I only know the Arzelà–Ascoli theorem for continuous functions on a compact topological space. However, in the context of characterizing weak convergence of probability measures on $C([0,\infty))$, I've seen that the following version is used (without proof):
If $N,\delta>0$ and $f\in C([0,\infty))$, let $$V^N(f,\delta):=\sup\left\{|f(t)-f(s)|:|t-s|\le\delta,s,t\le N\right\}.$$ Then $F\subseteq C([0,\infty))$ is relatively compact if and only if $\left\{f(0):f\in F\right\}$ is bounded and for all $N>0$, $$\lim_{\delta\to0+}\sup_{f\in F}V^N(f,\delta)=0\tag1.$$
Does any body know a reference for a version of the Arzelà–Ascoli theorem which captures this case?
Remark: Obviously, $(1)$ is equivalent to the uniform equicontinuity of $F$.
EDIT: There is the following version which can be found in Theorem 4.43 in the book of Folland:
Theorem 4.43: If $X$ is a compact Hausdorff space and $\mathcal F\subseteq C(X)$ is equicontinuous$^1$ and pointwise bounded$^2$, then $\mathcal F$ is totally bounded (with respect to the supremum metric) on $C(X)$ and relatively compact.
I guess the situation in the question can somehow be generalized in the following way: If $X$ is a Hausdorff space, $\mathcal F\subseteq C(X)$ and ${\mathcal F}_K:=\left\{\left.f\right|_K:f\in\mathcal F\right\}$ is equicontinuous and pointwise bounded for all compact $K\subseteq X$, then …
Maybe someone could elaborate on what exactly we can conclude and which additional assumption on $X$ we need. I know almost nothing about general topology, but I could imagine that we can consider $C(X)$ above as being equipped with the topology induced by the family of metrics $$d_{\infty,\:K}(f,g):=\sup_{x\in K}d(f(x),g(x))\;\;\;\text{for }f,g\in C(X)$$ for compact $K\subseteq X$, which should yield a sequentially complete space.
$^1$ i.e. for all $x\in X$ and $\varepsilon>0$, there is a neighborhood $N$ of $x$ with $f(N)\subseteq B_\varepsilon(f(x))$ for all $x\in X$.
$^2$ i.e. $\left\{f(x):f\in F\right\}$ is bounded for all $f\in\mathcal F$.
Best Answer
You can find the Arzelà–Ascoli in utmost generality (definitely more than you need) in Engelking's "General Topology", theorems 3.4.20 (p. 163) and 8.2.10 (p. 443). In your case, these theorems reduce to:
Minor nitpick: it is not clear if $N$ in your question is supposed to be natural; even if it were, you may easily replace it with real positive numbers, because every real number $x$ sits between its integer part $[x]$ and $[x]+1$, which are natural numbers. (In fact, $N$ plays the role of the interval $[0,N]$, which can be readily replaced with arbitrary compacts of $[0, \infty)$.)
Pick $x \in [0, \infty)$ arbitrary. In relation (1) take $N=x$. From the $\varepsilon - \delta$ definition of the concept of limit, taking $\varepsilon = 1$ there exists $\delta_1 > 0$ such that if $\delta < \delta_1$ then $\sup _{f \in F} V^x (f, \delta) \le 1$. Equivalently, $V^x (f, \delta) \le 1$ for all $f \in F$ and $\delta < \delta_1$. Explicitly,
$$\sup \{ |f(s) -f(t)| : 0 \le s, t \le x, \ |s-t| < \delta \} \le 1 \quad \forall f \in F \ .$$
In particular, taking $\delta = \frac {\delta_1} 2 < \delta_1$ in the above, we get that $|f(s) -f(t)| \le 1 $ for all pairs $0 \le s, t \le x$ with $|s-t| < \frac {\delta_1} 2$ and for all $f \in F$.
The interval $[0,x]$ can be covered with $n(x, \delta_1) := \left[ \frac {2x} {\delta_1} \right] + 1$ subintervals of length $\frac {\delta_1} 2$, whence (using the triangle inequality multiple times)
$$|f(x) - f(0)| \le \\ \le \left| f(x) - f \left( x - \frac {\delta_1} 2 \right) + f \left( x - \frac {\delta_1} 2 \right) - f \left( x - 2 \frac {\delta_1} 2 \right) + \dots + f \left( x - n(x, \delta_1) \frac {\delta_1} 2 \right) - f(0) \right| \le \\ \le \left| f(x) - f \left( x - \frac {\delta_1} 2 \right) \right| + \left| f \left( x - \frac {\delta_1} 2 \right) - f \left( x - 2 \frac {\delta_1} 2 \right) \right| + \dots + \left| f \left( x - n(x, \delta_1) \frac {\delta_1} 2 \right) - f(0) \right| \le \\ \le 1 + 1 + \dots + 1 = n(x, \delta_1) \ .$$
(Warning: my count above might be off by $\pm 1$ because of the endpoints, I'm never good at counting, but this doesn't change the end result: we have been able to find an upper bound for $|f(x) - f(0)|$ which is independent of $f \in F$.)
Finally, if $B = \{ |f(0)| \mid f \in F\}$ then
$$|f(x)| = |f(x) - f(0) + f(0)| \le |f(x) - f(0)| + |f(0)| \le n(x, \delta_1) + |f(0)| \in n(x, \delta_1) + B$$
and the right-hand side is obviously bounded, being the translate by the constant (with respect to $f \in F$) $n(x, \delta_1)$ of the bounded subset $B$.
(Notice that since $\{f(0) \mid f \in F\}$ was bounded, so will be $\{|f(0)| \mid f \in F\}$, trivially.)
Since $x$ was arbitrary, all the work above proves the pointwise boundedness of $F$.
The (uniform, but this is not needed) equicontinuity of $F$, on the other hand, comes practically for free, being encoded in (1), as you remark yourself in the question.
Since you have pointwise boundedness and equicontinuity, you have relative compactness in the topology of uniform convergence on compact subsets.