The definition of uniform continuity is wrong

Consider a map f\colon X\to Y where X and Y are metric spaces.

The standard (wrong) definition says: f is uniformly continuous if for every \epsilon>0 there exists \delta>0 such that d_X(x_1,x_2)<\delta \implies d_Y(f(x_1),f(x_2))<\epsilon.

According to this definition, every function f\colon \mathbb Z\to \mathbb R is uniformly continuous. What exactly is uniform about the function f(n)=2^{2^n}\sin n? I don't see it.

The right definition is: f is uniformly continuous if there exists a function \omega \colon [0,\infty)\to [0,\infty) such that \omega(0+)=0 and d_Y(f(x_1),f(x_2))\le \omega(d_X(x_1,x_2)) for all x_1,x_2\in X.

What is the difference, you ask? For nice spaces, such as intervals, there is none. But for uglier spaces: highly nonconvex, disconnected, etc the second definition is more restrictive, or, the way I put it, the first definition is overly permissive (e.g., toward the function f(n)=2^{2^n}).

There is a parallel with the definition of a quasisymmetric map, which is the version of “uniformly continuous” for relative distances. This definition was worked out after the concept of a general metric space was well understood, which is why mathematicians were able to get it right. Uniform continuity predates metric spaces: it was defined for functions on an interval, and then the same definition was applied to more general spaces whether it fit or not.

Here is a statement which is true with the right definition of uniform continuity and false with the wrong one.

Theorem. For a metric space X, the following are equivalent:

  1. X is compact
  2. every continuous function f\colon X\to\mathbb R is uniformly continuous
  3. every map f\colon X\to Y, for every metric space Y, is uniformly continuous

Proof. 1\implies 3. Given t\ge 0, consider the set E_t=\{(x_1,x_2)\in X\times X\colon d_X(x_1,x_2)\le t\}. This is a closed subset of the compact set X\times X, hence compact. Since the map G(x_1,x_2):=d(f(x_1),f(x_2)) is continuous, it follows that G(E_t) is a bounded subset of [0,\infty). Define \omega(t)=\sup G(E_t). One can show that \omega(0+)=0, for example, by covering the diagonal of X\times X with appropriate open neighborhoods.

3\implies 2 is trivial.

2\implies 1. We must show that if X is either not complete or not totally bounded, then there exists a continuous function f\colon X\to \mathbb R which is not uniformly continuous. First, suppose X is not complete. Pick p\in \overline X\setminus X where \overline X is the completion of X. The function f(x)=1/d_{\overline X}(x,p) is continuous but not bounded on \{x\in X\colon d_{\overline X}(x,p)\le 1\}. Since the right definition of uniform continuity requires f to be bounded on bounded sets, f is not uniformly continuous.

Next, suppose that X is not totally bounded. Then there exists \epsilon>0 and an infinite sequence x_n such that d_X(x_n,x_m)\ge \epsilon whenever n\ne m. Let f(x)=\sum_n n\max(\epsilon/2 - d(x,x_n), 0). This function is continuous on X but is not uniformly continuous.

At a first glance one might think than the difference between the two notions boils down to preserving bounded sets. This is not so, since any function f\colon \mathbb Z\to \mathbb R is bounded on bounded subsets of \mathbb Z, yet not every such function is uniformly continuous according to the right definition.

When \omega needs to be emphasized, one can say that f is \omega-continuous. In particular, f is Lipschitz continuous if it is \omega-continuous with \omega(t)=Ct for some C. Also, f is \alpha-Hölder continuous if it is \omega-continuous with \omega(t)=O(t^{\alpha}) as t\to 0. (I don’t think many people really want to impose the Hölder condition at large distances.) Finally, f is Dini-continuous if it is \omega-continuous with \int_0 \frac{\omega(t)}{t}\,dt<\infty.

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