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.
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