I'm working on the following analysis problem: Prove that if $f$ is differentiable on a, possibly infinite, interval $(a, b)$ and if $\lim_{x\to b}f'(x)=\infty$, then $f$ is not uniformly continuous on $(a, b)$.

I've tried several approaches, but the one that's gotten me the farthest so far is establishing that there is some interval $[m, b)$ on which $f$ is increasing, then introducing a non-decreasing Cauchy sequence $\{x_n\}$ in $[m, b)$ converging to $b$. I can then show that $\{f(x_n)\}$ is non-decreasing, and therefore has a limit.

Intuitively, I know that the limit should be $\infty$, and once I show that, it follows that $\{f(x_n)\}$ is not Cauchy. Then, since $\{x_n\}$ is a Cauchy sequence in $(a, b)$ but $\{f(x_n)\}$ is not Cauchy, it follows that $f$ is not uniformly continuous. However, I cannot seem to work out the details of showing that $\lim f(x_n)=\infty$. I'm assuming I need to use the Mean Value Theorem, since that is the main focus of this section in the book and also it's the only way I can think of relating the function to its derivative. Any help would be appreciated.

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    $\begingroup$ You won't necessarily have $\lim f(x_n) = \infty$. Consider for example $(a,b) = (0,1)$ and $f(x) = -\sqrt{1-x}$. Moreover, this function satisfies the hypothesis but is uniformly continuous on $(0,1)$, so the result you are trying to prove is false. $\endgroup$ – Bungo Aug 24 '17 at 21:58
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    $\begingroup$ @Bungo Your example seems to be uniformly continuous, which makes it a counterexample to the question :). $\endgroup$ – Erick Wong Aug 24 '17 at 22:01
  • $\begingroup$ @ErickWong Yep, I just noticed that :-) Edited my comment. $\endgroup$ – Bungo Aug 24 '17 at 22:02
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    $\begingroup$ Oh jeez. Dang it, Joseph L. Taylor, how could you do this to me. Well, I guess I'll move on then. Thank you. $\endgroup$ – themathandlanguagetutor Aug 24 '17 at 22:06

On an infinite interval, say $[a,\infty)$, if $\lim_{x \to \infty} f'(x) = + \infty$, then $f$ is not uniformly continuous.

By the MVT, for any $x <y$ in $[a,\infty)$ there exists $\xi$ with $x < \xi < y$ such that $|f(x) - f(y)| = |f'(\xi)|\, |x- y|$.

If $f'(x) \to +\infty$, then for any $\delta >0$, there exists $K> a$ such that if $x > K$ then $|f'(x)| > 2/\delta.$

Choosing any $x > K$ and $y> x$ such that $ |x - y| = \delta/2 < \delta$, we have

$$|f(x) - f(y)| = |f'(\xi)| \, |x - y| = |f'(\xi)| \frac{\delta}{2} > \frac{2}{\delta}\frac{\delta}{2} = 1.$$

Hence, $f$ is not uniformly continuous.

On a finite interval, the proposition is not true. This can be seen with the counterexample $f(x) = \sqrt{x}$ on $(0,1)$ where $f$ is continuous on the compact interval $[0,1]$ -- hence, uniformly continuous on $(0,1)$ -- but, $f'(x) = 1/\sqrt{x} \to +\infty$ as $x \to 0+$.

  • $\begingroup$ To the proposer: Note that the reason the method of proof for the domain $[a,\infty)$ fails for the case $f(x)=\sqrt x$ on $(0,1)$ is that in the domain $[a,\infty)$ we can fix a positive lower bound ( e.g. $\delta /2 $ ) on $|x-y|$ while letting $x, y$ go to \infty but we can't fix a positive lower bound on |$x-y|$ and have $x,y$ both go to $0.$ $\endgroup$ – DanielWainfleet Aug 25 '17 at 10:06
  • $\begingroup$ I'm simply illustrating that a function can be uniformly continuous on a finite interval yet the derivative can approach infinity as either endpoint is approached. The counterexample using the right endpoint has already been given in a comment. $\endgroup$ – RRL Aug 27 '17 at 18:44
  • $\begingroup$ It's a fine answer. I just wanted to point something out.It is instructive to examine proofs to see exactly where,or why, they fail for alternate cases. $\endgroup$ – DanielWainfleet Aug 29 '17 at 9:59
  • $\begingroup$ @DanielWainfleet: Point well taken. $\endgroup$ – RRL Aug 31 '17 at 7:28

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