I'm reading parts of Reed and Simon's Analysis of Operators and have come across a statement I find puzzling. They say that if $V$ is a bounded function of compact support on $\mathbb{R}^3$ then $-\Delta + V$ has only finitely many eigenvalues (discrete spectrum) in $(-\infty, -1]$.
Things I know:
- The Min-Max Principle
- $\sigma_{ess}(-\Delta) = [0,\infty)$, and for (relatively) compact $V$ then $\sigma_{ess}(-\Delta + V) = \sigma_{ess}(-\Delta)$.
- If $V \in R + (L^\infty)_\epsilon$ on $\mathbb{R}^3$ (where $R$ is the Rollnik class) then $\sigma_{ess}(-\Delta + V) = \sigma_{ess}(-\Delta)$.
My Thoughts:
It seems that $V \in L^\infty(\mathbb{R}^3)$ since it is globally bounded. At the same time by the Min-Max principle we must have $\mu_n(H)$ is either an eigenvalue below the bottom of the essential spectrum, or is the bottom of the essential spectrum for each $n$ (where $H = -\Delta + V$). Part of me wants to argue that in this instance we are guaranteed that for some $n, \; \mu_n(H) = \inf \sigma_{ess}(H) = 0$, but this may not be so. What if from the Min-Max principle we have that there are infinitely many eigenvalues counting multiplicity? Are we guaranteed to hit the bottom of the essential spectrum at some point using the Min-Max algorithm?
I feel like I'm asking a silly question, tell me if it is!