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.

  • $\begingroup$ 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. $\endgroup$ – nicomezi Nov 14 '20 at 16:58
  • $\begingroup$ @nicomezi Thanks for clearing things up. (I will delete those comments under the question is it's a bit tangential to this question). $\endgroup$ – Arctic Char Nov 14 '20 at 17:27
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    $\begingroup$ No problem, its not easy to be clear when talking about equicontinuity, so if on top of that you add some uniform condition ... Have a nice evening. $\endgroup$ – nicomezi Nov 14 '20 at 17:38

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