Principal Eigenvalue and Landscape Function of the Anderson Model on a Large Box

Abstract

We state a precise formulation of a conjecture concerning the product of the principal eigenvalue and the sup-norm of the landscape function of the discrete Anderson model restricted to a large box. We first provide the asymptotic of the principal eigenvalue as the size of the box grows, and then use it to give a partial proof of the conjecture. For the one dimensional case, we give a complete proof by means of Green function bounds.

1 Introduction and Results

The landscape function, introduced by Filoche and Mayboroda in [1], has been conjectured to capture the low eigenvalues of the Anderson model operator, discrete or continuous, restricted to a finite large box. We can find this conjecture loosely stated in [2, Equation 1.4] as: If 0 is the minimum of the support of the potential distribution then

$$\begin{aligned} \lambda _i L_i\approx 1+\frac{d}{4}, \qquad 1\le i\ll n^d, \end{aligned}$$

where $\{\lambda _i\}_{i}$ are the eigenvalues ordered increasingly, $\{L_i\}_{i}$ are the local maxima of the landscape function ordered decreasingly, d is the dimension, and n is the linear size on the box. Numerical experiments with Bernoulli and Uniform potential distributions support the conjecture (see [3, 4]), but to this moment there is no mathematical proof. In this article we give a precise formulation of the conjecture on the discrete setting for the case $i=1$, that is, for the product of the principal (smallest) eigenvalue and the sup-norm of the Landscape function on a large box. We claim such product converges almost surely to an explicit dimensional constant, different from $1+\frac{d}{4}$, as the size of the box goes to infinity and give the proof of the $\liminf$. For the special case $d=1$, we also give the proof of the $\limsup$.

We start with some definitions and notation. Given a finite set $A\subseteq {\mathbb {Z}}^d$ and a positive potential $W:A\rightarrow [0,\infty )$ we consider the Schrödinger operator

$$\begin{aligned} -\Delta _{A}+W&:\ell ^2(A)\longrightarrow \ell ^2(A),\\ (-\Delta _{A}+W)\phi (x)&{:}{=}\sum _{ \left| y-x\right| =1}\left[ \phi (x)-\phi (y)\right] +W(x)\phi (x), \end{aligned}$$

where $-\Delta _{A}$ has Dirichlet boundary conditions. From it, we define its principal eigenvalue and landscape function

$$\begin{aligned} \lambda _{A,W}{:}{=}\inf \sigma (-\Delta _{A}+W),\qquad L_{A,W}{:}{=}(-\Delta _{A}+W)^{-1}\mathbbm {1}_{A}. \end{aligned}$$

Notice that $\lambda _{A,W}>0$ and that $L_{A,W}$ is well defined on A, since $-\Delta _{A}>0$ and $W\ge 0$.

Let $V=\{V(x)\}_{x\in {\mathbb {Z}}^d}$ be an independent and identically distributed (i.i.d.) random non-negative potential whose probability measure and expectation we denote ${\mathbb {P}}$ and ${\mathbb {E}}$, and define for $n\in {\mathbb {N}}$ the box $\Lambda _n{:}{=}[-n,n]^d\cap {\mathbb {Z}}^d$. Our main objectives are the asymptotics of $\lambda _{\Lambda _n,V}$ and $\left\| L_{\Lambda _n,V}\right\| _\infty$ as $n\rightarrow \infty$, where, as customary, the restriction of V to $\Lambda _n$ is implicit.

In addition to V being non-negative (i.e., $\mathbb {P}\left[ V(0)\in (-\infty ,0)\right] =0$) we will always assume the distribution function $F(t)=\mathbb {P}\left[ V(0)\le t\right]$ satisfies one of the following mutually exclusive conditions:

(C1):

$0<F(0)<1$,   (e.g., Bernoulli(p))

(C2):

$F(t)=c\,t^\eta (1+o(1))$ as $t\downarrow 0$ for some $c,\,\eta >0$.   (e.g., Uniform(0, 1))

We write n instead of $\Lambda _n$ whenever convenient, for instance $-\Delta _n=-\Delta _{\Lambda _n}$ and $\lambda _{n,V}=\lambda _{\Lambda _n,V}$. We denote by $\omega _d$ and $\mu _d$ respectively, the volume of the unit ball in ${\mathbb {R}}^d$ and the principal eigenvalue of the continuous Laplacian ($-\sum _{i=1}^d\partial ^2/\partial x_i^2$) on such ball with Dirichlet boundary conditions.

We now state our conjecture and results. We are always assuming that V is non-negative and satisfies (C1) or (C2). We claim that:

Conjecture 0

$\displaystyle \lim _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{n,V}\right\| _\infty =\frac{\mu _d}{2d}$   ${\mathbb {P}}$-a.s.

The heuristic argument behind this conjecture is that both $\lambda _{n,V}$ and $\left\| L_{n,V}\right\| _\infty$ are controlled by the largest ball inside of $\Lambda _n$ with zero or very low potential. If the radius of such ball is r then, roughly, $\lambda _{n,V}$ is proportional to $r^{-2}$ and $\left\| L_{n,V}\right\| _\infty$ is proportional to $r^{2}$, making the product of “order one” in r. The appearance of the continuous constant $\frac{\mu _d}{2d}$ is another instance of the solution of a discrete problem converging to the solution of the corresponding continuous one. The disagreement between the dimensional constants $\frac{\mu _d}{2d}$ and $1+\frac{d}{4}$ is simply explained by the fact that $1+\frac{d}{4}$ was “guessed” from the numerical experiments, and the two constants are close to each other. For example, for $d=1$ we have $1+\frac{1}{4}=1.25$ and $\frac{\mu _1}{2}=\frac{\pi ^2}{8}\approx 1.23$.

Using the Min–Max Principle and our hypothesis on V it is straightforward to show that $\lambda _{n,V}$ is decreasing in n and converges to 0. Our first result is on the speed of this convergence; to state it we first need to define the deterministic sequences:

$$\begin{aligned} \epsilon _n{:}{=}{\left\{ \begin{array}{ll} 0,&{}{\textbf {(C1)}},\\ (\ln n)^{-2/d},&{}{\textbf {(C2)}}, \end{array}\right. } \qquad y_n{:}{=}\left( \frac{d\ln n}{\omega _d\left| \ln F(\epsilon _n)\right| }\right) ^{1/d}. \end{aligned}$$

Theorem 1

$\displaystyle \lim _{n\rightarrow \infty }y_n^2\lambda _{n,V} =\mu _d$   ${\mathbb {P}}$-a.s.

The proof of Theorem 1 is given in Sect. 2, and it is divided into the $\limsup$ and $\liminf$ bounds. The $\limsup$ bound follows from the Min–Max Principle and the previously mentioned heuristic of the largest ball with zero or very low potential. The $\liminf$ bound is more involved; it uses a Lifshitz tails result form [5] and the connection between the integrated density of states of the (infinite) Anderson model and the cumulative distribution function of $\lambda _{n,V}$.

Our second result is a partial proof of Conjecture 0, and a complete proof when $d=1$.

Theorem 2

• (i) $\displaystyle \varliminf _{n\rightarrow \infty }\lambda _{n,V} \left\| L_{n,V}\right\| _\infty \ge \frac{\mu _d}{2d}$   ${\mathbb {P}}$-a.s.

• (ii) If $d=1$ then $\displaystyle \lim _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{n,V}\right\| _\infty =\frac{\mu _1}{2}$   ${\mathbb {P}}$-a.s.

Remark

The article [6] has a proof of (ii) in the continuous setting for the (C1) case. Both proofs follow the heuristic of the largest ball with zero potential, but differ on how to obtain a lower bound of $\lambda _{n,V}$ and an upper bound of $L_{n,V}$.

We prove Theorem 2 in Sect. 3 after deriving some general properties of landscape functions. Most notable among these properties is Proposition 9, which states that $\lambda _{A,W}\left\| L_{A,W}\right\| _\infty$ is bounded from above and below by two dimensional constants uniformly on A and W. This is a consequence of an upper bound of the $\ell ^\infty \rightarrow \ell ^\infty$ norm of the semigroup generated by the Schrödinger operator, which we adapted from the book [7] to the discrete setting. The statement (i) of Theorem 2 follows from domain monotonicity of the landscape function and the asymptotic of $\lambda _{n,V}$ given in Theorem 1, while (ii) is based on the geometric resolvent identity and the restrictions of one dimensional geometry.

We tried to illustrate Theorem 2(ii) when $V(0)\overset{\text {d}}{=}$ Bernoulli(p) by plotting $\lambda _{n,V}\left\| L_{n,V}\right\| _\infty \,\text {v.s.}\,n$ for a single realization of the potential. However, the plot does not show any kind of accumulation up to $n=10^5$, suggesting the convergence is very slow. Instead, we draw the empirical distribution of $\lambda _{n,V}\left\| L_{n,V}\right\| _\infty -\frac{\mu _1}{2}$ from $10^5$ realizations for $n=10^2,10^3,10^4,10^5$. These are given in Fig. 1, from which we can see that the empirical distribution concentrates towards 0 as n increases.

In the proofs that follow, C or C(d) is a finite positive constant that may only depend on the dimension and can change from line to line. By $a_t\sim _t b_t$ we mean $\lim _{t\rightarrow \infty }\frac{a_t}{b_t}=1$.

Fig. 1

Empirical distribution of $\lambda _{n,V}\left\| L_{n,V}\right\| _\infty -\frac{\mu _1}{2}$ for $d=1$ and $V(0)\overset{\text {d}}{=}$ Bernoulli(0.3) computed form $10^5$ samples. The empirical mean (m) and empirical standard deviation (s) are shown in red and blue respectively

2 Principal Eigenvalue (Proof of Theorem 1)

2.1 The Lim Sup Bound

The goal of this subsection is to show

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }y_n^2\lambda _{n,V}\le \mu _d\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

(1)

As usual, getting a good upper bound on $\lambda _{n,V}$ just requires choosing a good test function and applying the Min–Max Principle.

Define $Y_n$ to be the radius of the largest open euclidean ball contained in $\Lambda _n$ in which V is uniformly bounded by $\epsilon _n$, that is,

$$\begin{aligned} Y_{n}{:}{=}\max \left\{ r\in {\mathbb {N}}\,\Big \vert \,\exists x\in \Lambda _n\,\text { such that }\,B(x,r)\cap {\mathbb {Z}}^d\subseteq \Lambda _n\cap V^{-1}\left( [0,\epsilon _n]\right) \right\} , \end{aligned}$$

where $B(x,r)=\{x'\in {\mathbb {R}}^d\,\vert \,\left| x-x'\right| <r\}$. Also, define $x_n\in \Lambda _n$ to be the center of a ball at which the maximum is attained (it may not be unique); and let $\phi \in \ell ^2(B(x_n,Y_n)\cap {\mathbb {Z}}^d)$ be the normalized eigenvector of $-\Delta _{B(x_n,Y_n)}$ associated to $\lambda _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}$, extended to $\Lambda _n$ by 0. Then, by the Min–Max Principle, we have

$$\begin{aligned} \lambda _{n,V}&\le \left\langle \phi ,\left( -\Delta _{n}+V\right) \phi \right\rangle _{\ell ^2(\Lambda _n)}\\&=\left\langle \phi ,\left( -\Delta _{B(x_n,Y_n)\cap {\mathbb {Z}}^d}+V\right) \phi \right\rangle _{\ell ^2(B(x_n,Y_n)\cap {\mathbb {Z}}^d)}\\&\le \lambda _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}+\epsilon _n. \end{aligned}$$

To conclude the proof of (1) we use that the event $\{\lim _{n\rightarrow \infty }Y_n/y_n=1\}$ has probability one (this will be proven in Proposition 3), together with translation invariance and the limit $\displaystyle \lim _{r\rightarrow \infty }r^2\lambda _{B(0,r)\cap {\mathbb {Z}}^d,0}=\mu _d$, to obtain

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }y_n^2\lambda _{n,V}&\le \lim _{n\rightarrow \infty }y_n^2\left( \lambda _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}+\epsilon _n\right) \\&=\lim _{n\rightarrow \infty }\frac{y_n^2}{Y_n^2}Y_n^2\lambda _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}=\mu _d\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

The limit $\displaystyle \lim _{r\rightarrow \infty }r^2\lambda _{B(0,r)\cap {\mathbb {Z}}^d,0}=\mu _d$ is a consequence of the discrete Laplacian converging to the continuous one, or random walk converging to Brownian motion. A proof following the latter approach can be found in [8, Proposition 8.4.2], where an extra factor d appears as a result of the probabilistic normalization of the Laplacian.

Proposition 3

$\displaystyle Y_n\sim _n y_n$   ${\mathbb {P}}$-a.s.

Proof

If $Y_n< y_n(1-\delta )^{1/d}$ for some $0<\delta <1$, then the inscribed ball of each of the $\left( \frac{2n}{2 y_n(1-\delta )^{1/d}}\right) ^d(1+o(1))$ disjoint cubes of side length $\big \lceil 2 y_n(1-\delta )^{1/d} \big \rceil$ that make up $\Lambda _n$ contains a point x with $V(x)>\epsilon _n$. Approximating the number of points in such balls by $\#(B(0,r)\cap {\mathbb {Z}}^d)\sim _r{\text {Vol}}(B(0,r))=\omega _d r^d$, we obtain for large n

$$\begin{aligned} \mathbb {P}\left[ Y_n< y_n(1-\delta )^{1/d}\right]&\le \left( 1-F(\epsilon _n)^{\omega _d y_n^d(1-\delta )(1+o(1))}\right) ^{\frac{n^d}{y_n^d(1-\delta )}(1+o(1))}\\&=\left( 1-\frac{1}{n^{d(1-\delta )(1+o(1))}}\right) ^{\frac{n^d\omega _d\left| \ln F(\epsilon _n)\right| }{d(\ln n)(1-\delta )}(1+o(1))}\\&\le \exp \left( -\frac{n^\delta \omega _d\left| \ln F(\epsilon _n)\right| }{2 d(\ln n)(1-\delta )}\right) , \end{aligned}$$

which is summable. Therefore, the Borel–Cantelli Lemma and sending $\delta \rightarrow 0$ give

$$\begin{aligned} 1\le \varliminf _{n\rightarrow \infty }y_n^{-1} Y_n\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

We show the $\limsup$ bound, first on an exponential sub-sequence, and then we extend it to the whole sequence. The extending argument requires a monotone sequence of random variables, which $Y_n$ may fail to be if (C2) holds. For this reason we introduce

$$\begin{aligned} Y_{n,n'}{:}{=}\max \left\{ r\in {\mathbb {N}}\,\Big \vert \,\exists x\in \Lambda _n\,\text { such that }\,B(x,r)\cap {\mathbb {Z}}^d\subseteq \Lambda _n\cap V^{-1}\big ([0,\epsilon _{n'}]\big )\right\} , \end{aligned}$$

which is increasing on n, decreasing on $n'$ and satisfies $Y_{n,n}=Y_n$. Since for $\delta >0$ and large m we have

$$\begin{aligned}&{\mathbb {P}}\left[ Y_{\left\lfloor e^{m+1} \right\rfloor ,\left\lfloor e^{m} \right\rfloor }> y_{\left\lfloor e^{m} \right\rfloor }(1+\delta )^{1/d}\right] \\&\qquad \qquad \le \sum _{x\in \Lambda _{\left\lfloor e^{m+1} \right\rfloor }}{\mathbb {P}}\left[ B(x,y_{\left\lfloor e^{m} \right\rfloor }(1+\delta )^{1/d})\cap {\mathbb {Z}}^d\subseteq V^{-1}[0,\epsilon _{\left\lfloor e^m \right\rfloor }]\right] \\&\qquad \qquad =\#\Lambda _{\left\lfloor e^{m+1} \right\rfloor } F(\epsilon _{\left\lfloor e^m \right\rfloor })^{\omega _d y_{\left\lfloor e^{m} \right\rfloor }^d(1+\delta )(1+o(1))}\\&\qquad \qquad = \frac{\#\Lambda _{\left\lfloor e^{m+1} \right\rfloor }}{\left\lfloor e^{m} \right\rfloor ^{d(1+\delta )(1+o(1))}}\\&\qquad \qquad \le C(d) e^{-m d \delta /2}, \end{aligned}$$

the Borel–Cantelli Lemma and the limit $\delta \rightarrow 0$ give

$$\begin{aligned} \varlimsup _{m\rightarrow \infty }y_{\left\lfloor e^{m} \right\rfloor }^{-1}Y_{\left\lfloor e^{m+1} \right\rfloor ,\left\lfloor e^{m} \right\rfloor }\le 1\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

For $n\in {\mathbb {N}}$ define $m(n)\in {\mathbb {N}}$ by $\left\lfloor e^{m(n)} \right\rfloor \le n<\left\lfloor e^{m(n)+1} \right\rfloor$. Since $y_n\sim _n y_{\left\lfloor e^{m(n)} \right\rfloor }$ and $Y_n\le Y_{\left\lfloor e^{m(n)+1} \right\rfloor ,\left\lfloor e^{m(n)} \right\rfloor }$, we conclude

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }y_n^{-1}Y_n&\le \varlimsup _{n\rightarrow \infty }y_n^{-1} Y_{\left\lfloor e^{m(n)+1} \right\rfloor ,\left\lfloor e^{m(n)} \right\rfloor }\\&=\varlimsup _{n\rightarrow \infty }y_{\left\lfloor e^{m(n)} \right\rfloor }^{-1}Y_{\left\lfloor e^{m(n)+1} \right\rfloor ,\left\lfloor e^{m(n)} \right\rfloor }\le 1\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

$\square$

2.2 The Lim Inf Bound

In this subsection we show that

$$\begin{aligned} \mu _d\le \varliminf _{n\rightarrow \infty }y_n^2\lambda _{n,V}\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

(2)

The main input for this is a result from [5] on the Lifshitz tail of the integrated density of states. We recall the integrated density of states of the Anderson model is a deterministic distribution function given by the ${\mathbb {P}}$-a.s. limit

$$\begin{aligned} I(t){:}{=}\lim _{n\rightarrow \infty }\frac{1}{\#\Lambda _n}\#\{\lambda \in \sigma (-\Delta _n+V)\,\vert \,\lambda \le t\},\qquad t\in {\mathbb {R}}, \end{aligned}$$

where the eigenvalues are counted with multiplicities. The central hypothesis of [5] is a scaling assumption on the cumulant-generating function of V(0), $H(t){:}{=}\ln \mathbb {E}\left[ e^{-t V(0)}\right]$, which we will check in the following proposition. To state it, we first need to define

$$\begin{aligned} (1,\infty )\ni t\longmapsto \alpha (t){:}{=}{\left\{ \begin{array}{ll}t^{1/(d+2)},&{} {\textbf {(C1)}},\\ \left( \frac{t}{\ln t}\right) ^{1/(d+2)},&{}{\textbf {(C2)}}, \end{array}\right. }\qquad {\widetilde{H}}{:}{=}{\left\{ \begin{array}{ll}\left| \ln F(0)\right| ,&{} {\textbf {(C1)}},\\ \frac{2\eta }{d+2},&{} {\textbf {(C2)}}. \end{array}\right. } \end{aligned}$$

Proposition 4

(Scaling assumption of [5]) For any compact $K\subseteq (0,\infty )$ we have

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\alpha ^{d+2}(t)}{t}H\left( \frac{t}{\alpha ^d(t)}y\right) =-{\widetilde{H}} \end{aligned}$$

uniformly on $y\in K$.

Proof

First assume (C1). In this case $\frac{\alpha ^{d+2}(t)}{t}=1$ and $\frac{t}{\alpha ^d(t)}=t^{2/(d+2)}$. Since for $t>0$ we have

$$\begin{aligned} \ln F(0)\le H(t)&=\ln \left( \mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)\le \frac{1}{\sqrt{t}}}\right] +\mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)> \frac{1}{\sqrt{t}}}\right] \right) \\&\le \ln \left( F\left( 1/\sqrt{t}\right) +e^{-\sqrt{t}}\right) , \end{aligned}$$

we conclude that

$$\begin{aligned}&\sup _{y\in K}\left| \frac{\alpha ^{d+2}(t)}{t}H\left( \frac{t}{\alpha ^d(t)}y\right) -\ln F(0)\right| \\&\qquad \qquad \le \sup _{y\in K}\ln \left( F\left( \frac{1}{t^{1/(d + 2)}\sqrt{y}}\right) +e^{-t^{1/(d + 2)}\sqrt{y}}\right) -\ln F(0)\\&\qquad \qquad =\ln \left( F\left( \frac{1}{t^{1/(d + 2)}\sqrt{\min K}}\right) +e^{-t^{1/(d + 2)}\sqrt{\min K}}\right) -\ln F(0)\xrightarrow [t\rightarrow \infty ]{} 0. \end{aligned}$$

Now assume (C2). In this case $\frac{\alpha ^{d+2}(t)}{t}=\frac{1}{\ln t}$ and $\frac{t}{\alpha ^d(t)}=t^{2/(d+2)}(\ln t)^{d/(d+2)}$. We introduce a parameter $0<\delta <1$ and observe that

$$\begin{aligned} H(t)=&\ln \left( \mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)\le t^{-\delta }}\right] +\mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)>t^{-\delta }}\right] \right) \\&\le \ln \left( F\left( t^{-\delta }\right) +e^{-t^{1-\delta }}\right) ,\qquad t>0, \end{aligned}$$

which implies

$$\begin{aligned}&\varlimsup _{t\rightarrow \infty }\sup _{y\in K}\frac{\alpha ^{d+2}(t)}{t}H\left( \frac{t}{\alpha ^d(t)}y\right) \\&\qquad \qquad \le \varlimsup _{t\rightarrow \infty }\sup _{y\in K}\frac{1}{\ln t}\ln \left( F\left( \left[ \frac{t}{\alpha ^d(t)}y\right] ^{-\delta }\right) +\exp \left( -\left[ \frac{t}{\alpha ^d(t)}y\right] ^{1-\delta }\right) \right) \\&\qquad \qquad = \varlimsup _{t\rightarrow \infty }\frac{1}{\ln t}\ln \left( F\left( \left[ \frac{t}{\alpha ^d(t)}\min K\right] ^{-\delta }\right) +\exp \left( -\left[ \frac{t}{\alpha ^d(t)}\min K\right] ^{1-\delta }\right) \right) \\&\qquad \qquad =-\frac{2\delta \eta }{d+2}\xrightarrow [\delta \rightarrow 1]{}-\frac{2\eta }{d+2}. \end{aligned}$$

For the $\displaystyle \varliminf _{t\rightarrow \infty }\inf _{y\in K}$ we use

$$\begin{aligned} H(t)&=\ln \left( \mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)\le t^{-1}}\right] +\mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)>t^{-1}}\right] \right) \\&\ge \ln \left( e^{-1}F(t^{-1})\right) ,\qquad t>0, \end{aligned}$$

to obtain

$$\begin{aligned} \varliminf _{t\rightarrow \infty }\inf _{y\in K}\frac{\alpha ^{d+2}(t)}{t}H\left( \frac{t}{\alpha ^d(t)}y\right)&\ge \varliminf _{t\rightarrow \infty }\inf _{y\in K}\frac{1}{\ln t}\ln \left( e^{-1}F\left( \left[ \frac{t}{\alpha ^d(t)}y\right] ^{-1}\right) \right) \\&\ge \varliminf _{t\rightarrow \infty }\frac{1}{\ln t}\ln \left( e^{-1}F\left( \left[ \frac{t}{\alpha ^d(t)}\max K\right] ^{-1}\right) \right) \\&=-\frac{2\eta }{d+2}. \end{aligned}$$

$\square$

Having checked the scaling assumption on H, we now have the Lifshitz tail result:

Theorem 5

[5, Theorem 1.3] Define the constant

$$\begin{aligned} \chi {:}{=}\inf _{g\in H^1({\mathbb {R}}^d),\,\left\| g\right\| _2=1}\left( \left\| \nabla g\right\| _2^2+{\widetilde{H}}{\text {Vol}}({\text {supp}}g)\right) , \end{aligned}$$

then

$$\begin{aligned} \lim _{t\downarrow 0}\frac{\ln I(t)}{t\alpha ^{-1}\left( t^{-1/2}\right) }=-2\,d^{d/2}\left( \frac{\chi }{d+2}\right) ^{(d+2)/2}. \end{aligned}$$

Remark

The function $t\mapsto \alpha (t)$ is eventually increasing so $\alpha ^{-1}(t)$ is well defined for large t. The original statement from [5] is far more general; our conditions on V make H fall into, what is there called, the $(\gamma =0)$-class.

The constant $\chi$ can be explicitly computed by means of the Faber–Krahn inequality:

Proposition 6

$\displaystyle \chi =(d+2)\left( \frac{{\widetilde{H}}\omega _d}{2}\right) ^{2/(d+2)}\left( \frac{\mu _d}{d}\right) ^{d/(d+2)}$.

Proof

Starting from $\chi =\inf _{g\in H^1({\mathbb {R}}^d),\,\left\| g\right\| _2=1}\left( \left\| \nabla g\right\| _2^2+D{\text {Vol}}({\text {supp}}g)\right)$ we see that we only need to consider the finite volume case. Hence

$$\begin{aligned} \chi&=\inf _{\begin{array}{c} A\subseteq {\mathbb {R}}^d,\\ {\text {Vol}}(A)<\infty \end{array}}\inf _{\begin{array}{c} g\in H^1({\mathbb {R}}^d),\,\left\| g\right\| _2=1,\\ {\text {supp}}g=A \end{array}}\left( \left\| \nabla g\right\| _2^2+{\widetilde{H}}{\text {Vol}}(A)\right) \\&=\inf _{\begin{array}{c} A\subseteq {\mathbb {R}}^d,\\ {\text {Vol}}(A)<\infty \end{array}}\left( \mu (A)+{\widetilde{H}}{\text {Vol}}(A)\right) , \end{aligned}$$

where $\mu (A)$ is the principal eigenvalue of the continuous Laplacian ($-\sum _{i=1}^d\partial ^2/\partial x_i^2$) defined on A with Dirichlet boundary conditions. The Faber-Krahn inequality states that over all domains of a given volume the one with the lowest principal eigenvalue is the ball, therefore, using $\mu (B(0,r))=\mu _d/r^2$ and ${\text {Vol}}(B(0,r))=\omega _d r^d$ we obtain

$$\begin{aligned} \chi =\inf _{0<r<\infty }\left( \frac{\mu _d}{r^2}+{\widetilde{H}}\omega _d r^d\right) . \end{aligned}$$

Evaluating at the only critical point $\displaystyle r=\left( \frac{2\mu _d}{{\widetilde{H}}\omega _d d}\right) ^{1/(d+2)}$ finishes the proof. $\square$

We now exploit the connection between I and the distribution of $\lambda _{n,V}$. This is a classic argument that can be found, for instance, in [9, Equation 4.46]. We present here a slightly modified version. Let $n\in {\mathbb {N}}$ and define a new potential

$$\begin{aligned} V'(x)&{:}{=}{\left\{ \begin{array}{ll} \infty ,&{} x\in \Gamma ,\\ V(x),&{} x\in {\mathbb {Z}}^d\setminus \Gamma , \end{array}\right. }\\ \Gamma&{:}{=}\left\{ x=(x_1,\ldots ,x_d)\in {\mathbb {Z}}^d\,\Big \vert \,x_i\in (2n+2){\mathbb {Z}} \text { for some } i=1,\ldots ,d\right\} . \end{aligned}$$

Clearly $V\le V'$ so for any $k\in {\mathbb {N}}$ and $t\in {\mathbb {R}}$ we have

$$\begin{aligned} \frac{\#\{\lambda \in \sigma (-\Delta _{(2n+2)k}+V)\,\vert \,\lambda \le t\}}{\#\Lambda _{(2n+2)k}}\ge \frac{\#\{\lambda \in \sigma (-\Delta _{(2n+2)k}+V')\,\vert \,\lambda \le t\}}{\#\Lambda _{(2n+2)k}}, \end{aligned}$$

where $-\Delta _{(2n+2)k}+V'$ has (by definition) Dirichlet boundary conditions at $\Gamma$. These Dirichlet boundary conditions at $\Gamma$ imply that $-\Delta _{(2n+2)k}+V'$ is a direct sum of $(2k)^d$ independent terms, all equal in distribution to $-\Delta _{n}+V$. Therefore, by taking the limit $k\rightarrow \infty$ on the above inequality and applying the Law of Large Numbers, we obtain

$$\begin{aligned} I(t)&\ge \left( \lim _{k\rightarrow \infty }\frac{(2k)^d}{\#\Lambda _{(2n+2)k}}\right) \mathbb {E}\left[ \#\{\lambda \in \sigma (-\Delta _{n}+V)\,\vert \,\lambda \le t\}\right] \\&\ge \left( \frac{1}{2n+2}\right) ^d\mathbb {P}\left[ \lambda _{n,V}\le t\right] . \end{aligned}$$

From the previous inequality, Theorem 5 and Proposition 6 we have

$$\begin{aligned} \mathbb {P}\left[ \lambda _{n,V}\le t\right] \le C(d) n^{d} I(t)\le C(d)\,n^{d}\exp \left[ -f(1/t)(1+o(1))\right] \quad \text {as } t\downarrow 0, \end{aligned}$$

(3)

where we have introduced $\displaystyle f(t){:}{=}\frac{{\widetilde{H}}\omega _d\mu _d^{d/2}\alpha ^{-1}(t^{1/2})}{t}$. To finish the proof we need the asymptotic of $f^{-1}(t)$ as $t\rightarrow \infty$:

Proposition 7

• (i) For (C1), $\displaystyle f^{-1}(t)=\frac{1}{\mu _d}\left( \frac{t}{\omega _d\left| \ln F(0)\right| }\right) ^{2/d}$.

• (ii) For (C2), $\displaystyle f^{-1}(t)\sim _t\frac{1}{\mu _d}\left( \frac{d\,t}{2\eta \omega _d\ln t}\right) ^{2/d}$.

Proof

For (C1) there is nothing to prove since $f(t)=\omega _d\left| \ln F(0)\right| \mu _d^{d/2}t^{d/2}$.

For (C2) we have

$$\begin{aligned} f(t)=\frac{k\alpha ^{-1}(t^{1/2})}{t}=k t^{d/2}\ln \alpha ^{-1}(t^{1/2}), \end{aligned}$$

with all the constants collected in $k=\frac{2\eta \omega _d\mu _d^{d/2}}{d+2}$. Since $\alpha$ is eventually increasing and has infinite limit, the same is true for f, in particular $f^{-1}(t)$ exists for large t. By solving for the $\alpha ^{-1}$ term in the first equality above, applying $\alpha$ and simplifying some exponents we arrive at $t=\left( \frac{f(t)}{k\ln \left[ t f(t)/k\right] }\right) ^{2/d}$. Replacing t by $f^{-1}(t)$ we have

$$\begin{aligned} f^{-1}(t)=\left( \frac{t}{k\ln \left[ t f^{-1}(t)/k\right] }\right) ^{2/d}. \end{aligned}$$

In order to deal with the $\ln [t f^{-1}(t)]$ term above, we multiplying the equality by t and then take the logarithm to obtain

$$\begin{aligned} \ln [t f^{-1}(t)]=\frac{d+2}{d}\ln t-\frac{2}{d}\ln \left[ k\ln \left[ t f^{-1}(t)/k\right] \right] . \end{aligned}$$

Since $t f^{-1}(t)\xrightarrow [t\rightarrow \infty ]{}\infty$, the last equality implies $\ln [t f^{-1}(t)]\sim _t \frac{d+2}{d}\ln t$ and therefore

$$\begin{aligned} f^{-1}(t)\sim _t\left( \frac{d\,t}{(d+2)k\ln t}\right) ^{2/d}=\frac{1}{\mu _d}\left( \frac{d\,t}{2\eta \omega _d\ln t}\right) ^{2/d}. \end{aligned}$$

$\square$

Going back to (3) with $n=\left\lfloor e^m \right\rfloor$ and $t=1/f^{-1}((1+\delta )d m)$ for some $m\in {\mathbb {N}}$ and $\delta >0$, we see that

$$\begin{aligned} \mathbb {P}\left[ \lambda _{\left\lfloor e^m \right\rfloor ,V}f^{-1}((1+\delta )d m)\le 1\right]&\le C(d) \left( \left\lfloor e^m \right\rfloor \right) ^{d}\exp \left[ -(1+\delta )d m(1+o(1))\right] \\ {}&\le C(d)\,e^{-m d\delta /2}, \end{aligned}$$

which is summable over $m\in {\mathbb {N}}$. Therefore, by the Borel–Cantelli Lemma we have

$$\begin{aligned} 1\le \varliminf _{m\rightarrow \infty } \lambda _{\left\lfloor e^m \right\rfloor ,V}f^{-1}((1+\delta )d m)=(1+\delta )^{2/d} \varliminf _{m\rightarrow \infty } \lambda _{\left\lfloor e^m \right\rfloor ,V}f^{-1}(d m)\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

As in the proof of Proposition 3, we define $m(n)\in {\mathbb {N}}$ by $\left\lfloor e^{m(n)} \right\rfloor \le n<\left\lfloor e^{m(n)+1} \right\rfloor$, so that $\ln n\sim _n (m(n)+1)$. Since $n\mapsto \lambda _{n,V}$ is monotone decreasing we have

$$\begin{aligned} \varliminf _{n\rightarrow \infty } \lambda _{n,V}f^{-1}(d \ln n)&\ge \varliminf _{n\rightarrow \infty } \lambda _{\left\lfloor e^{m(n)+1} \right\rfloor ,V}f^{-1}(d \ln n)\\ {}&=\varliminf _{n\rightarrow \infty } \lambda _{\left\lfloor e^{m(n)+1} \right\rfloor ,V}f^{-1}(d(m(n)+1))\\&\ge (1+\delta )^{-2/d}\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

The proof of (2) is finished by sending $\delta \rightarrow 0$ and noticing that Proposition 7 implies $f^{-1}(d \ln n)\sim _n\frac{y_n^2}{\mu _d}$ for both (C1) and (C2).

3 Landscape Function

We start this section by deriving some general properties of landscape functions.

For a finite set $A\subseteq {\mathbb {Z}}^d$ and a positive potential $W:A\rightarrow [0,\infty )$ we introduce the Green function with 0 as spectral parameter

$$\begin{aligned} G_{A,W}(x,y){:}{=}{\left\{ \begin{array}{ll} \left\langle \delta _x ,(-\Delta _A+W)^{-1}\delta _y\right\rangle _{\ell ^2(A)},&{}(x,y)\in A\times A,\\ 0,&{}(x,y)\in ({\mathbb {Z}}^d\times {\mathbb {Z}}^d)\setminus (A\times A). \end{array}\right. } \end{aligned}$$

This function is known to be symmetric, non-negative, and decreasing on the potential W. It is also known to satisfy the geometric resolvent identity:

$$\begin{aligned} G_{A,W}(x,y)=G_{A',W}(x,y)+\sum _{(i,j)\in \partial A'}G_{A',W}(x,i)G_{A,W}(j,y),\qquad A'\subseteq A, \end{aligned}$$

where $\partial A'{:}{=}\{(i,j)\in A'\times ({\mathbb {Z}}^d{\setminus } A')\,\vert \,\left| i-j\right| =1\}$ is the boundary of $A'$ (see [10, Section 5.3]). By extending the definition of $L_{A,W}$ to $L_{A,W}(x){:}{=}\sum _{y\in {\mathbb {Z}}^d}G_{A,W}(x,y)$ for all $x\in {\mathbb {Z}}^d$, the previously stated properties of $G_{A,W}$ translate into non-negativity, potential monotonicity and domain monotonicity of landscape functions:

• $L_{A,W}\ge 0$ and $L_{A,W}(x)=0$ if $x\in {\mathbb {Z}}^d\setminus A$.

• If $0\le W'\le W$ then $L_{A,W}\le L_{A,W'}$.

• If $A'\subseteq A$ then

  $$\begin{aligned} L_{A,W}(x)=L_{A',W}(x)+\sum _{(i,j)\in \partial A'}G_{A',W}(x,i)L_{A,W}(j)\ge L_{A',W}(x). \end{aligned}$$

  (4)

Our last general property is that $\lambda _{A,W}\left\| L_{A,W}\right\| _\infty$ is bounded from above and below by two positive constants uniformly on A and W. This is based on the following upper bound of the $\ell ^\infty \rightarrow \ell ^\infty$ norm of the semigroup, which can be found, for the continuous setting, in [7, Chapter 3, Theorem 1.2]. We could not find a proof in the literature for the discrete case, so we provide one in Appendix A.

Theorem 8

For a finite $A\subseteq {\mathbb {Z}}^d$ and $W:A\rightarrow [0,\infty )$ we have

$$\begin{aligned} \left\| \exp \left( -t[-\Delta _{A}+W]\right) \mathbbm {1}_A\right\| _\infty \le C(d)\left( 1+\left[ \lambda _{A,W}t\right] ^{d/2}\right) e^{-\lambda _{A,W}t},\qquad t\ge 0. \end{aligned}$$

As an an immediate corollary we obtain:

Proposition 9

For a finite $A\subseteq {\mathbb {Z}}^d$ and $W:A\rightarrow [0,\infty )$ we have

$$\begin{aligned} 1\le \lambda _{A,W}\left\| L_{A,W}\right\| _\infty \le C(d). \end{aligned}$$

Remark

The bound 1 is sharp. It is attained when A is a single point of ${\mathbb {Z}}^d$.

Proof

For the upper bound we use Theorem 8 and the substitution $u=\lambda _{A,W}t$:

$$\begin{aligned} \left\| L_{A,W}\right\| _\infty&= \left\| \int _0^\infty \exp \left( -t[-\Delta _{A}+W]\right) \mathbbm {1}_{A}\,\textrm{d}t\right\| _\infty \\&\le C(d)\int _0^\infty \left( 1+\left[ \lambda _{A,W}t\right] ^{d/2}\right) \exp \left( -\lambda _{A,W}t\right) \,\textrm{d}t\\&=\frac{C(d)}{\lambda _{A,W}}\int _0^\infty \left( 1+u^{d/2}\right) e^{-u}\,\textrm{d}u=\frac{C(d)}{\lambda _{A,W}}. \end{aligned}$$

For the lower bound we just need to notice that the positivity of $G_{A,W}$ implies

$$\begin{aligned} \left\| (-\Delta _A+W)^{-1}\right\| _{\ell ^\infty (A)\rightarrow \ell ^\infty (A)}= \left\| (-\Delta _A+W)^{-1}\mathbbm {1}_A\right\| _{\infty }=\left\| L_{A,W}\right\| _\infty , \end{aligned}$$

and therefore, denoting $\phi \in \ell ^2(A)$ an eigenvector of $-\Delta _A+W$ associated to $\lambda _{A,W}$, we obtain

$$\begin{aligned} \frac{1}{\lambda _{A,W}}=\frac{\left\| (-\Delta _A+W)^{-1}\phi \right\| _{\infty }}{\left\| \phi \right\| _\infty }\le \left\| (-\Delta _A+W)^{-1}\right\| _{\ell ^\infty (A)\rightarrow \ell ^\infty (A)}=\left\| L_{A,W}\right\| _\infty . \end{aligned}$$

$\square$

3.1 Proof of Theorem 2(i)

We start with the asymptotic of the sup-norm of the landscape function on balls with 0 potential.

Proposition 10

$\displaystyle \left\| L_{B(0,r)\cap {\mathbb {Z}}^d,0}\right\| _\infty \sim _r\frac{r^2}{2d}$.

Proof

Let $r>0$ and consider the function $\phi _r(x){:}{=}\frac{r^2-\left| x\right| ^2}{2d}$ defined on ${\mathbb {Z}}^d$. Clearly $-\Delta \phi _r(x)=1$ for all $x\in {\mathbb {Z}}^d$ and therefore $L_{B(0,r)\cap {\mathbb {Z}}^d,0}-\phi _r$ is harmonic in $B(0,r)\cap {\mathbb {Z}}^d$. By the Maximum Principle (see [8, Theorem 6.2.1]) we have

$$\begin{aligned} \left| \left\| L_{B(0,r)\cap {\mathbb {Z}}^d,0}\right\| _\infty -\frac{r^2}{2d}\right|&=\left| \sup _{x\in B(0,r)\cap {\mathbb {Z}}^d}L_{B(0,r)\cap {\mathbb {Z}}^d,0}(x)-\sup _{x\in B(0,r)\cap {\mathbb {Z}}^d}\phi _r(x)\right| \\&\le \sup _{x\in B(0,r)\cap {\mathbb {Z}}^d}\left| L_{B(0,r)\cap {\mathbb {Z}}^d,0}(x)-\phi _r(x)\right| \\ {}&=\sup _{x\in \partial ^+\left[ B(0,r)\cap {\mathbb {Z}}^d\right] }\left| L_{B(0,r)\cap {\mathbb {Z}}^d,0}(x)-\phi _r(x)\right| \\ {}&=\sup _{x\in \partial ^+\left[ B(0,r)\cap {\mathbb {Z}}^d\right] }\left| \phi _r(x)\right| \le C(d)\,r(1+o(1)) \end{aligned}$$

where $\partial ^+ A{:}{=}\left\{ x\in {\mathbb {Z}}^d{\setminus } A\,\Big \vert \,\exists \,y\in A\text { such that }\left| x-y\right| =1\right\}$ is the outer boundary of $A\subseteq {\mathbb {Z}}^d$. Dividing by $\frac{r^2}{2d}$ and taking the limit $r\rightarrow \infty$ give the proposition. $\square$

Recall the definitions of $Y_n$ and $x_n$ from Sect. 2.1. From domain monotonicity of landscape functions we have

$$\begin{aligned} L_{n,V}\ge L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,V}. \end{aligned}$$

For (C1), V is identically 0 in $B(x_n,Y_n)\cap {\mathbb {Z}}^d$ so Theorem 1, translation invariance and Propositions 3, 10 give

$$\begin{aligned} \varliminf _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{n,V}\right\| _\infty&\ge \lim _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}\right\| _\infty \\&=\lim _{n\rightarrow \infty }\frac{\mu _d}{y_n^2}\frac{Y_n^2}{2d}=\frac{\mu _d}{2d}\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

For (C2), we use the second resolvent identity, domain monotonicity of the eigenvalue, and Propositions 3, 9, 10 to obtain

$$\begin{aligned}&\lambda _{n,V}\left\| L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}-L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,V}\right\| _\infty \\&\qquad \qquad =\lambda _{n,V}\left\| (-\Delta _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0})^{-1}V L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,V}\right\| _\infty \\&\qquad \qquad \le \left\| (-\Delta _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0})^{-1}V \left[ \lambda _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,V}L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,V}\right] \right\| _\infty \\&\qquad \qquad \le C(d)\epsilon _n\left\| (-\Delta _{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0})^{-1}\mathbbm {1}_{B(x_n,Y_n)\cap {\mathbb {Z}}^d}\right\| _\infty \\&\qquad \qquad = C(d)\epsilon _n \left\| L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}\right\| _\infty \xrightarrow [n\rightarrow \infty ]{{\mathbb {P}}\text {-a.s.}}0, \end{aligned}$$

which implies

$$\begin{aligned} \varliminf _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{n,V}\right\| _\infty&\ge \lim _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,V}\right\| _\infty \\ {}&=\lim _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{B(x_n,Y_n)\cap {\mathbb {Z}}^d,0}\right\| _\infty =\frac{\mu _d}{2d}\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

This concludes the proof of Theorem 2(i).

3.2 Proof of Theorem 2(ii)

We assume from this point on that $d=1$. We set $\llbracket a , b\rrbracket {:}{=}[a,b]\cap {\mathbb {Z}}$ for any $a,b\in {\mathbb {Z}}$, $a\le b$. This proof is based on the following deterministic bound of the Green function in terms of the values of the potential.

Proposition 11

Let $a,b\in {\mathbb {Z}}$, $a\le b$ and $W:\llbracket a , b\rrbracket \rightarrow [0,\infty )$. For any $x\in \llbracket a , b\rrbracket$ we have

$$\begin{aligned} G_{\llbracket a , b\rrbracket ,W}(a,x)&\le \left( \sum _{j=0}^{x-a} (x-a+1-j) W(x-j)\right) ^{-1},\\ G_{\llbracket a , b\rrbracket ,W}(x,b)&\le \left( \sum _{j=0}^{b-y} (b-x+1-j) W(x+j)\right) ^{-1}. \end{aligned}$$

Proof

By translation invariance and isotropy of the Laplacian, it is enough to show the first inequality assuming that $a=1$, $b\ge 1$, $W:\llbracket 1 , b\rrbracket \rightarrow [0,\infty )$.

Fix some $x\in \llbracket 1 , b\rrbracket$. By potential monotonicity and Cramer’s rule we have

$$\begin{aligned} G_{\llbracket 1 , b\rrbracket ,W}(1,x)\le G_{\llbracket 1 , b\rrbracket ,W\mathbbm {1}_{\llbracket 1 , x\rrbracket }}(1,x)=\frac{\det \left( [-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket }]_{1\rightarrow \delta _x}\right) }{\det (-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket })}, \end{aligned}$$

where $[-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket }]_{1\rightarrow \delta _x}$ is the matrix obtained by replacing the first column of $-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket }$ by $\delta _x$. Computing the determinant on the numerator from its first column we see that

$$\begin{aligned} \det \left( [-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket }]_{1\rightarrow \delta _x}\right)&=(-1)^{x+1}\det \left( \begin{array}{c|c} T &{} 0 \\ \hline M &{} -\Delta _{\llbracket 1 , b-x\rrbracket } \end{array}\right) \\&=(-1)^{x+1}\det (T)\det (-\Delta _{\llbracket 1 , b-x\rrbracket })=b-x+1, \end{aligned}$$

since T is a lower triangular square matrix of size $x-1$ with $(-1)$ on all the diagonal, and $\det (-\Delta _{\llbracket 1 , k\rrbracket })=k+1$ for all $k\in {\mathbb {N}}$. Here we have used that the determinant of an empty matrix is 1.

Consider $\det (-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket })$ as a polynomial in $(W(j))_{j=1}^x$. It is clear that it does not contain squares, or greater powers, of any W(j). Moreover, a straightforward computation shows that the coefficient of $W(j_1)W(j_2)\cdots W(j_{k-1})W(j_k)$, with $1\le j_1<j_2<\cdots j_{k-1}<j_k\le x$ and $1\le k\le x$, is

$$\begin{aligned} \det (-\Delta _{\llbracket 1 , j_1-1\rrbracket })&\det (-\Delta _{\llbracket j_1+1 , j_2-1\rrbracket })\cdots \det (-\Delta _{\llbracket j_{k-1}+1 , j_k-1\rrbracket })\det (-\Delta _{\llbracket j_k+1 , b\rrbracket })\\&=j_1(j_2-j_1)\cdots (j_k-j_{k-1})(b-j_k+1). \end{aligned}$$

The remaining coefficient (the constant one) is $\det (-\Delta _{\llbracket 1 , b\rrbracket })=b+1$, which means all coefficients of $\det (-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket })$ are positive. Therefore, by keeping only the linear terms, we obtain

$$\begin{aligned} G_{\llbracket 1 , b\rrbracket ,W}(1,x)&\le \frac{b-x+1}{\det (-\Delta _{\llbracket 1 , b\rrbracket }+W\mathbbm {1}_{\llbracket 1 , x\rrbracket })}\le \frac{b-x+1}{\sum _{j=1}^x j(b-j+1)W(j)}\\&\le \left( \sum _{j=1}^x j W(j)\right) ^{-1}=\left( \sum _{j=0}^{x-1} (x-j) W(x-j)\right) ^{-1}. \end{aligned}$$

$\square$

With the previous proposition in mind we define for $\delta >0$ and $x\in {\mathbb {Z}}^d$

$$\begin{aligned} Z^+_\delta (x)&{:}{=}\min \left\{ n\in {\mathbb {N}}\,\Big \vert \,\sum _{j=1}^n(n+1-j)V(x+j)>\delta ^{-1}\right\} ,\\ Z^-_\delta (x)&{:}{=}\min \left\{ n\in {\mathbb {N}}\,\Big \vert \,\sum _{j=1}^n(n+1-j)V(x-j)>\delta ^{-1}\right\} ,\\ A_\delta (x)&{:}{=}\llbracket x-Z^-_\delta (x) , x+Z^+_\delta (x)\rrbracket . \end{aligned}$$

Notice that V(x) is not included in the definition of $Z^\pm _\delta (x)$ and therefore $Z^+_\delta (x)$ and $Z^-_\delta (x)$ are independent for all $x\in {\mathbb {Z}}$. Moreover, $Z^\pm _\delta (x)$ is equal in distribution to $Z^+_\delta (0)$ for all $x\in {\mathbb {Z}}$.

It follows from (4), the definitions above, potential monotonicity, and Propositions 9, 11 that

$$\begin{aligned} \lambda _{n,V}\left\| L_{n,V}\right\| _\infty&\le \lambda _{n,V}\max _{x\in \Lambda _n}\left[ L_{A_\delta (x),0}(x)+2\delta \left\| L_{n,V}\right\| _\infty \right] \\&\le \lambda _{n,V}\max _{x\in \Lambda _n}\left\| L_{A_\delta (x),0}\right\| _\infty +2\delta C. \end{aligned}$$

By domain monotonicity and translation invariance, the last maximum above is attained at the $x\in \Lambda _n$ that also maximises $\#A_\delta (x)=Z^+_\delta (x)+Z^-_\delta (x)+1$. Moreover, V being i.i.d. implies

$$\begin{aligned} \mathbb {P}\left[ \lim _{n\rightarrow \infty }\max _{x\in \Lambda _n}Z^+_\delta (x)+Z^-_\delta (x)=\infty \right] =1, \end{aligned}$$

and therefore Theorem 1 and Proposition 10 give

$$\begin{aligned} \varlimsup _{n\rightarrow \infty }\lambda _{n,V}\left\| L_{n,V}\right\| _\infty \le \frac{\mu _1}{2} \varlimsup _{n\rightarrow \infty }\left( \frac{\max _{x\in \Lambda _n}Z^+_\delta (x)+Z^-_\delta (x)}{2y_n}\right) ^2+2\delta C\quad {\mathbb {P}}\text {-a.s.} \end{aligned}$$

The proof of Theorem 2(ii) is finished with the next proposition followed by the limit $\delta \rightarrow 0$.

Proposition 12

For all $\delta >0$, $\displaystyle \varlimsup _{n\rightarrow \infty }\frac{1}{2y_n}\max _{x\in \Lambda _n}[Z^+_\delta (x)+Z^-_\delta (x)]\le 1\quad {\mathbb {P}}$-a.s.

Proof

We will prove this over an exponential subsequence by means of the Borel–Cantelli Lemma; the extension to the whole sequence is done as in the proof of Proposition 3 using the monotonicity of $n\mapsto \max _{x\in \Lambda _n}[Z^+_\delta (x)+Z^-_\delta (x)]$.

Assume (C1). For all $t>0$ we have

$$\begin{aligned} \mathbb {E}\left[ e^{-t V(0)}\right] =\mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)\le \frac{1}{\sqrt{t}}}\right] +\mathbb {E}\left[ e^{-t V(0)}\mathbbm {1}_{V(0)> \frac{1}{\sqrt{t}}}\right] \le F\left( 1/\sqrt{t}\right) +e^{-\sqrt{t}}. \end{aligned}$$

With this, we use the exponential Markov inequality and independence to obtain

$$\begin{aligned} \mathbb {P}\left[ Z_\delta ^+(0)>n\right]&=\mathbb {P}\left[ \sum _{j=1}^n(n+1-j)V(x+j)\le \delta ^{-1}\right] \\&\le e^{t/\delta }\prod _{j=1}^n\mathbb {E}\left[ \exp (-t j V(0))\right] \\&\le e^{t/\delta }\left( F\left( 1/\sqrt{t}\right) +e^{-\sqrt{t}}\right) ^n,\qquad n \in {\mathbb {N}}. \end{aligned}$$

Now we proceed with the distribution of $Z^+_\delta (0)+Z^-_\delta (0)$ as

$$\begin{aligned}&\mathbb {P}\left[ Z^+_\delta (0)+Z^-_\delta (0)>n\right] \\&\qquad \qquad =\mathbb {P}\left[ Z^+_\delta (0)>n-1\right] +\sum _{j=1}^{n-1}\mathbb {P}\left[ Z^+_\delta (0)=j\right] \mathbb {P}\left[ Z^-_\delta (0)>n-j\right] \\&\qquad \qquad \le 2\mathbb {P}\left[ Z^+_\delta (0)>n-1\right] +\sum _{j=2}^{n-1}\mathbb {P}\left[ Z^+_\delta (0)>j-1\right] \mathbb {P}\left[ Z^+_\delta (0)>n-j\right] \\&\qquad \qquad \le 2e^{t/\delta }\left( F\left( 1/\sqrt{t}\right) +e^{-\sqrt{t}}\right) ^{n-1}+(n-2)e^{2t/\delta }\left( F\left( 1/\sqrt{t}\right) +e^{-\sqrt{t}}\right) ^{n-1}\\&\qquad \qquad \le n e^{2t/\delta }\left( F\left( 1/\sqrt{t}\right) +e^{-\sqrt{t}}\right) ^{n-1}. \end{aligned}$$

For any $\epsilon >0$ define $t(\epsilon )$ by $\ln \left( F\left( 1/\sqrt{t(\epsilon )}\right) +e^{-\sqrt{t(\epsilon )}}\right) \le \frac{\ln F(0)}{1+\epsilon }$ so that

$$\begin{aligned}&\mathbb {P}\left[ \max _{x\in \Lambda _n}[Z^+_\delta (x)+Z^-_\delta (x)]>\left\lfloor (1+\epsilon )^2 2y_n \right\rfloor \right] \\&\qquad \qquad \le C n\,\mathbb {P}\left[ Z^+_\delta (0)+Z^-_\delta (0)>\left\lfloor (1+\epsilon )^2 2y_n \right\rfloor \right] \\&\qquad \qquad \le C e^{2t(\epsilon )/\delta } n \exp \left[ (1+\epsilon )^2 2 y_n\frac{\ln F(0)}{1+\epsilon }(1+o(1))\right] \\&\qquad \qquad =Ce^{2t(\epsilon )/\delta }n^{-\epsilon (1+o(1))}, \end{aligned}$$

which is summable over the exponential subsequence $n=\left\lfloor e^m \right\rfloor$, $m\in {\mathbb {N}}$.

Assume (C2). We follow the same steps as for (C1) above. To bound the Laplace transform of V(0) we consider the function $f(t){:}{=}a[F(t)]^{1/\eta }$ for some $a>0$. From (C2) follows that there exists $t_0\in (0,\infty )$ such that $F(t)\le 2\,c\,t^\eta$ for all $t\in [0,t_0]$. Therefore, by choosing $a{:}{=}(t_0^{-1}+(2c)^{1/\eta })^{-1}$ we obtain

$$\begin{aligned} 0\le f(t)\le {\left\{ \begin{array}{ll} a(2c)^{1/\eta }t\le t,&{}t\in [0,t_0],\\ a\le t_0\le t, &{}t\in (t_0,\infty ). \end{array}\right. } \end{aligned}$$

Moreover, since $\mathbb {P}\left[ f(V(0))\le t\right] =\left( \frac{t}{a}\right) ^\eta$ for $t\in [0,a]$, we have

$$\begin{aligned} \mathbb {E}\left[ \exp (-t V(0))\right]&\le \mathbb {E}\left[ \exp (-t f(V(0)))\right] =\frac{\eta }{(a)^{\eta }}\int _0^{a} e^{-t y} y^{\eta -1}\,\textrm{d}y\\ {}&\le \frac{\eta }{(a)^{\eta }}\int _0^{\infty } e^{-t y} y^{\eta -1}\,\textrm{d}y= \frac{\eta \Gamma (\eta )}{(a t)^\eta },\qquad t>0. \end{aligned}$$

The exponential Markov inequality at $t=n\eta \delta$, independence, and the Stirling bound $(n/e)^n\le n!$ lead us to

$$\begin{aligned} \mathbb {P}\left[ Z_\delta ^+(0)>n\right]&\le e^{t/\delta }\prod _{j=1}^n\mathbb {E}\left[ \exp (-t j V(0))\right] \le \left( \frac{\eta \Gamma (\eta )}{(at)^\eta }\right) ^n\frac{e^{t/\delta }}{(n!)^{\eta }}\\&\le \left( \frac{\eta ^{1-\eta }\Gamma (\eta )e^{2\eta }}{a^\eta \delta }\right) ^n n^{-2\eta n}{=}{:}K_\delta ^n \,n^{-2\eta n},\qquad n \in {\mathbb {N}}, \end{aligned}$$

from which follows

$$\begin{aligned}&\mathbb {P}\left[ Z^+_\delta (0)+Z^-_\delta (0)>n\right] \\&\qquad \qquad \le 2\mathbb {P}\left[ Z^+_\delta (0)>n-1\right] +\sum _{j=2}^{n-1}\mathbb {P}\left[ Z^+_\delta (0)>j-1\right] \mathbb {P}\left[ Z^+_\delta (0)>n-j\right] \\&\qquad \qquad \le 2 K_\delta ^{n-1}(n-1)^{-2\eta (n-1)}+K_\delta ^{n-1} \sum _{j=2}^{n-1} \left( j-1\right) ^{-2\eta (j-1)}(n-j)^{-2\eta (n-j)}. \end{aligned}$$

The function $[2,n-1]\ni j\mapsto \left( j-1\right) ^{-(j-1)}(n-j)^{-(n-j)}$ attains its unique maximum at $j=(n+1)/2$, therefore

$$\begin{aligned}&\mathbb {P}\left[ Z^+_\delta (0)+Z^-_\delta (0)>n\right] \\&\qquad \qquad \le 2 K_\delta ^{n-1}(n-1)^{-2\eta (n-1)}+(4^{\eta }K_\delta )^{n-1}(n-2)(n-1)^{-2\eta (n-1)}\\&\qquad \qquad \le (4^{\eta }K_\delta )^{n-1}n(n-1)^{-2\eta (n-1)}. \end{aligned}$$

Finally, for any $\epsilon >0$ we have

$$\begin{aligned}&\mathbb {P}\left[ \max _{x\in \Lambda _n}[Z^+_\delta (x)+Z^-_\delta (x)]>\left\lfloor (1+\epsilon )2y_n \right\rfloor \right] \\&\qquad \qquad \le C n\,\mathbb {P}\left[ Z^+_\delta (0)+Z^-_\delta (0)>\left\lfloor (1+\epsilon )2y_n \right\rfloor \right] \\&\qquad \qquad = C n \exp \left[ -(1+\epsilon )4\eta y_n(\ln y_n)(1+o(1))\right] \\&\qquad \qquad = C n \exp \left[ -(1+\epsilon )4\eta y_n(\ln \ln n)(1+o(1))\right] \\&\qquad \qquad =C n^{-\epsilon (1+o(1))}, \end{aligned}$$

which is summable over the exponential subsequence $n=\left\lfloor e^m \right\rfloor$, $m\in {\mathbb {N}}$. $\square$

Appendix A Proof of Theorem 8

Let $(X_t)_{t\ge 0}$ be a continuous time simple symmetric random walk on ${\mathbb {Z}}^d$ with jump intensity 1, and let $\textrm{P}_x$, $\textrm{E}_x$ be the associated probability measure and expectation conditioned on $X_0=x$. We remark that $(X_t)_{t\ge 0}$ is the Markov process of generator $-\Delta /(2d)$ on $\ell ^2({\mathbb {Z}}^d)$.

For a finite $A\subseteq {\mathbb {Z}}^d$ and $W:A\rightarrow [0,\infty )$, the Feynman-Kac formula lets us write the semigroup generated by $-\Delta _{A}+W$ acting on as

$$\begin{aligned} \exp \left( -\frac{t}{2d}[-\Delta _{A}+W]\right) \phi (x)=\textrm{E}_{x}\left[ \phi (X_{t})\exp \left( -\int _0^{t}\frac{W(X_s)}{2d}\,\textrm{d}s\right) \mathbbm {1}_{t<\tau _A}\right] , \end{aligned}$$

where $\tau _A{:}{=}\inf \{t\ge 0\,\vert \, X_t\notin A\}$ is the exit time of A. To simplify notation we set $\lambda =\frac{\lambda _{A,W}}{2d}$ and $K_t=\exp \left( -\frac{t}{2d}[-\Delta _{A}+W]\right)$. Depending on $\lambda$ we distinguish two cases.

Case 1: $\lambda \le \frac{1}{d}$.

Let $\overline{B_\infty }(x,r){:}{=}\{y\in {\mathbb {R}}^d\,\vert \,\left\| x-y\right\| _\infty \le r\}$. For $t\ge 1$, $x\in A$ and $r=2t\sqrt{\frac{e}{\lambda d}}$ we decompose $K_{t/\lambda }\mathbbm {1}_A(x)$ as

$$\begin{aligned} K_{t/\lambda }\mathbbm {1}_A(x)=K_{t/\lambda }\mathbbm {1}_{A\cap \overline{B_\infty }(x,r)}(x)+K_{t/\lambda }\mathbbm {1}_{A\setminus \overline{B_\infty }(x,r)}(x) \end{aligned}$$

and bound each term as follows.

For the first term we use that $t\mapsto K_t$ is a semigroup to obtain

$$\begin{aligned} K_{t/\lambda }\mathbbm {1}_{A\cap \overline{B_\infty }(x,r)}(x)&=\left\langle \delta _x ,K_{t/\lambda }\mathbbm {1}_{A\cap \overline{B_\infty }(x,r)}\right\rangle _{\ell ^2(A)}\\ {}&=\left\langle K_{1/\lambda }\delta _x ,K_{(t-1)/\lambda }\mathbbm {1}_{A\cap \overline{B_\infty }(x,r)}\right\rangle _{\ell ^2(A)}\\&\le \left\| K_{1/\lambda }\delta _x\right\| _{\ell ^2(A)}\left\| \mathbbm {1}_{A\cap \overline{B_\infty }(x,r)}\right\| _{\ell ^2(A)}\left\| K_{(t-1)/\lambda }\right\| _{\ell ^2(A)\rightarrow \ell ^2(A)}\\&\le C(d)\sqrt{K_{2/\lambda }\delta _x(x)}\,r^{d/2}e^{-(t-1)}\\&\le C(d)\sqrt{\textrm{P}_{0}\left[ X_{2/\lambda }=0\right] }\,r^{d/2}e^{-(t-1)}. \end{aligned}$$

We can estimate $\textrm{P}_{0}\left[ X_{2/\lambda }=0\right]$ using the characteristic function of $X_s$, which is $\phi _s(\theta )=\exp \left( -s+\frac{s}{d}\sum _{i=1}^d\cos (\theta _i)\right)$, by means of

$$\begin{aligned} \textrm{P}_{0}\left[ X_{s}=0\right] =\frac{1}{(2\pi )^d}\int _{[-\pi ,\pi ]^d}\phi _s(\theta )\,\textrm{d}\theta =\frac{e^{-s}}{(2\pi )^d}\left[ \int _{[-\pi ,\pi ]}\exp \left( \frac{s}{d}\cos \theta _1\right) \,\textrm{d}\theta _1\right] ^d. \end{aligned}$$

Laplace’s method applied to the integral inside the the square brackets yields $\textrm{P}_{0}\left[ X_{s}=0\right] s^{d/2}\xrightarrow [s\rightarrow \infty ]{}\left( \frac{d}{2\pi }\right) ^{d/2}$, and therefore $\textrm{P}_{0}\left[ X_{2/\lambda }=0\right] \le C(d)\lambda ^{d/2}$.

For the second term we use [11, Lemma 4.6] which states that

$$\begin{aligned} P[S_n>y]\le e^{-y^2/(2n)},\qquad y>0, \end{aligned}$$

where $S_n$ is a discrete time simple symmetric random walk on ${\mathbb {Z}}$ starting at 0, and P is its probability measure. Recalling that the first component of $X_{t}$, which we denote $X_t^1$, is a continuous time simple symmetric random walk on ${\mathbb {Z}}$ with jump intensity 1/d we have

$$\begin{aligned} K_{t/\lambda }\mathbbm {1}_{A\setminus \overline{B_\infty }(x,r)}(x)&\le \textrm{P}_{0}\left[ X_{t/\lambda }\notin \overline{B_\infty }(0,r)\right] \le 2d \textrm{P}_{0}\left[ X_{t/\lambda }^1>r\right] \\&\le 2d\sum _{n\ge 0}\frac{e^{-t/(\lambda d)}}{n!}\left( \frac{t}{\lambda d}\right) ^n P[S_n>r]. \end{aligned}$$

We split the series at $n=\frac{2e t}{\lambda d}$ and bound the two terms separately:

$$\begin{aligned} \sum _{n\le \frac{2e t}{\lambda d}}\frac{e^{-t/(\lambda d)}}{n!}\left( \frac{t}{\lambda d}\right) ^n P[S_n>r]&\le \sum _{n\le \frac{2e t}{\lambda d}}\frac{e^{-t/(\lambda d)}}{n!}\left( \frac{t}{\lambda d}\right) ^n e^{-r^2/(2n)}\\ {}&\le \exp \left( -\frac{r^2\lambda d}{4 e t}\right) \sum _{n\le \frac{2e t}{\lambda d}}\frac{e^{-t/(\lambda d)}}{n!}\left( \frac{t}{\lambda d}\right) ^n\\&\le \exp \left( -\frac{r^2\lambda d}{4 e t}\right) =e^{-t},\\ \sum _{n\ge \frac{2e t}{\lambda d}}\frac{e^{-t/(\lambda d)}}{n!}\left( \frac{t}{\lambda d}\right) ^n P[S_n>r]&\le e^{-t/(\lambda d)}\sum _{n\ge \frac{2e t}{\lambda d}}\frac{1}{n!}\left( \frac{t}{\lambda d}\right) ^n\\&\le e^{-t/(\lambda d)}\sum _{n\ge 0}\frac{1}{n!}\left( \frac{n}{2 e}\right) ^n\le C e^{-t}. \end{aligned}$$

With this we have shown $\left\| K_{t/\lambda }\mathbbm {1}_A\right\| _\infty \le C(d)\left( 1+t^{d/2}\right) e^{-t}$ for $t\ge 1$. Since $K_{t/\lambda }\mathbbm {1}(x)$ is always bounded by 1 we can add $\left( \inf _{0\le t\le 1}\left( 1+t^{d/2}\right) e^{-t}\right) ^{-1}$ to C(d), if necessary, to have

$$\begin{aligned} \left\| K_{t/\lambda }\mathbbm {1}_A\right\| _\infty \le C(d)\left( 1+t^{d/2}\right) e^{-t},\qquad t\ge 0. \end{aligned}$$

Replacing t by $2d\lambda t$ and gives the desired bound.

Case 2: $\lambda \ge \frac{1}{d}$.

This case follows from the heat kernel bound

$$\begin{aligned} \textrm{P}_{0}\left[ X_t=y\right] \le C(d) \exp \left[ -t-\left\| y\right\| _1\ln \left( \frac{\left\| y\right\| _1}{e t}\right) \right] ,\qquad \left\| y\right\| _1\ge e t, \end{aligned}$$

taken from [11, Theorem 5.17]. We proceed as before but now we use $\overline{B_1}(x,r){:}{=}\{y\in {\mathbb {R}}^d\,\vert \,\left\| x-y\right\| _1\le r\}$. For $t\ge 0$ and $r=\lambda t d e^2$ we have

$$\begin{aligned} K_{t}\mathbbm {1}_A(x)&=K_{t}\mathbbm {1}_{A\cap \overline{B_1}(x,r)}(x)+K_{t}\mathbbm {1}_{A\setminus \overline{B_1}(x,r)}(x)\\&\le \left\langle \delta _x ,K_{t}\mathbbm {1}_{A\cap \overline{B_1}(x,r)}\right\rangle _{\ell ^2(A)}+\textrm{P}_{0}\left[ X_t\notin \overline{B_1}(0,r)\right] \\&\le C(d) r^{d/2}e^{-\lambda t}+\textrm{P}_{0}\left[ X_t\notin \overline{B_1}(0,r)\right] . \end{aligned}$$

Clearly $r\ge e t$, so we can apply the heat kernel bound to obtain

$$\begin{aligned} \textrm{P}_{0}\left[ X_{t}\notin \overline{B_1}(0,r)\right]&\le C(d)\sum _{\begin{array}{c} y\in {\mathbb {Z}}^d\\ \left\| y\right\| _1>r \end{array}}\exp \left[ -t-\left\| y\right\| _1\ln \left( \frac{\left\| y\right\| _1}{e t}\right) \right] \\ {}&\le C(d)\sum _{\begin{array}{c} y\in {\mathbb {Z}}^d\\ \left\| y\right\| _1>r \end{array}}\exp \left[ -\left\| y\right\| _1\ln (e d)\right] \\ {}&\le C(d) e ^{-r}\sum _{y\in {\mathbb {Z}}^d}\exp \left[ -\left\| y\right\| _1\right] \\&= C(d) e^{-r}\le C(d) e^{-\lambda t}, \end{aligned}$$

and therefore $\left\| K_{t}\mathbbm {1}_A\right\| _\infty \le C(d)\left( 1+[\lambda t]^{d/2}\right) e^{-\lambda t}$. Replacing t by 2dt and gives the desired bound.