My question is, how can I prove that $\sin(5x)$ is not uniformly continuous?
Asked
Active
Viewed 156 times
0
-
1By Heine-Cantor, it is uniformly continuous on $[-\pi/5, \pi/5]$ and since it is periodic, this would imply it is uniformly continuous on $\mathbf{R}$ in a rather standard proof. – William M. Dec 16 '18 at 17:32
-
It is $5$-Lipschitz. – Aphelli Dec 16 '18 at 18:19