In one paper the author uses the statement without citation: 


Let $(M,g)$ be a Riemannian manifold. The gradient $\nabla F$ of a proper function $F: M\rightarrow \mathbb{R}$ is integrable vector field, i.e. its integral curves are defined for all times. 


In this particular case, $F$ is also bounded from below, and the Riemannian metric $g$ on $M$ is complete.


My question is 


1)Is this statement true in general?


2)If not, why is the statement true when additional assumptions are satisfied (function is bounded from below and metric is complete)?


I could not find a reference on this in the literature; if someone knows it - that will be enough.


Thanks for any help!



Gradient of a proper function is integrable
Unless there are some additional assumptions imposed, the claim is false. Consider $f(x)=x^4$ on $\mathbb{R}$. This function is bounded from below, $f'(x)=4x^3$, and solutions to $x'(t)=4x(t)^3$ have blowup in a finite time. 


