# Show that a function is uniformly continuous. Where to start?

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.

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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.

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Very fast way to have the conclusion (+1). Good to know that $f(x)=\sin(1/x)$ fails to be U.C. in $(-1,1)$. – Babak S. Oct 20 '12 at 19:14
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] ? – Joyeuse Saint Valentin Oct 20 '12 at 19:33
Yes, there is no problem with that. The "magic" happens at $x=0$ anyway. – Hagen von Eitzen Oct 20 '12 at 19:39
@Hempo: Yes, exactly. – Asaf Karagila Oct 20 '12 at 20:15
@Hempo: Is there a particular reason for which you unaccepted the answer? – Asaf Karagila Nov 3 '12 at 22:26

Define $f(1) = \sin(1)$ and $f(-1)=-\sin(-1)$

(1) The function is continuous on $x=0$.

Proof: $|f(x)-f(0)|=|x \sin(1/x)-0| \le |x|$. Given $\epsilon \gt 0$, set $\delta=\epsilon$, so that whenever $|x-0| = |x| \lt \delta$ it follows that $|f(x) - f(0)| \lt \epsilon$. Thus, f is continuous at $x=0$ $\square$

(2) The function is continuous everywhere when $x \not= 0$

Proof: $1/x$ is continuous when $x \not= 0$, while $\sin(u)$ is everywhere continuous. A composition of two continuous functions is continuous. $x$ is continuous everywhere. A multiplication of two continuous functions is continuous. So our function $x \sin(1/x)$ is continuous everywhere if $x \not = 0$ $\square$

(3) $f(x)=x \sin(1/x)$ is continuous on [-1,1]:

Proof: The combination of the these two facts means that the $f(x)$ is everywhere continuous. If a function is continuous everywhere, it's sure continuous on a subset [-1,1]. So we conclude f is continuous on [-1,1]. Because f is continuous on a compact set, we can conclude that f is uniformly continuous on [-1,1]. If a function is uniformly continuous on a compact set, we may conclude that it's also uniformly continuous on a subset of that compact set. (-1,1) is a subset of [-1,1]. Hence, f is uniformly continuous on (-1,1) $\square$

(Please improve my proof if you think it's necessary)

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