2
$\begingroup$

Given a function $f:\mathbb R \rightarrow \mathbb R $ such that $\lim _{ x\rightarrow\infty }{ f(x) }=\infty$, prove $f(x)\sin(x)$ is not uniformly continuous.

I tried to go by definition:

  • $\lim _{ x\rightarrow\infty }{ f(x) }=\infty$ $\ \Rightarrow$ $\ \forall M>0 \ \exists E>0\ \forall x>E:f(x)>M$
  • Proving the function is not uniformly continuous: $\ \exists \epsilon>0\ \forall\delta>0\ \exists x_1,x_2\ \in\mathbb R \ s.t. |x_1-x_2|<\delta:|f(x_1)-f(x_2)|\ \geq\epsilon$
  • $|f(x_1)\sin(x_1)-f(x_2)\sin(x_2)|> ?$

How can I continue from here?

$\endgroup$
6
$\begingroup$

Put $\epsilon = 1$.

Put $x_1 = 2 \pi n$, $x_2 = x_1 + \delta/2$.

$|f(x_1) \sin(x_1) - f(x_2) \sin(x_2)| = |f(x_2) \sin(\delta/2)|$.

For sufficiently large n, this quantity is greater than $\epsilon$

$\endgroup$
  • 1
    $\begingroup$ Concise and complete (+1) $\endgroup$ – Mark Viola Feb 17 '17 at 20:49

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.