3.51
By Exercise 3.49, there exists a linear functional
𝑓
such that
𝐻
=
{
𝑥
∈
𝑋
:
𝑓
(
𝑥
) = 0
}
Since
x
0
/
∈
𝐻
,
𝑓
(
x
0
)
∕
= 0. Without loss of generality, we can normalize so that
𝑓
(
x
0
) =
1. (If
𝑓
(
x
0
) =
𝑐
∕
= 1, then the linear functional
𝑓
′
=
1
/
c
𝑓
has
𝑓
′
(
x
0
) = 1 and kernel
𝑓
′
=
𝐻
.)
To show that
𝑓
is unique, suppose that
𝑔
is another linear functional with kernel
𝑔
=
𝐻
and
𝑔
(
x
0
) = 1. For any
x
∈
𝑋
, there exists
𝛼
∈ ℜ
such that
x
=
𝛼
x
0
+
v
with
𝑣
∈
𝐻
(Exercise 1.153). Since
𝑓
(
v
) =
𝑔
(
v
) = 0 and
𝑓
(
x
0
) =
𝑔
(
x
0
) = 1
𝑔
(
x
) =
𝑔
(
𝛼
x
0
+
v
) =
𝛼𝑔
(
𝑥
0
) =
𝛼𝑓
(
x
0
) =
𝑓
(
𝛼
x
0
+
v
) =
𝑓
(
x
)
3.52
Assume
𝑓
=
𝜆𝑔
,
𝜆
∕
= 0. Then
𝑓
(
𝑥
) = 0
⇐⇒
𝑔
(
𝑥
) = 0
Conversely, let
𝐻
=
𝑓
−
1
(0) =
𝑔
−
1
(0).
If
𝐻
=
𝑋
, then
𝑓
=
𝑔
=
0
.
Otherwise,
𝐻
is a hyperplane containing
0
.
Choose some
x
0
/
∈
𝐻
.
Every
x
∈
𝑋
has a unique
representation
x
=
𝛼
x
0
+
v
with
v
∈
𝐻
(Exercise 1.153) and
𝑓
(
x
) =
𝛼𝑓
(
x
0
)
𝑔
(
x
) =
𝛼𝑔
(
x
0
)
Let
𝜆
=
𝑓
(
x
0
)
/𝑔
(
x
0
) so that
𝑓
(
x
0
) =
𝜆𝑔
(
x
0
). Substituting
𝑓
(
x
) =
𝛼𝑓
(
x
0
) =
𝛼𝜆𝑔
(
x
0
) =
𝜆𝑔
(
x
)
3.53
𝑓
continuous implies that the set
{
𝑥
∈
𝑋
:
𝑓
(
𝑥
) =
𝑐
}
=
𝑓
−
1
(
𝑐
) is closed for every
𝑐
∈ ℜ
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 Fall '10
 Dr.DuMond
 Macroeconomics, Existence, All rights reserved, Quantification, Functional

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