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Unformatted text preview: Econ 204 Problem Set 6 Due Monday, August 17th Exercise 1 Consider f : R 2 ! R 2 such that f 2 C 3 ( R 2 ) . Now let F ( x;y;w;z ) = f ( x;y ) & ( w;z ) and suppose that F ( x;y;w;z ) = 0 has solutions in R 4 : Let S R 4 be the set of solutions to this system. Show that there exists a set B such that B c has measure zero and for ( x;y;w;z ) 2 S where ( w;z ) 2 B , there is a local implicit function h : W R 2 ! R 2 ( W open) such that F ( h ( w;z ) ;w;z ) = 0 for all ( w;z ) 2 W and h 2 C 3 ( R 2 ) : Exercise 2 Let f : [0 ; 1] ! [0 ; 1] be a correspondence de&ned as f ( x ) = f ; 1 = ( x + 1) g for x 6 = 0 and f (0) = f 1 = 2 g : Does f have a &xed point? If yes, &nd the point(s). Does any of the &xed point theorems you have learned apply here? Explain. Answer the same questions for f ( x ) = [ &; 1 = ( x +1)] for all x 2 [0 ; 1] where < & < 1 = 2 . Exercise 3 We say that a relation R on X is convex if whenever xRy and zRy then ( x + (1 & ) z ) Ry for all 2 (0 ; 1) . (if x...
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This note was uploaded on 10/04/2010 for the course ECON 204 taught by Professor Anderson during the Summer '08 term at University of California, Berkeley.
- Summer '08