Confusion about bounded derivative implies uniform continuity

Question

I'm currently learning about uniform continuity and was introduced to an intuitive way of "seeing" uniform continuity by observing the derivative of the function across an interval. Notably, if the derivative is bounded, this implies uniform continuity. That made sense for some examples but I've come across an example which confuses me. Let $f(x)=\sqrt{x}$. I've proven using the $\epsilon-\delta$ definition that this function is in fact uniformly continuous on its domain, $[0,\infty)$, but the derivative of $\sqrt{x}$ is tending to $+\infty$ as $x\rightarrow0^+$. To me, this isn't a bounded derivative and therefore that "intuitive" way of visualizing uniform continuity doesn't work here. What I'm hoping for is to understand why this is so (obviously my interpretation of this fact must be flawed), and secondly is this the best way to look at a function and to interpret uniform continuity?

Answer 1

Notice that you have found a one way relationship between derivatives and uniform continuity. If the derivative is bounded, then the function is uniformly continuous. The converse is certainly not true; the example you gave provides a nice counter example. It is NOT the case that if $f$ is uniformly continuous then $f'$ is bounded. This is all to say that there are some uniformly continuous functions that do not have bounded derivatives, while all functions with bounded derivatives are uniformly continuous.

In terms of your intuitive definition of uniform continuity: I would ask is there a point $x\in \text{dom} f$ for which the slope is the largest or smallest (possibly $\pm \infty$). For example, $$f:\left[0,\infty \right) \to \mathbf{R}$$ $$f(x)=\sqrt{x}$$ is uniformly continuous as its slope is "largest" at $x=0$ (very loosely speaking). Whereas, $$g:(0,\infty) \to \mathbf{R}$$ $$g(x)=\frac{1}{x}$$there is no $x\in (0,\infty)$ where the slope of $g$ is smallest; just scoot a little left to find a smaller slope.

Answer 2

My favorite interpretation of uniform continuity is through the notion of "continuity at infinitely close points". The next theorem illustrates this idea:

Theorem. Let $f : U \subseteq \mathbb{R} \to \mathbb{R}$. Then the followings are equivalent:

1. $f$ is uniformly continuous (on $U$).
2. For any sequences $(a_n), (b_n)$ in $U$ such that $|a_n - b_n| \to 0$, we have $|f(a_n) - f(b_n)| \to 0$.

In Item 2, the sequence $(a_n)$ plays the role of "points in $U$ that are infinitely close to the boundary of $U$" and/or "infinitely large points in $U$". Hence, Item 2 tells that $f$ is uniformly continuous if and only if it is "continuous" even at such points.

Alternatively, but still heuristically, imagine there is a way of enlarging $\mathbb{R}$ by adding infinitesimals and infinitely large numbers, so that all the conventional quantities, objects, relations, etc. in $\mathbb{R}$ extends to the enlargement, say $\mathbb{R}^*$. (Let us also denote the extension of each object $x$ to the realm of $\mathbb{R}^*$ by $x^*$.) Then $f : U \subseteq \mathbb{R} \to \mathbb{R}$ is uniformly continuous if and only if $f^* : U^* \to \mathbb{R}^*$ is continuous, in the sense that if $x, y \in U^*$ are infinitesimally close, then $f^*(x)$ and $f^*(y)$ are infinitesimally close as well. (This is not a daydream, though. There is a mathematically rigorous way of materializing this idea, through the hyperreal numbers and ultraproduct.)

The next two examples demonstrate this heuristics:

Example 1. Let $f : U = (0, \infty) \to \mathbb{R}$ be given by $f(x) = 1/x$. We show that $f$ "fails to be continuous at a point in $U$ infinitely close to $\partial U$". Indeed, set $a_n = \frac{1}{n}$ and $b_n = \frac{1}{n+1}$. Then

$$|a_n - b_n| \to 0 \qquad\text{but}\qquad |f(a_n) - f(b_n)| = 1 \not\to 0. $$

Heuristically, let $N$ be infinitely large positive number, and let $x = \frac{1}{N}$ and $y = \frac{1}{N+1}$. Since $x > 0$, we have $x \in U^*$. Then $x$ and $y$ are infinitely close, but $f^*(x) - f^*(y) = -1$ is not.

Example 2. Let $f : U = \mathbb{R} \to \mathbb{R}$ be given by $f(x) = x^2$. We then argue that $f$ "fails to be continuous at an infinitely large point in $U$". Indeed, set $a_n = n$ and $b_n = n + \frac{1}{n}$. Then

$$|a_n - b_n| \to 0 \qquad\text{but}\qquad |f(a_n) - f(b_n)| = 2 + \frac{1}{n^2} \not\to 0. $$

Likewise, letting $x = N$ and $y = N + \frac{1}{N}$ for infinitely large $N$ shows that $f^*$ is not continuous.