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I have been teaching real analysis for a while now, but I still feel as though I am not able to explain why we should care about uniform continuity.

At first, I simply passed it off as an upgrade of "normal" continuity. Basically, a "nice" continuity. This isn't untrue, but it isn't particularly interesting.

I've also explained it from the "uniform" perspective. Students can see the "delta which works for all epsilon" point that I make, but again, not especially motivating.

I assign as homework to prove that "uniformly continuous image of a Cauchy sequence is Cauchy", but again, it's tough for the weaker students to recognize why we care about this.

Later, I added a note to my lessons about two behaviors of continuous functions which indicate an ABSENCE of uniformly continuity:

  1. Being unbounded on a bounded domain, like $1/x$ on $(0,1]$.
  2. Having an unbounded slope as $x\to\pm\infty$.

This helped some, but it more illustrated what uniform continuity is NOT, rather than what it is.

Finally, I found this wonderful post which (among other things) helped me to recognize the importance of the theorem "$f:D\to\mathbb{R}$ is uniformly continuous if an only if it can be continuously extended to the closure of $D$."

While this last point is a useful fact and provides a quick visual check for uniform continuity as well (which I love), I still don't have a good idea how to motivate uniform continuity on $\mathbb{R}$.

For example, I can prove that $f(x)=x$ and $f(x)=\sqrt[3]{x}$ are uniformly continuous functions on $\mathbb{R}$, and that $f(x)=x^2$ is not. I can also relate that back to the idea that "Having unbounded slope as $x\to\pm\infty$" prevents a function from being uniformly continuous. Is there any way to revisit the "extends continuously" idea here? (I'm assuming no, since the closure of $\mathbb{R}$ is $\mathbb{R}$, of course.) Can I do more to intuitively understand the difference between these two functions that doesn't rely on the "uniform delta" idea?

AegisCruiser
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  • Allow me to point out that having unbounded slope as $x \to \pm \infty$ does not prevent a function from being uniformly continuous. – Adayah Oct 21 '24 at 18:12
  • @Adayah Can you give me a counterexample of this? I was under the impression that part (b) of this question was true: https://math.stackexchange.com/questions/118665/why-if-f-is-unbounded-then-f-isnt-uniformly-continuous – AegisCruiser Oct 21 '24 at 19:42
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    Part (b) is correct, but unbounded slope is more like (c), which is incorrect. Anyway, consider the function $f : [0, \infty) \to \mathbb{R}$ which is equal to $n$ on the interval $\left[n, n+\frac{1}{n^2}\right]$ for $n \in \mathbb{N}$ and $0$ elsewhere. Then $g(x) = \int_0^x f(t) , \text{d} t$ has unbounded slope, but is uniformly continuous. Also a modification of $\sqrt[3]{x}$ would work, but it's a bit cumbersome to describe. – Adayah Oct 22 '24 at 06:43

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You could initially introduce uniform continuity not as a standalone idea, but by proving the property holds for continuous functions on a compact interval, or by showing it is necessary and sufficient for being able to continuously extend a function on a bounded set to its closure. Then name that property 'uniform continuity', and give the general definition, and note that it can apply to functions on an unbounded domain as well.

rake
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