Is uniform continuity related to the rate of change of the function? I have been trying to understand how uniformly continuous functions differ from functions that are continuous but not uniformly continuous. Based on some examples and counter-examples, my feeling is that uniformly continuous functions cannot change "rapidly". Hopefully, the following examples will make it clear what I mean by that:


*

*$f(x) = 1/x$ is not uniformly continuous on $(0,1)$ because it "blows up" at $0$, that is, its slope becomes infinite very quickly. Similar functions would be $\log x$ on $(0,1)$ and $\tan x$ on $(0,\pi/2)$.

*$f(x) = x^2$ is not uniformly continuous on $(0,\infty)$ because it goes to infinity "rapidly" in the sense that the function's rate of change is greater than that of linear functions, for sufficiently large $x$.

*$f(x) = x^{1/3}$ is uniformly continuous on $(-1,1)$, because even though the slope does become infinite at $0$, it does not happen "rapidly" as in the previous two examples.

*$f(x) = \sin (1/x)$ is not uniformly continuous on $(0,1)$ because it oscillates "rapidly" as $x$ approaches $0$.

*$f(x) = x\sin(1/x)$ is uniformly continuous on $(-1,1)$ because, although the frequency of oscillation remains the same, the amplitude is "small enough" to make the function uniformly continuous; in some sense, the function is not changing rapidly enough for it to fail to be uniformly continuous.


This is admittedly a very naive way of looking at uniform continuity, but does it have any merit? Is there a way to describe uniform continuity by rigorously formulating the notion of rate of change of the function?
Some difficulties I already see are:


*

*There are continuous functions that are not differentiable, so the natural idea of rate of change is not applicable to them.

*There even exist continuous but nowhere differentiable functions on $\mathbb{R}$, and we know that any continuous function on a compact subset of $\mathbb{R}$ is uniformly continuous. So, the Weierstrass function restricted to $[0,1]$ is a uniformly continuous function. Surely, the function changes "rapidly" all over the place, but somehow it doesn't do so "rapidly enough"?


I have tagged this as a soft question because I know there is a lot of ambiguity in this post. It would be very helpful if anyone can give me insights about why this method of thinking of uniformly continuous functions (at least in the real one-variable case) can or cannot be fruitful.
Also, this is a related post that I found useful, especially the most upvoted answer: Why did mathematicians introduce the concept of uniform continuity?
 A: Rates of change without differentiability is indeed a problem. What you can measure, in a sense, is average rates of change. You can still define the average rate of change of $f$ over $[a, b]$ to be
$$\frac{f(b) - f(a)}{b - a}.$$
Note that there's no differentiability required here.
When $f$ is a Lipschitz-continuous function (stronger than uniform continuity), this is always bounded, regardless of the choice of $a$ and $b$ (in fact, this is the definition of being Lipschitz-continuous).
When $f$ is uniformly continuous, the situation is a little more subtle. It's not helpful to consider the single numerical value $\frac{f(b) - f(a)}{b - a}$, but instead view the largest possible allowance for $f(b) - f(a)$ as a function of the length of the interval, $b - a$. With a Lipschitz-continuous function, halving the largest allowable value of $f(b) - f(a)$ corresponds to halving the length $b - a$, but this isn't the case with a general uniformly continuous function. Instead, we have that $f(b) - f(a)$ tends to $0$, at some (possibly non-linear) rate as $b - a$ tends to $0$, regardless of the values that $a$ and $b$ actually take.
Let's take an example: $f(x) = \sqrt{x}$. Note that $f$ is not Lipschitz-continuous, as
$$\frac{f(b) - f(a)}{b - a} = \frac{\sqrt{b} - \sqrt{a}}{b - a} = \frac{1}{\sqrt{b} + \sqrt{a}},$$
and taking $b$ and $a$ arbitrarily small makes this value tend to $\infty$. However, if we fix $\varepsilon > 0$, we can force $|f(b) - f(a)| < \varepsilon$ by making $b - a$ small, regardless of the values of $a$ and $b$. That is, $f$ is uniformly continuous. To see this, first suppose $b - a < \varepsilon^2$. Then
$$b < \varepsilon^2 + a \le \varepsilon^2 + 2\sqrt{a}\varepsilon + a = (\varepsilon + \sqrt{a})^2,$$
hence $\sqrt{b} - \sqrt{a} < \varepsilon$. So, it's not a linear function of $\varepsilon$, but by forcing $b - a < \varepsilon^2$, we are forcing $f(b) - f(a)$ to be smaller than $\varepsilon$.
By taking smaller and smaller intervals, we can still have $\frac{f(b) - f(a)}{b - a}$ to be larger and larger, but when we fix $b - a$ to be at least some minimal length, we can make $f(b) - f(a)$ small.
A: It does, in the following sense: for a function $f$ to fail to be uniformly continuous, there has to be a pair of infinitely close points such that the difference in values of $f$ at those points is appreciable (hence not infinitesimal).  That's a tremendous rate of change and you don't even need to require the function to be differentiable for this to work.
