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Question: Let $X$ be a set of functions from $\mathbb{R}$ to itself. Consider the subset $X_0$ of the sequences $(f_n)_{n=1}^{\infty}\in X^{\mathbb{N}}$ for which $$ f_{\infty}(x) = \lim\limits_{n \to \infty} f_n(x), $$ is a well-defined function. Is there any literature studying the continuity of the map $$ \begin{aligned} \operatorname{Lim}: X_0 &\rightarrow \mathbb{R}^{\mathbb{R}}\\ (f_n)_{n=1}^{\infty} & \to f(\cdot):= \lim\limits_{n \to \infty} f_n(\cdot)? \end{aligned} $$

My question is in some sense motivated by the following problem:

Motivation: For each $n,k \in \mathbb{N}$ let $f_{n,k},f_{k}\in C(\mathbb{R},\mathbb{R})$ and suppose that for each $k \in \mathbb{N}$, $f_k$ is the uniform on compacts limit of $f_{n,k}$. Suppose also that there is a Borel function $f:\mathbb{R}\rightarrow \mathbb{R}$ for which $f(x)= \lim\limits_{k \to \infty}f_k\circ \dots \circ f_1(x)$.

Then, for each $x \in \mathbb{R}$, does

$$ \lim\limits_{(n,k)\to \infty}f_{n,k}\circ \dots \circ f_{n,1}(x) = f(x)? $$

In other words: when does $\lim\limits_{n \to\infty}\operatorname{Lim}((f_{n,k})_{k})=\operatorname{Lim}(f_{k})$?

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  • $\begingroup$ To give an example for Motivation: Let's build $f_{n,k}(x)$ such that it is equal to the identity outside of $[n+k-2, n+k]$, and it is linear on $[n+k-2, n+k-3/2]$ and on $[n+k-1/2,n+k]$, with $f_{n,k}(x) = n+k+1$ on the segment $[n+k-3/2,n+k-1/2]$. Clearly $f_{n,k}$ are continuous. For fixed $k$, $f_{n,k}$ converges uniformly on compacts to $x\mapsto x$. And so we have $f_k(x) = f(x) = x$. However given $n$ fixed, $f_{n,k} \circ \cdots f_{n,1}(x) = n+k + 1$ for every $x\in [n-1/2, n+1/2]$. So we don't have a limiting function $\mathrm{Lim}((f_{n,k})_k)$. $\endgroup$
    – Willie Wong
    Commented Jul 16, 2020 at 19:05
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    $\begingroup$ Actually, thinking about this more: if you let your functions to be constant functions, your question immediately reduce to the one about "when do limits commute for double sequences". So I think the fact that $f_*$ are functions are really a bit of a red herring here. $\endgroup$
    – Willie Wong
    Commented Jul 16, 2020 at 19:29
  • $\begingroup$ @WillieWong You're right I didn't think of that. In that case, do you know of a generalization of the moore-osgood theorem for function spaces? That would do the trick.. $\endgroup$
    – ABIM
    Commented Jul 16, 2020 at 19:31
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    $\begingroup$ Moore-Osgood works for general metric spaces. Using the notation of that link: let $Y$ be your desired function space (for example the space of bounded continuous functions with the $\sup$ norm), and let $X$ be either $\mathbb{N}$ (and you have to modify the theorem appropriately), or $\mathbb{R}$ (and use $1/n$ instead of $n$). $\endgroup$
    – Willie Wong
    Commented Jul 16, 2020 at 19:41
  • $\begingroup$ Interesting, then is it at-least possible to find a convergence subsequence after swapping the limits? $\endgroup$
    – ABIM
    Commented Jul 16, 2020 at 19:50

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