Macroeconomics Exam Review 117 - Solutions for Foundations...

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