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Let $(X, d)$ be a metric space, $C(X)$ the space of real-valued continuous functions on $Y$, and $\mathcal{F} \subset C(X)$. We say that

  • $\mathcal{F}$ is pointwise bounded if $\forall x \in X$, the set $\{f(x) \mid f \in \mathcal{F}\}$ is bounded.

  • $\mathcal{F}$ is uniformly bounded if the set $\{f(x) \mid f \in \mathcal{F}, x\in X\}$ is bounded.

  • $\mathcal F$ is pointwise equi-continuous if $\forall x\in X, \forall \varepsilon>0, \exists \delta>0, \forall f\in \mathcal F$, we get $d(x,y) < \delta \implies |f(y)-f(x)| < \varepsilon$.

  • $\mathcal F$ is uniformly equi-continuous if $\forall \varepsilon>0, \exists \delta>0, \forall f\in \mathcal F$, we get $d(x,y)< \delta \implies |f(y)-f(x)| < \varepsilon$.

Wikipedia's ver 1: Assume $X$ is compact. A subset $\mathcal F$ of $C(X)$ is relatively compact in $\| \cdot\|_\infty$ if and only if $\mathcal F$ is pointwise bounded and pointwise equi-continuous.

Carothers' ver 2: Assume $X$ is a compact. A subset $\mathcal{F}$ of $C(X)$ is compact in $\| \cdot\|_\infty$ if and only if $\mathcal{F}$ is closed, uniformly bounded, and uniformly equi-continuous.

I proved that above versions of Arzelà–Ascoli theorem are equivalent here. Now I want to apply them to get below result.

Theorem: Assume $X$ is compact and $(f_n) \subset C (X)$ such that $f_n \to f \in C(X)$. Then $(f_n)$ is uniformly equi-continuous.

Could you have a check on my attempt?

My attempt: We need below auxiliary result.

Lemma: Assume $x_n, a \in X$ such that $x_n \to a$. Then $S := \{a, x_1, x_2, \ldots\}$ is compact.

Proof of lemma: Let $(O_i)_{i\in I}$ be an open cover of $S$. There is $\hat i$ such that $a \in O_{\hat i}$. There is $N \in \mathbb N$ such that $x_n \in O_{\hat i}$ for all $n \ge N$. There is $i_n$ such that $x_n \in O_{i_n}$ for all $n \le N$. Then $\{O_i \mid i \in \{\hat i, i_1, \ldots, i_N\}\}$ is a finite subcover.

Now we come back to our Theorem. It follows from our Lemma that $\{f, f_1, f_2, \ldots\}$ is compact. Then $(f_n)$ is uniformly equi-continuous by Carothers' ver 2. This completes the proof.

Analyst
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    Sure but I mean (and this assumes $f_n \to f \in C(X)$ means $\sup_{x \in X} |f_n(x)-f(x)| \to 0$ as $n \to \infty$) without the theorem stated by Carother (which seems fair as they didn't prove the implication used in your argument) it really just comes down to 3 simple facts: (i) given $\varepsilon>0$ we have $|f_n(x)-f_n(y)|<|f(x)-f(y)|+2\varepsilon$ for all but finitely many $n \in \mathbb{N}$, (ii) any continuous function on a compact metric space is uniformly continuous, and (iii) any finite family of [uniformly] continuous functions is [uniformly] equicontinuous. – M A Pelto Jul 26 '22 at 01:41
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    @MAPelto Could you post your comment as an answer so that I can accept it? – Analyst Jul 26 '22 at 10:54

1 Answers1

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Let $\varepsilon>0$ be given and set $\varepsilon':=\varepsilon/4$.

(i) Since $\sup_{x \in X} |f_n(x)-f(x)| \to 0$ as $n \to \infty$, there is $N \in \mathbb{N}$ so that \begin{align}|f_n(x)-f_n(y)|&\leq |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)|\\&<|f(x)-f(y)|+2\varepsilon' \quad\quad\quad\quad\text{ whenever } n \geq N.\end{align} (ii) Since $f \in C(X)$ and $X$ is compact, there is $\delta_0>0$ so that $$|f(x)-f(y)|<2\varepsilon' \text{ whenever } d(x,y)<\delta_0.$$ Moreover, for each $k \in [N]:=\{k: 1 \leq k <N \}$ there is $\delta_k>0$ so that $$|f_k(x)-f_k(y)|<\varepsilon \text{ whenever } d(x,y)<\delta_k.$$ (iii) We set $\delta=\min\{\delta_0,\delta_1, \ldots, \delta_{N-1} \}$ to see that for all $n \in \mathbb{N}$ we now have $$|f_n(x)-f_n(y)|<\varepsilon \text{ whenever } d(x,y)<\delta.$$

M A Pelto
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