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I was struggling a little bit with the following question.

Let $A$ and $B$ be two non-empty sets and $f:A \times B \rightarrow \mathbb{R}$. Then it holds $\sup_{y\in Y}\left(\inf_{x\in X}f(x,y)\right)\leq \inf_{x\in X}\left(\sup_{y\in Y}f(x,y)\right)$.

Now, what I was wondering about is when exactly we have equality above, i.e. $\sup_{y\in Y}\left(\inf_{x\in X}f(x,y)\right)= \inf_{x\in X}\left(\sup_{y\in Y}f(x,y)\right)$.

Are there any theorems that give us conditions under which this holds?

I would appreciate your help.

Keen
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  • Hm, I'm a bit confused about $\mathrm{inf}_{x\in X} f(x,y)$. When we take the infimum, we obviously require the set to have a preorder, which is usually just the normal ordering on the real numbers. Since each slice $f_x(y)$ is not a real number, but in fact a function $B\to \mathbb{R}$, it is not immediately clear what the preorder we should be using is, and so I have no idea how this infimum is supposed to be defined. Usually if I wanted to talk about a preorder on functions I would do it in terms of their $L^\infty$ or $\sup$ norm, but then we'd be doing $\sup \inf\sup$. What did you mean? – Jack Crawford Jun 15 '19 at 10:53
  • @JackCrawford I interpreted that as simply constructing a function $g(y)$ such that $g(y) = \inf_{x\in X} f(x,y)$ for each given (fixed) $y\in Y$. Is there a problem with this approach? – cirpis Jun 15 '19 at 10:57
  • @JackCrawford Oh I see. I just assumed $g\leq h$ iff $g(y)\leq h(y)$ forall $y\in Y$. So like a pointwise preorder. – cirpis Jun 15 '19 at 11:04
  • @cirpis Oops, I accidentally deleted my last comment with the attempt to revise it but failed in copying it so I can't be bothered re-writing it. But hm, I'm not convinced this pointwise preorder you have suggested works. Suppose $f(x,y)$ is the function for a bell-curve (in $y$) translated across by $x$ units. Then for all $x_1,x_2\in A$, we neither have $f(x_1,y) > f(x_2,y)$ or $f(x_1,y) < f(x_2,y)$. But we also don't have $f(x_1,y) = f(x_2,y)$ for any $x_1$ or $x_2$, so... this seems like quite a confused relation. I'm not sure this gives the intended behaviour. – Jack Crawford Jun 15 '19 at 11:10
  • In this example, none of the functions ever completely dominate the other. They're all just bell-curves at different horizontal translations. Are they all the suprema and infima, simultaneously? – Jack Crawford Jun 15 '19 at 11:12
  • @JackCrawford Since when do we need a total preorder for suprema/infima? So as long as it forms a lattice, we good. In this case to find $\sup(f, g)$ just define $h(y) = \sup(f(y), g(y))$ for all $y\in Y$. Now $h\geq f$ and $h\geq g$. Furthermore, for all $u\geq f$ and $u\geq g$ we then have $u\geq h$, so it is indeed a supremum. – cirpis Jun 15 '19 at 11:14
  • @JackCrawford So in your bell curve example we would just construct a function which at every point $y$ would just be the highest (loosely) that any bell curve gets.

    Remember - a $\sup(A)$ need not be in $A$.

     – cirpis Jun 15 '19 at 11:17
  • @cirpis By that definition, wouldn't $\sup_{x\in X} f(x,y)$ always be a constant function, not actually depending on $y$? It looks like you're treating it like $\sup_{x\in X, y\in Y} f(x,y)$ when you say that the supremum over a set of bell-curves in $y$ translated by $x$ units is just the constant function defined by the highest point the bell-curve ever reaches (in $x$ or $y$). Edit: Actually, hm, I see what you mean. I am OK with this definition, now. It still seems less standard than ordering by the $\sup$-norm, but I will go with it. – Jack Crawford Jun 15 '19 at 11:24
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    @JackCrawford Not neccessarily, because $\sup$ needs to be the least upper bound. So for some families of curves it would be some funky shape, i.e. for any two distinct bell curves it yould just be the maximum of each of them at any given point.

    For the bell curve family you propose though, it would be the constant function, since the highest point gets dragged through by translation. But this is pure coincidence. May I suggest we more to a chatroom for extended discussion?

     – cirpis Jun 15 '19 at 11:29
  • Let us continue this discussion in chat. – Jack Crawford Jun 15 '19 at 11:31

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