Suppose I have some function $F(x,y):\mathbb{R}^2 \rightarrow \mathbb{R}$. Is it always the case that $$ \min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) = \min_{1 \leq y \leq M} (\min_{1 \leq x \leq N} |F(x,y)|) $$ holds?
Here I am interested in situations where $x,y \in \mathbb{R}$ or $\mathbb{N}$. If it's not always the case, under what conditions are they the same? Thank you!
On
Clearly, for any $y$ in $[1,M]$, we have that $$ \min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) \le (\min_{1 \leq x \leq N} |F(x,y)|) $$ since for any $y$ in $[1,M]$ we have that $ \min_{1 \leq y \leq M} |F(x,y)| \le |F(x,y)| $. In particular for the values of $y$ attaining the minimum of the right hand side. Hence $$ \min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) \le \min_{1 \leq y \leq M} (\min_{1 \leq x \leq N} |F(x,y)|). $$
The same argument interchanging $x$ and $y$ proves the reversed inequality.
This is an explanation for my comment. I will show the following
Let $L,R$ be the left and right hand side respectively. It is clear that for any $x\in X$, we have $L\le \inf_{y\in Y}f(x,y)$, and thus $$L\le\inf_{x\in X}\left(\inf_{y\in Y}f(x,y)\right)=R.$$
To show the converse, we first assume that $L\ne -\infty$. Then for any $\epsilon>0$, we can find $(x_0,y_0)\in X\times Y$ such that $f(x_0,y_0)<L+\epsilon$. Thus we have $$R\le f(x_0,y_0)<L+\epsilon.$$ Since this is true for any $\epsilon>0$, it must be that $R\le L$.
If $L=-\infty$, then for any $l\in\mathbb{R}$, we can find $(x_0,y_0)\in X\times Y$ such that $f(x_0,y_0)<l$, and thus $R\le f(x_0,y_0)<l$. This implies that $R=-\infty$ as well. QED.
Finally, by symmetry we also have
In particular, for compact sets $X=[1,N],Y=[1,M]$, if we assume that $F$ is continuous, then the infima are attained so we can write $$\min_{1 \leq x \leq N} (\min_{1 \leq y \leq M} |F(x,y)|) = \min_{1 \leq y \leq M} (\min_{1 \leq x \leq N} |F(x,y)|)$$