Let $V:\mathbb{R}^{n}\to\mathbb{R}$ smooth, such that $\lim\_{|x|\to\infty}V(x)=+\infty$.


What are conditions on $V$ that guarantee the existence of a function $U:\mathbb{R}^{n}\to\mathbb{R}$ such that for all $\alpha>0$


$$\liminf\_{|x|\to\infty} (\nabla V(x) \cdot \nabla U(x) - \alpha\, \Delta U(x)) > 0 \ ?$$


Of course choosing $U=V$ does the job if $\nabla V$ does not vanish at infinity and we assume $\limsup\_{|x|\to\infty}\Delta V(x)\leq0$. This condition seems too restrictive to me, any other idea?


***Edit:*** I've realized that what I actually need is the following: given any $\alpha>0$ prove the existence of a function $U$ (possibly $U=U(\alpha,x)$) satisfying the inequality (under suitable hypothesis on $V$). The fact that $U$ may depend on $\alpha$ should help.



A differential inequality involving gradient and laplacian
Suppose you know that $\Delta V=o(|\nabla V|^2)$ as $|x|\to\infty$. This implies that for any $\theta\in(0,1)$ you have eventually the inequality $\Delta V\le \theta|\nabla V|^2$. Then for any eventually increasing function $\phi$, if you take $U=\phi(V)$ your expression is bounded below by
$$
|\nabla V|^2((1-\alpha\theta)\phi'-\alpha \phi'').
$$
Then you can play with the choice of $\phi$. For instance, if $\phi''=o(\phi')$ and $\phi'>c>0$ you are done.


