TITLE: What is the intuition behind uniform continuity?
QUESTION [79 upvotes]: There’s another post asking for the motivation behind uniform continuity. I’m not a huge fan of it since the top-rated comment spoke about local and global interactions of information, and frankly I just did not get it. 
Playing with the definition, I want to say uniform continuity implies there’s a maximum “average rate of change”. Not literally a derivative, but the rate of change between two points is bounded in the domain. I’m aware that this is essentially Lipschitz continuity, and that Lipschitz implies uniform. This implies there’s more to uniform continuity than just having a bounded average rate of change. 
And also, how is it that $ f(x)=x$ is uniform yet $f(x)f(x)=g(x)=x^2$  is not? I understand why it isn’t, I can prove it. But I just don’t understand the motivation and importance of uniform continuity.

REPLY [5 votes]: One interpretation that I love is the one using non-standard analysis:

Let $f : E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Then the followings are equivalent:

$f$ is uniformly continuous.
Its hyperreal version ${{}^*\!f} : {{}^*\!E} \subseteq {}^*\mathbb{R} \to {}^*\mathbb{R}$ is continuous, i.e., whenever $x, y \in {{}^*\!E}$ are infinitely close, ${{}^*\!f}(x)$ and ${{}^*\!f}(y)$ are infinitely close as well.


Here, the 'hyperreal version' refers to the $*$-transform of $f$. The catch is that the domain ${{}^*\!E}$ of ${{}^*\!f}$ contains hyperreal numbers which are either infinitely close to $E$ or infinitely large. So the continuity needs to be tested around those numbers as well to establish uniform continuity.
This statement can be recast in terms of real numbers only:

Let $f : E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Then the followings are equivalent:

$f$ is uniformly continuous.

For any sequences $(a_n)$ and $(b_n)$ in $E$ such that $|a_n - b_n| \to 0$, we have $|f(a_n) - f(b_n)| \to 0$.



In this version, the role of infinitely close hyperreal numbers is replaced by a pair of sequences which become arbitrarily close. Again, we see that uniformly continuity is really about enforcing continuity infinitesimally beyond $E$. Here are some examples that demonstrate this idea:
Example 1. Let $f(x) = 1/x$ on $(0, \infty)$. If we pick two 'infinitesimals' $a_n = \frac{1}{n}$ and $b_n = \frac{1}{n+1}$, then they become arbitrarily close but $|f(a_n) - f(b_n)| = 1 \not\to 0$. So $f$ is not uniformly continuous.
Example 2. Let $f(x) = x^2$ on $\mathbb{R}$. If we pick two 'infinitely large' $a_n = n$ and $b_n = n+\frac{1}{n}$, then $a_n$ and $b_n$ become arbitrarily close, but $|f(a_n) - f(b_n)| = 2 + \frac{1}{n} \not\to 0$. So $f$ is not uniformly continuous.