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Macroeconomics Exam Review 148

# Macroeconomics Exam Review 148 - Solutions for Foundations...

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3.142 Clearly, if 𝑓 is convex, it is locally convex at every x 𝑆 , where 𝑆 is the required neighborhood. To prove the converse, assume to the contrary that 𝑓 is locally convex at every x 𝑆 but it is not globally convex. That is, there exists x 1 , x 2 𝑆 such that 𝑓 ( 𝛼 x 1 + (1 𝛼 ) x 2 ) > 𝛼𝑓 ( x 1 ) + (1 𝛼 ) 𝑓 ( x 2 ) Let ( 𝑡 ) = 𝑓 ( 𝑡 x 1 + (1 𝑡 ) x 2 ) Local convexity implies that 𝑓 is continuous at every x 𝑆 (Corollary 3.8.1), and therefore continuous on 𝑆 . Therefore, is continuous on [0 , 1]. By the continuous maximum theorem (Theorem 2.3), 𝑇 = arg max x [ x 1 , x 2 ] ( 𝑡 ) is nonempty and compact. Let 𝑡 0 = max 𝑇 . For every 𝜖 > 0, ( 𝑡 0 𝜖 ) ( 𝑡 0 ) and ( 𝑡 0 + 𝜖 ) < ℎ ( 𝑡 0 ) Let x 0 = 𝑡 0 x 1 + (1 𝑡 0 ) x 2 and x 𝜖 = ( 𝑡 0 + 𝜖 ) x 1 + (1 𝑡 0 𝜖 ) x 2 Every neighborhood 𝑉 of x 0 contains x 𝜖 , x 𝜖 [ x 1 , x 2 ] with 1 2 𝑓 ( x 𝜖 ) + 1 2 𝑓 ( x 𝜖 ) = 1 2 ( 𝑡 0 𝜖 ) + 1 2 ( 𝑡 0 + 𝜖 ) < ℎ ( 𝑡 0 ) = 𝑓 ( x 0 ) = 𝑓 ( 1 2 x 𝜖 + 1 2
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