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This is from Scheidemann, Complex Analysis.

Theorem (Arzela-Ascoli): Let $K$ be a compact separable metric space, $E$ a finite-dimensional Banach space and $(f_j)_{j\in\mathbb{N}}\subseteq C(K,E)$ a family of equicontinuous and uniformly bounded functions. Then $(f_j)_{j\in\mathbb{N}}$ contains a subsequence converging uniformly on $K$.

I don't understand the proof of the following corollary.

Corollary: Let $X$ be a locally compact separable metric space, $U\subseteq X$ open and $(f_j)_{j\in\mathbb{N}}\subseteq C(U,E)$ a family of locally equicontinuous and locally uniformly bounded functions. Then $(f_j)_{j\in\mathbb{N}}$ contains a subsequence converging locally uniformly on $U$.

"Proof": Since $X$ is locally compact every $a\in U$ possesses a relatively compact neighbourhood $V\subseteq U$. Apply Arzela-Ascoli on $\overline{V}$.

First of all, I think we can't just apply Arzela-Ascoli like that. Since $(f_j)_{j\in\mathbb{N}}$ is both locally equicontinuous and locally bounded, if $a\in U$ there exist $V_1,V_2\subseteq U$ open such that $(f_j\restriction_{V_1})_{j\in\mathbb{N}}$ is equicontinuous and $(f_j\restriction_{V_2})_{j\in\mathbb{N}}$ is bounded. Then let $V$ be a relatively compact neighbourhood of $a$ such that $V\subseteq\overline{V}\subseteq V_1\cap V_2$. By Arzela-Ascoli, $(f_j\restriction_{\overline{V}})_{j\in\mathbb{N}}$ has a subsequence converging uniformly on $\overline{V}$, say $(f_{j_k})_{k\in\mathbb{N}}$.

However, we still need to find a subsequence $(f_j)_{j\in\mathbb{N}}$ that converges locally uniformly on $U$, i.e., such a subsequence must converge uniformly on a neighbourhood of every $x\in U$. So, I think we can't just pick $(f_{j_k})_{k\in\mathbb{N}}$ since the construction of this subsequence was dependent of $a$.

What do you think?

Thank you.

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  • $\begingroup$ a compact metric space is always separable; no need to assume it $\endgroup$ – zhw. May 25 '16 at 19:36

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