Macroeconomics Exam Review 127 - c 2001 Michael Carter All rights reserved Solutions for Foundations of Mathematical Economics Therefore For every

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 > +