Math 829  The Arzela-Ascoli Theorem  Spring 1999

1 Introduction

Our setting is a compact metric space X which you can, if you wish, take to be a compact subset of ℝ^{n}, or even of the complex plane (with the Euclidean metric, of course). Let C(X) denote the space of all continuous functions on X with values in ℂ (equally well, you can take the values to lie in ℝ). In C(X) we always regard the distance between functions f and g in C(X) to be

dist (f, g) = max{|f(x) − g(x)| : x ∈ X}.

It is easy to check that “dist” is a metric (henceforth: the “max-metric”) on C(X), in which a sequence is convergent iff it converges uniformly on X. Similarly, a sequence in C(X) is Cauchy iff it is Cauchy uniformly on X. Thus the max-metric, which from now on we always assume to be part of the definition of C(X), makes that space complete. These notes prove the fundamental theorem about compactness in C(X):

1.1 The Arzela-Ascoli Theorem If a sequence {f_{n}}_{1}^{∞} in C(X) is bounded and equicontinuous then it has a uniformly convergent subsequence.

In this statement,

(a) “F ⊂ C(X) is bounded” means that there exists a positive constant M < ∞ such that |f(x)| ≤ M for each x ∈ X and each f ∈ F, and

(b) “F ⊂ C(X) is equicontinuous” means that: for every ε > 0 there exists δ > 0 (which depends only on ε) such that for x, y ∈ X:

d(x, y) < δ ⇒ |f(x) − f(y)| < ε ∀f ∈ F,

where d is the metric on X.

1.2 Exercise. The Arzela-Ascoli Theorem is the key to the following result: A subset F of C(X) is compact if and only if it is closed, bounded, and equicontinuous.

1.3 Exercise. You can think of ℝ^{n} as (real-valued) C(X) where X is a set containing n points, and the metric on X is the discrete metric (the distance between any two different points is 1). The metric thus induced on ℝ^{n} is equivalent to, but (unless n = 1) not the same as, the Euclidean one, and a subset of ℝ^{n} is bounded in the usual Euclidean way if and only if it is bounded in this C(X). Show that every bounded subset of this C(X) is equicontinuous, thus establishing the Bolzano-Weierstrass theorem as a generalization of the Arzela-Ascoli Theorem.

2 Proof of the Arzela-Ascoli Theorem.

Step I. We show that the compact metric space X is separable, i.e., has a countable dense subset S.

Given a positive integer n and a point x ∈ X, let

B(x, 1/n) = {y ∈ X : d(x, y) < 1/n},

the open ball of radius 1/n, centered at x. For a given n, the collection of all these balls as x runs through X is an open cover of x, so (because X is compact) there is a finite subcollection that also covers X. Let S_{n} denote the collection of centers of the balls in this finite subcollection. Thus S_{n} is a finite subset of X that is “1/n-dense” in the sense that every point of X lies within 1/n of a point of S_{n}. Clearly the union S of all the sets S_{n} is countable, and dense in X.

Step II. We find a subsequence of {f_{n}} that converges pointwise on S.

This is a standard diagonal argument. Let’s list the (countably many) elements of S as {x_{1}, x_{2}, …}. Then the numerical sequence {f_{n}(x_{1})}_{n=1}^{∞} is bounded, so by Bolzano-Weierstrass it has a convergent subsequence, which we’ll write using double subscripts: {f_{1,n}(x_{1})}_{n=1}^{∞}. Now the numerical sequence {f_{1,n}(x_{2})}_{n=1}^{∞} is bounded, so it has a convergent subsequence {f_{2,n}(x_{2})}_{n=1}^{∞}. Note that the sequence of functions {f_{2,n}}_{n=1}^{∞}, since it is a subsequence of {f_{1,n}}_{n=1}^{∞}, converges at both x_{1} and x_{2}. Proceeding in this fashion we obtain a countable collection of subsequences of our original sequence:

\begin{matrix}f_{1,1} & f_{1,2} & f_{1,3} & ⋯ \\ f_{2,1} & f_{2,2} & f_{2,3} & ⋯ \\ f_{3,1} & f_{3,2} & f_{3,3} & ⋯ \\ . & . & . & … \\ . & . & . & … \\ . & . & . & … \end{matrix}

where the sequence in the n-th row converges at the points x_{1}, …, x_{n}, and each row is a subsequence of the one above it.

Thus the diagonal sequence {f_{n,n}} is a subsequence of the original sequence {f_{n}} that converges at each point of S.

Step III. Completion of the proof.

Let {g_{n}} be the diagonal subsequence produced in the previous step, convergent at each point of the dense set S. Let ε > 0 be given, and choose δ > 0 by equicontinuity of the original sequence, so that d(x, y) < δ implies |g_{n}(x) − g_{n}(y)| < ε/3 for each x, y ∈ x and each positive integer n. Fix M > 1/δ so that the finite subset S_{M} ⊂ S that we produced in Step I is δ-dense in X. Since {g_{n}} converges at each point of S_{M}, there exists N > 0 such that

(∗) n, m > N ⇒ |g_{n}(s) − g_{m}(s)| < ε/3 ∀s ∈ S_{M}.

Fix x ∈ X. Then x lies within δ of some s ∈ S_{M}, so if n, m > M:

|g_{n}(x) − g_{m}(x)| ≤ |g_{n}(x) − g_{n}(s)| + |g_{n}(s) − g_{m}(s)| + |g_{m}(s) − g_{m}(x)|

The first and last terms on the right are < ε/3 by our choice of δ (which was possible because of the equicontinuity of the original sequence), and the same estimate holds for the middle term by our choice of N in (*). In summary: given ε > 0 we have produced N so that for each x ∈ X,

m, n > N ⇒ |g_{n}(x) − g_{m}(x)| < ε/3 + ε/3 + ε/3 = ε.

Thus on X the subsequence {g_{n}} of {f_{n}} is uniformly Cauchy, and therefore uniformly convergent. This completes the proof of the Arzela-Ascoli Theorem. □