Home > Analysis, Fixed Point > Brouwer fixed point theorem

Brouwer fixed point theorem

Let D be an open subset of \Bbb{R}^n such that \overline{D} is homeomorphic to the closed unit ball \overline{B} ( in \Bbb{R}^n). If \phi \in C(\overline{D}) (continuous function on \overline{D}) and \phi(\overline{D})\subset\overline{D}, then \phi has a fixed point in \overline{D}.

Proof: First reduce the problem to the case where D=B. To do this, consider the homeomorphism h:\overline{D} \to \overline{B}. Then \psi=h \circ \phi \circ h^{-1} maps \overline{B} to \overline{B}, and \psi(y)=y,\ y \in \overline{B} implies (since there exists x \in \overline{D} such that h(x)=y) we have h(\phi(x))=h(x), which means \phi(x)=x. This reduces the problem to consider our domain the unit ball in \Bbb{R}^n.

Now, if \psi(x_0)=x_0 for some x_0 \in \partial B then we are done. Suppose this does not happen. Consider the homotopy h_t(x)=x-t\psi(x), x \in \overline{B}, 0 \leq t\leq 1.  It is clear that if x \in \partial B and 0 \leq t <1, then t \psi(x) \in B, hence h_t(x)\neq 0, \forall x \in \partial B, 0\leq t<1. By hypothesis, o \notin h_1(\partial B). By invariance of topological degree of a mapping at a point which is not in the image of the boundary of the open set on which the mappings are is defined, under a homotopy, we have d(I-\psi,B,0)=d(I,B,0) (I is the identity mapping). But d(I,B,0)=1 since 0 \in B.

If the topology degree of a mapping \theta at a point is not 0, then there exists a point x in the domain of \theta such \theta(x)=0. Therefore, there exists an element x \in B such that x-\psi(x)=0, meaning that \psi, and therefore \phi has a fixed point.

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