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.



Existence of non-trivial solutions to Dirichlet problem with a potential lying between eigenvalues.
I assume $\Omega$ is a bounded open subset of $\mathbb{R}^n$ and, say, $V\in L^\infty(\Omega)$. What you say is correct, but of course you need to assume that the eigenvalues are also *consecutive* (otherwise e.g. $V$ itself could be another eigenvalue in between [edit: as you actually said]). The comparison principle you are seeking comes from the Courant–Fischer–Weyl variational characterization of the eigenvalues of $-\Delta + V,$ by which the $k$-th eigenvalue $\lambda\_k(-\Delta + V\\ )$ in the increasing order is expressed as a certain min-max of the Rayleigh quotient, which is monotone wrto $V:$ 
$$\frac{\int\_\Omega \left(|\nabla u|^2+V(x)u^2 \right)dx }{\int\_\Omega u^2 dx }$$
(for a precise statement see e.g. Courant-Hilbert, or Reed-Simon, or Gilbarg-Trudinger, &c).


So if we denote $\lambda\_k:=\lambda\_k(-\Delta)$ and assume $-\lambda\_{k+1} < V(x) < -\lambda\_{k}\\ ,$ it follows by the monotonicity
$$\lambda\_k(-\Delta+V)<\lambda\_k(-\Delta - \lambda\_k)=0 =\lambda\_{k+1}(-\Delta - \lambda\_{k+1})< \lambda\_{k+1}(-\Delta +V),$$
so that $0$ is not an eigenvalue of $-\Delta +V.$


