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By trying to extend certain limit properties of sequences from compact subsets to the entire set, I cam up with something that can be formed in the following question.

Let $a_{mn}$ be a double sequence of nonnegative real numbers. I want to be able to switch order of iterated limits in the following form $$\limsup_{m \to \infty}\limsup_{n \to \infty} a_{mn}=\limsup_{n \to \infty}\limsup_{m \to \infty} a_{mn}$$ under the following conditions

  • $a_{mn}$ is increasing in $m$ for every $n$

  • $\limsup_{m \to \infty} a_{mn}=\lim_{m \to \infty} a_{mn}=a_{n}, \forall n$

  • $\limsup_{n \to \infty} a_{mn} \leq a_m, \forall m$ where $a_m$ is increasing in $m$

I tried all of the ideas from here, but without any success. Any helpful ideas or references?

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  • $\begingroup$ Is the second hypothesis just stating that $\lim_{m \to \infty} a_{m n}$ exists and giving a name to it? $\endgroup$
    – LSpice
    Commented Feb 2, 2022 at 19:21
  • $\begingroup$ Exactly, and the limit can be dependent on $n$. $\endgroup$
    – Marko Rajkovic
    Commented Feb 2, 2022 at 19:25

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$a_{mn}=1(m\ge n)$ is a counterexample. (Here, $1(A)$ is the indicator of an assertion $A$. That is, $1(A):=1$ if $A$ is true and $1(A):=0$ if $A$ is false.)

If you insist on understanding "increasing" in the strict sense, then $a_{mn}=1(m\ge n)-1/m-1/n$ is a counterexample.

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    $\begingroup$ Thank you. I obviously was too distracted with the original complicated version of the problem that I missed this simple counterexample. $\endgroup$
    – Marko Rajkovic
    Commented Feb 2, 2022 at 19:57

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