# Proving that if a sequence of continuous real functions is uniformly convergent on a compact real subset then it is uniformly equicontinuous

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.

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.
• Your answer is correct, but as far as I can tell there have been more modifications then just replacing $x_0$ by $y$. And that is normal since the answer you refered to does not use uniform continuity, while you did. That was the missing argument I refered to, on my part. Hope we cleared things out. – nicomezi Nov 14 '20 at 16:58