Interchanging supremum with infimum Let $f:(0,a) \times (0,b) \to \mathbb{R}$ be a given function. Under what conditions is it true that
$$\sup_{x\in(0,a)} \inf_{y \in (0,b)} f(x,y) = \inf_{y \in (0,b)} \sup_{x\in(0,a)} f(x,y) \tag{1}$$
There is a related question saying that the inequality $"\leq"$ in $(1)$ holds true, however in general one cannot expect the converse inequality $"\geq"$. I'm interested in some additional assumptions that will make $(1)$ true. Any references will be much appreciated.
For example if we assume that $f$ is increasing in $y$ and decreasing in $x$, then $\sup_{x \in (0,a)}$ and $\inf_{y \in (0,b)}$ in $(1)$ can be replaced with $\lim_{x\to 0}$ and $\lim_{y\to 0}$ respectively. Then $(1)$ would hold for example if the double limit at $(0,0)$ exists. But these are very restrictive assumptions, anything more general?

For each $x$, let us write $C_x$ for the "column" $\{(x,y) : y \in (0,b)\}$. Similarly, for each $y$, let us write $R_y$ for the "row" $\{(x,y) : x \in (0,a) \}$. One pretty natural sufficient condition for (1) to hold is that there should exist $x$ and $y$ such that $f|_{R_y} \leq f|_{C_x}$ . This can be tweaked a bit to get a necessary and sufficient condition: (1) holds if and only if, for every $\epsilon > 0$, there exist $x$ and $y$ such that $f|_{R_y} \leq f|_{C_x} + \epsilon$, but to be honest this is not much different than just writing down the definitions of the left hand side and right hand side.
