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Definition: A family of functionals $\{F_n: X\to\bar{\mathbb R}\}$ on a metric space $X$ is said to be equi-coercive if, for every $\alpha \in \mathbb{R}$, there is a compact set $K_\alpha$ of $X$ such that for all $n$ we have $$\{F_n \leq \alpha\} \subseteq K_\alpha.$$

Question:

If $\{F_n\}$ is a family of equi-coercive functionals, how to show there is a non-empty compact set $K$ (independent of $n$) such that $$\inf_{x \in X} F_n(x) = \inf_{x \in K} F_n(x) $$ for all $n$?

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    $\begingroup$ What about $F_n(x)=n+|x-n|$ on $\mathbb R$? $\endgroup$
    – Jochen Wengenroth
    Commented Feb 5 at 11:55
  • $\begingroup$ Do not use $F_n$ and $\alpha$ where $f_n$ and $a$ will do. $\endgroup$
    – Iosif Pinelis
    Commented Feb 5 at 15:49

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Your question can be restated as follows:

Suppose that for each real $a$ there is a compact set $K_a$ such that $E_{a,n}:=F_n^{-1}((-\infty,a])\subseteq K_a$ for all $n$. Does then there necessarily exist a compact set $K$ such that $\inf_X F_n=\inf_K F_n$?

As noted by Jochen Wengenroth, this will not hold in such generality. However, this will be obviously true if it is additionally assumed that $E_{a_0,n}\ne\emptyset$ for some real $a_0$ and all $n$.

Indeed, then it is enough to take $K=K_{a_0}$ -- because then for each $n$ we will have $\inf_K F_n\le\inf_{E_{a_0,n}} F_n\le a_0$ and $F_n>a_0$ on the complement of $K$.

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