LECTURE NOTES ON DYADIC HARMONIC ANALYSIS 7

e2nix,

x E

[0,

1)),

such that

'Y( k)

=

jl( k)

=

JT

-zk df.L( z).

Such measure

f.L

is called

the

spectral measure of the process.

Moreover,

E(~k~n)

=

"f(k- n)

=

l

zn-zkdf.L(z)

=

(zn, zk)L2(J.L)·

Instead of studying the geometry of

{~n}nE7l

in

1{

we study the geometry of

{zn}nE7l

in L

2(f.L).

Let the

past

be denoted by

P

=

span{ zn : n

0} and the

future

by

:F

=

span{zn : n 2:

0}. When is the angle between the past and the future positive?

Remember that given two closed subspaces E 1,

E2

in a Hilbert space H then

cos(LE1E2)

=

sup{l(e1,e2)HI:

ei

E

Ei,

lleill

=

1}.

EXERCISE

1.6. Let

H, Et, E

2 be as above, assume that E1 nE2

=

{0}, E1

+E2

is dense in

H

then show that the following are equivalent: (I)

LE1E2

0, (II)

E

1

+

E2

=

H,

(III) The projection onto E1 parallel to

E2,

PE1 11E2

5

is bounded.

Moreover if

LE1E2

=

a

then

IIPEt~~E

2

II

=

1/

sin

a. Hint:

Use the Closed Graph

Theorem for the equivalences.

Now our question can be rephrased as: When is the

Riesz projection P+

bounded? Here

P+(~::::;

ckzk)

=

L:kO ckzk

is the projection onto the future par-

allel to the past, or the projection on.to the analytic part of the function. By the

next exercises, this is equivalent to asking when is

{zn}nE7l

a basis in L

2(f.L).

EXERCISE

1. 7. {

Xn}

is a basis on a Banach space X if and only if for all

x

E

X

there is a unique representation

x

=

I:

CkXk

(this is equivalent to asking that the

system {

Xn}

is complete and linearly independent). Show that {

Xn}

is a basis if and

only if IIPnll ~

C

for all

n,

where

Pn(l::ckxk)

=

2::~= 1

CkXk· Hint:

use Uniform

Boundedness Principle.

EXERCISE

1.8. In our setting

Pnf

=

l::~=O j(k)zk.

Show that

Pnf

=

P+f-

zn+l P+-zn+l

f.

Show also that if

P+

is bounded on

L2(f.L)

then

f.L

is absolutely

continuous with respect to Lebesgue measure:

df.L

=

w dm, w 2:

0.

There is an explicit formula for

P+f(z)

for lzl

1,

the Cauchy formula (the

projection onto the analytic part of the function). We already discussed the bound-

ary values when lzl

=

1:

1 1 -

P+f(z)

=

2f(z) +i2Hf(z),

lzl

=

1.

The behaviors of H and

fi

are similar as we mentioned before. Therefore the

question becomes: When is the Hilbert transform H bounded in L

2

(w)?

Where

w

is a

weight,

that is 0

~

w

E

Lfac·

This problem was solved in 1965 by Helson and

Szego, the necessary and sufficient conditions on the weight

w

are that

w

= eu+Hv,

for u,v

E

L

00

,

and llvlloo

7r/2

[HS]. They used complex analysis methods. In

1973, Hunt, Muckenhoupt and Wheeden found an equivalent condition using purely

real methods

[HMW]:

His bounded in

L

2

(w)

if and only if

wE

A

2

,

(1.3)

s~p

(

l~l1

w) (

~~~1

w-

1)

oo

A

2

-condition,

where the supremum is taken over all intervals

I.

We will say more about weights

and weighted inequalities in later lectures.

5If

x

E

E1 + E2,

then

x

=

XI +x2, Xi

E

Ei,

the decomposition is unique; define

PE1

11

E 2

x

=

XI.