A generalization of Brower's fixed point theorem says that any continuous map from compact, convex set in the plane $K$ to itself, $f:K \to K$, must have a fixed point.
It is easy to see that any set homeomorphic to the closed unit disk must have a Brower fixed point theorem. However, the above generalization includes a wider array of allowed sets than just those homeomorphic to the closed unit disk. For instance, it includes the set $\{0\} \times [0,1]$.
In some sense, that is not surprising as that just reduces to a lower dimensional case. But this makes me wonder, what additional sets are being added? Is it just lower dimensional cases? In other words, what are the conditions for a compact, convex set to be homeomorphic to the closed unit ball? Purely that it is 2 dimensional?
Best Answer
Yes, indeed, every nonempty compact convex subset of $R^n$ is homeomorphic to the closed unit ball of suitable dimension.
See my answer here. In the answer I was using the assumption that the set in question is symmetric and has nonempty interior; in your case it need not be symmetric so you use an asymmetric norm $p(x)$ instead. The point is that every convex subset $C$ with nonempty interior in the $n$-dimensional affine space $A^n$ is absorbing with respect to the origin that belongs to the interior of $C$ and, hence, defines a norm on the corresponding vector space. If you have a nonempty convex subset $C\subset R^n$ with empty interior, then $C$ will have nonempty interior in its affine hull, i.e. the smallest affine subspace of $R^n$ containing $C$, see for instance, Rahul's answer here.