Let $X$ be a metric space and $(f_n)$ be a sequence of continuous functions with $f_n: X \rightarrow \mathbb{R}$. I know that, in general, $\sup_{x \in X}\lim_{n\in \mathbb{N}}f_n(x)\ne \lim_{n\in \mathbb{N}} \sup_{x \in X}f_n(x)$, though equality holds when $(f_n)$ converges uniformly to a function $f$.
Is there an easy inequality direction though? My conjecture is that $\sup_{x \in X}\lim_{n\in \mathbb{N}}f_n(x)\leq \lim_{n\in \mathbb{N}} \sup_{x \in X}f_n(x)$ because depending on the mode of convergence, $\lim_{n\in \mathbb{N}}f_n(x)$ is not necessarily optimal for preserving supremums of $(f_n)$, say, if convergence is Lebesgue-almost everywhere and the points where the convergence does not happen are precisely where sup is attained. But is a very hand-wavy argument which I am not sure how to make precise.
Another concern that I have: if either $X$ is an open set, or $f_n$ are chosen from a non-compact subset $\mathcal{F}$, then it is possible that either $\lim_{n \in \mathbb{N}}f_n \notin \mathcal{F}$ or $\sup_{x \in X}f_n(x)\notin f_n(X)$. For example, if $\lim_{n \in \mathbb{N}}f_n \notin \mathcal{F}$, then when we take $\sup_{x \in X}\lim_{n\in \mathbb{N}}f_n(x)$ of it, we are dealing with a function that does not appear in $\lim_{n\in \mathbb{N}} \sup_{x \in X}f_n(x)$ at all. How would this affect the inequality?
