Martin Kernels for Markov Processes with Jumps

We prove the existence of boundary limits of ratios of positive harmonic functions for a wide class of Markov processes with jumps and irregular (possibly disconnected) domains of harmonicity, in the context of general metric measure spaces. As a corollary, we prove the uniqueness of the Martin kernel at each boundary point, that is, we identify the Martin boundary with the topological boundary. We also prove a Martin representation theorem for harmonic functions. Examples covered by our results include: strictly stable Lévy processes in Rd with positive continuous density of the Lévy measure; stable-like processes in Rd and in domains; and stable-like subordinate diffusions in metric measure spaces.


Introduction
The purpose of this article is to study boundary limits of ratios of positive functions which are harmonic in an arbitrary open set with respect to a Markov process with jumps. The proof of our main result, Theorem 2, relies on the boundary Harnack inequality for Markov Mateusz Kwaśnicki mateusz.kwasnicki@pwr.edu.pl Tomasz Juszczyszyn tomasz.juszczyszyn@pwr.edu.pl 1 Faculty of Pure and Applied Mathematics, Wroclaw University of Science and Technology, Wroclaw, Poland processes with jumps, proved recently in [12], and the oscillation reduction argument, developed in [6] and [11]. As an application, we obtain Martin representation of harmonic functions in Theorem 3. To explain the motivation for our research, we begin with a discussion of the classical case, where harmonicity has its usual meaning: f is harmonic in an open set D if Δf = 0 in D. The boundary Harnack inequality is a statement about positive harmonic functions in an open set, which are equal to zero on a part of the boundary. The result states that if D is regular enough (for example, a Lipschitz domain), x 0 is a boundary point of D, f and g are positive and harmonic in D ∩ B(x 0 , R), and both f and g converge to 0 on ∂D ∩ B(x 0 , R), then for every r ∈ (0, R) the ratio f/g has bounded relative oscillation in D ∩ B(x 0 , r): Here c = c(D, x 0 , r, R)−1 is a constant that depends only on the local geometric properties of D near x 0 , and B(x 0 , r) denotes the ball of radius r, centred at x 0 . The boundary Harnack inequality was first proved independently by A. Ancona ([5]), B. Dahlberg ([17]) and J.-M. Wu ([35]) for Lipschitz domains, and then extended by numerous authors to a wider class of domains and elliptic operators. We refer to [1][2][3][4]31] for further discussion and references. Under appropriate assumptions on the regularity of D, the estimate (1) turns out to be self-improving as r → 0 + , in the sense that the constant c in Eq. 1 converges to 1 as r → 0 + . Equivalently, the boundary limit exists. When D is a Lipschitz domain, then in fact c(D, x 0 , r, R) is of order r β as r → 0 + for some β > 0, which means that f/g extends to a Hölder continuous function at x 0 . A closely related concept of Martin representation of positive harmonic functions was introduced by R. S. Martin in his beautiful article [32], more than three decades before the boundary Harnack inequality became available. Given the existence of limits (2) (for example, if D is a Lipschitz domain), Martin's result asserts that there is a one-to-one correspondence between positive harmonic functions f in D and finite positive measures μ on the boundary of D. The two objects are linked by the formula where the Martin kernel is defined as the boundary limit of the ratio of Green functions: Herex ∈ D is an arbitrarily fixed reference point.
One of numerous equivalent definitions of harmonicity links harmonic functions with the Brownian motion: f is harmonic in D if and only if f has the mean-value property with respect to the distributions of the Brownian motion X t at first exit times: for all bounded open sets U such that the closure of U is contained in D.
Here E x denotes the expectation (and P x will denote the probability) corresponding to the Brownian motion process X t that starts at x, and τ U is the time of first exit from U : that would require rather lengthy and technical arguments. For this reason, unlike in [11], we refer to the general theory of Martin boundary. Our argument still requires extension of some elements of [11] for more general Markov processes, but the most involved part of the proof is avoided. For an excellent exposition of the general theory of Martin boundary, we refer to Chapter 14 of [16]. We conclude the introduction with a description of the structure of this article. The assumptions for the boundary Harnack inequality of [12] are briefly recalled in Section 2. We omit a detailed discussion of these conditions and refer the interested reader to the original paper. Instead, we present a number of examples right after the statement of Theorems 2 and 3 in Section 3. We also provide a counter-example, which shows that the boundary limits (2) typically fail to exist in irregular domains when the process X t has a non-trivial diffusion part. Finally, in Section 4 we prove Theorems 2 and 3.

Fundamental Assumptions for the Boundary Harnack Inequality
The formal statement of the assumptions for Theorem 2 requires some effort. We assume that (X, d, m) is a locally compact metric measure space in which all bounded closed sets are compact and m has full support, and that R 0 > 0 (possibly R 0 = ∞) is a localisation radius such that X \ B(x, r) = ∅ if x ∈ X and 0 < r < 2R 0 .
In [12] the following four conditions are introduced. A detailed discussion of these assumptions is beyond the scope of the present article, we refer the reader to [12] for more information. Here we only state the conditions, without explaining in a formal way the notions of semi-polar and polar sets, processes in duality X t andX t , their generators A andÂ, densities ν(x, y) andν(x, y) (with respect to the measure m) of the Lévy kernels of X t andX t , as well as their Green functions G D (x, y) =Ĝ D (y, x). We note that ν(x, y) describes the intensity of jumps from x to y and it is commonly used throughout the article. The Green function G D (x, y) is required for Theorem 3 only; informally, G D (x, y) is the average amount of time spent near y by the process X t , started at x, until τ D .

Assumption 1
The Hunt processes X t andX t are dual with respect to the measure m. The transition semigroups of X t andX t are both Feller and strong Feller. Every semi-polar set of X t is polar.

Assumption 4
If x 0 ∈ X, 0 < r < s < R < R 0 and B = B(x 0 , R), then We denote where the infimum is taken over all functions f described by the Assumption 2. If x 0 ∈ X and 0 < r < R < R 0 , then we denote Note that by Proposition 2.1 in [12], under Assumptions 1 through 3, C exit (x 0 , r) is finite. Following [6], we say that f is a regular harmonic function in an open set D if the meanvalue property (4) holds with U = D. By the strong Markov property, this implies that Eq. 4 holds for arbitrary open U ⊆ D, so in particular f is harmonic in D.
We use the short-hand notation f ≈ cg for the two inequalities c −1 g ≤ f ≤ cg, where c > 0 is a positive constant. The following theorem is a reformulation of the main result of [12]. [12]) Suppose that x 0 ∈ X, 0 < r 1 < r 2 < r 3 < r 6 < R 0 and a non-negative function f is a regular harmonic function in

Theorem 1 (Lemma 3.2 and Theorems 3.4 and 3.5 in
for some r 4 , r 5 , r 7 , r 8 such that 0 < r 1 < r 2 < r 3 < r 4 < r 5 < r 6 < r 7 < r 8 . Note that it is important that f is non-negative everywhere, not just in D. Theorem 1 implies the more classical statement of the boundary Harnack inequality (Theorem 3.5 in [12]): if f and g satisfy the assumptions of Theorem 1, then as in Eq. 1. We remark that although the original statement allows for an arbitrary sequence of radii, it will be sufficient for us to consider r 1 = r, r 2 = 2r, r 3 = 3r and r 6 = 4r, and we will commonly write C BHI = C BHI (x 0 , r) = C BHI (x 0 , r, 2r, 3r, 4r) in this case.

Main Results and Examples
For the existence of limits, we introduce one more definition. If x 0 ∈ X and 0 < r < R < R 0 , we let Theorem 2 Let D ⊆ X be open, x 0 ∈ ∂D and R > 0. Suppose that: (i) X t satisfies Assumptions 1 through 4; Suppose furthermore that non-negative functions f and g are regular harmonic functions in D ∩ B(x 0 , R) and are equal to zero in B(x 0 , R) \ D. Then either one of f and g is zero everywhere in D, or the finite, positive boundary limit Remark 1 Condition (ii) is required only for inaccessible boundary points x 0 , characterised by the property D∩B(x 0 ,R) E y τ D∩B(x 0 ,R) m(dy) < ∞. The result for accessible boundary points x 0 , for which the integral is infinite, holds under conditions (i) and (iii) through (v).
Remark 2 Theorem 2 also holds with g(x) = E x τ D∩B(x 0 ,R) . This is formally shown in Section 4.4, but the informal explanation is rather straightforward: g is essentially a regular harmonic function in D ∩ B(x 0 , R) (in sharp contrast with the case of continuous Markov processes). Indeed, suppose that X is unbounded, D is a bounded open set and that C Lévy (x 0 , r, R) converges to 1 as R → ∞. By Dynkin's formula (see Lemma 2 and estimate (14) below), is the limit of regular harmonic functions in D. Since the estimates in Theorem 2 are uniform in f and g, we obtain the desired result. (Note that the formal argument is completely different and requires no further assumptions on X and X t .) Remark 3 As remarked in the introduction, the limit in Eq. 10 exists if and only if the relative oscillation of f and g converges to one, that is, By inspecting the proof of Theorem 2, one immediately sees that, given D and x 0 , the boundary limits exist uniformly in f and g, in the sense that with the supremum taken over all f and g satisfying the assumptions of the theorem. We remark that in fact one can prove uniformity also in D, just as in [11], by appropriately modifying the final part of the proof. More formally, where the supremum is taken over all open sets D and f and g satisfying the assumptions of the theorem (here we let the ratio sup / inf be equal to 1 if D ∩ B(x 0 , r) is empty). The proof of this result is sketched in Section 4.4.

Remark 4
It is not necessary to assume that x 0 ∈ ∂D in Theorem 2. For x 0 / ∈ D the statement is void, but for x 0 ∈ D we obtain relative continuity of positive harmonic functions: if f and g are positive harmonic functions in D, then f/g is continuous in D. By Remark 3, the family of functions f/g is in fact relatively equicontinuous at x 0 , in the sense that the functions log(f/g) are equicontinuous at x 0 .
If the process is conservative, then the constant g(x) = 1 is harmonic. In the general case, P x (X(τ D ) = ∂) is continuous (this is proved as in [15]; with the notation of that article, Consequently, positive harmonic functions are relatively equicontinuous at x 0 . If in addition the characteristics of the process (that is, the constants in conditions (ii) through (v)) do not depend on x 0 , then positive functions harmonic in D are in fact uniformly relatively equicontinuous in every compact subset of D.
Before we discuss examples, we provide one application. Recall that the Green function G D (x, y) is the density of the mean occupation measure of X t up to τ D , that is, Under Assumptions 1 and 4, there is a version of G D (x, y) which is a harmonic function of x ∈ D \ {y}, and a co-harmonic (that is, harmonic for the dual process) function of y ∈ D \{x}. Hence, Theorem 2 (or, more precisely, its version for the dual process) immediately implies the existence of the Martin kernel for z = x 0 (this is exactly the same as the classical definition (3)

(b) The Martin kernel M D (x, z) is a harmonic function in D with respect to x if and only if z is an accessible boundary point:
where μ is a measure on ∂ m D, the set of accessible boundary points of D. (e) Conversely, given any non-negative function f and any measure μ on ∂ m D, the righthand side of Eq. 12 is either a harmonic function in D or infinity everywhere in D.

Remark 5
The terms accessible and inaccessible correspond to the probabilistic theory of Martin boundary. To be specific, the process X t killed at the time of first exit from D and conditioned in the sense of Doob by the Martin kernel M D (·, z) converges at its lifetime to z when z is accessible, and dies out in D when z is inaccessible. We refer to [16] for more information.
Remark 6 Unlike in the case of isotropic stable Lévy processes in [11], description of the infinite part of the Martin boundary of D for unbounded open sets is a completely different problem. This issue is addressed in a recent work of P. Kim, R. Song and Z. Vondraček ( [28,30]).

Remark 7
In order to apply the results of [16] about general theory of Martin representation, one requires the dual of the Green operatorĜ D to map bounded functions into bounded continuous ones (a strong Feller property for the Green operator, Hypothesis 13.42 in [16]). In particular,Ê xτD =Ĝ D 1(x) needs to be bounded in D. If X is unbounded, thenÊ xτD is bounded (this follows, for example, by the argument used in the proof of Proposition 2.1 in [12]). If, however, X is bounded (and hence compact), then one needs to assume boundedness ofÊ xτD explicitly (indeed, when X t is conservative and D = X, then clearlyτ D = ∞ with probability one).
Boundedness of E x τ D is assumed in order to keep perfect symmetry between X t andX t (which makes the proof easier to follow). Note, however, that this is a rather mild assumption. Indeed, it is rather easy to see that if X is compact and X \ D is not a polar set, then there is ε > 0 such that P x (τ D < 1) > ε andP x (τ D < 1) > ε for all x ∈ X, and therefore E x τ D andÊ xτD are bounded.
The boundary Harnack inequality stated in Theorem 1 was applied to a variety of Markov processes in Section 5 of [12]. The scale-invariant version of Theorem 1 under α-stable-like scaling discussed therein already asserts conditions (i), (iii) and (v) in Theorem 2. Verification of the remaining conditions (ii) and (iv) is typically straightforward, and we obtain several classes of processes for which Theorems 2 and 3 apply.
In our first example, we use the result of Example 5.5 in [12], where the boundary Harnack inequality for Lévy processes is considered. In the asymmetric case, equality of the notions of semi-polar and polar sets (in Assumption 1) is not trivial, and this was apparently overlooked in [12]. Fortunately, for all asymmetric Lévy processes listed therein, this condition is satisfied by Theorem 2 in [33].

Example 1 (Strictly stable Lévy processes) Let m be the Lebesgue measure in
Suppose that X t is a strictly α-stable Lévy process in R d , where d ≥ 1 and 0 < α < 2. Suppose, furthermore, that the Lévy measure of X t has a density function of the form ν(z) = ϕ(z/|z|)|z| −d−α , with ϕ continuous and positive on the unit sphere (for Lévy processes, ν(x, y) = ν(y − x)). It is easy to see that C Lévy (x 0 , r, R) converges to 1 as r → 0 + and that C Lévy-int (x 0 , r, R) = (R/r) α . By Example 5.5 in [12], X t satisfies the other assumptions of Theorem 2, and so we may use Theorems 2 and 3.
We remark that the above example can be extended to more general Lévy processes, including many subordinate Brownian motions and, more generally, unimodal isotropic Lévy processes. This is based on estimates obtained recently in [9,10,18,20] and will be studied in detail in [19]. Other extensions can be obtained by allowing the Lévy kernel to depend on x or restricting it to a domain, as described in the following two examples.
Example 2 (Stable-like processes) Let m be the Lebesgue measure in R d , R 0 = ∞. Suppose that 0 < α < 2 and where ϕ is symmetric (that is, ϕ(x, y) = ϕ(y, x)), bounded by positive constants, smooth, and has bounded partial derivatives of all orders. As in Example 5.6 in [12], in this case there is a pure-jump process X t with the Lévy kernel ν(x, y)m(dy), and the assumptions of Theorem 2 are satisfied.
Example 3 (Reflected stable processes) Let 0 < α < 2. Let X be the closure of either a Lipschitz domain in R d if α < 1 or a C 1,α+ε domain in R d if α ≥ 1 (with some ε > 0). Let m be the Lebesgue measure on X, and ν(x, y) = c|x − y| −d−α for some c > 0. Again as in Example 5.6 in [12], there is a pure-jump process X t with the Lévy kernel ν(x, y)m(dy), and the assumptions of Theorem 2 are satisfied for some R 0 > 0.
The state space X need not be Euclidean.
Example 4 (Stable-like subordinate diffusions) Let X be a sufficiently regular metric measure space in which there exists a diffusion process. For a rigorous definition, we refer to Example 5.7 in [12]; examples include Riemannian manifolds, Sierpiński gaskets or the Sierpiński carpet. Suppose that 0 < α < d w , where d w is the walk dimension of X (that is, an approximate scaling exponent for the diffusion process). Finally, let X t be a process subordinate to the diffusion process, corresponding to the (α/d w )-stable subordinator. In Example 5.7 in [12] it is shown that X t satisfies conditions (i), (iii) and (v) of Theorem 2, and one easily proves that C Lévy-int (x 0 , r, R) ≤ c(R/r) α for some c > 0. Verification of (ii) requires some work, especially when X is unbounded. For this reason, we only sketch the argument for compact X. For some c > 0 we have where q t (x, y) is the transition density of the diffusion process. Since for each t > 0, q t is Hölder continuous, it is easy to see that ν(x, y) is positive and uniformly continuous in x ∈ B(x 0 , r), y ∈ X \ B(x 0 , R), which clearly implies condition (ii). It follows that Theorems 2 and 3 apply to stable-like subordinate diffusions in compact metric measure spaces.
Surprisingly, Theorem 2 is not influenced by killing.
Example 5 (Processes with a multiplicative functional) Let M t be a strong continuous multiplicative functional such that M 0 = 1 with probability one for all starting points x ∈ X. Such a functional describes gradual killing of the process X t , and is typically obtained as the Feynman-Kac functional M t = exp(− t 0 V (X s )ds) for some non-negative function V . A function f is said to be harmonic with respect to the pair (X t , M t ) if it has the mean-value property instead of Eq. 4. As in Theorem 5.10 in [12], if the assumptions of Theorem 2 are satisfied by the process X t , then the conclusion also holds for functions harmonic with respect to the pair (X t , M t ).
Our final example shows that when X t has non-vanishing diffusion part, one cannot expect the existence of boundary limits (2) unless some geometric restrictions on D are imposed. For corresponding positive results in smooth domains, see [24].
Example 6 (Mixture of Brownian motion and stable process) Let X = R and let m be the Lebesgue measure. Let X t be a one-dimensional Lévy process which is the sum of two independent Lévy processes: the Brownian motion and the symmetric α-stable Lévy process for some α ∈ (1, 2). That is, the characteristic exponent of X t is given by c 1 ξ 2 + c 2 |ξ | α for some c 1 , c 2 > 0. Denote D = (−1, 1) \ {0}. Let p t (y − x) be the continuous version of the transition density of X t . Then the three functions are regular harmonic in D: for u this is just the martingale property of X t , for v (the compensated potential kernel of X t ) this is proved, for example, in [36], while for w it follows directly from the definition. Furthermore, for x ∈ R, with c 3 = c 3 (c 1 , c 2 , α) (see, for example, Lemma 2.14 in [20]). In particular, v(x) ≈ c 3 |x| for x ∈ D. Finally, by the boundary Harnack inequality given in Theorem 1 (see Examples 5.5 and 5.13 in [12] for a detailed discussion), we have c 2 , α). Let us define −1] (X(τ D ))).
Then f and g are non-negative, regular harmonic in D and equal to zero in (−1, 1) \ D = {0}, so that they satisfy the assumptions of Theorem 2. On the other hand, 1 2 . In particular, the limit of f (x)/g(x) as x → 0 does not exist.

Proofs of Main Results
In this section we prove Theorem 2. We will always assume that x 0 , R and D are fixed, where x 0 ∈ X, 0 < 2R < R 0 and D ⊆ B(x 0 , R) is an open set. It is also understood that x 0 ∈ ∂D, although, at least formally, the argument extends also to x 0 ∈ D and x 0 / ∈ D. Recall that the notation f ≈ cg stands for c −1 g ≤ f ≤ cg with c > 0. We Finally, we let s D (x) = E x τ D .
To simplify the notation, we drop D from the notation in subscripts whenever possible, and we write τ r = τ D r , τ r,s = τ D r,s , s r (x) = s D r (x), 1 r,s (x) = 1 D r,s (x) etc.
Our argument is based on the boundary Harnack inequality of [12], stated in Theorem 1. Under the assumptions of Theorem 2, the constant C BHI (x 0 , r, 2r, 3r, 4r) can be chosen so that it does not depend on r, as long as 0 < 4r ≤ R, and it will be denoted simply by C BHI (recall that x 0 and R are fixed). In a similar way, we denote C Lévy = C Lévy (x 0 , r, 2r) (with 0 < 2r < R 0 ) and C Lévy-int = C Lévy-int (x 0 , r, 2r) (with 0 < 2r < R 0 ), chosen independently of r. With one exception, we will only use constants C BHI , C Lévy and C Lévy-int with these parameters.
We prove Theorem 2 by considering separately two types of boundary points, which are called accessible and inaccessible in [11]. First, however, we introduce some further notation and prove preliminary estimates.

Decomposition of Harmonic Functions
From now on f and g are functions satisfying the assumptions of Theorem 2, and we assume that neither f nor g is equal to zero almost everywhere. Note that this implies that f and g are strictly positive in D. Whenever 0 < r < s ≤ R, we decompose f into the sum of two functions, f r,s andf r,s , which correspond to the process X t exiting D r near its boundary (into D r,s ) and away of its boundary (into D s,∞ ): ((f 1 r,s )(X(τ r ))),f r,s (x) = E x ((f 1 s,∞ )(X(τ r ))).
Not unexpectedly, a similar notation is used for the function g. Clearly, f = f r,s +f r,s , and both f r,s andf r,s are non-negative regular harmonic functions in D r which are equal to zero in B r \D r . Therefore, we can apply Theorem 1 to f 4r,s andf 4r,s whenever 0 < 4r < s ≤ R.
Note that by Theorem 1 with r = R 4 , we have whenever 0 ≤ r ≤ s ≤ R 4 . The next result states, in particular, that there is little difference whether we write s R/2 or s R in the above estimate.
Proof The first inequality is clear. For the other one, we use the strong Markov property and Theorem 1: X\B 3r E y s 8r (X(τ 4r ))ν(x 0 , y)m(dy) Furthermore, by Proposition 2.1 in [12] (combined with the last displayed formula in the proof of this result), .
It remains to use (9).
For convenience, we denote Our next result compares f 8r,s withf 8r,s . For f 8r,s , we will use Theorem 1, which states that in D 2r we have f 8r,s ≈ C BHI M 6r,∞ (f 8r,s )E x τ 4r . The same estimate can be written down forf 8r,s . However, M 6r,∞ (f 8r,s ) involves an integral off 8r,s over D 6r,8r , which is often problematic. A much better estimate forf 8r,s can be easily obtained from the following corollary of Dynkin's formula for X t .

Lemma 2 (formula (2.12) in [12]) Let D ⊆ X be open and bounded, and let f be a non-negative function equal to zero in D.
Then (14) for x ∈ D.

y)f (y)m(dy)dt
Using the definition off 8r,s and Eq. 5 to substitute ν(x 0 , y) for ν(X t , y) in Eq. 14, we havef Note that not only we have M s,∞ (f ) instead of M 6r,∞ (f 8r,s ), but also the constant C Lévy (x 0 , 8r, s) tends to 1 as r → 0 + .
Proof By Theorem 1, Finally, by Eq. 13, which is the desired upper bound. The lower bound is proved in a somewhat more complicated way. By Theorem 1 and estimate (15), Lévy (x 0 , 8r, 16r) in the second inequality because s ≥ 16r). By Lemma 1, E x τ 8r ≤ C τ E x τ 4r . Furthermore, by Theorem 1 (as in Eq. 13, but with R replaced by R/3) and again Lemma 1, On the other hand, by Eq. 13, We conclude that as desired.

Inaccessible Boundary Points
Throughout this part we assume that x 0 is inaccessible, that is, In this case f 8r,s and g 8r,s turn out to be negligible compared tof 8r,s andg 8r,s for sufficiently small r and s. Clearly, M 0,∞ (s R/2 ) ≤ M 0,∞ (s R ) < ∞. We remark that by Eq. 13, and M 0,∞ (g) < ∞ by the same argument, and hence one can pass to the limit separately in the numerator and the denominator of Eq. 10. Let 0 < ε < 1. By the upper bound in Lemma 3, there is s = s(ε) ≤ εR such that if 0 < 8r ≤ s, then for x ∈ D 2r . Furthermore, estimate (15) and the assumption lim for x ∈ D 8r . It follows that for x ∈ D 2r . The lower bound is proved in a similar manner, and we obtain for x ∈ D 2r . Since ε was arbitrary and s converges to 0 as ε → 0 + , we have We claim that if 0 < q < R 0 and η > 0, then there is p, which depends only on q, η and the characteristics of the process X t , such that 0 < p < q and inf x∈D q (f (x)/g(x)) − 1 (22) for all open sets D and all functions f and g as in Theorem 2 (this estimate is very similar to Eq. 21). By considering the supremum of both sides of Eq. 22 over all f , g and D, and then taking the upper limit as q → 0 + , we obtain the desired result: for arbitrary η > 0. Therefore, it remains to prove (22). Let 0 < q < 1 24 R 0 and η > 0. We consider two additional parameters δ, N > 0; the actual values of δ (small real) and N (large integer) are to be specified at the end of the argument. By the assumption lim r→0 + C Lévy (x 0 , r, R) = 1 one can construct a decreasing sequence of radii a 0 , a 1 , . . . , a N so that a 0 is the input radius q, 1 8 a N will be the output radius p, and we have 16a n+1 < a n and C Lévy (x 0 , 8a n+1 , a n ) ≤ 1 + δ for all n = 0, 1, . . . , N − 1.
In the other scenario, for each n the converse of Eq. 23 holds. Summing up these inequalities for n = 0, 1, . . . , N − 1 we obtain M a N ,a 0 (s R/2 ) ≥ Nδ(1 + M a 0 ,R/4 (s R/2 )), and we argue as in Section 4.3. Again by Lemma 3, we have Eq. 20 with r = 1 8 a N , s = a 0 and ε = C 3 BHI C Lévy C 3 τ (N δ) −1 . Inequality (21) follows. Dividing both sides of it by inf x∈D r (f (x)/g(x)) and using monotonicity of this expression in r, we obtain (22) for p = 1 8 a N , provided that ε(C BHI + 1) ≤ η. Since δ is now fixed, we may choose N large enough, so that this condition is satisfied. This completes the proof of the extension described in Remark 3.

Martin Representation
In this section we prove Theorem 3. We assume that the assumptions of Theorem 2 are satisfied in a uniform way for all x 0 ∈ D.
We note one important property of the Green function: if U is an open subset of D and y / ∈ ∂U , then (where, as usual, we assume that G U (x, y) = 0 whenever x / ∈ U or y / ∈ U ). In particular, G D (x, y) is a regular harmonic function in D \ B(y, r) for every r > 0. By a duality argument, G D (x, y) is a regular co-harmonic function in D \ B(x, r) for every r > 0. Furthermore, by the strong Markov property, (25) for any nonnegative function f . Note that if m(∂U ) = 0, then Eq. 25 follows from Eq. 24 and Fubini.

Proof of Theorem 3(a)
The assumptions are completely symmetric under duality, and hence we may apply Theorem 2 to both harmonic and co-harmonic functions. In particular, as already remarked before the statement of Theorem 3, the Martin kernel, defined as the boundary limit of co-harmonic functions exists for all boundary points z ∈ D (here and belowx ∈ D is a fixed reference point). In other words, the Martin boundary coincides with the Euclidean boundary.
The representation given in part (d) essentially follows now from the general theory of Martin boundary, together with some ideas developed in [11]. For simplicity, in the remaining part of the proof we simply write that a function is harmonic when we refer to harmonicity in D.

Proof of Theorem 3(b)
Following the proof of Theorem 2 in [11], we find that M D (x, x 0 ) is a harmonic function with respect to x if and only if x 0 is accessible. Indeed, for an inaccessible boundary point x 0 we have, by Eq. 10 in Theorem 2, for C = ( D ν(y, x 0 )G D (x, y)m(dy)) −1 > 0, and so the Martin kernel is not harmonic (to see this, simply use (25)). On the other hand, if x 0 is accessible and R > 0, then Recall that G D (x, y) is a regular harmonic function of x ∈ D \ B(x 0 , R) when y ∈ B(x 0 , R). By Fatou's lemma, and we claim that in fact equality holds, that is, we can exchange the limit with the expectation in Eq. 26. By Vitali's convergence theorem, it suffices to prove that that the ratio in the right-hand side of Eq. 26 is a uniformly integrable family of random variables for y ∈ D ∩ B(x 0 , r) for some r > 0. The argument is exactly the same as in the proof of formula (77) in [11]; for the convenience of the reader, we repeat it below.
Assume that 0 < 8r < R and that x,x / ∈ D ∩ B(x 0 , R). We will first prove that By the boundary Harnack inequality (Theorem 1) applied to G D (z, ·) and G D (x, ·), it suffices to consider a fixed y ∈ D ∩ B(x 0 , r), that is, to show that G D (·, y) is bounded in D \B(x 0 , 4r). This is relatively simple, but somewhat technical. Denote D 1 = D ∩B(x 0 , r), 4r). By Dynkin's formula (14), The supremum is finite by Assumption 3 and boundedness of D, and the integrals in the right-hand side are bounded by sup u∈D E u τ D and sup u∈DÊuτD , respectively. Furthermore, and the right-hand side is finite by Assumption 4. By adding the sides of these two bounds and using harmonicity of the Green function, we complete the proof of Eq. 28.
On the other hand, if we denote D = D \ B(x 0 , 8r) and D = D \ B(x 0 , R), then, again by Lemma 2, and, in a similar way,

G D (w, y)m(dw) .
It follows that where C does not depend on (sufficiently small) r > 0 and y ∈ D ∩ B(x 0 , r). Recall that G D\B(x 0 ,8r) (x, v) increases to G D (x, v) (because the corresponding exit times τ D\B(x 0 ,8r) increase to τ D ). By monotone convergence, the right-hand side converges to zero as r → 0 + . Together with Eq. 28, this completes the proof of uniform integrability of the right-hand side of Eq. 26. Part (b) follows, and in addition we see that for accessible boundary points z, the Martin kernel M D (x, z) is a regular harmonic function in D \ B(z, r) for every r > 0.
In order to apply the general theory of Martin boundary, we need to prove that the Green operator, which maps a measurable function f (x) to G D f (x) = D G D (x, y)f (y)m(dy), takes bounded functions into continuous ones. Let f be a bounded function on D, x 0 ∈ D and ε > 0. Clearly, |G D f (x)| ≤ f E x τ D for x ∈ D, so that G D f is bounded. Let r > 0 be small enough, so that E x τ B(x 0 ,r) < ε for x ∈ B(x 0 , r). By Eq. 25, The first term is continuous in B(x 0 , r) by Theorem 2 (see Remark 4). The other one is bounded by ε f , an arbitrarily small number. Therefore, G D f is continuous at x 0 .
The general theory of Martin boundary tells us now that if f satisfies the assumptions of Theorem 3 and f is equal to zero in the complement of D, then for some measure μ on the set of accessible boundary points ∂ m D, see Theorem 14.8 in [16]. Furthermore, if we show that for every z ∈ ∂ m D, M D (x, z) is a minimal harmonic function with respect to x, then the measure μ in the above representation is unique. Minimality of M D (x, z) is proved as in the final part of the proof of Lemma 14 in [11].

Proof of Theorem 3(c)
and that the measure μ in representation (29) is zero on ∂ m D ∩ B(x 0 , 4r) for some r > 0. Our goal is to prove that f is identically zero. This will imply that if f is harmonic and 0 ≤ f (x) ≤ M D (x, x 0 ) for all x ∈ X, then the measure μ in representation (29)  On the other hand, since f (x) ≤ M D (x, x 0 ), one easily finds that f is also a regular harmonic function in D \B(x 0 , r). This is exactly the same argument as in Lemma 9 in [11]; for the convenience of the reader, we provide the details at the end of this section. In particular, since f is bounded in D ∩ B(x 0 , 2r), it is bounded on D.
A sweeping argument, which is a simplified version of Lemma 10 in [11], proves then that f is a regular harmonic function in D: Let σ n be the sequence of consecutive exit times from alternately D ∩ B(x 0 , 4r) and D \ B(x 0 , r). That is, σ 0 = 0 and σ n+1 = σ n + τ V • ϑ σ n , where V = D ∩ B(x 0 , 4r) when n is even and V = D \ B(x 0 , r) when n is odd (and ϑ τ is the shift operator).
Clearly, σ n ≤ τ D < ∞. Since σ n is increasing, by quasi-left continuity, X(σ n ) has a limit as n → ∞. Therefore, it is impossible that σ n < τ D for infinitely many n. It follows that with probability one, eventually σ n = τ D .
Since f (x) = E x f (X(σ n )) and f is bounded, by dominated convergence we have f (x) = E x f (X(τ D )) = 0, as desired.
We have thus proved the representation (29) for harmonic functions f which are zero in the complement of D. The general case is handled as in Lemma 13 in [11]. The integrand in the right-hand side increases as n → ∞, and therefore by monotone convergence,

y)ν(y, z)m(dy) f (z)m(dz).
Let g(x) be equal to the right-hand side of Eq. 30 for x ∈ D, and to f (x) for x ∈ X\D. From Lemma 2 and the property (25)  .
Therefore, f − g is a non-negative harmonic function which is equal to zero in X \ D, and so it has a unique representation (29). Finally, the outer integral in Eq. 30 is finite, and so points at which the inner integral is infinite cannot contribute to the integral. It follows that we can change the outer integral to an integral over X \ (D ∪ ∂ m D). The proof of Eq. 12 is complete.

Proof of Theorem 3(e)
By the boundary Harnack inequality, if the right-hand side of Eq. 12 is finite at some x ∈ D, it is finite everywhere in D. Indeed, let f be given by Eq. 12. If f (x) = ∞ for some x ∈ D, by Theorem 1 f is infinite at every point of a ball B(x, r) contained in D. If y ∈ D, then again using Theorem 1 (for a ball centred at y), f is infinite at y.
Finally, harmonicity of the right-hand side of Eq. 12, whenever it is finite, follows from property (25) of the Green function, harmonicity of the Martin kernel and Fubini.
At the end of this section, we present the proof of Lemma 9 in [11], adapted to our setting. This result was used in the proof of Theorem 3(c).

Lemma 5 (Lemma 9 in [11]) Let U and D be open subsets of
for all x ∈ X, f and g are harmonic in D, g is a regular harmonic function in U and g(x) = 0 for x ∈ X \ D, then f is a regular harmonic function in U .
Proof Let D n be an ascending sequence of open sets such that D n ⊆ D and ∞ n=1 D n = D, and let U n = U ∩ D n . Then τ U n increases to τ U , and, by quasi-left continuity, X(τ U n ) converges to X(τ U ) with probability one. It follows that if X(τ U ) ∈ D \ U , then eventually τ U n = τ U for n large enough up to an event of probability zero. Hence, lim n→∞ E x (g1 D\U )(X(τ U n )) = E x (g1 D\U )(X(τ U )) = g(x).