Fourier Analysis on Local Fields. (MN-15) 9781400871339

Table of contents :
Cover
Table of Contents
Preface
Introduction
I. Introduction to Local Fields
II. Fourier Analysis on K, the One-Dimension Case
III. Fourier Analysis on Kn
IV. Regularization and the Theory of Regular and Sub-Regular Functions
V. The Littlewood-Paley Function and Some Applications
VI. Multipliers and Singular Integral Operators
VII. Conjugate Systems of Regular Functions and an F. and M. Riesz Theorem
VIII. Almost Everywhere Convergence of Fourier Series
Bibliography

Citation preview

FOURIER ANALYSIS ON LOCAL FIELDS BY M. H.

.~AIBLESON

PRINCETON UNIVERSITY PRESS AND

UNIVERSITY

OF TOKYO PRESS

PRINCETON, NEW JERSEY 1975

Copyright ~ 1975 by Princeton University Press Published by Princeton University Press, Princeton and London

All Rights Reserved L.C. Card: 74-32047 I.S.B.N.: 0-691-08165-4

Library of Congress Cataloging in Publication Data will be found on the last printed page of this book

~aE€~p)

QA2W'l

:12-~ \ q"lS c,,3 Published in Japan exclusively by University of Tokyo Press in other parts of the world by Princeton University Press

Printed in the United States of America

For Charlotte

FOVRI£R ANALYSIS ON LOCAL FIELDS

By M. H. Taibleson

Preface

These are tht lecture notes of a course given at Washington UniveLsity,

~aint

Louis during the Fall and Spring semesters 1972-73.

With the exception of the results on Fourier series in II §6, all the results described have appeared in a series of papers, some of whicb were authored singly, but orations

~hich

mo~tly

they are the result of happy collab-

started with Paul Sally.

Then I

~ent

on to work with

;tichard Hunt, Keith Phillips I yi

and

I x + yl < I xl.

max[lx+yl,lyll
V

and

$

'D .. {x E K:

is the unique maximal ideal in

{x

E K:

Ix! .s

1], f.j*,.. (xEK:

K is ultrametric,

D is the unique

is the group of units :tn ~.

There is apE

Ixl ... l},

~

K* , and

such that

- 10 -

'B .. b l:.

The residue space

characteristic the image of

If

p.

6.

*

is a finite field of

is the number of elements in

q

K* ~ (O,~)

subgroup of (0,00) on

I" D/'T).

under the valuation

generated by

ip I = q-l.

q.

I·! A

then

is the

Haa~ measure

is given by

K

More facts about K

o

A.

- ""

in

K.

converges commutatively. An easy consequence of the ultrametric inequality.

B.

1£!

m-

CaiJi.1 Then if

Proof.

We may assume

the relations

_ '\ N

x =L

be any fixed full set of coset representatives k

x E 'B , k E 2Z, x can be expressed uniquely as

k - O.

ctP

t

Ct are defined inductively by

The

N+l

J

mod(Ti

)•

.t.=O

C.

Let

A maps of

K (as a topological field).

A be an automorphism of D onto

C, 'D .2!!i2

.,.,

and has module

1

Then

as an automorphism

K+.

R!:22f. and so

Clearly the image of A C .. D.

Similarly

(; A~ ..

is a maximal compact subring of '1\.

Now note that

K

- 11 -

D.

A be as in

Let

Proof. so

C

From

l

then

~

--t

Consequently.l is dense in

•

All functions in .I have compact support since each

that property.

T ht

k

Thtk

has

That the functions in .I are continuous follows from

the ultrametric inequality. since

(1.11).)

k ' h E K, k Ell.

compact support that separates points.

.lli!ll.

and that

is the space of finite linear combinations of functions

Proposition (1.3).

Co

h + ~k ~ Thtk

Indeed if

is constant on cosets of

~k.

Th'it .I is an algebra that

separates points is left as a trivial exercise.

The rest of the

proposition follows from the Stone-Weierstrass theorem and the usual density arguments.

Proposition (1.4).

- 24 -

tk(x) ... , tk(S)Xx(S)d~" "K

of a character on

K+

to

~k

V (S)dt;.

•

0

and choose

The Fourier transform can be extended to finite Borel measures

~

as follows:

Remark.

~(x) = J ix(~)d~(S)

(1.1) is valid for

total variation of at the origin then

~),

.

~ E

..

~

but (1.6) fails (if 1).

1

(replace the L norm of ~

f

by the

has mass 1 concentrated

- Z5 -

We define convolution in the usual way. h = f* g

functions, then

If

f

and

g

are

is the function (when the defining integral

exists) : h(x)

=

(f*g)(x) ...

f(x-z)g(z:)dz'" J f(z)g(x-z)dz •

J

The following theorem is included for the sake of completeness. The proof is standard.

Theorem (1. 7) •

For

~,

f E LP , 1

1.!

S

p

S

CD

and

gEL

1

then

g E LP

h

a finite Borel measure, the convolution operator is

extended as follows: (1.1.* f) (x) ... , f(x-z)dj.l.(z)

•

" (1.7) extends to finite Borel measures

Remark. if

It

in the sense that

is a finite Borel measure (with total variation

f E LP , 1

case

~

p =

S P < co then ~

a>

*£

is replaced by:

In any case if

£

P E L

If

and

g

and

lila

f E Co

are in

£11 p -< 11fll p IIIJ.II M IJ.

*£

then

f

then

L1

111-ll1M)

E Co

and

The

and

* gEL1

and the

next result follows easily from the definitions: Theorem (1.8).

If

f ,g

1

E L then (f * g)

Borel measure then (jJ. * f)" - ~

f .

".

A

A

.. f g.

is a finite

- 26 -

We now consider the problem of inverting the Fourier transform. Formally we would expect: "f(x) ... always make sense since The example For

Ixl

is not necessarily in

f

but this does not

L1 when

f ELI.

1/Ixl)to(x) discussed early, is a case in point.

> 1, f(x) ~ (~n q/(l_q-l»lx\-l •

Definition.

If

= (tn

f(x)

Jf(S)Xx(S)dS" ,

If

g

g E Ll

is

locally integrable and

it is clear that

~g ~

r

.' g

let

k E LZ

as

k ~ "" •

Jg

It is also clear that this limit may exist even though

does not

exist, as a Lebesgue integral. ex>

Example.

Let)

8

'--'k"O

k

be any convergent, not absolutely convergent

series of complex numbers. 0

g(x)

Then

= {

a (l-q k

~g ~

-1

.

\' ~

)q

=1

-k

Define

, Ixl Ixl

ak ' but

~

g

1 k

.. q , k ~ 1

J'. Igl

co

\'

-/

K-1

Theorem (1.9) (Multiplication Theorem),

J

I'

by the rule:

f(x)g(x}dx .. .!, f(x) g(x) dx •

1kl 8

_

AjF1 .

We may assume that

~Tal >

such that

iFI

sUPa!Ta

l

such that

F c: U _ Tk k l

and

0>

IF1~ IU~_lTkl

- ')

~"l

ITkl.

The result is now immediate for any >",0 A}, and

For

'~S I fl > Als x I.

~

1

= (x:

11s1 : I fl

sup

S,sphere

f E L • A

If

is the function

xEs

Ix-zl~q-k

of finite measure. and

~

"

Theorem (1.13). Let

\f(z)ldz

f

fs) x~F

Sx

such that z E Sx

satisfies the conditions

A, 0 < A < 1 there is a finite sub-collection

Hence, given

,.,N

(Sk1:.l of mutually disjoint spheres such that

/ , I ski

> AI Fl.

Kal

Definition.

Let

be locally integrable.

f

be a regular point of fk(x) '"

l

Theorem (1.14).

If

regular point of

f.

f

Then

Thus,

x E K is said to

if

J

f(z)dz

~

f(x), as k

~'"

•

Ix_zl~q-k f

is locally integrable then a.e. x E K

!!-!

- 30 -

We may assume that

!J:22!.. (T x19 0) f.)

choose that

For

g E.I

fELL. (If

f

4 Ll

replace it with

g E "/, gk (x) .. g (x) for large k.

such that

Ilg-fll

l

< e.

Fix

e >0

For large values of

and we have

k

f- fk - (f- g) - (f-g)k' so that

The result follows if for each 0>0, E"{x:lim suplf(x)-f,;( (x)[>o)

is of measure zero. Ee::

lx:

But

If(x) -s(x)i >6/2] U Ix: lim supl(f-g)k(x)1 >o/Z}" Kl UE 2 • 1

1

IEli S 20- I1f- gill' [Ezi S 2 o - 1If- gill (the second of these by (1.13) so

IEl S

4611f- gill

= 46

E:.

e ~O

Let

lEI .. o.

and we see that

Corollary (1.15). particular, it converges at each point of continuity of Proof.

f.

An immediate corollary of (1.10) and (1.14).

Corollary (1.16).

11

f

!8£

f

are both integrable then

equal, a.e., to a continuous function.

~

f

i!

modified (on a set of

measure zero) to be continuous we obtain for all

f

x.

- 31 -

A

If

f

Jf

continuous function; namely that

is continuous

f

~(Xx~)

is integrable then

a.e.

Xx

converges to a

(l.l(b».

Modify

f

and

..f

By (1.15)

we see

and a set of measure zero

and we are done.

Corollary (1.17). By

(f-g)"=O.

In

1

E L

~

,.. f

f(x) ..

£(0)

particular~

?

that

..f

r· i

A

f(OdS.

~

Ixl

~

q

f

is continuous at

fELl.

Since

..

£

Since

f

0,

f.

is continuous

that

?

0

k

ELI •

Notation.

For

k,~ E ~, kv-t • max(k,t], k ".t • min[k,.t).

Definition.

For

k E;;Z

let

~.

at each regular point of

0 we obtain (from (1.10) and (1.14»

£(0) '" lim k-'"

f(x) .. g(x)

•

We only need to show that at

and

0

~. f(~)xx(S)dS

J f(S)dS

=

then

(1.16) (f-g)(x)-Oa.e.

Corollary (1.18). f

= g

R(x,k) .. q

-k '-k'

- 32 -

A

A

= 4i k ,

(1.19)

R(· ,k)

(1.20)

R(·,k)*R(. ,t} .. R(.,kvt)

(1.19) that

4i

k

.. R(' ,k)

is a restatement of (1.4).

= 'kt.t

(R(',k)*R(.,-t»"

= t

kvt

From (1.8) and (1.19) we see

'" R(.,kvt).

An application of

(1.17) completes the proof.

2

The L -theory

2.

Proof.

Let

g(x)

-

=>

(f*g)"cfg=!fI

2

f(-x). Then

.?o.

A

"

g" f.

Since

Since

1

1

f,g E L , f* gEL,

2 f,gEL ,f*g

is continuous.

To

see this we write

I (hg)(x+y)

But

IIf(o

-f*S(x)[ .. IJ(f(X+y-z) -f(x- z»g(z)dz[

+ y) - f(.)11 2

.. 0(1)

as

y ~ O.

We then apply (1.18) and we see that (f* g)" ..

Jlil 2

.. (f*g)(O) ..

J

g(-z)f(z)dz"

E L1

and

Iff" J'l f l 2 •

From this theorem we see that the map

on

lil2

f

~

"f

1 2 2 L n L, which is a dense subspace of L •

2

is an L -isometry We now extend the

- 33 -

Fourier transform to isometry on

L2.

From (2.1) it is easily seen to an

L2, and agrees with the Fourier transform on

Ll

on the

set

Definition.

If

f

E L2 ,

let

where the limit is taken in

2 L.

2 fJg E L

Theorem (2.2) (Multiplication).

for each

and

fk .. f t _ k

~

We use Schwartz's inequality and see that the left

k.

hand side converges to

and the right hand side of

The relations (1.2) and (1.5) hold for

Remark.

it is onto.

~

.f

is a 1 inear isometry.

2 If not there is agE L

g =:

O.

But 11&11

2

•

L2,

We on1 y need to show that such that

But (2.2) then implies that This implies that

r .. Jf g

f E L2.

The Fourier transform is unitary on

Theorem (2.3) • f

r .. .. Jr ..f g.

J fg

.. Ilgll

.f.

Jig .. 0

ffg

m

for all

0 for all f E

0 , a contradiction.

L~

- 34 -

Notation. f

~v.

--;I>

by

The inverse mapping to the Fourier transform is denoted

r

We denote the reflection operator for a function

on

f

K

(7(x) = f(-x».

In the proof of (2.1) we noted that L2

Similarly (i)

as usual.

f' --;::;-:

.. J f But

f

~

If

(2.4)

f

If

...

1

Theorem (2.7)

!.r.2.2i.

S

p

2•

.!i

g

f EL

I

, g E LP

,IS

From (1.8) the result holds for

transform and the fact that g

E L2.

P = 1.

* (gl +

f

1s continuous in

Thus

(f

* g)

* g1 + f * g2

g2) .. f

g

a.e.

•

2

From (1.7) we see that

I

L (f E L ),

sO

Since the Fourier transform is continuous in L2.

It will suffice to

From (1.8) and (2.6) we have

(f*(~_kg»A .. f(~_kg)" ~f

in

then

2,

P" 2, using the linearity of the Fourier

establish the result for

Suppose

S

P

A

,.

..

= f g

a. e.

f

* (~-kg) L2, (f

....,. f'lt g

* (I -kg»"

in

2 L •

~(f*g)"

- 36 -

3.

Distributions on ~

Let

K

be the class of functions defined in

§1.

This class can

be described alternatively as follows: ~

(3.1)

E~

~k

on cosets of

k,~

iff there are integers and is supported

such that

9

is constant

~t.

The following result is crucial and is the local field version of the usual situation where the "smoothness of the function" reflects the "rate of decrease at infinity" of its Fourier transform and conversely.

.ll cpE....., is constant on cosets of ~k and is

Theorem !3.2~. sUl!ported on

'11'-

supported on

'B

~.

-k

then

~E.I

We may assume that

~

q

k ~ t

and that

~

Theorem (3.3).

-k

,and hence so is

If

~"

The result for Clearly .I

w is a finite linear

Thtk, with

-k~ht_k which is constant on cosets of

supported on

and is

•

combination of functions of the form

(ThY~)A k

~-t

is constant on cosets of

~

~-t y

But

and is

.

E.I, h E K then

ThCP

E ~t.

h

T h cP

.....

, cP , cp

* 1\1

E.I •

is a triviality, as in the result for

is closed under multiplication of functions.

....

cpo

By (1.7) we

- 37 -

have (~*~)A = ~ ~ E~. iff ~ E ""')

(

.

A*

~

~ E ~* (or ~)

and

The context will

resolve the ambiguity. Let

wn

be the characteristic function of

characteristic function of

For

* f E L1 (K)

T!f(J() -

For

J

K*

dx

\"

""7"!.

I

I

00

'"""""k.. -CE>

q

n

be the

is called the Mellin transform.

the Mell in transform of

f(x).(x)

II

{'lr _ *: deg(It*) - •




0

rl(la)cp(a)da,

€~Ia 1~1t/~n for

deg(1t*)

-./tn q

q

11'.

That is,

iva

\)

r,." Cpr1 ; r,..* ,

We write

A

deg(1r*)

> 0)

E ,)1*',

Theorem (5.19).

Let

~.

*

Thus,

K.

5][;'1

n

.2:

n

Then

~

-1

cp E

< r 1/:,'fl > .. < :tJl X,cp >

=

* and is compactly supported on

~

lim

n-C:O

< :Dlf.X] ,cp > • n

f

...

*

deg(,.. )

If

A*

cp E ~.

and so

.*... 1

then,

1f

,..*

is ramified •

[X]

n

E,.;*

and

- 66 -

I"

J q

where

D n

and

-n~ lx i~n q

,...

D n

are the Dirichlet and conjugate Dirichlet kernels:

1 ,n " (x) Dn (x) .. - +' cos kx, "'Dn 2 Ll

.. \,n sin Ll

kx •

Well known facts from the

theory of Fourier series show that as distributions on the circle

~

Dn(a tn q)

(1l/J..n q)b

and

Dn(a tn q)

~P[t cot(a tn

q/2)1 •

A

little arithmetic shows that -q

io:-l

I

+ '2(1-q

-1

H 1-1 cot(o: tn

q/Z) 1

co

r 1 (ia)

!Ili[X]n d·1 io, ~ prl(iO:) + (.(l-q -l)/-{,n q)b ()*')

Thus,

so

r/F .. p r + (,,(1_q-l)/h q)o • 1 1

Definition. denoted

J

J{

For

n E "'* K ; u,v E K* ,

the Bessel function (of order .),

(u,v), is the value of the principal value integral

(5.20)

In Theorem (5.25) below we establish that J J[ (u,v) exists for all 1l E "'* K

and

* u,v E K.

If we use that fact several properties of the

Bessel function can be obtained by changing variables in (5.20).

- 67 -

LenJ:na (5.21).

(ii)

(1)

J (u,v) - J _I(v,u). 1t

It

= ~(v)J 1£ (v,u).

n(u)J (u,v) l{

= rc(-l)J

(iii) J (u,v) - J _I(-u,-v) It

It

.!!..

(iv)

If

~(-l) ..

1, J (u,u)

_l(u,v)

1s real-valued.

1t

= -1

re(-l)

l{

1s pure Unag1nary-va1ued.

,In(u,u)

k, a positive integer, n E A* K , v E K* we set

For

Fre(k,v)

(5.22)

=

1 ;'(x);(v/x)rc(x)!xl- dx

J

! xl =qk (5.23).

Le1Illl18

m

Ivi

Suppose

q

l:Sk 1,

m

1f

is ramified, deg(:II:)

/ 1

\uv\

, I uv/

o n:

I uv I

~

h? 1

m

_-l(u)F (m/2,uv); luvj

J (u,v) :II'

> 1,

In

In

> 1,

m odd

~

q , m

1f-1

m

luvl •

-1

S

qm, m? 2h, m even qm, m >Zh, m odd

'It

tC

:II'

(u)(F (h,uv) :II'

+

- qm, h < m < 2h, m even

F (m-h,uv)1; 'It

luvl - qm, h

Suppose

qh

(u)[ F (h, uv) + F (m-h, uv) + F (m/Z, uv) l;

I uvl tC

even

~

Jl

o

Sq

- q , m

,,(v)f(,,-l)+'It-l(u)f(1{); \uv\

Proof.

even

1

F (m/2,uv),

]J

m

> 1, m odd

1 +1]-2/q, (l-q -1 )[lOSqjuvj

(11)

q

luv! $ q, luvl ~ qm

mS 1 •

< m < 2h,

m odd.

- 70 -

J ~(uv/x)lt(x)jxl-ldx+1f-l(u)p J x(x)lf(x)lxj-1dx

J .. (u,v) • 71'-l(u)p

Ix\ >1

Ixl~ 1

r

• n(v) P

-

(x) IxI1-1 dx +

-1

X(~).

J

71'

-1

Ixl~luvl If

is ramified we see that

If

•

1s unramified set

(

r

Ivi a

-1

J (u,v) • 1I'(v)f(1f '11

+ \ U I -a

-1

) + ..

,,(x) - jx\a and recall that

-X (x) Ix I -ct-l dx

J

(' - (x) j x 1a-1 dx J:t.

a little calcUlation shows that it 18 equal to

:JI'(v)f(:JI'-l)

+

J .. (u,v) •

'1I-

If

.-l(v)f(1f).

1

luv\ -

a ..

0

we get

qm , m > 1.

Since

luv\

f ~(uv/x) .. (x)lxl-ldx

{u)P

Ixl~l

J

+ ,,-l(u)

;(uv/x)~(x),,(x)!x\-ldx

1< I x\1

If

J 1( u,v) -

f X(x)1f(X)lx\-l

(u)P

~(x).(x)lxl-ldX

> 1,

Then

- 71 -

I

• ,,(v)P

\xl~luvl +1C

-1

J ~(x),r(x)lxl-ldx

;:(x),,-l(x)lx\-l dx + n:-1(u)P

Ixl2:.1uvl

'\'

(u)

L

F,,(k,uv)

1-::;' k< m

If

is unramified we see that the first two terms are zero

1C

(by (5.2»

and the third (by (5.23)(1»

is

F,,(m/2,uv) if

m is even

and is zero otherwise. If

n:

1 < m $ h.

is ramified of degree From

(5.23) (ii)

h, h

?

1, consider first the case

the third term is zero and from the

definition of the gamma function the first two terms are and

•

-1

(U)f(1C) respectively.

m

If

>

h

If(v)f(n:

-1

)

then by (5.2) the first two

terms are zerO and the evaluation of the third term is established by reference to (5.23)(11). Remark.

J (u,v) J(

can be extended in obvious ways to characters

1C

that are not unitary to obtain Bessel functions of more general order. Remark.

The proof of (5.25) shows that for

J,,(U,v) --lI> J (u,v) 1

as

l(

(l-q

(5.26)

IJ

--lI> 1

-1

(l_q-l)

umramified,

and that

) (log

(u,v)l$ { l(

"

q

,

- 72 -

Corollary (5.27).

~

(1)

u,v E K* .

function of

J

1

E ..K*,

ft ,

that

(i)

(5.26).

luvl 7 0 • 1(

".

If

i1(u)! -

is bounded as a

u,v E K*, J (u,v) is bounded as a function of

For fixed

Proof.

1, J 1( (u,v)

(u,v) is bounded for (uv) bounded away from

as (ii)

I

+1

1(

I 1(v)i

a

"* EK

the result follows from (5.24) and the fact

1.

For

1(

= 1

The asymptotic formula for

the result is contained in

J (u,v) is immediate from

1

(5.2S)(i) (b). (ii)

Fix (u,v).

If

1(

is umramified then (l-q

-1

) I log

(l-q

If

11'

-1

q

lu~\

~

q

, Iuvl

>

q

)

is ramified then, iJiu,v)1 ~max[lr(lf)1 + ~

max[2 q

-i

Ir(1f- I )I,

, 3(l-q

-1

The next three theorems assert that the Mellin transform of

-

+ 1J + 2/q, luv!

I 1-1

p )(.(v/x)1I'(x) x

3(1_q-l)]

»). J.(u,v) can be regarded as

~(ux + v/x), the Fourier transform of

and as the inverse Mellin transform of

- 73 -

Theorem (5.28). I)I1f(",) .. J

~.

11'

Let

f(x)

v/x) , u,v ~ K*.

J

J{

~

r' J

We fix

I

f E ~I,

"-

~t

1

respectively, IV

x(ux+ v/x) (!!t"~) (x)

~

u,v E K*,

~

Choose a compact set

t he

..

,. E .,Pr.

C

support

f-

Since

f

For

I xl -1 dx:. j.. f'

J

K*

~

= C(u,v,hO'~)

0

~

~ E j*

we wish to

(u,v)9(1f)d1f •

is compactly supported we

*

C K

Then

( ",,-1m)"". ~'~

x(ux +v!x) ('m-1 ~) N (x) I x

1-1dx

C

-Jr -X(ux+v/x)0Jl-11fT)N (x) I x 1-1 dx K*

From (5.27) we Thus they are

A

and

-

C contains

x. ~.

(u,v) is bounded as a function of

distributions in show that

Then

(u,v) EJir',

X(ux + v/x) is bounded as a function of

see that

and

= ;(ux +

so that

- 74 -

Lemma (5.29).

.f(u)

=

K

Fix

n

A* , v E K* • EK

as a function of

J

:J(

(u,v) is locally integrable on

u, and so it 1s 4n

Ix 1-1 E v I • Let P X(v/x):n:(x)

choose a positive integer

~

h

a

~/.

such that

K+

We want to show next that

max[1,deg(n)1.

~k and 1s supported on 'l'-k • Now choose If

x E K+ •

J (u,v) E,,;' •

~.

f -

E ~' , as a function of

=f

X(V/X}K(X)!X! -1

p

~

Fix

cp E.I,

and

is constant on cosets of t

qt - maxtq

so

h1 v 1-1 ,qk 1.

n>.t then

(5.30)

-

J'

q -.tS

where in

~

f E

.I',

k

and

I x$q I k

~(v/x):n:(x)!xl -l~(x)dx

.t depend on :J(,v and cpo

we may choose a fixed pair and that

k

and

For

{cps}

a null sequence

.t and it is obvious that

- 75 -

For all

n

large enough we have

.. r w(u) K




!)

8=0

so


z IIG~I p will do. Ihl S 1.

We now suppose a G

and we know that

IIGa (. + h) _ Ga (')11

Since

IX+hl '"

Ixl

=

J

and for

Ixl

if

is radial (depends only on Ix[)

L

so if

>

Ihl > 1 Ihl

we see that

IGa(x+h)_Ga(x) IPdxJl/p,

1

S p < en

Ixlslhl

P

sup

l Ga (x+h) _GCt (x) !

, p

=

OX>

Ixlslhl Suppose fn(a)Ga(x) ..

Ix\

S

Ih1

Ifn(a>\

< 1

1IGct(.

p"

ox>.

Then

_qa-~o{x)

Re(a) > n

+ Ixla-,\o(x).

so

a

Since

+n.

Thus,

.O(x+h) = .o(x) for

(both terms are 1) we see that +

h) -

a G

(·)I1.,,:5

sup

I xis Ihl < I

Ilx+h 1a -

n

-

Ix ICt-n ) :E zlh\"

- 141 -

This completes the proof for NO'W assume that

1

p

Q

S. p < '" , a '"

a !rn(a)\ IIG (. + h)- Ga(")llp

J

=~

~

•

D ,

Ihi s. 1

•

!Ix+hl a - n_ !xla-nl p

jI/p

Ixl::;l hi

l::; P 0

lxl

I

>I

•

Aa (x) .. {1,lx G

Then we write

1-0}.

there are finite Borel measures

such that

I xl a.."~,a(X)

"-0

G

(x)

Proof.

Hence the theorem follows if finite measure.

(lxla,o}

is the Fourier transform of a 1

Actually it is the Fourier transform of an L function.

- 143 -

An easy calculation shows that

(\Xla,O}A(U)

~ut this function is in

- {(l.q-n)!(l_qa-n), f(a+n)lu\-(a+n)}. and

(jxla,o} is continuous.

From (1.16) it follows that

is the Fourier transform of an

Definit ion.

(Ixla,o)

Ll function.

a E C, 1 :5 P :5

For

Ll

00

,

let L~ .. (f : f- P g , g E LP }.

We norm is supplied with the

C

0,0:

Some Facts. with in

is isomorphic, as a Banach space with

CO' for all a. for all

,

Co a

oo

L norm.

Clearly

.,

is dense in

p

~, 1

P

L ,

Co ,a

:5 p

Re(a)

is Cauchy in

We can define

and there is a

is Cauchy in

LP ; its limit

Our considerations in this section show

I-a, Re (a) > 0

is

p

L

a

,

I :$ p


n«l/p) - (lIs»

~

0,

Re(a) - n«l/p) - (lIs» > 0, 1 < P < s < '" .

Then it is easy to see that for such a, jGa(x)i and independent of

p,r and

~ Aa i xi Re(a).n, Aa > 0

As a consequence of (4.9) we have, with

L11 fll s '" II Jagl\ s~ BJI IRe (a) gil s~ BaCpsllgllp =BaCpsllfllp a'

Thus,

Lemma (7.2).

If

LP(~) c LS(K~.

a

1 < P < s < '"', Re(a) .. n«l/p) - (l/s» The inclusion 1s continuous.

> 0 then

- 146 -

In the remainder of this section we will be concerned with showing rhe conclusion of (7.1) and (7.2) in the case Re(a)

= O.

1

s, then

00'

< s, then

we continue by

In this manner we get a countable collection

f(x), x ~ Ds s f2 (x) '" [l 1 r J w f (x) dx,

TW:T t

t

(iv) and (v) are trivially verified.

If

x E wt '

Wi

-

- 150 -

f(x)

r

1

-~

that

f(z)dz"

¢ Dg

x

then for all

5

(x)

We see

t

1 =~ IfIl ,

J

< qn s

f(x)dx

ill

t

U),

x E ill,

a.e. K ~ Ds ' f(x)

from (1.14) that for

5

(f(x) - f(z»dz.

1°\1'(1}

(vi) 15 valid. If

f

~ ~

(J)t

t

f

~ S

f(z)dz




II £;111 '" II fill

IIf;1I2Sllf~lIt

E L"',

'llf;lI!,

S qn 8

follows from (ii), (iii) and (iv).

is a trivial computation. and

1I£~PlSllf~!11

Note that,

+11£\\1'

Done.

s

- 151 -

Lemma (7.10).

For

i

a E C, a

0, n, k E z , k ~o

,

Ixl

[(l_q-n)/(l_qa-n)] qk(a-D)_ (qa-n/fn(a», Ga (x,k)

I Ia-n - qa-n )/rn(a), qk < Ix I (x

~

{

, I xl >

o

are also radial and continuous.

1

1

LP, 1 S p ~~.

By observation the functions are in

~.

~

~ qk

They

A trivial computation shows that the

functions have Fourier transforms

fl,\xl-a}tk(x)

so that they are

a

indeed, G (x,k).

Remarks (7.11). (a)

a

The cases

§

0

and

a

a

n

are easily supplied.

GO(x,k) - R(x,k)

,\x\ ~qk 1 (l-q-n)logq(q/lxl), qk < Ixl ~ 1 ft-k(l-q-n)

(b)

n

G (x,k) '"

l

Ixl

>1

£ -? J f

l:!l

.I' .!!! k ......,. -

L2 , Ga (. ,k) .. £ -? Pf

in

L2 .!.!

0

Suppose

Lemna (7.12).

a

,

Re(a) - 0,

*

(a)

1£

f E ,,/' , G (, ,k)

(b)

.!f

f E

(c)

If

fELl , Ga (. ,kh f

(J

k"""" -

A

>0

00, IIJa fI12"'llfI12'

converges a.e, {as k"""" -} to a function

fa (not necessarily a distribution of function type),

constant

co •

indepe.ndent of

f

~

s

>0

There is a

such that

- l52 -

fr22f. a

Thus

G (', k)

(Ga (., k)

*f

* f) ..

a

a

I

... J f ( • , k) ~ J f in

{I

.. ~ k (x) 1, x I-a)"f (x) •

Note that not only is

Ff E L2, IIPfli2 - Ilcafll2 = IIfl12 - !If\l2' Re(a) = O.

when

k ~ - "".

as

Furthermore

f E L1,

>0

is given.

GO(. ,k)

that

*f

... R(' ,k)

Itx:f(x)

*f

the result is

5-

1

Re(a) .. 0, a

We now suppose

•

That is

4 o.

f; E LZ,

(7.6) - (7.9).

Note that

exists

Ii (f;)diz .. IIPf;1I 2 -

/[x

a.e.,

:1 (f;>a(x)I

If(x)1

2

dx1Q ..J

f(x)? 0, fELL

~ediate,

... f(x,k) ~ f(x) a.e,

>5)1 ~iifill

Ixl> q-

k

is unitary

Write

o

co

f

,

f ... f~ + f~

Ilf~ll; ~ qn sllfl1 1 11f~llz.

and the fact

(by (4.11» f

•

as in

Thus, (f;)a(x)

Hence,

> s/2)\ :5il(f;)all~ 45- 2 :5 4qn IIflll s·1

s 1

a

(f *G (',k»(x)"

;0

j

and

since

s a f (x-z)G (z,k)dz, which is 1 defined and continuous for all x since fl8 E L1 , Ga (·,k) E L"" ,

Proof of Claim.

0(11

f(x) ~ :Pf(x) a.e.

We may assume that

a =0

If

J'

I

,,~

a G (. ,k) *

~.Pf

f

but

Thus, IIGa(.,k)* f - Pfll.} =

Now suppose s

(We used (3.19) and (4.11»,

J.

- 153 -

d:t(z,k) = 0

r

z ~

if

T~lUs, (£~*

r:.

s et '\ fl (x-z)G (z,k)dz ... L

J

Get(. ,k» (x)

s et fl (x-z)G (z,k)dz

[' ,I

t zE(x-!U )n('

zEn

t

0 ~ x - w " t

Let us note that for some

and

k

(x- W ) t

X-lOt c: D ,Of ){-w t ' and (x-lOt) x- ID

t

lI '

then

t

4:

K

D s.

There are

n D = ~ and the term is zero, or

n.o ..

x- ID

IzE(x-w

•

t

In the latter case, each

a

has the same norm, so

From the observation that

r £1s (z)dz

OEx-

x E (\It ' a contradiction since

then two possibilities:

element in

If

G (z,k) is constant on the set.

f~(x-z)dz t

)nl:

..

r

f~(x-z)dz

X-lO

t

we see that the remaining terms are also zero,

.. 0,

50

W

t

s

et

(G (',k)* £1) (x) = 0

for all x

! ~ Ds'

k 5 0, so

This proves the claim.

for

In particular

(f~)et(x) exists if

a.e. x ~

But

a.e. x.

Ds'

IDs I ..

x ~ Ds

0(1) as s ~ ""

and so £a(x) exists so

fet(x) exists for

Now we have,

s

n

... (1+4q)llfIi

1

5

-1

•

-1

• 154 -

11

Lemma (7.13).

E LP , 1 < P < "', Re(a)

f

Ap > 0 independent of

IIPfl1 p -
-

~

•

is a collection

that: (i) (x 5 + ~-k5,k S ) c ~ , k

7l 3

~.

E

sup{k:(x,k) E~) m(x) ==

:a kO E

if

S

= 1,2, ••• ,r;

= 0 , 1 , ••• ,r; and

is connected if for every pair (x,k) , (y,t) € ~ -k (x + ~ S k )Jr in ~ that connects the pair in 5

'

5

5=0

the sense that

Remark.

If

~

is a connected domain, then it is simple.

from the observation that if (x,k),

A function

f(x,k) on a domain

f(x,k) is constant on

(x

+ ~-k ,k).

(y,~)

~

in

ED

and

n K X?Z

~

This follows

is connected

is ~

if

- 185 -

A function

f(x,k) on a domain

f(x,k) is smooth and for all (x,k) E (3.1)

f (x,k)

q

r

-kn

£,

in

n K X7.1

is l:.egu1ar

if

t;. "" ~.r, ,

f(z,k-l)dz

x:'I\-k . A function

f(x,k) on a domain

~

in

Kn X ~

is sub-regular

="

(resp., super-regular) if it is smooth, real-valued and the" "~"

(3.1) is replaced by

=Remark. of

Ii

§l.

~

= Kn

X 7.l

Note also that

in

(respectively, " ;: ").

the notion of regularity coincides with that n

K XZ

is unbounded but is connected and

simple.

If

f(x,k) is a smooth function on a domain

f'

function

(x,k) may be determined as follows:

then

f/(x,k-l) = f(x,k-l) - f(x,k).

then

f'

(x,k-l) =0.

If

its derived

£

(x,k) E

If

(x,k-l) E ~

1) .... eli!

and (x,k) ~ ~

Our definition of regular, sub-regular and super-

regular may be trivially reformulated: Let its

f

be a smooth function on a domain

derived function.

f 1

sub-re;:;ular

f is

~

and let

f/(x,k) be

Then

(>0

~l

regular super regular

J

I

iff

V(x,k) E [., "'" ~.!9,

f' J

,I

x+'li

-k

f' (z,k-l)dz

1:: "

- 1.86 ..

Note that if

f is regular on a domain

z E x + 'U

known for

-k

(x, k) E

,

i>I "'"

0 iI

~

and

f'(z,k-l) is

and if anyone of the

[f(x,k),f(x+ak,k-l)} (where ~ak) are qn coset representativea s s k ",-k+l in 'B- ) is also known then all the other values are

values of

determined.

Thus,

Proposition (3.2). is given. then

t'

Proposition (3.3).

Since case where

~

&

If f

f

is regular on a connected domain

~

and

is determined up to an additive constant.

11

f(x,k) is sub-regular on a bounded domain

~,

is bounded it will suffice to consider the special

= (x +

~

-k ,k) U

(x

+ ~-k ,k-l), with

The result now follows from the definition of sub-regularity.

Proposition (3.4).

11

~

is a bounded domain and

regular functions that agree on f - g h j,

f

:I

h~

g

on

!n£

then they agree on

is regular so the supremum of

(take real and imaginary parts).

f

Similarly for

f - g

g

~

h

is taken on

g - f

and

gO

ii.

Proposition {3.S}.

(a)

If

f

is regular

I f\

is sub-regular.

(b) Linear combinations of regular functions are regular.

If

f

and

- 187 -

g f

are sub-regular, a, b v

g

is sub-regular.

~ 0

(c)

then a f + b g f

If

is sub-regular,

is sub-regular and

.!§....l!.

" f

f

is sub-regular.

The proofs are left as an exercise.

Proposition (3.6).

* f(· ,t)

R(· ,k)

1£

;? f(.,k

Since

f(x,k) is sub-regular on

The only case of interest is t .. k-l.

it for

then

vo{,).

is smooth, if

f

Kn X ~

That is,

k

~

~

k

t , R(',k)* f(o,t)

f(x,t).

a

t, and that will follow if we check

R(.,k}* f(.,k-l) ~ f(·,k).

But that is

the definition of sub-regularity.

If a regular function n X ~

m(x,k) majorizes the function

domain

~

~.

(h, a regular majorant of

that

If

in

K

we say that

on

m 1s a least regular malorant of

Convention. f(x,k).

If

f

is a regular majorant of

m f

E ~' we identify

f

f(x,k) on a

fJ)

f

on

(h ~ m) then we say

)

on

f

~.

and the regular function

In view of (105) this is reasonable.

Theorem (3.7). function on

Suppose

f

is a non-negative valued, sub-regular

n K X ~ and suppose that

supil f(' k

,k)11

P

..

A




for some p,

- 188 -

Then

(a)

if

1

(b)

if

p

Proof.


,IIMt fll p ,5 Ap,rJlfll p ' and for each s > 0

M.(,f(x) > s)\ ,5 Agllfll

is locally integrable, then

l

8-

1

n.t. lim(x,k)_zf(x,k)

= f(x).

- 191 -

Proof.

For

= 0,

t

(a) is (1.7) and (b) is III (1.13).

0,

(b) we see that is of weak type

and we get this from (a).

We construct a substitute for conformal mapping.

Definition. Suppose

~

Let

be a simple domain in

f(x,k) is regular on

n K X ~ and let m

I~, then the extension of

= m(~).

f, f(x,k) to

n

K X ~ is defined as follows:

f(x,k),

(x,k) E

f(x,t),

(x, t)

If

f(x,k)=

0

,

(x,k)

t,;

':: e

t£

Fact. on

and

or

(x,J,)

km

If £ is regular on the simple domain

~ X~ •

for some

k 0 such that

I\f(' ,k)I'

P -< M for all

k Ell.

This result fails for p = 1 as we see from examining the regular function R(x,k).

We note that

i\R(.,k)ll

l

== I but

R(x,k)

is not the regularization

of an Ll - function.

02.

Theorem (2.2) of this section gives some local field variants of

results in Taibleson [6,Ch.IIIl that are needed in the sequel. Lemma (2.1) is the discrete analogue of Hardy's inequality (see Hardy, Littlewood and Polya (l,pp.239-246), and the reader will note that the proof is totally trivial.

If one takes the trouble to write the

relations in Hardy's inequality as integrals on the multiplicative group: (O,~),dt/t}, we see that Hardy's inequality has the same trivial proof.

§3.

In this section we work out some more local field variants of the

properties of harmonic functions in euclidean half-spaces.

One of the

- 194 -

more important of these is (3.7) which provides sufficient conditions for a sub-regular function to have a least regular majorant.

For

probability buffs we point out that sub-regular functions correspond to sub-martingales as well as being analogues of SUb-harmonic functions. The definition of the extension of a function, regular on a simple domain to a regular function on

n K X ~, corresponds to the extension

of functions on certain sub-domains of the disc to the entire disc by conformal mapping, as used in Zygmund [l,vol.II,Ch.XIV,Sl). be used in Chapter V

~2

It will

to study the boundary behaviour of regular

functions. In the study of the "boundary behaviour" of martingales the corresponding extension is obtained by use of a

stopping time.

- 195 -

2h..;pter V.

The Littlewood-Paley function and some applications

In ~l we introduce the Littlewood-Paley functions g (';f), and p

study the relation between the

LP properties of

In §Z we introduce a truncated version of

f

and g (.;f). s

gZ(';f), Sf, and study the

local equivalence of the n.t. convergence of f(x,k), n.t. boundedness of f(x,k), and existence of Sf(x).

1.

(n.t. = "non-cangential")

The Littlewood-Paley function

Def init ion.

If

f (x, k) is regu lar on

n

>< 2'l we def ine the

K

Littlewood-Paley functions g (.jf) by P

g",(xjf) ..

SUP

k

E:2\f(x,k) - f(x,k-l)\

p g (x·f) " ' ; \ If(x,k) - f(x,k-l)!P-:l/ , 1:S P J p J ~ kEzz

If write

F

is the distribution to which

g (xiF) P

=

r "T (/

,

2

j \f(x,k-l)\ dx -

Jr

2 \f(x,k)\ dx.

2

\,T

Thus,

r

2"

:I

If(x,k) - f(x,k-l)1 )dx'"l" [.If(x,k-l)l dx-J1f(x,k)l dx]

~

t

! £(x,T) I Ldx

r

•

"

From IV (1.7), 1£(x,T)1 2 ~ (Mf(x»2 ELl. f(x,T)

~

0 as

J1'1 £(x,T),12 dx ilS

t

~

-

0:>

T

~O>.

~O

IV (1.10),

By the dominated convergence theorem,

T ~o:>.

as

By

By

J

IV (1.8), " I£(x,t-l)/ 2 dx ~!lfI122 '

•

I\T\ f(x,k)- £(x,k-1) 1 2-J1/2

From the monotone convergence theorem, LL~

t

converges a.e. to an L

8 (x ,f) .. 2

I" L

'"KEZ

2

2

function, and L -norm equal to

•

I£(x.,k) - f(x,k-l) \211/2 I , exists a.e., is in

I

L2

-1

for each

s

>

0

A

>

0

II

We use the decomposition of III (7.6) - (7.9) and the

argument of III (7.12)(c).

Thus,

.J

s

~.

\I f11 2

Thus, with

s > 0 fixed we have

and

- 197 -

x ~ D , where

if x

s

s

~

D

s

-1

if

and

We may argue as in III (7.12)(c) and &et that

g2(x;f) exists a,e.

and further that,

,

~ilf111s

Lemma (1.3).

If

1

there are constants

Proof.

The map

f

-1

+4q

fl

~

>0

I

sllfl1s

-2

f E LP , then

\. ~ Kn

-

lim t -~ T .... +""

"T < f' L, t l .J Kn

f(x,t-l)h(x,t-l)dx -

.

')

JK f(x,T)h(x,T)dx( n-

+'"

f(x)h(x)dx.

Theorem (1.8). such that

Let

f(x,k) be a regular function on

(i) f (x, k) -?- 0

(ii) 8 (·;f) E LP , 1 2

of a function

F E LP




0 independent of

(';011 P

2

the result follows almost exactly as in the

We only need to notice that

If(x,T)-f(x,t)1

< (T_t)l/p'g (x;f) E LP and using (1.7) obtain

-

p

I' -

I,',

1'-

f(x)h(x)dxl < !Ig (x;f)g ,(x;h)dx\ .

m

and

r

,~-m

--k=-'"

i

_

! f(x,k)-f(x,k-l) \

x

if

i 21:b J =

and

x E {) :

f(x,k) = f(x,k)

if

'< Sf(x)

- '"

:S M

-

< k ~ m,

and

0 •

0>

E~ , but (x,~) ~ ~ ~

.{, 5 k 5

(x,t) E

b~,

if

y E EM'

r \,m

2~

_

-

(x,t) E [1 (z)

(x,t) E rO(Y) for some _

t:

for some

m, but:, f(x,k) .. f(x,!,)

Since

For

Thus 0>

f(x,k)" f(x,k)

< k ~ .{,.

for some k? t,

:S

if

Thus,

Thus,

f(x,t)

= f(y,t). 2~

r \' m

~ ~ .._",I f (x,k)-f(x,k-l) 1 I ~ L ~~-+ll f(y,k)-f(y,k-l) 1

J

Sf(y):S M •

Thus, for

2l'

x E C, ","' ~L m 1-f(x,k)-f(x,k-1)\ ~ k=-'"

difference betwe~n this expression on

r'L;'~ 1-f(x,k)-f(x,k-l) 12rJ " ~m+l

f .c,

2~ ..J

Then f(x,k) .. 0,

-

--'1 0 J independent

A

v-

A

II p -< AI' :p \1 p

(Ink'!')

E L~, then

= (~)v*~,

Hence

LP we only need to

and that there is a constant

1

!:::

small enough.

k

m is a multiplier on

if we wish to show that show that

for

n

m E L"" k

•

11

c

L2

so

where the convolution is

defined as an integral in the ordinary manner.

Theorem (1.1). independent of

~

[xl =q'

P --'JIo co

dy I dx Im(x+y) -m(x) 12 I yI2n+€

m is a multiplier on

are constants ,~

and there are

B

> 0,

>

€

0,

tEll such that

J

Then

mEL'"

Suppose

A

> 0,

C

p

LP , 1

= Cp > 0

A depends

I

only on

< P 0,

-1

then (1.1) will follow. To see t:lis note that to establish (1.1) we need only show that there are constants k E 2Z

But

such that

Ii

A and 11 (mk)v*

C

p

ctJlI

as in (1.1) that are independent of

< A Cp 1!~1 p for all cp

p-

E" •

(m1 0 depends only on n

I'

1~(S)ldS

~

0

I ~(S) IdS -< Ane; B q -krJ2

Jr

1~12' where

e >

~

€.

=

'

A

nOD~)~

Since n

in ",-k-l

If

",-1 -

n cosets fill out

O,l, •.. ,k, 1 ~

We proceed by induction on

...,.",0

n

and so our result holds if

u(v)-u(n) E

k fixed and show that it holds for We may assume that r




t

< e/ (1 + A

P

(3.4).

For

115 n f-fll p < -

),

n liS

f .. b + g, where

Write

O.

and

g

E .I(D) where

large enough

bll

np

Corollary 0.6).

+ Ilbl!

If

< P < '" ~

P f E L , 1

..li

Corollary (3.5).


s1!

is independent of

(I),

f

s

>

g =

1

~

O.

rf,xE(1) Proof.

We may assume

Tn* (x;m)

Then

where

N

>0

S

lifll m;m =

*

2 Tng(x).

l.

Let

0, x

We look for the minimum of

at

and

f, P,

ill

IN p s

-1 1P, 2

Consider an expresbion of the form

exp(-l/eA)

F(p)

P = ileA.

s

:s [exp(-s/Ne)J!ill!

S

=

[exp(-s/NE)l




Iill I .

,

(Ap)P , A

Thus, if

2 e [exp(-s/Ne)11 w1

S

>0

:s p

we have

IE s I

Ul

p ~ 2, we have

For

is independent of

takes its minimum

4

> O.

s/eN

It ~ 2,

2~5

-

4.

Singular integral operators

We indicated at the end of

LP, 1 < P < ~

multipliers on 11

-

§Z

that an interesting class of m(x) ~ ~(x)J

i& obtained by letting

a unitary mul tipl icat ive charac ter.

If

is the ident ity then

J{

the multiplier map is convolution with the dirac delta.

implicit in that aevelopment, and the material in is ramified, and we set

(1

,

k - . '"

and

",!

«PK)

! zl ~

* Ci) ,.. (x) to

bounded on

=

r

* c) (x) = lim

«PK)

c; 1, co.}verge&

a.e. to the functLon.

In the proof of (3.6) we stated that the norm, b p , of the HardyLittlewood maximal operator satisfies the condition: p

~~.

for

~owever,

B P

fJ C.)

at.

in proving ~hat the maximal operator was bounded

1 < p

P

Po ,~

A

J

(1.2)

>

0,

!F(x,k)!Pdx:SA q

~

'" - - /

k f< -') "m"

Let

.t

k f(x-t)

dkf(x) == f(x,k)-f(x,k+l).

q

r

_(m+l)"\,q-l t( j) ~-. k J=

II

€

~_1

m

Then,

f(x-t)dt

~J +'B- m

- 246 -

(2.2)

1

"q-l

= --,{qf (J{)

since

I

t· I(

.

(eJ)f(x-e~,k)

~j=l

f(x-et,k +1) is contant over

j : 1,2, ••• ,q-l.

Also,

(2.2')

",-{k+l) Fix a c o s e tY:+ .> gO

= g~

..

r (2.3)

and

0

a

t ... 1, ••• ,q-2,

1

= O.

Let,

.. f(y,k+l)

~

... Ttf(y,k+l)

a{ '" T t dkf (y+€~)

= 1,2, ••• ,q-2;

Then for

(2.4)

1f(O) '" /'(0)

0

1. at t

and make the conventions,

x E y

j

j

+ Ek +

.. dkf (y+e:~)

= O,l, ••• ,q-l. ~

-k

,

(2.2) can be rewritten, for

-

Namely, if

2~

7 -

are given then

are obtained by a linear

t

t

i

j

transformation whose matrix has (i,j) entry (l/q r(~ »~ (e -e ). Since regularity and sub-regularity are defined in terms of the local behaviour of the

tail

(a~) we see that we are headed towards a

and

local definition of our conjugacy relations.

Zxamp1e: K, the 3-adic or 3-series number field. We have

q

= 3,

=2

q-l

unity; namely, ft(e) .. -1.

€

so K(e)

r (I()

~

I'

1

I

-1

)- x(-p

-1

»

(1/3) (x(P

-1

)- x(-P

is a primitive 3rd root of unity and x(-p

f(~) = (1I3)(± i/3), and so

If now a

al

We

n(x) x(x)dx

(lIq f(lf»

-1

-1

»

) is its square.

"'±

(l/i/3).

The choice is not essential, let us take 1/3 r(1t) .. l/i

then

= 1.

Ixl=3

(1/3) GCP

Thus

r'

J

rt(x) i(x) xl - dx .. (1/3)

Ixl=3

1

2 1(e )

= -1,

0, !f(e)

Thus, we have

[(If).

..

~()l- )

2nd root of

can be chosen to be -1.

(with our convention above), rt(O) evaluate

is a primitive

"

o

1 0 -1 = (ao,ao,a ) (we require only that

o

(ai ,a~,a;l)

a; '" (1/1/3

)(a~

is g1ven by

1 - a: ),

a~l

a~ =

.. (l/i/3

-

a~).

-1

0

a o + a o +ao - 0)

/3 ) (0:

(l/i

)(a~

1

/3 .

1

_

a~)

,

- 24'3 -

Notation.

q

e '" (e ,e ,··· ,c _ ) E C , let O 1 q 1

For

lie)) ..

r

,\q-l

' ! Lo ;

-

les!

2~11

J

2.

s"'O

Proposition (2.5).

o

1

at'" (aV a 1/ ~

(b)

(2.4).

q-l

,at

q

) E C , t = 1, ••• ,q-2

be defined as in

(2.3)

Then,

Ilat )! "Ilaoll, t .. 1,2, ••. ,q-2 ,q-1

(e)

•••

I

'-'1"0

i

1

a j at ..

O. whenever j

+

J,

is odd;

j,1 .. 0,1, ••• ,q-2.

f!22i. "q-l

(a) 1

a Lii",O

~

0

This either follows from regularity or we may start with from regularity, or by assuwpt1on.

We then let

at

be

0

defined by (2.4),

1,4

0, and the rest of the proof w1ll follow.

if t ~ 0 t

(l/q r(lt

»

,q-l"q-l

L..

L..,

j=O

'" (1/ q

'"

0,

t

r ('II' »

,q-l J'

'-'1=0

~

rr '(e

j

i -I';

1..0

a

i( \~q-l ) 0

~~ j=O

t 'II'

i

)0:

0

1)

j (€ - ~ )

rhus,

- 249 -

Note that

is a non-trivial character on

~

and is

[€j_€i}j:~ are a complete set of coset

ramified of degree 1 and representatives of

~*

~,recalling

in

our convention that

€o

0"

This proves (a). (b)

Let

be the restriction of

g

and (2.1) we see that ",- (k+l)

on y + -

y + 1'!\-(k+1) •

0kf to

T-tg '" T tdk f on y +~

- (k+l)

Using (2.2)

, and it 1s supported

.

Note that

Jg'" q

k \,q-l L,

j=O

a

j

'" 0,

0

and

-', 12·.~ __ k/21 \,q-l, 11 2 - *J g j q L I'-'j=O lifo J

I

I

This proves (b).

(c)

We use

above, but for simplicity we assume that

g

(This amounts to rnul tip lying g is supported on A

so

g

Since {

j

is supported on

g(O) '" 0,

~-(k+l)

k

~

Jj=l

and is

Thus,

by

~'y .)

and is constant on cosets of

and is constant on cosets of

we see that

k+l q-l

F._(k+l) + ~

A

g

y = O.

g ~ero

'!\

-k ,

'fjk+l.

is constant on each of the cosets otherwise.

- 250 -

'" r

T j8(X)T{,g(X)dX

'K

(Tjg)A(~) (T.f,g)"(-S)dS

co . :

K

easily that the sum is zero.

The next result shows the importance of the three properties exhibited in (Z.5). i j

Theorem (l.G). i

= 0, ••• , k-l; o J

Consider a k X(m+l) matrix (a ) with complex entries; '"' 0, •••

j

1

a. = (a , ,aj J

a

8

i

,m.

k-l

, ••• ,Ct ' J

k

) E C ,

,. ~-l i Zll/2 I\0: II .. : i Ia I I j

L.

W.

1.'"

0

j

~

'"

a

( a ,8 , ••• ,8 ) ,..~ ",m+l ~, o 1 n

Iiall" !\~,m /

L '-'1=0

18

,\2J1/2. J

- 251 -

Suppose

[O,l, ••• ,m}

disjoint.

~

= DUE,

D and

E

are non-empty and

Suppose now that,

(2.7)

O,l, ... ,m

(2.B)

1,2, .•. ,m

j

\*-1

(2.9)

i=O

i

aj

i a~

Then there is a

0

whenever

Po' 0

JED, t E W.

< Po < 1 such that

(2.10)

for all

~

P

Po ,where

Po depends only on

The proof is delayed.

Definition.

Let

k

and

It starts with the proof of Lemma (2.12).

F(x,k) '" (f (x,k),f1(x,k), ••• ,f (x,k» o

m

be a vector of smooth functions defined on a domain For each

(y,k+l) E v aj

a

m.

~

= fj(y,k+l)

b

~

1

11' .. dkf j (Y+f: ) k j

n

~ C K X ~ •

let 0 J 1, • •• ,m

j

c:l

i

= 0, ••.

n

,q -1

If the (a~) satisfy (2.7), (2.8) and (2.9) we say that is a conjugate System on

~.

F(x,k)

- 252 -

~.

(2.7) simply says that each

fj (x,k) is regular on

IF(x,k)~P

and the conclusion of (2.6); namely (2.10), say that is sub-regular.

Thus,

11

Corollary (2.11). a domain such that

~,

iJ,

F

=

(fO,fl, ••• ,f ) m

then there is a

I F(x,k) 1p

Po' 0 < Po

is sub-regular on

is a conjugate system on

< I , Po independent of F, j()

for all

p? Po •

Examples of Conjugate Systems. Using (2.4) we see that if dkT.r,f

dkf

is defined on that same domain, without reference to the

global existence of

Ttf.

Since

f (see IV §3), if the domain

of

determined up to a constant by connected domain ~,

on

~

dkf ~

dkf.

is connected then Thus, if

f

is

f is defined on a

then, taking this point of view,

(tj};=l

be a subset of

least one odd and one even integer. the even integers in

(T, f,I, f, ... ,T. f) '1

is just the derived function

Ttf

is defined

and is uniquely determined, up to an additive constant.

Thus, let

~

is defined on a domain, then

~2

{t j

D

is the odd integers and

then by (2.5) and (2.6),

is a conjugate system on

"-'m

connected domain in

)

If

[O,l, ••• ,q-2), with at

K X 'll •

K X :11., or any

- 253 -

In particular, (f,Tlf), (f,T f, .•• ,T _ f) are called the l q 2 principal and full conjugate systems, respectively.

If

(q-I)/2 is odd, then (f'~q_I)/2f) is a conjugate system, T(q_l)/2f

is a natural analogue of the Hilbert transform of

f.

Lemma (2.12).

be given as in (2.6) so that

(2.7), (2.8) and (2.9) are valid.

Then given

is a constant

A PI

>

iiCt

Proof.

~e may assume that

I'al', = 0, P

(2.13) ...".

l

II

>

(m-l)/m, there

such that (2.10) holds for all

0

provided

o

PI

P

> PI

< A Iiall •

-

> I

PI

or

P

llall

+0,

and

° < PI ~ P S 1,

> 1 the result is trivial.

for if

2S~

-

~

Using (2.8) and (2.9) we obtain,

( '\I \Re \'La. - a.i\ 2 ~~1

j

'i

'" L

1 '\' S \' I ~i

J

J

I '\' -&ja.i\ 2 I

i

J

c.."jED

-

i ajCt. J

I

"JED '~

+,

\

a

I

-'j~E j

i

I,\'

l

a .a ~i"j EE J j

I II i) 2 + ',' ( :s \''-'1 , ('\'JED ,a L j I a J. ! ~i I .

(2.14)

+

'\ I

--'j

0 is independent of

p.

If we enter these rebults in (2.15) we get

"k-l

0

:'1=0

-

l

= II,all P

r

.; k

"k-l Thus, (11k) ;' ·-"1"0

?

PI

>

p

+

(m-l)/m and

_ 0_ _

m.E..ll::.El

IIaJl2

2

(ila

i.

p

lIa 112

,.

lIa-+a11P > IlallP'ok + m+l

II

_0_

Iiall

Iia 112 _0_

Ila!12

2

{~m--'l \

\2r

lIa

II 1 'I, J J

I ; .!!!E. ( ~) b _0_ I 2 P m II all

lIa-+a~IP ~

Ilail P , provided,

(1Iaoll/lialJ) < A ,where - PI

This completes the proof of (2.12).

Proof of (2.6). \'

assume that

I

-"i

If

Ila-+a~1 > 0

lIali .. 0

0

of (2.12) we may

fix

(since,

the result holds trivially so we may kliali ~

\

I

-"i

i Ila-+a II).

PI' (m-I)/m < PI < 1 and assume that

lIa II > Alia!!. o PI Thus we may assume,

(2.16)

Also, in view

(11k) \L-lla-+aill ~1

= 1, Ila 1I > A 0

-

PI

!lall, A

PI

>0

- 257 -

Note that if

and

Ilao".

lia II .. 0

then

0

j, and so

lIajli

1 .. 0

a

is not: consistent with

= 0

0

co

for all

all i

a i .. 0

and

j

and thus

(2.16).

j

lIall .. (11k)

" I

For

for all

. Ila~lll

~i

and

0..

110: II > o

Let

-

A

PI

II all •

>0

But A PI

so

satisfying (2.7), (2.8), (2.9) and (2.16).

= 1,

I!all: 0, a contradiction.

be the collection of all vectors

B

i

b

= [a~i}~:~

is a compact set in

B

Ck(m+l) •

To prove the theorem it will be sufficient to show that there is

a O,O f

iff

0

is a component The

"if"

part

follows froc~ (2.11) and (l.l)(c), the "only if" part follows from a result of Chao then

f

E HP

(2) that for implies

n

any multiplicative character on

* C,

- 262 -

Chapter VIII.

Almost everywhere convergence of Fourier series

The results

1.

Theorem

(1. 1) •

then !lJ!f E LP

and there is a constant

llrro fll

such that