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Macroeconomics Exam Review 117

Macroeconomics Exam Review 117 - Solutions for Foundations...

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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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