Compactness Criteria In C(K,R^m)

Compact Closure in a Space of Continuous Functions

For a family of continuous functions, compactness in a function space is a statement about controlling whole functions at once. On a compact metric domain, the Arzelà–Ascoli theorem turns that control into two concrete tests: the family must be uniformly bounded, and its members must vary in a uniformly controlled way near each point. These tests concern the uniform topology, not merely the behavior of each function separately.

Let \((K,d)\) be a nonempty compact metric space, and give \(\mathbb{R}^m\) its Euclidean norm \(\|\cdot\|\) and standard topology. Write \(C(K,\mathbb{R}^m)\) for the set of continuous maps from \(K\) to \(\mathbb{R}^m\). For \(f,g\in C(K,\mathbb{R}^m)\), define the uniform metric by

This supremum is finite: each continuous function on the compact space \(K\) is bounded, so \(x\mapsto\|f(x)-g(x)\|\) is bounded. A family \(\mathcal{F}\subseteq C(K,\mathbb{R}^m)\) is relatively compact if its closure in this metric is compact. It is uniformly bounded if there is a finite \(M\geq0\) such that \(\|f(x)\|\leq M\) for every \(f\in\mathcal{F}\) and every \(x\in K\). It is equicontinuous if for every \(x\in K\) and every \(\varepsilon>0\), there is a \(\delta>0\) such that, for every \(f\in\mathcal{F}\) and every \(y\in K\), \(d(x,y)<\delta\) implies \(\|f(x)-f(y)\|<\varepsilon\). The same \(\delta\) must work for all members of the family.

Why Relative Compactness Requires Both Conditions

We first prove the direction that extracts the two conditions from compact closure. The argument uses finite uniform approximations: a compact set in a metric space can be covered by finitely many balls of any prescribed positive radius.

Proof. Suppose \(\overline{\mathcal{F}}\) is compact. If \(\mathcal{F}\) is empty, both conditions hold vacuously, so assume it is nonempty. To prove uniform boundedness, cover \(\overline{\mathcal{F}}\) by the open \(d_\infty\)-balls of radius \(1\) centered at its points. Compactness gives a finite subcover with centers \(g_1,\ldots,g_r\in\overline{\mathcal{F}}\). Each \(g_j\) is continuous on compact \(K\), hence bounded; let \(M_0=\max_{1\leq j\leq r}\sup_{x\in K}\|g_j(x)\|\), which is finite. For any \(f\in\mathcal{F}\), some \(j\) satisfies \(d_\infty(f,g_j)<1\). Thus, for every \(x\in K\), \(\|f(x)\|\leq\|g_j(x)\|+\|f(x)-g_j(x)\|<M_0+1\). This is a bound independent of \(f\) and \(x\).

Now fix \(x\in K\) and \(\varepsilon>0\). Cover \(\overline{\mathcal{F}}\) by uniform balls of radius \(\varepsilon/3\), and choose a finite subcover with centers \(h_1,\ldots,h_s\in\overline{\mathcal{F}}\). Each \(h_j\) is continuous at \(x\), so there is \(\delta_j>0\) such that \(d(x,y)<\delta_j\) implies \(\|h_j(x)-h_j(y)\|<\varepsilon/3\). Set \(\delta=\min_{1\leq j\leq s}\delta_j>0\). Given \(f\in\mathcal{F}\), choose a center \(h_j\) with \(d_\infty(f,h_j)<\varepsilon/3\). If \(d(x,y)<\delta\), the triangle inequality gives \[ \|f(x)-f(y)\| \leq \|f(x)-h_j(x)\|+\|h_j(x)-h_j(y)\|+\|h_j(y)-f(y)\| <\varepsilon. \] The choice of \(\delta\) is independent of \(f\), so \(\mathcal{F}\) is equicontinuous. \(\square\)

For the converse, the relevant form of the Arzelà–Ascoli theorem says that on a compact metric domain, an equicontinuous family whose values at each point have compact closure in the codomain has compact closure in the uniform metric. This is the standard Arzelà–Ascoli theorem established in the earlier material on that theorem. If \(\mathcal{F}\) is uniformly bounded by \(M\), then for each \(x\in K\), all values \(f(x)\) lie in the closed Euclidean ball of radius \(M\) in \(\mathbb{R}^m\). That ball is compact by the finite-dimensional Heine–Borel theorem. The pointwise compactness hypothesis therefore holds.

Proof. Suppose \(\mathcal{F}\) is uniformly bounded and equicontinuous. The preceding paragraph verifies that for each \(x\in K\), the set \(\{f(x):f\in\mathcal{F}\}\) has compact closure in \(\mathbb{R}^m\). The stated Arzelà–Ascoli theorem applies, since \(K\) is compact metric and \(\mathcal{F}\) is equicontinuous. It follows that \(\overline{\mathcal{F}}\) is compact in \(d_\infty\). \(\square\)

Thus the criterion is an if-and-only-if: uniform boundedness and equicontinuity are both necessary, and together they are sufficient. In particular, if \(\mathcal{F}\) is closed in \(C(K,\mathbb{R}^m)\), the criterion characterizes when \(\mathcal{F}\) itself is compact. Indeed, for a closed set, \(\overline{\mathcal{F}}=\mathcal{F}\); conversely, every compact subset of a metric space is closed and hence is relatively compact.

Applying the Criterion

A Common Mistake: Letting the Scale Depend on the Function

For the powers \(f_n(x)=x^n\), a tempting but invalid argument is: “For each fixed \(n\), continuity gives a \(\delta\) that controls \(f_n\), so the family is equicontinuous.” The \(\delta\) supplied by continuity may depend on \(n\); equicontinuity demands one \(\delta\) that works for every \(n\). The calculation above shows the failure concretely: for every \(\delta>0\), choose \(n\geq2\) with \(1/n<\delta\), then \(y=1-1/n\) satisfies \(|1-y|<\delta\) while \(|f_n(1)-f_n(y)|>1/2\). Pointwise continuity of every member does not supply uniform control across the family.

The distinction between the two conditions is also useful diagnostically. Constant functions can be equicontinuous while their sizes become unbounded, as in the second example. The powers are uniformly bounded while equicontinuity fails. The criterion rules out relative compactness in either case; passing only one test is not enough.