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Given, $f (x) = \sin (x \sin x)$ and we have to check if the function is uniformly continuous on $(0, \infty)$. So, far I haven't been successful but I tried to use this theorem " A function $f(x)$ is uniformly continuous on $(a,b)$ iff the extension of the function $f$ is continuous on $[a,b]$ "

Defining $f (x) = \sin (x \sin x)$, for $x$ in $(0, \infty)$ and $f (0) = 0$ I tried to show it continuous on $[0, \infty)$. But I think my procedure isn't quite right and the theorem could only be used for intervals of finite length.

Any suggestion regarding how to show it isnt uniformly continuous ?

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    $\begingroup$ @Marc That's not true! Say $f(0)=0$, $f(t)=t^2\sin(1/t^4)$ for $t\ne0$. Then $f$ is differentiable, $f'$ is unbounded on $[0,1]$, but $f$ is continuous on $[0,1]$, hence uniformly continuous on $[0,1]$. $\endgroup$ – David C. Ullrich Oct 2 at 12:17
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Hint: $\arcsin $ is a continuous function from $[-1,1]$ to $[0,2\pi]$. Continuous functions on a compact interval are uniformly continuous and composition of two uniformly continuous functions is uniformly continuous. Hence uniformly continuity of your function implies that of $x \sin x$. Can you show that $x \sin x$ is not uniformly continuous?

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  • $\begingroup$ Yes. It is a product of two uniformly continuous function and one of them is unbounded. $\endgroup$ – user487691 Oct 2 at 15:53

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