0
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ @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
0
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.