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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.

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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.

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  • $\begingroup$ Thank you for your answer. Taking $U=\phi(V)$ is great, but is it possible to weaken the hypothesis on $V$, for example $\Delta V \leq \theta |\nabla V|^2$ for some fixed $\theta$? $\endgroup$
    – tituf
    Commented Jul 1, 2020 at 15:40
  • $\begingroup$ It seems difficult, since you want your inequality to hold for arbitrarily large $\alpha$, but I can not exclude it $\endgroup$
    – Piero D'Ancona
    Commented Jul 1, 2020 at 21:46
  • $\begingroup$ If I weaken the request and ask the inequality to hold only for $0<\alpha<\alpha_0$, then $\Delta V\leq \theta |\nabla V|^2$ for some $\theta<\alpha_0^{-1}$ is a sufficient hypothesis. This can be seen by choosing $U=V$. Do you see any further possibility when only $0<\alpha<\alpha_0$ is considered? $\endgroup$
    – tituf
    Commented Jul 2, 2020 at 17:23
  • $\begingroup$ You could try with a multiplier of the form $\phi(x,V)$. The derivatives w.r.to $x$ should be of lower order, that is, small compared to the main terms. For instance of polynomial growth in $x$ $\endgroup$
    – Piero D'Ancona
    Commented Jul 2, 2020 at 18:22
  • $\begingroup$ If $U$ is allowed to depend on $\alpha$, is everything easier? $\endgroup$
    – tituf
    Commented Jul 2, 2020 at 21:59

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