Mathematics.

Compactness in Function Spaces

Equicontinuity and Arzelà–Ascoli Theorem

Real Analysis90 minDifficulty9 out of 10

You should know: uniform continuity, metric spaces, uniform convergence

Overview

The Arzela-Ascoli theorem characterises the compact subsets of C(K) — the space of continuous functions on a compact metric space K with the sup-norm. A set F of functions is relatively compact (has compact closure) if and only if it is uniformly bounded and equicontinuous. Equicontinuity is the condition that all functions in F are uniformly continuous with a single modulus of continuity. The theorem is the function-space analogue of the Bolzano-Weierstrass theorem (bounded sequences in R^n have convergent subsequences) and is a key tool in the existence theory for ODEs, PDEs, and integral equations.

Intuition

In \mathbb{R}^n, compactness = closed and bounded (Heine-Borel). For sequences of functions, boundedness alone is not enough — a bounded sequence of functions can oscillate wildly without a convergent subsequence. Equicontinuity provides the extra regularity needed: if the functions all vary at the same 'rate' (same modulus of continuity), then a bounded equicontinuous family behaves much like a bounded set in \mathbb{R}^n and admits convergent subsequences.

Formal Definition

Definition

Let K be a compact metric space and C(K) the space of continuous real-valued functions with sup-norm.

F⊆C(K) is equicontinuous at x0  ⟺  ∀ε>0, ∃δ>0, d(x,x0)<δ  ⟹  ∣f(x)−f(x0)∣<ε ∀f∈F.\mathcal{F} \subseteq C(K) \text{ is equicontinuous at } x_0 \iff \forall \varepsilon > 0,\ \exists \delta > 0,\ d(x,x_0) < \delta \implies |f(x)-f(x_0)| < \varepsilon\ \forall f \in \mathcal{F}.
Equicontinuity at a point
F is uniformly equicontinuous  ⟺  ∀ε>0, ∃δ>0, d(x,y)<δ  ⟹  ∣f(x)−f(y)∣<ε ∀f∈F.\mathcal{F} \text{ is uniformly equicontinuous} \iff \forall \varepsilon > 0,\ \exists \delta > 0,\ d(x,y) < \delta \implies |f(x)-f(y)| < \varepsilon\ \forall f \in \mathcal{F}.
Uniform equicontinuity
F is uniformly bounded  ⟺  ∃M, ∣f(x)∣≤M ∀f∈F, x∈K.\mathcal{F} \text{ is uniformly bounded} \iff \exists M,\ |f(x)| \leq M\ \forall f \in \mathcal{F},\ x \in K.
Uniform boundedness
Arzela-Ascoli: F‾ compact in C(K)  ⟺  F uniformly bounded and equicontinuous.\text{Arzela-Ascoli: } \overline{\mathcal{F}} \text{ compact in } C(K) \iff \mathcal{F} \text{ uniformly bounded and equicontinuous.}
Arzela-Ascoli theorem

Worked Examples

  1. Each f_n(x) = \sin(nx)/n is differentiable with f_n'(x) = \cos(nx).

    fn′(x)=cos⁡(nx).f_n'(x) = \cos(nx).

  2. So |f_n'(x)| \leq 1 for all x and all n. By the Mean Value Theorem: |f_n(x) - f_n(y)| \leq |x-y| for all n.

    ∣fn(x)−fn(y)∣≤sup⁡t∣fn′(t)∣⋅∣x−y∣≤∣x−y∣.|f_n(x)-f_n(y)| \leq \sup_t |f_n'(t)| \cdot |x-y| \leq |x-y|.

  3. Given \varepsilon > 0, take \delta = \varepsilon. Then |x-y| < \delta \implies |f_n(x)-f_n(y)| < \varepsilon for all n simultaneously — this is equicontinuity.

  4. Also \|f_n\|_\infty \leq 1/n \leq 1, so the family is uniformly bounded. By Arzela-Ascoli, \{f_n\} has a uniformly convergent subsequence.

Answer: The family \{\sin(nx)/n\} is equicontinuous (uniformly Lipschitz-1) and uniformly bounded, so has a uniformly convergent subsequence by Arzela-Ascoli.

Practice Problems

Difficulty 7/10

Is the family \mathcal{F} = \{f_n(x) = x^n : n \geq 1\} on [0,1] equicontinuous?

Difficulty 8/10

Prove the 'easy' direction of Arzela-Ascoli: if \overline{\mathcal{F}} is compact in C(K), then \mathcal{F} is uniformly bounded and equicontinuous.

Difficulty 9/10

Give an example showing that equicontinuity alone (without uniform boundedness) does not guarantee a convergent subsequence.

Common Mistakes

Common Mistake

A uniformly bounded sequence of continuous functions always has a uniformly convergent subsequence.

Uniform boundedness alone is not enough; equicontinuity is also needed. The sequence f_n(x) = \sin(nx) on [0,1] is uniformly bounded but has no uniformly convergent subsequence.

Common Mistake

Equicontinuity is the same as each function being uniformly continuous.

Each function may be uniformly continuous, but equicontinuity requires the same modulus of continuity (the same \delta) to work for all functions simultaneously.

Quiz

The Arzela-Ascoli theorem characterises compact subsets of C(K) as sets that are:
A family \mathcal{F} of functions is equicontinuous if:

Summary

  • Equicontinuity: a family \mathcal{F} \subseteq C(K) is equicontinuous if one \delta works for all f \in \mathcal{F} for each \varepsilon.
  • Arzela-Ascoli theorem: a family in C(K) (K compact metric) is relatively compact iff it is uniformly bounded and equicontinuous.
  • On compact K, equicontinuity is equivalent to uniform equicontinuity.
  • The theorem is the key existence tool in ODEs (Peano theorem), integral equations, and harmonic analysis.
  • Without either hypothesis (boundedness or equicontinuity), compactness can fail.

References

  1. BookRudin, W. — Principles of Mathematical Analysis (3rd ed.), McGraw-Hill, 1976, Chapter 7.
  2. WebsiteWikipedia — Arzela-Ascoli theorem