Harnack inequality for subordinate random walks

TL;DR

This paper establishes Harnack inequalities and estimates for transition probabilities and Green functions for a broad class of subordinate random walks on integer lattices, advancing understanding of their harmonic functions.

Contribution

It introduces new estimates and proves the Harnack inequality for subordinate random walks with Laplace exponents satisfying specific scaling conditions.

Findings

Derived transition probability estimates

Established Green function bounds

Proved Harnack inequality for harmonic functions

Abstract

In this paper, we consider a large class of subordinate random walks $X$ on integer lattice $\mathbb{Z}^d$ via subordinators with Laplace exponents which are complete Bernstein functions satisfying a certain lower scaling condition at zero. We establish estimates for one-step transition probabilities, the Green function and the Green function of a ball, and prove the Harnack inequality for non-negative harmonic functions.

Equations

$$\varphi (\lambda )\text{mathchar}58={\int }_{⟨0,\infty ⟩}\left(1-e^{-\lambda t}\right)\mu (dt)$$

$$c_m^{\varphi }\text{mathchar}58={\int }_{⟨0,\infty ⟩}\frac{t^m}{m!}{e}^{-t}\mu (dt)$$

$$\sum _{m=1}^{\infty }{c}_m^{\varphi }={\int }_{⟨0,\infty ⟩}(e^t-1)e^{-t}\mu (dt)={\int }_{⟨0,\infty ⟩}(1-e^{-t})\mu (dt)=\varphi (1)=1.$$

$$X_n\stackrel{d}{=}\sum _{k=1}^n{\xi }_k,n⩾0.$$

$$\mathbb{P}(X_1=x)$$

$$=\sum _{m=1}^{\infty }{\int }_{⟨0,\infty ⟩}\frac{t^m}{m!}{e}^{-t}\mu (dt)\mathbb{P}(Z_m=x),x\in {\mathbb{Z}}^d.$$

$$\frac{\varphi (R)}{\varphi (r)}⩾a_1{\left(\frac{R}{r}\right)}^{{\gamma }_1},\forall 0<r⩽R⩽1.$$

$$\frac{\varphi (R)}{\varphi (r)}⩽a_2{\left(\frac{R}{r}\right)}^{{\gamma }_2},\forall 0<r⩽R⩽1.$$

$$f(x)=Pf(x)=\sum _{y\in {\mathbb{Z}}^d}p(x,y)f(y),\forall x\in B.$$

$$f(x_1)⩽c_{a}f(x_2),\forall x_1,x_2\in B(x,an).$$

$$\underset{x\to \infty }{\lim }\frac{\Gamma (x+1,x)}{\Gamma (x+1)}=\frac{1}{2}.$$

$$\Gamma (x+1)\sim \sqrt{2\pi x}{x}^x{e}^{-x},x\to \infty$$

$$\Gamma (x+1,x)\sim \sqrt{2^{-1}\pi x}{x}^x{e}^{-x},x\to \infty$$

$$\underset{x\to \infty }{\lim }\frac{\Gamma (x+1,x)}{\Gamma (x+1)}=\underset{x\to \infty }{\lim }\frac{\sqrt{2^{-1}\pi x}{x}^x{e}^{-x}}{\sqrt{2\pi x}{x}^x{e}^{-x}}=\frac{1}{2}.$$

$${\int }_{{⟨-\delta ,\delta ⟩}^d}\mathrm{Re}\left(\frac{1}{1-{\phi }_{X_1}(\theta )}\right)d\theta <+\infty .$$

$${\phi }_{X_1}(\theta )$$

$$=\sum _{m=1}^{\infty }{\int }_{⟨0,+\infty ⟩}\frac{t^m}{m!}{e}^{-t}\mu (dt)\sum _{x\in {\mathbb{Z}}^d}{e}^{i\theta \cdot x}\mathbb{P}(Z_m=x)=\sum _{m=1}^{\infty }{\int }_{⟨0,+\infty ⟩}\frac{t^m}{m!}{e}^{-t}\mu (dt)(\phi (\theta ){)}^m$$

$$={\int }_{⟨0,+\infty ⟩}\left(e^{t\phi (\theta )}-1\right)e^{-t}\mu (dt)=\varphi (1)-\varphi (1-\phi (\theta ))=1-\varphi (1-\phi (\theta )).$$

$$\phi (\theta )=\frac{1}{d}\sum _{m=1}^d\cos ({\theta }_m),\theta =({\theta }_1,{\theta }_2,\dots ,{\theta }_m).$$

$${\int }_{{⟨-\delta ,\delta ⟩}^d}\mathrm{Re}\left(\frac{1}{1-{\phi }_{X_1}(\theta )}\right)d\theta ={\int }_{{⟨-\delta ,\delta ⟩}^d}\frac{1}{\varphi (1-\phi (\theta ))}d\theta .$$

$$|\phi (\theta )|=\phi (\theta )⩽1-\frac{1}{4d}|\theta {|}^2,\theta \in B(0,a).$$

$$\frac{1}{\varphi \left(1-\phi (\theta )\right)}⩽\frac{1}{\varphi \left(|\theta {|}^2/4d\right)},\theta \in B(0,a).$$

$${\int }_{{⟨-\delta ,\delta ⟩}^d}\frac{1}{\varphi (1-\phi (\theta ))}d\theta$$

$$⩽a_2(4d{)}^{{\gamma }_2}{\int }_{B(0,a)}\frac{1}{\varphi (|\theta {|}^2)}d\theta =c_1(4d{)}^{{\gamma }_2}{\int }_0^a\frac{r^{d-1}}{\varphi (r^2)}dr$$

$$=\frac{c_1(4d{)}^{{\gamma }_2}}{\varphi (a)}{\int }_0^a{r}^{d-1}\frac{\varphi (a)}{\varphi (r^2)}dr⩽\frac{c_1{a}_2(4ad{)}^{{\gamma }_2}}{\varphi (a)}{\int }_0^a{r}^{d-2{\gamma }_2-1}dr$$

$$g(r)=\frac{1}{r^d\varphi (r^{-2})}$$

$$j(r)=r^{-d}\varphi (r^{-2}).$$

$$g(r)=\frac{1}{\frac{r^d}{q^d}{q}^d\varphi (q^{-2})\frac{\varphi (r^{-2})}{\varphi (q^{-2})}}⩾\frac{1}{a_2}{\left(\frac{q}{r}\right)}^{d-2{\gamma }_2}g(q)⩾\frac{1}{a_2}g(q).$$

$$j(r)=\frac{r^{-d}}{q^{-d}}{q}^{-d}\varphi (q^{-2})\frac{\varphi (r^{-2})}{\varphi (q^{-2})}⩾a_1{\left(\frac{q}{r}\right)}^{d+2{\gamma }_1}j(q)⩾a_{1}j(q).$$

$$j(ar)⩽a^{-d-2}j(r),$$

$$g(ar)⩾a^{-d+2}g(r).$$

$$j(ar)⩾a^{-d-2}j(r).$$

$$g(ar)⩽\frac{g(r)}{a_1{a}^{d-2{\gamma }_1}}.$$

Full text

2010 Mathematics Subject Classification: Primary: 60J45; Secondary: 60G50, 60J10, 05C81

Keywords and phrases: random walk, subordination, Harnack inequality, harmonic function, Green function, Poisson kernel

1 Introduction

Let $(Y_{k})_{k\geqslant 1}$ be a sequence of independent, identically distributed random variables defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, taking values in the integer lattice $\mathbb{Z}^{d}$, with distribution $\mathbb{P}(Y_{k}=e_{i})=\mathbb{P}(Y_{k}=-e_{i})=1/2d$, $i=1,2,\ldots,d$, where $e_{i}$ is the $i$-th vector of the standard basis for $\mathbb{R}^{d}$. A simple symmetric random walk in $\mathbb{Z}^{d}$ $(d\geqslant 1)$ starting at $x\in\mathbb{Z}^{d}$ is a stochastic process $Z=(Z_{n})_{n\geqslant 0}$, with $Z_{0}=x$ and $Z_{n}=x+Y_{1}+\cdots+Y_{n}$.

Let $Z=(Z_{n})_{n\geqslant 0}$ be a simple symmetric random walk in $\mathbb{Z}^{d}$ starting at the origin. Further, let

[TABLE]

be a Bernstein function satisfying $\phi(1)=1$. Here $\mu$ is a measure on $\langle 0,\infty\rangle$ satisfying $\int_{\langle 0,\infty\rangle}{(1\wedge t)\mu(\operatorname{d\!}t)}<\infty$ called the Lévy measure. For $m\in\mathbb{N}$ denote

[TABLE]

and notice that

[TABLE]

Hence, we can define a random variable $R$ with $\mathbb{P}(R=m)=c_{m}^{\phi}$, $m\in\mathbb{N}$. Now we define the random walk $T=(T_{n})_{n\geqslant 0}$ on $\mathbb{Z}_{+}$ by $T_{n}\mathrel{\mathop{\mathchar 58\relax}}=\sum_{k=1}^{n}{R_{k}}$, where $(R_{k})_{k\geqslant 1}$ is a sequence of independent, identically distributed random variables with the same distribution as $R$ and independent of the process $Z$. Subordinate random walk is a stochastic process $X=(X_{n})_{n\geqslant 0}$ which is defined by $X_{n}\mathrel{\mathop{\mathchar 58\relax}}=Z_{T_{n}}$, $n\geqslant 0$. It is straightforward to see that the subordinate random walk is indeed a random walk. Hence, there exists a sequence of independent, identically distributed random variables $(\xi_{k})_{k\geqslant 1}$ with the same distribution as $X_{1}$ such that

[TABLE]

We can easily find the explicit expression for the distribution of the random variable $X_{1}$:

[TABLE]

We denote the transition matrix of the subordinate random walk $X$ with $P$, i.e. $P=(p(x,y)\mathrel{\mathop{\mathchar 58\relax}}x,y\in\mathbb{Z}^{d})$, where $p(x,y)=\mathbb{P}(x+X_{1}=y)$.

We will impose some additional constraints on the Laplace exponent $\phi$. First, $\phi$ will be a complete Bernstein function [13, Definition 6.1.] and it will satisfy the following lower scaling condition: there exist $0<\gamma_{1}<1$ and $a_{1}>0$ such that

[TABLE]

In dimension $d\leqslant 2$, we additionally assume that there exist $\gamma_{1}\leqslant\gamma_{2}<1$ and $a_{2}>0$ such that

[TABLE]

It is well known that, if $\phi$ is a Bernstein function, then $\phi(\lambda t)\leqslant\lambda\phi(t)$ for all $\lambda\geqslant 1$, $t>0$, implying $\phi(v)/v\leqslant\phi(u)/u$, $0<u\leqslant v$. Using these two facts, proof of the next lemma is straightforward.

Lemma 1.1**.**

If $\phi$ is a Bernstein function, then for all $\lambda,t>0$, $1\wedge\lambda\leqslant\phi(\lambda t)/\phi(t)\leqslant 1\vee\lambda$.

Using Lemma 1.1 we get $\phi(R)/\phi(r)\leqslant R/r$ for all $0<r\leqslant R\leqslant 1$ and this suffices for $d\geqslant 3$.

The main result of this paper is a scale invariant Harnack inequality for subordinate random walks. The proof will be given in the last section. Before we state the result, we define the notion of harmonic function with respect to subordinate random walk $X$.

Definition 1.2**.**

We say that a function ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ is harmonic in $B\subset\mathbb{Z}^{d}$, with respect to $X$, if

[TABLE]

This definition is equivalent to the mean-value property in terms of the exit from a finite subset of $\mathbb{Z}^{d}$: If $B\subset\mathbb{Z}^{d}$ is finite then ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ is harmonic in $B$, with respect to $X$, if and only if $f(x)=\mathbb{E}_{x}[f(X_{\tau_{B}})]$ for every $x\in B$, where $\tau_{B}\mathrel{\mathop{\mathchar 58\relax}}=\inf\{n\geqslant 1\mathrel{\mathop{\mathchar 58\relax}}X_{n}\notin B\}$.

For $x\in\mathbb{Z}^{d}$ and $r>0$ we define $B(x,r)\mathrel{\mathop{\mathchar 58\relax}}=\{y\in\mathbb{Z}^{d}\mathrel{\mathop{\mathchar 58\relax}}|y-x|<r\}$. We use shorthand notation $B_{r}$ for $B(0,r)$.

Theorem 1.3** (Harnack inequality).**

Let $X=(X_{n})_{n\geqslant 0}$ be a subordinate random walk in $\mathbb{Z}^{d}$, $d\geqslant 1$, with $\phi$ a complete Bernstein function satisfying (1.4) and in the case $d\leqslant 2$ also (1.5). For each $a<1$, there exists a constant $c_{a}<\infty$ such that if ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ is harmonic on $B(x,n)$, with respect to $X$, for $x\in\mathbb{Z}^{d}$ and $n\in\mathbb{N}$, then

[TABLE]

Notice that the constant $c_{a}$ is uniform for all $n\in\mathbb{N}$. That is why we call this result scale invariant Harnack inequality.

Some authors have already dealt with this problem and Harnack inequality was proved for symmetric simple random walk in $\mathbb{Z}^{d}$ [9, Theorem 1.7.2] and random walks with steps of infinite range, but with some assumptions on the Green function and some restrictions such as finite second moment of the step [3, 10].

Notion of discrete subordination was developed in [6] and it was discussed in details in [4], but under different assumptions on $\phi$ than the ones we have. Using discrete subordination we can obtain random walks with steps of infinite second moment, see Remark 3.3. Harnack inequality has not been proved so far for such random walks.

In Section 2 we state an important result about gamma function that we use later, we discuss under which conditions subordinate random walk is transient and we introduce functions $g$ and $j$ and examine their properties. The estimates of one-step transition probabilities of subordinate random walk are given in Section 3. In Section 4 we derive estimates for the Green function. This is very valuable result which gives the answer to the question posed in [5, Remark 1]. Using estimates developed in Section 3 and 4 and following [11, Section 4], in Section 5 we find estimates for the Green function of a ball. In Section 6 we introduce Poisson kernel and prove Harnack inequality.

Throughout this paper, $d\geqslant 1$ and the constants $a_{1},a_{2}$, $\gamma_{1},\gamma_{2}$ and $C_{i}$, $i=1,2,\ldots,9$ will be fixed. We use $c_{1},c_{2},\ldots$ to denote generic constants, whose exact values are not important and can change from one appearance to another. The labeling of the constants $c_{1},c_{2},\ldots$ starts anew in the statement of each result. The dependence of the constant $c$ on the dimension $d$ will not be mentioned explicitly. We will use “:=” to denote a definition, which is read as “is defined to be”. We will use $\operatorname{d\!}x$ to denote the Lebesgue measure in $\mathbb{R}^{d}$. We denote the Euclidean distance between $x$ and $y$ in $\mathbb{R}^{d}$ by $|x-y|$. For $a,b\in\mathbb{R}$, $a\wedge b\mathrel{\mathop{\mathchar 58\relax}}=\min\{a,b\}$ and $a\vee b\mathrel{\mathop{\mathchar 58\relax}}=\max\{a,b\}$. For any two positive functions $f$ and $g$, we use the notation $f\asymp g$, which is read as “$f$ is comparable to $g$”, to denote that there exist some constants $c_{1},c_{2}>0$ such that $c_{1}f\leqslant g\leqslant c_{2}f$ on their common domain of definition. We also use notation $f\sim g$ to denote that $\lim_{x\to\infty}f(x)/g(x)=1$.

2 Preliminaries

In this section we first state an important result about the ratio of gamma functions that is needed later. Secondly, we discuss under which conditions subordinate random walk is transient. At the end of this section we define functions $g$ and $j$ that we use later and we prove some of their properties.

2.1 Ratio of gamma functions

Lemma 2.1**.**

Let $\Gamma(x,a)=\int_{a}^{\infty}{t^{x-1}e^{-t}\operatorname{d\!}t}$. Then

[TABLE]

• Proof:

Using a well-known Stirling’s formula

[TABLE]

and [2, Formula 6.5.35] that states

[TABLE]

we get

[TABLE]

$\blacksquare$

2.2 Transience of subordinate random walks

Our considerations only make sense if the random walk that we defined is transient. In the case of a recurrent random walk, the Green function takes value $+\infty$ for every argument $x$. We will use Chung-Fuchs theorem to show under which condition subordinate random walk is transient. Denote with $\varphi_{X_{1}}$ the characteristic function of one step of a subordinate random walk. We want to prove that there exists $\delta>0$ such that

[TABLE]

The law of variable $X_{1}$ is given with (1). We denote one step of the simple symmetric random walk $(Z_{n})_{n\geqslant 0}$ with $Y_{1}$ and the characteristic function of that random variable with $\varphi$. Assuming $|\theta|<1$ we have

[TABLE]

From [9, Section 1.2, page 13] we have

[TABLE]

That is function with real values so

[TABLE]

From Taylor’s theorem it follows that there exists $a\leqslant 1$ such that

[TABLE]

Now we take $\delta$ such that ${\langle-\delta,\delta\rangle}^{d}\subset B(0,a)$. From (2.3), using the fact that $\phi$ is increasing, we get

[TABLE]

Hence,

[TABLE]

and the last integral converges for $d-2\gamma_{2}-1>-1$. So, subordinate random walk is transient for $\gamma_{2}<d/2$. Notice that in the case when $d\geqslant 3$ we have $\gamma_{2}<d/2$ even if $\gamma_{2}=1$. That is the reason why we do not need (1.5) in proving results in dimensions higher than or equal to $3$. If some particular result depends on the dimension, we will write its proof using (1.5) just to show that the result is true even in dimensions $1$ and $2$ when $\gamma_{2}<d/2$. In the case when $d\geqslant 3$, we can replace $a_{2}$ and $\gamma_{2}$ from (1.5) with $1$ and only use Lemma 1.1.

2.3 Properties of functions $g$ and $j$

Let ${g}\mathrel{\mathop{\mathchar 58\relax}}{\langle 0,+\infty\rangle}\to{\langle 0,+\infty\rangle}$ be defined by

[TABLE]

and let ${j}\mathrel{\mathop{\mathchar 58\relax}}{\langle 0,+\infty\rangle}\to{\langle 0,+\infty\rangle}$ be defined by

[TABLE]

We use these functions in numerous places in our paper. In this section we present some of their properties that we need later.

Lemma 2.2**.**

Assume (1.5) (if $d\leqslant 2$) and let $1\leqslant r\leqslant q$. Then $g(r)\geqslant a_{2}^{-1}g(q)$.

• Proof:

Using (1.5) and the fact that $d>2\gamma_{2}$ we have

[TABLE]

$\blacksquare$

We prove similar assertion for the function $j$.

Lemma 2.3**.**

Assume (1.4) and let $1\leqslant r\leqslant q$. Then $j(r)\geqslant a_{1}j(q)$.

• Proof:

Using (1.4) we have

[TABLE]

$\blacksquare$

Using (1.4), (1.5) and Lemma 1.1 we can easily prove a lot of different results about functions $g$ and $j$. We will state only those results that we need in the remaining part of our paper. For the first lemma we do not need any additional assumptions on the function $\phi$. For the second one we need (1.4) and for the third one we need (1.5).

Lemma 2.4**.**

Let $r\geqslant 1$. If $0<a\leqslant 1$ then

[TABLE]

[TABLE]

If $a\geqslant 1$ then

[TABLE]

Lemma 2.5**.**

Assume (1.4) and let $0<a\leqslant 1$ and $r\geqslant 1$ such that $ar\geqslant 1$. Then

[TABLE]

Lemma 2.6**.**

Assume (1.5) and let $r\geqslant 1$. If $0<a\leqslant 1$ such that $ar\geqslant 1$ then

[TABLE]

If $a\geqslant 1$ then

[TABLE]

3 Transition probability estimates

In this section, we will investigate the behavior of the expression $\mathbb{P}(X_{1}=z)$. We will prove that $\mathbb{P}(X_{1}=z)\asymp j(|z|)$, $z\neq 0$. First we have to examine the behavior of the expression $c_{m}^{\phi}$.

Lemma 3.1**.**

Assume (1.4) and let $c_{m}^{\phi}$ be as in (1.1). Then

[TABLE]

• Proof:

Since $\phi$ is a complete Bernstein function, there exists completely monotone density $\mu(t)$ such that

[TABLE]

From [8, Proposition 2.5] we have

[TABLE]

and

[TABLE]

Inequality (3.3) holds only if (1.4) is satisfied and for inequality (3.2) we do not need any scaling conditions. Using monotonicity of $\mu$, (2.1) and (3.3) we have

[TABLE]

for $m$ large enough. On the other side, using inequality (3.2), monotonicity of $\mu$ and Lemma 1.1, we get

[TABLE]

Hence, we have

[TABLE]

for $m$ large enough, but we can change constants and get (3.1). $\blacksquare$

We are now ready to examine the expression $\mathbb{P}(X_{1}=z)$.

Proposition 3.2**.**

Assume (1.4). Then

[TABLE]

• Proof:

Using (1) and the fact that $\mathbb{P}(Z_{m}=z)=0$ for $|z|>m$, we have

[TABLE]

To get the upper bound for the expression $\mathbb{P}(X_{1}=z)$ we will use [7, Theorem 2.1] which states that there are constants $C^{\prime}>0$ and $C>0$ such that

[TABLE]

Together with (3.1) this result yields

[TABLE]

Let us first examine $I_{1}(z)$. Using (1.4), we get

[TABLE]

Using Lemma 1.1 instead of (1.4) and replacing the upper limit in the integral by $\infty$, we get $I_{2}(z)\leqslant c_{5}\phi(|z|^{-2})$. Hence, $\mathbb{P}(X_{1}=z)\leqslant c_{6}|z|^{-d}\phi(|z|^{-2})$.

In finding the matching lower bound for $\mathbb{P}(X_{1}=z)$, periodicity of a simple random walk plays very important role. We write $n\leftrightarrow x$ if $n$ and $x$ have the same parity, i.e., if $n+x_{1}+x_{2}+\cdots+x_{d}$ is even. Directly from [9, Proposition 1.2.5], we get

[TABLE]

for $0\leftrightarrow z\leftrightarrow m$ and $|z|\leqslant m^{\alpha}$, $\alpha<2/3$. In the case when $1\leftrightarrow z\leftrightarrow m$ we have

[TABLE]

By combining (3.5) and (3.6), we can easily get

[TABLE]

We will find lower bound for $\mathbb{P}(X_{1}=z)$ when $z\leftrightarrow 0$ by using (3.5), the proof when $z\leftrightarrow 1$ being analogous using (3.7). If $z\leftrightarrow 0$ then $\mathbb{P}(Z_{m}=z)=0$ for $m=2l-1$, $l\in\mathbb{N}$. Hence,

[TABLE]

where in the last line we used Lemma 1.1. $\blacksquare$

Remark 3.3**.**

It follows immediately form Proposition 3.2 that the second moment of the step $X_{1}$ is infinite.

4 Green function estimates

The Green function of $X$ is defined by $G(x,y)=G(y-x)$, where

[TABLE]

Note that for $n\geqslant 1$

[TABLE]

Hence, for $y\neq 0$ we have

[TABLE]

where

[TABLE]

and $T_{n}$ is as before. Now we will investigate the behavior of the sequence $c(m)$. Since $\phi$ is a complete Bernstein function (hence special), we have

[TABLE]

for some non-increasing function ${u}\mathrel{\mathop{\mathchar 58\relax}}{\langle 0,\infty\rangle}\to{\langle 0,\infty\rangle}$ satisfying $\int_{0}^{1}{u(t)\operatorname{d\!}t}<\infty$, see [13, Theorem 11.3.].

Lemma 4.1**.**

Let $c(m)$ be as in (4.2). Then

[TABLE]

• Proof:

We follow the proof of [4, Theorem 2.3]. Define $M(x)=\sum_{m\leqslant x}{c(m)}$, $x\in\mathbb{R}$. The Laplace transformation $\mathcal{L}(M)$ of the measure generated by $M$ is equal to

[TABLE]

Now we calculate $\mathbb{E}[e^{-\lambda R_{1}}]$:

[TABLE]

where we used $\phi(1)=1$ in the last equality. Hence, $\mathcal{L}(M)(\lambda)=1/\phi(1-e^{-\lambda})$. Now we define $\Phi(\lambda)\mathrel{\mathop{\mathchar 58\relax}}=1/\phi(\lambda)$ and we want to show that

[TABLE]

It is easy to see that $\Phi^{(m)}(\lambda)=(-1)^{m}\int_{\langle 0,\infty\rangle}{t^{m}e^{-\lambda t}u(t)\operatorname{d\!}t}$. Hence,

[TABLE]

Since $\mathcal{L}(M)(\lambda)=1/\phi(1-e^{-\lambda})=\Phi(1-e^{-\lambda})$ from calculations (Proof:) and (4.6) we have

[TABLE]

The statement of this lemma follows by the uniqueness of the Laplace transformation. $\blacksquare$

Lemma 4.2**.**

Assume (1.4). Then

[TABLE]

• Proof:

Since $\phi$ is a complete Bernstein function, using (1.4) we can obtain

[TABLE]

where the upper bound is valid even without (1.4) (see [8, Corollary 2.4.]) and $u$ is as in Lemma 4.1. Using monotonicity of $u$, (2.1) and (4.7), we get that

[TABLE]

for $m$ large enough. Now we will find the upper bound for $c(m)$. Here we use that $t\mapsto t^{m}e^{-t}$ is unimodal with maximum at $m$. By splitting the integral and using (2.2), we have

[TABLE]

for $m$ large enough. Since $2^{-1}e^{1/2}<1$, $\int_{0}^{1}{u(t)\operatorname{d\!}t}<\infty$ and $\phi$ is increasing, we have

[TABLE]

for $m$ large enough, where we used (4.7) in the last inequality. We will estimate the integral $\int_{1}^{m/2}{u(t)\operatorname{d\!}t}$ by $mu(m/2)$. Using (4.7) and (1.4) we get

[TABLE]

Since $u(m/2)\geqslant c_{4}/((m/2)\phi(2m^{-1}))$ for $m\geqslant 2$, we have $1/\phi(2m^{-1})\leqslant(1/2c_{4})mu(m/2)$. Hence,

[TABLE]

Using this estimate and the fact that $2^{-1}e^{1/2}$ is less then $1$, we have

[TABLE]

for $m$ large enough. Now we have to show that $u(m/2)$ can be estimated by $u(m)$. Again, we will use (4.7) and (1.4):

[TABLE]

where we assumed that $t\geqslant 2$ because we need $2t^{-1}\leqslant 1$ so that we can use (1.4). Now, for $m$ large enough, we have

[TABLE]

Hence,

[TABLE]

for $m$ large enough. We can now change constants in such a way that the statement of this lemma is true for every $m\in\mathbb{N}$. $\blacksquare$

Theorem 4.3**.**

Assume (1.4) and, if $d\leqslant 2$, assume additionally (1.5). Then

[TABLE]

• Proof:

We assume $|x|\geqslant 1$ throughout the whole proof. In (4.1) we showed that $G(x)=\sum_{m=1}^{\infty}{c(m)p(m,x)}$, where $p(m,x)=\mathbb{P}(Z_{m}=x)$. Let $q(m,x)=2\left(d/(2\pi m)\right)^{\frac{d}{2}}e^{-\frac{d|x|^{2}}{2m}}$ and $E(m,x)=p(m,x)-q(m,x)$. By [9, Theorem 1.2.1]

[TABLE]

Since $p(m,x)=0$ for $m<|x|$, we have

[TABLE]

First we estimate

[TABLE]

By Lemma 4.2, (4.9) and (1.5)

[TABLE]

Now we have

[TABLE]

By Lemma 4.2, (1.4) and (1.5)

[TABLE]

where the last integral converges because of the condition $\gamma_{2}<d/2$. We estimate $J_{2}(x)$ using (3.4) and (1.4):

[TABLE]

Using $J_{11}(x)\geqslant(2c_{8})/(|x|^{d}\phi(|x|^{-2}))$ and $J_{12}(x)|x|^{d}\phi(|x|^{-2})\geqslant-c_{8}$ for $|x|$ large enough and for some constant $c_{8}>0$, we get

[TABLE]

On the other hand

[TABLE]

Here we used $J_{11}(x)\leqslant c_{9}/(|x|^{d}\phi(|x|^{-2}))$, $J_{2}(x)\leqslant c_{7}/(|x|^{d}\phi(|x|^{-2}))$ and $J_{12}(x)|x|^{d}\phi(|x|^{-2})\\ \leqslant c_{9}$ for $|x|$ large enough and for some constant $c_{9}>0$. So, we have $c_{8}\leqslant G(x)|x|^{d}\phi(|x|^{-2})\leqslant c_{10}$ for $|x|$ large enough. Now we can change the constants $c_{8}$ and $c_{10}$ to get

[TABLE]

$\blacksquare$

5 Estimates of the Green function of a ball

Let $B\subset\mathbb{Z}^{d}$ and define

[TABLE]

where $\tau_{B}$ is as before. A well-known result about Green function of a set is formulated in the following lemma.

Lemma 5.1**.**

Let $B$ be a finite subset of $\mathbb{Z}^{d}$. Then

[TABLE]

where $\sigma_{x}=\inf\{n\geqslant 1\mathrel{\mathop{\mathchar 58\relax}}X_{n}=x\}$.

Our approach in obtaining estimates for the Green function of a ball uses the maximum principle for the operator $A$ that we define by

[TABLE]

Since $\sum_{y\in\mathbb{Z}^{d}}{p(x,y)}=1$ and $p(x,y)=\mathbb{P}(X_{1}=y-x)$ we have

[TABLE]

Before proving the maximum principle, we will show that for the function $\eta(x)\mathrel{\mathop{\mathchar 58\relax}}=\mathbb{E}_{x}[\tau_{B_{n}}]$ we have $(A\eta)(x)=-1$, for all $x\in B_{n}$. Let $x\in B_{n}$. Then

[TABLE]

and this is obviously equivalent to $(A\eta)(x)=-1$, for all $x\in B_{n}$. It follows from the Definition 1.2 that $f$ is harmonic in $B\subset\mathbb{Z}^{d}$ if and only if $(Af)(x)=0$, for all $x\in B$.

Proposition 5.2**.**

Assume that there exists $x\in\mathbb{Z}^{d}$ such that $(Af)(x)<0$. Then

[TABLE]

• Proof:

If (5.2) is not true, then $f(x)\leqslant f(y)$, for all $y\in\mathbb{Z}^{d}$. In this case, we have

[TABLE]

This implies $(Af)(x)=(Pf)(x)-f(x)\geqslant 0$ which is in contradiction with the assumption that $(Af)(x)<0$. $\blacksquare$

We will now prove a series of lemmas and propositions in order to get the estimates for the Green function of a ball. In all those results we assume (1.4) and, if $d\leqslant 2$, we additionally assume (1.5). Throughout the rest of this section, we follow [11, Section 4].

Lemma 5.3**.**

There exist $a\in\langle 0,1/3\rangle$ and $C_{1}>0$ such that for every $n\in\mathbb{N}$

[TABLE]

• Proof:

From Lemma 5.1 we have

[TABLE]

We will first prove this lemma in the case when $x\neq y$. If we show that $\mathbb{E}_{x}[G(X_{\tau_{B_{n}}},y)]\leqslant c_{1}G(x,y)$ for some $c_{1}\in\langle 0,1\rangle$ we will have (5.3) with the constant $c_{2}=1-c_{1}$. Let $a\in\langle 0,1/3\rangle$ and $x,y\in B_{an}$. In that case, we have $|x-y|\leqslant 2an$. Since $X_{\tau_{B_{n}}}\notin B_{n}$, $x\neq y$ and $(1-a)/(2a)>1$ if and only if $a<1/3$, we have

[TABLE]

Using Theorem 4.3, (5.4), Lemma 2.2 and (2.11), we get

[TABLE]

Since $2a/(1-a)\longrightarrow 0$ when $a\rightarrow 0$ and $d>2\gamma_{2}$, if we take $a$ small enough and then fix it, we have $\mathbb{E}_{x}[G(X_{\tau_{B_{n}}},y)]\leqslant c_{1}G(x,y)$ for $c_{1}\in\langle 0,1\rangle$ and that is what we wanted to prove. Now we deal with the case when $x=y$. From Lemma 5.1 we have $G_{B_{n}}(x,x)=(\mathbb{P}(\tau_{B_{n}}<\sigma_{x}))^{-1}$ and from the definition of the function $G$ and the transience of random walk we get $G(x,x)=G(0)\in[1,\infty\rangle$. Now, we can conclude that

[TABLE]

If we define $C_{1}\mathrel{\mathop{\mathchar 58\relax}}=\min\{c_{2},(G(0))^{-1}\}$ we have (5.3). $\blacksquare$

Using Lemma 5.3 we can prove the following result:

Proposition 5.4**.**

There exists constant $C_{2}>0$ such that for all $n\in\mathbb{N}$

[TABLE]

where $a\in\langle 0,1/3\rangle$ is as in Lemma 5.3.

• Proof:

Let $x\in B_{\frac{an}{2}}$. In that case, we have $B(x,an/2)\subseteq B_{an}$. We set $b=a/2$ for easier notation. Notice that $\mathbb{E}_{x}[\tau_{B_{n}}]=\sum_{y\in B_{n}}G_{B_{n}}(x,y)$. Using this equality, Lemma 5.3, Theorem 4.3 and Lemma 1.1, we have

[TABLE]

for $n$ large enough. Hence, we can conclude that $\mathbb{E}_{x}[\tau_{B_{n}}]\geqslant C_{2}/\phi(n^{-2})$, for all $x\in B_{\frac{an}{2}}$, for $n$ large enough and for some $C_{2}>0$. As usual, we can adjust the constant to get the statement of this proposition for every $n\in\mathbb{N}$. Notice that this is true regardless of the dimension because here, we can always plug in $\gamma_{2}=1$. $\blacksquare$

Now we want to find the upper bound for $\mathbb{E}_{x}[\tau_{B_{n}}]$.

Lemma 5.5**.**

There exists constant $C_{3}>0$ such that for all $n\in\mathbb{N}$

[TABLE]

• Proof:

We define the process $M^{f}=(M_{n}^{f})_{n\geqslant 0}$ with

[TABLE]

where $f$ is a function defined on $\mathbb{Z}^{d}$ with values in $\mathbb{R}$, $A$ is defined as in (5.1) and $X=(X_{n})_{n\geqslant 0}$ is a subordinate random walk. By [12, Theorem 4.1.2], the process $M^{f}$ is a martingale for every bounded function $f$. Let $f\mathrel{\mathop{\mathchar 58\relax}}=\mathbbm{1}_{B_{2n}}$ and $x\in B_{n}$. By the optional stopping theorem, we have

[TABLE]

Hence

[TABLE]

We now investigate both sides of the relation (5.7). For every $k<\tau_{B_{n}}$, $X_{k}\in B_{n}$, and for every $y\in B_{n}$, using Proposition 3.2, (1.4) and (1.5), we have

[TABLE]

Using the above estimate, we get

[TABLE]

Using (5.7), (5.8) and $\mathbb{E}_{x}[f(X_{\tau_{B_{n}}})-f(X_{0})]=\mathbb{P}_{x}(X_{\tau_{B_{n}}}\in B_{2n})-1=-\mathbb{P}_{x}(X_{\tau_{B_{n}}}\in B_{2n}^{c})$, we get

[TABLE]

and this implies

[TABLE]

$\blacksquare$

In the next two results we develop estimates for the Green function of a ball. We define $A(r,s)\mathrel{\mathop{\mathchar 58\relax}}=\{x\in\mathbb{Z}^{d}\mathrel{\mathop{\mathchar 58\relax}}r\leqslant|x|<s\}$ for $r,s\in\mathbb{R}$, $0<r<s$.

Proposition 5.6**.**

There exists constant $C_{4}>0$ such that for all $n\in\mathbb{N}$

[TABLE]

where $\eta(y)=\mathbb{E}_{y}[\tau_{B_{n}}]$, $b=a/2$, and $a\in\langle 0,1/3\rangle$ is as in Lemma 5.3.

• Proof:

Let $x\in B_{\frac{bn}{2}}$ and $y\in A(bn,n)$. We define function $h(z)\mathrel{\mathop{\mathchar 58\relax}}=G_{B_{n}}(x,z)$. Notice that for $z\in B_{n}\setminus\{x\}$ we have

[TABLE]

Hence, $h$ is a harmonic function in $B_{n}\setminus\{x\}$. If we take $z\in B(x,bn/8)^{c}$ then $|z-x|\geqslant bn/8\geqslant 1$ for $n$ large enough. Using Lemma 2.2 and Theorem 4.3 we get

[TABLE]

Hence, $h(z)\leqslant kg(bn/8)$ for $z\in B(x,bn/8)^{c}$ and for some constant $k>0$. Notice that $A(bn,n)\subseteq B(x,bn/8)^{c}$, hence $y\in B(x,bn/8)^{c}$. Using these facts together with Proposition 3.2, we have

[TABLE]

where in the last line we used Lemma 2.3 together with $|v-y|\geqslant bn/8\geqslant 1$ for $v\in B(x,bn/8)$ and for $n$ large enough. Using (2.6) we get $j(bn/8)\leqslant(b/8)^{-d-2}j(n)$. Hence, using (5.6), we have

[TABLE]

for some $c_{2}>0$. On the other hand, using (2.9) and Proposition 5.4 we get

[TABLE]

Now we define $C_{4}\mathrel{\mathop{\mathchar 58\relax}}=(c_{2}\vee kc_{3})+1$ and using

[TABLE]

we get

[TABLE]

So, if we define $u(\cdot)\mathrel{\mathop{\mathchar 58\relax}}=C_{4}n^{-d}\eta(\cdot)-h(\cdot)\wedge kg(bn/8)$, we showed that $u$ is non-negative function on $B_{bn}$. It is obvious that it vanishes on $B_{n}^{c}$ and for $y\in A(bn,n)$ we have

[TABLE]

Since $u\geqslant 0$ on $B_{bn}$ and $u$ vanishes on $B_{n}^{c}$, if $\inf_{y\in\mathbb{Z}^{d}}u(y)<0$ then there would exist $y_{0}\in A(bn,n)$ such that $u(y_{0})=\inf_{y\in\mathbb{Z}^{d}}u(y)$. But then, by Proposition 5.2, $(Au)(y_{0})\geqslant 0$ which is in contradiction with $(Au)(y)<0$ for $y\in A(bn,n)$. Hence,

[TABLE]

and then, because $h(y)\leqslant kg(bn/8)$ for $y\in A(bn,n)$ we get

[TABLE]

$\blacksquare$

Now we will prove a proposition that will give us the lower bound for the Green function of a ball. We use the fact that $|B_{n}\cap\mathbb{Z}^{d}|\geqslant cn^{d}$ for some constant $c>0$, where $|\cdot|$ denotes the cardinality of a set.

Proposition 5.7**.**

There exist $C_{5}>0$ and $b\leqslant a/4$ such that for all $n\in\mathbb{N}$

[TABLE]

where $a$ is as in Lemma 5.3 and $\eta(y)=\mathbb{E}_{y}[\tau_{B_{n}}]$.

• Proof:

Let $a\in\langle 0,1/3\rangle$ as in Lemma 5.3. Then there exists $C_{1}>0$

[TABLE]

From Proposition 5.6 it follows that there exists constant $C_{4}>0$ such that

[TABLE]

From Lemma 5.5 we have

[TABLE]

for some constant $C_{3}>0$. By Theorem 4.3 and (2.4) there exists $c_{1}>0$ such that $G(x)\geqslant c_{1}g(|x|)$, $x\neq 0$. Now we take

[TABLE]

and fix it. Notice that $(C_{1}c_{1})/(a_{2}^{2}C_{3}b^{d-2\gamma_{2}})\geqslant 2C_{4}$. Let $x\in B_{bn}$, $v\in B(x,bn)$. Since $b\leqslant a/4$ we have $x,v\in B_{an}$. We want to prove that $G_{B_{n}}(x,v)\geqslant 2C_{4}n^{-d}\eta(v)$. We will first prove that assertion for $x\neq v$. In that case we have $1\leqslant|x-v|$. Since $v\in B(x,bn)$, we have $|x-v|\leqslant bn$ so we can use (5.12), Lemma 2.2 and (2.10) to get

[TABLE]

Using (5.14) and (5.15), we get $G_{B_{n}}(x,v)\geqslant 2C_{4}n^{-d}\eta(v)$ for $x\neq v$. Now we will prove that $G_{B_{n}}(x,x)\geqslant 2C_{4}n^{-d}\eta(x)$, for $x\in B_{bn}$ and for $n$ large enough. First note that

[TABLE]

since $d-2\gamma_{2}>0$. Therefore

[TABLE]

for $n$ large enough. Hence,

[TABLE]

Now we fix $x\in B_{bn}$ and define the function

[TABLE]

From (5.16) we have $h(v)\leqslant\frac{1}{2}G_{B_{n}}(x,v)$ for $v\in B(x,bn)$. Recall that $G_{B_{n}}(x,\cdot)$ is harmonic in $A(an/2,n)$. Using (5.13) we get $h(y)=G_{B_{n}}(x,y)$ for $y\in A(an/2,n)$. Hence, for $y\in A(an/2,n)$

[TABLE]

where we used Proposition 3.2 and Lemma 2.3 together with $1\leqslant|v-y|\leqslant 2n$. Using (5.15) and $|B_{n}\cap\mathbb{Z}^{d}|\geqslant c_{2}n^{d}$, we get

[TABLE]

Using (2.8) we get $j(2n)\geqslant 2^{-d-2}j(n)$. When we put this together with (Proof:) and (5.18), we get

[TABLE]

Define $u\mathrel{\mathop{\mathchar 58\relax}}=h-\kappa\eta$, where

[TABLE]

where $c_{5}>0$ will be defined later. For $y\in A(an/2,n)$

[TABLE]

For $x\in B_{bn}\subseteq B_{an/2}$, $v\in B_{an/2}$ we have $|x-v|\leqslant an\leqslant n$. We will first assume that $x\neq v$ so that we can use Theorem 4.3, Lemma 2.2 and (2.10). In this case, we have

[TABLE]

So, $G_{B_{n}}(x,v)\geqslant c_{5}n^{-d}\eta(v)$ for some constant $c_{5}>0$ and for $x\neq v$. If $x=v$ we can use the same arguments that we used when we were proving that $G_{B_{n}}(x,x)\geqslant 2C_{4}n^{-d}\eta(x)$ for $n$ large enough to prove that $G_{B_{n}}(x,x)\geqslant c_{5}n^{-d}\eta(x)$ for $n$ large enough. Hence, $G_{B_{n}}(x,v)\geqslant c_{5}n^{-d}\eta(v)$ for all $x\in B_{bn}$ and $v\in B_{an/2}$ and for $n$ large enough. Now we have

[TABLE]

Hence,

[TABLE]

Since $u(v)\geqslant 0$ for $v\in B_{an/2}$, $u(v)=0$ for $v\in B_{n}^{c}$ and $(Au)(v)<0$ for $v\in A(an/2,n)$ we can use the same argument as in Proposition 5.6 to conclude by Proposition 5.2 that $u(y)\geqslant 0$ for all $y\in\mathbb{Z}^{d}$. Since $G_{B_{n}}(x,y)\leqslant C_{4}n^{-d}\eta(y)$ for $x\in B_{an/4},y\in A(an/2,n)$ we have $h(y)=G_{B_{n}}(x,y)$. Using that, we have

[TABLE]

for $n$ large enough. As before, we can change the constant and get (5.11) for all $n\in\mathbb{N}$. $\blacksquare$

Using last two propositions, we have the next corollary.

Corollary 5.8**.**

Assume (1.4) and (1.5). Then there exist constants $C_{6},C_{7}>0$ and $b_{1},b_{2}\in\langle 0,\frac{1}{2}\rangle$, $2b_{1}\leqslant b_{2}$ such that

[TABLE]

• Proof:

This corollary follows directly from Proposition 5.6 and Proposition 5.7. We can set $b_{2}=a/2$ where $a\in\langle 0,1/3\rangle$ is as in Lemma 5.3 and $b_{1}=b$ where $b\leqslant a/4$ is as in Proposition 5.7. $\blacksquare$

6 Proof of the Harnack inequality

We start this section with the proof of the proposition that will be crucial for the remaining part of our paper.

Proposition 6.1**.**

Let ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}\times\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ be a function and $B\subset\mathbb{Z}^{d}$ a finite set. For every $x\in B$ we have

[TABLE]

• Proof:

We have

[TABLE]

Using (1.2), we get

[TABLE]

Hence,

[TABLE]

$\blacksquare$

Remark 6.2**.**

Formula (6.1) can be considered as a discrete counterpart of the continuous-time Ikeda-Watanabe formula. We will refer to it as discrete Ikeda-Watanabe formula.

It can be proved that if ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ is harmonic in $B$, with respect to $X$, then $\{f(X_{n\wedge\tau_{B}})\mathrel{\mathop{\mathchar 58\relax}}n\geqslant 0\}$ is a martingale with respect to the natural filtration of $X$ (proof is the same as [9, Proposition 1.4.1], except that we have a non-negative instead of a bounded function). Using this fact, we can prove the following lemma.

Lemma 6.3**.**

Let $B$ be a finite subset of $\,\mathbb{Z}^{d}$. Then ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ is harmonic in $B$, with respect to $X$, if and only if $f(x)=\mathbb{E}_{x}[f(X_{\tau_{B}})]$ for every $x\in B$.

• Proof:

Let us first assume that ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ is harmonic in $B$, with respect to $X$. We take arbitrary $x\in B$. By the martingale property $f(x)=\mathbb{E}_{x}[f(X_{n\wedge\tau_{B}})]$, for all $n\geqslant 1$. First, by Fatou’s lemma we have $\mathbb{E}_{x}[f(X_{\tau_{B}})]\leqslant f(x)$ so $f(X_{\tau_{B}})$ is a $\mathbb{P}_{x}$-integrable random variable. Since $B$ is a finite set, we have $f\leqslant M$ on $B$, for some constant $M>0$, and $\mathbb{P}_{x}(\tau_{B}<\infty)=1$. Using these two facts, we get

[TABLE]

Since the right hand side is $\mathbb{P}_{x}$-integrable, we can use the dominated convergence theorem and we get

[TABLE]

On the other hand, if $f(x)=\mathbb{E}_{x}[f(X_{\tau_{B}})]$, for every $x\in B$, then for $x\in B$ we have

[TABLE]

$\blacksquare$

Hence, if we take $B\subset\mathbb{Z}^{d}$ finite and ${f}\mathrel{\mathop{\mathchar 58\relax}}{\mathbb{Z}^{d}}\to{[0,\infty\rangle}$ harmonic in $B$, with respect to $X$, then by Lemma 6.3 and the discrete Ikeda-Watanabe formula, we get

[TABLE]

Let us define the discrete Poisson kernel of a finite set $B\subset\mathbb{Z}^{d}$ by

[TABLE]

If the function $f$ is non-negative and harmonic in $B_{n}$, with respect to $X$, from (6.2) we have

[TABLE]

Now we are ready to show that the Poisson kernel $K_{B_{n}}(x,z)$ is comparable to an expression that is independent of $x$. When we prove that, Harnack inequality will follow immediately.

Lemma 6.4**.**

Assume (1.4) and let $b_{1},b_{2}\in\langle 0,\frac{1}{2}\rangle$ be as in Corollary 5.8. Then $K_{B_{n}}(x,z)\asymp l(z)$ for all $x\in B_{b_{1}n}$, where

[TABLE]

• Proof:

Splitting the expression (6.3) for the Poisson kernel in two parts and using Proposition 3.2, we get

[TABLE]

Since $G_{B_{n}}(x,y)\asymp n^{-d}\mathbb{E}_{y}[\tau_{B_{n}}]$ for $x\in B_{b_{1}n}$, $y\in A(b_{2}n,n)$, for the second sum in the upper expression we have

[TABLE]

Now we look closely at the expression $\sum_{y\in B_{b_{2}n}}{G_{B_{n}}(x,y)j(|z-y|)}$. Using the fact that $y\in B_{b_{2}n}$, $b_{2}\in\langle 0,\frac{1}{2}\rangle$ and $|z|\geqslant n$ because $z\in B_{n}^{c}$, we have

[TABLE]

On the other hand

[TABLE]

Hence,

[TABLE]

Combining (6.6), (6.7) and using Lemma 2.3, we have

[TABLE]

Using (2.6), we get $j(\frac{1}{2}|z|)\leqslant 2^{d+2}j(|z|)=c_{1}j(|z|)$. Similarly, from (2.8), we get $j(2|z|)\geqslant 2^{-d-2}j(|z|)=c_{2}j(|z|)$. Hence, $a_{1}c_{2}j(|z|)\leqslant a_{1}j(2|z|)\leqslant j(|z-y|)\leqslant a_{2}^{-1}j\left(\frac{1}{2}|z|\right)\leqslant a_{2}^{-1}c_{1}j(|z|)$ for some $c_{1},c_{2}>0$. Therefore,

[TABLE]

Using (6.8) we have

[TABLE]

Now we want to show that $\sum_{y\in B_{b_{2}n}}{G_{B_{n}}(x,y)}\asymp 1/\phi\left(n^{-2}\right)$. Using the fact that $G_{B_{n}}$ is non-negative function and that $\mathbb{E}_{x}[\tau_{B_{n}}]\leqslant C_{3}/\phi\left(n^{-2}\right)$ for $x\in B_{n}$ we have

[TABLE]

To prove the other inequality we will use Lemma 5.3, Theorem 4.3, Lemma 2.2 together with $1\leqslant|x-y|\leqslant 2b_{2}n$, $|B_{n}\cap\mathbb{Z}^{d}|\geqslant c_{3}n^{d}$ and Lemma 1.1. Thus

[TABLE]

Hence,

[TABLE]

From (6.9) and (6.10) we have

[TABLE]

Finally, using (6.8) and (6.11) we have

[TABLE]

And now, from (6.12) and (6.5) we have the statement of the lemma. $\blacksquare$

Lemma 6.4 basically states that there exist constants $C_{8},C_{9}>0$ such that

[TABLE]

Now we are ready to prove our main result.

• Proof of Theorem 1.3:

Notice that, because of the spatial homogeneity, it is enough to prove this result for balls centered at the origin. We will prove the theorem for $a=b_{1}$, where $b_{1}$ is as in Corollary 5.8. General case follows using the standard Harnack chain argument. Let $x_{1},x_{2}\in B_{b_{1}n}$. Using (6.13) we get

[TABLE]

Now we can multiply both sides with $f(z)\geqslant 0$ and sum over all $z\notin B_{n}$ and we get

[TABLE]

If we look at the expression (6) we see that this means

[TABLE]

and that is what we wanted to prove. $\blacksquare$

Acknowledgement: This work has been supported in part by Croatian Science Foundation under the project 3526.