The proof I see for the Schauder fixed point theorem for Banach spaces is basically the following:

If $K$ is compact and convex and $A\colon K\to K$ is continuous, for each $\epsilon$ find a finite collection of points whose $\epsilon$-neighborhoods cover $K$, and take the convex hull $C$. There's a map $f\colon K\to C$ for which $\sup_K d(x,f(x))\leq\epsilon$. Since $C$ is homeomorphic to the unit ball of some dimension, $f\circ A |_C$ has a fixed point $x_\epsilon$ by the Brouwer fixed point theorem. A convergent subsequence as $\epsilon \to 0$ will be a fixed point of $A$.

I found this a little unsatisfying. A couple possible reasons: the proof is as much analysis as topology, and convexity is a rather un-topological property- it's easy to destroy with slight deformations which oughtn't change basic topological properties.

It'd seem more natural to me if we could say "here's a connectedness property which, along with compactness, implies the fixed point property, and we can easily see that convex subsets of Banach spaces are connected in that way." For instance, does a compact space which is both locally and globally contractible have the fixed point property? If so this would be a very tidy corollary.

Is anyone aware of a result along these lines, or a conjecture that people have seriously attempted to prove?

(Searching around I see that people in the 30s asked the question of whether every contractible compact space has the fixed point property. That was answered in the negative by someone named Kinoshita in 1953, with a counterexample which is contractible but not locally connected. BTW I understand just enough algebraic topology to have a vague idea what Lefschetz's fixed point theorem is about, but obviously an infinite dimensional space is not a finite simplicial complex.)

  • $\begingroup$ If a compact convex sub-set of a Banach space is an euclidean neighbourhood retract, then the version of Lefschetz-Hopf proven in Dold's Lectures on Algebraic Topology (1980) Proposition 6.22 would apply. However, I am not entirely confident that this is true. After some searching, I might ask this as a separate question and come back to this one once it when it gets an answer. $\endgroup$ – Ben Nov 8 '18 at 13:57
  • $\begingroup$ Of course not, the Hilbert cube is infinite-dimensional and ENRs are finite dimensional. $\endgroup$ – Ben Nov 8 '18 at 14:26

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