## Background

At the risk of greatly oversimplifying matters, let me state a heuristic from Granas and Dugundji's beautiful book: *fixed point theorems fall into two broad categories*. The first class is usually functional analytic and imposes strong conditions on the *map* $f:X \to X$ whereas the second class is usually algebraic topological and imposes strong conditions on the *space* $X$ itself.

A typical example of the first class of theorems is the fixed point theorem of Banach. While the spaces it applies to are fairly general (complete metric spaces), the function must have a Lipschitz constant strictly less than $1$. On the other hand, Brouwer's theorem falls into the second class. Any continuous map works, but the domain must be a compact and convex subset of Euclidean space (originally a disk). Of course, both these theorems have been *vastly* generalized from the versions that I am stating here.

## Question

One fundamental advantage of the Banach theorem is that it actually provides a recipe for converging to the fixed point as part of the standard proof: just start at an initial point and iterate. The proofs of the Brouwer theorem that I have seen do no such thing. The best known proof (I think) is the one by contradiction: assuming the domain is a disk, if $f(x)$ and $x$ are always distinct then the ray from $f(x)$ through $x$ to the boundary of said disk provides a deformation-retraction from the disk to its boundary, aha!

Here is my question:

Is there any way to actually find a fixed point when using Brouwer's theorem?

## A Possible Idea

One scheme that unfortunately fails is as follows. Consider the sequence of iterates $f^n(x)$ for $n \in \mathbb{N}$ and any initial $x$ in the domain. We have an infinite sequence in a compact set, and hence a convergent subsequence, so the limit point is a candidate. This won't work since **a**) we haven't used convexity at all, and **b**) one may just be converging to a periodic orbit of $f$.

anyapproach is not fast. Brouwer-completeness should be understood analogously to (it is different from) NP-completeness. $\endgroup$