I'm confused how to use the following theorem:

19.6 Theorem. Let $f$ be a continuous function on an interval $I$ [$I$ may be bounded or unbounded]. Let $I^◦$ be the interval obtained by removing from $I$ any endpoints that happen to be in $I$. If $f$ is differentiable on $I^◦$ and if $f′$ is bounded on $I^◦$, then $f$ is uniformly continuous on $I.$

So far, I have encountered examples. $f(x)= \sqrt{x}, g(x)= \frac{1}{x}, h(x)= x^2$ They are each on the interval $(0,\infty)$

I know $f$ is uniformly continuous, but $g$ and $h$ are not. However, the derivatives for each of these functions is unbounded on $(0,\infty)$

To show that a continuous function is not uniformly continuous on $(0, \infty)$, do I need to show the derivative is unbounded for every interval $[a, \infty )$ , where $a>0$?

If so, how would I prove the function is unbounded from $[a, \infty )$?

I would appreciate a worked out example with one of the functions above or one of your choosing.


If $f$ is differentiable on $I^\circ$ and $f'$ is bounded on $I^\circ$, then $f$ is uniformly continuous on $I$

is correct.

If $f$ is uniformly continuous on $I$, then $f$ is differentiable on $I^\circ$ and $f'$ is bounded on $I^\circ$

is not correct. The counterexample is exactly $f(x)=\sqrt x$.

To show that $f$ is not uniformly continous on $I$, it is enough to show that it is not uniformly continuous on $(0,1]$. And uniformly continous functions on a bounded interval are always bounded.

  • $\begingroup$ Could you explain how I could prove $g(x)$ and $h(x)$ are not uniformly continuous? I believe $g$ cannot be extended to a continuous function at $x=0$. How do you prove h is not uniformly continuous? $\endgroup$ – Zslice Oct 23 '14 at 8:14
  • $\begingroup$ @Zslice Look at the last paragraph I wrote. $\endgroup$ – 5xum Oct 23 '14 at 8:15
  • $\begingroup$ How does the last paragraph apply to $h(x)= x^2$? $\endgroup$ – Zslice Oct 23 '14 at 8:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.