Can 0 be an eigenvalue?

Question

Let $-\Delta $ be the positive Laplacian and consider the operator $$ -\Delta + V $$ on $L^2(\mathbb{R}^3)$ with domain the Sobolev space $W^{2,2}(\mathbb{R}^3)$. Here $V:\mathbb{R}^3\to \mathbb{R}$ is smooth, has compact support and acts by multiplication, $(Vf)(x)=V(x)f(x)$. Is it possible for $0$ to be an eigenvalue? That is, is it possible (for some choice of $V$), to find $f\in L^2(\mathbb{R}^3)$ such that $$ -\Delta f + Vf = 0 \qquad ? $$

Answer 1

Let me put an incomplete idea here in the hope that others might be able to improve on it.

Let $q\geq0$ be a compactly supported smooth function in $\mathbb{R}^3$, and let $u$ be its Newtonian potential, i.e., $$ \Delta u = q. $$ We know that $u$ is smooth, entirely subharmonic, and harmonic outside the support of $q$. We also know that $u$ decays at infinity, so by the maximum principle, $u$ is nonpositive. Moreover, because of the strong maximum principle, $u$ cannot attain the value $0$, hence $u<0$. Now we let $Q=q/u$, which is well defined because $u$ does not vanish anywhere. With this, we have $$ - \Delta u + Qu = 0. $$ The catch is that we cannot guarantee $u\in W^{2,2}(\mathbb{R}^3)$. In general all we can say is $u$ decays like $1/|x|$, which is not enough for $u\in L^2(\mathbb{R}^3)$. To get more decay, one must require $\int q=0$, and then we cannot guarantee $u\neq0$ everywhere. This makes the division $q/u$ problematic.