# When $\min \max = \max \min$?

Let $X \subset \mathbb{R}^n$ and $Y \subset \mathbb{R}^m$ be compact sets.

Consider a continuous function $f : X \times Y \rightarrow \mathbb{R}$.

Say under which condition we have

$$\min_{x \in X} \max_{y \in Y} f(x,y) = \max_{y \in Y} \min_{x \in X} f(x,y).$$

From this we have that $\max_{y \in Y} \min_{x \in X} f(x,y) \leq \min_{x \in X} \max_{y \in Y} f(x,y)$. So here we are looking for conditions on $f$ such that we have the equality.

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I would suggest you try $n=m=1, X=Y=[0,1]$ and some functions that are easy to handle, like a constant, $f(x,y)=x, f(x,y)=xy, f(x,y)=x+y$ and see if this gives you any ideas. – Ross Millikan Aug 31 '12 at 16:59
Typical sufficient conditions are $X,Y$ compact and convex, $f$ continuous, convex in $x$, concave in $y$. See Rockafellar, "Convex Analysis", Corollary 37.3.2. – copper.hat Aug 31 '12 at 17:05
In the context of Continuous Logic this is to ask when a formula $f(x,y)$ is such that $\exists x\forall y f(x,y)\iff\forall y\exists x f(x,y)$. – Asaf Karagila Aug 31 '12 at 17:09

take two examples:

$$f(x,y) = \cos(x+y)$$

and

$$f(x,y)=2xy(x-y)$$

In the first case $\min_x \max_y f(x,y)=1$ and $\max_y\min_x f(x,y)=-1$ for all $x,y$.

In the second one we have $y=0.5x$ and $x=0.5y$ since we have $f(x,y)^{''}>0$ for $\frac{d^2}{dx}f(x,y)$ and $f(x,y)^{''}<0$ for $\frac{d^2}{dy}f(x,y)$, they are minima and maxima respectively. As a result one gets $$\min_x \max_y f(x,y) = \max_y \min_x f(x,y)$$

for the second case.

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