Let $f:[0,1]\to\mathbb R$ be a continuous map such that the map $[0,1]\to \mathbb R^2,t\mapsto e^{2\pi if(t)}$ is smooth ($C^\infty$). Then is $f$ also smooth?
I tried to use the fact that $\arccos$ and $\arcsin$ are smooth if the domain is restricted to the interior. But I don't know what to do at the endpoints.
I figured it out. Note that both $\cos (2\pi f(t))$ and $\sin (2\pi f(t))$ are smooth. Near an arbitrary real number, at least one of $\arcsin$ and $\arccos$ does not cause an endpoint related problem. So we can use the one not causing the problem. That is, if $f(t_0)=0$, then we cannot use $\arccos$, but since $\arcsin$ is smooth near $0$, we see that $2\pi f(t)(=\arcsin \sin 2\pi f(t))$ is smooth near $t_0$.