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A student showed me the following exercise:

Is $h(x) = x \sin\left(\frac{1}{x}\right)$ uniformly continuous on $(0,1)$?

Admittedly, it has been a while since I have looked at problems involving uniform continuity and so I am not sure how to go about this problem.

My guess is that $h(x)$ is uniformly continuous on this interval but I am not sure how to go about picking an appropriate $\delta > 0$ such that $|x-c|< \delta$ implies $|h(x) - h(c)| < \epsilon$ for $x,c \in (0,1)$.

The student and I looked at $|x \sin(x) - c \sin(c)| \leq |x \sin(x)| + |c \sin(c)| \leq |x| + |c|$ although this feel like the wrong path to go down.

I appreciate any help! Thanks in advance!

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2 Answers 2

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$h(x)=x\sin(\frac{1}{x})$ can be extended to a continuous function on $[0,1]$ by setting $h(0)=0$. As $[0,1]$ is compact it follows that $h(x)$ is uniformly continuous on $[0,1]$, and therefore is uniformly continuous on $(0,1)$.

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    $\begingroup$ Wouldn't we have to first show that $h(x)$ is continuous on $(0,1)$? I am primarily stuck on showing continuity using the definition, unless there is another way to show that fact. $\endgroup$
    – user328442
    Commented Mar 16, 2017 at 23:08
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    $\begingroup$ $h(x)$ is continuous on $(0,1)$ because $x,\sin x$, and $\frac{1}{x}$ are. I wouldn't want to prove the continuity of $h$ directly from the definition. $\endgroup$
    – carmichael561
    Commented Mar 16, 2017 at 23:21
  • $\begingroup$ This makes a lot of sense. I didn't think of that. Thank you! :) $\endgroup$
    – user328442
    Commented Mar 16, 2017 at 23:24
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An idea: to use the equivalent definition of unif. continuity: let $\;\{x_n\}\;,\;\;\{y_n\}\;$ be any two sequences in $\;(0,1)\;$ s.t. $\;x_n-y_n\xrightarrow[n\to\infty]{}0\;$, then:

$$|h(x_n)-h(y_n)|=\left|x_n\sin\frac1{x_n}-y_n\sin\frac1{y_n}\right|\le|x_n-y_n|\left|\sin\frac1{x_n}\right|+y_n\left|\sin\frac1{x_n}-\sin\frac1{y_n}\right|=$$

$$=|x_n-y_n|\left|\sin\frac1{x_n}\right|+y_n\left|2\sin\left(\frac{\frac1{x_n}-\frac1{y_n}}2\right)\cos\left(\frac{\frac1{x_n}+\frac1{y_n}}2\right)\right|\xrightarrow[n\to\infty]{}0+0=0$$

since both summands above are of the form "function converging to zero times something bounded" .

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  • $\begingroup$ Forgive me but I don't follow on how $$y_n |2 \sin(\frac{\frac{1}{x_n}-\frac{1}{y_n}}{2}) \cos(\frac{\frac{1}{x_n}+\frac{1}{y_n}}{2})|$$ converges to 0. $\endgroup$
    – user328442
    Commented Mar 16, 2017 at 23:12
  • $\begingroup$ $\;x_n-y_n\to0\;$ and $\;\left|\sin\frac1{x_n}\right|\;$ is bounded, and $\;\sin\left(\frac{\frac1{x_n}-\frac1{y_n}}2\right)\to0\;$ and all the rest of the second summand is bounded... $\endgroup$
    – DonAntonio
    Commented Mar 16, 2017 at 23:14
  • $\begingroup$ Hmmm....Not so fast: there's something suspicious there. Lemme see if I can fix that later. $\endgroup$
    – DonAntonio
    Commented Mar 16, 2017 at 23:17

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