Let $f:[0,1]\times[0,1]\to\mathbb{R}$ be a continuous function. Define
$$
M\_x:=\max\limits\_{0\leq y\leq 1} f(x,y), \qquad m\_y:=\min\limits\_{0\leq x\leq 1} f(x,y).
$$
Is there a useful set of assumptions under which one can conclude that
$$
\inf\limits\_{0\leq x\leq 1}M\_x = \sup\limits\_{0\leq y\leq 1} m\_y ?
$$
As the example $f(x,y)=(x-y)^2$ shows, this is not true in general. On the other hand, I think (though I haven't checked the details carefully) that this is true provided that for any given $x$, the maximum of $f(x,y)$ is attained at a $unique$ $y$, and that for any given $y$, the minimum of $f(x,y)$ is attained at a $unique$ $x$. I would be grateful for any reference that discusses this question in detail.



Interchanging min and max for a continuous function of two variables
There are many results of this type, and in general they go under the name of minimax theorems. Often some sort of convexity / concavity assumption is made, as in [Sion's Minimax Theorem](http://en.wikipedia.org/wiki/Sion%2527s_minimax_theorem).


You are correct in stating that uniqueness of optimizers gives you equality when $f$ is continuous with domain $[0,1]\times [0,1]$. However, this can fail if the domain of $f$ is different. For example, let $f$ be the arc length metric on the circle. Then $M\_x = \pi$ for all $x$, $m\_y = 0$ for all $y$. The optima are achieved uniquely at $y = -x$ and $x=y$, respectively, but $\pi = \inf\_x M\_x \neq \sup\_y m\_y = 0$.


