Is $f(x)=x\sin x$ uniformly continuous in the interval $(0,a)$ when $a>0$?

Question

Is $f(x)=x\sin x$ uniformly continuous in the interval $(0,a)$ while $a>0$?

I have proven that its not uniformly continuous in the interval $[0,\infty)$ because the function "$x\sin x$" is continuous in general but its doesn't converge to any final limit. so its not uniformly continuous in the interval $[0,\infty)$ but what about the interval $(0,a)$ while $a>0$? any kind of help would be appreciated.

I mean I proved this in a previous follow up question that I solved on paper. "I didn't write it here" because I didn't think it will be relevant to the question itself. I thought mentioning that fact would help. — Firas Abd El Gani, Jan 02 '15 at 18:07
@John its $(0,a)$ an open interval. and if u meant the same thing, then can u explain why ? — Firas Abd El Gani, Jan 02 '15 at 18:09
$x\sin x$ is uniformly continuous on $[0, a]$, so it is uniformly continuous on $(0, a)$. — , Jan 02 '15 at 18:10
@john we can say that only because its not a closed interval according to the definition to uniform continuity? — Firas Abd El Gani, Jan 02 '15 at 18:12
Any continuous function is uniformly continuous over a (relatively) compact interval. — Jack D'Aurizio, Jan 02 '15 at 18:13
"Doesn't converge to any final limit" is misleading - for example $\sin x$ is uniformly continuous but doesn't converge to any value as $x\to \infty$ — DanZimm, Jan 02 '15 at 18:40
The fact that the function does not converge to any final limit does not imply that it is not uniformly continuous on the interval $[0,+\infty)$. — Ezequiel, Jan 02 '15 at 18:38
While this explains why OPs partly solution is wrong, it dosen't actually answer the question. — Alice Ryhl, Jan 02 '15 at 18:41
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. — kingW3, Jan 02 '15 at 18:59

Alice Ryhl · Accepted Answer · 2015-01-02T18:47:13.803

4

$x\sin x$ is continuous on the closed interval $[0,a]$, and is therefore uniformly continuous on that interval. Since $(0,a)$ is a subset of $[0,a]$, then $x\sin x$ is uniformly continuous on the open interval $(0,a)$

edited Jan 02 '15 at 18:47

answered Jan 02 '15 at 18:31

Alice Ryhl

8,001

Is $f(x)=x\sin x$ uniformly continuous in the interval $(0,a)$ when $a>0$?

1 Answers1