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.
If you look at this problem as a Primal Problem and its Dual Problem then you're basically asking when Strong Duality Holds.  

Basically when $ f\left( x, y \right) $ is Convex in $ x $ and Concave in $ y $.
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 $f(x,y)^{''}&gt;0$ for $\frac{d^2}{dx}f(x,y)$ and $f(x,y)^{''}&lt;0$ 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.