Let $f,f_1,f_2,\cdots$ be continuous real-valued functions on a compact metric space $E$, with $\lim_{n\to\infty} f_n=f$. Show that if $f_1(p)\le f_2(p)\le \cdots$ for all $p \in E$, then the sequence $f_1, f_2, \cdots$ is uniformly convergent.
Let $\varepsilon>0$. I want to show that there exists an $N$ such that for all $n \ge N$, $\sup(|f_n-f|)<\varepsilon.$ for all $x \in E.$ Equivalently, we want to show $(f_n-f)\to 0$. I let $g_n=f_n-f$ and notice that $||g_1||\ge ||g_2|| \ge \cdots$. The image $f(E)$ is also compact and attains a maximum.
I'm not too sure where to proceed here. Any help would be appreciated.Thanks!
You need the additional condition that $f$ is continuous.
Hint: Consider $G_n = \{x: f(x) -f_n(x) < \epsilon\}$. Show by compactness and the other hypotheses that $$E \subset \cup_{n=1}^\infty G_n \implies E\subset \cup_{k=1}^m G_{n_k} $$
and $f(x) - f_n(x) < \epsilon$ if $n > N = \max(n_1, \ldots n_m)$.