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 $f\in\mathcal F$.

  • $\begingroup$ I don't have a copy with me but I remember recently reading a version of Arzela-Ascoli that deals with $C(E)$ for $E$ a locally compact Hausdorff space in Folland. Try checking there. $\endgroup$ – Rhys Steele Apr 13 at 17:47
  • $\begingroup$ @RhysSteele Ah, there is a version for $\sigma$-locally compact Hausdorff spaces. But he's only assuming equicontinuity ... That's a bit surprising. Am I missing something? That would mean that we only need $$\lim_{\delta\to0+}V^N(f,\delta)=0\;\;\;\text{for all }f\in F$$ instead of $(1)$. And he's assuming (as actually usual) that $F$ is pointwise bounded. In the version in the question, boundedness is only required for $t=0$. Do the assumptions somehow imply that $F$ is actually bounded for all $t\ge0$? $\endgroup$ – 0xbadf00d Apr 13 at 18:03
  • $\begingroup$ Check the definition of equicontinuity again. $\lim_{\delta \to 0+} V^N(f,\delta) = 0$ says something like $f$ is uniformly continuous on $[0,N]$. Equicontinuity of a family $F$ essentially says that the family is uniformly uniformly continuous (which sounds stupid so we use a different word) in the sense that (on compacts) for every $\varepsilon > 0$ there is a $\delta$ such that for all $x,y$ with $d(x,y) < \delta$ we have for each $n$, $|f_n(x) - f_n(y)| < \varepsilon$. This should be the same as $(1)$. $\endgroup$ – Rhys Steele Apr 13 at 18:12
  • $\begingroup$ And yes, boundedness at a single point and equicontinuity together imply pointwise bounds at all points. This is easiest to see using the definition of equicontinuity I stated above and is a good exercise to check you understood it. $\endgroup$ – Rhys Steele Apr 13 at 18:13
  • $\begingroup$ @RhysSteele Yes, sorry. It precisely means uniform continuity on $[0,N]$. $\endgroup$ – 0xbadf00d Apr 13 at 18:14

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:

$F \subset C([0, \infty))$ is relatively compact in the topology of uniform convergence on compact subsets of $[0, \infty)$ if and only if $F$ is equicontinuous at every point $x \in [0, \infty)$ and $\{f(x) \mid f \in F\} \subseteq \mathbb R$ is bounded for every $x \in [0, \infty)$.

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.


EDIT This answer is incorrect. I used the wrong topology.

I think the way this theorem is stated is not true, if I didn't misread anything.

Consider a bump function $\varphi$ supported in $[0,1]$. Defined $\varphi_n$ by $$ \varphi_n(x) = \varphi(x-n), $$ then the family $\mathcal F=\{\varphi_n:n \in\Bbb N\}$ is bounded in $\|\cdot\|_\infty$ norm and satisfies the uniform continuity assumption. However, $(\varphi_n)$ has no converging subsequence so $\mathcal F$ is not relatively compact.

  • $\begingroup$ You've got the wrong topology. $\|\cdot\|_\infty$ isn't a norm on $C([0,\infty))$ since $[0,\infty)$ is only locally compact. The right notion of convergence on this space is that of uniform convergence on compacts and in this topology $\varphi_n \to 0$ since that sequence even has zero tails when restricted to any compact subset of $[0,\infty)$. $\endgroup$ – Rhys Steele Apr 13 at 17:46
  • $\begingroup$ The topology on $C([0,\infty))$ is the locally convex topology induced by the metric $$d(f,g):=\sum_{n\in\mathbb N}2^{-n}d_n(f,g),$$ $d_n$ being the metric on $C([0,n])$. $\endgroup$ – 0xbadf00d Apr 13 at 17:56
  • $\begingroup$ @RhysSteele Thank you, it seems like I didn't use the correct topology the OP intended. However, $\|\cdot\|_\infty$ is a norm on $C([0,\infty))$ though, just not the right one for this question (which I couldn't have known beforehand). $\endgroup$ – BigbearZzz Apr 13 at 18:01
  • $\begingroup$ @BigbearZzz Really? norms are by definition maps into $[0, \infty)$ with some extra properties. What is the $\sup$-norm of the continuous function $t \mapsto t$ on $[0,\infty)$? The topology OP has in mind is the standard one on this space (usually called the topology of uniform convergence on compacts), I agree it could have been stated but you also could have known. $\endgroup$ – Rhys Steele Apr 13 at 18:03
  • $\begingroup$ I see what's the problem now. Usually I am familiar with the notion that $(C(\Omega),\|\cdot\|_\infty)$ is the space consisting of only functions that are bounded on $\Omega$, but here your $C([0,\infty))$ includes unbounded functions as well. $\endgroup$ – BigbearZzz Apr 13 at 18:04

In the situation you described, the Arzela-Ascoli theorem can be applied to the functions restricted to intervals of the form [0,N]. Starting with any sequence in F, and applying a diagonalization argument to the subsequences which converge on [0,N], we get a subsequence converging on the positive half-line.

  • $\begingroup$ I've got some similar idea in a more general setting. Please take note of my edit. Do you think this holds in a more general set-up? $\endgroup$ – 0xbadf00d Apr 13 at 20:03
  • $\begingroup$ @0xbadf00d If the space X is $\sigma$-compact, then the same argument works. If not, then C(X) isn't metrizable and you'd have to use nets instead of sequences, which always confuses me. $\endgroup$ – Amichai Lampert Apr 13 at 20:38
  • $\begingroup$ @0xbadf00d This seems to be answered here math.stackexchange.com/questions/20670/… $\endgroup$ – Amichai Lampert Apr 13 at 20:47

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