{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Macroeconomics Exam Review 92

Macroeconomics Exam Review 92 - c 2001 Michael Carter All...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
= Let 𝐴 0 , 𝐴 1 , . . . , 𝐴 𝑛 be closed subsets of an 𝑛 dimensional simplex 𝑆 with vertices x 0 , x 1 , . . . , x 𝑛 such that conv { x 𝑖 : 𝑖 𝐼 } ⊆ 𝑖 𝐼 𝐴 𝑖 for every 𝐼 ⊆ { 0 , 1 , . . ., 𝑛 } . For 𝑖 = 0 , 1 , . . ., 𝑛 , let 𝑔 𝑖 ( x ) = 𝜌 ( x , 𝐴 𝑖 ) For any x 𝑆 with barycentric coordinates 𝛼 0 , 𝛼 1 , . . . , 𝛼 𝑛 , define 𝑓 ( x ) = 𝛽 0 x 0 + 𝛽 1 x 1 + ⋅ ⋅ ⋅ + 𝛽 𝑛 x 𝑛 where 𝛽 𝑖 = 𝛼 𝑖 + 𝑔 𝑖 ( x ) 1 + 𝑛 𝑗 =0 𝑔 𝑗 ( x ) 𝑖 = 0 , 1 , . . . , 𝑛 (2.45) By construction 𝛽 𝑖 0 and 𝑛 𝑖 =0 𝛽 𝑖 = 1. Therefore 𝑓 ( x ) 𝑆 . That is, 𝑓 : 𝑆 𝑆 . Furthermore 𝑓 is continuous. By Brouwer’s theorem, there exists a fixed point 𝑥 with 𝑓 ( x ) = x . That is 𝛽 𝑖 = 𝛼 𝑖 for 𝑖 = 0 , 1 , . . ., 𝑛 . Now, since the collection 𝐴 0 , 𝐴 1 , . . . , 𝐴 𝑛 covers 𝑆 , there exists some 𝑖 for which 𝜌 ( x , 𝐴 𝑖 ) = 0. Substituting 𝛽 𝑖 = 𝛼 𝑖 in (2.45) we have 𝛼 𝑖 = 𝛼 𝑖 1 + 𝑛 𝑗 =0 𝑔 𝑗 ( x ) which implies that 𝑔 𝑗 ( x ) = 0 for every 𝑗 . Since the
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}