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Unformatted text preview: y Since y / , y must be a boundary point of . By the previous exercise, there exists a supporting hyperplane at y , that is there exists a continuous linear functional such that ( y ) ( x ) for every x 3.185 1. ( ) . 2. ( ) is convex and hence an interval (Exercise 1.160. 3. ( ) is open in (Proposition 3.2). 3.186 is nonempty and convex and / . (Otherwise, there exists x and y such that = y + ( x ) which implies that x = y contradicting the assumption that = .) Thus there exists a continuous linear functional such that ( y x ) ( ) = 0 for every x , y so that ( x ) ( y ) for every x , y Let = sup x ( x ). Then ( x ) ( y ) for every x , y By Exercise 3.185, (int ) is an open interval in (...
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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