Let $\Omega$ be $\mathbb R^n$ or a complete (unbounded) open manifold, and $f$ be a smooth function on $\Omega$.

We consider a self-adjoint 2nd. elliptic operator $H$ on $L^2$ space(to simplify the case, one can regard $H$ as Lapalacian).

We assume that $f\to 0,~df \to0$ as $x\to \infty$.

Q Is this enough to show that $f\cdot\nabla$ is $H$-compact, i.e. $f\nabla(H+i\lambda)^{-1}$ is a compact operator, for some $\lambda\in\mathbb R_+$.

PS: I am not sure the problem was studied or not. It will also be greatly helpful, if anyone give some reference.


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