I'm consideirng the example of
$-\Delta u + V(x) u = 0$ in $\Omega$ with $u = 0$ on $\partial \Omega$. I'm trying to see if it's true that if $-\lambda_1 < V(x) < -\lambda_2 < 0$ on $\overline{\Omega}$ that we *do not* have existence of non-trivial solutions to this equation. Here $\lambda_1$ and $\lambda_2$ are two distinct eigenvalues of $-\Delta u$, both of which are of course positive.

In the special case that $V$ is constant, this is certainly true since otherwise $V$ would itself be an eigenvalue. I have tried to see if I can use a sort of comparison principle to sohw that $u \equiv 0$ if $-\Delta u + V(x)u = 0$ in the case that $V$ is not constant but I can't seem to establish this.

For simplicity of course assume that $V$ is as smooth and as bounded as you'd like.