One Continuity Bound for an Entire Family
Continuity of each function separately allows its choice of \(\delta\) to depend on the function. Equicontinuity asks for a shared choice: nearby inputs must have nearby outputs for every function in the family. This shared control is useful when studying a collection of functions at once. On a compact metric domain, local shared control at every point is enough to obtain one global choice of \(\delta\).
Let \((X,d)\) be a metric space with its metric topology, and let \(\mathcal{F}\) be a family of functions \(f:X\to\mathbb{R}\), where \(\mathbb{R}\) has its standard topology.
The order of the quantifiers matters. At a point \(x\), the chosen \(\delta\) may depend on \(x\), but it must work for every \(f\in\mathcal{F}\). Uniform equicontinuity requires a single \(\delta\) that works for all points as well as all functions. If \(\mathcal{F}\) is empty or \(X\) is empty, these universal conditions hold vacuously.
A common way to verify uniform equicontinuity is to establish a bound shared by the family. For example, if there is a constant \(L\geq 0\) such that \(|f(x)-f(y)|\leq Ld(x,y)\) for every \(f\in\mathcal{F}\) and every \(x,y\in X\), then \(\mathcal{F}\) is uniformly equicontinuous. If \(L>0\), choose \(\delta=\varepsilon/L\); if \(d(x,y)<\delta\), then \(|f(x)-f(y)|\leq Ld(x,y)<\varepsilon\). If \(L=0\), all these differences are zero, so any \(\delta>0\) works.
Shared Bounds in Practice
Worked Example: A Family with a Common Lipschitz Bound
On \(X=[0,1]\) with the Euclidean metric, let \(\mathcal{F}\) consist of all functions of the form \(f_a(x)=ax\), where \(a\in[-2,2]\). For any \(x,y\in[0,1]\) and any permitted \(a\),
Thus \(L=2\) is a shared Lipschitz bound. Given any \(\varepsilon>0\), choosing \(\delta=\varepsilon/2\) gives \(2|x-y|<\varepsilon\) whenever \(|x-y|<\delta\). For a numerical check, take \(\varepsilon=0.3\). The choice \(\delta=0.15\) gives \(|f_a(x)-f_a(y)|\leq2|x-y|<2(0.15)=0.3\). In particular, at \(x=0.2\), \(y=0.3\), and \(a=-2\), the actual difference is \(|-0.4-(-0.6)|=0.2\), which is less than \(0.3\). The bound and the direct calculation agree.
A shared estimate need not be exactly a Lipschitz bound. More generally, suppose there is a function \(\omega:[0,\infty)\to[0,\infty)\) such that \(\omega(0)=0\), \(\lim_{t\downarrow0}\omega(t)=0\), and \(|f(x)-f(y)|\leq\omega(d(x,y))\) for every \(f\in\mathcal{F}\) and every \(x,y\in X\). Then the family is uniformly equicontinuous. Indeed, given \(\varepsilon>0\), choose \(\delta>0\) so that \(0<t<\delta\) implies \(\omega(t)<\varepsilon\). If \(0<d(x,y)<\delta\), the shared estimate gives the required strict inequality. If \(d(x,y)=0\), then \(x=y\), so \(|f(x)-f(y)|=0<\varepsilon\). This separate zero-distance case is necessary: the limit condition alone does not constrain \(\omega(0)\).
Worked Example: Two Different Functions with One Bound
On \([0,1]\) with the Euclidean metric, consider the family \(\mathcal{F}=\{f,g\}\), where \(f(x)=x^2\) and \(g(x)=\sin x\). For \(x,y\in[0,1]\),
Here \(0\leq x+y\leq2\), which gives the inequality. The Mean Value Theorem applied to sine gives \(|\sin x-\sin y|=|\cos c||x-y|\) for some \(c\) between \(x\) and \(y\) when \(x\ne y\). Since \(|\cos c|\leq1\), \(|g(x)-g(y)|\leq|x-y|\); when \(x=y\), both sides are zero. Therefore both functions satisfy the shared bound \(2|x-y|\), and the family is uniformly equicontinuous with \(\delta=\varepsilon/2\).
For instance, with \(x=0.2\) and \(y=0.5\), the distance is \(0.3\). The two output differences are \(|0.2^2-0.5^2|=|0.04-0.25|=0.21\) and \(|\sin(0.2)-\sin(0.5)|\), which is at most \(0.3\) by the estimate just proved. For \(\varepsilon=0.7\), \(\delta=0.35\) applies to both functions and all pairs, since \(2(0.3)=0.6<0.7\). The common estimate verifies the family property without requiring the two functions to have identical formulas.
Compactness Upgrades Local Control
Proof. If \(X\) or \(\mathcal{F}\) is empty, uniform equicontinuity holds vacuously. Otherwise, fix \(\varepsilon>0\). At each \(z\in X\), equicontinuity at \(z\) gives a number \(r_z>0\) such that, for every \(f\in\mathcal{F}\), \(d(z,w)<r_z\) implies \(|f(z)-f(w)|<\varepsilon/2\). The open balls \(B(z,r_z/2)\), as \(z\) ranges over \(X\), cover \(X\). Compactness gives a finite subcover \(B(z_1,r_{z_1}/2),\ldots,B(z_m,r_{z_m}/2)\). Set \(\delta=\min_{1\leq i\leq m}r_{z_i}/2\), which is positive because it is the minimum of finitely many positive numbers. Let \(x,y\in X\) satisfy \(d(x,y)<\delta\), and choose \(i\) with \(d(x,z_i)<r_{z_i}/2\). Since \(\delta\leq r_{z_i}/2\), the triangle inequality gives \(d(y,z_i)\leq d(y,x)+d(x,z_i)<r_{z_i}/2+r_{z_i}/2=r_{z_i}\). Also \(d(x,z_i)<r_{z_i}\). Applying equicontinuity at \(z_i\) to \(x\) and to \(y\), for the same \(f\in\mathcal{F}\), yields \(|f(x)-f(z_i)|<\varepsilon/2\) and \(|f(z_i)-f(y)|<\varepsilon/2\). Therefore \(|f(x)-f(y)|\leq|f(x)-f(z_i)|+|f(z_i)-f(y)|<\varepsilon\). The choice of \(\delta\) works for all \(f,x,y\), proving uniform equicontinuity. Conversely, if the family is uniformly equicontinuous, its single \(\delta\) for \(\varepsilon\) works in particular when one of the two points is any fixed \(x\). Thus it is equicontinuous at \(x\). Since \(x\) was arbitrary, it is equicontinuous everywhere. \(\square\)
The compactness proof is a finite-cover argument. At each center \(z_i\), local control compares values with \(f(z_i)\). Two such comparisons, one for each point, give control of the difference between the values at the original points. Compactness supplies finitely many centers and hence one positive minimum radius. No estimate on the sizes of the function values is used.
Worked Example: Powers on the Compact Interval
Let \(X=[0,1]\) with the Euclidean metric, and let \(\mathcal{F}=\{f_n:n\in\mathbb{N}\}\), where \(f_n(x)=x^n\). This family is equicontinuous at each \(x_0<1\). To prove it, choose \(q\) with \(x_0<q<1\). If \(y\in[0,1]\) and \(|x_0-y|<q-x_0\), then \(y<q\), and both \(x_0\) and \(y\) belong to \([0,q]\). For \(u,v\in[0,q]\), factor the difference of powers to obtain, for \(n\geq1\),
The middle inequality holds because each of the \(n\) terms \(q^k\), for \(0\leq k\leq n-1\), is at least \(q^{n-1}\); the last uses the finite geometric sum, which is at most \(1/(1-q)\). Given \(\varepsilon>0\), choose \(\delta\) positive and smaller than both \(q-x_0\) and \(\varepsilon(1-q)\). Then \(|x_0-y|<\delta\) implies \(|f_n(x_0)-f_n(y)|\leq |x_0-y|/(1-q)<\varepsilon\) for every \(n\). Thus the family is equicontinuous at \(x_0\).
At \(x_0=1\), it is not equicontinuous. Fix \(\varepsilon=1/2\), and consider any proposed \(\delta>0\). Set \(y=1-\min(\delta/2,1/2)\). Then \(y\in[0,1)\) and \(|1-y|=\min(\delta/2,1/2)<\delta\). Because \(0\leq y<1\), its powers \(y^n\) tend to zero, so some \(n\in\mathbb{N}\) has \(y^n<1/2\). For that function, \(|f_n(1)-f_n(y)|=1-y^n>1/2\), contradicting the required inequality. Hence this family is not equicontinuous at 1 and so is not uniformly equicontinuous. This does not contradict the theorem: its premise requires equicontinuity at every point.
As a numerical check, at \(y=1/2\), \(f_1(y)=1/2\), \(f_2(y)=1/4\), and \(f_3(y)=1/8\), while each \(f_n(1)=1\). The corresponding differences are \(1/2\), \(3/4\), and \(7/8\). These values illustrate why a single pointwise choice at the endpoint cannot control all powers.
Why Compactness Matters
Worked Example: Local Equicontinuity Without a Uniform Choice
Consider \(\mathbb{R}\) with its standard metric, and define \(f_n(x)=\max(1-n|x-n|,0)\) for \(n\in\mathbb{N}\). Each function is zero outside \([n-1/n,n+1/n]\). The family is equicontinuous at every fixed \(x\in\mathbb{R}\), as follows. Choose an open interval \(J=(x-1/3,x+1/3)\). For all sufficiently large \(n\), the support interval \([n-1/n,n+1/n]\) does not meet \(J\): for example, once \(n\geq2\), its left endpoint is at least \(n-1/2\), which tends to infinity, while \(J\) is bounded. Thus only finitely many functions can be nonzero anywhere in \(J\). Each of those functions is Lipschitz with constant \(n\), since \(x\mapsto 1-n|x-n|\) has that bound and taking the maximum with zero does not increase differences. For the finitely many active indices, choose \(\delta>0\) smaller than \(1/3\) and than \(\varepsilon/n\) for each such index. If there are no active indices, any \(\delta<1/3\) works. For \(y\) with \(|x-y|<\delta\), both points lie in \(J\). Inactive functions vanish at both points, and each active function has difference at most \(n|x-y|<\varepsilon\). This proves equicontinuity at \(x\).
The family is nevertheless not uniformly equicontinuous. Given any \(\delta>0\), choose \(n\) so large that \(3/(4n)<\delta\), and set \(x=n\), \(y=n+3/(4n)\). Then \(f_n(n)=1\), whereas \(f_n(y)=\max(1-3/4,0)=1/4\). Thus \(|f_n(x)-f_n(y)|=3/4\), despite \(|x-y|=3/(4n)<\delta\). In particular, the uniform condition fails for \(\varepsilon=1/2\), for every proposed \(\delta\). The domain \(\mathbb{R}\) is not compact, so the compact-domain theorem does not apply.
Common pitfall. It is not enough to show that each function is continuous, or even that the family is equicontinuous separately at every point of a noncompact domain. Equicontinuity requires one local \(\delta\) for all functions at a given point; uniform equicontinuity requires one \(\delta\) for all points too. In the last example, each fixed point has a neighborhood where only finitely many functions matter, but the narrow supports move arbitrarily far out and their slopes increase. That is precisely why no global \(\delta\) works.
| Condition | What the choice of \(\delta\) must cover | Useful test |
|---|---|---|
| Equicontinuity at \(x\) | Every function, with \(x\) fixed | Find a shared local bound near \(x\) |
| Uniform equicontinuity | Every function and every pair of points | Find a shared global bound |
| Pointwise equicontinuity on compact \(X\) | Local choices at every point | Use a finite subcover and the compactness theorem |
Check Your Understanding
For each question, identify which quantifiers the proposed estimate must satisfy.
- Show that any family of constant functions on a metric space is uniformly equicontinuous.
- On \([0,1]\), test uniform equicontinuity of \(\{x\mapsto ax:|a|\leq5\}\), and give a valid \(\delta\) for a specified \(\varepsilon>0\).
- For \(f_n(x)=x^n\) on \([0,1]\), explain why the compact-domain theorem cannot be applied to conclude uniform equicontinuity.
- In the compactness theorem's proof, why is the minimum of the finitely many radii positive?
- Give an example of a bound \(\omega(t)\) with \(\lim_{t\downarrow0}\omega(t)=0\) but \(\omega(0)\ne0\). Why should the zero-distance case be handled separately?