In the one dimensional setting we have the following:
Assume $$f:\Bbb R\to\Bbb R$$ is differentiable and fulfills $$f'(x) > c >0 ,$$
at least outside of some compact set $K$. Then we have $$\lim_{x\to\pm\infty}f(x) = \pm \infty,$$
which may be seen for example by mean value theorem.
I am wondering how to generalize this to $$ f:\Bbb R^n\to\Bbb R^n.$$
I suspect we have $$\lim_{\|x\|\to \infty}\|f(x)\| = \infty,$$
under some conditions on the Jacobian $J_f$.
By the multivariate mean value theorem, one could get that
$$f_i(a) - f_i(b) = \langle J_f(z_i)_i,a-b\rangle ,$$
fore the component functions $f_i$ and some $z_i$ on the line segment between $a$ and $b$. At this point I am not sure how to put that into a condition on the Jacobian, for example if I use the matrix norm the inequality goes in the 'wrong' direction.
Edit:
Maybe a good starting point to think about this problem is strong convexity. Assume that $f$ is the gradient of some function $g$ : $$f(x) = \nabla_g(x).$$
$g$ is strongly convex, iff there exists some $m>0$ such that
$$\langle \nabla_g(x) - \nabla_g(y),x-y\rangle \geq m \Vert x-y\Vert_2^2 .$$
In particular, $g$ is strongly convex iff the Hessian of $g$, that is $J_f$, fullfills $$J_f \geq mI_n,$$
that is the smallest eigenvalue $\lambda_*$ of $J_f$ fulfills $\lambda_*\geq m$.
Then one can conclude by Cauchy-Schwarz:
$$ \Vert f(x) - f(y) \Vert \Vert x-y \Vert \geq \langle f(x) -f(y), x-y\rangle^2 \geq m^2 \Vert x-y\Vert^4 .$$
Which shows $\lim_{\|x\|\to \infty}\|f(x)\| = \infty.$
Of course, this line of argument requires that $f$ has a potential. Maybe one can find another condition without this assumption?