Show that the following function is uniformly continuous on $(-1,1)$

$$f(x) = \begin{cases} {x \sin \frac{1} {x}}, & \text{ } x\in(-1,0)\cup(0,1) \\ 0, & \text{ }x = 0. \end{cases} $$

We cannot use the theorem that a continuous function on a compact set K is continuous on K, because we don't have a compact set. I was told the following hint: "if a function is uniformly continuous on a set then it is also uniformly continuous on any subset of this set". I don't know exactly what to do with this information, can you help me ? :)

I know the definition of uniform continuity, I (should) know what open, closed, compact sets are.


Hint: Show that you can extend the definition to $[-1,1]$ and that it is continuous on the closed interval. Then use the theorem about uniform continuity.

  • $\begingroup$ Very fast way to have the conclusion (+1). Good to know that $f(x)=\sin(1/x)$ fails to be U.C. in $(-1,1)$. $\endgroup$ – mrs Oct 20 '12 at 19:14
  • $\begingroup$ So you suggest that I define f(1)= sin(1) and f(-1)= - sin(-1), then show that the function is continious. Continious functions on a compact set are uniformly continious, and then use the hint, namely a (-1,1) is a subset of [-1,1] ? $\endgroup$ – Applied mathematician Oct 20 '12 at 19:33
  • 2
    $\begingroup$ Yes, there is no problem with that. The "magic" happens at $x=0$ anyway. $\endgroup$ – Hagen von Eitzen Oct 20 '12 at 19:39
  • 1
    $\begingroup$ @Hempo: Yes, exactly. $\endgroup$ – Asaf Karagila Oct 20 '12 at 20:15

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.