3
FOURIER ANALYSIS
26
Note.
k
f
k
=
h
f, f
i
1
/
2
=
h
f, f
i
1
h
f, f
i
1
/
2
=
h
f, f
i
1
k
f
k
=
⌧
f,
f
k
f
k
=
h
f, g
i
where
g
=
f
k
f
k
and
k
g
k
=
f
k
f
k
= 1.
Definition.
A
Hilbert space
is a complete inner product space, i.e. a Banach space where the norm comes from
an inner product.
Example.
1.
C
n
is a Hilbert space.
2.
`
2
is a Hilbert space.
3.
L
2
(
E
) is a Hilbert space, by ReiszFisher theorem.
4.
C
[0
,
1] is not a Hilbert space, because it is dense in
L
2
with
k
·
k
L
2
. (It is complete with respect to the supnorm,
but the supnorm does not come from an inner product.)
3.2
Orthogonality
Let
H
be a Hilbert space.
Definition.
x, y
2
H
are orthogonal (denoted
x
?
y
) are orthogonal if
h
x, y
i
= 0.
Example.
1. (1
,
0
,
0)
?
(0
,
1
,
0) in
C
3
.
2.
χ
E
?
χ
E
C
in
L
2
(
R
).
3.
x
?
x
i
↵
x
= 0.
Definition.
We say
S
✓
H
is an
orthogonal set
if
x
?
y
whenever
x, y
2
S
and
x
6
=
y
. Moreover,
S
is
orthonormal
if it is orthogonal and
k
x
k
= 1
8
x
2
S
.
Note.
In an orthonormal set
S
,
h
x, y
i
=
(
0
x
6
=
y
1
x
=
y
.
Example.
Consider
H
=
`
2
. Put
e
n
have a 1 in the
n
th position and a 0 everywhere else.
S
=
{
e
n
}
n
is orthonormal.
Pythagorean Theorem.
If
x
?
y
, then
k
x
+
y
k
2
=
k
x
k
2
+
k
y
k
2
.
Proof.
k
x
+
y
k
2
=
h
x
+
y, x
+
y
i
=
h
x, x
i
+
h
x, y
i

{z
}
0
+
h
y, x
i

{z
}
0
+
h
y, y
i
=
k
x
k
2
+
k
y
k
2
.
Corollary.
More generally, if
{
x
k
k
N
1
is orthogonal, then
P
N
k
=1
x
k
2
=
P
N
k
=1
k
x
k
k
2
.
Theorem.
If
S
is orthogonal and consists of nonzero vectors, then
S
is a linearly independent set.
Proof.
Suppose
P
N
i
=1
a
i
x
i
= 0, where
x
i
2
S
and
a
i
2
C
. Then
0 =
N
sup
i
=1
a
i
x
i
2
=
X
k
a
i
x
i
k
2
=
X

a
i

2
k
x
i
k
2
)

a
i

2
k
x
i
k
2
= 0
8
i
and thus
x
i
= 0
8
i
.