# Interchanging min and max for a continuous function of two variables

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.

• A possible setting would be 1) guarantee that both the infsup and the supinf values are critical levels for $f$, and 2) require that $f$ has only one critical point. The fact that $f$ should be defined on the closed square , and not e.g. on $\mathbb R^2$, makes this less natural though Commented Dec 8, 2023 at 11:57

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 $$\pi = \inf_x M_x \neq \sup_y m_y = 0$$.