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Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous function, where $X \subset \mathbb{R}^n$ and $Y \subset \mathbb{R}^m$ are compact sets.

Say under which conditions we have that

$ \max_{x \in X} \max_{y \in Y} f(x,y) = \max_{y \in Y} \max_{x \in X} f(x,y) $

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    Well, $\sup_X \sup_Y f(x,y) = \sup_Y \sup_X f(x,y)$ and by continuity and compactness the extreme values are attained, so this must be true.2012-08-31

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Always :

Since $f$ is continuous on the compact set $X\times Y$, it has a maximum $m$, say at $(x_0,y_0)$.

Now, $y\mapsto \max_x f(x,y)$ is a function which takes value $m$ at $y_0$, so $\max_y \max_x f=m$. Using this argument the other way around, we get the wanted equality.

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    It holds whenever $f$ has indeed a maximum on $X\times Y$. This might not be the case if $f$ is non-continuous, even with $X\times Y$ compact (hence bounded). For instance we could define $f=0$ for all points except a sequence $(x_n, y_n)$ for which $f(x_n, y_n)=1-1/n$.2014-11-13