Must solutions to the time-independent Schrodinger equation that have discrete or negative eigenvalues be square-integrable?

Question

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 smooth function $\mathbb{R} \to \mathbb{R}$, which we assume 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:

1. $\psi(x)$ is square-integrable, i.e. $$\int \limits_{-\infty}^\infty dx\ |\psi(x)|^2 < \infty,$$
2. $\lambda < 0$, and
3. $\lambda$ lies in a discrete part of the set of eigenvalues of the differential operator in parentheses, i.e. there exists a proper real interval such that $\lambda$ is the only eigenvalue of the differential operator 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 set of eigenvalues 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 eigenvalues for the particular differential operator given in that example are 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.

Answer 1

(1) is perhaps better rephrased as: (1') $\lambda\in\sigma_p$ (in words: $\lambda$ is an eigenvalue) because then both (1') and (3) are statements about spectral properties. Also, in more compact notation, we can rewrite: (3) $\lambda\in\sigma_d$

As you pointed out, (1') $\implies$ (2) is false, and so is (1') $\implies$ (3), for the same reason. (2) $\implies$ (1') and (2) $\implies$ (3) are manifestly absurd.

(3) $\implies$ (2) is true because $V\to 0$ implies that $\sigma_{\textrm{ess}}=[0,\infty)$. (3) $\implies$ (1') is obvious since $\sigma_d\subseteq\sigma_p$.

Finally, indeed nothing changes in higher dimensions. A radially symmetric version of the von Neumann-Wigner potential still works as a counterexample to (1') $\implies\ldots$.