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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.

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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.

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  • $\begingroup$ 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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