This is a very straightforward question (although the answer may not be!) about a ubiquitous textbook physics situation. I assume that the answer is probably well-known, but I asked it on both Math and Physics SE, and to my surprise, I got many upvotes but not a single answer.
Consider the following version of the time-independent Schrodinger equation: $$ \left( -\frac{d^2}{dx^2} + V(x) \right) \psi(x) = \lambda\ \psi(x) $$ (where we have absorbed some unimportant physical constants into the function $\psi(x)$). The function $V(x)$ is a given real smooth function $\mathbb{R} \to \mathbb{R}$, which we assume to be continuous and to approach 0 at large arguments: $$ \lim_{|x| \to \infty} V(x) = 0. $$ The smooth complex-valued function $\psi:\mathbb{R} \to \mathbb{C}$ and the real constant $\lambda \in \mathbb{R}$ are to be determined. This equation is simply the eigenvalue equation for the linear second-order differential operator in parentheses. (We can loosen the smoothness requirements on $V(x)$ and $\psi(x)$, if doing so makes the problem more tractable.)
Non-rigorous physical heuristics suggest that these three statements are equivalent:
- $\psi(x)$ is square-integrable, i.e. $$\int \limits_{-\infty}^\infty dx\ |\psi(x)|^2 < \infty,$$
- $\lambda < 0$, and
- $\lambda$ lies in a discrete part of the eigenvalue spectrum of the differential operator in parentheses, i.e. there exists a proper real interval such that $\lambda$ is the only eigenvalue in the differential operator's spectrum that lies within that interval.
A related piece of "folk wisdom" considers the same eigenvalue equation in the case where $$ \lim_{|x| \to \infty} V(x) = +\infty, $$ and claims that in this case, (a) all eigenfunctions $\psi(x)$ must be square-integrable, and (b) the eigenvalue spectrum of the differential operator must be discrete.
But this "folk wisdom" is incorrect. This answer gives an explicit example of a function $V(x)$ and a square-integrable eigenfunction $\psi(x)$ with positive eigenvalue $\lambda$. Therefore, statement #1 above does not imply statement #2. (I do not know whether the spectrum for the particular differential operator given in that example is discrete or continuous around the relevant eigenvalue $\lambda = 1$, so I don't know the status of claim #3 for this example.)
What are the exact implications between the three statements above? Of the six possible implications, which have been proven to be true, which (other than $1 \implies 2$) have explicit known counterexamples, and which are still open problems?
I'd also like to know about the case of multiple spatial dimensions, although I assume that the answers are probably the same as for the 1D case.