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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.

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.

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 $E_j(L,a)\to E_j(a)$ (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.

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.

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.

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 $E_j(L,a)\to E_j(a)$ (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.