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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 understood 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.

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.

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?

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 understood 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.

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 $\alpha>0$

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

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

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?