Therefore
ˆ
𝑆
⊆
𝑆
. For every
x
∈
𝑋
(
𝑓
(
x
)
z
−
𝑓
(
z
)
x
)
𝑇
z
=
𝑓
(
x
)
z
𝑇
z
−
𝑓
(
z
)
x
𝑇
z
= 0
since
z
∈
𝑆
⊥
. Therefore
𝑓
(
x
) =
𝑓
(
z
)
∥
z
∥
2
x
𝑇
z
=
x
𝑇
(
z
𝑓
(
z
)
∥
z
∥
2
)
=
x
𝑇
y
where
y
=
z
𝑓
(
z
)
∥
z
∥
2
3.76
𝑋
∗
is always complete (Proposition 3.3). To show that it is a Hilbert space, we
have to that it has an inner product. For this purpose, it will be clearer if we use an
alternative notation
<
x
,
y
>
to denote the inner product of
x
and
y
. Assume
𝑋
is a
Hilbert space. By the Riesz representation theorem (Exercise 3.75), for every
𝑓
∈
𝑋
∗
there exists
y
𝑓
∈
𝑋
such that
𝑓
(
x
) =
<
x
,
y
𝑓
>
for every
x
∈
𝑋
Furthermore, if
y
𝑓
represents
𝑓
and
y
𝑔
represents
𝑔
∈
𝑋
∗
, then
y
𝑓
+
y
𝑔
represents
𝑓
+
𝑔
and
𝛼
y
𝑓
represents
𝛼𝑓
since
(
𝑓
+
𝑔
)(
x
) =
𝑓
(
x
) +
𝑔
(
x
) =
<
x
,
y
𝑓
>
+
<
x
,
y
𝑔
>
=
<
x
,
y
𝑓
+
y
𝑔
>
(
𝛼𝑓
)(
x
) =
𝛼𝑓
(
x
) =
𝛼 <
x
,
y
𝑓
>
=
<
x
, 𝛼
y
𝑓
>
Define an inner product on
𝑋
∗
by
< 𝑓, 𝑔 >
=
<
y
𝑔
,
y
𝑓
>
We show that it satisfies the properties of an inner product, namely
symmetry
< 𝑓, 𝑔 >
=
<
y
𝑔
,
y
𝑓
>
=
<
y
𝑓
,
y
𝑔
>
=
< 𝑔, 𝑓 >
additivity
< 𝑓
1
+
𝑓
2
, 𝑔 >
=
<
y
𝑔
,
y
𝑓
1
+
𝑓
2
>
=
<
y
𝑔
,
y
𝑓
1
+
y
𝑓
2
>
=
< 𝑓
1
, 𝑔 >
+
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- Fall '10
- Dr.DuMond
- Macroeconomics, Hilbert space, inner product
-
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