maximum of $F(x,y)$

Question

Consider some $F(x,y): \mathbb{R}^2 \to \mathbb{R}$. Under which assumptions is it true that: $\max_x \max_y F(x,y) = \max_y \max_x F(x,y) = \max_{(x,y)} F(x,y)$ ? If not, what are the conditions that will guarantee that? Is convexity of the domain enough? Please note that I did not make any other assumptions about f (if the max does not exist, let's consider the sup)

Answer 1

First of all: The $\max$ doesn't exist for all $F$ but the $\sup$ does so I assume your question is if it's always true that $$\sup_x \sup_y F(x,y) = \sup_y \sup_x F(x,y) = \sup_{(x,y)} F(x,y)$$ The answer is: Yes.

On the one hand we have: \begin{align*} &&F(x,y) &\le \sup_{(x,y)} F(x,y) \quad \forall x,y \\ &\Rightarrow &\sup_{x}F(x,y) &\le \sup_{(x,y)} F(x,y) \quad \forall y \\ &\Rightarrow &\sup_y \sup_xF(x,y) & \le \sup_{(x,y)} F(x,y) \end{align*}

On the other hand we have: \begin{align*} &&F(x,y) &\le \sup_y \sup_xF(x,y) \quad \forall x,y \\ &\Rightarrow& \sup_{(x,y)} F(x,y) &\le \sup_y \sup_xF(x,y)\end{align*}

Alltogether we get $$\sup_y \sup_xF(x,y) = sup_{(x,y)} F(x,y)$$ The proof for the other equation is similar.

Answer 2

Supposing a sort of simmetry for $F$: $F(-x,-y)=-F(x,y)$, it follows that formulation $\max_x \max_y F(x,y) = \max_y \max_x F(x,y)$ is equivalent to $\min_x \max_y F(-x,-y) = \max_y \min_x F(-x,-y)$.

Now, you are able to reply to your question verifying what are hypothesis under minimax Theorem of J. von Neumann.

Note that $\max_x F(x,y) = \max_x [-F(-x,-y)] = - \max_x F(-x,-y) = \min_x F(-x,-y) $ and $\max_x \max_y F(x,y) = \max_x \max_y [-F(-x,-y)] = - \max_x \max_x F(-x,-y) = \min_x \max_y F(-x,-y) $