Reference request on Min-Max theorem

Question

Consider the following min-max problem

$$\inf_{x\in M} \sup_{y\in N} F(x,y),$$

where $F: M\times N\to\mathbb R$ is Lipschitz and $y\mapsto F(x,y)$ is concave for all $x\in M$. Could we derive $\inf_{x\in M} \sup_{y\in N} F(x,y)=\sup_{y\in N} \inf_{x\in M}F(x,y)$ if $M\subset \mathbb R^m$ and $N\subset\mathbb R^n$ are both compact?

PS: To the best of my knowledge, the reference on Min-Max theorem is from M. Sion : https://msp.org/pjm/1958/8-1/pjm-v8-n1-p14-p.pdf However, the convexity of $x\mapsto F(x,y)$ is missing in my case. Any comments or references are highly appreciated!

PS2: Thank Nik Weaver for the counterexample and Iosif Pinelis for providing a helpful condition. The function above is defined as

$$F(x,y)~:=~\sum_{k=1}^n\int_{V_k(x,y)}\left\{|z-x_k|^2-y_k\right\}\rho(z)dx+\sum_{k=1}^n p_ky_k,$$

where $x=(x_1,\ldots, x_n)\in\Omega^n$, $y=(y_1,\ldots, y_n)\in\mathbb R^n$ and

$$V_k(x,y)~:=~\big\{z\in\Omega:~ |z-x_k|^2-y_k\le |z-x_{i}|^2-y_{i},~ \forall 1\le i\le n\big\}.$$

Here $\Omega\subset\mathbb R^d$ is compact, $\rho$ is a density function on $\Omega$ and $p_1,\ldots, p_n\in (0,1)$ are given weights satisfying

$$\int_{\Omega}\rho(z)dz ~=~ 1 ~=~ \sum_{k=1}^n p_k.$$

According to Iosif Pinelis, to show $\inf_{x\in\Omega^n}\sup_{y\in\mathbb R^n}F(x,y)=\sup_{y\in\mathbb R^n}\inf_{x\in\Omega^n}F(x,y)$, it suffices to show, for each $y\in\mathbb R^n$, there exists a unique $x_y\in\Omega^n$ s.t. $\inf_{x\in\Omega^n}F(x,y)=F(x_y,y)$.

It is known that $(V_k)_{1\le k\le n}$ is the weighted Voronoi tessellation (if $y=(0,\ldots, 0)$ it becomes the Voronoi tessellation, and the unique minimizer is given by the centroidal Voronoi tessellation).

Answer 1

It's false. Take $M = [0,1]$ and $N = \mathbb{R}$ and define $F(x,y) = 1 - |x-y|$. Taking $y_x = x$ satisfies condition (2). Here $\inf_M \sup_N F(x,y) = 1$ and $\sup_N \inf_M F(x,y) = 1/2$, achieved when $y = 1/2$.

Edit: this answers the original question. The new version of the question, with $N$ compact, is falsified by taking $N=[0,1]$ in the above example.

Answer 2

As pointed out by Nik Weaver, the minimax duality will not hold in general without assuming that $F(x,y)$ is convex in $x$.

However, if, for instance, for each $y$ the minimum of $F(x,y)$ in $x$ is attained at only one point, then under natural regularity conditions we have the minimax duality. See e.g. Theorem 1.1 or Theorem 1 or Theorem 1.

Remark 1: In Nik Weaver's counterexample, with $M=N=[0,1]$, the condition that the minimum of $F(x,y)$ in $x$ be attained at only one point was violated only for one value of $y$, namely $y=1/2$, and that was enough to bring the minimax duality down! On the other hand, as seen from the above citations, it is not necessary to require a unique minimizer in $x$ for each $y$ -- it is enough to require it just for one special $y$. In Nik Weaver's counterexample, $1/2$ would be precisely that special $y$ -- for which the uniqueness condition fails, though.

Remark 2: A necessary and sufficient condition for the minimax duality for generalized concave-convex functions $F$ is given here. This condition consists in the upper semi-continuity at $0$ of a certain functional constructed based on $F$.