# 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!

https://brainmass.com/math/graphs-and-functions/brouwers-fixed-point-theorem-18701

#### Solution Summary

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