# 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.
QED

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 $\int_0 \frac{\omega(t)}{t}\,dt<\infty$.