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    Brouwer's Fixed Point Theorem

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    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!

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    Solution Summary

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