# To check uniform continuity

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 ?

• @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 $[0,1]$. – David C. Ullrich Oct 2 at 12:17

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?