What do uniformly continuous functions look like?

Question

When I see a function, I want to be able to quickly determine whether it is uniformly continuous or not. Usually, this kind of skill comes after being exposed to many different examples that either do or do not have the desired feature, but the questions on this site about examples of functions that are continuous but not uniformly continuous all point to just one example ($f(x)=\frac 1 x$). What properties distinguish continuous functions from uniformly continuous functions?

An immediate answer to my question is that Lipschitz-continuous functions are uniformly continuous. Visually, the gradient of a Lipschitz-continuous function will always be bounded. Using this property, $f(x)=\frac 1 x$ has an unbounded gradient function, which implies that $f$ is not a Lipschitz-continuous function. But there are some functions that are uniformly continuous but not Lipschitz-continuous, so this is does not test whether a function is not uniformly continuous, only if a function is uniformly continuous.

Answer 1

My intuitive understanding of a uniformly continuous function is that it doesn't accelerate without bound and then stay there as the function itself increases (or decreases) without bound. So $x^2$ isn't uniformly continuous, because it gets faster and faster and faster and never slows down, but $\sqrt[3]{x}$ is, because despite being vertical $x=0$ it then slows down again.

Answer 2

Given any $\def\nn{\mathbb{N}}$$\def\rr{\mathbb{R}}$$n \in \nn$, any closed bounded subset of $\rr^n$ is compact.

Given any continuous function $f$ on a compact subset $S$ of $\rr^n$, we can prove that $f$ is uniformly continuous on $S$.

Thus any non-uniformly continuous function must have no (continuous) extension to a compact domain.

Two examples:

1. Let $f : \rr_{\ne 0} \to [0,1]$ such that $f(x) = \sin(\frac{1}{x})$ for any $x \in \rr_{\ne 0}$. Then $f$ is continuous on $\rr_{\ne 0}$ but not uniformly continuous. (The hole at $0$ cannot be patched.)

2. Let $f : \rr \to \rr$ such that $f(x) = x \sin(x)$ for any $x \in \rr$. Then $f$ is continuous on $\rr$ but not uniformly continuous. (An unbounded domain is never compact.)

Of course, some uniformly continuous functions do not have a domain that can be extended to a compact one, such as a step function.