0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}(J\,\overline{\rho^{-1}}/g^{1/2}+U_{1})+2\mathbf{p}\cdot\big{(}(J\,\overline{\rho^{-1}}/g^{1/2}+U_{1})\mathbf{q}\big{)}_{x}$
$\displaystyle\kern
20.00003pt+(\tilde{e}_{1}+U_{1}^{2}+2J\,\overline{\rho^{-1}}/g^{1/2}U_{1})\|\mathbf{q}\|^{2}_{x}+2(\mathbf{p}+U_{1}\mathbf{q}+J\,\overline{\rho^{-1}}/g^{1/2}\mathbf{q})\cdot\mathbf{p}_{x}=0\,.$
(A.3)

As a direct consequence of equation (3.0d) we obtain

$\mathbf{p}_{x}={\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}(U_{1}+J\,\overline{\rho^{-1}}/g^{1/2})-\big{(}(U_{1}+J\,\overline{\rho^{-1}}/g^{1/2})\mathbf{q}\big{)}_{x}\,.$
(A.4)

Using the above relation for $\mathbf{p}_{x}$ and eliminating $(U_{1})_{x}$
with the help of (A.1), one can obtain the simplified second wave conservation
equation (7.3) in terms of convective derivatives, where

$\displaystyle
W_{2}={\textstyle\frac{1}{2}}\big{[}(\tilde{e}_{1}+U_{1}^{2})\|\mathbf{q}\|_{x}+2U_{1}\mathbf{q}\cdot\big{(}2(\mathbf{U}_{\flat})_{x}-{\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}U_{1}+J\,\overline{\rho^{-1}}/g^{1/2}\mathbf{q}_{x}-\mathbf{D}_{\flat}S\big{)}$
$\displaystyle\kern
20.00003pt-\mathbf{q}\cdot\mathbf{D}_{\flat}(\tilde{e}_{1}+U_{1}^{2})-(U_{1}^{2}\|\mathbf{q}\|^{2})_{x}\Big{]}.$
(A.5)

Equation (A.4) yields

$J\,\overline{\rho^{-1}}/g^{1/2}\mathbf{q}_{x}-\mathbf{D}_{\flat}\big{(}J\,\overline{\rho^{-1}}/g^{1/2}\big{)}=\mathbf{D}_{\flat}U_{1}-\mathbf{p}_{x}-U_{1}\mathbf{q}_{x}\,.$
(A.6)

Using the above relation along with equations (A.0a), (A.0b), (7.0b), (7.0a)
and some tedious but straightforward algebra, one obtains (7.0b).

To express the averaged mass equation (7.11) in terms of convective
derivatives, first we replace $\mathbf{M}$ using (7.9) and obtain:

$D_{t}(\overline{\rho})+2\overline{\rho}\big{(}(U_{1})_{x}+{\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}\cdot\mathbf{U}_{\flat}\big{)}+2\big{(}(g\tilde{J})_{x}+{\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}\cdot(g\tilde{J}\mathbf{q})\big{)}=0\,,$
(A.7)

Simplifying further we obtain

$D_{t}(\overline{\rho})+2\overline{\rho}D_{x}U_{1}+2D_{x}(g\tilde{J})+2gM_{1}({\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}\cdot\mathbf{q})+2\overline{\rho}({\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}\cdot\mathbf{p})=0\,.$
(A.8)

To rewrite (A.8) in simpler form, we consider the combination (A.8)
$-~{}g(\overline{\rho}/g)$(7.2), which yields (7.1).

Finally, we consider the first component of the averaged momentum equation
(7.0a), using a similar approach as before one can rewrite it as follows:

$\displaystyle
D_{t}(gM_{1})+2g(M_{1}+\tilde{J})D_{x}U_{1}+2U_{1}D_{x}(g\tilde{J})+D_{x}(g\tilde{e}_{2})-\mathbf{q}\cdot\mathbf{D}_{\flat}(g\,(\overline{\rho}^{2}/g^{2}))$
$\displaystyle\kern
70.0001pt+2(\overline{\rho}^{2}/g^{2})g(\mathbf{q}\cdot\mathbf{q}_{x})+g\big{(}\tilde{e}_{2}-(\overline{\rho}^{2}/g^{2})+2U_{1}(M_{1}+\tilde{J})\big{)}{\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}\cdot\mathbf{q}+2M_{1}g{\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}\cdot\mathbf{p}=0\,.$
(A.9)

Next, taking the combination (A.3) $-~{}U_{1}$(A.8) yields

$\displaystyle
g(\overline{\rho}/g)D_{t}U_{1}+D_{t}(g\tilde{J})+4g\tilde{J}D_{x}U_{1}+D_{x}(g\tilde{e}_{2})-\mathbf{q}\cdot\mathbf{D}_{\flat}(g(\overline{\rho}^{2}/g^{2}))$
$\displaystyle\kern
90.00014pt+2g(\overline{\rho}^{2}/g^{2})(\mathbf{q}\cdot\mathbf{q}_{x})+g(\tilde{e}_{2}-(\overline{\rho}^{2}/g^{2})-2U_{1}\tilde{J})+2g\tilde{J}{\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}\cdot\mathbf{p}=0\,.$
(A.10)

To simplify this equation more we consider the combination
(7.0b)$-$(7.0a)$\mathbf{q}-$(7.11)$\mathbf{p}$ and obtain the following vector
equation:

$\displaystyle
M_{1}D_{t}\mathbf{q}+(\overline{\rho}/g)D_{t}\mathbf{p}+(2U_{1}\tilde{J}+\tilde{e}_{2})D_{x}\mathbf{q}+2\tilde{J}D_{x}\mathbf{p}$
$\displaystyle\kern
100.00015pt-(\overline{\rho}^{2}/g^{2})D_{x}\mathbf{q}+g\mathbf{D}_{\flat}((\overline{\rho}^{2}/g^{2}))+2(\overline{\rho}^{2}/g^{2})\mathbf{D}_{\flat}g=0\,.$
(A.11)

Finally, we consider the combination
(A.3)$-2g\tilde{J}\,$(7.2)$+\mathbf{q}\cdot\,$(A.3). Using (7.0a) and after
extensive simplifications, we obtain (7.2).

Next we show that, using the transformation to Riemann-type variables (7.1)
(7.2), (7.3), (7.1) and (7.2) yield (7.4). Note first that, using (7.1), we
have the following identities:

$\displaystyle U_{1}={\textstyle\frac{1}{2}}s_{1}\,,$ (A.0a)
$\displaystyle\tilde{e}_{1}=s_{2}-{\textstyle\frac{1}{4}}s_{1}^{2}\,,$ (A.0b)
$\displaystyle\tilde{e}_{2}=s_{4}+{\textstyle\frac{1}{16}}\,[-16s_{3}s_{1}-4s_{2}^{2}+8s_{1}^{2}s_{2}-s_{1}^{4}]\,,$
(A.0c)
$\displaystyle\tilde{J}={\textstyle\frac{1}{3}}s_{3}-{\textstyle\frac{1}{24}}\,s_{1}(6s_{2}-s_{1}^{2})\,,$
(A.0d)

[where again the $s_{n}$ are as in (5.1)], which allow us to express
$\tilde{e}_{1},\dots,\tilde{e}_{3}$ in terms of the Riemann invariants via
$s_{1},\dots,s_{4}$. The identity (A.0d) is especially important, since it
allows us to eliminate square roots from the modulation equations. Recall that
(2.12) only determines $J^{2}$, and
$J=\sigma(\lambda_{1}\lambda_{2}\lambda_{3})^{1/2}$, with $\sigma=\pm 1$. On
the other hand, (7.1) yields
$\sigma\lambda_{1}^{1/2}={\textstyle\frac{1}{2}}\,\sqrt{g}\,(r_{1}-r_{2}-r_{3}+r_{4})$,
where the sign $\sigma$ here is needed because but one needs
$\lambda_{3}\geqslant\lambda_{2}\geqslant\lambda_{1}\geqslant 0$ (cf. section
2.2), but $r_{1}-r_{2}-r_{3}+r_{4}$ can be either positive or negative
depending on the relative magnitude of $r_{1},\dots,r_{4}$. (In contrast, no
ambiguity arises for $\lambda_{2}^{1/2}$ and $\lambda_{3}^{1/2}$ when
$r_{1},\dots,r_{4}$ are well-ordered.) One can verify that, with this choices,
the sign of both the left-hand side and right-hand side of (A.0d) equal
$\sigma$. The following formulae are also useful:

$\displaystyle k_{1}=\frac{\sqrt{(r_{4}-r_{2})(r_{3}-r_{1})}}{2K_{m}}\,,$
(A.0a) $\displaystyle\bigg{(}{\frac{\partial k_{1}}{\partial
r_{1}}},\dots,{\frac{\partial k_{1}}{\partial
r_{4}}}\bigg{)}^{T}=\frac{\sqrt{(r_{4}-r_{2})(r_{3}-r_{1})}}{4K_{m}^{2}}\,\bigg{(}\frac{(r_{1}-r_{4})K_{m}+(r_{4}-r_{2})E_{m}}{(r_{2}-r_{1})(r_{4}-r_{1})},$
$\displaystyle\frac{(r_{3}-r_{2})K_{m}+(r_{1}-r_{3})E_{m}}{(r_{2}-r_{1})(r_{3}-r_{2})},\frac{(r_{3}-r_{2})K_{m}+(r_{2}-r_{4})E_{m}}{(r_{3}-r_{2})(r_{3}-r_{4})},\frac{(r_{4}-r_{1})K_{m}+(r_{1}-r_{3})E_{m}}{(r_{4}-r_{1})(r_{4}-r_{3})}\bigg{)}^{T}\,,$
(A.0b)

We are now ready to present the final steps of the derivation. We begin by
deriving (7.2), which is the simplest of the four equations. In this case we
simply need to express $D_{t}k_{1}$ in terms of the Riemann invariants, i.e.,

$D_{t}k=\sum_{j=1}^{4}\frac{\partial k_{1}}{\partial r_{j}}\,D_{t}r_{j}\,,$
(A.1)

which immediately yields (7.0a), with $W_{1}$ as in (7.0a). Next, equation
(7.3) simplifies due to the identity

$D_{t}(U_{1}+J\,\overline{\rho^{-1}}/g^{1/2})+(U_{1}-J\,\overline{\rho^{-1}}/g^{1/2})\,\frac{D_{t}k_{1}}{k_{1}}=\frac{2}{k_{1}}\sum_{j=1}^{4}r_{j}{\frac{\partial
k_{1}}{\partial r_{j}}}D_{t}r_{j}\,,$ (A.2)

and takes the form of (7.0b), with $W_{2}$ as in (7.0b). Next, taking the
combination (7.1)/2 + $gU_{1}/2\times$(7.3) +
$g(s_{2}-2U_{1}^{2})/4\times$(7.2) and using identities (A.3) and (A.2),
yields (7.0c), where $W_{3}$ is as in (7.0a). Finally, considering the linear
combination (7.2)/2 + $3U_{1}/2\times$(7.1) $g(s_{2}+2U_{1}^{2})/4\times$(7.3)
+ $g(3\tilde{J}+U_{1}s_{2}-2U_{1}^{3})/2\times$(7.2), and using identities
(A.3), (A.2) again, and after some tedious algebra, one finds (7.0d), with
$W_{4}$ as in (7.0b).

Our last task is to show that the compatibility relations
${\kern-0.83337pt\lower 0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower
0.20004pt\hbox{\kern 0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}\times\mathbf{k}={\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}\times\bar{\mathbf{u}}=\mathbf{0}$,
when written in terms of the Riemann-type variables
$\mathbf{r}=(r_{1},\dots,r_{4})^{T}$ as well as $\mathbf{q}$ and $\mathbf{p}$,
yield (7.1). To this end, we first use the definition of $\mathbf{q}$ as in
(7.0b) along with the compatibility condition ${\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}\times\mathbf{k}=0$. It can be
easily seen that

$k_{1}\mathbf{q}_{x}=(\mathbf{k}_{\flat})_{x}-k_{1,x}\,\mathbf{q}={\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}_{\flat}k_{1}-k_{1,x}\,\mathbf{q}=\mathbf{D}_{\flat}k_{1}\,$
(A.3)

(cf. the third equation in (7.2)), which yields the first half of (7.1). Next,
using the compatibility condition ${\kern-0.83337pt\lower
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\lower 0.20004pt\hbox{\kern
0.50003pt\hbox{$\nabla$}}\kern-8.33336pt\raise
0.20004pt\hbox{$\nabla$}\kern-8.33336pt\raise
0.20004pt\hbox{\kern-0.50003pt\hbox{$\nabla$}}}\times\bar{\mathbf{u}}=0$ with
the definition of $\bar{u}_{1}$ as in (7.0c) one can derive (7.5), namely,

${\mathbf{p}}_{x}=\mathbf{D}_{\flat}\big{(}U_{1}+J\,\overline{\rho^{-1}}/g^{1/2}\big{)}+\big{(}U_{1}-J\,\overline{\rho^{-1}}/g^{1/2}\big{)}\frac{\mathbf{D}_{\flat}k_{1}}{k_{1}}-2U_{1}\frac{\mathbf{D}_{\flat}k_{1}}{k_{1}}\,.$
(A.4)

Using the identity (A.2), one then obtains the second half of (7.1).

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# Positive eigenvalues of non-local Schrödinger operators with sign-changing
potentials

S. Molchanov 111Dept of Mathematics and Statistics, UNCC, NC 28223,
<EMAIL_ADDRESS>B. Vainberg 222Dept of Mathematics and Statistics, UNCC, NC
28223<EMAIL_ADDRESS>
###### Abstract

Simon’s results on the positive spectrum of recurrent Schrödinger operators
are extended to a wider class of operators and potentials. Some
counterexamples are provided for transient operators whose potential does not
satisfy the CLR estimate.

MSC2020: 35J10, 35P99, 35Q99, 47A10, 35R11

##  1 Introduction

Let

$H=\Delta+\sigma V(x),\quad x\in\mathbb{R}^{d},$ (1)

be a Schrödinger operator in $L^{2}(\mathbb{R}^{d})$ with a real-valued
potential $V$. If $V$ decays at infinity in an appropriate sense, then
$Sp_{ess}=(-\infty,0]$ and the positive spectrum is discrete. If $V(x)\geq 0$,
then, under some conditions on $V(x)$ (to be discussed later), the positive
eigenvalues exist for arbitrary $\sigma>0$ when the dimension $d$ is $1$ or
$2$, see [9, 1, 6]. The situation is absolutely different in higher
dimensions. If $d\geq 3$ and

$0<\int_{x:~{}V(x)>0}V(x)^{d/2}dx<\infty,$ (2)

then there is a critical value $\sigma_{cr}>0$ such that the operator $H$ does
not have positive spectrum when $\sigma\leq\sigma_{cr}$, and positive
eigenvalues exist (and their number $N(\sigma V)$ is finite) when
$\sigma>\sigma_{cr}$ [9]. The latter result is based on the CLR estimate [2,
4, 8, 5]. The same conditions ($d\geq 3$ and (2)) are needed for the validity
of quasi-classical asymptotics of $N(\sigma V)$ as $\sigma\to\infty$, [7].

This fundamental difference between the low dimensional ($d=1,2$) and higher
dimensional ($d\geq 3$) Schrödinger operators is related to the different
asymptotic behavior of the Markov semigroups $T_{t}=\exp(t\Delta)$ and the
corresponding Markov processes (Brownian motions) $b(t)$ as $t\to\infty$. When
$d=1,2$, the Brownian motion $b(t)$ is recurrent, and it is transient if
$d\geq 3$. The processes $b(t)$ is recurrent or transient if the integral

$\int_{1}^{\infty}p(t,x,x)dt$

diverges or converges, respectively. Here $p(t,x,y)$ is the integral kernel of
the semigroup $T_{t}=\exp(t\Delta)$, i.e., the fundamental solution of the
heat equation. Equivalently, the recurrent and transient cases can be
distinguished by the value of $\lim_{\lambda\to+0}G_{\lambda}(x),~{}x\neq 0$,
which is infinite in the recurrent case and finite in the transient one. Here
$G_{\lambda}(x-y)$ is the integral kernel of the resolvent
$(\Delta-\lambda)^{-1}$.

Two types of results are obtained in this paper. The recurrent operators are
studied in the first part of the paper. A couple of examples concerning
transient operators are given in the second part of the paper. In the
recurrent case, our main goal is to simplify the proofs and extend the
classical results [9] of Simon on the existence of positive eigenvalues to a
wider class of operators. Simon proved the existence of positive eigenvalues
for standard Schrödinger operators (1) with arbitrarily small $\sigma>0$ when
$d=1,2$ and the following conditions hold:

$\int_{\mathbb{R}^{d}}V(x)dx\geq 0,$ (3)
$\int_{\mathbb{R}}(1+|x|^{2})|V(x)|dx<\infty,\quad{\rm if}\quad d=1,$ (4)

and there are $\varepsilon,\delta>0$ such that

$\int_{\mathbb{R}^{2}}(1+|x|^{\varepsilon})|V(x)|dx<\infty,\quad\int_{\mathbb{R}^{2}}|V(x)|^{1+\delta}dx<\infty,\quad{\rm
if}\quad d=2.$ (5)

We will assume that $H=L+\sigma V(x)$, where the unperturbed operator $L$ may
be non-local and has the form of multiplication by a real-valued continuous
function $a(k)$ after the application of the Fourier transform:

$\widehat{L\psi}=a(k)\widehat{\psi}(k),$ (6)

where $a(k)$ has the following properties: $a(k)<0$ for $k\neq 0$, and
$~{}|a(k)|\leq C|k|^{d},\quad k\to 0.$ Additional restrictions on $V$ and $L$
will be imposed when the integral in (3) is zero.

The second condition on $a(k)$ is a manifestation of the recurrency of the
process with the generator $L$. Indeed, the integral kernel $G(\lambda,x-y)$
of the resolvent $(L-\lambda)^{-1}$ is given by

$G(\lambda,x-y)=(2\pi)^{-d/2}\int_{\mathbb{\mathbb{missing}}{R}^{d}}\frac{e^{i<k,x-y>}dk}{a(k)-\lambda}dk,\quad\lambda>0,$
(7)

and the second condition implies that
$\lim_{\lambda\to+0}G(\lambda,x)=\infty,~{}x\neq 0$.

Our conditions on $a(k)$ allow $L$ to be a generator of a symmetric
homogeneous in space and time Levy process. Function $a$ in this case has the
form

$a(k)=\int_{\mathbb{\mathbb{missing}}{R}^{d}}(\cos<k,z>-1)l(z)dz,$

where function $l$ (the density of the Levy measure associated with the
semigroup $\exp(tL)~{}$) has the properties

$l(z)=l(-z),~{}\quad l(z)\geq
0,~{}~{}\quad\int_{|z|>1}l(z)dz<\infty,~{}~{}\quad\int_{|z|<1}|z|^{2}l(z)dz<\infty.$

A typical example of such a generator $L$ is provided by a fractional power of
the Laplacian $L=-(-\Delta)^{\alpha/2},~{}0<\alpha\leq 2$. We will consider
this example only when $d=2,\alpha=2,$ or $d=1,~{}1\leq\alpha\leq 2,$ to
guarantee the recurrence of the process. The process is transient if
$d=1,~{}0<\alpha<1,$ or $d=2,~{}0<\alpha<2$.

Our first result concerns the general operator $H=L+\sigma V(x)$, where $L$ is
defined by (6), $V\in L_{1}$, and the integral of $V$ is strictly positive
(conditions (4), (5) are not imposed). A short proof ($12$ lines) of the
existence of positive eigenvalues is based on a study of the quadratic form of
$H$ in the Fourier space. When the integral in (3) is zero, our proof is a
little bit more technical compared to the case when the integral is positive.
We provide the proof for operator (1) when $d=2$ with slightly weaker
conditions than in (5) and for the perturbations of a fractional power of the
Laplacian when $d=1$. The last result of the first part of the paper is the
proof of the existence of positive eigenvalues for all $\sigma>0$ for
$1$-dimensional operator (1) on the half-axis $x>0$ with the Neumann boundary
condition at $x=0$ under Bargman’s condition on $V$:

$\int_{o}^{\infty}x|V(x)|dx<\infty.$ (8)

In particular, this result provides the existence of the positive eigenvalues
for all $\sigma>0$ for operators on the whole line $\mathbb{R}$ when the
potential is even. The latter case includes the class of positive-definite
potentials that appear in many important practical applications, where $V$ has
the form of a convolution: $V(x)=b(x)\ast b(-x)$. Hence $V$ is even and

$\int_{\mathbb{R}^{d}}V(x)dx=(\int_{\mathbb{R}^{d}}b(x)dx)^{2}\geq 0.$

Note that the Bargman condition (8) implies (see [9]) that the positive
eigenvalue is unique when $\sigma$ is small enough.

Two examples are provided in the second part of the paper. They concern the
transient classical Schrödinger operator (1) with $d\geq 3$ for which
condition (2) is slightly violated. The first example shows that it is
possible to have infinitely many positive eigenvalues when $V(x)$ decays
rather fast at infinity, but the integral in (2) diverges. The potential $V$
in this example has order $O(|x|^{-2}),~{}|x|\to\infty,$ on a sequence of non-
intersecting balls of increasing radii, and $V$ vanishes outside of the balls.
Note that a logarithmically stronger estimate $|V(x)|<C(1+|x|\ln(2+|x|)^{-2}$
implies (2) and the boundedness of $N(\sigma V)$. The second example shows
that $H$ may have no positive eigenvalues when a non-negative potential decays
slowly at infinity.

##  2 Perturbations of recurrent operators

######  Theorem 2.1.

Let $H=L+\sigma V(x)$ be an operator in $L^{2}(\mathbb{R}^{d}),$ where
operator $L$ is defined by (6) and function $a$ has the properties

$a(k)<0\quad~{}{\it when}\quad k\neq 0,\quad~{}|a(k)|\leq C|k|^{d},\quad k\to
0.$

If $V\in L_{1}(\mathbb{R}^{d})$ and the integral of $V$ is strictly positive:

$\int_{\mathbb{R}^{d}}V(x)dx>0,$

then operator $H=L+\sigma V$ has a positive eigenvalue
$\lambda=\lambda(\sigma)$ for arbitrary $\sigma>0$.

###### Proof.

For each $\psi\in L^{2}(\mathbb{R}^{d})$, we have

$<H\psi,\psi>=\int_{\mathbb{R}^{d}}a(k)|\widehat{\psi}(k)|^{2}dk+\sigma\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\widehat{V}(k-\xi)\widehat{\psi}(k)\overline{\widehat{\psi}(\xi)}dkd\xi:=I_{1}+\sigma
I_{2},$ (9)

where $\widehat{\psi},\widehat{V}$ are the Fourier transforms of functions
$\psi,V$, respectively. One needs only to show that, for each $\sigma>0$,
there is $\widehat{\psi}\in L^{2}(\mathbb{R}^{d})$ for which the quadratic
form above is positive.

Let us fix $\varepsilon_{1}>0$ such that $\widehat{V}(k-\xi)>v_{0}>0$ when
$|k|,|\xi|<\varepsilon_{1}$. Since $\widehat{V}(0)>0$ and $\widehat{V}(k)$ is
continuous, such $\varepsilon_{1}>0$ and $v_{0}$ exist. We define
$\psi(x)=\psi_{\varepsilon}(x)$ by using its Fourier image:

$\widehat{\psi}_{\varepsilon}(k)=\left\\{\begin{array}[]{c}|k|^{-d},~{}~{}0<\varepsilon\leq|k|\leq\varepsilon_{1}\\\
0,~{}~{}|k|\notin(\varepsilon,\varepsilon_{1})\\\ \end{array},\right.$

where $\varepsilon\in(0,\varepsilon_{1})$ will be chosen below. Then

$|I_{1}|\leq
C\int_{\varepsilon<|k|<\varepsilon_{1}}|k|^{-d}dk<C_{1}|\ln\varepsilon|,\quad\varepsilon\to
0,$
$I_{2}>v_{0}\int_{\varepsilon<|k|<\varepsilon_{1}}\widehat{\psi}_{\varepsilon}(k)dk\int_{\varepsilon<|\xi|<\varepsilon_{1}}\widehat{\psi}_{\varepsilon}(\xi)d\xi=C_{2}v_{0}|\ln\varepsilon|^{2}>0,\quad\varepsilon\to
0.$

Hence the form (9) is positive if we choose sufficiently small $\varepsilon$
(such that $\sigma|\ln\varepsilon|\gg 1$). ∎

Denote by $L$ the space of functions on $\mathbb{R}^{2}$ integrable with the
weight $\ln(2+|x|)$, and

$\|f\|_{L}=\int_{\mathbb{R}^{2}}\ln(2+|x|)|f(x)|dx.$

######  Theorem 2.2.

Let $d=2$ and suppose that the following assumptions hold for some $\delta>0$:

$\int_{\mathbb{R}^{2}}\ln^{2}(2+|x|)|V(x)|dx<\infty;~{}\quad\int_{\mathbb{R}^{2}}|V(x)|^{1+\delta}dx<\infty;\quad\int_{\mathbb{R}^{2}}V(x)dx=0.$

Then operator $H=\Delta+\sigma V$ has a positive eigenvalue
$\lambda=\lambda(\sigma)$ for arbitrary $\sigma>0$.

###### Proof.

Since the operator $\sigma^{-1}\Delta+V,~{}\sigma>0,$ increases monotonically
when $\sigma$ increases, it is enough to show the existence of a positive
eigenvalue of $H$ for small $\sigma$. This will imply its existence for all
$\sigma>0$.

We will look for eigenfunctions of $H$ with $\lambda>0$ in the form
$\psi=(\Delta-\lambda)^{-1}f,~{}f\in L$. Using the Fourier transform, one can
check that $\psi\in L^{2}(\mathbb{R}^{2})$. This function also satisfies the
equation $H\psi-\lambda\psi=0$ if $f$ is a solution of the equation

$f(x)+\sigma V(x)\int_{\mathbb{R}^{2}}G(\lambda,x-y)f(y)dy=0,\quad f\in
L,\quad\lambda>0,$ (10)

where $G$ is the integral kernel of the resolvent $(\Delta-\lambda)^{-1}$,
i.e., $G=\frac{-1}{2\pi}K_{0}(\sqrt{\lambda}|x-y|)$. Here $K_{0}$ is the
modified Bessel function of the second kind. Thus the theorem will be proved
if we show that, for arbitrarily small $\sigma>0$, there exists
$\lambda=\lambda(\sigma)$ for which equation (10) has a non-trivial solution
$f\in L$.

We fix a smooth function $h$ with a compact support such that
$\int_{\mathbb{R}^{2}}hdx=1$, and write $f$ in the form

$f=C_{f}h+f^{\bot},\quad{\rm
where}~{}~{}~{}C_{f}=\int_{\mathbb{R}^{2}}f(x)dx,\quad\int_{\mathbb{R}^{2}}f^{\bot}(x)dx=0.$

We single out the singularity of $G$ as $\lambda\to+0$:
$G=\frac{1}{4\pi}\ln\lambda+g(\lambda,|x-y|)$, where $g$ is continuous when
$\lambda\geq 0$ and there are constants $C,c$ such that

$|g|\leq C(|\ln|x-y||+1),\quad
g(0,x-y)=\frac{1}{2\pi}\ln|x-y|+c,\quad\lambda\geq 0.$ (11)

We put $G=\frac{1}{4\pi}\ln\lambda+g(\lambda,|x-y|)$ into (10) and rewrite it
as a system for $C_{f}$ and $f^{\bot}$:

$C_{f}+\sigma C_{f}B_{\lambda}h+\sigma B_{\lambda}f^{\bot}=0,$ (12)
$f^{\bot}(x)+C_{f}\frac{\sigma V(x)}{4\pi}\ln\lambda+\sigma
C_{f}[A_{\lambda}h-(B_{\lambda}h)h]+\sigma[A_{\lambda}f^{\bot}-(B_{\lambda}f^{\bot})h]=0,$
(13)

where

$\displaystyle A_{\lambda}:L\to L,\quad
A_{\lambda}f=V(x)\int_{\mathbb{R}^{2}}g(\lambda,x-y)f(y)dy,\quad\lambda\geq
0,$ (14) $\displaystyle B_{\lambda}:L\to\mathbb{R},\quad
B_{\lambda}f=\int_{\mathbb{R}^{2}}\int_{\mathbb{R}^{2}}V(x)g(\lambda,x-y)f(y)dydx,\quad\lambda\geq
0.$ (15)

Operators $A_{\lambda},B_{\lambda}$ are bounded uniformly in $\lambda\in[0,1]$
and strongly continuous when $\lambda\in[0,1]$. Indeed, the first estimate in
(11) implies that

$\displaystyle\|A_{\lambda}f\|_{L}\leq
C\int_{\mathbb{R}^{2}}\int_{\mathbb{R}^{2}}\ln(2+|x|)|V(x)|(|\ln|x-y||+1)f(y)|dydx$
$\displaystyle\leq C\int_{\mathbb{R}^{2}}(I_{1}(y)+I_{2}(y))|f(y)|dy,\quad
I_{i}=\int_{D_{i}}\ln(2+|x|)|V(x)|(|\ln|x-y||+1)dx,$ (16)

where $D_{1}=\\{x:~{}|x-y|<1\\},~{}D_{2}=\\{x:~{}|x-y|>1\\}$. We have

$\displaystyle I_{1}\leq
C\ln(2+|y|)\left(\int_{|x-y|<1}|V(x)|^{1+\delta}dx\right)^{\frac{1}{1+\delta}}\left(\int_{|x-y|<1}(|\ln|x-y||+1)^{\frac{1+\delta}{\delta}}dx\right)^{\frac{\delta}{1+\delta}}$
$\displaystyle\leq C_{1}\ln(2+|y|),$ $I_{2}\leq
C\int_{|x-y|>1}|V(x)|\ln^{2}(2+|x|)\ln(2+|y|)dx\leq C_{2}\ln(2+|y|).$

The estimates on $I_{i}$ together with (16) prove the uniform boundedness of
$A_{\lambda}$. The boundedness of $B_{\lambda}$ can be proved similarly. The
same arguments complemented by the continuity of $g$ and the dominated
convergence theorem can be used to prove the strong continuity of both
operators.

The boundedness of $A_{\lambda},B_{\lambda}$ allows us to rewrite (12), (13)
for small $\sigma\geq 0$ in a slightly shorter form:

$C_{f}[1+O(\sigma)]+\sigma B_{\lambda}f^{\bot}=0,$ (17)
$f^{\bot}(x)+\sigma[A_{\lambda}f^{\bot}-h(B_{\lambda}f^{\bot})]+C_{f}[\frac{\sigma
V(x)}{4\pi}\ln\lambda+O(\sigma)]=0,$ (18)

and then solve equation (18) for $f^{\bot}$:

$f^{\bot}=-C_{f}[I+\sigma(A_{\lambda}-hB_{\lambda})]^{-1}(\frac{\sigma
V(x)}{4\pi}\ln\lambda+O(\sigma))=-C_{f}[\frac{\sigma
V(x)}{4\pi}\ln\lambda(1+O(\sigma))+O(\sigma)].$

We put the latter expression into (17) and obtain

$C_{f}d=0,\quad{\rm where}\quad
d=d(\sigma,\lambda)=\frac{-\sigma^{2}m}{4\pi}\ln\lambda(1+o(1))+1+o(1),$ (19)

where the remainder terms vanish when $\sigma,\lambda\to+0$, and

$m=B_{0}V(x)=\int_{\mathbb{R}^{2}}\int_{\mathbb{R}^{2}}V(x)(\frac{\ln|x-y|}{2\pi}+c)V(y)dydx=\int_{\mathbb{R}^{2}}\int_{\mathbb{R}^{2}}V(x)\frac{\ln|x-y|}{2\pi}V(y)dydx.$
(20)

Since equation (10) is reduced to (19), the values of $\lambda$ for which
$d=0$ are the eigenvalues of operator $H$.

From (20) and condition $V\in L$ it follows that $|m|<\infty$. Since the
Fourier transform of function $\frac{1}{2\pi}\ln|x|$ is $-1/|k|^{2}$, the
Plancherel identity implies

$m=-\int_{\mathbb{R}^{2}}\frac{|\widehat{V}(k)|^{2}}{|k|^{2}}dk<0.$ (21)

We fix $\lambda=\lambda_{1}\in(0,1)$ so small that the remainder terms in (19)
satisfy $|o(1)|<1/2$ when $\sigma,\lambda\in[0,\lambda_{1}]$. Then we fix
$\sigma=\sigma^{\prime}\in(0,\lambda_{1})$ as small as we please for which
$d(\sigma^{\prime},\lambda_{1})>0$. Since
$\lim_{\lambda\to+0}d(\sigma^{\prime},\lambda)=-\infty$, there is
$\lambda=\lambda_{0}$ such that $d(\sigma^{\prime},\lambda_{0})<0$. Since the
strong continuity of operators $A_{\lambda},B_{\lambda}$ implies the
continuity of the remainder terms in (19), there exists a point
$\lambda=\lambda(\sigma^{\prime})\in(\lambda_{0},\lambda_{1})$ for which
$d(\sigma^{\prime},\lambda)=0$. This point is an eigenvalue of $H$.

∎

Consider now a one-dimensional version of Theorem 2.2 in the case when the
Laplacian is replaced by a more general operator. A very natural class of
these operators is presented by fractional powers $L=-(-\Delta)^{\alpha/2}$ of
the Laplacian. We assume that $\alpha\leq 2$, so that $L$ is a generator of a
symmetric stable process with parameter $\alpha$, and that $\alpha>1$, so that
process is recurrent. In fact, the process remains recurrent when $\alpha=1$,
but the statement and some estimates in the proof require changes, and we
decided to exclude this case from consideration.

######  Theorem 2.3.

Let $d=1,~{}H=-(-\Delta)^{\alpha/2}+V(x),~{}1<\alpha\leq 2$, and suppose that
the following assumptions hold:

$\int_{\mathbb{R}}(1+|x|^{2\alpha-2})|V(x)|dx<\infty,\quad\int_{\mathbb{R}^{d}}V(x)dx=0.$

Then the operator $H=\Delta+\sigma V$ has a positive eigenvalue
$\lambda=\lambda(\sigma)$ for each $\sigma>0$.

###### Proof.

The proof copies the one of Theorem 2.2 with some steps being simpler. As in
Theorem 2.2, it is enough to show the existence of a positive eigenvalue of
$H$ for small $\sigma$. Space $L$ now consists of functions integrable with
the weight $1+|x|^{\alpha-1}$ with

$\|f\|_{L}=\int_{\mathbb{R}}(1+|x|^{\alpha-1})|f(x)|dx.$

We reduce again the eigenvalue problem for operator $H$ to the one-dimensional
version of equation (10) with $G(\lambda,x-y)$ being the integral kernel of
the operator $(-(-\Delta)^{\alpha/2}-\lambda)^{-1}$.

We look at the singularity of $G$ as $\lambda\to+0$. Denote
$\nu=|x|\lambda^{1/\alpha}$. We have

$G(\lambda,x)=\frac{-1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{ikx}dk}{\lambda+|k|^{\alpha}}=\frac{-1}{\pi}\int_{0}^{\infty}\frac{\cos(kx)dk}{\lambda+|k|^{\alpha}}=\frac{-\lambda^{\frac{1-\alpha}{\alpha}}}{\pi}\int_{0}^{\infty}\frac{\cos(s\nu)ds}{1+s^{\alpha}}=$
$=\frac{-\lambda^{\frac{1-\alpha}{\alpha}}}{\pi}\int_{0}^{\infty}\frac{ds}{1+s^{\alpha}}+\frac{\lambda^{\frac{1-\alpha}{\alpha}}}{\pi}\int_{0}^{\infty}\frac{2\sin^{2}(s\nu/2)ds}{1+s^{\alpha}}:=-c_{1}(\alpha)\lambda^{\frac{1-\alpha}{\alpha}}+g(\lambda,x),\quad
c_{1}(\alpha)>0.$ (22)

We replace $1+s^{\alpha}$ in the last integrand by $s^{\alpha}$ and make the
substitution $s\nu\to s$. This leads to the estimate

$|g(\lambda,x)|\leq
C\lambda^{\frac{1-\alpha}{\alpha}}\nu^{\alpha-1}=C|x|^{\alpha-1},\quad\lambda>0.$
(23)

Function $g$ is continuous for $\lambda>0$ and can be extended by continuity
when $\lambda\to+0$ with

$g(0,x)=c_{2}(\alpha)|x|^{\alpha-1},\quad c_{2}(\alpha)>0.$ (24)

The latter statement follows from the relation

$g(\lambda,x)=\frac{\lambda^{\frac{1-\alpha}{\alpha}}}{\pi}\int_{0}^{\infty}\frac{2\sin^{2}(s\nu/2)ds}{s^{\alpha}}+\frac{\lambda^{\frac{1-\alpha}{\alpha}}}{\pi}\int_{0}^{\infty}\frac{2\sin^{2}(s\nu/2)ds}{s^{\alpha}(1+s^{\alpha})}$
$=\frac{|x|^{\alpha-1}}{\pi}\int_{0}^{\infty}\frac{2\sin^{2}(s/2)ds}{s^{\alpha}}+\frac{|x|^{\alpha-1}}{\pi}\int_{0}^{\infty}\frac{2\sin^{2}(s/2)ds}{s^{\alpha}(1+(s/\nu)^{\alpha})},$

where the first term is $c_{2}(\alpha)|x|^{\alpha-1}$ and the second one
vanishes when $x$ is fixed and $\lambda\to+0$.

The relations (23), (24) are the analogue of (11). Using (22)-(24) and the
representation $f=C_{f}h+f^{\bot},~{}C_{f}=\int_{\mathbb{R}}f(x)dx$, we
rewrite the one-dimensional version of equation (10) as the system (12)-(15)
with the singularity $\frac{\ln\lambda}{4\pi}$ of $G$ in (13) replaced by
$-c_{1}(\alpha)\lambda^{\frac{1-\alpha}{\alpha}}$ and the integration in (14),
(15) over $\mathbb{R}$ instead of $\mathbb{R}^{2}$. The remaining steps of the
proof are the same with the following analogue of formula (19) for function
$d$ defining the eigenvalues of $H$:

$d=d(\sigma,\lambda)=c_{1}(\alpha)\lambda^{\frac{1-\alpha}{\alpha}}\sigma^{2}m(1+o(1))+1+o(1),\quad
m=-\int_{\mathbb{R}^{2}}\frac{|\widehat{V}(k)|^{2}}{|k|^{\alpha}}dk<0.$

∎

##  3 ODE on the half-line.

Consider operator $H$ in $L^{2}(\mathbb{R}_{+})$ defined by the following
relations

$H\psi=\psi^{\prime\prime}(x)+\sigma V(x)\psi(x),\quad
x>0,\quad\psi^{\prime}(0)=0,\quad{\rm
where}\quad\int_{0}^{\infty}x|V(x)|dx<\infty.$ (25)

######  Theorem 3.1.

If $\int_{0}^{\infty}V(x)dx\geq 0$, then operator $H$ defined in (25) has a
positive eigenvalue $\lambda=\lambda(\sigma)$ for each $\sigma>0$.

We will need a couple of lemmas. The following statement can be found in [10]
under more restrictive assumption on $V$. It is referred to as the Schpat
theorem there.

######  Lemma 3.2.

Equation

$\psi^{\prime\prime}(x)+\sigma
V(x)\psi(x)=0,\quad\int_{0}^{\infty}x|V(x)|dx<\infty,$ (26)

has solutions $\psi_{1},\psi_{2}$ with the following asymptotic behavior at
infinity:

$\psi_{1}(x)=1+o(1),\quad\psi_{1}^{\prime}(x)=o(1),\quad\psi_{2}(x)=x(1+o(1)),\quad\psi_{2}^{\prime}(x)=1+o(1),\quad
x\to\infty.$ (27)

###### Proof.

Denote

$\alpha=|\sigma|\int_{0}^{\infty}x|V(x)|dx.$

Since the statement of the lemma does not depend on the choice of the origin
on the $x$-axis, without loss of the generality, we can assume that
$\alpha<1$.

We look for $\psi_{1}$ in the form $\psi_{1}=1+z(x)$. Then
$z^{\prime\prime}+\sigma V(x)z=-\sigma V(x)$, and this equation can be reduced
to the integral equation

$z(x)+\sigma Pz=-\sigma\int_{x}^{\infty}(\xi-x)V(\xi)d\xi,\quad x>0,\quad{\rm
where}\quad Pz=\int_{x}^{\infty}(\xi-x)V(\xi)z(\xi)d\xi.$ (28)

If $z$ is a solution of (28), then $\psi_{1}$ satisfies (26). We consider $P$
as an operator in the space $C$ of continuous bounded function on the semi-
axis $R_{+}$. Then $\|\sigma P\|\leq\alpha<1$, and therefore equation (28) is
uniquely solvable in $C$. Since $Pz$ vanishes at infinity for each $z\in C$,
from (28) it follows that $z$ vanishes at infinity. Differentiation of (28) in
$x$ implies that $z^{\prime}$ vanishes at infinity. Thus $\psi_{1}$ has the
desired asymptotic behavior.

Using the solution $\psi=\psi_{1}$ of (26), one can obtain the general
solution of (26) (by looking for it in the form $\psi=\psi_{1}z$). In
particular, the second solution can be chosen as

$\psi_{2}(x)=\psi_{1}(x)\int_{0}^{x}\psi_{1}^{-2}(\xi)d\xi.$

Then (27) holds.

∎

######  Lemma 3.3.

If $\int_{0}^{\infty}V(x)dx\geq 0$ and $\sigma>0$ is small enough, then the
solution $\psi=\psi_{0}$ of (26) with the initial data
$\psi_{0}(0)=1,~{}\psi_{0}^{\prime}(0)=0$ vanishes at some point $x_{0}>0$.

###### Proof.

We need initial data for $\psi_{1},\psi_{2}$ when $\sigma\to 0$ and $V$ is
fixed. Since the operator $(I+\sigma P)^{-1}$ is analytic in $\sigma$ at
$\sigma=0$, function $\psi_{1}$ as an element of $C$ is analytic at
$\sigma=0$. Hence

$\psi_{1}=1+z=1-\sigma\int_{x}^{\infty}(\xi-x)V(\xi)d\xi+\sigma^{2}P\int_{x}^{\infty}(\xi-x)V(\xi)d\xi+O(\sigma^{3}).$

Equation (28) implies that the Taylor series of $\psi_{1}$ in $\sigma$ can be
differentiated in $x$. Thus, after differentiation of (28) in $x$ followed by
integration by parts, we get

$\psi_{1}^{\prime}=\sigma\int_{x}^{\infty}V(\xi)d\xi-\sigma^{2}\int_{x}^{\infty}V(\eta)\int_{\eta}^{\infty}(\xi-\eta)V(\xi)d\xi
d\eta+O(\sigma^{3})$
$=\sigma\int_{x}^{\infty}V(\xi)d\xi-\sigma^{2}\int_{x}^{\infty}V(\xi)d\xi\int_{x}^{\infty}(\xi-x)V(\xi)d\xi+\sigma^{2}\int_{x}^{\infty}\left(\int_{\eta}^{\infty}V(\xi)d\xi\right)^{2}d\eta+O(\sigma^{3}),$

and therefore

$\psi_{1}(0)=1-\sigma a+O(\sigma^{2}),\quad\psi_{1}^{\prime}(0)=\sigma
b+\sigma^{2}(-ab+c^{2})+O(\sigma^{3}),$
$\quad\psi_{2}(0)=0,\quad\psi_{2}^{\prime}(0)=1+\sigma a+O(\sigma^{2}),$

where

$a=\int_{0}^{\infty}\xi V(\xi)d\xi,\quad b=\int_{0}^{\infty}V(\xi)d\xi,\quad
c^{2}=\int_{0}^{\infty}\left(\int_{\eta}^{\infty}V(\xi)d\xi\right)^{2}d\eta.$

We represent $\psi_{0}$ as $\psi_{0}=A\psi_{1}+B\psi_{2}$. Initial data for
these three solutions allow us to find $A$ and $B$:

$A=\frac{1}{1-\sigma a+O(\sigma^{2})},\quad B=-\frac{\sigma
b+\sigma^{2}(ab+c^{2})+O(\sigma^{3})}{1-\sigma^{2}a+O(\sigma^{3})}.$

Since $B<0$ for small $\sigma>0$, we have $\psi(x)\to-\infty$ as $x\to\infty$.
This and the condition $\psi_{0}(0)=1$ complete the proof.

∎

Proof of Theorem 3.1. Let $\phi=\psi_{0}$ for $x\leq x_{0},~{}\phi=0$ for
$x>x_{0}$. Then, for arbitrarily small $\sigma>0$,

$0=<H\phi,\phi>=-\|\phi^{\prime}\|_{L^{2}(\mathbb{R}_{+})}+\sigma\int_{0}^{\infty}V(x)\phi^{2}(x)dx.$

Since the first term on the right is negative, the second one is positive, and
therefore the quadratic form is positive when $\sigma$ is replaced by
$2\sigma$. Hence $H$ with $\sigma$ replaced by $2\sigma$ has a positive
eigenvalue. Since the arguments used in the first paragraph of the proof of
Theorem 2.2 imply that it is enough to justify the statement of Theorem 3.1
only for small $\sigma>0$, the proof is complete.

∎

##  4 Examples concerning the transient operators

Let

$H\psi=\Delta\psi+\sigma V(x)\psi,\quad x\in\mathbb{R}^{d},~{}~{}d\geq 3.$
(29)

Condition (2) is a standard requirement for CLR-type theorems that provide an
estimate from above on the number $N(\sigma V)$ of positive eigenvalues of $H$
and for the quasi-classical asymptotics of $N(\sigma V)$. Below we provide a
couple of examples concerning the situation when (2) does not hold.

Example 1 (Large separated spots). We consider operator (29), where $V(x)$ is
an arbitrary potential such that $V(x)\geq\frac{1}{|x|^{2}}$ on a sequence of
disjoint balls with fast-increasing radii. Then $N(\sigma V)$ can be infinite
for large enough $\sigma$.

Let us provide a particular example of this situation. We fix arbitrary points
$x_{n}\in\mathbb{R}^{d},~{}d\geq 3,$ such that $|x_{n}|=2^{n}$, choose
$R_{n}=\frac{1}{3}2^{n}$, and consider the balls
$B_{n}=\\{x:|x-x_{n}|<R_{n}\\}$. In order to show that the number of positive
eigenvalues is infinite, consider the test functions
$\psi_{n}(x)=\psi(\frac{|x-x_{n}|}{R_{n}})$, where $\psi=\psi(x)$ is a fixed
infinitely smooth function with the support in the unit ball $|x|<1$. Then

$<\Delta\psi_{n},~{}\psi_{n}>=\int_{\mathbb{R}^{d}}\Delta\psi_{n}(x)\psi_{n}(x)dx=R_{n}^{d-2}\int_{\mathbb{R}^{d}}\Delta\psi(x)\psi(x)dx=C_{1}R_{n}^{d-2},$
$<V\psi_{n},~{}\psi_{n}>=\int_{\mathbb{R}^{d}}V(x)|\psi_{n}(x)|^{2}dx\geq
R_{n}^{d-2}\int_{\mathbb{R}^{d}}|x|^{-2}|\psi(x)|^{2}dx=C_{2}R_{n}^{d-2},$

and therefore $<H\psi_{n},~{}\psi_{n}>$ is positive for all $n\geq 1$ if
$\sigma>|C_{1}|/C_{2}$. Then the variational principle implies that $N(\sigma
V)=\infty$.

Example 2 (Small sparse spots). Here we will provide an example of a non-
negative potential $V(x)$ for which operator (29) with $\sigma\in(0,1)$ does
not have positive eigenvalues in spite of the fact that $V$ decays at infinity
so slowly that

$\int_{\mathbb{R}^{d}}V^{\beta}(x)dx=\infty\quad{\rm for}~{}{\rm
each}~{}~{}~{}\beta>0.$ (30)

The potential $V$ will be a sum of sparse bumps.

We choose points $x_{n}$ as in the previous example (i.e., $|x_{n}|=2^{n}$).
Let $V_{0}\in C_{0}^{\infty}$ (elementary bump) have the support in the unit
ball $|x|<1$. We choose

$V(x)=\sum_{n_{0}}^{\infty}\frac{1}{\ln n}V_{0}(x-x_{n}),$ (31)

where $n_{0}\geq 3$ will be chosen later. Since $V(x)\to 0,~{}|x|\to\infty,$
the positive spectrum of $H$ is discrete. Hence, if the positive spectrum is
not empty, there is a ground state $\psi_{0}(x)>0$ that corresponds to the
largest eigenvalue $\lambda_{0}>0$. Obviously, (30) holds for potential (31),
and it remains to show that the positive spectrum of $H$ is empty for
$\sigma\in(0,1)$.

If $\psi_{0}(x)$ exists, then it satisfies the equation

$\psi_{0}(x)=\int_{\mathbb{R}^{d}}G_{\lambda_{0}}(x-y)\sigma
V(y-x_{n})\psi_{0}(y)dy=\sum_{n_{0}}^{\infty}\int_{|y-x_{n}|<1}G_{\lambda_{0}}(x-y)\frac{\sigma}{\ln
n}V_{0}(y-x_{n})\psi_{0}(y)dy,$ (32)

where $G_{\lambda_{0}}(x-y)$ is the integral kernel of the resolvent
$(\Delta-\lambda_{0})^{-1}$. Denote

$M_{n}=max_{x:|x-x_{n}|\leq 1}\psi_{0}(x).$

Since $0<G_{\lambda_{0}}(x)<c(d)/|x|^{d-2},~{}x\in\mathbb{R}^{d},$ equation
(32) with $\sigma\in(0,1)$ implies

$M_{m}\leq\frac{1}{\ln n_{0}}\sum\limits_{n\geq n_{0},~{}n\neq
m}\frac{v(d)c(d)M_{n}}{|2^{m}-2^{n}|^{d-2}-2}+\frac{c_{1}(d)}{\ln m}M_{m}$
$:=\frac{1}{\ln n_{0}}\sum\limits_{n\geq n_{0}~{}n\neq
m}\Gamma_{n,m}M_{n}+\gamma_{m}M_{m},\quad m\geq n_{0},$

where $v(d)$ is the volume of the unit ball in $\mathbb{R}^{d}$. We choose
$n_{0}$ so large that $\gamma_{m}<1/2$ for $m\geq n_{0}$. Then

$M_{m}\leq\frac{2}{\ln n_{0}}\sum\limits_{n\geq n_{0}~{}n\neq
m}\Gamma_{n,m}M_{n},\quad m\geq n_{0},$

where $\rho:=\max_{n}\sum_{n\geq 3}\Gamma_{n,m}<\infty$. Hence the norm of the
operator with the matrix elements $\Gamma_{n,m},~{}n,m\geq 3,~{}n\neq m,$ in
the space $l_{1}$ of summable sequences is bounded, and therefore the system
of inequalities for $M_{m}$ has only trivial solution if $n_{0}$ is large
enough. Thus the positive spectrum is empty.

Acknowledgments: The work of B. Vainberg was supported by the Simons
Foundation grant 527180.

## References

  * [1] M. Cranston, L. Koralov, S. Molchanov, B. Vainberg, Continuous Model for Homopolymers, J. Functional Analysis, 256 (2009) no. 8, 2656-2696.
  * [2] M. Cwikel, Weak type estimates for singular values and the number of bound states of Schrödinger operators, Ann. Math., (2) 106 (1977), 93-100.
  * [3] B. M. Levitan and I. S. Sargsjan, Introduction to Spectral Theory: Selfadjoint Ordinary Differential Operators, AMS, 1975.
  * [4] E. Lieb, Bounds on the eigenvalues of the Laplace and Schrödinger operators. Bull.Amer. Math. Soc., 82 (1976), no. 5, 751–753.
  * [5] S. Molchanov, B. Vainberg, On general Cwikel-Lieb-Rozenblum and Lieb-Thirring inequalities, in Around the research of Vladimir Maz’ya III, Editor A. Laptev, Int. Math. Ser. (N.Y.) 13, Springer, 2010, 201-246 .
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  * [8] G. Rozenblum, Distribution of the discrete spectrum of singular differential operators, (Russian) Dokl. Acad. Nauk SSSR, 202 (1972), 1012-1015; translation in Soviet Math. Dokl., 13 (1972), 245-249.
  * [9] B. Simon, The bound state of weakly coupled Schrödinger operators in one and two dimensions, Ann.of Phys.,97 (1976), 279-288.
  * [10] V.V. Stepanov, Course of Differential Equations (in Russian), Moscow, 1958.


Let $X=\ell_{p}$, $1<p<\infty$. The proof of Theorem 6.1 has shown that ${\rm
NA}(X)\cap\mathcal{S}(X)$ is $F_{\sigma}$ in
$(\mathcal{S}(X),\texttt{SOT}\mbox{${}^{*}$})$, hence in
$(\mathcal{S}(X),\|\,\cdot\,\|)$. On the other hand, by Bourgain’s result [B]
mentioned above, ${\rm NA}(X)\cap\mathcal{S}(X)$ is comeager in
$(\mathcal{S}(X),\|\,\cdot\,\|)$. So, we obtain that
$\mathcal{S}(X)\setminus{\rm NA}(X)$ is in fact _nowhere dense_ in
$(\mathcal{S}(X),\|\,\cdot\,\|)$. In other words, ${\rm
NA}(\ell_{p})\cap\mathcal{S}(\ell_{p})$ contains a norm-dense open subset of
$\mathcal{S}(\ell_{p})$; from which it follows that ${\rm NA}(\ell_{p})$
contains a norm-dense open subset of $\mathcal{B}(\ell_{p})$.

It follows from Theorems 6.1 and 3.4 that if $2<p<\infty$, a typical
$T\in\bigl{(}\mathcal{B}_{1}(\ell_{p}),\texttt{SOT}\bigr{)}$ does not attain
its norm. This leads to the following question:

######  Question 6.9.

Let $1<p<2$. Is it true that a typical
$T\in\bigl{(}\mathcal{B}_{1}(\ell_{p}),\texttt{SOT}\bigr{)}$ does not attain
its norm?

##  7\. Additional results

###  7.1. More on similarity of topologies

Let $\mathbf{M}$ be an abstract set. We have observed that two topologies on
$\mathbf{M}$ sharing the same comeager sets may not be similar. In the next
proposition, we show that if some extra assumptions on the topologies are
added, then the two properties become equivalent.

######  Proposition 7.1.

Let $\tau$ and $\tau^{\prime}$ be two topologies on $\mathbf{M}$. Assume that
$\tau\subseteq\tau^{\prime}$, that $\tau^{\prime}$ is Baire and second-
countable, and that the identity map
$\mathbf{i}_{\tau,\tau^{\prime}}:(\mathbf{M},\tau)\to(\mathbf{M},\tau^{\prime})$
is Borel $1$, i.e. any $\tau^{\prime}$\- open subset of $\mathbf{M}$ is
$\tau\,$-$\,F_{\sigma}$. Then $\tau$ and $\tau^{\prime}$ are similar if and
only if they have the same comeager sets.

###### Proof.

By Lemma 2.1, we only need to prove the “if” part. Assume that $\tau$ and
$\tau^{\prime}$ have the same comeager sets. Since
$\mathbf{i}_{\tau,\tau^{\prime}}$ is Borel $1$ and $\tau^{\prime}$ is second-
countable, we know that $\mathcal{C}(\tau,\tau^{\prime})$ is $\tau\,$\-
comeager. So $\mathcal{C}(\tau,\tau^{\prime})$ is $\tau^{\prime}\,$\- comeager
by our assumption, and hence $\tau^{\prime}\,$\- dense in $\mathbf{M}$ since
$\tau^{\prime}$ is Baire. By Lemma 2.3, it follows that $\tau$ and
$\tau^{\prime}$ are similar. (Note that it is not really necessary to assume
that $\tau\subseteq\tau^{\prime}$: it is enough to know that
$\mathcal{C}(\tau^{\prime},\tau)$ is $\tau\,$\- dense.) ∎

One gets the same conclusion with the following “symmetric” assumptions:
$\tau$ and $\tau^{\prime}$ are both Baire and second-countable, and the two
identity maps are Borel $1$.

We now prove a result that allows to give a few other characterisations of
similarity.

######  Proposition 7.2.

Let $\tau$ and $\tau^{\prime}$ be similar topologies on $\mathbf{M}$, and
assume that $\tau$ and $\tau^{\prime}$ are second-countable. Then
$\mathcal{C}(\tau,\tau^{\prime})\cap\mathcal{C}(\tau^{\prime},\tau)$ is
comeager (for both $\tau$ and $\tau^{\prime}$).

###### Proof.

It is enough to show that
$\mathcal{A}:=\mathcal{C}(\tau,\tau^{\prime})\cap\mathcal{C}(\tau^{\prime},\tau)$
is $\tau\,$\- comeager; and to this end we use the _Banach-Mazur game_
$\mathbf{G}(\mathcal{A})$ associated with $\mathcal{A}$ (see e.g. [Ke, p.
51]), in the topological space $(\mathbf{M},\tau)$. Recall that
$\mathbf{G}(\mathcal{A})$ is an infinite game with two players, denoted by I
and II. The two players play alternatively non-empty $\tau\,$\- open sets
$U_{0}\supseteq U_{1}\supseteq U_{2}\supseteq\cdots$, and player II wins the
run if $\bigcap_{n\geq 0}U_{n}\subseteq\mathcal{A}$. To prove that
$\mathcal{A}$ is $\tau\,$\- comeager, it is enough to show that player II has
a winning strategy in $\mathbf{G}(\mathcal{A})$.

Let $(O_{k})_{k\geq 0}$ and $(O^{\prime}_{k})_{k\geq 0}$ be countable bases
for $(\mathbf{M},\tau)$ and $(\mathbf{M},\tau^{\prime})$. Assume that player I
has just played a $\tau\,$\- open set $U_{2k}\neq\emptyset$. Then II answers
as follows. First, II chooses a $\tau^{\prime}\,$\- open set
$W^{\prime}_{k}\neq\emptyset$ such that $W^{\prime}_{k}\subseteq U_{2k}$ and,
moreover, $W^{\prime}_{k}\subseteq U_{2k}\cap O_{k}$ if $U_{2k}\cap
O_{k}\neq\emptyset$. This is possible since $\tau$ and $\tau^{\prime}$ are
similar. Then II plays a $\tau\,$\- open set $U_{2k+1}\neq\emptyset$ such that
$U_{2k+1}\subseteq W^{\prime}_{k}$ and, moreover, $U_{2k+1}\subseteq
W^{\prime}_{k}\cap O^{\prime}_{k}$ if $W^{\prime}_{k}\cap
O^{\prime}_{k}\neq\emptyset$. Again, this is possible since $\tau$ and
$\tau^{\prime}$ are similar. Let us check that this strategy is winning for
II.

Let $(U_{n})_{n\geq 0}$ be a run of the game $\mathbf{G}(\mathcal{A})$ where
II has followed the above strategy, and let $x\in\bigcap_{n\geq 0}U_{n}$. We
have to show that
$x\in\mathcal{A}=\mathcal{C}(\tau,\tau^{\prime})\cap\mathcal{C}(\tau^{\prime},\tau)$
i.e. that any $\tau\,$\- neighbourhood of $x$ is a $\tau^{\prime}\,$\-
neighbourhood and vice-versa. Let $V$ be a $\tau\,$\- neighbourhood of $x$.
Choose $k\geq 0$ such that $x\in O_{k}\subseteq V$. Then $x\in O_{k}\cap
U_{2k}$, so that in particular $O_{k}\cap U_{2k}\neq\emptyset$. With the above
notation, it follows that $W^{\prime}_{k}\subseteq O_{k}\cap U_{2k}$ and hence
$W^{\prime}_{k}\subseteq V$. Since $x\in U_{2k+1}\subseteq W^{\prime}_{k}$,
this shows that $V$ is a $\tau^{\prime}\,$\- neighbourhood of $x$. Now, let
$V^{\prime}$ be a $\tau^{\prime}\,$\- neighbourhood of $x$. Choose $k\geq 0$
such that $x\in O^{\prime}_{k}\subseteq V^{\prime}$. Then $x\in U_{2k+1}\cap
O^{\prime}_{k}\subseteq W^{\prime}_{k}\cap O^{\prime}_{k}$, so
$W^{\prime}_{k}\cap O^{\prime}_{k}\neq\emptyset$. Hence $U_{2k+1}\subseteq
W^{\prime}_{k}\cap O^{\prime}_{k}\subseteq V^{\prime}$; and since $x\in
U_{2k+1}$, it follows that $V^{\prime}$ is a $\tau\,$\- neighbourhood of $x$.
∎

######  Corollary 7.3.

Assume that $\tau$ and $\tau^{\prime}$ are Baire and second-countable. Then
the following assertions are equivalent.

  1. (i)

$\tau$ and $\tau^{\prime}$ are similar.

  2. (ii)

$\mathcal{C}(\tau,\tau^{\prime})\cap\mathcal{C}(\tau,\tau^{\prime})$ is
comeager for both $\tau$ and $\tau^{\prime}$.

  3. (iii)

$\mathcal{C}(\tau,\tau^{\prime})$ and $\mathcal{C}(\tau,\tau^{\prime})$ are
both $\tau\,$\- dense and $\tau^{\prime}\,$\- dense.

  4. (iv)

$\mathcal{C}(\tau,\tau^{\prime})$ is $\tau^{\prime}\,$\- dense and
$\mathcal{C}(\tau,\tau^{\prime})$ is $\tau\,$\- dense.

###### Proof.

This follows from Proposition 7.2 and Lemma 2.3. ∎

In order to state our second corollary, we need to recall a perhaps not so
well known notion. A subset $A$ of a topological space $E$ is said to be
_semi-open_ if $A$ is contained in the closure of its interior; equivalently,
if there exists an open set $O$ such that $O\subseteq
A\subseteq\overline{O\,}$. This notion was introduced in [Levine], and
independently in [Norway] (where semi-open sets are called _$\beta\,$ \-
sets_). The somewhat dual property of being contained in the interior of its
closure defines the class of _locally dense_ sets introduced in [CM]. There is
actually a rather abundant literature concerning various classes of
“generalised open sets”; see e.g. [Csa].

######  Corollary 7.4.

Assume that $\tau\subseteq\tau^{\prime}$, that $\tau$ is second-countable, and
that $\tau^{\prime}$ is Baire and second-countable. Then the following
assertions are equivalent.

  1. (a)

$\tau$ and $\tau^{\prime}$ are similar;

  2. (b)

$\mathcal{C}(\tau,\tau^{\prime})$ is $\tau^{\prime}$\- dense in $\mathbf{M}$;

  3. (c)

every $\tau^{\prime}\,$\- semi-open set is $\tau\,$\- semi-open;

  4. (d)

every $\tau^{\prime}\,$\- open set is $\tau\,$\- semi-open.

###### Proof.

By Proposition 7.2 and since $\tau^{\prime}$ is Baire, we know that
(a)$\implies$(b).

(b)$\implies$(c) Assume that $\mathcal{C}(\tau,\tau^{\prime})$ is
$\tau^{\prime}$\- dense in $\mathbf{M}$. Let $A\subseteq\mathbf{M}$ be
$\tau^{\prime}\,$\- semi-open, and choose a $\tau^{\prime}\,$\- open set
$O^{\prime}$ such that $O^{\prime}\subseteq
A\subseteq\overline{O^{\prime}}^{\,\tau^{\prime}}$. Denote by $O$ the
$\tau\,$\- interior of $O^{\prime}$, so that $O\subseteq O^{\prime}\subseteq
A$. Since $\mathcal{C}(\tau,\tau^{\prime})$ is $\tau^{\prime}$\- dense in
$\mathbf{M}$ and $O^{\prime}$ is $\tau^{\prime}\,$\- open,
$\mathcal{C}(\tau,\tau^{\prime})\cap O^{\prime}$ is $\tau^{\prime}\,$\- dense
in $O^{\prime}$; and since $\tau\subseteq\tau^{\prime}$, it follows that
$\overline{O^{\prime}}^{\,\tau^{\prime}}\subseteq\overline{\mathcal{C}(\tau,\tau^{\prime})\cap
O^{\prime}}^{\,\tau}$ Moreover, $\mathcal{C}(\tau,\tau^{\prime})\cap
O^{\prime}$ is contained in $O$ by the definition of
$\mathcal{C}(\tau,\tau^{\prime})$. So we see that $O\subseteq
A\subseteq\overline{O}^{\,\tau}$, and hence $A$ is $\tau\,$\- semi-open.

(c)$\implies$(d) is obvious.

(d)$\implies$(a) Assume (d). Since $\tau\subseteq\tau^{\prime}$, it is enough
to show that any set $A\subseteq\mathbf{M}$ with non-empty $\tau^{\prime}\,$\-
interior has non-empty $\tau\,$\- interior; which is obvious by (d). ∎

######  Remark 7.5.

The implications (b)$\implies$(c)$\implies$(d)$\implies$(a) do not require
$\tau$ and $\tau^{\prime}$ to be second-countable or Baire. Moreover, without
assuming that $\tau\subseteq\tau^{\prime}$, the proof of (b)$\implies$(c)
shows that if (b) holds true then, for any $\tau^{\prime}\,$\- semi-open set
$A\subseteq\mathbf{M}$, one can find a $\tau\,$\- open set $O$ such that
$O\subseteq A\subseteq\overline{O}^{\,\tau^{\prime}}$; and the proof of
(d)$\implies$(a) shows that if (d) holds true then any $\tau\,$\- dense set is
$\tau^{\prime}\,$\- dense.

######  Remark 7.6.

One cannot replace condition (c) in Corollary 7.4 above by the condition
“$\tau$ and $\tau^{\prime}$ have the same semi-open sets”. In fact, having the
same semi-open sets should be considered as a much stronger property than
similarity. For example, it follows from [Norway, Propositions 1 and 8] that
if $\tau$ and $\tau^{\prime}$ are two _regular_ topologies with the same semi-
open sets, then $\tau=\tau^{\prime}$. (Recall that a topology is regular if
every point of the space has a neighbourhood basis consisting of closed sets.)

###  7.2. More on points of WOT \- SOT continuity

In this sub-section, we add some examples of situations where one can conclude
that an operator $T\in{\mathcal{B}}_{1}(X)$ belongs to
$\mathcal{C}(\texttt{WOT},\texttt{SOT})$.

Our first example concerns isometries. Let us say that a projection operator
$P\in\mathcal{B}(X)$ is _uniformly strictly contractive_ if $\|P\|=1$ and,
whenever $(x_{n})$ is a sequence in $B_{X}$ such that $\|Px_{n}\|\to 1$, it
follows that $\|(I-P)x_{n}\|\to 0$. For example, any $L_{p}\,$\- projection,
$1\leq p<\infty$ is uniformly strictly contractive. (A projection
$P\in\mathcal{B}(X)$ is an $L_{p}\,$\- projection if $X={\rm
ran}(P)\oplus_{\ell_{p}}\ker(P)$; see [B&al].)

######  Example 7.7.

Let $X$ be reflexive. If $T\in{\mathcal{B}}_{1}(X)$ admits a left inverse
$L\in{\mathcal{B}}_{1}(X)$ such that the projection $P:=TL$ is uniformly
strictly contractive, then $T\in\mathcal{C}({\texttt{WOT}},{\texttt{SOT}})$.
In particular:

  1. (i)

any surjective isometry of $X$ belongs to
$\mathcal{C}({\texttt{WOT}},{\texttt{SOT}\mbox{${}^{*}$}})$;

  2. (ii)

if $E$ is any reflexive space and $1<p<\infty$, the forward shift $S$ on
$X:=\ell_{p}(\mathbb{Z}_{+},E)$ belongs to
$\mathcal{C}({\texttt{WOT}},{\texttt{SOT}})$.

###### Proof.

Note that the assumption made on $T$ implies that $T$ is an isometry since
$\|T\|\leq 1$ and $\|L\|\leq 1$.

Since $X$ is reflexive and separable, the set $\mathcal{C}(w,\|\,\cdot\,\|)$
is weakly dense in $B_{X}$; in particular, we have $\overline{\rm
span}\;\mathcal{C}(w,\|\,\cdot\,\|)=X$. Hence, by Proposition 2.13, it is
enough to show that
$T\bigl{(}\mathcal{C}(w,\|\,\cdot\,\|)\bigr{)}\subseteq\mathcal{C}(w,\|\,\cdot\,\|)$.

Let $z\in\mathcal{C}(w,\|\,\cdot\,\|)$, and let $(x_{n})$ be a sequence in
$B_{X}$ such that $x_{n}\xrightarrow{w}Tz$. Since $LTz=z$ and
$z\in\mathcal{C}(w,\|\,\cdot\,\|)$, we see that
$Lx_{n}\xrightarrow{\|\,\cdot\,\|}z$. Hence $\|TLx_{n}\|\to\|Tz\|=1$. Since
$P=TL$ is uniformly strictly contractive, it follows that $\|(I-TL)x_{n}\|\to
0$; and since $TLx_{n}\xrightarrow{\|\,\cdot\,\|}Tz$, we conclude that
$x_{n}\xrightarrow{\|\,\cdot\,\|}Tz$, which shows that
$Tz\in\mathcal{C}(w,\|\,\cdot\,\|)$.

Since the identity operator is a uniformly strictly contractive projection, we
see that any surjective isometry belongs to
$\mathcal{C}({\texttt{WOT}},{\texttt{SOT}})$, and (i) follows by duality. As
to (ii), take $L:=B$, the backward shift on $X=\ell_{p}(\mathbb{Z}_{+},E)$. ∎

Our second example follows from Proposition 2.14. Recall that a point $x\in
S_{X}$ is said to be a _strongly exposed point of $B_{X}$_ if there exists a
linear functional $x^{*}\in S_{X^{*}}$ such that $\langle x^{*},x\rangle=1$
and $x^{*}_{|B_{X}}$ has a strong maximum at $x$, i.e. every maximising
sequence $(x_{n})\subseteq B_{X}$ for $x^{*}$ converges in norm to $x$.

######  Example 7.8.

Let $X$ be reflexive with the Kadec-Klee property. If
$T\in{\mathcal{B}}_{1}(X)$ is such that $\|T\|=1$ and the weak closure of
$\mathcal{N}(T)$ contains the strongly exposed points of $B_{X}$, then
$T\in\mathcal{C}({\texttt{WOT}},{\texttt{SOT}})$.

###### Proof.

Since $X$ is reflexive, $B_{X}$ is the closed convex hull of its strongly
exposed points, by [Lind, Theorem 4]. So the result is clear by Proposition
2.14. ∎

######  Remark.

It is quite possible that $\mathcal{N}(T)$ contains all extreme points of
$B_{X}$ and yet $T$ is not an isometry. Here is a simple example. Let
$1<p<\infty$ and let $X:=X_{1}\oplus_{\ell_{p}}X_{2}$, where
$X_{1}:=(\mathbb{K}^{2},\|\,\cdot\,\|_{1})$ and $X_{2}$ is any reflexive
Banach space with the Kadec-Klee property. Then $X$ is reflexive with the
Kadec-Klee property, and the extreme points of $B_{X}$ are the points
$x=x_{1}\oplus x_{2}$, where $\|x_{1}\|^{p}+\|x_{2}\|^{p}=1$ and $x_{j}\in{\rm
Ext}\bigl{(}\|x_{j}\|B_{X_{j}}\bigr{)}$. Let $T:=A\oplus
I_{X_{2}}\in{\mathcal{B}}_{1}(X)$, where $A\in\mathcal{B}_{1}(X_{1})$ is any
non-isometric operator such that $\|Ae\|=\|Ae^{\prime}\|=1$, where
$(e,e^{\prime})$ is the canonical basis of $X_{1}$. Then $T$ is not an
isometry but $\mathcal{N}(T)\supseteq{\rm Ext}(B_{X})$.

We now prove a variant of Proposition 2.14 involving only “approximately
norming” vectors. Recall that the Banach space $X$ is said to have the
_Uniform Kadec-Klee property_ if the following holds true: for any
$\varepsilon>0$, there exists $\delta>0$ such that, whenever $(x_{n})$ is an
$\varepsilon\,$\- separated sequence in $B_{X}$ such that
$x_{n}\xrightarrow{w}x\in X$, it follows that $\|x\|<1-\delta$. (This property
was introduced by Huff [Hu].) For example, any uniformly convex space is UKK.
Also, if $T\in\mathcal{B}(X)$, let us say that a vector $z\in
X\setminus\\{0\\}$ is _$(1-\delta)\,$ \- norming_ for $T$ is
$\|Tz\|>(1-\delta)\|T\|\|z\|$. We denote by $\mathcal{N}_{\delta}(T)$ the set
of all $(1-\delta)\,$\- norming vectors for $T$.

######  Proposition 7.9.

Assume that $X$ is reflexive with the Uniform Kadec-Klee property, and let
$T\in{\mathcal{B}}_{1}(X)$. Assume that $\|T\|=1$ and that there exists some
constant $c>0$ such that, for every $\delta>0$, the set $\overline{\rm
conv}\,\bigl{(}\mathcal{N}_{\delta}(T)\cap S_{X}\bigr{)}$ contains the ball
$cB_{X}$. Then, $T\in\mathcal{C}(\emph{{WOT}},\emph{{SOT}})$.

###### Proof.

Let $(T_{n})$ be a sequence in ${\mathcal{B}}_{1}(X)$ such that
$T_{n}\xrightarrow{\texttt{WOT}}T$, and let $x\in X$. We have to show that
$T_{n}x\xrightarrow{\|\,\cdot\,\|}Tx$. So we fix $\alpha>0$, and we intend to
show that $\|T_{n}x-Tx\|\leq\alpha$ for all large enough $n$. Moreover, we may
assume that $\|x\|\leq 1$.

######  Claim 7.10.

Let $\varepsilon>0$, and let $\delta=\delta(\varepsilon)$ be given by the UKK
property. For any $z\in\mathcal{N}_{\delta}(T)\cap S_{X}$, we have
$\overline{\,\lim}\,\|T_{n}z-Tz\|\leq\varepsilon$.

###### Proof of Claim 7.10.

Since $z\in\mathcal{N}_{\delta}(T)\cap S_{X}$, we have $\|Tz\|>1-\delta$; and
since $T_{n}z\xrightarrow{w}Tz$, it follows that the sequence $(T_{n}z)$ has
no $\varepsilon\,$\- separated subsequence. From that, it is not hard to
deduce that one can find an increasing sequence of integers $(n_{k})_{k\geq
0}$ such that $\|T_{n_{k}}z-T_{n_{k^{\prime}}}z\|<\varepsilon$ for every
$k,k^{\prime}\geq 0$. Fixing $k$ and letting $k^{\prime}\to\infty$, we get
$\|T_{n_{k}}z-Tz\|\leq\varepsilon$ for all $k\geq 0$. So we have
$\underline{\,\lim}\,\|T_{n}-Tz\|\leq\varepsilon$; and since this can be
applied to any subsequence of $(T_{n})$, the result follows. ∎

Now, let us choose $\varepsilon:=\alpha c/3$, and let $\delta$ be given by the
UKK property. By assumption on $T$, one can find $u\in X$ which is a convex
combination of vectors $z_{1},\dots,z_{N}\in\mathcal{N}_{\delta}(T)\cap S_{X}$
such that $\|u-cx\|<\alpha c/3$. By Claim 7.10, we have
$\|T_{n}z_{i}-Tz_{i}\|\leq\varepsilon=\alpha c/3$ for $i=1,\dots,N$ if $n$ is
large enough; and then $\|T_{n}u-Tu\|\leq\alpha c/3$ by convexity. By the
triangle inequality and since $\|T\|,\|T_{n}\|\leq 1$, it follows that
$\|T_{n}(cx)-T(cx)\|\leq\alpha c$ for all large enough $n$; so
$\|T_{n}x-Tx\|\leq\alpha$, as required. ∎

######  Remark.

By the Hahn-Banach theorem and since $\mathcal{N}_{\delta}(T)$ is balanced for
any $\delta>0$, the assumption in Proposition 7.9 can be written as follows:

$\inf_{x^{*}\in S_{X^{*}}}\inf_{\delta>0}\sup_{z\in\mathcal{N}_{\delta}(T)\cap
S_{X}}|\langle x^{*},z\rangle|>0.$

We conclude this sub-section with a rather natural question. As observed in
[Kan, Proposition 2.6], if the Banach space $X$ is strictly convex then any
operator $T\in{\mathcal{B}}_{1}(X)$ such that $\|T\|=1$ and $\overline{{\rm
span}}\bigl{(}\mathcal{N}(T)\bigr{)}=X$ (in particular, any isometry) has to
be an _extreme point_ of ${\mathcal{B}}_{1}(X)$. In fact, without any
assumption on $X$, any $T\in{\mathcal{B}}_{1}(X)$ such that $\|T\|=1$,
$\mathcal{N}(T)\cap S_{X}\subseteq{\rm Ext}(B_{X})$ and $\overline{{\rm
span}}\bigl{(}\mathcal{N}(T)\bigr{)}=X$ is an extreme point of
${\mathcal{B}}_{1}(X)$. By Proposition 2.11, it follows that for $X=\ell_{2}$,
any $T\in\mathcal{C}(\texttt{WOT},\texttt{SOT})$ is an extreme point of
${\mathcal{B}}_{1}(X)$. Also, it is easy to see that
$\mathcal{C}(\texttt{WOT},\texttt{SOT})$ is an extreme subset of
${\mathcal{B}}_{1}(X)$, for any $X$: if $A,B\in{\mathcal{B}}_{1}(X)$ and
$\frac{A+B}{2}\in\mathcal{C}(\texttt{WOT},\texttt{SOT})$, then
$A\in\mathcal{C}(\texttt{WOT},\texttt{SOT})$ and
$B\in\mathcal{C}(\texttt{WOT},\texttt{SOT})$. This motivates the following
question.

######  Question 7.11.

For which reflexive spaces $X$ is it true that every
$T\in\mathcal{C}(\texttt{WOT},\texttt{SOT})$ is an extreme point of
${\mathcal{B}}_{1}(X)$?

###  7.3. Typical properties related to Fredholm theory

In this final sub-section, we gather a few remarks and some questions related
to Lemma 5.4. Since we consider spectral properties, we assume that
$\mathbb{K}=\mathbb{C}$. We begin by observing that the SOT∗ version of Lemma
5.4 holds true if $X$ is reflexive and has the MAP. In other words:

######  Fact 7.12.

If $X$ is reflexive and has the MAP then a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT}\mbox{${}^{*}$})$ is not upper semi-
Fredholm.

###### Proof.

By the proof of Lemma 5.4, it is in fact enough to show that the finite rank
contraction operators are SOT∗\- dense in ${\mathcal{B}}_{1}(X)$; and this
will follow if we can show that there exists a sequence of finite rank
operators $(R_{n})\subseteq{\mathcal{B}}_{1}(X)$ such that
$R_{n}\xrightarrow{\texttt{SOT}\mbox{${}^{*}$}}I_{X}$.

Since $X$ has the MAP, there is a sequence of finite rank operators
$(A_{k})\subseteq{\mathcal{B}}_{1}(X)$ such that
$A_{k}\xrightarrow{\texttt{SOT}}I_{X}$. Then
$A_{k}\xrightarrow{\texttt{WOT}}I_{X}$, hence
$A_{k}^{*}\xrightarrow{\texttt{WOT}}I_{X^{*}}$ since $X$ is reflexive. By
Mazur’s Theorem, it follows that $I_{X^{*}}$ belongs to the SOT \- closure of
the set $\bigl{\\{}A_{k}^{*};\;k\geq n\bigr{\\}}$, for any $n\in\mathbb{N}$.
So, by metrizability of $\bigl{(}\mathcal{B}_{1}(X^{*}),\texttt{SOT}\bigr{)}$,
one can find a sequence $(S_{n})$ of convex combinations of the $A_{k}^{*}$
such that $S_{n}\xrightarrow{\texttt{SOT}}I_{X^{*}}$ and $S_{n}\in{\rm
conv}\,\bigl{\\{}A_{k}^{*};\;k\geq n\bigr{\\}}$ for all $n\in\mathbb{N}$. The
sequence of finite-rank operators $(R_{n})$ defined by $R_{n}:=S_{n}^{*}$ is
then such that $R_{n}\xrightarrow{\texttt{SOT}}I_{X}$ (because
$A_{k}\xrightarrow{\texttt{SOT}}I_{X}$) and
$R_{n}^{*}\xrightarrow{\texttt{SOT}}I_{X^{*}}$. ∎

The next proposition shows in particular that Lemma 5.4 can be improved when
$X$ is an $\ell_{p}\,$\- direct sum or a $c_{0}\,$\- direct sum of finite-
dimensional spaces..

######  Proposition 7.13.

Assume that there exists a sequence $(P_{N})_{N\geq 0}$ of finite rank $M$\-
like projections on $X$ such that $P_{N}\xrightarrow{\emph{{SOT}}}I$. Then a
typical $T\in({\mathcal{B}}_{1}(X),\emph{{SOT}})$ is such that $T-\lambda I$
is not upper semi-Fredholm for any $\lambda\in\overline{\,\mathbb{D}}$. In
particular, the essential spectrum of a typical $T\in{\mathcal{B}}_{1}(X)$ is
equal to $\overline{\,\mathbb{D}}$.

###### Proof.

We use an improved version of Fact 5.5.

######  Fact 7.14.

Let $\lambda\in\overline{\,\mathbb{D}}$. A typical $T\in{\mathcal{B}}_{1}(X)$
has the following property: for any $\varepsilon>0$ and any $n\in\mathbb{N}$,
there exists a subspace $E\subseteq X$ with $\dim(E)>n$ such that
$\|(T-\lambda I)_{|E}\|<\varepsilon$.

###### Proof of Fact 7.14.

Let us denote by $\mathcal{G}$ the set of all $T\in{\mathcal{B}}_{1}(X)$ with
the above property. The set $\mathcal{G}$ is SOT -$\,G_{\delta}$; so we just
have to check that $\mathcal{G}$ is SOT \- dense in ${\mathcal{B}}_{1}(X)$.
Let $A\in{\mathcal{B}}_{1}(X)$ be arbitrary, and for any $N\geq 0$, set
$T_{N}:=P_{N}AP_{N}+\lambda(I-P_{N})$. Since the projection $P_{N}$ is $M$\-
like, we see that $T_{N}\in{\mathcal{B}}_{1}(X)$; and
$T_{N}\xrightarrow{\texttt{SOT}}A$. Moreover, we have $T_{N}\equiv\lambda I$
on $E:=\ker(P_{N})$, so $T\in\mathcal{G}$. ∎

It follows from Fact 7.14 that for any fixed
$\lambda\in\overline{\,\mathbb{D}}$, a typical $T\in{\mathcal{B}}_{1}(X)$ is
such that $T-\lambda I$ is not upper semi-Fredholm. Now, let $\Lambda$ be a
countable dense subset of $\overline{\,\mathbb{D}}$. By the Baire Category
Theorem, a typical $T\in{\mathcal{B}}_{1}(X)$ is such that $T-\lambda I$ is
not upper semi-Fredholm for any $\lambda\in\Lambda$. Since the set of all
$\lambda\in\mathbb{C}$ such that $T-\lambda I$ is not upper semi-Fredholm is
closed in $\mathbb{C}$, the result follows. ∎

######  Remark 7.15.

The SOT∗ version of Proposition 7.13 holds true (with exactly the same proof)
if one assumes that $P_{N}\xrightarrow{\texttt{SOT}\mbox{${}^{*}$}}I$. This
applies in particular to $X=c_{0}$ or $\ell_{p}$, $1<p<\infty$.

######  Remark 7.16.

Let $X=\ell_{1}$. It is proved in [GMM2] that a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ has the following two properties:
$T^{*}$ is an isometry, and every $\lambda\in\mathbb{D}$ is an eigenvalue of
$T$ of infinite multiplicity. The proof of the first property is rather
simple, but the proof of the second property given in [GMM2] is a bit
technical. By Proposition 7.13, we now see that the typicality of the second
property actually _follows_ from the typicality of the first one. Indeed, if
$T^{*}$ is an isometry, then $T-\lambda I$ is surjective for every
$\lambda\in\mathbb{D}$ since $(T-\lambda I)^{*}$ is an embedding, and hence
$\dim\ker(T-\lambda I)=\infty$ if $T-\lambda I$ is not Fredholm.

Assume that the Banach space $X$ has the MAP. As already pointed out, it
follows from Lemma 5.4 that if a typical $T\in{\mathcal{B}}_{1}(X)$ has closed
range, then a typical $T\in{\mathcal{B}}_{1}(X)$ has an infinite-dimensional
kernel. As the next remark shows, saying that a typical
$T\in{\mathcal{B}}_{1}(X)$ has closed range is in fact the same as saying that
a typical $T\in{\mathcal{B}}_{1}(X)$ is surjective.

######  Remark 7.17.

If $X$ has the MAP, then a typical
$T\in\bigl{(}{\mathcal{B}}_{1}(X),\texttt{SOT}\bigr{)}$ has dense range.
Consequently, a typical $T\in{\mathcal{B}}_{1}(X)$ has closed range if and
only if a typical $T\in{\mathcal{B}}_{1}(X)$ is surjective.

###### Proof.

For any $z\in X$, set
$\mathcal{G}_{z}:=\bigl{\\{}T\in{\mathcal{B}}_{1}(X);\;z\in\overline{\,{\rm
ran}(T)}\bigr{\\}}$. Let also $D$ be a countable dense subset of $X$. Then an
operator $T\in{\mathcal{B}}_{1}(X)$ has dense range if and only if
$T\in\bigcap_{z\in D}\mathcal{G}_{z}$. Moreover, each set $\mathcal{G}_{z}$ is
easily seen to be SOT -$\,G_{\delta}$ in ${\mathcal{B}}_{1}(X)$. So it is
enough to show that $\mathcal{G}_{z}$ is SOT \- dense in
${\mathcal{B}}_{1}(X)$, for any $z\in X$.

Let $\mathcal{U}$ be a non-empty open set in
$({\mathcal{B}}_{1}(X),\texttt{SOT})$. Since $X$ has the MAP, one can find a
finite rank operator $A$ such that $A\in\mathcal{U}$ and $\|A\|<1$. Since
$\dim(X)=\infty$, we can choose $x\in\ker(A)$ with $x\neq 0$, and then
$x^{*}\in X^{*}$ such that ${\langle x^{*},x\rangle}=1$. Let $\varepsilon>0$,
and set $T:=A+\varepsilon\,x^{*}\otimes z$. If $\varepsilon$ is small enough,
then $\|T\|\leq 1$ and $T\in\mathcal{U}$. Moreover, since $Tx=\varepsilon z$,
we have $z\in{\rm ran}(T)$ and hence $T_{N}\in\mathcal{G}_{z}$. ∎

We finish the paper with the following two questions:

######  Question 7.18.

Is a typical $T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ surjective when
$X=\ell_{p}$, $1<p<2$?

We have stated Question 7.18 for $1<p<2$ only because the answer is already
known for all other values of $p$. Indeed, the answer is “Yes” for
$X=\ell_{2}$ by [EM], and also “Yes” for $X=\ell_{1}$ by [GMM2, Theorem 4.1].
On the other hand, since a typical $T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$
is not invertible (by [GMM2, Proposition 2.1], see also Proposition 7.13), the
answer is “No” if $X=\ell_{p}$, $p>2$ or $c_{0}$, because a typical
$T\in({\mathcal{B}}_{1}(X),\texttt{SOT})$ is one-to-one by [GMM2, Theorem
2.3]. The answer is also “No” for the topology SOT∗ and any $1<p<\infty$, for
the same reason. It follows from Theorem 4.1 that the answer is “No” as well
for the topology SOT if $\mathbb{K}=\mathbb{R}$ and $p=3/2$.

######  Question 7.19.

Assume that $X$ has the MAP. Is it true that a typical
$T\in\bigl{(}{\mathcal{B}}_{1}(X),\texttt{SOT}\bigr{)}$ is such that
$T-\lambda I$ has dense range for every $\lambda\in\mathbb{C}$?

Competing interests: The authors declare none.

## References