3
FOURIER ANALYSIS
29
Proof.
(
)
) By part (a) of the theorem,
x
=
P
1
k
=1
h
x, e
k
i
e
k
. By the generalized Pythagorean Theorem,
k
x
k
2
=
1
X
k
=1
k h
x, e
k
i
e
k
k
2
=
1
X
k
=1

h
e
k
, x
i

2
.
(
(
) Suppose
{
e
k
}
is not complete. Get
x
6
= 0 such that
x
?
e
k
8
k
. By assumption,
k
x
k
2
=
X

h
x, e
k
i

2
=
X
0 = 0
)
x
= 0
.
Thus the result holds by contradiction.
Corollary.
The orthonormal set
{
e
k
}
is complete i
↵
T
:
H
!
`
2
:
x
7!
(
h
x, e
k
i
)
1
k
=1
is 11.
Proof.
(
)
) Suppose
T
(
x
) =
T
(
y
). Then
h
x, e
k
i
=
h
y, e
k
i
8
k
, so
x

y
?
e
k
8
e
k
, so
x
=
y
by completeness.
(
(
) Suppose
{
e
k
}
is not complete. Then
9
x
6
= 0 such that
x
?
e
k
8
k
. Then
T
(
x
) = 0 =
T
(0), so
T
is not 11.
Lecture 19: June 17
Summary.
The orthonormal set
{
e
k
}
1
k
=1
is complete i
↵
any of the following are true:
(0)
h
x, e
k
i
= 0
8
k
)
x
= 0. (The definition.)
(1)
H
=
sp
{
e
k
}
. (A basis for
H
.)
(2)
x
=
P
1
k
=1
h
x, e
k
i
e
k
8
x
2
H
.
(3)
k
x
k
2
=
P
1
k
=1

h
x, e
k
i

2
8
x
2
H
(Bessel’s inequality holds with equality.)
(4)
T
:
H
!
`
2
:
x
7!
(
h
x, e
k
i
)
1
k
=1
is 11.
We saw previously that
T
is onto: Given (
β
k
)
2
`
2
,
x
=
P
1
k
=1
β
k
e
k
2
H
and
β
k
=
h
x, e
k
i
. This combined with (3)
gives
k
T
(
x
)
k
2
`
2
=
1
X
k
=1

h
x, e
k
i

2
=
k
x
k
2
H
.
(5)
T
is an isometric (normpreserving) isomorphism (11, onto, linear).
Grand Theorem.
Every Hilbert space
H
has a basis. If
H
is separable, then any orthonormal set is countable (or
finite).
Important Fact.
L
2
[
a, b
] is separable because polynomials with rational coe
ffi
cients are dense.
Proof of Grand Theorem.
We first prove every Hilbert space
H
has a basis. Let
S
the the set of orthonormal sets
in
H
. Define a partial order on
S
by inclusion:
A
B
()
A
✓
B
. We need to show that every chain
C
has an
upper bound. Take
X
=
S
C
2
C
C
. Certainly
X
≥
C
8
C
2
C
, so we must show
X
2
S
. If
x
2
X
, then
x
2
C
for
some
C
2
C
, so
k
x
k
= 1.
Let
x
6
=
y
2
X
.
Since
C
is a chain, there is some
C
2
C
such that
x, y
2
C
.
But
C
is
an orthonormal set, so
x
?
y
. Thus
X
2
S
is an upper bound for
C
. By Zorn’s lemma,
S
has a maximal element
A
. If
9
x
6
= 0 such that
h
x, a
i
= 0
8
a
2
A
, then
A
[
{
x
k
x
k
}
would be an orthonormal set that properly contains
A
,
contradicting the maximality of
A
. Hence
A
is a complete orthonormal set, i.e. a basis for
H
.
Now suppose
H
is separable, and we shall prove
A
is countable. Notice that if
{
e
1
, e
2
}
is orthonormal, then
k
e
1

e
2
k
2
=
k
e
1
k
2
+
k 
e
2
k
2
= 2. Say
{
e
↵
}
is an uncountable orthonormal set. Let
{
f
n
}
1
1
be a countable dense subset of
H
.
Take
e
↵
and choose
n
such that
e
↵
2
B
(
f
n
,
1
2
). Since there are countably many such balls, and uncountably many
e
↵
s, there exists
n
and
e
↵
6
=
e
β
such that
e
↵
, e
β
2
B
(
f
n
,
1
2
). Then
p
2 =
k
e
↵

e
β
k k
e
↵

f
n
k
+
k
f
n

e
β

1
,