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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.

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2 Answers 2

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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.

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  • $\begingroup$ This is the best answer so far, but if new answers come in I will consider accepting them as the best. $\endgroup$
    – ahorn
    May 24, 2016 at 11:57
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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.

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  • $\begingroup$ If you are using the contrapositive, shouldn't you say that a continuous but non-uniformly continuous function always has a domain that is not compact? Isn't this a necessary but not sufficient condition? $\endgroup$
    – ahorn
    Apr 30, 2016 at 16:24
  • $\begingroup$ @ahorn: Yes that is what I meant by "non-uniformly continuous", and yes it is necessary but not sufficient as my final sentence made clear. $\endgroup$
    – user21820
    May 1, 2016 at 2:49
  • $\begingroup$ I have an issue with the wording "no extension to a non-compact domain." That is why I said "always has a domain that is not compact." If you use your wording, shouldn't you say, "have no extension to a compact domain?" $\endgroup$
    – ahorn
    May 1, 2016 at 7:06
  • $\begingroup$ @ahorn: Oops!!! Sorry I didn't realize I had the extra "not".. Editing now. $\endgroup$
    – user21820
    May 1, 2016 at 7:08
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    $\begingroup$ @ahorn: That's great! Really sorry about that careless mistake. My point is that I don't want to just say "has a non-compact domain" but also that it's not possible to extend it continuously to any compact domain, in other words the domain is really forced to be non-compact by the nature of the function. $\endgroup$
    – user21820
    May 1, 2016 at 7:11

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