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I have the following problem in hands, in the context of a broader investigation:

Let $V\in L^{n/2}$ compactly supported, where $n\geq 3$ is the dimension. I want to prove the following:

For any $\lambda>0$ (or at least for $\lambda$ big enough) there exists a domain $\Omega$ of class $\mathcal{C}^2$ surrounding the support of $V$ and such that $\lambda$ is not a Neumann eigenvalue for the operator $-\Delta+V$.

I think it makes sense to start by thinking just about just the operator $-\Delta$, as the boundary is away from the support of $V$.

I don't really need to construct this domain, just to prove the existence of it. In the case of Dirichlet eigenvalues, one can do it by using an argument based on the monotonicity of eigenvalues with respect to domain inclusion (Stefanov 1990). However, this monotonicity doesn't hold for Neumann eigenvalues (Freitas and Kennedy, 2023).

Currently I'm thinking of it like this: take a somewhat random domain $\Omega$. If $\lambda$ turns out to be a Neumann eigenvalue in $\Omega$, then construct a sequence of perturbations of the domain and prove somehow that $\lambda$ can't keep being a Neumann eigenvalue for all of them, maybe by some kind of orthogonality argument.

Looking at references (e.g. Arrieta 1991, 1997), I have found works in which authors investigate the convergence of the spectrum on perturbed domains to that of the original domain. What I am looking for is kind of the opposite, I want to change the domain in a way that I can assure that the eigenvalues vary.

If anyone has any clue, reference, insight or idea, even if vague, I would really appreciate it. This is the final piece left to complete a big (at least for me) work that compose my PhD thesis.

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    $\begingroup$ Is $V\geq 0$, or bounded below? If that is the case, what you suggested works well (I can write it as an answer if you like). If not, it might work as well, but it looks a bit more technical at first sight. $\endgroup$
    – username
    Commented Oct 16, 2023 at 18:14
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    $\begingroup$ It's quite annoying that this would be hard to prove since common sense of course suggests that it will be difficult to find a domain for which this is not true. $\endgroup$
    – Christian Remling
    Commented Oct 16, 2023 at 21:59
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    $\begingroup$ I completely agree with you Christian. Maybe the proof will be easy in hindsight, but I haven't been able to put the argument together so far. $\endgroup$
    – Manuel Cañizares
    Commented Oct 16, 2023 at 22:11
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    $\begingroup$ @ChristianRemling I like that idea, I will try it to make it rigurous, thank you! However, I deliberately omitted a detail of the problem in my question to make it easier and less specific to me. This is, I need the boundary of the domain to include a prescribed small open set (open in the $n-1$ dimensional topology). Therefore in principle one can't just consider concentric balls. Maybe the idea can be adapted though. $\endgroup$
    – Manuel Cañizares
    Commented Oct 19, 2023 at 6:29
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    $\begingroup$ I noticed that I assumed something on the distribution of eigenvalues of the laplacian so I asked if it was true. $\endgroup$
    – username
    Commented Oct 24, 2023 at 20:20

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