Harnack inequality for subordinate random walks
Ante Mimica Stjepan Šebek

3D Printing of Polymer Bonded Rare-Earth Magnets With a Variable Magnetic Compound Density for a Predefined Stray Field
    D. Suess
    December 30, 2023
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In this paper, we consider a large class of subordinate random walks X on integer lattice 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.




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

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




§ INTRODUCTION

Let (Y_k)_k ≥ 1 be a sequence of independent, identically distributed random variables defined on a probability space (Ω, , ), taking values in the integer lattice ^d, with distribution (Y_k = e_i) = (Y_k = -e_i) = 1/2d, i = 1, 2, …, d, where e_i is the i-th vector of the standard basis for ^d. A simple symmetric random walk in ^d (d ≥ 1) starting at x ∈^d is a stochastic process Z = (Z_n)_n ≥ 0, with Z_0 = x and Z_n = x + Y_1 + ⋯ + Y_n.

Let Z = (Z_n)_n ≥ 0 be a simple symmetric random walk in ^d starting at the origin. Further, let

    ϕ(λ) := ∫_⟨ 0, ∞⟩1 - e^-λ tμ ( t)

be a Bernstein function satisfying ϕ(1)=1. Here μ is a measure on ⟨ 0,∞⟩ satisfying ∫_⟨ 0,∞⟩(1 ∧ t) μ( t) < ∞ called the Lévy measure. For m ∈ denote

    c_m^ϕ := ∫_⟨ 0, ∞⟩t^m/m!e^-tμ (dt)

and notice that

    ∑_m = 1^∞c_m^ϕ = ∫_⟨ 0, ∞⟩(e^t - 1) e^-tμ( t) = ∫_⟨ 0, ∞⟩(1 - e^-t) μ( t) = ϕ(1) = 1.

Hence, we can define a random variable R with (R = m) = c_m^ϕ, m ∈. Now we define the random walk T = (T_n)_n ≥ 0 on _+ by T_n := ∑_k = 1^nR_k, where (R_k)_k ≥ 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 ≥ 0 which is defined by X_n := Z_T_n, n ≥ 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 (ξ_k)_k ≥ 1 with the same distribution as X_1 such that

    X_n d=∑_k = 1^n ξ_k,    n ≥ 0.

We can easily find the explicit expression for the distribution of the random variable X_1:

    (X_1 = x)
    	    =  (Z_T_1 = x) = (Z_R_1 = x) = ∑_m = 1^∞(Z_R_1 = x | R_1 = m) c_m^ϕ
        =  ∑_m = 1^∞∫_⟨ 0, ∞⟩t^m/m! e^-tμ(dt)(Z_m = x),    x ∈^d.

We denote the transition matrix of the subordinate random walk X with P, i.e. P = (p(x, y) : x, y ∈^d), where p(x, y) = (x + X_1 = y).

We will impose some additional constraints on the Laplace exponent ϕ. First, ϕ will be a complete Bernstein function <cit.> and it will satisfy the following lower scaling condition: there exist 0 < γ_1 < 1 and a_1 > 0 such that

    ϕ(R)/ϕ(r)≥ a_1 R/r^γ_1,   ∀   0 < r ≤ R≤ 1.

In dimension d ≤ 2, we additionally assume that there exist γ_1 ≤γ_2 < 1 and a_2 > 0 such that

    ϕ(R)/ϕ(r)≤ a_2 R/r^γ_2,   ∀   0 < r ≤ R≤ 1.

It is well known that,  if ϕ is a Bernstein function, then ϕ(λ t) ≤λϕ(t) for all λ≥ 1, t > 0, implying ϕ(v)/v ≤ϕ(u)/u, 0 < u ≤ v. Using these two facts, proof of the next lemma is straightforward.

	If ϕ is a Bernstein function, then for all λ, t > 0, 1 ∧λ≤ϕ(λ t) / ϕ(t) ≤ 1 ∨λ.

Using Lemma <ref> we get ϕ(R) / ϕ(r) ≤ R/r for all 0 < r ≤ R ≤ 1 and this suffices for d ≥ 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.

	We say that a function f^d[0, ∞⟩ is harmonic in B ⊂^d, with respect to X, if
	
    f(x) = Pf(x) = ∑_y ∈^dp(x, y) f(y),   ∀ x ∈ B.


This definition is equivalent to the mean-value property in terms of the exit from a finite subset of ^d: If B ⊂^d is finite then f^d[0, ∞⟩ is harmonic in B, with respect to X, if and only if f(x) = _x[f(X_τ_B)] for every x ∈ B, where τ_B := inf{n ≥ 1 : X_n ∉ B}.

For x ∈^d and r > 0 we define B(x, r) := {y ∈^d : y - x < r}. We use shorthand notation B_r for B(0, r).
[Harnack inequality]
	Let X = (X_n)_n ≥ 0 be a subordinate random walk in ^d, d ≥ 1, with ϕ a complete Bernstein function satisfying (<ref>) and in the case d ≤ 2 also (<ref>). For each a < 1, there exists a constant c_a < ∞ such that if f^d[0, ∞⟩ is harmonic on B(x, n), with respect to X, for x ∈^d and n ∈, then
	
    f(x_1) ≤ c_a f(x_2),   ∀ x_1, x_2 ∈ B(x, an).


Notice that the constant c_a is uniform for all 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 ^d <cit.> 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 <cit.>.

Notion of discrete subordination was developed in <cit.> and it was discussed in details in <cit.>, but under different assumptions on ϕ than the ones we have. Using discrete subordination we can obtain random walks with steps of infinite second moment, see Remark <ref>. Harnack inequality has not been proved so far for such random walks.

In Section <ref> 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 <ref>. In Section <ref> we derive estimates for the Green function. This is very valuable result which gives the answer to the question posed in <cit.>. Using estimates developed in Section <ref> and <ref> and following <cit.>, in Section <ref> we find estimates for the Green function of a ball. In Section <ref> we introduce Poisson kernel and prove Harnack inequality.

Throughout this paper, d≥ 1 and the constants a_1, a_2, γ_1, γ_2 and C_i, i = 1, 2, …, 9 will be fixed. We use c_1, c_2, … 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, … 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 x to denote the Lebesgue measure in ^d. We denote the Euclidean distance between x and y in ^d by x - y. For a, b ∈, a ∧ b := min{a, b} and a ∨ b := max{a, b}. For any two positive functions f and g, we use the notation f ≍ 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 ≤ g ≤ c_2 f on their common domain of definition. We also use notation f ∼ g to denote that lim_x →∞ f(x) / g(x) = 1.




§ 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.



 §.§ Ratio of gamma functions


	Let Γ(x, a) = ∫_a^∞t^x - 1 e^-t t. Then		
    lim_x →∞Γ(x + 1, x)/Γ(x + 1) = 1/2.



	Using a well-known Stirling's formula
		
    Γ(x + 1) ∼√(2 π x) x^x e^-x,    x →∞

	and <cit.> that states
	
    Γ(x + 1, x) ∼√(2^-1π x) x^x e^-x,    x →∞

	we get
	
    lim_x →∞Γ(x + 1, x)/Γ(x + 1) = lim_x →∞√(2^-1π x) x^x e^-x/√(2 π x) x^x e^-x = 1/2.





 §.§ 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 +∞ for every argument x. We will use Chung-Fuchs theorem to show under which condition subordinate random walk is transient. Denote with φ_X_1 the characteristic function of one step of a subordinate random walk. We want to prove that there exists δ > 0 such that

    ∫_⟨ -δ, δ⟩^d1/1 - φ_X_1 (θ)θ < +∞.

The law of variable X_1 is given with (<ref>). We denote one step of the simple symmetric random walk (Z_n)_n ≥ 0 with Y_1 and the characteristic function of that random variable with φ. Assuming θ < 1 we have

    φ_X_1 (θ)
    	    = e^i θ· X_1 = ∑_x ∈^de^i θ· x∑_m = 1^∞∫_⟨ 0, +∞⟩t^m/m! e^-tμ( t)(Z_m = x)
        = ∑_m = 1^∞∫_⟨ 0, +∞⟩t^m/m! e^-tμ( t)∑_x ∈^de^i θ· x(Z_m = x) = ∑_m = 1^∞∫_⟨ 0, +∞⟩t^m/m! e^-tμ( t) (φ(θ))^m
        = ∫_⟨ 0, +∞⟩e^t φ(θ) - 1 e^-tμ( t) = ϕ(1) - ϕ(1 - φ(θ)) = 1 - ϕ(1 - φ(θ)).

From <cit.> we have

    φ(θ) = 1/d∑_m = 1^dcos(θ_m),   θ = (θ_1, θ_2, …, θ_m).

That is function with real values so

    ∫_⟨ -δ, δ⟩^d1/1 - φ_X_1(θ)θ = ∫_⟨ -δ, δ⟩^d1/ϕ(1 - φ(θ))θ.

From Taylor's theorem it follows that there exists a ≤ 1 such that

    φ(θ) = φ(θ) ≤ 1 - 1/4dθ^2,   θ∈ B(0, a).

Now we take δ such that ⟨ -δ, δ⟩^d ⊂ B(0, a). From (<ref>), using the fact that ϕ is increasing, we get

    1/ϕ1 - φ(θ)≤1/ϕθ^2 / 4d,   θ∈ B(0, a).

Hence,

    ∫_⟨ -δ, δ⟩^d1/ϕ(1 - φ(θ))θ   ≤∫_⟨ -δ, δ⟩^d1/ϕθ^2 / 4dθ≤∫_B(0, a)ϕ(θ^2)/ϕθ^2 / 4d1/ϕ(θ^2)θ
       ≤ a_2 (4d)^γ_2∫_B(0, a)1/ϕ(θ^2)θ = c_1 (4d)^γ_2∫_0^ar^d - 1/ϕ(r^2) r
        = c_1 (4d)^γ_2/ϕ(a)∫_0^ar^d - 1ϕ(a)/ϕ(r^2) r≤c_1 a_2 (4ad)^γ_2/ϕ(a)∫_0^ar^d - 2 γ_2 - 1 r

and the last integral converges for d - 2 γ_2 - 1 > -1. So, subordinate random walk is transient for γ_2 < d/2. Notice that in the case when d ≥ 3 we have γ_2 < d/2 even if γ_2 = 1. That is the reason why we do not need (<ref>) 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 (<ref>) just to show that the result is true even in dimensions 1 and 2 when γ_2 < d/2. In the case when d ≥ 3, we can replace a_2 and γ_2 from (<ref>) with 1 and only use Lemma <ref>.



 §.§ Properties of functions g and j

Let g⟨ 0, +∞⟩⟨ 0, +∞⟩ be defined by

    g(r) = 1/r^d ϕ(r^-2)

and let j⟨ 0, +∞⟩⟨ 0, +∞⟩ be defined by

    j(r) = r^-dϕ(r^-2).

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


	Assume (<ref>) (if d ≤ 2) and let 1 ≤ r ≤ q. Then g(r) ≥ a_2^-1 g(q).


	Using (<ref>) and the fact that d > 2γ_2 we have
	
    g(r) = 1/r^d/q^d q^d ϕ(q^-2) ϕ(r^-2)/ϕ(q^-2)≥1/a_2q/r^d - 2γ_2 g(q) ≥1/a_2 g(q).


We prove similar assertion for the function j.

	Assume (<ref>) and let 1 ≤ r ≤ q. Then j(r) ≥ a_1 j(q).


	Using (<ref>) we have
	
    j(r) = r^-d/q^-d q^-dϕ(q^-2) ϕ(r^-2)/ϕ(q^-2)≥ a_1 q/r^d + 2γ_1 j(q) ≥ a_1 j(q).


Using (<ref>), (<ref>) and Lemma <ref> 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 ϕ. For the second one we need (<ref>) and for the third one we need (<ref>).

	Let r ≥ 1. If 0 < a ≤ 1 then
	
    j(ar) ≤ a^-d - 2 j(r),

	
    g(ar) ≥ a^-d + 2 g(r).

	If a ≥ 1 then
	
    j(ar) ≥ a^-d - 2 j(r).




	Assume (<ref>) and let 0 < a ≤ 1 and r ≥ 1 such that ar ≥ 1. Then
	
    g(ar) ≤g(r)/a_1 a^d - 2γ_1.




	Assume (<ref>) and let r ≥ 1. If 0 < a ≤ 1 such that ar ≥ 1 then
	
    g(ar) ≥g(r)/a_2 a^d - 2γ_2.
		
	If a ≥ 1 then
	
    g(ar) ≤a_2/a^d - 2γ_2 g(r).





§ TRANSITION PROBABILITY ESTIMATES

In this section, we will investigate the behavior of the expression (X_1 = z). We will prove that (X_1 = z) ≍ j(z), z ≠ 0. First we have to examine the behavior of the expression c_m^ϕ.

	Assume (<ref>) and let c_m^ϕ be as in (<ref>). Then
		
    c_m^ϕ≍ϕ(m^-1)/m,    m∈.



	Since ϕ is a complete Bernstein function, there exists completely monotone density μ(t) such that
	
    c_m^ϕ = ∫_0^∞t^m/m! e^-tμ(t) dt,    m∈.

	From <cit.> we have
	
    μ(t) ≤ (1 - 2e^-1)^-1 t^-1ϕ(t^-1) = c_1 t^-1ϕ(t^-1),    t > 0

	and
	
    μ(t) ≥ c_2 t^-1ϕ(t^-1),    t ≥ 1.

	Inequality (<ref>) holds only if (<ref>) is satisfied and for inequality (<ref>) we do not need any scaling conditions. Using monotonicity of μ, (<ref>) and (<ref>) we have

    c_m^ϕ≥μ(m)/m!∫_0^m t^m e^-t dt = μ(m) 1 - Γ(m + 1, m)/Γ(m + 1)≥1/4μ(m) ≥c_2/4ϕ(m^-1)/m

for m large enough. On the other side,  using inequality (<ref>), monotonicity of μ and Lemma <ref>, we get

    c_m^ϕ   ≤1/m!∫_0^m t^m e^-t c_1 ϕ(t^-1)/t dt + μ(m)/m!∫_m^∞t^m e^-t dt
       ≤c_1/m!ϕ(m^-1) ∫_0^m t^m - 1 e^-tϕ(t^-1)/ϕ(m^-1) dt + μ(m)/m!∫_0^∞t^m e^-t dt
       ≤ c_1 ϕ(m^-1) 1/Γ(m)∫_0^∞t^m - 2 e^-t dt + μ(m) = c_1 ϕ(m^-1) Γ(m - 1)/Γ(m) + μ(m) 
       ≤ c_1 ϕ(m^-1)/m + c_1 ϕ(m^-1)/m = 2c_1 ϕ(m^-1)/m.

Hence, we have

    c_2/4ϕ(m^-1)/m≤ c_m^ϕ≤ 2c_1 ϕ(m^-1)/m

for m large enough, but we can change constants and get (<ref>).

We are now ready to examine the expression (X_1 = z).

	Assume (<ref>). Then
	
    (X_1 = z) ≍z^-dϕ(z^-2),    z≠ 0.



Using (<ref>) and the fact that (Z_m = z) = 0 for z > m, we have

    (X_1 = z) = ∑_m ≥zc_m^ϕ(Z_m = z).

To get the upper bound for the expression (X_1 = z) we will use <cit.> which states that there are constants C' > 0 and C > 0 such that

    (Z_m = z) ≤ C' m^-d/2 e^-z^2/C m,   ∀  z ∈^d, ∀  m ∈.

Together with (<ref>) this result yields

    (X_1 = z)
    	   ≤∑_m ≥zc_1 ϕ(m^-1)/m C' m^-d/2 e^-z^2/C m≤ c_2 ∫_z^∞ϕ(t^-1) t^-d/2 - 1 e^-z^2/C t dt
        = c_2 ∫_0^z/Cϕ(C sz^-2) z^2/C s^-d/2 - 1 e^-s z^2/C s^2 ds
        = c_3 z^-d∫_0^1/Cϕ(C sz^-2) s^d/2 - 1 e^-s ds + ∫_1/C^z/Cϕ(C sz^-2) s^d/2 - 1 e^-s ds
        =: c_3 z^-d (I_1(z) + I_2(z)).

Let us first examine I_1(z). Using (<ref>), we get

    I_1(z) = ϕ(z^-2) ∫_0^1/Cϕ (C sz^-2)/ϕ(z^-2) s^d/2 - 1 e^-s ds≤ c_4 ϕ(z^-2).

Using Lemma <ref> instead of (<ref>) and replacing the upper limit in the integral by ∞, we get I_2(z) ≤ c_5 ϕ(z^-2). Hence, (X_1 = z) ≤ c_6 z^-dϕ(z^-2).

In finding the matching lower bound for (X_1 = z), periodicity of a simple random walk plays very important role. We write n ↔ x if n and x have the same parity, i.e., if n + x_1 + x_2 + ⋯ + x_d is even. Directly from <cit.>, we get

    (Z_m = z) ≥ c_7 m^-d/2 e^-d z^2/2m

for 0 ↔ z ↔ m and z≤ m^α, α < 2/3. In the case when 1 ↔ z ↔ m we have

    (Z_m = z) = 1/2d∑_i = 1^d [(Z_m - 1 = z + e_i) + (Z_m - 1 = z - e_i)].

By combining (<ref>) and (<ref>), we can easily get

    (Z_m = z) ≥ c_8 m^-d/2 e^-z^2/c m,   z≤ m^1/2, 1 ↔ z ↔ m.

We will find lower bound for (X_1 = z) when z ↔ 0  by using (<ref>), the proof when z ↔ 1 being analogous using (<ref>). If z ↔ 0 then (Z_m = z) =  0 for m = 2l - 1, l ∈. Hence,

    (X_1 = z)
    	   ≥∑_m ≥z^2, m = 2lc_9ϕ(m^-1)/m m^-d/2 e^--dz^2/2m = c_9∑_l ≥z^2 / 2ϕ((2l)^-1)/2l (2l)^-d/2 e^-dz^2/4l
       ≥ c_10∫_z^2 / 2^∞ϕ((2t)^-1)/2t (2t)^-d/2 e^-dz^2/4t t = c_10/2∫_z^2^∞ϕ(t^-1) t^-d/2 - 1 e^-dz^2/2t t
        = c_10/2∫_0^d/2ϕ2s/d z^2d z^2/2s^-d/2 - 1 e^-s d z^2/2s^2 s
        = c_11z^-dϕ(z^-2) ∫_0^d/2ϕ2s/dz^-2/ϕ (z^-2) s^d/2 - 1 e^-s s
       ≥ c_11z^-dϕ(z^-2) ∫_0^d/22/d s^d/2 e^-s ds = c_12z^-dϕ(z^-2),

where in the last line we used Lemma <ref>.


It follows immediately form Proposition <ref> that the second moment of the step X_1 is infinite.




§ GREEN FUNCTION ESTIMATES

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

    G(y) = [∑_n = 0^∞_{X_n = y}].

Note that for n ≥ 1

    (X_n = y)
    		    = (Z_T_n = y) = ∑_m = n^∞(Z_m = y) (T_n = m)
        = ∑_m = n^∞∑_m_1 + ⋯ + m_n = mc_m_1^ϕ⋯ c_m_n^ϕ(Z_m = y)

Hence, for y ≠ 0 we have

    G(y) = ∑_m = 1^∞∑_n = 1^m∑_m_1 + ⋯ + m_n = mc_m_1^ϕ⋯ c_m_n^ϕ(Z_m = y) = ∑_m = 1^∞c(m) (Z_m = y),

where

    c(m) = ∑_n = 1^m∑_m_1 + ⋯ + m_n = mc_m_1^ϕ⋯ c_m_n^ϕ = ∑_n = 0^∞(T_n = m),

and T_n is as before. Now we will investigate the behavior of the sequence c(m). Since ϕ is a complete Bernstein function (hence special), we have
	
    1/ϕ(λ) = ∫_⟨ 0, ∞⟩e^-λ t u(t)  t

	for some non-increasing function u⟨ 0, ∞⟩⟨ 0, ∞⟩ satisfying ∫_0^1u(t) t < ∞, see <cit.>.



	Let c(m) be as in (<ref>). Then

    c(m) = 1/m!∫_⟨ 0, ∞⟩t^m e^-t u(t)  t,    m ∈_0.



	We follow the proof of <cit.>. Define M(x) = ∑_m ≤ xc(m), x ∈. The Laplace transformation (M) of the measure generated by M is equal to	
    (M)(λ)
    		    = ∫_[0, ∞⟩e^-λ x M(x) = ∑_m = 0^∞c(m) e^-λ m = ∑_m = 0^∞e^-λ m∑_n = 0^∞(T_n = m)
        = ∑_n = 0^∞∑_m = 0^∞e^-λ m(T_n = m) = ∑_n = 0^∞[e^-λ T_n] = ∑_n = 0^∞[e^-λ R_1]^n = 1/1 - [e^-λ R_1].

	Now we calculate [e^-λ R_1]:
	
    [e^-λ R_1]
    		    = ∑_m = 1^∞e^-λ m∫_⟨ 0, ∞⟩t^m/m! e^-tμ( t) = ∫_⟨ 0, ∞⟩e^te^-λ - 1 e^-tμ( t) = 1 - ϕ(1 - e^-λ),

	where we used ϕ(1) = 1 in the last equality. Hence, (M)(λ) = 1 / ϕ(1 - e^-λ). Now we define Φ(λ) := 1 / ϕ(λ) and we want to show that
	
    Φ(1 - e^-λ) = ∑_m = 0^∞(-1)^m Φ^(m)(1)/m! e^-λ m.

	It is easy to see that
	Φ^(m)(λ) = (-1)^m ∫_⟨ 0, ∞⟩t^m e^-λ tu(t)  t. Hence,	
    ∑_m = 0^∞(-1)^m Φ^(m)(1)/m! e^-λ m = ∑_m = 0^∞(-1)^m/m! (-1)^m ∫_⟨ 0, ∞⟩t^m e^-t u(t)  t e^-λ m = Φ(1 - e^-λ).

	Since (M)(λ) = 1/ϕ(1 - e^-λ) = Φ(1 - e^-λ) from calculations (<ref>) and (<ref>) we have
	
    ∑_m = 0^∞c(m) e^-λ m = ∑_m = 0^∞1/m!∫_⟨ 0, ∞⟩t^m e^-t u(t)  t  e^-λ m.

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



	Assume (<ref>). Then
	
    c(m) ≍1/m ϕ(m^-1),    m ∈.



	Since ϕ is a complete Bernstein function, using (<ref>) we can obtain
		
    u(t) ≍1/t ϕ(t^-1),    t ≥ 1,

	where the upper bound is valid even without (<ref>) (see <cit.>) and u is as in Lemma <ref>. Using monotonicity of u, (<ref>) and (<ref>), we get that
	
    c(m) ≥ u(m) 1/m!∫_0^m t^m e^-t t = u(m) 1 - Γ(m + 1, m)/Γ(m + 1)≥1/4 u(m) ≥c_1/m ϕ(m^-1),

	for m large enough. Now we will find the upper bound for c(m). Here we use that t ↦ t^m e^-t is unimodal with maximum at m. By splitting the integral and using (<ref>), we have
	
    c(m)
    		    = 1/m!∫_0^m/2t^m e^-t u(t)  t + 1/m!∫_m/2^mt^m e^-t u(t)  t + 1/m!∫_m^∞t^m e^-t u(t)  t
       ≤c_2 2^-m m^m e^-m/2/√(2 π m) m^m e^-m∫_0^m/2u(t)  t + u(m/2) 1/m!∫_m/2^∞t^m e^-t t + u(m) 1/m!∫_m^∞t^m e^-t t
       ≤ c_2 (2^-1 e^1/2)^m/√(2 π m)∫_0^1u(t)  t + c_2 (2^-1 e^1/2)^m/√(2 π m)∫_1^m/2u(t)  t + u(m/2) + u(m),

	for m large enough. Since 2^-1 e^1/2 < 1, ∫_0^1u(t)  t < ∞ and ϕ is increasing, we have
	
    c_2 (2^-1 e^1/2)^m/√(2 π m)∫_0^1u(t)  t≤1/m≤1/m ϕ(m^-1)≤1/4c_1 u(m)

	for m large enough, where we used (<ref>) in the last inequality. We will estimate the integral ∫_1^m/2u(t)  t by mu(m/2). Using (<ref>) and (<ref>) we get
	
    ∫_1^m/2u(t) dt
    		   ≤ c_3 ∫_1^m/21/tϕ(t^-1) dt = c_3/ϕ(2m^-1)∫_1^m/2ϕ(2m^-1)/ϕ(t^-1) t^-1 dt 
       ≤c_3/ϕ(2m^-1)1/a_1(m/2)^γ_1∫_1^m/2t^γ_1 - 1 dt 
       ≤c_3/a_1ϕ(2m^-1)(m/2)^γ_1(m/2)^γ_1/γ_1 = c_3/a_1γ_11/ϕ(2m^-1).

	Since u(m/2) ≥ c_4 / ((m/2)ϕ(2m^-1)) for m ≥ 2, we have 1 / ϕ(2m^-1) ≤ (1 / 2c_4) mu(m/2). Hence,
	
    ∫_1^m/2u(t) dt ≤c_3/a_1γ_11/2c_4 mu(m/2) = c_5mu(m/2).

	Using this estimate and the fact that 2^-1 e^1/2 is less then 1, we have
	
    c_2 (2^-1 e^1/2)^m/√(2π m)∫_1^m/2u(t) dt ≤ c_6 (2^-1 e^1/2)^m/√(2π m) m u(m/2) = c_6(2^-1 e^1/2)^m m^1/2 u(m/2) ≤ u(m/2)

	for m large enough. Now we have to show that u(m/2) can be estimated by u(m). Again, we will use (<ref>) and (<ref>):
	
    u(t/2) ≤c_3/(t/2)ϕ(2t^-1)  = 2c_3/tϕ(t^-1)/ϕ(2t^-1)1/ϕ(t^-1)≤2c_3/a_1 2^γ_11/t ϕ(t^-1)≤2c_3/c_4 a_1 2^γ_1 u(t) = c_7 u(t),

	where we assumed that t ≥ 2 because we need 2t^-1≤ 1 so that we can use (<ref>). Now, for m large enough, we have
	
    c(m) ≤1/4c_1 u(m) + u(m/2) + u(m/2) + u(m) ≤1/4c_1 u(m) + 2c_7 u(m) + u(m) ≤c_8/m ϕ(m^-1).

	Hence,
	
    c_1/m ϕ(m^-1)≤ c(m) ≤c_8/m ϕ(m^-1)

	for m large enough. We can now change constants in such a way that the statement of this lemma is true for every m ∈.


	Assume (<ref>) and, if d ≤ 2, assume additionally (<ref>). Then
	
    G(x) ≍1/x^d ϕ(x^-2),   x≥ 1.



	We assume x≥ 1 throughout the whole proof. In (<ref>) we showed that G(x) = ∑_m = 1^∞c(m) p(m, x), where p(m, x) = (Z_m = x). Let q(m, x) = 2 d / (2 π m)^d/2 e^-d x^2/2m and E(m, x) = p(m, x) - q(m, x). By <cit.>
	
    E(m, x)≤ c_1 m^-d/2 / x^2.

	Since p(m, x) = 0 for m <x, we have
	
    G(x) = ∑_m > x^2c(m) p(m, x) + ∑_x≤ m ≤x^2c(m) p(m, x) =: J_1(x) + J_2(x).

	First we estimate
	
    J_1(x) = ∑_m > x^2c(m) q(m, x) + ∑_m > x^2c(m) E(m, x) =: J_11(x) + J_12(x).

	By Lemma <ref>, (<ref>) and (<ref>)
	
    J_12(x)   ≤ c_2 ∑_m > x^21/m ϕ(m^-1)m^-d/2/x^2 = c_2/x^2 ϕ(x^-2)∑_m > x^2ϕ(x^-2)/ϕ(m^-1) m^-d/2 - 1
       ≤c_3 x^-2 γ_2/x^2 ϕ(x^-2)∫_x^2^∞t^γ_2 - d/2 - 1 t = c_4/x^21/x^d ϕ(x^-2).

	Now we have
	
    lim_x→∞x^d ϕ(x^-2) J_12(x) = 0.

	By Lemma <ref>, (<ref>) and (<ref>)
	
    J_11(x)
    		   ≍∫_x^2^∞1/t ϕ(t^-1) t^-d/2 e^-d x^2/2 t t = 1/ϕ(x^-2)∫_x^2^∞ϕ(x^-2)/ϕ(t^-1) t^-d/2 - 1 e^-d x^2/2 t t
       ≍x^-2 γ_i/ϕ(x^-2)∫_x^2^∞t^γ_i - d/2 - 1 e^-d x^2/2 t t = 1/x^d ϕ(x^-2)∫_0^d/2s^d/2 -γ_i - 1 e^-s s≍1/x^d ϕ(x^-2),

	where the last integral converges because of the condition γ_2 < d/2. We estimate J_2(x) using (<ref>) and (<ref>):
	
    J_2(x)
    		   ≤ c_5  ∫_x^x^2t^-d/2 - 1/ϕ(t^-1) e^-x^2/C t t = c_5/ϕ(x^-2)∫_x^x^2ϕ(x^-2)/ϕ(t^-1) t^-d/2 - 1 e^-x^2/C t t
       ≤c_5 x^-2 γ_1/a_1 ϕ(x^-2)∫_x^x^2t^γ_1 - d/2 - 1 e^-x^2/C t t = c_5 x^-2 γ_1/a_1 ϕ(x^-2)∫_1/C^x/Cx^2/Cs^γ_1 - d/2 - 1 e^-sx^2/C s^2 s
       ≤c_6/x^d ϕ(x^-2)∫_0^∞s^d/2 - γ_1 - 1 e^-s s = c_7/x^d ϕ(x^-2).

	Using J_11(x) ≥ (2 c_8) / (x^dϕ(x^-2)) and J_12(x) x^d ϕ(x^-2) ≥ - c_8 for x large enough and for some constant c_8 > 0, we get
	
    G(x) x^d ϕ(x^-2) ≥ J_11(x) x^d ϕ(x^-2) + J_12(x) x^d ϕ(x^-2) ≥ 2c_8 - c_8 = c_8

	On the other hand
	
    G(x) x^d ϕ(x^-2) ≤ c_9 +  J_12(x) x^d ϕ(x^-2) + c_7 ≤ 2 c_9 + c_7 = c_10.

	Here we used J_11(x) ≤ c_9 / (x^d ϕ(x^-2)), J_2(x) ≤ c_7 / (x^d ϕ(x^-2)) and J_12(x) x^d ϕ(x^-2) 
≤ c_9 for x large enough and for some constant c_9 > 0. So, we have c_8 ≤ G(x) x^d ϕ(x^-2) ≤ c_10 for x large enough. Now we can change the constants c_8 and c_10 to get	
    G(x) ≍1/x^d ϕ(x^-2),   for all  x≥ 1.






§ ESTIMATES OF THE GREEN FUNCTION OF A BALL

Let B ⊂^d and define

    G_B (x, y) = _x ∑_n = 0^τ_B - 1_{X_n = y}

where τ_B is as before. A well-known result about Green function of a set is formulated in the following lemma.

	Let B be a finite subset of ^d. Then
	
    G_B (x, y)     = G(x, y) - _x G(X_τ_B, y),    x, y ∈ B, 
    
    		G_B (x, x)     = 1/_x (τ_B < σ_x),    x∈ B,

	where σ_x = inf{n ≥ 1 : 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

    (Af)(x) := ((P - I)f)(x) = (Pf)(x) - (If)(x) = ∑_y ∈^dp(x, y) f(y) - f(x).

Since ∑_y ∈^dp(x, y) = 1 and p(x, y) = (X_1 = y - x) we have

    (Af)(x) = ∑_y ∈^d(X_1 = y - x) (f(y) - f(x)).

Before proving the maximum principle, we will show that for the function η(x) := _x [τ_B_n] we have (Aη)(x) = -1, for all x ∈ B_n. Let x ∈ B_n. Then

    η(x) = ∑_y ∈^d_x[τ_B_n| X_1 = y] _x(X_1 = y) = ∑_y ∈^d(1 + _y[τ_B_n]) (X_1 = y - x) = 1 + (Pη)(x)

and this is obviously equivalent to (Aη)(x) = -1, for all x ∈ B_n. It follows from the Definition <ref> that f is harmonic in B ⊂^d if and only if (Af)(x) = 0, for all x ∈ B.

 Assume that there exists x ∈^d such that (Af)(x) < 0. Then
	
    f(x) > inf_y ∈^d f(y).
		


	If (<ref>) is not true, then f(x) ≤ f(y), for all y ∈^d. In this case, we have
	
    (Pf)(x) = ∑_y ∈^d(X_1 = y - x) f(y)≥ f(x) ∑_y ∈^d(X_1 = y - x) = f(x).

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

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 (<ref>) and, if d ≤ 2, we additionally assume (<ref>). Throughout the rest of this section, we follow <cit.>.

	There exist a ∈⟨ 0, 1/3 ⟩ and C_1 > 0 such that for every n ∈
	
    G_B_n(x, y) ≥ C_1 G(x, y),   ∀  x, y ∈ B_an.



	From Lemma <ref> we have
	
    G_B_n (x, y) = G(x, y) - _x[G(X_τ_B_n, y)].

	We will first prove this lemma in the case when x ≠ y. If we show that _x[G(X_τ_B_n, y)] ≤ c_1 G(x, y) for some c_1 ∈⟨ 0, 1 ⟩ we will have (<ref>) with the constant c_2 = 1 - c_1. Let a ∈⟨ 0, 1/3 ⟩ and x, y ∈ B_an. In that case, we have x - y≤ 2an. Since X_τ_B_n∉ B_n, x ≠ y and (1 - a) / (2a) > 1 if and only if a < 1/3, we have
	
    y - X_τ_B_n≥ (1 - a)n = 1 - a/2a 2an ≥1 - a/2ax - y≥ 1.

	Using Theorem <ref>, (<ref>), Lemma <ref> and (<ref>), we get
	
    G(X_τ_B_n, y)
    		   ≍ g(y - X_τ_B_n) ≤ a_2 g1 - a/2ax - y
       ≤ a_2^2 2a/1 - a^d - 2γ_2 g(x - y) ≍ a_2^2 2a/1 - a^d - 2γ_2 G(x, y).

	Since 2a / (1 - a) ⟶ 0 when a → 0 and d > 2 γ_2, if we take a small enough and then fix it, we have _x[G(X_τ_B_n, y)] ≤ c_1 G(x, y) for c_1 ∈⟨ 0, 1 ⟩ and that is what we wanted to prove. Now we deal with the case when x = y. From Lemma <ref> we have G_B_n(x, x) = ((τ_B_n < σ_x))^-1 and from the definition of the function G and the transience of random walk we get G(x, x) = G(0) ∈ [1, ∞⟩. Now, we can conclude that
	
    G_B_n(x, x) ≥ 1 = (G(0))^-1 G(0) = (G(0))^-1 G(x, x).

	If we define C_1 := min{c_2, (G(0))^-1} we have (<ref>).

Using Lemma <ref> we can prove the following result:

	There exists constant C_2 > 0 such that for all n ∈
	
    _x[τ_B_n] ≥C_2/ϕ(n^-2),   ∀  x ∈ B_an/2,

	where a ∈⟨ 0, 1/3 ⟩ is as in Lemma <ref>.


	Let x ∈ B_an/2. In that case, we have B(x, an/2) ⊆ B_an. We set b = a/2 for easier notation. Notice that _x[τ_B_n] = ∑_y ∈ B_n G_B_n(x, y). Using this equality, Lemma <ref>, Theorem <ref> and Lemma <ref>, we have
	
    _x[τ_B_n]
    		   ≥∑_y ∈ B(x, bn)G_B_n (x, y)≥∑_y ∈ B(x, bn) ∖{x}C_1 G(x, y)≍∑_y ∈ B(x, bn) ∖{x}g(x - y)
       ≍∫_1^bng(r) r^d - 1 r = ∫_1^bn1/r ϕ(r^-2) r = 1/ϕ(n^-2)∫_1^bn1/rϕ(n^-2)/ϕ(r^-2) r
       ≥1/a_2 ϕ(n^-2) n^2 γ_2∫_1^bnr^2 γ_2 - 1 r = 1/2a_2 γ_2 ϕ(n^-2)b^2 γ_2 - 1/n^2 γ_2≥b^2 γ_2/4 a_2 γ_2 ϕ(n^-2),

	for n large enough. Hence, we can conclude that _x[τ_B_n] ≥ C_2 / ϕ(n^-2), for all x ∈ B_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 ∈. Notice that this is true regardless of the dimension because here, we can always plug in γ_2 = 1.


Now we want to find the upper bound for _x [τ_B_n].

	There exists constant C_3 > 0 such that for all n ∈
	
    _x[τ_B_n] ≤C_3/ϕ(n^-2),   ∀  x ∈ B_n.



	We define the process M^f = (M_n^f)_n ≥ 0 with
	
    M_n^f := f(X_n) - f(X_0) - ∑_k = 0^n - 1(Af)(X_k)

	where f is a function defined on ^d with values in , A is defined as in (<ref>) and X = (X_n)_n ≥ 0 is a subordinate random walk. By <cit.>, the process M^f is a martingale for every bounded function f. Let f := _B_2n and x ∈ B_n. By the optional stopping theorem, we have
	
    _x [M_τ_B_n^f] = _x f(X_τ_B_n) - f(X_0) - ∑_k = 0^τ_B_n - 1(Af)(X_k) = _x [M_0^f] = 0.

	Hence
		
    _x f(X_τ_B_n) - f(X_0) = _x ∑_k = 0^τ_B_n - 1(Af)(X_k).

	We now investigate both sides of the relation (<ref>). For every k < τ_B_n, X_k ∈ B_n, and for every y ∈ B_n, using Proposition <ref>, (<ref>) and (<ref>), we have
	
    (Af)(y)
    		    = ∑_u ∈^d(X_1 = u - y) (f(u) - f(y))≍ -∑_u ∈ B_2n^cu - y^-dϕ(u - y^-2)
       ≍ -∫_n^∞r^-dϕ(r^-2) r^d - 1 r = -ϕ(n^-2) ∫_n^∞r^-1ϕ(r^-2)/ϕ(n^-2) r
       ≍ -ϕ(n^-2) ∫_n^∞r^-1n^2γ_i/r^2γ_i r = -ϕ(n^-2) n^2γ_in^-2γ_i/2γ_i≍ -ϕ(n^-2).

	Using the above estimate, we get	
    _x ∑_k = 0^τ_B_n - 1(Af)(X_k)≍_x -∑_k = 0^τ_B_n - 1ϕ(n^-2) = -ϕ(n^-2) _x[τ_B_n].

	Using (<ref>), (<ref>) and  _x [f(X_τ_B_n) - f(X_0)] = _x (X_τ_B_n∈ B_2n) - 1 = -_x(X_τ_B_n∈ B_2n^c), we get
	
    _x (X_τ_B_n∈ B_2n^c) ≍ϕ(n^-2) _x[τ_B_n]

	and this implies	
    _x [τ_B_n] ≤C_3 _x(X_τ_B_n∈ B_2n^c)/ϕ(n^-2)≤C_3/ϕ(n^-2).


In the next two results we develop estimates for the Green function of a ball. We define A(r, s) := {x ∈^d : r ≤x < s} for r, s ∈, 0 < r < s.

	There exists constant C_4 > 0 such that for all n ∈	
    G_B_n(x, y) ≤ C_4n^-dη(y),   ∀  x ∈ B_bn/2, y ∈ A(bn, n),

	where η(y) = _y[τ_B_n], b = a/2, and a ∈⟨ 0, 1/3 ⟩ is as in Lemma <ref>.


	Let x ∈ B_bn/2 and y ∈ A(bn, n). We define function h(z) := G_B_n (x, z). Notice that for z ∈ B_n ∖{x} we have
	
    h(z) = G_B_n(x, z) = G_B_n(z, x) = ∑_y ∈^d(X_1 = y - z) G_B_n (y, x) = ∑_y ∈^d(X_1 = y - z) h(y).

	Hence, h is a harmonic function in B_n ∖{x}. If we take z ∈ B(x, bn/8)^c then z - x≥ bn/8 ≥ 1 for n large enough. Using Lemma <ref> and Theorem <ref> we get
	
    g(bn/8) ≥ a_2^-1 g(z - x) ≍ G(x, z) ≥ G_B_n(x, z) = h(z).

	Hence, h(z) ≤ kg(bn/8) for z ∈ B(x, bn/8)^c and for some constant k > 0. Notice that A(bn, n) ⊆ B(x, bn/8)^c, hence y ∈ B(x, bn/8)^c. Using these facts together with Proposition <ref>, we have
	
    A
    		    (h ∧ kg(bn/8))(y) = A(h ∧ kg(bn/8) - h)(y) 
        = ∑_v ∈^d(X_1 = v - y) (h(v) ∧ kg(bn/8) - h(v) - h(y) ∧ kg(bn/8) + h(y)) 
       ≍∑_v ∈ B(x, bn/8)j(v - y) (h(v) ∧ kg(bn/8) - h(v))≥ -∑_v ∈ B(x, bn/8)j(v - y) h(v)
       ≥ -∑_v ∈ B(x, bn/8)a_1^-1 j(bn/8) h(v) = -a_1^-1 j(bn/8) ∑_v ∈ B(x, bn/8)G_B_n(x, v)≥ -a_1^-1 j(bn/8) η(x),

	where in the last line we used Lemma <ref> together with v - y≥ bn/8 ≥ 1 for v ∈ B(x, bn/8) and for n large enough. Using (<ref>) we get j(bn/8) ≤ (b/8)^-d-2 j(n). Hence, using (<ref>), we have
	
    A(h ∧ kg(bn/8))(y) ≥ -c_1 n^-dϕn^-2η(x) ≥ -c_1 n^-dϕn^-2 C_3ϕn^-2^-1 = -c_2 n^-d

	for some c_2 > 0. On the other hand, using (<ref>) and Proposition <ref> we get
	
    g(bn/8)
    		   ≤ a_1^-1 (bn/8)^-d + 2γ_1 g(n) ≤ (a_1 C_2)^-1 (bn/8)^-d + 2γ_1 n^-dη(z) = c_3n^-dη(z),   ∀ z ∈ B_bn.

	Now we define C_4 := (c_2 ∨ kc_3) + 1 and using
	
    h(z) ∧ kg(bn/8) ≤ kg(bn/8) ≤ kc_3n^-dη(z)

	we get
	
    C_4n^-dη(z) - h(z) ∧ kg(bn/8) ≥ (C_4 - kc_3)n^-dη(z) ≥ 0,    ∀  z ∈ B_bn

	So, if we define u(·) := C_4n^-dη(·) - h(·) ∧ 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 ∈ A(bn, n) we have
	
    (Au)(y) = C_4n^-d (Aη)(y) - A(h ∧ kg(bn/8))(y) ≤ -C_4n^-d + c_2n^-d < 0.

	Since u ≥ 0 on B_bn and u vanishes on B_n^c, if inf_y ∈^du(y) < 0 then there would exist y_0 ∈ A(bn, n) such that u(y_0) = inf_y ∈^du(y). But then, by Proposition <ref>, (Au)(y_0) ≥ 0 which is in contradiction with (Au)(y) < 0 for y ∈ A(bn, n). Hence,
	
    u(y) = C_4n^-dη(y) - h(y) ∧ kg(bn/8) ≥ 0,   ∀  y ∈^d

	and then, because h(y) ≤ kg(bn/8) for y ∈ A(bn, n) we get
	
    G_B_n(x, y) = h(y) ≤ C_4n^-dη(y),   ∀  x ∈ B_bn/2,  y ∈ A(bn, n).


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 ∩^d≥ c n^d for some constant c > 0, where · denotes the cardinality of a set.

	There exist C_5 > 0 and b ≤ a/4 such that for all n ∈	
    G_B_n(x, y) ≥ C_5n^-dη(y),   ∀  x ∈ B_bn, y ∈ A(an/2, n),

	where a is as in Lemma <ref> and η(y) = _y[τ_B_n]. 


	Let a ∈⟨ 0, 1/3 ⟩ as in Lemma <ref>. Then there exists C_1 > 0
	
    G_B_n (x, v) ≥ C_1G(x, v),    x, v ∈ B_an.

	From Proposition <ref> it follows that there exists constant C_4 > 0 such that
	
    G_B_n(x, v) ≤ C_4n^-dη(v),    x ∈ B_an/4, v ∈ A(an/2, n).

	From Lemma <ref> we have
	
    η(v) ≤C_3/ϕn^-2,    v∈ B_n,

	for some constant C_3 > 0. By Theorem <ref> and (<ref>) there exists c_1 > 0 such that G(x) ≥ c_1 g(x), x ≠ 0. Now we take
	
    b ≤mina/4, C_1 c_1/2 a_2^2 C_3 C_4^1/d - 2γ_2

	and fix it. Notice that (C_1 c_1) / (a_2^2 C_3 b^d - 2γ_2) ≥ 2C_4. Let x ∈ B_bn, v ∈ B(x, bn). Since b ≤ a/4 we have x, v ∈ B_an. We want to prove that G_B_n(x, v) ≥ 2C_4 n^-dη(v). We will first prove that assertion for x ≠ v. In that case we have 1 ≤x - v. Since v ∈ B(x, bn), we have x - v≤ bn so we can use (<ref>), Lemma <ref> and (<ref>) to get
	
    G_B_n(x, v) ≥ C_1 G(x, v) ≥C_1 c_1/a_2g(bn) ≥C_1 c_1/a_2^2 b^d - 2γ_2 g(n) ≥2C_3 C_4/n^d ϕ(n^-2).

	Using (<ref>) and (<ref>), we get G_B_n(x, v) ≥ 2C_4 n^-dη(v) for x ≠ v. Now we will prove that G_B_n(x, x) ≥ 2C_4 n^-dη(x), for x ∈ B_bn and for n large enough. First note that

    lim_n →∞n^d ϕ(n^-2) = lim_n →∞n^d ϕ(n^-2)/ϕ(1)≥lim_n →∞n^d 1/a_2 n^2 γ_2 = lim_n →∞1/a_2 n^d - 2γ_2 = ∞,

	since d - 2γ_2 > 0. Therefore 

    2 C_4 n^-dη(x) ≤2 C_4 C_3/n^d ϕ(n^-2)≤ 1 ≤ G_B_n(x, x)

for n large enough. Hence,
	
    C_4n^-dη(v) ≤1/2G_B_n(x, v),   ∀  x ∈ B_bn, v ∈ B(x, bn).

	Now we fix x ∈ B_bn and define the function
	
    h(v) := G_B_n(x, v) ∧C_4n^-dη(v).

	From (<ref>) we have h(v) ≤1/2G_B_n(x, v) for v ∈ B(x, bn). Recall that G_B_n(x, ·) is harmonic in A(an/2, n). Using (<ref>) we get h(y) = G_B_n(x, y) for y ∈ A(an/2, n). Hence, for y ∈ A(an/2, n)
	
    (Ah)(y)
    		    = A(h(·) - G_B_n(x, ·))(y) ≍∑_v ∈^dj(v - y) h(v) - G_B_n(x, v) - h(y) + G_B_n(x, y)
       ≤∑_v ∈ B(x, bn)j(v - y)h(v) - G_B_n(x, v)≤ -(a_1/2) j(2n) ∑_v ∈ B(x, bn)G_B_n(x, v),

	where we used Proposition <ref> and Lemma <ref> together with 1 ≤v - y≤ 2n. Using (<ref>) and B_n ∩^d≥ c_2 n^d, we get
	
    ∑_v ∈ B(x, bn)G_B_n(x, v)≥2C_3 C_4/n^d ϕn^-2B_bn∩^d≥2c_2 C_3 C_4/n^d ϕn^-2 (bn)^d = c_3/ϕn^-2.

	Using (<ref>) we get j(2n) ≥ 2^-d - 2j(n). When we put this together with (<ref>) and (<ref>), we get
	
    (Ah)(y) ≤ -c_4n^-d.

	Define u := h - κη, where
	
    κ := minc_4/2, c_5/2, C_4/2n^-d,

	where c_5 > 0 will be defined later. For y ∈ A(an/2, n)
	
    (Au)(y) = (Ah)(y) - κ(Aη)(y) ≤ -c_4n^-d + κ≤ -c_4n^-d + c_4/2 n^-d = -c_4/2n^-d < 0.

	For x ∈ B_bn⊆ B_an/2, v ∈ B_an/2 we have x - v≤ an ≤ n. We will first assume that x ≠ v so that we can use Theorem <ref>, Lemma <ref> and (<ref>). In this case, we have
	
    G_B_n(x, v)
    		   ≥ C_1G(x, v) ≍ g(x - v) ≥1/a_2g(an) ≥1/a_2^2 a^d - 2γ_2 g(n) ≥1/a_2^2 C_3 a^d - 2γ_2n^-dη(v).

	So, G_B_n(x, v) ≥ c_5 n^-dη(v) for some constant c_5 > 0 and for x ≠ v. If x = v we can use the same arguments that we used when we were proving that G_B_n(x, x) ≥ 2C_4 n^-dη(x) for n large enough to prove that G_B_n(x, x) ≥ c_5 n^-dη(x) for n large enough. Hence, G_B_n(x, v) ≥ c_5 n^-dη(v) for all x ∈ B_bn and v ∈ B_an/2 and for n large enough. Now we have
	
    h(v) = G_B_n(x, v) ∧C_4n^-dη(v)≥c_5n^-dη(v)∧C_4n^-dη(v) = (C_4 ∧ c_5) n^-dη(v).

	Hence,
	
    u(v) = h(v) - κη(v) ≥ (C_4 ∧ c_5) n^-dη(v) - C_4/2∧c_5/2 n^-dη(v) ≥ 0.

	Since u(v) ≥ 0 for v ∈ B_an/2, u(v) = 0 for v ∈ B_n^c and (Au)(v) < 0 for v ∈ A(an/2, n) we can use the same argument as in Proposition <ref> to conclude by Proposition <ref> that u(y) ≥ 0 for all y ∈^d. Since G_B_n(x, y) ≤ C_4 n^-dη(y) for x ∈ B_an/4, y ∈ A(an/2, n) we have h(y) = G_B_n(x, y). Using that, we have
	
    G_B_n(x, y) ≥κη(y) = C_5n^-dη(y),    x∈ B_bn, y ∈ A(an/2, n),

	for n large enough. As before, we can change the constant and get (<ref>) for all n ∈.

Using last two propositions, we have the next corollary.

	Assume (<ref>) and (<ref>). Then there exist constants C_6, C_7 > 0 and b_1, b_2 ∈⟨ 0, 1/2⟩, 2b_1 ≤ b_2 such that
	
    C_6 n^-d_y[τ_B_n] ≤ G_B_n(x, y) ≤ C_7 n^-d_y[τ_B_n],   ∀  x ∈ B_b_1n, y ∈ A(b_2n, n).



	This corollary follows directly from Proposition <ref> and Proposition <ref>. We can set b_2 = a/2 where a ∈⟨ 0, 1/3 ⟩ is as in Lemma <ref> and b_1 = b where b ≤ a/4 is as in Proposition <ref>.




§ 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.

	Let f^d ×^d[0, ∞⟩ be a function and B ⊂^d a finite set. For every x ∈ B we have
	
    _x f(X_τ_B - 1, X_τ_B) = ∑_y ∈ BG_B (x, y) f(y, y + X_1) _{y + X_1 ∉ B}.



	We have
	
    _x f(X_τ_B - 1, X_τ_B) = ∑_y ∈ B, z ∈ B^c_x (X_τ_B - 1 = y, X_τ_B = z) f(y, z).

	Using (<ref>), we get
	
    _x (X_τ_B - 1    = y, X_τ_B = z) = ∑_m = 1^∞_x (X_τ_B - 1 = y, X_τ_B = z, τ_B = m)
        = ∑_m = 1^∞ (x + X_m - 1 + ξ_m = z, x + X_m - 1 = y, X_1, …, X_m - 2∈ B - x)
        = ∑_m = 1^∞ (ξ_m = z - y)  (x + X_m - 1 = y, X_1, …, X_m - 2∈ B - x)
        =  (ξ_1 = z - y) ∑_m = 1^∞_x (X_m - 1 = y, X_1, …, X_m - 2∈ B)
        =  (X_1 = z - y) ∑_m = 1^∞_x (X_m - 1 = y, τ_B > m - 1) =  (X_1 = z - y) G_B (x, y).

	Hence,
	
    _x f(X_τ_B - 1, X_τ_B)    = ∑_y ∈ B, z ∈ B^cf(y, z) G_B (x, y)  (y + X_1 = z)
        = ∑_y ∈ BG_B (x, y) f(y, y + X_1) _{y + X_1 ∉ B}.



	Formula (<ref>) 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^d[0, ∞⟩ is harmonic in B, with respect to X, then {f(X_n ∧τ_B) : n ≥ 0} is a martingale with respect to the natural filtration of X (proof is the same as <cit.>, except that we have a non-negative instead of a bounded function). Using this fact, we can prove the following lemma.

	Let B be a finite subset of ^d. Then f^d[0, ∞⟩ is harmonic in B, with respect to X, if and only if f(x) = _x[f(X_τ_B)] for every x ∈ B.


	Let us first assume that f^d[0, ∞⟩ is harmonic in B, with respect to X. We take arbitrary x ∈ B. By the martingale property f(x) = _x[f(X_n ∧τ_B)], for all n ≥ 1. First, by Fatou's lemma we have _x[f(X_τ_B)] ≤ f(x) so f(X_τ_B) is a _x-integrable random variable. Since B is a finite set, we have f ≤ M on B, for some constant M > 0, and _x(τ_B < ∞) = 1. Using these two facts, we get
	
    f(X_n ∧τ_B) = f(X_n) _{n < τ_B} + f(X_τ_B) _{τ_B ≤ n}≤ M + f(X_τ_B).

	Since the right hand side is _x-integrable, we can use the dominated convergence theorem and we get
	
    f(x) = lim_n →∞_x[f(X_n ∧τ_B)] = _x[lim_n →∞f(X_n ∧τ_B)] = _x[f(X_τ_B)].

	
	On the other hand, if f(x) = _x[f(X_τ_B)], for every x ∈ B, then for x ∈ B we have
	
    f(x) = ∑_y ∈^d_x [ f(X_τ_B) | X_1 = y ] _x(X_1 = y) = ∑_y ∈^dp(x, y) _y[f(X_τ_B)] = ∑_y ∈^dp(x, y) f(y).


Hence, if we take B ⊂^d finite and f^d[0, ∞⟩ harmonic in B, with respect to X, then by Lemma <ref> and the discrete Ikeda-Watanabe formula, we get

    f(x) = _x f(X_τ_B) = ∑_y ∈ BG_B (x, y) f(y + X_1) _{y + X_1 ∉ B}.

Let us define the discrete Poisson kernel of a finite set B ⊂^d by

    K_B(x, z):= ∑_y ∈ BG_B (x, y) (X_1 = z - y) ,    x ∈ B, z ∈ B^c.

If the function f is non-negative and harmonic in B_n, with respect to X, from (<ref>) we have

    f(x)
    	    = ∑_y ∈ B_nG_B_n (x, y) ∑_z ∉ B_nf(y + X_1) _{y + X_1 ∉ B_n}| X_1 = z - y(X_1 = z - y)
        = ∑_z ∉ B_n∑_y ∈ B_nG_B_n (x, y) f(y + z - y) _{y + z - y ∉ B_n}(X_1 = z - y)
        = ∑_z ∉ B_nf(z) ∑_y ∈ B_nG_B_n (x, y) (X_1 = z - y) = ∑_z ∉ B_nf(z) K_B_n(x, z).

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.

	Assume (<ref>) and let b_1, b_2 ∈⟨ 0, 1/2⟩ be as in Corollary <ref>. Then K_B_n(x, z) ≍ l(z) for all x ∈ B_b_1n, where
	
    l(z) =  j(z)/ϕn^-2 + n^-d∑_y ∈ A(b_2n, n)_y[τ_B_n] j(z - y).



	Splitting the expression (<ref>) for the Poisson kernel in two parts and using Proposition <ref>, we get
	
    K_B_n(x, z) ≍∑_y ∈ B_b_2nG_B_n(x, y) j(z - y) + ∑_y ∈ A(b_2n, n)G_B_n(x, y) j(z - y).

	Since G_B_n(x, y) ≍ n^-d_y[τ_B_n] for x ∈ B_b_1n, y ∈ A(b_2n, n), for the second sum in the upper expression we have
	
    ∑_y ∈ A(b_2n, n)G_B_n(x, y) j(z - y)≍ n^-d∑_y ∈ A(b_2n, n)_y[τ_B_n] j(z - y).

	Now we look closely at the expression ∑_y ∈ B_b_2nG_B_n(x, y) j(z - y). Using the fact that y ∈ B_b_2n, b_2 ∈⟨ 0, 1/2⟩ and z≥ n because z ∈ B_n^c, we have
	
    z - y≤z + y≤z + b_2n ≤z + b_2z≤ (1 + b_2)z≤ 2z.

	On the other hand
	
    z≤z - y + y≤z - y + b_2n ≤z - y + b_2z.

	Hence,
	
    1/2z≤ (1 - b_2)z≤z - y.

	Combining (<ref>), (<ref>) and using Lemma <ref>, we have
	
    1/a_1 j1/2z≥ j(z - y) ≥ a_1 j(2z).

	Using (<ref>), we get j(1/2z) ≤ 2^d + 2j(z) = c_1 j(z). Similarly, from (<ref>), we get j(2z) ≥ 2^-d -2 j(z) = c_2 j(z). Hence, a_1 c_2 j(z) ≤ a_1 j(2z) ≤ j(z - y) ≤ a_2^-1 j1/2z≤ a_2^-1 c_1 j(z) for some c_1, c_2 > 0. Therefore,
	
    j(z - y) ≍ j(z),    y ∈ B_b_2n,  z ∈ B_n^c.

	Using (<ref>) we have
	
    ∑_y ∈ B_b_2nG_B_n(x, y) j(z - y)≍∑_y ∈ B_b_2nG_B_n(x, y) j(z) = j(z) ∑_y ∈ B_b_2nG_B_n(x, y).

	Now we want to show that ∑_y ∈ B_b_2nG_B_n(x, y)≍ 1 / ϕn^-2. Using the fact that G_B_n is non-negative function and that _x[τ_B_n] ≤ C_3 / ϕn^-2 for x ∈ B_n we have
	
    ∑_y ∈ B_b_2nG_B_n(x, y)≤∑_y ∈ B_nG_B_n(x, y) = _x[τ_B_n] ≤C_3/ϕn^-2.

	To prove the other inequality we will use Lemma <ref>, Theorem <ref>, Lemma <ref> together with 1 ≤x - y≤ 2b_2n, B_n ∩^d≥ c_3n^d and Lemma <ref>. Thus
	
    ∑_y ∈ B_b_2nG_B_n(x, y)   ≥ C_1∑_y ∈ B_b_2n∖{x}G(x, y)≍∑_y ∈ B_b_2n∖{x}g(x - y)
       ≥1/a_2 (B_b_2n∩^d - 1) g(2b_2n) ≥1/a_2c_3/2(b_2n)^d 1/2^d (b_2n)^d1/ϕn^-2ϕ(n^-2)/ϕ((2b_2n)^-2)
       ≥c_3/2 a_21/2^d ϕ(n^-2) (2b_2)^2 ≥c_3 (2b_2)^2/2^d + 1 a_21/ϕn^-2.

	Hence,
	
    ∑_y ∈ B_b_2nG_B_n(x, y)≥c_4/ϕn^-2.

	From (<ref>) and (<ref>) we have
	
    ∑_y ∈ B_b_2nG_B_n(x, y)≍1/ϕn^-2.

	Finally, using (<ref>) and (<ref>) we have
	
    ∑_y ∈ B_b_2nG_B_n(x, y) j(z - y)≍j(z)/ϕn^-2.

	And now, from (<ref>) and (<ref>) we have the statement of the lemma.

Lemma <ref> basically states that there exist constants C_8, C_9 > 0 such that

    C_8 l(z) ≤ K_B_n(x, z) ≤ C_9 l(z),    x ∈ B_b_1n,   z ∈ B_n^c.

Now we are ready to prove our main result.
[Proof of Theorem <ref>]
	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 <ref>. General case follows using the standard Harnack chain argument. Let x_1, x_2 ∈ B_b_1n. Using (<ref>) we get
	
    K_B_n(x_1, z) ≤ C_9 l(z) = C_9/C_8 C_8 l(z) ≤C_9/C_8 K_B_n(x_2, z).

	Now we can multiply both sides with f(z) ≥ 0 and sum over all z ∉ B_n and we get
	
    ∑_z ∉ B_nf(z) K_B_n(x_1, z)≤C_9/C_8∑_z ∉ B_nf(z) K_B_n(x_2, z).

	If we look at the expression (<ref>) we see that this means
	
    f(x_1) ≤C_9/C_8 f(x_2)

	and that is what we wanted to prove.


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

babamspl





Ante Mimica, 20-Jan-1981 - 9-Jun-2016, <https://web.math.pmf.unizg.hr/ amimica/>




Stjepan Šebek, Faculty of Electrical Engineering and Computing, University of Zagreb, 10000 Zagreb, Croatia

E-mail address: stjepan.sebek@fer.hr
