How to check whether $\sin (x \sin x)$ is uniformly continuous or not on $\mathbb R$.

My Try: I took two sequence namely $t_n = \frac{n\pi}{2} + \frac {\pi}{n}$ and $z_n = \frac{n\pi}{2}$ . Now $|z_n - t_n|$ converges to $0$ but $|\sin(z_n \sin z_n) - \sin (t_n\sin t_n)| \to \frac {(\pi)^2}{2}$. So we can say $\sin (x \sin x)$ is not uniformly continuous on $\mathbb R$.

Please correct me if I went wrong anywhere.

  • $\begingroup$ This method seems OK. An alternative one would be to show that the derivative is unbounded. $\endgroup$ – Michał Zapała Sep 11 '18 at 16:22
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    $\begingroup$ @MichałZapała Note that $f(x) = \sqrt{x}$ is uniformly continuous on $[0,+\infty)$ and $f'(x)$ is unbounded as $x \to 0^+$ $\endgroup$ – Robert Z Sep 11 '18 at 16:25
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    $\begingroup$ @MichałZapała Derivative unbounded would not show uniform continuity fails $\endgroup$ – zhw. Sep 11 '18 at 16:25
  • $\begingroup$ Erm, you're both right, of couse. How embarrasing $\endgroup$ – Michał Zapała Sep 11 '18 at 16:26
  • $\begingroup$ $\pi^2/2?$ No, that difference can be at most $2.$ $\endgroup$ – zhw. Sep 11 '18 at 16:45

Let $f(x)= \sin (x\sin x).$ Define $x_n=2\pi n,y_n = 2\pi n +1/(4n).$ Then $y_n-x_n\to 0,$ and $f(x_n)=0.$ Thus

$$ f(y_n)-f(x_n) = f(y_n) = \sin \left((2\pi n + 1/(4n))\sin(2\pi n + 1/(4n))\right)$$ $$\tag 1 = \sin ((2\pi n + 1/(4n))\sin(1/(4n)).$$

Now $\sin u = u +o(u)$ as $u\to 0.$ So $\sin(1/(4n)) = 1/(4n) +o(1/n)$ as $n\to \infty.$ Insert that into $(1)$ to see it equals $\sin (\pi/2 + o(1))\to 1$ as $n\to \infty.$ Therefore $f$ is not uniformly continuous on $\mathbb R.$


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