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The following question was part of my analysis assignment and I was unable to correctly solve it.

Is $\sin(1/x)$ uniformly continuous on $(0,1)$?

The following reasoning was given by my instructor : Is $\sin(1/x)$ can't be continuously extended to $0$. So, its not uniformly continuous on $(0,1)$.

But I have a question : why we need to extend Is $\sin(1/x)$ continuously to $0$ when asked domain is $(0,1)$?

I am not satisfied by the reasoning of my instrictor.

So, I am looking for a rigorious explanation here.

Thank you!!

balddraz
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    A continuous function on $(0, 1)$ is uniformly continuous if and only if it can be extended continuously to $[0, 1]$. See for example https://math.stackexchange.com/q/178266/42969. – Martin R Nov 08 '20 at 07:11
  • Since you require rigor (good for you) I am not allowed to post this response as an answer. Intuitively, the $\epsilon, \delta$ definition of continuity of $f(x)$ at a specific point $x_0$ is affected by $f'(x_0).$ The central difference between continuity at a point and uniform continuity throughout an interval is that you are forced to craft one specific $\epsilon,\delta$ algorithm that will be applied throughout the interval. That is okay as long as $|f'(x)|$ is finite (i.e. bounded). When it isn't bounded, you've got trouble. – user2661923 Nov 08 '20 at 07:44
  • @user2661923 This is not so precise. A function can be uniformly continuous whereas its derivative is not bounded. Think about $x \mapsto \sqrt{x}$. – TheSilverDoe Nov 08 '20 at 11:47
  • Your professor was not saying "this is the one and only way to show that $\sin 1/x$ is not uniformly continuous". It is quite possible to show it directly from the definition. What your professor is saying is that this is an easy way to see that $\sin 1/x$ cannot be uniformly continuous. Since (as Martin R says), you can show that any uniformly continuous function on $(0,1)$ can be extended continuously to $[0,1]$, but $\sin 1/x$ cannot be so extended, it cannot be uniformly continuous. – Paul Sinclair Nov 08 '20 at 17:08
  • @TheSilverDoe Very good point. My blind spot was stuck on the idea that the relationship between $\epsilon$ and $\delta$ has to be linear, which, as your example highlights, is not necessarily true. It would have been better for me to say, with respect to whether the function has uniform continuity, if $f'(x)$ is bounded, then you are okay. If it is not bounded, then you might have trouble. – user2661923 Nov 08 '20 at 18:10
  • @user2661923 Actually, the fact that the derivative is bounded is equivalent to the fact that the function is lipschitz (for $\mathcal{C}^1$ functions). Of course it implies that the function is uniformly continuous, but the converse is not true.So yes, the way you say it in your last sentence is the correct one :) – TheSilverDoe Nov 08 '20 at 18:13

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