What the Arzela–Ascoli Hypotheses Detect
Compactness of a family of continuous functions in the uniform metric rules out two different kinds of escape. The values at each fixed input cannot spread without bound, and the functions cannot develop sharper and sharper changes on smaller scales. The examples below isolate these failures: a spike family and an oscillation family retain pointwise relative compactness but lose equicontinuity, while a family of constant functions is equicontinuous but fails pointwise relative compactness.
Throughout, let \(I=[0,1]\) have its usual metric and topology. Write \(C(I,\mathbb{R})\) for the continuous maps from \(I\) to \(\mathbb{R}\), with the uniform metric \[ d_\infty(f,g)=\sup_{x\in I}|f(x)-g(x)|. \] A family is relatively compact in this function space if its closure in this metric is compact.
The quantifiers distinguish equicontinuity from continuity of each function separately. A continuous function on the compact interval \(I\) is uniformly continuous, but the resulting \(\delta\) may depend on the function. Equicontinuity requires one \(\delta\) to work simultaneously for the whole family.
The standard Arzela–Ascoli theorem for a compact metric domain and target \(\mathbb{R}\), established in the earlier material on that theorem, says that a family in \(C(I,\mathbb{R})\) is relatively compact in \(d_\infty\) if and only if it is equicontinuous and pointwise relatively compact. Here we examine what goes wrong when either condition is omitted.
Why Relative Compactness Requires Both Conditions
Proof. Let \(H\) be the compact closure of \(\mathcal{F}\) in \(C(I,\mathbb{R})\). Fix \(\varepsilon>0\). The open \(d_\infty\)-balls of radius \(\varepsilon/3\), centered at the members of \(H\), cover \(H\), so compactness gives a finite subcover with centers \(g_1,\ldots,g_k\in H\). Each \(g_i\) is continuous on the compact interval \(I\), hence uniformly continuous. Choose \(\delta_i>0\) so that \(|x-y|<\delta_i\) implies \(|g_i(x)-g_i(y)|<\varepsilon/3\), and put \(\delta=\min_{1\le i\le k}\delta_i>0\). For any \(f\in\mathcal{F}\), some \(g_i\) satisfies \(d_\infty(f,g_i)<\varepsilon/3\). Thus, if \(|x-y|<\delta\), then \[ |f(x)-f(y)|\le |f(x)-g_i(x)|+|g_i(x)-g_i(y)|+|g_i(y)-f(y)|<\varepsilon. \] This proves equicontinuity. For each fixed \(x\in I\), the evaluation map \(E_x:H\to\mathbb{R}\), defined by \(E_x(g)=g(x)\), is continuous because \(|E_x(g)-E_x(h)|\le d_\infty(g,h)\). A continuous image of a compact space is compact, so \(E_x(H)\) is compact and contains \(\{f(x):f\in\mathcal{F}\}\). Its closed subset \(\overline{\{f(x):f\in\mathcal{F}\}}\) is therefore compact. This proves pointwise relative compactness. \(\square\)
The two examples that follow show that pointwise relative compactness alone is not enough. In each, the functions are continuous and their values at every fixed input remain in a compact interval, but their variation cannot be controlled by one common \(\delta\).
Worked Example: Narrow Triangular Spikes
For \(n\in\mathbb{N}\), define \[ f_n(x)=\max\bigl(1-n|x-\tfrac12|,\,0\bigr),\qquad x\in I. \] Each \(f_n\) is continuous, since absolute value, scalar multiplication, and the maximum of two continuous real-valued functions preserve continuity. Also \(0\le f_n(x)\le1\) for every \(x\in I\). For any fixed \(x\), the set of values \(\{f_n(x):n\in\mathbb{N}\}\) lies in the compact interval \([0,1]\), so it has compact closure. Thus the family is pointwise relatively compact.
It is not equicontinuous. Given any \(\delta>0\), choose \(n\in\mathbb{N}\) with \(1/(2n)<\delta\), and set \(x=\tfrac12\), \(y=\tfrac12+\tfrac{1}{2n}\). Both points are in \(I\), and \[ |x-y|=\frac{1}{2n}<\delta,\qquad f_n(x)=1,\qquad f_n(y)=\max(1-\tfrac12,0)=\tfrac12. \] Hence \(|f_n(x)-f_n(y)|=\tfrac12\). Taking \(\varepsilon=\tfrac14\), no proposed \(\delta\) works for every \(n\).
This family is also not relatively compact in the uniform metric. At \(x=\tfrac12\), every \(f_n(x)=1\). At each fixed \(x\ne\tfrac12\), the distance \(r=|x-\tfrac12|\) is positive, and for every integer \(n\ge 1/r\), \(1-nr\le0\), so \(f_n(x)=0\). Therefore the pointwise limit is \(g(\tfrac12)=1\) and \(g(x)=0\) for \(x\ne\tfrac12\), which is discontinuous at \(\tfrac12\).
To justify why this rules out a uniformly convergent subsequence, recall the following direct fact. If continuous functions \(h_j\) converge uniformly to \(h\), then \(h\) is continuous: for \(x_0\in I\) and \(\varepsilon>0\), choose \(j\) with \(d_\infty(h_j,h)<\varepsilon/3\). By continuity of \(h_j\), choose \(\delta>0\) such that \(|x-x_0|<\delta\) gives \(|h_j(x)-h_j(x_0)|<\varepsilon/3\). Then \[ |h(x)-h(x_0)|\le |h(x)-h_j(x)|+|h_j(x)-h_j(x_0)|+|h_j(x_0)-h(x_0)|<\varepsilon. \] Every subsequence of \((f_n)\) has the same pointwise limit \(g\), so a uniformly convergent subsequence would have to converge to \(g\), contradicting this continuity fact. Since a compact metric space is sequentially compact, relative compactness would give a uniformly convergent subsequence. Thus the family is not relatively compact.
Oscillation: Bounded Values Do Not Control Change
Worked Example: Increasing-Frequency Sines
Define \(f_n(x)=\sin(2\pi n x)\) for \(x\in I\) and \(n\in\mathbb{N}\). Every value lies in \([-1,1]\), so at each fixed \(x\), the value set has compact closure by the compactness of this interval. The family is pointwise relatively compact.
To test equicontinuity, fix \(\varepsilon=\tfrac12\). Given \(\delta>0\), choose \(n\) with \(1/(4n)<\delta\) and take \(x=0\), \(y=1/(4n)\). Then \[ |x-y|=\frac{1}{4n}<\delta,\qquad f_n(0)=0,\qquad f_n(y)=\sin(2\pi n/(4n))=\sin(\pi/2)=1. \] Thus \(|f_n(y)-f_n(0)|=1>\tfrac12\), so the family is not equicontinuous.
We can also verify directly that no two distinct members are close in the uniform metric. For distinct \(n,m\in\mathbb{N}\), product-to-sum gives \[ \int_0^1 \bigl(f_n(x)-f_m(x)\bigr)^2\,dx =\frac12+\frac12-2\int_0^1\sin(2\pi n x)\sin(2\pi m x)\,dx=1. \] Indeed, each squared-sine integral is \(1/2\), and the cross integral is zero: the product-to-sum formula expresses it as half the difference of integrals of \(\cos(2\pi(n-m)x)\) and \(\cos(2\pi(n+m)x)\), both zero because \(n-m\) and \(n+m\) are nonzero integers. Since the interval has length \(1\), \[ \int_0^1 \bigl(f_n(x)-f_m(x)\bigr)^2\,dx \le d_\infty(f_n,f_m)^2. \] Consequently \(d_\infty(f_n,f_m)\ge1\) whenever \(n\ne m\). If the closure of the family were compact, its open balls of radius \(1/3\) would have a finite subcover. Each such ball could contain at most one \(f_n\), since two members in it would have distance less than \(2/3\) by the triangle inequality. A finite collection of these balls cannot cover all the \(f_n\). The family is not relatively compact.
The integral calculation gives a useful distinction: pointwise relative compactness controls the range at each fixed input, not the number of oscillations across the domain. The uniform metric detects those oscillations because it records the largest difference over all inputs.
Equicontinuity Without Pointwise Relative Compactness
Worked Example: Constant Functions Escaping to Infinity
For each \(n\in\mathbb{N}\), let \(f_n(x)=n\) for all \(x\in I\). Each function is continuous. For any \(x,y\in I\) and any \(n\), \[ |f_n(x)-f_n(y)|=|n-n|=0. \] Thus the family is equicontinuous: for every \(\varepsilon>0\), any \(\delta>0\) satisfies its defining condition.
At each fixed \(x\), however, the value set is \(\{1,2,3,\ldots\}\), which is not relatively compact in \(\mathbb{R}\). It is closed and unbounded; more explicitly, the open sets \((-N,N)\), \(N\in\mathbb{N}\), cover it, but no finite subcollection covers it. The family also fails to be relatively compact in \(d_\infty\), since \[ d_\infty(f_n,f_m)=|n-m|\ge1\quad\text{when }n\ne m. \] As with the sine family, a compact closure would have a finite cover by open balls of radius \(1/3\), and each ball could contain at most one of these functions. This is impossible for the infinite family.
This example shows why pointwise relative compactness cannot be replaced by continuity of each function or by equicontinuity. The family changes by zero across the domain, yet its values at every fixed input escape to infinity as \(n\) increases.
A Common Mistake: Choosing a Scale for Each Function
Common pitfall. For \(f_n(x)=\sin(2\pi n x)\), one can estimate \[ |f_n(x)-f_n(y)|\le 2\pi n|x-y|. \] It is tempting to conclude that the family is equicontinuous by choosing \(\delta=\varepsilon/(2\pi n)\). That calculation gives a scale depending on \(n\), not one \(\delta\) that works for all members of the family. For example, with \(\varepsilon=\tfrac12\), any proposed \(\delta>0\) admits \(n\) with \(1/(4n)<\delta\); at \(x=0\) and \(y=1/(4n)\), the function \(f_n\) changes by \(1\). The correction is to keep the quantifiers in order: choose \(\varepsilon\), then seek a single \(\delta\), and only then test every function in the family.
| Family | Pointwise relatively compact? | Equicontinuous? | Relatively compact in \(d_\infty\)? |
|---|---|---|---|
| Narrow triangular spikes | Yes: all values lie in \([0,1]\) | No: a fixed-size change occurs over distances tending to zero | No: any subsequence has a discontinuous pointwise limit |
| \(\sin(2\pi n x)\) | Yes: all values lie in \([-1,1]\) | No: the change from \(0\) to \(1\) occurs arbitrarily close to \(0\) | No: distinct members are at least distance \(1\) apart |
| Constant functions \(f_n(x)=n\) | No: the values are unbounded at each input | Yes: every function has zero variation | No: distinct members are at least distance \(1\) apart |
Check Your Understanding
For each question, identify which Arzela–Ascoli condition is relevant and verify the quantifiers or values involved.
- For the spike family, compute \(f_n(\tfrac12+\tfrac{1}{4n})\). How does this value help test equicontinuity?
- For the family \(g_n(x)=x^n\) on \(I\), determine the pointwise limit and decide whether the family is equicontinuous at \(x=1\).
- Explain why boundedness of \(\{f(x):f\in\mathcal{F}\}\) for each fixed \(x\) is not the same as equicontinuity.
- For the constant family \(f_n(x)=n\), give an open cover of the pointwise value set that has no finite subcover.
- In the sine example, why does a uniform lower bound on the distance between distinct family members rule out compact closure?