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Given any $f: X \times Y \rightarrow \mathbb{R} \cup \{ \infty, -\infty\}$, I was wondering

  1. if it is true that $\inf_{x \in X} \sup_{y \in Y} f(x,y) \geq \sup_{y \in Y} \inf_{x \in X} f(x,y)$?
  2. when it is true that $\inf_{x \in X} \sup_{y \in Y} f(x,y) = \sup_{y \in Y} \inf_{x \in X} f(x,y)$?
  3. if the answers to the above will be different if $\inf$ and/or $\sup$ be replaced with $\min$ and/or $\max$?
  4. if the answers to the above will be different if the codomain of $f$ is any totally ordered set? Suppose all exist.

Thank!

  • 3
    For each $x$, $\sup_y(f(x,y))\geq \sup_y\inf_z(f(z,y))$; because $f(x,y)$ is at least as large as the infimum. This holds for all $x$, so $\inf_x\sup_y f(x,y)\geq \sup_y\inf_x(f(x,y))$.2011-04-05
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    @Arturo: Nice proof! Thanks!2011-04-05

3 Answers 3

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$$\mathop{\inf\vphantom{p}}_{x\in X}\sup_{y\in Y}f(x,y)\ge\mathop{\inf\vphantom{p}}_{x\in X}\sup_{y\in Y}\mathop{\inf\vphantom{p}}_{x'\in X}f(x',y)=\sup_{y\in Y}\mathop{\inf\vphantom{p}}_{x'\in X}f(x',y)=\sup_{y\in Y}\mathop{\inf\vphantom{p}}_{x\in X}f(x,y)\;.$$

This proves $1$, and the same proof works if you replace inf by min and/or sup by max and/or replace $\mathbb{R}$ by any totally ordered set.

  • 0
    If anyone knows how to use phantoms to vertically align $x\in X$ with $y\in Y$, I'd be much obliged. I believe the problem is that "sup" has a descender and "inf" doesn't. I assume the solution must be something like \X{\inf\vphantom{\sup}}, but I don't know what \X should be.2011-04-05
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    Thanks! Just in case no one here knows, there is a SE site for Latex, http://tex.stackexchange.com/2011-04-05
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    OK, I found it; X is mathop.2011-04-05
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    @Tim: I think you've got the wrong quantifier there -- your comment is helpful not only if for all $p$ here, $p$ doesn't know, but if there exists $p$ here such that $p$ doesn't know :-)2011-04-05
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    My bad English. I actually wanted to say that in case no one knew the solution to your question, you could try the tex SE site. But it is great that you find the solution by yourself.2011-04-05
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    Looks like the second question in my post is meaningless, isn't it?2011-04-05
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    @Tim: I don't see why it should be meaningless, but I don't see an easy way of answering it -- it does feel like there should be a simple answer, though.2011-04-05
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    @joriki: I think an important caveat is that if we take a different total order then the $\sup$ might exist and the $\inf$ might not, which renders the question somewhat moot.2011-04-29
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2 is not true in general, but under certain assumptions, it is true. This is called Minimax theorem. See, for instance, Minimax Theorem.

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For 1 and 2, they might not exist. Try $f(x,y)=\frac{1}{y}$. Even more so for 3 and 4.

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    Often inf and sup are defined to be infinite if there is no bound. The question makes more sense if interpreted that way.2011-04-05
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    Thanks! You are right about the existence issue. What about when they exist? I just modified my questions for this.2011-04-05