If $f_{n}\longrightarrow f$ is a uniformly convergent sequence in a compact topological group $G$, then $\min(f_{n})\longrightarrow\min(f)$

Question

Let $G$ be a compact topological space and $f_{n}$ a uniformly convergent sequence in $C(G,\mathbb{R})$. Let $f$ be the limit of the $f_{n}$. I want to show that $\min(f_{n})\longrightarrow\min(f)$.

I've tried different approaches to this problem but I've only managed to prove that there exists a subsequence $f_{n_{k}}$ such that $\min(f_{n_{k}})\longrightarrow\min(f)$. If I could show that the whole sequence $\min(f_{n})$ has a limit then I would be pretty much done, but so far nothing has worked for me. I've read and tried to adapt the proofs in these two comments but some parts of the argument are hard to follow:

https://math.stackexchange.com/q/1828540

https://math.stackexchange.com/q/1823276

Any help will be greatly appreciated!

Answer 1

This is too elementary and does not require any theorem on topological groups.

Let $\epsilon >0$. Choose $n_0$ such that $|f_n(g)-f(g)| <\epsilon$ for all $g$ for $n >n_0$. Let $a_n=\inf_{g\in G} f_n(g)$ and $a=\inf_{g \in G} f(g)$. For $n >n_0$ we have $f_n(g) <f(g)+\epsilon$ so $a_n < f(g)+\epsilon$. Since this is true for all $g$ we get $a_n \leq a+\epsilon$. Similarly we get $a \leq a_n+\epsilon$ for. $n >n_0$. Since minimum is same as infimum we are done.