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$\newcommand{\div}{\operatorname{div}}$Suppose I have an elliptic operator $\mathcal{L} u = -\div (A \nabla u) $ on some open set $\Omega \subseteq \mathbb{R}^d$ where here $A$ is uniformly elliptic on $\Omega$. Assume that I can prove the existence of the Green's function for $ \mathcal{L}_v u = -\div (A \nabla u) + vu$ on $\Omega$ for any bounded, nonnegative $v$ on $\Omega$.

Is there any simple way to go from this to proving the existence of the Green's function $ \mathcal{L}_v u = -\div (A \nabla u) + vu$ on $\Omega$ for any $v$ real and bounded on $\Omega$? (that is, removing the nonnegative assumption).

The obvious point here being that for $E > 0$ large enough, $v + E$ is bounded and nonnegative on $\Omega$ and so we would know the existence of the Green's function for $ \mathcal{L}_{v, E} u = -\div (A \nabla u) + (v+E)u$ on $\Omega$. Can we use this knowledge to somehow get the Green's function for $\mathcal{L}_v?$

Even just the Schrödinger /fundamental solution case ($\Omega = \mathbb{R}^d$ and $A = I_{d \times d}$ so $\div (A \nabla u) = \Delta u$ ) would be very helpful.

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    $\begingroup$ A Green function means invertibility. But for negative $v$ you will sometimes hit and eigenvalue. Try in dimension 1... So no. $\endgroup$
    – username
    Commented Feb 18, 2022 at 20:18
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    $\begingroup$ Your use of \text{div} instead of \operatorname{div} was the reason for the lack of sufficient space between the minus sign and $\operatorname{div}.$ \text{} does not result in context-dependent spacing. $\endgroup$ Commented Feb 18, 2022 at 22:22
  • $\begingroup$ username, Thanks! Looking at eigenfunctions is trivial but clarifies things and kind of fits with the literature: While in the Schrodinger case, fundamental solutions exists for $v \geq 0$ with $v \in L_{\text{loc}} ^1(\mathbb{R}^n)$ according to S. Mayboroda and B. Poggi, "Exponential decay estimates...", for even bounded but not necessarily nonnegative v, the situation is more subtle and requires other conditions (for example $v \in L^{\frac{n}{2}, \infty} (\mathbb{R}^n)$ suffices, see G. Sakellaris "On scale invariant bounds for...") $\endgroup$ Commented Feb 22, 2022 at 19:11

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