In the context of differential geometry, we have a smooth vector field $X$ and an integral curve of $X$ called $\gamma$ such that $\exists\ t_0 \in \mathbb{R}$ with $\gamma'(t_0) = 0$. The exercise asks to prove that under those conditions, $\gamma$ is constant.
Invoking the Picard-Lindelöf theorem of existence and uniqueness of differential equations that can easily be proved.
I was wondering if there was another, more indirect and "by hand" way.
Thanks!
PS: I don't think this is a duplicate question, as it's not the same to have an integral curve into an arbitrary manifold as it is to have one to $\mathbb{R^n}$. Of course, given Anthony's answer, I can see that this specific question can be answered by reducing it to the other case. But there could also be another answer that takes a different approach.
Since $X$ is smooth, we can choose local coordinates with origin $\gamma(t_0)$ such that the components of $X$ satisfy a Lipschitz condition, i.e. $|X(x) - X(y)| \le C|x-y|$ for some constant $C$. In particular, substituting $x=0$ and $y=\gamma(t)$ and remembering $\gamma'(t_0) = X(0)=0$, we have $$|\gamma'(t)|=|X(\gamma(t))| \le C|\gamma(t)|.$$ Since $\frac d{dt} |\gamma(t)|^2 \le 2|\gamma(t)||\gamma'(t)|$, this yields the differential inequality $$\frac{df}{dt}\le 2Cf$$ for the differentiable function $f(t)=|\gamma(t)|^2$; so Gronwall's inequality along with our initial condition $f(t_0) = 0$ tells us that $f(t) \le 0$. Since $f$ is manifestly non-negative, we conclude that $f \equiv 0$ and thus $\gamma$ is constant.