Let $V:R^d\to R_+$ be with a compact support. The Schrödinger operator $H_a=-\Delta - a V$ acting in $L^2(R^d)$ has then (at most) finitely many negative eigenvalues. Denote the number of negative eigenvalues by $N(a)$, then one has clearly $N(a)\to+\infty$ for $a\to+\infty$, and there exists a sequence of critical coupling constants $(a_n)$ at which $a\mapsto N(a)$ has jumps.

My question is as follows:

Assume that $a=b$ is *not* critical, i.e. that $a\mapsto N(a)$ is constant for $a\in(b-\epsilon,b+\epsilon)$. Does it follow that the equation $(-\Delta-b V)u=0$ has no non-trivial bounded solutions? Maybe the "boundedness" should be understood is a suitable sense ("not growing very fast" or "suitably decaying decaying at infinity") depending on the dimension $d$.

Informally, the question is: how to characterize the critical values of $a$ in terms of the solutions of $(-\Delta-a V)u=0$. I am mostly interested in the case $d=2$.

I suspect that the answer should be well known but I do not manage to find a text saying it explicitly.

**Update.** For $d=1$ the answer by Christian Remling seems to be OK.

My impression that I have another method which works for $d=1,2$, at least if the potential is not very bad (see below).

Denote $h_a(u,u)=\int |\nabla u|^2 - a \int V u^2$ the quadratic form for $H_a$. Assume that for some $b$ we have two properties:

1) there are exactly $n$ negative eigenvalues (to be denoted by $\lambda_j$ with $u_j$ the respective eigenfunctions, which are supposed orthonormalized)

2) there is a non-zero bounded solution $u$ to $-\Delta u-bVu=0$.

We want to show that for every $a>b$ the operator $H_a$ has at least $n+1$ eigenvalues. By the min-max principle, it is sufficient to find $n+1$ functions $v_j$ such that the matrix $A_a=h_a(v_j,v_k)$ is negative definite.

We take $v_j=u_j$, $j=1,\dots,n$, and $v_{n+1}= \varphi u$, where $\varphi$ is a cut-off function whose support tends to the whole space, to be determined later. Then the matric $A_a$ takes the form $$ A_a=\begin{pmatrix} \lambda_1 & 0 &\dots & 2 \int u\nabla\varphi\nabla u_1+\int \Delta\varphi u u_1\\ 0 & \lambda_2 &\dots & 2 \int u\nabla\varphi\nabla u_2+\int \Delta\varphi u u_2\\ \dots\\ \dots&&& \int u^2 |\nabla \varphi|^2 \end{pmatrix} - (a-b) (\int V u_j u_k) $$

The functions $u_j$ are linearly idependent (in fact , $u_j$ and $u$ are linearly independent at any open set) and if the support of $\varphi$ covers that of $V$, and if $V$ is strictly positive on an open set, then the second matrix (which is substracted) is postivie definite (at it is a Gram matrix). Therefore, it is sufficient to find a cut-off function such that all $\varphi$-terms become very small, but this is quite standard:

For $d=2$: one takes a smooth $\chi$ on $R$ with $\chi(t)=1$ for $t\le 1$ and $\chi(t)=0$ for $t\ge 2$, and sets $\varphi(x)=\chi(\varepsilon \ln |x|)$ and sends $\varepsilon$ to $0$.

For $d=1$ one can take simply $\varphi(x)=\chi(\varepsilon |x|)$.