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Does Arzela-Ascoli hold in $C_0(X)$ (vanishing at infinity)?... namely a subset of $C_0(X)$ is relatively compact iff it is equicontinuous and bounded. In particular, if $C_0^1(X)$ is the Banach subspace of $C_0(X)$ of differentiable functions such that the derivative is in $C_0(X)$, it would imply that the closed unit ball of $C_0^1(X)$ is relatively compact in $C_0(X)$, and therefore that $C_0^1(X)$ is compactly embedded in $C_0(X)$.

I have seen in some class notes that it is true that Arzela-Ascoli holds in $C_0(X)$... but I have seen nowhere that $C_0^1(X)$ is compactly embedded in $C_0(X)$. Nevertheless, it seems that we can proove this by using the Alexandroff compactification of $X$ and defining $f(\infty):=0$.

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In order to save the Arzeli-Ascoli theorem in your context, you can add the condition that the functions in your set converge uniformly to zero at infinity. – jbc Sep 25 '12 at 14:58
thanks jbc. Yes, that's what I did. – nelson Oct 3 '12 at 17:56
up vote 0 down vote accepted

Consider $C_0(\mathbb{R})$ with the $\sup$ norm. Define $f_n(x) = \begin{cases} 1, && |x| <n \\ 1-(|x|-n), && n \leq |x| < n+1 \\ 0, && \text{otherwise}\end{cases}$

Let $F = \{ f_n \}_{n=1}^\infty \subset C_0(\mathbb{R}) $. $F$ is equicontinuous and uniformly bounded. However, $\|f_n - f_m \| = 1$ for all $n \neq m$, hence $F$ has no convergent subsequence and so cannot be relatively compact.

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Thanks for this example, I think the additional requirement needed to such a family is that it tends uniformly to 0 at infinity. – nelson Sep 25 '12 at 16:05

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