Is $\sin{(\log{x})}$ uniformly continuous on $(0,\infty)$?

Question

Is $\sin{(\log{x})}$ uniformly continuous on $(0,\infty)$?

Let $x,y \in (0,\infty)$. $$ |f(x)-f(y)| = |\sin{\log{x}} - \sin{\log{y}}| = \left|2 \cos{\frac{\log{xy}}{2}}\sin{\log{\frac{x}{y}}{2}} \right| \leq 2 \left|\sin{\frac{\log{\frac{x}{y}}}{2}} \right| \leq \left|\log{\frac{x}{y}} \right| = |\log{x} - \log{y}| $$

If we consider the interval is $(1,\infty)$, then it is easy to finish the above inequality. But now the interval is $(0,\infty)$, I guess it is not uniformly continuous and now I don't have any idea to show it is not uniformly continuous. Can anyone give a favor or some hints? Thank you!

Answer 1

It is not uniformly continuous on $(0,1)$. Let $x_n=e^{-n\pi}$, $y_n=e^{-n\pi-\pi/2}$. Then $x_n,y_n\in(0,1)$ for all $n\in\mathbb N$, and $$|x_n-y_n|=|e^{-n\pi}-e^{-n\pi-\pi/2}|=e^{-n\pi}|1-e^{-\pi/2}|\to 0,$$ but $$|\sin\log x_n-\sin\log y_n|=|\sin(-n\pi)-\sin(-n\pi-\pi/2)|=1,$$ which does not converge to $0$ as $n\to\infty$. Therefore $\sin\log x$ is not uniformly continuous on $(0,1)$, so it is not uniformly continuous on $(0,\infty)$.

Answer 2

No it is not uniformly continuous because $\log x$ is not uniformly continuous on $(0,1)$.

If we take the sequence of numbers $(e^{-n})$ then $|e^{-n} -e^{-n-1}|$ tends to zero but $|\log(e^{-n})-\log(e^{-n-1})|=1$