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The Arzela-Ascoli function basically says that a set of real-valued continuous functions on a compact domain is precompact under the uniform norm if and only if the family is pointwise bounded and equicontinuous.

Are there any analogs of this kind of result for spaces of noncontinuous functions? The specific set I have in mind is the càdlàg functions, which are right continuous and have left limits. Essentially, I want to know if there is any relationship between compactness and equicontinuity. Here, equicontinuity would obviously have to be relaxed to account for jumps, and compactness in the uniform topology could be relaxed to compactness in, say, the Skorokhod topology, or something weaker than uniform.

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5 Answers 5

In $L^p(\Omega)$ spaces ($1 \leq p < \infty$) , this theorem by A.Kolmogorov asserts that a family of functions $\mathcal{F}$ is compact in $L^p(\Omega)$ if and only if

1) $\mathcal{F}$ is bounded.

2) For every $\varepsilon > 0$, there exist $\delta > 0$ such that for each $f \in \mathcal{F}$ we have:

if $\| h \| < \delta $ then $ \| f(\cdot + h) - f(\cdot) \|_p < \varepsilon$

(this is equivalent to the equicontinuity condition but in p-mean).

3) For any $\varepsilon > 0$ , there exists a Borel set $B \subseteq \Omega$ such that for every $f \in A$ we have

$$ \left( \int_{\Omega \setminus B} | f(x) |^p dx \right)^{\frac{1}{p}} < \varepsilon$$

I don't have a reference since I've got this from my measure theory notes. The proof is somewhat technical and in the kernel of it the Arzelá Ascoli theorem is used.

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Helge Holden and I published an expository paper on the Kolmorogov theorem recently, see – there you can also find a link to the publised version in Expositiones Mathematicae. –  Harald Hanche-Olsen Nov 11 '10 at 14:03

For total boundedness in the supremum norm, replace equicontinuity with the condition that for each $\epsilon > 0$ there is a partition $a=t_0 < t_1<...<t_n=b$ of the interval $[a,b]$ s.t. for each $j$, the oscillation of each function in the collection on $[t_j,t_{j+1})$ is less than $\epsilon > 0$. (Notice that for collections of continuous functions on $[a,b]$, this condition is equivalent to equicontinuity.)

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Billingsley's book Convergence of Probability Measures has a section on the geometry of the Skorohod space $D=D[0,1]$, and in Theorem 14.3 gives an analogue of the Arzela-Ascoli theorem. I am using the first edition of this book.

A set $A$ has compact closure in the Skorohod topology if and only if $$\sup_{x\in A}\ \sup_t |x(t)|<\infty\ \mbox{ and }\ \lim_{\delta\to 0}\ \sup_{x\in > A}\ w^\prime_x(\delta)=0.$$

Here $$w^\prime_x(\delta)=\inf_{t_i}\ \max_{0 < i\leq r}\ \sup\{ |x(s)-x(t)|: s,t\in [ t_{i-1} , t_i) \} , $$ where the infimum extends over finite sets of points $\{ t_i\}$ such that $0=t_0 < t_1 < \cdots < t_r=1$ and $t_i-t_{i-1}>\delta\ $ for $1\leq i \leq r$.

Billingsley also gives a second characterization of compactness in Theorem 14.4 that he says is sometimes more convenient to work with.

I presume that this is the same as the solutions already given, but I thought another reference wouldn't hurt.

Here is another reference. In section 6 of chapter 3 of Ethier and Kurtz's Markov processes: Characterization and Convergence you will find similar criteria for relative compactness of subsets of $D_E[0,\infty)$, that is, $E$-valued Skorohod space, where $E$ is a metric space.

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Dunford & Schwartz (vol. 1, starting page 374) lists descriptions of compact sets in many normed spaces.

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Thanks for the reminder, Gerry. The cadlag space is $B(S,\Sigma)$ with $\Sigma$ the algebra generated by left closed, right open subintervals of $[a,b]$. The theorem in D-S that characterizes total boundedness in $B(S,\Sigma)$ is on page 260 in D-S v.1. For cadlag the condition is equivalent to the one I mentioned above. –  Bill Johnson Nov 10 '10 at 20:11

Consider a space $X$, whose points are right and left limiting points of $[a,b]$, i.e. points of the form $x+$ and $x-$, where $x\in(a,b)$ and points $a+,b-$. If $f\colon [a,b]\to \mathbb{R}$ is a Cadlag function, you can define $f(z)$ for $z\in X$ by formulas $f(x+)=\lim_{y\to x+} f(y)$, $f(x-)=\lim_{y\to x-} f(y)$. Moreover, you can define a topology on $X$ such that function on $X$ is continuous iff it is obtained by the procedure described in the previous sentence from Cadlag function on $[a,b]$. Space $X$ with this topology is compact and you can apply Arzela-Ascoli theorem. The result of this is the following:

Denote by $D$ a space of Cadlag functions endowed with topology, given by norm $\|f\|=\sup |f(x)|$ (it stronger than Skorokhod topology). Set $F\subset D$ is precompact if and only if the following conditions hold.
1. Functions in $F$ are uniformly bounded: $\sup\{|f(x)|, x\in [a,b], f\in F\}<\infty$.
2. Limits $\lim_{y\to x+} f(y)$ for $x\in [a,b)$ and $\lim_{y\to x-} f(y)$ for $x\in (a,b]$ are uniform with respect to the choice of $f\in F$.

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