Show $f(x) = x\sin{x}$ is not uniformly continuous on $\mathbb{R}$

Question

I know it's not Lipschitz continuous because its derivative is unbounded but I'm not sure if not Lipschitz continuous implies not uniformly continuous. Thanks

Answer 1

Lipschitz continuous implies uniformly continuous, but not the other way around. To show your function is not uniformly continuous, try working directly with the definition. For $\varepsilon >1$, can you locate, for every $\delta >0$ two points $x,y$ such that $|x-y|<\delta $ but $|f(x)-f(y)|\ge \varepsilon $? Can you see how the answer to this question solves the problem? Hint: the location of the points will depend on how small $\delta $ is.

Answer 2

If $x\sin(x)$ were uniformly continuous then given $\epsilon >0$ there will exist a $\delta>0$ such that $|x\sin(x)-y\sin(y)| < \epsilon$ whenever $|x-y|<\delta$. In particular, if you choose $x=2n\pi$ and $y=2n\pi +\frac{\delta}{2}$ then it will follow that $|(2n\pi+\frac{\delta}{2})\sin(\frac{\delta}{2})|<\epsilon$ for all natural numbers $n$, which is obviously not true and hence your function can not be uniformly continuous.