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.


  • 3
    $\begingroup$ 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 $\inf_x\sup_y f(x,y)\geq \sup_y\inf_x(f(x,y))$. $\endgroup$ – Arturo Magidin Apr 5 '11 at 4:46
  • $\begingroup$ @Arturo: Nice proof! Thanks! $\endgroup$ – Tim Apr 5 '11 at 14:35

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

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.

  • $\begingroup$ 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. $\endgroup$ – joriki Apr 5 '11 at 4:46
  • $\begingroup$ Thanks! Just in case no one here knows, there is a SE site for Latex, tex.stackexchange.com $\endgroup$ – Tim Apr 5 '11 at 4:53
  • $\begingroup$ OK, I found it; X is mathop. $\endgroup$ – joriki Apr 5 '11 at 4:59
  • $\begingroup$ @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 :-) $\endgroup$ – joriki Apr 5 '11 at 5:01
  • $\begingroup$ 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. $\endgroup$ – Tim Apr 5 '11 at 5:07

2 is not true in general, but under certain assumptions, it is true. This is called Minimax theorem. See, for instance, Minimax Theorem.


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

  • $\begingroup$ Often inf and sup are defined to be infinite if there is no bound. The question makes more sense if interpreted that way. $\endgroup$ – joriki Apr 5 '11 at 4:18
  • $\begingroup$ Thanks! You are right about the existence issue. What about when they exist? I just modified my questions for this. $\endgroup$ – Tim Apr 5 '11 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.