Form of function constant on a closed curve

Question

Suppose that a continuous function $f : (0,2\pi)^2 \to \mathbb{R}$ is constant along the closed curve $$ \gamma(t) = (\theta_0 + t \mod{2\pi}, \phi_0 + q t \mod{2\pi}) $$ where $q \in \mathbb{Q} - \{0\}$ is rational, that is $$ \partial_t f(\gamma(t)) = 0, $$ how can we show that $f$ has the form $$ f(\theta,\phi) = h(q\theta - \phi) $$ where $h : \mathbb{R} \to \mathbb{R}$ is continuous?

Answer 1

In general, $f$ doesn't have to be of that form, unless it is constant on all curves of the form that you've given (for any choice of $\theta_0$ and $\phi_0$).

In such a case, you can notice that for every point $(\theta,\phi)$ it lies on the same curve as point $(0, \phi-q\theta)$. You can therefore define $h(z)=f(0,-z)$. If $f$ is constant on each curve, we have $$f(\theta, \phi)=f(0, \phi-q\theta)=h(q\theta-\phi)$$.