Let $\{f_n\}$ be a sequence of real-valued continuous functions on $[a,b]$. I am trying to show that if $\{f_n\}$ is uniformly convergent, it is also uniformly equicontinuous.
My attempt at this seems like it must be too simple. I will outline it below. So, in class, we proved this theorem, followed by another proposition we proved:
Theorem: Let $\{f_n\}$ be a sequence of continuous real functions on a compact set $K\subset X$. Then $\{f_n\}$ is equicontinuous and pointwise convergent if and only if it is uniformly convergent.
Proposition: Let $K\subset X$ be compact. If $\mathcal{F}\subset C(K)$ is equicontinuous at every point in $K$, then $\mathcal{F}$ is uniformly equicontinuous in $K$.
So, my idea is that since $[a,b]$ is a compact set, the proof follows from simply combining these two statements. This feels too simple to be correct to me, so I was hoping if someone could tell me if this will suffice or if there is any other simple way to prove this without the need of these theorems? Thanks.
This is essentially answered here.
Let $\epsilon>0$ be given. Since $f_n$ converges uniformly, it is uniformly Cauchy: there is $n_0 \in \mathbb N$ so that $$ |f_n(x) - f_{n_0}(x)|<\epsilon/3, \ \ \ \forall n\ge n_0, x\in [a, b].$$ Since $f_1, \cdots, f_{n_0}$ are continuous on $[a, b]$, they are all uniform continuous. Thus there are $\delta_1, \cdots, \delta_{n_0}$ so that $$|f_k(x) - f_k(y)|<\epsilon/3$$ whenever $x, y \in [a, b]$ and $|x-y|<\delta_k$. Let $$\delta=\min_{k=1, \cdots, n_0} \delta_k.$$ Then for all $n\ge n_0$ and $|x-y|<\delta$,
$$|f_n(x)-f_n(y)|\le |f_n(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(y)|+|f_{n_0}(y)-f_n(y)|<\epsilon.$$
if $n \le n_0$ and $|x-y|<\delta$, then $|f_n(x) - f_n(y)|<\epsilon/3 < \epsilon$ by the choice of $\delta$.
Thus $$|f_n(x) - f_n(y)|<\epsilon$$ for all $n$ and for all $x, y\in [a, b]$ such that $|x-y|<\delta$. Thus $\{f_n\}$ is uniform equicontinuous.