# Minimax theorem on a non convex domain

A minimax theorem is a theorem which states that under certain conditions on $\mathcal{X}$, $\mathcal{Y}$ and $f$: $$\inf_{x \in \mathcal{X}}{\sup_{y \in \mathcal{Y}}{f(x,y)}} = \sup_{y \in \mathcal{Y}}{\inf_{x \in \mathcal{X}}{f(x,y)}}$$ All minimax theorems rely strongly on convexity: the sets $\mathcal{X}$ and $\mathcal{Y}$ are usually required to be convex subsets of vector spaces, and $f$ to be convex-concave.

My question is: of course one cannot expect much without some convexity, but is it possible to get rid of the convexity of the domain in any way?

For example, consider the simplest minimax problem: $\mathcal{X}$ is a subset of $\mathbb{R}^n$, $\mathcal{Y}$ a convex subset of $\mathbb{R}^n$ and $f(x,y) = \langle x,y \rangle$ is the usual scalar product.

Are there any minimax results in this setting (with of course additional assumptions on $\mathcal{X}$ or $\mathcal{Y}$, but $\mathcal{X}$ should not be assumed to be convex) ?

• I don't understand. You said "quasi convexity of the domain", so are you really speaking about the domain $\mathcal{X}$ (which is my question) or about the function $f$ (which is not really the point here) ? – Adrien Jan 26 '14 at 2:53
The situation were your objective $f(x,y)=x^T A y$ is bilinear corresponds to the case of zero-sum games (http://www.inf.ed.ac.uk/teaching/courses/agta/lec4.pdf). If you remove the convexity assumption in this framework, existence of equilibria is not guaranteed anymore. For instance, if you restrict your variables x, y to be integral then generally there is no equilibrium for the zero-sum game (and this is precisely the motivation for mixed strategies in Von Neumann's Theorem).