Bounded solutions for Schrödinger equation at the edge of the essential spectrum 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 $\varphi(x)=\chi(\varepsilon |x|)$.
 A: Update: To elaborate some on my discussion with Willie in the comments, I think I can now do this in $d=1$, and this is perhaps more interesting than the original answer.

$-u''-aVu=0$ has a bounded solution if and only if $N(a+)-N(a-)=1$.

To prove this, notice that a solution $u$ will be bounded precisely if $u'(-L)=u'(L)=0$, and here I can take any $L>0$ that is large enough so that $(-L,L)$ contains the support of $V$.
In other words, we obtain a bounded solution if and only if $0$ is an eigenvalue of $-d^2/dx^2-aV$ on $[-L,L]$ with Neumann boundary conditions. As we increase $a$, the Neumann eigenvalues $E_j(L,a)$ decrease strictly, so we already almost have what we want: there is a bounded solution if and only if $N_L(a)$ jumps at $a$, and here $N_L$ now counts the Neumann eigenvalues of the problem on $[-L,L]$.
To finish the proof, we need to show that $E_j(L,a)\to E_j(a)$ as $L\to\infty$ (and thus also $N_L\to N$). This is routine: we have that $E_j(L)\le E_j$, by min-max for the quadratic forms. To show that $E_j- E_j(L)\to 0$, we can also use min-max (observe that a normed solution with zero derivatives at $\pm L$ must be close to the decaying exponential outside the support of $V$, so a Neumann eigenfunction is a good test function for the whole line problem also after making it smooth near $\pm L$).

Original answer:
No. Let me do this in dimension $d=3$. Then $N(a)=0$ for all small $a>0$; see here. If $V$ is also spherically symmetric, then we can separate variables. A solution of the form $u=u(r)$ will satisfy $-(1/r^2)(r^2u')'+Vu=0$, and if we introduce $ru=y$, then this becomes $-y''+V(r)y=0$ (and now we need the solution with $y(0)=0$).
Once we're beyond the support of $V$, the general solution is $y=a+br$, so $u=y/r$ is bounded automatically.
