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Give an example of a bounded function $f: \Bbb R \to \Bbb R$ which is uniformly continuous but the function $g$ given by $g(x) = f(x^2)$ is not.

xavierm02
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user112741
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  • What are some bounded functions you know about? Have you tried any of them? There's a very simple one that should be on your list of favorite bounded functions. – dfeuer Dec 06 '13 at 19:49
  • @dfeuer. I know cos(x) is bounded but not sure how to verify using definition of uniform continuity – user112741 Dec 06 '13 at 19:54
  • The easiest way to prove that $\cos$ is uniformly continuous is using two facts: 1. A function that is continuous on a compact set (such as a closed interval) is uniformly continuous on that set, and 2. $\cos$ is periodic. – dfeuer Dec 06 '13 at 20:02

2 Answers2

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One such example is $$ f(x) = \sin(x). $$

Here is how I found this example:

The property of uniformly continuous means that the function has a maximal steepness at each fixed scale.

If you compare $f(x)$ with $f(x^2)$ then as $x$ moves towards infinity $x^2$ moves even faster. Therefore the function $f(x^2)$ seems to be changing faster and therefore becoming steeper and steeper than $f(x)$.

Therefore the task is to find a function $f$ that maintains at least some steepness. If you find such a function then $x^2$ will be constantly strengthening this steepness, ultimately breaking the uniform continuity.

However your function has to be bounded. That means it cannot maintain its steepness by growing just in one direction. It needs to oscillate. And at this point the sine function should automatically come to your mind.

If you want a proof that $f(x) = \sin(x^2)$ is indeed not uniformly continuous, then check out this answer.

Community
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Martin Drozdik
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    Yep, beat me to it. – JohnD Dec 06 '13 at 19:52
  • What does the OP learn from this? – dfeuer Dec 06 '13 at 19:52
  • @dfeuer How do you imagine a good answer? I think this is the best I can do, since posting a proof of why this is true would spoil the fun for OP – Martin Drozdik Dec 06 '13 at 19:54
  • I believe a good answer would work out some of the properties a function satisfying this requirement must have. For example, it can't be constant or monotone, but you could probably come up with some other good ones. – dfeuer Dec 06 '13 at 19:58
  • @MartinDrozdik: what did you mean by "at each fixed scale?". What I realize was that a uniformly continuous function must have a maximal slope. Is this what you mean? Thanks! – TheLast Cipher Sep 30 '19 at 05:10
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    @TheLastCipher Uniform continuity is a little more subtle than maximal slope. Maximal slope (Lipschitz continuity) implies uniform continuity, but not the other way around. For example $\sqrt{x}$ is uniform continuous, but does not have a maximal slope. The closer you go to zero, the steeper it gets. "At each fixed scale" maybe wasn't the best choice of words, but the wikipedia pages have really cool animated gifs to illustrate the point: https://en.wikipedia.org/wiki/Uniform_continuity https://en.wikipedia.org/wiki/Lipschitz_continuity – Martin Drozdik Sep 30 '19 at 12:10
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$f$ can't be constant, because if it were, $g$ would be uniformly continuous.

$f$ can't be monotone: this is actually interesting to prove, so I will leave it as a challenge to you. Hint: use the fact that a continuous function on a compact set is uniformly continuous.

It seems the simplest sort of bounded function $\Bbb R\to \Bbb R$ that is neither constant nor monotone is a periodic function. Can you prove that any continuous periodic function will do the trick?

dfeuer
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