0$, there exists a constant $C_p(E_0)$ so that
\[
\sum_{n=0}^{\infty} \left( \int_n^{n+1} |V(x)|\, dx \right)^{2p} \le C_p(E_0) \sum E_n^{p} ,
\]
provided that $E_1\le E_0$.
\end{Theorem}
This is reassuring, but we emphasize again that these inequalities do not really catch
the essence of our method. As outlined above, the behavior of $E_n$ governs the geometry
of the situation, and this part of the information gets lost when we pass to global
bounds as in Theorem~\ref{TILT}. This effect is also responsible for the additional
assumption that $\sup E_n \le E_0$ from part b): The intervals $(n,n+1)$ are not adapted
to the underlying geometry. The need for such a restriction is also apparent from the
fact that part b) is not invariant under the rescaling $V(x)\to g^2V(gx)$, $E\to g^2E$.
The estimates from part a) might be called \textit{inverse Lieb-Thirring inequalities.}
Lieb-Thirring inequalities are bounds of the type of part a) but with the opposite sign.
They hold for $p\ge 1/2$. In particular, for $p=1/2$, we have inequalities in both
directions, so $\int |V|$ and $\sum E_n^{1/2}$ are comparable. This is not a new result;
on the contrary, $p=1/2$ is essentially a sum rule, and the best constant is known
(\cite{GGM}, see also \cite{Schm}).
It is clear that we cannot have inverse Lieb-Thirring inequalities for $p>1/2$ because
$V$ can have local singularities so that $V \notin L_q$ for any $q>1$. It is also
important to work with \textit{Neumann} boundary conditions as there are non-zero
potentials $V \le 0$ with no Dirichlet eigenvalues. The whole-line analog of
Theorem~\ref{TILT} also holds and is perhaps more natural for precisely this reason.
As for the spectral properties, Theorem~\ref{TILT} has the following consequences:
\begin{Corollary}
\label{C1.1} Assume {\rm (}$\Sigma_{{\rm ess}}${\rm )}. Moreover, assume that $V\le 0$.
{\rm a)} If $\sum E_n^{1/2}<\infty$, then the spectrum is purely absolutely continuous on
$(0,\infty)$ for all boundary conditions.
{\rm b)} If $\sum E_n<\infty$, then there is absolutely continuous spectrum essentially
supported by $(0,\infty)$.
{\rm c)} If $\sum E_n^p<\infty$ for some $p<1$, then the solutions satisfy WKB-type
asymptotic formulae for Lebesgue almost all energies $E>0$.
\end{Corollary}
Part a) follows because Theorem~\ref{TILT}a) says that $V\in L_1$. As pointed out above,
this part of the corollary has been known before. Rybkin \cite[Theorem~1]{Ryb} has proved
that the assertion of part b) holds if $V \in \ell_2(L_1)$, that is, if
\[
\sum \left( \int_n^{n+1} |V(x)|\, dx \right)^2 < \infty .
\]
This is the ultimate form of a well-known theorem of Deift and Killip \cite{DeK} which
states that there is absolutely continuous spectrum essentially supported by $(0,\infty)$
if $V\in L_1+L_2$. So part b) of the corollary follows from Theorem~\ref{TILT}b).
The asymptotic formula alluded to in part c) reads
\[
y(x,E) = \exp \left( i\sqrt{E} x - \frac{i}{2\sqrt{E}}\int_0^x V(t)\, dt \right) + o(1)\quad\quad
(x\to\infty) .
\]
Christ and Kiselev \cite{CK} prove that this holds at almost all energies if
$V\in\ell_p(L_1)$ for some $p<2$, so Theorem~\ref{TILT}b) also implies part c) of the
corollary.
As above, these results are complemented by the following:
\begin{Theorem}
\label{TL1opt} Let $e_n>0$ be a non-increasing sequence with $\sum e_n=\infty$. Then
there exists a potential $V\le 0$ so that {\rm (}$\Sigma_{{\rm ess}}${\rm )} holds,
$E_n\le e_n$, and $\sigma_{{\rm ac}}=\emptyset$.
\end{Theorem}
We organize this paper in the obvious way: Section~2 gives a detailed discussion of our
general method. The subsequent sections are concerned with the applications of this to
the spectral theory of $H_{\pm}$, in the order suggested by this introduction. In the
final section, we present the examples announced in Theorems~\ref{Tdimopt} and
\ref{TL1opt}.
\medskip
\noindent\textit{Acknowledgments.} It is a pleasure to thank Rowan Killip and Barry Simon
for useful conversations. C.\ R.\ would like to express his gratitude for the hospitality
of Caltech, where this work was begun.
\section{A Method for Estimating $V$}
As in the previous section, let $H_{\pm}=-d^2/dx^2 \pm V(x)$. We write $H_{\sigma}$ if we
work with one of these operators, but do not want to specify which one. Boundary
conditions (where necessary) will \textit{always} be Dirichlet boundary conditions
($y=0$).
The following theorem may be viewed as the principal result of this paper. Things become
slightly easier in the whole-line setting because we avoid the somewhat artificial
technical problems associated with the effect that a boundary condition can screen part
of the potential. So we discuss this case first. The modifications needed to handle
half-line problems will be described after having completed the treatment of the
whole-line case.
\begin{Theorem}
\label{TWQ} Consider $H_{\pm}$ on $L_2(\mathbb R)$. Assume {\rm (}$\Sigma_{{\rm
ess}}${\rm )}. Then there exist a partition of $\mathbb R$ into intervals $J_n^{(k)}$
with disjoint interiors and a decomposition $V=W'+Q$ with the following properties:
{\rm a)} {\rm (basic properties of $W,Q$)} $W$ is absolutely continuous. If $\sum
E_n^{1/2}<\infty$, then $W\in L_2(\mathbb R)$ and $Q\in L_1(\mathbb R)$.
{\rm b)} {\rm(geometry of the intervals)} The indices $k,n$ vary over the following sets:
$n\in\mathbb N$ and $k\in\mathbb Z$, $-N_n-1 0$, we have two linearly independent solutions for which the limits
$\lim_{x\to\infty} R(x,k)$, $\lim_{x\to\infty} (\psi(x,k)-2kx)$ exist, then it follows
(by taking a suitable linear combination) that $E\notin S$. Thus it suffices to show that
for arbitrary but fixed initial
values $R(0,k)$, $\psi(0,k)$, these limits exist off a set of dimension
at most $4p$.
We will split the proof of this into two parts. This is not really necessary, but
it will help to make the presentation more transparent.
In the first step, we will show that the limits exist on the subsequence $a_n$ as long as we
stay off an exceptional set. In the second step, we will extend this to sequences tending to
infinity arbitrarily. Actually, rather similar arguments are applied in both steps, so the
second step will not be very difficult once we have completed the first step.
So in this first step, we are concerned with the series
\begin{equation}
\label{4.1}
\sum_{n=1}^{\infty} \left| \int_{a_{n-1}}^{a_n} W(x) e^{i\psi(x,k)}\, dx \right| .
\end{equation}
Indeed, real and imaginary parts of the integrals give us the leading terms from the
equations for $R$ and $\psi-2kx$, respectively; see \eqref{E:PRC}, \eqref{E:PTC}. So it
suffices to show that this series converges off an exceptional set of dimension $\le 4p$.
We will need control on the maximal function
\[
M_n(k) \equiv \max_{a_{n-1}\le c\le a_n} \left| \int_{a_{n-1}}^c W(x)e^{2ikx}\, dx \right| .
\]
Let $\mu$ be any (Borel) measure with finite $4p$ energy. Since $\|W\|_{L_2(a_{n-1},a_n)}
\lesssim L_n^{-1/2}$ by \eqref{5.1}, Lemma~\ref{Lcap} says that $\|M_n\|_{L_1(\mu)}
\lesssim L_n^{-2p}$. Now $\sum L_n^{-2p}<\infty$, so the Monotone Convergence Theorem
shows that $M_n(k)\in\ell_1$ for $\mu$ almost every $k$. Since $\mu$ is only assumed to
have finite $4p$ energy but is otherwise arbitrary, it follows that $\dim S_0\le 4p$,
where
\[
S_0 = \left\{ k>0: \sum_{n=1}^{\infty} M_n(k)= \infty \right\} .
\]
This conclusion is nothing but a standard relation between capacities and Hausdorff dimensions; for
example, one can argue as follows: Suppose that, contrary to our claim, $\dim S_0 > 4p$, and fix
$d\in (4p,\dim S_0)$. Then,
since $S_0$ is a Borel set of infinite $d$-dimensional Hausdorff measure, there exists a finite measure
$\mu\not= 0$ supported by $S_0$ with $\mu(I) \le C|I|^d$ for all intervals $I\subset\mathbb R$
\cite[Theorem 5.6]{Falc}. It is easily seen that $\mathcal{E}_{4p}(\mu) <\infty$ for such a $\mu$.
So what we have shown above now says that $M_n(k)\in\ell_1$ for $\mu$ almost every $k$, which clearly
contradicts the fact that $\mu$ is supported by $S_0$.
We now claim, more specifically, that \eqref{4.1} converges if $k\notin S_0$.
Write $\psi = 2kx +\varphi$, and consider one of the integrals
$\int_{a_{n-1}}^{a_n} W(x)e^{i\psi(x,k)}\, dx$ from \eqref{4.1}.
Integration by parts gives
\begin{multline*}
\int_{a_{n-1}}^{a_n} W(x)e^{i\psi(x,k)}\, dx = e^{i\varphi(a_n,k)}\int_{a_{n-1}}^{a_n} W(x)e^{2ikx}\, dx \\
- i \int_{a_{n-1}}^{a_n} dx\, e^{i\varphi(x,k)}(2W(x)\sin\psi(x,k)+\rho(x))
\int_{a_{n-1}}^x dt\, W(t) e^{2ikt} .
\end{multline*}
We have abbreviated the integrable term as $\rho=(1/k)(Q-W^2)(\cos\psi-1)$.
The first term on the right-hand side is summable for $k\notin S_0$, so
we must now show that the series over the second term on the right-hand side
is also absolutely convergent for these $k$.
This, however, is immediate from the bound
\[
\left( 2\|W\|_{L_1(a_{n-1},a_n)} + \|\rho\|_{L_1(a_{n-1},a_n)} \right) M_n(k)
\]
on this term since $\|W\|_{L_1(a_{n-1},a_n)}\lesssim 1$.
So we know now that $\lim_{n\to\infty} R(a_n,k)$, $\lim_{n\to\infty} (\psi(a_n,k)-2ka_n)$
exist for all $k\notin S_0$. As the second step of the proof of Theorem~\ref{Tdim}, we
need to extend this to sequences tending to infinity in an arbitrary way. This suggests
that we look at
\[
\widetilde{M}_n(k)\equiv \max_{a_{n-1} \le c\le a_n} \left| \int_{a_{n-1}}^c W(x)e^{i\psi(x,k)}\, dx
\right| .
\]
We claim that $\widetilde{M}_n(k)\to 0$ if $k\notin S_0$. This will complete the proof of
Theorem~\ref{Tdim} because it will then follow that $S\subset S_0$.
To prove our claim on $\widetilde{M}_n(k)$, we proceed as above and reduce matters to
the corresponding statement on $M_n(k)$, which we know is true. This does not require new ideas.
We just repeat the above computations, but with the upper limit $a_n$ now replaced by
$c=c(k)$. Everything goes through as before, and we have in fact proved the
stronger statement that $\widetilde{M}_n(k)\in \ell_1$ if $k\notin S_0$.\hfill$\Box$
\section{Sign-Definite Potentials}
We prove Theorem~\ref{TILT} here. We use the strategy from Section~2. The treatment
simplifies considerably because there is no need to resort to Lemma~\ref{LWQ}. Indeed, if
$H_+\ge -\epsilon$ on $I$, then
\begin{equation}
\label{6.1} -\int V\varphi^2 \le \epsilon\int\varphi^2 + \int\varphi'^2
\end{equation}
for all test functions $\varphi\in H_1(I)$ that vanish at the finite endpoints of $I$.
Since $V\le 0$ now, this may be used to bound the $L_1$ norm of $V$ over suitable
intervals. Therefore, the proof of Theorem~\ref{TWQ} now produces intervals $J_n^{(k)}$
with the same geometry as before, and $\int_J |V|\lesssim 1/|J|$. In particular, if $\sum
E_n^p<\infty$, then also $\sum |J_n^{(k)}|^{-2p}<\infty$, and this latter sum may be
estimated by a multiple of the first sum. The constant only depends on $p$. This follows
as usual from $|J_n^{(k)}|\gtrsim 2^{|k|}E_n^{-1/2}$ by first summing over $k$ and then
over $n$.
Theorem~\ref{TILT}a) for the whole-line problem is now immediate from H\"older's
inequality, which gives, for $0 0$, the operator $-d^2/dx^2 + V_g(x)$ on
$L_2(\mathbb R)$ has precisely one eigenvalue $-E$, and $E=E(g)=g^2+O(g^3)$.
{\rm b)} For small $g>0$, the operator $-d^2/dx^2 + W_g(x)$ on $L_2(\mathbb R)$ has
precisely one eigenvalue $-E$, and $E=E(g)=g^4/9+O(g^5)$.
\end{Lemma}
\begin{proof}[Sketch of the proof.]
It is clear that $-d^2/dx^2+V_g(x)$ has at most one eigenvalue for small $g>0$. This
follows from an elementary analysis of the solutions at zero energy (the number of zeros
on $(-N,\infty)$ of a solution $y$ with $y(-N)=0$ is the number of negative eigenvalues
of the operator on that interval). Since $W_g\ge V_g$, $-d^2/dx^2+W_g(x)$ has at most one
eigenvalue for small $g$. That the operators have at least one eigenvalue for $g>0$ is a
classical result for sign-definite potentials \cite[Theorem XIII.11]{RS4} and follows
from recent work \cite{dhks,dks} (and, incidentally, also from our Lemma~\ref{LWQ}) for
arbitrary potentials.
To approximately compute this eigenvalue $-E$, we note that the corresponding eigenfunction
$y$ must be a multiple of $e^{-E^{1/2}|x|}$ for $x\le -1$ and $x\ge 1$; the constant factors
may be different on these two half lines. So $-E$ is an eigenvalue precisely if there
exists $c\in\mathbb R$ so that
\begin{equation}
\label{9.1}
c \begin{pmatrix} 1 \\ -E^{1/2} \end{pmatrix} = T_g(1,-1;-E) \begin{pmatrix}
1 \\ E^{1/2} \end{pmatrix} .
\end{equation}
Here, $T_g$ is the transfer matrix, that is, the matrix that takes solution vectors
$(y,y')^t$ at $x=-1$ to their value at $x=1$. We of course have explicit formulae
for the transfer matrices for $V_g$ and
$W_g$, respectively, and a somewhat cumbersome but completely elementary discussion
of \eqref{9.1} then establishes the asserted asymptotics of $E$.
\end{proof}
Now consider a (half-line) potential $V$ of the form
\begin{equation}
\label{9.2}
V(x) = \sum_{n=1}^{\infty} V_{g_n}(x-x_n),
\end{equation}
with $g_n\to 0$. The $x_n$'s are typically very rapidly increasing so that the individual
bumps are well separated and thus almost independent of one another. To rigorously
analyze $V\!$, we build it up successively. The following lemma describes this situation.
\begin{Lemma}
\label{L9.2} Consider $H_a=-d^2/dx^2+Q(x)+V_g(x-a)$, where $Q$ has compact support and
$g$ is sufficiently small so that Lemma~\ref{Lsparse} applies. Suppose that
$-d^2/dx^2+Q(x)$ has precisely $N$ negative eigenvalues
$-\widetilde{E}_1,\ldots,-\widetilde{E}_N$ on $L_2(0,\infty)$. Then, for any
$\epsilon>0$, there exists $a_0$ so that for all $a\ge a_0$, the following holds: $H_a$
on $L_2(0,\infty)$ has precisely $N+1$ negative eigenvalues $-E_1,\ldots,-E_{N+1}$, and
\[
\left| E_i -\widetilde{E}_i\right|<\epsilon,\quad \left| E_{N+1} - E(g)\right|
<\epsilon .
\]
An analogous statement holds for $H_a$ on $L_2(0,2a)$.
\end{Lemma}
In other words, there is almost no interaction between $Q$ and $V_g$ and the
eigenvalues approximately behave like those of
the orthogonal sum of $-d^2/dx^2+Q(x)$ and $-d^2/dx^2+V_g(x)$.
\begin{proof}
Let $y$ be the solution of $-y''+(Q+V_g)y=0$ with $y(0)=0$, $y'(0)=1$. By taking $a$
large enough, we can make sure that $y$ has precisely $N$ zeros on $(0,a/2)$ (say) and
$y(a/2)y'(a/2)\ge 0$, $y(a/2)\not= 0$ (if $y$, $y'$ had different signs, there would be
another zero on $(a/2,\infty)$ for zero potential). Here we again use oscillation theory;
more precisely, the number of positive zeros of $y$ is exactly the number of negative
eigenvalues. An elementary discussion now shows that $y$ has exactly one more zero on
$(a/2,\infty)$ if $a$ is large enough. Thus $H_a$ has precisely $N+1$ negative
eigenvalues. Since this additional zero in fact lies in $(a+1,2a)$, this also holds for
$H_a$ on $L_2(0,2a)$.
It is now easy to approximately locate these eigenvalues: Cut off the eigenfunctions of
$-d^2/dx^2+Q(x)$ (on $L_2(0,\infty)$) and of $-d^2/dx^2+V_g(x-a)$ (on $L_2(\mathbb R)$)
by multiplying by suitable smooth functions which are equal to one except in a
neighborhood of, say, $a/2$ (and $2a$ for this latter eigenfunction). If $a$ is large
enough, we thus obtain functions $\varphi_i$ with
\[
\|(H_a+\widetilde{E}_i)\varphi_i\|
<\epsilon\|\varphi_i\| \quad (i=1,\ldots,N),\quad\quad \|(H_a+E(g))\varphi_{N+1}\|<\epsilon
\|\varphi_{N+1}\| .
\]
This completes the proof of the lemma.
\end{proof}
We now claim that given any sequence $\epsilon_n>0$, we can find $x_n$'s in \eqref{9.2},
so that $-d^2/dx^2+V(x)$ has eigenvalues $-E_n$, satisfying $|E_n-E(g_n)|<\epsilon_n$,
and no other negative eigenvalues. To prove this, we may assume that $\epsilon_n$
decreases. To simplify the book keeping, we in fact further assume that the intervals
$(E(g_n)-\epsilon_n,E(g_n)+\epsilon_n)$ are disjoint. This is to say, we assume that the
$g_n$'s are distinct and then further decrease the $\epsilon_n$'s if necessary.
Now apply Lemma~\ref{L9.2} with $Q=0$, $g=g_1$, and $\epsilon= 2^{-2}\epsilon_1$ to find
$x_1$. In the second step, apply the lemma with $Q=V_{g_1}(x-x_1)$, $g=g_2$, and
$\epsilon=2^{-3}\epsilon_2$ (which is also $\le 2^{-3}\epsilon_1$ by our assumption).
This gives $x_2$; we may also demand that $x_2-1>2x_1$. Continue in this way. The
construction ensures that $-d^2/dx^2+V(x)$ on $L_2(0,2x_n)$ has precisely $n$ negative
eigenvalues $E_i^{(n)}$, and these satisfy $|E_i^{(n)}-E(g_i)|\le\epsilon_i/2$. Moreover,
for fixed $i$, $E_i^{(n)}$ is a Cauchy sequence and hence convergent. Since every
eigenvalue of the half-line problem is an accumulation point of eigenvalues of the
problems on $(0,2x_n)$, only the limits $E_i\equiv\lim_{n\to\infty} E_i^{(n)}$ can be
eigenvalues of $-d^2/dx^2+V(x)$. On the other hand, the half-line problem has at least as
many eigenvalues $<-E$ as the corresponding problem on $(0,2x_n)$; thus, since the
$E_i$'s are from the disjoint intervals $(E(g_i)-\epsilon_i,E(g_i)+\epsilon_i)$ and the
problems on $(0,2x_n)$ have spectrum in these intervals for large $n$ by construction and
Lemma~\ref{L9.2}, the negative spectrum is precisely the set $\{ E_i\}$.
Let us now prove Theorem~\ref{TL1opt}. Given $e_n>0$ with $e_n\to 0$, $\sum e_n=\infty$,
pick $g_n>0$ so that $E(g_n)=e_n/2$, say. By slightly decreasing the $e_n$'s if
necessary, we may of course assume that the $e_n$'s and hence also the $g_n$'s are
distinct. By Lemma~\ref{Lsparse}a), the $g_n$'s will satisfy $g_n\sim (e_n/2)^{1/2}$ in
the sense that the ratio tends to one. Now choose $x_n$'s as above so that the operator
$-d^2/dx^2+V(x)$ with $V$ as in \eqref{9.2} has eigenvalues $E_n \le e_n$. Since $V\le
0$, $H_-=-d^2/dx^2-V(x)$ has no negative spectrum. When choosing the $x_n$'s, we can
further require that $x_n/x_{n+1}\to 0$. Since also $\sum g_n^2=\infty$, a well-known
result on sparse potentials applies and the spectrum is purely singular continuous on
$(0,\infty)$ (see \cite[Theorem 1.6(2)]{kls}).
The proof of Theorem~\ref{Tdimopt} uses $W_g$ and \cite[Theorem 4.2b)]{Remtams} instead
of $V_g$ and \cite[Theorem 1.6(2)]{kls}, respectively, but is otherwise analogous. First
of all, the analog of Lemma~\ref{L9.2} holds, with a similar proof. Let $e_n>0$ with
$\sum e_n^{1/4}=\infty$ be given. If $e_n$ is non-increasing, we also have that $\sum
e_{2n}^{1/4}=\infty$. Determine $g_n$'s so that $E(g_n)=e_{2n}/9$, where $E(g)$ now
refers to $W_g$. Lemma~\ref{Lsparse}b) shows that then $g_n\sim e_{2n}^{1/4}$. Define $V$
as in \eqref{9.2}, but with $W_g$ instead of $V_g$. We can then find $x_n$'s so that
$-d^2/dx^2+V(x)$ has eigenvalues $E_n\le e_{2n}$ in this case as well, following the same
arguments as above. Moreover, since $-d^2/dx^2-W_g(x)$ on $L_2(\mathbb R)$ has the same
eigenvalue $-E(g)$ as $-d^2/dx^2+W_g$, we can also arrange that $-d^2/dx^2-V(x)$ has
eigenvalues $E'_n \le e_{2n}$. Thus, after combining the eigenvalues of $H_{\pm}$ in one
sequence $E_n$, we have that $E_n\le e_n$, as desired. Finally, we can again require that
$x_n/x_{n+1}\to 0$. Since $\sum g_n=\infty$, Theorem 4.2b) from \cite{Remtams} applies
and shows that $\dim S=1$.
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