Continuity for a Whole Family
Continuity and uniform continuity describe one function at a time. When studying many functions together, a related question is whether their continuity can be controlled with the same choices of input scales. This shared control is called equicontinuity. Compactness connects the local version of this property to a global one, and it also helps characterize when a family of functions is compact in the uniform metric.
We will work with a nonempty compact metric space \((K,d)\) and real-valued continuous functions on \(K\). Write \(C(K)\) for the set of these functions. Since continuous real-valued functions on a compact space are bounded, the supremum metric \(d_\infty(f,g)=\sup_{x\in K}|f(x)-g(x)|\) is finite for \(f,g\in C(K)\). It measures the largest difference between two functions anywhere on the domain.
The distinction is where the scale may depend. Equicontinuity allows the radius \(\delta\) to depend on the point \(x\), but not on the function in the family. Uniform equicontinuity requires that radius to work across the whole domain as well. For a family containing just one function, equicontinuity is ordinary continuity, while uniform equicontinuity is uniform continuity.
Compactness Makes the Scale Uniform
On a compact domain, a finite-cover argument turns the local radii in equicontinuity into a single radius. This extends the same compactness principle used in the Heine-Cantor theorem: local control, combined with a finite subcover and a Lebesgue number, provides global control.
Proof. The claim is immediate if \(\mathcal{F}\) is empty, so suppose it is nonempty. Fix \(\varepsilon>0\). For each \(z\in K\), equicontinuity at \(z\) gives \(r_z>0\) such that, for every \(f\in\mathcal{F}\), \(d(w,z)<r_z\) implies \(|f(w)-f(z)|<\varepsilon/3\). The balls \(B_{r_z/2}(z)\), for \(z\in K\), cover \(K\). Compactness gives a finite subcover. By the Lebesgue Number Lemma, this cover has a Lebesgue number \(\lambda>0\): every subset of \(K\) with diameter less than \(\lambda\) is contained in one of these balls.
Take \(x,y\in K\) with \(d(x,y)<\lambda\). The set \(\{x,y\}\) has diameter less than \(\lambda\), so it lies in some \(B_{r_z/2}(z)\) from the finite subcover. In particular, \(d(x,z)<r_z\) and \(d(y,z)<r_z\). For every \(f\in\mathcal{F}\), the triangle inequality gives
The number \(\lambda\) works for all \(f,x,y\), so the family is uniformly equicontinuous. \(\square\)
The theorem has a useful converse in this setting: if a family is uniformly equicontinuous, then it is equicontinuous at every point, simply by using the same \(\delta\) there. Thus on compact metric domains the two properties are equivalent. The finite-cover argument is essential; on a noncompact domain, point-dependent continuity scales need not have a positive global lower bound.
Worked Example: A Bounded Family of Linear Functions
Let \(K=[-1,1]\), and for each \(a\in[-2,2]\) define \(f_a(x)=ax\). For any \(x,y\in K\),
Given \(\varepsilon>0\), choose \(\delta=\varepsilon/2\). Whenever \(|x-y|<\delta\), the displayed estimate gives \(|f_a(x)-f_a(y)|<\varepsilon\), for every \(a\in[-2,2]\). The family is therefore uniformly equicontinuous. Also, for each fixed \(x\), \(|f_a(x)|\leq 2\), so the family is pointwise bounded. The Arzelà-Ascoli theorem below will imply that this family has compact closure in the supremum metric.
The Arzelà-Ascoli Compactness Criterion
Equicontinuity controls how quickly functions can change across the domain. A second condition is needed to prevent their values from escaping without bound. A family \(\mathcal{F}\subseteq C(K)\) is pointwise bounded if, for every \(x\in K\), the set \(\{f(x):f\in\mathcal{F}\}\) is bounded in \(\mathbb{R}\). The bound is initially allowed to depend on \(x\).
Proof. First suppose that the closure of \(\mathcal{F}\) is compact in \(C(K)\). A compact subset of a metric space is bounded, so \(\mathcal{F}\) is bounded in the supremum metric. In particular, there is a finite \(M\) such that \(\sup_{x\in K}|f(x)|\leq M\) for every \(f\in\mathcal{F}\). This implies pointwise boundedness.
We next prove equicontinuity. Compact subsets of metric spaces are totally bounded, so for any \(\varepsilon>0\), finitely many functions \(g_1,\ldots,g_m\) in the compact closure of \(\mathcal{F}\) can be chosen so that every \(f\in\mathcal{F}\) satisfies \(d_\infty(f,g_i)<\varepsilon/3\) for some \(i\). Each \(g_i\) is continuous on compact \(K\), hence is uniformly continuous by the Heine-Cantor theorem. Because there are only finitely many \(g_i\), there is one \(\delta>0\) such that \(d(x,y)<\delta\) implies \(|g_i(x)-g_i(y)|<\varepsilon/3\) for every \(i\). For \(f\in\mathcal{F}\), choose a corresponding \(g_i\). Then
whenever \(d(x,y)<\delta\). Thus \(\mathcal{F}\) is uniformly equicontinuous, and in particular equicontinuous.
Conversely, suppose that \(\mathcal{F}\) is equicontinuous and pointwise bounded. If \(\mathcal{F}\) is empty, its closure is compact, so assume it is nonempty. We show first that \(\mathcal{F}\) is totally bounded in \(d_\infty\). Fix \(\varepsilon>0\). By the theorem just proved, \(\mathcal{F}\) is uniformly equicontinuous. Choose \(\delta>0\) such that, for all \(f\in\mathcal{F}\),
Since compact metric spaces are totally bounded, choose finitely many points \(p_1,\ldots,p_m\in K\) whose open \(\delta\)-balls cover \(K\). At each \(p_j\), pointwise boundedness gives a finite bound \(M_j\) with \(|f(p_j)|\leq M_j\) for every \(f\in\mathcal{F}\). Divide each interval \([-M_j,M_j]\) into finitely many subintervals of length at most \(\varepsilon/4\). Assign each function \(f\) to the finite list of subintervals containing its values \(f(p_1),\ldots,f(p_m)\). There are only finitely many such lists, and hence finitely many nonempty classes of functions.
Choose one representative from each nonempty class. If \(f\) and \(g\) belong to the same class, their values at every \(p_j\) differ by at most \(\varepsilon/4\). Given \(x\in K\), choose \(p_j\) with \(d(x,p_j)<\delta\). Then
Therefore each nonempty class lies within \(d_\infty\)-distance \(\varepsilon\) of its representative. The finite set of representatives is an \(\varepsilon\)-net, proving that \(\mathcal{F}\) is totally bounded.
The space \(C(K)\) is complete: \(K\) is a metric space and \(C(K)=C_b(K)\), so this follows from the completeness theorem for \(C_b(E)\). The closure of a totally bounded set is totally bounded, and the closure is complete as a closed subset of a complete space. By the Complete and Totally Bounded Metric Spaces Are Compact theorem, the closure of \(\mathcal{F}\) is compact. This proves the criterion. \(\square\)
Worked Examples: Testing the Two Conditions
Worked Example: Pointwise Bounded but Not Equicontinuous
On \(K=[0,1]\), consider \(f_n(x)=x^n\) for positive integers \(n\). For each \(x\in[0,1]\), \(0\leq f_n(x)\leq1\), so the family is pointwise bounded. It is not equicontinuous at \(x=1\). For \(n\geq2\), set \(x_n=1-1/n\). Then \(x_n\to1\), but
To verify the inequality, note that \(\left(1-\frac1n\right)^{-n}=\left(1+\frac1{n-1}\right)^n \geq 1+\frac{n}{n-1}>2\), using the first two terms of the binomial expansion. Thus \(\left(1-\frac1n\right)^n<1/2\). For every \(\delta>0\), one can choose \(n\) large enough that \(|x_n-1|=1/n<\delta\), while the corresponding function values differ by more than \(1/2\). This contradicts equicontinuity at \(1\) for \(\varepsilon=1/2\). Pointwise boundedness alone does not ensure compactness in the supremum metric.
Worked Example: Equicontinuous but Not Pointwise Bounded
Let \(K=[0,1]\), and take the family of constant functions \(c_n(x)=n\), for positive integers \(n\). For all \(x,y\in K\), \(|c_n(x)-c_n(y)|=0\), so the family is uniformly equicontinuous. At every fixed \(x\), however, the values \(\{c_n(x):n\geq1\}=\{1,2,3,\ldots\}\) are unbounded. Also, distinct functions in this family satisfy \(d_\infty(c_n,c_m)=|n-m|\geq1\). Thus no finite collection of supremum-metric balls of radius \(1/2\) can cover the family. It is not totally bounded and cannot have compact closure. Equicontinuity alone is not enough.
Why Both Conditions Matter
The finite-net proof explains the roles of the hypotheses. Equicontinuity lets us replace control over every point of \(K\) by control at finitely many nearby sample points. Pointwise boundedness then ensures that, at each of those sample points, only finitely many value ranges need to be considered. Together these produce finitely many representatives that approximate the entire family uniformly.
Pointwise boundedness is weaker in its wording than a single bound on all values of all functions. In this theorem, equicontinuity and compactness allow the finite-net argument to produce the needed uniform approximations without assuming such a global bound at the outset. By contrast, the constant-function example fails at every point, and no finite collection of functions can approximate all its members in the supremum metric.
The theorem concerns compactness of a family in the supremum metric, not merely continuity of its individual members. Every function in \(C(K)\) is continuous, but a collection of continuous functions may still fail to be equicontinuous or pointwise bounded. When applying the criterion, check both conditions on the family, and be clear whether the conclusion sought is compactness of the family itself or compactness of its closure.
Check Your Understanding
Use the definitions and compactness criterion above to answer the following questions.
- In the definition of equicontinuity, which choices may depend on the point, and which may not?
- Where does compactness enter the proof that an equicontinuous family is uniformly equicontinuous?
- Why do finitely many sample points suffice to build a finite net in the supremum metric?
- What property does the family of constant functions \(c_n(x)=n\) satisfy, and which Arzelà-Ascoli condition does it fail?
- Why does pointwise boundedness of \(x^n\) on \([0,1]\) not imply equicontinuity?