If $f_n(s) \rightarrow f(s)$ for all s. Is it correct to say that $\lim_{n\to\infty}(\min f_n)=\min f$?

Question

If $f_n(s) \rightarrow f(s)$ for all $s$, is it correct to say that $\lim_{n\to\infty}(\min f_n)=\min f$? Are minimums of $f_n$ converging to minimum of $f$?

Answer 1

No. For example, consider the following sequence in $\mathbb{R^R}$: $$ f_n(x)= \begin{cases} 1 & x\leq n\\ 0 & x\geq n+1\\ n+1-x & n\leq x\leq n+1 \end{cases} $$ converging pointwise to the constant function $1$, but $\min f_n=0\not\to 1$.

Answer 2

A simple counterexample is $$ f_n(x)=[x\le n] $$ where $[\dots]$ are Iverson Brackets.

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An analytic counterexample is $$ f_n(x)=e^{-x^2/n} $$