Browder's fixed point theorem in non-Hausdorff topological vector spaces

Question

Browder proved the following fixed point theorem in his 1968 Mathematische Annelen paper (Theorem 1):

Theorem. Let $K$ be a non-empty compact convex subset of a topological vector space $E$ (where we assume that $E$ is separated but not necessarily locally convex). Let $T$ be a mapping of $K$ into $2^K$, where for each $x$ in $K$, $T(x)$ is a non- empty convex subset of $K$. Suppose further that for each $y$ in $K$, $T^{-1}(y) = \{x | x \in K , y \in T(x)\}$ is open in $K$. Then there exists $x_0$ in $K$ such that $x_0\in T(xo)$.

In the paper he assumed Hausdorff separation axiom, and in the proof of the theorem he explicitly used it.

I wonder whether this theorem can be generalized to non-Hausdorff topological vector spaces. If not, is there a counterexample?

This question is of interest to economics since Browder's fixed point theorem has been used to show the existence of an equilibrium in games. And it is not rare to see papers that work with general topological vector spaces (they explicitly mention they do not assume Hausdorff separation axiom).

Answer 1

It is true indeed without assuming the Hausdorff separation axiom on the topological vector space $E$, for the reason that we can quotient over the closure of the origin, $\overline{\{0\}}$.

Let's denote $\pi:E\to \tilde E:=E/\overline{\{0\}}$ the quotient projection, and consider $\tilde K:=\pi (K)$, a compact convex subset of $\tilde E$.

For any $x\in K$ the set $T^{-1}(T(x))$ is a nbd of $x$ by assumption, hence it contains the closure of $x$ in $K$, that is all $u\in K$ such that $\pi x=\pi u$. By symmetry we conclude that $T(x)=T(u)$ whenever $\pi x=\pi u$. So the map $T$ factor through $\pi$, and there is a well-defined map $\tilde T:\tilde K\to 2^{\tilde K}$ such that for any $x\in K$, $\tilde T(\pi x)=\pi(T(x))$.

The multi $\tilde T$ satisfy the hypotheses of Browder's fixed point theorem for Hausdorff TVS, because for any $\tilde x\in \tilde K$, $\tilde T(\tilde x)$ is a non-empty convex subset of $\tilde K$ , and for any $y\in E$
$$\tilde T^{-1}(\pi y)=\bigcup_{v\in \overline y}\pi \big(T^{-1}(v)\big) \, , $$ an open subset of $K$ because $\pi:K\to\tilde K$ is an open map.

In consequence there is $u_0\in K$ such that $\pi u_0\in\pi\big(T(u_0)\big)$. This means that there is $x_0\in K$ such that $\pi u_0=\pi x_0$ and $x_0\in T(u_0)$: but as observed $T(u_0)=T(x_0)$ and $x_0$ is a fixed point of $T$.