Discrete harmonic functions in Lipschitz domains

We prove the existence and uniqueness of a discrete nonnegative harmonic function for a random walk satisfying finite range, centering and ellipticity conditions, killed when leaving a globally Lipschitz domain in $\mathbb{Z}^d$. Our method is based on a systematic use of comparison arguments and discrete potential-theoretical techniques.

Then, we let {S(n), n ∈ N} = (S n ) n∈N be the Markov chain on Z d defined by P[S n+1 = x + e/S n = x] = π(x, e); e ∈ Γ, x ∈ Z d , n = 0, 1, . . . (S n ) n∈N is a centered random walk with bounded increments which becomes spatially homogeneous if we assume the probabilities π(x, e) are independent of x. We shall assume that the set Γ contains all unit vectors in Z d , i.e. all the vectors e k = (0, . . . , 0, 1, 0, . . . 0) ∈ Z d , where the 1 is the k-th component. We shall impose to the random walk (S n ) n∈N to satisfy the following uniform ellipticity condition: (1.1) π(x, e) ≥ α, e ∈ Γ, x ∈ Z d , for some α > 0. We shall denote by: • C a globally Lipschitz domain of Z d that is, a domain C = D ∩ Z d where for some Lipschitz function on R d−1 satisfying for some A > 0, where |.| denote the Euclidean norm. We shall assume that ϕ(0) = 0.
• G y x , x, y ∈ C, the Green function defined by G y x = n∈N P x (S n = y, τ > n).
We are interested in positive functions h which are discrete harmonic for the random walk (S j ) j∈N killed at the boundary of C, i.e. in functions h : C → R + such that: i) For any x ∈ C, h(x) = e∈Γ π(x, e)h(x + e); ii) If x ∈ ∂C, then h(x) = 0; iii) If x ∈ C, then h(x) > 0; where C = ∂C ∪ C. The boundary of a set A ⊂ Z d is defined by ∂A = {x ∈ A c , x = z + e for some z ∈ A and e ∈ Γ} and A = A ∪ ∂A. In terms of the first exit time of the random walk from C, we have that Theorem 1.1. Let (S n ) n∈N be a centered random walk satisfying the above finite support and ellipticity conditions. Assume that C is a globally Lipschitz domain of Z d . Then, up to a multiplicative constant, there exists a unique positive function, harmonic for the random walk killed at the boundary.
The previous result has an important consequence on the Martin boundary theory attached to the random walk (S n ) n∈N killed on the boundary of C. Recall that for a transient Markov chain on a countable state space E, the Martin compactification of E is the unique smallest compactification E M of the discrete set E for which the Martin kernels y → k x y = G x y /G x 0 y (where x 0 is a given reference state in E) extend continuously for all x ∈ E. The minimal Martin boundary ∂ m E M is the set of all those γ ∈ ∂E M for which the function x → k x γ is minimal harmonic. Recall that a harmonic function h is minimal if 0 ≤ g ≤ h with g harmonic implies g = ch with some c > 0. By the Poisson-Martin boundary representation theorem, every nonnegative harmonic function h can be written as for a some positive Borel measure µ on ∂ m E M (cf. [12], [23], [25]).
An immediate consequence of Theorem 1.1 is the following. We conclude this introduction with some comments which may be helpful in placing the results of this paper in their proper perspective.
(i) The proof of Theorem 1.1 given in [8] uses in a crucial way the parabolic Harnack principle. We noted in [8] that a more satisfactory approach should dispense with parabolic information and restrict to elliptic tools. A way to get round the difficulties encountered in [8] is to use a lower estimate for superharmonic extensions of discrete positive harmonic functions derived by Kuo and Trudinger in [20]. This lower estimate encompasses three powerful ingredients: the Aleksandrov-Bakel'man-Pucci's maximum principle, a barrier technique and a Calderón-Zygmund covering argument. Going trough the superharmonic extension gives an alternative to the use of [8, Lemma 2.5] and provides a purely elliptic derivation of [8,Proposition 2.6] . An advantage of this approach is that it allows us to relax the assumptions 0 ∈ Γ and Γ = −Γ made in [8].
(ii) In case of homogeneous symmetric random walks on unbounded Lipschitz domains, the main results of this paper follows from [17]. Although the work of Gyrya and Saloff-Coste concerns diffusion on Dirichlet spaces, to derive the desired results for symmetric random walks, it suffices to consider the corresponding cable process (see [3, §2]). Since the harmonic functions for cable process and the random walk on the corresponding graph are essentially the same one has all the desired results (namely Theorem 1.1, Theorem 1.2, Theorem 2.2 and Theorem 2.3).
(iii) Spatially inhomogeneous random walks can be considered as the discrete analogues of diffusions generated by second-order differential operators in nondivergence form. As in [8], the main tools in this paper are discrete versions of Carleson estimate and boundary Harnack inequality (cf. [4], [5], [13], [14]).
(iv) We restrict ourselves in this paper to random walks in Lipschitz domains. However, the proofs given below should work for a larger class of domains, for instance uniform or inner uniform domains (cf. [1]).
2. Proof of Theorem 1.1 2.1. Harnack principle. We say that a function u : In addition to an obvious maximum principle, harmonic functions satisfy, when they are positive, a Harnack principle. For convenience this principle is formulated in balls. The discrete Euclidean ball of center y ∈ Z d and radius R ≥ 1 is denoted B R (y) and simply B R when y is clearly understood. We shall also have to use cubes. The cube of center y ∈ Z d and sides 2R, parallel to the coordinate axes is denoted Q R (y) and simply Q R when y is clear. The following theorem [20] is a centered version of Harnack principle established by Lawler [22] for random walks with symmetric bounded increments (as well homogeneous and inhomogeneous).
Theorem 2.1. (Harnack principle) Assume that u is a nonnegative harmonic function associated to a random walk satisfying centering, finite support and uniform ellipticity conditions in a ball B 2R (y). Then max The classical Carleson estimate [10] asserts that a positive harmonic function vanishing on a portion of the boundary is bounded, up to a smaller portion, by the value at a fixed point in the domain with a multiplicative constant independent of the function.
Theorem 2.2. Assume that u is a nonnegative harmonic function in C ∩ B 3R (y). Assume that u = 0 on ∂C ∩ B 2R (y). Then where C = C(d, α, Γ, A) > 0 is independent of y, R and u.
The proof of Theorem 1.1 relies on the following Proposition.
Proposition 2.1. Let y ∈ ∂C and R large enough. Let u be a nonnegative harmonic function Proof. To prove (2.2) we first observe that it suffices to show that Without loss of generality, we assume y = 0 and max u(x), x ∈ C ∩ Q 2R = 1. Then considering the function v : 3) reduces to the following lower estimate Since v is superharmonic in Q 2R (i.e Lv ≤ 0 in Q 2R ) we can use the estimate [20, 4.18] and deduce that where γ, δ > 0 are two positive constants depending on d, α and Γ and where the notation |S| is used to denote the cardinality of a subset S ⊂ Z d .
On the other hand, the fact that C is Lipschitz allows us to find a circular cone C ′ with vertex at the origin such that C ′ ⊂ C c . It follows then that there exists a positive constant µ (depending on A) such that for R large enough We conclude from (2.5) and (2.6) that which implies (2.4) and completes the proof of (2.3).

Proof of Theorem 2.2.
To prove the Carleson estimate (2.1) we first observe that the uniform ellipticity assumption implies that u(ξ) ≤ Ce C |ξ−ζ| u(ζ), ξ, ζ ∈ C ∩ B 3R (y); where C = C(d, α, Γ, A) > 0. This local Harnack principle allows us to assume that the distance of x from ∂C is sufficiently large. We shall denote by δ(x) (x ∈ C ∩ B 2R (y)) this distance and suppose that δ(x) ≥ C. The fact that C is Lipschitz combined with Harnack principle (Theorem 2.1) imply that where γ and C are positive constants depending on d, α, Γ and A. Let x ∈ C ∩ B 2R (y), and let us assume that where ρ is the constant obtained in (2.2) and γ the exponent that appears in (2.7). Let x 0 ∈ ∂C such that |x − x 0 | = δ(x). It follows easily from (2.8) and the fact that δ(x) is sufficiently large that B 3 √ dδ(x) (x 0 ) ⊂ B 2R (y). By Proposition 2.1 applied to the harmonic function u in the domain Hence, thanks to (2.8) It follows that and therefore, by (2.9) It remains to consider the case where In follows from (2.11) that and, thanks to (2.7), . Putting together (2.10) and (2.12) and taking the supremum over C ∩ B 2R (y), we deduce that Using the fact that (2R − |x − y|) ≈ R for x ∈ C ∩ B R (y) we deduce the estimate (2.1). ✷ where C = C(d, α, Γ, A, K) > 0.
The above formulation of the boundary Harnack principle follows the classical formulation but the proof of (2.13) which will be given below shows that the assumption v = 0 on ∂C ∩ B 2R (y) is not needed so that (2.1) constitutes a special case of (2.13).
The estimate (2.13) is an immediate consequence of the lower estimate contained in the following lemma.
For y ∈ ∂C and R ≥ r ≥ 1, we shall denote by For R ≥ r ≥ 1, the boundary of C R,r is the union of three sets: the "bottom" ∂C R,r ∩ C c , the "lateral side" ∂C R,r ∩ {x ∈ C, 0 ≤ δ(x) ≤ r} and the "top" ∂C R,r ∩ D R,r .
Proof of Theorem 2.3. In order to derive estimate (2.13) from (2.14), we first observe that it is always possible to assume that u(y + Re 1 ) = v(y + Re 1 ) = 1. For a large K, Carleson estimate (2.1) combined with Harnack principle imply that the function u is dominated by a positive constant c 0 in the region B KR (y) ∩ C. This constant c 0 (which depend on K) can be chosen so that by Harnack principle the lower estimate v ≥ 1 c 0 holds on D KR,R (y).
We have: where the second inequality follows from (2.14). We deduce then that which completes the proof of (2.13). ✷ Proof of Lemma 2.1. To prove estimate (2.14) it suffices to show that if u, v : C Kr,r (y) → R (where y ∈ ∂C, r ≥ 1 are fixed) satisfy x ∈ C Kr,r (y) on ∂C Kr,r (y) ∩ D Kr,r (y) then we have First we prove that under (2.15) the function u satisfies for appropriate constants α, β > 0. The proof of (2.18) relies on the following construction. We assume K large enough and we defineũ : Let U = (B M r (y) ∩ C) ∩ B 2r (ỹ) and w : U −→ R be defined by The boundary conditions satisfied byũ imply that 0 ≤ũ ≤ 1 in B M r (y) ∩ C. It follows from this that On the other hand assumption (2.19) ensures that the function w vanishes on ∂B M r (y) ∩ B 2r (ỹ). Letz = y + (M − 2)re 1 ∈ U. By the same argument used in the proof of the lower estimate (2.4) we see that where τ U is the exit time from U. It follows then that where the last inequality follows from (2.20). This means that 1−ũ(z) ≤ θ < 1 and thereforẽ u(z) ≥ 1 − θ > 0. On the other hand, since u ≥ 1 on ∂C Kr,r (y) ∩ D Kr,r (y), we deduce by the maximum principle thatũ ≥ u on B r (y) ∩ C. Combining with Harnack inequality we de deduce (2.18).
Let U j = B 2r (x j ) ∩ C (2j+1)r,r (y) and τ U j be the exit time from U j . By the same argument used in the proof of (2.4) we see that Using (2.16) (in particular, the fact that v ≤ 0 on ∂C Kr,r (y) ∩ (D Kr,r (y) ∪ C c )) we deduce Iterating this estimate we obtain which proves (2.22). It follows from (2.22) that provided that K is large enough. From the previous considerations it follows that u 1 = K β 2α u ≥ 0 in C r,r/K (y) with u 1 ≥ 1 in ∂C r,r/K (y) ∩ D r,r/K (y) thanks to (2.21) and, thanks to (2.23), with v 1 ≤ 0 on ∂C r,r/K (y) ∩ D r,r/K (y) ∪ C c .
In particular, we have ≥ 0 on C r,r (y) \ C r,r/K (y).
It follows that u 1 , v 1 satisfy the same assumptions as u, v with r replaced by r/K. We can then iterate and define u i , v i such that ≥ 0 on C r/K i ,r/K i (y) \ C r/K i ,r/K i+1 (y) i = 1, 2 . . .. We deduce that u − v ≥ 0 on S(y) = i≥0 C r/K i ,r/K i (y) \ C r/K i ,r/K i+1 (y).
Replacing K by K + 2 in the previous considerations we deduce that u ≥ v on S(x) which contains x. This show that u(x) ≥ v(x) and completes the proof of (2.17). ✷ Proof of Theorem 1.1. The proof is the same as [8]. Observe that instead of Carleson estimate, one can simply use the estimate u(ξ) ≤ C e C|ξ−ζ| u(ζ), ξ, ζ ∈ C, which follows from uniform ellipticity. The advantage of this estimate is that it works for all connected infinite domains, and not just for domains satisfying Carleson estimate. It is already enough to run the diagonal process argument used in [8].
As in [8] the uniqueness can be deduced by essentially the same method as in [1, Proof of Theorem 3] and [2, Lemma 6.2]. ✷