0

Assume you have a real valued function $f(x,y)$ defined on some domain $X\times Y$.

When people write $\sup\limits_{x \in X}\sup\limits_{y\in Y} |f(x,y)|$ is that actually the same as

$\sup\limits_{(x,y) \in X \times Y } |f(x,y)|$?

Such that

$\sup\limits_{y\in Y}\sup\limits_{x \in X} |f(x,y)|=\sup\limits_{x \in X}\sup\limits_{y\in Y} |f(x,y)|$

For instance in the Arzela Ascoli theorem one requires among a other condition for a set $\lbrace f_1,f_2,\ldots\rbrace $ to be relatively compact in $C(X, \mathbb{R})$ ($X$ for example compact metric space) that $\sup\limits_{n \in \mathbb{N}} \sup\limits_{x\in X} |f_n(x)|<\infty$.

Is that the same as $\sup\limits_{x\in X} \sup\limits_{n \in \mathbb{N}} |f_n(x)|<\infty$

  • What are your thoughts on this? What have you tried? What do you know about the same question with limits, for example? – Ted Shifrin Nov 28 '23 at 17:49
  • Well if it would be the same as $\sup\limits_{(x,y) \in X \times Y } |f(x,y)|$ then it would be true but unfortunately I think that it is not the same. So actually I can pull this question back since I think this is wrong in generall. – MackeyTopology Nov 28 '23 at 18:13
  • 1
    Perhaps this entry will help. – Ted Shifrin Nov 28 '23 at 18:24
  • @TedShifrin Thank you! – MackeyTopology Nov 28 '23 at 18:29
  • They need not be the same. Imagine a function that adds someone's height and weight. The person with the largest total sum need not be the heaviest person among the tallest people (nor the tallest person among the heaviest people). It could be someone who is not taller than everyone else and not heavier than everyone else, but just has a larger sum. And the heaviest among the tallest people need not have the same sum as the tallest among the heaviest people. – Arturo Magidin Nov 28 '23 at 19:31
  • @ArturoMagidin according to the proof https://math.stackexchange.com/questions/1734318/is-it-possible-to-interchange-order-of-supremum-and-supremum their always the same even when the range of $f$ is unbounded. Only prerequisit is that $X,Y$ are non empty. I first also thought that it wouldn't be true but now changed my mind. – MackeyTopology Nov 28 '23 at 19:33
  • I looked at the proof and couldn't find a mistake. But feel free to share youre opinions. Thank you! – MackeyTopology Nov 28 '23 at 19:33
  • Hmmm... yes, I see that I am thinking of a function that is defined on a subset of $X\times Y$, not on all of $X\times Y$. Sorry for the noise. – Arturo Magidin Nov 28 '23 at 19:36
  • no problem still interesting example though – MackeyTopology Nov 28 '23 at 19:38
  • 2
    If one defines $\sup \emptyset = -\infty$, then there is not even a prerequisite that $X$ and $Y$ are not empty. – Paul Sinclair Nov 29 '23 at 14:27

0 Answers0