem ([7])

For a real-valued locally integrable function (2.1) where

A

C[a](x,y) = i/{(x-y) + i(A(x) - A(y))},

is a primitive of

a.

We write simply by

operator defined by the kernel (2.1). Calder~n on

a

graph

Lreal~ = ~ >U 0 (See (1.23).)

Lreal,~

The norm

The operator

C[a]

C[a] = (-~)H

+

Tl[a] = T[a]

Meyer commutator (2.2)

= {a E L®;

the singular integral

We put a

is real-valued}.

showed

IIC[a]II2,2

is bounded if

a E Lreal"

is expressed formally in the following form Z (-i) n Tn[a], n=l

(the Calder~n commutator)

(n ~ 2), i.e., Tn[a ]

Tn[a](x,y)

C[a]

This is called the Cauchy transform of

{(x, A(x)); x E ~}.

Coifman-McIntosh-Meyer

Theorem B ([7]).

where

a, we define a kernel by

and

Tn[a]

is the n-th Coifman-

is an operator defined by

= (A(x) - A(y))n/(x-y) n+l.

Prior to this theorem, the following three theorems were shown. that

llTl[a]II2,2 ~ Const Ilall= (a E L~), Coifman-Meyer (2.3)

IITn[a]II2,2 ~ Const n! llalI~

CalderSn showed

[9] showed that

(a E L ~, n ~ 2)

and Calder~n showed that (2.4)

IIC[a]ll2,2

is bounded if

llall~ (a E ereal)

is small enough.

At present, there are three proofs of Theorem B; the original proof, a proof by the Tb

theorem [40] and a proof by perturbation.

contained proof by perturbation. Calder~n

[4] and David [17].

In this chapter, we show a self-

A proof by perturbation was first given by

Improving their methods and repeating a simple

perturbation method, we shall deduce Theorem B only from the boundedness of ([17], [42], [45]).

(See APPENDIX II.)

H

32

§2.2.

Two basic principles

(Zygmund

Here are two basic principles Coverin$ Lemma. IUxEA

~I < = •

Let

{~}X

[54]) in real analysis.

E A be a family of intervals in

Then there exists a sequence

{I~k}k=la

~

such that

of mutually disjoint

intervals such that

lIXkl.

iux~ A ~I-< 5 k=iE

The proof is as follows. larger than the supremum of

~i' "'''

~k-i

llxI

Now we show that

{j;

{IXk }

we have

> 211X. I, j # ~ . Let k

211kk I ,

the same midpoint as

If

{IXk }

be an interval such that

over all

X E Ak_ I,

where

(k ~ 2). (If

is the required sequence.

is

211~kl

A0 = A

is

and

Ak_ 1 = ~, we stop our

We first assume that

Since the intervals are mutually disjoint and

For

IX,

there exists

X ~ A.. Hence J be the smallest integer in the set.

IX. 3

which implies that

which gives that I~ k

211XII

Suppose that

according to the definition of our choice.

1% n Ixk # ~,

IUhE21 Ikl

IXk

lim k ~ ~ llXk I = 0.

[IxI

X ~ Aj}

IIxI ~

, we have

Let

X E A .

(i ~ j ~ k-l)}

is an infinite sequence.

IUk= I IXk I < ~

be an interval such that

over all

II~I

Ak_ 1 = {X E A ; Ik n IX. = ~ J induction at k-l.)

such that

IXI

have been chosen.

larger than the supremum of

{IXk }

Let

IX c IXk , where

and of length

--< I U IXk I --< 5 k=l

51

I"

IX k

Then

Since IXk

X ~

Ak,

is the interval of

Thus

% llXkl • k=l

is a finite sequence, each

IX

intersects with

in the same manner, we have the required inequality.

U

IXk"

Hence,

This completes the proof of

this lemma. Risin$ Sun Lemma. ~ a(x) ~ ~

for any

( ~ ~ T ~ ~ ),

Let

a

be a function in an interval

x E I, where

~ ~ O.

we define a function

the infimum is taken over all functions

B ~

Let in

I

such that

A by

I

such that

be a primitive of

a.

For

B(x) = inf ~(x), where ~ ~ A, ~' ~ Y

a.e. on

I.

33

Let

b = B'

and

components of

~ = {x 6 I; A(x) # B(x)} ~.

(2.5)

=

Uk= I I k ,

where

are the

{Ik}k= I

Then

Y _-< b(x) _-< ~

(2.6)

b(x) = Y

(2.7)

(a)l k

a.e. on

I,

(x 6 ~), I

(2.8)

_-< Y

I~I--< ~ -

((a)l k =

- (b)l T

~

III

fl k a(s)ds, k >-_ i), i "I~[ ~I b(s)ds).

((b)l =

I I

i

Inequalities

(2.5)-(2.7)

are easily seen.

(b) I III = 71 b(s)ds =

which gives (2.8). of Type i

71_62

We have

+ 7~

For the sake of convenience, we call this rising sun lemma RSL

(7-r~y,8-~e£~t);

an open set

z

we shall use later various rising sun lemmas.

~ , we denote by

{I~ ,k}k=l

its components.

For

The following two lemmas

are also the rising sun lemmas for integrable functions. Lemma 2.1 (The Calder~n-Zygmund k > 0.

decomposition

Then there exists an open set

~

[35, p. 12]).

=

in

(0,~)

such that

by

k

and

(k => i),

X

A(x) = f0 If(s) Ids

If(x) l < k

~c.

a.e. on

(x > 0), and define a function

B

B(x) = sup ~(x), where the supremum is taken over all functions

~ & A,

l~ll

f E LI

=

I~I < llflll/k, (Ifl)l~,k TO see this, we put

Let

such that

~

If(x) l ~

~' ~ k 1 ~

a.e. on

f~ If(s) Ids

a.e. on

Let

(Ifl) •

X

(0,~).

= ~ l~l,k

(0,®) - ~I "

~I = {x > 0; A(x) # B(x)}. (k ~ i),

Then

34

Considering Then

f(-x),

we obtain, in the same manner, an open set

~2

in

(0,~).

~i U {-x; x E ~2 } is the required open set. In the same manner, we have

Lemma 2.2. satisfy

Let

f

be an integrable function in an interval

k > (Ifl)l.

Then there exists an open set

i I~I _-< ~ 71 If(x)I ds, If(x) l < k

a.e.

M

is defined by

the supremum is taken over all intervals M

I

k > 0

such that

I

The (non-centered) maximal operator

denotes the norm of

in

and let

(k>= i),

(Ifl) I =< ~,k

on

~

I

I

Mf(x) = sup(Ifl) I,

containing

as an operator from

Lp

x.

For

to itself.

where

p > i,

NMIIp,p

The following lemma

is deduced from Covering Lemma. Lemma 2 . 3 ([35, p.7]). For

f ( L I,

IIMIIp,p~ Cp

X > O, we put

can choose an interval

Ix

(p > i). Ek = {x; Mf(x) > ~}.

containing

x

so that

For each

x ( EX, we

(Ifl) I > ~ . Covering Lemma X

shows that there exists a sequence that

IE~) ~ 5 %k= I Ilxk),

f ( L p and

fk(x) = 0

if

HM f[]

=

=

k~ I /i k if(s) ids ~

k > O, we define If(x) I ~ k/2. = Cp

f0

of mutually disjoint intervals such

which yields that

Ix; Mf(x) > ~I ~ ~5 For

{Ixk}k=l

fx

by

5 ~

[iflll "

fk(x) = f(x)

if

If(x) l > k/2

xP-l]x;

Mf(x) > k]dx

Cp 15

k p-I {Ix; Mfx(x)

Cp 75

xp-I ix ; MfN(x) > X / 2 )

dX

Cp /5

xp-2 llfkllI dX = Cp f~

xp-2 {fk/2 Ix;If(x) I > s I ds} dk

®

> X/21 + Ix; M(f-fk)(x) > k/2 I} dX

2s

Cp f0 Ix;If(x)] > sl { /0

which gives that

IIMIIp,p

and

Then

xp-2 d~} ds = Cp llfll ,

Cp.

At last we note John-Nirenbergts inequality, which was used in Chapter I. This is deduced from RSL. Lemma 2.4 ([32]).

Let

(For the proof of Theorem B, this is not necessary.)

f ( BMO

and

I

be an interval.

Then

35

Ix E I; If(x)-(f)l I >

-function

§2.3.

kI

=< exp(- Const >OIIl

(7~_> l).

([8], [35], [54])

In this section, we show a fundamental

the sake of simplicity, we deal with only kernels

K(x,y)

(See §i.9.)

in (1.22),

We use the notation

~(K), ff(l,K,f)

standard kernel, we define an operator =

K f(x)

I f l 1x '_' y

sup E > 0

K

For

inequality for standard kernels. satisfying

(1.21).

(1.24).

For a

by

> g K(x,y)f(y)dy I .

We show Lemma 2.5 ([35], p. 49). Then

Let

be a

K(x,y)

6-standard kernel (satisfying

(1.21)).

IIK I12,2 ~ Const o(K) + C 6 ~5(K). We begin by showing (2.9)

where

~(K ) ~ Const

~(K )

is the supremum of

and intervals I,

we put

if(K) + C 6 ~6(K),

I.

For

¢ > O,

(l/Ill) /I K (Xlf)(x)dx f E Lreal,l,

over all

an interval

J' = (x - s/2, x + s/2), J = (x - s, x + g), g =

h = XI_ J f. If

0 < ~ <

IiI,

we have, for any

Iflx-y I > s K(x'y)(XIf)(y)dyl

I

f E Lreal,l

and a point %1 fl J f

x

on

and

s E J' ,

= IKh(x) J ~ IKh(s) J + IKh(x)-Kh(s)J

IKh(s) I + C 5 ~5(K) ~ IK(Xlf)(s) I + IKg(s) l + C 6 ~6(K)

= IX ,(s)K(Xlf)(s)l

+ IKg(s) I + CO ~b(K),

I where

I

is the double of

I.

Taking first the square roots of the first

quantity and the last three quantities, respect to

s,

and taking next their means over

J'

with

we obtain

Iflx-Y I > s K(x'y)(XIf)(y)dyII/2 M(I X , K(XIf)II/2)(x) I If

S >= III, then

+ (IKgll/2)j, + C 6 ~6(K) I/2.

/Ix-yl > s K(x'Y)(XIf)(y)dy

= 0.

for all

Hence this inequality holds

s > 0, which shows that this inequality holds with the first quantity replaced by K * (Xif) (x)i/2 . Taking the squares of both sides of the resulting inequality,

and using Shwartz's inequality, we obtain

36

K (Xif)(x) _-< Const

M(IX , K(Xif)I1/2)(x) 2 I

1/2 2 + Const(iKg I )fl, + C6 ~6(K). Since

( M(IX I ,K(Xlf) iI/2) 2) I =< Const <

Const

~

{~(l,K,f) +

Const

{~(K) +

~i

fl* IK(X If) (x) idx

¢05(K) f , ( fl I -I

~

dx}

~6(K)}

and (IKgl

1/2.2 )j, ~ (IKgl)j, & (IK(kj,g) l)j, + (IK(Xj_j,g))j, ~(K) + Const

we have

(K*(XIf)) I ~ Const

Let

f E L 2,

K > 0.

~(K) + C 6 ~6(K),

I

of

*

Mf(x) > ~k

(2.11) (See §1.4). (x0

on

To prove this, it is sufficient to show that,

i

Mf(x) ~ ~Ikl =< - ~

~ E I,

llI.

=

g = Xjf

and

is the left endpoint of

1

-iO

i1 l '

h = Xjcf,

I).

where

IX E I; K g(x) > k I

+ Ix E I; K*h(x) > 2k I

(= L I + L2,

Note that

J = (x 0 - 2II I, x 0 + 2III)

Then we have

Ix E I; K*f(x) > 3X 1 <

L 2.

Assuming that

we prove

Ix E I; K*f(x) > 3X} <

First we estimate

inequality:

Ix; K*f(x) > k I

I, this inequality evidently holds.

for some

Let

k

{x; K f(x) > X} ,

Ix E I; K f(x) > 3X, If

I@f(x) =< ~k I ~_ - ~

is determined later.

for each component

M f(<) ~ ~k

i

Ix; K f(x) > 3k,

~ > 0

which implies (2.9).

We show the following good

*

(2.10) where

~5(K),

say).

K h(x 0) ~ >~. For ~ > 0

and

x E I, we have

37

Iflx-y I > e K(x'y)h(y)dy - 7 1 x 0 _ Y 1 > g K(x0'Y)h(y)dy I ~

IK(x,y) - K(x0,Y)IIN(Y)I

dy + Const

oos(K) M f ~ )

=< C6 oos(K) Mf(~) =< C5 oo6(K) ~ . Since

g > 0

is arbitrary, we have, with a constant

(2.12)

depending only on

5,

K*h(x) <= K*h(x 0) + C6,1 c°5(K) ~X -<_ {i + CS,IOOs(K)~}k

This shows that

L2 = 0

if

C5 ,I oo5 (K) D

Lemma 2.1, there exists an open set

I~I <

C8,1

~ =

(x E I). < i.

Uk= 1 Ik (I k = ~ , k )

Ilglll/(100DX), (Igl)l k = 1 0 0 ~ X Ig(x)I _<- i00~)~

We define a function

g

a.e. on

Next we estimate

L I.

By

such that

(k > I),

~c

by

f J g(x)

(x ~ ~e),

\ L Then llgll~

i00~.

Put

I~*I < 21~I For

s > 0

and

(x ~ I k, k ~ i).

(g) lk

~* = Uk= ~ I Ik* , where

Ik*

is the double of

=< llglll/(50 "qX) --<Mf(~)IJl/(50 ~X) =<

x k = (the midpoint of

(x - s, x + s).

Then

111/15"

x ~ ~ *c , there exist at most two intervals (saying

which intersect with the boundary of

Ik.

II

and

12 )

We have, with

Ik),

I71x-y I > s K(x,y)(g(y) - g(y))dy I = If(i IN 12) n (x-s,x+g) c K(x,y)(g(y)-g(y))dY +

=<

% I k c (x-s, x+s) e

{K(x,y)-K(X,Xk)} flk

Const oo6(K) {(Igl)ll + (Igl)12}

(g(Y) - g(y))dy I

38

+ C0 co6(K)

where

&(x) :

Z k=l

(Igl) (Ix-Xkl

+ Ilkl) I~5

~ C6 co6(K) ~

(i + A(x)).

Ik

Sk= I "'[IkIl'~/(IX-Xk I + ''Ilkl)I+6.

Since

s > 0

is arbitrary, we

have K g(x) ~ K Since J

of

supp(g) c J

g(x) + Cs~o6(K) q%(l + &(x)) and

(x e C'c).

I~l ~ III, the support of

g

is contained in the double

J. Hence (2.9) shows that f

, K

g(x)dx =< f , K (X , g)(x)dx J J

I-~

--< ~(K ) llgll~ IJ*l =< {Const o(K) + C0 ~8(K)} n~ Ill. We have easily I

, {C 6 ~8(K) NX (i + 8(x)) } dx I-~

-<- c6~8( K ) ~ { l ~ i Consequently,

we have, with an absolute constant

depending only on (2.13)

+ I~i} ~_ c0 ~8(K)~l~ l~J. CO

and a constant

C6, 2

6,

L I --< Ix E I - @ ; i

K

g(x) > k I +

I~*I

*

/

, K g(x)dx + iII/15 I-~

Ti 7

, {K*~g(x) + c 0 ~8(K)n~(l + M x ) ) }

dx + I~I/15

I-~ {(C O c(K) + C6, 2 ~8(K))~

+ (1/15)}

III .

Let = min {(2 C6,1~6(K))-I , (30 C O c(K) + 30 C8, 2 ~6(K)) -I} • Then (2,12) and (2.13) show that •

1

lx E I; K f(x) > 3X l ~ L 1 + L 2 = L 1 ~ Thus (2.11) holds, which implies

To

III*

(2.10).

In the same manner as in §1.4, (2.10) yields that llK*fll2 ~ (Const/~)

IIMfll2 ~ {Const ~(K) + C 5 ~8(K)} llfH2,

3g

which implies the required inequality in our lemma. ~2.4.

A-priori estimates In this section, we show some inequalities which play important roles later.

For an operator

T

from

L2

to itself, we put

(2.14)

~0(T) = sup { ~

$(I, T, XI); I

interval},

(2.15)

$(T) = sup

{

$(I, T, f); f ~ Lreal,l, I

(2.16)

~(T) = sup

{~

~(I, T, f); 0 =< f <= i,

(2.17)

~(I, T, f) = fl

IT(XI f)(x)I2 dx,

(2.18)

$(I, T, f) = fl

IT(Xlf)(x)I 2 f(x)dx.

I

interval}, interval},

where

For an open set (2.19) For a

~

with

I~I < ~ , we put

~(T;~) = sup {

8-standard kernel

~(l,T,f);

K(x,y),

f ( Lreal,l,

we have easily

I component of

~} .

o(K) ~ ~(K) I/2 ~- IIKII2,2 ,

and hence, by Lemma 2.5, (2.20)

~(K) I/2 -_< IIKII2,2 ~_ Const

For a non-negative measure respect to

~ ,

i.e.,

~

on

(f,g)

~(K) I/2 + C 8 0~8(K).

~, we denote by = f~

fg

Lebesgue measure, we omit the suffix.)

d~.

(.,.)

the inner product with

(In the case of the 1-dimension

Here is an inequality necessary for the

proof of Theorem B. Lemma 2.6.

Let

open set in

I

I

be an (open) interval,

and let

K(x,y), T(x,y)

K(x,y) = T(x,y)

for any

supported on

and a non-negative measure

~

l(Ku,v)

~ = Uk= 1 I k (I k = I2,k)

x, y ~ I - ~, x # y.

I =< I(ru,v)

I

+ C8(co8(K) + cos(T))

+

be an

be two 8-standard kernels such that

~

Then, for any with

u, v ~ L 2

d~/dx ~ Lreal,l,

y~ I((K-r)(Xlk u), XlkV) k=l

1

HuPI~2 Ilvll,~2 ,

where

" (2.21) Proof.

Since

llwll,~2 = IIwll2 +

{ E k=l

]IXI~

XI_ ~ (K-T)(XI_~U) = 0,

2 }1/2 wll2 we have

(w = u,v; I k is the double of Ik).

40 I (Ku,v)~I

--< I (Tu,v)~I

+

I ((K-T)(X~u), X~v)~I

+ ] ((K-T)(X~ u), Xl_~V)g[ ( = I (ru,v)~l

+

[ ((K-T) 0(I_~ u), X~v)~[

+ L I + L 2 + L3,

say).

Without loss of generality, we may assume that with

Xk = Xlk

IIii _>- 1121 ->- . . . .

We have,

(k > i),

L 1 =< Z ]((K-T)(XkU), XkV)~l k=l ®

+

Z k=l

(= Let

xk

~

l((K-T)(XkU), Xjv)~ I +

Z

j=k+l

Z ]((K-T)(XkU), XkV)~] k=l

be an endpoint of

*¢ y ( (Ik ) N Ij,

k < j,

Ik

k-I

Z

Z

k=2

j=I

+ LII + LI2,

such that

xk ( I -~

l((K-T)~kU), Xjv)~I

say). (k >-- i).

If

x ( Ik,

then

IK(x,y) - T(x,y) I = IK(x,y) - K(Xk,X j) + T(Xk,X j) - T(x,y) l C8 llklS/[x-ylI+6

* c . If xj ( (Ik)

(Here we assume IK(x,y) - T(x,y) l ~ LII

+

C8 llklS/Ix-yll+8.)

Z

Z

k=2

j=k+l

f

Hence

*c I(K-T) (XkU) (x)v(x) ]dx (Ik ) N Ij

Z fl {fie k= 2 ke C8(¢o8(K) + cos(T))

+

then we have evidently

--< Z f , l(K-T)(XkU)(X)V(x)Idx k=l Ik-I k

=< C8(¢o8(K) + ~6(T))

+

* xj ( Ik,

Z k=2

fI k

[ Z k=l IIk]6 ix_yll+8 {

>~ k=l

lu(Y) l Mv(y)dy}

f * Ik-I k

{flk

dy}

Iv(x)l

lu(y) Idy} Iv(x) l dx ] f , IH(X k u)(x)v(x)I dx Ik-I k ~- C6(es(K) + ¢o8(T))

_-
lluII2 llvIl,~2

dx

41

Using the adjoint kernel of

K(x,y) - T(x,y), we have, in the same manner,

LI2 ~ C8(~8(K) + ~8(T)) If

x E Ik,

*e y ~ (Ik ) n (I - ~),

HulI~2llvll~2 then

IK(x,y) - T(x,y) l = IK(x,y) - K(Xk,Y ) + r(xk,Y) - T(x,y) l

IIkIs/Ix-yI 1+8

c8

Hence

we have, in the same manner as in L2

+

a

Z k=l

Z k=l

LII,

f , [(K-T)(XkU)(X)V(x)Idx Ik-I k f

* c l(K-T)(XkU)(X)V(X) I dx (Ik) n (i-~)

C8( ~8(K) + ~8(T)) lluIl,~2 IIvII,~2 . using the adjoint kernel, we have also n 3 ~ C8(~8(K)+ ~8(T)) lluN,~2 Ilvll*~2 • Q.E.D.

Thus the required inequality holds. The following three lemmas are corollaries of Lemma 2.6. Lemma 2.7. for any

Let

I, ~, K(x,y) and

T(x,y)

be the same as in Lemma 2.6.

Then,

f E Lreal,l, ~(I,K,f) 5 (~(I,T,f) +

Z

~(Ik,K,f)

k=l + C8(~(T;~)I/2 + oos(K) + o)8(T)) Ill. Proof.

using Lemma 2.6 with

u = klf, v = X I K(XIf)/IK(XIf) I

and

d~

= dx,

we have ~(l,K,f) = l(Ku,v) l <-- I(Tu,v) I

+

[((K-T)(XkU), XkV) I

Z

k=l +

C8(~08(K) + ~os(T)) III

_-<~(l,T,f) +

Z k=l

~(l,K,f)

+ C8(~(T;~)I/2+~8(K) +~8(T)) lIl. Lemma 2.8.

Let

anti-symmetric

I, ~

Q.E.D.

be the same as in Lemma 2.6.

8-standard kernels such that

Let

K(x,y), T(x,y)

K(x,y) = T(x,y)

be two

42

(x, y 6 1 - ~, x # y).

Then, for any

(l,K,f) < ~(l,T,f) + Z

f 6 Lreal,l

with

0-<- f < i,

~(Ik,K,f) + C5 AI(K,T;~)III

k='i

where AI(K,T;~) =

{~(T~52)I/2 +~os(K) +~os(T)} {a(K) +a(T) +o~8(K) +~Os(T)}

Proof. Without loss of generality we may assume that supp(f) c I. Lemma 2.6 with u = f, v = XIKf and d ~ = f dx, we have ~(l,K,f) +

< l(rf, XiKf)fdxI

+

C8(~8(K ) + c08(T)) [Ifll,~2

Using

Z ]((K-T)(Xkf), XkKf)fd x ] k=l

11XiKfIl,~2

(= L 1 + L 2 + L3,

say).

We have easily L 3 =< C8( ~6(K) + ~8(T)) IIKII2,2 _-<

t I]

C6(~o6(K) + ~08(T))(a(K ) +¢o6(K))II I .< C8 AI(K,T;~) ] I [ .

(See (2.22)). Since K(x,y), T(x,y) are anti-syn~netric, (Xkf(K-T)~k f))Ik = 0 (k ~ i). Hence we have , with x k' = (the midpoint of Ik), L2 =

<=

Z k=l z

k=l

Iflk f(x)(K-T)(Xkf)(x) K{(X k + X , + X *c )f} (x)dx 1 Ik-I k Ik If

f(x)(K-T)(Xkf)(x) K(Xkf)(x) dxl Ik I ( f. Ik- Ik

+

~o8(K) k=iZ flk l(K-r)(Xkf)(x)

+

k=IZ Iflk f(x)(K-r)(Xkf)(x) { K(X

*e

~ )

f) (x) - K(X

Ik

=<

Z

$(Ik,K,f) +

k=l

+

~6(K)

Z

(~)

}

dx

llXkT(Xkf)II2 llK(Xkf)ll2

k=l

Z

ll(K-r)(Xkf)II2 {fie ( f ,

k=l

+

Ik ef )

dx

--~)2dx}i/2

Ik-I k

C8 ~o6(K) k=iZ flk I(K-T) (Xkf) (x) I dx

--< Z

~(Ik,K,f) + Cs(~(T;a) I/2 + ~8(K)) (]IKII2,2

k=l

--< Z k=l

~(Ik,K,f) + C8 AI(K,T;~) Ill.

+ IPTIF2, 2) I~I

1

43 Using Lemma 2.6 with

u = f, v = EiTf

L I = l(Kf, XiTf)fdxl

and

~ ~(l,T,f) +

d~ = f dx,

we have

Z I((K-T)(Xkf), k=l

XkTf)fd x 1

+ C6(~Os(K) + ~o6(T)) llfll,~2 IIXITflI,~2 (= ~(l,T,f) + Ell + LI2,

say).

We have LI2 -<-C8(~06(K) + ~5(T)) IITII2,2 II! :< C 6 AI(K,T;~) In the same manner as in

Ii~. ~

L 2,

Z Ifik f(x)(K-T)(Xkf)(x ) T{(X k + X , + X ,c)f}(x) dx p k=l Ik-I k Ik

LII =

=<

k=iZ I71k f(x)(K-T)(Xkf)(x) T(Xkf)(x) dx 1 + C 8 AI(K,T;~) (IIKII2,2 + liTIl2,2) ~(T;~) I/2 III + C 8 AI(K,T;~) c 8 AI(K,T;~)

III

III. Q.E.D.

Thus the required inequality holds. Lemma 2.9.

Let

E c [, we put

I, ~, K(x,y)

K E = ~KME, where

define inductively define

T~ k)

Ill

and ME

T(x,y)

be the same as in Lemma 2.8.

is a multiplier: g ÷ XEg.

K~ k) = ~ 4 k - l ) ( k

i; K~ 0)

in the same manner as

K~ k).

@(l'K~2)'f){1 ~ ~(I, TI(2), f) +

For

We

is the identity operator).

Then, for any

We

f ( LTeal,l,

Z $(I k, K (2) f) + C 6 A3(K,T;~), k=l Ik '

where A3(K,T;~ ) = {$0(K) I/2 + $0(T) I/2 + ~(T;9) I/2 + ~6(K) +

ms(T)}

× {o(K) + o(T) + ~8(K) + ~8(T)} 3. Proof.

We divide the proof into several steps.

(First Step). (2.22) For any

We begin by showing that fIXI(J) fll,~2 =< C6(~(x ) + ~8(x))JlI 1 g ( L2

supported on

I,

we have, with

(i -<_j ~ 3, X = K, T). Ik

= (the double of

I k) ,

44

fl~

IXg(x) I2 dx ~ 3 / , !X~ ** g)(x)I 2 dx Ik Ik

+ 3 f , IX~ **c g)(x) - (X~ **c g))Ik I2 dx Ik Ik Ik + 3 Ii~I l(X(Xlk**C g))Ik I2 =< +

C6 ~6(X) 2

3 IlxiI~,2 IiXlk** glI~

 Ik

)ikl2

and

IXkl I(X(X**c g))Ik12 ~

211kl

l(xg)Xkl2+

Ik =< 2 [Ikl l(Xg)ikl2 + C6(~(K ) +~6(K)) 2

211kl [(X(X ** g))ik 12 Ik fIX ** gll~" Ik

Thus % k=l

f , Ik

IXg(x)I 2 dx

=< Const

0~6(K))2

+ CS(~(K) +

=< C6((~(K) + ~o6(K))2

% Ilkl l(Xg)ikl2 k=l

(ilkk ~g]122 +

% k =I

{Hgt122 +

% k=l

IlK ** g]I22 ) Ik

Ilk ** gll~ }, Ik

which shows that HgiI,~2 _-< C6(a(K) + ~6(K)) llgIl**~2 , where put

II'II**~2 g = Xif.

is defined by (2.21) with

If

Ik

replaced by

Then this inequality gives (2.22).

use the above argument. with

I~

To estimate

g = Xi(2)f

j = 2, we put

j = i, we g = Xif

and

IIXifll**~2 , we use again the above argument

replaced by the double of Ik .

j = 3, we put

If

Ik . If

Then we obtain consequently (2.22).

and use the above argument 3 times.

Then we obtain

(2.22). (Second Step). (2.23)

We show t h a t , for any E k=l

I(XkU, Yiv) I ~

u, v E L2

% I(Xku, YkV) l + C 6 { ~6(Y)(~(X) + k=l

+ $0(xll/2(~(Y) + o~6(Y))} Ilull2 where

X k = Xik ' Yk = Yik. We have

supported on I ,

Ilvli,~ 2

(X = K, T;

0~6(X))

Y = K, T),

45

+

Put

Z k=l

(XkU, YIV) I ~

Z k=l

(XkU, Y(X *c v))I Ik


We have, with L2 =

(k ~ i).

(=

(the midpoint of

Z k=l

Illk (XkU(X) - ~k )

l(XkU, Y(X

, Ik-I k

v))l

{~

~I k lu(Y) l2 dy} I/2

(k ~ i)

Ik) {f *c (Y(x,y) - Y(x{,y))v(y)dy} dx Ik

Y(X *c v)(x) dx I ik

--
Z k=l

Z I(XkU, YkV) I + L 1 + L 2, say). k=l

$0(X) I/2

xk

fl k

l +

Then

= I(uX k Xk)ikl ~

+ ~k

+

l(XkU, YkV)

Z k=l

Z (IIXk(XkU) ~vllI + -}~kl k=l

l
fixk MvllI)

[l×kY(× , v)HI) Ik-I k

=< C6 c°6(Y) (IIXli2,2 + ~o(X) I/2) ilull2 Iivll2 + C 6 $0(X) I/2 (NY[]2,2 + c06(Y)) [lull2 J[vH,e2 C6 { ~b(Y)(o(X) + cob(X)) + ~0(X)I/2(o(Y) + cob(Y))} IIull2 !Ivli,~22 and LI

e6(Y)

= Const

Z k=l

dy) dx

Ilk IXkU(X) l ( I , Ik-I k

e6(Y)

Z k=l

I ,

Iv(Y) llHIXkUl(Y)IdY

Ik-I k

C5 ~6(Y)(o(X) + ~b(X)) llull2 Ilv![,~2 • Hence (2.23) holds. (Third Step). (2.24)

We show that . (3):.

l l(Kk f, t
z) i -<-

Z k=l

~ (2)

%(Ik' ~k

' f)

where A3(K) = ($0(K) I/2 + ~6(K))(~(K) + ~6(K)) 3.

+

C 6 A3(K)

Ill,

46

Using (2.23) with

v = .~I(2)f ' X = Y = K

u = Xlf,

and using (2.22)

we have

Z I (Kk f, K~3)f) 1 -< Z" I (Kkf, K k K~2)f) I k=l k=l +

%

($0(K) I/2 +~o6(K))(~(K)

Z I (Kk(2)f, K I(2)f)l k=l

+

+o08(K))

C6 A3(K)

fixif!I2 IIK~2) fll~2 2

III

eo

Using (2.23) with

Z

u = Zk= 1 Kkf,

v = Klf , X = Y = K,

I(Kk(2)f, K~2) f) I <

k=l

Using (2.23)

I(Kk(3)f, K I f)l

+ C8 A3(K)IIl-

k=l

with %

l

we have

u =

Zk= I 42)f,

I (~k " (3)~~' Klf)l

v = Xlf,

-<- %

k=l

X = Y = K,

~(Ik' 4 2)'f) +

we have

C8 A3(K)

III"

k=l

Thus (2.24) holds. (Fourth Step).

j = i,

we use (2.23) with

Z I(ZkTI f,K~2)f) I = k=l +

(i ~ j =< 3, Z = K, T).

Z l(ZkT~J)f, K~3-J)f) I =< C 8 A3(K,T;~)III k=l

(2.25)

If

We show that

u = K~2)f,{

v =

X = Z, Y

Xlf,

% I(ZkK~2)f, Tlf)I k=l

=<

T.

Then

% l(ZkK~ 2)f, Tkf) l k=l

C6{~6(T) (~(Z) + ~6(Z)) + ~0(Z)I/2(~(T) + ~8(T))} ]IK~2)fll 2 I]fll,~2 C6 a(T;e)I/2(~(Z) + ~o6(Z))(a(K) + ~Os(K))2 IIl

If

+

C8 {~os(T)(o(Z) + o~8(Z)) + $o(Z)I/2(o(T) + ~8(T))} (<~(K) + ~o8(K))2 III

<

C8

AB(K, T; e) III.

j = 2, we use (2.23) with Z l(ZkT~2)f, Klf) I k=l =<

u = Kif, =

v = Tif,

X = Z, Y = T.

Z l(ZkKlf, T~2)f)l k=l

Z I(ZkKI f, TkTlf) I + k=l

C 8 AN(K , T; ~) III.

Then

47

Using (2.23) with

u = Ek= I ZkKlf , v = f,

Z I(Zk Klf, TkTlf)I k=l _-<

If

=

Y [(rkZkKlf , Tkf) [ + k=l

j = 3, we use (2.23) 3 times.

(Final Step).

we have

Z I(Tk Z k Elf, rlf) 1 k=l C6 A3(K,T;~ ) III =< C6A3(K,T; ~) [I[.

Then, in the same manner, we obtain (2.25).

We now show the required inequality in our lemma.

generality, we may assume that v = K$3)f," &

X = Y = T,

d~ = dx

supp(f) c I.

Without loss of

Using Lemma 2.6 with

u = f,

and using (2.22), (2.24), we have

(2.26)

+

Z IC C K k - Tk)f, K~3)f)[ k=l I(rl f, K~3)f) I +

% I(Kkf,K~B)f)I k=l

=< l(rlf ' K~3)f) I +

Using Lemma 2.6 with

+

Z

k=l

+ C8(~6(K) + ~6(T)) NfH,~ 2 l]K~3)fli,~2

~ ~~(Ik, ~ (2) ,f) + k=l

u =Tlf , v = K~2)f

Ill

C 5 A3(K,T;~)

[I[.

and using (2.22), (2.25), we have

I((Kk - Tk)Tlf , K~2)f)l

~l(T~2)f, K~2) f) I +

+ C 6 A3(K,T;~)

+

C 8 A3(K,T;~)

[I I

C 5 A3(K,T;~)l!I.

Repeating this argument 2 times~ we obtain

(2.27) i(T2)f, K 2)f)i

<

2), f) ÷ c6 A3(K,T )

Thus (2.26) and (2.27) give the required inequality in our lermna. §2.5.

Proof of Theorem A by perturbation ([45]) In this section, we deduce Theorem A from the boundedness of the Hilbert

transform and Lemma 2.9. Fixing

0 < s < 1/2,

(We do not use Cotlar's lemma nor the Fourier transform.)

we define

S[a](x,y) =

ks(x-y ) T[a](x,y),

48

where

(IsIl~)

( ~ <

- i

g (s) = (i - s l s l )

We shall show that gives Theorem A. Since

Isl --< 2~)

(2S <

!s I --< ll(2s))

(1/(2s)

<

IIS[a]ll2,2 $ Const flail, .

Isl

<_- l/s).

Once this is known, Fatou's lemma

Without loss of generality, we may assume that

~l(S[a]) ~ Const , Lemma 2.5 shows that

a ( Lreal,l.

Iis[a]II2,2 ~ Const{o(S[a]) + I}

Hence it is sufficient to show that (2.28)

~S = sup {o(S[a]);

a ( Lreal,l } <

Const .

Let -

OS(2) where

S[a]~ 2)

=

i

For a, f ( Lreal,l

(2.29) Considering

i~ I -a

(-1/3-&.,l-a.)

~(I, S[a]i

,

f);

is defined in the same manner as

Shwartz's inequality shows that is finite.

(2)

sup { ~["

K~ 2)

in Lemma 2.9.

~

<~ From the definition of = S(2)" and an interval I, we show that

~(I, S[a]~ 2), f)

shows that there exists

Then S[.],

@ (2) S

2 )4 3 ~ 3 =< {( ~ + 7 } ~S(2) + Const {~S + i}.

if necessary, we may assume that

= {x ( I; A(x) # B(x)}

a, f ( Lreal,l , I interval},

(a) I $ O.

b E Lreal,l

RSL

of Type 1

such that, with

(B(x) = A(x0) + f~O b(s)ds,

x0

is the left endpoint of

I),

(2.30)

-1/3 ~ b(x) ~ 1

(2.31)

I~I ~

(The function

b

l-(b)I 1+(1/3)

a.e.

Ill

(x ( I - ~ ) ,

Lemma 2.9 with

I,

l-(a)I i+(i-7~

b(x) = -1/3

we have

I,

we may put

I.

6 = i,

~(I, S[a]~2),f)=< $(I, S[b]~ 2), f)

(x (IC).)

(x, y E I - ~ ) .

we have

+

Z

~(I k, S[a]~2k), f)

k=l

+ Const

~,

Since (2.30), (2.31) are

b(x) = 1

S[a](x,y) = S[b](x,y)

K = S[a], T = S[b],

on

3 Ill ~ 7 III"

obtained from RSL is defined only on

independent of the behavior outside A(x) = B(x)

~

on

A3(S[a] , S[b]; ~) [I I,

Since Using

49

where

~ = Uk= I Ik

dominated by

The second quantity in the above inequality is

(Ik = l~,k)-

= b - (1/3).

(3/4) ~S(2) Ill . Put

S[D] = S[b] + (~/3)H,

~(I, =~

Then

II~II~ ~ 2/3

and

and hence

S[b]~2),f)

=<- ~(I,

S["~]~ 2) , f) + Const

IIsEE]ll~, 2 III

3 (2)4~s(2) i! I +Const {%+1} IIl.

Thus $(I, S[a]~ 2)

,

2 )4 3 {( ~ + ~}

f) ~

3 {oS + i}III

+ Const

$

+ Const

S(2)

A3(S[a], S[b]; ~) IIl.

A3(S[a], S[b]; ~). x E J,

It is necessary to estimate for any interval

J

and

k (x-y) ~x-y

]SIal Xj(x) I -<- Ifj

[I[

Integration by parts shows that,

a(y)dy I + Const.

Lemma 2.5 shows that fj

k (x-y) g x-y

ifj

a(y)dy I dx ~ Const NH*(Xja)II2 IJl I12

Const (~(H) + 1) IJl E Const IJl. Hence we have

(i/IJ I) ~(Ji S[a], Xj) ~ Const.

~0(S[a]) & Const . b(x) = -1/3

on

In the same manner, ~,

Taking the supremum over all

~0(S[b]) ~ Const.

Since

we have, for g ~ Lreal,l ,

~(I k, S[b], g) = Const $(I k, H, g) & Const Ilkl and hence

J,

~(S[b]; ~) ~ Const.

Consequently,

AB(S[a], S[b]; ~) ~ Const (~S + i)

(k a i),

3

Const {~ (S[a]) + ~(S[D]) + 1} 3

,

which gives (2.29). Taking the supremum of and all intervals

(i/IIl) 5(I, S[a]~ 2), f)

over all

I, we obtain, by (2.29),

2 )4 ¼ ~ 8S(2) ~ {( ~ + } ~S(2)

+ Const

3 {oS + i}.

a, f 6 Lreal,l

50

Since

(2/3) 4 + (3/4) < I

and

~ ~ ~ ~S(2) ,

this inequality yields (2.28).

This completes the proof of Theorem A.

§2.6.

Proof of Theorem B by perturbation ([17]) In this section, we deduce Theorem B from the boundedness of

H.

Here are

two lemmas necessary for the proof. Lemma 2.10 (Calder~n [4]). Proof.

Fixing

IITn[a]ll2,2 $ (Const) n llalI~

0 < s < 1/2,

we put

Sn[a](x,y ) = Since

el(Sn[a]) ~ Const n

(n ~ i).

X g (x-y) Tn[a](x,y)

JPalI~,it

is sufficient to show that n+l

(2.32)

~S

=

sup {~(Sn[a]);

a ( ereal,l } ~ C O

n for some absolute constant So[-] = (-~)H. (0 ~ k ~ n-l).

CO

Then We d e f i n e

(which will be determined later). Let

~So =

~S(2) n

T = Sn[b] , 6 = i,

(2.33)

k+l ffSk ~ C O

Suppose that

in the same manner as

4 Let a, b, f, I, ~ <~Sn -<-~S (2) " n K = Sn[a],

~(-[ H) ~ ~.

~S(2)"

be the same as in §2.5.

Using Lemma 2.9 with

we have

~(I, Sn[a]~ 2), f ) =< ~(I, Sn[b]~ 2), f ) +

Z

~(I k, Sn[a]~2), f)

k=l + Const Put

b

= (3/2)(b-

A3(Sn[a] , Sn[b];~ ) l l I .

(1/3)).

Then

2 b* i Sn[b ] = Sn[ 3 + ~] =

I]b I[~ =< 1 n

and

(2)k

i n-k

Sk[b* I

k=0

By Lemma 2.5 and the assumption of our induction, we have IISk[d]II2,2 ~ Const Const (C~ + n)

Then

{~Sk + ~l(Sk[d])} (d ( e~eal,l,

0 ~ k ~ n-l).

Hence, in the same manner as in §2.5, we obtain, by (2.33),

51

~(I, Sn[a] 2 )

f)

~ { (~)4n + 4 } S (2) n

+ Const (C~ + n ) ~ S

+ n) 3 + Const (C~ + n) 4. n

Since

4 ~S

and

I

are arbitrary,

@S(2)

.

Since

a, f E Lreal,l

quantity replaced by

this inequality holds with the first

(2/3) 4n + (3/4) ~ 0.99

and

n ~ ~ n

=< C 0' (C

we have S (2) '

+ n)

for some absolute constant

C O' •

Let

Sn

n

C O = max {2 C~, ,} •

Then

~S

~ Gn _n+l •

This shows that (2.32) holds for all

n

n ~ 0.

This c o m p l e t e s t h e p r o o f o f Lemma 2 . 1 0 .

Remark 2.11.

Q.E.D.

It is known that IITn[a]II~,BM 0 ~ Const {IiTn[a]II2, 2 + ~l(Tn[a])} ¢onst

{I!Tn[a]!12,2 + (n+l)}

(n ~ O,

a E Lreal,l),

where

is the norm of Tn[a ] from L ~ to IITn[a]II~,BMO The proof of Lemma 2.10 by Theorem 1.12 is as follows. Let

and

n ~ i.

Integration by parts shows that

BMO.

(See Lemma 2.5.)

a E Lreal,l

Tn[a]l = Tn_l[a]a.

Hence

Theorem I shows that IITn[a]H2, 2 ~ Const {IITn[a]IIIBM0 + n} = Const {liTn_l[a]aIIBM O + n} ~

Const {IITn_I[a]II~,BM O + n}

Const {Hrn_l[a]II2,2 + n} , which gives Lemma 2.10. For

a E Lreal,

(2.34) where

A

we define a kernel

E[a](x,y) =

is a primitive of

Coifman-McIntosh-Meyer Lemma 2.12.

[7].

a.

1 x-y

exp {i A(x)-A(y)_} x-y

The following lemma was first shown by

A proof by perturbation was given by David

There exists an absolute constant NO IIE[a]II2,2 ~ Const(l + llall=)

Proof.

Since

,

NO

such that

(a E L~eal).

~l(E[a]) ~ Const (i + llall~), it is sufficient NO ~(E[a~) ~ Const (i + llall )

[17].

to show that

52

for some absolute constant (2.35)

N O ~ i.

We put

( ~ > 0)

aE(~) = sup {o(E[a]); a ( ereal,6}

and show that NO

(2.36)

~E(~) -<_ Const(l + 6)

Lemma 2.10 shows that, for any

( ~ > 0).

a ( Lreal, in

~(E[a]) = -<- Z

~(

Z n=O

nl

=

Tn[a])

i

=< Z n=O

~-f ~(Tn[a])

i n! (C°nst)n 11alln <= exp {Const(l + Iiaii®)} •

n=0 Hence > i,

OE(~) < ~

for all

a ( L~eal,~

,

~ > 0

f ( Lreal,l

Type i (-~/3- A.,B- a.)

and

~E(1) ~ Const.

and

I

Let

be an interval.

shows that there exists

b ( Lreal

RSL of such that, with

= {x ( I; A(x) # B(x)}, - 613 ~ b(x)

6 -

Using Lemma 2.7 with

~

6

a.e.

(b) I

on

I,

b(x)

= - ~13

on

~

,

3

we have

K = E[a], T = E[b], 5 = i,

~(I, E[a], f) _~ a(l, E[b], f) +

Z k=l

O(Ik, E[a], f)

+ Const {$(E[b];e) I/2 + ~01(E[a]) + ~Ol(E[b])} Ill, where

I k = I~, k

~(I, E[b],f) = Since

(k >_- i).

~(I, E[~], f).

b(x) = -~/3

on

Put

~ = b - (6/3). We have

Then

II{II~ =< 2~/3

~, we have

D(E[b]; ~) =

~2~(H;~) =< Const.

Thus

~(I, E[a], f) ~ OE (

) +

~E(~) + Const 6 ,

which yields that ~E(6) _-< OE( that is,

) +~

and

001(E[a]) + O~l(E[b]) =< Const ~ .

aE(~) + Const ~ ,

53

(2.37) Let

n

OE(~) < 4 ~E(23 ~)

+ Const ~ .

be the minimum of integers

n =< (log ~)/(log 3/2) + Const.

k->- 1

such that

Inequality

OE(~) =< 4 n ~ E ( ( 2 ) ~ )

(2/3)k~ ~ I.

Then

(2.37) shows that n-i % k=0

+ Const

4 k (2)k3

-_< 4 n {~E(1) + Const (2)n ~} _-< Const where

4n_-< Const (i + ~) N0 ,

N O = (log 4)/(log 3/2).

Thus (2.36) holds.

This completes the proof of

Lemma 2.12.

Q.E.D.

We now give the proof of Theorem B. (x ~ ~),

Since

i/(I + ix) = f0

e-iXs e-S ds

we have C[a] = f~

E[-sa]e -s ds.

By Lemma 2.12, we have IIC[a]I12,2 =< f0 IIE[-sa]ll2,2 e-s ds NO Const f0 (I + sllall )

NO e -s ds _-< Const (i + IIall~)

This completes the proof of Theorem B. §2.7.

Estimates of norms of

E[.]

and

C[.]

In this section, we show Theorem C ([44]).

For any real-valued

(2.38)

NE[a]II2,2 ~ Const (i + IIalIBMO),

(2.39)

IIC[a]I12,2~ Const (i +~IIalIBMO).

In [7], (2.38) was given with was given with

__~MO

replaced by

[44] via [42], [43], [50]. use RSL repeatedly.

a(x), I, ~, ~

and

we define analogously aretan

y.

Then

IIalIBMO

in

BM0,

9 IIalIBMO,

and (2.39)

This theorem was established in

In this note, we deduce Theorem C from (2.36).

The lower bound of A(x) B(x)

a

replaced by

8 IIaIIBMO.

(The sun also rises!)

Type i (y - r.,$ - a.) Let

function

We

RSL in §2.2 is called of a(x)

is independent of (2.5)-(2.8).

be the same as in RSL in §2.2.

For

0 ~ T ~ ~ ,

by using the sun at the right upper infinity of angle

54

~ b(x) ~ 7 (a)ik ~ T

a.e. on (k g i),

I, I~I ~

b(x) = 7

(x ( I k, k ~ i),

(-~) * (b)l

III.

(-~) + 7 This RSL is called of Type 2 ( y - h a y , ~ - d g 6 e e ~ ) . independent

of the above estimates.

For

0 ~ 7 ~ ~ ,

where the supremum is taken over all functions a.e. on

!.

We define

b, ~

--<-b(x) --< 7

In this case

@

in the same manner.

a.e. on

I,

~

we define

such that

is B(x) = sup @(x),

• ~ A,

~' ~ T

Then

b(x) = 7

(x ( I k, k >_- i),

(-~) + (b) (a)i k > y

(k ~ i),

I~I =<

This RSL is called of Type 3 ( y - r . , of the above estimates.

For

~-d.)

~ ~ 7 ~ 0,

(-~) + 7

.

I Ill.

In this case

we define

B(x),

as in Type 3, by using the sun at the lower right infinity

~

is independent

in the same manner

of angle

- arctan

171'

Then 7 ~ b(x) ~ ~

(a)ik ~ T

a.e. on

(k ~ i),

This RSL is called of Type 4

(Y-~.,B

I,

I~I ~

- a.).

b(x) = 7 - (b)l ~ _ 7

(x ( I k, k ~ i),

llI.

In this case,

~

is independent

of the above estimates.

Type 1

Type 2

Type 3

Type 4

RSL of Type j is reduced to RSL of Type i by a suitable affine transformation.

55 §2.8.

Proof of (2.38)

(First Step). (2.35).

We divide the proof into three steps.

~E(~) = sup {~(E[a]);a E Lreal,~} (See (2.16).)

~E(~)

defined by

e

0 ~ f ~ i

(~ > 0).

In this step, we show that, for

(2.40) where

Recall

Put

~E(~) ~ ~E(@~) + ~

(0 < e < i) and

~(I, E[a], f).

~E~T ^ .l+e ~) + (Cs/e)

is determined

an interval

Since

I,

~ ~ i,

we

later. study

For

0 < 5 ~ i,

~5{~E(~)

+

a E Lreal,~,

an a-priori

~(l,E[a],f) = ~(l,E[-a],f),

estimate

~5}

f E Lreal, with

respect

(a)ik ~ Using Lemma 2.8 with

b(x)

=< ~

e~

a.e.

on

(k => i),

I,

I~I

K = E[a], T = E[b],

b(x)

Since

~o6(E[a]) _< C5 ~5,

III

(x E I k,

k >

b ~ L ~real

i),

( ~ = (b)l)-

we have

~(I, E[a], f) -<_ ~(I, m[b], f) + +

= - e~

=< ~ +- ~

to

we may assume that

> 0. RSL of Type i (- e~- ~., ~ - ~.) shows that there exists (a) I = such that, with ~ = {x E I; A(x) # B(x)} = Ok= I Ik (Ik = l~,k). - 6~ ~

,

% k=l

~(Ik, E[a], f)

C 6 AI(E[a] , E[b]; ~) IIl. 0~6(E[b] ) < C5 ~6

$(E[b]; ~)i/2

and

= Const $(H;~) I/2 < Const =

we have AI(E[a ],E[b];~) = {&(E[b];~)I/2 + ~5(E[a]) + ~5(E[b]) } × {~(E[a]) +

~(E[b]) +

~5(E[a]) + ~5(E[b])}

_<_ C5 F5 {~E(F) + ~5 } , and hence ~(I,E[a],f) < ~(I,E[b],f) +

% $(ik,E[a ],f) + C5 ~5{~E(~) + ~5} k=l

ii1.

56

For each

Ik,

there exist

we use~ RSL of Type 2 (e~ -&.,-6-d.).~ bk ~ L r e a l

and an open s e t

-~ & bk(X) ~ e~

(a) I

k,g

a.e.

¢ e~

( ~

on

Let

bk = bk -

-(-~) + e~

(i -e)~/2).

~(Ik,E[%k],f) = $(Ik,E[bk],f).

g(ik,E[a],f) < + % +

Then

(a)~

~k = U g = l l k , g

Ik,

- e~

<

in

Ik

such t h a t

(x E ~k ),

bk(X) = 8~

I),

-(-6 )+(bk) Ik =<

Since

I Ikl <

~+ (a)ik ~+ e~

i- e

IXkl --< "i + e IXk I"

ll%k[l= =< (l +e)~/2

and

Hence, by Lermma 2.8, we have

~(Ik,E[bk],f) +

Y $(Ik, g, E[a], f) g=l l+e AI(E[a],E[bk]; g~k) llkl =< ~E ( T ~ ) Ilkl

ixkl '

~ g(Ik,g, E[a] ,f) + C5 ~5 {OE(~) + ~5} g=l

which yields that g(l,E[al,f) N $(l,E[b],f) + ~E ( Tl+O +

For each

~)

z l~kt k=$

k=l% g=i% o(Ik' g, E[a],f) + C5 ~5 {~E(~) + ~6}{ii I + k=lY llkl}.

Ik,~, we use RSL of Type i (-e~-&.,~-4.) ~k,~ = U~=I Ik,g,m

we obtain an open set (a)

Ik,g, m

~

- e~,

Z m= 1

llk,~,ml

Since

(a)

Ik,g such that

in

Ik,~

< =

1 - e I + e

llk,g

>--eF,

I "

In the same manner as above, i+@ ~(l,E[a],f) ~ ~(l,E[b],f) + ~E ( T ~)

{ £ llkl + E Z llk,~l} k=l k=l ~=i

^

% % k=Ig=l

% ~(Ik,g,m, E[a], f) m=l

+ C5 ~5{aE(~) + ~5} {lll+

% llkl+ k=l

% k=l

Z Ilk,~I} • g=l

57 ^

Since

^ in

-<- - 8 ~ ,

(a) I

c~(Ik,g,m, E[a], f)

is estimated in the same manner as

k,& ,m

~(Ik,E[a],f).

Repeating this argument, .i+@ l-e i-8 2 E ( ~ - ~) I~I {1 + i ~ ÷ (i~) +

~(l,E[a] ,f) =< ~(l,E[b] ,f) + + C 8 p8 {~E(~) + ~ 8 } {JZj

+ I~I ( I + ~

^ l+e~. ~ - n ~(l,E[b],f) + OE(2-~-p) ~ + 8~ ^ ^ ,l+e = o(l,E[b] ,f) + OE<--~- ~)

To estimate

l-e

I-8 2 + (i-~-) + ...)}

l+e JiJ + (C5/8)~5{aE(~) + ~6} 28

~ - n 2e~

c ( Lr eal

~' = Oj=l ~ Ij'

and an open set

-8~ -_< C(X) K e8

a.e.

-(-e~) + (c) x

on

in

I

to b(x)

and

I.

such that

I ,

P~ + n

Jxj _-<

]e'I-<- ./-'(-e~) + e~

Jxl

-I~I + (%/e) ~8{C~E(~) + ~8} I~I.

~(l,E[b],f), we use RSL of Type 3 (85-r.,-e~-d.)

There exist

"}

jlJ.

2ep

In the same manner as above, ~(l,E[b],f) =< ~(l,E[c],f) + ^ ,l+e

Z j=l

e~ +

~E (e~)Ill + ~E<--2- ~)

~(l],E[b],f) + C8 ~8 {OE(~) + ~8} iiI

q,

2e~

Consequently, 7 Ui

$(l,E[a],f) <= {~E(e~) + SE (--2- ~)

+ { ~~E <),i+8 T^

~28~-~

1+8 SE(e~) + - ~ -

2e~

+ C8 ~8(~E(~) + ~6)}

+ (C8/8)~5(~E(~) + ~5)} ^ .l+e O E i _ T ~ ) + (C8/8) ~6{~E(~) + ~8},

which gives (2.40). (Second Step). (2.41)

In this step, from (2.40), we deduce ~E(~) ~ Const

~

( ~ g i).

Let 1+8

hX(e) = e) ~ + - ~ -

l+e)k

(----f--

(o< e < 1,

~.> o).

58 Then

~(e) > i

for any

0<

e
k > 2

i = h2(i/3 ) = Hence we choose

e = 1/3

an interval

we have

I,

in (2.40).

min h2(0 ). 0<0<1

For any

f + 2~ ~(l,E[a],f) _-< 3 ~(l,E[a], - - 3

^ f + 2~ Const {q(l,E[a], ~

qE(~) ~ Const ~E(~) I12.

(2.42)

)

Ceal,~'

a 6

f E Lreal,l

and

2 ) + 3 ~(l,E[a], ~ X I )

1/2

^

2 ~i)i/2}

1/2

III

+ a(l, E[a], ~

DE(7)I/21II,

Const

and hence

and

SE(~) =< ~E (

Inequality (2.40), 0 = 113

+ 2 DE(

) + C5 ~8

shows that

+ ~5}

( ~ ->_ i).

By (2,36), (2.43) where

( ~ ~ i),

~E(~) ~ Const {~E(~) + ~}2 ~_ Const ~N

N = 2N 0 + 2.

Suppose that

N ->_ 3.

We put

N+I ~m = sup{ ~E(~) ~ Then

-c3 -<_ Const.

For any

; i ~ ~ =< (3)m }

2

m >_- 4

and

(3/2) m-I

(m = 3,4 .... ).

< ~ < (3/2) m, we have,

by (2.42), N+I ~E(~ )

~

2

N+I

_<_ ~

2

N+I

N+I

2

2

{~E(~3) + 2 ~E(23 -~)

+

C8( ~

N+I +

2

+

~28)}

N

~:m-1 + c8 (~ 2 +8

+

3

N+I _<,{( ) 2

N+I

~m-i + C5 ~ Hence

~m ~ ~m-I + C 8

N+I

-(li2)

+ 2( ) 2 } .~m_i +

C6(~

p28) }

)}

+

8-(1/2) 2 (m-i)((i/2)-8) ~ Tm-i + C8(3)

(2/3) (m-I)((1/2)-8)

~m <= ~3 + Const

m-i Z k=3

We choose (_~)k/4°3<= Const,

8

1/4.

Then

59

which gives that

Const

~E(~) i.e., (2.43) holds with

N

~

N+I 2

( ~_>-i),

replaced by

(N+I)/2.

Repeating this argument, we

obtain

~E(~) =<

Const 9 3

( ~ ~ i).

Put ~'m = sup {~ E(~) ~-2; 1 =< 9 =< (~)m } Then

' < Const. Since ~3 =

(1/3) 2 + 2(2/3) 2

=

i,

(m = 3,4 .... )

we have, in the same manner as

above, ~'m -<-~tm_l + C6

T'm_l + C6

We put

6 = 1/4.

Then

{(2)(m-I)((I/2)-6)

(Q)(m-l)((I/2)-6)9~

~t ~ Const, m

~E(~) ~ Const Since

~E(~) ~ Const ~E(9) I/2,

(Final Step).

+ (3)(m-I)(2-6)}

(m >= 4).

which yields that

92

( ~ a i)°

this shows (2.41).

At last, we deduce (2.38) from (2.41).

Let

a E Lreal.

We show

that (2.44)

~5(E[a]) ~ C 6 (i + ~25)

1/Ix-y]

We have evidently

IE[a](x,y) I ~

Ix-x'l ~

We may assume that

Ix-yl/2

( ~ = IlalJBM0, 0 < 5 ~ 1/2) Let

x, x', y ~ IR satisfy

y < x < x'.

fyX la(s ) _ (a)(x,x,) ] ds =< Const

We have

~(x-y) log x'-y

which gives that

I A(x)-A(Y)x-y yA(y) - A(x')x'i X'-y

f:' (a(s)

-

I =

I (~_y~(~,_y) X' - X

(a)(x,xt))ds ]

/ yX

(a(s) -(a)(x,x,))ds

80

Xt--X

Const

{i + log ~

x-y

}

xt_X

~

Const

~(x'-x)i/2/(x-y) I/2.

Thus IE[a](x,y) - E[a](x',y) l =<

1 x'-y

+

x'

-

I exp { i

x

A(x),A(y) x-y 1

x' - x (x-y)(x'-y)

i + C6 ~

x' - x

Const

I

+ C6 ~26

(x_y)2

For

f E

a = a - (a) I. an open set

Lreal,l Then

A(x)-A(y) x-y

(Ik = IE, k)

in

I

2-~~i I~(s) Ids ~ III/2' i

I a(x) l =< 2~

a.e.

on

_

}I

{i A(X'x!-A(J~)}I25

A(x')-A(y) x'-y

I 25

C5(I + ~28)(x'-x)6/(x-y) I+6 .

I, we estimate

o(l,E[a],f) = ~(l,E[a],f).

=<

{i A(x')-A(y) xV-y

- exp

(x' - x) 6 (x_y)l + 6

and an interval

~ = Uk= I Ik

I~I

}

I exp {i A(x)-A(Y-~)} e X P x - -y

(x-y)(x'-y) + C6 < =

X ! -- X

"(x-y)(x''y~

o(l,E[a],f).

Let

Le~=na 2.2 shows that there exists

such that

(IaI )i k ~ 2~

(k ~ i),

I - ~ .

We put

b(x) =

Then

b E

e =real,2~

~)

~(x)

(x ~ I-

~(a0)ik

(x l )~ ~). Ik' ( xk >

Using Lemma 2.7 with

K = E[a], T = E[b],

we have

~(l,E[a],f) = o(l,E[a],f) <= o(l,E[b],f) +

~-

For each have

Ik,

~E(2~)

III +

Z ~(Ik,E[a],f) + C 6 (i + ~25) ii I k=l k=Z I ~(Ik,E[a],f)

we use Lemma 2.2 with

a - (a)ik.

+

C8

(I + ~26) iii.

Repeating this discussion, we

81

i 1 ]I I {i + ~ + ~ +

~(I, E[a], f) =< qE(2~)

i {i + ~ + ~

+ C8 (i + ~ 26) III

i

... }

+ ... } =<_ {2 ~E(2~) + C8 (i + ~ 28) } IiI

Thus o(E[a]) ~ 2 ~E(2~) + C8 (i + ~28) Const (i+ ~) = Const(l + IIaIIBMO). Consequently, Lem~na 2.5 and (2.44) yield (2.38) in the ease where a

be a real-valued function in

an(X) =

(x) -

Since

a n ~ Lreal ,

BMO.

a E Lreal.

Let

We put

iax 1 Ia(x) I =< n

a(x)

<

(n->_ I).

-n

we have

IiE[an]II2,2 ~ Const(l + IlanllBMO) ~ Const(l + IIalIBMO). Letting §2.9.

n

tend to infinity,

we obtain (2.38).

Proof of (2.39) Put $C(~,~) = sup {~(C[a]); a ~ Lreal, ~ = a -<_ ~}

We show that, for

~ ~ I,

(e _-< 0,

~ _-> 0).

0 < 8 ~ i,

^

(2.45)

~C(-~,~) ~ 2 ~C (- ~ , ~) + C 8

~8{~C(~) + ~8}

where ~C(~) = sup {o(C[a]); a E LTeal,~} • (Tchamitehian

[51] showed that

¢~C(-~,~) -< 2 OC (- ~2' ~2) +

C6 B6

Inequality (2.45) is an improvement of his inequality.) for the proof of (2.45).

(~ ->_ i). Here is a lemma necessary

62 Lemma 2.13. Proof.

<~c(O,~)= ~C(--~,O) _-<- C5 ~5 {Crc(~ ) + ~5}

The first equality is evident. We define

6 = log 2/log(i/4q). For interval I, we estimate Type 3 (4qB-/t.,0-d.).

(# >_-- i , 0 < 6 -< i ) . q > 0

by

a E Lreal, 0 _-< a--< [3, f ~ eal,l' 0 -< f =< 1 ~(l,C[a],f). If (a)l < 2q~ , we use RSL of

There exists

b 6 L~eal

and an

such that, with

= {x 6 I; A(x) ~ B(x)} , 0 =< b(x) =< 4Q#

on

a.e.

I,

I~1 <

=

0+2r]~3 { i I =< i i i / 2 .

o+~{3

Lemma 2.8 shows that ^

~(l,C[a],f) _-< $(l,C[b],f)

+

Z k=l

~(Io, k,~

C[a], f)

+ c~ ~8 {~c(~) + ~} IT I _< {C-c(O, 4q#) +y$ C.c(0,{3 ) + c6~(ac(#)

+ {3~)} Izl

If (a)l -> 2q~ , we use RSL of Type i (qB-A.,{3-a.). such that~ with = {x E I; A(x) # B(x)} , ~

=< b(x) =< ~

a.e.

There exists

- 2~

on

tI]

=

1-

b ~ Lreal

~

tit.

Lemma 2.8 shows that ee

~(l,C[a],f)_~$(l,C[b],f)

+

Z ~(I~, k, C[a],f) k=l

+c6 ~6{~c(~) +~8} II -

l_~q

We have ~(l,C[b],f) ~71 If I

= fB(1)

IfB(1)

(Const/~)

i (x-y)+ i(B(xiiBiy))

I (B-l(s)-B-l(t)) ~(B(1),C[(B-I)'],

f(y)dyl 2 dx

+ i(s-t)

f=B-l(t) dtl2 B'°B-I(t)

ds B'°B-I(s)

foB-i/B'oB -I)

NO (Const/~) Const {i +If(B-I) ' XB(1)II ~ } II(foB-I/B'°B-I)XB(1)IIIIB(1)I C6 ~-31B(1) I ~ C6 ~-2 i1 I ~ C6111 ,

63 where

B(1) = {B(x); x ~ I}

and

NO

is the absolute constant in Lemma 2.12.

Hence, in this case,

~(i, C[a],f) =< l-zq l-'q

~c(O,~)lll

+ c5 p~{~c(~) + ~ } I I l

•

Thus we have, in both cases, i

~(I, C[a]

f)

--< maX{~c(0'4~) + 215 C (0'~)' z_n_2_2n i-~ ~c(O,~)}

+ c5p6{~c(~) + ~5} ,

which yields that

$C (0'13) --< max{~;c(O'4"q~3) + "2i ~c(O,~), ~l-q + c~{~c(~) If the second quantity in

~c(O,~)}

+ ~}.

max {.,.}

is larger than the first quantity, then

~c(O,~ ) ~ z_,_-_-n c8 ~{~c(~)

+ ~6} = ca ~5{~c(~) + ~5} .

If the first quantity is larger than the second quantity, then (2.46)

$C(0,~) ~ 2 ~C(0,4~) + C6~6{~C(~) + ~6} .

Thus, in both cases, (2.46) holds. k

such that

(4~])k ~ =< i.

Then

Let

m

be the smallest integer of

m _-< {log ~/log(I/4~)}

+ i.

Inequality (2.46)

yields that ~C(0,~) --< 2m ~c(O,(4~)m~) + C 6

m-i Z 2k(4~])kS~8{~C((4~)k~) + (4~)k8 ~6} k=0

2m $C(0,i) + C5 {I + 2m(4~) m6} ~6 {aC(~) +

~5}

C5(~8 + 2TM) {OC(~) + ~6} ~ C6 ~6 {OC(~) + ~6 } . We now prove (2.45). interval

I,

we estimate

For

a ~ e~eal,~, f E e~eal,

$(l,C[a],f).

Since

0 ~ f ~ i

(a) I ~ 0.

RSL of Type i (- e~-~.,6- ~.)

that there exists

b E L~eal

such that, with

(Ik = l~,k),

and an

$(l,C[a],f) = $(l,C[-a],f), we

may assume that

= U~= 1 Ik

Q.E.D.

(e = 1/2)

~ = {x E I; A(x) # B(x)}

shows

64 -

~

~

b(x)

~

(a)ik ~ - 8~

~

a.e.

on

I,

b(x)

=

-

85

- (b) I i~ I =< ~ ~ @~ ii I ~-

(k => i) '

(x

(~),

(e = I/2).

II~I, l+e

Lemma 2.8 shows that g(l,C[a],f) ~ ~(l,C[b],f) + gC(-6~,~)Ill +

Z $(Ik,C[a],f) + C5~8{~C(B) + ~5} ii I k=l

Z g(Ik,C[a],f) + C5~5{ffC(~) + ~8}III . k=l

For each k ~ I, we use RSL of Type 3 (0-&.,-~- d.). Since exist b k 6 Lreal such that

and an open set

- ~ ~ bk(X) ~ 0

a.e.

(a)l ~ - @~,

~

there

k

~k = U~=I Ik,~ (Ik,g = l~k~~) in Ik

on

Ik,

bk(X) = 0

(x ( ~k ),

+ (bk)ik (a)ik, ~ 0 (~ ~ I), B + 0

Lemmas 2.8 and 2.13 show that ~(Ik,C[a],f) ~ ~(Ik,C[bk],f) + --<~C(-~,0) Ilkl +

Z $(Ik,g,C[a],f) + C8~5{~C(~) + ~5} llkl g=l

Z ~(Ik,g,C[a],f) + C5~5{~C(~) + ~5} llkl ~=i

--< Z $(ik,e,C[a],f) + C5 ~5 {OC(~) + ~5} iikl ' g=l Thus ~(l,C[a],f) ~ $C(-8~,~)II] + Z Z $(Ik,g,C[a],f) k=l ~=I + (a)ik,~ => 0 Since

(a) I ~ k,g

csFS{Oc(F) + ~8}II j, (k,~ >_-i), k=1% ~=IZ Ilk,~l =<

ii +- 8e iii.

0, we can apply the above argument to Ik, 3 ° Repeating this,

we have 1

Tff

$(l,C[a],f) ~ {$C(-@~,~) + c8~5(~c(~) + ~6)}{I + il-e ~+

~!~

~c(_ep,F ) + c5~5 {~c(~) + ~5},

~.l-e.2 # + ...}

65

which gives

(2.47)

~c(_~,~) =< z+i2e ~c (-e~'~) + c~ ~

To estimate -

8~a

~ ~,

(a) I ~ 0.

~C(-8~,~), we study

f £ L~eal,

0 & f & i

~(l,C[a],f)

such that, with

for

and an interval

RSL of Type 3 (8~-r.,-@$-a.) ~ = {x 6 I; A(x) # B(x)}

- e~ = b(x) =< e~

I.

a ~ Lreal, First we assume that

shows that there exists = Uk= I Ik

a.e. on

I,

a.

(k . > .1)

e~ .

b(x) = e~

I~I < e~ + g

b 6 Lreal

(Ik = l~,k),

e~ + (b) I (a)l k

(e = ~/2).

{Oc (~) + ~ }

(x ~ ~), e~_

I~I ~ e~+e~

½ I~I =

I~I'

Lemma 2.8 shows that E $(ik,C[a],f) + C8~8{~C(~) + ~8} k=l

$(I,C[a],f) ~ $(l,C[b],f) +

~c(-el3,el3) I~1 + For each

k ~ i,

there exist such that

~(Ik,C[a],f) + C858{~C(~) + ~8}II I. k=l

we use RSL of Type I (O-r.,~-a.).

bk 6 Lreal

0 ~ bk(X) & ~

ii I

and an open s e t

a.e. on

Ik,

Since

~k = Ug=l Ik,~

bk(X) = 0

(ak)ik a e~,

(Ik,g = I~k,g))

(x 6 ~k ), (a) I

~ 0

(9~=>l),

k,i

- (bk) Ik

II

~

~ - e~

llkl

= (l-e) llkl.

Lemmas 2.8 and 2.13 show that ~(ik,C[a],f ) _<_$(Ik,C[bk],f ) + + Thus

Z ~(Ik,&, C[a],f) g=l

C8~8 {C~C(~) + ~8}llkl --< Z ~(Ik,~,C[a],f) + C8~8{~C(~) + ~8}llkl g=l

•

66

$(l,C[a],f) ~ $C(-8~,@#)

(a) I ~ 0 k,g Since

(a) I

~ 0,

Ill

+

(k,g ~ i),

z k=l

% k=l

z ~(Ik, g, C[a], f) g=l

Z ilk,gl ~ g=l

we can apply the above argumen~ to

i_~8 ill.

Ik, ~.

Repeating this,

k,g we have

(2.48) ~ 1 G(I,C[a],f) _-_-{~-C(-@l~,e#)+C6~35(0C({3)+~6)}{1+ t2----~8+ (~)2+...} <

=

Next we assume that there exists

2

~c(_e#,e#) + csp~{Cc(~) + #6} .

1+0

(a) I > 0.

RSL of Type4 (0- A.,6-G.) shows that

such that, with

b E Lreal

~ = {x E I; A(x) # B(x)} = Uk= I Ik

(Ik : l~,k), 0 ~ b(x) ~ ~

a.e. on

I,

(a)Ik ~ 0

(k ~ 0).

Lemmas 2.8 and 2.12 show that $(l,C[a],f) =< ~(l,C[b],f) +

--<

Since

(a) I

% ~(ik,C[a],f) + C5~6{GC(~) + ~5} ]I] k=l

Z $(Ik,C[a],f) + C5~5 {OC(~) + ~6}]I[. k=l

~ 0, we can apply the argument in the case of

(a)

I

~ 0

to the

estimate of k $(ik,C[a],f) ; we have 2 $C(_@~,@~) Ilk I + C5~5 {GC(~) + ~5} llkl " ~(Ik'C[a]'f) ~ i-~ Thus, in this ease also, (2.48) holds.

Consequently,

2

~c (-e~'~) ~i-~ $c (-°~'e~) + c6~{~c(~) + ~}

(e = 1/2).

This estimate and (2.47) immediately yield (2.45). In the same manner as in (Second Step) of the proof of (2.38), (2.45) shows that

G(C[a]) ~ Const(l +~/ilaIl~) (a ~ LTeal).

From this inequality, we

now deduce (2.49)

G(C[a]) ~ Const(l + H~aI~BMO)

(a

is real-valued).

67

For

a E L~eal,

f E L~eal,l

and an interval

Let 8 = IIaNBMO. Inequality (2.44) holds with Lemma 2.2 shows that there exists an open set

I, we study

~(l,C[a],f).

E[.] replaced by C[-]. ~ = U"k=l Ik (Ik = I~, k) in

such that 121

~

--~

fI la(s) - (a)II ds ~ III/2, (a - (a)i)Ik ~ 2~

la(x) - (a)iI

~ 2~

aoe.

on

(k a i),

I - ~.

We put

b(x) =

(x { Ik, k >_- i) (a) I

(x ~ R).

Then ~(l,C[a],f) =< ~(l,C[b],f) +

+

Z ~(Ik,C[a],f) k=l

c5(i + 825 ) Ill.

IITn[']ll2,2 --= i) for some absolute constant C O . (See Lemma 2.10). If I(a)iI ~ 4 c O 8, then IIbll.~ (2 + 4C0)~, and hence

Recall that

~(l,C[a],f) ~ ~C(I + (2 + 4C0) ~) Ill ~ If

l(a)ll > 4c0~,

we put

~(l,C[b],f) =

I .....li-i(a)iI -~

{~ +

b = b - (a) I.

fi ]I I

fl

Then

Const (i +v~)

Ill.

II~II. ~ 2 ~.

We have

f
dyldx

-i - - ) n T n [I~ ] ( X l f ) ( x ) I d x l(-~)H(Xlf)(x) + n=1Z (l_i(a)

Z (i + l(a)l12) -n/2 c0n ii~ii~} ii I =< Const llI. n=l

Thus ~(l,C[a],f) _~ Const(l + V~) We can apply this argument to

Ik.

III +

Z k=l

~(Ik,C[al,f).

Repeating this, we have

1 -[-+q-a(l, C[a], f) =< Const (i +V~),

88

which gives (2.49).

Lemma 2.5 and (2.49) yield (2.39).

This completes the proof

of (2.39).

§2.10.

Application of (2.38) As is well-known, Theorems B and C are applicable to the higher dimensional

Neumann problem~ pseudo-differential operators and the estimate of analytic capacity ([6], [I0]).

(See Chapter III.)

application of (2.38). ¢,

LP(F)

denotes the Banach space of functions

IIfIILP(F)

The Cauchy(-Hilbert)

We say that

lim S ~0

f l<-zl > s

LP(F)

Fixing

p

(lap<-).

f(~) <-z

Id~l

•

M

if,

where

~(z,~)

is the length of

F

We show

Let

F

z 0 6 F,

r= Let

with norm

to itself is denoted by

be a chord-arc curve with constant

IIHFIIL2(F),L2(F) a Const Proof.

F

F is a chord-arc curve with constant

g(z,~) a Mlz-~l,

and ~ .

Corollary 2.14.

in the complex plane

•

z, ~ 6 F, z

on

F

is defined by

H F as an operator from

IIHFII LP(F),LP(F)

f

f(z)IPldzl} I/p

H F f(z) = 1~

The norm of

for any

= { IF

transform on

(2.50)

between

In this section, we show an immediate

For a locally rectifiable curve

we parametrize

{z(t); t E m }

be an even function in

CO

,

F

M.

Then

M 2. so that

~(z0,z(t)) = Jtf.

such that 3

p(x) = 1

(la xaM),

~

Hp(k)rl =< Const,

k=O supp(p) c [-M-l, -1/2] U [1/2, M+l]. Put

h(z) = p(Izl)/z

we have, with

(z 6 ¢).

Since

Is-tl/M a Iz(s)-z(t)I

~ Is-tl (s,t 6 ~),

a(t) = Mz'(t)

1 z(s)-z(t)

= ~s-t

M s-t

h(s-~it fts

(Mz(s)-Mz(t))-I s-t

a(u)du)

( = M T¢[a,h](s,t), say).

69

We have

Iz'(t) I = 1

a.e.

and

HFf(z(s)) = (-~)

lira f l s _ ~ > g s -~ 0

= (-~) M T¢[a,h] {(faz)z'}(s) Hence

II~IIL2(F),L2(F )

= n M IIT~[a,h]II2,2.

F h(s + it) = f~= / ~

f(z(t))z'(t)dt

z(s)-z(t)

a.e.

Let

e-i(sx+tY)h(x + iy)dx dy •

Then Tc[a,h ] = Const ~_~ / ~ Fh(s+it) E[Re {a(s-it)}] ds dt = Const ~_~ f_=Fh(s+it) oo

since

~_~ /_

{E[Re {a(s-it) } ] + ~ H}

ds dr,

oo

Fh(s+it)ds dt = Const h(0) = 0. II E[Re {a(s-it)} ] + ~ H!I2, 2

Lemma 2.10 and (2.38) show that

Const fiRe a(s-it)llBMO Const

M Is+itl ,

and hence llTc[a,h ] N2,2 ~ Const M /_~ f_~ I Fh(s+it) I Is+itl ds dt. Integration by parts shows that, for

n = 2, 4,

I Fh(s+it) I = l(_is)-n f_~ f_~e-i(sx+ty)

n B n 8x

h(x+iy) dx dy I

Const/Isl n . In the same manner, [ Fh(s + it) l ~

I Fh(s + it) I &

Const/Is + itl n

® f ® IFh(s+it)[

Is+itj ds dt =

Const {ffls+itl ~ 1 Consequently,

Const/It[ n

(n = 2, 4).

Thus

(n = 2,4 ) . We have

Ills+itl ~

ds dt + f f Is+itl Is+itpl

i + ffls+itl > i ds dt Is+itl3 }

~

Const.

70

lIHr

llL2(r),L2(r)

= ~ M lIT¢[a,h]II2

Const M 2 f_Z f_Z

IFh(s+it) I Is+itl as at

This completes the proof of Corollary 2.24,

Const M 2 . Q.E.D.

CHAPTER III.

§3.1.

Relation between

ANALYTIC CAPACITIES

~'(.) and

H

In this chapter, we study analytic integralgeometry

OF CRANKS

capacity

y(.)

from the point of view of

and the Cauchy transform on graphs. We shall estimate

so-called cranks.

For a compact set

the Banach space of bounded analytic supremum norm

l!'il ~.

E

in the complex plane

functions

in

~

The analytic capacity of

E

~,

7(.)

H'(E c)

U {=} - E( = E c)

of

denotes

with

is defined by

H

(3.1)

T(E) = sup{If'(~) I ; Ilfll

~ l, f E H®(EC)},

H" where

f'(®) = lim z ~ ~ z(f(z) - f(=)),

the Taylor expansion at

~

i.e.,

([29, p.6]).

f'(')

If

g(z) = (f(z) - f(®))/(l - f(®) f(z)) satisfies

g(~) = 0,

IigIl ~ & i

is the 1/z-coefficient

f E H~(EC), II fll ~ ~ i, H

of

then

( E H=(Ee))

and

H

Ig'(®)I Hence, ~.

= I f'(~)I /(i - I f(~)l 2) _>_ If,(= ) I.

to estimate

y('), we can restrict our attention

The Cauchy transform of a complex measure i 2~i

C ~(Z) =

f~

~

d~(~) ----q-"z-~ i

in

¢

to functions vanishing at is defined by

(z ~ supp(~)).

We put (3.2)

y+(E)

i = sup{-~-

/ d~;

IIC~H ~

i,

~ ~ 0, s u p p ( ~ ) c

E} .

H

Since

(C~)'(')

i 2~i the open disk of center z IEIg = 2 inf

{D(Zk,rk)}k= 1 is defined by

Zk= I of

rk, E

f d~

we have

and of radius

Y(E) g T+(E). r.

For

Let

D(z,r)

denote

s > 0, we put

where the infimum is taken over all coverings

with radii less than

IEI = lim s ~ oIEls"

If

equals its 1-dimenslon Lebesgue measure.

s.

The generalized

length of

E c e, then the generalized We shall compare

T(-),

E

length of

Y+(-)

E

and

If. A set any

z £ F,

constant

M.

F c ¢

is called a locally chord-arc

there exists

E > 0

A locally chord-arc

a locally chord-arc

such that

Fn

curve with constant

D(z,e)

is a chord-arc curve with

curve is not, in general,

compact curve with constant

i00.

M, if, for

connected.

We define

Let

F

be

72

(3.3)

p(F) = inf y(E)/IEI,

p+(r) = inf y+(E)/IEI,

where the infimums are taken over all compact sets N'II (i ~ p < ~) L p (r)

be the same as in §2.10.

r

with supremum norm

on

F

li"IIL~(F)

and let

E

Let

L (r)

on

L~(F)

F .

Let

be the

LP(F), L"

space on

be the space of functions

f

with norm

NflIL(F)

=

If()l>

sup

The Cauchy transform

HF

on

operator from

to

LI(F)

LI(F)

F

>

;

is defined by (2.50).

HHFIIL1(F) ,L~(F) "

is denoted by

w relations among

p(F), o+(F)

HF

The norm of

as an

Here are

and IIHFIILI(F),LI(F)"

Theorem D.

(3.4)

C°nst/IIHFIrLZ(F),L~(F)

~ p+(F) ~ Const/llHrHLl(F), L~(F),

(3.5)

p+(F) ~ p(r) ~ Const p+(F)

1/3

.

We begin by showing the second inequality in (3.4). llfH

= i.

k > 0,

For

Let

f E L2(F),

- ~ < e ~ ~, we put

Ll(r) Ek, e = {z ~ F; HF f ( z ) ~ D(ke i e , k / 4 ) } , There exists a compact set

Fk~ 8

in

exists a non-negative measure

EX, ~

on

FX, e

d~ ~ H Since IIC~II

IFx,el ~ IEx,el/2.

such that

There

such that

y+(Fx,e)/2.

k,e

~ i,

d~ = hldz I

we can write

with

h 6 L~(F),

0 <_- h <_- 2~.

We

H ~

show that

IIHFhN

~ Const.

L~(F) For

zO ~ F

choose first i00.

such that

¢ > 0

Choose next

0 < e' < s

IHF h(z 0) where so that

so that

H F h(zo)

exists (in the sense of (2.50)), we

F n D(Zo,¢ )

is a chord-arc curve with constant

so that, for any

z E F c n D(Zo,¢'),

2i C~(z)[ g ]HF h(z O) - 2i C(hId~])(z)[

is the restriction of

h

to

F n D(Zo,g).

+ i,

Choose at last

0 < ¢" < g'

73

7,(,o)-~-

IHr Since

~"

F n D(z0,¢)

such that

[(<) r n D(z0,~") c

I
~ - z0

<-- J--

is a chord-arc curve with constant

s "'4 ~ =< I~o - z~'l<,=< ~"i2

i00, there exists

" r n D(%,i0 -i° ~') = ~.

and

IHr h(z O) - 2i C ~(z~) I ~ IHr ~(z o) - 2i c<~Id~l)l --<

I ~- m

=<

2 s

r n D(Zo,¢")C

+

IdOl

~ - z0

~-

rr

~ - z~

t=o - =6t r n (D(=o,~)-D(~o,~")) t ~ - =oi I~ - %t 1

21

z 0' ~ ¢

Thus we have

+ i

I<11+2

Idol

Id~l + 2 _<-Const.

F N D(Zo,~") Since

IIC~II =< i,

we have

IHF h(zo) [ -<- Const.

Since

~F h(z)

H" r,

TIHr h}i

~ Const. L~(F)

Sinc e 1

7+(Fk,e)

1 -~-

~

I Fk,e

hldzl - T + ( F k ' 9 ) '

we have k ~

Y+(Fx, e)

=< ~

I IF

k e i0 hldzll k,e

=<

i ~--

iiFk,@ (HF f) h Idzll

~2n

IIF f(H F

h)Idzll

Const IIfIILl + (r)

+

+ k 7

~

IFk,@ hldz 1

%(Fx,e)

k ~ T+(Fk,e) ,

which gives y+(Fk, e) ~ Const llfIILi(F)/k ~ Const/k . Since

IEk,@l ~ 21Fk,eI we have

~ 2 y+(Fk,e)/p+(r)

~ Const/(k p + ( r ) ) ,

exists a.e. on

74

Iz ~ r;

l~r f(z) l > k I ~_

~ {const/(X p+(r))}

Z

I U n=0

.4)n (g

=

i00 u E k=l

I (5/4)nk, 2nk/100

Const/(k p+(F)).

n=O

Since

k > 0

is arbitrary,

flIL~(r )~

IIHr

Const/ o+(r)

(f ~

L2(F),llfNLl(r) = i).

A standard argument shows that this inequality holds with g E LI(F),

IIglILI(F)

= i.

f

replaced by any

Hence the second inequality in (3.4) holds.

The proof of the first inequality in (3.4) was essentially given by Davie [21], Marshall

[37] and Davie-¢ksendal

[22].

Here is a tool for the proof.

Lemma 3.1 (the separation theorem [53, p. 108]). convex sets in the Banach space

II'IIL.(F). Then

C(F)

Let

P, Q

be two compact

of continuous functions on

there exists a complex measure

~ on

F

F

with norm

such that, for any

f E P, g E Q, Re fF f d~ >

Re fF g d~.

Since F is a locally chord-arc compact curve with constant i00, there exists F N D(z,s 0) is a chord-arc curve with constant i00 for any ~0 > O such that 8 z E F. Let H F (0 < s < SO/2) be an operator defined by

H r~ f(z) = ~i f

f(~l ~ - z

r n D(z,c) c

Id~I

We show that there exists an absolute constant C O (3.6)

Let

M F2g

s IIH FII ~ ~ LI(F) ,e~(F)

C0

F(z,~) = F N D(z,~). IIMISlILI(F ) i

For

such that (= C0m 0, say).

be a maximal operator defined by

MF2g f(z) = 0 < nsup~ 2s where

llHrllnl(r) ,L~(F)

(f e El(F))

f ~ LI(F)

with

~

,n (F)

~

~

i

/F(z,n)

if(~) i id~l

Then Const.

IIftlLl(F) ~ i,

we put

E = {z ~ F; IHF f(z) I =< k ' M2~F f
(f E LI(F)),

75 Then

l r - El ~ {tlHrltLl(r),L!(D

}/x NM~lld(r),L~(r)

+

(m0 + Const)/k Let

z0 E E.

Then, for any

IH~

IH~ f(z) I ~

< 1

z ~ E N D(ZO,e/2),

f(z) -

l

ifF(zo,8 )

Hrf(z)1 + IHr

f(z) I

Id~II + k

f(~)

=< 1w[ ]fF(Zo,g)-f(~)~- z -< ~i

Const mO/k.

~ f(<) -z

IdYll + Const M2Fg f(z)

+k

]d~] I + Const k .

Hence we have

[H~ f(z)[= lHr(~0,~) where

fo

Since

F(zo,g)

is the restriction of

fo(Z)[ f

+ C~ k

to

F(zo,S)

and

is a chord-arc cu=ve with constant

IJHF(zo,s)H LI(F),L~(F ) ~ Const.

Iz ~ E n r(z0,~12);

We choose a finite covering and

IH~ f(z)I

C O'

is an absolute constant.

i00, Corollary 2.14 shows that

> (% + l)xl

n {D(Zk,g/2)}k=l

llZk= n 1 XF(Zk,~)IIL.(£) =<

characteristic function of

n r(z0,~/2)),

(See also (2.10).) Thus

Iz E E 0 £(ZO,S/2); ]HF(zo,s)fo(Z)l

(i ~ k ~ n)

(z E E

F(Zk,S)°

g

Iz ~r; IHr f(zll > ( % + i ) ~- Iz ~E;PH~ f(z)I > ( % + l ) X l

> kI ~ of

Const,

r

(Const/k)IIfOllel(r(Zo,~))

so that where

zk E F XF(zk,g )

Then

xI +

n =~ Z ]z ( E fl F(Zk,S/2);IH F f(z) 1 > k=l

IF - E l

(% + 1)~J + Const mo/k

n _<- (Const/k) k=IZ IF(zk,e ) If(~)l Id~l +

Const mo/k

which gives (3.6). Given

0 < ~ < gO/2

is the

and a compact set

E c F,

we put

&

Const mo/k,

76

F = {f E L'(F); 0 ~ f(z) ~ i, IE f(z)IdzI ~ IEl/2,

where

P = {Hi f; f ( F},

Q

mo = NHFIILI(F),L]w(F)

and

P n Q # @.

Suppose that

supp(f) c E} ,

[Igile.(r )~

= {g E C(F); CO

P N Q = @.

d~ >

Taking the supremum of

Re

Since

Re IF g d~

IF

g d~

Como},

is the constant in (3.6). P, Q

We show that

are compact and convex in

C(F), Lemma 3.1 shows that there exists a measure Re /F HFs f

3

~

on

F

such that

(f E ~ g (Q)

over all

g E Q,

Re IF H F~ f d~ ~ 3 C0m 0 IFid~I

we have

(f E F)

which implies that - Re IF f go

IdzI { 3 Com 0

(f E F),

where

g0(z) =

(7 i r Id~l) -I fr, I< - zp > ~

By (3.6), we have, for any

h E LI(F)

Iz ( F;IH ~ h(z) I ~ Since the kernel of

H Fe

2 Como/IEiI ~

Iz E F; Igo(z)I ~

Then

~

with

IIhIILI(F) ~ i,

IEI/2.

fFld~I ~

2 Como/iEiI ~

F = {z E E; Igo(z) I ~

function.

d~(<).

is uniformly bounded, this inequality holds with

h Idzl replaced by any measure

Let

with

1

< - z

2 Como/iEl}

XF E F .

I.

Hence

XF

be its characteristic

IEI/2. and let

Hence we have

3 Com 0 _-< - Re fF XF go

fdzI --< IF Igo (z) I Idzl

2 Com 0 --< ~

fF Idz] =< 2 Como,

which is a contradiction. Since

P N Q # ~,

IIHF f¢ile®(F ) <= 3 tom O. ~i >-- e2 >-- "'''

Thus

P N Q # ~.

there exists Let

limn -~ ~ Sn

{On}n= I =

0

and

f~ E L'(F)

such that

fs E

F,

be a sequence of positive numbers such that {f

gn

IdzI}~=l

converges weakly (as a

77 sequence of measures). Idzl; we write by

IIHr

fOllL~(r )~

Const m O.

0 < 2s e < e k < SO/2.

Sk St 1% f %)I

+

Let

ISr, _ F(Zo,CoI2)

st fk

ek, e e

Id~lt

Z

1

(

1

1

1

~ - z

IdYll

fF_F(Zo,ao/2) f ~

-

ee

)

+ Const

(~)

IdYll

) fse(~)

fee(K)IdYll sO c~ M F f (Zo)

Idol

IdYll + Const {i + (ek/t~)Irl}.

fee(~) - r(zo,Ck)

~ - z

~ - z0

fee(<) ~ - z

F -F(z0,c k)

satisfy

z E F(ZO,ek/2),

e k < I~ - Z o l < eO/2 ( ~ - Z0

If

If r

and let

--

!

!

We show that

Imr_F
~ !

imr,

~

fO 6 ~

z0 E F

Then, for any

2 Const (ek / s0)

Let

We have

i

! -

Then the limit is absolutely continuous with respect to

fOidz I.

z

denote the restriction of

fee

to

F(Zo,ek).

Then, for any

z E F(ZO,ek/2), et

Id~ll = IHr f

!~ ifr- r(Zo,~k) ~ - z 3 Com 0 +

st

st

(z) - H r fk(z) l

~e ~e IHF fk (z) I

which shows that Ck t ~ I HF f (Zo) l ~

s~ s~ IHF fk (z) l

+

+

3 Com 0

Const {i + (%l~)Irl}

(z E F(Zo,Sk/2)).

By (3.6), we have st st Iz E r; IHr fk (z) I g

10-3 Ir(Zo,~k) I

103 Com 0 ] < 10 -3 st = rlfk NLI(F )

1 7 Ir(ZO'Sk/2)I "

This shows that the generalized length of is larger than or equal to

{z E F(z0,gk/2);

s t st IHF fk (z) I < 103Como }

IP(Zo,Sk/2) I/2. Hence the mean of

=

78 ¢g sg IHF fk (z) l

[Hr

over this set is dominated by

Const mo, which shows that

(Zo)l -<_ Const m0 + 3 C0m0 + Const {1 + ( 8 k / S 2 ) [ F I }

f

= /2 < Const {m0 + (¢k 80 ) Irl } " Since

z0 ( F

is arbitrary,

ii. " f"Jl Letting first

Const {m 0 + ( S k / g ~ ) ) F I } .

8

tend to infinity, and letting next

k

tend to infinity, we have

[[HF f 0 IIL.(F) ~ Const m 0. Now let

d~ 0 = f01dz [.

Since

f0( F,

[[Hr f0[[L.(F) ~ Const mo,

the

maximum modulus principle shows that

[[c~°lt H ® --
C1

~1

is an absolute constant. 00

I~

[EI/Cl'

fE d~° ~-

->_ O, supp(~ 00) c E,

Let

00

tlc~°°ll

supp(~ °) c E,

= O/(Clm0).

Then

< 1. =

H® Hence d, °° ,

which shows that

T+(E)/IE 1 ~ Const m 0.

Taking the infimum over all compact sets

E c F, we obtain the first inequality in (3.4). The first inequality in (3.5) is evident. inequality in (3.5).

Let

At last

f ~ L2(F), llfIILl(r) = i.

For

we show the second k > 0, - ~ < O ~ ~,

we put

EX, e = {z ~ F; H F f(z) ( D(ke i0, pk/4)} There exists a compact set exists

C

g (H

(Fk, 0)

Ilgll. H

We can write

~ z,

Since

in

EX, 0

such that

tFk,ol ~ IEk,o I/2"

such that

g ( - ) = 0,

g = C(hldzl)

llhllL~(r)< 2~,

FX, 0

(o = ~(r)).

with

Ig'(-)l

~ y(Fk,0)/2.

h (L~(F)

I1~'r hllL~(r) ~ Const,

satisfying supp(h) c Fk, 0

There

79 1

= ~2~ IfF>~,0 hldztI

Y(Fk, e) -~ [g'(~)l

=< Y(Fk,@)'

we have

k -<

1 2"--~ l [fFk,o

IfFk,e Xe ie hldzll

(Hrf) h Idzll

+ p% fFk, 0 lhlJdzl

--< 2T

Jfr

-< Const

llfllLl(r) + E 7(Fk,e),

which gives

f

HF h Idzll +

IFk, 0 1

k

y(Fk,o) ~ Const plflILl(r)/k = Const/k. IEx,0] ~

we have, with

21Fk,ol

~

Since

y(Fk,0)/p ~ Const/(pk),

q = (the integral part of

Iz ~ r; IHr f(z)l > xl

[U

103/p),

q U

n=0 k=l

E{l+(p/4)}n%,2~k/ql

Const/(p3k), which gives that IIH F ]ILl(F) ,L~(F)

=< Const/p(F) 3

This inequality and the first inequality in (3.4) immediately yield the second inequality in (3.5). §3.2.

Vitushkin's example, Garnett's example, Calder~n's problem and extremal problems

([5], [28], [46], [52])

Painlev~ showed that the analytic capacity of a compact set of zero generalized

l e n g t h i s e q u a l to z e r o ,

generalized

l e n g t h , we can c h o o s e a f i n i t e

that

Znk=l r k -< IEI"

1f' (~) ] =

Then, f o r any

covering

f ( H~(E c)

n Z k=l

rk _-
E of f i n i t e p o s i t i v e n {D(Zk,rk) }k=l of E so

with

i If f (~) d~ [ 2--~ 0 {Uk=ID(Zk,rk) } Const

which gives

For a compact s e t

IlfllH~ ~

1,

80

(3.7)

Y(E) =< ConstlE 1 .

This inequality immediately yields Palnleve's theorem. an example follows.

P Let

such that

Y(P®) = 0, IP I > 0.

P0 = [0,I].

[0,1/2], [i/2,1].

We divide

P0

Vitushkin [52] constructed

The set

P

is defined as

into two non-overlapping closed segments

Fixing their midpoints, we rotate these two segments so that

the resulting two segments are perpendicular to the x-axis. resulting segments are on divide each segment length.

(of

~.)

Let

P1 )

PI

discussion, we define

that

P

y(Q ) = O,

length

Pn;

into two non-overlapping closed segments of equal

Pn

nn= 0 Uk=nP k.

1/4

Let

The set

QI

in the four corners of

4 -n

2n

closed segments of length

Q~

is defined as follows.

Q~

2-n.

such

Let

be the union of four closed squares with sides of

with each component of

length

is a union of

Repeating this

Garnett [28] also constructed an example

IQ~I > 0.

Q0 = [0,I] x [0,i].

Qn-i

We

Fixing their midpoints, we rotate these four segments so that the

resulting four segments are perpendicular to the y-axis.

We put

(The midpoints of the

be the union of these two segments.

Qn-i

Q0"

In the same manner,

in the four corners of the component.

There are several proofs of Ahlfors function [29, p. 18] of

Qn

is defined from

replaced by four closed squares with sides of

y(Q ) = 0. Q

We put

Q~

= Nn=0 Qn"

Supposing that the non-trivial

exists, Garnett [28] showed a contradiction.

Using Besicovitch's set theory [i], Mattila [38] also gave an indirect proof.

A

direct proof from the point of view of the construction of Garabedian functions [29, p. 19] is given in [46].

This method is applicable to estimate

Y(')

of

various sets. It is sufficient to construct a sequence compact sets

Rn

and

f n (H=(R~)

so that

{(R n,fn )}n=10000~ R n o Qn'

of pairs of

fn(~) = i

and

78Rn [fn(Z) l Idzl ~ Const/(log n).

If such a sequence exists, we have, for any

g (H~(Q~),

IIgll

~ i, g(~) = 0,

H~

Ig'(~)l

=

2~i

i ~2-~ fDR

IfORn g(z) dz I

Ifn(Z) lldzl

2--~i I fSRn g(Z)fn(Z) dz I

~ Const/(log n),

n which shows that

y(Qn ) ~ Const/(log n) o

is constructed as follows. For a closed square

Q

We denote by

Thus m

y(Q ) = 0.

The pair

the integral part of

with sides parallel to the coordinate axes,

denotes its lower left corner and

g(Q)

Q+

Q

denote two closed squares in

(Rn,f n)

(log n)/2.

denotes the length of a side.

~(Q) Let

Q_,

with sides parallel to the coordinate axes

81

such that

2(Q_) = £(Q+) = 4-mg(Q),

g(Q_) = ¢(Q)

and the upper right corners of 4k Qk = U j=l Qk,j with

Q+ and Q are identical. For k > 0, we can write k components {Qk,j}4_1. Note that %(;Qk,j) = 4_k. We define inductively j-1 4m compact sets {Vk}k= 0 by V 0 = (Q0)_U (Q0)+, Vk = U where

E~ ~k (Qkm'~)-U

(Qkm'~)+

~k = {i -< ~ -< 4 km," Q k m , ~ N 4m (3.8)

(i < k < 4m),

(U k-i j=0 V j) = ~} . Let 4TM

Rn = {k=0 ~ V k} 0 Q(4m+l) m,

4m + 1

~0 = {i} . We now put

fn (z) = k=0~ %E~kHu(4km(z - E(Qkm,%))),

where u(z) = exp[e i~/4 4 -m { - - 1 + 1 }]~ z_(e i~/4 4-m/~) z_l_i+(e i~/4 4-m//~) Then

(Rn,fn)

satisfies the required conditions. (See [46].)

82

Calder~n [5] suggests to study

C[a]

for

a ~ L°° real"

This problem seems very hard;

in effect, Theorem D and Garnett's example immediately yield = o~

(See Remark 3.16.) Let

Qn.)

!

Then

Qn

is a union of

QnI 4n

=

{(2-i)z/3; z 6 Qn }. squares

V

4n

{Qn,j }j=l

{IIC[a]II2,2; a E Lreal}

(We rotate and contract with sides of length

(/5/3)4-n; the sides of Qn,j are not parallel to the coordinate axes. Let the projection of

Qn,j

to ~R and let

Qn,j

be the segment in

Qn,j

In, j be

whose n

projection to

~

colnsides with

In, j (i ~ j ~ 4n).

The intervals

{In,j}4=l

are mutually non-overlapping, the length of each interval is 4-n and their 4n ,,, [0,i]. Let F n = Uj= I Qn,j , where Qn~,j is the closed sub-segment

union equals of

Qn,j

of the same midpoint as

Qn,j

such that

is a locally chord-arc compact curve with constant IFn I = vr~-/6

and

Then

Fn

Since

y(Qn ) = (vr~/3)Y(Qn) , Theorem D shows that

IIHFnIL

l(rn) ,Llw(F n)

(3.9)

IQn,jl = IQn,jl/2. i.

~

Const/p (Fn) -_> Const/p(F n)

_a Const IFnl/Y(Fn) = Const/y(Fn) >= Const/Y(Qn) = Const/Y(Qn). We see that llHr IIi < n L (rn) ,Lwl(rn ) (See (3.18).)

Const {l]HFnlIL2(rn) ,L2(Fn ) ,,, 4 n {Qn,j}j= I

Since the projections of

we can define a graph {(x, An(X)); x 6 ~} A'n (= an) ~ Lreal." Since

an(X) = 1/3

IIHFnIIL2(Fn),L2(Fn ) ~

containing

+ i} .

to ~ Fn

are mutually disjoint, such that

a.e. on the projection of

Fn

to

~R,

Const llC[an]II2,2 .

Thus (3.9) shows that I/Y(Q n) ~ Since

Const{IiC[an]II2, 2 + i}.

lim n ~ Y ( Q n ) = y(Q ) = 0,

this gives that

{IIc[a]II2,2; a E L~eal}

=-.

It is very important to give various reasonable grounds to Vitushkin-Garnett's examples. Let

From this point of view, we consider the following extremal problem.

I 0 = [0,I).

For

s I, ..., s n E ~, we define

TSl ..... sn(X,y) = i/{(x-y) + i(Asl ..... an(X) - Asl ..... sn(Y) )} , where

83

(x ¢ I 0) As I ..... sn(X)

= I 0 sk

k-i

k

(--~-_~

x < n .

1 <- k < n) .

.

.

Our extremal problem is the following: ex(~(n) = max {~(Tsl ,...,sn); Sl, ..., sn ( m}

(n _~ i).

We see that (n ~ i).

Const iV~og(n+l) & ex (n) ~ Const ~ o g ( n + l ) (See Appendix I.) A~(x) = E 10m k= I

A0

We define a function

Sk(X)

(x 6 I0),

where

on

~

by

A~(x) = 0

(x ~ I0)

and

n

is the integral part of

m

(log n)/(log i0)

and 0 gk(X) =

Let

TAO(X,y)

(j-l)10 -k ~ x < j i0 -k, i ~ J ~ i0 k

i0-k be the kernel associated with

j

is

odd

j

is even.

A O.n Then

n

~(T 0 ) ~ Const I/~ = Const ~og(10m). An (See (3.14).)

Hence

r 0 = {(x, A~(x)); x ~ I0 }

is one of the worst graphs with

n

respect to

ex ( n ) . ~

The g r a p h

F0 n

is

similar

to

Qm"

Hence Theorem D suggests

that Problem 3.2.

Const T(F~) ~ min {T(Fsl .... ,Sn); s I ..... Sn6 ~} Const

where

§3.3.

T(F~)

(n g i),

Fsl,...,s n = {(x, A Sl"'''Sn (x)); x ( I 0} •

The Cauchy transform on cranks As a first step of harmonic analysis on discontinuous graphs, it is natural to

begin with worst graphs. M > 0

such that, for any

We say that a set

E c ¢

is thick, if there exists

z ( E, r > 0,

I/M ~ IE n D(z,r) I/r ~ M. The 1-dimension Caldergn-Zygmund decomposition is applicable to thick sets. thick sets are also natural objects.

From the point of view of §3.2 and

Hence

84

"thick sets", we define (thick) cranks. An interval

i

in

I 0 = [0,i)

is called a dyadic interval if

I

is

expressed in the form

I = [(j-l)2 -g, j2 -g) with integers g >-_ 0, 1 =< j =< 2g. A m R = {Ik}k= 1 of mutually disjoint dyadic intervals is called a m I0) if I 0 = Uk= 1 I k, For a positive integer q and two coverings

finite sequence covering (of n

~

R' = {lj}j=l,

m

= {Ik}k=l,

we write by

R'<

q

R

if each

I'~ J

is expressed as a

union of at least 2 q elements of R of same length. A segment I 0 is called a (thick) crank of degree

0.

For a positive integer

is called a (thick) crank of degree and

n

functions

(3.10)

AI, ..., A n

I 0
RI

~q

on

... 2

(ql .... , qn )

n, a graph

F = {(x, ~ ( x ) ) ; x

n, if there exist I0

~ qn

n

coverings

n

tuple

AF

(3.12)

on each element

For two positive integers

R

for some n

I

of

Rk,

Ak

is a constant and

(i ~ k ~ n). n, q

and two real numbers

=, ~

less than or equal to

i, we define a crank F(n,q,e,~) = {(x, ~(n,q,a,~)(x));

x E I0 }

by ~(n,q,~,~)

= A(q'~'~) + .... + A(q'a'~) n (x E [(j-l)2 -qk, j 2-qk), j

odd)

A(q,a,~)(x ) = I ~2-qk k

(~2_qk

(x E [(j-l)2 -qk, j 2-qk),

j

even).

We show

(3.13)

F

be a crank of degree

IIHFNL2(F)

s Const ,L2(F)

There exists an absolute constant (3.14)

n

such that

= A I + ...+ An,

)~(x) I ~ Ill

Let

R

of positive integers,

(3.11)

Theorem E.

E I0 }

R I, ...

lIHF(n)llL2(F(n )

~0

Then

V~.

such that, if

),L2(F(n))

(r(n) = r(n,q,~,~),

n.

~

Const V~n

n ~ i).

I~ - ~I 2 ~ D0/q,

then

85

In this section, we give the proof of the first half of Theorem E. For a crank

F

= {(x, AF(x)); x ~ I0}, we put

AF(x) = 0

outside

I0

and define a

kernel (x # y, x, y ~ ) .

TF(X,y) = I/{(x-y) + i(AF(x) - AF(Y)) } For

f E L2(F), we have H F f(x + i AF(X)) =

1

- ~

1

p.v.

f0 TF(x'Y)f(Y + i AF(y))dy

a.e. on

I 0,

and hence

(3.15)

IIHFIIL2(F) ~ ,L2(F)

!

lITFI12,2.

Here are three lemmas necessary for the proof of (3.13). Le~mma 3.3.

Let

F

be a crank of degree

NTFII2, 2 _-< Const Proof.

{~(TF) + I} .

ITF(x,Y) I =< i/Ix-yl

For any dyadic interval

(3.17)

ITF(x,Y) - TF(x',Y) I --
This is shown as follows. and let

.

I, we have

(x, x' 6 I, y ~ I

AI, ..., An

Then, for any

Let

k

R I, ..., Rn

(i =< k ~

I)).

= (the double of

be the coverings associated with

be the functions associated with

the smallest integer of I.

Then

Evidently, (3.16)

in

n.

n)

such that

AF. Rk

We denote by

has an element contained

x, x ~ 6 I, <

J

0

(i ~ k < m I)

=

I

2ml-k+l • III

(ml~ k

IAk(X ) - Ak(X')l ~ n).

Thus

ITr(x,y) - rr(x',y) 1 ~ Constl(x + i At(x)) - (x' + i Ar(x')) ! /Ix-yl 2 n

<= Const{Ix-x' I +

Z k=m I

IAk(X) - Ak(X')l} /Ix-yl 2

F

mI

ConstllI/Ix-yl 2.

86

The proof of this lemma is analogous play the same role as

el(.).

to the proof of Lemma 2.5;

(3.16) and (3.17)

In the same manner as in the proof of (2.9), we

obtain ~(T~) ~ Const {~(TF) + i}. J' = (x - (~/2), x + (e/2))

(Replace containing

x.)

(3.18)

where

Ix; T~ f(x) > 3 k, <

i

=

i00

J'

J~ f(x) ~ ~ k I

I x; T~ f(x) > k I

~ = C O {~(TF) + i}

{Ik}k= 1

by the largest dyadic interval in

Using this inequality, we obtain

and

CO

(f E L 2,

is a suitable constant.

by a suitable sequence of dyadic intervals.)

immediately yields

k > O) (Replace

Inequality

(3.18) Q.E.D.

IITFII2,2 ~ Const {~(T F) + i}.

For a non-negative

integer

o(n) = sup {~(TF);

F

we put

crank of degree

~(n) = sup {$(I O, TF,f); F

n,

& n} ,

f E Lreal, 0 ~ f a i,

crank of degree

~ n} .

We show Lemma 3.4.

For two positive integers

(3.19) Proof.

Let

satisfying

Let

(3.10)

F

with

g ~ n-l,

Without loss of generality we may assume that

be a crank of degree

and let

A I, ...

A '

(3.12).

g

~(n) ~ $(g) + ~(n -g - i) + Const ~(n). f E Lreal, 0 ~ f ~ i.

supp(f) c I 0.

n,

be

n, n

R 1 . . . . , Rn be

A F, = A 1 + ...+ An_g_l

,

m

Rn_ g Then

F'

= {Ik}k= I.

is a crank of degree

n - ~ - i.

We have

$(I 0, T F, f) = (TFf , TFf)fd x = ~(I0, TF,, f) + ((T r - Tr,)f, TFf)fd x + (TF,f, and

coverings

functions satisfying

n

Put F' = {(x, AF,(X)); x ~ I0} ,

n

(T r - TF,)f)fd x

(3.11)

87

((T F - TF,)f , TFf)fd x m

=

Z flk(T F - Tr,)(X ik f)(x) TF(Xlkf)(x) f(x)dx k=l m

Z flk(TF - TF')~Ik f)(x) T F ~ c f)(x) f(x)dx k=l Ik m E f c (rr - TF')(Xlkf)(x) rrf(x) f(x)dx k=l Ik ( = L I + L 2 + L3, Since

say).

TF,(x,y) = i/(x-y)

(x,y E I k, i _-< k =< m)

and

~(n) ->- Const, we have, by

Lemma 3.3, m

ILl1 ~

m

Z ~(I k, TF, f) + ~ % flk IH(Xlkf)(x) k=l k=l

Tr(Xlkf)(x)Idx

m

~(I k, TF, f) + Const IITFII2,2

Z k=l m

^

~(Ik, TF, f) + Const o(n).

E k=l

Extending coordinates, we see that, for each 0 ~ fk ~ i

and a crank

Fk

of degree

g

Ik,

there exist

fk ~ L ~real'

such that

$(I k, T F, f) = llkl ~(I 0 , TFk , fk ). Hence m

ILl1 ~

% llkl ~(I0, fk ) + Const o(n) k= I TF k ' ~(~) + Const a(n).

Recall (3.16) and (3.17). have, with

Since

x k = (the midpoint of

TF(X,y ) - TF,(x,y)

m

IL21 =

is anti-symmetric, we

Ik) , I *k = (the double of

I Z 7!k(T F - TF,)(Xlkf)(x) k=l

× {TF(× *c f)(x) - TF(× *c f)(xk)} Ik Ik

f(x) dx

Ik)

88 m

+

Z flk(TF - TF,)(Xlkf)(x) k=l

TF(× ,

Ik-Ik

f)(x) f(x)dx I

m Z Ilk I(TF - rF,)(Xlkf)(x)l Mf(x) dx

=<- Const

k=l m

+ k=iZ Ilk I(TF - TF,)(Xlkf)(x) I (/ik_lk ~

) dx

_-< Const lITF - TF,II2,2 =< Const ~(n). We have IL31 ~ j,k;j#kZ Iij {flk

ITr(x,y) - Tr,(x,Y) idY} ITFf(X) I dx

fi~cN lj {71, c j n Ik

z

j,k;j#k

ITF(X,y) - TF, (x,y)IdY}ITFf(x) I dx

+

Z {I * c I TF(x,y)-TF,(x,Y) IdY} ITFf(x) Idx j ,k;j#k $1kC*N I.J (lj -lj) N Ik

+

Z I , {Ilk ITF(X,Y) - TF,(x,y) IdY} ITFf(X)Idx j,k;j#k (Ik-Ik) N lj

= L31 + L32 + L33,

[L331

m

z f,

~

{flk ITF(X'Y) - Tr,(x,y) Idy} ITFf(x) idx

k=l Ik-Ik m

2

2

Z f , (flk ~ ) k=l Ik-Ik m Z k=l

ITFf(x)idx

~ ) 3 {f ,

Ik-Ik

(flk

dx}i/3

ITFf(x) I3/2 dx} 2/3 {I .

Ik-Ik

m

Const

Z Ilk II/3 {7 , k=l Ik

ITFf(x)I3/2 dx} 2/3

Const ( Zm llkl)1/3 { mE I . ITFf(x) I3/2 dx}2/3 k=l k=l Ik m llk 12 Const { I_~ ( Z ) ITFf(x) l3/2 dx} 2/3 k=l (X-Xk)2 +llk 12

89

m

( Z k=l

IIk 12 )4}I/6 {7 ~ITFf(x) l2 dx} I/2 T llk 12 (X-Xk)

Const

{f

Const

I[Trf][2 =< Const ][TF][2~2 =< Const ~(n), m

IL3,21 ~

Z {7 , ITF(x,Y) - TF,(x,y) IdY} ITrf(x) Idx j=l flj lj-lj m

z 7i j (7 , j=l I.-i.

=< 2

] ~dy

) ITrf(x) idx

J ]

Const IITFfll2 ~ Const o(n) and

IL311 Z

j,k;j#k

fI~

l~(x) - Ar,(x) I + IAF(y)-AF,(y) I dy}ITFf(x) Idx

{f *c c N lj ~,] n ~k

m Z 71. j=l 3

Ix - yl 2

n I Z A (x) l (71,c ~=n-g ~ . J

m

dy ) iTFf(x) idx Ix-yl 2

n

Z fik k= I c

{Tik I Z A (y) I/Ix-yl 2 dy} ITrf(x) Idx ~=n-g

m

Const

j=l flj

lljl (fl,c • ]

dy ) irFf(x) idx Ix_yl2

m

+ Const

Z J"Ik*c IIkl (7Ik k=l

dy 2 ) ITrf (x) idx Ix-Yl

m Const /i 0 ITFf(X)]dx + Const f ~ ( Z - k= I Const

llk 12 + Ilk 12 iX_Xkl 2

)ITFf(x) Idx

IITrfll2 ~ Const ~(n).

Thus IL31 ~ Censt ~(n). Consequently, l((r r _ rF,)f, rFf)fdxl

~ ILII + Ie21 + IL31 ~ ~(g) + Const ~(n).

90

In the same manner, we have

I(TF,f, (Tr - TF,)f)fdx[ m Ik=IZ Ilk Since

TF,(Xlkf)(x) (T F - TF,)(Xlkf)(x) f(x)dxl

+ Const ~(n).

TF,(x,y) = i/(x-y) (x,y E Ik, 1 ~ k ~ m), the first quantity in the right-

hand side is dominated by

Const ~(n).

Thus

3(10, Tr, f) ~ ~(Io, TF, , f) + $(g) + Const ~(n) ~(n - g -

I) + ~(g) + Const c(n), Q.E.D.

which gives (3.19). Lemma 3.5. Proof.

a(n) 2 ~ Const ~(n)

We see that, for any crank

(See(Second Step) in §2.8.) (3.20) Since

sup {$(Tr);

F ,

and let

a crank

FI

F

crank of degree

f E L~eal, 0 ~ f ~ i.

of degree

~ n

and

~(I, TF, f) = III $(I0, TFI,fI ). non-dyadic interval

such that

n} ~ Const $(n).

n = 0.

Let F

fI ~ LTeal' Hence

be a crank of degree

For any dyadic interval 0 ~ fI ~ 1

I, there exist

such that

(i/IIl) $(I, T F, f) ~ $(n).

For any

I c I0, there exist mutually disjoint two dyadic intervals

Jill

= tI21

~--~i$(I, TF, f) = ~ i ~

~(TF) 2 ~ Const $(TF).

Hence it is sufficent to show that

$(0) ~ Const, (3.20) holds for

n g 1

I I, 12

(n $ 0).

~2JII, ~ I

IUI

2

Then

3(11 U 12, T r, Xlf)

{3(11' T r, Xlf) + $(I 2, T r, Xlf)

+ 7i I (if2

dy )2 dx + 7i 2(7I 1

ix_y I

id)

ix_y I

2

dx}

Const {~(n) + i} ~ Const $(n). For any interval

I c ~, we have

1 $(I, TF,f) ~ ill Thus

1 $(I 0 N I, TF,f) + Const ~ Const ~(n).

$(TF) ~ Const ~(n), which gives (3.20). In the same manner as in Lemmas 3.4 and 3.5, we have

Q.E.D.

91

Lemma 3.6.

o(n) ~ ~(n-l) + Const

(n ~ i).

We now give the proof of (3.13). 3.6 show that

~(n) < -

any non-negative integer

for all

Since

n ~ i.

~(0) ~ Const,

Lemmas 3.4 and

By Lemmas 3.4 and 3.5, we have, for

m,

~(2 m) ~ ~(2 m-l) + $(2 m-I - i) + Const ~(2 m) _<- 2 $(2 m-l) + Const ~(2 m) =<... m =< 2m $(i) + Const Z 2m-k ~(2 k) =< 2TM $(i) + Const 2m ~(2 TM) k=l _<- Const 2m ~(2m) I/2, which shows that

$(2 m) -<_ Const 2 2m,

~(2 m) -_< 2m ~(i) + Const

Hence

mE 2m_k ~(k) I/2 k=l

m % 2m-k 2 k =< Const(m+l)2 TMk=l

-_< 2m $(I) + Const Consequently,

~(2 TM) <= 2m $(i) + Const

<_- 2m ~(i) + Const

For an integer 2m =< n < 2m+l .

m 2m_k ~(k) 1/2 Z k=l

m% 2 m - k ~ / - ~ k=l

2k/2 -<_ Const 2m.

n g i, we choose a non-negative integer

m

so that

Then

$(n) ~ $(2 m+l) & Const 2m+l ~ Const n, Consequently, Lemmas 3.3 and 3.5 yield (3.13).

§3.4.

Proof of the latter half of Theorem E The following idea is essentially due to David [18, Chap. III].

q, ~, ~,

we write F(n) = F(n, q, e, ~)

A n = A (q'~'~) n

We put T (x,y) = TF(n)(X,y) ~0 (n) = ~(I0' T0n'

XIO)

x-y

'

(ng

i).

(n e i). -

Fixing

92 Here are two lemmas necessary for the proof.

~0(I) ~ [~ - ~12/1o0.

Lemma 3.7. Proof.

Let

x ( [(k0-l)2-q, k02-q) (k0 (~)[

0

=

I T 1 ×i0

1710

- #12 -q

- ~12- q

=

f

Z k even Z

Then

1 { (x-y) + i(Al(X)-Al(Y))

Z k even >= ~ - ~12 -q I Re

is odd).

i ~} dy I x-y

dy [(k-l)2-q,k2 -q) {(x-y) + i(a-~)2-q}(x-y)

I

I

I [(k-l)2-q,k2 -q)

7

dy

k even

[(k_l)2-q,k2-q)

f2-q +I 2- q

dy y2 + (~_6)2' 2-2q

In the same manner, we have, for any IT~ XIo(X) I a

Is - ~I/lO.

~

0

......

(x_y)2 + (~_~)2 2-2q > I S _ BI/10. =

x ( [(ko-l)2-q, k02-q) (k0

is even),

Thus ~°(I) = $(I°' t1'

Lemma 3.8.

(~

= I[o

For two positive integers

(3.21) Proof.

XI0) >

)2 dx > Is - ~12/I00.

lO

=

n, g

with

Q.E.D.

g ~ n - i,

T0(n) e ~0(~) + ~0(n - g) - Const q-l~-~-.

We write Ik = [(k-l)2 -qg, k2 -qg )

(i ~ k ~ 2qg).

We have ~o(n)

=

Izo Ir~ ×~o(X) t 2 dx

+ fl0(T~- T ~ ) (XI x )0 = LI =

= ~o(~)

T n0 XI 0 (x) dx + fl 0 TZ0 Xl0(X) (T~ - T~) XI0(X) dx

T0(g) + L 1 + L 2, 2qg 0 T~(x,y)dy} dx k=iZ /ik {/ik(T~(x,y) - Tg(x,y))dy}{flk

03

2qg Z 0 flk {flk (T~(x,y) - T~(x,y))dy} {flo_lk T~(x,y)dy}dx k=l 2 q&

k=IE fl0-1 k {flk (Tn0(x'y) - TO(x'y))dY}

T0n XI O(x) dx

= LII + LI2 + LI3.

Note that

0 T~(x,y) = 0

(x, y EIk, 1 ~ k =< 2q~)n

T (x,y) = [(x-y) + i

Z = g+l

and

(A (x) - A (y))]-i

1 x-y

(x, y E Ik, i ~ k ----2qg) Hence, extending the coordinate axes, we have 2qg

2q&

LII = k=iZ flk [flk T~(x,y)dy] 2 ^

Let

p be the integer such that

integer. where Let

dx = k=iZ Ilk] ~0(n - L) = ~0(n - £).

~k

q4 < p ~ 2q4 and (log p)/log 2 is an For each i ~ k ~ 2qg, we write Ik = Ik, 1 U... U I , 2 k,p 2 P 1 are mutually disjoint dyadic intervals of length p-2 2q~ {Ik,j}j= denote the closed interval of the same midpoint as

Ik and of length

(i + p-l) Ilkl, and let Ik,j denote the closed interval of the same midpoint as Ik,j and of length (i + p-4) llk,jl . We have, with x k = (the midpoint of Ik) , Xk, j = (the midpoint of Ik,j), 2qg LI2 = k=iZ flk {fiE T~(x,y)dy}

{fl0_lk T~(x,y)dy} dx

2qg

k=iE

fl k {flk T~(x,y)dy}

{f(l0 N ~k)-Ik T~(x,y)dy}dx

2qg z k=l

% .(T~(x,y)-T~(Xk,Y))dy}dx flk {Ilk T~(x,y)dy} {fl0_(10 N ik)

= LI21 + LI22 ,

ILl211

2qg 2 k=iZ fie Iflk Tn0(X,y)dyI (fik_l k

dy ~

) dx

04 2q&

2 q8

Z ~(Ik ' TO )}i/2 { Z flk (f~k_lk ~ k=I n' XIk k=I

{f~

Const Const

)2 dx}i/2

log2
p-i/2 ~o(n _ ~)i/2

and 2qZ p2 Z Z SI {71 LI2 2 = k=l j=i k,j k,j 2q~ p2 Z j=l E k=l

rO(x,y)dy){fiO_(iOnik)(TOn(x,y)-TOn(~k,Y))aY> dx

f(lk N ZIk,j)-Ik,j {/Ik,j } {fI0-(l0 Nik)}

2qg p2

Z j=l Z k=l

f Ik 0 ~ck,j {fIk '3}

{fI0-(~0 n ik)}

= L1221 + L1222 + L1223 • Since

n 2 IX-Xk,j I +t ~ (A~(x>-A~(Xk,j)) 1 ~=~+I ...... ]TnO(x,y) - T0n(Xk,j, Y)] _-< [x_yI IXk,j - y] Cons¢ {p-2 2-q8 + 2-q(8+i)} / ]Xk,j _ y12 Const p-2 2-q%/iXk,j _ yi2

(x ( Ik,j, y £ ~),

(3.13) shows that 2qg p2 IL12211 = I Z Z

k=l

j=i

fik,j{fIk,j T~(x,y)dy}

× {fi0_(i0 fl Ik) 2q8 p2

Const

2q~ p2

Const

-2

P 2-qg dy) dx Z Z fI Ifik,jT~(x,y)dy] (f~ k=l j=l k,j ~ IXk,j-Y]2 p-i Z Z fIk,j]fI T~ (x'y)dyl k=l j =i k,j

95 Const p-i NT~H2,2 ~ Const p-i ff . Since ITnO(x,y) - TOn(xk,Y) I ~ Const p

2-q~llxk-y[2

(x E I k, y E ~k ) ,

[TO(x,y)] =< Const 2-q(g+l)/Ix-yl2<_- Const 2-q(e+l) p8 llk,j [-2 =< Const pl0 2-q/llk,jl

(x 6 leo ~C Ik,j, y ~ Ik,j)

we have [L1222] =< Const

2qg P2 ~-~ E Z f gIk,j-lk,j (flk'j )(flk k=l j=l

~_ Const p2 fol

P 2-q~ dy)dx IXk-y]2

log(l + p-4s-l) ds ~ Const/p

and 2qg p2 [L1223 [ =< Const Z Y k=l j=l =~ Const p14 2-q

f

~c (Const pl02-q)(f~ Ik 0 k,j k

p 2 -q~ 2 dy) dx

IXk-Y I

2q~ E flk dx =< Const p14 2-q . k=l

Thus, by (3.13), ILl21 ~ILI211 + IL12211 + IL12221 + IL1223I Const {p-i/2 ~0(n - ~)i/2 + p-i/~ + p-i + p14 2-q } =< Const p-i/2 {~o(n ~ - e)i/2 + V~ } ~ Const~/q. Since iT0n(X,y)

-

0 rg(x,y) I _~ Const 2-q(g+l)/ Ix-yl2

Const p2 2-q llkl/{iX_Xkl2 + llk12 } (x E I~Ck, y EIk), we have, in the same manner as in the estimate of 2qg ILl31 _-
( fie ~ )

IL331 in §3.3,

IT0nXlo(X) lax

96

+ Const p2 2-q

sz0

2q~ {%

k=l

ilkl 2

Ix-x~I2 + tlkl2

2qg Const { Z ~ ( ~ k=l fIk-I k IIk

n } LT°×I0 (x)ldx

)3 dx}i/3

2qg × { Z f~k IT0 X10(x) 13/2 dx}2/3 k= 1 n +

2qg Const p2 2-q {flo ( Z k=l Const

+

llkl 2 ) 2 2 2 dx}i/2{/io ITn~IO(x) 12dx}I/2 IX-Xkl + llkl

i, ~ ~i/3 { Z2q~ 7~k IT0 {fl0/P log3(l + s)aS~ n X10(x) 13/2 dx} 2/3 k=l

Const p2 2-q 1IT,ll2,2 Const (p-i/4 + p2 2-q)NT~II2,2 =< Const q-iVan

Thus L I ~ LII - ILl2 [ - ILl31 ~ ~o(n - g) - Const q-l~/~. In the same manner, 2qg

IL21 _~ Ik=iZ/ik {flk TO(x,y)dy} {flk(TO(x,y) - TO(x,y))dy} dxl + Const q-l~-~n. Since

T~(x,y) ~ 0

(x, y 6 Ik, 1 ~ k & 2q&), IL21 & Const q-l~nn.

Consequently,

~0(n) ~ ~O(g) + L I - IL21 ~ ~O(g) + ~o(n - g) - Const q-l~-~.

Q.E.D.

We now give the proof of (3.14). Lemmas 3.7 and 3.8 show that, for any positive integer m, ~0(2 m) ~ 2 ~0(2m-l) - Const q-i 2m/2 ~ ... 2m ~0(i) - Const q-I m~l k=0

2k 2(m_k)/2

2m {I~ - ~12/i00 - COnst/q} .

97

For any integer n ->_ 2, 2m- i < n ~ 2m . Then

we choose a positive integer

~0(n) >__ ~0(2 m-l) + ~o(n - 2m-l) - Const q

2m-I {I= - ~12/i00 - Const/q} ~V~{I~

- ~12/200 - Const/q}

Thus (3.14) holds if

~0

-i

m

so that

V~

- Const q - i V ~

.

is large enough.

This completes the proof of the

latter half of Theorem E. Corollary 3.9. If

For any crank

Is - fll2 ~ q0/q,

then

(F(n) = r(n,q,~,~), Proof.

n g i),

For any crank

F

F

of degree

where

~0

of degree

NHrII1

n ~ I,

1

~

is the constant in Theorem E.

n ~ i, we have

const HTFfILl

L (F),Lw(r) Since

~

p(F) ~ p+(r) ~ Const/~-~.

p(r(n)) I/3 ~ Const p+(r(n)) ~ Const/~-

,L~(~)" (~)

is a locally chord-arc compact curve with constant

valid for

F .

inequalities.

i, Theorem D is

Thus Theorem D, Lemma 3.3 and (3.13) yield the required

The latter half is also deduced from Theorem D, Lemma 3.3 and

(3.14).

Q.E.D.

David [18, Chap. III] showed that (2.39) is best possible in the following sense: (3.22)

sup {IIC[a]II2,2; a 6 Lreal,M }

This is deduced from Theorem E as follows.

IITF0(n) Xl0112 ->_ C o n s t V ~ where

IAnl ~ Const n.

A n = {(x, f0

(n > i) , ~0 ) , i, 0).

Fo(n), we define an arc

An

Adding some segments with endpoints

There exists a Lipsehitz graph

b (s)ds); x ~ I0}

IFo(n) U Amn - Fo(n) n A*n I ~ s, 0 < g < 1/2

(M ->_ 1).

We showed that

Fo(n) = F(n, i + (the integral part of

parallel to the y-axis to Then

>= Const I/M

is determined later.

such that

IA

where

is an absolute constant and

CO

We have

Sio Ib~(~)ldx ~ IA~I ----c o n

~ CO n

and

0

and

1.

98 Lemma 2.2 shows that there exists an open set

~ = Uk= 1 Ik

(I k = I~, k)

in

I0

such that

[~J ~ ~,

(Ib~l)ik & Con/8

Ib~(x) I ~ Con/8

(k a i),

a.e. on

I0 - ~ .

We put

C(x>= I and

An

* Ák (bn)

= {(x, f0 bn (s)ds); x ( I0}.

projection of

F0(n) N A n

to ~

:x( :o' we have

(x ( Ik, k ~ i)

IF1 ~ 2s.

Then

and let

(s> s

libn I]. N C 0 n/g . Let F - I 0 - E.

:o C (s>ds) = ),i

E

be the

Since

X

From the definition,

..,

C[b n ](x,y) = TF0(n)(X,y)

(x,y 6 E).

Hence T h e o r e m E s h o w s t h a t

{rE IC[b~*] ×E(x)]2dx}l/2

=

{rE Irr0(n> ×E(x)12 dx}l/2

=> {rE ITFo(n)×IO(x) 12 dx}i/2 - {fIoITpo(n) XF (x) 12 dx}i/2 {Const n - /F ITF0(n) Xl0(X) 12 dx}i/2 - Const V ~ V ~ . By (3.13) and (3.18), we have IITF0 (n )if4,4

~

Const{o( TF0 (n )) + l} ~ Const V~,

and hence,

7F ITFo(n) ×lo(X) l2 dx _-
{(i - ConstV~c )I/2 -

sufficiently small, we have

bn** (Lreal,C0n/e)

** ]112,2 >= Constq-f llC[bn

(n => i),

Const VT}.

99

which yields (3.22). §3.5.

Analytic capacities of fat cranks For

With

p > 0, z 6 ¢

0 ~_ ~ -<- i/i00

and

E c ¢, we write

and a segment

the closed segment

J(~)

J c ¢

[pE + z] = { ~ + z;

of the same midpoint as

and of length

(i + ~) IJl.

and a segment

J c C

~ 6 E}.

parallel to the x-axis, we associate J, parallel to the x-axis

With a positive integer

q ~_ 2,

0 ~ ~ --< I/I00

parallel to the x-axis, we associate 2q-i

J(q,~) = 2q {Jk}k= 1

where

U k=l

{[J2k_l(~) + i 2-qlJl]

are mutually non-overlapping segments on

these are ordered from left to right. segments of length of type

O.

2-q(l + ~) IJl.

For a finite sequence

less than or equal to

i/i00,

The set

({Jk}k=l£

F = for some 2.

J(q,~)

J

of length

is a union of

2q

2-qlJl; closed

A segment F 0 = [0,i] is called a crank n = {~j }j=0' ~0 0 of non-negative numbers

a finite union

F

of closed segments parallel to n {~j }j=O if there exists a crank

the x-axis is called a (fat) crank of type F' = Uk=l~ Jk

U J2k(@)} ,

are components of

F')

of type

n-i {~J }j=0

such that

U Jk(qk , ~n ) k=l

g-tuple

of positive integers larger than or equal to

(ql' "''' q~)

We write simply

T .

F'[(ql ' ...

, q&;~n ) Proposition 3.10.

(3.23)

Let

F

be a crank of type n Z

T(T) ~ Const {

U H

~=I j=0 At present, the estimate of

T(F)

{~j}~=O '

1

~0 = 0.

}-I.

(l+~j) from above is unknown.

[29, p. 87] and (3.8) do not yield satisfactory inequalities. proof of (3.23) is standard.

r0

We put

[ (. ;~i)

rI

Let

Then

n 0 {F }~=

[ ... [ (. ;~2) (. ;+n)

be

n+l

r n

The method in

The following

cranks such that

100

h (z) = j=0 (i + ~j) ' g (z) = Im ~I" hp(z) P h (~)

1

Id~I

= --~ Im p.v. /F where

Im

denotes the imaginary part.

Then

(z E r , 0 =< ~ ~ n),

go ~ 0.

We show that, for any

I--< ~ =< n,

(3.24)

ilg~II N L (F) -

F _ I = Uk= 1 Jk

We can write can write

F

fig i l l ~- L (Fp_l)

+

and let

z0

I ~i

{Jk}k= 1

~-tuple

be the point on

Ig (z0) - g~_l(Z~)

~ j=0 (i + ~j)

with components

= Uk= 1 Jk(qk , ~ ) with some

z0 E Jk0(qk0,~ )

const

Jk0

.

of

F

(ql' "''' q~)" nearest to

Im fJk0(qk0,~ )

and

i'

Let Then

z0.

~h (~)z0

h~_l(~)

i

,

Im fjk 0

Id~l I

- z0 1

+I 7

E k#k 0

{Im 7jk(qk,@~ )

= L (I) + L (2),

h(~)

h

~ _ z0-- ]d~I - Im 7Jk

~-±- z~

i(~)

Id~l~l

say.

We can write G0 U yj j=l

Jk0(qk0 '~) =

G0 {Tj }j=l ;

with its components

qk (G0 = 2 0)

these components are ordered so that the

x-coordinate of their midpoints are increasing. may assume that

cY0/2 z0 ~ Uj=I Y2j-I h Im fj

we have

Since

G0/2 {Y2j}j=I

i (~) -

k0

"

Without loss of generality, we

~-

Id~I z 0

are disjoint and

h (~)

.........

= Im /U°0/2j 1

I

Y2j-i

Z 0

Id~I

=

0,

101

h(~)

1

L (I)

--

IdIL

IIm I ~012

~ - z0

Uj=I T2j -i (~0 IJkol

f~

=< ! =

2

dx

-

- (x-Re z0) + (~011Jk01) 2

~

j=O

1

(i

+ ® _ j)

"

For 1 ~ k ~ g, 0 ~ v ~ ~ -i, 7k(V) denotes the component of Fv generating Jk" In particular, Yk(~-l) = Jk (i ~ k ~ g). We put

L(2)v = k Z( G

fJk

h (~) ~ -~ Zo--- Id~l

!fJk(qk'~b ). hb-l(~)-, Id~II ~ - z0

(i ~ v ~ W-I),

where Gv = {i ~ k & ~; k # k0, Yk(V-l) = Yk0(V-l), Tk(V) # 7k0(V)}. Then h (~) L (2) =< k#k0Z Ifjk(qk,@# > _ b_ z0 We now estimate L(2)'v Note that observation s h o w s t h a t , f o r a n y k (3.25)

h_l(~) Id~l . ./Jk . . ~ -----*--z 0 Id~ll = ~Zl v=l L(2)~"

ITk(V)I = IYk0(V) I ( G

dis(J k U Jk(qk,@ ), Jk0 U Jk0(qk0,@ )) ~ diS(Yk(V), Tk 0(v))

- 2 l~k0(~)l

{(1 + ~v+l )

2-2 +

+ (l+ ¢~,+l)(l+ @v+ 2)

3 => diS(Yk(V ), Yk0(~)) - ~

2-4 ~f

...

Tl (I + %j) 2-2(~-v) } j=v+l

=> diS(Yk(V), Yko(V)) - 2 IYko(V) I

Since

(k (Gv) . A geometric

,

IYko(V) l

1.01 { --7-

.i.01.2 + ~--#-~

+ "'" }

102

/jk(qk,¢~ ) h (~) Jd~ I = Ijk we have, with

zk = (the midpoint of

L (2) = Z v k (G

{

Z k ~G ~-i N j=O

Const

i * zk - z 0

Id~l

Qk = Jk U Jk(qk,@ ),

Ifjk(qk,¢~ ) { ~ -i Zo

IJk Const

Jk )'

i , zk - z 0

v

+

=

h_l(~)

} h

1

} h (~)Id~I

l(~ )

idol 1

~ - z0

{(JJkl + IJk01) dis(Q k, Qk0 )-2 7jk

i (i + Cj)

Z k E

G

h~-l(~)

{(IJk] + ]Jk01)IJkl

Id~l}

diS(Qk,Qk0)-2}. !

The segment length

Tk0(V - i)

generates

IYk0(V) I , where

We may assume that

' 2qv

components

2qv {km}m= 1

of

rv

q~ = log {(i + ~v) IYk0(V-l) I/ IYk0(V) l} /log 2.

k I = 7ko(V).

Let q' (2 ~ m ~- 2 v ) "

G v,m = {k ~ G v ; km = Yk(V) } q~

Then

Gv

2 v = Urn=2

Z k(G

G v,m .

We have

(IJkl + IJkol)IJkl v~m

~_ Ikil 2-2(~-i-v)

N (i + ¢j) j -< - ~-i

v<

=

Ik112 2-2(~-I-~)

~

Z IJkl k ( G v,m

(i + @j)2 _-< ikli2 2-(~-I-~) ,

v < j _ - < ~-i where

N

(i + ~j) v<

denotes

j _~ ~-i

observation and (3.25) show

that

i

if

v = D -i.

Hence a geometric

of

103

v

2qv L (2) =< Const

H

j=0

1 (i + ~j)

%

(IJkJ + IJk0l)IJkl diS(Qk,Qk0)-2

Z

m= 2

k #G

v,m

v

~-i -<_ Const

j=O

1 (i + @j)

~-i

1

j=0

(i + ~j)

Const ~-i H j=O

(i +i @j)

--< Const

--< Const

~-i H j=O

2qv % m=2

diS(km,)~l)-2

Z k ~G

2q~

i)VlI 2 2-(~-l-v )

y,

(IJkl + IJko I) IJkl v,m

dis(Xm,kl)-2

m=2

I~i 12 2 -(~-l-v)

1

Z k=l

(

1 IXll k

)2

2_(~_i_~).

( i + ~j)

Tbus *

Ig (z0) I ~-I g _l(zO) I + L

(i)

+

L(2)

E v= I

V

1 =< IIg~_IIIL~(F~ 1 )

+

_-< IIg~_lll ~ L (r _ I)

+ Const

_

Since

z0 ~ F

n

j=0

(l + ~j)

H j=0

+ Const

~-i H j=0

1 (i + ~j)

(i + ~j) "

is arbitrary, we obtain (3.24).

By (3.24) and llIm H F

go m 0,

we have

hnll n L°°(rn)

~

Const

n E ~=i

H 1 j=0 (i + ~j)

(= Const

~n'

say).

Evidently, fF

hn(~) n

Id~I

=

i,

IIhnli L°~(Fn)

Hence we can define a non-negative function

hn

on

rn

so that

104

Ir n

hn(<)

I d<[

80/
=

'

llhn*NL'(rn) + lllm"r h$1L~(r# S l, h:(<) = 0

is c o n t i n u o u s l y

h*(<) n where

60

at endpoints of each component of

is an absolute constant. ^* hn(Z) =

i 2~i

u (z) = Im n (3.26)

Then

f

n

fF

h:(z)

Vn(Z) = Re

'

Fn ,

Let

h (<) n ~ - z

Ii

along

Id
fn(Z) = {I - exp(h:(z))} /{i + exp(~:(z))}

is analytic outside :

Ifn (')I The

differentiable

F , n

12~

nontangential limit of

F fF

(z ~ Fn).

and

n

h:(<)

{d~ 1 :

n

60/(2
fUn(Z) I at each point on

Fn

is dominated by

llIm HFn h~llL~O(Fn) _< 1.

Since sup

fUn(Z) I

is subharmonic in

F c fUn(Z) [ =< i. z (

Fc n

Hence, for any

and continuous on

U {~},

we have

z ~ F n,

n 1 + exp(2 Vn(Z)) - 2 exp(Vn(Z))COS(Un(Z)) i,

Ifn(Z) I2 1 + exp(2 v (z)) + 2 exp(vn(Z))COS(Un(Z)) n which shows that

T(r n) a

IifnllH~~ i.

If'(-)I ~ n

Consequently,

601(2 ~n )

= Const{

This completes the proof of Proposition 3.10.

n Z ~=i

H I j=0 (i + _~j)

-i.

105

§3.6.

Analytic Let

equation NE(r,@)

capacity and integralgeometric

L(r,@)

(r > 0, -~ < e ~ ~)

x cos @ + y sin @ = r. = #{E N L(r,@)},

elements of

For

less

than

e.

Cr (E) =

and

we write

is the (cardinal)

0 < a ~ i,

finite

line defined by the

E C ~,

number of

we put

Na{u~=lD(Zk,rk)}(r,@)~

where t h e infimum i s t a k e n o v e r a l l radii

set

#{E A L(r,O)} e > 0

Cr(e)(E)~ = i n f f _ ~ { f ~

denote the straight

For a compact

where

E n g(r,8).

quantities

dr} dS,

coverings

{D(Zk'rk)

k=l

of

E

with

We p u t

lim

Cr(e)(E)

(0 < ~. < i),

~ + 0 Bu(E) = If

lim lim c÷Oa÷O

E c D(0,1),

then

is the probability I01 ! ~)

Bu(E)/2~

intersecting

to compare a compact

is called the Buffon needle probability;

(measured by

curve. Then Crofton's (3.7), we have

Cr ( ¢ ) ( E ) .

with

dr d@/2~)

E.

formula

Suppose

~(.) E

with

Cr (.)

such that

that

E

L(r,O)

is a locally

[49, p.13] shows that

T(E) ~ Const CrI(E ).

set

of needles

(and with

Bu(.)).

and

chord-arc

compact

CrI(E ) = Coust I E I.

From this point of view,

y(E) = 1

this

(0 < r < i,

By

it is interesting

It is known that there exists

Bu(E) = 0

(cf. Jones-Mural

[34]).

We shall show Theorem F. y(E) = 1

For any and

Kakutani,

The author expresses

(3.28),

Galton-Watson

and Professors

process

various

set

E

such that

Coifman,

integralgeometric

time of the sun's rays integralgeometric

= l)

P r o b ( X ~ = 0) = 1 - B H

to use the to Professor since

RSL is the first stopping to try to give an

C,

Cr (.).

For

= -1)

0 < ~ < I,

let

(X~}~= I

on the standard probabillty

such that = erob(X~

According

Hence it is interesting

random variables

(F 0 = [0,1])

Cr (.).

cancers outside bodies.

we study

sequence of independent

eroh(X~

Steger who suggested

formulae are used for surgeries,

(or needles).

proof of Theorem

In this section,

his thanks to Professor Kakutani who

[30] for the estimate of

X-rays react to, for example,

(PO,S,Prob)

there exists a compact

Cr (E) = 0.

Acknowledgement. communicated

0 < ~ < 1/2,

=

(n a 1 ) .

/2,

be a

space

106

We put S~0 = 0'

S~n = X~l + "''+ X~n

This is a model of random walks. (3.27)

Then

y0~(x) = i,

Prob(y~n >_-0) = i

(n > I).

We define a Galton-Watson process

y~n(X)= y~n_l(X)+ S~y~n_l(x)(X)

(n >_-0).

c (n) = 2~+I

{Yn}n= 0

(n ->-i, x E F0).

We put

fO/2 { Z

k s b~n)(t(e))}cos Od@

(n >__l, 0 ~ <~ < i),

k=l where b(kn)(6) = Prob(Y~n = k), t(8)

= ~

tan e

(0 ~ e < arctan i)

Itane- 2j I

(arctan (2j-l) i e < arctan(2j+l), j ~ i).

First we show Lemma 3.11.

ca(n ) _-
(n ~ i, 0 < ~ S 1).

The proof of this lemma is standard. {bk(J)(6)}k=O

is defined by

(3.28)

The generating function of Then

P~(x) = Zk= 0 b(J)(~) x k.

P~ (x) = P~j-I( ~2 + ( i - 6)x + ~2 x2 ).

In effect, (3.27) shows that, for any

k_a0, w

(3.29)

b(J)(~) = Prob(y~ = k) = =

where

Z(~)

Z b~(j-l) (6) ~=0

xk-eoeffieient of P~_I ( 3 +

P~(x) is J ( i - 6)x + ~ x 2 )

(3.28) holds.

Z Prob(y~_ I J =g) Prob(S p~ = k - g) g=O

z(~)

~'

Let

(~)h(1-~)~2(-~)~3,

el! g2 ! g3 !

is the summation taken over all triples

non-negative integers such that

by

(~i' g2' g3 ) of

~1 + g2 + g3 = g ' g2 + 2 g3 = k.

b~J)(x)

and the

The

xk-coefficient of

is equal to the last quantity in (3.29).

Thus

107

vj(~) = I 1 (i - x) -a 8-- P~ j(x) dx ~j(~) ~x where

(

n

{~j ~)}j=0

(0 ~ j ~ n)

are inductively defined by

~j(~) = ~ +(i-9)~j_l(X)+

40(~) = 0,

~ ~j_l(X)2

Since

fo1 a_

(i - x)-a(k x k-l) dx a Const

k~

i - ~

1 fl-(i/k) (i - x)-~(k x k-l) dx

(k ~_ 2),

we have Z ks bln)(~) ~ Const (i - a) v0(~). k=l Since

(l-~)-a = {i- (~ + ( 1 - ~ ) x +

~x 2)}-=

(o=~i),

(3.28) shows that vj(~) =

f~j(~) ( i - x)-~{(l-~) + ~x} ~

P~n_j_l(~2+(l-~)x + ~2 x2)dx

+ ~ ~-°~(~-~+ ~

~l ~l-(~ + ~-~ ~j (~)

~n-j-l~2~ +(~-~+ ~ d ~

= Vj+l(~), and

hence v0(~) ~ Vl(~) ~

...

~

1 = f~n(~ ) (i- x)-a ~%We have easily

Vn(~) p~0(x)dx = i i _

I1 - ~n(~) I -<- Const/( ~ n).

(i - ~n(~)) I-~.

Thus

Z k a b(kn)(~) <= Const/(~n) l-a k=l Consequently,

we have, with

e. = arctan j J

(j ~ 0),

108

c (n) _-
f 0/2 t(@) ~-I cos6

Const n cr-I { I081 (tan Q)a-lcos

Const n ~-I {

II

"

i

Z j=l

This completes Let

....

@ d@ +

" @2j+l j fl I 82j-I

Itan 8- 2j

I~-1

cos @ d@}

Y dy (i + y2) 3/2

0

+

d@

y~-i

I_ I

3/2 dy

~

Const/(~ nl-~).

{i + (y + 2j) 2}

the proof of Lemma 3.11.

F'

and

(the degree of

F

be two cranks such that

F') ~ i.

n = (the degree of

If there exist an n-tuple

(A I, ..., A n )

r)

-

of cranks

such that r' = A I

[

A2

[

(0-1; ") for some

(0`i' .... ~ ) '

n-tuple

integers, we write

---

F' [[ F

t

Cry(F)

=

= {q(~) }k~l g

For

crank

of degree

F* n (3.30)

where Proof. we put

of positive

and

,

i _-< ~ < n} .

F ,

0 ~ g ~ i, 0 ~ ~ ~ i, n

gO ~ 1

(0 ~ ~ ~ i). and

n ~ i,

there exists a

such that

r'n)=> g0'

~(F0'

r

of finite sequences

2~ I_~~ {I~ NF(r,0)~ dr} dG

Lemma 3.12.

An =

(Qn; ")

~(F',F) = gin {q~);{ 1 s k -< g Note that, for any crank

[

(0`2; ")

ICr~(r*n)-

C~ (n) l < C ,

F 0 = [0,i]. For a finite increasing sequence gap(~

0` = {qk}gk=l

= min {qk - qk-l; 1 =< k =< g} ,

~n = (0`1. . . . .

h),

~

= {qi~) }k~l

where

of positive integers, qo

= O.

Let

be an n-tuple of finite increasing

sequences of positive integers such that q~) ~i = i,

we put

~

= 2

gap(~n) = m i n { g a p ( ~ ) ;

q(~) + 2 2

q~) + ...+ 2

i ~ ~ ~ n} .

With

~-i

~n'

(2 --< ~ --< n),

we associate

n

cranks

109

r~ *

=

F(Q 1 . . . . .

Q)

*

(i)

(1 & ~ & n) ,0)

r I = FI(QI) = r0(q 1

has

gk

Let

Suppose that

components

r _i = Uk~

expressed in the form {ji~-1))~-i "k=l ;

( F 0 = [0,i]).

rk

been defined so that

as follows.

j ~1)

rl, r2, ..., r _ I

(1 N k N g - l ) .

Then

have

F _1

is

with its components

these are ordered so that the x-coordinates of their midpoints

are increasing. We put

,

,

r = r(%

. . . . .

%) =

g~-i

ji~_l) (qi~_l),o) "

u

k=l The set with

rn(~n)

is a crank of degree

0j = arctan j,

0 ~ 0j < ~/2

2-~ Cr~(rn*(@n)) "

Zj=l

( =

{I O.J 0j -i

" f-n{fo

I0

+I

n.

We now s~udy

Cr~(rn(~n)).

We have,

(j ~ 0),

Nr*(~n)n

(r,O) ~ dr} de

-eJ-i 70 } + i

s]

-Oj

Z {dj(~n ) + d j(~n )} j=l.

+ do(~n) , say).

i

tan 0 4

1-tan 0 tan 0 4 4

1-tan 0 4

L(x cos e,e)

For

0 < 6 < @i'

we put

Ix ~ r0;

~0)(@)

~,

= 0,

(x c o s e , e )

r(Q I . . . . . 0.~)

= kl

(k_>- 0, i < ~ =<- n).

110

Then

bk(~)(e) = 0

(k>= 2~ + 1 , 0 ~_ ~ ~ n).

We have

b~0)(e) = 1 = Prob(y0an 8 = i)

Let

bk(1)(e) = Prob(ylan @ = k)

( 0 --< k ~- 2).

2 ~ ~ ~ n.

V

1 ~ j ~ 2~-I,

there correspond

j

To a component

of

{x E F0; N ,

(x cosS, e)=j},

components

j(~-l) , ..-, j(~-l) of F* F~-I vI vj ~-I L(x cos 8,8) for all x E V; these are ordered so that the

which intersect with

x-coordinates of their midpoints are increasing.

If

(~)

is sufficiently

%1

large, then

Ix ~ V; # {j(~-l) ((~) -

,0) fl L ( x cos e , e ) }

=

k

%1.

~i

IV I Prob(Slan 8 = k _ l ) 1

is sufficiently small for all

0 =< k--< 2.

If

(~)

q(~) -

%l' ~2

(~)

%1

are

sufficiently large, then

-IvI

Prob(S~ an e = k - 2 )

I

is sufficiently small for all

0 ~ k & 4.

if

qVj

~

' ~V 2

- qv 1

.....

Repeating this argument, we see that,

- q(~)

are sufficiently large, then

~-i

.stanj8 = k-j)l llxv; {rnL<xcos0,e)=kl-lvIProb is sufficiently small for all

0 ~ k ~ 2j.

Hence, if

gap(Q)

is sufficiently

)

large, then , (x cos @,e) llX 6 F0; N F ~-i

j, N , F

(x cos

8,8)

kl

- bj~(~-l)(8) Prob(S tan3 8 = k - J)l is sufficiently small for all

0 -_< k =< 2j.

If

gap(Q)

is sufficiently large,

then I x ~ F0; NF, ~-I

(x cos e,e) = 0,

N , (x cos 8,e) >= i I F

111

( = JJx ~ £0; N , (x cos 8,8) = 0, F ~-i _ ~-i)(8

) Prob(s;an 0 >__ l) j)

is sufficiently small. if

gap(%) ~ p (@),

Jlx ~ F0;

Thus there exists a positive integer

, (x cos O,O) F_ 1

0 _~ k -~ 2j, 0 -~ j =< 2g-l. 2~-i z j=O

~tbk~)(e) _ K

p~(e)

such that,

then j,

k I

N , (x cos @,O) F

b(~-l)(0). Prob(S tan. @ = k - j)J 3 3

-

for all

N

N , (x cos 8,0) >_ 1 J £

=< g 2-3n3

This yields that

~(~-i) (e).

erob(stan O = k - J)Jl

]

I

3

2~-I j Z {ix E r0; N , (x cos 8,0) = j, N , (x cos 8,0) = kJ j=0 £ ~-i r _ ~-i)(%)

Prob(s~an 8 = k - j)}J

=< s(2 g-I + i)2 -3n3 -_< Const Put

p(O) = max{p (8); 1 _~ ~ =< n} . 2n Z k=0

If

Z

k=0

2 n-I J z j 0

(0 _<- k -<_ 2~).

gap(~ n) ->- p(8),

Ibk(n)(0) - bk(n)(tan 0) I = 2n

<=

s 2-2n3

then

2n E Ib(n)(o) - Prob(Yntan 8 = k) t k=0

~!n-l)(e) Prob(S.tan O = k - j )

2n-i Z Prob(Yn_itan8 = j)Prob(sjan 8 = k-j)

+ Const ~ (2 n + i)2 -2n3

j=0 2n-i _-< (2n+l)

~_...=<

~(n-l) _ , . tan 0 =j) Jbj (8) - wroD[Yn_ 1

Z j=0 n H ~=2

(2 ~ + i) n

+

Const ~=2

2 Z j =0

(2 ~ + i) .

+ Const 6(2n+I) 2-2n3

,. tan ~!l)(o) _ ~r°DiYl 3

. n +. i)} . (2

2_2n3

e

= j)j

3 < Const ~ 2-n ,

112 which

shows

that

If~ NF~(~n) (r,@) a

dr

-

Z

k ~ " (n)(tan @)cos

k=l

@ I

bk

2n

Iro"

N , (x cos @,9) g COS @

dx

kg

Z

F

b~n)(tan"e)eose

I

k=l n

2n = I

Z k=l

k~

~k~(n)(@)

- b~n)(tan

@)}

cos @I

3 =< There

Const

exists

s 2 ~n 2-n

a positive

~

integer

F = {0 < @ < @i; P(@) > Pl }

Idl(Qn)- fO01 @i If0

Const Pl

~ .

such that the measure s 2 -n"

is less than

Z" k=l

k ~ b. k(n)(tan

{/O N , (r,@) ~ dr - % F k=l

If

of

gap(~n)

then

~- PI'

@) cos @ d@ I

k s bk(n) (tan @) cos @} d@ I

n

l{f F + /(0,01)_ F} { Const For an integer integer

pj

2 ~n IFI + Const

s

j # I, we can choose,

such that,

if

8_1 = 0.

&

Const

that if

as above,

a positive

then

Z k ~ " (n)(t(@)) k=l ok

This shows

s .

in the same manner

gap(~ n) ~ pj,

81Jl Idj ( ~ ) - T e l j l _ l

where

} de I

gap(~n)

cos

@ del

< =

s /(i + j2),

is sufficiently

large,

then

(3.30)

holds. §3.7.

Q.E.D. Proof

of Theorem

F ([34])

Here are three lemmas necessary 0<

~<

Lemma

for the proof.

From now on, we fix

1/2. 3.13.

For

and a non-negative component

of

F n,

t O => 1 function

and

n >= i, w

n

on

there exist a crank F

n

such that

w

n

Fn

of degree

is a constant

n

on each

113

~(r0, r[) ~ ~0' cr(r[) ~ ClI(~ n~-~), (3.31) llWnll i * L (Fn)

I,

llWnll~ ~ L (r[)

where

C1

Proof.

Leamms 3 . 1 1 a n d 3 . 1 2 show t h a t

satisfying

C I, film H , w~lle.(F~ ) ~ Cn~'n~, Fn

is an absolute constant.

the first

there

two i n e q u a l i t i e s

in

exists

(3.31).

II~rn,

~

ConstV~-,

tlHFn,

~

Const V~.

IIL2(Fn),L2(Fn)

a crank

F

of degree n n ( 3 , 1 3 ) shows t h a t

Inequality

which yields

fiLl(m) ,L~(Fn )

Thus, in the same manner as in the proof of Theorem D, we obtain a non-negative f u n c t i o n on satisfying F* n mean o v e r e a c h c o m p o n e n t o f Lemma 3 . 1 4 .

and

w

Let

~0 ~ 1

and

three

m

~j = 0

such that

Fm.

n ~ 1.

in

the required

Let

F

m

(3.31).

function

Taking the w • n

be a c r a n k o f t y p e

rm

such that

Then there exists a crank

(ra+l N j N ~ - n )

Wm+ n

inequalities

we o b t a i n

be a non-negative function on

on each component of with

the last F , n

and a n o n - n e g a t i v e

mw

function

{ 6 j } ~0=_

is a constant

Fm+ n

is a constant on each component of

Q .E.D ,

•

of type wm+n

.m+n

16j~j= 0

on

Fm+n

Fm+ n,

rm [[ rm+n' c(rm' Fm+n) >= gO' Cr (Fm+n) -<- C 1 IFml/(~ n l-a ),

IlWm+nllL 1

(3.32) tl~m

=

(rm+n)

Hr

m+n where

C1

IlWml]L1 "

,

IlWm+nll

(Fm)

~ lllm

win+nil L'(Fm+n)

~ C 1 llWmll . L'(Fm+n)

L

HFmWmllL®(Fm)+ C2g-~ llwmll

is the constant in Lemma 3.13 and

C2

(rm)

'

. L (Fm)

is an absolute constant.

Proof. We can write g I Jk with its components {Jk}k= I. rm = Uk= g be the left endpoints of {j k}k=l, respectively. We put {Zk}k= 1

Let

,

114

g Fro+n =

U k=l

A k,

A k = [IJkIF n + Zk],

Wm+n(Z ) = Wn*((Z-Zk)/IJkl) Wm(Zk) where

F

n respectively.

and

w

n

Then

i -<_ k =< ~),

are the crank and the function in Lemma 3.13, m+n Fm+ n is a crank of type {~j }j=0 such that

F m [[ Fm+n, L(Fm, Fm+n) m gO"

Cr (rm+ n) -~

(z E A k,

The second inequality in (3.31) shows that

Z Cr (Ak) = k=l

Z IJkl Cr (rn*) --< C l Irml/(a nl-~). k=l Q.E.D.

In the same manner as in Proposition 3.10, we have (3.32). In the same manner, we have Lemma 3,15.

Let

g0 ~ i, n ~ i, rm

m {6j}j= 0 , w m

be a crank of type

be a

non-negative function on Fm such that w m is a constant on each component m+n of Fm, and let {~j}j=m+l be non-negative numbers less than or equal to m+n i/I00, Then there exist a crank of type {~j}j=0 and a non-negative function of

Wm+ n

Fm+n'

on

Fm+ n

Fm [[ Fm+n'

such that

Wm+ n

is a constant on each component

~(Fm' Fm+n) ~ g0'

llWm+nIILl(rm+n)

llWmllLl (Fm) ' m+n

llWm+nIIe.(Fm+n)

IIIm HFm+n

(I + ~

=<-IIWmI[e,(Fm) / #=m+l H

Wm+nll ~ -~ llIm H r Wmll ~ L (Fm+ n) m L (Fm) m+n

+ Elwll .

j

z

L (Fm)

{i/

j=m+l

~

and a positive integer

~0(i - e) > i.

Let

define a sequence

Pm

mk+ I = non~ + and define a sequence

E.

Choose a positive number

~0/2 < i - (i/n 0 ) and ~m be the integral part of (i01/i00) 0 (m >_- I).

{mk}k= 0

n o ->_ 2

(I+%)}

~=m+l

We now construct the required compact set ~0

),

so that

of non-negative integers by

Pn0mk

{~j}~=0

We

m 0 = 0, m I = n o

(k ~ i), of non-negatlve numbers by

115

Cj

Let

[ = {~k}k=i

(O < j -<- m I)

i/i00

(mk <

0

(n0n k < j =< ink+l, k>= i).

Using Lemma 3.14 with rml

rml,

and a non-negative function and

Wn0ml,

gO = ~i' n = m l,

and a non-negative functlOnn m

gO = &!' n = (nO - l)ml,

rn0ml

j _~ nomk) k->_ i)

be an increasing sequence of positive integers which will be

determined later. obtain a crank

0

wml Wn0ml

and

0 1 {¢j}j=ml+l,

we obtain a crank

rm2

{wmk}k=l

and

w 0 m i,

we obtain a crank t0

=

~2

)

n

rnoml = Pn0ml,

and a non-negative function {Fmk}~=l

we

Using Lemma 3.15 with

Using Lemma 3.14 with

•

Repeating this argument, we obtain a sequence sequence

wml"

F0

Wm2.

of cranks and a

of non-negative functions such that, for

k a 2,

r~ [[ r~+l, ~(r~, r~+ l) >-_~k' llwk IlLl(r~) IIIm HF~

i,

]i®

ilwmk L

wmk IiL'(r~)

k-i +

co / n (rk) -< ~=0

(]_ + ¢~),

,

=< C0 V~I

nomv Z j =mv+l

v CO

Z v=l

nomk, 1

k

=

k-i

j 11

{L/

(i + ¢ ) }

FL=O

+

nomv

v+lDv-6-----. ~

Z C0 ~=i

Pnomv/

(l

+

~=i

n0mk_ I Cra(r~) =< C O

where

£~ = rmk

and show that

m

nomv_ I

and

H ~=0

We put

C O = max{Cl, C2}.

T(E([)) ~ Const.

(v ~ 2)

(I + 4~ )/(~ PnOmk 1 ),

and

Let

m I = no,

k $ 2.

we have

Then mv

n0mk_ 1

Ilwm k I1. , L (rk) k =

Since

C0 (

~

k/ n (i + ¢9 c0 ~= mk-l+l

I01 )-(no-l)mk-i ~-~ ~

Const.

E([) = Nj= 1

uLj

Ilwmkllel(r~) = i.

Since

v no (v ~ l), and hence

¢~),

116

nom ~ ~Pnom---~/

n

.lOl.~onomv/2 uonst(l--~ )

(i + ~)

.i01 -(no-l)mv ~)

~=0 .i01. -{(l-(i/no))-(~O/2)}nomv Const (I--~)

(v >= i),

we have IIIm HF~ wmkllL®(F ~) ~ Const.

Thus, in the same manner as in (3.26), we obtain an analytic function

fk ~ H~(FI c)

such that

llfkllH~ ~ i, Since

k ~ 2

lfi(~)l

e

Const.

is arbitrary, we obtain an analytic function

f (H~(E([) c)

such that llfll _<- l, H~ which shows that

If'(-)l ->_ Const,

Y(E([)) ~ Const.

Let .101.(1-(l-a)~o)n0mk ~i--OO)

gk = (C0/~) Then

limk ~ ® gk = 0 Cr (F~)

(k ~ 2).

and i01 n0mk-i

(2c0/~)(i~)

.101.-(l-~)~0n0mk-i

~

2 gk_ 1 . We can inductively choose

~0 = {gk}k= I 0®

Cr(i/k)(E([0))~ ~ 2 gk-l'

which shows that

satisfies

Y(E) > 0

Remark 3.16.

and

Throughout

so that, for any Cr (E([0)) = 0.

the note, we use Theorem D to estimate

below.

Here is a weaker inequality than Theorem D. Then

(cf. [29, p. 19]).

I 7r f

dzI2/{Irfll 2(F)

E = E([ O)

Let

F

Y(')

from

be a locally chord-arc

+ llHr(f dz/Idzl)[l 2(r)

This is also useful to estimate

we can deduce (3.23) and

Thus

Cr (E) = 0.

curve.

Y(F) => Const

k ~ 2,

{IIC[a]I12,2; a ( ereal}

= ~

Y(r)

from below.

In effect,

from this inequality.

APPENDIX

For

I.

A N EXTREMAL PROBLEM

s I . . . . , s n E ~, w e define

Ts I,

..., s n

(x,y) = i/{(x-y)

+ i(Asl '

..., s n

(x) - A

s I'

--., s n

(y)},

where 0

A

x ~ I0 :

[0,i)

(x) s I, ..., s n

k-i k ( - -n =< x < -n'

Sk

1 =< k =< n).

Put

(4.1)

ex (n) = max {~(Tsl ' ..., s ); Sl'

.... s n E ~ }

•

n

(See (1.22).) Theorem G.

We show

Const

~TOg(n+l)

The first inequality a positive E c U~=_

integer

n,

[I 0 + ik/n],

to

I0

are mutually

by

x + iAE(X ) E E

the projection

of

TE(X,y)

~ ex (n) ~ C o n s t ~ g ( n + l ) is shown in §3.4.

F n E

denotes

For

(x E pr(E)) to

I O.

Let

such that

Proof.

and

T E''

and their projections

we define a function

AE(X)

= 0

AE(X)

(x ~ pr(E)), where

on

pr(E)

is

E E Fn on

for the proof.

and let WI

W I, W 2

and

be two disjoint

AE(X) ~ 0

f W 1 IrE(XW2f)(x)I 2 dx ~ Const

We define an operator

g Let

E E Fn,

For

such that

= i/{(x-y) + i(AE(X ) - AE(Y)) } .

AE(X) ~ 0

(4.2)

E c ~

We define a kernel by

Here are three lemmas necessary Lemma 4.1.

W e prove the second inequality.

the totality of sets

has a finite number of components

disjoint,

E

(n ~ i).

T E'

~ f_~ g(y)/{(x-y)

denote the adjoint

on

W 2.

Then,

IIXw2fiI~ .

by

- i A(y)} dy.

operator of

T E'

subsets of

Then we have

for any

pr(E) f E L2 ,

118

I T"g(x)I

< H*g(x) + Const Mg(x), nT~ llp ,P =< C P

which shows that

AE(X) - AE(Y) ~ AE(X) ~ 0

(p > i).

Hence

(x 6 W I, y 6 W2),

lIT~Ilp,p =< C P

we have,

(p > i).

Since

in the same manner as in

the proof of (2.9), I TE(Xw2 f)(x) I =< Const {M(T~f)(x) (x £ Wl),

+ IIrE[14/3,4/3' M(I Xw2fl 4/3)(x)3/4} which gives

(4.2).

Put 1

co

~(n) = sup {T'/UT-6TT'Ip~tLjl ~(E,f);

E E u~-' f 6 Lreal, 0 ! f <- i},

where ~(E,f) Lemma 4.2. Proof.

= /pr(E)

For any

For

n ~ I,

E £ FI,

(D = O, ±i,

...).

ITE(Xpr(E) f)(x)I2 ~(n) < ~.

we put

Then,

f(x)dx.

G = pr(E)

for any

'

G' = pr(E N {Im z = ~})

f 6 Lreal, 0 ! f ! I,

f(Y) dy - Z i I TE(XG f)(x) - fG~ x - y ~=-~ 1 + i(k-~) /G'~ f(Y) dyl Const

Z ~=-~ (k-~) 2 + 1

(x 6 G~, k = 0, ±i,

...).

Hence ~(E,f) ~ Const ( Z

fG

+

i

z

k=-~

IG

Const which shows that

llz

IH(x G f)(x)i 2 dx fG' f(y)dy] 2 +

~ {IGkl + IfG, f(y)dyl 2 k=-~

+

1 G~I }

E E Fn, f ~ Lreal,

0 ! f $ I,

we put

E k = {z - k; z 6 [n El, k ~ Re z < k+l}, ~.x + k. fk(x) = ~[--~---) Then

--< C o n s t

z U=- °° (k-~) 2 + 1

I G1,

~(i) ~ Const. co

For

z k =-oo

~=-~

(0 <_ k _< n-I).

G' = [n pr(E)],

}

119

~(E,f) = _111fG' I fG' n 1 { ~

<= n

2 f(~) dx f(y/n) ,d~ (x - y) + i(n AE(--Xn) - n AE(Y))

~(Ek,f k) + C n }.

k=l co

Since

E k E FI' fk E Lreal, 0 < f < i, The following Lemma is analogous

For

E E F22 n,

F =

U

F E F

(the projection of

.

(n ~ i).

we define

to the line Then

Q .E .D.

~(n) < co.

to Lemma 3.4.

~(22n) ~ 2 ~(2 n) + Const ~(22n) I/2

Lemma 4,3. Proof.

we have

E N {(k-l)2 -n ~ Im z < k 2 -n}

Im z = (k-l)2-n).

Let

2n G = pr(E), G. = pr(E [~ {(j-l)2 -n _< Re z < j 2-n}), j Gj, k = pr(E n {(j-l)2 -n _<_ Re z < j 2 -n, (k-l)2 -n < I m (j = 1 . . . . . We have, for

2 n, k = O, +i,

f 6 Lreal, 0 ~ f $ i,

~(E,f) = ~(F,f) + I G (T E - TF)(XGf)(x ) TE(XGf)(x)

f(x)dx

+ I G TF(XGf)(x)

f(x)dx

= ~(F,f)

(T E - TF)(XGf)(x)

+ L (I) + L (2)

and 2n L (I) = j=l 2n +Z

j=i

z < k 2-n})

IG (T E - TF)(XG f)(x) TE(XG f)(x) f(x)dx J 3 ] fG (TE - TF)(XG f)(x) TE(XG_G f)(x ) f(x)dx J J J

...).

120

2n Z j=l

IG_G. (TE -TF)(XG.f)(x) TE(XGf)(x) f(x)dx J J

L I + L 2 + L 3. For

i <_-j _<- 2n,

such that

there exists

E!j ( F2n, F'j E FI

and

f.3 E L "real, 0 =< fj =< 1

IEli = IF]I = 2niGjl,

IG" ITE(XG. f) (x) 12 f(x)dx = 2-n <(E],fj) J J fG. ITF(XG. f)(x) 12 f(x)dx = 2-n ~(F],fj). ] J Hence 2n _-< Z j=l

ILII

+

2n z j=l

IG ITE(XG.f)(x)I 2 f(x)dx j j

{IG. ITF(XG f)(x)i2f(X)dx}i/2{IG.ITE(XG.f)(x)i2f(x) dx} I/2 J J J J 2n % ¢(E',fj)j + j =i

2-n

2n Z j =i

<(2 n)

2-n

2n Z j =I

<(F~,fj) I/2 <(E],fj) I/2

IgjI + ~(i) I/2 ~(2n) I/2

2n Z j =i

IGjI

= IGI {~(2 n) + ~(i) I/2 <(2n) I/2} .

Since

F2n c F22n,

we have

<(2 n) ~ <(22n).

Thus Lemma 4.2 yields that

}LII ~ IGI {~(2 n) + Const <(22n) I/2} . Let

xj = (j-l)2 -n

Then, for any

and

Gj = Gj_ 1 U Gj U Gj+ I

(i ~ j ~ 2n),

x E G - G., J

I (T E -

2-n T F)(X Gjf)(X) I --< Const fG.3 (x_y)2 + 2-2n 2-niGj I _-< Const

For any

where

g E L 2, we have

(x-xj)2 + 2-2n

-.

f(y)dy

G O =~.

121

I~Z

2n Z j=l

2-nlGjl g(x)dxl (x-xj) 2 + 2-2n 2n ~ fG. j=l 3

Const

2-n {f ~ -

(x-Y)2 +

Const ~G M g(x) dx ~ Const ~

Ig(x) Idx} dy

2-2n llgll2 ,

which shows that 2n ii Z j=l

2-nl G4 j a 2_2n. iI2 _<_Const ~ (--xj)2 +

.

Thus 2n IL31-_<

{7G - ~ j

Z j=l

~ } I(TE_TF)(XG f)(X)TE(XGf)(x)l f(x)dx fGj-Gj j

+

2n

<= Const

j=l

2-nlcj j (x-xj) 2 + 2-2n

7G-Gj

iTE(XGf)(x)if(x)dx

2n +

2

%

+

2

2n % j=l

"

2n { Z j=l

-<_ Const fG

{fG.-G. 3 3

=< Const ~ ( E , f ) _-< Const ~ V ~

dy_ ) ITE(XGf)(x)I f(x)dx Ix-Y[

(~G3

~G.-G. J J

j=l

2-nlGj I ITE(XGf) (x) If (x)dx 2 (x-xj) + 2-2n-}

(fG

--~ J

I/2 + Const

)2dx}i/2 {fG,ITE(Xgf)(x)12 f(x)dx}i/2 3 2n E ~IGjl { ~ ITm(×Gf)(x)12f( x)dx}I/2 j=l 3

~(E,f) I/2 + Const ~

~(E,f) I/2

_<- Const IGI ~(22n) I/2 Let

G.3 ,k

= Gj

U

,k-i

Gj

,k

U Cj,k+ I

(j = 1 .... , 2n, k = 0, -+i, ...).

Then 2I% IL21 =< 2n + IZ j=l

I E

j=i

Z k=-~ fGj ,k

(TE-TF)(X G

f)(x) TE(X G ~ f)(x) f(x)dxl j ,k - J

f)(x) TE(XG_~ f)(x) f(x)dxl Z ~gj (TE-TF)(XG. U gj k=-" ,k 3 ,k-i ,k+l 3

122

2n IS j=l

Z fG (TE-TF)(XG _~" f)(x) TE(XG_~ f)(x) f(x)dx I k=-- j,k J ],k J

2n lj=iS /Gj (TE-TF)(XGjf)(x) TE(X~ f _G ) (Jx ) j

f(x)dx I

L21 + L22 + L23 + L24° We have 2n IL241 =< 2 Z j=l

{7G. I(TE-TF)(XG f)(x)l 2 f(xldx} I/2 3 3 dy

× {fg. (fG.-G. 3 J J Note that

)2 dx}i/2 ~ Const IGI ~(22n) I/2.

~(2 2n) ->_~(i) > Const.

Since

I(TE - TF)(XG _~" f)(x)l J J,k

_-< Const

" S g=-~

2nlGj ,~I 2 (k-u) + i

=< Const

(x ~ Gj ,k),

we have IL23 I ~

=

2n S j=i

Const

Const

, S ~G k=-j ,k

ITE(XG_~ f)(x) I f(x)dx j

2n E fg. ITE {(XG - XG.-G. - XG.)f}(x)l f(x)dx k=I J 3 J J

=< Const Igl

{~(22n) I/2 + i} -<_Const IGl ~(22n) I/2.

Lemma 4.1 shows that 2n IL221

~

{ E j=l

E fGj I(TE-TF)(XG UGj k=-,k j,k-i ,k+l

2n x { Z

Const

f)(x)l 2 dx} I/2

fG. ITE(XG-G. f)(x)12 f(x)dx}i/2

j=l

j

2n { Z j=l

S IGj U Gj f(x)2 dx}i/2 k=-,k-i ,k+l

× {~(E,f) +

2n Z j=l

j

f~, ITE(X~ f)(x)12f(x)dx} I/2 3 3

123

Const IGI ~(22n) I/2. Since

(TE-TF)(x,y)

fGj ,k For

is anti-symmetric, we have

(TE-TF) (XG

f) (x)f (x)dx = 0. j ,k

Gj, k # ~, we choose a point xj, k on G.j,k. Then IL211 = I Z (TE-TF)(XC. f)(x) j,k;Gj,k# ~ fGj,k j,k x {TE(XG_~j f) (x) - TE(XG_~.jf)(xj,k)} f(x)dx Const

Z /Gj,k I(TE-TF)(XG f)(x)If(x)dx j,k;Gj,k# ~ j,k

Const

IGl ~(22n) I/2.

Consequently, IL(1) I ~ Igl {~(2n) + Const ~(22n) I/2} . In the same manner, IL(2) I ~

2n j=l% 7GJ ITF(XGjf)(X)TE(XGjf)(x)If(x)dx + Const IGI ~(22n) I/2

Since the first quantity in the left hand side of the above inequality is dominated by Const IGI ~(22n) I/2, we obtain IL(2) l =< Const IGI ~(22n) I/2 Thus ~(E,f) ~ ~(F,f) + IGI

IGI {~(2n) + Const ~(22n) I/2}

{2 ~(2n) + Const ~(22n) I/2} ,

which yields the required inequality.

Q.E.D.

We now prove the second inequality in Theorem G. Lemmas 4.2 and 4.3 show that n ~(22n-i) n ~(22 ) =< 2 + Const ~(22 )1/2 2n

~(220)

Const

n-i 2k 2n-k 1/2 E ~(2 ) k=O n-i 2k ~(22n-k) i/2 } {2n + 7 k=O + Const

124

which yields that

~(22n) < Const 2 n.

(See the proof of (2.43).)

Let n

denote the integer satisfying

2 n

-<_ n < 2 n.

Then

n ~(2 n) =< ~(22

) =< Const 2 n _-< Const n. -g

For

E E Fn

and

f E Lreal, 0 ~ f ~ i,

g(x) = f(x 2 n/n).

Then

we put

F E F

F = [n2

n E]

and

Hence we have 2gn "

~(E,f)

-2- n ~(F,g) ~ Const

=

IF1 <(2 gn)

n

Const

IEI gn ~ Const

IEI log(n+l),

which gives that (4.3) For

~(n) ~ Const log(n+l)

s I, ..., sn E ~

(ng

i).

we put n

^

E(s I . . . . .

s n) =

U

k-1

{x + Sk, -n -

k

=< x < n },

k=i

where

~k = (the integral part of

I c I0

and

Then we have, for an interval

f E Lreal, 0 ~ f ~ i,

~(I, Tsl . . . . . Hence

nsk)/n.

Sn ,f) ~

(4.3) shows that

Const

{3(1, TE(~I ' ... , Sn ), f) + l l I }

$(I, T

,f) ~ Const

[I[ log(n+l).

• Since

s I, ..., s n Tsl . . . . .

sn(X,y) = i/(x-y)

(x, y ~ I0) ,

this inequality gives

^

~(Tsl ' .--, Sn) ~ Const log(n+l).

Consequently,

) ~ Const $(T °(Tsl, Since

..., s n

s I, ..., s n E ~

)1/2 ~ C o n s t ~ l o g ( n + l ) . s I, ..., s n

are arbitrary,

the second inequality in Theorem G holds.

This completes the proof of Theorem G. Let

BMO(F)

denote the Banach space of functions

f

on a finite union

F

of segments, modulo constants, with norm IIfIIBMo(£) = sup ~ £

1 N D(z,2r) I-

/£ N D(z,r)

If(z) - (f)r N D(z,r) IidzI

is the mean of f over r n D(z,r) where (f)r N D(z,r) and the supremum is taken over all z E ¢, r > O° Put

with respect to

'

Idzl

125

r

Sl' .... Sn

= {(x, A

Corollary 4.4.

Sl' .-., sn(X)); x E

II Sl, ..., s n

; s I, ..., s n ~ ~ } L (FSl '

), BM0(Fsl ' °°'~

=< Const ~log(n+l)

BM0(F).

Theorem G immediately yields

.

Const'~og(n+l)

max{IIH r

where

I 0)

IIHpII

SN

) °''~

S n

(n ~ i) ,

, L (F),BM0(F)

is the norm of

HF

as an operator from

L'(F)

to

APPENDIX II.

PROOF OF THEOREM B BY

P. W. JONES-$. SEMMES

Quite recently, P. w. Jones-S.Semmes gave a proof of Theorem B by complex variable methods.

The following note is their lecture in M~y, 1987.

(The author

expresses his thanks to P. W. Jones-S.Semmes who permitted the author to write here their proof (cf. P.W. Jones [33]). Lemma 4.5.

Let

F = {x + iA(x); x ( ~}

= {z 6 ~; Im z > A(Re z)} . extension, say simply

(4.4)

where

da

Here is a fact obtained by C. Kenig. be a Lipschitz graph and

Then, for any

g(z) (z ( ~ ) ,

to

g (L2(F)

having an analytic

~ ,

IIglIL2(F) ~ CM {ifzig'(z)I 2 dis(z,F)d~(z)} I/2,

is the area element and

CM

is a constant depending only on

M = IA'II. For

z ( ~,

z * = z - 2i(Im z - A(Re z)).

we write

Cf(z) = C(f d~IF)(z) (z £ ~) 1 2hi

~f(z) = For at

z ( F, z.

f(~) ~ - z

IF

we write by

~f(z)

z 6 F U ~,

we have

For

~f(z) = - i ~ 1 2~i

Cf(~)

(~ ~ ~ )

(Cf)"(z + i t ) t dt d~} dt

(Cf)'(w) --- - {Im w - A(Re w)} {~ + iA'(Re w)} do(w) (z - w*) 2

be a sequence of mutually disjoint cubes (with sides parallel to

the coordinate axes) such that 6(Qk )

the nontangentlal limit of

I~ {I F (If)'(< + it/2)t (z+(it/2) - ~)2

2

{Qk}i= 1

d~.

(Cf)i(z + i t ) d r = ~

~-y IIi Let

f 6 L2(F) , we put

For

, i.e.,

is the length of a side of

-

SSQk

~ = Ok=l Qk' Qk

and

dk/CM ~ Z(Qk ) ~ CMdk (k { i), where

d k = diS(Qk,F).

(Cf)'(w)

(= - w*) 2

Then

{Im w - A(Re w)}{~ + iA'(Re w)} do(w),

IfQk Ilm w - A(Re w) I do(w) ~ CM diS(Qk,F) 3

(k ! i).

127

Hence ~ " dk 2 ][CfIIL2(F) --
(4.5)

co

where the supremum is taken over all sequences [Ckl ~ d k, zk E Qk

(k ~ i).

Let

For two sequences function h(~) on

{Ck}k= I, {Zk}k= 1 C by

h(~) = Ck(Cf)'(z k) dk I/2

{Ck}k= 1, {Zk}k= 1 such t h a t denote the 2~dimension maximal operator.

M2

satisfying the above condition, we define a

( ~ E Qk' k ~ i), h(~) = 0

( ~ E ¢ - ~).

Then Lemma 4.5 shows that

"

(4.6)

dk

l[ Z C k ( ~ f ) ' ( z k) ( -- , k=l z - zk

=< CM

{ff~ ]ff¢h(~)] IN

CM

{ff~M2 h(z) 2

IIL2(F)

2 --dk * 3 12 dis(z,F) do(z)} I/2 (z - zk)

{ff~ Ik=El Ck(Cf)'(Zk)

CM

)2

~]I/211m ]z - ~*[3 z~I/2

{ffc lh(z) 12 d~(z)}I/2

d~(z)}i/2 ~ CM

CM { Z" l(~f),(Zk)i 2 a, 3k .)1 / 2 • k=l

CM { "Z ICk(Cf)'(z k) ]2 dk}i/2 ~ k=l Let G Then

denote the totality of sequences

da(~)12 da(z)}i/2

{~k}k= I

such that

k=l ~k~2dk = i.

.3~i/2 [ Z l(Cf)'(z k) 12 ak# k=l

(4.7)

=

2j '

sup {I Z (Cf)'(Zk) ak de k=l

Lemma 4.4 shows that, for any

{~k}k=l E G} .

{~k}k=l E G, 2 dk

(4.8) I k=iZ (Cf)'(Zk) ek dk2 I <

=

i 2~

IIftl

L2(F)

II z

k= I

< CM llf[IL2(r) { ff¢_f~

-

2~i [ f r

f(z)

Z k=l

~k

(Z - zk)

2

dz I

2 dk ek

(z

IkZ=l ~k

-

Zk )2

II L2(F)

2 dk (z - Zk )3

12 dis(z,F) do(z) }1/2

128

CMIlfllc2(r) {k=lZ

I~k 12 dk}l/2

Ilcfll

Thus (4.5)-(4.8> show that Theorem B.

ct~

<

=

~ cM Ilfll L2(F)

IlfllL~(n. ,

which yields

L2(F)

The proof of Lemma 4.5 by P.W.Jones-S.Se~nes is as follows. Put A = Ilgll~2(r), B = ff~ Ig'(z)I 2 dis(z,F) do,z).

Let

to one mapping form the upper half plane

~.

U

to

~(z)

be a conformal one

Then

A = f ~ Ig ° ~(x)1 2 I~'(x) i dx S = flu Ig'° ¢(z) 12 dis(¢(z),r) Koebe's i/4-theorem shows that

flu Since

l¢'(z) l2 do(z)

dis(~(z),F) ~ l~'(z) l(Im z)/4

(z 6 U),

and hence

Ig'° ~(z) 12 l~'(z) I3 y do(z) ~ Const B.

larg ~'(x) I ~ g/2 - (i/M)

(x ~ ~),

Green's formula shows that

A ~ CM [f_~ Ig o ~(x) I2 ~'(x) dx I

= CM [ffu A(Ig {flu

i CM

+ ffu

V E BMO

since

[g'° ~(z) 12 [O'(z) 13 y do(z)

](g'o ¢)(z)(g o

=< CM[B + We can write

o O(z) I2 ~'(z)) y do(z) I

Bl/2{fYuig °

~'(z) = e V(z) Im V ~ L~.

{flu Thus

~(z) 12[~'(z)l ~ (z) 2 y do(z) I/2].

with an analytic function

V(z)

in

U.

Then

Thus

I¢"(z)/¢'(z)l 2 y dy dx is a Carleson measure in

¢)(z)]l¢'(z)[ 2 ]¢"(z)] y do(z)}

U.

( = Iv'(s)I 2 y dx dy )

Since

g o @(z) eV(z)/2

is analytic in

~"(z) 2 }1/2 AI/2. [g ° ~(z) 12]~'(z)I ~ ( z ) y do(z) ~ CM

A g CM(B + BI/2AI/2),

which yields the required inequality.

U,

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1980.

SUBJECT INDEX

Ahlfors function analytic capacity Area integral BMO

Interpolation

80 y(E)

71

John-Nirenberg's inequality

2

locally chord-arc LP( -)

1

Buffon needle probability Bu(E)

21

105

T[a]

68

Lreal 31 L(r,0) i05

105

Calderdn commutator

71

L~(.) 72

1

Calder6n's problem

82

maximal operator

M

34

Calder@n's theorem

1

Mclntosh expression

13

Calderdn-Zygmund decomposition Carleson measure

33

6

C

chord-arc curve

HF

68

9

T-atom

Coifman-Meyer-Stein's theorem

Covering Lemma

32

crank

105

E

1.1

15

tent space

ii

T1 theorem

16

31

Vitushkin's example

Cra(E)

¥+(E)

105

108

I(F',F)

21

~(E,f)

83

o fat crank

99

Galton-Watson process Garabedian function Garnett's example

106

80 71

Good % inequalities

4

Hilbert transform

~(n)

118

p(F)

72 72

o-function

generalized length

Green's formula

118

p+(F)

80

aC(g)

OE(B)

61 52

~E(~) 55

7

integralgeometric quantity

35

61

$C(~,B)

5

105

80

71

8-standard kernel

51

ex (n)

74

T-atomic decomposition

Tn[a]

126

E[a]

ii

Tb theorem

83

Crofton's formula

3

105

separation theorem

Coifman-Meyer expression

17

Prob

Rising Sun Lemma 32

71

68

Cotlar's lemma

105

Poisson kernel

Cauchy-Hilbert transform Cauchy transform

NE(r,6)

o(l,K,f)

25

15

13

34

133

$(I,T,f)

39

@(I,T,f)

39

~(K)

25

~(n)

86

$0(T) 39 ~(T) 39 @(T)

39

3(T ;~-~) 39 ~0(n) 91 @(n) 86 co6(K)

16