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Unformatted text preview: 3.142 Clearly, if is convex, it is locally convex at every x , where is the required neighborhood. To prove the converse, assume to the contrary that is locally convex at every x but it is not globally convex. That is, there exists x 1 , x 2 such that ( x 1 + (1 ) x 2 ) > ( x 1 ) + (1 ) ( x 2 ) Let ( ) = ( x 1 + (1 ) x 2 ) Local convexity implies that is continuous at every x (Corollary 3.8.1), and therefore continuous on . Therefore, is continuous on [0 , 1]. By the continuous maximum theorem (Theorem 2.3), = arg max x [ x 1 , x 2 ] ( ) is nonempty and compact. Let = max . For every > 0, ( ) ( ) and ( + ) < ( ) Let x = x 1 + (1 ) x 2 and x = ( + ) x 1 + (1 ) x 2 Every neighborhood of x contains x , x...
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This note was uploaded on 01/16/2012 for the course ECO 2024 taught by Professor Dr.dumond during the Fall '10 term at FSU.
- Fall '10