The transition operator of a random walk perturbated by sparse
potentials111private notes .

Takuya Mine222Faculty of Arts and Sciences, Kyoto Institute of Technology,
Matsugasaki, Sakyo-ku, Kyoto 606-8585, Japan and Nobuo Yoshida333Graduate
School of Mathematics, Nagoya University, Furocho, Chikusaku, Nagoya 464-8602,
Japan

###### Abstract

We consider an operator $P_{V}=(1+V)P$ on $\ell^{2}({\mathbb{Z}}^{d})$, where
$P$ is the transition operator of a symmetric irreducible random walk, and $V$
is a “sparse” potential. We first characterize the essential spectra of this
operator. Secondly, we prove that all the eigenfunctions which correspond to
discrete spectra decay exponentially fast. Thirdly, we give a sufficient
condition for this operator to have an absolute spectral gap at the right edge
of the spectra. Finally, as an application of the absolute spectral gap and
the exonential decay of the eigenfunctions, we prove a limit theorem for the
random walk under the Gibbs measure associated to the potential $V$.

##  1 Introduction

In this article, we investigate the spectral properties of an operator
$P_{V}=(1+V)P$ on $\ell^{2}({\mathbb{Z}}^{d})$, where $P$ is the transition
operator of a symmetric irreducible random walk, and $V$ is a nonnegative
bounded function. Here, $V$ is supposed to be a so called “sparse” potential,
of which a typical example is that with the property

$\min\\{|x-y|\;;\;x,y\in{\rm supp}V,\;x\neq y,\;|x|\geq r,\;|y|\geq
r\\}\stackrel{{\scriptstyle r\rightarrow\infty}}{{\longrightarrow}}\infty,$
(1.1)

Before introducing the contents of this article, we start by explaining the
probabilistic backgrounds which brought us to the study of the spectra of
$P_{V}$.

Let $(S_{n})_{n\in{\mathbb{N}}}$ be a random walk on a probability space
$(\Omega,{\cal F},P)$ which is associated with the transition operator $P$. We
suppose that $S_{0}=0$. Then, the semigroup $P_{V}^{n}$ ($n\in{\mathbb{N}}$)
is expressed by the discrete version of the Feynmann-Kac formula.

$P_{V}^{n}f(x)=E\left[f(x+S_{n})\prod^{n-1}_{j=0}(1+V(x+S_{j}))\right],\;\;f\in\ell^{2}({\mathbb{Z}}^{d}),\;\;x\in{\mathbb{Z}}^{d}.$
(1.2)

Similarly, if we restrict $P_{V}$ to $\ell^{2}({\mathbb{N}}^{d})$ by imposing
Dirichlet boundary condition, and denote the restriction by $P^{+}_{V}\in{\cal
B}(\ell^{2}({\mathbb{N}}^{d}))$, then, we obtain

$\displaystyle(P_{V}^{+})^{n}f(x)$ $\displaystyle=$ $\displaystyle
E\left[f(x+S_{n}){\bf
1}\\{S_{j}\in{\mathbb{N}}^{d},\;j=1,\ldots,n\\}\prod^{n-1}_{j=0}(1+V(x+S_{j}))\right],$
(1.3) $\displaystyle f\in\ell^{2}({\mathbb{N}}^{d}),\;\;x\in{\mathbb{N}}^{d}.$

Let
$\partial{\mathbb{N}}^{d}=\bigcup_{\alpha=1}^{d}\\{x\in{\mathbb{N}}^{d}\;;\;x_{\alpha}=0\\}$,
and $\beta>0$ be a positive parameter. In connection with statistical physics,
the following sequence $\mu_{N}$, $N\geq 1$ of measures on $(\Omega,{\cal F})$
are studied.

$\mu_{N}(d\omega)={1\over Z_{N}}E\left[{\bf
1}\\{S_{j}\in{\mathbb{N}}^{d},\;j=1,\ldots,N\\}\exp\left(\beta\sum_{j=1}^{N-1}{\bf
1}_{\partial{\mathbb{N}}^{d}}(S_{j})\right):d\omega\right],$ (1.4)

where $Z_{N}$ is the normalizing constant. The above measure is related with
the operator $P_{V}^{+}$ with $V=\exp(\beta{\bf
1}_{\partial{\mathbb{N}}^{d}})-1$ via the formula (1.3). Under the measure
$\mu_{N}$, two competing effects coexist: The paths of the random walk is
attracted to $\partial{\mathbb{N}}^{d}$ by the potential to lower the eneregy:

$-\beta\sum_{j=1}^{N-1}{\bf 1}_{\partial{\mathbb{N}}^{d}}(S_{j}).$

On the other hand, paths of the random walk are pushed away from
$\partial{\mathbb{N}}^{d}$ by the entropic repulsion. These competing effects
cause the phase transition, called wetting transition, which is expressed as
follows, cf. [3, 7]. There exists $\beta_{c}\in(0,\infty)$ such that

$\displaystyle\lim_{N\rightarrow\infty}$ $\displaystyle Z_{N}^{1/N}$
$\displaystyle=\max\\{|\lambda|\;;\;\lambda\in\sigma(P^{+}_{V})\\}$ (1.7)
$\displaystyle\left\\{\begin{array}[]{ll}=1,&\mbox{if $\beta\leq\beta_{c}$
(delocalized phase)},\\\ >1,&\mbox{if $\beta>\beta_{c}$ (localized
phase)}.\end{array}\right.$

Intuitively, if $N$ is large, the random walk under the measure $\mu_{N}$
behaves as if there were no potentials in the delocalized phase, while it is
localized near $\partial{\mathbb{N}}^{d}$ in the localized phase. In fact, for
$d=1$, these intuitive pictures are justified mathematically in [3]. In
particular, it was shown there that, in the localized phase, the measure
$\mu_{N}$ converges as $N\rightarrow\infty$ to the law of a positively
recurrent Markov chain.

Technically, a crutial step in the proof of the limit theorem in [3] referred
to above is the existence of the absolute spectral gap of $P^{+}_{V}$ (cf.
(1.11) below), and it is here that the particularity of one dimension comes
into play. Indeed, if $d=1$, the operator $P^{+}_{V}$ is a compact
perturbation of $P^{+}$, from which the existence of the absolute spectral gap
follows via Weyl’s essential spectrum theorem. Unfortunately,
$P^{+}_{V}-P^{+}$ is no longer a compact in higher dimensions. The present
article comes out as a partial progress in the effort to carry the results in
[3] over to higher dimensions and therefore, to noncompact perturbation cases.

We now explain contents of this paper a little more in detail. For simplicity,
we consider the whole lattice ${\mathbb{Z}}^{d}$, rather than its first
quadrant ${\mathbb{N}}^{d}$.

Let $p:{\mathbb{Z}}^{d}\rightarrow[0,\infty)$ be a transition probability of a
symmetric irreducible random walk. We define
$P:\ell^{2}({\mathbb{Z}}^{d})\rightarrow\ell^{2}({\mathbb{Z}}^{d})$ by

$Pf(x)=\sum_{y\in{\mathbb{Z}}^{d}}p(x-y)f(y),\;\;f\in\ell^{2}({\mathbb{Z}}^{d}),\;\;x\in{\mathbb{Z}}^{d}.$
(1.8)

We consider a perturbation
$P_{V}:\ell^{2}({\mathbb{Z}}^{d})\rightarrow\ell^{2}({\mathbb{Z}}^{d})$ of the
operator $P$ of the form

$P_{V}=(1+V)P,$ (1.9)

where $V:{\mathbb{Z}}^{d}\rightarrow[0,\infty)$ is a bounded function. Here,
as usual, $V$ is regarded as the multiplication operator. In the context of
probability theory, this type of perturbation is quite natural as we have
already seen.

In this article, we investigate the structure of spectra $\sigma(P_{V})$, as
well as its consequence on the long time behavior of the semigruop (1.2). In
this context, an important quantity is the right edge

$r(P_{V})=\max\sigma(P_{V})$

of the spectra. More precisely, we are interested in the following properties

$\displaystyle r(P_{V})$ $\displaystyle>$
$\displaystyle\sup\\{\lambda\;;\;\lambda\in\sigma(P_{V}),\;\lambda\neq
r(P_{V})\\}\;\;\mbox{({\bf spectral gap})},$ (1.10) $\displaystyle r(P_{V})$
$\displaystyle>$
$\displaystyle\sup\\{|\lambda|\;;\;\lambda\in\sigma(P_{V}),\;|\lambda|\neq
r(P_{V})\\}\;\;\mbox{({\bf absolute spectral gap}).}$ (1.11)

As it turns out in the sequel, the following quantity plays an important role
in studying these properties.

$v_{0}=\inf_{n\in{\mathbb{N}}}\sup_{|x|\geq n}V(x).$ (1.12)

For example, as is discussed earlier in [3], properties (1.10) and (1.11) are
relatively easy to obtain when $v_{0}=0$, i.e., $V$ decays at infinity.
Indeed, the multiplication operator $V$ is compact in this case, and hence, by
Weyl’s essential spectrum theorem, cf. [2, p.358, Proposition 4.2 (e)], [6,
p.112, Theorem XIII.14], the set of essential spectra $\sigma_{\rm
ess}(P_{V})$ is the same as $\sigma_{\rm ess}(P)=\sigma(P)=[\ell(P),1]$
($-1\leq\ell(P)<1$). Thus, one immediately obtains (1.10) as soon as one knows
that $r(P_{V})>1$. Then, it can be improved to (1.11) under reasonable
additional assumptions on the transition function $p$.

In this article, we are mainly interested in the case of $v_{0}>0$, where the
multiplication operator $V$ is no longer compact. To compensate the lack of
the decay of $V$ at infinity, we will assume that the support of $V$ is sparse
enough, to ensure that the perturbation is not too large to control, see
(1.26) below for the precise formulation of the sparseness. To the best of our
knowledge, research in this direction was initiated in [5], where the
Schrödinger operator $-{d^{2}\over dx^{2}}+V$ on the real line is discussed.

Firstly, we characterize in Theorem 1.2.1 the set $\sigma_{\rm ess}(P_{V})$ of
essential spectra of $P_{V}$. Here, we adapt the method in [4] to the present
setting. Theorem 1.2.1 has a corollary, which tells us that the excess
$\sigma_{\rm ess}(P_{V})\backslash\sigma(P)$ is nonempty if, e.g., $d\leq 2$
and $v_{0}>0$, cf. (1.12). For $d\geq 3$, we have the same conclusion if
$v_{0}$ is sufficiently large. Thus, the spectral aspect of the operator
$P_{V}$ is indeed different from the compact perturbation case.

Secondly, we prove in Theorem 1.2.3 that that all the eigenfunctions which
correspond to discrete spectra decay exponentially fast.

Thirdly, we establish the spectral gap (1.10) in Theorem 1.2.4, and then,
improve it to the absolute spectral gap (1.11) in Corollary 1.2.5.

Finally, as an application of the absolute spectral gap (1.11) obtained in
Corollary 1.2.5 and the exponential decay of the eigenfunctions (Theorem
1.2.3), we prove a limit theorem for the random walk in the potential $V$.
More precisely, we consider a Gibbs measures $\mu_{N}$ ($N\in{\mathbb{N}}$) on
$(\Omega,{\cal F})$ given by

$\mu_{N}(d\omega)={1\over
Z_{N}}E\left[\prod^{N-1}_{j=0}(1+V(S_{j})):d\omega\right],$ (1.13)

where $Z_{N}$ is the normalizing constant. Then, we will prove that the
measure $\mu_{N}$ converges as $N\rightarrow\infty$ to a positively recurrent
Markov chain on ${\mathbb{Z}}^{d}$, cf. Theorem 1.3.1 below.

The rest of this article is organized as follows.

###### Contents

  1. 1 Introduction
    1. 1.1 General notations
    2. 1.2 Spectra of the operator $P_{V}$
    3. 1.3 Limit of the random walk in potential $V$
    4. 1.4 An example in dimension one
  2. 2 Proof of Theorem 1.2.1 and Corollary 1.2.2
    1. 2.1 Outline
    2. 2.2 Lemmas
  3. 3 Proof of Theorem 1.2.3
    1. 3.1 Outline
    2. 3.2 Lemmas
  4. 4 Proof of Theorem 1.2.4, Corollary 1.2.5 and Proposition 1.2.6
    1. 4.1 Outline
    2. 4.2 Lemmas

###  1.1 General notations

$\bullet$ For a normed vector space $X$, we denote the totality of bounded
linear operators $T:X\rightarrow X$ by ${\cal B}(X)$. For $T\in{\cal B}(X)$,
$\|T\|$, $\sigma(T)$ and $\rho(T)$ stands respectively for its operator norm,
the totality of its spectra, and that of its resolvents. If $X$ is a Hilbert
space and $T\in{\cal B}(X)$ is self-adjoint, we write

$r(T)=\max\sigma(T),\;\;\;\ell(T)=\min\sigma(T).$ (1.14)

$\bullet$ The Banach space $\ell^{p}({\mathbb{Z}}^{d})$ will be abbreviated by
$\ell^{p}$. For a subset $S\subset{\mathbb{Z}}^{d}$, $\ell^{p}(S)$ is
identified with the totality of $f\in\ell^{p}$ which vanish outside $S$.

$\bullet$ For $T\in{\cal B}(\ell^{2})$, we write its kernel by
$T(x,y)\stackrel{{\scriptstyle\rm
def}}{{=}}\langle\>\delta_{x},T\delta_{y}\>\rangle$
($x,y\in{\mathbb{Z}}^{d}$).

$\bullet$ For $u\in\ell^{p}$ and $u\in\ell^{q}$ ($p,q\in[1,\infty]$, ${1\over
p}+{1\over q}=1$) let
$\langle\>u,v\>\rangle=\sum_{x\in{\mathbb{Z}}^{d}}u(x)v(x)^{*}$, where $c^{*}$
denotes the complex conjugate of $c\in{\mathbb{C}}$.

###  1.2 Spectra of the operator $P_{V}$

Let $p:{\mathbb{Z}}^{d}\rightarrow[0,\infty)$ be a transition probability of a
random walk, i.e., $\sum_{x\in{\mathbb{Z}}^{d}}p(x)=1$. Additionally, we
assume that

$\displaystyle\mbox{({\bf symmetry}) $p(x)=p(-x)$ for all
$x\in{\mathbb{Z}}^{d}$};$ (1.15) $\displaystyle\mbox{({\bf finite range}) $p$
is supported on a finite set};$ (1.16) (irreducibility) for all
$x\in{\mathbb{Z}}^{d}$, there exists $n\in{\mathbb{N}}$ such that
$p_{n}(x)>0$, (1.17)

where $p_{n}$ denotes the $n$-fold convolution. Then, we define $P\in{\cal
B}(\ell^{2})$ and $P_{V}\in{\cal B}(\ell^{2})$ respectively by (1.8) and
(1.9).

Let $\widehat{p}(\theta)=\sum_{x\in{\mathbb{Z}}^{d}}p(x)\exp({\bf
i}\theta\cdot x)$, $\theta\in[-\pi,\pi]^{d}$. Then, the set $\sigma(P)$ of the
spectra of $P$ is the interval $[\ell(P),1]$, where
$\ell(P)=\min\widehat{p}\in[2p(0)-1,1)$. For
$\lambda\in{\mathbb{C}}\backslash\sigma(P)$, let $G_{\lambda}\in{\cal
B}(\ell^{2})$ be the resolvent operator

$G_{\lambda}=(\lambda-P)^{-1}.$ (1.18)

We then define

$g_{\lambda}(x)\stackrel{{\scriptstyle\rm def}}{{=}}\lambda
G_{\lambda}(0,x),\;\;\lambda\in{\mathbb{R}}\backslash\sigma(P),\;x\in{\mathbb{Z}}^{d}.$
(1.19)

In the sequel, we need to deal with essential spectra of operators which are
not self-adjoint. To do so, we adopt the following definition via the theory
of Fredholm operator. Let $X$ be a Banach space and $T\in{\cal B}(X)$. We say
that $T$ is a Fredholm operator if

$\mbox{${\rm Ran}T$ is closed, $\dim{\rm Ker}T<\infty$, and $\dim\left(X/{\rm
Ran}T\right)<\infty$}.$ (1.20)

We denote the totality of Fredholm operator by ${\rm F}(X)$. We then define
the set $\sigma_{\rm ess}(T)\subset{\mathbb{C}}$ of essential spectra of $T$
by

$\lambda\in\sigma_{\rm ess}(T)\;\;\Longleftrightarrow\;\;\lambda-T\not\in{\rm
F}(X).$ (1.21)

If $X$ is a Hilbert space and $T$ is self-adjoint, then,

$T\in{\rm F}(X)\;\;\Longleftrightarrow\;\;\dim{\rm
Ker}T<\infty\;\mbox{and}\;0\not\in\overline{\sigma(T)\backslash\\{0\\}}.$
(1.22)

cf. [2, p.359, Proposition 4.6].

In what follows, we will exploit the following characterization of the
essential spectra of the multiplication operator $V$.

$v\in\sigma_{\rm
ess}(V)\;\;\Longleftrightarrow\;\;\sharp\\{x\in{\mathbb{Z}}^{d}\;;\;|V(x)-v|<\varepsilon\\}=\infty\;\;\mbox{for
all $\varepsilon>0$},$ (1.23)

where $\sharp$ stands for the cardinality. Moreover,

$\max\sigma_{\rm ess}(V)=v_{0},$ (1.24)

where $v_{0}$ is defined by (1.12). Let us consider an inner product and the
associated norm.

$\langle\>f,g\>\rangle_{V}\stackrel{{\scriptstyle\rm
def}}{{=}}\langle\>(1+V)^{-1}f,g\>\rangle,\;\;\|f\|_{V}=\sqrt{\langle\>f,f\>\rangle_{V}},\;\;f,g\in\ell^{2}.$
(1.25)

We note that the norms $\|\cdot\|$ and $\|\cdot\|_{V}$ are equivalent. The
Hilbert space $(\ell^{2},\|\cdot\|_{V})$ will be denoted by $\ell^{2}_{V}$.
Then, the operator $P_{V}\in{\cal B}(\ell^{2}_{V})$ is self-adjoint. As a
consequence, all the spectra of $P_{V}$ are real numbers.

We first identify the set $\sigma_{\rm ess}(P_{V})$ of the essential spectra.

###### Theorem 1.2.1

Suppose that $V:{\mathbb{Z}}^{d}\rightarrow[0,\infty)$ is a bounded function
such that

$a_{\varepsilon}(x)\stackrel{{\scriptstyle\rm
def}}{{=}}\sum_{y\in{\mathbb{Z}}^{d}\atop y\neq
x}\sqrt{V(x)V(y)}\exp(-\varepsilon|x-y|)\stackrel{{\scriptstyle|x|\rightarrow\infty}}{{\longrightarrow}}0\;\;\mbox{for
any $\varepsilon>0$}.$ (1.26)

Then,

$\sigma_{\rm ess}(P_{V})=\sigma(P)\cup\Lambda_{V},$ (1.27)

where

$\Lambda_{V}=\\{\lambda\in{\mathbb{R}}\backslash\sigma(P)\;;\;\mbox{there
exists $v\in\sigma_{\rm ess}(V)\backslash\\{0\\}$ such that
$g_{\lambda}(0)=1+v^{-1}$}\\}.$ (1.28)

Remark Condition (1.26) is satisfied, not only when
$V(x)\stackrel{{\scriptstyle|x|\rightarrow\infty}}{{\longrightarrow}}0$, but
also when the support of $V$ is sufficiently sparse. For example, the
condition (1.26) follows from (1.1). Indeed, suppose that $|x|\geq 2r$. Then,
$|x-y|\geq r$ if $|y|\leq r$. Thus, letting ${\rm d}(r)$ denote the minimum on
the left-hand side of (1.1), we have

$\displaystyle a_{\varepsilon}(x)$ $\displaystyle\leq$
$\displaystyle\|V\|_{\infty}\left(\sum_{|y|\leq r}+\sum_{|y|>r\atop y\neq
x}\right)\exp(-\varepsilon|x-y|)$ $\displaystyle\leq$ $\displaystyle
Cr^{d}\exp(-\varepsilon r)+C\exp\left(-{\varepsilon{\rm d}(r)\over
2}\right)\sum_{y\in{\mathbb{Z}}^{d}\atop y\neq
0}\exp\left(-{\varepsilon|y|\over 2}\right)\stackrel{{\scriptstyle
r\rightarrow\infty}}{{\longrightarrow}}0.$

Theorem 1.2.1 has the following corollary, which says that, unlike the compact
perturbation case, we may find essential spectra of $P_{V}$ outside
$\sigma(P)$.

###### Corollary 1.2.2

a)

$\sigma_{\rm ess}(P_{V})\cap(1,\infty)\neq\emptyset$, if and only if

$\mbox{$v_{0}>0$ and $1+v_{0}^{-1}=g_{\lambda_{0}}(0)$ for some
$\lambda_{0}\in(1,\infty)$},$ (1.29)

where $v_{0}$ is defined by (1.12). Moreover, (1.29) implies that
$\lambda_{0}=\max\sigma_{\rm ess}(P_{V})$.

b)

$\sigma_{\rm ess}(P_{V})\cap(-\infty,\ell(P))\neq\emptyset$, if and only if

$\mbox{$\ell(P)<0$, $v_{0}>0$ and $1+v_{0}^{-1}=g_{\lambda_{0}}(0)$ for some
$\lambda_{0}\in(-\infty,\ell(P))$}.$ (1.30)

Remark It follows from (2.12) that the condition (1.29) is satisfied whenever
$\sigma_{\rm ess}(V)\backslash\\{0\\}\neq\emptyset$ if $d\leq 2$. Similarly,
we see from (2.12) that the condition (1.30) is satisfied whenever
$\sigma_{\rm ess}(V)\backslash\\{0\\}\neq\emptyset$ if $d\leq 2$ and
$\ell(P)<0$.

The next result (Theorem 1.2.3) deals with discrete spectra of $P_{V}$. More
precisely, it states that the corresponding eigenfunctions decay exponentially
fast. Together with the absolute spectral gap (Corollary 1.2.5), this theorem
plays an important role in the proof of Theorem 1.3.1 below.

###### Theorem 1.2.3

Suppose that the condition (1.26) holds true,
$\lambda\in\sigma(P_{V})\backslash\sigma_{\rm ess}(P_{V})$ and that a function
$\varphi\in\ell^{2}$ satisfies $(\lambda-P_{V})\varphi=0$. Then, there exist
constants $\alpha,C\in(0,\infty)$ such that

$|\varphi(x)|\leq C\exp(-\alpha|x|),\;\;\mbox{for all
$x\in{\mathbb{Z}}^{d}$}.$ (1.31)

The next result deals with the right edge $r(P_{V})$ of the spectra of
$P_{V}$, cf. (1.14).

###### Theorem 1.2.4

In addition to conditions (1.26) and (1.29), suppose that

$V({\mathbb{Z}}^{d})\cap[v_{0},\infty)\neq\emptyset.$ (1.32)

Then, there is a spectral gap at $r(P_{V})$, i. e., (1.10) holds. Moreover,

there exists a strictly positive, normalized function $\varphi\in\ell^{2}$
such that $\displaystyle{\rm Ker}(r(P_{V})-P_{V})={\mathbb{C}}\varphi.$ (1.33)

In the following corollary to Theorem 1.2.4, we improve the spectral gap
(1.10) to the absolute spectral gap (1.11). An operator $T\in{\cal
B}(\ell^{2})$ is said to be bipartite w.r.t.
$J\in\\{-1,1\\}^{{\mathbb{Z}}^{d}}$ if $T(x,y)=0$ for all
$(x,y)\in({\mathbb{Z}}^{d})^{2}$ such that $J(x)J(y)=1$.

###### Corollary 1.2.5

Suppose that (1.10) and (1.33) hold true, which is the case if (1.26), (1.29)
and (1.32) are satisfied. Then:

a)

There is an absolute spectral gap at $r(P_{V})$, i. e. (1.11) holds if and
only if

$-r(P_{V})<\ell(P_{V})\;\;\mbox{or $P$ is bipartite w.r.t. some
$J\in\\{-1,1\\}^{{\mathbb{Z}}^{d}}$.}$ (1.34)

b)

Assume (1.34). Then, there exists $\varepsilon\in(0,1)$ such that

$\|r(P_{V})^{-n}P^{n}_{V}(f-\Pi_{V}f)\|_{V}\leq\varepsilon^{n}\|f-\Pi_{V}f\|_{V}\;\;\;\mbox{for
all $f\in\ell^{2}$ and $n\in{\mathbb{N}}$},$ (1.35)

where, with $\varphi$ from (1.33),

$\Pi_{V}f\stackrel{{\scriptstyle\rm
def}}{{=}}\left\\{\begin{array}[]{ll}\langle\>f,\varphi\>\rangle_{V}\varphi,&\mbox{if
$-r(P_{V})<\ell(P_{V})$},\\\
\langle\>f,\varphi\>\rangle_{V}\varphi+\langle\>f,J\varphi\>\rangle_{V}J\varphi,&\mbox{if
$P$ is bipartite w.r.t.
$J\in\\{-1,1\\}^{{\mathbb{Z}}^{d}}$},\end{array}\right.$ (1.36)

Finally, we provide a sufficient condition in terms of the transition
probability $p$ for the condition (1.34).

###### Proposition 1.2.6

Let $A=\\{x\in{\mathbb{Z}}^{d}\;;\;\sum_{\alpha\in I}x_{\alpha}\in
2{\mathbb{Z}}\\}$ with $\emptyset\neq I\subset\\{1,\ldots,d\\}$. Suppose that
either $p$ vanishes on $A$, or

$p(0)>\sum_{x\in A\backslash\\{0\\}}p(x).$ (1.37)

Then, the condition (1.34) is satisfied for any nonnegative
$V\in\ell^{\infty}({\mathbb{Z}}^{d})$. More precisely,

If $p$ vanishes on $A$, then, $P$ is bipartite w.r.t. $J={\bf 1}_{A}-{\bf
1}_{A^{\sf c}}$. If (1.37) holds, then, $-r(P_{V})<\ell(P_{V})$ for any
nonnegative $V\in\ell^{\infty}({\mathbb{Z}}^{d})$.

###  1.3 Limit of the random walk in potential $V$

We introduce a random walk $(S_{n})_{n\in{\mathbb{N}}}$ on a probability space
$(\Omega,{\cal F},P)$ such that $S_{0}=0$, $P(S_{1}=\cdot)=p$. We define a
probability measure $\mu_{N}$ on $(\Omega,{\cal F})$ by (1.13).

We assume (1.26) and (1.29) in what follows. We will have a positively
recurrent Markov chain as the limit process of $\mu_{N}$. To describe the
Markov chain obtained as the limit, we introduce an operator
$P_{V,\varphi}\in{\cal B}(\ell^{2})$ by

$P_{V,\varphi}(x,y)=r(P_{V})^{-1}\varphi(x)^{-1}P_{V}(x,y)\varphi(y),$ (1.38)

where $\varphi$ is from Theorem 1.2.4. Then, $P_{V,\varphi}$ is a transition
probability for a Markov chain in ${\mathbb{Z}}^{d}$, which we denote by
$(\\{S_{n}\\}_{n\in{\mathbb{N}}},\\{\nu^{x}\\}_{x\in{\mathbb{Z}}^{d}})$. It is
easy to check that the Markov chain $\\{\nu^{x}\\}_{x\in{\mathbb{Z}}^{d}}$ is
reversible with respect to the probability measure $m$ on ${\mathbb{Z}}^{d}$
defined by

$\langle\>f,m\>\rangle=\frac{\langle\>f,\varphi^{2}\>\rangle_{V}}{\langle\>{\bf
1},\varphi^{2}\>\rangle_{V}},\;\;\;f\in\ell^{\infty}.$ (1.39)

In particular, $\\{\nu^{x}\\}_{x\in{\mathbb{Z}}}$ is positively recurrent.

###### Theorem 1.3.1

Assume the same hypothesis as Corollary 1.2.5. Then, there are constants
$C=C(p,V)>0$ and $\varepsilon=\varepsilon(p,V)\in(0,1)$ as follows; if $n\geq
k+C$, $f:{\mathbb{Z}}^{k}\rightarrow{\mathbb{R}}$ is polynomialy bounded and
$F(\omega)=f(S_{1}(\omega),\ldots,S_{k}(\omega))$, $\omega\in\Omega$, then

$\left|\int_{\Omega}Fd\mu_{n}-\int_{\Omega}Fd\nu\right|\leq
B(f)\varepsilon^{n-k},$ (1.40)

where $B(f)$ is a constant which depends only on $p$, $V$ and $f$.

Proof: Given Theorem 1.2.3 and Corollary 1.2.5, the proof of Theorem 1.3.1 is
identical to that of [3, Theorem 1.3], hence is omitted. $\Box$

###  1.4 An example in dimension one

We provide a simple example for $d=1$. We define
$p:{\mathbb{Z}}\rightarrow[0,1)$ such that $p(0)=q\in[0,1)$
$p(1)=p(-1)=(1-q)/2\in(0,1/2]$. Then, for
$\lambda\in{\mathbb{R}}\backslash[2q-1,1]$, the function $g_{\lambda}(x)$ is
computed explicitly.

$g_{\lambda}(x)={\lambda\over\sqrt{\delta(\lambda)}}\times\left\\{\begin{array}[]{ll}\varphi(\lambda)^{|x|},&\mbox{if
$\lambda>1$},\\\ -\varphi(\lambda)^{-|x|},&\mbox{if
$\lambda<2q-1$},\end{array}\right.$ (1.41)

where $\delta(\lambda)=(\lambda-1)(\lambda-(2q-1))>0$ and

$\varphi(\lambda)={\lambda-q-\sqrt{\delta(\lambda)}\over
1-q}\in\left\\{\begin{array}[]{ll}(0,1),&\mbox{if $\lambda>1$},\\\
(-\infty,-1),&\mbox{if $\lambda<2q-1$}.\end{array}\right.$
$\displaystyle\bullet$ $g_{\lambda}(0)$ is strictly decreasing in
$\lambda\in(1,\infty)$, $g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\searrow
1}}{{\longrightarrow}}\infty$,
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\nearrow\infty}}{{\longrightarrow}}1$.
(1.45) $\displaystyle\bullet$ if $q=0$, then $g_{\lambda}(0)$ is strictly
increasing in $\lambda\in(-\infty,-1)$,
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\searrow-\infty}}{{\longrightarrow}}1$,
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\nearrow-1}}{{\longrightarrow}}\infty$.
$\displaystyle\bullet$ if $0<q<1/2$, then, $g_{\lambda}(0)$ is strictly
decreasing in $\lambda\in(-\infty,{2q-1\over q})$, strictly increasing in
$\lambda\in({2q-1\over q},2q-1)$,
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\searrow-\infty}}{{\longrightarrow}}1$,
$g_{2q-1\over q}(0)={\sqrt{1-2q}\over 1-q}$, $\displaystyle\mbox{$g_{2q-1\over
2q}(0)=1$, $g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\nearrow
2q-1}}{{\longrightarrow}}\infty$},$ $\displaystyle\bullet$
$\displaystyle\mbox{ if $1/2\leq q<1$, then, $g_{\lambda}(0)$ is strictly
decreasing in $\lambda\in(-\infty,2q-1)$},$
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\searrow-\infty}}{{\longrightarrow}}1$,
$g_{0}(0)=0$, and if $1/2<q<1$,
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\nearrow
2q-1}}{{\longrightarrow}}-\infty$.

For $v>0$, we set

$\lambda_{\pm}(v)\stackrel{{\scriptstyle\rm
def}}{{=}}c(v)\left(q\pm\sqrt{q^{2}-(2q-1)c(v)^{-1}}\right),\;\;\mbox{where}\;\;c(v)\stackrel{{\scriptstyle\rm
def}}{{=}}{(v+1)^{2}\over 2v+1}.$ (1.46)

Suppose that $V:{\mathbb{Z}}\rightarrow[0,\infty)$ is a bounded function which
satisfies (1.26).

$\Lambda_{V}=\left\\{\begin{array}[]{ll}\\{\lambda_{\pm}(v)\;;\;v\in\sigma_{\rm
ess}(V)\backslash\\{0\\}\\},&\mbox{if $0\leq q<1/2$},\\\
\\{\lambda_{+}(v)\;;\;v\in\sigma_{\rm ess}(V)\backslash\\{0\\}\\},&\mbox{if
$1/2\leq q<1$}.\end{array}\right.$ (1.47)

To see this, we observe that for $v>0$ and
$\lambda\in{\mathbb{R}}\backslash[2q-1,1]$,

$\displaystyle g_{\lambda}(0)=1+v^{-1}$ $\displaystyle\Longrightarrow$
$\displaystyle{\lambda^{2}\over\delta(\lambda)}=(1+v^{-1})^{2}$
$\displaystyle\Longleftrightarrow$
$\displaystyle\lambda^{2}-2qc(v)\lambda+(2q-1)c(v)=0$
$\displaystyle\Longleftrightarrow$ $\displaystyle\lambda=\lambda_{\pm}(v).$

Moreover,

$\lambda_{-}(v)<2q-1<1<\lambda_{+}(v)=\lambda_{-}(v)+2c(v)\sqrt{q^{2}-(2q-1)c(v)^{-1}}.$
(1.48)

Taking (1.45)–(1.48) into account (In particular, if $0<q<1/2$, then,
${2q-1\over 2q}<\lambda_{-}(v)<2q-1$ and hence $g_{\lambda}(0)>1$ at
$\lambda=\lambda_{-}(v)$ by (1.45)), we have

$\lambda\in{\mathbb{R}}\backslash[2q-1,1],\;g_{\lambda}(0)=1+v^{-1}\;\Longleftrightarrow\;\lambda=\left\\{\begin{array}[]{ll}\lambda_{\pm}(v),&\mbox{if
$0\leq q<1/2$},\\\ \lambda_{+}(v),&\mbox{if $1/2\leq q<1$}.\end{array}\right.$

This, together with Theorem 1.2.1, implies (1.47).

##  2 Proof of Theorem 1.2.1 and Corollary 1.2.2

###  2.1 Outline

Proof of Theorem 1.2.1 Step1 We prove that $\sigma_{\rm
ess}(P_{V})\backslash\sigma(P)=\Lambda_{V}$. Here, we adapt the method in [4]
to the present setting. Let $\lambda\in{\mathbb{R}}\backslash\sigma(P)$. Then,
We introduce the modified Birman-Schwinger operator $G_{V,\lambda}\in{\cal
B}(\ell^{2})$ by

$G_{V,\lambda}=V^{1/2}(\lambda G_{\lambda}-1)V^{1/2}.$ (2.1)

We first show in Lemma 2.2.3 that

$\lambda\in\sigma_{\rm ess}(P_{V})\;\;\Longleftrightarrow\;\;1\in\sigma_{\rm
ess}(G_{V,\lambda}).$ (2.2)

We then prove in Lemma 2.2.4 that

$G_{V,\lambda}=(g_{\lambda}(0)-1)V+H_{V,\lambda},$ (2.3)

where $H_{V,\lambda}$ is a compact operator. In fact, this is where the
condition (1.26) is used. Then, we see from (2.3) that

$\sigma_{\rm ess}(G_{V,\lambda})=(g_{\lambda}(0)-1)\sigma_{\rm ess}(V),$ (2.4)

via Weyl’s essential spectrum theorem, cf. [2, p.358, Proposition 4.2 (e)],
[6, p.112, Theorem XIII.14]. By (2.4),

$1\in\sigma_{\rm ess}(G_{V,\lambda})\;\;\Longleftrightarrow\;\;\exists
v\in\sigma_{\rm
ess}(V)\backslash\\{0\\},\;g_{\lambda}(0)-1=v^{-1}\;\;\Longleftrightarrow\;\;\lambda\in\Lambda_{V}.$
(2.5)

Thus, we obtain (1.27) from (2.2) and (2.5).
Step2 We prove that $\sigma(P)\subset\sigma_{\rm ess}(P_{V})$. This step is
the subject of Lemma 2.2.6 below. $\Box$

Proof of Corollary 1.2.2: a) Suppose that the condition (1.29) holds. Then,
$\lambda_{0}\in\Lambda_{V}\cap(1,\infty)$. Since $\sigma_{\rm
ess}(V)\cap(1,\infty)=\Lambda_{V}\cap(1,\infty)$ by Theorem 1.2.1, it follows
that $\sigma_{\rm ess}(V)\cap(1,\infty)\neq\emptyset$. Moreover, by (2.12),
$g_{\lambda}(0)$ is strictly decreasing in $\lambda\in(1,\infty)$. Therefore.
(1.29) implies that $\lambda_{0}=\max\sigma_{\rm ess}(P_{V})$.
Suppose on the other hand that $\sigma_{\rm
ess}(V)\cap(1,\infty)\neq\emptyset$. Then, since $\sigma_{\rm
ess}(V)\cap(1,\infty)=\Lambda_{V}\cap(1,\infty)$ by Theorem 1.2.1, there exist
$\lambda\in(1,\infty)$ and $v\in\sigma_{\rm ess}(V)\backslash\\{0\\}$ such
that $g_{\lambda}(0)=1+v^{-1}$. Then, $v_{0}\geq v>0$. Moreover, since
$0<v_{0}^{-1}\leq v^{-1}$ and by (2.12),
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\rightarrow\infty}}{{\longrightarrow}}1$,
it follows from the mean value theorem that there exists
$\lambda_{0}\in(1,\infty)$ such that $g_{\lambda_{0}}(0)=1+v_{0}^{-1}$.
b) The proof is similar as above. $\Box$

###  2.2 Lemmas

Properties of the function $\lambda\mapsto g_{\lambda}(0)$, which we need in
this article are summarized in the following

###### Lemma 2.2.1

$\displaystyle\bullet$ $g_{\lambda}(0)$ is strictly decreasing in
$\lambda\in(1,\infty)$,
$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\rightarrow\infty}}{{\longrightarrow}}1$.
(2.12) Moreover, $g_{1+}(0)\left\\{\begin{array}[]{ll}=\infty,&\mbox{if $d\leq
2$},\\\ \in(0,\infty),&\mbox{if $d\geq 3$}.\end{array}\right.$
$\displaystyle\bullet$ if $\ell(P)<0$, then,
$g_{\lambda}(0)>{|\lambda|\over|\lambda|+p(0)}$ for
$\lambda\in(-\infty,\ell(P))$.
$\displaystyle\mbox{$g_{\lambda}(0)\stackrel{{\scriptstyle\lambda\rightarrow-\infty}}{{\longrightarrow}}1$,
Moreover, $g_{\ell(P)-}(0)=\infty$ if $d\leq 2$},$ $\displaystyle\bullet$
$\displaystyle\mbox{ if $\ell(P)\geq 0$, then, $g_{\lambda}(0)$ is strictly
decreasing in $\lambda\in(-\infty,\ell(P))$},$
$g_{\lambda}(0)\in\left\\{\begin{array}[]{ll}(0,1),&\mbox{for
$\lambda\in(-\infty,0)$},\\\ (-\infty,0),&\mbox{for
$\lambda\in(0,\ell(P))$}.\end{array}\right.$

Proof: The behavior (2.12) of $g_{\lambda}(0)$ for $\lambda>1$ is well-known
in the context of the random walk. Thus, we omit the proofs.

On the other hand, the following integral formula is well-known.

$g_{\lambda}(0)={\lambda\over(2\pi)^{d}}\int_{[-\pi,\pi]^{d}}{d\theta\over\lambda-\widehat{p}(\theta)},\;\;\;\lambda\in{\mathbb{R}}\backslash\sigma(P).$
(2.13)

We derive properties (2.12)–(2.12) from this formula. We take (2.12) for
example. Since $\lambda$ is negative,

$\displaystyle g_{\lambda}(0)$ $\displaystyle=$
$\displaystyle{|\lambda|\over(2\pi)^{d}}\int_{[-\pi,\pi]^{d}}{d\theta\over|\lambda|+\widehat{p}(\theta)}$
$\displaystyle>$
$\displaystyle{|\lambda|\over|\lambda|+(2\pi)^{-d}\int_{[-\pi,\pi]^{d}}\widehat{p}(\theta)d\theta}={|\lambda|\over|\lambda|+p(0)},$

where we have used Jensen inequality to the convex function $x\mapsto 1/x$
($x>0$) to proceed from the first line to the second.

To show that $g_{\ell(P)-}(0)=\infty$ if $d\leq 2$, we take
$\theta_{0}\in[-\pi,\pi]^{d}$ such that $\ell(P)=\widehat{p}(\theta_{0})$.
Since ${\partial\widehat{p}\over\partial\theta_{\alpha}}(\theta_{0})=0$ for
all $\alpha=1,\ldots,d$, we have

$\displaystyle\widehat{p}(\theta_{0}+\theta)-\ell(P)$ $\displaystyle=$
$\displaystyle\widehat{p}(\theta_{0}+\theta)-\widehat{p}(\theta_{0})$
$\displaystyle=$ $\displaystyle{1\over
2}\sum_{\alpha,\beta=1}^{d}\theta_{\alpha}\theta_{\beta}\int^{1}_{0}(1-t){\partial^{2}\widehat{p}\over\partial\theta_{\alpha}\partial\theta_{\beta}}(\theta_{0}+t\theta)dt$
$\displaystyle\leq$ $\displaystyle C|\theta|^{2}\;\;\;\mbox{for
$\theta\in[-\pi,\pi]^{d}$}.$

By applying the monotone convergence theorem to the integral
$\int_{[-\pi,\pi]^{d}}{d\theta\over\widehat{p}(\theta)-\lambda}$ as
$\lambda\nearrow\ell(P)$, we have

$\displaystyle g_{\ell(P)-}(0)$ $\displaystyle=$
$\displaystyle{|\ell(P)|\over(2\pi)^{d}}\int_{[-\pi,\pi]^{d}}{d\theta\over\widehat{p}(\theta)-\ell(P)}={|\ell(P)|\over(2\pi)^{d}}\int_{[-\pi,\pi]^{d}}{d\theta\over\widehat{p}(\theta_{0}+\theta)-\ell(P)}$
$\displaystyle\geq$
$\displaystyle{|\ell(P)|\over(2\pi)^{d}C}\int_{[-\pi,\pi]^{d}}{d\theta\over|\theta|^{2}}=\infty.$

$\Box$

Remark If $\ell(P)<0$, $g_{\lambda}(0)$ is not necessarily monotone in
$\lambda\in(-\infty,\ell(P))$. See (1.45) for a counterexample.

###### Lemma 2.2.2

Let $A,B\in{\cal B}(X)$ on a Banach space $X$. Then,

a)

$1\in\sigma_{\rm ess}(BA)$ if and only if $1\in\sigma_{\rm ess}(AB)$.

b)

$1\in\rho(BA)$ if and only if $1\in\rho(AB)$. Moreover, if these conditions
hold true, then

$(1-BA)^{-1}=1+B(1-AB)^{-1}A.$ (2.14)

Proof: a) Since the roles of $A$ and $B$ are exchangeable, it is enough to
verify only the ”if” part, which is equivalently stated as $1-AB\in{\rm F}(X)$
$\Rightarrow$ $1-BA\in{\rm F}(X)$. To prove this, we use Atkinson’ s theorem
which says the following for $T\in{\cal B}(X)$, cf. [1, p.161, Theorem 4.46].

A1)

Suppose that $T\in{\rm F}(X)$ with $n_{1}=\dim{\rm Ker}T$ and
$n_{2}=\dim(X/{\rm Ran}T)$. Then, there exist $S,K_{1},K_{2}\in{\cal B}(X)$
with ${\rm rank}K_{j}=n_{j}$ ($j=1,2$) such that

$ST=1+K_{1}\;\;\mbox{and}\;\;TS=1+K_{2}.$ (2.15)

A2)

Conversely, suppose that there exist $S,K_{1},K_{2}\in{\cal B}(X)$ of which
$K_{1}$ and $K_{2}$ are compact such that (2.15) holds true. Then, $T\in{\rm
F}(X)$.

Comming back to the proof of the lemma, suppose that
$T\stackrel{{\scriptstyle\rm def}}{{=}}1-AB\in{\rm F}(X)$. Then, by Atkinson’s
theorem, there exist $S,K_{1},K_{2}\in{\cal B}(X)$ as are stated in A1) above.
Then, $1+BSA\in{\cal B}(X)$, ${\rm rank}BK_{j}A\leq n_{j}$ ($j=1,2$) and

$(1+BSA)(1-BA)=1+BK_{1}A,\;\;(1-BA)(1+BSA)=1+BK_{2}A.$ (2.16)

Therefore, $1-BA\in{\rm F}(X)$, by A2) above.
b) The equivalence of $1\in\rho(BA)$ and $1\in\rho(AB)$ can be regarded as a
special case of part a), where $K_{1}=K_{2}=0$ in the proof above. Moreover,
suppose that $1\in\rho(BA)$, or equivalently, $1\in\rho(AB)$. Then,
$S=(1-AB)^{-1}$ in (2.16). Thus, the equality (2.14) follows from (2.16).
$\Box$

###### Lemma 2.2.3

Let $\lambda\in{\mathbb{R}}\backslash\sigma(P)$, and $G_{V,\lambda}$ be the
operator defined by (2.1). Then,

a)

$\lambda\in\sigma_{\rm ess}(P_{V})$ if and only if $1\in\sigma_{\rm
ess}(G_{V,\lambda})$.

b)

$\lambda\in\sigma(P_{V})$ if and only if $1\in\sigma(G_{V,\lambda})$.
Moreover, if $\lambda\in\rho(P_{V})$, or equivalently,
$1\in\rho(G_{V,\lambda})$, then,

$(\lambda-
P_{V})^{-1}=G_{\lambda}+G_{\lambda}V^{1/2}(1-G_{V,\lambda})^{-1}V^{1/2}PG_{\lambda}.$
(2.17)

Proof: a) We need to verify the equivalence. $\lambda-P_{V}\in F(\ell^{2})$
$\Longleftrightarrow$ $1-G_{V,\lambda}\in F(\ell^{2})$. We decompose this task
into the following two steps.

$\displaystyle\lambda-P_{V}\in F(\ell^{2})\;\;$
$\displaystyle\Longleftrightarrow$ $\displaystyle\;\;1-G_{\lambda}PV\in
F(\ell^{2}),$ (2.18) $\displaystyle 1-G_{\lambda}PV\in F(\ell^{2})\;\;$
$\displaystyle\Longleftrightarrow$ $\displaystyle\;\;1-G_{V,\lambda}\in
F(\ell^{2}).$ (2.19)

Note that

$\lambda-P_{V}=\lambda-P-VP=(\lambda-P)(1-G_{\lambda}VP).$ (2.20)

or equivalently,

$1-G_{\lambda}VP=G_{\lambda}(\lambda-P_{V}).$ (2.21)

Note that $\lambda-P,G_{\lambda}\in F(\ell^{2})$. Recall also that
$F(\ell^{2})$ is closed under the composition cf. [1, p.158, Theorem 4.43].
Then, (2.18) follows from (2.20) and (2.21).

The equivalence (2.19) follows from Lemma 2.2.2 a), since for
$A\stackrel{{\scriptstyle\rm def}}{{=}}G_{\lambda}V^{1/2}$ and
$B\stackrel{{\scriptstyle\rm def}}{{=}}V^{1/2}P$,

$\displaystyle AB$ $\displaystyle=$ $\displaystyle
G_{\lambda}V^{1/2}V^{1/2}P=G_{\lambda}VP,$ $\displaystyle BA$ $\displaystyle=$
$\displaystyle V^{1/2}PG_{\lambda}V^{1/2}=V^{1/2}(\lambda
G_{\lambda}-1)V^{1/2}=G_{V,\lambda}.$

b) The proof is similar as above. We use Lemma 2.2.2 b) instead of Lemma 2.2.2
a). Suppose that $\lambda\in\rho(P_{V})$. Then, it follows from (2.21) that

$(\lambda-P_{V})^{-1}=(1-G_{\lambda}VP)^{-1}G_{\lambda}.$

On the other hand, by plugging $A=V^{1/2}P$ and $B=G_{\lambda}V^{1/2}$ into
(2.14),

$\displaystyle(1-G_{\lambda}VP)^{-1}$ $\displaystyle=$ $\displaystyle
1+G_{\lambda}V^{1/2}(1-V^{1/2}PG_{\lambda}V^{1/2})^{-1}V^{1/2}P$
$\displaystyle=$ $\displaystyle
1+G_{\lambda}V^{1/2}(1-G_{V,\lambda})^{-1}V^{1/2}P.$

Combining these, we obtain (2.17). $\Box$

###### Lemma 2.2.4

With $g_{\lambda}(0)$, $G_{V,\lambda}$ defined respectively by (1.19) and
(2.1), the following operator $H_{V,\lambda}\in{\cal B}(\ell^{2})$ is compact.

$H_{V,\lambda}\stackrel{{\scriptstyle\rm
def}}{{=}}G_{V,\lambda}-(g_{\lambda}(0)-1)V.$

Proof: We decompose $G_{V,\lambda}$ into diagonal and off diagonal components
as follows.

$G_{V,\lambda}=V^{1/2}(\lambda
G_{\lambda}-1)V^{1/2}=(g_{\lambda}(0)-1)V+H_{V,\lambda},$

where the operator $H_{V,\lambda}$ is given by the kernel

$H_{V,\lambda}(x,y)=\lambda\sqrt{V(x)V(y)}G_{\lambda}(x,y){\bf 1}_{x\neq y}.$

We will show that

$\|H_{V,\lambda}-H_{V,\lambda}^{(N)}\|\stackrel{{\scriptstyle
N\rightarrow\infty}}{{\longrightarrow}}0,$ (2.22)

where $K_{N}$ is a finite rank operator defined by the kernel

$H_{V,\lambda}^{(N)}(x,y)=\lambda\sqrt{V(x)V(y)}G_{\lambda}(x,y){\bf 1}_{x\neq
y,|x|\leq N}.$

Thus, (2.22) shows that $H_{V,\lambda}$ is a compact operator.

By a standard estimate,

$\|H_{V,\lambda}-H_{V,\lambda}^{(N)}\|\leq|\lambda|\sqrt{A_{N}B_{N}},$

where

$A_{N}=\sup_{|x|\geq N}\sum_{y\in{\mathbb{Z}}^{d}\atop x\neq
y}\sqrt{V(x)V(y)}|G_{\lambda}(x,y)|,\;\;\;B_{N}=\sup_{x\in{\mathbb{Z}}^{d}}\sum_{y\in{\mathbb{Z}}^{d}\atop
y\neq x,|y|\geq N}\sqrt{V(x)V(y)}|G_{\lambda}(x,y)|.$

By (1.16), there exist $C=C(\lambda)\in(0,\infty)$ and
$\varepsilon=\varepsilon(\lambda)\in(0,\infty)$ such that

$|G_{\lambda}(x,y)|\leq C\exp(-\varepsilon|x-y|).$ (2.23)

Thus, (2.22) follows from (1.26) and

$\sup_{x\in{\mathbb{Z}}^{d}}b_{N,\varepsilon}(x)\stackrel{{\scriptstyle
N\rightarrow\infty}}{{\longrightarrow}}0,\;\;\mbox{where}\;\;b_{N,\varepsilon}(x)=\sum_{y\in{\mathbb{Z}}^{d}\atop
y\neq x,|y|>N}\sqrt{V(x)V(y)}\exp(-\varepsilon|x-y|).$ (2.24)

Therefore, it is enough to show that (1.26) implies (2.24). Moreover, it
follows immediately from (1.26) that $\sup_{|x|>\lfloor
N/2\rfloor}b_{N,\varepsilon}(x)\stackrel{{\scriptstyle
N\rightarrow\infty}}{{\longrightarrow}}0$. Hence, it is enough to verify that

$\sup_{|x|\leq\lfloor N/2\rfloor}b_{N,\varepsilon}(x)\stackrel{{\scriptstyle
N\rightarrow\infty}}{{\longrightarrow}}0.$ (2.25)

If $|x|\leq\lfloor N/2\rfloor$ and $|y|\geq N$, then $|y-x|\geq N/2$. Thus,

$\displaystyle\sup_{|x|\leq\lfloor N/2\rfloor}b_{N,\varepsilon}(x)$
$\displaystyle\leq$ $\displaystyle\exp(-\varepsilon
N/2)\|V\|_{\infty}\sup_{|x|\leq\lfloor
N/2\rfloor}\sum_{y\in{\mathbb{Z}}^{d}}\exp(-\varepsilon|x-y|/2)$
$\displaystyle=$ $\displaystyle\exp(-\varepsilon
N/2)\|V\|_{\infty}\sum_{y\in{\mathbb{Z}}^{d}}\exp(-\varepsilon|y|/2)\stackrel{{\scriptstyle
N\rightarrow\infty}}{{\longrightarrow}}0,$

which proves (2.25). $\Box$

In what follows, we use the following notation. For
$x=(x_{\alpha})_{\alpha=1}^{d}\in{\mathbb{Z}}^{d}$,

$|x|_{\infty}=\max_{1\leq\alpha\leq d}|x_{\alpha}|.$ (2.26)

For $c\in{\mathbb{Z}}^{d}$ and a positive integer $\ell$,

$Q(c,\ell)=\\{x\in{\mathbb{Z}}^{d}\;;\;|x-c|_{\infty}\leq\ell\\}.$ (2.27)

Suppose that the condition (1.26) holds true and that $v_{0}>0$. Then the
following lemma shows that there are infinitely many disjoint cubes
$Q(c,\ell)$ in which $V$ takes the value close to $v_{0}$ at the center $c$,
while the value $V(x)$ for the other points of the cube are close to zero.

###### Lemma 2.2.5

Suppose that the condition (1.26) holds true and that $v_{0}>0$. Then, for any
$L,\ell\in(0,\infty)$ and $\varepsilon\in(0,1)$, there exists
$c\in{\mathbb{Z}}^{d}$ such that

$\displaystyle Q(0,L)\cap Q(c,\ell)$ $\displaystyle=$
$\displaystyle\emptyset,$ (2.28) $\displaystyle V(c)$ $\displaystyle>$
$\displaystyle(1-\varepsilon)v_{0},$ (2.29) $\displaystyle\sum_{x\in
Q(c,\ell)\backslash\\{c\\}}V(x)$ $\displaystyle<$ $\displaystyle\varepsilon.$
(2.30)

Proof: Take $\delta\in(0,1)$ such that
$\delta^{2}\exp(2\ell)<\varepsilon(1-\varepsilon)v_{0}$. By assumption
$v_{0}>0$, the set

$H_{\varepsilon}=\\{x\in{\mathbb{Z}}^{d}\;;\;V(x)>(1-\varepsilon)v_{0}\\}$

is unbounded. Therefore, by (1.26), we can find $c\in H_{\varepsilon}$ such
that

$|c|>L+\ell,\;\;\mbox{and}\;\;A(c)\stackrel{{\scriptstyle\rm
def}}{{=}}\sum_{x\in{\mathbb{Z}}^{d}\atop x\neq
c}\sqrt{V(c)V(x)}\exp(-|c-x|_{\infty})<\delta.$ (2.31)

The first inequality of (2.31) is equivalent to (2.28), while the second
inequality implies (2.30) as follows.

$\displaystyle\sum_{x\in Q(c,\ell)\backslash\\{c\\}}V(x)$ $\displaystyle\leq$
$\displaystyle\left(\sum_{x\in
Q(c,\ell)\backslash\\{c\\}}\sqrt{V(x)}\right)^{2}\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{high2})}}}{{\leq}}A(c)^{2}{\exp(2\ell)\over(1-\varepsilon)v_{0}}\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{high4})}}}{{<}}\delta^{2}{\exp(2\ell)\over(1-\varepsilon)v_{0}}<\varepsilon.$

$\Box$

###### Lemma 2.2.6

Suppose that the condition (1.26) holds true. Then,
$\sigma(P)\subset\sigma_{\rm ess}(P_{V})$.

Proof: Let $\lambda\in\sigma(P)$ be arbitrary. We will prove that $\lambda-
P_{V}\not\in{\rm F}(\ell^{2})$, or equivalently, $\lambda-P_{V}^{*}\not\in{\rm
F}(\ell^{2})$, where $P_{V}^{*}$ is the adjoint operator of $P_{V}$ on
$\ell^{2}$, therefore, $P_{V}^{*}=P+PV$.

For this purpose, we will use the following criterion for an operator $T$ on a
Hilbert space $X$ not to be a Fredholm operator. $T\not\in{\rm F}(X)$ if there
exists a normalized sequence $\\{u_{n}\\}\subset X$ such that

$u_{n}\stackrel{{\scriptstyle n\rightarrow\infty}}{{\longrightarrow}}0$ weakly
and $Tu_{n}\stackrel{{\scriptstyle n\rightarrow\infty}}{{\longrightarrow}}0$
strongly, (2.32)

The converse is also true if $T$ is self-adjoint, cf. [2, p. 350, Theorem
2.3]. A sequence with above property is called a Weyl sequence.

We will construct a Weyl sequence $u_{n}\in\ell^{2}$ for $\lambda-P_{V}^{*}$,
that is, $u_{n}\in\ell^{2}$ is normalized, weakly convergent to zero, and

$\|(\lambda-P-PV)u_{n}\|\stackrel{{\scriptstyle
n\rightarrow\infty}}{{\longrightarrow}}0.$ (2.33)

Step1: Let $r>0$ be an integer such that ${\rm supp}[p]\subset Q(0,r)$, We
first show that, there exist $c_{n}\in{\mathbb{Z}}^{d}$, $n\in{\mathbb{N}}$
such that

$\displaystyle Q(c_{m},m+r)\cap Q(c_{n},n+r)$ $\displaystyle=$
$\displaystyle\emptyset\;\;\;\;\mbox{if $m\neq n$},$ (2.34)
$\displaystyle\max_{x\in Q(c_{n},n+r)}V(x)$ $\displaystyle<$
$\displaystyle(n+2)^{-1}\;\;\mbox{for all $n\in{\mathbb{N}}$}.$ (2.35)

Suppose that $v_{0}=0$. Then, for any $\varepsilon>0$, there exists $R>0$ such
that $V(x)<\varepsilon$ if $|x|>R$. Thus, it is easy to find such $c_{n}$ by
an inductive procedure.

Suppose on the contrary that $v_{0}>0$. We then proceed inductively with the
help of Lemma 2.2.5 as follows. We start by taking $L=1$, $\ell=3r$ and
$\varepsilon=1/2$ in Lemma 2.2.5, so that we can find
$b_{0}\in{\mathbb{Z}}^{d}$ such that

$\max_{x\in Q(b_{0},3r)\backslash\\{b_{0}\\}}V(x)<1/2.$

Thus, by choosing $c_{0}$ such that $Q(c_{0},r)\subset
Q(b_{0},3r)\backslash\\{b_{0}\\}$, we obtain (2.35) for $n=0$. Next, we take
$L=|c_{0}|+1+r$, $\ell=3+3r$ and $\varepsilon=1/3$ in Lemma 2.2.5, so that we
can find $b_{1}\in{\mathbb{Z}}^{d}$ such that

$Q(c_{0},r)\cap Q(b_{1},3+3r)=\emptyset,\;\;\mbox{and}\;\;\max_{x\in
Q(b_{1},3+3r)\backslash\\{b_{1}\\}}V(x)<1/3.$

Thus, by choosing $c_{1}$ such that $Q(c_{1},1+r)\subset
Q(b_{1},3+3r)\backslash\\{b_{1}\\}$, we obtain (2.35) for $n=1$. By repeating
this procedure, we obtain $c_{n}$, $n\in{\mathbb{N}}$ as desired.

Step2: We now construct the Weyl sequence $u_{n}$, $n\in{\mathbb{N}}$. Let
$\theta\in[-\pi,\pi]^{d}$ be such that $\lambda=\widehat{p}(\theta)$ and set
$e_{\lambda}(x)=\exp({\bf i}\theta\cdot x)$. Then by Step1, there exist
$c_{n}\in{\mathbb{Z}}^{d}$, $n\in{\mathbb{N}}$ which satisfy (2.34) and
(2.35). We set

$\varphi_{n}=e_{\lambda}{\bf
1}_{Q(c_{n},n+r)}\;\;\mbox{and}\;\;u_{n}=\varphi_{n}/\|\varphi_{n}\|.$ (2.36)

We will show that $u_{n}$, $n\in{\mathbb{N}}$ is the Weyl sequence which we
look for. By (2.34), $u_{n}$, $n\in{\mathbb{N}}$ are orthonormal. Moreover, we
see from (2.35) that

$0\leq V\leq(n+2)^{-1}{\bf 1}_{Q(c_{n},n+r)},$

and hence $\|Vu_{n}\|\leq(n+2)^{-1}\stackrel{{\scriptstyle
n\rightarrow\infty}}{{\longrightarrow}}0$. Thus, it only remains to verify
that

$\|(\lambda-P)u_{n}\|\stackrel{{\scriptstyle
n\rightarrow\infty}}{{\longrightarrow}}0.$ (2.37)

To see this, we observe that

$(\lambda-P)e_{\lambda}=0.$ (2.38)

Note also that for $f,g:{\mathbb{Z}}^{d}\rightarrow{\mathbb{C}}$ and
$x\in{\mathbb{R}}^{d}$,

$\mbox{$f=g$ on $Q(x,r)$}\;\;\Longrightarrow\;\;Pf(x)=Pg(x).$

This, together with (2.36) and (2.38), implies that

$\mbox{$(\lambda-P)\varphi_{n}=0$ on $Q(c_{n},n)$}.$ (2.39)

If we set $h_{n}\stackrel{{\scriptstyle\rm def}}{{=}}{\bf
1}_{Q(c_{n},n+r)\backslash Q(c_{n},n)}$, then

$\|h_{n}\|^{2}=\sharp(Q(c_{n},n+r)\backslash
Q(c_{n},n))=(2n+2r+1)^{d}-(2n+1)^{d}=O(n^{d-1}),$

and hence

$\|(\lambda-P)\varphi_{n}\|\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{sub*ess*ev2})}}}{{=}}\|h_{n}(\lambda-P)\varphi_{n}\|\leq\|h_{n}\|\|(\lambda-P)\varphi_{n}\|_{\infty}=O(n^{(d-1)/2}).$
(2.40)

This implies (2.37), since $\|\varphi_{n}\|=\|{\bf
1}_{Q(c_{n},n+r)}\|=(2n+2r+1)^{d/2}$. $\Box$

##  3 Proof of Theorem 1.2.3

###  3.1 Outline

For $\alpha>0$, we denote by $\ell^{\infty,\alpha}$ the Banach space of
exponentially decaying functions $u:{\mathbb{Z}}^{d}\rightarrow{\mathbb{C}}$
with the exponent at least $\alpha$, more precisely, the functions which
satisfy

$\|u\|_{\infty,\alpha}\stackrel{{\scriptstyle\rm
def}}{{=}}\sup_{x\in{\mathbb{Z}}^{d}}\exp(\alpha|x|)|u(x)|<\infty.$ (3.1)

We also recall the estimate (2.23) on the exponential decay of the kernel
$G_{\lambda}(x,y)$.

Suppose that the condition (1.26) holds true. Let
$\lambda\in{\mathbb{C}}\backslash\sigma_{\rm ess}(P_{V})$ and
$\alpha\in(0,\varepsilon)$, where $\varepsilon>0$ is from (2.23). For
$K\subset{\mathbb{Z}}^{d}$, we define

$V_{K}=V{\bf 1}_{{\mathbb{Z}}^{d}\backslash K}.$ (3.2)

In the sequel, we write $K\Subset{\mathbb{Z}}^{d}$, when $K$ is a finite
subset of ${\mathbb{Z}}^{d}$. We will prove in Lemma 3.2.4 below that there
exists $K\Subset{\mathbb{Z}}^{d}$ such that $\lambda\in\rho(P_{V_{K}})$ and
$(\lambda-P_{V_{K}})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$.

Suppose additionally that $\lambda\in\sigma(P_{V})$ and that a function
$\varphi\in\ell^{2}$ satisfies $(\lambda-P_{V})\varphi=0$. Then, with the set
$K$ from Lemma 3.2.4, we rewrite $(\lambda-P_{V})\varphi=0$ as

$(\lambda-P_{V_{K}})\varphi={\bf 1}_{K}VP\varphi.$

Since $\lambda\in\rho(P_{V_{K}})$ by Lemma 3.2.4, it follows from the above
display that

$\varphi=(\lambda-P_{V_{K}})^{-1}{\bf 1}_{K}VP\varphi.$ (3.3)

Since the function ${\bf 1}_{K}VP\varphi$ is supported on the finite set $K$
and $(\lambda-P_{V_{K}})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$ by Lemma
3.2.4, we obtain (1.31) from (3.3). $\Box$

###  3.2 Lemmas

Let $\lambda\in{\mathbb{C}}\backslash\sigma_{\rm ess}(P_{V})$ and
$\alpha\in(0,\varepsilon)$, where $\varepsilon>0$ is from (2.23). As is
discussed in section 3.1, it is enough to prove that there exists
$K\Subset{\mathbb{Z}}^{d}$ such that $\lambda\in\rho(P_{V_{K}})$ and
$(\lambda-P_{V_{K}})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$ (Lemma 3.2.4). We
will implement this by dealing with the modified Birman-Schwinger operator
with the potential $V_{K}$:

$G_{V_{K},\lambda}=V_{K}^{1/2}(\lambda G_{\lambda}-1)V_{K}^{1/2},$

cf. (2.1). As in the proof of Lemma 2.2.4, we decompose $G_{V_{K},\lambda}$
into diagonal and off diagonal components as follows.

$G_{V_{K},\lambda}=\gamma
V_{K}+H_{V_{K},\lambda},\;\;\mbox{where}\;\;\gamma=g_{\lambda}(0)-1.$ (3.4)

We first look at the diagonal component $\gamma V_{K}$ of the above
decomposition.

###### Lemma 3.2.1

Suppose that $\lambda\not\in\sigma(P)\cup\Lambda_{V}$. Then, there exists
$K_{0}\Subset{\mathbb{Z}}^{d}$ such that if $K_{0}\subset
K\Subset{\mathbb{Z}}^{d}$,

$\varepsilon_{0}\stackrel{{\scriptstyle\rm
def}}{{=}}\inf_{x\in{\mathbb{Z}}^{d}}|1-\gamma V_{K}(x)|>0.$ (3.5)

Proof: It follows from the assumption $\lambda\not\in\Lambda_{V}$ that either

$\mbox{{\bf i)} $\gamma=0$}\;\;\mbox{or}\;\;\mbox{{\bf ii)} $\gamma\neq 0$ and
$\gamma^{-1}\not\in\sigma_{\rm ess}(V)$}.$

If $\gamma=0$, then, (3.5) is clearly true with $K=\emptyset$ and
$\varepsilon_{0}=1$. Suppose ii) above. Then, there exists $\delta>0$ such
that

$K_{0}\stackrel{{\scriptstyle\rm
def}}{{=}}\\{x\in{\mathbb{Z}}^{d}\;;\;|1-\gamma
V(x)|<\gamma\delta\\}=\\{x\in{\mathbb{Z}}^{d}\;;\;|\gamma^{-1}-V(x)|<\delta\\}\Subset{\mathbb{Z}}^{d}.$

Let $K_{0}\subset K\Subset{\mathbb{Z}}^{d}$. Then, it is clear that

$|1-\gamma V_{K}(x)|=|1-\gamma V(x)|\geq\gamma\delta\;\;\mbox{for all
$x\in{\mathbb{Z}}^{d}\backslash K$},$

which implies (3.5). $\Box$

The following lemma deals with the off diagonal part $H_{V_{K},\lambda}$ of
the operator $G_{V_{K},\lambda}$, cf. (3.4) which is given by the kernel.

$H_{V_{K},\lambda}(x,y)=\lambda\sqrt{V_{K}(x)V_{K}(y)}G_{\lambda}(x,y){\bf
1}_{x\neq y}.$ (3.6)

###### Lemma 3.2.2

Suppose that the condition (1.26) holds true. Then, for all
$\alpha\in(0,\varepsilon)$ and $\beta\in(0,\infty)$, where $\varepsilon$ is
from (2.23), there exists $K_{1}\Subset{\mathbb{Z}}^{d}$ such that if
$K_{1}\subset K\Subset{\mathbb{Z}}^{d}$, then

$\displaystyle\|H_{V_{K},\lambda}\|_{{\cal B}(\ell^{2})}\leq\beta,$ (3.7)
$\displaystyle H_{V_{K},\lambda}\in{\cal
B}(\ell^{\infty,\alpha})\;\;\mbox{with}\;\;\|H_{V_{K},\lambda}\|_{{\cal
B}(\ell^{\infty,\alpha})}\leq\beta.$ (3.8)

Proof: Set $\delta\stackrel{{\scriptstyle\rm def}}{{=}}\varepsilon-\alpha>0$.
Then, by (1.26), there exists a $K_{1}\Subset{\mathbb{Z}}^{d}$ such that

$\sup_{x\in{\mathbb{Z}}^{d}\backslash K_{1}}\sum_{y\in{\mathbb{Z}}^{d}\atop
y\neq x}\sqrt{V(x)V(y)}\exp(-\delta|x-y|)\leq{\beta\over|\lambda|C},$ (3.9)

where the constant $C$ is from (2.23). Let $K_{1}\subset
K\Subset{\mathbb{Z}}^{d}$. By (2.23) and (3.6),

$H_{V_{K},\lambda}(x,y)\leq|\lambda|C\sqrt{V_{K}(x)V_{K}(y)}\exp(-\varepsilon|x-y|){\bf
1}_{x\neq y}.$ (3.10)

Since $H_{V_{K},\lambda}:\ell^{2}\rightarrow\ell^{2}$ is symmetric, we have by
a standard estimate that

$\displaystyle\|H_{V_{K},\lambda}\|$ $\displaystyle\leq$
$\displaystyle\sup_{x\in{\mathbb{Z}}^{d}}\sum_{y\in{\mathbb{Z}}^{d}}|H_{V_{K},\lambda}(x,y)|$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{H*W*ker*2})}}}{{\leq}}$
$\displaystyle|\lambda|C\sup_{x\in{\mathbb{Z}}^{d}\backslash
K}\sum_{y\in{\mathbb{Z}}^{d}\atop y\neq
x}\sqrt{V(x)V(y)}\exp(-\varepsilon|x-y|)\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{H*W*K})}}}{{\leq}}\beta,$

which shows (3.7).

Suppose that $u\in\ell^{\infty,\alpha}$ and $x\in{\mathbb{Z}}^{d}$. Then,
noting that $|x|\leq|x-y|+|y|$,

$\displaystyle\exp(\alpha|x|)|(H_{V_{K},\lambda}u)(x)|$ $\displaystyle\leq$
$\displaystyle\exp(\alpha|x|)\sum_{y\in{\mathbb{Z}}^{d}}|H_{V_{K},\lambda}(x,y)||u(y)|$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{H*W*ker*2})}}}{{\leq}}$
$\displaystyle{\bf 1}_{{\mathbb{Z}}^{d}\backslash
K}(x)|\lambda|C\sum_{y\in{\mathbb{Z}}^{d}\atop y\neq
x}\sqrt{V(x)V(y)}\exp(-\delta|x-y|)\exp(\alpha|y|)|u(y)|$ $\displaystyle\leq$
$\displaystyle{\bf 1}_{{\mathbb{Z}}^{d}\backslash
K}(x)|\lambda|C\|u\|_{\infty,\alpha}\sum_{y\in{\mathbb{Z}}^{d}\atop y\neq
x}\sqrt{V(x)V(y)}\exp(-\delta|x-y|)\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{H*W*K})}}}{{\leq}}\beta\|u\|_{\infty,\alpha},$

which shows (3.8). $\Box$

Combining Lemma 3.2.1 and Lemma 3.2.2, we obtain the following

###### Lemma 3.2.3

Suppose that (1.26) holds true. Then, for
$\lambda\in{\mathbb{C}}\backslash\sigma_{\rm ess}(P_{V})$ and
$\alpha\in(0,\varepsilon)$, where $\varepsilon$ is from (2.23), there exists
$K\Subset{\mathbb{Z}}^{d}$ such that $1\in\rho(G_{V_{K},\lambda})$ and that
$(1-G_{V_{K},\lambda})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$.

Proof: We have $\lambda\not\in\sigma_{\rm
ess}(P_{V})=\sigma(P)\cup\Lambda_{V}$, by the choice of $\lambda$ and Theorem
1.2.1. Thus, we may apply Lemma 3.2.1 and take $\varepsilon_{0}\in(0,1]$ and
$K_{0}\Subset{\mathbb{Z}}^{d}$ from there. We then apply Lemma 3.2.2 with
$\beta<\varepsilon_{0}$. As a consequence, we can find
$K_{i}\Subset{\mathbb{Z}}^{d}$ ($i=0,1$) with which (3.5), (3.7) and (3.8)
hold true, where $\beta<\varepsilon_{0}$. We now suppose that $K_{0}\cup
K_{1}\subset K\Subset{\mathbb{Z}}^{d}$. Then, it follows from (3.5) that the
operator $1-\gamma V_{K}$ has its inverse in ${\cal B}(\ell^{2})$ with the
norm at most $1/\varepsilon_{0}$. Then, setting

$R_{K}=(1-\gamma V_{K})^{-1}H_{V_{K},\lambda}$

for simplicity, we have

$\|R_{K}\|_{{\cal B}(\ell^{2})}\leq\|(1-\gamma V_{K})^{-1}\|_{{\cal
B}(\ell^{2})}\|H_{V_{K},\lambda}\|_{{\cal
B}(\ell^{2})}\leq\varepsilon_{0}^{-1}\cdot\beta<1,$ (3.11)

and therefore, the following Neumann series converges:

$(1-R_{K})^{-1}=\sum_{n=0}^{\infty}R_{K}^{n}.$ (3.12)

Since

$1-G_{V_{K},\lambda}=1-\gamma V_{K}-H_{V_{K},\lambda}=(1-\gamma
V_{K})(1-R_{K}),$

the operator $1-G_{V_{K},\lambda}$ is invertible in ${\cal B}(\ell^{2})$, and
herefore, $1\in\rho(G_{V_{K},\lambda})$.

By repeating the same argument as above, with ${\cal B}(\ell^{2})$ replaced by
${\cal B}(\ell^{\infty,\alpha})$, we see that
$(1-G_{V_{K},\lambda})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$. Indeed, it is
clear that $1-\gamma V_{K}$ has the inverse in ${\cal
B}(\ell^{\infty,\alpha})$ with the norm at most $\varepsilon_{0}^{-1}$.
Moreover, by (3.8), $\|H_{V_{K},\lambda}\|_{{\cal
B}(\ell^{\infty,\alpha})}\leq\beta<\varepsilon_{0}$. Consequently, we see that
$\|R_{K}\|_{{\cal B}(\ell^{\infty,\alpha})}<1$ and hence that the Neumann
series (3.12) converges in ${\cal B}(\ell^{\infty,\alpha})$. This implies that
$(1-G_{V_{K},\lambda})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$. $\Box$

As is discussed in section 3.1, Theorem 1.2.3 follows from the following

###### Lemma 3.2.4

Suppose that (1.26) holds true. Then, for
$\lambda\in{\mathbb{C}}\backslash\sigma_{\rm ess}(P_{V})$ and
$\alpha\in(0,\varepsilon)$ where $\varepsilon$ is from (2.23), there exists
$K\Subset{\mathbb{Z}}^{d}$ such that $\lambda\in\rho(P_{V_{K}})$ and that
$(\lambda-P_{V_{K}})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$.

Proof: The first half of the lemma follows from Lemma 3.2.2, since,
$\lambda\in\rho(P_{V_{K}})$ if and only if $1\in\rho(G_{V_{K},\lambda})$ by
Lemma 2.2.3. As for the latter half, by applying (2.17) with $V$ there
replaced by $V_{K}$, we have

$(\lambda-
P_{V_{K}})^{-1}=G_{\lambda}+G_{\lambda}V_{K}^{1/2}(1-G_{V_{K},\lambda})^{-1}V_{K}^{1/2}PG_{\lambda}.$

It is clear that $V_{K}^{1/2}$ and $P$ belong to ${\cal
B}(\ell^{\infty,\alpha})$. It is also clear from the proof of Lemma 3.2.2 that
$G_{\lambda}\in{\cal B}(\ell^{\infty,\alpha})$. Moreover,
$(1-G_{V_{K},\lambda})^{-1}\in{\cal B}(\ell^{\infty,\alpha})$ by Lemma 3.2.3.
Combining these, we conclude that $(\lambda-P_{V_{K}})^{-1}\in{\cal
B}(\ell^{\infty,\alpha})$. $\Box$

##  4 Proof of Theorem 1.2.4, Corollary 1.2.5 and Proposition 1.2.6

###  4.1 Outline

Proof of Theorem 1.2.4: We first prove that

$1\leq\dim{\rm
Ker}(r(P_{V})-P_{V})<\infty\;\;\mbox{and}\;\;r(P_{V})\not\in\overline{\sigma(P_{V})\backslash\\{r(P_{V})\\}}.$
(4.1)

For this purpose, we prove via a “lower bound by a delta potential” (Lemma
4.2.1) that there exists $u_{0}\in\ell^{2}$ such that

$\mbox{$\|u_{0}\|_{V}=1$ and
$\langle\>P_{V}u_{0},u_{0}\>\rangle_{V}>\max\sigma_{\rm ess}(P_{V})$},$ (4.2)

which implies (4.1) as follows. It follows from (4.2) that

$r(P_{V})=\sup_{\|u\|_{V}=1}\langle\>P_{V}u,u\>\rangle_{V}\geq\langle\>P_{V}u_{0},u_{0}\>\rangle_{V}>\max\sigma_{\rm
ess}(P_{V}).$

Thus, $r(P_{V})\not\in\sigma_{\rm ess}(P_{V})$, or equivalently,

$\dim(r(P_{V})-P_{V})<\infty\;\;\mbox{and}\;\;r(P_{V})\not\in\overline{\sigma(P_{V})\backslash\\{r(P_{V})\\}},$

(cf. (1.21) and (1.22)). Since $r(P_{V})\in\sigma(P_{V})$, $r(P_{V})$ is an
eigenvalue and therefore, we obtain (4.1).

Now that we have established (4.1), the rest of the proposition follows
immediately from an extension of the Perron-Frobenius theorem to infinite
dimensional setting (See Lemma 4.2.3 below). Indeed,

$P_{V}(x,y)=(1+V(x))P(x,y)\geq P(x,y)\;\;\mbox{for all
$x,y\in{\mathbb{Z}}^{d}$},$

which shows that $P_{V}$ is positive and irreducible. Moreover, by Lemma 4.2.2
below, the operator norm of $P_{V}:\ell^{2}_{V}\rightarrow\ell^{2}_{V}$ equals
to $r(P_{V})$. $\Box$

Proof of Corollary 1.2.5 $P_{V}$ is positive and irreducible. Moreover,
$P_{V}$ is bipartite w.r.t. $J\in\\{-1,1\\}^{{\mathbb{Z}}^{d}}$ if and only if
$P$ is bipartite w.r.t. $J$. We can pass from Theorem 1.2.4 to Corollary 1.2.5
by applying Lemma 4.2.4 below to $T=P_{V}$. $\Box$

Proof of Proposition 1.2.6: Case1: Suppose that $p$ vanishes on $A$. We prove
that $P$ is bipartite w.r.t. $J={\bf 1}_{A}-{\bf 1}_{A^{\sf c}}$. Indeed, it
follows from assumptions that ${{\rm supp\>}}[p]\subset A^{\sf c}$, and
therefore by definition of $A$ that $(x,y)\not\in A^{2}\cup(A^{\sf c})^{2}$ if
$y-x\in{{\rm supp\>}}[p]$. This implies that $P$ is bipartite w.r.t. $J={\bf
1}_{A}-{\bf 1}_{A^{\sf c}}$.

Case2: Suppose that (1.37) holds. We prove that $-r(P_{V})<\ell(P_{V})$ for
any nonnegative $V\in\ell^{\infty}({\mathbb{Z}}^{d})$. For this purpose, we
use the following inequality, cf. Lemma 4.2.5 below.

$-r(P_{V})+2\ell({\bf 1}_{A}P{\bf 1}_{A})\leq\ell(P_{V}).$ (4.3)

On the other hand, we have

$\ell({\bf 1}_{A}P{\bf 1}_{A})=\min_{\theta\in[-\pi,\pi]^{d}}\sum_{x\in
A}p(x)\exp({\bf i}\theta\cdot x)\geq p(0)-\sum_{x\in
A\backslash\\{0\\}}p(x)>0.$

Then, it follows from (4.3) that $-r(P_{V})<\ell(P_{V})$. $\Box$

###  4.2 Lemmas

We start by explaining the “lower bound by a delta potential” referred to in
the proof of Theorem 1.2.4.

###### Lemma 4.2.1

a)

Suppose that $v_{0}>0$. Suppose also that $\lambda\in(1,\infty)$, $v>0$,
$g_{\lambda}(0)=1+v^{-1}$, and
$V({\mathbb{Z}}^{d})\cap[v,\infty)\neq\emptyset$. Then, there exists a
$u_{\lambda}\in\ell^{2}$ such that

$\mbox{$\|u_{\lambda}\|_{V}=1$, and
$\langle\>P_{V}u_{\lambda},u_{\lambda}\>\rangle_{V}>\lambda$}.$ (4.4)

b)

Under hypothesis of Theorem 1.2.4, there exists a $u_{0}\in\ell^{2}$ such that
(4.2) holds.

Proof: a) We take $c\in{\mathbb{Z}}^{d}$ such that $V(c)\geq v$. Then, by
(1.17), $\varphi_{\lambda}(x)\stackrel{{\scriptstyle\rm
def}}{{=}}G_{\lambda}(x,c)>0$ for all $x\in{\mathbb{Z}}^{d}$. We will show
that the function $u_{\lambda}\stackrel{{\scriptstyle\rm
def}}{{=}}\varphi_{\lambda}/\|\varphi_{\lambda}\|_{V}$ satisfies (4.4). To see
this, we make an auxiliary use of the potential $U=v\delta_{c}$ to give a
lower bound for $V$. We first observe that

$P_{U}u_{\lambda}=\lambda u_{\lambda}.$ (4.5)

Indeed, noting the equality $PG_{\lambda}=\lambda G_{\lambda}-1$, we have for
all $x\in{\mathbb{Z}}^{d}$ that

$\displaystyle P_{U}\varphi_{\lambda}(x)$ $\displaystyle=$
$\displaystyle(1+v\delta_{c}(x))PG_{\lambda}\delta_{c}(x)$ $\displaystyle=$
$\displaystyle(1+v\delta_{c}(x))(\lambda G_{\lambda}(x,c)-\delta_{c}(x))$
$\displaystyle=$ $\displaystyle\lambda
G_{\lambda}(x,c)-\delta_{c}(x)+v(\lambda G_{\lambda}(c,c)-1)\delta_{c}(x)$
$\displaystyle=$ $\displaystyle\lambda
G_{\lambda}(x,c)=\lambda\varphi_{\lambda}(x),$

which implies (4.5).

Now, we use (4.5) to see (4.4) as follows. It follows from $v\leq V(c)$ that
$U(x)\leq V(x)$ for all $x\in{\mathbb{Z}}^{d}$. Moreover, since $v_{0}>0$,
$V(x)>0$ for infinitely many $x$’s, and hence $U(x)<V(x)$ for at least an $x$
(in fact, for infinitely many $x$’s) . These, together with the strict
positivity of $u_{\lambda}$ implies that

$1=\|u_{\lambda}\|_{V}=\|(1+V)^{-1/2}u_{\lambda}\|<\|(1+U)^{-1/2}u_{\lambda}\|=\|u_{\lambda}\|_{U}.$
(4.6)

Moreover, since $\langle\>Pu_{\lambda},u_{\lambda}\>\rangle>0$,

$\langle\>P_{V}u_{\lambda},u_{\lambda}\>\rangle_{V}=\langle\>Pu_{\lambda},u_{\lambda}\>\rangle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{||u||<1})}}}{{>}}{\langle\>Pu_{\lambda},u_{\lambda}\>\rangle\over\|u_{\lambda}\|_{U}}={\langle\>P_{U}u_{\lambda},u_{\lambda}\>\rangle_{U}\over\|u_{\lambda}\|_{U}}\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{vp*ev})}}}{{=}}\lambda.$

b) Here, by Corollary 1.2.2, $\lambda_{0}=\max\sigma_{\rm ess}(P_{V})$ and
$v_{0}$ are related as

$g_{\lambda_{0}}(0)=1+v_{0}^{-1}.$

Therefore, part b) of the present lemma follows from part a). $\Box$

We present a series of lemmas in the following more general framework. Let
$(S,{\cal A},m)$ be a $\sigma$-finite measure space. We denote the norm and
the the inner product of the Hilbert space $L^{2}(m)$ resectively by
$\|\cdot\|$ and $\langle\>\cdot,\cdot\>\rangle$. The totality of $f\in
L^{2}(m)$ such that $f\geq 0$, $m$-a.e. is denoted by $L^{2}_{+}(m)$. An
operator $T\in{\cal B}(L^{2}(m))$ is said to be positive if $Tf\in
L^{2}_{+}(m)$ for all $f\in L^{2}_{+}(m)$.

###### Lemma 4.2.2

Suppose that $T\in{\cal B}(L^{2}(m))$ is positive. Then, for all $f\in
L^{2}(m)$,

$T|f|-|Tf|\in L^{2}_{+}(m),\;\;\mbox{and
thus,}\;\;|\langle\>f,Tf\>\rangle|\leq\langle\>|f|,T|f|\>\rangle.$ (4.7)

In particular, if $T$ is positive and self-adjoint, then,

$|\ell(T)|\leq r(T)=\|T\|.$

Proof: Let $f\in L^{2}(m)$ be arbitrary. Since $T$ maps a real-valued function
to a real-valued function, we have

$T(\mathop{\rm Re}\nolimits f)=\mathop{\rm Re}\nolimits(Tf).$ (4.8)

Moreover, for each $z\in{\mathbb{C}}$,

$|z|=\sup_{\theta\in{\mathbb{R}}}\mathop{\rm Re}\nolimits(\exp({\bf
i}\theta)z)=\sup_{\theta\in{\mathbb{Q}}}\mathop{\rm Re}\nolimits(\exp({\bf
i}\theta)z).$ (4.9)

Let $x\in S$ and $\theta\in{\mathbb{Q}}$ be fixed. Then, by applying (4.9) to
$z=f(x)$, and to $z=Tf(x)$, we see that

$\displaystyle g_{\theta}(x)$ $\displaystyle\stackrel{{\scriptstyle\rm
def}}{{=}}$ $\displaystyle|f(x)|-\mathop{\rm Re}\nolimits(\exp({\bf
i}\theta)f(x))\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{|z|=})}}}{{\geq}}0,$
(4.10) $\displaystyle T|f|(x)-|Tf(x)|$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{|z|=})}}}{{=}}$
$\displaystyle T|f|(x)-\sup_{\theta\in{\mathbb{Q}}}\mathop{\rm
Re}\nolimits(\exp({\bf i}\theta)Tf(x))$ (4.11) $\displaystyle=$
$\displaystyle\inf_{\theta\in{\mathbb{Q}}}\left(T|f|(x)-\mathop{\rm
Re}\nolimits(\exp({\bf i}\theta)Tf(x))\right)$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{TRe=ReT})}}}{{=}}$
$\displaystyle\inf_{\theta\in{\mathbb{Q}}}Tg_{\theta}(x).$

By (4.10), we have $Tg_{\theta}\in L^{2}_{+}(m)$. Then, combining this with
(4.11), we have $T|f|-|Tf|\in L^{2}_{+}(m)$. In particular, if $T$ is positive
and self-adjoint, then,

$-\langle\>f,Tf\>\rangle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{201*1})}}}{{\leq}}\langle\>|f|,T|f|\>\rangle\leq
r(T)\|f\|^{2}.$

Taking the supremum of the left-hand side of the above inequality over all
normalized functions $f$ in $L^{2}(m)$, we obtain $-\ell(T)\leq r(T)$, which,
together with the obvious inequality $\ell(T)\leq r(T)$, implies that
$|\ell(T)|\leq r(T)$. As a consequence, we obtain the equality $r(T)=\|T\|$.
$\Box$

Let $T\in{\cal B}(L^{2}(m))$. $T$ is said to be irreducible if, for all
functions $f,g\in L^{2}_{+}(m)$ which are not identically zero $m$-a.e., there
exists $n\in{\mathbb{N}}$ such that $\langle\>f,T^{n}g\>\rangle>0$. If there
exists a measurable function $J:S\rightarrow\\{-1,1\\}$ such that $JT=-TJ$,
i.e., $JTf=-T(Jf)$ for all $f\in L^{2}(m)$, then, $T$ is said to be bipartite
and $J$ is called the sign of $T$. $T$ is simply said to be bipartite if $T$
is bipartite w.r.t. some sign $J$.

We consider the following conditions for an operator $T\in{\cal B}(L^{2}(m))$,
in connection with its irreducibility and bipartiteness.

$\displaystyle{\rm Ker}(\|T\|+T)\neq\\{0\\}.$ (4.12) $\displaystyle{\rm
Ker}(\|T\|-T)\neq\\{0\\}.$ (4.13) There exists a normalized function
$\varphi\in L^{2}(m)$ such that $\displaystyle\varphi>0\;\mbox{a.e. and}\;{\rm
Ker}(\|T\|-T)={\mathbb{C}}\varphi.$ (4.14)

###### Lemma 4.2.3

Suppose that $T\in{\cal B}(L^{2}(m))$ is positive. Then,

$\mbox{(\ref{vp*202*1}) $\Longleftrightarrow$ (\ref{vp*202*0}) and $T$ is
irreducible}.$ (4.15)

Suppose in addition that $T$ is irreducible. Then,

(4.12) $\Longleftrightarrow$ (4.14) and that $T$ is bipartite. (4.16)

Moreover, the converse part of the above equivalence entails the following
relation. If $T$ is bipartite w.r.t. the sign $J$, then, (4.12) holds with

${\rm Ker}(\|T\|+T)={\mathbb{C}}J\varphi.$ (4.17)

Proof: Let us make a preliminary observation which applies to all $T\in{\cal
B}(L^{2}(m))$. Suppose that $u\in L^{2}(m)$ is normalized and that
$\langle\>u,Tu\>\rangle=\sigma\|T\|$, where $\sigma$ is either $1$ or $-1$.
Then,

$Tu=T^{*}u=\sigma\|T\|u.$ (4.18)

Indeed, since $\langle\>u,Tu\>\rangle=\sigma\|T\|\in{\mathbb{R}}$,

$\|\sigma\|T\|u-Tu\|^{2}=\|T\|^{2}-2\sigma\|T\|\langle\>u,Tu\>\rangle+\|Tu\|^{2}=-\|T\|^{2}+\|Tu\|^{2}\leq
0.$

Thus, $Tu=\sigma\|T\|u$. On the other hand,
$\langle\>u,Tu\>\rangle=\sigma\|T\|$ implies that
$\langle\>u,T^{*}u\>\rangle=\sigma\|T^{*}\|$. Hence by letting $T^{*}$ play
the role of $T$ above, we obtain $T^{*}u=\sigma\|T\|u$ as well.
(4.15): This equivalence is due to [6, p.202, Theorem XIII.43].
(4.16) ($\Rightarrow$): Suppose that $\psi\in L^{2}(m)$ is a nonzero function
which satisfies the equation $T\psi=-\|T\|\psi$. Since the same equation is
satisfied by the real and imaginary parts of $\psi$, we may assume that $\psi$
is real-valued. Moreover, by normalization, we may assume that $\|\psi\|=1$.
Therefore, we see from the preliminary observation (4.18) that

$T\psi=T^{*}\psi=-\|T\|\psi.$ (4.19)

On the other hand, applying (4.7) to $f=\psi$, we obtain

$\|T\|=-\langle\>\psi,T\psi\>\rangle\leq\langle\>|\psi|,T|\psi|\>\rangle\leq\|T\|.$

and hence, $\langle\>|\psi|,T|\psi|\>\rangle=\|T\|$. Therefore, we see from
the preliminary observation (4.18) that $T|\psi|=\|T\||\psi|$, Thus, we have
proved (4.13), which, together with the irreducibility of $T$, implies (4.14).
Then, it follows from (4.14) that $|\psi|=\varphi$, and by applying (4.18), we
have

$T\varphi=T^{*}\varphi=\|T\|\varphi.$ (4.20)

Since $|\psi|=\varphi$, the function $J\stackrel{{\scriptstyle\rm
def}}{{=}}\psi/\varphi$ is defined a.e. and takes values in $\\{-1,1\\}$
there. We will prove that $T=-JTJ$. By linearlity, it is enough to prove that
$g\stackrel{{\scriptstyle\rm def}}{{=}}Tf+JTJf=0$ a.e. for all $f\in
L^{2}_{+}(m)$. Since $f\pm Jf\in L^{2}_{+}(m)$, we have $Tf\pm TJf\in
L^{2}_{+}(m)$, and hence $g\in L^{2}_{+}(m)$. Therefore, it is enough to show
that $\langle\>\varphi,g\>\rangle=0$, which can be done by noting that
$\varphi J=\psi$, and that $\psi J=\varphi$ as follows.

$\displaystyle\langle\>\varphi,g\>\rangle$ $\displaystyle=$
$\displaystyle\langle\>\varphi,Tf\>\rangle+\langle\>\psi,TJf\>\rangle=\langle\>T^{*}\varphi,f\>\rangle+\langle\>T^{*}\psi,Jf\>\rangle$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{202*obs*1}).
(\ref{202*obs*2})}}}{{=}}$
$\displaystyle\|T\|\langle\>\varphi,f\>\rangle-\|T\|\langle\>\psi,Jf\>\rangle=\|T\|\langle\>\varphi,f\>\rangle-\|T\|\langle\>\varphi,f\>\rangle=0.$

(4.16) ($\Leftarrow$): By the relation $TJ=-JT$, the eigenspaces ${\rm
Ker}(\|T\|\pm T)$ are mapped to each other bijectively by $J$. Thus, (4.14)
implies (4.17). $\Box$

###### Lemma 4.2.4

Suppose that an operator $T\in{\cal B}(L^{2}(m))$ is positive, self-adjoint,
and satisfies (4.14) and

$r(T)>\sup\\{\lambda\;;\;\lambda\in\sigma(T),\;\lambda\neq r(T)\\}.$ (4.21)

Then,

a)

The following conditions are equivalent.

$r(T)>\sup\\{|\lambda|\;;\;\lambda\in\sigma(T),\;|\lambda|\neq r(T)\\}$.
(4.22) Either $-r(T)<\ell(T)$ or $T$ is bipartite. (4.23)

b)

Suppose that one of the conditions (4.22) and (4.23) holds true (therefore
that both do). Then, there exists $\varepsilon\in(0,1)$ such that

$\|r(T)^{-n}T^{n}(f-\Pi f)\|\leq\varepsilon^{n}\|f-\Pi f\|\;\;\;\mbox{for all
$f\in L^{2}(m)$ and $n\in{\mathbb{N}}$},$ (4.24)

where, with $\varphi$ from (4.14),

$\Pi f\stackrel{{\scriptstyle\rm
def}}{{=}}\left\\{\begin{array}[]{ll}\langle\>f,\varphi\>\rangle\varphi,&\mbox{if
$-r(T)<\ell(T)$},\\\
\langle\>f,\varphi\>\rangle\varphi+\langle\>f,J\varphi\>\rangle
J\varphi,&\mbox{if $T$ is bipartite w.r.t. the sign $J$}.\end{array}\right.$
(4.25)

Proof: a) (4.22) $\Rightarrow$ (4.23): Suppose that (4.23) fails. Then, it
follows from Lemma 4.2.3 a) that $\ell(T)=-r(T)$ is not an eigenvalue, and
therefore by (1.22), $-r(T)\in\overline{\sigma(T)\backslash\\{-r(T)\\}}$.
Hence (4.22) fails.
(4.23) $\Rightarrow$ (4.22): If $-r(T)<\ell(T)$, then (4.22) holds, since

$\sigma(T)\cap[-r(T),\ell(T))=\emptyset.$ (4.26)

On the other hand, if $T$ is bipartite, then, (4.22) follows from (4.21) and
the symmety of $\sigma(T)$ with respect to the origin.
b) Let $T=\int_{\sigma(T)}\lambda dE_{\lambda}$ denote the spectral
decomposition of the self-adjoint operator $T$ on $\ell^{2}(m)$. We first
verify that

$\Pi=\int_{\sigma(T)\atop|\lambda|=r(T)}dE_{\lambda}.$ (4.27)

We treat each of the cases in condition (4.23) separately. Let us temporarily
denote the orthogonal projection on the right-hand side of (4.27) by $\Pi^{\rm
RHS}$.

Case1: Suppose that $-r(T)<\ell(T)$. Then, it follows from (4.26) that

$\\{\lambda\in\sigma(T)\;;\;|\lambda|=r(T)\\}=\\{r(T)\\}.$

We see from this and (4.14) that

${\rm Ran}(\Pi^{\rm RHS})={\rm Ker}(r(T)-T)={\mathbb{C}}\varphi,$

which implies (4.27).

Case2: Suppoes that $T$ is bipartite the sign $J$. Then, it follows from the
symmetry of $\sigma(T)$ with respect to the origin that

$\\{\lambda\in\sigma(T)\;;\;|\lambda|=r(T)\\}=\\{\pm r(T)\\}.$

Putting this, (4.14) and (4.17) together, we have

${\rm Ran}(\Pi^{\rm RHS})={\rm Ker}(r(T)-T)\oplus{\rm
Ker}(-r(T)-T)={\mathbb{C}}\varphi\oplus{\mathbb{C}}J\varphi,$

which implies (4.27).

Now, (4.24) follows easily from condition (4.22). Let

$r=\sup\\{|\lambda|\;;\;\lambda\in\sigma(T),\;|\lambda|\neq r(T)\\}$

and let $f^{\perp}=f-\Pi f$. Then, it follows from (4.27) that

$\|T^{n}f^{\perp}\|_{V}^{2}=\int_{\sigma(T)\atop|\lambda|\leq
r}\lambda^{2n}d\langle\>E_{\lambda}f^{\perp},f^{\perp}\>\rangle\leq
r^{2n}\|f^{\perp}\|^{2},$

which proves (4.24). $\Box$

###### Lemma 4.2.5

Suppose that $p:{\mathbb{Z}}^{d}\rightarrow[0,\infty)$ is a transition
probability of a symmetric random walk, and that
$A=\\{x\in{\mathbb{Z}}^{d}\;;\;\sum_{\alpha\in I}x_{\alpha}\in
2{\mathbb{Z}}\\}$ with $\emptyset\neq I\subset\\{1,\ldots,d\\}$. Then,

$-r(P_{V})+2\ell({\bf 1}_{A}P{\bf 1}_{A})\leq\ell(P_{V}).$ (4.28)

Proof: We already know from Lemma 4.2.2 that $-r(P_{V})\leq\ell(P_{V})$. Thus,
we may assume that $\ell({\bf 1}_{A}P{\bf 1}_{A})\geq 0$. Since

$-\ell(P_{V})=\sup\\{-\langle\>P_{V}u,u\>\rangle_{V}\;;\;\|u\|_{V}=1\\},$

it is enough to show that

$-\langle\>P_{V}u,u\>\rangle_{V}\leq(-2\ell({\bf 1}_{A}P{\bf
1}_{A})+r(P_{V}))\|u\|^{2}_{V}\;\;\mbox{for all $u\in\ell^{2}$}.$ (4.29)

To see this, let $u\in\ell^{2}$ be arbitrary and $J={\bf 1}_{A}-{\bf
1}_{A^{\sf c}}$. We first verify that

$-\langle\>Pu,u\>\rangle\leq-2\ell({\bf 1}_{A}P{\bf
1}_{A})\|u\|_{2}^{2}+\langle\>PJu,Ju\>\rangle.$ (4.30)

Let $P_{+}={\bf 1}_{A}P{\bf 1}_{A}+{\bf 1}_{A^{\sf c}}P{\bf 1}_{A^{\sf c}}$
and $P_{-}={\bf 1}_{A}P{\bf 1}_{A^{\sf c}}+{\bf 1}_{A^{\sf c}}P{\bf 1}_{A}$.
Then,

$P_{\pm}J=\pm JP_{\pm},$ (4.31)

as is easily be seen. Moreover, for fixed $c\in A^{\sf c}$, the map $x\mapsto
x+c$ ($A$ $\longrightarrow$ $A^{\sf c}$) is a bijection. Hence

$\displaystyle\langle\>P_{+}u,u\>\rangle$ $\displaystyle=$
$\displaystyle\left(\sum_{x,y\in A}+\sum_{x,y\not\in
A}\right)p(y-x)u(x)u(y)^{*}$ (4.32) $\displaystyle=$
$\displaystyle\sum_{x,y\in A}p(y-x)u(x)u(y)^{*}$
$\displaystyle+\sum_{x^{\prime},y^{\prime}\in
A}p(y^{\prime}-x^{\prime})u(x^{\prime}+c)u(y^{\prime}+c)^{*}$
$\displaystyle\geq$ $\displaystyle\ell({\bf 1}_{A}P{\bf 1}_{A})\sum_{x\in
A}\left(|u(x)|^{2}+|u(x+c)|^{2}\right)=\ell({\bf 1}_{A}P{\bf
1}_{A})\|u\|^{2}.$

Thus, we obtain (4.30) as follows.

$\displaystyle\langle\>PJu,Ju\>\rangle$ $\displaystyle=$
$\displaystyle\langle\>P_{+}Ju,Ju\>\rangle+\langle\>P_{-}Ju,Ju\>\rangle$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{edge5})}}}{{=}}$
$\displaystyle\langle\>JP_{+}u,Ju\>\rangle-\langle\>JP_{-}u,Ju\>\rangle$
$\displaystyle=$
$\displaystyle\langle\>P_{+}u,u\>\rangle-\langle\>P_{-}u,u\>\rangle=2\langle\>P_{+}u,u\>\rangle-\langle\>Pu,u\>\rangle$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{edge6})}}}{{\geq}}$
$\displaystyle 2\ell({\bf 1}_{A}P{\bf
1}_{A})\|u\|_{2}^{2}-\langle\>Pu,u\>\rangle.$

We then use (4.30) to prove (4.29) as follows. Note that $\ell({\bf
1}_{A}P{\bf 1}_{A})\|u\|^{2}\geq\ell({\bf 1}_{A}P{\bf 1}_{A})\|u\|_{V}^{2}$,
since $\|u\|\geq\|u\|_{V}$ and we have assumed that $\ell({\bf 1}_{A}P{\bf
1}_{A})\geq 0$. Therefore,

$\displaystyle-\langle\>P_{V}u,u\>\rangle_{V}$ $\displaystyle=$
$\displaystyle-\langle\>Pu,u\>\rangle$
$\displaystyle\stackrel{{\scriptstyle\scriptsize\mbox{(\ref{edge4})}}}{{\leq}}$
$\displaystyle-2\ell({\bf 1}_{A}P{\bf
1}_{A})\|u\|^{2}+\langle\>PJu,Ju\>\rangle$ $\displaystyle=$
$\displaystyle-2\ell({\bf 1}_{A}P{\bf
1}_{A})\|u\|^{2}+\langle\>P_{V}Ju,Ju\>\rangle_{V}$ $\displaystyle\leq$
$\displaystyle(-2\ell({\bf 1}_{A}P{\bf 1}_{A})+r(P_{V}))\|u\|^{2}_{V}.$

$\Box$

Acknowledgments The second author was supported by JSPS KAKENHI Grant Number
25400136

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