McShane’s Extension Theorem Revisited

Abstract

A fundamental extension theorem of McShane states that a bounded real-valued uniformly continuous function defined on a nonempty subset A of a metric space 〈X, d〉 can be extended to a uniformly continuous function on the entire space. In the first half of this note, we obtain McShane’s Extension Theorem from the simpler fact that a real-valued Lipschitz function defined on a nonempty subset of the space has a Lipschitz constant preserving extension to the entire space. In the second half of the note, we use this theorem to give an elementary proof of the equivalence of the most important characterizations of metric spaces in which the real-valued uniformly continuous functions form a ring. These characterizations of such a basic property, due to Cabello-Sánchez and separately Bouziad and Sukhacheva, are remarkably recent.

1 Introduction

In the sequel, all metric spaces will be assumed to contain at least two points. If 〈X, d〉 is a metric space, we denote the continuous real-valued functions on X by $C(X,\mathbb {R})$ and the uniformly continuous real-valued functions on X by $UC(X,\mathbb {R})$. If A is a nonempty closed subset of 〈X, d〉 and $f: A \rightarrow \mathbb {R}$ is continuous, then the Tietze extension theorem [10, p. 149] guarantees that f has a continuous extension to X. However, if f is uniformly continuous, no uniformly continuous extension need exist: letting $\mathbb {N}$ be the positive integers equipped the usual metric of $\mathbb {R}$, clearly $f:\mathbb {N}\rightarrow \mathbb {R}$ defined by f(n) = n² has no uniformly continuous extension to the entire line. McShane’s Extension Theorem [17] says that if f is real-valued, uniformly continuous and bounded on an arbitrary nonempty subset A of the metric space, then f has a uniformly continuous (bounded) extension to X. This was later proved in the more general setting of uniform spaces by Katetov [15, Theorem 3] in the same article in which he proved his famous “sandwich theorem” in the context of normal spaces.

Actually, McShane showed that such an extension exists provided the uniformly continuous function f satisfies the following linear growth condition on A: there exist nonnegative constants m and b such that for each t ≥ 0

$$ \sup\{|f(a_{1}) - f(a_{2})|:~\{a_{1},a_{2}\} \subseteq A ~\text{ and }~ d(a_{1},a_{2}) \leq t\} \leq mt + b $$

and the extension will satisfy the same growth condition [17, Theorem 2].

Each real-valued Lipschitz function f on A satisfies such a growth condition where m is a Lipschitz constant for the function and b = 0 and thus it has a Lipschitz constant preserving extension. But there is a much easier well-known way to prove the existence of such a Lipschitz extension. In the first part of our note, we derive McShane’s result for bounded uniformly continuous functions as a consequence of the Lipschitz extension theorem. We do so by showing that a bounded uniformly continuous function on A is actually Lipschitz with respect to a second metric ρ on X that is uniformly equivalent to d. Because of the importance of McShane’s result and as our approach is so transparent, we include all details for pedagogical purposes. That said, a key part of our argument uses a construction of McShane.

In the second part of the note, we heavily use McShane’s result to derive some recent characterizations of metric spaces on which the real-valued uniformly continuous functions are stable under pointwise products, that is, for which $UC(X,\mathbb {R})$ forms a ring under the operations of pointwise addition and pointwise multiplication. The first internal characterization seems to be due to Cabello-Sánchez [8], who showed that $UC(X,\mathbb {R})$ is a ring if and only if each subset of X either is Bourbaki bounded or contains an infinite uniformly isolated subset. Shortly thereafter, Beer, Garrido and Meroño [4] showed that this condition amounts to the agreement of the bornology of Bourbaki bounded subsets with a larger bornology $\mathscr{B}$ that they called the bornology of infinitely nonuniformly isolated subsets. This bornology is somehow intrinsic to global uniform continuity: in the same way that a globally continuous function is one that is uniformly continuous restricted to each relatively compact set, and a function maps Cauchy sequences to Cauchy sequences if and only if it is uniformly continuous restricted to each totally bounded set [5, Proposition 4.11], a function is globally uniformly continuous if and only if it is uniformly continuous when restricted to each infinitely nonuniformly isolated subset [4, Proposition 3.7].

Within a year of the second article, Bouziad and Sukhacheva [7] came up with several additional technical internal characterizations plus two characterizations involving the structure of $C(X,\mathbb {R})$ and $UC(X,\mathbb {R})$ that surely deserve the attention of all analysts:

• $\forall f \in UC(X,\mathbb {R}),~\exists k \in \mathbb {N}$ such that {x ∈ X : |f(x)| > k} is uniformly isolated;

• $\forall f \in UC(X,\mathbb {R}),~\forall g \in C(\mathbb {R},\mathbb {R})$, we have $g \circ f \in UC(X,\mathbb {R})$.

We feel that analysts would be well-served by a self-contained proof of the equivalence of the above two conditions of Bouziad and Sukhacheva with the internal characterization of Cabello-Sánchez that does not reference their other characterizations, and which is provided totally within the metric framework, rather than in the setting of uniform spaces that are described in terms of a topological game in which Bouziad and Sukhacheva chose to work.

2 Preliminaries

We denote a set X with at least two points equipped with a metric d by 〈X, d〉. Two metrics d₁ and d₂ on a set X are declared equivalent if their induced topologies agree; this means that the identity function id_X between 〈X, d₁〉 and 〈X, d₂〉 is continuous in both directions. If the identity function is uniformly continuous in both directions, the metrics are declared uniformly equivalent.

A function $f: \langle X,d \rangle \rightarrow \mathbb {R}$ is called Lipschitz if for some λ > 0 and all x₁, x₂ ∈ X, we have |f(x₁) − f(x₂)|≤ λd(x₁, x₂). In this case we say that λ is a Lipschitz constant for f or simply that f is λ-Lipschitz. A function that is 1-Lipschitz is called nonexpansive. Each Lipschitz function is uniformly continuous.

That $f: X \rightarrow \mathbb {R}$ is λ-Lipschitz on X can obviously be expressed in this alternative manner: whenever x₁, x₂ ∈ X,

$$ f(x_{1}) \leq f(x_{2}) + \lambda d(x_{1},x_{2}). $$

From this formulation, one easily gets this well-known fact (see, e.g., [12, p. 43]).

Lemma 1

Let {f_j : j ∈ J} be a family ofλ-Lipschitzreal-valued functions on a metric space 〈X, d〉 such that for somex₀ ∈ X, infj∈Jf_j(x₀) > −∞. Thenx↦infj∈Jf_j(x) is a real-valuedλ-Lipschitzfunction on X.

The Lipschitz extension theorem is an immediate consequence of our lemma.

Theorem 1

LetA be a nonempty subset of a metric space〈X, d〉. If$f: \langle A,d \rangle \rightarrow \mathbb {R}$isλ-Lipschitz,then f has aλ-Lipschitzextension to X.

Proof

For each a ∈ A, define $f_{a}:X \rightarrow \mathbb {R}$ by f_a(x) = f(a) + λd(x, a). For each p ∈ A, we have f(p) = inf_a∈Af_a(p). By Lemma 1, inf_a∈Af_a is the desired extension. □

We denote the open ball of radius ε about a point p of a metric space 〈X, d〉 by S_d(p, ε). A subset A of 〈X, d〉 is called bounded if it is contained in a ball, while A is called totally bounded if ∀ε > 0 there exists a finite subset F of X such that A ⊆∪_x∈FS_d(x, ε). We denote the set of limit points of a subset A of the metric space by A^′. A subset A of 〈X, τ〉 is called relatively compact if it has compact closure.

If A ≠ ∅ is a subset of the metric space 〈X, d〉 and p ∈ X, the distance ofpfromA is defined by d(p, A) := inf{d(p, a) : a ∈ A}. The isolation functionalI_d : X → [0, ∞) for the metric space (see, e.g., [4, 5, 8]) is defined by

$$ I_{d}(x) := d(x, X\backslash \{x\}). $$

Note that I_d(x) = 0 if and only if x ∈ X^′. If A ⊆ X and there exists λ > 0 such that ∀a ∈ A, I_d(a) ≥ λ, we call A a uniformly isolated subset of X. It is a standard exercise to show that distance functionals are nonexpansive. Isolation functionals are also nonexpansive. As the proof is surprisingly subtle, we include it as a courtesy to the reader.

Proposition 1

Let〈X, d〉 be a metric space. Then the isolation functionalI_d(⋅) for the space is nonexpansive.

Proof

We need to show that when x ≠ w, we have |I_d(x) − I_d(w)|≤ d(x, w). For the proof, we consider two separate cases for x ≠ w:

• (a) either d(x, w) = I_d(x) or d(x, w) = I_d(w);

• (b) max{I_d(x), I_d(w)} < d(x, w).

In case (a) we can assume without loss of generality that I_d(x) = d(x, w). Since I_d(w) ≤ d(x, w) we have |I_d(x) − I_d(w)| = I_d(x) − I_d(w) ≤ I_d(x) = d(x, w). In case (b), choose ε > 0 with ε < d(x, w) − max{I_d(x), I_d(w)} and then p ∈ X with d(x, p) < I_d(x) + ε. By the choice of ε, p ≠ w and we get

$$ I_{d}(w) \leq d(w,p) \leq d(w,x) + d(x,p) < d(w,x) + I_{d}(x) + \varepsilon, $$

and letting ε go to zero, this yields I_d(w) − I_d(x) ≤ d(w, x). In the same way, I_d(x) − I_d(w) ≤ d(w, x) so that that the isolation functional is nonexpansive. □

A family of subsets $\mathscr{B}$ of 〈X, d〉 that contains the singletons, that is stable under taking subsets and that is stable under finite unions is called a bornology (see, e.g, [4, 5, 14, 20]). Obviously, the smallest bornology is the family of finite subsets of X. The bornology of bounded subsets contains the bornology of totally bounded subsets which in turn contains the bornology of relatively compact subsets. The bornology of Bourbaki bounded subsets [3, 4, 6, 11, 20], which we next describe, lies between the bounded subsets and the totally bounded subsets.

Let ε > 0; we call a finite list of (not necessarily distinct) points x₀, x₁, x₂,…, x_n of 〈X, d〉 an ε-chain of lengthn from a to b if x₀ = a, x_n = b and whenever j ∈{1,2,…, n}, we have d(x_j− 1, x_j) < ε.

Definition 1

A subset A of a metric space 〈X, d〉 is called Bourbaki bounded if for each ε > 0 there exists a finite subset F of X and $n \in \mathbb {N}$ such that each point of A can be connected to some point of F by a ε-chain of length n.

We now state the key fact linking Bourbaki boundedness to uniform continuity [3, 13].

Proposition 2

A nonempty subsetA of a metric space〈X, d〉 is Bourbaki bounded if and only if the restriction of each memberof$UC(X,\mathbb {R})$toA is a bounded function.

Finally, we make precise the notion of infinitely nonuniformly isolated subset mentioned at the end of the Introduction and was first considered in [4].

Definition 2

A subset A of a metric space 〈X, d〉 is called infinitely nonuniformly isolated if whenever E is an infinite subset of A, we have infe∈EI_d(e) = 0.

It is easy to check that the family of such sets forms a bornology that contains each Bourbaki bounded set [4, Proposition 3.5]. Further the Cabello-Sánchez criterion can be understood in the following way: if a subset A of the metric space fails to be Bourbaki bounded, then it fails to be infinitely nonuniformly isolated. Taken together, these two statements say that the two bornologies coincide.

3 Uniformly Continuous Extensions from Lipschitz Extensions

Let $C \subseteq \mathbb {R}$ be a nonempty convex subset. We say that $g:C \rightarrow \mathbb {R}$ is concave if whenever t₁ and t₂ lie in C and α ∈ [0,1], we have

$$ \alpha g(t_{1}) + (1 - \alpha)g(t_{2}) \leq g(\alpha t_{1} + (1 - \alpha)t_{2}). $$

By a convex function$h:C \rightarrow \mathbb {R}$, we mean a function such that − h is concave. There is a much broader literature on convex functions. Facts about convex functions can be dualized to obtain corresponding facts about concave functions. For example, $h:C \rightarrow \mathbb {R}$ is convex if and only if its epigraph$\{(t,\alpha ) \in C \times \mathbb {R} : \alpha \geq h(t)\}$ is a convex set [18, p. 80], while $g:C \rightarrow \mathbb {R}$ is concave if and only if its hypograph$\{(t,\alpha ) \in C \times \mathbb {R} : \alpha \leq g(t)\}$ is a convex set.

Actually, we will only be looking at nonnegative nonconstant concave functions defined on [0, ∞). We collect some facts about these (all known) in the following result. We call g : [0, ∞) → [0, ∞) nondecreasing if whenever t₂ > t₁ ≥ 0, we have g(t₂) ≥ g(t₁).

Proposition 3

Letg : [0, ∞) → [0, ∞) be concave and nonconstant. Then the following statements are valid:

• 1. g is a subadditive function;

• 2. g is a nondecreasing function;

• 3. whenevert > 0, we haveg(t) > 0.

Proof

For the first statement, we are asked to show that whenever t₁ ≥ 0 and t₂ ≥ 0 then g(t₁ + t₂) ≤ g(t₁) + g(t₂). By nonnegativity of values, this is clearly true if t₁ = 0 = t₂. Otherwise, t₁ + t₂ > 0 and we compute

$$ \begin{array}{@{}rcl@{}} g(t_{1}) = g\left( \frac{t_{1}}{t_{1} + t_{2}}(t_{1} + t_{2}) + \frac{t_{2}}{t_{1} + t_{2}}(0)\right) &\geq& \frac{t_{1}}{t_{1} + t_{2}}g(t_{1} + t_{2}) + \frac{t_{2}}{t_{1} + t_{2}}g(0)\\ &\geq& \frac{t_{1}}{t_{1} + t_{2}}g(t_{1} + t_{2}). \end{array} $$

In the same way, $g(t_{2}) \geq \frac {t_{2}}{t_{1} + t_{2}}g(t_{1} + t_{2})$. Adding these inequalities yields our subadditivity.

For the second statement, suppose g(t₂) < g(t₁) where t₂ > t₁. Put γ = g(t₁) − g(t₂). By concavity it is routine to show using mathematical induction that for each $n \in \mathbb {N}$,

$$ g(t_{1} + n(t_{2} - t_{1})) \leq g(t_{1}) - n\gamma. $$

As a result, g(t₁ + n(t₂ − t₁)) will be negative for large n which is impossible.

For the third statement, suppose t₁ > 0 and g(t₁) = 0. Since g is nondecreasing and nonnegative, g is zero throughout [0, t₁]. If g(t₂) > 0 for some t₂ > t₁, then concavity fails, as

$$ g(t_{1}) = g\left( \frac{t_{2} - t_{1}}{t_{2}}(0) + \frac{t_{1}}{t_{2}}(t_{2})\right) < \frac{t_{2} - t_{1}}{t_{2}}g(0) + \frac{t_{1}}{t_{2}}g(t_{2}) = \frac{t_{1}}{t_{2}}g(t_{2}). $$

We conclude that g is the zero function on [0, ∞), a contradiction. □

We now let Δ denote the family of nonnegative concave functions g on [0, ∞) that are globally continuous and for which g^− 1({0}) = {0}. Of course, g(t) = arctan (t) belongs to Δ as does g(t) = ln(t + 1). As an example of a member of Δ that is eventually constant, consider g defined by $g(t) = \sqrt {t}$ for t ∈ [0,1] and g(t) = 1 otherwise.

Our next result is a stripped-down version of the lemma preceding Theorem 2 of [17] that is adequate for our purposes.

Proposition 4

LetA be a nonempty subset of 〈X, d〉 and let$f: A \rightarrow \mathbb {R}$bebounded, nonconstant and uniformly continuous. Then there existsg_f ∈Δ that majorizesh_f : [0, ∞) → [0, ∞) definedby

$$ h_{f}(t) = \sup\{|f(a_{1}) - f(a_{2})| : \{a_{1},a_{2}\} \subseteq A ~\text{ and }~ d(a_{1},a_{2}) \leq t\}. $$

Proof

Clearly h_f(0) = 0 and h_f is nondecreasing. Let g_f : [0, ∞) → [0, ∞) be the infimum of the family of all nondecreasing affine functions that majorize h_f on [0, ∞). This infimum is a well-defined nonnegative real-valued function because 0 ≤ h_f ≤ 2M where M = sup{|f(a)| : a ∈ A}. Evidently g_f is concave [18, p. 80] and g_f majorizes h_f. Since f is nonconstant, there exists t₁ > 0 such that h_f(t₁) > 0 so that g_f(t₁) > 0 as well. To show that $g_{f} \in C([0,\infty ),\mathbb {R})$, it suffices to show that g_f is continuous at t = 0 as a concave function is automatically continuous at each interior point of its domain [18, p. 4]. We will show that g_f(0) = 0 at the same time.

Let ε ∈ (0,2M) be otherwise arbitrary. By uniform continuity of f on A, the function h_f is continuous at the origin. This means that for some t₂ > 0 and all t ∈ [0, t₂] we have h_f(t) < ε. Let v be the affine function whose graph contains (0, ε) and (t₂,2M). Clearly, v is nondecreasing and majorizes h_f and thus majorizes g_f. From these two facts, we have both g_f(0) ≤ ε and $\limsup _{t \rightarrow 0} g_{f}(t) \in [0,\varepsilon ]$. Letting ε tend to zero gives $g_{f}(0) = 0 = \lim _{t \rightarrow 0} g_{f}(t)$. Finally, as g_f(t₁) ≥ h_f(t₁) > 0 we see that g_f is nonconstant and so by Proposition 3, 0 is the only t-value for which g_f(t) = 0. We have now shown that g_f belongs to Δ. □

Theorem 2

LetA be a nonempty subset of 〈X, d〉 and let$f: A \rightarrow \mathbb {R}$bebounded, nonconstant and uniformly continuous. Letg_fbe the function of the last proposition. Theng_f ∘ d : X × X → [0, ∞) is a metric uniformly equivalent to d with respect to which f is a nonexpansivefunction on A.

Proof

Put ρ = g_f ∘ d. Since $g_{f}^{-1}(\{0\}) = \{0\}$ and d is a metric, ρ(x₁, x₂) = 0 if and only if x₁ = x₂. Clearly, ρ(x₁, x₂) = ρ(x₂, x₁) for all x₁, x₂ in X. The triangle inequality follows from the subadditivity of g_f and its nondecreasing nature as established in Proposition 3: given x₁, x₂ and x₃ in X

$$ \rho(x_{1},x_{3}) = g_{f}(d(x_{1},x_{3})) \leq g_{f} (d(x_{1},x_{2}) + d(x_{2},x_{3})) \leq \rho(x_{1},x_{2}) + \rho (x_{2},x_{3}). $$

To show uniform equivalence of d and ρ, we must show that the identity function on X is uniformly continuous in both directions. Let ε > 0; by the continuity of g_f at t = 0 there exists δ > 0 such that if t = |t − 0| < δ then g_f(t) < ε. As a result, if d(x₁, x₂) < δ, then ρ(x₁, x₂) < ε. For uniform continuity in the other direction, again let ε > 0, and put δ = g_f(ε) > 0. If ρ(x₁, x₂) < δ then since g_f is nondecreasing, we have d(x₁, x₂) < ε.

It remains to show that $f: A \rightarrow \mathbb {R}$ is 1-Lipschitz with respect to ρ. Suppose a₁ and a₂ are arbitrary points of A. We have

$$ |f(a_{1}) - f(a_{2})| \leq h_{f}(d(a_{1},a_{2})) \leq g_{f}(d(a_{1},a_{2})) = \rho(a_{1},a_{2}) $$

as required. □

Examining our argument that g_f ∘ d is a metric uniformly equivalent to d, we only used these properties of g_f: (i) g_f is nondecreasing; (ii) g_f is subadditive; (iii) $g_{f}^{-1}(\{0\}) = \{0\}$; (iv) g_f is continuous at t = 0. In the end, concavity was not called on at all nor was continuity of g_f at any positive number. We do not regard what we have done here as really new because a problem in the classical monograph on general topology by Kelley [16, p. 131] asks the reader to show that if g : [0, ∞) → [0, ∞) is continuous, nondecreasing, subadditive, and g(t) = 0 if and only if t = 0, then if d is any metric, g ∘ d will be a metric equivalent to d. We refer the interested reader to the expository article of Corazza [9] that considers general functions g : [0, ∞) → [0, ∞) such that for any metric d, g ∘ d is a metric or better yet, g ∘ d is a metric equivalent to the initial metric.

We can now easily derive McShane’s Extension Theorem from the Lipschitz extension theorem.

Theorem 3

LetA be a nonempty subset of a metric space〈X, d〉 and suppose$f: A \rightarrow \mathbb {R}$isa bounded uniformly continuous function on A. Then f can be extended to a uniformly continuous function on〈X, d〉.

Proof

This is trivial if f is a constant function. Otherwise, by Theorem 2, there is a metric ρ on X that is uniformly equivalent to d such that with respect to ρ, $f:A \rightarrow \mathbb {R}$ is 1-Lipschitz. By Theorem 1, f has a 1-Lipschitz extension $\widehat {f}$ to the entire space. Since $\widehat {f}$ is uniformly continuous with respect to ρ and id_X : 〈X, d〉→〈X, ρ〉 is uniformly continuous, the extension $\widehat {f}$ is uniformly continuous on X with respect to d. □

Note that in our proof, we did not use the full uniform equivalence of d and ρ, only the fact that id_X is uniformly continuous in one direction.

4 When Do the Uniformly Continuous Functions Form a Ring?

Of course, $UC(X,\mathbb {R})$ is stable under pointwise product if either (i) each member of $UC(X,\mathbb {R})$ is bounded, that is, X itself is Bourbaki bounded, or (ii) $UC(X,\mathbb {R}) = C(X,\mathbb {R})$. Members of the second class of spaces are called UC-spaces or Atsuji spaces in the literature. Atsuji in his seminal article [1] first showed that X is Bourbaki bounded iff each member of $UC(X,\mathbb {R})$ is a bounded function and separately gave a number of characterizations of UC-spaces which have been added to over the years (see, e.g., [4, Theorem 2.2 and Theorem 3.13]). But there are simple metric spaces that are neither Bourbaki bounded nor are UC-spaces for which $UC(X,\mathbb {R})$ is a ring, e.g., $(-1,0) \cup \mathbb {N}$ equipped with the usual metric of the line. We invite the reader to check that $UC((-1,0) \cup \mathbb {N},\mathbb {R})$ is a ring using the Cabello-Sánchez internal criterion, just to see how easily applicable it is.

As promised in the Introduction, we intend to give an elementary metric space proof of the equivalence of the Cabello-Sánchez criterion with the function space conditions of Bouziad and Sukhacheva. Although coincidence of the bornology of infinitely nonuniformly bounded subsets with the Bourbaki bounded subsets can be used to prove that $UC(X,\mathbb {R})$ is a ring in a very attractive way (see the proof of [4, Theorem 3.9]), we do not include it as an additional criterion. All the implications of substance in our proof rely on McShane’s Extension Theorem for bounded uniformly continuous real-valued functions.

Theorem 4

Let 〈X, d〉 be a metric space. The following conditions are equivalent:

• 1. $UC(X,\mathbb {R})$isstable under pointwise products and so forms a ring;

• 2. whenever$f \in UC(X,\mathbb {R})$, thenf²is uniformly continuous as well;

• 3. $\forall f \in UC(X,\mathbb {R}),~\exists k \in \mathbb {N}$suchthat {x ∈ X : |f(x)| > k} is uniformly isolated;

• 4. $\forall f \in UC(X,\mathbb {R}),~\forall g \in C(\mathbb {R},\mathbb {R})$, we have$g \circ f \in UC(X,\mathbb {R})$;

• 5. each subset of X is either Bourbaki bounded or contains an infiniteuniformly isolated subset.

Proof

The equivalence of conditions (1) and (2) follows from the identity

$$ fg = \frac{1}{2}\left( (f + g)^{2} - f^{2} - g^{2}\right) $$

and the fact that $UC(X,\mathbb {R})$ is always a vector space. We intend to establish this implication string: (2) ⇒ (3) ⇒ (4) ⇒ (5) ⇒ (2).

(2) ⇒ (3). We prove that the contrapositive holds. Suppose condition (3) fails for some uniformly continuous function f. Without loss of generality, by replacing f by |f|, we may assume that the values of f are nonnegative. By uniform continuity, there exists a decreasing positive sequence 〈δ_k〉 with $\lim _{k\rightarrow \infty } \delta _{k} = 0$ such that $\forall k \in \mathbb {N}$,

$$ d(x,w) < \delta_{k} \Rightarrow |f(x) - f(w)| < \frac{1}{3k} \quad (x,w \in X). $$

Since (3) fails, {x : f(x) > 1} is not uniformly isolated, and we can choose x₁ and w₁ such that 0 < d(x₁, w₁) < δ₁ and f(x₁) > 1. Again by the failure of (3), {x : f(x) > f(x₁) + 1} is not uniformly isolated, and we can choose x₂ and w₂ such that 0 < d(x₂, w₂) < δ₂ and f(x₂) > f(x₁) + 1. Continuing, choose for each positive integer k, x_k and w_k such that 0 < d(x_k, w_k) < δ_k and f(x_k+ 1) > f(x_k) + 1. By construction, it is easy to verify that the following estimates all hold:

• (i) $\forall k \in \mathbb {N}, f(x_{k}) > k$;

• (ii) $\forall k \in \mathbb {N},~ f(x_{k}) + \frac {1}{3k} > f(w_{k}) > f(x_{k}) - \frac {1}{3k} > 0$;

• (iii) whenever j ≠ k, d(x_j, x_k) ≥ δ₁, d(w_j, w_k) ≥ δ₁, and d(x_j, w_k) ≥ δ₁.

We only verify in (iii) that d(w_j, w_k) ≥ δ₁ for j ≠ k. Without loss of generality, we may assume that j > k. It suffices to show that f(w_j) − f(w_k) ≥ 1/3. This follows immediately from the following three facts:

• (a) f(x_j) − f(x_k) > j − k ≥ 1;

• (b) $f(w_{j}) > f(x_{j}) - \frac {1}{3j} \geq f(x_{j}) - \frac {1}{3}$;

• (c) $f(w_{k}) < f(x_{k}) + \frac {1}{3k} \leq f(x_{k}) + \frac {1}{3}$.

Define $h : \{x_{k}: k \in \mathbb {N}\} \cup \{w_{k}: k \in \mathbb {N}\} \rightarrow \mathbb {R}$ by h(x_k) = 0 and $h(w_{k}) = \frac {1}{k}$ for all k. By estimate (iii) above, h so defined is a bounded uniformly continuous function, so that by McShane’s Extension Theorem this function has a uniformly continuous extension to X which, for convenience, we shall denote by h as well. Of course, $f + h \in UC(X,\mathbb {\mathbb {R}})$. We will show that (f + h)² fails to be uniformly continuous by proving that for each $k, (f + h)^{2}(w_{k}) > (f + h)^{2}(x_{k}) + \frac {4}{3}$. Fix $k \in \mathbb {N}$; we compute using (i) and (ii) and the fact that h(x_k) = 0 that

$$ \begin{array}{@{}rcl@{}} (f + h)^{2}(w_{k}) &=& f^{2}(w_{k}) + 2f(w_{k})\frac{1}{k} + \frac{1}{k^{2}}\\ &>& \left( f(x_{k}) - \frac{1}{3k}\right)^{2} + \frac{2}{k}\left( f(x_{k}) - \frac{1}{3k}\right)+ \frac{1}{k^{2}}\\ &>& f^{2}(x_{k}) - \frac{2}{3k}f(x_{k}) + \frac{2}{k}f(x_{k})\\ &=& f^{2}(x_{k}) + \frac{4}{3k}f(x_{k}) > (f + h)^{2}(x_{k}) + \frac{4}{3}. \end{array} $$

(3) ⇒ (4). Suppose $g \in C(\mathbb {R},\mathbb {R})$ and by condition (3) choose $k \in \mathbb {N}$ such that for some λ > 0, |f(x)| > k ⇒ I_d(x) > λ. Since g restricted to [−k, k] is uniformly continuous, there exists δ₁ > 0 such that whenever {α, β}⊆ [−k, k] with |α − β| < δ₁, we have |g(α) − g(β)| < ε. By uniform continuity of f, pick δ < λ such that d(x, w) < δ ⇒|f(x) − f(w)| < δ₁. It is easy to check that

$$ d(x,w) < \delta \quad\Longrightarrow\quad |(g\circ f)(x) - (g\circ f)(w)| < \varepsilon \quad (x,w \in X). $$

(4) ⇒ (5). Suppose the Cabello-Sánchez criterion (5) fails. Then we can find a subset A of X which fails to be Bourbaki bounded and is simultaneously infinitely nonuniformly isolated; note that A must be infinite as each finite set is Bourbaki bounded. Let $f \in UC(X,\mathbb {R})$ fail to be bounded on A. Without loss of generality we can assume that f has nonnegative values. As a result, for each positive integer k, {a ∈ A : f(a) > k} is infinite and so inf{I_d(a) : a ∈ A and f(a) > k} = 0. Proceeding exactly as in the proof of (2) ⇒ (3) we can produce sequences 〈x_n〉 and 〈w_n〉 with exactly the same properties (and where each x_n lies in A—a fact that we will not use).

By McShane’s Extension Theorem, we can find $g \in UC(X,\mathbb {R})$ such that for all $n \in \mathbb {N}$, (a) g(x_n) = 0, (b) g(w_n) = 0 if f(x_n)≠f(w_n), and (c) $g(w_{n}) = \frac {1}{3n}$ if f(x_n) = f(w_n). By construction, the sequence of reals

$$ (f + g)(x_{1}), (f + g)(w_{1}), (f + g)(x_{2}), (f + g)(w_{2}), (f + g)(x_{3}), \ldots $$

has distinct terms, and since

$$ \lim\limits_{n\rightarrow\infty} (f + g)(x_{n}) = \infty = \lim\limits_{n\rightarrow\infty} (f + g)(w_{n}), $$

the set of terms of this sequence is a closed discrete set. In particular, any real function defined on the set of terms equipped with the relative topology of the line is continuous, so by the Tietze extension theorem, there exists $h \in C(\mathbb {R},[0,1])$ such that for each $n \in \mathbb {N}$, h((f + g)(x_n)) = 0 and h((f + g)(w_n)) = 1. This shows that h ∘ (f + g) is not uniformly continuous so that condition (4) fails.

(5) ⇒ (2). We again prove the contrapositive. Suppose $f \in UC(X,\mathbb {R})$ while f² fails to be uniformly continuous. Then for some ε > 0 we can find for each $n \in \mathbb {N}$ points x_n and w_n in X such that 0 < d(x_n, w_n) < 1/n but |f²(x_n) − f²(w_n)|≥ ε. Since f² is globally continuous, by passing to a subsequence we can assume that the terms of 〈x_n〉 are distinct. Now the restriction of f to $\{x_{n} : n \in \mathbb {N}\}$ cannot be bounded, else by uniform continuity its restriction to $\{w_{n} : n \in \mathbb {N}\}$ would be bounded. Thus f² would be uniformly continuous on $\{x_{n} : n \in \mathbb {N}\} \cup \{w_{n} : n \in \mathbb {N}\}$ which is impossible. By Proposition 2, $\{x_{n} : n \in \mathbb {N}\}$ is not Bourbaki bounded. By construction $\{x_{n}: n \in \mathbb {N}\}$ has no infinite uniformly isolated subset. Thus, the internal criterion of Cabello-Sánchez fails. □

5 Concluding Remarks

The results of Section 3 suggest that Lipschitz analysis might inform our understanding of certain results in general mathematical analysis that seem to have little to do with Lipschitz functions. McShane’s Extension Theorem is not the only such result; the weak Ekeland Variational Principle for lower bounded lower semicontinuous functions is better understood through Lipschitz analysis as well. Beer and Ceniceros [2] showed that the principle is valid precisely in those spaces 〈X, d〉 in which whenever $f: X \rightarrow \mathbb {R}$ is λ-Lipschitz for some positive λ and lower bounded, there exists ε ∈ (0, λ) and x₀ ∈ X such that for all x ∈ X, we have f(x₀) ≤ f(x) + εd(x, x₀). Intuitively, this means at some x₀ ∈ X, we can improve on the Lipschitz constant λ. Of course, this property of lower-bounded Lipschitz constants must be characteristic of completeness of the metric d (see [19]).

It is hoped that some readers will be motivated to look for Lipschitzian behavior lying unexpectedly behind other important results in general mathematical analysis.