From a Shared Estimate to Compactness
A single continuous function can behave very differently at different points, but a family of functions may be controlled by one estimate that applies to every member. On a compact metric domain, a uniform bound controls the possible values at each point, while a shared Lipschitz estimate prevents the functions from changing too rapidly. Together these conditions give the compactness conclusion of Arzela–Ascoli.
Let \((K,d)\) be a compact metric space and let \(C(K,\mathbb{R}^m)\) denote the continuous functions from \(K\) to \(\mathbb{R}^m\). We use the uniform metric \(d_\infty(f,g)=\sup_{x\in K}\|f(x)-g(x)\|\). For compact \(K\), this metric gives the usual compact-open topology on this function space. A family \(F\subseteq C(K,\mathbb{R}^m)\) is uniformly bounded if there is \(M<\infty\) such that \(\|f(x)\|\le M\) for every \(f\in F\) and \(x\in K\). It is equicontinuous if for every \(\varepsilon>0\) and every \(x\in K\), there is \(\delta>0\) such that \(d(x,y)<\delta\) implies \(\|f(x)-f(y)\|<\varepsilon\) for every \(f\in F\).
In finite-dimensional Euclidean space, a bounded set is relatively compact by the Heine–Borel theorem. Thus a uniform bound supplies the pointwise relative compactness condition. The remaining work is to verify equicontinuity and, if actual compactness rather than relative compactness is wanted, closedness.
Uniformly Bounded Lipschitz Families
Proof. If \(K=\varnothing\), there is exactly one function from \(K\) to \(\mathbb{R}^m\), so every family of such functions is either empty or a singleton and is compact. Assume now that \(K\ne\varnothing\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon/(L+1)>0\). If \(d(x,y)<\delta\), then for every \(f\in F\), \(\|f(x)-f(y)\|\le Ld(x,y)<L\varepsilon/(L+1)<\varepsilon\) when \(L>0\); if \(L=0\), the left side is zero and is also less than \(\varepsilon\). Hence \(F\) is equicontinuous. For each fixed \(x\), all values \(f(x)\) lie in the closed ball of radius \(M\) in \(\mathbb{R}^m\), which is compact by Heine–Borel. The Arzela–Ascoli criterion therefore says \(F\) is relatively compact. Since \(F\) is closed, it equals its intersection with its compact closure and is a closed subset of a compact space, hence is compact. \(\square\)
The constraints themselves also define a closed family. Indeed, if \(f_n\) satisfies them and \(f_n\to f\) uniformly, then for each \(x\), \(\|f(x)\|=\lim_n\|f_n(x)\|\le M\). For each \(x,y\), continuity of the norm and convergence at both points give \(\|f(x)-f(y)\|=\lim_n\|f_n(x)-f_n(y)\|\le Ld(x,y)\). Thus \(f\) satisfies both constraints. This verifies the “in particular” assertion, rather than assuming closedness.
Worked Example: Anchored Lipschitz Functions on the Unit Interval
Let \(K=[0,1]\) with its usual metric and let \(F\) consist of all continuous \(f:[0,1]\to\mathbb{R}\) such that \(f(0)=0\) and \(|f(x)-f(y)|\le 2|x-y|\) for all \(x,y\). Show that \(F\) is compact in the uniform metric.
First compute a uniform bound. Taking \(y=0\) gives \(|f(x)|=|f(x)-f(0)|\le 2|x-0|=2x\le2\) for every \(x\in[0,1]\). Thus \(M=2\) works, and the shared Lipschitz constant is \(L=2\). If \(f_n\in F\) and \(f_n\to f\) uniformly, then \(f(0)=\lim_n f_n(0)=0\), and for any \(x,y\), \(|f(x)-f(y)|=\lim_n|f_n(x)-f_n(y)|\le2|x-y|\). So \(F\) is closed. The theorem now gives compactness.
Check: The function \(f(x)=2x\) belongs to the family: \(f(0)=0\), \(|f(x)|=2x\le2\), and \(|f(x)-f(y)|=2|x-y|\). For \(x=3/4\) and \(y=1/4\), the difference is \(|3/2-1/2|=1\), while \(2|3/4-1/4|=2(1/2)=1\), so the Lipschitz inequality holds with equality at these points.
Replacing the Lipschitz Bound by a Common Modulus
A common Lipschitz constant is one particular way to control how fast all functions can vary. More generally, a common modulus of continuity is a function \(\omega:[0,\infty)\to[0,\infty)\) with \(\omega(0)=0\) and \(\lim_{r\downarrow0}\omega(r)=0\), such that every function in the family satisfies \(\|f(x)-f(y)\|\le\omega(d(x,y))\). The function \(\omega\) need not be linear. Its limit at zero ensures that sufficiently close inputs give uniformly close outputs.
Proof. If \(K=\varnothing\), the conclusion follows as in the preceding theorem. Otherwise fix \(\varepsilon>0\). Since \(\omega(r)\to0\) as \(r\downarrow0\), there is \(\delta>0\) such that \(0\le r<\delta\) implies \(\omega(r)<\varepsilon\). Thus \(d(x,y)<\delta\) gives \(\|f(x)-f(y)\|\le\omega(d(x,y))<\varepsilon\) simultaneously for every \(f\in F\). So \(F\) is equicontinuous. Uniform boundedness and Heine–Borel give relative compactness of the values at each point. Arzela–Ascoli implies relative compactness of \(F\); its assumed closedness then gives compactness. \(\square\)
The constraint imposed by a fixed modulus is preserved by uniform limits: if \(f_n\to f\) uniformly and every \(f_n\) satisfies the estimate, then for each \(x,y\), \(\|f(x)-f(y)\|=\lim_n\|f_n(x)-f_n(y)\|\le\omega(d(x,y))\). The same pointwise limit argument preserves any fixed uniform bound. Consequently, the family of all functions satisfying a fixed bound and a fixed common-modulus estimate is closed, so it is compact.
Worked Example: A Square-Root Modulus
On \(K=[0,1]\), consider all continuous functions \(f:[0,1]\to\mathbb{R}\) satisfying \(|f(x)|\le1\) and \(|f(x)-f(y)|\le\sqrt{|x-y|}\). This family is compact by the common-modulus theorem, with \(\omega(r)=\sqrt r\).
The modulus condition is equicontinuity in explicit form: given \(\varepsilon>0\), choose \(\delta=\varepsilon^2\). If \(|x-y|<\delta\), then \(|f(x)-f(y)|\le\sqrt{|x-y|}<\sqrt{\delta}=\varepsilon\), for every function in the family. The bound \(1\) gives pointwise relative compactness in \(\mathbb{R}\), and the constraints are closed under uniform limits by the preceding argument.
Check that the family is nonempty: take \(f(x)=\sqrt{x}\). For \(x\ge y\ge0\), both sides below are nonnegative and \((\sqrt{x}-\sqrt{y})^2\le(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y\). Taking square roots gives \(|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}\). Also \(0\le\sqrt{x}\le1\) on \([0,1]\). At \(x=1/4,y=1\), the left side is \(|1/2-1|=1/2\), and the right side is \(\sqrt{3/4}=\sqrt3/2\); indeed \(1/2\le\sqrt3/2\), since squaring gives \(1/4\le3/4\).
Worked Example: A Vector-Valued Lipschitz Family
Let \(K=[0,1]\), and let \(G\) be the family of all continuous \(g:[0,1]\to\mathbb{R}^2\) with \(g(0)=(0,0)\) and \(\|g(x)-g(y)\|\le\sqrt5\,|x-y|\). The bound follows by setting \(y=0\): \(\|g(x)\|\le\sqrt5\,x\le\sqrt5\). The family is closed because uniform limits preserve the value at zero and the Lipschitz inequality. The theorem with \(M=L=\sqrt5\) gives compactness.
For a concrete member, set \(g(x)=(x,-x^2)\). Its value at zero is \((0,0)\), and \(\|g(x)\|^2=x^2+x^4\le2\le5\), so it obeys the uniform bound. For \(x,y\in[0,1]\), \(\|g(x)-g(y)\|^2=(x-y)^2+(x^2-y^2)^2=(x-y)^2(1+(x+y)^2)\le5(x-y)^2\), because \(x+y\le2\). Taking square roots verifies the required estimate.
Check: at \(x=3/4\) and \(y=1/4\), \(g(x)=(3/4,-9/16)\) and \(g(y)=(1/4,-1/16)\). Their difference is \((1/2,-1/2)\), whose norm is \(\sqrt{1/4+1/4}=\sqrt{1/2}\). The allowed bound is \(\sqrt5\,|3/4-1/4|=\sqrt5/2\); squaring both nonnegative quantities reduces the comparison to \(1/2\le5/4\), which holds.
A Common Estimate to Avoid
A frequent mistake is to treat a square-root modulus as a Lipschitz estimate with constant \(1\). At \(x=1/4\) and \(y=0\), the proposed inequality \(|\sqrt{x}-\sqrt{y}|\le|x-y|\) reads \(1/2\le1/4\), which is false. The valid estimate is \(|\sqrt{x}-\sqrt{y}|\le\sqrt{|x-y|}\), as verified above. This distinction matters: equicontinuity requires a common bound tending to zero with the distance, but it does not require that bound to be linear.
In the Lipschitz case, one can take \(\omega(r)=Lr\), so the common-modulus result includes the Lipschitz result. The key steps in either case are to establish a uniform bound, verify one shared estimate that gives equicontinuity, and check closedness when claiming compactness of the family itself. Without closedness, Arzela–Ascoli gives relative compactness, not necessarily compactness.
Check Your Understanding
For each question, identify the bound, common estimate, and closedness condition that are relevant.
- For functions \(f:[0,1]\to\mathbb{R}\) with \(f(0)=0\) and Lipschitz constant at most \(3\), find a uniform bound and state why the family is compact.
- Show that \(\omega(r)=r^{1/3}\) is a common modulus and specify a \(\delta\) that gives output differences less than \(\varepsilon\).
- Why does a uniform bound on \(f(x)\) imply pointwise relative compactness when the codomain is \(\mathbb{R}^m\)? Name the result used.
- Suppose \(f_n\) all satisfy \(\|f_n(x)-f_n(y)\|\le 4d(x,y)\) and \(f_n\to f\) uniformly. Write the limit argument that shows \(f\) satisfies the same estimate.
- For the square-root modulus, test the claimed Lipschitz-constant-\(1\) inequality at \(x=1/9\), \(y=0\), and give the correct modulus estimate.