Let $f_{k,m}:\mathbb{R}^n\to \mathbb{R}, \forall k, m \in \mathbb{N}$ and suppose that $\lim_{m\to\infty}f_{k,m} = f_{k,-}$ uniformly in $m$ and $\lim_{k\to\infty}f_{k,m} = f_{-,m}$ at least pointwise. Can I then combine the double limit $\lim_{k\to\infty}\lim_{m\to\infty}f_{k,m}$ to $\lim_{k\to\infty}\lim_{m\to\infty}f_{k,m} = \lim_{k\to\infty}f_{k,k}$? I already tried to search my analysis notes and books, but as this sort of limit combining is quite rare at least in standard curriculum (I cannot reside any that important theorem that combines limits), so I could not find anything useful. I am aware that under these assumptions conditions one can change the order of the limits of $\lim_{n\to\infty}\lim_{m\to\infty}f_{k,m}$ to $\lim_{m\to\infty}\lim_{k\to\infty}f_{k,m}$.

Edit: A guiding motivation to my question relates to convergence discussed in a problem such as this: Dense subspace of the space of functions vanishing at infinity, where one acceptable approach is this. If I understood that answer correctly, we are showing that we can approximate a continuous function $\tilde{f}_k$ with a compact support in the infinity norm with a sequence of smooth compactly supported functions $g_{k,m}$. We are taking it as granted that we can approximate a continuous function $f$ vanishing at infinity with such $\tilde{f}_k$s, so in the end we are approximating $f$ with a sequence that itself is approximated with some other functions. Therefore once wrote out, the entire expression is something like

$$\lim_{k\to \infty}||f - \tilde{f}_k||_\infty = \lim_{k\to \infty}||f - \lim_{m\to\infty}g_{k,m}||_\infty$$

  • 2
    $\begingroup$ Better not use $n$ both as an index of the sequence and as the dimension of the space :) $\endgroup$
    – Martin R
    Commented Nov 9, 2022 at 10:01
  • $\begingroup$ @MartinR Ah, good catch! It is now fixed. Although I would be intrigued to know what you could say for example about such a function whose index is tied to the dimension of the ambient space! $\endgroup$ Commented Nov 9, 2022 at 10:04
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    $\begingroup$ Does the “uniform” refer to $x$? Then this math.stackexchange.com/a/631292/42969 is a counterexample, consider the constant functions $f_{k, m}(x) = 1$ if $k \le m$ and $f_{k, m}(x) = 0$ otherwise. $\endgroup$
    – Martin R
    Commented Nov 9, 2022 at 10:14
  • $\begingroup$ I guess (for tied dimensions) you could identify functions from $\mathbb{R}^n$ to $\mathbb{R}$ to the set of functions from $\mathbb{R}^\mathbb{N}$ to $\mathbb{R}$ constant relative to all the components of index $\geq n+1$ as a starting point? Not that I have any clue for what you'd do after that hmm $\endgroup$
    – Bruno B
    Commented Nov 9, 2022 at 10:14
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    $\begingroup$ In ”$\lim_{m\to\infty}f_{k,m}(x) = f_{k,-}(x)$ uniformly” the “uniformly” can refer to $x$, or to $k$, or to both. You should clarify what exactly your assumptions are. $\endgroup$
    – Martin R
    Commented Nov 9, 2022 at 10:40


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