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Brouwer's Fixed Point Theorem

Please see the attached file for the fully formatted problems.

Prove that if D is the closed disc |x| =< 1 in R2, then any map f E C2[D --> D]
has a fixed point: f(x) = x. The proof is by contradiction, and uses Stokes theorem. Follow the steps outlined below.
(1) Define a new map F(x) = ...
Show that F has no fixed points if r is small enough.
(2) Draw the ray from F(x) to x (these are distinct) and note where it cuts the circle C : |x| = 1. This point G(x) = (cos , sin ) depends smoothly on x, i.e.  2 C2(D); moreover, it reduces to the identity on C.
(3) Now compute
(4) Explain why the above is a contradiction?
Since this is an analysis problem, please be sure to be rigorous, and include as much detail as possible so that I can understand. Please also state if you are making use of some fact or theorem. Thanks!


Solution Summary

Problems involving Brouwer's fixed Point Theorem are solved. The solution is detailed and well-presented.