Let $V: \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ vector field and suppose that $\int_{1}^{\infty}\frac{dr}{B(r)}=\infty$ where $$ B(r) = \sup(|V(x)|: |x|<r ). $$ Prove that the maximal solutions of the differential equation $\dot x = V(x)$ live on all $\mathbb{R}$.
Let $\sigma_p:(\alpha, \omega) \to \mathbb{R}^n$ be the maximal solution with $\sigma_p(0)=p$, I want to prove that $\omega = \infty$ (I imagine that $\alpha=-\infty$ can be proved similarly).
I've tried reasoning by contradiction so that $t_R = \inf(0\le t \lt \omega: |\sigma_p(t)|\ge R)$ is finite for every $R>0$, but the best I could came up with is $B(R) \ge |V(\sigma_p(t_R))|$ and I don't know if $$ \int_{1}^{\infty} \frac{1}{|V(\sigma_p(t_R))|}dR< +\infty $$ i.e. if it converges.
I've also some tried other things like $$ |\sigma_p(t)-p|\le \int_{0}^{t}|V(\sigma_p(s))|ds $$ and got $R\le t_R B(R)$, but nothing again.
Can you give me a hint?
There needs to be an additional condition $B(1) \neq 0$, otherwise the integral may diverge because of the singularity at $1$ even though the field $V$ is not complete.
The idea is to consider the growth rate of $r = |x|$. Since $\sup_{|\xi| < r} |V(\xi)| = \sup_{|\xi| \leq r} |V(\xi)|$ for a continuous function $V$, we have $$\frac d {dt} r^2 = \frac d {dt} (x \cdot x) = 2 x \cdot V(x) \leq 2 r \left| V(x) \right| \leq 2 r B(r), \\ \dot r \leq B(r).$$ Then $r$ is dominated by the solution of $\dot \rho = B(\rho)$ with an initial condition $\rho(0) = \rho_0 \geq \max(r(0), 1)$. The domain of the solution $\rho$ extends to $\infty$, which can be shown as here. Therefore the domain of $x$ also extends to $\infty$, since $|x|$ is bounded on any finite interval where $x$ exists.
By reversing $t$, we show in the same way that the domain of $x$ extends to $-\infty$.