A sequence of functions decreasing to 0 is equicontinuous in a compact metric space.

Question

How to proof that?

Let M be a compact metric space and $ \{f_n\} \subset C(M,\mathbb{R})$, so that $\{f_n\}$ is decreasing and $ lim f_n(x)=0 $, then $\{f_n\}$ is equicontinuous

Answer 1

Let $\varepsilon >0$. By Dini's theorem, the sequence converge uniformly to $0$. Let $N$ s.t. $$\sup_{x\in M}|f_n(x)|\leq \frac{\varepsilon}{2},$$ for all $n\geq N$.

The $f_n$ are uniformly continuous on $M$. Let $\delta_n$ s.t. $$|x-y|\leq \delta_n\implies |f_n(x)-f_n(y)|\leq \varepsilon.$$ Set $\delta=\min_{1\leq n\leq N} \delta_n$. Then, $$|x-y|\leq \delta\implies \forall n,\ |f_n(x)-f_n(y)|<\varepsilon,$$ what prove the claim.